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

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.

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Li, Y. , Liu, J. and Zhu, X. (2020) Blow-Up for a Periodic Two-Component Camassa-Holm Equation with Generalized Weakly Dissipation. Journal of Applied Mathematics and Physics, 8, 2223-2240. doi: 10.4236/jamp.2020.810167.

1. Introduction

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

$\left\{\begin{array}{l}{u}_{t}-{u}_{xxt}+k{u}_{x}+3u{u}_{x}-2{u}_{x}{u}_{xx}-u{u}_{xxx}+\lambda \left(u-{u}_{xx}\right)+\sigma \rho {\rho }_{x}=0,\text{ }\text{ }t>0,x\in R,\\ {\rho }_{t}+{\left(\rho u\right)}_{x}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}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

$\left\{\begin{array}{l}{u}_{t}-{u}_{xxt}+k{u}_{x}+3u{u}_{x}-2{u}_{x}{u}_{xx}-u{u}_{xxx}-\rho {\rho }_{x}=0,\text{ }t>0,x\in R\\ {\rho }_{t}+{\left(\rho u\right)}_{x}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0,x\in R\end{array}$ (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 ${‖\text{ }\text{ }\cdot \text{ }‖}_{X}$ 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 ${H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$.

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

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

${‖\left(A\left(y\right)-A\left(z\right)\right)w‖}_{X}\le {\mu }_{1}{‖y-z‖}_{X}{‖w‖}_{Y}$ (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

${‖\left(B\left(y\right)-B\left(z\right)\right)w‖}_{X}\le {\mu }_{2}{‖y-z‖}_{Y}{‖w‖}_{X}$ (2.3)

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

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

${‖f\left(y\right)-f\left(z\right)‖}_{Y}\le {\mu }_{3}{‖y-z‖}_{Y}$ (2.4)

${‖f\left(y\right)-f\left(z\right)‖}_{X}\le {\mu }_{4}{‖y-z‖}_{X}$ (2.5)

where $z,y\in Y$, ${\mu }_{1},{\mu }_{2},{\mu }_{3},{\mu }_{4}$ are constants which only depending $\left\{{‖y‖}_{Y},{‖z‖}_{Y}\right\}$.

If the 1), 2), 3) hold, given ${u}_{0}\in Y$, there is a maximal $T>0$ depending only on ${‖{u}_{0}‖}_{Y}$ 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 $g\left(x\right):=\frac{\mathrm{cosh}\left(x-\left[x\right]-\frac{1}{2}\right)}{2\mathrm{sinh}\frac{1}{2}}$, $x\in R$, ${\left(1-{\partial }_{x}^{2}\right)}^{-1}f=g\ast f$ 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

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

Theorem 2.2 Let ${z}_{0}=\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}×{H}^{s-1}$ 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 $z=\left(\begin{array}{l}u\\ \rho \end{array}\right)$, $T={H}^{s}×{H}^{s}$, $X={H}^{s-1}×{H}^{s-1}$, $\wedge ={\left(1-{\partial }_{x}^{2}\right)}^{\frac{1}{2}}$, $Q=\left(\begin{array}{cc}\wedge & 0\\ 0& \wedge \end{array}\right)$, 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 ${H}^{s}×{H}^{s}$ onto ${H}^{s-1}×{H}^{s-1}$. 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 $z\in {H}^{s}×{H}^{s}$, $s>\frac{3}{2}$ belongs to $G\left({L}^{2}×{L}^{2},1,\beta \right)$.

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

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

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

where $y,z,w\in {H}^{s}×{H}^{s}$.

