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\affiliation{Mohamed Houasni$^{1}$ and Salah Zitouni$^{2}$}
{$^{1}$ Facult\'{e} des Sciences et de la Technologie, Universit\'{e} DBKM,\\
Khemis miliana, Alg\'{e}rie\\
$^{2}$ Department of Mathematics and Informatics  \\
Souk Ahras Univ, P.O. Box 1553,
Souk Ahras, 41000, Algeria}



\begin{document}

\title{Exponential stability for a delayed flexible structure }
\author{}
\maketitle

\begin{abstract}
In this paper, we consider a non-uniform and delayed flexible structure with
temperature and micro-temperature effects. We prove the well-posed of the
problem using semi-groups theory, as well as an exponential stability using
the multiplier method under a small condition on time delay.
\end{abstract}

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\section{Introduction}

In this paper, we aim to study the following inhomogeneous flexible
structure system with micro-temperature effect:
\begin{equation}
\left\{
\begin{array}{l}
m(x)u_{tt}-(p(x)u_{x}+2\delta (x)u_{xt})_{x}+dw_{x}+\eta \theta _{x}+\mu
u_{t}\left( x,t-\tau _{0}\right) =0, \\
c\theta _{t}-k\theta _{xx}+\eta u_{tx}+k_{1}w_{x}=0, \\
\tau w_{t}-k_{3}w_{xx}+k_{2}w+k_{1}\theta _{x}+du_{tx}=0,%
\end{array}%
\right.  \label{11m.}
\end{equation}%
where $u(x,t)$ is the displacement of a particle at position $x\in (0,L)$
and time $t>0,$ $\theta $ and $w$ are the temperature of the body and the
micro-temperature vector respectively. $\eta >0$ is the coupling constant,
that accounts for the heating effect, and $k,k_{1},k_{2},k_{3},c,d,\tau >0.$
$m(x),$ $\delta (x)$ and $p(x)$ are responsible for the non-uniform
structure of the body, and, respectively, denote mass per unit length of
structure, coefficient of internal material damping and a positive function
related to the stress acting on the body at a point $x$. We consider the
following initial and boundary conditions:%
\begin{equation}
\begin{array}{l}
u\left( .,0\right) =u_{0}\left( x\right) ,\text{ }u_{t}\left( .,0\right)
=u_{1}\left( x\right) ,\text{ }\theta \left( .,0\right) =\theta _{0}\left(
x\right) ,w\left( .,0\right) =w_{0}\left( x\right) ,\text{ \ }\forall x\in
\lbrack 0,L] \\
u\left( 0,t\right) =u\left( L,t\right) =\theta \left( 0,t\right) =\theta
\left( L,t\right) =w_{x}\left( 0,t\right) =w_{x}\left( L,t\right) =0,\text{ }%
\forall t\geq 0.%
\end{array}
\label{12m.}
\end{equation}

In the presence of second sound, Alves et al. \cite{1} concerned with the
system%
\begin{equation}
\left\{
\begin{array}{l}
m(x)u_{tt}-(p(x)u_{x}+2\delta (x)u_{xt})_{x}+\eta \theta _{x}=0, \\
\theta _{t}+kq_{x}+\eta u_{tx}=0, \\
\tau q_{t}+\beta q+k\theta _{x}=0,%
\end{array}%
\right.  \label{1334.}
\end{equation}%
They established the well-posedness of the system and proved its stability
exponential and polynomial under suitable boundary conditions. We know that
the dynamic systems with delay terms have become a major research subject in
differential equation since the $1970s$ of the last century. The delay
effect that is similar to memory processes is important in the research of
applied mathematics such as physics, non-instant transmission phenomena and
biological motivations. Moreover, Datko et al. \cite{dd1} in 1986 showed
that the presence of the delay may not only destabilize a system which is
asymptotically stable in the absence of the delay but may also lead to
ill-posedness (see also \cite{2444} and \cite{Ri}). Li et al. \cite{lii}
considered (\ref{1334.}) with a delay term of the form $\mu u_{t}\left(
x,t-\tau _{0}\right) $ in its first equation,
\begin{equation}
\left\{
\begin{array}{l}
m(x)u_{tt}-(p(x)u_{x}+2\delta (x)u_{xt})_{x}+\eta \theta _{x}+\mu
u_{t}\left( x,t-\tau _{0}\right) =0, \\
\theta _{t}+kq_{x}+\eta u_{tx}=0, \\
\tau q_{t}+\beta q+k\theta _{x}=0,%
\end{array}%
\right.
\end{equation}%
they proved that the system is exponential decay under a "small" condition
on time delay. For more details discussion on the subject see \cite{11,2g}
and the references therein.

