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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 212, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/212\hfil Existence and exponential decay]
{Existence and exponential decay of solutions for transmission
problems with delay}

\author[A. Benseghir \hfil EJDE-2014/212\hfilneg]
{Aissa Benseghir}  % in alphabetical order

\address{Aissa Benseghir \newline
Universit\'e Ferhat Abbas de S\'etif, Algeria}
\email{aissa5919@yahoo.fr}

\thanks{Submitted July 14, 2014. Published October 14, 2014.}
\subjclass[2000]{35B37, 35L55, 74D05, 93D15, 93D20}
\keywords{Wave equation; transmission problem; delay term; 
\hfill\break\indent exponential stability}

\begin{abstract}
 In this article we consider a transmission problem in a bounded domain with a
 delay term in the first equation. Under suitable assumptions on the
 weight of the damping and the weight of the delay, we prove the
 existence and the uniqueness of the solution using the semigroup theory.
 Also we show the exponential stability of the solution 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}

In this article, we consider the transmission problem with a delay term,
\begin{equation}  \label{Main_system_Trasmission}
\begin{gathered}
u_{tt}(x,t) -au_{xx}(x,t) +\mu _1u_{t}(
x,t) +\mu _2u_{t}(x,t-\tau ) =0, \quad  (x,t) \in
\Omega \times (0,+\infty ), \\
v_{tt}(x,t) -bv_{xx}(x,t) =0, \quad (x,t)\in (L_1,L_2)\times (0,+\infty ),
\end{gathered}
\end{equation}
where $0<L_1<L_2<L_3$, $\Omega =] 0,L_1[ \cup ]L_2,L_3[ $, $a,b,\mu _1,\mu _2$ 
are positive constants, and $\tau >0$ is the delay.

System \eqref{Main_system_Trasmission} is subjected to the following
boundary  and transmission conditions:
\begin{equation}  \label{Boundary_Condition}
\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{Initial_Conditions}
\begin{gathered}
u(x,0) =u_{0}(x),\quad u_{t}(x,0)
=u_1(x) , \quad  x\in \Omega,  \\
u(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}
\end{equation}
For $\mu_2=0$, system \eqref{Main_system_Trasmission}-\eqref{Initial_Conditions} 
has been investigated in \cite{BstRapo_2007}; for
$\Omega=[0,L_1]$, the authors showed the well-posedness and exponential
stability of the total energy. Mu\~noz Rivera and Oquendo \cite{RO00}
studied the wave propagations over materials consisting of elastic and
viscoelastic components; that is,
\begin{equation}
\begin{gathered}
\rho _1u_{tt}-\alpha _1u_{xx}=0, \quad  x\in ] 0,L_{0}[ ,\; t>0, \\
\rho _2v_{tt}-\alpha _2v_{xx}+\int_{0}^{t}g(
t-s) v_{xx}(s) ds=0, \quad  x\in ] L_{0},L[ ,\; t>0,
\end{gathered} \label{Rivera_Oquendo}
\end{equation}
with the boundary and initial conditions:
\begin{equation}
\begin{gathered}
u(0,t) =v(L,t), \quad u(L_{0},t) =v(L_{0},t) , \quad t>0,  \\
\alpha _1u_{x}(L_{0},t) =\alpha _2v_{x}(L_{0},t)
-\int_{0}^{t}g(t-s) v_{x}(s) ds, \quad t>0,
 \\
u(x,0) =u_{0}(x) ,\quad u_{t}(x,0)=u_1(x) , \quad x\in [0,L_0],  \\
v(x,0) =v_{0}(x) ,\quad v_{t}(x,0)=v_1(x) , \quad x\in [L_0,L],
\end{gathered}  \label{Boundary_Rivera_Oquendo}
\end{equation}
where $\rho _1$ and $\rho _2$ are densities of the materials and $\alpha
_1,\alpha _2$ are elastic coefficients and $g$ is positive exponential
decaying function. They showed that the dissipation produced by the
viscoelastic part is strong enough to produce an exponential decay of the
solution, no matter how small is its size. Ma and Oquendo \cite{MaOquen_2006}
considered transmission problem involving two Euler-Bernoulli equations
modeling the vibrations of a composite beam. By using just one boundary
damping term in the boundary, they showed the global existence and decay
property of the solution. Marzocchi et al \cite{MRN021} investigated
a 1-D semi-linear transmission problem in classical thermoelasticity and
showed that a combination of the first, second and third energies of the
solution decays exponentially to zero, no matter how small the damping
subdomain is. A similar result has sheen shown by Messaoudi and Said-Houari
\cite{MS09}, where a transmission problem in thermoelasticity of type III
has been investigated. See also Marzocchi et al \cite{MRN022} for  a
multidimensional linear thermoelastic transmission problem.

For $\mu_2>0$, problem \eqref{Main_system_Trasmission} has a delay term
in the internal feedback. This delay term may destabilize system
\eqref{Main_system_Trasmission}-\eqref{Initial_Conditions} that is exponentially
stable in the absence of delays \cite{BstRapo_2007}. The effect of the delay
in the stability of hyperbolic systems has been investigated by many people.
See for instance \cite{Dat91,DLP86}.

