\documentclass[reqno]{amsart}
\usepackage{hyperref}


\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 123, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/123\hfil Two-component Camassa-Holm system]
{Cauchy problem for a generalized weakly dissipative periodic two-component \\
Camassa-Holm system}

\author[W. Chen,  L. Tian, X. Deng \hfil EJDE-2014/123\hfilneg]
{Wenxia Chen, Lixin Tian, Xiaoyan Deng}  % in alphabetical order

\address{Wenxia Chen \newline
Nonlinear Scientific Research Center, Faculty of Science,
Jiangsu University, Zhenjiang, Jiangsu 212013, China}
\email{swp@ujs.edu.cn}

\address{Lixin Tian \newline
Nonlinear Scientific Research Center, Faculty of Science,
Jiangsu University, Zhenjiang, Jiangsu 212013, China}

\address{Xiaoyan Deng \newline
Nonlinear Scientific Research Center, Faculty of Science,
Jiangsu University, Zhenjiang, Jiangsu 212013, China}

\thanks{Submitted September 27, 2013. Published May 14, 2014.}
\subjclass[2000]{35B44, 35B30, 35G25}
\keywords{Wave-breaking; weakly dissipative; blow-up rate;
\hfill\break\indent periodic two-component Camassa-Holm equation; 
 global solution}

\begin{abstract}
 In this article, we study a generalized weakly dissipative periodic
 two-component Camassa-Holm system. We show that this system can exhibit the
 wave-breaking phenomenon and determine the exact blow-up rate of strong solution to
 the system. In addition, we establish a sufficient condition for having a global 
 solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks



\section{Introduction}


In recent years, the Camassa-Holm equation \cite{c1},
\begin{equation} \label{e1.1}
u_t-u_{txx}+3uu_x=2u_xu_{xx}+uu_{xxx}, \quad t > 0, x \in\mathbb{R}
\end{equation}
which models the propagation of shallow water waves has attracted
considerable attention from a large number of researchers, and two
remarkable properties of \eqref{e1.1} were found.
The first one is that the equation
possesses the solutions in the form of peaked solitons or
`peakons' \cite{c1,c5}. The peakon $u(t,x) = ce^{ - | {x - ct}|}$, $c \neq 0$
is smooth except at its crest and the tallest
among all waves of the fixed energy. It is a feature observed for the
traveling waves of largest amplitude which solves the governing
equations for water waves \cite{c6,c7,t2,z1}.
The other remarkable property is that the
equation has breaking waves \cite{c1,c8};
 that is, the solution remains
bounded while its slope becomes unbounded in finite time. After wave
breaking the solutions can be continued uniquely as either global
conservative \cite{b1} or global dissipative solutions \cite{b2}.

The Camassa-Holm equation also admits many integrable multicomponent
generalizations. The most popular one is
\begin{equation} \label{e1.2}
\begin{gathered}
 m_t - Au_x + um_x + 2u_x m + \rho \rho _x = 0 \\
 \rho _t + (\rho u)_x = 0 \\
 m = u - u_{xx}
 \end{gathered}
\end{equation}
Notice that the C-H equation can be obtained via the obvious
reduction $\rho \equiv 0$ and $A = 0$. System \eqref{e1.2} was derived
in \cite{s1}, where $\rho (t,x)$ is related to the free surface
elevation from the equilibrium (or scalar density), and $A \ge 0$
characterizes a linear underlying shear flow. Recently,
Constantin-Ivanov \cite{c9} and Ivanov \cite{i1} established a rigorous
justification of the derivation of system \eqref{e1.2}. Mathematical
properties of the system have been also studied further in many
works, for example \cite{a1,c3,c4,e1,f1,g4,h2,p1,t1}.
Chen, Liu and Zhang \cite{c3} established a
reciprocal transformation between the two-component Camassa-Holm
system and the first negative flow of the AKNS hierarchy. Escher,
Lechtenfeld, and Yin \cite{e1} investigated local well-posedness for the
two-component Camassa-Holm system with initial data $(u_0 ,\rho _0 -
1) \in H^s\times H^{s - 1}$ with $s \ge 2$ by applying Kato's theory
\cite{k1} and provided some precise blow-up scenarios for strong
solutions to the system. The local well-posedness is improved by Gui
and Liu \cite{g5} to the Besov Spaces (especially in the Sobolev space
$H^s\times H^{s - 1}$ with $s > 3/2)$, and they showed that
the finite time blow-up is determined by either the slope of the
first component $u$ or the slope of the second component $\rho$
\cite{c5,e1}. The blow-up criterion is made more precise in 
\cite{l1} where Liu and Zhang showed that the wave breaking in finite time
only depends on the slope of $u$. This blow-up criterion is
improved to the lowest Sobolev spaces
$H^s\times H^{s - 1}$ with $s > 3/2$ \cite{g4}.

In general, it is difficult to avoid energy dissipation mechanisms
in a real world. We are interested in the effect of the weakly
dissipative term on the two-component Camassa-Holm equation. Wu,
Escher and Yin have investigated the blow-up phenomena, the blow-up rate
of the strong solutions of the weakly dissipative CH equation \cite{w2}
and DP equation \cite{w1}. Inspired by the above results, in this paper,
we investigate the following generalized weakly dissipative
two-component Camassa-Holm system
\begin{equation} \label{e1.3}
\begin{gathered}
 u_t - u_{txx} - Au_x + 3uu_x - \sigma (2u_x u_{xx} + uu_{xxx} ) + \lambda
(u - u_{xx} ) + \rho \rho _x = 0, \\
 t > 0,\; x \in\mathbb{R},\\
 \rho _t + (\rho u)_x = 0, \quad  t > 0,\; x \in\mathbb{R}, \\
 u(0,x) = u_0 (x), \rho (0,x) = \rho _0 (x), \quad x \in\mathbb{R},\\
  u(t,x) = u(t,x+1),\rho(t,x) = \rho(t,x+1), \quad t\geq0,\; x \in\mathbb{R},
\end{gathered}
\end{equation}
or equivalently,
\begin{equation} \label{e1.4}
\begin{gathered}
 m_t - Au_x + \sigma (um_x + 2u_x m) + 3(1 - \sigma )uu_x + \lambda m + \rho
\rho _x = 0, \\
 \rho _t + (\rho u)_x = 0, \\
 m = u - u_{xx}, \\
\end{gathered}
\end{equation}
where $\lambda m = \lambda (I-\partial_{xx})u$ is the weakly
dissipative term, $\lambda \ge 0$ and $A$ are constants, and $\sigma $
is a new free parameter. When $A = 0$, $\lambda =0$ and $\rho = 1$,
Guan and Yin have obtained a new result of the existence of the
strong solution and some new blow-up results \cite{g1}. Meanwhile,
they have proved the global existence of the weak solution about the
two-component CH equation \cite{g2}. Henry investigates
the infinite propagation speed of the solution for a two-component
CH equation \cite{h1}.

Similar to \cite{c9,e1}, we can use the method of Besov spaces together
with the transport equation theory to show that system \eqref{e1.4} is
locally well-posedness in $H^s\times H^{s - 1}$ with
$s > 3/2$. The two equations for $u$ and $\rho $ are of a
transport structure $\partial _t f + v\partial _x f = g$. It is well
known that most of the available estimates require $v$ to have some
level of regularity. Roughly speaking, the regularity of the initial
data is expected to be preserved as soon as $v$ belongs to
$L^1(0,T;Lip)$. More specially, $u$ and $\rho $ are ``transported''
along directions of $\sigma u$ and $u$ respectively. Then, the
solution can be estimated in a Gronwall way involving $\| {u_x
} \|_{L^\infty } $. Hence, one can use these estimates to
derive a criterion which says if
 $\int_0^T {\| {u_x (\tau )}\|_{L^\infty } d\tau } < \infty $,
then solutions can be
extended further in time. Compared with the result in \cite{c2}, we find
that the equation \eqref{e1.4} has the same blow-up rate when the blow-up
occurs. This fact shows that the blow-up rate of equation \eqref{e1.4} is
not affected by the weakly dissipative term. But the occurrence of
blow-up of equation \eqref{e1.4} is affected by the dissipative parameter
$\lambda$.

The basic elementary framework is as follows. Section 2 gives the
local well-posedness of system \eqref{e1.4} and a wave-breaking
criterion, which implies that the wave breaking only depends on the
slope of $u$, not the slope of $\rho$. Section 3 improves the
blow-up criterion with a more precise conditions. Section 4
determine the exact blow-up rate of strong solutions of system
\eqref{e1.4}. Finally, section 5 provides a sufficient condition for global
solutions.

\subsection*{Notation}
 Throughout this paper, we identity periodic function spaces
over the unit $S$ in $\mathbb{R}^2$, i.e. $S=R/Z$.

\section{Formation of singularities for $\sigma \neq 0$}

We consider the following generalized weakly dissipative two -
component Camassa - Holm system:
\begin{equation} \label{e2.1}
\begin{gathered}
 u_t - u_{txx} - Au_x + 3uu_x - \sigma (2u_x u_{xx} + uu_{xxx} ) + \lambda
(u - u_{xx} ) + \rho \rho _x = 0, \\
 t > 0,\;x \in\mathbb{R},\\
 \rho _t + (\rho u)_x = 0, \quad t > 0,\; x \in\mathbb{R},\\
 u(0,x) = u_0 (x), \quad \rho (0,x) = \rho _0 (x),\\
  u(t,x) = u(t,x+1),\quad \rho(t,x) = \rho(t,x+1),
\end{gathered}
\end{equation}
 where $\lambda \ge 0$ and $A$ are constants, and $\sigma$ is a
new free parameter.