The linear theory of thermo-elasticity with micro-temperatures for materials
with inner structure whose particles, in addition to the classical
displacement and temperature fields, possess micro-temperatures was
constructed by Ie\c{s}an and Quintanilla \cite{iq,isan2}. The work is
motivated by increasing use of materials which possess thermal variation at
a microstructure level. The same authors proved an existence theorem and
established the continuous dependence of solutions of the initial data and
body loads. Originally, Grot \cite{grot} was the first to take into
consideration the inner structure of a body in order to develop a
thermodynamic theory for thermo-elastic materials where micro-elements, in
addition to classic micro-deformations, possess micro-temperatures. While,
the fundamental solution of the equations of the theory of thermo-elasticity
with micro-temperatures was constructed by Svanadze \cite{sva}. Riha \cite%
{riha,riha2} developed a further study concerning heat conduction in
thermo-elastic materials with inner structure. We note that the concept of
micro-temperature was just used in the theory of thermodynamics for elastic
materials with microstructure. In addition to micro-deformations of the
string, the micro-elements of the continuum possess micro-temperatures which
represent the variation of the temperature within a micro-volume. We refer
the interested readers to \cite%
{ap,cas,ch2,ch,dri,hach,isan6,isan3,isan4,mag,saci,sca} for details
discussion on the theory.

Motivated by works mentioned above, we investigate (\ref{11m.})-(\ref{12m.})
under suitable condition and establish the well-posedness of the problem
using semi-group theory, as well as the stability result of the solution
using the multiplier method. Our purpose in the present manuscript is to
obtain an exponential decay rate estimates of the energy function of (\ref%
{11m.}) with a "small" condition on time delay. Introducing the
microtemperature damping instead of the second sound together with the delay
term in the internal feedback of flexible structure makes our problem
different from those considered so far in the literature.

This paper is organized as follows; In the second section, we introduce some
assumptions needed in our work then prove the well-posedness of the system (%
\ref{11m.})-(\ref{12m.}). In the last section we state and prove our
stability result.

\section{Well-posedness of the problem}

In this section, we present some assumptions and give the existence and
uniqueness result of system (\ref{11m.})-(\ref{12m.}) using the semigroup
theory. Throughout this paper, $c^{\prime }$ represents a generic positive
constant and is different in various occurrences.

Taking the following new variable
\begin{equation*}
z(x,\rho ,t)=u_{t}\left( x,t-\rho \tau _{0}\right) ,\text{ in }\left(
0,L\right) \times \left( 0,1\right) \times \left( 0,\infty \right) ,
\end{equation*}%
then we obtain%
\begin{equation*}
\left\{
\begin{array}{l}
sz_{t}(x,\rho ,t)+z_{\rho }(x,\rho ,t)=0, \\
z(x,0,t)=u_{t}(x,t).%
\end{array}%
\right.
\end{equation*}%
Consequently, problem (\ref{11m.})-(\ref{12m.}) is equivalent to%
\begin{equation}
\left\{
\begin{array}{l}
\begin{array}{l}
m(x)u_{tt}-(p(x)u_{x}+2\delta (x)u_{xt})_{x}+dw_{x}+\eta \theta _{x}+\mu
z(x,1,t)=0, \\
c\theta _{t}-k\theta _{xx}+\eta u_{tx}+k_{1}w_{x}=0, \\
\tau w_{t}-k_{3}w_{xx}+k_{2}w+k_{1}\theta _{x}+du_{tx}=0, \\
sz_{t}(x,\rho ,s,t)+z_{\rho }(x,\rho ,s,t)=0,%
\end{array}
\\
\begin{array}{l}
u\left( .,0\right) =u_{0}\left( x\right) ,\text{ }u_{t}\left( .,0\right)
=u_{1}\left( x\right) ,\text{ }\theta \left( .,0\right) =\theta _{0}\left(
x\right) ,w\left( .,0\right) =w_{0}\left( x\right) , \\
u\left( 0,t\right) =u\left( L,t\right) =\theta \left( 0,t\right) =\theta
\left( L,t\right) =w_{x}\left( 0,t\right) =w_{x}\left( L,t\right) =0,\text{ }%
\forall t\geq 0, \\
z(x,0,t)=u_{t}(x,t) \\
z(x,\rho ,0)=f_{0}(x,-\rho \tau _{0})%
\end{array}%
\end{array}%
\right.  \label{11m..}
\end{equation}%
where $f_{0}$ is the history function.