In \cite{NP06} the authors examined a system of wave equations with a linear
boundary damping term with a delay:
\begin{equation}
\begin{gathered}
u_{tt}-\Delta u=0, \quad  x\in \Omega ,\; t>0 \\
u(x,t)=0, \quad  x\in \Gamma _{0},\ t>0 \\
 \frac{\partial u}{\partial \nu }(x,t)=\mu_1u_{t}(x,t)+
\mu_2u_{t}(x,t-\tau ) \quad  x\in \Gamma _1,\ t>0 \\
u(x,0)=u_{0}(x), \quad  x\in \Omega, \\
u_{t}(x,0)=u_1(x) \quad  x\in \Omega, \\
u_{t}(x,t-\tau)=g_{0}(x,t-\tau) \quad  x\in \Omega,\; \tau\in(0,1)
\end{gathered} \label{delay_1}
\end{equation}
and  under the assumption
\begin{equation}
\mu _2<\mu _1  \label{coeff}
\end{equation}
they proved that the solution is exponentially stable.
On the contrary, if \eqref{coeff}
does not hold, they found a sequence of delays for which the corresponding
solution of \eqref{delay_1} will be unstable. We also recall the result by
Xu et al \cite{XYL06}, where the authors proved the same result as
in \cite{NP06} for the one space dimension by adopting the spectral analysis
approach.

The aim of this article is to study the well-posedness and asymptotic
stability of system \eqref{Main_system_Trasmission}-\eqref{Initial_Conditions}
 provided that \eqref{coeff} is satisfied. The paper is organized as
follows. The well-posedness of the problem is analyzed in Section
\ref{Section_2} using the semigroup theory. In Section \ref{Section_3}, we prove
the exponential decay of the energy when time goes to infinity.

\section{Well-posedness of the problem}\label{Section_2}

In this section, we prove the  existence and the uniqueness of a local
solution of system \eqref{Main_system_Trasmission}-\eqref{Initial_Conditions} by using the semi-group theory. So let us
introduce the following new variable \cite{NP06}
\begin{equation}\label{Change_Variables}
y(x,\rho ,t)=u_{t}(x,t-\tau \rho ).
\end{equation}
Then, we obtain
\begin{equation}\label{Third_Equation}
\tau y_{t}(x,\rho ,t)+y_{\rho }(x,\rho ,t)=0,\quad \text{in }\Omega \times
(0,1)\times (0,+\infty ).
\end{equation}
Therefore,  problem \eqref{Main_system_Trasmission} is equivalent to
\begin{equation}\label{Main_system_2}
\begin{gathered}
u_{tt}(x,t) -au_{xx}(x,t) +\mu _1u_t(x,t) +\mu _2y( x,1,t) =0,
\quad (x,t)\in \Omega \times ] 0,+\infty [ \\
v_{tt}(x,t) -bv_{xx}(x,t) =0, \quad (x,t) \in ] L_1,L_2[ \times ] 0,+\infty [ \\
\tau y_{t}(x,\rho ,t)+y_{\rho }(x,\rho ,t)=0,\quad \text{in }\Omega \times (0,1)\times (0,+\infty )
\end{gathered}
\end{equation}
which together with \eqref{Initial_Conditions} can be rewritten as
\begin{equation}\label{First_Order_system}
\begin{gathered}
 U'=\mathscr{A}U, \\
U(0)=(u_{0},v_{0},u_1,v_1,f_{0}(.,-.\tau ))^{\text{T}},
\end{gathered}
\end{equation}
where the operator $\mathscr{A}$ is defined by
\begin{equation}
\mathscr{A}\begin{pmatrix}
u \\
v \\
\varphi \\
\psi \\
y
\end{pmatrix}
 =\begin{pmatrix}
\varphi \\
\psi \\
au_{xx}-\mu _1\varphi-\mu _2y(.,1) \\
bv_{xx} \\
-\frac{1}{\tau }y_{\rho }
\end{pmatrix}
 \label{e2.5}
\end{equation}
with the domain
\begin{equation*}
D(\mathscr{A})=\big\{
(u,v,\varphi ,\psi ,y)^{\text{T}}\in \mathscr{H};
y(.,0)=\varphi \text{ on }\Omega \big\} ,
\end{equation*}
where
\[
\mathscr{H}=\big\{\big(H^{2}(\Omega )\times H^2(L_1,L_2)\big)\cap X_\ast\big\}
\times H^{1}(\Omega )\times H^{1}(L_1,L_2)\times
L^{2}(0,1,H^{1}(\Omega )).
\]
Here the space $X_\ast$ is defined by
\begin{align*}
X_\ast=\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),\; au_{x}(L_{i},t) =bv_{x}( L_{i},t),\; i=1,2
 \Big\}.
\end{align*}
Now the energy space is defined by
\begin{equation*}
\mathscr{K}= X_\ast \times L^{2}(\Omega )\times
L^{2}(L_1,L_2)\times L^{2}((\Omega )\times (0,1)).
\end{equation*}
Let
\[
U=(u,v,\varphi ,\psi ,y)^{\text{T}},\quad \bar{U}=(\bar{u},\bar{v},\bar{
\varphi },\bar{\psi },\bar{y})^{\text{T}}.
\]
Then, for a positive constant $\zeta $ satisfying
\begin{equation}\label{zeta}
\tau\mu_2\leq \zeta\leq \tau(2\mu_1-\mu_2),
\end{equation}
we define the inner product in $\mathscr{K}$ as
follows:
\[
(U,\bar{U})_{\mathscr{K}}
=\int_{\Omega }\{\varphi \bar{\varphi }+au_{x}
\bar{u}_{x}\}\,dx+\int_{L_1}^{L_2}\{\psi \bar{\psi }+bv_{x}
\bar{v_{x}}\}\,dx+\zeta \int_{\Omega }\int_{0}^{1}y(x,\rho )\bar{y}
(x,\rho )\,d\rho\,dx\,.
\]
The existence and uniqueness result is stated as follows.

