Microscopic Hamiltonian dynamics perturbed by a conservative noise

Transcription

1 Microscopic Hamiltonian dynamics perturbed by a conservative noise CNRS, Ens Lyon April 2008

2 Introduction

3 Introduction Fourier s law : Consider a macroscopic system in contact with two heat baths with different temperatures T l T r. When the system reaches its steady state < > ss, one expects Fourier s law holds: < J(q) > ss = κ(t (q)) (T (q)), q macroscopic point

4 Introduction Fourier s law : Consider a macroscopic system in contact with two heat baths with different temperatures T l T r. When the system reaches its steady state < > ss, one expects Fourier s law holds: < J(q) > ss = κ(t (q)) (T (q)), q macroscopic point J(q) is the energy current; T (q) the local temperature; κ(t ) the conductivity.

5 Introduction Fourier s law : Consider a macroscopic system in contact with two heat baths with different temperatures T l T r. When the system reaches its steady state < > ss, one expects Fourier s law holds: < J(q) > ss = κ(t (q)) (T (q)), q macroscopic point J(q) is the energy current; T (q) the local temperature; κ(t ) the conductivity. If system has (microscopic) size N, finite conductivity means < J > ss N 1.

6 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains

7 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains H = { p 2 x + W (q x) + } V (q x q y ), Λ Z d 2m x 2 4 x Λ y x

8 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains H = { p 2 x + W (q x) + } V (q x q y ), Λ Z d 2m x 2 4 x Λ y x W : pinning potential; V : interaction potential

9 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains H = { p 2 x + W (q x) + } V (q x q y ), Λ Z d 2m x 2 4 x Λ y x W : pinning potential; V : interaction potential If V (r) = α r 2, W (q) = ν q 2 (harmonic chain), < > ss is an explicit Gaussian measure and < J > ss 1 : Fourier s law is false (Lebowitz, Lieb, Rieder 67)

10 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains H = { p 2 x + W (q x) + } V (q x q y ), Λ Z d 2m x 2 4 x Λ y x W : pinning potential; V : interaction potential If V (r) = α r 2, W (q) = ν q 2 (harmonic chain), < > ss is an explicit Gaussian measure and < J > ss 1 : Fourier s law is false (Lebowitz, Lieb, Rieder 67) Non linearity is extremely important to have normal heat conduction.

11 Hamiltonian microscopic models : Fermi-Pasta-Ulam chains H = { p 2 x + W (q x) + } V (q x q y ), Λ Z d 2m x 2 4 x Λ y x W : pinning potential; V : interaction potential If V (r) = α r 2, W (q) = ν q 2 (harmonic chain), < > ss is an explicit Gaussian measure and < J > ss 1 : Fourier s law is false (Lebowitz, Lieb, Rieder 67) Non linearity is extremely important to have normal heat conduction. But it is not sufficient : It has been observed experimentally and numerically for nonlinear chains that if d 2 and momentum is conserved ( W = 0, unpinned) then conductivity is still infinite (finite otherwise).

12 Motivations/Goal

13 Motivations/Goal Give a rigorous derivation of Fourier s law from the microscopic model.

14 Motivations/Goal Give a rigorous derivation of Fourier s law from the microscopic model. If Fourier s law does not hold, κ N N δ, universality of the diverging order δ of the conductivity?

15 Motivations/Goal Give a rigorous derivation of Fourier s law from the microscopic model. If Fourier s law does not hold, κ N N δ, universality of the diverging order δ of the conductivity? Numerical simulations are not conclusive (δ [0.25; 0.47] for the same models) and subject of intense debate.

16 The Models FPU chains are mathematically very difficult to study. We perturb the Hamiltonian dynamics by a stochastic noise. These stochastic perturbations simulate (qualitatively) the long time (chaotic) effect of the deterministic nonlinear model.