The aim of this section is to prove that system (\ref{11m.}) is well-posed.
From Equation (\ref{11m.})$_{3}$ and the boundary conditions, we have%
\begin{equation*}
\frac{d}{dt}\int_{0}^{L}w\left( x,t\right) dx+\frac{k_{2}}{\tau }%
\int_{0}^{L}w\left( x,t\right) dx=0,\text{ }\forall t\geq 0,
\end{equation*}%
thus%
\begin{equation*}
\int_{0}^{L}w\left( x,t\right) dx=\left( \int_{0}^{L}w_{0}dx\right) \exp
\left( \frac{-t}{\tau }k_{2}\right) ,\text{ }\forall t\geq 0,
\end{equation*}%
So, if we set%
\begin{equation*}
\widetilde{w}\left( x,t\right) =w\left( x,t\right) -\frac{1}{L}\left(
\int_{0}^{L}w_{0}dx\right) \exp \left( \frac{-t}{\tau }k_{2}\right),
\end{equation*}
\begin{equation*}
\text{ } t\geq 0,\text{ }x\in \lbrack 0,L],
\end{equation*}
then, $(u,u_{t},\theta ,\widetilde{w},z)$ satisfies Equation (\ref{11m.}),
and
\begin{equation*}
\int_{0}^{L}\widetilde{w}\left( x,t\right) dx=0,
\end{equation*}%
for all $t\geq 0$. In the sequel, we shall work with $\widetilde{w}$ but we
write $w$ for simplicity.

The energy functional associated to (\ref{11m..}), namely,%
\begin{equation}
\mathcal{E}(t)=\frac{1}{2}\int_{0}^{L}\left\{
p(x)u_{x}^{2}+m(x)u_{t}^{2}+c\theta ^{2}+\tau w^{2}\right\} dx+\frac{%
\left\vert \mu \right\vert \tau _{0}}{2}\int_{0}^{L}\int_{0}^{1}z^{2}(x,\rho
,t)d\rho dx,  \label{ett.}
\end{equation}%
we denote $\mathcal{E}(t)=\mathcal{E}\left( t,u,u_{t},\theta ,w,z\right) $
and $\mathcal{E}(0)=\mathcal{E}\left( 0,u_{0},u_{1},\theta
_{0},w_{0},f_{0}\right) $ for simplicity of notations. Then the energy $E$
is decreasing function and satisfies, for and all $t\geq 0,$%
\begin{eqnarray*}
\mathcal{E}^{\prime }(t) &=&-2\int_{0}^{L}\delta \left( x\right)
u_{xt}^{2}dx-k_{2}\int_{0}^{L}w^{2}dx-k_{3}\int_{0}^{L}w_{x}^{2}dx-k%
\int_{0}^{L}\theta _{x}^{2}dx \\
&&-\mu \int_{0}^{L}u_{t}z_{1}(x,1,t)dx \\
&\leq &\left( \left\vert \mu \right\vert -2\delta (\xi _{6})/l\right)
\int_{0}^{L}u_{t}^{2}dx-k_{2}\int_{0}^{L}w^{2}dx-k_{3}%
\int_{0}^{L}w_{x}^{2}dx-k\int_{0}^{L}\theta _{x}^{2}dx.
\end{eqnarray*}%
To obtain precise decay rates of $E(t)$ as $t\rightarrow +\infty ,$ we
assume that

\begin{equation}
m,\text{ }\delta ,\text{ }p\in W^{1,\infty }(0,L),\text{ }m(x),\text{ }p(x)%
\text{, }\delta (x)>0,\text{ \ }\forall x\in \lbrack 0,L].  \label{h2.}
\end{equation}