System \eqref{e2.1} can be written in the  ``transport'' form
\begin{equation} \label{e2.2}
\begin{gathered}
u_t + \sigma uu_x = - \partial _x G \ast
( -Au + \frac{3 - \sigma}{2}u^2 + \frac{\sigma}{2}u_x^2 + \frac{1}{2}\rho^2)
- \lambda u \quad t > 0,\; x \in\mathbb{R}\\
\rho _t + (\rho u)_x = 0 \quad t > 0,\; x \in\mathbb{R} \\
  u(0,x) = u_0 (x), \rho (0,x) = \rho _0 (x),\quad x \in\mathbb{R}\\
  u(t,x) = u(t,x+1),\rho(t,x) = \rho(t,x+1),\quad t\geq 0,\; x \in\mathbb{R}
\end{gathered}
\end{equation}
where $G(x): = \frac{\cosh(x-[x]-\frac{1}{2})}{2\sinh(1/2)}$,
$x \in S$, and $(1-\partial _x^2)^{-1} f=G \ast f$
for all $f \in L^2 (S)$.

Applying the transport equation theory combined with the method of Besov
spaces, one may follow the similar argument as in \cite{g5} to obtain the
following local well-posedness result for the system \eqref{e2.1}. The
proof is very similar to that of \cite[Theorem 1.1]{g5} and  is
omitted.

\begin{theorem} \label{thm2.1}
Assume $( u_0, \rho_0-1) \in H^s(S) \times H^{s-1}(S)$ with
$s > 3/2$, then there exist a maximal time
$T = T (\|(u_0, \rho_0-1)\|_{H^s \times H^{s-1}}) >0$ and a unique
solution $( u, \rho -1)$ of equation \eqref{e2.1} in $C([0,T); H^s
\times H^{s-1}) \cap C^1([0,T); H^{s-1} \times H^{s-2})$
with initial data $( u_0, \rho_0)$. Moreover, the solution depends
continuously on the initial data, and $T$ is independent of $s$.
\end{theorem}

\begin{lemma}[\cite{p1}] \label{lem2.1}
Let $0 < s < 1$. Suppose
that $f_0 \in H^s$, $g \in L^1([0,T];H^s)$,
$v, v_x \in L^1([0,T];L^\infty)$, and that
$f \in L^\infty([0,T];H^s) \cap C([0,T); S')$
solves the one-dimensional linear transport equation
\begin{gather*}
\partial _t f + v\partial _x f = g\\
f(0,x) = f_0(x) \\
\end{gather*}
then $f \in C( [0,T];H^s)$. More precisely, there exists a
constant $C$ depending only on $s$ such that
\begin{equation*}
\|f(t)\|_{H^s} \leq \|f_0\|_{H^s} +
C\Big(\int_0^t{\|g(\tau)\|_{H^s} d\tau} +
\int_0^t{\|f(\tau)\|_{H^s} V'(\tau) d\tau}\Big),
\end{equation*}
then
\begin{equation*}
\|f(t)\|_{H^s} \leq e^{CV(t)}(\|f_0\|_{H^s} +
C\int_0^t{\|g(\tau)\|_{H^s} d\tau}),
\end{equation*}
where $V(t) = \int_0^t{(\|v(\tau)\|_{L^\infty} +
\|v_x(\tau)\|_{L^\infty})}d\tau$.
\end{lemma}

We may use \cite[Lemma 2.1]{g4} to handle the regularity
propagation of solutions to \eqref{e2.1}. In addition, Lemma \ref{lem2.1} was proved
using the Littlewood-Paley analysis for the transport equation and
Moser-type estimates. Using this result and performing the same
argument as in \cite{g4}, we can obtain the following blow-up criterion.

\begin{theorem} \label{thm2.2}
Let $\sigma \neq 0$, $( u, \rho)$ be the solution of \eqref{e2.1} with initial data
 $( u_0, \rho_0-1) \in H^s(S) \times H^{s-1}(S)$ with
$s > 3/2$, and $T$ be the maximal time of existence. Then
\begin{equation} \label{e2.3}
T < \infty \Rightarrow \int_0^t{\|u_x(\tau)\|_{L^\infty} d\tau} = \infty.
\end{equation}
\end{theorem}

Regarding the finite time blow-up, we consider the trajectory
equation of the system \eqref{e2.1},
\begin{equation} \label{e2.4}
\begin{gathered}
\frac{dq(t,x)}{dt} = u(t,q(t,x)), \quad t \in [0,T) \\
q(0,x) = x, \quad x \in S ,
\end{gathered}
\end{equation}
where $u \in C^1([0,T);H^{s - 1})$ is the first component
of the solution $(u,\rho )$ to \eqref{e2.1} with initial data
$(u_0 ,\rho _0 ) \in H^s(S)\times H^{s - 1}(S)$ with $s > 3/2$, and
 $T >0$ is the maximal time of the existence. Applying Theorem \ref{thm2.1}, we
know that $q(t, \cdot ):S \to S$ is the diffeomorphism for every
$t\in [0,T)$, and
\begin{equation} \label{e2.5}
 q_x(t,x) = \exp \Big(\int_0^t{u_x(\tau,q(\tau,x))d\tau}\Big) >0,
\quad \forall(t,x) \in [0,T)\times S.
\end{equation}
Hence, the $L^\infty$-norm of any function $v(t,\cdot)\in
L^\infty, t \in [0,T)$ is preserved under the diffeomorphism
$q(t,\cdot)$ with $t \in [0,T)$; that is,
 $\|v(t,\cdot)\|_{L^\infty} = \|v(t,q(t,\cdot))\|_{L^\infty}$.

\begin{lemma}[\cite{c8}] \label{lem2.2}
 Let $T > 0$  and $v \in C^1([0,T);H^1(R))$, then for every $t \in [0,T)$,
 there exists at least one point $\xi (t) \in\mathbb{R}$ with
$m(t): = \inf_{x \in\mathbb{R}} [ {v_x (t,x)} ] = v_x (t,\xi (t))$.
The function $m(t)$ is absolutely continuous on $(0,T)$ with
\[
\frac {dm(t)}{dt} = v_{tx}(t,\xi(t)) \quad \text{a.e.  on } (0,T).
\]
\end{lemma}

\begin{lemma} \label{lem2.3}
Assume $(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$ with $s > 3/2$,
and $(u,\rho )$ is the solution of system \eqref{e2.1}, then
$\| {(u,\rho - 1)} \|_{H^1\times L^2}^2 \le \| {(u_0 ,\rho _0 - 1)}
\|_{H^1\times L^2}^2 $.
\end{lemma}

\begin{proof}
 Multiplying the first equation in \eqref{e2.1} by
$u$ and using integration by parts gives
\[
\frac{d}{dt}\int_S {(u^2 + u_x ^2)dx}
+ 2\lambda \int_S {(u^2 + u_x ^2)dx} + 2\int_S {u\rho \rho _x dx} = 0
\]
Rewriting the second equation in \eqref{e2.1} in the form
$(\rho - 1)_t + \rho _x u + \rho u_x = 0$, and multiplying by
$(\rho - 1)$ and using integration by parts, we have
\[
\frac{d}{dt}\int_S {(\rho - 1)^2dx} + 2\int_S {u\rho \rho _x dx} -
2\int_S {u\rho _x dx} + 2\int_S {u_x \rho ^2dx} - 2\int_S {u_x \rho
dx} = 0.
\]
Combining the above equalities, we have
\begin{gather*}
\frac{d}{dt}\int_S {(u^2 + u_x ^2 + (\rho - 1)^2)dx} + 2\lambda
\int_S {(u^2 + u_x ^2)dx} = 0,\\
\frac{d}{dt}\int_S {( {u^2 + u_x ^2 + (\rho - 1)^2 + 2\lambda
\int_0^t {(u^2 + u_x ^2)d\tau } } )dx} = 0.
\end{gather*}
So we have
\begin{align*}
&\int_S {( {u^2 + u_x ^2 + (\rho - 1)^2 + 2\lambda \int_0^t
{(u^2 + u_x ^2)d\tau } } )dx} \\
&= \int_S {(u_0 ^2 + u_{0x} ^2 +
(\rho _0 - 1)^2)dx} = \| {(u_0 ,\rho _0 - 1)}
\|_{H^1\times L^2}^2.
\end{align*}
Since $2\lambda \int_0^t {(u^2 + u_x ^2) d\tau } \ge 0$,
we obtain
\[
\| {(u,\rho - 1)} \|_{H^1\times L^2}^2 = \int_S {(u^2 +
u_x ^2 + (\rho - 1)^2)dx} \le \| {(u_0 ,\rho _0 - 1)}
\|_{H^1\times L^2}^2 .
\]
The proof is complete.
\end{proof}

\begin{lemma}[\cite{y1}] \label{lem2.4}
(1) For all $f \in H^1(S)$, we have
\begin{equation*}
\max_{x\in [0,1]} f^2(x) \leq \frac{e+1}{2(e-1)}\| f \|_1^2,
\end{equation*}
where  $\frac{ e+1}{ 2(e-1)}$ is the best constant.

