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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 128, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/128\hfil Energy decay in thermoelasticity]
{Energy decay in thermoelasticity type III with viscoelastic damping 
 and delay term}

\author[T. A. Apalara, S. A. Messaoudi , M. I. Mustafa \hfil EJDE-2012/128\hfilneg]
{Tijani A. Apalara, Salim A. Messaoudi, Muhammad I. Mustafa}  % in alphabetical order

\address{Tijani A. Apalara  \newline
Department of Mathematics and Statistics\\
King Fahd University of Petroleum and Minerals\\
Dhahran 31261, Saudi Arabia}
\email{tijani@kfupm.edu.sa}

\address{Salim A. Messaoudi \newline
Department of Mathematics and Statistics\\
King Fahd University of Petroleum and Minerals\\
Dhahran 31261, Saudi Arabia}
\email{messaoud@kfupm.edu.sa}

\address{Muhammad I. Mustafa \newline
Department of Mathematics and Statistics\\
King Fahd University of Petroleum and Minerals\\
Dhahran 31261, Saudi Arabia}
\email{mmustafa@kfupm.edu.sa}

\thanks{Submitted June 11, 2012. Published August 15, 2012.}
\subjclass[2000]{35B37, 35L55, 74D05,  93D15, 93D20}
\keywords{Damping; delay; relaxation function; thermoelasticity; viscoelasticity.}

\begin{abstract}
 In this article, we consider a thermoelastic system of type III with a
 viscoelastic damping and internal delay.  We use the multiplier method
 to prove, under suitable assumptions, general energy decay results
 from which the exponential and polynomial types of decay are only 
 special cases.
\end{abstract}

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

\section{Introduction}

 In this article, we consider the problem
\begin{equation}\label{E:First}
\begin{gathered}
\begin{aligned}
&u_{tt}(x,t)-\mu \Delta u(x,t)-(\mu+\lambda)\nabla(\operatorname{div}u( x,t))
+\beta\nabla \theta(x, t)\\
&+\int_0^{t}g( s)\Delta u( x,t-s)ds +\mu _1u_t( x,t)+\mu _2 u_t( x,t-\tau)=0,
\quad  x\in\Omega,\; t > 0
\end{aligned}\\
\theta_{tt}(x,t)-\kappa \Delta\theta(x,t)-\delta\Delta\theta_{t}(x,t)
+\beta \operatorname{div}u_{tt}( x,t),=0, \quad  x\in\Omega,\; t > 0
\\
u(x,0)=u_0(x),\quad  u_{t}(x, 0)=u_1(x), \quad  x\in\Omega, \\
\theta(x, 0)=\theta_0(x), \quad   \theta_{t}(x, 0)=\theta_1(x), \quad  x\in\Omega,
\\
 u_{t}(x, -t)=f_0(x, t), \quad x\in\Omega,\; t\in(0, \tau)\\
 u(x,t)=\theta(x,t)=0,\quad x\in\partial\Omega,\; t \ge 0
\end{gathered}
\end{equation}
where $\Omega$ is a bounded domain of $\mathbb{R}^n (n\geq2)$ with a boundary
$\partial\Omega$ of class $C^{2}$, $ u=u(x,t)\in\mathbb{R}^n$  is the displacement
vector, $\theta (x,t) $ is the difference temperature, the relaxation function
 $g$ is positive and decreasing, the coefficients
$\mu, \lambda,\beta, \mu_1, \kappa, \delta$ are positive  constants,
$\mu_2$ is a real number, and $\tau >0$ represents the time delay.
This is a (type III) thermoelastic system with the presence of a viscoelastic
damping and constant internal delay supplemented by initial data
$ u_0, u_1, \theta_0, \theta_1$ and a history function $f_0$.

 Time delays so often arise in many physical, chemical, biological, thermal
and economical phenomena. In recent years, the control of PDEs with time
delay effects has become an active area of research, see for
 example \cite{Abdallah,Suh} and the references therein.
The presence of delay may be a source of instability. See, for example
 \cite{Datko,Nicaise,Xu}, where it was proved that an arbitrarily small
delay may destabilize a system which is uniformly asymptotically stable
in the absence of delay unless additional conditions or control terms have
 been used.

Consider  the system
\begin{equation}\label{E:Scd}
\begin{gathered}
u_{tt}(x,t)-\Delta u(x,t)=0, \quad  x\in\Omega,\; t > 0\\
u(x,t)=0,\quad x\in\Gamma_0,\; t > 0\\
\frac{\partial u}{\partial \nu}{(x, t)}=-\mu _1u_t( x,t)-\mu _2u_t( x,t-\tau), \quad
 x\in\Gamma_1,\; t > 0.
\end{gathered}
\end{equation}
It is well known that in the absence of delay $(\mu_2=0, \mu_1>0)$,
system \eqref{E:Scd} is exponentially stable, see
\cite{Komonik}--\cite{Lasiecka 2}, \cite{Zuazua}. Whereas,
in the presence of delay $(\mu_2>0)$, Nicaise and Pignotti \cite{Nicaise} proved,
under the assumption $\mu_2<\mu_1$, that the energy is exponentially stable.
However, for the opposite case $(\mu_2\ge \mu_1),$ they were able to construct
a sequence of delays for which the corresponding solution is unstable.
The same results were obtained for the case when both the damping and the delay
act internally in the domain, see also \cite{Ait} for the treatment of this
problem in more general abstract form. Nicaise and Pignotti \cite{Nicaise 2}
treated the situation when the constant delay in system \eqref{E:Scd} is replaced
with a distributed delay of the form
\begin{equation*}
\int_{\tau_1}^{\tau_2}\mu_2(s)u_{t}(x, t- s)ds
\end{equation*}
and established an exponential stability result similar to the one
in \cite{Nicaise} under the condition that
\begin{equation*}
\int_{\tau_1}^{\tau_2}\mu_2(s)ds<\mu_1.
\end{equation*}
Kirane and Said-Houari \cite{Kirane} considered a viscoelastic wave equation
of the form
\begin{equation*}
u_{tt}(x,t)-\Delta u(x,t)+\int_0^{t}g(t-s)\Delta u( x,s)ds
+\mu _1u_t( x,t)+\mu _2u_t( x,t-\tau)=0,
\end{equation*}
for  $x\in\Omega,\; t > 0$,
together with initial and Dirichlet boundary conditions.
 They established general energy decay results under the condition that
$\mu_2\leq\mu_1$.
In fact, the presence of a viscoelastic damping together with a frictional
damping allowed $\mu_2=\mu_1$.

  Recently, Pignotti \cite{Pignotti} considered the equation
\[
u_{tt}(x,t)-\Delta u(x,t)+a\chi _{\omega }u_{t}(x,t)+ku_{t}(x,t-\tau
)=0,\quad\text{in }\Omega \times (0,\infty )
\]
for $a,\tau >0$ and $k$ a real number. She established, under some
geometry condition on the domain, a well posedness of the problem and an
exponential decay result for $| k | < a$.


