\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 112, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/112\hfil Stabilization of a semilinear wave equation] {Stabilization of a semilinear wave equation with variable coefficients and a delay term in the boundary feedback} \author[J. Li, H. Feng, J. Wu \hfil EJDE-2013/112\hfilneg] {Jing Li, Hongyinping Feng, Jieqiong Wu} % in alphabetical order \address{Jing Li \newline School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China. \newline Tel +86-351-7010555 fax +86-351-7010979} \email{mathlj@sxu.edu.cn} \address{Hongyinping Feng \newline School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China} \email{fhyp@sxu.edu.cn} \address{Jieqiong Wu \newline School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China} \email{jieqiong@sxu.edu.cn} \thanks{Submitted January 28, 2013. Published April 30, 2013.} \subjclass[2000]{35L05 58J45} \keywords{Delay feedback; Riemannian geometric method; variable coefficients; \hfill\break\indent semilinear wave equation} \begin{abstract} We study the uniform stabilization of a semilinear wave equation with variable coefficients and a delay term in the boundary feedback. The Riemannian geometry method is applied to prove the exponential stability of the system by introducing an equivalent energy function. \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} Let $\Omega$ be a bounded domain in $\mathbb{R}^n\ (n\geq2)$ with smooth boundary $\partial\Omega=\Gamma_0\bigcup\Gamma_1$. Assume that $\Gamma_0$ is nonempty and relatively open in\ $\partial\Omega$ and $\overline{\Gamma}_0\cap\overline{\Gamma}_1=\emptyset$. Define \begin{equation} \mathcal{A }u=-\operatorname{div} (A(x)\nabla u )\quad\text{for } u\in H^1(\Omega),\label{e1.1} \end{equation} where $\operatorname{div}(X)$ denote the divergence of the vector field $X$ in the Euclidean metric, $A(x)=(a_{ij}(x))$ is a matrix function with $a_{ij}=a_{ji}$ of class $C^1$, satisfying \begin{equation} \begin{gathered} \lambda \sum^n_{i=1}\xi_{i}^2\leq\sum^n_{i,j=1}a_{ij}(x)\xi_{i}\xi_j \leq\Lambda \sum^n_{i=1}\xi_{i}^2\quad \forall x\in \Omega,\\ 0\neq\xi=(\xi_1,\xi_2\cdots\xi_n)^T\in\mathbb{R}^n, \end{gathered} \label{e1.2} \end{equation} for some positive constants $\lambda, \Lambda$. We consider the initial boundary value problem \begin{equation} \begin{gathered} u_{tt}(x,t)+\mathcal{A }u(x,t)+h(\nabla u)+f(u)=0 \quad\text{in }\Omega\times (0,+\infty), \\ u=0 \quad\text{on } \Gamma_0\times (0,+\infty), \\ \frac{\partial u}{\partial\nu_A}=-\mu_1u_t(x,t)-\mu_2u_t(x,t-\tau) \quad\text{on } \Gamma _1\times (0,+\infty), \\ u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x)\quad\text{in }\Omega, \\ u_t(x,t-\tau)=g_0(x,t-\tau)\quad\text{on } \Gamma_1\times [0,\tau], \end{gathered} \label{e1.3} \end{equation} where $$ \frac{\partial u}{\partial\nu_A}=\sum^n_{i,j=1}a_{ij} \frac{\partial u}{\partial x_j}\nu_{i}, $$ and $ \nu(x)=(\nu_1,\nu_2,\cdots \nu_n)^T$ denotes the outside unit normal vector of the boundary, $\nu_A=A\nu$. $f:R\to \mathbb{R}$ and $h:\mathbb{R}^n\to \mathbb{R}$ are continuous nonlinear functions satisfying some assumptions (see (A1), (A2)). Here, $\tau>0$ is a time delay, $\mu_1,\mu_2$ are positive real numbers, and the initial values $(u_0,u_1,g_0)$ belong to suitable spaces. The problem of uniform stabilization for the solution to the wave equation has been widely investigated. We refer the reader to \cite{c3,g1,g3,l1,l2}. The system \eqref{e1.3} was claimed to be a nondissipative wave system in the literature. The stability of a nondissipative system is a important mathematical problem and has attracted much attention in recent years. On the other hand, delay effects arise in many applications and practical problems and it is well-known that an arbitrarily small delay may destabilize a system which is uniformly asymptotically stable in absence of the delay, see \cite{d1,n1,n2,x1}. Consequently, we consider the stabilization for a nondissipative wave system with a delay term in the boundary feedback. When $A(x)\equiv I$, we say that the system \eqref{e1.3} is of constant coefficients. In this case, many results on such problems are available in the literature, see \cite{d1,g1,l1,n1,n2,x1}. The coefficients matrix $A(x)$ is related to the material in applications. Our main goal is to dispense with the restriction $A(x)\equiv I$, and we consider the variable coefficients case. The main tool is the Riemannian geometric method which was first introduced in \cite{y1} to obtain the observability inequality. This method was then applied to established the controllability and stabilization in \cite{c1,c2,g4,w2,y2} for second-order hyperbolic equations with the variable coefficients principal part. For a survey on the Riemannian geometric method, we refer the reader to \cite{g2}. We will show that the nondissipative system \eqref{e1.3} is essentially a dissipative system by introducing an equivalent energy function of the system. A similar nondissipative system with variable coefficients has been studied in \cite{g3}. However, the delay term was not considered. The appearance of the delay term often brings great difficulty. We will select a new equivalent energy function, which is different from the equivalent energy function in \cite{g3}, to obtain the exponential stability of the solution to \eqref{e1.3}. Our paper is organized as follows. In Section 2, some necessary notation is introduced and the main results are presented. In Section 3, some preliminary results and the main theorem are proved. The proof of the existence theorem of the solution is presented in the Appendix. \section{Notation and statement of results} All definitions and notation are standard and classical in the literature, see \cite{w1}. Set \begin{equation} G(x)=(g_{ij}(x))=A^{-1}(x).