\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 93, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/93\hfil Uniform decay of solutions]
{Uniform decay of solutions to Cauchy viscoelastic problems
with density}

\author[M. Kafini\hfil EJDE-2011/93\hfilneg]
{Mohammad Kafini}

\address{Mohammad Kafini \newline
Department of Mathematics and Statistics\\
KFUPM, Dhahran 31261 \\
Saudi Arabia}
\email{mkafini@kfupm.edu.sa}

\thanks{Submitted May 25, 2011. Published July 18, 2011}
\subjclass[2000]{35B05, 35L05, 35L15, 35L70}
\keywords{Cauchy problem; relaxation
function; viscoelastic; weighted spaces}

\begin{abstract}
 In this article we consider the decay of solutions to
 a linear Cauchy viscoelastic problem with density.
 This study includes the exponential and polynomial
 rates as particular cases.  To compensate for the lack
 of Poincare's inequality in the whole space, we consider
 the solutions in spaces weighted by the density.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

 In this article we are concerned with the initial-value
problem
\begin{equation} \label{e1}
\begin{gathered}
\rho (x)u_{tt}-\Delta u(x)+\int_0^{t}g(t-s)\Delta u(x,s)ds=0,\quad
 x\in \mathbb{R}^n,\; t>0, \\
 u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \mathbb{R}^n,
\end{gathered}
\end{equation}
where $u_0$, $u_1$ are initial data chosen in suitable spaces
and $g$ is the relaxation function subjected to some conditions
to be specified later.
The density $\rho (x)$ satisfying the following conditions
\begin{itemize}
\item[(H1)] $\rho :\mathbb{R}^n\to \mathbb{R}$,
$n\geq 2$, $\rho (x)>0$,
$\rho (x)\in C^{0,\gamma }(\mathbb{R}^n)$
with $\gamma \in (0,1)$ and
$\rho \in L^{n/2}(\mathbb{R}^n)\cap L^{\infty }(\mathbb{R}^n)$.
\end{itemize}

In the whole space case, Poincare's inequality and some Lebesgue
and Sobolev embedding inequalities are not valid.
To overcome this difficulty in this case, we exploit the density
to introduce weighted spaces for solutions
of our problem.

The work with weighted spaces was studied by many authors.
Papadopoulos and Stavarakakis \cite{p1} established existence of a global
solutions and blow up results for the  non local quasilinear
hyperbolic problem of Kirchhoff type
\begin{gather*}
u_{tt}-\phi (x)\| \nabla u(t)\| ^2\Delta u+\delta
u_t=| u| ^{a}u,\\
 x\in \mathbb{R}^n,\; t\geq 0,
\end{gather*}
in the case where $n\geq 3$, $\delta \geq 0$ and
$\rho (x)=(\phi (x))^{-1}$ is a positive function lying in
$L^{n/2}(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$.
Brown and Stavarakakis \cite{b1} proved the existence of
positive solutions for the semilinear elliptic equation
\begin{gather*}
-\Delta u(x)=\lambda g(x)f(u(x)),\quad 0<u<1,\; x\in \mathbb{R}\\
\lim_{|x| \to +\infty }u(x)=0
\end{gather*}
and for $g\in L^{n/2}(\mathbb{R}^n)$ for $n=1,2,3$.
Karachalios and Stavarakakis \cite{k3} proved a local existence of
solutions and global attractor in the energy space
$D^{1,2}(\mathbb{R}^n)\times L_g^2(\mathbb{R}^n)$ for a
semilinear hyperbolic problem
\[
u_{tt}-\phi (x)\Delta u+\delta u_t+\lambda f(u)=\eta (x),
\quad x\in\mathbb{R}^n,\;t>0,
\]
in the case where $\delta >0$, $n\geq 3$ and
$\rho (x)=(\phi (x))^{-1}$ lies
in $L^{n/2}(\mathbb{R}^n)$.