17 The Models FPU chains are mathematically very difficult to study. We perturb the Hamiltonian dynamics by a stochastic noise. These stochastic perturbations simulate (qualitatively) the long time (chaotic) effect of the deterministic nonlinear model. FPU chains conserve total energy H. If the system is unpinned (W = 0), it conserves also total momentum x p x.

18 The Models FPU chains are mathematically very difficult to study. We perturb the Hamiltonian dynamics by a stochastic noise. These stochastic perturbations simulate (qualitatively) the long time (chaotic) effect of the deterministic nonlinear model. FPU chains conserve total energy H. If the system is unpinned (W = 0), it conserves also total momentum x p x. Two different noises:

19 The Models FPU chains are mathematically very difficult to study. We perturb the Hamiltonian dynamics by a stochastic noise. These stochastic perturbations simulate (qualitatively) the long time (chaotic) effect of the deterministic nonlinear model. FPU chains conserve total energy H. If the system is unpinned (W = 0), it conserves also total momentum x p x. Two different noises: Noise 1 = only energy conservative

20 The Models FPU chains are mathematically very difficult to study. We perturb the Hamiltonian dynamics by a stochastic noise. These stochastic perturbations simulate (qualitatively) the long time (chaotic) effect of the deterministic nonlinear model. FPU chains conserve total energy H. If the system is unpinned (W = 0), it conserves also total momentum x p x. Two different noises: Noise 1 = only energy conservative Noise 2 = energy and momentum conservative

21 The Models The generator L (adjoint of the Fokker-Planck operator) has two terms L = A + γs

22 The Models The generator L (adjoint of the Fokker-Planck operator) has two terms L = A + γs A is the Liouville operator A = { H qx H } px x px qx

23 The Models The generator L (adjoint of the Fokker-Planck operator) has two terms L = A + γs A is the Liouville operator A = { H qx H } px x px qx S is a diffusion on the shell of constant kinetic energy (noise 1) or of constant kinetic energy and constant momentum (noise 2).

24 The Models The generator L (adjoint of the Fokker-Planck operator) has two terms L = A + γs A is the Liouville operator A = { H qx H } px x px qx S is a diffusion on the shell of constant kinetic energy (noise 1) or of constant kinetic energy and constant momentum (noise 2). γ > 0 regulates the strength of the noise.

25 Construction of the noise Example : Noise 1, energy conserving, m x = 1, d=1

26 Construction of the noise Example : Noise 1, energy conserving, m x = 1, d=1 For every nearest neigbor atoms x and x + 1, surface of constant kinetic energy e S 1 e = {(p x, p x+1 ) R 2 ; p 2 x + p 2 x+1 = e}

27 Construction of the noise Example : Noise 1, energy conserving, m x = 1, d=1 For every nearest neigbor atoms x and x + 1, surface of constant kinetic energy e S 1 e = {(p x, p x+1 ) R 2 ; p 2 x + p 2 x+1 = e} The following vector field X x,x+1 is tangent to S 1 e X x,x+1 = p x+1 px p x px+1 so X 2 x,x+1 generates a diffusion on S1 e (Brownion motion on the circle).

28 Construction of the noise Example : Noise 1, energy conserving, m x = 1, d=1 For every nearest neigbor atoms x and x + 1, surface of constant kinetic energy e S 1 e = {(p x, p x+1 ) R 2 ; p 2 x + p 2 x+1 = e} The following vector field X x,x+1 is tangent to S 1 e X x,x+1 = p x+1 px p x px+1 so Xx,x+1 2 generates a diffusion on S1 e (Brownion motion on the circle). We define S = 1 2 N 2 x=1 X 2 x,x+1

29 Construction of the noise Example : Noise 1, energy conserving, m x = 1, d=1 For every nearest neigbor atoms x and x + 1, surface of constant kinetic energy e S 1 e = {(p x, p x+1 ) R 2 ; p 2 x + p 2 x+1 = e} The following vector field X x,x+1 is tangent to S 1 e X x,x+1 = p x+1 px p x px+1 so Xx,x+1 2 generates a diffusion on S1 e (Brownion motion on the circle). We define S = 1 2 N 2 x=1 X 2 x,x+1 Noise 2, d 1... are of the same type.

