The Cauchy problem for the seventh-order dispersive equation in Sobolev space

Abstract

This paper is devoted to the Cauchy problem for the higher-order dispersive equation $u_t+{\partial }_x^{7}u={\partial }_x^2(u^2)$, $x,t\in \mathbf{R}$. The local well-posedness of the associated Cauchy problem is established in Sobolev space $H^s(\mathbf{R})$ with $s>-\frac{7}{4}$ with the aid of the Fourier restriction norm method.

MSC:35K30.

1 Introduction

In this paper, we are concerned with the Cauchy problem for the following seventh-order dispersive equation:

$$u_t+{\partial }_x^{7}u={\partial }_x^2\left(u^2\right),x,t\in \mathbf{R},$$

(1.1)

$$u(x,0)=u_0(x).$$

(1.2)

Kenig et al. [1] established that

$$u_t+{\partial }_x^{2j+1}u+P(u,{\partial }_{x}u,\dots ,{\partial }_x^{2j}u)=0,j\in \mathbf{N},x,t\in \mathbf{R},$$

(1.3)

$$u(x,0)=u_0(x),$$

(1.4)

is locally well-posed in some weighted Sobolev spaces for small initial data and for arbitrary initial data. Recently, Pilod [2] studied the following higher-order nonlinear dispersive equation:

$$u_t+{\partial }_x^{2j+1}u=\sum _{0\le j_1+j_2\le 2j}{a}_{j_1,j_2}{\partial }_x^{j_1}u{\partial }_x^{j_2}u,$$

(1.5)

where $x,t\in \mathbf{R}$ and u is a real- (or complex-) valued function and proved it is locally well-posed in weighted Besov and Sobolev spaces for small initial data and proved ill-posedness results when $a_{0,k}\ne 0$ for some $k>j$ in the sense that (1.5) cannot have its flow map $C^2$ at the origin in $H^s(R)$. Very recently, Guo et al. [3] studied the Cauchy problem for

$$u_t+{\partial }_x^5+c_1{\partial }_{x}u{\partial }_x^{2}u+c_{2}u{\partial }_x^{3}u=0,$$

(1.6)

and he proved that it is locally well-posed in $H^s(\mathbf{R})$ with $s\ge \frac{5}{4}$ with the aid of a short time Bourgain space.

In this paper, inspired by [1–5], by using the Fourier restriction norm method, we establish that (1.1)-(1.2) is locally well-posed in Sobolev space $H^s$ with $s>-\frac{7}{4}$.

Now we give some notations and definitions. Throughout this paper, we always assume that ψ is a smooth function, ${\psi }_{\delta }(t)=\psi (\frac{t}{\delta })$, satisfying $0\le \psi \le 1$, $\psi =1$ when $t\in [0,1]$, $supp\psi \subset [-1,2]$ and $\sigma =\tau -{\xi }^7$, ${\sigma }_k={\tau }_k-{\xi }_k^7$ ($k=1,2$),

$$\begin{array}{c}W(t)u_0={\int }_{\mathbf{R}}{e}^{i(x\xi -t{\xi }^7)}{\mathcal{F}}_x{u}_0(\xi )d\xi ,\hfill \\ {\parallel f\parallel }_{L_t^q{L}_x^p}={\left({\int }_{\mathbf{R}}{\left({\int }_{\mathbf{R}}\right|f(x,t){|}^{p}dx)}^{\frac{q}{p}}dt\right)}^{\frac{1}{q}},{\parallel f\parallel }_{L_t^p{L}_x^p}={\parallel f\parallel }_{L_{xt}^p}.\hfill \end{array}$$