It is also worth mentioning the work of  Zhou \cite{z1} and
Cavalcanti cite{c2}.
In this work, the following nonlinear wave equation with damping
and source terms of the form
\[
u_{tt}-\phi (x)\Delta u+a| u_t| ^{m-1}u_t=f(x,u),\quad
x\in \mathbb{R}^n,\; t>0,
\]
was considered. Where the author proved, in the linear damping
case, that the solution blows up in finite time even for vanishing
initial energy. Criteria to guarantee blow up of solutions with
positive initial energy were established for both linear and
nonlinear cases. Global existence and large time behavior also
proved.

In \cite{m3}, a class of abstract viscoelastic systems of the form
\begin{equation} \label{e2}
\begin{gathered}
u_{tt}(t)+\mathcal{A}u(t)+\beta u(t)-(g*\mathcal{A}^{\alpha }u)(t)=0 \\
u(0)=u_0,\quad u_t(0)=u_1,
\end{gathered}
\end{equation}
for $0\leq \alpha \leq 1$, $\beta \geq 0$, were investigated.
The main focus was on the case when $0<\alpha <1$ and the main result
was that solutions for \eqref{e2} decay polynomially even if the
kernel $g$ decay exponentially. This
result is sharp (see \cite[Theorem 12]{m3}. This result has been
improved by Rivera \textit{et al.} \cite{m4}, where the authors studied
a more general abstract problem than \eqref{e2} and established a
necessary and sufficient condition to obtain an exponential decay.
In the case of lack of exponential
decay, a polynomial decay has been proved.

 Kafini and Messaoudi \cite{k1} looked into the following Cauchy
viscoelastic problem
\begin{gather*}
u_{tt}-\Delta u+\int_0^{t}g(t-s)\Delta u(x,s)ds=0,\quad x\in \mathbb{R}
^n,\; t>0 \\
u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \mathbb{R}^n
\end{gather*}
 and showed that, for compactly supported initial data $u_0$,
$u_1$ and for an exponentially decaying relaxation function $g$,
the decay of the first energy of solution is polynomial.
The finite-speed propagation is used to compensate for the lack
of Poincar\'{e}'s inequality in $\mathbb{R}^n$.
 For nonexistence, the same authors \cite{k2} established a blow-up
result to the following Cauchy viscoelastic problem
\begin{gather*}
u_{tt}-\Delta u+\int_0^{t}g(t-s)\Delta u(x,s)ds+u_t=| u|
^{p-1}u,\quad x\in \mathbb{R}^n,\; t>0 \\
u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \mathbb{R}^n,
\end{gather*}
under the conditions
\begin{gather*}
\int_0^{t}g(s)ds)<\frac{2p-2}{2p-1},\quad g'(t)\leq 0,\\
E(0)=\frac{1}{2}\| u_1\| _2^2+\frac{1}{2}\| \nabla
u_0\| _2^2-\frac{1}{p+1}\| u_0\| _{p+1}^{p+1}\leq 0.
\end{gather*}
This result extends the one of \cite{z2}, established for the wave
equation in bounded domain.

In this article, we will extend the result in \cite{z1} to our viscoelastic
problem. We aim to study the effect of the density to the decay rates. In
the case $\rho (x)=1$ (as in \cite{k1}), the best decay obtained is polynomial.
Here, we establish a general decay result for solutions where the
exponential and polynomial are only special cases. This result does not
contradict the past results in \cite{k1,m3,m4}. The choice of spaces
of solutions and the density one make it possible to get an exponential
decay rate. Where we have $\rho (x)$ is a continuous and $L^{n/2}(
\mathbb{R}^n)\cap L^{\infty }(\mathbb{R}^n)$ function that make most of
its contribution concentrate at the early time in the contrast of the late
time ($t\to \infty $). Obviously, $\rho (x)$ cannot be a constant
here. In our proof, we use the multiplier method together with the Lyapunov
functional method as in \cite{m1} with some necessary modification due to the
nature of the problem. The paper organized as follows. In Section 2, we
define our function space and the assumptions on the kernel $g$. In section
3, we state and prove our main result.

