\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 88, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/88\hfil Exponential decay of solutions] {Exponential decay of solutions to a viscoelastic equation with nonlinear localized damping} \author[S. Berrimi, S. A. Messaoudi\hfil EJDE-2004/88\hfilneg] {Said Berrimi, Salim A. Messaoudi} % in alphabetical order \address{Said Berrimi\hfill\break Math Department\\ University of Setif\\ Setif, Algeria} \email{berrimi@yahoo.fr} \address{Salim A. Messaoudi \hfill\break Mathematical Sciences Department\\ KFUPM, Dhahran 31261, Saudi Arabia} \email{messaoud@kfupm.edu.sa} \date{} \thanks{Submitted May 17, 2004. Published June 29, 2004.} \subjclass[2000]{35B35, 35L20, 35L70} \keywords{Exponential decay; global existence; nonlinear localized damping} \begin{abstract} In this paper we consider the nonlinear viscoelastic equation $$ u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)\,d\tau +a(x)|u_{t}|^{m}u_{t}+b|u|^{\gamma }u=0, $$ in a bounded domain. Without imposing geometry restrictions on the boundary, we establish an exponential decay result, under weaker conditions than those in \cite{c2}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Cavalcanti \textit{et al} \cite{c2} studied the equation \begin{equation} \label{e1.1} \begin{gathered} u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau +a(x)u_{t}+|u|^{\gamma }u=0,\ \mbox{\ in }\Omega \times (0,\infty) \\ u(x,t)=0,\quad x\in \partial \Omega,\;t\geq 0 \\ u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),\quad x\in \Omega , \end{gathered} \end{equation} where $\Omega $ is a bounded domain of $\mathbb{R}^{n}$ ($n\geq 1$) with a smooth boundary $\partial \Omega $, $\gamma >0$, $g$ is a positive function, and $a:\Omega \to \mathbb{R}^{+}$ is a function, which may be null on a part of $\Omega $. Under the condition that $a(x)\geq a_{0}>0$ on $\omega \subset \Omega $, with $\omega $ satisfying some geometry restrictions and \[ -\xi_{1}g(t)\leq g'(t)\leq -\xi_{2}g(t),\quad t\geq 0, \] such that $\|g\|_{L^{1}((0,\infty))}$ is small enough, the authors obtained an exponential rate of decay. This work extended the result of Zuazua \cite{z1}, in which he considered \eqref{e1.1} with $g=0$ and the linear damping is localized. Cavalcanti \textit{et al} \cite{c3} considered the equation \[ u_{tt}-k_{0}\Delta u+\int_{0}^{t}\mathop{\rm div}[a(x)g(t-\tau) \nabla u(\tau)]d\tau +b(x)h(u_{t})+f(u)=0,\ \mbox{\ in }\Omega \times (0,\infty), \] under similar conditions on the relaxation function $g$ and $a(x)+b(x)\geq \delta >0$, for all $x\in \Omega $. They improved the result in \cite{c2} by establishing exponential stability for $g$ decaying exponentially and $h$ linear and polynomial stability for $g$ decaying polynomially and $h$ nonlinear. Their proof, based on the use of piecewise multipliers, is similar to the one in \cite{c2}. Another problem, where the damping induced by the viscosity is acting on the domain and a part of the boundary, was also discussed by Cavalcanti \textit{et al} \cite{c4} and existence and uniform decay rate results were established. In the same direction, Cavalcanti \textit{et al} \cite{c1} have also studied, in a bounded domain, the equation \[ |u_{t}|^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau -\gamma \Delta u_{t}=0, \] with $x\in \Omega$, $t>0$, $\rho >0$. They proved a global existence result for $\gamma \geq 0$ and an exponential decay for $\gamma >0$. This last result has been extended to a situation, where a source term is competing with the strong damping mechanism and the one induced by the viscosity, by Messaoudi and Tatar \cite{m2}. There, the authors combined well known methods with perturbation techniques to show that a solution with positive but small energy exist globally and decay to the rest state exponentially. Messaoudi \cite{m3} considered the equation \[ u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau +au_{t}|u_{t}|^{m}=b|u|^{\gamma }u,\quad \mbox{in }\Omega \times (0,\infty) \] and showed, under suitable conditions on $g$, that solutions with negative energy blow up in finite time if $\gamma >m$, and continue to exist if $m\geq \gamma $. We also should mention the work of Kavashima and Shibata \cite{k2}, in which a global existence and exponential stability of small solutions to a nonlinear viscoelastic problem has been established. In the absence of the viscoelastic term ($g=0$), the problem has been extensively studied and many results concerning global existence and nonexistence have been proved. For instance, for the problem \begin{equation} \label{e1.2} \begin{gathered} u_{tt}-\Delta u+au_{t}|u_{t}|^{m}=b|u|^{\gamma }u,\quad \mbox{in }\Omega \times (0,\infty) \\ u(x,t)=0,\quad x\in \partial \Omega ,\; t\geq 0 \\ u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),\quad x\in \Omega , \end{gathered} \end{equation} with $m,\gamma \geq 0$, it is well known that, for $a=0$, the source term $bu|u|^{\gamma }$, ($\gamma >0$) causes finite time blow up of solutions with negative initial energy (see \cite{b1,k1}) and for $b=0$, the damping term $au_{t}|u_{t}|^{m}$ assures global existence for arbitrary initial data (see \cite{h1,k3}). The interaction between the damping and the source terms was first considered by Levine \cite{l1,l2} in the linear damping case ($m=0$). He showed that solutions with negative initial energy blow up in finite time. Georgiev and Todorova \cite{g1} extended Levine's result to the nonlinear damping case ($m>0$). In their work, the authors introduced a different method and determined suitable relations between $m$ and $\gamma $, for which there is global existence or alternatively finite time blow up. Precisely; they showed that solutions with negative energy continue to exist globally 'in time' if $m\geq \gamma $ and blow up in finite time if $\gamma >m$ and the initial energy is sufficiently negative. Without imposing the condition that the initial energy is sufficiently negative, Messaoudi \cite{m1} extended the blow up result of \cite{g1} to