The local well-posedness for nonlinear fourth-order Schrödinger equation with mass-critical nonlinearity and derivative

Abstract

We study the Cauchy problem of the nonlinear fourth-order Schrödinger equation with mass-critical nonlinearity and derivative: $i{u}_t+a{u}_{xxxx}+b{u}^2{\overline{u}}_{xx}+c{|u|}^{8}u=0$, $x\in R$, $t\in R$, where a, b, and c are real numbers. We obtain the local well-posedness for the Cauchy problem with low regularity initial value data by the Fourier restriction norm method.

1 Introduction

In [1], Fibich et al. discussed the following fourth-order Schrödinger equation:

$$\{\begin{array}{l}i{u}_t+a{\mathrm{\Delta }}^{2}u+c{|u|}^{\alpha }u=0,x\in R^n,t\in R,\\ u(x,0)=u_0(x),x\in R^n.\end{array}$$

(1.1)

They gave the sufficient conditions of the existence of solutions in the space $H^2(R^n)$. For the case $\alpha =\frac{8}{n}$, the mass is still invariant under the scaling $u(x,t)\mapsto {\lambda }^{-\frac{4}{\alpha }}u(\frac{x}{\lambda },\frac{t}{{\lambda }^4})$. We call this case mass-critical.

Meanwhile, in [1], Fibich et al. mentioned the physical motivation of (1.1). With nonparaxial effects, one obtained the perturbed nonlinear Schrödinger equation:

$$\{\begin{array}{l}i{u}_t(x,y,t)+\mathrm{\Delta }u+\epsilon {\mathrm{\Delta }}^{2}u\\ +\epsilon ({|u|}^{2}u+{|u|}^{4}u+{|u|}^2\mathrm{\Delta }u+u\mathrm{\nabla }u\cdot \mathrm{\nabla }\overline{u}+\overline{u}\mathrm{\nabla }u\cdot \mathrm{\nabla }u)=0,\\ u(x,y,0)=u_0(x,y).\end{array}$$

(1.2)

Furthermore, with vectorial effects, one obtained the following equation:

$$\{\begin{array}{l}i{u}_t(x,y,t)+\mathrm{\Delta }u+\epsilon {\mathrm{\Delta }}^{2}u+\epsilon ({|u|}^{2}u+{|u|}^{4}u+{|u|}^2\mathrm{\Delta }u+u\mathrm{\nabla }u\cdot \mathrm{\nabla }\overline{u}+\overline{u}\mathrm{\nabla }u\cdot \mathrm{\nabla }u)\\ +\epsilon ({|u_x|}^{2}u+{(u_x)}^2\overline{u}+{|u|}^2{u}_{xx}+u^2{\overline{u}}_{xx})=0,\\ u(x,y,0)=u_0(x,y).\end{array}$$

(1.3)

Evidently, the nonlinearities with derivatives appear. It is well known that nonlinearities with derivatives bring about more difficulties to solve the problem for us. Especially, there are so many nonlinearities with derivatives in (1.2) and (1.3).

So, in this paper we will study the following nonlinear fourth-order Schrödinger equation with mass-critical nonlinearity and two-order derivative in one dimension:

$$\{\begin{array}{l}i{u}_t+a{\mathrm{\Delta }}^{2}u+b{u}^2{\overline{u}}_{xx}+c{|u|}^{8}u=0,x\in R,t\in R,\\ u(x,0)=u_0(x),x\in R,\end{array}$$

(1.4)

where $u(x,t)$ are complex-valued function, $\overline{u}(x,t)$ is the complex conjugate quantity of $u(x,t)$. a, b, and c are real numbers. We are interested in obtaining the well-posedness for the Cauchy problem of (1.1) with initial value data under low regularity (which means $u_0(x)\in H^s(R)$, $s<2$). Tao et al. obtained the global well-posedness for the Schrödinger equations with derivative (${({|u|}^{2}u)}_x$) by the I-method (see [2, 3]). Bourgain obtained the well-posedness for the nonlinear Schrödinger equation (${|u|}^{p}u$) by the Fourier restriction norm method (see [4]). The character of (1.4) lies in the coexistence of the mass-critical nonlinearity and the two-order derivative. We will discuss the local well-posedness for the fourth-order Schrödinger equation by the Fourier restriction norm method.

For the complicated case, we will discuss it in another paper.

First, we introduce the following notations. We define the Sobolev norms $H^s$ by

$${\parallel f\parallel }_{H^s}:={\parallel {〈D_x〉}^{s}f\parallel }_{L_x^2}={\parallel {〈\xi 〉}^s\hat{f}(\xi )\parallel }_{L_{\xi }^2},$$

where $〈\cdot 〉=(1+|\cdot |)$, and $\hat{f}$ denotes the Fourier transformation of $f(x)$.

