Local Well-Posedness of a Two-Component Novikov System in Critical Besov Spaces

Abstract

In this paper, we establish the local well-posedness for a two-component Novikov system in the sense of Hadamard in critical Besov spaces $B_{p,1}^{1+\frac{1}{p}}(\mathbb{R})\times B_{p,1}^{1+\frac{1}{p}}(\mathbb{R}),1\le p<\infty$. We first provide a uniform bound for the approximate solutions constructed by iterative scheme, then we show the convergence and regularity; afterwards, based on the Lagrangian coordinate transformation techniques, we prove the uniqueness result; finally, we show that the the solution map is continuous.

1. Introduction

In what follows, we are concerned with the local well-posedness of the Cauchy problem for the Novikov system:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{q}_t=-(u^2+v^2)q_x-3(u{u}_x+v{v}_x)q+r(u{v}_x-u_{x}v),\hfill & t>0,x\in \mathbb{R},\hfill \\ r_t=-(u^2+v^2)r_x-3(u{u}_x+v{v}_x)r+q(u{v}_x-u_{x}v),\hfill & t>0,x\in \mathbb{R},\hfill \\ q=u-u_{xx},r=v-v_{xx},\hfill \\ u(0,x)=u_0(x),v(0,x)=v_0(x),\hfill & x\in \mathbb{R}.\hfill \end{array}\right.\end{array}$$

(1)

Equation (1) was first proposed in [1,2], it can be derived as the reduction of the following multi-component Novikov system

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{M}_t=-2U_x^{T}VM-U^T{V}_{x}M-U^{T}V{M}_x-M^{T}V{U}_x+M^T{V}_{x}U,\hfill \\ N_t=-2U^T{V}_{x}N-U_x^{T}VN-U^{T}V{N}_x-N^{T}U{V}_x+N^T{U}_{x}V,\hfill \\ M=U-U_{xx},N=V-V_{xx}\hfill \end{array}\right.\end{array}$$

(2)

by setting $M=N={(q,r)}^T,U=V={(u,v)}^T$.

Equation (2) was derived from the zero-curvature equation for the vector prolongation of Lax pair representation for the Geng–Xue equation [3]. It can be reformulated as a bi-Hamiltonian system with the following Hamiltonian functionals:

$$\begin{array}{ccc}\hfill H_0& =& \frac{1}{2}\int \langle M,V\rangle \langle U_x,V\rangle -\langle N,U\rangle \langle V_x,U\rangle +(\langle N,U_x\rangle -\langle M,V_x\rangle )\langle U,V\rangle \mathrm{d}x,\hfill \\ \hfill H_1& =& \frac{1}{2}\int \langle M,V\rangle +\langle N,U\rangle \mathrm{d}x.\hfill \end{array}$$

While $M=q,U=u,N=V=1$, Equation (2) is reduced to the Degasperis-Procesi (DP) equation [4]

$$m_t+u{m}_x+3u_{x}m=0,m=u-u_{xx}.$$

It is a model for nonlinear shallow water dynamics [5]. The DP equation was proved to be formally integrable by constructing a Lax pair and the direct and inverse scattering approach [6]. Moreover, it was also shown that it has a bi-Hamiltonian structure and an infinite number of conservation laws [7] and admits exact peakon solutions, which are analogous to Camassa–Holm peakons. It is noted that the peakons replicate a feature that is characteristic for the waves of great heights or waves with largest amplitudes, which are exact solutions for the governing equations of irrotational water waves. The local well-posedness of DP equation in Sobolev and Besov spaces has been investigated in many papers, see [8,9,10].

While $M=N=q,V=U=u$, Equation (2) is reduced to the Novikov equation

$$m_t+3u{u}_{x}m+u^2{m}_x=0,m=u-u_{xx}.$$

The Novikov equation shares similar properties with the Camassa–Holm equation, such as a Lax pair in the matrix form, a bi-Hamiltonian structure, infinitely many conserved quantities, and peakon solutions given by the formula $u(x,t)=\sqrt{c}{e}^{-|x-ct|}$ [11].

While $M=q,N=r,U=u,V=v$, Equation (2) is reduced to the Geng–Xue equation [3]

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{m}_t+3u{v}_{x}m+uv{m}_x=0,\hfill \\ n_t+3u_{x}vn+uv{n}_x=0,\hfill \\ m=u-u_{xx},n=v-v_{xx}.\hfill \end{array}\right.\end{array}$$

The integrability [3,12], dynamics and structure of the peaked solitons of the Geng–Xue system [13,14] were considered recently. In [15], the well-posedness and wave breaking phenomena of the Cauchy problem were discussed.

The well-posedness and ill-posedness of Camassa–Holm type equations in Besov spaces have recently been intensively explored [16,17,18,19]. The main difference between the Novikov equation and the Camassa–Holm equation is that the Novikov equation contains terms of cubic nonlinearity, while the Camassa–Holm equation only contains quadratic terms. In [20], the local well-posedness of the Novikov equation was researched in Besov spaces $B_{p,r}^s$ for $s>max\{\frac{3}{2},1+\frac{1}{p}\}$ with $p\in [1,\infty )$ and $r\in [1,\infty )$; meanwhile, the ill-posedness of the Novikov equation was also considered in $B_{2,\infty }^{\frac{3}{2}}$ by using peakon solutions. The ill-posedness result was lately studied by Li et al. [21] in supercritical Besov spaces $B_{p,\infty }^s$ with $s>max\{1+\frac{1}{p},\frac{3}{2}\}$ and $1\le p\le \infty$, in the sense that the solution map starting from $u_0$ is discontinuous at $t=0$ in the metric of $B_{p,\infty }^s$. For the local well-posedness of Novikov equation in critical Beosov spaces, it is worth mentioning that Ni and Zhou [22] obtained local well-posedness in $B_{2,1}^{\frac{3}{2}}$, later, the local well-posedness result was extended to the broader range of Besov spaces $B_{p,1}^{1+\frac{1}{p}}$ with $1\le p<\infty$ based on the Lagrangian coordinate transformation by Ye et al. [23]. The continuous dependence of the solution to Novikov equation on the initial data in the space $\mathcal{C}([0,T];B_{p,r}^s)$ with $r<\infty$ was also researched in [18], and the solution map was proved to be weakly continuous with $r=\infty$.

For the two-component Novikov system (1), the local well-posedness result in super critical Besov spaces $B_{p,r}^s\times B_{p,r}^s$ for $s>max\{1+\frac{1}{p},\frac{3}{2}\}$ with $p\in [1,\infty )$ and $r\in [1,\infty )$ has already been established [24]. However, whether (1) is well-posed or not in the critical Besov spaces $B_{p,1}^{1+\frac{1}{p}}\times B_{p,1}^{1+\frac{1}{p}}$ with $p\in [1,\infty )$ is still open to debate, therefore, it would be reasonable to study the local well-posedness of Equation (1) in such spaces, we aim to construct the local well-posedness result in the present paper, the methods adopted in this paper are also applicable to two-component Camassa–Holm equations and Geng–Xue system etc.

The motivation to use the methods adopted in each step can be stated as follows: We construct a sequence of approximate solutions using an iterative scheme for the purpose of showing the existence of the solution. In the first step, we demonstrate the uniform boundedness of the solution sequence via induction; then, in step 2, we extract two subsequences of the above mentioned approximate solution sequence, the norm of the difference between which in the space $B_{p,1}^{\frac{1}{p}}\times B_{p,1}^{\frac{1}{p}}$ is then shown to converge to zero, thus we obtain the compactness of the approximate sequence and get a solution $(u,v)$ which solves Equation (3) in the sense of distributions. The Moser-type inequality is always needed in the proof of uniqueness; however, the critical index restriction $s=1+\frac{1}{p}$ makes it impossible to apply the Moser-type inequality any longer. Fortunately, inspired by [23], we devise the Lagrangian scale of (1) to overcome this difficulty and succeed in proving the uniqueness in step 3. In order to prove the continuous dependence of the solution on the initial data in step 4, we decompose the solution components, and afterwards, by making use of transport theories and the energy method, we manage to show that if the initial data sequence converges to a fixed initial data, then the corresponding approximate solution sequence converges to the solution corresponding to the fixed initial data.

For $T>0$ and $1\le p<\infty$, we define

$$\begin{array}{c}\hfill E_{p,1}^{1+\frac{1}{p}}(T)=C([0,T];B_{p,1}^{1+\frac{1}{p}})\bigcap C^1([0,T];B_{p,1}^{\frac{1}{p}}).\end{array}$$

Before proceeding any further, we need to reformulate (1) into a workable form of transport equation-type. By applying the inverse of Helmholtz operator ${(1-{\mathsf{\partial }}_x^2)}^{-1}$ to both sides of the equations in (1), we obtain

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle u_t+(u^2+v^2)u_x=-{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\left(u^3+u{v}^2+\frac{3}{2}u{u}_x^2+u_{x}v{v}_x+\frac{1}{2}u{v}_x^2\right)}\hfill \\ {\displaystyle -\frac{1}{2}{(1-{\mathsf{\partial }}_x^2)}^{-1}\left(u_x^3+u_x{v}_x^2\right)\triangleq F(u,v),}\hfill \\ {\displaystyle v_t+(u^2+v^2)v_x=-{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\left(v^3+u^{2}v+\frac{3}{2}v{v}_x^2+u{u}_x{v}_x+\frac{1}{2}{u}_x^{2}v\right)}\hfill \\ {\displaystyle -\frac{1}{2}{(1-{\mathsf{\partial }}_x^2)}^{-1}\left(u_x^2{v}_x+v_x^3\right)\triangleq G(u,v),}\hfill \\ u(0,x)=u_0(x),v(0,x)=v_0(x).\hfill \end{array}\right.\end{array}$$

(3)

We introduce our main result regarding local well-posedness of (3) in critical Besov spaces as follows.

Theorem 1.