\section{Preliminaries}

 To achieve our result, we assume the following assumptions on the
relaxation function $g$:
\begin{itemize}
\item[(G1)] $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is a
differentiable function such that
\[
g(0)>0,\quad 1-\int_0^{\infty }g(s)ds=l>0.
\]

\item[(G2)] There exists a nonincreasing differentiable function
$\xi :\mathbb{R}_{+}\to \mathbb{R}_{+}$ such that
\[
g'(t)\leq -\xi (t)g(t),\quad t\geq 0, \quad
\int_0^{\infty }\xi (t)dt=+\infty .
\]
\end{itemize}
 There are many functions satisfying (G1) and (G2), for example
\begin{gather*}
g_1(t) =\frac{\alpha }{(1+t)^{\nu }},\quad \nu >1 \\
g_2(t) =\alpha e^{-\beta (t+1)^p},\quad 0<p\leq 1 \\
g_{3}(t) =\frac{\alpha }{(1+t)[\ln (1+t)]^{\nu }},\quad \nu >1
\end{gather*}
where $\alpha $ and $\beta $ are chosen properly.

\begin{remark} \label{rmk2.1} \rm
Condition (G1) is necessary to guarantee the
hyperbolicity of the system.
\end{remark}

We define the function space of our problem and its norm,
as in \cite{b1,p1}, as follows:\\
(A) The function space for \eqref{e1} is
$X=D^{1,2}(\mathbb{R}^n)\times L_{\rho }^2(\mathbb{R}^n)$,
with
\[
D^{1,2}(\mathbb{R}^n)=\{ f\in L^{\frac{2n}{n-2}}(\mathbb{R}
^n):\nabla f\in L^2(\mathbb{R}^n)\} .
\]
(B) The space $L_{\rho }^2(\mathbb{R}^n)$ is defined
to be the closure of $C_0^{\infty }(\mathbb{R}^n)$ functions with
respect to the inner product
\[
(f,h)_{L_{\rho }^2(\mathbb{R}^n)}=\int_{\mathbb{R}^n}\rho fh\,dx.
\]
One can easily check that $L_{\rho }^2(\mathbb{R}^n)$ is a
separable Hilbert space and
\[
\| f\| _{L_{\rho }^2(\mathbb{R}^n)}^2=(f,f)_{L_{\rho }^2(
\mathbb{R}^n)}.
\]
(C) For $1<p<\infty $, if $f$ is a measurable function on
$\mathbb{R}^n$, we define
\[
\| f\| _{L_{\rho }^p}=\Big( \int_{\mathbb{R}^n}\rho |
f| ^pdx\Big) ^{1/p}
\]
and let $L_{\rho }^p(\mathbb{R}^n)$ consist of all $f$ for
which $\| f\| _{L_{\rho }^p(\mathbb{R}^n)}<\infty $.

 For the weighted space $L_{\rho }^p(\mathbb{R}^n)$, we have
the following lemma

\begin{lemma}[\cite{k3}] \label{lem2.2}
Let $\rho $ satisfies {\rm (H1)},
 then for any $u\in D^{1,2}(\mathbb{R}^n)$,
\[
\| u\| _{L_{\rho }^{q}}\leq \| \rho \| _{L^{s}}\|
\nabla u\| _{L^2}, \quad\text{with}\quad
s=\frac{2n}{2n-qn+2q},\quad 2\leq q\leq \frac{2n}{n-2}.
\]
\end{lemma}

\begin{corollary} \label{coro2.3}
If $q=2$, then Lemma \ref{lem2.2}. yields
\[
\| u\| _{L_{\rho }^2}\leq \| \rho \| _{L^{n/2}}\| \nabla u\| _{L^2},
\]
where we can assume $\| \rho \| _{L^{n/2}}=C_0>0$  to get
\begin{equation} \label{e3}
\| u\| _{L_{\rho }^2(\mathbb{R}^n)}\leq C_0\| \nabla
u\| _{L^2}.
\end{equation}
\end{corollary}