We also define the spaces $X_{s,b}(R\times R)$ (see [5]) on $R\times R$ by

$${\parallel u\parallel }_{X_{s,b}}:={\parallel {〈\xi 〉}^s{〈\tau -a{\xi }^4〉}^b\hat{u}(\xi ,\tau )\parallel }_{L_{\tau }^2{L}_{\xi }^2},$$

where $\hat{u}(\xi ,\tau )$ denotes the Fourier transformation of $u(x,t)$.

We denote by $U(t)$ ($t\in R$) the fundamental solution operator of the fourth-order Schrödinger equation, i.e.,

$$U(t)\phi (x)=F^{-1}(e^{iat{\xi }^4}\hat{\phi }(\xi ))\text{for }\phi \in S^{\prime }(R),$$

where $\hat{\phi }$ denotes the Fourier transformation of φ, and $F^{-1}$ represents the inverse Fourier transformation.

We use C to denote various constants which may be different from in particular cases of use throughout.

The main result of this paper is the following theorem.

Theorem 1.1 Let $s\ge \frac{1}{2}$, $\frac{1}{2}<b<\frac{5}{8}$. Then the system (1.4) is locally solvable in $H^s(R)$, i.e., for any $u_0(x)\in H^s(R)$, there exists a corresponding $T>0$ such that the system (1.4) has a unique solution in the class

$$C([0,T];H^s(R))\cap X_{s,b}.$$

Moreover, the mapping $u_0(x)\to u(x,t)$ is Lipschitz continuous from $H^s(R)$ to $C([0,T];H^s(R))$.

In [6], Cui et al. obtained the local well-posedness with the initial condition satisfying $u_0(x)\in H^s(R)$, $s\ge 0$ for $b=0$ in (1.4).

Thus from the above theorem, we can see the following result.

Remark 1.1 When mass-critical nonlinearity and nonlinearity with derivative appear at the same time, nonlinearity with second-order derivative plays more important role. This property is consistent with the classical Schrödinger equation which has both mass-critical nonlinearity and first-order derivative nonlinearity.

2 The preliminary estimates

Definition 2.1 For two integers $8\le q\le \mathrm{\infty }$ and $2\le r<\mathrm{\infty }$, we say that $(q,r)$ is an admissible pair if the following condition is satisfied:

$$\frac{2}{q}=\frac{1}{4}(1-\frac{2}{r}).$$

We have the following Strichartz estimate (see [6]): For any admissible pair $(q,r)$

$${\parallel U(t)\phi (x)\parallel }_{L_x^r{L}_t^q}\le C{\parallel \phi \parallel }_{L^2}.$$

(2.1)

Lemma 2.1 Assume that $(q,r)$ is an admissible pair. Let $b>\frac{1}{2}$. We have

$${\parallel u\parallel }_{L_x^r{L}_t^q}\le C{\parallel u\parallel }_{0,b}.$$

(2.2)

For any $f\in L_{\tau }^2{L}_{\xi }^2$, we have

$${\parallel D_x^{-\frac{1}{4}}{F}_b\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}\le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2},$$

(2.3)

$${\parallel D_x^{\frac{3}{2}}{F}_b\parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}\le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2},$$

(2.4)

where ${\hat{F}}_b(\xi ,\tau )=\frac{f(\xi ,\tau )}{{(1+|\tau -a{\xi }^4|)}^b}$.

Proof Firstly, we prove the inequality (2.2).

For any $u(x,t)\in S(R^2)$, we have

$$u(x,t)=C{\int }_R{e}^{it\lambda }{\int }_R{e}^{ix\xi -ita{\xi }^4}\hat{u}(\xi ,\lambda +a{\xi }^4)d\xi d\lambda =C{\int }_R{e}^{it\lambda }U(t)u_{\lambda }(x)d\lambda ,$$

where ${\hat{u}}_{\lambda }(\xi )=\hat{u}(\xi ,\lambda +a{\xi }^4)$.

