\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 27, 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 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/27\hfil Stabilization of the wave equation] {Stabilization of the wave equation with variable coefficients and a dynamical boundary control} \author[Z. Zhang \hfil EJDE-2016/27\hfilneg] {Zhifei Zhang} \address{Zhifei Zhang \newline Department of Mathematics, Huazhong University of Science and Technology, Wuhan, 430074, China} \email{zhangzf@hust.edu.cn} \thanks{Submitted December 9, 2015. Published January 15, 2016.} \subjclass[2010]{34H05, 35L05, 49K25, 93C20, 93D15} \keywords{Exponential decay; wave equation with variable coefficients; \hfill\break\indent dynamical boundary control} \begin{abstract} In this article we consider the boundary stabilization of the wave equation with variable coefficients and a dynamical Neumann boundary control. The dynamics on the boundary comes from the acceleration terms which can not be ignored in some physical applications. It has been known that addition of dynamics to the boundary may change drastically the stability properties of the underlying system. In this paper by applying a boundary feedback control we obtain the exponential decay for the solutions. Our proof relies on the Geometric multiplier skills and the energy perturbed approach. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $\Omega$ be a bounded domain in $\mathbb{R}^N$ $(N\geq 2)$ with smooth boundary $\Gamma$. We assume $\Gamma=\Gamma_0\cup\Gamma_1$ with $\overline{\Gamma_0}\cap\overline{\Gamma_1}=\emptyset$. We consider the following wave equation with dynamical Neumann boundary condition \begin{equation}\label{w} \begin{gathered} u_{tt}-\operatorname{div}A(x)\nabla u=0,\quad\text{in }\Omega\times(0,\infty),\\ u(x,t)=0,\quad \text{on }\Gamma_0\times(0,\infty),\\ m(x)u_{tt}(x,t)+\partial_{\nu_A}u(x,t)=C(t), \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, \end{gathered} \end{equation} where $\operatorname{div} X$ denotes the divergence of the vector field $X$ in the Euclidean metric, $A(x) = (a_{ij}(x))$ are symmetric and positive definite matrices for all $x\in\mathbb{R}^N$ and $a_{ij}(x)$ are smooth functions on $\mathbb{R}^N$. Let $\partial_{\nu_A}u=\sum_{i,j=1}^na_{ij}(x)\partial_{x_j}u\nu_i$, where $\nu=(\nu_1,\nu_2,\cdots,\nu_n)^T$ denotes the outward unit normal vector of the boundary and $\nu_A=A\nu$. Here $C(t)$ is the boundary feedback control. We say that equation \eqref{w} is with dynamical boundary conditions when $m(x)\neq0$. In some physical applications one has to take the acceleration terms into account on the boundary. Usually in this case to describe what really happened we need the models with dynamical boundary conditions. They are not only important theoretically but also have strong backgrounds for physical applications. There are numerous of these applications in the bio-medical domain (\cite{10,VD}) as well as in applications related to noise suppression and control of elastic structures (\cite{Ak,BST,MI,XL}). Moreover, in