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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 226, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/226\hfil Energy decay]
{Energy decay of a variable-coefficient wave equation with nonlinear
time-dependent localized damping}

\author[J. Wu, F. Feng, S. Chai \hfil EJDE-2015/226\hfilneg]
{Jieqiong Wu, Fei Feng, Shugen Chai}

\address{Jieqiong Wu (corresponding author)\newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{jieqiong@sxu.edu.cn}

\address{Fei Feng \newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{fengfei599@126.com}

\address{Shugen Chai \newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{sgchai@sxu.edu.cn}

\thanks{Submitted May 28, 2015. Published September 2, 2015.}
\subjclass[2010]{35L05, 35L70, 93B27}
\keywords{Energy decay; time-dependent and space-dependent damping;
\hfill\break\indent localized damping;  Riemannian geometry method; 
variable coefficient}

\begin{abstract}
 We study the energy decay for the Cauchy problem of the wave equation with
 nonlinear time-dependent and space-dependent damping.
 The damping is localized in a bounded domain and near infinity, and
 the principal part of the wave equation has a variable-coefficient.
 We apply the multiplier method for variable-coefficient equations, and
 obtain an energy decay that depends on the property of the coefficient
 of the damping term.
\end{abstract}

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

\section{Introduction}

Let $n\geq 2$ be an integer.  We consider the energy decay for  the solution to the
Cauchy problem of the wave equation with time-dependent and space-dependent damping,
 \begin{gather}
u_{tt}-\operatorname{div}(A(x)\nabla u) +u+\sigma(t)\rho(x, u_t)=0, \quad
 x\in \mathbb{R}^n,\;  t>0,\label{1.1}
\\
 u(0,x)=u_0(x), \quad   u_t(0,x)=u_1(x),  \quad  x\in \mathbb{R}^n,\label{1.2}
\end{gather}
where  $A(x)=(a_{ij}(x))$ is a symmetric,
positive matrix  for each $x\in \mathbb{R}^n$;
$\sigma:\mathbb{R}^+\to \mathbb{R}^+$ is a non-increasing function  of
class $C^1$. Let  $\Omega$ and $\Omega_1$ be two bounded domains
such that  $\Omega_1\subset\Omega$. We assume that
\begin{equation}
 \rho(x, u_t)=\begin{cases}
a(x)u_t, & x\in \mathbb{R}^n\setminus \Omega_1\\
a(x)h(u_t), &  x\in \Omega_1
\end{cases} \label{1.3}
\end{equation}
where $h:\mathbb{R}\to \mathbb{R} $ is a nonlinear continuous nondecreasing function satisfying
$h(s)s>0$ for all $s\neq 0$ and $a(\cdot)\in L^\infty(\mathbb{R}^n)$
is a nonnegative function satisfying
\begin{equation}
 a(x)\geq \varepsilon_0>0,\quad x\in \Omega_1\cup \Omega^c.\label{1.4}
\end{equation}
 When $x\in \Omega_1$,
\begin{gather}
 c_1|u_t|^m\leq |h(u_t)|\leq c_2|u_t|^{1/m},\quad |u_t|\leq 1,\label{1.5}\\
 c_3|u_t|\leq |h(u_t)|\leq c_4|u_t|,\quad |u_t|\geq 1\label{1.6}
\end{gather}
where $m\geq 1$ and $c_i>0$ $(i=1,2,3,4)$ are given numbers.

 Since \eqref{1.4} implies that the dissipation may vanish in
$\Omega\setminus\Omega_1$, we call the damping $\sigma(t)\rho(x, u_t)$
a time-dependent localized damping.


 Many studies concerning energy decay of wave equations in  the bounded domain
are available in the literature.  We refer the readers to
 \cite{ da-1, na-2,na-3,zu-1}. The author of \cite{na-2} derived precise
energy decay estimates for the initial-boundary value problem to the wave
equation with a localized nonlinear dissipation which depended on the
time as well as the space variable.  The result of \cite{na-2} is a
generalization of  the work \cite{na-3}, where the dissipation was independent
of the time. A wave equation with time-dependent but space-independent
damping was investigated in \cite{da-1} and the decay rate was obtained by
solving a nonlinear ODE.

For the energy decay of wave equation in the exterior domain or on the whole space,
many results have been  obtained for the case of constant coefficients,
that is $A(x)\equiv I$ (see   \cite{da-2,be-1,da-3,na-4,na-5,na-6,zu-2}).
The damping in \cite{da-3,na-5,zu-2} were time-independent and localized in
a sub-domain of $\mathbb{R}^n$.
   Energy decay of the  Cauchy problem for the wave equation with time-dependent
damping was studied in \cite{be-1} where the coefficient of the damping did not
depend on the space variable.
It is interesting  to study the case where the damping is time-dependent
and  exists on a local domain of $\mathbb{R}^n$.
 \cite{da-2} addressed the decay for a wave equation on an exterior domain
 where the damping is time-dependent and  effective only in a ball of
$\mathbb{R}^n$. The dimension of the space variable was restricted to be odd
 since the Lax-Phillips' scattering theory was used in \cite{da-2}.
Nakao \cite{na-4} considered the Cauchy problem of wave equation with a
dissipation effective outside of a given ball of $\mathbb{R}^n$.

