03 - SN Partial Differential Equations and Applications Webinar

Springer Nature

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Join Julio D. Rossi of Universidad de Buenos Aires as he discusses the interplay between partial differential equations and probability. Also discussed are approximations using game theory to viscosity solutions to an elliptic system governed by two different operators. Don’t miss analysis of a game that combines Tug-of-War with Random Walks in two different boards with a positive probability of jumping from one board to the other and more.

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00:05

So good morning jewel home. Good morning feeds that evening

00:11

to all.

00:12

Good goodnight for some it is a pleasure.

00:16

Great delight for all of us to introduce professor who

00:21

Rosie from University of Buenos Ritis.

00:25

Holio is a leading expert on elliptic PDE's,

00:30

pre boundary problems, regularity theory and also very well known

00:36

to be.

00:37

An excellent speaker so.

00:42

We are delighted to have him here.

00:45

He just published a very beautiful paper on spring nurture

00:49

PD an applications and also an expository paper on the

00:53

notices of the Ms,

00:55

which I strongly advise you all to check.

00:58

So another beautiful.

01:02

So another beautiful article from Julio,

01:05

so with no further ado,

01:06

I will pass the.

01:09

The floor to Julio and.

01:12

Thank you again for you.

01:15

OK, so thank you very much.

01:16

Thank you for the introduction.

01:19

Apps I will try to share my screen now.

01:24

If you.

01:27

Allow me to continue that said.

01:34

Here we are.

01:39

OK, so thank you again for the introduction and thanks

01:43

to the organizers.

01:44

Thanks young for the organization of this seminar.

01:48

For me it's a pleasure to be here today.

01:51

Well, it's it's a pleasure because I am at home.

01:54

I'd like to be at home,

01:56

but it's a pleasure to give this seminar today.

01:59

So that's the title of the talk and I'm going

02:02

to talk about the relation between game theory of probability

02:06

and piddies.

02:07

That is by now well known stuff.

02:09

But I will recall something at the beginning and at

02:13

the end I will talk about our results for nonlinear

02:17

system.

02:18

The results are showing work with Alfredo Miranda and Alfredo

02:21

Miranda is in.

02:22

That is here in the audience as well.

02:24

So you can ask questions so you can say hi

02:27

to him.

02:28

Yeah OK, let us start with the the plan of

02:31

the talk.

02:32

What is the idea of the talk that plan of

02:35

the talk is first to present some simple relation between

02:40

the random walk?

02:41

And the Laplacian that usual Laplacian and next enter into

02:46

nonlinear stuff.

02:47

I'm trying to arrive arrive at the end.

02:50

Two and a linear system to find solutions.

02:54

To a non linear systems of piddies of partial differential

02:57

equations using let me say,

02:59

again theoretical approximation, so using probability and values of gains.

03:05

So let us start with the simplest thing that is

03:08

the Laplacian and the linear.

03:10

Part of the talk. So you have a domain in

03:13

our two dart 2.

03:14

Just for simplicity you can take RN if you want

03:17

a nice smooth domain in our two you start at

03:21

some point X inside Omega and you ask the following

03:25

question. You move completely at random.

03:29

And you have a final payoff.

03:32

In the exterior of the domain and when you exit

03:35

the domain,

03:36

you get this amount of money given by F capital

03:39

F of the exiting point.

03:41

So when you start moving around and you exit the

03:44

domain,

03:45

you get some payment and you ask yourself,

03:48

OK, what is the expected expectation because you're moving at

03:52

random according to the probability that you are moving off

03:57

the final payoff and you call this stuff U.

04:00

Of X and you ask yourself,

04:02

OK, you affect solves what?

04:06

OK, we are going to assume two things because of

04:09

random walks in random walks.

04:11

We assume two things. You assume that you walk.

04:14

Completely blind, so you are blind and you don't know

04:18

any preferred direction,

04:19

so you don't know where you're going in your walk

04:23

and you are also completely drunk,

04:25

which is a sort of crazy thing for me at

04:27

this hour in the morning because for me it's 10:00

04:31

o'clock in the morning,

04:32

but you can assume that I am drunk.

04:35

And so I have no memory of what I've done

04:38

in my previous steps,

04:39

so each movement is independent of the of the history

04:43

of what I've done until this time.

04:46

OK, to be more concrete,

04:47

fix an Epsilon positive some number which measure the length

04:52

of the steps that you are making,

04:54

and you have four possibilities because you are in R2,

04:58

you can move one step forward,

05:00

one step backwards, one step to your left,

05:03

or one step to your right.

05:05

On all these four positions,

05:08

next positions of the pure.

05:11

Of your token of your of,

05:13

you have the same probability one fold and this is

05:17

because you don't have any prefer direction,

05:20

so you can move one step forward with the same

05:24

probability to move one step to the right.

05:27

One step to the left.

05:29

OK, if you say that you epsilon.

05:33

Is the spectation of the final payoff when you exit

05:37

the domain,

05:38

you can compute formula. You can compute an equation for

05:42

you Epsilon and it's the following.

05:45

You apply conditional expectations that says that the expected amount

05:50

of money that I'm going to get starting at X

05:53

is the probability of making one step forward times the

05:57

expected amount of money that I get.

06:00

Making one step forward plus the probability.

06:03

Of making one step backwards times the probability the expected

06:08

amount of money that I will get.

06:10

Making one step backwards and so on.

06:12

And you get this formula.

06:14

And if you rearrange them,

06:15

you multiply by 4 and you subtract you and there's

06:19

two sides.

06:20

Then you get these things so you have silent is

06:24

a solution to this equation and in these equations you

06:28

can see easily a second order discretization of the second

06:32

derivative in the direction of X and a second order

06:36

that reduce station of the second derivative in the direction

06:41

of Y.

06:42

So when you try to pass to the limit,

06:44

asylum goes to zero. You end up with the viscosity

06:47

solution with the solution that we say of the Laplacian

06:51

equals to 0.

