\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 71, 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} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/71\hfil Exponential decay of solutions] {Exponential decay of solutions to a fourth-order viscoelastic evolution equation \\ in $\mathbb{R}^n$} \author[M. Kafini\hfil EJDE-2010/71\hfilneg] {Mohammad Kafini} \address{Mohammad Kafini \newline Department of Mathematics and Statistics, KFUPM-DCC, Dhahran 31261, Saudi Arabia} \email{mkafini@kfupm.edu.sa} \thanks{Submitted March 3, 2010. Published May 17, 2010.} \subjclass[2000]{35B05, 35L05, 35L15, 35L70} \keywords{Decay; Cauchy problem; relaxation function; viscoelastic} \begin{abstract} In this article, we consider a Cauchy problem for a viscoelastic wave equation of fourth order. Under suitable conditions on the initial data and the relaxation function, we show that the rate of decay is exponential. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} In this work concerns the Cauchy problem $$\label{e1} \begin{gathered} u_{tt}-\Delta u+u+\int_0^{t}g(t-s)(\Delta u(s)-u(s)) ds-\Delta u_{tt}=0,\quad x\in \mathbb{R}^n,\; t>0, \\ u(x,0)=u_0(x),\quad u_{t}(x,0)=u_1(x),\quad x\in \mathbb{R}^n, \end{gathered}$$ where $u_0$, $u_1$ are initial data and $g$ is the relaxation function subjected to some conditions to be specified later. This type of evolution equations of fourth order arises in the study of strain solitary waves \cite{c4,z1} and in the theory of viscoelasticity when the material density depends on $u_{t}$, see \cite{f1,r1}. Hrusa and Nohel \cite{h1} studied the one-dimensional nonlinear viscoelastic equation $$\label{e2} u_{tt}=(\phi (u_{x}(x,t))) _{x} -\int_0^{t}a'(t-s)(\psi (u_{x}(x,s))) _{x}ds=0$$ in $\mathbb{R}^n$. They proved, under reasonable conditions on $\phi$, $\psi$ and smallness condition on the initial data, the existence of a unique global classical solution. They also established an asymptotic result but no rate of decay was given. Dassios and Zafiropoulus \cite{d2} showed that for the same kernel the decay is of order $t^{-3/2}$, if the material is occupying the whole space $\mathbb{R}^3$. Mu\~noz \cite{m1} extended the result of Dassios and Zafiropoulus to $\mathbb{R}^n$. Precisely, he showed that if the kernel is decaying exponentially then the solution decays exponentially for material occupying bounded domains whereas the decay is of the order $t^{-n/2}$ for material occupying the whole $n$-dimensional space. For nonexistence and formation of singularities, we mention the work by Dafermos \cite{d1} in 1985. Recently, Kafini and Messaoudi \cite{k1} considered the Cauchy problem $$\label{e3} \begin{gathered} u_{tt}-\Delta u+\int_0^{t}g(t-s)\Delta u(x,s)ds+u_{t}=|u|^{p-1}u,\quad x\in \mathbb{R}^n,\; t>0 \\ u(x,0)=u_0(x),\quad u_{t}(x,0)=u_1(x),\quad x\in \mathbb{R}^n\,. \end{gathered}$$ They showed that if the initial energy is negative and $\int_0^{\infty }g(s)ds<\frac{2p-2}{2p-1},\quad \int_{\mathbb{R} ^n}u_0u_1dx\geq 0,$ then the solution blows up in finite time. Also, in \cite{k2}, the same authors showed their blow-up result for the coupled