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Mean field limit for interacting particles

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Mean field limit for interacting particles

P-E Jabin (University of Nice) Joint works with J. Barr´e and M. Hauray

(2)

Interacting particles

ConsiderN particles,identicaland interacting two by two through the kernelK. Denote Xi(t)∈Πd andVi(t)∈Rd the position and velocity of the i-th particle. Then

d

dtXi =Vi, d

dtVi = 1 N

X

j6=i

K(Xi−Xj). (1)

The N1 is a renormalization to get the correct time scale.

The most important case isCoulomb interaction K(x) =−∇Φ, Φ(x) = α

|x|d−2 + regular.

The caseα >0 corresponds to the repulsive/electrostatic case (plasmas...) andα <0 to the attractive/gravitational case (cosmology...).

Other kernels of interest exist however...

(3)

Well posedness for the dynamics

IfK is regular (Lipschitz), then Eq. (1) has aunique solutionfor all initial data thanks to Cauchy-Lipschitz.

IfK is singular, but the potential is repulsivethen the same result trivially holds. For example, Coulomb case theenergy

E(t) = 1 N

X

i

|Vi|2+ α N2

X

i

X

j6=i

1

|Xi −Xj| =E(0), gives

|Xi −Xj| ≥ C

N2, ∀i 6=j.

This shows that the force is in fact Lipschitz. But notice that this estimate isuselessas N→+∞.

Inother cases, one may obtain existence/uniqueness for a.e.initial conditions, see DiPerna-Lions, Ambrosio or Hauray.

(4)

Formal limit

AsN →+∞, Eq. (1) is even more cumbersome to use and some limit to a PDE is expected.

Definition of the problem : Take a sequence of initial data

ZN,0 = (X1N,0, . . . ,XNN,0, V1N,0, . . . ,VNN,0).

Consider the sequence of solutions

ZN(t,ZN,0) = (X1N, . . . ,XNN, V1N, . . . ,VNN) to (1) with corresponding initial data

XiN(0) =XiN,0, ViN(0) =ViN,0, i = 1. . .N.

Define the empirical measure (cf P.L. Lions’ course) fN(t,x,v) = 1

N

N

X

i=1

δ(x−XiN(t))⊗δ(v−ViN(t)).

Can we get an equation for some limitf offN?

(5)

Thenif K ∈C(Πd) orXiN(t)6=XjN(t) for everyt,i 6=j, we get the Vlasov equation (with the conventionK(0) = 0)

tfN+v· ∇xfN+ (K ?xρN)· ∇vfN = 0, ρN(t,x) =

Z

R3

fN(t,x,v)dv. (2)

AsN → ∞, one expectsfN −→f in w − ∗topology, withf a solution to (2) with

f(t = 0) = lim

N fN(t = 0).

Even forCoulombpotential, Eq. (2) is well posed (existence+uniqueness, at least in 3d) if for example

f(t = 0)∈L1∩L with compact support in velocity (see Horst, Lions-Perthame, Pfaffelmoser, Schaeffer,...).

Butif K 6∈C(Πd) then the limitfN →f is extremely difficult because of the product

(K ?xρN)fN.

(6)

Macroscopic models

Thesame questionmay be asked for the dynamics of particles in the purelyphysical space, namely

d

dtXi = 1 N

X

j6=i

µiµjK(Xi −Xj), (3) with theµi of order 1. Then defining

ρN(t,x) =

N

X

i=1

µiδ(x−Xi(t)), one expects as the limit forρN

tρ+∇x(K ? ρ ρ) = 0. (4)

The main example is in2dwith K =x/|x|2,|µi|= 1 and then (4) is just the incompressibleEuler equation.

(7)

The derivation of (4) is usuallyeasier than the one of (2). Indeed theforceis typically regular provided that

dmin(t) = min

i6=j |Xi(t)−Xj(t)|,

is large enough (for exampleorderN1/d). And if, for some locally boundedF and frx close to xi, an estimate like

k1 N

X

j6=i

µiK(x−Xj)kW1,∞ ≤F(dmin/N1/d), is available, then as for anyt, there exists i,k such that dmin=|Xi−Xk|. Assuming µik

d

dtdmin= d

dt|Xi −Xk|

≥ −1 N

X

j6=i,k

µj(K(Xi −Xj)−K(Xk −Xj))

+ small

≥ −k1 N

X

j6=i

µjK(x−Xj)kW1,∞dmin.

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One deduces then that d

dtdmin≥ −dminF(dmin/N1/d).

