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1 The PDE associated with local interaction

This section is more di¢ cult than previous ones, hence we sketch its content.

First, we reformulate the model with contact interaction in a form which is suitable for the investigation of its macroscopic limit, 1.1.

Then, since the macrocopic limit problem is very di¢ cult and innovative, we change the problem and freeze the scaling parameter in the interaction potential, 1.2.

With this trick we may apply Mean Field theory and deduce a PDE, parametrized by the freezed parameter, say M .

Then we send M to in…nity, to …nd a new PDE, called Porous Media equation, which in a sense corresponds to in…nitely many particles and in…nitely rescaled potential, 1.3.

However, we have not done the limit together, with M = N , as it should be. We have

…rst taken N ! 1, then M ! 1. Is it the same? We do not know the answer. We present a few simulations that indicate that the two results are at least numerically very close, for certain classes of problems, 1.4.

Encouraged by the numerical results, we describe in detail a partial result in the direc- tion of showing that the two limits are the same. It holds true for a modi…ed version of the rescaling, called intermediate or moderate interaction theory, 2-3.

1.1 Reformulation of the SDE 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) 1r 1, r 0, G (r) = (1 r)2=2 1r 1. We consider V as a "potential". Writing V0 for rV , we have

V0(x) = g (jxj) x jxj: Hence we may write equation

dXti;N = N1=d XN j=1

g N1=d Xti;N Xtj;N Xti;N Xtj;N Xti;N Xtj;N

dt + dBti

in the form

dXti;N = N1=d XN j=1

V0 N1=d Xti;N Xtj;N dt + dBti: Setting further

VN(x) = N V N1=dx

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we have

VN0 (x) = N N1=dV0 N1=dx hence we may write the previous SDE as

dXti;N = 1 N

XN j=1

VN0 Xti;N Xtj;N dt + dBti:

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 V is

dXti;N = 1 N

XN j=1

VN0 Xti;N Xtj;N dt + dBti: (1)

1.2 The Mean Field equation with a concentrated potential

Let us decuple the number of particles from the scaling of the potential and consider the equation

dXti;N;M = 1 N

XN j=1

VM0 Xti;N;M Xtj;N;M dt + dBti (2) where now we have two parameters: N is the number of particles, M is a parameter which de…nes the rescaling

VM(x) = M V M1=dx :

Obviously the true contact-interaction problem is to take M = N and send N ! 1.

But we already know the result of sending N ! 1 if we keep M …xed (it is the Mean Field theory): the empirical measure StN;M = N1 PN

i=1 Xti;N;M weakly converges to the measure-valued solution tof the Mean Field equation

@ t

@t =

2

2 t+ div t VM0 t :

Assume (we need it below) that t has a density ut(x). More precisely, both tand ut(x) depend on M ; so let us write uMt (x). This function is a weak solution of the PDE

@uMt

@t =

2

2 uMt + div uMt VM0 uMt : (3) We ask two questions: does uMt converge to the solution ut of some PDE? Is this PDE the correct one for the limit of the empirical measure StN = StN;N? We may answer the

…rst question, but the second one is open.

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1.3 Taking the limit in the Mean Field PDE

Let us …rst investigate the limit, as M ! 1, of solutions uM of equation (3). We do not want to be rigorous here (it is possible to state a rigorous result) and thus assume that uM 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 C01,

uMt ; = uM0 ; +

2

2 Z t

0

uM; ds Z t

0

VM0 uM; uMr ds

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

hut; i = hu0; i +

2

2 Z t

0 hu; i ds lim

M !1

Z t 0

VM0 uM; uMr ds:

In addition, if we prove that VM0 uM ! v in a suitable sense, we get hut; i = hu0; i +

2

2 Z t

0 hu; i ds Z t

0 hv; ur i ds:

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 C01, VM0 uM; = uM; VM0 ( ) : Moreover,

VM0 ( ) (x) = Z

VM0 (y x) (y) dy = r Z

VM(y x) (y) dy:

Assume now that VM is proportional to a sequence of molli…ers: this is true by the formula VM(x) = M V M1=dx if V is proportional to a smooth probability density

V (x) = 21f (x)

where 21 > 0 and f is a pdf. In this case, since is smooth and compact support, RVM(y x) (y) dy ! 21 (x), hence (VM0 ( ) ) (x) ! 21r (x) and …nally

VM0 uM; ! 21hu; r i = 21hru; i : We have found

v = 21ru and therefore the limit PDE

@u

@t =

2

2 u +

2 1

2 u2: (4)

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

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Remark 3 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 uM of the Mean Field equation converges to the solution u of the Porous Media equation.

1.4 Open problem and a few simulations

Let us repeat the open problem. We have a few objects: the particle system with contact interaction (1) and its empirical measure StN; the particle system with mean …eld interac- tion VM0 (2) and its empirical measure StN;M; the solution uMt of the mean …eld PDE (3);

the solution ut of the porous media PDE (4). We know that

N !1lim StN;M = uMt lim

M !1uMt = ut in suitable senses. Do we have also

N !1lim StN = ut?

The answer is not known. In this section we give some evidence of the fact that the answer, although theoretically uncertain, numerically is very close to be positive, in the case of repulsive interaction with integrable potential V .

The aim of the following simulation is to see, at the particle level, what happens if we consider N particles with potential VM(x), for di¤erent choices of M . We consider the case, in d = 1,

V0(x) = 1jxj 1(1 jxj) x jxj and we rescale it as above:

VM0 (x) = M M1=dV0 M1=dx

= M2V0(M x) :

The following functions de…ne part of V0 with a generic rescaling, where in the sequel we shall take R = M 1, = M2.

norma=function(x) abs(x) H=function(r) (sign(r)+1)/2

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g=function(r,R,alp) alp*H(1-r/R)*(1-r/R)

The code is now

M1=1; M2=100

N=200; Nloc=10; n=100000; dt=0.00001; sig=1; h=sqrt(dt); T=20; L=5; L0 = 1

e= matrix(1/N,N,1) alp1=M1^2; R1=1/M1 alp2=M2^2; R2=1/M2

X1=matrix(0,N,n); X2=matrix(0,N,n);

X1[,1]=rnorm(N,0,L0); X2[,1]=X1[,1];

plot(c(-L,L), c(0,0.6), type="n") for(t in 1:(n-1)){

DX1.0= matrix(X1[,t],N,N); DX1=DX1.0-t(DX1.0) DX2.0= matrix(X2[,t],N,N); DX2=DX2.0-t(DX2.0)

Kx1=g(norma(DX1),R1,alp1)*DX1/(norma(DX1)+ 0.000001) Kx2=g(norma(DX2),R2,alp2)*DX2/(norma(DX2)+ 0.000001) X1[,t+1]=X1[,t] + dt*Kx1%*%e + h*sig*rnorm(N)

X2[,t+1]=X2[,t] + dt*Kx2%*%e + h*sig*rnorm(N) if(t%%T==0 )

{polygon(c(-L,L,L,-L),c(0,0,0.6,0.6), col="white", border=NA) lines(density(X1[,t+1],bw=sd(X1[,t+1])/4),col="green")

lines(density(X2[,t+1],bw=sd(X2[,t+1])/4),col="red") }

}

Remark 4 The previous code, thanks to the help of participants, has been improved with respect to previous ones, by taking a matrix-approach to the computation of interactions between particles. This speeds-up considerably the computation.

Remark 5 Taking values like 2-3 instead of 4-5 in bw=sd(X2[,t+1])/4) smooths-out the pro…le but one then suspects that fatness is due to smoothing and not to di¤ usion.

The result seems to be that M does not matter: the di¤usion is essentially the same.