system $$\label{e4} \begin{gathered} u_{tt}-\Delta u+\int_0^{t}g(t-s)\Delta u(x,s)ds=f_1(u,v), \quad \text{in }\mathbb{R}^n\times (0,\infty ) \\ v_{tt}-\Delta v+\int_0^{t}h(t-s)\Delta v(x,s)ds=f_2(u,v), \quad \text{in }\mathbb{R}^n\times (0,\infty ) \\ u(x,0)=u_0(x),\quad u_{t}(x,0)=u_1(x),\quad x\in \mathbb{R}^n \\ v(x,0)=v_0(x),\quad v_{t}(x,0)=v_1(x),\quad x\in \mathbb{R}^n. \end{gathered}$$ For more results related to stability and asymptotic behavior of viscoelastic equations, we refer the reader to the books by Renardy et al. \cite{r2}, Mu\~noz and Oquendo \cite{m2}, Fabrizio and Morro \cite{f2}, and Baretto et al. \cite{b1}. Most of the works \cite{b2,b3,c1,c2,c3} concerning the linear case of viscoelastic wave equations use assumptions of the form $$\label{e5} 1-\int_0^{\infty }g(s)ds=l>0,$$ and, for $a>0$, $$\label{e6} g'(t)\leq -ag^{p}(t),\quad 1\leq p<3/2,\quad t\geq 0.$$ Lately, a few papers \cite{f3,m3,m4,t1} appeared with alternative conditions. For instance, Furati and Tatar \cite{f3} proved that for sufficiently small $g$ and $g'$ can give also an exponential decay. Namely, they assumed $g(t)e^{\alpha t}$ and $g'(t)e^{\alpha t}$ have small $L^{1}$-norms. Conditions like \eqref{e5} or \eqref{e6} are not imposed. In particular, $g$ is not necessarily always negative. Recently, Messaoudi and Tatar \cite{m5} improved some earlier results concerning the exponential decay. They showed that the weak dissipation induced by the convolution term is sufficient to drive the system to rest with an exponential rate. Precisely, they established their result under the conditions $$\label{e7} g'(t)\leq 0\quad \text{and}\quad \int_0^{\infty }g(t)e^{\alpha t}dt<+\infty$$ for some large positive constant $\alpha$. Our aim, in this paper, is to establish a rate of exponential decay for the energy of solutions to \eqref{e1}, under the same conditions on $g$ and $g'$ as in \cite{m5} but in $\mathbb{R}^n$. Unlike in the bounded domain case, Poincar\'{e}'s inequality and some embedding inequalities are no longer valid. To overcome this difficulty, more recently, Kafini and Messaoudi \cite{k3}, exploited the nature of the wave propagation. In our problem, we want to achieve our goal without using such property. We will define functionals with special type that are equivalent to the energy functional. We remark that our proof is also valid for bounded domains $(\Omega \subset \mathbb{R}^n)$. Only it is needed to add the condition $u=0$ on $\partial \Omega \times (0,\infty )$ to the original system. This paper is organized as follows. In section 2, we state the conditions needed on $g$, and present, without proof, a global existence result. Section 3 starts with five technical lemmas before the statement and proof of the main result. \section{Preliminaries} In this section we present some material needed for the proof of our result. For this goal, we use the assumptions: \begin{itemize} \item[(G1)] $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is a differentiable function such that $1-\int_0^{\infty }g(s)ds=l>0,\quad t\geq 0.