06:51

So harmonic function with boundary conditions equals 2F in the

06:55

boundary of Omega,

06:56

because here is your final pay offices file at the

07:00

exterior and the boundary of our domain.

07:02

OK, this this simple thing of.

07:06

Of approximating solutions to the PDE to the harmonic to

07:10

harmonic function with boundary conditions.

07:14

As for the digit problem for the Laplacian,

07:17

you have a simple way of approximating this by a

07:20

sequence of the same values of games related to probability,

07:25

and this is the starting point.

07:27

The first step into Monte Carlo method cause you can

07:31

simulate.

07:32

This random walks in the lattice because you here we

07:36

have the lattice because we are moving only in the

07:40

coordinate axis in the direction of the coordinate axis.

07:44

So you can simulate this in a computer and this

07:47

is very fast.

07:48

This is very good because this give me the opportunity

07:51

to make a connection with the previous talk but are

07:55

moved Jensen in this same seminar.

07:57

So he talk about this Monte Carlo methods and machine

08:00

learning for solving PD is this is the starting point

08:04

of this,

08:04

but we're going to do it.

08:07

For nonlinear equations and I am not going to say

08:11

anything concerning the computation of this of this functions website

08:17

because I don't know how to do it.

08:19

I am not a numerical study.

08:23

So now for the Infinity lab reaction,

08:26

the other operator that will came into into into account

08:29

in this talk is the Infinity depression.

08:32

What is the Infinity Laplacian?

08:34

The Infinity depression is this combination of 1st and 2nd

08:39

derivatives.

08:40

Ann is essentially given by the matrix.

08:42

This is the hessian.

08:44

Of the function you times the direction of the gradient

08:48

transit direction of the grand entrance pose.

08:51

So this is the second derivative in the direction of

08:55

the gradient of the function itself.

08:58

This this operator is not linear.

09:00

Clearly it's not linear because you have this.

09:03

This multiplication of 2nd derivative by first derivatives.

09:07

Isn't diversions form not in diversiones form because you cannot

09:11

integrate by parts,

09:13

it is elliptic. Something that is nice.

09:17

Because, uh, OK, it is monotone in the hessian.

09:23

Decision that is monotone in the matrix that is involved

09:26

and is strongly degenerate because you can only control one

09:29

of the second derivatives and the direction in which you

09:33

control the second derivative of your function changes with the

09:36

gradient of the function itself.

09:38

So to compute this operator in a smooth function,

09:41

let me say that I have a smooth function with

09:44

gradient that is different from zero.

09:46

Then you take took the direction of the gradient and

09:49

you compute the second derivative in this direction and the

09:53

direction is changing from one point to another.

09:55

Because the gradient is changing from one point to another,

09:59

and of course you have to say something when the

10:02

gradient vanish is,

10:03

but I will not enter into details and let me

10:06

say that the operator is well defined even if the

10:09

gradient vanish is in the sense that.

10:11

Thereafter, semi continuous envelope on the lower semi continuous envelope

10:16

of U.

10:16

Verify some equations in the viscosity sense.

10:19

OK, what is the solution?

10:22

So this operator equal to 0.

10:25

Oh God, so you finish the harmonic functions.

10:28

The simplest one is a linear function and affine function.

10:31

A plane is a solution because the Hessian banishes.

10:34

But of course this is let me say the simplest

10:38

one.

10:38

They they tricky one or the nice solution is this

10:42

one.

10:42

The X to the four cells.

10:44

Why to the four cells it is infinitely harmonic in

10:48

the whole our tool.

10:49

And as you can see this function is not see

10:53

two,

10:53

so you cannot compute. The Hessian this matrix of the

10:57

second derivative is you cannot compute it.

11:01

A little after, as long as you computed along the

11:04

Axis,

11:05

this is forbidden. So.

11:09

OK, This Is This is the key of enter into

11:12

viscosity solutions.

11:13

This is a viscosity solution to this equation.

11:16

I am not going to present you the definition of

11:19

what is a disco city solution,

11:21

but let me say that in an informal way this

11:24

is a simple way of understanding what is what is

11:28

a solution to PD,

11:29

just taking derivative on touching functions.

11:32

You touch the graph of your solution from above or

11:35

from below with smooth functions and you compute.

11:39

The derivative in your touching function from above or below.

11:44

OK, so in here you can see that this nonlinear

11:47

operator is strongly the generate elliptic,

11:50

so you have solutions that are not as smooth as

11:54

you want,

11:54

like for depression. For harmonic functions are analytic,

11:59

but these are not in general.

12:01

So if you want to approximate or you want to

12:04

put some numerical scheme or something to solve this equation,

12:09

this Infinity Laplacian you have to take care that the

12:13

error term.

12:14

Is not going to be in general of 2nd order

12:18

because you will not have a C2 function and in

12:23

the limit.

12:24

So you cannot compute the error,

12:26

just making Taylor expansions because you don't have enough regularity

12:30

to make the Taylor expansions.

12:32

That's what I mean. OK,

12:34

so concerning existence of this equation,

12:37

that impression of you close to zero with boundary data

12:41

F.

12:41

At this the digit problem for this.

12:43

For this operator you have existence of a viscosity solution

12:47

and this this solution is related to the optimal lipschitz

12:51

extension problem,

12:52

so you assume that your boundary data capital F is

12:56

a Lipschitz function and you ask yourself,

12:58

OK, can I extend it in Alicia's way so you

13:01

can finally choose function inside?

13:04

Omega that takes this boundary data and the answer is

13:08

yes,

13:08

and the optimal way of doing this.

13:11

That is the solution. The function that extends F that

13:15

is left is under minimizes the Lipschitz constant in every

13:19

possible sub domain in Omega is exactly the solution to

13:23

this PD. To this Infinity Laplacian of you equal to

13:27

0 Infinity Harmonic and boundary data equals 12.

13:31

So this is the geometrical problem that is behind this

13:35

nonlinear brain.