\begin{theorem}\label{Theorem_1}
For any $U_{0}\in \mathscr{K}$ there exists a unique solution
$U\in C([0,+\infty [ ,\mathscr{K})$ of problem \eqref{First_Order_system}. Moreover, if $U_{0}\in D(\mathscr{A})$, then
\[
U\in C([0,+\infty [ ,D(\mathscr{A}))\cap C^{1}([0,+\infty [ ,\mathscr{K}).
\]
\end{theorem}

\begin{proof}
To prove the result stated in Theorem \ref{Theorem_1}, we use the semigroup
theory, that is, we show that the operator $\mathscr{A}$ generates a
$C_{0}$-semigroup in $\mathscr{K}$. In this step, we concern ourselves to
prove that the operator $\mathscr{A}$ is dissipative. Indeed, for
$U=(u,v,\varphi ,\psi ,y)^{\text{T}}\in D(\mathscr{A})$,
where $\varphi (L_2)=\psi (L_2)$ and $\zeta $ is a positive constant, we have
\begin{equation} \label{Scalar_Product}
\begin{aligned}
(\mathscr{A}U,U)_{\mathscr{K}}
&= a\int_{\Omega }u_{xx}\varphi
\,dx+b\int_{L_1}^{L_2}v_{xx}\psi \,dx-\mu _1\int_{\Omega }\varphi^2 \,dx
 \\
&\quad -\mu _2\int_{\Omega }y(.,1)\varphi \,dx-\frac{\zeta }{\tau }\int_{\Omega
}\int_{0}^{1}y(x,\rho )y_{\rho }(x,\rho )\,d\rho\,dx\\
&\quad +a\int_\Omega u_x\varphi_x\,dx+b\int_{L_1}^{L_2}v_x\psi_x\,dx\,.
\end{aligned}
\end{equation}
Looking now at the last term of the right-hand side of \eqref{Scalar_Product},
we have
\begin{equation} \label{y_identity}
\begin{aligned}
\zeta \int_{\Omega }\int_{0}^{1}y(x,\rho )y_{\rho }(x,\rho )\,d\rho\,dx
&= \zeta \ \int_{\Omega }\frac{1}{2}\frac{\partial }{\partial \rho }
y^{2}(x,\rho )\,d\rho\,dx   \\
&= \frac{\zeta }{2}\ \int_{\Omega }(y^{2}(x,1)-y^{2}(x,0))\,dx.
\end{aligned}
\end{equation}
Integrating by parts in \eqref{Scalar_Product}, keeping in mind the
fact that $y(x,0,t)=\varphi(x,t)$ and using \eqref{y_identity}, we have
from \eqref{Scalar_Product}
\begin{equation} \label{Product_1}
\begin{aligned}
(\mathscr{A}U,U)_{\mathscr{K}}
&=  a[u_x\varphi]_{\partial\Omega} +b[v_x\psi]_{L_1}^{L_2}
-\big(\mu _1-\frac{\zeta}{2\tau}\big)\int_{\Omega }\varphi^2 \,dx\\
&\quad -\mu _2\int_{\Omega }y(.,1)\varphi \,dx
 -\frac{\zeta }{2\tau }\int_{\Omega}y^{2}(x,1)\,dx.
\end{aligned}
\end{equation}
Using Young's inequality, \eqref{Boundary_Condition},
and the equality $\varphi (L_2)=\psi (L_2)$,  from \eqref{Product_1}, we obtain
\begin{equation} \label{Product_2}
\big(\mathscr{A}U,U\big)_{\mathscr{K}}
 \leq-\big(\mu _1-\frac{\zeta}{2\tau}-\frac{\mu_2}{2}\big)
\int_{\Omega }\varphi^2 \,dx-\big(\frac{\zeta }{2\tau }-\frac{\mu_2}{2}\big)
\int_{\Omega }y^{2}(x,1)\,dx.
\end{equation}
Consequently, using \eqref{zeta},  we deduce that
$(\mathscr{A}U,U)_{\mathscr{K}}\leq 0$. Thus, the operator $\mathscr{A}$
is dissipative.