solutions with negative initial energy only. For results of same nature, we refer the reader to Levine and Serrin \cite{l3} and Levine and Park \cite{l4}, Vitillaro \cite{v1}. In the present work, we are concerned with \begin{equation} \label{e1.3} \begin{gathered} u_{tt}-\Delta u+\int_{0}^{t}g(t-\tau)\Delta u(\tau)d\tau +a(x)u_{t}|u_{t}|^{m}+|u|^{\gamma }u=0,\quad \mbox{in }\Omega \times (0,\infty) \\ u(x,t)=0,\quad x\in \partial \Omega ,\; t\geq 0 \\ u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),\quad x\in \Omega , \end{gathered} \end{equation} for $m\geq 0$. We will prove an exponential decay result under weaker conditions on both $a$ and $g$. In fact we will allow $a$ to vanish on any part of $\Omega $ (including $\Omega $ itself). As a consequence, the geometry restriction imposed on a part of $\partial \Omega $ by Cavalcanti \textit{et al} \cite{c2} is dropped. Although this present work and \cite{c3} both improve [3], they have different nature and use different approaches. Our method of proof is based on the use of the perturbed energy technique. Our choice of the Lyaponov functional made our proof easier than the one in \cite{c2,c3}. This paper is organized as follows. In Section 2, We present some notation and material needed for our work and we state the global existence theorem in \cite{c2}. Section 3 contains the statement and the proof of our main result. \section{Preliminaries} In this section, we shall prepare some material needed in the proof of our result and state, without proof, a global existence result, which may be proved by repeating the argument of \cite{c2}. We use the standard Lebesgue space $L^{p}(\Omega)$ and Sobolev space $H_{0}^{1}(\Omega)$ with their usual scalar products and norms. The symbols $\nabla $ and $\Delta $ will stand for the gradient and the Laplacian respectively and the subscript $t$ will denote the time differentiation. For the relaxation function $g(t)$ we assume \begin{itemize} \item[(G1)] $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is a bounded $\mathcal{C}^{1}$ function such that $g(0)>0$ and \[1-\int_{0}^{\infty }g(s)ds=l>0. \] \item[(G2)] There exists a positive constant $\xi $ such that $g'(t)\leq -\xi g(t)$, for $t\geq 0$. \end{itemize} \begin{proposition} \label{prop1} Let $(u_{0},u_{1}) \in H_{0}^{1}(\Omega) \times L^{2}(\Omega) $. Assume that $g$ satisfies (G1) and \begin{equation} \label{e2.1} \begin{gathered} 0 \leq \gamma \leq \frac{2}{n-2}, \quad n\geq 3 \\ \gamma \geq 0,\quad n=1,2\,. \end{gathered} \end{equation} Then problem \eqref{e1.3} has a unique global solution, \begin{equation} \label{e2.2} \begin{gathered} u \in C([ 0,\quad \infty) ;H_{0}^{1}(\Omega)) \\ u_{t} \in C([ 0,\quad \infty) ;L^{2}(\Omega )) \cap L_{a}^{m+2}(\Omega \times (0,\quad \infty )) , \end{gathered} \end{equation} where $L_{a}^{m+2}$ is the weighted Lebesgue space. \end{proposition} \begin{remark} \label{rmk2.1} \rm Condition \eqref{e2.1} is needed so that the nonlinearity is Lipschitz from $H^{1}(\Omega)$ to $L^{2}(\Omega)$. Condition (G1) is necessary to guarantee the hyperbolicity of the system \eqref{e1.3}. \end{remark} Now, we introduce the energy \begin{equation} \label{e2.3} \mathcal{E}(t):=\frac{1}{2}\Big(1-\int_{0}^{t}g(s)ds\Big) \|\nabla u(t)\|_{2}^{2}+\frac{1}{2}\|u_{t}\|_{2}^{2}+\frac{1}{2}(g\circ \nabla u)(t)+\frac{1}{\gamma +2}\|u\|_{\gamma +2}^{\gamma +2}, \end{equation} where \begin{equation} \label{e2.4} (g\circ v)(t)=\int_{0}^{t}g(t-\tau)\|v(t)-v(\tau)\|_{2}^{2}d\tau . \end{equation} \begin{remark} \label{rmk2.2} \rm Multiplying equation \eqref{e1.3} by $u_{t}$ and integrating over $\Omega $, then using integration by parts and hypotheses (G1) and (G2), we obtain, after some manipulations, \begin{equation} \label{e2.5} \begin{aligned} \mathcal{E}'(t) &\leq -\Big(\int_{\Omega }a(x)|u_{t}|^{m+2}dx-\frac{1}{2}(g'\circ \nabla u)(t)+\frac{1}{2} g(t)\|\nabla u(t)\|^{2}\Big) \\ &\leq -\int_{\Omega }a(x)|u_{t}|^{m+2}dx+\frac{1}{2}(g'\circ \nabla u)(t)\leq 0. \end{aligned} \end{equation} This implies that ``modified'' energy is uniformly bounded (by $\mathcal{E}(0)$) and is decreasing in $t$. \end{remark} We will also use the embedding $H_{0}^{1}(\Omega)\hookrightarrow L^{q}(\Omega)$ for $2\leq q\leq 2n/(n-2)$ if $n\geq 3$ or $q\geq 2$ if $ n=1,2$ and $L^{r}(\Omega)\hookrightarrow L^{q}(\Omega)$, for $\ q1$, in which case \begin{equation} \label{e3.3} \int_{\Omega }a(x)|u|^{m+2}dx\leq \|a\|_{\infty }\|u\|_{m+2}^{m+2}\leq \|a\|_{\infty }\|u\|_{m+2}^{\gamma +2}\leq C_{p}^{\gamma +2}\|a\|_{\infty }\|u\|_{\gamma +2}^{\gamma +2}\,. \end{equation} If $m>\gamma $ then \begin{align*} \int_{\Omega }a(x)|u|^{m+2}dx &\leq C_{p}^{m+2}\|a\|_{\infty }\|\nabla u\|_{2}^{m+2}\leq C_{p}^{m+2}\|a\|_{\infty }\|\nabla u\|_{2}^{2}\big(\frac{2\mathcal{E}(t)}{l}\big) ^{m/2} \\ &\leq C_{p}^{m+2}\|a\|_{\infty } \big(\frac{2\mathcal{E}(0)}{l}\big)^{m/2}\|\nabla u\|_{2}^{2} \end{align*} Combining \eqref{e3.2}, \eqref{e3.3} with the above inequality, we complete the proof. \end{proof} \begin{theorem} \label{thm3.2} Let $(u_{0},u_{1}) \in H_{0}^{1}(\Omega) \times L^{2}(\Omega) $. Assume that $g$ satisfies (G1) and (G2), such that \[ 0\leq \max \{m,\gamma \}\leq \frac{2}{n-2}, \quad n\geq 3. \] Then there exist positive constants $k$ and $K$, such that the solution given by \eqref{e2.2} satisfies $\mathcal{E}(t)\leq Ke^{-kt}$ for all $t\geq 0$. \end{theorem} \begin{proof} We define the function \begin{equation} \label{e3.6} F(t):=\mathcal{E}(t)+\varepsilon_{1}\Psi (t)+\varepsilon_{2}\chi (t) \end{equation} where $\varepsilon_{1}$ and $\varepsilon_{2}$ are positive constants to be specified later and \begin{gather*} \Psi (t) :=\int_{\Omega }uu_{t}\,dx \\ \chi (t) :=-\int_{\Omega }u_{t}\int_{0}^{t}g(t-\tau )(u(t)-u(\tau))d\tau \,dx\,. \end{gather*} It is straightforward to see that for $\varepsilon_{1}$ and $\varepsilon_{2}$ small, we have \begin{equation} \label{e3.7} \alpha_{1}F(t)\leq \mathcal{E}(t)\leq \alpha_{2}F(t), \end{equation} holds for two positive constants $\alpha_{1}$ and $\alpha_{2}$. In