Blow-Up for a Periodic Two-Component Camassa-Holm Equation with Generalized Weakly Dissipation

Yang Li^*, Jingyi Liu, Xincheng Zhu

Nonlinear Scientific Research Center, Jiangsu University, Zhenjiang, China.

DOI: 10.4236/jamp.2020.810167                

Abstract

In this paper we study a periodic two-component Camassa-Holm equation with generalized weakly dissipation. The local well-posedness of Cauchy problem is investigated by utilizing Kato’s theorem. The blow-up criteria and the blow-up rate are established by applying monotonicity. Finally, the global existence results for solutions to the Cauchy problem of equation are proved by structuring functions.

Keywords

Periodic Two-Component Camassa-Holm Equation, Local Well-Posedness, Blow-Up, Global Existence, Monotonicity

1. Introduction

In this paper, we consider the Cauchy problem of periodic two-component Camassa-Holm equation with a generalized weakly dissipation:

$\{\begin{array}{l}{u}_t-u_{xxt}+k{u}_x+3u{u}_x-2u_x{u}_{xx}-u{u}_{xxx}+\lambda \left(u-u_{xx}\right)+\sigma \rho {\rho }_x=0,t>0,x\in R,\\ {\rho }_t+{\left(\rho u\right)}_x=0,t>0,x\in R,\\ u\left(0,x\right)=u_0\left(x\right);\rho \left(0,x\right)={\rho }_0\left(x\right),x\in R,\\ u\left(t,x\right)=u\left(t,x+1\right);\rho \left(t,x\right)=\rho \left(t,x+1\right),t\ge 0,x\in R,\end{array}$ (1.1)

where $\lambda \ge 0$ and k is a fixed constant; $\sigma$ is a free parameter.

It is well known that the two-component integrable Camassa-Holm equation is

$\{\begin{array}{l}{u}_t-u_{xxt}+k{u}_x+3u{u}_x-2u_x{u}_{xx}-u{u}_{xxx}-\rho {\rho }_x=0,t>0,x\in R\\ {\rho }_t+{\left(\rho u\right)}_x=0,t>0,x\in R\end{array}$ (1.2)

which is a model for wave motion on shallow water, where $u\left(t,x\right)$ standing for the fluid velocity at time $t\ge 0$ in the spatial x direction [1], $\rho \left(t,x\right)$ is in connection with the horizontal deviation of the surface from equilibrium (i.e. amplitude). Equation (1.2) possesses a bi-Hamiltonian structure [2] and the solution interaction of peaked travelling waves and wave breaking [1] [2] [3]. It is completely integrable [3] and becomes the Camassa-Holm equation when $\rho =0$.

Equation (1.2) was derived physically by Constantin and Ivanov [4] in the context of shallow water theory. As soon as this equation was put forward, it attracted attention of a large number of researchers. Escher et al. [5] established the local well-posedness and present the blow-up scenarios and several blow-up results of strong solutions to Equation (1.2). Constantin and Ivanov [6] investigated the global existence and blow-up phenomena of strong solutions of Equation (1.2). Guan and Yin [7] obtained a new global existence result for strong solutions to Equation (1.2) and several blow-up results, which improved the results in [6]. Gui and Liu [8] established the local well-posedness for Equation (1.2) in a range of the Besov spaces, they also characterized a wave breaking mechanism for strong solutions. Hu and Yin [9] [10] studied the blow-up phenomena and the global existence of Equation (1.2).

Dissipation is an inevitable phenomenon in real physical word. It is necessary to study periodic two-Camassa-Holm equation with a generalized weakly dissipation. Hu and Yin [11] study the blow-up of solutions to a weakly dissipative periodic rod equation. Hu considered global existence and blow-up phenomena for a weakly dissipative two-component Camassa-Holm system [12] [13]. The purpose of this paper is to study the blow-up phenomenon of the solutions of Equation (1.1). The results show that the behavior of solutions to the periodic two-component Camassa-Holm equation with a generalized weakly dissipation is similar to Equation (1.2) and the blow-up rate of Equation (1.1) is not affected by the dissipative term when $\sigma >0$.

The paper is organized as follows. Section 2 gives the local well-posedness of the Cauchy problem associated with Equation (1.1). The blow-up criteria for solutions and two conditions for wave breaking in finite time are given in Section 3. Furthermore, we also learn the blow-up rate of solutions. In Section 4, we address the global existence of Equation (1.1).

2. Local Well-Posedness

Let us introduce some notations, the $S=R/Z$ is the circle of unit length, the $\left[x\right]$ stands for the integer part of $x\in R$, the $\ast$ stands for the convolution, the ${\Vert \cdot \Vert }_X$ is used to represent the norm of Banach space X.

In this section, we investigate the local well-posedness for the Cauchy problem of Equation (1.1) by applying Kato’s theory [14] in $H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$.

For convenience we recall the Kato’s theorem in the suitable form for our purpose. Consider the following abstract quasilinear evolution equation:

$\{\begin{array}{l}\frac{\text{d}z}{\text{d}t}+A\left(z\right)z=f\left(z\right),t\ge 0\\ z\left(0\right)=z_0\end{array}$ (2.1)

There are two Hilbert’s spaces X and Y, Y is continuously and densely embedded in X and $Q:Y\to X$ is a topological isomorphism, the $L\left(Y,X\right)$ stands for the space of all bounded linear operator from Y to X.

Theorem 2.1 [14] 1) $A\left(y\right)\in L\left(Y,X\right)$, for $\forall y\in X$ with

${\Vert \left(A\left(y\right)-A\left(z\right)\right)w\Vert }_X\le {\mu }_1{\Vert y-z\Vert }_X{\Vert w\Vert }_Y$ (2.2)

where $z,y,w\in Y$, $A\left(y\right)\in G\left(X,1,\beta \right)$, i.e. $A\left(y\right)$ is quasi-m-accretive, uniformly on bounded sets in Y.

2) $QA\left(y\right)Q^{-1}=A\left(y\right)+B\left(y\right)$, where $B\left(y\right)\in L\left(X\right)$ is uniformly bounded on a bounded sets in Y

${\Vert \left(B\left(y\right)-B\left(z\right)\right)w\Vert }_X\le {\mu }_2{\Vert y-z\Vert }_Y{\Vert w\Vert }_X$ (2.3)

where $z,y\in Y$, $w\in X$.

3) $f:Y\to Y$ is a bounded map on bounded sets in Y

${\Vert f\left(y\right)-f\left(z\right)\Vert }_Y\le {\mu }_3{\Vert y-z\Vert }_Y$ (2.4)

${\Vert f\left(y\right)-f\left(z\right)\Vert }_X\le {\mu }_4{\Vert y-z\Vert }_X$ (2.5)

where $z,y\in Y$, ${\mu }_1,{\mu }_2,{\mu }_3,{\mu }_4$ are constants which only depending $\left\{{\Vert y\Vert }_Y,{\Vert z\Vert }_Y\right\}$.

If the 1), 2), 3) hold, given $u_0\in Y$, there is a maximal $T>0$ depending only on ${\Vert u_0\Vert }_Y$ and a unique solution u of Equation (2.1) such that

$u=u\left(\cdot ,u_0\right)\in C\left(\left[0,T\right);Y\right)\cap C^1\left(\left[0,T\right);X\right)$ (2.6)

Moreover, the map $u\to u\left(\cdot ,u_0\right)$ is continuous from Y to $C\left(\left[0,T\right);Y\right)\cap C^1\left(\left[0,T\right);X\right)$.