In \cite{Mustafa}, Mustafa studied a thermoelastic system with boundary
time-varying delay in one dimensional space and showed that the damping
effect through heat conduction is still strong enough to uniformly stabilize
the system even in the presence of boundary time-varying delay.
For more results concerning time delay in one dimensional as well as
multi-dimensional space, we refer the reader to
\cite{Fridman}, \cite{Nicaise 3}--\cite{Nicaise 5}.

We also recall some results regarding thermoelastic systems of type III.
In one space dimension, Quintanilla and Racke \cite{Quintanilla} considered
the equation
\begin{gather*}
u_{tt}-\alpha u_{xx}+\beta\theta_{x}=0, \quad   \text{in } [0, \infty)\times(0, L)\\
\theta_{tt}-\delta\theta_{xx}+\gamma u_{ttx}-\kappa \theta_{txx}=0, \quad
 \text{in } [0, \infty)\times(0, L)
\end{gather*}
and used the spectral analysis method and the energy method to obtain the
exponential stability for various boundary conditions;
(Dirichlet-Dirichlet or Dirichlet-Neuman). Furthermore, they proved an energy
decay result for the radially symmetric situations in the multi-dimensional
case $(n=2,3)$.
Zhang and Zuazua \cite{Zhang} analyzed the long time behavior of the solution
of the n-dimensional system \eqref{E:First}, when $g=\mu_1=\mu_2=0$,
 and showed that (i) for most domains the energy of the system does not
decay uniformly, (ii) under suitable conditions on the domain that may be
described in terms of Geometric Optics, the energy of the system decays
exponentially, and (iii) for most domains in two space dimensions, the
energy of smooth solutions decays polynomially.
Messaoudi and Soufyane \cite{Messaoudi 3} considered the system
\begin{gather*}
u_{tt}-\mu \Delta u-(\mu+\lambda)\nabla(\operatorname{div}u)
+\beta\nabla \theta =0, \quad  \text{in } \Omega \times  \mathbb{R}^+
\\
\theta_{tt}-\kappa\Delta\theta-\delta\Delta\theta_{t}
 + \beta \operatorname{div}u_{tt}=0, \quad
 \text{in}\ \Omega \times \mathbb{R}^+
\end{gather*}
subject to a boundary feedback of viscoelastic type that acts on a part
of the boundary and established exponential and polynomial stability results.
This result was later generalized by Messaoudi and Al-Shehri \cite{Messaoudi}
by taking a wider class of relaxation functions. They proved a more general
decay result, from which the exponential and polynomial decay estimates
are only special cases.

Recently, Qin and Ma \cite{Qin} considered the system
\begin{gather*}
u_{tt}- \Delta u +\int_0^{t}g( t-s)\Delta u(s)ds+\nabla \theta =0, \quad
  x\in\Omega,\; t > 0\\
\theta_{tt}-\Delta\theta_{t}-\Delta\theta + \operatorname{div}u_{tt}=0, \quad
  x\in\Omega,\; t > 0\\
\theta =0,  \quad  x\in\partial\Omega,\; t > 0\\
u=0,  \quad  x\in\Gamma_0,\; t > 0\\
\frac{\partial u}{\partial \nu}-\int_0^{t}g( t-s)\Delta u(s)ds+H(u_{t})=0, \quad
x\in\Gamma_1,\; t > 0
\end{gather*}
and established a general decay result depending on both $g$ and $H$.
This result extends the decay result obtained by Messaoudi and
Mustafa \cite{Messaoudi 2} obtained earlier for wave equations. For more
results on Thermoelasticity type III, we refer the reader to
\cite{Liu,Messaoudi 4,Messaoudi 5,Quintanilla} and references therein.

In this article, we investigate system \eqref{E:First} under suitable assumptions
on the weight of the delay term and prove general decay result from which
the exponential and polynomial types of decay are only special cases.
This work extends the result obtained by Kirane and Said-Houari \cite{Kirane}
for a viscoelastic wave equation  to the thermoviscoelastic system with a delay.
We should mention here that, to the best of our knowledge, there is no result
concerning systems of thermoelasticty of type III with the presence of delays.
The rest of our paper is organized as follows. In section 2, we introduce some
transformations and assumptions needed in our work. Some technical lemmas and
the statement with proof of our main results will be given in section 3 and
section 4 respectively. Finally, we give some examples to illustrate our results.

\section{Assumptions and Transformations}

In this section, we present some materials needed in the proof of our results.
 We use the standard Lebesgue space $ L^2(\Omega)$  and the Sobolev space
 $H_0^1(\Omega)$ with their usual scalar products and norms. Throughout this paper,
$c$ is used to denote a generic positive constant.

For the relaxation function g, we assume the following:
\begin{itemize}
\item[(A1)] $g: \mathbb{R}^+\to \mathbb{R}^+$ is a $C^{1}$ function satisfying
$$
g(0) > 0, \quad  \mu-\int_0^{\infty}g(s)ds=l > 0.
$$

\item[(A2)] There exists a positive non-increasing differentiable function
$\eta: \mathbb{R}^+\to \mathbb{R}^+$ satisfying
$$
g' (t) \leq -\eta (t) g(t), \quad  t\ge 0.
$$
\end{itemize}

\begin{remark} \label{remark 2.1} \rm
There are many functions that satisfy (A1) and (A2).
Below are three examples of such functions with the assumptions that
$a, b > 0$ and $a < \mu b$.

\begin{enumerate}
\item If $g(t) = ae^{-bt}$, then $g' (t) = -\eta (t) g(t)$, where $\eta (t) = b$.

\item If $g(t) = \frac{a}{(1+t)^{b+1}}$, then $g' (t) = -\eta (t) g(t)$,
 where $\eta (t) = \frac{b+1}{1+t}$.

\item If $g(t)=\frac{a}{(e+t) [ln(e+t)]^{b+1}}$, then $g' (t) = -\eta (t) g(t)$,
where 
$$
\eta (t)=\frac{1}{e+t}+\frac{b+1}{(e+t) \ln (e+t)}.
$$
\end{enumerate}
\end{remark}