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

${‖\left(B\left(y\right)-B\left(z\right)\right)w‖}_{{H}^{s-1}×{H}^{s-1}}\le {\mu }_{2}{‖y-z‖}_{{H}^{s}×{H}^{s}}{‖w‖}_{{H}^{s-1}×{H}^{s-1}}$ (2.10)

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

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

1) ${‖f\left(y\right)-f\left(z\right)‖}_{{H}^{s}×{H}^{s}}\le {\mu }_{3}{‖y-z‖}_{{H}^{s}×{H}^{s}}$, $y,z\in {H}^{s}×{H}^{s}$ (2.11)

2) ${‖f\left(y\right)-f\left(z\right)‖}_{{H}^{s-1}×{H}^{s-1}}\le {\mu }_{4}{‖y-z‖}_{{H}^{s-1}×{H}^{s-1}}$, $y,z\in {H}^{s}×{H}^{s}$ (2.12)

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

$\begin{array}{l}{‖f\left(y\right)-f\left(z\right)‖}_{{H}^{s}×{H}^{s}}\\ \le {‖-{\partial }_{x}{\left(1-{\partial }_{x}^{2}\right)}^{-1}\left[\left({y}_{1}^{2}-{u}^{2}\right)+\frac{1}{2}\left({y}_{1x}^{2}-{u}_{x}^{2}\right)+k\left({y}_{1}-u\right)+\frac{\sigma }{2}\left({y}_{2}^{2}-{\rho }^{2}\right)\right]‖}_{{H}^{s}}\\ \text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{‖\lambda \left({y}_{1}-u\right)‖}_{{H}^{s}}+{‖{u}_{x}\rho -{y}_{1x}{y}_{2}‖}_{{H}^{s}}\end{array}$

$\begin{array}{l}\le {‖\left({y}_{1}^{2}-{u}^{2}\right)+\frac{1}{2}\left({y}_{1x}^{2}-{u}_{x}^{2}\right)+k\left({y}_{1}-u\right)‖}_{{H}^{s-1}}+\frac{|\sigma |}{2}{‖{y}_{2}^{2}-{\rho }^{2}‖}_{{H}^{s-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+|\lambda |{‖{y}_{1}-u‖}_{{H}^{s}}+{‖\left({u}_{x}-{y}_{1x}\right)\rho ‖}_{{H}^{s}}+{‖{y}_{1x}\left(\rho -{y}_{2}\right)‖}_{{H}^{s}}\end{array}$

$\begin{array}{l}\le {‖\left({y}_{1}-u\right)\left({y}_{1}+u\right)‖}_{{H}^{s-1}}+\frac{1}{2}{‖\left({y}_{1x}-{u}_{x}\right)\left({y}_{1x}+{u}_{x}\right)‖}_{{H}^{s-1}}+|k|{‖{y}_{1}-u‖}_{{H}^{s-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{ }+|\lambda |{‖{y}_{1}-u‖}_{{H}^{s}}+\frac{|\sigma |}{2}{‖{y}_{2}-\rho ‖}_{{H}^{s-1}}{‖{y}_{2}+\rho ‖}_{{H}^{s-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+{‖{y}_{1}-u‖}_{{H}^{s}}{‖\rho ‖}_{{H}^{s}}+{‖{y}_{1}‖}_{{H}^{s}}{‖\rho -{y}_{2}‖}_{{H}^{s}}\end{array}$

$\begin{array}{l}\le {‖\left({y}_{1}-u\right)\left({y}_{1}+u\right)‖}_{{H}^{s-1}}+\frac{1}{2}{‖\left({y}_{1x}-{u}_{x}\right)\left({y}_{1x}+{u}_{x}\right)‖}_{{H}^{s-1}}+|k|{‖{y}_{1}-u‖}_{{H}^{s-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{ }+|\lambda |{‖{y}_{1}-u‖}_{{H}^{s}}+\frac{|\sigma |}{2}{‖{y}_{2}-\rho ‖}_{{H}^{s-1}}{‖{y}_{2}+\rho ‖}_{{H}^{s-1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+{‖{y}_{1}-u‖}_{{H}^{s}}{‖\rho ‖}_{{H}^{s}}+{‖{y}_{1}‖}_{{H}^{s}}{‖\rho -{y}_{2}‖}_{{H}^{s}}\end{array}$

