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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 12, 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 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/12\hfil Blow-up and general decay of solutions]
{Blow-up and general decay of solutions
for a nonlinear viscoelastic equation}

\author[W. Chen, Y. Xiong \hfil EJDE-2013/12\hfilneg]
{Wenying Chen, Yangping Xiong}  % in alphabetical order

\address{Wenying Chen \newline
College of Mathematics and Statistics,
Chongqing Three Gorges University,
Chongqing 404000, China}
\email{wenyingchenmath@yahoo.com}

\address{Yangping Xiong \newline
Department of Mathematics,
Zhejiang Normal University, Jinhua 321004, China}
\email{xiongyangping@gmail.com}


\thanks{Submitted November 19, 2012. Published January 14, 2013.}
\subjclass[2000]{35B45, 35B65, 35Q30, 76D05}
\keywords{Blow-up; decay; viscoelastic equation}

\begin{abstract}
 In this article we investigate a nonlinear viscoelastic equation
 that admits blow-up and decay. First, we establish blow-up
 results for this equation, even for vanishing initial energy.
 Then, we show that the solutions decay under suitable conditions.
\end{abstract}

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

\def\p{{\partial}}
\def\n{\nabla}
\def\L{{\Lambda}}
\def\D{{\Delta}}

\section{Introduction}

In this article, we consider the viscoelastic equation
\begin{equation}\label{1.1}
 \begin{gathered}
 u_{tt}-\Delta u+\int^t_0 g(t-\tau)\Delta u(\tau)d\tau
 +u_t=u|u|^{p-1}, \quad (x,t)\in\Omega\times(0,\infty), \\
 u(x,t)=0, \quad x\in\partial\Omega, t\geq 0,\\
 u(x,0)=u_0(x), \quad u_t(x,0)=u_1(x), \quad x\in\Omega.
\end{gathered}
\end{equation}
where $\Omega$ is a bounded domain of $\mathbb{R}^n$ $(n\geq 1)$
with a smooth boundary $\partial\Omega$, $p>1$, and $g$ is
a positive nonincreasing function.

There have been extensive studies on some special cases of this
equation and the physical background is also given in these works;
see \cite{B,C,J,L,L1,L2,M4,V,X,B1,Z2} and references
therein. For instance, the equation without $u_t$ is studied in
\cite{B}, the local existence theorem is established, and for
certain initial data and suitable conditions on $g$ and $p$, that
this solution is global with energy which decays exponentially or
polynomially depending on the rate of the decay of the relaxation
function $g$. In the absence of the viscoelastic term $(g = 0)$, for
instance, the equation
\begin{equation}
u_{tt}-\Delta u+a u_t|u_t|^m=b
u|u|^{\gamma}, \quad (x,t)\in\Omega\times(0,\infty),
\end{equation}
 we know that the source term $b u|u|^{\gamma}(\gamma>0)$ causes
finite-time blow-up of solutions with negative initial energy when
$a=0$, cf. \cite{J}. The interaction between the damping and the
source terms was first considered by Levine \cite{L1,L2} for the
linear damping case $(m=0)$. He showed that solutions with negative
initial energy blow up in finite time. Recently, In \cite{Z2}, it is
proved that the solution blows up in finite time even for vanishing
initial energy. Another case with time dependent damping $b(t)u_t$
is studied in \cite{N3}. Georgiev and Todorova \cite{V} extended
Levine¡¯s result to the nonlinear damping case $(m>0)$. In \cite{C},
it is showed that the solution blows up in finite time even for
vanishing initial energy. We mention the work of Liu and Zhou
\cite{L}, the equation
\begin{equation}
 u_{tt}-\Delta u=a^{-k} |u|^{\gamma}, \quad
(x,t)\in \mathbb{R}^n\times(0,\infty),
\end{equation}
is studied, it is proved that the solutions blow up in finite time with more
relaxed initial data and extended index $\gamma$.

For the problem \eqref{1.1} in $\mathbb{R}^n$, Mohammad Kafinia and
Salim Messaoudib in \cite{K} give a finite-time blow-up result
under suitable conditions on the initial data and the relaxation
function, this work extend the result of \cite{Z1}, established for
the wave equation, to the problem \eqref{1.1} in $\mathbb{R}^n$. In
this paper we improve the result of blow-up in \cite{K}, and discuss
the phenomenon of decay for the solution of equation \eqref{1.1}.
This is an important breakthrough, since it is only well known that
the solution blows up in finite time if the initial energy is
negative from all the previous literature.

Now, we list some notation that will be
used in our paper. Use $\|\cdot\|_p$ to denote the $L^p(\mathbb{R}^n)$ norm.
Throughout this paper, $C$ denotes a generic positive constant (generally large),
it may be different from line to line.

The remainder of the paper will be organized as follows. In the next
section, we review some preliminaries that will be used in the proof of
our main theorems.
Then, the blow-up phenomenon will be considered in
Section \ref{section3}. In the last Section, we discuss the
decay of the solution to equation \eqref{1.1}.