Noting that $b>\frac{1}{2}$, using the Strichartz estimate, we obtain

$$\begin{array}{rl}{\parallel u(x,t)\parallel }_{L_x^r{L}_t^q}& \le C{\int }_R{\parallel U(t)u_{\lambda }(x)\parallel }_{L_x^r{L}_t^q}d\lambda \le C{\int }_R{\parallel u_{\lambda }(x)\parallel }_{L_x^2}d\lambda \\ \le C{\parallel {〈\lambda 〉}^b{\hat{u}}_{\lambda }(\xi )\parallel }_{L_{\xi }^2{L}_{\tau }^2}=C{\parallel u\parallel }_{0,b}.\end{array}$$

Next we prove (2.3), we only need to prove that for $b>\frac{1}{2}$ we have

$${\parallel F_b(x,t)\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}\le C{\parallel f\parallel }_{L_{\tau }^2{\dot{H}}_{\xi }^{\frac{1}{4}}}.$$

(2.5)

$${\parallel U(t)\phi \parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}\le C{\parallel \phi \parallel }_{{\dot{H}}^{\frac{1}{4}}}\mathrm{\forall }\phi \in {\dot{H}}^{\frac{1}{4}}(R).$$

(2.6)

Changing the variable to $\tau =\lambda +a{\xi }^4$, we obtain

$$\begin{array}{rl}{F}_b(x,t)& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{i(x\xi +t\tau )}\frac{f(\xi ,\tau )}{{(1+|\tau -a{\xi }^4|)}^b}d\xi d\tau \\ ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{it\lambda }\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{ix\xi +iat{\xi }^4}f(\xi ,\lambda +a{\xi }^4)d\xi \right)\frac{1}{{(1+|\lambda |)}^b}d\lambda .\end{array}$$

(2.7)

From (2.6) and (2.7), we can obtain

$$\begin{array}{rl}{\parallel F_b(x,t)\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}& \le {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{it\lambda }{\parallel {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{ix\xi +iat{\xi }^4}f(\xi ,\lambda +a{\xi }^4)d\xi \parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}\frac{1}{{(1+|\lambda |)}^b}d\lambda \\ \le C{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\parallel f(\xi ,\lambda +a{\xi }^4)\parallel }_{{\dot{H}}_{\xi }^{\frac{1}{4}}}\frac{1}{{(1+|\lambda |)}^b}d\lambda \\ \le C{\parallel f\parallel }_{L_{\tau }^2{\dot{H}}_{\xi }^{\frac{1}{4}}},\end{array}$$

so that (2.5) holds.

Similarly, (2.4) follows from the following inequality:

$${\parallel D_x^{\frac{3}{2}}U(t)\phi \parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}\le C{\parallel \phi \parallel }_{L^2}.$$

This inequality has been proved in Theorem 4.1 of [7]. We omit its detailed proof here. □

Lemma 2.2 If $b>\frac{5}{8}\frac{r-2}{r}$ for $2\le r\le 10$. Then we have

$${\parallel F_b\parallel }_{L_x^r{L}_t^r}\le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2},$$

(2.8)

where $F_b(x,t)$ is as same as in Lemma 2.1.

Proof Noting that $(10,10)$ is an admissible pair, so we have

$${\parallel U(t)\phi (x)\parallel }_{L_x^{10}{L}_t^{10}}\le C{\parallel \phi \parallel }_{L^2(R)}.$$

(2.9)

Therefore, using (2.7), Minkowski’s inequality, (2.9), and taking $b>\frac{1}{2}$, we can obtain

$${\parallel F_b\parallel }_{L_x^{10}{L}_t^{10}}\le C{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\parallel f(\xi ,\lambda +a{\xi }^4)\parallel }_{L_{\xi }^2}\frac{d\lambda }{{(1+|\lambda |)}^b}\le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}.$$

(2.10)

By interpolation [8] between (2.10) and the following relation:

$${\parallel F_0\parallel }_{L_x^2{L}_t^2}={\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2},$$

we immediately obtain for all $b>\frac{5}{8}\frac{r-2}{r}$

$${\parallel F_b\parallel }_{L_x^r{L}_t^r}\le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}.$$

 □

We take a function $\psi \in C_0^{\mathrm{\infty }}(R)$ with $\psi =1$ on $[-1,1]$ and $supp\psi \subset [-2,2]$. We denote ${\psi }_{\delta }(\cdot )=\psi ({\delta }^{-1}(\cdot ))$.

Similar to the proof of Lemma 3.2 in [9] (or see [[10], Lemmas 3.1-3.3], [[11], Lemma 2.3], [[12], Lemma 2.7]), we have the following estimates.

Lemma 2.3 For any real s, $0<\delta <1$, $\frac{1}{2}<b<1$, and $\frac{1}{2}<b_1<b_2<1$. We have

$$\begin{array}{c}{\parallel {\psi }_{\delta }(t)U(t)w_0\parallel }_{X_{s,b}}\le C{\delta }^{\frac{1}{2}-b}{\parallel w_0\parallel }_{H^s},\hfill \\ {\parallel {\psi }_{\delta }(t){\int }_0^{t}U(t-\tau )f(\tau )d\tau \parallel }_{X_{s,b}}\le C{\delta }^{\frac{1}{2}-b}{\parallel f\parallel }_{X_{s,b-1}},\hfill \\ {\parallel {\psi }_{\delta }(t)F\parallel }_{X_{s,b_1-1}}\le C{\delta }^{b_2-b_1}{\parallel F\parallel }_{X_{s,b_2-1}}.\hfill \end{array}$$