Gronwall lemma thencontrolsdmin, at least for a short time.

However for the case inphase space, the forceis stillcontrolled by dmin the minimal distance in the physical space but it itself only boundsthe minimaldistance in phase space

dminv = inf

i6=j(|Xi −Xj|+|Vi −Vj|).

Hence it is not possible toclose the estimate...

For Euler, see Goodman, Hou and Lowengrub, or Schochet.

(9)

Regular case

Theeasiestway of obtaining the limit is to take enough regularity onK to be able to pass to the limitin (K ? ρN)fN.

Theorem

Assume thatK is continuousand that ∃R s.t.|Vi(0)| ≤R, ∀i.

Thenthere exists a subsequence σ s.t.

1)fσ(N)−→f in w − ∗L(R+, M1d×Rd)) 2)ρσ(N)−→ρ=R

Rdf dv in w− ∗L(R+, M1d)) 3) f is asolution to (2)in the sense of distribution.

Comments :

1) This provides theexistence of measure valued solutionsto (2).

2) There isno uniqueness theory for (2) under such aweak assumptionfor K : no convergence of the full fN.

3) The particles could be verypoorly distributed (all concentrated at 0 for ex.).

(10)

Quick proof

•For any t

Z

Πd×Rd

fN(t,x,v)dx dv = 1.

By weak-* compactness ofL(R+, M1d×Rd)) (dual space of L1(R+, C0d×Rd))), one has σ s.t. fσ(N) −→f.

•One has

|Vi(t)| ≤ |Vi(0)|+1 N

X

j6=i

Z t

0

|K(Xi−Xj)|ds ≤R+tkKk, ∀i.

Therefore ρσ(N)=R

fσ(N)dv converges weak-* in L(R+, M1d)) toρ=R

f dv.

•Witht fixed,K ?xρσ(N) isequicontinuous in x (same modulous of continuity asK).

•In the sense of distribution, integrating (2) in velocity

tρσ(N)+ divx Z

Rd

v fσ(N)dv

= 0.

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ThereforeK ?x ρσ(N) is equicontinuous inx andt.

•By Ascoli’s theorem, K ?xρσ(N) converges uniformly(in L([0, T]×Πd) for any T) to K?xρ.

Consequently in the sense of distributions

(K ?xρσ(N))fσ(N)−→(K ?xρ)f

Passing to the limit in the other terms of (2) is straightforward.

(12)

Well posedness

IfK is more regular, the following stability holds (Dobrushin, Braun and Hepp, Spohn...)

Theorem

AssumeK ∈W1,∞ and f1, f2∈L(R+, M1d×Rd))are two solutions to (2) with compact support then

kf1(t)−f2(t)kW−1,1d×Rd)≤C kf1(0)−f2(0)kW−1,1d×Rd)

×exp(Ck∇KkL t).

Comments :

1)Well posedness of measures solutionsto (2) and not only convergence.

2) Theexponentialgrowth rate is probably not optimal.

3) Controls theconcentrationof particles.

4) The constantC only depends on the total mass of both solutions.

(13)

Idea of the proof Define thecharacteristicsfor each solution

d

dtXγ(t,x,v) =Vγ(t,x,v), d

dtVγ(t,x,v) =K ?x ργ(t,Xγ), X(0,x,v) =x, V(0,x,v) =v, γ = 1,2.

This is well defined as K ? ργ is Lipschitz (K is C1) and it mimics the dynamics of the particles (1). Moreoever

|∇Xγ|+|∇Vγ| ≤eC tk∇Kk. Then denoting

L={φ∈C1d×Rd), kφk≤1 andk∇φk≤1}, one has

kf1(t)−f2(t)kW−1,1= sup

φ∈L

Z

Πd×Rd

φ(x,v) (f1(t,x,v)−f2(t,x,v))

= sup

φ∈L

Z

(φ(X1,V1)f1(0,x,v)−φ(X2,V2)f2(0,x,v))dx dv

(14)

Therefore

kf1(t)−f2(t)kW−1,1 ≤sup

φ∈L

Z

Πd×Rd

φ(X1,V1)(f1(0)−f2(0)) + sup

φ∈L

Z

Πd×Rd

|φ(X1,V1)−φ(X2,V2)|f2(0,x,v), so

kf1(t)−f2(t)kW−1,1 ≤ k∇(X1,V1)kkf1(0)−f2(0)kW−1,1

+k(X1,V1)−(X2,V2)k Z

f2(0,x,v)dx dv. And consequently it only remains to bound

k(X1,V1)−(X2,V2)k. First of all

d

dt|X1−X2| ≤ |V1−V2|.