\section{Preliminaries}

In this section we review some preliminaries that will be used in the proof
of our main theorems. Throughout this paper,
\[
\frac{n+2}{[n-2]_+}= \begin{cases}
 \infty, & n=1, 2, \\
 \frac{n+2}{n-2}, & n\geq3.
 \end{cases}
\]
The relaxation function $g$ satisfies:
\begin{itemize}
\item[(H1)] $g:\mathbb{R}_+\to \mathbb{R}_+$ is a differentiable
 function such that
\[
g(0)>0, \quad 1-\int^\infty_0 g(\tau)d\tau=l>0, \quad t\geq 0.
\]
\item[(H2)] There exists a positive differentiable function $\xi(t)$ such that
\[ %\label{2.0}
g'(t)\leq-\xi(t)g(t), \quad t\geq 0.
\]
and
\[
\big|\frac{\xi'(t)}{\xi(t)}\big|\leq k, \xi(t)>0, \xi'(t)\leq 0, t>0.
\]
\end{itemize}

\begin{remark} \label{rmk2.1} \rm
Since $\xi$ is nonincreasing, then $\xi(t)\leq\xi(0)=M$.
\end{remark}

The embedding $H_0^1(\Omega)\hookrightarrow L^q(\Omega)$ for
 $2\leq q\leq \frac{2n}{n-2}$, if $n\geq 3$ and $q\geq 2$,
if $n=1,2$; $L^r(\Omega)\hookrightarrow L^q(\Omega)$ for $q<r$; that is to say,
there exists a constant $C_e$, such that
\begin{equation} \label{em}
\|u\|_q\leq C_e\|\nabla u\|_2, \quad \|u\|_q\leq C_e\|u\|_r.
\end{equation}
These two inequalities will be used frequently in this article.

We define the  energy corresponding to problem \eqref{1.1} as
\begin{equation} \label{2.10}
E(t)=\frac{1}{2}\|u_t\|^2_2+\frac{1}{2}
\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+\frac{1}{2}(g\circ\n u)(t)-\frac{1}{p+1}\|u\|^{p+1}_{p+1},
\end{equation}
here
\[
(g\circ v)(t)=\int^t_0 g(t-\tau)\|v(t)-v(\tau)\|^2_2d\tau\,.
\]
By a direct calculation we obtain
\begin{equation} \label{2.2}
E'(t)=\frac{1}{2}(g'\circ\n u)(t)-\frac{1}{2}g(t)\|\n u\|^2_2
-\|u_t\|^2_2\leq\frac{1}{2}(g'\circ\n u)(t)\leq 0.
\end{equation}
Hence, we can deduce that $E(t)\leq E(0)$ for $t\geq 0$.

\begin{remark} \label{rmk2.2} \rm
The largest $T$ for which the solution exists on $[0,T)\times\mathbb{R}^n$
is called the lifespan of the solution of \eqref{1.1}.
The supremum of the $T$'s is denoted by  $T^*$ .
If $T^*=\infty$, we say the solution is global
while $T^*<\infty$ we say that solution blows up in finite time.
\end{remark}

\begin{lemma} \label{lem2.3}
If $p$ satisfies $p<\frac{n+2}{[n-2]_+}$, then there exists a positive
constant $C>1$, such that
\begin{equation} \label{2.1}
\|u\|_{p+1}^s\leq C\left(\|\n u\|_2^2+\|u\|^{p+1}_{p+1}
\right)\quad \text{with } 2\leq s\leq p+1,
\end{equation}
for any $u$ being a solution of \eqref{1.1} on $[0,T)$. Consequently,
\begin{equation} \label{2.5}
\|u\|_{p+1}^s \leq C\left(H(t)+\|u_t\|^2_2+(g\circ\n u)(t)+\|\n u\|_2^2
\right)\quad \text{with }2\leq s\leq p+1,
\end{equation}
on $[0,T)$ and here $H(t):=-E(t)$.
\end{lemma}

\begin{proof}
If $\|u\|_{p+1} \leq 1$, the estimate
$\|u\|_{p+1}^s \leq \|u\|_{p+1}^2 \leq B^2\|\n u\|_2^2 $ is true.

If $\|u\|_{p+1}>1$, we have
$\|u\|_{p+1}^s \leq \|u\|_{p+1}^{p+1}$. Combining the two inequalities we obtain \eqref{2.1}.

Note that \eqref{2.5} follows from \eqref{2.1} and the
definition of energy corresponding to the solution.
\end{proof}

\section{Blow-up phenomenon}\label{section3}

\begin{theorem} \label{theorem31}
Assume that {\rm (H1), (H2)} hold, $1<p<\frac{n+2}{[n-2]_+}$,
$\int^\infty_0 g(\tau)d\tau<\frac{(p+1)(p-1)
}{1+(p+1)(p-1)}$ and $E(0)< 0$. Then the solution blows up
in finite time.
\end{theorem}

\begin{proof}
From the definition of $H(t)$, we have
\[
H'(t)=-\frac{1}{2}(g'\circ\n u)(t)
+\frac{1}{2}g(t)\|\n u\|^2_2+\|u_t\|^2_2\geq 0,
\]
and
\[
0<H(0)\leq H(t)\leq\frac{1}{p+1}\|u\|^{p+1}_{p+1}.
\]
Moreover, we define
\[
L(t)=H^{1-\alpha}(t)+\epsilon\int_\Omega uu_t\,dx,
\]
here $\epsilon$ small to be chosen later, $0<\alpha\leq
\frac{p-1}{2(p+1)}$.