 However,  in the case of variable coefficient, few results  about
the energy decay of  wave equation in the exterior domain or on the
 whole space can be seen  (see\cite{li,ya-1}).  Yao \cite{ya-1}
studied the energy decay for the Cauchy problem of the variable-coefficient
wave  equation with a linear damping. The authors of \cite{li} considered
the energy decay of variable-coefficient wave equation with nonlinear
damping in an exterior domain.

 The purpose of this paper is to derive the energy decay for
\eqref{1.1}-\eqref{1.2} with the  assumptions \eqref{1.3}-\eqref{1.6}.
Our first main goal is to dispense with the restriction $A(x)\equiv I$.
We will use the Riemannian geometry method  which was  introduced by
Yao  \cite{ya} (see also \cite{ys}) to deal with controllability for the
wave equation with variable-coefficient principal part and has been viewed
as an important method for variable-coefficient models.

  Our second goal is to generalize the work \cite{be-1}, where the damping
exists on the whole space. In this paper,  we assume that  the damping is
localized near the origin point and near infinity and may disappear in a
large area.   We try to explain how  the time-dependence of the damping affect
the decay when the dissipation is effective in a localized domain.
Energy decay is obtained under some conditions on $A(x)$ and the coefficient
of the damping.

We introduce a Riemanian metric on $\mathbb{R}^n$ by
\begin{equation}
g(x)=A^{-1}(x)\quad\mbox{for}\quad x\in\mathbb{R}^n.\label{1.7}
\end{equation}
We denote by $\nabla f$ and $\nabla_g f$ the gradients of $f$ in the
standard metric of the Euclidean space $\mathbb{R}^n$ and in the metric $g$,
respectively.

   The energy of the model \eqref{1.1}-\eqref{1.2}  is defined by
\begin{equation}
E(t)=\frac{1}{2}\int_{\mathbb{R}^n}(u_t^2+|\nabla_gu|_g^2+u^2)dx.\label{1.8}
\end{equation}
where
 $\nabla_g u=A(x)\nabla u$ and
$$
|\nabla_gu|_g^2=\langle \nabla_g u, \nabla_g u\rangle_g
=\langle A(x)\nabla u, \nabla u\rangle
 =\sum\limits_{ij=1}^na_{ij}(x)\frac{\partial u}{\partial x_j}
\frac{\partial u}{\partial x_i}.
$$
 We refer the readers to  \cite{ya} and \cite{ys} for more information about the metric $g(x)$.

We use the following assumption in this article
\begin{itemize}
\item[(H1)] There is a vector field $H$ on
$(\mathbb{R}^n, g)$  such that
\begin{equation}
 D_gH(X,X)\geq \bar{\sigma}|X|_g^2,\quad X\in \mathbb{R}^n_x,\; x\in \bar{\Omega},
\label{1.9}
\end{equation}
where $\bar{\sigma}>0$ is a constant and $D_g$ is  the Levi-Civita
connection of the metric $g$.
\end{itemize}
Our main result read as follows.

\begin{theorem} \label{thm1.1}
Let $(u_0, u_1)\in H^1(\mathbb{R}^n)\times L^2(\mathbb{R}^n)$.
In addition to (H1) assume that $\sigma:\mathbb{R}^+\to \mathbb{R}^+$ is a
 non-increasing $C^1$ function and $\sigma(t)\geq \sigma_0>0$ for all
$t\geq 0$. Then  \eqref{1.1}-\eqref{1.2} admits a unique solution $u$ satisfying
\[
E(t)\leq C E(0)\Big(\frac{1}{\int_0^t\sigma(s)ds}\Big)^{\frac{2}{m-1}},\quad
 \forall t>0\quad \text{if } m>1,
\]
and
\[
E(t)\leq C E(0)\exp(-\omega\int_0^t\sigma(s)ds), \quad \forall t>0 \quad
\text{if } m=1,
\]
for some $\omega>0$ and $C>0$.
\end{theorem}

 \section{Proof of main result}

To prove our main result we use the following lemma.

\begin{lemma}[\cite{ma}] \label{lem2.1}
Let $E:\mathbb{R}^+\to \mathbb{R}^+$ be a non-increasing function and
$\phi:\mathbb{R}^+\to\mathbb{R}^+$ a strictly increasing $C^1$ function such that
$$
\phi(0)=0,\quad \phi(t)\to+\infty\quad\text{as } t\to+\infty.
$$
Assume that there exist $r\geq 0$ and  $\omega>0$ such that
$$
\int_S^{+\infty} E^{1+r}(t)\phi'(t)dt
\leq\frac{1}{\omega}E^r(0)E(S),\quad  0\leq S<+\infty.
$$
Then
\begin{gather*}
E(t)\leq E(0)\Big(\frac{1+r}{1+\omega r \phi(t)}\Big)^{1/r} \quad 
\text{for all $t\geq 0$,  if $r>0$},\\
E(t)\leq CE(0)e^{1-\omega\phi(t)}\quad\text{for all $t\geq 0$,  if $r=0$}.
\end{gather*}
\end{lemma}

