The Cauchy problem for the modified Novikov equation

Abstract

In this paper, we are concerned with the Cauchy problem for the modified Novikov equation. By using the transport equation theory and Littlewood-Paley decomposition as well as nonhomogeneous Besov spaces, we prove that the Cauchy problem for the modified Novikov equation is locally well posed in the Besov space $B_{p,r}^s$ with $1\le p,r\le +\mathrm{\infty }$ and $s>max\{1+\frac{1}{p},\frac{3}{2}\}$ and show that the Cauchy problem for the modified Novikov equation is locally well posed in the Besov space $B_{2,1}^{3/2}$ with the aid of Osgood lemma.

MSC: 35G25, 35L05, 35R25.

1 Introduction

Recently, Zhao and Zhou [1] considered the exact traveling wave solution to the following modified Novikov equation:

$$u_t-u_{txx}+4u^4{u}_x=3u{u}_x{u}_{xx}+u^2{u}_{xxx}.$$

(1.1)

We recall that the Novikov equation

$$u_t-u_{txx}=u^2{u}_{xxx}+3u{u}_x{u}_{xx}-4u^2{u}_x$$

(1.2)

was discovered by Vladimir Novikov [2] and it possesses the bi-Hamiltonian structure, infinite conservation laws. The well-posedness and blow-up of the Cauchy problem for the Novikov equation in Sobolev spaces and Besov spaces have been investigated by some authors [3]–[7]. The weak solution of the Cauchy problem for the Novikov equation has been investigated by some authors [4], [5], [8]. Recently, Li and Yan [9] considered the Cauchy problem for the KdV equation with higher dispersion.

We define $P_1(D)={\partial }_x{(1-{\partial }_x^2)}^{-1}$ and $P_2(D)={(1-{\partial }_x^2)}^{-1}$. By using the fact that $G(x)=\frac{1}{2}{e}^{-|x|}$ and $G(x)\ast f={(1-{\partial }_x^2)}^{-1}f$ for all $f\in L^2(\mathbf{R})$ and $G\ast y=u$, we can rewrite (1.1) as follows:

$$u_t+u^2{u}_x+P_1(D)[\frac{4}{5}{u}^5-\frac{1}{3}{u}^3+\frac{3}{2}u{u}_x^2]+P_2(D)\left[\frac{1}{2}{u}_x^3\right]=0,t>0.$$

Now we consider the following problem:

$$u_t+u^2{u}_x+P_1(D)[\frac{4}{5}{u}^5-\frac{1}{3}{u}^3+\frac{3}{2}u{u}_x^2]+P_2(D)\left[\frac{1}{2}{u}_x^3\right]=0,t>0,$$

(1.3)

$$u(x,0)=u_0(x),x\in \mathbf{R}.$$

(1.4)

To the best of our knowledge, the well-posedness and blow-up of the Cauchy problem for (1.3) and (1.4) in Besov spaces are open up to now. More precisely, in this paper, motivated by [10], [11], using Littlewood-Paley decomposition and nonhomogeneous Besov spaces, we prove that the Cauchy problem for (1.4) is locally well posed in the Besov space $B_{p,r}^s$ with $s>max\{1+\frac{1}{p},\frac{3}{2}\}$ and we give a blow-up criterion.

To introduce the main results, we define

$$\begin{array}{c}{E}_{p,r}^s(T)=C([0,T];B_{p,r}^s)\cap C^1([0,T];B_{p,r}^{s-1})\text{if }r<\mathrm{\infty },\hfill \\ E_{p,\mathrm{\infty }}^s(T)=L^{\mathrm{\infty }}(0,T;B_{p,\mathrm{\infty }}^s)\cap Lip([0,T];B_{p,r}^{s-1}).\hfill \end{array}$$

The main results of this paper are as follows.

Theorem 1.1

Let$1\le p,r\le \mathrm{\infty }$and$s>max(\frac{3}{2},1+\frac{1}{p})$and$u_0\in B_{p,r}^s$. Then there exists a time$T>0$such that problem (1.3) and (1.4) has a unique solution$u$in$E_{p,r}^s(T)$. The map$u_0⟶u$is continuous from a neighborhood of$u_0$in$B_{p,r}^s$into$C([0,T];B_{p,r}^{s^{\prime }})\cap C^1([0,T];B_{p,r}^{s^{\prime }-1})$for every$s^{\prime }<s$. When$r<\mathrm{\infty }$, the solution to problem (1.3) and (1.4) is continuous in$E_{p,r}^s(T)$.

Theorem 1.2

When$u_0\in B_{2,1}^{3/2}$, (1.3) and (1.4) is locally well posed in the sense of Hadamard.

The remainder of this paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we establish local well-posedness of the Cauchy problem for the generalized Camassa-Holm equation in Besov spaces. In Section 4, we prove Theorem 1.2.

2 Preliminaries

In this section, the nonhomogeneous Besov spaces and the theory of transport equation which can be seen in [10]–[13] are presented.