(2) For all $f \in H^3(S)$, we have
\begin{equation*}
\max_{x\in [0,1]} f^2(x) \leq c \| {f} \|_1^2,
\end{equation*}
where the possible best constant $c \in (1,\frac{13}{12} ]$, and the best
constant is $\frac{ e+1}{2(e-1)}$.
\end{lemma}

\begin{lemma} \label{lem2.5}
If $ f \in H^3(S)$, then
 \begin{equation*}
\max_{x\in [0,1]} f_x^2(x) \leq \frac{1}{12} \| {f} \|_{H^2(S)}^2.
\end{equation*}
\end{lemma}

\begin{proof}
 From \cite[Theorem 2.1]{y1},
 the Fourier expansion of $f(x)$  can be written as
\begin{equation*}
f(x) = \frac{a_0}{2} + \sum _{n=1}^\infty a_n \cos(2\pi nx).
\end{equation*}
Then
\begin{equation*}
f_x(x) = - \sum _{n=1}^\infty (2n \pi a_n \sin(2\pi nx)).
\end{equation*}
Using that $\sum _{n=1}^\infty 1/ n^2=\pi^2/6$, we have
\begin{align*}
\max_{x\in S}f_x^2(x)
&\leq \Big(\sum _{n=1}^\infty | 2n \pi a_n |\Big)^2\\
&=\Big(\sum _{n=1}^\infty (2n \pi)^2|  a_n | \frac{1}{2n \pi}\Big)^2\\
&\leq \sum _{n=1}^\infty((2n \pi)^2|  a_n |)^2\sum _{n=1}^\infty
 (\frac{1}{2n \pi})^2 \\
&\leq\frac{1}{24}\sum _{n=1}^\infty(16n^4\pi^4a_n^2)\\
&=\frac{1}{12}\sum _{n=1}^\infty(8n^4\pi^4a_n^2)\\
&=\frac{1}{12}\int_S{f_{xx}^2 dx} \leq \frac{1}{12} \| {f} \|_{H^2(S)}^2.
\end{align*}
The proof is complete.
\end{proof}

Applying the above lemmas and the method of characteristics, we may carry
out the estimates along the characteristics $q(t,x)$ which captures
$\sup _{x \in S} u_x (t,x)$ and $\inf_{x \in S} u_x (t,x)$.

\begin{lemma} \label{lem2.6}
Let $\sigma \ne 0$ and $(u,\rho )$
be the solution of \eqref{e2.1} with initial data
$(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$, $s> 3/2$, and $T$
be the maximal time of existence.

(1) When $\sigma > 0$, we have
\begin{equation} \label{eq9}
\sup _{x \in S} u_x (t,x) \le \|
{u_{0x} } \|_{L^\infty } + \sqrt {\frac{\lambda ^2}{\sigma ^2}
+ \frac{\| {\rho _0 } \|_{L^\infty }^2 + C_1^2 }{\sigma}};
\end{equation}

(2) When $\sigma < 0$, we have
\begin{equation} \label{e2.7}
\inf _{x \in S} u_x (t,x) \ge - \| {u_{0x} }
\|_{L^\infty } - \sqrt {\frac{\lambda ^2}{\sigma ^2} -
\frac{C_2^2 }{\sigma }};
\end{equation}
where the constants are defined as follows:
\begin{gather} \label{e2.8}
C_1 = \sqrt {\frac{5(e+1)}{2(e-1)}+(\frac{1+A^2}{2}+\frac{(e+1)
| 3-\sigma |}{e-1})\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2},\\
 \label{e2.9}
C_2 = = \sqrt {\frac{5(e+1)}{2(e-1)}+(\frac{A^2}{2}
+\frac{(5-\sigma)e+3-\sigma}{2(e-1)})\| {(u_0 ,\rho _0 - 1)}
\|_{H^1\times L^2}^2}.
\end{gather}
\end{lemma}

\begin{proof}
The local well-posedness theorem and a
density argument imply that it suffices to prove the desired
estimates for $s \ge 3$. Thus, we take $s = 3$ in the proof. Here we
may assume that $u_0 \ne 0$. Otherwise, the results become trivial.

Differentiating the first equation in \eqref{e2.2} with respect
to $x$ and using the identity $ - \partial _x^2 G * f = f - G * f$,
we have
\begin{equation} \label{e2.10}
u_{tx} + \sigma uu_{xx} + \frac{\sigma }{2}u_x^2 = \frac{1}{2}\rho
^2 + \frac{3 - \sigma }{2}u^2 + A\partial _x^2 G * u - G *
(\frac{\sigma }{2}u_x^2 + \frac{3 - \sigma }{2}u^2 + \frac{1}{2}\rho
^2) - \lambda u_x.
\end{equation}

(1) When $\sigma > 0$, using Lemma \ref{lem2.2} and the fact that
\[
\sup _{x \in S} [v_x (t,x)] = - \inf _{x \in S} [ - v_x (t,x)],
\]
we can consider $\bar {m}(t)$ and $\eta (t)$ as
\begin{equation} \label{e2.11}
\bar {m}(t): = u_x (t,\eta (t)) = \sup _{x \in S}
(u_x (t,x)), \quad t \in [0,T).
\end{equation}
This gives
\begin{equation} \label{e2.12}
u_{xx} (t,\eta (t)) = 0   \quad \text{a.e. on } t \in [0,T)
\end{equation}
Take the trajectory $q(t,x)$ defined in \eqref{e2.4}.
We know that $q(t, \cdot ):S \to S$ is a diffeomorphism for every
$t \in [0,T)$,then there exists $x_1 (t) \in S$ such that
\begin{equation} \label{e2.13}
q(t,x_1 (t)) = \eta (t),\quad  t \in [0,T).
\end{equation}
Let
\begin{equation} \label{e2.14}
\bar {\zeta }(t) = \rho (t,q(t,x_1 )), \quad t \in [0,T).
\end{equation}
Then along the trajectory $q(t,x_1 (t))$, equation \eqref{e2.10}
and the second equation of \eqref{e2.1} become
\begin{equation} \label{e2.15}
\begin{gathered}
 \bar{m}'(t) = - \frac{\sigma }{2}\bar {m}^2(t) - \lambda \bar {m}(t) +
\frac{1}{2}\bar {\zeta }^2(t) + f(t,q(t,x_1 )) \\
 \bar{\zeta }'(t) = - \bar {\zeta }(t)\bar {m}(t),
\end{gathered}
\end{equation}
 where
\begin{equation} \label{e2.16}
f = \frac{3 - \sigma }{2}u^2 + A\partial _x^2 G * u - G *
\Big(\frac{\sigma }{2}u_x^2 + \frac{3 - \sigma }{2}u^2 +
\frac{1}{2}\rho ^2\Big).
\end{equation}
Since $\partial _x^2 G * u = \partial _x G * \partial _x u$, we have
\begin{align*}
f &= \frac{3 - \sigma }{2}u^2 + A\partial _x G * \partial _x u - G *
(\frac{\sigma }{2}u_x^2 + \frac{3 - \sigma }{2}u^2) - \frac{1}{2}G *
1 - G * (\rho - 1) \\
&\quad - \frac{1}{2}G * (\rho - 1)^2 \\
&\leq \frac{3 - \sigma }{2}u^2 + A\partial _x G * \partial _x u - G *
( \frac{3 - \sigma }{2}u^2) - \frac{1}{2}G * 1 - G * (\rho - 1) \\
&\le \frac{| 3 - \sigma | }{2}u^2 + A| {\partial _x G * \partial _x u}
| + | {G * (\frac{3 - \sigma
}{2}u^2)} | + \frac{1}{2}| G * 1 | + | {G * (\rho - 1)} |.
\end{align*}
Based on the following formulas:
\begin{gather*}
\frac{| 3 - \sigma | }{2}u^2 \leq \frac{| 3 - \sigma | }{2} \cdot
\frac{e+1}{2(e-1)}\| {u} \|_{H^1}^2,
\\
A| {\partial _x G * \partial _x u} | \le A\| {G_x }
\|_{L^2} \| {u_x } \|_{L^2}
\leq \frac{e+1}{2(e-1)}+\frac{1}{4}A^2\| {u_x }\|_{L^2}^2,
\\
| {G * (\frac{\sigma }{2}u_x^2}) | \le \| {G_x }
\|_{L^\infty} \| {\frac{\sigma }{2}u_x^2} \|_{L^1}
\leq \frac{e+1}{2(e-1)}\cdot\frac{\sigma}{2}\| {u_x }\|_{L^2}^2,
\\
| {G * (\frac{3-\sigma }{2}u^2}) | \le \| {G_x }
\|_{L^\infty} \| {\frac{3-\sigma }{2}u^2} \|_{L^1}
\leq \frac{e+1}{2(e-1)}\cdot\frac{| 3-\sigma |}{2}\| {u}\|_{L^2}^2,
\\
\frac{1}{2}| {G * 1} |  \leq \frac{1}{2}\| {G }
\|_{L^\infty} \leq \frac{e+1}{4(e-1)},
\\
| {G * (\rho - 1)} | \le \| G \|_{L^2} \|
{\rho - 1} \|_{L^2} \leq \frac{e+1}{2(e-1)}+\frac{1}{4}\|
{\rho - 1} \|_{L^2}^2,
\\
\frac{1}{2}| {G * (\rho - 1)^2} | \le \frac{1}{2}\|
G \|_{L^\infty} \| {(\rho - 1)^2} \|_{L^1} \leq
\frac{e+1}{4(e-1)}\| {\rho - 1} \|_{L^2}^2,
\end{gather*}
from the above inequalities and Lemma \ref{lem2.3} we obtain an
upper bound of $f$,
\begin{equation} \label{e2.17}
\begin{aligned}
f &\le \frac{5(e+1)}{4(e-1)}+\frac{1}{4}\| {\rho - 1}
\|_{L^2}^2+(\frac{A^2}{4}+\frac{(e+1)| 3-\sigma
|}{2(e-1)})\| {u} \|_{H^1}^2 \\
&\le \frac{5(e+1)}{4(e-1)}+(\frac{A^2+1}{4}+\frac{(e+1)| 3-\sigma
|}{2(e-1)})\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times
L^2}^2 \\
&= \frac{1}{2}C_1^2.
\end{aligned}
\end{equation}
Similarly, we obtain a lower bound of $f$,
\begin{equation} \label{e2.18}
\begin{aligned}
 - f
&\le \frac{\sigma - 3}{2}u^2 + A| {\partial _x G * \partial _x u}
| + | {G * (\frac{\sigma }{2}u_x^2 + \frac{3 - \sigma
}{2}u^2)} | + \frac{1}{2}| {G * 1} | \\
&\quad + | {G *(\rho - 1)} | + \frac{1}{2}G * (\rho - 1)^2
\\
& \le \frac{5(e+1)}{4(e-1)}+\frac{e}{2(e-1)}\| {\rho- 1} \|_{L^2}^2
+(\frac{A^2}{4}+\frac{(e+1)(| \sigma | +2 | 3-\sigma|)}{4(e-1)})\| {u} \|_{H^1}^2
\\
&\le \frac{5(e+1)}{4(e-1)}+(\frac{A^2}{4}+\frac{2e+(e+1)(| \sigma |
+2 | 3-\sigma|)}{4(e-1)}) \| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2.
\end{aligned}
\end{equation}
Combining \eqref{e2.17} and \eqref{e2.18}, we obtain
\begin{equation} \label{e2.19}
| f | \le \frac{5(e+1)}{4(e-1)}+(\frac{A^2}{4}
+\frac{2e+(e+1)(| \sigma | +2 | 3-\sigma|)}{4(e-1)})
\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2.
\end{equation}
Since $s \ge 3$, it follows that $u \in C_0^1 (S)$ and
\begin{equation} \label{e2.20}
\inf _{x \in S} u_x (t,x) \le 0_{,} \quad \sup _{x \in S} u_x (t,x) \ge 0,
 \quad t \in [0,T).
\end{equation}
 Hence, we obtain
\begin{equation} \label{e2.21}
\bar {m}(t) > 0    \quad  \text{for } t \in [0,T).
\end{equation}
From the second equation in \eqref{e2.15}, we have
\begin{gather} \label{e2.22}
\bar {\zeta }(t) = \bar {\zeta }(0)e^{ - \int_0^t {\bar {m}(\tau
)d\tau }},\\
| {\rho (t,q(t,x_1 ))} | = | {\bar {\zeta }(t)}
| \le | {\bar {\zeta }(0)} | \le \| {\rho _0 }
\|_{L^\infty }. \nonumber
\end{gather}
For any given $x \in S$, we define
\[
P_1 (t) = \bar {m}(t) - \| {u_{0x} } \|_{L^\infty } -
\sqrt {\frac{\lambda ^2}{\sigma ^2} + \frac{\| {\rho _0 }
\|_{L^\infty }^2 + C_1^2 }{\sigma }}.
\]

