\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 23, pp. 1--21.\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-2016/23\hfil Well-posedness and decay of solutions]
{Well-posedness and  decay of solutions for a
transmission problem with history and delay}

\author[G. Li, D. Wang, B. Zhu \hfil EJDE-2016/23\hfilneg]
{Gang Li, Danhua Wang, Biqing Zhu}

\address{Gang Li \newline
College of Mathematics and Statistics,
Nanjing University of Information Science and Technology,
 Nanjing 210044, China}
\email{ligang@nuist.edu.cn}

\address{Danhua Wang \newline
College of Mathematics and Statistics,
Nanjing University of Information Science and Technology,
Nanjing 210044, China}
\email{matdhwang@yeah.net}

\address{Biqing Zhu \newline
College of Mathematics and Statistics,
Nanjing University of Information Science and Technology,
Nanjing 210044, China}
\email{brucechu@163.com}

\thanks{Submitted March 20, 2015. Published January 13, 2016.}
\subjclass[2010]{35B37, 35L55, 74D05, 93D15, 93D20}
\keywords{Wave equation; transmission problem; past history; delay term}

\begin{abstract}
 In this article, we consider a transmission problem in the presence of
 history and delay terms. Under appropriate hypothesis on the relaxation
 function and the relationship between the weight of the damping and
 the weight of the delay, we prove well-posedness by using the semigroup
 theory. Also we establish a decay result by introducing a suitable 
 Lyaponov functional.
\end{abstract}

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

\section{Introduction}\label{s1}

In this article,  we study the  following transmission system with
a past history and a delay term
\begin{equation}
\begin{gathered}
\begin{aligned}
&u_{tt}(x,t)-au_{xx}(x,t)+\int_0^{\infty}g(s)u_{xx}(x,t-s){\rm d}s\\
&+\mu_1u_{t}(x,t)+\mu_2u_{t}(x,t-\tau)=0, \quad(x,t)\in\Omega\times(0,+\infty),
\end{aligned}\\
 v_{tt}(x,t)-bv_{xx}(x,t)=0,\quad (x,t)\in(L_1,L_2)\times(0,+\infty),
\end{gathered}\label{1.1}
\end{equation}
under the boundary and transmission conditions
\begin{equation}
\begin{gathered}
u(0,t)=u(L_3,t)=0,\\
u(L_{i},t)=v(L_{i},t),\quad i=1,2,\\
au_{x}(L_{i},t)-\int_0^{\infty}g(s)u_{x}(L_{i},t-s){\rm d}s=bv_{x}(L_{i},t),\quad
i=1,2,
\end{gathered}\label{1.2}
\end{equation}
and the initial conditions
\begin{equation}
\begin{gathered}
u(x,-t)=u_0(x,t),\quad u_{t}(x,0)=u_1(x), \quad x\in\Omega,\\
u_{t}(x,t-\tau)=f_0(x,t-\tau), \quad x\in\Omega,\; t\in(0,\tau),\\
v(x,0)=v_0(x),\quad v_{t}(x,0)=v_1(x), \quad x\in(L_1,L_2),
\end{gathered}\label{1.3}
\end{equation}
where $0<L_1<L_2<L_3$, $\Omega=(0,L_1)\cup(L_2,L_3)$, $a$, $b$, $\mu_1$,
 $\mu_2$ are positive constants, $u_0$ is given history,
and $\tau>0$ represents the delay.

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.7\textwidth]{fig1}
\caption{The configuration for problem \eqref{1.1}--\eqref{1.3}.} \label{figure1}
\end{center}
\end{figure}


Transmission problems arise in several applications of physics and biology.
We note that problem \eqref{1.1}-\eqref{1.3} is related to the wave
propagation over a body which consists of two different type of materials:
the elastic part and the viscoelastic part that has the past history and
time delay effect.

For  wave equations with various dissipations, many results concerning
stabilization of solutions have been proved.
Recently, wave equations with viscoelastic damping have been investigated
by many authors, see
\cite{bm2006,c2001,cds2002,l2014,lc2014,ls2014,tp2013,w2013,w20132}
and the references therein.
It is showed that the dissipation produced by the viscoelastic part
can produce the decay of the solution. For example,
  Cavalcanti et al \cite{cds2002} studied the  equation
$$
u_{tt}-\Delta u+\int_0^{t}g(t-\tau)\Delta u(\tau){\rm d}\tau
+a(x)u_{t}+|u|^{\gamma}u=0,\quad \text{in } \Omega\times(0,\infty),
$$
where $a: \Omega\to \mathbb{R}_{+}$. Under the conditions
that $a(x)\geq a_0>0$ on $\omega\subset\Omega$, with $\omega$
satisfying some geometry restrictions and
$$
-\xi_1g(t)\leq g'(t)\leq-\xi_2g(t),\quad t\geq 0,
$$
the authors showed the exponential decay.
 Berrimi and Messaoudi \cite{bm2006}  considered the equation
$$
u_{tt}-\Delta u+\int_0^{t}g(t-\tau)\Delta u(\tau){\rm d}\tau
=|u|^{\gamma}u,\quad \text{in } \Omega\times(0,\infty)
$$
with only the viscoelastic dissipation and proved
that the solution energy  decays
exponentially or polynomially depending on the rate of the decay of
the relaxation function $g$.
Guesmia \cite{g2011}  considered  the  asymptotic behavior of solutions
to an abstract linear dissipative integrodifferential equation with infinite
memory (past history) and introduced a new approach which allows a larger
class of past-history kernels and consequently a more general decay result
for a class of hyperbolic problems with past history is obtained.
 For other past (infinite) history problems, see
\cite{amq2013,gm2012, mma2013,qr2013,q2014} and the references therein.

In recent years, the control of PDEs with time delay effects has become an
active area of research. The presence of delay may be a source of instability.
For example, it was proved in \cite{dl1986,l20132,np2006} that an arbitrarily
small delay may destabilize a system which is uniformly asymptotically stable
in the absence of delay unless additional conditions or control terms were used.
Kirane and Said-Houari \cite{ks2011} considered the viscoelastic wave equation
 with a delay
\begin{align*}
&u_{tt}(x,t)-\Delta u(x,t)+\int_0^{t}g(t-s)\Delta u(x,t-s){\rm d}s\\
&+\mu_1u_{t}(x,t)+\mu_2u_{t}(x,t-\tau)=0,\quad \text{in } \Omega\times(0,\infty),
\end{align*}
where $\mu_1$ and $\mu_2$ are positive constants.
They established an energy decay result under the condition
that $0\le \mu_2\le \mu_1$.
Later, Liu \cite{l2013} improved this result by considering the equation
with a time-varying delay term, with not necessarily positive coefficient
$\mu_2$ of the delay term.

Transmission  problems
related to \eqref{1.1}-\eqref{1.3} have also been extensively studied.
Bastos and Raposo \cite{br2007}
investigated the transmission problem with frictional damping
and  showed the well-posedness and exponential stability of the total energy.
 Mu\~noz Rivera and Portillo
Oquendo \cite{mp2000} considered the transmission problem of viscoelastic waves
and proved that    the dissipation produced by the viscoelastic part
can produce exponential decay of the solution, no matter how small its size is.
Bae \cite{b2010} studied the transmission problem,
which one component is clamped and the other is in a viscoelastic
fluid producing a dissipative mechanism on the boundary, and established a
decay result which depends on the rate of the decay of the relaxation function.
The effect of the delay in the stability of transmission system \eqref{1.1}
in the absence of the past history term has been investigated
by Benseghir in \cite{b2014}. In \cite{wlz2016}, the present authors studied
the well-posedness and  decay of solution for a
transmission problem in a bounded domain with a viscoelastic term
$\int_0^{t}g(t-s)u_{xx}(x,s){\rm d}s$ and
a delay term.

Motivated by the above results, we intend to study in this paper the
well-posedness and the decay result of
problem \eqref{1.1}-\eqref{1.3},  in which the infinite memory (past
history) term $\int_0^{\infty}g(s)u_{xx}(x,t-s){\rm d}s$ is involved.
The main difficulty we encounter here arises from
the simultaneous appearance of the past history and the delay term.
We need also pay more attention to the influence of the transmission boundary.
To attain our goal, we use the semigroup theory to prove the well-posedness,
and introduce a suitable Lyaponov functional to establish the  decay result.

This article is organized as follows. In Section \ref{s2},
 we give some materials needed for our work and state our main results.
In Section \ref{s3},we prove the well-posedness of the
 problem. The decay result is proved in Section
\ref{s4}.