Note that $g\left(x\right):=\frac{\cosh \left(x-\left[x\right]-\frac{1}{2}\right)}{2\sinh \frac{1}{2}}$, $x\in R$, ${\left(1-{\partial }_x^2\right)}^{-1}f=g\ast f$ for all $f\in L^2\left(S\right)$ and $g\ast \left(u-u_{xx}\right)=u$. Then Equation (1.1) can be rewritten as

$\{\begin{array}{l}{u}_t+u{u}_x=-{\partial }_{x}g\ast \left(u^2+\frac{1}{2}{u}_x^2+ku+\frac{\sigma }{2}{\rho }^2\right)-\lambda u\\ {\rho }_t+{\left(\rho u\right)}_x=0\\ u\left(0,x\right)=u_0\left(x\right);\rho \left(0,x\right)={\rho }_0\left(x\right)\\ u\left(t,x\right)=u\left(t,x+1\right);\rho \left(t,x\right)=\rho \left(t,x+1\right)\end{array}$ (2.7)

Theorem 2.2 Let $z_0=\left(u_0,{\rho }_0-1\right)\in H^s\times H^{s-1}$ with $s\ge 2$, there exists a maximal time $T>0$ which is independent on s and exists a unique solution $\left(u,\rho \right)$ of Equation (1.1) in the interval $\left[0,T\right)$ with initial data $z_0$, such that the solution depends continuously on the initial data.

The remainder of this section is devoted the proof of Theorem 2.2. Let $z=\left(\begin{array}{l}u\\ \rho \end{array}\right)$, $T=H^s\times H^s$, $X=H^{s-1}\times H^{s-1}$, $\wedge ={\left(1-{\partial }_x^2\right)}^{\frac{1}{2}}$, $Q=\left(\begin{array}{cc}\wedge & 0\\ 0& \wedge \end{array}\right)$, and

$A\left(z\right)=\left(\begin{array}{cc}u{\partial }_x& 0\\ 0& u{\partial }_x\end{array}\right)$ (2.8)

The [15] shows that Q is an isomorphism from $H^s\times H^s$ onto $H^{s-1}\times H^{s-1}$. It is sufficiently to verify $A\left(z\right)$, $B\left(z\right)$, $f\left(z\right)$ satisfy 1), 2), 3) to prove the theorem 2.2. For this purpose, the following lemmas are necessary.

Lemma 2.1 [15] The operator $A\left(z\right)$ is defined in (2.8) with $z\in H^s\times H^s$, $s>\frac{3}{2}$ belongs to $G\left(L^2\times L^2,1,\beta \right)$.

Lemma 2.2 [15] The operator $A\left(z\right)$ is defined in (2.8) with $z\in H^s\times H^s$, $s>\frac{3}{2}$ belongs to $G\left(H^{s-1}\times H^{s-1},1,\beta \right)$.

Lemma 2.3 [15] The operator $A\left(z\right)$ is defined in (2.8) with $z\in H^s\times H^s$, $s>\frac{3}{2}$ belongs to $L\left(H^s\times H^s,H^{s-1}\times H^{s-1}\right)$, moreover,

${\Vert \left(A\left(y\right)-A\left(z\right)\right)w\Vert }_{H^{s-1}\times H^{s-1}}\le {\mu }_1{\Vert y-z\Vert }_{H^s\times H^s}{\Vert w\Vert }_{H^s\times H^s}$ (2.9)

where $y,z,w\in H^s\times H^s$.

Lemma 2.4 [15] Let $B\left(z\right)=QA\left(z\right)Q^{-1}-A\left(z\right)$ with $z\in H^s\times H^s$, $s>\frac{3}{2}$, then the operator $B\left(z\right)\in L\left(H^{s-1}\times H^{s-1}\right)$ and

${\Vert \left(B\left(y\right)-B\left(z\right)\right)w\Vert }_{H^{s-1}\times H^{s-1}}\le {\mu }_2{\Vert y-z\Vert }_{H^s\times H^s}{\Vert w\Vert }_{H^{s-1}\times H^{s-1}}$ (2.10)

for $y,z\in H^s\times H^s$, and $w\in H^{s-1}\times H^{s-1}$.

Lemma 2.5 Let $z\in H^s\times H^s$, $s>\frac{3}{2}$, and

$f\left(z\right)=-\left(\begin{array}{c}{\partial }_x{\left(1-{\partial }_x^2\right)}^{-1}\left(u^2+\frac{1}{2}{u}_x^2+ku+\frac{\sigma }{2}{\rho }^2\right)+\lambda u\\ \rho u_x\end{array}\right)$

Then f is bounded on bounded sets in $H^s\times H^s$ and satisfies

1) ${\Vert f\left(y\right)-f\left(z\right)\Vert }_{H^s\times H^s}\le {\mu }_3{\Vert y-z\Vert }_{H^s\times H^s}$, $y,z\in H^s\times H^s$ (2.11)

2) ${\Vert f\left(y\right)-f\left(z\right)\Vert }_{H^{s-1}\times H^{s-1}}\le {\mu }_4{\Vert y-z\Vert }_{H^{s-1}\times H^{s-1}}$, $y,z\in H^s\times H^s$ (2.12)

Proof: For any $z,y\in H^s\times H^s$, $s>\frac{3}{2}$,

$\begin{array}{l}{\Vert f\left(y\right)-f\left(z\right)\Vert }_{H^s\times H^s}\\ \le {\Vert -{\partial }_x{\left(1-{\partial }_x^2\right)}^{-1}\left[\left(y_1^2-u^2\right)+\frac{1}{2}\left(y_{1x}^2-u_x^2\right)+k\left(y_1-u\right)+\frac{\sigma }{2}\left(y_2^2-{\rho }^2\right)\right]\Vert }_{H^s}\\ +{\Vert \lambda \left(y_1-u\right)\Vert }_{H^s}+{\Vert u_x\rho -y_{1x}{y}_2\Vert }_{H^s}\end{array}$

$\begin{array}{l}\le {\Vert \left(y_1^2-u^2\right)+\frac{1}{2}\left(y_{1x}^2-u_x^2\right)+k\left(y_1-u\right)\Vert }_{H^{s-1}}+\frac{\left|\sigma \right|}{2}{\Vert y_2^2-{\rho }^2\Vert }_{H^{s-1}}\\ +\left|\lambda \right|{\Vert y_1-u\Vert }_{H^s}+{\Vert \left(u_x-y_{1x}\right)\rho \Vert }_{H^s}+{\Vert y_{1x}\left(\rho -y_2\right)\Vert }_{H^s}\end{array}$

$\begin{array}{l}\le {\Vert \left(y_1-u\right)\left(y_1+u\right)\Vert }_{H^{s-1}}+\frac{1}{2}{\Vert \left(y_{1x}-u_x\right)\left(y_{1x}+u_x\right)\Vert }_{H^{s-1}}+\left|k\right|{\Vert y_1-u\Vert }_{H^{s-1}}\\ +\left|\lambda \right|{\Vert y_1-u\Vert }_{H^s}+\frac{\left|\sigma \right|}{2}{\Vert y_2-\rho \Vert }_{H^{s-1}}{\Vert y_2+\rho \Vert }_{H^{s-1}}\\ +{\Vert y_1-u\Vert }_{H^s}{\Vert \rho \Vert }_{H^s}+{\Vert y_1\Vert }_{H^s}{\Vert \rho -y_2\Vert }_{H^s}\end{array}$