fact \begin{equation} \label{e3.8} \begin{aligned} F(t)&\leq \mathcal{E}(t)+(\varepsilon_{1}/2) \int_{\Omega }|u_{t}|^{2}dx+(\varepsilon_{1}/2) \int_{\Omega }|u| ^{2}dx \\ &\quad+(\varepsilon_{2}/2) \int_{\Omega }|u_{t}|^{2}dx+(\varepsilon_{2}/2) \int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(u(t)-u(\tau))d\tau\Big) ^{2}dx \\ &\leq \mathcal{E}(t)+(\varepsilon_{1}/2) \int_{\Omega }|u_{t}|^{2}dx+(\varepsilon_{1}/2) C_{p}\int_{\Omega }| \nabla u| ^{2}dx \\ &\quad +(\varepsilon_{2}/2) \int_{\Omega }|u_{t}|^{2}dx +(\varepsilon_{2}/2) C_{p}(1-l)(g\circ \nabla u)(t)\\ &\leq \frac{1}{\alpha _1}\mathcal{E}(t) \end{aligned} \end{equation} where $\mathcal{E}(t)$ is the energy, and \begin{align*} F(t)&\geq \mathcal{E}(t)-(\varepsilon_{1}/2) \int_{\Omega}|u_{t}|^{2}dx-(\varepsilon_{1}/2) \int_{\Omega }|u| ^{2}dx \\ &\quad -(\varepsilon_{2}/2) \int_{\Omega }|u_{t}|^{2}dx -(\varepsilon_{2}/2) C_{p}(1-l)(g\circ \nabla u)(t) \\ &\geq \frac{1}{2}l\|\nabla u(t)\|_{2}^{2}+\frac{1}{2}\|u_{t}\|_{2}^{2} +\frac{1}{2}(g\circ \nabla u)(t)+\frac{1}{\gamma +2}\|u\|_{\gamma +2}^{\gamma +2}\\ &\quad - \frac{\varepsilon_{1}+\varepsilon_{2}}2 \int_{\Omega }|u_{t}|^{2}dx -(\frac{\varepsilon_{1}}2)C_{p}\int_{\Omega }| \nabla u| ^{2}dx -(\frac{\varepsilon_{2}}2) C_{p}(1-l)(g\circ \nabla u)(t)\\ &\geq \frac{1}{\alpha _1} \mathcal{E}(t) \end{align*} for $\varepsilon_{1}$ and $\varepsilon_{2}$ small enough. Using equation \eqref{e1.3}, we easily see that \begin{equation} \label{e3.10} \begin{aligned} \Psi '(t) &=\int_{\Omega}(uu_{tt}+u_{t}^{2})dx\\ &=\int_{\Omega }u_{t}^{2}dx-\int_{\Omega}|\nabla u|^{2}dx +\int_{\Omega }\nabla u(t)\int_{0}^{t}g(t-\tau)\nabla u(\tau)d\tau\,dx\\ &\quad -\int_{\Omega }|u|^{\gamma +2}dx-\int_{\Omega}a(x)|u_{t}|^{m}u_{t}udx \end{aligned} \end{equation} We now estimate the third term in the right-hand side of \eqref{e3.10} as follows: \begin{align*} &\int_{\Omega }\nabla u(t).\int_{0}^{t}g(t-\tau)\nabla u(\tau)d\tau\,dx\\ &\leq \frac{1}{2}\int_{\Omega }|\nabla u(t)|^{2}dx+\frac{1}{2} \int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)|\nabla u(\tau)|d\tau\Big) ^{2}dx \\ &\leq \frac{1}{2}\int_{\Omega }|\nabla u(t)|^{2}dx+\frac{1}{2} \int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(|\nabla u(\tau) -\nabla u(t)|+|\nabla u(t)|)d\tau\Big) ^{2}dx\,. \end{align*} Using Cauchy-Schwarz and Young's inequality, and $\int_{0}^{t}g(\tau)d\tau \leq \int_{0}^{\infty }g(\tau )d\tau =1-l$, we obtain that for any $\eta >0$, \begin{align*} &\int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(|\nabla u(\tau) -\nabla u(t)|+|\nabla u(t)|)d\tau\Big) ^{2}dx\\ &\leq \int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(|\nabla u(\tau) -\nabla u(t)|d\tau\Big) ^{2}dx+\int_{\Omega }\Big(\int_{0}^{t}g(t-\tau) |\nabla u(t)|d\tau\Big) ^{2}dx \\ &\quad +2\int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(|\nabla u(\tau) -\nabla u(t)|d\tau\Big) \Big(\int_{0}^{t}g(t-\tau)|\nabla u(t)|d\tau\Big) dx \\ &\leq (1+\eta)\int_{\Omega }\Big(\int_{0}^{t}g(t-\tau )|\nabla