${〈\xi 〉}^s={(1+{\xi }^2)}^{\frac{s}{2}}$ for any $\xi \in \mathbf{R}$, and $\mathcal{F}u$ denotes the Fourier transformation of u with respect to its all variables. ${\mathcal{F}}^{-1}u$ denotes the Fourier inverse transformation of u with respect to its all variables. ${\mathcal{F}}_{x}u$ denotes the Fourier transformation of u with respect to its space variable. ${\mathcal{F}}_x^{-1}u$ denotes the Fourier inverse transformation of u with respect to its space variable. $\mathcal{S}({\mathbf{R}}^n)$ is the Schwarz space and ${\mathcal{S}}^{\prime }({\mathbf{R}}^n)$ is its dual space. $H^s(\mathbf{R})$ is the Sobolev space with norm ${\parallel f\parallel }_{H^s(\mathbf{R})}\stackrel{\mathrm{△}}{=}{\parallel {〈\xi 〉}^s{\mathcal{F}}_{x}f\parallel }_{L_{\xi }^2(\mathbf{R})}$. For any $s,b\in \mathbf{R}$, $X_{s,b}({\mathbf{R}}^2)$ is the Bourgain space with phase function $\varphi (\xi )=-{\xi }^7$. That is, a function $u(x,t)$ in ${\mathcal{S}}^{\prime }({\mathbf{R}}^2)$ belongs to $X_{s,b}({\mathbf{R}}^2)$ iff

$${\parallel u\parallel }_{X_{s,b}({\mathbf{R}}^2)}\stackrel{\mathrm{△}}{=}{\parallel {〈\xi 〉}^s{〈\sigma 〉}^b\mathcal{F}u(\xi ,\tau )\parallel }_{L_{\tau }^2(\mathbf{R})L_{\xi }^2(\mathbf{R})}<\mathrm{\infty }.$$

For any given interval L, $X_{s,b}(\mathbf{R}\times L)$ is the space of the restriction of all functions in $X_{s,b}({\mathbf{R}}^2)$ on $\mathbf{R}\times L$, and for $u\in X_{s,b}(\mathbf{R}\times L)$ its norm is

$${\parallel u\parallel }_{X_{s,b}(\mathbf{R}\times L)}=inf\{{\parallel U\parallel }_{X_{s,b}({\mathbf{R}}^2)};U{|}_{\mathbf{R}\times L}=u\}.$$

When $L=[0,T]$, $X_{s,b}(\mathbf{R}\times L)$ is abbreviated as $X_{s,b}^T$.

The main result of this paper is as follows.

Theorem 1.1 Assume that $u_0(x)\in H^s(\mathbf{R})$ with $s>-\frac{7}{4}$. Then the Cauchy problem for (1.1) is locally well-posed.

The remainder of paper is arranged as follows. In Section 2, we make some preliminaries. In Section 3, we give an important bilinear estimate. In Section 4, we establish Theorem 1.1.

2 Preliminaries

Lemma 2.1 Let $b>\frac{1}{2}$. Then

$${\parallel u\parallel }_{L_{xt}^4}\le C{\parallel u\parallel }_{X_{0,\frac{4b}{7}}},$$

(2.1)

$${\parallel D_x^{\frac{5}{4}}u\parallel }_{L_t^4{L}_x^{\mathrm{\infty }}}\le C{\parallel u\parallel }_{X_{0,b}},$$

(2.2)

$${\parallel u\parallel }_{L_t^4{L}_x^2}\le C{\parallel u\parallel }_{X_{0,\frac{1}{2}b}},$$

(2.3)

$${\parallel u\parallel }_{X_{0,-\frac{1}{2}b}}\le C{\parallel u\parallel }_{L_t^{\frac{4}{3}}{L}_x^2},$$

(2.4)

$${\parallel D_x^{\frac{5}{8}}u\parallel }_{L_{xt}^4}\le C{\parallel u\parallel }_{X_{0,\frac{3}{4}b}}.$$

(2.5)

Proof For the proof of (2.1)-(2.5), we refer the readers to Lemma 2.1 of [5].

