Logarithmic Sobolev inequalities and spectral concentration for the cubic Schrödinger equation

Abstract

The nonlinear Schrödinger equation NLSE(p, β), −iut = −uxx + β|u| p−2 u = 0, arises from a Hamiltonian on infinite-dimensional phase space L 2 (T). For p ≤ 6, Bourgain (Comm. Math. Phys. 166 (1994), 1-26) has shown that there exists a Gibbs measure µL 2 ≤ N } in phase space such that the Cauchy problem for NLSE(p, β) is well posed on the support of µ β N , and that µ β N is invariant under the flow. This paper shows that µ β N satisfies a logarithmic Sobolev inequality for the focussing case β < 0 and 2 ≤ p ≤ 4 on Ω N for all N > 0; also µ β satisfies a restricted LSI for 4 ≤ p ≤ 6 on compact subsets of Ω N determined by Hölder norms. Hence for p = 4, the spectral data of the periodic Dirac operator in L 2 (T; C 2 ) with random potential φ subject to µ β N are concentrated near to their mean values. The paper concludes with a similar result for the spectral data of Hill's equation when the potential is random and subject to the Gibbs measure of KdV.

Quantitative bounds on the rate of approach to equilibrium for some one-dimensional stochastic nonlinear Schrödinger equations

“…The Holley-Stroock Lemma has been used for related models by Gordon Blower [1]; see also [2]. Combining this Theorem with the results of Caffarelli, Feyel andÜstünel we are able to carry out a convexity comparison directly in the infinite-dimensional setting and to avoid sharp cut-offs or finite-dimensional approximations.…”

“…We define P n to be the projector onto the span of the functions {e −i2πkx/L : −n ≤ k ≤ n} in L 2 . In what follows a decomposition into low-frequency and high-frequency modes is crucial.…”

Hill's spectral curves and the invariant measure of the periodic KdV equation

“…Proof. By Lemma 4.2, F n : Ω N → R is Lipschitz with constant κ 2 C(N )n 4 . By Corollary 2 of [3], any Lipschitz function on (Ω N , ν β N ) satisfies a Gaussian concentration of measure inequality as in (4.11).…”

“…are Lipschitz continuous on Ω N . As in [4], we introduce the circles C(n 2 /4, 1/4) and apply Cauchy's integral formula to obtain…”