A KAM Theorem for the Time Quasi-periodic Reversible Perturbations of Linear Wave Equations Beyond Brjuno Conditions

Abstract

This paper is concerned with the existence of quasi-periodic response solutions (i.e., solutions that are quasi-periodic with the same frequencies as forcing term) for a class of forced reversible wave equations with derivative nonlinearity. The forcing frequency $\omega \in \mathbb {R}^2$ would be Liouvillean which is weaker than the usual Diophantine and Brjuno conditions. The derivative nonlinearity in the equation also leads to some difficulty in measure estimate. To overcome it, we also use the Töplitz–Lipschitz property of vector field. The proof is based on an infinite dimensional Kolmogorov–Arnold–Moser theorem for reversible systems.

Appendix.  Proofs of Some Technical Lemmas

1.1 Proof of Inequality (5.25) in Proposition 5.1

Proof

Below to estimate $\Vert A^{-1}_{ij}\Vert _{\mathcal {O}}$, we first estimate

$$\begin{aligned} \frac{|\frac{\partial }{\partial \xi }(\Omega _{ij}+[B_{ij}])|}{(\langle k, \omega \rangle -(\Omega _{ij}+[B_{ij}])^2}. \end{aligned}$$

For $i\ne j,$ on one hand, since $\Omega _{j}(\xi )=d(\xi )j+\widetilde{\Omega }_{j}(\xi ),$

$$\begin{aligned}{} & {} |\partial _\xi \Omega _{ij}|= |\partial _\xi d(i-j)+\partial _\xi (\widetilde{\Omega }_{i}-\widetilde{\Omega }_{j})|\le (A_2+2|\widetilde{\Omega }|_{\mathcal {O}})|i-j|.\\{} & {} |\partial _\xi [B_{ij}]| \le 2\Vert B\Vert _{\infty ,D(r)\times \mathcal {O}} \le 2\zeta |i-j|. \end{aligned}$$

Then

$$\begin{aligned} |\partial _\xi (\Omega _{ij}+[B_{ij}])|\le (A_2+\frac{3A_1}{4})|i-j|. \end{aligned}$$

On the other hand,

$$\begin{aligned} |\partial _\xi (\Omega _{ij}+[B_{ij}])|\ge |\partial _\xi \Omega _{ij}|-|\partial _\xi [B_{ij}]| \ge \frac{A_1}{4}|i-j|, \end{aligned}$$

thus

$$\begin{aligned} \frac{A_1}{4}|i-j|\le |\partial _\xi (\Omega _{ij}+[B_{ij}])|\le (A_2+\frac{3A_1}{4})|i-j|. \end{aligned}$$

Note that

$$\begin{aligned} \begin{aligned} |\langle k,\omega \rangle -(\Omega _{ij}+[B_{ij}])|&\ge |\Omega _{ij}|-|[B_{ij}]|-|\langle k,\omega \rangle |\\&\ge (A_0-\frac{3A_1}{4})|i-j|-K. \end{aligned} \end{aligned}$$

Then when $|i-j|> \mathcal {C}K,$

$$\begin{aligned} \frac{|\partial _\xi (\Omega _{ij}+[B_{ij}])}{(\langle k, \omega \rangle -(\Omega _{ij}+[B_{ij}]))^2} \le \frac{(A_2+\frac{3A_1}{4})}{\mathcal {C}K}. \end{aligned}$$

When $|i-j|\le \mathcal {C}K,$ due to $\Omega (\xi )+[B]\in \mathcal{M}\mathcal{C}_\omega (\gamma , \tau , K, \mathcal {O}),$

$$\begin{aligned} \frac{|\partial _\xi (\Omega _{ij}+[B_{ij}])}{(\langle k, \omega \rangle -(\Omega _{ij}+[B_{ij}]))^2} \le \frac{(A_2+\frac{3A_1}{4})\mathcal {C}K^{2\tau +1}}{\gamma ^{2}}. \end{aligned}$$