this article we assume that \[ m(x)\in L^{\infty}(\Gamma_1);\quad m(x)\geq m_0>0,\quad x\in\Gamma_1. \] Our motivating example is the hybrid wave equation \begin{equation}\label{mo} \begin{gathered} u_{tt}-\Delta u=0,\quad\text{in }\Omega\times(0,\infty),\\ u(x,t)=0,\quad \text{on }\Gamma_0\times(0,\infty),\\ mu_{tt}(x,t)+\partial_{\nu}u(x,t)=-\partial_{\nu}u_t, \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. \end{gathered} \end{equation} It is shown that the above system with $m>0$ is not uniformly stable, while the case when $m=0$ the system is exponentially stable under suitable geometrical conditions on $\Omega$. That is to say, the dynamical terms on the boundary may change the stability property of the system. For details, see \cite{FL,Lag,LL}. The purpose of this paper is to study how the dynamic boundary conditions on $\Gamma_1$ affect the decay of the system. In this article, we shall design a collocated boundary feedback to obtain the exponential stabilization of the system \eqref{w}. We set \begin{equation} \label{fd} C(t)=-\beta u_t(x,t)-\gamma \partial_{\nu_A}u_t\,, \end{equation} where the constants $\beta$ and $\gamma$ are positive numbers such that $\beta\gamma0$ such that for all $x\in\overline{\Omega}$, for all $X\in M_x$(the tangent space at $x$): \begin{equation} \label{a1} DH(X, X)\equiv \langle D_XH, X\rangle_g \geq\rho |X|^2. \end{equation} \item[(A2)] \begin{equation}\label{a2} H\cdot\nu\leq 0 \quad\text{on } \Gamma_0\,. \end{equation} \end{itemize} \begin{remark} \rm For any Riemannian manifold $M$, the existence of such a vector field $H$ in (A1) has been proved in \cite{Y3}, where some examples are given. See also \cite{Y4}. For the Euclidean metric, taking the vector field $H=x-x_0$ and we have $DH(X,X)=|X|^2$, which means assumption (A1) always holds true with $\rho=1$ for the Euclidean case. \end{remark} \subsection*{Energy of the system \eqref{w}} Before we go to the stabilization of the system, we should first define an energy connected with the natural energy of the hybrid system. We set \[ \eta(x,t)=mu_t(x,t)+\gamma\partial_{\nu_A}u,\ x\in\Gamma_1\,. \] Let $u$ be a regular solution of system \eqref{w1}. Then we associate to system \eqref{w1} the energy functional E(t) as $$ E(t)=\int_{\Omega}\big(u_t^2+|\nabla_gu|_g^2\big)dx +\int_{\Gamma_1}\frac{1}{m-\beta\gamma}\eta^2d\Gamma \,. $$ The main result of this paper reads as follows. \begin{theorem}\label{th12} Let the geometrical assumptions {\rm (A1)} and {\rm (A2)} hold. Then there exist constants $C>0$ and $\omega>0$ such that \begin{equation}\label{re} E(t)\leq Ce^{-\omega t}E(0)\,,\quad t\geq 0\,. \end{equation} \end{theorem} This article is organized as follows. In the next section, we discuss the well-posedness of the nonlinear close-loop system by semigroup theory. Section 3 devotes to the proof of the exponential stability. We construct two auxiliary functions to estimate the energy, thus to obtain the main result finally. \section{Well-posedness of the closed loop system} In this section, we study well-posedness