url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.convertunits.com/from/gigahertz/to/degree/hour | 1,701,450,722,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00575.warc.gz | 798,718,914 | 12,760 | ## Convert gigahertz to degree/hour
gigahertz degree/hour
How many gigahertz in 1 degree/hour? The answer is 7.716049382716E-16. We assume you are converting between gigahertz and degree/hour. You can view more details on each measurement unit: gigahertz or degree/hour The SI derived unit for frequency is the hertz. 1 hertz is equal to 1.0E-9 gigahertz, or 1296000 degree/hour. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between gigahertz and degrees/hour. Type in your own numbers in the form to convert the units!
## Quick conversion chart of gigahertz to degree/hour
1 gigahertz to degree/hour = 1.296E+15 degree/hour
2 gigahertz to degree/hour = 2.592E+15 degree/hour
3 gigahertz to degree/hour = 3.888E+15 degree/hour
4 gigahertz to degree/hour = 5.184E+15 degree/hour
5 gigahertz to degree/hour = 6.48E+15 degree/hour
6 gigahertz to degree/hour = 7.776E+15 degree/hour
7 gigahertz to degree/hour = 9.072E+15 degree/hour
8 gigahertz to degree/hour = 1.0368E+16 degree/hour
9 gigahertz to degree/hour = 1.1664E+16 degree/hour
10 gigahertz to degree/hour = 1.296E+16 degree/hour
## Want other units?
You can do the reverse unit conversion from degree/hour to gigahertz, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Gigahertz
The SI prefix "giga" represents a factor of 109, or in exponential notation, 1E9.
So 1 gigahertz = 109 .
The definition of a hertz is as follows:
The hertz (symbol Hz) is the SI unit of frequency. It is named in honour of the German physicist Heinrich Rudolf Hertz who made some important contributions to science in the field of electromagnetism.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 622 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.85265 |
http://mathhelpforum.com/discrete-math/177589-prove-exists-m-among-ai-whose-sum-least-m.html | 1,481,370,483,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543035.87/warc/CC-MAIN-20161202170903-00323-ip-10-31-129-80.ec2.internal.warc.gz | 175,150,531 | 9,469 | ## Prove that exists m among ai whose sum is at least m
Let n be the element of N and let a1,a2,...,an be positive real numbers such that
a1+a2+a3+...+an= (1/(a1)^2)+(1/(a2)^2)+...+(1/(an)^2)
Prove that for any m=1,2,...,n there exist m numbers among ai whose sum is at least m. | 99 | 280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-50 | longest | en | 0.852322 |
https://www.airmilescalculator.com/distance/mde-to-svi/ | 1,709,068,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00125.warc.gz | 618,031,854 | 38,290 | # How far is San Vicente Del Caguán from Rionegro?
The distance between Rionegro (José María Córdova International Airport) and San Vicente Del Caguán (Eduardo Falla Solano Airport) is 279 miles / 450 kilometers / 243 nautical miles.
The driving distance from Rionegro (MDE) to San Vicente Del Caguán (SVI) is 474 miles / 763 kilometers, and travel time by car is about 12 hours 35 minutes.
279
Miles
450
Kilometers
243
Nautical miles
1 h 1 min
## Distance from Rionegro to San Vicente Del Caguán
There are several ways to calculate the distance from Rionegro to San Vicente Del Caguán. Here are two standard methods:
Vincenty's formula (applied above)
• 279.394 miles
• 449.640 kilometers
• 242.786 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 280.897 miles
• 452.060 kilometers
• 244.093 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Rionegro to San Vicente Del Caguán?
The estimated flight time from José María Córdova International Airport to Eduardo Falla Solano Airport is 1 hour and 1 minutes.
## Flight carbon footprint between José María Córdova International Airport (MDE) and Eduardo Falla Solano Airport (SVI)
On average, flying from Rionegro to San Vicente Del Caguán generates about 66 kg of CO2 per passenger, and 66 kilograms equals 146 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Rionegro to San Vicente Del Caguán
See the map of the shortest flight path between José María Córdova International Airport (MDE) and Eduardo Falla Solano Airport (SVI).
## Airport information
Origin José María Córdova International Airport
City: Rionegro
Country: Colombia
IATA Code: MDE
ICAO Code: SKRG
Coordinates: 6°9′52″N, 75°25′23″W
Destination Eduardo Falla Solano Airport
City: San Vicente Del Caguán
Country: Colombia
IATA Code: SVI
ICAO Code: SKSV
Coordinates: 2°9′7″N, 74°45′58″W | 596 | 2,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.748939 |
http://physics.stackexchange.com/questions/43929/equivalent-spring-equations-for-non-helical-coil-shapes | 1,469,716,503,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828283.6/warc/CC-MAIN-20160723071028-00149-ip-10-185-27-174.ec2.internal.warc.gz | 190,993,749 | 19,336 | # Equivalent spring equations for non-helical coil shapes?
The compression spring equations are generally given for helical coil. What are the equivalent equations for alternative coil shapes, like oval?
-
Just to get the Nomenclature straight, the spring is still a helical spring, just the cross section can change between circular, ovate, square or trapezoidal. If the profile of the outside diameter changes along the spring, then it is called a beehive spring. – ja72 Nov 13 '12 at 0:25
I'm not referring to cross section, but coil shape. – ThePeopleWantToKnow Nov 26 '12 at 6:36
So what you are looking really is deflection of curved beam loaded perpendicular to the coil? – ja72 Nov 26 '12 at 14:28
The mechanics of all spring problems is derived from Hooke's law which models the properties of springs for small changes in length. Even for helical springs this law is a (very good) approximation. You can consider different spring geometries, different materials, larger changes in length... and this approximation will obviously worsen. Of course one can try to create more complicated models to take this matters into account, but it's generally not worth it and, almost certainly, you are never going to see different equations for spring dynamics.
-
What would you recommend as a good translation factor from a circular coil to an oval? – ThePeopleWantToKnow Nov 26 '12 at 6:38
I'm guessing that the "equations that are generally given" you refer to the ones calculating deflection of a helical spring, based on the torsional deformation of the coil, such as what is found here. The following picture is taken from there:
As you can see, any cross section of the coil is subject to shear, as well as to torsion. If your coil is not perfectly circular, it will have the same shear, but different torsion, depending on the point around the coil you are considering, $T=Fr$, with $T$ the torsion torque, $F$ the axial load, and $r$ the effective radius of the coil.
In the reference above, they compute the deformation from the total strain energy, which is made up of a shear term and a torsional term. In their case it is relatively easy, since the torsion is constant throughout the coil. You will have to integrate around the coil based on the shape of your coil. The exercises here should help in figuring out how to do that.
If you get stuck, shout in the comments, and I can guide you through the calculaitons for a particular case.
Or you can go the quick, dirty way, and copy the formulas found in this discussion.
-
I understand the issues with cross section--NASA has a good paper on the benefits of an egg shape. My problem is with coil shape. Think rectangular / magazine springs--which are typically oval in form today. – ThePeopleWantToKnow Nov 26 '12 at 6:40
So the link I gave deals with round helical springs of non-circular cross-section, did not ready it carefully enough. But my explanation should allow you to calculate the deflection for a non circular helicoid. It's the same method proposed by in ja72's answer, by the way. – Jaime Nov 27 '12 at 0:32
Start by reading the deflection of curved beams, especially the part on how to solve this with strain energy method.
http://www.codecogs.com/reference/engineering/materials/curved_beams.php
Next apply this method to combined torsion, shear and bending for your geometry. The strain energy is $$U = \int \frac{T^2}{2 G J}\,{\rm d}l + \int \frac{F^2}{2 G A}\,{\rm d}l + \int \frac{M^2}{2 E I} {\rm d}l$$ where $T$ is torque, $F$ shear force, $G$ shear modulus, $J$ polar second moment of area, $A$ area, $M$ is bending moment, $E$ modulus of elasticity, $I$ second moment of area and ${\rm d}l$ a small section of the spring.
Typically for helical spring the moment component is negligible.
Once you have the total strain energy equation as a function of the applied loading $P$ the the deflection along the load axis is $$\delta = \frac{{\rm d}U}{{\rm d}P}$$.
- | 965 | 3,962 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-30 | latest | en | 0.946097 |
https://rallycoding.com/problems/314/comments | 1,679,869,536,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00784.warc.gz | 547,056,219 | 3,600 | a year ago
``````const solve = (workers, tasks) => {
};
``````
2 years ago
``````const solve = (workers, tasks) => {
return Math.ceil(tasks.reduce((acc, cur) => acc + cur, 0) / 2);
};``````
2 years ago
Here is how I solved mine. Basically just resorting the workers array so that the most available worker always took the most work first.
const solve = (workers, tasks) => { let workerArray = new Array(workers).fill(0) let sortedTasks = tasks.sort((a,b) => a > b ? -1 : 1) while(sortedTasks.length){ workerArray[0] += sortedTasks.shift() workerArray = workerArray.sort((a,b) => a > b ? 1: -1) }
return Math.max(...workerArray) }; | 187 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-14 | latest | en | 0.774763 |
https://www.physicsforums.com/threads/two-mechanics-questions-projectile-and-inclined-plane.134191/ | 1,544,680,837,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00630.warc.gz | 1,015,368,753 | 13,419 | # Homework Help: Two Mechanics Questions: Projectile and Inclined Plane
1. Sep 30, 2006
### Thomas154321
1) Two particles, A and B, of masses m and 2m, respectively,are placed on a line of greatest slope, l, of a rough inclined plane which makes an angle of 30 degrees with the horizontal. The coefficient of friction between A and the plane is (√3)/6 and the coefficient of frictuon between B and the plane is (√3)/3. The particles are at rest with B higher up l than A and are connected by a light inextensible string which is taut. A force P is applied to B.
a) Show that the least magnitude of P for which the two particles move upwards along l is (11√3)mg/8 and give, in this case, the direction in which P acts.
B) Find the least magnitude of P for which the particles do not slip downwards along l.
http://img243.imageshack.us/img243/8022/mechanics1uy8.png [Broken]
I think it's all in the diagram except the angle of inclination of 30. The angle between P and the plane is θ.
1: Resolving parallel to the slope at B gives: Pcosθ - FB - T - 2mgsin30 = 2ma.
2: Resolving perpendicular to the plane at B gives: RB = Psinθ + 2mg cos30.
3: Resolving parallel to the slope at A gives: T - FA - mg sin30 = ma.
4: Resolving perpendicular to the plane at A gives: RA = mg cos30.
FB = (√3)/3*RB
= (√3)/3*(Psinθ + 2mg cos30).
FA = (√3)/6*RA
= mg/4.
From eq. 3: T = ma + FA + mg/2
= ma + 3mg/4.
Now I'm a bit confused - I have 8 unknowns - a, P, θ, FA, FB, RA, RB, T - and only 6 equations. For part b) I can put a=0 and F=uR, but in part a), a =/ 0. I don't know what to do next. Any help please?
2) The points A and B are 180 metres apart and lie on horizontal ground. A missile is launched from A at speed 100m/s and at an acute angle of elevation to the line AB of arcsin (3/5). A time T seconds later, an anti-missile missile is launched from B, at speed 200m/s and at an acute angle of elevation to the line BA of arcsin (4/5). The motion of both missiles takes place in the vertical plane containing A and B, and the missiles collide.
Taking g=10m/s and ignoring air resistance, find T.
I'm really not too sure on how to solve this, so I've just calculated the heights of each and said that these must be equal.
Vertically at A: a=-g, u=100*(3/5), s=yA
Using s=ut + 1/2at^2: yA = 100*(3/5)t -5t^2.
Vertically at B: a=-g, u=200*(4/5), s=yB
Using s=ut + 1/2at^2: yB = 200*(4/5)(t-T) - 5(t-T)^2
100*(3/5)t -5t^2 = 200*(4/5)(t+T) - 5(t+T)^2
=> 60t - 5t^2 = 160(t+T) - 5(t+T)^2
=> 60t - 5t^2 = 160t + 160T - 5t^2 -10tT -5T^2
=> 5T^2 + (10t-160)T -100t = 0.
This equation doesn't seem to get me anywhere. Can someone explain what I should be trying to do? Thanks.
Last edited by a moderator: May 2, 2017
2. Sep 30, 2006
### HallsofIvy
"Mechanics" is physics, not mathematics. I moving this to elementary physics.
3. Oct 1, 2006
### Thomas154321
Are you going to move it then? | 962 | 2,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-51 | latest | en | 0.922935 |
http://clay6.com/qa/12039/find-the-maximum-and-minimum-values-if-any-of-the-following-functions-given | 1,600,475,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00276.warc.gz | 29,519,558 | 6,164 | # Find the maximum and minimum values, if any, of the following functions given by?$(iv)\;f(x)=x^3+1$
$\begin{array}{1 1} Minimum=0 \\ No \;finite\;Maximum\;and\;minimum \\ Maximum =1 \\ Maximum=1\;Minimum =0 \end{array}$ | 76 | 221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-40 | latest | en | 0.389343 |
https://www.unitconverters.net/energy/btu-it-to-gram-force-meter.htm | 1,670,109,652,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00563.warc.gz | 1,100,590,804 | 3,556 | Home / Energy Conversion / Convert Btu (IT) to Gram-force Meter
# Convert Btu (IT) to Gram-force Meter
Please provide values below to convert Btu (IT) [Btu (IT), Btu] to gram-force meter [gf*m], or vice versa.
From: Btu (IT) To: gram-force meter
### Btu (IT) to Gram-force Meter Conversion Table
Btu (IT) [Btu (IT), Btu]Gram-force Meter [gf*m]
0.01 Btu (IT), Btu1075.857558547 gf*m
0.1 Btu (IT), Btu10758.57558547 gf*m
1 Btu (IT), Btu107585.7558547 gf*m
2 Btu (IT), Btu215171.51170941 gf*m
3 Btu (IT), Btu322757.26756411 gf*m
5 Btu (IT), Btu537928.77927351 gf*m
10 Btu (IT), Btu1075857.558547 gf*m
20 Btu (IT), Btu2151715.1170941 gf*m
50 Btu (IT), Btu5379287.7927351 gf*m
100 Btu (IT), Btu10758575.58547 gf*m
1000 Btu (IT), Btu107585755.8547 gf*m
### How to Convert Btu (IT) to Gram-force Meter
1 Btu (IT), Btu = 107585.7558547 gf*m
1 gf*m = 9.2949107626341E-6 Btu (IT), Btu
Example: convert 15 Btu (IT), Btu to gf*m:
15 Btu (IT), Btu = 15 × 107585.7558547 gf*m = 1613786.3378205 gf*m | 406 | 993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.432489 |
https://lfe.io/books/sicp/ch1/exercises-10.html | 1,660,816,748,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00564.warc.gz | 350,566,194 | 7,507 | ### Exercises
#### Exercise 1.34
Suppose we define the function
``````(defun f (g)
(funcall g 2))
``````
Then we have
``````> (f #'square/1)
4
> (f (lambda (z) (* z (+ z 1))))
6
``````
What happens if we (perversely) ask the interpreter to evaluate the combination `(f #'f/1)`? Explain. | 102 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.656337 |
http://clay6.com/qa/30843/the-system-shown-find-the-angular-frequency-w-oscillation-for-s-h-m- | 1,481,012,776,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00133-ip-10-31-129-80.ec2.internal.warc.gz | 54,702,721 | 27,793 | Browse Questions
Oscillations
# The system shown find the angular frequency (w) oscillation for S.H.M .
$(a)\;\sqrt{\large\frac{K_{2} K_{1}}{(K_{1}+K_{2})\;m}}\qquad(b)\;\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}\qquad(c)\;\sqrt{\large\frac{K_{2} K_{1}}{( 2K_{2}+K_{1})\;m}}\qquad(d)\;None\;of\;these$
Answer : (b) $\;\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}$
Explanation :
Let us strech the spring x
Let the spring $\;K_{2}\;$ move by $\;x_{2}\;$ and $\;K_{1}\;$ move by $\;x_{1}\;$ then the mass (m) moves $\;2x_{1}+x_{2}=x$
As the system is in equilibrium forces are same and let the force be T
So
$2T=K_{1} x_{1} \quad \; T=K_{2}x_{2}$
$x_{1}=\large\frac{2T}{K_{1}} \quad \; x_{2}=\large\frac{T}{K_{2}}$
Therefore , $2x_{1}+x_{2}=x$
$\large\frac{4T}{K_{1}}+\large\frac{T}{K_{2}}=x$
$T(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})=x$
$T=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$
$T=ma$
$ma=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$
$a=\large\frac{K_{2}K_{1}}{(4K_{2}+K_{1})}\times\large\frac{x}{m}$
$w=\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}$
edited Mar 9, 2014 by yamini.v | 546 | 1,150 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-50 | longest | en | 0.415069 |
https://www.convertunits.com/from/kiloton/to/British+thermal+unit | 1,632,231,763,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00507.warc.gz | 742,792,679 | 13,136 | ››Convert kiloton [explosive] to British thermal unit
kiloton British thermal unit
How many kiloton in 1 British thermal unit? The answer is 2.5216441204589E-10.
We assume you are converting between kiloton [explosive] and British thermal unit.
You can view more details on each measurement unit:
kiloton or British thermal unit
The SI derived unit for energy is the joule.
1 joule is equal to 2.3900573613767E-13 kiloton, or 0.00094781707774915 British thermal unit.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilotons and Btus.
Type in your own numbers in the form to convert the units!
››Quick conversion chart of kiloton to British thermal unit
1 kiloton to British thermal unit = 3965666653.3024 British thermal unit
2 kiloton to British thermal unit = 7931333306.6049 British thermal unit
3 kiloton to British thermal unit = 11896999959.907 British thermal unit
4 kiloton to British thermal unit = 15862666613.21 British thermal unit
5 kiloton to British thermal unit = 19828333266.512 British thermal unit
6 kiloton to British thermal unit = 23793999919.815 British thermal unit
7 kiloton to British thermal unit = 27759666573.117 British thermal unit
8 kiloton to British thermal unit = 31725333226.42 British thermal unit
9 kiloton to British thermal unit = 35690999879.722 British thermal unit
10 kiloton to British thermal unit = 39656666533.024 British thermal unit
››Want other units?
You can do the reverse unit conversion from British thermal unit to kiloton, or enter any two units below:
Enter two units to convert
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››Definition: Btu
The British thermal unit (BTU or Btu) is a non-metric unit of energy, used in the United States and, to a lesser extent, the UK (where it isgenerally only used for heating systems). The SI unit is the joule (J), which is used by most other countries.
››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 600 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-39 | latest | en | 0.660906 |
https://worksheets.tutorvista.com/understanding-algebraic-concepts-worksheet.html | 1,563,728,774,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00045.warc.gz | 606,036,488 | 12,121 | Understanding Algebraic Concepts Worksheet | Problems & Solutions
# Understanding Algebraic Concepts Worksheet
Understanding Algebraic Concepts Worksheet
• Page 1
1.
The income of Bryan is 2 times the income of Kristen. What is the income of Bryan if Kristen has an income of $$x$? a.$($x$ - 2) b. $($x$ ÷ 2) c.$($x$ + 2) d. \$(2 × $x$)
2.
Rashmi is $n$ years old. How old is Linda, if Rashmi is 5 years younger than Linda?
a. $n$ - 5 b. 5 - $n$ c. - $n$ - 5 d. $n$ + 5
3.
Peter is $p$ years old, Robert is 30 years elder than Peter. Which number sentence would help you to find Robert's age?
a. $p$ - 30 b. $p$ × 30 c. $p$ + 30 d. $p$ ÷ 30
4.
Nathan scored 7 less than Joe in Zoology. Find Nathan′s score, if Joe′s score is $m$.
a. $m$ - 7 b. 7 - $m$ c. $m$ × 7 d. $m$ + 7
5.
Joseph ate twice the number of biscuits that Martin ate. How many biscuits did Martin eat, if Joseph ate $x$ biscuits?
a. 2 - $x$ b. $x$ + 2 c. $x$ - 2 d. $\frac{x}{2}$
6.
Bryan had 20 pencils and he placed an equal number of pencils in 5 boxes. Which number sentence helps you find the number of pencils ($n$) he put in each box?
a. $n$ = 20 × 5 b. $n$ = 20 ÷ 5 c. $n$ = 20 + 5 d. 5 + $n$ = 20
7.
Peter distributed 75 marbles equally among 5 children. Which number sentence helps you find the number of marbles with each child?
a. 5 ÷ $x$ = 75 b. 5 + $x$ = 75 c. $x$ = 75 ÷ 5 d. 75 - 5 = $x$
8.
Francis distributed 5 bags of corn chips equally among 5 children. Which number sentence helps you find the number of bags($n$) each child would get?
a. $n$ = 5 + 5 b. $n$ = 5 - 5 c. $n$ = 5 ÷ 5 d. $n$ = 5 × 5
9.
Michael scored 39 in Botany. John scored 5 more than Michael. Which number sentence helps you find John′s score?
a. $s$ = 39 - 5 b. $s$ = 39 ÷ 5 c. $s$ = 39 × 5 d. $s$ = 39 + 5
Ashley started reading a book that contains 200 pages. She finished reading 76 pages. Which number sentence helps you find the number of pages($n$) that she needs to read to complete the book?
a. $n$ = 200 + 76 b. $n$ = 200 × 76 c. $n$ = 200 - 76 d. $n$ = 200 ÷ 76 | 758 | 2,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 48, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-30 | latest | en | 0.808009 |
http://slideplayer.com/slide/4052600/ | 1,508,576,468,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824675.67/warc/CC-MAIN-20171021081004-20171021101004-00639.warc.gz | 302,548,744 | 24,047 | Stoichiometry Chemistry Chapter 12
Presentation on theme: "Stoichiometry Chemistry Chapter 12"— Presentation transcript:
Stoichiometry Chemistry Chapter 12
"That which you persist in doing becomes easy to do - not that the nature of the thing has changed, but your power and ability to do has increased." -- H.J. Grant
Chemistry & Baking Chemistry is a lot like baking. What do you need to bake chocolate chip cookies? Recipe If you need more cookies, what are you going to do to the recipe? In this example what are the reactants & the products? A balanced chemical equation provides the same kind of quantitative information that a recipe does!
Using Balanced Chemical Equations
Chemist use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. If you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created. The calculation of chemical quantities in chemical reactions is known as stoichiometry.
Interpreting Chemical Equations
A balanced chemical equation can be interpreted in terms of different quantities, such as: # of atoms - # and types of atoms # of molecules - Coefficients of a balanced equation indicate the relative # of molecules Moles – Coefficients of a balanced equation indicate the relative # of moles Mass – can be calculated using mole / mass relationship; law of conservation of mass. Volume – can be calculate using mole / volume relationship.
From Balanced Equations to Moles
A balanced equation is essential for all calculations involving amounts of reactants & products. N2(g) + 3H2(g) -- 2NH3(g) The most important interpretation of this equation is that 1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia. Based on this interpretation you can write ratios that relate moles of reactants to moles of product.
Mole Ratios A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation N2(g) + 3H2(g) -- 2NH3(g) 3 mol of H2 2 mol of NH3 1 mol N2 3 mol of H2 2 mol of NH3 1 mol of N2
Mole ratio is a conversion factor
Using Mole Ratios In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product moles of reactants moles of products. Mole ratio is a conversion factor
Steps to Solve Stoichiometric Calculations
In a typical stoichiometric problem: Balance the equation Convert given quantity to moles (if needed) Then the mole ratio is used to convert from moles of given to moles of wanted Finally, the wanted moles are converted to specified unit of measurement, as required by the problem.
More on Cooking & Chemistry
When baking, after getting your recipe, what do you do? Check ingredients What happens if the cookie recipe calls for 4 eggs and you only have 2? Your recipe will be limited by the fact that you only have 2 eggs; the eggs are the limiting ingredients. A chemist often faces similar situations.
Limiting & Excess Reagents
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. Limiting reagents – the reagent that determines the amount of product that is formed by a reaction. (makes the least) The reaction occurs until the limiting reagent is used up! Excess reagent – the reactant that is not used up.
What do we have to do next???
Now that we have moles, what next??
What is mole ratio of Cu to Cu2S??
Knows Limiting reagent is Cu at 1.26 mol (from previous problem) 1 mol of Cu2S has a molar mass of 159.1g Next ??? What is mole ratio of Cu to Cu2S??
Now we have everything we need to calculate how much product will be formed. So what’s next ??
We should have started with 80.0g of Cu, however, we had already converted the 80.0g to moles (in previous problem).
Percent Yield When an equation is used to calculate the amount of product that will be formed during a reaction, the calculated value represents the theoretical yield or the max product that could be produced from given amounts of reactants. The amount of product actually formed during the reaction (in lab) is called the actual yield. The percent yield is a ratio of the actual to theoretical yield expressed as a percent.
Percent Yield (cont) The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.
So how are we going to solve it??
What is mole ratio of CaCO3 to CaO ??
We have everything we need so let’s solve it !
Now let’s calculate the percent yield.
We have what we need so let’s solve it ! | 1,038 | 4,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-43 | longest | en | 0.916198 |
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In conclusion, you need to make the most of the Polynomial course Assist Service make life easier on your own. As a student, you must comprehend the value of fixing such problems and use Gamma Function to aid with your issue. | 1,836 | 10,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.96485 |
https://hq.holyfiregames.com/hq/topic/d551be-na2o2-%2B-co2 | 1,621,135,129,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00274.warc.gz | 336,131,313 | 5,930 | Substitute immutable groups in chemical compounds to avoid ambiguity. 2Na2O2(s) + 2CO2(g) →2Na2CO3(s) + O2(g) What mass (in grams) of Na2O2 is necessary to consume 1.00 L CO2 at STP? The answer will appear below, Always use the upper case for the first character in the element name and the lower case for the second character. 2Na 2 O 2 + 2CO 2 → 2Na 2 CO 3 + O 2 [ Check the balance ] Hydrogen peroxide react with carbon dioxide to produce sodium carbonate and oxygen. Сoding to search: 2 Na2O2 + 2 H2O = 4 NaOH + O2. Please register to post comments. If you do not know what products are enter reagents only and click 'Balance'. Error: equation CO2+Na2O2=Na2CO3 is an impossible reaction Please correct your reaction or click on one of the suggestions below: CO2 + Na2O2 = Na2CO3 + O2 Instructions and examples below may help to solve this problem You can always ask for help in the forum By using this website, you signify your acceptance of, calcium hydroxide + carbon dioxide = calcium carbonate + water, Enter an equation of a chemical reaction and click 'Balance'. Our channel. Find an answer to your question Read the given equation. Na2O2 + CO2 -> O2 + Na2CO3 R = 0.083145 L bar K-1 mol-1 I got 1.57L Chemistry. Compound states [like (s) (aq) or (g)] are not required. This particular compound is sodium peroxide.. You're right that usually oxygen has a charge of -2, but in this case, there's no way that each \$\ce{Na}\$ can have an oxidation state of +2.. Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. Enter either the number of moles or weight for one of the compounds to compute the rest. It Works By Reacting With CO2 In The Air To Produce Sodium Carbonate (Na2CO3) And O2. Become a Patron! Sodium peroxide is hydrolyzed by water to form sodium hydroxide plus hydrogen peroxide according to the reaction: Na2O2 + 2 H2O → 2 NaOH + H2O2. This reaction takes place at a temperature of 450-550°C. You place a piece of paper on a balance and find that its mass is 0.63 g. Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it reacts with exhaled CO2 to form Na2CO3 and O2. In many cases a … If you do not know what products are enter reagents only and click 'Balance'. Reaction stoichiometry could be computed for a balanced equation. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide, To enter an electron into a chemical equation use {-} or e. To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}. Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. 2 Na2O2 (s) + 2 CO2 (g) →2 Na2CO3 (s) + O2 (g) What Volume (in Liters) Of CO2 Can Be Consumed At STP By 695 G Na2O2? If you do not know what products are enter reagents only and click 'Balance'. What is the volume of the oxygen produced when 1.00L carbon dioxide reacts with 20.0 g of Na2O2 based on the following unbalanced reaction at 25C and 2.00 bar? Sodium oxide react with carbon dioxide. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide, To enter an electron into a chemical equation use {-} or e. To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}. Na2O2 + CO2 -> O2 + Na2CO3 R = 0.083145 L bar K-1 mol-1 I got 1.57L . In many cases a complete equation will be suggested. Sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide according to the reaction In many cases a … Na 2 O + CO 2 Na 2 CO 3 [ Check the balance ] Sodium oxide react with carbon dioxide to produce sodium carbonate. It works by reacting with CO2 in the air to produce sodium carbonate (Na2CO3) and O2. If you do not know what products are enter reagents only and click 'Balance'. Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. So you then work backwards, deciding if it's \$\ce{Na+}\$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom. Add / Edited: 28.09.2014 / Evaluation of information: 5.0 out of 5 / number of votes: 1. Sodium peroxide, Na2O2, is the normal product when sodium is burned. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. Na2O2 + CO2 → Na2CO3 + O2 What volume of O2 gas is produced from 4.60 liters of CO2, assuming all value… The answer will appear below, Always use the upper case for the first character in the element name and the lower case for the second character. In many cases a complete equation will be suggested. By using this website, you signify your acceptance of, Instructions and examples below may help to solve this problem, calcium hydroxide + carbon dioxide = calcium carbonate + water, Enter an equation of a chemical reaction and click 'Balance'. What is the volume of the oxygen produced when 1.00L carbon dioxide reacts with 20.0 g of Na2O2 based on the following unbalanced reaction at 25C and 2.00 bar? Enter either the number of moles or weight for one of the compounds to compute the rest. Reaction stoichiometry could be computed for a balanced equation. Find another reaction. Substitute immutable groups in chemical compounds to avoid ambiguity. ChemiDay you always could choose go nuts or keep calm with us or without. Compound states [like (s) (aq) or (g)] are not required. In this video we'll balance the equation Na2CO3 = Na2O + CO2 and provide the correct coefficients for each compound. Sodium Peroxide (Na2O2) Is Used To Remove Carbon Dioxide From (and Add Oxygen To) The Air Supply In Spacecrafts. The reaction proceeds at room temperature. How many liters of respired air can react with 88.5 g Na2O2 if each liter of respired air contains 0.0720 g CO2? It is a strong oxidizer. the reaction of sodium peroxide Na2O2 with CO2 is used in space vehicles to remove CO2 from the air and generate O2 for breathing: 2Na2O2 + 2CO2 ----> 2Na2CO3+O2 a)assuming that air is breathed at an average rate of 4.50L/min (25C and 735 mmHg) and that the concentration of CO2 in expelled air is 3.4% by volume, how many grams of CO2 are produced in 24hr? Oxygen can take multiple oxidation states. | 1,826 | 6,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.865734 |
https://www.vanessabenedict.com/how-many-oz-is-a-gram/ | 1,725,933,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00046.warc.gz | 972,389,691 | 14,272 | # How much does 1 oz weigh in grams?
#### ByVanessa
Jun 9, 2022
Untitled Document
## How much does 1 oz weigh in grams
In fact, 1 ounce equals approximately 28.35 grams.
## Is a gram bigger than 1 8 oz
One ounce equals 28.3495 g. A one ounce cannabis purchase usually has an actual weight of around 28 ounces, and sometimes even. Similarly, buying on the 9th is not a true 1/8 ounce (3.G) 54688, but rather a mathematically more convenient weight of 3.5g.
Untitled Document
## What is the difference between Gram positive and Gram negative organisms when referring to Gram staining ie what makes Gram positive purple and Gram negative pink
Cells with a thick movable or mobile wall appear blue (Gram positive) because crystal violet is retained in the exact cells and hence the red dye is not visible. Cells that have a thin cell wall and are therefore discolored appear red (gram negative).
## Which is are true regarding features of PESA Act 1996 1 Gram Sabha shall identify beneficiaries under poverty alleviation programs 2 the recommendations of the Gram Sabha is mandatory prior to grant of prospecting license for minor minerals 3 Gram Sabha
1) The Gram Sabha should no more define beneficiaries than poverty reduction programs. 2) Gram Is Sabha concepts are required before approving exploration for smaller minerals. 4) Every panchayat at the village level must obtain a certificate from the Gram Sabha regarding the use of the funds.
## How does the Gram staining procedure differentiate between gram negative and Gram-positive bacteria quizlet
Gram-positive microorganisms have a lot of peptidoglycan in their cell wall, which allows them to retain the crystal violet dye, so they begin to stain blue-violet. Gram-negative bacteria require less peptidoglycan in their stable vertical cell structure, so the crystals cannot retain their purple color, so they turn red-pink.
## How does the Gram staining procedure differentiate between Gram negative and gram positive bacteria
Currently, Gram-positive bacteria have cell walls containing thick layers of peptidoglycan (90% of the wall). They turn purple. Gram-negative bacteria retain cell walls with thin cell peptidoglycan (10% wall) and high fat content. They turn pink.
Untitled Document | 493 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.894472 |
http://www.studymode.com/essays/Unity-Is-Strenght-1443165.html | 1,529,666,732,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00437.warc.gz | 507,145,099 | 25,398 | # Unity Is Strenght
Topics: Likert scale, Guttman scale, Thurstone scale Pages: 4 (988 words) Published: February 21, 2013
A Guttman scale presents a number of items to which the person is requested to agree or not agree. This is typically done in a 'Yes/No' dichotomous format. It is also possible to use a Likert scale, although this is less commonly used.Questions in a Guttman scale gradually increase in specificity. The intent of the scale is that the person will agree with all statements up to a point and then will stop agreeing.The scale may be used to determine how extreme a view is, with successive statements showing increasingly extremist positions.If needed, the escalation can be concealed by using intermediate questions. Concealed example (hardening attitude towards crime), using Likert scale: | Strongly
agree| Tend to
agree| Neither
agree
nor
disagree| Tend to
disagree| Strongly
disagree|
Criminals should be punished| [ ]| [ ]| [ ]| [ ]| [ ]| Litter is a problem in the street| [ ]| [ ]| [ ]| [ ]| [ ]| Sentences for many crimes should be longer| [ ]| [ ]| [ ]| [ ]| [ ]| Streets in this town are not well lit| [ ]| [ ]| [ ]| [ ]| [ ]| More criminals deserve the death penalty| [ ]| [ ]| [ ]| [ ]| [ ]| Question selection
Generate a list of possible statements
Get a set of judges to score the statements with a Yes or No, depending on whether they agree or disagree with them. Draw up a table with the respondent in rows and statements in columns, showing whether they answered Yes or No. Sort the columns so the statement with the most Yes's is on the left. Sort the rows so the respondent with the most Yes's is at the top. Select a set of questions that have the least set of 'holes' (No's between 'Yes's).
A Thurstone scale has a number of statements to which the respondent is asked to agree or disagree. There are three types of scale that Thurstone described:
* Equal-appearing intervals method
* Successive intervals method
* Paired comparisons method
| Agree| Disagree|
I like going...
Please join StudyMode to read the full document | 544 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-26 | latest | en | 0.901131 |
https://www.developerfaqs.com/6323/if-statement-using-io-int-haskell | 1,553,205,345,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.7/warc/CC-MAIN-20190321213516-20190321235516-00049.warc.gz | 754,646,208 | 10,288 | # If statement using IO Int haskell
• A+
Category：Languages
I have a game , user vs computer and I want to randomly choose who starts the game. I have
``a = getStdRandom \$ randomR (0, 1) ``
This gets a random number 0 or 1. However it is a `IO Int`, so I can't have an if statement comparing it to a number like
``if a == 0 then userStarts else computerStarts ``
I have tried to compare `IO Int` with `IO Int` and it doesn't work, and I have also tried
Converting IO Int to Int
I am very new to Haskell, not sure how to approach this. Code details requested:
``randomNumber = getStdRandom \$ randomR (0, length symbols - 5) -- this will be 0 or 1 randomNumber2 = getStdRandom \$ randomR (0, length symbols - 5) -- according to -- the solution I need another function returning IO int. a = do x <- randomNumber randomNumber2 \$ pureFunction x ``
Error I get:
``• Couldn't match expected type ‘t0 -> IO b with actual type ‘IO Int’ • The first argument of (\$) takes one argument, but its type ‘IO Int’ has none In a stmt of a 'do' block: randomNumber2 \$ pureFunction x In the expression: do x <- randomNumber randomNumber2 \$ pureFunction x • Relevant bindings include a :: IO b (bound at Path:87:1) randomNumber2 \$ pureFunction x Path:89:20: error: Variable not in scope: pureFunction :: Int -> t0 randomNumber2 \$ pureFunction x ``
When you say `a = getStdRandom \$ randomR (0,1)` you are saying "let a be the action of getting a random value between 0 and 1". What you want is within some function's do block `a <- getStdRandom \$ randomR (0,1)` which is "let a be the result of running the action of getting a random value between 0 and 1".
As such:
``import System.Random main :: IO () main = do a <- getStdRandom \$ randomR (0, 1 :: Int) if a == 0 then userStarts else computerStarts -- Placeholders for completeness userStarts, computerStarts :: IO () userStarts = putStrLn "user" computerStarts = putStrLn "computer" ``
N.B. I specified the `1` is an int or else the compiler won't know if you want a random int, int64, double, float, or something else entirely.
EDIT: @monocell makes a good point that generating an int in a range just to get a boolean is somewhat indirect. You can just directly generate a boolean result and this requires no range:
`` a <- getStdRandom random if a then userStarts else computerStarts `` | 644 | 2,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | longest | en | 0.769334 |
http://thefiringline.com/forums/showpost.php?p=5249351&postcount=21 | 1,394,516,442,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011131391/warc/CC-MAIN-20140305091851-00015-ip-10-183-142-35.ec2.internal.warc.gz | 175,395,590 | 5,693 | View Single Post
October 10, 2012, 06:02 AM #21 HiBC Senior Member Join Date: November 13, 2006 Posts: 3,366 You are getting good info. If you divide the sight radius into the range you are shooting you will get a number that is the multiplier of any sight adjustment you make.Of course,you must convert to the same units..I'll give you an example(I do not have my calculator handy,so I'll use simple numbers) Suppose you have an 18 in sight radius.We can call that 0.5 yd.If you are shooting at 50 yds,0.5 yds goes into 50 100 times.So,with these numbers,if you need to move the group 5 in at 50 yds,it would be 0.050 at the sight. I'm telling you this so you can have an idea how much to move it.The brass punch and tapping hammer are correct.Of course,you do the tapping on the dovetail part of the sight,not the leaf spring the sight blade is on. Before you start whacking,however,I would consider this.Those dog leg stock older guns with lower velocity rounds are a bit "dynamic" when fired.Recoil is starting before the bullet leaves the barrel. So,as has been suggested,a benchrest group point of impact may not be the same as the sort of shot you might typically take with a 30-30. You migfht,for fun,see how it shoots standing on your hind legs at 50 yds.Shoot 10 real careful and find the center of the group.See if it is still off that much. Its not real easy to get a good spot weld on that rifle on a bench.You are also dealing with a slow lock time,and open vs peep sights.Try sight black on your gold bead,focus on the front sight but really concentrate on a perfect sight picture as you squeeze,then follow through and call the shots. If the barrel is in good shape,that old 30-30 will likely give "minute of battle rifle" 3 inch groups or so at 100 yds. IMO,they are about a 150 yd rifle. Be a little patient.Its not an M-4.Have fun!
Page generated in 0.04403 seconds with 7 queries | 488 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2014-10 | longest | en | 0.949115 |
https://www.mathgrapher.com/representation-of-a-square-wave/ | 1,721,288,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00876.warc.gz | 751,014,596 | 8,955 | Iterations
Calculate algebraic series such as e = 1+ 1/2! + 1/3! + …, a square wave, Fibonacci numbers. Study iterative maps, e.g. the (one-dimensional) Logistic map: (see below) or more complicated multi-dimensional maps. The logistic map is perhaps one of the simplest mathematical system showing many characteristics of the development of chaotic behavior.
Several analytical tools are valuable to study the results of Iterations and ODE’s such as:
Time series | Power spectra | 2D projections | Fixed points | Lyapunov exponents
A special window has been added in Mathgrapher v2 to allow detailed presentation of 2D orbits at the pixel level and to study the stability of the orbits (see the Examples and Demonstrations for the Henon map, Standard map and Mandelbrot and Julia sets.
Henon map: Definition | 2D orbit | Region of Stability
Mandelbrot and Julia sets: Definition | Mandelbrot: vary parameters | Julia: vary initial conditions
Series representation of square wave
Define F1=F2 and F2=F1+sin((2n+1)x)/(2n-1) with initial values 0.
Vary one parameter in the Iteration type panel. Vary x and do 50 iterations (N=50) for 200 values of x ranging from -5 to 15. This yields the following graph (this is also shown in one of the demonstrations) | 308 | 1,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.845741 |
https://www.salon-beautycare.de/cone-crusher/3020-calculations.html | 1,638,329,875,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00330.warc.gz | 1,035,381,147 | 8,568 | Cone Crusher Calculations Mill Chargecalculations Of Cone Crusher
# Calculations Mill Chargecalculations Of Cone Crusher
### Technical Notes 5 Crushers Mineral Tech
The chief difference between cone and gyratory or jaw crushers is the nearly parallel arrangement of the mantle and the cone at the discharge end in the cone crusher this is illustrated in figure 52 reduction ratios in the following ranges are common for cone crushers 61 81 for secondaries 41 61 for tertiary and quaternary crushing
### Design And Analysis Of Impact Crushers
Mechanism usedbased on the mechanism used crushers are of three types namely cone crusherjaw crusher and impact crusher our objective is to design various components of an impact crusher like drive mechanism shaft rotor hammers casing feed and discharge mechanism which will be useful in minimizing weight cost and maximizing the capacity
### Dynamic Modeling And Simulation Of Sag Mill Circuits
Agsag mill is inefficient in grinding particles of a certain size typically in the range of 2555 mm ie pebbles therefore cone crushers are often used as pebble crushers and integrated into agsag mill circuits to break the critical size particles that accumulate in the mill and to increase the performance of the primary grinding circuits
### Crusher No Load Power Calculation
Cone crusher power calculation crusher machineor crushing machineincludes jaw crushercone crusher circulating load in roll mill calculation basalt crusher ball mill circulating load ball mill calculation of power draw of dry multicopartment ball mills the size and ball read more
### Vibrating Screen Efficiency Calculation Theory
Vibrating screen efficiency calculation theory protable plant ball mill efficiency calculations crusher machine ball mill charge calculations stone crusher machine mill charge load calculation stone crushing machine warman centrifugal slurry pumps ndash excellent minerals solutions calculation of the systemvibrating screen efficiency basically
### Calculation Of Capacity Of Cone Crusher Binq Mining
Mar 04 2013 raymond mill capacity calculations crusher mills cone crusher 22003000 hot sale high capacity airswept coal ball mill 22003000 hot sale high capacity airswept coal ball mill calculation in aluminium industry by luoyang more detailed
### Design Of A Crushing System That Improves The Crushing
Fine crushing ex gravity stamp mill following are the crushers under the scope of our syllabus jaw crushers gyratory crushers cone crushers 251 gyratory crusher gyratory crushers are the most efficient primary crushers for dealing with blasted hard rock in ore and natural stone mining
### Reduction Ratio Calculations Of Briquette Machine
Crusher reduction ratio formula tembaletu trust reduction ratio formula for crusher za review on design and analysis of jaw plate of jawcrusher ijirse a jaw crusher is a type of size reduction machine which is widely used in for crusher is 254 mm width of jaw is w 508 mm reduction ratio 51 minimum 45 mm from rose and english formula
### Grinding Equations In Roller Mills Pdf 4ft Standard Cone
Grinding equations in roller mills pdf 4ft standard cone ball mill capacity calculation 4ft standard cone crusher we are a largescale manufacturer specializing in producing various mining machines including different types of sand and gravel equipment milling equipment mineral processing equipment and building materials equipment
### Ball Mill Charge Calculations Vyspolecz
Mill steel charge volume calculation ball mills have been the traditional method of comminution in the mineral ball mill charge grinding classifiion circuits metallurgist jun 17 2018 the crop load of ball mill it ore mill medium water but usually it was used in a percentage formation so how to calculate this percentage
### Critical Speed Calculation Formula Of Ball Mill
How to calculate critical speed of a ball mill cone crusher 4 days ago ball mill rpm calculation 3 john gregorysteel grinding balls this formula calculates the critical speed of any ball mill calculations for mill motor power mill speed and media charge calculations for mill motor power mill speed and media charge advantages
### Crusher Size Reduction Ratio Calculation Method
The vertical shaft impact crusher size reduction ratio is 48 to 1 the vertical roller mill size reduction ratio is 225 to 1 the hammer crusher size reduction ratio is 20 to 1 pre articlecrusher plays an important role in artificial sand production line next articledevelopment of the worlds cone crusher industry related machine
### How To Calculate Sag Mill Ball Charge
Mar 01 2021 sag is an acronym for semiautogenous grindingsag mills are autogenous mills but use grinding balls like a ball mill a sag mill is usually a primary or first stage grinder technical notes 8 grinding r p king 83 centrifugal force outward fc mp 2 dm 2 81 is the angular velocity mp is the mass of any particle media or charge in the mill
### Estimate Jaw Crusher Capacity 911 Metallurgist
Feb 17 2016 metallurgical contentcapacities and horsepower of jaw crusher tonshrcapacities of a gyratory crushers tonshrtypical capacities of twinroll crushers tonshrtypical capacities of cone crusherstypical capacities of hammermills example capacity calculation of a 10 x 20 250 mm x 500 mm pp 2800 28 sg e 02 halfway between dolomite and sandstone a 2501000 x
### O I Skarin N O Tikhonov Calculation Of The
Drum shell and cone is presented the mathematical procedure is based on the calculation of the overall potential and kinetic energy of the ball charge during its motion inside the rotating mill the concluding section of the article offers empirical formulas to calculate idling and aggregate capacity of the mill
### Comminution Calculationspdf Find The Bulk Density Of
Ore cone crusher rod mill ball mill product 4 101 600 m p80 12700 m p80 20 mesh 841 m p80 140 mesh 105 m answers p 1 7167 kw
### Construction Working And Maintenance Of Crushers
Generally gyratory crushers jaw crushers high speed double roll crushers low speed sizers impactors and hammer mills are used as primary crushers in the secondary and subsequent stages the material is further reduced and refined for proper size and shape mostly based on specifications most secondary crushers are cone crushers and
### Design And Analysis Of A Horizontal Shaft
Process used in cone or jaw crushers 7 an impact crusher can be further classified as horizontal impact crusher hsi and vertical shaft impact crusher vsi based on the type of arrangement of the impact rotor and shaft horizontal shaft impact crusher these break rock by impacting the rock with hammersblow bars that are fixed upon the outer
### Determine Charge Volume In A Ball Mill
Mill steel charge volume calculation mar 19 2017 we can calculate the steel charge volume of a ball or rod mill and express it as the of the volume within the liners that is filled with grinding media while the mill is stopped the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill
### Ball Mill Charge Calculations Mining
Calculation of ball mill charge volume mtm calculate ball mill charge how to calculate for ball mill media charge to calculate grinding media charge for continuous type ball mill m 0000676 x d2 x l example to calculate grinding media charge for a 180 cm dia x 180 cm long batch type ball mill with duralox 50 mm thick bricks formula to be used is read the rest ball charges calculators
### Eccentric Shaft An Overview Sciencedirect Topics
The cone crusher is a modified gyratory crusher the essential difference is that the shorter spindle of the cone crusher is not suspended as in the gyratory but is supported in a curved universal bearing below the gyratory head or cone figure 82 power is transmitted from the source to the countershaft to a vbelt or direct drive
### How To Calculate Capacity Of Hammer Millindustry News
Hammer mill is suitable for crushing various materials of brittleness such as coal gangue coke slay shale and loose lime stone etc whose compressive resistance does not exceed 10mpa and the surface moisture content should not be greater than 8 how to calculate capacity of hammer mill different types the capacity is also different there are three models of dbm hammer mill pc400875
### Bauxite Ore Crusher Machine For Sale
Bauxite ore crusher wholesale crushers suppliers alibabacom offers 451 bauxite ore crusher products about 68 of these are crusher 1 are other mining machines a wide variety of bauxite ore crusher options are available to you such as cone crusher jaw crusher and roller crusher
### Amazoncom Freetophome Herb Grinder Electric Spice
This item freetophome herb grinderelectric spice crushersmall usb recharge grinderssmart control millmini automatic portable cranklarge bladealuminiumblack836inch cigamate herb grinder spice grinder rainbow grinder hand cranked clear top grinder 25 inch 4 pieces herb grinder with drawer 248 x 311 inch
### About The Mine And Mill Equipment Cost Calculator
If you value your time if you value the quality of your estimates if you want to make your job easier the cost calculator will work for you once you have determined your mine or mill equipment requirements the rest is easy in a matter of minutes you can determine capital and hourly operating costs for the equipment needed for your project
### Reduction Ratio Calculations Of Briquette Machine
Crusher reduction ratio formula tembaletu trust reduction ratio formula for crusher za review on design and analysis of jaw plate of jawcrusher ijirse a jaw crusher is a type of size reduction machine which is widely used in for crusher is 254 mm width of jaw is w 508 mm reduction ratio 51 minimum 45 mm from rose and english formula
### Calculation Of Ball Mill Charge Volume
Ball mill charge sensor logan sainlez mill steel charge volume calculation we can calculate the steel charge volume of a ball or rod mill and express it as the of the volume within the liners that is filled with grinding media while the mill is stopped the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top
### Pdf New Developments In Cone Crusher Performance
New developments in cone crusher performance optimization crusher design evolutioncompression cone crusher designs today have evolved from the simple cone crusher first developed in the mid1920s by edgar b symons to the modern high performance crushers early crushers used springs for tramp iron protection and were manually adjusted
### Technical Notes 5 Crushers Mineral Tech
The chief difference between cone and gyratory or jaw crushers is the nearly parallel arrangement of the mantle and the cone at the discharge end in the cone crusher this is illustrated in figure 52 reduction ratios in the following ranges are common for cone crushers 61 81 for secondaries 41 61 for tertiary and quaternary crushing
### Comminution Calculationspdf Find The Bulk Density Of
Ore cone crusher rod mill ball mill product 4 101 600 m p80 12700 m p80 20 mesh 841 m p80 140 mesh 105 m answers p 1 7167 kw
### Jaw Crusher Force Calculation Grinding Mill China
Posted atjune 26 2013 47 2790 ratings allis chalmers cone crusher for sale stone requirement in india learn more calculation of impact force of impact crusher crusher usa about calculation of impact force of impact crusherrelated informationhealth care budget deficit calculator the u s health care system is possibly the most
### Design Of A Crushing System That Improves The Crushing
Fine crushing ex gravity stamp mill following are the crushers under the scope of our syllabus jaw crushers gyratory crushers cone crushers 251 gyratory crusher gyratory crushers are the most efficient primary crushers for dealing with blasted hard rock in ore and natural stone mining
### Dynamic Modeling And Simulation Of Sag Mill Circuits
Agsag mill is inefficient in grinding particles of a certain size typically in the range of 2555 mm ie pebbles therefore cone crushers are often used as pebble crushers and integrated into agsag mill circuits to break the critical size particles that accumulate in the mill and to increase the performance of the primary grinding circuits
### Construction Working And Maintenance Of Crushers
Generally gyratory crushers jaw crushers high speed double roll crushers low speed sizers impactors and hammer mills are used as primary crushers in the secondary and subsequent stages the material is further reduced and refined for proper size and shape mostly based on specifications most secondary crushers are cone crushers and
### Cement Grinding Tube Mill Significance Of Jaw Crusher
Grinding efficiency crusher mills cone crusher jaw ball mill grinding efficiency can be increased by making three changes in operating practice ball mill media charge calculation pdf nip angle of jaw crusher meaning advanced grinding mill used in place of vertical roller mill method of crusher raw material for cement
### Derive The Expression Of Critical Speed For A Standard
Derive the expression of critical speed for a standard ball mill products as a leading global manufacturer of crushing grinding and mining equipments we offer advanced reasonable solutions for any sizereduction requirements including derive the expression of critical speed for a standard ball mill quarry aggregate and different kinds of minerals
### Energy And Mass Balance Crusher In Mumbai Sale
Mass balance for crusher in cement plant mass balance for crusher in cement plant mass balance of a kiln system cement co2 protocol the following diagram illustrates an example of the mass flows in a cement plant and the mass balance of a kiln system from raw meal rm to clinker get price energy and mass balance ball mill in mumbai
### A Dry Grinding Ball Mill In Closed Circuit Is To B
A dry grinding ball mill in closed circuit is to be fed with 100 metric tonneshour of a material with a work index of 15 kwhhr and a size distribution of 80 passing 08cm the feed was prepared in a cone crusher the required product size is 80 passing 60um calculate the power requirement
### Calculation Ball Crusher
Crusher no load power calculation determining circulating load around 2 screens circulating load formula in ball millcrusher circulating load calculationsbimorgin here is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone asball mill load calculation youtubemay 23 2018 ball mill circulating load calculation pakistan crusher
### Ball Mill Charge Calculation Per Size
Ballmillfinish mill ball charge calculationget price mill grinding wikipedia w is the work index measured in a laboratory ball mill kilowatthours per metric or short ton p 80 is the mill
### Concrete Geith Concrete Crusher For Sale
A concrete crusher for example can crush rocks and mix them with concrete and asphalt for roadside construction projects whether you need a used rock crusher for sale a concrete crusher or something for iron ore sandstone and other similar materials the
### Ball Mill Designpower Calculation
Dec 12 2016 ball mill power calculation example a wet grinding ball mill in closed circuit is to be fed 100 tph of a material with a work index of 15 and a size
### Crusher Efficiency Calculations 911 Metallurgist
Sep 15 2014 the 1in screen is a second deck for the 38 tph from the jaw crusher so the deck correction factor is 090 and that screen capacity is 21 x 09 189 tphsq ft the screen area needed under the jaw crusher is 38189 201 sq ft
### Impact Crushers Design And Calculations
Impact crushers design and calculations impact crusher in the production process the main parameters of the equipment has an important significance for normal operation generally includes three basic parameters the rotor speed productivity and motor power so for these impact crusher parameter selection and calculation method of everybody
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Jun 19 2015 the basic parameters used in ball mill design power calculations rod mill or any tumbling mill sizing are material to be ground characteristics bond work index bulk density specific density desired mill tonnage capacity dtph operating solids or pulp density feed size as f80 and maximum chunk size product size as p80 and maximum and finally the type of circuit openclosed | 3,644 | 18,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.821333 |
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Topic: Simple analytical properties of n/d
Replies: 20 Last Post: Mar 11, 2013 11:01 PM
Messages: [ Previous | Next ]
William Elliot Posts: 2,637 Registered: 1/8/12
Re: Simple analytical properties of n/d
Posted: Mar 3, 2013 10:22 PM
On Sun, 3 Mar 2013, Ross A. Finlayson wrote:
> Survey: does anybody find that:
> lim_d->oo lim_n->d n/d = 1
>
> It's clear that it does, for all values of d e N.
>
> Then, as a function f = n/d from N to R[0,1], d e N, n <= d E N, is it
> not constant monotone increasing?
Is that f(d) = n/d or f(n) = n/d?
What's deN and dEN?
> If not increasing, how is lim_n->d n/d = 1?
lim(x->d) x/d = 1, because since f(x) = x/d is continuous,
lim(x->d) x/d = d/d = 1
For integer variables n, lim(n->d) f(n) is meaningless.
Give a definition for it. Wouldn't it be the same as f(d)? | 331 | 1,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-22 | latest | en | 0.897082 |
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## Description
In this new ground-breaking book, the author has uncovered the mathematical formula that defines how easily The Force can be used in every day life right now, not a long time ago and right here on Earth and not in some galaxy far, far away. Those of us who may want to use The Force for physical violence will be greatly disappointed, however, since The Force that created the universe and all things within it is not capable of being abused by mere mortals and there is no Dark Side of The Force. There are people who are on the Dark Side of something, to be sure, but it is not The Force. The book details how the Force Equation is as simple to understand as Einstein's famous equation of E = mc2. With Einstein's simple formula of how matter is converted into energy, we were able to construct the Atomic bomb. In the case of The Force Equation we learn how to convert Time into a greater power than mankind has ever known or will ever know and which shall make all atomic power pale in comparison. When we learn how to do this, we are able to achieve things that we would never be able to without the use of The Force. Most of us, luckily experience moments of clarity and even accomplish great things at random times in our lives, however, we're oblivious to the true nature of The Force that can be used to do this more and more often, no longer relying on dumb luck. For today, there is a mathematical equation that makes it possible to use The Force at will. In this ground-breaking new book you will learn how E = mc2 has now morphed into F = tc3 . . . where 'F' is The Force. 'T' is Time and 'c3' is the speed of light cubed. This amounts to an almost infinite amount of power that is always available - just under the radar - to each and every thinking individual on planet Earth today. More importantly reader will learn how to apply this formula in their every day life. This information will instruct you on how to completely live in the moment because in each and every moment of time, even as little as a microsecond, as the Force Equation shows, there lies hidden enough power to completely change the world any of us are experiencing or even the total experience that we enjoy collectively on this planet. Of course, it is not done with laser swords, jedi-knights, fancy space-ships, or legions of clone warriors. It's all done in much more subtle ways. And, it can be done in much more spiritual ways. And, it can be done in many systemic ways. This book is the blueprint for all of them in the order of importance to each and every one of us living today. If you were as intrigued by the Star Wars movies as I was but wanted to know more about The Force - in this book you will learn the actual laws of Physics that relate to the magical and mystical Force as used by the Jedi Knights of this most famous saga to defeat the Empire. The book shows in plain and simple language The Force and how it relates to all other known Laws of Physics. We're so convinced that this is the most important scientific breakthrough of all time that the publishers want to give any reader who will leave a REVIEW on Amazon or Audible with their impressions of this book, all of the following free benefits that can save the average family hundreds of dollars, if not thousands over time. The details on how to have them sent to you are in the final chapter of the book. A Free Grocery Club Membership where you get Free Grocery Coupons every month - (Total Value over \$1,000 Per Year) Free Restaurant Meals - (Total Value over \$2,000 Per Year) Free Travel Discount Club - saves hundreds over all other Online Travel discounts (My family saved over \$1,000 last year using this plan.) Free Discount Health Care Membership - (Everything including Dental included and saves the average family over \$1,000 Per Year in Dental alone. And - Free TV (No Cable charges) - over 1,000 Channels. All the latest movies and TV Shows, music, video games etc. - alshow more
## Product details
• Paperback | 74 pages
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• 1508442797
• 9781508442790 | 948 | 4,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-39 | latest | en | 0.959866 |
klobouk.fsv.cvut.cz | 1,432,905,604,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930109.71/warc/CC-MAIN-20150521113210-00322-ip-10-180-206-219.ec2.internal.warc.gz | 141,866,337 | 13,655 | PEXESO library
PEXESO is the C++ Library of Optimizing Methods, created as a marginal result of about one year research about this topics at the Faculty of Civil Engineering of the Czech Technical University in Prague. It is intended to a Linux environment and to be compiled and used as a Shared Library (libpexeso.so).
Index of this document:
1. What is that - the Optimization Methods
Generally, Optimization is a task that has finding the best value of some function as its aim. For example, let's have a function
f(x)=x^2-5x+8.
Than, let's find a minimum of this function. We can write down a derivative of this function:
f'(x)=2x-5,
and look, where this derivative is equal to zero. Obviously, f'(x)=0 for x=2.5. Now, let's evaluate our function in the point x=2.5: we get a value 1.75, which is the global minimum of our function. This task we've just solved is calling 'minimization' which is a special case of 'optimization' (because 'optimization' could also mean 'maximization'.
The example we've got above is very simple. But the real-world tasks where there is some interest to find the global minimum or maximum of some function, can be very complicated. The first difficulty that can be met is that the function we like to optimize has not an analytic expression or it is too complicated to be derivated. Another thing is, if the function is not continuous or have more than one or two parameters (I mean these 'x').
For the more complicated problems, numeric methods are usually used. These methods can be divided into several groups:
gradient methods, simulated annealing, multilayered global optimalizators, evolutionary computation.
Of course, this document cannot deal with all these methods in detail. I think the best place where You can obtain lots of information about the Global Optimization is this site: http://solon.cma.univie.ac.at/~neum/glopt.html. Also, You can buy some books in the bookstore.
`2. Idea of the PEXESO library
PEXESO library provides functionality that covers several types of already discovered optimization methods of the evolutionary type. The main idea is, that the one, who tries to solve some optimization problem, is able to compose and tune all the evolutinary algorithm by himself. There is a possibility to choose among several types of selection strategies ('steady state', 'trial vectors', 'tournament selection' or 'total offspring') and compose own combination of evolutionary recombination operators ('simplified differential', 'hard mutation, 'local mutation' and a set of Rainer Storn's differential operators). Another selection strategies and another recombination operators will be added in the near future. Seems that now this library is able to compose evolution strategies, differential evolution, differential genetic algorithms and many other combinations of operators and selection strategies that don't have a name until now.
Except the possibilty of composing the evolutionary method by yourself, the one can also choose from three pre-composed optimization strategies:
PEXESO is a C++ library. This approach, whatever unpleasant for many developers, is considered by me as the most simple, after all. There are two main classes, that model the optimization task and the optimization method.
Class px_task models the optimized task in global manner: it contents the optimized function itself, the domain on which the function is defined as long as the stopping criteria and the way how to display information about the progress of the computation and the results. The one who wants to optimize his own task by the PEXESO library, helas, must inherit its own class 'my_optimization_task' from this one, something like:
and so on. There are five methods that must be written each time, and many other, that can be left as they are or redefined if necessary.
On the other hand, class px_strategy stands as an ancestor of any selection strategy and also holds pointers to the genetic (recombination) operators. There is no need to inherit anything in this part of using PEXESO, only the developer must choose between several particular classes that are inherited from px_strategy:
the easiest way is to use a function px_create_any, for example:
px_strategy *s=px_create_any( "Tournament",10,t ) ;
(parameters will be discussed later). Any of these strategies than should be filled by (up to 8) recombination operators using the functions:
px_strategy::set_main_operator(...); px_strategy::set_marginal_operator(...);
The idea is that we've got one operator which is called main (it must be present everytime) and several operators that are optional. The optional (marginal) operators have their 'probabilities'. When the algorithm creates a new generation, for each chromosome (trial vector) it performs a random experiment to determine, if any of these marginal operators are going to be used; otherwise, the main operator is used. So, if we have a marginal operator with its own probability of 5% and a main operator, its mean, that from 100 new chromosomes 5 will be created using the marginal operator and 95 using the main operator.
Important note: the PEXESO library is only able to operate on the real domains.
Let's note some links to sites that concern the Global Optimization, Evolutionary Computation and so on.
• The site on Global Optimization: http://solon.cma.univie.ac.at/~neum/glopt.html
• The homepage of Differential Evolution by Rainer Storn: http://www.icsi.berkeley.edu/~storn/code.html
• The papers of Zbyszek Michalewitz: http://www.coe.uncc.edu/~zbyszek/papers.html
• Using the Differential Evolution on the Reinforced Concrete Beam Optimization: http://klobouk.fsv.cvut.cz/~ondra/mr.beam/mr.beam.html
• 4. Algorithmic model of PEXESO library
Every evolutionary computation method can be modelled by the same algorithmic scheme, which can be written in consequent form:
void EVOLUTIONARY_METHOD ( task_to_optimize *ot ) { double bsf ; FIRST_GENERATION( ot ) ; bsf=EVALUATE_CHROMOSOMES( ot,0,number_of_selected ) ; int stop=0,generation=0 ; do { GENERATE_OFFSPRING( ot ) ; bsf=EVALUATE_CHROMOSOMES( ot,nember_of_selected_chromosomes,number_of_all_chromosomes ) ; PERFORM_SELECTION() ; if ( TO_BE_STOPPED( bsf,generation )) stop=1 ; generation++ ; } while ( !stop ) ; }
In this code, bsf is a variable that contains the best result reached so far (Best So Far). Let's now discuss each sub-functions one by one:
FIRST_GENERATION - generates the first set of chromosomes EVALUATE_CHROMOSOMES - assigns a value of the optimized function to each of the chromosomes GENERATE_OFFSPRING - generates the new (trial) set of chromosomes as the recombinations of the current ones; this function calls all the recombination operators PERFORM_SELECTION - this function selects the servivors from both the sets of current and new chromosomes; this function is the one that mainly differs in different classes inherited from px_strategy TO_BE_STOPPED - checks whether the optimum has been reached or the number of generations allowed has been overrun (of course, what exactly this functions checks can differ for different optimization tasks)
5. Description of Selection Strategies
Four types of Selection Strategies are implemented. Important functions used by them are: create_new_one (using the recombination operators creates new chromosome) and select_better (that selects which of two chromosomes is better and if it is a new one, replaces the current one with it).
• Steady state: in each generation only one new chromosome is generated; than another chromosome is changed from the population and these two are compared; if the new one is better, then replaces the current one.
• Trial vectors: for each of the chromosomes the potential substitutor is generated and if is better than the current one, the current one is going to be replaced by it.
• Tournament: at first, the whole new population is generated and than, the survivors are selected both from new and the current populations this way: each time two chromosomes are selected randomly and worse one is going to be killed.
• Total offspring: works int the same manner as the previous one except that the new generations is two times bigger than current and the survivors are selected only from the new chromosomes.
• 6. Description of Recombination Operators
Now, let's introduce all types of differential operators. Each of them is given by an equation, where:
ch[i] is an i-th chromosome, ch[i][j] is the j-th coordinate of i-th chromosome, ch' is always a chromosome from the offspring rand_ch is completely new random chromosome best_ch is the best chromosome in current generation rand( a,b ) is a random double between a and b range[j] is the range of the domain for coordinate j n is the number of coordinates Iexp(CR) is a subset of {1,2,...,n} for parameter CR in exponencial sense Ibin(CR) is a subset of {1,2,...,n} for parameter CR in binomial sense p,q,r,s,t are random indexes
• Simplified Differential (parameter CR):
• ch'[i]=ch[i]+CR*( ch[p]-ch[q] ),
• Hard Mutation (parameter MR):
• ch'[i]=ch[i]+MR*( rand_ch-ch[i] ),
• Local Mutation (parameter LR):
• ch'[i][j]=ch[i][j]+LR*rand( -1,1 )*range[j],
• Best One Exponencial (parameters F,CR):
• ch'[i][j]=best_ch[j]+F*( ch[p][j]-ch[q][j] ) for each j in Iexp(CR),
• Rand One Exponencial (parameters F,CR):
• ch'[i][j]=ch[p][j]+F*( ch[q][j]-ch[r][j] ) for each j in Iexp(CR),
• Best Two Exponencial (parameters F,CR):
• ch'[i][j]=best_ch[j]+F*( ch[p][j]-ch[q][j] )+F*( ch[r][j]-ch[s][j] ) for each j in Iexp(CR),
• Rand Two Exponencial (parameters F,CR):
• ch'[i][j]=ch[p][j]+F*( ch[q][j]-ch[r][j] )+F*( ch[s][j]-ch[t][j] ) for each j in Iexp(CR),
• Rand To Best One Exponencial (parameters F,CR):
• ch'[i][j]=ch[i][j]+F*( best_ch[j]-ch[i][j] )+F*( ch[p][j]-ch[q][j] ) for each j in Iexp(CR),
• Best One Binomial (parameters F,CR):
• ch'[i][j]=best_ch[j]+F*( ch[p][j]-ch[q][j] ) for each j in Ibin(CR),
• Rand One Binomial (parameters F,CR):
• ch'[i][j]=ch[p][j]+F*( ch[q][j]-ch[r][j] ) for each j in Ibin(CR),
• Best Two Binomial (parameters F,CR):
• ch'[i][j]=best_ch[j]+F*( ch[p][j]-ch[q][j] )+F*( ch[r][j]-ch[s][j] ) for each j in Ibin(CR),
• Rand Two Binomial (parameters F,CR):
• ch'[i][j]=ch[p][j]+F*( ch[q][j]-ch[r][j] )+F*( ch[s][j]-ch[t][j] ) for each j in Ibin(CR),
• Rand To Best One Binomial (parameters F,CR):
• ch'[i][j]=ch[i][j]+F*( best_ch[j]-ch[i][j] )+F*( ch[p][j]-ch[q][j] ) for each j in Ibin(CR).
7. The most simple example
Writing an optimization application using the PEXESO library contents of these steps:
2. use some of PEXESO's precomposed optimizing algorithms or compose your own one
3. write main()
4. compile it using -lpexeso
In our example, we'll try to solve the Chebychev trial polynomial problem with the Differential Evolution. It's the same trial task as you can find on Rainer Storn's page on Differential Evolution.
So, let's define our class chebychev:
#include #define Dim 9 #define M 60 class chebychev : public px_task { public: double fitness ( double *ox ) ; double get_desired_extreme ( void ) ; int get_dimension ( void ) ; void new_point ( double *ox ) ; void get_limits ( int odir , double &omin , double &omax , double &oprec ) ; } ;
Note: in this example, we declare and define only the mathods, that are absolutely necessary (and that are pure virtual in px_task). Constants Dim and M will be used in the code consequently.
So, we need these functions:
fitness (it is the optimized function itself; this method returns the evaluation of given chromosome - potential solution vector ox) get_desired_extreme (this is a method, that is beeing used by another methods to determine, whether it is worth to continue computations; if you don't know the best value, return here the biggest number that could be expected) get_dimension (simply returns the number of parameters of the optimized function) new_point (create here random new point in the domain and store it in the vector ox) get_limits (assign here variables omin and omax, which are the most far limits of the domain in the direction of odir, and oprec, which is the desired precision; if no such, fill this with zero)
Now, let's code these methods:
double chebychev::fitness ( double *ox ) { int i,j ; double px,x=-1,dx=M,result=0 ; dx=2.0/dx ; for ( i=0 ; i<=M ; i++ ) { px=ox[0] ; for ( j=1 ; j1.0 )) result+=( 1.0-px )*( 1.0-px ) ; x+=dx ; } px=ox[0] ; for ( j=1 ; j
Note: PEXESO does maximization, so if you minimize something, please return -value.
double chebychev::get_desired_extreme ( void ) { return 0.0 ; } int chebychev::get_dimension ( void ) { return Dim ; } void chebychev::new_point ( double *ox ) { int j ; for ( j=0 ; j
We want to optimize this function using the Differential Evolution, which belongs to the precomposed strategies of PEXESO, so we don't need to compose an optimalization algorithms by ourselves. We'll simply use the function px_setup_as_storn. So, now let's proceed to write the main:
int main ( void ) { px_task *T=new chebychev ; px_strategy *S=px_setup_as_storn( "RandToBestOneExponencial",0.85,1.0,T ) ; S->run( T ) ; delete S ; delete T ; return 1 ; }
Do you want to discuss the main step by step? So,
1. You have to instanciate the chebychev class you've defined:
2. You have to create an optimization strategy. Because we want the Differential Evolution, we should use a function px_setup_as_storn. "RandToBestOneExponencial" is the name of the operator, 0.85 is F and 1.0 is CR.
px_strategy *S=px_setup_as_storn( "RandToBestOneExponencial",0.85,1.0,T ) ;
3. Let's run the optimization:
S->run( T ) ;
4. What was allocated, must be also unallocated:
delete S ; delete T ;
OK, we've written this. Now, let's compile it:
g++ -o chebychev chebychev.c -lpexeso
and run it:
./chebychev
Well, it's running, even it is not writing anything to screen. If you wait some time, it will end and print the results to the screen.
This example is in full in doc/example1.c.
8. If we want to see some progress
The thing we don't like most on the first run, is that we don't see, what is going on during the computation. If we want to see the progress, we can redefine the method px_task::info_about_status:
void px_task::info_about_status ( double **osolutions , double *oforces , int obtg , int ogeneration , int opopulation ) ;
which has these parameters:
osolutions: are all the vectors (chromosomes) that are in the population oforces: are all the forces of the chromoseomes obtg: is an index of the best chromosome in the current generation ogeneration: is a number of current generation opopulation: is a number of chromosomes in the population
So, let's code this:
class chebychev : public px_task { ... void info_about_status ( double **osolutions , double *oforces , int obtg , int ogeneration , int opopulation ) ; ... } ; void chebychev::info_about_status ( double **osolutions , double *oforces , int obtg , int ogeneration , int opopulation ) { printf( "in generation (%d) the best value is (%f)\n",ogeneration,oforces[obtg] ) ; }
This will print the value of the best chromosome for each generations. Everything else can stay the same. Look for the example2.c.
9. Why running so long
Now, we can see what is going on during the computation, and this is still not good. It is running too long and for last about 9000 of about 10000 generations it is reporting that the best value is zero, which is the defined result. Also, at the and we could see something like this: stagnation by ( 9958 ). It means that nothing has changed for last 9958 generations. Why it is so? Well, there is a function to_be_stopped in the px_task, which tells the algorithms, when to stop. Now, this function only checks if the number of generations in not bigger than 10240, and if so, then stops the algorithm.
So, if we are looking for a reasonable behaviour of the algorithm, we should redefine this method. Our new one should check, if the best value is close enough (for example par 1.0E-7) to zero and than stop (stop means return 1, continue means return 0):
class chebychev : public px_task { ... int to_be_stopped ( double obsf , int ogeneration , int oiteration , int ostagnation ) ; ... } ; int chebychev::to_be_stopped ( double obsf , int ogeneration , int oiteration , int ostagnation ) { if ( obsf>-1.0E-7 ) return 1 ; return 0 ; }
OK, try it or see example3.c.
10. Non-rectangular domain
Well, most of the optimization tasks that come to focus, are defined on some n-dimensional rectangular domain. If it is always, we could go with only a simple function get_limits (as described above) but because of certain cases where this is not true, it is necessary to deal somehow with this thing. Let's try to solve the consequent example: let our function be too simple:
f(x[0],x[1])=x[1],
but defined on non-rectangular domain like this:
(OK, I know that we could simple solve it in the polar coordinates, but I just want to explain something to you, so don't treat me as a stupid cow, please.)
We will solve it on the whole rectangle (1.0-1.05)x(1.0-1.05), so our chromosomes can fly far from domain and we have to catch them and return them to the domain - see it on the picture above, please. We, for example, can conserve an angle and reduce the distance from the center...
We will need a different function new_point, that will generate points only in the desired domain:
void thin_arc::new_point ( double *ox ) { double ro,fi ; ro=(( double )rand()/RAND_MAX )*0.05+1.0 ; fi=(( double )rand()/RAND_MAX )*M_PI_2 ; ox[0]=ro*cos( fi ) ; ox[1]=ro*sin( fi ) ; }
And also, we must make this returning to domain. So, there are 2 redefinable methods in class px_task, that are called somewhere from the algorithm. One of them, check_domain tells the algorithms, if it should check the chromosomes if there are or are not in the domain, and the second, return_to_domain, does this thing itself-it returns chromosomes into the domain. Normally, as it is made in px_task, check_domain returns 0 and return_to_domain does nothing and is never called, because check_domains returns zero, which means that the algorithms will not call it. We simply have to redefine these two methods:
int thin_arc::check_domain ( void ) { return 1 ; } int thin_arc::return_to_domain ( double *ox ) { double ro,fi ; fi=atan( ox[1]/ox[0] ) ; ro=sqrt( ox[0]*ox[0]+ox[1]*ox[1] ) ; if ( fi<0.0 ) fi=0.0 ; if ( fi>M_PI_2 ) fi=M_PI_2 ; if ( ro<1.0 ) ro=1.0 ; if ( ro>1.05 ) ro=1.05 ; ox[0]=ro*cos( fi ) ; ox[1]=ro*sin( fi ) ; }
Now, let's see the definition of class thin_arc and the other methods:
#include #include #include class thin_arc : public px_task { public: double fitness ( double *ox ) ; double get_desired_extreme ( void ) ; int get_dimension ( void ) ; void new_point ( double *ox ) ; void get_limits ( int odir , double &omin , double &omax , double &oprec ) ; int check_domain ( void ) ; int return_to_domain ( double *ox ) ; int to_be_stopped ( double obsf , int ogeneration , int oiteration , int ostagnation ) ; void info_about_status ( double **osolutions , double *oforces , int obtg , int ogeneration , int opopulation ) ; } ; double thin_arc::fitness ( double *ox ) { return ox[1] ; } double thin_arc::get_desired_extreme ( void ) { return 1.05 ; } int thin_arc::get_dimension ( void ) { return 2 ; } void thin_arc::get_limits ( int odir , double &omin , double &omax , double &oprec ) { omin=0.0 ; omax=1.05 ; oprec=0.0 ; } int thin_arc::to_be_stopped ( double obsf , int ogeneration , int oiteration , int ostagnation ) { if ( obsf>1.0499999 ) return 1 ; return 0 ; } void thin_arc::info_about_status ( double **osolutions , double *oforces , int obtg , int ogeneration , int opopulation ) { printf( "in generation (%d) the best value is (%f)\n",ogeneration,oforces[obtg] ) ; } int main ( void ) { px_task *T=new thin_arc ; px_strategy *S=px_setup_as_storn( "RandToBestOneExponencial",0.85,1.0,T ) ; S->run( T ) ; delete S ; delete T ; return 1 ; }
OK, we have example4.c.
11. If you have KRESLITKO...
Well, if you have KRESLITKO installed on your linux box, you can try to compile and run example5.c. You get a graphical experience how does the PEXESO library work if optimizing a task like examlpe4 (the previous one) was:
Here, you can see a red thin-arc, which is our domain, blue crosses which are the chromosomes and blacj rectangle, which locates the average coordnates of all the chromosomes in the population. Have a nice time with that.
So far, enough with tutorials. Now let's describe the function reference of the two main classes, firstly the px_task.
Of course, you don't need to wory about most of these methods. You need to redefined only five of them, the others may stay as they are, and you can redefined them anytime you need it. Most of them are intended more to the development environment that is beeing used by us (called NAVARRO).
13. Class px_strategy function reference
Now let's describe the methods themselves of class px_strategy.
px_strategy ( px_task *ot , int opop_limit=0 ) ; a constructor; send him please your optimization task as ot and if you feel a need, you can limit the number of population, but unless you are not sure what are you doing please leave this alone ~px_strategy ( void ) ; a destructor void set_main_operator ( char *otype , double *orate ) ; sets one operator as a main one; pass here the name from this set: "SimplifiedDifferential", "HardMutation", "LocalMutation", "BestOneExponencial", "RandOneExponencial", "BestTwoExponencial", "RandTwoExponencial", "RandToBestOneExponencial", "BestOneBinomial", "RandOneBinomial", "BestTwoBinomial", "RandTwoBinomial", "RandToBestOneBinomial" as otype and the vector of its parameters os orate void set_marginal_operator ( char *otype , double oprobability , double *orates ) ; sets one of marginal operators; the difference from the previous is that marginal operator must have a probability, that should be passed as oprobability void clear_pool ( void ) ; this method is intended for repeated computations in some kind of benchmark test; is simply clears the values that have been found in previous run of computation, so the algorithm can be restarted from knowing nothing void run ( px_task *ot ) ; this method runs the computation and does everything around
We provide three precomposed strategies:
And a function that return empty strategy of certain type (empty means that it should be yet filled by the operators):
px_strategy *px_create_any ( char *oselection , int opopulation_rate , px_task *ot ) ; this creates empty strategy with selection type given by oselection and with population rate given by opopulation_rate; pass also your optimization task as ot
15. Assembling your own evolutionary method
Let's now try to compose our own optimization method. It should be, for example, of the steady_state type selection, with the local_mutation as a main operator and hard_mutation with probability 10% as a marginal operator. We can than run this method with a problem of thin_arc and see, how will this poor method try to solve that trivial task. At first, let's write down a function, that creates our desired method and returns a pointer to it:
px_strategy *create_our_own ( px_task *ot ) { px_strategy *s ; s=px_create_any( "SteadyState",20,ot ) ; double rate[1] ; rate[0]=0.0025 ; s->set_main_operator( "LocalMutation",rate ) ; rate[0]=0.25 ; s->set_marginal_operator( "HardMutation",0.1,rate ) ; return s ; }
Now, let's replace a line that loads the method in main:
px_strategy *S=create_our_own( T ) ;
Procees as is usual. Can see the example6.c. You may start this computation and let it run until you or the computer are bored to death.
Latest release is: PEXESO-0014
Date: May 24 2001
17. How to compile & install PEXESO
The PEXESO library is anything-on-the-world-independent. So, after you unpack the pexeso-xxxx.tgz by:
tar xzvf pexeso-xxxx
you can immediately run 'make' and than 'make install' as root; if you don't like it, you may run 'make uninstall' also as root. Compiling the application is standard, only you should use a library option: -lpexeso.
18. Release history
Archive name: Date of release: What's new: pexeso-0014.tgz May 24, 2001 the first release ever
19. References to the literature
Author Title Where published Storn, R. & Price, K. Differentisal Evolution - a Simple and Efficient Hueristic for Global Optimization over Continuous Spaces Journal of Global Optimization, 11, 341-359 Michalewicz, Z. Genetic Algorithms + Data Structures = Evolution Programs Springer-Verlag 1992 Michalewicz, Z. & Hinterding, R. & Michalewicz, M Evolutionary Algorithms Chapter 2 in Fuzzy Evolutionary Computation, W. Pedrycz (editor), Kluwer Academic, 1997 Hrstka, O. & Kucerova, A. Search for optimization method on multidimensional real domains CTU Reports, Contribution to Mechanics of Materials and Structures, 4, 87-104, 2000
20. Index of Optimizing Tasks that have been recently solved by PEXESO
Problem description Where to find a documentation Periodic unit cell problem solving none Finding a parameters of a retention graph of soil none Reinforced concrete beam optimization http://klobouk.fsv.cvut.cz/~ondra/mr.beam/mr.beam.html FSWE Type0 (the study of computational complexity with respect to a problem dimension) http://klobouk.fsv.cvut.cz/~ondra/sade/sade.html FSWE Type1 (the study of behavoiour of evolutionary methods for the strongly multi-modal functions) none Test set of functions http://klobouk.fsv.cvut.cz/~ondra/about_ga/about_ga.html The Chebychev trial polynomial problem none | 6,694 | 26,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2015-22 | longest | en | 0.914742 |
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1 Movement and Position
2 Syllabus points: 1.2 plot and interpret distance-time graphs 1.3 know and use the relationship between average speed, distance moved and 1.4 describe experiments to investigate the motion of everyday objects such as toy cars or tennis balls 1.5 know and use the relationship between acceleration, velocity and time: 1.6 plot and interpret velocity-time graphs 1.7 determine acceleration from the gradient of a velocity-time graph 1.8 determine the distance travelled from the area between a velocity-time graph and the time axis.
3 distance (m) distance (m) Syllabus points: 1.2 plot and interpret distance-time graphs You need to be able to plot accurate distance-time graphs. Skills tested here will include: plotting points accurately labelling axes with quantities and units selecting an appropriate scale (filling over half the page, avoiding weird scales) Distance-time graphs and velocity-time graphs are some of the only graphs in physics where you connect the points dot-to-dot. The shape of a distance-time graph helps describe the motion of the object. At rest A flat line shows that the object s distance is not changing; i.e. they are not moving/they are stationary. The gradient of a distance time graph tells you the average speed between the two points in the time selected. This is because the of a flat line is zero, the speed is zero.. As the gradient Constant speed A straight line shows that the object is travelling the same amount of distance each second; i.e. they are moving at a constant speed. The gradient of a distance time graph tells you the average speed between the two points in the time selected. This is because the. As the gradient of a straight line is constant, the speed is constant. A steeper gradient means a greater (i.e. faster) speed.
4 distance (m) Increasing speed A curved line represents a change in speed. This could be an increase (acceleration) or a decrease (deceleration). If the gradient of the line is increasing (shallow to steep) then the speed is increasing (acceleration); if the gradient of the line is decreasing (steep to shallow) then the speed is decreasing (deceleration). The gradient of a distance time graph tells you the average speed between the two points in the time selected. This is because the. 1.3 know and use the relationship between average speed, distance moved and Speed is defined as how far an object travels in a given time. You are probably used to the speed of cars being measured in miles per hour (i.e. how many miles you would travel if you moved at that speed for a whole hour). So travelling at 60 mph you would move 30 miles in half an hour and 120 miles in 2 hours. In physics we use the metric system, so the standard units for the quantities are: distance moved; metres, m time taken; seconds; s average speed; metres per second; m/s There are two categories of speed; average speed and instantaneous speed. Both have a very similar equation;
5 distance (m) This makes a little more sense when comparing 2 objects moving the same distance in the same time Constant speed Slow then fast In the graph above the blue line (top line) has the same gradient throughout and shows the object moving as a constant speed. We can find the speed of this by finding the gradient. In this case 50 m/10 s = 5 m/s. The red line (lower line) has 2 distinct gradients; for the first 5 seconds the object travels 10 m so has an instantaneous speed of 10 m/5 s = 2 m/s. In the next 5 s it travels 40 m so has an instantaneous speed of 40 m/5 s = 8 m/s. However, they have still travelled a total of 50 m in 5 s so their average speed is 50 m/ 10s = 5 m/s. 1.4 describe experiments to investigate the motion of everyday objects such as toy cars or tennis balls How would you go about investigating the motion of an everyday object? This could be a toy car on a slope or a ball rolling along a surface. 1) You could measure out the distance between 2 points using a ruler, mark these points with some sort of marker. Then start the stopwatch when the object passes the first marker. Stop the stopwatch when it goes past the second marker. You should repeat this a couple of times to ensure that you are collecting consistent data. To find the average speed of everyday object you just need to divide the distance between markers by the average time measured on the stopwatch. 2) The next method uses light gates and works in a similar way; set up two light gates so that the everyday object passes through both. Light gates consist of a beam of light and a detector. When the beam of light is broken by something this sends a signal to a computer or datalogger. Measure the distance between these light gates with a ruler. When the everyday object passes through the first light gate the computer or datalogger it is conneted to will start a stopwatch. When the everyday object passes through the second light gate the stopwatch will be stopped. Again you can find the average speed by dividing the distance between light gates by the average time taken.
6 3) The next method is to use a high speed camera and markers in the scene. Measure out a distance and mark with cone markers again. Video the everyday object moving past the two cones. Watch back the video and look at the time when the everyday object goes past the first cone. Let the video continue playing and then look at the time when it reaches the second cone. Then do the distance measured divided by time taken to get average speed. 4) The final method is to use a ticker tape timer. This device marks a dot every 0.02 s on a piece of ticker tape (thin piece of paper). The other end of the ticker tape is attached to the everyday object. Once the everyday object has moved the ticker tape is processed. This involves marking a dot and then moving on 5 more dots; this represents 0.1 s of travel (5 x 0.02 s = 0.1 s). The speed of the everyday object is equal to the length covered by 5 dots divided by 0.1 s. 5cm Above, each section of 5 dots (which represent 0.1 s of travel) are the same length; this means the everyday object is travelling at a constant speed. In this example the everyday object is moving at 5 cm/0.1 s = 50 cm/s Above, the distance between dots has increased; however, the time between dots is still the same. This means the everyday object is moving further in the same amount of time; i.e. it is accelerating. Above, the distance between the dots has decreased; however, the time between dots is the same. This means the everyday object is moving less far in the same amount of time; i.e. it is decelerating. 1.5 know and use the relationship between acceleration, velocity and time: Acceleration is the rate of change of velocity (remember that rate in physics means per unit time so acceleration is the change in velocity per second). The unit for acceleration is m/s 2 or m/s/s. This is because it is the change in velocity (m/s) per second (/s) i.e. metres per second per second. Velocity is essentially speed in a given direction (for example a velocity of 5 m/s would mean moving down at a speed of 5 m/s). We ll look more at this in another set of notes. For now, don t worry too much about it.
7 velocity (m/s) velocity (m/s) Change in velocity is equal to the final velocity (v) minus the initial velocity (u). For example if a car accelerates from 10 m/s to 20 m/s in 5s the acceleration will be:. If the change in velocity is negative then the acceleration will be negative i.e. a deceleration. For example if a car brakes from 30 m/s to 10 m/s in 4 s the acceleration will be. This is equal to a deceleration of 5 m/s plot and interpret velocity-time graphs You need to be able to plot accurate velocity-time graphs. Skills tested here will include: plotting points accurately labelling axes with quantities and units selecting an appropriate scale (filling over half the page, avoiding weird scales) Distance-time graphs and velocity-time graphs are some of the only graphs in physics where you connect the points dot-to-dot. The shape of a velocity-time graph helps describe the motion of the object. Stationary This velocity-time graph shows the velocity remains at zero. This means the object is not moving. time Constant velocity A flat line shows that the object s velocity is not changing; i.e. they are moving with a constant velocity. The gradient of a velocity-time graph tells you the acceleration between the two points in time selected. This is because the is zero, the acceleration is zero. time. As the gradient of a flat line
8 velocity (m/s) velocity (m/s) Constant acceleration A straight line shows that the objects velocity is changing at a constant rate; i.e. they are accelerating with a constant acceleration. The gradient of a velocity-time graph tells you the acceleration between the two points in time selected. This is because the. As the gradient of a straight line is constant, the acceleration is constant. Increasing acceleration A curved line shows that the object s acceleration is changing. If the gradient of the line is increasing (shallow to steep)_then the rate of acceleration is increasing; in the much more common case of the gradient of the line decreasing (steep to shallow) then the acceleration is decreasing. The gradient of a velocity-time graph tells you the acceleration between the two points in time selected. This is because the.