Now to show that the operator $\mathscr{A}$ is
maximal monotone, it is sufficient to show that the operator
$\lambda I-\mathscr{A}$ is surjective for  a fixed $\lambda >0$.
Indeed, given $ (f_1,g_1,f_2,g_2,h)^{\text{T}}\in \mathscr{K}$, we
seek $U=(u, v,\varphi ,\psi,y)^{\text{T}}\in D(\mathscr{A})$ solution of
\begin{equation}\label{Equation_system}
\begin{pmatrix}
\lambda u-\varphi  \\
\lambda v-\psi  \\
\lambda \varphi -au_{xx}+\mu _1y(.,0)+\mu _2y(.,1) \\
\lambda \psi -bv_{xx} \\
\lambda y+\frac{1}{\tau }y_{\rho }
\end{pmatrix}
= \begin{pmatrix}
f_1 \\
g_1 \\
f_2 \\
g_2 \\
h
\end{pmatrix}
\end{equation}
suppose we have find $(u,v)$ with the appropriate regularity, then
\begin{equation} \label{u_v_solution}
\begin{gathered}
\varphi  = \lambda u-f_1 \\
\psi  = \lambda v-g_1 .
\end{gathered}
\end{equation}
It is clear that $\varphi \in H^{1}(\Omega )$ and
$\psi \in H^{1}(L_1,L_2)$, furthermore, by \eqref{Equation_system},
we can find $y$ as $y(x,0)=\varphi (x)$, $x\in \Omega $, using the approach
as in Nicaise and Pignotti \cite{NP06}, we obtain, by using the equation
in \eqref{Equation_system}
\[
y(x,\rho )=\varphi (x)e^{-\lambda \rho \tau }+\tau e^{-\lambda \rho \tau
}\int_{0}^{\rho }h(x,\sigma )e^{\lambda \sigma \tau }d\sigma\,.
\]
From \eqref{u_v_solution}, we obtain
\[
y(x,\rho )=\lambda u(x)e^{-\lambda \rho \tau }-f_1(x)e^{-\lambda \rho \tau
}+\tau e^{-\lambda \rho \tau }\int_{0}^{\rho }h(x,\sigma )e^{\lambda \sigma
\tau }d\sigma\,.
\]
By using \eqref{Equation_system} and \eqref{u_v_solution},
the functions $u,v$ satisfy the
following equations:
\begin{equation} \label{u_v_equation_2}
\begin{gathered}
\lambda ^{2}u-au_{xx}+\mu _1y(.,0)+\mu _2y(.,1) = f_2+\lambda f_1 \\
\lambda ^{2}v-bv_{xx} = g_2+\lambda g_1\,.
\end{gathered}
\end{equation}
Since
\begin{align*}
y(x,1) &= \varphi (x)e^{-\lambda \tau }
+\tau e^{-\lambda \tau}\int_{0}^{1}h(x,\sigma )e^{\lambda \tau }d\sigma  \\
&= \lambda ue^{-\lambda \tau }+y_{0}(x),
\end{align*}
for $x\in \Omega $, we have
\begin{equation}
y_{0}(x)=-f_1(x)+\tau e^{-\lambda \tau }\int_{0}^{1}h(x,\sigma )e^{\lambda
\tau }d\sigma   \label{e2.14}
\end{equation}
The problem \eqref{u_v_equation_2} can be reformulated  as
\begin{equation} \label{Lax_Milgram_form}
\begin{gathered}
\begin{aligned}
&\int_{\Omega }(\lambda ^{2}u-au_{xx}+\mu _1\lambda u++\mu _2\lambda
ue^{-\lambda \tau })\omega _1\,dx   \\
&= \int_{\Omega }(f_2+\lambda f_1-\mu _2\lambda y_{0}(x))\omega _1\,dx,
\end{aligned}  \\
\int_{L_1}^{L_2}(\lambda ^{2}v-bv_{xx})\omega _2\,dx
= \int_{L_1}^{L_2}(g_2+\lambda g_1)\omega _2\,dx,
\end{gathered}
\end{equation}
for any $(\omega_1,\omega_2)\in X_\ast$.

Integrating the first equation in \eqref{Lax_Milgram_form}  by parts, we obtain
\begin{equation}\label{Lax_Milgram_2}
\begin{aligned}
&\int_{\Omega }(\lambda ^{2}u-au_{xx}+\mu _1u +\mu _2\lambda
ue^{-\lambda \tau })\omega _1\,dx   \\
&= \int_{\Omega }\lambda ^{2}u\omega _1\,dx
 -a\int_{\Omega }u_{xx}\omega_1\,dx
+\mu _1\int_{\Omega }\lambda u\,dx
 +\mu _2\int_{\Omega }\lambda ue^{-\lambda \tau }\omega _1\,dx   \\
&= \int_{\Omega }\lambda ^{2}u\omega _1\,dx
 +a\int_{\Omega }u_{x}(\omega _1)_{x}\,dx
 -[au_{x}\omega _1]_{\partial\Omega }\\
&\quad +\mu _1\int_{\Omega }\lambda u\,dx
 +\mu _2\int_{\Omega }\lambda ue^{-\lambda \tau }\omega _1\,dx
\\
&= \int_{\Omega }(\lambda ^{2}+\mu _1\lambda +\mu _2\lambda e^{-\lambda
\tau })u\omega _1\,dx+a\int_{\Omega }u_{x}(\omega _1)_{x}\,dx
-[au_{x}\omega _1]_{\partial\Omega }\,.
\end{aligned}
\end{equation}
Integrating the second equation in \eqref{Lax_Milgram_form} by parts, we obtain
\begin{equation} \label{Lax_Milgram_3}
\int_{L_1}^{L_2}(\lambda ^{2}v-bv_{xx})\omega_2\,dx
=\int_{L_1}^{L_2}\lambda ^{2}v \omega_2\,dx
+b\int_{L_1}^{L_2}v_{x}(\omega _2)_{x}\,dx-[bv_{x}\omega_2]_{L_1}^{L_2}\,.
\end{equation}
Using \eqref{Lax_Milgram_2} and \eqref{Lax_Milgram_3}, the problem
\eqref{Lax_Milgram_form} is equivalent to the problem
\begin{equation}\label{Lax_Milgram_4}
\Phi ((u,v),(\omega _1,\omega _2))=l(\omega _1,\omega _2)
\end{equation}
where the bilinear form $\Phi :{(X_\ast\times X_\ast)}\to\mathbb{R}$
 and the linear form $l: {X_\ast}\to \mathbb{R}$ are defined by
\begin{align*}
\Phi ((u,v),(\omega _1,\omega _2))
&= \int_{\Omega }(\lambda ^{2}+\mu _1\lambda
 +\mu _2\lambda e^{-\lambda \tau })u\omega _1\,dx+a\int_{\Omega}u_{x}(\omega _1)_{x}\,dx
-[au_{x}\omega _1]_{\partial\Omega } \\
&\quad +\int_{L_1}^{L_2}\lambda ^{2}v\ \omega
_2\,dx+b\int_{L_1}^{L_2}v_{x}(\omega _2)_{x}\,dx-[bv_{x}\omega_2]_{L_1}^{L_2}
\end{align*}
and
\[
l(\omega _1,\omega _2) = \int_{\Omega }(f_2+\lambda f_1-\mu
_2\lambda y_{0}(x))\omega _1\,dx+\int_{L_1}^{L_2}(g_2+
\lambda g_1)\omega _2\,dx\,.
\]
Using the properties of the space $X_\ast$, it is clear that  $\Phi $
is continuous and coercive, and $l$ is continuous. So applying the Lax-Milgram
theorem, we deduce that for all $(\omega _1,\omega _2)\in X_\ast$,
problem \eqref{Lax_Milgram_4}  admits a unique solution $(u,v)\in X_\ast$.
It follows from \eqref{Lax_Milgram_2} and \eqref{Lax_Milgram_3} that
$(u,v)\in \{\big(H^{2}(\Omega )\times H^2(L_1,L_2)\big)\cap X_\ast\}$.
Therefore, the operator $\lambda I-\mathscr{A}$ is dissipative
for any $\lambda >0$. Then the result in Theorem \ref{Theorem_1}
follows from the Hille-Yoshida theorem.
\end{proof}