Notice that $P_1 (t)$ is a $C^1$-function
in $[0,T)$ and satisfies
\[
P_1 (0) = \bar {m}(0) - \| {u_{0x} } \|_{L^\infty } -
\sqrt {\frac{\lambda ^2}{\sigma ^2} + \frac{\| {\rho _0 }
\|_{L^\infty }^2 + C_1^2 }{\sigma }} \le \bar {m}(0) - \|
{u_{0x} } \|_{L^\infty } \le 0.
\]
Next, we claim that
\begin{equation} \label{e2.23}
P_1 (t) \le 0 \quad \text{for }  t \in [0,T).
\end{equation}
If not, then suppose that there is a $t_0 \in [0,T)$ such that
$P_1 (t_0 ) > 0$. Define $t_1 = \max \{t < t_0 :P_1 (t) = 0\}$, then
$P_1 (t_1 ) = 0$, $P_1 ' (t_1 ) \ge 0$. That is,
\[
\bar {m}(t_1 ) = \| {u_{0x} } \|_{L^\infty } + \sqrt
{\frac{\lambda ^2}{\sigma ^2} + \frac{\| {\rho _0 }
\|_{L^\infty }^2 + C_1^2 }{\sigma }} , \quad
{\bar{m}}'(t_1 ) = {P}'_1 (t_1 ) \ge 0.
\]
 On the other hand, we have
\begin{align*}
{\bar {m}}'(t_1 )
&= - \frac{\sigma }{2}\bar {m}^2(t_1 ) - \lambda
\bar {m}(t_1 ) + \frac{1}{2}\bar {\zeta }^2(t_1 ) + f(t_1 ,q(t_1,x_1 ))\\
& \le - \frac{\sigma }{2}\Big(\| {u_{0x} } \|_{L^\infty } + \sqrt
{\frac{\lambda ^2}{\sigma ^2} + \frac{\| {\rho _0 }
\|_{L^\infty }^2 + C_1^2 }{\sigma }} + \frac{\lambda }{\sigma
}\Big)^2 \\
&\quad + \frac{\lambda ^2}{2\sigma } + \frac{1}{2}\| {\rho _0} \|_{L^\infty }^2
+ \frac{1}{2}C_{_1 }^2 < 0.
\end{align*}
This yields a contraction. Thus, $P_1 (t) \le 0$ for $t \in [0,T)$. Since $x$
is chosen arbitrarily, we obtain \eqref{eq9}).


(2) When $\sigma < 0$ , we have a finer estimate
\begin{equation} \label{e2.24}
\begin{aligned}
 - f
&\le -A(\partial _x G * \partial _x u) +G * \frac{3 - \sigma
}{2}u^2 + \frac{1}{2}(G * 1) + G *
(\rho - 1) + \frac{1}{2}G * (\rho - 1)^2
\\
&\le A| \partial _x G * \partial _x u  | + | G * \frac{3 - \sigma
}{2}u^2 | + \frac{1}{2}| G * 1 | + | G *
(\rho - 1) | + \frac{1}{2}| G * (\rho - 1)^2|
\\
&\le \frac{5(e+1)}{4(e-1)}+(\frac{A^2}{4}+\frac{(5-\sigma)e+3-\sigma}{4(e-1)})
\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2=\frac{1}{2}C_2^2.
\end{aligned}
\end{equation}
We consider the functions $m(t)$ and $\xi (t)$ in Lemma \ref{lem2.2},
\begin{equation} \label{e2.25}
m(t): = \inf _{x \in S} [ {u_x (t,x)} ], \quad t \in [0,T)
\end{equation}
Then $u_{xx} (t,\xi (t)) = 0$  a.e. on $t \in [0,T)$.
Choose $x_2 (t) \in S$, such that $q(t,x_2 (t)) = \xi (t)$,
$t \in [0,T)$.
Let $\zeta (t) = \rho (t,q(t,x_2 )), t \in [0,T)$. Along
the trajectory $q(t,x_2 )$, equation \eqref{e2.10} and the second equation
of \eqref{e2.1} become
\begin{gather*}
 {m}'(t) = - \frac{\sigma }{2}m^2(t) - \lambda m(t) + \frac{1}{2}\zeta ^2(t)
+ f(t,q(t,x_2 )) \\
 {\zeta }'(t) = - \zeta (t)m(t).
 \end{gather*}
Let $P_2 (t) =  m(t) + \| {u_{0x} }\|_{L^\infty }
+ \sqrt {\frac{\lambda ^2}{\sigma ^2} -\frac{C_2^2 }{\sigma }},
 \quad \forall x \in\mathbb{R}$. Then $P_2 (t)$ is a
$C^1$-function in $[0,T)$ and satisfies
\[
P_2 (0) = m(0) + \| {u_{0x} } \|_{L^\infty } + \sqrt
{\frac{\lambda ^2}{\sigma ^2} - \frac{C_2^2 }{\sigma }} \ge m(0) +
\| {u_{0x} } \|_{L^\infty } \ge 0.
\]
Now we claim that
\begin{equation} \label{e2.26}
P_2 (t) \ge 0\quad  \text{for } t \in [0,T).
\end{equation}
 Assume that there is a $\bar {t}_0 \in [0,T)$ such that
 $P_2 (\bar {t}_0 ) < 0$. Define
$t_2 = \max \{t < \bar {t}_0 :P_2 (t) = 0\}$, then $P_2 (t_2 ) = 0$,
$P_2 ' (t_2 ) \le 0$. That is,
\[
m(t_2 ) = - \| {u_{0x} } \|_{L^\infty } - \sqrt
{\frac{\lambda ^2}{\sigma ^2} - \frac{C_2^2 }{\sigma }}, \quad
{m}'(t_2 ) = {P}'_2 (t_2 ) \le 0.
\]
In addition, we have
\begin{align*}
{m}'(t_2 )
&= - \frac{\sigma }{2}m^2(t_2 ) - \lambda m(t_2 ) +
\frac{1}{2}\zeta ^2(t_2 ) + f(t_2 ,q(t_2 ,x_2 ))\\
& \ge - \frac{\sigma }{2}( { - \| {u_{0x} } \|_{L^\infty } -
\sqrt {\frac{\lambda ^2}{\sigma ^2} - \frac{C_2^2 }{\sigma }} +
\frac{\lambda }{\sigma }} )^2 + \frac{\lambda ^2}{2\sigma } -
\frac{1}{2}C_2^2 > 0.
\end{align*}
This is a contradiction.
Then we have $P_2 (t) \ge 0$ for $t \in [0,T)$,
since $x$ is chosen arbitrarily.
\end{proof}

Now, we present the following estimates for
$\| \rho \|_{L^\infty (S)} $, if $\sigma u_x $ is bounded from below.

\begin{lemma}[\cite{c2}] \label{lem2.7}
Let $\sigma \ne 0$ and $(u,\rho )$ be the solution of \eqref{e2.1} with initial
data $(u_0 ,\rho_0 - 1) \in H^s(S)\times H^{s - 1}(S),s > 3/2$, and $T$ be
the maximal time of the existence. If there is a $M \ge 0$ such that
$\inf _{(t,x) \in [0,T)\times S} \sigma u_x \ge - M$, Then we have
following two statements.

(1) If $\sigma > 0$, then $\| {\rho (t, \cdot )}
\|_{L^\infty (S)}
\le \| {\rho _0 } \|_{L^\infty (S)} e^{Mt / \sigma }$.