By differentiating the above equality, we have
\begin{equation} \label{3.30}
\begin{aligned}
L'(t)&= (1-\alpha)H^{-\alpha}(t)H'(t)+\epsilon\int_\Omega |u_t|^2dx
+\epsilon\int_\Omega uu_{tt}dx \\
&= (1-\alpha)H^{-\alpha}(t)\Big(-\frac{1}{2}(g'\circ\n u)(t)
+\frac{1}{2}g(t)\|\n u\|^2_2+\|u_t\|^2_2\Big)\\
&\quad +\epsilon\|u_t\|^2_2-\epsilon\|\n u\|^2_2
 +\epsilon\int_\Omega \n u(t)\int^t_0 g(t-\tau)\n u(\tau)\,d\tau\,dx\\
&\quad -\epsilon\int_\Omega uu_tdx+\epsilon\|u\|^{p+1}_{p+1}.
\end{aligned}
\end{equation}
Using Young and Schwarz inequality, we obtain
\begin{gather} \label{3.31}
\begin{aligned}
&\int_\Omega\n u(t)\int^t_0 g(t-\tau)\n u(\tau)\,d\tau\,dx\\
&\geq -\delta\|\n u\|^2_2-\frac{1}{4\delta}\Big(\int^t_0 g(\tau)d\tau\Big)
(g\circ\n u)(t)
+\Big(\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2\,,
\end{aligned}\\
\label{3.32}
\int_\Omega uu_tdx\leq\frac{\delta^2}{2}\|u\|^2_2
+\frac{\delta^{-2}}{2}\|u_t\|^2_2
\end{gather}
Inserting \eqref{3.31} and \eqref{3.32} into \eqref{3.30}, we deduce
\begin{align*}
L'(t)&\geq (1-\alpha)H^{-\alpha}(t)\Big(-\frac{1}{2}(g'\circ\n u)(t)
+\frac{1}{2}g(t)\|\n u\|^2_2+\|u_t\|^2_2\Big)+\epsilon\|u_t\|^2_2 \\
&\quad +\Big(-1-\delta+\int^t_0 g(\tau)d\tau\Big)\epsilon\|\n u\|^2_2
-\frac{\epsilon}{4\delta}\Big(\int^t_0 g(\tau)d\tau\Big)(g\circ\n u)(t) \\
&\quad -\frac{\epsilon\delta^2}{2}\|u\|^2_2
  -\frac{\epsilon\delta^{-2}}{2}\|u_t\|^2_2 +\epsilon\|u\|^{p+1}_{p+1}.
\end{align*}
If we set $\delta^2=kH^\alpha$, $\delta^{-2}=k^{-1}H^{-\alpha}$, $k>0$
and we have
\[
H^\alpha(t)\|u\|^2_2\leq C(\frac{1}{p+1})^\alpha \|u\|^{2+\alpha (p+1)}
_{p+1}.
\]
Then
\begin{align*}
L'(t)&\geq
\Big(1-\alpha-\frac{\epsilon}{2k}\Big)H^{-\alpha}(t)\|u_t\|^2_2
+\Big[p+1-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha\Big]\epsilon H(t) \\
&\quad +\Big[\frac{p-1}{2}\Big(1-\int^t_0g(\tau)d\tau\Big)-\delta
-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha\Big]\epsilon\|\n u\|^2_2 \\
&\quad +\Big[\frac{p+1}{2}-\frac{1}{4\delta}\int^t_0 g(\tau)d\tau
-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha\Big]\epsilon(g\circ\n u)(t) \\
&\quad +\Big[\frac{p+3}{2}-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha\Big]
\epsilon\|u_t\|^2_2.
\end{align*}
According to the hypothesis in Theorem 3.1 and take $k$ and $\delta$
to be small enough such that
\begin{gather*}
\frac{p-1}{2}\Big(1-\int^t_0g(\tau)d\tau\Big)-\delta
-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha> 0,\\
\frac{p+1}{2}-\frac{1}{4\delta}\int^t_0 g(\tau)d\tau
-\frac{kC}{2}\Big(\frac{1}{p+1}\Big)^\alpha>0.
\end{gather*}
Choose $\epsilon$ ($k$ is fixed) small enough such that
\[
1-\alpha-\frac{\epsilon}{2k}\geq 0, \quad
L(0)=H^{1-\alpha}(0)+\epsilon \int_\Omega u_0 u_1dx > 0.
\]
Then, we can deduce that
\[
L'(t) \geq C[H(t)+\|u_t\|^2_2+\|\n u\|^2_2+(g \circ\n u)(t)].
\]
Thanks to H\"{o}lder and Young inequality, we obtain
\begin{equation} \label{3.33}
\begin{split}
\Big|\int_\Omega uu_tdx\Big|^{1/(1-\alpha)}
&\leq \|u\|^{1/(1-\alpha)}_2 \|u_t\|^{1/(1-\alpha)}_2
\leq C\|u\|^{1/(1-\alpha)}_{p+1} \|u_t\|^{1/(1-\alpha)}_2 \\
&\leq C(\|u\|^s_{p+1}+ \|u_t\|^2_2) \\
&\leq C\left(H(t)+\|u_t\|^2_2+(g\circ\n u)(t)+\|\n u\|_2^2\right),
\end{split}
\end{equation}
where $2\leq s= \frac{2}{1-2\alpha}\leq p+1$. Hence,
\begin{align*}
L^{1/(1-\alpha)}(t)
&= \Big(H^{1- \alpha }(t)+ \epsilon\int _\Omega
uu_t dx\Big)^{1/(1-\alpha)}\\
&\leq 2^{1/(1-\alpha)}\Big(H(t) +\Big|\int _\Omega uu_t dx\Big|^{1/(1-\alpha)}
\Big) \\
&\leq C\left(H(t)+\|u_t\|^2_2+(g\circ\n u)(t)+\|\n u\|_2^2\right),
\end{align*}
which implies that $L'(t)\geq \lambda L^{1/(1-\alpha)}(t)$,
where $\lambda$ is a constant depending on $C$, $p$, $\alpha$ and $\epsilon$.
 Therefore
\[
L(t)\geq(L^{\frac{-\alpha}{1-\alpha}}(0)
+ \frac{-\alpha}{1-\alpha}\lambda t)^{-\frac{1-\alpha}{\alpha}}.
\]
So $L(t)$ approaches infinite as $t$ tends to
$(1-\alpha)/\big(\alpha \lambda L^{\frac{\alpha}{1-\alpha}}(0)\big)$.
This completes the proof.
\end{proof}