\section{Exponential decay of solutions}
\label{Section_3}

In this section we study the asymptotic behavior of the system
\eqref{Main_system_Trasmission}-\eqref{Initial_Conditions}. For any regular
solution of \eqref{Main_system_Trasmission}-\eqref{Initial_Conditions}, we
define the energy as
\begin{gather}  \label{Energy_E_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{Energy_E_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}
The total energy is defined as
\begin{equation}  \label{Total_Energy}
E(t)=E_1(t)+E_2(t) +\frac{\zeta}{2}\int_{\Omega}\int_{0}^{1}y^2(x,
\rho,t)\,d\rho\,dx
\end{equation}
where $\zeta $ is the positive constant defined by \eqref{zeta}.
Our decay result reads as follows.

\begin{theorem}\label{Theorem_2}
Let $(u,v)$ be the solution of
\eqref{Main_system_Trasmission}-\eqref{Initial_Conditions}. Assume that
$\mu_2>\mu_1$ and
\begin{equation}  \label{Assumption_L_1_L_2}
\frac{a}{b}<\frac{L_3+L_1-L_2}{2(L_2-L_1)}.
\end{equation}
Then there exist two positive constants $C$ and $d$, such that
\begin{equation}  \label{Energy_Exponential}
E(t)\leq Ce^{-d t},\quad \forall t\geq 0.
\end{equation}
\end{theorem}

\begin{remark}\label{Remark_Conditions} \rm
Assumption  \eqref{Assumption_L_1_L_2} gives the relationship
between the boundary regions and the transmission permitted. It can be also
seen as a restriction on the wave speeds of the two equations and the damped
 part of the domain.   It is known that for Timoshenko systems \cite{Souf99}
and Bresse systems \cite{ARA10} that the wave speeds always control the decay
rate of the solution. It is an interesting open question to show the behavior
of the solution if \eqref{Assumption_L_1_L_2} is not satisfied.
\end{remark}

For the proof of Theorem \ref{Theorem_2} we use the following lemmas.

\begin{lemma}\label{Lemma_Decay}
Let $(u,v,y)$ be the solution of \eqref{Main_system_2},
\eqref{Initial_Conditions}. Assume that $\mu _1\geq \mu _2$. Then we have
the  inequality
\begin{equation}   \label{dE_dt}
\frac{dE(t)}{dt}\leq \big(-\mu_1+\frac{\mu_2}{2}+\frac{\zeta }{2\tau }
\big)\int_{\Omega }y^{2}(x,0,t)\,dx+\big(\frac{\mu _2}{2}-\frac{\zeta }{
2\tau }\big)\int_{\Omega }y^{2}(x,1,t)\,dx.
\end{equation}
\end{lemma}

\begin{proof}
From \eqref{Total_Energy} we have
\begin{equation}  \label{dE_1_dt}
\frac{dE_1(t)}{dt}=\int_{\Omega }u_{tt}(x,t)u_{t}(x,t)\,dx+a\int_{\Omega
}u_{xt}(x,t)u_{x}(x,t)\,dx\,.
\end{equation}
Using system \eqref{Main_system_2}, and integrating by parts, we obtain
\begin{equation}  \label{dE_1_dt_1}
\frac{dE_1(t)}{dt} = a[u_xu_t]_{\partial\Omega}-\mu _1\int_{\Omega
}u_{t}^2(x,t)-\mu _2\int_{\Omega }u_{t}(x,t)y(x,1,t) )\,dx\,.
\end{equation}
On the other hand,
\begin{equation}  \label{dE_2}
\frac{dE_2(t)}{dt}=b[v_{x}v_{t}]_{L_1}^{L_2}.
\end{equation}
Using the fact that
\begin{equation}  \label{rho_term}
\begin{aligned}
\frac{d}{dt}\frac{\zeta}{2}\int_{\Omega}\int_{0}^{1}y^2(x,\rho,t)\,d\rho\,dx
&= \zeta\int_{\Omega}\int_{0}^{1}y(x,\rho,t)y_t(x,\rho,t)\,d\rho\,dx
\\
&=-\frac{\zeta}{\tau}\int_{\Omega}\int_0^1y_{\rho}(x,\rho,t)y(x,\rho,t)d
\rho \,dx   \\
&=-\frac{\zeta}{2\tau}\int_{\Omega}\int_0^1\frac{d}{d\rho}
y^2(x,\rho,t)\,d\rho\,dx   \\
&=-\frac{\zeta}{2\tau}\int_{\Omega}(y^2(x,1,t)-y^2(x,0,t))\,dx\,,
\end{aligned}
\end{equation}
collecting \eqref{dE_1_dt_1}, \eqref{dE_2}, \eqref{rho_term}, using
\eqref{Boundary_Condition} and applying Young's inequality, we show that
\eqref{dE_dt} holds. The proof is complete.
\end{proof}