We have completed the proof of Lemma 2.1. □

Lemma 2.2 Assume that $b=\frac{1}{2}+ϵ$. Then

$${\parallel I^{\frac{1}{2}}(u_1,u_2)\parallel }_{L_{xt}^2}\le C\prod _{k=1}^2{\parallel u_k\parallel }_{X_{0,b}},$$

(2.6)

where

$$\mathcal{F}{I}^{\frac{1}{2}}(u_1,u_2)(\xi ,\tau )={\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}|{\xi }_1^6-{\xi }_2^6{|}^{\frac{1}{2}}\mathcal{F}{u}_1({\xi }_1,{\tau }_1)\mathcal{F}{u}_2({\xi }_2,{\tau }_2)d{\xi }_{1}d{\tau }_1.$$

Lemma 2.2 is the case of $n=3$ of Lemma 3.1 of [5].

Lemma 2.3 For any $0<\delta <\frac{1}{2}$, and $s\in \mathbf{R}$, for $b>0$, we have

$${\parallel {\psi }_{\delta }(t)W(t)u_0\parallel }_{X_{s,b}}\le C{\delta }^{\frac{1}{2}-b}{\parallel u_0\parallel }_{H^s}.$$

(2.7)

For $-\frac{1}{2}<b^{\prime }\le 0\le b^{\prime }+1$, we have

$${\parallel {\psi }_{\delta }(t){\int }_0^{t}W(t-\tau )ud\tau \parallel }_{X_{s,b}}\le C{\delta }^{1+b^{\prime }-b}{\parallel u\parallel }_{X_{s,b^{\prime }}}.$$

(2.8)

Lemma 2.3 can be found as Lemma 2.4 of [6].

3 Bilinear estimates

In this section, we will give an important bilinear estimate.

We give an important relation before proving the bilinear estimate.

$$|\sigma -{\sigma }_1-{\sigma }_2|=|{\xi }^7-{\xi }_1^7-{\xi }_2^7|\sim {\xi }_{\min }{\xi }_{\max }^6,$$

where

$$\begin{array}{r}{\xi }_{\min }=min\{|\xi |,|{\xi }_1|,|{\xi }_2|\},\\ {\xi }_{\max }=max\{|\xi |,|{\xi }_1|,|{\xi }_2|\}.\end{array}$$

(3.1)

Lemma 3.1 Let $s\ge -\frac{7}{4}+21ϵ$, $b=\frac{1}{2}+ϵ$, where $0\ll ϵ\le 1$, $b^{\prime }=-\frac{1}{2}+2ϵ$. Then

$${\parallel {\partial }_x^2\prod _{k=1}^2(u_k)\parallel }_{X_{s,b^{\prime }}}\le C\prod _{k=1}^2{\parallel u_k\parallel }_{X_{s,b}}.$$

(3.2)

Proof Let

$$\begin{array}{c}{F}_k({\xi }_k,{\tau }_k)={〈{\xi }_k〉}^s{〈{\sigma }_k〉}^b\mathcal{F}{u}_k({\xi }_k,{\tau }_k),\hfill \\ F(\xi ,\tau )={〈\xi 〉}^{-s}{〈\sigma 〉}^{b^{\prime }}\mathcal{F}u(\xi ,\tau ),\hfill \\ \sigma =\tau -{\xi }^7,{\sigma }_k={\tau }_k-{\xi }_k^7,k=1,2.\hfill \end{array}$$

To establish (3.2), it is sufficient to derive the following inequality:

$${\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )|F|\prod _{k=1}^2|F_k|d{\xi }_{1}d{\tau }_{1}d\xi d\tau \le C{\parallel F\parallel }_{L_{\xi \tau }^2}{\parallel F_1\parallel }_{L_{\xi \tau }^2}{\parallel F_2\parallel }_{L_{\xi \tau }^2},$$

(3.3)

where

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )=\frac{{|\xi |}^2{〈\sigma 〉}^{b^{\prime }}{〈\xi 〉}^s}{{\prod }_{k=1}^2{〈{\xi }_k〉}^s{〈{\sigma }_k〉}^b}.$$

(3.4)

Without loss of generality, we assume that $F\ge 0$, $F_k\ge 0$ ($k=1,2$). To derive (3.3), it suffices to prove that

$${\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \le C{\parallel F\parallel }_{L_{\xi \tau }^2}{\parallel F_1\parallel }_{L_{\xi \tau }^2}{\parallel F_2\parallel }_{L_{\xi \tau }^2}.$$