Let $(u_0,v_0)\in B_{p,1}^{1+\frac{1}{p}}\times B_{p,1}^{1+\frac{1}{p}}$ with $1\le p<\infty$, then there exists a time $T>0$ such that the Cauchy problem (3) has a unique solution $(u,v)\in E_{p,1}^{1+\frac{1}{p}}(T)\times E_{p,1}^{1+\frac{1}{p}}(T)$, and the solution map $(u_0,v_0)\to (u,v)$ is continuous from a neighborhood of $B_{p,1}^{1+\frac{1}{p}}\times B_{p,1}^{1+\frac{1}{p}}$ into $E_{p,1}^{1+\frac{1}{p}}(T)\times E_{p,1}^{1+\frac{1}{p}}(T)$.

The rest of this paper is organized as follows. The properties of Besov spaces, as well as the transport theories, which provide a basis of the study of the associated local well-posedness problem is given in Section 2. In Section 3, we split the proof of Theorem 1 into four steps, which cover uniform boundedness, convergence and regularity, uniqueness, and continuous dependence, respectively.

Notation 1.

For given Banach space X, we denote its norm by ${\parallel ·\parallel }_X$. Since all function spaces are over $\mathbb{R}$, we drop $\mathbb{R}$ in our notations of function spaces for simplicity if there is no ambiguity.

2. Preliminaries

In this section, we shall recall some properties of the Besov spaces and the transport theories.

Definition 1

([25]). Assume that $s\in \mathbb{R}$, $p,r\in [1,\infty ]$, $u\in {\mathcal{D}}^{\prime }(\mathbb{T})$. The Besov space is defined as follows:

$$B_{p,r}^s=\{u\in {\mathcal{D}}^{\prime }{(\mathbb{T}): \parallel u\parallel }_{B_{p,r}^s}<\infty \},$$

where

$$\begin{array}{c}\hfill {\parallel u\parallel }_{B_{p,r}^s}\doteq \left\{\begin{array}{cc}({\displaystyle \sum _{q\ge -1}}(2^{sq}\parallel {\mathsf{\Delta }}_q{u\parallel }_{L^p}{{)}^r)}^{\frac{1}{r}},\hfill & \mathrm{when}1\le r<\infty ,\hfill \\ \underset{q\ge -1}{sup}{2}^{sq}{\parallel {\mathsf{\Delta }}_{q}u\parallel }_{L^p},\hfill & \mathrm{when}r=\infty .\hfill \end{array}\right.\end{array}$$

The following lemma summarizes some useful properties of the Besov space $B_{p,r}^s$.

Lemma 1

([25]). Let $s\in \mathbb{R},1\le p,r,p_j,r_j\le \infty ,j=1,2$, then

(1)

Algebraic properties: $\forall s>0,B_{p,r}^s$ is a Banach algebra if and only if $s>\frac{1}{p}$ or $s\ge \frac{1}{p}$ and $r=1$.

(2)

1-D Moser-type estimates:

(i) 

For $s>0$,

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

(ii) 

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

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

(3)

Fatou lemma: If ${\{v_n\}}_{n\in \mathbb{N}}$ is bounded in $B_{p,r}^s$ and $v_n\to v$ in ${\mathcal{D}}^{\prime }(\mathbb{T})$, then $v\in B_{p,r}^s$ and

$${\parallel v\parallel }_{B_{p,r}^s}\le \underset{n\to \infty }{lim\; inf}{\parallel v_n\parallel }_{B_{p,r}^s}.$$

We also prepare some useful lemmas on the following transport equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_{t}g+v{\mathsf{\partial }}_{x}g=h,\hfill \\ {g|}_{t=0}=g_0(x).\hfill \end{array}\right.\end{array}$$

(4)

Lemma 2

([26]). Assume that $1\le p,r\le \infty$, $\theta >-min(\frac{1}{p},\frac{1}{p^{\prime }})$. If $g_0\in B_{p,r}^{\theta },h\in L^1(0,T;B_{p,r}^{\theta })$, $v\in L^{\rho }(0,T;B_{\infty ,\infty }^{-M})$, where $\rho >1,M>0$, and

$$\begin{array}{c}\hfill {\mathsf{\partial }}_{x}v\in L^1(0,T;B_{p,r}^{\theta -1}),if\theta =1+\frac{1}{p},r=1.\end{array}$$

Then there exists a unique solution $g\in C([0,T];B_{p,r}^{\theta })$ to Equation (4) if $r<\infty$.

Lemma 3

([26]). Assume that $1\le p\le \infty ,1\le r\le \infty ,\theta >-min(\frac{1}{p},\frac{1}{p^{\prime }})$. Then there exists a constant C such that for all solutions $g\in L^{\infty }(0,T;B_{p,r}^{\theta })$ of (4) with initial data $g_0$ in $B_{p,r}^{\theta }$ and $h\in L^1(0,T;B_{p,r}^{\theta })$, we have for a.e. $t\in [0,T]$,

$$\begin{array}{c}\hfill {\parallel g(t)\parallel }_{B_{p,r}^{\theta }}\le \parallel g_0{\parallel }_{B_{p,r}^{\theta }}+{\int }_0^t\parallel h(t^{\prime }){\parallel }_{B_{p,r}^{\theta }}d{t}^{\prime }+{\int }_0^t{V}^{\prime }(t^{\prime }){\parallel g(t^{\prime })\parallel }_{B_{p,r}^{\theta }}d{t}^{\prime }\end{array}$$

or

$$\begin{array}{c}\hfill {\parallel g(t)\parallel }_{B_{p,r}^{\theta }}\le e^{CV(t)}(\parallel g_0{\parallel }_{B_{p,r}^{\theta }}+{\int }_0^t{e}^{-CV(t^{\prime })}\parallel h(t^{\prime }){\parallel }_{B_{p,r}^{\theta }}d{t}^{\prime })\end{array}$$

with

$$\begin{array}{c}\hfill V^{\prime }(t)={\parallel {\mathsf{\partial }}_{x}v(t)\parallel }_{B_{p,r}^{\theta -1}},if\theta =1+\frac{1}{p},r=1.\end{array}$$

Lemma 4

([27]). (Existence and uniqueness) Suppose $1\le p\le p_1\le \infty ,1\le r\le \infty ,\sigma \ge -min(\frac{1}{p},1-\frac{1}{p})$ with strict inequality if $r<\infty$. Suppose in addition that $g_0\in B_{p,r}^{\sigma },h\in L^1(0,T;B_{p,r}^{\sigma })$, $v\in L^k(0,T;B_{\infty ,\infty }^{-M})$, where $k>1,M>0$, and

$$\begin{array}{c}\hfill {\mathsf{\partial }}_{x}v\in L^1(0,T;B_{p,r}^{\sigma -1}),if\sigma =1+\frac{1}{p},r=1\end{array}$$

Then (4) has a unique solution $f\in C([0,T];B_{p,r}^{\sigma })$.

Lemma 5

([26]). Suppose that $1\le p\le \infty ,1\le r\le \infty ,s>1+\frac{1}{p}$ or $s=1+\frac{1}{p},1\le p<\infty r=1$. For $n\in \overline{\mathbb{N}}$, denote $b^n$ be the solution of

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_t{b}^n+B^n{\mathsf{\partial }}_x{b}^n=G,\hfill \\ b^n{|}_{t=0}=b_0(x),\hfill \end{array}\right.\end{array}$$

with $G(t,x)\in L^{\infty }(0,T;B_{p,r}^{s-1}),b_0(x)\in B_{p,r}^{s-1}$. Assume that

$$\begin{array}{c}\hfill \underset{n\in \overline{\mathbb{N}}}{sup}{\parallel B^n(t)\parallel }_{B_{p,r}^s}\le \beta (t)forsome\beta \in L^1(0,T)\end{array}$$

and $B^n\to B^{\infty }$ in $L^1(0,T;B_{p,r}^{s-1})$. Then the sequence ${\{b^n\}}_{n\in \overline{\mathbb{N}}}$ converges to $b^{\infty }$ in $C([0,T];B_{p,r}^{s-1})$.

Proposition 1

([27]). Assume that $1\le p,r\le \infty$, s is a real number.

(i)

For any given $u\in B_{p,r}^s,\varphi \in B_{p^{\prime },r^{\prime }}^{-s}$,

$$(u,\varphi )\to \sum _{|j-i|\le 1}\langle {\mathsf{\Delta }}_{j}u,{\mathsf{\Delta }}_i\varphi \rangle$$

defines a continuous bilinear functional on $B_{p,r}^s\times B_{p^{\prime },r^{\prime }}^{-s}$.

(ii)

Let $Q_{p^{\prime },r^{\prime }}^{-s}$ be the set of functions $\varphi \in \mathcal{S}$ satisfying ${\parallel \varphi \parallel }_{p^{\prime },r^{\prime }}^{-s}\le 1$. If u is in ${\mathcal{S}}^{\prime }$, then

$${\parallel u\parallel }_{B_{p,r}^s}\le C\underset{\varphi \in Q_{p^{\prime },r^{\prime }}^{-s}}{sup}\langle u,\varphi \rangle .$$

3. Proof of Theorem 1

We split the proof of the main theorem into four steps. First we prove the uniform boundedness of the approximate solutions, then we show its convergence; afterward we prove the uniqueness of solution; finally, we establish the continuous dependence of the solution on the initial data.

Step 1. Uniform boundedness of approximate solutions constructed with an iterative scheme.