Let ${\mu }_{3}=\frac{5+|\sigma |}{2}{‖y‖}_{{H}^{s}×{H}^{s}}+\frac{3+|\sigma |}{2}{‖z‖}_{{H}^{s}×{H}^{s}}+|k|+|\lambda |$, then

${‖f\left(y\right)-f\left(z\right)‖}_{{H}^{s}×{H}^{s}}\le {\mu }_{3}{‖y-z‖}_{{H}^{s}×{H}^{s}}$, $y,z\in {H}^{s}×{H}^{s}$

Making $y=0$ in the above inequality, it shows that f is bounded on bounded sets in ${H}^{s}×{H}^{s}$, 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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}×{H}^{s-1}$, $s>\frac{3}{2}$, T is the maximal time of existence of the solution, then

$T<\infty ⇒{\int }_{0}^{T}{‖{u}_{x}\left(\tau \right)‖}_{{L}^{\infty }}\text{d}\tau =\infty$ (3.1)

Consider the following equation of trajectory:

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

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

${q}_{x}\left(t,x\right)={\text{e}}^{{\int }_{0}^{t}{u}_{x}\left(\tau ,q\left(\tau ,x\right)\right)\text{d}\tau }>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall \left(t,x\right)\in \left[0,T\right)×S$ (3.3)

Hence

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

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

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 ${z}_{0}=\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$ 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

${‖u‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\le {‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}$ (3.5)

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

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

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

According to (3.6) and (3.7)

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

Then

$\begin{array}{l}{\int }_{S}\left({u}^{2}+{u}_{x}^{2}+\sigma {\left(\rho -1\right)}^{2}+2\lambda {\int }_{0}^{t}\left({u}^{2}+{u}_{x}^{2}\right)\text{d}\tau \right)\text{d}x\\ ={\int }_{S}\left({u}_{0}^{2}+{u}_{0x}^{2}+\sigma {\left({\rho }_{0}-1\right)}^{2}\right)\text{d}x={‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\end{array}$

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

$\begin{array}{c}{‖u‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}={\int }_{S}\left({u}^{2}+{u}_{x}^{2}+\sigma {\left(\rho -1\right)}^{2}\right)\text{d}x\\ \le {‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\end{array}$

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

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

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

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

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

$\underset{x\in S}{\mathrm{sup}}{u}_{x}\left(t,x\right)\le {‖{u}_{0x}‖}_{{L}^{\infty }}+\sqrt{{\lambda }^{2}+\sigma {‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+{C}_{1}^{2}}$

where ${C}_{1}=\sqrt{\frac{\left(3\sigma +2\right)\left(e+1\right)}{2\left(e-1\right)}+\left(\frac{e+1}{e-1}+\frac{{k}^{2}+1}{2}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\right)}$.

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

$\stackrel{¯}{m}\left(t\right)={u}_{x}\left(t,\eta \left(t\right)\right)=\underset{x\in S}{\mathrm{sup}}\left({u}_{x}\left(t,x\right)\right),m\left(t\right)=\underset{x\in S}{\mathrm{inf}}\left({u}_{x}\left(t,x\right)\right)$ (3.12)

From the Fermat’s lemma, we know

${u}_{xx}\left(t,\eta \left(t\right)\right)\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}a.e.\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,T\right)$

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

$\stackrel{¯}{\zeta }\left(t\right)=\rho \left(t,q\left(t,{x}_{1}\right)\right)$, $t\in \left[0,T\right)$ (3.14)

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

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

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

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

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

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

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

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

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

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

Therefore we get the upper bound of f

$\begin{array}{l}f\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{{k}^{2}}{4}\right){‖u‖}_{{H}^{1}}^{2}+\frac{1}{4}\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\\ \text{ }\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{{k}^{2}+1}{4}\right)\left({‖u‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)\\ \text{ }\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{e+1}{2\left(e-1\right)}+\frac{{k}^{2}+1}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\right)\\ \text{ }=\frac{1}{2}{C}_{1}^{2}\end{array}$ (3.16)