Let us introducing the vector function $U=(u,v,\theta ,w,z)^{T}$, where $%
v=u_{t}$, using the standard Lebesgue space $L^{2}(0,L)$ and the Sobolev
space $H_{0}^{1}(0,L)$ with their usual scalar products and norms for define
the spaces:%
\begin{equation*}
\mathcal{H}:=H_{0}^{1}(0,L)\times \left[ L^{2}(0,L)\right] ^{2}\times
L_{\ast }^{2}(0,L)\times L^{2}((0,L)\times (0,1)),
\end{equation*}%
and%
\begin{equation*}
H_{\ast }^{2}(0,L)=\left\{ w\in H^{2}(0,L):w_{x}\left( L\right) =w_{x}\left(
0\right) =0\right\} ,
\end{equation*}%
where%
\begin{equation*}
L_{\ast }^{2}(0,L)=\left\{ w\in L^{2}(0,L):\int_{0}^{L}w\left( s\right)
ds=0\right\} .
\end{equation*}%
We equip $\mathcal{H}$ with the inner product%
\begin{eqnarray*}
(U,\widetilde{U})_{\mathcal{H}} &=&\int_{0}^{L}p\left( x\right) u_{x}%
\widetilde{u}_{x}dx+\int_{0}^{L}m\left( x\right) v\widetilde{v}%
dx+c\int_{0}^{L}\theta \widetilde{\theta }dx+\tau \int_{0}^{L}w\widetilde{w}%
dx \\
&&+\tau _{0}\left\vert \mu \right\vert \int_{0}^{L}\int_{0}^{1}z\widetilde{z}%
(x,\rho ,t)d\rho dx.
\end{eqnarray*}

Next, the system (\ref{11m..}) can be reduced to the following abstract
Cauchy problem:%
\begin{equation}
\left\{
\begin{array}{ll}
U^{\prime }(t)+\left( \mathcal{A+B}\right) U(t)=0, & t>0 \\
U(0)=U_{0}=(u_{0},u_{1},\theta _{0},w_{0},f_{0})^{T}, &
\end{array}%
\right.  \label{proo.}
\end{equation}%
where the operator $\mathcal{A}:D(\mathcal{A})\rightarrow \mathcal{H}$ is
defined by%
\begin{equation*}
\mathcal{A}U=\left(
\begin{array}{c}
-v \\
-\frac{1}{m(x)}\left[ (p(x)u_{x}+2\delta (x)v_{x}-\eta \theta -dw)_{x}-\mu
z(x,1,t)-\left\vert \mu \right\vert v\right] \\
\frac{1}{c}\left( -k\theta _{xx}+\eta u_{tx}+k_{1}w_{x}\right) \\
\frac{1}{\tau }(-k_{3}w_{xx}+k_{2}w+k_{1}\theta _{x}+du_{tx})%
\end{array}%
\right) ,
\end{equation*}%
\begin{equation*}
\mathcal{B}U=\left(
\begin{array}{c}
0 \\
-\frac{\left\vert \mu \right\vert v}{m(x)} \\
0 \\
0%
\end{array}%
\right) ,
\end{equation*}%
The domain of $\mathcal{A}$ is then%
\begin{equation*}
D(\mathcal{A})=\left\{
\begin{array}{l}
U\in \mathcal{H}\mid u\in H^{2}(0,L)\cap H_{0}^{1}(0,L),v\in H_{0}^{1}(0,L),%
\text{ }\theta \in H^{2}(0,L), \\
w\in L_{\ast }^{2}(0,L)\cap H_{\ast }^{2}(0,L),\text{ }z,z_{\rho }\in
L^{2}((0,L)\times (0,1)),z(x,0)=u_{t}(x)%
\end{array}%
\text{ }\right\} ,
\end{equation*}%
which is dense in $\mathcal{H}$.

\begin{proposition}
\label{Propo} Let $U_{0}\in \mathcal{H}$ be given. Problem (\ref{proo.})
possesses then a unique solution satisfying $U\in C\left(
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+};\mathcal{H}\right) .$ If $U_{0}\in D(\mathcal{A}),$ then $U\in
C^{1}\left(
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+};\mathcal{H}\right) \cap C\left(
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+};D(\mathcal{A})\right) .$
\end{proposition}

\begin{proof}
Using semi-group theory. We prove that $\mathcal{A}$ is a maximal monotone
operator and $\mathcal{B}$ is Lipschitz continuous in $\mathcal{H}.$
Consequently, $\mathcal{A+B}$ is the infinitesimal generator of a linear
contraction C$_{0}$-semigroup on $\mathcal{H}$.
\end{proof}