13:36

OK, concerning uniqueness for this equation,

13:39

uniqueness was proved by Jensen.

13:42

This was the first provide uniqueness.

13:45

There is a simple proof by Burleson Bouska in 2001,

13:49

and there is a survey concerning this equation by Crandall,

13:53

an engineer, and concerning regularity.

13:56

Regularity is a simple issue here because as you as

14:00

you see in this this slide this X to the

14:03

4th third Y to the 4th third is not.

14:06

See two but it is.

14:07

See one 1/3. So it is a little better than

14:14

C1.

14:14

Did the gradient discolor continues?

14:17

But it's not going. Let's not sit too.

14:20

So what? What is known is that the gradient is

14:23

Holder continuous is C.

14:24

1A in dimension 2, but only in dimension 2 and

14:28

in dimension bigger than two.

14:29

So it dimension bigger or equal than three.

14:32

The solution is differential, so you have a gradient.

14:36

But up to now you have any models of continuity

14:39

of the garden.

14:40

So this is a tricky stuff and I have to

14:43

confess that I tried to prove C-11 third,

14:45

but of course I failed.

14:48

On the with Eduardo Teixeira we have some results with

14:52

relative shader and we have some results of regularity of

14:57

solutions to equations like this,

14:59

but close to a free boundary point,

15:02

not everywhere, not inside the domain.

15:05

We say so regularity. Let me say that just saying

15:08

that regularity's are tricky stuff and I am not going

15:11

to enter into too much details about this,

15:13

but. Try to focus on the relation between these kind

15:18

of equations with game theory with probability.

15:21

OK, what are the game?

15:23

What is the game that is related to these equations?

15:26

So I have to present you the rules of the

15:29

games and I'm going to let me say spend a

15:32

couple of minutes in this slide.

15:35

So what are the rules?

15:36

This is a two persons there,

15:38

some games. Two people play the game and they not

15:41

do not cooperate.

15:42

They are in contest and the total amount of money

15:45

that one of the player is getting is exactly the

15:47

total amount of money that the other players paying.

15:51

So here you don't have any state of any taxes

15:53

or any casino rules that says that some part of

15:56

the money is keeping by the casino.

15:58

Not here is a private arrangement between two people that

16:02

are playing this game.

16:04

OK, the first player. Tries to maximize the expected outcome

16:09

and because they are contest the second player try to

16:13

minimize the outcome.

16:15

Because the player one is getting the money and player

16:19

2 is paying that money to pay that amount of

16:22

money you play inside the domain.

16:24

Omega in Iran.

16:26

This is about the domain smooth bounded domain and you

16:30

assume that the boundary is included in some strip around

16:34

the boundary gamma.

16:35

In the exterior and in demand,

16:37

you have a final payoff function exactly as before.

16:41

So when you exit the domain,

16:43

the first player gets F that amount of money given

16:47

by F of this point and the second player pays

16:50

this amount of money and whatever the rules of the

16:54

games you start at some point it's not inside Omega,

16:58

and then you flip a coin at,

17:00

let me say 1/2 one half probability coin and the

17:04

winner.

17:04

If you get heads. The first player chooses if you

17:08

get tails.

17:09

The second player chooses the winner of the coin toss,

17:11

chooses the new position of the game.

17:14

At any point inside the ball of radius of silent,

17:17

here is the same as step that you are making

17:20

the size of the step it silent that you were

17:22

making.

17:23

The random walk is here again.

17:24

This is them silent. But now you can choose the

17:27

first player can choose to proceed the new position of

17:30

the game and the second player can choose the new

17:33

position of the game. You continue playing until you exit

17:37

the domain and when you exit the domain the first

17:39

player get this amount of money and the second player

17:42

pays this amount of money.

17:44

This is because this minus.

17:46

Yeah.

17:47

So this is a 0 sum game because the sum

17:51

of the two is equal to 0.

17:53

OK, So what are those?

17:54

So this is a sketch of what's going on.

17:57

Maybe it is too dark to see something,

18:00

but OK, that's the picture.

18:02

So you started, it's not.

18:03

You flip the coin with probability 1/2 and half.

18:08

One of the player juices that we say that player

18:11

one chooses the new position of the game and you

18:14

restart again.

18:15

Here with probability 1/2, one half you take the ball

18:18

of radius of silent.

18:19

You flip the coin and now it's player two who

18:22

chooses and now is player one player,

18:24

one player, one, player 1 until you exit and when

18:27

you exit in this first time you exit the domain

18:31

and this extor then player 1.

18:33

Gets this amount of money F of X2 and the

18:36

other place paying this amount of money.

18:39

OK, those are the rules of the game.

18:42

Now. If you take the sequence of states of the

18:47

game,

18:47

so the sequence of positions.

18:50

That your Bitcoin, at the token that you are moving

18:56

inside the domain.

18:58

It takes until you exit until this extra that you

19:01

exit has some probability that depends on the coin tosses,

19:05

of course, but depends on the starting point.

19:08

The starting position is not on the strategy of the

19:12

players.

19:13

As one honest or what is the strategy of the

19:15

players?

19:16

This is a new ingredient that appears here and that

19:18

you don't have in the random walk.

19:20

The strategies of the players is the decision.

19:23

Of each of the two players of.

19:26

What point they're going to choose?

19:28

The first one, the first player and the second player

19:32

inside the bowl of Radius Epsilon at each movement of

19:36

the game,

19:37

so both players has to make a decision.

19:40

At every point in every stage of the game.

19:46

And the decision gives you the probability 1/2 because of

19:51

the coin tosses or jumping to the position chooses by

19:55

play.

19:56

1 1/2 of the jump to the position.

19:59

Choose it by player 2.

20:02

So these these strategies of this thing of strategies is

20:05

not present in the random walk in the usual random

20:08

walk.

20:09

When you start in the random walk,

20:11

everything is a chance. If everything is random,

20:14

you're not going to make any decision because you are

20:17

blind and completed drunk.