Now, as in \cite{Zhang}, we introduce the new variable
\begin{equation}\label{E:3}
v(x, t) = \int_0^{t}\theta (x, s)ds+\chi(x),
\end{equation}
where $\chi(x)$ is the solution of
\begin{equation}\label{E:4}
\begin{gathered}
-\kappa\Delta \chi = \delta\Delta\theta_0-\theta_1-\beta \operatorname{div} u_1,
 \quad  \text{in }  \Omega, \\
\chi = 0, \quad   \text{on } \partial\Omega,
\end{gathered}
\end{equation}
Then, integrating the second equation in \eqref{E:First} with respect to $t$
and using \eqref{E:3} and \eqref{E:4}, we have
$$
v_{tt}-\kappa\Delta v- \delta\Delta v_{t}+\beta \operatorname{div} u_{t}=0.
$$
By introducing as in \cite{Nicaise}, another new dependent variable
$$
z(x, \rho, t) = u_{t} (x,t-\tau\rho), \quad
x\in \Omega, \; \rho \in (0,1), \;  t > 0.
$$
problem \eqref{E:First}  takes the form
\begin{equation}\label{E:5}
\begin{gathered}
\begin{aligned}
&u_{tt}(x,t)-\mu \Delta u(x,t)-(\mu+\lambda)\nabla(\operatorname{div}u( x,t))
+\beta \nabla v_{t}(x, t)\\
&+\int_0^{t}g(t-s)\Delta u( x,s)ds
+\mu _1u_{t}( x,t)+\mu _2 z( x, 1, t)=0, \quad
 x\in\Omega,\; t > 0
\end{aligned}\\
v_{tt}(x,t)-\kappa\Delta v(x,t)-\delta\Delta v_{t}(x,t)
+ \beta \operatorname{div}u_{t}( x,t)=0, \quad  x\in\Omega,\; t > 0
\\
\tau z_{t}(x, \rho, t) + z_{\rho}(x, \rho, t) = 0, \quad
x\in \Omega, \rho \in (0,1), \;  t > 0
\\
z(x, 0, t) = u_{t}(x, t), \quad  x\in\Omega,\; t > 0
\\
 u(x,0)=u_0(x),\quad  u_{t}(x, 0)=u_1(x),\quad x\in\Omega,\\
v(x, 0)=v_0(x),\quad  v_{t}(x, 0) = v_1(x),  \quad   x\in\Omega,
\\
z(x, \rho, 0) = f_0(x, \tau \rho), \quad x\in\Omega,\; \rho\in(0, 1)
\\
 u(x,t) = v(x,t) = 0,\quad x\in\partial\Omega,\; t \ge 0
\end{gathered}
\end{equation}
Thus, we will consider problem \eqref{E:5} instead of \eqref{E:First}.
In what follows, we consider $(u,v,z)$ to be a solution of system \eqref{E:5}
with the regularity needed to justify the calculations in this paper.
By repeating the arguments of \cite{Kirane}, one can easily prove the existence
and uniqueness of strong and weak solutions.

Next, we assume that $| \mu_2| \le \mu_1$ and that $\xi$ is a positive
constant satisfying
\begin{equation}\label{E:6}
\begin{gathered}
\tau | \mu_2| < \xi < \tau (2\mu_1-| \mu_2|), \quad  \text{if }
 |\mu_2| < \mu_1, \\
\xi = \tau \mu_1, \quad   \text{if } \mu_1 = | \mu_2|,
\end{gathered}
\end{equation}
The energy associated with problem \eqref{E:5} is
\begin{equation}\label{E:7}
\begin{aligned}
E(t) &= \frac{1}{2}\int_{\Omega}{| u_{t}|}^{2}dx
 + \frac{1}{2}\int_{\Omega} v_{t}^{2}dx + \frac{1}{2}
 \Big(\mu - \int_0^{t}g(s)ds\Big)\int_{\Omega}{| \nabla u |}^{2}dx
 +\frac{\kappa}{2}\int_{\Omega}{| \nabla v|}^{2}dx\\
&\quad + \frac{(\mu + \lambda)}{2}\int_{\Omega}{| \operatorname{div} u|}^{2}dx
+ \frac{1}{2}(g\circ \nabla u)(t)+ \frac{\xi}{2}
 \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)\,ds\,dx,
\end{aligned}
\end{equation}
where
$$
(g \circ \nabla u)(t) = \int_{\Omega}\int_0^{t}g(t-s){\bigm| \nabla u(x, t)
- \nabla u(x, s)\bigm|}^2\,ds\,dx.
$$

\section{Technical Lemmas}

In this section we establish several lemmas needed for the proof of our main
result.

\begin{lemma}\label{lemma 1}
Let $(u,v,z)$ be the solution of \eqref{E:5}. Then the energy functional,
defined by \eqref{E:7}, satisfies
\begin{equation}\label{E:8}
\begin{aligned}
&E' (t) \le -m_0\Big(\int_{\Omega}{| u_{t}|}^{2}dx
 + \int_{\Omega}z^{2}(x, 1, t)dx \Big) + \frac{1}{2}(g' \circ  \nabla u)(t)\\
&\quad -\frac{1}{2}g(t)\int_{\Omega}{| \nabla u |}^{2}dx
- \delta \int_{\Omega}{| \nabla v_{t} |}^{2}dx \le 0, \quad \forall t \ge 0,
\end{aligned}
\end{equation}
\end{lemma}
for some constant $m_0$, where $m_0>0$ if $| \mu_2| < \mu_1$ and
 $m_0=0$ if $\mu_1= | \mu_2|$. 

\begin{proof}
A multiplication of the first and the second equation in \eqref{E:5} by $u_t$  and
$v_t$ respectively, and integration over $\Omega$, using integration by parts and
the boundary conditions, yield
\begin{equation} \label{E:9}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt} \Bigl\{\int_{\Omega}{| u_{t}|}^{2}dx
+ \int_{\Omega} v_{t}^{2}dx + \Big(\mu - \int_0^{t}g(s)ds\Big)
\int_{\Omega}{| \nabla u |}^{2}dx\\
&+\kappa\int_{\Omega}{| \nabla v|}^{2}dx
+ (\mu + \lambda)\int_{\Omega}{| \operatorname{div} u|}^{2}dx
+ (g\circ \nabla u)(t) \Bigr\}
\\
&= \frac{1}{2}(g' \circ  \nabla u)(t) -\delta\int_{\Omega}{| \nabla v_{t}|}^{2}dx
-\frac{1}{2}g(t)\int_{\Omega}{| \nabla u |}^{2}dx\\
&\quad-\mu_1\int_{\Omega}{| u_{t}|}^{2}dx
 -\mu_2 \int_{\Omega}u_{t}\cdot z(x, 1, t)dx.
\end{aligned}
\end{equation}
Now, multiplying the third equation in \eqref{E:5}  by $\xi z$ and integrating
over $\Omega \times (0,1)$, we obtain
\begin{equation} \label{E:10}
 \frac{\xi}{2}\frac{d}{dt}\int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx
= -\frac{\xi}{2\tau}\int_{\Omega}z^{2}(x, 1, t)dx
+  \frac{\xi}{2\tau}\int_{\Omega}{| u_{t}|}^{2}dx.
\end{equation}
A combination of \eqref{E:9} and \eqref{E:10}, leads to
\begin{align*}
E' (t) &= \frac{1}{2}(g' \circ  \nabla u)(t)
 -\frac{1}{2}g(t)\int_{\Omega}{| \nabla u |}^{2}dx
 -\delta\int_{\Omega}{|\nabla v_{t}|}^{2}dx
 - \big(\mu_1 -\frac{\xi}{2\tau}\big)\int_{\Omega}{|  u_{t} |}^{2}dx\\
&\quad - \mu_2 \int_{\Omega}u_{t}\cdot z(x, 1, t)dx
 -\frac{\xi}{2\tau}\int_{\Omega}z^{2}(x, 1, t)dx.
\end{align*}
Then by Young's inequality, we have
\begin{align*}
E' (t) &\le \frac{1}{2}(g' \circ  \nabla u)(t)-\frac{1}{2}g(t)
 \int_{\Omega}{| \nabla u |}^{2}dx-\delta\int_{\Omega}{|\nabla v_{t}|}^{2}dx \\
&\quad - \big(\mu_1 -\frac{\xi}{2\tau}-\frac{| \mu_2|}{2}\big)
\int_{\Omega}{|  u_{t} |}^{2}dx-\big(\frac{\xi}{2\tau}-\frac{| \mu_2|}{2}\big)
\int_{\Omega} z^{2}(x, 1, t)dx.
\end{align*}
Consequently, using \eqref{E:6}, estimate \eqref{E:8} follows.
\end{proof}

