The Gaussian free field, Gibbs measures and NLS on planar domains

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1 The Gaussian free field, Gibbs measures and on planar domains N. Burq, joint with L. Thomann (Nantes) and N. Tzvetkov (Cergy) Université Paris Sud, Laboratoire de Mathématiques d Orsay, CNRS UMR 8628 LAGA, Paris-13, Séminaire intercontinental, feb 20th, 2014

2 Purpose Study the dynamics of solutions of non linear Schrödinger equations in a planar domain, M R 2, with Dirichlet or Neumann boundary conditions, (i t + )u u 2 u = 0, u t=0 = u 0, u M = 0, (resp. ν u M = 0), from a statistical point of view, i.e. u 0 is a random variable or (equivalently, endow the space of initial data with a probability measure, µ 0. Well posedness on the support of the measure (almost sure WP): definition of a flow Φ(t). Statistical properties of the measure propagated by the flow, µ(t) = Φ(t) (µ 0 ): continuity, recurence, growth of Sobolev norms,...

3 The Gaussian free field Consider a sequence of (complex) independent Gaussian random variables g k N (0, 1) (of law 1 π e z 2 dz )and for e k, ( + 1)e k = λ 2 k e k, e k M = 0, n N, the random variable u 0 = g k e k (x). λ n N k equivalently, consider the map ω (Ω, p) u = g k (ω) e k (x), λ n N k and endow the space of distributions D (M), with the image of p by this map.

4 The GFF continued equivalently, write u = n N u k e k (x), identify D (M) with the space of sequences U = (u k ) C N (satisfying some temperance growth conditions), and endow C N with the probability measure + λ 2 k k=1 π e λ2 k u k 2 du k. Lemma The GFF is for any ɛ > 0 supported by H ɛ (M): µ 0 (H ɛ (M) = 1, but µ 0 (L 2 (M)) = 0.

5 Wick re-ordering, Bourgain s result If u is a solution to, then v = e it( u 2 L 2 1) u is a solution to (i t + 1)v ( v 2 2 v 2 L 2)v = 0, v t=0= u 0. (R) Theorem (Bourgain 1996) There exists δ > 0 such that for µ 0 -almost every u 0, there exists a unique (global in time) solution of (R) on the torus T 2, u = Φ(t)u 0 in e it u 0 + X δ,1/2+. Furthermore there exists a function G L 1 (dµ 0 ), positive on the support of µ 0 such that the measure dν = G(u)dµ is invariant by the flow Φ(t). ν is a Gibbs measure. Recall u X s,b = D t b e it u L 2 t ;Hx s

6 Our result Let M R 2 smooth bounded domain, and : u 4 L 4 := u 2 2 u 2 L 2 2 L 2 4 u 4 L 2. Theorem (N.B. L. Thomann, N. Tzvetkov) e : u 4 L 4 is µ 0 a.s. finite, and in L 1 (dµ 0 ). Furthermore, there exists for µ 0 a.e. initial data u 0 a global in time solution to (R), with Dirichlet (resp. Neumann) bdry conditions, and the flow Φ(t) satisfies Φ(t) (e : u 4 L 4 dµ 0 ) = e : u 4 L 4 dµ 0. Rk 1: No uniqueness in our result: weak solutions. But contrarily to usual (deterministic) weak solutions some kind of uniqueness remains: for any such flow get same invariant measure Rk 2 Work in progress: hope to get strong solutions.

