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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 22, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/22\hfil Existence and decay of solutions]
{Existence and decay of solutions to a viscoelastic plate equation}

\author[S. A. Messaoudi, S. E. Mukiawa \hfil EJDE-2016/22\hfilneg]
{Salim A. Messaoudi, Soh Edwin Mukiawa}

\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{Soh Edwin Mukiawa \newline
Department of Mathematics and Statistics\\
King Fahd University of Petroleum and Minerals\\
Dhahran 31261, Saudi Arabia}
\email{g201206120@kfupm.edu.sa}

\thanks{Submitted October 12, 2015. Published January 13, 2016.}
\subjclass[2010]{35L35, 37B25, 34D20, 74H20, 74H25}
\keywords{Existence; decay; plate viscoelastic; fourth order}

\begin{abstract}
 In this article we study the fourth-order viscoelastic plate equation
 \[
 u_{tt} + \Delta^2 u -\int_0^t g(t-\tau)\Delta^2u(\tau)d\tau = 0
 \]
 in the bounded domain $\Omega = (0,\pi)\times(-\ell,\ell)\subset\mathbb{R}^2$
 with non traditional boundary conditions. We establish the well-posedness
 and a decay result.
\end{abstract}

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

\section{Introduction}

This article is devoted to the well-posedness and the decay rate of the 
energy functional for the fourth-order  viscoelastic  plate  problem
\begin{equation}  \label{p1}
\begin{gathered}
u_{tt} + \Delta^{2} u -\int_0^t g(t-\tau)\Delta^2u(\tau)d\tau
= 0 ,\quad \Omega\times (0,T) \\
u(0,y,t) = u_{xx}(0, y, t) = 0  ,\quad\text{for }
 (y,t)\in (-\ell, \ell)\times (0, T) \\
u(\pi, y, t) = u_{xx}(\pi, y, t) = 0 ,\quad\text{for }(y,t)\in
 (-\ell, \ell)\times(0,T) \\
u_{yy}(x, \pm l, t) + \sigma u_{xx}(x, \pm l, t) = 0, \quad\text{for }
 (x,t)\in  (0,\pi)\times (0,T) \\
u_{yyy}(x, \pm l, t) + (2-\sigma)u_{xxy}(x, \pm l, t) = 0 ,
\quad\text{for }(x,t)\in  (0,\pi)\times (0,T)\\
u(x, y, 0) = u_0(x, y),\quad u_t(x, y, 0 )=  u_1(x, y) ,
\quad\text{in } \Omega
\end{gathered}
\end{equation}
where $\Omega  =(0,\pi)\times(-\ell,\ell)$, $0 < \sigma < \frac{1}{2}$  
and $g$ is a positive and nonincreasing function. This type of problems 
models  the motion of a  viscoelastic plate. The fundamental work of 
Ferrero and  Gazzola \cite{Gaz} in $2013$, where they modeled a suspension 
bridge as a rectangular plate with  the same  boundary conditions as \eqref{p1}, 
suggests  the investigation of the viscoelastic   material used in construction.
 Al-Gwaiz et al \cite{gwa} also investigated the  bending and stretching 
energies of the  rectangular plate model suggested in \cite{Gaz}. 
Contributions on the analysis of a suspension bridge have  also come from  
Mckenna and Walter \cite{mckenna}, Mckenna et al \cite{mckenna1}, 
Ma and Zhong \cite{maa2} and Bochicchio et al \cite{bo2}.

The  existence, decay and blow up properties of viscoelastic problems has 
attracted  a lot of attention since the pioneer work by Dafermos 
\cite{da,daa} in $1970$. Hence, a considerable number of  results for models 
similar to \eqref{p1}, for both second and fourth order have been established. 
We begin  with the result of Messaoudi \cite{SA}, where he considered
\begin{equation} \label{mm}
u_{tt} - \Delta u + \int_0^t g(t-\tau)\Delta u(\tau)d\tau = 0, \quad\text{in }
	\Omega\times(0,+\infty)
\end{equation}
with general conditions on the relaxation function $g$, and  proved a general 
decay result that is not necessarily  of exponential or polynomial type. 
His result generalized and improved many results in literature such as 
\cite{raaaa,raaa,raa,ra,ma}. 
Rivera et al \cite{RI} considered the  fourth-order equation
\begin{equation}
u_{tt} + \gamma\Delta u_{tt} + \Delta^{2} u -\int_0^t g(t-\tau)\Delta^2u(\tau)d\tau 
= 0,\quad\text{in }	\Omega\times(0,T)
\end{equation}
together with initial and dynamical boundary conditions and proved that the 
sum of the first and second energies decays exponentially (polynomially) 
if the kernel $g$ decays exponentially (polynomially). 
Mustafa and Ghassan \cite{New5} considered the plate equation
\begin{equation}
u_{tt} + \Delta^{2} u = 0,\quad\text{in }\Omega\times(0,+\infty)
\end{equation}
with viscoelastic damping localized on a part of the boundary  and 
established a decay result.
For more results related to the plate equation, we refer the reader 
to Messaoudi \cite{edwin},  Kang \cite{tih}, Santos and Junior \cite{tip}, 
Lagnese \cite{jj}, Horn and Lasiecka \cite{hh}, Lasiecka \cite{rr}, 
and Lasiecka et al \cite{rr2}, Cabanillas et al \cite{raaaaa}
 Lasiecka et al \cite{lesi}.

    The aim of this work is to take advantage of the techniques used in 
\cite{SA} and the new model in \cite{Gaz} to establish a global existence 
and general decay results for problem \eqref{p1}. We organize this work as 
follows. In section 2, we present some important and fundamental materials 
to be used in  establishing our main  results. 
In section 3, we state and prove the global  existence result. 
Finally, in section 4 we state and prove the general decay result.