Similarly, we turn to the lower bound of f

$\begin{array}{c}-f\le {u}^{2}+k|{\partial }_{x}g\ast {\partial }_{x}u|+|g\ast \left({u}^{2}+\frac{1}{2}{u}_{x}^{2}\right)|+\frac{\sigma }{2}|g\ast 1|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sigma |g\ast \left(\rho -1\right)|+\frac{\sigma }{2}|g\ast {\left(\rho -1\right)}^{2}|\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\frac{e+1}{e-1}{‖u‖}_{{L}^{2}}^{2}+\left(\frac{3\left(e+1\right)}{4\left(e-1\right)}+\frac{{k}^{2}}{4}\right){‖{u}_{x}‖}_{{L}^{2}}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\frac{e+1}{4\left(e-1\right)}+\frac{1}{4}\right)\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{{k}^{2}+1}{4}\right)\left({‖u‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)\\ \le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{{k}^{2}+1}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\right)\end{array}$ (3.17)

According to (3.16) and (3.17)

$|f|\le \frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{{k}^{2}+1}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\right)$ (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

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

hence

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

From the second Equation of (3.15), we have

$\stackrel{¯}{\zeta }\left(t\right)=\stackrel{¯}{\zeta }\left(0\right){\text{e}}^{-{\int }_{0}^{t}\stackrel{¯}{m}\left(\tau \right)\text{d}\tau }$

then

$|\rho \left(t,q\left(t,{x}_{1}\right)\right)|=|\stackrel{¯}{\zeta }\left(t\right)|\le |\stackrel{¯}{\zeta }\left(0\right)|\le {‖{\rho }_{0}‖}_{{L}^{\infty }}$

For any given $x\in S$, define

${P}_{1}\left(t\right)=\stackrel{¯}{m}\left(t\right)-{‖{u}_{0x}‖}_{{L}^{\infty }}-\sqrt{{\lambda }^{2}+\sigma {‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+{C}_{1}^{2}}$ (3.21)

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

$\begin{array}{c}{P}_{1}\left(0\right)=\stackrel{¯}{m}\left(0\right)-{‖{u}_{0x}‖}_{{L}^{\infty }}-\sqrt{{\lambda }^{2}+\sigma {‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+{C}_{1}^{2}}\\ \le \stackrel{¯}{m}\left(0\right)-{‖{u}_{0x}‖}_{{L}^{\infty }}\le 0\end{array}$

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 ${t}_{1}=\mathrm{max}\left\{t<{t}_{0}:{P}_{1}\left(t\right)=0\right\}$, we have

${P}_{1}\left({t}_{1}\right)=0,{{P}^{\prime }}_{1}\left({t}_{1}\right)\ge 0$

then

$\stackrel{¯}{m}\left({t}_{1}\right)={‖{u}_{0x}‖}_{{L}^{\infty }}+\sqrt{{\lambda }^{2}+\sigma {‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+{C}_{1}^{2}}$.

From (3.21), we know

${\stackrel{¯}{m}}^{\prime }\left({t}_{1}\right)={{P}^{\prime }}_{1}\left({t}_{1}\right)\ge 0$ (3.22)

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

$\begin{array}{l}{\stackrel{¯}{m}}^{\prime }\left({t}_{1}\right)=-\frac{1}{2}{\stackrel{¯}{m}}^{2}\left({t}_{1}\right)-\lambda \stackrel{¯}{m}\left({t}_{1}\right)+\frac{\sigma }{2}{\stackrel{¯}{\zeta }}^{2}\left({t}_{1}\right)+f\left({t}_{1},q\left({t}_{1},{x}_{1}\right)\right)\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\le -\frac{1}{2}{\left(m\left({t}_{1}\right)+\lambda \right)}^{2}+\frac{1}{2}{\lambda }^{2}+\frac{\sigma }{2}{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+\frac{1}{2}{C}_{1}^{2}\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }\le -\frac{1}{2}{\left({‖{u}_{0x}‖}_{{L}^{\infty }}+\sqrt{{\lambda }^{2}+\sigma {‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+{C}_{1}^{2}}+\lambda \right)}^{2}+\frac{\sigma }{2}{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}+\frac{1}{2}{C}_{1}^{2}\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{ }\text{ }\text{ }\text{ }<0\end{array}$