Suppose that $u^{(0)}=v^{(0)}=0$, we define by induction a solution sequence ${(u^{(n)},v^{(n)})}_{n\in \mathbb{N}}$ of the following transport equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_t{u}^{(n+1)}+[{(u^{(n)})}^2+{(v^{(n)})}^2]{\mathsf{\partial }}_x{u}^{(n+1)}=F(u^{(n)},v^{(n)}),\hfill \\ {\mathsf{\partial }}_t{v}^{(n+1)}+[{(u^{(n)})}^2+{(v^{(n)})}^2]{\mathsf{\partial }}_x{v}^{(n+1)}=G(u^{(n)},v^{(n)}),\hfill \\ u^{(n+1)}{{|}_{t=0}=u_0,v^{(n+1)}|}_{t=0}=v_0.\hfill \end{array}\right.\end{array}$$

(5)

Assume that ${(u^{(n)},v^{(n)})}_{n\in \mathbb{N}}$ belongs to $E_{p,1}^{1+\frac{1}{p}}(T)\times E_{p,1}^{1+\frac{1}{p}}(T)$ for all $T>0$, we know from Lemma 1 that $B_{p,1}^{1+\frac{1}{p}},B_{p,1}^{\frac{1}{p}}$ are algebras and the embedding $B_{p,1}^{1+\frac{1}{p}}\hookrightarrow B_{p,1}^{\frac{1}{p}}\hookrightarrow L^{\infty }$ holds. Thus, we have

$$\begin{array}{c}\hfill F(u^{(n)},v^{(n)})\le C(\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}{{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})}^3.\end{array}$$

It can be deduced from Lemma 2 that there exists a unique solution $(u^{(n+1)}(t),v^{(n+1)}(t))$$\in E_{p,1}^{1+\frac{1}{p}}(T)\times E_{p,1}^{1+\frac{1}{p}}(T)$ of (5). Moreover, by applying Lemma 3, one obtains

$$\begin{array}{c}\hfill \parallel u^{(n+1)}{(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\le e^{CV(t)}\left(\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+C{\int }_0^t{e}^{-CV(\tau )}(\parallel u^{(n)}{(\tau )\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}(\tau ){{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})}^3\mathrm{d}\tau \right).\end{array}$$

(6)

where $V(t)={\int }_0^t(\parallel u^{(n)}(t^{\prime }){\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^n(t^{\prime }){{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})}^2\mathrm{d}{t}^{\prime }$.

Similarly, we can infer that

$$\begin{array}{c}\hfill \parallel v^{(n+1)}{(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\le e^{CV(t)}\left(\parallel v_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+C{\int }_0^t{e}^{-CV(\tau )}(\parallel u^{(n)}{(\tau )\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}(\tau ){{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})}^3\mathrm{d}\tau \right).\end{array}$$

(7)

Let $X^{(n+1)}(t)=\parallel u^{(n+1)}{(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n+1)}{(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}},X_0=\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+{\parallel v_0\parallel }_{B_{p,1}^{1+\frac{1}{p}}}$, it follows by adding (6) and (7) that

$$\begin{array}{c}\hfill X^{(n+1)}(t)\le e^{CV(t)}\left(X_0+C{\int }_0^t{e}^{-CV(\tau )}{(X^{(n+1)}(\tau ))}^3\mathrm{d}\tau \right).\end{array}$$

(8)

Fix a $T>0$ such that $T<\frac{1}{4C(\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v_0{{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})}^2}$, and suppose by induction that

$$\begin{array}{c}\hfill X^{(n)}(t)\le {(1-4C{X}_0^{2}t)}^{-\frac{1}{2}}.\end{array}$$

(9)

Since

$$C(v(t)-v(\tau ))=C{\int }_{\tau }^t{(X^{(n)}(t^{\prime }))}^2\mathrm{d}{t}^{\prime }\le \frac{1}{4}{\int }_{\tau }^t\frac{4C{X}_0^2}{1-4C{X}_0^2{t}^{\prime }}\mathrm{d}{t}^{\prime }=\frac{1}{4}ln(1-4C{X}_0^2\tau )-\frac{1}{4}ln(1-4C{X}_0^{2}t),$$

substituting the above inequality into (8) yields

$$\begin{array}{ccc}\hfill X^{(n+1)}(t)& \le & {(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{X}_0+C{(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{\int }_0^t{(1-4C{X}_0^2\tau )}^{-\frac{5}{4}}\mathrm{d}\tau \hfill \\ & =& {(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{X}_0+{(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{X}_0{\int }_0^{t}C{X}_0^2{(1-4C{X}_0^2\tau )}^{-\frac{5}{4}}\mathrm{d}\tau \hfill \\ & =& {(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{X}_0+{(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}{X}_0[{(1-4C{X}_0^{2}t)}^{-\frac{1}{4}}-1]\hfill \\ & =& {(1-4C{X}_0^{2}t)}^{-\frac{1}{2}}{X}_0.\hfill \end{array}$$

Which is equivalent to

$$\begin{array}{c}\hfill \parallel u^{(n+1)}{(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+{\parallel v^{(n+1)}(t)\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\le \frac{\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+{\parallel v_0\parallel }_{B_{p,1}^{1+\frac{1}{p}}}}{[1-4Ct(\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}{{)}^2]}^{\frac{1}{2}}}\end{array}$$

Therefore, ${(u^{(n)},v^{(n)})}_{n\in \mathbb{N}}$ is uniformly bounded in $L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})\times L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})$.

Step 2. Convergence and regularity of solutions.

Next, we will prove that ${(u^{(n)},v^{(n)})}_{n\in \mathbb{N}}$ is a Cauchy sequence in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$. It follows from (5) that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_t(u^{(n+m+1)}-u^{(n+1)})+[{(u^{(n+m)})}^2+{(v^{(n+m)})}^2]{\mathsf{\partial }}_x(u^{(n+m+1)}-u^{(n+1)})\hfill \\ =F_1(u^{(n+m)},v^{(n+m)},u^{(n+1)},u^{(n)},v^{(n)}),\hfill \\ {\mathsf{\partial }}_t(v^{(n+m+1)}-v^{(n+1)})+[{(u^{(n+m)})}^2+{(v^{(n+m)})}^2]{\mathsf{\partial }}_x(v^{(n+m+1)}-v^{(n+1)})\hfill \\ =G_1(u^{(n+m)},v^{(n+m)},u^{(n+1)},u^{(n)},v^{(n)}),\hfill \\ (u^{(n+m+1)}-u^{(n+1)}){{|}_{t=0}=(v^{(n+m+1)}-v^{(n+1)})|}_{t=0}=0.\hfill \end{array}\right.\end{array}$$

Among which

$$\begin{array}{ccc}& & F_1(u^{(n+m)},v^{(n+m)},u^{(n+1)},u^{(n)},v^{(n)})\hfill \\ & =& [(u^{(n)}-u^{(n+m)})(u^{(n)}+u^{(n+m)})+(v^{(n)}-v^{(n+m)})(v^{(n)}+v^{(n+m)})]u_x^{(n+1)}\hfill \\ & & -{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\{\frac{3}{2}{(u_x^{(n+m)})}^2(u^{(n+m)}-u^{(n)})+\frac{3}{2}{u}^{(n)}(u_x^{(n+m)}-u_x^{(n)})(u_x^{(n+m)}+u_x^{(n)})\hfill \\ & & +(u_x^{(n+m)}-u_x^{(n)})v_x^{(n+m)}{v}^{(n+m)}+u_x^{(n)}{v}_x^{(n+m)}(v^{(n+m)}-v(n))+u_x^{(n)}{v}^{(n)}(v_x^{(n+m)}-v_x^{(n)})\hfill \\ & & +(u^{(n+m)}-u^{(n)})[{(u^{(n+m)})}^2+u^{(n+m)}{u}^{(n)}+{(u^{(n)})}^2]+(u^{(n+m)}-u^{(n)}){(v^{(n)})}^2\hfill \\ & & +u^{(n)}(v^{(n+m)}-v^{(n)})(v^{(n+m)}+v^{(n)})+\frac{1}{2}(u^{(n+m)}-u^{(n)}){(v_x^{(n+m)})}^2\hfill \\ & & +\frac{1}{2}{u}^{(n)}(v_x^{(n+m)}-v_x^{(n)})(v_x^{(n+m)}+v_x^{(n)})\}-\frac{1}{2}{(1-{\mathsf{\partial }}_x^2)}^{-1}\{(u_x^{(n+m)}-u_x^{(n)})[{(u_x^{(n+m)})}^2\hfill \\ & & +u_x^{(n+m)}{u}_x^{(n)}+{(u_x^{(n)})}^2]+(u_x^{(n+m)}-u_x^{(n)}){(v_x^{(n)})}^2+u_x^{(n)}(v_x^{(n+m)}-v_x^{(n)})(v_x^{(n+m)}+v_x^{(n)})\}\hfill \end{array}$$

$$\begin{array}{ccc}& & G_1(u^{(n+m)},v^{(n+m)},v^{(n+1)},u^{(n)},v^{(n)})\hfill \\ & =& [(u^{(n)}-u^{(n+m)})(u^{(n)}+u^{(n+m)})+(v^{(n)}-v^{(n+m)})(v^{(n)}+v^{(n+m)})]v_x^{(n+1)}\hfill \\ & & -{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\{\frac{3}{2}{(v_x^{(n+m)})}^2(v^{(n+m)}-v^{(n)})+\frac{3}{2}{v}^{(n)}(v_x^{(n+m)}-v_x^{(n)})(v_x^{(n+m)}+v_x^{(n)})\hfill \\ & & +(u^{(n+m)}-u^{(n)})v_x^{(n+m)}{u}_x^{(n+m)}+u^{(n)}{v}_x^{(n+m)}(u_x^{(n+m)}-u_x^{(n)})+u_x^{(n)}{u}^{(n)}(v_x^{(n+m)}-v_x^{(n)})\hfill \\ & & +(v^{(n+m)}-v^{(n)})[{(v^{(n+m)})}^2+v^{(n+m)}{v}^{(n)}+{(v^{(n)})}^2]+(v^{(n+m)}-v^{(n)}){(u^{(n+m)})}^2\hfill \\ & & +v^{(n)}(u^{(n+m)}-u^{(n)})(u^{(n+m)}+u^{(n)})+\frac{1}{2}(v^{(n+m)}-v^{(n)}){(u_x^{(n)})}^2\hfill \\ & & +\frac{1}{2}{v}^{(n)}(u_x^{(n+m)}-u_x^{(n)})(u_x^{(n+m)}+u_x^{(n)})\}-\frac{1}{2}{(1-{\mathsf{\partial }}_x^2)}^{-1}\{(v_x^{(n+m)}-v_x^{(n)})[{(v_x^{(n+m)})}^2\hfill \\ & & +v_x^{(n+m)}{v}_x^{(n)}+{(v_x^{(n)})}^2]+(v_x^{(n+m)}-v_x^{(n)}){(u_x^{(n+m)})}^2+v_x^{(n)}(u_x^{(n+m)}-u_x^{(n)})(u_x^{(n+m)}+u_x^{(n)})\}.\hfill \end{array}$$