$\begin{array}{l}\le {\Vert \left(y_1-u\right)\left(y_1+u\right)\Vert }_{H^{s-1}}+\frac{1}{2}{\Vert \left(y_{1x}-u_x\right)\left(y_{1x}+u_x\right)\Vert }_{H^{s-1}}+\left|k\right|{\Vert y_1-u\Vert }_{H^{s-1}}\\ +\left|\lambda \right|{\Vert y_1-u\Vert }_{H^s}+\frac{\left|\sigma \right|}{2}{\Vert y_2-\rho \Vert }_{H^{s-1}}{\Vert y_2+\rho \Vert }_{H^{s-1}}\\ +{\Vert y_1-u\Vert }_{H^s}{\Vert \rho \Vert }_{H^s}+{\Vert y_1\Vert }_{H^s}{\Vert \rho -y_2\Vert }_{H^s}\end{array}$

Let ${\mu }_3=\frac{5+\left|\sigma \right|}{2}{\Vert y\Vert }_{H^s\times H^s}+\frac{3+\left|\sigma \right|}{2}{\Vert z\Vert }_{H^s\times H^s}+\left|k\right|+\left|\lambda \right|$, then

${\Vert f\left(y\right)-f\left(z\right)\Vert }_{H^s\times H^s}\le {\mu }_3{\Vert y-z\Vert }_{H^s\times H^s}$, $y,z\in H^s\times H^s$

Making $y=0$ in the above inequality, it shows that f is bounded on bounded sets in $H^s\times H^s$, the proof of 1) is complete.

Similarly, the inequality (2.12) also can be proved.

Proof of Theorem 2.2: The 1) is true for $A\left(z\right)$ from the inequality (2.9), the 2) is true for $B\left(z\right)$ from the inequality (2.10), the 3) is true for $f\left(z\right)$ from the inequalities (2.11) (2.12). According to the Theorem 2.1, the proof of the Theorem 2.2 is complete.

3. Blow-Up

This section will establish a blow-up criterion for solution of Equation (1.1) when $\sigma >0$.

Theorem 3.1 [8] [16] Let $\sigma \ne 0$ and $\left(u,\rho \right)$ be the solution of (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\times H^{s-1}$, $s>\frac{3}{2}$, T is the maximal time of existence of the solution, then

$T<\infty \Rightarrow {\displaystyle {\int }_0^T{\Vert u_x\left(\tau \right)\Vert }_{L^{\infty }}\text{d}\tau }=\infty$ (3.1)

Consider the following equation of trajectory:

$\{\begin{array}{ll}\frac{\text{d}q\left(t,x\right)}{\text{d}t}=u\left(t,q\left(t,x\right)\right),\hfill & t\in \left[0,T\right)\hfill \\ q\left(0,x\right)=x,\hfill & x\in S\hfill \end{array}$ (3.2)

The (3.2) shows $q\left(t,\cdot \right):S\to S$ is the differential homeomorphism for every $t\in \left[0,T\right)$

$q_x\left(t,x\right)={\text{e}}^{{\displaystyle {\int }_0^t{u}_x\left(\tau ,q\left(\tau ,x\right)\right)\text{d}\tau }}>0,\forall \left(t,x\right)\in \left[0,T\right)\times S$ (3.3)

Hence

${\Vert v\left(t,\cdot \right)\Vert }_{L^{\infty }}={\Vert v\left(t,q\left(t,\cdot \right)\right)\Vert }_{L^{\infty }}$ (3.4)

Lemma 3.1 [17] Let $T>0$ and $v\in C^1\left(\left[0,T\right);H^1\left(R\right)\right)$, then for every $t\in \left[0,T\right)$, there exists at least one point $\xi \left(t\right)\in R$ with

$m\left(t\right):=\underset{x\in R}{\inf }\left[v_x\left(t,x\right)\right]=v_x\left(t,\xi (\; t\; )\right)$

The function $m\left(t\right)$ is absolutely continuous in $\left(0,T\right)$ with

$\frac{\text{d}m\left(t\right)}{\text{d}t}=v_{tx}\left(t,\xi \left(t\right)\right)$ a.e. in $\left(0,T\right)$.

Lemma 3.2 Let $z_0=\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$ with $s\ge 2$, there exist a maximal time $T>0$ and a unique solution $\left(u,\rho \right)$ of Equation (1.1) with initial data $z_0$, then we have

${\Vert u\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\le {\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2$ (3.5)

Proof: Multiply the first equation of Equation (1.1) by u and integrate

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S\left(u^2+u_x^2\right)\text{d}x}+2\lambda {\displaystyle {\int }_S\left(u^2+u_x^2\right)\text{d}x}+2\sigma {\displaystyle {\int }_S\rho {\rho }_{x}u\text{d}x}=0$ (3.6)

The second equation of Equation (1.1) can be rewritten as

${\left(\rho -1\right)}_t+{\rho }_{x}u+\rho u_x=0$

Multiply the above equation by $\left(\rho -1\right)$ and integrate

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S{\left(\rho -1\right)}^2\text{d}x}+2{\displaystyle {\int }_{S}u\rho {\rho }_x\text{d}x}-2{\displaystyle {\int }_{S}u{\rho }_x\text{d}x}+2{\displaystyle {\int }_S{u}_x{\rho }^2\text{d}x}-2{\displaystyle {\int }_S{u}_x\rho \text{d}x}=0$ (3.7)

According to (3.6) and (3.7)

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S\left(u^2+u_x^2+\sigma {\left(\rho -1\right)}^2+2\lambda {\displaystyle {\int }_0^t\left(u^2+u_x^2\right)\text{d}\tau }\right)\text{d}x}=0$

Then

$\begin{array}{l}{\displaystyle {\int }_S\left(u^2+u_x^2+\sigma {\left(\rho -1\right)}^2+2\lambda {\displaystyle {\int }_0^t\left(u^2+u_x^2\right)\text{d}\tau }\right)\text{d}x}\\ ={\displaystyle {\int }_S\left(u_0^2+u_{0x}^2+\sigma {\left({\rho }_0-1\right)}^2\right)\text{d}x}={\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\end{array}$

Notice that $2\lambda {\displaystyle {\int }_0^t\left(u^2+u_x^2\right)\text{d}x}\ge 0$, then

$\begin{array}{c}{\Vert u\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2={\displaystyle {\int }_S\left(u^2+u_x^2+\sigma {\left(\rho -1\right)}^2\right)\text{d}x}\\ \le {\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\end{array}$

Lemma 3.3 [18] [19] 1) For every $f\in H^1\left(S\right)$, we have

$\underset{x\in \left[0,1\right]}{\max }{f}^2\left(x\right)\le \frac{e+1}{2\left(e-1\right)}{\Vert f\Vert }_{H^1}^2$ (3.8)

where the constant $\frac{e+1}{2\left(e-1\right)}$ is the best constant.

2) For every $f\in H^3\left(S\right)$, we have

$\underset{x\in \left[0,1\right]}{\max }{f}^2\left(x\right)\le c{\Vert f\Vert }_{H^1}^2$ (3.9)

where the best constant c is $\frac{e+1}{2\left(e-1\right)}$.

