\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 73, 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 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/73\hfil Decay results for viscoelastic diffusion equations] {Decay results for viscoelastic diffusion equations in absence of instantaneous elasticity} \author[M. Kafini \hfil EJDE-2012/73\hfilneg] {Mohammad Kafini} % in alphabetical order \address{Mohammad Kafini \newline Department of Mathematics and Statistics\\ KFUPM, Dhahran 31261\\ Saudi Arabia} \email{mkafini@kfupm.edu.sa} \thanks{Submitted December 18, 2011. Published May 10, 2012.} \subjclass[2000]{35B05, 35L05, 35L15, 35L70} \keywords{Diffusion equation; instantaneous elasticity; exponential decay; \hfill\break\indent relaxation function; viscoelastic} \begin{abstract} We study the diffusion equation in the absence of instantaneous elasticity \begin{equation*} u_t-\int_0^{t}g(t-\tau )\Delta u(\tau )\,d\tau =0,\quad (x,t)\in \Omega \times (0,+\infty ), \end{equation*} where $\Omega \subset \mathbb{R}^n$, subjected to nonlinear boundary conditions. We prove that if the relaxation function $g$ decays exponentially, then the solutions is exponential stable. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} A diffusion equation in the absence of instantaneous elasticity has the form $$\label{e1} u_t-\int_0^{t}g(t-\tau )\Delta u(\tau )\,d\tau =0,\quad (x,t)\in \Omega \times (0,+\infty ).$$ When the fluid is enclosed in a region $\Omega \subset \mathbb{R}^n$ the above equation is supplemented by conditions at $\partial \Omega$, the boundary of $\Omega$. For instance, one can consider the nonlinear boundary condition: $$\label{e2} \partial _{\nu }u+f(u)=0,\quad \partial \Omega \times (0,+\infty ).$$ Let us assume that $$\label{e3} u(x,0)=u_0(x),\quad x\in \Omega .$$ Denoting \begin{equation*} (g\ast \varphi )(t)=\int_0^{t}g(t-s)\varphi (s)\,ds \end{equation*} and differentiating equation \eqref{e1}, with respect to $t$, we arrive at the Volterra equation $$\label{e4} \frac{1}{g(0)}u_{tt}=\Delta u+\frac{1}{g(0)}(g'\ast \Delta u).$$ Considering the Volterra inverse operator we obtain $$\label{e5} u_{tt}-g(0)\Delta u+k\ast u_{tt}=0,$$ where the resolvent kernel satisfies \begin{equation*} k+\frac{1}{g(0)}g'\ast k=-\frac{1}{g(0)}g'. \end{equation*} Thus \eqref{e5} becomes $$\label{e6} u_{tt}-g(0)\Delta u+k(0)u_t-k(t)u_t(0)+k'\ast u_t=0.$$ Reciprocally, supposing in a natural way that $u_t(0)=0$, the identity \eqref{e6} implies \eqref{e1}. Since we are interested in relaxation functions of exponential type and \eqref{e6} involves the resolvent kernel $k$, we want to know if $k$ has the same properties. The following lemma answers this question. Let $h$ be a relaxation function and $k$ its resolvent kernel; that is, $$\label{e7} k(t)-k\ast h(t)=h(t).