results for system \eqref{w1} using semigroup theory. Denote $H^1_{\Gamma_0}=\{u\in H^1(\Omega)\big| u|_{\Gamma_0}=0\}$. We consider the unknown \[ U=\big(u, v=u_t|_{\Omega}, \eta\big)^T\,, \] in the state space, denoted by \begin{equation} \label{state} {\Upsilon}:= H^1_{\Gamma_0}(\Omega)\times L^2(\Omega)\times L^2(\Gamma_1)\,, \end{equation} with the norm defined by \begin{equation}\label{norm} \|U\|^2=\|(u,v,\eta)^T\|^2 =\int_{\Omega}\big(|\nabla_g u|^2_g+v^2\big)dx +\int_{\Gamma_1}\frac{1}{m-\beta\gamma}\eta^2d\Gamma\,. \end{equation} The closed-loop system \eqref{w1} can be rewritten in the abstract form \begin{equation}\label{abstr} \begin{gathered} U'=\mathcal{A}U,\\ U_0=\big(u_0, u_1, \eta_0\big)^T, \end{gathered} \end{equation} where the operator $\mathcal{A}$ is defined by $$ \mathcal{A}\begin {pmatrix} u\\ v\\ \eta\end{pmatrix} = \begin{pmatrix} v\\ \operatorname{div}\nabla_g u \\ -\frac{1}{\gamma}\eta+(\frac{m}{\gamma}-\beta)v(x,t) \end{pmatrix} $$ with domain \begin{equation} \begin{aligned} D(\mathcal{A}):=\big\{&(u,v,\eta)^T \in H^1(\Omega) \times L^2(\Omega) \times L^2(\Gamma_1): \\ \operatorname{div}\nabla_g u\in L^2(\Omega),\eta=mv|_{\Gamma_1} +\gamma\partial_{\nu_A}u\big\}\,. \end{aligned} \end{equation} We will show that $\mathcal{A}$ generates a $C_0$ semigroup on $\Upsilon$. Now we state the well-posedness result. \begin{theorem} \label{thm2.1} For any initial datum $U_0\in \Upsilon$, there exists a unique solution $U\in C([0,\infty),\Upsilon)$ of system \eqref{abstr}. Moreover, if $U_0\in {D}(\mathcal A)$, then $U\in C([0,\infty),{D}(\mathcal A))\cap C^1([0,\infty),{D}(\mathcal A))$. \end{theorem} \begin{proof} \textbf{Step 1.} We prove that $\mathcal A$ is dissipative. We know $\Upsilon$ is a Hilbert space equipped with the adequate scalar product $\langle\cdot,\cdot\rangle_{\Upsilon}$ and norm $\|U\|$ defined by \eqref{norm}. For $U\in D(\mathcal A)$, a simple computation leads to \begin{equation} \begin{aligned} \langle \mathcal{A}U,U\rangle_{\Upsilon} &= \frac12\frac{d}{dt}\|U\|^2 \\ &= \int_{\Gamma}u_t\partial_{\nu_A}ud\Gamma+\frac{1}{m-\beta\gamma}\int_{\Gamma_1} \eta\eta_td\Gamma \\ &= \int_{\Gamma_1}u_t\partial_{\nu_A}ud\Gamma-\frac{1}{(m-\beta\gamma)\gamma} \int_{\Gamma_1}{\eta}^2 d\Gamma+\int_{\Gamma_1}\frac{1}{\gamma}\eta u_t(x,t)d\Gamma \\ &= \int_{\Gamma_1}u_t\partial_{\nu_A}ud\Gamma-\frac{1}{2m\gamma} \int_{\Gamma_1}\eta^2d\Gamma \\ &\quad -\Big(\frac{1}{(m-\beta\gamma)\gamma} -\frac{1}{2m\gamma}\Big)\int_{\Gamma_1}{\eta}^2 d\Gamma +\int_{\Gamma_1}\frac{1}{\gamma}\eta u_t(x,t)d\Gamma \\ &= -\int_{\Gamma_1}\big(\frac{m}{2\gamma}u^2_t +\frac{\gamma}{2m}\partial^2_{\nu_A}u\big)d\Gamma \\ &\quad-\Big(\frac{1}{(m-\beta\gamma)\gamma}-\frac{1}{2m\gamma}\Big) \int_{\Gamma_1}{\eta}^2 d\Gamma+\int_{\Gamma_1}\frac{1}{\gamma} \eta u_t(x,t)d\Gamma. \end{aligned} \label{den} \end{equation} Now we handle the items in \eqref{den} by applying H\"older inequality \begin{equation} \int_{\Gamma_1}\frac{1}{\gamma}\eta u_t(x,t)d\Gamma \leq \frac{1}{k_1m\gamma}\int_{\Gamma_1}\eta^2d\Gamma+\frac{k_1m}{4\gamma} \int_{\Gamma_1}u_t^2d\Gamma\,,\label{ueta1} \end{equation} where $k_1>0$ is to be chosen later. Substituting the inequality \eqref{ueta1} to \eqref{den} yields \begin{equation}\label{edot} \begin{aligned} \langle \mathcal{A}U,U\rangle_{\Upsilon} &\leq -\int_{\Gamma_1}\frac{\gamma}{2m}\partial^2_{\nu_A}ud\Gamma -\int_{\Gamma_1}\big(\frac{m}{2\gamma} -\frac{k_1m}{4\gamma}\big)u^2_td\Gamma\\ &\quad -\big(\frac{1}{(m-\beta\gamma)\gamma}-\frac{1}{2m\gamma} -\frac1{k_1m\gamma}\big)\int_{\Gamma_1}{\eta}^2 d\Gamma \\ &\leq -\int_{\Gamma_1}\frac{\gamma}{2m}\partial^2_{\nu_A}ud\Gamma -\frac{2\beta\gamma}{m+\beta\gamma}\int_{\Gamma_1}u^2_td\Gamma -\frac{\beta(m+\beta\gamma)}{2m^2(m-\beta\gamma)} \int_{\Gamma_1}{\eta}^2 d\Gamma\,, \end{aligned} \end{equation} by choosing $k_1=\frac{2m}{m+\beta\gamma}$, which is negative noticing that $\beta\gamma0$. Given $(\bar a, \bar b, \bar c)^T\in \Upsilon$, we seek a solution $U=(u, v,\eta)^T\in D(\mathcal A)$ of $$ ({\lambda I-\mathcal A}) \begin{pmatrix} u\\ v\\ \eta\end{pmatrix} =\begin {pmatrix} \bar a\\ \bar b \\ \bar c\end{pmatrix}\,; $$ that is, satisfying \begin{equation}\label{w3} \begin{gathered} \lambda u-v=\bar a, quad \text{in } \Omega,\\ \lambda v-\operatorname{div} A(x)\nabla u=\bar b,\quad \text{in } \Omega,\\ \lambda\eta+\frac{1}{\gamma}\eta+(\beta-\frac{m}{\gamma})v(x,t)=\bar c, \quad \text{on } \Gamma_1,\\ \eta=mv|_{\Gamma_1}+\gamma\partial_{\nu_A}u,\ \text{on }\Gamma_1\,, \end{gathered} \end{equation} where we take $t$ as a parameter for granted. Suppose that we have found $u$ with the appropriate regularity, then from the equation \eqref{w3}-1, \eqref{w3}-3 and \eqref{w3}-4 we have $$ v:=\lambda u-\bar a\,,\ \eta=(\lambda+\frac1\gamma)^{-1} \big(\bar c-(\beta-\frac{m}{\gamma})(\lambda u-\bar a)\big). $$ Now we state the process on how to get $u$. Eliminating $v$ and noticing $\eta=mv|_{\Gamma_1}+\gamma\partial_{\nu_A}u(x,t)$, we find that the function $u$ satisfies \begin{equation}\label{4u} \begin{gathered} \lambda^2 u-\operatorname{div} A(x)\nabla u =\lambda\bar a+\bar b,\quad \text{in } \Omega,\\ u=0, \quad \text{on }\Gamma_0, \\ \partial_{\nu_A}u=\frac{1}{\gamma}\eta-\frac m\gamma (\lambda u-\bar a),\quad \text{on } \Gamma_1. \end{gathered} \end{equation} We obtain a weak formulation of system \eqref{4u} by multiplying the equation by $\psi$ and using Green's formula \begin{equation} \begin{aligned} &\int_{\Omega}(\lambda^2u\psi+\langle A(x)\nabla u,\nabla\psi\rangle)dx +\int_{\Gamma_1}\frac{\lambda(m\lambda+\beta)}{\lambda\gamma+1} u\psi \\ &= \int_{\Omega}(\bar b+\lambda\bar a)\psi dx +\int_{\Gamma_1}(\frac{1}{\lambda\gamma+1}\bar c +\frac{m\lambda+\beta}{\lambda\gamma+1}\bar a)\psi\,, \end{aligned} \label{4u1} \end{equation} for any $\psi\in H^1_{\Gamma_0}(\Omega)=\{\psi\in H^1(\Omega)|\psi|_{\Gamma_0}=0\}$. As the left hand side of \eqref{4u1} is coercive on $H^1(\Omega)$, Lax-Milgram Theorem guarantees the existence and uniqueness of a solution $u\in H^1(\Omega)$ of \eqref{4u}. \smallskip \noindent\textbf{Step 3.