3) For every $f\in H^3\left(S\right)$, we have

$\underset{x\in \left[0,1\right]}{\max }{f}_x^2\left(x\right)\le \frac{1}{12}{\Vert f\Vert }_{H^2}^2$ (3.10)

Lemma 3.4 Suppose $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T be the maximal time of existence, then

$\underset{x\in S}{\sup }{u}_x\left(t,x\right)\le {\Vert u_{0x}\Vert }_{L^{\infty }}+\sqrt{{\lambda }^2+\sigma {\Vert {\rho }_0\Vert }_{L^{\infty }}^2+C_1^2}$

where $C_1=\sqrt{\frac{\left(3\sigma +2\right)\left(e+1\right)}{2\left(e-1\right)}+\left(\frac{e+1}{e-1}+\frac{k^2+1}{2}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\right)}$.

Proof: The theorem 2.2 and a density argument imply that it is sufficient to prove the desired estimates for $s=3$.

Differentiate the first equation of Equation (2.7) with respect to x

$u_{tx}=u^2-\frac{1}{2}{u}_x^2-\lambda u_x+\frac{\sigma }{2}{\rho }^2-k{\partial }_x^{2}g\ast u-g\ast \left(u^2+\frac{1}{2}{u}_x^2+\frac{\sigma }{2}{\rho }^2\right)-u{u}_{xx}$ (3.11)

Define

$\bar{m}\left(t\right)=u_x\left(t,\eta \left(t\right)\right)=\underset{x\in S}{\sup }\left(u_x\left(t,x\right)\right),m\left(t\right)=\underset{x\in S}{\inf }\left(u_x\left(t,x\right)\right)$ (3.12)

From the Fermat’s lemma, we know

$u_{xx}\left(t,\eta \left(t\right)\right))=0,a.e.t\in \left[0,T\right)$

there exists $x_1\left(t\right)\in S$ such that

$q\left(t,x_1\left(t\right)\right)=\eta \left(t\right),t\in \left[0,T\right)$ (3.13)

Set

$\bar{\zeta }\left(t\right)=\rho \left(t,q\left(t,x_1\right)\right)$, $t\in \left[0,T\right)$ (3.14)

From (3.11) and the second equation of Equation (1.1), we obtain

$\{\begin{array}{l}{\bar{m}}^{\prime }\left(t\right)=-\frac{1}{2}{\bar{m}}^2\left(t\right)-\lambda \bar{m}\left(t\right)+\frac{\sigma }{2}{\bar{\zeta }}^2\left(t\right)+f\left(t,q\left(t,x_1\right)\right)\\ {\bar{\zeta }}^{\prime }\left(t\right)=-\bar{\zeta }\left(t\right)\bar{m}\left(t\right)\end{array}$ (3.15)

where $f=u^2-k{\partial }_x^{2}g\ast u-g\ast \left(u^2+\frac{1}{2}{u}_x^2+\frac{\sigma }{2}{\rho }^2\right)$.

Notice that ${\partial }_x^{2}g\ast u={\partial }_{x}g\ast {\partial }_{x}u$, then

$\begin{array}{l}f=u^2-k{\partial }_x^{2}g\ast u-g\ast \left(u^2+\frac{1}{2}{u}_x^2\right)-\frac{\sigma }{2}g\ast \left({\rho }^2\right)\\ =u^2-k{\partial }_x^{2}g\ast u-g\ast \left(u^2+\frac{1}{2}{u}_x^2\right)-\frac{\sigma }{2}g\ast 1-\sigma g\ast \left(\rho -1\right)-\frac{\sigma }{2}g\ast {\left(\rho -1\right)}^2\\ \le u^2+k\left|{\partial }_{x}g\ast {\partial }_{x}u\right|+\frac{\sigma }{2}\left|g\ast 1\right|+\sigma \left|g\ast \left(\rho -1\right)\right|\end{array}$

From (3.8) (3.9) and (3.10), we have

$u^2\le \frac{e+1}{2\left(e-1\right)}{\Vert u\Vert }_{H^1}^2$

$k\left|{\partial }_{x}g\ast {\partial }_{x}u\right|\le k{\Vert g_x\Vert }_{L^2}{\Vert u_x\Vert }_{L^2}\le \frac{e+1}{2\left(e-1\right)}+\frac{1}{4}{k}^2{\Vert u_x\Vert }_{L^2}^2$

$\left|g\ast \left(u^2+\frac{1}{2}{u}_x^2\right)\right|\le \frac{e+1}{2\left(e-1\right)}{\Vert u\Vert }_{L^2}^2+\frac{e+1}{4\left(e-1\right)}{\Vert u_x\Vert }_{L^2}^2$

$\frac{\sigma }{2}\left|g\ast 1\right|\le \frac{\sigma }{2}{\Vert g\Vert }_{L\infty }\le \frac{\sigma \left(e+1\right)}{4\left(e-1\right)}$

$\sigma \left|g\ast \left(\rho -1\right)\right|\le \sigma {\Vert g\Vert }_{L^2}{\Vert \rho -1\Vert }_{L^1}\le \frac{\sigma \left(e+1\right)}{2\left(e-1\right)}+\frac{\sigma }{4}{\Vert \rho -1\Vert }_{L^2}^2$

$\frac{\sigma }{2}\left|g\ast {\left(\rho -1\right)}^2\right|\le \frac{\sigma }{2}{\Vert g\Vert }_{L^{\infty }}{\Vert \left(\rho -1\right)\Vert }_{L^1}\le \frac{\sigma \left(e+1\right)}{4\left(e-1\right)}{\Vert \rho -1\Vert }_{L^2}^2$

Therefore we get the upper bound of f

$\begin{array}{l}f\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{k^2}{4}\right){\Vert u\Vert }_{H^1}^2+\frac{1}{4}\sigma {\Vert \rho -1\Vert }_{L^2}^2\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{k^2+1}{4}\right)\left({\Vert u\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{k^2+1}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\right)\\ =\frac{1}{2}{C}_1^2\end{array}$ (3.16)

Similarly, we turn to the lower bound of f

$\begin{array}{c}-f\le u^2+k\left|{\partial }_{x}g\ast {\partial }_{x}u\right|+\left|g\ast \left(u^2+\frac{1}{2}{u}_x^2\right)\right|+\frac{\sigma }{2}\left|g\ast 1\right|\\ +\sigma \left|g\ast \left(\rho -1\right)\right|+\frac{\sigma }{2}\left|g\ast {\left(\rho -1\right)}^2\right|\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\frac{e+1}{e-1}{\Vert u\Vert }_{L^2}^2+\left(\frac{3\left(e+1\right)}{4\left(e-1\right)}+\frac{k^2}{4}\right){\Vert u_x\Vert }_{L^2}^2\\ +\left(\frac{e+1}{4\left(e-1\right)}+\frac{1}{4}\right)\sigma {\Vert \rho -1\Vert }_{L^2}^2\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{k^2+1}{4}\right)\left({\Vert u\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{k^2+1}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\right)\end{array}$ (3.17)

According to (3.16) and (3.17)

$\left|f\right|\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{k^2+1}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\right)$ (3.18)

From Sobolev’s embedding theorem, we have $u\in C_0^1\left(S\right)$, due to the periodic of Equation (1.1), then

$\underset{x\in S}{\inf }{u}_x\left(t,x\right)\le 0,\underset{x\in S}{\sup }{u}_x\left(t,x\right)\ge 0,t\in \left[0,T\right)$ (3.19)

hence

$\bar{m}\left(t\right)\ge 0,t\in \left[0,T\right)$ (3.20)