The result is similar for large densities:

M1=1; M2=100

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N=200; Nloc=10; n=100000; dt=0.00001; sig=1; h=sqrt(dt); T=20; L=5; L0 = 0.2

e= matrix(1/N,N,1) alp1=M1^2; R1=1/M1 alp2=M2^2; R2=1/M2

X1=matrix(0,N,n); X2=matrix(0,N,n);

X1[,1]=rnorm(N,0,L0); X2[,1]=X1[,1];

plot(c(-L,L), c(0,2), type="n") for(t in 1:(n-1)){

DX1.0= matrix(X1[,t],N,N); DX1=DX1.0-t(DX1.0) DX2.0= matrix(X2[,t],N,N); DX2=DX2.0-t(DX2.0)

Kx1=g(norma(DX1),R1,alp1)*DX1/(norma(DX1)+ 0.000001) Kx2=g(norma(DX2),R2,alp2)*DX2/(norma(DX2)+ 0.000001) X1[,t+1]=X1[,t] + dt*Kx1%*%e + h*sig*rnorm(N)

X2[,t+1]=X2[,t] + dt*Kx2%*%e + h*sig*rnorm(N) if(t%%T==0 )

{polygon(c(-L,L,L,-L),c(0,0,2,2), col="white", border=NA) lines(density(X1[,t+1],bw=sd(X1[,t+1])/4),col="green") lines(density(X2[,t+1],bw=sd(X2[,t+1])/4),col="red") }

}

Replacing the potential with the following one

V0(x) = 10 1jxj 1(1 jxj)10 x jxj does not change the picture:

norma=function(x) abs(x) H=function(r) (sign(r)+1)/2

g=function(r,R,alp) 10*alp*H(1-r/R)*(1-r/R)^10

Obviously we should go to M = N , but it seems that nothing changes. The conclusion seems to be that taking the limits N ! 1 …rst and then M ! 1 gives a very similar result as imposing M = N and taking N ! 1.

Some geometric intuition, not described here, seems to con…rm this result for interme- diate values of the density. When the density is very large, the case M = N could have

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a stronger di¤usion, but this is not biologically relevant. When the density is very small the di¤usion of the Mean Field equation should prevail, but the Brownian di¤usion in that case becomes dominant.

Overall, we think that the Porous Media equation is a very good model for contact repulsive interaction, when V is integrable.

Exercise 6 Compare the simulation of the Porous Media PDE (4) with the particle system with true contact interaction (1).

For this reason, let us examine in detail the following case.

2 Intermediate interaction: preparation

2.1 Intermediate regime

The open problem above, namely whether the limit of the contact-interaction case is equal to taking …rst N ! 1 and then M ! 1, has a full rigorous (and positive) solution if we rescale the interaction potential in a weaker way. Karl Oelschäger identi…ed this intermediate problem, between mean …eld and contact interactions, called intermediate (or moderate) regime, where it is possible to prove that StN converges to the solution of the porous media equation. It is the case when

VN(x) = N V N =dx

with 2 (0; 0), for a suitable 0< 1. For = 0 we have mean …eld interaction; for = 1 we have contact interaction. The regime 2 (0; 0) is thus more local than mean …eld but not so realistic as contact interaction.

The restriction to the intermediate regime will be used to prove tightness. But also the problem of passage to the limit is very di¢ cult, as explained below. In this case, it is not the restriction on to play a role but an assumption on the structure of V as the convolution of two kernels, assumption again identi…ed by Oelschäger. Let us discuss …rst the issue of convergence.

2.2 Taking the limit As in a previous section we have:

Lemma 7 The empirical measure StN satis…es the identity d StN; t = StN;@ t

@t dt StN; r t VN0 StN dt + dMt;N+

2

2 StN; t dt

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for all 2 Cb1:2 [0; T ] Rd , where

Mt;N = 1 N

XN i=1

Z t

0 r s Xsi;N dBsi:

In order to understand the di¢ culty to take the limit in this identity, assume for a second that StN weakly converges to some measure t. We can easily pass to the limit in the terms D

StN;@@ttE

and StN; t ; the martingale term Mt;N goes to zero in mean square (the proof is easy; we shall see below similar computations). But the limit of the nonlinear term

StN; r t VN0 StN is much more di¢ cult.