$ \item[(G2)] $g'(t)\leq 0$ and $\int_0^{\infty }g(t)e^{\alpha t}dt<+\infty$, for some large positive $\alpha$. \end{itemize} \begin{proposition} \label{prop2.1} Assume that {\rm (G1), (G2)} hold, $u_0\in H^{1}(\mathbb{R}^n)$, and $u_1\in L^{2}(\mathbb{R}^n)$, with compact support. Then \eqref{e1} has a unique local solution $u\in C([0,\infty );H^{1}(\mathbb{R}^n)),\quad u_{t}\in C([0,\infty );L^{2}(\mathbb{R}^n))\cap L^{2}([0,\infty )\times \mathbb{R}^n).$ \end{proposition} Now, we introduce the modified'' energy functional \begin{align*} E(t) &= \frac{1}{2}\Big[ \int_{\mathbb{R}^n}( | u_{t}| ^{2}+| \nabla u_{t}| ^{2}) dx+\Big(1-\int_0^{t}g(s)ds\Big) \int_{\mathbb{R}^n}| \nabla u| ^{2}dx \\ &\quad +\Big(1-\int_0^{t}g(s)ds\Big) \int_{\mathbb{R}^n} | u| ^{2}dx+(g\boxdot u)(t)\Big] \end{align*} where $(g\boxdot u) (t)=\int_0^{t}g(t-s)\int_{\mathbb{R}^n} [ | \nabla u(t)-\nabla u(s)| ^{2}+| u(t)-u(s)| ^{2}] \,dx\,ds.$ \begin{lemma} \label{lem2.2} If $u$ is a solution of \eqref{e1}, then the modified'' energy satisfies $$\label{e8} E'(t)=\frac{1}{2}(g'\boxdot u)-\frac{1}{2}g(t)\| \nabla u\| _2^{2}\leq \frac{1}{2}(g'\boxdot u)\leq 0.$$ \end{lemma} \begin{proof} By multiplying the equation in \eqref{e1} by $u_{t}$ and integrating over $\mathbb{R}^n$, using integration by parts and repeating the same computations as in \cite{m5}, we obtain the result. \end{proof} In this paper, we use the notation $\overline{f}=\int_0^{\infty }| f(s)| ds.$ \section{Decay of solutions} In this section, we establish four lemmas, and then we state and prove our main result. to this end, we introduce the following functionals: $$\label{e9} \Phi _1(t):=\int_{\mathbb{R}^n}\int_0^{t}G(t-s) \left[ | \nabla u(t)-\nabla u(s)| ^{2}+| u(t)-u(s)| ^{2}\right] \,ds\,dx$$ with $G(t):=e^{-\alpha t}\int_{t}^{\infty }e^{\alpha s}g(s)ds$, \begin{gather} \label{e10} \Phi _2(t):=\Big(\int_{\mathbb{R}^n}uu_{t}dx+\int_{\mathbb{R}^n} \nabla u.\nabla u_{t}dx\Big) , \\ \label{e11} \begin{aligned} \Phi _3(t) &:=-\Big[ \int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad +\int_{\mathbb{R}^n}u_{t}\int_0^{t}g(t-s) (u(t)-u(s)) \,ds\,dx\Big] \end{aligned} \\ \label{e12} F(t):=E(t)+\sum_{i=1}^3\gamma _i\Phi _i(t), \quad t\geq 0. \end{gather} \begin{lemma} \label{lem3.1} Assume {\rm (G1), (G2)} hold. Then, for small enough $\gamma _2$ and $\gamma _3$, there exist two positive constants $\xi _1,\xi _2$ such that $$\label{e13} \xi _1E(t)\leq F(t)\leq \xi _2[E(t)+\Phi _1(t)].$$ \end{lemma} \begin{proof} We estimate the terms in the above functionals using Young's inequality as follows \begin{gather} \label{e14} \int_{\mathbb{R}^n}uu_{t}dx\leq \| u\| _2^{2}+ \frac{1}{4}\| u_{t}\| _2^{2}, \\ \label{e15} \int_{\mathbb{R}^n}\nabla u.