To apply Lemma \ref{lem2.1}, we need some estimates on  the energy terms, 
which are based on the identities below.
Multiply  \eqref{1.1} by $u_t$ and integrate by parts
 over $\mathbb{R}^n$ with respect to the variable $x$ to obtain
\begin{equation}
\frac{d}{dt}E(t)=-\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx.\label{2.1}
\end{equation}

 Let   $H$ be a  vector field on $\mathbb{R}^n$.  
Multiply \eqref{1.1} by $H(u)$ and integrate by parts
 over $\mathbb{R}^n$  with respect to the variable $x$ to obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\int_{\mathbb{R}^n}u_tH(u)dx+\int_{\mathbb{R}^n}
D_gH(\nabla_gu, \nabla_gu)dx \\
&+\int_{\mathbb{R}^n}H(u)u\,dx
 +\frac{1}{2}\int_{\mathbb{R}^n}(u_t^2-|\nabla_gu|_g^2) \mathrm{div}H dx\\
&=- \int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)H(u)dx.
\end{aligned}\label{2.2}
\end{equation}

 Let $q\in C^1(\mathbb{R}^n)$ be a function. Multiply \eqref{1.1} by $qu$ 
and  integrate by parts  over $\mathbb{R}^n$ with respect to the variable
 $x$ to obtain
 \begin{equation}
\begin{aligned}
&\frac{d}{dt}\int_{\mathbb{R}^n} quu_t\,dx
+\int_{\mathbb{R}^n} q \big(|\nabla_g u|^2_g-u_t^2\big)\,dx
-\int_{\mathbb{R}^n}u^2 \operatorname{div}A\nabla q\, dx
+\int_{\mathbb{R}^n}qu^2dx\\
&=-\int_{\mathbb{R}^n}qu\sigma(t)\rho(x,u_t)dx.
\end{aligned}\label{2.3}
\end{equation}