Lemma 2.1

(Littlewood-Paley decomposition)

There exists a couple of smooth radial functions$(\chi ,\varphi )$valued in$[0,1]$such that$\chi$is supported in the ball$B=\{\xi \in {\mathbf{R}}^n,|\xi |\le \frac{4}{3}\}$and$\varphi$is supported in the ring$C=\{\xi \in {\mathbf{R}}^n,\frac{3}{4}\le |\xi |\le \frac{8}{3}\}$. Moreover,

and

$$\begin{array}{c}Supp\varphi (2^{-q}\cdot )\cap Supp\varphi (2^{-q^{\prime }}\cdot )=\mathrm{\varnothing }\mathit{\text{if }}|q-q^{\prime }|\ge 2,\hfill \\ Supp\chi (\cdot )\cap Supp\varphi (2^{-q}\cdot )=\mathrm{\varnothing }\mathit{\text{if }}|q|\ge 1.\hfill \end{array}$$

Then, for$u\in {\mathcal{S}}^{\prime }(\mathbf{R})$, the nonhomogeneous dyadic blocks are defined as follows:

$$\begin{array}{c}{\mathrm{\Delta }}_{q}u=0\mathit{\text{if }}q\le -2,\hfill \\ {\mathrm{\Delta }}_{-1}u=\chi (D)u={\mathcal{F}}_x^{-1}\chi {\mathcal{F}}_{x}u,\hfill \\ {\mathrm{\Delta }}_{q}u=\varphi \left(2^{-q}D\right)={\mathcal{F}}_x^{-1}\varphi \left(2^{-q}\xi \right){\mathcal{F}}_{x}u\mathit{\text{if }}q\ge 0.\hfill \end{array}$$

Thus$u={\sum }_{q\in \mathbf{Z}}{\mathrm{\Delta }}_{q}u\mathit{\text{in}}{\mathcal{S}}^{\prime }(\mathbf{R})$.

Remark

The low frequency cut-off $S_q$ is defined by

$$S_{q}u=\sum _{p=-1}^{q-1}\mathrm{\Delta }u=\chi \left(2^{-q}D\right)u={\mathcal{F}}_x^{-1}\chi \left(2^{-q}\xi \right){\mathcal{F}}_{x}u,\mathrm{\forall }q\in N.$$

It is easily checked that

$$\begin{array}{c}{\mathrm{\Delta }}_p{\mathrm{\Delta }}_{q}u\equiv 0\text{if }|p-q|\ge 2,\hfill \\ {\mathrm{\Delta }}_q(S_{p-1}u{\mathrm{\Delta }}_{p}v)\equiv 0\text{if }|p-q|\ge 5,\mathrm{\forall }u,v\in {\mathcal{S}}^{\prime }(\mathbf{R})\hfill \end{array}$$

as well as

$${\parallel {\mathrm{\Delta }}_{q}u\parallel }_{L^p}\le {\parallel u\parallel }_{L^p},{\parallel S_{q}u\parallel }_{L^p}\le C{\parallel u\parallel }_{L^p},\mathrm{\forall }1\le p\le +\mathrm{\infty }$$

with the aid of Young’s inequality, where $C$ is a positive constant independent of $q$.

Definition

(Besov spaces)

Let $s\in \mathbf{R}$, $1\le p\le +\mathrm{\infty }$. The nonhomogeneous Besov space $B_{p,r}^s({\mathbf{R}}^n)$ is defined by

$$B_{p,r}^s\left({\mathbf{R}}^n\right)=\{f\in {\mathcal{S}}^{\prime }(\mathbf{R}):{\parallel f\parallel }_{B_{p,r}^s}={\parallel 2^{qs}{\mathrm{\Delta }}_{q}f\parallel }_{l^r(L^p)}={\parallel {\left(2^{qs}{\parallel {\mathrm{\Delta }}_{q}f\parallel }_{L^p}\right)}_{q\ge -1}\parallel }_{l^r}<\mathrm{\infty }\}.$$

In particular, $B_{p,r}^{\mathrm{\infty }}={\bigcap }_{s\in \mathbf{R}}{B}_{p,r}^s$. Let $T>0$, $s\in \mathbf{R}$ and $1\le p\le \mathrm{\infty }$. Define $E_{p,r}^s={\bigcap }_{T>0}{E}_{p,r}^s(T)$.

Lemma 2.2

Let$s\in \mathbf{R}$, $1\le p,r,p_j,r_j\le \mathrm{\infty }$, $j=1,2$, then:

• (1) Topological properties: $B_{p,r}^s$ is a Banach space which is continuously embedded in ${\mathcal{S}}^{\prime }(\mathbf{R})$.

• (2) Density: $C_c^{\mathrm{\infty }}$ is dense in $B_{p,r}^s\iff 1\le p,r<\mathrm{\infty }$.