To obtain another blow-up result we first give the following lemma.

\begin{lemma} \label{lem3.2}
Assume that {\rm (H1), (H2)} hold, additionally, assume that
\[
\|u_0\|_{p+1}>\lambda_0\equiv B_0^{\frac{-2}{p-1}}, \quad
E(0)<E_0=\Big(\frac{1}{2}-\frac{1}{p+1}\Big)B_0^{\frac{-2(p+1)}{p-1}}.
\]
Then
\[
\|u\|_{p+1}>\lambda_0, \quad
\|\n u\|_2 >B_0^{\frac{-(p+1)}{p-1}},\quad\text{for all }t \geq 0,
\]
where $B_0=\frac{B}{l^{1/2}}$ for $\|u\|_{p+1}\leq\ B\|\n u\|_2$.
\end{lemma}

\begin{proof}
From \eqref{2.10} and the hypothesis, we know that
\begin{align*}
E(t)&=\frac{1}{2} \|u_t\|^2_2+ \frac{1}{2}
\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+\frac{1}{2}(g\circ\n u)(t)-\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&\geq \frac{1}{2}\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
-\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&\geq \frac{l}{2}\|\n u\|^2_2-\frac{1}{p+1}\|u\|^{p+1}_{p+1}
\geq\frac{1}{2B^2_0}\|u\|^2_{p+1}-\frac{1}{p+1}\|u\|^{p+1}_{p+1}.
\end{align*}
Set $h(\xi)= \frac{1}{2B_0^2}\xi^2- \frac{1}{p+1}\xi^{p+1}$, $\xi \geq 0$.
Then $h(\xi)$ satisfies
 \begin{itemize}
\item $h(\xi)$ is strictly increasing on $[0,\lambda_0)$;

\item $h(\xi)$ takes its maximum value
$(\frac{1}{2}-\frac{1}{p+1})B_0^{\frac{-2(p+1)}{p-1}}$ at $\lambda_0$;

\item $ h(\xi)$ is strictly decreasing on $(\lambda_0, \infty)$.
\end{itemize}
Since $E_0 > E(0)\geq E(t) \geq h(\|u\|_{p+1})$ for all $t\geq0$, there is
no time $t^*$ such that $\|u(\cdot,t^*)\|_{p+1}=\lambda_0$. By the continuity
of the $\|u(\cdot,t)\|_{p+1}-norm$ with respect to the time variable, one has
\[
\|u(\cdot,t)\|_{p+1} > \lambda_0=B_0^{\frac{-2}{p-1}}\quad
 \text{for all $t \geq 0$},
\]
and consequently,
\[
\|\n u(\cdot,t)\|_2 \geq \frac{1}{l^{1/2}B_0}\|u(\cdot,t)\|_{p+1}
>\frac{1}{l^{1/2}}B_0^{\frac{-(p+1)}{p-1}}>B_0^{\frac{-(p+1)}{p-1}}.
\]
This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.3}
Suppose that{\rm (H1), (H2)} hold, $1<p<\frac{n+2}{[n-2]_+}$,
$$
\int^\infty_0 g(\tau)d\tau<\frac{(p+1)(p-1)}{1+(p+1)(p-1)},
$$
 $\|u_0\|_{p+1}>\lambda _0$ and $ E(0)\leq E_0$.
Then the solution of \eqref{1.1} blows up in finite time.
\end{theorem}