\begin{theorem}[\cite{z1}] \label{thm2.4}
Suppose that {\rm (H1)} holds and $g$ satisfies {\rm (G1)}.
Assume that $1\leq p\leq \frac{n+2}{n-2}$  if
$n\geq 2$ or $1\leq p$ if $n=2$.
Then for any initial data
\[
u_0\in \emph{D}^{1,2}(\mathbb{R}^n)\quad \text{and}\quad u_1\in
L_{\rho }^2(\mathbb{R}^n),
\]
 problem \eqref{e1} has a unique solution
\[
u\in C([0,T);D^{1,2}(\mathbb{R}^n)) \quad \text{and}\quad
u_t\in C([0,T);L_{\rho }^2(\mathbb{R}^n)),
\]
for $T$ small enough.
\end{theorem}

We now introduce the `modified' energy functional
\begin{equation} \label{e4}
E(t)=\frac{1}{2}\| u_t\| _{L_{\rho }^2}^2+\frac{1}{2}\Big(
1-\int_0^{t}g(s)ds\Big) \| \nabla u\| _2^2+\frac{1}{2}
(g\circ \nabla u),
\end{equation}
where
\[
(g\circ v)(t)=\int_0^{t}g(t-s)\| v(t)-v(s)\| _2^2ds,\quad
\forall v\in L^2(\mathbb{R}^n).
\]

\section{Decay of solutions}

In this section, we state and prove our main result. For this
purpose, we set
\begin{equation} \label{e5}
F(t):=E(t)+\varepsilon _1\Psi (t)+\varepsilon _2\chi (t),
\end{equation}
where $\varepsilon _1$ and $\varepsilon _2$ are positive
constants and
\begin{gather}
\Psi (t) :=\int_{\mathbb{R}^n}\rho uu_t\,dx, \label{e6}\\
\chi (t) :=-\int_{\mathbb{R}^n}\rho u_t\int_0^{t}g(t-s)\big(
u(t)-u(s)\big) \,ds\,dx. \label{e7}
\end{gather}

\begin{lemma} \label{lem3.1}
Along the solution of \eqref{e1}, the `modified' energy satisfies
\begin{equation} \label{e8}
E'(t)=\frac{1}{2}(g'\circ \nabla u)-\frac{1}{2}g(t)\|
\nabla u\| _2^2\leq \frac{1}{2}(g'\circ \nabla u)\leq 0.
\end{equation}
\end{lemma}

\begin{proof}
 By multiplying \eqref{e1} by $u_t$ and integrating over $
\mathbb{R}^n$, using integration by parts, hypotheses (G1) and (G2)
and some manipulations as in \cite{c1,c2,m2}, we reach the result.
\end{proof}

\begin{lemma} \label{lem3.2} For any
$\varepsilon _1$  and $\varepsilon _2$ small enough,
\begin{equation} \label{e9}
\alpha _1F(t)\leq E(t)\leq \alpha _2F(t),
\end{equation}
holds for two positive constants $\alpha _1$  and
$\alpha_2$.
\end{lemma}