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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of the solution. If there exists $M\ge 0$ such that

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

then

${‖\rho \left(t,\cdot \right)‖}_{{L}^{\infty }\left(S\right)}\le {‖{\rho }_{0}‖}_{{L}^{\infty }\left(S\right)}{\text{e}}^{Mt}$ (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

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

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

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

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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$, 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

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

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

${u}_{x}\left(t,x\right)\ge -M,\text{\hspace{0.17em}}\forall \left(t,x\right)\in \left[0,T\right)×S$

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

On the other hand, Sobolev embedding theorem ${H}^{s}↦{L}^{\infty }$ 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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If there exists ${x}_{0}\in S$ satisfies

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

and

$\begin{array}{l}{‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\\ \le \left(\frac{\left(8\sigma -1\right)\left(e+1\right)}{36\left(e-1\right)}-\frac{1}{2}{\lambda }^{2}\right)\frac{4\left(e+1\right)}{3\left(e+1\right)+18\left({k}^{2}+1\right)\left(e-1\right)}\end{array}$ (3.27)

then the corresponding solution u of Equation (1.1) blows up in finite time when $0, where

$\begin{array}{c}{T}^{\ast }=\frac{2\left(1+|{u}_{0x}\left({x}_{0}\right)|\right)}{\frac{\left(8\sigma -1\right)\left(e+1\right)}{18\left(e-1\right)}+\left(\frac{3\left(e+1\right)+18\left({k}^{2}+1\right)\left(e-1\right)}{2\left(e-1\right)}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖{\rho }_{0}-1‖}_{{L}^{2}}^{2}\right)-{\lambda }^{2}\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2}{1-\lambda }\end{array}$

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

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

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

$\begin{array}{c}{\stackrel{¯}{m}}^{\prime }\left(t\right)=-\frac{1}{2}{m}^{2}\left(t\right)-\lambda m\left(t\right)+{u}^{2}\left(t,\xi \left(t\right)\right)-k{\partial }_{x}g\ast {\partial }_{x}u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-g\ast \left({u}^{2}+\frac{1}{2}{u}_{x}^{2}+\frac{\sigma }{2}{\rho }^{2}\right)\left(t,\xi \left(t\right)\right)\\ =-\frac{1}{2}{m}^{2}\left(t\right)-\lambda m\left(t\right)+{f}^{\ast }\left(t,q\left(t,{x}_{2}\right)\right)\\ =-\frac{1}{2}{\left(m\left(t\right)+\lambda \right)}^{2}+\frac{1}{2}{\lambda }^{2}+{f}^{\ast }\left(t,q\left(t,{x}_{2}\right)\right)\end{array}$ (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:

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

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

The similar process to (3.16) leads to

${f}^{\ast }\le \frac{\left(1-8\sigma \right)\left(e+1\right)}{36\left(e-1\right)}+\frac{3\left(e+1\right)+18\left(1+{k}^{2}\right)\left(e-1\right)}{4\left(e-1\right)}\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)=-{C}_{2}$

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

${t}_{1}=\frac{2\left(1+|{u}_{0x}\left({x}_{0}\right)|\right)}{2{C}_{2}-{\lambda }^{2}}$

integrating (3.32) over $\left[0,{t}_{1}\right]$ yields

$m\left({t}_{1}\right)=m\left(0\right)+{\int }_{0}^{{t}_{1}}{m}^{\prime }\left(t\right)\text{d}t\le |{u}_{0x}\left({x}_{0}\right)|+\left(\frac{1}{2}{\lambda }^{2}-{C}_{2}\right){t}_{1}=-1$ (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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If there exists ${x}_{0}\in S$ satisfies

${\rho }_{0}\left({x}_{0}\right)=0,{u}_{0x}\left({x}_{0}\right)=\underset{x\in R}{\mathrm{inf}}{u}_{0x}\left(x\right)$ (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,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), 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

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

Hence

$0.