9 velocity (m/s) velocity (m/s) You may be asked to describe the shape of a velocity time graph; it is important that you focus on the shape. Consider the following example: The gradient is initially steep. Then it flattens out to a horizontal line. Finally the gradient increases again but to a lesser gradient than during the first part of the graph. The motion that this represents is an object accelerating with a high constant acceleration, then moves at a constant velocity, then accelerates again but with a lower acceleration. 1.7 determine acceleration from the gradient of a velocity-time graph As discussed above, A positive (sloping upwards) gradient shows an acceleration. A negative (sloping downwards) gradient shows deceleration. 1.8 determine the distance travelled from the area between a velocity-time graph and the time axis. The area between a velocity-time graph and the time axis is equal to velocity time which is the distance travelled. This is sometimes known as the area under a graph. The orange line is the velocity-time graph line. The blue shaded area represents the area under the graph / area between the velocity-time graph and the time axis. This means the area of the blue shapes is equal to the distance travelled.
10 velocity (m/s) part 1 part In the example above it is easier to split the objects journey into 2 parts: In part one the object is accelerating with a constant acceleration of (as this is the gradient of the line). The area under the graph is equal to ½ x 40 s x 20 m/s (area of a triangle ½ base height). This means that during the accelerating stage of the journey the object has moved 400 m. In part two the object is moving with a constant velocity (flat line, no acceleration). The area under the graph is equal to 20 m/s 20 s = 400 m. (Care needs to be taken here as the time between 40 s and 60 s is 20 s). The total distance travelled is 800 m. The average speed for this journey would be 800 m/60 s = 13.3 m/s. This means an object travelling at 13.3 m/s would arrive at the end point at the same time as the object in the graph.
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Name: Date: 1. Carl Lewis set a world record for the 100.0-m run with a time of 9.86 s. If, after reaching the finish line, Mr. Lewis walked directly back to his starting point in 90.9 s, what is the magnitude | 9,336 | 38,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-26 | latest | en | 0.912955 |
https://www.airmilescalculator.com/distance/cns-to-ptj/ | 1,611,551,680,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00411.warc.gz | 663,040,871 | 43,537 | Distance between Cairns (CNS) and Portland (PTJ)
Flight distance from Cairns to Portland (Cairns Airport – Portland Airport (Victoria)) is 1499 miles / 2412 kilometers / 1302 nautical miles. Estimated flight time is 3 hours 20 minutes.
Driving distance from Cairns (CNS) to Portland (PTJ) is 1899 miles / 3056 kilometers and travel time by car is about 37 hours 40 minutes.
Map of flight path and driving directions from Cairns to Portland.
Shortest flight path between Cairns Airport (CNS) and Portland Airport (Victoria) (PTJ).
How far is Portland from Cairns?
There are several ways to calculate distances between Cairns and Portland. Here are two common methods:
Vincenty's formula (applied above)
• 1498.562 miles
• 2411.702 kilometers
• 1302.215 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1503.393 miles
• 2419.476 kilometers
• 1306.412 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Cairns Airport
City: Cairns
Country: Australia
IATA Code: CNS
ICAO Code: YBCS
Coordinates: 16°53′8″S, 145°45′18″E
B Portland Airport (Victoria)
City: Portland
Country: Australia
IATA Code: PTJ
ICAO Code: YPOD
Coordinates: 38°19′5″S, 141°28′15″E
Time difference and current local times
The time difference between Cairns and Portland is 1 hour. Portland is 1 hour ahead of Cairns.
AEST
AEDT
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 179 kg (395 pounds).
Frequent Flyer Miles Calculator
Cairns (CNS) → Portland (PTJ).
Distance:
1499
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
1499
Round trip? | 492 | 1,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.826591 |
https://www.arxiv-vanity.com/papers/2009.04641/ | 1,618,883,176,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038921860.72/warc/CC-MAIN-20210419235235-20210420025235-00208.warc.gz | 745,942,868 | 26,178 | # A note on pure braids and link concordance
Miriam Kuzbary School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332 http://people.math.gatech.edu/ mkuzbary3/
###### Abstract.
The knot concordance group can be contextualized as organizing problems about 3- and 4-dimensional spaces and the relationships between them. Every 3-manifold is surgery on some link, not necessarily a knot, and thus it is natural to ask about such a group for links. In 1988, Le Dimet constructed the string link concordance groups and in 1998, Habegger and Lin precisely characterized these groups as quotients of the link concordance sets using a group action. Notably, the knot concordance group is abelian while, for each , the string link concordance group on strands is non-abelian as it contains the pure braid group on strands as a subgroup. In this work, we prove that even the quotient of each string link concordance group by its pure braid subgroup is non-abelian.
## 1. Introduction
A knot is a smooth embedding of into and an -component link is a smooth embedding of disjoint copies of into . The set of knots modulo a -dimensional equivalence relation called concordance forms the celebrated knot concordance group using the operation connected sum, while the set of -component links modulo concordance does not have a well-defined connected sum (even when the links are ordered and oriented) [FM66]. As a result, multiple notions of link concordance group have arisen in the literature with somewhat different structural properties [Hos67] [LD88] [DO12].
We focus on the -strand string link concordance group of Le Dimet as it is the only link concordance group which is non-abelian. This is in stark contrast to the original knot concordance group which is abelian, and thus is perhaps a further indicator of the complexity of the link concordance story. While each link has an infinite number of distinct string link representatives, the correspondence between links and their string link representatives was characterized by Habegger and Lin using the action of on [HL98]. In particular, a string link is trivial in this group exactly when the link in it represents is concordant to the unlink.
The string link concordance group is known to be non-abelian as it contains , the pure braid group on strands, as a proper subgroup [LD88]. Therefore, it is not clear whether being non-abelian is merely an inherited property from its pure braid subgroup. This question can be viewed as in some sense complementary to the question of classifying links up to link homotopy answered in [HL90], as every string link is link homotopic to a pure braid. Understanding how sits inside is made more difficult by the fact that is not a normal subgroup for [KLW98]. Therefore, we must instead consider the quotient of by the normal closure . Despite this difficulty, we have proven the following theorem.
###### Theorem 1.1.
is non-abelian for all .
It is natural to further ask about the structure of the quotient group and we have made the following observation.
###### Proposition 1.2.
has a central subgroup isomorphic to .
Notice that Theorem 1.1 leads to questions about how close this quotient is to being abelian. More precisely, is the quotient solvable? It is important to note that the pure braid group itself is residually nilpotent.
###### Conjecture 1.3.
is not solvable and neither is .
There are many natural questions arising from these results, including:
• Does contain elements of finite order?
• What is the abelianization of ?
• How do the images of the -solvable filtration and bipolar filtration behave in this quotient?
• What can be said about the structure of itself and, more precisely, which properties of are preserved in this closure?
## 2. Proofs and tools used in proofs
In this section, we carefully define -component string links and the equivalence relation on them required to form a group. We then outline the main invariants used to obstruct the relevant quotient being abelian and present proofs of the results.
###### Definition 2.1 (n-component string link).
Let be the unit disk, the unit interval, and be points in the interior of . An -component (pure) string link is a smooth proper embedding such that
σ|Ii(0)= {pi}×{0} σ|Ii(1)= {pi}×{1}
As is typical in the study of knots and links, we will often abuse notation and use to refer to both the map itself and to its image embedded in . Notice that a -component string link is an -strand pure braid when each slice is the -punctured disk. Just as we can take the closure of a braid, we can take the closure of a string link and get a link in . Note that not every link is the closure of a pure braid, however, every link is the closure of a pure string link in the following way. First, introduce an embedded disk intersecting each component of once positively. Then, cut open along this disk to get a collection of knotted arcs in . In order to multiply string links and , glue to by the identity map on .
Geometrically, this multiplication is exactly what one would obtain from taking two closed -component links and in , choosing basing disks and for each link (which would give and if one were to cut open along them), and then band summing the strand of with the strand of exactly at the points and . Note that the set of component pure string links merely forms a monoid and not a group, and therefore to define inverses we consider the following notion.
###### Definition 2.2 (String link concordance).
Two -component string links and are concordant if there is a smooth embedding which is transverse to the boundary such that
H|(⨆mI×{0})= σ1 H|(⨆mI×{1})= σ2 H|(⨆m∂I×I)= j0×idI
with .
Now, we can see the inverse of a string link is simply reflected across . More precisely, if is the reflection map , , and , then .
###### Theorem 2.3 ([Ld88]).
The set of -component string links modulo string link concordance with the above binary operation forms a group which we will notate as .
It is clear through taking string link closures that concordant string links close to form concordant links in , and furthermore that is isomorphic to the usual knot concordance group .
### 2.1. Milnor’s Invariants
The main result in this work was proven using Milnor’s invariants. These integer-valued concordance invariants first defined by Milnor in 1957 can be interpreted as higher order linking numbers and computed in many different ways. Roughly, these invariants detect how deep the longitudes of an -component link lie in the lower central series of the link group of the link group . The mechanism for doing so involves comparing the nilpotent quotients to those of the free group on letters and using combinatorial group theoretic techniques to analyze the longitudes in these quotients. In a sense, we can view this process as taking the “Taylor expansion” in non-commuting variables of each longitude using Fox Calculus; the Milnor’s invariants are exactly the coefficients of the resulting polynomial modulo a subset of lower order coefficients. It is certainly not obvious that such things would be concordance invariants. It is a deep result that the entire nilpotent quotients themselves are link concordance invariants and that Milnor’s invariants extract concordance data from these quotients.
These invariants are difficult to compute; however, due to Cochran we have an algorithm for computing the non-vanishing Milnor’s invariants of lowest weight (which should be thought of as the lowest degree coefficients of the aforementioned polynomials) using surface systems. The main idea relies on the celebrated theorem of Turaev and Porter which relates Milnor’s invariants to Massey products on the link exterior. Cochran’s process dualizes the Massey product defining system machinery to use intersections of surfaces instead of cup products of cochains. For details of the construction, see [Coc90].
Recall that one of the unique properties of is that it is nonabelian for [LD88]. This is because it contains the pure braid group as a subgroup; therefore, it is not clear whether this non-abelian property is merely inherited from the pure braid group. This question is made far more difficult by the following fact.
###### Theorem 2.4 (Kirk-Livingston-Wang [Klw98]).
is not a normal subgroup of for .
### 2.2. Proofs
See 1.1
###### Proof.
First notice that the Milnor invariant with for any pure braid ; this is because this Milnor’s invariant only depends on the -component sublinks of (namely, the link whose components are the and components of P) and two component pure braids are simply full twists. Therefore, the only non-vanishing is exactly the linking number .
Now, let with all pairwise linking numbers zero for reasons which will soon be clear. We see that where and for all . Now, by Theorem 8.12 in [Coc90] we see that .
Again, will depend only on the -component sublinks of each string link made up of the and components of . Each of these sublinks will be a product of the and components of , , and . We will show what happens for a specific . Call these components , , and
As is clear in Figure 1, is a product of full twists and thus we can slide along the twisting region to obtain the word . We see this string link is concordant to and thus . Finally, this tells us .
What we have now shown is that any with vanishing pairwise linking numbers also has vanishing Sato-Levine invariants. Hence to prove the theorem it will suffice to show a commutator in has a non-vanishing Sato-Levine invariant. Note that any commutator of string links has pairwise linking number and there is no indeterminacy in computing the where is the closure.
Consider the link in Figure 2. It is the commutator of a representative of the Hopf link with a representative of the Whitehead link. To build a surface system, we will first label the components and and think of these curves as indexed by meridional generators and in . The closure, , bounds surfaces and as in Figure 3.
These surfaces and intersect in the curve in Figure 4 with two components, and one can think of this curve as being indexed by the commutator . In order to be careful and consistent with orientations as in [Coc90], we will consider this curve as the intersection and orient it so the triple gives the right-handed orientation of .
All linkings of , , and with their pushoffs of weight 3 vanish, and the only non-trivial linking of weight is . Thus, by [Coc90]. This proves the result for . Now, consider the link whose first two strands are isotopic to and whose remaining strands are trivial. This link is also a commutator, and we immediately see therefore the result follows.
See 1.2
###### Proof.
The result follows from a series of simple observations. First, note that any string link of the form which is split with strand isotopic to a knot cut open at a point and all other strands trivial will commute with any string link by simply sliding along the strand of . Furthermore, note that is if and, if , it is the split link with component isotopic to split open at a point and component isotopic to split open at a point. Therefore, the set generates . Furthermore, no element in this set is in as each component of an element of must have slice components (this is clear as pure braids have unknotted components).
### 2.3. Acknowledgments
The author was supported by an NSF Graduate Research Fellowship under Grant No. 1450681 as well as an AAUW American Fellowships Dissertation Fellowship. Additionally, the author was partially supported by NSF grant DMS-1745583 as a postdoc. The author is deeply grateful to her advisor, Shelly Harvey, for her guidance. The author would further like to express her appreciation for Tim Cochran for giving her a roadmap of where to look to understand Milnor’s invariants and telling her to physically make links using household objects to check her more unwieldy linking number computations. The author would like to thank Jennifer Hom for her mentorship and generous feedback and would also like to thank her REU student Benjamin Pagano for pointing out a computational error in an earlier version of this paper. | 2,723 | 12,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | latest | en | 0.915094 |
https://community.qlik.com/thread/281094 | 1,539,874,970,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00203.warc.gz | 641,703,333 | 23,022 | 5 Replies Latest reply: Nov 9, 2017 10:08 AM by Sunny Talwar
# Calculate average lead time
Hi!
I have got a table similar to the format below in which I need to calculate the average lead time (finish date - start date) in a certain period of time (activities finished in the last 7 days, in the last 14 days and so on).
IDSTART DATEFINISH DATE
130/10/201706/11/2017
201/11/201708/11/2017
201/11/2017
08/11/2017
303/11/201707/11/2017
403/11/2017
505/11/201710/11/2017
505/11/201710/11/2017
Please note 2 points:
• the id might repeat
• only the ids already finished (finish date is not null) should be taken into account
Can someone help me?
• ###### Re: Calculate average lead time
May be this
Avg({<[FINISH DATE] = {'*'}>} Aggr([FINISH DATE] - [START DATE], ID))
• ###### Re: Calculate average lead time
Sunny, thks for the answer! What if I wanted to set a time variable and calculated the average of all the activities that finished after that variable?
E.G: FINISH DATE >= 01/01/2018
• ###### Re: Calculate average lead time
May be this
Avg({<[FINISH DATE] = {"\$(=Date(MakeDate(2018, 1, 1), 'DD/MM/YYYY'))"}>} Aggr([FINISH DATE] - [START DATE], ID))
• ###### Re: Calculate average lead time
Sorry for insisting, but does it mean that all the IDs with finish date after this date would be taken into account? If it doesn´t, how can we do it?
Cheers!
• ###### Re: Calculate average lead time
My bad.... this should be the right expression
Avg({<[FINISH DATE] = {"\$(='>=' & Date(MakeDate(2018, 1, 1), 'DD/MM/YYYY'))"}>} Aggr([FINISH DATE] - [START DATE], ID)) | 484 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-43 | latest | en | 0.79454 |
https://markxu.com/prediction-markets | 1,701,356,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00359.warc.gz | 432,646,520 | 3,332 | # Artificially Intelligent
Any mimicry distinguishable from the original is insufficiently advanced.
# Revenge of the Prediction Market
| 642 words
Recommended reading: Prediction Markets: Tales from the Election
Suppose I wanted to know the probability of some future event. How might I do this?
One way would be to pay forecasters from the Good Judgment Project to forecast the event. These forecasters are generally pretty good at what they do, so this would probably work pretty well. However, this would cost a bit of money, and you would have less than 100 eyes on the problem. Can we do better?
One way would be to use a prediction aggregation platform like Metaculus, with some tweaks like weighting forecasts by the forecaster’s historical accuracy. Ideally, we would want to weight forecasts based on how confident the forecasters were. If we want people working on it full time we might have to add large monetary incentives. Can we do better?
One way would be to use one of humanity’s best knowledge aggregation mechanisms: markets. A prediction market works by defining two assets: one worth $1 if the event happens and$0 otherwise and another with the opposite behavior. $1 buys you one copy of both assets, which can be traded on the market. Assuming that market participants are profit maximizers, the assets’ price should both sum to $1 and be equal to the confidence-weighted aggregate probability given to the event by market participants. To see why this is true if the sum of the price is more than $1, I can buy a copy of both assets for$1 and sell them both, netting a profit and lowering the price. If the price sums to less than $1, then I can buy both assets and weight for the event to either happen or not happen, netting$1 in both cases. If my subjective probability is greater than the current price, buying the asset nets me profit in expectation and raises the price. If my subjective probability estimate is lower than the price, buying the other asset nets me profit in expectation and raises the other asset’s price. The amount of asset I buy is, roughly speaking, how confident I am in my subjective probability estimate. Thus the equilibrium price of the asset will be the confidence-weighted aggregate subjective probability of market participants.
Prediction markets still have several problems, like not being very good investments compared to other assets for long time-horizons or low probabilities, but at least the incentives align.
So how does this work out in practice? The recent Augur market on the 2020 presidential election has been the largest prediction market to date. On January 10th, an asset that would be worth $1 if Trump lost the election was trading at $0.87. Even before the election, there was reason to be alive that the market was priced improperly.
It’s hard to describe how I feel about this. On the one hand, the market had pretty high barriers to entry, competed with other crypto assets, and was politically charged; all reasons you wouldn’t expect it to be efficient. On the other hand… fuck. How do I say this? Prediction markets were supposed to be the chosen one. Sigh. Instead we just found out that they had their own problems.
The markets get better as time passes. Those good at predicting make money, so they will control a larger share of future markets. Those bad at predicting lose money, so they will control a lesser share of future markets. As time rachets forward, the equilibrium price will become close to the best guess given the available information.
But still. Why does the market, which is shaped by the invisible hand, which is efficient, contain so much inefficiency? Why does the prediction, which is determined by the market, which is efficient, look so stupid? Why do the trades, which are made by traders, who are profit maximizers, lose so much money? | 801 | 3,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.951472 |
https://yourconverterworld.com/spring-pitch-calculator/ | 1,718,628,363,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00487.warc.gz | 940,005,810 | 17,206 | ## Introduction to Spring Pitch Calculator:
A spring pitch calculator is a useful tool that helps engineers and designers determine the pitch of a helical compression spring. The pitch of a spring is a crucial parameter in spring design, as it directly affects the spring’s behavior and performance. This calculator simplifies the process of calculating the spring pitch by taking into account the wire diameter, number of coils, and mean diameter of the spring.
## Working of Spring Pitch Calculator:
### Gather Input Data:
#### Wire Diameter (d):
Input the diameter of the wire used to make the spring.
#### Number of Coils (n):
Enter the total number of coils or turns in the spring.
#### Mean Diameter (D):
Provide the mean diameter of the spring, which is essentially the average diameter of the coil.
#### Input Validation:
The calculator checks whether the values entered for wire diameter, number of coils, and mean diameter are valid numeric values.
## Calculate Spring Pitch:
The spring pitch (P) is calculated using the following formula:
`Spring Pitch (P) = Mean Diameter (D) / Number of Coils (n)`
## Display Result:
• The calculated spring pitch (P) is displayed in the “Spring Pitch (P)” field on the calculator interface.
## Interpretation:
The result represents the pitch of the helical compression spring in the specified units (e.g., millimeters or inches). The pitch is the distance between adjacent coil centers.
## Formula for Calculating Spring Pitch:
The formula to calculate the spring pitch (P) is:
`Spring Pitch (P) = Mean Diameter (D) / Number of Coils (n)`
• Spring Pitch (P): The distance between adjacent coil centers.
• Mean Diameter (D): The average diameter of the spring.
• Number of Coils (n): The total number of coils or turns in the spring.
## Summary:
In summary, the Spring Pitch Calculator simplifies the process of determining the pitch of a helical compression spring, which is a critical parameter in spring design. Engineers and designers can use this tool to ensure that the spring meets their specific requirements for a wide range of applications, from automotive to industrial machinery. | 440 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.832832 |
https://jp.mathworks.com/matlabcentral/cody/problems/2677-find-out-next-state-ns-of-t-flip-flop/solutions/530188 | 1,581,987,241,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00305.warc.gz | 432,296,822 | 15,651 | Cody
# Problem 2677. Find out next state (NS) of T Flip-Flop.
Solution 530188
Submitted on 18 Nov 2014 by rifat
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% Q_PS=0; T=0; y_correct = 0; assert(isequal(T_FF(Q_PS,T),y_correct))
2 Pass
%% Q_PS=1; T=0; y_correct = 1; assert(isequal(T_FF(Q_PS,T),y_correct))
3 Pass
%% Q_PS=0; T=1; y_correct = 1; assert(isequal(T_FF(Q_PS,T),y_correct))
4 Pass
%% Q_PS=1; T=1; y_correct = 0; assert(isequal(T_FF(Q_PS,T),y_correct)) | 206 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.608005 |
http://hitchhikersgui.de/Line_search | 1,548,209,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583879117.74/warc/CC-MAIN-20190123003356-20190123025356-00315.warc.gz | 105,997,291 | 9,804 | # Line search
In optimization, the line search strategy is one of two basic iterative approaches to find a local minimum ${\displaystyle \mathbf {x} ^{*}}$ of an objective function ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$. The other approach is trust region.
The line search approach first finds a descent direction along which the objective function ${\displaystyle f}$ will be reduced and then computes a step size that determines how far ${\displaystyle \mathbf {x} }$ should move along that direction. The descent direction can be computed by various methods, such as gradient descent, Newton's method and Quasi-Newton method. The step size can be determined either exactly or inexactly.
## Example use
Here is an example gradient method that uses a line search in step 4.
1. Set iteration counter ${\displaystyle \displaystyle k=0}$, and make an initial guess ${\displaystyle \mathbf {x} _{0}}$ for the minimum
2. Repeat:
3. Compute a descent direction ${\displaystyle \mathbf {p} _{k}}$
4. Choose ${\displaystyle \displaystyle \alpha _{k}}$ to 'loosely' minimize ${\displaystyle h(\alpha )=f(\mathbf {x} _{k}+\alpha \mathbf {p} _{k})}$ over ${\displaystyle \alpha \in \mathbb {R} _{+}}$
5. Update ${\displaystyle \mathbf {x} _{k+1}=\mathbf {x} _{k}+\alpha _{k}\mathbf {p} _{k}}$, and ${\displaystyle \displaystyle k=k+1}$
6. Until ${\displaystyle \|\nabla f(\mathbf {x} _{k})\|}$ < tolerance
At the line search step (4) the algorithm might either exactly minimize h, by solving ${\displaystyle h'(\alpha _{k})=0}$, or loosely, by asking for a sufficient decrease in h. One example of the former is conjugate gradient method. The latter is called inexact line search and may be performed in a number of ways, such as a backtracking line search or using the Wolfe conditions.
Like other optimization methods, line search may be combined with simulated annealing to allow it to jump over some local minima.
## Algorithms
### Direct search methods
In this method, the minimum must first be bracketed, so the algorithm must identify points x1 and x2 such that the sought minimum lies between them. The interval is then divided by computing ${\displaystyle f(x)}$ at two internal points, x3 and x4, and rejecting whichever of the two outer points is not adjacent to that of x3 and x4 which has the lowest function value. In subsequent steps, only one extra internal point needs to be calculated. Of the various methods of dividing the interval,[1] golden section search is particularly simple and effective, as the interval proportions are preserved regardless of how the search proceeds:
${\displaystyle {\frac {1}{\phi }}(x_{2}-x_{1})=x_{4}-x_{1}=x_{2}-x_{3}=\phi (x_{2}-x_{4})=\phi (x_{3}-x_{1})=\phi ^{2}(x_{4}-x_{3})}$ where ${\displaystyle \phi ={\frac {1}{2}}(1+{\sqrt {5}})\approx 1.618}$ | 763 | 2,835 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 17, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-04 | latest | en | 0.778091 |
https://community.tableau.com/thread/204047 | 1,542,621,198,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745522.86/warc/CC-MAIN-20181119084944-20181119110944-00179.warc.gz | 580,364,134 | 25,157 | 2 Replies Latest reply on Apr 3, 2016 8:48 AM by Manideep Bhattacharyya
# Calculation help
Each time a Case ID is sequenced to a unique assigned to group that case ID is considered to have been escalated. I want to be able to display a number that represents the number of escalated case IDs by each group. I am all out of ideas. Any advice/ideas? Thank you.
For example: This Case ID has been escalated 2 times.
• ###### 1. Re: Calculation help
If I'm understanding what you need correctly, then I'm thinking that a LOD expression might provide the solution you are looking for in this situation. For example, the following might work for you.
{fixed [Case ID] : countd([Assigned to Group]) } - 1
The "Fixed" part tells Tableau to do the calculation at the level of Case ID. The CountD will count then count the number of unique groups assigned for each Case ID, and finally -1 since the first assigned group doesn't count as an escalation.
Hope this helps
-Marc
• ###### 2. Re: Calculation help
Dear Stephen - Your workbook contain SQL, that won't open in my desktop. Please try to simulate things in the Superstore dataset and post the problem. This will help you to get solution faster. However if I understood you correct then assigning any activity to "ALT CCI CB MYCOUNTY/MY ADMIN" means escalation.
Create a Calculated field "No. of Escalation"
{fixed [Case ID] : SUM(if Assigned to Group] = "ALT CCI CB MYCOUNTY/MY ADMIN" THEN 1 ELSE 0 END)}
This will return 2 for the case HD0000023119947
Make that Field discrete and drag that field next to the Case will give you required result.
Thanks -
Manideep | 397 | 1,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-47 | latest | en | 0.905525 |
https://forums.developer.nvidia.com/t/waht-is-meaning-of-normal-dot-vel-in-intargralboundaryconstraint-in-aneurysm-py/296899 | 1,721,724,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00716.warc.gz | 219,787,065 | 6,070 | # Waht is meaning of normal_dot_vel in IntargralBoundaryConstraint in aneurysm.py
I tried to understand the code of `aneurysm.py`. However, there is one issue which I cannot understand. What is the meaning of `"normal_dot_vel": 2.540`? The detailed code is below.
I think this code refers to the boundary condition of the outlet region and `normal_dot_vel` represents the volume flow. Is this correct? However, I cannot understand how the flow rate value of 2.540 is calculated.
If someone has knowledge of this code, please provide information. Thank you for your cooperation.
Shusaku
# Integral Continuity 1
``````integral_continuity = IntegralBoundaryConstraint(
nodes=nodes,
geometry=outlet_mesh,
outvar={"normal_dot_vel": 2.540},
batch_size=1,
integral_batch_size=cfg.batch_size.integral_continuity,
lambda_weighting={"normal_dot_vel": 0.1},
)
``````
Hi Shusaku,
yes, you are right. The `normal_dot_vel` represents the volumetric flow rate.
Volumetric flow rate is calculated with inlet area and the mean velocity at the inlet:
`vfr = inlet_area * mean_inlet_vel`
With a velocity magnitude of the parabolic velocity profile at the inlet: `inlet_vel = 1.5` you can approximate the mean velocity with: `mean_inlet_vel = 0.5 * inlet_vel`
The scaled inlet area is: `inlet_area = 21.1284 * (0.4**2)`
This results in: ` vfr = 2.54`
I hope this helps.
Best Daniel
That makes sense! I have one more question.
The variable `normal_dot_vel` has both positive and negative values. What does a negative value signify?
# Integral Continuity 1
``````integral_continuity = IntegralBoundaryConstraint(
nodes=nodes,
geometry=outlet_mesh,
outvar={"normal_dot_vel": 2.540},
batch_size=1,
integral_batch_size=cfg.batch_size.integral_continuity,
lambda_weighting={"normal_dot_vel": 0.1},
)
# Integral Continuity 2
integral_continuity = IntegralBoundaryConstraint(
nodes=nodes,
geometry=integral_mesh,
outvar={"normal_dot_vel": -2.540},
batch_size=1,
integral_batch_size=cfg.batch_size.integral_continuity,
lambda_weighting={"normal_dot_vel": 0.1},
) | 525 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.799818 |
https://www.scribd.com/document/307601280/Ultimate-Ect-Reviewer | 1,568,840,352,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00030.warc.gz | 986,095,862 | 63,492 | You are on page 1of 2
# Which of the following form a moral philosophy,
## which associated with mutual interest among
men, constitutes the foundation of ethics?
A. 2LA
C. 2kLA
D. LA (2 + k)
## A 2 F capacitor with initial charge q0 = 100
C, is connected across a 100 resistor at t =
0. Calculate the time in which the transient
voltage across the resistor drops from 40 to 10
V.
## D. Honesty, courage, and integrity
For a carbon composition resistor color coded
with yellow, violet, orange, and silver stripes,
from left to right, the resistance and tolerance
are:
A. 740ohms @5%
B. 7400ohms @1%
C. 4700ohms @10%
D. 47,000ohms @10%
it?
A. Selectivity
B. Precession
++C. Rigidity
D. Sensitivity
A. 2.77 s
## What system in industrial electronics has the
ability to monitor certain variable in the
industrial processes, the same can perform
self-correcting action?
++B. 0.277 ms
A. Coal-slurry system
C. 2.77 ms
## ++B. Closed-loop system
C. 0.277 s
C. Open-loop system
## What do you call a small D.C generator built
into alternators to provide excitation current to
field windings?
A. prime mover
## In a series RLC circuit with R= 200 L= 0.1 H
and C= 5F, find the transient current after the
switch is closed at t=0 applying a 200V source.
A. 3.3kohms
++B. excitor
A. 2e-1000tsin 1000t
B. 33kohms
C. commutator
B. 2e-1000tcos 1000t
C. 330kohms
C. 2e-100tsin 100t
D. 3.03kohms
What is the condition of the diode in a parallellimiter when the output is developed?
## If 20V are applied across a resistor and there
are 6.06mA of current, the resistance is
## According to our Code of Ethics, Engineers
should not injure the reputation of their fellow
engineers. However, if one has proof that a
professional colleague is involved in illegal
activities, ethically he should
B. Conducting
## A dc current of 50 Amp flows through a long
straight conductor. Determine the force on a
magnetic pole of 50 unit pole strength placed 5
cm from the conductor
C. Cut-off
D. Shorted
B. 1000 dynes
## Which of the following robot is primarily used
for machining processes?
C. 100pi dynes
C. 6.93 ms
D. 6.93 s
A. Cincinatti T1
## What is the hysteresis loss at a frequency of 60
cycles per second of an audio transformer
magnetic core which has a maximum flux
density of 10,000 gausses? The volume of solid
iron of this core is 66.8 cubic cm and the
hysteresis coefficient (a constant that depends
upon grade and quality of iron) id 5.35 x 10-4.
++B. Cincinatti T3
## A. blow the wistle
B. Inform in writing
A. 0.5386 W
B. 53.86 W
++C. 5.386 W
D. 5386 W
When two coils of equal inductances are
connected in series with coefficient of coupling
and their fields in phase, find the total
inductance of two coils.
++A. Shunted
D. 2000 dynes
C. Cincinatti T2
## A solenoid has a magnetic reluctance of 2.2
x10-3. it has 300 turns and a core area of 5 sq.
cm. What is the flux density when the current
flowing is 1 Amp?
D. Cincinatti T4
A. 26,300 Gauss
## The Kuka spot welding robot has how many
degrees of freedom?
B. 12,200 Gauss
A. 4
++B. 6
C. 5
D.7
What is the property of a gyro so that the axis
of rotation or spin axis tends to remain in a
D. 21,200 Gauss | 1,019 | 3,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-39 | latest | en | 0.892453 |
https://freetech4teach.teachermade.com/2023/03/a-round-up-of-pi-day-resources/ | 1,712,993,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00240.warc.gz | 246,307,588 | 20,393 | ### A Round-up of Pi Day Resources
Pi Day is on Tuesday. Last week I shared a few resources for teaching and learning about pi. This post is a summary of those resources and a few more.
OPEN Phys Ed offers five free physical education lesson plans centered around Pi Day. The lesson plans are designed to be used in elementary school and middle school. The five Pi Day lesson plans offered by OPEN are:
• Pi Toss
• Pi Day Races
• Pi Day Dice Relay
• Cake or Pi?
• Who Wants Pi?
To access the lesson plans you do need to register for a free OPEN Phys Ed account. Once you have an account you can download the lesson plans for free as PDFs and Word documents.
On Drawings Of… Lillie Marshall is offering three free printable coloring pages for Pi Day. The first is a visual explanation from the “Pi Queen” and the other two are sheets of little pi-themed cards that students can color and give to each other. Get the Pi Day printables here
Numberphile has a few good videos about pi and Pi Day. Pi with real pies is a three minutes and fourteen seconds video that explains Pi and how it can be calculated.
After showing the video above, you might want to follow up with this video, How Pi Was Nearly Changed to 3.2.
A Mile of Pi, as you might guess, is about a mile of digits.
Exploratorium’s Science Snacks site has three hands-on activities that you can do on Pi Day (or any other day of the year).
• Pi Toss is an activity in which students toss tooth picks is a physical recreation of Buffon’s Needle Problem.
• Pi Graph is an activity in which students graph the diameter and circumference of a series of objects in order to see the linear relationship between any circle’s diameter and circumference.
• Cutting Pi is an activity in which students use string to measure the circumference of an object and then attempt to cut the diameter of the object from the string as many times as possible. In other words, it’s a physical way to divide the circumference by the diameter.
Tynker is a service that offers programming lessons for elementary school and middle school students. For Pi Day Tynker has a free lesson plan in which students practice their programming skills by making art based on Pi. The free lesson plan has students use Tynker’s block programming interface to create art and animations featuring the digits of Pi.
Pi Skyline is an art project that has a Pi Day theme. In the project students shade graph paper to correspond to the digits in pi. Then they cut out the graph and place it on a shaded background to create a city skyline effect. Watch this one minute video to see how the project comes together.
Finally, if you want to give your students a Pi Day ear worm, play the Pi Day Song for them.
##### Archives
Thank You Readers for 14 Amazing Years! | 606 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.921476 |
https://cstheory.stackexchange.com/questions/46209/almost-all-objects-have-property-p-vs-it-is-easy-to-test-whether-an-object-h | 1,656,462,666,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00637.warc.gz | 248,668,020 | 67,880 | # "Almost all objects have property P" vs. "It is easy to test whether an object has property P"
I am interested in any relation between "almost all objects(from a universe) possessing a particular property P" versus "testing whether an object has property P being poly. time decidable".
My guess is that they are completely separate (that is, one doesn't imply the other). Am I missing something?
(Note: Almost all in the sense of probability)
PS: I am not sure whether the tag probability is appropriate here, sorry.
• Almost all reals are irrational. But testing whether "$0$ if Golbach's Conjecture is true and $\sqrt 2$ if it is not" is irrational is not easy Jan 20, 2020 at 21:50
• @HagenvonEitzen How is that a property? Jan 21, 2020 at 0:34
• Almost all strings are not valid C programs that loop forever. But it's very, very hard to test this property. Jan 21, 2020 at 4:32
• @Acccumulation being rational is a property. H von E wants to test this property on a number that has a somewhat annoying definition and finds that it is hard. So one thing we can conclude that how hard it is to test a property of a given object depends at least in part on how the object is presented to you Jan 21, 2020 at 13:12
• @DanielWagner Isn't that halting problem? Jan 21, 2020 at 16:10
They are separate (assuming $$P \ne NP$$). Consider the following property $$P(x)$$: $$x$$ is a $$2n$$-bit string, where either the first $$n$$ bits are not all zeros, or the last $$n$$ bits are a yes-instance of 3SAT. It's clear that testing whether $$x$$ satisfies $$P$$ is NP-hard, yet almost all strings satisfy it: the density $$\to 1$$ as $$n \to \infty$$.
• Nice... anyway you don't need to assume P!=NP, just replace 3SAT by any problem known to not be in P, say an EXPTIME-complete problem Jan 20, 2020 at 8:12
• This proves one direction. The implication in other direction is intuitively wrong; So, this answer completes the picture. Jan 20, 2020 at 12:11
• @CyriacAntony For an explicit counterexample to the other direction, consider the set of strings of even length - easy to recognize, but neither it nor its complement is "almost everything." (And this can be easily tweaked to a set of strings whose limsup frequency is 1 and whose liminf frequency is 0.) Jan 20, 2020 at 17:56
A (counter)example from the recent research literature: almost every simply typed $$\lambda$$-calculus term has a long $$\beta$$-reduction sequence (Asada et al., 2019), but this property is very hard to test, even if P = NP!
Asymptotically, almost every STLC term of order $$k$$ and length $$n$$ has reduction sequence length $$(2\uparrow \uparrow (k - 1))^{\Theta(n)}$$, where $$2 \uparrow \uparrow n = 2^{2^{2^{...}}}$$ is the exponential function iterated $$n$$ times. However, to test this property, the only way is to $$\beta$$-reduce this term and check the reduction sequence length. STLC is strongly normalizing, so it is certainly decidable, but apparently this will take at least $$O((2\uparrow \uparrow (k-1))^{\Theta(n)})$$ time in the worst case, assuming that each reduction step takes $$O(1)$$ time. Deciding this property is apparently not in P. In fact, it is in $$k$$-EXPTIME, so it is not in P even if P=NP!
In the other direction, it's trivial to show that the implication doesn't hold: it is easy to check if a STLC term has a polynomial-length reduction sequence, but almost no term has such a short reduction sequence.
It's quite common for properties split numbers into "almost all" and "almost none" sets. For instance, almost none of Turing machines halt. Almost all real numbers are normal. Almost none of real number are algebraic.
• This does not seem to answer the OP's question. Can you modify the answer to relate to the property of being testable in polynomial time? Jan 21, 2020 at 8:13
• And how do we know that almost no Turing machines halt? One of the known normal numbers is Chaitin constant - and if almost no Turing machines halt that constant would be almost zero, which isn't normal. Jan 21, 2020 at 16:58
• @HansOlsson: I used a notation for Turing machines for awhile for which most corruptions of interesting Turing machines looped forever. Jan 21, 2020 at 21:12 | 1,102 | 4,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-27 | longest | en | 0.913195 |
https://mathexamination.com/class/toeplitz-matrix.php | 1,619,120,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00565.warc.gz | 491,258,621 | 7,017 | ## Take My Toeplitz Matrix Class
A "Toeplitz Matrix Class" QE" is a standard mathematical term for a generalized constant expression which is utilized to fix differential equations and has solutions which are routine. In differential Class resolving, a Toeplitz Matrix function, or "quad" is used.
The Toeplitz Matrix Class in Class type can be expressed as: Q( x) = -kx2, where Q( x) are the Toeplitz Matrix Class and it is a crucial term. The q part of the Class is the Toeplitz Matrix consistent, whereas the x part is the Toeplitz Matrix function.
There are four Toeplitz Matrix functions with appropriate option: K4, K7, K3, and L4. We will now take a look at these Toeplitz Matrix functions and how they are solved.
K4 - The K part of a Toeplitz Matrix Class is the Toeplitz Matrix function. This Toeplitz Matrix function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To resolve for K4 we multiply it by the correct Toeplitz Matrix function: k( x) = x2, y2, or x-y.
K7 - The K7 Toeplitz Matrix Class has a solution of the type: x4y2 - y4x3 = 0. The Toeplitz Matrix function is then multiplied by x to get: x2 + y2 = 0. We then need to multiply the Toeplitz Matrix function with k to get: k( x) = x2 and y2.
K3 - The Toeplitz Matrix function Class is K3 + K2 = 0. We then multiply by k for K3.
K3( t) - The Toeplitz Matrix function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Toeplitz Matrix function which offers: K2( t) = K( t) times k.
The Toeplitz Matrix function is also known as "K4" because of the initials of the letters K and 4. K implies Toeplitz Matrix, and the word "quad" is pronounced as "kah-rab".
The Toeplitz Matrix Class is among the main approaches of fixing differential formulas. In the Toeplitz Matrix function Class, the Toeplitz Matrix function is first multiplied by the suitable Toeplitz Matrix function, which will give the Toeplitz Matrix function.
The Toeplitz Matrix function is then divided by the Toeplitz Matrix function which will divide the Toeplitz Matrix function into a real part and an imaginary part. This provides the Toeplitz Matrix term.
Finally, the Toeplitz Matrix term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right hand side and the term "q".
The Toeplitz Matrix Class is an essential concept to comprehend when resolving a differential Class. The Toeplitz Matrix function is simply one approach to resolve a Toeplitz Matrix Class. The techniques for fixing Toeplitz Matrix formulas include: singular worth decomposition, factorization, optimal algorithm, numerical service or the Toeplitz Matrix function approximation.
## Pay Me To Do Your Toeplitz Matrix Class
If you would like to end up being familiar with the Quartic Class, then you need to first start by checking out the online Quartic page. This page will reveal you how to utilize the Class by utilizing your keyboard. The explanation will also show you how to produce your own algebra formulas to help you study for your classes.
Prior to you can comprehend how to study for a Toeplitz Matrix Class, you need to first understand making use of your keyboard. You will discover how to click the function keys on your keyboard, along with how to type the letters. There are 3 rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.
By pressing Alt and F2, you can increase and divide the value by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power.
When you press Alt and F3, you will enter the number you are attempting to multiply and divide. To increase a number by itself, you will press Alt and X, where X is the number you want to increase. When you press Alt and F3, you will key in the number you are attempting to divide.
This works the very same with the number 6, except you will just key in the two digits that are six apart. Finally, when you press Alt and F3, you will utilize the fourth power. Nevertheless, when you press Alt and F4, you will utilize the real power that you have actually discovered to be the most proper for your problem.
By utilizing the Alt and F function keys, you can multiply, divide, and after that use the formula for the third power. If you need to increase an odd number of x's, then you will need to enter an even number.
This is not the case if you are attempting to do something complex, such as increasing 2 even numbers. For instance, if you wish to increase an odd number of x's, then you will require to go into odd numbers. This is particularly true if you are trying to figure out the response of a Toeplitz Matrix Class.
If you wish to transform an odd number into an even number, then you will need to press Alt and F4. If you do not know how to increase by numbers by themselves, then you will need to use the letters x, a b, c, and d.
While you can multiply and divide by utilize of the numbers, they are a lot easier to use when you can take a look at the power tables for the numbers. You will have to do some research study when you first start to use the numbers, however after a while, it will be force of habit. After you have actually developed your own algebra equations, you will have the ability to develop your own multiplication tables.
The Toeplitz Matrix Solution is not the only method to resolve Toeplitz Matrix formulas. It is very important to discover trigonometry, which uses the Pythagorean theorem, and after that use Toeplitz Matrix formulas to fix issues. With this approach, you can understand about angles and how to fix problems without needing to take another algebra class.
It is necessary to attempt and type as quickly as possible, due to the fact that typing will help you understand about the speed you are typing. This will assist you compose your answers faster.
## Pay Someone To Take My Toeplitz Matrix Class
A Toeplitz Matrix Class is a generalization of a linear Class. For example, when you plug in x=a+b for a given Class, you obtain the value of x. When you plug in x=a for the Class y=c, you obtain the values of x and y, which offer you an outcome of c. By using this basic principle to all the formulas that we have attempted, we can now solve Toeplitz Matrix equations for all the worths of x, and we can do it quickly and efficiently.
There are numerous online resources readily available that provide free or affordable Toeplitz Matrix equations to resolve for all the worths of x, including the cost of time for you to be able to take advantage of their Toeplitz Matrix Class assignment aid service. These resources normally do not require a membership fee or any sort of investment.
The answers provided are the outcome of complex-variable Toeplitz Matrix equations that have been fixed. This is also the case when the variable used is an unknown number.
The Toeplitz Matrix Class is a term that is an extension of a linear Class. One advantage of using Toeplitz Matrix equations is that they are more general than the linear formulas. They are simpler to resolve for all the values of x.
When the variable utilized in the Toeplitz Matrix Class is of the kind x=a+b, it is simpler to resolve the Toeplitz Matrix Class since there are no unknowns. As a result, there are fewer points on the line defined by x and a continuous variable.
For a right-angle triangle whose base indicate the right and whose hypotenuse indicate the left, the right-angle tangent and curve graph will form a Toeplitz Matrix Class. This Class has one unknown that can be discovered with the Toeplitz Matrix formula. For a Toeplitz Matrix Class, the point on the line defined by the x variable and a constant term are called the axis.
The presence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a Toeplitz Matrix Class, are an offered, we can find all the values of x and they will sum to the given worths. This is attained when we use the Toeplitz Matrix formula.
The aspect of being a continuous element is called the system of equations in Toeplitz Matrix equations. This is sometimes called the central Class.
Toeplitz Matrix equations can be fixed for other worths of x. One way to resolve Toeplitz Matrix equations for other worths of x is to divide the x variable into its element part.
If the variable is given as a positive number, it can be divided into its aspect parts to get the normal part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a constant. In such a case, the formula is a third-order Toeplitz Matrix Class.
If the variable x is unfavorable, it can be divided into the very same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order Toeplitz Matrix Class.
Solution aid service in resolving Toeplitz Matrix formulas. When using an online service for solving Toeplitz Matrix equations, the Class will be resolved immediately. | 2,054 | 9,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.881531 |
http://www.ece.lsu.edu/ee4720/1997/hw08_sol.html | 1,544,912,859,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00014.warc.gz | 369,388,195 | 3,019 | # EE 4720 Computer Architecture - HW 8 Solution
### Problem 1
The original program with line numbers added:
```1: add r1, r2, r3
2: sub r6, r7, r8
3: lw r10, 0(r20)
5: sub r14, r1, r9
7: sub r16, r17, r18
9: sw 0(r20), r6
```
The instructions are rearranged and placed in groups of 3 for 3VDLX. The first three instructions can be, and are, grouped together because they don't use any values they produce. The next three, on lines 4, 5, and 6, cannot be grouped together because the value produced on line 5 is needed on line 6. The inefficient solution is to keep the instructions in the same order and use nops to avoid hazards, starting with the load latency:
```add r1, r2, r3; sub r6, r7, r8; lw r10, 0(r20)
nop; nop; nop
add r11, r10, r12; sub r14, r1, r9; nop
add r1, r14, r15; sub r16, r17, r18; add r19, r21, r22
sw 0(r20), r6 nop; nop
```
The first set of nops is needed because of the load instruction. The nop in the third 3VDLX instruction is needed because of the true dependency between lines 5 and 6 in the original program. The nops in the last 3VDLX instruction would be needed if we had no further instructions (which is unlikely when the last instruction is not a CTI, but that's where the problem ends).
The instructions can be rearranged for efficient execution. The instructions at lines 5, 7 and 8 can be placed in the second 3VDLX instruction. The remaining can be placed in the third 3VDLX instruction yielding:
```add r1, r2, r3; sub r6, r7, r8; lw r10, 0(r20)
sub r16, r17, r18; add r19, r21, r22; sub r14, r1, r9
```
Since the VLIW instructions execute as a unit, there is no need to distinguish between the separate parts of a pipeline stage, as is done for superscalar.
```add sub lw IF ID EX MEM WB
sub add sub IF ID EX MEM WB
```
### Problem 2
Each pipeline stage has room for three instructions, superscripts are used to distinguish the parts. To save space, M is used for the MEM stage.
The first three instructions can start executing without delay. Of the next three, only the one on line 5 can start, the add on line 4 must wait for the load and the add on line 6 must wait for the sub on line 5. With two instructions in ID the last group of three instructions are stalled at time 3.
```Time 0 1 2 3 4 5 6 7
add IF1 ID1 EX1 M1 WB1
sub IF2 ID2 EX2 M2 WB2
lw IF3 ID3 EX3 M3 WB3
add IF1 ID1 EX1 M1 WB1
sub IF2 ID2 EX2 M2 WB2
add IF3 ID3 EX3 M3 WB3
sub IF1 ID1 EX1 M1 WB1
add IF2 ID2 EX2 M2 WB2
sw IF3 ID3 EX3 M3 WB3
```
### Problem 3
Reservation stations, as described in class, add an extra cycle of latency because of the complexity of using the CDB (in the WB "stage"). The reservation station numbers associated with the functional units are:
```Int 1 2
L/S 3 4
Int 5 6
L/S 7 8
Int 9 10
L/S 11 12
```
where Int refers to an integer execution unit and L/S refers to a load/store unit. Note that reservation stations are dedicated to functional units. If ready instructions were waiting in reservation stations 1 and 2, only one could start in a cycle, even if the other two integer execution units were free.
The pipeline notation is the same used in class, for example, 5:EX3 indicates that execution unit 3 is executing an instruction from reservation station 5.
Load and store instructions need an ALU to compute addresses, in this solution a load/store unit has its own ALU, that stage is indicated by AD.
Execution is given below:
```Time 0 1 2 3 4 5 6 7
sub IF2 ID2 5:EX2 5:WB2
lw IF3 ID3 11:AD3 11:M3 11:WB3
add IF1 ID1 2:RS 2:RS 2:EX1 2:WB1
sub IF2 ID2 6:RS 6:EX2 6:WB2
add IF3 ID3 9:RS 9:RS 9:RS 9:EX3 9:WB3
sub IF1 ID1 1:EX1 1:WB1 | 1,261 | 3,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-51 | latest | en | 0.87314 |
http://www.fixya.com/support/t1418764-more_watts_needed | 1,493,223,118,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121453.27/warc/CC-MAIN-20170423031201-00354-ip-10-145-167-34.ec2.internal.warc.gz | 524,047,820 | 37,085 | # More watts needed
Can a pre amp or something be added to a home theater setup to increase the wattage to 2 floorstanding jbl 250 watt fronts?
I have a Onkyo tx- sr601 receiver..
Can the front channels on that receiver be vbidged?
Thanks,
Justin
Posted by on
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• Contributor
The TX-SR601 does not have pre-amp outputs. Unless you know how to splice it at the circuit level and install your own RCA plugs, I don't think you can accomplish this task with the 601.
Posted on Jan 11, 2010
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Add another power amp to line outs from your main amp to your speakers this will give more power(current) to your main speakers as you have decreased load on main amp
Posted on Jan 08, 2009
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Posted on Jan 02, 2017
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## Related Questions:
### Given 1hp motor how to compute the exact amount of amphere use in circuit breaker
You're not going to be able to do this with just a known Horse Power.
There are 3 elements to the equation, with any two, you can work out the third.
If you want to know how the amperage, you will need to know the voltage and wattage of the motor. I imagine that you already know the voltage (It's going to be 220V or 110 volt)
Watts divided by volts = Amps
Examples:
A 220v 1000 watt motor (1000 divided by 220) will draw 4.55 amps
A 110v 800 watt motor (800 divided by 110) will draw 7.27 amps
Bear in mind that most washing machines have a couple of windings for wash and spin. As an average, the was winding will usually be about 500 watts to spin and about 250 watts to wash. ALSO, bear in mind that if you are using this data for a WASHING MACHINE, then there is a water heating element in there too and that draws about 2Kw (2000 watts)
Dont just take this as read, you DO need to check wattages, but, working on what I have just said, the max consumption on a 220V machine will look like this:
At Spin, with a 500 Watt consumption: (500/220) = 2.3 amps
While Washing with a 250 watt consumption: (250/220) = 1.14 amps
Consider that the WASH and HEAT may be running at the same time.
2Kw heating (2000/220) = 9.1 amps PLUS 1.14 amps for the motor - Total wattage 10.24 amps
Watts / Volts = Amps
Amps x Volts = Watts
Watts divided by amps = Volts
Aug 25, 2011 | Washing Machines
### I have the phillips hts5580w. how much wattage output does it have? an does this system have a built in amplifier as well?
Anything that drives speakers must be or have amplifiers. They are rated in watts.
http://www.retrevo.com/support/Philips-HTS5580W-Home-Theater-Systems-manual/id/23889dj591/t/2/
It says the power output, presumably in the phoniest of terms - all channels summed together with THIRTY PRECENT DISTORTION which would seriously risk any speaker system and the health of tha amps themselves - is 1000 watts. This is a joke, a lie, marekting at its worst, especially when the totoal power consumption of the unit (that power whci it drwas from the wall outlet) is ONLY 135 watts.
Philips must be onto something no one else has ever seen - they get six times as much sound energy out of the unit as the amount of electricity that goes into it, plus all that heat. That, my friend, would be a miracle and would put all the reputable amp maufacturers, who rate their amps in honest watts per channel at certain (vanishingly low) distortion figures, to shame.
If you push this unit into sustained 30% distortion it will self destruct.
Aug 04, 2011 | Philips HTS5580W Theater System
### 1 question can Cutler Hammer breakers have the amperage feed through the breakers. To the breaker box and from the box grid to the load. Does it matter which way the amperage is fed through the breaker?...
There seem to be several questions mingled in with those 2 questions.
1) Amps can go either way
2) Breakers trip when heat exceeds certain level. Heat is caused by amps.
3) You want to add a subpanel in garage.
I do not know the code in your area for installing a subpanel.
Your plan will work by connecting 6 gauge wire to main breaker, and new 60 amp breaker will protect wires between main box and subpanel.
Remember, wires going into main breaker cannot be turned off without pulling meter.
Some areas require license to pull meter.
http://waterheatertimer.org/How-to-replace-circuit-breaker.html
4) To work around license and meter-pull, simply replace a 240V breaker in main panel with your new 60 amp breaker.
Then move breakers around to match new set-up.
New subpanel can accommodate two new 240V breakers and one new 120 Breaker
http://waterheatertimer.org/How-to-install-a-subpanel.html
5) 6 gauge wire is correct for 60 amp breaker.
6) I'm not sure what you mean by the box rails?
7) Do you need more amperage on main service?
This means adding a larger service panel with 150 Watt or 200 Watt main breaker.
Some areas require service upgrade when remodeling or adding circuits.
8) How to figure total amp draw at your house.
Add up total watts being used.
For example you have 1/2 Hp motor
754 watts per Hp
1/2 Hp = 377Watts
Volts x Amps = Watts
Amps = Watts divided by Volts.
377 Watts divided by 240Volts = 1.57 amps (plus a bit more amperage when motor starts)
40 watt light bulb divided by 120Volts = .33 amps
Double oven has label located inside door that shows upper and lower wattage.
Dishwasher, dryer, big screen TV, satellite receiver all have labels that show wattage.
Water heater has label that shows wattage of each element.
http://waterheatertimer.org/Figure-Volts-Amps-Watts-for-water-heater.html
http://waterheatertimer.org/See-inside-main-breaker-box.html
Add a comment for more help
Feb 18, 2011 | Cutler Hammer 100 Amp Main Breaker...
### Car subwoofer just past half volume cuts off the amp
Assuming there is no short, or lose ground wire somewhere inside the box, diodes( in the amp) are fine, you might be overheating it by not allowing it to get good ventilation. This can be a big deal on some low end Chinese amps. Or the speakers are blown , or your Resistors (in the amp) are bad, but for arguments sake.
Most people don't use the proper size and or conductivity types of wiring, for pushing high amperage Think about it like this, Watts (power) =.Amps (energy) x Voltage (pressure) Resistance or Ohm's (amount) determines how much Wattage will go in or out, Ohms = Volts x2 / Watts.
What gauge wire are you using on your setup?
Jan 09, 2010 | Car Audio & Video
### Does onkyo HT-s3100 supports 4 Ohms speakers from JBL(control 1)
Palaboy is incorrect. JBL Control 1 speakers are 4 Ohm 150 watt speakers. He's probably thinking of the Control 1x speakers which are 8 ohm 80 watt speakers.
If you hook your Control 1s to that Onkyo, it won't sound great and could cause the receiver to overheat and trip the protection circuit if played at high volume.
Aug 11, 2009 | Onkyo HT-S3100 Receiver
### Wattage for Kirby Sentria G10D
The conversion of Amps to Watts is governed by the equation
Watts = Amps x Volts
For example 1 amp @ 110 volts = 110 watts
your case 7 amp @ 120 volts = 840 watts
unless you are running it at 230v but I don't think you can
I would look for one rated for Volt-Amps instead of watts
It's best to round up to be safe add 40 - 60 watts
Don
Apr 25, 2009 | Kirby Ultimate G Bagged Upright Vacuum
### Turntable to Receiver
Yes you will need a pre-amp to connect a turntable to this unit. For the AUX input you will need something that can output 400mV into 47k ohm, somewhere like RadioShack should have what you need. Please update the question & let us know if the information given was useful to you - Good Luck!
Aug 31, 2007 | Audio Players & Recorders
### Combining an integrated receiver and an external amp.
yes, you are "forgoing the 100 Watts of my integrated receiver and now just using the 100 watts from the external amp"
Jun 03, 2007 | Home Theater Direct MA-1235 Amplifier
### 1000 watt amp sony xplod
how do you have your system hooked? What speakers do you have? There is a lot more too it then just wattage.
Jan 23, 2007 | Sony Xplod Car Audio Amplifier XM-DS1300P5...
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Level 3 Expert | 2,328 | 8,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-17 | longest | en | 0.94825 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/192/C4xQ8s2S3.html | 1,585,475,553,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494064.21/warc/CC-MAIN-20200329074745-20200329104745-00484.warc.gz | 604,325,848 | 7,687 | Copied to
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## G = C4×Q8⋊2S3order 192 = 26·3
### Direct product of C4 and Q8⋊2S3
Series: Derived Chief Lower central Upper central
Derived series C1 — C12 — C4×Q8⋊2S3
Chief series C1 — C3 — C6 — C2×C6 — C2×C12 — C2×D12 — C2×Q8⋊2S3 — C4×Q8⋊2S3
Lower central C3 — C6 — C12 — C4×Q8⋊2S3
Upper central C1 — C2×C4 — C42 — C4×Q8
Generators and relations for C4×Q82S3
G = < a,b,c,d,e | a4=b4=d3=e2=1, c2=b2, ab=ba, ac=ca, ad=da, ae=ea, cbc-1=ebe=b-1, bd=db, cd=dc, ece=b-1c, ede=d-1 >
Subgroups: 328 in 122 conjugacy classes, 55 normal (39 characteristic)
C1, C2 [×3], C2 [×2], C3, C4 [×2], C4 [×2], C4 [×5], C22, C22 [×4], S3 [×2], C6 [×3], C8 [×3], C2×C4 [×3], C2×C4 [×5], D4 [×3], Q8 [×2], Q8, C23, Dic3, C12 [×2], C12 [×2], C12 [×4], D6 [×4], C2×C6, C42, C42, C22⋊C4, C4⋊C4, C4⋊C4 [×2], C2×C8 [×2], SD16 [×4], C22×C4, C2×D4, C2×Q8, C3⋊C8 [×2], C3⋊C8, C4×S3 [×2], D12 [×2], D12, C2×Dic3, C2×C12 [×3], C2×C12 [×2], C3×Q8 [×2], C3×Q8, C22×S3, C4×C8, D4⋊C4, Q8⋊C4, C4.Q8, C4×D4, C4×Q8, C2×SD16, C2×C3⋊C8 [×2], C4⋊Dic3, D6⋊C4, Q82S3 [×4], C4×C12, C4×C12, C3×C4⋊C4, C3×C4⋊C4, S3×C2×C4, C2×D12, C6×Q8, C4×SD16, C4×C3⋊C8, C12.Q8, C6.D8, Q82Dic3, C4×D12, C2×Q82S3, Q8×C12, C4×Q82S3
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], S3, C2×C4 [×6], D4 [×2], C23, D6 [×3], SD16 [×2], C22×C4, C2×D4, C4○D4, C4×S3 [×2], C3⋊D4 [×2], C22×S3, C4×D4, C2×SD16, C4○D8, Q82S3 [×2], S3×C2×C4, C4○D12, C2×C3⋊D4, C4×SD16, C4×C3⋊D4, C2×Q82S3, Q8.13D6, C4×Q82S3
Smallest permutation representation of C4×Q82S3
On 96 points
Generators in S96
(1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16)(17 18 19 20)(21 22 23 24)(25 26 27 28)(29 30 31 32)(33 34 35 36)(37 38 39 40)(41 42 43 44)(45 46 47 48)(49 50 51 52)(53 54 55 56)(57 58 59 60)(61 62 63 64)(65 66 67 68)(69 70 71 72)(73 74 75 76)(77 78 79 80)(81 82 83 84)(85 86 87 88)(89 90 91 92)(93 94 95 96)
(1 33 51 11)(2 34 52 12)(3 35 49 9)(4 36 50 10)(5 44 64 48)(6 41 61 45)(7 42 62 46)(8 43 63 47)(13 80 66 72)(14 77 67 69)(15 78 68 70)(16 79 65 71)(17 31 75 23)(18 32 76 24)(19 29 73 21)(20 30 74 22)(25 82 56 60)(26 83 53 57)(27 84 54 58)(28 81 55 59)(37 92 86 96)(38 89 87 93)(39 90 88 94)(40 91 85 95)
(1 26 51 53)(2 27 52 54)(3 28 49 55)(4 25 50 56)(5 39 64 88)(6 40 61 85)(7 37 62 86)(8 38 63 87)(9 81 35 59)(10 82 36 60)(11 83 33 57)(12 84 34 58)(13 74 66 20)(14 75 67 17)(15 76 68 18)(16 73 65 19)(21 79 29 71)(22 80 30 72)(23 77 31 69)(24 78 32 70)(41 95 45 91)(42 96 46 92)(43 93 47 89)(44 94 48 90)
(1 43 23)(2 44 24)(3 41 21)(4 42 22)(5 76 12)(6 73 9)(7 74 10)(8 75 11)(13 60 86)(14 57 87)(15 58 88)(16 59 85)(17 33 63)(18 34 64)(19 35 61)(20 36 62)(25 96 80)(26 93 77)(27 94 78)(28 95 79)(29 49 45)(30 50 46)(31 51 47)(32 52 48)(37 66 82)(38 67 83)(39 68 84)(40 65 81)(53 89 69)(54 90 70)(55 91 71)(56 92 72)
(5 18)(6 19)(7 20)(8 17)(9 35)(10 36)(11 33)(12 34)(13 92)(14 89)(15 90)(16 91)(21 41)(22 42)(23 43)(24 44)(25 82)(26 83)(27 84)(28 81)(29 45)(30 46)(31 47)(32 48)(37 80)(38 77)(39 78)(40 79)(53 57)(54 58)(55 59)(56 60)(61 73)(62 74)(63 75)(64 76)(65 95)(66 96)(67 93)(68 94)(69 87)(70 88)(71 85)(72 86)
G:=sub<Sym(96)| (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,33,51,11)(2,34,52,12)(3,35,49,9)(4,36,50,10)(5,44,64,48)(6,41,61,45)(7,42,62,46)(8,43,63,47)(13,80,66,72)(14,77,67,69)(15,78,68,70)(16,79,65,71)(17,31,75,23)(18,32,76,24)(19,29,73,21)(20,30,74,22)(25,82,56,60)(26,83,53,57)(27,84,54,58)(28,81,55,59)(37,92,86,96)(38,89,87,93)(39,90,88,94)(40,91,85,95), (1,26,51,53)(2,27,52,54)(3,28,49,55)(4,25,50,56)(5,39,64,88)(6,40,61,85)(7,37,62,86)(8,38,63,87)(9,81,35,59)(10,82,36,60)(11,83,33,57)(12,84,34,58)(13,74,66,20)(14,75,67,17)(15,76,68,18)(16,73,65,19)(21,79,29,71)(22,80,30,72)(23,77,31,69)(24,78,32,70)(41,95,45,91)(42,96,46,92)(43,93,47,89)(44,94,48,90), (1,43,23)(2,44,24)(3,41,21)(4,42,22)(5,76,12)(6,73,9)(7,74,10)(8,75,11)(13,60,86)(14,57,87)(15,58,88)(16,59,85)(17,33,63)(18,34,64)(19,35,61)(20,36,62)(25,96,80)(26,93,77)(27,94,78)(28,95,79)(29,49,45)(30,50,46)(31,51,47)(32,52,48)(37,66,82)(38,67,83)(39,68,84)(40,65,81)(53,89,69)(54,90,70)(55,91,71)(56,92,72), (5,18)(6,19)(7,20)(8,17)(9,35)(10,36)(11,33)(12,34)(13,92)(14,89)(15,90)(16,91)(21,41)(22,42)(23,43)(24,44)(25,82)(26,83)(27,84)(28,81)(29,45)(30,46)(31,47)(32,48)(37,80)(38,77)(39,78)(40,79)(53,57)(54,58)(55,59)(56,60)(61,73)(62,74)(63,75)(64,76)(65,95)(66,96)(67,93)(68,94)(69,87)(70,88)(71,85)(72,86)>;
G:=Group( (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,33,51,11)(2,34,52,12)(3,35,49,9)(4,36,50,10)(5,44,64,48)(6,41,61,45)(7,42,62,46)(8,43,63,47)(13,80,66,72)(14,77,67,69)(15,78,68,70)(16,79,65,71)(17,31,75,23)(18,32,76,24)(19,29,73,21)(20,30,74,22)(25,82,56,60)(26,83,53,57)(27,84,54,58)(28,81,55,59)(37,92,86,96)(38,89,87,93)(39,90,88,94)(40,91,85,95), (1,26,51,53)(2,27,52,54)(3,28,49,55)(4,25,50,56)(5,39,64,88)(6,40,61,85)(7,37,62,86)(8,38,63,87)(9,81,35,59)(10,82,36,60)(11,83,33,57)(12,84,34,58)(13,74,66,20)(14,75,67,17)(15,76,68,18)(16,73,65,19)(21,79,29,71)(22,80,30,72)(23,77,31,69)(24,78,32,70)(41,95,45,91)(42,96,46,92)(43,93,47,89)(44,94,48,90), (1,43,23)(2,44,24)(3,41,21)(4,42,22)(5,76,12)(6,73,9)(7,74,10)(8,75,11)(13,60,86)(14,57,87)(15,58,88)(16,59,85)(17,33,63)(18,34,64)(19,35,61)(20,36,62)(25,96,80)(26,93,77)(27,94,78)(28,95,79)(29,49,45)(30,50,46)(31,51,47)(32,52,48)(37,66,82)(38,67,83)(39,68,84)(40,65,81)(53,89,69)(54,90,70)(55,91,71)(56,92,72), (5,18)(6,19)(7,20)(8,17)(9,35)(10,36)(11,33)(12,34)(13,92)(14,89)(15,90)(16,91)(21,41)(22,42)(23,43)(24,44)(25,82)(26,83)(27,84)(28,81)(29,45)(30,46)(31,47)(32,48)(37,80)(38,77)(39,78)(40,79)(53,57)(54,58)(55,59)(56,60)(61,73)(62,74)(63,75)(64,76)(65,95)(66,96)(67,93)(68,94)(69,87)(70,88)(71,85)(72,86) );
G=PermutationGroup([(1,2,3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16),(17,18,19,20),(21,22,23,24),(25,26,27,28),(29,30,31,32),(33,34,35,36),(37,38,39,40),(41,42,43,44),(45,46,47,48),(49,50,51,52),(53,54,55,56),(57,58,59,60),(61,62,63,64),(65,66,67,68),(69,70,71,72),(73,74,75,76),(77,78,79,80),(81,82,83,84),(85,86,87,88),(89,90,91,92),(93,94,95,96)], [(1,33,51,11),(2,34,52,12),(3,35,49,9),(4,36,50,10),(5,44,64,48),(6,41,61,45),(7,42,62,46),(8,43,63,47),(13,80,66,72),(14,77,67,69),(15,78,68,70),(16,79,65,71),(17,31,75,23),(18,32,76,24),(19,29,73,21),(20,30,74,22),(25,82,56,60),(26,83,53,57),(27,84,54,58),(28,81,55,59),(37,92,86,96),(38,89,87,93),(39,90,88,94),(40,91,85,95)], [(1,26,51,53),(2,27,52,54),(3,28,49,55),(4,25,50,56),(5,39,64,88),(6,40,61,85),(7,37,62,86),(8,38,63,87),(9,81,35,59),(10,82,36,60),(11,83,33,57),(12,84,34,58),(13,74,66,20),(14,75,67,17),(15,76,68,18),(16,73,65,19),(21,79,29,71),(22,80,30,72),(23,77,31,69),(24,78,32,70),(41,95,45,91),(42,96,46,92),(43,93,47,89),(44,94,48,90)], [(1,43,23),(2,44,24),(3,41,21),(4,42,22),(5,76,12),(6,73,9),(7,74,10),(8,75,11),(13,60,86),(14,57,87),(15,58,88),(16,59,85),(17,33,63),(18,34,64),(19,35,61),(20,36,62),(25,96,80),(26,93,77),(27,94,78),(28,95,79),(29,49,45),(30,50,46),(31,51,47),(32,52,48),(37,66,82),(38,67,83),(39,68,84),(40,65,81),(53,89,69),(54,90,70),(55,91,71),(56,92,72)], [(5,18),(6,19),(7,20),(8,17),(9,35),(10,36),(11,33),(12,34),(13,92),(14,89),(15,90),(16,91),(21,41),(22,42),(23,43),(24,44),(25,82),(26,83),(27,84),(28,81),(29,45),(30,46),(31,47),(32,48),(37,80),(38,77),(39,78),(40,79),(53,57),(54,58),(55,59),(56,60),(61,73),(62,74),(63,75),(64,76),(65,95),(66,96),(67,93),(68,94),(69,87),(70,88),(71,85),(72,86)])
48 conjugacy classes
class 1 2A 2B 2C 2D 2E 3 4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 4K 4L 4M 4N 6A 6B 6C 8A ··· 8H 12A 12B 12C 12D 12E ··· 12P order 1 2 2 2 2 2 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 6 6 6 8 ··· 8 12 12 12 12 12 ··· 12 size 1 1 1 1 12 12 2 1 1 1 1 2 2 2 2 4 4 4 4 12 12 2 2 2 6 ··· 6 2 2 2 2 4 ··· 4
48 irreducible representations
dim 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 4 4 type + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C2 C2 C4 S3 D4 D6 D6 D6 SD16 C4○D4 C3⋊D4 C4×S3 C4○D8 C4○D12 Q8⋊2S3 Q8.13D6 kernel C4×Q8⋊2S3 C4×C3⋊C8 C12.Q8 C6.D8 Q8⋊2Dic3 C4×D12 C2×Q8⋊2S3 Q8×C12 Q8⋊2S3 C4×Q8 C2×C12 C42 C4⋊C4 C2×Q8 C12 C12 C2×C4 Q8 C6 C4 C4 C2 # reps 1 1 1 1 1 1 1 1 8 1 2 1 1 1 4 2 4 4 4 4 2 2
Matrix representation of C4×Q82S3 in GL4(𝔽73) generated by
27 0 0 0 0 27 0 0 0 0 1 0 0 0 0 1
,
1 0 0 0 0 1 0 0 0 0 1 3 0 0 48 72
,
1 0 0 0 0 1 0 0 0 0 61 55 0 0 4 12
,
72 1 0 0 72 0 0 0 0 0 1 0 0 0 0 1
,
0 1 0 0 1 0 0 0 0 0 1 0 0 0 48 72
G:=sub<GL(4,GF(73))| [27,0,0,0,0,27,0,0,0,0,1,0,0,0,0,1],[1,0,0,0,0,1,0,0,0,0,1,48,0,0,3,72],[1,0,0,0,0,1,0,0,0,0,61,4,0,0,55,12],[72,72,0,0,1,0,0,0,0,0,1,0,0,0,0,1],[0,1,0,0,1,0,0,0,0,0,1,48,0,0,0,72] >;
C4×Q82S3 in GAP, Magma, Sage, TeX
C_4\times Q_8\rtimes_2S_3
% in TeX
G:=Group("C4xQ8:2S3");
// GroupNames label
G:=SmallGroup(192,584);
// by ID
G=gap.SmallGroup(192,584);
# by ID
G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-3,253,232,58,1684,851,102,6278]);
// Polycyclic
G:=Group<a,b,c,d,e|a^4=b^4=d^3=e^2=1,c^2=b^2,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c^-1=e*b*e=b^-1,b*d=d*b,c*d=d*c,e*c*e=b^-1*c,e*d*e=d^-1>;
// generators/relations
×
𝔽 | 5,948 | 9,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | longest | en | 0.323376 |
https://atplquestions.com/atplh-ir-atplh-question/general-navigation/12 | 1,709,232,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00200.warc.gz | 103,512,479 | 7,284 | 12 / 20
An aircraft departs points A (50°N, 012°E) on a westerly track and flies for 300 NM to point B. From point B, the aircraft flies a southerly track of 300 NM to point C. From point C, the aircraft finally flies a westerly track of 300 NM to point D. The coordinates of point D are:
• A
64°N, 010°48’W.
• B
45ºN, 002º51’W
• C
45°N, 010°48’E.
• D
64°N, 033°06’E.
To solve this type of question, it is important to be familiar with:
• Departure (NM) = Change of Longitude (min) x 60 x cos Lat
• One minute of latitude equals one nautical mile and degrees of latitude are 60 nm apart.
(1) From 50ºN 012ºE => Aircraft flies westwards for 300 NM
Start by using the Departure formula to find the change of longitude:
300 NM = Change of longitude (min) x 60 x cos (50º)
Change of longitude (º) = 300 / [cos (50º) x 60] = 7.78 (Convert it to minutes)
Change of longitude (º) = 7.47º westerly
- After the first leg (westwards), the aircraft will be at 012ºE - 007.47º = 4.13º E
• 50ºN 004.13ºE
(2) From 50ºN 003.7ºE => Aircraft flies southwards for 300 NM
1º Latitude = 60 NM
Therefore, 300 NM / 60 NM = 5º southerly
- After the second leg (southwards), the aircraft will be at 50ºN - 5ºS = 45ºN
• 45ºN 004.13ºE
(3) From 45ºN 003.7ºE => Aircraft flies westwards for 300 NM
Start by using the Departure formula to find the change of longitude:
300 NM = Change of longitude (min) x 60 x cos (45º)
Change of longitude (º) = 300 / [cos (45º) x 60] = 7.07 (Convert it to minutes)
Change of longitude (º) = 7.04º westerly
- After the third leg (westwards), the aircraft will be at 004.13ºE – 7.04º = 2.51ºW
• 45ºN 002º51’W
=> Aircraft's final position: 45ºN 002º51’W
Your Notes (not visible to others)
This question has appeared on the real examination, you can find the related countries below. | 620 | 1,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-10 | latest | en | 0.723817 |
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A098371 #{m: GCD(n,a(m)) = Max{GCD(n,a(k)): 1<=k
1, 1, 2, 1, 4, 2, 6, 1, 1, 4, 10, 1, 12, 7, 1, 3, 16, 2, 18, 1, 1, 10, 22, 1, 2, 13, 1, 1, 28, 2, 30, 1, 1, 16, 2, 1, 36, 19, 1, 3, 40, 1, 42, 1, 1, 21, 46, 2, 4, 4, 9, 1, 52, 2, 1, 1, 1, 27, 58, 1, 60, 29, 2, 2, 2, 1, 66, 12, 1, 2, 70, 1, 72, 37, 2, 1, 2, 1, 78, 1, 1, 40, 82, 1, 7, 42, 2, 2, 88, 2, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS a(n) = n-1 iff n = 1 or n is prime; a(A098373(n)) = n and a(m) <> n for m < A098373(n); a(A098374(n)) = 1; a(n) = #{m: GCD(n,a(m)) = A098372(n)}. LINKS CROSSREFS Cf. A096216, A000010. Sequence in context: A217916 A057923 A147763 * A300234 A070777 A173614 Adjacent sequences: A098368 A098369 A098370 * A098372 A098373 A098374 KEYWORD nonn AUTHOR Reinhard Zumkeller, Sep 05 2004 STATUS approved
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Last modified July 17 14:44 EDT 2019. Contains 325106 sequences. (Running on oeis4.) | 586 | 1,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.586897 |
https://nl.mathworks.com/matlabcentral/cody/players/4529133-marc-jakobi/solved | 1,591,354,209,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00576.warc.gz | 468,522,107 | 22,345 | Cody
# Marc Jakobi
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1 – 50 of 228 | 1,084 | 3,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-24 | latest | en | 0.711148 |
https://www.coursehero.com/file/6686969/Lab-9-Measurement-of-the-Index-of-Refraction/ | 1,490,850,025,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00247-ip-10-233-31-227.ec2.internal.warc.gz | 883,237,158 | 22,288 | Lab 9 Measurement of the Index of Refraction
# Lab 9 Measurement - Lab 9 Measurement of the Index of Refraction 2‘ Lab 9 Report(100 points Name-’" PHYS 1214 Section,2 Day and Time,1 L 14
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Unformatted text preview: Lab 9 Measurement of the Index of Refraction 2‘ Lab 9 Report (100 points) Name: -’ " PHYS 1214 Section: ,2 Day and Time: ,1 L/ 14 Station Check-in: ‘5 4 Table 1. Data for the Measurement of the Prism An Angle Readings Angle between rays 2 a-b and b’-a' Angle of Minimum Deviation (D*) Yellow line (579.2 nm) Violet line (404.8 nm) Exercises and Questions 1. From the values of A and D)k that were measured, calculate the corresponding values of the index of refraction for each cfigil’or. Show your work, including equations. (12 pts) H ‘w 7‘ ‘ Vests/m V\ f ; S Q " ' . “I a. 2 am: + H0‘ A We? j\ s! ' 1 a i i \j i m a d; x : ; it: m " B “2:. i / 2. The reference value for the violet line is nviolet=1.6522, and the reference value for the yellow line is nyeuow=1.6213. Calculate the percent error for both lines. Show your work for one ; in: Lazigawm " " o i may 14m I H ,g, M ’7 «r 1 ' - 0 11mm “00 " ’0 46M ’ 115 fleézz K/OO ' [glen/:Z‘gwwi f E Lab 9 Measurement of the Index of Refraction - 3. Based solely on your data, which color of light is deviated M ghost gand least) in passing through the prism? Explain your reasoning. (10 pts) 9~ 1% K 5‘ k." r ( A; 7 . R I, ,‘ v. 9 g , . _ i W “"3” fliQViii“? "J i ,1' V37 2,49” :1 I I 4. Based solely on your data, which c olor of light has the greatest (and least) speed in glass? Explain your reasoning. (10 pts) s '3‘ j ’ 31v ». ‘ ‘ ’ i} f ( § a“ a ’“ r {Where W «W/ «~— 52 Q we; '57; a w z v write” 0 c1 5. Based on your data, calculate the sp one calculation, including equations and units. (10 pts) ...
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## This note was uploaded on 01/10/2012 for the course PHYS 2114 taught by Professor Bandy during the Spring '08 term at Oklahoma State.
### Page1 / 2
Lab 9 Measurement - Lab 9 Measurement of the Index of Refraction 2‘ Lab 9 Report(100 points Name-’" PHYS 1214 Section,2 Day and Time,1 L 14
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Ask a homework question - tutors are online | 790 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-13 | longest | en | 0.765444 |
https://www.crazy-numbers.com/en/2574 | 1,708,663,084,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00569.warc.gz | 748,623,286 | 6,506 | Discover a lot of information on the number 2574: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 2574
Is 2574 a prime number? No
Is 2574 a perfect number? No
Number of divisors 24
List of dividers 1, 2, 3, 6, 9, 11, 13, 18, 22, 26, 33, 39, 66, 78, 99, 117, 143, 198, 234, 286, 429, 858, 1287, 2574
Sum of divisors 6552
Prime factorization 2 x 32 x 11 x 13
Prime factors 2, 3, 11, 13
## How to write / spell 2574 in letters?
In letters, the number 2574 is written as: Two thousand five hundred and seventy-four. And in other languages? how does it spell?
2574 in other languages
Write 2574 in english Two thousand five hundred and seventy-four
Write 2574 in french Deux mille cinq cent soixante-quatorze
Write 2574 in spanish Dos mil quinientos setenta y cuatro
Write 2574 in portuguese Dois mil quinhentos setenta e quatro
## Decomposition of the number 2574
The number 2574 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
Other ways to write 2574
In letter Two thousand five hundred and seventy-four
In roman numeral MMDLXXIV
In binary 101000001110
In octal 5016
In US dollars USD 2,574.00 (\$)
In euros 2 574,00 EUR (€)
Some related numbers
Previous number 2573
Next number 2575
Next prime number 2579
## Mathematical operations
Operations and solutions
2574*2 = 5148 The double of 2574 is 5148
2574*3 = 7722 The triple of 2574 is 7722
2574/2 = 1287 The half of 2574 is 1287.000000
2574/3 = 858 The third of 2574 is 858.000000
25742 = 6625476 The square of 2574 is 6625476.000000
25743 = 17053975224 The cube of 2574 is 17053975224.000000
√2574 = 50.734603575863 The square root of 2574 is 50.734604
log(2574) = 7.8532163881561 The natural (Neperian) logarithm of 2574 is 7.853216
log10(2574) = 3.4106085425684 The decimal logarithm (base 10) of 2574 is 3.410609
sin(2574) = -0.86017703936403 The sine of 2574 is -0.860177
cos(2574) = -0.50999554993249 The cosine of 2574 is -0.509996
tan(2574) = 1.6866363627641 The tangent of 2574 is 1.686636 | 883 | 2,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-10 | latest | en | 0.753503 |
https://community.wolfram.com/groups/-/m/t/2369756?sortMsg=Recent | 1,638,922,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00469.warc.gz | 247,107,474 | 31,327 | # Help fitting and selecting data?
Posted 2 months ago
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Kindly find the data:https://drive.google.com/file/d/1Zw_GBnhFYbEglEKzTXnCv4o5xEMMT9n1/view?usp=sharingI want to fit my data. After using drop and select, I am getting blank {}. I want to fit my data of XZ Plane and potential. Kindly help, as soon as possible
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Posted 2 months ago
Add whatever filtering on values you want (* In DataxzStein, x is column 1 and z is column 2 *) DataxzSteinSubset = Select[DataxzStein, Between[#[[1]], {-0.32, 0.32}] && Between[#[[2]], {-0.22, 0.22}] &] MinMax /@ Transpose@DataxzSteinSubset (* {{-0.315, 0.315}, {-0.215, 0.215}, {0.447852, 0.889891}} *) Use DataxzSteinSubset to fit.
Posted 2 months ago
I mean to say. I want to command Mathematica that Please select data points with a range from ( -0.32 to +0.32) for x column(1st), a range from ( -0.22 to +0.22) for z column (3rd) along with its corresponding potential data in the 4th columns. Then I will use those data points for fitting.
Posted 2 months ago
Something like this? (* MinMax for x, y, z, p *) MinMax /@ Transpose@DataxyzStein (* {{-0.495, 0.495}, {-0.495, 0.495}, {-0.305, 0.305}, {0.299106, 0.998006}} *) (* Select positive x values *) DataxzSteinPosX = Select[DataxzStein, #[[1]] >= 0 &] MinMax /@ Transpose@DataxzSteinPosX (* {{0.005, 0.495}, {-0.305, 0.305}, {0.302262, 0.998001}} *)
Posted 2 months ago
Hi Rohit,Thank you so much for your time and consideration. But let's say I want to extract a specific range of data from my column data points for fitting; how to do that?Sorry for the silly question. For a beginner, I am struggling to learn. It would be helpful if you answer this. Regards, Soumyaranjan
Posted 2 months ago
Extract columns 1, 3, 4 DataxzStein = DataxyzStein[[All, {1, 3, 4}]] FittXZStein = NonlinearModelFit[ DataxzStein, {ModelxzStein}, {Subscript[T, 0], Subscript[T, 2], Subscript[T, 4], Subscript[T, 6], Subscript[T, 8], Subscript[T, 10]}, {x, z}] FittXZStein["RSquared"] (* 0.9845 *) FittXZStein["ParameterTable"] | 703 | 2,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-49 | latest | en | 0.720072 |
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# MTH101 GDB No1. May 29, 2015 to Monday, June 01, 2015
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Thora Difficul Hai Plz Koe Esay sa Btao na :D
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3 | 902 | 3,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-05 | longest | en | 0.889406 |
https://math.pro/db/thread-1483-1-14.html | 1,653,567,790,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00597.warc.gz | 439,822,044 | 7,213 | # 2007TRML
## 2007TRML
#### 附件
2007individual.pdf (93.49 KB)
2012-8-11 16:16, 下載次數: 4908
2007team.pdf (80.1 KB)
2012-8-11 16:16, 下載次數: 4835
TOP
## 回復 1# may513 的帖子
I-6
[解答]
I-8
[解答]
$$a_{1} , a_{2} , a_{2}-a_{1}, a_{4}=-a_{1} , a_{5}=-a_{2} , a_{6}=-a_{3} \Rightarrow a_{k+3}=-a_{k}\Rightarrow$$ 六個一循環且連續 6 項之和為 0
$$2005=6\cdot334+1\Rightarrow a_{1}=a_{2005}=2006$$
$$\sum\limits _{k=1}^{2007}a_{k}=a_{1}+a_{2}+a_{3}=2a_{2}=192$$
Team 6
[解答]
$$\Rightarrow n\cdot2^{n-3}\leq(n+1)\cdot n\Rightarrow2^{n-3}\leq n+1$$
110.8.12thepiano補充
(A)1 (B)2 (C)3 (D)4 (E)5 | 329 | 577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | longest | en | 0.216732 |
https://www.coursehero.com/file/7724044/P-13-2-s-ketch-t-he-signals-a-f-t-b-f-t6-c/ | 1,490,906,297,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218203515.32/warc/CC-MAIN-20170322213003-00361-ip-10-233-31-227.ec2.internal.warc.gz | 884,043,849 | 21,968 | Signal Processing and Linear Systems-B.P.Lathi copy
# Signal Processing and Linear Systems-B.P.Lathi copy
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Unformatted text preview: cted in Fig. P1.4-2a. ( d) ( h) b (t - 2) sin 1rtdt e (x-l) cos 5)]8(x - 3) dx 1 . 7 -2 CD (a) (b) 2 f (t) ~~ + (sin t )y(t) = ~ + 2 f(t) ( g) 2 t f (t) + 2y(t) = ~~ + 2y(t) = f(t)~ ( h) y (t) = 2 ( a) y(t) -1 -1 ~ED = f (t - ( b) y(t) 1 = f ( - t) ( e) y (t) = = f lat) ( f) y (t) F ig. P l.4-7 1 .4-7 Using t he g eneralized function definition, show t hat 8(t) is a n even function of t. f (r) d r 2) ( d) y (t) = t f(t - 2) ISs f(r)dr = (~) 2 F ind a nd s ketch J~oo f (x) dx for t he signal f (t) i llustrated in Fig. P l.4-7. 1 .4-8 loo For t he s ystems described by t he e quations below, w ith t he i nput f (t) a nd o utput y (t), d etermine which of t he s ystems are time-invariant p arameter s ystems a nd which are time-varying p arameter s ystems. ( c) y (t) o + 2 = f (t) ( f) ~~ + y2(t) = f (t) ( d) 1 .4-6 ( c) 3 y(t) = f (t) Hint: S tart w ith Eq. (1.24a) as t he definition of 6(t). Now change variable t show t hat I: 1 .4-9 P rove t hat = - x to q ,(t)6(-t)dt = q,(O) 1 8(at) = ~6(t) Hint: Show t hat 1 1 00 - 00 . . 4 -10 Show t hat q,(t)8(at) dt I: = ~q,(0) 8(t)q,(t) dt = -4>(0) 1 . 7 -3 F or a certain LTI system with t he i nput f (t), t he o utput y (t) a nd t he two initial conditions Xl(O) a nd X2(O), following observations were made: f (t) o o u (t) 1 e2t cos 3t ( d) e - 2t ( e) e2t ( f) 5. e -tu(t) 2 e - t (3t -1 -1 + 2)u(t) 2u(t) D etermine y (t) when b oth t he i nitial conditions are zero a nd t he i nput f (t) is as shown in Fig. P l.7-3. Hint: T here a re t hree causes: t he i nput a nd each of t he two initial conditions. Because of linearity property, if a cause is increased b y a f actor k, t he r esponse t o t hat cause also increases by t he s ame factor k. Moreover, if causes are added, t he c orresponding responses add. w here q,(t) a nd 4>(t) a re continuous a t t = 0, a nd q,(t) - > 0 a s t - > ± oo. T his integral defines 8( t) a s a generalized function. Hint: Use i ntegration by p arts . .. 4 -11 A sinusoid eat cos wt c an be expressed as a s um o f exponentials est a nd e -· t (Eq. ( l.30c) w ith complex frequencies s = a + j w a nd s = a - jw. L ocate in t he complex plane t he frequencies of t he following sinusoids: ( a) cos 3t ( b) e - 3t cos 3t ( c) y (t) Xl(O) -5 1 F ig. P 1.7-3 f {t) t __ ---~--------- 102 1 .1-4 1 I ntroduction t o S ignals a nd S ystems 1 .7-8 A system is specified by its i nput-output relationship as = loo ( c) yet) J(T)dT = ret) n , integer ( d) yet) = cos [J(t)] ( b) yet) = J(3t - 6) Show t hat t he circuit in Fig. P1.7-5 is zero-state linear b ut is not zero-input linear. Assume all diodes to have identical (matched) characteristics. Hint: In zero s tate (when the initial capacitor voltage vc(O) = 0), t he circuit is linear. If t he i nput J (t) = 0, a nd vc(O) is nonzero, t he c urrent yet) does not exhibit linearity with respect to its cause vc(O). y (t) For t he systems described by t he e quations below, with the i nput J(t) a nd o utput yet), d etermine which of t he s...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
Ask a homework question - tutors are online | 1,333 | 3,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-13 | longest | en | 0.755986 |
http://stackoverflow.com/questions/3269529/i-dont-understand-this-huffman-algorithm-implementation/3269546 | 1,448,826,011,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398459333.27/warc/CC-MAIN-20151124205419-00262-ip-10-71-132-137.ec2.internal.warc.gz | 216,822,217 | 19,705 | # I don't understand this Huffman algorithm implementation
`````` template<class T>
void huffman(MinHeap<TreeNode<T>*> heap, int n)
{
for(int i=0;i<n-1;i++)
{
TreeNode<T> *first = heap.pop();
TreeNode<T> *second = heap.pop();
TreeNode<T> *bt = new BinaryTreeNode<T>(first, second, first.data, second.data);
heap.push(bt);
}
}
``````
In my Fundamentals of Data Structures in C++ textbook, it gave a 2 page definition of Huffman coding, and the code above. To me, the book wasn't enough detailed, so I've done the googling and I learned how the process of Huffman coding works. The textbook claims that at the end of the code above, a Huffman tree is made. But to me it seems wrong, because a Huffman tree, is not necessary a complete tree, but the code above seems to always give a complete tree because of the `heap.push()`. So can someone explain to me how this piece of code is not wrong?