\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let  the vector field $H$ be such that  the assumption (H1) holds. 
Let $\hat{\Omega}\subset\mathbb{R}^n$ and 
$\hat{\hat{\Omega}} \subset\mathbb{R}^n$ be two bounded domains such that
$\bar{\Omega} \subset\hat{\Omega}\subset\bar{\hat{\Omega}}\subset \hat{\hat{\Omega}}$. 
Let $\varphi$ and $\psi$ be two $C_0^\infty$ functions and
$ 0\leq \varphi\leq 1$, $0\leq \psi\leq 1$, and
\begin{equation}
\varphi=\begin{cases}
1,& x\in \Omega,\\
0,& x\in \hat{\Omega}^c=\mathbb{R}^n\setminus\hat{\Omega},
\end{cases}
\quad
\psi=\begin{cases}
1,& x\in \hat{\Omega},\\
0,& x\in \mathbb{R}^n\setminus\hat{\hat{\Omega}}.
\end{cases} \label{2.4}
\end{equation}
Let $q_0=\frac{\operatorname{div} (\varphi H)}{2}-\bar{\sigma}\varphi$.
Replacing $H$ by $\varphi H$ in \eqref{2.2} and replacing $q$ by $q_0$
in \eqref{2.3}, respectively, we can obtain two  identities.
Then we add them up and  obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\int_{\mathbb{R}^n}\big(u_t\varphi H(u)+q_0uu_t\big)dx
+\int_{\mathbb{R}^n}D_g(\varphi H)(\nabla_gu, \nabla_gu)dx\\
&+\int_{\mathbb{R}^n}\varphi H(u)u\,dx
+\int_{\mathbb{R}^n}\bar{\sigma}\varphi(u_t^2-|\nabla_gu|_g^2) dx\\
&-\int_{\mathbb{R}^n} |u|^2\operatorname{div}(\nabla_g q_0)dx
+\int_{\mathbb{R}^n} q_0u^2dx\\
&=-\int_{\mathbb{R}^n} q_0u\sigma(t)\rho(x,u_t)dx
-\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)\varphi H(u)dx.
\end{aligned} \label{2.5}
\end{equation}
Let $k$ be a large constant determined later. Combining \eqref{2.1}
 with \eqref{2.5}, we have
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\int_{\mathbb{R}^n}\big(u_t\varphi H(u)+q_0uu_t\big)dx+kE'(t)
+\int_{\mathbb{R}^n} D_g(\varphi H)(\nabla_gu, \nabla_gu)dx\\
&+\int_{\mathbb{R}^n}\varphi H(u)u\,dx
 +\int_{\mathbb{R}^n}\bar{\sigma}\varphi(u_t^2-|\nabla_gu|_g^2) dx
 -\int_{\mathbb{R}^n} |u|^2\operatorname{div}(\nabla_g q_0)dx\\
& +\int_{\mathbb{R}^n} q_0u^2dx \\
&=-\int_{\mathbb{R}^n} q_0u\sigma(t)\rho(x,u_t)dx
 -\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)\varphi H(u)dx\\
&\quad -k\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx.
\end{aligned}\label{2.6}
\end{equation}
Set
$$
Y(t)=\int_{\mathbb{R}^n} D_g(\varphi H)(\nabla_gu, \nabla_gu)dx
+\int_{\mathbb{R}^n}\bar{\sigma}\varphi(u_t^2-|\nabla_gu|_g^2) dx
+k\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx.
$$
Set $\sigma_1=\sup_{\bar{\hat{\Omega}}}|D_g(\varphi H)(X, X)|$.
Noticing \eqref{1.3},\eqref{1.4},\eqref{1.9} and \eqref{2.4},
 we have the following estimates on $Y(t)$:
\begin{equation} \begin{aligned}
Y(t)
&\geq \int_{\hat{\Omega}\setminus\Omega}
 D_g(\varphi) H(\nabla_gu, \nabla_gu)dx
+\int_{\mathbb{R}^n\setminus\Omega}\bar{\sigma}\varphi u_t^2dx
- \int_{\hat{\Omega}\setminus\Omega}\bar{\sigma}\varphi|\nabla_gu|_g^2\, dx\\
&\quad +\frac{k}{4}\int_{\mathbb{R}^n\setminus\Omega}\epsilon_0\sigma(t)u_t^2dx
+\int_{\Omega}\bar{\sigma}u_t^2dx
+\frac{3k}{4}\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx
\\
&\geq (-\sigma_1-\bar{\sigma})\int_{\hat{\Omega}\setminus\Omega}|\nabla_g u|^2_gdx
+\int_{\mathbb{R}^n\setminus\Omega}\big(\frac{k}{4}\varepsilon_0\sigma(t)
 +\bar{\sigma}\varphi\big)u_t^2dx \\
&\quad +\int_{\Omega}\bar{\sigma}u_t^2dx
+\frac{3k}{4}\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx\,.
\end{aligned} \label{2.7}
\end{equation}
Noticing $\sigma(t)\geq \sigma_0>0$, we choose $k$ large enough so that
$\frac{k\sigma_0 \varepsilon_0}{4}>\bar{\sigma}$. Then we have
\begin{equation}
Y(t)\geq \int_{\mathbb{R}^n}\bar{\sigma}|u_t|^2dx
-\int_{\hat{\Omega}\setminus\Omega}(\sigma_1+\bar{\sigma})|\nabla_g u|^2_g
 +\frac{3k}{4}\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx\,. \label{2.8}
\end{equation}
Now we estimate the last term of the right side of \eqref{2.8}.
\begin{equation}\begin{aligned}
&\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\,dx \\
&\geq \int_{\mathbb{R}^n}\psi \sigma(t)\rho(x,u_t)u_t\,dx
=\int_{\hat{\hat{\Omega}}}\psi \sigma(t)\rho(x,u_t)u_t\,dx\\
&=\int_{\hat{\hat{\Omega}}\setminus\Omega_1}\psi \sigma(t)a(x)u_t^2dx
 +\int_{\Omega_1}\psi \sigma(t)\rho(x,u_t)u_t\,dx
\\
&=\int_{\hat{\hat{\Omega}}}\psi \sigma(t)a(x)u_t^2\,dx
-\int_{\Omega_1}\sigma(t)a(x)u_t^2\,dx
+\int_{\Omega_1}\psi \sigma(t)\rho(x,u_t)u_t\,dx\,.
\end{aligned}\label{2.9}
\end{equation}
For the first term on the right of \eqref{2.9}, noticing
$\sigma(t)\geq \sigma_0$ and using \eqref{1.4}, \eqref{2.4} and \eqref{2.3}
for $q=\sigma_0\psi a(x)$, we have
\begin{equation}\begin{aligned}
\int_{\hat{\hat{\Omega}}}\psi \sigma(t)a(x)u_t^2\,dx
&\geq \sigma_0\int_{\hat{\hat{\Omega}}}\psi a(x)u_t^2\,dx
=\int_{\mathbb{R}^n}\sigma_0\psi a(x)u_t^2dx\\
&\geq \frac{d}{dt}\int_{\mathbb{R}^n} \sigma_0a(x)\psi u u_t\,dx
+\sigma_0 \varepsilon_0\int_{\hat{\Omega}\setminus\Omega} |\nabla_g u|^2_g\,dx \\
&\quad -\sigma_0\int_{\mathbb{R}^n}u^2 \operatorname{div}A\nabla\big(a(x)\psi\big)dx
  +\sigma_0\int_{\mathbb{R}^n}a(x)\psi u^2dx \\
&\quad  +\sigma_0\int_{\mathbb{R}^n}a(x)\psi \sigma(t)\rho(x,u_t)u\,dx.
\end{aligned}\label{2.10}
\end{equation}
since $\operatorname{supp} \psi\subset \hat{\hat{\Omega}}$ and
$a(x)\geq \varepsilon_0>0$ for $x\in \Omega^c$.