(3.5)

By using the symmetry between $|{\xi }_1|$ and ${\xi }_2$, without loss of generality, we can assume that $|{\xi }_1|\le |{\xi }_2|$. Obviously,

$$\{\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_2|\ge |{\xi }_1|\}\subset \bigcup _{k=1}^6{\mathrm{\Omega }}_k,$$

where

$$\begin{array}{c}{\mathrm{\Omega }}_1=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_1|\le |{\xi }_2|\le 18\},\hfill \\ {\mathrm{\Omega }}_2=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_2|\ge 18,|{\xi }_2|\ge 4|{\xi }_1|,|{\xi }_1|\le 1\},\hfill \\ {\mathrm{\Omega }}_3=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_2|\ge 18,|{\xi }_2|\ge 4|{\xi }_1|,|{\xi }_1|\ge 1\},\hfill \\ {\mathrm{\Omega }}_4=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi =\sum _{k=1}^2{\xi }_k,\tau =\sum _{k=1}^2{\tau }_k,|{\xi }_2|\ge 18,\hfill \\ \phantom{{\mathrm{\Omega }}_4=}|{\xi }_1|\le |{\xi }_2|\le 4|{\xi }_1|,|\xi |\le \frac{1}{2}|{\xi }_1|,{\xi }_1{\xi }_2\le 0\},\hfill \\ {\mathrm{\Omega }}_5=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_2|\ge 18,|{\xi }_1|\le |{\xi }_2|\le 4|{\xi }_1|,|{\xi }_2|\ge \frac{|\xi |}{2}\},\hfill \\ {\mathrm{\Omega }}_6=\{({\xi }_1,{\tau }_1,\xi ,\tau )\in {\mathbf{R}}^4,\xi ={\xi }_1+{\xi }_2,\tau ={\tau }_1+{\tau }_2,|{\xi }_2|\ge 18,|{\xi }_1|\le |{\xi }_2|\le 4|{\xi }_1|,{\xi }_1{\xi }_2\ge 0\}.\hfill \end{array}$$

We will denote the integrals in (3.5) corresponding to ${\mathrm{\Omega }}_k$ ($1\le k\le 6$) by $J_k$ ($1\le k\le 6$), respectively. Let $f={\mathcal{F}}^{-1}\frac{F}{{〈\sigma 〉}^{-b^{\prime }}}$, $f_k={\mathcal{F}}^{-1}\frac{F_k}{{〈{\sigma }_k〉}^b}$, $k=1,2$.

• (1) Subregion ${\mathrm{\Omega }}_1$. Since $|{\xi }_1|\le |{\xi }_2|\le 18$, we have $|\xi |\le 36$, which yields

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le \frac{C}{{〈\sigma 〉}^{-b^{\prime }}{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

Then, by the Plancherel identity, the Hölder inequality, and $\frac{4}{7}b<b$, we derive

$$\begin{array}{rcl}{J}_1& \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{F{\prod }_{k=1}^2{F}_k}{{〈\sigma 〉}^{-b^{\prime }}{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \le C{\int }_{{\mathbf{R}}^2}{f}_1{f}_{2}fdxdt\\ \le & C{\parallel f\parallel }_{L_{xt}^2}\prod _{k=1}^2{\parallel f_k\parallel }_{L_{xt}^4}\le C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel f_k\parallel }_{X_{0,\frac{4}{7}b}}\le C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel F_k\parallel }_{L_{\xi \tau }^2}.\end{array}$$

• (2) Subregion ${\mathrm{\Omega }}_2$. In this subregion, obviously, $|{\xi }_2|\sim |\xi |$.