\nabla u_{t}dx\leq \| \nabla u\| _2^{2}+\frac{1}{4}\| \nabla u_{t}\| _2^{2}, \\ \label{e16} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}g(t-s) (\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\leq \| \nabla u_{t}\| _2^{2}+\frac{\overline{g}}{4} \int_0^{t}g(t-s)\int_{\mathbb{R}^n}| \nabla u(t)-\nabla u(s)| ^{2}\,dx\,ds, \end{aligned} \\ \label{e17} \begin{aligned} &\int_{\mathbb{R}^n}u_{t}\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\leq \| u_{t}\| _2^{2}+\frac{\overline{g}}{4} \int_0^{t}g(t-s)\int_{\mathbb{R}^n}| u(t)-u(s)| ^{2}\,dx\,ds. \end{aligned} \end{gather} By inserting \eqref{e14}-\eqref{e17} in \eqref{e12}, we obtain for some positive constant $\xi _2$, \label{e18} \begin{aligned} F(t) &\leq \gamma _1\Phi _1(t)+(\frac{1}{2}+\frac{\gamma _2}{4} +\gamma _3) \int_{\mathbb{R}^n}|u_{t}| ^{2}dx \\ &\quad +(\frac{l}{2}+\gamma _2) \int_{\mathbb{R} ^n}| \nabla u| ^{2}dx+(\frac{l}{2}+\gamma_2) \int_{\mathbb{R}^n}| u| ^{2}dx \\ &\quad +\big[ \frac{1}{2}+\frac{\gamma _2}{4}+\gamma _3\big] \int_{\mathbb{R}^n}| \nabla u_{t}| ^{2}dx +\big[ \frac{1}{2}+\frac{\overline{g}\gamma _3}{4}\big] (g\boxdot u)\\ &\leq \xi _2[E(t)+\Phi _1(t)]. \end{aligned} Moreover, the same estimates give \begin{align*} F(t) &\geq (\frac{1}{2}-\frac{\gamma _2}{4}-\gamma _3) \int_{\mathbb{R}^n}| u_{t}| ^{2}dx \\ &\quad +(\frac{l}{2}-\gamma _2) \int_{\mathbb{R} ^n}| \nabla u| ^{2}dx+(\frac{l}{2}-\gamma _2) \int_{\mathbb{R}^n}| u| ^{2}dx \\ &\quad +[ \frac{1}{2}-\frac{\gamma _2}{4}-\gamma _3] \int_{\mathbb{R}^n}| \nabla u_{t}| ^{2}dx+[ \frac{1}{2}-\frac{ \overline{g}\gamma _3}{4}] (g\boxdot u). \end{align*} By taking $\gamma _2$ and $\gamma _3$ small enough, we arrive, for some positive constant $\xi _1$, $$\label{e19} F(t)\geq \xi _1E(t).$$ Combining of \eqref{e18} and \eqref{e19}, the result follows. \end{proof} \begin{lemma} \label{lem3.2} If {\rm (G1), (G2)} hold. Then $\Phi_1(t)\emph{\ satisfies}$, for any $\delta _1,\delta _2>0$, $$\label{e20} \Phi _1'(t)\leq -\big(\alpha -\frac{2\overline{G}}{\delta _1}- \frac{2\overline{G}}{\delta _2}\big) \Phi _1(t)-(g\boxdot u)+\delta _1\| \nabla u_{t}\| _2^{2}+\delta _2\| u_{t}\| _2^{2}.$$ \end{lemma} \begin{proof} We obtain the result by differentiating \eqref{e9} and using Young's inequality as follows \begin{align*} \Phi _1'(t) &= -\alpha \Phi _1(t)-(g\boxdot u) +2\int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}G(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad +2\int_{\mathbb{R}^n}u_{t}\int_0^{t}G(t-s)( u(t)-u(s)) \,ds\,dx \\ &\leq -\alpha \Phi _1(t)-(g\boxdot u)+\delta _1\| \nabla u_{t}\| _2^{2}+\frac{2}{\delta _1}\overline{G}\Phi _1(t)+\delta _2\| u_{t}\| _2^{2}+\frac{2}{\delta _2}\overline{G}\Phi _1(t), \end{align*} where $\overline{G}=\int_0^{\infty }G(s)ds =\int_0^{\infty }\Big( e^{-\alpha t}\int_{t}^{\infty }e^{\alpha s}g(s)ds\Big) dt \leq \frac{1}{\alpha }\int_0^{\infty }e^{\alpha s}g(s)ds<\infty .$ \end{proof} \begin{lemma} \label{lem3.3} Assume {\rm (G1), (G2)} hold. Then along the solution of \eqref{e1}, for any $\delta_3,\delta _4>0$, the function $\Phi_2(t)$ satisfies $$\label{e21} \Phi _2'(t)\leq \| u_{t}\| _2^{2}+\| \nabla u_{t}\| _2^{2}-(l-\delta _3) \| \nabla u\| _2^{2}-(l-\delta _4) \| u\| _2^{2}+\frac{ \overline{g}}{4\delta _{5}}(g\boxdot u).$$ \end{lemma} \begin{proof} By differentiating \eqref{e10}, we have $$\label{e22} \Phi _2'(t) = \int_{\mathbb{R}^n}| u_{t}| ^{2}dx+\int_{\mathbb{R}^n}uu_{tt}dx +\int_{\mathbb{R}^n}| \nabla u_{t}| ^{2}dx+\int_{ \mathbb{R}^n}\nabla u.\nabla u_{tt}dx.$$ Along \eqref{e1}, we find \begin{align*} &\int_{\mathbb{R}^n}uu_{tt}dx+\int_{\mathbb{R} ^n}\nabla u.