\begin{lemma}\label{lemma 2}
Suppose that {\rm (A1)} and {\rm (A2)} hold, and let $(u,v,z)$ be the solution
of \eqref{E:5}. Then the functional
$$
F_1(t):=\int_{\Omega}u_{t}\cdot u dx
$$
satisfies the following estimate, for some positive constant $m_1$,
\begin{equation}\label{E:11}
\begin{aligned}
F_1' (t) &\leq c\Big(\int_{\Omega}{| u_{t}|}^{2}dx + \int_{\Omega}v_{t}^{2}dx
+ \int_{\Omega}z^2 (x, 1, t)dx + (g \circ\nabla u)(t)\Big)\\
&\quad - m_1\Big(\int_{\Omega}{| \nabla u |}^{2}dx
\int_{\Omega}{| \operatorname{div} u|}^{2}dx\Big).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Direct computations using the first equation in \eqref{E:5}, yield
\begin{align*}
F_1' (t)
&= \int_{\Omega}{| u_{t}|}^{2}dx -\mu \int_{\Omega}{ | \nabla u |}^{2}dx
- (\mu +\lambda)\int_{\Omega}{| \operatorname{div} u |}^2dx
 +\beta \int_{\Omega}v_{t}\cdot \operatorname{div}u dx\\
&\quad +\int_{\Omega}{ \nabla u}\cdot \int_0^{t}g(t-s)\nabla u (s) \,ds\,dx
- \mu_1\int_{\Omega}u \cdot u_{t}dx - \mu_2 \int_{\Omega}z(x, 1, t)\cdot u\,dx.
\end{align*}
Using Young's and Poincar\' e's inequalities,for $\delta_1 > 0$, we have
\begin{equation}\label{E:12}
\begin{aligned}
& F_1' (t) \le -\Big(\frac{ \mu}{2}-\delta_1(\mu_1+| \mu_2|)\Big)
 \int_{\Omega}{ | \nabla u |}^{2}dx+\frac{1}{2\mu}\int_{\Omega}
\Big(\int_0^{t}g(t-s)\nabla u (s)ds\Big)^{2}dx\\
& +\big(1+\frac{c\mu_1}{4\delta_1}\big)\int_{\Omega}{| u_{t}|}^{2}dx
 - (\mu +\lambda-\delta_1)\int_{\Omega}{| \operatorname{div}u|}^2dx
 + \frac{1}{4\delta_1}\int_{\Omega}v_{t}^{2}dx\\
& +\frac{c| \mu_2|}{4\delta_1}\int_{\Omega}z ^{2}(x, 1, t)dx.
\end{aligned}
\end{equation}
The second term in the right-hand side of \eqref{E:12} is estimated as follows:
\begin{align*}
&\int_{\Omega}\Big(\int_0^{t}g(t-s)  |\nabla u (s)| ds\Big)^{2}dx\\
&\le \int_{\Omega}\Big(\int_0^{t}g(t-s) (| \nabla u (s)-\nabla u(t)|
 +|\nabla u(t) | )ds\Big)^{2}dx \\
& = \int_{\Omega}\Big(\int_0^{t} g(t-s) |\nabla u (s)- \nabla u(t)| ds\Big)^{2}dx
 + \int_{\Omega}\Big(\int_0^{t}g(t-s) | \nabla u (t)| ds\Big)^{2}dx\\
&\quad +2\int_{\Omega} \Big(\int_0^{t}g(t-s)| \nabla u(s)-\nabla u(t)| ds\Big)
\Big(\int_0^{t}g(t-s)| \nabla u(t)| ds\Big)dx.
\end{align*}
A simple calculation, using Cauchy-Schwarz and Young's inequalities, for $\eta>0$,
gives
\begin{equation}\label{E:13}
\begin{aligned}
&\int_{\Omega}\Big(\int_0^{t}g(t-s)  |\nabla u (s)| ds\Big)^{2}dx\\
&\le (\mu-l)^{2}(1+\eta)\int_{\Omega}{| \nabla u |}^{2}dx
  +(\mu-l)\big(1+\frac{1}{\eta}\big)(g \circ\nabla u)(t).
\end{aligned}
\end{equation}
By inserting \eqref{E:13} into \eqref{E:12}  and choosing $\eta=\frac{l}{\mu-l}$,
we arrive at
\begin{align*}
F_1' (t)
&\le \big(1+\frac{c\mu_1}{4\delta_1}\big)
\int_{\Omega}{| u_{t}|}^{2}dx -\big(\frac{l}{2}-\delta_1(\mu_1+| \mu_2|)\big)
 \int_{\Omega}{ | \nabla u |}^{2}dx\\
&\quad  - (\mu +\lambda-\delta_1)\int_{\Omega}{| \operatorname{div}u|}^2dx
 +\frac{1}{4\delta_1}\int_{\Omega}v_{t}^{2}dx+\frac{(\mu-l)}{2l}(g \circ\nabla u)(t)\\
&\quad +\frac{c| \mu_2|}{4\delta_1}\int_{\Omega}z ^{2}(x, 1, t)dx.
\end{align*}
By taking $\delta_1$ small enough, \eqref{E:11} follows.
\end{proof}

\begin{lemma}\label{lemma 3}
let $(u,v,z)$ be the solution of \eqref{E:5}. Then the functional
$$
F_2(t):=\int_{\Omega}v_{t}v dx+\beta\int_{\Omega}v\operatorname{div}u dx
+\frac{\delta}{2}\int_{\Omega}{| \nabla u | }^{2}dx
$$
satisfies the following estimate, for any positive constant $\delta_2$,
\begin{equation}\label{E:14}
F_2' (t) \le \big(1+ \frac{\beta}{4\delta_2}\big)
\int_{\Omega}v_{t}^{2}dx +\beta\delta_2
\int_{\Omega}{| \operatorname{div} u |}^{2}dx-\kappa\int_{\Omega}{| \nabla v| }^2 dx.
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $F_2 (t)$ and using the second equation 
in \eqref{E:5}, it  follows that
$$
F_2' (t)=\int_{\Omega}v_{t}^{2}dx+\kappa\int_{\Omega}v \Delta vdx
+\delta\int_{\Omega}v \Delta v_{t}dx
+\beta\int_{\Omega} v_{t}\operatorname{div}u dx
+\delta\int_{\Omega}\nabla v \cdot \nabla v_{t}dx.
$$
Use of Green's formula and the boundary conditions lead to
$$
F_2' (t)=\int_{\Omega}v_{t}^{2}dx-\kappa\int_{\Omega} {| \nabla v |}^{2}dx
+\beta\int_{\Omega} v_{t}\operatorname{div}u dx.
$$
By exploiting Young's inequality for $\delta_2>0$, estimate \eqref{E:14}
is established.
\end{proof}