\label{e2.1} \end{equation} For each $x\in\mathbb{R}^n$ in the tangent space $\mathbb{R}^n_{x}=\mathbb{R}^n$, we denote the inner product and the norm as \begin{equation} \label{e2.2} g(X,Y)=\langle X,Y\rangle_{g}=\sum^n_{i,j=1}g_{ij}(x) \alpha _{i}\beta_j,\quad |X|_{g}=\langle X,X\rangle_{g}^{1/2} \end{equation} for all $$ X=\sum^n_{i=1}\alpha_{i} \frac{\partial }{\partial x_{i}},\quad Y=\sum^n_{i=1}\beta_{i} \frac{\partial }{\partial x_{i}}\in\mathbb{R}^n_{x},\quad x\in\mathbb{R}^n. $$ From \cite[Lemma 2.1]{y1}, it holds that \begin{equation} \langle X(x),A(x)Y(x)\rangle_{g}= X(x)\cdot Y(x)\quad x\in\mathbb{R}^n,\label{e2.3} \end{equation} where the central dot denotes the Euclidean product of $\mathbb{R}^n$. It is easy to check that $(\mathbb{R}^n,g)$ is a Riemannian manifold with the metric $g$. Denote as $D$ the Levi-Civita connection in the Riemannian metric $g$. Let $H$ be a vector field on $( \mathbb{R}^n,g)$. Then the covariant differential $DH$ of $H$ determines a bilinear form on $ \mathbb{R}^n_{x}\times \mathbb{R}^n_{x}$ for each $x\in\mathbb{R}^n$, by \begin{equation} DH(X,Y)=\langle D_{Y}H,X\rangle_{g}\quad \forall X,Y\in\mathbb{R}^n_{x}, \label{e2.4} \end{equation} where $D_{Y}H$ stands for the covariant derivative of vector field $H$ with respect to $Y$. Denote as $\nabla_{g}u$ the the gradient of $u$ in the Riemannian metric $g$. It follows from \cite[Lemma 2.1]{y1} that \begin{equation} \nabla_{g}u=\sum^n_{i=1}\Big(\sum^n_{j=1}a_{ij}(x) \frac{\partial u }{\partial x_j}\Big) \frac{\partial}{\partial x_{i}},\quad |\nabla _{g}u|^2_{g}=\sum^n_{i,j=1}a_{ij} \frac{\partial u }{\partial x_{i}} \frac{\partial u }{\partial x_j}. \label{e2.5} \end{equation} We refer the reader to \cite{y1} for further relationships. The following assumptions are needed for proving our results. \begin{itemize} \item[(A1)] $f:\mathbb{R}\to \mathbb{R}$ is a $C^1$-function deriving from a potential \begin{equation} F(s)= \int^{s}_0f(\tau)d\tau\geq 0\quad \forall s\in\mathbb{R},\label{e2.6} \end{equation} and satisfies \begin{equation} |f(s)|\leq b_1|s|^{\rho}+b_2,\quad |f'(s)|\leq b_1|s|^{\rho-1}+b_2,\label{e2.7} \end{equation} where $ b_1,b_2$ are positive constants and the parameter $\rho$ satisfies \begin{equation} 1\leq\rho\leq \begin{cases} 2,& n\leq3,\\ \frac{n }{n-2},& n\geq4. \end{cases} \label{e2.8} \end{equation} \item[(A2)] $h:\mathbb{R}^n\to \mathbb{R}$ is a $C^1$-function and there exist two constants $\beta>0$ and $L>0$ such that \begin{equation} |h(\xi)|\leq\beta\sqrt{\lambda}|\xi|,\quad |\nabla h(\xi)|\leq L\quad \forall\xi\in\mathbb{R}^n.\label{e2.9} \end{equation} \item[(A3)] There exists a vector field $H$ on the Riemannian manifold $( \mathbb{R}^n,g)$ such that \begin{equation} DH(X,X)= c(x)|X|_{g}^2\quad \forall x\in\overline{\Omega}, \; X \in\mathbb{R}^n_{x}.\label{e2.10} \end{equation} Let $b=\min_{\overline{\Omega}}c(x)>0 $ and $B=\max_{\overline{\Omega}}c(x)$ such that \begin{equation} B<\min\big\{b+ \frac{2b-3\varepsilon_0}{n},r\big(b- \frac{\varepsilon_0}{n}\big) \big\} \quad\text{for some $\varepsilon_0\in(0,b)$ and $r>1$}.\label{e2.11} \end{equation} Moreover, \begin{equation} \text{$H\cdot\nu\leq0$ on $\Gamma_0$ and $H\cdot\nu\geq\delta>0$ on $\Gamma_1$ for some constant $\delta$}.\label{e2.12} \end{equation} \end{itemize} Note that (A3) implies that \begin{equation} nb\leq \operatorname{div}(H)\leq nB.\label{e2.13} \end{equation} A number of examples of such a vector field $H$ on $(\mathbb{R}^n,g)$ for which the condition \eqref{e2.10} is satisfied without any constraints on $B$ are presented in \cite{y1}. When $A(x)\equiv I$, condition \eqref{e2.10} is automatically satisfied by choosing $H=x-x_0$. If $\mu_2=0$, that is, in absence of the delay term, the energy of the system \eqref{e1.3} is exponentially decaying to zero, see \cite{g3}. On the contrary, if $\mu_1=0$, that is, there exists only the delay part in the boundary condition on $\Gamma_1$, the system \eqref{e1.3} becomes unstable. See, for instance \cite{d2}. So it is interesting to seek a stabilization result when both $\mu_1$ and $\mu_2$ are nonzero. In this case, the boundary feedback is composed of two parts and only one of them has a delay. The stability of a linear wave equation with constant coefficients and a delay in the boundary feedback has been studied in \cite{n1}. There it is shown that if $\mu_1=\mu_2$, then there exists a sequence of arbitrary small (and large) delays such that instabilities occur, if $\mu_2>\mu_1$; delays which destabilize the system were also obtained. In this article, in agreement with \cite{n1}, we assume that \begin{equation} \mu_2<\mu_1.\label{e2.14} \end{equation} Set $$ V=\{v\in H^1(\Omega)|v=0 \quad \text{on }\Gamma_0\},\quad W=H^2(\Omega)\cap V. $$ \begin{theorem} \label{thm1} Under assumptions {\rm (A1), (A2)} and \eqref{e2.14}, for any given initial values $(u_0,u_1)\in W\times W$, $g_0\in C^1([-\tau,0]; L^2(\Gamma_1))$, satisfying \begin{equation} \frac{\partial u_0}{\partial\nu_A}=-\mu_1u_1-\mu_2g_0(x,-\tau)\quad \text{on } \Gamma _1\label{e2.15} \end{equation} and $T>0$, system \eqref{e1.3} admits a unique strong solution $u$ on $(0,T)$ such that \begin{equation} u\in L^{\infty}(0,T; V),\quad u _t\in L^{\infty}(0,T; V), \quad u _{tt}\in L^{\infty}(0,T; L^2(\Omega)).