Then for all $i\ne j,$

$$\begin{aligned} \frac{|\partial _\xi (\Omega _{ij}+[B_{ij}])|}{(\langle k, \omega \rangle -(\Omega _{ij}+[B_{ij}]))^2}\le \frac{(A_2+\frac{3A_1}{4})\mathcal {C}K^{2\tau +1}}{\gamma ^{2}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&\Vert A^{-1}_{ij}\Vert _{\mathcal {O}}\\&\quad =\max \limits _{|k|\le K}\sup \limits _{\xi \in \mathcal {O}}\left( \frac{1}{|\langle k, \omega \rangle -(\Omega _{ij}(\xi )+[B_{ij}(\theta )])|}+\frac{|\frac{\partial }{\partial \xi }(\Omega _{ij}+[B_{ij}])|}{(\langle k, \omega \rangle -(\Omega _{ij}+[B_{ij}])^2}\right) \\&\quad \le \frac{(1+A_2+\frac{3A_1}{4})\mathcal {C}K^{2\tau +1}}{\gamma ^{2}} =:\frac{C_0K^{2\tau +1}}{\gamma ^{2}}. \end{aligned} \end{aligned}$$

$\square$

1.2 Proof of Lemma 6.1

Proof

• (1) In fact, by Lemma 3.1, one has $Q_{n+1}\ge Q_{n}^{\mathcal {A}}.$

  $$\begin{aligned}{} & {} \tilde{r}_{m}\le \tilde{r}_{0}=2r_{+},\\{} & {} (r-\tilde{r}_{m})^3\ge (r-2r_{+})^3\ge (\frac{r_{0}}{8Q_{n}^{4}})^3, \end{aligned}$$

  and

  $$\begin{aligned} \begin{aligned} \frac{360\tilde{r}_{m}Q_{n+1}\zeta }{(r-\tilde{r}_{m})^3}&\le 360\tilde{r}_{m}Q_{n+1}\zeta (\frac{8Q_{n}^{4}}{r_0})^3\\&\le 360\frac{2r_0}{4Q_{n+1}^{4}}Q_{n+1}2\varepsilon _{0}^{\frac{1}{2}} \frac{512Q_{n}^{12}}{r_{0}^{3}}\\&=C\frac{Q_{n}^{12}\varepsilon _{0}^{\frac{1}{2}}}{Q_{n+1}^{3}r_{0}^{2}} \le 1. \end{aligned} \end{aligned}$$

  (9.1)

  Thus we have our conclusion $360\tilde{r}_{m}Q_{n+1}\zeta \le (r-\tilde{r}_{m})^3$.

• (2) From Lemma 3.1, we have $Q_{n+1}\ge Q_{n}^{\mathcal {A}}$ and $\ln Q_{n+1}\le Q^U_n.$ Using these and (6.7),

  $$\begin{aligned} \frac{256}{(r-\tilde{r}_m)^2}\textrm{e}^{-\frac{r-\tilde{r}_m}{2}Q_{n+1}}\zeta\le & {} 256(\frac{8Q_{n}^{4}}{r_0})^{2} \textrm{e}^{-\frac{r-\tilde{r}_m}{2}Q_{n+1}}2\varepsilon _{0}^{\frac{1}{2}},\\ \frac{256\cdot 64\cdot 2}{r_{0}^{2}}\varepsilon _{0}^{\frac{1}{2}}\le & {} \frac{1}{2},\\ Q_{n}^{8}\textrm{e}^{-(r-\tilde{r}_m)Q_{n+1}}\le & {} \textrm{e}^{-\frac{r_0}{8Q_{n}^{4}}Q_{n+1}}Q_{n}^{8} \le \textrm{e}^{-\frac{Q_{n+1}}{Q_{n}^{5}}}Q_{n}^{8}\\\le & {} \textrm{e}^{-\frac{Q_{n+1}^{\frac{1}{2}}\ln Q_{n+1}}{Q_{n}^{5}}}\textrm{e}^{Q_{n}\ln Q_{n+1}} \le \textrm{e}^{-\ln Q_{n+1}(Q_{n}^{\frac{\mathcal {A}}{2}-5}-Q_{n})}\\\le & {} \textrm{e}^{-(\ln Q_{n+1})Q_{n}^{\frac{\mathcal {A}}{2}-5-1}} \le Q_{n+1}^{-(n+2-n_{0})2^{n+2-n_{0}}c\tau U}\\= & {} \left( \frac{\varepsilon }{\varepsilon _{-}}\right) ^{n+2-n_{0}} \le \varepsilon \le \tilde{\zeta }_m, \end{aligned}$$

  thus

  $$\begin{aligned} \frac{256}{(r-\tilde{r}_m)^2}e^{-\frac{r-\tilde{r}_m}{2}Q_{n+1}}\zeta \le \frac{1}{2}\tilde{\zeta }_m^{\frac{1}{2}}. \end{aligned}$$