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 $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$, $s\ge 2$, and T is the maximal time of existence of the solution. If $T<\infty$, then

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

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

Set

$M=\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+{k}^{2}}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)$ (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 $\underset{t\to {T}^{-}}{\mathrm{lim}}\left(m\left(t\right)+\lambda \right)=-\infty$, 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)

$\frac{1}{2}-\epsilon \le \frac{\text{d}}{\text{d}t}\left(\frac{1}{m\left(t\right)+\lambda }\right)\le \frac{1}{2}+\epsilon ,\text{\hspace{0.17em}}t\in \left({t}_{0},T\right)$

Integrating (3.39) over $\left(t,T\right)$ with respect to $t\in \left[{t}_{0},T\right)$, notice that $\underset{t\to {T}^{-}}{\mathrm{lim}}\left(m\left(t\right)+\lambda \right)=-\infty$, then

$\left(\frac{1}{2}-\epsilon \right)\left(T-t\right)\le -\left(\frac{1}{m\left(t\right)+\lambda }\right)\le \left(\frac{1}{2}+\epsilon \right)\left(T-t\right),\text{\hspace{0.17em}}t\in \left({t}_{0},T\right)$

Since $\epsilon$ is arbitrary, so

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

That is $\underset{t\to {T}^{-}}{\mathrm{lim}}m\left(t\right)\left(T-t\right)=-2$, 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 $0<\sigma <2$, $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$ 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 $\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)>0$, then

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

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)}{C}_{4}{\text{e}}^{{C}_{3}t}$

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_{4}^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{{C}_{3}t}{2-\sigma }}$

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

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_{4}^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{{C}_{3}t}{2-\sigma }}$

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)}{C}_{4}{\text{e}}^{{C}_{3}t}$

where

${C}_{3}=1+\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+{k}^{2}}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)$

${C}_{4}=1+{‖{u}_{0x}‖}_{{L}^{\infty }}^{2}+{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}$

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

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

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

$\left\{\begin{array}{l}{m}^{\prime }\left(t\right)=-\frac{1}{2}{m}^{2}\left(t\right)-\lambda m\left(t\right)+\frac{\sigma }{2}{\zeta }^{2}\left(t\right)+f\left(t,q\left(t,{x}_{2}\right)\right)\\ {\zeta }^{\prime }\left(t\right)=-\zeta \left(t\right)\stackrel{¯}{m}\left(t\right)\end{array}$ (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 ${w}_{1}\left( t \right)$

$\begin{array}{c}{{w}^{\prime }}_{1}\left(t\right)=\zeta \left(0\right){\zeta }^{\prime }\left(t\right)+\frac{\zeta \left(0\right)}{{\zeta }^{2}\left(t\right)}\left(1+{m}^{2}\left(t\right)\right){\zeta }^{\prime }\left(t\right)+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right){m}^{\prime }\left(t\right)\\ =-\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{{\zeta }^{2}\left(t\right)}\left(1+{m}^{2}\left(t\right)\right)\zeta \left(t\right)m\left(t\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)\left(-\frac{1}{2}{m}^{2}\left(t\right)-\lambda m\left(t\right)+\frac{\sigma }{2}{\zeta }^{2}\left(t\right)+f\right)\\ =\left(\sigma -1\right)\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)-\frac{2\lambda \zeta \left(0\right)}{\zeta \left(t\right)}{m}^{2}\left(t\right)+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)f\end{array}$