-
The heap's tree structure does not necessarily match the resulting Huffman tree -- rather, the heap contains a forest of partial Huffman trees, initially each consisting of a single symbol node. The loop then repeatedly takes the two nodes with the least weight, combines them into one node, and puts the resulting combined node back. At the end of the process, the heap contains one finished tree.
-
Then what I don't get is that if the heap is not necessarily the huffman tree, how am I able traverse or use the heap to find the correct leaf node. – user299648 Jul 17 '10 at 0:09
The heap is a temporary auxiliary data structure used to improve efficiency of the operation of finding the two Huffman nodes with the least weight. In the end, the heap contains only one element, a single `BinaryTreeNode<T>`, which can be popped out and is then the root of the finished Huffman tree; the heap can then be destroyed because it is no longer needed. – Jeffrey Hantin Jul 17 '10 at 0:20
thank you so much! you were really helpful. – user299648 Jul 17 '10 at 0:26
Huffman encoding works by taking the two lowest value items at each step. When you first call the function (since your MinHeap is sorted by value) the two lowest value items are popped off and "combined" into a decision node which is then put back into the heap. That node is scored by the sum of its child scores and put back into the heap. Inserting it back into the heap puts it into the right place based on its score; if it's still lower than any other items it'll be first, otherwise it'll be somewhere else.
So this algorithm is building the tree from the bottom up, and by the time you empty the heap you'll have a complete tree. I don't understand what the 'n' is for, though; the loop should be `while (heap.size() > 1)`. Regardless, the tree is not "full", different branches will be different lengths depending on how the frequencies of the items in the initial heap are scored.
- | 670 | 2,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2015-48 | longest | en | 0.900478 |
https://www.physicsforums.com/threads/question-about-element-of-and-subset-symbols.540807/ | 1,531,900,160,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590069.15/warc/CC-MAIN-20180718060927-20180718080927-00391.warc.gz | 975,123,190 | 13,690 | # Question about element of and subset symbols
1. Oct 15, 2011
### fishingspree2
Question about "element of" and "subset" symbols
I've always thought that ∈ is defined when talking about elements in a set. For example, if A is a set and x is an element, then x ∈ A is defined. It wouldn't make sense to say x $\subseteq$ A
In the same way, if A and B are sets and A is contained in B, then it is incorrect to say A ∈ B. We should use A$\subseteq$ B.
Question is: say a set which contains the empty set: {Ø}
I would think we should write Ø $\subseteq$ {Ø} because Ø is a set itself.
But at the same time Ø ∈ {Ø} looks like it could also make sense... I am slightly confused.
Can anyone shed some light on the matter? Thank you very much.
2. Oct 16, 2011
### Stephen Tashi
Re: Question about "element of" and "subset" symbols
You go to the trouble of distinguishing between "element of" and "subset of" and then you use the ambiguous term "contains". Tisk, tisk.
3. Oct 16, 2011
### HallsofIvy
Re: Question about "element of" and "subset" symbols
Specifically, fishingspree2, you are being ambiguous when you say "if A and B are sets and A is contained in B" without distinguishing the two meanings of "contained in". In "naive set theory" it is perfectly possible to have a set whose members are sets. Given that $\{\Phi\}$ is the set whose only member is the empty set, it is correct to say that the empty set is a member of that set. And, since the empty set is a subset of any set, both $\Phi\subset \{\Phi\}$ and $\Phi\in\{\Phi\}$ are both valid.
4. Oct 16, 2011
### disregardthat
Re: Question about "element of" and "subset" symbols
In set theory every element of a set is a set. There are no mathematical objects but sets in set theory.
5. Oct 26, 2011
### xxxx0xxxx
Re: Question about "element of" and "subset" symbols
Both are true, in the one case,
$$x \mbox{ is a set} \Rightarrow ( \emptyset \subseteq x)$$
and the other,
$$x \in \{ \emptyset \} \Leftrightarrow x = \emptyset$$. | 559 | 2,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-30 | latest | en | 0.91547 |
https://www.numbersaplenty.com/3580092 | 1,726,755,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00807.warc.gz | 841,568,727 | 3,342 | Search a number
3580092 = 223333149
BaseRepresentation
bin1101101010000010111100
320201212222000
431222002330
51404030332
6204422300
742300405
oct15520274
96655860
103580092
11202585a
121247990
139846c9
146929ac
154aab7c
hex36a0bc
3580092 has 24 divisors (see below), whose sum is σ = 9282000. Its totient is φ = 1193328.
The previous prime is 3580091. The next prime is 3580103. The reversal of 3580092 is 2900853.
3580092 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (27).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3580091) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 16467 + ... + 16682.
It is an arithmetic number, because the mean of its divisors is an integer number (386750).
Almost surely, 23580092 is an apocalyptic number.
It is an amenable number.
3580092 is an abundant number, since it is smaller than the sum of its proper divisors (5701908).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3580092 is a wasteful number, since it uses less digits than its factorization.
3580092 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 33162 (or 33154 counting only the distinct ones).
The product of its (nonzero) digits is 2160, while the sum is 27.
The square root of 3580092 is about 1892.1131044417. The cubic root of 3580092 is about 152.9788513636.
The spelling of 3580092 in words is "three million, five hundred eighty thousand, ninety-two". | 524 | 1,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.885721 |
https://www.iitianacademy.com/sat-maths-formula-sheet/ | 1,721,744,111,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00037.warc.gz | 705,037,763 | 33,745 | # SAT – Maths : Formula Sheet
Area of a Circle
\begin{aligned} & A=\pi r^2 \\ & C=2 \pi r \end{aligned}
$A=\pi r^2$
$$\pi$$ is a constant that can, for the purposes of the SAT, be written as 3.14 (or 3.14159 )
$$r$$ is the radius of the circle (any line drawn from the center point straight to the edge of the circle)
Circumference of a Circle
$C=2 \pi r(\text { or } C=\pi d)$
$$d$$ is the diameter of the circle. It is a line that bisects the circle through the midpoint and touches two ends of the circle on opposite sides. It is twice the radius.
Area of a Rectangle
$$A=l w$$
$$l$$ is the length of the rectangle
$$w$$ is the width of the rectangle
Area of a Triangle
\begin{aligned} A & =\frac{1}{2} b h \\ \end{aligned}
$$b$$ is the length of the base of triangle (the edge of one side)
$$h$$ is the height of the triangle
In a right triangle, the height is the same as a side of the 90-degree angle. For non-right triangles, the height will drop down through the interior of the triangle, as shown above (unless otherwise given).
The Pythagorean Theorem
###
$a^2+b^2=c^2$
In a right triangle, the two smaller sides ( $$a$$ and $$b$$ ) are each squared. Their sum is the equal to the square of the hypotenuse (c, longest side of the triangle).
Properties of Special Right Triangle: Isosceles Triangle
An isosceles triangle has two sides that are equal in length and two equal angles opposite those sides.
An isosceles right triangle always has a 90-degree angle and two 45 degree angles.
The side lengths are determined by the formula: $$x, x, x \sqrt{2}$$, with the hypotenuse (side opposite 90 degrees) having a length of one of the smaller sides * $$\sqrt{2}$$.
E.g., An isosceles right triangle may have side lengths of 12,12 , and $$12 \sqrt{2}$$.
Properties of Special Right Triangle: 30, 60, 90 Degree Triangle
• A 30, 60, 90 triangle describes the degree measures of the triangle’s three angles.
• The side lengths are determined by the formula: $$x, x \sqrt{3}$$, and $$2 x$$
• The side opposite 30 degrees is the smallest, with a measurement of $$x$$.
• The side opposite 60 degrees is the middle length, with a measurement of $$x \sqrt{3}$$.
• The side opposite 90 degrees is the hypotenuse (longest side), with a length of $$2 x$$.
• For example, a 30-60-90 triangle may have side lengths of $$5,5 \sqrt{3}$$, and 10 .
Volume of a Rectangular Solid
$V=l w h$
$$l$$ is the length of one of the sides.
$$h$$ is the height of the figure.
$$w$$ is the width of one of the sides.
Volume of a Cylinder
###
$V=\pi r^2 h$
$$r$$ is the radius of the circular side of the cylinder.
$$h$$ is the height of the cylinder.
Volume of a Sphere
$V=\left(\frac{4}{3}\right) \pi r^3$
$$r$$ is the radius of the sphere.
Volume of a Cone
###
$V=\left(\frac{1}{3}\right) \pi r^2 h$
$$r$$ is the radius of the circular side of the cone.
$$h$$ is the height of the pointed part of the cone (as measured from the center of the circular part of the cone).
Volume of a Pyramid
$V=\left(\frac{1}{3}\right) l w h$
$$l$$ is the length of one of the edges of the rectangular part of the pyramid.
$$h$$ is the height of the figure at its peak (as measured from the center of the rectangular part of the pyramid).
$$w$$ is the width of one of the edges of the rectangular part of the pyramid.
Law: the number of degrees in a circle is 360
Law: the number of radians in a circle is $$2 \pi$$
Law: the number of degrees in a triangle is 180
Scroll to Top | 990 | 3,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-30 | latest | en | 0.837488 |
https://www.kaysonseducation.co.in/questions/p-span-sty_129 | 1,656,683,172,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00472.warc.gz | 883,039,844 | 11,207 | If a, b , c are non-zero real numbers, then : Kaysons Education
If A, B , C Are Non-zero Real Numbers, Then
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Solution
Correct option is
SIMILAR QUESTIONS
Q1
If the system of the linear equations x + y + z = 6, x + 2y + 3z = 14 and 2 x + 5y + λz = μ, (λ, μ ∈ R) has a unique solution, then
Q2
Q3
find the maximum value of f (x).
Q4
Find the number of values of t for which the system of equations
has non-trivial solution.
Q5
Find the number of real roots of the equation
Q6 | 282 | 1,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | longest | en | 0.7808 |
https://elearning.reb.rw/course/view.php?id=506§ion=1 | 1,708,620,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00456.warc.gz | 228,253,320 | 30,454 | • ### Unit1: SOUND WAVES
Fig.1. 1: People listening to music
Key unit Competence
By the end of the unit I should be able to analyze the effects of sound waves in elastic medium
My goals
• Describe how sound propagates through a substance
• Give the characteristics of sound.
• Relate loudness and pitch to amplitude and frequency
• Carry out calculations relating decibels and intensity
• Establish relationship between characteristics of notes and sound waves
• Explain beats and establish beat frequency
• Explain Doppler – Fizeau effect.
• Give examples of musical pipe instruments.
• Establish the fundamental frequency and 2nd harmonic, 3rd harmonic, in vibrating strings and in pipes
• Explain Doppler – Fizeau effect. • Give examples of musical pipe instruments.
• Establish the fundamental frequency and 2nd harmonic, 3rd harmonic, in vibrating strings and in pipes
Introductory activity
What properties explain most the behavior of sound?
1. Most people like to listen to music, but hardly anyone likes to listen to noise. What is the difference between a musical sound and noise?
2. A guitarist plays any note. The sound is made by the vibration of the guitar string and propagates as a wave through the air and reaches your ear. Which of the following statement is the right?
• The vibration on the string and the vibration in the air have the same wavelength.
• They have the same frequency.
• They have the same speed.
• None of the above is the same in the air as it is on the string.
1.1 CHARACTERISTICS AND PROPERTIES OF SOUND WAVES
Activity 1.1: Properties of sound
On an interview for Physics placement in a certain school in Rwanda, Claudette a S.6 leaver who had applied for the job was asked about sound waves during the interview.
She was asked to state the properties of sound waves. Confidently, she responded that the properties are reflection, refraction, diffraction and interference. This was enough to make Claudette pass the first level of the interview.
However, in the second step, she was required to discuss different media in which sound waves can propagate. Claudette started discussing these different media. What surprised the interviewer was Claudette’s ability to relate sound waves to other kinds of waves stating that these waves behaves the same way when they pass from one medium to another. Looking at Claudette’s face, the interviewer asked her to discuss the laws governing reflection and refraction of sound waves. With a smile, she started by saying that since sound waves have the same properties as for light; these laws therefore do not change.
As she was attempting to state them, the interviewer stopped her and congratulated her upon her confidence and bravery she showed in the room. She was directly told that she was successful and she was given the job. Claudette is now working as assistant S2 Physics teacher and doubles as a Physics laboratory attendant.
1.1.1 Properties of sound waves
Most of us start our lives by producing sound waves! We spend much of our life surrounded by objects which produce sound waves. Most machines in use vibrate and produce sound so the only sure way to silence them would be to put them in vacuum where there would be no surrounding medium for the vibrating surfaces of the machine to push against, hence no sound waves. Some physiologists are concerned with how speech is produced, how speech impairment might be corrected, how hearing loss can be alleviated.
Sound is associated with our sense of hearing and, therefore, with the physiology of our ears that intercept the sound and the psychology of our brain which interprets the sensations that reach our ears. Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases.
As the sound wave propagates, many interactions can occur, including reflection, refraction, diffraction and interference. When a sound wave hits a surface, a part of the energy gets scattered while a part of it is absorbed. Absorption is the phenomenon of the wave where the energy of sound wave gets transformed from one form to another. The high frequency sound waves are more easily absorbed than low frequency sounds. It happens most with the soft materials.
1.1.2 Characteristics of sound waves
Activity 1.2: Characteristics of sound waves
1. How to calculate the speed of sound waves in different materials.
2. How to calculate the intensity of a sound wave.
3. From the Fig.1.2, can you hear the ultrasound waves that a bat uses for echolocation? Why or why not?
Fig.1. 2: Range of frequencies heard by various animals and human (Randall & Knight.,Physics for scientists and engineers: Stategic approach., 2008)
Usually, the characteristics used to describe waves are period, frequency, wavelength, and amplitude.
a. Frequency ranges
Any periodic motion has a frequency, which is the number of complete cycles in a second and a period which is the time used to complete one cycle. While the frequency is measured in Hertz (Hz), the period is measured in seconds (s). For a wave, the frequency is the number of wave cycles that pass a point in a second. A wave’s frequency equals the frequency of the vibrating source producing the wave. Sound waves are classified into three categories that cover different frequency ranges:
Audible soundlies within the range of sensitivity of the human ear. They can be generated in a variety of ways, such as musical instruments, human voices, or loudspeakers. It is almost impossible to hear sounds outside the range of 20 Hz to 20 kHz. These are the limits of audibility for human beings but the range decreases with age.
Infrasonic waves have frequencies below the audible range. They are sound waves with frequencies that are below 20 Hz limit. Some animals such as elephants can use infrasonic waves to communicate with each other, even when they are separated by many kilometers. Rhinoceros also use infrasonic as low as 5 Hz to call one another.
Ultrasonic waves have frequencies above the audible range. They are sound waves whose frequencies are higher than 20 KHz. You may have used a “silent” whistle to retrieve your dog. The ultrasonic sound emitted by that device is easily heard by dogs, although humans cannot detect it at all. Ultrasonic waves are also used in medical imaging. Many animals hear a much wider range of frequencies than human beings do. For example, dog whistles vibrate at a higher frequency than the human ear can detect, while evidence suggests that dolphins and whales communicate at frequencies beyond human hearing (ultrasound) (Cutnell & Johnson, 2006).
b. Wavelength
Wavelength is the distance covered by a wave in a period. It is represented by the separation between a point on one wave and a similar point on the next cycle of the wave. For a transverse wave, wavelength is measured between adjacent crests or between adjacent troughs. For a longitudinal wave such as sound wave, wavelength is the distance between adjacent compressions or rarefaction.
c. Speed of sound
For a periodic wave, the shape of the string at any instant is a repeating pattern. The length of one complete wave pattern is the distance from one crest to the next or from one trough to the next or from any point to the corresponding point on the next repetition of the wave shape. We call this distance the wavelength of the wave, denoted by the Greek letter lambda (λ). The wave pattern travels with constant speed and advances a distance of one wavelength in a time interval of one period T. So the wave speed is given by
(1.01)
where f is the frequency of the wave. Sound travels faster in liquids and solids than in gases, since the particles in liquids and solids are closer together and can respond more quickly to the motion of their neighbors. As examples, the speed of sound is 331 m/s in air,1500 m/s in water and 5000 m/s in iron (though these can change depending on temperature and pressure). Sound does not travel in vacuum.
Example 1.1 Wavelength of a musical sound
1. Sound waves can propagate in air. The speed of the sound depends on temperature of the air; at 200 C it is 344 m/s it is. What is the wavelength of a sound wave in air if the frequency is 262 Hz (the approximate frequency of middle C on a piano)?
Using Equation of wave (1.01): λ=V/f = (344m/s)/262Hz=1.31m
Factors which affect the velocity of sound in air
• The speed of sound waves in a medium depends on the compressibility and density of the medium. If the medium is a liquid or a gas and has a bulk modulus Band density ρ , the speed of sound waves in that medium is given by:
(1.02)
It is interesting to compare this expression with the equation applicable to transverse waves on a string. In both cases, the wave speed depends on an elastic property of the medium (bulk modulus B or tension in the string T) and on an inertial property of the medium (the density ρ or linear mass µ ). In fact, the speed of all mechanical waves follows an expression of the general form (1.03)
For longitudinal sound waves in a solid rod of material, for example, the speed of sound depends on Young’s modulus Y and the density ρ • Changes of pressure have no effect on the velocity of sound in air. Sir Isaac Newton showed that:
(1.04)
In accordance with Boyle’s law, if the pressure of a fixed mass of air is doubled, the volume will be halved. Hence the density will be doubled. Thus at constant temperature, the ratio P/ρ will always remain constant no matter how the pressure may change. The speed of sound increases with temperature If the air temperature increases at constant pressure the air will expand according to Charles’ law, and therefore become less dense.
The ratio P/ρ will therefore increase, and hence the speed of sound increases with temperature.
For sound traveling through air, the relationship between wave speed and medium temperature is (1.05)
d. Amplitude
The amplitude of a wave is the maximum displacement of the medium from its rest position. The amplitude of a transverse wave is the distance from the rest position to a crest or a trough. The more energy a wave has, the greater is its amplitude.
1.1.3 Checking my progress
1. The correct statement about sound waves is that:
a. They are transverse waves
b. They can be polarized
c. They require material medium to propagate
2. Sound travels in
a. Air
b. Water
c. Iron
d. All of these
3. Two men talk on the moon. Assuming that the thin layer of gases on the moon is negligible, which of the following is the right answer:
a. They hear each other with lower frequency
b. They hear each other with higher frequency
c. They can hear each other at such frequency
d. They cannot hear each other at all
4. Do you expect an echo to return to you more quickly on a hot day or a cold day?
a. Hot day.
b. Cold day.
c. Same on both days.
5.A sound wave is different than a light wave in that a sound wave is:
a. Produced by an oscillating object and a light wave is not.
b. Not capable of traveling through a vacuum.
c. Not capable of diffracting and a light wave is.
d. Capable of existing with a variety of frequencies and a light wave has a single frequency.
6. A spider of mass 0.30 g waits in its web of negligible mass see Fig. below. A slight movement causes the web to vibrate with a frequency of about 15 Hz.
Fig.1. 3 A spider of mass waits in its web
a. Estimate the value of the spring stiffness constant k for the web assuming simple harmonic motion.
b. At what frequency would you expect the web to vibrate if an insect of mass 0.10 g were trapped in addition to the spider?
1.2 PRODUCTION OF STATIONARY SOUND WAVES
Fig.1. 4: A guitarist.
Activity 1.3: Production of stationary sound waves.
Look at the Fig.1.4 of guitarist and then answer the following question.
1. How do vibrations cause sound?
2. What determines the particular frequencies of sound produced by an organ or a flute?
3. How resonance occurs in musical instruments?
4. How to describe what happens when two sound waves of slightly different frequencies are combined?
1.2.1 Sound in pipes
The source of any sound is vibrating object. Almost any object can vibrate and hence be a source of sound. For musical instruments, the source is set into vibration by striking, plucking, bowing, or blowing. Standing waves (also known as stationary waves are superposition of two waves moving in opposite directions, each having the same amplitude and frequency) are produced and the source vibrates at its natural resonant frequencies.
The most widely used instruments that produce sound waves make use of vibrating strings, such as the violin, guitar, and piano or make use of vibrating columns of air, such as the flute, trumpet, and pipe organ. They are called wind instruments.
We can create a standing wave:
• In a tube, which is open on both ends. The open end of a tube is approximately a node in the pressure (or an antinode in the longitudinal displacement).
• In a tube, which is open on one end and closed on the other end. The closed end of a tube is an antinode in the pressure (or a node in the longitudinal displacement). In both cases a pressure node is always a displacement antinode and vice versa.
a. Tube of length L with two open ends
An open pipe is one which is open at both ends. The length of the pipe is the distance between consecutive antinodes. But the distance between consecutive
Quick check 1.1: Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if the pipe is open at both ends.
b. Tube of length L with one open end and one closed end
The longest standing wave in a tube of length L with one open end and one closed end has a displacement antinode at the open end and a displacement node at the closed end. This is the fundamental.
Another way to analyze the vibrations in a uniform tube is to consider a description in terms of the pressure in the air. Where the air in a wave is compressed, the pressure is higher, whereas in a wave expansion (or rarefaction), the pressure is less than normal.We call a region of increased density a compression; a region of reduced density is a rarefaction.
The wavelength is the distance from one compression to the next or from one rarefaction to the next.
Quick check 1.2: Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if the pipe is closed at the left end and open at the right end.
1.2.2 Vibrating strings
The string is a tightly stretched wire or length of gut. When it is struck, bowed or plucked, progressive transverse waves travel to both ends, which are fixed, where they are reflected to meet the incident waves. A stationary wave pattern is formed for waves whose wavelengths fit into the length of the string, i.e. resonance occurs. If you shake one end of a cord (slinky) and the other end is kept fixed, a continuous wave will travel down to the fixed end and be reflected back, inverted. The frequencies at which standing waves are produced are the natural frequencies or resonant frequencies of the cord. A progressive sound wave (i.e. a longitudinal wave) is produced in the surrounding air with frequency equal to that of the stationary transverse wave on the string.
Now let consider a cord stretched between two supports that is plucked like a guitar or violin string. Waves of a great variety of frequencies will travel in both directions along the string, will be reflected at the ends, and will be travel back in the opposite direction. The ends of the string, since they are fixed, will be nodes.
Consider a string of length L fixed at both ends, as shown in Fig.1.12. Standing waves are set up in the string by a continuous superposition of wave incident on and reflected from the ends.
Note that there is a boundary condition for the waves on the string. The ends of the string, because they are fixed, must necessarily have zero displacement and are, therefore, nodes by definition.
The normal modes of vibration form a harmonic series: (b) the fundamental note (first harmonic); (c) First overtone (second harmonic); (d) the second overtone (third harmonic) (Halliday, Resneck, & Walker, 2007).
1.2.3. Wave Interference and Superposition
a. Wave interference
Up to this point we’ve been discussing waves that propagate continuously in the same direction. But when a wave strikes the boundaries of its medium, all or part of the wave is reflected
When you yell at a building wall or a cliff face some distance away, the sound wave is reflected from the rigid surface and you hear an echo. When you flip the end of a rope whose far end is tied to a rigid support, a pulse travels the length of the rope and is reflected back to you. In both cases, the initial and reflected waves overlap in the same region of the medium. This overlapping of waves is called interference
In general, the term “interference” refers to what happens when two or more waves pass through the same region at the same time Fig.1.14 shows an example of another type of interference that involves waves that spread out in space.
Two speakers, driven in phase by the same amplifier, emit identical sinusoidal sound waves with the same constant frequency. We place a microphone at point P in the figure, equidistant from the speakers. Wave crests emitted from the two speakers at the same time travel equal distances and arrive at point P at the same time; hence the waves arrive in phase, and there is constructive interference.
The total wave amplitude at P is twice the amplitude from each individual wave, and we can measure this combined amplitude with the microphone.
Now let’s move the microphone to point Q, where the distances from the two speakers to the microphone differ by a half-wavelength. Then the two waves arrive a half-cycle out of step, or out of phase; a positive crest from one speaker arrives at the same time as a negative crest from the other. Destructive interference takes place, and the amplitude measured by the microphone is much smaller than when only one speaker is present. If the amplitudes from the two speakers are equal, the two waves cancel each other out completely at point Q, and the total amplitude there is zero.
Activity 1.4:
Problem 1
The Adventures of Marvin the Mouse: You and your friend are walking down by the pool when you hear a cry for help. Poor Marvin the Mouse has fallen into the pool and needs your help. The sides of the pool are to slippery for Marvin to climb out but there is an inner tube anchored in the center of the pool. Oh no! The sides of the inner tube are too slippery and high for Marvin to climb. He’s getting tired and can’t swim to the sides; he has just enough energy to float by the inner tube. Having studied about waves, you and your friend take up positions on opposite sides of the pool. How did you help Marvin get safely onto the inner tube?
Problem 2: Dance club designer
You are the designer of a new Dance Club. You have been informed that you need to design the club in such a way that the telephone is placed in a location that allows the customers to hear the people on the other side. The phone company states that they can only put the phone line in at a point 20 m from the stage. Develop a model which allows the customers to use the phone with the least amount of trouble given that the phone must be placed at a distance of 20 m, (2/3 the room size), from the stage. This will be an area where there will be virtually no sound.
c. Resonance of sound
We have seen that a system such as a taut string is capable of oscillating in one or more normal modes of oscillation. If a periodic force is applied to such a system, the amplitude of the resulting motion is greater than normal when the frequency of the applied force is equal to or nearly equal to one of the natural frequencies of the system. This phenomenon is known as resonance. Although a block–spring system or a simple pendulum has only one natural frequency, standing-wave systems can have a whole set of natural frequencies. Because oscillating systems exhibits large amplitude when driven at any of its natural frequencies, these frequencies are often referred to as resonance frequencies. Fig.1.15 shows the response of an oscillating system to various driving frequencies, where one of the resonance frequencies of the system is denoted by fo.
One of our best models of resonance in a musical instrument is a resonance tube. This is a hollow cylindrical tube partially filled with water and forced into vibration by a tuning fork (Fig.1.16). The tuning fork is the object that forced the air, inside the resonance tube, into resonance.
As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency. Yet, in the absence of resonance, the sound of these vibrations is not loud enough to discern.
Resonance only occurs when the first object is vibrating at the natural frequency of the second object. So if the frequency at which the tuning fork vibrates is not identical to one of the natural frequencies of the air column inside the resonance tube, resonance will not occur and the two objects will not sound out together with a loud sound. But the location of the water level can be altered by raising and lowering a reservoir of water, thus decreasing or increasing the length of the air column.
So by raising and lowering the water level, the natural frequency of the air in the tube could be matched to the frequency at which the tuning fork vibrates. When the match is achieved, the tuning fork forces the air column inside of the resonance tube to vibrate at its own natural frequency and resonance is achieved. The result of resonance is always a big vibration - that is, a loud sound. A more spectacular example is a singer breaking a wine glass with her amplified voice. A good-quality wine glass has normal-mode frequencies that you can hear by tapping it.
If the singer emits a loud note with a frequency corresponding exactly to one of these normal-mode frequencies, large-amplitude oscillations can build up and break the glass (Fig. 1.18)
d. Beats and its phenomena
Beats occur when two sounds-say, two tuning forks- have nearly, but not exactly, the same frequencies interfere with each other. A crest may meet a trough at one instant in time resulting in destructive interference. However at later time the crest may meet a crest at the same point resulting in constructive interference. To see how beats arise, consider two sound waves of equal amplitudes and slightly different frequencies as shown on the figure below.
In 1.00 s, the first source makes 50 vibrations whereas the second makes 60. We now examine the waves at one point in space equidistant from the two sources. The waveforms for each wave as a function of time, at a fixed position, are shown on the top graph of Fig. 1.19; the red line represents the 50 Hz wave, and the blue line represents the 60 Hz wave. The lower graph in Fig. 1.18 shows the sum of the two waves as a function of time. At the time the two waves are in phase they interfere constructively and at other time the two waves are completely out of phase and interfere destructively. Thus the resultant amplitude is large every 0.10 s and drops periodically in between. This rising and falling of the intensity is what is heard as beats. In this case the beats are 0.10 s apart. The beat frequency is equal to the difference in frequencies of the two interfering waves.. Consider two sound waves of equal amplitude traveling through a medium with slightly different frequencies f1 and f2atchosen point x = 0:
Quick check 1.4: A tuning fork produces a steady 400 Hz tone. When this tuning fork is struck and held near a vibrating guitar string, twenty beats are counted in five seconds. What are the possible frequencies produced by the guitar string?
9. Why is a pulse on a string considered to be transverse?
10. A guitar string has a total length of 90 cm and a mass of 3.6 g. From the bridge to the nut there is a distance of 60 cm and the string has a tension of 520 N. Calculate the fundamental frequency and the first two overtones
1.3 CHARACTERISTICS OF MUSICAL NOTES
Activity 1.5: Characteristics of musical notes
The physical characteristics of a sound wave are directly related to the perception of that sound by a listener. Before you read this section answer these questions. As you read this section answer again these questions. Compare your answer.
1. What is the difference between the sound of whistle and that of drum?
2. Can you tell which musical instrument is played if the same note is played on different instrument without seeing it? Explain
3. How can you calculate the intensity of a sound wave?
A musical note is produced by vibrations that are regular and repeating, i.e. by periodic motion. Non-periodic motion results in noise which is not pleasant to the ear. Many behaviors of musical note can be explained using a few characteristics: intensity and loudness, frequency and pitch, and quality or timber.
1.3.1. Pitch and frequency
The sound of a whistle is different from the sound of a drum. The whistle makes a high sound. The drum makes a low sound. The highness or lowness of a sound is called its pitch. The higher the frequency, the higher is the pitch. The frequency of an audible sound wave determines how high or low we perceive the sound to be, which is known as pitch.
Frequency refers to how often something happens or in our case, the number of periodic, compression-rarefaction cycles that occur each second as a sound wave moves through a medium -- and is measured in Hertz (Hz) or cycles/second. The term pitch is used to describe our perception of frequencies within the range of human hearing.
If a note of frequency 300 Hz and note of 600 Hz, are sounded by a siren, the pitch of the higher note is recognized to be an upper octave of the lower note. The musical interval between two notes is an upper octave if the ratio of their frequencies is 2:1. It can be shown that the musical interval between two notes depends on the ratio of their frequencies, and not on the actual frequencies.
Whether a sound is high-pitched or low-pitched depends on how fast something vibrates. Fast vibrations make high-pitched sounds. Slow vibrations make low pitched sounds.
Do not confuse the term pitch with frequency. Frequency is the physical measurement of the number of oscillations per second. Pitch is a psychological reaction to sound that enables a person to place the sound on a scale from high to low, or from treble to bass. Thus, frequency is the stimulus and pitch is the response. Although pitch is related mostly to frequency, they are not the same. A phrase such as “the pitch of the sound” is incorrect because pitch is not a physical property of the sound.The octave is a measure of musical frequency.
1.3.2 Intensity and amplitude
A police siren makes a loud sound. Whispering makes a soft sound. Whether a sound is loud or soft depends on the force or power of the sound wave. Powerful sound waves travel farther than weak sound waves. To talk to a friend across the street you have to shout and send out powerful sound waves. Your friend would never hear you if you whispered.
A unit called the decibel measures the power of sound waves. The sound waves of a whisper are about 10 decibels. Loud music can have a level of 120 decibels or more. Sounds above 140 decibels can actually make your ears hurt. The energy carried by a sound wave is proportional to the square of its amplitude. The energy passing a unit area per unit time is called the intensity of the wave.
The intensity of spherical sound wave at a place p is defined as the energy per second per m2, or power per m2 flowing normally through an area at X. i.e
where r is the distance from the source for a spherical wave
Quick check 1.4: A point source emits sound waves with an average power output of 80.0 W.
a. Find the intensity 3.00 m from the source.
b. Find the distance at which the sound level is 40 dB.
Activity 1.6: Noise or music
Most people like to listen to music, but hardly anyone likes to listen to noise.
1. What is the physical difference between musical sound and noise?
2. What is the effect of noise to human being?
The physical characteristics of a sound wave are directly related to the perception of that sound by a listener. For a given frequency the greater the pressure amplitude of a sinusoidal sound wave, the greater the perceived loudness.
The loudness or softness of sound depends on the intensity of the sound wave reaching the person concerned. Loudness is a subjective quantity unlike intensity. Sound that is not wanted or unpleasant to the ear is called noise. High intensity can damage hearing.The higher the intensity, the louder is the sound. Our ears, however, do not respond linearly to the intensity. A wave that carries twice the energy does not sound twice as loud.
1.3.3 Quality or timbre
If the same note is sounded on the violin and then on the piano, an untrained listener can tell which instrument is being used, without seeing it. We would never mistake a piano for flute. We say that the quality or timbre of note is different in each case. The manner in which an instrument is played strongly influences the sound quality. Two tones produced by different instruments might have the same fundamental frequency (and thus the same pitch) but sound different because of different harmonic content. The difference in sound is called tone color, quality, or timbre. A violin has a different timbre than a piano.
1.3.4 Checking my progress
1. Complete each of the following sentences by choosing the correct term from the word bank: loudness, pitch, sound quality, echoes, intensity and noise
a. The ------------ of a sound wave depends on its amplitude
b. Reflected sound waves are called --------------------------
c. Two different instruments playing the same note sound different because of -----------------
2. Plane sound wave of frequency 100 Hz fall normally on a smooth wall. At what distances from the wall will the air particles have: a. Maximum b. Minimum amplitude of vibration? Give reasons for your answer. The speed of sound in air may be taken as 340 m/s
3. A boy whistles a sound with the power of 0.5x10-4w . What will be his sound intensity at a distance of 5m?
4. Calculate the intensity level equivalent to an intensity 1 nW/m2
5. If the statement is true, write true. If it is false, change the underlined word or words to make the statement true.
a. Intensity is mass per unit volume.
b. Loudness is how the ear perceives frequency
c. Music is a set of notes that are pleasing
1.4 APPLICATIONS OF SOUND WAVES
Activity1.7: Doppler Effect and uses of sound waves
1. Why does the pitch of a siren change as it moves past you?
2. How is Doppler’s effect used in communication with satellites?
3. Explain how is the Doppler’s effect used in Astronomy?
4. People use sound for other things other than talking and making music. In your own word, give more examples and explanations to support this statement.
1.4.1 The Doppler Effect
Doppler’s effect is the apparent variation in frequency of a wave due to the relative motion of the source of the wave and the observer.
Fig.1. 19 C.J.Doppler (Douglass, PHYSICS, Principles with applications., 2014)
The effect takes its name from the Austrian Mathematician Christian Johann Doppler (1803-1853), who first stated the physical principle in 1842. Doppler’s principle explains why, if a source of sound of a constant pitch is moving toward an observer, the sound seems higher in pitch, whereas if the source is moving away it seems lower. This change in pitch can be heard by an observer listening to the whistle of an express train from a station platform or another train.
The upper signs apply if source and/or observer move toward each other. The lower signs apply if they are moving apart. The word toward is associated with an increase
in observed frequency. The words away from are associated with a decrease in observed frequency.
Although the Doppler’s effect is most typically experienced with sound waves, it is a phenomenon that is common to all waves. For example, the relative motion of source and observer produces a frequency shift in light waves. The Doppler’s effect is used in police radar systems to measure the speeds of motor vehicles. Likewise, astronomers use the effect to determine the speeds of stars, galaxies, and other celestial objects relative to the Earth.
Quick check 1.5: Middle C on the musical scale has a frequency of 264 Hz. What is the wavelength of the sound wave?
1.4.2 Uses of Ultrasonic
a. Echolocation
Some marine mammals, such as dolphin, whales, and porpoises use sound waves to locate distant objects. In this process, called echolocation, a dolphin produces a rapid train of short sound pulses that travel through the water, bounce off distant objects, and reflect back to the dolphin. From these echoes, dolphins can determine the size, shape, speed, and distance of their potential prey. Experiments have shown that at distance of 114 m, a blindfolded dolphin can locate a stainless-steel sphere with a diameter of 7.5 cm and can distinguish between a sheet of aluminum and a sheet of copper (Cutnell & Johnson, 2006).
The Ultrasonic waves emitted by a dolphin enable it to see through bodies of other animals and people (Fig.1.20). Skin muscles and fat are almost transparent to dolphins, so they see only a thin outline of the body but the bones, teeth and gas-filled cavities are clearly apparent. Physical evidence of cancers, tumors, heat attacks, and even emotional shake can all be seen by dolphin. What is more interesting, the dolphin can reproduce the sonic signals that paint the mental image of its surroundings, and thus the dolphin probably communicates its experience to other dolphins. It needs no words or symbol for fish, for example, but communicates an image of the real thing.
Fig.1. 21: The Ultrasonic waves emitted by a dolphin enable it to see through bodies of other animals and people.
Bats also use echo to navigate through air.Bats use ultrasonic with frequencies up to 100 kHz to move around and hunt (Fig.1.23).
Fig.1. 22 Bats use ultrasonic with frequencies up to 100 kHz to move around and hunt
The waves reflect off objects and return the bat’s ears. The time it takes for the sound waves to return tells the bat how far it is from obstacles or prey. The bat uses the reflected sound waves to build up a picture of what lies ahead. Dogs, cats and mice can hear ultrasound frequencies up to 450 kHz. Some animals not only hear ultrasound but also use ultrasonic to see in dark.
b. In medicine
The sonogram is device used in medicine and exploits the reflected ultrasound to create images. This pulse-echo technique can be used to produce images of objects inside the body and is used by Physicians to observe fetuses. Ultrasound use a high frequency in the range of 1 MHz to 10 MHz that is directed into the body, and its reflections from boundaries or interfaces between organs and other structures and lesions in the body are then detected. (Michael, Loannis, & Martha, 2006) Tumors and other abnormal growths can be distinguished; the action of heart valves and the development of a foetus (Fig.1.24) can be examined; and information about various organs of the body, such as the brain, heart, liver, and kidneys, can be obtained.
Although ultrasound does not replace X-rays, for certain kinds of diagnosis it is more helpful. Some tissues or fluid are not detected in X-ray photographs, but ultrasound waves are reflected from their boundaries. Echoes from ultrasound waves can show what is inside the body. Echo is a reflection of sound off the surface of an object.
Fig.1. 23: Ultrasound image as an example of using high-frequency sound waves to see within the human body (Douglass, PHYSICS, Principles with applications., 2014).
In medicine, ultrasonic is used as a diagnostic tool, to destroy diseased tissue, and to repair damaged tissue.Ultrasound examination of the heart is known as echocardiography
c. Sonar
The sonar or pulse-echo technique is used to locate underwater objects and to determine distance. A transmitter sends out a sound pulse through the water, and a detector receives its reflection, or echo, a short time later. This time interval is carefully measured, and from it the distance to the reflecting object can be determined since the speed of sound in water is known. The depth of the sea and the location of sunken ships, submarines, or fish can be determined in this way. Sonar also tells how fast and what direction things are moving. Scientists use sonar to make maps of the bottom of the sea.
An analysis of waves reflected from various structures and boundaries within the Earth reveals characteristic patterns that are also useful in the exploration for oil and minerals.
Radar used at airports to track aircraft involves a similar pulse-echo technique except that it uses electromagnetic (EM) waves, which, like visible light, travel with a speed of 3 ×108 m/s. One reason for using ultrasound waves, other than the fact that they are inaudible, is that for shorter wavelengths there is less diffraction so the beam spreads less and smaller objects can be detected.
1.4.3 Uses of infrasonic
Elephants use infrasonic sounds waves to communicate with one another. Their large ears enable them to detect these low frequency sound waves which have relatively long wavelengths. Elephants can effectively communicate in this way even when they are separated by many kilometers. Some animals, such as this young bateared fox, have ears adapted for the detection of very weak sounds.
Fig.1. 24: Some animals, such as this young bat-eared fox, have ears adapted for the detection of very weak sounds.
1.4.4 Checking my progress
For question 1 to 2: Choose the letter of the best answer
1. Choose the best answer: Bats can fly in the dark without hitting anything because
a. They are flying mammals
b. Their night vision is going
c. They are guided by ultrasonic waves produced by them
d. Of no scientific reason
2. Bats and dolphins use echolocation to determine distances and find prey.
What characteristic of sound waves is most important for echolocation?
a. Sound waves reflect when they hit a surface
b. Sound waves spread out from a source
c. Sound waves diffract around corner
d. Sound waves interfere when they overlap
3. Discuss application of sound waves in medicine and navigation
4. Explain how sonar is used to measure the depth of a sea
5. a. What is meant by Doppler Effect?
b. A police car sound a siren of 1000 Hz as it approaches a stationary
observer at a speed of 33.5 m/s. What is the apparent frequency of the siren as heard by the observer if the speed of sound in air is 340 m/s.
c. Give one application of the Doppler Effect.
1.5 END UNIT ASSESSMENT
1.5.1 Multiple choices question
For question 1 to 6, choose the letter of the best answer
1. Which of the following affects the frequency of wave?
a. Reflection
b. Doppler effect
c. Diffraction d. All of the above
2. Consider the following statements:
I. Recording of sound on tapes was first invented by Valdemar Poulsen.
II. Audio tapes have magnetic property.
III. The tapes may also be made of PVC (Polyvinyl-chloride)Of these statements:
a. I, II, and III all are correct.
b. I, II, and III all are wrong
c. I and II are correct, III is wrong
d. I and II are wrong, III is correct
3. Nodes are
a. Positions of maximum displacement
b. Positions of no displacement
c. A position between no displacement and maximum displacement
d. None of these
4. Sound waves are
a. Transverse waves characterized by the displacement of air molecules.
b. Longitudinal waves characterized by the displacement of air molecules.
c. Longitudinal waves characterized by pressure differences.
d. Both (B) and (C). e. (A), (B), and (C).
5. In which of the following is the wavelength of the lowest vibration mode the same as the length of the string or tube?
a. A string.
b. A tube closed at one end.
c. All of the above.
d. An open tube.
e. E. None of the above.
6. When a sound wave passes from air into water, what properties of the wave will change?
a. Frequency.
b. Wave speed.
c. Both frequency and wavelength.
d. Wavelength.
e. Both wave speed and wavelength.
1.5.2 Structured questions
1. Does the phenomenon of wave interference apply only to sinusoidal waves? Explain.
2. As oppositely moving pulses of the same shape (one upward, one downward) on a string pass through each other, there is one instant at which the string shows no displacement from the equilibrium position at any point. Has the energy carried by the pulses disappeared at this instant of time? If not, where is it?
3. Can two pulses traveling in opposite directions on the same string reflect from each other? Explain.
4. When two waves interfere, can the amplitude of the resultant wave be greater than the amplitude of any of the two original waves? Under which conditions?
5. When two waves interfere constructively or destructively, is there any gain or loss in energy? Explain.
6. Explain why your voice seems to sound better than usual when you sing in the shower.
7. An airplane mechanic notices that the sound from a twin-engine aircraft rapidly varies in loudness when both engines are running. What could be causing this variation from loud to soft?
8. Explain how a musical instrument such as a piano may be tuned by using the phenomenon of beats.
9. Fill in the gap
a. As a sound wave or water ripple travels out from its source, its ------------- decreases.
b. The vibrating air in a/an ----------------------------- has displacement antinodes at both ends.
c. For a /an ……………., the fundamental corresponds to a wavelength four times the length of the tube.
d. The ……………….. refers to the change in pitch of a sound due to the motion either of the source or of the observer. If source and observer are approaching each other, the perceived pitch is …….. If they are moving apart, the perceived pitch is ……………. 10. A bat, moving at 5.00 m/s, is chasing a flying insect. If the bat emits a 40.0 kHz chirp and receives back an echo at 40.4 kHz, at what speed is the insect moving toward or away from the bat? (Take the speed of sound in air to be v = 340 m/s.)
11. If you hear the horn of the car whose frequency is 216 Hz at a frequency of 225 Hz, what is their velocity? Is it away from you or toward you? The speed of sound is 343 m/s
12. You run at 12.5 m/s toward a stationary speaker that is emitting a frequency of 518 Hz. What frequency do you hear? The speed of sound is 343 m/s
13. If you are moving and you hear the frequency of the speaker at 557 Hz, what is your velocity? Is it away from or toward the speaker? The speed of sound is 343 m/s
1.5.3 Essay type question | 9,520 | 43,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-10 | latest | en | 0.940691 |
https://ctse5040.wordpress.com/spreadsheets-all-areas-of-life/ | 1,550,264,393,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479159.2/warc/CC-MAIN-20190215204316-20190215230316-00367.warc.gz | 544,665,076 | 13,795 | # Spreadsheets: All Areas of Life
What makes a subject worth studying? At the end of the day, a subject is only as significant as it is practical in everyday life. If math had no applications in the real world, there would be no point in learning it. This fact drives students’ hunger for life application math problems, because this type of problem serves as a window through which students relate math to their life. The task of an effective math teacher, then, is to pose practical problems that challenge students to use a variety of technological tools and critical thinking skills, as this process most nearly mimics situations they will face later in life.
When formulating any problem that will involve technology, it is critical for teachers to make sure that the technology to be used promotes problem analyzation, strategizing, connections across representations, and solution reflection (Dick and Hollebrands, 2011). Part of this process includes students making decisions about which technology or tool will be most effective in problem solving, so an ideal problem can be solved with multiple methods. This analysis will focus on a real-life application problem which can be solved using spreadsheet software and/or the graphing application Desmos.
Consider the following problem:
Farmer Fran is building a rectangular corral. She has 120 meters of fencing.
• What are the dimensions of the largest corral she can build?
• What if she builds her corral adjacent to her barn, so that the barn serves as one side of the corral?
It is worth noting that this problem could be modified in order to appeal to different types of students. For example, this problem could involve 120 cm worth of icing around a cookie cake or 120 yards worth of chalk around a soccer field.
Continue to the next page to see a solving strategy for this problem. | 381 | 1,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | latest | en | 0.94865 |
http://www.math.yorku.ca/Who/Faculty/kunquan/1310/quiz/q6soln/ | 1,508,192,155,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00095.warc.gz | 627,897,977 | 1,622 | MATH 1310.03F A Quiz 6 Thursday, November 2, 2000
NAME:
Student Number: No.Marks
(25 Marks) Discuss the convergence of the following improper integral.
Solution.
1. When q=1, . Hence, we have
and . This implies and diverges.
2. When q>1, we have Let . Then
So, we have
Hence, . The integral converges.
Kunquan Lan
Thu Nov 2 13:07:16 EST 2000 | 114 | 351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | latest | en | 0.798352 |
https://statshelponline.com/statistical-inference-assignment-help-19353 | 1,519,061,933,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00414.warc.gz | 785,366,362 | 14,173 | Select Page
# Statistical Inference Assignment Help
There are various system of treatments are readily available for inference and induction. Inferential Data are commonly utilized in screening predefined hypotheses and likewise utilized to make evaluations based on sample information.Statistical inference through its subjects such as Regression Analysis, Checking of Hypothesis, Possibility circulations-- Regular, Binomial, Poisson, Hyper geometric and Self-confidence Periods has actually turned into one of the complicated and crucial locations in Data. Our gifted swimming pool of Stats professionals, Stats assignment tutors and Stats research tutors can accommodate your whole requirements in the location of Statistical inference theory such as Assignment Help, Research Help, Task Paper Help and Test Preparation Help. With well annotated uses of notes and literature evaluations, our online stats tutors provide you the superior quality services.
In specific, the statistical reasonings are playing important part in taking a look at numerous system of treatments which work for created appropriate conclusions originated from a dataset collected from systems which are normally impacted by random variation such as observational mistakes, random tasting, or random experimentation.
The random variation consists of observational mistakes, random tasting, or random experimentation. Numerous systems of treatments are readily available for inference and induction.In specific, the statistical reasonings are playing a vital function in evaluating different system of procedures that work in order to create proper choices originated from a dataset gathered from systems that are normally affected by random variation for instance observational mistakes, random tasting, or haphazard exploring.
Statistical inference is the treatment for presuming homes of an intrinsic circulation by assessment of information. Inferential statistical analysis presumes homes about a population: this consists of obtaining approximations and screening theories.
There many system of procedures are readily available for Statistical inference is that branch of statistics, which has an interest in making use of possibility concept to manage unpredictability in decision-making. The field of statistical inference in fact had a satisfying improvement, provided the latter half of the 19th century.
Computational Inference Assignment Help
More especially, the treatment of statistical inference used in revealing various structures or treatments.It has mathematical services that discuss connections in between random variables and requirements. It makes anticipations about the random variables, and sometimes, requirements. The Computational Statistical Inference has an interest in advancing the theory, method, and algorithmic development of simulation-based approaches to statistical inference and unpredictability metrology. The statistical inference was a handy tool in understanding the info offered to us by the research study hall from Alaska.
In addition, the result/conclusion/answers gotten for such an unique scenario has to be fundamental sufficient to be asked for many circumstances. Inferential Stats are thoroughly made use of in evaluating hypotheses in addition to for making assessments based upon sample info. Bear in mind, a statistical inference concentrates on learning characteristics of the population from a sample; the population qualities are requirements and sample qualities are statistics. The random variation includes observational errors, random screening, or random experimentation. Many systems of treatments are provided for inference and induction. Initial requirements of this system are to produce an useful conclusion/answer when it is utilized to a specific circumstance. Statistical inference is proper for a great deal of circumstances in order to deciding from the similar dataset and a number of methods which are used to carry out such inference. A statistical style is a representation of a complex phenomenon that produced the info.
Statistical Inference Assignment Help
According to the very first requirement of the system, one can produce affordable decisions/responses by utilizing such system of procedures in nearly any distinct circumstance. Inferential Stats are popular in screening theories that were predefined as well as utilized to make quotes based on sample information. The statistical thinkings are playing a vital part in examining various system of treatments, which works for producing suitable conclusions came from a dataset gathered from systems, which are generally affected by random variation such as observational errors, random tasting, or random experimentation. The choice of a statistical inference is a statistical proposal. A point approximation, i.e. a special worth that finest estimates some specification of interest; a trustworthy time, i.e. a set of worths consisting of 95% of posterior belief; rejection of a theory; classification or clustering of information points into groups.
A fundamental type: details = style + residuals
Style should explain most of the variation in the info
Residuals are a representation of a lack-of-fit, which is of the part of the details uncommon by the style. Statistical inference is based upon the laws of possibility, and makes it possible for professionals to presume conclusions about a provided population based upon outcomes observed through random tasting. 2 of the important terms in statistical inference are requirement and figure: A figure is a number, which may be determined from the details observed in a random sample, without needing using any unknown requirements, such as a sample mean At UAE online Statistical Inference-II assignment help service; we just work with skilled Statistical Inference-II assignment professionals throughout the world. When you purchase a Statistical Inference-II assignment help, you are guaranteed of getting your Statistical Inference-II paper on-time.
Statistical inference makes propositions about a population, using info drawn from the population with some sort of screening. Used a hypothesis about a population, for which we wish to draw thinkings, statistical inference consists of (to start with) picking a statistical style of the treatment that develops the info, and (second of all) deducing propositions from the style. At UAE online Statistical Inference-II assignment help service; we just work with skilled Statistical Inference-II assignment specialists around the world. At our Statistical Inference-II assignment help online service, our group is just devoted to supplying the best time analysis assignment responses to all orders put on our site. You are guaranteed of getting your Statistical Inference-II paper on-time when you buy a Statistical Inference-II assignment help. A requirements is a number describing a population, such as a part or portion.
https://youtu.be/vSf6i7O1z-w | 1,194 | 6,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-09 | longest | en | 0.924248 |
https://wiki-helper.com/question/equilateral-triangles-are-drawn-on-the-three-sides-of-a-right-angled-triangle-show-that-the-area-40374575-30/ | 1,675,410,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00340.warc.gz | 638,319,596 | 14,367 | ## Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenus
Question
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to sum of the area of triangles on the other two sides.
in progress 0
2 years 2021-06-22T16:58:06+00:00 1 Answers 0 views 0 | 106 | 410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.856511 |
https://fornoob.com/need-help-on-how-to-expand-xh4/ | 1,670,586,621,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00179.warc.gz | 292,329,665 | 18,856 | # Need help on how to expand (X+H)^4!!?
Hello,
I’m trying to work out {(x+h)^4 -x^4}/h
I would gratefully like the answer with explantion because I am having trouble working it out. Plus (X+h)^4, I need help working it out and somewhere I have seen a sort of general formulas for bracket sums with powers.
Thank you very much.
• (x + h)⁴ = x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴
the powers of the x decrease and the powers of the h increase
the coefficients come from pascals triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
etc
These can also be found using combinations
nCr
(a + b)^n = ∑(r = 0 to n) nCr a^(n – r) b^r
• X H 4
• Memorize Pascal’s triangle it will help a lot, multiplying all the terms out takes far too long. I’ve included a link so you can see what I’m talking about.
Expanding (x+h)^4 results in
(x^4+4x^3h+6x^2h^2+4xh^3+h^4-x^4)/h
Simplifies to (4x^3h+6x^2h^2+4xh^3+h^4)/h after subtracting the x^4 term.
Dividing by h results in 4x^3+6x^2h+4xh^2+h^3.
• (x+h)^4=[(x+h)^2]*[(x+h)^2]
This is the long way to do it, but it works with any number of terms inside the parentheses and any power.
The general formula for binomial powers comes from combinatorial calculus.
Look up Pascal’s triangle for a simple way to derive the coefficients.
• For the best answers, search on this site shorturl.im/awozz
Hi, (x + 4)(x + 3) = x(x + 3) + 4(x + 3) = x² + 3x + 4x + 12 = x² + 7x + 12 <==ANSWER I hope that helps!! 🙂
• try looking up the pascals triangle, it will help you a lot | 549 | 1,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-49 | latest | en | 0.853848 |
https://www.aafp.org/pubs/afp/issues/2003/0901/p937.html | 1,726,880,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00337.warc.gz | 552,474,023 | 15,964 | Am Fam Physician. 2003;68(5):937-938
## Clinical Question
What is the probability of group A beta-hemolytic streptococcal (GABHS) pharyngitis (strep throat) in a patient who presents with sore throat?
Applying the Evidence: Justin is a 13-year-old boy who has had sore throat, fever, and swollen glands for two days. On examination, he has exudative tonsillitis and tender anterior cervical nodes, but no posterior cervical adenopathy. What is his probability of having strep throat?
Answer: Justin has a strep score of 5, which indicates a 52 percent risk of GABHS pharyngitis. It would be reasonable to treat him empirically. As another option, the physician could obtain a specimen for throat culture and call the patient in two days to discontinue the antibiotics if the culture is negative; however, the physician should consider the possibility of false-negative culture results when the pretest probability of strep throat is sufficiently high. A third option is to base treatment on a rapid step test for high-risk patients, although the physician should keep in mind that this test has the same risk of false-negative results as the throat culture.
## Evidence Summary
The probability of GABHS pharyngitis as the cause of sore throat is greatest in children younger than 15 years, especially those younger than 10. In a typical outpatient setting, 30 percent of children five to nine years of age with sore throat have GABHS pharyngitis; the probability declines to 15 to 20 percent in preadolescents and adolescents (ages 10 to 19 years). Only 5 to 10 percent of adults with sore throat have GABHS pharyngitis.13
Signs and symptoms that help rule in GABHS pharyngitis include tonsillar exudates (positive likelihood ratio [LR+]: 3.4), pharyngeal exudates (LR+: 2.1), and exposure to strep throat in the previous two weeks (LR+: 1.9). The absence of tender anterior cervical nodes (negative likelihood ratio [LR-]: 0.6), the absence of tonsillar enlargement (LR-: 0.6), and the absence of exudate (LR-: 0.7) are the most helpful findings for ruling out GABHS pharyngitis.4 The presence of a scarlatiniform rash or palatine petechiae is uncommon but very specific for the diagnosis of GABHS pharyngitis.4
A number of clinical decision rules have been developed and validated to help physicians more accurately estimate the probability of GABHS pharyngitis. Some have been tested only in adults, some only in children, some only in primary care settings, and some only in emergency departments or college health centers. The spectrum of disease may differ in each setting. The best clinical decision rule, created by McIsaac and colleagues,3 is included in the accompanying evidence-based patient encounter form for the management of sore throat. This tool is simple to use and takes the patient's age, tonsillar exudates, absence of cough, cervical adenopathy, and fever into account. It has been found to be clinically accurate in a group of more than 600 adults and children who presented to family physicians with sore throat.3
This guide is one in a series that offers evidence-based tools to assist family physicians in improving their decision-making at the point of care.
This series is coordinated by Mark H. Ebell, MD, MS, deputy editor for evidence-based medicine.
A collection of Point-of-Care Guides published in AFP is available at https://www.aafp.org/afp/poc. | 785 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.925453 |
https://socratic.org/questions/how-do-you-differentiate-x-cosx-sinx-x-using-the-quotient-rule | 1,611,730,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00243.warc.gz | 553,278,352 | 5,778 | # How do you differentiate (x-cosx)/ (sinx+x) using the quotient rule?
Jul 4, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + x \left(\sin x - \cos x\right) + \sin x + \cos x}{\sin x + x} ^ 2$
#### Explanation:
Let $y = \frac{x - \cos x}{\sin x + x}$ Using Quotient Rule for Diffn., we have,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sin x + x\right) \left(x - \cos x\right) ' - \left(x - \cos x\right) \left(\sin x + x\right) '}{\sin x + x} ^ 2$
$= \frac{\left(\sin x + x\right) \left(1 + \sin x\right) - \left(x - \cos x\right) \left(\cos x + 1\right)}{\sin x + x} ^ 2$
$= \frac{{\sin}^{2} x + x \sin x + \sin x + x + \left(\cos x - x\right) \left(\cos x + 1\right)}{\sin x + x} ^ 2$
$= \frac{{\sin}^{2} x + x \sin x + \sin x + x + {\cos}^{2} x - x \cos x + \cos x - x}{\sin x + x} ^ 2$
$= \frac{1 + x \left(\sin x - \cos x\right) + \sin x + \cos x}{\sin x + x} ^ 2$ | 386 | 880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-04 | latest | en | 0.225233 |
https://www.freeastroscience.com/2021/09/what-is-singularity-theory.html | 1,675,004,659,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00668.warc.gz | 772,144,908 | 31,669 | ## Thursday, September 9, 2021
### What is the Singularity theory
Singularity theory studies spaces that are almost manifolds, but not quite. A string can serve as an example of a one-dimensional manifold, if one neglects its thickness. A singularity can be made by balling it up, dropping it on the floor, and flattening it.
In some places the flat string will cross itself in an approximate “X” shape. The points on the floor where it does this are one kind of singularity, the double point: one bit of the floor corresponds to more than one bit of string. Perhaps the string will also touch itself without crossing, like an underlined “U”.
This is another kind of singularity. Unlike the double point, it is not stable, in the sense that a small push will lift the bottom of the “U” away from the “underline”.
Vladimir Arnold defines the main goal of singularity theory as describing how objects depend on parameters, particularly in cases where the properties undergo sudden change under a small variation of the parameters. These situations are called perestroika bifurcations or catastrophes.
Classifying the types of changes and characterizing sets of parameters which give rise to these changes are some of the main mathematical goals. Singularities can occur in a wide range of mathematical objects, from matrices depending on parameters to wavefronts.
In singularity theory the general phenomenon of points and sets of singularities is studied, as part of the concept that manifolds (spaces without singularities) may acquire special, singular points by a number of routes.
Projection is one way, very obvious in visual terms when three-dimensional objects are projected into two dimensions (for example in one of our eyes); in looking at classical statuary the folds of drapery are amongst the most obvious features.
Singularities of this kind include caustics, very familiar as the light patterns at the bottom of a swimming pool.
Other ways in which singularities occur is by degeneration of manifold structure. The presence of symmetry can be good cause to consider orbifolds, which are manifolds that have acquired “corners” in a process of folding up, resembling the creasing of a table napkin.
At about the same time as Hironaka’s work, the catastrophe theory of RenĂ© Thom was receiving a great deal of attention. This is another branch of singularity theory, based on earlier work of Hassler Whitney on critical points.
Roughly speaking, a critical point of a smooth function is where the level set develops a singular point in the geometric sense. This theory deals with differentiable functions in general, rather than just polynomials.
To compensate, only the stable phenomena are considered. One can argue that in nature, anything destroyed by tiny changes is not going to be observed; the visible is the stable. Whitney had shown that in low numbers of variables the stable structure of critical points is very restricted, in local terms.
Thom built on this, and his own earlier work, to create a catastrophe theory supposed to account for discontinuous change in nature. | 638 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | latest | en | 0.937185 |
https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2008-February/032608.html | 1,563,566,549,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526337.45/warc/CC-MAIN-20190719182214-20190719204214-00439.warc.gz | 459,349,388 | 3,493 | # [gmx-users] Harmonic dihedral restraints
Justin A. Lemkul jalemkul at vt.edu
Tue Feb 26 21:28:56 CET 2008
```Quoting Robert Johnson <bobjohnson1981 at gmail.com>:
> Not sure if anyone saw this...I had a question about how to include
> the dihedral restraints in the topology file. Unfortunately, this info
> doesn't seem to be in the manual. There isn't a [ dihedral_restraints
> ] section listed in Table 5.4. I presume this section should be
> formatted as:
>
> [ dihedral_restraints ]
> i j k l 1 phi_0 k delta_phi
I've never done this before, but I recall a discussion about dihedral restraints
not too long ago. There is some detailed information on the Wiki:
http://wiki.gromacs.org/index.php/Dihedral_Restraints
Maybe it will be of use.
-Justin
>
> Is that correct?
>
> Thanks,
> Bob
>
> On Tue, Feb 26, 2008 at 11:37 AM, David Mobley <dmobley at gmail.com> wrote:
> > Bob,
> >
> > The other way of putting what Mark said is that phi is only meaningful
> > on some range (-pi to pi, or 0 to 2pi, depending on how you define it)
> > and so what you require is that the potential be harmonic for the
> > region in which phi is meaningful. You don't care what happens outside
> > that. Or, in this case, you handle the issue by mapping phi values
> > outside the allowed range back into that allowed range.
> >
> > David
> >
> >
> > On Mon, Feb 25, 2008 at 2:24 PM, Robert Johnson
> >
> >
> > <bobjohnson1981 at gmail.com> wrote:
> > > Well, the potential is of the form V=k(x1-x2)^2, but I don't see how
> > > it's harmonic. What you would want is the x1 and x2 to refer to
> > > dihedral angles. However, the potential in equation 4.70 has this
> > > weird phi-phi_0 MOD 2pi term and this delta_phi parameter. It's just
> > > not obvious to me how equation 4.70 can be expanded or rearranged to
> > > look like a harmonic potential in the dihedral angles. Am I just not
> > > seeing something correctly?
> > > Bob
> > >
> > >
> > >
> > >
> > > On Mon, Feb 25, 2008 at 2:47 PM, David Mobley <dmobley at gmail.com>
> wrote:
> > > > Robert,
> > > >
> > > > I am not sure which manual you are looking at, but in the GROMACS
> 3.3
> > > > manual, equation 4.70 gives the dihedral restraint as harmonic.
> > > >
> > > > David
> > > >
> > > >
> > > >
> > > >
> > > > On Mon, Feb 25, 2008 at 10:10 AM, Robert Johnson
> > > > <bobjohnson1981 at gmail.com> wrote:
> > > > > Hello everyone,
> > > > > I am trying to calculate the absolute free energy of binding
> between a
> > > > > DNA base and a nanotube. To do this, I am first calculating the
> free
> > > > > energy associated with restraining the base in the correct
> binding
> > > > > geometry in accordance with Boresch et. al. J. Phs. Chem. B.,
> 107,
> > > > > 2003. In this paper, all restraints (1 distance, 2 angles, 3
> > > > > dihedrals) are assumed to be harmonic. In Gromacs, there already
> > > > > exists a harmonic distance restraint. Technically, the angle
> restraint
> > > > > (Equation 4.67 in the manual) is not harmonic. However, for small
> > > > > angle displacements it can be approximated as harmonic, so that's
> not
> > > > > a problem either.
> > > > >
> > > > > However, there is no harmonic dihedral restraint. A reasonable
> > > > > solution would be to use an improper dihedral (Equation 4. 59 in
> the
> > > > > manual) for the restraint. Is this alright, or are there any
> problems
> > > > > that could arise from using this? To my knowledge, exclusions are
> > > > > defined by bonds. Thus, I don't think I have to worry about the
> > > > > improper dihedral affecting the exclusions. Is this correct?
> > > > >
> > > > > Thanks,
> > > > > Bob
> > > > > _______________________________________________
> > > > > gmx-users mailing list gmx-users at gromacs.org
> > > > > http://www.gromacs.org/mailman/listinfo/gmx-users
> > > > > Please search the archive at http://www.gromacs.org/search before
> posting!
> > > > > Please don't post (un)subscribe requests to the list. Use the
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> > > > >
> > > > _______________________________________________
> > > > gmx-users mailing list gmx-users at gromacs.org
> > > > http://www.gromacs.org/mailman/listinfo/gmx-users
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> > > >
> > > _______________________________________________
> > > gmx-users mailing list gmx-users at gromacs.org
> > > http://www.gromacs.org/mailman/listinfo/gmx-users
> > > Please search the archive at http://www.gromacs.org/search before
> posting!
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> > > www interface or send it to gmx-users-request at gromacs.org.
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> > >
> > _______________________________________________
> > gmx-users mailing list gmx-users at gromacs.org
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> > Please search the archive at http://www.gromacs.org/search before posting!
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> > www interface or send it to gmx-users-request at gromacs.org.
> > Can't post? Read http://www.gromacs.org/mailing_lists/users.php
> >
> _______________________________________________
> gmx-users mailing list gmx-users at gromacs.org
> http://www.gromacs.org/mailman/listinfo/gmx-users
> Please search the archive at http://www.gromacs.org/search before posting!
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>
========================================
Justin A. Lemkul | 1,962 | 6,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-30 | latest | en | 0.916526 |
https://jukebox.esc13.net/interactiveGlossary/HTML_files/origin.html | 1,632,160,413,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00135.warc.gz | 365,409,075 | 2,212 | Origin
My Definition
Key Characteristics
The following are true of the origin:
• It is a point on the coordinate plane.
• It is located at the intersection of the x-axis and the y-axis.
• It is represented by the ordered pair, (0, 0).
• It is often labeled with the capital letter, O.
• It is where you begin counting to find a point.
Example
Non-example
Point A is located at (2, 3).
TEKS: 5(8)(A), 8(3)(C) | 115 | 419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.944797 |
http://zugidapyr.pev.pl/mortgage-refinance-calculator-extra-principal.html | 1,632,297,466,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00658.warc.gz | 124,563,284 | 5,554 | Extra Mortgage Payments Calculator
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http://www.myhomeloantools.com/extra_payments.php | 890 | 4,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.87386 |
https://in.mathworks.com/help/radar/ref/sarminaperture.html | 1,675,814,032,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00259.warc.gz | 324,787,435 | 20,539 | # sarminaperture
Lower bound on antenna area for SAR
## Syntax
``aac = sarminaperture(r,lambda,v,grazang)``
``aac = sarminaperture(r,lambda,v,grazang,dcang)``
## Description
example
````aac = sarminaperture(r,lambda,v,grazang)` returns the lower bound on antenna area based on synthetic aperture radar (SAR) constraints.```
example
````aac = sarminaperture(r,lambda,v,grazang,dcang)` specifies the Doppler cone angle that identifies the direction towards the scene relative to the direction of motion of the array.```
## Examples
collapse all
Estimate the antenna area constraint of a side-looking airborne SAR operating in broadside at 16.7 GHz with a sensor velocity of 100 m/s for a target range of 10 km. Assume a nominal grazing angle of ${30}^{\circ }$.
```fc = 16.7e9; lambda = freq2wavelen(fc); grazang =30; v = 100; R = 10e3;```
Compute the antenna area constraint.
`area = sarminaperture(R,lambda,v,grazang)`
```area = 4.1486e-04 ```
## Input Arguments
collapse all
Range from target to antenna in meters, specified as a positive real scalar or a vector.
Data Types: `double`
Radar wavelength in meters, specified as a positive real scalar or a vector.
Data Types: `double`
Sensor velocity in meters per second, specified as a positive real scalar.
Data Types: `double`
Grazing angle in degrees, specified as a scalar in the range [`0`, `90`].
Data Types: `double`
Doppler cone angle in degrees, specified as a scalar in the range [`0`, `180`]. This argument identifies the direction toward the scene relative to the direction of motion of the array.
Data Types: `double`
## Output Arguments
collapse all
Upper bound on area coverage rate in square meters per second, returned as a matrix. The rows of `aac` correspond to the range values in `r`. The columns of `aac` correspond to the wavelength values in `lambda`.
## Version History
Introduced in R2021a | 489 | 1,897 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | longest | en | 0.613596 |
https://link.springer.com/article/10.1007/s00440-022-01148-7 | 1,717,026,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00866.warc.gz | 292,509,337 | 158,227 | ## 1 Introduction
The aim of this paper is to study the local and global well posedness of evolution equations for Borel probability measures driven by a suitable notion of probability vector fields in an Eulerian framework.
For the sake of simplicity, let us consider here a finite dimensional Euclidean space $${\textsf {X} }$$ with scalar product $$\langle \cdot ,\cdot \rangle$$ and norm $$|\cdot |$$ (our analysis however will not be confined to finite dimension and will be carried out in a separable Hilbert space) and the space $$\mathcal {P}({\textsf {X} })$$ (resp. $$\mathcal {P}_b({\textsf {X} })$$) of Borel probability measures in $${\textsf {X} }$$ (resp. with bounded support).
### 1.1 A Cauchy-Lipschitz approach, via vector fields
A first notion of vector field can be described by maps $${\varvec{b}}:\mathcal {P}_b({\textsf {X} })\rightarrow \mathrm C({\textsf {X} };{\textsf {X} })$$, typically taking values in some subset of continuous vector fields in $${\textsf {X} }$$ (as the locally Lipschitz ones of $$\mathrm {Lip}_{loc}({\textsf {X} };{\textsf {X} })$$), and satisfying suitable growth-continuity conditions. In this respect, the evolution driven by $${\varvec{b}}$$ can be described by a continuous curve $$t\mapsto \mu _t\in \mathcal {P}_b({\textsf {X} })$$, $$t\in [0,T]$$, starting from an initial measure $$\mu _0\in \mathcal {P}_b({\textsf {X} })$$ and satisfying the continuity equation
in the distributional sense, i.e.
\begin{aligned} \int _0^T\int _{\textsf {X} }\Big (\partial _t \zeta +\langle \nabla \zeta ,{\varvec{v}}_t\rangle \Big )\,\mathrm d\mu _t\,\mathrm dt=0,\quad {\varvec{v}}_t={\varvec{b}}[\mu _t], \quad \text {for every } \zeta \in \mathrm {C}^1_c((0,T)\times {\textsf {X} }).\nonumber \\ \end{aligned}
(1.2)
If $${\varvec{b}}$$ is sufficiently smooth, solutions to (1.1c,d) can be obtained by many techniques. Recent contributions in this direction are given by the papers [5, 10, 26, 27], we also mention [28, 29] for the analysis in presence of sources. In particular, in [5] the aim of the authors is to develop a suitable Cauchy-Lipschitz theory in Wasserstein spaces for differential inclusions which generalizes (1.1b) to multivalued maps $${\varvec{b}}:\mathcal {P}_b({\textsf {X} })\rightrightarrows \mathrm {Lip}_{loc}({\textsf {X} };{\textsf {X} })$$ and requires (1.1b), (1.2) to hold for a suitable measurable selection of $${\varvec{b}}$$. As it occurs in the classical finite-dimensional case, the differential-inclusion approach is suitable to describe the dynamics of control systems, when the velocity vector field involved in the continuity equation depends on a control parameter.
### 1.2 The Explicit Euler method
It seems natural to approximate solutions of (1.1c,d) by a measure-theoretic version of the Explicit Euler scheme. Choosing a step size $$\tau >0$$ and a partition $$\{0,\tau ,\ldots ,n\tau ,\ldots , N\tau \}$$ of the interval [0, T], with $$N:=\left\lceil T/\tau \right\rceil$$, we construct a sequence $$M^n_\tau \in \mathcal {P}_b({\textsf {X} })$$, $$n=0,\ldots , N,$$ by the algorithm
\begin{aligned} M^0_\tau :=\mu _0,\quad M^{n+1}_\tau :=({\varvec{i}}_{\textsf {X} }+\tau {\varvec{b}}^n_\tau )_\sharp M^n_\tau ,\quad {\varvec{b}}^n_\tau \in {\varvec{b}}[M^n_\tau ], \end{aligned}
(1.3)
where $${\varvec{i}}_{\textsf {X} }(x):=x$$ is the identity map and $${\varvec{r}}_\sharp \mu$$ denotes the push forward of $$\mu \in \mathcal {P}({\textsf {X} })$$ induced by a Borel map $${\varvec{r}}:{\textsf {X} }\rightarrow {\textsf {X} }$$ and defined by $${\varvec{r}}_\sharp \mu (B):=\mu ({\varvec{r}}^{-1}(B))$$ for every Borel set $$B\subset {\textsf {X} }$$. If $${\bar{M}}_\tau$$ is the piecewise constant interpolation of the discrete values $$(M^n_\tau )_{n=0}^N$$, one can then study the convergence of $$\bar{M}_\tau$$ as $$\tau \downarrow 0$$, hoping to obtain a solution to (1.1c,d) in the limit.
It is then natural to investigate a few relevant questions:
$$\langle {\mathrm{E}}.1\rangle$$:
What is the most general framework where the Explicit Euler scheme can be implemented?
$$\langle {\mathrm{E}}.2\rangle$$:
What are the structural conditions ensuring its convergence?
$$\langle {\mathrm{E}}.3\rangle$$:
How to characterize the limit solutions and their properties?
Concerning the first question $$\langle {\mathrm{E}}.1\rangle$$, one immediately realizes that each iteration of (1.3) actually depends on the probability distribution on the tangent bundle $$\mathsf {TX}={\textsf {X} }\times {\textsf {X} }$$, where the second component plays the role of velocity, in the sense that
\begin{aligned} \Phi ^n_\tau :=({\varvec{i}}_{\textsf {X} },{\varvec{b}}^n_\tau )_\sharp M^n_\tau \in \mathcal {P}(\mathsf {TX}) \end{aligned}
whose first marginal is $$M^n_\tau$$. If we denote by $${\textsf {x} },{\textsf {v} }:\mathsf {TX}\rightarrow {\textsf {X} }$$ the projections
\begin{aligned} {\textsf {x} }(x,v):=x,\qquad {\textsf {v} }(x,v):=v, \end{aligned}
and by $$\textsf {exp} ^\tau : \mathsf {TX}\rightarrow {\textsf {X} }$$ the exponential map in the flat space $${\textsf {X} }$$, defined by
\begin{aligned} \textsf {exp} ^\tau (x,v):=x+\tau v, \end{aligned}
we recover $$M^{n+1}_\tau$$ by a single step of “free motion” driven by $$\Phi ^n_\tau$$ and given by
\begin{aligned} M^{n+1}_\tau = \textsf {exp} ^\tau _\sharp \Phi ^n_\tau =({\textsf {x} }+\tau {\textsf {v} })_\sharp \Phi ^n_\tau . \end{aligned}
This operation does not depend on the fact that $$\Phi ^n_\tau$$ is concentrated on the graph of a map (in this case $${\varvec{b}}^n_\tau \in {\varvec{b}}[M^n_\tau ]$$): one can more generally assign a multivalued map $${\varvec{\mathrm {F}}}:\mathcal {P}_b({\textsf {X} })\rightrightarrows \mathcal {P}_b(\mathsf {TX})$$ such that for every $$\mu \in \mathcal {P}_b({\textsf {X} })$$, every measure $$\Phi \in {\varvec{\mathrm {F}}}[\mu ]\in \mathcal {P}_b(\mathsf {TX})$$ has first marginal $$\mu ={\textsf {x} }_\sharp \Phi$$. We call $${\varvec{\mathrm {F}}}$$ a multivalued probability vector field (MPVF in the following), which is in good analogy with the Riemannian interpretation of $$\mathcal {P}_b(\mathsf {TX})$$. The disintegration $$\Phi _x\in \mathcal {P}_b({\textsf {X} })$$ of $$\Phi$$ with respect to $$\mu$$ provides a (unique up to $$\mu$$-negligible sets) Borel family of probability measures on the space of velocities such that $$\Phi =\int _{\textsf {X} }\Phi _x\,\mathrm d\mu (x)$$. In particular, $$\Phi$$ is induced by a vector field $${\varvec{b}}$$ only if $$\Phi _x=\delta _{{\varvec{b}}(x)}$$ is a Dirac mass for $$\mu$$-a.e. x. In the general case, (1.3) reads as
\begin{aligned} M^0_\tau :=\mu _0,\quad M^{n+1}_\tau := \textsf {exp} ^\tau _\sharp \Phi ^n_\tau = ({\textsf {x} }+\tau {\textsf {v} })_\sharp \Phi ^n_\tau ,\quad \Phi ^n_\tau \in {\varvec{\mathrm {F}}}[M^n_\tau ]. \end{aligned}
(1.4)
In addition to its greater generality, this point of view has other advantages: working with the joint distribution $${\varvec{\mathrm {F}}}[\mu ]$$ instead of the disintegrated vector field $${\varvec{b}}[\mu ]$$ potentially allows for the weakening of the continuity assumption with respect to $$\mu$$. This relaxation corresponds to the introduction of Young’s measures to study the limit behaviour of weakly converging maps [13]. Adopting this viewpoint, the classical discontinuous example in $${\mathbb {R}}$$ (see [16]), where $${\varvec{b}}(x)=-\mathrm {sign} (x)$$, admits a natural closed realization as MPVF given by
\begin{aligned} \Phi \in {\varvec{\mathrm {F}}}[\mu ] \quad \Leftrightarrow \quad \Phi _x={\left\{ \begin{array}{ll}\delta _{{\varvec{b}}(x)}&{}\text {if }x\ne 0\\ (1-\theta )\delta _{-1}+\theta \delta _1&{}\text {if }x=0 \end{array}\right. } \quad \text {for some }\theta \in [0,1]. \end{aligned}
In particular, $${\varvec{\mathrm {F}}}[\delta _0]=\left\{ \delta _0\otimes \left( (1-\theta )\delta _{-1}+ \theta \delta _{1}\right) \mid \theta \in [0,1]\right\}$$ (see also [9, Example 6.2]).
The study of measure-driven differential equations/inclusions is not new in the literature [15, 34]. However, these studies, devoted to the description of impulsive control systems [8] and mainly motivated by applications in rational mechanics and engineering, have been used to describe evolutions in $${\mathbb {R}}^d$$ rather than in the space of measures.
A second advantage in considering a MPVF is the consistency with the theory of Wasserstein gradient flows generated by geodesically convex functionals introduced in [3] (Wasserstein subdifferentials are particular examples of MPVFs) and with the multivalued version of the notion of probability vector fields introduced in [26, 27], whose originating idea was indeed to describe the uncertainty affecting not only the state of the system, but possibly also the distribution of the vector field itself.
A third advantage is to allow for a more intrinsic geometric viewpoint, inspired by Otto’s non-smooth Riemannian interpretation of the Wasserstein space: probability vector fields provide an appropriate description of infinitesimal deformations of probability measures, which should be measured by, e.g., the $$L^2$$-Kantorovich-Rubinstein-Wasserstein distance
\begin{aligned} W_2^2(\mu ,\nu ):=\min \left\{ \int _{{\textsf {X} }\times {\textsf {X} }}|x-y|^2\,\mathrm d\varvec{\gamma }(x,y): \varvec{\gamma }\in \Gamma (\mu ,\nu )\right\} , \end{aligned}
(1.5)
where $$\Gamma (\mu ,\nu )$$ is the set of couplings with marginals $$\mu$$ and $$\nu$$ respectively. It is well known [3, 32, 35] that if $$\mu ,\nu$$ belong to the space $$\mathcal {P}_2({\textsf {X} })$$ of Borel probability measures with finite second moment
\begin{aligned} {\textsf {m} }_2^2(\mu ):=\int _{\textsf {X} }|x|^2\,\mathrm d\mu (x)<\infty , \end{aligned}
then the minimum in (1.5) is attained in a compact convex set $$\Gamma _o(\mu ,\nu )$$ and $$(\mathcal {P}_2({\textsf {X} }),W_2)$$ is a complete and separable metric space. Adopting this viewpoint and proceeding by analogy with the theory of dissipative operators in Hilbert spaces, a natural class of MPVFs for evolutionary problems should at least satisfy a $$\lambda$$-dissipativity condition, with $$\lambda \in {\mathbb {R}}$$, such as
\begin{aligned}&W_2(\textsf {exp} ^\tau _\sharp \Phi ,\textsf {exp} ^\tau _\sharp \Psi ) \le (1+\lambda \tau ) W_2(\mu ,\nu )+o(\tau )\nonumber \\&\quad \text {as }\tau \downarrow 0, \text {for every }(\Phi ,\Psi )\in {\varvec{\mathrm {F}}}[\mu ]\times {\varvec{\mathrm {F}}}[\nu ], \mu \ne \nu . \end{aligned}
(1.6)
### 1.3 Metric dissipativity
Condition (1.6) in the simple case $$\lambda =0$$ has a clear interpretation in terms of one step of the Explicit Euler method: it is an asymptotic contraction as the time step goes to 0. By using the properties of the Wasserstein distance, we will first compute the right derivative of its square along the deformation $$\textsf {exp} ^\tau$$ as follows
\begin{aligned} \left[ \Phi , \Psi \right] _{r}:= & {} \frac{1}{2}\frac{\mathrm d}{\mathrm d\tau } W_2^2(\textsf {exp} ^\tau _\sharp \Phi ,\textsf {exp} ^\tau _\sharp \Psi )\Big |_{\tau =0+} \nonumber \\= & {} \min \left\{ \int _{\mathsf {TX}\times \mathsf {TX}}\langle w-v,y-x\rangle \,\mathrm d\varvec{\Theta }(x,v;y,w): \varvec{\Theta }\in \Gamma (\Phi ,\Psi ),\ ({\mathsf {x}},{\mathsf {y}})_\sharp \varvec{\Theta }\in \Gamma _o(\mu ,\nu )\right\} \nonumber \\ \end{aligned}
(1.7)
and we will show that (1.6) admits the equivalent characterization
\begin{aligned} \left[ \Phi , \Psi \right] _{r}\le \lambda W_2^2(\mu ,\nu )\quad \text { for every }(\Phi ,\Psi )\in {\varvec{\mathrm {F}}}[\mu ]\times {\varvec{\mathrm {F}}}[\nu ]. \end{aligned}
(1.8)
If we interpret the left hand side of (1.8) as a sort of Wasserstein pseudo-scalar product of $$\Phi$$ and $$\Psi$$ along the direction of an optimal coupling between $$\mu$$ and $$\nu$$, (1.8) is in perfect analogy with the canonical definition of $$\lambda$$-dissipativity (also called one-sided Lipschitz condition) for a multivalued map $${\mathrm F}:{\textsf {X} }\rightrightarrows {\textsf {X} }$$, which reads as
\begin{aligned} \langle w-v,y-x\rangle \le \lambda |x-y|^2\quad \text { for every }(v,w)\in {\mathrm F}[x]\times {\mathrm F}[y]. \end{aligned}
(1.9)
It turns out that the (opposite of the) Wasserstein subdifferential $$\varvec{\partial }{\mathcal {F}}$$ [3, Sect. 10.3] of a geodesically $$(-\lambda )$$-convex functional $$\mathcal F:\mathcal {P}_2({\textsf {X} })\rightarrow (-\infty ,+\infty ]$$ is a MPVF and satisfies a condition equivalent to (1.6) and (1.8). We also notice that (1.8) reduces to (1.9) in the particular case when $$\Phi =\delta _{(x,v)},\Psi =\delta _{(y,w)}$$ are Dirac masses in $$\mathsf {TX}$$.
### 1.4 Conditional convergence of the Explicit Euler method
Contrary to the Implicit Euler method, however, even if a MPVF satisfies (1.8), every step of the Explicit Euler scheme (1.4) affects the distance by a further quadratic correction according to the formula
\begin{aligned} W_2^2(\textsf {exp} ^\tau _\sharp \Phi ,\textsf {exp} ^\tau _\sharp \Psi )&\le W_2^2(\mu ,\nu )+2\tau \left[ \Phi , \Psi \right] _{r}+\tau ^2\Big ( |\Phi |_2^2+|\Psi |_2^2\Big ),\\ |\Phi |_2^2&:=\int _\mathsf {TX}|v|^2\,\mathrm d\Phi (x,v), \end{aligned}
which depends on the order of magnitude of $$\Phi$$ and $$\Psi$$, and thus of $${\varvec{\mathrm {F}}}$$, at $$\mu$$ and $$\nu$$.
Our first main result (Theorems 6.5 and 6.7), which provides an answer to question $$\langle {\mathrm{E}}.2\rangle$$, states that if $${\varvec{\mathrm {F}}}$$ is a $$\lambda$$-dissipative MPVF according to (1.8) then every family of discrete solutions $$({\bar{M}}_\tau )_{\tau >0}$$ of (1.4) in an interval [0, T] satisfying the abstract stability condition
\begin{aligned} |\Phi ^n_\tau |_2\le L\quad \text {if }0\le n\le N:=\left\lceil T/\tau \right\rceil , \end{aligned}
(1.10)
is uniformly converging to a Lipschitz continuous limit curve $$\mu :[0,T]\rightarrow \mathcal {P}_2({\textsf {X} })$$ starting from $$\mu _0$$, with a uniform error estimate
\begin{aligned} W_2(\mu _t,{\bar{M}}_\tau (t))\le CL\sqrt{\tau (t+\tau )}\mathrm e^{\lambda _+ t} \end{aligned}
(1.11)
for every $$t\in [0,T]$$, and a universal constant $$C\le 14$$. Apart from the precise value of C, the estimate (1.11) is sharp [31] and reproduces in the measure-theoretic framework the celebrated Crandall-Liggett estimate [12] for the generation of dissipative semigroups in Banach spaces. We derive it by adapting to the metric-Wasserstein setting the relaxation and doubling variable techniques of [23], strongly inspired by the ideas of Kružkov [21] and Crandall-Evans [11].
This crucial result does not require any bound on the support of the measures and no local compactness of the underlying space $${\textsf {X} }$$, so that we will prove it in a general Hilbert space, possibly with infinite dimension. Moreover, if $$\mu ,\nu$$ are two limit solutions starting from $$\mu _0,\nu _0$$ we show that
\begin{aligned} W_2(\mu _t,\nu _t)\le W_2(\mu _0,\nu _0)\mathrm e^{\lambda t}\quad \text { for every } t\in [0,T], \end{aligned}
as it happens in the case of gradient flows of $$(-\lambda )$$-convex functions. Once one has these building blocks, it is not too difficult to construct a local and global existence theory, mimicking the standard arguments for ODEs.
### 1.5 Metric characterization of the limit solution
As we stated in question $$\langle {\mathrm{E}}.3\rangle$$, a further important point is to get an effective characterization of the solution $$\mu$$ obtained as limit of the approximation scheme.