It is easily checked that

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^2{〈\sigma 〉}^{b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

Consequently, by the Cauchy-Schwarz inequality and Lemma 2.2, we have

$$\begin{array}{rcl}{J}_2& \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{F{\prod }_{k=1}^2{F}_k{|{\xi }_1^6-{\xi }_2^6|}^{\frac{1}{2}}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\parallel {\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{|{\xi }_1^6-{\xi }_2^6|}^{\frac{1}{2}}{\prod }_{k=1}^2{F}_k}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}d{\xi }_{1}d{\tau }_1\parallel }_{L_{\xi \tau }^2}{\parallel F\parallel }_{L_{\xi \tau }^2}\\ \le & C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel F_k\parallel }_{L_{\xi \tau }^2}.\end{array}$$

• (3) Subregion ${\mathrm{\Omega }}_3$. In this subregion, we derive $|\xi |\sim |{\xi }_2|$. Thus,

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^2{|{\xi }_1|}^{-s}{〈\sigma 〉}^{b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

• (i) Case $|\sigma |=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$. By (3.1), we derive

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{2+6b^{\prime }}{|{\xi }_1|}^{-s+b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

If $-s+b^{\prime }\le 0$, then

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{2+6b^{\prime }}{|{\xi }_1|}^{-s+b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

If $-s+b^{\prime }\ge 0$, then

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{2-s+7b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

This case can be proved similarly to ${\mathrm{\Omega }}_2$.

• (ii) Case $|{\sigma }_1|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$. Since ${〈\sigma 〉}^{b+b^{\prime }}\le {〈{\sigma }_1〉}^{b+b^{\prime }}$, we have

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^2{|{\xi }_1|}^{-s}{〈{\sigma }_1〉}^{b^{\prime }}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}\le C\frac{{|\xi |}^{2+6b^{\prime }}{|{\xi }_1|}^{-s+b^{\prime }}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}.$$

If $-s+b^{\prime }\le 0$, we have

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|{\xi }_2|}^{5/4}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b},$$

consequently, by using the Cauchy-Schwarz inequality and (2.5) and (2.4), we have

$$\begin{array}{c}{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{|{\xi }_2|}^{\frac{2n-1}{4}}F{\prod }_{j=1}^2{F}_j}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \hfill \\ \le {\parallel {〈\sigma 〉}^{-b}{\int }_{{\mathbf{R}}^2}{|{\xi }_2|}^{\frac{5}{4}}{〈\sigma 〉}^{-b}\prod _{j=1}^2{F}_{j}d{\xi }_{1}d{\tau }_1\parallel }_{L_{\xi \tau }^2}{\parallel F\parallel }_{L_{\xi \tau }^2}\hfill \\ \le C{\parallel {\mathcal{F}}^{-1}(F_1)\left(D_x^{\frac{5}{4}}{f}_2\right)\parallel }_{L_t^{\frac{4}{3}}{L}_x^2}{\parallel F\parallel }_{L_{\xi \tau }^2}\hfill \\ \le C{\parallel {\mathcal{F}}^{-1}(F_1)\parallel }_{L_{xt}^2}{\parallel D_x^{5/4}{f}_2\parallel }_{L_t^4{L}_x^{\mathrm{\infty }}}{\parallel F\parallel }_{L_{\xi \tau }^2}\hfill \\ \le C{\parallel F_1\parallel }_{L_{\xi \tau }^2}{\parallel F\parallel }_{L_{\xi \tau }^2}{\parallel f_2\parallel }_{X_{0,b}}\hfill \\ \le C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{j=1}^2{\parallel F_j\parallel }_{L_{\xi \tau }^2}.\hfill \end{array}$$

If $-s+b^{\prime }\ge 0$, since $s\ge -\frac{7}{4}+21ϵ$, we have

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{3-s+7b^{\prime }}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}\le C\frac{{|{\xi }_2|}^{5/4}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}.$$

This case can be proved similarly to the above case.

• (iii) Case $|{\sigma }_2|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$. This case is similar to (ii) case $|{\sigma }_1|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$.