(2) If $\sigma < 0$,  then $\| {\rho (t, \cdot )}
\|_{L^\infty (S)}\le \| {\rho _0 } \|_{L^\infty (S)} e^{Nt}$,

\noindent where $N = \| {u_{0x} } \|_{L^\infty } +
(C_2 / \sqrt { - \sigma } )$ and $C_2 $ is given in \eqref{e2.24}.
\end{lemma}

\begin{proof}
The proof of Lemma \ref{lem2.7} is similar to that of
\cite[Proposition 3.8]{c2}, so we omit it here.
\end{proof}

From the above results, we can get the necessary and sufficient
conditions for the blow-up of solutions.

\begin{theorem}[Wave-breaking criterion for $\sigma \ne 0$] \label{thm2.3}
 Let
$\sigma \ne 0$ and $(u,\rho )$ be the solution of \eqref{e2.1} with initial
data $(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$, $s >3/2$, and $T$
be the maximal time of existence. Then the
solution blows up in finite time if and only if
\begin{equation} \label{e2.27}
\lim _{t \to T^-} \inf _{x \in S} \sigma u_x(t,x) = -\infty.
\end{equation}
\end{theorem}

\begin{proof}
Assume that $T < \infty $ and \eqref{e2.27} is not
valid, then there is some positive number $M > 0$, such that
$\sigma u_x (t,x) \ge - M$, $\forall (t,x) \in [0,T)\times S$.
From the above lemmas, we have $| {u_x (t,x)} | \le C$, where
$C = C(A,M,\sigma ,\lambda ,\| {(u_0 ,\rho _0 - 1)}
\|_{H^s\times H^{s - 1}} )$.  Thus, Theorem \ref{thm2.2} implies
that the maximal existence time $T = \infty $, which contradicts the
assumption $T < \infty $.

On the other hand, the Sobolev embedding theorem
$H^s \hookrightarrow L^\infty $ with $s > 1/2$ implies that if
\eqref{e2.27} holds, the corresponding solution blows up in finite time.
The proof is complete.
\end{proof}

\section{Blow-up scenarios}

\begin{theorem} \label{thm3.1}
Let $\sigma > 0$ and $(u,\rho )$ be the solution of
\eqref{e2.1} with initial data $(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s -
1}(S),s> 3/2$, and $T$ be the maximal time of existence. Assume that there
is some $x_0 \in S$ such that $\rho _0 (x_0 ) = 0_{,} u_{0x} (x_0 )
= \inf _{x \in S} u_{0x} (x)$ and
\begin{equation} \label{e3.1}
\begin{aligned}
&\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2 \\
&< \Big(\frac{8e-10}{18(e-1)} - \frac{\lambda ^2}{2\sigma }\Big)\frac{4(e-1)}
{(18A^2+19)e-(18A^2+17)+(2 | 3-\sigma | + \sigma)(e+1)},
\end{aligned}
\end{equation}
then the corresponding solution to system \eqref{e2.1} blows up
in finite time in the following sense: there exists a $T$ such that
\begin{equation} \label{e3.2}
\begin{aligned}
0 < T 
&\le \frac{2}{\sigma - \lambda } + \Big(72\sigma(e-1)(1+| u_{0x}(x_0)|)\Big)\\
&\quad \div\Big(\sigma(32e-40-324e-324A^2e+324A^2+306)
-36\lambda^2(e-1)\\
&\quad +(2 | 3-\sigma | + \sigma)(e-1)\| {(u_0 ,\rho
_0 - 1)} \|_{H^1\times L^2}^2 \Big)
\end{aligned}
\end{equation}
and that $\liminf_{t \to T^-}(\inf_{x \in S}u_x(t,x)) = -\infty$.
\end{theorem}

\begin{proof}
Here we also consider $s \geq 3$. We still
consider along the trajectory $q (t,x_2)$ defined as before. In this
way, we can write the transport equation of $\rho $ in \eqref{e2.1} along
the trajectory of $q (t,x_2)$ as
\begin{equation} \label{e3.3}
\frac {d\rho (t,\xi(t))}{dt} = -\rho (t,\xi(t))u_x (t,\xi(t)) .
\end{equation}
By the assumption, we have
\[
m(0) = u_x (0,\xi (0)) = \inf _{x \in S} u_{0x} (x)= u_{0x} (x_0 ).
\]
Choose $\xi (0) = x_0 $ and then $\rho_0 (\xi (0))
= \rho _0 (x_0 ) = 0$. Then by \eqref{e3.3}, we derive
\begin{equation} \label{e3.4}
\rho (t,\xi (t)) = 0, \quad \forall t \in [0,T).
\end{equation}
Evaluating the result at $x = \xi (t)$ and combining \eqref{e3.4}
with $u_{xx} (t,\xi (t)) = 0$, we have
\begin{equation} \label{e3.5}
\begin{aligned}
{m}'(t)
&= - \frac{\sigma }{2}m^2(t) - \lambda m(t) + \frac{3 -
\sigma }{2}u^2(t,\xi (t)) + A(G_x * u_x )(t,\xi (t))\\
&\quad - G * (\frac{\sigma }{2}u_x^2 + \frac{3 - \sigma }{2}u^2 + \frac{1}{2}\rho
^2)(t,\xi (t))\\
&= - \frac{\sigma }{2}m^2(t) - \lambda m(t) + f(t,q(t,x_2 )) \\
&= - \frac{\sigma }{2}( {m(t) + \frac{\lambda }{\sigma }} )^2
+ \frac{\lambda ^2}{2\sigma } + f(t,q(t,x_2 )).
\end{aligned}
\end{equation}
 We modify the estimates:
\begin{gather*}
A| {G_x * u_x } | \le A\| {G_x } \|_{L^2}\| {u_x } \|_{L^2}
\le \frac{1}{18}\cdot \frac{ e+1}{ 2(e-1)} +
\frac{9}{2}A^2\| {u_x } \|_{L^2}^2,
\\
| G*(\rho -1) | \leq \| {G} \|_{L^2}\| {\rho -1} \|_{L^2}
\leq \frac{1}{18}\cdot \frac{ e+1}{ 2(e-1)} +
\frac{9}{2}\| {\rho -1 }\|_{L^2}^2.
\end{gather*}
Similarly, we obtain the upper bound of $f$ as
\begin{align*}
f &\le
 \frac{10-8e}{18(e-1)} + \frac{ (18A^2+19)e
 -(18A^2+17) +(2 | 3-\sigma | + \sigma)(e+1)}{ 4(e-1)}\\
&\quad\times \| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2 : = - C_3.
\end{align*}
By  assumption \eqref{e3.1}, we obtain
$\frac{\lambda ^2}{2\sigma } - C_3 < 0$ and
\begin{equation} \label{e3.6}
{m}'(t) \le - \frac{\sigma }{2}\big( {m(t) + \frac{\lambda }{\sigma
}} \big)^2 + \frac{\lambda ^2}{2\sigma } - C_3 \le \frac{\lambda
^2}{2\sigma } - C_3 < 0, \quad t \in [0,T).
\end{equation}
So $m(t)$ is strictly decreasing in $[0,T)$. If the
solution $(u,\rho )$ of \eqref{e2.1} exists globally in time, that is,
$T =\infty $, we will show that it leads to a contradiction.

 Let $t_1 = \frac{ 2\sigma (1 +| {u_{0x} (x_0 )} |)}{ 2\sigma C_3 - \lambda
^2}$.  Integrating \eqref{e3.6} over $[0,t_1 ]$ gives
\begin{equation} \label{e3.7}
m(t_1 ) = m(0) + \int_0^{t_1 } {{m}'(t)dt} \le | {u_{0x} (x_0
)} | + (\frac{\lambda ^2}{2\sigma } - C_3 )t_1 = - 1.
\end{equation}
 For $t \in [t_1 ,T)$, we have $m(t) \le m(t_1 )\le - 1$.
From \eqref{e3.6}, we have
\begin{equation} \label{e3.8}
{m}'(t) \le - \frac{\sigma }{2}\big(
{m(t) + \frac{\lambda }{\sigma }}\big)^2.
\end{equation}
Integrating  over $[t_1 ,T)$, by \eqref{e3.7}, yields
\begin{gather*}
  -\frac{1}{m(t) + \frac{\lambda }{\sigma }} + \frac{1}{\frac{\lambda
}{\sigma } - 1} \le - \frac{1}{m(t) + \frac{\lambda }{\sigma }} +
\frac{1}{m(t_1 ) + \frac{\lambda }{\sigma }} \le - \frac{\sigma
}{2}(t - t_1 ), \quad t \in [t_1 ,T),
\\
m(t) \le
\frac{1}{\frac{\sigma
}{2}(t - t_1 ) + \frac{\sigma
}{\lambda - \sigma }} - \frac{\lambda
}{\sigma } \to - \infty, \quad \text{as } t \to t_1 +
\frac{2}{\sigma - \lambda }.
\end{gather*}
So, $T \le  t_1 + \frac{2}{\sigma - \lambda}$, which is a contradiction to
$T = \infty$. Consequently, the proofis complete.
\end{proof}

\begin{theorem} \label{thm3.2}
 Let $\sigma \ne 0$ and $(u,\rho )$ be the solution of
\eqref{e2.1} with initial data $(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s -
1}(S)$, $s> 3/2$, and $T$ be the maximal time of the existence.