Combining \eqref{2.8} with \eqref{2.9} and \eqref{2.10}, we have
\begin{equation}
\begin{aligned}
Y(t)
&\geq\Big(\frac{k\sigma_0\varepsilon_0}{4}-(\sigma_1+\bar{\sigma})\Big)
\int_{\hat{\Omega}\setminus\Omega} |\nabla_g u|^2_g\,dx
 + \int_{\mathbb{R}^n}\bar{\sigma} u_t^2\, dx \\
&\quad +\frac{k}{2}\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u_t\, dx
 +\frac{\sigma_0 k}{4}\Big[\frac{d}{dt}\int_{\mathbb{R}^n} a(x)\psi u u_t\,dx\\
&\quad -\int_{\mathbb{R}^n}u^2 \operatorname{div}A\nabla\big(a(x)\psi\big)dx
+\int_{\mathbb{R}^n}a(x)\psi u^2dx \\
&\quad +\int_{\mathbb{R}^n}a(x)\psi \sigma(t)\rho(x,u_t)udx\Big]
-\frac{k}{4}\int_{\Omega_1}\sigma(t)a(x) u^2_t\,dx \\
&\quad  +\frac{k}{4}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx
\end{aligned}\label{2.11}
\end{equation}
Choose $k$ large enough so that
$\frac{k\sigma_0 \varepsilon_0}{4}>\bar{\sigma}+\sigma_1$.
Then it follows from   \eqref{2.11} and   \eqref{2.6} that
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\Big(\int_{\mathbb{R}^n}\big(u_t\varphi H(u)+q_0uu_t\big)dx+kE(t)
+\frac{\sigma_0 k }{4}\int_{\mathbb{R}^n}a(x)\psi u u_t\,dx\Big)\\
&+\int_{\mathbb{R}^n}\bar{\sigma } u_t^2\,dx
 +\frac{k}{2}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx
 -\frac{k}{4}\int_{\Omega_1}\sigma(t)a(x)\psi u^2_t\,dx\\
& +\frac{k}{4}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx 
 +\frac{\sigma_0 k}{4}\Big[\int_{\mathbb{R}^n}
 \big(a\psi-\operatorname{div}A\nabla\big(a(x)\psi)\big)u^2\,dx\\
& +\int_{\mathbb{R}^n}a(x)\psi \sigma(t)\rho(x,u_t)udx\Big]
+\int_{\mathbb{R}^n}\varphi H(u)u\,dx \\
&+\int_{\mathbb{R}^n} \big(q_0-\operatorname{div}\nabla_g q_0\big)u^2\,dx
 +\int_{\mathbb{R}^n} q_0u\sigma(t)\rho(x,u_t)dx \\
& +\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)\varphi H(u)dx\leq 0.
\end{aligned}\label{2.12}
\end{equation}
We may have  equality by setting  $q=\frac{\bar{\sigma}}{2}$ in \eqref{2.3}.
 Then add that equality  to \eqref{2.12} to yield
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\Big(\int_{\mathbb{R}^n}\big(u_t\varphi H(u)
+q_0uu_t\big)dx+kE(t)+\frac{\sigma_0 k }{4}\int_{\mathbb{R}^n}a(x)\psi u u_t\,dx\\
&+\int_{\mathbb{R}^n}\frac{\bar{\sigma }}{2}u u_t\,dx\Big)
 +\int_{\mathbb{R}^n}\frac{\bar{\sigma }}{2}\big(|\nabla_gu|_g^2
 +u_t^2+u^2+\sigma(t)\rho(x,u_t)u\big)dx \\
&+\frac{k}{2}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx
 -\frac{k}{4}\int_{\Omega_1}\sigma(t)a(x)\psi u^2_t\,dx\\
& +\frac{k}{4}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx 
 +\frac{\sigma_0 k}{4}\Big[\int_{\mathbb{R}^n}
 \Big(a\psi-\operatorname{div}A\nabla\big(a(x)\psi\big)\Big)u^2dx \\
& +\int_{\mathbb{R}^n}a(x)\psi \sigma(t)\rho(x,u_t)udx\Big]
+\int_{\mathbb{R}^n}\varphi H(u)u\,dx\\
&+\int_{\mathbb{R}^n} \big(q_0-\operatorname{div}\nabla_g q_0\big)u^2\,dx 
+\int_{\mathbb{R}^n} q_0\sigma(t)\rho(x,u_t)u \,dx\\
&+ \int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)\varphi H(u)dx\leq 0.
\end{aligned}\label{2.13}
\end{equation}
Set
$$
X(t)=\int_{\mathbb{R}^n}\big(u_t\varphi H(u)
+q_0uu_t\big)dx+kE(t)+\frac{\sigma_0 k}{4}\int_{\mathbb{R}^n}a(x)\psi u u_t\,dx
+\int_{\mathbb{R}^n}\frac{\bar{\sigma} }{2}u u_t\,dx.
$$
From \eqref{2.13} and the definition of $E(t)$, we have
\begin{equation}
\begin{aligned}
&\bar{\sigma}E(t)\\
&\leq -\frac{d}{dt}X(t)
 -\int_{\mathbb{R}^n}\frac{\bar{\sigma} }{2}\sigma(t)\rho(x,u_t)u\, dx
 -\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)\varphi H(u)\,dx\\
&\quad -\int_{\mathbb{R}^n} q_0\sigma(t)\rho(x,u_t)u dx
 -\frac{k}{2}\int_{R^n}\sigma(t)\rho(x,u_t)u_t\,dx \\
&\quad  -\frac{k}{4}\int_{\Omega_1}\sigma(t)\rho(x,u_t)u_t\,dx 
 -\frac{\sigma_0 k}{4}\Big[\int_{\mathbb{R}^n}
 \Big(a\psi-\operatorname{div}A\nabla\big(a(x)\psi\big)\Big)u^2\,dx\\
&\quad +\int_{\mathbb{R}^n}a(x)\psi \sigma(t)\rho(x,u_t)u\,dx\Big]
 +\frac{k}{4}\int_{\Omega_1}\sigma(t)a(x)\psi u^2_t\,dx\\
&\quad -\int_{\mathbb{R}^n}\varphi H(u)u\,dx
 -\int_{\mathbb{R}^n}\big(q_0-\operatorname{div}\nabla_g q_0\big)u^2\,dx \\
&\leq -\frac{d}{dt}X(t)+\sigma_4\Big|\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u\,dx\Big|
-\frac{k}{2}E'(t)\\
&\quad +(\frac{\sigma_0 k\sigma_2}{4}+\sigma_3) 
  \int_{\hat{\hat{\Omega}}}|u|^2\,dx 
 +\frac{k}{4}\int_{\Omega_1}\sigma(t)a(x) u^2_t\,dx
 +\sigma_5\int_{\hat{\Omega}}|\nabla_gu|_gu\,dx \\
&\quad +\sigma_5\int_{\mathbb{R}^n}\sigma(t)|\rho(x,u_t)||\nabla_g u|_g\,dx
\end{aligned}\label{2.14}
\end{equation}
where
\begin{gather*}
\sigma_2=\sup_{x\in \bar{\hat{\hat{\Omega}}}}
 |a\psi-\operatorname{div}\nabla_g(a\psi)|,\quad
\sigma_3=\sup_{x\in \bar{\hat{\Omega}}}|q_0-\operatorname{div}\nabla_gq_0|,\\
\sigma_4=\max\big\{\frac{\sigma_0k|a|_\infty}{4}, \frac{\bar{\sigma}}{2},
\sup_{\bar{\hat{\Omega}}}|q_0|\big\},\quad
\sigma_5=\sup_{\bar{\hat{\Omega}}}|\varphi H|.
\end{gather*}
For the second term, the sixth term and the last term on the right of \eqref{2.14},
we use Young's inequality to obtain
\begin{equation}
\begin{aligned}
\sigma_4\big|\int_{\mathbb{R}^n}\sigma(t)\rho(x,u_t)u dx\big|
&\leq \sigma_4\int_{\mathbb{R}^n}\big(C_\varepsilon|\sigma(t)\rho(x,u_t)|^2
+\varepsilon u^2 \big)dx \\
&\leq \sigma_4\int_{\mathbb{R}^n}C_\varepsilon|\sigma(t)\rho(x,u_t)|^2
 +2\varepsilon C_4E(t),
\end{aligned}\label{2.15}
\end{equation}
\begin{equation}
\begin{aligned}
\sigma_5\int_{\hat{\Omega}}|\nabla_gu|_gu dx
&\leq \sigma_5\int_{\mathbb{R}^n} \varepsilon |\nabla_g u|^2_g \,dx
+C_\varepsilon\sigma_5\int_{\hat{\Omega}}u^2\,dx, \\
&\leq 2\varepsilon C_5E(t)+C_\varepsilon\sigma_5\int_{\hat{\Omega}}u^2\,dx,
\end{aligned}\label{2.16}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\sigma_5\int_{\mathbb{R}^n}\sigma(t)|\rho(x,u_t)||\nabla_g u|_g\,dx\\
&\leq \varepsilon \sigma_5\int_{\mathbb{R}^n}|\nabla_g u|_g^2\,dx
+C_\varepsilon\sigma_5\int_{\mathbb{R}^n}|\sigma(t)\rho(x,u_t)|^2\,dx\\
& \leq  2\varepsilon C_5E(t)
 +C_\varepsilon\sigma_5\int_{\mathbb{R}^n}|\sigma(t)\rho(x,u_t)|^2dx
\end{aligned}\label{2.17}
\end{equation}
where $\varepsilon>0$ is small and $C_\varepsilon$ is a constant dependent
on $\varepsilon$.