$\begin{array}{l}\le \left(\sigma -1\right)\zeta \left(0\right)\zeta \left(t\right)m\left(t\right)+\frac{\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)+\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)f\\ =\frac{2\zeta \left(0\right)}{\zeta \left(t\right)}m\left(t\right)\left(\frac{1}{2}+f+\frac{\sigma -1}{2}{\zeta }^{2}\left(t\right)\right)\\ \le \frac{\zeta \left(0\right)}{\zeta \left(t\right)}\left(1+{m}^{2}\left(t\right)\right)\left(1+|f|\right)\\ \le {C}_{3}{w}_{1}\left(t\right)\end{array}$ (4.4)

where ${C}_{3}=1+\frac{\left(3\sigma +2\right)\left(e+1\right)}{4\left(e-1\right)}+\left(\frac{7\left(e+1\right)}{4\left(e-1\right)}+\frac{1+{k}^{2}}{4}\right)\left({‖{u}_{0}‖}_{{H}^{1}}^{2}+\sigma {‖\rho -1‖}_{{L}^{2}}^{2}\right)$.

Then

$\begin{array}{l}{w}_{1}\left(t\right)\le {{w}^{\prime }}_{1}\left(0\right){\text{e}}^{{C}_{3}t}=\left({\zeta }^{2}\left(0\right)+1+{m}^{2}\left(0\right)\right){\text{e}}^{{C}_{3}t}\\ \text{ }\text{ }\text{ }\text{ }\text{ }\le \left(1+{‖{u}_{0x}‖}_{{L}^{\infty }}^{2}+{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}\right){\text{e}}^{{C}_{3}t}={C}_{4}{\text{e}}^{{C}_{3}t}\end{array}$ (4.5)

where ${C}_{4}=1+{‖{u}_{0x}‖}_{{L}^{\infty }}^{2}+{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}$.

From (4.3), we have

$\zeta \left(0\right)\zeta \left(t\right)\le {w}_{1}\left(t\right),\text{\hspace{0.17em}}|\zeta \left(0\right)|m\left(t\right)\le {w}_{1}\left(t\right)$ (4.6)

then

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|=|m\left(t\right)|\le \frac{{w}_{1}\left(t\right)}{|\zeta \left(0\right)|}\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)}{C}_{4}{\text{e}}^{{C}_{3}t},t\in \left[0,T\right)$

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 ${w}_{2}\left( t \right)$

$\begin{array}{l}{{w}^{\prime }}_{2}\left(t\right)=\frac{2{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}m\left(t\right)\left(\left(\sigma -1\right){\zeta }^{2}\left(t\right)+\frac{\sigma -1}{2}{m}^{2}\left(t\right)-\lambda m\left(t\right)+f+\frac{\sigma }{2}\right)\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\le \frac{{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}\left(1+{m}^{2}\left(t\right)\right)\left(|f|+\frac{\sigma }{2}\right)\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\le \frac{{\zeta }^{\sigma }\left(0\right)}{{\zeta }^{\sigma }\left(t\right)}\left(1+{m}^{2}\left(t\right)\right)\left(|f|+1\right)\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\le {C}_{3}{w}_{2}\left(t\right)\end{array}$ (4.8)

then

$\begin{array}{c}{w}_{2}\left(t\right)\le {w}_{2}\left(0\right){\text{e}}^{{C}_{3}t}=\left({\zeta }^{2}\left(0\right)+1+{m}^{2}\left(0\right)\right){\text{e}}^{{C}_{3}t}\\ \le \left(1+{‖{u}_{0x}‖}_{{L}^{\infty }}^{2}+{‖{\rho }_{0}‖}_{{L}^{\infty }}^{2}\right){\text{e}}^{{C}_{3}t}\\ ={C}_{4}{\text{e}}^{{C}_{3}t}\end{array}$ (4.9)

