\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 198, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/198\hfil Stabilization of wave equations]
{Stabilization of wave equations with variable coefficient
and delay in the dynamical boundary feedback}

\author[Dandan Guo, Zhifei Zhang \hfil EJDE-2017/198\hfilneg]
{Dandan Guo, Zhifei Zhang}

\address{Dandan Guo \newline
School of Mathematics and Statistics,
Huazhong University of Science and Technology,
 Wuhan 430074, China.\newline
Hubei Key Laboratory of Engineering Modeling and Scientific Computing,
Huazhong University of Science and Technology, Wuhan 430074, China}
\email{m201570062@hust.edu.cn}

\address{Zhifei Zhang (corresponding author)\newline
School of Mathematics and Statistics,
Huazhong University of Science and Technology,
Wuhan 430074, China.\newline
Hubei Key Laboratory of Engineering Modeling and Scientific Computing,
Huazhong University of Science and Technology, Wuhan 430074, China}
\email{zhangzf@hust.edu.cn}

\dedicatory{Communicated by Goong Chen}

\thanks{Submitted June 13, 2017. Published August 9, 2017.}
\subjclass[2010]{34H05, 35L05, 49K25, 93C20, 93D15}
\keywords{Exponential decay; wave equation with variable coefficient;
\hfill\break\indent  dynamical boundary control; time delay}

\begin{abstract}
 In this article we consider the boundary stabilization of a wave
 equation with variable coefficients. This equation has an acceleration
 term and a delayed velocity term on the boundary.
 Under suitable geometric conditions, we obtain the exponential
 decay for the solutions. Our proof relies on the geometric multiplier
 method and the Lyapunov approach.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\newcommand{\R}{{\mathbb R}}
\newcommand{\C}{{\mathbb C}}
\newcommand{\A}{{\mathcal A}}
\newcommand{\X}{{\mathcal X}}

\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),\;u_t(x,0)={u}_1(x),\quad  \text{in}\ \Omega,
\end{gathered}
\end{equation}
where $\operatorname{div} X$ denote 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 $\mathbb{R}^N$.
$\partial_{\nu_A}u=\sum_{i,j=1}^na_{ij}(x)\partial_{x_j}u\nu_i$, where 
$\nu=(\nu_1,\nu_2,\dots,\nu_n)^T$ denotes the outward unit normal vector of the
boundary and $\nu_A=A\nu$. $C(t)$ is the boundary feedback control.

We call the equation \eqref{w} is with dynamical boundary conditions if 
$m(x)\neq0$, which means the system has an acceleration term on part 
of the boundary. This is what happened in some physical applications 
when one has to take the acceleration terms into account on the boundary. 
Actually we need the models with dynamical boundary conditions, 
see \cite{CDS,FL,MK} and many others. 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,BR,BST,XL}. 
For the above reasons in this article we assume
\begin{equation} \label{al}
 m(x)\in L^{\infty}(\Gamma_1)\quad\text{and}\quad m(x)> \alpha>0,
\end{equation}
where $\alpha$ is a positive constant.
Here we denote $\epsilon_1(x)$ a nonnegative function as
\begin{equation} \label{eps}
\epsilon_1(x) =\frac{m(x)}{\alpha}-1> 0,\quad x\in\Gamma_1.
\end{equation}
On the other hand, time delay effects arise in many practical problems
in science and engineering. Most phenomena
depends on not only the present state but also the history of the
system in a very complicated
way. For instance the practical systems often suffer from the actuator
saturation and sometimes the control input delay.
It is well known that delay effects might turn a well-behaved system
into a wild one by inducing some instabilities, see \cite{Dat1,Freitas,NP1}.
Recently boundary feedbacks are designed to overcome the negative effect
of time delays and stabilize the system; see \cite{ars,LC,litter,ning2,xu}
 and the references therein.

In this article, we discuss the stabilization of wave equations subject to 
dynamical boundary conditions with acceleration terms(i.e. $u_{tt}$) and 
time delayed velocity terms.
We shall design a collocated boundary feedback
with time delay effect in the velocity input to obtain the exponential 
stabilization of the system. That is, $C(t)$ is a feedback law with input time delay
\begin{equation} \label{fd}
C(t)=-\alpha u_t(x,t)-\beta u_t(x,t-\tau)-\partial_{\nu_A}u_t\,,
\end{equation}
where $\tau$ is the time delay, $\alpha$ is the positive constant given
in \eqref{al} and the constant $\beta>0$ denotes the effect of time delay
in the velocity input. $\beta=0$ means the absence of input delay.
For the discussion of stabilization results in the case of absence of delay,
see \cite{chen1,CG,CDS,Lag,LT.3,YL} and our recent paper \cite{zhang1}.
Here in this article we assume that, for $x\in\Gamma_1$, we have
\begin{gather} \label{mu}
0<\frac{\beta}{\alpha}<\sqrt{\frac{\epsilon_1(x)}{1+\epsilon_1(x)}},\\
\label{eps1}
\epsilon_1(x)\in{R^{+}\setminus[\frac{\beta^2}{2\alpha(\alpha+\beta)},
\frac{\beta^2}{2\alpha(\alpha-\beta)}]},
\end{gather}
where $\epsilon_1(x)$ is given in \eqref{eps}.
If $t<\tau$, then $u_t(x,t-\tau)$ is determined by the datum in the past
and we need an initial value in the past.
We thus give the initial condition
\begin{equation} \label{ini}
u_t(x,t-\tau)=f_0(x,t-\tau),\quad (x,t)\in\Gamma_1\times(0,\tau).
\end{equation}