Recall, from general facts of analysis, that weak convergence of functions is not su¢ cient to pass to the limit in nonlinear terms: the sequence fn(x) = sin (nx), x 2 [0; 1] converges weakly to zero, but fn2 does not. Hence the sole property of weak convergence of the measures StN cannot be su¢ cient.

Another general fact we know from analysis is that if StN weakly converge to t and fN is a sequence of functions which converges uniformly to f , then

N !1lim fN; StN = hf; ti But uniform convergence is essential.

Example 8 If n = xn, = x, xn ! x, then n * . If fn converges uniformly to f , we see that hfn; ni = fn(xn) converges to f (x). But weaker convergences do not imply the same result: there is no reason why fn(xn) should converge to f (x) if we have only pointwise convergence, or Lp convergence, and so on.

Can we say that VN0 StN converges uniformly to some limit? This is a very di¢ cult question. The only easy thing we can say is that it weakly converges:

VN0 StN; ! 21

Z

r (x) t(dx) = 21 Z

(x) rut(x) dx in the case when t has a di¤erentiable density ut(x). Indeed

VN0 StN; = StN; VN( ) r ! 21

Z

r (x) t(dx) = 21 Z

(x) rut(x) dx:

Implicitly we have understood that, in this problem, sooner or later we have to prove that t has a density. One way is to investigate the convergence of the molli…ed empirical measure hNt (x) = WN StN (x).

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After these preliminary remarks on the di¢ culty of taking the limit in the non-linear term, let us describe the brilliant trick devised by Karl Oelschläger, under an additional assumption. Assume that (take 21 = 1 here for simplicity of exposition)

VN = WN WN

where WN are classical molli…ers and, from now on, we write WN (x) = WN ( x) : Then we have

StN; r t VN0 StN = r tStN; WN WN0 StN

= WN r tStN ; r WN StN where we have used the property

WN f; g = hf; WN gi easy to prove. Now, assume we can prove that the quantity

WN r tStN ; r WN StN r t WN StN ; r WN StN

is small for large N (namely that commuting WN with the pointwise product r t is irrelevant, in the limit). If so, we have to deal with

r t WN StN ; r WN StN

and here we just need weak L2-convergence of r WN StN and strong L2-convergence of WN StN. Recall indeed that if fn! f in L2 Rd and gn* g in L2 Rd , then

jhfn; gni hf; gij jhfn; gni hf; gnij + jhf; gni hf; gij kfn f k kgnk + jhf; gni hf; gij

and now the …rst term goes to zero by strong convergence of fn to f and boundedness of gn; the second term goes to zero by weak convergence of gn to g.

In the next section we describe a modi…ed model where it is possible to prove such properties and then to pass to the limit rigorously.

2.3 Bounds on particle position

Our ultimate aim is to prove tightness of the laws QN of the empirical measures SN. [In fact, at the end, we follow a di¤erent approach, namely we prove tightness of molli…ed empirical measure, but it is convenient for pedagogical reasons to argue about tightness of

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the laws QN of SN] To reach this result, the …rst and more important estimate, following the computations made for the Mean Field case, would be to prove that

E

"

sup

t2[0;T ]

Xti;N

# C:

But now

Xti;N X0i;N + 1 N

XN j=1

Z t 0

VN0 Xsi;N Xsj;N ds + Bti

and VN0 is unbounded with respect to N , so we cannot repeat the estimates of the mean

…eld case. Notice that 1 N

X

j

VN0 Xti Xtj dt = ruN t; Xti

where

uN(t; x) = VN StN (x) = 1 N

X

j

VN x Xtj : Hence the interacting particle system can be written in the form

dXti;N = ruNt Xti dt + dBti: Thus

Xti;N X0i;N + Z t

0 ruNs Xsi ds + Bti : If we prove suitable estimates on ruNt , we have a tool to prove Eh

supt2[0;T ] Xti;N i C.