} \ Finally, the well-posedness result follows from Lummer-Phillips Theorem. \end{proof} \section{Exponential Stability} In this section, we show the exponential stability of the system by energy perturbed approach. Here we define two auxiliary functions \begin{gather*} V_1(t)=\int_{\Omega}H(u)u_tdx\,,\\ V_2(t)=\frac12\int_{\Omega}(\operatorname{div}_0H-\rho)uu_t. \end{gather*} To estimate the functions, we need some lemmas from \cite{Y4} and \cite{LR}. \begin{lemma}[{\cite[Theorem 2.1]{Y4}}] \label{l1} Suppose that $u(x,t)$ is a solution of the equation $u_{tt}+\operatorname{div}A\nabla u=0$. Then \[ \dot{V}_1(t)= B_1+I_1\,, \] where we denote the boundary term \begin{equation}\label{b1} B_1(\Gamma)=\int_{\Gamma}\partial_{\nu_{A}}uH(u)d\Gamma +\frac12\int_{\Gamma}(u_t^2-|\nabla_gu|_g^2)H\cdot\nu d\Gamma, \end{equation} and the internal term \[ I_1=-\int_{\Omega}D_gH(\nabla_gu,\nabla_gu)dx -\frac12\int_{\Omega}(u_t^2-|\nabla_gu|_g^2)\operatorname{div}_0 H\,dx, \] here $\operatorname{div}_0H$ is the divergence of $H$ in the Euclidean metric of $R^n$. \end{lemma} \begin{lemma}[{\cite[Theorem 2.2]{Y4}}] \label{l2} Suppose that $u(x,t)$ is a solution of the equation $u_{tt}+\operatorname{div}A\nabla u=0$. Then \[ \dot{V}_2(t)= B_2+I_2\,, \] where we denote the boundary term by \[ B_2(\Gamma)=-\frac14\int_{\Gamma}u^2\partial_{\nu_A} (\operatorname{div}_0H)d\Gamma+\frac12\int_{\Gamma} (\operatorname{div}_0H-\rho)u\partial_{\nu_A}ud\Gamma, \] and the internal term \[ I_2=\frac12\int_{\Omega}(\operatorname{div}_0H-\rho) (u_t^2-|\nabla_gu|_g^2)dx +\frac14\int_{\Omega}u^2\operatorname{div}(\partial_{\nu_A}\operatorname{div}_0H)dx\,. \] \end{lemma} \begin{lemma}[{\cite[Lemma 7.2]{LR}}] \label{l3} Let $\varepsilon>0$ be given small. Let $u$ solves the problem \eqref{w}. Then \begin{equation}\label{321} \int_{\varepsilon}^{T-\varepsilon}\int_{\Gamma_1}|\nabla_gu|_g^2d\Gamma dt \leq C_{T,\varepsilon}\big\{\int_0^T\int_{\Gamma_1}\big((\partial_{\nu_A}u)^2 +u_t^2\big)d\Gamma dt+\|u\|_{H^{\frac12+\varepsilon}(\Omega\times(0,T))}\big\}\,. \end{equation} \end{lemma} According to Lemmas \ref{l1}, \ref{l2} and \ref{l3} we obtain: \begin{lemma}\label{l4} Suppose that the geometrical assumptions {\rm (A1)} and {\rm (A2)} hold. Let $u$ solve \eqref{w}. Then there exist constants $c_1,c_2,c_3>0$ such that \begin{gather} \frac{\rho}{2}E(t)+\dot{V}_1(t)+\dot{V}_2(t) \leq \frac{\rho}{2}\int_{\Gamma_1}\frac{1}{m-\beta\gamma} \eta^2d\Gamma+c_1\int_{\Gamma_1}(u_t^2+|\nabla_gu|_g^2)d\Gamma +c_1\int_{\Omega}u^2dx\,,\label{l34}\\ |V_1(t)|\leq c_2E(t),\ |V_2(t)|\leq c_3E(t)\,. \label{l341} \end{gather} \end{lemma} \begin{proof} Obviously the estimate \eqref{l341} is true. Now we prove the inequality \eqref{l34}. First we estimate the boundary terms $B_1$, $B_2$ given in Lemma \ref{l1} and Lemma \ref{l2}. Since $u|_{\Gamma_0}=0$, we have \begin{equation} \nabla_gu=\partial_{\nu_A}u\frac{\nu_A}{|\nu_A|^2_g}\,, \end{equation} which induces \begin{equation} \label{z1} |\nabla_gu|_g^2=|\partial_{\nu_A}u|^2_g\frac{1}{|\nu_A|_g^2}. \end{equation} Similarly we have \begin{equation} \label{z2} H(u)=\langle