• (3) Embedding: $B_{p_1,r_1}^s\hookrightarrow B_{p_2,r_2}^{s-n(\frac{1}{p_1}-\frac{1}{p_2})}$ if $p_1\le p_2$ and $r_1\le r_2$.

  $$B_{p,r_2}^{s_2}\hookrightarrow B_{p,r_1}^{s_1}\mathit{\text{ locally compact}}\mathit{\text{if }}{s}_1<s_2.$$

• (4) Algebraic properties: $\mathrm{\forall }s>0$, $B_{p,r}^s\cap L^{\mathrm{\infty }}$ is a Banach algebra. $B_{p,r}^s$ is a Banach algebra ⇔ $B_{p,r}^s\hookrightarrow L^{\mathrm{\infty }}$ ⇔ $s>\frac{1}{p}$ or ($s\ge \frac{1}{p}$ and $r=1$). In particular, $B_{2,1}^{1/2}$ is continuously embedded in $B_{2,\mathrm{\infty }}^{1/2}\cap L^{\mathrm{\infty }}$ and $B_{2,\mathrm{\infty }}^{1/2}\cap L^{\mathrm{\infty }}$ is a Banach algebra.

• (5) 1-D Moser-type estimates:

• (i) For $s>0$,

  $${\parallel fg\parallel }_{B_{p,r}^s}\le C({\parallel f\parallel }_{B_{p,r}^s}{\parallel g\parallel }_{L^{\mathrm{\infty }}}+{\parallel f\parallel }_{L^{\mathrm{\infty }}}{\parallel g\parallel }_{B_{p,r}^s}).$$

• (ii) $\mathrm{\forall }{s}_1\le \frac{1}{p}<s_2$ ($s_2\ge \frac{1}{p}$ if $r=1$) and $s_1+s_2>0$, we have

  $${\parallel fg\parallel }_{B_{p,r}^{s_1}}\le C{\parallel f\parallel }_{B_{p,r}^{s_1}}{\parallel g\parallel }_{B_{p,r}^{s_2}}.$$

• (6) Complex interpolation:

  $${\parallel f\parallel }_{B_{p,r}^{\theta s_1+(1-\theta )s_2}}\le {\parallel f\parallel }_{B_{p,r}^{s_1}}^{\theta }{\parallel g\parallel }_{B_{p,r}^{s_2}}^{1-\theta },\mathrm{\forall }f\in B_{p,r}^{s_1}\cap B_{p,r}^{s_2},\mathrm{\forall }\theta \in [0,1].$$

• (7) Real interpolation: $\mathrm{\forall }\theta \in (0,1)$, $s_1>s_2$, $s=\theta s_1+(1-\theta )s_2$, there exists a constant $C$ such that

  $${\parallel u\parallel }_{B_{p,1}^s}\le \frac{C(\theta )}{s_1-s_2}{\parallel u\parallel }_{B_{p,\mathrm{\infty }}^{s_1}}^{\theta }{\parallel u\parallel }_{B_{p,\mathrm{\infty }}^{s_2}}^{1-\theta },\mathrm{\forall }u\in B_{p,\mathrm{\infty }}^{s_1}.$$

In particular, for any$0<\theta <1$, we have that

$${\parallel u\parallel }_{B_{2,1}^{1/2}}\le {\parallel u\parallel }_{B_{2,1}^{\frac{3}{2}-\theta }}\le C(\theta ){\parallel u\parallel }_{B_{2,\mathrm{\infty }}^{1/2}}^{\theta }{\parallel u\parallel }_{B_{2,\mathrm{\infty }}^{3/2}}^{1-\theta }.$$

(2.1)

• (8) Fatou lemma: if is bounded in $B_{p,r}^s$ and $u_n⟶u$ in ${\mathcal{S}}^{\prime }(\mathbf{R})$, then $u\in B_{p,r}^s$ and

  $${\parallel u\parallel }_{B_{p,r}^s}\le \underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel u_n\parallel }_{B_{p,r}^s}.$$

• (9) Let $m\in \mathbf{R}$ and $f$ be an $S^m$-multiplier (i.e., $f:{\mathbf{R}}^n\to \mathbf{R}$ is smooth and satisfies that $\mathrm{\forall }\alpha \in N^n$, ∋ a constant $C_{\alpha }$, s.t. $|{\partial }_{\alpha }f(\xi )|\le C_{\alpha }{(1+|\xi |)}^{m-|\alpha |}$ for all $\xi \in {\mathbf{R}}^n$). Then the operator $f(D)$ is continuous from $B_{p,r}^s$ to $B_{p,r}^{s-m}$. Notice that $P_1(D)$ is continuous from $B_{p,r}^s$ to $B_{p,r}^{s-1}$ and $P_2(D)$ is continuous from $B_{p,r}^s$ to $B_{p,r}^{s-2}$.

• (10) The usual product is continuous from $B_{2,1}^{-1/2}\times (B_{2,\mathrm{\infty }}^{1/2}\cap L^{\mathrm{\infty }})$ to $B_{2,\mathrm{\infty }}^{-1/2}$.