20:18

Here you are not blind,

20:20

you are not drunk. You are smart player that tries

20:23

to maximize or to minimize boarding.

20:25

If your player one or player two,

20:27

you try to maximize or minimize the expected outcome and

20:30

you have to make a decision at every step.

20:33

But you still have some randomness.

20:35

You have still have something to compute an expected value

20:40

because you have this coin tosses with probability 1/2 and

20:44

half.

20:45

So if you fix the toy strategies,

20:47

you can compute the expected value starting at X.

20:51

Note with these two strategies of the final payoff,

20:54

and this gives you some number some value.

20:57

And now since players know what we're doing.

21:01

The player one tries to maximize the suspected result and

21:04

player two tries to minimize,

21:06

so there are two extreme cases.

21:09

The first one is that your computer in FEMA and

21:13

then the supremum.

21:14

So player 2 is trying to minimize computes the infamous

21:18

among all possible strategies.

21:20

Of the expected value and then player one compute the

21:25

supremum and II extreme values are reversed in Freeman this

21:29

supremum.

21:30

The first one is the best possible thing that player

21:35

2 May hope cause against any strategy of player 1.

21:39

He minimizes 1st and then player one tries to save

21:43

what he can and compute the supremum and the other

21:47

way around.

21:48

In the second extreme cases OK,

21:50

the definition the clear definition from game theory that says

21:55

that the game has a value.

21:57

And we call it you Epsilon because we have to

22:01

emphasize what is the size of the step that we're

22:05

making.

22:06

If and only if the two extreme values coincide and

22:09

this is just reversing film on the supremum,

22:12

and this is something delicate in general because you cannot

22:17

reverse in film with supremum without without taking care of

22:21

what you're doing,

22:22

but the general theorem by Parasram Sheffield Wilson says that

22:27

this game has a value,

22:28

so this whole game that is called in the literature

22:32

has some value Epsilon.

22:34

What is the equation verified by Epsilon?

22:37

The equation verified by the silent that is called the

22:40

dynamic programming principle is the following.

22:43

The expected value of the game at some point X

22:47

is given by the probability of player 1 to win.

22:52

Times the expected value provided that player one wins.

22:58

And expected value provided that player one wins.

23:01

Since player one wants to maximize.

23:05

The expected outcome. He will choose a point in the

23:08

bull riders accident that is close to the supremum of

23:12

the expected values.

23:13

And you have to add the other possibility that is

23:17

the probability of player 2 to win the coin toss.

23:21

Times the expected value provided that player two wins and

23:25

he will choose something the other player will choose something

23:29

close to the infamous.

23:31

In the bowl of fire subside.

23:33

If you rearrange them, that is,

23:35

you multiply by two and you subtract.

23:38

In the other side of the equation,

23:40

you get this form. Of writing.

23:44

The equation verified by the value function of this game

23:49

that you Epsilon.

23:50

And in this, in this equation you can notice 1st

23:53

that the equation is nonlinear.

23:55

Because you have this supremum infinum inside your equation,

24:00

so this is a nonlinear.

24:02

Equation and if you have a smooth function,

24:05

let me say that uses mood and you compute the

24:08

maximum the closed bowl and the minimum the clothes balls

24:12

obtained.

24:13

At some point XM Capital M little M.

24:16

The dynamic programming principle reads us the supremum for a

24:20

smooth function is the maximum,

24:22

so this is you at the maximum point infineum.

24:26

Is given by the minimum.

24:28

This is you at the minimum point minus.

24:32

To to your fix.

24:35

And now you say OK,

24:37

the maximum the point at which the maximum is attained

24:41

is close to the direction of the gradient of you.

24:46

Because for a smooth function,

24:47

the function grows faster in the direction of the gradient.

24:53

And the minimum point. Is close to minus the direction

24:57

of the gradient because its function is growing direction of

25:01

the gradient then is decreasing faster in the direction of

25:05

minus the gradient. So your formula your previous.

25:10

Formula if you extend minus you little M.

25:14

Minus two, your fix is the same as saying You

25:18

at X Plus Epsilon.

25:19

The direction of the gradient plus VU of X Minus

25:24

Epsilon the direction of the gradient minus to your ex.

25:29

And this when you divide by Epsilon Square is a

25:32

discretization of the 2nd order derivative in the direction of

25:37

the graph.

25:38

So what do you expect?

25:39

Is that the dynamic programming principle?

25:42

This dynamic programming principle computed?

25:45

At this function, you Epsilon is a sort of 2nd

25:48

order discretization of the second derivative in the direction of

25:53

the gradient,

25:54

and this gives you the Infinity Laplacian.

25:57

So it is not very surprising.

25:59

Of course, this computation you have to make it rigorously

26:04

and you can do it taking viscosity arguments.

26:08

But the key point is to prove that this sequence

26:11

of functions,

26:12

this sequence of value functions of your game convergence uniformly.

26:17

In Omega Bar to some function to some continuous function

26:21

on this continuous function is going to be the unique

26:25

viscosity solution to the infinite depression equals to zero.

26:29

With boundary data F. So in this way better from

26:34

Sheffield and Wilson proved an alternative way of show of

26:38

showing existence and uniqueness.

26:41

Of Infinity Harmonic functions and it's more than that because

26:45

you can play this game in much more general ambience

26:48

than the Euclidean case,

26:49

but I will stick to the Euclidean case in all

26:53

the talk during the whole talk.

26:55

So let me recap. It'll it'll,

26:57

it'll what are the rules of the game,

26:59

the rules of the games are the following you have

27:02

to players.

27:03

You start with some point X,

27:05

not inside Omega. You flip a coin an according of

27:08

the result of the talk.

27:10

If you get heads. You move player one chooses the

27:14

new position and you get tails.

27:16

Player 2 chooses the new position inside the ball radio

27:21

subside on around your previous position and.