(15)

And now

d

dt|V1−V2| ≤|K ? ρ1(t,X1)−K ? ρ2(t,X2)|

≤ |K ? ρ1(t,X1)−K? ρ1(t,X2)|

+|K ?(ρ1−ρ2)(t,X2)|

≤ |X1−X2| k∇Kk Z

ρ1dx +k∇Kk1−ρ2kW−1,1. Putting all estimates together

d

dtkf1(t)−f2(t)kW−1,1 ≤Ck∇Kkkf1(t)−f2(t)kW−1,1, and we conclude by Gronwall lemma.

(16)

Conclusion

The caseK ∈C1 is completely solved. However it is very far from the kernels that are used in practice.

The caseK ∈C is superficially easybut in fact not satisfactory.

The well posedness is most probably out of reach (uniqueness of X˙ =b(x) with b only continuous ?).

The interesting cases areK ∼ |x|−α. We don’t even haveK ∈C but in factK ∈C1d\ {0}).

(17)

The problem

Consider the dynamics forK ∼ |x|−α d

dtXi =Vi, d

dtVi = 1 N

X

j6=i

K(Xi−Xj). (1) Define

fN = 1 N

N

X

i=1

δ(x−Xi(t))δ(v−Vi(t)).

Can we prove thatfN −→f with f solution to

tf +v· ∇xf + (K ?xρ)· ∇vf = 0, ρ(t,x) =

Z

Rd

f(t,x,v)dv. (2)

(18)

Weakly singular case

Still using Gronwall type estimates, we have (Hauray-Jabin) Theorem

AssumeK ∼ |x|1α with α <1. For any sequence of initial data ZN,0 with uniform compact support and such that

dmin(0) = min

i6=j(|XiN,0−XjN,0|+|ViN,0−VjN,0|) ≥cN−1/(2d). Thenfor any t we have c0 such thatdmin(t)≥c0N−1/(2d) and the sequencefull fN converges toward the unique solution f to (2)with f(t = 0) = limNfN(t = 0)in L1∩L and compactly supported.

(19)

Comments :

1) The conditionα <1 is probably close tooptimal even though it is quitefarfrom theCoulombian case. It is used to control the integral of the force along trajectories, even when 2 particles are close :

Z

t

dt

|X +Vt|α <∞

2) If theinitialpositions and velocities are chosen randomlythen theprobabilityto satisfy the condition ondmin vanishes. Therefore this isnot satisfying from astatistical physics point of view but quiteall right fornumerical purposes.

In fact this assumption even tells a lot on the limitf. For example if

dmin(0)≥cN1/(2d), andfN(0)−→f, then automaticallyf ∈L.

3) Thecompact supportassumption corresponds to the usual hypothesis foruniqueness on (2) and is rather natural.

(20)

The macroscopic equivalent

For macroscopic models, the result is better (see Hauray) Theorem

AssumeK ∼ |x|1α with α <d−1. Take any sequence of initial data XN,0 such that

dmin(0) = min

i6=j(|XiN,0−XjN,0|) ≥cN−1/d, consider the dynamics (1) with µi = 1,

d

dtXi = N1 P

j6=iK(Xi −Xj). Then for any t≤T we have c0 such thatdmin(t)≥c0N−1/(d) and the sequencefull ρN converges toward the unique solutionρ to (4). If div K = 0, then T =∞.

Comments :

1) The conditionα <d −1 is now very reasonableas 2d Euler is just the limit case. Moreoverno symmetry is needed onK contrary to the previous derivations.

2) All other remarks concerningdmin unfortunately still hold.

(21)

A partial proof

Let us only show the estimate on the minimal distance. Of course d

dt|Xi−Xk| ≥ − 1 N

X

j6=i,k

|K(Xi −Xj)−K(Xk −Xj)|

+ 1

N(|K(Xi−Xk)|+|K(Xk −Xi)|).

IfK ∼ |x|α then

|K(Xi−Xk)| ≤ C

|Xi−Xk|α ≤ C dminα , and on the other hand

|K(Xi−Xj)−K(Xk−Xj)| ≤|Xi −Xk|

×

C

|Xi −Xj|α+1 + C

|Xk −Xj|α+1

. The main point is to control

1 N

X

j6=i

C

|Xi −Xj|α+1.

(22)

We simply mimic the usual convolution estimates. Denote Nk =|{j, j 6=i and 2kdmin ≤ |Xi −Xj| ≤2k+1dmin}.

