\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 174, pp. 1--13.\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-2017/174\hfil Transmission problem with distributed delay]
{Well-posedness and exponential decay of solutions for a
transmission problem with distributed delay}

\author[G. Liu \hfil EJDE-2017/174\hfilneg]
{Gongwei Liu}

\address{Gongwei Liu \newline
College of Science,
Henan University of Technology,
Zhengzhou 450001, China}
\email{gongweiliu@126.com}


\thanks{Submitted April 10, 2017. Published July 10, 2017.}
\subjclass[2010]{35B37, 35L55, 74D05, 93D15, 93D20}
\keywords{Transmission problem; distributed delay; exponential decay}

\begin{abstract}
 In this article, we consider a transmission problem in a bounded domain with
 a distributed delay in the first equation.  Using a semigroup theorem,
 we prove the existence and uniqueness of global solution under suitable
 assumptions on the weight of damping and the weight of distributed delay.
 Also we establish the exponential stability of the solution by introducing
 a suitable Lyapunov functional.
\end{abstract}

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

\section{Introduction}

In this article, we study the  transmission problem with a distributed delay,
\begin{equation}\label{1.1}
\begin{gathered}
u_{tt}(x, t)-au_{xx}(x, t)+\mu_1u_t(x, t)
 +\int_{\tau_1}^{\tau_2}\mu_2(s)u_t(t-s)ds=0,
 \quad  x\in \Omega, \; t>0,\\
v_{tt}(x,t)-bv_{xx}(x,t)=0 ,\quad  x\in (L_1, L_2),\;  t\geq0,
\end{gathered}
\end{equation}
under the boundary and the transmission conditions
\begin{equation}\label{1.2}
\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)=bv_x(L_i, t),\quad i=1,2
\end{gathered}
\end{equation}
and the initial conditions
\begin{equation}\label{1.3}
\begin{gathered}
u(x, 0)=u_0(x),\quad u_t(x, 0)=u_1(x),\quad x\in \Omega,\\
v(x, 0)=v_0(x),\quad v_t(x, 0)=v_1(x),\quad  x\in (L_1, L_2),\\
u_t(x, -t)=f_0(x, -t),\quad x\in \Omega,\quad t\in (0, \tau_2)
\end{gathered}
\end{equation}
where $0<L_1<L_2<L_3$, $\Omega=(0,L_1)\cup (L_2, L_3)$, $a, b, \mu_1$
are positive constants, and the initial data $(u_0, u_1, v_0, v_1, f_0)$
belongs to suitable space. Moreover,  $\mu_2: [\tau_1, \tau_2]\to \mathbb{R}$
 is a bounded function, where $\tau_1$ and $\tau_2$ are two real number
satisfying $0\leq \tau_1<\tau_2$.

It is  known that transmission problems  happen frequently
in applications where the domain is occupied by two or several materials, whose
elastic properties are different, joined together over the whole of a surface.
From the mathematical point of view, a transmission problem for wave propagation
consists on a hyperbolic equation for which the corresponding elliptic operator
 has discontinuous coefficients, see \cite{BaRa, Lions1990}.

In absence of delay ($\mu_2(s)=0$), the system \eqref{1.1}-\eqref{1.3} has been
investigated in \cite{BaRa} by Bastaos and Raposo; for $\Omega=[0, L_1]$,
they showed that the well-posedness and exponential stability of the total energy.
 Rivera and  Oquendo \cite{RiOq2000} studied the transmission problem of
viscoelastic waves and established that the dissipation produced by the
viscoelastic part is strong enough to produce the exponential stability,
no matter small its size is.
Interested readers are referred to
\cite{MaOp2006, MaRi2002MMAS, MaRi2002IMA, MeSH2009}, for more results concerning
 other types of transmission problems.

Introducing the delay term  makes the problem different from those considered in
the literatures. Delay effect  arises in many applications  depending not only
on the present state but also on some past occurrences.
It may turn a well-behaved system into a wild one. The
presence of delay may be a source of instability. For example, it was shown
in \cite{Dakto1986, Dakto1988, Guesmia2013,  Nicaise2006, Nicaise2008, Xu2006}  that
an arbitrarily small delay may destabilize a system that is uniformly asymptotically
stable in the absence of delay unless additional control terms have been used.
Here we mention the some interesting results on the relation
between the delay term and source term
\cite{LiuZhang2016, LiuYZ2017, Feng2017, Nicaise2015}.

Nicaise and Pignotti \cite{Nicaise2008} considered the wave equation with
liner frictional damping and internal distributed delay
\begin{equation*}
  u_{tt}-\Delta u+\mu_1u_t+a(x)\int_{\tau_1}^{\tau_2}\mu_2(s)u_{t}(t-s)ds=0
  \end{equation*}
in $\Omega\times (0, \infty)$, with initial and mixed Dirichlet-Neumann
boundary conditions and $a$ is a suitable function.
They obtained exponential decay of  the solution under the assumption that
\begin{equation*}
  \|a\|_{\infty}\int_{\tau_1}^{\tau_2}\mu_2(s)ds<\mu_1.
\end{equation*}
The authors also obtained the same result when the distributed delay acted on
the part of the boundary. Mustafa and Kafini \cite{MuKa2013} considered a
thermoelastic system with internal distributed delay, they obtained exponential
stability under suitable condition; for the boundary distributed delay,
similar result was obtained by \cite{Mu2014}. Here we also mention the work
on Timoshenko system with second sound and internal distributed delay
in \cite{Ap2014} by Apalara, and wave equation with strong distributed
delay \cite{Me2016JMP} by Messsaoudi et al.