(1) When $\sigma > 0$, assume that there is an $x_0 \in S$ such
that $\rho _0 (x_0 ) = 0$, $u_{0x} (x_0 ) = \inf _{x \in S} u_{0x} (x)$ and
$u_{0x} (x_0 ) < - \sqrt {\frac{\lambda ^2}{\sigma ^2} + \frac{C_1^2 }{\sigma }} -
\frac{\lambda }{\sigma }$, where $C_1 $ is defined in \eqref{e2.8}. Then
the corresponding solution to system \eqref{e2.1} blows up in finite time
in the following sense: there exists a $T_1 $ such that
\[
0 < T_1 \le - \frac{2(\lambda + \sigma u_{0x} (x_0))}
{(\lambda + \sigma u_{0x} (x_0 ))^2 - (\lambda ^2 + \sigma C_1^2 )},
\]
and
\[
\liminf_{t \to T_1^ - } \{\inf _{x \in S} u_x (t,x)\} = - \infty.
\]

 (2) When $\sigma < 0$, assume that there is some $x_0 \in S$ such
that $u_{0x} (x_0 ) > \sqrt {\frac{\lambda^2}{\sigma ^2}
- \frac{ C_2^2}{\sigma }} - \frac{\lambda}{\sigma }$, where $C_2 $
is defined in \eqref{e2.9}. Then the
corresponding solution to system \eqref{e2.1} blows up in finite time in
the following sense: there exists a $T_2 $ such that
\[
0 < T_2 \le -
\frac{2(\lambda + \sigma u_{0x} (x_0
))}{(\lambda + \sigma u_{0x} (x_0 ))^2 - (\lambda ^2 -
\sigma C_2^2 )},
\]
and
\[
\liminf_{t \to T_2^ - } \{\sup _{x \in S} u_x (t,x)\} = \infty.
\]
\end{theorem}

\begin{proof}
 (1) When $\sigma > 0$, using the upper
bound of $f$ in \eqref{e2.17} and \eqref{e3.4}, we have
\[
{m}'(t) \le - \frac{\sigma }{2}\Big(m(t) + \frac{\lambda }{\sigma
}\Big)^2 + \frac{\lambda ^2}{2\sigma } + \frac{1}{2}C_1^2 , \quad t
\in [0,T).
\]
By the assumption $m(0) = u_{0x} (x_0 ) < -
\sqrt {\frac{\lambda ^2}{\sigma ^2} + \frac{C_1^2 }{\sigma }} -
\frac{\lambda }{\sigma }$, we have that ${m}'(0) < 0$ and $m(t)$ is
strictly decreasing over $[0,T)$.
Set
$$
\delta = \frac{1}{2} - \frac{1}{\sigma
(u_{0x} (x_0 ) + \frac{\lambda }{\sigma })^2}\Big(\frac{\lambda
^2}{2\sigma } + \frac{1}{2}C_1^2 \Big)  \in (0,\frac{1}{2}).
$$
Since $m(t) < m(0) = u_{0x} (x_0 ) < -
\frac{\lambda }{\sigma }$, it holds
\[
{m}'(t) \le - \frac{\sigma }{2}\Big( {m(t) + \frac{\lambda }{\sigma
}} \Big)^2 + \frac{\lambda ^2}{2\sigma } + \frac{1}{2}C_1^2 \le -
\delta \sigma \Big(m(t) + \frac{\lambda }{\sigma }\Big)^2.
\]
By a similar argument as in the proof of Theorem \ref{thm3.1}, we obtain
\[
m(t) \le \frac{\lambda + \sigma u_{0x}
(x_0 )}{\sigma + (\delta \sigma ^2u_{0x} (x_0 ) +
\lambda \delta \sigma )t} - \frac{\lambda
}{\sigma } \to - \infty  \quad\text{as }t \to -
\frac{1}{\lambda \delta + \delta \sigma
u_{0x} (x_0 )}.
\]
Thus, we have $0 < T_1 \le -\frac{1}{\lambda \delta 
+ \delta \sigma u_{0x} (x_0)}$.

(2) when $\sigma < 0$, we consider the functions 
$\bar {m}(t)$ and $\eta (t)$ as defined in \eqref{e2.11} and take the trajectory
$q(t,x_1 )$ with $x_1 $ defined in \eqref{e2.13}, then
\begin{equation} \label{e3.9}
\begin{aligned}
{\bar {m}}'(t) 
&= - \frac{\sigma }{2}\bar {m}^2(t) - \lambda \bar
{m}(t) + \frac{1}{2}\rho ^2(t,\eta (t)) + f(t,q(t,x_1 ))\\
& \ge - \frac{\sigma }{2}\Big(\bar {m}(t) + \frac{\lambda }{\sigma }\Big)^2 +
\frac{\lambda ^2}{2\sigma } + f(t,q(t,x_1 )).
\end{aligned}
\end{equation}
From the lower bound of $f$ in \eqref{e2.24}, we obtain
$$
{\bar {m}}'(t) \ge - \frac{\sigma }{2}\Big(\bar {m}(t) +
\frac{\lambda }{\sigma }\Big)^2 + \frac{\lambda ^2}{2\sigma } -
\frac{1}{2}C_2^2, \quad t \in [0,T).
$$
 By the assumption $\bar {m}(0) \ge u_{0x} (x_0 ) >
\sqrt {\frac{\lambda ^2}{\sigma ^2} - \frac{C_2^2
}{\sigma }} - \frac{\lambda }{\sigma }$, we have that $\bar {m}'(0) > 0$
and $\bar {m}(t)$ is strictly increasing over
$[0,T)$.

Set 
$$
\theta = \frac{(\sigma u_{0x} (x_0 ) +
\lambda )^2 - (\lambda ^2 - \sigma C_2^2 )}{2(\sigma u_{0x} (x_0 ) +
\lambda )^2} \in (0,\frac{1}{2}).
$$
Since $\bar {m}(t) > \bar {m}(0) \ge u_{0x} (x_0 ) > -
\frac{\lambda }{\sigma }$, we obtain
\[
{\bar {m}}'(t) \ge - \frac{\sigma }{2}\Big(\bar {m}(t) +
\frac{\lambda }{\sigma }\Big)^2 + \frac{\lambda ^2}{2\sigma } -
\frac{1}{2}C_2^2 \ge - \theta \sigma \Big(\bar {m}(t) +
\frac{\lambda }{\sigma }\Big)^2.
\]
Similarly, we obtain
\[
\bar {m}(t) \ge \frac{\lambda + \sigma u_{0x} (x_0
)}{\sigma + (\theta \sigma ^2u_{0x} (x_0 ) + \lambda \theta \sigma
)t} - \frac{\lambda }{\sigma } \to \infty \quad 
\text{as } t \to - \frac{1}{\lambda \theta + \theta \sigma u_{0x} (x_0
)}.
\]
Therefore, $0 < T_2 \le - \frac{1}{\lambda
\theta + \theta \sigma u_{0x} (x_0 )}$. The proof is complete.
\end{proof}

\noindent\textbf{Remark.} 
If $\sigma = 3$ and $A = 0$, then all
solutions of system \eqref{e2.1} with initial data $(u_0 ,\rho _0 - 1) \in
H^s(S)\times H^{s - 1}(S)$ with $s > 3/2$  satisfying $u_0 \ne 0$
and $\rho _0 (x_0 ) = 0$ for some $x_0 \in S$, blow up in
finite time.


\section{Blow-up rate}

\begin{theorem} \label{thm4.1} 
Let $\sigma \ne 0$. If $T < \infty $ is the blow-up time
of the solution $(u,\rho )$ to \eqref{e2.1} with initial data 
$(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$, $s > 3/2$ satisfying
the assumptions of Theorem \ref{thm3.2}.  Then
\begin{gather} \label{e4.1}
\mathop {\lim}_{t \to T^-}\{\inf _{x \in
S}u_x(t,x)(T-t)\} = -\frac{2}{\sigma}, \quad \sigma > 0,\\
\label{e4.2}
\lim_{t \to T^-}\{\mathop{\sup}_{x \in S}u_x(t,x)(T-t)\}
 = -\frac{2}{\sigma},
\quad \sigma < 0.
\end{gather}
\end{theorem}

\begin{proof} 
We assume that $s=3$ to prove the theorem.

(1) when $\sigma > 0$, from \eqref{e3.5} we have
\begin{equation} \label{e4.3}
m'(t) = - \frac{\sigma }{2}\Big(m(t) + \frac{\lambda }{\sigma
}\Big)^2 + \frac{\lambda ^2}{2\sigma } + f(t,q(t,x)).
\end{equation}
 From \eqref{e2.19},  note that
\begin{equation} \label{e4.4}
M = \frac{5(e+1)}{4(e-1)}+(\frac{A^2}{4}
+\frac{2e+(e+1)(| \sigma | +2 | 3-\sigma|)}{4(e-1)})
\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2,
\end{equation}
Then
\begin{equation} \label{e4.5}
 - \frac{\sigma }{2}( {m(t) + \frac{\lambda }{\sigma }} )^2 -
\frac{\lambda ^2}{2\sigma } - M \le {m}'(t) \le - \frac{\sigma
}{2}( {m(t) + \frac{\lambda }{\sigma }} )^2 +
\frac{\lambda ^2}{2\sigma } + M.
\end{equation}

Choose $\varepsilon \in (0,\frac{\sigma }{2})$, since
 $\lim _{t \to T^-}\big(m(t)+\frac{\lambda}{\sigma}\big) = -\infty$,
there is some $t_0 \in (0,T)$, such that
$m(t_0)+\frac{\lambda}{\sigma} < 0$   and
 $\big(m(t_0)+\frac{\lambda}{\sigma}\big)^2 
> \frac{1}{\varepsilon}\big(\frac{\lambda^2}{2\sigma}+M\big)$. 
Since $m$ is locally Lipschitz, it follows that $m$ is absolutely
continuous. We deduce that $m$ is decreasing on  $[t_0,T)$ and
\begin{equation} \label{e4.6}
\Big(m(t)+\frac{\lambda}{\sigma}\Big)^2 >
\frac{1}{\varepsilon}\Big(\frac{\lambda^2}{2\sigma}+M\Big), \quad t
\in [t_0,T).
\end{equation}
Combining \eqref{e4.5} with \eqref{e4.6}, we have
\begin{equation} \label{e4.7}
\frac{\sigma}{2} - \varepsilon \leq
\frac{d}{dt}\Big(\frac{1}{m(t)+\frac{\lambda}{\sigma}}\Big) \leq
\frac{\sigma}{2} + \varepsilon, \quad t \in [t_0,T).
\end{equation}
Integrating over $(t,T)$ with $t \in [t_0,T)$  and
noticing that $\lim _{t \to T^-}\big(m(t)+\frac{\lambda}{\sigma}\big) = -\infty$,
we obtain
\[
(\frac{\sigma}{2} - \varepsilon)(T - t) \leq
-\frac{1}{m(t)+\frac{\lambda}{\sigma}} \leq (\frac{\sigma}{2} +
\varepsilon)(T - t).
\]
Since $\varepsilon \in (0,\frac{\sigma}{2})$
is arbitrary, in view of the definition of $m(t)$, we have
\[
\lim _{t \to T^-}\{m(t)(T-t)+\frac{\lambda}{\sigma}(T-t)\} =
-\frac{2}{\sigma};
\]
 that is, $\lim _{t \to T^-}\{\mathop{inf}_{x \in S}u_x(t,x)(T-t)\} 
= -\frac{2}{\sigma}$.