\begin{lemma}\label{lemma 4}
let $(u,v,z)$ be the solution of \eqref{E:5}. Then the functional
$$
F_3(t):=\tau \int_{\Omega}\int_0^{1}e^{-\tau \rho}z^{2}(x, \rho, t) d \rho dx,
$$
satisfies the following  estimate, for some positive constant $m_2$,
\begin{equation}\label{E:15}
F_3'\le - m_2\Big(\int_{\Omega}z^{2}(x, 1, t)dx
+\tau \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t) d \rho dx\Big)
+\int_{\Omega}{| u_{t}|}^{2}dx.
\end{equation}
\end{lemma}

\begin{proof}
By differentiating $F_3 (t)$ and using the third equation in \eqref{E:5}, we obtain
\begin{align*}
F_3' (t)
&= -2\int_{\Omega}\int_0^{1}e^{-\tau \rho}z(x, \rho, t) z_{\rho}(x, \rho, t)d \rho dx \\
&=  -\frac{d}{d\rho} \int_{\Omega}\int_0^{1}e^{-\tau \rho}z^{2}(x, \rho, t) d \rho dx -\tau \int_{\Omega}\int_0^{1}e^{-\tau \rho}z^{2}(x, \rho, t) d \rho dx\\
&=  -\int_{\Omega}[e^{-\tau}z^{2}(x, 1, t)-z^{2}(x, 0, t)] dx-\tau \int_{\Omega}\int_0^{1}e^{-\tau \rho}z^{2}(x, \rho, t) d \rho dx\\
&\le - m_2\Big(\int_{\Omega}z^{2}(x, 1, t)dx
 +\tau \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t) d \rho dx\Big)
+\int_{\Omega}{| u_{t}|}^{2}dx.
\end{align*}
which gives \eqref{E:15}.
\end{proof}

\begin{lemma}\label{lemma 5}
Suppose that {\rm (A1)} and {\rm (A2)} hold and let $(u,v,z)$ be the solution
of \eqref{E:5}. Then for $\mu_1=| \mu_2|$ and for any $t_0>0,$ the functional
 $$
F_4(t):=-\int_{\Omega}u_{t}\cdot \int_0^{t}g(t-s)(u(t)-u(s))\,ds\,dx,
$$
satisfies the following estimate, for some positive constant $m_3$,
and for any positive $\delta_3$, $\delta_4$, $\delta_5$,
\begin{equation}\label{E:16}
\begin{aligned}
F_4' (t)
&\le -m_3\int_{\Omega}{| u_{t}|}^{2}dx
 +\frac{\beta}{2} \int_{\Omega}{| \nabla v_{t}| }^{2}dx
 +\delta_3c\int_{\Omega}{| \nabla u| }^{2}dx\\
& \quad +\delta_4(\mu + \lambda)\int_{\Omega}{| \operatorname{div} u|}^{2}dx 
 + C_{\delta}(g \circ\nabla u)(t)+\delta_5\mu_1\int_{\Omega}z^2 (x, 1, t)dx\\
&\quad -c(g'\circ \nabla u)(t), \quad \forall t\ge t_0 > 0.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Differentiation of $F_4 (t)$, using \eqref{E:5} and integrating by parts
 together with the boundary conditions, yield
\begin{equation}\label{17}
\begin{aligned}
&F_4' (t)
= \mu \int_{\Omega}\nabla u \cdot 
 \Big(\int_0^{t}g(t-s)(\nabla u(s)-\nabla u(t))ds\Big)dx\\
&\quad +(\mu + \lambda)\int_{\Omega}(\operatorname{div} u)\cdot 
\Big(\int_0^{t}g(t-s)(\operatorname{div}u(s)-\operatorname{div}u(t))ds\Big)dx\\
&\quad -\beta\int_{\Omega}\nabla v_{t} \cdot 
\Big(\int_0^{t}g(t-s)( u(s)- u(t))ds\Big) dx\\
&\quad -\int_{\Omega}\Big(\int_0^{t}g(t-s)\nabla u(s)ds\Big) \cdot 
\Big(\int_0^{t}g(t-s)(\nabla u(s)- \nabla u(t))ds\Big)dx\\
&\quad +\mu_1 \int_{\Omega}u_{t} \cdot \int_0^{t}g(t-s)( u(s)- u(t))ds dx\\
&\quad +\mu_2 \int_{\Omega}z(x, 1, t) \cdot \int_0^{t}g(t-s)( u(s)- u(t))ds dx
-\Big(\int_0^{t}g(s)ds\Big)\int_{\Omega}{| u_{t}|}^{2}dx\\
&\quad -\int_{\Omega}u_{t} \cdot \int_0^{t}g'(t-s)( u(s)- u(t))\,ds\,dx.
\end{aligned}
\end{equation}
Now, we estimate the terms in the right hand side of \eqref{17} using 
Young's, Cauchy-Schwarz, and Poincar\' e's inequalities.
 So,  for $\delta_3, \delta_4, \delta_5, \delta_{6} > 0$, we  obtain
\begin{equation}\label{E:18}
\begin{aligned}
I_1&=\int_{\Omega}\nabla u \cdot 
\Big(\int_0^{t}g(t-s)(\nabla u(s)-\nabla u(t))ds\Big)dx\\
&\le \delta_3\int_{\Omega}{| \nabla u |}^{2}dx+\frac{1}{4\delta_3}\int_{\Omega}
\Big(\int_0^{t}g(t-s)| \nabla u(s)-\nabla u(t)| ds\Big)^{2}dx\\
&\le \delta_3\int_{\Omega}{| \nabla u |}^{2}dx+\frac{1}{4\delta_3}\int_{\Omega}
\Big(\int_0^{t}g(s)ds\Big)\Big(\int_0^{t}g(t-s){| \nabla u(s)-\nabla u(t)|}^{2} 
ds\Big)dx\\
&\le \delta_3\int_{\Omega}{| \nabla u |}^{2}dx+\frac{\mu-l}{4\delta_3}
(g\circ \nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:19}
\begin{aligned}
I_2&=\int_{\Omega}(\operatorname{div} u)\cdot
\Big(\int_0^{t}g(t-s)(\operatorname{div}u(s)-\operatorname{div}u(t))ds\Big)dx\\
&\le \delta_4\int_{\Omega}{| \operatorname{div}u |}^{2}dx
+\frac{1}{4\delta_4}\int_{\Omega}
\Big(\int_0^{t}g(t-s) \left( \operatorname{div}u(s)-\operatorname{div}u(t)\right)
 ds\Big)^{2}dx\\
&\le \delta_4\int_{\Omega}{| \operatorname{div}u |}^{2}dx
 +\frac{\mu -l}{4\delta_4}\int_{\Omega}\int_0^{t}g(t-s){| \operatorname{div}u(s)
 -\operatorname{div}u(t)|}^{2} \,ds\,dx\\
&\le \delta_4\int_{\Omega}{| \operatorname{div}u |}^{2}dx
 +\frac{\mu -l}{2\delta_4}(g \circ\nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:20}
\begin{aligned}
I_3
&= -\int_{\Omega}\nabla v_{t} \cdot \Big(\int_0^{t}g(t-s)( u(s)- u(t))ds\Big) dx\\
&\le\frac{1}{2}\int_{\Omega}{| \nabla v_{t}|}^{2}dx+\frac{c(\mu-l)}{2}
(g \circ\nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:21}
\begin{aligned}
I_4
&=-\int_{\Omega}\Big(\int_0^{t}g(t-s)\nabla u(s)ds\Big) \cdot 
\Big(\int_0^{t}g(t-s)(\nabla u(s)- \nabla u(t))ds\Big)dx\\
&\le \delta_3\int_{\Omega}\Big(\int_0^{t}g(t-s)| \nabla u(s) | ds\Big)^{2}dx\\
&\quad +\frac{1}{4\delta_3}\int_{\Omega}
\Big(\int_0^{t}g(t-s)| \nabla u(s)-\nabla u(t)| ds\Big)^{2}dx\\
&\le 2{(\mu -l)}^{2}\delta_3\int_{\Omega}{| \nabla u|}^{2}dx
+(\mu -l)\left(2\delta_3+\frac{1}{4\delta}\right)(g \circ\nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:22}
\begin{aligned}
I_5&=  \int_{\Omega}u_{t} \cdot \int_0^{t}g(t-s)( u(s)- u(t))ds dx\\
&\le \delta_{6}\int_{\Omega}{| u_{t}|}^{2}dx
+\frac{c(\mu -l)}{4\delta_{6}}(g \circ\nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:23}
\begin{aligned}
I_{6}&= \int_{\Omega}z(x, 1, t) \cdot \int_0^{t}g(t-s)( u(s)- u(t))ds dx\\
&\le \delta_5\int_{\Omega}z^{2}(x, 1, t)dx
+\frac{c(\mu -l)}{4\delta_5}(g \circ\nabla u)(t).
\end{aligned}
\end{equation}