\begin{proof} By applying Young's inequality to \eqref{e5} and using
\eqref{e6} and \eqref{e7}, we obtain
\begin{align*}
F(t) &\leq E(t)+\varepsilon _1\big( \delta \| u_t\|
_{L_{\rho }^2}^2+\frac{1}{4\delta }\| u\| _{L_{\rho
}^2}^2\big) \\
&\quad+ \varepsilon _2\Big( \delta \| u_t\| _{L_{\rho }^2}^2+
\frac{1}{4\delta }\| \int_0^{t}g(t-s)\big( u(t)-u(s)\big)
ds\| _{L_{\rho }^2}^2\big) \\
&\leq E(t)+\varepsilon _1\Big( \delta \| u_t\| _{L_{\rho
}^2}^2+\frac{C_0}{4\delta }\| \nabla u\| _2^2\Big)
+\varepsilon _2\Big( \delta \| u_t\| _{L_{\rho }^2}^2+
\frac{C_0}{4\delta }(1-l)(g\circ \nabla u)\Big) \\
&\leq E(t)+\delta (\varepsilon _1+\varepsilon _2)\| u_t\|
_{L_{\rho }^2}^2+\frac{\varepsilon _1C_0}{4\delta }\| \nabla
u\| _2^2+\frac{\varepsilon _2C_0}{4\delta }(1-l)(g\circ \nabla
u) \\
&\leq E(t)+\beta E(t)\leq \alpha _2F(t).
\end{align*}
Also, for $\varepsilon _1$ and $\varepsilon _2$ small enough,
we have
\begin{align*}
F(t) &\geq E(t)-\varepsilon _1\Psi (t)-\varepsilon _2\chi (t) \\
&\geq E(t)-\delta (\varepsilon _1+\varepsilon _2)\| u_t\|
_{L_{\rho }^2}^2-\frac{\varepsilon _1C_0}{4\delta }\| \nabla
u\| _2^2-\frac{\varepsilon _2C_0}{4\delta }(1-l)(g\circ \nabla
u) \\
&\geq E(t)-\beta E(t)\geq \alpha _1F(t).
\end{align*}
Consequently, \eqref{e9} follows.
\end{proof}

\begin{lemma} \label{lem3.3}
Assume {\rm (H1), (G1), (G2)}.
Along the solution of \eqref{e1}, the functional
\[
\Psi (t):=\int_{\mathbb{R}^n}\rho uu_t\,dx,
\]
satisfies
\begin{equation} \label{e10}
\Psi '(t)\leq \| u_t\| _{L_{\rho }^2}^2-(
l-\delta ) \| \nabla u\| _2^2+\frac{1}{4\delta }(
1-l) ( g\circ \nabla u) (t).
\end{equation}
\end{lemma}

\begin{proof} From the definition of $\Psi (t)$ in \eqref{e6}
 we have
\begin{equation} \label{e11}
\Psi '(t)=\int_{\mathbb{R}^n}\rho u_t^2dx+\int_{\mathbb{R}
^n}\rho uu_{tt}dx.
\end{equation}

 To estimate the last term in \eqref{e11}, we multiply \eqref{e1}
 by $u$ and integrate by parts over $\mathbb{R}^n$. So, we obtain
\begin{equation} \label{e12}
\int_{\mathbb{R}^n}\rho uu_{tt}dx=\int_{\mathbb{R}^n}\nabla u(t)\cdot
\int_0^{t}g(t-s)\nabla u(s)\,ds\,dx-\int_{\mathbb{R}^n}| \nabla
u(t)| ^2dx.
\end{equation}
The first term in \eqref{e12} can be estimated as follows
\begin{equation} \label{e13}
\begin{split}
&\int_{\mathbb{R}^n}\nabla u(t)\cdot \int_0^{t}g(t-s)\nabla u(s)\,ds\,dx
 \\
&= \int_{\mathbb{R}^n}\nabla u(t)\cdot \int_0^{t}g(t-s)( \nabla
u(s)-\nabla u(t)+\nabla u(t)) \,ds\,dx \\
&= \Big( \int_0^{t}g(s)ds\Big) \int_{\mathbb{R}^n}| \nabla
u(t)| ^2dx+\int_{\mathbb{R}^n}\nabla u(t)\cdot
\int_0^{t}g(t-s)( \nabla u(s)-\nabla u(t)) \,ds\,dx   \\
&\leq \Big( \int_0^{t}g(s)ds\Big) \int_{\mathbb{R}^n}| \nabla
u(t)| ^2dx+\delta \int_{\mathbb{R}^n}| \nabla u(t)|
^2dx+\frac{1}{4\delta }\Big( \int_0^{t}g(s)ds\Big) ( g\circ
\nabla u) (t).
\end{split}
\end{equation}
 Recalling  that
\[
\int_0^{t}g(s)ds\leq \int_0^{\infty }g(s)ds=1-l,
\]
we obtain the result.
\end{proof}

