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Example 1 g (r) = (1 r) 1 r 1 , r 0, G (r) = (1 r) 2 =2 1 r 1 . We consider V as a "potential". Writing V 0 for rV , we have

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1 The PDE associated with local interaction

Given the function g (r) 0, r 0, of the previous lecture, let G (r) 0, r 0, be a primitive of g (r) and let

V (x) = G (jxj) :

Example 1 g (r) = (1 r) 1 r 1 , r 0, G (r) = (1 r) 2 =2 1 r 1 . We consider V as a "potential". Writing V 0 for rV , we have

V 0 (x) = g (jxj) x jxj : Hence we may write equation

dX t i;N = N 1=d X N j=1

g N 1=d X t i;N X t j;N X t i;N X t j;N X t i;N X t j;N

dt + dB t i

in the form

dX t i;N = N 1=d X N j=1

V 0 N 1=d X t i;N X t j;N dt + dB t i : Setting further

V N (x) = N V N 1=d x we have

V N 0 (x) = N N 1=d V 0 N 1=d x hence we may write the previous SDE as

dX t i;N = 1 N

X N j=1

V N 0 X t i;N X t j;N dt + dB t i :

With these new notations, recall the result:

Corollary 2 The correct rescaled dynamics to investigate the macroscopic limit of a sys- tem with local interaction g is

dX t i;N = 1 N

X N j=1

V N 0 X t i;N X t j;N dt + dB t i :

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In order to conjecture the associated macroscopic PDE we have two routes: the more rigorous one is to write the equation for the empirical measure S t N and investigate its limit.

The second one, which a priori could even lead to a wrong conjecture, is to take the Mean Field equation

@u N

@t =

2

2 u N + div u N V N 0 u N (1)

and investigate its limit.

Remark 3 It is very important to notice that there is no reason, based on the rigorous Mean Field theory to believe that S t N is close to u N . What is rigorously true is the weak convergence of S t N;M to u M as N ! 1, for every …xed M 2 N, where S t N;M is the empirical measure associated to the system

dX t i;N;M = 1 N

X N j=1

V M 0 X t i;N;M X t j;N;M dt + dB t i

Pretending that S t N is close to u N is like pretending that the Mean Field limit lim N !1 S t N;M = u M is su¢ ciently uniform in M , a fact that is completely outside the results we have proved so far. Indeed, when M increases, the uniform and Lipschitz constants of V M 0 deteriorate and we had to assume a bounded Lipschitz continuous kernel in the Mean Field theory, with all estimates depending on its constants.

1.1 Taking the limit in the Mean Field PDE

Let us …rst investigate the limit, as N ! 1, of solutions u N of equation (1). We do not want to be rigorous here and thus assume that u N converges, in a suitable sense, to a function u. We want to …nd the equation satis…ed by u. We have, for every test function

2 C 0 1 ,

u N t ; = u N 0 ; +

2

2 Z t

0

u N ; ds Z t

0

V N 0 u N ; u N r ds

hence, under relatively weak convergence properties, the …rst three terms converge, hence we have

hu t ; i = hu 0 ; i +

2

2 Z t

0 hu; i ds lim

N !1

Z t 0

V N 0 u N ; u N r ds:

In addition, if we prove that V N 0 u N ! v in a suitable sense, we get hu t ; i = hu 0 ; i +

2

2 Z t

0 hu; i ds Z t

0 hv; ur i ds:

(3)

This is the weak form of the PDE

@u

@t =

2

2 u + div (uv) :

We have only to identify v. We have, for every test function 2 C 0 1 , V N 0 u N ; = u N ; V N 0 ( ) :

Moreover,

V N 0 ( ) (x) = Z

V N 0 (y x) (y) dy = r Z

V N (y x) (y) dy:

Assume now that V N is proportional to a sequence of molli…ers: this is true by the formula V N (x) = N V N 1=d x if V is proportional to a smooth probability density

V (x) = 2 1 f (x)

where 2 1 > 0 and f is a pdf. In this case, since is smooth and compact support, R V N (y x) (y) dy ! 2 1 (x), hence (V N 0 ( ) ) (x) ! 2 1 r (x) and …nally

V N 0 u N ; ! 2 1 hu; r i = 2 1 hru; i : We have found

v = 2 1 ru and therefore the limit PDE

@u

@t =

2

2 u +

2 1

2 u 2 :

This equation is known in the literature as Porous Media equation.