The effect of the delay term  $u_t(x, t-\tau)$ in the transmission system has
been investigated by Benseghir \cite{Be2014}. Recently, the well-posedness
and the decay of solution for a transmission problem in a bounded domain
with a viscoelastic term and a delay term $u_{t}(x, t-\tau)$  have been studied
in \cite{Li2016EJDE, Wang2016JNSA}.

In this  work we consider the transmission system \eqref{1.1}-\eqref{1.3},
and prove the well-posedness and the exponential stability.
Our work extends the stability results in \cite{BaRa, Be2014} to the
transmission system with distributed delay.

 The plan of this paper is as follows.
In section 2, we present some notations and assumptions needed for our work,
and then establish the well-posedness of our problem by virtue of
 the semigroup methods. In section 3, we state and prove the stability result
by introducing a suitable Lyapunov function.


\section{Well-posedness of the problem}

 Throughout this paper, $c$ and  $c_i$ are used to denote the generic positive
constant. From now on, we shall omit $x$ and $t$ in all functions of $x$
and $t$ if there is no ambiguity.

 As in \cite{Nicaise2008}, we introduce the new variable
 \begin{equation*}
   z(x, \rho, t, s)=u_t(x, t-\rho s), \quad x\in \Omega,\;\rho\in(0,1),\; t>0,\;
 s\in(\tau_1, \tau_2).
 \end{equation*}
Then the above variable $z$ satisfies
\begin{equation}\label{2.1}
  sz_{t}(x, \rho, t, s)+z_{\rho}(x, \rho, t, s)=0, \quad
 x\in \Omega,\;\rho\in(0,1),\; t>0,\; s\in(\tau_1, \tau_2).
\end{equation}
Consequently, system \eqref{1.1} is equivalent to
\begin{equation}\label{2.2}
\begin{gathered}
u_{tt}(x, t)-au_{xx}(x, t)+\mu_1u_t(x, t)+\int_{\tau_1}^{\tau_2}
 \mu_2(s)z(x, 1, t, s)ds=0,  \\\  x\in \Omega, \; t>0,\\
 v_{tt}(x,t)-bv_{xx}(x,t)=0 ,\quad   x\in (L_1, L_2),\;  t\geq0,\\
  sz_{t}(x, \rho, t, s)+z_{\rho}(x, \rho, t, s)=0, \quad
 x\in \Omega,\;\rho\in(0,1),\; t>0,\; s\in(\tau_1, \tau_2).
\end{gathered}
\end{equation}
Defining $U=(u, v, \varphi, \psi, z)^T$, we formally get that $U$
satisfies
\begin{equation}\label{2.3}
\begin{gathered}
U'=\mathcal{A}U,\\
U(0)=U_0=(u_0, v_0, u_1, v_1, f_0),
\end{gathered}
\end{equation}
where the operator $\mathcal{A}$ is defined as
\[
 \mathcal{A}\begin{pmatrix}u\\ v \\ \varphi \\ \psi \\ z \end{pmatrix}
=\begin{pmatrix}\varphi\\ \psi \\ au_{xx}-\mu_1\varphi
-\int_{\tau_1}^{\tau_2}\mu_2(s)z(x, 1, t, s)ds \\
bv_{xx} \\ -\frac{1}{s}z_{\rho}(x, \rho, t, s)
\end{pmatrix}.
\]
Introducing the  space
\begin{align*}
  X_{*}=\big\{ &(u, v)=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),\; au_x(L_i, t)=bv_x(L_i, t),\; i=1, 2 \big\},
\end{align*}
we define the energy space as
\begin{equation*}
  \mathcal{H}=X_{*}\times L^2(\Omega) \times L^2(L_1, L_2)
\times L^2\big(\Omega\times(0,1)\times(\tau_1, \tau_2) \big)
\end{equation*}
equipped with the inner product
\begin{align*}
  \langle U, \tilde{U}\rangle_{\mathcal{H}}
&=\int_{\Omega}(\varphi \tilde{\varphi}+au_x\tilde{u}_x)dx
 +\int_{L_1}^{L_2}(\psi\tilde{\psi}+bv_x\tilde{v}_x)dx \\
&\quad +\int_{\Omega}\int_0^1\int_{\tau_1}^{\tau_2}
 s|\mu_2(s)|z(x,\rho,s)\tilde{z}(x,\rho,s)\,ds\,d\rho\,\,dx.
\end{align*}
The domain of $\mathcal{A}$ is
\begin{align*}
  D(\mathcal{A})
=\big\{ &(u, v, \varphi, \psi, z)^{T}\in \mathcal{H}:
 (u, v)\in (H^2(\Omega)\times H^2(L_1, L_2))\cap X_{*}, \\
& \varphi \in H^1(\Omega ),\;\psi \in H^1(L_1, L_2),
 z(x, 0, s)=\varphi,\\
& z, z_{\rho}\in  L^2\big(\Omega\times(0,1)\times(\tau_1, \tau_2) \big)  \big\}.
\end{align*}
Clearly, $D(\mathcal{A})$ is dense in $\mathcal{H}$.