 \section{Preliminaries}
In this section, we present some fundamental materials needed for the proof of
 our main results. For this, we assume the following conditions on the 
relaxation function $g$.
\begin{itemize}
\item[(A1)]  $g:\mathbb{R_+}\to \mathbb{R_+}$ is a differentiable 
function such that
\begin{equation}
g(0) > 0, \quad 1 - \int_0^{+\infty}g(s)ds = l_0 > 0.
\end{equation}

\item[(A2)]  There exists a differentiable function $\xi$ satisfying
\begin{equation}
\begin{gathered}
 g'(t)\leq -\xi(t)g(t),\quad t\geq 0\\
\xi(t)> 0,\quad \xi'(t)\leq 0,\quad \forall t > 0,\quad
\int_0^{+\infty}\xi(s)ds = +\infty
\end{gathered}
\end{equation}
\end{itemize}

The following three functions  satisfy (A1)--(A2).
\begin{gather*}
g_1(t)=\frac{ae^{-t}}{(1+t)}, \quad a>0, \\
g_2(t)= \frac{a}{(1+t)^p},\quad p>1, \; a>0,
g_3(t)= ae^{-b(1+t)^p},\quad  0<p\leq 1, \; a,\; b >0.
\end{gather*}
We introduce the   space
\begin{equation}
H_{\ast}^{2}(\Omega) = \{w\in H^{2}(\Omega): w  =0 \text{ on }
\{0, \pi \} \times (-\ell,\ell) \},
\end{equation}
 with the inner product
\begin{equation} \label{p2}
(u,v)_{H_{\ast}^2}=\int_{\Omega}[(\Delta u\Delta v 
+(1-\sigma)(2u_{xy}v_{xy}-u_{xx}v_{yy}-u_{yy}v_{xx})]\,dx\,dy,
\end{equation}
and  set  $\mathcal{H}(\Omega) $ as the dual  of  $ H_{\ast}^2(\Omega)$.
For  completeness, we state some results from Ferrero and  Gazzola \cite{Gaz}.

\begin{lemma}[\cite{Gaz}] \label{p3}
Assume $0 < \sigma <1/2$. Then the  norm $\|\cdot\|_{H_{\ast}^2(\Omega)}$ 
given by $\|\cdot\|_{H_{\ast}^2(\Omega)}^2= (\cdot,\cdot)_{H_*^2}$ 
is equivalent to the usual  $H^2(\Omega)$-norm. Moreover, 
$H_*^2(\Omega)$ is a Hilbert space  when endowed with the scalar product 
$(u,v)_{H_*^2}$. 
\end{lemma}

\begin{theorem}[\cite{Gaz}]
Assume $0 < \sigma <1/2$ and let $f \in L^2(\Omega)$. 
Then there exists a unique function $u \in H_{\ast}^2(\Omega)$ such that
\begin{equation}\label{555}
\int_{\Omega} [\Delta u \Delta v +(1-\sigma)(2u_{xy} v_{xy} 
-u_{xx} v_{yy}-u_{yy} v_{xx} )]\,dx\,dy = \int_{\Omega} fv,
\end{equation}
for all $v\in H_*^2(\Omega)$.
\end{theorem}

The function $u \in H_{\ast}^2(\Omega)$ satisfying \eqref{555} is called the weak solution of  the stationary problem
\begin{equation} \label{pp}
\begin{gathered}
 \Delta^2 u = f,\\
u(0,y)= u_{xx}(0,y)= u(\pi,y)= u_{xx}(\pi,y) = 0,\\
u_{yy}(x,\pm l) + \sigma u_{xx}(x,\pm l)
= u_{yyy}(x,\pm l) + (2-\sigma)u_{xxy}(x,\pm l) = 0.
\end{gathered}
\end{equation}

\begin{lemma}[\cite{wang}] \label{lem}
Let  $u \in H_{\ast}^2(\Omega)$ and assume $1 \leq  p < +\infty$. 
Then, there exists a positive constant $C_e = C_e(\Omega,p)> 0$ such that 
$$
\|u\|_p^p \leq C_e\|u\|_{H_{\ast}^2(\Omega)}^p.
$$
\end{lemma}

Let us also  introduce  the  energy functional associated to problem 
\eqref{p1},
\begin{equation} \label{p5}
E(t)= \frac{1}{2}\int_{\Omega} u_t^2 + \frac{1}{2}
\Big(1 - \int_0^t g(s)ds\Big)\|u\|_{H_{\ast}^2(\Omega)}^2 
+ \frac{1}{2}\int_0^t g(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau.
\end{equation}

\section{Well-posedness}

In this section, we show that problem \eqref{p1} has a unique global weak solution.

\begin{definition} \label{def3.1} \rm
A function
\begin{equation} 
u\in C( [0,T), H_{\ast}^2(\Omega))\cap
 C^1( [0,T), L^2(\Omega))\cap C^2([0,T),\mathcal{H}(\Omega))
\end{equation}
is called a weak solution of \eqref{p1}  if
\begin{equation}\label{e1}
\begin{gathered}
\int_{\Omega} u_{tt}w + (u,w)_{H_{\ast}^2(\Omega)}
-\int_0^t g(t-\tau)(u(\tau),w)_{H_{\ast}^2(\Omega)} d\tau = 0,
\quad \forall w\in H_{\ast}^2(\Omega),
\\
u(0) = u_0, \quad u_t(0) = u_1.
\end{gathered}
\end{equation}
\end{definition}

\begin{theorem} \label{thm3.1}
Let $(u_0,u_1)\in H_{\ast}^2(\Omega)\times L^2(\Omega)$. 
Assume that {\rm (A1), (A2)} hold. Then problem \eqref{p1}
 has a unique weak global solution
\begin{equation}
u\in C( [0,T), H_{\ast}^2(\Omega)),\quad
u_t\in C([0,T), L^2(\Omega)),\quad
u_{tt}\in C([0,T),\mathcal{H}(\Omega))
\end{equation}
\end{theorem}