\begin{lemma} \label{lem3.4}
Assume {\rm (G1),  (G2)}.
Along the solution of \eqref{e1}, for any $\delta>0$,
the functional
\[
\chi (t):=-\int_{\mathbb{R}^n}\rho u_t\int_0^{t}g(t-s)(
u(t)-u(s)) \,ds\,dx
\]
satisfies
\begin{equation} \label{e14}
\begin{split}
\chi '(t) &\leq \delta ( 1+2(1-l)^2) \| \nabla
u\| _2^2+(1-l)\big[ ( 2\delta +\frac{1}{2\delta }) +
\frac{1}{4\delta }\big] ( g\circ \nabla u) (t)   \\
&\quad \frac{g(0)}{4\delta }C_0( -(g'\circ \nabla u)(t))
-\Big( \int_0^{t}g(s)ds-\delta \Big) \int_{\mathbb{R}^n}\rho
u_t^2dx.
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
 From the definition of $\chi (t)$ in \eqref{e7}, we have
\begin{equation} \label{e15}
\begin{split}
\chi '(t)
&= -\int_{\mathbb{R}^n}\rho u_{tt}\int_0^{t}g(t-s)( u(t)-u(s))
 \,ds\,dx \\
&\quad -\int_{\mathbb{R}^n}\rho u_t\int_0^{t}g'(t-s)(
u(t)-u(s)) \,ds\,dx
-\Big( \int_0^{t}g(s)ds\Big) \int_{\mathbb{R}
^n}\rho u_t^2dx.
\end{split}
\end{equation}
To simplify the first term in \eqref{e15}, we multiply \eqref{e1}
by $\int_0^{t}g(t-s)( u(t)-u(s)) ds$ and integrate by parts over
$\mathbb{R}^n$. So we obtain
\begin{equation} \label{e16}
\begin{split}
&\int_{\mathbb{R}^n}\rho u_{tt}\int_0^{t}g(t-s)( u(t)-u(s))
\,ds\,dx   \\
&= \int_{\mathbb{R}^n}\Delta u(x)\int_0^{t}g(t-s)( u(t)-u(s))
\,ds\,dx   \\
&-\int_{\mathbb{R}^n}( \int_0^{t}g(t-s)( u(t)-u(s))
ds\int_0^{t}g(t-s)\Delta u(s)ds) dx.
\end{split}
\end{equation}
 The first term in the right side of \eqref{e16} is estimated as follows
\begin{equation} \label{e17}
\begin{split}
&\int_{\mathbb{R}^n}\Delta u(x)\int_0^{t}g(t-s)( u(t)-u(s))
\,ds\,dx   \\
&\leq -\int_{\mathbb{R}^n}\nabla u(x)\cdot \int_0^{t}g(t-s)(\nabla
u(t)-\nabla u(s))\,ds\,dx   \\
&\leq \int_{\mathbb{R}^n}\nabla u(x)\cdot \int_0^{t}g(t-s)(\nabla
u(s)-\nabla u(t))\,ds\,dx   \\
&\leq \delta \int_{\mathbb{R}^n}| \nabla u(t)| ^2dx+\frac{1}{
4\delta }( \int_0^{t}g(s)ds) ( g\circ \nabla u) (t)
 \\
&\leq \delta \int_{\mathbb{R}^n}| \nabla u(t)| ^2dx+\frac{1}{
4\delta }(1-l)( g\circ \nabla u) (t),
\end{split}
\end{equation}
 while the second term becomes, as in \eqref{e13},
\begin{equation} \label{e18}
\begin{split}
&-\int_{\mathbb{R}^n}( \int_0^{t}g(t-s)( u(t)-u(s))
ds\int_0^{t}g(t-s)\Delta u(x,s)ds) dx   \\
&= \int_{\mathbb{R}^n}( \int_0^{t}g(t-s)\nabla u(s)ds) \cdot
( \int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s))ds) dx   \\
&\leq \delta \int_{\mathbb{R}^n}| \int_0^{t}g(t-s)\nabla
u(s)ds| ^2dx+\frac{1}{4\delta }\int_{\mathbb{R}^n}|
\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s))ds| ^2dx   \\
&\leq ( 2\delta +\frac{1}{4\delta }) \int_{\mathbb{R}^n}(
\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s))ds) ^2dx   \\
&\quad 2\delta (1-l)^2\int_{\mathbb{R}^n}| \nabla u(t)| ^2dx
 \\
&\leq ( 2\delta +\frac{1}{4\delta }) (1-l)( g\circ \nabla
u) (t)+2\delta (1-l)^2\int_{\mathbb{R}^n}| \nabla u(t)|
^2dx.
\end{split}
\end{equation}
 Back to \eqref{e15}, the second term can be estimated as follows
\begin{equation} \label{e19}
\begin{split}
&-\int_{\mathbb{R}^n}\rho u_t\int_0^{t}g'(t-s)(
u(t)-u(s)) \,ds\,dx   \\
&\leq \delta \int_{\mathbb{R}^n}\rho u_t^2dx+\frac{1}{4\delta }
\| \int_0^{t}-g'(t-s)( u(t)-u(s)) ds\|
_{L_{\rho }^2}^2   \\
&\leq \delta \int_{\mathbb{R}^n}\rho u_t^2dx+\frac{g(0)}{4\delta }
C_0( -(g'\circ \nabla u)(t)) .
\end{split}
\end{equation}
By combining \eqref{e15}-\eqref{e19}, the assertion of the
lemma is established.
\end{proof}