\nabla u_{tt}dx \\ &= -\int_{\mathbb{R}^n}| \nabla u| ^{2}dx+\int_{ \mathbb{R}^n}\nabla u.\int_0^{t}g(t-s)\nabla u(s)\,ds\,dx \\ &\quad-\int_{\mathbb{R}^n}| u| ^{2}dx-\int_{\mathrm{I \hskip-2ptR}^n}u\int_0^{t}g(t-s)u(s)\,ds\,dx; \end{align*} thus \eqref{e22} becomes \label{e23} \begin{aligned} \Phi _2'(t) &= \int_{\mathbb{R}^n}| u_{t}| ^{2}dx+\int_{\mathbb{R}^n}| \nabla u_{t}| ^{2}dx-\int_{\mathbb{R}^n}| \nabla u|^{2}dx-\int_{\mathbb{R}^n}| u| ^{2}dx \\ &\quad +\int_{\mathbb{R}^n}\nabla u.\int_0^{t}g(t-s)\nabla u(s)\,ds\,dx-\int_{\mathbb{R}^n}u\int_0^{t}g(t-s)u(s)\,ds\,dx. \end{aligned} Using the estimates \label{e24} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u(t).\int_0^{t}g(t-s)\nabla u(s)\,ds\,dx \\ &\leq \delta _3\| \nabla u\| _2^{2}+\frac{\overline{g}}{ 4\delta _3}\int_{\mathbb{R}^n}\int_0^{t}g(t-s)| \nabla u(t)-\nabla u(s)| ^{2}\,ds\,dx+\overline{g}\| \nabla u\| _2^{2} \end{aligned} and \label{e25} \begin{aligned} &-\int_{\mathbb{R}^n}u(t)\int_0^{t}g(t-s)u(s)\,ds\,dx \\ &\leq \delta _4\| u\| _2^{2}+\frac{\overline{g}}{4\delta _4 }\int_{\mathbb{R}^n}\int_0^{t}g(t-s)| u(t)-u(s)| ^{2}\,ds\,dx+\overline{g}\| u\| _2^{2}. \end{aligned} Adding \eqref{e24} and \eqref{e25}, for $\delta _{5}=\min \{ \delta_3,\delta _4\}$, yields \label{e26} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u(t).\int_0^{t}g(t-s)\nabla u(s)\,ds\,dx-\int_{\mathbb{R} ^n}u(t)\int_0^{t}g(t-s)u(s)\,ds\,dx \\ &\leq \delta _3\| \nabla u\| _2^{2}+\overline{g}\| \nabla u\| _2^{2}+\delta _4\| u\| _2^{2}+\overline{g} \| u\| _2^{2}+\frac{\overline{g}}{4\delta _{5}}(g\boxdot u). \end{aligned} Inserting \eqref{e26} in \eqref{e23} gives the desired result \eqref{e21}. \end{proof} \begin{lemma} \label{lem3.4} Suppose {\rm (G1), (G2)} hold. Then along the solution of \eqref{e1}, for any $\delta_6,\delta _7,\delta _9,\delta _{10},\delta _{12}, \delta _{13}>0$, the function $\Phi _3(t)$ satisfies \label{e27} \begin{aligned} \Phi _3'(t) &\leq -\Big(\int_0^{t}g(s)ds-\delta _9\Big) \| \nabla u_{t}\| _2^{2}-\Big( \int_0^{t}g(s)ds-\delta _{10}-\delta _{13}\Big) \| u_{t}\| _2^{2} \\ &\quad +(\delta _6+\delta _{12}) \| \nabla u\| _2^{2}+\delta _7\| u\| _2^{2}+\Big(\frac{1}{4\delta _{8}}+ \frac{1}{4\delta _{14}}+1\Big) \overline{g}(g\boxdot u)\\ &\quad -\frac{g(0)}{4\delta _{11}}(g'\boxdot u). \end{aligned} \end{lemma} \begin{proof} Differentiation of \eqref{e11} yields \begin{aligned} \label{e28} \Phi _3'(t) &= -\int_{\mathbb{R}^n}\nabla u_{tt}.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad -\int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}g'(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad -\int_{\mathbb{R}^n}u_{tt}\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\quad -\int_{\mathbb{R}^n}u_{t}\int_0^{t}g'(t-s)(u(t)-u(s)) \,ds\,dx \\ &\quad -\Big(\int_0^{t}g(s)ds\Big) \| u_{t}\| _2^{2}-\Big(\int_0^{t}g(s)ds\Big) \| \nabla u_{t}\| _2^{2}. \end{aligned} Along \eqref{e1}, we find \label{e29} \begin{aligned} &\int_{\mathbb{R}^n}u_{tt}\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\quad +\int_{\mathbb{R}^n}\nabla u_{tt}.