u(t)|d\tau\Big) ^{2}dx \\ &\quad +(1+\frac{1}{\eta })\int_{\Omega }\Big(\int_{0}^{t}g(t-\tau )(|\nabla u(\tau)-\nabla u(t)|d\tau\Big) ^{2}dx \\ &\leq (1+\frac{1}{\eta })\int_{\Omega }\int_{0}^{t}g(t-\tau )d\tau \int_{0}^{t}g(t-\tau)|\nabla u(\tau)-\nabla u(t)|^{2}d\tau\,dx\\ &\quad +(1+\eta)\int_{\Omega }|\nabla u(t)|^{2} \Big(\int_{0}^{t}g(t-\tau)d\tau\Big) ^{2}dx \\ &\leq (1+\eta)(1-l)^{2}\int_{\Omega }|\nabla u(t)|^{2}dx \\ &\quad +(1+\frac{1}{\eta })(1-l)\int_{\Omega }\int_{0}^{t}g(t-\tau )|\nabla u(\tau)-\nabla u(t)|^{2}d\tau\, dx\,. \end{align*} For the fifth term of the right-hand side of \eqref{e3.10}, we use Young's inequality and Lemma \ref{lm3.1} to get \begin{equation} \label{e3.13} \begin{aligned} \int_{\Omega }a(x)|u_{t}|^{m}u_{t}udx &\leq \delta \int_{\Omega }a(x)|u|^{m+2}dx +c(\delta)\int_{\Omega}a(x)|u_{t}|^{m+2}dx \\ &\leq c(\delta)\int_{\Omega }a(x)|u_{t}|^{m+2}dx+\delta C\{\|\nabla u\|_{2}^{2}+\|u\|_{\gamma +2}^{\gamma +2}\} \end{aligned} \end{equation} By combining \eqref{e3.10}--\eqref{e3.13}, we have % \label{e3.14} \begin{align*} &\Psi '(t)\\ &\leq \int_{\Omega}u_{t}^{2}dx-\int_{\Omega }|\nabla u|^{2}dx -\int_{\Omega}|u|^{\gamma +2}dx+\frac{1}{2}\int_{\Omega }|\nabla u(t)|^{2}dx \\ &\quad +\frac{1}{2}(1+\eta)(1-l)^{2}\int_{\Omega }|\nabla u(t)|^{2}dx+c(\delta)\int_{\Omega }a(x)|u_{t}|^{m+2}dx +\delta C_{p}\|\nabla u\|_{2}^{2} \\ &\quad +\delta C_{p}\|u\|_{\gamma +2}^{\gamma +2}\}+\frac{1}{2}(1+\frac{1}{\eta } )(1-l)\int_{\Omega }\int_{0}^{t}g(t-\tau)|\nabla u(\tau )-\nabla u(t)|^{2}d\tau\,dx \\ &\leq \int_{\Omega }u_{t}^{2}dx-\int_{\Omega }|\nabla u|^{2}dx-\int_{\Omega }|u|^{\gamma +2}dx +\frac{1}{2}[ 1+(1+\eta)(1-l)^{2}] \int_{\Omega }|\nabla u(t)|^{2}dx \\ &\quad +\frac{1}{2}(1+\frac{1}{\eta })(1-l)(g\circ \nabla u)(t)+c(\delta )\int_{\Omega }a(x)|u_{t}|^{m+2}dx+\delta C\{\|\nabla u\|_{2}^{2}+\|u\|_{\gamma +2}^{\gamma +2}\} \end{align*} By choosing $\eta =l/(1-l)$ and $\delta =l$ $/4C$, the above inequality becomes \begin{equation} \label{e3.15} \begin{aligned} \Psi '(t) &\leq \int_{\Omega }u_{t}^{2}dx-\frac{l}{4} \int_{\Omega }|\nabla u|^{2}dx-\frac{4-l}{4}\int_{\Omega }|u|^{\gamma +2}dx+\frac{1-l}{2l}(g\circ \nabla u)(t) \\ &\quad +c(\delta)\int_{\Omega }a(x)|u_{t}|^{m+2}dx. \end{aligned} \end{equation} Next we estimate \begin{equation} \begin{aligned} \label{e3.16} \chi '(t) &=-\int_{\Omega }u_{tt}\int_{0}^{t}g(t-\tau)(u(t) -u(\tau))d\tau\,dx \\ &\quad -\int_{\Omega }u_{t}\int_{0}^{t}g'(t-\tau )(u(t)-u(\tau))d\tau\,dx-(\int_{0}^{t}g(s)ds) \int_{\Omega }u_{t}^{2}dx \\ &=\int_{\Omega }\nabla u(t).(\int_{0}^{t}g(t-\tau )(\nabla u(t)-\nabla u(\tau))d\tau) dx \\ &\quad -\int_{\Omega }(\int_{0}^{t}g(t-\tau)\nabla u(\tau)d\tau ).(\int_{0}^{t}g(t-\tau)(\nabla u(t)-\nabla u(\tau))d\tau)dx \\ &\quad +\int_{\Omega }a(x)u_{t}(t)\int_{0}^{t}g(t-\tau) (u(t)-u(\tau))d\tau\,dx \\ &\quad +\int_{\Omega }|u|^{\gamma }u\int_{0}^{t}g(t-\tau )(u(t)-u(\tau))d\tau\,dx \\ &\quad -\int_{\Omega }u_{t}\int_{0}^{t}g'(t-\tau )(u(t)-u(\tau))d\tau\,dx-(\int_{0}^{t}g(s)ds) \int_{\Omega }u_{t}^{2}dx \end{aligned} \end{equation} Similarly to \eqref{e3.10}, we estimates the right-hand side terms of the above inequality. So for $\delta >0$, we have: For the first term, \begin{equation} \label{e3.17} -\int_{\Omega }\nabla u(t).