\begin{proof}
Set $G(t)=E_0+H(t)$, then
\[
G'(t)=-\frac{1}{2}(g'\circ\n u)(t)+\frac{1}{2}g(t)\|\n u\|^2_2+\|u_t\|^2_2\geq 0,
\]
from which we obtain
\begin{align*}
0< G(t)
&= E_0+H(t)=\big(\frac{1}{2}-\frac{1}{p+1}\big)B_0^
{\frac{-2(p+1)}{p-1}}+H(t) \\
&< \big(\frac{1}{2}-\frac{1}{p+1}\big)\|\n u\|^2_2+H(t)
<C(\|\n u\|^2_2+H(t))
\end{align*}
and
\begin{align*}
0&< G(t)\\
&= E_0-\frac{1}{2}\|u_t\|^2_2-\frac{1}{2}
\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
-\frac{1}{2}(g\circ\n u)(t)+\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&\leq E_0-\frac{1}{2}\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&\leq \big(\frac{1}{2}-\frac{1}{p+1}\big)B_0^
{\frac{-2(p+1)}{p-1}}-\frac{l}{2}\big(\frac{1}{l^{1/2}}\big)^2B_0^
{\frac{-2(p+1)}{p-1}}+\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&\leq \frac{1}{p+1}\|u\|^{p+1}_{p+1}.
\end{align*}
Let
\[
Q(t)=G^{1-\alpha}(t)+\epsilon\int_\Omega uu_tdx,
\]
with $\epsilon$ small to be chosen later, $0<\alpha\leq
\frac{p-1}{2(p+1)}$.

By the same process as in the proof of Theorem \ref{theorem31},
deduce that
\[
 Q'(t)\geq C[H(t)+\|u_t\|^2_2+\|\n u\|^2_2+(g \circ\n u)(t)].
\]
Thanks to \eqref{3.33}, we obtain
\begin{align*}
Q^{1/(1-\alpha)}(t)
&=  \Big(G^{1- \alpha }(t)+ \epsilon\int
_\Omega uu_t dx\Big)^{1/(1-\alpha)}\\
&\leq 2^{1/(1-\alpha)}\Big(G(t)
+\Big|\int _\Omega uu_t dx\Big|^{1/(1-\alpha)}\Big)
\\
&\leq  C\left(H(t)+\|u_t\|^2_2+(g\circ\n u)(t)+\|\n
u\|_2^2\right),
\end{align*}
which implies that $Q'(t)\geq \lambda Q^{1/(1-\alpha)}(t)$, where
$\lambda$ is a constant depending
on $C$, $p$, $\alpha$ and $\epsilon$. Therefore
\[
Q(t)\geq(Q^{\frac{-\alpha}{1-\alpha}}(0)+
\frac{-\alpha}{1-\alpha}\lambda t)^{- \frac{1-\alpha}{\alpha}}.
\]
 So $Q(t)$ approaches infinite as $t$ tends to
$\frac{1-\alpha}{\alpha \lambda Q^{\frac{\alpha}{1-\alpha}}(0)}$.
This completes the proof.
\end{proof}

\section{Decay solutions}

The purpose of this section is to give a decay result of the solution.
Set
\[
I(t)=\Big(1-\int^t_0 g(\tau)d\tau\Big)
\|\n u\|^2_2+(g\circ\n u)(t)-\|u\|^{p+1}_{p+1}.
\]
As in \cite{M4},  to give our decay result, we first prove
the following lemmas.

\begin{lemma} \label{lem4.1}
Suppose that {\rm (H1), (H2)} hold, $p<\frac{n+2}{[n-2]_+}$,
and $(u_0,u_1)\in H_0^1(\Omega)\times L^2(\Omega)$ such that
\begin{equation} \label{2.3}
\beta=\frac{C_e^{p+1}}{l}\Big(\frac{2(p+1)E(0)}{(p-1)l}\Big)^{(p-1)/2}<1,
I(u_0)>0,
\end{equation}
then $I(u(t))>0$, for all $t>0$. Here $C_e$ is given in \eqref{em}.
\end{lemma}