$$ \begin{lemma}[\cite{m1,r1,r2}] \label{lem1.1} If $h$ is a positive continuous function, then $k$ is also a positive continuous function. Moreover, if there exist positive constants $c_0<\gamma$, such that \begin{equation*} h(t)\leq c_0e^{-\gamma t}, \end{equation*} then the function $k$ satisfies \begin{equation*} k(t)\leq \frac{c_0(\gamma -\epsilon )}{\gamma -\epsilon -c_0} e^{-\epsilon t}, \end{equation*} for all $0<\epsilon <\gamma -c_0$. \end{lemma} \begin{proof} Note that $k(0)=h(0)>0$. Now, we take \begin{equation*} t_0=\text{inf}\{t\in \mathbb{R}^{+}:k(t)=0\}, \end{equation*} so $k(t)>0$ for all $t\in [ 0,t_0[$. If $t_0\in \mathbb{R}^{+}$, from equation \eqref{e7} we obtain that $-k\ast h(t_0)=h(t_0)$ but this is contradictory. Therefore, $k(t)>0$ for all $t\in \mathbb{R}^{+}$. Now, let us fix $\epsilon$, such that $0<\epsilon <\gamma -c_0$ and denote \begin{equation*} k_{\epsilon }(t):=e^{\epsilon t}k(t),\quad h_{\epsilon }(t):=e^{\epsilon t}h(t). \end{equation*} Multiplying equation \eqref{e7} by $e^{\epsilon t}$ we obtain \begin{equation*} k_{\epsilon }(t)=h_{\epsilon }(t)+k_{\epsilon }\ast h_{\epsilon }(t), \end{equation*} hence \begin{align*} \underset{s\in [ 0,t]}{\sup }k_{\epsilon }(s) &\leq \sup h_{\epsilon }(s) +\Big\{ \int_0^{\infty }c_0e^{(\epsilon -\gamma )s}ds\Big\} \sup_{s\in [ 0,t]} k_{\epsilon }(s) \\ &\leq c_0+\frac{c_0}{( \gamma -\epsilon ) }\sup_{s\in[ 0,t]} k_{\epsilon }(s). \end{align*} Therefore, \begin{equation*} k_{\epsilon }(t)\leq \frac{c_0(\gamma -\epsilon )}{\gamma -\epsilon -c_0}, \end{equation*} which is the desired result. \end{proof} Thanks to Lemma \ref{lem1.1}, we can use equation \eqref{e6} instead of \eqref{e1}. Usually when $g$ is such that $\operatorname{Re}(\hat{g})>0$ ($\hat{g}$ is the Fourier transform of $g$), we can, by the Laplace transformation, reduce equation \eqref{e1} to an elliptic problem. By the variational method, we can resolve such an equation (see Raynal \cite{r2}). In what follows we shall adopt a different procedure in order to establish the well-posedness of problem \eqref{e1}. So, from the above comments, we can consider equation \eqref{e6}, instead of equation \eqref{e1}, supplemented by the initial data \eqref{e3}, the compatibility condition $u_t(0)=0$, and the boundary conditions $$\label{e8} \begin{gathered} \partial _{\nu }u+\beta u_t+|u|^{\rho }u=0,\quad{on }\Gamma _1\times (0,\infty ) \\ u=0,\quad \Gamma _0\times (0,+\infty ), \end{gathered}$$ assuming that $k\in W^{2,1}(0,+\infty )$, $\beta >0$ and $0<\rho <2/(n-2)$ if $n\geq 3$ or $\rho >0$ if $n=1,2$. We shall assume that $\Omega$ is a bounded domain of $\mathbb{R}^n$, $n\geq 1$, with a smooth boundary $\Gamma =\Gamma _0\cup \Gamma_1$. Here, $\Gamma _0\neq \emptyset$; $\Gamma _0$ and $\Gamma _1$ are closed and disjoint and $\nu$ represents the unit outward normal to $\Gamma$. The variational formulation associated with problem \eqref{e6} is \begin{align*} &(u_{tt}(t),v)_{\Omega }-g(0)(\Delta u(t),v)_{\Omega}\\ &+k(0)(u_t(t),v)_{\Omega }+\int_0^{t}k'(t-\tau )(u'(\tau ),v)_{\Omega }\,d\tau +g(0)(f(u),v)_{\Gamma _1}=0, \end{align*} for all $v\in H_{\Gamma _0}^1(\Omega ):=\{u\in H^1(\Omega );u=0 \text{ on }\Gamma _0\}$. We can easily obtain the existence and uniqueness of global \emph{regular solutions} making use, for instance, of the Faedo-Galerkin method. Evidently the additional term given by $\beta u_t$ plays an essential role by allowing us to control the nonlinear term on the boundary. This is strongly necessary because of Lopatinski condition is lost. Thus, it is clear and it has been recognized a long time ago, that well-posedness theory with semilinear boundary nonlinearity and finite energy solutions must rely on and take advantage of the boundary dissipation. See, for instance, Cavalcanti et al \cite{c2}, Lasiecka and Tataru \cite{l1} and references therein. However, from the physical point of view to have two dissipations, namely, $k(0)u_t$ (internal) and $\beta u_t$ (on the boundary) is too much to establish the exponential decay. So, by considering the techniques employed in Lasiecka and Tataru \cite{l1} or in Cavalcanti, Cavalcanti, and Soriano \cite{c2}, it is possible to obtain the existence of weak solutions to \eqref{e6} subject to the boundary conditions $$\label{e9} \partial _{\nu }u+|u|^{\rho }u=0,\quad \text{on }\Gamma _1\times (0,\infty ).$$ Unfortunately, because of the nonlinear boundary condition \eqref{e9}, the uniqueness is lost. Indeed, let $A$ be the operator whose domain is defined by \begin{equation*} D(A)=\left\{ (u,v)\in H_{\Gamma _0}^1(\Omega )\times H_{\Gamma _0}^1(\Omega );u-\mathcal{N}[g_1(\gamma _0v)+f_1(\gamma _0u)]\in D(-\Delta )\right\} \end{equation*} and the operator by \begin{equation*} A\begin{pmatrix} u \\ v\end{pmatrix} =\begin{pmatrix} -v \\ \Delta (u-\mathcal{N}[g_1(\gamma _0v)+f_1(\gamma _0u)]) \end{pmatrix}. \end{equation*} We are assuming that $$\label{e10} g_1(s)=\beta s,\quad f_1 \text{ is a Lipschitz continuous function on } \mathbb{R},$$ and \begin{align*} D(-\Delta ) &= \{ v\in H_{\Gamma _0}^1(\Omega );\Delta v\in L^2(\Omega )\} \\ &= \big\{ v\in H_{\Gamma _0}^1(\Omega )\cap H^2(\Omega );\frac{ \partial v}{\partial \nu }=0\text{ on }\Gamma _1\big\} , \end{align*} and $\mathcal{N}:H^{s}(\Gamma _1)\to H_{\Gamma _0}^{s+3/2}(\Omega )$, $s\in \mathbb{R}$, is the Neumann map defined by \begin{equation*} \mathcal{N}p=q\Leftrightarrow \left\{ \begin{gathered} -\Delta q=0\quad \text{in } \Omega \\ q=0\quad \text{on }\Gamma _0 \\ \frac{\partial q}{\partial \nu }=p\quad \text{on }\Gamma _1 \end{gathered} \right. \end{equation*} We observe that $$\label{e11} (u,v)\in D(A)\Leftrightarrow \left\{ \begin{gathered} (u,v)\in [ H_{\Gamma _0}^1(\Omega )] ^2, \\ u-\mathcal{N}[g_1(\gamma _0v)+f_1(\gamma _0u)]\in H_{\Gamma_0}^1(\Omega ), \\ \Delta (u-\mathcal{N}[g_1(\gamma _0v)+f_1(\gamma _0u)])\in L^2(\Omega ). \end{gathered} \right.$$ By the nonlinear semigroup theory \cite[Theorem 2.1]{l1}, the operator $A$ is $\omega$-accretive on the space $E:=H_{\Gamma_0}^1(\Omega )\times L^2(\Omega )$, for some $\omega$ suitably large. Moreover, $A+\omega I$ is maximal monotone and $$\label{e12} D(A)\text{ is dense in }H_{\Gamma _0}^1(\Omega )\times L^2(\Omega ).