\begin{proof}
We  use the Galerkin approximation method. 
Let  $\{ w_j\} _{j=1}^{\infty}$ be a basis of the separable space 
$H_{\ast}^2(\Omega)$ and
$V_m = \operatorname{span}\{ w_1,w_2,\dots,w_m \} $ be a finite subspace 
of $H_{\ast}^2(\Omega)$ spanned by the first $m$ vectors.
Let
$$
u_0^m(x,y)= \sum_{j=1}^{m} a_jw_j(x,y)\quad\text{and}\quad
u_1^m(x,y)= \sum_{j=1}^{m} b_jw_j(x,y) 
$$
 be sequences in $ H_{\ast}^2(\Omega)$ and $L^2(\Omega)$  respectively, such that
\begin{equation} \label{666}
u_0^m \to u_0\text{ in }H_{\ast}^2(\Omega),\quad
u_1^m\to u_1\text{ in }L^2(\Omega).
\end{equation}
We seek a solution of the form
$$
 u^m(x,y,t)=\sum_{j=1}^{m} c_j(t)w_j(x,y), 
$$ 
which satisfies the approximate problem
\begin{equation} \label{e2}
\begin{gathered}
\begin{aligned}
&\int_{\Omega} u_{tt}^m(x,y,t)w_j + (u^m(x,y,t),w_j)_{H_{\ast}^2(\Omega)} \\
& -\int_0^t g(t-\tau)(u^m(x,y,\tau),w_j)_{H_{\ast}^2(\Omega)} d\tau =0,\quad
\forall w_j\in V_m,\; j=1,2,\dots,m.
\end{aligned}\\
u^m(0) = u_0^m, \quad u_t^m(0) = u_1^m\,.
\end{gathered}
\end{equation}
We note that \eqref{e2} leads to  system of ODEs with $m$ unknown functions 
$c_j$, $j=1,2,\dots,m$. Thus, using ODE theory (see \cite{cavalcanti}), 
we obtain functions
$$ 
c_j :[0,t_m)\to \mathbb{R}, \ j=1,2,\dots,m,
$$ 
which satisfy \eqref{e2} for almost every $t\in (0, t_m),\ 0 < t_m < T$. 
Therefore, we obtain a local solution $u^m$ of $\eqref{e2}$ in a maximal interval 
$[0,t_m),t_m\in(0,T]$. Next, we show that $t_m = T$ and that the local solution 
is uniformly bounded independent of $m$ and $t$.
For this, we multiply  $\eqref{e2}$ by $c'_j(t)$ and sum over 
$j=1,2,\dots,m$, to obtain
\begin{align*}
&\frac{d}{dt}\Big[\frac{1}{2}\|u_t^m\|^2_{L^2(\Omega)} 
+ \frac{1}{2}(1 - \int_0^t g(s)ds)\|u^m\|_{H_{\ast}^2(\Omega)}^2 \\
&\quad + \frac{1}{2}\int_0^t g(t-\tau)
 \|u^m(t)-u^m(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau \Big] \\
&= \frac{1}{2}\int_0^t g'(t-\tau)\|u^m(t)- u^m(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau 
- \frac{1}{2}g(t)\|u^m\|_{H_{\ast}^2(\Omega)}^2 
\end{align*}
It follows from \eqref{p5} that
\begin{equation}
\label{e3}
\frac{d}{dt}E^m(t)
= \frac{1}{2}\int_0^t g'(t-\tau)\|u^m(t)- u^m(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau 
- \frac{1}{2}g(t)\|u^m\|_{H_{\ast}^2(\Omega)}^2 \leq 0,
\end{equation}
by assumptions (A1) and (A2). 
Integrating \eqref{e3} over $(0,t)$, $t\in(0,t_m)$ and noting that 
$(u_0^m)$ and $(u_1^m)$ are bounded   in $ H_{\ast}^2(\Omega)$ and $L^2(\Omega)$  
respectively (as convergent sequences \eqref{666}), we obtain
\begin{equation}
E^m(t)\leq E^m(0)=\frac{1}{2}\|u_1^m\|^2_{L^2(\Omega)} 
+ \frac{1}{2}\|u_0^m\|_{H_{\ast}^2(\Omega)}^2 \leq C
\end{equation}
where $C$ is a positive constant independent of $m$ and $t$. Therefore,
\begin{align*}
&\frac{1}{2}\|u_t^m\|^2_{L^2(\Omega)} 
 + \frac{1}{2}(1 - \int_0^t g(s)ds)\|u^m\|_{H_{\ast}^2(\Omega)}^2 \\
&+ \frac{1}{2}\int_0^t g(t-\tau)\|u^m(t)-u^m(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau 
\leq C.
\end{align*}
This implies
\begin{equation}\label{777}
\frac{1}{2}\sup_{t\in (0,t_m)}\|u_t^m\|^2_{L^2(\Omega)} 
+ \frac{l_0}{2}\sup_{t\in (0,t_m)}\|u^m\|_{H_{\ast}^2(\Omega)}^2 \leq C.
\end{equation}
 So, the approximate solution is uniformly bounded independent of $m$ and $t$. 
Therefore, we can extend $t_m$ to $T$. Moreover, we obtain from \eqref{777} that
\begin{equation} \label{e4}
\begin{gathered}
 (u^m)\text{is a bounded sequence in } L^{\infty}((0,T),H_{\ast}^2(\Omega)),\\
(u^m_t)\text{ is a bounded sequence  in } L^{\infty}((0,T),L^2(\Omega)).
\end{gathered}
\end{equation}
Thus, there exists a subsequence $(u^k)$  of $(u^m)$ such that
\begin{equation}  \label{e5}
\begin{gathered}
u^k\rightharpoonup u \text{ weakly star in 
$L^{\infty}((0,T),H_{\ast}^2(\Omega))$  and  weakly  in }
L^2((0,T),H_{\ast}^2(\Omega))\\
u_t^k\rightharpoonup u_t \text{ weakly  star in 
$L^{\infty}((0,T),L^2(\Omega))$  and weakly in }L^2((0,T),L^2(\Omega))
\end{gathered}
\end{equation}
Using  that $H_{\ast}^2(\Omega)$ is compactly embedded in $L^2(\Omega)$ 
(remember that $\Omega$ is bounded and $H_{\ast}^2(\Omega) \subset H^2(\Omega))$, 
we can extract  a subsequence $(u^l)$  of $(u^k)$ such that
\begin{gather*}
u^l\to u\text{ strongly  in } L^2((0,T),L^2(\Omega)), \\
u^l\to u\text{ a.e in }\Omega\times(0,T).
\end{gather*}
Now, replacing $(u^m)$ by $(u^l)$ in \eqref{e2} and integrating over $(0,t)$ 
we obtain
\begin{equation} \label{e6}
\begin{aligned}
&\int_{\Omega} u_t^lw_j + \int_0^t(u^l,w_j)_{H_{\ast}^2(\Omega)}dt 
- \int_0^t\int_0^s g(s-\tau)(u^l(\tau),w_j)_{H_{\ast}^2(\Omega)} d\tau ds\\
& = \int_{\Omega} u_1^lw_j,\quad \forall j\leq l.
\end{aligned}
\end{equation}
Letting $l\to +\infty$, we obtain
\begin{equation}\label{e7}
\begin{aligned}
&\int_{\Omega} u_tw_j + \int_0^t(u,w_j)_{H_{\ast}^2(\Omega)}dt 
- \int_0^t\int_0^s g(s-\tau)(u(\tau),w_j)_{H_{\ast}^2(\Omega)} d\tau ds\\