\section{Preliminaries and statement of main results}\label{s2}

In this section, we present some materials that shall be used for
proving our main results. For the relaxation function $g$, we have the folloing
assumptions:
\begin{itemize}
\item[(A1)] $g: \mathbb{R}_{+}\to\mathbb{R}_{+}$ is a $C^1$
function satisfying
$$
g(0)>0,\quad a-\int_0^{\infty}g(s){\rm d}s=a-g_0=l>0.
$$

(A2) There exists a non-increasing differentiable function
$\xi(t): \mathbb{R}_{+}\to\mathbb{R}_{+}$ such that
$$
g'(t)\leq -\xi(t) g(t), \quad \forall t\geq 0 \quad \text{and}\quad
\int_0^{\infty}\xi(t){\rm d}t=+\infty.
$$
\end{itemize}
As in \cite{np2006}, we introduce the  variable
$$
z(x,\rho,t)=u_{t}(x,t-\tau\rho),\quad (x,\rho,t)\in\Omega\times(0,1)\times(0,\infty).
$$
Then
$$
\tau z_{t}(x,\rho,t)+z_{\rho}(x,\rho,t)=0, \quad
(x,\rho,t)\in\Omega\times(0,1)\times(0,\infty).
$$
Following the ideal in \cite{d19701}, we set
\begin{equation}\label{1.10}
\eta^{t}(x,s)=u(x,t)-u(x,t-s),\quad (x,t,s)\in\Omega\times\mathbb{R}_{+}
\times\mathbb{R}_{+}.
\end{equation}
Then
$$
\eta^{t}_{t}(x,s)+\eta^{t}_{s}(x,s)=u_{t}(x,t), \quad
(x,t,s)\in\Omega\times\mathbb{R}_{+}\times\mathbb{R}_{+}.
$$
Thus, system \eqref{1.1} becomes
\begin{equation}
\begin{gathered}
\begin{aligned}
&u_{tt}(x,t)-lu_{xx}(x,t)-\int_0^{\infty}g(s)\eta_{xx}^{t}(x,s){\rm d}s\\
&+\mu_1u_{t}(x,t)+\mu_2z(x,1,t)=0,\quad (x,t)\in\Omega\times(0,+\infty),
\end{aligned}\\
v_{tt}(x,t)-bv_{xx}(x,t)=0,\quad (x,t)\in(L_1,L_2)\times(0,+\infty), \\
\tau z_{t}(x,\rho,t)+z_{\rho}(x,\rho,t)=0,\quad
 (x,\rho,t)\in\Omega\times(0,1)\times(0,+\infty),\\
\eta^{t}_{t}(x,s)+\eta^{t}_{s}(x,s)=u_{t}(x,t),\quad
(x,s,t)\in\Omega\times(0,+\infty)\times(0,+\infty),
 \end{gathered}\label{2.5}
\end{equation}
 the boundary and transmission conditions \eqref{1.2}  become
\begin{equation}
\begin{gathered}
u(0,t)=u(L_3,t)=0,\\
u(L_{i},t)=v(L_{i},t),\quad i=1,2,\; t\in(0,+\infty),\\
l u_{x}(L_{i},t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_{i},s){\rm d}s=bv_{x}(L_{i},t),
\quad i=1,2,\; t\in(0,+\infty),
 \end{gathered}\label{2.6}
\end{equation}
 and the initial conditions \eqref{1.3} become
\begin{equation}
\begin{gathered}
u(x,-t)=u_0(x,t),\quad u_{t}(x,0)=u_1(x),\; x\in\Omega,\\
z(x,0,t)=u_{t}(x,t),\quad z(x,1,t)=f_0(x,t-\tau),\quad
 (x,t)\in\Omega\times(0,+\infty),\\
v(x,0)=v_0(x),\quad v_{t}(x,0)=v_1(x),\quad x\in(L_1,L_2),
 \end{gathered}
\end{equation}
It is clear that
\begin{equation}
\begin{gathered}
\eta^{t}(x,0)=0,\quad \text{for all } x>0,\\
\eta^{t}(0,s)=\eta^{t}(L_3,s)=0,\quad \text{for all } s> 0,\\
\eta^{0}(x,s)=\eta_0(s),\quad \text{for all } s> 0.
  \end{gathered}
\end{equation}
Let $V:=(u,v,\varphi,\psi,z,\eta^{t})^T$, then $V$ satisfies the problem
\begin{equation}
\begin{gathered}
V_{t}=\mathscr{A}V(t),\quad t>0,\\
 V(0)=V_0,
 \end{gathered}\label{2.7}
\end{equation}
where $V_0:=(u_0(\cdot,0),v_0,u_1,v_1,f_0(\cdot,-\tau),\eta_0)^T$
and the operator $\mathscr{A}$ is defined by
\[
\mathscr{A}=\begin{pmatrix}
u\\
 v\\
 \varphi\\
 \psi\\
 z\\
 w\\
\end{pmatrix}=
\begin{pmatrix}
\varphi\\
 \psi\\
 lu_{xx} +\int_0^{+\infty}g(s)w_{xx}(s){\rm d}s-\mu_1\varphi-\mu_2z(.,1)\\
 bv_{xx}\\
 -\frac{1}{\tau}z_{\rho}\\
 -w_{s}+\varphi\\
\end{pmatrix}.
\]
where
\begin{align*}
X_{*}=\Big\{&(u,v)\in H^1(\Omega)\cap H^1(L_1,L_2): u(0,t)=u(L_3,t)=0,
 u(L_{i},t)=v(L_{i},t),\\
&l u_{x}(L_{i},t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_{i},s){\rm d}s
=bv_{x}(L_{i},t),i=1,2\Big\}
\end{align*}
and
$L_{g}^2(\mathbb{R}_{+},H^1(\Omega))$ denotes the Hilbert space of $H^1$-valued
functions on $\mathbb{R}_{+}$,
endowed with the inner product
$$
(\phi,\vartheta)_{L_{g}^2\left(\mathbb{R}_{+},H^1(\Omega)\right)}
=\int_{\Omega}\int_0^{+\infty}g(s)\phi_{x}(s)\vartheta_{x}(s){\rm d}s{\rm d}x.
$$
Set
$$
V=(u,v,\varphi,\psi,z,w)^T,\quad \bar{V}=(\bar{u},\bar{v},\bar{\varphi},\bar{\psi},\bar{z},\bar{w})^T.
$$

We define the inner product in $\mathscr{H}$,
\begin{align*}
\langle V,\bar{V}\rangle_{\mathscr{H}}
&=\int_{\Omega}\varphi\bar{\varphi}{\rm d}x+\int_{L_1}^{L_2}\psi\bar{\psi}{\rm d}x+
\int_{\Omega}lu_{x}\bar{u}_{x}{\rm d}x+\int_{L_1}^{L_2}bv_{x}\bar{v}_{x}{\rm d}x\\
&\quad +\int_{\Omega}\int_0^{+\infty}g(s)w_{x}(s)\bar{w}_{x}(s){\rm d}s{\rm d}x
+\zeta\int_{\Omega}\int_0^1z\bar{z}{\rm d}\rho{\rm d}x.
\end{align*}
The domain of $\mathscr{A}$ is
\begin{align*}
D(\mathscr{A})=\Big\{
&(u,v,\varphi,\psi,z,w)^T\in\mathscr{H}: u\in H^{2}(\Omega)\cap H^1(\Omega),\\
&v\in H^2(L_1,L_2)\cap H^1(L_1,L_2),\varphi\in H^1(\Omega),\psi\in H^1(L_1,L_2),\\
&w\in L_{g}^2\left(\mathbb{R}_{+},H^{2}(\Omega)\cap H^1(\Omega)\right),
w_{s}\in\left(\mathbb{R}_{+},H^1(\Omega)\right),\\
&z_{\rho}\in
L^2((0,1),L^2(\Omega)),w(x,0)=0,z(x,0)=\varphi(x) \Big\}.
\end{align*}
The well-posedness of problem \eqref{2.5}-\eqref{2.6} is ensured by
the following theorem.

\begin{theorem}\label{th2.1}
Assume that $\mu_2\leq\mu_1$, {\rm (A1)} and {\rm (A2)} hold.
Let $V_0\in \mathscr{H}$, then there exists a unique weak solution
$V\in C\left(\mathbb{R}_{+},\mathscr{H}\right)$ of problem
\eqref{2.7}. Moreover, if $V_0\in D(\mathscr{A})$, then
$$
V\in C\left(\mathbb{R}_{+},D(\mathscr{A})\right)
\cap C^1\left(\mathbb{R}_{+},\mathscr{H}\right).
$$
\end{theorem}

For a solution $u$ of \eqref{1.1}-\eqref{1.3}, we define the energy
\begin{equation} \label{3.1}
\begin{aligned}
E(t)&=\frac{1}{2}\int_{\Omega}[u_{t}^2(x,t)+lu_{x}^2(x,t)]{\rm d}x
+\frac{1}{2}\int_{L_1}^{L_2}[v_{t}^2(x,t)+bv_{x}^2(x,t)]{\rm d}x \\
&+\frac{1}{2}\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x+\frac{\zeta}{2}\int_{\Omega}\int_0^1z^2(x,\rho,t){\rm d}\rho{\rm d}x,
\end{aligned}
\end{equation}
where $\zeta$ is the positive constant satisfying
\begin{equation}
\begin{gathered}
 \zeta \mu_2<\zeta<\tau(2\mu_1-\mu_2), \quad \mu_2<\mu_1, \\
\zeta=\tau \mu_2, \quad \mu_2=\mu_1.
\end{gathered} \label{2.9}
\end{equation}
Our decay result reads as follows.

\begin{theorem}\label{th3.1}
Let $(u,v)$ be the solution of \eqref{1.1}-\eqref{1.3}. Assume
that {\rm (A1), (A2)} hold,
that $\mu_2\leq\mu_1$, that for some $m_0\geq0$,
\begin{equation}\label{3.40}
\int_{\Omega}u_{0x}^2(x,s){\rm d}x\leq m_0,\quad \forall s >0
\end{equation}
and that
\begin{equation}\label{3.30}
a>\frac{8(L_2-L_1)}{L_1+L_3-L_2}l,\quad b>\frac{8(L_2-L_1)}{L_1+L_3-L_2}l
\end{equation}
hold, then there exists constants $\gamma_0,\gamma_2>0$ such that,
for all $t\in\mathbb{R}_{+}$ and for all $\gamma_1\in(0,\gamma_0)$,
\begin{equation}\label{3.26}
E(t)\leq \gamma_2\Big(1+\int_0^{t}(g(s))^{1-\gamma_1}{\rm d}s\Big)
e^{-\gamma_1\int_0^{t}\xi(s){\rm d}s}+\gamma_2\int_{t}^{+\infty}g(s){\rm d}s.
\end{equation}
\end{theorem}

\begin{remark} \label{rmk2.3} \rm
Here we consider some examples to illustrate our estimate \eqref{3.26}.