By applying the algebraic property of Besov spaces $B_{p,1}^{1+\frac{1}{p}}$ and $B_{p,1}^{\frac{1}{p}}$, we obtain

$$\begin{array}{ccc}& & \parallel F_1(u^{(n+m)},v^{(n+m)},u^{(n+1)},u^{(n)},v^{(n)}){\parallel }_{B_{p,1}^{\frac{1}{p}}}\le C(\parallel u^{(n+m)}-u^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel v^{(n+m)}-v^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}})\hfill \\ & & \times [\parallel u^{(n+1)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}(\parallel u^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})\hfill \\ & & +\parallel u^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2].\hfill \end{array}$$

Similarly,

$$\begin{array}{ccc}& & \parallel G_1(u^{(n+m)},v^{(n+m)},v^{(n+1)},u^{(n)},v^{(n)}){\parallel }_{B_{p,1}^{\frac{1}{p}}}\le C(\parallel u^{(n+m)}-u^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel v^{(n+m)}-v^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}})\hfill \\ & & \times [\parallel v^{(n+1)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}(\parallel u^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})\hfill \\ & & +\parallel u^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2].\hfill \end{array}$$

Thanks to Lemmas 2 and 3, we see that for any $t\in [0,T]$,

$$\begin{array}{ccc}& & \parallel (u^{(n+m+1)}-u^{(n+1)})(t){\parallel }_{B_{p,1}^{\frac{1}{p}}}+{\parallel (v^{(n+m+1)}-v^{(n+1)})(t)\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & \le & C{e}^{C{\int }_0^t{\parallel [{(u^{(n+m)})}^2+{(v^{(n+m)})}^2](t^{\prime })\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\mathrm{d}{t}^{\prime }}\hfill \\ & \times & {\int }_0^t{e}^{-C{\int }_0^{\tau }{\parallel [{(u^{(n+m)})}^2+{(v^{(n+m)})}^2]({\tau }^{\prime })\parallel }_{B_{p,1}^{\frac{1}{p}}}\mathrm{d}{\tau }^{\prime }}\times (\parallel (u^{(n+m)}-u^{(n)})(t){\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel (v^{(n+m)}-v^{(n)})(t){\parallel }_{B_{p,1}^{\frac{1}{p}}})\hfill \\ & \times & (\parallel u^{(n+1)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n+1)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel u^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n+m)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2)\mathrm{d}\tau .\hfill \end{array}$$

Therefore, by making use of the induction, one can see a positive constant $C_T^{\prime }$ exists, which is independent of n and m, such that

$$\begin{array}{ccc}& & \parallel u^{(n+m+1)}-u^{(n+1)}{\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+{\parallel v^{(n+m+1)}-v^{(n+1)}\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}\hfill \\ & \le & \frac{{(T{C}_T)}^{n+1}}{(n+1)!}(\parallel u^{(m)}{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}+\parallel v^{(m)}{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})})\le C_T^{\prime }{2}^{-n},\hfill \end{array}$$

which indicates that the sequence ${(u^{(n)},v^{(n)})}_{n\in \mathbb{N}}\in ([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$ is a Cauchy sequence and converges to $(u,v)\in C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$. Therefore, via Fatou’s Lemma, we obtain

$$\begin{array}{c}\hfill {\parallel u\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+{\parallel v\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}\le \frac{\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+{\parallel v_0\parallel }_{B_{p,1}^{1+\frac{1}{p}}}}{[1-4CT(\parallel u_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v_0{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}{{)}^2]}^{\frac{1}{2}}}\end{array}$$

Next, we need to show that $(u,v)$ solves the Equation (3) using the following steps.

In view of (5), we consider the following system

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle {\int }_0^t(\langle u^{(n+1)},{\mathsf{\partial }}_t\varphi (\tau )\rangle +\langle u^{(n+1)}[{(u^{(n)})}^2+{(v^{(n)})}^2](\tau ),{\mathsf{\partial }}_x\varphi (\tau )\rangle +{\langle u^{(n+1)}{[{(u^{(n)})}^2+v^{(n)})}^2]}_x(\tau )}\hfill \\ -F(u^{(n)},v^{(n)}),\varphi (\tau )\rangle )\mathrm{d}\tau =\langle u^{(n+1)}(t),\varphi (t)\rangle -\langle u^{(n+1)}(0),\varphi (0)\rangle ,\hfill \\ {\displaystyle {\int }_0^t(\langle v^{(n+1)},{\mathsf{\partial }}_t\varphi (\tau )\rangle +\langle v^{(n+1)}[{(u^{(n)})}^2+{(v^{(n)})}^2](\tau ),{\mathsf{\partial }}_x\varphi (\tau )\rangle +{\langle v^{(n+1)}{[{(u^{(n)})}^2+v^{(n)})}^2]}_x(\tau )}\hfill \\ -G(u^{(n)},v^{(n)}),\varphi (\tau )\rangle )\mathrm{d}\tau =\langle v^{(n+1)}(t),\varphi (t)\rangle -\langle v^{(n+1)}(0),\varphi (0)\rangle ,\hfill \end{array}\right.\end{array}$$

(10)

where $\varphi \in C^1([0,T];\mathcal{S})$ is a test function. By virtue of Proposition 1, we can take the limit as $n\to \infty$ in the first equation of (10) to check its convergence

$$\begin{array}{ccc}& & \left|{\displaystyle {\int }_0^t\langle u^{(n+1)}(\tau )[{(u^{(n)})}^2+{(v^{(n)})}^2](\tau ),{\mathsf{\partial }}_x\varphi (\tau )\rangle \mathrm{d}\tau -{\int }_0^t\langle u(\tau )(u^2+v^2)(\tau ),{\mathsf{\partial }}_x\varphi \rangle \mathrm{d}\tau }\right|\hfill \\ & =& \left|{\displaystyle {\int }_0^t\langle u^{(n+1)}[{(u^{(n)})}^2+{(v^{(n)})}^2]-u(u^2+v^2),{\mathsf{\partial }}_x\varphi \rangle \mathrm{d}\tau }\right|\hfill \\ & \le & {\displaystyle {\int }_0^t\left|\langle (u^{(n+1)}-u)[{(u^{(n)})}^2+{(v^{(n)})}^2],{\mathsf{\partial }}_x\varphi \rangle \right|\mathrm{d}\tau }\hfill \\ & & {\displaystyle +{\int }_0^t\left|\langle u[{(u^{(n)})}^2+{(v^{(n)})}^2-(u^2+v^2)],{\mathsf{\partial }}_x\varphi \rangle \right|\mathrm{d}\tau }\hfill \\ & \le & T(\parallel (u^{(n+1)}-u)[{(u^{(n)})}^2+{(v^{(n)})}^2]{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}\parallel {\mathsf{\partial }}_x\varphi {\parallel }_{L_T^{\infty }(B_{p^{\prime },\infty }^{-1-\frac{1}{p}})}\hfill \\ & & +\parallel u[{(u^{(n)})}^2+{(v^{(n)})}^2-(u^2+v^2)]{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}\parallel {\mathsf{\partial }}_x\varphi {\parallel }_{L_T^{\infty }(B_{p^{\prime },\infty }^{-1-\frac{1}{p}})})\hfill \\ & \le & T(\parallel u^{(n+1)}-u{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}\parallel {(u^{(n)})}^2+{(v^{(n)})}^2{\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}\hfill \\ & & +{\parallel u\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}\parallel {(u^{(n)})}^2+{(v^{(n)})}^2-(u^2+v^2){\parallel }_{L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})}{)\parallel \varphi \parallel }_{L_T^{\infty }(B_{p^{\prime },\infty }^{-\frac{1}{p}})},\hfill \end{array}$$

which converges to 0 as $n\to \infty$, where $p^{\prime }=\frac{p}{p-1}$, deduced from which we find that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{\int }_0^t\langle u^{(n+1)}[{(u^{(n)})}^2+{(v^{(n)})}^2],{\mathsf{\partial }}_x\varphi \rangle \mathrm{d}\tau ={\int }_0^t\langle u(u^2+v^2),{\mathsf{\partial }}_x\varphi \rangle \mathrm{d}\tau .\end{array}$$

The convergence of other terms can be computed in a similar way. Thus we can derive for any $t\in [0,T]$,

$$\begin{array}{ccc}& & {\int }_0^t(\langle u,{\mathsf{\partial }}_t\varphi \rangle +\langle u(u^2+v^2),{\mathsf{\partial }}_x\varphi \rangle +\langle u{\mathsf{\partial }}_x(u^2+v^2),\varphi \rangle -\langle F(u,v),\varphi \rangle )\mathrm{d}\tau \hfill \\ & =& \langle u,\varphi \rangle -\langle u(0),\varphi (0)\rangle .\hfill \end{array}$$

Similarly, for the second equation of (10), we can derive that

$$\begin{array}{ccc}& & {\int }_0^t(\langle v,{\mathsf{\partial }}_t\varphi \rangle +\langle v(u^2+v^2),{\mathsf{\partial }}_x\varphi \rangle +\langle v{\mathsf{\partial }}_x(u^2+v^2),\varphi \rangle -\langle G(u,v),\varphi \rangle )\mathrm{d}\tau \hfill \\ & =& \langle v,\varphi \rangle -\langle v(0),\varphi (0)\rangle .\hfill \end{array}$$

Consequently, $(u,v)$ solves the Equation (3), Lemma 4 then shows that $(u,v)\in C([0,T];B_{p,1}^{1+\frac{1}{p}})\times C([0,T];B_{p,1}^{1+\frac{1}{p}})$. Notice that $u_t=F(u,v)-(u^2+v^2){\mathsf{\partial }}_{x}u\in C([0,T];B_{p,1}^{\frac{1}{p}})$ and $v_t=G(u,v)-(u^2+v^2){\mathsf{\partial }}_{x}v\in C([0,T];B_{p,1}^{\frac{1}{p}})$, we conclude that $(u,v)\in E_{p,1}^{1+\frac{1}{p}}(T)\times E_{p,1}^{1+\frac{1}{p}}(T)$.