& =\int_{\Omega} u_1 w_j,\quad \forall j\geq 1.
\end{aligned}
\end{equation}
This implies
\begin{equation}\label{e8}
\begin{aligned}
\int_{\Omega} u_tw 
&= - \int_0^t(u,w)_{H_{\ast}^2(\Omega)}dt 
 + \int_0^t \int_0^s g(s-\tau)(u(\tau),w)_{H_{\ast}^2(\Omega)} d\tau ds\\
&\quad + \int_{\Omega} u_1w,\quad \forall w\in H_{\ast}^2(\Omega).
\end{aligned}
\end{equation}
Now, observe that the terms in the right-hand side of \eqref{e8} are 
absolutely continuous since they are functions of $t$ defined by integrals 
over $(0,t)$, hence differentiable almost everywhere. 
Thus, differentiating  \eqref{e8}, we obtain that for a.e  $t\in (0,T)$,
\begin{equation} \label{e9}
\int_{\Omega} u_{tt}w + (u,w)_{H_{\ast}^2(\Omega)}
 - \int_0^t g(t-\tau)(u(\tau),w)_{H_{\ast}^2(\Omega)} d\tau  = 0 
\end{equation}
for all $w\in L^2((0,T), H_{\ast}^2(\Omega))$.
To handle the  initial conditions, we note that
\begin{equation}
\begin{gathered}
u^l\rightharpoonup u \text{ weakly in }L^2((0,T),H_{\ast}^2(\Omega))\\
u_t^l\rightharpoonup u_t \text{ weakly in }L^2((0,T),L^2(\Omega))
\end{gathered}
\end{equation}
Thus, using Lions' Lemma \cite{jean}, we obtain
\begin{equation}
u^l \to u \quad\text{in } C([0,T),L^2(\Omega)).
\end{equation}
Therefore, $u^l(x,y,0)$ makes sense and
$u^l(x,y,0)\to u(x,y,0)$  in $L^2(\Omega)$.
Also we have that
$$
u^l(x,y,0)=u_0^l(x,y)\to u_0(x,y) \quad\text{in }H_*^2(\Omega).
$$
Hence
\begin{equation}\label{ic1}
u(x,y,0) = u_0(x,y).
\end{equation}
As in \cite{Gaz,marie}, let $\phi\in C_0^{\infty}(0,T)$  and  replacing 
$(u^m)$ by $(u^l)$, we obtain from $\eqref{e2}$  and  for any $j\leq l$ that
\begin{equation}
\begin{aligned}
&-\int_0^T( u^l_t(t),w_j)_{L^2(\Omega)}\phi'(t)dt \\
&= -\int_0^T(u^l(t),w_j )_{H_*^2(\Omega)}\phi(t)dt 
+ \int_0^T\int_0^t g(t-\tau)(u^l(\tau),w_j)_{H_{\ast}^2(\Omega)} \phi(t)d\tau dt.
\end{aligned}
\end{equation}
As $l\to +\infty$, we obtain that
\begin{align*}
&-\int_0^T( u_t(t),w_j)_{L^2(\Omega)}\phi'(t)dt \\
&= -\int_0^T(u(t),w_j )_{H_*^2(\Omega)}\phi(t)dt 
+ \int_0^T\int_0^t g(t-\tau)(u(\tau),w_j)_{H_{\ast}^2(\Omega)} \phi(t)d\tau dt,
\end{align*}
for all $j\geq 1$. This implies
\begin{align*}
&-\int_0^T( u_t(t),w)_{L^2(\Omega)}\phi'(t)dt \\
&= -\int_0^T(u(t),w )_{H_*^2(\Omega)}\phi(t)dt 
+ \int_0^T\int_0^t g(t-\tau)(u(\tau),w)_{H_{\ast}^2(\Omega)} 
\phi(t)d\tau dt,
\end{align*}
for all $w\in H_*^2(\Omega)$.
This means $u_{tt}\in L^2([0,T),\mathcal{H}(\Omega))$.
Thus,
 \begin{equation}
 u_t \in L^2([0,T),L^2(\Omega)),\quad
u_{tt}\in L^2([0,T),\mathcal{H}(\Omega))\Longrightarrow 
u_t \in C([0,T),\mathcal{H}(\Omega)).
 \end{equation}
So, $u_t^l(x,y,0)$ makes sense (see \cite[p.116]{marie}). 
It follows that
 $$
u_t^l(x,y,0)\to u_t(x,y,0)\quad\text{in } \mathcal{H}(\Omega).
$$
But  
$$
u_t^l(x,y,0)=u_1^l(x,y)\to u_1(x,y)\quad\text{in }L^2(\Omega).
$$
Hence
\begin{equation} \label{ic2}
u_t(x,y,0) = u_1(x,y).
\end{equation}
For the uniqueness, suppose  $u$ and $\bar{u}$  satisfy \eqref{e9}, \eqref{ic1} 
and \eqref{ic2}. Then  $v = u-\bar{u}$  satisfies
\begin{equation} \label{e11}
\begin{gathered}
\int_{\Omega} v_{tt}w + (v,w)_{H_{\ast}^2(\Omega)} 
- \int_0^t g(t-\tau)(v(\tau),w)_{H_{\ast}^2(\Omega)} d\tau  = 0 ,\\
 \forall w\in L^2((0,T), H_{\ast}^2(\Omega)),\\
v(0) = v_t(0) = 0.
\end{gathered}
\end{equation}
Replacing $w$  by $v_t$ in \eqref{e11}, we obtain
\begin{equation} \label{e12}
\frac{d}{dt}\Big[\frac{1}{2}\int_{\Omega} v_t^2 
+ \frac{1}{2}\|v\|_{H_{\ast}^2(\Omega)}^2 \Big] 
-\int_0^t g(t-\tau)(v(\tau),v_t(t))_{H_{\ast}^2(\Omega)}d\tau = 0.
\end{equation}
We have that
\begin{equation} \label{p11}
\begin{aligned}
J_1 &=  -\int_0^t g(t-\tau)(v(\tau),v_t(t))_{H_{\ast}^2(\Omega)}d\tau\\
&=  \int_0^t g(t-\tau)(v_t(t),v(t)-v(\tau))_{H_{\ast}^2(\Omega)}d\tau
- \int_0^t g(s)ds(v_t(t),v(t))_{H_{\ast}^2(\Omega)}\\
&=  \int_0^t g(t-\tau)\frac{d}{dt}\frac{1}{2}\|v(t)
 - v(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau
- \int_0^t g(s)ds\frac{d}{dt}\frac{1}{2}\|v(t)\|_{H_{\ast}^2(\Omega)}^2 \\
&= \frac{d}{dt}\frac{1}{2} \int_0^t g(t-\tau)\|v(t)- v(\tau)\|_{H_{\ast}^2(\Omega)}^2 
  d\tau \\
&\quad -\frac{1}{2} \int_0^t g'(t-\tau)\|v(t)- v(\tau)\|_{H_{\ast}^2(\Omega)}^2 
  d\tau\\
&\quad -\frac{d}{dt}\frac{1}{2}\int_0^t g(s)ds\|v(t)\|_{H_{\ast}^2(\Omega)}^2 
 + \frac{1}{2}g(t)\|v(t)\|_{H_{\ast}^2(\Omega)}^2.
\end{aligned}
\end{equation}
Inserting \eqref{p11} into \eqref{e12} and taking note of \eqref{p5}, we obtain
\begin{equation}\label{p12}
\frac{d\tilde{E}(t)}{dt} 
= \frac{1}{2} \int_0^t g'(t-\tau)\|v(t)- v(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau 
- \frac{1}{2}g(t)\|v(t)\|_{H_{\ast}^2(\Omega)}^2 \leq 0,
\end{equation}
by (A1) and (A2).
Integrating \eqref{p12} over $(0,t)$, we obtain
\begin{equation}
\tilde{E}(t)\leq \tilde{E}(0)= 0.
\end{equation}
This implies
$$  
\|v_t\|^2_{L^2(\Omega)} + \|v\|_{H_{\ast}^2(\Omega)}^2 = 0. 
$$
Therefore, $u = \bar{u}$. The proof is complete.
\end{proof}