Remark 4 Notice that the equation does not depend on the details of the interaction po- tential V , only on

2 1 =

Z

V (x) dx:

This looks strange, from the viewpoint of the particle system.

The result guessed so far is correct from the viewpoint of the PDEs: it is a theorem

that the solution u N of the Mean Field equation converges to the solution u of the Porous

Media equation. What is not true is the fact that S t N converges to u.

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1.2 The equation for the empirical measure

As in a previous section we have:

Lemma 5 The empirical measure S t N satis…es the identity d S t N ; t = S t N ; @ t

@t dt S t N ; r t V N 0 S t N dt + dM t ;N +

2

2 S t N ; t dt for all 2 C b 1:2 [0; T ] R d , where

M t ;N = 1 N

X N i=1

Z t

0 r s X s i;N dB s i :

First, let us argue formally: assume that S t N weakly converge to some measure t . We can pass to the limit in the terms D

S t N ; @ @t

t

E

, M t ;N (it goes to zero in mean square), S t N ; t . Concerning

S t N ; r t V N 0 S t N the analysis of convergence is much more di¢ cult.

The …nal answer will not be unique and general, so for pedagogical reasons let us discuss the di¢ culties, before we give some partial result.

A true general fact is that (if S t N weakly converge to t )

N !1 lim f N ; S t N = hf; t i

when f N converges uniformly to f . But uniform convergence is essential.

Example 6 If n = x

n

, = x , x n ! x, then n * . If f n converges uniformly to f , we see that hf n ; n i = f n (x n ) converges to f (x). But weaker convergences do not imply the same result: there is no reason why f n (x n ) should converge to f (x) if we have only pointwise convergence, or L p convergence, and so on.

Can we say that V N 0 S t N converges uniformly to some limit? We have, as in a compu- tation already done above,

V N 0 S t N ; = S t N ; V N 0 ( ) = Z

r Z

V N (y x) (y) dy S t N (dx) : We have r R

V N (y x) (y) dy ! 2 1 r (x) uniformly, hence V N 0 S t N ; ! 2 1

Z

r (x) t (dx) :

(5)

We are faced with two main di¢ culties: …rst t has to have a di¤erentiable density u t (x), such that

2 1

Z

r (x) t (dx) = 2 1 Z

(x) ru t (x) dx:

Second, even so, the convergence of V N 0 S t N to 2 1 ru t would be only weak, not uniform as required.

The fact that ru t (x) appears in the limit is not avoidable: even if we do not have all pieces of proof, it is di¢ cult to believe that no term like ru t (x) will appear. Thus we have to prove that S t N weakly converges to a measure t having di¤erentiable density. This is an interesting problem in itself. The idea which allows to solve this problem is to analyze the molli…ed empirical measure

h N t (x) = W N S t N (x) :

If this converges to a di¤erentiable function (by means of some compactness result for functions), then we can solve at least the problem of ru t (x).

1.3 Intermediate regime

In the sequel we shall try to understand the macroscopic limit when the potential V is rescaled, a problem which is much more di¢ cult than the Mean Field problem. However, the case of Contact interaction, which corresponds to

V N (x) = N V N 1=d x

will turn out to be too di¢ cult. Karl Oelschäger identi…ed an intermediate problem, between Mean Field and Contact interactions, which can be studied in some cases, called Intermediate (or Moderate) regime. It is the case

V N (x) = N V N =d x

with 2 [0; 1]. For = 0 we have Mean Field interaction; for = 1 we have Contact interaction. Nothing changes apparently in the other formulae written above, except for the fact that certain results can be proved (at present) only for certain values of and also, more important, that certain quantitative results for < 1 seem to be di¤erent from the those for = 1.