As a first property, considered in [26, 27] in the case of a single-valued PVF, one could hope that $$\mu$$ satisfies the continuity equation (1.1a) coupled with the barycentric condition, thus replacing (1.1b) with
\begin{aligned} {\varvec{v}}_t(x)=\int _\mathsf {TX}v\,\mathrm d\Phi _t(x,v),\quad \Phi _t\in {\varvec{\mathrm {F}}}[\mu _t]. \end{aligned}
(1.12)
This is in fact true, as shown in [26, 27] in the finite dimensional case, if $${\varvec{\mathrm {F}}}$$ is single valued and satisfies a stronger Lipschitz dependence w.r.t. $$\mu$$ (see (H1) in Sect. 7.5).
In the framework of dissipative MPVFs, we will replace (1.12) with its relaxation à la Filippov (see e.g. [36, Chapter 2] and [2, Chapter 10]) given by
\begin{aligned} {\varvec{v}}_t(x)=\int _\mathsf {TX}v\,\mathrm d\Phi _t(x,v)\quad \text {for some}\quad \Phi _t\in \overline{{\text {co}}}({\text {cl}}({\varvec{\mathrm {F}}})[\mu _t]), \end{aligned}
where $${\text {cl}}({\varvec{\mathrm {F}}})$$ is the sequential closure of the graph of $${\varvec{\mathrm {F}}}$$ in the strong-weak topology of $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ and $$\overline{{\text {co}}}({\text {cl}}({\varvec{\mathrm {F}}})[\mu ])$$ denotes the closed convex hull of the given section $${\text {cl}}({\varvec{\mathrm {F}}})[\mu ]$$. We refer to [25] and Sect. 2.2 for more details on the mentioned strong-weak topology; in fact, a more restrictive “directional” closure could be considered, see Sect. 5.5 and in particular Theorem 5.27.
However, even in the case of a single valued map, (1.12) is not enough to characterize the limit solution, as it has been shown by an interesting example in [9, 27] (see also the gradient flow of Example 7.7).
From a Wasserstein viewpoint, one could consider the differential inclusion
\begin{aligned} ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t^W)_\sharp \mu _t\in {\varvec{\mathrm {F}}}[\mu _t],\quad \text {for a.e. }\,t\in [0,T], \end{aligned}
(1.13)
where $${\varvec{v}}^W$$ is the Wasserstein metric velocity field associated to $$\mu$$ (see Theorem 2.10). However, while this property was appropriate to characterize limit solution $$\mu$$ in the case of gradient flows, it is not reasonable for a general MPVF $${\varvec{\mathrm {F}}}$$. Indeed, the given MPVF $${\varvec{\mathrm {F}}}$$, even if regular, could have no relation with the tangent space $${{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2({\textsf {X} })$$ where $${\varvec{v}}_t^W$$ lies.
In order to address the problem of characterizing the limit solution $$\mu$$, here we follow the metric viewpoint adopted in [3] for gradient flows and we will characterize the limit solutions by a suitable Evolution Variational Inequality satisfied by the squared distance function from given test measures. As a byproduct (see Theorem 5.4), this interpretation will be reflected in a relaxed formulation of the inclusion (1.13) with respect to a suitable extension $${\hat{{\varvec{\mathrm {F}}}}}$$ of $${\varvec{\mathrm {F}}}$$ introduced in Sect. 4.3. This approach is also strongly influenced by the Bénilan notion of integral solutions to dissipative evolutions in Banach spaces [4]. The main idea is that any differentiable solution to $$\dot{x}(t)\in {\mathrm F}[x(t)]$$ driven by a $$\lambda$$-dissipative operator in a Hilbert space as in (1.9) satisfies
\begin{aligned} \begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}|x(t)-y|^2&= \langle \dot{x}(t),x(t)-y\rangle \\ {}&= \langle \dot{x}(t)-w,x(t)-y\rangle +\langle w,x(t)-y\rangle \\ {}&\le \lambda |x(t)-y|^2-\langle w,y-x(t)\rangle \end{aligned} \end{aligned}
for every $$w\in {\mathrm F}[y]$$. In the framework of $$\mathcal {P}_2({\textsf {X} })$$, we replace $$w\in {\mathrm F}[y]$$ with $$\Psi \in {\varvec{\mathrm {F}}}[\nu ]$$ and the scalar product $$\langle w,y-x(t)\rangle$$ with
\begin{aligned} \left[ \Psi , \mu _t\right] _{r}:=\min \left\{ \int _{\mathsf {TX}\times {\textsf {X} }}\langle w,y-x\rangle \,\mathrm d\varvec{\Theta }(y,w;x): \varvec{\Theta }\in \Gamma (\Psi ,\mu _t),\ ({\mathsf {y}},{\mathsf {x}})_\sharp \varvec{\Theta }\in \Gamma _o(\nu ,\mu _t)\right\} , \end{aligned}
as in (1.7). According to this formal heuristic, we consider the $$\lambda$$-EVI characterization of a limit curve $$\mu$$ as
As for Bénilan integral solutions, we can considerably relax the apriori smoothness assumptions on $$\mu$$, just imposing that $$\mu$$ is continuous and ($$\lambda$$-EVI) holds in the sense of distributions in (0, T). In this way, we obtain a robust characterization, which is stable under uniform convergence (cf. Proposition 5.6) and also allows for solutions taking values in the closure of the domain of $${\varvec{\mathrm {F}}}$$. This is particularly important when $${\varvec{\mathrm {F}}}$$ involves drift terms with superlinear growth (see Example 7.5).
The crucial point of this approach relies on a general error estimate, which extends the validity of (1.11) to a general $$\lambda$$-EVI solution $$\mu$$ and therefore guarantees its uniqueness, whenever the Explicit Euler method is solvable, at least locally in time (see Sect. 5.3).
Combining local in time existence with suitable global confinement conditions (see e.g. Theorem 5.32) we can eventually obtain a robust theory for the generation of a $$\lambda$$-flow, i.e. a semigroup $$(\mathrm S_t)_{t\ge 0}$$ in a suitable subset D of $$\mathcal {P}_2({\textsf {X} })$$ such that $$\mathrm S_t[\mu _0]$$ is the unique $$\lambda$$-EVI solution starting from $$\mu _0$$ and for every $$\mu _0,\mu _1\in D$$
\begin{aligned} W_2(\mathrm S_t[\mu _0],\mathrm S_t[\mu _1])\le W_2(\mu _0,\mu _1)\mathrm e^{\lambda t}\quad \text { for every } t\ge 0, \end{aligned}
as in the case of Wasserstein gradient flows of geodesically $$(-\lambda )$$-convex functionals.
### 1.6 Explicit vs Implicit Euler method
In the framework of contraction semigroups generated by $$\lambda$$-dissipative operators in Hilbert or Banach spaces, a crucial role is played by the Implicit Euler scheme, which has the advantage to be unconditionally stable, and thus avoids any apriori restriction on the local bound of the operator, as we did in (1.10). In Hilbert spaces, it is well known that the solvability of the Implicit Euler scheme is equivalent to the maximality of the graph of the operator.
In the case of a Wasserstein gradient flow of a geodesically convex $$\mathcal {F}:\mathcal {P}_2({\textsf {X} })\rightarrow (-\infty ,+\infty ]$$, every step of the Implicit Euler method (also called JKO/Minimizing Movement scheme [3, 20]) can be solved by a variational approach: $$M^{n+1}_\tau$$ has to be selected among the solutions of
(1.14)
Notice, however, that in this case the MPVF $$\varvec{\partial }{\mathcal {F}}$$ is defined implicitely in terms of $${\mathcal {F}}$$ and each step of (1.14) provides a suitable variational selection in $$\varvec{\partial }{\mathcal {F}}$$, leading in the limit to the minimal selection principle.
In the case of more general dissipative evolutions, it is not at all clear how to solve the Implicit Euler scheme, in particular when $${\varvec{\mathrm {F}}}[\mu ]$$ is not concentrated on a map, and to characterize the maximal extension of $${\varvec{\mathrm {F}}}$$ (in the Hilbertian case the maximal extension of a dissipative operator $${\mathrm F}$$ is explicitly computable at least when the domain of $${\mathrm F}$$ has not empty interior, see the Theorems of Robert and Bénilan in [30]). The analogy with the Hilbertian theory does not extend to some properties: in particular, a dissipative MPVF $${\varvec{\mathrm {F}}}$$ in $$\mathcal {P}_2({\textsf {X} })$$ is not locally bounded in the interior of its domain (see Example 7.3) and maximality may fail also for single-valued continuous PVFs (see Example 7.4). Even more remarkably, in the Hilbertian case a crucial equivalent characterization of dissipativity reads as
\begin{aligned} |x-y|\le |(x-\tau v)-(y-\tau w)|\quad \text { for every }(v,w)\in {\mathrm F}[x]\times {\mathrm F}[y], \end{aligned}
which implies that the resolvent operators $$({\varvec{i}}_{\textsf {X} }-\tau {\mathrm F})^{-1}$$ – and thus every single step of the Implicit Euler scheme – are contractions on $${\textsf {X} }$$. On the contrary, if we assume the forward characterizations (1.6) and (1.8) of dissipativity in $$\mathcal {P}_2({\textsf {X} })$$ (with $$\lambda =0$$) we cannot conclude in general that
\begin{aligned} W_2(\mu ,\nu )\le W_2(\textsf {exp} ^{-\tau }_\sharp \Phi , \textsf {exp} ^{-\tau }_\sharp \Psi )\quad \text { for every }(\Phi ,\Psi )\in {\varvec{\mathrm {F}}}[\mu ]\times {\varvec{\mathrm {F}}}[\nu ], \end{aligned}
(1.15)
since the squared distance map $$f(t):=W^2_2(\textsf {exp} ^{t}_\sharp \Phi , \textsf {exp} ^{t}_\sharp \Psi )$$, $$t\in {\mathbb {R}}$$, is not convex in general (see e.g. [3, Example 9.1.5]) and the fact that its right derivative at $$t=0$$ (corresponding to $$\left[ \Phi , \Psi \right] _{r}$$) is $$\le 0$$ according to (1.8) does not imply that $$f(0)\le f(t)$$ for $$t<0$$ (corresponding to (1.15) for $$t=-\tau$$).
For these reasons, we decided to approach the investigation of dissipative evolutions in $$\mathcal {P}_2({\textsf {X} })$$ by the Explicit Euler method, and we defer the study of the implicit one to a forthcoming paper.
### 1.7 Plan of the paper
As already mentioned, our theory works for a general separable Hilbert space $${\textsf {X} }$$, and we recollect some preliminary material concerning the Wasserstein distance in Hilbert spaces and the properties of strong-weak topology for $$\mathcal {P}_2(\mathsf {TX})$$ in Sect. 2.
In Sect. 3, we will study the semi-concavity properties of $$W_2$$ along general deformations induced by the exponential map $$\textsf {exp} ^\tau$$ and we introduce and study the pairings $$\left[ \cdot , \cdot \right] _{r}$$, $$\left[ \cdot , \cdot \right] _{l}$$. We will apply such tools to derive the precise expressions of the left and right derivatives of $$W_2$$ along absolutely continuous curves in $$\mathcal {P}_2({\textsf {X} })$$ in Sect. 3.2.
In Sect. 4, we will introduce and study the notion of $$\lambda$$-dissipative MPVF, in particular its behaviour along geodesics (Sect. 4.2) and its extension properties (Sect. 4.3).
Sections 5 and 6 contain the core of our results. Section 5 is devoted to the notion of $$\lambda$$-EVI solutions and to their properties: local uniqueness, stability and regularity in Sect. 5.3, global existence in Sect. 5.4 and barycentric characterizations in Sect. 5.5. Section 6 contains the main estimates for the Explicit Euler scheme: the Cauchy estimates between two discrete solutions corresponding to different step sizes in Sect. 6.2 and the uniform error estimates between a discrete and a $$\lambda$$-EVI solution in Sect. 6.3.
Finally, a few examples are collected in Sect. 7.
## 2 Preliminaries
In this section, we introduce the main concepts and results of Optimal Transport theory that will be extensively used in the rest of the paper. We start by listing the adopted notation.
$${\varvec{b}}_{\Phi }$$ the barycenter of $$\Phi \in \mathcal {P}(\mathsf {TX})$$ as in Definition 3.1 $$\mathrm B_X(x,r)$$ the open ball with radius $$r>0$$ centered at $$x\in X$$ $$\mathrm {C}(X;Y)$$ the set of continuous functions from X to Y $$\mathrm {C}_b(X)$$ the set of bounded continuous real valued functions defined in X $$\mathrm {C}_c(X)$$ the set of continuous real valued functions with compact support $${{\,\mathrm{Cyl}\,}}({\textsf {X} })$$ the space of cylindrical functions on $${\textsf {X} }$$, see Definition 2.9 $${\text {cl}}({\varvec{\mathrm {F}}}),{\text {co}}({\varvec{\mathrm {F}}})[\mu ]$$ the sequential closure and convexification of $${\varvec{\mathrm {F}}}$$, see Sect. 4.3 $$\overline{{\text {co}}}({\varvec{\mathrm {F}}})[\mu ],{\hat{{\varvec{\mathrm {F}}}}}$$ sequential closure of convexification and extension of $${\varvec{\mathrm {F}}}$$, see Sect. 4.3 $${\frac{\mathrm d}{\mathrm dt}}^{+}\zeta , {\frac{\mathrm d}{\mathrm dt}}_{+}\zeta$$ the right upper/lower Dini derivatives of $$\zeta$$, see (5.3) $$\mathrm {D}({\varvec{\mathrm {F}}})$$ the proper domain of a set-valued function as in Definition 4.1
$${\mathscr {E}}(\mu _0,\tau ,T,L), {\mathscr {M}}(\mu _0,\tau ,T,L)$$ the sets associated to the Explicit Euler scheme (EE) defined in (5.12) $$f_\sharp \nu$$ the push-forward of $$\nu \in \mathcal {P}(X)$$ through the map $$f:X\rightarrow Y$$ $$\Gamma (\mu ,\nu )$$ the set of admissible couplings between $$\mu ,\nu$$, see (2.1) $$\Gamma _o(\mu ,\nu )$$ the set of optimal couplings between $$\mu ,\nu$$, see Definition 2.5 $$\Gamma _o^{i}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}),\,i=0,1$$ the set of optimal couplings conditioned to $${\varvec{\mathrm {F}}}$$, see (4.12) $${\mathcal {I}}$$ an interval of $${\mathbb {R}}$$ $${\varvec{i}}_X(\cdot )$$ the identity function on a set X $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$ the set of time instants t s.t. $${\textsf {x} }^t_\sharp \varvec{\mu }$$ belongs to $$\mathrm {D}({\varvec{\mathrm {F}}})$$, see (4.7) $$\lambda _+$$ the positive part of $$\lambda \in {\mathbb {R}}$$, given by $$\lambda _+=\max \{\lambda ,0\}$$ $$\Lambda ,\Lambda _o$$ the sets of couplings as in Definition 3.8 and Theorem 3.9 $${\mathcal {L}}$$ the 1-dimensional Lebesgue measure $${\textsf {m} }_2(\nu )$$ the 2-nd moment of $$\nu \in \mathcal {P}(X)$$ as in Definition 2.5 $$|\Phi |_2$$ the 2-nd moment of $$\Phi \in \mathcal {P}(\mathsf {TX})$$ as in (3.2) $$|{\varvec{\mathrm {F}}}|_2(\mu )$$ the 2-nd moment of $${\varvec{\mathrm {F}}}$$ at $$\mu$$ as in (5.20) $$|{\dot{\mu }}_t|$$ the metric derivative at t of a locally absolutely continuous curve $$\mu$$ $$\mathcal {P}(X)$$ the set of Borel probability measures on the topological space X $$\mathcal {P}_b(X)$$ the set of Borel probability measures with bounded support $$\mathcal {P}_2(X)$$ the subset of measures in $$\mathcal {P}(X)$$ with finite quadratic moments $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$ the space $$\mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ endowed with a weaker topology as in Definition 2.14 $$\mathcal {P}_{}(\mathsf {TX}|\mu )$$ the subset of $$\mathcal {P}_2(\mathsf {TX})$$ with fixed first marginal $$\mu$$ as in (3.3) $$\left[ \cdot , \cdot \right] _{r}$$, $$\left[ \cdot , \cdot \right] _{l}$$ the pseudo scalar products as in Definition 3.5 $$\left[ \Phi _t, \varvec{\vartheta }\right] _{b,t}$$,$$[\Phi _t,\varvec{\vartheta }]_{r,t}$$, $$[\Phi _t,\varvec{\vartheta }]_{l,t}$$ the duality pairings as in Definition 3.18 $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}$$, $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}$$ the duality pairings as in Definition 4.8 $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+},[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}$$ the limiting duality pairings as in Definition 4.11 $${{\,\mathrm{supp}\,}}(\nu )$$ the support of $$\nu \in \mathcal {P}(X)$$ $${{\,\mathrm{Tan}\,}}_{\mu }\mathcal {P}_2(X)$$ the tangent space defined in Theorem 2.10 $$W_2(\mu ,\nu )$$ the $$L^2$$-Wasserstein distance between $$\mu$$ and $$\nu$$, see Definition 2.5 $${\textsf {X} }$$ a separable Hilbert space $$\mathsf {TX}$$ the tangent bundle to $${\textsf {X} }$$, usually endowed with the strong-weak topology $${\textsf {x} },{\textsf {v} },\textsf {exp} ^t,{\textsf {s} }$$ the projection, exponential and reversion maps defined in (3.1) and (3.26) $${\textsf {x} }^t$$ the evaluation map defined in (3.4) $$\left\lfloor \cdot \right\rfloor ,\,\left\lceil \cdot \right\rceil$$ the floor and ceiling functions, see (5.8)
In the present paper we will mostly deal with Borel probability measures defined in (subsets of) some separable Hilbert space endowed with the strong or a weaker topology. The convenient setting is therefore provided by Polish/Lusin and completely regular topological spaces.
Recall that a topological space X is Polish (resp. Lusin) if its topology is induced by a complete and separable metric (resp. is coarser than a Polish topology). We will denote by $$\mathcal {P}(X)$$ the set of Borel probability measures on X. If X is Lusin, every measure $$\mu \in \mathcal {P}(X)$$ is also a Radon measure, i.e. it satisfies
\begin{aligned} \forall \, B\subset X\text { Borel, }\forall \, \varepsilon >0 \quad \exists \, K\subset B \text { compact s.t. } \mu (B {\setminus } K) < \varepsilon . \end{aligned}
X is completely regular if it is Hausdorff and for every closed set C and point $$x\in X{\setminus } C$$ there exists a continuous function $$f: X \rightarrow [0,1]$$ s.t. $$f(x)=0$$ and $$f(C)=\{1\}$$.
Given X and Y Lusin spaces, $$\mu \in \mathcal {P}(X)$$ and a Borel function $$f: X \rightarrow Y$$, there is a canonical way to transfer the measure $$\mu$$ from X to Y through f. This is called the push forward of $$\mu$$ through f, denoted by $$f_{\sharp }\mu$$ and defined by $$(f_{\sharp }\mu )(B) := \mu (f^{-1}(B))$$ for every Borel set B in Y, or equivalently
\begin{aligned} \int _{Y} \varphi \,\mathrm d(f_{\sharp }\mu ) = \int _X \varphi \circ f \,\mathrm d\mu \end{aligned}
for every $$\varphi$$ bounded (or nonnegative) real valued Borel function on Y. A particular case occurs if $$X=X_1 \times X_2$$, $$Y=X_i$$ and $$f=\pi ^i$$ is the projection on the i-th component, $$i=1,2$$. In this case, f is usually denoted with $$\pi ^i$$ or $$\pi ^{X_i}$$, and $$\pi ^{X_i}_{\sharp }\mu$$ is called the i-th marginal of $$\mu$$.
This notation is particularly useful when dealing with transport plans: given $$X_1$$ and $$X_2$$ two completely regular spaces and $$\mu \in \mathcal {P}(X_1)$$, $$\nu \in \mathcal {P}(X_2)$$, we define
\begin{aligned} \Gamma (\mu , \nu ) := \left\{ \varvec{\gamma }\in \mathcal {P}(X_1 \times X_2) \mid \pi ^{1}_{\sharp } \varvec{\gamma }= \mu \, , \, \pi ^{2}_{\sharp } \varvec{\gamma }= \nu \right\} , \end{aligned}
(2.1)
i.e. the set of probability measures on the product space having $$\mu$$ and $$\nu$$ as marginals.
On $$\mathcal {P}(X)$$ we consider the so-called narrow topology which is the coarsest topology on $$\mathcal {P}(X)$$ s.t. the maps $$\mu \mapsto \int _X \varphi \,\mathrm d\mu$$ are continuous for every $$\varphi \in \mathrm {C}_b(X)$$, the space of real valued and bounded continuous functions on X. In this way a net $$(\mu _\alpha )_{\alpha \in \mathbb A} \subset \mathcal {P}(X)$$ indexed by a directed set $${\mathbb {A}}$$ is said to converge narrowly to $$\mu \in \mathcal {P}(X)$$, and we write $$\mu _\alpha \rightarrow \mu$$ in $$\mathcal {P}(X)$$, if
\begin{aligned} \lim _{\alpha } \int _X \varphi \,\mathrm d\mu _\alpha = \int _X \varphi \,\mathrm d\mu \quad \text { for every }\varphi \in \mathrm {C}_b(X). \end{aligned}
We recall the well known Prokhorov’s theorem in the context of completely regular topological spaces (see [33, Appendix]).
### Theorem 2.1
(Prokhorov) Let X be a completely regular topological space and let $$\mathcal {F} \subset \mathcal {P}(X)$$ be a tight subset i.e.
\begin{aligned} \text {for all }\, \varepsilon >0\; \text { there exists }\, K_{\varepsilon } \subset X \text { compact s.t. } \sup _{\mu \in \mathcal {F}}\mu (X {\setminus } K_{\varepsilon }) < \varepsilon . \end{aligned}
Then $$\mathcal {F}$$ is relatively compact in $$\mathcal {P}(X)$$ w.r.t. the narrow topology.
It is then relevant to know when a given $$\mathcal {F} \subset \mathcal {P}(X)$$ is tight. If X is a Lusin completely regular topological space, then the set $$\mathcal {F} = \{ \mu \}\subset \mathcal {P}(X)$$ is tight. Another trivial criterion for tightness is the following: if $$\mathcal {F} \subset \mathcal {P}(X_1 \times X_2)$$ is s.t. $$\mathcal {F}_i := \{ \pi _{\sharp }^i \varvec{\gamma }\mid \varvec{\gamma }\in \mathcal {F} \} \subset \mathcal {P}(X_i)$$ are tight for $$i=1,2$$, then also $$\mathcal {F}$$ is tight. We also recall the following useful proposition (see [3, Remark 5.1.5]).
### Proposition 2.2
Let X be a Lusin completely regular topological space and let $$\mathcal {F} \subset \mathcal {P}(X)$$. Then $$\mathcal {F}$$ is tight if and only if there exists $$\varphi : X \rightarrow [0, + \infty ]$$ with compact sublevels s.t.
\begin{aligned} \sup _{\mu \in \mathcal {F}} \int _{X} \varphi \,\mathrm d\mu < + \infty . \end{aligned}
We recall the so-called disintegration theorem (see e.g. [3, Theorem 5.3.1]).
### Theorem 2.3
Let $${\mathbb {X}}, X$$ be Lusin completely regular topological spaces, $$\varvec{\mu }\in \mathcal {P}(\mathbb {X})$$ and $$r:\mathbb {X}\rightarrow X$$ a Borel map. Denote with $$\mu =r_{\sharp }\varvec{\mu }\in \mathcal {P}(X)$$. Then there exists a $$\mu$$-a.e. uniquely determined Borel family of probability measures $$\{\varvec{\mu }_x\}_{x\in X}\subset \mathcal {P}(\mathbb {X})$$ such that $$\varvec{\mu }_x(\mathbb {X}{\setminus } r^{-1}(x))=0$$ for $$\mu$$-a.e. $$x\in X$$, and
\begin{aligned} \int _{\mathbb {X}}\varphi ({\varvec{x}})\,\mathrm d\varvec{\mu }({\varvec{x}})=\int _X\left( \int _{r^{-1}(x)}\varphi ({\varvec{x}})\,\mathrm d\varvec{\mu }_x({\varvec{x}})\right) \,\mathrm d\mu (x) \end{aligned}
for every bounded Borel map $$\varphi :\mathbb {X}\rightarrow {\mathbb {R}}$$.
### Remark 2.4
When $$\mathbb {X}=X_1\times X_2$$ and $$r=\pi ^1$$, we can canonically identify the disintegration $$\{\varvec{\mu }_x\}_{x\in X_1} \subset \mathcal {P}(\mathbb {X})$$ of $$\varvec{\mu }\in \mathcal {P}(X_1\times X_2)$$ w.r.t. $$\mu =\pi ^1_\sharp \varvec{\mu }$$ with a family of probability measures $$\{\mu _{x_1}\}_{x_1\in X_1} \subset \mathcal {P}(X_2)$$. We write $$\varvec{\mu }= \int _{X_1}\mu _{x_1}\,\mathrm d\mu (x_1)$$.
### 2.1 Wasserstein distance in Hilbert spaces
Let X be a separable (possibly infinite dimensional) Hilbert space. We will denote by $$X^s$$ (respt. $$X^w$$) the Hilbert space endowed with its strong (resp. weak) topology. Notice that $$X^w$$ is a Lusin completely regular space. The spaces $$X^s$$ and $$X^w$$ share the same class of Borel sets and therefore of Borel probability measures, which we will simply denote by $$\mathcal {P}(X)$$, using $$\mathcal {P}(X^s)$$ and $$\mathcal {P}(X^w)$$ only when we will refer to the corresponding topology. Finally, if $$X$$ has finite dimension then the two topologies coincide.
We now list some properties of Wasserstein spaces and we refer to [3, § 7] for a complete account of this matter.
### Definition 2.5
Given $$\mu \in \mathcal {P}(X)$$ we define
\begin{aligned} {\textsf {m} }_2^2(\mu ):= \int _X|x|^2 \,\mathrm d\mu (x)\qquad \text {and}\qquad \mathcal {P}_2(X) := \Bigl \{ \mu \in \mathcal {P}(X) \mid {\textsf {m} }_2(\mu )< + \infty \Bigr \}. \end{aligned}
The $$L^2$$-Wasserstein distance between $$\mu , \mu ' \in \mathcal {P}_2(X)$$ is defined as
\begin{aligned} W_2^2(\mu , \mu ')&:= \inf \left\{ \int _{X\times X} |x-y|^2 \,\mathrm d\varvec{\gamma }(x,y) \mid \varvec{\gamma }\in \Gamma (\mu , \mu ') \right\} . \end{aligned}
(2.2)
The set of elements of $$\Gamma (\mu , \mu ')$$ realizing the infimum in (2.2) is denoted with $$\Gamma _o(\mu , \mu ')$$. We say that a measure $$\varvec{\gamma }\in \mathcal {P}_2(X\times X)$$ is optimal if $$\varvec{\gamma }\in \Gamma _o(\pi ^1_\sharp \varvec{\gamma },\pi ^2_\sharp \varvec{\gamma })$$.
We will denote by $$\mathrm B(\mu ,\varrho )$$ the open ball centered at $$\mu$$ with radius $$\varrho$$ in $$\mathcal {P}_2(X)$$. The metric space $$(\mathcal {P}_2(X), W_2)$$ enjoys many interesting properties: here we only recall that it is a complete and separable metric space and that $$W_2$$-convergence (sometimes denoted with $$\overset{W_2}{\longrightarrow }$$) is stronger than the narrow convergence. In particular, given $$(\mu _n)_{n\in {\mathbb {N}}}\subset \mathcal {P}_2(X)$$ and $$\mu \in \mathcal {P}_2(X)$$, we have [3, Remark 7.1.11] that
\begin{aligned} \mu _n\overset{W_2}{\rightarrow }\mu ,\text { as }n\rightarrow +\infty \quad \Longleftrightarrow \quad {\left\{ \begin{array}{ll}\mu _n\rightarrow \mu \text { in }\mathcal {P}(X^s),\\ {\textsf {m} }_2(\mu _n)\rightarrow {\textsf {m} }_2(\mu ), \end{array}\right. } \text { as }n\rightarrow +\infty . \end{aligned}
(2.3)
Finally, we recall that sequences converging in $$(\mathcal {P}_2(X), W_2)$$ are tight. More precisely we have the following characterization of compactness in $$\mathcal {P}_2(X)$$.
### Lemma 2.6
(Relative compactness in $$\mathcal {P}_2(X)$$) A subset $${\mathcal {K}}\subset \mathcal {P}_2(X)$$ is relatively compact w.r.t. the $$W_2$$-topology if and only if
1. (1)
$${\mathcal {K}}$$ is tight w.r.t. $$X^s$$,
2. (2)
$${\mathcal {K}}$$ is uniformly 2-integrable, i.e.
\begin{aligned} \lim _{k\rightarrow \infty }\sup _{\mu \in {\mathcal {K}}}\int _{|x|\ge k}|x|^2\,\mathrm d\mu =0. \end{aligned}
(2.4)
### Proof
Tightness is clearly a necessary condition; concerning (2.4) let us notice that the maps
\begin{aligned} F_k:\mathcal {P}_2(X)\rightarrow [0,\infty ),\quad F_k(\mu ):=\int _{|x|\ge k}|x|^2\,\mathrm d\mu \end{aligned}
are upper semicontinuous, are decreasing w.r.t. k, and converge pointwise to 0 for every $$\mu \in \mathcal {P}_2(X)$$. Therefore, if $${\mathcal {K}}$$ is relatively compact, they converge uniformly to 0 thanks to Dini’s Theorem.
In order to prove that (1) and (2) are also sufficient for relative compactness, it is sufficient to check that every sequence $$(\mu _n)_{n\in {\mathbb {N}}}$$ in $${\mathcal {K}}$$ has a convergent subsequence. Applying Prokhorov Theorem 2.1, we can find $$\mu \in \mathcal {P}(X)$$ and a convergent subsequence $$k\mapsto \mu _{n_k}$$ such that $$\mu _{n_k}\rightarrow \mu$$ in $$\mathcal {P}(X^s)$$. Since $${\textsf {m} }_2(\mu _n)$$ is uniformly bounded, then $$\mu \in \mathcal {P}_2(X)$$. Applying [3, Lemma 5.1.7], we also get
\begin{aligned} \lim _{k\rightarrow \infty }{\textsf {m} }_2(\mu _{n_k})={\textsf {m} }_2(\mu ) \end{aligned}
so that, by (2.3), we conclude
\begin{aligned} \lim _{k\rightarrow \infty }W_2(\mu _{n_k},\mu )=0. \end{aligned}
$$\square$$
### Definition 2.7
(Geodesics) A curve $$\mu :[0,1]\rightarrow \mathcal {P}_2(X)$$ is said to be a (constant speed) geodesic if for all $$0\le s\le t\le 1$$ we have
\begin{aligned} W_2(\mu _s,\mu _t)=(t-s)W_2(\mu _0,\mu _1), \end{aligned}
where $$\mu _t$$ denotes the evaluation at time $$t\in [0,1]$$ of $$\mu$$. We also say that $$\mu$$ is a geodesic from $$\mu _0$$ to $$\mu _1$$.
We say that $$A\subset \mathcal {P}_2(X)$$ is a geodesically convex set if for any pair $$\mu _0,\mu _1\in A$$ there exists a geodesic $$\mu$$ from $$\mu _0$$ to $$\mu _1$$ such that $$\mu _t\in A$$ for every $$t\in [0,1]$$.
We recall also the following useful properties of geodesics (see [3, Theorem 7.2.1, Theorem 7.2.2]).
### Theorem 2.8
(Properties of geodesics) Let $$\mu _0,\mu _1\in \mathcal {P}_2(X)$$ and $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$. Then $$\mu :[0,1]\rightarrow \mathcal {P}_2(X)$$, defined by
\begin{aligned} \mu _t := ({\textsf {x} }^t)_{\sharp } \varvec{\mu },\quad t\in [0,1], \end{aligned}
(2.5)
is a (constant speed) geodesic from $$\mu _0$$ to $$\mu _1$$, where $${\textsf {x} }^t:X^2\rightarrow X$$ is given by
\begin{aligned} {\textsf {x} }^t(x_0,x_1):=(1-t)x_0+tx_1. \end{aligned}
Conversely, any (constant speed) geodesic $$\mu$$ from $$\mu _0$$ to $$\mu _1$$ admits the representation (2.5) for a suitable plan $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$.
Finally, if $$\mu$$ is a geodesic connecting $$\mu _0$$ to $$\mu _1$$, then for every $$t \in (0,1)$$ there exists a unique optimal plan between $$\mu _0$$ and $$\mu _t$$ (resp. between $$\mu _t$$ and $$\mu _1$$) and it is concentrated on a map.
We define the counterpart of $$\mathrm {C}^{\infty }_c({\mathbb {R}}^d)$$ when we have $$X$$ in place of $${\mathbb {R}}^d$$.
### Definition 2.9
($${{\,\mathrm{Cyl}\,}}(X)$$) We denote by $$\Pi _d(X)$$ the space of linear maps $$\pi :X\rightarrow {\mathbb {R}}^d$$ of the form $$\pi (x)=(\langle x,e_1\rangle ,\ldots ,\langle x,e_d\rangle )$$ for an orthonormal set $$\{e_1,\ldots ,e_d\}$$ of $$X$$. A function $$\varphi : X\rightarrow {\mathbb {R}}$$ belongs to the space of cylindrical functions on $$X$$, $${{\,\mathrm{Cyl}\,}}(X)$$, if it is of the form
\begin{aligned} \varphi = \psi \circ \pi \end{aligned}
where $$\pi \in \Pi _d(X)$$ and $$\psi \in \mathrm C^\infty _c({\mathbb {R}}^d)$$.
We recall the following result (see [3, Theorem 8.3.1, Proposition 8.4.5 and Proposition 8.4.6]) characterizing locally absolutely continuous curves in $$\mathcal {P}_2(X)$$ defined in a (bounded or unbounded) open interval $${\mathcal {I}}\subset {\mathbb {R}}$$. We use the notation $$\mu _t$$ for the evaluation at time $$t\in {\mathcal {I}}$$ of a map $$\mu :{\mathcal {I}}\rightarrow \mathcal {P}_2(X)$$.
### Theorem 2.10
(Wasserstein velocity field) Let $$\mu :{\mathcal {I}}\rightarrow \mathcal {P}_2(X)$$ be a locally absolutely continuous curve defined in an open interval $${\mathcal {I}}\subset {\mathbb {R}}$$. There exists a Borel vector field $${\varvec{v}}:{\mathcal {I}}\times X\rightarrow X$$ and a set $$A(\mu ) \subset {\mathcal {I}}$$ with $$\mathcal {L}({\mathcal {I}}{\setminus } A(\mu ))=0$$ such that the following hold
1. (1)
$$\displaystyle {\varvec{v}}_t\in {{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2(X) :={} \overline{\{ \nabla \varphi \mid \varphi \in {{\,\mathrm{Cyl}\,}}(X) \}}^{L^2_{\mu _t}(X;X)}$$, for every $$t\in A(\mu )$$;
2. (2)
$$\int _{X} |{\varvec{v}}_t|^2\,\mathrm d\mu _t=|{{\dot{\mu }}}_t|^2:=\lim _{h\rightarrow 0}\frac{W_2^2(\mu _{t+h},\mu _t)}{h^2}$$, for every $$t\in A(\mu )$$;
3. (3)
the continuity equation
\begin{aligned} \partial _t\mu _t+\nabla \cdot ({\varvec{v}}_t\mu _t)=0 \end{aligned}
holds in the sense of distributions in $${\mathcal {I}}\times X$$.
Moreover, $${\varvec{v}}_t$$ is uniquely determined in $$L^2_{\mu _t}(X;X)$$ for $$t\in A(\mu )$$ and
\begin{aligned} \lim _{h \rightarrow 0} \frac{W_2(({\varvec{i}}_X+h{\varvec{v}}_t)_{\sharp }\mu _t, \mu _{t+h})}{|h|} =0 \quad \text {for every }t \in A(\mu ). \end{aligned}
(2.6)
We conclude this section with a useful property concerning the upper derivative of the Wasserstein distance, which in fact holds in every metric space.
### Lemma 2.11
Let $$\mu : {\mathcal {I}}\rightarrow \mathcal {P}_2(X)$$, $$\nu \in \mathcal {P}_2(X)$$, $$t \in {\mathcal {I}}$$, $$\varvec{\sigma }_t \in \Gamma _o(\mu _t, \nu )$$, and consider the constant speed geodesic $$\nu ^t:[0,1]\rightarrow \mathcal {P}_2(X)$$ defined by $$\nu _s^t: = ({\textsf {x} }^s)_{\sharp }\varvec{\sigma }_t$$ for every $$s \in [0,1]$$. The upper right and left Dini derivatives $$b^{\pm }:(0,1] \rightarrow \mathbb {R}$$ defined by
\begin{aligned} \begin{aligned} b^+(s):=&\frac{1}{2s} \limsup _{h \downarrow 0} \frac{W_2^2(\mu _{t+h},\nu _s^t) - W_2^2(\mu _t, \nu _s^t)}{h},\\ b^-(s):=&\frac{1}{2s} \limsup _{h \downarrow 0} \frac{W_2^2(\mu _{t}, \nu _s^t) - W_2^2(\mu _{t-h}, \nu _s^t)}{h} \end{aligned} \end{aligned}
are respectively decreasing and increasing in (0, 1].
### Proof
Take $$0\le s'<s\le 1$$. Since $$\nu ^t:[0,1]\rightarrow \mathcal {P}_2(X)$$ is a constant speed geodesic from $$\mu _t$$ to $$\nu$$, we have
\begin{aligned} W_2(\mu _t, \nu _s^t) = W_2(\mu _t, \nu _{s'}^t) + W_2(\nu _{s'}^t, \nu _s^t), \end{aligned}
then, by triangular inequality
\begin{aligned} W_2(\mu _{t+h}, \nu _s^t) - W_2(\mu _t, \nu _s^t)&\le W_2(\mu _{t+h}, \nu _{s'}^t) + W_2(\nu _{s'}^t, \nu _s^t) - W_2(\mu _t, \nu _s^t) \\&= W_2(\mu _{t+h}, \nu _{s'}^t) - W_2(\mu _t, \nu _{s'}^t). \end{aligned}
Dividing by $$h>0$$ and passing to the limit as $$h\downarrow 0$$ we obtain that the function $$a:[0,1] \rightarrow \mathbb {R}$$ defined by
\begin{aligned} a^+(s):= \limsup _{h \downarrow 0} \frac{W_2(\mu _{t+h}, \nu _s^t) - W_2(\mu _t, \nu _s^t)}{h} \end{aligned}
is decreasing. It is then sufficient to observe that for $$s>0$$
\begin{aligned} b^+(s) = a^+(s) \frac{W_2(\mu _t, \nu _s^t)}{s}= a^+(s) \,W_2(\mu _t, \nu ). \end{aligned}
The monotonicity property of $$b^-$$ follows by the same argument. $$\square$$
### 2.2 A strong-weak topology on measures in product spaces
Let us consider the case where $$X={\textsf {X} }\times {\textsf {Y} }$$ where $${\textsf {X} },{\textsf {Y} }$$ are separable Hilbert spaces. The space $$X$$ is naturally endowed with the product Hilbert norm and $$\mathcal {P}_2(X)$$ with the corresponding topology induced by the $$L^2$$-Wasserstein distance. However, it will be extremely useful to endow $$\mathcal {P}_2(X)$$ with a weaker topology which is related to the strong-weak topology on $$X$$, i.e. the product topology of $${\textsf {X} }^s\times {\textsf {Y} }^w$$. We follow the approach of [25], to which we refer for the proofs of the results presented in this section.
In order to define the topology, we consider the space $$\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} })$$ of test functions $$\zeta :{\textsf {X} }\times {\textsf {Y} }\rightarrow {\mathbb {R}}$$ such that
\begin{aligned}&\zeta \text { is sequentially continuous in } {\textsf {X} }^s\times {\textsf {Y} }^w, \end{aligned}
(2.7)
\begin{aligned}&\forall \,\varepsilon >0\ \exists \,A_\varepsilon \ge 0: |\zeta (x,y)|\le A_\varepsilon (1+|x|_{\textsf {X} }^2)+\varepsilon |y|_{\textsf {Y} }^2\quad \forall \,(x,y)\in {\textsf {X} }\times {\textsf {Y} }. \end{aligned}
(2.8)
Notice in particular that functions in $$\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} })$$ have quadratic growth. We endow $$\mathrm {C}^{sw}_{2}(X)$$ with the norm
\begin{aligned} \Vert \zeta \Vert _{\mathrm {C}^{sw}_{2}(X)}:=\sup _{(x,y)\in X}\frac{|\zeta (x,y)|}{1+|x|_{\textsf {X} }^2+|y|_{\textsf {Y} }^2}. \end{aligned}
### Remark 2.12
When $${\textsf {Y} }$$ is finite dimensional, (2.7) is equivalent to the continuity of $$\zeta$$.
### Lemma 2.13
$$(\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} }),\Vert \cdot \Vert _{\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} })})$$ is a Banach space.
### Definition 2.14
(Topology of $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$, [25]) We denote by $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$ the space $$\mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ endowed with the coarsest topology which makes the following functions continuous
\begin{aligned} \varvec{\mu }\mapsto \int \zeta (x,y)\,\mathrm d\varvec{\mu }(x,y),\quad \zeta \in \mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} }). \end{aligned}
It is obvious that the topology of $$\mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ is finer than the topology of $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$ and the latter is finer than the topology of $$\mathcal {P}({\textsf {X} }^{s}\times {\textsf {Y} }^w)$$. It is worth noticing that any bounded bilinear form $$B:{\textsf {X} }\times {\textsf {Y} }\rightarrow {\mathbb {R}}$$ belongs to $$\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} })$$, so that for every net $$(\varvec{\mu }_\alpha )_{\alpha \in {\mathbb {A}}} \subset \mathcal {P}({\textsf {X} }\times {\textsf {Y} })$$ indexed by a directed set $${\mathbb {A}}$$, we have
\begin{aligned} \lim _{\alpha \in {\mathbb {A}}}\varvec{\mu }_\alpha =\varvec{\mu }\quad \text {in }\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })\quad \Rightarrow \quad \lim _{\alpha \in {\mathbb {A}}}\int B \,\mathrm d\varvec{\mu }_\alpha = \int B \,\mathrm d\varvec{\mu }. \end{aligned}
(2.9)
The following proposition justifies the interest in the $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$-topology.
### Proposition 2.15
1. (1)
Assume that $$(\varvec{\mu }_\alpha )_{\alpha \in {\mathbb {A}}}\subset \mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ is a net indexed by the directed set $${\mathbb {A}}$$, $$\varvec{\mu }\in \mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ and they satisfy
1. (a)
$$\varvec{\mu }_\alpha \rightarrow \varvec{\mu }$$ in $$\mathcal {P}({\textsf {X} }^s\times {\textsf {Y} }^w)$$,
2. (b)
$$\displaystyle \lim _{\alpha \in {\mathbb {A}}}\int |x|_{\textsf {X} }^2\,\mathrm d\varvec{\mu }_\alpha (x,y)=\int |x|_{\textsf {X} }^2\,\mathrm d\varvec{\mu }(x,y)$$,
3. (c)
$$\displaystyle \sup _{\alpha \in {\mathbb {A}}} \int |y|_{\textsf {Y} }^2\,\mathrm d\varvec{\mu }_\alpha (x,y)<\infty$$,
then $$\varvec{\mu }_\alpha \rightarrow \varvec{\mu }$$ in $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$. The converse property holds for sequences: if $${\mathbb {A}}={\mathbb {N}}$$ and $$\varvec{\mu }_n\rightarrow \varvec{\mu }$$ in $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$ as $$n\rightarrow \infty$$ then properties (a), (b), (c) hold.
2. (2)
For every compact set $${\mathcal {K}}\subset \mathcal {P}_2({\textsf {X} }^s)$$ and every constant $$c<\infty$$ the sets
\begin{aligned} {\mathcal {K}}_c:=\left\{ \varvec{\mu }\in \mathcal {P}_2({\textsf {X} }\times {\textsf {Y} }):\pi ^{{\textsf {X} }}_\sharp \varvec{\mu }\in {\mathcal {K}},\quad \int |y|_{\textsf {Y} }^2\,\mathrm d\varvec{\mu }(x,y)\le c\right\} \end{aligned}
are compact and metrizable in $$\mathcal {P}_2^{sw}({\textsf {X} }\times {\textsf {Y} })$$ (in particular they are sequentially compact).
It is worth noticing that the topology $$\mathcal {P}_2^{ws}({\textsf {X} }\times {\textsf {Y} })$$ is strictly weaker than $$\mathcal {P}_2({\textsf {X} }\times {\textsf {Y} })$$ even when $${\textsf {Y} }$$ is finite dimensional. In fact, $$\mathrm {C}^{sw}_{2}({\textsf {X} }\times {\textsf {Y} })$$ does not contain the quadratic function $$(x,y)\mapsto |y|_{\textsf {Y} }^2$$, so that convergence of the quadratic moment w.r.t. y is not guaranteed.
## 3 Directional derivatives and probability measures on the tangent bundle
From now on, we will denote by $${\textsf {X} }$$ a separable Hilbert space with norm $$|\cdot |$$ and scalar product $$\langle \cdot ,\cdot \rangle$$. We denote by $$\mathsf {TX}$$ the tangent bundle to $${\textsf {X} }$$, which is identified with the set $${\textsf {X} }\times {\textsf {X} }$$ with the induced norm $$|(x,v)|:=\big (|x|^2+|v|^2\big )^{1/2}$$ and the strong-weak topology of $${\textsf {X} }^s \times {\textsf {X} }^w$$(i.e. the product of the strong topology on the first component and the weak topology on the second one). We will denote by $${\textsf {x} },{\textsf {v} }:\mathsf {TX}\rightarrow {\textsf {X} }$$ the projection maps and by $$\textsf {exp} ^t:\mathsf {TX}\rightarrow {\textsf {X} }$$ the exponential map defined by
\begin{aligned} {\textsf {x} }(x,v):=x,\quad {\textsf {v} }(x,v):=v,\quad \textsf {exp} ^t(x,v):=x+tv. \end{aligned}
(3.1)
The set $$\mathcal {P}(\mathsf {TX})$$ is defined thanks to the identification of $$\mathsf {TX}$$ with $${\textsf {X} }\times {\textsf {X} }$$ and is endowed with the narrow topology induced by the strong-weak topology in $$\mathsf {TX}$$. For $$\Phi \in \mathcal {P}(\mathsf {TX})$$ we define
\begin{aligned} |\Phi |_2^2:= \int _{\mathsf {TX}} |v|^2 \,\mathrm d\Phi (x,v). \end{aligned}
(3.2)
We denote by $$\mathcal {P}_2(\mathsf {TX})$$ the subset of $$\mathcal {P}(\mathsf {TX})$$ of measures for which $$\int \big (|x|^2+|v|^2\big )\,\mathrm d\Phi <\infty$$ endowed with the topology of $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ as in Sect. 2.2. If $$\mu \in \mathcal {P}({\textsf {X} })$$ we will also consider
\begin{aligned} \mathcal {P}_{}(\mathsf {TX}|\mu ):= \left\{ \Phi \in \mathcal {P}(\mathsf {TX}) \mid {\textsf {x} }_{\sharp }\Phi = \mu \right\} ,\quad \mathcal {P}_{2}(\mathsf {TX}|\mu ):= \left\{ \Phi \in \mathcal {P}_{}(\mathsf {TX}|\mu ): |\Phi |_2<\infty \right\} .\nonumber \\ \end{aligned}
(3.3)
When we deal with the product space $${\textsf {X} }^2$$, we will use the notation
\begin{aligned} {\textsf {x} }^t:{\textsf {X} }^2\rightarrow {\textsf {X} },\quad {\textsf {x} }^t(x_0,x_1):=(1-t)x_0+tx_1,\quad t\in [0,1]. \end{aligned}
(3.4)
If $${\varvec{v}}\in L^2_\mu ({\textsf {X} };{\textsf {X} })$$ we can consider the probability measure
\begin{aligned} \Phi =({\varvec{i}}_{\textsf {X} },{\varvec{v}})_\sharp \mu \in \mathcal {P}_{2}(\mathsf {TX}|\mu ). \end{aligned}
(3.5)
In this case we will say that $$\Phi$$ is concentrated on the graph of the map $${\varvec{v}}$$. More generally, given a Borel family of probability measures $$(\Phi _x)_{x\in {\textsf {X} }}\subset \mathcal {P}_2({\textsf {X} })$$ satisfying
\begin{aligned} \int \Big (\int |v|^2\,\mathrm d\Phi _x(v)\Big )\,\mathrm d\mu (x)<\infty \end{aligned}
(3.6)
we can consider the probability measure
\begin{aligned} \Phi = \int _{{\textsf {X} }}\Phi _x\,\mathrm d\mu (x) \in \mathcal {P}_{2}(\mathsf {TX}|\mu ). \end{aligned}
(3.7)
Conversely, every $$\Phi \in \mathcal {P}_{2}(\mathsf {TX}|\mu )$$ can be disintegrated into a Borel family $$(\Phi _x)_{x\in {\textsf {X} }}\subset \mathcal {P}_2({\textsf {X} })$$ satisfying (3.6) and (3.7). The measure $$\Phi$$ can be associated with a vector field $${\varvec{v}}\in L^2_\mu ({\textsf {X} };{\textsf {X} })$$ if and only if for $$\mu$$-a.e. $$x\in {\textsf {X} }$$ we have $$\Phi _x=\delta _{{\varvec{v}}(x)}$$. Recalling the disintegration Theorem 2.3 and Remark 2.4, we give the following definition.
### Definition 3.1
Given $$\Phi \in \mathcal {P}_{2}(\mathsf {TX}|\mu )$$, the barycenter of $$\Phi$$ is the function $${\varvec{b}}_{\Phi }\in L^2_\mu ({\textsf {X} };{\textsf {X} })$$ defined by
\begin{aligned} {\varvec{b}}_{\Phi }(x):=\int _{\textsf {X} }v\,\mathrm d\Phi _x(v) \quad \text {for }\mu \text {-a.e. }x\in {\textsf {X} }, \end{aligned}
where $$\{\Phi _x\}_{x\in {\textsf {X} }}\subset \mathcal {P}_2({\textsf {X} })$$ is the disintegration of $$\Phi$$ w.r.t. $$\mu$$.
### Remark 3.2
Notice that, by the linearity of the scalar product, we get the following identity which will be useful later
\begin{aligned} \int _{{\textsf {X} }} \langle \varvec{\zeta }(x), {\varvec{b}}_{\Phi }(x)\rangle \,\mathrm d\mu (x) = \int _{\mathsf {TX}} \langle \varvec{\zeta }(x), v\rangle \,\mathrm d\Phi (x,v) \end{aligned}
(3.8)
for all $$\varvec{\zeta }\in L^2_\mu ({\textsf {X} };{\textsf {X} })$$.
### 3.1 Directional derivatives of the Wasserstein distance and duality pairings
Our starting point is a relevant semi-concavity property of the function
\begin{aligned} f(s,t):= \frac{1}{2} W_2^2(\textsf {exp} ^s_{\sharp } \Phi _0, \textsf {exp} ^t_\sharp \Phi _1),\quad s,t\in {\mathbb {R}}, \end{aligned}
(3.9)
with $$\Phi _0,\Phi _1\in \mathcal {P}_2(\mathsf {TX})$$. We first state an auxiliary result, whose proof is based on [3, Proposition 7.3.1].
### Lemma 3.3
Let $$\Phi _0, \Phi _1 \in \mathcal {P}_2(\mathsf {TX})$$, $$s,t \in {\mathbb {R}}$$, and let $$\varvec{\vartheta }^{s,t} \in \Gamma (\textsf {exp} ^s_{\sharp }\Phi _0, \textsf {exp} ^t_{\sharp } \Phi _1)$$. Then there exists $$\varvec{\Theta }^{s,t} \in \Gamma (\Phi _0, \Phi _1)$$ such that $$(\textsf {exp} ^s, \textsf {exp} ^t)_{\sharp }\varvec{\Theta }^{s,t}= \varvec{\vartheta }^{s,t}$$.
### Proof
Define, for every $$r,s,t \in {\mathbb {R}}$$,
\begin{aligned}&\Sigma ^r : \mathsf {TX}\rightarrow \mathsf {TX},\quad \Sigma ^r(x, v) := (\textsf {exp} ^r(x, v), v); \\&\Lambda ^{s,t}: \mathsf {TX}\times \mathsf {TX}\rightarrow \mathsf {TX}\times \mathsf {TX},\quad \Lambda ^{s,t}:=(\Sigma ^s,\Sigma ^t). \end{aligned}
Consider the probabilities $$(\Sigma ^s)_{\sharp }\Phi _0, (\Sigma ^t)_{\sharp }\Phi _1$$ and $$\varvec{\vartheta }^{s,t}$$. They are constructed in such a way that there exists $$\varvec{\Psi }^{s,t} \in \mathcal {P}(\mathsf {TX}\times \mathsf {TX})$$ s.t.
\begin{aligned} ({\textsf {x} }^0,{\textsf {v} }^0)_{\sharp } \varvec{\Psi }^{s,t} = (\Sigma ^s)_{\sharp }\Phi _0, \quad ({\textsf {x} }^1,{\textsf {v} }^1)_{\sharp } \varvec{\Psi }^{s,t} = (\Sigma ^t)_{\sharp }\Phi _1, \quad ({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp } \varvec{\Psi }^{s,t}= \varvec{\vartheta }^{s,t}, \end{aligned}
where we adopted the notation $${\textsf {x} }^i(x_0,v_0,x_1,v_1):=x_i$$ and $${\textsf {v} }^i(x_0,v_0,x_1,v_1):=v_i$$, $$i=0,1$$. We conclude by taking $$\varvec{\Theta }^{s,t}:= (\Lambda ^{-s,-t})_{\sharp } \varvec{\Psi }^{s,t}$$. $$\square$$
### Proposition 3.4
Let $$\Phi _0,\Phi _1 \in \mathcal {P}_2(\mathsf {TX})$$ with $$\mu _1={\textsf {x} }_\sharp \Phi _1$$ and $$\varphi ^2:=|\Phi _0|_2^2+|\Phi _1|_2^2$$, let $$f:{\mathbb {R}}^2\rightarrow {\mathbb {R}}$$ be the function defined by (3.9) and let $$h,g:{\mathbb {R}}\rightarrow {\mathbb {R}}$$ be defined by
\begin{aligned} h(s)&:=f(s,s)= \frac{1}{2} W_2^2(\textsf {exp} ^s_{\sharp }\Phi _0,\textsf {exp} ^s_\sharp \Phi _1),\nonumber \\ g(s)&:=f(s,0)= \frac{1}{2} W_2^2(\textsf {exp} ^s_{\sharp } \Phi _0, \mu _1),\quad s\in {\mathbb {R}}. \end{aligned}
(3.10)
1. (1)
The function $$(s,t)\mapsto f(s,t)-\frac{1}{2}\varphi ^2( s^2+t^2)$$ is concave, i.e. it holds
\begin{aligned} \begin{aligned}&f((1-\alpha )s_0 + \alpha s_1, (1-\alpha )t_0 + \alpha t_1) \ge (1-\alpha )f(s_0,t_0) + \alpha f(s_1,t_1) \\&\quad - \frac{1}{2}\alpha (1-\alpha )\Big [(s_1-s_0)^2 +(t_1-t_0)^2 \Big ]\varphi ^2 \end{aligned} \end{aligned}
(3.11)
for every $$s_0, s_1 ,t_0,t_1\in {\mathbb {R}}$$ and every $$\alpha \in [0,1]$$.
2. (2)
The function $$s\mapsto h(s)-\varphi ^2 s^2$$ is concave.
3. (3)
the function $$s\mapsto g(s)-\frac{1}{2}s^2|\Phi _0|_2^2$$ is concave.
### Proof
Let us first prove (3.11). We set $$s:= (1-\alpha )s_0 + \alpha s_1$$, $$t:=(1-\alpha )t_0+\alpha t_1$$ and we apply Lemma 3.3 to find $$\varvec{\Theta }\in \Gamma (\Phi _0,\Phi _1)$$ such that $$(\textsf {exp} ^s,\textsf {exp} ^t)_\sharp \varvec{\Theta }\in \Gamma _o(\textsf {exp} ^s_{\sharp }\Phi _0, \textsf {exp} ^t_\sharp \Phi _1)$$. Then, recalling the Hilbertian identity
\begin{aligned} |(1-\alpha )a + \alpha b|^2 = (1-\alpha ) |a|^2+\alpha |b|^2-\alpha (1-\alpha ) |a-b|^2, \quad a,b\in {\textsf {X} }, \end{aligned}
we have
\begin{aligned}&W_2^2( \textsf {exp} ^{s}_{\sharp } \Phi _0, \textsf {exp} ^t_\sharp \Phi _1) \\&\quad = \int |x_0+sv_0-(x_1+tv_1)|^2 \,\mathrm d\varvec{\Theta }\\ {}&\quad = \int |(1-\alpha )(x_0+s_0v_0)+\alpha (x_0+s_1v_0)- (1-\alpha )(x_1+t_0v_1)-\alpha (x_1+t_1v_1)|^2 \,\mathrm d\varvec{\Theta }\\ {}&\quad =(1-\alpha )\int |x_0+s_0v_0-(x_1+t_0v_1)|^2\,\mathrm d\varvec{\Theta }+\alpha \int |x_0+s_1v_0-(x_1+t_1v_1)|^2\,\mathrm d\varvec{\Theta }\\ {}&\qquad -\alpha (1-\alpha ) \int |(s_1-s_0)v_0+(t_1-t_0)v_1|^2 \,\mathrm d\varvec{\Theta }\\&\quad \ge (1-\alpha ) W_2^2(\textsf {exp} _{\sharp }^{s_0} \Phi _0,\textsf {exp} ^{t_0}_\sharp \Phi _1) + \alpha W_2^2(\textsf {exp} _{\sharp }^{s_1} \Phi _0, \textsf {exp} ^{t_1}_\sharp \Phi _1) \\ {}&\qquad - \alpha (1-\alpha )\Big ((s_1-s_0)^2+(t_1-t_0)^2\Big ) \Big (\int |v_0|^2 \,\mathrm d\Phi _0+ \int |v_1|^2\,\mathrm d\Phi _1\Big ). \end{aligned}
which is the thesis. Claims (2) and (3) follow as particular cases when $$t=s$$ or $$t=0$$. $$\square$$
Semi-concavity is a useful tool to guarantee the existence of one-sided partial derivatives at (0, 0): for every $$\alpha ,\beta \in {\mathbb {R}}$$ we have (see e.g. [19, Ch. VI, Prop. 1.1.2]) that
\begin{aligned} f_r'(\alpha ,\beta )&=\lim _{\varrho \downarrow 0}\frac{f(\alpha \varrho ,\beta \varrho )-f(0,0)}{\varrho }= \sup _{\varrho>0}\frac{f(\alpha \varrho ,\beta \varrho )-f(0,0)}{\varrho }-\frac{\varrho \varphi ^2}{2}(\alpha ^2+\beta ^2),\\ f_l'(\alpha ,\beta )&=\lim _{\varrho \downarrow 0}\frac{f(0,0)-f(-\alpha \varrho ,-\beta \varrho )}{\varrho }\\&= \inf _{\varrho >0}\frac{f(0,0)-f(-\alpha \varrho ,-\beta \varrho )}{\varrho }+\frac{\varrho \varphi ^2}{2}(\alpha ^2+\beta ^2). \end{aligned}
$$f_r'$$ (resp. $$f'_l$$) is a concave (resp. convex) and positively 1-homogeneous function, i.e. a superlinear (resp. sublinear) function. They satisfy
\begin{aligned}&f_r'(-\alpha ,-\beta )=-f'_l(\alpha ,\beta )\quad \text {for every }\alpha ,\beta \in {\mathbb {R}}, \end{aligned}
(3.12)
\begin{aligned}&f_l'(\alpha ,\beta )\ge f_r'(\alpha ,\beta )\quad \text {for every }\alpha ,\beta \in {\mathbb {R}}, \end{aligned}
(3.13)
\begin{aligned}&f_r'(\alpha ,\beta )\ge \alpha f_r'(1,0)+\beta f_r'(0,1)\quad \text {for every }\alpha ,\beta \ge 0,\nonumber \\&f(s,t)\le f(0,0)+f'_r(s,t)-\frac{\varphi ^2}{2}(s^2+t^2)\quad \text {for every }s,t\in {\mathbb {R}}. \end{aligned}
(3.14)
Notice moreover that
\begin{aligned} f_r'(1,0)=g'_r(0)=\lim _{\varrho \downarrow 0}\frac{g(\varrho )-g(0)}{\varrho }\end{aligned}
where g is the function defined in (3.10); a similar representation holds for $$f_l'(1,0)$$. We introduce the following notation for $$f'_r$$, $$f'_l$$, $$g'_r$$ and $$g'_l$$.
### Definition 3.5
Let $$\mu _0,\mu _1 \in \mathcal {P}_2({\textsf {X} })$$, $$\Phi _0 \in \mathcal {P}_{2}(\mathsf {TX}|\mu _0)$$ and $$\Phi _1\in \mathcal {P}_{2}(\mathsf {TX}|\mu _1)$$. We define
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r}&:= \lim _{s \downarrow 0} \frac{W_2^2(\textsf {exp} ^s_{\sharp }\Phi _0, \mu _1) - W_2^2(\mu _0, \mu _1)}{2s}, \\ \left[ \Phi _0, \mu _1\right] _{l}&:= \lim _{s \downarrow 0} \frac{W_2^2(\mu _0, \mu _1)- W_2^2(\textsf {exp} ^{-s}_{\sharp }\Phi _0, \mu _1)}{2s}, \end{aligned}
and analogously
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}&:= \lim _{t \downarrow 0} \frac{W_2^2(\textsf {exp} ^t_{\sharp }\Phi _0, \textsf {exp} ^t_{\sharp }\Phi _1) - W_2^2(\mu _0, \mu _1)}{2t},\\ \left[ \Phi _0, \Phi _1\right] _{l}&:=\lim _{t \downarrow 0} \frac{W_2^2(\mu _0, \mu _1)-W_2^2(\textsf {exp} ^{-t}_{\sharp }\Phi _0, \textsf {exp} ^{-t}_{\sharp }\Phi _1)}{2t}. \end{aligned}
Recalling the definitions of f and g given by (3.9) and (3.10), with $$\Phi _0$$ and $$\Phi _1$$ as above, we notice that
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r}&= g'_r(0)=f_r'(1,0),\\ \left[ \Phi _0, \mu _1\right] _{l}&=g'_l(0)=f_l'(1,0),\\ \left[ \Phi _0, \Phi _1\right] _{r}&=f'_r(1,1),\\ \left[ \Phi _0, \Phi _1\right] _{l}&= f_l'(1,1). \end{aligned}
### Remark 3.6
Notice that $$\left[ \Phi _0, \mu _1\right] _{r} = \left[ \Phi _0, \Phi _{\mu _1}\right] _{r}$$ and $$\left[ \Phi _0, \mu _1\right] _{l} = \left[ \Phi _0, \Phi _{\mu _1}\right] _{l}$$, where
\begin{aligned} \Phi _{\mu _1} = ({\varvec{i}}_{\textsf {X} },0)_{\sharp }\mu _1 \in \mathcal {P}_2(\mathsf {TX}). \end{aligned}
Moreover, given $$\Phi \in \mathcal {P}(\mathsf {TX})$$ and using the notation
\begin{aligned} -\Phi :=J_\sharp \Phi , \quad \text {with } J(x,v):=(x,-v), \end{aligned}
(3.15)
we have
\begin{aligned} \left[ -\Phi _0, -\Phi _1\right] _{r}=-\left[ \Phi _0, \Phi _1\right] _{l},\quad \text { and } \quad \left[ -\Phi _0, \mu _1\right] _{r}=-\left[ \Phi _0, \mu _1\right] _{l}. \end{aligned}
In particular, the properties of $$\left[ \cdot , \cdot \right] _{l}$$ (in $$\mathcal {P}_2(\mathsf {TX}) \times \mathcal {P}_2(\mathsf {TX})$$ or $$\mathcal {P}_2(\mathsf {TX}) \times \mathcal {P}_2({\textsf {X} })$$) and the ones of $$\left[ \cdot , \cdot \right] _{r}$$ in $$\mathcal {P}_2(\mathsf {TX}) \times \mathcal {P}_2({\textsf {X} })$$ can be easily derived by the corresponding ones of $$\left[ \cdot , \cdot \right] _{r}$$ in $$\mathcal {P}_2(\mathsf {TX}) \times \mathcal {P}_2(\mathsf {TX})$$.
Recalling (3.14) and (3.12) we obtain the following result.
### Corollary 3.7
For every $$\mu _0, \mu _1 \in \mathcal {P}_2({\textsf {X} })$$ and for every $$\Phi _0\in \mathcal {P}_{2}(\mathsf {TX}|\mu _0)$$, $$\Phi _1\in \mathcal {P}_{2}(\mathsf {TX}|\mu _1)$$, it holds
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r} + \left[ \Phi _1, \mu _0\right] _{r} \le \left[ \Phi _0, \Phi _1\right] _{r} \quad \text { and }\quad \left[ \Phi _0, \mu _1\right] _{l} + \left[ \Phi _1, \mu _0\right] _{l} \ge \left[ \Phi _0, \Phi _1\right] _{l}. \end{aligned}
Let us now show an important equivalent characterization of the quantities we have just introduced. As usual we will denote by $${\textsf {x} }^0,{\textsf {v} }^0,{\textsf {x} }^1:\mathsf {TX}\times {\textsf {X} }\rightarrow {\textsf {X} }$$ the projection maps of a point $$(x_0,v_0,x_1)$$ in $$\mathsf {TX}\times {\textsf {X} }$$ (and similarly for $$\mathsf {TX}\times \mathsf {TX}$$ with $${\textsf {x} }^0,{\textsf {v} }^0,{\textsf {x} }^1, {\textsf {v} }^1$$).
First of all we introduce the following sets.
### Definition 3.8
For every $$\Phi _0 \in \mathcal {P}(\mathsf {TX})$$ with $$\mu _0={\textsf {x} }_\sharp \Phi _0$$ and $$\mu _1 \in \mathcal {P}_2({\textsf {X} })$$ we set
\begin{aligned} \Lambda (\Phi _0, \mu _1):= \left\{ \varvec{\sigma }\in \Gamma (\Phi _0, \mu _1) \mid ({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp }\varvec{\sigma }\in \Gamma _o(\mu _0, \mu _1) \right\} . \end{aligned}
Analogously, for every $$\Phi _0, \Phi _1 \in \mathcal {P}(\mathsf {TX})$$ with $$\mu _0={\textsf {x} }_\sharp \Phi _0$$ and $$\mu _1={\textsf {x} }_\sharp \Phi _1$$ in $$\mathcal {P}_2({\textsf {X} })$$ we set
\begin{aligned} \Lambda (\Phi _0, \Phi _1):= \left\{ \varvec{\Theta }\in \Gamma (\Phi _0, \Phi _1) \mid ({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp } \varvec{\Theta }\in \Gamma _o(\mu _0, \mu _1) \right\} . \end{aligned}
In the following proposition and subsequent corollary, we provide a useful characterization of the pairings $$\left[ \cdot , \cdot \right] _{r}$$ and $$\left[ \cdot , \cdot \right] _{l}$$. Similar results with analogous proofs can be found also in [18, Theorem 4.2] and [14, Corollary 3.18] where $${\textsf {X} }$$ is a smooth compact Riemannian manifold.
### Theorem 3.9
For every $$\Phi _0, \Phi _1 \in \mathcal {P}_2(\mathsf {TX})$$ and $$\mu _1 \in \mathcal {P}_2({\textsf {X} })$$ we have
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r}&= \min \left\{ \int _{\mathsf {TX}\times {\textsf {X} }} \langle x_0-x_1, v_0\rangle \,\mathrm d\varvec{\sigma }\mid \varvec{\sigma }\in \Lambda (\Phi _0, \mu _1) \right\} , \end{aligned}
(3.16)
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}&= \min \left\{ \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }\mid \varvec{\Theta }\in \Lambda (\Phi _0, \Phi _1) \right\} . \end{aligned}
(3.17)
We denote by $$\Lambda _o(\Phi _0, \mu _1)$$ (resp. $$\Lambda _o(\Phi _0, \Phi _1)$$) the subset of $$\Lambda (\Phi _0,\mu _1)$$ (resp. $$\Lambda (\Phi _0,\Phi _1)$$) where the minimum in (3.16) (resp. (3.17)) is attained.
### Proof
First, we recall that the minima in the right hand side are attained since $$\Lambda (\Phi _0,\mu _1)$$ and $$\Lambda (\Phi _0,\Phi _1)$$ are compact subsets of $$\mathcal {P}_2(\mathsf {TX}\times {\textsf {X} })$$ and $$\mathcal {P}_2(\mathsf {TX}\times \mathsf {TX})$$ respectively by Lemma 2.6 and the integrands are continuous functions with quadratic growth. Thanks to Remark 3.6, we only need to prove the equality (3.17). For every $$\varvec{\Theta }\in \Lambda (\Phi _0, \Phi _1)$$ and setting $$\mu _0={\textsf {x} }_\sharp \Phi _0$$, $$\mu _1={\textsf {x} }_\sharp \Phi _1$$, we have
\begin{aligned}&W_2^2(\textsf {exp} ^s_{\sharp }(\Phi _0), \textsf {exp} ^s_{\sharp }(\Phi _1))\\&\quad \le \int _{\mathsf {TX}\times \mathsf {TX}} | (x_0-x_1) +s(v_0-v_1)|^2 \,\mathrm d\varvec{\Theta }\\&\quad = \int _{{\textsf {X} }^2} |x_0-x_1|^2 \,\mathrm d({\textsf {x} }^0, {\textsf {x} }^1)_{\sharp }\varvec{\Theta }\\&\qquad + 2s \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }+ s^2 \int _{{\textsf {X} }^2} |v_0-v_1|^2 \,\mathrm d\varvec{\Theta }\\&\quad = W_2^2(\mu _0,\mu _1) + 2s \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }+ s^2 \int _{{\textsf {X} }^2} |v_0-v_1|^2 \,\mathrm d\varvec{\Theta }. \end{aligned}
and this immediately implies
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}\le \min \left\{ \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }\mid \varvec{\Theta }\in \Lambda (\Phi _0, \Phi _1) \right\} . \end{aligned}
In order to prove the converse inequality, thanks to Lemma 3.3, for every $$s>0$$ we can find $$\varvec{\Theta }_s \in \Gamma (\Phi _0, \Phi _1)$$ s.t.
\begin{aligned} (\textsf {exp} ^s, \textsf {exp} ^s)_{\sharp }\varvec{\Theta }_s \in \Gamma _o(\textsf {exp} ^s_{\sharp }\Phi _0, \textsf {exp} ^s_{\sharp }\Phi _1). \end{aligned}
Then
\begin{aligned} \frac{W_2^2(\textsf {exp} ^s_{\sharp }\Phi _0, \textsf {exp} ^s_{\sharp }\Phi _1) - W_2^2(\mu _0,\mu _1)}{2s}&\ge \frac{1}{2s}\int _{\mathsf {TX}\times \mathsf {TX}} |(x_0-x_1)+s(v_0-v_1)|^2 \,\mathrm d\varvec{\Theta }_s \nonumber \\&\quad -\frac{1}{2s} \int _{\mathsf {TX}\times \mathsf {TX}} |x_0-x_1|^2 \,\mathrm d\varvec{\Theta }_s\nonumber \\&\ge \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }_s. \end{aligned}
(3.18)
Since $$\Gamma (\Phi _0, \Phi _1)$$ is compact in $$\mathcal {P}_2(\mathsf {TX}\times \mathsf {TX})$$, there exists a vanishing sequence $$k\mapsto s_k$$ and $$\varvec{\Theta }\in \Gamma (\Phi _0, \Phi _1)$$ s.t. $$\varvec{\Theta }_{s_k} \rightarrow \varvec{\Theta }$$ in $$\mathcal {P}_2(\mathsf {TX}\times \mathsf {TX})$$. Moreover it holds $$(\textsf {exp} ^{s_k}, \textsf {exp} ^{s_k})_{\sharp }\varvec{\Theta }_{s_k} \rightarrow ({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp }\varvec{\Theta }$$ in $$\mathcal {P}(\mathsf {TX}\times \mathsf {TX})$$ so that $$({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp }\varvec{\Theta }\in \Gamma _o(\mu _0, \mu _1)$$, and therefore $$\varvec{\Theta }\in \Lambda (\Phi _0, \Phi _1)$$. The convergence in $$\mathcal {P}_2(\mathsf {TX}\times \mathsf {TX})$$ yields
\begin{aligned} \lim _k \int _{\mathsf {TX}\times \mathsf {TX}}\langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }_{s_k} = \int _{\mathsf {TX}\times \mathsf {TX}}\langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }, \end{aligned}
so that, passing to the limit in (3.18) along the sequence $$s_k$$, we obtain
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}\ge \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }\end{aligned}
for some $$\varvec{\Theta }\in \Lambda (\Phi _0,\Phi _1)$$. $$\square$$
### Corollary 3.10
Let $$\Phi _0, \Phi _1 \in \mathcal {P}_2(\mathsf {TX})$$ and $$\mu _1\in \mathcal {P}_2({\textsf {X} })$$, then
\begin{aligned} \left[ \Phi , \mu _1\right] _{l}&= \max \left\{ \int _{\mathsf {TX}\times {\textsf {X} }} \langle x_0-x_1, v_0\rangle \,\mathrm d\varvec{\sigma }\mid \varvec{\sigma }\in \Lambda (\Phi _0, \mu _1) \right\} ,\nonumber \\ \left[ \Phi _0, \Phi _1\right] _{l}&= \max \left\{ \int _{\mathsf {TX}\times \mathsf {TX}} \langle x_0-x_1, v_0-v_1\rangle \,\mathrm d\varvec{\Theta }\mid \varvec{\Theta }\in \Lambda (\Phi _0, \Phi _1) \right\} .\qquad \end{aligned}
(3.19)
### 3.2 Right and left derivatives of the Wasserstein distance along a.c. curves
Let us now discuss the differentiability of the map $${\mathcal {I}}\ni t \mapsto \frac{1}{2}W_2^2(\mu _t, \nu )$$ along a locally absolutely continuous curve $$\mu : {\mathcal {I}}\rightarrow \mathcal {P}_2({\textsf {X} })$$, with $${\mathcal {I}}$$ an open interval of $${\mathbb {R}}$$ and $$\nu \in \mathcal {P}_2({\textsf {X} })$$.
### Theorem 3.11
Let $$\mu :{\mathcal {I}}\rightarrow \mathcal {P}_2({\textsf {X} })$$ be a locally absolutely continuous curve and let $${\varvec{v}}:{\mathcal {I}}\times {\textsf {X} }\rightarrow {\textsf {X} }$$ and $$A(\mu )$$ be as in Theorem 2.10. Then, for every $$\nu \in \mathcal {P}_2({\textsf {X} })$$ and every $$t \in A(\mu )$$, it holds
\begin{aligned} \lim _{h \downarrow 0} \frac{W_2^2(\mu _{t+h}, \nu )-W_2^2(\mu _t, \nu )}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r},\nonumber \\ \lim _{h \uparrow 0} \frac{W_2^2(\mu _{t+h}, \nu )-W_2^2(\mu _t, \nu )}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{l}, \end{aligned}
(3.20)
so that the map $$s \mapsto W_2^2(\mu _s, \nu )$$ is left and right differentiable at every $$t \in A(\mu )$$. In particular,
1. (1)
if $$t \in A(\mu )$$ and $$\nu \in \mathcal {P}_2({\textsf {X} })$$ are s.t. there exists a unique optimal transport plan between $$\mu _t$$ and $$\nu$$, then the map $$s \mapsto W_2^2(\mu _s, \nu )$$ is differentiable at t;
2. (2)
there exists a subset $$A({\mu ,\nu })\subset A(\mu )$$ of full Lebesgue measure such that $$s\mapsto W_2^2(\mu _s,\nu )$$ is differentiable in $$A({\mu ,\nu })$$ and
\begin{aligned} \begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}W_2^2(\mu _t,\nu )&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r}= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{l}\\&=\int \langle {\varvec{v}}_t(x_1),x_1-x_2\rangle \,\mathrm d\varvec{\mu }(x_1,x_2) \end{aligned} \end{aligned}
for every $$\varvec{\mu }\in \Gamma _o(\mu _t,\nu )$$, $$t\in A({\mu ,\nu })$$.
### Proof
Let $$\nu \in \mathcal {P}_2({\textsf {X} })$$ and for every $$t\in {\mathcal {I}}$$ we set $$\Phi _t := ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t\in \mathcal {P}_2(\mathsf {TX})$$. By Theorem 3.9, we have
\begin{aligned} \lim _{h \downarrow 0} \frac{W_2^2(\textsf {exp} _{\sharp }^h\Phi _t, \nu )-W_2^2(\mu _t, \nu )}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r},\\ \lim _{h \uparrow 0} \frac{W_2^2(\textsf {exp} _{\sharp }^h\Phi _t, \nu )-W_2^2(\mu _t, \nu )}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{l}. \end{aligned}
Since $$\textsf {exp} _{\sharp }^h \Phi _t = ({\varvec{i}}_{\textsf {X} }+h {\varvec{v}}_t)_{\sharp }\mu _t$$, then thanks to Theorem 2.10 we have that the above limits coincide respectively with the limits in the statement, for all $$t \in A(\mu )$$.
Claim (1) comes by the characterizations given in Theorem 3.9 and Corollary 3.10. Indeed, if there exists a unique optimal transport plan between $$\mu _t$$ and $$\nu$$, then $$\left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r}=\left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{l}$$.
Claim (2) is a simple consequence of the fact that $$s\mapsto W_2^2(\mu _s,\nu )$$ is differentiable a.e. in $${\mathcal {I}}$$. $$\square$$
### Remark 3.12
In Theorem 3.11 we can actually replace $${\varvec{v}}$$ with any Borel velocity field $${\varvec{w}}$$ solving the continuity equation for $$\mu$$ and s.t. $$\Vert {\varvec{w}}_t\Vert _{L^2_{\mu _t}} \in L^1_{loc}({\mathcal {I}})$$. Indeed, we notice that by [3, Lemma 5.3.2],
\begin{aligned} \Lambda (({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t,\nu )&=\{({\textsf {x} }^0 , {\varvec{v}}_t\circ {\textsf {x} }^0,{\textsf {x} }^1 )_{\sharp }\varvec{\gamma }\mid \varvec{\gamma }\in \Gamma _o(\mu _t,\nu )\},\\\Lambda (({\varvec{i}}_{\textsf {X} }, {\varvec{w}}_t)_{\sharp }\mu _t,\nu )&=\{({\textsf {x} }^0 , {\varvec{w}}_t\circ {\textsf {x} }^0,{\textsf {x} }^1 )_{\sharp }\varvec{\gamma }\mid \varvec{\gamma }\in \Gamma _o(\mu _t,\nu )\}, \end{aligned}
so that, by [3, Proposition 8.5.4], we get
\begin{aligned} \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r}&=\left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{w}}_t)_{\sharp }\mu _t, \nu \right] _{r},\\ \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{l}&=\left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{w}}_t)_{\sharp }\mu _t, \nu \right] _{l}. \end{aligned}
### Remark 3.13
In general, if $$\mu :{\mathcal {I}}\rightarrow \mathcal {P}_2({\textsf {X} })$$ is a locally absolutely continuous curve and $$\nu \in \mathcal {P}_2({\textsf {X} })$$, then the map $${\mathcal {I}}\ni s \mapsto W_2^2(\mu _s, \nu )$$ is locally absolutely continuous and thus differentiable in a set of full measure $$A({\mu ,\nu }) \subset {\mathcal {I}}$$ which, in principle, depends both on $$\mu$$ and $$\nu$$. What Theorem 3.11 shows is that, independently of $$\nu$$, there is a full measure set $$A(\mu )$$, depending only on $$\mu$$, where this map is left and right differentiable. If moreover $$\nu$$ and $$t \in A(\mu )$$ are such that there is a unique optimal transport plan between them, we can actually conclude that such a map is differentiable at t. We refer in particular to Appendix A for a concrete example showing the optimality of the result stated in Theorem 3.11.
### Theorem 3.14
Let $$\mu ^1,\mu ^2:{\mathcal {I}}\rightarrow \mathcal {P}_2({\textsf {X} })$$ be locally absolutely continuous curves and let $${\varvec{v}}^1,{\varvec{v}}^2: {\mathcal {I}}\times {\textsf {X} }\rightarrow {\textsf {X} }$$ be the corresponding Wasserstein velocity fields satisfying (2.6) in $$A({\mu ^1})$$ and $$A({\mu ^2})$$ respectively. Then, for every $$t \in A({\mu ^1})\cap A({\mu ^2})$$, it holds
\begin{aligned} \lim _{h \downarrow 0} \frac{W_2^2(\mu ^1_{t+h}, \mu ^2_{t+h})-W_2^2(\mu ^1_t, \mu ^2_t)}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^1_t)_{\sharp }\mu ^1_t, ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^2_t)_{\sharp }\mu ^2_t\right] _{r},\\ \lim _{h \uparrow 0} \frac{W_2^2(\mu ^1_{t+h}, \mu ^2_{t+h})-W_2^2(\mu ^1_t, \mu ^2_t)}{2h}&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^1_t)_{\sharp }\mu ^1_t, ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^2_t)_{\sharp }\mu ^2_t\right] _{l}. \end{aligned}
In particular, there exists a subset $$A\subset A({\mu ^1})\cap A({\mu ^2})$$ of full Lebesgue measure such that $$s \mapsto W_2^2(\mu ^1_s, \mu ^2_s)$$ is differentiable in A and
\begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}W_2^2(\mu ^1_t,\mu ^2_t)&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^1_t)_{\sharp }\mu ^1_t, ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^2_t)_{\sharp }\mu ^2_t\right] _{r}\nonumber \\&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^1_t)_{\sharp }\mu ^1_t, ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}^2_t)_{\sharp }\mu ^2_t\right] _{l} \nonumber \\&= \int \langle {\varvec{v}}^1_t-{\varvec{v}}^2_t,x_1-x_2\rangle \,\mathrm d\varvec{\mu }(x_1,x_2) \end{aligned}
(3.21)
for every $$\varvec{\mu }\in \Gamma _o(\mu ^1_t,\mu ^2_t)$$, $$t\in A$$.
The proof of Theorem 3.14 follows by the same argument of the proof of Theorem 3.11.
### 3.3 Convexity and semicontinuity of duality parings
We want now to investigate the semicontinuity and convexity properties of the functionals $$\left[ \cdot , \cdot \right] _{r}$$ and $$\left[ \cdot , \cdot \right] _{l}$$.
### Lemma 3.15
Let $$(\Phi _n )_{n\in {\mathbb {N}}} \subset \mathcal {P}_2(\mathsf {TX})$$ be converging to $$\Phi$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$, and let $$(\nu _n )_{n\in {\mathbb {N}}} \subset \mathcal {P}_2({\textsf {X} })$$ be converging to $$\nu$$ in $$\mathcal {P}_2({\textsf {X} })$$. Then
\begin{aligned} \liminf _n \left[ \Phi _n, \nu _n\right] _{r} \ge \left[ \Phi , \nu \right] _{r}\quad \text { and }\quad \limsup _n \left[ \Phi _n, \nu _n\right] _{l} \le \left[ \Phi , \nu \right] _{l}. \end{aligned}
(3.22)
Finally, if $$(\Phi _n^i)_{n\in {\mathbb {N}}}$$, $$i=0,1$$, are sequences converging to $$\Phi ^i$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ then
\begin{aligned} \liminf _{n\rightarrow \infty }\left[ \Phi ^0_n, \Phi ^1_n\right] _{r}\ge \left[ \Phi ^0, \Phi ^1\right] _{r},\qquad \limsup _{n\rightarrow \infty }\left[ \Phi ^0_n, \Phi ^1_n\right] _{l}\ge \left[ \Phi ^0, \Phi ^1\right] _{l}. \end{aligned}
(3.23)
### Proof
We just consider the proof of the first inequality (3.22); the other statements follow by similar arguments and by Remark 3.6.
We can extract a subsequence of $$(\Phi _n)_{n \in {\mathbb {N}}}$$ (not relabeled) s.t. the $$\liminf$$ is achieved as a limit. We have to prove that
\begin{aligned} \lim _n \left[ \Phi _n, \nu _n\right] _{r} \ge \left[ \Phi , \nu \right] _{r}. \end{aligned}
(3.24)
For every $$n\in {\mathbb {N}}$$ take $$\varvec{\sigma }_n \in \Lambda _o(\Phi _n, \nu _n)$$ and $${\bar{\varvec{\vartheta }}}_n=({\textsf {x} }^0,{\textsf {x} }^1)_\sharp \varvec{\sigma }_n$$. Since the marginals of $${\bar{\varvec{\vartheta }}}_n$$ are converging w.r.t. $$W_2$$, the family $$({\bar{\varvec{\vartheta }}}_n)_{n\in {\mathbb {N}}}$$ is relatively compact in $$\mathcal {P}_2({\textsf {X} }^2)$$. Hence, $$(\varvec{\sigma }_n)_{n\in {\mathbb {N}}}$$ is relatively compact in $$\mathcal {P}_2^{sws}(\mathsf {TX}\times {\textsf {X} })$$ by Proposition 2.15, since the moments $$\int |v_0|^2\,\mathrm d\varvec{\sigma }_n(x_0,v_0,x_1)=|\Phi _n|^2_2$$ are uniformly bounded by assumption. Thus, possibly passing to a further subsequence, we have that $$(\varvec{\sigma }_n)_{n\in {\mathbb {N}}}$$ converges to some $$\varvec{\sigma }$$ in $$\mathcal {P}_2^{sws}(\mathsf {TX}\times {\textsf {X} })$$. In particular $$\varvec{\sigma }\in \Lambda (\Phi , \nu )$$ since optimality of the $${\textsf {X} }^2$$ marginals is preserved by narrow convergence. Indeed, it sufficies to use [3, Proposition 7.1.3] noting that
\begin{aligned} \int |x_0-x_1|^2\,\mathrm d\varvec{\sigma }_n\le 2\,{\textsf {m} }_2^2({\textsf {x} }_\sharp \Phi _n)+2\,{\textsf {m} }_2^2(\nu _n)\le K, \end{aligned}
for some $$K\ge 0$$.
The relation in (2.9) then yields
\begin{aligned} \lim _{n\rightarrow \infty } \left[ \Phi _n, \nu _n\right] _{r} = \lim _{n\rightarrow \infty } \int \langle v_0, x_0-x_1\rangle \,\mathrm d\varvec{\sigma }_n = \int \langle v_0, x_0-x_1\rangle \,\mathrm d\varvec{\sigma }\end{aligned}
which yields (3.24) since the r.h.s. is larger than $$\left[ \Phi , \nu \right] _{r}$$ by Theorem 3.9. $$\square$$
### Remark 3.16
Notice that in the special case in which $$\Lambda (\Phi , \nu )$$ is a singleton, then the limit exists and it holds
\begin{aligned} \lim _{n\rightarrow \infty } \left[ \Phi _n, \nu _n\right] _{r} = \left[ \Phi , \nu \right] _{r},\qquad \lim _{n\rightarrow \infty } \left[ \Phi _n, \nu _n\right] _{l} = \left[ \Phi , \nu \right] _{l}. \end{aligned}
### Lemma 3.17
For every $$\mu ,\nu \in \mathcal {P}_2({\textsf {X} })$$ the maps $$\Phi \mapsto \left[ \Phi , \nu \right] _{r}$$ and $$(\Phi ,\Psi )\mapsto \left[ \Phi , \Psi \right] _{r}$$ (resp. $$\Phi \mapsto \left[ \Phi , \nu \right] _{l}$$ and $$(\Phi ,\Psi )\mapsto \left[ \Phi , \Psi \right] _{l}$$) are convex (resp. concave) in $$\mathcal {P}_{2}(\mathsf {TX}|\mu )$$ and $$\mathcal {P}_{2}(\mathsf {TX}|\mu )\times \mathcal {P}_{2}(\mathsf {TX}|\nu )$$.
### Proof
We prove the convexity of $$(\Phi ,\Psi )\mapsto \left[ \Phi , \Psi \right] _{r}$$ in $$\mathcal {P}_{2}(\mathsf {TX}|\mu )\times \mathcal {P}_{2}(\mathsf {TX}|\nu )$$; the argument of the proofs of the other statements are completely analogous.