\section{Decay of solutions}

In this section, we discuss the stability of  solution of problem \eqref{p1}. 
Let us begin by defining the Lyapunov functional
\begin{equation} \label{p6}
	F(t)= E(t) + \epsilon_1\Psi(t) + \epsilon_2\chi(t),
\end{equation}
where $\epsilon_1$ and $\epsilon_2$ are positive constants to be specified later and
\begin{equation} \label{p7}
\begin{gathered}
\Psi(t)= \int_{\Omega} uu_t,\\
\chi(t)= -\int_{\Omega} u_t\int_0^t g(t-\tau)(u(t)-u(\tau))d\tau \,dx\,dy.
\end{gathered}
\end{equation}

\begin{lemma} \label{lem4.1}
Assume {\rm (A1),  (A2)} hold. Then  the  energy  functional, defined 
in \eqref{p5}, satisfies
\begin{equation}
\label{p8}
\frac{dE(t)}{dt}  
= \frac{1}{2}\int_0^t g'(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau 
- \frac{1}{2}g(t)\|u\|_{H_{\ast}^2(\Omega)}^2 \leq 0.
\end{equation}
\end{lemma}

\begin{proof}
By using $\eqref{e9}$ and  the density of $H_{\ast}^2(\Omega)$ in $L^2(\Omega)$
 we obtain
\begin{equation} \label{f2}
\int_{\Omega} u_{tt}w + (u,w)_{H_{\ast}^2(\Omega)} 
- \int_0^t g(t-\tau)(u(\tau),w)_{H_{\ast}^2(\Omega)} d\tau  = 0 
\end{equation}
 for all $w\in L^2([0,T), L^2(\Omega))$.
Repeating exactly the same arguments as in \eqref{e12}-\eqref{p12}, 
we obtain the result.
\end{proof}