Step 3. Uniqueness with respect to the initial data.

To prove the uniqueness result, we follow the structure of [23] and introduce the associated Lagrangian scale of (3) as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{ccc}{\displaystyle \frac{\mathrm{d}q}{\mathrm{d}t}=H(u,v)(t,q(t,\sigma ))\triangleq (u^2+v^2)(t,q(t,\sigma )),t>0,}\hfill & & \sigma \in \mathbb{R},\hfill \\ q(0,\sigma )=\sigma ,\hfill & & \sigma \in \mathbb{R}.\hfill \end{array}\right.\end{array}$$

By introducing the new variable $U(t,\sigma )=u(t,q(t,\sigma )),V(t,\sigma )=u(t,q(t,\sigma ))$, then (3) becomes

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{U}_t=(F(u,v))(t,q(t,\sigma ))\triangleq \tilde{F}(U,V,q),\hfill \\ V_t=(G(u,v))(t,q(t,\sigma ))\triangleq \tilde{G}(U,V,q),\hfill \\ {U(t,\sigma )|}_{t=0}=U_0{(\sigma ),V(t,\sigma )|}_{t=0}=U_0(\sigma ),\sigma \in \mathbb{R}.\hfill \end{array}\right.\end{array}$$

(11)

By Denoting $U_i:=u_i(t,q(t,\sigma )),V_i:=v_i(t,q(t,\sigma )),i=1,2$, we may derive from (11) that

$$\begin{array}{ccc}\hfill \frac{d}{\mathrm{d}t}{U}_i& =& \tilde{F}(U_i,V_i,q_i)\hfill \\ & =& \frac{1}{2}{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_i(t,\sigma )-x)e^{-|q_i(t,\sigma )-x|}(u_i^3+u_i{v}_i^2+\frac{3}{2}{u}_i{u}_{ix}^2+u_{ix}{v}_i{v}_{ix}+\frac{1}{2}{u}_{ix}^2{v}_i)\hfill \\ & & -\frac{1}{4}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(u_{ix}^3+u_{ix}{v}_{ix}^2)\mathrm{d}x,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill \frac{d}{\mathrm{d}t}{V}_i& =& \tilde{G}(U_i,V_i,q_i)\hfill \\ & =& \frac{1}{2}{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_i(t,\sigma )-x)e^{-|q_i(t,\sigma )-x|}(v_i^3+u_i^2{v}_i+\frac{3}{2}{v}_i{v}_{ix}^2+u_i{u}_{ix}{v}_{ix}+\frac{1}{2}{u}_{ix}^2{v}_i)\hfill \\ & & -\frac{1}{4}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(v_{ix}^3+u_{ix}^2{v}_{ix})\mathrm{d}x,\hfill \end{array}$$

(13)

and

$$\begin{array}{ccc}& & \frac{\mathrm{d}}{\mathrm{d}t}{U}_{i\sigma }={(\tilde{F}(U_i,V_i,q_i))}_{\sigma }=U_i^3{q}_{i\sigma }+U_i{V}_i^2{q}_{i\sigma }+\frac{3}{2}\frac{U_i{U}_{i\sigma }^2}{q_{i\sigma }}+\frac{U_{i\sigma }{V}_{i\sigma }{V}_i}{q_{i\sigma }}\hfill \\ & +& \frac{1}{2}\frac{U_i{V}_{i\sigma }^2}{q_{i\sigma }}-\frac{q_{i\sigma }}{2}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(u_i^3+u_i{v}_i^2+\frac{3}{2}{u}_i{u}_{ix}^2+u_{ix}{v}_i{v}_{ix}+\frac{1}{2}{u}_i{v}_{ix}^2)\mathrm{d}x\hfill \\ & -& \frac{1}{2q_{i\sigma }^2}{U}_{i\sigma }(U_{i\sigma }^2+V_{i\sigma }^2)+\frac{q_{i\sigma }}{4}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(u_{ix}^3+u_{ix}{v}_{ix}^2)\mathrm{d}x,\hfill \end{array}$$

(14)

$$\begin{array}{ccc}& & \frac{\mathrm{d}}{\mathrm{d}t}{V}_{i\sigma }={(\tilde{G}(U_i,V_i,y_i))}_{\sigma }=V_i^3{q}_{i\sigma }+U_i^2{V}_i{q}_{i\sigma }+\frac{3}{2}\frac{V_i{V}_{i\sigma }^2}{q_{i\sigma }}+\frac{U_i{U}_{i\sigma }{V}_{i\sigma }}{q_{i\sigma }}\hfill \\ & +& \frac{1}{2}\frac{U_{i\sigma }^2{V}_i}{q_{i\sigma }}-\frac{q_{i\sigma }}{2}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(v_i^3+u_i^2{v}_i+\frac{3}{2}{v}_i{v}_{ix}^2+u_i{u}_{ix}{v}_{ix}+\frac{1}{2}{u}_{ix}^2{v}_i)\mathrm{d}x\hfill \\ & -& \frac{1}{2q_{i\sigma }^2}{V}_{i\sigma }(U_{i\sigma }^2+V_{i\sigma }^2)+\frac{q_{i\sigma }}{4}{\int }_{-\infty }^{+\infty }{e}^{-|q_i(t,\sigma )-x|}(v_{ix}^3+u_{ix}^2{v}_{ix})\mathrm{d}x.\hfill \end{array}$$

(15)

Similar to the proof of Theorem 1.1 in [23], we can find that $(U_i(t,\sigma ),V_i(t,\sigma ))\in L_T^{\infty }(W^{1,p}\cap W^{1,\infty })\times L_T^{\infty }(W^{1,p}\cap W^{1,\infty })$, $q_i(t,\sigma )-\sigma \in L_T^{\infty }(W^{1,p}\cap W^{1,\infty })$ and $\frac{1}{2}\le q_{i\sigma }\le C_{(u_0,v_0)}$ for sufficient small T.

We estimate $\parallel \tilde{F}(U_1,V_1,q_1)-\tilde{F}(U_2,V_2,q_2){\parallel }_{L^p\cap L^{\infty }}$ next. The main difficulty lies in estimates of coupling terms containing derivatives. For the terms containing the sign function, we choose to estimate the following term as an example

$$\begin{array}{c}\hfill {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_{1x}{v}_1{v}_{1x}\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_{2x}{v}_2{v}_{2x}\mathrm{d}x.\end{array}$$

Notice that $q_i(i=1,2)$ is monotonically increasing, therefore $\mathrm{sign}(q_i(\sigma )-q_i(\tau ))=\mathrm{sign}(\sigma -\tau )$, and thus,

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_{1x}{v}_1{v}_{1x}\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_{2x}{v}_2{v}_{2x}\mathrm{d}x\hfill \\ & =& {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-q_1(\tau ))e^{-|q_1(\sigma )-q_1(\tau )|}\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & & -{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-q_2(\tau ))e^{-|q_2(\sigma )-x|}\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_2\mathrm{d}\tau \hfill \\ & =& {\int }_{-\infty }^{+\infty }\mathrm{sign}(\sigma -\tau )(e^{-|q_1(\sigma )-q_1(\tau )|}-e^{-|q_2(\sigma )-q_2(\tau )|})\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & & +{\int }_{-\infty }^{+\infty }\mathrm{sign}(\sigma -\tau )e^{-|q_2(\sigma )-q_2(\tau )|}(\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1-\frac{U_{2\tau }{V}_{2\tau }}{q_{2\tau }}{V}_2)\mathrm{d}\tau \triangleq I_1+I_2.\hfill \end{array}$$

(16)

If $\sigma >\tau$(or $\sigma <\tau$), then $q_i(\sigma )>q_i(\tau )$(or $q_i(\sigma )<q_i(\tau ))$, therefore we can derive

$$\begin{array}{ccc}\hfill I_1& =& {\int }_{-\infty }^{\sigma }(e^{-(q_1(\sigma )-q_1(\tau ))}-e^{-(q_2(\sigma )-q_2(\tau ))})\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & & -{\int }_{\sigma }^{+\infty }(e^{-(q_1(\sigma )-q_1(\tau ))}-e^{-(q_2(\sigma )-q_2(\tau ))})\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & =& {\int }_{-\infty }^{\sigma }{e}^{\sigma -\tau }(e^{-{\int }_0^{t}H(U_1,V_1)(\sigma )-H(U_1,V_1)(\tau )\mathrm{d}{t}^{\prime }}-e^{-{\int }_0^{t}H(U_2,V_2)(\sigma )-H(U_2,V_2)(\tau )\mathrm{d}{t}^{\prime }})\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & & -{\int }_{\sigma }^{+\infty }{e}^{\sigma -\tau }(e^{-{\int }_0^{t}H(U_1,V_1)(\sigma )-H(U_1,V_1)(\tau )\mathrm{d}{t}^{\prime }}-e^{-{\int }_0^{t}H(U_2,V_2)(\sigma )-H(U_2,V_2)(\tau )\mathrm{d}{t}^{\prime }})\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau \hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})({\int }_{-\infty }^{\sigma }{e}^{-(\sigma -\tau )}\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau +{\int }_{\sigma }^{+\infty }{e}^{\sigma -\tau }\frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau )\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})\hfill \\ & & \times [1_{\ge 0}(x)e^{-\left|x\right|}\ast \frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau +1_{\le 0}(x)e^{-\left|x\right|}\ast \frac{U_1{V}_{1\tau }}{q_{1\tau }}{V}_1\mathrm{d}\tau ].\hfill \end{array}$$

(17)

By using a similar technique and applying the inverse trigonometric inequality, we have

$$\begin{array}{ccc}\hfill I_2& \le & C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(18)