Concerning the weight of the distributed delay, we assume that
\begin{equation}\label{2.4}
  \int_{\tau_1}^{\tau_2}|\mu_2(s)|ds \leq \mu_1.
\end{equation}

The well-posedness of the system \eqref{2.2}, \eqref{1.2}and \eqref{1.3}
is ensured by the following theorem.

\begin{theorem}\label{thm2.1}
 Under the assumption \eqref{2.4}, for any $U_0\in \mathcal{H}$, there exists
a unique weak solution $U\in C(\mathbb{R}^+, \mathcal{H})$ of problem \eqref{2.3}.
Moreover, if $U_0\in D(\mathcal{A})$, then
$U\in C(\mathbb{R}^+, D(\mathcal{A}))\cap C(\mathbb{R}^+, \mathcal{H})$.
\end{theorem}

\begin{proof}
We use the semigroup approach and the Hille-Yosida theorem to prove the
well-posedness  of the problem. First, we prove that the operator
$\mathcal{A}$ is dissipative.
Indeed, for $U=(u, v, \varphi, \psi, z)\in D(\mathcal{A})$, where
$\varphi(L_i)=\psi(L_i), i=1, 2$, we have
\begin{equation}\label{2.5}
 \begin{split}
&\langle \mathcal{A}U, U\rangle_{\mathcal{H}}\\
&=\int_{\Omega}\big(au_{xx}-\mu_1\varphi-\int_{\tau_1}^{\tau_2}\mu_2(s)z(x, 1, t, s)ds\big)\varphi\,dx+a\int_{\Omega}u_x\varphi_xdx\\
&\quad +\int_{L_1}^{L_2}bv_{xx}\psi\,dx+\int_{L_1}^{L_2}bv_x\psi_x\,dx
+\int_{\Omega}\int_{\tau_1}^{\tau_2}\int_0^1|\mu_2(s)|zz_{\rho}d \rho \,ds\,dx.
\end{split}
\end{equation}
For the last term of the right hand side of \eqref{2.5}, we have

\begin{equation}\label{2.6}
  \begin{split}
&\int_{\Omega}\int_{\tau_1}^{\tau_2}\int_0^1|\mu_2(s)|zz_{\rho}d \rho \,ds\,dx\\
&=\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}\int_0^1|\mu_2(s)
 |\frac{d}{d\rho}|z(x,\rho,t,s)|^2d\rho \,ds\,dx\\
&\quad + \frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,s)ds
x-\frac{1}{2}\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\int_{\Omega}z^2(x,0,s)dx.
\end{split}
\end{equation}
Integrating by parts in \eqref{2.5}, and noticing the fact
$z(x, 0, t, s)=\varphi(x, t)$, from \eqref{2.6}, we have
\begin{align*}
\langle \mathcal{A}U, U\rangle_{\mathcal{H}}
&=[au_x\varphi]_{\partial\Omega}+[bv_x\psi]_{L_1}^{L_2}
-\Big(\mu_1-\frac{1}{2}\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\Big)
 \int_{\Omega}\varphi^2dx\\
&\quad +\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x, 1, s)\,ds\,dx
-\int_{\Omega}\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1,s)\varphi \,ds\,dx.
\end{align*}
Using Young's inequality, and the equality $\varphi(L_i)=\psi(L_i)$, $i=1, 2$,
 from \eqref{1.2} and \eqref{2.6} we have
\begin{align*}
\langle \mathcal{A}U, U\rangle_{\mathcal{H}}
&\leq- (\mu_1-\int_{\tau_1}^{\tau_2}|\mu_2(s)|)\int_{\Omega}\varphi^2dx
 -\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x, 1, s)\,ds\,dx\\
&\quad +\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x, 1, s)\,ds\,dx
\\
&\leq - \Big(\mu_1-\int_{\tau_1}^{\tau_2}|\mu_2(s)|\Big)
\int_{\Omega}\varphi^2dx\leq0,
\end{align*}
by \eqref{2.4}. Hence, the operator $\mathcal{A}$ is dissipative.