30 Evaluation of the conductivity

31 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ).

32 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?.

33 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?. Problem : If T l T r, we don t know < > ss

34 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?. Problem : If T l T r, we don t know < > ss LINEAR RESPONSE THEORY (GREEN-KUBO):

35 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?. Problem : If T l T r, we don t know < > ss LINEAR RESPONSE THEORY (GREEN-KUBO): Non-rigorous perturbative arguments predict: lim lim κ N T l,t r T N is given by the Green-Kubo formula κ GK (T ) (space-time variance of the current at equilibrium)

36 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?. Problem : If T l T r, we don t know < > ss LINEAR RESPONSE THEORY (GREEN-KUBO): Non-rigorous perturbative arguments predict: lim lim κ N T l,t r T N is given by the Green-Kubo formula κ GK (T ) (space-time variance of the current at equilibrium) Advantage : κ GK requires to consider the dynamics at equilibrium. We know the equilibrium measure (Gibbs measure with temperature T ).

37 Evaluation of the conductivity NON EQUILIBRIUM SETTING : system in contact with two heat baths (T l T r ). κ N is of same order as N < J > ss. κ N N δ, δ =?. Problem : If T l T r, we don t know < > ss LINEAR RESPONSE THEORY (GREEN-KUBO): Non-rigorous perturbative arguments predict: lim lim κ N T l,t r T N is given by the Green-Kubo formula κ GK (T ) (space-time variance of the current at equilibrium) Advantage : κ GK requires to consider the dynamics at equilibrium. We know the equilibrium measure (Gibbs measure with temperature T ). Inconvenient : The quantity to evaluate is a dynamical quantity.

38 Green Kubo formula

39 Green Kubo formula Energy e x of atom x is (m x = 1 to simplify) e x = p2 x 2 + W (q x) + V (q x q y ) 2 2 y x

40 Green Kubo formula Energy e x of atom x is (m x = 1 to simplify) e x = p2 x 2 + W (q x) + V (q x q y ) 2 2 y x Local conservation of energy: d e x (t) e x (0) = (J x ek,x(t) J x,x+ek (t)) k=1

41 Green Kubo formula Energy e x of atom x is (m x = 1 to simplify) e x = p2 x 2 + W (q x) + V (q x q y ) 2 2 y x Local conservation of energy: d e x (t) e x (0) = (J x ek,x(t) J x,x+ek (t)) k=1 The current of energy J has the decomposition: J x,x+ek (t) = t 0 j x,x+ek (s)ds + M x,x+ek (t)

42 Green Kubo formula Energy e x of atom x is (m x = 1 to simplify) e x = p2 x 2 + W (q x) + V (q x q y ) 2 2 y x Local conservation of energy: d e x (t) e x (0) = (J x ek,x(t) J x,x+ek (t)) k=1 The current of energy J has the decomposition: J x,x+ek (t) = t 0 j x,x+ek (s)ds + M x,x+ek (t) with M martingale (mean 0 w.r.t. any initial condition) j x,x+ek = 1 2 ( V )(q x+e k q x ) (p x+ek + p x ) γ ek p 2 x

43 Consider the system of length N. Linear response theory predicts the following expression for the conductivity:

44 Consider the system of length N. Linear response theory predicts the following expression for the conductivity: [κ GK ] 1,1 (T ) = lim lim t N 2 1 2T 2 E eq. 1 J x,x+e1 (t) tn d x T d N E eq. is the expectation w.r.t. EQUILIBRIUM, meaning the uniform measure on the shell {H = x T d e x = N d T }. N

45 Consider the system of length N. Linear response theory predicts the following expression for the conductivity: [κ GK ] 1,1 (T ) = lim lim t N 2 1 2T 2 E eq. 1 J x,x+e1 (t) tn d x T d N E eq. is the expectation w.r.t. EQUILIBRIUM, meaning the uniform measure on the shell {H = x T d e x = N d T }. N If the conductivity is infinite, how to obtain the diverging order of the conductivity κ N of the system of length N?