Let $\phi$ be a function satisfying the conditions in Lemma \ref{lem2.1}. 
Let $r\geq 0 $ be a constant  determined later.
Multiplying \eqref{2.14} by $E^r\phi'$ and  integrating on $[S, T]$ 
with respect to $t$,  from \eqref{2.15} -\eqref{2.17} we have 
\begin{equation}
\begin{aligned}
&\int_S^TE^{r+1}(t)\phi'\bar{\sigma}dt\\
&\leq -\int_S^TE^r\phi'X'(t)dt+(\sigma_4C_\varepsilon
+\sigma_5 C_\varepsilon)\int_S^TE^r\phi'\int_{\mathbb{R}^n}|
\sigma(t)\rho(x,u_t)|^2\,dx\,dt\\
&\quad +(2\varepsilon\sigma_4+4\varepsilon\sigma_5)\int^T_SE^{r+1}\phi'dt
 -\frac{k}{2}\int^T_SE^r\phi' E'dt \\
&\quad +(\frac{k\sigma_0\sigma_2}{4}+\sigma_3
 +C_\varepsilon\sigma_5)\int^T_SE^{r}\phi'\int_{\hat{\hat{\Omega}}}u^2\,dx\,dt\\
&\quad +\frac{k}{4}\int^T_SE^{r}\phi' \int_{\Omega_1}\sigma(t)a(x)u_t^2\,dx\,dt.
\end{aligned}\label{2.18}
\end{equation}