Lemma 2.4 [12]

If f, $f_1$, $f_2$, and $f_3$ belong to a Schwartz space on $R^2$, then we have

$${\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\hat{\overline{f}}(\tau ,\xi ){\hat{f}}_1({\tau }_1,{\xi }_1){\hat{f}}_2({\tau }_2,{\xi }_2){\hat{f}}_3({\tau }_3,{\xi }_3)d\delta ={\int }_R{\int }_R\overline{f}{f}_1{f}_2{f}_{3}dxdt,$$

where

$$\begin{array}{c}{\mathrm{\Gamma }}_{\tau }=\{({\tau }_1,{\tau }_2,{\tau }_3)\in R^3;{\tau }_1+{\tau }_2+{\tau }_3=\tau \},{\mathrm{\Gamma }}_{\xi }=\{({\xi }_1,{\xi }_2,{\xi }_3)\in R^3;{\xi }_1+{\xi }_2+{\xi }_3=\xi \},\hfill \\ d\delta =d\tau d{\tau }_{1}d{\tau }_{2}d{\tau }_{3}d\xi d{\xi }_{1}d{\xi }_{2}d{\xi }_3.\hfill \end{array}$$

Lemma 2.5 Let $s\ge \frac{1}{2}$, $\frac{1}{2}<b<\frac{11}{16}$, $b^{\prime }>\frac{1}{2}$. Then we have

$${\parallel u_1{u}_2{\overline{u}}_{3xx}\parallel }_{X_{s,b-1}}\le C{\parallel u_1\parallel }_{X_{s,b^{\prime }}}{\parallel u_2\parallel }_{X_{s,b^{\prime }}}{\parallel u_3\parallel }_{X_{s,b^{\prime }}}.$$

(2.11)

Proof By the definition of $X_{s,b}$ and duality, the inequality (2.11) is reduced to the following estimate:

$$\begin{array}{rl}\bigwedge & ={\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^{s}i{|{\xi }_3|}^2\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2},\end{array}$$

for all $\overline{f}\in L^2$, where

$$\begin{array}{c}\sigma =\tau -a{\xi }^4,{\sigma }_1={\tau }_1-a{\xi }_1^4,{\sigma }_2={\tau }_2-a{\xi }_2^4,{\sigma }_3={\tau }_3+a{\xi }_3^4,\hfill \\ f_1({\tau }_1,{\xi }_1)={〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{\hat{u}}_1,f_2({\tau }_2,{\xi }_2)={〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{\hat{u}}_2,f_3({\tau }_1,{\xi }_1)={〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}{\hat{\overline{u}}}_3.\hfill \end{array}$$

Let ${\hat{F}}_{\rho }^j({\tau }_j,{\xi }_j)=\frac{f_j({\tau }_j,{\xi }_j)}{{(1+|{\tau }_j-a{\xi }_j^4|)}^{\rho }}=\frac{f_j({\tau }_j,{\xi }_j)}{{〈{\sigma }_j〉}^{\rho }}$, $j=1,2$, ${\hat{F}}_{\rho }^3({\tau }_3,{\xi }_3)=\frac{f_3({\tau }_3,{\xi }_3)}{{(1+|{\tau }_3+a{\xi }_3^4|)}^{\rho }}=\frac{f_3({\tau }_3,{\xi }_3)}{{〈{\sigma }_3〉}^{\rho }}$.

Without loss of generality, we can assume that $\overline{f}\ge 0$, $f_j\ge 0$ for $j=1,2,3$.

We split the domain of integration into two cases $|\xi |\ge 3$ and $|\xi |\le 3$.

Case I. Assume that $|\xi |\le 3$.

Noting that $s\ge \frac{1}{2}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.1, we obtain

$$\begin{array}{rl}\bigwedge & \le {\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^{\frac{3}{2}}{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{|{\xi }_1|}^{\frac{1}{4}}{〈{\sigma }_1〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\sigma }_2〉}^{b^{\prime }}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_R{\int }_R|{\overline{F}}_0|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\right|\cdot \left|D_x^{\frac{3}{2}}{F}_{b^{\prime }}^3\right|dxdt\\ \le C{\parallel F_0\parallel }_{L_x^2{L}_t^2}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{\frac{3}{2}}{F}_{b^{\prime }}^3\parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

Case II. Assume that $|\xi |\ge 3$.

Subcase 1. Assume that $|{\xi }_3|\le 1$. Then we have $|\xi |\le 3max(|{\xi }_1|,|{\xi }_2|)$.