7 Wick re-ordering on the torus For u = e it( ) u 0 = n Z 2 u 2 u = n 1,n 2,n 3 g n n ein x n 2t, n = n 2 + 1, g n1g n2g n3 n 1 n 2 n 3 ei(n 1 n 2 +n 3 ) x ( n 1 2 n n 3 2 )t g n 2 g m = +2 eim x ( m 2 )t n 2 m n 1 n 2,n 3 n 2 n,m n g n 2 g n n 3 e in x ( n 2 )t ( u 2 2 u 2 L 2)u g n1g n2g n3 = n 1 n 2 n 3 ei(n 1 n 2 +n 3 ) x n n 1 n 2,n 3 n 2 g n 2 g n e in x n 3

8 Wick re-ordering on T 2 : analysis Last term is a.s in H 2 : E( n g n 2 g n n 3 e in x 2 H δ ) = n Z 2 E( g n 6 ) n 6 2δ < + if 6 2δ > 2 First term is a.s. in X 1/2, 1/2+. Indeed, E( 2 X s, b) n 1 n 2,n 3 n 2 ( 1 = E (1 + n 1 2 n n 3 2 k 2 ) b k = k n 1 n 2,n 3 n 2,n 1 n 2 +n 3 =k ( E n 1 n 2,n 3 n 2 g n1 g n2 g n3 2) n 1 n 2 n 3 k s m 1 m 2,m 3 m 2 g n1 g n2 g n3 g m1 g m2 g m3 )

9 Since Gaussian are independent, complex and have mean equal to 0, and since n 1 n 2, n 3 n 2, expectancy vanishes unless (n 1, n 2, n 3 ) = (m 1, m 2, m 3 ) or (n 1, n 2, n 3 ) = (m 3, m 2, m 1 ) E( n Z 2 1 n 2 n 1 n 2,n 3 n 2 2 X s, b) n 1 n 2,n 3 n 2 k=n 1 n 2 +n 3 k 2s n 1 2 n 2 2 n (1 + n 1 2 n n 3 2 k 2 ) 2b diverges logarithmically. Factor 1 (1 + n 1 2 n n 3 2 k 2 ) 2b gives additional convergence which can be exchanged to compensate the k 2s term, hence end up in X s,b, for some s > 0 a level at which Cauchy theory well posed! This (plus quite some work) proves Bourgain s Theorem.

10 The problem on a manifold Wick Reordering is in the folklore of quantum field theory. Best (most efficient) approach seems to be via Fock representation Fock representation seems to be not so well suited to X s,b analysis (possible development?) Take boundary into account (should not be a serious problem in the context of Fock representation though) Keep the elementary approach and understand at that level the compensations which allow for Wick re-ordering

11 Back to Wick re-ordering: The case of S 2 Take e n Hilbert base of spherical harmonics, eigenvalues λ 2 n. u 2 u = n 1,n 2,n 3 g n1g n2g n3 n 1 n 2 n 3 e i(λ2 n1 λ2 n2 +λ2 n3 )t e n1 (x)e n2 (x)e n3 (x) g n 2 g m = + 2 n 2 m e iλ2 m t e n 2 (x)e m (x) n 1 n 2,n 3 n 2 n,m n g n 2 g n e iλ2 n t e n 3 n 2 (x)e n (x) As in the previous analysis, last term OK. Main issues: e n 2 (x) 1 e n1 e n2 e n3 is not an eigenfunction

12 The Weyl formula on S 2 Let E k = Vect {e n ; λ 2 n = k(k + 1)} an eigenspace, and e n1, e n2k+1 any orthonormal basis of E k. Proposition Let e(x, y, k) = 2k+1 p=1 e n p (x)e np (y). x S 2, e(x, x, k) = 2k + 1 Vol (S 2 ). In a mean value sense, the eigenfunctions are constant on S 2 Proof: e(x, y, k) is the kernel of the orthogonal projector on E k. Hence for any isometry T, e(x, Ty) = e(t 1 x, y). Theorem (Van der Kam, Zelditch, N.B-G. Lebeau) There exists orthonormal basis (e n,k ) having for any p < + uniformly bounded L p norms (actually most ONB are such)

13 Back to Wick re-ordering, analysis of 2nd term n,m g n 2 g m n 2 m e iλ2 m t e n 2 (x)e m (x) = ( g n 2 1)g m e iλ2 n 2 m t e n 2 (x)e m (x) m n,m + g m t n 2 m e iλ2 m e n 2 (x)e m (x) k n,e n E k,m = I + g m n 2 m e iλ2 m t e m (x) k n,e n E k,m = I + g n 2 g m t n 2 m e iλ2 m e m (x) n,m + n,m (1 g n 2 )g m e iλ2mt e n 2 m (x) = I + II + III m

14 and n Indeed, since I + III = n ( g n 2 1) n 2 II = e it u 2 L 2eit u ( g n 2 1) ( e n 2 n 2 (x) 1)e it u ( e n 2 (x) 1) is a.s finite (renormalizable). n m E ( ( g n 2 1)( g m 2 1) ) = E( g n 2 1)E( g m 2 1) = 0, ( E n ( g n 2 1) ( e n 2 n 2 (x) 1) 2 ) = ( E n n ( g n 2 1) 2 n 4 ( e n 2 (x) 1) 2 ) n 1 n 4 < +

15 Back to Wick reordering: analysis of the first term With γ(n 1, n 2, n 3, p)) = e S 2 n1 e n2 e n3 e p dx, ( E g n1 g n2 g ) n3 n1 n n1 n2 1 n 2 n 3 e i(λ2 λ2 n2 +λ2 n3 )t e n1 (x)e n2 (x)e n3 (x) 2 X s,b n 3 n 2 = E( k,p p n 1 n 2 n 3 n 2 p 2s k 2b n 1 n 2,n 3 n 2 k λ 2 n 1 +λ 2 n 2 λ 2 n 3 [0,1) g n1 g n2 g n3 n 1 n 2 n 3 γ(n 1, n 2, n 3, p) 2 ) p 2s λ 2 n 1 + λ 2 n 2 λ 2 n 3 2b n 1 2 n 2 2 n 3 2 γ(n 1, n 2, n 3, p)) 2 Difference with T 2 : additional sum (in p), but presence of γ 2, which is invariant wrt permutations and in l 1 : choice of bases, γ(n 1, n 2, n 3, p)) 2 = e n1 e n2 e n3 2 L 2 (M) C < + p

16 General manifolds Use Weyl formula (Volume of manifold normalized to 1) e(x, λ, µ) = e n 2 (x) µ λ n<λ Theorem (Hormander Th ) Let d(x) be the distance of the point x M to the boundary M. There exists C > 0 such that for any λ > 1 and x M satisfying d(x, M) λ 1/2, any d [0, 1], we have e(x, λ + dλ 1/2, λ) d 2π λ3/2 Cλ, (1) Theorem (Sogge) There exists C > 0 such that for any λ > 1 and x M e(x, λ + 1, λ) Cλ, (2)

17 The strategy of proof Regularize the system by cutting in the non linearity the frequencies higher than N (λ n > N) and get global in time solutions. Get a flow Φ N (t). Define a family ν N = e 1 2 : u N 4 L 4 : dµ 0 of probability measures invariant by the flows Φ N. Show that these measures converge (in a sense to be precised) to a limit measure ν 0 = e 1 2 : u 4 L 4 : dµ 0 Pass to the limit (in a sense to be precised) in the family of solutions Φ N (t)u 0 Φ(t)u 0 Show that the flow Φ(t) solves (R) and leaves the measure ν 0 invariant Strategy quite standard in parabolic settings (see e.g. Da Prato-Debussche),

18 The approximating systems Let Π N be the orthogonal projector on the space Let Φ N (t)u 0 be the solution of Vect (e n ; λ n N). (i t + )u Π N ( ( ΠN (u) 2 2 Π N (u) 2 L 2)Π N(u) ) = 0, u t=0 = u 0, u M = 0, (resp. ν u M = 0) Hamiltonian system with Hamiltonian 1 H = 2 xu 2 dx Π N(u) 4 L Π N(u) 4 L 2 M