Substituting estimates (17) and (18) back into (16) yields

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_{1x}{v}_1{v}_{1x}\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_{2x}{v}_2{v}_{2x}\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast \frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1+1_{\le 0}(x)e^{-\left|x\right|}\ast \frac{U_{1\tau }{V}_{1\tau }}{q_{1\tau }}{V}_1]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(19)

Similar to (19), we can derive the estimates of other terms containing the sign functions as follows:

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_1^3\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_2^3\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (U_1^3{q}_{1\tau })+1_{\le 0}(x)e^{-\left|x\right|}\ast (U_1^3{q}_{1\tau })]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(20)

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_1{v}_1^2\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_2{v}_2^2\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (U_1{V}_1^2{q}_{1\tau })+1_{\le 0}(x)e^{-\left|x\right|}\ast (U_1{V}_1^2{q}_{1\tau })]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|V_1-V_2|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(21)

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_1{u}_{1x}^2\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_2{u}_{2x}^2\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (U_1\frac{U_{1\tau }^2}{q_{1\tau }})+1_{\le 0}(x)e^{-\left|x\right|}\ast (U_1\frac{U_{1\tau }^2}{q_{1\tau }})]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|U_{1\tau }-U_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|U_{1\tau }-U_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(22)

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }\mathrm{sign}(q_1(\sigma )-x)e^{-|q_1(\sigma )-x|}{u}_1{v}_{1x}^2\mathrm{d}x-{\int }_{-\infty }^{+\infty }\mathrm{sign}(q_2(\sigma )-x)e^{-|q_2(\sigma )-x|}{u}_2{v}_{2x}^2\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (U_1\frac{V_{1\tau }^2}{q_{1\tau }})+1_{\le 0}(x)e^{-\left|x\right|}\ast (U_1\frac{V_{1\tau }^2}{q_{1\tau }})]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_1-U_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(23)

For the other two terms which do not contain the sign function, we estimate them as follows:

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-x|}(u_{1x}^3)\mathrm{d}x-{\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-x|}(u_{2x}^3)\mathrm{d}x\hfill \\ & =& {\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-q_1(\tau )|}\frac{U_{1\tau }^3}{q_{1\tau }^2}\mathrm{d}\tau -{\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-x|}\frac{u_{2\tau }^3}{q_{2\tau }^2}\mathrm{d}\eta \hfill \\ & =& {\int }_{-\infty }^{+\infty }(e^{-|q_1(\sigma )-q_1(\tau )|}-e^{-|q_2(\sigma )-q_2(\tau )|})\frac{U_{1\tau }^3}{q_{1\tau }^2}\mathrm{d}\tau +{\int }_{-\infty }^{+\infty }{e}^{-|q_2(\sigma )-q_2(\tau )|}(\frac{U_{1\tau }^3}{q_{1\tau }^2}-\frac{U_{2\tau }^3}{q_{2\tau }^2})\mathrm{d}\tau \hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (\frac{U_{1\tau }^3}{q_{1\tau }^2})+1_{\le 0}(x)e^{-\left|x\right|}\ast (\frac{U_{1\tau }^3}{q_{1\tau }^2})]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(24)

Similarly

$$\begin{array}{ccc}& & {\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-x|}(u_{1x}{v}_{1x}^2)\mathrm{d}x-{\int }_{-\infty }^{+\infty }{e}^{-|q_1(\sigma )-x|}(u_{2x}{v}_{2x}^2)\mathrm{d}x\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})[1_{\ge 0}(x)e^{-\left|x\right|}\ast (\frac{U_{1\tau }{V}_{1\tau }^2}{q_{1\tau }^2})+1_{\le 0}(x)e^{-\left|x\right|}\ast (\frac{U_{1\tau }{V}_{1\tau }^2}{q_{1\tau }^2})]\hfill \\ & & +C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)].\hfill \end{array}$$

(25)

Combining (19)–(25), we find that

$$\begin{array}{ccc}& & |\tilde{F}(U_1,V_1,q_1)-\tilde{F}(U_2,V_2,q_2)|\hfill \\ & \le & C[1_{\ge 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\eta }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (|U_{1\tau }-U_{2\tau }|+|V_1-V_2|+|V_{1\tau }-V_{2\tau }|+|q_{1\tau }-q_{2\tau }|)]\hfill \\ & & +C(\parallel U_1-U_2{\parallel }_{L^{\infty }}+\parallel V_1-V_2{\parallel }_{L^{\infty }})\hfill \\ & & \times [1_{\ge 0}(x)e^{-\left|x\right|}\ast (U_1^3{q}_{1\tau }+U_1{V}_1^2{q}_{1\tau }+\frac{U_1{U}_{1\tau }^2}{q_{1\tau }}+\frac{U_1{V}_{1\tau }^2}{q_{1\tau }}+\frac{U_{1\tau }^3}{q_{1\tau }^2}+\frac{U_{1\tau }{V}_{1\tau }{V}_1}{q_{1\tau }}+\frac{U_{1\tau }{V}_1^2}{q_{1\tau }^2})\hfill \\ & & +1_{\le 0}(x)e^{-\left|x\right|}\ast (U_1^3{q}_{1\tau }+U_1{V}_1^2{q}_{1\tau }+\frac{U_1{U}_{1\tau }^2}{q_{1\tau }}+\frac{U_1{V}_{1\tau }^2}{q_{1\tau }}+\frac{U_{1\tau }^3}{q_{1\tau }^2}+\frac{U_{1\tau }{V}_{1\tau }{V}_1}{q_{1\tau }}+\frac{U_{1\tau }{V}_{1\tau }^2}{q_{1\tau }^2})].\hfill \end{array}$$

Which implies that

$$\begin{array}{ccc}& & \parallel \tilde{F}(U_1,V_1,q_1)-\tilde{F}(U_2,V_2,q_2){\parallel }_{L^{\infty }\cap L^p}\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }\cap L^p}+\parallel U_{1\sigma }-U_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}+\parallel V_1-V_2{\parallel }_{L^{\infty }\cap L^p}\hfill \\ & & +\parallel V_{1\sigma }-V_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}+\parallel q_{1\sigma }-q_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}).\hfill \end{array}$$

(26)

Similarly, we can also obtain

$$\begin{array}{ccc}& & \parallel {(\tilde{F}(U_1,V_1,q_1))}_{\sigma }-{(\tilde{F}(U_2,V_2,q_2))}_{\sigma }{\parallel }_{L^{\infty }\cap L^p}\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^{\infty }\cap L^p}+\parallel U_{1\sigma }-U_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}+\parallel V_1-V_2{\parallel }_{L^{\infty }\cap L^p}\hfill \\ & & +\parallel V_{1\sigma }-V_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}+\parallel q_{1\sigma }-q_{2\sigma }{\parallel }_{L^{\infty }\cap L^p}).\hfill \end{array}$$

(27)

Combining (26) and (27), we have

$$\begin{array}{ccc}& & \parallel \tilde{F}(U_1,V_1,q_1)-\tilde{F}(U_2,V_2,q_2){\parallel }_{W^{1,\infty }\cap W^{1,p}}\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel q_1-q_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}).\hfill \end{array}$$

(28)

Next, we are ready to prove uniqueness. Suppose that $u_1,v_1$ and $(u_2,v_2)$ are two solutions to (3), then $(U_i(t,\sigma )=u_i(t,q(t,\sigma )),V_i(t,\sigma )=v_i(t,q(t,\sigma )))$ satisfies (3) and the following equations for $i=1,2$

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{U}_{it\sigma }={(\tilde{F}(U_i,V_i,q_i))}_{\sigma },\hfill \\ V_{it\sigma }={(\tilde{G}(U_i,V_i,q_i))}_{\sigma }.\hfill \end{array}\right.\end{array}$$

With the help of (28), we see

$$\begin{array}{ccc}& & \parallel U_1-U_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+{\parallel q_1-q_2\parallel }_{W^{1,\infty }\cap W^{1,p}}\hfill \\ & \le & C(\parallel U_1(0)-U_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1(0)-V_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel q_1(0)-q_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}})\hfill \\ & & +C{\int }_0^T{\parallel \tilde{F}(U_1,V_1,q_1)-\tilde{F}(U_2,V_2,q_2)\parallel }_{W^{1,\infty }\cap W^{1,p}}\mathrm{d}t\hfill \\ & \le & C(\parallel U_1(0)-U_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1(0)-V_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}})\hfill \\ & & +C{\int }_0^T(\parallel U_1-U_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel q_1-q_2{\parallel }_{W^{1,\infty }\cap W^{1,p}})\mathrm{d}t,\hfill \end{array}$$

(29)

where we use the fact that $q_1(0)=q_2(0)=\sigma$.