In equation \eqref{w} we adopt the feedback law given in \eqref{fd} 
and the initial datum given in \eqref{ini} to obtain the following closed 
loop system:
\begin{equation}\label{w1}
\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)=-\alpha u_t(x,t)-\beta u_t(x,t-\tau)
 -\partial_{\nu_A}u_t , \quad \text{on } \Gamma_1\times(0,\infty),
\\ 
u(x,0)={u}_0(x),\;u_t(x,0)={u}_1(x),\quad \text{in }  \Omega,
\\ 
u_t(x,t-\tau)=f_0(x,t-\tau),\quad (x,t) \in \Gamma_1\times(0,\tau).
\end{gathered}
\end{equation}

We define
\[
 g=A^{-1}(x),\quad  x\in {\Omega}
\]
as a Riemannian metric on ${\Omega}$ and consider the couple $({\Omega},g)$ as 
a Riemannian manifold. Let $D$ denote the Levi-Civita connection of 
the metric $g$. For each $x\in{\Omega}$, the metric $g$ induces an inner product 
on $R_x^n$ by
\[
\langle X,Y\rangle_g=\langle A^{-1}(x)X,Y\rangle,\quad
 |X|_g^2=\langle X,X\rangle_g\,,\quad  X,Y\in R^n_x\,,
\]
where $\langle\cdot,\cdot\rangle$ denotes the standard metric of the 
Euclidean space $R^n$.

To obtain the stabilization of problem \eqref{w1}, 
the following geometric hypotheses are assumed:

 There exists a vector field $H$ on Riemannian manifold $({\Omega},g)$ such that 
the following properties hold:
\begin{itemize}
\item[(A1)] $DH(\cdot,\cdot)$ is strictly positive definite on
$\overline{\Omega}$: there exists a constant $\kappa>0$ 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\kappa |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} \label{rmk1.1} \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 
 with $\kappa=1$ for the Euclidean case. 
\end{remark}

 Before we go to the stabilization of the system, we should first define 
an energy connected with the natural energy of the hybrid system \eqref{w}.
We set 
\begin{equation}\label{etaa}
\eta(x,t)=m(x)u_t(x,t)+\partial_{\nu_A}u,\ x\in\Gamma_1\,.
\end{equation}
Let $u$ be a regular solution of system \eqref{w1}. 
Then we associate with system \eqref{w1} the energy functional
\begin{equation} \label{ree}
\begin{aligned}
E(t)&=\int_{\Omega}\big(u_t^2+|\nabla_gu|_g^2\big)dx
+\int_{\Gamma_1}\frac{1}{m(x)-\alpha}\eta^2d\Gamma
&\quad +\xi\int_0^1\int_{\Gamma_1}u^2_t(x,t-\rho\tau)d\Gamma d\rho\,,
\end{aligned}
\end{equation}
for some constant $\xi>0$ which will be determined later.

The main result of this paper is the following:

\begin{theorem}\label{th12} 
Let the geometric 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 an appropriate Lyapunov functional to obtain the main result.

\section{Well-posedness of the closed loop system}

In this section, we shall study well-posedness results for system 
\eqref{w1} using semigroup theory.
Let
$$ 
z(x,\rho,t)=u_t(x,t-\rho\tau),\quad  x\in\Gamma_1, \rho\in(0,1),\quad t>0\,.
$$
Then the closed loop system \eqref{w1} is equivalent to the  system
\begin{equation}\label{w2}
\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),\\ 
\tau z_t(x,\rho,t)+z_{\rho}(x,\rho,t)=0,\quad \text{in }
 \Gamma_1\times(0,1)\times(0,\infty),\\ 
\eta_t(x,t)=-\eta+(m-\alpha)u_t(x,t)-\beta u_t(x,t-\tau),\quad \text{in }
 \Gamma_1\times(0,\infty), \\ 
u(x,0)={u}_0(x),\quad u_t(x,0)={u}_1(x),\quad x\in\Omega, \\ 
\eta(x,0)=mu_1(x)+\partial_{\nu_A}u_0(x),\quad \text{on } \Gamma_1\,,\\ 
z(x,0,t)=u_t(x,t),\quad \text{on } \Gamma_1\times(0,\infty),\\ 
z(x,\rho,0)=f_0(x,-\rho\tau),\quad \text{on }  \Gamma_1\times(0,1)\,,
\end{gathered}
\end{equation}
where $\eta$ is given by \eqref{etaa}.
We consider the unknown
$$ 
U=\big(u, w=u_t|_{\Omega}, \eta, z\big)^T\,,
$$ 
in the state space, denoted by
\begin{equation} \label{state}
{\Upsilon}:= H^1(\Omega)\times L^2(\Omega)\times L^2(\Gamma_1)
\times L^2(\Gamma_1\times(0,1))\,,
\end{equation}
 with the norm defined by
\begin{equation}\label{norm}
\begin{aligned}
&\|U\|^2\\
&= \|(u,w,\eta,z)^T\|^2\\
&= \int_{\Omega}\big(\langle A(x)\nabla u,\nabla u\rangle+w^2\big)dx
+\int_{\Gamma_1}\frac{1}{m(x)-\alpha}\eta^2d\Gamma
+ \xi\int_0^1\int_{\Gamma_1}z^2d\rho d\Gamma,
\end{aligned}
\end{equation}
where $\xi>0$ is the constant in \eqref{ree}.