\begin{theorem} \label{thm3.5}
Let $u_0\in D^{1,2}(\mathbb{R}^n)$
 and $u_1\in L_{\rho }^2(\mathbb{R}^n)$ be given.
Assume that {\rm (H1)} holds and $g$  satisfies {\rm (G1)} and
{\rm (G2)}. Then, for each $t_0>0$, there exist strictly positive
constants $C_1$ and $C_2$  such that the energy of solution given
by \eqref{e1} satisfies, for all $t\geq t_0$,
\begin{equation} \label{e20}
E(t)\leq C_1E( t_0) e^{-C_2\int_{t_0}^{t}\xi (s)ds},\quad
\forall t>t_0.
\end{equation}
\end{theorem}

\begin{proof}
Since $g$ is positive and $g(0)>0$, then for any $t_0>0$ we have
\[
\int_0^{t}g(s)ds\geq \int_0^{t_0}g(s)ds=g_0>0,\quad \forall t\geq
t_0.
\]
Differentiation of \eqref{e3} using \eqref{e8}, \eqref{e10} and
\eqref{e14}, yields
\begin{equation} \label{e21}
\begin{split}
F'(t)
&= E'(t)+\varepsilon _1\Psi '(t)+\varepsilon _2\chi '(t)   \\
&\leq -( \varepsilon _2(g_0-\delta )-\varepsilon _1) \int_{
\mathbb{R}^n}\rho u_t^2{}dx+[ \frac{1}{2}-\varepsilon _2\frac{
g(0)}{4\delta }C_0] ( (g'\circ \nabla u)(t))
 \\
&\quad -[ \varepsilon _1( l-\delta ) -\varepsilon _2\delta
( 1+2(1-l)^2) ] \| \nabla u\| _2^2 \\
&\quad \big\{ \frac{\varepsilon _1}{4\delta }( 1-l) +\varepsilon
_2(1-l)[ \big( 2\delta +\frac{1}{2\delta }\big)
+\frac{1}{4\delta } ] \big\} ( g\circ \nabla u) (t)
\end{split}
\end{equation}
 At this point we choose $\delta $ so small that
\begin{gather*}
\max \{ g_0-\delta ,l-\delta \} > \frac{1}{2}g_0, \\
\delta ( 1+2(1-l)^2) < \frac{1}{4}g_0.
\end{gather*}
Whence $\delta $ is fixed, the choice of any two positive constants
$\varepsilon _1$ and $\varepsilon _2$ satisfying
\begin{equation} \label{e22}
\frac{1}{4}g_0\varepsilon _2<\varepsilon _1<\frac{1}{2}
g_0\varepsilon _2,
\end{equation}
will make
\begin{gather*}
k_1 = \varepsilon _2(g_0-\delta )-\varepsilon _1>0, \\
k_2 = \varepsilon _1( l-\delta ) -\varepsilon _2\delta
(1+2(1-l)^2)>0.
\end{gather*}
Then we pick $\varepsilon _1$ and $\varepsilon _2$ so small
that \eqref{e9} and \eqref{e22} remain valid and
\[
\frac{1}{2}-\varepsilon _2\frac{g(0)}{4\delta }C_0>0.
\]
Therefore,  for some positive constants $\beta ,\beta
_1$ and $\beta _2$, we have
\begin{equation} \label{e23}
\begin{split}
F'(t)