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

$\begin{array}{l}\frac{{w}_{2}\left(t\right)}{{\zeta }^{\sigma }\left(0\right)}={\left({\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}\right)}^{\frac{2}{\sigma }}+{\left(\frac{{\left(1+{m}^{2}\right)}^{\frac{2-\sigma }{2}}}{{\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}}\right)}^{\frac{2}{2-\sigma }}\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \frac{\sigma }{2}{\left({\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}\right)}^{\frac{2}{\sigma }}+\frac{2-\sigma }{2}{\left(\frac{{\left(1+{m}^{2}\right)}^{\frac{2-\sigma }{2}}}{{\zeta }^{\frac{\sigma \left(2-\sigma \right)}{2}}}\right)}^{\frac{2}{2-\sigma }}\\ \text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\ge {\left(1+{m}^{2}\right)}^{\frac{2-\sigma }{2}}\ge {|m\left(t\right)|}^{2-\sigma }\end{array}$

Hence

$|\underset{x\in S}{\mathrm{inf}}{u}_{x}\left(t,x\right)|\le {\left(\frac{{w}_{2}\left(t\right)}{|{\zeta }^{\sigma }\left(0\right)|}\right)}^{\frac{1}{2-\sigma }}\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_{4}^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{{C}_{3}t}{2-\sigma }},t\in \left[0,T\right)$

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

Similarly,

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

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

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

From (3.20) and (4.8), we obtain ${{\stackrel{¯}{w}}^{\prime }}_{1}\left(t\right)\le {C}_{3}{\stackrel{¯}{w}}_{1}\left(t\right)$, then ${\stackrel{¯}{w}}_{1}\left(t\right)\le {C}_{4}{\text{e}}^{{C}_{3}t}$.

Similarly, we get

$\frac{{\stackrel{¯}{w}}_{1}\left(t\right)}{{\stackrel{¯}{\zeta }}^{\sigma }\left(0\right)}\ge {|\stackrel{¯}{m}\left(t\right)|}^{2-\sigma }$

then

$|\underset{x\in S}{\mathrm{sup}}{u}_{x}\left(t,x\right)|\le {\left(\frac{{\stackrel{¯}{w}}_{1}\left(t\right)}{|{\stackrel{¯}{\zeta }}^{\sigma }\left(0\right)|}\right)}^{\frac{1}{2-\sigma }}\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}^{\frac{\sigma }{2-\sigma }}\left(x\right)}{C}_{4}^{\frac{1}{2-\sigma }}{\text{e}}^{\frac{{C}_{3}t}{2-\sigma }},t\in \left[0,T\right)$

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

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

From (3.20) and (4.4), we have ${{\stackrel{¯}{w}}^{\prime }}_{2}\left(t\right)\le {C}_{3}{\stackrel{¯}{w}}_{2}\left(t\right)$, then ${\stackrel{¯}{w}}_{2}\left(t\right)\le {C}_{4}{\text{e}}^{{C}_{3}t}$.

Hence

$|\underset{x\in S}{\mathrm{sup}}{u}_{x}\left(t,x\right)|=|\stackrel{¯}{m}\left(t\right)|\le \frac{{\stackrel{¯}{w}}_{2}\left(t\right)}{|\stackrel{¯}{\zeta }\left(0\right)|}\le \frac{1}{\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)}{C}_{4}{\text{e}}^{{C}_{3}t},t\in \left[0,T\right)$

Theorem 4.2 Let $0<\sigma <2$, $\left({u}_{0},{\rho }_{0}-1\right)\in {H}^{s}\left(S\right)×{H}^{s-1}\left(S\right)$ 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 $\underset{x\in S}{\mathrm{inf}}{\rho }_{0}\left(x\right)>0$, 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

${{\int }_{0}^{T}|{u}_{x}\left(t,x\right)|}_{{L}^{\infty }}\text{d}t=\infty$ (4.12)

The assumptions and the Theorem 4.1 show

$|{u}_{x}\left(t,x\right)|<\infty$

For all $\left(t,x\right)\in \left[0,T\right)×S$, 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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