Following \cite{ANP10}, we define  the functional
\begin{equation*}
I(t)=\int_{\Omega }\int_{t-\tau }^{t}e^{s-t}u_{t}^{2}(x,s)\,ds\,dx\,,
\end{equation*}
and state the following lemma.

\begin{lemma}\label{Lemma_I}
 Let $(u,v)$ be the solution of
\eqref{Main_system_Trasmission}-\eqref{Initial_Conditions}. Then
\begin{equation}  \label{dI_dt_1}
\frac{dI(t)}{dt}\leq \int_{\Omega }u_{t}^{2}(x,t)\,dx-e^{-\tau }\int_{\Omega
}u_{t}^{2}(x,t-\tau )\,dx-e^{-\tau }\int_{\Omega }\int_{t-\tau
}^{t}u_{t}^{2}(x,s)\,ds\,dx.
\end{equation}
\end{lemma}

The proof of the above Lemma  is straightforward, so we omit it.
Now, we define the functional $\mathscr{D}(t)$ as follows
\begin{equation}  \label{Functional_D}
\mathscr{D}(t)=\int_\Omega uu_t\,dx+\frac{\mu_1}{2}\int_\Omega
u^2\,dx+\int_{L_1}^{L_2} vv_t \,dx.
\end{equation}
Then, we have the following estimate.

\begin{lemma}\label{Lemma_D}
The functional $\mathscr{D}(t)$ satisfies 
\begin{equation}  \label{D_dt_estimate}
\begin{aligned}
\frac{d}{dt}\mathscr{D}(t)
&\leq -(a-\epsilon_0c_0^2)\int_\Omega
u_x^2\,dx-b\int_{L_1}^{L_2} v_x^2\,dx   \\
&\quad +\int_\Omega u_t^2\,dx+\int_{L_1}^{L_2} v_t ^2\,dx
+C(\epsilon_0)\int_{\Omega }y^{2}(x,1,t )\,dx
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathscr{D}(t)$ with respect to $t$ and using
\eqref{Main_system_Trasmission}, we find that
\begin{equation}  \label{dD_dt_1}
\begin{aligned}
\frac{d}{dt}\mathscr{D}(t)
&= \int_\Omega u_t^2\,dx+\int_{L_1}^{L_2} v_t
^2\,dx-a\int_\Omega u_x^2\,dx-b\int_{L_1}^{L_2} v_x^2\,dx   \\
&\quad -\mu_2\int_{\Omega }u(x,t)y(x,1,t)\,dx+[au_xu]_{\partial\Omega}+
[bv_xv]_{L_1}^{L_2}.
\end{aligned}
\end{equation}
Applying young's inequality and using the boundary conditions
\eqref{Boundary_Condition}, we have
\begin{equation}
\begin{aligned}{}
[au_xu]_{\partial\Omega}+[bv_xv]_{L_1}^{L_2}
&= au_x(L_1,t)u(L_1,t)-au_x(L_2,t)u(L_2,t)   \\
&\quad +bv_x(L_2,t)v(L_2,t)-bv_x(L_1,t)v(L_1,t)=0.  
\end{aligned}
\end{equation}
On the other hand, we have by Poincar\'e's inequality and Young's inequality,
\begin{equation}
\mu_2\int_{\Omega }u(x,t)y(x,1,t)\,dx\leq \epsilon_0c^2_0\int_\Omega
u_x^2\,dx+C(\epsilon_0)\int_{\Omega }y^{2}(x,1,t )\,dx
\end{equation}
where $\epsilon_0$ is a positive constants and $c_0$ is the Poinca\'e's
constant. Consequently, plugging the above estimates into \eqref{dD_dt_1},
we find \eqref{D_dt_estimate}.
\end{proof}

Now, inspired by \cite{MRN021}, we introduce the functional
\begin{equation}  \label{q_function}
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_2-L_3-L_1}{2(L_2-L_1) }(x-L_1) +
\frac{L_1}{2},      & x\in [ L_1,L_2]
\end{cases}
\end{equation}
Next, we define the functionals
\[
\mathscr{F} _1(t)=-\int_{\Omega }q(x)u_{x}u_{t}\,dx,\quad
\mathscr{F} _2(t)=-\int_{L_1}^{L_2}q(x)v_{x}v_{t}\,dx.
\]
Then, we have the following estimates.