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &= -\int_{\mathbb{R}^n}\nabla u.\int_0^{t}g(t-s)( \nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad -\int_{\mathbb{R}^n}u\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\quad +\int_{\mathbb{R}^n}\Big( \int_0^{t}g(t-s)\nabla u(s)ds.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) ds\Big) dx \\ &\quad -\int_{\mathbb{R}^n}\Big( \int_0^{t}g(t-s)u(s)ds.\int_0^{t}g(t-s)( u(t)-u(s)) ds\Big) dx. \end{aligned} The first two terms in the right side of \eqref{e29} can be estimated as follows \begin{gather} \label{e30} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u.\int_0^{t}g(t-s)( \nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\leq \delta _6\| \nabla u\| _2^{2}+\frac{\overline{g}}{ 4\delta _6}\int_{\mathbb{R}^n}\int_0^{t}g(t-s)| \nabla u(t)-\nabla u(s)| ^{2}\,ds\,dx, \end{aligned} \\ \label{e31} \begin{aligned} &\int_{\mathbb{R}^n}u\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\leq \delta _7\| u\| _2^{2}+\frac{\overline{g}}{4\delta _7 }\int_{\mathbb{R}^n}\int_0^{t}g(t-s)| u(t)-u(s)| ^{2}\,ds\,dx, \end{aligned} \end{gather} from these two estimates, for $\delta _{8}=\min \{ \delta _6,\delta _7\}$, we have \label{e32} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx +\int_{\mathbb{R}^n}u\int_0^{t}g(t-s)(u(t)-u(s)) \,ds\,dx \\ &\leq \delta _6\| \nabla u\| _2^{2}+\delta _7\| u\| _2^{2}+\frac{\overline{g}}{4\delta _{8}}(g\boxdot u). \end{aligned} Using $\int_{\mathbb{R}^n}\big| \int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) ds\big| ^{2}dx \leq \overline{g}\int_{\mathbb{R}^n}\int_0^{t}g(t-s) | \nabla u(t)-\nabla u(s)| ^{2}\,ds\,dx,$ we estimate the last two terms in \eqref{e29}, for $\delta _{14}=\min\{ \delta _{12},\delta _{13}\}$, as follows \begin{align*} %33 &\int_{\mathbb{R}^n}\Big( \int_0^{t}g(t-s)\nabla u(s)ds.\int_0^{t}g(t-s)(\nabla u(t)-\nabla u(s)) ds\Big) dx \\ &\quad -\int_{\mathbb{R}^n}\Big( \int_0^{t}g(t-s)u(s)ds.\int_0^{t}g(t-s)( u(t)-u(s)) ds\Big) dx \\ &\leq \int_{\mathbb{R}^n}\nabla u(t).\int_0^{t}g(t-s) (\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\quad +\int_{\mathbb{R}^n}u(t)\int_0^{t}g(t-s)( u(t)-u(s)) \,ds\,dx \\ &\quad +\int_{\mathbb{R}^n}\Big| \int_0^{t}g(t-s)( \nabla u(t)-\nabla u(s)) ds\Big| ^{2}dx \\ &\quad +\int_{\mathbb{R}^n}\Big| \int_0^{t}g(t-s)( u(t)-u(s)) ds\Big| ^{2}dx \\ &\leq \delta _{12}\| \nabla u\| _2^{2}+\delta _{13}\| u_{t}\| _2^{2}+\frac{\overline{g}}{4\delta _{14}}(g\boxdot u)+ \overline{g}(g\boxdot u). \end{align*} Similarly, the second and the fourth term of \eqref{e28} can be handled as follows \begin{gather} \label{e34} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}g'(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx \\ &\leq \delta _9\| \nabla u_{t}\| _2^{2}-\frac{g(0)}{4\delta _9}\int_{\mathbb{R}^n}\int_0^{t}g'(t-s)| \nabla u(t)-\nabla u(s)| ^{2}\,ds\,dx, \end{aligned} \\ \label{e35} \begin{aligned} &\int_{\mathbb{R}^n}u_{t}\int_0^{t}g'(t-s)(u(t)-u(s)) \,ds\,dx \\ &\leq \delta _{10}\| u_{t}\| _2^{2}-\frac{g(0)}{4\delta _{10}} \int_{\mathbb{R}^n}{}\int_0^{t}g'(t-s)| u(t)-u(s)| ^{2}\,ds\,dx, \end{aligned} \end{gather} from the \eqref{e34} and \eqref{e35}, for $\delta _{11}=\min \{ \delta _9,\delta _{10}\}$, we have \label{e36} \begin{aligned} &\int_{\mathbb{R}^n}\nabla u_{t}.