Let $$\Phi _k\in \mathcal {P}_{2}(\mathsf {TX}|\mu )$$, $$\Psi _k\in \mathcal {P}_{2}(\mathsf {TX}|\nu )$$, and let $$\beta _k\ge 0$$, with $$\sum _k\beta _k=1$$, $$k=1,\ldots ,K$$. We set $$\Phi =\sum _{k=1}^K\beta _k\Phi _k$$, $$\Psi =\sum _{k=1}^K\beta _k\Psi _k$$, For every k let us select $$\varvec{\Theta }_{k}\in \Lambda (\Phi _k,\Psi _k)$$ such that
\begin{aligned} \left[ \Phi _k, \Psi _k\right] _{r}= \int \langle v_1-v_0,x_1-x_0\rangle \,\mathrm d\varvec{\Theta }_{k}. \end{aligned}
It is not difficult to check that $$\varvec{\Theta }:=\sum _{k}\beta _{k}\varvec{\Theta }_{k}\in \Lambda (\Phi ,\Psi )$$ so that
\begin{aligned} \left[ \Phi , \Psi \right] _{r}\le & {} \int \langle v_1-v_0,x_1-x_0\rangle \,\mathrm d\varvec{\Theta }\\= & {} \sum _{k}\beta _{k}\int \langle v_1-v_0,x_1-x_0\rangle \, \mathrm d\varvec{\Theta }_{k}=\sum _k \beta _{k}\left[ \Phi _k, \Psi _k\right] _{r}. \end{aligned}
$$\square$$
### 3.4 Behaviour of duality pairings along geodesics
We have seen that the duality pairings $$\left[ \cdot , \cdot \right] _{r}$$ and $$\left[ \cdot , \cdot \right] _{l}$$ may differ when the collection of optimal plans $$\Gamma _o(\mu _0,\mu _1)$$ contains more than one element. It is natural to expect a simpler behaviour along geodesics. We will introduce the following definition, where we use the notation
\begin{aligned} {\textsf {x} }^t(x_0,x_1):=(1-t)x_0+tx_1,\quad {\textsf {v} }^0(x_0,v_0,x_1):=v_0 \end{aligned}
for every $$(x_0,v_0,x_1)\in \mathsf {TX}\times {\textsf {X} }$$, $$t\in [0,1]$$.
### Definition 3.18
For $$\varvec{\vartheta }\in \mathcal {P}_2({\textsf {X} }\times {\textsf {X} })$$, $$t\in [0,1]$$, $$\vartheta _t={\textsf {x} }^t_\sharp \varvec{\vartheta }$$ and $$\Phi _t\in \mathcal {P}_{2}(\mathsf {TX}|\vartheta _t)$$, we set
\begin{aligned} \Gamma _t(\Phi _t,\varvec{\vartheta }):= \left\{ \varvec{\sigma }\in \mathcal {P}_2(\mathsf {TX}\times {\textsf {X} }) \mid ({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp }\varvec{\sigma }=\varvec{\vartheta }\,\text { and }\, ({\textsf {x} }^t \circ ({\textsf {x} }^0,{\textsf {x} }^1),{\textsf {v} }^0)_\sharp \varvec{\sigma }=\Phi _t\right\} ,\nonumber \\ \end{aligned}
(3.25)
which is not empty since $$\vartheta _t={\textsf {x} }^t_\sharp \varvec{\vartheta }={\textsf {x} }_\sharp \Phi _t$$. We set
\begin{aligned} \left[ \Phi _t, \varvec{\vartheta }\right] _{b,t}&:= \int \Big \langle x_0-x_1,{\varvec{b}}_{\Phi _t}({\textsf {x} }^t (x_0,x_1))\Big \rangle \,\mathrm d\varvec{\vartheta }(x_0,x_1),\\ [\Phi _t,\varvec{\vartheta }]_{r,t}&:= \min \left\{ \int \langle x_0-x_1, v_0\rangle \,\mathrm d\varvec{\sigma }(x_0,v_0,x_1) \mid \varvec{\sigma }\in \Gamma _t(\Phi _t,\varvec{\vartheta }) \right\} , \\ [\Phi _t,\varvec{\vartheta }]_{l,t}&:= \max \left\{ \int \langle x_0-x_1, v_0\rangle \,\mathrm d\varvec{\sigma }(x_0,v_0,x_1) \mid \varvec{\sigma }\in \Gamma _t(\Phi _t,\varvec{\vartheta }) \right\} . \end{aligned}
If moreover $$\Phi _0\in \mathcal {P}_2(\mathsf {TX}|\vartheta _0)$$, $$\Phi _1\in \mathcal {P}_2(\mathsf {TX}|\vartheta _1)$$, $$\varvec{\vartheta }\in \Gamma (\vartheta _0,\vartheta _1)$$, we define
\begin{aligned}{}[\Phi _0,\Phi _1]_{r,\varvec{\vartheta }}&:= [\Phi _0,\varvec{\vartheta }]_{r,0}- [\Phi _1,\varvec{\vartheta }]_{l,1},\\ [\Phi _0,\Phi _1]_{l,\varvec{\vartheta }}&:= [\Phi _0,\varvec{\vartheta }]_{l,0}- [\Phi _1,\varvec{\vartheta }]_{r,1}. \end{aligned}
Notice that, if $$(\Phi _t)_x$$ is the disintegration of $$\Phi _t$$ with respect to $$\vartheta _t={\textsf {x} }_\sharp \Phi _t$$, we can consider the barycentric coupling $$\varvec{\sigma }_t:= \int _{{\textsf {X} }\times {\textsf {X} }}(\Phi _t)_{{\textsf {x} }^t}\,\mathrm d\varvec{\vartheta }\in \Gamma _t(\Phi _t,\varvec{\vartheta })$$, i.e.
\begin{aligned} \int \psi (x_0,v_0,x_1)\,\mathrm d\varvec{\sigma }_t= \int \Big [\int \psi (x_0,v_0,x_1)\,\mathrm d(\Phi _t)_{(1-t)x_0+tx_1}(v_0)\Big ]\,\mathrm d\varvec{\vartheta }(x_0,x_1) \end{aligned}
so that $$\left[ \Phi _t, \varvec{\vartheta }\right] _{b,t}=\int \langle {v_0},x_0-x_1\rangle \,\mathrm d\varvec{\sigma }_t$$ and
\begin{aligned}{}[\Phi _t,\varvec{\vartheta }]_{r,t}\le \left[ \Phi _t, \varvec{\vartheta }\right] _{b,t}\le [\Phi _t,\varvec{\vartheta }]_{l,t}. \end{aligned}
If we define the reversion map
\begin{aligned} {\textsf {s} }:{\textsf {X} }^2\rightarrow {\textsf {X} }^2,\quad {\textsf {s} }(x_0,x_1):=(x_1,x_0), \end{aligned}
(3.26)
with a similar definition for $$\mathsf {TX}\times {\textsf {X} }$$, given by $$\mathsf s(x_0,v_0,x_1):=(x_1,v_0,x_0)$$, it is easy to check that
\begin{aligned} \varvec{\sigma }\in \Gamma _t(\Phi _t,\varvec{\vartheta })\quad \Leftrightarrow \quad {\textsf {s} }_{\sharp }\varvec{\sigma }\in \Gamma _{1-t}(\Phi _t,{\textsf {s} }_\sharp \varvec{\vartheta }) \end{aligned}
so that
\begin{aligned}{}[\Phi _t,\varvec{\vartheta }]_{r,t}=- [\Phi _t,{\mathsf {s}}_\sharp \varvec{\vartheta }]_{l,1-t},\quad [\Phi _t,\varvec{\vartheta }]_{l,t}=- [\Phi _t,{\textsf {s} }_\sharp \varvec{\vartheta }]_{r,1-t}. \end{aligned}
(3.27)
We point out that (3.16) and (3.19) have simpler versions in two particular cases, which will be explained in the next remark.
### Remark 3.19
(Particular cases) Suppose that $$\varvec{\vartheta }\in \mathcal {P}_2({\textsf {X} }^2)$$, $$t\in [0,1]$$, $$\vartheta _t={\textsf {x} }^t_\sharp \varvec{\vartheta }$$, $$\Phi _t\in \mathcal {P}_{2}(\mathsf {TX}|\vartheta _t)$$ and $${\textsf {x} }^t:{\textsf {X} }^2\rightarrow {\textsf {X} }$$ is $$\varvec{\vartheta }$$-essentially injective so that $$\varvec{\vartheta }$$ is concentrated on a Borel map
\begin{aligned} (X^0_t,X^1_t):{\textsf {X} }\rightarrow {\textsf {X} }\times {\textsf {X} },\text { i.e. }\varvec{\vartheta }=(X^0_t,X^1_t)_\sharp \vartheta _t. \end{aligned}
In this case $$\Gamma _t(\Phi _t,\varvec{\vartheta })$$ contains a unique element given by $$(X^0_t\circ {\textsf {x} },{\textsf {v} },X^1_t\circ {\textsf {x} })_\sharp \Phi _t$$ and
\begin{aligned}{}[\Phi _t,\varvec{\vartheta }]_{r,t}= & {} [\Phi _t,\varvec{\vartheta }]_{l,t}= \left[ \Phi _t, \varvec{\vartheta }\right] _{b,t}= \int \langle v,X^0_t(x)-X^1_t(x)\rangle \,\mathrm d\Phi _t(x,v)\nonumber \\= & {} \int \langle {\varvec{b}}_{\Phi _t},X^0_t-X^1_t\rangle \,\mathrm d\vartheta _t, \end{aligned}
(3.28)
where in the last formula we have applied the barycentric reduction (3.8). When $$t=0$$ and $$\varvec{\vartheta }$$ is the unique element of $$\Gamma _o(\vartheta _0,\vartheta _1)$$ then $$X^0_t(x)=x$$ and we obtain
\begin{aligned} \left[ \Phi _t, \vartheta _1\right] _{r}= & {} \left[ \Phi _t, \vartheta _1\right] _{l}=[\Phi _t,\varvec{\vartheta }]_{r,0}=[\Phi _t,\varvec{\vartheta }]_{l,0}\\= & {} \int \langle v,x-X^1_t(x)\rangle \,\mathrm d\Phi _t(x,v)= \int \langle {\varvec{b}}_{\Phi _t},x-X^1_t(x)\rangle \,\mathrm d\vartheta _0(x). \end{aligned}
Another simple case is when
\begin{aligned} \Phi _t=({\varvec{i}}_{\textsf {X} },{\varvec{w}})_\sharp \vartheta _t \end{aligned}
for some vector field $${\varvec{w}}\in L^2_{\vartheta _t}({\textsf {X} };{\textsf {X} })$$ as in (3.5) (i.e. its disintegration $$\Phi _x$$ w.r.t. $$\vartheta _t$$ takes the form $$\delta _{{\varvec{w}}(x)}$$ and $${\varvec{w}}={\varvec{b}}_{\Phi _t}$$). We have
\begin{aligned}{}[\Phi _t,\varvec{\vartheta }]_{r,t}= [\Phi _t,\varvec{\vartheta }]_{l,t}= \int \Big \langle {\varvec{w}}((1-t)x_0+tx_1),x_0-x_1\Big \rangle \,\mathrm d\varvec{\vartheta }(x_0,x_1). \end{aligned}
In particular we get
\begin{aligned} \left[ \Phi _t, \vartheta _1\right] _{r}= \min \Bigg \{\int \langle {\varvec{w}}(x),x_0-x_1\rangle \,\mathrm d\varvec{\vartheta }(x_0,x_1)\mid \varvec{\vartheta }\in \Gamma _o(\vartheta _0,\vartheta _1)\Bigg \}. \end{aligned}
An important case in which the previous Remark 3.19 applies is that of geodesics in $$\mathcal {P}_2({\textsf {X} })$$.
### Lemma 3.20
Let $$\mu _0, \mu _1 \in \mathcal {P}_2({\textsf {X} })$$, $$\mu :[0,1]\rightarrow \mathcal {P}_2({\textsf {X} })$$ be a constant speed geodesic induced by an optimal plan $$\varvec{\mu }\in \Gamma _o(\mu _0, \mu _1)$$ by the relation
\begin{aligned} \mu _t = {\textsf {x} }^t_{\sharp } \varvec{\mu },\quad t\in [0,1],\quad \text {where}\quad {\textsf {x} }^t(x_0,x_1)=(1-t)x_0+tx_1. \end{aligned}
If $$t \in (0,1)$$, $$\Phi _t\in \mathcal {P}_2(\mathsf {TX}|\mu _t)$$, $${{\hat{\varvec{\mu }}}}={\textsf {s} }_\sharp \varvec{\mu }\in \Gamma _o(\mu _1,\mu _0)$$, with $${\textsf {s} }$$ the reversion map in (3.26), then
\begin{aligned} \frac{1}{1-t} \left[ \Phi _t, \mu _1\right] _{r}&= \frac{1}{1-t} \left[ \Phi _t, \mu _1\right] _{l} \nonumber \\&= [\Phi _t,\varvec{\mu }]_{r,t}\nonumber \\&=[\Phi _t,\varvec{\mu }]_{l,t}\nonumber \\&= -\frac{1}{t} \left[ \Phi _t, \mu _0\right] _{r}\nonumber \\&= -\frac{1}{t} \left[ \Phi _t, \mu _0\right] _{l}\nonumber \\&= - [\Phi _t,{{\hat{\varvec{\mu }}}}]_{r,1-t}\nonumber \\&=-[\Phi _t,{{\hat{\varvec{\mu }}}}]_{l,1-t}. \end{aligned}
(3.29)
### Proof
The crucial fact is that $${\textsf {x} }^t : {\textsf {X} }^2 \rightarrow {\textsf {X} }$$ is injective on $${{\,\mathrm{supp}\,}}(\varvec{\mu })$$ and thus a bijection on its image $${{\,\mathrm{supp}\,}}(\mu _t)$$. Indeed, take $$(x_0, x_1), (x_0', x_1') \in {{\,\mathrm{supp}\,}}(\varvec{\mu })$$, then
\begin{aligned} \left| {\textsf {x} }^t(x_0, x_1) - {\textsf {x} }^t(x_0', x_1') \right| ^2&= (1-t)^2|x_0-x_0'|^2 + t^2|x_1-x_1'|^2 \\&\quad + 2t(1-t)\langle x_0-x_0', x_1-x_1'\rangle \\&\ge (1-t)^2|x_0-x_0'|^2 + t^2|x_1-x_1'|^2 \end{aligned}
thanks to the cyclical monotonicity of $${{\,\mathrm{supp}\,}}(\varvec{\mu })$$ (see [3, Remark 7.1.2]).
Then, for every $$x \in {{\,\mathrm{supp}\,}}(\mu _t)$$, there exists a unique couple $$(x_0, x_1) =(X^0_t(x),X^1_t(x))\in {{\,\mathrm{supp}\,}}(\varvec{\mu })$$ s.t. $$x=(1-t)x_0 + tx_1$$, where we refer to Remark 3.19 for the definitions of $$X^0_t, X^1_t$$ (cf. also [32, Theorem 5.29]). Hence, in the following diagram all maps are bijections:
where $$\varvec{\mu }_{t1} =({\textsf {x} }^t,{\textsf {x} }^1)_\sharp \varvec{\mu }=({\varvec{i}}_{\textsf {X} }, X^1_t)_{\sharp }\mu _t$$ is the unique element of $$\Gamma _o(\mu _t, \mu _1)$$ and $$\varvec{\mu }_{t0}=({\textsf {x} }^t,{\textsf {x} }^0)_\sharp \varvec{\mu }=({\varvec{i}}_{\textsf {X} }, X^0_t)_{\sharp }\mu _t= ({\textsf {x} }^{1-t},{\textsf {x} }^1)_\sharp {{\hat{\varvec{\mu }}}}$$ is the unique element of $$\Gamma _o(\mu _t, \mu _0)$$ (see Theorem 2.8). Since
\begin{aligned} \frac{x-X^1_t(x)}{1-t} = \frac{x-x_1}{1-t}= x_0-x_1 = -\frac{x-x_0}{t} = -\frac{x-X^0_t(x)}{t}, \end{aligned}
and $$\Lambda (\Phi _t, \mu _1) = \{ ({\varvec{i}}_{\mathsf {TX}} , X^1_t \circ {\textsf {x} })_{\sharp } \Phi _t \}$$ thanks to Theorem 2.8, by Theorem 3.9 and Corollary 3.10 we have
\begin{aligned} \left[ \Phi _t, \mu _1\right] _{r} = \left[ \Phi _t, \mu _1\right] _{l} = \int _{\mathsf {TX}} \langle v, x-X^1_t(x)\rangle \,\mathrm d\Phi _t(x,v). \end{aligned}
Analogously, $$\Lambda (\Phi _t, \mu _0) = \{ ({\varvec{i}}_{\mathsf {TX}} , X^0_t \circ {\textsf {x} })_{\sharp } \Phi _t \}$$. Hence
\begin{aligned} \left[ \Phi _t, \mu _0\right] _{r} = \left[ \Phi _t, \mu _0\right] _{l} = \int _{\mathsf {TX}} \langle v, x-X^0_t(x)\rangle \,\mathrm d\Phi _t(x,v). \end{aligned}
Also recalling (3.27) and (3.28) we conclude. $$\square$$
## 4 Dissipative probability vector fields: the metric viewpoint
### Definition 4.1
(Multivalued Probability Vector Field - MPVF) A multivalued probability vector field $${\varvec{\mathrm {F}}}$$ is a nonempty subset of $$\mathcal {P}_2(\mathsf {TX})$$ with domain $$\mathrm {D}({\varvec{\mathrm {F}}}) := {\textsf {x} }_\sharp ({\varvec{\mathrm {F}}})= \{ {\textsf {x} }_\sharp \Phi :\Phi \in {\varvec{\mathrm {F}}}\}$$. Given $$\mu \in \mathcal {P}_2({\textsf {X} })$$, we define the section $${\varvec{\mathrm {F}}}[\mu ]$$ of $${\varvec{\mathrm {F}}}$$ as
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ] := ({\textsf {x} }_\sharp )^{-1}(\mu )\cap {\varvec{\mathrm {F}}}= \left\{ \Phi \in {\varvec{\mathrm {F}}}\mid {\textsf {x} }_{\sharp }\Phi = \mu \right\} . \end{aligned}
A selection $${\varvec{\mathrm {F}}}'$$ of $${\varvec{\mathrm {F}}}$$ is a subset of $${\varvec{\mathrm {F}}}$$ such that $$\mathrm {D}({\varvec{\mathrm {F}}}')=\mathrm {D}({\varvec{\mathrm {F}}})$$. We call $${\varvec{\mathrm {F}}}$$ a probability vector field (PVF) if $${\textsf {x} }_\sharp$$ is injective in $${\varvec{\mathrm {F}}}$$, i.e. $${\varvec{\mathrm {F}}}[\mu ]$$ contains a unique element for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$. A MPVF $${\varvec{\mathrm {F}}}$$ is a vector field if for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, the section $${\varvec{\mathrm {F}}}[\mu ]$$ contains a unique element $$\Phi$$ concentrated on a map, i.e. $$\Phi =({\varvec{i}}_{\textsf {X} }, {\varvec{b}}_{\Phi })_\sharp \mu$$.
### Remark 4.2
We can equivalently formulate Definition 4.1 by considering $${\varvec{\mathrm {F}}}$$ as a multifunction, as in the case, e.g., of the Wasserstein subdifferential $$\varvec{\partial }{\mathcal {F}}$$ of a function $${\mathcal {F}}:\mathcal {P}_2({\textsf {X} })\rightarrow (-\infty ,+\infty ]$$, see [3, Ch. 10] and the next Sect. 7.1. According to this viewpoint, a MPVF is a set-valued map $${\varvec{\mathrm {F}}}:\mathcal {P}_2({\textsf {X} })\supset \mathrm {D}({\varvec{\mathrm {F}}}) \rightrightarrows \mathcal {P}_2(\mathsf {TX})$$ such that $${\textsf {x} }_{\sharp }\Phi =\mu$$ for all $$\Phi \in {\varvec{\mathrm {F}}}[\mu ]$$. In this way, each section $${\varvec{\mathrm {F}}}[\mu ]$$ is nothing but the image of $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ through $${\varvec{\mathrm {F}}}$$. In this case, probability vector fields correspond to single valued maps: this notion has been used in [27] with the aim of describing a sort of velocity field on $$\mathcal {P}({\textsf {X} })$$, and later in [26] dealing with Multivalued Probability Vector Fields (called Probability Multifunctions).
### Definition 4.3
(Metrically $$\lambda$$-dissipative MPVF) A MPVF $${\varvec{\mathrm {F}}}\subset \mathcal {P}_2(\mathsf {TX})$$ is (metrically) $$\lambda$$-dissipative, with $$\lambda \in {\mathbb {R}}$$, if
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r} \le \lambda W_2^2(\mu _0,\mu _1)\quad \text { for every }\Phi _0,\Phi _1\in {\varvec{\mathrm {F}}},\,\mu _i={\textsf {x} }_\sharp \Phi _i. \end{aligned}
(4.1)
We say that $${\varvec{\mathrm {F}}}$$ is (metrically) $$\lambda$$-accretive if $$-{\varvec{\mathrm {F}}}=\{-\Phi :\Phi \in {\varvec{\mathrm {F}}}\}$$ (recall (3.15)) is $$-\lambda$$-dissipative, i.e.
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{l} \ge \lambda W_2^2(\mu _0,\mu _1)\quad \text { for every }\Phi _0,\Phi _1\in {\varvec{\mathrm {F}}},\,\mu _i={\textsf {x} }_\sharp \Phi _i. \end{aligned}
In Sect. 7 we collect explicit examples of $$\lambda$$-dissipative MPVFs.
### Remark 4.4
Notice that (4.1) is equivalent to asking for the existence of a coupling $$\varvec{\Theta }\in \Lambda (\Phi _0,\Phi _1)$$ (thus $$({\textsf {x} }^0,{\textsf {x} }^1)_\sharp \varvec{\Theta }$$ is optimal between $$\mu _0={\textsf {x} }_\sharp \Phi _0$$ and $$\mu _1={\textsf {x} }_\sharp \Phi _1$$) such that
\begin{aligned} \int \langle v_1-v_0,x_1-x_0\rangle \,\mathrm d\varvec{\Theta }\le \lambda W_2^2(\mu _0,\mu _1)= \lambda \int |x_1-x_0|^2\,\mathrm d\varvec{\Theta }. \end{aligned}
As anticipated in the Introduction, dealing with (1.6) and (1.8), the $$\lambda$$-dissipativity condition (4.1) has a natural metric interpretation: if $$\Phi _0,\Phi _1\in {\varvec{\mathrm {F}}}$$ with $$\mu _0={\textsf {x} }_\sharp \Phi _0$$, $$\mu _1={\textsf {x} }_\sharp \Phi _1$$, performing a first order Taylor expansion of the map
\begin{aligned} t\mapsto \frac{1}{2}W_2^2(\textsf {exp} ^t\Phi _0,\textsf {exp} ^t\Phi _1) \end{aligned}
at $$t=0$$, recalling Definition 3.5, we have
\begin{aligned} W_2^2(\textsf {exp} ^t\Phi _0,\textsf {exp} ^t\Phi _1)\le (1+2\lambda t)W_2^2(\mu _0,\mu _1)+o(t)\quad \text {as }t\downarrow 0. \end{aligned}
### Remark 4.5
Thanks to Corollary 3.7, (4.1) implies the weaker condition
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r}+\left[ \Phi _1, \mu _0\right] _{r} \le \lambda W_2^2(\mu _0, \mu _1)\quad \text {for every }\Phi _0,\Phi _1 \in {\varvec{\mathrm {F}}},\,\mu _i={\textsf {x} }_\sharp \Phi _i.\qquad \end{aligned}
(4.2)
It is clear that the inequality of (4.2) implies the inequality of (4.1) whenever $$\Gamma _o(\mu _0,\mu _1)$$ contains only one element. More generally, we will see in Corollary 4.13 that (4.2) is in fact equivalent to (4.1) when $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is geodesically convex (according to Definition 2.7).
As in the standard Hilbert case, $$\lambda$$-dissipativity can be reduced to dissipativity (meaning 0-dissipativity) by a simple transformation as shown in Lemma 4.6. Let us introduce the map
\begin{aligned} L^\lambda :\mathsf {TX}\rightarrow \mathsf {TX},\quad L^\lambda (x,v):=(x,v-\lambda x). \end{aligned}
### Lemma 4.6
$${\varvec{\mathrm {F}}}$$ is a $$\lambda$$-dissipative MPVF (resp. satisfies (4.2)) if and only if $${\varvec{\mathrm {F}}}^\lambda :=L^\lambda _\sharp ({\varvec{\mathrm {F}}})= \{L^\lambda _\sharp \Phi \mid \Phi \in {\varvec{\mathrm {F}}}\}$$ is dissipative (resp. satisfies (4.2) with $$\lambda =0$$).
### Proof
Let us first check the case of (4.2). If $$\varvec{\sigma }\in \mathcal {P}_2(\mathsf {TX}\times {\textsf {X} })$$ with $$({\textsf {x} }^i)_\sharp \varvec{\sigma }=\mu _i$$, $$i=0,1$$, the transformed plan $$\varvec{\sigma }^\lambda :=(L^\lambda ,{\varvec{i}}_{\textsf {X} })_\sharp \varvec{\sigma }$$ satisfies
\begin{aligned} \int \langle v_0,x_0-x_1\rangle \,\mathrm d\varvec{\sigma }^\lambda&= \int \langle v_0-\lambda x_0,x_0-x_1\rangle \,\mathrm d\varvec{\sigma }\nonumber \\&= \int \langle v_0,x_0-x_1\rangle \,\mathrm d\varvec{\sigma }-\frac{\lambda }{2}\int |x_0-x_1|^2\,\mathrm d\varvec{\sigma }\nonumber \\&\quad +\frac{\lambda }{2}\Big ({\textsf {m} }_2^2(\mu _1)-{\textsf {m} }_2^2(\mu _0)\Big ). \end{aligned}
(4.3)
Since $$\varvec{\sigma }\in \Lambda _o(\Phi _0,\mu _1)$$ if and only if $$\varvec{\sigma }^\lambda \in \Lambda _o(L^\lambda _\sharp \Phi _0,\mu _1)$$, (4.3) yields
\begin{aligned} \int \langle v_0,x_0-x_1\rangle \,\mathrm d\varvec{\sigma }^\lambda = \int \langle v_0,x_0-x_1\rangle \,\mathrm d\varvec{\sigma }-\frac{\lambda }{2}\Big ({\textsf {m} }_2^2(\mu _0)-{\textsf {m} }_2^2(\mu _1)+W_2^2(\mu _0,\mu _1)\Big ) \end{aligned}
and therefore
\begin{aligned} \left[ L^\lambda _\sharp \Phi _0, \mu _1\right] _{r}= \left[ \Phi _0, \mu _1\right] _{r}-\frac{\lambda }{2}\Big ({\textsf {m} }_2^2(\mu _0)-{\textsf {m} }_2^2(\mu _1)+W_2^2(\mu _0,\mu _1)\Big ). \end{aligned}
(4.4)
Using the corresponding identity for $$\left[ L^\lambda _\sharp \Phi _1, \mu _0\right] _{r}$$ we obtain that $${\varvec{\mathrm {F}}}^\lambda$$ is dissipative.
Similarly, if $$\varvec{\Theta }\in \mathcal {P}_2(\mathsf {TX}\times \mathsf {TX})$$ with $${\textsf {x} }^i_\sharp \varvec{\Theta }=\mu _i$$, the plan $$\varvec{\Theta }^\lambda :=(L^\lambda ,L^\lambda )_\sharp \varvec{\Theta }$$ satisfies
\begin{aligned} \int \langle v_0-v_1,x_0-x_1\rangle \,\mathrm d\varvec{\Theta }^\lambda&= \int \langle v_0-v_1-\lambda (x_0-x_1),x_0-x_1\rangle \,\mathrm d\varvec{\Theta }\nonumber \\&= \int \langle v_0-v_1,x_0-x_1\rangle \,\mathrm d\varvec{\Theta }-\lambda \int |x_0-x_1|^2\,\mathrm d\varvec{\Theta }. \end{aligned}
(4.5)
Reasoning with a similar argument as for the case of assumption (4.2), using the identity (4.5), we get the equivalence between the $$\lambda$$-dissipativity of $${\varvec{\mathrm {F}}}$$ and the dissipativity of $${\varvec{\mathrm {F}}}^\lambda$$. $$\square$$
Let us conclude this section by showing that $$\lambda$$-dissipativity can be deduced from a Lipschitz like condition similar to the one considered in [27] (see Sect. 7.5).
### Lemma 4.7
Suppose that the MPVF $${\varvec{\mathrm {F}}}$$ satisfies
\begin{aligned} \mathcal {W}_2({\varvec{\mathrm {F}}}[\nu ], {\varvec{\mathrm {F}}}[\nu ']) \le L W_2(\nu , \nu ')\quad \text { for every } \nu , \nu ' \in \mathrm {D}({\varvec{\mathrm {F}}}), \end{aligned}
where $$\mathcal {W}_2:\mathcal {P}_2(\mathsf {TX})\times \mathcal {P}_2(\mathsf {TX})\rightarrow [0,+\infty )$$ is defined by
\begin{aligned} \mathcal {W}_2^2(\Phi _0,\Phi _1) = \inf \left\{ \int _{\mathsf {TX}\times \mathsf {TX}} |v_0 - v_1|^2 \,\mathrm d\varvec{\Theta }(x_0,v_0,x_1,v_1) : \varvec{\Theta }\in \Lambda (\Phi _0,\Phi _1) \right\} , \end{aligned}
with $$\Lambda (\cdot ,\cdot )$$ as in Definition 3.8. Then $${\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.1), for $$\lambda :=\frac{1}{2}(1+L^2)$$
### Proof
Let $$\nu ',\nu ''\in \mathrm {D}({\varvec{\mathrm {F}}})$$, then by Theorem 3.9 and Young’s inequality, we have
\begin{aligned} \begin{aligned} \left[ {\varvec{\mathrm {F}}}[\nu '], {\varvec{\mathrm {F}}}[\nu '']\right] _{r}&=\min \left\{ \int _{\mathsf {TX}\times \mathsf {TX}}\langle x'-x'', v'-v''\rangle \,\mathrm d\varvec{\Theta }\,:\,\varvec{\Theta }\in \Lambda ({\varvec{\mathrm {F}}}[\nu '],{\varvec{\mathrm {F}}}[\nu ''])\right\} \\&\le \frac{1}{2}\left( W_2^2(\nu ',\nu '')+\mathcal {W}_2^2({\varvec{\mathrm {F}}}[\nu '],{\varvec{\mathrm {F}}}[\nu ''])\right) \\ {}&\le \frac{L^2+1}{2}\,W_2^2(\nu ',\nu ''). \end{aligned} \end{aligned}
$$\square$$
### 4.2 Behaviour of $$\lambda$$-dissipative MPVF along geodesics
Let us now study the behaviour of a MPVF $${\varvec{\mathrm {F}}}$$ along geodesics. Recall that in the case of a dissipative map $${\mathrm F}:{\mathsf {H}}\rightarrow {\mathsf {H}}$$ in a Hilbert space $${\mathsf {H}}$$, it is quite immediate to prove that the real function
\begin{aligned} f(t) := \langle F(x_t), x_0-x_1\rangle ,\quad x_t = (1-t)x_0 + tx_1, \quad t \in [0,1] \end{aligned}
(4.6)
is monotone increasing. This property has a natural counterpart in the case of measures.
Let $${\varvec{\mathrm {F}}}\subset \mathcal {P}_2(\mathsf {TX})$$, $$\mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$. In order to compute the measure-theoretic analogue of the scalar product in (4.6), we need to define the set
\begin{aligned} \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}):={}\left\{ t\in [0,1]:{\textsf {x} }^t_\sharp \varvec{\mu }\in \mathrm {D}({\varvec{\mathrm {F}}})\right\} , \end{aligned}
(4.7)
since we can evaluate the MPVF $${\varvec{\mathrm {F}}}$$ along geodesics only for time instants $$t\in [0,1]$$ at which they lie inside the domain.
### Definition 4.8
Let $${\varvec{\mathrm {F}}}\subset \mathcal {P}_2(\mathsf {TX})$$ be a MPVF. Let $$\mu _0, \mu _1 \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$ and let $$\mu _t:={\textsf {x} }^t_{\sharp } \varvec{\mu }$$, $$t\in [0,1]$$. For every $$t\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$ we define
\begin{aligned}{}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t} := \sup \Big \{ [\Phi _t,\varvec{\mu }]_{r,t} \mid \Phi _t \in {\varvec{\mathrm {F}}}[\mu _t] \Big \}, \quad [{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} := \inf \Big \{ [\Phi _t,\varvec{\mu }]_{l,t} \mid \Phi _t \in {\varvec{\mathrm {F}}}[\mu _t] \Big \}. \end{aligned}
### Theorem 4.9
Let us suppose that the MPVF $${\varvec{\mathrm {F}}}$$ satisfies (4.2), let $$\mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, and let $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$. Then the following properties hold
1. (1)
$$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}$$ for every $$t \in (0,1)\cap \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$;
2. (2)
$$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,s} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}+\lambda (t-s)\,W_2^2(\mu _0,\mu _1)$$ for every $$s,t\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, $$s < t$$;
3. (3)
the maps
\begin{aligned} t\mapsto [{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}+\lambda t\,W_2^2(\mu _0,\mu _1)\quad \text {and}\quad t\mapsto [{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}+\lambda t\,W_2^2(\mu _0,\mu _1) \end{aligned}
are increasing respectively in $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}){\setminus } \{1\}$$ and in $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}){\setminus } \{0\}$$;
4. (4)
if $$t_0$$ is a right accumulation point of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, then
\begin{aligned} \lim _{t\downarrow t_0}\,[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t} = \lim _{t\downarrow t_0}\,[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} \end{aligned}
(4.8)
and these right limits exist. If, instead, $$t_0$$ is a left accumulation point of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, the same holds with the right limits in (4.8) replaced by the left limits at $$t_0$$;
5. (5)
$$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} = [{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}$$ at every interior point t of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$ where one of them is continuous.
### Proof
Throughout all the proof we set
\begin{aligned} f_r(t):=[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}\quad \text {and}\quad f_l(t):=[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}. \end{aligned}
(4.9)
Thanks to Lemma 4.6 and in particular to (4.4), it is easy to check that it is sufficient to consider the dissipative case $$\lambda =0$$.
1. (1)
It is a direct consequence of Lemma 3.20 and the definitions of $$f_r$$ and $$f_l$$.
2. (2)
We prove that for every $$\Phi _s \in {\varvec{\mathrm {F}}}[\mu _s]$$ and $$\Phi '_t \in {\varvec{\mathrm {F}}}[\mu _t]$$ it holds
\begin{aligned}{}[\Phi _s,\varvec{\mu }]_{r,s} \le [\Phi '_t,\varvec{\mu }]_{l,t}. \end{aligned}
(4.10)
The thesis will follow immediately passing to the $$\sup$$ over $$\Phi _s \in {\varvec{\mathrm {F}}}[\mu _s]$$ in the l.h.s. and to the $$\inf$$ over $$\Phi '_t \in {\varvec{\mathrm {F}}}[\mu _t]$$ in the r.h.s.. It is enough to prove (4.10) in case at least one between st belongs to (0, 1). Let us define the map $$L: \mathcal {P}_2(\mathsf {TX}\times {\textsf {X} }) \rightarrow {\mathbb {R}}$$ as
\begin{aligned} L(\gamma ):= \int _{\mathsf {TX}\times {\textsf {X} }} \langle v_0, x_0-x_1\rangle \,\mathrm d\gamma (x_0,v_0,x_1) \quad \gamma \in \mathcal {P}_2(\mathsf {TX}\times {\textsf {X} }). \end{aligned}
Observe that, since it never happens that $$s=0$$ and $$t=1$$ at the same time, the map $$T_{s,t}:\Gamma _s(\Phi _s, \varvec{\mu })\rightarrow \Lambda (\Phi _s,\mu _t)$$, with $$\Gamma _s(\cdot ,\cdot )$$ as in (3.25) and $$\Lambda (\cdot ,\cdot )$$ as in Definition 3.8, defined as
\begin{aligned} T_{s,t}(\varvec{\sigma }) := \left( {\textsf {x} }^s \circ ({\textsf {x} }^0,{\textsf {x} }^1),{\textsf {v} }^0, {\textsf {x} }^t \circ ({\textsf {x} }^0,{\textsf {x} }^1)\right) _{\sharp } \varvec{\sigma }\end{aligned}
is a bijection s.t. $$(t-s)L(\varvec{\sigma })=L(T_{s,t}(\varvec{\sigma }))$$ for every $$\varvec{\sigma }\in \Gamma _s(\Phi _s, \varvec{\mu })$$. This immediately gives that
\begin{aligned} (t-s) [\Phi _s,\varvec{\mu }]_{r,s} = \left[ \Phi _s, \mu _t\right] _{r}. \end{aligned}
In the same way we can deduce that
\begin{aligned} (s-t) [\Phi '_t,\varvec{\mu }]_{l,t} = \left[ \Phi '_t, \mu _s\right] _{r}. \end{aligned}
Thanks to the dissipativity assumption (4.2) of $${\varvec{\mathrm {F}}}$$, we get
\begin{aligned} (t-s)[\Phi _s,\varvec{\mu }]_{r,s} - (t-s)[\Phi '_t,\varvec{\mu }]_{l,t} = \left[ \Phi _s, \mu _t\right] _{r} +\left[ \Phi '_t, \mu _s\right] _{r} \le 0. \end{aligned}
3. (3)
Combining (1) and (2) we have that for every $$s,t\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$ with $$0< s<t <1$$ it holds
\begin{aligned} f_l(s) \le f_r(s) \le f_l(t) \le f_r(t), \end{aligned}
(4.11)
with $$f_r, f_l$$ as in (4.9). This implies that both $$f_l$$ and $$f_r$$ are increasing in $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})\cap (0,1)$$. Observe that, again combining (1) and (2), it also holds
\begin{aligned} f_r(0)&\le f_l(t) \le f_r(t),\\ f_l(t)&\le f_r(t) \le f_l(1) \end{aligned}
for every $$t \in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}){\setminus } \{0,1\}$$, and then $$f_r$$ is increasing in $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}){\setminus }\{1\}$$ and $$f_l$$ is increasing in $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}}){\setminus } \{0\}$$.
4. (4)
It is an immediate consequence of (4.11).
5. (5)
It is a straightforward consequence of (4).
$$\square$$
Thanks to Theorem 4.9(4), we have
\begin{aligned} \lim _{t\downarrow 0}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}&=\lim _{t\downarrow 0}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t},\\ \lim _{t\uparrow 1}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}&= \lim _{t\uparrow 1}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}, \end{aligned}
and those limits exist whenever the starting time $$t_0=0$$ and the final time $$t_1=1$$ are accumulation points of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, respectively. Due to the importance played by these objects in Sect. 5, we give the following definitions. These are intended to weaken the requirement for the operator’s domain $$\mathrm {D}({\varvec{\mathrm {F}}})$$ to be open or geodesically convex.
### Definition 4.10
Let $${\varvec{\mathrm {F}}}\subset \mathcal {P}_2(\mathsf {TX})$$, $$\mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$. We define the sets
\begin{aligned} \Gamma _o^{i}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}):={}&\left\{ \varvec{\mu }\in \Gamma _o(\mu _0,\mu _1): i\text { is an accumulation point of }\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})\right\} , i=0,1 \end{aligned}
(4.12)
\begin{aligned} \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}):={}&\Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\cap \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}). \end{aligned}
(4.13)
Notice that these sets depend on $${\varvec{\mathrm {F}}}$$ just through $$\mathrm {D}({\varvec{\mathrm {F}}})$$. In particular, if $$\mu _0,\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$ and $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is open or geodesically convex according to Definition 2.7 then $$\Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\ne \emptyset$$.
By the previous discussion, the next definition is well posed.
### Definition 4.11
Let us suppose that the MPVF $${\varvec{\mathrm {F}}}$$ satisfies (4.2), let $$\mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$.
\begin{aligned} \text {If }\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\text { we set} \quad [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}&:= \lim _{t\downarrow 0}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t} =\lim _{t\downarrow 0}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} \\ \text {If }\varvec{\mu }\in \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\text { we set} \quad [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}&:= \lim _{t\uparrow 1}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}= \lim _{t\uparrow 1}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t}. \end{aligned}
In the following statements, we make use of the objects introduced in Definition 4.10 in order to get refined dissipativity conditions involving the limiting pseudo-scalar products of Definition 4.11. These results will be useful in the sequel: in Proposition 4.17 they allow to get a dissipativity property of a suitable notion of extension $${\hat{{\varvec{\mathrm {F}}}}}$$ of $${\varvec{\mathrm {F}}}$$; in Sect. 5 (see in particular Lemma 5.3) they are relevant to study the properties of so-called $$\lambda$$-EVI solutions for a $$\lambda$$-dissipative MPVF $${\varvec{\mathrm {F}}}$$.
### Corollary 4.12
Let us keep the same notation of Theorem 4.9 and let $$s\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})\cap (0,1)$$ with $$\Phi \in {\varvec{\mathrm {F}}}[\mu _s]$$.
1. (1)
If $$\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$, we have that
\begin{aligned}{}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}&\le [\Phi ,\varvec{\mu }]_{l,s}+\lambda s W^2= [\Phi ,\varvec{\mu }]_{r,s}+\lambda s W^2; \end{aligned}
(4.14)
if moreover $$\Phi _0\in {\varvec{\mathrm {F}}}[\mu _0]$$ then
\begin{aligned} \left[ \Phi _0, \mu _1\right] _{r}\le [\Phi _0,\varvec{\mu }]_{r,0}\le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}. \end{aligned}
(4.15)
2. (2)
If $$\varvec{\mu }\in \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$, we have that
\begin{aligned}{}[\Phi ,\varvec{\mu }]_{l,s}-\lambda (1-s) W^2 = [\Phi ,\varvec{\mu }]_{r,s}-\lambda (1-s) W^2 \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}; \end{aligned}
if moreover $$\Phi _1\in {\varvec{\mathrm {F}}}[\mu _1]$$ then
\begin{aligned}{}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}\le [\Phi _1,\varvec{\mu }]_{l,1}\le -\left[ \Phi _1, \mu _0\right] _{r} \end{aligned}
(4.16)
3. (3)
In particular, for every $$\Phi _0\in {\varvec{\mathrm {F}}}[\mu _0]$$, $$\Phi _1\in {\varvec{\mathrm {F}}}[\mu _1]$$ and $$\varvec{\mu }\in \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$ we obtain
\begin{aligned}{}[\Phi _0,\Phi _1]_{r,\varvec{\mu }}\le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}- [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}\le \lambda W^2_2(\mu _0,\mu _1). \end{aligned}
(4.17)
(4.17) immediately yields the following property.
### Corollary 4.13
Suppose that a MPVF $${\varvec{\mathrm {F}}}$$ satisfies
\begin{aligned} \text {for every } \mu _0,\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}}) \text { the set } \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}) \text { of }(4.13)\text { is not empty} \end{aligned}
(4.18)
(e.g. if $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is open or geodesically convex), then $${\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.13) if and only if it satisfies (4.2).
### Proposition 4.14
Let $${\varvec{\mathrm {F}}}\subset \mathcal {P}_2(\mathsf {TX})$$ be a MPVF satisfying (4.2), let $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ and let $$\Phi \in \mathcal {P}_{2}(\mathsf {TX}|\mu _0)$$. Consider the following statements
1. (P1)
$$\left[ \Phi , \mu \right] _{r}+\left[ \Psi , \mu _0\right] _{r}\le \lambda W_2^2(\mu _0,\mu )$$ for every $$\Psi \in {\varvec{\mathrm {F}}}$$ with $$\mu ={\textsf {x} }_\sharp \Psi$$;
2. (P2)
for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exists $$\Psi \in {\varvec{\mathrm {F}}}[\mu ]$$ s.t. $$\left[ \Phi , \mu \right] _{r}+\left[ \Psi , \mu _0\right] _{r}\le \lambda W_2^2(\mu _0,\mu )$$;
3. (P3)
$$[\Phi ,\varvec{\mu }]_{r,0} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}$$ for every $$\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$;
4. (P4)
$$[\Phi ,\varvec{\mu }]_{r,0} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}$$ for every $$\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$;
5. (P5)
$$[\Phi ,\varvec{\mu }]_{r,0} \le \lambda W_2^2(\mu _0,\mu _1)+ [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}$$ for every $$\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$\varvec{\mu }\in \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$;
6. (P6)
$$[\Phi ,\varvec{\mu }]_{r,0} \le \lambda W_2^2(\mu _0,\mu _1)+ [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}$$ for every $$\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\varvec{\mu }\in \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$.
Then the following hold
1. (1)
(P1) $$\Rightarrow$$ (P2) $$\Rightarrow$$ (P3) $$\Rightarrow$$ (P4);
2. (2)
(P1) $$\Rightarrow$$ (P2) $$\Rightarrow$$ (P5) $$\Rightarrow$$ (P6);
3. (3)
if for every $$\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$ $$\Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\ne \emptyset$$, then (P4) $$\Rightarrow$$ (P1) (in particular, (P1), (P2), (P3), (P4) are equivalent);
4. (4)
if for every $$\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$ $$\Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\ne \emptyset$$, then (P6) $$\Rightarrow$$ (P1) (in particular, (P1), (P2), (P5), (P6) are equivalent).
### Proof
We first prove that (P2) $$\Rightarrow$$ (P3),(P5). Let us choose an arbitrary $$\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$; by the definition of $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t}$$ and arguing as in the proof of Theorem 4.9(2), for all $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$ and $$t\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$ there exists $$\Psi _t\in {\varvec{\mathrm {F}}}[\mu _{t}]$$ such that
\begin{aligned}{}[\Phi ,\varvec{\mu }]_{r,0}&=\frac{1}{t}\left[ \Phi , \mu _t\right] _{r}\\ {}&\le -\frac{1}{t} \left[ \Psi _t, \mu _0\right] _{r}+t\lambda W_2^2(\mu _0,\mu _1)\\ {}&= [\Psi _t,\varvec{\mu }]_{r,t}+t\lambda W_2^2(\mu _0,\mu _1)\\&\le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t} +t\lambda W_2^2(\mu _0,\mu _1), \end{aligned}
where we also used (3.29). If $$\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$, by passing to the limit as $$t\downarrow 0$$ we get (P3).
In the second case, assuming that $$\varvec{\mu }\in \Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$, we can pass to the limit as $$t\uparrow 1$$ and we get (P5).
We now prove item (3). Let $$\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\Psi \in {\varvec{\mathrm {F}}}[\mu _1]$$, $$\varvec{\mu }\in \Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$, $$s\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})\cap (0,1)$$, $$\Phi _s\in {\varvec{\mathrm {F}}}[\mu _s]$$, with $$\mu _s={\textsf {x} }^s_\sharp \varvec{\mu }$$. Assuming (P4) and using (4.15), (4.14), (3.29) and (4.2), we have
\begin{aligned} \left[ \Phi , \mu _1\right] _{r}\le [\Phi ,\varvec{\mu }]_{r,0} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}&\le [\Phi _s,\varvec{\mu }]_{r,s}+\lambda s W_2^2(\mu _0,\mu _1)\\&=\frac{1}{1-s}\left[ \Phi _s, \mu _1\right] _{r}+\lambda sW_2^2(\mu _0,\mu _1)\\ {}&\le -\frac{1}{1-s}\left[ \Psi , \mu _s\right] _{r}+\lambda (1+s)W_2^2(\mu _0,\mu _1). \end{aligned}
By Lemma 3.15, letting $$s\downarrow 0$$ we get (P1). Item (4) follows by (4.15), (4.16). $$\square$$
### 4.3 Extensions of dissipative MPVF
Let us briefly study a few simple properties about extensions of $$\lambda$$-dissipative MPVFs. The first one concerns the sequential closure in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ (the sequential closure may be smaller than the topological closure, but see Proposition 2.15): given $$A\subset \mathcal {P}_2(\mathsf {TX})$$, we will denote by $${\text {cl}}(A)$$ its sequential closure defined by
\begin{aligned} {\text {cl}}(A):=\left\{ \Phi \in \mathcal {P}_2(\mathsf {TX}): \exists \,(\Phi _n)_{n\in {\mathbb {N}}}\subset A:\Phi _n\rightarrow \Phi \ \text {in }\mathcal {P}_2^{sw}(\mathsf {TX})\right\} . \end{aligned}
### Proposition 4.15
If $${\varvec{\mathrm {F}}}$$ is a $$\lambda$$-dissipative MPVF according to (4.1), then its sequential closure $${\text {cl}}({\varvec{\mathrm {F}}})$$ is $$\lambda$$-dissipative as well according to (4.1).
### Proof
If $$\Phi ^i$$, $$i=0,1$$, belong to $${\text {cl}}({\varvec{\mathrm {F}}})$$, we can find sequences $$(\Phi ^i_n)_{n\in {\mathbb {N}}}\subset {\varvec{\mathrm {F}}}$$ such that $$\Phi ^i_n\rightarrow \Phi ^i$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ as $$n\rightarrow \infty$$, $$i=0,1$$. It is then sufficient to pass to the limit in the inequality
\begin{aligned} \left[ \Phi ^0_n, \Phi ^1_n\right] _{r}\le \lambda W_2^2(\mu ^0_n,\mu ^1_n),\quad \mu ^i_n={\textsf {x} }_\sharp \Phi ^i_n \end{aligned}
using the lower semicontinuity property (3.23) and the fact that convergence in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ yields $$\mu ^i_n\rightarrow {\textsf {x} }_\sharp \Phi ^i$$ in $$\mathcal {P}_2({\textsf {X} })$$ as $$n\rightarrow \infty$$. $$\square$$
A second result concerns the convexification of the sections of $${\varvec{\mathrm {F}}}$$. For every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ we set
\begin{aligned} {\text {co}}({\varvec{\mathrm {F}}})[\mu ]:=&\text {the convex hull of }{\varvec{\mathrm {F}}}[\mu ]= \Bigg \{\sum _k\alpha _k\Phi _k:\Phi _k\in {\varvec{\mathrm {F}}}[\mu ], \alpha _k\ge 0,\sum _k\alpha _k=1\Bigg \}, \end{aligned}
(4.19)
\begin{aligned} \overline{{\text {co}}}({\varvec{\mathrm {F}}})[\mu ]:=&{\text {cl}}({\text {co}}({\varvec{\mathrm {F}}})[\mu ]). \end{aligned}
(4.20)
Notice that if $${\varvec{\mathrm {F}}}[\mu ]$$ is bounded in $$\mathcal {P}_2(\mathsf {TX})$$ then $$\overline{{\text {co}}}({\varvec{\mathrm {F}}})[\mu ]$$ coincides with the closed convex hull of $${\varvec{\mathrm {F}}}[\mu ]$$.
### Proposition 4.16
If $${\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.1), then $${\text {co}}({\varvec{\mathrm {F}}})$$ and $$\overline{{\text {co}}}({\varvec{\mathrm {F}}})$$ are $$\lambda$$-dissipative as well according to (4.1).
### Proof
By Proposition 4.15 and noting that $$\overline{{\text {co}}}({\varvec{\mathrm {F}}})\subset {\text {cl}}({\text {co}}({\varvec{\mathrm {F}}}))$$, it is sufficient to prove that $${\text {co}}({\varvec{\mathrm {F}}})$$ is $$\lambda$$-dissipative. By Lemma 4.6 it is not restrictive to assume $$\lambda =0$$. Let $$\Phi ^i\in {\text {co}}({\varvec{\mathrm {F}}})[\mu _i]$$, $$i=0,1$$; there exist positive coefficients $$\alpha ^i_k$$, $$k=1,\ldots ,K$$, with $$\sum _k\alpha ^i_k=1$$, and elements $$\Phi ^i_k\in {\varvec{\mathrm {F}}}[\mu ^i]$$, $$i=0,1$$, such that $$\Phi ^i=\sum _{k=1}^K\alpha ^i_k\Phi ^i_k$$. Setting $$\beta _{h,k}:=\alpha ^0_h\alpha ^1_k$$, we can apply Lemma 3.17 and we obtain
\begin{aligned} \left[ \Phi ^0, \Phi ^1\right] _{r} = \Big [{\sum _{h,k}\beta _{h,k}\Phi ^0_h},{\sum _{h,k}\beta _{h,k}\Phi ^1_k} \Big ]_r \le \sum _{h,k}\beta _{h,k}\left[ \Phi ^0_h, \Phi ^1_k\right] _{r}\le 0. \square \end{aligned}
We recall that in the Hilbertian case (cf. e.g. [7]), a fundamental role is played by the notion of maximality for a dissipative operator $${\mathrm F}\subset {\textsf {H} }\times {\textsf {H} }$$. Indeed, this notion enables to extablish the existence and uniqueness of solutions of the corresponding evolution equation and to get crucial properties of the resolvent operator. Moreover, if $${\mathrm F}$$ is maximal, in order to prove that an element $$(x,v) \in {\textsf {H} }\times {\textsf {H} }$$ belongs to $${\mathrm F}$$ it is enough to verify that it satisfies the dissipativity inequality
\begin{aligned} \langle v-w, x-y \rangle \le 0 \quad \text { for every } (y,w) \in {\mathrm F}. \end{aligned}
(4.21)
For these reasons, if $${\mathrm F}$$ is not maximal it is important to study its maximal extension, whose elements (xv) must satisfy (4.21).
By analogy with the Hilbertian framework, it is interesting to study the properties of the extended MPVF defined by
\begin{aligned} {\hat{{\varvec{\mathrm {F}}}}}:=\left\{ \Phi \in \mathcal {P}_2(\mathsf {TX}):\begin{array}{l}\mu ={\textsf {x} }_\sharp \Phi \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\\ \left[ \Phi , \nu \right] _{r}+\left[ \Psi , \mu \right] _{r} \le \lambda W_2^2(\mu ,\nu ) \quad \forall \,\Psi \in {\varvec{\mathrm {F}}},\ \nu ={\textsf {x} }_\sharp \Psi \end{array} \right\} .\nonumber \\ \end{aligned}
(4.22)
This notion of extension $${{\hat{{\varvec{\mathrm {F}}}}}}$$ of a MPVF $${\varvec{\mathrm {F}}}$$ will be involved later in Sect. 5 dealing with differential inclusions in Wasserstein spaces, in particular in Theorem 5.4 and in Sect. 5.5.
It is obvious that $${\varvec{\mathrm {F}}}\subset {{\hat{{\varvec{\mathrm {F}}}}}}$$; if the domain of $${\varvec{\mathrm {F}}}$$ satisfies the geometric condition (4.24), the following result shows that $${{\hat{{\varvec{\mathrm {F}}}}}}$$ provides the maximal $$\lambda$$-dissipative extension of $${\varvec{\mathrm {F}}}$$.
### Proposition 4.17
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1).
1. (a)
If $${\varvec{\mathrm {F}}}'\supset {\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.1), with $$\mathrm {D}({\varvec{\mathrm {F}}}')\subset \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, then $${\varvec{\mathrm {F}}}'\subset {{\hat{{\varvec{\mathrm {F}}}}}}$$. In particular $$\overline{{\text {co}}}({\text {cl}}({\varvec{\mathrm {F}}}))\subset {{\hat{{\varvec{\mathrm {F}}}}}}$$.
2. (b)
$$\widehat{{{\text {cl}}({\varvec{\mathrm {F}}})}}={{\hat{{\varvec{\mathrm {F}}}}}}$$ and $$\widehat{{{\text {co}}({\varvec{\mathrm {F}}})}}={{\hat{{\varvec{\mathrm {F}}}}}}$$.
3. (c)
$${{\hat{{\varvec{\mathrm {F}}}}}}$$ is sequentially closed and $${\hat{{\varvec{\mathrm {F}}}[\mu ]}}$$ is convex for every $$\mu \in \mathrm {D}({{\hat{{\varvec{\mathrm {F}}}}}})$$.
4. (d)
If $$\mathrm {D}({\varvec{\mathrm {F}}})$$ satisfies (4.18), then the restriction of $${{\hat{{\varvec{\mathrm {F}}}}}}$$ to $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is $$\lambda$$-dissipative according to (4.1) and for every $$\mu _0,\mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$ it holds
\begin{aligned}{}[{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}= [{{\hat{{\varvec{\mathrm {F}}}}}},\varvec{\mu }]_{0+},\quad [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}= [{{\hat{{\varvec{\mathrm {F}}}}}},\varvec{\mu }]_{1-} \quad \text {for every } \varvec{\mu }\in \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}).\nonumber \\ \end{aligned}
(4.23)
5. (e)
If $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\ \mu _1\in \mathrm {D}({\varvec{\mathrm {F}}})$$ and $$\Gamma _o^{1}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\ne \emptyset$$ then
\begin{aligned} \Phi _i\in {\hat{{\varvec{\mathrm {F}}}[\mu _i]\quad \Rightarrow \quad }} \left[ \Phi _0, \Phi _1\right] _{r}\le \lambda W_2^2(\mu _0,\mu _1). \end{aligned}
6. (f)
If
\begin{aligned} \text {for every } \mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})} \text { the set } \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}}) \text { is not empty,} \end{aligned}
(4.24)
then $${{\hat{{\varvec{\mathrm {F}}}}}}$$ is $$\lambda$$-dissipative as well according to (4.1) and for every $$\mu _0,\mu _1\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ (4.23) holds.
### Proof
Claim (a) is obvious since every $$\lambda$$-dissipative extension $${\varvec{\mathrm {F}}}'$$ of $${\varvec{\mathrm {F}}}$$ in $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ satisfies $${\varvec{\mathrm {F}}}'\subset {{\hat{{\varvec{\mathrm {F}}}}}}$$.
(b) Let us prove that if $$\Phi \in {{\hat{{\varvec{\mathrm {F}}}}}}$$ then $$\Phi \in \widehat{{\text {cl}}({\varvec{\mathrm {F}}})}$$. If $$\Psi \in {\text {cl}}({\varvec{\mathrm {F}}})$$ we can find a sequence $$(\Psi _n)_{n\in {\mathbb {N}}}\subset {\varvec{\mathrm {F}}}$$ converging to $$\Psi$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ as $$n\rightarrow \infty$$. We can then pass to the limit in the inequalities
\begin{aligned} \left[ \Phi , \nu _n\right] _{r}+\left[ \Phi _n, \mu \right] _{r}\le \lambda W_2^2(\mu ,\nu _n),\quad \mu ={\textsf {x} }_\sharp \Phi ,\ \nu _n={\textsf {x} }_\sharp \Psi _n, \end{aligned}
using the lower semicontinuity results of Lemma 3.15. We conclude since $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}=\overline{\mathrm {D}({\text {cl}}({\varvec{\mathrm {F}}}))}$$.
In order to prove that $$\Phi \in {\hat{{\varvec{\mathrm {F}}}\ }} \Rightarrow \ \Phi \in {\widehat{{\text {co}}({\varvec{\mathrm {F}}})}}$$ we take $$\Psi =\sum \alpha _k\Psi _k\in {\text {co}}({\varvec{\mathrm {F}}})$$; for some $$\Psi _k\in {\varvec{\mathrm {F}}}[\nu ]$$, $$\nu ={\textsf {x} }_\sharp \Psi \in \mathrm {D}({\varvec{\mathrm {F}}})$$, and positive coefficients $$\alpha _k$$, $$k=1,\ldots ,K$$, with $$\sum _{k}\alpha _k=1$$. Taking a convex combination of the inequalities
\begin{aligned} \left[ \Phi , \nu \right] _{r}+\left[ \Psi _k, \mu \right] _{r}\le \lambda W_2^2(\mu ,\nu ),\quad \text {for every }k=1,\ldots ,K, \end{aligned}
and using Lemma 3.17 we obtain
\begin{aligned} \left[ \Phi , \nu \right] _{r}+\left[ \Psi , \mu \right] _{r}\le \sum _k\alpha _k\Big (\left[ \Phi , \nu \right] _{r}+\left[ \Psi _k, \mu \right] _{r}\Big ) \le \lambda W_2^2(\mu ,\nu ). \end{aligned}
The proof of claim (c) follows by a similar argument.
(d) Let $$\mu _i\in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\Phi _i\in {\hat{{\varvec{\mathrm {F}}}[\mu _i]}}$$, $$i=0,1$$, and $$\varvec{\mu }\in \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$. The implication (P1)$$\Rightarrow$$(P4) of Proposition 4.14 applied to $$\varvec{\mu }$$ and to $${\textsf {s} }_\sharp \varvec{\mu }$$, with $${\textsf {s} }$$ the reversion map in (3.26), yields
\begin{aligned}{}[\Phi _0,\varvec{\mu }]_{r,0}\le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+},\quad [\Phi _1,{\textsf {s} }_\sharp \varvec{\mu }]_{r,0}\le [{\varvec{\mathrm {F}}},{\textsf {s} }_\sharp \varvec{\mu }]_{0+}= -[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-} \end{aligned}
so that (4.17) yields
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}\le [\Phi _0,\varvec{\mu }]_{r,0}+ [\Phi _1,{\textsf {s} }_\sharp \varvec{\mu }]_{r,0} \le [{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}- [{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}\le \lambda W_2^2(\mu _0,\mu _1). \end{aligned}
In order to prove (4.23) we observe that $${\varvec{\mathrm {F}}}\subset \hat{\varvec{\mathrm {F}}}$$ so that, for every $$\varvec{\mu }\in \Gamma _o^{01}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})$$ and every $$t\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, we have $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{r,t} \le [{{\hat{{\varvec{\mathrm {F}}}}}},\varvec{\mu }]_{r,t}$$ and $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{l,t} \ge [{{\hat{{\varvec{\mathrm {F}}}}}},\varvec{\mu }]_{l,t}$$, hence (4.23) is a consequence of Definition 4.11 and Theorem 4.9.
The proof of claim (f) follows by the same argument.
In the case of claim (e), we use the implication (P1)$$\Rightarrow$$(P6) of Proposition 4.14 applied to $$\varvec{\mu }$$ and the implication (P1)$$\Rightarrow$$(P3) applied to $${\textsf {s} }_\sharp \varvec{\mu }$$, obtaining
\begin{aligned}{}[\Phi _0,\varvec{\mu }]_{r,0}\le \lambda W_2^2(\mu _0,\mu _1) +[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-},\quad [\Phi _1,{\textsf {s} }_\sharp \varvec{\mu }]_{r,0}\le [{\varvec{\mathrm {F}}},{\textsf {s} }_\sharp \varvec{\mu }]_{0+}= -[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-} \end{aligned}
and then
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}\le [\Phi _0,\varvec{\mu }]_{r,0}+ [\Phi _1,{\textsf {s} }_\sharp \varvec{\mu }]_{r,0} \le \lambda W_2^2(\mu _0,\mu _1). \end{aligned}
$$\square$$
## 5 Solutions to measure differential inclusions
### 5.1 Metric characterization and EVI
Let $${\mathcal {I}}$$ denote an arbitrary (bounded or unbounded) interval in $${\mathbb {R}}$$.
The aim of this section is to study a suitable notion of solution to the following differential inclusion in the $$L^2$$-Wasserstein space of probability measures
\begin{aligned} {\dot{\mu }}_t \in {\varvec{\mathrm {F}}}[\mu _t],\qquad t\in {\mathcal {I}}, \end{aligned}
(5.1)
driven by a MPVF $${\varvec{\mathrm {F}}}$$ as in Definition 4.1. In particular, we will address the usual Cauchy problem when (5.1) is supplemented by a given initial condition.
Measure Differential Inclusions have been introduced in [26] extending to the multi-valued framework the theory of Measure Differential Equations developed in [27]. In these papers, the author aims to describe the evolution of curves in the space of probability measures under the action of a so called probability vector field $${\varvec{\mathrm {F}}}$$ (see Definition 4.1 and Remark 4.2). However, as exploited also in [9], the definition of solution to (5.1) given in [9, 26, 27] is too weak and it does not enjoy uniqueness property which is recovered only at the level of the semigroup through an approximation procedure.
From the Wasserstein viewpoint, the simplest way to interpret (5.1) is to ask for a locally absolutely continuous curve $$\mu :{\mathcal {I}}\rightarrow \mathcal {P}_2({\textsf {X} })$$ to satisfy
\begin{aligned} ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp } \mu _t \in {{\varvec{\mathrm {F}}}}[\mu _t] \quad \text {for a.e. }t \in {\mathcal {I}}, \end{aligned}
(5.2)
where $${\varvec{v}}$$ is the Wasserstein metric velocity vector associated to $$\mu$$ (see Theorem 2.10). Even in the case of a regular PVF, however, (5.2) is too strong, since there is no reason why a given $${\varvec{\mathrm {F}}}[\mu _t]$$ should be associated to a vector field of the tangent space $${{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2({\textsf {X} })$$. Starting from (5.2), we thus introduce a weaker definition of solution to (5.1), modeled on the so-called EVI formulation for gradient flows, which will eventually suggest, as a natural formulation of (5.1), the relaxed version of (5.2) as a differential inclusion with respect to the extension $${{\hat{{\varvec{\mathrm {F}}}}}}$$ of $${\varvec{\mathrm {F}}}$$ introduced in (4.22).
We start from this simple remark: whenever $${\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.1), recalling Theorem 3.11 and Remark 4.5, one easily sees that every locally absolutely continuous solution according to the above definition (5.2) also satisfies the Evolution Variational Inequality ($$\lambda$$-EVI)
for every $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ and every $$\Phi \in {\varvec{\mathrm {F}}}[\nu ]$$, where $$\left[ \cdot , \cdot \right] _{r}$$ is the functional pairing in Definition 3.5 and the writing $$\mathscr {D}'\big ({\text {int}}\left( {\mathcal {I}}\right) \big )$$ means that the expression has to be understood in the distributional sense over $${\text {int}}\left( {\mathcal {I}}\right)$$ (in fact, ($$\lambda$$-EVI) holds a.e. in $${\mathcal {I}}$$). This provides a heuristic motivation for the following definition.
### Definition 5.1
($$\lambda$$-EVI solution) Let $${\varvec{\mathrm {F}}}$$ be a MPVF and let $$\lambda \in {\mathbb {R}}$$. We say that a continuous curve $$\mu : {\mathcal {I}}\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is a $$\lambda$$-EVI solution to (5.1) for the MPVF $${\varvec{\mathrm {F}}}$$ if ($$\lambda$$-EVI) holds for every $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ and every $$\Phi \in {\varvec{\mathrm {F}}}[\nu ]$$.
A $$\lambda$$-EVI solution $$\mu$$ is said to be a strict solution if $$\mu _t\in \mathrm {D}({\varvec{\mathrm {F}}})$$ for every $$t\in {\mathcal {I}}$$, $$t > \inf {\mathcal {I}}$$.
A $$\lambda$$-EVI solution $$\mu$$ is said to be a global solution if $$\sup {\mathcal {I}}=+\infty$$.
In Example 7.5 we will clarify the interest of imposing no more than continuity in the above definition.
Recall that the right upper and lower Dini derivatives of a function $$\zeta :{\mathcal {I}}\rightarrow {\mathbb {R}}$$ are defined for every $$t \in {\mathcal {I}}$$, $$t < \sup {\mathcal {I}}$$ by
\begin{aligned} {\frac{\mathrm d}{\mathrm dt}}^{+}\zeta (t):= \limsup _{h \downarrow 0} \frac{\zeta (t+h)-\zeta (t)}{h}, \qquad {\frac{\mathrm d}{\mathrm dt}}_{+}\zeta (t):= \liminf _{h \downarrow 0} \frac{\zeta (t+h)-\zeta (t)}{h}.\nonumber \\ \end{aligned}
(5.3)
### Remark 5.2
Arguing as in [22, Lemma A.1] and using the lower semicontinuity of the map $$t\mapsto \left[ \Phi , \mu _t\right] _{r}$$, the distributional inequality of ($$\lambda$$-EVI) can be equivalently reformulated in terms of the right upper or lower Dini derivatives of the squared distance function and requiring the condition to hold for every $$t\in {\text {int}}\left( {\mathcal {I}}\right)$$:
A further equivalent formulation [22, Theorem 3.3] involves the difference quotients: for every $$s,t\in {\mathcal {I}}$$, $$s<t$$
Finally, if $$\mu$$ is also locally absolutely continuous, then ($$\lambda$$-EVI$$_1$$) and ($$\lambda$$-EVI$$_2$$) are also equivalent to
\begin{aligned} \begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt} W_2^2(\mu _t,\nu )&\le \lambda W_2^2(\mu _t,\nu )- \left[ \Phi , \mu _t\right] _{r} \end{aligned} \quad \text {for a.e. } t\in {\mathcal {I}}\text { and every}\ \Phi \in {\varvec{\mathrm {F}}},\ \nu ={\textsf {x} }_\sharp \Phi . \end{aligned}
The following lemma discusses further properties of $$\lambda$$-EVI solutions. We refer respectively to (4.7), (4.12) and Definition 4.11 for the definitions of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, $$\Gamma _o^{i}({\cdot },{\cdot }|{\varvec{\mathrm {F}}})$$, with $$i=0,1$$, and for the definitions of $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}$$ and $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{1-}$$.
### Lemma 5.3
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$\mu : {\mathcal {I}}\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ be a continuous $$\lambda$$-EVI solution to (5.1). We have
\begin{aligned} \frac{1}{2}{\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t, \nu )&\le [{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{0+} \quad \text {for every } \nu \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\, t\in {\text {int}}\left( {\mathcal {I}}\right) ,\, \varvec{\mu }_t\in \Gamma _o^{0}({\mu _t},{\nu }|{\varvec{\mathrm {F}}}), \end{aligned}
(5.4a)
\begin{aligned} \frac{1}{2}{\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t, \nu )&\le \lambda W_2^2(\mu _t,\nu )+[{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{1-} \quad \nonumber \\&\quad \text {for every } \nu \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\, t\in {\text {int}}\left( {\mathcal {I}}\right) ,\, \varvec{\mu }_t\in \Gamma _o^{1}({\mu _t},{\nu }|{\varvec{\mathrm {F}}}). \end{aligned}
(5.4b)
If moreover $$\mu$$ is locally absolutely continuous with Wasserstein velocity field $${\varvec{v}}$$ satisfying (2.6) for every t in the subset $$A(\mu )\subset {\mathcal {I}}$$ of full Lebesgue measure, then
\begin{aligned} \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r}&\le \lambda W_2^2(\mu _t,\nu )- \left[ \Phi , \mu _t\right] _{r}&\text {if } t\in A(\mu ),\,\, \Phi \in {\varvec{\mathrm {F}}},\ \nu ={\textsf {x} }_\sharp \Phi , \end{aligned}
(5.5a)
\begin{aligned}{}[({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t,\varvec{\mu }_t]_{r,0}&\le [{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{0+}&\text {if } t\in A(\mu ), \nu \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\ \varvec{\mu }_t\in \Gamma _o^{0}({\mu _t},{\nu }|{\varvec{\mathrm {F}}}), \end{aligned}
(5.5b)
\begin{aligned}{}[({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t,\varvec{\mu }_t]_{r,0}&\le \lambda W_2^2(\mu _t,\nu )+ [{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{1-}&\text {if } t\in A(\mu ),\, \nu \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})},\ \varvec{\mu }_t\in \Gamma _o^{1}({\mu _t},{\nu }|{\varvec{\mathrm {F}}}). \end{aligned}
(5.5c)
### Proof
In order to check (5.5a) it is sufficient to combine (3.20) of Theorem 3.11 with ($$\lambda$$-EVI$$_1$$). (5.5b) and (5.5c) then follow applying Proposition 4.14. Let us now prove (5.4a): fix $$\nu \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ and $$t \in {\text {int}}\left( {\mathcal {I}}\right)$$. Take $$\varvec{\mu }_t \in \Gamma _o(\mu _t,\nu )$$ and define the constant speed geodesic $$\nu ^t:[0,1]\rightarrow \mathcal {P}_2({\textsf {X} })$$ by $$\nu _{s}^t: = ({\textsf {x} }^s)_{\sharp }\varvec{\mu }_t$$, thus in particular $$\nu _{0}^t=\mu _t$$ and $$\nu _{1}^t=\nu$$. Then by Lemma 2.11, for every $$s\in \mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})\cap (0,1)$$ and $$\Phi _s\in {\varvec{\mathrm {F}}}(\nu _s^t)$$ we have
\begin{aligned} \begin{aligned} \frac{1}{2}{\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t, \nu )&\le \frac{1}{2s} {\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t, \nu _s^t) \\ {}&\le - \frac{1}{s} \left[ \Phi _s, \mu _t\right] _{r} + \frac{\lambda }{s} W_2^2( \mu _t,\nu _{s}^t) \\ {}&\le [{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{r,s}+\lambda s W_2^2(\mu _t,\nu ), \end{aligned} \end{aligned}
where the second inequality comes from ($$\lambda$$-EVI$$_1$$). Taking $$\varvec{\mu }_t \in \Gamma _o^{0}({\mu _t},{\nu }|{\varvec{\mathrm {F}}})$$ and passing to the limit as $$s\downarrow 0$$ we get (5.4a). Analogously for (5.4b). $$\square$$
We can now give an interpretation of absolutely continuous $$\lambda$$-EVI solutions in terms of differential inclusions.
### Theorem 5.4
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$\mu : {\mathcal {I}}\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ be a locally absolutely continuous curve.
1. (1)
If $$\mu$$ satisfies the differential inclusion (5.2) driven by any $$\lambda$$-dissipative extension of $${\varvec{\mathrm {F}}}$$ in $${\mathrm {D}({\varvec{\mathrm {F}}})}$$, then $$\mu$$ is also a $$\lambda$$-EVI solution to (5.1) for $${\varvec{\mathrm {F}}}$$.
2. (2)
$$\mu$$ is a $$\lambda$$-EVI solution of (5.1) for $${\varvec{\mathrm {F}}}$$ if and only if
\begin{aligned} ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp } \mu _t \in {{{\hat{{\varvec{\mathrm {F}}}}}}}[\mu _t] \quad \text {for a.e. }t \in {\mathcal {I}}. \end{aligned}
(5.6)
3. (3)
If $$\mathrm {D}({\varvec{\mathrm {F}}})$$ satisfies (4.18) and $$\mu _t\in \mathrm {D}({\varvec{\mathrm {F}}})$$ for a.e. $$t\in {\mathcal {I}}$$, then the following properties are equivalent:
• $$\mu$$ is a $$\lambda$$-EVI solution to (5.1) for $${\varvec{\mathrm {F}}}$$.
• $$\mu$$ satisfies (5.5b).
• $$\mu$$ is a $$\lambda$$-EVI solution to (5.1) for the restriction of $${{\hat{{\varvec{\mathrm {F}}}}}}$$ to $$\mathrm {D}({\varvec{\mathrm {F}}})$$.
4. (4)
If $${\varvec{\mathrm {F}}}$$ satisfies (4.24) then $$\mu$$ is a $$\lambda$$-EVI solution to (5.1) for $${\varvec{\mathrm {F}}}$$ if and only if it is a $$\lambda$$-EVI solution to (5.1) for $${{\hat{{\varvec{\mathrm {F}}}}}}$$.
### Proof
(1) It is sufficient to apply Theorem 3.11 and the definition of $$\lambda$$-dissipativity.
The left-to-right implication $$\Rightarrow$$ of (2) follows by (5.5a) of Lemma 5.3 and the definition of $$\hat{\varvec{\mathrm {F}}}$$.
Conversely, if $$\mu$$ satisfies (5.6), $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\Phi \in {\varvec{\mathrm {F}}}[\nu ]$$, then Theorem 3.11 and the definition of $${{\hat{{\varvec{\mathrm {F}}}}}}$$ yield
\begin{aligned} \frac{1}{2}\frac{\,\mathrm d}{\,\mathrm dt} W_2^2(\mu _t, \nu )&= \left[ ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp } \mu _t, \nu \right] _{r}\le \lambda W_2^2(\mu _t, \nu ) - \left[ \Phi , \mu _t\right] _{r} \quad \text {a.e.~in }{\mathcal {I}}. \end{aligned}
Claim (3) is an immediate consequence of Lemma 5.3, Proposition 4.17(d) and Proposition 4.14.
Claim (4) is a consequence of Proposition 4.17(f) and the $$\lambda$$-dissipativity of $${{\hat{{\varvec{\mathrm {F}}}}}}$$. $$\square$$
The result stated in Theorem 5.4 suggests a compatibility between the notion of EVI solution for a dissipative MPVF and the notion of gradient flow for a convex functional in $$\mathcal {P}_2({\textsf {X} })$$. This correspondence is analysed in Sect. 7.1, where we consider the particular case where the MPVF is the opposite of the Fréchet subdifferential of a proper, lower semicontinuous and convex functional $$\mathcal {F}: \mathcal {P}_2({\textsf {X} }) \rightarrow (-\infty , + \infty ]$$ (see Proposition 7.2).
We derive a further useful a priori bound for $$\lambda$$-EVI solutions.
### Proposition 5.5
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$T \in (0, + \infty ]$$. Every $$\lambda$$-EVI solution $$\mu :[0,T) \rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ with initial datum $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ satisfies the a priori bound
\begin{aligned} W_2(\mu _t,\mu _0)\le 2 |{\varvec{\mathrm {F}}}|_2(\mu _0) \int _0^t\mathrm e^{\lambda s}\,\mathrm ds \end{aligned}
(5.7)
for all $$t \in [0, T)$$, where
\begin{aligned} |{\varvec{\mathrm {F}}}|_2(\mu ):={}\inf \left\{ |\Phi |_2:\Phi \in {\varvec{\mathrm {F}}}[\mu ]\right\} \end{aligned}
for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$.
### Proof
Let $$\Phi \in {\varvec{\mathrm {F}}}(\mu _0)$$. Then ($$\lambda$$-EVI) with $$\nu :=\mu _0$$ yields
\begin{aligned} {\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t,\mu _0)-2\lambda W_2^2(\mu _t,\mu _0)\le -2\left[ \Phi , \mu _t\right] _{r}\le 2|\Phi |_2\,W_2(\mu _t,\mu _0) \end{aligned}
for every $$t\in [0,T)$$. We can then apply the estimate of Lemma B.1 to obtain
\begin{aligned} \mathrm e^{-\lambda t} W_2(\mu _t,\mu _0)\le 2|\Phi |_2\int _0^t \mathrm e^{-\lambda s}\,\mathrm ds \end{aligned}
for all $$t\in [0,T)$$, which in turn yields (5.7). $$\square$$
We conclude this section with a result showing the robustness of the notion of $$\lambda$$-EVI solution.
### Proposition 5.6
If $$\mu _n: {\mathcal {I}}\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is a sequence of $$\lambda$$-EVI solutions locally uniformly converging to $$\mu$$ as $$n\rightarrow \infty$$, then $$\mu$$ is a $$\lambda$$-EVI solution.
### Proof
$$\mu$$ is a continuous curve defined in $${\mathcal {I}}$$ with values in $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$. Using pointwise convergence, the lower semicontinuity of $$\mu \mapsto \left[ \Phi , \mu \right] _{r}$$ of Lemma 3.15, and Fatou’s Lemma, it is easy to pass to the limit in the equivalent characterization ($$\lambda$$-EVI$$_3$$) of $$\lambda$$-EVI solutions, written for $$\mu _n$$. $$\square$$
### 5.2 Local existence of $$\lambda$$-EVI solutions by the Explicit Euler Scheme
In order to prove the existence of a $$\lambda$$-EVI solution to (5.1), our strategy is to employ an approximation argument through an Explicit Euler scheme as it occurs for ODEs.
In the following $$\left\lfloor \cdot \right\rfloor$$ and $$\left\lceil \cdot \right\rceil$$ denote the floor and the ceiling functions respectively, i.e.
\begin{aligned} \left\lfloor t \right\rfloor :=\max \left\{ m\in {\mathbb {Z}}\mid m\le t\right\} \quad \text {and}\quad \left\lceil t \right\rceil :=\min \left\{ m\in {\mathbb {Z}}\mid m\ge t\right\} , \end{aligned}
(5.8)
for any $$t\in {\mathbb {R}}$$.
### Definition 5.7
(Explicit Euler Scheme) Let $${\varvec{\mathrm {F}}}$$ be a MPVF and suppose we are given a step size $$\tau >0$$, an initial datum $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$, a bounded interval [0, T], corresponding to the final step $${\mathrm N(T,\tau )}:=\left\lceil T/\tau \right\rceil ,$$ and a stability bound $$L>0$$. A sequence $$(M^n_\tau ,\Phi _\tau ^n)_{0\le n\le {\mathrm N(T,\tau )}}\subset \mathrm {D}({\varvec{\mathrm {F}}})\times {\varvec{\mathrm {F}}}$$ is a L-stable solution to the Explicit Euler Scheme in [0, T] starting from $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ if
\begin{aligned} \left\{ \begin{aligned} M_{\tau }^0&= \mu _0 ,\\ \Phi _\tau ^n&\in {\varvec{\mathrm {F}}}[M_{\tau }^n],\ |\Phi _\tau ^n|_2 \le L&0\le n<{\mathrm N(T,\tau )},\\ M_{\tau }^{n}&= (\textsf {exp} ^{\tau })_{\sharp } \Phi _\tau ^{n-1}&1\le n\le {\mathrm N(T,\tau )}. \end{aligned} \right. \end{aligned}
(EE)
We define the following two different interpolations of the sequence $$(M^n_\tau ,\Phi _\tau ^n)$$:
• the affine interpolation:
\begin{aligned} M_{\tau }(t) := (\textsf {exp} ^{t-n\tau })_{\sharp } \Phi _\tau ^n \quad \text { if } t \in [n\tau , (n+1)\tau ] \text { for some } n \in {\mathbb {N}},\ 0\le n<{\mathrm N(T,\tau )},\nonumber \\ \end{aligned}
(5.9)
• the piecewise constant interpolation:
\begin{aligned} \bar{M}_{\tau }(t)&:= M^{\left\lfloor t/\tau \right\rfloor }_{\tau }, \quad t\in [0,T], \end{aligned}
(5.10)
\begin{aligned} {{\varvec{F}}}_{\tau }(t)&:= \Phi _\tau ^{\left\lfloor t/\tau \right\rfloor }, \quad t\in [0,T]. \end{aligned}
(5.11)
We define the following (possibly empty) sets
\begin{aligned} {\mathscr {E}}(\mu _0,\tau ,T,L):= & {} \left\{ (M_\tau ,{{\varvec{F}}}_\tau )\mid M_\tau , {{\varvec{F}}}_\tau \text { are the curves given by }(5.9),(5.11)\text { respectively}\right\} \nonumber \\ {\mathscr {M}}(\mu _0,\tau ,T,L):= & {} \left\{ M_\tau \mid M_\tau \text { is the curve given by }(5.9)\right\} . \end{aligned}
(5.12)
### Remark 5.8
We immediately notice that, if $$(M_\tau ,{{\varvec{F}}}_\tau )\in \mathscr {E}(\mu _0,\tau ,T,L)$$ and $$\bar{M}_{\tau }(\cdot )$$ is as in (5.10), then the following holds for any $$0\le s\le t\le T$$:
1. (1)
the affine interpolation can be trivially written as
\begin{aligned} M_\tau (t)=\left( \textsf {exp} ^{t-\left\lfloor t/\tau \right\rfloor \tau }\right) _{\sharp }\left( {{\varvec{F}}}_\tau (t)\right) ; \end{aligned}
2. (2)
$$M_\tau$$ satisfies the uniform Lipschitz bound
\begin{aligned} W_2(M_\tau (t),M_\tau (s))\le L|t-s|; \end{aligned}
(5.13)
3. (3)
we have the following estimate
\begin{aligned} W_2(\bar{M}_{\tau }(t), M_{\tau }(t))= W_2\left( M_\tau \left( \left\lfloor \frac{t}{\tau } \right\rfloor \tau \right) ,M_\tau (t)\right) \le L\tau . \end{aligned}
(5.14)
The estimate (5.14) shows that the stability and convergence results stated for the affine interpolation (see Theorem 5.9) can be easily adapted to the piecewise constant one.
Notice that, since in general $${\varvec{\mathrm {F}}}[\mu ]$$ is not reduced to a singleton, the sets $${\mathscr {E}}(\mu _0,\tau ,T,L)$$ and $$\mathscr {M}(\mu _0,\tau ,T,L)$$ may contain more than one element (or may be empty). Stable solutions to the Explicit Euler scheme generated by a $$\lambda$$-dissipative MPVF exhibit a nice behaviour, which is clarified by the following important result, which will be proved in Sect. 6 (see Proposition 6.3 and Theorems 6.46.5 and 6.7), with explicit estimates of the error constants $$A(\delta )$$. We stress that in the next statement $$A(\delta )$$ solely depend on $$\delta$$ (in particular, it is independent of $$\lambda , L, T,\tau ,\eta , M_\tau ,M_\eta$$).
### Theorem 5.9
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1).
1. (1)
For every $$\mu _0,\mu _0' \in \mathrm {D}({\varvec{\mathrm {F}}})$$, every $$M_\tau \in \mathscr {M}(\mu _0,\tau ,T,L)$$, $$M_\tau '\in {\mathscr {M}}(\mu _0',\tau ,T,L)$$ with $$\tau \lambda _+\le 2$$ we have
\begin{aligned} W_2(M_\tau (t),M_\tau '(t)) \le \mathrm e^{\lambda t}W_2(\mu _0,\mu _0') +8L\sqrt{t\tau }\,\Big (1+|\lambda |\sqrt{t\tau }\Big )\mathrm e^{\lambda _+ t} \end{aligned}
(5.15)
for every $$t\in [0,T]$$.
2. (2)
For every $$\delta >1$$ there exists a constant $$A(\delta )$$ such that if $$M_\tau \in \mathscr {M}(M^0_\tau ,\tau ,T,L)$$ and $$M_\eta \in {\mathscr {M}}(M^0_\eta ,\eta ,T,L)$$ with $$\lambda _+(\tau +\eta )\le 1$$ then
\begin{aligned} W_2(M_\tau (t),M_\eta (t))\le \Big (\delta \, W_2(M^0_\tau ,M^0_\eta )+ A(\delta ) L\sqrt{(\tau +\eta )(t+\tau +\eta )}\Big )\mathrm e^{\lambda _+\, t} \end{aligned}
for every $$t\in [0,T]$$.
3. (3)
For every $$\delta >1$$ there exists a constant $$A(\delta )$$ such that if $$\mu :[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is a $$\lambda$$-EVI solution and $$M_\tau \in {\mathscr {M}}(M^0_\tau ,\tau ,T,L)$$ then
\begin{aligned} W_2(\mu _t, M_{\tau }(t))\le \Big (\delta \, W_2(\mu _0,M^0_\tau )+ A(\delta )L\sqrt{\tau (t+\tau )}\Big ) \mathrm e^{\lambda _+ t} \end{aligned}
(5.16)
for every $$t\in [0,T]$$.
4. (4)
If $$n\mapsto \tau (n)$$ is a vanishing sequence of time steps, $$(\mu _{0,n})_{n\in {\mathbb {N}}}$$ is a sequence in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ converging to $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ in $$\mathcal {P}_2({\textsf {X} })$$ and $$M_n\in {\mathscr {M}}(\mu _{0,n},\tau (n),T,L)$$, then $$M_n$$ is uniformly converging to a Lipschitz continuous limit curve $$\mu :[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ which is a $$\lambda$$-EVI solution starting from $$\mu _0$$.
### Definition 5.10
(Local and global solvability of (EE)) We say that the Explicit Euler Scheme (EE) associated to a MPVF $${\varvec{\mathrm {F}}}$$ is locally solvable at $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ if there exist strictly positive constants $$\varvec{\tau },T,L$$ such that $${\mathscr {E}}(\mu _0,\tau ,T,L)$$ is not empty for every $$\tau \in (0,\varvec{\tau })$$. We say that (EE) is globally solvable at $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ if for every $$T>0$$ there exist strictly positive constants $$\varvec{\tau },L$$ such that $${\mathscr {E}}(\mu _0,\tau ,T,L)$$ is not empty for every $$\tau \in (0,\varvec{\tau })$$.
If we assume that the Explicit Euler scheme is locally solvable, Theorem 5.9 provides a crucial tool to obtain local existence and uniqueness of $$\lambda$$-EVI solutions.
Let us now state the main existence result for $$\lambda$$-EVI solutions. Given $$T \in (0, + \infty ]$$ and $$\mu :[0,T)\rightarrow \mathcal {P}_2({\textsf {X} })$$ we denote by $$|{{\dot{\mu }}}_t|_+$$ the right upper metric derivative
\begin{aligned} |{{\dot{\mu }}}_t|_+:= \limsup _{h\downarrow 0}\frac{W_2(\mu _{t+h},\mu _t)}{h}. \end{aligned}
### Theorem 5.11
(Local existence and uniqueness) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1).