From the second Equation of (3.15), we have

$\bar{\zeta }\left(t\right)=\bar{\zeta }\left(0\right){\text{e}}^{-{\displaystyle {\int }_0^t\bar{m}\left(\tau \right)\text{d}\tau }}$

then

$\left|\rho \left(t,q\left(t,x_1\right)\right)\right|=\left|\bar{\zeta }\left(t\right)\right|\le \left|\bar{\zeta }\left(0\right)\right|\le {\Vert {\rho }_0\Vert }_{L^{\infty }}$

For any given $x\in S$, define

$P_1\left(t\right)=\bar{m}\left(t\right)-{\Vert u_{0x}\Vert }_{L^{\infty }}-\sqrt{{\lambda }^2+\sigma {\Vert {\rho }_0\Vert }_{L^{\infty }}^2+C_1^2}$ (3.21)

then $P_1\left(t\right)$ is $C^1$ -function in $\left[0,T\right)$ and satisfies

$\begin{array}{c}{P}_1\left(0\right)=\bar{m}\left(0\right)-{\Vert u_{0x}\Vert }_{L^{\infty }}-\sqrt{{\lambda }^2+\sigma {\Vert {\rho }_0\Vert }_{L^{\infty }}^2+C_1^2}\\ \le \bar{m}\left(0\right)-{\Vert u_{0x}\Vert }_{L^{\infty }}\le 0\end{array}$

Next, we will show $P_1\left(t\right)\le 0,t\in \left[0,T\right)$.

By contradictory arguement, there exists $t_0\in \left[0,T\right)$ such that $P_1\left(t_0\right)>0$. Making $t_1=\max \left\{t<t_0:P_1\left(t\right)=0\right\}$, we have

$P_1\left(t_1\right)=0,{P^{\prime }}_1\left(t_1\right)\ge 0$

then

$\bar{m}\left(t_1\right)={\Vert u_{0x}\Vert }_{L^{\infty }}+\sqrt{{\lambda }^2+\sigma {\Vert {\rho }_0\Vert }_{L^{\infty }}^2+C_1^2}$.

From (3.21), we know

${\bar{m}}^{\prime }\left(t_1\right)={P^{\prime }}_1\left(t_1\right)\ge 0$ (3.22)

On the other hand, from the first Equation of (3.15), we have

$\begin{array}{l}{\bar{m}}^{\prime }\left(t_1\right)=-\frac{1}{2}{\bar{m}}^2\left(t_1\right)-\lambda \bar{m}\left(t_1\right)+\frac{\sigma }{2}{\bar{\zeta }}^2\left(t_1\right)+f\left(t_1,q\left(t_1,x_1\right)\right)\\ \le -\frac{1}{2}{\left(m\left(t_1\right)+\lambda \right)}^2+\frac{1}{2}{\lambda }^2+\frac{\sigma }{2}{\Vert {\rho }_0\Vert }_{L^{\infty }}^2+\frac{1}{2}{C}_1^2\\ \le -\frac{1}{2}{\left({\Vert u_{0x}\Vert }_{L^{\infty }}+\sqrt{{\lambda }^2+\sigma {\Vert {\rho }_0\Vert }_{L^{\infty }}^2+C_1^2}+\lambda \right)}^2+\frac{\sigma }{2}{\Vert {\rho }_0\Vert }_{L^{\infty }}^2+\frac{1}{2}{C}_1^2\\ <0\end{array}$

It yields a contradiction, then the proof of the Lemma 3.4 is complete.

Lemma 3.5 Suppose $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of the solution. If there exists $M\ge 0$ such that

$\underset{\left(t,x\right)\in \left[0,T\right)\times S}{\inf }{u}_x\ge -M$ (3.23)

then

${\Vert \rho \left(t,\cdot \right)\Vert }_{L^{\infty }\left(S\right)}\le {\Vert {\rho }_0\Vert }_{L^{\infty }\left(S\right)}{\text{e}}^{Mt}$ (3.24)

Proof: For any given $x\in S$, define

$U\left(t\right)=u_x\left(t,q\left(t,x_1\right)\right),\gamma \left(t\right)=\rho \left(t,q\left(t,x\right)\right)$

the second equation of Equation (1.1) becomes

${\gamma }^{\prime }\left(t\right)=-\gamma U$

then

$\gamma \left(t\right)=\gamma \left(0\right){\text{e}}^{-{\displaystyle {\int }_0^{t}U\left(\tau \right)\text{d}\tau }}$.

From (3.23), we know $U\left(t\right)\ge -M,t\in \left[0,T\right)$. Hence

$\left|\rho \left(t,q\left(t,x\right)\right)\right|=\left|\gamma \left(t\right)\right|\le \left|\gamma \left(0\right)\right|{\text{e}}^{-{\displaystyle {\int }_0^{t}U\left(\tau \right)\text{d}\tau }}\le \left|\gamma \left(0\right)\right|{\text{e}}^{Mt}\le {\Vert {\rho }_0\Vert }_{L^{\infty }}{\text{e}}^{Mt}$

which together with (3.4), then the proof of lemma 3.5 is complete.

Theorem 3.2 Suppose $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution, then the solution of Equation (1.1) blows up in finite time if and only if

$\underset{t\to T^{-}}{\lim }\underset{x\in S}{\inf }{u}_x\left(t,x\right)=-\infty$ (3.25)

Proof: Suppose that $T<\infty$ and (3.25) is invalid, then there exists $M>0$ satisfies

$u_x\left(t,x\right)\ge -M,\forall \left(t,x\right)\in \left[0,T\right)\times S$

The Lemma 3.4 shows that $u_x\left(t,x\right)$ is bounded on $\left[0,T\right)$, i.e. $\left|u_x\left(t,x\right)\right|\le C$, where $C=C\left(k,M,\sigma ,\lambda ,{\Vert \left(u_0,{\rho }_0-1\right)\Vert }_{H^s\times H^{s-1}}\right)$. Then from the Theorem 3.1, we have $T=\infty$, which contradicts the assumption $T<\infty$.

On the other hand, Sobolev embedding theorem $H^s\mapsto L^{\infty }$ with $s>\frac{1}{2}$ implies that if (3.25) holds, then the corresponding solution blows up in finite time, the proof of Theorem 3.2 is complete.

Next we give two blow-up conditions in finite time.

Theorem 3.3 Suppose $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If there exists $x_0\in S$ satisfies

${\rho }_0\left(x_0\right)=0,u_{0x}\left(x_0\right)=\underset{x\in R}{\inf }{u}_{0x}\left(x\right)$ (3.26)

and

$\begin{array}{l}{\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2\\ \le \left(\frac{\left(8\sigma -1\right)\left(e+1\right)}{36\left(e-1\right)}-\frac{1}{2}{\lambda }^2\right)\frac{4\left(e+1\right)}{3\left(e+1\right)+18\left(k^2+1\right)\left(e-1\right)}\end{array}$ (3.27)

then the corresponding solution u of Equation (1.1) blows up in finite time when $0<T<T^{\ast }$, where

$\begin{array}{c}{T}^{\ast }=\frac{2\left(1+\left|u_{0x}\left(x_0\right)\right|\right)}{\frac{\left(8\sigma -1\right)\left(e+1\right)}{18\left(e-1\right)}+\left(\frac{3\left(e+1\right)+18\left(k^2+1\right)\left(e-1\right)}{2\left(e-1\right)}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert {\rho }_0-1\Vert }_{L^2}^2)-{\lambda }^2\right)}\\ +\frac{2}{1-\lambda }\end{array}$