\begin{lemma} \label{p13}
For every $u\in H_{\ast}^2(\Omega)$, we have
\begin{equation}\label{p14}
\begin{aligned}
&\int_{\Omega}\Big( \int_0^t g(t-\tau)(u(t)-u(\tau))d\tau\Big)^2 \,dx\,dy \\
&\leq C_e(1-l_0) \int_0^t g(t-\tau)\|u(t)- u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau,
\end{aligned}
\end{equation}
where $C_e >0$ is the embedding constant introduced in Lemma \ref{lem}.
\end{lemma}

\begin{proof}
Since $g$ is positive, we have
\[
\int_{\Omega}\Big( \int_0^t g(t-\tau)(u(t)-u(\tau))d\tau\Big)^2  
=\int_{\Omega}\Big( \int_0^t \sqrt{ g(t-\tau)}\sqrt{ g(t-\tau)}(u(t)-u(\tau))d\tau
\Big)^2
\]
By applying Cauchy-Schwarz, (A1) and  Lemma \ref{lem}, we obtain
\begin{align*}
&\int_{\Omega}( \int_0^t g(t-\tau)(u(t)-u(\tau))d\tau)^2\\
&\leq \int_{\Omega}( \int_0^t g(s)ds)( \int_0^t g(t-\tau)(u(t)- u(\tau))^2 d\tau)\\
&\leq C_e(1-l_0) \int_0^t g(t-\tau)\|u(t)- u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau .
\end{align*}
\end{proof}

\begin{lemma}
For $\epsilon_1$ and $\epsilon_2$  small enough, there exists two positive 
constants $\alpha_1$ and $\alpha_2$ such that
\begin{equation} \label{p15}
\alpha_1 F(t)\leq E(t)\leq \alpha_2 F(t)
\end{equation}
\end{lemma}

The proof of the above lemma uses similar techniques as in 
\cite[Lemma 3.3]{SA}; we omit it here.

\begin{lemma} \label{p18}
Under  assumptions {\rm (A1), (A2)}, the functional
$$ 
\Psi(t)= \int_{\Omega} uu_t
$$ satisfies, along the solution of \eqref{p1},
\begin{equation} \label{p19}
\Psi'(t)\leq \int_{\Omega}u_t^2 - \frac{l_0}{2}\|u\|_{H_{\ast}^2(\Omega)}^2 
+ \frac{1-l_0}{2l_0}\int_0^t g(t-\tau)\| u(t)
- u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau.
\end{equation}
\end{lemma}

\begin{proof}
By using  \eqref{e9} and replacing $w$ by $u$, direct differentiations yield
\begin{equation} \label{p20}
\Psi'(t)=  \int_{\Omega}u_t^2 - \|u\|_{H_{\ast}^2(\Omega)}^2 
+ \int_0^t g(t-\tau)( u(t), u(\tau))_{H_{\ast}^2(\Omega)}d\tau.
\end{equation}
By using Cauchy-Schwarz and Young's inequalities, we estimate the third term
$$
J_2 =\int_0^t g(t-\tau)( u(t),u(\tau))_{H_{\ast}^2(\Omega)}d\tau,
$$
for any $\eta>0$, as follows
\begin{align*}
J_2 &\leq \int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)}
\|u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau\\
&\leq \frac{1}{2}\|u(t)\|_{H_{\ast}^2(\Omega)}^2 
 + \frac{1}{2}\Big(\int_0^t g(t-\tau)(\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}
 +\|u(t)\|_{H_{\ast}^2(\Omega)})d\tau\Big)^2\\
&= \frac{1}{2}\|u(t)\|_{H_{\ast}^2(\Omega)}^2 
  + \frac{1}{2}\Big(\int_0^t g(t-\tau)\|u(t)- u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau\Big)^2\\
&\quad +\frac{1}{2}(\int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)})d\tau)^2\\
&\quad +\Big(\int_0^t g(t-\tau)\|u(t)- u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau\Big)
\Big(\int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)}d\tau\Big).
\end{align*}
By using  Lemma \ref{p13}, we obtain
\begin{equation} \label{p21}
\begin{aligned}
J_2 &\leq \frac{1}{2}(1 +(1-l_0)^2 )\|u\|_{H_{\ast}^2(\Omega)}^2
+ \frac{1}{2}(1-l_0)\int_0^t g(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau\\
&\quad +\frac{\eta}{2}(\int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)}d\tau)^2
 + \frac{1}{2\eta}(\int_0^t g(t-\tau)\|u(t)- u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau)^2\\
&\leq \frac{1}{2}(1 +(1-l_0)^2(1+\eta))\|u\|_{H_{\ast}^2(\Omega)}^2 \\
&\quad +\frac{1}{2}(1-l_0)(1+\frac{1}{\eta})\int_0^t g(t-\tau)
 \|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau.
\end{aligned}
\end{equation}
Now, substituting \eqref{p21} in \eqref{p20}, we obtain
\begin{equation} \label{p22}
\begin{aligned}
\Psi'(t)&\leq \int_{\Omega}u_t^2 + \frac{1}{2}((1-l_0)^2(1+\eta) - 1 )
\|u\|_{H_{\ast}^2(\Omega)}^2\\
&\quad +\frac{1}{2}(1-l_0)(1+\frac{1}{\eta})\int_0^t g(t-\tau)
 \|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau,\quad \forall \eta >0.
\end{aligned}
\end{equation}
 We choose $\eta = \frac{l_0}{1-l_0}$ and obtain the result.
 \end{proof}