The system \eqref{w2} 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,f_0(\cdot,-\cdot\tau)\big)^T,
\end{gathered}
\end{equation}
where the operator $\mathcal{A}$ is defined by
$$
\mathcal{A}\begin{pmatrix}
u\\ w\\ \eta\\z\end{pmatrix}
=\begin{pmatrix}
 w\\ \operatorname{div} A(x)\nabla u
\\ -\eta+\big(m(x)-\alpha\big)w(x,t)-\beta z(x,1,t)\\ -\tau^{-1}z_{\rho}
\end{pmatrix}
$$
with domain
\begin{equation}
\begin{aligned}
D(\mathcal{A}):=\Big\{&(u,w,\eta,z)^T \in H^1(\Omega) \times L^2(\Omega) 
\times L^2(\Gamma_1)\times L^2({\Gamma}_1;H^1(0,1)): \\
&\operatorname{div}A(x)\nabla u\in L^2(\Omega),\eta=m(x)w|_{\Gamma_1}
+\partial_{\nu_A}u,z(x,0,t)=w(x,t)\\
& \text{on } \Gamma_1\times(0,\infty)\Big\}\,.
\end{aligned}
\end{equation}

We will show that $\mathcal{A}$ generates a $C_0$ semigroup on $\Upsilon$
 under the assumption \eqref{mu}. 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} {\bf 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
 +\int_{\Gamma_1}\frac{1}{m(x)-\alpha}\eta\eta_td\Gamma
 +\xi\int^1_0\int_{\Gamma_1}z(x,\rho,t)z_{t}(x,\rho,t) d\rho d\Gamma\\
&= \int_{\Gamma_1}u_t\partial_{\nu_A}ud\Gamma
 -\int_{\Gamma_1}\frac{1}{m(x)-\alpha}{\eta}^2 d\Gamma
 +\int_{\Gamma_1}\eta u_t(x,t)d\Gamma\\
&\quad -\int_{\Gamma_1}\frac{\beta}{m(x)-\alpha}z(x,1,t)\eta d\Gamma
 +\xi\int^1_0\int_{\Gamma_1}z(x,\rho,t)z_{t}(x,\rho,t) d\rho d\Gamma\\
&= \int_{\Gamma_1}u_t\partial_{\nu_A}ud\Gamma-\int_{\Gamma_1}\frac{1}{2m(x)}
\eta^2d\Gamma\\
&\quad -\int_{\Gamma_1}\big(\frac{1}{m(x)-\alpha}
-\frac{1}{2m(x)}\big){\eta}^2 d\Gamma+\int_{\Gamma_1}\eta u_t(x,t)d\Gamma\\
&\quad-\int_{\Gamma_1}\frac{\beta}{m(x)-\alpha}z(x,1,t)\eta d\Gamma
 +\xi\int^1_0\int_{\Gamma_1}z(x,\rho,t)z_{t}(x,\rho,t) d\rho d\Gamma\\
&= -\int_{\Gamma_1}\big(\frac{m(x)}{2}u^2_t+\frac{1}{2m(x)}
 \partial^2_{\nu_A}u\big)d\Gamma\\
&\quad -\int_{\Gamma_1}\big(\frac{1}{m(x)-\alpha}
 -\frac{1}{2m(x)}\big){\eta}^2 d\Gamma
 +\int_{\Gamma_1}\eta u_t(x,t)d\Gamma\\
&\quad -\int_{\Gamma_1}\frac{\beta}{m(x)-\alpha}z(x,1,t)\eta d\Gamma
 +\frac{\xi}{2\tau}\int_{\Gamma_1}\big(u_t^2(x,t)-z^2(x,1,t)\big)d\Gamma\,,
\end{aligned} \label{den}
\end{equation}
where we used that
\begin{equation}\label{ueta0}
\begin{aligned}
&\xi\int^1_0\int_{\Gamma_1}z(x,\rho,t)z_{t}(x,\rho,t)d\rho d\Gamma\\
&= -\tau^{-1}\xi\int^1_0\int_{\Gamma_1}z_{\rho}(x,\rho,t)z(x,\rho,t)d\rho d\Gamma\\
&=  -\frac12\tau^{-1}\xi\int_{\Gamma_1}\int_0^1\frac{dz^2}{d\rho}d\rho d\Gamma \\ 
&= \frac{\xi}{2\tau}\int_{\Gamma_1}\big(u_t^2-z^2(x,1,t)\big)d\Gamma\,.
\end{aligned}
\end{equation}
Now we handle the items in \eqref{den} by applying H\"older's inequality
\begin{gather}\label{ueta1}
\int_{\Gamma_1}\eta u_t(x,t)d\Gamma
\leq \frac{1}{k_1}\int_{\Gamma_1}\frac{\eta^2}{m(x)}d\Gamma
+\frac{k_1}{4}\int_{\Gamma_1}m(x)u_t^2d\Gamma\,, \\
\label{ueta2}
-\int_{\Gamma_1}\frac{\beta}{m(x)-\alpha}z(x,1,t)\eta d\Gamma
 \leq \int_{\Gamma_1}\frac{\beta}{m(x)-\alpha}
\big(\frac1{2k_2}\eta^2+\frac{k_2}{2}z^2(x,1,t)\big )d\Gamma\,,
\end{gather}
where $k_1>0$ and $k_2>0$ can be chosen later.
Substituting  \eqref{ueta0}, \eqref{ueta1} and \eqref{ueta2}
 in \eqref{den} yields
\begin{equation} \label{pan}
\begin{aligned}
 \langle \mathcal{A}U,U\rangle_{\Upsilon}
&\leq -\int_{\Gamma_1}\big(\frac{m(x)}{2}-\frac{k_1m(x)}{4}
-\frac{\xi}{2\tau}\big)u^2_td\Gamma \\
&\quad -\int_{\Gamma_1}\big(\frac{1}{m(x)-\alpha}(1-\frac{\beta}{2k_2})
-\frac{1}{2m(x)}-\frac1{k_1m(x)}\big){\eta}^2 d\Gamma\\
&\quad +\int_{\Gamma_1}\big(\frac{\beta}{m(x)-\alpha}\frac{k_2}{2}
-\frac{\xi}{2\tau}\big)z^2(x,1,t)d\Gamma
-\int_{\Gamma_1}\frac{1}{2m(x)}\partial^2_{\nu_A}ud\Gamma.
\end{aligned}
\end{equation}