46 Consider the system of length N. Linear response theory predicts the following expression for the conductivity: [κ GK ] 1,1 (T ) = lim lim t N 2 1 2T 2 E eq. 1 J x,x+e1 (t) tn d x T d N E eq. is the expectation w.r.t. EQUILIBRIUM, meaning the uniform measure on the shell {H = x T d e x = N d T }. N If the conductivity is infinite, how to obtain the diverging order of the conductivity κ N of the system of length N? κ N κ GK with truncation of the time up to time t N = N (phononic picture).

47 Consider the system of length N. Linear response theory predicts the following expression for the conductivity: [κ GK ] 1,1 (T ) = lim lim t N 2 1 2T 2 E eq. 1 J x,x+e1 (t) tn d x T d N

48 Consider the system of length N. Linear response theory predicts the following expression for the conductivity: [κ GK ] 1,1 (T ) = lim lim t N 2 1 2T 2 E eq. 1 J x,x+e1 (t) tn d The current of energy J has the decomposition: J x,x+ek (t) = with M martingale and t 0 x T d N j x,x+ek (s)ds + M x,x+ek (t) j x,x+ek = 1 2 ( V )(q x+e k q x ) (p x+ek + p x ) γ ek p 2 x

49 [κ GK ] 1,1 (T ) = lim lim t N 1 2T 2 N d t E eq [ x t 0 j x,x+e1 (s)ds ] 2 + γ d (Martingale contribution)

50 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 1: energy conserving

51 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Momentum is NOT conserved. Noise 1: energy conserving

52 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 1: energy conserving Momentum is NOT conserved. Homogenous chain : m x = 1

53 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 1: energy conserving Momentum is NOT conserved. Homogenous chain : m x = 1 Theorem (B., Olla, JSP 05) κ GK is finite (pinned or unpinned) in any dimension. Fourier s law holds and linear response theory is correct: System of length N in contact with two Langevin baths at temperature T l and T r in its steady state lim N < j x,x+1 > ss = κ GK (T r T l ) N

54 Results : Harmonic case, random masses

55 Results : Harmonic case, random masses Numerical simulations and rigorous results (Dahr, Lebowitz, O Connor, Rubin-Greer...), non-equilibrium setting, disordered chain (without noise)

56 Results : Harmonic case, random masses Numerical simulations and rigorous results (Dahr, Lebowitz, O Connor, Rubin-Greer...), non-equilibrium setting, disordered chain (without noise) Harmonic chain without pinning: κ N N δ with δ [ 1/2, 1/2] depending on the spectral properties of the baths!!!

57 Results : Harmonic case, random masses Numerical simulations and rigorous results (Dahr, Lebowitz, O Connor, Rubin-Greer...), non-equilibrium setting, disordered chain (without noise) Harmonic chain without pinning: κ N N δ with δ [ 1/2, 1/2] depending on the spectral properties of the baths!!! Harmonic chain with harmonic pinning: κ N e cn.

58 Results : Harmonic case, random masses Numerical simulations and rigorous results (Dahr, Lebowitz, O Connor, Rubin-Greer...), non-equilibrium setting, disordered chain (without noise) Harmonic chain without pinning: κ N N δ with δ [ 1/2, 1/2] depending on the spectral properties of the baths!!! Harmonic chain with harmonic pinning: κ N e cn. Anharmonic chain (with or without pinning): κ N = O(1).