(1) Let
$$
g(t)=k_1e^{-k_2(1+t)^{q}}, \quad 0<q<1, \; k_1>0,\; k_2>0,
$$
then it is clear that (A2) holds for $\xi(t)=k_2q(1+t)^{q-1}$.
Consequently, applying \eqref{3.26}, we obtain
the decay result
$$
E(t)\leq\tilde{c_1}e^{-\tilde{c_2}k_2(1+t)^{q}},
$$
where $\tilde{c_1}$ and $\tilde{c_2}$ are positive constants.

(2) Let
$$
g(t)=k_3e^{-k_4[\ln(1+t)]^{p}},\quad k_3>0,\; k_4>0,
$$
then our assumption (A2) holds with $\xi(t)=\frac{k_4p[\ln(1+t)]^{p-1}}{1+t}$.
\eqref{3.26} gives us
$$
E(t)\leq\tilde{c_3}e^{-\tilde{c_4}k_4[\ln(1+t)]^{p}},
$$
where $\tilde{c_3}$ and $\tilde{c_4}$ are positive constants.
\end{remark}

\section{Well-posedness of the problem}\label{s3}

In this section, by combining the frameworks of \cite{b2014} with some necessary
 modifications due to the problem treated here,
we use the semigroup approach and the Hille-Yosida theorem to prove the
existence and uniqueness of problem
\eqref{1.1}-\eqref{1.3}.

\begin{proof}[Proof of Theorem \ref{th2.1}]
First, we  prove that the operator $\mathscr{A}$ is dissipative. In fact, for
$(u,v,\varphi,\psi,z,w)^T\in D(\mathscr{A})$, where
$\varphi(L_{i})=\psi(L_{i}),i=1,2$ and $\zeta$ is a positive
constant satisfying \eqref{2.9}, we have
\begin{equation} \label{2.8}
\begin{aligned}
\langle\mathscr{A}V,V\rangle_{\mathscr{H}}
&=\int_{\Omega}lu_{xx}\varphi{\rm d}x+\int_{\Omega}
\Big(\int_0^{+\infty}g(s)w_{xx}(s){\rm d}s-\mu_1\varphi-\mu_2z(.,1)\Big)
 \varphi{\rm d}x \\
&\quad+\int_{\Omega}lu_{x}\varphi_{x}{\rm d}x+\int_{L_1}^{L_2}bv_{x}\psi_{x}{\rm d}x
 +\int_{L_1}^{L_2}bv_{xx}\psi{\rm d}x \\
&\quad+\int_{\Omega}\int_0^{+\infty}g(s)w_{x}(s)(-w_{s}+\varphi)_{x}{\rm d}s{\rm d}x
  \\
&\quad-\frac{\zeta}{\tau}\int_{\Omega}\int_0^1zz_{\rho}(x,\rho){\rm d}\rho{\rm d}x.
\end{aligned}
\end{equation}
For the last term of the right side of \eqref{2.8}, we obtain
\begin{align*}
\zeta\int_{\Omega}\int_0^1zz_{\rho}(x,\rho){\rm d}\rho{\rm d}x
= \zeta\int_{\Omega}\int_0^1\frac{1}{2}\frac{\partial}{\partial\rho}z^2(x,\rho)
{\rm d}\rho{\rm d}x
= \frac{\zeta}{2}\int_{\Omega}(z^2(x,1)-z^2(x,0)){\rm d}x.
\end{align*}
Noticing that $z(x,0,t)=\varphi(x,t)$, $w(x,0)=0$ and
$\varphi(L_{i})=\psi(L_{i}),i=1,2$, we obtain
\begin{align*}
\langle\mathscr{A}V,V\rangle_{\mathscr{H}}
&= \Big[lu_{x}\varphi+\int_0^{+\infty}g(s)w_{x}(s){\rm d}s\varphi
 \Big]_{\partial\Omega}+[bv_{x}\psi]_{L_1}^{L{2}} \\
&\quad +\int_{\Omega}
(-\mu_1\varphi-\mu_2z(.,1))\varphi{\rm d}x
 -\Big[\frac{1}{2}\int_0^{+\infty}g(s)|w_{x}(x,s)|^2{\rm d}s\Big]_{\partial\Omega} \\
&\quad +\frac{1}{2}\int_{\Omega}\int_0^{+\infty}g'(s)|w_{x}(x,s)|^2{\rm d}s{\rm d}x
 -\frac{\zeta}{\tau}\int_{\Omega}\int_0^1zz_{\rho}{\rm d}\rho{\rm d}x \\
&\leq -\big(\mu_1-\frac{\zeta}{2\tau}\big)\int_{\Omega}\varphi^2(x){\rm d}x
 -\frac{\zeta}{2\tau}\int_0^1z^2(x,1){\rm d}x \\
&-\mu_2\int_{\Omega}z(x,1)\varphi(x){\rm d}x
 +\frac{1}{2}\int_{\Omega}\int_0^{+\infty}g'(s)|w_{x}(x,s)|^2{\rm d}s{\rm d}x,
\end{align*}
where we have used  that
\begin{align*}
&[lu_{x}\varphi+\int_0^{+\infty}g(s)w_{x}(s){\rm d}s\varphi]_{\partial\Omega} \\
&=\Big(lu_{x}(L_1,t)+\int_0^{+\infty}g(s)w_{x}(L_1,s){\rm d}s\Big)\varphi(L_1,t) \\
&\quad -\Big(lu_{x}(L_2,t)+\int_0^{+\infty}g(s)w_{x}(L_2,s){\rm d}s\Big)
 \varphi(L_2,t) \\
&=-[bv_{x}\psi]_{L_1}^{L{2}}
\end{align*}
Using Young's inequality, we have
\begin{align*}
\langle\mathscr{A}V,V\rangle_{\mathscr{H}}
&=-\Big(\mu_1-\frac{\zeta}{2\tau}-\frac{\mu_2}{2}\Big)
\int_{\Omega}\varphi^2(x){\rm d}x
 -\Big(\frac{\zeta}{2\tau}-\frac{\mu_2}{2}\Big)\int_0^1z^2(x,1){\rm d}x \\
&\quad +\frac{1}{2}\int_{\Omega}\int_0^{+\infty}g'(s)|w_{x}(x,s)|^2{\rm d}s{\rm d}x.
\end{align*}
Consequently, taking \eqref{2.9} and (A2) into account, we conclude that
\begin{align*}
\langle\mathscr{A}V,V\rangle_{\mathscr{H}}\leq 0;
\end{align*}
that is, $\mathscr{A}$ is dissipative.

Next, we prove that $-\mathscr{A}$ is maximal. Actually, let
$F=(f_1,f_2,f_3,f_4,f_5,f_6)^T\in \mathscr{H}$, we prove that there
exists $V=(u,v,\varphi,\psi,z,w)^T\in D(\mathscr{A})$ satisfying
\begin{equation}\label{2.10}
(\lambda I-\mathscr{A})V=F,
\end{equation}
which is equivalent to
\begin{equation}
\begin{gathered}
\lambda u-\varphi=f_1,\\
\lambda v-\psi=f_2,\\
 \lambda\varphi-lu_{xx}-\int_0^{\infty}g(s)w_{xx}(s){\rm d}s
+\mu_1\varphi+\mu_2z(.,t)=f_3,\\
 \lambda\psi-bv_{xx}=f_4,\\
\lambda z+\frac{1}{\tau}z_{\rho}=f_5,\\
\lambda w+w_{s}-\varphi=f_6.
 \end{gathered}\label{2.11}
\end{equation}
Assume that with the suitable regularity we have found $u$ and $v$, then
\begin{equation}
\begin{gathered}
\varphi=\lambda u-f_1,\\
\psi=\lambda v-f_2.
\end{gathered}\label{2.12}
\end{equation}
So we have $\varphi\in H^1(\Omega)$ and $\psi\in H^1(L_1,L_2)$. Moreover,
we can find $z$ with
$$
z(x,0)=\varphi(x),\quad \text{for } x\in \Omega.
$$
Following the same approach in \cite{np2006} and using the equation
in \eqref{2.11}, we obtain
$$
z(x,\rho)=\varphi(x)e^{-\lambda\rho\tau}
 +\tau e^{-\lambda\rho\tau}\int_0^{\rho}f_5(x,\sigma)
 e^{\lambda\sigma\tau}{\rm d}\sigma.
$$
From \eqref{2.12}, we obtain
\begin{equation}\label{2.14}
z(x,\rho)=\lambda ue^{-\lambda\rho\tau}-f_1e^{-\lambda\rho\tau}\tau
+\tau e^{-\lambda\rho\tau}\int_0^{\rho}f_5(x,\sigma)
e^{\lambda\sigma\tau}{\rm d}\sigma.
\end{equation}