Next we find positive numbers $k_1$, $k_2$ and $\xi$ that
 guarantee the negativity of $\langle \mathcal{A}U,U\rangle_{\Upsilon}$. 
That is, we need, for all $x\in\Gamma_1$,	
\begin{gather}\label{11}
  \frac{m(x)}{2}- \frac{k_1m(x)}{4}- \frac{\xi}{2\tau}>0\,, \\
\label{12}
   \frac{1}{m(x)-\alpha}(1-\frac{\beta}{2k_2})-\frac{1}{2m}-\frac{1}{k_1m(x)}>0\,,\\
\label{13}
   \frac{\beta}{m(x)-\alpha}(\frac{k_2}{2})-\frac{\xi}{2\tau}<0\,.
\end{gather}
Inequalities \eqref{11} and \eqref{13} imply
\begin{equation}
  \frac{\beta}{m(x)-\alpha}(\frac{k_2}{2})<\frac{m(x)}{2}- \frac{k_1m(x)}{4}\,,
\end{equation}
that is,
\begin{equation} \label{k21}
  k_2<\frac{(2-k_1)m(x)}{2}\frac{m(x)-\alpha}{\beta}\,.
\end{equation}
We can easily show that  inequality \eqref{12} is equivalent to
\begin{equation}
    {k_2}(\frac{1}{m(x)-\alpha}-\frac{k_1+2}{2m(x)k_1})
-\frac{\beta}{2(m(x)-\alpha)}>0\,,
\end{equation}
from which we find that
\begin{equation}
  2m(x)k_1-(k_1+2)\big(m(x)-\alpha\big)>0\,,
\end{equation}
that is,
\begin{equation}\label{k11}
  k_1>2\frac{m(x)-\alpha}{m(x)+\alpha}\,,
\end{equation}
and
\begin{equation}\label{k22}
  k_2>\frac{\beta m(x)k_1}{2m(x)k_1-(k_1+2)\big(m(x)-\alpha\big)}\,,
\end{equation}
for all $x\in\Gamma_1$.

Now we can determine the constants $k_1$ and $k_2$ according to 
 inequalities \eqref{k21}, \eqref{k11} and \eqref{k22}.

Firstly we aim to get $k_1$.
The inequalties \eqref{k21} and \eqref{k22} yield
\[
  \frac{\beta m(x)k_1}{2m(x)k_1-(k_1+2)\big(m(x)-\alpha\big)}
<\frac{(2-k_1)m(x)\big(m(x)-\alpha\big)}{2\beta}\,,
\]
that is,
\begin{align*}
&k_1^2\Big(m(x)\big(m(x)-\alpha\big)\big(m(x)+\alpha\big)\Big)
+k_1(2m(x))\Big(\beta^2-2m(x)\big(m(x)-\alpha\big)\Big)\\
&+4m(x)\big(m(x)-\alpha\big)^2<0\,.
\end{align*}
For the quadratic form of $k_1$, we compute the discriminant
\begin{align*}
  \Delta(x)
&= 4m^2\big({\beta}^2-2m(m-\alpha)\big)^2-16m^2(m-\alpha)^2(m^2-{\alpha}^2)\\
&= 4m^2\big({\beta}^4+4(m-\alpha)(m\alpha^2-\alpha^3-m{\beta}^2)\big)>0\,,
\end{align*}
for all $x\in\Gamma_1$, where we used assumption \eqref{mu}.
Thus we get a possible candidate 
\begin{equation}\label{k_1}
  k_1=\min_{x\in\Gamma_1}\big\{\frac{2m(x)}{m(x)+\alpha}
-\frac{\beta^2}{m(x)^2-\alpha^2}\big\}\,.
\end{equation}
It's easy to verify that $0<k_1<2$ and inequality \eqref{k11} holds.

