Wave Turbulence and Complete Integrability

Abstract

The most famous completely integrable PDEs—KdV and 1D cubic NLS—do not display any kind of wave turbulence, since the conservation laws control high regularity, making impossible the long time appearance of small scales. In this course, I will discuss a new type of integrable infinite-dimensional system, posed on functions of one variable, allowing dramatic growth of high Sobolev norms. The analysis is connected to the solution of an inverse spectral problem for Hankel operators from classical analysis. I will also discuss how this phenomenon can be exported to some Hamiltonian PDEs which can be seen as perturbations of this new type of integrable system.

Appendix A: The L ∞ Estimate and Its Consequences

In this section, we show how the lax pair structure leads to the following a priori estimate for solutions of the cubic Szegő equation.⁴

Theorem A.15 ([2])

Assume $u_0\in H^s_+$ for some s > 1. Then the corresponding solution u of the cubic Szegő equation satisfies

$$\displaystyle \begin{aligned}\sup_{t\in {\mathbb R}}\Vert u(t)\Vert _{L^\infty}<+\infty \ .\end{aligned}$$

The proof of this theorem relies on the following proposition.

Proposition A.1

Given $u\in H^{\frac 12}_+$, we denote by {s _j(u)}_j≥1 the sequence of singular values of H _u, repeated according to multiplicity. The following double inequality holds:

$$\displaystyle \begin{aligned}\frac 12\sum _{n=0}^\infty \vert \hat u(n)\vert \le \sum _{j=1}^\infty s_j(u)\le \sum _{n=0}^\infty \left (\sum _{\ell =0}^\infty \vert \hat u(n+\ell )\vert ^2\right )^{\frac 12}\end{aligned}$$

Furthermore, the right hand side is controlled by

$$\displaystyle \begin{aligned} C_s\Vert u\Vert _{H^s}\end{aligned}$$

for every s > 1.

Remark A.3

This proposition can be interpreted as a double inequality for the trace norm of the Hankel operator H _u. A more complete characterization of functions u such that the trace norm of H _u is finite is given in Peller [13]. Here we provide an elementary proof.

Assuming the proposition, let us show the theorem. From the second equality in the proposition, we have

$$\displaystyle \begin{aligned}\sum_{j=1}^\infty s_j(u_0)<+\infty .\end{aligned}$$

Since each s _j(u) is a conservation law, we have, for every $t\in {\mathbb R}$,

$$\displaystyle \begin{aligned}\sum_{j=1}^\infty s_j(u(t))=\sum_{j=1}^\infty s_j(u_0).\end{aligned}$$

Finally, by the first inequality in the proposition,

$$\displaystyle \begin{aligned}\sup_{t\in {\mathbb R}}\sum_{n\ge 0}\vert \hat u(t,n)\vert \le 2\ \sup_{t\in{\mathbb R}}\sum_{j=1}^\infty s_j(u(t))=2\sum_{j=1}^\infty s_j(u_0).\end{aligned}$$

The proof is completed by the elementary observation that

$$\displaystyle \begin{aligned}\Vert u\Vert_{L^\infty}\le \sum_{n\ge0}\vert \hat u(n)\vert .\end{aligned}$$

We now pass to the proof of the proposition.

Proof of Proposition A.1

We denote by {e _j}_j≥1 an orthonormal basis of $\overline { \operatorname {\mathrm {Ran}}(H_u)}=\overline { \operatorname {\mathrm {Ran}}(H_u^2)}$ such that $H_u^2e_j=s_j^2u$. Such a basis exists because $H_u^2$ is a compact selfadjoint operator. Notice that

$$\displaystyle \begin{aligned}\Vert H_u(e_j)\Vert ^2=(H_u^2(e_j)\vert e_j)=s_j^2.\end{aligned}$$

We set, for every n ≥ 0, ε _n(x) = e ^inx.

Let us prove the first inequality. Observe that

$$\displaystyle \begin{aligned}\hat u(2n)=(H_u(\varepsilon _n)\vert \varepsilon _n),\ \hat u(2n+1)=K_u(\varepsilon _n\vert \varepsilon_n).\end{aligned}$$

On the other hand,

$$\displaystyle \begin{aligned}(H_u(\varepsilon _n)\vert \varepsilon_n)=\sum _j(H_u(\varepsilon_n)\vert e_j)(e_j\vert \varepsilon_n),\end{aligned}$$

and (H _u(ε _n)|e _j) = (H _u(e _j)|ε _n), so that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{n=0}^\infty \vert \hat u(2n)\vert &\displaystyle \le &\displaystyle \sum_{n,j}\vert (H_u(e_j)\vert \varepsilon_n)(e_j\vert \varepsilon_n)\vert \\ &\displaystyle \le &\displaystyle \sum_j\left (\sum_{n=0}^\infty \vert (H_u(e_j)\vert \varepsilon_n)\vert^2 \right )^{\frac 12}\left (\sum_{n=0}^\infty \vert (e_j\vert \varepsilon_n)\vert^2 \right )^{\frac 12}\\ &\displaystyle =&\displaystyle \sum_j\Vert H_u(e_j)\Vert \Vert e_j\Vert =\sum_js_j . \end{array} \end{aligned} $$

Arguing similarly with K _u, we obtain

$$\displaystyle \begin{aligned}\sum_{n=0}^\infty \vert \hat u(2n+1)\vert \le \sum_ks^{\prime}_k .\end{aligned}$$