59 Results : Harmonic case, random masses Numerical simulations and rigorous results (Dahr, Lebowitz, O Connor, Rubin-Greer...), non-equilibrium setting, disordered chain (without noise) Harmonic chain without pinning: κ N N δ with δ [ 1/2, 1/2] depending on the spectral properties of the baths!!! Harmonic chain with harmonic pinning: κ N e cn. Anharmonic chain (with or without pinning): κ N = O(1). Theorem (B., 08) Harmonic system with random masses and energy conservative noise 1.The conductivity defined by Green-Kubo formula is strictly positive and bounded above: 0 < c κ KG C + < +

60 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 2: energy/momentum conserving

61 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 2: energy/momentum conserving Theorem (Basile, B., Olla, PRL 06) ( ) C 1,1 (t) = lim N j x,x+e1 (t), j 0,e1 (0) C 1,1 (t) = T 2 4π 2 d x [0,1] d ( k 1ω(k)) 2 e tγψ(k) dk where ω is the dispertion relation of the harmonic chain ω(k) = (ν + 4α eq. d sin 2 (πk j )) 1/2, ψ(k) k 2 j=1

62 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 Noise 2: energy/momentum conserving Corollary C 1,1 (t) t d/2 in the unpinned case (ν = 0) C 1,1 (t) t d/2 1 in the pinned case (ν > 0)

63 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2

64 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 κ N = 1 2T 2 N d t N E eq [ x tn 0 j x,x+e1 (s)ds ] 2 + γ d, t N = N

65 Results : Harmonic case : V (r) = αr 2, W (q) = νq 2 κ N = 1 2T 2 N d t N E eq [ x tn 0 j x,x+e1 (s)ds ] 2 + γ d, t N = N Corollary If the system is unpinned (ν = 0) then truncated Green-Kubo formula for κ N gives: { κ N N 1/2 if d = 1 κ N log N if d = 2 In all other cases κ N is bounded in N and converges to κ GK.

66 Results : Anharmonic case, Canonical version of GK

67 Results : Anharmonic case, Canonical version of GK Theorem (Basile,B.,Olla 08) For d 3, if W > 0 is general or if W = 0 and 0 < c V C + < + then κ N C. For d = 2, if W = 0 and 0 < c V C + < κ N C(log N) 2. For d = 1, if W = 0 and 0 < c V C + <, then κ N C N. In any dimension, if V are quadratic and W > 0 is general then κ N C.

68 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ.

69 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ. FPU (α) : δ = 0.35 for γ small to δ = 0.48 for γ larger.

70 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ. FPU (α) : δ = 0.35 for γ small to δ = 0.48 for γ larger. FPU (β): δ = 0.41 for γ small to δ = 0.47 for γ larger.

71 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ. FPU (α) : δ = 0.35 for γ small to δ = 0.48 for γ larger. FPU (β): δ = 0.41 for γ small to δ = 0.47 for γ larger. Thermal conductivity increases with strength of the noise!!!

72 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ. FPU (α) : δ = 0.35 for γ small to δ = 0.48 for γ larger. FPU (β): δ = 0.41 for γ small to δ = 0.47 for γ larger. Thermal conductivity increases with strength of the noise!!! Theoretical arguments are controversial. For FPU (β):

73 Anharmonic case : Numerical simulations, (EPJ 08), Basile, Delfini, Lepri, Livi, Olla, Politi Simulations (d = 1) for unpinned systems with energy/momentum conservative noise 2. The strength of the noise is regulate by γ. Then κ N N δ. FPU (α) : δ = 0.35 for γ small to δ = 0.48 for γ larger. FPU (β): δ = 0.41 for γ small to δ = 0.47 for γ larger. Thermal conductivity increases with strength of the noise!!! Theoretical arguments are controversial. For FPU (β): Kinetic approach : δ = 2/5 (Perverzev, Lukkarinen and Spohn) MCT approach : δ = 1/2 (Delfini, Lepri,Livi, Politi)

74 Proof of the harmonic case

75 Proof of the harmonic case [ ( E eq. dte λt 0 x ) ] j x,x+e1 (t) j 0,e1 (0)

76 Proof of the harmonic case [ ( E eq. dte λt 0 x = 0 dte λt ( x ) ] j x,x+e1 (t) j 0,e1 (0) ) e tl j x,x+e1, j 0,e1 eq.