\begin{lemma}\label{Lemma_J_1_J_2}
For any $\epsilon_2>0$, we have the estimates:
\begin{equation}  \label{J_1_estimate}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_1(t)
&\leq C(\epsilon_2)\int_{\Omega }u_{t}^2\,dx
+\big(\frac{a}{2}+\epsilon_2\big)\int_{\Omega
}u_{x}^2\,dx+C(\epsilon_2)\int_{\Omega }y^2(x,1,t)\,dx   \\
&\quad -\frac{a}{4}[(L_3-L_2)u_x^2(L_2,t)+L_1u_x^2(L_1,t)]
\end{aligned}
\end{equation}
and
\begin{equation}  \label{J_2_estimate}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_2(t) 
&\leq \frac{L_2-L_3-L_1}{4( _2-L_1) }(\int_{L_1 }^{L_2}v_{t}^2\,dx+\int_{L_1
}^{L_2}bv_{x}^2\,dx)   \\
&\quad +\frac{b}{4}\big((L_3-L_2)v_x^2(L_2,t)+L_1v_x^2(L_1,t)\big).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Taking the derivative of $\mathscr{F}_1(t)$ with respect to $t$ and using
 \eqref{Main_system_Trasmission}, we obtain
\begin{equation}  \label{dF_1_dt_1}
\begin{aligned}
&\frac{d}{dt}\mathscr{F}_1(t)\\
&= -\int_{\Omega }q(x)u_{tx}u_{t}\,dx-\int_{\Omega}q(x)u_{x}u_{tt}\,dx   \\
&= -\int_{\Omega }q(x)u_{tx}u_{t}\,dx
-\int_{\Omega }q(x)u_{x}(au_{xx}(x,t) -\mu _1u_t(x,t) -\mu _2y(x,1,t))\,dx.
\end{aligned}
\end{equation}
Integrating by parts, 
\begin{equation}  \label{First_term}
\int_{\Omega }q(x)u_{tx}u_{t}\,dx=-\frac{1}{2}\int_{\Omega
}q'(x)u_{t}^2\,dx+\frac{1}{2}[ q(x)u_t^2]_{\partial\Omega}.
\end{equation}
On the other hand,
\begin{equation}  \label{Second_term}
\int_{\Omega }aq(x)u_{xx}u_{x}\,dx=-\frac{1}{2}\int_{\Omega
}aq'(x)u_{x}^2\,dx+\frac{1}{2}[ aq(x)u_x^2]_{\partial\Omega}.
\end{equation}
Substituting  \eqref{First_term} and \eqref{Second_term} in \eqref{dF_1_dt_1},
we find that
\begin{equation}  \label{dF_1_dt_2}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_1(t)
&= \frac{1}{2}\int_{\Omega }q'(x)u_{t}^2\,dx+
\frac{1}{2}\int_{\Omega }aq'(x)u_{x}^2\,dx-\frac{1}{2}[ q(x)u_t^2
]_{\partial\Omega}\\
&\quad -\frac{1}{2}[ aq(x)u_x^2]_{\partial\Omega}
 +\int_{\Omega }q(x)u_{x}\big(\mu _1u_t(x,t) +\mu _2y(x,1,t)\big)\,dx\,.
\end{aligned}
\end{equation}
Using Young's inequality and  \eqref{q_function}, equation
\eqref{dF_1_dt_2} becomes 
\begin{equation} \label{dF_1_dt_3}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_1(t)
&\leq C(\epsilon_2)\int_{\Omega }u_{t}^2\,dx+\big(
\frac{a}{2}+\epsilon_2\big)\int_{\Omega }u_{x}^2\,dx-\frac{1}{2}[
q(x)u_t^2]_{\partial\Omega}\\
&\quad -\frac{a}{2}[ q(x)u_x^2]_{\partial\Omega}  
 +C(\epsilon_2)\int_{\Omega }y^2(x,1,t)\,dx.
\end{aligned}
\end{equation}
for any $\epsilon_2>0$. Since $q(L_1)>0$ and $q(L_2)<0$,  by using
the boundary conditions \eqref{Boundary_Condition}, we have
\begin{equation}  \label{Boundary_1}
\frac{1}{2}[ q(x)u_t^2]_{\partial\Omega}\geq 0.
\end{equation}
Also, we have
\begin{equation}  \label{Boundary_2}
\begin{aligned}
-\frac{a}{2}[ q(x)u_x^2]_{\partial\Omega}
&=-\frac{aL_1}{4}[ u_x^2(L_1,t)+u_x^2(0,t)]\\
&\quad -\frac{a(L_3-L_2)}{4}[ u_x^2(L_3,t)+u_x^2(L_2,t)].
\end{aligned}
\end{equation}
Taking into account \eqref{Boundary_1} and \eqref{Boundary_2}, then
\eqref{dF_1_dt_3} gives \eqref{J_1_estimate}.

By the same method, taking the derivative of $\mathscr{F}_2(t)$ with respect
to $t$, we obtain
\begin{equation}  \label{F_2_1}
\begin{aligned}
\frac{d}{dt}\mathscr{F}_2(t)
&= -\int_{L_1}^{L_2}q(x)v_{tx}v_{t}\,dx-\int_{L_1}^{L_2}q(x)v_{x}v_{tt}   \\
&=  \frac{1}{2}\int_{L_1 }^{L_2}q'(x)v_{t}^2\,dx-\frac{1}{2}[
q(x)v_t^2]_{L_1}^{L_2}+\frac{1}{2}\int_{L_1}^{L_2}bq' (x)v_{x}^2\,dx
 -\frac{b}{2}[ q(x)v_x^2]_{L_1}^{L_2}   \\
&\leq \frac{L_2-L_3-L_1}{4(L_2-L_1) }\Big(\int_{L_1 }^{L_2}v_{t}^2\,dx
+\int_{L_1 }^{L_2}bv_{x}^2\,dx\Big)   \\
&\quad +\frac{b}{4}\big((L_3-L_2)v_x^2(L_2,t)+L_1v_x^2(L_1,t)\big).
\end{aligned}
\end{equation}
which is exactly \eqref{J_2_estimate}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Theorem_2}]
We define the Lyapunov functional
\begin{equation}  \label{Functional_L}
\mathscr{L} (t)=NE(t)+I(t)+\gamma_2\mathscr{D}(t)+\gamma_3 \mathscr{F}
_1(t)+\gamma _{4}\mathscr{F} _2(t),
\end{equation}
where $N,\gamma _2,\gamma_3$ and $\gamma _{4}$ are positive constants that
will be fixed later.