\begin{proof}
Since $I(u_0)>0$,  there exists $T_m<T$, such that
\[
I(u(t))>0, \quad \forall t\in[0,T_m],
\]
which gives
\begin{align*}
&\frac{1}{2}\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+\frac{1}{2}(g\circ\n u)(t)-\frac{1}{p+1}\|u\|^{p+1}_{p+1} \\
&= \frac{p-1}{2(p+1)}\Big[\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+(g\circ\n u)(t)\Big]+\frac{1}{p+1}I(t) \\
&\geq \frac{p-1}{2(p+1)}\Big[\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
+(g\circ\n u)(t)\Big].
\end{align*}
So we have
\begin{equation} \label{2.7}
l\|\n u\|^2_2\leq\Big(1-\int^t_0 g(\tau)d\tau\Big)\|\n u\|^2_2
\leq\frac{2(p+1)}{p-1}E(t)\leq\frac{2(p+1)}{p-1}E(0).
\end{equation}
By using (H1), \eqref{2.3} and \eqref{2.7}, we obtain
\[
\|u\|^{p+1}_{p+1}\leq C_e^{p+1}\|\n u\|^{p+1}_2\leq \beta l\|\n u\|^2_2
<\left(1-\int^t_0 g(\tau)d\tau\right)\|\n u\|^2_2
\]
Hence,
\[
I(t)=\left(1-\int^t_0 g(\tau)d\tau\right)\|\n u\|^2_2
+(g\circ\n u)(t)-\|u\|^{p+1}_{p+1}>0,\quad \forall t\in [0,T_m].
\]
By repeating this process, and using  that
\[
\lim_{t\to T_m}\frac{C_e^{p+1}}{l}\Big(\frac{2(p+1)E(u,u_t)}{(p-1)l}\Big)^{(p-1)/2}\leq\beta<1,
\]
we show that $T_m$ is extended to $T$.
\end{proof}

To establish the decay rate, we use the functional
\begin{equation} \label{3.1}
F(t)=E(t)+\epsilon_1\Psi(t)+\epsilon_2\Phi(t),
\end{equation}
where $\epsilon_1$ and $\epsilon_2$ are positive constants and
\[
\Psi(t)=\xi(t)\int_\Omega uu_tdx, \quad
\Phi(t)=-\xi(t)\int_\Omega u_t\int_0^tg(t-\tau)(u(t)-u(\tau))\,d\tau\,dx.
\]
This functional, for $\xi(t)=1$, was first introduced in
\cite{B} and \cite{B1}. Now, let us consider some useful properties
of this functional.

\begin{lemma} \label{lem4.2}
Assume that $u(x, t)$ is the solution of \eqref{1.1} and that \eqref{2.3} holds.
Then there exists $k_1<1$ and $k_2>1$ such that
\begin{equation} \label{4.2}
k_1E(t)\leq F(t)\leq k_2E(t).
\end{equation}
\end{lemma}

\begin{proof}
Using  Young, Schwarz and Poincar\'e inequality, we obtain
\begin{gather} \label{3.4}
\int_\Omega uu_tdx\leq\frac{C_*^2}{2}\|\n u\|^2_2+\frac{1}{2}\|u_t\|^2_2,\\
 \label{3.5}
\int_\Omega u_t\int_0^tg(t-\tau)(u(t)-u(\tau))\,d\tau\,dx
\leq \frac{1}{2}\|u_t\|^2_2+\frac{1}{2}(1-l)C_*^2(g\circ\n u)(t).
\end{gather}
Using \eqref{3.4} and \eqref{3.5}, we have
\begin{align*}
k_2E(t)-F(t)
&\geq \Big[\Big(\frac{k_2-1}{2}-\frac{k_2-1}{p+1}
\Big)l-\frac{\epsilon_1C_*^2M}{2}\Big]\|\n u\|^2_2 \\
&\quad +\frac{1}{2}\{k_2-1-(\epsilon_1+\epsilon_2)M\}\|u_t\|^2_2
+\frac{k_2-1}{p+1}I(t) \\
&\quad+\Big[\frac{k_2-1}{2}-\frac{k_2-1}{p+1}-\frac{\epsilon_2(1-l)
C_*^2M}{2}\Big](g\circ\n u)(t).
\end{align*}
Similarly,
\begin{align*}
F(t)-k_1E(t)
&\geq \Big[\Big(\frac{1-k_1}{2}-\frac{1-k_1}{p+1}
\Big)l-\frac{\epsilon_1C_*^2M}{2}\Big]\|\n u\|^2_2 \\
&\quad +\frac{1}{2}[1-k_1-(\epsilon_1+\epsilon_2)M]\|u_t\|^2_2+\frac{1-k_1}{p+1}
I(t) \\
&\quad+\Big[\frac{1-k_1}{2}-\frac{1-k_1}{p+1}-\frac{\epsilon_2(1-l)
C_*^2M}{2}\Big](g\circ\n u)(t).
\end{align*}
 By choosing $\epsilon_1$ and $\epsilon_2$ small enough,
such that $k_2E(t)-F(t)\geq 0$ and
$F(t)-k_1E(t)\geq 0$, we complete the proof.
\end{proof}