\section{Exponential stability}

In this section, we introduce some lemmas allow us to achieve our goal,
which is the proof of the stability result.

\begin{lemma}
\cite{Miil} (Poincar\'{e} type Scheeffer's inequality): Let $h\in $ $%
H_{0}^{1}(0,L)$. Then it holds%
\begin{equation}
\int_{0}^{L}\left\vert h\right\vert ^{2}dx\leq \text{ }l\int_{0}^{L}\left\vert h_{x}\right\vert ^{2}dx,\text{ \ }l=\frac{L^{2}}{\pi ^{2}}.
\label{Muu.}
\end{equation}
\end{lemma}

\begin{lemma}
\cite{1,3m} \label{mean.}: Let $(u,u_{t},\theta ,w)$ be the solution to
system (\ref{11m.})-(\ref{12m.}), with an initial datum in $D(\mathcal{A})$.
Then, for any $t>0$, there exists a sequence of real numbers (depending on
t), denoted by $\xi _{i}\in \lbrack 0,L](i=1,...,6)$, such that:%
\begin{equation*}
\int_{0}^{L}p\left( x\right) u_{x}^{2}dx=p(\xi _{1})\int_{0}^{L}u_{x}^{2}dx,%
\text{ }\int_{0}^{L}\delta \left( x\right) u^{2}dx=\delta (\xi
_{4})\int_{0}^{L}u^{2}dx,
\end{equation*}%
\begin{equation*}
\text{ \ }\int_{0}^{L}m\left( x\right) u_{t}^{2}dx=m(\xi
_{2})\int_{0}^{L}u_{t}^{2}dx,\text{ }\int_{0}^{L}\delta \left( x\right)
u_{x}^{2}dx=\delta (\xi _{5})\int_{0}^{L}u_{x}^{2}dx,
\end{equation*}%
\begin{equation*}
\int_{0}^{L}m\left( x\right) u^{2}dx=m(\xi _{3})\int_{0}^{L}u^{2}dx,\text{ }%
\int_{0}^{L}\delta \left( x\right) u_{xt}^{2}dx=\delta (\xi
_{6})\int_{0}^{L}u_{xt}^{2}dx.
\end{equation*}
\end{lemma}

\begin{lemma}
Let $\left( u,v,\theta ,w,z\right) $ be the solution of (\ref{11m..}), then
the energy $\mathcal{E}$ is non-increasing function and satisfies, for all $%
t\geq 0,$%
\begin{eqnarray}
\mathcal{E}^{\prime }(t) &=&-2\int_{0}^{L}\delta \left( x\right)
u_{xt}^{2}dx-k_{2}\int_{0}^{L}w^{2}dx-k_{3}\int_{0}^{L}w_{x}^{2}dx-k%
\int_{0}^{L}\theta _{x}^{2}dx-\mu \int_{0}^{L}u_{t}z_{1}(x,1,t)dx  \notag \\
&\leq &\left( \left\vert \mu \right\vert -2\delta (\xi _{6})/l\right)
\int_{0}^{L}u_{t}^{2}dx-k_{2}\int_{0}^{L}w^{2}dx-k_{3}%
\int_{0}^{L}w_{x}^{2}dx-k\int_{0}^{L}\theta _{x}^{2}dx.  \label{ET''}
\end{eqnarray}
\end{lemma}

\begin{proof}
Multiplying the equations in (\ref{11m..})$_{1}$,(\ref{11m..})$_{2}$, (\ref%
{11m..})$_{3}$ and (\ref{11m..})$_{4}$ by $u_{t}$, $\theta ,$ $w$ and $%
\left\vert \mu \right\vert z$, respectively, integrate over $(0,L)$ and $%
(0,L)\times (0,1)$ and using (\ref{Muu.}), we obtain (\ref{ET''}).
\end{proof}

\begin{lemma}
The functional%
\begin{equation}
I_{1}\left( t\right) :=\int_{0}^{L}\left( \delta
(x)u_{x}^{2}+m(x)u_{t}u\right) dx,  \label{I1m.}
\end{equation}%
satisfies%
\begin{eqnarray}
I_{1}^{\prime }\left( t\right)  &\leq &-\left( p(\xi _{1})-\mu ^{2}l-\left(
\eta +d\right) \epsilon _{1}\right) \int_{0}^{L}u_{x}^{2}dx+m(\xi
_{2})\int_{0}^{L}u_{t}^{2}dx+\frac{\eta }{4\epsilon _{1}}\int_{0}^{L}\theta
^{2}dx  \notag \\
&&+\frac{d}{4\epsilon _{1}}\int_{0}^{L}w^{2}dx+\frac{1}{4}%
\int_{0}^{L}z^{2}(x,1,t)dx,  \label{I1'}
\end{eqnarray}%
for any $\varepsilon _{1}>0.$
\end{lemma}