It is easy to see that the last equation in \eqref{2.11} with $w(x,0)=0$
has a unique solution
\begin{align}\label{2.15}
w(x,s)
&=\Big(\int_0^{s}e^{\lambda y}(f_6(x,y)+\varphi(x)){\rm d}y\Big)e^{-\lambda s} \\
&=\Big(\int_0^{s}e^{\lambda y}(f_6(x,y)+\lambda u(x)-f_1(x)){\rm d}y\Big)
e^{-\lambda s}.
\end{align}
By using \eqref{2.11}, \eqref{2.12} and \eqref{2.15}, the functions $u$ and
$v$ satisfy
\begin{equation}
\begin{gathered}
 \left(\lambda^2+\mu_1\lambda+\mu_2\lambda e^{-\lambda\tau}\right)u
-\tilde{l}u_{xx}=\tilde{f},\\
\lambda^2v-bv_{xx}=f_4+\lambda f_2,
\end{gathered}\label{2.16}
\end{equation}
where
$$
\tilde{l}=l+\lambda\int_0^{\infty}g(s)e^{-\lambda s}
\Big(\int_0^{s}e^{\lambda y}{\rm d}y\Big){\rm d}s
$$
and
\begin{align*}
\tilde{f}
&= \int_0^{\infty}g(s)e^{-\lambda s}\Big(\int_0^{s}e^{\lambda y}
 (f_6(x,y)-f_1(x,y))_{xx}{\rm d}y\Big){\rm d}s\\
&\quad -\mu_2\tau e^{-\lambda\tau}
\int_0^1f_5(x,\sigma)e^{\lambda\sigma\tau}{\rm d}\sigma
+\left(\lambda+\mu_1+\mu_2e^{-\lambda\tau}\right)f_1+f_3.
\end{align*}
We just need to prove that \eqref{2.16} has a solution $(u,v)\in X_{*}$
and replace in \eqref{2.12}, \eqref{2.14} and \eqref{2.15}
to get $V=(u,v,\varphi,\psi,z,w)^T\in D(\mathscr{A})$
satisfying \eqref{2.10}. Consequently, problem \eqref{2.16} is
equivalent to the problem
\begin{equation}\label{2.17}
\Phi((u,v),(\omega_1,\omega_2))=l(\omega_1,\omega_2),
\end{equation}
where the bilinear form $\Phi: \left(X_{*},X_{*}\right)\to\mathbb{R}$ and
the linear form $l: X_{*}\to\mathbb{R}$ are defined by
\begin{align*}
\Phi((u,v),(\omega_1,\omega_2))
&=\int_{\Omega}
\big[\left(\lambda^2+\mu_1\lambda+\mu_2\lambda e^{-\lambda\tau}\right)
u\omega_1+\tilde{l}u_{x}(\omega)_{x}\big]{\rm d}x
-[\tilde{l}u_{x}\omega_1]_{\partial\Omega}\\
&\quad +\int_{L_1}^{L_2}\left(\lambda^2v\omega_2
 +bv_{x}(\omega_2)_{x}\right){\rm d}x-[bv_{x}\omega_2]_{L_1}^{L_2}
\end{align*}
and
$$
l(\omega_1,\omega_2)=\int_{\Omega}\tilde{f}\omega_1{\rm d}x
+\int_{L_1}^{L_2}(f_4+\lambda f_2)\omega_2{\rm d}x.
$$
Using the properties of the space $X_{*}$, it is easy to see that $\Phi$
is continuous and coercive, and $l$ is continuous.
Applying the Lax-Milgram theorem, we infer that for all
$(\omega_1,\omega_2)\in X_{*}$, problem \eqref{2.17} has
a unique solution $(u,v)\in X_{*}$. It follows from \eqref{2.16} that
$(u,v)\in \left\{\left(H^{2}(\Omega)\times H^{2}(L_1,L_2)\right)\cap X_{*}\right\}$.
 Thence, the operator
$\lambda I-\mathscr{A}$ is surjective for any $\lambda>0$. At last, the result
of Theorem \ref{th2.1} follows from
the Hille-Yosida theorem.
\end{proof}


\section{Decay of the solution}\label{s4}


In this section, we consider a decay result of problem
\eqref{1.1}-\eqref{1.3}. We will discuss two cases,
the case $\mu_2<\mu_1$ and $\mu_2=\mu_1$. We will separate two cases
since the proof are slightly different.

\subsection{The case $\mu_2<\mu_1$}
For the proof of Theorem \ref{th3.1}, we need some lemmas.

\begin{lemma}\label{le3.1}
Let $(u,v,z)$ be the solution of \eqref{2.5}-\eqref{2.6}.
 Assume that $\mu_2<\mu_1$ and $E(t)$ satisfies \eqref{3.1}.
Then we have the inequality
\begin{equation}\label{3.1'}
\begin{aligned}
\frac{d}{dt}E(t)
&\leq-c_1\Big[\int_{\Omega}u_{t}^2(x,t){\rm d}x
+\int_{\Omega}z^2(x,1,t){\rm d}x\Big] \\
&\quad +\frac{1}{2}\int_{\Omega}\int_0^{\infty}g'(s)|\eta_{x}^{t}(x,s)|^2
{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Multiplying the first equation in \eqref{2.5} by $u_{t}$, the second
equation by $v_{t}$, using the fourth equation in \eqref{2.5},
integrating by parts and using \eqref{2.6}, we obtain
\begin{equation} \label{3.2}
\begin{aligned}
&\int_{\Omega}\Big(u_{t}u_{tt}+lu_{x}u_{xt}
 +\int_0^{\infty}g(s)\eta_{x}^{t}(x,s)u_{xt}{\rm d}s
 +\mu_1u_{t}^2+\mu_2z(x,1,t)u_{t}\Big){\rm d}x \\
&-\Big[\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s \Big) u_{t}
 \Big]_{\partial\Omega}-[bv_{x}v_{t}]_{L_1}^{L_2}
 +\int_{L_1}^{L_2}(v_{t}v_{tt}+bv_{x}v_{xt}){\rm d}x \\
&=\frac{1}{2}\frac{d}{dt}\Big[\int_{\Omega}[u_{t}^2(x,t)+lu_{x}^2(x,t)]
 {\rm d}x\Big]+\frac{1}{2}\frac{d}{dt}
 \Big[\int_{L_1}^{L_2}[v_{t}^2(x,t)+bv_{x}^2(x,t)]{\rm d}x\Big] \\
&\quad +\frac{1}{2}\frac{d}{dt}\int_{\Omega}\int_0^{\infty}g(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
 +\Big[\frac{1}{2}\int_{\Omega}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}x\Big]_0^{\infty} \\
&\quad +\mu_1\int_{\Omega}u_{t}^2(x,t){\rm d}x+\mu_2\int_{\Omega}u_{t}(x,t)z(x,1,t)
 {\rm d}x\\
&\quad -\frac{1}{2}\int_{\Omega}\int_0^{\infty}g'(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x 
=0,
\end{aligned}
\end{equation}
where we have used  that
\begin{align*}
&-\Big[\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s
 \Big) u_{t}\Big]_{\partial\Omega} \\
&=\Big(lu_{x}(L_2,t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_2,s){\rm d}s
 \Big) u_{t}(L_2,t) \\
&\quad -\Big(lu_{x}(L_1,t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_1,s){\rm d}s \Big)
 u_{t}(L_1,t) \\
&=[bv_{x}v_{t}]_{L_1}^{L_2},
\end{align*}
and
$$
\Big[\frac{1}{2}\int_{\Omega}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}x\Big]_0^{\infty}=0.
$$

Multiplying the third equation in \eqref{2.5} by $\zeta z/\tau$,
integrating the result over $\Omega\times(0,1)$ with
respect to $x$ and $\rho$ respectively, we have
\begin{equation}\label{3.3}
\frac{\zeta}{2}\frac{d}{dt}\int_{\Omega}\int_0^1z^2(x,\rho,t){\rm d}\rho{\rm d}x
=-\frac{\zeta}{2\tau}\int_{\Omega}(z^2(x,1)-z^2(x,0)){\rm d}x.
\end{equation}
From \eqref{3.1}, \eqref{3.2} and \eqref{3.3}, we arrive at
\begin{equation} \label{3.4}
\begin{aligned}
\frac{d}{dt}E(t)
&=-\Big(\mu_1-\frac{\zeta}{2\tau}\Big)\int_{\Omega}u_{t}^2(x,t){\rm d}x
 -\frac{\zeta}{2\tau}\int_{\Omega}z^2(x,1,t){\rm d}x \\
&-\mu_2\int_{\Omega}u_{t}(x,t)z(x,1,t){\rm d}x
 +\frac{1}{2}\int_{\Omega}\int_0^{\infty}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
Young's inequality gives us
\begin{align*}
\frac{d}{dt}E(t)
&\leq-\Big(\mu_1-\frac{\zeta}{2\tau}-\frac{\mu_2}{2}\Big)
\int_{\Omega}u_{t}^2(x,t){\rm d}x
-\Big(\frac{\zeta}{2\tau}-\frac{\mu_2}{2}\Big)\int_{\Omega}z^2(x,1,t){\rm d}x \\
&\quad +\frac{1}{2}\int_{\Omega}\int_0^{\infty}g'(s)|\eta_{x}^{t}
 (x,s)|^2{\rm d}s{\rm d}x.
\end{align*}
Thanks to \eqref{2.9} and \eqref{3.1}, our conclusion holds.
The proof is complete.
\end{proof}

Now we define the functional
$$
\mathscr{D}(t)=\int_{\Omega}uu_{t}{\rm d}x
 +\frac{\mu_1}{2}\int_{\Omega}u^2{\rm d}x+\int_{L_1}^{L_2}vv_{t}{\rm d}x.
$$
Then we have the following lemma.

