Gibbs measures associated to the integrals of motion of the periodic derivative nonlinear Schrödinger equation

Abstract

We study the one-dimensional periodic derivative nonlinear Schrödinger equation. This is known to be a completely integrable system, in the sense that there is an infinite sequence of formal integrals of motion ${\textstyle \int }h_k$, $k\in {\mathbb {Z}}_{+}$. In each ${\textstyle \int }h_{2k}$ the term with the highest regularity involves the Sobolev norm $\dot{H}^{k}({\mathbb {T}})$ of the solution of the DNLS equation. We show that a functional measure on $L^2({\mathbb {T}})$, absolutely continuous w.r.t. the Gaussian measure with covariance $({\mathbb {I}}+(-\varDelta )^{k})^{-1}$, is associated to each integral of motion ${\textstyle \int }h_{2k}$, $k\ge 1$.

Appendix: Gaussian measures in Sobolev spaces: a toolbox

We are here interested in giving a succinct but self-contained survey on the theory of Gaussian measures in Hilbert Sobolev spaces. For a complete treatment we refer to [2, 33].

1.1 Concentration of measure in $\dot{H}^{k}({\mathbb {T}})$

Here we study the concentration property of the Gaussian measure with covariance $({\mathbb {I}}+(-\varDelta )^{k})^{-1}$. The main feature is that the measure is concentrated on functions in $L^2({\mathbb {T}})$ having slightly less then $k-\frac{1}{2}$ weak derivatives as regularity. This is stated precisely in the following.

Proposition 6.1

For every $k\ge 0$ we have $\gamma _k(\bigcap \nolimits _{\varepsilon >0} \dot{H}^{k-\frac{1}{2}-\varepsilon })=1$.

We will proceed by steps. At first we prove

Lemma 6.2

$\gamma _k(\dot{H}^{k-\frac{1}{2}+\varepsilon })=0$ for every $\varepsilon \ge 0$.

Proof

We take any function $\varphi \in \dot{H}^{s}({\mathbb {T}})$ with $s\ge k-\frac{1}{2}$. We have that $\Vert \varphi _N\Vert _{\dot{H}^s}$ is finite uniformly in N, where we recall $\varphi _N$ is the projection on the Fourier modes $|n|\le N$ defined by (1.5) and (1.6). We show that for all $\lambda >0$

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^s}\le \lambda \right) \rightarrow 0,\quad {\text{ as }}\quad N\rightarrow \infty . \end{aligned}$$

To do so, we make use of the Markov inequality: For every $\mu >0$

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^s}\le \lambda \right)\le & {} e^{\frac{\mu \lambda }{2}} \int \prod _{|n|\le N}\left( \frac{1+n^{2k}}{\sqrt{2\pi }}\hbox {d}\varphi _n \hbox {d}\bar{\varphi }_n\right) e^{-\frac{1}{2}\sum _n\left( 1+n^{2k}\right) |\varphi _n|^2}e^{-\frac{\mu }{2}\sum _n n^{2s}|\varphi _n|^2}\\\le & {} e^{\frac{\mu \lambda }{2}} \int \frac{1}{(2\pi )^{2N+1}}\hbox {d}\varphi ^{\prime }_n \hbox {d}\bar{\varphi }^{\prime }_n e^{-\frac{1}{2}\sum _n|\varphi ^{\prime }_n|^2}e^{-\mu \sum _n n^{2(s-k)}|\varphi ^{\prime }_n|^2}\\= & {} \exp \left[ \frac{\mu \lambda }{2}-\frac{1}{2}\sum _{\begin{array}{c} |n|\le N, \\ n\ne 0 \end{array}}\ln \left( 1+\frac{\mu }{|n|^{\kappa }}\right) \right] , \end{aligned}$$

where we have performed the change of variables $\varphi ^{\prime }_n=\sqrt{1+n^{2k}}\varphi _n$ and set $-2(k-s)=:\kappa$. Let us first consider negative $\kappa$. In this case

$$\begin{aligned} \sum _{\begin{array}{c} |n|\le N, \\ n\ne 0 \end{array}}\ln \left( 1+\frac{\mu }{|n|^{\kappa }}\right) \ge 2N\ln (1+\mu ), \end{aligned}$$

and so we have an exponential decay in N for every choice of positive $\mu$:

$$\begin{aligned} \gamma _p\left( \Vert \varphi _N\Vert _{\dot{H}^{s}}\le \lambda \right) \lesssim e^{\frac{\mu \lambda }{2}} e^{-2N \ln (1+\mu )},\quad (s>k). \end{aligned}$$

(6.1)

For $\kappa \in [0,1)$ the series $\sum \nolimits _n\ln \left( 1+\frac{\mu }{n^{\kappa }}\right)$ diverges as $N^{1-\kappa }$. Hence,