Applying the Gronwall Lemma to (29) yields

$$\begin{array}{ccc}& & \parallel U_1-U_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,\infty }\cap W^{1,p}}+{\parallel q_1-q_2\parallel }_{W^{1,\infty }\cap W^{1,p}}\hfill \\ & \le & C(\parallel U_1(0)-U_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel V_1(0)-V_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}})\hfill \\ & =& C(\parallel u_1(0)-u_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel v_1(0)-v_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}).\hfill \end{array}$$

(30)

Therefore, it follows from (30) that

$$\begin{array}{ccc}& & \parallel u_1-u_2{\parallel }_{L^p}+{\parallel v_1-v_2\parallel }_{L^p}\hfill \\ & \le & C(\parallel u_1\circ q_1-u_2\circ q_1{\parallel }_{L^p}+\parallel v_1\circ q_1-v_2\circ q_1{\parallel }_{L^p})\hfill \\ & \le & C(\parallel u_1\circ q_1-u_2\circ q_2+u_2\circ q_2-u_2\circ q_1{\parallel }_{L^p}+\parallel v_1\circ q_1-v_2\circ q_2+v_2\circ q_2-v_2\circ q_1{\parallel }_{L^p})\hfill \\ & \le & C(\parallel u_1\circ q_1-u_2\circ q_2{\parallel }_{L^p}+\parallel u_2\circ q_2-u_2\circ q_1{\parallel }_{L^p}+\parallel v_1\circ q_1-v_2\circ q_2{\parallel }_{L^p}+\parallel v_2\circ q_2-v_2\circ q_1{\parallel }_{L^p})\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^p}+\parallel u_{2x}{\parallel }_{L^{\infty }}\parallel q_2-q_1{\parallel }_{L^p}+\parallel V_1-V_2{\parallel }_{L^p}+\parallel v_{2x}{\parallel }_{L^{\infty }}\parallel q_2-q_1{\parallel }_{L^p})\hfill \\ & \le & C(\parallel U_1-U_2{\parallel }_{L^p}+\parallel V_1-V_2{\parallel }_{L^p}+\parallel q_2-q_1{\parallel }_{L^p}).\hfill \end{array}$$

(31)

Similarly, we can obtain

$$\begin{array}{c}\hfill \parallel {\mathsf{\partial }}_x(u_1-u_2){\parallel }_{L^p}+\parallel {\mathsf{\partial }}_x(v_1-v_2){\parallel }_{L^p}\le C(\parallel U_1-U_2{\parallel }_{W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,p}}+\parallel q_2-q_1{\parallel }_{W^{1,p}}).\end{array}$$

(32)

(31) together with (32) implies that

$$\begin{array}{c}\hfill \parallel u_1-u_2{\parallel }_{W^{1,p}}+\parallel v_1-v_2{\parallel }_{W^{1,p}}\le C(\parallel U_1-U_2{\parallel }_{W^{1,p}}+\parallel V_1-V_2{\parallel }_{W^{1,p}}+\parallel q_2-q_1{\parallel }_{W^{1,p}}).\end{array}$$

Then, from the embedding theorem, it follows that

$$\begin{array}{ccc}\hfill \parallel u_1-u_2{\parallel }_{B_{p,\infty }^1}+{\parallel v_1-v_2\parallel }_{B_{p,\infty }^1}& \le & C(\parallel u_1-u_2{\parallel }_{W^{1,p}}+\parallel v_1-v_2{\parallel }_{W^{1,p}})\hfill \\ & \le & C(\parallel u_1(0)-u_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}}+\parallel v_1(0)-v_2(0){\parallel }_{W^{1,\infty }\cap W^{1,p}})\hfill \\ & \le & C(\parallel u_1(0)-u_2(0){\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v_1(0)-v_2(0){\parallel }_{B_{p,1}^{1+\frac{1}{p}}}).\hfill \end{array}$$

Thus, if $u_1(0)=u_2(0),v_1(0)=v_2(0)$, we can prove the uniqueness result immediately.

Step 4. Continuous dependence on the initial data

Let $(u_0^{(n)},v_0^{(n)})\to (u_0^{\infty },v_0^{\infty })$ in $B_{p,1}^{1+\frac{1}{p}}\times B_{p,1}^{1+\frac{1}{p}}$ as $n\to \infty$, and $(u^{(n)},v^{(n)}),(u^{\infty },v^{\infty })$ are the solutions with initial data $(u_0^{(n)},v_0^{(n)}),(u_0^{\infty },v_0^{\infty })$, respectively. By the above discussion, we find that $(u^{(n)},v^{(n)}),(u^{\infty },v^{\infty })$ are uniformly bounded in $L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})\times L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})$ and

$$\begin{array}{c}\hfill \parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,\infty }^0}+\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,\infty }^0}\le C(\parallel u_0^{(n)}-u_0^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v_0^{(n)}-v_0^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}),\end{array}$$

which indicates that $(u^{(n)},v^{(n)})\to (u^{\infty },v^{\infty })$ in $C([0,T];B_{p,\infty }^0)\times C([0,T];B_{p,\infty }^0)$. By applying the interpolation inequality we can find

$$\begin{array}{c}\hfill (u^{(n)},v^{(n)})\to (u^{\infty },v^{\infty })in{B}_{p,1}^{\frac{1}{p}}\times B_{p,1}^{\frac{1}{p}}.\end{array}$$

(33)

Next, in order to prove the continuous dependence result, we just need to prove

$$\begin{array}{c}\hfill ({\mathsf{\partial }}_x{u}^{(n)},{\mathsf{\partial }}_x{v}^{(n)})\to ({\mathsf{\partial }}_x{u}^{\infty },{\mathsf{\partial }}_x{v}^{\infty })\end{array}$$

in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$.

For this purpose, we decompose the solution components as follows:

$$\begin{array}{c}\hfill {\mathsf{\partial }}_x{u}^{(n)}=w_1^{(n)}+z_1^{(n)},{\mathsf{\partial }}_x{v}^{(n)}=w_2^{(n)}+z_2^{(n)},\end{array}$$

where $(w_1^{(n)},w_2^{(n)})$ solves the following problem:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_t{w}_1^{(n)}+H(u^{(n)},v^{(n)}){\mathsf{\partial }}_x{w}_1^{(n)}=\tilde{F}(u^{\infty },v^{\infty }),\hfill \\ {\mathsf{\partial }}_t{w}_2^{(n)}+H(u^{(n)},v^{(n)}){\mathsf{\partial }}_x{w}_2^{(n)}=\tilde{G}(u^{\infty },v^{\infty }),\hfill \\ w_1^{(n)}(0,x)={\mathsf{\partial }}_x{u}_0^{\infty },w_2^{(n)}(0,x)={\mathsf{\partial }}_x{v}_0^{\infty }.\hfill \end{array}\right.\end{array}$$

and $(z_1^{(n)},z_2^{(n)})$ solves the following problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathsf{\partial }}_t{z}_1^{(n)}+H(u^{(n)},v^{(n)}){\mathsf{\partial }}_x{z}_1^{(n)}=\tilde{F}(u^{(n)},v^{(n)})-\tilde{F}(u^{\infty },v^{\infty }),\hfill \\ {\mathsf{\partial }}_t{z}_2^{(n)}+H(u^{(n)},v^{(n)}){\mathsf{\partial }}_x{z}_2^{(n)}=\tilde{G}(u^{(n)},v^{(n)})-\tilde{G}(u^{\infty },v^{\infty }),\hfill \\ z_1^{(n)}(0,x)={\mathsf{\partial }}_x{u}_0^{(n)}-{\mathsf{\partial }}_x{u}_0^{\infty },z_2^{(n)}(0,x)={\mathsf{\partial }}_x{v}_0^{(n)}-{\mathsf{\partial }}_x{v}_0^{\infty }.\hfill \end{array}\right.\end{array}$$

$\tilde{F}(u,v)$ and $\tilde{G}(u,v)$ are defined as follows:

$$\begin{array}{ccc}\hfill \tilde{F}(u,v)& =& -{(u^2+v^2)}_x{u}_x-{\mathsf{\partial }}_x^2{(1-{\mathsf{\partial }}_x^2)}^{-1}{f}_1-{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}{f}_2,\hfill \end{array}$$

(34)

$$\begin{array}{ccc}\hfill \tilde{G}(u,v)& =& -{(u^2+v^2)}_x{v}_x-{\mathsf{\partial }}_x^2{(1-{\mathsf{\partial }}_x^2)}^{-1}{g}_1-{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}{g}_2,\hfill \end{array}$$

(35)

among which,

$$\begin{array}{c}\hfill f_1(u,v)=u^3+u{v}^2+\frac{3}{2}u{u}_x^2+u_{x}v{v}_x+\frac{1}{2}u{v}_x^2,f_2=\frac{1}{2}(u_x^3+u_x{v}_x^2),\\ \hfill g_1(u,v)=v^3+u^{2}v+\frac{3}{2}v{v}_x^2+u{u}_x{v}_x+\frac{1}{2}{u}_x^{2}v,g_2=\frac{1}{2}(v_x^3+u_x^2{v}_x).\end{array}$$

Note that ${(u^{(n)},v^{(n)})}_{n\in \overline{\mathbb{N}}}$ is bounded in $L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})\times L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})$, we can deduce that ${({\mathsf{\partial }}_x{u}^{(n)},{\mathsf{\partial }}_x{v}^{(n)})}_{n\in \overline{\mathbb{N}}}$ and ${(F(u^{(n)},v^{(n)}),G(u^{(n)},v^{(n)}))}_{n\in \overline{\mathbb{N}}}$ are also bounded in $L_T^{\infty }(B_{p,1}^{\frac{1}{p}})\times L_T^{\infty }(B_{p,1}^{\frac{1}{p}})$. As $(u^{(n)},v^{(n)})\to (u^{\infty },v^{\infty })$ in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$, Lemma 5 guarantees that $(w_1^{(n)},w_2^{(n)})\to (w_1^{\infty },w_2^{\infty })$ in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$.