2 Energy estimates

Our ultimate aim in this section would be to prove tightness of the laws Q N of the empirical

measures S N . To reach this result, the …rst and more important estimate, following the

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computations made for the Mean Field case, would be to prove that

E

"

sup

t2[0;T ]

X t i;N

# C:

But now

X t i;N X 0 i;N + 1 N

X N j=1

Z t 0

V N 0 X s i;N X s j;N ds + B t i

and V N 0 is unbounded, being

V N 0 (x) = N N 1=d V 0 N 1=d x

it growth as N N 1=d . The following energy-type estimates provide a solution, under certain restrictions on .

Notice that

1 N

X

j

V N 0 X t i X t j dt = ru N t; X t i where

u N (t; x) = V N S t N (x) = 1 N

X

j

V N x X t j : Hence the interacting particle system can be written in the form

dX t i;N = ru N t; X t i dt + dB t i : Thus

X t i;N X 0 i;N + Z t

0 ru N s; X s i ds + B i t : If we prove suitable estimates on ru N , we have a tool to prove E h

sup t2[0;T ] X t i;N i C.

2.1 Energy estimates on the Porous Media equation

We …rst describe an a priori estimate for the Porous Media equation since, we think, it was the underlying idea behind the next computation on S t N . For simplicity of notations, consider the equation

@u

@t =

2

2 u + div (uru) :

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For this equation, the standard energy estimate reads d

dt 1 2

Z

u 2 dx = Z

u @u

@t dx = Z

u

2

2 u + div (uru) dx

=

2

2 Z

jruj 2 dx Z

u jruj 2 dx hence

1 2

Z

u 2 t dx + Z t

0 2

2 Z

jruj 2 dx + Z

u jruj 2 dx ds = 1 2

Z u 2 0 dx where we stress the remarkable fact that

Z

u div (uru) dx = Z

ru (uru) dx = Z

u jruj 2 dx:

This provides estimates on ru. Recall that at the end of the previous subsection we identi…ed as a possible tool exactly the control of ru N . Thus we shall try now to repeat these energy computations on u N .

2.2 Energy estimates on the empirical measure Recall Lemma 5:

d S t N ; t = S t N ; @ t

@t dt S t N ; r t V N 0 S t N dt + dM t ;N +

2

2 S t N ; t dt:

In order to write an energy estimate for S N t we have to mollify it, we need a function. Let us apply the convolution with a molli…er W N and set

w N (t; x) = W N S t N (x) = 1 N

X

j

W N x X t j :

We shall discover below if we have to choose W N = V N or not. Replace t (y) by W N (x y), with x given as a parameter, and think S t N ; t as an integration in the y variable, so that

S t N ; t = S t N ; W N (x ) = w N (t; x) and so on for the other terms:

S N t ; W N (x ) = W N S t N (x)

S t N ; rW N (x ) V N 0 S t N = Z

r y W N (x y) V N 0 S t N (y) S N t (dy)

= div

Z

W N (x y) V N 0 S t N (y) S t N (dy)

= div W N V N 0 S t N S t N :

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We get

dw N (t; x) = div W N V N 0 S t N S N t (x) dt + dM t N (x) +

2

2 w N (t; x) dt where

M t N (x) = 1 N

X N i=1

Z t

0 rW N x X s i;N dB i s :

Let us apply Itô formula to w 2 N (t; x), with x treated again as a parameter. We have dw 2 N (t; x) = 2w N (t; x) dw N (t; x) + d [w N ( ; x)] t

= 2w N (t; x) div W N V N 0 S t N S t N (x) dt + 2 w N (t; x) w N (t; x) dt +2w N (t; x) dM t N + d M N (x) t

and

d M N (x) t =

2

N 2 X N i=1

rW N x X t i;N 2 dt:

Think to the previous identity as integrated in time, as it is rigorously. Then we integrate in dx, namely we compute the di¤erential of the "energy" 1 2 R

R

d

w N 2 (t; x) dx:

1 2

Z

R

d

w 2 N (t; x) dx = 1 2

Z

R

d

w 2 N (0; x) dx +

Z

R

d

Z t 0

w N (s; x) div W N V N 0 S s N S s N (x) ds dx +

2

2 Z

R

d

Z t 0

w N (s; x) w N (s; x) ds dx +

Z

R

d

Z t 0

w N (s; x) dM s N dx

+ Z

R

d

2

N 2 X N

i=1

Z t

0 rW N x X s i;N 2 ds

! dx:

Let us exchange integration (the functions are integrable under minor assumptions on W N ).