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^{s}}\le \lambda \right) \lesssim e^{-\mu N^{1-2(k-s)}},\quad \left( k\ge s>k-\frac{1}{2}\right) . \end{aligned}$$

(6.2)

Finally, for $\kappa =1$ we have a logarithmic divergence at exponent and therefore

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^{s}}\le \lambda \right) \le \left( \frac{e^{\frac{\lambda }{2}}}{N}\right) ^{\mu },\quad \left( s=k-\frac{1}{2}\right) , \end{aligned}$$

(6.3)

for arbitrary $\mu >0$. We obtain the statement by taking $N\rightarrow \infty$ in (6.1)–(6.3). $\square$

Remark 6.3

The same strategy can be also used to show the stronger statement

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^s}\le \ln N \right) \rightarrow 0{ \text{ as } }N\rightarrow \infty ,\qquad \left( s\ge k-\frac{1}{2}\right) . \end{aligned}$$

Lemma 6.4

We have that for every $s<k-\frac{1}{2}$ and $\lambda >0$

$$\begin{aligned} \gamma _k\left( \Vert \varphi \Vert _{\dot{H}^s}\ge \lambda \right) \lesssim e^{-\lambda {/}4}. \end{aligned}$$

(6.4)

Proof

Let us take a function $\varphi \in \dot{H}^{s}$ for some $s<k-\frac{1}{2}$. We look at its truncation $\varphi _N$ and again it is $\Vert \varphi _N\Vert _{\dot{H}^s}$ finite uniformly in N. We exploit the reverse Chernoff bound at finite N: For every $\mu \in (0,1)$ and $\lambda >0$, we get

$$\begin{aligned} \gamma _k\left( \Vert \varphi _N\Vert _{\dot{H}^s}\ge \lambda \right)\le & {} e^{-\frac{\mu \lambda }{2}} \int \prod _{|n|\le N}\left( \frac{1+n^{2k}}{\sqrt{2\pi }}\hbox {d}\varphi _n \hbox {d}\bar{\varphi }_n\right) e^{-\frac{1}{2}\sum \nolimits _n(1+n^{2k})|\varphi _n|^2}e^{\frac{\mu }{2}\sum _n n^{2s}|\varphi _n|^2}\\\le & {} e^{-\frac{\mu \lambda }{2}} \int \frac{1}{(2\pi )^{2N+1}}\hbox {d}\varphi ^{\prime }_n \hbox {d}\bar{\varphi }^{\prime }_n e^{-\frac{1}{2}\sum _n\left| \varphi ^{\prime }_n\right| ^2}e^{\mu \sum _n n^{2(s-k)}|\varphi _n|^2}\\= & {} \exp \left[ -\frac{\mu \lambda }{2}-\frac{1}{2}\sum _{\begin{array}{c} |n|\le N, \\ n\ne 0 \end{array}}\ln \left( 1-\frac{\mu }{|n|^{\kappa }}\right) \right] , \end{aligned}$$

where again we have used the same change of variables as before. Note that now it is $\kappa >1$. Since $\frac{1}{2}\sum \nolimits _n\ln (1-\frac{\mu }{n^{\kappa }})$ is convergent for all $\mu <1$ and $\kappa >1$, we can choose $\mu \in (0,1)$ and take the limit $N\rightarrow \infty$. We get (6.4) by setting $\mu =1{/}2$. $\square$

Equation (6.4) implies that $\Vert u\Vert _{H^{q}}$ is bounded with probability 1 for every $k<s-\frac{1}{2}$. This is sufficient to complete the proof of Proposition 6.1.

1.2 Quadratic forms

Then we present some results about quadratic forms of Gaussian random variables, used in the paper.

Proposition 6.5

Let $k \ge 2$ and Q be a $(2N+1)\times (2N+1)$ matrix such that

$$\begin{aligned} \sup _{l,h } \frac{|Q_{l h}|}{\sqrt{1+h^{2k}}} =: T_k < + \infty \end{aligned}$$

(6.5)

Then for $\lambda >0$

$$\begin{aligned} \gamma _k\left( (\varphi ,Q\varphi )\ge \lambda \right) \lesssim e^{-\lambda {/}4T_k}. \end{aligned}$$

(6.6)

Proof

To begin with, we exploit the Markov inequality: For any $\mu >0$

$$\begin{aligned} \gamma _k\left( (\varphi ,Q\varphi )\ge \lambda \right) \le e^{-\mu \lambda }\mathbb {E}{e^{\mu (\varphi ,Q\varphi )}}. \end{aligned}$$

(6.7)