1. (a)
If the Explicit Euler Scheme is locally solvable at $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$, then there exists $$T>0$$ and a unique Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ starting from $$\mu _0$$, satisfying
\begin{aligned} t\mapsto \mathrm e^{-\lambda t}|{{\dot{\mu }}}_t|_+ \quad \text {is decreasing in }[0,T). \end{aligned}
(5.17)
If $$\mu ':[0,T']\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is any other $$\lambda$$-EVI solution starting from $$\mu _0$$ then $$\mu _t=\mu '_t$$ if $$0\le t\le \min \{T, T'\}$$.
2. (b)
If the Explicit Euler Scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ and
\begin{aligned}&\text {for any local } \lambda -\mathrm{EVI} \text { solution } \mu \text { starting from } \mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})\nonumber \\&\quad \text {there exists }\delta >0:\quad t\in [0,\delta ]\quad \Rightarrow \quad \mu _t\in \mathrm {D}({\varvec{\mathrm {F}}}), \end{aligned}
(5.18)
then for every $$\mu _0 \in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exist a unique maximal time $$T\in (0,\infty ]$$ and a unique strict locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T)\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ starting from $$\mu _0$$, which satisfies (5.17) and
\begin{aligned} T<\infty \quad \Rightarrow \quad \lim _{t\uparrow T}\mu _t\not \in \mathrm {D}({\varvec{\mathrm {F}}}). \end{aligned}
(5.19)
Any other $$\lambda$$-EVI solution $$\mu ':[0,T')\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ starting from $$\mu _0$$ coincides with $$\mu$$ in $$[0,\min \{T,T'\})$$.
### Proof
(a) Let $$\varvec{\tau }, T,L$$ positive constants such that $$\mathscr {E}(\mu _0,\tau ,T,L)$$ is not empty for every $$\tau \in (0,\varvec{\tau })$$. Thanks to Theorem 5.9(2), the family $$M_\tau \in {\mathscr {E}}(\mu _0,\tau ,T,L)$$ satisfies the Cauchy condition in $$\mathrm C([0,T];\mathcal {P}_2({\textsf {X} }))$$ so that there exists a unique limit curve
\begin{aligned} \mu =\lim _{\tau \downarrow 0}M_\tau \end{aligned}
which is also Lipschitz in time, thanks to the a-priori bound (5.13). Theorem 5.9(4) shows that $$\mu$$ is a $$\lambda$$-EVI solution starting from $$\mu _0$$ and the estimate (5.16) of Theorem 5.9(3) shows that any other $$\lambda$$-EVI solution in an interval $$[0,T']$$ starting from $$\mu _0$$ should coincide with $$\mu$$ in the interval $$[0,\min \{T',T\}]$$.
Let us now check (5.17): we fix st such that $$0\le s<t<T$$ and $$h\in (0,T-t)$$, and we set
\begin{aligned} s_\tau :=\tau \left\lfloor s/\tau \right\rfloor \quad \text {and}\quad h_\tau :=\tau \left\lfloor h/\tau \right\rfloor . \end{aligned}
The curves
\begin{aligned} r\mapsto M_\tau (s_\tau +r)\quad \text {and}\quad r\mapsto M_\tau (s_\tau +h_\tau +r) \end{aligned}
belong to $$\mathscr {M}(M_\tau (s_\tau ),\tau ,t-s,L)$$ and $$\mathscr {M}(M_\tau (s_\tau +h_\tau ),\tau ,t-s,L)$$, so that (5.15) yields
\begin{aligned} W_2(M_\tau (s_\tau +t-s),M_\tau (s_\tau +h_\tau +(t-s)))\le \mathrm e^{\lambda (t-s)} W_2(M_\tau (s_\tau ),M_\tau (s_\tau +h_\tau ))+B\sqrt{\tau }, \end{aligned}
for $$B=B(\lambda , L, \varvec{\tau },T)$$. Passing to the limit as $$\tau \downarrow 0$$ we get
\begin{aligned} W_2(\mu _t,\mu _{t+h})\le \mathrm e^{\lambda (t-s)} W_2(\mu _s,\mu _{s+h}). \end{aligned}
Dividing by h and passing to the limit as $$h\downarrow 0$$ we get (5.17).
(b) Let us call $${\mathcal {S}}$$ the collection of $$\lambda$$-EVI solutions $$\mu :[0,S)\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ starting from $$\mu _0$$ with values in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ and defined in some interval [0, S), $$S=S(\mu )$$. Thanks to (5.18) and the previous claim the set $${\mathcal {S}}$$ is not empty.
It is also easy to check that two curves $$\mu ',\mu ''\in {\mathcal {S}}$$ coincide in the common domain [0, S) with
\begin{aligned} S:=\min \left\{ S(\mu '), S(\mu '')\right\} . \end{aligned}
Indeed, the set
\begin{aligned} \left\{ t\in [0,S):\mu '_r=\mu ''_r \text { if }0\le r\le t\right\} \end{aligned}
contains $$t=0$$, is closed since $$\mu ',\mu ''$$ are continuous, and it is also open since, if $$\mu '=\mu ''$$ in [0, t], then the previous claim and the fact that $$\mu '_t=\mu ''_t\in \mathrm {D}({\varvec{\mathrm {F}}})$$ show that $$\mu '=\mu ''$$ also in a right neighborhood of t. Since [0, S) is connected, we conclude that $$\mu '=\mu ''$$ in [0, S).
We can thus define
\begin{aligned} T:=\sup \left\{ S(\mu ):\mu \in {\mathcal {S}}\right\} , \end{aligned}
obtaining that there exists a unique $$\lambda$$-EVI solution $$\mu$$ starting from $$\mu _0$$ and defined in [0, T) with values in $$\mathrm {D}({\varvec{\mathrm {F}}})$$.
If $$T<\infty$$, since $$\mu$$ is Lipschitz in [0, T) thanks to (5.17), we know that there exists the limit
\begin{aligned} {\bar{\mu }}:=\lim _{t\uparrow T}\mu _t \end{aligned}
in $$\mathcal {P}_2({\textsf {X} })$$. If $${\bar{\mu }}\in \mathrm {D}({\varvec{\mathrm {F}}})$$ we can extend $$\mu$$ to a $$\lambda$$-EVI solution with values in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ and defined in an interval $$[0,T')$$ with $$T'>T$$, which contradicts the maximality of T. $$\square$$
Recall that a set A in a metric space X is locally closed if every point of A has a neighborhood U such that $$A\cap U = \bar{A} \cap U$$. Equivalently, A is the intersection of an open and a closed subset of X. In particular, open or closed sets are locally closed.
We refer to Definition 5.1 for the notion of strict EVI solutions, used in the following.
### Corollary 5.12
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) for which the Explicit Euler Scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$. If $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is locally closed then for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exists a unique maximal strict and locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T)\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$, $$T \in (0, + \infty ]$$, satisfying (5.19).
Let us briefly discuss the question of local solvability of the Explicit Euler scheme. The main constraints of the Explicit Euler construction relies on the a priori stability bound and in the condition $$M_\tau ^n\in \mathrm {D}({\varvec{\mathrm {F}}})$$ for every step $$0\le n\le {\mathrm N(T,\tau )}$$. This constraint is feasible if at each measure $$M^n_\tau$$, $$0\le n<{\mathrm N(T,\tau )}$$, the set $${{\,\mathrm{Adm}\,}}_{\tau ,L}(M^n_\tau )$$ defined by
\begin{aligned} {{\,\mathrm{Adm}\,}}_{\tau ,L}(\mu ):=\left\{ \Phi \in {\varvec{\mathrm {F}}}[\mu ]: |\Phi |_2 \le L \quad \text {and}\quad \textsf {exp} _{\sharp }^{\tau } \Phi \in \mathrm {D}({\varvec{\mathrm {F}}}) \right\} \end{aligned}
is not empty. If $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is open and $${\varvec{\mathrm {F}}}$$ is locally bounded, then it is easy to check that the Explicit Euler scheme is locally solvable (see Lemma 5.13). We will adopt the following notation:
\begin{aligned} |{\varvec{\mathrm {F}}}|_2(\mu ):={}&\inf \left\{ |\Phi |_2:\Phi \in {\varvec{\mathrm {F}}}[\mu ]\right\} \quad \text {for every }\mu \in \mathrm {D}({\varvec{\mathrm {F}}}), \end{aligned}
(5.20)
and we will also introduce the upper semicontinuous envelope $$|{\varvec{\mathrm {F}}}|_{2\star }$$ of the function $$|{\varvec{\mathrm {F}}}|_2$$: i.e.
\begin{aligned} \begin{aligned} |{\varvec{\mathrm {F}}}|_{2\star }(\mu ):={}&\inf _{\delta >0}\sup \left\{ |{\varvec{\mathrm {F}}}|_2(\nu ): \nu \in \mathrm {D}({\varvec{\mathrm {F}}}),\ W_2(\nu ,\mu )\le \delta \right\} \\={}&\sup \left\{ \limsup _{k\rightarrow \infty }|{\varvec{\mathrm {F}}}|_2(\mu _k):\mu _k\in \mathrm {D}({\varvec{\mathrm {F}}}),\ \mu _k\rightarrow \mu \text { in }\mathcal {P}_2({\textsf {X} })\right\} . \end{aligned} \end{aligned}
### Lemma 5.13
If $${\varvec{\mathrm {F}}}$$ is a $$\lambda$$-dissipative MPVF according to (4.1), $$\mu _0\in \mathrm {Int}(\mathrm {D}({\varvec{\mathrm {F}}}))$$ and $${\varvec{\mathrm {F}}}$$ is bounded in a neighborhood of $$\mu _0$$, i.e. there exists $$\varrho >0$$ such that $$|{\varvec{\mathrm {F}}}|_2$$ is bounded in $$\mathrm B(\mu _0,\varrho )$$, then the Explicit Euler scheme is locally solvable at $$\mu _0$$ and the locally Lipschitz continuous solution $$\mu$$ given by Theorem 5.11(a) satisfies
\begin{aligned} |{{\dot{\mu }}}_t|_+ \le e^{\lambda t} |{\varvec{\mathrm {F}}}|_{2\star }(\mu _0) \quad \text {for all } \, t \in [0,T). \end{aligned}
(5.21)
In particular, if $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is open and $${\varvec{\mathrm {F}}}$$ is locally bounded, for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exists a unique maximal locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T)\rightarrow \mathcal {P}_2({\textsf {X} })$$ satisfying (5.19) and (5.21).
### Proof
Let $$\mu _0\in \mathrm {Int}(\mathrm {D}({\varvec{\mathrm {F}}}))$$ and let $$\varrho , L>0$$ so that $$|{\varvec{\mathrm {F}}}|_2(\mu )<L$$ for every $$\mu \in \mathrm B(\mu _0,\varrho )$$. We set
\begin{aligned} T:=\varrho /(2L)\quad \text {and}\quad \varvec{\tau }:=\min \{T, 1\} \end{aligned}
and we perform a simple induction argument to prove that
\begin{aligned} W_2(M^n_\tau ,\mu _0)\le L n\tau <\varrho \end{aligned}
if $$n\le {\mathrm N(T,\tau )}$$, so that we can always find an element $$\Phi ^n_\tau \in {{\,\mathrm{Adm}\,}}_{\tau ,L}(M^n_\tau )$$. In fact, if $$W_2(M^n_\tau ,\mu _0)<Ln\tau$$ and $$n<{\mathrm N(T,\tau )}$$ then
\begin{aligned} W_2(M^{n+1}_\tau ,\mu _0)\le W_2(M^{n+1}_\tau ,M^n_\tau )+ W_2(M^n_\tau ,\mu _0) \le L(n+1)\tau . \end{aligned}
The property in (5.17) shows that $$|{{\dot{\mu }}}_t|_+\le L\mathrm e^{\lambda t}$$ for every $$L>|{\varvec{\mathrm {F}}}|_{2\star }(\mu _0)$$, so that we obtain (5.21). $$\square$$
More refined estimates will be discussed in the next sections. Here we will show another example, tailored to the case of measures with bounded support.
### Proposition 5.14
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Assume that $$\mathrm {D}({\varvec{\mathrm {F}}})\subset \mathcal {P}_\mathrm{b}({\textsf {X} })$$ and for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exist $$\varrho >0$$, $$L>0$$ such that, for every $$\mu \in \mathcal {P}_\mathrm{b}({\textsf {X} })$$ with $${{\,\mathrm{supp}\,}}(\mu )\subset {{\,\mathrm{supp}\,}}(\mu _0)+\mathrm B_{\textsf {X} }(\varrho )$$, there exists $$\Phi \in {\varvec{\mathrm {F}}}[\mu ]$$ such that
\begin{aligned} {{\,\mathrm{supp}\,}}({\textsf {v} }_\sharp \Phi )\subset \mathrm B_{\textsf {X} }(L). \end{aligned}
Then for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exists $$T\in (0,+\infty ]$$ and a unique maximal strict and locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T)\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ satisfying (5.19).
### Proof
Arguing as in the proof of Lemma 5.13, it is easy to check that setting $$T:=\varrho /4L$$, $$\varvec{\tau } = \min \{T, 1\}$$, we can find a discrete solution $$(M_\tau ,{{\varvec{F}}}_\tau )\in \mathscr {E}(\mu _0,\tau ,T,L)$$ satisfying the more restrictive condition
\begin{aligned} {{\,\mathrm{supp}\,}}(M^n_\tau )\subset {{\,\mathrm{supp}\,}}(\mu _0)+\mathrm B_{\textsf {X} }(Ln\tau )\subset {{\,\mathrm{supp}\,}}(\mu _0)+\mathrm B_{\textsf {X} }(\varrho /2),\quad \text {and}\quad {{\,\mathrm{supp}\,}}({\textsf {v} }_\sharp \Phi ^n_\tau )\subset \mathrm B_{\textsf {X} }(L). \end{aligned}
So that the Explicit Euler scheme is locally solvable and $$M_\tau$$ satisfies the uniform bound
\begin{aligned} {{\,\mathrm{supp}\,}}(M_\tau (t))\subset {{\,\mathrm{supp}\,}}(\mu _0)+\mathrm B_{\textsf {X} }(\varrho /2) \end{aligned}
(5.22)
for every $$t\in [0,T]$$. Theorem 5.11 then yields the existence of a local solution, and Theorem 5.9(3) shows that the local solution satisfies the same bound (5.22) on the support, so that (5.18) holds. $$\square$$
### 5.3 Stability and uniqueness
In the following theorem we prove a stability result for $$\lambda$$-EVI solutions of (5.1), as it occurs in the classical Hilbert case. We distinguish three cases: the first one assumes that the Explicit Euler scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$.
### Theorem 5.15
(Uniqueness and Stability) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) such that the Explicit Euler scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$, and let $$\mu ^1, \mu ^2: [0,T) \rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$T\in (0, + \infty ]$$, be $$\lambda$$-EVI solutions to (5.1). If $$\mu ^1$$ is strict, then
\begin{aligned} W_2(\mu ^1_t,\mu ^2_t)\le W_2(\mu ^1_0,\mu ^2_0)\,\mathrm e^{\lambda _+ \,t}\quad \text { for every }t\in [0,T). \end{aligned}
(5.23)
In particular, if $$\mu ^1_0=\mu ^2_0$$ then $$\mu ^1\equiv \mu ^2$$ in [0, T).
If $$\mu ^1,\mu ^2$$ are both strict, then
\begin{aligned} W_2(\mu _t^1, \mu _t^2) \le W_2(\mu ^1_0, \mu ^2_0)\, \mathrm e^{\lambda t}\quad \text { for every }t\in [0,T). \end{aligned}
(5.24)
### Proof
In order to prove (5.23), let us fix $$t\in (0,T)$$. Since the Explicit Euler scheme is locally solvable and $$\mu ^1_t\in \mathrm {D}({\varvec{\mathrm {F}}})$$, there exist $$\varvec{\tau },\delta ,L$$ such that $${\mathscr {M}}(\mu ^1_t,\tau ,\delta ,L)$$ is not empty for every $$\tau \in (0,\varvec{\tau })$$. If $$M^1_\tau \in \mathscr {M}(\mu ^1_t,\tau ,\delta ,L)$$, then (5.16) yields
\begin{aligned} \begin{aligned} W_2(\mu ^1_{t+h},\mu ^2_{t+h})&\le W_2(M^1_{\tau }(h),\mu ^2_{t+h})+ W_2(M^1_{\tau }(h),\mu ^1_{t+h}) \\ {}&\le \delta \,W_2(\mu ^1_t,\mu ^2_t)\mathrm e^{\lambda _+ h}+ B\sqrt{\tau }\quad \text {if }0\le h\le \delta , \end{aligned} \end{aligned}
for $$B=B(\lambda , L, \varvec{\tau },\delta )$$ Passing to the limit as $$\tau \downarrow 0$$ we obtain
\begin{aligned} W_2(\mu ^1_{t+h},\mu ^2_{t+h})\le \delta \, W_2(\mu ^1_t,\mu ^2_t) \mathrm e^{\lambda _+ h} \end{aligned}
and a further limit as $$\delta \downarrow 1$$ yields
\begin{aligned} W_2(\mu ^1_{t+h},\mu ^2_{t+h})\le W_2(\mu ^1_t,\mu ^2_t) \mathrm e^{\lambda _+ h} \end{aligned}
for every $$h\in [0,\delta ]$$, which implies that the map $$t\mapsto \mathrm e^{-\lambda _+ t}W_2(\mu ^1_t,\mu ^2_t)$$ is decreasing in $$[t,t+\delta ]$$. Since t is arbitrary, we obtain (5.23).
In order to prove the estimate (5.24) (which is better than (5.23) when $$\lambda <0$$), we argue in a similar way, using (5.15).
As before, for a given $$t\in (0,T)$$, since the Explicit Euler scheme is locally solvable and $$\mu ^1_t,\mu ^2_t\in \mathrm {D}({\varvec{\mathrm {F}}})$$, there exist $$\varvec{\tau },\delta ,L$$ such that $$\mathscr {M}(\mu ^1_t,\tau ,\delta ,L)$$ and $${\mathscr {M}}(\mu ^2_t,\tau ,\delta ,L)$$ are not empty for every $$\tau \in (0,\varvec{\tau })$$. If $$M^i_\tau \in {\mathscr {M}}(\mu ^i_t,\tau ,\delta ,L)$$, for $$i=1,2$$, (5.15) and (5.16) then yield
\begin{aligned} \begin{aligned} W_2(\mu ^1_{t+h},\mu ^2_{t+h})&\le W_2(\mu ^1_{t+h},M^1_\tau (h))+ W_2(M^1_{\tau }(h),M^2_\tau (h))+ W_2(\mu ^2_{t+h},M^2_\tau (h)) \\ {}&\le \mathrm e^{\lambda h} W_2(\mu ^1_t,\mu ^2_t)+B\sqrt{\tau }\end{aligned} \end{aligned}
if $$0\le h \le \delta$$, with $$B=B(\lambda , L, \varvec{\tau },\delta )$$. Passing to the limit as $$\tau \downarrow 0$$ we obtain
\begin{aligned} W_2(\mu ^1_{t+h},\mu ^2_{t+h})\le \mathrm e^{\lambda h} W_2(\mu ^1_t,\mu ^2_t) \end{aligned}
which implies that the map $$t\mapsto \mathrm e^{-\lambda t}W_2(\mu ^1_t,\mu ^2_t)$$ is decreasing in (0, T). $$\square$$
It is possible to prove (5.24) by a direct argument depending on the definition of $$\lambda$$-EVI solution and a geometric condition on $$\mathrm {D}({\varvec{\mathrm {F}}})$$. The simplest situation deals with absolutely continuous curves.
### Theorem 5.16
(Stability for absolutely continuous solutions) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$\mu ^1, \mu ^2: [0,T) \rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$T\in (0, + \infty ]$$, be locally absolutely continuous $$\lambda$$-EVI solutions to (5.1). If $$\Gamma _o^{0}({\mu ^1_t},{\mu ^2_t}|{\varvec{\mathrm {F}}}) \ne \emptyset$$ for a.e. $$t \in (0,T)$$, then (5.24) holds. In particular, if $$\mu ^1_0=\mu ^2_0$$ then $$\mu ^1\equiv \mu ^2$$ in [0, T).
### Proof
Since $$\mu ^1,\mu ^2$$ are locally absolutely continuous curves, we can apply Theorem 3.14 and find a subset $$A\subset A({\mu ^1})\cap A({\mu ^2})$$ of full Lebesgue measure such that (3.21) holds and $$\Gamma _o^{0}({\mu ^1_t},{\mu ^2_t}|{\varvec{\mathrm {F}}}) \ne \emptyset$$ for every $$t \in A$$. Selecting $$\varvec{\mu }_t\in \Gamma _o^{0}({\mu ^1_t},{\mu ^2_t}|{\varvec{\mathrm {F}}})$$, we have
\begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}W_2^2(\mu ^1_t,\mu ^2_t) = \int \langle {\varvec{v}}_t^1(x_1),x_1-x_2\rangle \,\mathrm d\varvec{\mu }_t(x_1,x_2)+ \int \langle {\varvec{v}}_t^2(x_2),x_2-x_1\rangle \,\mathrm d\varvec{\mu }_t(x_1,x_2). \end{aligned}
Note that
\begin{aligned}&\Gamma _0\left( ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t^1)_{\sharp }\mu _t^1,\varvec{\mu }_t\right) =\Lambda \left( ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t^1)_{\sharp }\mu _t^1,\mu _t^2\right) =\left\{ ({\textsf {x} }^0,{\varvec{v}}_t^1\circ {\textsf {x} }^0,{\textsf {x} }^1)_\sharp \varvec{\mu }_t\right\} ,\\&\Gamma _0\left( ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t^2)_{\sharp }\mu _t^2,{\textsf {s} }_\sharp \varvec{\mu }_t\right) =\Lambda \left( ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t^2)_{\sharp }\mu _t^2,\mu _t^1\right) =\left\{ ({\textsf {x} }^1,{\varvec{v}}_t^2\circ {\textsf {x} }^1,{\textsf {x} }^0)_\sharp \varvec{\mu }_t\right\} \end{aligned}
by [3, Lemma 5.3.2], where $$\Gamma _0(\cdot ,\cdot )$$ is the set defined in (3.25) with $$t=0$$ and $$\Lambda (\cdot ,\cdot )$$ is defined in Definition 3.8. Hence, using (5.5b), (5.5c) and recalling the definition of reversion map $${\textsf {s} }$$ in (3.26), for every $$t\in A$$ we get
\begin{aligned} \frac{1}{2}\frac{\mathrm d}{\mathrm dt}W_2^2(\mu ^1_t,\mu ^2_t)&= [({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t^1)_{\sharp }\mu _t^1,\varvec{\mu }_t]_{r,0}+[({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t^2)_{\sharp }\mu _t^2,{\textsf {s} }_\sharp \varvec{\mu }_t]_{r,0}\\&\le [{\varvec{\mathrm {F}}},\varvec{\mu }_t]_{0+}+ \lambda W_2^2(\mu ^1_t,\mu ^2_t) +[{\varvec{\mathrm {F}}},{\textsf {s} }_\sharp \varvec{\mu }_t]_{1-}\\&= \lambda W_2^2(\mu ^1_t,\mu ^2_t), \end{aligned}
where we also used the property
\begin{aligned}{}[{\varvec{\mathrm {F}}},{\textsf {s} }_\sharp \varvec{\mu }_t]_{1-}=-[{\varvec{\mathrm {F}}}, \varvec{\mu }_t]_{0+}. \end{aligned}
$$\square$$
The last situation deals with the comparison between an absolutely continuous and a merely continuous $$\lambda$$-EVI solution. The argument is technically more involved and takes inspiration from the proof of [23, Theorem 1.1]: we refer to the Introduction of [23] for an explanation of the heuristic idea.
### Theorem 5.17
(Refined stability) Let $$T>0$$ and $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Let
1. (i)
$$\mu ^1:[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ be an absolutely continuous $$\lambda$$-EVI solution for $${\varvec{\mathrm {F}}}$$, with $$\mu ^1_0\in \mathrm {D}({\varvec{\mathrm {F}}})$$;
2. (ii)
$$\mu ^2:[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ be $$\lambda$$-EVI solution for $${\varvec{\mathrm {F}}}$$.
If at least one of the following properties hold:
1. (1)
$$\Gamma _o^{0}({\mu ^1_r},{\mu ^2_s}|{\varvec{\mathrm {F}}}) \ne \emptyset \text { for every } s \in (0,T) \text { and }r\in [0,T) {\setminus } N$$ with $$N\subset (0,T),\ {\mathcal {L}}(N)=0$$;
2. (2)
$$\mu ^1$$ satisfies (5.2),
then
\begin{aligned} W_2(\mu ^1_t, \mu ^2_t) \le e^{\lambda t} W_2(\mu ^1_0, \mu ^2_0)\quad \text { for every }t\in [0,T]. \end{aligned}
### Proof
We extend $$\mu ^1$$ in $$(-\infty , 0)$$ with the constant value $$\mu ^1_0$$, denote by $${\varvec{v}}$$ the Wasserstein velocity field associated to $$\mu ^1$$ (and extended to 0 outside $$A(\mu ^1)$$) and define the functions $$w,f,h:(-\infty ,T]\times [0,T]\rightarrow {\mathbb {R}}$$ by
\begin{aligned} w(r,s)&:= W_2(\mu ^1_r, \mu ^2_s)\\ f(r,s)&:= {\left\{ \begin{array}{ll} 2|{\varvec{\mathrm {F}}}|_2(\mu _0^1) w(0,s) \quad &{}\text { if } r<0, \\ 0 &{}\text { if } r \ge 0, \end{array}\right. }\qquad h(r,s):= {\left\{ \begin{array}{ll} 0\quad &{}\text { if } r<0, \\ 2\left[ ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_r)_\sharp \mu ^1_r, \mu ^2_s\right] _{r} &{}\text { if } r \ge 0. \end{array}\right. } \end{aligned}
Theorem 3.11 yields
\begin{aligned} \frac{\partial }{\partial r} w^2(r,s) = h(r,s) \quad&\text{ in } \mathscr {D}'(-\infty ,T), \text { for every } s \in [0,T]. \end{aligned}
(5.25)
In case (1) holds, writing (5.4b) for $$\mu ^2$$ with $$\nu =\mu ^1_r$$ and $$r\in (-\infty ,T]{\setminus } N$$, then for every $$\varvec{\mu }_{rs} \in \Gamma _o^{0}({\mu ^1_r},{\mu ^2_s}|{\varvec{\mathrm {F}}})$$ we obtain
\begin{aligned} {\frac{\mathrm d}{\mathrm ds}}^{+}w^2(r,s) \le 2\lambda w^2(r,s) -2[{\varvec{\mathrm {F}}},\varvec{\mu }_{rs}]_{0+} \quad&\text { for } s\in (0,T)\text { and } r \in (-\infty ,T){\setminus } N. \end{aligned}
(5.26)
On the other hand (5.5b) yields
\begin{aligned} \begin{aligned} -2[{\varvec{\mathrm {F}}},\varvec{\mu }_{rs}]_{0+}&\le -2 [({\varvec{i}}_{\textsf {X} },{\varvec{v}}_r)_\sharp \mu ^1_r,\varvec{\mu }_{rs}]_{r,0} \\&\le -2\left[ ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_r)_\sharp \mu ^1_r, \mu ^2_s\right] _{r} \quad \text {for every }r\in A(\mu ^1){\setminus } N,\\ -2[{\varvec{\mathrm {F}}},\varvec{\mu }_{rs}]_{0+}&\le 2|{\varvec{\mathrm {F}}}|_2(\mu _0^1) w(0,s)=f(r,s) \quad \text {for every }r<0. \end{aligned} \end{aligned}
(5.27)
Combining (5.26) and (5.27) we obtain
\begin{aligned} {\frac{\mathrm d}{\mathrm ds}}^{+}w^2(r,s)\le 2\lambda w^2(r,s) +f(r,s)-h(r,s)\quad \text {for }s\in (0,T),\ r\in (-\infty ,0]\cup A(\mu ^1){\setminus } N. \end{aligned}
Since, recalling Theorem 2.10, we have $$|h(r,s)|\le 2 |{{\dot{\mu }}}^1_r|\,w(r,s)$$, then applying Lemma B.4 we get
\begin{aligned} \frac{\partial }{\partial s} w^2(r,s)\le 2\lambda w^2(r,s) +f(r,s)-h(r,s)\quad \text {in }{\mathscr {D}}'(0,T),\text { for a.e.~}r\in (-\infty ,T].\nonumber \\ \end{aligned}
(5.28)
The expression in (5.28) can also be deduced in case (2) using (5.2).
By multiplying both inequalities (5.25) and (5.28) by $$e^{-2\lambda s}$$ we get
\begin{aligned} \frac{\partial }{\partial r} \Big (e^{-2\lambda s}w^2(r,s)\Big ) = e^{-2\lambda s}h(r,s)\quad&\text{ in } \mathscr {D}'(-\infty ,T) \text { and every } s \in [0,T], \\ \frac{\partial }{\partial s} \Big (e^{-2\lambda s}w^2(r,s) \Big )\le e^{-2\lambda s}\big (f(r,s)-h(r,s)\big ) \quad&\text{ in } \mathscr {D}'(0,T) \text { and a.e.~} r \in (-\infty ,T]. \end{aligned}
We fix $$t\in [0,T]$$ and $$\varepsilon >0$$ and we apply the Divergence theorem in [23, Lemma 6.15] on the two-dimensional strip $$Q_{0,t}^{\varepsilon }$$ as in Fig. 1,
\begin{aligned} Q_{0,t}^{\varepsilon } := \{ (r,s) \in {\mathbb {R}}^2 \mid 0\le s \le t \, , \, s-\varepsilon \le r \le s \}, \end{aligned}
(5.29)
and we get
\begin{aligned} \int _{t-\varepsilon }^t e^{-2\lambda t} w^2(r,t)\,\mathrm dr \le \int _{-\varepsilon }^0 w^2(r,0) \,\mathrm dr + \iint _{Q^{\varepsilon }_{0,t}} e^{-2\lambda s} f(r,s)\,\mathrm dr \,\mathrm ds. \end{aligned}
Using
\begin{aligned} w(t,t) \le \int _{r}^t |{\dot{\mu }}^1_u| \,\mathrm du+ w(r,t) \le \int _{t-\varepsilon }^t |{\dot{\mu }}^1_u| \,\mathrm du + w(r,t) \quad \text { if } t-\varepsilon \le r \le t, \end{aligned}
then, for every $$\delta , \delta _{\star }>1$$ conjugate coefficients ($$\delta _{\star }=\delta /(\delta -1)$$), we get
\begin{aligned} w^2(t,t)\le \delta w^2(r,t)+\delta _{\star }\left( \int _{t-\varepsilon }^t |{\dot{\mu }}^1_u| \,\mathrm du \right) ^2. \end{aligned}
(5.30)
Integrating (5.30) w.r.t. r in the interval $$(t-\varepsilon ,t)$$, we obtain
\begin{aligned} e^{-2\lambda t} w^2(t,t) \le \frac{\delta }{\varepsilon } \int _{t-\varepsilon }^t e^{-2\lambda t} w^2(r,t)\,\mathrm dr + \delta _{\star }\left( \int _{t-\varepsilon }^t |{\dot{\mu }}^1_u| \,\mathrm du \right) ^2 \max \{1, e^{2|\lambda |T}\}.\nonumber \\ \end{aligned}
(5.31)
Finally, we have the following inequality
\begin{aligned} \varepsilon ^{-1} \iint _{Q^{\varepsilon }_{0,t}} e^{-2 \lambda s} f(r,s) \,\mathrm dr \,\mathrm ds \le 2|{\varvec{\mathrm {F}}}|_2(\mu _0^1) \int _0^{\varepsilon } e^{-2\lambda s}w(0,s)\,\mathrm ds. \end{aligned}
(5.32)
Summing up (5.31) and (5.32) we obtain
\begin{aligned} e^{-2\lambda t} w^2(t)\le & {} \delta \left( w^2(0) + 2|{\varvec{\mathrm {F}}}|_2(\mu _0^1) \int _0^{\varepsilon } e^{-2\lambda s}w(0,s)\,\mathrm ds \right) \\&+ \delta _{\star }\left( \int _{t-\varepsilon }^t |{\dot{\mu }}^1_u| \,\mathrm du \right) ^2 \max \{1, e^{2|\lambda |T}\}. \end{aligned}
where we have used the notation $$w(s)= w(s,s)$$. Taking the limit as $$\varepsilon \downarrow 0$$ and $$\delta \downarrow 1$$, we obtain the thesis. $$\square$$
### Corollary 5.18
(Local Lipschitz estimate) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$\mu :(0,T) \rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$, $$T\in (0, + \infty ]$$, be a $$\lambda$$-EVI solution to (5.1). If at least one of the following two conditions holds
1. (a)
$$\mu$$ is strict and (EE) is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$,
2. (b)
$$\mu$$ is locally absolutely continuous and (4.24) holds,
then $$\mu$$ is locally Lipschitz and
\begin{aligned} t\mapsto \mathrm e^{-\lambda t} |{{\dot{\mu }}}_t|_+ \quad \text {is decreasing in }(0,T). \end{aligned}
(5.33)
### Proof
Since for every $$h>0$$ the curve $$t\mapsto \mu _{t+h}$$ is a $$\lambda$$-EVI solution, (5.24) yields
\begin{aligned} \mathrm e^{-\lambda (t-s)}W_2(\mu _{t+h},\mu _t) \le W_2(\mu _{s+h},\mu _s) \end{aligned}
for every $$0<s<t$$. Dividing by h and taking the limsup as $$h\downarrow 0$$, we get (5.33), which in turn shows the local Lipschitz character of $$\mu$$. $$\square$$
### 5.4 Global existence and generation of $$\lambda$$-flows
We collect here a few simple results on the existence of global solutions and the generation of a $$\lambda$$-flow. A first result can be deduced from the global solvability of the Explicit Euler scheme.
### Theorem 5.19
(Global existence) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). If the Explicit Euler Scheme is globally solvable at $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$, then there exists a unique global and locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,\infty )\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ starting from $$\mu _0$$.
### Proof
We can argue as in the proof of Theorem 5.11(a), observing that the global solvability of (EE) allows for the construction of a limit solution on every interval [0, T], $$T>0$$. $$\square$$
Let us provide a simple condition ensuring global solvability, whose proof is deferred to Sect. 6.
### Proposition 5.20
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Assume that for every $$R>0$$ there exist $$M=\mathrm M(R)>0$$ and $${\bar{\tau }}={\bar{\tau }}(R)>0$$ such that, for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ with $${\textsf {m} }_2(\mu )\le R$$ and every $$0<\tau \le {\bar{\tau }}$$,
\begin{aligned} \text {there exists } \,\Phi \in {\varvec{\mathrm {F}}}[\mu ]\,\text { s.t. }\, |\Phi |_2\le \mathrm M(R)\,\text { and }\, \textsf {exp} ^\tau _\sharp \Phi \in \mathrm {D}({\varvec{\mathrm {F}}}). \end{aligned}
(5.34)
Then the Explicit Euler scheme is globally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$.
Global existence of $$\lambda$$-EVI solution is also related to the existence of a $$\lambda$$-flow.
### Definition 5.21
We say that the $$\lambda$$-dissipative MPVF $${\varvec{\mathrm {F}}}$$, according to (4.1), generates a $$\lambda$$-flow if for every $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ there exists a unique $$\lambda$$-EVI solution $$\mu =\mathrm S[\mu _0]$$ starting from $$\mu _0$$ and the maps $$\mu _0\mapsto \mathrm S_t[\mu _0]=(\mathrm S[\mu _0])_t$$ induce a semigroup of Lipschitz transformations $$(\mathrm S_t)_{t\ge 0}$$ of $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ satisfying
\begin{aligned} W_2(\mathrm S_t[\mu _0],\mathrm S_t[\mu _1])\le \mathrm e^{\lambda t}W_2(\mu _0,\mu _1)\quad \text { for every }t\ge 0. \end{aligned}
(5.35)
### Theorem 5.22
(Generation of a $$\lambda$$-flow) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). If at least one of the following properties is satisfied:
1. (a)
the Explicit Euler Scheme is globally solvable for every $$\mu _0$$ in a dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$;
2. (b)
the Explicit Euler Scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ and, for every $$\mu _0$$ in a dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$, there exists a strict global $$\lambda$$-EVI solution starting from $$\mu _0$$;
3. (c)
the Explicit Euler Scheme is locally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ and $$\mathrm {D}({\varvec{\mathrm {F}}})$$ is closed;
4. (d)
for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\mu _1 \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ we have $$\Gamma _o^{0}({\mu _0},{\mu _1}|{\varvec{\mathrm {F}}})\ne \emptyset$$ and, for every $$\mu _0$$ in a dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$, there exists a locally absolutely continuous strict global $$\lambda$$-EVI solution starting from $$\mu _0$$;
5. (e)
for every $$\mu _0$$ in a dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$, there exists a locally absolutely continuous solution of (5.2) starting from $$\mu _0$$,
then $${\varvec{\mathrm {F}}}$$ generates a $$\lambda$$-flow.
### Proof
1. (a)
Let D be the dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$ for which (EE) is globally solvable. For every $$\mu _0\in D$$ we define $$\mathrm S_t[\mu _0]$$, $$t\ge 0$$, as the value at time t of the unique $$\lambda$$-EVI solution starting from $$\mu _0$$, whose existence is guaranteed by Theorem 5.19. If $$\mu _0,\mu _1\in D$$, $$T>0$$, we can find $$\varvec{\tau }, L$$ such that $${\mathscr {M}}(\mu _0,\tau ,T,L)$$ and $$\mathscr {M}(\mu _1,\tau ,T,L)$$ are not empty for every $$\tau \in (0,\varvec{\tau })$$. We can then pass to the limit in the uniform estimate (5.15) for every choice of $$M^i_\tau \in {\mathscr {M}}(\mu _i,\tau ,T,L)$$, $$i=0,1$$, obtaining (5.35) for every $$\mu _0,\mu _1\in D$$. We can then extend the map $$\mathrm S_t$$ to $$\overline{D}=\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ still preserving the same property. Proposition 5.6 shows that for every $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ the continuous curve $$t\mapsto \mathrm S_t[\mu _0]$$ is a $$\lambda$$-EVI solution starting from $$\mu _0$$. Finally, if $$\mu :[0,T')\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is any $$\lambda$$-EVI solution starting from $$\mu _0$$, we can apply (5.16) to get
\begin{aligned} W_2(\mu _t,M^1_\tau (t))\le \Big (2W_2(\mu _0,\mu _1)+C(\varvec{\tau }, L,T)\sqrt{\tau }\Big )\mathrm e^{\lambda _+ t} \end{aligned}
(5.36)
for every $$t\in [0,T]$$, $$T<T'$$ and $$\tau <\varvec{\tau }$$, where $$C(\varvec{\tau }, L,T)>0$$ is a suitable constant. Passing to the limit as $$\tau \downarrow 0$$ in (5.36) we obtain
\begin{aligned} W_2(\mu _t,\mathrm S_t[\mu _1])\le 2W_2(\mu _0,\mu _1)\mathrm e^{\lambda _+ t}\quad \text { for every }t\in [0,T]. \end{aligned}
(5.37)
Choosing now a sequence $$\mu _{1,n}$$ in D converging to $$\mu _0$$ and observing that we can choose arbitrary $$T<T'$$, we eventually get $$\mu _t=\mathrm S_t[\mu _0]$$ for every $$t\in [0,T')$$.
2. (b)
Let D be the dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$ such that there exists a global strict $$\lambda$$-EVI solution starting from D. By Theorem 5.15 such a solution is unique and the corresponding family of solution maps $$\mathrm S_t:D\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ satisfy (5.35). Arguing as in the previous claim, we can extend $$\mathrm S_t$$ to $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ still preserving (5.35) and the fact that $$t\mapsto \mathrm S_t[\mu _0]$$ is a $$\lambda$$-EVI solution. If $$\mu$$ is $$\lambda$$-EVI solution starting from $$\mu _0$$, Theorem 5.15 shows that (5.37) holds for every $$\mu _1\in D$$. By approximation we conclude that $$\mu _t=\mathrm S_t[\mu _0]$$.
3. (c)
Corollary 5.12 shows that for every initial datum $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exists a global $$\lambda$$-EVI solution. We can then apply Claim (b).
4. (d)
Let D be the dense subset of $$\mathrm {D}({\varvec{\mathrm {F}}})$$ such that there exists a locally absolutely continuous strict global $$\lambda$$-EVI solution starting from D. By Theorem 5.16 such a solution is the unique locally absolutely continuous solution starting from $$\mu _0$$ and the corresponding family of solution maps $$\mathrm S_t:D\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ satisfy (5.35). Arguing as in the previous claim (b), we can extend $$\mathrm S_t$$ to $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ still preserving (5.35) (again thanks to Theorem 5.16) and the fact that $$t\mapsto \mathrm S_t[\mu _0]$$ is a $$\lambda$$-EVI solution. If $$\mu$$ is a $$\lambda$$-EVI solution starting from $$\mu _0 \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ and $$(\mu _0^n)_{n \in {\mathbb {N}}} \subset D$$ is a sequence converging to $$\mu _0$$, we can apply Theorem 5.17(1) and conclude that $$\mu _t=\mathrm S_t[\mu _0]$$.
5. (e)
The proof follows by the same argument of the previous claim, eventually applying Theorem 5.17(2).$$\square$$
By Lemma 5.13 we immediately get the following result.
### Corollary 5.23
If $${\varvec{\mathrm {F}}}$$ is locally bounded $$\lambda$$-dissipative MPVF according to (4.1), with $$\mathrm {D}({\varvec{\mathrm {F}}})=\mathcal {P}_2({\textsf {X} })$$, then for every $$\mu _0\in \mathcal {P}_2({\textsf {X} })$$ there exists a unique global $$\lambda$$-EVI solution starting from $$\mu _0$$.
We conclude this section by showing a consistency result with the Hilbertian theory, related to the example of Sect. 7.2.
### Corollary 5.24
(Consistency with the theory of contraction semigroups in Hilbert spaces) Let $$F \subset {\textsf {X} }\times {\textsf {X} }$$ be a dissipative maximal subset generating the semigroup $$(R_t)_{t \ge 0}$$ of nonlinear contractions [7, Theorem 3.1]. Let $${\varvec{\mathrm {F}}}$$ be the dissipative MPVF according to (4.1), defined by
\begin{aligned} {\varvec{\mathrm {F}}}:= \left\{ \Phi \in \mathcal {P}_2(\mathsf {TX}) \mid \Phi \text { is concentrated on } F \right\} . \end{aligned}
The semigroup $$\mu _0\mapsto \mathrm S_t[\mu _0]:=(R_t)_{\sharp }\mu _0$$, $$t\ge 0$$, is the 0-flow generated by $${\varvec{\mathrm {F}}}$$ in $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$.
### Proof
Let D be the set of discrete measures $$\frac{1}{n}\sum _{j=1}^n\delta _{x_j}$$ with $$x_j\in \mathrm {D}(F)$$. Since every $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is supported in $$\overline{\mathrm {D}(F)}$$, D is dense in $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$. Our thesis follows by applying Theorem 5.22(e) if we show that for every $$\mu _0^n=\frac{1}{n}\sum _{j=1}^n\delta _{x_{j,0}}\in D$$ there exists a locally absolutely continuous solution $$\mu ^n:[0,\infty )\rightarrow D$$ of (5.2) starting from $$\mu _0^n$$.
It can be directly checked that
\begin{aligned} \mu _t^n:=(R_t)_\sharp \mu _0^n= \frac{1}{n}\sum _{j=1}^n\delta _{x_{j,t}},\quad x_{j,t}:=R_t(x_{j,0}) \end{aligned}
satisfies the continuity equation with Wasserstein velocity vector $${\varvec{v}}_t$$ (defined on the finite support of $$\mu _t^n$$) satisfying
\begin{aligned} {\varvec{v}}_{t}(x_{j,t})=\dot{x}_{j,t}=F^\circ (x_{j,t})\quad \text {and}\quad |{\varvec{v}}_t(x_{j,t})|\le |F^\circ (x_{j,0})| \end{aligned}
for every $$j=1,\ldots , n$$, and a.e. $$t>0$$, where $$F^\circ$$ is the minimal selection of F. It follows that
\begin{aligned} ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t^n \in {\varvec{\mathrm {F}}}[\mu _t^n]\quad \text { for a.e. }t>0, \end{aligned}
so that $$\mu ^n$$ is a Lipschitz EVI solution for $${\varvec{\mathrm {F}}}$$ starting from $$\mu _0^n$$. We can thus conclude observing that the map $$\mu _0\mapsto (R_t)_\sharp \mu _0$$ is a contraction in $$\mathcal {P}_2({\textsf {X} })$$ and the curve $$\mu _t^n=(R_t)_\sharp \mu _0^n$$ is continuous with values in $$\overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$. $$\square$$
### 5.5 Barycentric property
If we assume that the MPVF $${\varvec{\mathrm {F}}}$$ is a sequentially closed subset of $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ with convex sections, we are able to provide a stronger result showing a particular property satisfied by the solutions of (5.1) (see Theorem 5.27). This is called barycentric property and it is strictly connected with the weaker definition of solution discussed in [9, 26, 27].
We first introduce a directional closure of $${\varvec{\mathrm {F}}}$$ along smooth cylindrical deformations. We set
\begin{aligned} \mathrm {exp}^{\varphi }(x):=x+\nabla \varphi (x) \end{aligned}
for every $$\varphi \in {{\,\mathrm{Cyl}\,}}({\textsf {X} })$$, and
\begin{aligned} {\overline{{\varvec{\mathrm {F}}}}}[\mu ]:={} \left\{ \Phi \in \mathcal {P}_2({\textsf {X} })\,\Bigg |\,\begin{array}{l}\exists \,\varphi \in {{\,\mathrm{Cyl}\,}}({\textsf {X} }),\ (r_n)_{n\in {\mathbb {N}}}\subset [0,+\infty ),\ r_n\downarrow 0,\\ \Phi _n\in {\varvec{\mathrm {F}}}[\mathrm {exp}^{r_n\varphi }_\sharp \mu ]:\, \Phi _n\rightarrow \Phi \text { in }\mathcal {P}_2^{sw}(\mathsf {TX})\end{array}\right\} . \end{aligned}
(5.38)
### Definition 5.25
(Barycentric property) Let $${\varvec{\mathrm {F}}}$$ be a MPVF. We say that a locally absolutely continuous curve $$\mu : {\mathcal {I}}\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ satisfies the barycentric property (resp. the relaxed barycentric property) if for a.e. $$t \in {\mathcal {I}}$$ there exists $$\Phi _t \in {\varvec{\mathrm {F}}}[\mu _t]$$ (resp. $$\Phi _t \in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}}[\mu _t])$$) such that
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt} \int _{\textsf {X} }\varphi (x) \,\mathrm d\mu _t (x)= \int _{\mathsf {TX}} \langle \nabla \varphi (x), v\rangle \,\mathrm d\Phi _t(x,v) \quad \text { for every }\varphi \in {{\,\mathrm{Cyl}\,}}({\textsf {X} }).\quad \end{aligned}
(5.39)
Notice that $${\varvec{\mathrm {F}}}\subset {\overline{{\varvec{\mathrm {F}}}}}\subset {\text {cl}}({\varvec{\mathrm {F}}})$$ and $${\overline{{\varvec{\mathrm {F}}}}}={\varvec{\mathrm {F}}}$$ if $${\varvec{\mathrm {F}}}$$ is sequentially closed in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$. Recalling Proposition 4.17(a) we also get
\begin{aligned} \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}})\subset {{\hat{{\varvec{\mathrm {F}}}}}}, \end{aligned}
so that the relaxed barycentric property implies the corresponding property for the extended MPVF $${{\hat{{\varvec{\mathrm {F}}}}}}$$ defined in (4.22). In particular, considering the directional closure $${\overline{{\varvec{\mathrm {F}}}}}$$ in place of the sequential closure $${\text {cl}}({\varvec{\mathrm {F}}})$$ not only allows us to obtain a finer result, but it could be easier to compute when one considers specific examples, being $$\bar{{\varvec{\mathrm {F}}}}$$ the closure of $${\varvec{\mathrm {F}}}$$ along regular directions.
### Remark 5.26
If $${\textsf {X} }= {\mathbb {R}}^d$$, the property stated in Definition 5.25 coincides with the weak definition of solution to (5.1) given in [26].
The aim is to prove that the $$\lambda$$-EVI solution of (5.1) enjoys the barycentric property of Definition 5.25, under suitable mild conditions on $${\varvec{\mathrm {F}}}$$. This is strictly related to the behaviour of $${\varvec{\mathrm {F}}}$$ along the family of smooth deformations induced by cylindrical functions. Let us denote by $$\mathbf {pr}_\mu$$ the orthogonal projection in $$L^2_\mu ({\textsf {X} };{\textsf {X} })$$ onto the tangent space $${{\,\mathrm{Tan}\,}}_\mu \mathcal {P}_2({\textsf {X} })$$ and by $${\varvec{b}}_{\Phi }$$ the barycenter of $$\Phi$$ as in Definition 3.1.
### Theorem 5.27
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Assume that for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ there exist constants $$M,\varepsilon >0$$ such that
\begin{aligned} \mathrm {exp}^{\varphi }_\sharp \mu \in \mathrm {D}({\varvec{\mathrm {F}}})\quad \text {and}\quad |{\varvec{\mathrm {F}}}|_2(\mathrm {exp}^{\varphi }_\sharp \mu )\le M \end{aligned}
(5.40)
for every $$\varphi \in {{\,\mathrm{Cyl}\,}}({\textsf {X} })$$ such that $$\displaystyle \sup _{\textsf {X} }|\nabla \varphi |\le \varepsilon$$. If $$\mu : {\mathcal {I}}\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ is a locally absolutely continuous $$\lambda$$-EVI solution of (5.1) with Wasserstein velocity field $${\varvec{v}}$$ satisfying (2.6) for every t in the subset $$A(\mu )\subset {\mathcal {I}}$$ of full Lebesgue measure, then
\begin{aligned} \text {for every }t \in A(\mu )\text { there exists }\Phi _t \in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}})[\mu _t]\text { such that}\quad {\varvec{v}}_t = \mathbf {pr}_{\mu _t} \circ {\varvec{b}}_{\Phi _t}.\qquad \end{aligned}
(5.41)
In particular, $$\mu$$ satisfies the relaxed barycentric property.
If moreover $${\overline{{\varvec{\mathrm {F}}}}}={\varvec{\mathrm {F}}}$$ and, for every $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, the section $${\varvec{\mathrm {F}}}[\nu ]$$ is a convex subset of $$\mathcal {P}_2(\mathsf {TX})$$, i.e.
\begin{aligned} {\varvec{\mathrm {F}}}[\nu ]= {\text {co}}({\varvec{\mathrm {F}}})[\nu ], \end{aligned}
then $$\mu$$ satisfies the barycentric property (5.39).
### Proof
We divide the proof of (5.41) into two steps.
### Claim 1
Let $$t\in A(\mu )$$ and $$M=M_t$$ be the constant associated to the measure $$\mu _t$$ in (5.40). Then $${\varvec{v}}_t\in \overline{{\text {co}}}(K_t)$$, where
\begin{aligned} K_t := \left\{ \mathbf {pr}_{\mu _t}({\varvec{b}}_{\Phi })\,:\, \Phi \in {\overline{{\varvec{\mathrm {F}}}}}[\mu _t], \, |\Phi |_2 \le M_t\right\} \subset {{\,\mathrm{Tan}\,}}_{\mu _t} \mathcal {P}_2({\textsf {X} }). \end{aligned}
(5.42)
### Proof of Claim 1
For every $$\zeta \in {{\,\mathrm{Cyl}\,}}({\textsf {X} })$$ there exists $$\delta =\delta (\zeta )>0$$ such that $$\nu ^{\zeta }:=\mathrm {exp}^{-\delta \zeta }_\sharp \mu _t \in \mathrm {D}({\varvec{\mathrm {F}}})$$ and $$\varvec{\sigma }^\zeta :=({\varvec{i}}_{\textsf {X} }, \mathrm {exp}^{-\delta \zeta })_\sharp \mu _t\in \Gamma _o^{01}({\mu _t},{\nu ^\zeta }|{\varvec{\mathrm {F}}})$$ is the unique optimal transport plan between $$\mu _t$$ and $$\nu ^{\zeta }$$.
Thanks to Theorem 3.11, the map $$s \mapsto W_2^2(\mu _s, \nu ^{\zeta })$$ is differentiable at $$s=t$$, moreover by employing also (5.5b), it holds
\begin{aligned} \delta \int _{{\textsf {X} }} \langle {\varvec{v}}_t(x), \nabla \zeta (x)\rangle \,\mathrm d\mu _t(x) = \frac{\,\mathrm d}{\,\mathrm dt} \frac{1}{2}W_2^2(\mu _t, \nu ^{\zeta }) \le [{\varvec{\mathrm {F}}},\varvec{\sigma }^\zeta ]_{0+} = \lim _{s \downarrow 0}\, [{\varvec{\mathrm {F}}},\varvec{\sigma }^\zeta ]_{l,s}.\qquad \end{aligned}
(5.43)
We can choose a decreasing vanishing sequence $$(s_k)_{k \in {\mathbb {N}}} \subset (0,1)$$, measures $$\nu _k^\zeta :={\textsf {x} }^{s_k}_\sharp \varvec{\sigma }^\zeta$$ and $$\Phi _k^\zeta \in {\varvec{\mathrm {F}}}[\nu _k^\zeta ]$$ such that $$\sup _k|\Phi _k^\zeta |_2\le M_t$$ and $$\Phi _k^\zeta \rightarrow \Phi ^\zeta$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$. Then, by (5.16), we get $$\Phi ^{\zeta } \in {\overline{{\varvec{\mathrm {F}}}}}[\mu _t]$$ with $$|\Phi ^{\zeta }|_2 \le M_t$$ and by (5.43) and the upper semicontinuity of $$\left[ \cdot , \cdot \right] _{l}$$ (see Lemma 3.15) we get
\begin{aligned} \delta \int _{{\textsf {X} }} \langle {\varvec{v}}_t(x), \nabla \zeta (x)\rangle \,\mathrm d\mu _t(x) \le \left[ \Phi ^{\zeta }, \nu ^{\zeta }\right] _{l} = \delta \int _{\mathsf {TX}}\langle v, \nabla \zeta (x)\rangle \,\mathrm d\Phi ^{\zeta }(x,v). \end{aligned}
(5.44)
Indeed, notice that, by [3, Lemma 5.3.2], we have $$\Lambda (\Phi ^{\zeta }, \nu ^\zeta )=\{\Phi ^{\zeta }\otimes \nu ^\zeta \}$$ with $$({\textsf {x} }^0,{\textsf {x} }^1)_{\sharp }(\Phi ^{\zeta }\otimes \nu ^\zeta )=\varvec{\sigma }^\zeta$$.
By means of the identity highlighted in Remark 3.2, the expression in (5.44) can be written as follows
\begin{aligned} \langle {\varvec{v}}_t, \nabla \zeta \rangle _{L^2_{\mu _t}({\textsf {X} };{\textsf {X} })} \le \langle {\varvec{b}}_{\Phi ^{\zeta }}, \nabla \zeta \rangle _{L^2_{\mu _t}({\textsf {X} };{\textsf {X} })} = \langle \mathbf {pr}_{\mu _t}({\varvec{b}}_{\Phi ^{\zeta }}), \nabla \zeta \rangle _{L^2_{\mu _t}({\textsf {X} };{\textsf {X} })} \end{aligned}
so that
\begin{aligned} \langle {\varvec{v}}_t, \nabla \zeta \rangle _{L^2_{\mu _t}({\textsf {X} };{\textsf {X} })} \le \sup _{{\varvec{b}} \in K_t}\,\langle {\varvec{b}}, \nabla \zeta \rangle _{L^2_{\mu _t}({\textsf {X} };{\textsf {X} })} \end{aligned}
for all $$\zeta \in {{\,\mathrm{Cyl}\,}}({\textsf {X} })$$, with $$K_t$$ as in (5.42). Applying Lemma B.3 in $${{\,\mathrm{Tan}\,}}_{\mu _t} \mathcal {P}_2({\textsf {X} })\subset L^2_{\mu _t}({\textsf {X} };{\textsf {X} })$$ we obtain that $${\varvec{v}}_t \in \overline{{\text {co}}}(K_t)$$.
### Claim 2
For every $${\varvec{w}}\in \overline{{\text {co}}}(K_t)$$ there exists $$\Psi \in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}})[\mu _t]$$ such that $${\varvec{w}}=\mathbf {pr}_{\mu _t} \circ {\varvec{b}}_{\Psi }$$.
### Proof of Claim 2
Notice that an element $${\varvec{w}}\in {{\,\mathrm{Tan}\,}}_\mu \mathcal {P}_2({\textsf {X} })$$ coincides with $$\mathbf {pr}_{\mu }({\varvec{b}}_{\Psi })$$ for $$\Psi \in \mathcal {P}_{2}(\mathsf {TX}|\mu )$$ if and only if
\begin{aligned} \int \langle {\varvec{w}},\nabla \zeta \rangle \,\mathrm d\mu = \int \langle v,\nabla \zeta \rangle \,\mathrm d\Psi (x,v) \end{aligned}
(5.45)
for every $$\zeta \in \mathrm {Cyl}({\textsf {X} })$$. It is easy to check that any element $${\varvec{w}}\in {\text {co}}(K)_t$$ can be represented as $$\mathbf {pr}_{\mu _t}({\varvec{b}}_{\Psi })$$ (and thus as in (5.45)) for some $$\Psi \in {\text {co}}({\overline{{\varvec{\mathrm {F}}}}}[\mu _t])$$. If $${\varvec{w}}\in \overline{{\text {co}}}(K_t)$$ we can find a sequence $$(\Psi _n)_{n\in {\mathbb {N}}}\subset {\text {co}}({\overline{{\varvec{\mathrm {F}}}}}[\mu _t])$$ such that $$|\Psi _n|_2\le M_t$$ and $${\varvec{w}}_n=\mathbf {pr}_{\mu _t}({\varvec{b}}_{\Psi _n})\rightarrow {\varvec{w}}$$ in $$L^2_{\mu _t}({\textsf {X} };{\textsf {X} })$$. Since the sequence $$(\Psi _n)_{n \in {\mathbb {N}}}$$ is relatively compact in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ by Proposition 2.15(2), we can extract a (not relabeled) subsequence converging to a limit $$\Psi$$ in $$\mathcal {P}_2^{sw}(\mathsf {TX})$$, as $$n\rightarrow +\infty$$. By definition $$\Psi \in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}}[\mu _t])$$ with $$|\Psi |_2\le M_t$$. We can eventually pass to the limit in (5.45) written for $${\varvec{w}}_n$$ and $$\Psi _n$$ thanks to $$\mathcal {P}_2^{sw}(\mathsf {TX})$$ convergence, obtaining the corresponding identity for $${\varvec{w}}$$ and $$\Psi$$ in the limit.
The thesis (5.41) follows by Claim 1 and Claim 2.
Finally, being $$\mu$$ locally absolutely continuous, it satisfies the continuity equation driven by $${\varvec{v}}$$ in the sense of distributions (see Theorem 2.10), so that by (5.41) we have
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt} \int _{\textsf {X} }\zeta (x) \,\mathrm d\mu _t(x)= & {} \int _{{\textsf {X} }} \langle \nabla \zeta (x), {\varvec{v}}_t(x)\rangle \,\mathrm d\mu _t(x) \\= & {} \int _{\mathsf {TX}} \langle \nabla \zeta (x), v\rangle \,\mathrm d\Phi _t(x,v) \quad \text {for all }\zeta \in \text {Cyl}({\textsf {X} }), \end{aligned}
for all $$t\in A(\mu )$$. $$\square$$
### Remark 5.28
We notice that it is always possible to estimate the value of $$M_t$$ in (5.42) by $$|{\varvec{\mathrm {F}}}|_{2\star }(\mu _t)$$.
### Remark 5.29
Using a standard approximation argument (see for example the proof of Lemma 5.1.12(f) in [3]) it is possible to show that actually the barycentric property (5.39) holds for every $$\varphi \in \mathrm {C}^{1,1}({\textsf {X} }; {\mathbb {R}})$$ (indeed, in this case, $$\nabla \varphi \in {{\,\mathrm{Tan}\,}}_{\mu } \mathcal {P}_2({\textsf {X} })$$ for every $$\mu \in \mathcal {P}_2({\textsf {X} })$$).
### Remark 5.30
We point out that the result stated in Theorem 5.27 is still valid if we replace the convex hull of $${\varvec{\mathrm {F}}}$$ defined in (4.19) using the “flat” structure of $$\mathcal {P}_2(\mathsf {TX})$$, with the following one which makes use of plan interpolations
\begin{aligned} \widetilde{\text {co}}({\varvec{\mathrm {F}}})(\nu ):=\left\{ \bigg ({\textsf {x} },\sum _{k_1}^N\alpha _k{\textsf {v} }_k\bigg )_\sharp \varvec{\Phi }\,\Bigg |\,\begin{array}{l}\varvec{\Phi }\in \mathcal {P}({\textsf {X} }^{N+1}),\,({\textsf {x} },{\textsf {v} }_k)_\sharp \varvec{\Phi }=\Phi _k,\,\Phi _k\in {\varvec{\mathrm {F}}}[\nu ],\\ \alpha _k\ge 0,\,k=1,\dots ,N,\,\sum _{k=1}^N\alpha _k=1, \, N \in {\mathbb {N}}\end{array}\right\} , \end{aligned}
for any $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, where
\begin{aligned} {\textsf {x} }(x,v_1,\dots ,v_N)=x\quad \text {and}\quad {\textsf {v} }_k(x,v_1,\dots ,v_N)=v_k,\,\,k=1,\dots ,N. \end{aligned}
Indeed, $${\text {co}}({\varvec{\mathrm {F}}})(\nu )$$ and $$\widetilde{\text {co}}({\varvec{\mathrm {F}}})(\nu )$$ share the same barycentric projection. However, while $${\text {co}}({\varvec{\mathrm {F}}})$$ preserves dissipativity as proved in Proposition 4.16, $$\widetilde{\text {co}}({\varvec{\mathrm {F}}})(\nu )$$ does not satisfy this property in general, as highlighted in the following example: let $${\textsf {X} }={\mathbb {R}}$$ and consider the PVF $${\varvec{\mathrm {F}}}$$, with domain $$\mathrm {D}({\varvec{\mathrm {F}}})=\left\{ \delta _0,\,\frac{1}{2}\delta _1+\frac{1}{2} \delta _0\right\}$$, defined by
\begin{aligned} {\varvec{\mathrm {F}}}[\delta _0]:=\frac{1}{2}\delta _{(0,3)}+\frac{1}{2}\delta _{(0,-3)},\qquad {\varvec{\mathrm {F}}}\left[ \frac{1}{2}\delta _1+\frac{1}{2} \delta _0\right] :=\frac{1}{2}\delta _{(1,2)}+\frac{1}{2}\delta _{(0,1)}. \end{aligned}
Then $${\varvec{\mathrm {F}}}$$ is dissipative, indeed
\begin{aligned} \left[ {\varvec{\mathrm {F}}}[\delta _0], {\varvec{\mathrm {F}}}\left[ \frac{1}{2}\delta _1+\frac{1}{2} \delta _0\right] \right] _{r}\le -1\le 0. \end{aligned}
However, $$\widetilde{\text {co}}({\varvec{\mathrm {F}}})$$ is not dissipative, indeed, if we take $$\delta _{(0,0)}\in \widetilde{\text {co}}({\varvec{\mathrm {F}}})[\delta _0]$$, we have
\begin{aligned} \left[ \delta _{(0,0)}, {\varvec{\mathrm {F}}}\left[ \frac{1}{2}\delta _1+\frac{1}{2} \delta _0\right] \right] _{r}= 2>0. \end{aligned}
As a complement to the studies investigated in this section, we prove the converse characterization of Theorem 5.27 in the particular case of regular measures or regular vector fields. We refer to [3, Definitions 6.2.1, 6.2.2] for the definition of $$\mathcal {P}_2^r({\textsf {X} })$$, that is the space of regular measures on $${\textsf {X} }$$. When $${\textsf {X} }={\mathbb {R}}^d$$ has finite dimension, $$\mathcal {P}_2^r({\textsf {X} })$$ is just the subset of measures in $$\mathcal {P}_2({\textsf {X} })$$ which are absolutely continuous w.r.t. the d-dimensional Lebesgue measure $${\mathcal {L}}^d$$.
### Theorem 5.31
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Let $$\mu : {\mathcal {I}}\rightarrow \mathrm {D}({\varvec{\mathrm {F}}})$$ be a locally absolutely continuous curve satisfying the relaxed barycentric property of Definition 5.25. If for a.e. $$t\in {\mathcal {I}}$$ at least one of the following properties holds:
1. (1)
$$\mu _t\in \mathcal {P}_2^r({\textsf {X} })$$,
2. (2)
$${\overline{{\varvec{\mathrm {F}}}}}[\mu _t]$$ contains a unique element $$\Phi _t$$ concentrated on a map, i.e. $$\Phi _t=({\varvec{i}}_{\textsf {X} }, {\varvec{b}}_{\Phi _t})_\sharp \mu _t$$
then $$\mu$$ is $$\lambda$$-EVI solution of (5.1).
### Proof
Take $$\varphi \in {{\,\mathrm{Cyl}\,}}({\textsf {X} })$$ and observe that, since $$\mu$$ has the relaxed barycentric property, then for a.e. $$t\in {\mathcal {I}}$$ (recall Theorem 3.11) there exists $$\Phi _t\in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}}[\mu _t])$$ such that
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\int _{\textsf {X} }\varphi (x)\,\mathrm d\mu _t(x)=\int _{\mathsf {TX}}\langle \nabla \varphi (x), v\rangle \,\mathrm d\Phi _t =\int _{\textsf {X} }\langle \nabla \varphi , \mathbf {pr}_{\mu _t} \circ {\varvec{b}}_{\Phi _t}\rangle \,\mathrm d\mu _t= \int _{\textsf {X} }\langle {\varvec{v}}_t, \nabla \varphi \rangle \,\mathrm d\mu _t, \end{aligned}
hence $$\mu$$ solves the continuity equation $$\partial _t\mu _t+\text {div}({\varvec{v}}_t\mu _t)=0$$, with $${\varvec{v}}_t=\mathbf {pr}_{\mu _t} \circ {\varvec{b}}_{\Phi _t}\in {{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2({\textsf {X} })$$. By Theorem 3.11, we also know that
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\frac{1}{2}W_2^2(\mu _t,\nu )=\int _{{\textsf {X} }^2}\langle {\varvec{v}}_t(x_0), x_0-x_1\rangle \,\mathrm d\varvec{\gamma }_t(x_0,x_1) \end{aligned}
(5.46)
for any $$t\in A(\mu ,\nu )$$, $$\varvec{\gamma }_t\in \Gamma _o(\mu _t,\nu )$$, $$\nu \in \mathcal {P}_2({\textsf {X} })$$. Possibly disregarding a Lebesgue negligible set, we can decompose the set $$A(\mu ,\nu )$$ in the union $$A_1\cup A_2$$, where $$A_1, A_2$$ correspond to the times t for which the properties (1) and (2) hold.
If $$t \in A_1$$ and $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, then by [3, Theorem 6.2.10], since $$\mu _t\in \mathcal {P}_2^r({\textsf {X} })$$, there exists a unique $$\varvec{\gamma }_t\in \Gamma _o(\mu _t,\nu )$$ and $$\varvec{\gamma }_t=({\varvec{i}}_{\textsf {X} }, {\varvec{r}}_t)_\sharp \mu _t$$ for some map $${\varvec{r}}_t$$ s.t. $${\varvec{i}}_{{\textsf {X} }}-{\varvec{r}}_t\in {{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2({\textsf {X} })\subset L^2_{\mu _t}({\textsf {X} };{\textsf {X} })$$ (recall [3, Proposition 8.5.2]), so that
\begin{aligned} \int _{{\textsf {X} }^2}\langle {\varvec{v}}_t(x_0), x_0-x_1\rangle \,\mathrm d\varvec{\gamma }_t(x_0,x_1)= & {} \int _{\textsf {X} }\langle {\varvec{v}}_t(x_0), x_0-{\varvec{r}}_t(x_0)\rangle \,\mathrm d\mu _t(x_0)\nonumber \\= & {} \int _{\textsf {X} }\langle {\varvec{b}}_{\Phi _t}, x_0-{\varvec{r}}_t(x_0)\rangle \,\mathrm d\mu _t(x_0) \nonumber \\= & {} \int _{\mathsf {TX}}\langle v, x-{\varvec{r}}_t(x)\rangle \,\mathrm d\Phi _t(x,v)\nonumber \\= & {} \left[ \Phi _t, \nu \right] _{r}, \end{aligned}
(5.47)
where we also applied Theorem 3.9 and Remark 3.19, recalling that in this case $$\Lambda (\Phi _t,\nu )$$ is a singleton.
If $$t\in A_2$$ we can select the optimal plan $$\varvec{\gamma }_t\in \Gamma _o(\mu _t,\nu )$$ along which
\begin{aligned} \left[ \Phi _t, \nu \right] _{r}= [\Phi _t,\varvec{\gamma }_t]_{r,0}= \int _{\textsf {X} }\langle {\varvec{b}}_{\Phi _t}(x_0), x_0-x_1\rangle \,\mathrm d\varvec{\gamma }_t(x_0,x_1). \end{aligned}
If $${\varvec{r}}_t$$ is the barycenter of $$\varvec{\gamma }_t$$ with respect to its first marginal $$\mu _t$$, recalling that $${\varvec{i}}_{{\textsf {X} }}-{\varvec{r}}_t\in {{\,\mathrm{Tan}\,}}_{\mu _t}\mathcal {P}_2({\textsf {X} })$$ (see also the proof of [3, Thm. 12.4.4]) we also get
\begin{aligned} \int _{{\textsf {X} }^2}\langle {\varvec{v}}_t(x_0), x_0-x_1\rangle \,\mathrm d\varvec{\gamma }_t(x_0,x_1)= & {} \int _{\textsf {X} }\langle {\varvec{v}}_t(x_0), x_0-{\varvec{r}}_t(x_0)\rangle \,\mathrm d\mu _t(x_0) \nonumber \\= & {} \int _{\textsf {X} }\langle {\varvec{b}}_{\Phi _t}(x_0), x_0-{\varvec{r}}_t(x_0)\rangle \,\mathrm d\mu _t(x_0)\nonumber \\= & {} \int _{\textsf {X} }\langle {\varvec{b}}_{\Phi _t}(x_0), x_0-x_1\rangle \,\mathrm d\varvec{\gamma }_t(x_0,x_1)\nonumber \\= & {} \left[ \Phi _t, \nu \right] _{r}, \end{aligned}
(5.48)
where we still applied Theorem 3.9 and Remark 3.19.
Combining (5.46) with (5.47) and (5.48) we eventually get
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\frac{1}{2}W_2^2(\mu _t,\nu )&=\left[ \Phi _t, \nu \right] _{r} \le -\left[ \Psi , \mu _t\right] _{r}+\lambda W_2^2(\mu _t,\nu )\quad \text { for every }\Psi \in {\varvec{\mathrm {F}}}[\nu ], \end{aligned}
by definition of $${{\hat{{\varvec{\mathrm {F}}}}}}$$ and the fact that $$\overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}})[\mu _t]\subset {{\hat{{\varvec{\mathrm {F}}}}}}[\mu _t]$$. $$\square$$
Thanks to Theorem 5.31, we can apply to barycentric solutions the uniqueness and approximation results of the previous Sections. We conclude this section with a general result on the existence of a $$\lambda$$-flow for $$\lambda$$-dissipative MPVFs, which is the natural refinement of Proposition 5.14
### Theorem 5.32
(Generation of $$\lambda$$-flow) Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Assume that $$\mathcal {P}_b({\textsf {X} })\subset \mathrm {D}({\varvec{\mathrm {F}}})$$ and for every $$\mu _0\in \mathcal {P}_b({\textsf {X} })$$ there exist $$\varrho >0$$ and $$L>0$$ such that, for every $$\mu$$ with $${{\,\mathrm{supp}\,}}(\mu )\subset {{\,\mathrm{supp}\,}}(\mu _0)+\mathrm B_{\textsf {X} }(\varrho )$$,
\begin{aligned} \text {there exists }\, \Phi \in {\varvec{\mathrm {F}}}[\mu ]\,\text { s.t. }\, {{\,\mathrm{supp}\,}}({\textsf {v} }_\sharp \Phi )\subset \mathrm B_{\textsf {X} }(L). \end{aligned}
(5.49)
Let $${\varvec{\mathrm {F}}}_b := {\varvec{\mathrm {F}}}\cap \mathcal {P}_b(\mathsf {TX})$$. If there exists $$a\ge 0$$ such that for every $$\Phi \in {\varvec{\mathrm {F}}}_b$$
\begin{aligned} {{\,\mathrm{supp}\,}}(\Phi )\subset \left\{ (x,v)\in \mathsf {TX}: \langle v,x\rangle \le a(1+|x|^2)\right\} , \end{aligned}
(5.50)
then $${\varvec{\mathrm {F}}}$$ generates a $$\lambda$$-flow.
### Proof
It is enough to prove that $${\varvec{\mathrm {F}}}_b$$ generates a $$\lambda$$-flow. Applying Proposition 5.14 to the MPVF $${\varvec{\mathrm {F}}}_b$$, we know that for every $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}}_b)$$ there exists a unique maximal strict and locally Lipschitz continuous $$\lambda$$-EVI solution $$\mu :[0,T)\rightarrow \mathcal {P}_b({\textsf {X} })$$ driven by $${\varvec{\mathrm {F}}}_b$$ and satisfying (5.19). We argue by contradiction, and we assume that $$T<+\infty$$. Notice that by (5.49) $${\varvec{\mathrm {F}}}$$ satisfies (5.40), so that $$\mu$$ is a relaxed barycentric solution for $${\varvec{\mathrm {F}}}_b$$. Since $$\mu _0\in \mathcal {P}_b({\textsf {X} })$$, we know that $${{\,\mathrm{supp}\,}}(\mu _0)\subset \mathrm B_{\textsf {X} }(r_0)$$ for some $$r_0>1$$.
It is easy to check that (5.50) holds also for every $$\Phi \in \overline{{\text {co}}}({\overline{{\varvec{\mathrm {F}}}}}_b)$$. Moreover, setting $$b:=2a$$, condition (5.50) yields
\begin{aligned} \langle v,x\rangle \le b|x|^2\quad \text {for every}\quad (x,v)\in {{\,\mathrm{supp}\,}}\Phi \in {\varvec{\mathrm {F}}}_b,\ |x|\ge 1. \end{aligned}
(5.51)
Let $$\phi (r):{\mathbb {R}}\rightarrow {\mathbb {R}}$$ be any smooth increasing function such that $$\phi (r)=0$$ if $$r\le r_0$$ and $$\phi (r)=1$$ if $$r\ge r_0+1$$, and let $$\varphi (t,x):=\phi (|x|\mathrm e^{-b t})$$. Clearly $$\varphi \in \mathrm C^{1,1}({\textsf {X} }\times [0,+\infty ))$$, with
\begin{aligned} \nabla \varphi (t,x)&=\frac{x}{|x|}\phi '(|x|\mathrm e^{-b t})\mathrm e^{-b t}\, \text { if }\,x\ne 0,\\ \nabla \varphi (t,0)&=0,\\ \partial _t \varphi (t,x)&=-b\phi '(|x|\mathrm e^{-b t})|x|\mathrm e^{-bt}. \end{aligned}
We thus have for a.e. $$t\in [0,T)$$
\begin{aligned} \frac{\mathrm d}{\mathrm dt}\int _{\textsf {X} }\varphi (t,x)\,\mathrm d\mu _t&= \mathrm e^{-b t}\int _\mathsf {TX}\Big (-b\phi '(|x|\mathrm e^{-b t})|x|+ \langle v,x\rangle |x|^{-1}\phi '(|x|\mathrm e^{-b t})\Big )\mathrm d\Phi _t(v,x) \\ {}&\le \mathrm e^{-b t}\int _\mathsf {TX}\Big (-b\phi '(|x|\mathrm e^{-b t})|x|+ b|x|\phi '(|x|\mathrm e^{-b t})\Big )\mathrm d\Phi _t(v,x)= 0 \end{aligned}
where in the last inequality we used (5.51) and the fact that the integrand vanishes if $$|x|\le 1$$. We get
\begin{aligned} \int _{\textsf {X} }\varphi (t,x)\,\mathrm d\mu _t=0\quad \text {in }[0,T); \end{aligned}
this implies that $${{\,\mathrm{supp}\,}}(\mu _t)\subset \mathrm B_{\textsf {X} }((r_0+1)\mathrm e^{bt})$$ so that the limit measure $$\mu _T$$ belongs to $$\mathcal {P}_b({\textsf {X} })$$ as well, leading to a contradiction with (5.19) for $${\varvec{\mathrm {F}}}_b$$.
We deduce that $$\mu$$ is a global strict $$\lambda$$-EVI solution for $${\varvec{\mathrm {F}}}_b$$. We can then apply Theorem 5.22(b) to $${\varvec{\mathrm {F}}}_b$$. $$\square$$
## 6 Explicit Euler scheme
In this section, we collect all the main estimates concerning the Explicit Euler scheme (EE) of Definition 5.7. For the sequel, we recall the notations
\begin{aligned} M_{\tau }(\cdot )\quad \text {and}\quad \bar{M}_\tau (\cdot ) \end{aligned}
for the affine and piecewise constant interpolations, respectively, of the sequence $$(M^n_\tau ,\Phi _\tau ^n)$$ in (EE). We also recall the notations
\begin{aligned} {\mathscr {E}}(\mu _0,\tau ,T,L)\quad \text {and}\quad {\mathscr {M}}(\mu _0,\tau ,T,L) \end{aligned}
for the (possibly empty) set of all the curves $$(M_\tau ,{{\varvec{F}}}_\tau )$$ and $$M_\tau$$, respectively, arising from the solution of (EE).
### 6.1 The Explicit Euler scheme: preliminary estimates
Our first step is to prove simple a priori estimates and a discrete version of ($$\lambda$$-EVI) as a consequence of Proposition 3.4.
### Proposition 6.1
Every solution $$(M_\tau ,{{\varvec{F}}}_\tau )\in {\mathscr {E}}(\mu _0,\tau ,T,L)$$ of (EE) satisfies
\begin{aligned} W_2(M_{\tau }(t),\mu _0)\le & {} L t,\quad |{{\varvec{F}}}_\tau (t)|_2 \le L \quad \text {for every }t \in [0,T], \end{aligned}
(6.1)
\begin{aligned} W_2(M_{\tau }(t), M_{\tau }(s))\le & {} L|t-s| \quad \text {for every } s,t \in [0,T], \end{aligned}
(6.2)
and
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt} \frac{1}{2} W_2^2(M_{\tau }(t), \nu ) \le \left[ {{\varvec{F}}}_{\tau }(t), \nu \right] _{r} + \tau |{{\varvec{F}}}_{\tau }(t)|_2^2 \le \left[ {{\varvec{F}}}_{\tau }(t), \nu \right] _{r} + \tau L^2 \end{aligned}
(IEVI)
for every $$t\in [0,T]$$ and $$\nu \in \mathcal {P}_2({\textsf {X} })$$, with possibly countable exceptions. In particular
\begin{aligned} \frac{1}{2}W_2^2(M_{\tau }^{n+1}, \nu )- \frac{1}{2}W_2^2(M_{\tau }^{n}, \nu )\le \tau \left[ \Phi _\tau ^n, \nu \right] _{r}+ \frac{1}{2} {\tau ^2 } L^2 \end{aligned}
(6.3)
for every $$0\le n<{\mathrm N(T,\tau )}$$ and $$\nu \in \mathcal {P}_2({\textsf {X} })$$.
### Proof
The second inequality of (6.1) is a trivial consequence of the definition of $${\mathscr {E}}(\mu _0,\tau ,T,L)$$, the first inequality is a particular case of (6.2). The estimate (6.2) is immediate if $$n\tau \le s<t\le (n+1)\tau$$ since
\begin{aligned} W_2 (M_{\tau }(s), M_{\tau }(t))&= W_2( (\textsf {exp} ^{s-n\tau })_{\sharp }\Phi _\tau ^n, (\textsf {exp} ^{t-n\tau })_{\sharp }\Phi _\tau ^n) \\&\le \sqrt{\int _{\mathsf {TX}} |(t-s) v)|^2 \,\mathrm d\Phi _\tau ^n} \\&=(t-s) \sqrt{\int _{\mathsf {TX}} |v|^2 \,\mathrm d\Phi _\tau ^n}\\&\le (t-s)L. \end{aligned}
This implies that the metric velocity of $$M_\tau$$ is bounded by L in [0, T] and therefore $$M_\tau$$ is L-Lipschitz.
Let us recall that for every $$\nu \in \mathcal {P}_2({\textsf {X} })$$ and $$\Phi \in \mathcal {P}_2(\mathsf {TX})$$ the function $$g(t):= \frac{1}{2} W_2^2(\textsf {exp} ^t_{\sharp } \Phi ,\nu )$$ satisfies
\begin{aligned}&t\mapsto g(t)-\frac{1}{2} t^2|\Phi |_2^2\text { is concave},\quad g'_r(0) = \left[ \Phi , \nu \right] _{r},\nonumber \\&g'(t)\le \left[ \Phi , \nu \right] _{r}+t|\Phi |_2^2\quad \text {for }\,t\ge 0, \end{aligned}
(6.4)
by Definition 3.5 and Proposition 3.4. In particular, the concavity yields the differentiability of g with at most countable exceptions. Thus, taking any $$n\in {\mathbb {N}}$$, $$0\le n<{\mathrm N(T,\tau )}$$, $$t\in [n\tau ,(n+1)\tau )$$ and $$\Phi = \Phi _\tau ^n$$ so that $$\textsf {exp} ^t_{\sharp } \Phi =M_\tau (t)$$, (6.4) yields (IEVI). The inequality in (6.3) follows by integration in each interval $$[n\tau ,(n+1)\tau ]$$. $$\square$$
In the following, we prove a uniform bound on curves $$M_\tau \in {\mathscr {M}}(\mu _0,\tau ,T,L)$$ which is useful to prove global solvability of the Explicit Euler scheme, as stated in Proposition 5.20. We will use the Gronwall estimates of Lemma B.1 and Lemma B.2.
### Proposition 6.2
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). Assume that for every $$R>0$$ there exist $$M=\mathrm M(R)>0$$ and $${\bar{\tau }}={\bar{\tau }}(R)>0$$ such that, for every $$\mu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ with $${\textsf {m} }_2(\mu )\le R$$ and every $$0<\tau \le {\bar{\tau }}$$,
\begin{aligned} \text {there exists } \,\Phi \in {\varvec{\mathrm {F}}}[\mu ]\,\text { s.t. }\, |\Phi |_2\le \mathrm M(R)\,\text { and }\, \textsf {exp} ^\tau _\sharp \Phi \in \mathrm {D}({\varvec{\mathrm {F}}}). \end{aligned}
(6.5)
Then the Explicit Euler scheme is globally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$. More precisely, if for a given $$\mu _0\in \mathrm {D}({\varvec{\mathrm {F}}})$$ with $$\Psi _0\in {\varvec{\mathrm {F}}}[\mu _0]$$, $${\textsf {m} }_0:={\textsf {m} }_2(\mu _0),$$ and we set
\begin{aligned} R:= {\textsf {m} }_0 +\Big (|\Psi _0|_2+1\Big ) \sqrt{2T}\mathrm e^{(1+2\lambda _+)T},\quad L:=\mathrm M(R), \quad \varvec{\tau }=\min \left\{ \frac{1}{L^2}, {\bar{\tau }}(R), T\right\} , \nonumber \\ \end{aligned}
(6.6)
then for every $$\tau \in (0,\varvec{\tau }]$$ the set $$\mathscr {E}(\mu _0,\tau ,T,L)$$ is not empty.
### Proof
We want to prove by induction that for every integer $$N\le {\mathrm N(T,\tau )}$$, (EE) has a solution up to the index N satisfying the upper bound
\begin{aligned} {\textsf {m} }_2(M^{N}_\tau )\le R, \end{aligned}
(6.7)
corresponding to the constants RL given by (6.6). For $$N=0$$ the statement is trivially satisfied. Assuming that $$0\le N<{\mathrm N(T,\tau )}$$ and elements $$(M^n_\tau ,\Phi _\tau ^n)$$, $$0\le n< N$$, $$M^N_\tau$$, are given satisfying (EE) and (6.7), we want to show that we can perform a further step of the Euler Scheme so that (EE) is solvable up to the index $$N+1$$ and $${\textsf {m} }_2(M^{N+1}_\tau )\le R$$.
Notice that by the induction hypothesis, for $$n=0, \dots , N-1$$, we have $$|\Phi _\tau ^n|_2\le L$$; since $${\textsf {m} }_2(M^N_\tau )\le R$$, by (6.5) we can select $$\Phi _\tau ^N\in {\varvec{\mathrm {F}}}[M^N_\tau ]$$ with $$|\Phi _\tau ^N|_2\le L$$ such that $$M^{N+1}_\tau =\textsf {exp} ^\tau _\sharp \Phi _\tau ^N\in \mathrm {D}({\varvec{\mathrm {F}}})$$. Using (6.3) with $$\nu =\mu _0$$, the $$\lambda$$-dissipativity with $$\Psi _0\in {\varvec{\mathrm {F}}}[\mu _0]$$
\begin{aligned} \left[ \Phi _\tau ^n, \mu _0\right] _{r}\le \lambda W_2^2(M_{\tau }^{n},\mu _0)-\left[ \Psi _0, M_{\tau }^{n}\right] _{r}, \end{aligned}
and the bound
\begin{aligned} -\left[ \Psi _0, M_{\tau }^n\right] _{r}\le \frac{1}{2}W_2^2(M_{\tau }^n,\mu _0)+\frac{1}{2}|\Psi _0|_2^2, \end{aligned}
we end up with
\begin{aligned} \frac{1}{2}W_2^2(M_{\tau }^{n+1}, \mu _0)- \frac{1}{2}W_2^2(M_{\tau }^{n}, \mu _0) \le \frac{\tau ^2}{2} L^2 + \tau \left( \frac{1}{2} + \lambda _+ \right) \, W_2^2(M_{\tau }^{n}, \mu _0) + \frac{\tau }{2} |\Psi _0|_2^2 , \end{aligned}
for every $$n\le N$$. Using the Gronwall estimate of Lemma B.2 we get
\begin{aligned} W_2(M_{\tau }^n, \mu _0)&\le \sqrt{T+\tau } \Big ( |\Psi _0|_2+\sqrt{\tau }L\Big ) \mathrm e^{(\frac{1}{2}+\lambda _+)\, (T+\tau )} \le \sqrt{2T}\Big ( |\Psi _0|_2+1\Big ) \mathrm e^{(1+2\lambda _+) T} \end{aligned}
for every $$n\le N+1$$, so that
\begin{aligned} {\textsf {m} }_2(M^{N+1}_\tau )\le {\textsf {m} }_0+ \sqrt{2T}\Big ( |\Psi _0|_2+1\Big ) \mathrm e^{(1+2\lambda _+) T}\le R. \end{aligned}
$$\square$$
We conclude this section by proving the stability estimate (5.15) of Theorem 5.9. We introduce the notation
\begin{aligned} I_\kappa (t):=\int _0^t \mathrm e^{\kappa r}\,\mathrm dr= \frac{1}{\kappa }(\mathrm e^{\kappa t}-1)\quad \text {if }\kappa \ne 0;\quad I_0(t):=t. \end{aligned}
Notice that for every $$t\ge 0$$
\begin{aligned} I_\kappa (t)\le t\mathrm e^{\kappa t}\quad \text {if }\kappa \ge 0. \end{aligned}
(6.8)
### Proposition 6.3
Let $$M_\tau \in {\mathscr {M}}(\mu _0,\tau ,T,L)$$ and $$M_\tau '\in {\mathscr {M}}(\mu _0',\tau ,T,L)$$. If $$\lambda _+\tau \le 2$$ then
\begin{aligned} W_2(M_\tau (t),M_\tau '(t))\le W_2(\mu _0,\mu _0')\mathrm e^{\lambda t}+ 8L\sqrt{t\tau }\Big (1+|\lambda |\sqrt{t\tau }\Big )\mathrm e^{\lambda _+ t} \end{aligned}
for every $$t\in [0,T]$$.