Noting that $1-b>\frac{5}{16}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.2, we obtain

$$\begin{array}{rl}\bigwedge & \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\sigma }_1〉}^{b^{\prime }}{〈{\sigma }_2〉}^{b^{\prime }}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_R{\int }_R|{\overline{F}}_{1-b}|\cdot \left|F_{b^{\prime }}^1\right|\cdot \left|F_{b^{\prime }}^2\right|\cdot \left|F_{b^{\prime }}^2\right|dxdt\\ \le C{\parallel F_{1-b}\parallel }_{L_x^4{L}_t^4}{\parallel F_{b^{\prime }}^1\parallel }_{L_x^4{L}_t^4}{\parallel F_{b^{\prime }}^2\parallel }_{L_x^4{L}_t^4}{\parallel F_{b^{\prime }}^3\parallel }_{L_x^4{L}_t^4}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

Subcase 2. Assume that $|{\xi }_3|\ge 1$. We split this domain of integration in several pieces.

${\mathbf{1}}^{\circ }$ Assume that $|\xi |\le 3max(|{\xi }_1|,|{\xi }_2|,|{\xi }_3|)=3|{\xi }_1|$.

Similar to the proof of subcase 1, noting that $s\ge \frac{1}{2}$, $1-b>\frac{5}{16}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.2, we obtain

$$\begin{array}{rl}\bigwedge & \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\sigma }_1〉}^{b^{\prime }}{〈{\sigma }_2〉}^{b^{\prime }}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

${\mathbf{2}}^{\circ }$ Assume that $|\xi |\le 3max(|{\xi }_1|,|{\xi }_2|,|{\xi }_3|)=3|{\xi }_2|$.

In this case, the proof is similar to that of the above case, so here we omit the detailed proof.

${\mathbf{3}}^{\circ }$ Assume that $|\xi |\le 3max(|{\xi }_1|,|{\xi }_2|,|{\xi }_3|)=3|{\xi }_3|$.

In this case, we easily get

$$|\xi |\sim |{\xi }_3|.$$

(2.12)

By a straightforward calculation, we can obtain

$$\begin{array}{rl}4max(|\sigma |,|{\sigma }_1|,|{\sigma }_2|,|{\sigma }_3|)& \ge |\sigma |+|\sigma |+|{\sigma }_2|+|{\sigma }_3|\ge |\sigma -{\sigma }_1-{\sigma }_2-{\sigma }_3|\\ =2|a||\xi {\xi }_3+{\xi }_1{\xi }_2||2{\xi }_1^2+3{\xi }_1{\xi }_2+2{\xi }_2^2+\xi {\xi }_3|\\ \ge C{〈\xi 〉}^2{〈{\xi }_3〉}^2.\end{array}$$

(2.13)

3.0: Assume that $|\sigma |=max(|\sigma |,|{\sigma }_1|,|{\sigma }_2|,|{\sigma }_3|)$, so $〈\sigma 〉\ge C{〈\xi 〉}^4$ holds.

Noting that $b<\frac{11}{16}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.1, we obtain

$$\begin{array}{rl}\bigwedge & ={\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^{s}i{|{\xi }_3|}^2\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\xi 〉}^{4(1-b)}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi ){|{\xi }_1|}^{\frac{1}{4}}{f}_1({\tau }_1,{\xi }_1){|{\xi }_2|}^{\frac{1}{4}}{f}_2({\tau }_2,{\xi }_2){|{\xi }_3|}^{\frac{3}{2}}{|{\xi }_3|}^{\frac{1}{2}}{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\xi 〉}^{4(1-b)}{〈{\xi }_1〉}^s{|{\xi }_1|}^{\frac{1}{4}}{〈{\sigma }_1〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^{\frac{3}{2}}{f}_3({\tau }_3,{\xi }_3)d\delta }{{|{\xi }_1|}^{\frac{1}{4}}{〈{\sigma }_1〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\sigma }_2〉}^{b^{\prime }}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_R{\int }_R|F_0|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\right|\cdot \left|D_x^{\frac{3}{2}}{F}_{b^{\prime }}^3\right|dxdt\\ \le C{\parallel F_0\parallel }_{L_x^2{L}_t^2}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{\frac{3}{2}}{F}_{b^{\prime }}^3\parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

3.1: Assume that $|{\sigma }_1|=max(|\sigma |,|{\sigma }_1|,|{\sigma }_2|,|{\sigma }_3|)$. By (2.10) and (2.11), we immediately obtain $〈{\sigma }_1〉\ge C{〈{\xi }_3〉}^4$.