\begin{equation}\label{E:24}
\begin{aligned}
I_{7}&=-\int_{\Omega}u_{t} \cdot \int_0^{t}g'(t-s)( u(s)- u(t))\,ds\,dx\\
&\le \delta_{6}\int_{\Omega}{| u_{t}|}^{2}dx+\frac{1}{4\delta_{6}}
 \int_{\Omega}\Big(\int_0^{t}g' (t-s) (u(s)-u(t))ds\Big)^{2} dx\\
&\le \delta_{6}\int_{\Omega}{| u_{t}|}^{2}dx+\frac{1}{4\delta_{6}}
 \int_{\Omega}\Big(\int_0^{t}-g' (s)ds\Big)
 \Big(\int_0^{t}-g' (t-s) {| (u(s)-u(t)|}^{2} ds\Big) dx\\
&\le \delta_{6}\int_{\Omega}{| u_{t}|}^{2}dx-\frac{cg(0)}{4\delta_{6}}
(g' \circ \nabla u)(t).
\end{aligned}
\end{equation}
Since the function $g$ is positive, continuous and $g(0) > 0$, then for any
 $t \ge t_0 > 0$, we have
\begin{equation}\label{25}
\int_0^{t}g(s)ds \ge \int_0^{t_0}g(s)ds=g_0.
\end{equation}
A combination of \eqref{17}$-$\eqref{25}, bearing in mind that
 $\mu_1=| \mu_2|$ leads to
\begin{align*}
F_4' (t)
&\le -[g_0-\delta_{6}(1+\mu_1)]\int_{\Omega}{| u_{t}|}^{2}dx 
 +\delta_5\mu_1\int_{\Omega}z^2 (x, 1, t)dx
 -\frac{cg(0)}{4\delta_{6}}(g'o\nabla u)(t)\\
&\quad +\frac{\beta}{2} \int_{\Omega}{| \nabla v_{t}| }^{2}dx 
 +\delta_3[\mu +2(\mu -l)^{2}]\int_{\Omega}{| \nabla u| }^{2}dx
 +\delta_4(\mu + \lambda)\int_{\Omega}{| \operatorname{div} u|}^{2}dx\\
& \quad + (\mu -l)\big[\frac{\mu +1}{4\delta_3}+\frac{\mu +\lambda}{2\delta_4}
 +2\delta_3+\frac{c\mu_1}{4}\big(\frac{1}{\delta_5}
 +\frac{1}{\delta_{6}}\big)+\frac{c\beta}{2}\big](g \circ\nabla u)(t), 
\end{align*}
for all $t\ge t_0$.
Next, we choose $\delta_6$ small enough to obtain \eqref{E:16}.
\end{proof}


\section{Asymptotic Stability}

This section is divided into two parts.  In the first subsection, 
we discuss the case where $| \mu_2| < \mu_1$ and in the second,
 we discuss the case where $\mu_1=| \mu_2|$.

	
\subsection{General stability for $| \mu_2| < \mu_1$}
For $\varepsilon > 0$, to be chosen appropriately later, we  let
\begin{equation}\label{E:26}
\mathcal {L}(t):=E(t)+\varepsilon F_1(t)+\varepsilon F_2(t)+\varepsilon F_3(t).
\end{equation}

\begin{lemma}\label{lemma 6}
There exist two positive constants $\alpha_1$ and $\alpha_2$ such that
\begin{equation}\label{E:27}
\alpha_1E(t) \le L(t) \le \alpha_2E(t), \quad  \forall t \ge 0,
\end{equation}
for $\varepsilon$ small enough
\end{lemma}

\begin{proof}
Let 
$$
\mathcal{G}(t)=\varepsilon F_1(t)+\varepsilon F_2(t)+\varepsilon F_3(t).
$$
By using Young's and Poincar\'e's inequalities, we obtain
\begin{align*}
| \mathcal{G}(t)| 
&\le \frac{\varepsilon}{2}\int_{\Omega}
\left({| u_{t} |}^{2}+v_{t}^{2}+c{| \nabla u|}^{2}
+\left(c(1+\beta)+\delta\right){| \nabla v|}^{2}
+{| \operatorname{div}u|}^{2}\right)dx \\
&\quad +\varepsilon \tau \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx\\
&\quad \le \varepsilon cE(t).
\end{align*}
Consequently, $| \mathcal{L}(t)-E(t)| \le \varepsilon cE(t)$,
which yields 
$$
(1-\varepsilon c)E(t)\le \mathcal{L}(t)\le (1+\varepsilon c)E(t).
$$
By choosing $\varepsilon$ small enough, \eqref{E:27} follows.
\end{proof}