\begin{lemma}
The functional $\mathscr{D}(t)$ satisfies
\begin{equation} \label{3.5}
\begin{aligned}
\frac{d}{dt}\mathscr{D}(t)
&\leq\int_{\Omega}u_{t}^2{\rm d}x+\int_{L_1}^{L_2}v_{t}^2{\rm d}x
 +(L^2\varepsilon+\varepsilon-l)\int_{\Omega}u_{x}^2{\rm d}x
 -\int_{L_1}^{L_2}bv_{x}^2{\rm d}x \\
&\quad+\frac{g_0}{4\varepsilon}\int_{\Omega}\int_0^{\infty}g(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
 +\frac{\mu_2^2}{4\varepsilon}\int_{\Omega}z^2(x,1,t){\rm d}x.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathscr{D}(t)$ with respect to $t$ and
using \eqref{2.5}, we have
\begin{equation} \label{3.6}
\begin{aligned}
\frac{d}{dt}\mathscr{D}(t)
&=\int_{\Omega}u_{t}^2{\rm d}x-l\int_{\Omega}u_{x}^2{\rm d}x
 -\mu_2\int_{\Omega}z(x,1,t)u{\rm d}x+[bv_{x}v]_{L_1}^{L_2}
 +\int_{L_1}^{L_2}v_{t}^2{\rm d}x \\
&\quad +\Big[\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s
 \Big)u\Big]_{\partial\Omega}\\
&\quad -\int_{\Omega}u_{x}(x,t)\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s{\rm d}x
 -\int_{L_1}^{L_2}bv_{x}^2{\rm d}x \\
&=\int_{\Omega}u_{t}^2{\rm d}x-l\int_{\Omega}u_{x}^2{\rm d}x
 -\mu_2\int_{\Omega}z(x,1,t)u{\rm d}x+\int_{L_1}^{L_2}v_{t}^2{\rm d}x\\
&\quad -\int_{L_1}^{L_2}bv_{x}^2{\rm d}x
 -\int_{\Omega}u_{x}(x,t)\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s{\rm d}x,
\end{aligned}
\end{equation}
where we  used  that
\begin{align*}
\Big[\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big)u
 \Big]_{\partial\Omega} 
&=\Big(lu_{x}(L_1,t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_1,s){\rm d}s \Big) u(L_1,t) \\
&\quad -\Big(lu_{x}(L_2,t)+\int_0^{\infty}g(s)\eta_{x}^{t}(L_2,s){\rm d}s
  \Big) u(L_2,t) \\
&=-[bv_{x}v_{t}]_{L_1}^{L_2}.
\end{align*}
By the boundary condition \eqref{1.2}, we have
\begin{gather*}
u^2(x,t)=\Big(\int_0^{x}u_{x}(x,t){\rm d}x\Big)^2
\leq L_1\int_0^{L_1}u_{x}^2(x,t){\rm d}x,\quad x\in [0,L_1], \\
u^2(x,t)\leq (L_3-L_2)\int_{L_2}^{L_3}u_{x}^2(x,t){\rm d}x,\quad x\in [L_2,L_3],
\end{gather*}
which implies
\begin{equation}\label{1.6}
\int_{\Omega}u^2(x,t){\rm d}x\leq L^2\int_{\Omega}u_{x}^2{\rm d}x,\quad
x\in\Omega,
\end{equation}
where $L=\max\{L_1,L_3-L_2\}$.
By making use of Young's inequality and \eqref{1.6},  for any
$\varepsilon>0$, we obtain
\begin{equation}\label{3.7}
\mu_2\int_{\Omega}z(x,1,t)u{\rm d}x
\leq\frac{\mu_2^2}{4\varepsilon}\int_{\Omega}z^2(x,1,t){\rm d}x
+L^2\varepsilon\int_{\Omega}u_{x}^2{\rm d}x.
\end{equation}
Young's inequality, H$\ddot{o}$lder's inequality and (A2) imply that
\begin{equation}\label{3.8}
\int_{\Omega}u_{x}(x,t)\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s{\rm d}x
\leq\varepsilon\int_{\Omega}u_{x}^2(x,t){\rm d}x+\frac{g_0}{4\varepsilon}
\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{equation}
Inserting the estimates \eqref{3.7} and \eqref{3.8} into \eqref{3.6},
then \eqref{3.5} is fulfilled.
\end{proof}

Next, enlightened by \cite{mmn2002}, we introduce the functional
\[
q(x)=\begin{cases}
x-\frac{L_1}{2},& x\in[0,L_1],\\
\frac{L_1}{2}-\frac{L_1+L_3-L_2}{2(L_2-L_1)}(x-L_1),&x\in(L_1,L_2),\\
x-\frac{L_2+L_3}{2},&x\in[L_2,L_3].\\
\end{cases}
\]
It is easy to see that $q(x)$ is bounded:  $|q(x)|\leq M$,
 where $ M=\max\{\frac{L_1}{2},\frac{L_3-L_2}{2}\}$.

We define the functionals
$$
\mathscr{F}_1(t)=-\int_{\Omega}q(x)u_{t}
\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x,\quad
\mathscr{F}_2(t)=-\int_{L_1}^{L_2}q(x)v_{x}v_{t}{\rm d}x.
$$
Then we have the following results.

\begin{lemma}\label{le3.5}
The functionals $\mathscr{F}_1(t)$ and $\mathscr{F}_2(t)$ satisfy
\begin{equation} \label{3.9}
\begin{aligned}
&\frac{d}{dt}\mathscr{F}_1(t)\\
&\leq \Big(\frac{l+g_0}{2}+\frac{M^2\mu_1^2}{4\varepsilon_1}
 +\varepsilon_1M^2\Big)\int_{\Omega}u_{t}^2{\rm d}x
 +\left(l^2+2l^2\varepsilon_1\right)\int_{\Omega}u_{x}^2{\rm d}x \\
&\quad +\frac{M^2\mu_2^2}{4\varepsilon_1}\int_{\Omega}z^2(x,1,t){\rm d}x
 +\left(g_0+2g_0\varepsilon_1\right)
 \int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&-\frac{g(0)}{4\varepsilon_1}\int_{\Omega}\int_0^{\infty}g'(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
 -\Big[\frac{l+g_0}{2}q(x)u_{t}^2\Big]_{\partial\Omega} \\
&-\Big[\frac{q(x)}{2}\Big(lu_{x}(x,t)+\int_0^{\infty}g(s)
 \eta_{x}^{t}(x,s){\rm d}s\Big)^2\Big]_{\partial\Omega}
\end{aligned}
\end{equation}
and
\begin{equation} \label{3.10}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_2(t)
&\leq -\frac{L_1+L_3-L_2}{4(L_2-L_1)}
 \Big(\int_{L_1}^{L_2}v_{t}^2{\rm d}x+\int_{L_1}^{L_2}bv_{x}^2{\rm d}x\Big)
 +\frac{L_1}{4}v_{t}^2(L_1) \\
&\quad +\frac{L_3-L_2}{4}v_{t}^2(L_2)+\frac{b}{4}
 \left((L_3-L_2)v_{x}^2(L_2,t)+L_1v_{x}^2(L_1,t)\right).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathscr{F}_1(t)$ with respect to $t$ and using
\eqref{2.5}, we obtain
\begin{equation} \label{3.11}
\begin{aligned}
&\frac{d}{dt}\mathscr{F}_1(t)\\
&=-\int_{\Omega}q(x)u_{tt}\Big(lu_{x}
 +\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&\quad -\int_{\Omega}q(x)u_{t}\Big(lu_{xt}+\int_0^{\infty}g(s)
 \eta_{xt}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&=-\int_{\Omega}q(x)\Big(lu_{xx}+\int_0^{\infty}g(s)\eta_{xx}^{t}(x,s){\rm d}s
 \Big)\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&\quad +\int_{\Omega}q(x)(\mu_1u_{t}(x,t)+\mu_2z(x,1,t))
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&\quad -\int_{\Omega}q(x)u_{t}
 \Big(lu_{xt}+\int_0^{\infty}g(s)\eta_{xt}^{t}(x,s){\rm d}s\Big){\rm d}x.
\end{aligned}
\end{equation}
We pay attention to
\begin{equation} \label{3.12}
\begin{aligned}
&-\int_{\Omega}q(x)\Big(lu_{xx}+\int_0^{\infty}g(s)\eta_{xx}^{t}(x,s){\rm d}s\Big)
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&=\frac{1}{2}\int_{\Omega}q'(x)
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big)^2{\rm d}x \\
&\quad-\Big[\frac{q(x)}{2}\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s)
 {\rm d}s\Big)^2\Big]_{\partial\Omega}.
\end{aligned}
\end{equation}
The last term in \eqref{3.11} can be treated as follows
\begin{equation} \label{3.13}
\begin{aligned}
&-\int_{\Omega}q(x)u_{t}
 \Big(lu_{xt}+\int_0^{\infty}g(s)\eta_{xt}^{t}(x,s){\rm d}s\Big){\rm d}x \\
&= -l\int_{\Omega}q(x)u_{t}u_{xt}{\rm d}x
 -\int_{\Omega}q(x)u_{t} \int_0^{\infty}g(s)\eta_{xt}^{t}(x,s){\rm d}s{\rm d}x \\
&= \Big[-\frac{l}{2}q(x)u_{t}^2\Big]_{\partial\Omega}
 +\frac{l}{2}\int_{\Omega}q'(x)u_{t}^2{\rm d}x\\
&\quad -\int_{\Omega}q(x)u_{t}\int_0^{\infty}g(s)\left(u_{t}-\eta_{s}^{t}\right)_{x}{\rm d}s{\rm d}x \\
&= \Big[-\frac{l}{2}q(x)u_{t}^2\Big]_{\partial\Omega}
 +\frac{l}{2}\int_{\Omega}q'(x)u_{t}^2{\rm d}x
 -g_0\int_{\Omega}q(x)u_{t}u_{tx}{\rm d}x \\
&\quad +\int_{\Omega}q(x)u_{t}\int_0^{\infty}g(s)\eta_{sx}^{t}(x,s){\rm d}s{\rm d}x \\
&= \Big[-\frac{l+g_0}{2}q(x)u_{t}^2\Big]_{\partial\Omega}
 +\frac{l+g_0}{2}\int_{\Omega}q'(x)u_{t}^2{\rm d}x\\
&\quad -\int_{\Omega}q(x)u_{t}\int_0^{\infty}g'(s)\eta_{x}^{t}{\rm d}s{\rm d}x,
\end{aligned}
\end{equation}
where we  used  that
$$
-\Big[\int_{\Omega}q(x)u_{t}g(s)\eta_{x}^{t}(x,s){\rm d}x\Big]_0^{\infty}=0.
$$
Inserting \eqref{3.12} and \eqref{3.13} in \eqref{3.11}, we arrive at
\begin{equation} \label{3.14}
\begin{aligned}
&\frac{d}{dt}\mathscr{F}_1(t)\\
&= -\Big[\frac{q(x)}{2}\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s)
 {\rm d}s\Big)^2\Big]_{\partial\Omega}
 -\Big[\frac{l+g_0}{2}q(x)u_{t}^2\Big]_{\partial\Omega} \\
&\quad +\frac{1}{2}\int_{\Omega}q'(x)
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big)^2{\rm d}x
 +\int_{\Omega}q(x)(\mu_1u_{t}(x,t) \\
&\quad +\mu_2z(x,1,t))\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big)
 {\rm d}x \\
&\quad +\frac{l+g_0}{2}\int_{\Omega}q'(x)u_{t}^2{\rm d}x
 -\int_{\Omega}q(x)u_{t}\int_0^{\infty}g'(s)\eta_{x}^{t}{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
Using Minkowski and Young's inequalities, we have
\begin{equation}\label{3.15}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}\Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s)
 {\rm d}s\Big)^2{\rm d}x \\
& \leq l^2\int_{\Omega}u_{x}^2{\rm d}x
 +g_0\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
Young's inequality gives us that for any $\varepsilon_1>0$,
\begin{equation} \label{3.16}
\begin{aligned}
&\Big|\mu_1\int_{\Omega}q(x)u_{t}(x,t)
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x\Big| \\
&\leq \frac{M^2\mu_1^2}{4\varepsilon_1}\int_{\Omega}u_{t}^2(x,t){\rm d}x
 +l^2\varepsilon_1\int_{\Omega}u_{x}^2(x,t){\rm d}x \\
&\quad +g_0\varepsilon_1\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
\end{aligned}
\end{equation}
and
\begin{equation} \label{3.17}
\begin{aligned}
&\Big|\mu_2\int_{\Omega}q(x)z(x,1,t)
 \Big(lu_{x}+\int_0^{\infty}g(s)\eta_{x}^{t}(x,s){\rm d}s\Big){\rm d}x\Big| \\
&\leq \frac{M^2\mu_2^2}{4\varepsilon_1}\int_{\Omega}z^2(x,1,t){\rm d}x
 +l^2\varepsilon_1\int_{\Omega}u_{x}^2(x,t){\rm d}x \\
&\quad +g_0\varepsilon_1\int_{\Omega}\int_0^{\infty}g(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
It is clear that
\begin{equation} \label{3.18}
\begin{aligned}
&\left|\int_{\Omega}q(x)u_{t}\int_0^{\infty}g'(s)
 \eta_{x}^{t}{\rm d}s{\rm d}x\right| \\
&\leq \varepsilon_1M^2\int_{\Omega}u_{t}^2{\rm d}x
 -\frac{g(0)}{4\varepsilon_1}\int_{\Omega}
 \int_0^{\infty}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
Inserting \eqref{3.15}-\eqref{3.18} into \eqref{3.14}, we obtain \eqref{3.9}.