Secondly we should determine the constant $k_2$.
Now  inequalities \eqref{k21} and \eqref{k22} become
\begin{equation}
k_2<\frac{m(x)\big(m(x)-\alpha\big)}{2\beta}(2-k_1)=\frac{2m(x)\alpha\big(m(x)-\alpha\big)+m(x)\beta^2}{2\beta\big(m(x)+\alpha\big)}\,,
\end{equation}
and
 \[
   k_2>\frac{\beta m(x)k_1}{2m(x)k_1-(k_1+2)\big(m(x)-\alpha\big)}
=\frac{\beta m(x)}{m(x)+\alpha}\frac{2m(x)
\big(m(x)-\alpha\big)-\beta^2}{2\alpha\big(m(x)-\alpha\big)-\beta^2}\,,
 \]
that is,
 \begin{equation}\label{k20}
  \frac{\beta m(x)}{m(x)+\alpha}\frac{2m(x)\big(m(x)-\alpha\big)
-\beta^2}{2\alpha\big(m(x)-\alpha\big)-\beta^2}< k_2
<\frac{2m(x)\alpha\big(m(x)-\alpha\big)+m(x)\beta^2}{2\beta\big(m(x)+\alpha\big)}\,.
 \end{equation}
To obtain $k_2$ we only have to verify that
 \begin{align*}
 &\frac{2m(x)\alpha\big(m(x)-\alpha\big)+m(x)\beta^2}{2\beta\big(m(x)+\alpha\big)}
-\frac{\beta m(x)}{m(x)+\alpha}\frac{2m(x)\big(m(x)-\alpha\big)
 -\beta^2}{2\alpha\big(m(x)-\alpha\big)-\beta^2}\\
&= \Big(\big(2m(x)\alpha\big(m(x)-\alpha\big)+m(x)\beta^2\big)\big(2\alpha\big(m(x)
 -\alpha\big)-\beta^2\big)\\
&\quad -2\beta^2m(x)\Big(2m(x)\big(m(x)-\alpha\big)-\beta^2\big)\Big)
\big/ \Big(2\beta\big(m(x)+\alpha\big)\big(2\alpha\big(m(x)-\alpha\big)
-\beta^2\big)\Big)\\
& >0\,.
\end{align*}
In fact, we have
\begin{align*}
& \big(2m\alpha(m-\alpha)+m\beta^2\big)\big(2\alpha(m-\alpha)-\beta^2\big)
 -2\beta^2m\big(2m(m-\alpha)-\beta^2\big)\\
&=4m^3\alpha^2+4m\alpha^4-8m^2\alpha^3+m\beta^4-4\beta^2m^3+4\beta^2m^2\alpha\\
&=(1+\epsilon_1)\big(4\alpha^3\epsilon_1^2(\alpha^2-\beta^2)-4\alpha^3
 \beta^2\epsilon_1+\alpha\beta^4\big)>0\,,
\end{align*}
where we used assumption \eqref{eps1}.

Thus by \eqref{k20} we can take
\begin{equation}\label{k2}
\begin{aligned}
  k_2&= \frac{1}{2}\Big(\min_{x\in\Gamma_1}\frac{2m(x)\alpha\big(m(x)
-\alpha\big)+m(x)\beta^2}{2\beta\big(m(x)+\alpha\big)} \\
&\quad +\max_{x\in\Gamma_1}\frac{\beta m(x)}{m(x)+\alpha}
 \frac{2m(x)\big(m(x)-\alpha\big)-\beta^2}{2\alpha\big(m(x)-\alpha\big)-\beta^2}
 \Big)\,.
\end{aligned}
\end{equation}
Finally, we can take a constant $\xi$ to satisfy
\[
  \frac{\beta k_2}{m-\alpha}<\frac{\xi}{\tau}<\frac{m(2-k_1)}{2}\,,
\]
after $k_1$ and $k_2$ are given in \eqref{k_1} and \eqref{k2}.
Thus we show that the operator $\mathcal{A}$ is dissipative.
\smallskip