19 Formally, the GFF µ 0 is, in the coordinate system given by the identification u = n u ne n (x) given by dµ 0 = + λ 2 + k k=1 π e λ2 k u k 2 du k = e λ2 k u k 2 + λ 2 k k=1 π du k k=1 = e P k λ2 k u k 2 + λ 2 k k=1 π du x 2 k = e L 2 + λ 2 k k=1 π du k, and ν N = e 1 2 : u N 4 L 4 : dµ 0 = e 2H(u) + λ 2 k k=1 π du k is (at least formally) invariant (because the Hamiltonian itself is invariant and a Hamiltonian system can be seen as an ODE with a divergence free vector field hence the (infinite product of) Lebesgue measures is also invariant

20 Passing to the limit ν N ν Definition S separable complete metric space, (ρ N ) N 1 probability measures on Borel σ algebra B(S). (ρ N ) on (S, B(S)) is tight if ε > 0, K ε S compact such that ρ N (K ε ) 1 ε for all N 1. Theorem (Prokhorov) (ρ N ) N 1 is tight iff it is weakly compact, i.e. there is a subsequence (N k ) k 1 and a limit measure ρ such that for every bounded continuous function f : S R, f (x)dρ Nk (x) = f (x)dρ (x). lim k S Rk. Weak convergence implies convergence in law S

21 A tight sequence of probability measures Let T > 0 and define a probability measure on the space of (space time) functions f (t, x) p N by the image measure of ν N by the map u 0 Φ N (t)u 0. Proposition The sequence p N is for any ɛ > 0 tight on C 0 ([0, T ]; H ɛ (M)). Proof: Fix 0 < ɛ < ɛ. the measure ν 0 is supported by H ɛ which embeds compactly in H ɛ. This gains compactness in space. Then use equation to gain compactness in time.

22 Passing to the limit Φ N (t) Φ(t) Setting: we have a family of r.v. X Nk = Φ Nk (t)u 0 such that the laws of X Nk, p Nk are weakly convergent to a probability measure p. We d like to deduce that the r.v. X Nk are a.s. convergent. This is False (take X Nk = ( 1) N k X). However Theorem (Skorohod) Assume that S is a separable metric space. Let (ρ N ) N 1 and ρ be probability measures on S. Assume that ρ N ρ weakly. Then there exists a probability space on which there are S valued random variables (Y N ) N 1, Y such that L(Y N ) = ρ N for all N 1, L(Y ) = ρ and Y N Y a.s.

23 Passing to the limit in the equation I Need first to check that the r.v. Y N satisfy the same equation as X N = Φ N (t): Consider the r.v. Z N = ((i t + )Y N Π N ( ( ΠN (Y N ) 2 2 Π N (Y N ) 2 L 2)Π N(Y N ) ) All the functions appearing in the r.h.s. are continuous from C 0 ((0, T ); H ɛ ) to S = H 1 ((0, T ); H 2 ɛ ). Hence L(Z N ) ( = L ((i t + )Y N Π N ( ΠN (Y N ) 2 2 Π N (Y N ) 2L 2)Π N(Y N ) )) ( = L ((i t + )X N Π N ( ΠN (X N ) 2 2 Π N (X N ) 2L 2)Π N(X N ) )) Hence Z N = 0 a.s. = L(0) = δ 0

24 Passing to the limit in the equation II Need to pass to the limit in Π N ( ( ΠN (X N ) 2 2 Π N (X N ) 2 L 2)Π N(X N ) ). Idea is: almost sure convergence in H ɛ, and estimate in probability of a stronger term namely : u 4 L 4 :. Gives convergence in probability of the non linear term. Hence convergence a.s. for a subsequence.

25 Toward a generalization of Bourgain s result on any smooth bounded domain? Need a nice local Cauchy theory at the level of regularity where the measure is supported (rest = more or less automatic) Standard (X s,b ) WP thresholds for cubic : T 2 rational s > 0 (Bourgain) S 2 s > 1 4 (N. B, Gérard, Tzvetkov) T 2 irrational s > 1 3 (Catoire Wang) Compact surfaces (without bdry) s > 1 2 (N. B-Gérard-Tzvetkov) Compact surfaces (with bdry) s > 2 3 (Anton, Blair-Smith-Sogge) Control 1st iteration in X s,1/2+0, s > s c Perform X s,b analysis in the Fock representation

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