(\int_{0}^{t}g(t-\tau )(\nabla u(t)-\nabla u(\tau))d\tau) dx \leq \delta \int_{\Omega }|\nabla u|^{2}dx+\frac{1-l}{4\delta } (g\circ \nabla u)(t). \end{equation} For the second term, \begin{align*} %\label{e3.18} &\int_{\Omega }\Big(\int_{0}^{t}g(t-s)\nabla u(s)ds\Big) \Big(\int_{0}^{t}g(t-s)(\nabla u(t)-\nabla u(s)) ds\Big) dx \\ &\leq \delta \int_{\Omega }\big| \int_{0}^{t}g(t-s)\nabla u(s)ds\big| ^{2}dx+\frac{1}{4\delta }\int_{\Omega }\big| \int_{0}^{t}g(t-s)(\nabla u(t)-\nabla u(s)) ds\big| ^{2}dx \\ &\leq \delta \int_{\Omega }\Big(\int_{0}^{t}g(t-s)(| \nabla u(t)-\nabla u(s)| +| \nabla u(t)|) ds\Big) ^{2}dx \\ &\quad +\frac{1}{4\delta }(\int_{0}^{t}g(t-s)ds) \int_{\Omega }\int_{0}^{t}g(t-s)| \nabla u(t)-\nabla u(s)| ^{2}ds\,dx\,. \\ &\leq \delta \int_{\Omega }\Big(\int_{0}^{t}g(t-s)| \nabla u(t)-\nabla u(s)| ds\Big) ^{2}dx \\ &\quad +2\delta (1-l) ^{2}\int_{\Omega }| \nabla u| ^{2}dx+\frac{1}{4\delta }(1-l) (g\circ \nabla u)(t) \\ &\leq (2\delta +\frac{1}{4\delta }) (1-l)(g\circ \nabla u)(t)+2\delta (1-l)^{2}\int_{\Omega }| \nabla u| ^{2}dx\,. \end{align*} For the third term, \[ %\label{e3.19} \int_{\Omega }a(x)u_{t}(t)\int_{0}^{t}g(t-\tau)(u(t)-u(\tau))d\tau\,dx \leq \delta \|a\|_{\infty }\int_{\Omega }u_{t}^{2}dx +\frac{C_{p}(1-l)}{4\delta }(g\circ \nabla u)(t). \] For the fourth term, \begin{equation} \label{e3.20} \begin{aligned} &\int_{\Omega }|u|^{\gamma }u\int_{0}^{t}g(t-\tau)(u(t)-u(\tau))d\tau\,dx \\ &\leq \delta \int_{\Omega }|u|^{2(\gamma +1)}dx+\frac{1}{4\delta } \int_{\Omega }\Big(\int_{0}^{t}g(t-\tau)(u(t)-u(\tau))d\tau\Big) ^{2}dx \end{aligned} \end{equation} We use \eqref{e2.1} \eqref{e2.3} and \eqref{e2.5} to obtain \begin{equation} \label{e3.21} \int_{\Omega }|u|^{2(\gamma +1)}dx\leq C_{p}\|\nabla u\|_{2}^{2(\gamma +1)}\leq C_{p}(\frac{\mathcal{E}(0)}{l}) ^{2\gamma }\|\nabla u\|_{2}^{2} \end{equation} By inserting \eqref{e3.21} in \eqref{e3.20}, we get \[ %\label{e3.22} \int_{\Omega }|u|^{\gamma }u\int_{0}^{t}g(t-\tau)(u(t)-u(\tau))d\tau\,dx \leq \delta C_{p}(\frac{\mathcal{E}(0)}{l}) ^{2\gamma }\|\nabla u\|_{2}^{2}+\frac{C_{p}(1-l)}{4\delta }(g\circ \nabla u)(t)\,. \] For the fifth term, \begin{equation} \label{e3.23} \begin{aligned} &-\int_{\Omega }u_{t}\int_{0}^{t}g'(t-\tau)(u(t)-u(\tau))d\tau\,dx \\ &\leq \delta \int_{\Omega }| u_{t}| ^{2}dx +\frac{g(0)}{4\delta }C_{p}\int_{\Omega }\int_{0}^{t}-g^{'}(t-s) | \nabla u(t)-\nabla u(s)| ^{2}ds\,dx\,. \end{aligned} \end{equation} Combining \eqref{e3.16}--\eqref{e3.23} yields \begin{equation} \label{e3.24} \begin{aligned} &\chi '(t) \\ &\leq \delta \{1+2(1-l)^{2}+C_{p}(\frac{\mathcal{E} (0)}{l}) ^{2\gamma }\}\|\nabla u\|_{2}^{2} \big[\frac{1-l}{2\delta }+2\delta (1-l)+\frac{C_{p}(1-l)}{2\delta }\big] (g\circ \nabla u)(t) \\ &\quad +\frac{g(0)}{4\delta }C_{p}(-(g'\circ \nabla u)(t)) +\big[ \delta (1+\|a\|_{\infty })-\int_{0}^{t}g(s)ds\big] \int_{\Omega }u_{t}^{2}dx\,. \end{aligned} \end{equation} Since $g(0)>0$ then there exists $t_{0}>0$ such that \begin{equation} \label{e3.25} \int_{0}^{t}g(s)ds\geq \int_{0}^{t_{0}}g(s)ds=g_{0}>0,\quad \forall t\geq t_{0}. \end{equation} Using \eqref{e3.6}, \eqref{e3.15}, \eqref{e3.24}, and \eqref{e3.25}, we obtain \begin{equation} \label{e3.26} \begin{aligned} &F'(t) \\ &\leq -(1-\varepsilon_{1}c(\delta)) \int_{\Omega }a(x)|u_{t}|^{m+2}dx \\ &\quad -\big[ \varepsilon_{2}\{g_{0}-\delta (1+\|a\|_{\infty }) \}-\varepsilon_{1}\big] \int_{\Omega }u_{t}^{2}dx-\varepsilon_{1} \frac{4-l}{4}\int_{\Omega }|u|^{\gamma +2}dx \nonumber \\ &\quad-\big[ \frac{\varepsilon_{1}l}{4}-\varepsilon_{2}\delta \{1+2(1-l)^{2}+C_{p}(\frac{\mathcal{E}(0)}{l}) ^{2\gamma }\}\big] \|\nabla u\|_{2}^{2}\\ &\quad +\big[ \frac{1}{2}-\frac{\varepsilon_{1}(1-l)}{2\xi l} -\varepsilon_{2}\{\frac{g(0)}{4\delta }C_{p}+\frac{(1-l)C_{p}}{ 2\delta \xi }+\frac{1-l}{2\delta \xi }+\frac{2\delta (1-l)}{\xi }\}\big] (g'\circ \nabla u)(t). \end{aligned} \end{equation} Now, we choose $\delta $ so small that \begin{gather*} g_{0}-\delta (1+\|a\|_{\infty }) >\frac{1}{2}g_{0} \\ \frac{4}{l}\delta \{1+2(1-l)^{2}+C_{p}(\frac{\mathcal{E}(0)}{l}) ^{2\gamma }\} <\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{e3.27} \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 (1+\|a\|_{\infty }) \}-\varepsilon_{1}>0 \\ k_{2} =\frac{\varepsilon_{1}l}{4}-\varepsilon_{2}\delta \{1+2(1-l)^{2}+C_{p}(\frac{\mathcal{E}(0)}{l}) ^{2\gamma }\}>0. \end{gather*} We then pick $\varepsilon_{1}$ and $\varepsilon_{2}$ so small that \eqref{e3.7} and \eqref{e3.27} remain valid and \begin{gather*} 1-\varepsilon_{1}c(\delta)>0, \\ \frac{1}{2}-\frac{\varepsilon_{1}(1-l)}{2\xi l}-\varepsilon_{2}\{\frac{g(0) }{4\delta }C_{p}+\frac{(1-l)C_{p}}{2\delta \xi }+\frac{1-l}{2\delta \xi }+ \frac{2\delta (1-l)}{\xi }\}>0 \end{gather*} Therefore, we arrive at $F'(t)\leq -\beta \mathcal{E}(t)$ for all $t\geq t_{0}$. This inequality and \eqref{e3.7} yield $F'(t)\leq -\beta \alpha_{1}F(t)$, for all $t\geq t_{0}$. A simple integration leads to \[ F(t)\leq F(t_{0})e^{\beta \alpha_{1}t_{0}}e^{-\beta \alpha_{1}t},\quad \forall t\geq t_{0}. \] This inequality and \eqref{e3.7} yields \[ \mathcal{E}(t)\leq \alpha_{2}F(t_{0})e^{\beta \alpha_{1}t_{0}}e^{-\beta \alpha_{1}t},\quad \forall t\geq t_{0}, \] which completes the proof. \end{proof} \begin{remark} \label{rmk3.1} \rm Note that our result is proved without imposing any restriction on the size of $\|g\|_{L^{1}}$. Also note that the function $a$ may vanish on the whole domain $\Omega $. In other words, contrary to \cite{c2}, measure $(\omega)$ can be zero. As a consequence, no geometry restriction on the boundary has been assumed. \end{remark} \subsection*{Acknowledgment} The authors would like to express their sincere thanks to King Fahd University of Petroleum and Minerals for its support. 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