(2) When $\sigma < 0$, we consider the functions
$\bar{m}(t)$ and $\eta(t)$ as defined in \eqref{e2.11}.
From \eqref{e3.9} and \eqref{e4.4}, we have 
$\bar{m}' (t) \geq -\frac{\sigma}{2}\Big(\bar
{m}(t)+\frac{\lambda}{\sigma}\Big)^2 + \frac{\lambda^2}{2\sigma}-M$.

 Because $\bar{m} (t) \to \infty$ as $t \to T^-$, there
is a $t_1 \in (0,T)$, such that $\bar{m}(t_1) >
\sqrt{\frac{\lambda^2}{\sigma^2}-\frac{2M}{\sigma}}-\frac{\lambda}{\sigma}
> 0$. Thus, we have that $\bar{m}' (t) > 0$ and
$\bar{m}(t)$ is strictly increasing on $[t_1,T)$, and
\begin{equation} \label{e4.8}
\bar{m}(t) > \bar{m}(t_1) > 0.
\end{equation}
By the transport equation for $\rho$, we have 
\begin{equation*}
\frac{d\rho (t,\eta (t))}{dt} = - \bar{m}(t)\rho (t,\eta (t)).
\end{equation*}
 Then
\begin{equation} \label{e4.9}
\rho (t,\eta (t)) = \rho (t_1 ,\eta (t_1 )) e^{ -
\int_{t_1}^t \bar{m}(\tau)d\tau}, \quad t \in [t_1 ,T).
\end{equation}
Combining \eqref{e4.8} with \eqref{e4.9} yields
\begin{equation} \label{e4.10}
\rho ^2(t,\eta (t)) \le \rho ^2(t_1 ,\eta (t_1 )), \quad t \in [t_1
,T)
\end{equation}
From \eqref{e3.9} and \eqref{e4.10}, we have
\begin{equation} \label{e4.11}
\begin{aligned}
& - \frac{\sigma }{2}\Big(\bar m + \frac{\lambda }{\sigma }\Big)^2 +
\frac{\lambda ^2}{2\sigma } - \frac{1}{2}\rho ^2(t_1 ,\eta (t_1 )) -
M \\
& \le \bar m ' \le - \frac{\sigma }{2}\Big(\bar m +
\frac{\lambda }{\sigma }\Big)^2 - \frac{\lambda ^2}{2\sigma } +
\frac{1}{2}\rho ^2(t_1 ,\eta (t_1 )) + M.
\end{aligned}
\end{equation}
Choose $\varepsilon \in (0, - \frac{\sigma}{2})$, and  pick a 
$t_2 \in [t_1 ,T)$, such that
\begin{equation} \label{e4.12}
\Big(\bar m (t_2 ) + \frac{\lambda }{\sigma }\Big)^2 >
\frac{1}{\varepsilon }\Big(\frac{1}{2}\rho ^2(t_1 ,\eta (t_1 )) + M
- \frac{\lambda ^2}{2\sigma }\Big).
\end{equation}
From \eqref{e4.11} and \eqref{e4.12}, we have
\begin{equation} \label{e4.13}
\frac{\sigma }{2} - \varepsilon \le \frac{d}{dt}\Big(\frac{1}{\bar m
(t) + \frac{\lambda }{\sigma }}\Big) \le \frac{\sigma }{2} +
\varepsilon, \quad t \in [t_2 ,T).
\end{equation}
Integrating \eqref{e4.13} over $[t,T)$ with $t \in [t_2 ,T)$ and
$\lim_{t \to T^-}\bar{m}(t) = \infty$ gives
\[
(\frac{\sigma }{2} - \varepsilon )(T - t) \le - \frac{1}{\bar m (t)
+ \frac{\lambda }{\sigma }} \le (\frac{\sigma }{2} + \varepsilon )(T- t).
\]
Since $\varepsilon \in (0, - \frac{\sigma }{2})$ is arbitrary, 
in view of the definition of $\bar m (t)$, we have
\[
\lim_{t \to T^-}\{\sup_{x \in S}u_x(t,x)(T-t)\} =-\frac{2}{\sigma}.
\]
This completes the proof of Theorem \ref{thm4.1}.
\end{proof}

\section{Existence of a global solution}

In this section, we provide a sufficient condition for the
global solution of system \eqref{e2.1} in the case when $0<\sigma<2$.

\begin{lemma} \label{lem5.1}
Let $0<\sigma<2$ and $(u,\rho )$
be the solution of \eqref{e2.1} with initial data 
$(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$, $s> 3/2$, and $T$ be 
the maximal time of existence. Assume that
$\inf_{x \in S}\rho_0(x)>0$.


(1) When $0<\sigma\leq1$, it holds
\begin{gather*}
| \inf _{x \in S}u_x(t,x) |
\leq \frac{ 1}{ \inf _{x \in S}\rho_0(x)}C_4e^{C_3t},
\\
| \sup _{x \in S}u_x(t,x) |
\leq \frac{ 1}{ \inf _{x \in S}\rho_0^{\frac{\sigma}{2-\sigma}}(x)}
C_4^{\frac{1}{2-\sigma}}e^{\frac{C_3t}{2-\sigma}}.
\end{gather*}


(2) When $1<\sigma<2$, it holds
\begin{gather*}
| \inf _{x \in S}u_x(t,x) |
\leq \frac{ 1}{ \inf _{x \in S}\rho_0^{\frac{\sigma}{2-\sigma}}(x)}
C_4^{\frac{1}{2-\sigma}}e^{\frac{C_3t}{2-\sigma}},
\\
| \sup _{x \in S}u_x(t,x) |
\leq \frac{ 1}{ \inf _{x \in S}\rho_0(x)}
C_4e^{C_3t},
\end{gather*}
where constants $C_3$ and $C_4$ are defined as follows:
\begin{gather*}
C_3 = 1+ \frac{5(e+1)}{4(e-1)}+(\frac{A^2}{4}+\frac{2e+(e+1)(| \sigma | +2 | 3-\sigma|)}{4(e-1)})
\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2,
\\
C_4 = 1+ \| {u_{0x}} \|_{L^\infty}^2+\| {\rho _0} \|_{L^\infty}^2.
\end{gather*}
\end{lemma}

\begin{proof} 
A density argument indicates that it suffices
to prove the desired results for $ s \geq 3 $.
Since $s \ge 3$, we have $u \in C_0^1 (S)$ and
\begin{equation*}
\inf _{x \in S} u_x (t,x) < 0 ,  \quad 
\sup _{x \in S} u_x (t,x) > 0, \quad t \in [0,T).
\end{equation*}

 (1) First we will derive the estimate for $| \inf _{x \in S}u_x(t,x) |$.
 Define $m(t)$ and $\xi(t)$ as in \eqref{e2.25}, and consider along the 
characteristics $q(t,x_2(t))$. Then 
\begin{equation} \label{e5.1}
m(t) \leq 0   \quad   \text{for }   t \in [0,T).
\end{equation}
 Let $\zeta(t)=\rho(t,\xi(t))$ and evaluating \eqref{e2.10} and
the second equation of system \eqref{e2.1} at $(t,\xi(t))$, we have
\begin{equation} \label{e5.2}
\begin{gathered}
 {m}'(t) = - \frac{\sigma }{2}m^2(t) - \lambda m(t) + \frac{1}{2}\zeta ^2(t)
+ f(t,q(t,x_2 )) \\ 
 {\zeta }'(t) = - \zeta (t)m(t), \\
 \end{gathered}
\end{equation}
where $f$ is defined in \eqref{e2.16}. The second equation above implies that
$\zeta(t)$ and $\zeta(0)$ are of the same sign.

Next we construct a Lyapunov function for our system as in \cite{c10}.
Since here we have a free parameter $\sigma$, we could not find a
uniform Lyapunov function. Instead, we split the case $0 < \sigma
\leq 1$ and the case $1 < \sigma < 2$. From the assumption of the
theorem, we know that $\zeta(0)=\rho(0,\xi(0))>0$.