### Proof
Let us set $$w(t):=W_2(M_\tau (t),M_\tau '(t))$$. Since by Proposition 3.4(2), in every interval $$[n\tau ,(n+1)\tau ]$$ the function $$t\mapsto w^2(t)-4L^2 (t-n\tau )^2$$ is concave, with
\begin{aligned} \frac{\mathrm d}{\mathrm dt}w^2(t)\bigg |_{t=n\tau +}= 2\left[ {{\varvec{F}}}_\tau (t), {{\varvec{F}}}_\tau '(t)\right] _{r} \le 2\lambda W_2^2(\bar{M}_\tau (t),{\bar{M}}_\tau '(t)), \end{aligned}
we obtain
\begin{aligned} \frac{\mathrm d}{\mathrm dt} w^2(t)\le 2\lambda W_2^2({\bar{M}}_\tau (t),\bar{M}_\tau '(t))+8L^2\tau \end{aligned}
for every $$t\in [0,T]$$, with possibly countable exceptions. Using the identity
\begin{aligned} a^2-b^2=2b(a-b)+|a-b|^2 \end{aligned}
with $$a=W_2({\bar{M}}_\tau (t),{\bar{M}}_\tau '(t))$$ and $$b=W_2(M_\tau (t),M_\tau '(t))$$ and observing that
\begin{aligned} |a-b|\le W_2({\bar{M}}_\tau (t),M_\tau (t))+W_2({\bar{M}}_\tau '(t),M_\tau '(t))\le 2L\tau , \end{aligned}
we eventually get
\begin{aligned} \frac{\mathrm d}{\mathrm dt} w^2(t)&\le 2\lambda w^2(t) +8L^2\tau + 8|\lambda |L\tau w(t) +\lambda _+ 8L^2\tau ^2 \\ {}&\le 2\lambda w^2(t) +8|\lambda |L\tau w(t) +24 L^2\tau , \end{aligned}
since $$\lambda _+\tau \le 2$$ by assumption. The Gronwall estimate in Lemma B.1 and (6.8) yield
\begin{aligned} w(t)&\le \Big (w^2(0) \mathrm e^{2\lambda t}+24L^2\tau \mathrm I_{2\lambda }(t)\Big )^{1/2}+8|\lambda | L\tau \mathrm I_{\lambda }(t) \\ {}&\le w(0) \mathrm e^{\lambda t}+8L\sqrt{t\tau }\Big (1+|\lambda |\sqrt{t\tau }\Big ) \mathrm e^{\lambda _+ t}. \end{aligned}
$$\square$$
### 6.2 Error estimates for the Explicit Euler Scheme
This subsection is devoted to the proof of the core of Theorem 5.9. In particular, we prove a Cauchy estimate for the affine interpolant of the Explicit Euler Scheme under different step sizes and a uniform (optimal, see [31]) error estimate between the affine interpolation and the $$\lambda$$-EVI solution for $${\varvec{\mathrm {F}}}$$. We stress that the results obtained for the affine interpolant of the sequence generated by the Explicit Euler Scheme in Definition 5.7 can be adjusted for the piecewise contant interpolant $$\bar{M}_\tau (\cdot )$$ thanks to (5.14).
### Theorem 6.4
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). If $$M_\tau \in {\mathscr {M}}(M^0_\tau ,\tau ,T,L)$$, $$M_\eta \in \mathscr {M}(M^0_\eta ,\eta ,T,L)$$ with $$\lambda \sqrt{T(\tau +\eta )}\le 1$$, then for every $$\delta >1$$ there exists a constant $$C(\delta )$$ such that
\begin{aligned} W_2(M_\tau (t),M_\eta (t))\le \Big (\sqrt{\delta } W_2(M^0_\tau ,M^0_\eta )+ C(\delta ) L \sqrt{(\tau +\eta )(t+\tau +\eta )}\Big )\mathrm e^{\lambda _+\, t} \end{aligned}
for every $$t \in [0,T]$$.
### Proof
We argue as in the Proof of Theorem 5.17. Since $$\lambda$$-dissipativity implies $$\lambda '$$-dissipativity for $$\lambda '\ge \lambda$$, it is not restrictive to assume $$\lambda > 0$$. We set $$\sigma :=\tau +\eta$$. We will extensively use the a priori bounds (6.1) and (6.2); in particular,
\begin{aligned} W_2(M_{\tau }(t), {\bar{M}}_{\tau }(t))\le L\tau ,\quad W_2(M_{\eta }(t), {\bar{M}}_{\eta }(t))\le L\eta . \end{aligned}
We will also extend $$M_\tau$$ and $${\bar{M}}_\tau$$ for negative times by setting
\begin{aligned} M_\tau (t)={\bar{M}}_\tau (t)=M^0_\tau ,\quad {{\varvec{F}}}_\tau (t)=M^0_\tau \otimes \delta _0\quad \text {if }t<0. \end{aligned}
(6.9)
The proof is divided into several steps.
1. Doubling variables.
We fix a final time $$t\in [0,T]$$ and two variables $$r,s\in [0,t]$$ together with the functions
\begin{aligned} w(r,s):={}&W_2(M_\tau (r),M_\eta (s)),\quad&w_\tau (r,s):={}&W_2({\bar{M}}_\tau (r),M_\eta (s)),\nonumber \\ w_\eta (r,s):={}&W_2(M_\tau (r),{\bar{M}}_\eta (s)),\quad&w_{\tau ,\eta }(r,s):={}&W_2({\bar{M}}_\tau (r),{\bar{M}}_\eta (s)), \end{aligned}
(6.10)
observing that
\begin{aligned} \max \left\{ |w-w_\tau |, |w_\eta -w_{\tau ,\eta }|\right\} \le L\tau ,\quad \max \left\{ |w-w_\eta |, |w_\tau -w_{\tau ,\eta }|\right\} \le L\eta .\qquad \end{aligned}
(6.11)
By Proposition 6.1, we can write (IEVI) for $$M_{\tau }$$ and get
and for $$M_{\eta }$$ obtaining
Apart from possible countable exceptions, ($$\text {IEVI}_{\tau }$$) holds for $$r\in (-\infty ,t]$$ and ($$\text {IEVI}_{\eta }$$) for $$s \in [0,t]$$. Taking $$\nu _1 = \bar{M}_{\eta }(s)$$, $$\nu _2=\bar{M}_{\tau }(r)$$, $$\Phi = {{\varvec{F}}}_\tau (\max \{r, 0\})\in {\varvec{\mathrm {F}}}[{\bar{M}}_\tau (r)]$$, summing the two inequalities $$(\text {IEVI}_{\tau ,\eta })$$, setting
\begin{aligned} f(r,s):= {\left\{ \begin{array}{ll} 2L W_2({\bar{M}}_\eta (s),M_\tau (0))=2Lw_\eta (0, s)&{} \text {if }r<0,\\ 0&{}\text {if }r\ge 0, \end{array}\right. } \end{aligned}
using (6.1) and the $$\lambda$$-dissipativity of $${\varvec{\mathrm {F}}}$$, we obtain
\begin{aligned} \frac{\partial }{\partial r} w_\eta ^2(r,s) + \frac{\partial }{\partial s} w_\tau ^2(r,s) \le 2\lambda w_{\tau ,\eta }^2(r,s) + 2L^2\sigma +f(r,s) \end{aligned}
in $$(-\infty ,t]\times [0,t]$$ (see also [23, Lemma 6.15]). By multiplying both sides by $$e^{-2\lambda s}$$, we have
\begin{aligned} \begin{aligned}&\frac{\partial }{\partial r} \mathrm e^{-2\lambda s} w_\eta ^2 + \frac{\partial }{\partial s} \mathrm e^{-2\lambda s} w_\tau ^2 \le \Big (2\lambda \left( w_{\tau ,\eta }^2 -w_\tau ^2 \right) + f +2L^2\sigma \Big )\mathrm e^{-2\lambda s}. \end{aligned} \end{aligned}
(6.12)
Using (6.11), the inequalities
\begin{aligned}&w_{\tau ,\eta }+w_\tau =w_{\tau ,\eta }-w_\tau +2(w_\tau -w)+2w\le 2L\sigma +2w,\\&|w(r,s)-w(s,s)|\le L|r-s| \end{aligned}
and the elementary inequality $$a^2-b^2\le |a-b||a+b|,$$ we get
\begin{aligned} 2\big (w_{\tau ,\eta }^2(r,s)-w_\tau ^2(r,s)\big )\le R_{r,s},\quad \text {if }r,s\le t, \end{aligned}
where $$R_{r,s}:=4L^2\sigma (\sigma +|r-s|)+4L\sigma w(s,s)$$. Thus (6.12) becomes
\begin{aligned} \frac{\partial }{\partial r} \mathrm e^{-2\lambda s} w_\eta ^2 + \frac{\partial }{\partial s} \mathrm e^{-2\lambda s} w_\tau ^2 \le Z_{r,s}, \end{aligned}
(6.13)
where $$Z_{r,s}:=\Big (R \lambda +f+ 2L^2\sigma \Big ) \mathrm e^{-2\lambda s}$$.
2. Penalization.
We fix any $$\varepsilon >0$$ and apply the Divergence Theorem to the inequality (6.13) in the two-dimensional strip $$Q^{\varepsilon }_{0,t}$$ as in (5.29) and we get
\begin{aligned}&\int _{t-\varepsilon }^t \mathrm e^{-2\lambda t} w_\tau ^2(r,t) \,\mathrm dr \le \int _{-\varepsilon }^0 w^2_\tau (r,0)\,\mathrm dr +\nonumber \\&\quad + \int _0^t \mathrm e^{-2\lambda s} \left( w_\tau ^2(s,s)-w_\eta ^2(s,s) \right) \,\mathrm ds + \int _0^t \mathrm e^{-2\lambda s} \left( w_\eta ^2(s-\varepsilon ,s)- w_\tau ^2(s-\varepsilon ,s) \right) \,\mathrm ds\nonumber \\&\quad +\iint _{Q_{0,t}^\varepsilon } Z_{r,s}\,\mathrm dr \mathrm ds. \end{aligned}
(6.14)
3. Estimates of the r.h.s..
We want to estimate the integrals (say $$I_0,I_1,I_2,I_3)$$ of the right hand side of (6.14) in terms of
\begin{aligned} w(s):=w(s,s)\quad \text {and}\quad W(t):=\sup _{0\le s\le t}\mathrm e^{-\lambda s}w(s). \end{aligned}
We easily get
\begin{aligned} I_0=\int _{-\varepsilon }^0 w_\tau ^2(r,0)\,\mathrm dr=\varepsilon w^2(0). \end{aligned}
(6.11) yields
\begin{aligned} |w_\tau (s,s)-w_\eta (s,s)| \le L(\tau +\eta )=L\sigma \end{aligned}
and
\begin{aligned} |w^2_\tau (s,s)-w^2_\eta (s,s)| \le L\sigma \Big ( L\sigma +2w(s) \Big ); \end{aligned}
after an integration,
\begin{aligned} \begin{aligned} I_1&\le {L^2 \sigma ^2 t} +2L\sigma \int _0^t e^{-2\lambda s}w(s) \,\mathrm ds\le L^2\sigma ^2t +2L\sigma tW(t). \end{aligned} \end{aligned}
Performing the same computations for the third integral term at the r.h.s. of (6.14) we end up with
\begin{aligned} \begin{aligned} I_2&=\int _0^t e^{-2\lambda s} \left( w_\eta ^2(s-\varepsilon ,s)-w_\tau ^2(s-\varepsilon ,s) \right) \,\mathrm ds \\&\le {L^2 t \sigma ^2} + 2L\sigma \int _0^t e^{-2\lambda s} w(s-\varepsilon ,s) \,\mathrm ds\\&\le L^2\sigma ^2t+2L^2\sigma \varepsilon t+ 2L\sigma \int _0^t e^{-2\lambda s} w(s) \,\mathrm ds\\&\le L^2\sigma ^2t+2L^2\sigma \varepsilon t+ 2L\sigma t W(t). \end{aligned} \end{aligned}
Eventually, using the elementary inequalities,
\begin{aligned} \iint _{Q^\varepsilon _{0,t}} \lambda \mathrm e^{-2\lambda s}\,\mathrm dr\,\mathrm ds\le \frac{\varepsilon }{2},\quad \iint _{Q^\varepsilon _{0,t}} \mathrm e^{-2\lambda s}w(s,s)\,\mathrm dr\,\mathrm ds= \varepsilon \int _0^t \mathrm e^{-2\lambda s} w(s)\,\mathrm ds, \end{aligned}
and $$f(r,s)\le 2L^2(\eta +s)+2Lw(s)$$ for $$r<0$$ and $$f(r,s)=0$$ for $$r\ge 0$$, we get
\begin{aligned} I_3&=\iint _{Q^\varepsilon _{0,t}}Z_{r,s}\,\mathrm dr \mathrm ds \le 2L^2\sigma \varepsilon (\sigma +\varepsilon )+ 4L\lambda \sigma \varepsilon \int _0^t \mathrm e^{-2\lambda s} w(s)\,\mathrm ds+ 2L^2\sigma \varepsilon t \\&\quad + 2\iint _{Q^\varepsilon _{0,\min \{\varepsilon , t\}}}(L^2(\eta +s)+Lw(s))\mathrm e^{-2\lambda s}\,\mathrm dr \mathrm ds \\&\le 2L^2\sigma \varepsilon (\sigma +\varepsilon )+ 2L^2\varepsilon ^2 (\sigma +\varepsilon )+ 2L^2\sigma \varepsilon t + 4L\lambda \sigma \varepsilon tW(t)+ 2L\varepsilon ^2 W(\min \{t, \varepsilon \}). \end{aligned}
We eventually get
\begin{aligned} \sum _{k=0}^3 I_k\le & {} \varepsilon w^2(0)+2L^2\sigma ^2 t+4L^2\sigma \varepsilon t+ 2L^2\varepsilon (\sigma +\varepsilon )^2\nonumber \\&+ 4L\sigma (1+\lambda \varepsilon ) tW(t) +2L\varepsilon ^2 W(\min \{t, \varepsilon \}). \end{aligned}
(6.15)
4. L.h.s. and penalization
We want to use the first integral term in (6.14) to derive a pointwise estimate for w(t); (6.2) and (6.10) yield
\begin{aligned} w(t)=w(t,t)\le L(t-r)+w(r,t)\le L(\tau +|t-r|) +w_\tau (r,t). \end{aligned}
(6.16)
We then square (6.16), use the Young inequality (i.e. $$2ab\le \frac{a^2}{\vartheta }+\vartheta b^2$$ for any $$a,b\ge 0$$, $$\vartheta >0$$), multiply the resulting inequality by $$\frac{e^{-2\lambda t}}{\varepsilon }$$ and integrate over the interval $$(t-\varepsilon ,t)$$. So that, for every $$\delta ,\delta _\star >1$$ conjugate coefficients, we get
\begin{aligned} \mathrm e^{-2\lambda t} w^2(t)\le & {} \frac{\delta }{\varepsilon }\int _{t-\varepsilon }^t e^{-2\lambda t} w_\tau ^2(r,t) \,\mathrm dr + \delta _\star L^2 (\tau +\varepsilon )^2\\\le & {} \frac{\delta }{\varepsilon }(I_0+I_1+I_2+I_3) + \delta _\star L^2 (\tau +\varepsilon )^2, \end{aligned}
with $$I_0, I_1, I_2, I_3$$ as in step 3. Using (6.15) yields
\begin{aligned} \mathrm e^{-2\lambda t} w^2(t)\le & {} (2\delta +\delta _\star ) L^2(\sigma +\varepsilon )^2+ \delta \Big (w^2(0)+ 2L^2\sigma ^2t/\varepsilon +4 L^2\sigma t \Big ) \\&+ \frac{4L(1+\lambda \varepsilon )\sigma \delta }{\varepsilon } tW(t)+ 2L\varepsilon \delta W(\min \{t, \varepsilon \}). \end{aligned}
5. Conclusion.
Choosing $$\varepsilon :=\sqrt{\sigma \,\max \{\sigma , t\}}$$ and assuming $$\lambda \sqrt{T \sigma }\le 1$$, we obtain
\begin{aligned} \mathrm e^{-2\lambda t} w^2(t)&\le \delta w^2(0) + (14\delta +4\delta _\star ) L^2\,\sigma \,\max \{\sigma , t\} +10\delta L\sqrt{\sigma \,\max \{\sigma , t\}} W(t). \end{aligned}
(6.17)
Since the right hand side of (6.17) is an increasing function of t, (6.17) holds even if we substitute the left hand side with $$\mathrm e^{-2\lambda s}w^2(s)$$ for every $$s\in [0,t]$$; we thus obtain the inequality
\begin{aligned} W^2(t) \le \delta w^2(0) + (14\delta +4\delta _\star ) L^2\,\sigma \,\max \{\sigma , t\} +10\delta L\sqrt{\sigma \,\max \{\sigma , t\}} W(t). \end{aligned}
Using the elementary property for positive ab
\begin{aligned} W^2\le a+2bW\quad \Rightarrow \quad W\le b+\sqrt{b^2+a}\le 2b+\sqrt{a}, \end{aligned}
(6.18)
we eventually obtain
\begin{aligned} \mathrm e^{-\lambda t}w(t)&\le \Big (\delta w^2(0)+ (14\delta +4\delta _\star )L^2\,\sigma \,\max \{\sigma , t\}\Big )^{1/2} +10\delta L\sqrt{\sigma \,\max \{\sigma , t\}} \\ {}&\le \sqrt{\delta } w(0)+ C(\delta )L\sqrt{\sigma \,\max \{\sigma , t\}}, \end{aligned}
with $$C(\delta ):=(14\,\delta +4\,\delta _\star )^{1/2}+ 10\,\delta$$. $$\square$$
### Theorem 6.5
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1). If $$\mu :[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ is a $$\lambda$$-EVI solution and $$M_\tau \in \mathscr {M}(M^0_\tau ,\tau ,T,L)$$, then for every $$\delta >1$$ there exists a constant $$C(\delta )$$ such that
\begin{aligned} W_2(\mu _t, M_{\tau }(t))\le \Big (\sqrt{\delta }\, W_2(\mu _0,M^0_\tau )+C(\delta ) L\sqrt{\tau (t+\tau )}\Big ) \mathrm e^{\lambda _+ t} \end{aligned}
for every $$t\in [0,T]$$.
### Remark 6.6
When $$\mu _0=M^0_\tau$$ and $$\lambda \le 0$$ we obtain the optimal error estimate
\begin{aligned} W_2(\mu _t, M_{\tau }(t))\le 13 L\sqrt{\tau (t+\tau )}. \end{aligned}
### Proof
We repeat the same argument of the previous proof, still assuming $$\lambda >0$$, extending $$M_\tau ,{\bar{M}}_\tau , {{\varvec{F}}}_\tau$$ as in (6.9) and setting
\begin{aligned} w(r,s):=W_2(M_\tau (r),\mu _s),\quad w_\tau (r,s):=W_2(\bar{M}_\tau (r),\mu _s). \end{aligned}
We use ($$\lambda$$-EVI) for $$\mu _s$$ with $$\nu ={\bar{M}}_\tau (r)$$ and $$\Phi ={{\varvec{F}}}_\tau (\max \{r, 0\})$$ and (IEVI) for $$M_{\tau }(r)$$ with $$\nu =\mu _s$$ obtaining
\begin{aligned} \frac{\partial }{\partial r} \frac{\mathrm e^{-2\lambda s}}{2} W_2^2(M_{\tau }(r), \mu _s)&\le \mathrm e^{-2\lambda s}\Big (\tau |{{\varvec{F}}}_\tau (r) |_2^2 + \left[ {{\varvec{F}}}_{\tau }(r) , \mu _s\right] _{r}\Big )&s\in [0,T], r \in (-\infty , T) \\ \frac{\partial }{\partial s} \frac{\mathrm e^{-2\lambda s}}{2} W_2^2(\mu _s, {\bar{M}}_\tau (r))&\le -\mathrm e^{-2\lambda s} \left[ {{\varvec{F}}}_\tau (\max \{r, 0\}) , \mu _s\right] _{r}&\text {in } {\mathscr {D}}'(0,T), r\in (-\infty ,T). \end{aligned}
Using [23, Lemma 6.15] we can sum the two contributions obtaining
\begin{aligned} \frac{\partial }{\partial r} \mathrm e^{-2\lambda s} w^2(r,s)+ \frac{\partial }{\partial s} \mathrm e^{-2\lambda s} w_\tau ^2(r,s) \le Z_{r,s}, \end{aligned}
where $$Z_{r,s}:=(2L^2 \tau +2f(r,s))\mathrm e^{-2\lambda s}$$, and
\begin{aligned} f(r,s):= {\left\{ \begin{array}{ll} L W_2(M_\tau (0),\mu _s)=Lw(0, s)&{} \text {if }r<0,\\ 0&{}\text {if }r\ge 0. \end{array}\right. } \end{aligned}
Let $$t \in [0,T]$$ and $$\varepsilon >0$$. Applying the Divergence Theorem in $$Q_{0,t}^\varepsilon$$ (see (5.29) and Figure 1), we get
\begin{aligned}&\int _{t-\varepsilon }^t \mathrm e^{-2\lambda t} w_\tau ^2(r,t) \,\mathrm dr \le \int _{-\varepsilon }^0 w_\tau ^2(r,0)\,\mathrm dr\nonumber \\&\quad + \int _0^t \mathrm e^{-2\lambda s} \left( w_\tau ^2(s,s)-w^2(s,s) \right) \,\mathrm ds + \int _0^t \mathrm e^{-2\lambda s} \left( w^2(s-\varepsilon ,s)- w_\tau ^2(s-\varepsilon ,s) \right) \,\mathrm ds\nonumber \\&\quad + \iint _{Q_{0,t}^\varepsilon } Z_{r,s} \,\mathrm dr\mathrm ds. \end{aligned}
(6.19)
Using
\begin{aligned} w(t,t)\le w(r,t)+L(t-r)\le w_\tau (r,t)+L(\tau +\varepsilon )\quad \text {if }t-\varepsilon \le r\le t, \end{aligned}
we get for every $$\delta ,\delta _\star >1$$ conjugate coefficients ($$\delta _\star =\delta /(\delta -1)$$)
\begin{aligned} \mathrm e^{-2\lambda t} w^2(t) \le \frac{\delta }{\varepsilon }\int _{t-\varepsilon }^t e^{-2\lambda t} w_\tau ^2(r,t) \,\mathrm dr + \delta _\star L^2 (\tau +\varepsilon )^2. \end{aligned}
(6.20)
Similarly to (6.11) we have
\begin{aligned} |w_\tau (s,s)-w(s,s)| \le L\tau ,\quad |w^2_\tau (s,s)-w^2(s,s)| \le L\tau \Big ( L\tau +2w(s) \Big ) \end{aligned}
and, after an integration,
\begin{aligned} \begin{aligned}&\int _0^t e^{-2\lambda s} \left( w^2_\tau (s,s)-w^2(s,s) \right) \,\mathrm ds \le {L^2 t \tau ^2} +2L\tau \int _0^t e^{-2\lambda s}w(s) \,\mathrm ds. \end{aligned} \end{aligned}
(6.21)
Performing the same computations for the third integral term at the r.h.s. of (6.19) we end up with
\begin{aligned} \begin{aligned} \int _0^t e^{-2\lambda s} \left( w^2(s-\varepsilon ,s)-w_\tau ^2(s-\varepsilon ,s) \right) \,\mathrm ds&\le {L^2t \tau ^2} + 2L\tau \int _0^t e^{-2\lambda s} w(s-\varepsilon ,s) \,\mathrm ds\\&\le L^2 t \tau (\tau +2 \varepsilon ) + 2L\tau \int _0^t e^{-2\lambda s} w(s) \,\mathrm ds. \end{aligned} \end{aligned}
(6.22)
Finally, since if $$r<0$$ we have $$f(r,s)=Lw(0,s)\le L^2s+Lw(s,s)$$, then
\begin{aligned} \varepsilon ^{-1}\iint _{Q^\varepsilon _{0,t}}Z_{r,s}\,\mathrm dr \mathrm ds&\le 2L^2t\tau +\varepsilon ^{-1}\iint _{Q^\varepsilon _{0,\min \{\varepsilon , t\}}}2f(r,s)\mathrm e^{-2\lambda s}\,\mathrm dr \mathrm ds \nonumber \\&\le 2L^2t\tau +L^2 \varepsilon ^2+2L\varepsilon \sup _{0\le s\le \min \{\varepsilon , t\}}\mathrm e^{-\lambda s}w(s). \end{aligned}
(6.23)
Using (6.21), (6.22), (6.23) in (6.19), we can rewrite the bound in (6.20) as
\begin{aligned} \begin{aligned} \mathrm e^{-2\lambda t} w^2(t)&\le \delta _\star L^2(\tau +\varepsilon )^2+ \delta \Big (w^2(0)+ 2L^2t\tau ^2/\varepsilon +2L^2t\tau +L^2\varepsilon ^2\\&\quad + 2L\varepsilon \sup _{0\le s\le \min \{\varepsilon , t\}}\mathrm e^{-\lambda s}w(s) \Big )\\&\quad +\frac{4\delta L\tau }{\varepsilon } \int _0^t e^{-2\lambda s} w(s) \,\mathrm ds. \end{aligned} \end{aligned}
Choosing $$\varepsilon :=\sqrt{\tau \,\max \{\tau , t\}}$$ we get
\begin{aligned} \mathrm e^{-2\lambda t} w^2(t)\le & {} 4\,\delta _\star L^2\,\tau \,\max \{\tau , t\}+ \delta \Big (w^2(0)+ 5L^2\,\tau \,\max \{\tau , t\}\Big ) \\&+6\,\delta L\sqrt{\tau \,\max \{\tau , t\}} \sup _{0\le s\le t}\mathrm e^{-\lambda s}w(s). \end{aligned}
A further application of (6.18) yields
\begin{aligned} \mathrm e^{-\lambda t} w(t)&\le \Big (\delta w^2(0)+ (5\delta +4\delta _\star ) L^2\,\tau \,\max \{\tau , t\}\Big )^{1/2} +6\delta L\sqrt{\tau \,\max \{\tau , t\}} \\ {}&\le \sqrt{\delta }w(0)+ C(\delta )L\sqrt{t+\tau }\sqrt{\tau }, \end{aligned}
with $$C(\delta ):=(5\delta +4\delta _\star )^{1/2}+6\delta$$. $$\square$$
As proved in the following, the limit curve of the interpolants $$(M_{\tau })_{\tau >0}$$ of the Euler Scheme defined in (5.9) is actually a $$\lambda$$-EVI solution of (5.1).
### Theorem 6.7
Let $${\varvec{\mathrm {F}}}$$ be a $$\lambda$$-dissipative MPVF according to (4.1) and let $$n\mapsto \tau (n)$$ be a vanishing sequence of time steps, let $$(\mu _{0,n})_{n\in {\mathbb {N}}}$$ be a sequence in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ converging to $$\mu _0\in \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ in $$\mathcal {P}_2({\textsf {X} })$$ and let $$M_n\in {\mathscr {M}}(\mu _{0,n},\tau (n),T,L)$$. Then $$M_n$$ is uniformly converging to a Lipschitz continuous limit curve $$\mu :[0,T]\rightarrow \overline{\mathrm {D}({\varvec{\mathrm {F}}})}$$ which is a $$\lambda$$-EVI solution starting from $$\mu _0$$.
### Proof
Theorem 6.4 shows that $$M_n$$ is a Cauchy sequence in $$\mathrm C([0,T];\overline{\mathrm {D}({\varvec{\mathrm {F}}})})$$, so that there exists a unique limit curve $$\mu$$ as $$n\rightarrow \infty$$. Moreover, $$\mu$$ is also L-Lipschitz and, recalling (5.14), we have that $$\mu$$ is also the uniform limit of $$\bar{M}_{\tau (n)}$$.
Let us fix a reference measure $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ and $$\Phi \in {\varvec{\mathrm {F}}}[\nu ]$$. The (IEVI) and the $$\lambda$$-dissipativity of $${\varvec{\mathrm {F}}}$$ yield
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt} \frac{1}{2} W_2^2(M_{n}(t), \nu )&\le \tau (n) | {{\varvec{F}}}_{\tau (n)}(t)|_2^2 + \left[ {{\varvec{F}}}_{\tau (n)}, \nu \right] _{r}\\&\le \tau (n)\, L^2 + \lambda W_2^2(\bar{M}_{\tau (n)}(t), \nu ) -\left[ \Phi , \bar{M}_{\tau (n)}(t)\right] _{r} \end{aligned}
for a.e. $$t \in [0,T]$$. Integrating the above inequality in an interval $$(t,t+h)\subset [0,T]$$ we get
\begin{aligned}&\frac{W_2^2(M_n(t+h),\nu )- W_2^2(M_n(t),\nu ) }{2h} \le \tau (n) L^2\nonumber \\&\quad +\frac{1}{h}\int _t^{t+h} \Big (\lambda W_2^2(\bar{M}_{\tau (n)}(s), \nu ) -\left[ \Phi , \bar{M}_{\tau (n)}(s)\right] _{r}\Big )\,\mathrm ds. \end{aligned}
(6.24)
Notice that as $$n \rightarrow + \infty$$, by (5.14), we have
\begin{aligned} \liminf _{n \rightarrow + \infty }\left[ \Phi , {\bar{M}}_{\tau (n)}(s)\right] _{r}\ge \left[ \Phi , \mu _s\right] _{r} \end{aligned}
for every $$s \in [0,T]$$, together with the uniform bound given by
\begin{aligned} \left| \left[ \Phi , \bar{M}_{\tau (n)}(s)\right] _{r} \right| \le \frac{1}{2}W_2^2(\bar{M}_{\tau (n)}(s), \nu ) + \frac{1}{2} |\Phi |_2^2 \end{aligned}
for every $$s \in [0,T]$$. Thanks to Fatou’s Lemma and the uniform convergence given by Theorem 6.4, we can pass to the limit as $$n \rightarrow + \infty$$ in (6.24) obtaining
\begin{aligned} \frac{W_2^2(\mu _{t+h},\nu )- W_2^2(\mu _t,\nu ) }{2h} \le \frac{1}{h}\int _t^{t+h} \Big (\lambda W_2^2(\mu _s, \nu ) -\left[ \Phi , \mu _s\right] _{r}\Big )\,\mathrm ds. \end{aligned}
A further limit as $$h\downarrow 0$$ yields
\begin{aligned} \frac{1}{2}{\frac{\mathrm d}{\mathrm dt}}^{+}W_2^2(\mu _t,\nu )\le \lambda W_2^2(\mu _t,\nu )- \left[ \Phi , \mu _t\right] _{r} \end{aligned}
which provides ($$\lambda$$-EVI). $$\square$$
## 7 Examples of $$\lambda$$-dissipative MPVFs and $$\lambda$$-flows
In the first part of this section, we present significant examples of $$\lambda$$-dissipative MPVFs which are interesting for applications. In Sect. 7.4, we give some examples of MPVFs generating $$\lambda$$-flows with particular properties. We then conclude with Sect. 7.5, where we compare our framework with that developed in [27], revisiting in particular the splitting particle example in Example 7.11.
### 7.1 Subdifferentials of $$\lambda$$-convex functionals
Recall that a functional $$\mathcal {F}:\mathcal {P}_2({\textsf {X} })\rightarrow (-\infty ,+\infty ]$$ is $$\lambda$$-(geodesically) convex on $$\mathcal {P}_2({\textsf {X} })$$ (see [3, Definition 9.1.1]) if for any $$\mu _0,\mu _1$$ in the proper domain $$D(\mathcal {F}):=\{\mu \in \mathcal {P}_2({\textsf {X} })\mid \mathcal {F}(\mu )<+\infty \}$$ there exists $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$ such that
\begin{aligned} \mathcal {F}(\mu _t)\le (1-t)\mathcal {F}(\mu _0)+t\mathcal {F}(\mu _1)-\frac{\lambda }{2}t(1-t) W_2^2(\mu _0,\mu _1) \end{aligned}
for every $$t\in [0,1]$$, where $$\mu :[0,1]\rightarrow \mathcal {P}_2({\textsf {X} })$$ is the constant speed geodesic induced by $$\varvec{\mu }$$, i.e. $$\mu _t={\textsf {x} }^t_{\sharp }\varvec{\mu }$$.
The Fréchet subdifferential $$\varvec{\partial }\mathcal {F}$$ of $$\mathcal {F}$$ [3, Definition 10.3.1] is a MPVF which can be characterized [3, Theorem 10.3.6] by
\begin{aligned}&\Phi \in \varvec{\partial }\mathcal {F}[\mu ] \quad \Leftrightarrow \quad \mu \in D(\mathcal {F}), \mathcal {F}(\nu )-\mathcal {F}(\mu )\ge -\left[ \Phi , \nu \right] _{l} +\frac{\lambda }{2}W_2^2(\mu ,\nu )\\&\quad \text {for every }\nu \in D(\mathcal {F}). \end{aligned}
According to the notation introduced in (3.15), we set
\begin{aligned} -\varvec{\partial }\mathcal {F}[\mu ]= J_\sharp \varvec{\partial }\mathcal {F}[\mu ],\quad \text {with } J(x,v):=(x,-v), \end{aligned}
(7.1)
and we have the following result.
### Theorem 7.1
If $$\mathcal {F}:\mathcal {P}_2({\textsf {X} })\rightarrow (-\infty ,+\infty ]$$ is a proper, lower semicontinuous and $$\lambda$$-convex functional, then $$-\varvec{\partial }\mathcal {F}$$ is a $$(-\lambda )$$-dissipative MPVF according to (4.1).
In the following proposition, we prove a correspondence between gradient flows for $$\mathcal {F}$$ and $$(-\lambda )$$-EVI solutions for the MPVF $$-\varvec{\partial } \mathcal {F}$$. We refer respectively to (4.7), (4.12) and Definition 4.11 for the definitions of $$\mathrm I(\varvec{\mu }|{\varvec{\mathrm {F}}})$$, $$\Gamma _o^{0}({\cdot },{\cdot }|{\varvec{\mathrm {F}}})$$ and $$[{\varvec{\mathrm {F}}},\varvec{\mu }]_{0+}$$.
### Proposition 7.2
Let $$\mathcal {F}: \mathcal {P}_2({\textsf {X} }) \rightarrow (-\infty , + \infty ]$$ be a proper, lower semicontinuous and $$\lambda$$-convex functional and let $$\mu :{\mathcal {I}}\rightarrow \mathrm {D}(\varvec{\partial }\mathcal {F})$$ be a locally absolutely continuous curve, with $${\mathcal {I}}$$ a (bounded or unbounded) interval in $${\mathbb {R}}$$. Then
1. (1)
if $$\mu$$ is a Gradient Flow for $$\mathcal {F}$$ i.e.
\begin{aligned} ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp }\mu _t \in -\varvec{\partial } \mathcal {F}(\mu _t) \quad \text { a.e. } t \in {\mathcal {I}}, \end{aligned}
then $$\mu$$ is a $$(-\lambda )$$-EVI solution of (5.1) for the MPVF $$-\varvec{\partial } \mathcal {F}$$ as in (7.1);
2. (2)
if $$\mu$$ is a $$(-\lambda )$$-EVI solution of (5.1) for the MPVF $$-\varvec{\partial } \mathcal {F}$$ and the domain of $$\varvec{\partial } \mathcal {F}$$ satisfies
\begin{aligned} \text { for a.e. } t \in {\mathcal {I}}, \, \Gamma _o^{0}({\mu _t},{\nu }|\varvec{\partial } \mathcal {F}) \ne \emptyset \quad \text {for every }\, \nu \in \mathrm {D}(\varvec{\partial }\mathcal {F}), \end{aligned}
then $$\mu$$ is a Gradient Flow for $$\mathcal {F}$$.
### Proof
The first assertion is a consequence Theorem 5.4(1). We prove the second claim; by (5.5b) we have that for a.e. $$t \in {\mathcal {I}}$$ it holds
\begin{aligned} \left[ ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r} \le [({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t)_{\sharp }\mu _t,\varvec{\mu }_t]_{r,0} \le [-\varvec{\partial } \mathcal {F},\varvec{\mu }_t]_{0+} \end{aligned}
for every $$\nu \in \mathrm {D}(\mathcal {F})$$ and $$\varvec{\mu }_t \in \Gamma _o^{0}({\mu _t},{\nu }|\varvec{\partial } \mathcal {F})$$. We show that for every $$\nu _0, \nu _1 \in \mathrm {D}(\varvec{\partial } \mathcal {F})$$ and every $$\varvec{\nu }\in \Gamma _o^{0}({\nu _0},{\nu _1}|{\varvec{\mathrm {F}}})$$
\begin{aligned}{}[-\varvec{\partial } \mathcal {F},\varvec{\nu }]_{0+} \le \mathcal {F}(\nu _1) - \mathcal {F}(\nu _0) - \frac{\lambda }{2}W_2^2(\nu _0, \nu _1). \end{aligned}
(7.2)
To prove that, we take $$s \in \mathrm I(\varvec{\nu }|\varvec{\partial } \mathcal {F})\cap (0,1)$$ and $$\Phi _s \in - \varvec{\partial } \mathcal {F}(\nu _s)$$, where $$\nu _s := {\textsf {x} }^{s}_{\sharp }\varvec{\nu }$$. By definition of subdifferential we have
\begin{aligned} \left[ \Phi _s, \nu _1\right] _{r} \le \mathcal {F}(\nu _1)-\mathcal {F}(\nu _s)- \frac{\lambda }{2} W_2^2(\nu _s, \nu _1). \end{aligned}
Dividing by $$(1-s)$$, using (3.29) and passing to the infimum w.r.t. $$\Phi _s \in - \varvec{\partial } \mathcal {F}(\nu _s)$$ we obtain
\begin{aligned}{}[-\varvec{\partial } \mathcal {F},\varvec{\nu }]_{r,s} \le \frac{1}{1-s}\left( \mathcal {F}(\nu _1) - \mathcal {F}(\nu _s) \right) - \frac{\lambda (1-s)}{2}W_2^2(\nu _0, \nu _1). \end{aligned}
Passing to the limit as $$s \downarrow 0$$ and using the lower semicontinuity of $$\mathcal {F}$$ lead to the result. Once that (7.2) is established we have that for a.e. $$t \in {\mathcal {I}}$$ it holds
\begin{aligned} \left[ ({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t)_{\sharp }\mu _t, \nu \right] _{r} \le \mathcal {F}(\nu ) - \mathcal {F}(\mu _t) - \frac{\lambda }{2}W_2^2(\mu _t, \nu )\quad \text { for every }\nu \in \mathrm {D}(\varvec{\partial }\mathcal {F}). \end{aligned}
(7.3)
To conclude it is enough to use the lower semicontinuity of the l.h.s. (see Lemma 3.15) and the fact that $$\mathrm {D}(\varvec{\partial }\mathcal {F})$$ is dense in $$\mathrm {D}(\mathcal {F})$$ in energy: indeed we can apply [25, Corollary 4.5] and [3, Lemma 3.1.2] to the proper, lower semicontinuous and convex functional $$\mathcal {F}^{\lambda }: \mathcal {P}_2({\textsf {X} }) \rightarrow (-\infty , + \infty ]$$ defined as
\begin{aligned} \mathcal {F}^{\lambda }(\nu )=\mathcal {F}(\nu ) - \frac{\lambda }{2}{\textsf {m} }_2^2(\nu ) \end{aligned}
to get the existence, for every $$\nu \in \mathrm {D}(\mathcal {F})$$, of a family $$(\nu ^\tau )_{\tau >0} \subset \mathrm {D}(\mathcal {F}^{\lambda }) = \mathrm {D}(\mathcal {F})$$ s.t.
\begin{aligned} \nu ^{\tau } \rightarrow \nu , \quad \mathcal {F}^{\lambda }(\nu ^{\tau }) \rightarrow \mathcal {F}^{\lambda }(\nu ) \quad \text { as } \tau \downarrow 0. \end{aligned}
Of course $$\mathcal {F}(\nu ^{\tau }) \rightarrow \mathcal {F}(\nu )$$ as $$\tau \downarrow 0$$ and, applying [3, Lemma 10.3.4], we see that $$\nu ^{\tau } \in \mathrm {D}(\varvec{\partial }\mathcal {F}^{\lambda })$$. However $$\varvec{\partial }\mathcal {F}^{\lambda } = L^{\lambda }_{\sharp } \varvec{\partial }\mathcal {F}$$ (see (4.4)) so that $$\nu ^\tau \in \mathrm {D}(\varvec{\partial }\mathcal {F})$$. We can thus write (7.3) for $$\nu ^{\tau }$$ in place of $$\nu$$ and pass to the limit as $$\tau \downarrow 0$$, obtaining that, by definition of subdifferential, $$({\varvec{i}}_{\textsf {X} },{\varvec{v}}_t)_{\sharp }\mu _t \in -\varvec{\partial }\mathcal {F}(\mu _t)$$ for a.e. $$t\in {\mathcal {I}}$$. $$\square$$
Referring to [3], here we list interesting and explicit examples of $$(-\lambda )$$-dissipative MPVFs, according to (4.1), induced by proper, lower semicontinuous and $$\lambda$$-convex functionals, focusing on the cases when $$\mathrm {D}(\varvec{\partial }\mathcal {F})=\mathcal {P}_2({\textsf {X} }).$$
1. (1)
Potential energy. Let $$P:{\textsf {X} }\rightarrow {\mathbb {R}}$$ be a l.s.c. and $$\lambda$$-convex functional satisfying
\begin{aligned} |\partial ^o P(x)| \le C(1+|x|) \quad \text {for every }x \in {\textsf {X} }, \end{aligned}
for some constant $$C>0$$, where $$\partial ^o P(x)$$ is the element of minimal norm in $$\partial P(x)$$. By [3, Proposition 10.4.2] the PVF
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ]:=({\varvec{i}}_{{\textsf {X} }} , -\partial ^o P)_{\sharp } \mu , \quad \mu \in \mathcal {P}_2({\textsf {X} }), \end{aligned}
is a $$(-\lambda )$$-dissipative selection of $$-\varvec{\partial }{\mathcal {F}}_P$$ for the potential energy functional
\begin{aligned} {\mathcal {F}}_P(\mu ) := \int _{\textsf {X} }P \,\mathrm d\mu , \quad \mu \in \mathcal {P}_2({\textsf {X} }). \end{aligned}
2. (2)
Interaction energy. If $$W:{\textsf {X} }\rightarrow [0,+\infty )$$ is an even, differentiable, and $$\lambda$$-convex function for some $$\lambda \in {\mathbb {R}}$$, whose differential has a linear growth, then, by [3, Theorem 10.4.11], the PVF
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ]:= \left( {\varvec{i}}_{{\textsf {X} }} , (-\nabla W *\mu )\right) _{\sharp }\mu , \quad \mu \in \mathcal {P}_2({\textsf {X} }), \end{aligned}
is a $$(-\lambda )$$-dissipative selection of $$-\varvec{\partial }{\mathcal {F}}_W$$, the opposite of the Wasserstein subdifferential of the interaction energy functional
\begin{aligned} \mathcal {F}_W(\mu ) := \frac{1}{2} \int _{{\textsf {X} }^2} W(x-y) \,\mathrm d(\mu \otimes \mu )(x,y) , \quad \mu \in \mathcal {P}_2({\textsf {X} }). \end{aligned}
3. (3)
Opposite Wasserstein distance. Let $$\bar{\mu } \in \mathcal {P}_2({\textsf {X} })$$ be fixed and consider the functional $$\mathcal {F}_{\text {Wass}} : \mathcal {P}_2({\textsf {X} }) \rightarrow {\mathbb {R}}$$ defined as
\begin{aligned} \mathcal {F}_{\text {Wass}}(\mu ) := - \frac{1}{2} W_2^2(\mu , \bar{\mu }), \quad \mu \in \mathcal {P}_2({\textsf {X} }), \end{aligned}
which is geodesically $$(-1)$$-convex [3, Proposition 9.3.12]. Setting
the PVF
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ]:=({\varvec{i}}_{\textsf {X} }, {\varvec{i}}_{\textsf {X} }-{\varvec{b}}(\mu ) )_{\#}\mu ,\quad \mu \in \mathcal {P}_2({\textsf {X} }) \end{aligned}
is a selection of $$-\varvec{\partial }\mathcal {F}_{\text {Wass}}(\mu )$$ and it is therefore 1-dissipative according to (4.1).
### 7.2 MPVF concentrated on the graph of a multifunction
The previous example of Sect. 7.1 has a natural generalization in terms of dissipative graphs in $${\textsf {X} }\times {\textsf {X} }$$ [1, 2, 7]. We consider a (non-empty) $$\lambda$$-dissipative set $${\mathrm F}\subset {\textsf {X} }\times {\textsf {X} }$$, i.e. satisfying
\begin{aligned} \langle v_0-v_1, x_0-x_1\rangle \le \lambda |x_0-x_1|^2\quad \text { for every }(x_0,v_0), (x_1,v_1)\in {\mathrm F}. \end{aligned}
The corresponding MPVF defined as
\begin{aligned} {\varvec{\mathrm {F}}}:=\left\{ \Phi \in \mathcal {P}_2(\mathsf {TX})\mid \Phi \text { is concentrated on }{\mathrm F}\right\} \end{aligned}
is $$\lambda$$-dissipative as well, according to (4.1). In fact, if $$\Phi _0,\Phi _1\in {\varvec{\mathrm {F}}}$$ with $$\nu _i={\textsf {x} }_\sharp \Phi _i$$, $$i=0,1$$, and $$\varvec{\Theta }\in \Lambda (\Phi _0,\Phi _1)$$ then $$(x_0,v_0,x_1,v_1)\in {\mathrm F}\times {\mathrm F}$$ $$\varvec{\Theta }$$-a.e., so that
\begin{aligned} \int _{\mathsf {TX}\times \mathsf {TX}} \langle v_0-v_1, x_0-x_1\rangle \,\mathrm d\varvec{\Theta }(x_0,v_0,x_1,v_1) \le \lambda \int _{\mathsf {TX}\times \mathsf {TX}}|x_0-x_1|^2\,\mathrm d\varvec{\Theta }=\lambda W_2^2(\nu _0,\nu _1). \end{aligned}
since $$({\textsf {x} }^0,{\textsf {x} }^1)_\sharp \varvec{\Theta }\in \Gamma _o(\nu _0,\nu _1)$$. Taking the supremum w.r.t. $$\varvec{\Theta }\in \Lambda (\Phi _0,\Phi _1)$$ we obtain $$\left[ \Phi _0, \Phi _1\right] _{l}\le \lambda W_2^2(\nu _0,\nu _1)$$ which is even stronger than $$\lambda$$-dissipativity. If $$\mathrm {D}({\mathrm F})={\textsf {X} }$$ then $$\mathrm {D}({\varvec{\mathrm {F}}})$$ contains $$\mathcal {P}_\mathrm{c}({\textsf {X} })$$, the set of Borel probability measures with compact support. If $${\mathrm F}$$ has also a linear growth, then it is easy to check that $$\mathrm {D}({\varvec{\mathrm {F}}})=\mathcal {P}_2({\textsf {X} })$$ as well.
Despite the analogy just shown with dissipative operators in Hilbert spaces, there are important differences with the Wasserstein framework, as highlighted in the following examples. In particular, in Sect. 4.2 we showed how dissipativity allows to deduce relevant properties when the MPVF $${\varvec{\mathrm {F}}}$$ is tested against optimal directions. On the contrary, whenever $${\textsf {v} }_\sharp {\varvec{\mathrm {F}}}[\mu ]$$ is orthogonal to $${{\,\mathrm{Tan}\,}}_\mu \mathcal {P}_2({\textsf {X} })$$, we are not able to deduce informations through the dissipativity assumption, as shown in Example 7.3 and Example 7.4.
### Example 7.3
Let $${\textsf {X} }={\mathbb {R}}^2$$, let $$B:=\{ x \in {\mathbb {R}}^2 \mid |x| \le 1\}$$ be the closed unit ball, let $${\mathcal {L}_B}$$ be the (normalized) Lebesgue measure on B, and let $${{\varvec{r}}}:{\mathbb {R}}^2\rightarrow {\mathbb {R}}^ 2$$, $${{\varvec{r}}}(x_1,x_2)=(x_2,-x_1)$$ be the anti-clockwise rotation of $$\pi /2$$ degrees. We define the MPVF
\begin{aligned} {\varvec{\mathrm {F}}}[\nu ] = {\left\{ \begin{array}{ll} ({\varvec{i}}_{{\mathbb {R}}^2}, 0)_{\sharp } \nu , \quad &{}\text { if } \nu \in \mathcal {P}_2({\mathbb {R}}^2) {\setminus } \{ {\mathcal {L}_B}\}, \\ \left\{ ({\varvec{i}}_{{\mathbb {R}}^2}, a{{\varvec{r}}})_{\sharp }{\mathcal {L}_B}\mid a \in {\mathbb {R}}\right\} , \quad &{}\text { if } \nu = {\mathcal {L}_B}. \end{array}\right. } \end{aligned}
Observe that $$\mathrm {D}({\varvec{\mathrm {F}}}) = \mathcal {P}_2({\mathbb {R}}^2)$$ and $${\varvec{\mathrm {F}}}$$ is obviously unbounded at $$\nu = {\mathcal {L}_B}$$, i.e.
\begin{aligned} \sup \left\{ |\Phi |_2\,:\,\Phi \in {\varvec{\mathrm {F}}}[{\mathcal {L}_B}]\right\} =+\infty . \end{aligned}
The MPVF $${\varvec{\mathrm {F}}}$$ is also dissipative with $$\lambda =0$$ according to (4.1): indeed, thanks to Remark 3.6 it is enough to check that
\begin{aligned} \left[ ({\varvec{i}}_{{\mathbb {R}}^2}, a{{\varvec{r}}})_{\sharp } {\mathcal {L}_B}, \nu \right] _{r} =0 \quad \text { for every } \nu \in \mathcal {P}_2({\mathbb {R}}^2), \, a \in {\mathbb {R}}. \end{aligned}
(7.4)
To prove (7.4), we notice that the optimal transport plan from $${\mathcal {L}_B}$$ to $$\nu$$ is concentrated on a map which belongs to the tangent space $${{\,\mathrm{Tan}\,}}_{{\mathcal {L}_B}}\mathcal {P}_2({\mathbb {R}}^2)$$ [3, Prop. 8.5.2]; by Remark 3.19 we have just to check that
\begin{aligned} \int _{{\mathbb {R}}^2} \langle {{\varvec{r}}}(x), \nabla \varphi (x)\rangle \,\mathrm d{\mathcal {L}_B}(x) =0 \quad \text { for every } \varphi \in \mathrm {C}^{\infty }_c({\mathbb {R}}^2), \end{aligned}
that is a consequence of the Divergence Theorem on B. This example is in contrast with the Hilbertian theory of dissipative operators according to which an everywhere defined dissipative operator is locally bounded (see [7, Proposition 2.9]).
### Example 7.4
In the same setting of the previous example, let us define the MPVF
\begin{aligned} {\varvec{\mathrm {F}}}[\nu ] = ({\varvec{i}}_{{\mathbb {R}}^2}, {{\varvec{r}}})_{\sharp } \nu , \quad {{\varvec{r}}}(x_1,x_2)=(x_2,-x_1),\quad \nu \in \mathcal {P}_2({\mathbb {R}}^2). \end{aligned}
It is easy to check that $${\varvec{\mathrm {F}}}$$ is dissipative according to (4.1) and Lipschitz continuous (as a map from $$\mathcal {P}_2({\mathbb {R}}^2)$$ to $$\mathcal {P}_2(\mathrm {T}{\mathbb {R}}^2)$$). Moreover, arguing as in Example 7.3, we can show that $$({\varvec{i}}_{{\mathbb {R}}^d}, 0)_{\sharp } {\mathcal {L}_B}\in {{\hat{{\varvec{\mathrm {F}}}}}} [{\mathcal {L}_B}]$$, where $${{\hat{{\varvec{\mathrm {F}}}}}}$$ is defined in (4.22). This is again in contrast with the Hilbertian theory of dissipative operators, stating that a single valued, everywhere defined, and continuous dissipative operator coincides with its maximal extension (see [7, Proposition 2.4]).
### 7.3 Interaction field induced by a dissipative map
Let us consider the Hilbert space $${\textsf {Y} }={\textsf {X} }^n$$, $$n\in {\mathbb {N}}$$, endowed with the scalar product $$\langle {\varvec{x}},\varvec{y}\rangle :=\frac{1}{n} \sum _{i=1}^n \langle x_i,y_i\rangle$$, for every $${\varvec{x}}=(x_i)_{i=1}^n,\ {\varvec{y}}=(y_i)_{i=1}^n\in {\textsf {X} }^n$$. We identify $$\mathsf {TY}$$ with $$(\mathsf {TX})^n$$ and we denote by $${\textsf {x} }^i,{\textsf {v} }^i$$ the i-th coordinate maps. Every permutation $$\sigma :\{1,\ldots ,n\}\rightarrow \{1,\ldots ,n\}$$ in $$\mathrm {Sym}(n)$$ operates on $${\textsf {Y} }$$ by the obvious formula $$\sigma (\varvec{x})_i=x_{\sigma (i)}$$, $$i=1,\ldots ,n$$, $${\varvec{x}}\in {\textsf {Y} }$$.
Let $$G: {\textsf {Y} }\rightarrow {\textsf {Y} }$$ be a Borel $$\lambda$$-dissipative map bounded on bounded sets (this property is always true if $${\textsf {Y} }$$ has finite dimension) and satisfying
\begin{aligned} {{\varvec{x}}}\in \mathrm {D}(G)\quad \Rightarrow \quad \sigma ({{\varvec{x}}})\in \mathrm {D}(G),\ G(\sigma ({{\varvec{x}}}))=\sigma (G({{\varvec{x}}}))\quad \text {for every permutation }\sigma .\nonumber \\ \end{aligned}
(7.5)
Denoting by $$(G^1,\ldots ,G^n)$$ the components of G, by $${\textsf {x} }^i$$ the projections from $${\textsf {Y} }$$ to $${\textsf {X} }$$ and by $$\mu ^{\otimes n}=\bigotimes _{i=1}^n\mu$$, we have that the MPVF
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ]:=({\textsf {x} }^1,G^1)_\sharp \mu ^{\otimes n} \quad \text {with domain } \mathrm {D}({\varvec{\mathrm {F}}}):=\mathcal {P}_b({\textsf {X} }) \end{aligned}
is $$\lambda$$-dissipative as well according to (4.1). Indeed, let $$\mu ,\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$, $$\varvec{\gamma }\in \Gamma _o(\mu ,\nu )$$ and let
\begin{aligned} \Phi =({\textsf {x} }^1,G^1)_\sharp \mu ^{\otimes n}\quad \text {and}\quad \Psi = ({\textsf {x} }^1,G^1)_\sharp \nu ^{\otimes n}. \end{aligned}
We can consider the plan $$\varvec{\beta }:= P_\sharp \varvec{\gamma }^{\otimes n}\in \Gamma (\mu ^{\otimes n},\nu ^{\otimes n})$$, where
\begin{aligned} P((x_1,y_1),\ldots ,(x_n,y_n) ):= ((x_1,\ldots ,x_n),(y_1,\ldots , y_n)). \end{aligned}
Considering the map $$H^1({{\varvec{x}}},{{\varvec{y}}}):=(x_1,G^1({{\varvec{x}}}),y_1,G^1({{\varvec{y}}}))$$ we have $$\varvec{\Theta }:=H^1_\sharp \varvec{\beta }\in \Lambda (\Phi ,\Psi )$$, so that
\begin{aligned} \left[ \Phi , \Psi \right] _{r}&\le \int \langle v_1-w_1,x_1-y_1\rangle \,\mathrm d\varvec{\Theta }(x_1,v_1,y_1,w_1)\\&= \int \langle G^1({{\varvec{x}}})-G^1({{\varvec{y}}}),x_1-y_1\rangle \,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}}) \\ {}&=\frac{1}{n}\sum _{k=1}^n \int \langle G^k({{\varvec{x}}})-G^k({{\varvec{y}}}),x_k-y_k\rangle \,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}})\\&= \int \langle G({{\varvec{x}}})-G({{\varvec{y}}}),{{\varvec{x}}}-{{\varvec{y}}}\rangle \,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}}), \end{aligned}
where we used (7.5) and the invariance of $$\varvec{\beta }$$ with respect to permutations. The $$\lambda$$-dissipativity of G then yields
\begin{aligned} \int \langle G({{\varvec{x}}})-G({{\varvec{y}}}),{{\varvec{x}}}-{{\varvec{y}}}\rangle \,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}})&\le \lambda \int |{{\varvec{x}}}-{{\varvec{y}}}|^2_{\textsf {Y} }\,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}})\\&= \lambda \frac{1}{n}\sum _{k=1}^n \int |x_k-y_k|^2_{\textsf {Y} }\,\mathrm d\varvec{\beta }({{\varvec{x}}},{{\varvec{y}}})\\&= \lambda \frac{1}{n}\sum _{k=1}^n \int |x_k-y_k|^2_{\textsf {Y} }\,\mathrm d\varvec{\varvec{\gamma }}(x_k,y_k) =\lambda W_2^2(\mu ,\nu ). \end{aligned}
A typical example when $$n=2$$ is provided by
\begin{aligned} G(x_1,x_2):=(A(x_1-x_2),A(x_2-x_1)) \end{aligned}
where $$A:{\textsf {X} }\rightarrow {\textsf {X} }$$ is a Borel, locally bounded, dissipative and antisymmetric map satisfying $$A(-z)=-A(z)$$. We easily get
\begin{aligned} \langle&G({{\varvec{x}}})-G({{\varvec{y}}}),{{\varvec{x}}}-{{\varvec{y}}}\rangle \\ {}&= \frac{1}{2} \Big (\langle A(x_1-x_2)-A(y_1-y_2),x_1-y_1\rangle - \langle A(x_1-x_2)-A(y_1-y_2),x_2-y_2\rangle \Big ) \\ {}&= \frac{1}{2} \langle A(x_1-x_2)-A(y_1-y_2),x_1-x_2-(y_1-y_2)\rangle \le 0. \end{aligned}
In this case
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ]=({\varvec{i}}_{\textsf {X} },{\varvec{a}}[\mu ])_\sharp \mu ,\quad {\varvec{a}}[\mu ](x)=\int _{\textsf {X} }A(x-y)\,\mathrm d\mu (y) \quad \text {for every }x\in {\textsf {X} }. \end{aligned}
### 7.4 A few borderline examples
In this subsection, we collect a few examples which reveal the importance of some of the technical tools we developed in Sect. 5. First of all we exhibit an example of dissipative MPVF generating a 0-flow, for which solutions starting from given initial data are merely continuous. In particular, the nice regularizing effect of gradient flows (see [6] for the Hilbert case and [3, Theorem 4.0.4, Theorem 11.2.1] for the general metric and Wasserstein settings), according to which a solution belongs to the domain of the functional for any $$t>0$$ even if the initial datum merely belongs to its closure, does not hold for general dissipative evolutions. This also clarifies the interest in a definition of continuous, not necessarily absolutely continuous, solution given in Definition 5.1.
### Example 7.5
(Lifting of dissipative evolutions and lack of regularizing effect) Let us consider the situation of Corollary 5.24, choosing the Hilbert space $${\textsf {X} }=\ell ^2({\mathbb {N}})$$. Following [31, Example 3] we can easily find a maximal linear dissipative operator $$A: \mathrm {D}(A)\subset \ell ^2({\mathbb {N}}) \rightarrow \ell ^2({\mathbb {N}})$$ whose semigroup does not provide a regularizing effect. We define A as
\begin{aligned} A(x_1, x_2, \dots , x_{2k-1}, x_{2k}, \dots ) = (-x_2, x_1, \dots , -kx_{2k}, kx_{2k-1}, \dots ), \quad x\in \mathrm {D}(A), \end{aligned}
with domain
\begin{aligned} \mathrm {D}(A):= \Bigg \{x\in \ell ^2({\mathbb {N}}):\sum _{k=1}^\infty k^2 |x_k|^2<\infty \Bigg \}, \end{aligned}
so that there is no regularizing effect for the semigroup $$(R_t)_{t \ge 0}$$ generated by (the graph of) A: evolutions starting outside the domain $$\mathrm {D}(A)$$ stay outside the domain and do not give raise to locally Lipschitz or a.e. differentiable curves. Corollary 5.24 shows that the 0-flow $$(S_t)_{t \ge 0}$$ generated by $${\varvec{\mathrm {F}}}$$ on $$\mathcal {P}_2(X)$$ is given by
\begin{aligned} S_t[\mu _0] = (R_t)_{\sharp }\mu _0 \quad \text { for every } \mu _0 \in \overline{\mathrm {D}({\varvec{\mathrm {F}}})} = \mathcal {P}_2({\textsf {X} }) \end{aligned}
so that there is the same lack of regularizing effect on probability measures.
In the next example we show that a constant MPVF generates a barycentric solution.
### Example 7.6
(Constant PVF and barycentric evolutions) Given $$\theta \in \mathcal {P}_2({\textsf {X} })$$, we consider the constant PVF
\begin{aligned} {\varvec{\mathrm {F}}}[\mu ] := \mu \otimes \theta . \end{aligned}
$${\varvec{\mathrm {F}}}$$ is dissipative according to (4.1): in fact, if $$\Phi _i=\mu _i\otimes \theta$$, $$i=0,1$$, $$\varvec{\mu }\in \Gamma _o(\mu _0,\mu _1)$$, and $${\varvec{r}}:{\textsf {X} }\times {\textsf {X} }\times X\rightarrow \mathsf {TX}\times \mathsf {TX}$$ is defined by $${\varvec{r}}(x_0,x_1,v):=(x_0,v;x_1,v)$$, then
\begin{aligned} \Theta ={\varvec{r}}_\sharp (\varvec{\mu }\otimes \theta )\in \Lambda (\Phi _0,\Phi _1) \end{aligned}
so that (3.17) yields
\begin{aligned} \left[ \Phi _0, \Phi _1\right] _{r}\le \int \langle x_0-x_1,v-v\rangle \,\mathrm d(\varvec{\mu }\otimes \theta )(x_0,x_1,v)=0. \end{aligned}
Applying Proposition 5.20 and Theorem 5.19 we immediately see that $${\varvec{\mathrm {F}}}$$ generates a 0-flow $$(\mathrm S_t)_{t\ge 0}$$ in $$\mathcal {P}_2({\textsf {X} })$$, obtained as a limit of the Explicit Euler scheme. It is also straightforward to notice that we can apply Theorem 5.27 to $${\varvec{\mathrm {F}}}$$ so that for every $$\mu _0 \in \mathcal {P}_2({\textsf {X} })$$ the unique EVI solution $$\mu _t=\mathrm S_t\mu _0$$ satisfies the continuity equation
\begin{aligned} \partial _t \mu _t+\nabla \cdot ({\varvec{b}}\mu _t)=0,\quad {\varvec{b}}=\int _{\textsf {X} }v\,\mathrm d\theta (v). \end{aligned}
Since $${\varvec{b}}$$ is constant, we deduce that $$\mathrm S_t$$ acts as a translation with constant velocity $${\varvec{b}}$$, i.e.
\begin{aligned} \mu _t=({\varvec{i}}_{\textsf {X} }+t{\varvec{b}})_\sharp \mu _0, \end{aligned}
so that $$\mathrm S_t$$ coincides with the semigroup generated by the PVF $${\varvec{\mathrm {F}}}'[\mu ]:=({\varvec{i}}_{\textsf {X} },{\varvec{b}})_\sharp \mu$$.
We conclude this subsection with a 1-dimensional example of a curve which satisfies the barycentric property but it is not an EVI solution.
### Example 7.7
Let $${\textsf {X} }= {\mathbb {R}}$$. It is well known (see e.g. [24]) that $$\mathcal {P}_2({\mathbb {R}})$$ is isometric to the closed convex subset $$\mathcal {K} \subset L^2(0,1)$$ of the (essentially) increasing maps under the action of the isometry $$\mathcal {J}: \mathcal {P}_2({\mathbb {R}}) \rightarrow \mathcal {K}$$ which maps each measure $$\mu \in \mathcal {P}_2({\mathbb {R}})$$ into the pseudo inverse of its cumulative distribution function.
It follows that for every $${\bar{\nu }}\in \mathcal {P}_2({\mathbb {R}})$$ the functional $$\mathcal {F}: \mathcal {P}_2({\mathbb {R}}) \rightarrow {\mathbb {R}}$$ defined as
\begin{aligned} \mathcal {F}(\mu ) :=\frac{1}{2} W_2^2(\mu , \bar{\nu }) \end{aligned}
is 1-convex, since it satisfies $$\mathcal {F}(\mu )=\mathcal {G}(\mathcal {J}(\mu ))$$ where $$\mathcal {G}: L^2(0,1) \rightarrow {\mathbb {R}}$$ is defined as
\begin{aligned} \mathcal {G}(u) :={\left\{ \begin{array}{ll} \frac{1}{2}\Vert u-\mathcal {J}(\bar{\nu }) \Vert ^2&{}\text {if }u\in \mathcal {K},\\ +\infty &{}\text {otherwise}.\end{array}\right. } \end{aligned}
Thus $$\mathcal {F}$$ generates a gradient flow $$(\mathrm S_t)_{t \ge 0}$$ which is a semigroup of contractions in $$\mathcal {P}_2({\mathbb {R}})$$; for every $$\mu _0\in \mathcal {P}_2({\mathbb {R}})$$, the map $$\mathrm S_t[\mu _0]$$ is the unique $$(-1)$$-EVI solution for the MPVF $$-\varvec{\partial }\mathcal {F}$$ starting from $$\mu _0 \in \mathcal {P}_2({\mathbb {R}})$$ (see Proposition 7.2). Since the notion of gradient flow is purely metric, the gradient flow of $$\mathcal {G}$$ starting from $$\mathcal {J}(\mu _0)$$ is just the image through $$\mathcal {J}$$ of the gradient flow of $$\mathcal {F}$$ starting from $$\mu _0 \in \mathcal {P}_2({\mathbb {R}})$$. Indeed: let $$\mu$$ be the gradient flow for $$\mathcal {F}$$ starting from $$\mu _0\in \mathcal {P}_2({\mathbb {R}})$$, then by e.g. [3, Theorem 11.1.4] we have that $$\mu$$ satisfies
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\frac{1}{2}W_2^2(\mu _t,\nu )\le \mathcal {F}(\nu )-\mathcal {F}(\mu _t)-\frac{1}{2}W_2^2(\mu _t,\nu )\quad \text { for a.e. }t>0,\,\text {for every }\nu \in \mathcal {P}_2({\mathbb {R}}), \end{aligned}
so that we get
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\frac{1}{2}\Vert \mathcal {J}(\mu _t)-\mathcal {J}(\nu )\Vert ^2\le \mathcal {G}(\mathcal {J}(\nu ))-\mathcal {G}(\mathcal {J}(\mu _t))-\frac{1}{2}\Vert \mathcal {J}(\mu _t)-\mathcal {J}(\nu )\Vert ^2, \end{aligned}
which, recalling the characterization of gradient flows in Hilbert spaces, gives that $$u(t):=\mathcal {J}(\mu _t)$$ is the gradient flow of $$\mathcal {G}$$ starting from $$\mathcal {J}(\mu _0)$$. It is easy to check that
\begin{aligned} u(t) := \mathrm e^{-t} \mathcal {J}(\mu _0) + (1-\mathrm e^{-t})\mathcal {J}(\bar{\nu }) \end{aligned}
is the gradient flow of $$\mathcal {G}$$ starting from $$u_0=\mathcal {J}(\mu _0)$$. Note that u(t) is the $$L^2(0,1)$$ geodesic from $$\mathcal {J}(\bar{\nu })$$ to $$\mathcal {J}(\mu _0)$$ evaluated at the rescaled time $$e^{-t}$$, so that $$\mathrm S_t[\mu _0]$$ must coincide with the evaluation at time $$\mathrm e^{-t}$$ of the (unique) geodesic connecting $$\bar{\nu }$$ to $$\mu _0$$ i.e.
\begin{aligned} \mathrm S_t[\mu _0] = {\textsf {x} }^{s}_{\sharp } \varvec{\gamma }, \quad s=\mathrm e^{-t}\in (0,1], \end{aligned}
where $$\varvec{\gamma }\in \Gamma _o(\bar{\nu },\mu _0)$$.
Let us now consider the particular case $$\bar{\nu }=\frac{1}{2} \delta _{-a} + \frac{1}{2} \delta _a$$, where $$a>0$$ is a fixed parameter and $$\mu _0 = \delta _0$$. It is straightforward to see that
\begin{aligned} \mu _t=\mathrm S_t[\delta _0] = \frac{1}{2} \delta _{a(1-\mathrm e^{-t})} + \frac{1}{2} \delta _{a(\mathrm e^{-t}-1)}, \quad t \ge 0\end{aligned}
so that
\begin{aligned} ({\varvec{i}}_{\textsf {X} }, {\varvec{v}}_t)_{\sharp } \mu _t= \frac{1}{2} \delta _{((1-e^{-t})a, e^{-t}a)} + \frac{1}{2} \delta _{((e^{-t}-1)a, -e^{-t}a)} \in -\varvec{\partial }\mathcal {F}(\mu _t), \quad \text { a.e. } t>0, \end{aligned}
where $${\varvec{v}}$$ is the Wasserstein velocity field of $$\mu _t$$. On the other hand, [3, Lemma 10.3.8] shows that
\begin{aligned} \delta _0 \otimes \left( \frac{1}{2}\delta _{-a} + \frac{1}{2} \delta _{a} \right) \in - \varvec{\partial }\mathcal {F}(\delta _0) \end{aligned}
so that the constant curve $${\bar{\mu }}_t := \delta _0$$ for $$t\ge 0$$ has the barycentric property for the MPVF $$-\varvec{\partial }\mathcal {F}$$ but it is not a EVI solution for $$-\varvec{\partial }\mathcal {F}$$, being different from $$\mu _t=\mathrm S_t[\delta _0]$$.
### 7.5 Comparison with [27]
In this section, we provide a brief comparison between the assumptions we required in order to develop a strong concept of solution to (5.1) and the hypotheses assumed in [27]. We remind that the relation between our solution and the weaker notion studied in [27] was exploited in Sect. 5.5. Here, we conclude with a further remark coming from the connections between our approximating scheme proposed in (EE) and the schemes proposed in [9] and [27].
We consider a finite time horizon [0, T] with $$T>0$$, the space $${\textsf {X} }={\mathbb {R}}^d$$ and we deal with measures in $$\mathcal {P}_b({\mathbb {R}}^d)$$ and in $$\mathcal {P}_b({\mathsf {T}}{\mathbb {R}}^d)$$, i.e. compactly supported. We also deal with single-valued probability vector fields (PVF) for simplicity, which can be considered as everywhere defined maps $${\varvec{\mathrm {F}}}:\mathcal {P}_b({\mathbb {R}}^d)\rightarrow \mathcal {P}_b({\mathsf {T}}{\mathbb {R}}^d)$$ such that $${\textsf {x} }_{\sharp }{\varvec{\mathrm {F}}}[\nu ]=\nu$$. This is indeed the framework examined in [27].
We start by recalling the assumptions required in [27] for a PVF $${\varvec{\mathrm {F}}}:\mathcal {P}_b({\mathbb {R}}^d)\rightarrow \mathcal {P}_b({\mathsf {T}}{\mathbb {R}}^d)$$.
1. (H1)
there exists a constant $$M>0$$ such that for all $$\nu \in \mathcal {P}_b({\mathbb {R}}^d)$$,
\begin{aligned} \sup _{(x,v) \in {{\,\mathrm{supp}\,}}({\varvec{\mathrm {F}}}[\nu ])}|v| \le M\left( 1 + \sup _{x \in {{\,\mathrm{supp}\,}}(\nu )}|x|\right) ; \end{aligned}
2. (H2)
$${\varvec{\mathrm {F}}}$$ satisfies the following Lipschitz condition: there exists a constant $$L\ge 0$$ such that for every $$\Phi ={\varvec{\mathrm {F}}}[\nu ],\ \Phi '={\varvec{\mathrm {F}}}[\nu ']$$ there exists $$\varvec{\Theta }\in \Lambda (\Phi ,\Phi ')$$ satisfying
\begin{aligned} \int _{{\mathsf {T}}{\mathbb {R}}^d\times {\mathsf {T}}{\mathbb {R}}^d} |v_0 - v_1|^2 \,\mathrm d\varvec{\Theta }(x_0,v_0,x_1,v_1) \le L^2 W_2^2(\nu ,\nu '), \end{aligned}
with $$\Lambda (\cdot ,\cdot )$$ as in Definition 3.8.
### Remark 7.8
Condition (H1) is (H:bound) in [27], while (H2) corresponds to (H:lip) in [27] in case $$p=2$$ (see also Remark 5 in [27]).
We stress that actually in [27] condition (H2) is local, meaning that L is allowed to depend on the radius R of a ball centered at 0 and containing the supports of $$\nu$$ and $$\nu '$$. Thanks to assumption (H1), it is easy to show that for every final time T all the discrete solutions of the Explicit Euler scheme and of the scheme of [27] starting from an initial measure with support in $$\mathrm B(0,R)$$ are supported in a ball $$\mathrm B(0,R')$$ where $$R'$$ solely depends on R and T. We can thus restrict the PVF $${\varvec{\mathrm {F}}}$$ to the (geodesically convex) set of measures with support in $$\mathrm B(0,R')$$ and act as L does not depend on the support of the measures.
### Proposition 7.9
If $${\varvec{\mathrm {F}}}:\mathcal {P}_b({\mathbb {R}}^d)\rightarrow \mathcal {P}_b({\mathsf {T}}{\mathbb {R}}^d)$$ is a PVF satisfying (H2), then $${\varvec{\mathrm {F}}}$$ is $$\lambda$$-dissipative according to (4.1) for $$\lambda =\frac{L^2+1}{2}$$, the Explicit Euler scheme is globally solvable in $$\mathrm {D}({\varvec{\mathrm {F}}})$$, and $${\varvec{\mathrm {F}}}$$ generates a $$\lambda$$-flow, whose trajectories are the limit of the Explicit Euler scheme in each finite interval [0, T].
### Proof
The $$\lambda$$-dissipativity comes from Lemma 4.7. We prove that (5.34) holds. Let $$\nu \in \mathrm {D}({\varvec{\mathrm {F}}})$$ and take $$\varvec{\Theta }\in \Lambda ({\varvec{\mathrm {F}}}[\nu ],{\varvec{\mathrm {F}}}[\delta _0])$$ such that
\begin{aligned} \int _{{\mathsf {T}}{\mathbb {R}}^d\times {\mathsf {T}}{\mathbb {R}}^d}|v'-v''|^2\,\mathrm d\varvec{\Theta }\le L^2W_2^2(\nu ,\delta _0)=L^2 {\textsf {m} }_2^2(\nu ). \end{aligned}
Since $${\varvec{\mathrm {F}}}[\delta _0] \in \mathcal {P}_c({\mathsf {T}}{\mathbb {R}}^d)$$ by assumption, there exists $$D>0$$ such that $${{\,\mathrm{supp}\,}}({\textsf {v} }_\sharp {\varvec{\mathrm {F}}}[\delta _0])\subset B_D(0)$$. Hence, we have
\begin{aligned} L^2{\textsf {m} }_2^2(\nu )&\ge \int _{{\mathsf {T}}{\mathbb {R}}^d\times {\mathsf {T}}{\mathbb {R}}^d}|v'-v''|^2\,\mathrm d\varvec{\Theta }\\ {}&\ge \int _{{\mathsf {T}}{\mathbb {R}}^d\times {\mathsf {T}}{\mathbb {R}}^d}[|v'|-D]_+^2\,\mathrm d\varvec{\Theta }\\&\ge \int _{{\mathsf {T}}{\mathbb {R}}^d}|v'|^2\,\mathrm d{\varvec{\mathrm {F}}}[\nu ]-2D\int _{{\mathsf {T}}{\mathbb {R}}^d}|v'|\,\mathrm d{\varvec{\mathrm {F}}}[\nu ], \end{aligned}
where $$[\,.\,]_+$$ denotes the positive part. By the trivial estimate $$|v'|\le D+\frac{|v'|^2}{4D}$$, we conclude
\begin{aligned} |{\varvec{\mathrm {F}}}[\nu ]|_2^2\le 2\left( 2D^2+L^2{\textsf {m} }_2^2(\nu )\right) . \end{aligned}
Hence (5.34) and thus the global solvability of the Explicit Euler scheme in $$\mathrm {D}({\varvec{\mathrm {F}}})$$ by Proposition 5.20. To conclude it is enough to apply Theorem 5.22(a) and Theorem 6.7. $$\square$$
It is immediate to notice that the semi-discrete Lagrangian scheme proposed in [9] coincides with the Explicit Euler Scheme given in Definition 5.7. In particular, we can state the following comparison between the limit obtained by the Explicit Euler scheme (EE) (leading to the $$\lambda$$-EVI solution of (5.1)) and that of the approximating LASs scheme proposed in [27] (leading to a barycentric solution to (5.1) in the sense of Definition 5.25).
### Corollary 7.10
Let $${\varvec{\mathrm {F}}}$$ be a PVF satisfying (H1)-(H2), $$\mu _0\in \mathcal {P}_{b}({\mathbb {R}}^d)$$ and let $$T\in (0,+\infty )$$. Let $$(n_k)_{k\in {\mathbb {N}}}$$ be a sequence such that the LASs scheme $$(\mu ^{n_k})_{k\in {\mathbb {N}}}$$ of [27, Definition 3.1] converges uniformly-in-time and let $$(M_{\tau _k})_{k\in {\mathbb {N}}}$$ be the affine interpolants of the Explicit Euler Scheme defined in (5.9), with $$\tau _k=\frac{T}{n_k}$$. Then $$(\mu ^{n_k})_{k\in {\mathbb {N}}}$$ and $$(M_{\tau _k})_{k\in {\mathbb {N}}}$$ converge to the same limit curve $$\mu :[0,T]\rightarrow \mathcal {P}_b({\mathbb {R}}^d)$$, which is the unique $$\lambda$$-EVI solution of (5.1) in [0, T].
### Proof
By Proposition 7.9, $${\varvec{\mathrm {F}}}$$ is a $$\left( \frac{L^2+1}{2}\right)$$-dissipative MPVF according to (4.1) s.t. $${M}(\mu _0, \tau , T, {\tilde{L}}) \ne \emptyset$$ for every $$\tau >0$$, where $${\tilde{L}}>0$$ is a suitable constant depending on $$\mu _0$$ and $${\varvec{\mathrm {F}}}$$. Thus by Theorem 6.7, $$(M_{\tau _k})_{k \in {\mathbb {N}}}$$ uniformly converges to a $$\lambda$$-EVI solution $$\mu :[0,T]\rightarrow \mathcal {P}_2({\mathbb {R}}^d)$$ which is unique since $${\varvec{\mathrm {F}}}$$ generates a $$\left( \frac{L^2+1}{2}\right)$$-flow. Since we start from a compactly supported $$\mu _0$$, the semi-discrete Lagrangian scheme of [9] and our Euler Scheme actually coincide. To conclude we apply [9, Theorem 4.1] obtaining that $$\mu$$ is also the limit of the LASs scheme. $$\square$$
We conclude that among the possibly not-unique (see [9]) barycentric solutions to (5.1) - i.e. the solutions in the sense of [27]/Definition 5.25 - we are selecting only one (the $$\lambda$$-EVI solution), which turns out to be the one associated with the LASs approximating scheme.
In light of this observation, we revisit an interesting example studied in [27, Sect 7.1] and [9, Sect. 6].
### Example 7.11
(Splitting particle) For every $$\nu \in \mathcal {P}_b({\mathbb {R}})$$ define:
\begin{aligned} B(\nu ):=\sup \left\{ x:\nu (]-\infty ,x])\le \frac{1}{2}\right\} ,\qquad \eta (\nu ):=\nu (]-\infty ,B(\nu )]) - \frac{1}{2}, \end{aligned}
so that $$\nu (\{B(\nu )\})=\eta (\nu )+\frac{1}{2}-\nu (]-\infty ,B(\nu )[)$$. We define the PVF $${\varvec{\mathrm {F}}}[\nu ]:= \int {\varvec{\mathrm {F}}}_x[\nu ]\,\mathrm d\nu (x)$$, by
\begin{aligned} {\varvec{\mathrm {F}}}_x[\nu ]:=\left\{ \begin{array}{ll} \delta _{-1} &{} \text {if}\ x<B(\nu )\\ \delta _{1} &{} \text {if}\ x>B(\nu )\\ \frac{1}{\nu (\{B(\nu )\})} \left( \eta \delta _{1}+ \left( \frac{1}{2}-\nu (]-\infty ,B(\nu )[)\right) \delta _{-1}\right) &{} \text {if}\ x=B(\nu ), \nu (\{B(\nu )\})>0. \end{array} \right. \end{aligned}
By [27, Proposition 7.2], $${\varvec{\mathrm {F}}}$$ satisfies assumptions (H1)-(H2) with $$L=0$$ and the LASs scheme admits a unique limit. Moreover, the solution $$\mu :[0,T]\rightarrow \mathcal {P}_b({\mathbb {R}})$$ obtained as limit of LASs, is given by
\begin{aligned} \mu _t(A)= & {} \mu _0((A\cap ]-\infty ,B(\mu _0)-t[)+t) + \mu _0((A\cap ]B(\mu _0)+t,+\infty [)-t)\nonumber \\&+\frac{1}{\mu _0(\{B(\mu _0)\})} \left( \eta \delta _{B(\mu _0)+t}(A)+ (\frac{1}{2}-\mu _0(]-\infty ,B(\mu _0)[))\delta _{B(\mu _0)-t}(A)\right) .\qquad \end{aligned}
(7.6)
By Corollary 7.10, (7.6) is the (unique) $$\lambda$$-EVI solution of (5.1). In particular:
1. (i)
if $$\mu _0=\frac{1}{b-a}{\mathcal {L}}\llcorner _{[a,b]}$$, i.e. the normalized Lebesgue measure restricted to [ab], we get $$\mu _t=\frac{1}{b-a}{\mathcal {L}}\llcorner _{[a-t,\frac{a+b}{2}-t]} +\frac{1}{b-a} {\mathcal {L}}\llcorner _{[\frac{a+b}{2}+t,b+t]}$$;
2. (ii)
if $$\mu _0=\delta _{x_0}$$, we get $$\mu _t=\frac{1}{2} \delta _{x_0+t}+\frac{1}{2} \delta _{x_0-t}$$.
Notice that, in case (i), since $$\mu _t\ll \mathcal {L}$$ for all $$t\in (0,T)$$, i.e. $$\mu _t\in \mathcal {P}_2^r({\mathbb {R}})$$, we can also apply Theorem 5.31 to conclude that $$\mu$$ is the $$\lambda$$-EVI solution of (5.1) with $$\mu _0=\frac{1}{b-a}\mathcal L\llcorner _{[a,b]}$$. Moreover, take $$\varepsilon >0$$, and consider case (i) where we denote by $$\mu ^\varepsilon _0$$ the initial datum and by $$\mu ^\varepsilon$$ the corresponding $$\lambda$$-EVI solution to (5.1) with $$a=x_0-\varepsilon$$, $$b=x_0+\varepsilon$$. We can apply (5.35) with $$\mu _0=\mu _0^{\varepsilon }$$ and $$\mu _1=\delta _{x_0}$$ in order to give another proof that, for all $$t\in [0,T]$$, the $$W_2$$-limit of $$S_t[\mu _0^{\varepsilon }]$$ as $$\varepsilon \downarrow 0$$, that is $$S_t[\delta _{x_0}]=\frac{1}{2} \delta _{x_0+t}+\frac{1}{2} \delta _{x_0-t}$$, is a $$\lambda$$-EVI solution starting from $$\delta _{x_0}$$. Thus we end up with (ii).
Dealing with case (ii), we recall that, if $$\mu _0=\delta _{x_0}$$ then also the stationary curve $${\bar{\mu }}_t=\delta _{x_0}$$, for all $$t\in [0,T]$$, satisfies the barycentric property of Definition 5.25 (see [9, Example 6.1]), thus it is a solution in the sense of [27]. However, $${\bar{\mu }}$$ is not a $$\lambda$$-EVI solution since it does not coincide with the curve given by (ii). This fact can also be checked by a direct calculation as follows: we find $$\nu \in \mathcal {P}_b({\mathbb {R}})$$ such that
\begin{aligned} \frac{\,\mathrm d}{\,\mathrm dt}\frac{1}{2}W_2^2({\bar{\mu }}_t,\nu )>\lambda W_2^2({\bar{\mu }}_t,\nu )-\left[ {\varvec{\mathrm {F}}}[\nu ], {\bar{\mu }}_t\right] _{r}\qquad t\in (0,T), \end{aligned}
(7.7)
where $$\lambda =\frac{1}{2}$$ is the dissipativity constant of the PVF $${\varvec{\mathrm {F}}}$$ coming from the proof of Proposition 7.9. Notice that the l.h.s. of (7.7) is always zero since $$t\mapsto {\bar{\mu }}_t=\delta _0$$ is constant. Take $$\nu =\mathcal L\llcorner _{[0,1]}$$ so that we get $${\varvec{\mathrm {F}}}[\nu ]= \int {\varvec{\mathrm {F}}}_x[\nu ]\,\mathrm d\nu (x)$$, with $${\varvec{\mathrm {F}}}_x[\nu ]=\delta _1$$ if $$x>\frac{1}{2}$$, $${\varvec{\mathrm {F}}}_x[\nu ]=\delta _{-1}$$ if $$x<\frac{1}{2}$$. Noting that $$\Lambda ({\varvec{\mathrm {F}}}[\nu ],\delta _0)=\{{\varvec{\mathrm {F}}}[\nu ]\otimes \delta _0\}$$, by using the characterization in Theorem 3.9 we compute
\begin{aligned} \left[ {\varvec{\mathrm {F}}}[\nu ], \delta _0\right] _{r}= & {} \int _{\mathsf {TX}}\langle x, v\rangle \,\mathrm d{\varvec{\mathrm {F}}}[\nu ]\\= & {} \int _0^{1/2}\langle x, v\rangle \,\mathrm d{\varvec{\mathrm {F}}}_x[\nu ](v)\,\mathrm dx\\&+\int _{1/2}^1\langle x, v\rangle \,\mathrm d{\varvec{\mathrm {F}}}_x[\nu ](v)\,\mathrm dx=\frac{1}{4}. \end{aligned}
Since $$W_2^2(\delta _0,\nu )={\textsf {m} }_2^2(\nu )=\frac{1}{3}$$, we have
\begin{aligned} \lambda W_2^2({\bar{\mu }}_t,\nu )-\left[ {\varvec{\mathrm {F}}}[\nu ], {\bar{\mu }}_t\right] _{r}=\frac{1}{6}-\frac{1}{4}<0, \end{aligned}
and thus we obtain the desired inequality (7.7) with $$\nu ={\mathcal {L}}\llcorner _{[0,1]}$$. | 112,479 | 293,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.714407 |
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/12%3A_Vectors_in_Space | 1,716,582,641,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00387.warc.gz | 325,608,090 | 30,737 | # 12: Vectors in Space
• Gilbert Strang & Edwin “Jed” Herman
• OpenStax
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A quantity that has magnitude and direction is called a vector. Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.
• 12.0: Prelude to Vectors in Space
• 12.1: Vectors in the Plane
When measuring a force, such as the thrust of the plane’s engines, it is important to describe not only the strength of that force, but also the direction in which it is applied. Some quantities, such as or force, are defined in terms of both size (also called magnitude) and direction. A quantity that has magnitude and direction is called a vector.
• 12.2: Vectors in Three Dimensions
To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into three dimensions.
• 12.3: The Dot Product
In this section, we develop an operation called the dot product, which allows us to calculate work in the case when the force vector and the motion vector have different directions. The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector. The dot product can also help us measure the angle formed by a pair of vectors and the position of a vector relative to the coordinate axes.
• 12.4: The Cross Product
In this section, we develop an operation called the cross product, which allows us to find a vector orthogonal to two given vectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in the section.
• 12.5: Equations of Lines and Planes in Space
To write an equation for a line, we must know two points on the line, or we must know the direction of the line and at least one point through which the line passes. In two dimensions, we use the concept of slope to describe the orientation, or direction, of a line. In three dimensions, we describe the direction of a line using a vector parallel to the line. In this section, we examine how to use equations to describe lines and planes in space. | 1,189 | 4,161 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.348404 |
https://homework.cpm.org/cpm-homework/homework/category/CCI_CT/textbook/Calc3rd/chapter/Ch5/lesson/5.1.1/problem/5-7 | 1,585,801,473,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506580.20/warc/CC-MAIN-20200402014600-20200402044600-00354.warc.gz | 384,729,649 | 16,189 | ### Home > CALC3RD > Chapter Ch5 > Lesson 5.1.1 > Problem5-7
5-7.
You know that the first derivative, $f^\prime$, tells us the slope and the rate of change of $f$. Homework Help ✎
1. What does the second derivative, $f^{\prime\prime}$, tell you about $f^\prime$? What does $f^{\prime\prime}$ tell you about $f$ ?
$f''(x)$ tells us the same thing about $f′(x)$ that $f′(x)$ tells us about $f(x)$.