Noting that $\frac{1}{2}<b<\frac{11}{16}$, $b^{\prime }>\frac{1}{2}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.1, we obtain

$$\begin{array}{rl}\bigwedge & ={\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^{s}i{|{\xi }_3|}^2\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le {\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{1-b}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{1-b}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{〈{\xi }_1〉}^s{〈{\xi }_3〉}^{4(1-b)}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{|\xi |}^{\frac{3}{2}}{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1){|{\xi }_2|}^{\frac{1}{4}}{f}_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{\xi }_3{|}^{\frac{1}{4}}{f}_3({\tau }_3,{\xi }_3)d\delta }{{|\xi |}^{\frac{3}{2}}{〈\sigma 〉}^{b^{\prime }}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{|{\xi }_2|}^{\frac{1}{4}}{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{|{\xi }_3|}^{\frac{1}{4}}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{|\xi |}^{\frac{3}{2}}\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1){|{\xi }_2|}^{\frac{1}{4}}{f}_2({\tau }_2,{\xi }_2){|{\xi }_3|}^{\frac{1}{4}}{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\sigma }_2〉}^{b^{\prime }}{|{\xi }_3|}^{\frac{1}{4}}{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le C{\int }_R{\int }_R\left|D_x^{\frac{3}{2}}{F}_{b^{\prime }}\right|\cdot \left|F_0^1\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^3\right|dxdt\\ \le C{\parallel D_x^{\frac{3}{2}}{F}_{b^{\prime }}\parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}{\parallel F_0^1\parallel }_{L_x^2{L}_t^2}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^3\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

3.2: Assume that $|{\sigma }_2|=max(|\sigma |,|{\sigma }_1|,|{\sigma }_2|,|{\sigma }_3|)$.

The proof is similar to that of the case 3.1, so here omit the detailed proof.

3.3: Assume that $|{\sigma }_3|=max(|\sigma |,|{\sigma }_1|,|{\sigma }_2|,|{\sigma }_3|)$. By (2.10) and (2.11), we immediately obtain $〈{\sigma }_3〉\ge C{〈{\xi }_3〉}^4$.

Noting that $\frac{1}{2}<b<\frac{11}{16}$, $b^{\prime }>\frac{1}{2}$, by Lemma 2.4, the Hölder inequality, and Lemma 2.1, we obtain

$$\begin{array}{rl}\bigwedge & ={\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^{s}i{|{\xi }_3|}^2\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{1-b}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{b^{\prime }}}\\ \le {\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\sigma }_3〉}^{1-b}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈\xi 〉}^s\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\xi }_3〉}^{4(1-b)}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{〈{|\xi |}^{\frac{3}{2}}\xi 〉}^s\overline{f}(\tau ,\xi ){|{\xi }_1|}^{\frac{1}{4}}{f}_1({\tau }_1,{\xi }_1){|{\xi }_2|}^{\frac{1}{4}}{f}_2({\tau }_2,{\xi }_2){|{\xi }_3|}^2{f}_3({\tau }_3,{\xi }_3)d\delta }{{|\xi |}^{\frac{3}{2}}{〈\sigma 〉}^{b^{\prime }}{|{\xi }_1|}^{\frac{1}{4}}{〈{\xi }_1〉}^s{〈{\sigma }_1〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\xi }_2〉}^s{〈{\sigma }_2〉}^{b^{\prime }}{〈{\xi }_3〉}^s{〈{\xi }_3〉}^{4(1-b)}}\\ \le C{\int }_{{\mathrm{\Gamma }}_{\tau }}{\int }_{{\mathrm{\Gamma }}_{\xi }}\frac{{|\xi |}^{\frac{3}{2}}\overline{f}(\tau ,\xi )f_1({\tau }_1,{\xi }_1)f_2({\tau }_2,{\xi }_2)f_3({\tau }_3,{\xi }_3)d\delta }{{〈\sigma 〉}^{b^{\prime }}{|{\xi }_1|}^{\frac{1}{4}}{〈{\sigma }_1〉}^{b^{\prime }}{|{\xi }_2|}^{\frac{1}{4}}{〈{\sigma }_2〉}^{b^{\prime }}}\\ \le C{\int }_R{\int }_R\left|D_x^{\frac{3}{2}}{F}_{b^{\prime }}\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\right|\cdot \left|D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\right|\cdot \left|F_0^3\right|dxdt\\ \le C{\parallel D_x^{\frac{3}{2}}{F}_{b^{\prime }}\parallel }_{L_x^{\mathrm{\infty }}{L}_t^2}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^1\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel D_x^{-\frac{1}{4}}{F}_{b^{\prime }}^2\parallel }_{L_x^4{L}_t^{\mathrm{\infty }}}{\parallel F_0^3\parallel }_{L_x^2{L}_t^2}\\ \le C{\parallel f\parallel }_{L_{\tau }^2{L}_{\xi }^2}{\parallel f_1\parallel }_{L_{{\tau }_1}^2{L}_{{\xi }_1}^2}{\parallel f_2\parallel }_{L_{{\tau }_2}^2{L}_{{\xi }_2}^2}{\parallel f_3\parallel }_{L_{{\tau }_3}^2{L}_{{\xi }_3}^2}.\end{array}$$