Using a similar argument as in \cite{ya-1} and \cite{na-1}, 
we can prove that if $T-S$ is large enough, then 
\begin{equation}
\begin{aligned}
\int^T_SE^{r}\phi'\int_{\hat{\hat{\Omega}}}u^2\,dx\,dt
&\leq C_\eta\int^T_SE^{r}\phi'\Big(\int_{\mathbb{R}^n}|\sigma(t)\rho(x,u_t)|^2\,dx\\
&\quad +\int_{\Omega_1}\sigma(t)u_t^2dx\Big)\,dt
+\eta\int_S^TE^{r+1}(t)\phi'\,dt
\end{aligned}\label{2.19}
\end{equation}
holds for any $\eta>0$.

For the rest of this article, let $\phi(t)=\int_0^t\sigma(s)ds$. 
It is clear that $\phi(t)$ satisfies the conditions of Lemma \ref{lem2.1}. 
 Noticing that $|X(t)|\leq CE(t)$ and $E'(t)\leq 0$,
we have by  integrating  by parts with respect to $t$,
\begin{equation}
\begin{aligned}
&-\int_S^TE^r\phi'X'(t)dt\\
&=-E^r(t)\phi'(t)X(t)\Big|^T_S+\int_S^T(E^r\phi')'X(t)dt\\
& \leq C\sigma(0)E^{r+1}(S)+\int_S^T|rE^{r-1}E'\phi'X(t)|dt
 +\int_S^T|E^{r}\phi''X(t)|dt\\
& \leq C\sigma(0)E^{r+1}(S)+C\sigma(0)\int_S^TrE^{r}(-E')dt
 +\int_S^TE^{r+1}(-\phi'')dt \\
& \leq CE^{r+1}(S).
\end{aligned}\label{2.20}
\end{equation}
We also have
\begin{equation}
 -\frac{k}{2}\int^T_SE^r\phi' E'dt\leq CE^{r+1}(S).\label{2.21}
\end{equation}
Now we estimate the second term on the right of \eqref{2.18}.
Set $\Omega_1^+=\{x\in \Omega_1; |u_t|\geq1\}$ and
$\Omega_1^-=\{x\in \Omega_1; |u_t|<1\}$.
Then $\Omega_1=\Omega_1^+\cup \Omega_1^-$.
Noticing \eqref{1.5}, using H$\ddot{o}$lder inequality and Young's
inequality, we have
\begin{equation}
\begin{aligned}
&\int_S^TE^r\phi'\int_{\Omega_1^-}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt\\
&=\int_S^TE^r\phi'\int_{\Omega_1^-}\sigma^2(t)|a(x)h(u_t)|^2\,dx\,dt \\
&\leq c_2\int_S^TE^r\phi'\sigma^2(t)\int_{\Omega_1^-}a^2(x)\big(h(u_t)u_t
 \big)^{\frac{2}{m+1}}\,dx\,dt \\
&=c_2\int_S^TE^r\phi'\int_{\Omega_1^-}\big(\sigma(t)a(x)h(u_t)u_t
 \big)^{\frac{2}{m+1}}\big(a(x)\sigma(t)\big)^{2-\frac{2}{m+1}}\,dx\,dt\\
&\leq C|a|_\infty^{2-\frac{2}{m+1}}\int_S^TE^r\phi'\sigma(t)^{2-\frac{2}{m+1}}
 \Big(\int_{\Omega_1^-}\sigma(t)a(x)h(u_t)u_t\big)dx\Big)^{\frac{2}{m+1}}dt\\
&\leq C|a|_\infty^{2-\frac{2}{m+1}}\sigma^2(0)\Big[\varepsilon_1
 \int_S^T\big(E^r\phi'\sigma(t)^{-\frac{2}{m+1}}\big)^{\frac{m+1}{m-1}}dt\\
&\quad +C_{\varepsilon_1}\int_S^T\int_{\Omega_1^-}\sigma(t)a(x)h(u_t)u_t\,dx\,dt\Big].
\end{aligned}\label{2.22}
\end{equation}
Now, we set $r=\frac{m-1}{2}$. Thus, we have from \eqref{2.22} and \eqref{2.1}
\begin{equation}
\begin{aligned}
&\int_S^TE^r\phi'\int_{\Omega_1^-}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt\\
&\leq C|a|_\infty^{2-\frac{2}{m+1}}\sigma^2(0)
\Big[\varepsilon_1\int_S^TE^{\frac{m+1}{2}}\phi'dt
+C_{\varepsilon_1}E(S)\big].
\end{aligned}\label{2.23}
\end{equation}
Noticing \eqref{1.6} and \eqref{2.1}, it is easy to obtain
\begin{equation}
\begin{aligned}
\int_S^TE^r\phi'\int_{\Omega_1^+}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt
&\leq c_4\sigma(0)|a|_\infty  \int_S^TE^r\phi'(-E'(t)) dt\\
&\leq CE^{r+1}(S).
\end{aligned}\label{2.24}
 \end{equation}
On the other hand, it is easy to obtain
\begin{equation}
\int_S^TE^r\phi'\int_{\mathbb{R}^n\setminus\Omega_1}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt
\leq CE^{r+1}(S)\label{2.25}
\end{equation}
 since $\rho(x,u_t)=a(x)u_t$  when  $x\in\mathbb{R}^n\setminus\Omega_1$.