We have

Z

R

d

Z t 0

w N (s; x) w N (s; x) ds dx

= Z t

0

Z

R

d

w N (s; x) w N (s; x) dx ds

=

Z t 0

Z

R

d

jrw N (s; x)j 2 dx ds

(9)

that we shall write on the left-hand-side of the identity above. Moreover, and this is the main point of the computation,

Z

R

d

Z t 0

w N (s; x) div W N V N 0 S s N S s N (x) ds dx

= Z t

0

Z

R

d

w N (s; x) div W N V N 0 S s N S s N (x) dx ds

=

Z t 0

Z

R

d

rw N (s; x) W N V N 0 S N s S s N (x) dx ds and, by expanding the convolution, one can easily check that

Z

R

d

rw N (s; x) W N V N 0 S s N S s N (x) dx

= Z

R

d

(W N ( ) rw N (s)) (x) V N 0 S s N (x) S s N (dx) : If we choose W N in such a way that

W N ( ) rw N (s) = V N 0 S s N (2)

then we have

Z

R

d

Z t 0

w N (s; x) div W N V N 0 S s N S s N (x) ds dx

=

Z t 0

Z

R

d

V N 0 S s N (x) 2 S s N (dx) ds which similar to the term R

u jruj 2 dx in the PDE estimate above and can be put on the left-hand-side of the identity above. Unfortunately, given V N , it does not seem possible to choose always a molli…er W N such that () above holds true. Thus we put this as an assumption.

Lemma 7 If there is W N such that

W N ( ) W N = V N

then we have the energy estimate 1

2 Z

R

d

w N 2 (t; x) dx +

2

2 Z t

0

Z

R

d

jrw N (s; x)j 2 dx ds + Z t

0

Z

R

d

V N 0 S s N (x) 2 S s N (dx) ds

= 1 2

Z

R

d

w N 2 (0; x) dx + Z

R

d

Z t 0

w N (s; x) dM s N dx + Z

R

d

2

N 2 X N i=1

Z t

0 rW N x X s i;N 2 ds

!

dx:

(10)

In order to use the estimate on ru N = V N 0 S s N provided by this lemma, we need to con- trol the terms on the right-hand-side of this identity. The martingale term R t

0 w N (s; x) dM s N has zero average, so we hope to control it. Let us understand the last term. We have

Z

R

d

2

N 2 X N i=1

Z t

0 rW N x X s i;N 2 ds

! dx

=

2

N 2 X N

i=1

Z t 0

Z

R

d

rW N x X s i;N 2 dx ds

=

2

N 2 X N

i=1

Z t 0

Z

R

d

jrW N (x)j 2 dx ds by the change of variable x ! x X s i;N in the integral in dx,

=

2 t N

Z

R

d

jrW N (x)j 2 dx:

If this quantity is bounded, we control ru N , otherwise not, with this approach. We have Z

W N ( x + y) W N (y) dy = V N (x) = N V N =d x :

Assume W N (x) of the form N W N =d x as it should be if we want it to be a molli…er.

Then Z

W N ( x + y) W N (y) dy = N Z

N W N =d (y x) W N =d y dy

= N Z

W y N =d x W (y) dy

= N (W ( ) W ) N =d x : We deduce that we need

W ( ) W = V and = , namely

W N (x) = N W N =d x : Made these remarks, we have

1 N

Z

R

d

jrW N (x)j 2 dx = N N

Z

R

d

N N =d rW N =d x 2 dx

= N

N N 2 =d Z

R

d

N rW N =d x 2 dx

= N

N N 2 =d Z

R

d

jrW (x)j 2 dx:

(11)

If we assume R

R

d

jrW (x)j 2 dx < 1 and + 2

d 1

namely

d d + 2 we have the estimate.

Theorem 8 Under the previous assumptions, in particular if d+2 d , and if E R

R

d

w 2 N (0; x) dx C, E X 0 i;N 2 C, the family of laws Q N of the empirical measure processes S N are tight on C [0; T ] ; Pr 1 R d .