\begin{proof}
Differentiating Equation (\ref{I1m.}) with respect to $t$ and using
Equations (\ref{11m.})$_{1}$. By using Young's inequality (see \cite{bre} p.
92) and Lemma \ref{mean.}, we complete the proof.
\end{proof}

\begin{lemma}
The functional%
\begin{equation}
I_{2}(t):=\tau c\int_{0}^{L}\theta \left( \int_{0}^{x}w(y)dy\right) dx,
\label{f44.}
\end{equation}%
satisfies%
\begin{eqnarray}
I_{2}^{\prime }\left( t\right)  &\leq &\left( -k_{1}c+3\varepsilon
_{2}\right) \int_{0}^{L}\theta ^{2}dx+\frac{1}{2\varepsilon _{2}}%
\int_{0}^{L}u_{t}^{2}dx+\frac{1}{4\varepsilon _{2}}\int_{0}^{L}\theta
_{x}^{2}dx  \notag \\
&&+\left( k_{1}\tau +2\varepsilon _{2}c^{\prime }+c^{\prime }\right)
\int_{0}^{L}w^{2}dx+\frac{1}{4\varepsilon _{2}}\int_{0}^{L}w_{x}^{2}dx,
\label{f4m.}
\end{eqnarray}%
for any $\varepsilon _{2}>0.$
\end{lemma}

\begin{proof}
Taking the derivative of (\ref{f44.}) and using (\ref{11m.})$_{2}$ and (\ref%
{11m.})$_{3}$, Integration by parts and using Young's inequality, we infer (%
\ref{f4m.}).
\end{proof}

\begin{lemma}
The functional%
\begin{equation}
I_{3}(t)=\tau _{0}\int_{0}^{L}\int_{0}^{1}e^{-2\tau _{0}\rho }z^{2}(x,\rho
,t)d\rho dx  \label{r}
\end{equation}

satisfies%
\begin{equation}
I_{3}^{^{\prime }}(t)\leq -2I_{3}(t)-c^{\prime
}\int_{0}^{L}z_{1}^{2}(x,1,t)dx+\int_{0}^{L}\varphi _{t}^{2}dx  \label{dr}
\end{equation}

for $c^{\prime }>0.$

\begin{proof}
By differentiating $I_{3}$, then using (\ref{11m..})$_{5}$ and integrating
by parts.
\end{proof}
\end{lemma}

Next, we define a Lyapunov functional $\mathcal{L}$ and show that it is
equivalent to the energy functional.

\begin{lemma}
For $N$ sufficiently large, the functional defined by%
\begin{equation}
\mathcal{L}(t):=N\mathcal{E}\left( t\right)
+I_{1}(t)+N_{1}I_{2}(t)+N_{2}I_{3}(t).  \label{F(t)m.}
\end{equation}%
where $N$ and $N_{1}$ are positive real numbers to be chosen appropriately
later, satisfies%
\begin{equation}
c_{1}^{\prime }\mathcal{E}(t)\leq \mathcal{L}(t)\leq c_{2}^{\prime }\mathcal{%
E}(t),\text{ \ \ }\forall t\geq 0.  \label{equm.}
\end{equation}%
where $c_{1}^{\prime }$ and $c_{2}^{\prime }$ are positive constants.
\end{lemma}

Now, we are ready to state and prove the main result of this section.