When $0 < \sigma \leq 1$, we define the  Lyapunov function
\[
\omega_1(t)=\zeta(0)\zeta(t)+\frac{ \zeta(0)}{ \zeta(t)}(1+m^2(t)),
\]
 which is always positive for $t \in [0,T)$. Differentiating 
$\omega_1(t)$ and using \eqref{e5.2} gives
\begin{equation} \label{e5.3}
\begin{aligned}
\omega'_1(t)
&=\zeta(0)\zeta'(t)-\frac{ \zeta(0)}{ \zeta^2(t)}(1+m^2(t))
\zeta'(t)+\frac{ 2\zeta(0)}{ \zeta(t)}m(t)m'(t)
\\
&=-\zeta(0)\zeta(t)m(t)-\frac{ \zeta(0)}{ \zeta^2(t)}(1+m^2(t))(-\zeta(t)m(t))\\
&\quad +\frac{ 2\zeta(0)}{ \zeta(t)}m(t)(-\frac{\sigma}{2}m^2(t)-\lambda m(t)
+\frac{1}{2}\zeta^2(t)+f)
\\
&=(1-\sigma)\frac{ \zeta(0)}{ \zeta(t)}m^3(t)
+\frac{ \zeta(0)}{ \zeta(t)}m(t)
-\frac{ 2\lambda\zeta(0)}{ \zeta(t)}m^2(t)
+\frac{ 2\zeta(0)}{ \zeta(t)}m(t)f
\\
&\leq \frac{ \zeta(0)}{ \zeta(t)}m(t)
+\frac{ 2\zeta(0)}{ \zeta(t)}m(t)f\\
&\leq \frac{ \zeta(0)}{ \zeta(t)}(1+m^2(t))
(1+| f |)\leq C_3\omega_1(t),
\end{aligned}
\end{equation}
where 
\[
C_3 = 1+ \frac{5(e+1)}{4(e-1)}
+(\frac{ A^2}{4}+\frac{2e+(e+1)(| \sigma | +2 | 3-\sigma|)}{ 4(e-1)})
\| {(u_0 ,\rho _0 - 1)} \|_{H^1\times L^2}^2.
\]
 This gives
\begin{equation} \label{e5.4}
\begin{aligned}
\omega_1(t)
&\leq \omega_1(0)e^{C_3t}=(\zeta^2(0)+1+m^2(0))e^{C_3t}\\
&\leq (1+\| {u_{0x}} \|_{L^\infty}^2+\| {\rho_{0}} \|_{L^\infty}^2)e^{C_3t}
=: C_4e^{C_3t},
\end{aligned}
\end{equation}
where $C_4=1+\| {u_{0x}} \|_{L^\infty}^2+\| {\rho_{0}} \|_{L^\infty}^2$.

Recalling that $\zeta(t)$ and $\zeta(0)$ are of the same sign,
the definition of $\omega_1(t)$ implies $\zeta(t)\zeta(0)\leq \omega_1(t)$
and $| \zeta(0)| | m(t) |\leq \omega_1(t)$. By \eqref{e5.4}, we obtain
\[
| \inf _{x \in S}u_x(t,x) |
= | m(t) | \leq \frac{ \omega_1(t)}{{| \zeta(0)|}}
\leq\frac{ 1}{ \inf _{x \in S}\rho_0(x)}
C_4e^{C_3t}, \quad \text{for } t\in [0,T).
\]

When $1<\sigma<2$, we define the  Lyapunov function
\begin{equation} \label{e5.5}
\omega_2(t)=\zeta^\sigma(0)\frac{ \zeta^2(t)+1+m^2(t)}{ \zeta^\sigma(t)}.
\end{equation}
Then
\begin{equation} \label{e5.6}
\begin{aligned}
\omega'_2(t)&=\frac{ 2\zeta^\sigma(0)}{ \zeta^\sigma(t)}m(t)
(\frac{ \sigma-1}{ 2}\zeta^2(t)-\lambda m(t)+f+\frac{ \sigma}{ 2})
\\
&\leq \frac{ \zeta^\sigma(0)}{ \zeta^\sigma(t)}
(1+m^2(t))(| f | +\frac{\sigma}{2})
\leq \frac{ \zeta^\sigma(0)}{ \zeta^\sigma(t)}
(1+m^2(t))(| f | +1)
\leq C_3\omega_2(t).
\end{aligned}
\end{equation}
Thus, we obtain
\begin{align*}
\omega_2(t)&\leq\omega_2(0)e^{C_3t}=(\zeta^2(0)+1+m^2(0))e^{C_3t}\\
&\leq (1+\| {u_{0x}} \|_{L^\infty}^2+\| {\rho_{0}} \|_{L^\infty}^2)e^{C_3t}
= C_4e^{C_3t}.
\end{align*}
Applying Young's inequality $ab\leq \frac{ a^p}{ p}+\frac{ b^q}{ q}$
to \eqref{e5.5} with $p=\frac{2}{\sigma}$ and $q=\frac{2}{2-\sigma}$ yields
\begin{align*}
\frac{\omega_2(t)}{\zeta^\sigma(0)}
&=\Big(\zeta^{\frac{\sigma(2-\sigma)}{2}}\Big)^{\frac{2}{\sigma}}
+\Big(\frac{(1+m^2)^{\frac{2-\sigma}{2}}}{\zeta^{\frac{\sigma(2-\sigma)}{2}}}\Big)^
{\frac{2}{2-\sigma}}\\
&\geq \frac{\sigma}{2}\Big(\zeta^{\frac{\sigma(2-\sigma)}{2}}\Big)^{\frac{2}{\sigma}}
+\frac{2-\sigma}{2}
\Big(\frac{(1+m^2)^{\frac{2-\sigma}{2}}}{\zeta^{\frac{\sigma(2-\sigma)}{2}}}\Big)
^{\frac{2}{2-\sigma}}
\\
&\geq (1+m^2)^{\frac{2-\sigma}{2}}
\geq {| m(t) |}^{2-\sigma}.
\end{align*}
So we have
\[
| \inf _{x \in S}u_x(t,x) |
\leq \Big(\frac{\omega_2(t)}{\zeta^\sigma(0)}\Big)^{\frac{1}{2-\sigma}}
\leq \frac{ 1}{ \inf _{x \in S}\rho_0^{\frac{\sigma}{2-\sigma}}(x)}
C_4^{\frac{1}{2-\sigma}}e^{\frac{ C_3t}{2-\sigma}}.
\]

 (2) Now, we  estimate $| \sup _{x \in S}u_x(t,x) |$.
Consider $\bar m(t),\eta(t),q(t,x_1)$ as in \eqref{e2.11} and \eqref{e2.13}, and
\begin{equation} \label{e5.7}
\begin{gathered}
 {\bar m}'(t) = - \frac{\sigma }{2}\bar m^2(t) - \lambda \bar m(t) 
+ \frac{1}{2}\bar\zeta ^2(t) + f(t,q(t,x_1 )) \\ 
 {\bar\zeta }'(t) = - \bar\zeta (t)\bar m(t) 
 \end{gathered}
\end{equation}
for $t\in [0,T)$, where $\bar\zeta(t)=\rho(t,\eta(t))$.
We know that
\begin{equation} \label{e5.8}
\bar m(t)\geq 0 \quad \text{for } t\in [0,T).
\end{equation}

When $0<\sigma\leq 1$, we define the Lyapunov function
\begin{equation} \label{e5.9}
\bar\omega_1(t)=\bar\zeta^\sigma(0)\frac{ \bar\zeta^2(t)+1
+\bar m^2(t)}{ \bar\zeta^\sigma(t)}.
\end{equation}
Then from \eqref{e5.6} and \eqref{e5.8}, we have
$\bar\omega'_1(t) \leq C_3\bar\omega_1(t)$,  then 
$\bar\omega_1(t) \leq C_4e^{C_3t}$.
Hence, by a similar argument as before, we obtain
\[
\frac{\bar\omega_1(t)}{\bar\zeta^\sigma(0)}\geq
{| \bar m(t) |}^{2-\sigma} .
\]
Then 
\[
| \sup _{x \in S}u_x(t,x) |
\leq (\frac{\bar\omega_1(t)}{\bar\zeta^\sigma(0)})^{\frac{
1}{ 2-\sigma}} \leq\frac{ 1}{
\inf _{x \in S}\rho_0^{\frac{\sigma}{2-\sigma}}(x)}
C_4^{\frac{1}{2-\sigma}}e^{\frac{C_3t}{2-\sigma}}, \quad t \in
[0,T).
\]

When $1<\sigma<2$, consider the  Lyapunov function
\begin{equation} \label{e5.10}
\bar\omega_2(t)=\bar\zeta(0)\bar\zeta(t)
+\frac{ \bar\zeta(0)}{ \bar\zeta(t)}(1+\bar m^2(t)).
\end{equation}
From \eqref{e5.3} and \eqref{e5.8}, we have
$\bar\omega'_2(t) \leq C_3\bar\omega_2(t)$ and 
$\bar\omega_2(t) \leq C_4e^{C_3t}$.
Therefore,
\[
| \sup _{x \in S}u_x(t,x) |
=| \bar m(t) | \leq \frac{\bar\omega_2(t)}{ \bar\zeta(0)}
\leq \frac{ 1}{ \inf _{x \in S}\rho_0(x)}
C_4e^{C_3t}, \quad t \in [0,T).
\]
The proof is complete.
\end{proof}

\begin{theorem} \label{thm5.1} 
Let $0<\sigma <2$ and $(u,\rho )$ be the solution of
\eqref{e2.1} with initial data 
$(u_0 ,\rho _0 - 1) \in H^s(S)\times H^{s - 1}(S)$, $s> 3/2$, and $T$ 
be the maximal time of existence. If $\inf _{x \in S}\rho_0(x)>0$,
then $T=+\infty$ and the solution $(u,\rho )$ is global.
\end{theorem}

\begin{proof} 
Assume on the contrary that $T<+\infty$ and
the solution blows up in finite time. It then follows from Theorem \ref{thm2.2}, 
that
\begin{equation} \label{e5.11}
\int_0^T{\|u_x(t)\|_{L^\infty}
dt} = \infty.
\end{equation}
 However, from the assumptions of the theorem and Lemma \ref{lem5.1}, we have
$| u_x(t,x) | < \infty$ for all $(t,x)\in [0,T)\times S$. 
This is a contradiction to \eqref{e5.11}. So $T=+\infty$, 
and it means that the solution $(u,\rho)$ is global.
\end{proof}


\subsection*{Acknowledgements}
We like to thank the anonymous referees for their valuable comments and 
suggestions. This work was supported by the National
Natural Science Foundation of China [11101190, 11171135, 11201186], 
and Natural Science Foundation of Jiangsu Province [BK2012282],
China Postdoctoral Science Foundation funded project 
(Nos: 2012M511199, 2013T60499) and Jiangsu University foundation 
grant [11JDG117].

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