Next, we prove the operator $\mathcal{A}$ is maximal. It is sufficient to show
that the operator $\lambda I-\mathcal{A}$ is surjective for a fixed $\lambda>0$.
Indeed, given $\mathcal{F}=(f_1, f_2, f_3, f_4, f_5)\in \mathcal{H}$,
we  prove that there exists $U=(u, v, \varphi, \psi, z)\in D(\mathcal{A})$ satisfying
\begin{equation}\label{2.7}
 (\lambda I- \mathcal{A})U=\mathcal{F},
\end{equation}
that is
\begin{equation}\label{2.8}
  \begin{gathered}
   \lambda u-\varphi=f_1,\\
\lambda v-\psi=f_2,\\
\lambda \varphi-au_{xx}+\mu_1\varphi+\int_{\tau_1}^{\tau_2}|\mu_2(s)|z(x,1,t,s)ds=f_3,\\
\lambda \psi-bv_{xx}=f_4,\\
\lambda sz+z_{\rho}=sf_5.
  \end{gathered}
\end{equation}
Suppose we have obtained $(u, v)$ with the suitable regularity, then
\begin{equation}\label{2.9}
  \begin{gathered}
   \varphi=\lambda u-f_1,\\
\psi=\lambda v-f_2,
  \end{gathered}
\end{equation}
so we have $\varphi \in H^1(\Omega)$ and $\psi \in H^1(L_1, L_2)$.
Moreover, using the approach as in Nicaise and Pignotti \cite{Nicaise2006},
 we obtain that the last equation in \eqref{2.8} with $z(x, 0, s)$ has a unique
solution
\begin{equation*}
  z(x,\rho, s)=\varphi(x)e^{-\lambda \rho s}+se^{\lambda \rho s}\int_0^{\rho}e^{\lambda \sigma s}f_5(x, \sigma, s)d\sigma.
\end{equation*}
It follows from \eqref{2.9} that
\begin{equation}\label{2.10}
  z(x,\rho, s)=\lambda ue^{-\lambda \rho s}-f_1e^{-\lambda \rho s}
+se^{\lambda \rho s}\int_0^{\rho}e^{\lambda \sigma s} f_5(x, \sigma, s)d\sigma,
\end{equation}
in particular, $z(x, 1, s)=\lambda ue^{-\lambda s}+z_0(x, s)$ with
$z_0\in L^{2}(\Omega\times (\tau_1, \tau_2))$ defined by
\begin{equation*}
  z_0(x, s)= -f_1e^{-\lambda s}
+se^{\lambda  s}\int_0^{1}e^{\lambda \sigma s} f_5(x, \sigma, s)d\sigma.
\end{equation*}
By \eqref{2.8} and \eqref{2.9}, the functions $(u, v)$ satisfy the  equations
\begin{equation}\label{2.11}
  \begin{gathered}
     \tilde{k}u-au_{xx}=\tilde{f},\\
\lambda^2v-bv_{xx}=f_2+\lambda f_4,
   \end{gathered}
\end{equation}
where
\begin{gather*}
  \tilde{k}=\lambda^2+\lambda \mu_1
+\int_{\tau_1}^{\tau_2}\lambda |\mu_2(s)|e^{-\lambda s}ds>0, \\
 \tilde{f}=f_3+(\lambda+\lambda\mu_1)f_1
-\int_{\tau_1}^{\tau_2}|\mu_2(s)|z_0(x,s)ds\in L^2(\Omega),
\end{gather*}
which can be reformulated as
\begin{equation}\label{2.12}
  \begin{gathered}
    \int_{\Omega}( \tilde{k}u-au_{xx})w_1dx=\int_{\Omega}\tilde{f}w_1dx,\\
\int_{L_1}^{L_2}(\lambda^2v-bv_{xx})w_2dx=\int_{L_1}^{L_2}(f_2+\lambda f_4)w_2dx,
   \end{gathered}
\end{equation}
for any $(w_1, w_2)\in X_{*}$.

Integrating by parts in \eqref{2.12}, we obtain that the variational formulation
corresponding to \eqref{2.11} takes the form
\begin{equation}\label{2.13}
  \Phi\big((u, v), (w_1, w_2)\big)=l(w_1, w_2),
\end{equation}
where the bilinear form $\Phi: (X_{*}, X_{*})\to \mathbb{R}$ and the linear
form $l: X_{*}\to \mathbb{R}$ are defined by
\begin{align*}
 \Phi\big((u, v), (w_1, w_2)\big)
&=\int_{\Omega}\tilde{k}uw_1dx+\int_{\Omega}au_xw_{1x}
 -[au_xw_1]_{\Omega}+\int_{L_1}^{L_2}\lambda^2vw_2dx\\
&\quad +\int_{L_1}^{L_2}v_xw_{2x}dx-[bv_xw_2]_{L_1}^{L_2},
\end{align*}
and
\begin{equation*}
  l(w_1, w_2)=\int_{\Omega}\tilde{f}w_1dx+\int_{L_1}^{L_2}(f_2+\lambda f_4)w_2dx.
\end{equation*}
By 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 deduce that problem l \eqref{2.13}
admits a unique solution $(u, v)\in X_*$ for all $(w_1, w_2)\in X_*$.
It follows from \eqref{2.11} that
$(u, v)\in \big((H^2(\Omega)\times H^2(L_1, L_2))\big)\cap X_*$.
Thus, the operator $\lambda I-\mathcal{A}$ is surjective for any $\lambda>0$.
Hence the Hille-Yosida theorem guarantees the existence of a unique solution
to the problem \eqref{2.7}. This completes the proof.
\end{proof}

\section{Exponential stability}

In this section, we state and prove the stability result for the energy of
the system \eqref{1.1}-\eqref{1.3}.
For the regular solution of the system \eqref{1.1}-\eqref{1.3}, we define
the energy as (see \cite{Be2014})
\begin{gather}\label{3.1}
  E_1(t)=\frac{1}{2}\int_{\Omega}u_t^2(x,t)dx+\frac{a}{2}\int_{\Omega}u_x^2(x,t)dx,\\
\label{3.2}
  E_2(t)=\frac{1}{2}\int_{L_1}^{L_2}v_t^2(x,t)dx
+\frac{b}{2}\int_{L_1}^{L_2}v_x^2(x,t)dx.
\end{gather}
 And the total energy is defined as
\begin{equation}\label{3.3}
  E(t)=E_1(t)+E_2(t)+\frac{1}{2}\int_{\Omega}
\int_0^1\int_{\tau_1}^{\tau_2}s|\mu_2(s)|z^2(x,\rho, t, s)\,ds\,d\rho\,dx.
\end{equation}