\begin{lemma} \label{lem4.3}
Let {\rm (H1) and (H2)} hold, and $p\leq\frac{n+2}{[n-2]_+}$.
Assume that $(u_0,u_1)\in H_0^1(\Omega)\times L^2(\Omega)$ and
$u$ is the solution of \eqref{1.1}.
Then
\begin{equation} \label{3.8}
\begin{split}
\Psi'(t)
&\leq \Big(1+\frac{(1-k)(1+k)C_*^2}{l}\Big)\xi(t)\|u_t\|^2_2
+\frac{1-l}{2l}\xi(t)(g\circ\n u)(t) \\
&\quad-\frac{l}{4}\xi(t)\|\n u\|^2_2+\xi(t)\|u\|^{p+1}_{p+1}
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
By a direct computation, we have
\begin{equation} \label{3.9}
\begin{split}
\Psi'(t)
&= \xi(t)\Big(\|u_t\|^2_2+\|u\|^{p+1}_{p+1}-\|\n u\|^2_2
+\int_\Omega\n u(t)\int_0^tg(t-\tau)\n u(\tau)\,d\tau\,dx \\
&\quad -\int_\Omega uu_tdx\Big)+\xi'(t)\int_\Omega uu_tdx \\
&:=\xi(t)\left(\|u_t\|^2_2+\|u\|^{p+1}_{p+1}-\|\n u\|^2_2+A_1-A_2
\right)+\xi'(t)A_2.
\end{split}
\end{equation}
By Young, Schwarz and Poincar\'{e} inequality, we have
\begin{gather}
A_1\leq \frac{1}{2}\|\n u\|^2_2+\frac{1}{2}\big(1+\frac{1}{\eta}\big)
(1-l)(g\circ\n u)(t)
 +\frac{1}{2}(1+\eta)(1-l)^2\|\n u\|^2_2, \label{3.10}
\\
A_2 \leq \alpha C_*^2\|\n u\|^2_2+\frac{1}{4\alpha}\|u_t\|^2_2.\label{3.11}
\end{gather}
Combining \eqref{3.9} and \eqref{3.10} with \eqref{3.11} yields
\begin{align*}
\Psi'(t)
&\leq \big(1+\frac{1-k}{4\alpha}\big)\xi(t)\|u_t\|^2_2
+\frac{1}{2}\big(1+\frac{1}{\eta}\big)(1-l)\xi(t)(g\circ\n u)(t) \\
&\quad -\Big[\frac{1}{2}-\frac{(1+\eta)(1-l)^2}{2}-(1+k)\alpha C_*^2\Big]
\xi(t)\|\n u\|^2_2+\xi(t)\|u\|^{p+1}_{p+1}.
\end{align*}
We choose $\eta=l/(1-l)$ and $\alpha=l/(4(1+k)C_*^2)$;
then \eqref{3.8} is true.
\end{proof}

\begin{lemma} \label{lem4.4}
Let {\rm (H1)} and {\rm (H2)} hold, $p\leq\frac{n+2}{[n-2]_+}$,
$(u_0,u_1)\in H_0^1(\Omega)\times L^2(\Omega)$ and $u$ is the solution
of \eqref{1.1}.
Then
\begin{equation} \label{3.14}
\begin{split}
\Phi'(t)
&\leq \delta\Big[1+2(1-l)^2+C_e^{2p}
\Big(\frac{2(p+1)E(0)}{l(p-1)}\Big)^{p-1}\Big]\xi(t)\|\n u\|^2_2 \\
&\quad-\frac{g(0)C_*^2}{4\delta}\xi(t)(g'\circ\n u)(t)
+\Big[\big(2\delta +\frac{1}{2\delta}\big)(1-l)
+\frac{(2+k)(1-l)C_*^2}{4\delta}\Big] \\
&\quad \times\xi(t)(g\circ\n u)(t)+\Big[\delta(2+k)-\int_0^tg(\tau)d\tau
\Big]\xi(t)\|u_t\|^2_2.
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
Straightforward computations show that
\begin{equation} \label{3.15}
\begin{split}
\Phi'(t)&= \xi(t)\int_\Omega\n u\int_0^tg(t-\tau)(\n u(t)-\n u(\tau))
\,d\tau\,dx \\
&\quad -\xi(t)\int_\Omega\Big(\int_0^tg(t-\tau)\n u(\tau)d\tau\Big)
\Big(\int_0^tg(t-\tau)(\n u(t)-\n u(\tau))d\tau\Big)dx \\
&\quad +\xi(t)\int_\Omega u_t\int_0^tg(t-\tau)(u(t)-u(\tau))\,d\tau\,dx \\
&\quad -\xi(t)\int_\Omega u|u|^{p-1}\int_0^tg(t-\tau)(u(t)-u(\tau))\,d\tau\,dx \\
&\quad -\xi(t)\int_\Omega u_t\int_0^tg'(t-\tau)(u(t)-u(\tau))\,d\tau\,dx
-\xi(t)\Big(\int_0^tg(\tau)d\tau\Big)\|u_t\|^2_2 \\
&\quad -\xi'(t)\int_\Omega u_t\int_0^tg(t-\tau)(u(t)-u(\tau))\,d\tau\,dx \\
&:=\xi(t)\Big[I_1+I_2+I_3+I_4+I_5-\Big(\int_0^tg(\tau)d\tau\Big)\|u_t\|^2_2\Big]
-\xi'(t)I_3.
\end{split}
\end{equation}
By Young and Poincar\'{e} inequality, we have
\begin{gather}
I_1\leq \delta\|\n u\|^2_2+\frac{1-l}{4\delta}(g\circ\n u)(t),\label{3.16}\\
I_2\leq \Big(2\delta+\frac{1}{4\delta}\Big)(1-l)(g\circ\n u)(t)
+2\delta(1-l)^2\|\n u\|^2_2,\label{3.17}\\
I_3\leq \delta\|u_t\|^2_2+\frac{C_*^2(1-l)}{4\delta}(g\circ\n u)(t),\label{3.18}\\
I_4\leq \delta C_e^{2p}\Big(\frac{2(p+1)E(0)}{l(p-1)}\Big)^{p-1}\|\n u\|^2_2
+\frac{(1-l)C_*^2}{4\delta}(g\circ\n u)(t),\label{3.19}\\
I_5\leq \delta\|u_t\|^2_2-\frac{g(0)C_*^2}{4\delta}(g'\circ\n u)(t).\label{3.20}
\end{gather}
Combining \eqref{3.15}-\eqref{3.20}, we have the required estimate \eqref{3.14}.
\end{proof}