27:24

The game ends when you exit the domain and when

27:27

you exit the domain you have a final payoff.

27:30

And if you do that,

27:32

the value of the game that is this in few

27:35

moments,

27:35

Supreme among the strategies. Converges uniformly.

27:41

To some function you. On this function do is turns

27:44

out to be the unique viscosity solution to the PD

27:48

to the limit between two things that I have to

27:51

mention is that in general this sequence of functions you

27:55

Epsilon are not continuous.

27:57

Because you have a discontinuity given by the size silent

28:02

that you are making the cause.

28:04

If you are very close to the boundary.

28:07

If you are, let me say blank silent.

28:11

Distance Epsilon to the boundary then you are not able

28:14

to jump in one step out,

28:16

but you are. You are a little closer than you

28:19

are able to jump only one step,

28:22

and the difference is this 1/2 and half.

28:25

So you need to win two times to exit the

28:28

domain or you have to win only one time for

28:31

points that are very,

28:33

very close. So this creates a discontinuity in EU Epsilon.

28:37

So this uniform convergence is delicate because you cannot use.

28:42

Let me say arguments like Arcelus quality items directly becausw

28:47

this.

28:48

You epsilons are discontinuous in general.

28:53

On the well, the fact that it is a solution

28:56

of viscosity solution to the limit P is not surprising

29:00

after this.

29:01

But that you can make.

29:05

OK, if you have a questions or comments,

29:08

please go ahead and mute yourself and ask.

29:10

Do whatever you want.

29:13

OK, concerning our results, we're going to deal with the

29:17

system and our motivation.

29:19

Here is how to present a again theoretical approximation.

29:23

For a system of Piddies in which you have two

29:26

different operators,

29:27

so you have this Infinity laplacian that I before and

29:31

the usual a pleasure with the random walk associated with

29:34

a random walk.

29:35

This is associated with this type of all game that

29:39

I showed you before.

29:40

This is this first equation.

29:42

This is the second equation,

29:43

so this is a system of equations.

29:45

You have two. Boundary Data F&G defined on the boundary

29:49

of the domain that I will assume that there are

29:52

both Lipschitz functions.

29:54

This Lipschitz regularity is not needed,

29:57

but helps a lot for the proof,

30:00

so uniform continuity and modulus of continuity is OK.

30:05

But Lipschitz Continuity is much simpler to do the proofs.

30:08

The domain is smooth, bounded at the smooth,

30:11

we don't need so much smoothness of the domain with

30:14

an exterior bowl condition.

30:16

It's OK, you can do it.

30:18

But assume that it is smooth and that's fine.

30:22

They the only thing that I have to to emphasize

30:26

here is that the coupling between the two equations given

30:30

by this minus V of X at this minus U

30:33

of X and the coupling is not any coupling.

30:36

Any possible coupling, you cannot coefficients.

30:41

Here in front of you be of X on your

30:43

of X depending on X,

30:45

but for example I don't know how to do it

30:48

for V Square and due to the power 3 for

30:50

example.

30:51

So you I don't know how to put powers in

30:54

the coupling.

30:55

With this game theoretical approximation.

30:59

And also this plus U of X + v of

31:04

X.

31:04

Are there and I don't know how to do it

31:08

with you square and be square here or there?

31:12

So the coupling on this extra terms that appears in

31:16

the operators.

31:17

Have to be there, in our case an we don't

31:21

know how to deal with the more general nonlinear.

31:25

Couplings.

31:27

But the Infinity Laplacian is nonlinear,

31:29

so this. This is a nonlinear system.

31:33

In at the end, and the main motivation for us

31:36

is to take two different operators.

31:39

OK, this system is not variational,

31:42

so you don't have an energy to minimize,

31:45

so you are far from.

31:46

Let me say that the variational theory in Sobolev spaces

31:51

here you have to tackle the problem using viscosity solutions

31:56

with positive stuff.

31:58

OK, what are the rules of the game that that

32:02

we figure it out?

32:03

How to approximate solutions to this system using a game

32:08

you have again a two person 0 sum game.

32:11

Now the game is played into two different boards,

32:15

so you have two copies.

32:18

Of the same set, Omega one up and one down

32:22

the first board and the second board 2 final payoff

32:27

functions F&G.

32:28

And what are the rules of the game in the

32:31

first board with probability 1 minus Epsilon Square and this

32:35

is the exact scale at which you have to choose

32:38

the probability. You did the players play games at the

32:42

screen before,

32:43

so with probability 1 minus Epsilon squared,

32:46

they toss a coin with probability 1/2,

32:49

one half, and with probability 1/2 one half wanted.

32:52

Player one chooses the new position,

32:54

the player 2 chooses a new position.

32:57

And with the rest of the probability they subside on

33:00

the square with the sum of the two is equal

33:03

to 1 with the Epsilon Square the position of the

33:06

game jumps to the second board at the same point

33:09

exactly at the same point inside.

33:11

Only in the second board with probability 1 minus Epsilon

33:15

squared.

33:15

Again, you move at random inside the ball radio,

33:18

subsided with uniform probability. So here you don't have to

33:23

make any decision.

33:24

You don't flip a coin,

33:26

or if you want to flip a coin according to

33:29

uniform probability in the polar sub silent and with probability

33:33

Epsilon Square you can go back to the 1st to

33:36

the 1st part. So here you have a picture.

33:41

And the scheme of what you're doing in the first

33:44

board.

33:44

You flip a coin with probability 1 minus upside on

33:47

the square,

33:47

and player one chooses or player 2 chooses the new

33:51

position of the game inside the polarizers of silent.

33:54

And with probability Epsilon squared,

33:56

you jump. To the second board in the second ball

34:00

with probability 1 minus Epsilon squared,

34:04

you choose any point completely at random inside the ball

34:08

in this shaded area inside the ball rise of silent

34:11

and with probability of selling square you go back.

34:15

You have some final payoff given by F&G outside.