\noindent\textbf{Step 2.}
 We will show that $\lambda I-\mathcal{A}$ is surjective for a fixed $\lambda>0$.
Given $(\bar a, \bar b, \bar c,\bar f)^T\in \Upsilon$, we seek a solution 
$U=(u, w,\eta,z)^T\in D(\mathcal A)$ of
$$
({\lambda I-\mathcal A})\begin{pmatrix}
u\\ w\\ \eta\\z\end{pmatrix}
=\begin{pmatrix} \bar a\\ \bar b
 \\ \bar c \\ \bar f\end{pmatrix}
$$
that is, satisfying
\begin{equation}\label{w3}
\begin{gathered} 
\lambda u-w=\bar a,\quad \text{in } \Omega,\\
\lambda w-\operatorname{div} A(x)\nabla u=\bar b,\quad \text{in } \Omega, \\
\lambda\eta+\eta-(m-\alpha) w(x,t)+\beta z(x,1,t)=\bar c,\quad \text{on }  \Gamma_1, \\
\lambda z+\tau^{-1}z_{\rho}=\bar f, \quad \text{on } \Gamma_1\times(0,1), \\
\eta=mw|_{\Gamma_1}+\partial_{\nu_A}u(x,t),z(x,0,t)=w(x),z(x,\rho,0)
=f_0(x,-\rho\tau)\quad \text{on } \Gamma_1.
\end{gathered}
\end{equation}
where we take $t$ as a parameter.
Suppose that we have found $u$ with the appropriate regularity, 
then from  equation \eqref{w3} we have
$ w:=\lambda u-\bar a$.
Therefore we have the initial value problem for $z$,
\begin{gather*} 
\lambda z+\tau^{-1}z_{\rho}=\bar f, \\
z(x,0)=w(x)\quad \text{on } \Gamma_1\,.
\end{gather*}
We can easily see that
$$ 
z(x,\rho)=w(x)e^{-\lambda\rho\tau}
+\tau e^{-\lambda\rho\tau}\int_0^{\rho}\bar f(x,\theta)
e^{\lambda\theta\tau}d\theta\,.
$$
In particular we have
\begin{equation} 
z(x,1)= (\lambda u-\bar a)e^{-\lambda\tau}
 +\tau e^{-\lambda\tau}\int_0^{1}\bar f(x,\theta)e^{\lambda\theta\tau}d\theta
=: \lambda u e^{-\lambda\tau}+z_0\,,  \label{z1}
\end{equation}
where $z_0=-\bar ae^{-\lambda\tau}+\tau e^{-\lambda\tau}
\int_0^{1}\bar f(x,\rho)e^{\lambda\rho\tau}d\rho$.
So once we know~$u$,we can get~$w=\lambda u-\bar a$,and we already know
$z(x,1)$, so we obtain
\begin{equation}
  \eta=\frac{\bar c}{\lambda+1}+\frac{m-\alpha}{\lambda+1}w
-\frac{\beta}{\lambda+1}z(x,1)
\end{equation}
from \eqref{w3}(3).
Now we lay the emphasis on how to get $u$.
Eliminating $w$  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,\ \text{in}\ \Omega
\\ u=0, \quad \text{on } \Gamma_0 \\ 
\partial_{\nu_A}u = \frac{\bar c}{\lambda+1}-\frac{\lambda m+\alpha}{\lambda+1}
  (\lambda u-\bar a)-\frac{\beta}{\lambda+1}z(x,1),\quad \text{in } \Gamma_1
\end{gathered}
\end{equation}
where $z(x,1)$ is given in \eqref{z1}.
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}u\psi\frac{\lambda(\lambda m+\alpha)}{\lambda+1}d\Gamma
 +\int_{\Gamma_1}u\psi\frac{\lambda\beta e^{-\lambda\tau}}{\lambda+1}d\Gamma\\
&=\int_{{\Omega}}(\lambda\bar a+\bar b)\psi dx+\int_{\Gamma_1}\psi
 \frac{\bar c}{\lambda+1}d\Gamma+\int_{\Gamma_1}\psi
 \frac{\bar a(\lambda m+\alpha)}{\lambda+1}d\Gamma
 -\int_{\Gamma_1}\psi\frac{\beta z_0}{\lambda+1}d\Gamma\,,
\end{aligned}\label{4u1}
\end{equation}
for any $\psi\in H^1({\Omega})$, where $z_0$ is given in \eqref{z1}.
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{Energy decay}

In this section, we obtain the exponential stability of the system by 
energy perturbed approach. At first we rewrite the energy $E(t)$ defined 
in \eqref{ree} as
\[
E(t)=\mathcal{E}(t)+E_{\tau}(t)\,,
\]
where $\mathcal{E}(t)$ is the standard energy regardless of the time delay
\begin{gather*} 
\mathcal{E}(t)=\int_{\Omega}\big(u_t^2+|\nabla_gu|_g^2\big)dx
+\int_{\Gamma_1}\frac{1}{m(x)-\alpha}\eta^2d\Gamma\,,\\
 E_{\tau}(t)=\int_0^1\int_{\Gamma_1}u^2_t(x,t-\rho\tau)d\Gamma d\rho\,.
\end{gather*}
To obtain the estimate of the energy $E(t)$ we define three auxiliary functions:
\begin{gather*}
V_1(t)= \int_{\Omega}H(u)u_tdx\,,\\
  V_2(t)= \frac{1}{2}\int_\Omega(\operatorname{div}H-\kappa) uu_tdx\,,\\
   V_3(t)= \xi\int_{\Gamma_1}\int_0^1e^{-2\tau\rho}u^2_t(x,t-\rho\tau)d\rho d\Gamma\,.
\end{gather*}

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 
$u_{tt}-\operatorname{div}A\nabla u=0$. 
We have
$\dot{V}_1(t)= B_1+I_1$,
where  the boundary term is
\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
\[
 I_1=-\int_{\Omega}D_gH(\nabla_gu,\nabla_gu)dx
-\frac12\int_{\Omega}(u_t^2-|\nabla_gu|_g^2)\operatorname{div} Hdx\,.
\]
\end{lemma}

\begin{lemma}[{\cite[Theorem 2.2.]{Y4}}] \label{l22}
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 the boundary term is
\[
  B_2(\Gamma)=-\frac14\int_{\Gamma}u^2\partial_{\nu_A}(\operatorname{div} H)d\Gamma
+\frac12\int_{\Gamma}(\operatorname{div} H
-\kappa)u\partial_{\nu_A}ud\Gamma,
\]
 and the internal term
\[
 I_2=\frac12\int_{\Omega}(\operatorname{div} H-\kappa)(u_t^2
-|\nabla_gu|_g^2)dx+\frac14\int_{\Omega}u^2\operatorname{div}A(x)
\nabla (\operatorname{div}H)dx\,.
\]
\end{lemma}

\begin{lemma}[{\cite[Prosition 3.1.]{NP1}}] \label{l5}
Let $u$ solve  problem \eqref{w1}. We have
\begin{equation}
\dot{V_3}(t)=\frac{\xi}{\tau}\int_{\Gamma_1}u_t^2d\Gamma
-\frac{\xi}{\tau}\int_{\Gamma_1}e^{-2\tau}z^2(x,1,t)d\Gamma
-2\xi\int_{\Gamma_1}\int_0^1e^{-2\rho\tau}z^2d\rho d\Gamma\,.
\end{equation}
\end{lemma}