For the energy decay result, we assume a restriction on the weight of the
distribute delay and the damping as
\begin{equation}\label{3.4}
  \int_{\tau_1}^{\tau_2}|\mu_2(s)|ds<\mu_1.
\end{equation}
The  stability result reads as follows.

\begin{theorem} \label{thm3.1}
Let $(u, v, z)$ be the solution of the system \eqref{2.2}, \eqref{1.2}
and \eqref{1.3}. Assume \eqref{3.4} and
\begin{equation}\label{3.5}
  \frac{a}{b}<\frac{L_1+L_3-L_2}{2(L_2-L_1)},\quad L_3>3(L_2-L_1).
\end{equation}
Then there exist two positive constants $K$ and $\kappa$, such that
\begin{equation}\label{3.6}
E(t)\leq Ke^{-\kappa t}, \quad \forall t\geq0.
\end{equation}
\end{theorem}

 The proof will be established through the following Lemmas.

\begin{lemma} \label{lem3.1}
Let  assumption \eqref{3.4} holds. Then the energy functional defined by
\eqref{3.3}, satisfies the estimate
\begin{equation}\label{3.7}
 E'(t)\leq -\Big(\mu_1-\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\Big)
\int_{\Omega}u_t^2(x, t)dx\leq0.
\end{equation}
\end{lemma}

\begin{proof}
By differentiating \eqref{3.1}, using the first equation in \eqref{2.2},
and integrating by parts, we obtain
\begin{equation*}
  E_1'(t)=[au_xu_t]_{\Omega}-\mu_1\int_{\Omega}u_t^2(x,t)dx
-\int_{\Omega}\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1, t, s)u_t(x, t)\,ds\,dx.
\end{equation*}
Similarly,
\begin{equation*}
  E_2'(t)=[bv_xv_t]_{L_1}^{L_2}.
\end{equation*}
Noticing that $z(x, 0, t, s)=u_t(x, t)$, from \eqref{2.2},  we obtain
\begin{align*}
 &\frac{1}{2}\frac{d}{dt}\int_{\Omega}
 \int_0^1\int_{\tau_1}^{\tau_2}s|\mu_2(s)|z^2(x, \rho, t, s)\,ds\,d\rho\,dx\\
&=-\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx
 +\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|u_t^2(x, t)\,ds\,dx.
\end{align*}
Meanwhile, using Young's inequality, we have
\begin{align*}
&-\int_{\Omega}\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1, t, s)u_t(x, t)\,ds\,dx\\
&\leq\frac{1}{2}\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\int_{\Omega}u_t^2(x,t)dx
 +\frac{1}{2}\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx.
\end{align*}

Combining the above equalities and using \eqref{3.4}, we show that
\eqref{3.7} holds, where we also use the fact
$[au_xu_t]_{\partial\Omega}=[bv_xv_t]_{L_1}^{L_2}$ from \eqref{1.2}.
\end{proof}

As in \cite{Me2016JMP}, we define the functional
\begin{equation*}
  I(t)=\int_{\Omega}\int_0^1\int_{\tau_1}^{\tau_2}se^{-\rho s}|\mu_2(s)|z^2
(x,\rho, t, s)\,ds\,d\rho\,dx,
\end{equation*}
then we have the following estimate.

\begin{lemma} \label{lem3.2}
The functional $I(t)$ satisfies the  estimate
\begin{equation}\label{3.8}
  \begin{split}
   I'(t)&\leq -e^{-\tau_2}\int_{\Omega}\int_{\tau_1}^{\tau_2}
 |\mu_2(s)|z^2(x, 1, t, s)\,ds\,dx\\
&\quad +\Big(\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\Big)\int_{\Omega}u_t^2(x, t)dx\\
&\quad -e^{-\tau_2}\int_{\Omega}\int_0^1\int_{\tau_1}^{\tau_2}s|\mu_2(s)
 |z^2(x, \rho, t, s)\,ds\,d\rho\,dx.
  \end{split}
\end{equation}
\end{lemma}

\begin{proof}
By differentiating $I(t)$ and using the third equation in \eqref{2.2}, we obtain
\begin{align*}
  I'(t)&= -\int_{\Omega}\int_{\tau_1}^{\tau_2}\int_0^1e^{-\rho s}
 |\mu_2(s)|\frac{d}{d\rho}z^2(x,\rho, t,s)d\rho \,ds\,dx\\
&=-\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|\int_0^1\frac{d}{d\rho}
 (e^{-\rho s}z^2(x,\rho,t,s))d\rho \,ds\,dx \\
&\quad -\int_{\Omega}\int_0^1\int_{\tau_1}^{\tau_2}s|\mu_2(s)|e^{-\rho s}
 z^2(x, \rho, t, s)\,ds\,d\rho\,dx.
  \end{align*}
Hence
\begin{align*}
I'(t)&=-\int_{\Omega}\int_{\tau_1}^{\tau_2}e^{-s}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx
+\Big(\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds\Big)\int_{\Omega}u_t^2(x, t)dx\\
&\quad -\int_{\Omega}\int_{\tau_1}^{\tau_2}s|\mu_2(s)|
\int_0^1e^{-\rho s}z^2(x,rho,t,s)\,d\rho \,ds\,dx.
\end{align*}
Recalling $e^{-s}\leq e^{-\rho s}\leq1$, for all $\rho\in[0, 1]$, and
$-e^{-s}\leq -e^{-\tau_2}$, for all $s\in[\tau_1, \tau_2]$, we obtain \eqref{3.8}.
\end{proof}