\label{e2.16} \end{equation} Moreover, if $(u_0,u_1)\in V\times L^2(\Omega)$, $g_0\in L^2(-\tau,0; L^2(\Gamma_1))$, then \eqref{e1.3} possess at least a weak solution in the space $C([0,T]; V)\cap C^1([0,T]; L^2(\Omega))$. \end{theorem} The Galerkin's approximation will be used for proving Theorem \ref{thm1}. Under assumption \eqref{e2.14}, define the energy of \eqref{e1.3} as \begin{equation} E(t)= \frac{1}{2} \int_{\Omega}[|u_t|^2+|\nabla_{g}u|_{g}^2+2F(u)]dx + \frac{\xi}{2} \int^1_0 \int_{\Gamma_1}u_t^2(x,t-\tau\rho) d\Gamma d\rho,\label{e2.17} \end{equation} where $\xi$ is a strictly positive constant satisfying \begin{equation} \tau\mu_2\leq\xi\leq\tau(2\mu_1- \mu_2).\label{e2.18} \end{equation} Denote $E_{s}(t)$ as \begin{equation} E_{s}(t)= \frac{1}{2} \int_{\Omega}[|u_t|^2 +|\nabla_{g}u|_{g}^2+2F(u)]dx.\label{e2.19} \end{equation} Our main result is the following theorem. \begin{theorem} \label{thm2} Let $u$ be a (strong or weak) solution of \eqref{e1.3}. Suppose that {\rm (A1)--(A3)} and \eqref{e2.14} hold. In addition assume that $f$ satisfies \begin{equation} 2rF(s)\leq sf(s)\quad \text{for some constant $r>1$, and all $s\in\mathbb{R}$}. \label{e2.20} \end{equation} If $\beta$ in \eqref{e2.9} is sufficiently small, then there exist positive constants $C $ and $\omega $ independent of initial values such that \begin{equation} E(t)\leq CE(0)\exp \{-\omega t\}\quad \forall t\geq0. \label{e2.21} \end{equation} \end{theorem} \section{Proof of Theorem \ref{thm2}} \begin{proposition} \label{prop1} Let $u$ be a (strong or weak) solution to the system \eqref{e1.3}, the following estimate holds: \begin{equation} \frac{dE(t)}{dt}\leq-C_1 \int_{\Gamma_1}[u_t^2(x,t)+u_t^2(x,t-\tau)]d\Gamma +\beta E_{s}(t),\label{e3.1} \end{equation} with $C_1$ is a positive constant to be specified later. \end{proposition} \begin{proof} Differentiating \eqref{e2.17}, we obtain \begin{equation} \begin{aligned} \frac{dE(t)}{dt}\ &= \int_{\Omega}\left[u_tu_{tt} +\langle\nabla_{g}u,\nabla_{g}u_t\rangle_{g}+f(u)u_t\right]dx\\ &\quad +\xi \int^1_0 \int_{\Gamma_1}u_t(x,t-\tau\rho)u_{tt}(x,t-\tau\rho) d\Gamma d\rho \\ &= \int_{\Gamma_1} \frac{\partial u}{\partial\nu_A}u_td\Gamma - \int_{\Omega}u_th(\nabla u)dx\\ &\quad +\xi \int^1_0 \int_{\Gamma_1}u_t(x,t-\tau\rho) u_{tt}(x,t-\tau\rho) d\Gamma d\rho. \end{aligned}\label{e3.2} \end{equation} Now, let $y(x,\rho)=u(x,t-\tau\rho)$. So we have \begin{equation} u_t=- \frac{1}{\tau} y_{\rho}, \quad u_{tt}= \frac{1}{\tau^2}y_{\rho\rho}.\label{e3.3} \end{equation} Therefore, \begin{equation} \int^1_0 \int_{\Gamma_1}u_t(x,t-\tau\rho)u_{tt}(x,t-\tau\rho)d\Gamma d\rho =- \frac{1}{\tau^{3}} \int^1_0 \int_{\Gamma_1}y_{\rho}(x,\rho) y_{\rho\rho}(x,\rho)d\Gamma d\rho. \label{e3.4} \end{equation} Integrating by parts in $\rho$, we obtain \begin{equation} \begin{aligned} & \int^1_0 \int_{\Gamma_1}y_{\rho}(x,\rho) y_{\rho\rho}(x,\rho)d\Gamma d\rho\\ &=\Big( \int_{\Gamma_1}y_{\rho}(x,\rho) y_{\rho}(x,\rho)d\Gamma\Big) \Big|^1_0 - \int^1_0 \int_{\Gamma_1}y_{\rho\rho}(x,\rho) y_{\rho}(x,\rho)d\Gamma d\rho \\ &= \int_{\Gamma_1}\left[ y_{\rho}^2(x,1)- y_{\rho}^2(x,0)\right] d\Gamma - \int^1_0 \int_{\Gamma_1}y_{\rho\rho}(x,\rho) y_{\rho}(x,\rho) d\Gamma d\rho. \end{aligned} \label{e3.5} \end{equation} That is $$ \int^1_0 \int_{\Gamma_1}y_{\rho}(x,\rho) y_{\rho\rho}(x,\rho) d\Gamma d\rho = \frac{1}{2} \int_{\Gamma_1}\left[ y_{\rho}^2(x,1)- y_{\rho}^2(x,0)\right] d\Gamma. $$ Therefore, \begin{equation} \begin{aligned} &\int^1_0 \int_{\Gamma_1}u_t(x,t-\tau\rho)u_{tt}(x,t-\tau\rho)d\Gamma d\rho\\ &= - \frac{1}{2\tau^{3}} \int_{\Gamma_1}\left[ y_{\rho}^2(x,1) - y_{\rho}^2(x,0)\right] d\Gamma\\ &= \frac{1}{2\tau} \int_{\Gamma_1}\left[ u_t^2(x,t)- u_t^2(x,t-\tau)\right] d\Gamma \end{aligned} \label{e3.6} \end{equation} which, together with the boundary condition of \eqref{e1.3} on $\Gamma_1$ and \eqref{e3.2}, leads to \begin{equation} \begin{aligned} \frac{dE(t)}{dt} &= -\mu_1 \int_{\Gamma_1}u_t^2(x,t)d\Gamma-\mu_2 \int_{\Gamma_1}u_t(x,t) u_t(x,t-\tau)d\Gamma- \int_{\Omega}u_t(x,t)h(\nabla u)dx\\ &\quad + \frac{\xi}{2\tau} \int_{\Gamma_1}\left[u_t^2(x,t)-u_t^2(x,t-\tau)\right] d\Gamma. \end{aligned}\label{e3.7} \end{equation} Applying the Cauchy-Schwarz inequality to \eqref{e3.7}, from \eqref{e2.9} and the fact $F(s)\geq0$, we have \begin{equation} \begin{aligned} \frac{dE(t)}{dt} &\leq \Big(-\mu_1+ \frac{\mu_2}{2}+ \frac{\xi}{2\tau}\Big) \int_{\Gamma_1}u_t^2(x,t) d\Gamma+\Big( \frac{\mu_2}{2} - \frac{\xi}{2\tau}\Big) \int_{\Gamma_1} u_t^2(x,t-\tau)d\Gamma\\ &\quad + \frac{\beta}{2} \int_{\Omega}[|u_t|^2+|\nabla_{g}u|_{g}^2]dx\\ &\leq \Big(-\mu_1+ \frac{\mu_2}{2}+ \frac{\xi}{2\tau}\Big) \int_{\Gamma_1}u_t^2(x,t) d\Gamma+\Big( \frac{\mu_2}{2} - \frac{\xi}{2\tau}\Big) \int_{\Gamma_1} u_t^2(x,t-\tau)d\Gamma\\ &\quad +\beta E_{s}(t), \end{aligned} \label{e3.8} \end{equation} which implies \begin{equation} \frac{dE(t)}{dt}\leq-C_1 \int_{\Gamma_1}\left[u_t^2(x,t)+u_t^2(x,t-\tau)\right] d\Gamma+\beta E_{s}(t), \label{e3.9} \end{equation} with $$ C_1=\min \{\mu_1- \frac{\mu_2}{2}- \frac{\xi}{2\tau}, \frac{\xi}{2\tau} - \frac{\mu_2}{2}\}. $$ Due to \eqref{e2.18}, we have $C_1>0$. The proof is complete. \end{proof} \begin{remark} \label{rmk3.1}\rm From inequality \eqref{e3.9}, it seems that the system \eqref{e1.3} is not dissipative. However, this is a wrong impression. Actually, by introducing an equivalent energy function, we will find that the system \eqref{e1.3} is essentially dissipative under some suitable conditions. \end{remark} \begin{lemma} \label{lem1} Let $H$ be a vector field on $\overline{\Omega}$. For any (strong or weak) solution to \eqref{e1.3} we have \begin{equation} \frac{\partial u}{\partial\nu_A}H(u)=|\nabla_{g}u|_{g}^2(H\cdot\nu)\quad \text{on }\Gamma_0.