Proof. We have proved that 1

2 E Z

R

d

w N 2 (t; x) dx +

2

2 E Z t

0

Z

R

d

jrw N (s; x)j 2 dx ds +E

Z t 0

Z

R

d

jru N (s; x)j 2 S s N (dx) ds C

(recall that u N = V N S s N ). Recall the proof of tightness made for the mean …eld problem.

We have …rst to prove that E

"

sup

t2[0;T ]

1 N

X N i=1

X t i;N

# C:

From

X t i;N X 0 i;N + Z t

0 ru N s; X s i ds + B t i we have

1 N

X N i=1

X t i;N 1 N

X N i=1

X 0 i;N + Z t

0

1 N

X N i=1

ru N s; X s i;N ds + 1 N

X N i=1

B t i

= 1

N X N i=1

X 0 i;N + Z t

0

Z

R

d

jru N (s; x)j S s N (dx) ds + 1 N

X N i=1

B i t 1

N X N i=1

X 0 i;N + Z t

0

Z

R

d

jru N (s; x)j 2 S s N (dx)

1=2

ds + 1 N

X N i=1

B t i 1

N X N i=1

X 0 i;N + C T

Z t 0

Z

R

d

jru N (s; x)j 2 S s N (dx) ds

1=2

+ 1

N X N i=1

B t i

(12)

hence

E

"

sup

t2[0;T ]

1 N

X N i=1

X t i;N

#

1 N

X N i=1

E h

X 0 i;N i

+ C T E

" Z T 0

Z

R

d

jru N (s; x)j 2 S s N (dx) ds

1=2 #

+ E

"

sup

t2[0;T ]

1 N

X N i=1

B t i

# :

The …rst term is bounded by assumption. The second one is bounded by the estimate above. The third one is easily bounded as we have done other times. The …rst estimate on the empirical measure is proved.

Then we have to estimate W 1 S t N ; S s N 1

N X N

i=1

X t i X s i 1

N X N

i=1

Z t

s ru N r; X r i;N dr + N

X N i=1

B t i B s i

= Z t

s

Z

R

d

jru N (s; x)j S r N (dx) dr + N

X N i=1

B t i B s i

(t s) 1=2 Z t

s

Z

R

d

jru N (s; x)j 2 S r N (dx) dr

1=2

+ N X N

i=1

B i t B s i : Therefore

E h

W 1 S t N ; S s N 2 i

2 (t s) E Z t

s

Z

R

d

jru N (s; x)j 2 S r N (dx) dr + 2

2

N X N

i=1

E h

B t i B s i 2 i C (t s) :

This estimate is not su¢ cient to apply our classical theorem, since we need C (t s) 1+ . We omit the proof of this improvement since it is quite technical, being satis…ed with the idea of the full proof. The technical argument re-starts from the identity of Lemma 7, estimates in the average higher powers of the terms on the left-hand-side (but martingale inequalities are needed here), and then goes back to the previous estimate on W 1 S t N ; S s N but being able to handle a power higher than 2.

With due little e¤ort, the bound Z t

0

Z E h

r W N S t N 2 i

dxdt C

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gives us the information that the limit measure t of S N t has a density u t w.r.t. Lebesgue

measure and Z t

0

Z E h

jru t (x)j 2 i

dxdt C:

This is a …rst step in the direction of taking the limit.

2.3 Appendix

We have used the following simple facts.

Lemma 9 i) f g = g f ii)

(f g) h = f (g h) hence we may write

f g h without ambiguities

iii) Z

(f g) h = Z

g f h

where f (x) = f ( x).

Proof.

((f g) h) (x) = Z

(f g) (x y) h (y) dy

= Z Z

f (x y z) g (z) h (y) dzdy

= Z Z

f x y 0 g (z) h y 0 z dzdy 0

= Z

f x y 0 dy 0 Z

g (z) h y 0 z dz:

Z

(f g) (x) h (x) dx = Z Z

f (x y) g (y) dyh (x) dx

= Z Z

f (x y) h (x) dxg (y) dy:

Let us formalize a few details outlined above, also to show how to deal with the sto-

chastic case, 6= 0.

Riferimenti

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