\begin{theorem}
Let $\left( u,v,\theta ,w,z\right) $ be the solution of (\ref{11m..}), then
the energy $\mathcal{E}$ satisfies, for all $t\geq 0$,%
\begin{equation*}
\mathcal{E}(t)\leq c_{1}e^{-c_{2}t},
\end{equation*}%
where $c_{1}$ and $c_{2}$ are positive constants.
\end{theorem}

\begin{proof}
We differentiate (\ref{F(t)m.}), and recall (\ref{ET''}), (\ref{I1'}), and (%
\ref{f4m.}), we obtain%
\begin{eqnarray*}
\mathcal{L}^{\prime }(t) &\leq &\left\{ N\left( \left\vert \mu \right\vert
-2\delta (\xi _{6})/l\right) +\frac{N_{1}}{2\varepsilon _{2}}+m(\xi
_{2})+N_{2}\right\} \int_{0}^{L}u_{t}^{2}dx \\
&&+\left\{ \left( \mu ^{2}l-p(\xi _{1})\right) +\left( \eta +d\right)
\epsilon _{1}\right\} \int_{0}^{L}u_{x}^{2}dx \\
&&+\left\{ -Nk_{2}+N_{1}\left( k_{1}\tau +2\varepsilon _{2}c^{\prime
}+c^{\prime }\right) +\frac{d}{4\epsilon _{1}}\right\} \int_{0}^{L}w^{2}dx \\
&&+\left\{ N_{1}\left( -k_{1}c+3\varepsilon _{2}\right) +\frac{\eta }{%
4\epsilon _{1}}\right\} \int_{0}^{L}\theta ^{2}dx+\left\{ -Nk+\frac{N_{1}}{%
4\varepsilon _{2}}\right\} \int_{0}^{L}\theta _{x}^{2}dx \\
&&+\left\{ -Nk_{3}+\frac{N_{1}}{4\varepsilon _{2}}\right\}
\int_{0}^{L}w_{x}^{2}dx+\left\{ \frac{1}{4}-c^{\prime }N_{2}\right\}
\int_{0}^{L}z^{2}(x,1,t)dx
\end{eqnarray*}

At this point, we choose $\epsilon _{2}$ small enough such that%
\begin{equation*}
-k_{1}c+3\varepsilon _{2}\,<0,
\end{equation*}

then, we choose $\left\vert \mu \right\vert $ small enough such that%
\begin{equation*}
\left\vert \mu \right\vert -2\delta (\xi _{6})/l<0,\text{ }\mu ^{2}l-p(\xi
_{1})<0
\end{equation*}

Now, we take $\epsilon _{1}$ small enough such that%
\begin{equation*}
-p(\xi _{1})+\mu ^{2}l+\left( \eta +d\right) \epsilon _{1}<0
\end{equation*}

then, choosing $N_{1}$ large enough so that%
\begin{equation*}
N_{1}\left( -k_{1}c+3\varepsilon _{2}\right) +\frac{\eta }{4\epsilon _{1}}<0.
\end{equation*}

Once $N_{1}$ is fixed, we then choose $N_{2}$ large enough so that%
\begin{equation*}
\frac{1}{4}-c^{\prime }N_{2}<0.
\end{equation*}%
Finally, we then choose $N$ large enough so that
\begin{equation*}
\begin{array}{l}
-Nc^{\prime }+\frac{N_{1}}{2\varepsilon _{2}}+m(\xi _{2})+N_{2}<0, \\
-Nk_{2}+N_{1}\left( k_{1}\tau +2\varepsilon _{2}c^{\prime }+c^{\prime
}\right) +\frac{d}{4\epsilon _{1}}<0, \\
-Nk+\frac{N_{1}}{4\varepsilon _{2}}<0,\text{ }-Nk_{3}+\frac{N_{1}}{%
4\varepsilon _{2}}<0.%
\end{array}%
\end{equation*}

Thus, using (\ref{ett.}), we arrive at%
\begin{equation}
\mathcal{L}^{\prime }(t)\leq -c\mathcal{E}(t),\text{ \ \ }\forall t>0.
\label{53.}
\end{equation}

A combination of (\ref{equm.}) and (\ref{53.}) gives%
\begin{equation}
\mathcal{L}^{\prime }(t)\leq -c_{2}\mathcal{L}(t),\text{ \ \ }\forall t>0.
\label{65.}
\end{equation}

where $c_{2}=c/c_{2}^{\prime },$ a simple integration of (\ref{65.}) over $%
(0,t)$ yields%
\begin{equation*}
c_{1}^{\prime }\mathcal{E}(t)\leq \mathcal{L}(t)\leq \mathcal{L}%
(0)e^{-c_{2}t}\text{ ,\ \ }\forall t>0.
\end{equation*}

Taking $c_{1}=\mathcal{L}(0)/c_{1}^{\prime }$ which completes the proof.
\end{proof}

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\end{document}