\label{e3.10} \end{equation} \end{lemma} \begin{proof} Let $x\in\Gamma_0$. We decompose $\nabla_{g}u$ into a direct sum in $(\mathbb{R}^n_{x},g)$ \begin{equation} \nabla_{g}u(x)=\big\langle\nabla_{g}u(x), \frac{\nu_{A}(x)}{|\nu_A|_{g}} \big\rangle_{g} \frac{\nu_{A}(x)}{|\nu_A|_{g}}+Y(x),\label{e3.11} \end{equation} where $Y(x)\in\mathbb{R}^n_{x}$ with $\langle Y(x),\nu_{A}(x)\rangle_{g}=0$. Taking \eqref{e2.3} into account, we obtain \begin{equation} Y(x)\cdot\nu(x)=\langle Y(x),A(x)\nu(x)\rangle_{g} =\langle Y(x),\nu_{A}(x)\rangle_{g}=0,\label{e3.12} \end{equation} which imply $Y(x)\in\Gamma_{0x}$, the tangent space of $\Gamma_0$ at $x$. Since $u=0$ on $\Gamma_0$, it follows from \eqref{e3.11} and \eqref{e3.12} that \begin{equation} \begin{aligned} |\nabla_{g}u|_{g}^2 &=\nabla_{g}u(u) = \frac{1}{|\nu_A|_{g}^2}\langle\nabla_{g}u(x),\nu_{A}(x)\rangle_{g}^2+Y(u)\\ &= \frac{1}{|\nu_A|_{g}^2}\Big{|} \frac{\partial u}{\partial\nu_A}\Big{|}^2. \end{aligned} \label{e3.13} \end{equation} Similarly, $H$ can be decomposed into a direct sum \begin{equation} H=\langle H(x), \frac{\nu_{A}(x)}{|\nu_A|_{g}}\rangle_{g} \frac{\nu_{A}(x)}{|\nu_A|_{g}}+Z(x),\label{e3.14} \end{equation} where $Z(x)\in \Gamma_{0x}$. Recalling that $u=0$ on $\Gamma_0$, from \eqref{e2.3} and \eqref{e3.14}, we obtain \begin{equation} H(u)= \frac{\langle H(x),\nu_{A}(x)\rangle_{g}}{|\nu_A|_{g}^2} \big( \frac{\partial u}{\partial\nu_A}\big) = \frac{ H(x)\cdot\nu(x)}{|\nu_A|_{g}^2} \big(\frac{\partial u}{\partial\nu_A}\big)\label{e3.15} \end{equation} which, together with \eqref{e3.13}, leads to \eqref{e3.10}. The proof is complete. \end{proof} Let \begin{equation} P(t)= \int_{\Omega}\left[2H(u)+(nb-\varepsilon_0)u\right]u_tdx \quad \text{for some } \varepsilon_0\in(0,b).\label{e3.16} \end{equation} \begin{proposition} \label{prop2} Let $u$ be a (strong or weak) solution of \eqref{e1.3}, under the assumptions of Theorem \ref{thm2}, there exist two positive constants $\theta$ and $N$ such that \begin{equation} \frac{dP(t)}{dt}\leq-2\theta E_{s}(t) +N \int_{\Gamma_1}[u_t^2(x,t)+u_t^2(x,t-\tau)] d\Gamma.\label{e3.17} \end{equation} \end{proposition} \begin{proof} Differentiating \eqref{e3.16} with respect to $t$ we obtain \begin{equation} \begin{aligned} \frac{dP(t)}{dt} &= \int_{\Omega}u_t[2H(u_t)+(nb-\varepsilon_0)u_t]dx - \int_{\Omega}\mathcal{A}u[2H(u)+(nb-\varepsilon_0)u]dx\\ &\quad - \int_{\Omega}h(\nabla u)[2H(u)+(nb-\varepsilon_0)u]dx - \int_{\Omega}f(u)[2H(u)+(nb-\varepsilon_0)u]dx\\ &= I_1(t)+I_2(t)+I_3(t)+I_4(t), \end{aligned}\label{e3.18} \end{equation} where \begin{gather*} I_1(t)= \int_{\Omega}u_t[2H(u_t)+(nb-\varepsilon_0)u_t]dx,\\ I_2(t)=- \int_{\Omega}\mathcal{A}u[2H(u)+(nb-\varepsilon_0)u]dx,\\ I_3(t)=- \int_{\Omega}h(\nabla u)[2H(u)+(nb-\varepsilon_0)u]dx,\\ I_4(t)=- \int_{\Omega}f(u)[2H(u)+(nb-\varepsilon_0)u]dx. \end{gather*} Now we estimate $I_{i}(t)$, $(i=1,2,3,4)$. Noting that $u=0$ on $\Gamma_0$, we have \begin{align*} I_1(t)&= \int_{\Omega}H(u_t^2)dx + \int_{\Omega}(nb-\varepsilon_0)u_t^2dx\\ &= \int_{\Gamma_1}u_t^2(H\cdot\nu)d\Gamma - \int_{\Omega}[\operatorname{div}(H)-nb]u_t^2dx -\varepsilon_0 \int_{\Omega}u_t^2dx, \end{align*} where $\operatorname{div}(H)$ denote the divergence of the vector field $H$ in the Euclidean metric. Denoting $M=\max_{\overline{\Omega}}|H|_{g}$, from \eqref{e2.13} we obtain \begin{equation} I_1(t)\leq M \int_{\Gamma_1}u_t^2d\Gamma -\varepsilon_0 \int_{\Omega}u_t^2dx.\label{e3.19} \end{equation} Next, we estimate $I_2(t)$. \begin{equation} \begin{aligned} I_2(t) &= 2 \int_{\partial\Omega} \frac{\partial u}{\partial\nu_A}H(u)d\Gamma -2 \int_{\Omega}\langle\nabla_{g}u,\nabla_{g}(H(u))\rangle_{g}dx + \int_{\partial\Omega}(nb-\varepsilon_0)u \frac{\partial u}{\partial\nu_A}d\Gamma\\ &\quad - \int_{\Omega}(nb-\varepsilon_0)|\nabla_{g}u|_{g}^2dx\\ &= -2 \int_{\Omega }DH(\nabla_{g}u,\nabla_{g}u)dx+ \int_{\Omega } \left[\operatorname{div}(H)-nb+\varepsilon_0\right]|\nabla_{g}u|_{g}^2dx\\ &\quad + \int_{\Gamma_1}\Big[2 \frac{\partial u}{\partial\nu_A}H(u) -|\nabla_{g}u|_{g}^2(H\cdot\nu)+(nb-\varepsilon_0)u \frac{\partial u}{\partial\nu_A}\Big]d\Gamma\\ &\quad + \int_{\Gamma_0}|\nabla_{g}u|_{g}^2(H\cdot\nu)d\Gamma, \end{aligned}\label{e3.20} \end{equation} where the validity of the last step comes from the fact $u=0$ on $\Gamma_0$ and \eqref{e3.10}. Since \begin{equation} \int_{\Gamma_1}2 \frac{\partial u}{\partial\nu_A}H(u)d\Gamma \leq \int_{\Gamma_1}\Big[\delta|\nabla_{g}u|_{g}^2 + \frac{M^2}{\delta}| \frac{\partial u}{\partial\nu_A}|^2\Big]d\Gamma, \label{e3.21} \end{equation} from \eqref{e2.10}, \eqref{e2.12}, \eqref{e2.13}, \eqref{e3.20}, \eqref{e3.21} we obtain \begin{equation} \begin{aligned} I_2(t) &\leq \int_{\Omega}\left[\operatorname{div}(H)-(n+2)b+\varepsilon_0\right] |\nabla_{g}u|_{g}^2dx\\ &\quad + \int_{\Gamma_1}\Big[\delta|\nabla_{g}u|_{g}^2 + \frac{M^2}{\delta}\big|\frac{\partial u}{\partial\nu_A}\big|^2 -\delta|\nabla_{g}u|_{g}^2+(nb-\varepsilon_0)u \frac{\partial u}{\partial\nu_A}\Big]d\Gamma\\ &\leq \left[nB-(n+2)b+\varepsilon_0\right] \int_{\Omega}|\nabla_{g}u|_{g}^2dx\\ &\quad + \int_{\Gamma_1}\Big[ \frac{M^2}{\delta}\big|\frac{\partial u}{\partial\nu_A}\big|^2+(nb-\varepsilon_0)u \frac{\partial u}{\partial\nu_A}\Big]d\Gamma. \end{aligned}\label{e3.22} \end{equation} Using the trace theorem, $$ \int_{\Gamma_1}|v|^2d\Gamma\leq \widetilde{C} \int_{\Omega}| \nabla_{g}v|_{g}^2dx $$ for some constant $\widetilde{C}>0$, for all $v\in V$, and the boundary condition of \eqref{e1.3} on $\Gamma_1$, we estimate the last term on the right-hand side of \eqref{e3.22} as \begin{equation} \begin{aligned} & \int_{\Gamma_1}\Big[ \frac{M^2}{\delta} \big| \frac{\partial u}{\partial\nu_A}\big|^2+(nb-\varepsilon_0)u \frac{\partial u}{\partial\nu_A}\Big]d\Gamma\\ &\leq \int_{\Gamma_1} \frac{M^2}{\delta}\big|\frac{\partial u}{\partial\nu_A}\big|^2d\Gamma+(nb-\varepsilon_0) \int_{\Gamma_1}\Big[\eta|u|^2+ \frac{1}{4\eta}\big|\frac{\partial u}{\partial\nu_A}\big|^2\Big]d\Gamma\\ &\leq \widetilde{C}(nb-\varepsilon_0)\eta \int_{\Omega}|\nabla_{g}u|_{g}^2dx +\Big( \frac{nb-\varepsilon_0}{4\eta}+ \frac{M^2}{\delta}\Big) \int_{\Gamma_1}\big|\frac{\partial u}{\partial\nu_A}\big|^2d\Gamma\\ &\leq \widetilde{C}(nb-\varepsilon_0)\eta \int_{\Omega}|\nabla_{g}u|_{g}^2dx\\ &\quad +C_2\Big( \frac{nb-\varepsilon_0}{4\eta} + \frac{M^2}{\delta}\Big) \int_{\Gamma_1}\left[u_t^2(x,t) +u_t^2(x,t-\tau)\right]d\Gamma\\ &= \varepsilon_0 \int_{\Omega}|\nabla_{g}u|_{g}^2dx+M_1 \int_{\Gamma_1} \left[u_t^2(x,t) +u_t^2(x,t-\tau)\right]d\Gamma, \end{aligned}\label{e3.23} \end{equation} where $\eta= \frac{\varepsilon_0}{\widetilde{C}(nb-\varepsilon_0)}$, $ M_1=C_2\big( \frac{nb-\varepsilon_0}{4\eta}+ \frac{M^2}{\delta}\big)$ were used in the last step and $C_2$ is a positive constant. Substitute \eqref{e3.23} into \eqref{e3.22} to obtain \begin{equation} I_2(t)\leq\left[nB-(n+2)b+2\varepsilon_0\right] \int_{\Omega}|\nabla_{g}u|_{g}^2dx+M_1 \int_{\Gamma_1}[u_t^2(x,t) +u_t^2(x,t-\tau)]d\Gamma.\label{e3.24} \end{equation} Applying the Cauchy inequality and recalling \eqref{e2.9}, we can obtain the estimation of $I_3(t)$ as follows: \begin{equation} \begin{aligned} I_3(t)&\leq 2\beta M \int_{\Omega}|\nabla_{g}u|_{g}^2dx +\beta(nb-\varepsilon_0) \int_{\Omega}|\nabla_{g}u|_{g}|u|dx\\ &\leq \beta \left[2M+ \frac{nb-\varepsilon_0}{2}(1+\overline{C})\right] \int_{\Omega}|\nabla_{g}u|_{g}^2dx, \end{aligned} \label{e3.25} \end{equation} where $\overline{C}$ is a positive constant satisfying $ \int_{\Omega}|u|^2\leq \overline{C} \int_{\Omega}|\nabla_{g}u|_{g}^2dx$ for all $u\in V$. Finally, we estimate $I_4(t)$ . By \eqref{e2.12}, \eqref{e2.13}, \eqref{e2.20}, the nonnegativity of $F, F(0)=0$, $u=0$ on $\Gamma_0$, we have \begin{equation} \begin{aligned} I_4(t) &\leq -(nb-\varepsilon_0)r \int_{\Omega}2F(u)dx-2 \int_{\Omega}H(F(u))dx\\ &= - \int_{\Omega}[(nb-\varepsilon_0)r-\operatorname{div}(H)]2F(u)dx - \int_{\Gamma_1}2F(u)(H\cdot\nu)d\Gamma\\ &\leq [nB-(nb-\varepsilon_0)r] \int_{\Omega}2F(u)dx. \end{aligned}\label{e3.26} \end{equation} Let $$ 0<\beta< \frac{\varepsilon_0}{2M+ \frac{(nb-\varepsilon_0)}{2}(1+\overline{C})}. $$ Combine \eqref{e3.18}, \eqref{e3.19}, \eqref{e3.24}, \eqref{e3.25} and \eqref{e3.26} to obtain \eqref{e3.17}, where \begin{equation} \begin{gathered} \theta:=\min \{(n+2)b-nB-3\varepsilon_0,(nb-\varepsilon_0)r-nB,\varepsilon_0\},\\ N:=M_1+M. \end{gathered}\label{e3.27} \end{equation} By \eqref{e2.11} and the values of $M$ and $M_1$, we have $\theta>0,N>0$. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2}] Define \begin{equation} S(t):= \int^t_{t-\tau} \int_{\Gamma_1}e^{s-t}u_t^2(x,s)d\Gamma ds. \label{e3.28} \end{equation} We can easily estimate \begin{equation} \begin{aligned} \frac{dS(t)}{dt} &= \int_{\Gamma_1}u_t^2(x,t)d\Gamma - \int_{\Gamma_1}e^{-\tau}u_t^2(x,t-\tau)d\Gamma - \int^t_{t-\tau} \int_{\Gamma_1}e^{s-t}u_t^2(x,s)d\Gamma ds\\ &\leq \int_{\Gamma_1}u_t^2(x,t)d\Gamma-e^{-\tau} \int_{\Gamma_1}u_t^2(x,t-\tau)d\Gamma-e^{-\tau} \int^t_{t-\tau} \int_{\Gamma_1}u_t^2(x,s)d\Gamma ds. \end{aligned}\label{e3.29} \end{equation} Let us define a new energy function for \eqref{e1.3} as \begin{equation} L(t):=E(t)+\gamma_1P(t)+\gamma_2S(t),\label{e3.30} \end{equation} where $\gamma_1,\gamma_2$ are suitable positive small constants that will be specified later on. Note that $L(t)$ is equivalent to the energy $E(t)$ if $\gamma_1,\gamma_2$ are small enough. In particular, there exist a positive constant $C_3$ and suitable positive constants $\alpha_1,\alpha_2$ such that \begin{equation} \alpha_1E(t)\leq L(t)\leq\alpha_2E(t)\quad \forall 0\leq \gamma_1,\gamma_2\leq C_3.