• (4) Subregion ${\mathrm{\Omega }}_4$. In this subregion, $|{\xi }_1|\sim |{\xi }_2|$, and it is easy to obtain

  $$|{\xi }_1^6-{\xi }_2^6|\ge C|\xi |{|{\xi }_1|}^5,|{\xi }^6-{\xi }_1^6|\ge C{|{\xi }_1|}^6,|{\xi }^6-{\xi }_2^6|\ge C{|{\xi }_2|}^6.$$

• (i) Case $|\sigma |=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$. By using, $|{\xi }_1|\sim |{\xi }_2|$, when $s\ge 0$, we have

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^3{〈\sigma 〉}^{b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}{〈\sigma 〉}^{b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

This case can be proved similarly to Subregion ${\mathrm{\Omega }}_2$. When $s\le 0$, we have

$$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^3{|{\xi }_1|}^{-2s}}{{〈\sigma 〉}^{-b^{\prime }}{\prod }_{j=1}^2{〈{\sigma }_j〉}^b}.$$

If $|\sigma |=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$, since $-\frac{7}{4}+21ϵ\le s\le 0$, then

$$\begin{array}{rcl}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )& \le & C\frac{{|\xi |}^{3+b^{\prime }}{|{\xi }_1|}^{-2s+6b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\\ \le & C\frac{{|\xi |}^{1/2}{|{\xi }_1|}^{5/2}{|\xi |}^{\frac{5}{2}+b^{\prime }}{|{\xi }_1|}^{-2s+6b^{\prime }-5/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\\ \le & C\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.\end{array}$$

By using the Cauchy-Schwarz inequality, we have

$$\begin{array}{rcl}{J}_4& \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\parallel {\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{|{\xi }_1^6-{\xi }_2^6|}^{1/2}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_1\parallel }_{L_{\xi \tau }^2}{\parallel F\parallel }_{L_{\xi \tau }^2}\\ \le & C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{j=1}^2{\parallel F_j\parallel }_{L^2}.\end{array}$$

• (ii) Case $max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}=|{\sigma }_1|$. Since ${〈\sigma 〉}^{b+b^{\prime }}\le {〈{\sigma }_1〉}^{b+b^{\prime }}$, by using $-\frac{7}{4}+21ϵ\le s\le 0$, we have

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^3{|{\xi }_1|}^{-2s}}{{〈\sigma 〉}^b{〈{\sigma }_2〉}^b{〈{\sigma }_1〉}^{-b^{\prime }}}\le C\frac{{|\xi |}^{3+b^{\prime }}{|{\xi }_1|}^{-2s+6b^{\prime }}}{{〈\sigma 〉}^b{〈{\sigma }_2〉}^b}\le C\frac{{|{\xi }^6-{\xi }_2^6|}^{1/2}}{{〈\sigma 〉}^b{〈{\sigma }_2〉}^b}.$$

This case can be proved similarly to $max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}=|\sigma |$.

• (iii) Case $max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}=|{\sigma }_2|$.

This case can be proved similarly to $max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}=|{\sigma }_1|$.

• (5) Subregion ${\mathrm{\Omega }}_5$. In this region $|\xi |\sim |{\xi }_1|\sim |{\xi }_2|$, thus, we have

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{3-s}{〈\sigma 〉}^{b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

• (i) If $|\sigma |=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$, by using (3.1) and $s\ge -\frac{9}{4}+\frac{21}{2}ϵ$, we have

  $$K_1({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{3-s+7b^{\prime }}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}\le C\frac{{\prod }_{k=1}^2{|{\xi }_k|}^{\frac{5}{8}}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}.$$

By using the Plancherel identity, the Hölder inequality, and $\frac{3}{4}b<b$ as well as (2.5), we have

$$\begin{array}{rcl}{J}_5& \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{\prod }_{k=1}^2{|{\xi }_k|}^{\frac{5}{8}}}{{\prod }_{k=1}^2{〈{\sigma }_k〉}^b}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\mathcal{F}}^{-1}F\prod _{k=1}^2{D}_x^{\frac{5}{8}}{f}_{k}dxdt\\ \le & C{\parallel {\mathcal{F}}^{-1}F\parallel }_{L_{xt}^2}\prod _{k=1}^2{\parallel D_x^{\frac{5}{8}}{f}_k\parallel }_{L_{xt}^4}\\ \le & C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel f_k\parallel }_{X_{0,\frac{3}{4}b}}\le C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel F_k\parallel }_{L_{\xi \tau }^2}.\end{array}$$