Now we define the functional
\begin{equation*}
 \mathcal{D}(t)=\int_{\Omega}uu_tdx+\frac{\mu_1}{2}\int_{\Omega}u^2dx
+\int_{L_1}^{L_2}vv_tdx.
\end{equation*}
Then we have the following estimate.

\begin{lemma} \label{lem3.3}
The functional $\mathcal{D}(t)$ satisfies
\begin{equation}\label{3.9}
\begin{split}
\mathcal{D}'(t)
&\leq -(a-\varepsilon_0 C_0^2)\int_{\Omega}u_x^2dx-b\int_{L_1}^{L_2}v_x^2dx
 +\int_{\Omega}u_t^2dx+\int_{L_1}^{L_2}v_t^2dx\\
&\quad +\frac{1}{4\varepsilon_0}\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
 \int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx.
\end{split}
\end{equation}
\end{lemma}

\begin{proof} Taking the derivative of  $\mathcal{D}(t)$ with respect to $t$,
using \eqref{2.2}, we obtain
\begin{equation}\label{3.10}
  \begin{split}
    \mathcal{D}'(t)
&=\int_{\Omega}u_t^2dx+\int_{L_1}^{L_2}v_t^2dx-a\int_{\Omega}u_x^2dx
 -\int_{L_1}^{L_2}v_x^2dx\\
&\quad -\int_{\Omega}\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1,t,s)u(x,t)\,ds\,dx
 +[au_xu]_{\partial\Omega}+[bv_xv]_{L_1}^{L_2}.
  \end{split}
\end{equation}
It follows from the boundary condition \eqref{1.2} that
\begin{equation*}
[au_xu]_{\partial\Omega}+[bv_xv]_{L_1}^{L_2}=0.
\end{equation*}
Using the boundary condition \eqref{1.2}, we obtain
\begin{gather*}
  u^2(x,t)=\big(\int_0^xu_x(x,t)dx\big)^2\leq L_1\int_0^{L_1}u_x^2(x,t)dx,\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)dx,\,\,\,x\in[L_2, L_3],
\end{gather*}
which imply the following Poincar\'e's inequality
\begin{equation}\label{3.11}
  \int_{\Omega}u^2(x,t)dx\leq C_0^2\int_{\Omega}u_x^2(x,t)dx,\quad
x\in \Omega,
\end{equation}
where $C_0=\max\{L_1, L_3-L_2\}$ is the Poincar\'e's constant.
Using Young's inequality and \eqref{3.11}, we have
\begin{align*}
  &-\int_{\Omega}\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1,t,s)u(x,t)\,ds\,dx\\
&\leq \varepsilon_0C_0^2\int_{\Omega}u_x^2(x,t)dx
+\frac{1}{4\varepsilon_0}\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
\int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx,
\end{align*}
for any $\varepsilon_0>0$. Inserting the above estimates in \eqref{3.10},
then \eqref{3.9} is fulfilled.
\end{proof}

Inspired by \cite{MaRi2002MMAS}, we introduce the functional
\begin{equation}\label{3.12}
 q(x)=\begin{cases}
x-\frac{L_1}{2}, &x\in[0,L_1],\\
x-\frac{L_2+L_3}{2}, &x\in[L_2,L_3],\\
\frac{L_1}{2}+\frac{L_2-L_3-L_1}{2(L_2-L_1)}(x-L_1), &x\in[L_1,L_2].
\end{cases}
\end{equation}
We define the  two functionals
\begin{equation*}
  \mathcal{F}_1(t)=-\int_{\Omega}q(x)u_xu_t\,dx,\quad
\mathcal{F}_2(t)=-\int_{L_1}^{L_2}q(x)v_xv_t\,dx.
\end{equation*}
Then, we have the following estimates.