\label{e3.31} \end{equation} Therefore, $L(t)$ is an equivalent energy function of\eqref{e1.3} for small $\gamma_1,\gamma_2$. Differentiating the function $L(t)$ and recalling \eqref{e3.1}, \eqref{e3.17}, \eqref{e3.29} we deduce \begin{equation} \begin{aligned} \frac{dL(t)}{dt} &= \frac{dE(t)}{dt}+\gamma_1 \frac{dP(t)}{dt}+\gamma_2 \frac{dS(t)}{dt}\\ &\leq (-2\gamma_1\theta+\beta)E_{s}(t)+(-C_1+\gamma_1N+\gamma_2) \int_{\Gamma_1}u_t^2(x,t)d\Gamma\\ &\quad +(-C_1+\gamma_1N-\gamma_2e^{-\tau}) \int_{\Gamma_1}u_t^2(x,t-\tau) d\Gamma\\ &\quad -\gamma_2e^{-\tau} \int^t_{t-\tau} \int_{\Gamma_1}u_t^2(x,s)d\Gamma ds. \end{aligned}\label{e3.32} \end{equation} Note that \begin{equation} \begin{aligned} E(t)&=E_{s}(t)+ \frac{\xi}{2} \int^1_0 \int_{\Gamma_1}u_t^2(x,t-\tau\rho)d\Gamma d\rho\\ &= E_{s}(t)+ \frac{\xi}{2\tau} \int^t_{t-\tau} \int_{\Gamma_1}u_t^2(x,s)d\Gamma ds. \end{aligned}\label{e3.33} \end{equation} Choosing $\gamma_1,\gamma_2$ sufficiently small such that $-C_1+\gamma_1N+\gamma_2<0$, $-C_1+\gamma_1N-\gamma_2e^{-\tau}<0$ and choosing $\beta>0$ small enough such that $-2\gamma_1\theta+\beta<0$, from \eqref{e3.32} and \eqref{e3.33}, we have \begin{equation} \frac{dL(t)}{dt}\leq-\widehat{C}E(t),\label{e3.34} \end{equation} with $\widehat{C}$ is a positive constant. Applying the second inequality of \eqref{e3.31}, from \eqref{e3.34}, we have \begin{equation} \frac{dL(t)}{dt}\leq- \frac{\widehat{C}}{\alpha_2}L(t).\label{e3.35} \end{equation} Then, we easily obtain \begin{equation} L(t)\leq L(0)\exp (-\omega t)\quad \forall t\geq0,\label{e3.36} \end{equation} with $\omega$ is a positive constant. Using \eqref{e3.31} again, we deduce the estimate \eqref{e2.21}. The proof is complete. \end{proof} \section{Appendix: Proof of Theorem \ref{thm1}} As in \cite{g3}, we use Galerkin approximations to prove the well-posedness of \eqref{e1.3}. The change of variable \begin{equation} v(x,t)=u(x,t)-\phi(x,t),\label{A.1} \end{equation} where \begin{equation} \phi(x,t)=u_0(x)+tu_1(x)\quad (x,t)\in Q:=\Omega\times(0,T),\label{A.2} \end{equation} gives the following problem, which is equivalent to \eqref{e1.3}, \begin{equation} \begin{gathered} v_{tt}-\operatorname{div} (A(x)\nabla v)+h(\nabla v+\nabla \phi)+f(v+\phi) =\mathcal{F} \quad\text{in }\Omega\times (0,T), \\ v=0 \quad\text{on } \Gamma_0\times (0,T), \\ \frac{\partial v}{\partial\nu_A}=-\mu_1[v_t(x,t)+u_1] -\mu_2[v_t(x,t-\tau)+u_1]+\mathcal{B}\quad \text{on } \Gamma _1\times (0,T),\\ v(x,0)= v_t(x,0)=0\quad\text{in }\Omega, \\ v_t(x,t-\tau)=g_0(x,t-\tau)-u_1\quad\text{on } \Gamma_1\times [0,\tau], \end{gathered} \label{A.3} \end{equation} where $\mathcal{F}=\operatorname{div} (A(x)\nabla \phi)$, $\mathcal{B}=- \frac{\partial \phi}{\partial\nu_A} $ and $\operatorname{div}(X)$ denote the divergence of the vector field $X$ in the Euclidean metric. Let $\{w_{i}\}_{i\in N}$ be a basis for $W$ that is orthonormal in $L^2(\Omega)$, and let $V_m$ be the space spanned by $w_1\cdots w_m$. When $g_0\in C^1([-\tau,0]; L^2(\Gamma_1))$, we choose a sequence $g_{0m}\to g_0$ strongly in $ C^1([-\tau,0]; L^2(\Gamma_1))$. Now we define the approximation $$ v_m(t)=\sum^{m}_{j=1}\gamma_j(t)w_j, $$ where $v_m(t)$ are solutions to the Cauchy problem \begin{equation} \begin{gathered} \begin{aligned} &\int_{\Omega}v_{mtt}(t)wdx+ \int_{\Omega}\langle\nabla_{g}v_m(t), \nabla_{g}w\rangle_{g}dx + \int_{\Omega}h(\nabla v_m(t)+\nabla \phi(t))wdx\\ & + \int_{\Omega}f(v_m(t)+\phi(t))wdx + \int_{\Gamma_1}[\mu_1(v_{mt}(t)+u_1) +\mu_2(v_{mt}(t-\tau)+u_1)]wd\Gamma\\ &= \int_{\Omega}\mathcal{F}(t)wdx+ \int_{\Gamma_1}\mathcal{B}wd\Gamma, \end{aligned} \\ v_m(0)= v_{mt}(0)=0, \\ v_{mt}(x,t)=g_{0m}(x,t)-u_1 \quad \text{on } \Gamma_1\times [-\tau,0], \end{gathered}\label{A.4} \end{equation} for all $w\in V_m$. According to the standard theory of ordinary differential equations, the finite dimensional problem \eqref{A.4} has solutions $v_m(t)$ defined on some interval $[0,T_m)$. The a priori estimates that follow imply that $T_m=T$. \noindent\textbf{Step 1:} The first-order estimate of $v_m$. Replacing $w$ by $v_{mt}(t)$ in \eqref{A.4} leads to \begin{equation} \begin{aligned} & \frac{1}{2} \frac{d}{dt}\Big( \int_{\Omega}[|v_{mt}(t)|^2 +|\nabla_{g}v_m(t)|_{g}^2+2F(v_m(t)+\phi(t))]dx\Big)\\ &+ \int_{\Omega}h(\nabla v_m(t)+\nabla \phi(t))v_{mt}(t)dx - \int_{\Omega}f(v_m(t)+\phi(t))u_1dx\\ &= \int_{\Omega}\mathcal{F}(t)v_{mt}(t)dx+ \frac{d}{dt} \Big( \int_{\Gamma_1}\mathcal{B}(t)v_m(t)d\Gamma\Big) - \int_{\Gamma_1}\mathcal{B}_t(t)v_m(t)d\Gamma\\ &\quad - \int_{\Gamma_1}[\mu_1(v_{mt}(t)+u_1) +\mu_2(v_{mt}(t-\tau)+u_1)][v_{mt}(t)+u_1]d\Gamma\\ &\quad + \int_{\Gamma_1}[\mu_1(v_{mt}(t)+u_1)+\mu_2(v_{mt}(t-\tau) +u_1)]u_1d\Gamma. \end{aligned}\label{A.5} \end{equation} Using the Sobolev imbedding theorem, H\"{o}lder's inequality, (A1) and the regularities of the initial values, we infer that \begin{equation} \begin{aligned} \int_{\Omega}f(v_m(t)+\phi(t))u_1dx &\leq C \int_{\Omega}|v_m(t)+\phi(t)|^{\rho}|u_1|dx +C \int_{\Omega}|u_1|dx\\ &\leq C\Big( \int_{\Omega}|v_m(t)|^{\rho}|u_1|dx + \int_{\Omega}|\phi(t)|^{\rho}|u_1|dx\Big)+C\\ &\leq C\Big( \int_{\Omega}|v_m(t)|^{2\rho}dx\Big)^{1/2} \Big( \int_{\Omega}|u_1|^2dx\Big)^{1/2}\\ &\quad + C\Big( \int_{\Omega}|u_0|^{\rho}|u_1|+t^{\rho}|u_1|^{\rho+1}\Big)dx +C\\ &\leq C\Big( \int_{\Omega}|\nabla_{g}v_m(t)|_{g}^2dx\Big)^{\rho/2}+Ct^{\rho}+C. \end{aligned}\label{A.6} \end{equation} Here and in what follows, we use the constant $C > 0$ to denote some constants independent of functions involved although it may have different values in different contexts. By (A2), it holds \begin{equation} \begin{aligned} &\int_{\Omega}h(\nabla v_m(t)+\nabla \phi(t))v_{mt}(t)dx\\ &\leq \frac{\beta^2}{2} \int_{\Omega}|\nabla_{g} v_m(t) +\nabla_{g} \phi(t)|_{g}^2dx + \frac{1}{2} \int_{\Omega}| v_{mt}(t)|^2dx. \end{aligned}\label{A.7} \end{equation} Combining \eqref{A.5}--\eqref{A.7}, recalling the trace theorem, $$ \int_{\Gamma_1}|v|^2d\Gamma \leq \widetilde{C} \int_{\Omega}|\nabla_{g}v|_{g}^2dx $$ for some constant $\widetilde{C}>0$ and all $v\in V$, it follows that \begin{align*} & \frac{1}{2} \frac{d}{dt}\Big( \int_{\Omega}[|v_{mt}(t)|^2 +|\nabla_{g}v_m(t)|_{g}^2+2F(v_m(t)+\phi(t))]dx\Big)\\ &\leq C\Big( \int_{\Omega}|\nabla_{g}v_m(t)|_{g}^2dx\Big)^{\rho/2} +Ct^{\rho} + \frac{\beta^2}{2} \int_{\Omega}|\nabla_{g} v_m(t)+\nabla_{g} \phi(t)|_{g}^2dx\\ &\quad + \frac{1}{2} \int_{\Omega}| v_{mt}(t)|^2dx + \frac{1}{2} \int_{\Omega}|\mathcal{F}(t)|^2dx + \frac{1}{2} \int_{\Omega}| v_{mt}(t)|^2dx\\ &\quad + \frac{d}{dt}\Big( \int_{\Gamma_1}\mathcal{B}(t)v_m(t)d\Gamma\Big) + \frac{\widetilde{C}}{2} \int_{\Gamma_1}|\mathcal{B}_t(t)|^2d\Gamma + \frac{1}{2} \int_{\Omega}|\nabla_{g} v_m(t)|_{g}^2dx+C\\ &\quad - \int_{\Gamma_1}[\mu_1(v_{mt}(t)+u_1) +\mu_2(v_{mt}(t-\tau)+u_1)][v_{mt}(t)+u_1]d\Gamma\\ &\quad + \int_{\Gamma_1}[\mu_1(v_{mt}(t)+u_1) +\mu_2(v_{mt}(t-\tau)+u_1)]u_1d\Gamma\,. \end{align*} Integrating the obtained result over the interval $(0,t)$, noticing $ v_m(0)= v_{mt}(0)=0$, $\frac{\rho}{2}\leq 1$ and applying the trace theorem, we obtain \begin{equation} \begin{aligned} & \int_{\Omega}\left[|v_{mt}(t)|^2 +|\nabla_{g}v_m(t)|_{g}^2+2F(v_m(t)+\phi(t))\right]dx \\ &\leq (C+2\beta^2+1) \int^t_0 \int_{\Omega}|\nabla_{g}v_m(s)|_{g}^2dxds+Ct^{\rho+1} +2 \int^t_0 \int_{\Omega}| v_{ms}(s)|^2dxds\\ &\quad +2\beta^2 \int^t_0 \int_{\Omega}|\nabla_{g}\phi(s)|_{g}^2dxds + \int^t_0 \int_{\Omega}|\mathcal{F}(s)|^2dxds+2 \int_{\Gamma_1}\mathcal{B}(t)v_m(t)d\Gamma\\ &\quad +\widetilde{C}t \int_{\Gamma_1}\Big{|} \frac{\partial u_1}{\partial\nu_A}\Big{|}^2d\Gamma+Ct+C\\ &\quad - \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1)+\mu_2(v_{ms}(s-\tau)+u_1)][v_{ms}(s)+u_1]d\Gamma ds\\ &\quad + \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1)+\mu_2(v_{ms}(s-\tau)+u_1)]u_1d\Gamma ds\\ &\leq (C+2\beta^2+1) \int^t_0 \int_{\Omega}|\nabla_{g}v_m(s)|_{g}^2dxds +2 \int^t_0 \int_{\Omega}| v_{ms}(s)|^2dxds\\ &\quad +\zeta \int_{\Omega}|\nabla_{g}v_m(t)|_{g}^2dx +C(t^{\rho+1}+t+t^{3})+C\\ &\quad - \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)][v_{ms}(s)+u_1]d\Gamma ds\\ &\quad + \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)]u_1d\Gamma ds, \end{aligned}\label{A.8} \end{equation} where $\zeta>0$ is a sufficiently small constant that will be specified later on. Using the Cauchy-Schwartz inequality, we deduce \begin{equation} \begin{aligned} &- \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)][v_{ms}(s)+u_1]d\Gamma ds\\ &\leq \int^t_0 \int_{\Gamma_1}\left[\big( \frac{\mu_2}{2}-\mu_1\big)|v_{ms}(s) +u_1|^2 + \frac{\mu_2}{2}|v_{ms}(s-\tau)+u_1|^2\right]d\Gamma ds. \end{aligned}\label{A.9} \end{equation} Now, using the history values about $v_{mt}(t)$ $t\in[-\tau,0]$, the second term in the right-hand side of \eqref{A.9} can be rewritten as \begin{equation} \begin{aligned} &\int^t_0 \int_{\Gamma_1}|v_{ms}(s-\tau)+u_1|^2d\Gamma ds\\ &= \int^{t-\tau}_{-\tau} \int_{\Gamma_1}|v_{m\rho}(\rho)+u_1|^2d\Gamma d\rho\\ &= \int^{0}_{-\tau} \int_{\Gamma_1}|v_{m\rho}(\rho)+u_1|^2d\Gamma d\rho + \int^{t-\tau}_0 \int_{\Gamma_1}|v_{m\rho}(\rho)+u_1|^2d\Gamma d\rho\\ &= \int^{0}_{-\tau} \int_{\Gamma_1}|g_{0m}(\rho)|^2d\Gamma d\rho + \int^{t-\tau}_0 \int_{\Gamma_1}|v_{m\rho}(\rho)+u_1|^2d\Gamma d\rho\\ &\leq C_0+ \int^t_0 \int_{\Gamma_1}|v_{m\rho}(\rho)+u_1|^2d\Gamma d\rho, \end{aligned}\label{A.10} \end{equation} where $C_0$ is a positive constant. From \eqref{A.9} and \eqref{A.10}, we deduce \begin{equation} \begin{aligned} &- \int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)][v_{ms}(s)+u_1]d\Gamma ds\\ &\leq \int^t_0 \int_{\Gamma_1}(\mu_2-\mu_1)|v_{ms}(s)+u_1|^2d\Gamma ds+C. \end{aligned}\label{A.11} \end{equation} On the other hand, taking the Cauchy-Schwartz inequality, the inequality \eqref{A.10} and the regularities of the initial values, we deduce \begin{equation} \begin{aligned} &\int^t_0 \int_{\Gamma_1}[\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)]u_1d\Gamma ds\\ &\leq \eta \int^t_0 \int_{\Gamma_1}|\mu_1(v_{ms}(s)+u_1) +\mu_2(v_{ms}(s-\tau)+u_1)|^2d\Gamma ds \\ &\quad +C(\eta) \int^t_0 \int_{\Gamma_1}|u_1|^2d\Gamma ds\\ &\leq 2\eta \int^t_0 \int_{\Gamma_1}|\mu_1(v_{ms}(s)+u_1)|^2d\Gamma ds\\ &\quad +2\eta \int^t_0 \int_{\Gamma_1}|\mu_2(v_{ms}(s-\tau) +u_1)|^2d\Gamma ds+C'(\eta)\\ &\leq 2(\mu_1^2+\mu_2^2)\eta \int^t_0 \int_{\Gamma_1}|v_{ms}(s)+u_1|^2d\Gamma ds +C+C'(\eta)\quad t\in[0,T], \end{aligned}\label{A.12} \end{equation} where $\eta>0$ is a sufficiently small constant that will be specified later on and $C(\eta),C'(\eta)$ are positive constants. Substituting \eqref{A.11}, \eqref{A.12} into \eqref{A.8} and choosing $\zeta>0$ small enough, we obtain \begin{equation} \begin{aligned} & \int_{\Omega}\left[|v_{mt}(t)|^2 +|\nabla_{g}v_m(t)|_{g}^2+2F(v_m(t)+\phi(t))\right]dx\\ & +[\mu_1-\mu_2-2(\mu_1^2+\mu_2^2)\eta] \int^t_0 \int_{\Gamma_1}|v_{ms}(s)+u_1|^2d\Gamma ds\\ &\leq (C+2\beta^2+1) \int^t_0 \int_{\Omega}|\nabla_{g}v_m(s)|_{g}^2dxds +2 \int^t_0 \int_{\Omega}| v_{ms}(s)|^2dxds\\ &\quad +C(t^{\rho+1}+t+t^{3})+C. \end{aligned}\label{A.13} \end{equation} Finally, noting the fact $\mu_2<\mu_1$, $F(s)\geq0$ for all $s\in\mathbb{R}$, choosing $\eta>0$ sufficiently small, by Gronwall's lemma, we obtain the first-order estimate of $v_m$ \begin{equation} \begin{aligned} &\int_{\Omega}\left[|v_{mt}(t)|^2 +|\nabla_{g}v_m(t)|_{g}^2+2F(v_m(t)+\phi(t))\right]dx\\ &\quad + \int^t_0 \int_{\Gamma_1}|v_{ms}(s)+u_1|^2d\Gamma ds \leq C_4, \end{aligned}\label{A.14} \end{equation} where $C_4>0$ is a