Proof: Without loss of generality, assume $s=3$, and choose $x_2\left(t\right)$ such that $q\left(t,x_2\left(t\right)\right)=\xi \left(t\right)$, $t\in \left[0,t\right)$, along the trajectory $q\left(t,x_2\right)$, we rewrite the transport Equation of $\rho$ in (2.7) as

$\frac{\text{d}\rho \left(t,\xi \left(t\right)\right)}{\text{d}t}=-\rho \left(t,\xi \left(t\right)\right)u_x\left(t,\xi \left(t\right)\right)$ (3.28)

From (3.26), we have

$m\left(0\right)=u_x\left(0,\xi \left(0\right)\right)=\underset{x\in S}{\inf }{u}_{0x}\left(x\right)=u_{0x}(\; x\; 0\; )$

Let $x_0=\xi \left(0\right)$, then ${\rho }_0\left(\xi \left(0\right)\right)={\rho }_0\left(x_0\right)$, from (3.28)

$\rho \left(t,\xi \left(t\right)\right)=0,\forall t\in \left[0,T\right)$ (3.29)

From (3.11), (3.29) and $u_{xx}\left(t,\xi \left(t\right)\right)=0$, we obtain

$\begin{array}{c}{\bar{m}}^{\prime }\left(t\right)=-\frac{1}{2}{m}^2\left(t\right)-\lambda m\left(t\right)+u^2\left(t,\xi \left(t\right)\right)-k{\partial }_{x}g\ast {\partial }_{x}u\\ -g\ast \left(u^2+\frac{1}{2}{u}_x^2+\frac{\sigma }{2}{\rho }^2\right)\left(t,\xi \left(t\right)\right)\\ =-\frac{1}{2}{m}^2\left(t\right)-\lambda m\left(t\right)+f^{\ast }\left(t,q\left(t,x_2\right)\right)\\ =-\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2+\frac{1}{2}{\lambda }^2+f^{\ast }\left(t,q\left(t,x_2\right)\right)\end{array}$ (3.30)

where

$f^{\ast }=u^2\left(t,\xi \left(t\right)\right)-k{\partial }_{x}g\ast {\partial }_{x}u-g\ast \left(u^2+\frac{1}{2}{u}_x^2+\frac{\sigma }{2}{\rho }^2\right)\left(t,\xi \left(t\right)\right)$ (3.31)

Modify the estimates:

$k\left|{\partial }_{x}g\ast {\partial }_{x}u\right|\le k{\Vert g_x\Vert }_{L^2}{\Vert u_x\Vert }_{L^2}\le \frac{e+1}{36\left(e-1\right)}+\frac{9}{2}{k}^2{\Vert u_x\Vert }_{L^2}^2$

$\sigma \left|g\ast \left(\rho -1\right)\right|\le \sigma {\Vert g\Vert }_{L^2}{\Vert \rho -1\Vert }_{L^2}\le \left(\frac{e+1}{36\left(e-1\right)}+\frac{9}{2}\right)\sigma {\Vert \rho -1\Vert }_{L^2}^2$

The similar process to (3.16) leads to

$f^{\ast }\le \frac{\left(1-8\sigma \right)\left(e+1\right)}{36\left(e-1\right)}+\frac{3\left(e+1\right)+18\left(1+k^2\right)\left(e-1\right)}{4\left(e-1\right)}\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)=-C_2$

From the above inequality and (3.27), we have $\frac{1}{2}{\lambda }^2-C_2<0$, then

$m^{\prime }\left(t\right)\le -\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2+\frac{1}{2}{\lambda }^2-C_2\le \frac{1}{2}{\lambda }^2-C_2<0,t\in \left[0,T\right)$ (3.32)

So $m\left(t\right)$ is strictly decreasing in $\left[0,T\right)$.

If there exist global solutions, we will show that this leads to a contradiction. Let

$t_1=\frac{2\left(1+\left|u_{0x}\left(x_0\right)\right|\right)}{2C_2-{\lambda }^2}$

integrating (3.32) over $\left[0,t_1\right]$ yields

$m\left(t_1\right)=m\left(0\right)+{\displaystyle {\int }_0^{t_1}{m}^{\prime }\left(t\right)\text{d}t}\le \left|u_{0x}\left(x_0\right)\right|+\left(\frac{1}{2}{\lambda }^2-C_2\right)t_1=-1$ (3.33)

Hence we know $m\left(t\right)\le m\left(t_1\right)\le -1,t\in \left[t_1,T\right)$.

From (3.32), we have

$m^{\prime }\left(t\right)\le -\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2$ (3.34)

Integrating (3.34) over $\left[t_1,T\right)$ and knowing $m\left(t_1\right)\le -1$, we get

$-\frac{1}{m\left(t\right)+\lambda }+\frac{1}{\lambda -1}\le -\frac{1}{m\left(t\right)+\lambda }+\frac{1}{\lambda +m\left(t_1\right)}\le -\frac{1}{2}\left(t-t_1\right),t\in \left[t_1,T\right)$

then

$m\left(t\right)\le \frac{1}{\frac{1}{2}\left(t-t_1\right)+\frac{1}{\lambda -1}}-\lambda \to -\infty$, as $t\to t_1+\frac{2}{1-\lambda }$

Thus $T\le t_1+\frac{2}{1-\lambda }$ is a contradiction with $T=\infty$.

The proof of the Theorem 3.3 is complete.

Theorem 3.4 Let $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If there exists $x_0\in S$ satisfies

${\rho }_0\left(x_0\right)=0,u_{0x}\left(x_0\right)=\underset{x\in R}{\inf }{u}_{0x}\left(x\right)$ (3.35)

and

$u_{0x}\left(x_0\right)<-\sqrt{{\lambda }^2+C_1^2}-\lambda$ (3.36)

then the corresponding solution u of Equation (1.1) blows up in finite time when $0<T<T^{\ast \ast }$,where

$T^{\ast \ast }=-\frac{2\left(\lambda +u_{0x}\left(x_0\right)\right)}{{\left(\lambda +u_{0x}\left(x_0\right)\right)}^2-\left({\lambda }^2+C_1^2\right)}$

Proof: From (3.16), we have

$m^{\prime }\left(t\right)\le -\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2+\frac{1}{2}\left({\lambda }^2+C_1^2\right),t\in \left[0,T\right)$

From (3.36), we have $m^{\prime }\left(0\right)<0$, $m\left(t\right)$ is strictly decreasing on $\left[0,T\right)$ and set

$\delta =\frac{1}{2}-\frac{1}{{\left(\lambda +u_{0x}\left(x_0\right)\right)}^2}\left(\frac{1}{2}{\lambda }^2+\frac{1}{2}{C}_1^2\right)\in \left(0,\frac{1}{2}\right)$.

Because $m\left(t\right)<m\left(0\right)=u_{0x}\left(x_0\right)<-\lambda$, then

$m^{\prime }\left(t\right)\le -\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2+\frac{1}{2}\left({\lambda }^2+C_1^2\right)\le -\delta {\left(m\left(t\right)+\lambda \right)}^2$

Similar discussion of the Theorem 3.3

$m\left(t\right)\le \frac{\lambda +u_{0x}\left(x_0\right)}{1+\delta t\left(\lambda +u_{0x}\left(x_0\right)\right)}-\lambda \to -\infty$, $t\to -\frac{1}{\lambda \delta +\delta u_{0x}(\; x\; 0\; )}$

Hence

$0<T<-\frac{2\left(\lambda +u_{0x}\left(x_0\right)\right)}{{\left(\lambda +u_{0x}\left(x_0\right)\right)}^2-\left({\lambda }^2+C_1^2\right)}$.