2. Write equations for $f′(x)$ and $f''(x)$ if $f(x)=x^3+3x^2-9x+2$.
$f′(x)$: $3x^2+6x-9$
$f''(x)$:$6x + 6$
3. Is $f$ getting steeper or less steep at $x = 1$? At $x = -2$? Use your derivatives from part (b) to explain your reasoning.
Consider this: Even though a slope of $−2$ is steeper than a slope of $−1$, if the slope changes from $−1$ to $−2$, then the slope is decreasing.
4. The values of $f''(1)$ and $f''(-2)$ can be used to determine concavity at $x = 1$ and $x = -2$. Where is $f$ concave up? Where is $f$ concave down?
Positive values of the 2nd-derivative indicate that the function is concave up while negative values indicate that the function is concave down. | 372 | 1,083 | {"found_math": true, "script_math_tex": 30, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | latest | en | 0.791019 |
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# Problem 690. Remove the two elements next to NaN value
Solution 177219
Submitted on 14 Dec 2012 by Mehmet OZC
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [6 10 5 8 9 NaN 23 9 7 3 21 43 NaN 4 6 7 8]; y_correct = [6 10 5 8 9 7 3 21 43 7 8]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
%% x = [25 NaN 1 3]; y_correct = 25; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
%% x = [ NaN 15 15 17 ] y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
x = NaN 15 15 17 | 231 | 626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.588434 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/judith677/split-each-cell-value-and-expand-it-according-to-the-specified-rule-2op6 | 1,726,864,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00440.warc.gz | 430,337,689 | 23,149 | ## DEV Community
Judith-Excel-Sharing
Posted on
# Split Each Cell Value And Expand It According To The Specified Rule
Problem description & analysis:
The following table records someone’s answers to a set of questions:
An answer generally consists of options separated by a semicolon. If it is a string “All of the Above”, it has all options under the same question number in dictionary table Sheet2.
`````` A B
1 1 A
2 1 b
3 1 c
4 1 d
5 2 a
6 2 b
7 2 c
8 2 d
9 3 a
10 3 b
11 3 c
12 3 d
13 4 a
14 4 b
15 4 c
16 4 d
17 4 e
``````
We need to split each answer into individual options, as shown below:
`````` A B
1 Question What I want
2 1 A
3 1 b
4 1 c
5 1 d
6 2 A
7 2 B
8 2 C
9 3 B
10 3 C
11 4 a
12 4 b
13 4 c
14 4 d
15 4 e
``````
Solution:
Use SPL XLL to enter the following formula:
``````=spl("=dt=?1,dc=?2,E@b(dt.news(if(~(2)==\$[All of the Above],dc.select(~(1)==dt.~(1)).(~(2)), ~(2).split(\$[;]));dt.~(1),~))",D2:E5,Sheet2!A2:Sheet2!B18)
``````
As shown in the picture below:
Explanation:
E@b converts an Excel table to a sequence. ~(1) represents the 1st child member of the current member in a sequence; \$[] represents a string. | 490 | 1,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.752251 |
http://nobel.scas.bcit.ca/chem0010/unit8/8.5_calculations.htm | 1,550,457,556,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00407.warc.gz | 200,871,265 | 5,211 | UNIT # 8 Introduction Objectives 8.18.2 8.3 Problems 1 | 2 | 3 | 4
CHEMICAL REACTIONS
8.5 - Calculations based on Chemical Equations
In section 8.2, we learned that in balancing chemical equations, coefficients can be inserted into the chemical equations so that
the number of atoms of each type on the LEFT-HAND-SIDE = the number of atoms of each type on the RIGHT-HAND-SIDE
With these coefficients, we can write a ratio of any two or more substances in the chemical equation.
Let's take a look at the balanced chemical equation for the synthesis of water:
2 H2 (g) + O2 (g) 2 H20 (l) *
* The coefficient in front of O2 (g) is by default "1". Coefficient of "1" is usually not written in.
Here are some ratios that we can write from the above balanced chemical equation:
1. 2 molecules of H2 (g) requires 1 molecule of O2 (g) to produce 2 molecules of H20 (l).
The important thing is, the ratio is 2:1:2 for the substances H2 : O2 : H20 . In any combination of any two substances, the coefficients in the balanced chemical equation tells us: the ratio between H2 : O2 is always 2:1. the ratio between H2 : H20 is always 2:2, or 1:1. the ratio between O2 : H20 is always 1:2.
2. If we scale up 2 times, then:
4 molecules of H2 (g) requires 2 molecules of O2 (g) to produce 4 molecules of H20 (l).
This ratio of 2:1:2 is always maintained no matter how much we scale up.
1. If we scale up 12 times, then:
2 dozens molecules of H2 (g) requires 1 dozens molecules of O2 (g) to produce 2 dozens molecules of H20 (l).
This ratio of 2:1:2 is always maintained no matter how much we scale up.
2. If we scale up to a mole, then:
2
moles of H2 (g) requires 1 mole of O2 (g) to produce 2 moles of H20 (l).
This ratio of 2:1:2 is always maintained no matter how much we scale up.
I repeat: The important thing is, the ratio is always 2:1:2 for the substances H2 : O2 : H20 . The coefficients give the mole ratio of the reactants and products. In any combination of any two substances, the coefficients in the balanced chemical equation tells us: the ratio between H2 : O2 is always 2:1. the ratio between H2 : H20 is always 2:2, or 1:1. the ratio between O2 : H20 is always 1:2.
This is known as stoichiometric ratio. In unit 7, we learned that moles and grams can be interconverted by a compound's molar mass (section 7.5.1). By knowing the mole ratio of the substances in the reaction, it enables us to calculate the mass of the reactants consumed and the mass of products formed.
We must master calculations involving the different types of stoichiometric problems that arise from a balanced chemical equation. They are:
1. Mole-mole Calculations (section 8.5.1)
2. Mass-mass Calculations (section 8.5.2)
3. Mass-mole Calculations (section 8.5.3)
4. Excess reactant and limiting reagent problems (section 8.5.4)
When you are going through these sections, you need to go slowly and thoroughly to understand the logic in each step. Refer to your textbook for more examples of each type of stoichiometric problems.
Section 10.1
Information Obtained from a Balanced Chemical Equation ..p249
Back
All contents copyrighted © 1996-2006 British Columbia Institute of Technology Chemistry Department - 3700 Willingdon Avenue Burnaby, B.C. Canada V5G 3H2 | 898 | 3,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-09 | latest | en | 0.867853 |
https://learn.careers360.com/ncert/question-if-a-b-c-are-three-points-on-a-line-such-that-ab-5-cm-bc-3-cm-and-ac-8-cm-which-one-of-them-lies-between-the-other-two/ | 1,712,948,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00636.warc.gz | 325,646,250 | 34,512 | #### 4. If A,B,C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
AB = 5 cm
BC = 3 cm
AC = 8 cm
Therefore AB + BC = AC
Therefore point B lies between points A and C. | 80 | 240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.946163 |
http://www.mangahigh.com/en-us/math_games/algebra/solving_equations_using_graphs/use_graphs_to_solve_equations | 1,519,317,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814140.9/warc/CC-MAIN-20180222160706-20180222180706-00162.warc.gz | 495,158,918 | 3,819 | MANGAHIGH
# Algebra
## Use graphs to solve equations
### Solving equations using graphs
#### Algebra
Find the roots or solve f(x)=0 as an alternative way of asking for a solution to the trinomial equal to zero. Solve trinomials of type e.g. x²+2x+2 = 4 after having drawny = x²+2x+2 or y = x2+2x+1 or y = x²+x-3 by drawing a linear graph or reading off points. Also solve cubics, reciprocals and exponentials equations with the aid of graphs
Play Now | 140 | 455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-09 | latest | en | 0.805417 |
https://discourse.mcneel.com/t/2-pivot-points-on-one-object/16148 | 1,653,055,418,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00127.warc.gz | 269,396,953 | 7,158 | # 2 pivot points on one object?
So I’m stuck trying to make something work, not sure the best way to go about it. I want to create the sketch below, where the purple is the “track” and the white is a rigid bar with wheels at each end which rides in the track. I’ve constrained the lower wheel to the blue line, but i cant figure out how to constrain the upper wheel to the same line and have it work the way I would expect. Is there a way to have multiple pivot points? The bar is a child to the lower wheel so it rotates with it but i cant figure out how to get the upper wheel to track AND keep that end of the bar in the right place. Any ideas?
Much appreciated!
I don’t know how to do this in Bongo, but it’s relatively easy in Grasshopper. The vertical pivot is constrained to the vertical line and the horizontal pivot is constrained to the intersection of the horizontal axis and a circle centered on the vertical pivot with radius the length of the bar. I suppose you can add these kinds of constraints in Bongo as well…
Short video…
–Mitch
1 Like
Mitch, thanks for taking the time to respond and to make that video! I assumed grasshopper would be another way to go but this is only part of a larger mechanism. There are additional components attached to the moving piece that change positions as well depending on the angle of the bar, and more from there. I could do it all in grasshopper but I was hoping bongo would eliminate a lot of work since that patch would get fairly large quickly and also it would be nice to be able to animate automatically.
I assume it can be done with Bongo but i’m missing something. Everything else seems to be working. If I can’t get it working ill probably switch to GH, in which case I may have a more questions!
Thanks again!
You should create a structure like this:
The lower wheel can be made moving horizontal by means of keyframes (Position transform). When you want it to move by means of a Simple constraint ‘To Path’ (using the blue line as a path) you’ll have to add an extra proxy (in the center of the wheel) as the ‘head’ of the chain, because a Simple constrained object can’t be part of an IK chain.
The upper wheel must be constrained as ‘Part of an IK chain’ to the blue line.
The lower wheel must be a Hinge, hence allowing the whole to rotate.
Succes.
1 Like
a demo.
rollbar.3dm (32.7 KB)
1 Like
Luc, thanks for your response. I think i figured it out based on that! Lots of other stuff to work out now, but this issue is resolved. I wasn’t doing the IK chain right - and I had the upper wheel as a joint & constraint. Now that I have a better idea of how it needs to be set up the rest of it should go much smoother. Thanks!!
My pleasure. Any more issues: keep on posting.
Maybe you can have a look at a little PDF tutorial I once made:
[http://bongo.rhino3d.com/forum/topics/the-basic-why-s-of-IK][1]
[1]: http://bongo.rhino3d.com/forum/topics/the-basic-why-s-of-ik.
I must rectify this; an Simple constrained object can sure be part of an IK chain but it cannot be a joint (Hinge, …) simultaneously.
And how about 3 pivot points on one object?
I’m trying to animate a object like this without results.
I don’t understand how to allow every bar to rotate in 3 points: at the ends and in the middle.
Any ideas?
Please have a look at this old post with instructions:
It should give you an idea on how to get it animated. | 799 | 3,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-21 | latest | en | 0.954634 |
https://wiki.hydrogenaud.io/index.php?title=Joint_stereo&oldid=15737 | 1,632,232,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00568.warc.gz | 648,136,124 | 7,498 | # Joint stereo
It has been suggested that Mid-side stereo be merged into this article or section. (Discuss)
Joint stereo coding methods try to increase the coding efficiency when encoding stereo signals by exploiting commonalities between the left and right channel signal. In the context of MP3 encoding, there are actually 2 totally different method of Joint Stereo: Mid-side Stereo and Intensity Stereo.
## Mid-side Stereo
Mid-side stereo coding calculates a "mid"-channel by addition of left and right channel, and a "side"-channel, i.e.:
$Left = L \qquad Right = R\,$
$Middle=\frac{L+R}{2} \qquad Side=\frac{L-R}{2}$
$Left=Middle+Side \qquad Right=Middle-Side$
Whenever a signal is concentrated in the middle of the stereo image (i.e. more mono-like), mid-side stereo can achieve a significant saving in bitrate, since one can use less bitrate to encode the side-channel. Even more important is the fact that by applying the inverse matrix in the decoder the quantization noise becomes correlated and falls in the middle of the stereo image where it is masked by the signal.
Unlike intensity stereo which destroys phase information, mid-side coding keeps the phase information pretty much intact. Correctly implemented mid-side stereo does very little or no damage to the stereo-image and increases compression efficiency either by reducing size or increasing overall quality.
## Intensity Stereo
Intensity stereo coding is a method that achieves a saving in bitrate by replacing the left and the right signal by a single representing signal plus directional information. This replacement is psychoacoustically justified in the higher frequency range since the human auditory system is insensitive to the signal phase at frequencies above approximately 2kHz.
Intensity stereo is by definition a lossy coding method thus it is primarily useful at low bitrates. For coding at higher bitrates only mid-side stereo should be used.
Some more details about joint stereo & mid-side coding:
• Bugs and/or not-optimized encoders may implement mid-side coding incorrectly, making mid-side coding sounds worse than simple stereo, while in reality (see the formulas above) there should be no difference in quality between mid-side stereo and simple stereo.
• Modern/optimized encoders will use mid-side coding or simple stereo coding as necessary, depending on the correlation between the Left and Right channels. | 484 | 2,420 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.894624 |
https://murtele.cf/688680-price-per-unit-converter.shtml | 1,586,063,339,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00030.warc.gz | 582,143,873 | 4,359 | # Price per unit converter
##### 2020-04-05 06:08
It's easy to choose the right amount of film for your project. Use our calculator for conversion formulas, metric equivalents and more. Or leave the math to us and contact oneHelp improve this site by sending anonymous usage data (privacy policy) price per unit converter
Do this input before the input of the price in the lighter fields on the left. The input must be a positive number. Then type the price to convert into the field behind which you find the corresponding amount. Move the mouse over a unit or click on it to read its full name. Click on any empty space of the window or on the calculate button.
You can conclude that the per apple price unit rate is 0. 601. Ryan paid a unit price of 0. 60 per apple (60 cents per 1 apple. 601). The pottery store can make 176 coffee mugs in an 8 hour day. How many mugs can they make in one hour? We want to know the number of mugs made per hour unit so we set up a ratio with hours in the denominator. Thanks to our price quantity calculator, you know that the unit price for the smaller package is 1, while for the larger package it is 90. The You Save box tells you how much more you'd have paid if you bought the same quantity at the price per unit of a smaller item in our example 15.price per unit converter Dec 06, 2018 The cost per unit is: (30, 000 Fixed costs 50, 000 variable costs) 10, 000 units 8 cost per unit. In the following month, ABC produces 5, 000 units at a variable cost of 25, 000 and the same fixed cost of 30, 000. The cost per unit is: (30, 000 Fixed costs 25, 000 variable costs) 5, 000 units 11unit. Related Courses
## Price per unit converter free
Price per oz app Cost per ounce You are faced with the following dilemma: You want to buy a pretzel, and you have three choices: 2. 5oz for 0. 99; 6oz for 2. 29; or a large 1pounder for 6. 79. Which choice will give you the most pretzel for your money, and how to calculate the price per ounce? price per unit converter Computes the cost per piece unit of food and general goods. Free calculator helps choose cheapest product per item. Converts price from x pieces to price per piece. The unit price for 2 liters is 1. 5 per liter, while the 1. 5 liter is 1. 66 per liter. Therefore the 2 liters is the best option price wise as its 16 cents cheaper per liter. Definition: In retail, it is the price divided by the number of units, which can be liters, ounces etc. To use the per unit method, we normalize all the system impedances (and admittances) within the network under consideration to a common base. These normalized impedances are known as per unit impedances. Any per unit impedance will have the same value on both the primary and secondary of a transformer and is independent of voltage level. Oct 29, 2018 Total manufacturing price per unit (Direct materials direct labor manufacturing overhead)number of units manufactured: (2, )360. 6, . 33.
Rating: 4.97 / Views: 950 | 739 | 2,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-16 | latest | en | 0.857694 |
https://www.daytodaygk.com/reasoning-quiz-sbi-po-ibps-209/ | 1,723,332,129,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00235.warc.gz | 579,378,744 | 16,467 | # Reasoning Quiz for SBI PO | IBPS – 209
1) Statements:
Some cycles are buses.
All cars are buses.
Some buses are trains.
Conclusions:
I. All cars are cycles.
II. Some trains are buses.
III. Some trains are cars.
a) None follows
b) Only I and II follow
c) Only land III follow
d) Only II and III follow
e) None of these
Answer d) Only II and III follow
2) Statements:
All pencils are sticks.
Some sticks are notes.
All diaries are notes.
Conclusions:
I. Some notes are diaries.
II. Some sticks are pencils.
III. Some diaries are sticks.
a) All follow
b) Only I follow
c) Only I and II follow
d) Only II follows
e) None of these
3) Statements:
Some buds are leaves.
No leaf is fruit.
Some fruits are buds.
Conclusions:
I. Some fruits are leaves.
II. All buds are fruits.
III. Some leaves are buds.
a) Only I or II follows
b) Only III follows
c) Only II follows
d) None follows
e) None of these
4) Statements:
Some birds are animals.
All animals are rivers.
Some rivers are lions.
Conclusions:
I. Some lions are animals
II. Some rivers are birds
III. No animal is lion
a) Only II follows
b) Only either I or III follows
c) I and II follows
d) Only either II or III follow
e) None of these
Answer c) I and II follows
5) Statements:
All boxes are pans.
Some boxes are jugs.
Some jugs are glasses.
Conclusions:
I. Some glasses are boxes
II. No glass is box
III. Some jugs are pans
IV. No jug is pan
a) Only I and II follows
b) Either I or II and III follows
c) Only III follows
d) Either I or II , and either III or IV follow
e) None of these
6) Ramesh starting from a fixed point goes 15 km towards North and then after turning to his right he goes 15 km. then he goes 10, 15 and 15 metres after turning to his left each time. How far is he from his starting point?
a) 5 metres
b) 10 metres
c) 20 metres
d) 15 metres
e) Cannot be determined
7) Sonalika goes 12 km towards North from a fixed point and then she goes 8 km towards South from there. In the end she goes 3 km towards east. How far and in what direction is she from her starting point?
a) 7 km East
b) 5 km West
c) 7 km West
d) 5 km North-East
e) None of these
8) Sunita goes 30 km towards North from a fixed point, then after turning to her right she goes 15 km. After this she goes 30 km after turning to her right. How far and in what direction is she from her starting point?
a) 45 km, East
b) 15 km, East
c) 45 km, West
d) 45 Km, North
e) None of these
9) Kanchan goes 5 m towards east from a fixed point N and then 35 km after turning to her left. Again she goes 10 metres after turning to her right. After this she goes 35 m after turning to her right. How far is she from N?
a) 40 m
b) At N
c) 10 m
d) 15 m
e) None of these
10) Shri Prakash walked 40 metres facing towards North. From there he walked 50 metres after turning to his left. After this he walked 40 metres after turning to his left. How far and in what direction is he now from his starting point?
a) 40 m, North
b) 50 m, West
c) 10 m, East
d) 10 m, West
e) None of these | 883 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-33 | latest | en | 0.920092 |
https://1library.net/document/q06j09gq-application-of-linear-transformation-in-numerical-calculation.html | 1,656,862,206,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00369.warc.gz | 119,262,190 | 38,726 | Application of linear transformation in numerical calculation
Full text
(1)
*
Qinggong College, Heibei United University, Tangshan, China
_____________________________________________________________________________________________
ABSTRACT
The linear transformation is the most simple and the most basic transformation. Such as, the linear function is the most simple and the most basic function. The linear transformation is a main research object of linear algebra. This text uses a series of related knowledge of linear transformation thought in linear algebra, makes study and analysis finely to the position and application of the linear transformation thought in mathematics, and sieves several typical example.
Keywords: Linear transformation, Matrix, Numerical Calculation, The application
_____________________________________________________________________________________________
INTRODUCTION
Linear transformation is one of the core content of linear algebra. It plays an important role in between the whole structure and linear vector space memory link. Concept of linear transformation is the transformation of coordinates in analytic geometry, some transformation in mathematical analysis to replace the generalization and abstraction. Its theory and methods in analytic geometry, differential equations and many other fields, and it has widespread application.
LINEAR TRANSFORMATION
Linear mapping
Definition 1 Suppose
be a mapping from
to
W
. If you meet the following conditions, called
is a mapping from
to
.
(1)For arbitrary
(2)For arbitrary
Some basic properties of linear mapping:
Condition (1) and condition (2), in the definition 1, is the condition (3) equivalent conditions. (3)For arbitrary
and
In fact, if the mapping
W
satisfies the condition (1) and (2), for any
and
,
Conversely, suppose (3) was established, made
1
,get condition (1); and
0
,get the condition (2). In the condition (2),
get
0
(2)
In other words, the linear mapping will be zero vectors onto the zero vectors. By (3), as the mathematical induction on
,easy to launch:
1 1
n n
1
1
n
n
For arbitrary
1
n
and
n
V
are established.
Linear transformation
Let
V
be a vector space of a number field
P
.
Some basic properties of linear transformation,
1. Suppose
A
be a linear transformation of
,
and
.
2. Linear transformation keeps linear combination and linear equation invariant. If
is a linear combination of
1
2
r,
r r
1 1
2 2
Then after the linear transformation
,
is a linear combination of
1
2
r :
1 1 2 2
r
r
If there is a linear relationship between
1
2
r
2 2 1
1
r r
So
1
1 2
2 r
r
3. Linear transformation can change linear correlation vectors into linear dependent set of vectors.
Example [1]
Case 1 Suppose
be a nonzero vector in geometry space, each vector
to it in the
on the injection of transformation is a linear transformation,with
expressed, formula is
Here
)
denotes the inner product.
Case 2 In the linear space
or
]
n , differential quotient is a linear transformation. This transformation usually useDrepresentative,
D
=
)
MATRIX OF A LINEAR TRANSFORMATION
The definition of linear transformation matrix [2]
Definition 2 Suppose
1
2
n is a set of basis in n -dimensional linear space
V
which Belongs to the number field
P
. AndAis a linear transformation of
V
.So the basis vectors can be linearly expressed as
1 11 1 21 2 1
2 12 1 22 2 2
1 1 2 2
n n
n n
n n n nn n
(3)
1 2 1 2 1 2
n
n
n
(1)
Among them
11 12 1
21 22 2
1 2
n
n
n n nn
Matrix
A
is called the matrix of linear transformationAon the base of
1
2
n.
The related theorem [3]
Theorem 1 Suppose
1
2
n is a set of basis in linear space
,
1
2
nis
n
vector arbitrary in
V
. There is a linear transformation ofA
A
i
i ,
Theorem 2Suppose
1
2
n is a set of cardinality in n-dimensional linear space
V
which Belongs to the number field
P
. In this base, every linear transformation according to the formula (1) corresponding to a
matrix. This corresponds with the following properties.
(1)Linear transformation sum corresponding to the matrix sum.
(2)The product of linear transformation corresponding to the matrix product.
(3)The number product of linear transformation corresponding to the matrix number product. (4)Invertible Linear Transformation corresponding to invertible matrix.
Theorem 3 Suppose matrix
A
is the matrix of linear transformationAon the base of
1
2
n. On the base of
1
2
n, vector
coordinate is
1
2
n
.So theA
coordinate, on the base of
1
2
n, is
1
2
n
.And
1 1
2 2
n n
Theorem 4Suppose matrix of linear transformation ofA, which in linear space
V
, under the two bases of (2) and (3) are respectively
and
B
.From the base (2) to (3) the transition matrix is
,and
1 .
n
1
2
(2)
1
2 n
(3)
Example[3]
Case 1 Suppose
1
2
m is a set of basis in subspace
W
which belongs to n -dimensional linear space
)
.It expanded a set of basis in
as
1
2
n.Specifies a linear transformAmeet
i i
i
(4)
A2=A
Matrix of projectionAunder the base of
1
2
nis
By determining a set of basis, established a mapping from linear transformation in n -dimensional linear space
V
which Belongs to the number field
to the
matrix.
Case 2 Suppose the matrix of linear transformationA,which in three dimensional linear spaces
V
, under the base of
3 2 1
is
11 12 13
21 22 23
31 32 33
(1)Seeking the matrix ofAunder the base of
3
2
1.
(2)Seeking the matrix ofAunder the base of
1
2
3,among them
and
.
Solving:
A
1
11
1
21
2
31
3 A
2
12
1
22
2
32
3 A
3
13
1
23
2
33
3
So
A
3
33
3
23
2
13
3
A
2
32
3
22
2
12
1
A
1
31
3
21
2
11
1
The matrix ofAunder the base of
3
2 1
is
11 12 13
21 22 23
31 32 33
In the same way,
A 1 11 1
21
2 31
3
A
2
12
1
22
2
32
3
A 3 13 1
23
2 33
3
(5)
The matrix ofAunder the base of
1
2
3is
33 32 31 23 22 21 13 12 11
Case 3 Suppose
V
is a Two-dimensional linear space in the number field
, and
1
2 is a set of basis. The matrix of linear transformationAunder the base of
1
2is
2
Calculating the matrix ofAunder another base of
1
2,among them
1 2
1 2
So, the matrix ofAunder the base of
1
2is that 1
Case 4 Suppose
1
2
3
4 is a set of basis in four dimensional linear spaces
V
,known the matrix ofAunder the base of
1
2
3
4is
1
Seeking the matrix of A under the base of
1
1
2
4
2
2
3
4
3
3
4 ， 4
4
. Solving:
1
2
3
4
1
2
3
4
(6)
Case 5Two bases of
3 of given [4]
1
1
2
2
3
3
)
The definition of linear transformationA: A
i
i
3
(1) Write out the transition matrix from
1
2
3to
1
2
3.
(2) Write out the matrix of
A
under the base of
1
2
3. (3) Write out the matrix of
A
under the base of
1
2
3. Solving:
1
Because of
A
i
i
i
So
1
2
3
1
2
3
(7)
1
APPLICATION OF LINEAR TRANSFORMATION IN DEFORMATION MEASUREMENT
Digital camera belongs to the non-metric camera. It has no fiducial mark and orientation equipment; when it goes on field photography can not make camera horizontally, therefore the initial inner and outer orientation elements of value can not be obtained when the camera at the moment of photography. So the past commonly used method is no longer applicable, we must adopt new data processing method [5].
In this study using the direct linear transformation (DLT) method, the method doesn’t need fiducial mark and the initial inner and outer orientation elements of value. Therefore this method is suitable for industrial and civil Photogrametry. The principle is
1 2 3 4 9 10 11
5 6 7 8 9 10 11
(4)
Considering the effects of camera lens distortion measurement of non quantity, in the formula (4) is introduced corresponding correction terms
.
2 0 1
2 0 1
Obtained 2
0 1 1 2 3 4 9 10 11
2
0 1 5 6 7 8 9 10 11
(5)
Among them
K
1 is the coefficient symmetry lens distortion,
0
,
0is the main points of the image coordinatograph coordinates. The image diameters is
1
2 2 2
0 0
Inverse calculating the principal point coordinates of image
0 1 9 2 10 3 11 2 2 2
0 5 9 6 10 7 11 2 2 2
(6)
Inverse calculating the principal distance of photograph
2 2 2 2 2 2 2 2
0 1 2 3 9 10 11
2 2 2 2 2 2 2 2
0 5 6 7 9 10 11
x
z
(8)
Among them
x z
F
Direct linear transformation formula is a set of nonlinear equations; it is calculated with the method of least squares iterative method. The algorithm consists of two steps,
9 10 11 1 2 3 4
9 10 11 5 6 7 8
x
z
(8)
The error equation of positive operator
1 2 2 0
2 0
11 1
(9)
By (9) can be calculated for 11
L
coefficients and a distortion coefficient
K
1,the above formula written as
B L W
The Method equation is
T T
The solution is
1
2 11
1
T T T
W
(10)
This operation step is an iterative process; the difference of two adjacent operations is less than 0.01mm as a limit cycle iterative solution method. Coefficient of each
is calculated by
L
value. And once again put the coefficient of
A
into the error equation, calculation of
L
coefficient, so the cycle until less than setting limits 0.01mm [6].
When was solved, we can obtain the coordinatograph coordinates, 2
0 1
2
0 1
(11)
Because each photograph has respective coefficient
L
,the process of positive operator should be carried out with piecewise.
4
1 9 2 10 3 11
5 9 6 10 7 11 8
（12）
For each photograph of each image point can have type (12), In order to obtain the
, ,
, at least two photograph. The formula is
N S Q
The Method equation is
T T
(9)
The solution is
T
T
1 T
Q
(13)
The above calculations take the point by point manner, to accelerate the computation speed, reducing the occupied memory capacity [7].
CONCLUSION
This paper introduces the concept, basic knowledge and correlation theory of linear and linear transformation matrix, and given many numerical examples, embodies the importance of linear transformation thought in numerical calculation. Finally introduces linear transform in the application of deformation measurement.
REFERENCES
[1] Peking University Department of Mathematics algebraic group in geometry and algebra Research Laboratory, higher algebra[M], Higher Education Press,2001
[2]Li Xin, Matrix theory and its application[M], Chongqing University press,2005
[3]Chen Wendeng, Huang Xiankai, Mathematics Review Guide[M], World Book Inc,2004
[4]Yu Chengxin, Zhang Xiangdong, Mou Yuzhi, Song Chuanzeng, Quan Jin, Ding Ning, Li Xianli, The application study of the direct linear transform in deformation observation[J],Journal of Shandong institute of ARCH and ENG,2002:23-28
[5] Lipschutz S., Theory and Problems of Linear Algebra[M], McGraw-Hill，1991
[6] Griffel D H., Linear Algebra and its Applications[M], Marcei Dekker, 1998
Updating...
Updating... | 3,307 | 11,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-27 | latest | en | 0.850943 |
https://www.sapling.com/12156745/calculate-military-retirement-va-compensation | 1,601,284,566,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00644.warc.gz | 1,041,059,910 | 63,449 | # How to Calculate Military Retirement & VA Compensation
Calculating concurrent military retirement and VA compensation benefits is a highly personal and specialized calculation.
If you have served in the military, the United States government rewards your sacrifice with retirement benefits. If you became disabled while serving in the military, the Department of Veterans Affairs provides VA disability compensation or disability pension depending on the your age. You may be eligible to receive these benefits at the same time. This practice was forbidden until 2004, but it is allowed now, provided that you meet the eligibility requirements. Concurrent benefits are highly personalized based on your specific situation, so there are no tables or generic ways to determine your benefits. However, there is a process by which you can predict your likely concurrent pay based on your situation.
## Step 1
Figure out your retirement compensation. This will be determined according to your retirement plan and will be based on your disability or your years of service. You may have received a statement of your military retirement benefits that either provides your actual benefit or your projected benefit. Otherwise you must calculate your estimated retirement compensation based on the formula listed in your military retirement-plan document.
## Step 2
Subtract the Current Baseline Offset. This will be equal to the lesser of your retirement compensation or your VA disability compensation or disability pension. This number is called your net retired pay.
## Step 3
Determine your Concurrent Retirement and Disability Pay table rate. This rate is based on your VA Disability Rating -- your percentage of disability from 50 to 100 percent.
## Step 4
Subtract your table rate from the lesser of your retirement compensation or your VA disability compensation or disability pension. This number will be your remaining offset.
## Step 5
Calculate your Concurrent Retirement and Disability Pay Phase-in Percentage. This percentage will be based on the current year. At the time of publication, the Concurrent Retirement and Disability Pay Phase-in percentage is 98.18 percent.
## Step 6
Multiply the Concurrent Retirement and Disability Pay Phase-in Percentage by the remaining offset. This amount is your phase-in amount.
## Step 7
Add the Concurrent Retirement and Disability Pay table rate from above to the phase-in amount; that is the amount of your estimated concurrent benefits.
### Tip
In order to receive concurrent benefits, you must be a retiree of the military with 20-plus years of service and have a service-related disability rating of 50 percent or more.
Concurrent Retirement and Disability Pay is an automatic process, so you will not need to apply in order to receive it. Provided you are eligible, you should see the increase in your benefits.
### Things You'll Need
• VA Disability Rating
• Copy of your military retirement plan
• Statement of military retirement benefits, if any
### Warning
Although you will be receiving concurrent benefits, the aggregate amount of the military retirement benefit and the VA disability compensation or disability pension will not exceed the amount of the benefits to which you are entitled from the military and the VA.
references | 607 | 3,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | longest | en | 0.93888 |
http://www.ck12.org/book/Basic-Speller-Student-Materials/r1/section/4.2/ | 1,493,533,240,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124299.47/warc/CC-MAIN-20170423031204-00191-ip-10-145-167-34.ec2.internal.warc.gz | 487,716,979 | 30,210 | # 4.2: When It's -s and When It's -es
Difficulty Level: At Grade Created by: CK-12
## When It's -s and When It's -es
1. In the last lesson you found these two groups of singular nouns:
Sort these twelve singular nouns into this matrix. Remember that the letter <x> at the end of words spells the combination of sounds [ks]. When you get done, two of the squares should still be empty:
2. When you want to refer to more than one of something with a singular noun that ends in the sounds ______, ______, ______, or _____, you add -es.
3. Now you can write a more useful rule for choosing -s and -es: When you want to refer to more than one of something with a noun that ends in the sounds _____, _____, _____, or ______, you add -es, but with most other nouns you add ________.
Word Changes
1. Write the word catch in the blank: ... _________
2. Add the suffix that means “more than one”: ... _________
3. Change the \begin{align*}<\text{s}>\end{align*} to the letter that comes right in front of it in the alphabet: ... _________
4. Change the first letter of the word to <w> and change the last letter to \begin{align*}<\text{s}>\end{align*}: ... _________
5. Change the first vowel in the word to <l>: ... _________
6. Change the first letter in the word to the letter that comes between <o> and <q> in the alphabet and change the \begin{align*}<\text{s}>\end{align*} back to <r>: ... _________
7. Add the suffix that means “more than one”: ... ________
8. Take away the \begin{align*}<\text{p}>\end{align*} and the <t>. Then move the <r> up to the front of the word: ... ________
9. Change the last letter of the word back to an <r>: ... _________
Riddle: A baseball player who makes a lot of money might be called a \begin{align*}\frac{}{\text{Word} \ \#9} \ \frac{}{\text{Word} \ \#6}.\end{align*}
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https://www.teachoo.com/5137/1343/Example-28---Show-that-set-of-letters-needed-to-spell-CATARACT/category/Equal-sets/ | 1,685,839,241,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00283.warc.gz | 1,127,597,111 | 32,791 | Equal sets
Chapter 1 Class 11 Sets
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Example 23 Show that the set of letters needed to spell “ CATARACT ” and the set of letters needed to spell “ TRACT” are equal. Let X be the set of letter in “CATARACT”. ∴ X = {C, A, T, R} Let Y be the set of letter in “TRACT”. ∴ Y = {T, R, A, C} Since every element of X and Y are equal Hence, X = Y | 133 | 443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-23 | longest | en | 0.869907 |
https://fykos.org/year29/problems/series6 | 1,657,001,692,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00386.warc.gz | 315,845,941 | 11,975 | # 6. Series 29. Year
### 1. It's about what's inside of us
In the year 2015, a Nobel prize for Physics was given for an experimental confirmation of the oscillation of neutrinos. You have probably already heard about neutrinos and maybe you know that they interact with matter very weakly so they can pass without any deceleration through Earth and similar large objects. Try to find out, using available literature and Internet sources, how many neutrinos are at any instant moment in an average person. Don't forget to reference the sources.
### 2. Optometric
Pikos' friend wears glasses. When she puts them on, her eyes seem to be smaller. Is she shortsighted or farsighted? Justify your answer.
### 3. Going downhill
We are going up and down the same hill with the slope $α$, driving at the same speed $v$ and having the same gear (and therefore the same RPM of the engine), in a car with mass $M$. What is the difference between the power of the engine up the hill (propulsive power) and down the hill (breaking power)?
### 4. Fire in the hole
Neutral particle beams are used in various fusion devices to heat up plasma. In a device like that, ions of deuterium are accelerated to high energy before they are neutralized, keeping almost the initial speed. Particles coming out of the neutralizer of the COMPASS tokamak have energy 40 keV and the current in the beam just before the neutralization is 12 A. What is the force acting on the beam generator? What is its power?
### 5. Particle race
Two particles, an electron with mass $m_{e}=9,1\cdot 10^{-31}\;\mathrm{kg}$ and charge $-e=-1,6\cdot 10^{-19}C$ and an alpha particle with mass $m_{He}=6,6\cdot 10^{-27}\;\mathrm{kg}$ and charge 2$e$, are following a circular trajectory in the $xy$ plane in a homogeneous magnetic field $\textbf{B}=(0,0,B_{0})$, $B_{0}=5\cdot 10^{-5}T$. The radius of the orbit of the electron is $r_{e}=2\;\mathrm{cm}$ and the radius of the orbit of the alpha particle is $r_{He}=200\;\mathrm{m}$. Suddenly, a small homogeneous electric field $\textbf{E}=(0,0,E_{0})$, $E_{0}=5\cdot 10^{-5}V\cdot \;\mathrm{m}^{-1}$ is introduced. Determine the length of trajectories of these particles during in the time $t=1\;\mathrm{s}$ after the electric field comes into action. Assume that the particles are far enough from each other and that they don't emit any radiation.
### P. iApple
Think up and describe a device that can deduce its orientation relative to gravitational acceleration and convert this information to an electrical signal. Come up with as many designs as you can. (An accelerometer-like device that is in most smart phones.)
### E. Malicious coefficient of restitution
If we drop a bouncing ball or any other elastic ball on an appropriate surface, it starts to bounce. During every hit on the surface some kinetic energy of the ball is dissipated (into heat, sound, etc.) and the ball doesn't return to its initial height. We define the coefficient of restitution as the ratio of the kinetic energy after and before the hit. Is there any dependence between the coefficient of restitution and the height which the ball fell from? Choose one suitable ball and one suitable surface (or several if you want) for which you determine the relation between the coefficient of restitution and the height of the fall. Describe the experiment properly and perform a sufficient number of measurements.
### S. A closing one
• Find, in literature or online, the change of enthalpy and Gibbs free energy in the following reaction
$$2\,\;\mathrm{H}_2 \mathrm{O}_2\longrightarrow2\,\mathrm{H}_2\mathrm{O},$$
where both the reactants and the product are gases at standard conditions. Find the change of entropy in this reaction. Give results per mole.
• Power flux in a photon gas is given by
$j=\frac{3}{4}\frac{k_\;\mathrm{B}^4\pi^2}{45\hbar^3c^3}cT^4$.
Substitute the values of the constants and compare the result with the Stefan-Boltzmann law.
• Calculate the internal energy and the Gibbs free energy of a photon gas. Use the internal energy to write the temperature of a photon gas as a function of its volume for an adiabatic expansion (a process with $δQ=0)$.
Hint: The law for an adiabatic process with an ideal gas was derived in the second part of this series (Czech only).
• Considering a photon gas, show that if $δQ⁄T$ is given by
$$\delta Q / T = f_{,T} \;\mathrm{d} T f_{,V} \mathrm{d} V\,,$$
then functions $f_{,T}$ and $f_{,V}$ obey the necessary condition for the existence of entropy, that is
$$\frac{\partial f_{,T}(T, V)}{\partial V} = \frac{\partial f_{,V}(T, V)}{\partial T}$$ | 1,176 | 4,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-27 | longest | en | 0.887314 |
https://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-4-section-4-3-multiplying-and-dividing-fractions-exercise-set-page-247/21 | 1,534,416,773,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210615.10/warc/CC-MAIN-20180816093631-20180816113631-00210.warc.gz | 891,236,226 | 15,498 | ## Prealgebra (7th Edition)
$\frac{4}{9}$
$(-\frac{2}{3})^2=(-\frac{2}{3})\times(-\frac{2}{3})=\frac{2\times2}{3\times3}=\frac{4}{9}$ | 67 | 134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-34 | longest | en | 0.278426 |
https://www.performancetrading.it/Documents/KsIntroduction/KsI_Formulae.htm | 1,713,892,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00072.warc.gz | 848,853,218 | 3,776 | Home > Doc > Introduction to Implied, Local and Stochastic Volatility > Formulae derivation for Heston Volatility
# Formulae derivation for Heston Volatility
The "General Heston Model" is:
We abbreviate the second of these in the form:
In doing so, we arrive to the General PDE for stochastic volatility:
(46)
We can see that the last PDE {46} and the original Black-Scholes PDE have a lot of similarity. Applying a similar set of transformations and using T equal to the maturity of the contract:
τ = T − t
x = log(S) + (r − D)(T − t)
V = U (x, ν, τ )e−r(T −t)
and after some routine calculus using the chain rule, leads to a PDE for "U" in the form:
Now, if we introduce the Fourier transform in the following form:
(47)
then at maturity, where τ = T − t = 0, we have:
which is the Fourier Transform of the payoff expressed in terms of the logarithm of the asset price. So the differentiation w.r.t. "x" is equal to the multiplication by (−iw) in the transform:
(48)
European option
The payoff of a call European option is max[S −K, 0] in terms of our original variables. In terms of logarithmic variables, we have:
V (x, ν, 0) = max [ex − K, 0]
so the Fourier transform of the payoff is:
The Vanilla Call
We need to check when this integral actually converges and bear in mind that "w" can be any complex number. We need the exponential to decay as "x" becomes large, so that the integral converges. This ONLY happens if Im(w) > 1 and, when this is true, we can evaluate the integral with some simplification of which gives:
The Vanilla Put
Here conditions remain the same, except that this time the integral converges only if Im(w) < 0. When this is true, we get an identical transform:
Digital Calls and Puts
For a digital call, the transformed payoff is:
For a digital put, we have:
Prof. Klaus Schmitz
Summary: Index | 475 | 1,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.844546 |
http://borenson.com/tabid/1003/Default.aspx | 1,506,073,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688932.49/warc/CC-MAIN-20170922093346-20170922113346-00238.warc.gz | 52,936,626 | 6,776 | Powerful Whole Brain Learning
Dr. Borenson's Balance Model for Solving Equations
## Demystifying the Learning of Algebra
Hands-On Equations is a supplementary program that can be used with any math curriculum to provide students with a concrete foundation for algebra. It uses the visual and kinesthetic instructional approach developed by Dr. Henry Borenson to demystify abstract algebraic concepts. This hands-on, intuitive approach enhances student self-esteem and interest in mathematics.
## What are the benefits of using Hands-On Equations?
• No algebraic prerequisites are required
• It is a game-like approach that fascinates students
• The gestures or "legal moves" used to solve the equations reinforce the concepts at a deep kinesthetic level
• The program can be used as early as the 3rd grade with gifted students, 4th grade with average students and 5th grade with LD students; it also serves as an excellent component of a middle-school pre-algebra program
• Students attain a high level of success with the program (see research studies section)
• The program provides students with a strong foundation for later algebraic studies
• The concepts and skills presented are essential for success in an Algebra 1 class
## Algebra concepts your student will learn in only seven lessons!
• the concept of an unknown
• how to evaluate an expression
• how to combine like terms
• the relational meaning of the equal sign (both sides have the same value)
• the meaning of an algebra equation
• how to balance algebra equations (using the subtraction property of equality)
• the concept of the check of an equation
• the ability to solve one and two-step algebra equations
• solving equations with unknowns on both sides (see video)
• how to work with a multiple of a parenthetical expression
## In Levels II and III, students learn:
• the concept of the opposite of an unknown
• how to evaluate algebraic expressions involving x and (-x).
• the additive property of inverses
• the addition property of equality
• the concept that subtracting an entity gives the same result as adding its opposite
• addition and subtraction of integers
"A strong foundation in algebra should be in place by the end of eighth grade...".
Principles and Standards for School Mathematics, NCTM | 473 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-39 | longest | en | 0.914502 |
https://welovehousesitting.com/how-to-get-277v-from-480v-2/ | 1,679,718,564,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00314.warc.gz | 694,564,766 | 18,959 | # How To Get 277V From 480V
277 volts is a common voltage found in the United States. It can be achieved by taking 480 volts and stepping it down to 277 volts with a transformer.
## How To Get 277V From 480V
There are a few ways to convert 480v to 277v. One way is to use a transformer, which will step down the voltage. Another way is to use a rectifier, which will convert the AC voltage to DC voltage. Finally, you can use a voltage regulator, which will stabilize the voltage output.
The necessary tools for this conversion are a 480v to 277v transformer, a line cord, and an outlet box. The transformer can be mounted inside or outside the outlet box. The line cord is connected between the transformer and the outlet box. The outlet box is then connected to the 277v supply.
• Calculate the percentage of power loss. determine how many transformers are needed. calcul
• Find the voltage of the input
• Find the voltage of the output
on ‘480v ac supply’ – How to get 277v from 480v: – 1) Convert 480v ac to 277v ac by using a transformer with a step-down ratio of 2:1. – 2) Use the 277v ac output to power devices that require 277v ac.
## Frequently Asked Questions
### What Voltage Is One Leg Of 480?
There are 480 volts between legs of a 480 volt circuit.
### Why Is It 277 480?
The number 277 480 is the result of a mathematical calculation involving the Fibonacci sequence.
### Does 277 Volts Have A Neutral?
277 volts does not have a neutral. 277 volts is a three-phase voltage and has three hots, or line-to-line voltage. There is no neutral wire in a 277 volt system.
## In Summary
There are several ways to get 277v from 480v. One way is to use a transformer to step down the voltage. Another way is to use a rectifier to convert the voltage to DC, and then use a DC-to-AC inverter to convert it back to AC. | 458 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-14 | latest | en | 0.857554 |
https://community.airtable.com/t5/formulas/nested-if-statement/td-p/129198 | 1,718,635,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00271.warc.gz | 154,956,095 | 50,832 | # Nested IF Statement
Topic Labels: Formulas
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4 - Data Explorer
Hello,
I’m new to Airtable & I have a formula that an Airtable support doc says should work & I can get to work in Excel, but keeps returning an error in my Airtable base.
What I want to accomplish is this:
IF the Jan 22 Commission Field is not blank, Jan 22 Prof Calc should equal that.
Otherwise, IF Jan 22 Actual is blank, Jan Prof Calc should equal zero. IF it’s not blank, Jan 22 Prof Calc should equal Jan 22 Remaining.
1 Solution
Accepted Solutions
16 - Uranus
A couple things:
• No Airtable formula starts with “=”. Remove it.
• `<>""` should be `!=` and that is asking “does not equal a blank string”, and you’re comparing number fields which won’t produce strings in the first place. A better way to check if a field is not blank, string or otherwise, is `IF({field name})`. If you want to check if it is blank or false you could also do `NOT({field name})`
• If you ever want you outcome to be a proper number, never put any possible outcome from the nested IFs in quotes. `"0"` is a string, `0` is a number.
So the proper formula would be:
``````IF(
{Jan 22 Commission},
{Jan 22 Commission},
IF(
NOT({Jan 22 Actual}),
0,
{Jan 22 Remaining}
)
)
``````
3 Replies 3
8 - Airtable Astronomer
You need to remove the “=” from before the IF (Airtable syntax isn’t quite the same as Excel) and use != for “does not equal/is not”. So your formula will be:
`IF({Jan 22 Commission}!="" {Jan 22 Commission}, IF({Jan 22 Actual}="", "0", {Jan 22 Remaining}))`
16 - Uranus
A couple things:
• No Airtable formula starts with “=”. Remove it.
• `<>""` should be `!=` and that is asking “does not equal a blank string”, and you’re comparing number fields which won’t produce strings in the first place. A better way to check if a field is not blank, string or otherwise, is `IF({field name})`. If you want to check if it is blank or false you could also do `NOT({field name})`
• If you ever want you outcome to be a proper number, never put any possible outcome from the nested IFs in quotes. `"0"` is a string, `0` is a number.
So the proper formula would be:
``````IF(
{Jan 22 Commission},
{Jan 22 Commission},
IF(
NOT({Jan 22 Actual}),
0,
{Jan 22 Remaining}
)
)
``````
4 - Data Explorer
Thanks so much! I appreciate the formula & the explanation. | 634 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-26 | latest | en | 0.856976 |
https://in.mathworks.com/matlabcentral/answers/2124851-how-to-plot-isosurfaces-using-3-vector-columns-of-data | 1,726,830,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00072.warc.gz | 270,453,470 | 37,894 | How to plot isosurfaces using 3 vector columns of data?
18 views (last 30 days)
Bacha Munir on 3 Jun 2024
Commented: Bacha Munir on 25 Jun 2024
I have 4 columns. The first three are the x,y, and z coordinates while the last one is the iso level. How can I plot them in Matlab?. thanks
Mathieu NOE on 3 Jun 2024
what are we supposed to do with the iso level - to do a countour plot ?
you can first plot your x,y,z data as a scatter plot
figure
scatter3(data(:,1), data(:,2), data(:,3))
Mathieu NOE on 3 Jun 2024
Edited: Mathieu NOE on 3 Jun 2024
maybe this ?
pretty much what you could obtain with contour (in principe ,but I am not sure how to make contour work on your data)
NB that I extracted the "level" Z data using a certain tolerance (tolZ), then from the raw coordinates I made a closed smoothed curve using smoothn available from the Fex : smoothn - File Exchange - MATLAB Central (mathworks.com)
hope it helps !
x = data(:,1);
y = data(:,2);
z = data(:,3);
%
figure
scatter3(x, y, z)
hold on
% get level data (here we are dealing with one value only)
level = unique(data(:,4));
level(isnan(level)) = [];
% set a tolerance on the z coordinates
tolZ = level/30;
ind = abs(data(:,3) - level)<tolZ;
xa = data(ind,1);
ya = data(ind,2);
za = data(ind,3);
% select points and re order with theta in ascending order
centroid_x = mean(xa);
centroid_y = mean(ya);
[th,r] = cart2pol(xa-centroid_x,ya-centroid_y);
[th,ia,ic] = unique(th);
r = r(ia);
za = za(ia);
% closing the curve
r(end+1) = r(1);
th(end+1) = th(1)+2*pi;
za(end+1) = za(1);
[xa,ya] = pol2cart(th,r);
xa = xa + centroid_x;
ya = ya + centroid_y;
plot3(xa, ya, za,'*g')
% create smoothed closed curve
A = smoothn({xa, ya, za},10);
xs = A{1};
ys = A{2};
zs = A{3};
% force the smoothed curve to be closed
xs(end+1) = xs(1);
ys(end+1) = ys(1);
zs(end+1) = zs(1);
plot3(xs,ys,zs,'r','linewidth',3)
Mathieu NOE on 24 Jun 2024
hello
I don't thinl smoothn is appropriate here for scattered data representing a closed surface - I got a bad result here - this is how I changed a bit your code :
clc
clearvars
fid=fopen('surface plot 1.txt');
data = [Z{1} Z{2} Z{3}];
fclose(fid);
% Remove rows containing NaN values
data = data(all(~isnan(data), 2), :);
% try do some smoothing
xx=data(:,1);yy=data(:,2);zz=data(:,3);
Zr=smoothn({xx,yy,zz});
data = [Zr{1} Zr{2} Zr{3}];
fid=fopen('geometry.txt');
geo = [Z{1} Z{2} Z{3}];
fclose(fid);
% Remove rows containing NaN values
geo = geo(all(~isnan(geo), 2), :);
%% Run program
[t,tnorm]=MyRobustCrust(data(:,1:3));
%% plot of the output triangulation
figure(1)
grayColor = [.7 .7 .7];
title('Output Triangulation','fontsize',14)
p = trisurf(t,data(:,1),data(:,2),data(:,3));
p.EdgeColor = 'none';
p.FaceColor = 'm';
view(-40,24)
box on
camlight(40,40)
camlight(-20,-10)
% now add the geometry dots
%
hold on
scatter3(geo(:,1),geo(:,2),geo(:,3),5,'filled');
hold off
this also may interest you if you need to smooth the surface
Bacha Munir on 25 Jun 2024
Ok brother. Thank you so much. | 974 | 2,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.803428 |
https://worlddatabaseofhappiness.eur.nl/correlational-findings/28277/ | 1,722,824,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00870.warc.gz | 508,038,292 | 12,184 | Correlational findings
Study OECD (2017): study FR 2015
Public
15-16 aged, secundary school students, France, 2017
Survey name
INT-PISA 2015
Sample
Respondents
N = 6108
Non Response
Assessment
Questionnaire: Conputer Assisted Web Interview (CAWI)
Correlate
Authors's Label
Time per week spend learning in regular lessons
Our Classification
Remarks
Learning time in test language regular lessons was computed by multiplying the number of minutes on average in the test language class by numberof test language class periods per week. Comparable indices were computed for mathematics and science. Learning time in total was computed using information about the average minutes in a class period in relation to information about the number of class periods per week attended in total. For convenience purposes, the information on learning time has been transforme into hours
Distribution
quartiles presented,
M hours= 27
SD= 7
Operationalization
Selfreport on 3 question about 4 subjects:
- How many class periods per week are you typically required to attend for: language / mathematics/ science?
- In a normal, full week at school, how may class periods per week are you required to attend in total? (slider range: 0 - '80 or more')
- How many minutes, on average, are there in a class period? (slider range: 0 - '120 or more')
Observed Relation with Happiness
Happiness Measure Statistics Elaboration / Remarks O-SLW-c-sq-n-11-b = - ns 4 M= 7,59
3 M= 7,62
2 M= 7,68
1 M= 7,62
4-1 difference -0,02 | 379 | 1,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.873802 |
http://www.artemisia.na.it/p70dd/exponential-distribution-equation-4d0fc4 | 1,709,089,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474690.22/warc/CC-MAIN-20240228012542-20240228042542-00671.warc.gz | 45,643,751 | 4,413 | obtainNote can be approximated by a normal distribution with mean We observe the first An exponential distribution compounded with a gamma distribution on the mean yields a Burr distribution. the. By continuing, you consent to the use of cookies. independent, the likelihood function is equal to property; for example, the arrival rate of cosmic ray alpha particles or Given a Poisson distribution with rate of change , the distribution of waiting times between successive changes (with ) is (1) (2) (3) and the probability distribution function is (4) It is implemented in the Wolfram Language as ExponentialDistribution[lambda]. $$. Geiger counter tics. Exponential Distribution. setting it equal to zero, we$$ in a given period). populations? Bounds on Time The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to … Functions for computing exponential PDF values, CDF values, and for producing isBy The exponential distribution is the only distribution to For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. \mbox{CDF:} & F(t) = 1-e^{-\lambda t} \\ to, The score functionwhere \begin{array}{ll} It may be possible to pass the CRE exam knowing one formula. can only belong to the support of the distribution, we can only positive values (and strictly so with probability & \\ is the parameter that needs to be estimated. first order condition for a maximum is View our, probability density, cumulative density, reliability and hazard functions, Probability and Statistics for Reliability, Discrete and continuous probability distributions, « Preventive Maintenance Goals and Activities, https://accendoreliability.com/standby-redundancy-equal-failure-rates-imperfect-switching/. "Exponential distribution - Maximum Likelihood Estimation", Lectures on probability theory and mathematical statistics, Third edition. is. rate during the respective time durations. When these events trigger failures, the exponential the distribution and the rate parameter The normal distribution, commonly known as the bell curve, occurs throughout statistics. \mbox{PDF:} & f(t, \lambda) = \lambda e^{-\lambda t} \\ life distribution model will naturally apply. Knowing the exponential distribution reliability function is one that you should memorize. any model by piecewise exponential distribution segments patched together. asymptotic normality of maximum likelihood estimators are satisfied. Cookies Policy, Rooted in Reliability: The Plant Performance Podcast, Product Development and Process Improvement, Musings on Reliability and Maintenance Topics, Equipment Risk and Reliability in Downhole Applications, Innovative Thinking in Reliability and Durability, 14 Ways to Acquire Reliability Engineering Knowledge, Reliability Analysis Methods online course, Reliability Centered Maintenance (RCM) Online Course, Root Cause Analysis and the 8D Corrective Action Process course, 5-day Reliability Green Belt ® Live Course, 5-day Reliability Black Belt ® Live Course, This site uses cookies to give you a better experience, analyze site traffic, and gain insight to products or offers that may interest you. is 0.6321. The exponential model works well for inter arrival The cumulative hazard function for the exponential is just the integral of the failure rate or … A quantity is subject to exponential decay if it decreases at a rate proportional to its current value. the asymptotic variance In this lecture, we derive the maximum likelihood estimator of the parameter Our first question was: Why is λ * e^(−λt) the PDF of the time until the next event occurs? Taboga, Marco (2017). Your email address will not be published. A generic term of the The definition of exponential distribution is the probability distribution of the time *between* the events in a Poisson process.. Histogram form with corresponding exponential PDF drawn through the histogram.
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Probability And Statistics Formulas And Examples, Steady Progress Mtg, Safety Sensor Working Principle, Is Calcium Carbonate Crystalline, Kabul House Phone Number, French Verb Conjugation Practice Audio, 2008 Ktm 690 Smc, | 844 | 4,226 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.845008 |
https://www.coursehero.com/file/85875861/Chapter-08-Study-Problem-02-Plumbing-and-Mechanical-Requirements-Answerspdf/ | 1,638,903,356,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00218.warc.gz | 788,932,559 | 50,368 | # Chapter 08 - Study Problem 02 - Plumbing and Mechanical Requirements Answers.pdf
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Chapter 8: Study Problem #2 Answers 1. 9 female, 9 male Explanation: An elementary school is considered an Educational (E) occupancy in IPC Table 403.1. If you look for “Educational” under the column titled “Classification,” there are no subclassifications. The ratio for water closets is “1 per 50. As required by the IPC, you must first divide the total occupant load in half to determine the required number of fixtures for each gender. You are given the occupant load of 820. 820 divided by 2 = 410 for each gender. Then, take the 410 and divide it by the ratio of 50 to obtain 8.2 water closets. Note that “a” at the bottom of the table indicates that you must round up for any fraction of a number of persons. Therefore, rounding up, this school requires 9 water closets for females and 9 water closets for males. (Note: If you had not first halved the total occupant load as the code now requires, you would have obtained a different result of 8 water closets for each sex.) 2. 12; the code requires 4 water closets for the visitors and 8 water closets for employees. Explanation: A nursing home is listed in IPC Table 403.1 as an Institutional (I 2) occupancy. First, find “Institutional” under the column titled “Classification.” Then locate “I 2” under the “Occupancy” column. Adjacent to I 2 you can see that there is a separate categories for the residents, for “Employees, other than residential care” and “Visitors, other than residential care” in the “Description” column. First looking at the requirement for employees, you see that the water closet ratio is “1 per 25” occupants. You are given the total occupant load of 155, which | 431 | 1,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | latest | en | 0.917197 |
http://parkingpalblog.com/freebooks/student-resource-manual-to-accompany-contemporary-college-algebra-and-trigonometry-a-graphing | 1,508,673,580,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00898.warc.gz | 260,412,666 | 13,385 | # Read Student Resource Manual to accompany Contemporary College Algebra and Trigonometry. A graphing approach. Foss/Palmer/Hamilton. Published by Harourt. 2001 Edition PDF, azw (Kindle), ePub, doc, mobi
Format: Paperback
Language:
Format: PDF / Kindle / ePub
Size: 13.01 MB
As a distant second it should prepare you for the ACT math sections. The pace is faster than a high school class in precalculus, we aim for greater command of the material, especially to extend what we have learned to new situations. In the first book of the Sphaerica, there is the first known conception and definition of a spherical triangle (Heath 262). For example, x1 would be x[1]. [/frame]Square Roots – Type the radicand (the number inside the square root symbol) inside the parenthesis.
Pages: 0
Publisher: Harcourt (2001)
ISBN: B002P20X8A
PLANE TRIGONOMETRY AND TABLES
A Treatise On Trigonometry
College Algebra
Algebra & Trigonometry, Volume II, INSTRUCTOR'S SOLUTIONS MANUAL, 6TH, pb, 2002
Student Study and Solutions Manual for Larson/Hostetler's Precalculus: A Concise Course, 2nd
Plane Trigonometry - Part I. - Elementary Course Excluding The Use Of Imaginary Quantities
Plane and Spherical Trigonometry; An Elementary Text-Book
Rudiments of Plane Geometry, Including Geometrical Analysis, and Plane Trigonometry
***RE-PRINT*** Elements of plane and spherical trigonometry
e-Study Guide for: Algebra and Trigonometry by Gary K. Rockswold, ISBN 9780321568014
Trigonometry: A Complete Introduction: Teach Yourself
Algebra and Trigonometry Enhanced with Graphing Utilities (6th Edition)
Analytic trigonometry
PRENTICE HALL MATH GEOMETRY TENNNESSEE TEST PREPARATION 2004C
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Plane trigonometry with tables, (His Mathematics for technical training. [2])
The Principles of Plane Trigonometry, Mensuration, Navigation and Surveying: Adapted to the Method of Instruction in the American Colleges
College Trigonometry
TI-83 Graphing Calculator Manual for Trigonometry, 4th
Trigonometry: Plane and Spherical (Revised & Enlarged Edition)
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Course of Civil Engineering: Comprising Plane Trigonometry, Surveying, and Levelling. with Their Application to the Construction of Common Roads, Railways, Canals ... | 711 | 2,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-43 | latest | en | 0.776995 |
https://www.physicsforums.com/threads/will-these-two-brushed-dc-motors-reach-the-same-rpm.892666/ | 1,566,356,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00469.warc.gz | 928,420,782 | 23,985 | # Will these two brushed DC motors reach the same RPM
#### William123
Hi
I'm working on a project in school (or at home) about brushed DC motors. I have recently built a prototype myself but instead of using permanent magnets around the armature I'm using an electromagnet from another motor that died. I intend to build another motor using the same armature but this time with permanent (neodymium) magnets and I'm going to measure the RPM of the two motors when they draw the same amount of current. What are your predictions? I read somewhere on the internet once that voltage determines the speed (this is probably not the whole truth) and according to this statement the first prototype should run much faster because the resistance is higher in that motor, which means higher voltage to reach the same amp draw. This doesn't really make sense to me though, I feel like they should reach about the same RPM.
I also noticed that the back emf generated by this first prototype wasn't all that much. It was noticeable but I'd like to see a bigger difference in the current draw when the motor is running vs when it is stalled. At 40-45V the amp draw was 1.8 when stalled and 1.6 when running maybe (i can double check this later). Does this have anything to do with the fact that I'm not using permanent magnets or is it all based on the friction inside the motor?
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#### davenn
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I intend to build another motor using the same armature but this time with permanent (neodymium) magnets and I'm going to measure the RPM of the two motors when they draw the same amount of current. What are your predictions?
There's a lot of variables, the main one being the variations in the magnetic field strength
I read somewhere on the internet once that voltage determines the speed (this is probably not the whole truth) and according to this statement the first prototype should run much faster because the resistance is higher in that motor,
but if it's the same armature, then the resistance should be the same
#### William123
That's a good platform for experimenting.
Flux speed and current determine back emf.
Flux and current determine torque.
With a separate field you can vary flux.
Look up "generator saturation curve"
for starters try http://www.electrical4u.com/magnetization-curve-of-dc-generator/
Thank you!
There's a lot of variables, the main one being the variations in the magnetic field strength
but if it's the same armature, then the resistance should be the same
I don't think I mentioned it but the resistance of the stator alone is about 15 ohms maybe and it's wired in series with the armature, so when I replace that with permanent magnets the resistance will be only 4 ohms (the resistance of the armature). I'm going to build version 2 as soon as I can and see which version reaches the highest RPM (but really this will be a race between the electromagnetic stator and Nd magnets).
#### jim hardy
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There are two equations that define external characteristics of a dynamo.
Flux is fundamental. With permanent magnets for a field it's constant, with wound field it's proportional to field current.
Symbol for flux is Φ .
1. Counter EMF = K X Φ X RPM , removing the X's gives KΦRPM . You determine constant K by open circuit test , that's your saturation curve as in the link i gave you.
With permanent magnets you'll measure KΦ as a single constant.
With wound field you control Φ by field amps.
2. Torque = same KΦ X 7.04 X Iarmature . That gives torque in foot pounds , if you use Newton-meters it's an easy units conversion.
Observe this about behavior when when field is wired in series with armature .
Apply voltage and large current flows because there's no counter EMF.
Large current hrough field makes high flux. So there's a lot of torque.
Machine accelerates. Counter EMF appears reducing current. So torque decreases,
but so does field current hence flux, meaning more RPM is required to maintain counter EMF.
So machine speeds up some more.
If there's no external mechanical load the only thing limiting speed is friction of bearings and air.
It is possible to overspeed a series motor so be careful.
When that happens the wires fly out of the armature by centrifugal force and it wrecks the machine.
old jim
#### William123
There are two equations that define external characteristics of a dynamo.
Flux is fundamental. With permanent magnets for a field it's constant, with wound field it's proportional to field current.
Symbol for flux is Φ .
1. Counter EMF = K X Φ X RPM , removing the X's gives KΦRPM . You determine constant K by open circuit test , that's your saturation curve as in the link i gave you.
With permanent magnets you'll measure KΦ as a single constant.
With wound field you control Φ by field amps.
2. Torque = same KΦ X 7.04 X Iarmature . That gives torque in foot pounds , if you use Newton-meters it's an easy units conversion.
Observe this about behavior when when field is wired in series with armature .
Apply voltage and large current flows because there's no counter EMF.
Large current hrough field makes high flux. So there's a lot of torque.
Machine accelerates. Counter EMF appears reducing current. So torque decreases,
but so does field current hence flux, meaning more RPM is required to maintain counter EMF.
So machine speeds up some more.
If there's no external mechanical load the only thing limiting speed is friction of bearings and air.
It is possible to overspeed a series motor so be careful.
When that happens the wires fly out of the armature by centrifugal force and it wrecks the machine.
old jim
This information will be valuable later when I start writing the report so thanks again.
I discovered that the rotor turns faster when I put 2 neodymium magnets on the stator. I think the next step is to improve the commutator and brushes to reduce the sparking and friction because it's running surprisingly well right now. A bit hot maybe but it's all good.
#### CWatters
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I discovered that the rotor turns faster when I put 2 neodymium magnets on the stator.
In general a DC permanent magnet motor will accelerate until the back EMF equals the applied voltage. The back emf is proportional to the flux/strength of the magnetic field so a motor with weak magnets will tend to spin faster than one with stronger magnets. That may seem counter intuitive but it doesn't mean a motor with weak magnets is "better". It just means the motor constant is different. Typically the efficiency will be worse. The peak power may also be lower.
By attaching these two magnets you probably altered the field in the motor and made it weaker. If you want the motor to go faster on a particular voltage then removing turns from the rotor should give better results (lower winding losses).
#### William123
In general a DC permanent magnet motor will accelerate until the back EMF equals the applied voltage. The back emf is proportional to the flux/strength of the magnetic field so a motor with weak magnets will tend to spin faster than one with stronger magnets. That may seem counter intuitive but it doesn't mean a motor with weak magnets is "better". It just means the motor constant is different. Typically the efficiency will be worse. The peak power may also be lower.
By attaching these two magnets you probably altered the field in the motor and made it weaker. If you want the motor to go faster on a particular voltage then removing turns from the rotor should give better results (lower winding losses).
I had no idea about this. If I can find a good source on this I can include this in my report aswell. You can basically choose between torque/efficiency and higher RPM by altering the strength of the field then? (depending on what you're going to use the motor for)
And thank you for the answer, I appreciate it.
#### CWatters
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Gold Member
Yes. Usually motors are designed to produce a specified rpm when operating from a specified voltage. The motor constant k = rpm/voltage. So a 9v 9000 rpm motor has a motor constant of 1000 rpm per volt. You can achieve that using ferrite magnets and a lot of turns or stronger rare earth magnets and fewer turns. A motor with fewer turns has lower winding resistance and hence lower losses due to I^2R. So rare earth magnets motors can be said to have lower "copper losses" compared to ferrite magnet motors.
#### jim hardy
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I don't think I mentioned it but the resistance of the stator alone is about 15 ohms maybe and it's wired in series with the armature, s
figure out how it's wired.
Then play with those two equations.
Terminal voltage is the sum of counter-emf and IXR product of armature current and resistance .
High armature resistance means you can't make much torque. IXR hogs the available voltage leaving not much to do useful work.
#### CWatters
Homework Helper
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You can basically choose between torque/efficiency and higher RPM by altering the strength of the field then? (depending on what you're going to use the motor for)
Yes, although there are a lot of factors that effect the decision on what type of magnets to use. If there are few constraints then Ferrite Magnets are normally used because they cost less to make than rare earth magnets. Rare Earth magnets can also loose their strength more easily if overheated.
I've been struggling to find good references that cover the subject at the right level but perhaps see..
Page 5..
http://www.gearseds.com/files/Lesson3_Mathematical Models of Motors.pdf
Theoretically the motor armature should continue to accelerate to a higher and higher speed unless there is a force that works in opposition to the battery voltage. Clearly the motor armature does not continue to accelerate but instead reaches a finite top speed.
The motor shaft will not continue to rotate faster because the spinning of the armature coils within the permanent magnetic field of the motor generates a voltage or a back emf that opposes the applied voltage.
https://en.wikipedia.org/wiki/Motor_constants
Motor velocity constant, back EMF constant
snip
The field flux may also be integrated into the formula:[9]
Eb = Kωϕω
where Eb is back EMF, Kωis the [motor] constant, ϕ is the flux, and ω is the angular speed
snip
By Lenz's law, a running motor generates a back-EMF proportional to the RPM. Once the motor's rotational velocity is such that the back-EMF is equal to the battery voltage (also called DC line voltage), the motor reaches its limit speed
So the stronger the magnets, the greater the Back EMF at any given rpm and the lower rpm at which the Back EMF equals the supply voltage.
#### William123
figure out how it's wired.
Then play with those two equations.
Terminal voltage is the sum of counter-emf and IXR product of armature current and resistance .
High armature resistance means you can't make much torque. IXR hogs the available voltage leaving not much to do useful work.
So this is why you would want a big counter-emf, right? I ordered a non-contact tachometer this week for the purpose of testing this. I have a question about that second formula though. If Torque = KΦ X 7.04 X I and I'm using an electromagnet instead of permanent magnets, the torque doesn't increase linearly, does it? Because as I increase the current, Φ grows too, doesn't it? If I would use permanent magnets on the other hand, Φ would stay the same all the time and the current through the armature would be the only thing changing. Is this correct?
So the stronger the magnets, the greater the Back EMF at any given rpm and the lower rpm at which the Back EMF equals the supply voltage.
This makes sense. I also realised that maybe I don't need a source to tell me that:
a motor with weak magnets will tend to spin faster than one with stronger magnets.
I could prove it myself mathematically and through testing. Thank you for the gearseds source though, it will definitely come in handy! :)
Edit: And thank you for telling me about Lenz's law. Did you mean to add I*R to the back-emf when you said "Once the motor's rotational velocity is such that the back-EMF is equal to the battery voltage", because as Jim said, terminal voltage is equal to I*R + back-emf.
Edit #2: I found another source confirming that Vt = I*RArmature + Eb
Last edited:
#### jim hardy
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So this is why you would want a big counter-emf, right?
Yes !
I have a question about that second formula though. If Torque = KΦ X 7.04 X I and I'm using an electromagnet instead of permanent magnets, the torque doesn't increase linearly, does it? Because as I increase the current, Φ grows too, doesn't it?
If it's series wound, you are correct. Armature and field amps are the same so torque equation has amps in it twice,
once in Iarmature and once in Φ .
That's why automobile starters are series wound they make immense torque to overcome static friction in the engine and to push those pistons up against the compressed gas in the cylinders . Listen to yours and you'll hear it labor to do that as each cylinder makes its compression stroke. When you've trained your ear you can tell by that sound when your battery is nearing end of life and you have time to look for a sale.
If I would use permanent magnets on the other hand, Φ would stay the same all the time and the current through the armature would be the only thing changing. Is this correct?
Yes.
It is very heartening to see somebody figuring things out and improving his understanding.
Apply your new insights to everyday experience and you'll grow more of them quickly.
Thanks for sharing. It helps us old guys feel maybe we're helping out.
DC machinery was one of my favorite courses . We're surrounded by them in everyday life so we owe it to ourselves to understand them.
Payback is you're not helpless when an electric window on your car quits working.
We live in fascinating times. Have fun with the technology.
old jim
#### CWatters
Homework Helper
Gold Member
Do you have a way to apply a variable load and measure the torque? If you do then you can calculate the output power and efficiency. It's relatively easy to make a motor that delivers the required power at some specified rpm but the ultimate test of how good it is would be to compare the efficiency under load.
"Will these two brushed DC motors reach the same RPM"
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• Solo and co-op problem solving | 3,251 | 14,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-35 | longest | en | 0.960051 |
https://allenfrostline.com/2018/05/06/literature-review-optimal-order-execution-4/ | 1,563,377,589,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00510.warc.gz | 299,303,808 | 5,489 | Today we implement the order placement strategy in Almgren and Chriss (2000) s.t. for a certain order size $Q$, we can estimate the probability to perform the optimal strategy in the paper within time horizon of $T$.
# Mathematical Formulation
It is tolerable[1] in HFT that we assume stock price evolves according to the discrete time arithmetic Brownian motion:
$\begin{cases} dS(t) = \mu dt + \sigma dW(t),\\\\ dQ(t) = -\dot{Q}(t)dt \end{cases}$
where $Q(t)$ is the quantity of stock we still need to order at time $t$. Now let $\eta$ denote the linear coefficient for temporary market impact, and let $\lambda$ denote the penalty coefficient for risks. To minimize the cost function
$C = \eta \int_0^T \dot{Q}^2(t) dt + \lambda\sigma\int_0^T Q(t) dt$
we have the unique solution given by
$Q^*(t) = Q\cdot \left(1 - \frac{t}{T^*}\right)^2$
where $Q\equiv Q(0)$ is the total and initial quantity to execute, and the optimal liquidation horizon $T^*$ is given by
$T^* = \sqrt{\frac{4Q\eta}{\lambda\sigma}}.$
Here, $\eta$ and $\lambda$ are exogenous parameters and $\sigma$ is estimated from the price time series (see the previous post) within $K$ time units, given by
$\hat{\sigma}^2 = \frac{\sum_{i=1}^n (\Delta_i - \hat{\mu}_{\Delta})^2}{(n-1)\tau}$
where $\\{\Delta_i\\}$ are the first order differences of the stock price using $\tau$ as sample period, $n\equiv\lfloor K / \tau\rfloor$ is the length of the array, and
$\hat{\mu}_{\Delta} = \frac{\sum_{i=1}^n \Delta_i}{n}.$
Notice that $\hat{\sigma}^2$ is proved asymptotically normal with variance
$Var(\hat{\sigma}^2) = \frac{2\sigma^4}{n}.$
Now that we know
$\hat{\sigma}^2 \equiv \frac{16Q^2\eta^2}{\lambda^2 \hat{T}^4} \overset{d}{\to} \mathcal{N}\left(\sigma^2, \frac{2\sigma^4}{n}\right)$
which yields
$\frac{16Q^2\eta^2}{\lambda^2\hat{\sigma}^2\hat{T}^4}\overset{d}{\to}\mathcal{N}\left(1, \frac{2}{n}\right),$
to keep consistency of parameters, with $n\equiv \lfloor K/\tau\rfloor \to\infty$ we can also write
$\frac{16Q^2\eta^2}{\lambda^2\hat{\sigma}^2\hat{T}^4}\overset{d}{\to}\mathcal{N}\left(1, \frac{2\tau}{K}\right).$
with which we can estimate the probability of successful strategy performance. Specifically, the execution strategy is given above, and the expected cost of trading is
$C^* = \eta \int_0^{T^*} \left(\frac{2Q}{T}\left(1 - \frac{t}{T^*}\right)\right)^2 dt + \lambda\sigma\int_0^{T^*} Q\cdot \left(1 - \frac{t}{T^*}\right) dt = \frac{4\eta Q^2}{3T^*} + \frac{\lambda \sigma QT^*}{3} = \frac{4}{3}\sqrt{\eta\lambda\sigma Q^3}.$
# Implementation
(1.465147881156472, 0.8431842483948604)
which means there's a probability of 84.3% that we can perform our order placement strategy of size 10 within 3.6405 time units and minimize the trading cost of 1.47 at optimum.
1. 1.Over long-term investment time scales or in extremely volatile markets it is important to consider geometric rather than arithmetic Brownian motion; this corresponds to letting $\sigma$ scale with $S$ But over the short-term trading time horizons of interest the total fractional price changes are small and the difference between arithmetic and geometric Brownian motions is negligible. (Almgren and Chriss, 2000) ↩︎ | 1,094 | 3,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-30 | latest | en | 0.708301 |
https://math.stackexchange.com/questions/1578667/on-volume-forms-and-norms-on-exterior-powers | 1,716,714,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00080.warc.gz | 331,174,228 | 37,341 | # On volume forms and norms on exterior powers
Let $V$ be a $n$-dimensional vector space. Given an inner product on $V$ one may define an inner product on the simple $k$-vectors of $\Lambda^k(V)$ by $$\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k\rangle_{\Lambda^k(V)} := \operatorname{det}\left(\langle v_i, w_j \rangle_V\right)$$ and extend it bilinearly. As usual this induces a norm on $\Lambda^k(V)$.
Burago/Ivanov claim in [Lemma 2.4, p. 6] that an oriented volume form $\omega\in \Lambda^2(V^\ast) \cong \left(\Lambda^2(V)\right)^\ast$ on $V$ determines a linear isometry $J: V \to V^\ast$ in "a standard way".
I don't understand the "isometry"-part. Here is what I have so far:
Define the mapping $J:V \to V^\ast$ by $J(u)(v) := \omega(u \wedge v), v\in V$. I can show that this is an isomorphism. I can define a somewhat "dual volume form" $\omega^\ast \in \Lambda^2(V) \cong \Lambda^2(V^{\ast\ast}) \cong \left(\Lambda^2(V^\ast)\right)^\ast$ by means of $$\omega^\ast(l\wedge g) := \omega\left(J^{-1}(l)\wedge J^{-1}(g)\right)$$ Thus, $$\omega^\ast\left(J(u)\wedge J(u')\right) = \omega(u\wedge u').$$ This looks quite promising already. (I am able to generalise this to $\Lambda^n(V)$ via $\widetilde J: \Lambda^{n-1}(V) \to V^\ast, \sigma \mapsto \omega(\sigma \wedge \cdot)$ and Hodge dual)
The way Burago/Ivanov use the "isometry"-part in their paper is $\left|J(v) \wedge J(v')\right| = \left|v \wedge v'\right|$ though (where the norms are on the respective exterior powers).
Is there a relationship between the induced norm on the exterior power and a corresponding volume form? Maybe by choosing an orthonormal basis for $V$ and taking the standard volume form $\varepsilon^1 \wedge \cdots \wedge \varepsilon^n$ determined by the dual basis $\{\varepsilon^i\}$ ?
• I don't get it. I mean, if $V$ is $2$-dimensional, then most exterior powers of $V$ are trivial, right? Feb 16, 2016 at 13:48
• Yes. The only non-trivial ones are $V=\Lambda^1(V)$ and $\Lambda^2(V)$ where the latter is again isomorphic to $V$ by the Hodge dual. But does this make my question trivial? That would be great because as you can see below it took me a bit to prove the stated question Feb 16, 2016 at 16:10
I will answer my own question in some more generality.
Let $V$ be an $n$-dimensional real inner product space. On $\Lambda^k(V)$ we define the inner product \begin{align} \left\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k\right\rangle_{\Lambda^k(V)} := \operatorname{det}\left(\langle v_i, w_j \rangle_V\right) \end{align} and extend it bilinearly. This induces a norm $\left\|\sigma\right\|_{\Lambda^k(V)} := \sqrt{\left\langle \sigma, \sigma\right\rangle_{\Lambda^k(V)}}.$
Given an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ for $V$ one can easily calculate that $e_1\wedge e_2\wedge \cdots\wedge e_n$ is a unit $n$-vector in the one-dimensional real vector space $\Lambda^n(V)$.
Introduce the basis $\{\varepsilon^1,\varepsilon^2,\ldots,\varepsilon^n\}$ for $V^\ast$ dual to $\{e_1,e_2,\ldots,e_n\}$, that is, $\varepsilon^i(e_j) = \delta^i_j$. The Riesz representation theorem gives an inner product on $V^\ast$ and the dual basis is orthonormal with respect to this inner product. Then analoguously, $\varepsilon^1\wedge \varepsilon^2\wedge \ldots\wedge \varepsilon^n$ is a unit $n$-form in $\Lambda^n(V^\ast)$.
Now let $\omega\in \Lambda^n(V^\ast)$ be the standard volume form given by $\omega = \varepsilon^1\wedge \varepsilon^2\wedge \ldots\wedge \varepsilon^n$. Then $\omega(e_1\wedge e_2\wedge \cdots\wedge e_n)=1$.
Any $n$-vector $v_1\wedge v_2\wedge \cdots\wedge v_n\Lambda^n(V)$ is a multiple $c$ of the basis $n$-vector $e_1\wedge e_2\wedge \cdots\wedge e_n$.
Lemma 1
If $v_1\wedge v_2\wedge \cdots\wedge v_n = c e_1\wedge e_2\wedge \cdots\wedge e_n$ then \begin{align} \left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)} = \left|\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)\right|. \end{align}
Proof: \begin{align} \left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)} &= \frac{\left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)}} {\left\|e_1\wedge e_2\wedge \cdots\wedge e_n\right\|_{\Lambda^n(V)}}\\ &= |c| \\ &= \left|\frac{\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)} {\omega(e_1\wedge e_2\wedge \cdots\wedge e_n)}\right| \\ &= \left|\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)\right| \end{align}
Let us define the map $\widetilde J: \Lambda^{n-1}(V) \to V^\ast$ by $\widetilde J(\sigma)(v) := \omega(\sigma \wedge v)$. This map is an isomorphism which can be seen by a straight forward calculation and I won't do it here.
The Hodge dual gives an isomorphism $\star: \Lambda^k(V) \to \Lambda^{n-k}(V)$ which is characterised by \begin{align} (\star \lambda) \wedge \theta = \left\langle\lambda,\theta\right\rangle_{\Lambda^{k}(V)} e_1\wedge e_2\wedge \cdots\wedge e_n \end{align} where $\lambda, \theta \in \Lambda^k(V)$ are arbitrary.
The composite map $J := \widetilde J\circ\star: V \to V^\ast$ is therefore also an isomorphism.
Define an $n$-vector $\omega^\ast := J_\ast \omega \in \Lambda^n(V) \cong \left(\Lambda^n(V^\ast)\right)^\ast$ as the pushforward of the volume form $\omega$ by $J$, that is, \begin{align} \omega^\ast(\phi^1\wedge \phi^2\wedge\cdots\wedge \phi^n) := \omega\left(J^{-1}(\phi^1)\wedge J^{-1}(\phi^2)\wedge\cdots\wedge J^{-1}(\phi^n)\right). \end{align} for $\phi^1\wedge \phi^2\wedge\cdots\wedge \phi^n\in \Lambda^n(V^\ast)$. This is well-defined because $J$ is an isomorphism and nonzero because $\omega$ is. Therefore, $\omega^\ast$ is a volume form on $V^\ast$ (what I called dual volume form in the question)
Lemma 2
For all $i=1,2,\ldots,n$ it holds that $J(e_i) = \epsilon^i$ and thus \begin{align*} \omega^\ast(\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge\epsilon^n) &= 1. \end{align*}
Proof: By the equation for the Hodge dual we have \begin{align*} J(e_i)(e_j) &= (\widetilde J \circ \star)(e_i)(e_j) = \widetilde J (\star e_i)(e_j) \\ &= \omega(\star e_i \wedge e_j) \\ &= \omega\left(\left\langle e_i,e_j\right\rangle_{V} e_1\wedge e_2\wedge\cdots\wedge e_n\right) \\ &= \delta^i_j \omega\left(e_1\wedge e_2\wedge\cdots\wedge e_n\right) = \delta^i_j \end{align*} The second assertation follows from the definition of $\omega^\ast$.
Finally, we can prove the following result.
Proposition
Let $\{e_1,e_2,\ldots,e_n\}$ be an orthonormal basis for $V$ and $\omega$ the standard volume form. The isomorphism $J$ as above is isometric in the following sense \begin{align*} \left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)} = \left\|v_1\wedge v_2\wedge\cdots\wedge v_n\right\|_{\Lambda^n(V)} \end{align*} for all $v_1\wedge v_2\wedge\cdots\wedge v_n\in \Lambda^n(V)$.
Proof: Lemma 1 and 2 yield \begin{align*} \left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)} &= \frac{\left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)}} {\left\|\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge \epsilon^n\right\|_{\Lambda^n(V^\ast)}} \\ &= \left|c\right| \\ &= \left|\frac{\omega^\ast\left(J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right)} {\omega^\ast\left(\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge \epsilon^n\right)}\right| \\ &= \left|\omega^\ast\left(J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right)\right| \\ &= \left|\omega(v_1\wedge v_2\wedge \cdots \wedge v_n)\right| \\ &= \left\|v_1\wedge v_2\wedge \cdots \wedge v_n\right\|_{\Lambda^n(V)}. \end{align*} | 2,788 | 7,511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.741906 |
http://options-try-platform-best-binary.pw/binary-hexadecimal-octal-decimal-calculator-8171.php | 1,544,655,550,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824180.12/warc/CC-MAIN-20181212225044-20181213010544-00568.warc.gz | 218,111,333 | 5,234 | ## Binary to Decimal to Hexadecimal Converter
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The binary to octal conversion can be done binary hexadecimal octal decimal calculator grouping of bits method. Follow the below steps to perform such conversions manually. Separate the digits of a given binary number into groups from right to left side, each binary hexadecimal octal decimal calculator 4 bits. Add 0's to the left, if the last group doesn't contain 3 digits. Find the equivalent octal number for each group. Write the all groups octal numbers together, maintaining the group order provides the equivalent octal number for the given binary.
Solved Example Problem The below solved example problem may useful to understand how to perform binary to octal number conversion. Problem Convert Binary number 2 to its octal equivalent. Octal to Binary Conversion This conversion can be done by finding the binary equivalent for an each digit of the octal number, combining them together in the same order. The below steps may useful to know how to perform octal to binary number conversion. Separate the digits of the given octal number, if it contains more than 1 digit.
Find the equivalent binary number for each digit of octal number. Add 0's to the left if any of the binary binary hexadecimal octal decimal calculator is shorter than 3 bits. Write the all group's binary numbers together, maintaining the same group order provides the equivalent binary for the given octal number.
Solved Example Problem The below solved example problem may useful to understand how to perform octal to binary number conversion. Problem Convert the octal 8 to its binary equivalent. Numbers Conversion Table Decimal Binary Octal Hex 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 10 8 9 11 9 10 12 A 11 13 B 12 14 C 13 15 D 14 16 E 15 17 F.
Binary - Octal Conversion. Binary to Octal Octal to Binary. BCD to Decimal Converter. Binary hexadecimal octal decimal calculator - Decimal Converter.
Binary - Gray Code Converter. Binary - Hex Converter. Decimal - Octal Converter. | 477 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-51 | latest | en | 0.790525 |
http://stackoverflow.com/questions/5355592/bayesian-network-independance-and-conditional-independance | 1,454,822,232,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701148475.34/warc/CC-MAIN-20160205193908-00311-ip-10-236-182-209.ec2.internal.warc.gz | 205,450,883 | 18,397 | Bayesian Network: Independance and Conditional Independance
I'm having some misunderstanding concerning Bayesian network. My main misunderstanding are independence and conditional independence!!
If e.g. I have to calculate `P(Burglary|Johncall)`, is it `P(Burglary|Johncalls)=P(Burglary)` because i'm seeing that Burglary is independent of Johncalls??
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I'm voting to close this question as off-topic because it is not directly about programming. – Pang Feb 7 '15 at 1:57
Burglary is independent from JohnCalls given Alarm. So P(B|A,J) = P(B|A).
Explaining the example
The idea is, that John can only tell you that there is an alarm. But if you already know that there is an alarm, then the phone call from John will tell you nothing new about the possibility of a burglary. Yes, you know that John heard the alarm, but that's not what you're interested in when asking for Burglary.
Conditional Independence
In school, you've probably learned about unconditional independence, given when P(A|B) = P(A)*P(B). Unconditional independence makes things easy to calculate but happens pretty rarely - inside the belief network unconditionally independent nodes would be unconnected.
Conditional independence on the other hand is a bit more complicated but happens more often. It means that the probability of two events becomes independent of each other when another "separating" fact is learned.
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does the markov blanket has anything to do with Conditional Independence ?? – Noor Mar 19 '11 at 6:21
@Noor A variable is independent of everything else, given its markov blanket. You don't need to know of anything that is "behind" the blanket. This is similar to markov processes, where the blanket is the last state (t-1), which makes the current state (t) independent of all other past states (< t-1). – ziggystar Mar 19 '11 at 15:38
Been a while, sorry for resurrecting this. Is the inverse conclusion valid as well? Is a variable A always dependent of everything inside its markov blanket? – Hendrik Wiese Mar 24 '15 at 13:31
@HendrikWiese Given one node `A`, all other nodes are always a valid Markov Blanket–so the answer is no. But your statement holds for a minimal Markov Blanket, i.e. one where you cannot remove a variable without invalidating the Markov blanket. – ziggystar Mar 24 '15 at 13:38
I see. Thanks for the explanation, @ziggystar. – Hendrik Wiese Mar 24 '15 at 15:35 | 587 | 2,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-07 | latest | en | 0.930077 |
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