\begin{theorem}\label{thrm 4.2}
let $(u,v,z)$ be the solution of \eqref{E:5}.
 Assume $| \mu_2| < \mu_1$ and {\rm (A1), (A2)} hold. Then, there exist
two positive constants $c_0$  and $c_1$ such that the energy functional 
for the system \eqref{E:5} satisfies
\begin{equation}\label{E:28}
E(t)\le c_0e^{-c_1\int_0^{t}\eta (s)ds},  \quad \forall t\ge 0.
\end{equation}
\end{theorem}

\begin{proof}
By differentiating \eqref{E:26} and using \eqref{E:8}, \eqref{E:11}, \eqref{E:14} 
and \eqref{E:15}, and Poincar\'e's inequality, we obtain
\begin{align*}
\mathcal{L}'(t)
&\le- [ m_0-\varepsilon c]\int_{\Omega}{| u_{t} |}^{2}dx
 -\varepsilon m_1\int_{\Omega}{| \nabla u|}^{2}dx
 -\varepsilon \kappa\int_{\Omega}{| \nabla v|}^{2}dx\\
&\quad -\varepsilon \left[m_1 -\beta\delta_2 \right]\int_{\Omega}{| \operatorname{div}u|}^{2}dx-\varepsilon m_2\tau \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx\\
&\quad +\varepsilon c(g \circ\nabla u)(t)
 -\big[\delta-\varepsilon c \big(c+\frac{\beta}{4\delta_2}\big)\big]
 \int_{\Omega}{| \nabla v_{t}| }^{2}dx\\
&\quad -[(m_0-\varepsilon c)+\varepsilon m_2]\int_{\Omega}z^{2}(x, 1, t)dx.
\end{align*}
At this point, we choose $\delta_2$ small enough such that 
$(m_1 -\beta\delta_2) > 0$.
Next, by picking
$$
\varepsilon <\ min\{\frac{m_0}{c} , \frac{\delta}{c ( c+\frac{\beta}{4\delta_2})}\},
$$
we obtain
\begin{align*}
\mathcal{L}'(t)
&\le k_1(g \circ\nabla u)(t)-k_2\Bigl\{ \int_{\Omega}{| u_{t} |}^{2}dx
+\int_{\Omega}{| \nabla u|}^{2}dx+\int_{\Omega}{| \nabla v|}^{2}dx\\
&\quad +\int_{\Omega}{| \operatorname{div}u|}^{2}dx
+\int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx
+\int_{\Omega}{| \nabla v_{t}| }^{2}dx\Bigl\},
\end{align*}
for positive constants $k_1$ and $ k_2$.
Then, using Poincar\'e's inequality and \eqref{E:7}, we obtain
\begin{equation}\label{E:29}
\mathcal{L}'(t)\le -k_0E(t)+k_1(g \circ\nabla u)(t), \quad \forall t \ge 0,
\end{equation}
for a positive constant $k_0$.
By multiplying \eqref{E:29} by $\eta (t)$ and using (A2)  and \eqref{E:8}, 
we arrive at
$$
\eta (t)\mathcal{L}'(t)\le- k_0\eta (t) E(t)-2k_1 E'(t), \quad \forall t \ge 0,
$$
which can be rewritten as
$$
\left(\eta(t)\mathcal{L}(t)+2k_1 E(t)\right)'-\eta'(t)\mathcal{L}(t)
\le - k_0\eta(t)E(t),  \; \forall t\ge 0.
$$
Using the fact that $\eta'(t) \le 0, \forall t \ge 0,$ we have
$$
\left(\eta(t)\mathcal{L}(t)+2k_1 E(t)\right)'
\le - k_0\eta(t)E(t),  \quad  \forall t\ge 0.
$$
By exploiting \eqref{E:27}, it can easily be shown that
\begin{equation}\label{E:30}
\mathcal{R}(t) =\eta (t) \mathcal {L}(t)+2k_1E(t)\sim E(t).
\end{equation}
Consequently, for some positive constant $c_1$, we obtain
\begin{equation}\label{E:31}
\mathcal{R}'(t)\le -c_1\eta(t)\mathcal{R}(t), \quad  \forall t\ge 0.
\end{equation}
A simple integration of \eqref{E:31} over $(0,t)$ leads to
\begin{equation}\label{E:32}
\mathcal{R}(t)\le \mathcal{R}(0)e^{-c_1\int_0^{t}\eta(s)ds}, \quad \forall t\ge 0.
\end{equation}
The conclusion of the theorem follows by combining \eqref{E:30} and \eqref{E:32}.
\end{proof}

\subsection{General stability for $| \mu_2|=\mu_1$}
By recalling \eqref{E:6}, we have $\xi =\tau \mu_1.$ Hence, \eqref{E:8} 
takes the form
\begin{equation}\label{E:33}
E' (t) \le  \frac{1}{2}(g' \circ  \nabla u)(t)
-\frac{1}{2}g(t)\int_{\Omega}{| \nabla u |}^{2}dx 
-\delta \int_{\Omega}{| \nabla v_{t} |}^{2}dx \le 0, \quad \forall t \ge 0.
\end{equation}
We then use \eqref{E:11}, \eqref{E:14}, and \eqref{E:15} with $\mu_1 = | \mu_2|$ 
and define another Lyapunov functional
\begin{equation}\label{E:34}
\tilde{\mathcal {L}}(t) := NE(t)+\varepsilon_1F_1(t)
+F_2(t)+\varepsilon_2F_3(t)+F_4(t),
\end{equation}
where $N, \varepsilon_1$ and $\varepsilon_2$ are positive real numbers which
 will be chosen properly later.

\begin{lemma}\label{lemma 7}
For $N$ large enough, $\tilde{\mathcal {L}}(t)$ and $E(t)$ satisfy
\begin{equation}\label{E:35}
\alpha_3E(t) \le \tilde{\mathcal {L}}(t) \le \alpha_4E(t), \quad \forall t \ge 0,
\end{equation}
for two positive constants $\alpha_3$ and $\alpha_4$.
\end{lemma}

The inequality in the above lemma is established with similar steps as in the proof 
of Lemma \ref{lemma 6}.