34:19

The domain Omega, and when you exit the domain as,

34:22

you can expect, the last thing is that when you

34:26

exit the domain with the position.

34:29

Exit the domain Omega, then the first player gets this

34:33

amount of money.

34:34

On the second player, places amount of money if you

34:38

exit.

34:38

In the first board an if your position of the

34:41

game exit being in the second board,

34:44

then the final payoff is given by G.

34:48

OK, you have a pair of expected values because now.

34:53

They they spectation depends in your starting point is the

34:58

first board or in the second board.

35:01

If your initial position it's not is in the first

35:05

board.

35:05

Then the expected outcome. I call it you Epsilon and

35:08

if you are in the second board expected outcome,

35:11

I call it appear upside and the two are different,

35:14

are going to be different because it is not the

35:17

same starting here or there.

35:18

You can imagine that the probability of jumping from one

35:22

board to the other is so small.

35:24

Silent square is a small that if you're close to,

35:27

for example, if you are close to the boundary,

35:29

this this is the main step of the proof.

35:32

In fact, that I will recall at the end if

35:34

you are close to the boundary in the 1st.

35:37

The. The first board, then you will end the game

35:42

close to this point.

35:44

Without jumping with very high probability.

35:49

So the value function starting at this point close to

35:52

the boundary in the first board is quite different from

35:55

the value function starting the second board close to the

35:58

boundary, because here you have final payoff that is different

36:02

in general from this final paper.

36:07

OK, what is the equations that the two?

36:13

Value functions that the pair of value function for this

36:17

game verify is the following you have you epsilon is

36:20

given by the probability of jumping for the second bowl

36:24

times the expected value of you jump on the expected

36:28

value in the second word is given by V Epsilon.

36:32

Plus one minus Epsilon, the probability of playing.

36:36

And the result of playing the Tao of war game.

36:39

So this flip the coin one half 1/2 player one

36:42

chooses the new position,

36:43

player 2 chooses the new position of the game.

36:47

And in the second board,

36:49

expected payoff is given by upside and square the probability

36:53

of jumping to the first board at the same position

36:56

plus one minus Epsilon squared.

36:58

Playing and playing is just by random,

37:01

so you just integrate your Epsilon.

37:03

Here you can replace the integral by this one,

37:06

4th one 4th 1/4. This mean value of the random

37:10

walk is going to be the same at the end

37:13

the limit is going to be the same.

37:15

OK, outside the domain you have the final payoff given

37:19

by F&G.

37:20

So these two equations hold inside for X inside Omega

37:24

and this this boundary conditions or exterior conditions.

37:29

Are are given for X outside domain OK this this

37:33

system is particular system of equations has a unique solution

37:38

and you can look at this particular system of equations

37:42

as a sort of discretization at level epsilon of the

37:46

equations in this system.

37:48

In this system of Piddies,

37:50

because here you have second order derivatives,

37:54

here you don't have any derivatives,

37:57

but you have a sort of differences.

38:00

Nonlinear differences that gives you the Infinity depression,

38:03

and here you have a difference,

38:05

a linear. Linear stuff because the interval is linear.

38:09

That gives you the Laplacian,

38:11

the lid, and there you can see that when you

38:14

divide by Epsilon Square everything.

38:16

In order to obtain the 2nd order discretization,

38:20

here you get this coupling stuff.

38:23

This is why this is silent,

38:25

has to be Epsilon Square.

38:28

2. To keep the dimensions of the of the system

38:33

of the problem.

38:34

OK, and of course this is coupled because here you

38:37

have the upside on it.

38:39

Here you have the website,

38:40

you cannot coefficient. That depends on X saying that the

38:44

probability of jumping from one board to another depends on

38:47

the point in consideration.

38:49

But I don't know how to do it with powers,

38:51

so I don't know how to take here.

38:54

For example P Square. So repeat this and I emphasize

38:58

that the coupling in this system is quite particular.

39:02

OK, what is? What does the term says that error

39:05

says the following thing that the huge silent converges uniformly

39:09

to some limit V Epsilon canoe converges uniformly as well.

39:12

Do some limit B and despair.

39:14

UV is characterized as being the unique viscosity solution to

39:19

the system,

39:20

proving that the system has a unique solution is a

39:24

pure PD proof,

39:25

so you prove a comparison principle for this system and

39:30

this is just a.

39:31

Let me say using viscosity solutions because it is tough.

39:36

So this uniqueness came from a poor PD argument.

39:39

Conversions is delicate and the main idea to prove conversions

39:44

is to use a sort of asymptotic our sellers palamon

39:47

which you control the solutions.

39:50

The value functions you upside down and get silent,

39:54

but in general this continues again.

39:57

But you you want to control the measure of the

40:01

oscillations of the jumps,

40:03

and to do that you have to do the following

40:06

thing.

40:07

You have to place yourself first,

40:09

close to a point, why on the boundary of Omega

40:13

an if you start close to appoint wine in the

40:17

boundary of Omega then.

40:19

Each of the two players has strategies that forces the

40:24

game to end at some point that is quite close

40:28

to Y without changing board with very high probability.

40:33

Unfor that. With you, you need to control the expected

40:38

number of times that you play,

40:40

and you need that expected number of times that you

40:44

play,

40:44

provided that you start close to the boundary close to

40:48

a boundary point.

40:50

Is a let me say less or equal than a

40:53

small quantity over Epsilon Square and this is the key

40:58

part of the proof.

41:00

That you have to control becausw you want the probability

41:06

of changing boards.

41:07

When you are close to the boundary is is very

41:10

small and you want this for two reasons.

41:13

First to prove a uniform conversions and the second thing

41:17

is that you want at the limit U&V is solution

41:20

to the PD,

41:21

so you want them to verify the boundary conditions and

41:25

that the boundary conditions are attained with continuity.