\int_0^{t}g'(t-s)(\nabla u(t)-\nabla u(s)) \,ds\,dx +\int_{\mathbb{R}^n}u_{t}\int_0^{t}g'(t-s)(u(t)-u(s)) \,ds\,dx \\ &\leq \delta _9\| \nabla u_{t}\| _2^{2}+\delta _{10}\| u_{t}\| _2^{2}-\frac{g(0)}{4\delta _{11}}(g'\boxdot u). \end{aligned} Combining \eqref{e28}-\eqref{e36}, the result follows. \end{proof} \begin{theorem} \label{thm3.6} Assume {\rm (G1), (G2)} hold for large $\alpha$. Then, for any $t_0>0$, there exist two positive constants $K$ and $k$ such that $E(t)\leq Ke^{-kt}.$ \end{theorem} \begin{proof} Differentiating \eqref{e12} and using \eqref{e8} yields $$\label{e37} F'(t)=E'(t)+\sum_{i=1}^3\gamma _i\Phi _i'(t) \leq \frac{1}{2}(g'\boxdot u)+\sum_{i=1}^3\gamma _i\Phi_i'(t).$$ Since $g$ is continuous and $g(0)>0$ then, for any $t\geq t_0>0$, we have $\int_0^{t}g(s)ds\geq \int_0^{t_0}g(s)ds=g_0>0.$ By inserting \eqref{e20}, \eqref{e21} and \eqref{e27} in \eqref{e37}, we obtain \label{e38} \begin{aligned} F'(t) &\leq -\big(\alpha -\frac{2\overline{G}}{\delta _1}-\frac{ 2\overline{G}}{\delta _2}\big) \gamma _1\Phi _1(t) +[ \frac{1}{2}-\frac{\gamma _3g(0)}{4\delta _{11}}] (g'\boxdot u) \\ &\quad -\big[ \gamma _1-\overline{g}(\frac{\gamma _2}{4\delta _{5}} +\gamma _3(\frac{1}{4\delta _{8}}+\frac{1}{4\delta _{14}}+1) ) \big] (g\boxdot u) \\ &\quad -[ \gamma _2(l-\delta _3) -\gamma _3(\delta _6+\delta _{12}) ] \| \nabla u\| _2^{2} \\ &\quad -[ \gamma _3(g_0-\delta _9) -\gamma _2-\gamma _1\delta _1] \| \nabla u_{t}\| _2^{2} \\ &\quad -[ \gamma _3(g_0-\delta _{10}-\delta _{13}) -\gamma _2-\gamma _1\delta _2] \| u_{t}\| _2^{2} -[ \gamma _2(l-\delta _4) -\gamma _3\delta _7] \| u\| _2^{2}. \end{aligned} At this point, we fix $\delta _3=\delta _4 0 \\ \gamma _2(l-\delta _3) -\gamma _3(\delta _6+\delta _{12}) > 0 \\ \gamma _3(g_0-\delta _9) -\gamma _2 = k_1>0 \\ \gamma _3(g_0-\delta _{10}-\delta _{13}) -\gamma _2 = k_2>0. \end{gather*} So we choose$\delta _6+\delta _7+\delta _{12}<\lambda $and$\gamma _3$small enough so that \eqref{e13} and \eqref{e39} remain valid and $\frac{1}{2}-\gamma _3\Big(\frac{g(0)}{4\delta _{11}}\Big) >0.$ Then we pick$\gamma _1$large enough so that $\gamma _1-\overline{g}\Big(\frac{\gamma _2}{4\delta _{5}}+\gamma _3\big(\frac{1}{4\delta _{8}}+\frac{1}{4\delta _{14}}+1\big) \Big) >0,$ and$\delta _1,\delta _2$small enough so that $k_1-\gamma _1\delta _1 >0,\quad k_2-\gamma _1\delta _2 >0.$ Therefore if$\alpha $is large enough so that$\alpha -\frac{2 \overline{G}}{\delta _1} -\frac{2\overline{G}}{\delta _2}>0$, then ,for all$t\geq t_0$, \eqref{e38} becomes $F'(t)\leq -c[ E(t)+\Phi _1(t)] \leq \frac{-c}{\xi _2}F(t).$ Integrating over$(t_0,t)$yields $F(t)\leq F(t_0)e^{ct_0/\xi _2}e^{-ct/\xi _2}.$ The equivalence in \eqref{e13} completes the proof for$K=\frac{F(t_0)}{ \xi _1}e^{ct_0/\xi _2}$and$k=c/\xi _2$. \end{proof} \subsection*{Acknowledgments} Author would like to thank the King Fahd University of Petroleum and Minerals for its support. \begin{thebibliography}{00} \bibitem{b1} R. 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