The proof of the theorem 3.4 is complete.

Next we will show the blow-up rate of solutions and the result shows: the blow-up rate is not affected by the weakly dissipation.

Theorem 3.5 (blow-up rate) Let $\sigma >0$, and $\left(u,\rho \right)$ be the solution of Equation (1.1) with initial data $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If $T<\infty$, then

$\underset{t\to T^{-}}{\lim }\left(\underset{x\in S}{\inf }{u}_x\left(t,x\right)\left(T-t\right)\right)=-2$

Proof: Without loss of generality, assume $s=3$.

Set

$M=\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+k^2}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)$ (3.37)

From (3.30), we have

$-\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2-\frac{1}{2}{\lambda }^2-M\le m^{\prime }\left(t\right)\le -\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^2+\frac{1}{2}{\lambda }^2+M$ (3.38)

Because of $\underset{t\to T^{-}}{\lim }\left(m\left(t\right)+\lambda \right)=-\infty$, there exists $t_0\in \left(0,T\right)$ satisfies $m\left(t_0\right)+\lambda <0$ and ${\left(m\left(t_0\right)+\lambda \right)}^2>\frac{1}{\epsilon }\left({\lambda }^2+M\right),\epsilon \in \left(0,\frac{1}{2}\right)$. Since m is locally Lipschitz, m is absolutely continuous. We deduce that m is decreasing in $\left[t_0,T\right)$ and

${\left(m\left(t\right)+\lambda \right)}^2>\frac{1}{\epsilon }\left({\lambda }^2+M\right)$ (3.39)

According to (3.38) and (3.39)

$\frac{1}{2}-\epsilon \le \frac{\text{d}}{\text{d}t}\left(\frac{1}{m\left(t\right)+\lambda }\right)\le \frac{1}{2}+\epsilon ,t\in \left(t_0,T\right)$

Integrating (3.39) over $\left(t,T\right)$ with respect to $t\in \left[t_0,T\right)$, notice that $\underset{t\to T^{-}}{\lim }\left(m\left(t\right)+\lambda \right)=-\infty$, then

$\left(\frac{1}{2}-\epsilon \right)\left(T-t\right)\le -\left(\frac{1}{m\left(t\right)+\lambda }\right)\le \left(\frac{1}{2}+\epsilon \right)\left(T-t\right),t\in \left(t_0,T\right)$

Since $\epsilon$ is arbitrary, so

$\underset{t\to T^{-}}{\lim }\left\{m\left(t\right)\left(T-t\right)+\lambda \left(T-t\right)\right\}=-2$

That is $\underset{t\to T^{-}}{\lim }m\left(t\right)\left(T-t\right)=-2$, the blow-up rate of solutions of Equation (1.1) is not effected by the weakly dissipation.

4. Global Existence

In this section, we provide a sufficient condition for the global solution of Equation (1.1) in the case $0<\sigma <2$.

Theorem 4.1 Let $0<\sigma <2$, $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$ with $s>\frac{3}{2}$, there exist a maximal time $T>0$ and a unique solution $\left(u,\rho \right)$ of Equation (1.1) with initial data. Assume that $\underset{x\in S}{\inf }{\rho }_0\left(x\right)>0$, then

1) when $0<\sigma \le 1$,

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0\left(x\right)}{C}_4{\text{e}}^{C_{3}t}$

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_4^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{C_{3}t}{2-\sigma }}$

2) when $1<\sigma <2$,

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_4^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{C_{3}t}{2-\sigma }}$

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0\left(x\right)}{C}_4{\text{e}}^{C_{3}t}$

where

$C_3=1+\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+k^2}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)$

$C_4=1+{\Vert u_{0x}\Vert }_{L^{\infty }}^2+{\Vert {\rho }_0\Vert }_{L^{\infty }}^2$

Proof: It is sufficient to prove the desired results for $s=3$.

1) We will estimate the $\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|$.

From (3.22), we have

$m\left(t\right)\le 0,t\in \left[0,T\right)$ (4.1)

Let $\zeta \left(t\right)=\rho \left(t,\xi \left(t\right)\right)$, thus we have

$\{\begin{array}{l}{m}^{\prime }\left(t\right)=-\frac{1}{2}{m}^2\left(t\right)-\lambda m\left(t\right)+\frac{\sigma }{2}{\zeta }^2\left(t\right)+f\left(t,q\left(t,x_2\right)\right)\\ {\zeta }^{\prime }\left(t\right)=-\zeta \left(t\right)\bar{m}\left(t\right)\end{array}$ (4.2)

where f is defined as (3.15). The second Equation of (3.15) shows that $\zeta \left(t\right)$ and $\zeta \left(0\right)$ have the same sign. Hence $\zeta \left(0\right)=\rho \left(0,\xi \right)\left(0\right)>0$.

Suppose $0<\sigma \le 1$, define the function

$w_1\left(t\right)=\zeta \left(0\right)\zeta \left(t\right)+\frac{\zeta \left(0\right)}{\zeta \left(t\right)}\left(1+m^2\left(t\right)\right)$ (4.3)

which is positive on $t\in \left[0,T\right)$.

Differentiate $w_1(\; t\; )$

$\begin{array}{c}{w^{\prime }}_1\left(t\right)=\zeta \left(0\right){\zeta }^{\prime }\left(t\right)+\frac{\zeta \left(0\right)}{{\zeta }^2\left(t\right)}\left(1+m^2\left(t\right)\right){\zeta }^{\prime }\left(t\right)+\frac{2\zeta (0)}{\zeta \left(t\right)}m\left(t\right)m^{\prime }\left(t\right)\\ =-\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{{\zeta }^2\left(t\right)}\left(1+m^2\left(t\right)\right)\zeta \left(t\right)m\left(t\right)\\ +\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)\left(-\frac{1}{2}{m}^2\left(t\right)-\lambda m\left(t\right)+\frac{\sigma }{2}{\zeta }^2\left(t\right)+f\right)\\ =\left(\sigma -1\right)\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)-\frac{2\lambda \zeta \left(0\right)}{\zeta \left(t\right)}{m}^2\left(t\right)+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)f\end{array}$

$\begin{array}{l}\le \left(\sigma -1\right)\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)f\\ =\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)\left(\frac{1}{2}+f+\frac{\sigma -1}{2}{\zeta }^2\left(t\right)\right)\\ \le \frac{\zeta \left(0\right)}{\zeta \left(t\right)}\left(1+m^2\left(t\right)\right)\left(1+\left|f\right|\right)\\ \le C_3{w}_1\left(t\right)\end{array}$ (4.4)

where $C_3=1+\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+k^2}{4}\right)\left({\Vert u_0\Vert }_{H^1}^2+\sigma {\Vert \rho -1\Vert }_{L^2}^2\right)$.

Then

$\begin{array}{l}{w}_1\left(t\right)\le {w^{\prime }}_1\left(0\right){\text{e}}^{C_{3}t}=\left({\zeta }^2\left(0\right)+1+m^2\left(0\right)\right){\text{e}}^{C_{3}t}\\ \le \left(1+{\Vert u_{0x}\Vert }_{L^{\infty }}^2+{\Vert {\rho }_0\Vert }_{L^{\infty }}^2\right){\text{e}}^{C_{3}t}=C_4{\text{e}}^{C_{3}t}\end{array}$ (4.5)

where $C_4=1+{\Vert u_{0x}\Vert }_{L^{\infty }}^2+{\Vert {\rho }_0\Vert }_{L^{\infty }}^2$.