\begin{theorem}\label{thrm 4.4}
let $(u,v,z)$ be the solution of \eqref{E:5}. Assume $| \mu_2| = \mu_1$ and
{\rm (A1), (A2)} hold. Then, for any $t_0 > 0$, there exist positive constants 
$c_2$  and $c_3$ independent of $t$ such that the energy functional 
of the system \eqref{E:5} satisfies
\begin{equation}\label{E:36}
E(t)\le c_2e^{-c_3\int_{t_0}^{t}\eta (s)ds},  \quad \forall t\ge t_0.
\end{equation}
\end{theorem}

\begin{proof}
Differentiating $\tilde{\mathcal {L}}(t)$ and using \eqref{E:11}, \eqref{E:14}, 
\eqref{E:15}, \eqref{E:16}, \eqref{E:33} and Poincar\'e's inequality, we obtain
\begin{align*}
\tilde{\mathcal {L}}'(t)
&\le -[ m_3-\varepsilon_1c-\varepsilon_2]\int_{\Omega}{| u_{t} |}^{2}dx
-[\varepsilon_1 m_1-\delta_3c]\int_{\Omega}{| \nabla u|}^{2}dx
- \kappa\int_{\Omega}{| \nabla v|}^{2}dx\\
&\quad -\varepsilon_2m_2\tau \int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx
-\left[\varepsilon_1m_1-\beta\delta_2-\delta_4(\mu +\lambda)\right]
 \int_{\Omega}{| \operatorname{div}u|}^{2}dx\\
&\quad -\big[N\delta-\frac{\beta c}{2}-c\big(1+\frac{\beta}{4\delta_2}
 +\varepsilon_1c\big)\big]\int_{\Omega}{| \nabla v_{t}| }^{2}dx
 +[\frac{N}{2}-c](g'\circ \nabla u)(t) \\
&\quad -\left[\varepsilon_2m_2-\varepsilon_1c-\delta_5\mu_1 \right]
\int_{\Omega}z^{2}(x, 1, t)dx+[\varepsilon_1c+C_{\delta}](g \circ\nabla u)(t).
\end{align*}
Now, we let
$$
\varepsilon_2=\frac{m_{}3}{2},\quad
\delta_3=\frac{\varepsilon_1m_1}{2c}, \quad
\delta_4=\frac{\varepsilon_1m_1}{2(\mu+\lambda)}.
$$
Next, we choose $\varepsilon_1$ small enough so that
$$
\tilde{k}_1:=[\frac{m_3}{2}-\varepsilon_1c] > 0 ,\quad
\tilde{k}_2:=[\frac{m_2m_3}{2}-\varepsilon_1c ] > 0.
$$
Once $\varepsilon_1$ is fixed, we then take
$\delta_5=\tilde{k}_2/(2\mu_1)$
and choose $\delta_2$ small enough so that
$$
\tilde{k}_3:=[\frac{\varepsilon_1m_1}{2}-\beta\delta_2] > 0
.$$
Finally, we choose $N$ so large that \eqref{E:35} remains valid and, furthermore,
$$
\tilde{k}_4:=\big[N\delta-\frac{\beta c}{2}
 -c\big(1+\frac{\beta}{4\delta_2}+\varepsilon_1c\big)\big]> 0, \quad
[\frac{N}{2}-c ] > 0.
$$
Hence, we arrive at
\begin{align*}
\tilde{\mathcal {L}'}(t)
&\le -\tilde{k}_1\int_{\Omega}{| u_{t} |}^{2}dx-\frac{\varepsilon_1m_1}{2} 
 \int_{\Omega}{| \nabla u|}^{2}dx- \kappa\int_{\Omega}{| \nabla v|}^{2}dx\\
&\quad -\tilde{k}_4\int_{\Omega}{| \nabla v_{t}| }^{2}dx
 -\tilde{k}_3\int_{\Omega}{| \operatorname{div}u|}^{2}dx
 +\tilde{k}_5(g \circ\nabla u)(t)\\
&\quad -\frac{m_2m_3\tau}{2}\int_{\Omega}\int_0^{1}z^{2}(x, \rho, t)d\rho dx.
\end{align*}
Using Poincar\'e's inequality, we obtain
\begin{equation}\label{E:37}
\tilde{\mathcal {L}'}(t)\le -\tilde{k}_0E(t)+ \tilde{k}_5(g \circ\nabla u)(t), \quad
 \forall t\ge t_0,
\end{equation}
where $\tilde{k}_0$ and $\tilde{k}_5$ are two positive constants.

By multiplying \eqref{E:37} by $\eta (t)$ and using $(A_2 )$ and \eqref{E:33}, 
we obtain
\begin{gather*}
\eta(t) \tilde{\mathcal {L}'}(t) \le - \tilde{k}_0\eta(t)E(t)-2\tilde{k}_5E'(t),
  \quad  \forall t\ge t_0,
\\
\left(\eta(t)\tilde{\mathcal {L}}(t)+2 \tilde{k_5} E(t)\right)'\le
- \tilde{k_0}\eta(t)E(t),  \quad \forall t\ge t_0.
\end{gather*}
If we set
\begin{equation}\label{38}
\tilde{\mathcal {R}}(t) = \eta(t)\tilde{\mathcal {L}}(t)+2 \tilde{k_5} E(t)
\sim E(t),
\end{equation}
and follow the same steps as in Theorem \ref{thrm 4.2}, we arrive at
\begin{equation}\label{E:39}
\tilde{\mathcal{R}}(t)\le \tilde{\mathcal{R}}(t_0)
e^{-\tilde{c}_3\int_0^{t}\eta(s)ds}, \quad \forall t\ge t_0.
\end{equation}
Consequently,\eqref{E:36} is established by virtue of \eqref{38} and \eqref{E:39}.

Note that Estimate \eqref{E:36} also holds for $ t \in [0, t_0]$ 
by the continuity  and boundedness of  $E(t)$  and $\eta (t)$.
\end{proof}

Now, we give some examples to illustrate the energy decay rates 
obtained by Theorem \ref{thrm 4.2} which is also valid for Theorem \ref{thrm 4.4}.
We consider the three examples under Remark \ref{remark 2.1} with the same
assumptions on $a$ and $b$ as stated before.
\begin{enumerate}
\item If $g(t) = ae^{-bt}$, then 
$$
E(t) \le c_0e^{-bc_1t}, \quad \forall t \ge 0.
$$

\item If $g(t) = \frac{a}{(1+t)^{b+1}}$, then 
$$
E(t) \le \frac{c_0}{(1+t)^{(b+1)c_1}}, \quad \forall t \ge 0.
$$
\item If $g(t)=\frac{a}{(e+t) [ln(e+t)]^{b+1}}$, then
 $$
E(t) \le \frac{c_0e^{c_1}}{\{(e+t) [ ln (e+t)]^{b+1}\}^{c_1}}, \quad \forall t \ge 0.
$$
\end{enumerate}

\begin{remark}\rm
As in  Pignotti \cite{Pignotti}, we do not require that $\mu_2$ be positive.
Our result extends, in a way, the result of Kirane and Said-Houari \cite{Kirane},
 where $\mu_2$ is taken to be positive.
\end{remark}

\subsection*{Acknowledgements}
The authors thank KFUPM for its continuous support and the referee for pointing
 out a valuable reference which improved this work a lot. This work has been 
funded by KFUPM under Project \# FT 111002.

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\end{document}