Now, it is clear from the boundary conditions \eqref{Boundary_Condition},
that
\begin{equation}  \label{u_v}
a^2u_x^2(L_i,t)=b^2v^2_x(L_i,t), \quad i=1,2.
\end{equation}
Taking the derivative of \eqref{Functional_L} with respect to $t$ and making
use of \eqref{dE_dt}, \eqref{dI_dt_1}, \eqref{D_dt_estimate},
\eqref{J_1_estimate}, \eqref{J_1_estimate} and taking into account \eqref{u_v},
we obtain
\begin{equation}  \label{dL_dt_main}
\begin{aligned}
\frac{d}{dt}\mathscr{L} (t)
&\leq \big\{N\big(-\mu _1+\frac{\mu _2}{2}
+\frac{\zeta }{2\tau })+1+\gamma _2+\gamma _3C(\epsilon _2)\big\}
\int_{\Omega }u_t^{2}\,dx   \\
&\quad +\big\{N\big(\frac{\mu _2}{2}-\frac{\zeta }{2\tau })-e^{-\tau
}+\gamma _2C(\epsilon _{0})+C(\epsilon_2)\gamma _3\big\}
\int_{\Omega}y^{2}(x,1,t)\,dx   \\
&\quad+\big\{\gamma_2(-a+\epsilon _{0}c_{0}^{2})+\gamma _3\epsilon_2+\frac{
\gamma _3a}{2}\big\}\int_{\Omega }u_{x}^{2}\,dx \\
&\quad +\big\{b\frac{L_2-L_3-L_1}{4(_2-L_1)}\gamma _{4}-\gamma
_2b\big\}\int_{L_1}^{L_2}v_{x}^{2}\,dx   \\
&\quad +\big\{\frac{L_2-L_3-L_1}{4(L_2-L_1)}\gamma _{4}+\gamma
_2\big\}\int_{L_1}^{L_2}v_{t}^{2}\,dx-e^{-\tau }\int_{\Omega
}\int_{t-\tau }^{t}u_{t}^{2}(x,s)\,ds\,dx   \\
&\quad -(\gamma_3-\frac{a}{b}\gamma_4)\frac{a(L_3-L_2)}{4}
u^2_x(L_2,t)-(\gamma_3-\frac{a}{b}\gamma_4)\frac{aL_1}{4}
u_x^2(L_1,t).
\end{aligned}
\end{equation}
At this point, we choose our constants in \eqref{dL_dt_main}, carefully,
such that all the coefficients in \eqref{dL_dt_main} will be negative.
Indeed, under the assumption \eqref{Assumption_L_1_L_2}, we can always find
$\gamma_2,\gamma_3$ and $\gamma_4$ such that
\begin{equation}  \label{Coefficient}
\frac{L_2-L_3-L_1}{4(L_2-L_1)}\gamma _{4}+\gamma _2<0,\quad
\gamma_3>\frac{a}{b}\gamma_4,\quad \gamma_2>\frac{\gamma_3}{2}.
\end{equation}
Once the above constants are fixed, we may choose $\epsilon_2$ and
$\epsilon_0$ small enough such that
\begin{equation*}
\epsilon_0c_0^2+\gamma_3\epsilon_2<a(\gamma_2-\gamma_3/2).
\end{equation*}
Finally, keeping in mind \eqref{zeta} and choosing $N$ large enough such
that the first and the second coefficients in \eqref{dL_dt_main} are
negatives.

Consequently, from the above, we deduce that there exist a positive constant $
\eta _1, $ such that \eqref{dL_dt_main} becomes
\begin{equation}  \label{dL_dt_main_3}
\begin{aligned}
\frac{d\mathscr{L} (t)}{dt}
&\leq -\eta _1\int_{\Omega
}(u_{t}^{2}(x,t)+u_{x}^{2}(x,t)+u_{t}^{2}(x,t-\tau ))\,dx   \\
&\quad -\eta _1\int_{L_1}^{L_2}(v_{t}^{2}(x,t)+v_{x}^{2}(x,t))\,dx-\eta
_1\int_{\Omega }\int_{t-\tau }^{t}u_{t}^{2}(x,s)\,ds\,dx \,.
\end{aligned}
\end{equation}
Consequently, recalling \eqref{Total_Energy},  we deduce that there
exist also $\eta _2>0$, such that
\begin{equation}  \label{L_E_estimate}
\frac{d\mathscr{L} (t)}{dt}\leq -\eta _2E(t),\quad \forall t\geq 0 .
\end{equation}
On the other hand, it is not hard to see that from \eqref{Functional_L} and
for $N$ large enough, there exist two positive constants $\beta _1$ and
$\beta _2$ such that
\begin{equation}  \label{Equiv_L_E}
\beta _1\ E(t)\leq \mathscr{L} (t)\leq \beta _2E(t),\quad \forall t\geq
0.
\end{equation}
Combining \eqref{L_E_estimate} and \eqref{L_E_estimate}, we deduce that
there exists $\Lambda >0$ for which the estimate
\begin{equation}  \label{L_estimate_last}
\frac{d\mathscr{L} (t)}{dt}\leq -\Lambda \mathscr{L} (t),\quad \forall
t\geq 0,
\end{equation}
holds. Integrating \eqref{L_E_estimate} over $(0,t)$ and using (\eqref
{L_E_estimate} once again, then \eqref{Energy_Exponential} holds. Then, the
proof is complete.
\end{proof}

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\end{document}