Now we compute

$$\begin{aligned} \mathbb {E}{e^{\mu (\varphi , Q \varphi )}}= & {} \int \prod _{|n|\le N}\left( \frac{1+n^{2k}}{\sqrt{2\pi }}\hbox {d}\varphi _n \hbox {d}\bar{\varphi }_n\right) \exp \left[ -\frac{1}{2}\sum _{i,j}\bar{\varphi }_i \left( (1+j^{2k})\delta _{ij}-2\mu Q_{ij}\right) \varphi _j\right] \nonumber \\= & {} \frac{1}{(2\pi )^{N}} \int \hbox {d}\varphi ^{\prime }_{-N}\ldots \hbox {d}\varphi ^{\prime }_N\bar{\varphi }^{\prime }_{-N}\ldots \hbox {d}\bar{\varphi }^{\prime }_N\exp \left[ -\frac{1}{2}\sum _{i,j}\bar{\varphi }^{\prime }_i\left( \delta _{ij}-2\mu Q_{ij}(k)\right) \varphi ^{\prime }_j\right] \nonumber \\= & {} e^{-\frac{1}{2}\ln \det ({\mathbb {I}}-2\mu Q(k))}. \end{aligned}$$

(6.8)

where we have performed the change of variables $\varphi ^{\prime }_j=\sqrt{1+j^{2k}}\varphi _j$, $\bar{\varphi }^{\prime }_j=\sqrt{1+j^{2k}}\bar{\varphi }_j$ and we have introduced $Q_{ij}(k):=Q_{ij}{/}\sqrt{(1+j^{2k})(1+i^{2k})}$. We claim that

$$\begin{aligned} | {{\mathrm{Tr}}}((Q_{ij}(k))^{m}) | \lesssim T_{k}^{m}, \quad m \in {\mathbb {Z}}_{+}, \end{aligned}$$

(6.9)

so the expansion of the determinant

$$\begin{aligned} -\ln \det ({\mathbb {I}} - 2 \mu Q(k)) = \sum _{m=1}^{+\infty } \frac{(2 \mu )^{m} {{\mathrm{Tr}}}\left( ( Q_{ij}(k))^{m}\right) }{m} \end{aligned}$$

is convergent provided that $\mu < \frac{1}{2 T_{k}}$. We choose $\mu = \frac{1}{4T_{k}}$, so that (6.7, 6.8) imply the desired inequality. It remains to show the (6.9).

$$\begin{aligned} {{\mathrm{Tr}}}\left( (Q(k) )^{m}\right)= & {} \sum _{i_{1}, \ldots , i_{m+1}} Q_{i_{1} i_{2}}(k) \ldots Q_{i_{m} i_{m+1}}(k) \delta _{i_{1} i_{m+1}} \\= & {} \sum _{i_{1}, \ldots , i_{m+1}} \frac{ Q_{i_{1} i_{2}} \ldots Q_{i_{m} i_{m+1}} \delta _{i_{1} i_{m+1}} }{ \sqrt{ \left( 1+i_{1}^{2k}\right) \left( 1+i_{2}^{2k}\right) \left( 1+i_{2}^{2k}\right) \ldots \left( 1+i_{m}^{2k}\right) \left( 1+i_{m}^{2k}\right) \left( 1+i_{m+1}^{2k}\right) } } \\\le & {} T_{k}^{m}\sum _{i_{1},\ldots , i_{m+1}} \frac{ \delta _{i_{1}, i_{m+1}} }{ \sqrt{ \left( 1+i_{1}^{2k}\right) \ldots \left( 1+i_{m}^{2k}\right) } } \\= & {} T_{k}^{m}\sum _{i_{1}, \ldots , i_{m}} \frac{ 1 }{ \sqrt{\left( 1+i_{1}^{2k}\right) \ldots \left( 1+i_{m}^{2k}\right) } } \\= & {} T_{k}^{m}\left( \sum _{i} \frac{ 1 }{ \sqrt{ \left( 1+i^{2k}\right) } } \right) ^{m} \lesssim T_{k}^{m}, \end{aligned}$$

where we have used the assumption (6.5) in the first inequality and $k \ge 2$ in the last inequality. $\square$

Remark 6.6

We observe that we can make different assumptions on the matrix Q and obtain similar inequalities. For instance, if the trace norm of Q(k) is finite uniformly in N, we have (see, for instance, Lemma 3.3 in [35])

$$\begin{aligned} -\ln \det \left( {\mathbb {I}} - 2 \mu Q(k)\right) \le \Vert Q(k)\Vert _{{{\mathrm{Tr}}}} \end{aligned}$$

and so for every N

$$\begin{aligned} \gamma _k\left( (\varphi , Q \varphi )\ge \lambda \right) \lesssim e^{-\lambda {/}\Vert Q(k) \Vert _{{{\mathrm{Tr}}}}}, \end{aligned}$$

(6.10)

by the same argument of the last proposition. If we assume the Hilbert-Schmidt norm of Q(k) to be finite uniformly in N, we obtain the Hanson–Wright inequality (see [16] and more recently [32]), holding for any N

$$\begin{aligned} \gamma _k\left( {{\mathrm{Var}}}(\varphi ,Q\varphi )\ge \lambda ^2\right) \lesssim e^{-c\min \left( \lambda ^2{/}\Vert Q(k)\Vert _{HS}, \lambda {/}\Vert Q(k)\Vert \right) }, \end{aligned}$$

(6.11)

where $\Vert A\Vert$ denotes the operator norm of A and c is a positive constant.