The next section shows that there is hope to prove an estimates on ruNt . To compare the empirical function uNt used here with the function ut of the next section, notice that uNt (x) = VN StN (x), if it converges, it should converge to the solution utof the Porous Media equation, by the general conjecture underlying this section.

2.4 Energy estimates on the Porous Media equation

We …rst describe an a priori estimate for the Porous Media equation since, we think, it is conceptually the idea behind the next computation on StN. 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

u2dx = Z

u@u

@tdx = Z

u

2

2 u + div (uru) dx

=

2

2 Z

jruj2dx Z

u jruj2dx hence

1 2

Z

u2tdx + Z t

0 2

2 Z

jruj2dx + Z

u jruj2dx ds = 1 2

Z u20dx where we stress the remarkable fact that

Z

u div (uru) dx = Z

ru (uru) dx = Z

u jruj2dx:

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 ruN. Thus we shall try now to repeat these energy computations on uN.

3 Rigorous results on intermediate interaction

In this section we prove the following result. We consider the SDE dXti;N = 1

N XN j=1

VN0 Xti;N Xtj;N dt + dBti with

VN(x) = N V N =dx

with 2 (0; 0), for a suitable 0 < 1 stated by the next theorem. Recall that we write W (x) = W ( x).

Theorem 9 Assume that

V = 21W W

where W is a smooth compact support probability density function and 21 is a positive constant. Set VN(x) = N V N =dx , WN(x) = N W N =dx (so that VN = 21WN WN) and set wNt (x) := WN StN (x). Assume

d d + 2: Finally, assume that the initial conditions X0i;N satisfy

E Z

Rd

wN0 (x)2dx C; E X0i;N 2 C

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and S0N; ! hu0; i in probability, for every 2 Cc1, with u02 L2 Rd , 0 u0 1.

Then the family of laws eQN of the functions wNt (x) are tight on the space Y described below, which includes L2loc [0; T ] Rd .

They have the unique limit eQ = u, and ut is the unique weak solution of the Porous Media equation

@u

@t =

2

2 u +

21

2 u2 with initial condition u0.

The proof of this theorem requires several steps that we divide in the next subsections.

3.1 Energy estimates on the molli…ed empirical measure Recall Lemma 7:

d StN; t = StN;@ t

@t dt StN; r t VN0 StN dt + dMt;N+

2

2 StN; t dt:

and the de…nitions

wNt (x) = WN StN (x) = 1 N

X

j

WN x Xtj uNt (x) = VN StN (x) :

In this section we prove the following two results.

Lemma 10 For the molli…ed empirical measure wtN(x), under the assumption V = 21W W , we have

1 2

Z

Rd

wNt (x)2dx +

2

2 Z t

0

Z

Rd rwNs (x)2dx ds + Z t

0

Z

Rd

VN0 SsN (x)2SsN(dx) ds

= 1 2

Z

Rd

wN0 (x)2dx + Z

Rd

Z t 0

wsN(x) dMsN dx + Z

Rd 2

N2 XN

i=1

Z t

0 rWN x Xsi;N 2ds

! dx:

Corollary 11 Under the further assumption d+2d , there is a constant C > 0 such that 1

2E Z

Rd

wNt (x) 2dx +

2

2 E Z t

0

Z

Rd rwNs (x)2dx ds + E Z t

0

Z

Rd ruNs (x) 2SsN(dx) ds 1

2 Z

Rd

wN0 (x)2dx + C:

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Let us prove the lemma. Replace t(y) by WN(x y), with x given as a parameter, and think StN; t as an integration in the y variable, so that

StN; t = StN; WN(x ) = wNt (x) and so on for the other terms:

StN; WN(x ) = WN StN (x) = wtN(x)

StN; rWN(x ) VN0 StN = Z

ryWN(x y) VN0 StN (y) SNt (dy)