H, \nabla_gu\rangle_g=\partial_{\nu_A}u\frac{1}{|\nu_A|^2_g} H\cdot\nu. \end{equation} Substituting the equalities \eqref{z1} and \eqref{z2} to \eqref{b1} yields \begin{equation*} B_1(\Gamma_0)=\frac12\int_{\Gamma}(u_t^2-|\nabla_gu|_g^2)H\cdot\nu d\Gamma\leq 0\,, \end{equation*} where we notice the geometrical assumption (A2) is true. It is obvious that \begin{equation} \label{bb1} B_1(\Gamma)=B_1(\Gamma_0)+B_1(\Gamma_1)\leq c_1\int_{\Gamma_1}(u_t^2+|\nabla_gu|_g^2)d\Gamma\,. \end{equation} Since $u|_{\Gamma_0}=0$, we have $B_2(\Gamma_0)=0$. And we have \begin{equation} \label{bb2} B_2(\Gamma)=B_2(\Gamma_0)+B_2(\Gamma_1)\leq c_1\int_{\Gamma_1}|\nabla_gu|_g^2d\Gamma. \end{equation} Next, we estimate the internal terms $I_1$ and $I_2$. By the geometrical assumption (A1), we have \begin{equation} \label{i1} I_1\leq-\rho\int_{\Omega}|\nabla_gu|_g^2dx -\frac12\int_{\Omega}(u_t^2-|\nabla_gu|_g^2)\\operatorname{div}_0H\,dx. \end{equation} It is obvious that \begin{equation} \label{i2} I_2\leq\frac12\int_{\Omega}(\operatorname{div}_0H-\rho)(u_t^2-|\nabla_gu|_g^2)dx +c_1\int_{\Omega}u^2dx. \end{equation} Combining the above inequalities \eqref{bb1}, \eqref{bb2}, \eqref{i1} and \eqref{i2} we complete the proof. \end{proof} The following is the observability inequality for the system \eqref{w}. \begin{lemma} \label{lem3.5} Suppose that the geometrical assumptions {\rm (A1)} and {\rm (A2)} hold. Let $u$ solve problem \eqref{w}. Then for any given $\varepsilon>0$, there exists a time $T_0>0$ and a positive constant $C_{T,\varepsilon,\rho}$ such that \begin{equation}\label{yd} E(0)\leq C_{T,\varepsilon,\rho}\Big\{\int_0^T\int_{\Gamma_1} \big(u_t^2+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt +\|u\|_{H^{1/2+\varepsilon}(\Omega\times(0,T))}\Big\}\,, \end{equation} for all $T>T_0$. \end{lemma} \begin{proof} For any $\varepsilon$ small enough, integrating the inequality \eqref{l34} on the interval $(\varepsilon,T-\varepsilon)$ yields \begin{align*} &\frac{\rho}{2}\int_{\varepsilon}^{T-\varepsilon}E(t)dt +V_1(T-\varepsilon)-V_1(\varepsilon)+V_2(T-\varepsilon)-V_2(\varepsilon) \\ &\leq \frac{\rho}{2}\int_{\varepsilon}^{T-\varepsilon} \int_{\Gamma_1}\frac{1}{m-\beta\gamma}\eta^2d\Gamma +c_1\int_{\varepsilon}^{T-\varepsilon} \int_{\Gamma_1}(u_t^2+|\nabla_gu|_g^2)d\Gamma +c_1\int_{\varepsilon}^{T-\varepsilon}\int_{\Omega}u^2dx\,. \end{align*} Then we use inequality \eqref{321} in Lemma \ref{l3} and inequality \eqref{l341} in Lemma \ref{l4} to obtain \begin{equation} \begin{aligned} \int_{\varepsilon}^{T-\varepsilon}E(t)dt &\leq C_{T,\varepsilon,\rho}\Big\{\int_0^T\int_{\Gamma_1}(\eta^2+u_t^2 +|\partial_{\nu_A}u|_g^2)d\Gamma\\ &\quad +\|u\|_{H^{1/2+\varepsilon}(\Omega\times(0,T))}\Big\}dx +c_0\big(E(T-\varepsilon)+E(\varepsilon)\big)\,, \end{aligned} \label{c1} \end{equation} where the constant $C_{T,\varepsilon,\rho}$ depends on $c_1$, $\rho$, $\frac{1}{m-\beta\gamma}$, $meas(\Omega)$ and the constant $c_0=\frac{4}{\rho}\max\{c_2,c_3\}$. Here $c_2$, $c_3$ are the constants given in inequality \eqref{l341}. We notice that \begin{equation} \begin{aligned} &E(0)+c_0\big(E(T-\varepsilon)+E(\varepsilon)\big) \\ &= \int_{\varepsilon}^{2c_0+\varepsilon+1}E(t)dt +\int_{\varepsilon}^{2c_0+\varepsilon+1}\big(E(0)-E(t)\big)dt\\ &\quad +c_0\big(E(\varepsilon)-E(0)\big)+c_0\big(E(T-\varepsilon)-E(0)\big) \\ &= \int_{\varepsilon}^{2c_0+\varepsilon+1}E(t)dt -\int_{\varepsilon}^{2c_0+\varepsilon+1}\big(\int_0^t\dot{E}(\tau)d\tau\big)dt +c_0\int_0^{\varepsilon}\dot{E}(\tau)d\tau \\ &\quad +c_0\int_{0}^{T-\varepsilon}\dot{E}(\tau)d\tau \\ &\leq \int_{\varepsilon}^{2c_0+\varepsilon+1}E(t)dt +3c_4\int_0^{\max\{T-\varepsilon,2c_0+\varepsilon+1\}} \int_{\Gamma_1}\big(u^2_t+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt\,, \end{aligned} \label{ke} \end{equation} where we used the following inequality known from \eqref{edot}, \[ \dot{E}(t)=\langle \mathcal{A}U,U\rangle_{\Upsilon} \leq c_4\int_{\Gamma_1}\big(u^2_t+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt\,. \] Now we shall take $T_0=2c_0+2\varepsilon+1$ to guarantee that $T-\varepsilon>2c_0+\varepsilon+1$, for all $T>T_0$. Substituting \eqref{ke} in \eqref{c1} completes the proof. \end{proof} In what follows we use the compactness-uniqueness argument to absorb the lower order term in \eqref{yd}. We list the lemma and omit the proof, which could be found in \cite{LLT.1,LT.3,ning1,Y,ZY} and many others. \begin{lemma}\label{l6} Suppose that the geometrical assumptions {\rm (A1)} and {\rm (A2)} hold. Let $u$ solve problem \eqref{w}. Then for any $T>T_0$, there exists a positive constant $C$ depending on $T$, $\varepsilon$, $\rho$, $\operatorname{meas}(\Omega)$ such that \begin{equation*} E(0)\leq C\int_0^T\int_{\Gamma_1}\big(u_t^2+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt\,. \end{equation*} \end{lemma} \begin{proof}[Proof of Theorem \ref{th12}] From \eqref{edot} we know that \begin{equation} \label{edot2} \begin{aligned} -\dot{E}(t) &=-\langle \mathcal{A}U,U\rangle_{\Upsilon}\\ &\geq \int_{\Gamma_1}\frac{\gamma}{2m}\partial^2_{\nu_A}ud\Gamma +\frac{2\beta\gamma}{m+\beta\gamma}\int_{\Gamma_1}u^2_td\Gamma +\frac{\beta(m+\beta\gamma)}{2m^2(m-\beta\gamma)}\int_{\Gamma_1}{\eta}^2 d\Gamma \\ &\geq c_5\int_{\Gamma_1}\big(u_t^2+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt\,, \end{aligned} \end{equation} where \[ c_5=\min\{\frac{\gamma}{2m},\frac{2\beta\gamma}{m+\beta\gamma}, \frac{\beta(m+\beta\gamma)}{2m^2(m-\beta\gamma)}\}. \] By Lemma \ref{l6} and the above inequality \eqref{edot2}, we have, that for all $T>T_0$, \[ E(0) \leq C\int_0^T\int_{\Gamma_1}\big(u_t^2+(\partial_{\nu_A}u)^2+\eta^2\big)d\Gamma dt \leq -\frac{C}{c_5}\int_0^T\dot{E}(t)=-\frac{C}{c_5}(E(T)-E(0))\,, \] which yields \[ % \label{re1} E(T)\leq \frac{{C}-c_5}{{C}} E(0)\,. \] The exponential decay result \eqref{re} follows from the above inequality. \end{proof} \subsection*{Acknowledgments} This research was supported by the National Natural Science Foundation of China under grant nos. 61473126, 61573342 and 11571128. \begin{thebibliography}{00} \bibitem{Ak} K. 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