&\leq -\beta ( \int_{\mathbb{R}^n}\rho
u_t{}^2dx+\| \nabla u\| _2^2) +\beta _1(g\circ
\nabla u)(t),   \\
&\leq -\beta E(t)+\beta _2(g\circ \nabla u)(t)\quad \forall t\geq t_0.
\end{split}
\end{equation}
Multiplying \eqref{e23} by $\xi (t)$ gives
\[
\xi (t)F'(t)\leq -\beta \xi (t)E(t)+\beta _2\xi (t)(g\circ \nabla
u)(t),\quad \forall t\geq t_0.
\]
The last term can be estimated, using (H2), as follows
\[
\beta _2\xi (t)(g\circ \nabla u)(t)\leq -\beta _2(g'\circ
\nabla u)(t)\leq -2\beta _2E'( t) .
\]
Thus, \eqref{e23} becomes
\begin{equation} \label{e24}
\begin{gathered}
\xi (t)F'(t) \leq -\beta \xi (t)E(t)-2\beta _2E'(t)   \\
\xi (t)F'(t)+2\beta _2E'( t) \leq -\beta \xi (t)E(t).
\end{gathered}
\end{equation}
It is clear that
\[
F_1(t)=\xi (t)F(t)+2\beta _2E( t) \sim E( t) .
\]
 Therefore, using \eqref{e24} and the fact that $\xi '(t)\leq
0, $ we arrive at
\begin{equation} \label{e25}
F_1'(t)=( \xi (t)F(t)+2\beta _2E( t) )
'\leq -\beta \xi (t)E( t) .
\end{equation}
Integration over $(t_0,t)$ leads to, for some constant $C_2>0$ such
that
\[
F_1(t)\leq F_1( t_0) e^{-C_2\int_{t_0}^{t}\xi
(s)ds},\quad \forall t>t_0.
\]
Recalling \eqref{e9}, estimate \eqref{e15} yields the desired
result \eqref{e20}.
\end{proof}


\begin{remark} \label{rmk3.1} \rm
Our result is established without using the condition
$\int_0^{\infty }\xi (s)ds=+\infty $, which is crucial for obtaining
 uniform stability.

Exponential decay is obtained for $\xi(t)\equiv a$, and
polynomial decay for $\xi (t)=a(1+t)^{-1}$, where $a$
is a positive constant.

 Estimate \eqref{e20} is also true for $t\in [0,t_0]$ by
virtue of continuity and boundedness of $E(t)$ and $\xi (t)$.
\end{remark}

\subsection{Acknowledgments}
The author would like to express his sincere thanks to King Fahd
University of Petroleum and Minerals for its support.

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\end{document}


In addition, the following paper in connection with
the main subject of the present paper should be quoted, namely:



The above reference treats the wave equation subject to a density
term $\rho(x)\geq 0$ and it
has been published before the reference [4] quoted in the present article.