41:30

If everything is Lipsius and the domain is as hold,

41:34

like in this case like the thing that I am

41:37

presented here,

41:39

the limit you get and slightly better regularity of the

41:43

limit you get at the limit UV are given by

41:46

a pair of Lipschitz functions,

41:48

so the Lipschitz continuity of the data G&V extends to

41:53

the limit and you get Lipschitz continuity of U&P.

41:57

With the modulus of continuity that is given by the

42:01

worst of the two models of continuity of the tulips

42:05

is constant of G&B of G&F.

42:07

So, so you and we are going to be Lipschitz

42:09

functions.

42:10

But another thing that I don't know how to do

42:13

it.

42:13

Is to prove for the system.

42:17

Yeah, better regularity. Let me say some continuity of the

42:21

gradient.

42:21

For example, in dimension 2,

42:23

continuity of the gradient for the pair UV,

42:27

I don't know how to how to tackle this stuff.

42:32

OK, so those are the things on at the end.

42:37

When you have the estimates at points that are close

42:41

to the boundary,

42:42

then you want to extend this estimates to the whole

42:46

to the whole domain Omega and this is done doing

42:50

a coupling strategy between two games.

42:53

But this is just a probability argument.

42:57

All these estimates this estimates for a points that are

43:02

close to the boundary are let me say I've given

43:06

in terms of probabilistic arguments because you want to use.

43:11

They the game approximation. And the game approximation shows how

43:16

to deal with the.

43:17

Let me say strategies for one player or the other

43:20

player in such a way that you get 7 Super

43:23

Solutions.

43:24

For this system, this is essentially what you are doing.

43:27

So the system of equations that is verified by the

43:31

value functions of the of the game when you specify

43:35

some strategies,

43:36

you stay in Southern Supper solutions to this thing and

43:40

you can go on.

43:41

I'm broke something OK concerning references for this for this.

43:48

For this kind of things.

43:51

I will mention this ones that are the closest ones.

43:55

For for what I'm presenting today,

43:58

the first one and quickest reference our notes on this

44:02

kossity solution.

44:03

So these two are just general references for Victor City

44:07

Solutions Para Sheffield and Parrish.

44:10

Ramshackle and Wilson are the first stuff of tug of

44:14

war games,

44:14

and later with the Manfredi and Peruvian and we probably

44:19

unk.

44:19

We made some progress into nonlinear operators as well.

44:24

You can you can take more general operators here.

44:26

Let me mention that you can take more general operators

44:30

that are not Infinity,

44:31

depression or depression. For example,

44:33

you can take the pillow plush and Acula Plush and

44:36

normalize Phillip Lation and normalized killer pleasure in the first

44:40

equation and the second equation you can do any kind

44:43

of operators as long as you have a game for

44:46

which you have this.

44:48

Expected amount of times that you're playing with a nice

44:52

spectation and you compute something there.

44:56

OK, extra references. Concerning game theory,

45:00

you have a bunch of names here.

45:02

I am not going to refer to every for everyone

45:06

and I am sure that in this list.

45:08

You have some omissions. Of course,

45:12

on the.

45:14

But but you can tackle.

45:15

As I told you other other operators,

45:17

more general operators, and there are some names for the

45:21

viscosity solutions to the pill oppression and regularity or the

45:25

pillow.

45:26

Plush and a consultant discussed solution.

45:28

I already mentioned things and concerning the relation between game

45:33

theory in this kind of game theory.

45:36

And B is. There is another bunch of names and

45:39

let me mention at the end that there are mean

45:42

value formulas.

45:43

Related to this game genetical stuff that says that,

45:47

for example, a harmonic function function is harmonic if and

45:50

only if it verifies the Min value formula.

45:53

So the value at some point is the mean value

45:56

of the function inside the bowl of radius R.

45:59

At any other producer. OK,

46:01

there are some in value formulas for nonlinear equations as

46:04

well,

46:05

and those are asymptotic mean value formulas.

46:08

And there you have names related to this asymptotic mean

46:13

value form.

46:14

And for the P Laplacian for example,

46:17

you have an asymptotic mean value form.

46:20

And this was a work with Manfredi and Barbian.

46:25

OK so I will stop here,

46:27

so thank you very much for your attention and I

46:30

think that my time is OK.

46:32

I have a still 3 minutes but I will stop

46:35

here so thanks a lot for your attention.

46:39

Woof, that's it.

46:45

Thank you so much Julio for this beautiful talk.

46:50

So now I would like to open the floor for

46:54

eventough questions.

46:59

And if you have a question,

47:01

you can either place it on on the chat or

47:04

mute yourself and ask.

47:14

Not my hardcover collecting in the information in the test.

47:27

Hulu

47:30

but what is it not at mixmax mean,

47:33

but a Nash equilibrium. OK this games.

47:37

They give them value or the value for this game

47:43

is not.

47:43

A Nash Equilibrium is just.

47:46

This infamous supremum.

47:51

This Supreme female of infamous Supreme coincide and this is

47:55

the key Librium value of the game.

47:57

This is the value of the game.

48:03

OK, and then this is unique as well.

48:05

So you Epsilon the value of the game is unique.

48:08

There is a unique solution to this dynamic programming principle

48:12

with the boundary conditions.

48:14

And the same thing happens with the system for the

48:17

system.

48:18

You have uniqueness of this DP,

48:20

so there is a unique equilibrium.

48:22

Let me say for the game there is a unique

48:26

pair of value functions that is suspected amount of money

48:31

that you're going to make.

48:33

OK.

48:38

When is the constant Kappa gains from the constant Kappa?

48:42

But they didn't mention the constant copper that is here.

48:45

You can OK. You can choose that constant Kappa making

48:50

that the game when you play at random in the

48:53

pool.

48:54

Alright, this sub silent you play at random in the

48:57

ball of radius.

48:58

Let me say B type silent so you change the

49:01

radius of the ball and you can select the constant

49:04

Kappa and this constant Kappa Canes Ascentia Lee from.