• (ii) If $|{\sigma }_1|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$, then ${〈\sigma 〉}^{b^{\prime }}{〈{\sigma }_1〉}^{-b}\le {〈{\sigma }_1〉}^{b^{\prime }}{〈\sigma 〉}^{-b}$. By using (3.1), we have

  $$K_2({\xi }_1,{\tau }_1,\xi ,\tau )\le C\frac{{|\xi |}^{2-s}{〈{\sigma }_1〉}^{b^{\prime }}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}\le C\frac{{|\xi |}^{2-s+7b^{\prime }}}{{〈{\sigma }_2〉}^b{〈\sigma 〉}^b}\le C\frac{{|\xi |}^{\frac{5}{8}}{|{\xi }_2|}^{\frac{5}{8}}}{{〈\sigma 〉}^b{〈{\sigma }_2〉}^b}.$$

By using the Plancherel identity, the Hölder inequality, (2.5) and $\frac{3}{4}b<b$, we have

$$\begin{array}{rcl}{J}_5& \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}{K}_1({\xi }_1,{\tau }_1,\xi ,\tau )F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}{\int }_{\begin{array}{c}\xi ={\xi }_1+{\xi }_2\\ \tau ={\tau }_1+{\tau }_2\end{array}}\frac{{|\xi |}^{\frac{5}{8}}{|{\xi }_2|}^{\frac{5}{8}}}{{〈{\sigma }_2〉}^b{〈{\sigma }_2〉}^b}F\prod _{k=1}^2{F}_{k}d{\xi }_{1}d{\tau }_{1}d\xi d\tau \\ \le & C{\int }_{{\mathbf{R}}^2}\left({\mathcal{F}}^{-1}{F}_1\right)\left(D_x^{\frac{5}{8}}{\mathcal{F}}^{-1}\left(\frac{F}{{〈\sigma 〉}^b}\right)\right)D_x^{\frac{5}{8}}{f}_{2}dxdt\\ \le & C{\parallel {\mathcal{F}}^{-1}{F}_1\parallel }_{L_{xt}^2}{\parallel D_x^{\frac{5}{8}}{f}_2\parallel }_{L_{xt}^4}{\parallel D_x^{\frac{5}{8}}{\mathcal{F}}^{-1}\left(\frac{F}{{〈\sigma 〉}^b}\right)\parallel }_{L_{xt}^4}\\ \le & C{\parallel F_1\parallel }_{L_{\xi \tau }^2}{\parallel f_2\parallel }_{X_{0,\frac{3}{4}b}}{\parallel \frac{F}{{〈\sigma 〉}^b}\parallel }_{X_{0,\frac{3}{4}b}}\le C{\parallel F\parallel }_{L_{\xi \tau }^2}\prod _{k=1}^2{\parallel F_k\parallel }_{L_{\xi \tau }^2}.\end{array}$$

• (iii) If $|{\sigma }_2|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$.

This case can be proved similarly to the case $|{\sigma }_1|=max\{|\sigma |,|{\sigma }_1|,|{\sigma }_2|\}$.

• (6) Subregion ${\mathrm{\Omega }}_6$. In this region, we have $|\xi |\sim |{\xi }_1|\sim |{\xi }_2|$.

This case can be proved similarly to the Subregion ${\mathrm{\Omega }}_5$.

We have completed the proof of Lemma 3.1. □

4 Proof of Theorem 1.1

The system (1.1)-(1.2) is equivalent to the following integral equation:

$$u(t)=W(t)u_0+{\int }_0^{t}W(t-\tau ){\partial }_x^2\left(u^2\right)d\tau .$$

(4.1)

We define

$$\mathrm{\Phi }(u)=\mathrm{\Psi }(t)W(t)u_0+{\mathrm{\Psi }}_{\delta }(t){\int }_0^{t}W(t-\tau ){\partial }_x^2\left(u^2\right)d\tau .$$

(4.2)

Combining Lemmas 2.3 and 3.1 with the fixed point theorem, we easily obtain Theorem 1.1.