 □

Lemma 2.6 Let $s\ge \frac{1}{2}$, $\frac{1}{2}<b<\frac{5}{8}$. Then we have the following inequality:

$${\parallel \prod _{j=1}^9{u}_j\parallel }_{X_{s,b-1}}\le C\prod _{j=1}^9{\parallel u_j\parallel }_{X_{s,b}}.$$

(2.14)

Proof Firstly, we prove (2.14) holds for $s=0$.

By the definition, the Hölder inequality, and the Sobolev inequality, we have

$$\begin{array}{rl}{\parallel u\parallel }_{0,b-1}& =\underset{{\parallel v\parallel }_{0,1-b}\le 1}{sup}|〈u,v〉|\le \underset{{\parallel v\parallel }_{0,1-b}\le 1}{sup}{\parallel u\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}{\parallel v\parallel }_{L_x^2{L}_t^{\frac{2}{2b-1}}}\\ \le C\underset{{\parallel v\parallel }_{0,1-b}\le 1}{sup}{\parallel u\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}{\parallel v\parallel }_{L_x^2{\dot{W}}_t^{1-b,2}}\le C\underset{{\parallel v\parallel }_{0,1-b}\le 1}{sup}{\parallel u\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}{\parallel v\parallel }_{0,1-b}\\ \le C{\parallel u\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}.\end{array}$$

(2.15)

That means

$${\parallel \prod _{j=1}^9{u}_j\parallel }_{0,b-1}\le C{\parallel \prod _{j=1}^9{u}_j\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}.$$

(2.16)

Again using the Hölder inequality, the Sobolev inequality in the variable x, and Lemma 2.1, we obtain

$${\parallel \prod _{j=1}^9{u}_j\parallel }_{L_x^2{L}_t^{\frac{2}{3-2b}}}\le \prod _{j=1}^9{\parallel u_j\parallel }_{L_x^{\overline{p_j}}{L}_t^{q_j}}\le C\prod _{j=1}^9{\parallel u_j\parallel }_{{\dot{W}}_x^{s_j,p_j}{L}_t^{q_j}}\le C\prod _{j=1}^9{\parallel u_j\parallel }_{s_j,b},$$

(2.17)

where ${\sum }_{j=1}^9\frac{1}{q_j}=\frac{3}{2}-b$, ${\sum }_{j=1}^9\frac{1}{\overline{p_j}}=\frac{1}{2}$, $\frac{1}{p_j}-\frac{1}{\overline{p_j}}=s_j$, ${\sum }_{j=1}^9{s}_j=4b-2$, $(q_j,p_j)$ is admissible pair.

Combining (2.16) and (2.17), we obtain

$${\parallel \prod _{j=1}^9{u}_j\parallel }_{0,b-1}\le C\prod _{j=1}^9{\parallel u_j\parallel }_{s_j,b}.$$

(2.18)

Secondly, using the Leibniz rule for fractional power, in a similar way, we obtain

$$\begin{array}{rl}{\parallel D_x^s\left(\prod _{j=1}^9{u}_j\right)\parallel }_{0,b-1}& \le C\sum _{k=1}^9\prod _{j=1}^9{\parallel u_j\parallel }_{L_x^{\overline{p_j^k}}{L}_t^{q_j^k}}\le C\sum _{k=1}^9\prod _{j=1}^9{\parallel u_j\parallel }_{{\dot{W}}_x^{s_j^k,p_j^k}{L}_t^{q_j^k}}\\ \le C\sum _{k=1}^9\prod _{j=1}^9{\parallel u_j\parallel }_{s_j^k,b},\end{array}$$

(2.19)

where ${\sum }_{j=1}^9\frac{1}{q_j^k}=\frac{3}{2}-b$, ${\sum }_{j=1}^9\frac{1}{\overline{p_j^k}}=\frac{1}{2}$, $\frac{1}{p_j^k}-\frac{1}{\overline{p_j^k}}=s_j$, ${\sum }_{j=1}^9{s}_j^k=s+4b-2$, $(q_j^k,p_j^k)$ is admissible pair for $k=1,\dots ,9$.