\begin{lemma} \label{p23}
Assume  conditions $(A1)$ and $(A2)$ hold. Then the functional
\begin{equation}
\label{p24}
\chi(t)= -\int_{\Omega} u_t\int_0^t g(t-\tau)(u(t)-u(\tau))d\tau \,dx\,dy
\end{equation}
satisfies, along the solution of \eqref{p1},
\begin{equation} \label{p25}
\begin{aligned}
\chi'(t)&\leq \Big(\frac{\delta}{2} -\int_0^t g(s)ds \Big)
\int_{\Omega}u_t^2  +  \frac{\delta}{2}( 1+2(1-l_0)^2)\|u\|_{H_{\ast}^2(\Omega)}^2\\
&\quad - \frac{C_eg(0)}{2\delta}\int_0^t g'(t-\tau)
\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau \\
&\quad + (1-l_0)(\delta + \frac{1}{\delta})\int_0^t
 g(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}^2 d\tau,\quad \forall
 \delta >0.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 By differentiating \eqref{p24}  and using \eqref{e9}, with $u$ instead of $w$, 
we obtain
\begin{equation} \label{p26}
\begin{aligned}
\chi'(t)&=  -\Big( \int_0^t g(s)ds\Big)\int_{\Omega}u_t^2 
- \int_{\Omega} u_t\int_0^t g'(t-\tau)(u(t)-u(\tau))\,d\tau \,dx\,dy\\
&\quad - \int_{\Omega} u_{tt}\int_0^t g(t-\tau)(u(t)-u(\tau))\,d\tau \,dx\,dy\\
&= -\Big( \int_0^t g(s)ds\Big)\int_{\Omega}u_t^2 
 - \int_{\Omega} u_t\int_0^t g'(t-\tau)(u(t)-u(\tau))d\tau \,dx\,dy\\
&\quad + \Big( u(t),\int_0^t g(t-\tau)(u(t)-u(\tau))d\tau\Big)_{H_{\ast}^2(\Omega)} \\
&\quad - \int_0^t g(t-\tau)\Big(u(\tau), \int_0^t g(t-\tau)(u(t)-u(\tau))d\tau
\Big)_{H_{\ast}^2(\Omega)}d\tau.
\end{aligned}
\end{equation}
By using Cauchy-Schwarz inequality,  Young's inequality  and Lemma \ref{p13}
 for $-g'$ instead of $g$, we estimate the terms in the right-hand side of 
\eqref{p26}. Thus, for the term
$$ 
J_3 =- \int_{\Omega} u_t\int_0^t g'(t-\tau)(u(t)-u(\tau))d\tau \,dx\,dy ,
$$
we have that for any $\delta > 0$,
\begin{equation} \label{p27}
\begin{aligned}
J_3&\leq \frac{\delta}{2}\int_{\Omega}u_t^2 
 + \frac{1}{2\delta}\int_{\Omega}
\Big(\int_0^t -g'(t-\tau)(u(t)-u(\tau))d\tau \Big)^2\,dx\,dy \\
&\leq \frac{\delta}{2}\int_{\Omega}u_t^2
  + \frac{1}{2\delta}\int_{\Omega}\Big( \int_0^t -g'(s)ds\Big)
\Big(\int_0^t -g'(t-\tau)(u(t)-u(\tau))^2d\tau \Big)\,dx\,dy\\
&\leq \frac{\delta}{2}\int_{\Omega}u_t^2 
 - \frac{C_eg(0)}{2\delta}\int_0^t g'(t-\tau)
\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau .
\end{aligned}
\end{equation}
 For the term
 $$
J_4 =( u(t),\int_0^t g(t-\tau)(u(t)-u(\tau)d\tau)_{H_{\ast}^2(\Omega)},
$$
 we have
 \begin{equation}  \label{p28}
\begin{aligned}
 J_4&\leq  \|u(t)\|_{H_{\ast}^2(\Omega)}\int_0^t g(t-\tau)
\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau\\
&\leq \frac{\delta}{2}\|u(t)\|_{H_{\ast}^2(\Omega)}^2 
 + \frac{1}{2\delta}( \int_0^t g(t-\tau)
 \|   u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau ) ^2\\
&\leq \frac{\delta}{2}\|u(t)\|_{H_{\ast}^2(\Omega)}^2 
+ \frac{(1-l_0)}{2\delta}\int_0^t g(t-\tau)
\|   u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau.
\end{aligned}
\end{equation}
Similarly, for the term
$$
J_5 =-\int_0^t g(t-\tau)(u(\tau), \int_0^t g(t-\tau)
(u(t)-u(\tau))d\tau)_{H_{\ast}^2(\Omega)}d\tau,  
$$
we obtain
\begin{equation} \label{p29}
\begin{aligned}
J_5 
&\leq \Big(\int_0^t g(t-\tau)\|u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau \Big)
\Big(\int_0^t g(t-\tau)\|   u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau  \Big) \\
&\leq  \frac{\delta}{2}\Big(\int_0^t g(t-\tau)( \|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}
+ \|u(t)\|_{H_{\ast}^2(\Omega)}) d\tau\Big)^2 \\
&\quad +\frac{1}{2\delta}\Big(\int_0^t g(t-\tau)
 \|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)}d\tau \Big)^2\\
&\leq \frac{\delta}{2}\Big(\int_0^t g(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)} 
 d\tau\Big)^2 \\
&\quad +\frac{\delta}{2}\Big(\int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)}d\tau\Big)^2\\
&\quad + \delta (\int_0^t g(t-\tau)\|u(t)-u(\tau)\|_{H_{\ast}^2(\Omega)} d\tau)
\Big(\int_0^t g(t-\tau)\|u(t)\|_{H_{\ast}^2(\Omega)}d\tau\Big)\\
&\quad +\frac{(1-l_0)}{2\delta}\int_0^t g(t-\tau)\|u(t)-u(\tau)
 \|^2_{H_{\ast}^2(\Omega)}d\tau\\
&\leq  (\delta + \frac{1}{2\delta})(1-l_0)
\int_0^t g(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau \\
&\quad +\delta (1- l_0)^2 \|u\|_{H_{\ast}^2(\Omega)}^2.
\end{aligned}
\end{equation}
By substituting \eqref{p27}--\eqref{p29} in \eqref{p26}, we obtain 
\eqref{p25}, for any $\delta >0$.
\end{proof}