Summing up, we have proved

$$\displaystyle \begin{aligned}\sum _{n=0}^\infty \vert \hat u(n)\vert \leq \sum_js_j+\sum_ks^{\prime}_k\le 2\sum_js_j .\end{aligned}$$

We now pass to the second inequality. Notice that

$$\displaystyle \begin{aligned}(H_u(e _j), H_u(e _{j'}))=(H_u^2(e _{j'}), e _j)=s _{j}^2\delta _{jj'}.\end{aligned}$$

In other words, the sequence {H _u(e _j)∕s _j} is orthonormal. We then define the following antilinear operator on $L^2_+$,

$$\displaystyle \begin{aligned}\Omega _u(h)=\sum _j (e _j, h)\frac{H_u(e_j)}{s_j}.\end{aligned}$$

Notice that, due to the orthonormality of both systems {e _j} and {H _u(e _j)∕s _j}, ∥ Ω_u(h)∥≤∥h∥. Similarly, we define

$$\displaystyle \begin{aligned}^t\Omega _u(h)=\sum _j \frac{(H_u(e _j), h)}{s_j}e_j,\end{aligned}$$

so that

$$\displaystyle \begin{aligned}\forall h,h'\in L^2_+,\quad (\Omega _u(h)\vert h')=({}^t\Omega _u(h')\vert h).\end{aligned}$$

We next observe that

$$\displaystyle \begin{aligned} \begin{array}{rcl} s_j&\displaystyle =&\displaystyle (H_u(e_j)\vert \Omega _u(e_j))=\sum_{n=0}^\infty (H_u(e_j)\vert \varepsilon_n)(\varepsilon_n\vert \Omega _u(e_j))\\ &\displaystyle =&\displaystyle \sum_{n=0}^\infty \sum_{\ell =0}^\infty \hat u(n+\ell)(\varepsilon_\ell \vert e_j)(\varepsilon_n\vert \Omega _u(e_j))\\ \end{array} \end{aligned} $$

But using the transpose of Ω_u, we get

$$\displaystyle \begin{aligned}\sum_j(\varepsilon_\ell\vert e_j)(\varepsilon_n\vert \Omega_u(e_j))=\sum_j(\varepsilon_\ell\vert e_j)(e_j\vert ^t\Omega _u(\varepsilon_n))=(\varepsilon_\ell \vert ^t\Omega_u(\varepsilon_n))=(\varepsilon_n \vert \Omega _u(\varepsilon_\ell)).\end{aligned}$$

Consequently,

$$\displaystyle \begin{aligned}\sum_js_j=\sum_{n,\ell\ge 0}\hat u(n+\ell)(\varepsilon_n \vert \Omega _u(\varepsilon_\ell)).\end{aligned}$$

Applying the Cauchy–Schwarz inequality to the sum on ℓ, we infer

$$\displaystyle \begin{aligned}\sum_js_j\le \sum _{n=0}^\infty \Vert \Omega _u(\varepsilon _\ell)\Vert \left (\sum _{\ell =0}^\infty \vert \hat u(k+\ell )\vert ^2\right )^{\frac 12},\end{aligned}$$

and the claim follows from the fact that ∥ Ω_u(ε _ℓ)∥≤∥ε _ℓ∥ = 1.

We finally need to control the right hand side of this last inequality. By the Cauchy–Schwarz inequality in the n sum, we have, for every s > 1,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum _{n=0}^\infty \left (\sum _{\ell =0}^\infty \vert \hat u(n+\ell )\vert ^2\right )^{\frac 12}&\displaystyle \le &\displaystyle \left (\sum _{n=0}^\infty (1+n)^{1-2s}\right )^{\frac 12} \left (\sum _{n,\ell \ge 0}(1+n)^{2s-1}\vert \hat u(n+\ell )\vert ^2\right )^{\frac 12} \\ &\displaystyle \le &\displaystyle \left (\frac{s}{s-1}\right )^{\frac 12} \left (\sum _{n,\ell \ge 0}(1+n+\ell )^{2s-1}\vert \hat u(n+\ell )\vert ^2\right )^{\frac 12}\\ &\displaystyle \le &\displaystyle C_s\Vert u\Vert_{H^s}, \end{array} \end{aligned} $$

and Proposition A.1 is proved. □

Corollary A.2

For s > 1 and $u_0\in H_+^s$, the corresponding solution t↦u(t) of the Szegő equation ⁵ satisfies

$$\displaystyle \begin{aligned} \|u(t)\|{}_{H^s}\leq C_se^{C^{\prime}_s|t|},\quad \forall t\in\mathbb{R},\end{aligned}$$

where C _s, $C^{\prime }_s$ are positive constants which only depend on s and $\|u_0\|{ }_{H^s}$.

Proof

We compute $\frac {d}{dt}\|D^su(t)\|{ }_{L^2}^2$ and use the boundedness of the L ^∞ norm to write

$$\displaystyle \begin{aligned} \left| \frac{d}{dt}\|D^su(t)\|{}_{L^2}^2\right| \leq C \|u\|{}_{H^s}^2,\end{aligned}$$

for an appropriate constant C depending on the norm of u ₀ in $H^s_+$. A Gronwall inequality then completes the proof of the corollary. □