• (11) There exists a constant $C>0$ such that the following interpolation inequality holds:

  $${\parallel f\parallel }_{B_{2,1}^{1/2}}\le C{\parallel f\parallel }_{B_{2,\mathrm{\infty }}^{1/2}}ln(e+\frac{{\parallel f\parallel }_{B_{2,\mathrm{\infty }}^{3/2}}}{{\parallel f\parallel }_{B_{2,\mathrm{\infty }}^{1/2}}}).$$

Lemma 2.3

(A priori estimates in Besov spaces)

Let$1\le p,r\le \mathrm{\infty }$and$s>-min(\frac{1}{p},1-\frac{1}{p})$. Assume that$f_0\in B_{p,r}^s$, $F\in L^1(0,T;B_{p,r}^s)$and${\partial }_{x}v$belongs to$L^1(0,T;B_{p,r}^{s-1})$if$s>1+\frac{1}{p}$or to$L^1(0,T;B_{p,r}^{1/p}\cap L^{\mathrm{\infty }})$otherwise. If$f\in L^{\mathrm{\infty }}(0,T;B_{p,r}^s)\cap C([0,T];{\mathcal{S}}^{\prime }(\mathbf{R}))$solves the following 1-D linear transport equation:

$$f_t+v{f}_x=F,$$

(2.2)

$$f(x,0)=f_0,$$

(2.3)

then there exists a constant$C$depending only on$s$, $p$, $r$such that the following statements hold:

• (1) If $r=1$ or $s\ne 1+\frac{1}{p}$, then

  $${\parallel f\parallel }_{B_{p,r}^s}\le {\parallel f_0\parallel }_{B_{p,r}^s}+{\int }_0^t{\parallel F(\tau )\parallel }_{B_{p,r}^s}d\tau +C{\int }_0^t{V}^{\prime }(\tau ){\parallel f(\tau )\parallel }_{B_{p,r}^s}d\tau$$

or hence,

$${\parallel f\parallel }_{B_{p,r}^s}\le e^{CV(t)}({\parallel f_0\parallel }_{B_{p,r}^s}+{\int }_0^t{e}^{-CV(\tau )}{\parallel F(\tau )\parallel }_{B_{p,r}^s}d\tau )$$

(2.4)

with$V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{B_{p,r}^{1/p}\cap L^{\mathrm{\infty }}}d\tau$if$s<1+\frac{1}{p}$and$V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{B_{p,r}^{s-1}}d\tau$else.

• (2) If $s\le 1+\frac{1}{p}$, $f_0^{\prime }\in L^{\mathrm{\infty }}$ and $f_x\in L^{\mathrm{\infty }}((0,T)\times \mathbf{R})$ and $F_x\in L^1(0,T;L^{\mathrm{\infty }})$, then

  $$\begin{array}{c}{\parallel f(t)\parallel }_{B_{p,r}^s}+{\parallel f_x(t)\parallel }_{L^{\mathrm{\infty }}}\hfill \\ \le e^{CV(t)}({\parallel f_0\parallel }_{B_{p,r}^s}+{\parallel f_{0x}\parallel }_{L^{\mathrm{\infty }}}+{\int }_0^t{e}^{-CV(\tau )}[{\parallel F(\tau )\parallel }_{B_{p,r}^s}+{\parallel F_x(\tau )\parallel }_{L^{\mathrm{\infty }}}]d\tau )\hfill \end{array}$$

with

$$V(t)={\int }_0^t{\parallel {\partial }_{x}v(\tau )\parallel }_{B_{p,r}^{1/p}\cap L^{\mathrm{\infty }}}.$$

• (3) If $f=v$, then for all $s>0$, (1) holds true when $V(t)={\int }_0^t{\parallel v_x(\tau )\parallel }_{L^{\mathrm{\infty }}}d\tau$.

• (4) If $r<\mathrm{\infty }$, then $f\in C([0,T];B_{p,r}^s)$. If $r=\mathrm{\infty }$, then $f\in C([0,T];B_{p,1}^{s^{\prime }})$ for all $s^{\prime }<s$.

Lemma 2.4

(Existence and uniqueness)

Let$p$, $r$, $s$, $f_0$and$F$be as in the statement of Lemma  2.3. Assume that$v\in L^{\rho }(0,T;B_{\mathrm{\infty },\mathrm{\infty }}^{-M})$for some$\rho >1$and$M>0$and$v_x\in L^1(0,T;B_{p,r}^{s-1})$if$s>1+\frac{1}{p}$or$s=1+\frac{1}{p}$and$r=1$and$v_x\in L^1(0,T;B_{p,\mathrm{\infty }}^{1/p}\cap L^{\mathrm{\infty }})$if$s<1+\frac{1}{p}$. Then problem (2.2) and (2.3) has a unique solution$f\in L^{\mathrm{\infty }}(0,T;B_{p,r}^s)\cap ({\bigcap }_{s^{\prime }<s}C([0,T];B_{p,1}^{s^{\prime }}))$and the inequalities of Lemma  2.3can hold true. Moreover, if$r<\mathrm{\infty }$, then$f\in C([0,T];B_{p,r}^s)$.

3 Proof of Theorem 1.1

By using the following six steps, we will complete the proof of Theorem 1.1.