It can also be derived from Lemmas 2 and 3 that, for any $t\in [0,T]$, $z_1^{\infty }=z_2^{\infty }=0$. In view of (34):

$$\begin{array}{ccc}& & \tilde{F}(u^{(n)},v^{(n)})-\tilde{F}(u^{\infty },v^{\infty })\hfill \\ & =& -2\{(u^{(n)}-u^{\infty }){(u_x^{(n)})}^2+u^{\infty }{(u^{(n)}-u^{\infty })}_x{(u^{(n)}+u^{\infty })}_x+u_x^{(n)}{v}_x^{(n)}(v^{(n)}-v^{\infty })\hfill \\ & & +v^{\infty }{v}_x^{(n)}{(u^{(n)}-u^{\infty })}_x+u_x^{\infty }{v}^{\infty }{(v^{(n)}-v^{\infty })}_x\}\hfill \\ & & -{\mathsf{\partial }}_x^2{(1-{\mathsf{\partial }}_x^2)}^{-1}\{(u^{(n)}-u^{\infty })({(u^{(n)})}^2+u^{(n)}{u}^{\infty }+{(u^{\infty })}^2)+(u^{(n)}-u^{\infty }){(v^{(n)})}^2\hfill \\ & & +u^{\infty }(v^{(n)}+v^{\infty })(v^{(n)}-v^{\infty })+\frac{3}{2}(u^{(n)}-u^{\infty }){(u_x^{(n)})}^2+\frac{3}{2}{u}^{\infty }{(u^{(n)}-u^{\infty })}_x{(u^{(n)}+u^{\infty })}_x\hfill \\ & & +u_x^{(n)}{v}_x^{(n)}(v^{(n)}-v^{\infty })+v^{\infty }{v}_x^{(n)}{(u^{(n)}-u^{\infty })}_x+u_x^{\infty }{v}^{\infty }{(v^{(n)}-v^{\infty })}_x+\frac{1}{2}(u^{(n)}-u^{\infty }){(v_x^{(n)})}^2\hfill \\ & & +\frac{1}{2}{u}^{\infty }{(v^{(n)}+v^{\infty })}_x{(v^{(n)}-v^{\infty })}_x\}\hfill \\ & & -\frac{1}{2}{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\{{(u^{(n)}-u^{\infty })}_x({(u_x^{(n)})}^2+u_x^{(n)}{u}_x^{\infty }+{(u_x^{\infty })}^2)+{(u^{(n)}-u^{\infty })}_x({(v_x^{(n)})}^2\hfill \\ & & +u_x^{\infty }{(v^{(n)}+v^{\infty })}_x{(v^{(n)}-v^{\infty })}_x\}.\hfill \end{array}$$

$$\begin{array}{ccc}& & \tilde{G}(u^{(n)},v^{(n)})-\tilde{G}(u^{\infty },v^{\infty })\hfill \\ & =& -2\{(v^{(n)}-v^{\infty }){(v_x^{(n)})}^2+v^{\infty }{(v^{(n)}-v^{\infty })}_x{(v^{(n)}+v^{\infty })}_x+v_x^{(n)}{u}_x^{(n)}(u^{(n)}-u^{\infty })\hfill \\ & & +u^{\infty }{u}_x^{(n)}{(v^{(n)}-v^{\infty })}_x+v_x^{\infty }{u}^{\infty }{(u^{(n)}-u^{\infty })}_x\}\hfill \\ & & -{\mathsf{\partial }}_x^2{(1-{\mathsf{\partial }}_x^2)}^{-1}\{(v^{(n)}-v^{\infty })({(v^{(n)})}^2+v^{(n)}{v}^{\infty }+{(v^{\infty })}^2)+(v^{(n)}-v^{\infty }){(u^{(n)})}^2\hfill \\ & & +v^{\infty }(u^{(n)}+u^{\infty })(u^{(n)}-u^{\infty })+\frac{3}{2}(v^{(n)}-v^{\infty }){(v_x^{(n)})}^2+\frac{3}{2}{v}^{\infty }{(v^{(n)}-v^{\infty })}_x{(v^{(n)}+v^{\infty })}_x\hfill \\ & & +v_x^{(n)}{u}_x^{(n)}(u^{(n)}-u^{\infty })+u^{\infty }{u}_x^{(n)}{(v^{(n)}-v^{\infty })}_x+v_x^{\infty }{u}^{\infty }{(u^{(n)}-u^{\infty })}_x+\frac{1}{2}(v^{(n)}-v^{\infty }){(u_x^{(n)})}^2\hfill \\ & & +\frac{1}{2}{v}^{\infty }{(u^{(n)}+u^{\infty })}_x{(u^{(n)}-u^{\infty })}_x\}\hfill \\ & & -\frac{1}{2}{\mathsf{\partial }}_x{(1-{\mathsf{\partial }}_x^2)}^{-1}\{{(v^{(n)}-v^{\infty })}_x({(v_x^{(n)})}^2+v_x^{(n)}{v}_x^{\infty }+{(v_x^{\infty })}^2)+{(v^{(n)}-v^{\infty })}_x({(u_x^{(n)})}^2\hfill \\ & & +v_x^{\infty }{(u^{(n)}+u^{\infty })}_x{(u^{(n)}-u^{\infty })}_x\}.\hfill \end{array}$$

Notice that ${(u^{(n)},v^{(n)})}_{n\in \overline{\mathbb{N}}}$ is bounded in $L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})\times L_T^{\infty }(B_{p,1}^{1+\frac{1}{p}})$, by applying the algebraic property of Besov space $B_{p,1}^{\frac{1}{p}}\times B_{p,1}^{\frac{1}{p}}$, we have

$$\begin{array}{ccc}& & \parallel \tilde{F}(u^{(n)},v^{(n)})-\tilde{F}(u^{\infty },v^{\infty }){\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & \le & C\{\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\parallel u^{(n)}+u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\hfill \\ & & +\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel v^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}{\parallel u^{(n)}-u^{\infty }\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\hfill \\ & & +\parallel v^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\parallel u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}{\parallel v^{(n)}-v^{\infty }\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\hfill \\ & & +(\parallel u^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}^2+\parallel u^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}\parallel u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}^2)\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & & +\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\parallel v^{(n)}+v^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\}\hfill \\ & \le & C(\parallel u^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{(n)}{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2+\parallel v^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}^2)(\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}})\hfill \\ & \le & C(\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel w_1^{(n)}-w_1^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel z_1^{(n)}-z_1^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel w_2^{(n)}-w_2^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & & +\parallel z_2^{(n)}-z_2^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}).\hfill \end{array}$$

Similarly we obtain

$$\begin{array}{ccc}& & \parallel \tilde{G}(u^{(n)},v^{(n)})-\tilde{G}(u^{\infty },v^{\infty }){\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & \le & C(\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel w_1^{(n)}-w_1^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel z_1^{(n)}-z_1^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel v^{(n)}-v^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel w_2^{(n)}-w_2^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & & +\parallel z_2^{(n)}-z_2^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}).\hfill \end{array}$$

It follows that for all $n\in \mathbb{N}$,

$$\begin{array}{ccc}& & \parallel z_1^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}+{\parallel z_2^{(n)}\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & \le & e^{C{\int }_0^t{\parallel [{(u^{(n)})}^2+{(u^{(n)})}^2](t^{\prime })\parallel }_{B_{p,1}^{\frac{1}{p}}}\mathrm{d}{t}^{\prime }}\hfill \\ & & \times (\parallel {\mathsf{\partial }}_x{u}_0^{(n)}-{\mathsf{\partial }}_x{u}_0^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel {\mathsf{\partial }}_x{v}_0^{(n)}-{\mathsf{\partial }}_x{v}_0^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+{\int }_0^t\parallel \tilde{F}(u^{(n)},v^{(n)})-\tilde{F}(u^{\infty },v^{\infty }){\parallel }_{B_{p,1}^{\frac{1}{p}}}\hfill \\ & & +\parallel \tilde{G}(u^{(n)},v^{(n)})-\tilde{G}(u^{\infty },v^{\infty }){\parallel }_{B_{p,1}^{\frac{1}{p}}}\mathrm{d}\tau )\hfill \\ & \le & C\{\parallel {\mathsf{\partial }}_x{u}_0^{(n)}-{\mathsf{\partial }}_x{u}_0^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel {\mathsf{\partial }}_x{v}_0^{(n)}-{\mathsf{\partial }}_x{v}_0^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+{\int }_0^t(\parallel u^{(n)}-u^{\infty }{\parallel }_{B_{p,1}^{1+\frac{1}{p}}}+{\parallel v^{(n)}-v^{\infty }\parallel }_{B_{p,1}^{1+\frac{1}{p}}}\hfill \\ & & +\parallel w_1^{(n)}-w_1^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel w_2^{(n)}-w_2^{\infty }{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel z_1^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}}+\parallel z_2^{(n)}{\parallel }_{B_{p,1}^{\frac{1}{p}}})\mathrm{d}\tau \}.\hfill \end{array}$$

In view of the fact that $({\mathsf{\partial }}_x{u}_0^{(n)},{\mathsf{\partial }}_x{v}_0^{(n)})\to ({\mathsf{\partial }}_x{u}_0^{\infty },{\mathsf{\partial }}_x{v}_0^{\infty })$ in $B_{p,1}^{\frac{1}{p}}\times B_{p,1}^{\frac{1}{p}}$, $(u^{(n)},v^{(n)})\to (u^{\infty },v^{\infty })$ in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$, $(w_1^{(n)},w_2^{(n)})\to (w_1^{\infty },w_2^{\infty })$ in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$, by applying the Gronwall Lemma, one can deduce that $(z_1^{(n)},z_2^{(n)})\to (0,0)$ in $C([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}})$. Therefore,

$$\begin{array}{ccc}& & \parallel {\mathsf{\partial }}_x{u}^{(n)}-{\mathsf{\partial }}_x{u}^{\infty }{\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+{\parallel {\mathsf{\partial }}_x{v}^{(n)}-{\mathsf{\partial }}_x{v}^{\infty }\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}\hfill \\ & \le & \parallel w_1^{(n)}-w_1^{\infty }{\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+\parallel z_1^{(n)}-z_1^{\infty }{\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+\parallel w_2^{(n)}-w_2^{\infty }{\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})}+{\parallel z_2^{(n)}-z_2^{\infty }\parallel }_{L_T^{\infty }(B_{p,1}^{\frac{1}{p}})},\hfill \end{array}$$

tends to 0 as $n\to \infty$, which implies

$$\begin{array}{c}\hfill ({\mathsf{\partial }}_x{u}^{(n)},{\mathsf{\partial }}_x{v}^{(n)})\to ({\mathsf{\partial }}_x{u}^{\infty },{\mathsf{\partial }}_x{v}^{\infty })inC([0,T];B_{p,1}^{\frac{1}{p}})\times C([0,T];B_{p,1}^{\frac{1}{p}}).\end{array}$$

(36)

Combining (33) with (36), we find

$$\begin{array}{c}\hfill (u^{(n)},v^{(n)})\to (u^{\infty },v^{\infty })inC([0,T];B_{p,1}^{1+\frac{1}{p}})\times C([0,T];B_{p,1}^{1+\frac{1}{p}}).\end{array}$$

Theorem 1 is a direct consequence of step 1–step 4.