constant independent of $m\in N$ and $t\in[0,T]$. \noindent\textbf{Step 2:} The second-order estimate of $v_m$. We estimate the term $\|v_{mtt}(0)\|_{L^2(\Omega)}$. Take $t=0$ in \eqref{A.4} and notice the fact $v_m(0)=v_{mt}(0)=0$, to obtain \begin{align*} & \int_{\Omega}v_{mtt}(0)wdx+ \int_{\Omega}h(\nabla u_0)wdx + \int_{\Omega}f(u_0)wdx + \int_{\Gamma_1}[\mu_1u_1+\mu_2g_{0m}(-\tau)]wd\Gamma\\ &= \int_{\Omega}\operatorname{div} (A(x)\nabla u_0)wdx + \int_{\Gamma_1}\Big(- \frac{\partial u_0}{\partial\nu_A}\Big)wd\Gamma\quad \forall w\in V_m \end{align*} which, together with \eqref{e2.15}, leads to \begin{align*} &\int_{\Omega}v_{mtt}(0)wdx+ \int_{\Omega}h(\nabla u_0)wdx + \int_{\Omega}f(u_0)wdx + \int_{\Gamma_1}[\mu_2g_{0m}(-\tau)]wd\Gamma\\ &= \int_{\Omega}\operatorname{div}( A(x)\nabla u_0)wdx + \int_{\Gamma_1}\mu_2g_0(-\tau)wd\Gamma\quad \forall w\in V_m, \end{align*} which, together with (A1), (A2) and the regularities of the initial values, lead to $$ \|v_{mtt}(0)\|_{L^2(\Omega)}\leq C_5, $$ where $C_5>0$ is a constant independent of $m\in N$. Next, differentiate \eqref{A.4} with respect to $t$ and replace $w$ by $v_{mtt}$, to obtain \begin{equation} \begin{aligned} &\frac{1}{2} \frac{d}{dt} \Big[ \int_{\Omega}\left(|v_{mtt}(t)|^2 +|\nabla_{g}v_{mt}(t)|_{g}^2\right)dx\Big]\\ & + \int_{\Omega}\nabla h(\nabla v_m(t)+\nabla \phi(t)) (\nabla v_{mt}(t)+\nabla u_1)v_{mtt}(t)dx\\ & + \int_{\Omega}f'(v_m(t)+\phi(t))(v_{mt}(t)+u_1)v_{mtt}(t)dx\\ & + \int_{\Gamma_1}[\mu_1v_{mtt}(t)+\mu_2v_{mtt}(t-\tau)]v_{mtt}(t)d\Gamma\\ &= \int_{\Omega}\mathcal{F}_t(t)v_{mtt}(t)dx+ \frac{d}{dt} \Big( \int_{\Gamma_1}\mathcal{B}_t(t)v_{mt}(t)d\Gamma\Big). \end{aligned} \label{A.15} \end{equation} Taking (A2) into account, we infer that \begin{equation} \begin{aligned} &\int_{\Omega}\nabla h(\nabla v_m(t)+\nabla \phi(t))(\nabla v_{mt}(t) +\nabla u_1)v_{mtt}(t)dx\\ &\leq C\Big(1+ \int_{\Omega}|\nabla_{g}v_{mt}(t)|_{g}^2dx + \int_{\Omega}|v_{mtt}(t)|^2dx\Big). \end{aligned}\label{A.16} \end{equation} We use H\"{o}lder's inequality, the Sobolev imbedding theorem, and the trace theorem, by noticing (A1), \eqref{A.14} and the regularities of the initial values, to obtain \begin{equation} \begin{aligned} &\int_{\Omega}f'(v_m(t)+\phi(t))(v_{mt}(t)+u_1)v_{mtt}(t)dx\\ &\leq C \int_{\Omega}(|v_m(t)|^{\rho-1}+|\phi(t)|^{\rho-1}+C)(|v_{mt}(t)|+|u_1|)|v_{mtt}(t)|dx\\ &\leq C \int_{\Omega}(|v_m(t)|^{2(\rho-1)}|v_{mt}(t)|^2dx+C \int_{\Omega}|\phi(t)|^{2(\rho-1)}|v_{mt}(t)|^2dx\\ &\quad +C \int_{\Omega}|v_{mtt}(t)|^2dx+C\\ &\leq C\Big( \int_{\Omega}(|v_m(t)|^{2(\rho-1)\cdot\frac{n}{2}}dx\Big)^{2/n} \Big( \int_{\Omega}(|v_{mt}(t)|^{2\cdot\frac{n}{n-2}}dx\Big)^{(n-2)/n}\\ &\quad +C \int_{\Omega}|\phi(t)|^{2(\rho-1)}|v_{mt}(t)|^2dx +C \int_{\Omega}|v_{mtt}(t)|^2dx+C\\ &\leq C \int_{\Omega}\left(|\nabla_{g}v_{mt}(t)|_{g}^2+|v_{mtt}(t)|^2\right)dx+C \end{aligned} \label{A.17} \end{equation} and \begin{gather} \int_{\Omega}\mathcal{F}_t(t)v_{mtt}(t)dx \leq C \int_{\Omega}|v_{mtt}(t)|^2dx+C,\label{A.18}\\ \int_{\Gamma_1}\mathcal{B}_t(t)v_{mt}(t)d\Gamma \leq C \frac{\widetilde{C}}{4\xi} +\xi \int_{\Omega}|\nabla_{g}v_{mt}(t)|_{g}^2dx,\label{A.19} \end{gather} where $\xi>0$ is a sufficiently small constant that will be specified later on. Finally, combining \eqref{A.16}--\eqref{A.19}, integrating \eqref{A.15} over $(0,t)$, choosing $\xi>0$ sufficiently small and recalling $\|v_{mtt}(0)\|_{L^2(\Omega)}\leq C_5$, we obtain \begin{equation} \begin{aligned} &\int_{\Omega}\left(|v_{mtt}(t)|^2+|\nabla_{g}v_m(t)|_{g}^2\right)dx + \int^t_0 \int_{\Gamma_1}[\mu_1v_{\rm mss}(s)+\mu_2v_{\rm mss}(s-\tau)]v_{\rm mss}(s)d\Gamma ds\\ &\leq C \int^t_0 \int_{\Omega}\left(|v_{\rm mss}(s)|^2+|\nabla_{g}v_{ms}(s)|_{g}^2 \right)\,dx\,ds+Ct+C. \end{aligned}\label{A.20} \end{equation} Note that \begin{equation} \begin{aligned} &\int^t_0 \int_{\Gamma_1}|v_{\rm mss}(s-\tau)|^2d\Gamma ds\\ &= \int^{t-\tau}_{-\tau} \int_{\Gamma_1}|v_{m\rho\rho}(\rho)|^2d\Gamma d\rho\\ &= \int^{0}_{-\tau} \int_{\Gamma_1}|g_{0m\rho}(\rho)|^2d\Gamma d\rho + \int^{t-\tau}_0 \int_{\Gamma_1}|v_{m\rho\rho}(\rho)|^2d\Gamma d\rho\\ &\leq C_0'+ \int^t_0 \int_{\Gamma_1}|v_{m\rho\rho}(\rho)|^2d\Gamma d\rho, \end{aligned}\label{A.21} \end{equation} where $C_0'$ is a positive constant. From \eqref{A.21}, we infer \begin{equation} \begin{aligned} &\int^t_0 \int_{\Gamma_1}[\mu_1v_{\rm mss}(s)+\mu_2v_{\rm mss}(s-\tau)] v_{\rm mss}(s)\,d\Gamma ds\\ &\geq \int^t_0 \int_{\Gamma_1}\left[\big(\mu_1- \frac{\mu_2}{2}\big) |v_{\rm mss}(s)|^2 - \frac{\mu_2}{2}|v_{\rm mss}(s-\tau)|^2\right]d\Gamma ds\\ &\geq \int^t_0 \int_{\Gamma_1}(\mu_1-\mu_2)|v_{\rm mss}(s)|^2d\Gamma ds-C \end{aligned}\label{A.22} \end{equation} which, together with \eqref{A.20}, leads to \begin{equation} \begin{aligned} & \int_{\Omega}\left(|v_{mtt}(t)|^2+|\nabla_{g}v_m(t)|_{g}^2\right)dx + \int^t_0 \int_{\Gamma_1}(\mu_1-\mu_2)|v_{\rm mss}(s)|^2d\Gamma ds\\ &\leq \int^t_0 \int_{\Omega}\left(|v_{\rm mss}(s)|^2 +|\nabla_{g}v_m(s)|_{g}^2\right)\,dx\,ds+Ct+C. \end{aligned} \label{A.23} \end{equation} Recalling the fact $\mu_2<\mu_1,$ by Gronwall's lemma, we obtain the second-order estimate of $v_m$, $$ \int_{\Omega}\left(|v_{mtt}(t)|^2+|\nabla_{g}v_m(t)|_{g}^2\right)dx + \int^t_0 \int_{\Gamma_1}(\mu_1-\mu_2)|v_{\rm mss}(s)|^2d\Gamma ds \leq C_6, % A.24 $$ where $C_6$ is a positive constant independent of $m\in N$ and $t\in[0,T]$. 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