From (4.3), we have

$\zeta \left(0\right)\zeta \left(t\right)\le w_1\left(t\right),\left|\zeta \left(0\right)\right|m\left(t\right)\le w_1\left(t\right)$ (4.6)

then

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|=\left|m\left(t\right)\right|\le \frac{w_1\left(t\right)}{\left|\zeta \left(0\right)\right|}\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0\left(x\right)}{C}_4{\text{e}}^{C_{3}t},t\in \left[0,T\right)$

Suppose $1<\sigma <2$, define the function

$w_2\left(t\right)={\zeta }^{\sigma }\left(0\right)\frac{{\zeta }^2\left(t\right)+1+m^2\left(t\right)}{{\zeta }^{\sigma }\left(t\right)}$ (4.7)

Differentiate $w_2(\; t\; )$

$\begin{array}{l}{w^{\prime }}_2\left(t\right)=\frac{2{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}m\left(t\right)\left(\left(\sigma -1\right){\zeta }^2\left(t\right)+\frac{\sigma -1}{2}{m}^2\left(t\right)-\lambda m\left(t\right)+f+\frac{\sigma }{2}\right)\\ \le \frac{{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}\left(1+m^2\left(t\right)\right)\left(\left|f\right|+\frac{\sigma }{2}\right)\\ \le \frac{{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}\left(1+m^2\left(t\right)\right)\left(\left|f\right|+1\right)\\ \le C_3{w}_2\left(t\right)\end{array}$ (4.8)

then

$\begin{array}{c}{w}_2\left(t\right)\le w_2\left(0\right){\text{e}}^{C_{3}t}=\left({\zeta }^2\left(0\right)+1+m^2\left(0\right)\right){\text{e}}^{C_{3}t}\\ \le \left(1+{\Vert u_{0x}\Vert }_{L^{\infty }}^2+{\Vert {\rho }_0\Vert }_{L^{\infty }}^2\right){\text{e}}^{C_{3}t}\\ =C_4{\text{e}}^{C_{3}t}\end{array}$ (4.9)

Here we apply Young’s inequality $ab\le \frac{a^p}{p}+\frac{b^q}{q}$, for $p=\frac{2}{\sigma }$, $q=\frac{2}{2-\sigma }$.

$\begin{array}{l}\frac{w_2\left(t\right)}{{\zeta }^{\sigma }\left(0\right)}={\left({\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}\right)}^{\frac{2}{\sigma }}+{\left(\frac{{\left(1+m^2\right)}^{\frac{2-\sigma }{2}}}{{\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}}\right)}^{\frac{2}{2-\sigma }}\\ \ge \frac{\sigma }{2}{\left({\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}\right)}^{\frac{2}{\sigma }}+\frac{2-\sigma }{2}{\left(\frac{{\left(1+m^2\right)}^{\frac{2-\sigma }{2}}}{{\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}}\right)}^{\frac{2}{2-\sigma }}\\ \ge {\left(1+m^2\right)}^{\frac{2-\sigma }{2}}\ge {\left|m\left(t\right)\right|}^{2-\sigma }\end{array}$

Hence

$\left|\underset{x\in S}{\inf }{u}_x\left(t,x\right)\right|\le {\left(\frac{w_2\left(t\right)}{\left|{\zeta }^{\sigma }\left(0\right)\right|}\right)}^{\frac{1}{2-\sigma }}\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_4^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{C_{3}t}{2-\sigma }},t\in \left[0,T\right)$

2) Next we control $\left|\underset{x\in S}{\sup }{u}_x\left(t,x\right)\right|$.

Similarly,

$\{\begin{array}{l}{\bar{m}}^{\prime }\left(t\right)=-\frac{1}{2}{\bar{m}}^2\left(t\right)-\lambda \bar{m}\left(t\right)+\frac{\sigma }{2}{\bar{\zeta }}^2\left(t\right)+f\left(t,q\left(t,x_1\right)\right)\\ {\bar{\zeta }}^{\prime }\left(t\right)=-\bar{\zeta }\left(t\right)\bar{m}(\; t\; )\end{array}$

Suppose $0<\sigma \le 1$, define the function

${\bar{w}}_1\left(t\right)={\bar{\zeta }}^{\sigma }\left(0\right)\frac{{\bar{\zeta }}^2\left(t\right)+1+{\bar{m}}^2\left(t\right)}{{\bar{\zeta }}^{\sigma }\left(t\right)}$ (4.10)

From (3.20) and (4.8), we obtain ${{\bar{w}}^{\prime }}_1\left(t\right)\le C_3{\bar{w}}_1\left(t\right)$, then ${\bar{w}}_1\left(t\right)\le C_4{\text{e}}^{C_{3}t}$.

Similarly, we get

$\frac{{\bar{w}}_1\left(t\right)}{{\bar{\zeta }}^{\sigma }\left(0\right)}\ge {\left|\bar{m}\left(t\right)\right|}^{2-\sigma }$

then

$\left|\underset{x\in S}{\sup }{u}_x\left(t,x\right)\right|\le {\left(\frac{{\bar{w}}_1\left(t\right)}{\left|{\bar{\zeta }}^{\sigma }\left(0\right)\right|}\right)}^{\frac{1}{2-\sigma }}\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_4^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{C_{3}t}{2-\sigma }},t\in \left[0,T\right)$

Suppose $1<\sigma <2$, define the function

${\bar{w}}_2\left(t\right)=\bar{\zeta }\left(0\right)\bar{\zeta }\left(t\right)+\frac{\bar{\zeta }\left(0\right)}{\bar{\zeta }\left(t\right)}\left(1+{\bar{m}}^2\left(t\right)\right)$ (4.11)

From (3.20) and (4.4), we have ${{\bar{w}}^{\prime }}_2\left(t\right)\le C_3{\bar{w}}_2\left(t\right)$, then ${\bar{w}}_2\left(t\right)\le C_4{\text{e}}^{C_{3}t}$.

Hence

$\left|\underset{x\in S}{\sup }{u}_x\left(t,x\right)\right|=\left|\bar{m}\left(t\right)\right|\le \frac{{\bar{w}}_2\left(t\right)}{\left|\bar{\zeta }\left(0\right)\right|}\le \frac{1}{\underset{x\in S}{\inf }{\rho }_0\left(x\right)}{C}_4{\text{e}}^{C_{3}t},t\in \left[0,T\right)$

Theorem 4.2 Let $0<\sigma <2$, $\left(u_0,{\rho }_0-1\right)\in H^s\left(S\right)\times H^{s-1}\left(S\right)$ with $s\ge 2$, there exist a maximal time $T>0$ and a unique solution $\left(u,\rho \right)$ of Equation (1.1) with initial data. If $\underset{x\in S}{\inf }{\rho }_0\left(x\right)>0$, then $T=\infty$ and the the solution $\left(u,\rho \right)$ is global.

Proof: By contradictory arguement, assume $T<\infty$ and the solution blows up. The Theorem 3.1 shows

${{\displaystyle {\int }_0^T\left|u_x\left(t,x\right)\right|}}_{L^{\infty }}\text{d}t=\infty$ (4.12)

The assumptions and the Theorem 4.1 show

$\left|u_x\left(t,x\right)\right|<\infty$

For all $\left(t,x\right)\in \left[0,T\right)\times S$, that is a contradiction to (4.12).

The proof of Theorem 4.2 is complete.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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