77 Proof of the harmonic case [ ( E eq. dte λt 0 x = 0 dte λt ( x ( = (λ L) 1 x ) ] j x,x+e1 (t) j 0,e1 (0) ) e tl j x,x+e1, j 0,e1 j x,x+e1 ), j 0,e1 eq. eq.

78 Proof of the harmonic case [ ( E eq. dte λt 0 x = 0 dte λt ( x ( = (λ L) 1 x ) ] j x,x+e1 (t) j 0,e1 (0) ) e tl j x,x+e1, j 0,e1 j x,x+e1 ), j 0,e1 eq. eq. Solve the resolvent equation (λ L)h λ = x j x,x+e 1 compute lim N < j 0,e1 h λ > eq.. and

79 Proof of the harmonic case [ ( E eq. dte λt 0 x = 0 = dte λt ( x (λ L) 1 ( x j x,x+e1 (t) e tl j x,x+e1 ) j x,x+e1 ) ) j 0,e1 (0), j 0,e1, j 0,e1 Solve the resolvent equation (λ L)h λ = x j x,x+e 1 and compute lim N < j 0,e1 h λ > eq.. Function h λ is local in the energy conservative case and non-local in the energy/momentum case. For d 3 or ν > 0, the decay of h λ is sufficient to assure a finite conductivity. ] eq. eq.

80 First order correction to local equilibrium What is h 0 = lim λ 0 h λ?

81 First order correction to local equilibrium What is h 0 = lim λ 0 h λ? Consider the system in contact with two reservoirs with temperature T l = β 1 l and T r = βr 1.

82 First order correction to local equilibrium What is h 0 = lim λ 0 h λ? Consider the system in contact with two reservoirs with temperature T l = β 1 l and T r = βr 1. If T l = T r = β 1, NESS < > ss is the Gibbs mesure Z 1 exp ( βh)

83 First order correction to local equilibrium What is h 0 = lim λ 0 h λ? Consider the system in contact with two reservoirs with temperature T l = β 1 l and T r = βr 1. If T l = T r = β 1, NESS < > ss is the Gibbs mesure Z 1 exp ( βh) Let f ss the density of < > ss w.r.t. the local equilibrium state < > le = Z 1 exp ( β(x/n)h), β(q) = β l + (β r β l )q

84 First order correction to local equilibrium What is h 0 = lim λ 0 h λ? Consider the system in contact with two reservoirs with temperature T l = β 1 l and T r = βr 1. If T l = T r = β 1, NESS < > ss is the Gibbs mesure Z 1 exp ( βh) Let f ss the density of < > ss w.r.t. the local equilibrium state < > le = Z 1 exp ( β(x/n)h), β(q) = β l + (β r β l )q If δt = T r T l is small f ss = 1 + δth 0 + o((δt ) 2 )

85 Anharmonic case : upper bounds

86 Anharmonic case : upper bounds E eq [ ] 2 N j x,x+e1 (s)ds 0 x

87 Anharmonic case : upper bounds E eq [ ] 2 N j x,x+e1 (s)ds ( ) ( 10N j x,x+e1, (N 1 L) 1 x x 0 x j x,x+e1 ) eq

88 Anharmonic case : upper bounds E eq [ ] 2 N j x,x+e1 (s)ds ( ) ( 10N j x,x+e1, (N 1 γs) 1 x x 0 x j x,x+e1 ) eq

89 Anharmonic case : upper bounds E eq [ ] 2 N j x,x+e1 (s)ds ( ) ( 10N j x,x+e1, (N 1 γs) 1 x x 0 x j x,x+e1 ) eq ( ) (N 1 γs) 1 j x,x+e1 = x d G N (x y)pxv j (q j y+e 1 qy j ) j=1 where G N (z) is the solution of the resolvent equation N 1 G N (z) 2γ( G N )(z) = 1 2 [δ 0(z) + δ e1 (z)] x,y