$$ Let us assume that $\{ u^{0},u^1\} \in H_{\Gamma _0}^1(\Omega )\times L^2(\Omega )$ and consider, in view of \eqref{e12}, $\{ u_{\mu }^{0},u_{\mu }^1\} \subset D(A)$ such that $$\label{e13} u_{\mu }^{0}\to u^{0}\text{ in }H_{\Gamma _0}^1(\Omega )\quad \text{and}\quad u_{\mu }^1\to u^1\text{ in }L^2(\Omega )\quad \text{as }\mu \to +\infty .$$ Thus, $\{ u_{\mu }^{0},u_{\mu }^1\}$ satisfies, for all $\mu \in \mathbb{N}$ the compatibility conditions \begin{equation*} \frac{\partial u_{\mu }^{0}}{\partial \nu }+\frac{1}{\mu }u_{\mu}^1 +f_{1,\mu }(u_{\mu }^{0})=0, \end{equation*} where $\beta$ is chosen equal to $1/\mu$ and $f_{1,\mu }(s)$ is the sequance of Lipschitz continuous (truncated) functions defined by $$\label{e14} f_{1,\mu }(s):=\begin{cases} |s| ^{\rho }s,\quad | s| <\mu \\ |\mu| ^{\rho }\mu ,\quad s\geq \mu \\ |-\mu| ^{\rho }(-\mu ),\quad s\leq -\mu . \end{cases}$$ Initially, we consider regular solutions to the auxiliary problem $$\label{e15} \begin{gathered} u_{tt}^n-\alpha \Delta u^n+k(0)u_t^n+\int_0^{t}k'(t-s)u_t^n(x,s)ds=0 \quad \text{in }\Omega \times (0,+\infty ) \\ u^n(x,t)=0,\quad x\in \Gamma _0 \\ \frac{\partial u^n}{\partial \nu }+\frac{1}{n}u_t^n+f_{1,n}(u^n)=0 \quad \text{on }\Gamma _1\times (0,+\infty ) \\ u^n(x,0)=u_0^n(x),\quad u_t^n(x,0)=u_1^n(x),\quad x\in \Omega . \end{gathered}$$ We obtain a sequence of regular solutions to problem \eqref{e15} which will converge, as $n$ approaches infinity, to a desired weak solution ($\{u^{0},u^1\}\in H_{\Gamma _0}^1(\Omega )\times L^2(\Omega )$). The procedure described above can be followed \emph{verbatim} as considered in \cite{c1} and therefore will be omitted. It is important to be mentioned that while problem \eqref{e15} possesses a unique solution, the uniqueness of the limit problem, namely $$\label{e16} \begin{gathered} u_{tt}-\alpha \Delta u+k(0)u_t+\int_0^{t}k'(t-s)u_t(x,s)ds=0\quad \text{in }\Omega \times (0,+\infty ) \\ u(x,t)=0,\quad x\in \Gamma _0 \\ \frac{\partial u}{\partial \nu }+| u| ^{\rho }u=0\quad \text{on }\Gamma _1\times (0,+\infty ) \\ u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \Omega . \end{gathered}$$ is lost because of the nonlinear boundary term $| u|^{\rho }u$. Of course for Dirichlet or Neumann homogeneous boundary conditions the uniqueness is recovered. The main task of this work, is to prove the exponential stability of problem \eqref{e15}. Namely, we would like to have $$\label{e17} E_{n}(t)\leq CE_{n}(0)e^{-\omega \,t},$$ where $E_{n}(t)$ is the energy associated with \eqref{e15} and the constants $C$ and $\omega$ do not depend on $n$. So, using denseness arguments as considered in \cite{c1,l1}, we can pass to the limit in \eqref{e17} to obtain the desired exponential decay rate for those weak solutions that are limit of regular solutions of problem \eqref{e15}. Evidently the procedure is valid for any weak solution if $u=0$ or $\partial_{\nu }u=0$ on $\Gamma$. \section{Preliminaries} In this section we present some material needed in the proof of our result. We will us the following assumptions: \begin{itemize} \item[(G1)] $k',k''':[0,\infty )\to \mathbb{R}^{+}$ with $k(0)>0$. \item[(G2)] $k'':[0,\infty )\to \mathbb{R}^{-}$ with $k''(0)<0$. \item[(G3)] There exist two positive constants $\zeta _1$ and $\zeta_2$ such that \begin{equation*} k''\leq -\zeta _1k'\quad \text{and}\quad k'''\geq -\zeta _2k''. \end{equation*} \end{itemize} An example of function $k$ satisfying (G1)--(G3) is $k(t)=a-e^{-bt}$, where $a>1$, $b>0$. \begin{lemma}[Poincar\'e] \label{lem2.1} There exists a positive constant $\beta (\Omega )$ such that \begin{equation*} | u| _2^2\leq \beta | \nabla u|_2^2,\quad \forall u\in H_{\Gamma _0}^1(\Omega ). \end{equation*} \end{lemma} Our main task is concerned with the asymptotic behavior of solutions to the problem $$\label{e18} \begin{gathered} u_{tt}-\alpha \Delta u+k(0)u_t+\int_0^{t}k'(t-s)u_t(x,s)ds=0,\quad x\in \Omega , \;t>0, \\ u(x,t)=0,\quad x\in \Gamma _0, \\ \frac{\partial u}{\partial \nu }+\frac{1}{n}u_t+b| u| ^{\rho }u=0\quad \text{ on }\Gamma _1, \\ u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \Omega . \end{gathered}$$ where $\Omega$ is a bounded domain in $\mathbb{R}^n$ with a smooth boundary $\partial \Omega =\Gamma _0\cup \Gamma _1$, $\alpha >0$ and $k(t)\in C^{3}(\mathbb{R}^{+})$ satisfying {\rm (G1)--(G3)}. Once exponential stability for \eqref{e18} is established, then the same technique is used for viscoelastic diffusion problem $$\label{e19} \begin{gathered} u_t-\int_0^{t}g(t-\tau )\Delta u(\tau )\,d\tau =0,\quad x\in \Omega ,\; t>0, \\ u(x,t)=0,\quad x\in \Gamma _0, \\ \frac{\partial u}{\partial \nu }+b| u| ^{\rho }u=0 \quad \text{on }\Gamma _1, \\ u(x,0)=u_0(x),\quad u_t(x,0)=0,\quad x\in \Omega , \end{gathered}$$ obtained by denseness arguments, having in mind the comments established in section 1. Of course this is not true for all weak solutions of \eqref{e19} unless we have $u=0$ or $\partial _{\nu }u=0$ on $\Gamma$. After integrating by parts the last term in \eqref{e18}, we obtain $$\label{e20} \begin{gathered} u_{tt}-\alpha \Delta u+k(0)u_t+k'(0)u(t)-k'(t)u_0 \\ +\int_0^{t}k''(t-s)u(x,s)ds=0,\quad x\in \Omega , \; t>0, \\ u(x,t)=0,\quad x\in \Gamma _0, \\ \frac{\partial u}{\partial \nu }+\frac{1}{n}u_t+b|u| ^{\rho }u=0 \quad \text{on }\Gamma _1, \\ u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x),\quad x\in \Omega . \end{gathered}$$ The modified energy functional associated with \eqref{e20} is \begin{align*} E(t) &= \frac{1}{2}\int_{\Omega }u_t^2dx+\frac{\alpha }{2}\int_{\Omega }| \nabla u| ^2dx+\frac{1}{2}k'(t)\int_{\Omega }\left( u(x,t)-u_0(x)\right) ^2dx \\ &\quad -\frac{1}{2}\int_{\Omega }\int_0^{t}k''(t-s)\left( u(x,s)-u(x,t)\right) ^2\,ds\,dx +\frac{\alpha b}{\rho +2}\int_{\Gamma_1}| u| ^{\rho +2}dx. \end{align*} \section{Decay of solutions} In this section we state and prove our main result. For this purpose, we set $$\label{e21} F(t)=E(t)+\varepsilon \varphi (t),\quad t\geq 0,$$ where $\varepsilon$ is a positive constant and $$\label{e22} \varphi (t)=\int_{\Omega }u_tu\,dx+\frac{k(0)}{2}\int_{\Omega }u^2dx+\frac{ \alpha }{2n}\int_{\Gamma _1}u^2dx,$$ \begin{lemma} \label{lem3.1} The modified energy satisfies, along the solution of \eqref{e20}, \begin{align*} E'(t) &= -k(0)\int_{\Omega }u_t^2dx+\frac{1}{2}k''(t)\int_{\Omega }\left( u(x,t)-u_0(x)\right) ^2\,ds\,dx \\ &\quad -\frac{1}{2}\int_{\Omega }\int_0^{t}k^{'''}(t-s) \left( u(x,s)-u(x,t)\right) ^2\,ds\,dx-\frac{\alpha }{n}\int_{\Gamma _1}| u_t| ^2dx\leq 0. \end{align*} \end{lemma} \begin{proof} Multiplying \eqref{e20} by $u_t$ and integrating over $\Omega$, using integration by parts, hypotheses (G1) and (G2) and some manipulations as in \cite{c1} , we obtain the result for any regular solution. This result remains valid for any limit weak'' solution (not for all weak solutions) by a simple denseness argument. \end{proof} \begin{lemma} \label{lem3.2} For $\varepsilon >0$ small enough, we have \begin{equation*} | F(t)-E(t)| \leq \varepsilon \lambda E(t),\quad t\geq 0, \end{equation*} where $\lambda$ is a constant independent of $\varepsilon$ and $n$. \end{lemma} \begin{proof} Using the Poincar\'{e} and the Cauchy-Schwarz inequalities, we obtain \begin{align*} | \varepsilon \varphi (t)| &\leq \frac{\varepsilon }{2} \int_{\Omega }u_t^2dx+\frac{\varepsilon }{2}( 1+k(0)) \int_{\Omega }u^2dx++\frac{\varepsilon \alpha }{2n}\int_{\Gamma _1}u^2dx \\ &\leq \frac{\varepsilon }{2}\int_{\Omega }u_t^2dx+\frac{\varepsilon \left( 1+k(0)+\frac{\alpha }{n}\right) \beta }{2}\int_{\Omega }| \nabla u| ^2dx \\ &\leq \frac{\varepsilon }{2}\int_{\Omega }u_t^2dx+\frac{\varepsilon \left( 1+k(0)+\alpha \right) \beta }{2}\int_{\Omega }| \nabla u| ^2dx\leq \lambda \varepsilon E(t), \end{align*} where $\beta$ is the Poincar\'{e} constant and \begin{equation*} \lambda =\max \left\{ 1,\frac{\left( 1+k(0)+\alpha \right) \beta }{\alpha } \right\} . \end{equation*} Then from \eqref{e21}, it follows that $$\label{e23} | F(t)-E(t)| \leq \varepsilon \lambda E(t),\quad t\geq 0.$$ \end{proof} \begin{lemma} \label{lem3.3} Under assumptions{\rm (G1)--(G2)}, the functional $\varphi (t)$ satisfies, along the solution of \eqref{e20} and for any $\delta >0$, \label{e24} \begin{aligned} \varphi '(t) &\leq \int_{\Omega }u_t^2dx-\left( \alpha -\delta \beta [ k'(t)+1]\right) \int_{\Omega }| \nabla u| ^2dx-\alpha b\int_{\Gamma _1}| u| ^{\rho +2}dx \\ &\quad -\frac{k'(0)}{4\delta }\int_{\Omega }\int_0^{t}k''(t-s) \left( u(s)-u(t)\right) ^2\,ds\,dx+\frac{k'(t)}{4\delta } \int_{\Omega }\left( u(t)-u_0(x)\right) ^2dx. \end{aligned} \end{lemma} \begin{proof} Differentiation of \eqref{e22}, using \eqref{e20}, yields $$\label{e25} \begin{split} \varphi '(t) &= \int_{\Omega }u_t^2dx-\alpha \int_{\Omega }| \nabla u| ^2dx-k'(0) \int_{\Omega}u^2dx-\alpha b\int_{\Gamma _1}| u| ^{\rho +2}dx \\ &\quad +k'(t)\int_{\Omega }u_0(x)u(t)dx-\int_{\Omega}u(t)\int_0^{t} k''(t-s)u(s)\,ds\,dx. \end{split}$$ Using Young's, Cauchy-Schwarz's, Poincar\'{e}'s and H \"{o}lder's inequalities, the last two terms in \eqref{e25} can be estimated as follows \label{e26} \begin{aligned} &k'(t)\int_{\Omega }u_0(x)u(t)dx \\ &= k'(t)\int_{\Omega}\left( u_0(x)-u(t)+u(t)\right) u(t)dx \\ &= k'(t)\int_{\Omega }\left( u_0(x)-u(t)\right) u(t)dx+k'(t)\int_{\Omega }u^2(t)dx \\ &\leq \frac{k'(t)}{4\delta }\int_{\Omega }\left(u(t)-u_0(x)\right) ^2dx +\delta k'(t)\int_{\Omega }u^2(t)dx+k'(t)\int_{\Omega }u^2dx \\ &\leq \frac{k'(t)}{4\delta }\int_{\Omega }\left(u(t)-u_0(x)\right) ^2dx +\left( \delta +1\right) k'(t)\int_{\Omega }u^2dx \end{aligned} and \label{e27} \begin{aligned} &-\int_{\Omega }u(t)\int_0^{t}k''(t-s)u(s)\,ds\,dx \\ &= -\int_{\Omega }u(t)\int_0^{t}k''(t-s)[ \big( u(s)-u(t)\big) +u(t)] \,ds\,dx \\ &= -\int_{\Omega }u(t)\int_0^{t}k''(t-s)\left( u(s)-u(t)\right) \,ds\,dx-\int_{\Omega }u^2(t)\int_0^{t}k''(t-s)\,ds\,dx \\ &\leq \delta \int_{\Omega }u^2(t)dx+\frac{1}{4\delta }\int_{\Omega }\Big( \int_0^{t}k''(t-s)\left( u(s)-u(t)\right) ds\Big)^2dx\\ &\quad -\int_{\Omega }u^2(t)\int_0^{t}k''(t-s)\,ds\,dx \\ &\leq \delta \int_{\Omega }u^2(t)dx+\frac{k'(0)}{4\delta } \int_{\Omega }\int_0^{t}-k''(t-s)\left( u(s)-u(t)\right) ^2\,ds\,dx\\ &\quad +k'(0)\int_{\Omega }u^2dx-k'(t)\int_{\Omega}u^2dx \\ &\leq -\frac{k'(0)}{4\delta }\int_{\Omega }\int_0^{t}k''(t-s) \left( u(s)-u(t)\right) ^2\,ds\,dx+[ \delta +k'(0)-k'(t)] \int_{\Omega }u^2dx. \end{aligned} Combining \eqref{e25}-\eqref{e27}, the result in \eqref{e24} follows. \end{proof} At this point, we state and prove our main result. \begin{theorem} \label{thm3.1} Assume that {\rm (G1)--(G3)} hold, and let $(u_0,u_1)\in H_{\Gamma _0}^1(\Omega )\times L^2(\Omega )$. Then, there exist two positive constants $C$ and $\omega$, independent of $n$, such that the limit weak solution of \eqref{e20} satisfies, for all $t\geq 0$, \begin{equation*} E(t)\leq CE(0)e^{-\omega t}. \end{equation*} \end{theorem} \begin{proof} Using Lemmas \ref{lem3.1} and \ref{lem3.3}, we have \label{e28} \begin{aligned} F'(t) &= E'(t)+\varepsilon \varphi '(t) \\ &\leq -\left( k(0)-\varepsilon \right) \int_{\Omega }u_t^2dx-\varepsilon \left( \alpha -\delta \beta [ k'(t)+1]\right) \int_{\Omega }| \nabla u| ^2dx \\ &\quad -\varepsilon \frac{k'(0)}{4\delta }\int_{\Omega }\int_0^{t}k''(t-s)\left( u(s)-u(t)\right) ^2\,ds\,dx \\ &\quad -\frac{1}{2}\int_{\Omega }\int_0^{t}k'''(t-s)\left( u(s)-u(t)\right) ^2\,ds\,dx-\varepsilon \alpha b\int_{\Gamma _1}| u| ^{\rho +2}dx \\ &\quad +\left( \frac{k''(t)}{2}+\varepsilon \frac{k'(t)}{ 4\delta }\right) \int_{\Omega }\left( u(t)-u_0(x)\right) ^2dx-\frac{ \alpha }{n}\int_{\Gamma _1}| u_t| ^2dx. \end{aligned} Using (G3), \eqref{e28}, and $k'(t)\leq k'(0)$, and dropping the last term, we arrive at \label{e29} \begin{aligned} F'(t) &\leq -\left( k(0)-\varepsilon \right) \int_{\Omega }u_t^2dx-\varepsilon \left( \alpha -\delta \beta [ k'(0)+1]\right) \int_{\Omega }| \nabla u| ^2dx \\ &\quad +\left( \frac{\zeta _2}{2}-\varepsilon \frac{k'(0)}{4\delta } \right) \int_{\Omega }\int_0^{t}k''(t-s)\left( u(s)-u(t)\right) ^2\,ds\,dx \\ &\quad -k'(t)\left( \frac{\zeta _1}{2}-\varepsilon \frac{1}{4\delta } \right) \int_{\Omega }\left( u(t)-u_0(x)\right) ^2dx-\varepsilon \alpha \int_{\Gamma _1}| u| ^{\rho +2}dx. \end{aligned} Now, we choose $\delta$ such that \begin{equation*} \delta <\frac{\alpha }{\beta \left( k'(0)+1\right) }. \end{equation*} Whence $\delta$ is fixed, we select $\varepsilon$ satisfying \begin{equation*} \varepsilon <\min \{ k(0),\frac{2\delta \zeta _2}{k'(0)} ,2\delta \zeta _1,\frac{1}{\lambda }\} , \end{equation*} hence \eqref{e29} yields, for some $c>0$, $$\label{e30} F'(t)\leq -cE(t),\quad \forall t\geq 0.