Taking $s_k^k=s$, $p_k^k=\mathrm{\infty }$, $q_k^k=8$; and for some $k_0\ne k$, $s_{k_0}^k=4b-2$, $p_{k_0}^k=\frac{1}{4b-2}$, $q_{k_0}^k=\frac{8}{5-8b}$; $s_j^k=0$, $p_j^k=14$, $q_j^k=\frac{28}{3}$, for $j\ne k,k_0$ in (2.19), noting $s\ge \frac{1}{2}$ and $\frac{1}{2}<b<\frac{5}{8}$, we obtain

$$\begin{array}{rl}{\parallel D_x^s\left(\prod _{j=1}^9{u}_j\right)\parallel }_{0,b-1}& \le C\sum _{k=1}^9{\parallel u_k\parallel }_{s,b}{\parallel u_{k_0}\parallel }_{4b-2,b}\prod _{j=1,j\ne k,k_0}^9{\parallel u_j\parallel }_{0,b}\\ \le C\sum _{k=1}^9{\parallel u_k\parallel }_{s,b}{\parallel u_{k_0}\parallel }_{s,b}\prod _{j=1,j\ne k,k_0}^9{\parallel u_j\parallel }_{0,b}\\ \le C\prod _{j=1}^9{\parallel u_j\parallel }_{s,b},\end{array}$$

(2.20)

which completes the proof. □

3 Proof of main result

Proof of Theorem 1.1 We will prove Theorem 1.1 by using the Banach fixed point theorem. Let $T<\delta$. We rewrite (1.4) in integral form:

$$u(t)=U(t)u_0-{\int }_0^{t}U(t-t^{\prime })(b{u}^2{\overline{u}}_{xx}+c{|u|}^{8}u)\left(t^{\prime }\right)d{t}^{\prime }.$$

(3.1)

We denote

$$Z=\{u\in X_{s,b}:{\parallel u\parallel }_{X_{s,b}}\le 2C_0{\parallel u_0\parallel }_{H^s}\},$$

and define a mapping S as follows: For $u\in X_{s,b}$,

$$Su(t)=\psi (t)U(t)u_0-\psi (t){\int }_0^{t}U(t-t^{\prime }){\psi }_{\delta }\left(t^{\prime }\right)[b{u}^2{\overline{u}}_{xx}+c{|u|}^{8}u]\left(t^{\prime }\right)d{t}^{\prime }.$$

In the sequel we will prove that S is well defined and it is a contraction map on Z.

By Lemma 2.3 and Lemmas 2.5-2.6, for $\frac{1}{2}<b<b^{\prime }<\frac{5}{8}$, we obtain

$$\begin{array}{rl}{\parallel Su\parallel }_{X_{s,b}}& \le C_0{\parallel u_0\parallel }_{H^s}+C_1{\delta }^{b^{\prime }-b}({\parallel u\parallel }_{X_{s,b}}^3+{\parallel u\parallel }_{X_{s,b}}^9)\\ \le C_0{\parallel u_0\parallel }_{H^s}+C_1{\delta }^{b^{\prime }-b}(2^3{C}_0^3{\parallel u_0\parallel }_{H^s}^3+2^9{C}_0^9{\parallel u_0\parallel }_{H^s}^9).\end{array}$$

If we take δ such that $C_1{\delta }^{b^{\prime }-b}(2^3{C}_0^2{\parallel u_0\parallel }_{H^s}^2+2^9{C}_0^8{\parallel u_0\parallel }_{H^s}^8)<1$, then $SZ\subset Z$.

Similarly, for $u_1,u_2\subset Z$, in an analogous way to the above, we can obtain

$$\begin{array}{rl}{\parallel S{u}_1-S{u}_2\parallel }_{X_{s,b}}& \le C_2{\delta }^{b^{\prime }-b}({\parallel u_1\parallel }_{X_{s,b}}^2+{\parallel u_2\parallel }_{X_{s,b}}^2+{\parallel u_1\parallel }_{X_{s,b}}^8+{\parallel u_2\parallel }_{X_{s,b}}^8){\parallel u_1-u_2\parallel }_{X_{s,b}}\\ \le 2C_2{\delta }^{b^{\prime }-b}(2^2{C}_0^2{\parallel u_0\parallel }_{H^s}^2+2^8{C}_0^8{\parallel u_0\parallel }_{H^s}^8){\parallel u_1-u_2\parallel }_{X_{s,b}}.\end{array}$$

Furthermore, if we take δ such that $C_2{\delta }^{b^{\prime }-b}(2^3{C}_0^2{\parallel u_0\parallel }_{H^s}^2+2^9{C}_0^8{\parallel u_0\parallel }_{H^s}^8)<1$ then S is a contraction mapping of Z into itself. The desired result immediately follows from Banach’s fixed point theorem. That means that there is a unique solution which solves the Cauchy problem (1.4) for $T<\delta$. The Lipschitz continuousness from $H^s(R)$ to $C([0,T];H^s(R))$ is easily obtained from the above proof process. □