By the same method, taking the derivative of $\mathscr{F}_1(t)$ with respect to $t$, 
we obtain
\begin{align*}
\frac{d}{dt}\mathscr{F}_2(t)
&= -\int_{L_1}^{L_2}q(x)v_{xt}v_{t}{\rm d}x
 -\int_{L_1}^{L_2}q(x)v_{x}v_{tt}{\rm d}x\\
&= \Big[-\frac{1}{2}q(x)v_{t}^2\Big]_{L_1}^{L_2}
 +\frac{1}{2}\int_{L_1}^{L_2}q'(x)v_{t}^2{\rm d}x+\frac{1}{2}\int_{L_1}^{L_2}bq'(x)v_{x}^2{\rm d}x\\
&\quad +\Big[-\frac{b}{2}q(x)v_{x}^2\Big]_{L_1}^{L_2}\\
&\leq -\frac{L_1+L_3-L_2}{4(L_2-L_1)}
 \Big(\int_{L_1}^{L_2}v_{t}^2{\rm d}x
 +\int_{L_1}^{L_2}bv_{x}^2{\rm d}x\Big)+\frac{L_1}{4}v_{t}^2(L_1)\\
&\quad +\frac{L_3-L_2}{4}v_{t}^2(L_2)
 +\frac{b}{4}\left((L_3-L_2)v_{x}^2(L_2,t)+L_1v_{x}^2(L_1,t)\right).
\end{align*}
Thus, the proof of Lemma \ref{le3.5} is complete.


As in \cite{amm2012}, we define the functional
$$
\mathscr{F}_3(t)=\tau\int_{\Omega}\int_0^1e^{-\tau\rho}z^2(x,\rho,t){\rm d}\rho{\rm d}x,
$$
then we have the following estimate.

\begin{lemma}[\cite{amm2012}]
The functionals $\mathscr{F}_3(t)$ satisfies
$$
\frac{d}{dt}\mathscr{F}_3(t)\leq-c_2
\Big(\int_{\Omega}z^2(x,1,t){\rm d}x
+\tau\int_{\Omega}\int_0^1z^2(x,\rho,t){\rm d}\rho{\rm d}x\Big)
+\int_{\Omega}u_{t}^2(x,t){\rm d}x.
$$
\end{lemma}

\begin{proof}[Proof of Theorem \ref{th3.1}]
 We define the Lyapunov functional
\begin{equation}\label{3.20}
L(t)=N_1E(t)+N_2\mathscr{D}(t)+\mathscr{F}_1(t)+N_4\mathscr{F}_2(t)+\mathscr{F}_3(t),
\end{equation}
where $N_1, N_2, N_4$ are positive constants that will be fixed later.

Taking the derivative of \eqref{3.20} with respect to $t$ and taking 
advantage of the above lemmas, we have
\begin{equation} \label{3.21}
\begin{aligned}
\frac{d}{dt}L(t)
&\leq -\Big\{N_1c_1-1-N_2-\Big(\frac{l+g_0}{2}+\frac{M^2\mu_1^2}{4\varepsilon_1}
 +\varepsilon_1M^2\Big)\Big\}
 \int_{\Omega}u_{t}^2{\rm d}x \\
&\quad -\Big\{N_1c_1+c_2-\frac{\mu_2^2N_2}{4\varepsilon}
 -\frac{M^2\mu_2^2 }{4\varepsilon_1}\Big\}
  \int_{\Omega}z^2(x,1,t){\rm d}x \\
&\quad -\big\{N_2(l-L^2\varepsilon-\varepsilon)
 -(l^2+2l^2\varepsilon_1)\big\}\int_{\Omega}u_{x}^2{\rm d}x \\
&\quad -\Big\{\frac{b(L_1+L_3-L_2)}{4(L_2-L_1)}N_4+N_2b\Big\}
 \int_{L_1}^{L_2}v_{x}^2{\rm d}x \\
&\quad -\Big\{\frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4-N_2\Big\}
 \int_{L_1}^{L_2}v_{t}^2{\rm d}x \\
&\quad -(b-N_4)\frac{b}{4}\left((L_3-L_2)v_{x}^2(L_2,t)+L_1v_{x}^2(L_1,t)\right) \\
&\quad -(a-N_4)\Big[\frac{L_1}{4}v_{t}^2(L_1,t)
+\frac{L_3-L_2}{4}v_{t}^2(L_2,t)\Big] \\
&\quad +c(N_2)\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&\quad +\big(\frac{N_1}{2}-\frac{g(0)}{4\varepsilon_1}\big)
\int_{\Omega}\int_0^{\infty}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
At this moment, we wish all coefficients except the last two in \eqref{3.21} will be negative.
We want to choose $N_2$ and $N_4$ to ensure that
\begin{gather*}
 a-N_4\geq0,\quad 
  b-N_4\geq0,\\
  \frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4-N_2>0.
 \end{gather*}
For this purpose, since $\frac{8l(L_2-L_1)}{L_1+L_3-L_2}<\min\{a,b\}$
we first choose $N_4$ satisfying
\[
\frac{8l(L_2-L_1)}{L_1+L_3-L_2}<N_4\leq\min\{a,b\}.
\]
Once $N_4$ is fixed, we pick $N_2$ satisfying
$$
2l<N_2<\frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4.
$$
Once the above constants are fixed, we take $\varepsilon<\frac{l}{8(L^2+1)}$ 
and $\varepsilon_1<\frac{1}{4}$ such that
$$
N_2(l-L^2\varepsilon-\varepsilon)-\left(l^2+2l^2\varepsilon_1\right)>0.
$$
Finally, choosing $N_1$ large enough such that 
the last coefficient in \eqref{3.21} is positive.
From the above, we deduce that there exists two positive constants $\alpha_1$ 
and $\alpha_2$ such that \eqref{3.21} becomes
\begin{equation}\label{3.41}
\frac{d}{dt}L(t)\leq-\alpha_1E(t)
+\alpha_2\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{equation}

On the other hand, by the definition of the functionals $\mathscr{D}(t)$, 
$\mathscr{F}_1(t)$, $\mathscr{F}_2(t)$, $\mathscr{F}_3(t)$ and $E(t)$, 
for $N_1$ large enough, there exists a positive constant $\beta$ satisfying
$$
|N_2\mathscr{D}(t)+\mathscr{F}_1(t)+N_4\mathscr{F}_2(t)+\mathscr{F}_3(t)|\leq\beta E(t),
$$
which implies 
$$
(N_1-\beta)E(t)\leq L(t)\leq(N_1+\beta)E(t).
$$

To finish the proof of the stability estimates, we need to estimate the 
last term in \eqref{3.41}. For the convenience of reading, we briefly 
repeat the process of \cite{gm2014}.