\begin{theorem} \label{thm4.1}
Let $(u_0 , u_1)\in H_{\ast}^2(\Omega)\times L^2(\Omega)$. Assume $g$ and $\xi$ 
satisfy {\rm (A1)} and {\rm (A2)}. Then, for any $t_0>0$, there exist positive 
constants $K$ and $\lambda$ such that the solution of \eqref{p1} satisfies
\begin{equation}\label{p30}
E(t)\leq Ke^{-\lambda\int_{t_0}^t \xi(s)ds}, \quad \forall t\geq t_0.
\end{equation}
\end{theorem}

\begin{proof}
Since $g$ is positive, continuous, and $g(0)>0$, then for any $t\geq t_0$ we have
$$
\int_0^t g(s)ds \geq \int_0^{t_0} g(s)ds = g_0 >0 .
$$
Combination of \eqref{p8}, \eqref{p19} and \eqref{p25}, gives that 
for any $t\geq t_0$,
\begin{equation} \label{p31}
\begin{aligned}
&F'(t)\\
&\leq  -\Big( \epsilon_2( g_0 -\frac{\delta}{2})-\epsilon_1\Big) 
\int_{\Omega} u_t^2 - \Big( \frac{\epsilon_1l_0}{2}- \epsilon_2 
\frac{\delta}{2}(1+2(1-l_0)^2)\Big) \|u\|_{H_{\ast}^2(\Omega)}^2\\
&\quad +\Big( \frac{1}{2}- \epsilon_2\frac{C_eg(0)}{2\delta}\Big)
 \int_0^t g'(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau\\
&\quad + \Big( \frac{\epsilon_1(1-l_0)}{2l_0}+ \epsilon_2(\delta 
+ \frac{1}{\delta})(1-l_0)\Big) 
\int_0^t g(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau.
\end{aligned}
\end{equation}
Now, we choose $\delta$ small enough such that
\begin{equation}
\label{p32}
 g_0 -\frac{\delta}{2} > \frac{g_0}{2}, \quad
 \frac{4\delta}{l_0}(1 + 2(1-l_0 )^2 ) < \frac{g_0}{4}.
\end{equation}
By using \eqref{p32}, we easily check that any $\epsilon_1$ and $\epsilon_2$, 
satisfying
\begin{equation}
\label{p34}
\frac{\epsilon_2g_0}{16} < \epsilon_1 < \frac{\epsilon_2g_0}{2},
\end{equation}
will make
$$
\beta_1 = \Big( \epsilon_2( g_0 -\frac{\delta}{2})-\epsilon_1\Big)>0,\quad
\beta_2 = \Big( \frac{\epsilon_1l_0}{2}-
  \epsilon_2 \frac{\delta}{2}(1+2(1-l_0)^2)\Big)>0. 
$$
Next, we pick  $\epsilon_1$ and $\epsilon_2$ small enough such that \eqref{p15}
 and \eqref{p34} remain valid and further we have
$$
 \frac{1}{2}- \epsilon_2\frac{C_eg(0)}{2\delta}>0,\quad
 \frac{\epsilon_1(1-l_0)}{2l_0}+ \epsilon_2(\delta 
+ \frac{1}{\delta})(1-l_0)>0.
$$
Thus, \eqref{p31} becomes
\begin{equation} \label{p35}
\begin{aligned}
F'(t)
&\leq-\beta_1 \int_{\Omega} u_t^2 -\beta_2\|u\|_{H_{\ast}^2(\Omega)}^2 
 + \tilde{C}\int_0^t g(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau \\
&\leq -\beta E(t) + C\int_0^t g(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}
 d\tau,\quad\forall t\geq t_0.
\end{aligned}
\end{equation}
 Multiplying \eqref{p35} by $\xi(t)$ and using the facts that $\xi$
 is decreasing and
 \begin{equation}
 g'(t)\leq -\xi(t)g(t),\quad
 E'(t)\leq \frac{1}{2}\int_0^t 
g'(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau,
 \end{equation}
 we arrive at
 \begin{align*}
\xi(t)F'(t)
&\leq-\beta\xi(t) E(t) + C\xi(t)\int_0^t g(t-\tau)
 \|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau\\
&\leq -\beta\xi(t) E(t) + C\int_0^t \xi(t-\tau)
 g(t-\tau)\|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau\\
&\leq -\beta\xi(t) E(t) + C\int_0^t -g'(t-\tau)
 \|u(t)-u(\tau)\|^2_{H_{\ast}^2(\Omega)}d\tau\\
&\leq-\beta\xi(t) E(t) - CE'(t), \quad \forall t\geq t_0.
\end{align*}
 This gives
 $$ 
(\xi(t)F(t) +  CE(t) )' -\xi'(t)F(t)
\leq  -\beta\xi(t) E(t),\,\,\,\, \forall t\geq t_0.
$$
 Consequently,
 \begin{equation}  \label{p37}
 ( \xi(t)F(t) +  CE(t) )'\leq  -\beta\xi(t) E(t),\quad  \forall t\geq t_0.
 \end{equation}
  Let
  \begin{equation}  \label{p39}
  L = \xi F +  CE \sim E,
  \end{equation}
since $F\sim E$ and $0\leq \xi(t)\leq \xi(0)$.
  Then \eqref{p37} and \eqref{p39} lead to
  \begin{equation}
  L'(t)\leq -\lambda\xi(t)L(t),\quad \forall t\geq t_0.
  \end{equation}
A simple integration in $(t_0,t)$ yields
  \begin{equation}   \label{p40}
  L(t)\leq L(t_0) e^{-\lambda\int_{t_0}^t \xi(s)ds} ,\quad \forall t\geq t_0.
  \end{equation}
Again, recalling \eqref{p39}, we obtain
  \begin{equation}
  E(t)\leq Ke^{-\lambda\int_{t_0}^t \xi(s)ds} ,\quad \forall t\geq t_0.
  \end{equation}
  This completes the proof.
  \end{proof}

\subsection*{Acknowledgements}
The authors thank KFUPM for its continuous  support and  also thank 
Prof. Kirane for his careful reading and precious suggestions.

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\end{thebibliography}

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