First step: Approximate solution. We will construct a solution with the aid of a standard iterative process. Starting from $u^{(0)}:=0$, by the inductive method and solving the following linear transport equation (3.1) and (3.2), we derive a sequence of smooth functions

$$\begin{array}{c}[{\partial }_t+{\left(u^{(n)}\right)}^2{\partial }_x]u^{(n+1)}=-P_1(D)[\frac{4}{5}{\left(u^{(n)}\right)}^5-\frac{1}{3}{\left(u^{(n)}\right)}^3+\frac{3}{2}{u}^{(n)}{\left(u_x^{(n)}\right)}^2]\hfill \\ \phantom{[{\partial }_t+{\left(u^{(n)}\right)}^2{\partial }_x]u^{(n+1)}=}-P_2(D)\left[\frac{1}{2}{\left(u_x^{(n)}\right)}^3\right],\hfill \end{array}$$

(3.1)

$$u^{(n+1)}(x,0)=u_0^{(n+1)}=S_{n+1}{u}_0.$$

(3.2)

It is easily checked that $S_{n+1}{u}_0\in B_{p,r}^{\mathrm{\infty }}$, by using Lemma 2.4 and the inductive method, for all $n\in N$, we have that (3.1) and (3.2) has a global solution which belongs to $C({\mathbf{R}}^{+},B_{p,r}^{\mathrm{\infty }})$.

Second step: Uniform bounds. We will prove

$$\begin{array}{rcl}{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^s}& \le & e^{C{\int }_0^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{2}d\tau }{\parallel u_0\parallel }_{B_{p,r}^s}+\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{2}d\tau }{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{3}d\tau \\ +\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{2}d\tau }{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^{5}d\tau \end{array}$$

(3.3)

for all $n\in \mathbf{N}$.

Combining (2.4) of Lemma 2.3 with (3.1), we have

$$\begin{array}{rcl}{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^s}& \le & e^{C{\int }_0^t{\parallel ({(u^{(n)})}^2)(t^{\prime })\parallel }_{B_{p,r}^s}d{t}^{\prime }}{\parallel u_0\parallel }_{B_{p,r}^s}\\ +{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel ({(u^{(n)})}^2)(t^{\prime })\parallel }_{B_{p,r}^s}d{t}^{\prime }}{\parallel F(u^{(n)},u_x^{(n)})\parallel }_{B_{p,r}^s}d\tau ,\end{array}$$

(3.4)

where

$$F(u^n,u_x^n)=P_1(D)[\frac{4}{5}{\left(u^{(n)}\right)}^5-\frac{1}{3}{\left(u^{(n)}\right)}^3+\frac{3}{2}{u}^{(n)}{\left(u_x^{(n)}\right)}^2]+P_2(D)\left[\frac{1}{2}{\left(u_x^{(n)}\right)}^3\right].$$

(3.5)

When $s>max\{1+\frac{1}{p},\frac{3}{2}\}$, by using (4) in Lemma 2.2, we have

$${\parallel \left({\left(u^{(n)}\right)}^2\right)\left(t^{\prime }\right)\parallel }_{B_{p,r}^s}\le C{\parallel \left(u^{(n)}\right)\left(t^{\prime }\right)\parallel }_{B_{p,r}^s}^2,$$

(3.6)

$${\parallel F(u^{(n)},u_x^{(n)})\parallel }_{B_{p,r}^s}\le C[{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^3+{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^5].$$

(3.7)

Combining (3.6)-(3.7) with (3.4), we have (3.3).

Let $T>0$ satisfy

$$T<min\{\frac{1}{16C{\parallel u_0\parallel }_{B_{p,r}^s}^2},\frac{1}{C}\},$$

(3.8)

$${\parallel u^{(n)}(t)\parallel }_{B_{p,r}^s}\le \frac{2{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{1/2}}.$$

(3.9)

By using (3.9), we have

$$e^{C{\int }_{\tau }^t{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^2(t^{\prime })d{t}^{\prime }}\le e^{-\frac{1}{4}{\int }_{\tau }^t\frac{d(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^2{t}^{\prime })}{1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^2{t}^{\prime }}}={\left(\frac{1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^2\tau }{1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t}\right)}^{\frac{1}{4}}.$$

(3.10)

When $\tau =0$ in (3.10), we have

$$e^{C{U}^n(t)}\le {\left(\frac{1}{1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t}\right)}^{1/4}.$$

(3.11)

By using (3.10) and (3.9), we have

$$\begin{array}{c}\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}(t^{\prime })\parallel }_{B_{p,r}^s}^{2}d{t}^{\prime }}{\parallel u^{(n)}(\tau )\parallel }_{B_{p,r}^s}^{5}d\tau \hfill \\ \le {(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{-\frac{1}{4}}{\int }_0^t\frac{16C{\parallel u_0\parallel }_{B_{p,r}^s}^5}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^2\tau )}^{\frac{9}{4}}}d\tau \hfill \\ =\frac{4}{5}{\parallel u_0\parallel }_{B_{p,r}^s}^3{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{-\frac{1}{4}}[{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{-\frac{5}{4}}-1].\hfill \end{array}$$

(3.12)

With the aid of the mean value theorem, we have

$$[{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{-\frac{5}{4}}-1]=20C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t{(1-\xi )}^{-9/4},$$