From \eqref{2.23}-\eqref{2.25}, we have
\begin{equation}
\begin{aligned}
&\int_S^TE^r\phi'\int_{\mathbb{R}^n}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt\\
&\leq C\varepsilon_1|a|_\infty^{2-\frac{2}{m+1}}\sigma^2(0)
 \int_S^TE^{\frac{m+1}{2}}\phi'dt+C_{\varepsilon_1}E(S)+CE^{r+1}(S).
\end{aligned} \label{2.26}
\end{equation}
Similarly, for the last term on the right side of \eqref{2.18}, we have
\begin{equation}
\begin{aligned}
& \int^T_SE^{r}\phi' \int_{\Omega_1}\sigma(t)a(x)u_t^2\,dx\,dt\\
&\leq C\varepsilon_1|a|_\infty^{\frac{m-1}{m+1}}\sigma(0)
\int_S^TE^{\frac{m+1}{2}}\phi'dt+C_{\varepsilon_1}E(S)+CE^{r+1}(S).
\end{aligned} \label{2.27}
\end{equation}
At this point, we choose $\varepsilon$ and $\eta$ small enough so that
\[
 2\varepsilon\sigma_4+4\varepsilon\sigma_5+\eta(\frac{k\sigma_0\sigma_2}{4}
+\sigma_3+C_\varepsilon\sigma_5)<\frac{\bar{\sigma}}{2}.
\]
Then using \eqref{2.19}-\eqref{2.21},  \eqref{2.18} becomes
\begin{equation}
\begin{aligned}
&\frac{\bar{\sigma}}{2}\int_S^TE^{r+1}(t)\phi'dt\\
&\leq CE^{r+1}(S) +\big(\sigma_4C_\varepsilon+\sigma_5 C_\varepsilon
+C_\eta(\frac{k\sigma_0\sigma_2}{4}+\sigma_3+C_\varepsilon\sigma_5)\big) \\
&\times \int_S^TE^r\phi'\int_{\mathbb{R}^n}|\sigma(t)\rho(x,u_t)|^2\,dx\,dt
-\frac{k}{2}\int^T_SE^r\phi' E'dt \\
&\quad +\big(\frac{k}{4}+C_\eta(\frac{k\sigma_0\sigma_2}{4}+\sigma_3
 +C_\varepsilon\sigma_5)\big)\int^T_SE^{r}\phi'
\int_{\Omega_1}\sigma(t)a(x)u_t^2\,dx\,dt.
\end{aligned}\label{2.28}
\end{equation}
Once $\varepsilon$ and $\eta$ are  fixed, we pick $\varepsilon_1$ small
enough so that
$$
C\varepsilon_1|a|_\infty^{\frac{m-1}{m+1}}\big(\sigma(0)+\sigma^2(0)\big)
(\sigma_4C_\varepsilon+\sigma_5 C_\varepsilon+\frac{k}{4}
+C_\eta(\frac{k\sigma_0\sigma_2}{4}+\sigma_3
+C_\varepsilon\sigma_5)\big)\leq \frac{\bar{\sigma}}{4}.
$$
Then using \eqref{2.26}-\eqref{2.27}  in \eqref{2.28},   we have
\begin{equation}
 \int_S^TE^{\frac{m+1}{2}}\phi'dt\leq CE(S).\label{2.29}
\end{equation}
Let $T\to\infty$ in \eqref{2.29}.
Then Theorem \ref{thm1.1} follows from Lemma \ref{lem2.1}  and \eqref{2.29}
with $r=\frac{m-1}{2}$ and $\phi(t)=\int_0^t\sigma(s)ds$.
\end{proof}

\subsection*{Acknowlegments}
This work was supported by the National Science Foundation of China for
the Youth (61104129,11401351),
by the Youth Science Foundation of Shanxi Province (2011021002-1),
by the National Science Foundation of China (11171195, 61174082 and 61174083).


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 \end{document}