• (3) We prove $\tilde{K}_m\le K^{(m)}$ and $\tilde{K}_m\le K.$ We first prove $\tilde{K}_m\le K^{(m)}.$ Owing to $\frac{2^{n+2-n_{0}}c\tau U}{2(24\tau +36)}\ge 5$ and

  $$\begin{aligned} m\le L-1=1+\left\lfloor \frac{2^{n+2-n_{0}}c\tau U \ln Q_{n+1}}{2(24\tau +36)\ln \frac{5}{2}}\right\rfloor , \end{aligned}$$

  then we have the inequalities

  $$\begin{aligned} \begin{aligned} \tilde{K}_m&\le \frac{2}{\tilde{\sigma }_m}\ln \frac{1}{\tilde{\varepsilon }_{m-1}} \le \frac{5\cdot 2^{m+2}\cdot 4Q_{n+1}^{4}}{r_0}\ln \frac{1}{\varepsilon ^{(\frac{5}{4})^{m-1}}}\\&\le \frac{160(\frac{5}{2})^{m-1}Q_{n+1}^{4}}{r_0}\ln \frac{1}{\varepsilon } \le \frac{160(\frac{5}{2})^{\lfloor \frac{2^{n+2-n_{0}}c\tau U \ln Q_{n+1}}{2(24\tau +36) \ln \frac{5}{2}}\rfloor }Q_{n+1}^{4}}{r_0}\ln \frac{1}{\varepsilon }\\&\le \frac{160Q_{n+1} ^{\frac{2^{n+2-n_{0}}c\tau U}{2(24\tau +36)}+4}}{r_0}\ln \frac{1}{\varepsilon } \le Q_{n+1}^{\frac{2^{n+2-n_{0}}c\tau U}{(24\tau +36)}}\ln \frac{1}{\varepsilon }\\&\le \left( \frac{1}{\varepsilon }\right) ^{\frac{1}{24\tau +36}}\left( \frac{1}{\varepsilon }\right) ^{\frac{1}{24\tau +36}} = \left( \frac{1}{\varepsilon }\right) ^{\frac{1}{12\tau +18}}, \end{aligned} \end{aligned}$$

  (9.2)

  and

  $$\begin{aligned}{} & {} 2^{L+2}\le \left( \frac{5}{2}\right) ^{\lfloor \frac{2^{n+2-n_{0}}c\tau U \ln Q_{n+1}}{2(24\tau +36) \ln \frac{5}{2}}\rfloor }\le Q_{n+1}^{\frac{2^{n+2-n_{0}}c\tau U}{2(24\tau +36)}}, \\{} & {} \frac{1}{2^{m+2}}\ge \frac{1}{2^{L+2}}\ge Q_{n+1}^{-\frac{2^{n+2-n_{0}}c\tau U}{2(24\tau +36)}}\ge \varepsilon ^{\frac{1}{2(24\tau +36)}}. \end{aligned}$$