(3.13)

where

$$16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t<\xi <1.$$

Combining (3.13) with (3.12), we have that

$$\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}(t^{\prime })\parallel }_{B_{p,r}^s}^{2}d{t}^{\prime }}{\parallel u^{(n)}(\tau )\parallel }_{B_{p,r}^s}^{5}d\tau \le \frac{16{\parallel u_0\parallel }_{B_{p,r}^s}^{5}Ct}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{5/2}}.$$

(3.14)

Inserting (3.10)-(3.14) into (3.3) leads to

$${\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^s}\le \frac{2{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{1/2}}.$$

(3.15)

Consequently, ${(u^{(n)})}_{n\in N}$ is uniformly bounded in $C([0,T];B_{p,r}^s)$. By using the fact that $B_{p,r}^{s-1}$ with $s>max\{1+\frac{1}{p},\frac{3}{2}\}$ is an algebra and $B_{p,r}^s\hookrightarrow B_{p,r}^{s-1}$ as well as the definition of the Besov spaces $B_{p,r}^s$, we derive that

$$\begin{array}{rcl}{\parallel {\left(u^{(n)}\right)}^2{u}_x^{(n+1)}\parallel }_{B_{p,r}^{s-1}}& \le & C{\parallel {\left(u^{(n)}\right)}^2\parallel }_{B_{p,r}^{s-1}}{\parallel u_x^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\\ \le & C{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^2{\parallel u^{(n+1)}\parallel }_{B_{p,r}^s}\\ \le & \frac{8C{\parallel u_0\parallel }_{B_{p,r}^s}^3}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{3}{2}}}.\end{array}$$

(3.16)

Since $s>max\{1+\frac{1}{p},\frac{3}{2}\}$, which leads to that $B_{p,r}^{s-1}$ is an algebra, by using the $S^{-1}$-multiplier property of $P_1(D)$ and the $S^{-2}$-multiplier property of $P_2(D)$ as well as (3.8), we have

$$\begin{array}{c}{\parallel F(u^{(n)},u_x^{(n)})\parallel }_{B_{p,r}^{s-1}}\le \frac{C}{2}[{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^3+{\parallel u^{(n)}\parallel }_{B_{p,r}^s}^5]\hfill \\ \le 4C[\frac{{\parallel u_0\parallel }_{B_{p,r}^s}^3}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{3}{2}}}+\frac{4{\parallel u_0\parallel }_{B_{p,r}^s}^5}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{5}{2}}}].\hfill \end{array}$$

(3.17)

Consequently, combining (3.1) with (3.16) and (3.17), we derive that

$${\parallel u_t^{(n+1)}\parallel }_{B_{p,r}^{s-1}}\le 4C[\frac{3{\parallel u_0\parallel }_{B_{p,r}^s}^3}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{3}{2}}}+\frac{4{\parallel u_0\parallel }_{B_{p,r}^s}^5}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{5}{2}}}],$$

(3.18)

which yields ${(u^{(n)})}_n\in C([0,T];B_{p,r}^s)\cap C^1([0,T];B_{p,r}^{s-1})$.

Third step: Convergence. We will derive that ${(u^{(n)})}_n$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$.

For $m,n\in N$, from (3.1), we have

$${(u^{(n+m+1)}-u^{(n+1)})}_t+{\left(u^{(n+m)}\right)}^2{(u^{(n+m+1)}-u^{(n+1)})}_x=\sum _{k=1}^5{T}_k,$$

(3.19)

where

$$\begin{array}{c}{T}_1=-\frac{4}{5}{P}_1(D)[{\left(u^{(n+m)}\right)}^5-{\left(u^{(n)}\right)}^5],\hfill \\ T_2=\frac{1}{3}{P}_1(D)[{\left(u^{(n+m)}\right)}^3-{\left(u^{(n)}\right)}^3],\hfill \\ T_3=-\frac{3}{2}{P}_1(D)\left[(u^{n+m}-u^n){\left(u_x^{(n+m)}\right)}^2\right],\hfill \\ T_4=-\frac{3}{2}{P}_1(D)[{\left(u_x^{(n+m)}\right)}^2-{\left(u_x^{(n)}\right)}^2],\hfill \\ T_5=-\frac{1}{2}{P}_2(D)[{\left(u^{(n+m)}\right)}_x^3-{\left(u^{(n)}\right)}_x^3],\hfill \\ T_6=[{\left(u^{(n+m)}\right)}^2-{\left(u^{(n)}\right)}^2]u_x^{(n+1)}.\hfill \end{array}$$

When $s>max\{1+\frac{1}{p},\frac{3}{2}\}$, by using the $S^{-1}$ multiplier property of $P_1(D)$, the $S^{-2}$ multiplier property of $P_2(D)$ and $B_{p,r}^{s-1}\hookrightarrow B_{p,r}^{s-2}$, we have

$${\parallel T_j\parallel }_{B_{p,r}^{s-1}}\le C{\parallel u^{(n+m)}-u^{(n)}\parallel }_{B_{p,r}^{s-1}},$$