$$ Also \eqref{e23} leads to \begin{equation*} ( 1-\lambda \varepsilon ) E(t)\leq F(t)\leq ( 1+\lambda \varepsilon ) E(t),\quad \forall t\geq 0. \end{equation*} Consequently, for any $0<\gamma \leq$ $1-\lambda \varepsilon$, we have $$\label{e31} \gamma E(t)\leq F(t)\leq \left( 2-\gamma \right) E(t),\quad \forall t\geq 0.$$ Inserting \eqref{e31} in \eqref{e30}, we obtain \begin{equation*} F'(t)\leq -\frac{c}{2-\gamma }F(t)=-\omega F(t),\quad \forall t\geq 0, \end{equation*} where $\omega =\frac{c}{2-\gamma }$. A direct integration yields \begin{equation*} F(t)\leq F(0)e^{-\omega t},\quad \forall t\geq 0. \end{equation*} Using \eqref{e31} again gives \begin{equation*} E(t)\leq \frac{1}{\gamma }F(t)\leq \frac{1}{\gamma }F(0)e^{-\omega t}\leq \frac{2-\gamma }{\gamma }E(0)e^{-\omega t}=CE(0)e^{-\omega t},\quad \forall t\geq 0. \end{equation*} This completes the proof. \end{proof} \subsection*{Acknowledgments} The author wants to express his sincere thanks to King Fahd University of Petroleum and Minerals for its support. \begin{thebibliography}{00} \bibitem{b1} J. T. Beale; \emph{Spectral properties of an acoustic boundary condition}, Indiana Univ. Math. J. 25(1976), 895-917. \bibitem{c1} M. M. Cavalcanti, V. N. Domingos Cavalcanti, I. Lasiecka; \emph{Well-posedness and optimal decay rates for the wave equation with nonlinear boundary damping-source interaction}. J. Differential Equations \textbf{236} no. \textbf{2} (2007), 407-459. \bibitem{c2} M. M. Cavalcanti, V. N. Domingos Cavalcanti, J. A. Soriano J.A.; \emph{Global solvability and asymptotic stability for the wave equation with nonlinear feedback and source term on the boundary,} Adv. Math. Sci. Appl. \textbf{16} no. \textbf{2} (2006), 661-696. \bibitem{l1} I. Lasiecka, D. Tataru; \emph{Uniform Boundary Stabilization of Semilinear Wave Equations with Nonlinear Boundary Damping}, Differential and Integral Equations \textbf{6 }no. \textbf{3} (1993), 507-533. \bibitem{m1} Jaime E. Mu\~{n}oz Rivera, Doherty Andrade; \emph{Exponential decay of non-linear wave equation with a viscoelastic boundary condition}. Math. Methods Appl. Sci. 23 (2000), no. 1, 41--61. \bibitem{p1} Pr\"{u}ss; \emph{Evolutionary Integral Equations and Applications}, Birkha\"{u}ser Verlag, Basel, 1993. \bibitem{r1} R. Racke; \emph{Lectures on nonlinear evolution equations. Initial value problems}. Aspect of Mathematics E19. Friedr. Vieweg \& Sohn, Braunschweig/Wiesbaden (1992). \bibitem{r2} Raynal M. L.; \emph{Sur un probl\`{e}me de diffusion non lin\'{e}aire}, C.R. Acad. Sci., Paris, S\'{e}r. A \textbf{280} (1975), 785-787. \end{thebibliography} \end{document}