Using (A2) and \eqref{3.1'}, we have
\begin{equation} \label{3.42}
\begin{aligned}
 \xi(t)\int_{\Omega}\int_0^{t}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
&\leq \int_{\Omega}\int_0^{t}\xi(s)g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&\leq -\int_{\Omega}\int_0^{t}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&\leq -\int_{\Omega}\int_0^{+\infty}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&\leq -2\frac{d}{dt}E(t).
\end{aligned}
\end{equation}
Moreover, (A2) and the definition of $E(t)$ give us
$$
\int_{\Omega}u_{x}^2(x,t){\rm d}x\leq\frac{2}{l}E(t)
\leq\frac{2}{l}E(0),\quad \forall t\in\mathbb{R}_{+}.
$$
Using \eqref{1.10}, \eqref{3.1}, \eqref{3.40} and \eqref{3.1'}, we arrive at
\begin{align*}
 \xi(t)\int_{\Omega}|\eta_{x}^{t}(x,s)|^2{\rm d}x
&=  \xi(t)\int_{\Omega}(u_{x}(x,t)-u_{x}(x,t-s))^2{\rm d}x\\
&\leq 2 \xi(t)\int_{\Omega}u_{x}^2(x,t){\rm d}x
 +2 \xi(t)\int_{\Omega}u_{x}^2(x,t-s){\rm d}x\\
&\leq \frac{8}{l}E(0) \xi(t)+2m_0 \xi(t),\quad \forall t,s\in\mathbb{R}_{+}.
\end{align*}
Then, we infer that for all $t\in\mathbb{R}_{+}$,
\begin{equation}\label{3.43}
 \xi(t)\int_{\Omega}\int_{t}^{+\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
\leq\big(\frac{8}{l}E(0)+2m_0\big)\xi(t)\int_{t}^{+\infty}g(s){\rm d}s.
\end{equation}
Multiplying \eqref{3.41} by $\xi(t)$ and using \eqref{3.42} and
\eqref{3.43}, we obtain
\begin{equation}\label{3.60}
\xi(t)\frac{d}{dt}L(t)+\beta_1\frac{d}{dt}E(t)
\leq-\alpha_1\xi(t)E(t)+\beta_2\xi(t)\int_{t}^{+\infty}g(s){\rm d}s,
\end{equation}
where $\beta_1=2\alpha_2$ and $\beta_2=\alpha_2\big(\frac{8}{l}E(0)+2m_0\big)$.

Now, we define functionals $\mathscr{L}(t)$ and $h(t)$ as
$$
\mathscr{L}(t)=\xi(t)L(t)+\beta_1E(t)\quad \text{and}\quad
 h(t)=\xi(t)\int_{t}^{+\infty}g(s){\rm d}s.
$$
The fact that $L(t)$ and $E(t)$ are equivalent and (A2) imply that 
for some positive constants $\eta_1$ and $\eta_2$,
\begin{equation}\label{3.44}
\eta_1E(t)\leq\mathscr{L}(t)\leq\eta_2E(t)\,.
\end{equation}
Using \eqref{3.60}, \eqref{3.44} and (A2), we obtain
$$
\frac{d}{dt}\mathscr{L}(t)\leq-\gamma_0\xi(t)\mathscr{L}(t)+\beta_2h(t),
$$
where $\gamma_0=\alpha_1/\eta_2$.
We conclude that, for any $\gamma_1\in(0,\gamma_0)$,
$$
\frac{d}{dt}\mathscr{L}(t)\leq-\gamma_1\xi(t)\mathscr{L}(t)+\beta_2h(t).
$$
By integrating over $[0,T]$ with $T\geq 0$, we obtain
$$
\mathscr{L}(T)\leq e^{-\gamma_1\int_0^T\xi(s){\rm d}s}
\Big(\mathscr{L}(0)+\beta_2\int_0^Te^{\gamma_1
\int_0^{t}\xi(s){\rm d}s}h(t){\rm d}t\Big).
$$
Using \eqref{3.44}, we have
\begin{equation}\label{3.46}
E(T)\leq\frac{1}{\eta_1}e^{-\gamma_1\int_0^T\xi(s){\rm d}s}
\Big(\mathscr{L}(0)+\beta_2\int_0^Te^{\gamma_1\int_0^{t}\xi(s){\rm d}s}h(t){\rm d}t
\Big).
\end{equation}
We notice that
$$
e^{\gamma_1\int_0^{t}\xi(s){\rm d}s}h(t)=\frac{1}{\gamma_1}
\Big(e^{\gamma_1\int_0^{t}\xi(s){\rm d}s}\Big)'\int_{t}^{+\infty}g(s){\rm d}s.
$$
So integration by parts gives us
\begin{align*}
&\int_0^Te^{\gamma_1\int_0^{t}\xi(s){\rm d}s}h(t){\rm d}t\\
&= \frac{1}{\gamma_1}\Big(e^{\gamma_1\int_0^T\xi(s){\rm d}s}
 \int_{T}^{+\infty}g(s){\rm d}s-\int_0^{+\infty}g(s){\rm d}s
+\int_0^Te^{\gamma_1\int_0^{t}\xi(s){\rm d}s}g(t){\rm d}t\Big).
\end{align*}
Consequently, combining with \eqref{3.46}, we have
\begin{equation}\label{3.47}
\begin{aligned}
E(T)&\leq \frac{1}{\eta_1}
\Big(\mathscr{L}(0)e^{-\gamma_1\int_0^T\xi(s){\rm d}s}
 +\frac{\beta_2}{\gamma_1}\int_{T}^{+\infty}g(s){\rm d}s\Big) \\
&\quad +\frac{\beta_2}{\eta_2\gamma_1}e^{-\gamma_1\int_0^T\xi(s){\rm d}s}
\int_0^Te^{\gamma_1\int_0^{t}\xi(s){\rm d}s}g(t){\rm d}t.
\end{aligned}
\end{equation}
On the other hand, thanks to (A2), we have
$$
\Big(e^{\gamma_1\int_0^{t}\xi(s){\rm d}s}(g(t))^{\gamma_1}\Big)'
\leq 0,\quad \forall t\in\mathbb{R}_{+};
$$
then
$$
e^{\gamma_1\int_0^{t}\xi(s){\rm d}s}(g(t))^{\gamma_1}\leq(g(0))^{\gamma_1}.
$$
Thus
\begin{equation}\label{3.48}
\int_0^Te^{\gamma_1\int_0^{t}\xi(s){\rm d}s}g(t){\rm d}t
\leq(g(0))^{\gamma_1}\int_0^T(g(t))^{1-\gamma_1}{\rm d}t.
\end{equation}
Finally, \eqref{3.47} and \eqref{3.48} imply that for the solution 
of \eqref{1.1}-\eqref{1.3} with
$$
\gamma_2=\frac{1}{\eta_1}\max
\Big\{\mathscr{L}(0),\frac{\beta_2}{\gamma_1}, 
 \frac{\beta_2}{\gamma_1}(g(0))^{\gamma_1}\Big\},
$$
\eqref{3.26} holds. The proof is complete.
\end{proof}

\subsection{Case $\mu_2=\mu_1$}
In this subsection, we assume that $\mu_1=\mu_2=\mu$ and prove the
decay result of problem \eqref{1.1}-\eqref{1.3}.
By \eqref{2.9}, we choose $\zeta=\tau\mu$, then we obtain the following 
consequence of Lemma \ref{le3.1}.

\begin{lemma}
Let $(u,v,z)$ be the solution of \eqref{2.5}-\eqref{2.6}. 
Assume that $\mu_1=\mu_2=\mu$ and $E(t)$ satisfies \eqref{3.1}. 
Then we have the inequality
\begin{equation}\label{5.4'}
\frac{d}{dt}E(t)\leq\frac{1}{2}\int_{\Omega}\int_0^{\infty}
g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{equation}
\end{lemma}

When $\mu_1=\mu_2=\mu$, we need negative term $-\int_{\Omega}u_{t}^2{\rm d}x$ to get 
$-cE(t)$.
For this purpose, we define the functional
$$
\mathscr{F}_4(t)=-\int_{\Omega}u_{t}\int_0^{\infty}g(s)(u(t)
-u(t-s)){\rm d}s{\rm d}x.
$$
Then we have the following estimate.