From Lemmas \ref{l1},  \ref{l22} and  \ref{l5}, we obtain the following result.

\begin{lemma}\label{l4} 
Suppose that the geometric assumptions {\rm (A1)} and {\rm (A2)} hold. 
Let $u$ solves  problem \eqref{w1}. There exist constants 
$c_1,c_2,c_3,c_4,c_5>0$ such that
\begin{gather} \label{l34}
\begin{aligned}
&\frac{\kappa}{2}{E}(t)+\dot{V}_1(t)+\dot{V}_2(t)
 +\frac{\kappa}{4\xi}e^{2\tau}\dot{V}_3(t)\\
&\leq \frac{\kappa}{2}\int_{\Gamma_1}\frac{1}{m(x)-\alpha}\eta^2d\Gamma
 +c_1\int_{\Gamma_1}(u_t^2+|\nabla_gu|_g^2)d\Gamma+c_2\int_{{\Omega}}u^2dx\,,
\end{aligned}\\
|V_1(t)| \leq c_3\mathcal{E}(t),\quad |V_2(t)|\leq c_4\mathcal{E}(t), \quad
 |V_3(t)|\leq c_5E_{\tau}(t) \label{l341}\,.
\end{gather}
\end{lemma}

\begin{proof}
 Obviously the estimate \eqref{l341} is true. Now we prove inequality 
\eqref{l34}. First we estimate the boundary terms $B_1$, $B_2$ given 
in Lemma \ref{l1} and Lemma \ref{l22}. Since $u|_{\Gamma_0}=0$, we have on 
$\Gamma_0$ that
\[ 
\nabla_gu=\partial_{\nu_A}u\frac{\nu_A}{|\nu_A|^2_g}\,,
\]
which implies
\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 equality \eqref{z2} in \eqref{b1} yields
\begin{equation*}
B_1(\Gamma_0)=\frac12\int_{\Gamma_0}(u_t^2+|\nabla_gu|_g^2)H\cdot\nu d\Gamma
\leq 0\,,
\end{equation*}
where we noticed the geometric assumption (A2) is true.
It is obvious that
\begin{equation} \label{bb1}
B_1(\Gamma)=B_1(\Gamma_0)+B_1(\Gamma_1)\leq \frac{c_1}4\int_{\Gamma_1}(u_t^2
+|\nabla_gu|_g^2)d\Gamma\,.
\end{equation}
Since $u|_{\Gamma_0}=0$, we have $B_2(\Gamma_0)=0$. Also  we have
\begin{equation} \label{bb2}
B_2(\Gamma)=B_2(\Gamma_0)+B_2(\Gamma_1)\leq \frac{c_1}4\int_{\Gamma_1}|\nabla_gu|_g^2d\Gamma.
\end{equation}
Next, we estimate the internal terms $I_1$ and $I_2$.
From to the geometric assumption (A1), we have
\begin{equation} \label{i1}
 I_1\leq-\kappa\int_{\Omega}|\nabla_gu|_g^2dx
-\frac12\int_{\Omega}(u_t^2-|\nabla_gu|_g^2)\operatorname{div} Hdx.
\end{equation}
It is obvious that
\begin{equation}
I_1+I_2\label{i2}\leq-\frac{\kappa}2\int_{\Omega}(u_t^2+|\nabla_gu|_g^2)dx
+c_2\int_{\Omega}u^2dx.
\end{equation}
Combining  inequalities \eqref{bb1}, \eqref{bb2}, \eqref{i1}
and \eqref{i2} we obtain
\begin{equation} \label{zm1}
\begin{aligned}
&\frac{\kappa}{2}\mathcal{E}(t)+\dot{V}_1(t)+\dot{V}_2(t) \\
&\leq \frac{\kappa}{2}\int_{\Gamma_1}\frac{1}{m-\alpha}\eta^2d\Gamma
+\frac{c_1}2\int_{\Gamma_1}(u_t^2+|\nabla_gu|_g^2)d\Gamma+c_2\int_{{\Omega}}u^2dx\,.
\end{aligned}
\end{equation}
Using Lemma \ref{l5} we know that
\begin{equation}\label{zm2}
\begin{aligned}
&\frac{\kappa}{2}E_{\tau}(t)+\frac{\kappa}{4\xi}e^{2\tau}\dot{V}_3(t) \\
&\leq \frac{\kappa}{2}E_{\tau}(t)+\frac{\kappa}{4\tau}e^{2\tau}
\int_{\Gamma_1}u_t^2d\Gamma-\frac{\kappa}{2}\int_{\Gamma_1}
\int_0^1e^{2\tau-2\rho\tau}z^2d\rho d\Gamma\\
&\leq \frac{\kappa}{4\tau}e^{2\tau}\int_{\Gamma_1}u_t^2d\Gamma\,.
\end{aligned}
\end{equation}
Combining inequalities \eqref{zm1} and \eqref{zm2} we complete the proof.
\end{proof}