  We conclude from all these inequalities that

  $$\begin{aligned} \frac{K^{(m)}}{\tilde{K}_m}\ge & {} \left( \frac{\gamma ^{2}\tilde{\sigma }_m^{2}}{2C_0\tilde{\zeta }_{m}^{\frac{1}{2}}}\right) ^{\frac{1}{2\tau +1}} \cdot \varepsilon ^{\frac{1}{12\tau +18}}\\\ge & {} \left( \frac{\gamma _{0}^{2}r_{0}^{2}}{(20\cdot 2^{m+2}Q_{n+1}^{4})^{2}\cdot 2C_{0}\cdot 2\varepsilon ^{\frac{1}{2}}}\right) ^{\frac{1}{2\tau +1}}\cdot \varepsilon ^{\frac{1}{12\tau +18}}\\\ge & {} \left( \frac{\gamma _{0}^{2}r_{0}^{2}}{C_{0}Q_{n+1}^{8}}\right) ^{\frac{1}{2\tau +1}} \cdot \left( \frac{1}{\varepsilon }\right) ^{\frac{1}{2(2\tau +1)}-\frac{1}{12\tau +18}-\frac{1}{2(24\tau +36)}}\\\ge & {} \left( \frac{\gamma _{0}^{2}r_{0}^{2}}{C_{0}Q_{n+1}^{8}}\right) ^{\frac{1}{2\tau +1}}\cdot Q_{n+1}^{2^{n+2-n_{0}} c\tau U\left( \frac{1}{2(2\tau +3)}-\frac{1}{12\tau +18}-\frac{1}{2(24\tau +36)}\right) }\\\ge & {} \left( \frac{\gamma _{0}^{2}r_{0}^{2}}{C_{0}Q_{n+1}^{8}}\right) ^{\frac{1}{2\tau +1}}\cdot Q_{n+1}^{2^{n+2-n_{0}}8} \ge 1, \end{aligned}$$

  i.e., $\tilde{K}_m\le K^{(m)}.$ In (9.2), we have obtained $\tilde{K}_m\le (\frac{1}{\varepsilon })^{\frac{1}{12\tau +18}}$. This together with $K\ge \left( \frac{\gamma ^{2}}{2\varepsilon ^{\frac{1}{2}}}\right) ^{\frac{1}{4\tau +6}}$ shows that

  $$\begin{aligned} \begin{aligned} \frac{\tilde{K}_m}{K} \le&\frac{\left( \frac{1}{\varepsilon }\right) ^{\frac{1}{12\tau +18}}}{\left( \frac{\gamma ^{2}}{2\varepsilon ^{\frac{1}{2}}}\right) ^{\frac{1}{4\tau +6}}} \le \left( \frac{1}{\varepsilon }\right) ^{\frac{1}{12\tau +18}} \left( \frac{2\varepsilon ^{\frac{1}{2}}}{\gamma ^{2}}\right) ^{\frac{1}{4\tau +6}}\\ \le&\left( \frac{2\varepsilon ^{\frac{1}{6}}}{\gamma ^{2}}\right) ^{\frac{1}{4\tau +6}} \le 1, \end{aligned} \end{aligned}$$

  (9.3)

  i.e., $\tilde{K}_m\le K.$ Thus we complete the proof of $\tilde{K}_m\le \min \{K, K^{(m)}\}.$

$\square$

1.3 Some Basic Inequalities

Lemma 9.1

(Cauchy’s estimate, [29]) Suppose  $0<\delta <r.$ $f(\theta , z, \bar{z})$ is real analytic on D(r, s),  then

$$\begin{aligned}{} & {} \left\| \frac{\partial f}{\partial \theta _b}\right\| _{D(r-\delta ,s)}\le \frac{c}{\delta }\Vert f\Vert _{D(r,s)}, \\{} & {} \left\| \frac{\partial f}{\partial z^{\pm }_i}\right\| _{D(r,s/2)}\le \frac{c}{s}\Vert f\Vert _{D(r,s)}\textrm{e}^{p|i|}, \end{aligned}$$

here c is a constant.

Lemma 9.2

([5]) Let $g: \mathcal {I}\rightarrow \mathbb {R}$ be $b+3$ times differentiable, and assume that

• (1) $\forall \sigma \in \mathcal {I}$ there exists $s\le b+2$ such that $g^{(s)}(\sigma )>B$.

• (2) There exists A such that $|g^{(s)}(\sigma )|\le A$ for $\forall \sigma \in \mathcal {I}$ and $\forall s$ with $1\le s\le b+3$.

Define

$$\begin{aligned} \mathcal {I}_h\equiv \{\sigma \in \mathcal {I}: |g(\sigma )|\le h\}, \end{aligned}$$

then

$$\begin{aligned} \frac{\textrm{meas}(\mathcal {I}_h)}{\textrm{meas}(\mathcal {I})}\le \frac{A}{B}2\left( 2+3+\cdots +(b+3)+ 2B^{-1}\right) h^{\frac{1}{b+3}}. \end{aligned}$$