49:08

The fact that when you subtract the mean value and

49:12

the ball right subside on,

49:14

you subtract V Epsilon here.

49:17

The the mean value of the upside down at Y

49:20

minus we have silent X in the bottom right.

49:22

Sub silent is it's not given by the Laplacian or

49:25

be upside not 2nd order is given by a constant

49:28

time celebration and the constant is just.

49:31

Then they say the measure of the ball of radius

49:34

one in our end.

49:35

So you have something there that depends only in the

49:40

dimension.

49:40

But you you can adjust and take any positive constant.

49:43

If you don't think any positive constant that the coupling

49:47

has the reverse sign and with the River sign.

49:50

I don't know how to do it.

49:52

So the coupling is this coupling and not any any

49:55

possible coupling that you can imagine.

50:02

Let me open the chat again.

50:10

They set the hypothesis on the boundary on the boundary

50:14

for us.

50:15

Is a domain with exterior bowl conditions.

50:17

So you have a uniform exterior board condition at any

50:22

point in the boundary.

50:23

You have a bowl that touches.

50:26

The domain at that point and that has a uniform

50:30

radius around the boundary.

50:32

This is enough for the argument.

50:34

If you want some Lipschitz regularity,

50:37

let me say that the boundary has to be a

50:40

little more smoother.

50:41

Let me say that for a C2 boundary,

50:43

then you can pass to the limit and obtain lips,

50:46

regularly delivered, but to pass to the limit in a

50:49

uniform way,

50:50

you only need an exterior ball condition.

50:55

Holy, there is another question regarding eventual.

50:59

No local analogue of these games.

51:04

Oh

51:06

well, for for the part of the Laplacian you can

51:10

imagine that instead of taking the meaning,

51:13

the ball of radius of silent you take something that

51:17

is a convolution,

51:19

kernel, convolution with some kernel with a fixed radius,

51:23

and there something can be done.

51:26

But for the non local version of the Infinity Laplacian

51:30

there is also paper for that,

51:32

that is Caffarelli Bullion Figoli.

51:35

But I don't know how to do the estimates.

51:39

Close to a boundary point that we made here.

51:42

So this this I don't know.

51:47

OK, can we get an equilibrium for Kappa negative?

51:51

I don't know. I don't know either as I told

51:55

you the coupling here is.

51:57

Quite a special and we need this minus in the

52:00

coupling term,

52:01

so the coupling terms that appear in the game are

52:05

minus two and minus B in the other equation.

52:09

So I don't know how to change the sign of

52:12

that.

52:18

I can I can share my slides,

52:20

so thank you very much.

52:23

Julio, just maybe maybe one more question regarding instead of

52:29

choosing in a bowl if you choose an ellipse.

52:33

So probably adding some coefficients too.

52:36

Probably you will get a second order here.

52:40

Here you will get a second order operator,

52:44

2nd order elliptic operator like the sum of AJ second

52:49

derivative of BJ.

52:51

Yeah you can do it as long as the.

52:54

The city that this ellipse is that you are taking

52:57

the mean values that do not generate close to the

53:00

boundary or something like this.

53:02

You need the estimates close to a boundary point and

53:06

then to extend the estimates to the interior.

53:08

This works OK, but close to a boundary point.

53:12

You need that your game ends very quickly in such

53:15

a way that the probability of jumping to the other

53:19

board is very small and you need to end very

53:22

quickly and close to this point so you are going

53:25

to be there if you are,

53:27

let me say fighting boxing against Muhammad Ali in one

53:31

corner of the ring you are going to be dead

53:34

there very quickly and with a very small probability of

53:37

reaching the other. The other end of the ring.

53:41

But that's the kind of game that that you need

53:45

to prove all that the kind of estimates that you

53:48

need to prove to pass to the limit in in.

53:51

Since, like the this system of equations.

53:56

Nice. Maybe one more question,

54:00

what happens if you have source term water where the

54:05

probabilistic interpretation of source terms like instead of having the

54:12

system equal to 0 in the here,

54:15

here, here you have F and yeah or HH&WW.

54:21

You can do that adding a running payoff to this

54:24

thing and saying that in the first board each time

54:27

that the players play,

54:29

player 1 gets an amount given by Epsilon Square Time

54:34

H of this point.

54:35

And if you do that.

54:38

You did the.

54:41

In this. Equations you get something like + on Square

54:48

H of X and in the second plus Epsilon Square

54:55

W.

54:55

Of X and if you pass to the limit,

54:57

I see you obtain this thing with with the right

55:02

hand side.

55:03

That I mentioned depends only on the point on X

55:06

and does not depend on you as well.

55:08

This is not a nonlinear function of the expected value

55:11

itself,

55:11

so this is much more delicate.

55:13

And for this.

55:17

This wrong hand size you have to ask for something

55:20

we have.

55:20

We don't make the computation but we.

55:23

You need to ask for something in order to get

55:26

the good estimates near the boundary.

55:29

Maybe for continuous functions it's OK for a uniform modulus

55:34

of continuity of the right hand side should be enough.

55:38

To pass to Delhi.

55:40

But should be enough. Makes an unclear proposition because you

55:45

have to make the computations.

55:49

Of course.

55:54

Yeah, more questions in the chat.

55:56

Let me OK. I think we don't have any further

56:00

questions so I would like to thank you holy once

56:03

more for this beautiful talk.

56:06

Very insightful talk and I hope to see you all

56:10

next week.

56:10

In another spring nature PD and applications webnar thank you

56:17

again guys.

56:18

OK, so thank you very much and thank you again

56:21

for the invitation for the whole organization of this thing

56:25

and thank you Eduardo for being friendly.

56:27

Stream Lee. Nice chairman, let me say.

56:32

Thank you very much. Thank you both very much.

56:34

Then yeah, I hope to see many of you again

56:37

next week for a talk by by Tomas Batch on.

56:40

An elliptic piddies nonlinear, once thanks a lot.

56:42

Bye bye everybody.