\begin{lemma}\label{le5.6}
The functional $\mathscr{F}_4(t)$ satisfies
\begin{equation} \label{5.19}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_4(t)
&\leq -(g_0-\delta_2-\delta_2\mu)\int_{\Omega}u_{t}^2{\rm d}x
 +\delta_2l^2\int_{\Omega}u_{x}^2{\rm d}x
+\delta_2\mu\int_{\Omega}z^2(x,1,t){\rm d}x \\
&\quad +\Big(g_0+\frac{g_0}{4\delta_2}+\frac{\mu g_0L^2}{2\delta_2}\Big)
 \int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&\quad -\frac{g(0)L^2}{\delta_2}\int_{\Omega}\int_0^{\infty}g'(s)
|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathscr{F}_4(t)$ with respect to $t$ and using 
\eqref{2.5}, we have
\begin{align}
&\frac{d}{dt}\mathscr{F}_4(t) \nonumber \\
&= -\int_{\Omega}\Big(lu_{xx}+\int_0^{\infty}g(s)\eta_{xx}^{t}(x,s){\rm d}s
 -\mu u_{t}-\mu z(x,1,t)\Big) \nonumber \\
&\quad \times\int_0^{\infty}g(s)(u(t)-u(t-s)){\rm d}s{\rm d}x
 -\int_{\Omega}u_{t}\int_0^{\infty}g(s)(u_{t}(t)-u_{t}(t-s)){\rm d}s{\rm d}x 
\nonumber \\
&= \int_{\Omega}lu_{x}\int_0^{\infty}g(s)(u_{x}(t)-u_{x}(t-s)){\rm d}s{\rm d}x
 -g_0\int_{\Omega}u_{t}^2{\rm d}x \nonumber \\
&\quad +\int_{\Omega}u_{t}\int_0^{\infty}g(s)\eta_{s}^{t}(s){\rm d}s{\rm d}x
 +\int_{\Omega}\Big(\int_0^{\infty}g(s)(u_{x}(t)-u_{x}(t-s)){\rm d}s\Big)^2{\rm d}x 
 \nonumber\\
&\quad +\int_{\Omega}\mu u_{t}\int_0^{\infty}g(s)(u(t)-u(t-s)){\rm d}s{\rm d}x
 \nonumber \\
&\quad +\int_{\Omega}\mu z(x,1,t)\int_0^{\infty}g(s)(u(t)-u(t-s)){\rm d}s{\rm d}x.
 \label{5.191}
\end{align}
Using Young's inequality and \eqref{1.6}, we obtain for any $\delta_2>0$,
\begin{gather} \label{5.192}
\begin{aligned}
&\int_{\Omega}lu_{x}\int_0^{\infty}g(s)(u_{x}(t)-u_{x}(t-s)){\rm d}s{\rm d}x \\
&\leq \delta_2l^2\int_{\Omega}u_{x}^2{\rm d}x+\frac{g_0}{4\delta_2}
\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x,
\end{aligned} \\
\label{5.193}
\begin{aligned}
&\int_{\Omega}\mu u_{t}\int_0^{\infty}g(s)(u(t)-u(t-s)){\rm d}s{\rm d}x \\
&\leq \delta_2\mu\int_{\Omega}u_{t}^2{\rm d}x
 +\frac{\mu g_0L^2}{4\delta_2}\int_{\Omega}\int_0^{\infty}g(s)
 |\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x,
\end{aligned} \\
\label{5.196}
\begin{aligned}
&\int_{\Omega}\mu z(x,1,t)\int_0^{\infty}g(s)(u(t)-u(t-s)){\rm d}s{\rm d}x \\
&\leq \delta_2\mu\int_{\Omega}z^2(x,1,t){\rm d}x
+\frac{\mu g_0L^2}{4\delta_2}\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{gather}
We notice that
\begin{equation} \label{5.194}
\begin{aligned}
&\int_{\Omega}\Big(\int_0^{\infty}g(s)(u_{x}(t)-u_{x}(t-s)){\rm d}s\Big)^2{\rm d}x \\
&\leq \int_{\Omega}\int_0^{\infty}g(s){\rm d}s
 \Big(\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s\Big){\rm d}x \\
&\leq g_0\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x
\end{aligned}
\end{equation}
and
\begin{equation} \label{5.195}
\begin{aligned}
\int_{\Omega}u_{t}\int_0^{\infty}g(s)\eta_{s}^{t}(s){\rm d}s{\rm d}x
&= -\int_{\Omega}u_{t}\int_0^{\infty}g'(s)\eta^{t}(s){\rm d}s{\rm d}x \\
&\leq \delta_2\int_{\Omega}u_{t}{\rm d}x
 -\frac{g(0)L^2}{4\delta_2}\int_0^{\infty}g'(s)\eta_{x}^{t}(s){\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
Inserting the estimates \eqref{5.192}-\eqref{5.195} into \eqref{5.191},
we obtain \eqref{5.19}. The proof is complete.
\end{proof}

Now, we define the Lyapunov functional
\begin{equation}\label{5.20}
G(t)=N_1E(t)+N_2\mathscr{D}(t)+\mathscr{F}_1(t)+N_4\mathscr{F}_2(t)
+N_5\mathscr{F}_3(t)+N_6\mathscr{F}_4(t),
\end{equation}
where $N_1, N_2, N_4, N_5$ and $N_6$ are positive constants that will be fixed later.

Taking the derivative of \eqref{5.20} with respect to $t$ and taking advantage 
of the above lemmas, we have
\begin{equation} \label{5.21}
\begin{aligned}
\frac{d}{dt}G(t)
&\leq -\Big\{N_6(g_0-\delta_2-\delta_2\mu)-N_2
 -\Big(\frac{l+g_0}{2}+\frac{M^2\mu^2}{4\varepsilon_1}
+\varepsilon_1M^2\Big)\\
&\quad -N_5\Big\}\int_{\Omega}u_{t}^2{\rm d}x \\
&\quad -\Big\{N_5c_2-\frac{N_2\mu^2}{4\varepsilon}-\frac{M^2\mu^2 }{4\varepsilon_1}
 -N_6\delta_2\mu\Big\} \int_{\Omega}z^2(x,1,t){\rm d}x \\
&-\big\{N_2(l-L^2\varepsilon-\varepsilon)-\big(l^2+2l^2\varepsilon_1\big)
 -N_6\delta_2l^2\big\}\int_{\Omega}u_{x}^2{\rm d}x \\
&\quad -\big\{\frac{b(L_1+L_3-L_2)}{4(L_2-L_1)}N_4+N_2b\big\}
 \int_{L_1}^{L_2}v_{x}^2{\rm d}x \\
&\quad -\Big\{\frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4-N_2\Big\}
 \int_{L_1}^{L_2}v_{t}^2{\rm d}x \\
&\quad -(b-N_4)\frac{b}{4}\left((L_3-L_2)v_{x}^2(L_2,t)+L_1v_{x}^2(L_1,t)\right) \\
&\quad -(a-N_4)\big[\frac{L_1}{4}v_{t}^2(L_1,t)
+\frac{L_3-L_2}{4}v_{t}^2(L_2,t)\big] \\
&\quad +c(N_2,N_6)\int_{\Omega}\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x \\
&+\Big(\frac{N_1}{2}-\frac{g(0)}{4\varepsilon_1}-\frac{N_6g(0)L^2}{4\delta_2}\Big)
\int_{\Omega}\int_0^{\infty}g'(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{aligned}
\end{equation}
At this moment, we wish all coefficients except the last two in \eqref{5.21}
will be negative. We want to choose $N_2$ and $N_4$ to ensure that
\begin{gather*}
 a-N_4\geq0,\quad
  b-N_4\geq0,\\
 \frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4-N_2>0.
\end{gather*}.
For this purpose, since $\frac{8l(L_2-L_1)}{L_1+L_3-L_2}<\min\{a,b\}$
we first choose $N_4$ satisfying
\begin{align*}
\frac{8l(L_2-L_1)}{L_1+L_3-L_2}<N_4\leq\min\{a,b\}.
\end{align*}
Once $N_4$ is fixed, we pick $N_2$ satisfying
$$
2l<N_2<\frac{L_1+L_3-L_2}{4(L_2-L_1)}N_4.
$$
Then we take $\varepsilon<\frac{l}{8(L^2+1)}$ and $\varepsilon_1<\frac{1}{8}$ such that
$$
N_2(l-L^2\varepsilon-\varepsilon)-2l^2\varepsilon_1>\frac{3}{2}l^2.
$$
Once $\varepsilon$ and $\varepsilon_1$ are fixed, we take $N_5$ satisfying
$$
N_5>\max\Big\{\frac{2N_2\mu^2}{\varepsilon c_2},
\frac{2M^2\mu^2}{\varepsilon_1 c_2}\Big\}
$$
such that
$$
N_5c_2-\frac{N_2\mu^2}{4\varepsilon}-\frac{M^2\mu^2 }{4\varepsilon_1}
>\frac{3}{4}N_5c_2.
$$
Further, we choose $N_6$ satisfying
$$
N_6>\max\Big\{\frac{8N_2}{g_0},\frac{4(l+g_0)}{g_0}
+\frac{2M^2\mu^2}{\varepsilon_1g_0}+\frac{8\varepsilon_1M^2}{g_0},
\frac{8N_5}{g_0}\Big\}
$$
such that
$$
\frac{7}{8}N_6g_0-N_2-\Big(\frac{l+g_0}{2}+\frac{M^2\mu^2}{4\varepsilon_1}
+\varepsilon_1M^2\Big)-N_5>\frac{N_6g_0}{2}.
$$
Then, we pick $\delta_2$ satisfying
$$
\delta_2<\min\Big\{\frac{g_0}{8(1+\mu)},\frac{N_5c_2}{8N_6\mu},\frac{1}{4N_6}\Big\}
$$
such that
$$
\frac{3}{4}N_5c_2-N_6\delta_2\mu>\frac{5}{8}N_5c_2,\quad
\frac{1}{8}N_6g_0-\delta_2-\delta_2\mu>0,\quad
\frac{l^2}{2}-N_6\delta_2l^2>\frac{l^2}{4}.
$$
Finally, choosing $N_1$ large enough such that the first two coefficients
in \eqref{5.21} are negative and the last coefficient in \eqref{5.21}
 is positive.
From the above, we deduce that there exists two positive constants
$\alpha_3$ and $\alpha_4$ such that \eqref{5.21} becomes
\begin{equation}\label{5.41}
\frac{d}{dt}G(t)\leq-\alpha_3E(t)+\alpha_4\int_{\Omega}
\int_0^{\infty}g(s)|\eta_{x}^{t}(x,s)|^2{\rm d}s{\rm d}x.
\end{equation}
The remaining part of the proof of Theorem \ref{th3.1} can be finished,
following the same steps as in the previous proof.
\end{proof}

\subsection*{Acknowledgments}
The authors express sincere thanks to Dr. Wenjun Liu, the editors and 
anonymous reviewers for their constructive comments
and suggestions that helped to improve this article. 
This work was supported by the project of Chinese
Ministry of Finance (Grant No. GYHY200906006), the 
National Natural Science Foundation of China
(Grant No. 11301277), and the Natural Science Foundation of 
Jiangsu Province (Grant No. BK20151523).

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\end{document}