(3.20)

where $1\le j\le 6$, $j\in \mathbf{N}$. Since $\mathrm{\forall }n\in N$, we have

$${\parallel u^{(n)}\parallel }_{B_{p,r}^s}\le \frac{2{\parallel u_0\parallel }_{B_{p,r}^s}}{{(1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t)}^{\frac{1}{2}}}.$$

By using (3.20), we have

$$\begin{array}{rcl}{\parallel \sum _{j=1}^6{T}_j\parallel }_{B_{p,r}^{s-1}}& \le & \sum _{j=1}^6{\parallel T_j\parallel }_{B_{p,r}^{s-1}}\\ \le & C{\parallel u^{(n+m)}-u^{(n)}\parallel }_{B_{p,r}^{s-1}}\frac{4{\parallel u_0\parallel }_{B_{p,r}^s}^2}{1-16C{\parallel u_0\parallel }_{B_{p,r}^s}^{2}t}.\end{array}$$

(3.21)

When $s\ne 2+\frac{1}{p}$, from Lemma 2.4 and (3.20), we have

$$\begin{array}{c}{\parallel (u^{(n+m+1)}-u^{(n+1)}(t)\parallel }_{B_{p,r}^{s-1}}\hfill \\ \le e^{C{W}^{(n+m)}(t)}{\parallel (u^{(n+m+1)}-u^{(n+1)})(\cdot ,0)\parallel }_{B_{p,r}^{s-1}}\hfill \\ +C{\int }_0^t{e}^{C{W}^{(n+m)}(t)-C{W}^{(n+m)}(\tau )}\sum _{j=1}^6{\parallel T_j\parallel }_{B_{p,r}^{s-1}}d\tau ,\hfill \end{array}$$

(3.22)

where

$$W^{n+m}(t)={\int }_0^t{\parallel {\partial }_x{\left(u^{(n+m)}\right)}^2(\tau )\parallel }_{B_{p,r}^{\frac{1}{p}}\cap L^{\mathrm{\infty }}}d\tau$$

(3.23)

if $s-1<1+\frac{1}{p}$ and

$$W^{n+m}(t)={\int }_0^t{\parallel {\partial }_x{\left(u^{(n+m)}\right)}^2(\tau )\parallel }_{B_{p,r}^{s-2}}d\tau$$

(3.24)

if $s-1>1+\frac{1}{p}$. From (3.23), if $s-1<1+\frac{1}{p}$, by using $B_{p,r}^{s-1}\hookrightarrow L^{\mathrm{\infty }}$ with $s>1+\frac{1}{p}$, we have

$$W^{n+m}(t)\le C{\int }_0^t{\parallel u^{(n+m)}(\tau )\parallel }_{B_{p,r}^s}^{2}d\tau .$$

(3.25)

From (3.25), if $s-1>1+\frac{1}{p}$, we have

$$W^{n+m}(t)\le C{\int }_0^t{\parallel u^{(n+m)}(\tau )\parallel }_{B_{p,r}^s}^{2}d\tau .$$

(3.26)

It is easily showed that

$${\parallel \sum _{q=n+1}^{n+m}{\mathrm{\Delta }}_q{u}_0\parallel }_{B_{p,r}^{s-1}}\le C{2}^{-n}{\parallel u_0\parallel }_{B_{p,r}^{s-1}}.$$

(3.27)

Inserting (3.25)-(3.27) into (3.22), we have

$$\begin{array}{c}{\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{B_{p,r}^{s-1}}\hfill \\ \le C_T(2^{-n}+{\int }_0^t{\parallel (u^{(n+m)}-u^{(n)})(\tau )\parallel }_{B_{p,r}^{s-1}}d\tau ).\hfill \end{array}$$

(3.28)

We define

$$W_{n,k}(t)={\parallel (u^{(n+m)}-u^{(n)})(t)\parallel }_{B_{p,r}^{s-1}},$$

(3.29)

$$W_n(t)=\underset{m\in \mathbf{N}}{sup}{W}_{n,m}(t),$$

(3.30)

$$\tilde{W}(t)=\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}{W}_n(t).$$

(3.31)

Combining (3.28) with (3.29)-(3.30), we have that

$$W_{n+1}(t)\le C{\int }_0^t{W}_n(\tau )d\tau .$$

(3.32)

From (3.31) and (3.32), by using the Fatou lemma, we have that

$$\tilde{W}(t)\le C{\int }_0^t\tilde{W}(\tau )d\tau .$$

(3.33)

Applying the Gronwall inequality to (3.33), we have that

$$\tilde{W}(t)\le e^C\tilde{W}(0).$$

(3.34)

From (3.31), we have that $\tilde{W}(0)=0$. Thus, $\tilde{W}(t)=0$. Consequently, ${(u^{(n)})}_n$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$; moreover, ${(u^{(n)})}_n$ is convergent to some limit function $u\in C([0,T];B_{p,r}^{s-1})$.