We are ready to give our decay result.

\begin{theorem} \label{thm4.5}
Suppose that {\rm (H1), (H2)} and \eqref{2.3} hold,
$p\leq\frac{n+2}{[n-2]_+}$,
$(u_0,u_1)\in H_0^1(\Omega)\times L^2(\Omega)$.
Then there exists positive constants
$\alpha$ and $\lambda$ such that the solution of \eqref{1.1} satisfies
\[
E(t)\leq \alpha e^{-\lambda\int_{t_0}^t \xi(\tau)d\tau},\quad t\geq t_0.
\]
\end{theorem}

\begin{proof}
Since $g$ is positive, continuous and $g(0)>0$, then for any $t_0>0$, we have
\begin{equation} \label{3.21}
\int_0^tg(\tau)d\tau\geq\int_0^{t_0}g(\tau)d\tau=g_0>0, \quad\forall t\geq t_0.
\end{equation}
Combining \eqref{2.2}, \eqref{3.1}, \eqref{3.8}, \eqref{3.14} and \eqref{3.21},
 for $t\geq t_0$, we have
\begin{equation} \label{4.17}
\begin{split}
F'(t)
&\leq -\Big\{\epsilon_2[g_0-\delta(2+k)]-\epsilon_1
\Big(1+\frac{(1-k)(1+k)C_*^2}{l}\Big)\Big\}\xi(t)\|u_t\|^2_2 \\
&\quad -\Big\{\frac{\epsilon_1l}{4}-\epsilon_2\delta
\Big[1+2(1-l)^2+C_e^{2p} \Big(\frac{2(p+1)E(0)}{l(p-1)}\Big)^{p-1}\Big]\Big\}
 \xi(t)\|\n u\|^2_2 \\
&\quad +\Big\{\frac{\epsilon_1(1-l)}{2l}+\epsilon_2
\Big[\Big(2\delta+\frac{1}{2\delta}\Big)(1-l)
+\frac{(2+k)(1-l)C_*^2}{4\delta}\Big]\Big\}\xi(t)(g\circ\n u)(t) \\
&\quad +\Big(\frac{1}{2}-\frac{\epsilon_2g(0)C_*^2M}{4\delta}\Big)(g'\circ\n u)(t)
+\epsilon_1\xi(t)\|u\|^{p+1}_{p+1} \\
&:=-J_1\xi(t)\|u_t\|_2^2-J_2\xi(t)\|\n u\|_2^2+J_3\xi(t)(g\circ\n u)(t) \\
&\quad +J_4(g'\circ\n u)(t)+\epsilon_1\xi(t)\|u\|^{p+1}_{p+1}.
\end{split}
\end{equation}
We choose suitable constants $\epsilon_1$ and $\epsilon_2$ satisfying
\[
\frac{\epsilon_1\big(1+\frac{(1-k)(1+k)C_*^2}{l}\big)}{g_0-\delta(2+k)}
<\epsilon_2<\frac{\epsilon_1l}{4\delta\big[1+2(1-l)^2+C_e^{2p}
\big(\frac{2(p+1)E(0)}{l(p-1)}\big)^{p-1}\big]}
\]
and $\delta$, $\epsilon_1$ small enough,
such that
\[
g_0-(2+k)\delta>\frac{1}{2}g_0,\quad J_1>0,\quad J_2>0,\quad
k_3:=J_4-J_3>0,
\]
which imply
\[
J_4(g'\circ\n u)(t)+J_3\xi(t)(g\circ\n u)(t)\leq-k_3\xi(t)(g\circ\n u)(t).
\]
Applying \eqref{4.2} and \eqref{4.17} yields
\[
F'(t)\leq-\gamma\xi(t)E(t)\leq\frac{-\gamma}{k_2}\xi(t)F(t).
\]
Therefore, after integrating the above inequality and using \eqref{4.2} again,
we obtain the desire result.
\end{proof}

\subsection*{Acknowledgments}
This work is partially supported by grant  T200905 from the
Zhejiang Innovation Project, and grant 11226176 from the NSFC.
The authors want thank the anonymous referee for the careful reading of
the original manuscript and the helpful suggestions.

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\end{document}