\begin{lemma} \label{lem3.4}
For any $\varepsilon_1>0$, the functionals $\mathcal{F}(t)$ and
$\mathcal{F}_2(t)$  satisfy
\begin{equation}\label{3.13}
\begin{split}
\mathcal{F}'_1(t)
&\leq C(\varepsilon_1)\int_{\Omega}u_t^2dx+(\frac{a}{2}+\varepsilon_1)
 \int_{\Omega}u_x^2dx\\
&\quad + C(\varepsilon_1)\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
 \int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^{2}(x,1,t,s)\,ds\,dx\\
&\quad -\frac{a}{4}[(L_3-L_2)u_x^2(L_2,t)+L_1u_x^2(L_1,t)]
 -\frac{1}{4}[L_1u_t^2(L_1,t)+(L_3-L_2)u_t^2(L_2,t)],
\end{split}
\end{equation}
and
\begin{equation}\label{3.14}
\begin{split}
\mathcal{F}'_2(t)
&= -\frac{L_1+L_3-L_2}{4(L_2-L_1)}\big(\int_{L_1}^{L_2}v_t^2dx
 +\int_{L_1}^{L_2}bv^2_xdx\big)+\frac{L_1}{4}v_t^2(L_1,t)\\
&\quad +\frac{L_3-L_2}{4}v_t^2(L_2, t)+\frac{b}{4}[(L_3-L_2)v_x^2(L_2, t)
 +L_1v_x^2(L_1, t)].
\end{split}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathcal{F}_1(t)$ with respect to $t$ and
using \eqref{2.2}, we have
\begin{equation}\label{3.15}
\begin{split}
\mathcal{F}'_1(t)
&=-\int_{\Omega}q(x)u_xu_{tt}-\int_{\Omega}q(x)u_{xt}u_t\,dx\\
&=-\int_{\Omega}q(x)u_x\big(au_{xx}-\mu_1u_t
 -\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1,t,s)ds\big)\,dx\\
&\quad -\int_{\Omega}q(x)u_{xt}u_t\,dx.
\end{split}
\end{equation}
Integrating by parts, we have
\begin{gather*}
  \int_{\Omega}q(x)u_{xt}u_tdx=-\frac{1}{2}\int_{\Omega}q'(x)u_t^2dx
 +\frac{1}{2}[q(x)u_t^2]_{\partial\Omega}, \\
 \int_{\Omega}q(x)au_xu_{xx}dx=-\frac{1}{2}\int_{\Omega}aq'(x)u_x^2dx
+\frac{1}{2}[aq(x)u_x^2]_{\partial\Omega}.
\end{gather*}
Inserting the above two equalities into \eqref{3.15}, and noticing
 \eqref{3.12} and Young's inequality, we obtain
\begin{equation}\label{3.16}
\begin{split}
\mathcal{F}'_1(t)
&=\frac{1}{2}\int_{\Omega}u_t^2dx+\frac{1}{2}\int_{\Omega}u_x^2dx
 -\frac{1}{2}[aq(x)u_x^2]_{\partial\Omega}\\
&\quad -\frac{1}{2}[q(x)u_t^2]_{\partial\Omega}+\int_{\Omega}q(x)u_x\big(\mu_1u_t
 +\int_{\tau_1}^{\tau_2}\mu_2(s)z(x,1,t,s)ds\big)dx\\
&\leq C(\varepsilon_1)\int_{\Omega}u_t^2dx+(\frac{a}{2}+\varepsilon_1)
 \int_{\Omega}u_x^2dx-\frac{1}{2}[aq(x)u_x^2]_{\partial\Omega}
 -\frac{1}{2}[q(x)u_t^2]_{\partial\Omega}\\
&\quad +C(\varepsilon_1)\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
 \int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^{2}(x,1,t,s)\,ds\,dx,
\end{split}
\end{equation}
for any $\varepsilon_1>0$. On the other hand, by the boundary conditions
\eqref{1.2}, we have
\begin{gather*}
\frac{1}{2}[q(x)u_t^2]_{\partial\Omega}
=\frac{1}{4}[L_1u_t^2(L_1,t)+(L_3-L_2)u_t^2(L_2,t)]\geq0, \\
 \frac{1}{2}[aq(x)u_x^2]_{\partial\Omega}=\frac{a}{4}[(L_3-L_2)u_x^2(L_2,t)
+L_1u_x^2(L_1,t)].
\end{gather*}
Inserting the above two equalities into \eqref{3.16}, then \eqref{3.16}
gives \eqref{3.13}.

By the same method, taking the derivative of  $\mathcal{F}_2(t)$ with respect
to $t$, we have
\begin{align*}
\mathcal{F}'_2(t)
&=-\int_{L_1}^{L_2}q(x)v_{xt}v_tdx-\int_{L_1}^{L_2}q(x)v_xv_{tt}dx\\
&=\frac{1}{2}\int_{L_1}^{L_2}q'(x)v_t^2dx-\frac{1}{2}[q(x)v_t^2]_{L_1}^{L_2}
 +\frac{1}{2}\int_{L_1}^{L_2}bq'(x)v_x^2dx-\frac{1}{2}[bq(x)v_x^2]_{L_1}^{L_2}\\
&= -\frac{L_1+L_3-L_2}{4(L_2-L_1)}\big(\int_{L_1}^{L_2}v_t^2dx
 +\int_{L_1}^{L_2}bv^2_xdx\big)+\frac{L_1}{4}v_t^2(L_1,t)\\
&+\frac{L_3-L_2}{4}v_t^2(L_2, t)+\frac{b}{4}[(L_3-L_2)v_x^2(L_2, t)
 +L_1v_x^2(L_1, t)].
\end{align*}
Hence, the proof  is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm3.1}]
 We define the Lyapunov functional
\begin{equation}\label{3.17}
  L(t)=N_1E(t)+N_2I(t)+\gamma_1\mathcal{F}_1(t)+\gamma_2\mathcal{F}_2(t)
+\gamma_3\mathcal{D}(t),
\end{equation}
where $N_1, N_2, \gamma_1, \gamma_2, \gamma_3$ are positive constants that
will be chosen later.