= div

Z

WN(x y) VN0 StN (y) StN(dy)

= div WN VN0 StN StN : We get

dwtN(x) = div WN VN0 StN SNt (x) dt + dMtN(x) +

2

2 wNt (x) dt where

MtN(x) = 1 N

XN i=1

Z t

0 rWN x Xsi;N dBis:

Let us apply Itô formula to wtN(x) 2, with x treated again as a parameter. We have d wNt (x) 2 = 2wtN(x) dwN(t; x) + d wN(x) t

= 2wtN(x) div WN VN0 StN StN (x) dt + 2wtN(x) wNt (x) (t; x) dt +2wtN(x) dMtN + d MN(x) t

and

d MN(x) t=

2

N2 XN i=1

rWN x Xti;N 2dt:

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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" 12R

Rd wNt (x)2dx:

1 2

Z

Rd

wNt (x)2dx = 1 2

Z

Rd

w0N(x) 2dx +

Z

Rd

Z t 0

wsN(x) div WN VN0 SsN SsN (x) ds dx +

2

2 Z

Rd

Z t 0

wNs (x) wsN(x) ds dx +

Z

Rd

Z t 0

wsN(x) dMsN dx

+ Z

Rd 2

N2 XN

i=1

Z t

0 rWN x Xsi;N 2ds

! dx:

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

We have

Z

Rd

Z t 0

wsN(x) wNs (x) ds dx

= Z t

0

Z

Rd

wsN(x) wNs (x) dx ds

=

Z t 0

Z

Rd rwNs (x)2dx ds

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

Z

Rd

Z t 0

wsN(x) div WN VN0 SsN SsN (x) ds dx

= Z t

0

Z

Rd

wsN(x) div WN VN0 SsN SsN (x) dx ds

=

Z t 0

Z

RdrwNs (x) WN VN0 SsN SsN (x) dx ds and, by a rule already used above,

= Z t

0

Z

Rd

WN rwN(s) (x) VN0 SsN (x) SsN(dx) ds:

From the assumption V = 21W W it follows VN = 21WN WN (we check it below) and thus

WN rwsN(x) = WN rWN SsN = rWN WN SsN

= r VN SsN = VN0 SsN:

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(we have used the properties (f g) h = f (g h) and f rg = rf g, easy to prove).

We deduce

Z

Rd

Z t 0

wsN(x) div WN VN0 SsN SsN (x) ds dx

=

Z t 0

Z

Rd

VN0 SsN (x)2SsN(dx) ds:

Let us prove that VN = 21WN WN:

2

1 WN WN (x) = 21 Z

WN( x + y) WN(y) dy = 21N Z

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

= 21N Z

W y N =dx W (y) dy = 21N W W N =dx

= N V N =dx = VN(x) : The lemma is proves.

Let us now prove the corollary. We have to prove that the last term in the identity of the lemma is bounded by a constant, in expected value (the martingale termRt

0 wN(s; x) dMsN has zero average). The required bound will follow from the following result. Notice that we shall perform just identities and even without expected value, hence the result, in terms of range of , is optimal.

Lemma 12 If d+2d , there is a constant C > 0 such that, for all s t in [0; T ], Z

Rd 2

N2 XN i=1

Z t

s rWN x Xri;N 2dr

!

dx C (t s) : Proof. We have

Z

Rd 2

N2 XN i=1

Z t

s rWN x Xri;N 2dr

! dx

=

2

N2 XN i=1

Z t s

Z

Rd rWN x Xri;N 2dx dr

=

2

N2 XN i=1

Z t s

Z

RdjrWN(x)j2dx dr

because, by the change of variable x ! x Xri;N in the integral in dx, we have Z

Rd rWN x Xri;N 2dx = Z

RdjrWN(x)j2dx:

Riferimenti

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An important and less trivial consequence of the rank formula is the following. 1.28 Theorem (Rank of

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Since we are comparing two polynomials, it follows that the identity holds for all x

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