Remark 6.7

For any linear operator $A\varphi :=\sum \nolimits _{n=1}^N a_i \varphi _i$, with

$$\begin{aligned} {\mathcal {T}}_k:=\sum _{|i|\le N} \frac{|a_i|^2}{(1+i^{2k})}<\infty \quad {\text{ uniformly } \text{ in }}\,N, \end{aligned}$$

by using $\gamma _k(|A\varphi |\ge \lambda )=\gamma _k(|A\varphi |^2\ge \lambda ^2)$ we can infer

$$\begin{aligned} \gamma _k\left( |A\varphi |\ge \lambda \right) \lesssim e^{-\lambda ^2{/}{\mathcal {T}}_k}. \end{aligned}$$

(6.12)

Note that if $A\varphi =\varphi ^{(s)}$ we have ${\mathcal {T}}_k<\infty$ uniformly in N for $s<k-\frac{1}{2}$. In this way we can improve Lemma 6.4, obtaining a sub-Gaussian decay.

Proposition 6.8

Let Q(x) be a $N\times N$ matrix as before. Moreover, we assume Q(x) to be Hölder continuous w.r.t. $x\in {\mathbb {T}}$ with exponent $\alpha$ and constant $L_N$, i.e.,

$$\begin{aligned} \left| (\varphi ,Q(x)\varphi )-(\varphi ,Q(y)\varphi )\right| \le L_N|x-y|^\alpha , \quad \mathrm{for\,every}\,\varphi \in {\mathbb {R}}^N. \end{aligned}$$

(6.13)

Then for any $\varepsilon >0$ and $\lambda \ge 2L_N \varepsilon ^\alpha$

$$\begin{aligned} \gamma _k\left( \sup _{x}(\varphi ,Q\varphi )\ge \lambda \right) \lesssim \frac{e^{-\lambda {/}4T_k}}{\varepsilon }. \end{aligned}$$

(6.14)

Proof

We exploit Proposition 6.5 along with an $\varepsilon$-net argument. For $\varepsilon >0$ we divide the interval ${\mathbb {T}}$ in $1{/}\varepsilon$ points at distance $\varepsilon$. We denote by $x_j$ a point in the jth segment, and by $x^*$ the point in which the maximum is attained. By Proposition 6.5 for each $x\in {\mathbb {T}}$ we obtain for $\lambda >0$

$$\begin{aligned} \gamma _k\left( (\varphi ,Q(x)\varphi )\ge \lambda \right) \lesssim e^{-\lambda {/}4T_k}. \end{aligned}$$

(6.15)

Let $j_0$ be such that $|x_{j_0}-x^*|\le \varepsilon$. Therefore, it has to be

$$\begin{aligned} \left| \left( \varphi ,Q\left( x^*\right) \varphi \right) -\left( \varphi ,Q\left( x_{j_0}\right) \varphi \right) \right| \le L_N\varepsilon ^\alpha , \quad \text{ for } \text{ every } \,\varphi \in {\mathbb {R}}^N. \end{aligned}$$

Then we use the union bound for the probabilities:

$$\begin{aligned} \gamma _k\left( \left| Q\left( x^*\right) \right| \ge \lambda \right)\le & {} \sum _j \gamma _k\left( \left| Q\left( x^*\right) \right| \ge \lambda \, \Big |\, |x^*-x_j|\le \varepsilon \right) \\\le & {} \sum _j \gamma _k\left( \left| Q(x_j)\right| \ge \frac{\lambda }{2}\right) \\&+\,\sum _j \gamma _k\left( \left| Q(x_j)-Q\left( x^*\right) \right| \ge \frac{\lambda }{2}\Big | \left| x^*-x_j\right| \le \varepsilon \right) . \end{aligned}$$

We immediately see by (6.13) that the second addendum in the last formula is zero as soon as $\lambda \ge 2L_N \varepsilon ^\alpha$. Therefore, we bound the first addendum by the total number of terms in the sum, which is $\varepsilon ^{-1}$, times the estimate (6.15), so obtaining (6.14). $\square$