It follows from the boundary conditions \eqref{1.2} that
\begin{equation}\label{3.18}
 a^2u_x^2(L_i, t)=bv_x^2(L_i, t),\quad i=1, 2.
\end{equation}
Taking the derivative of \eqref{3.17} with respective to $t$, using the above
lemmas and \eqref{3.18}, we have
\begin{equation}\label{3.19}
\begin{split}
&L'(t) \\
&\leq-\Big\{N_1(\mu_1-\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds)
 -N_2\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
 -\gamma_1C(\varepsilon_1)-\gamma_3\Big\}\int_{\Omega}u_t^2dx\\
&\quad -\Big\{N_2e^{-\tau_2}-\gamma_1C(\varepsilon_1)
 \int_{\tau_1}^{\tau_2}|\mu_2(s)|ds-\gamma_3
 \frac{\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds}{4\varepsilon_0}\Big\} \\
&\quad\times \int_{\Omega}\int_{\tau_1}^{\tau_2}|\mu_2(s)|z^2(x,1,t,s)\,ds\,dx\\
&\quad -\big\{(a-\varepsilon_0C_0^2)\gamma_3-(\frac{a}{2}+\varepsilon_1)\gamma_1\big\}
 \int_{\Omega} u_x^2dx \\
&\quad -\big\{\frac{L_1+L_3-L_2}{4(L_2-L_1)}\gamma_2+\gamma_3\big\}
 \int_{L_1}^{L_2}bv_x^2dx\\
&-\big\{\frac{L_1+L_3-L_2}{4(L_2-L_1)}\gamma_2-\gamma_3\big\}
\int_{L_1}^{L_2}v_t^2dx \\
&\quad -N_2e^{-\tau_2}\int_{\Omega}\int_0^1
 \int_{\tau_1}^{\tau_2}s|\mu_2(s)|z^2(x,\rho,t,s)\,ds\,d\rho\,dx\\
&\quad -\big\{\gamma_1-\gamma_2\big\}
\big(\frac{L_1}{4}u_t^2(L_1,t)+\frac{L_3-L_2}{4}u_t^2(L_2, t)\big)\\
&\quad -\big\{\gamma_1-\frac{a}{b}\gamma_2\big\}
 \big(\frac{a}{4}[L_1u_x^2(L_1,t)+(L_3-L_2)u_x^2(L_2,t)] \big).
\end{split}
\end{equation}
At this point we will choose all the constants, carefully, such that all
the coefficients in \eqref{3.19} will be negative. In fact,
it follows from the assumption \eqref{3.5} that we can always choose
$\gamma_1, \gamma_2$ and $\gamma_3$ such that
\begin{equation*}
  \frac{L_1+L_3-L_2}{4(L_2-L_1)}\gamma_2-\gamma_3>0,\quad
\gamma_1>\frac{a}{b}\gamma_2, \quad \gamma_1>\gamma_2,
\quad \gamma_3>\frac{\gamma_1}{2}.
\end{equation*}
Once the above constants $\gamma_1, \gamma_2, \gamma_3$ are fixed,
we may choose $\varepsilon_0$ and $\varepsilon_1$ sufficiently small such that
\begin{equation*}
  \gamma_3\varepsilon_0C_0^2+\gamma_1\varepsilon_1<a(\gamma_3-\frac{\gamma_1}{2}).
\end{equation*}
Then we can take $N_2$ sufficiently large such that
\begin{equation*}
  N_2e^{-\tau_2}-\gamma_1C(\varepsilon_1)\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds
-\gamma_3\frac{\int_{\tau_1}^{\tau_2}|\mu_2(s)|ds}{4\varepsilon_0}>0.
\end{equation*}
Finally, noticing the assumption \eqref{3.4}, we can always choose $N_1$
sufficiently large such that the first coefficient in \eqref{3.19} is negative.

Thus, we obtain  that there exists a positive constant $\alpha$ such
that \eqref{3.19} yields
\begin{align*}
  L'(t)&\leq-\alpha\Big(\int_{\Omega}u_t^2dx+\int_{\Omega}au_{xx}^2dx
 +\int_{L_1}^{L_2}v_t^2dx \\
&\quad +\int_{L_1}^{L_2}bv_{xx}^2dx
 +\int_{\Omega}\int_0^1\int_{\tau_1}^{\tau_2}s|\mu_2(s)|
 z^2(x,\rho, t, s)\,ds\,d\rho\,dx\Big),
\end{align*}
recalling \eqref{3.3}, which implies
\begin{equation}\label{3.20}
 L'(t)\leq -\frac{\alpha}{2}E(t), \quad \forall \geq0.
\end{equation}

On the hand, it is not hard to see that $L(t)\sim E(t)$, i.e.
there exist two positive constants $\beta_1$ and $\beta_2$ such that
\begin{equation}\label{3.21}
 \beta_1E(t)\leq L(t)\leq\beta_2E(t),\quad t\geq0.
\end{equation}
Combining \eqref{3.20} and \eqref{3.21}, we obtain that
\begin{equation*}
  L'(t)\leq -\kappa L(t),\quad t\geq0
\end{equation*}
for the positive constant $\kappa=\alpha/\beta_2$.
Integration over $(0, t)$ gives
\begin{equation*}
  L(t)\leq L(0)e^{-\kappa t}, \quad t\geq0,
\end{equation*}
recall \eqref{3.21} again, then \eqref{3.6} holds. Hence,
the proof is complete.
\end{proof}


\subsection*{Acknowledgments}
The authors would like to thank the referees for the careful reading
of this paper and for the valuable suggestions to improve the
presentation and the style of the paper. This project is supported by
Key Scientific Research Foundation of the Higher
Education Institutions of Henan Province, China (Grant No.15A110017 )and
the Basic Research Foundation of Henan University of Technology,
China (No.2013JCYJ11).


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\end{document}