By the definition ofdmin we of course haveNk = 0 for anyk <0 and as we are in Πd,Nk = 0 for k >k0 =−lndmin/ln 2.

Decomposing we get 1

N X

j6=i

1

|Xi −Xj|α+1 ≤ Cd N

k0

X

k=0

2−k(α+1)dmin−α−1 Nk. Again by the definition ofdmin

Nk ≤Cd2kd, so asα+ 1<d

1 N

X

j6=i

1

|Xi−Xj|α+1 ≤ Cd

N

k0

X

k=0

2k(d−α−1)dmin−α−1

≤ Cd

N 2k0(d−α−1)dmin−α−1≤Cd N−1 dmind .

(23)

Putting all the estimates together, one gets d

dt|Xi −Xk| ≥ −|Xi−Xk|Cd N−1

dmind −C dmin−α N . Therefore taking the i andk such that|Xi−Xk|=dmin

d

dtdmin ≥ −dminCd N−1

dmind

−C dmin−α N

≥ −dmin N−1 dmind

Cd+C dmind−α ,

and we conclude by Gronwall lemma.

(24)

An almost everywhere approach

We would like to still derive astability estimate,i.e. showing that

|XiN(t,ZN,0)−XiN(t,ZN,0+δ)|remains of orderδ for shiftsδ that may go to 0.

However as we do not want to use Cauchy-Lipschitz/Gronwall like estimates, this can only betrue for almost all initial data.

More precisely (following Crippa-De Lellis) one is looking for an estimate like

Z

ΠdN×RdN

P(ZN,0) log

1 + 1 N|δ|

N

X

i=1

(|XiN(t,ZN,0)−XiN(t,ZN,0+δ)|

+|ViN(t,ZN,0)−ViN(t,ZN,0+δ)|)

≤C (1 +t).

(5) The functionP of the initial configuration determines thenotion of almost everywhereand so must be chosen with care...

(25)

Idea of the proof

Differentiate in time and for a givent use the change of variables ZN,0 −→ZN(t,ZN,0),

which has jacobian 1. Then, as the measuree−HN is invariant under the flow, one has to control quantities like

Z

ΠdN×RdN

Pt(ZN,0)

|X10−X20|α+1 dZN,0,

with Pt the image by the flow of at timet of P. This should be all right ifα+ 1<d.

If one is not careful, one ends up instead with Z

ΠdN×RdN

Pt(ZN,0) max

i

1 N

X

j6=i

1

|Xi0−Xj0|α+1

dZN,0 = +∞...

(26)

The choice of P

From the details of the proof one gets the followingcondition onP Z

Πd(N−k)×RdN

Pt(ZN,0)dXN−k0 . . .dXNdV10. . .dVN0 ≤Ck. The simplest (and more or less only reasonable) way of satisfying this is to chooseP invariant for the flow. For instance if

K =−∇Φ with Φ≥0

P(ZN,0) =βNe−HN(ZN,0), Where the hamiltonianHN can be one of the two

HN =EN = 1 N

N

X

i=1

|ViN|2+ 1 N2

N

X

i=1

X

j6=i

Φ(XiN−XjN) (easy).

or

HN =N EN =

N

X

i=1

|ViN|2+1 N

N

X

i=1

X

j6=i

Φ(XiN−XjN) (much harder).

(27)

Interpretation of the estimate

TakeK =−∇Φ with Φ≥0and K ∼ |x|−α with α <d −1.

•First of all one can replaceZN(t,ZN,0+δ) in (5) byZN,δ the solution to the dynamics with aregularized kernelKδ. Hence (5) controls

Z

Π3N×R3N

P(ZN,0) log 1 + 1

|δ|kfN(t)−fδ(t)kW−1,1 ,

with fδ the solution to (2) with the regularizedkernel Kδ.

=⇒Convergence to the solutions of (2) withf0 = limNfN(0).

•But in general one expects some sort of concentration of the measureP so that the possible limitsf0 are very limited.

•For HN =EN one has almostalways fN(0)−→0.

(28)

ForHN =N EN one has almostalways fN(0)−→ρ(x)e−|v|2, with ρ theminimizer of

Z

Π6

1

2Φ(x−y)ρ(x)ρ(y)dx dy + Z

Π3

ρ(x) logρ(x)dx.

•For the moment this gives only a stabilityof the two stationary solutionsthroughperturbationbyDirac masses.

=⇒Interest of having non invariant measures instead of HN.

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