When $s=2+\frac{1}{p}$, by using (6) of Lemma 2.2, we derive that

$$\begin{array}{c}{\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{L_T^{\mathrm{\infty }}{B}_{p,r}^{s-1}}\hfill \\ ={\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{L_T^{\mathrm{\infty }}{B}_{p,r}^{1+\frac{1}{p}}}\hfill \\ \le {\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{L_T^{\mathrm{\infty }}{B}_{p,r}^{s_1}}^{\theta }{\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{L_T^{\mathrm{\infty }}{B}_{p,r}^{s_2}}^{1-\theta }\hfill \\ \le {\parallel (u^{(n+m+1)}-u^{(n+1)})(t)\parallel }_{B_{p,r}^{1+\frac{1}{p}}}^{\theta }{[{\parallel u^{(n+m+1)}\parallel }_{B_{p,r}^{2+\frac{1}{p}}}+{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^{2+\frac{1}{p}}}]}^{1-\theta }\hfill \\ \le {\left(C_T^{\prime }\right)}^{\theta }{2}^{-\theta n}{[{\parallel u^{(n+m+1)}\parallel }_{B_{p,r}^{2+\frac{1}{p}}}+{\parallel u^{(n+1)}(t)\parallel }_{B_{p,r}^{2+\frac{1}{p}}}]}^{1-\theta },\hfill \end{array}$$

(3.35)

where

$$s_1\in (max(1+\frac{1}{p},\frac{3}{2})-1,1+\frac{1}{p}),s_2\in (1+\frac{1}{p},2+\frac{1}{p}).$$

Consequently, ${(u^{(n)})}_n$ is a Cauchy sequence in $C([0,T];B_{p,r}^{s-1})$ and ${(u^{(n)})}_n$ converges to some limit function $u\in C([0,T];B_{p,r}^{s-1})$.

Fourth step: Existence of solution in$E_{p,r}^s(T)$. Existence of solution $E_{p,r}^s(T)$ can be proved similarly to [11].

Fifth step: Uniqueness of solution. We consider case $s\ne 2+\frac{1}{p}$ and case $s=2+\frac{1}{p}$, respectively. In fact, this can be proved similarly to [14].

Sixth step: Continuity with respect to the initial data. Continuity with respect to the initial data can be proved similarly to [6].

4 Proof of Theorem 1.2

Since $B_{2,1}^{3/2}$ and $B_{2,1}^{1/2}$ are Banach algebras, by using a proof similar to (3.3), we can prove that

$$\begin{array}{rcl}{\parallel u^{(n+1)}(t)\parallel }_{B_{2,1}^{3/2}}& \le & e^{C{\int }_0^t{\parallel u^{(n)}\parallel }_{B_{2,1}^{3/2}}^{2}d\tau }{\parallel u_0\parallel }_{B_{2,1}^{3/2}}+\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}\parallel }_{B_{2,1}^{3/2}}^{2}d\tau }{\parallel u^{(n)}\parallel }_{B_{2,1}^{3/2}}^{3}d\tau \\ +\frac{C}{2}{\int }_0^t{e}^{C{\int }_{\tau }^t{\parallel u^{(n)}\parallel }_{B_{2,1}^{3/2}}^{2}d\tau }{\parallel u^{(n)}\parallel }_{B_{2,1}^{3/2}}^{5}d\tau \end{array}$$

(4.1)

for all . We assume that

$$T<min\{\frac{1}{16C{\parallel u_0\parallel }_{B_{2,1}^{3/2}}^2},\frac{1}{C}\},$$

(4.2)

$${\parallel u^{(n)}(t)\parallel }_{B_{2,1}^{3/2}}\le \frac{2{\parallel u_0\parallel }_{B_{2,1}^{3/2}}}{{(1-16C{\parallel u_0\parallel }_{B_{2,1}^{3/2}}^{2}t)}^{1/2}}.$$

(4.3)

By a proof similar to (3.15), we can prove that

$${\parallel u^{(n+1)}(t)\parallel }_{B_{2,1}^{3/2}}\le \frac{2{\parallel u_0\parallel }_{B_{2,1}^{3/2}}}{{(1-16C{\parallel u_0\parallel }_{B_{2,1}^{3/2}}^{2}t)}^{1/2}}.$$

(4.4)

Thus, ${(u^{(n)})}_n$ is uniformly bounded in $B_{2,1}^{3/2}$. From (3.1), by using (4) in Lemma 2.2, we can prove that ${(u_t^{(n)})}_n$ is uniformly bounded with respect to $n$ in $B_{2,1}^{1/2}$.

Consequently, ${(u^{(n)})}_n\in C([0,T];B_{2,1}^{3/2})\cap C^1([0,T];B_{2,1}^{1/2})$.

We define

$${\rho }_{n,k}(t)={\parallel (u^{(n+m)}-u^{(n)})(t)\parallel }_{B_{2,\mathrm{\infty }}^{1/2}},$$

(4.5)

(4.7)

$$\tilde{\rho }(t)=\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}{\rho }_n(t).$$

(4.7)

By a proof similar to [15], we derive that

$$\tilde{\rho }(t)=0,$$

(4.8)

and with the aid of (2.1), we can prove that ${(u^{(n)})}_n$ is a Cauchy sequence in $C([0,T];B_{2,1}^{1/2})$.

The rest of Theorem 1.2 can be proved similarly to [13], [15].