To elinate the tangential part of the derivative $\nabla_g u$ we need 
the following lemma from \cite{LR}.

\begin{lemma}p\cite[Lemma 7.2.]{LR}] \label{l3}
 Let $\varepsilon>0$ be given small. Let $u$ solves the problem \eqref{w1}. Then
\begin{equation}\label{321}
\begin{aligned}
&\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{aligned}
\end{equation}
\end{lemma}

The following is the observability inequality for the system \eqref{w1}.

\begin{lemma} 
Suppose that the geometric assumptions {\rm (A1)} and {\rm (A2)} hold. 
Let $u$ solve problem \eqref{w1}. 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}
\begin{aligned}
E(0)&\leq C_{T,\varepsilon,\rho}\Big\{\int_0^T\int_{\Gamma_1}
\big(u_t^2+(\partial_{\nu_A}u)^2+z^2(x,1,t)+\eta^2\big)d\Gamma dt\\
&\quad +\|u\|_{H^{1/2+\varepsilon}(\Omega\times(0,T))}\Big\}\,,
\end{aligned}
\end{equation}
for all $T>T_0$.
\end{lemma}

\begin{proof}
For a given $\varepsilon$ small enough, integrating inequality \eqref{l34} 
on the interval $(\varepsilon,T-\varepsilon)$ yields
\begin{align*}
&\frac{\kappa}{2}\int_{\varepsilon}^{T-\varepsilon}E(t)dt
 +V_1(T-\varepsilon)-V_1(\varepsilon)
 +V_2(T-\varepsilon)-V_2(\varepsilon) \\
&+\frac{\kappa e^{2\tau}}{4\xi}V_3(T-\varepsilon)
 -\frac{\kappa e^{2\tau}}{4\xi}V_3(\varepsilon) \\
&\leq \frac{\kappa}{2}\int_{\varepsilon}^{T-\varepsilon}
 \int_{\Gamma_1}\frac{1}{m-\alpha}\eta^2d\Gamma
 +c_1\int_{\varepsilon}^{T-\varepsilon}\int_{\Gamma_1}(u_t^2
 +|\nabla_gu|_g^2)d\Gamma 
+c_2\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 $C_{T,\varepsilon,\kappa}$ depends on $c_1$,
 $\kappa$, $\frac{1}{m-\alpha}$, $\operatorname{meas}(\Omega)$ 
and the constant $C_{T,\varepsilon}$ given in inequality \eqref{321}.

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 \\
&\quad +c\int_0^{{2c_0+\varepsilon+1}}\int_{\Gamma_1}\big(u^2_t
+(\partial_{\nu_A}u)^2+z^2(x,1,t)+\eta^2\big)d\Gamma dt\,,
\end{aligned}\label{ke}
\end{equation}
where we used $\dot{E}(t)\leq 0$ and the fact which is known from \eqref{den} 
that 
\begin{equation}
 -\dot{E}(t)=-\langle \mathcal{A}U,U\rangle_{\Upsilon}
\leq c\int_{\Gamma_1}\big(u^2_t+(\partial_{\nu_A}u)^2+z^2(x,1,t)
+\eta^2\big)d\Gamma dt\,.
\end{equation}

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} to inequality \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{LT.3,ning2,Y,ZY} and many others.

\begin{lemma}\label{l6} 
Suppose that the geometric 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+z^2(x,1,t)
+\eta^2\big)d\Gamma dt\,.
\end{equation*}
\end{lemma}

\begin{proof}[Proof of Theorem \ref{th12}]
 From \eqref{pan}, we have
 \begin{equation}\label{E1}
   \dot{E}\leq-C_1\int_{\Gamma_1}[u_t^2+\eta^2+z(x,1,t)^2+\partial^2_{\nu_A}u]dx\,,
 \end{equation}
 where we denote
\begin{align*}
C_1&=\min\Big((\frac{m}{2}-\frac{k_1m}{4}-\frac{\xi}{2\tau}\big),
(\frac{1}{m-\alpha}(1-\frac{\beta}{2k_2})
-\frac{1}{2m}-\frac1{k_1m}\big) , \\
&\quad -(\frac{\beta}{m-\alpha}\frac{k_2}{2}-\frac{\xi}{2\tau})\Big)\,.
\end{align*}
Thus from Lemma \ref{l6} we have that for all $T>T_0$,
\begin{align*}
   E(0)&\leq C\int_0^T\int_{\Gamma_1}\big(\eta^2+\partial_\nu A^2u+u_t^2+z^2(x,1,t)
 \big)d\Gamma dt \\  
&\leq-\frac{C}{C_1}\int_0^T\dot{E}dt
   =-\frac{C}{C_1}(E(T)-E(0))\,;
\end{align*}
that is, 
\[
 E(T)\leq\frac{C-C_1}{C}E(0)\,,
\]
from which the exponential decay result follows.
\end{proof}

\subsection*{Acknowledgements}
This work was supported by the National Natural Science
Foundation of China, Grants No. 61473126
and by the Fundamental Research Funds for the Central Universities.


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