\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 129, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/129\hfil Stabilization of laminated beams]
{Stabilization of laminated beams with \\ interfacial slip}

\author[A. Lo, N.-e. Tatar \hfil EJDE-2015/129\hfilneg]
{Assane Lo, Nasser-eddine Tatar}

\address{Assane Lo \newline
King Fahd University of Petroleum and Minerals,
Department of Mathematics and Statistics,
Dhahran 31261, Saudi Arabia}
\email{assane@kfupm.edu.sa}

\address{Nasser-eddine Tatar \newline
King Fahd University of Petroleum and Minerals,
Department of Mathematics and Statistics,
Dhahran 31261, Saudi Arabia}
\email{tatarn@kfupm.edu.sa}

\thanks{Submitted February 24, 2015. Published May 7, 2015.}
\subjclass[2010]{34B05, 34D05, 34H05}
\keywords{Exponential stabilization; vibration reduction; Timoshenko system;
 \hfill\break\indent slip; boundary control; multiplier technique}

\begin{abstract}
 We study a laminated beam consisting of two identical beams of uniform
 thickness, which is modeled as Timoshenko beams.
 An adhesive of small thickness is bonding the two layers and creating
 a restoring force producing a damping. It has been shown that the
 interfacial slip between the layers alone is not enough to stabilize the
 system exponentially to its equilibrium state.
 Some boundary control has been used in the literature for that
 purpose. In this paper, we show that for viscoelastic material there 
 is no need for any kind of internal or boundary control.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks


\section{Introduction}

Many structures in mechanical engineering, electrical engineering, civil
engineering and aerospace engineering are formed by a single beam or a
number of beams. We can cite for instance, robot arms, rotor turbine and
helicopter blades, turbo-machineries, electronic equipment, antennas,
missiles, panels, pipelines, buildings, bridges, etc.\ There are mainly
three important theories. The first one is named after Euler and Bernoulli
and the second one after Rayleigh. To alleviate the shortcomings in these
two theories, Timoshenko came up with a new theory which is better suited
for engineering practice and is nowadays widely used for moderately thick
beams. Both, rotatory inertia and the effect of shear forces are taken into
account. In his theory, Timoshenko also assumed that the plane
cross-sections perpendicular to the beam centerline remain plane but could
become oblique after deformation. An additional kinematics variable is added
in the displacement assumptions. Internal and external forces like the
weight of the beam, heavy loads, wind, earthquakes and interaction with
other bodies or materials are examples of some sources causing high stresses
accompanying unwanted vibration. These stresses not only bring some
discomfort, reduce the fatigue-life of the material and produce annoying
noise but also are harmful to the structure as they may cause significant
damage or complete destruction of the machine or equipment. Therefore, some
ways and devices capable of enhancing dynamic stability must accompany these
structures. To this end various devices and energy dissipation mechanisms
have been designed either in the material itself such as smart materials
(piezoelectric, pietzoceramic, viscoelastic), on its surface (viscoelastic
layers, sandwich plates,\ldots ) or at the boundary (or part of the
boundary). Some well-known dampers are: friction dampers, sensors and
actuators, special loads, viscoelastic dampers, tuned mass dampers, tuned
liquid dampers and tuned mass liquid dampers. Sometimes they are classified
into active, semi-active and passive control methods. In this paper, we
would like to investigate the case of two identical beams with an adhesive
layer in the interface creating a restoring force. It has been already shown
that when this restoring force is proportional to the amount of slip the
created frictional damping is unable by itself to stabilize the system
exponentially. The first investigators have been forced to control the
system by an additional boundary feedback. We intend to seek other ways and
means, preferably less costly, less demanding and easy to implement, to
stabilize the system exponentially.

\subsection*{Statement of the problem}

The original structure consists of a two-layered beam with an adhesive layer
bonding the two adjoining surfaces. The adhesive layer creates a restoring
force which is assumed proportional to the amount of slip. Therefore, we are
in the presence of a structural damping due to interfacial slip. Moreover,
we assume that the adhesive layer is of negligible thickness and mass so
that the contribution of its mass to the kinetic energy of the structure can
be ignored. The equations of motion modeling the system are derived using
Timoshenko theory and a third equation is coupled with the first two
describing the dynamic of the slip and containing the internal frictional
(Kelvin-Voigt) damping. Namely, we have the  system
\begin{gather*}
\rho w_{tt}+G(\psi -w_x)_x=0, \\
I_{\rho }(3s_{tt}-\psi _{tt})-G(\psi -w_x)
-D(3s_{xx}-\psi _{xx})=0, \\
3I_{\rho }s_{tt}+3G(\psi -w_x)+4\gamma s+4\beta
s_t-3Ds_{xx}=0,
\end{gather*}
supplemented by the initial data
\[
(w,\psi ,s)(x,0)=(w_0,\psi _0,s_0),\quad
(w_t,\psi _t,s_t)(x,0)=(w_1,\psi _1,s_1)
\]
and cantilever boundary conditions.

Here $w,\psi ,\rho , G,\ I_{\rho }, D, \gamma , \beta $ are
transverse displacement, rotation angle, density, shear stiffness, mass
moment of inertia, flexural rigidity, adhesive stiffness, adhesive damping
parameter and $s$ is proportional to the amount of slip along the
interface. The expression $\xi :=3s-\psi $ is the effective rotation angle.

It has been shown in \cite{w1} that the frictional damping created by the
interfacial slip alone is not enough to stabilize the system exponentially
to its equilibrium state. Therefore, a natural question that can be asked
is: what are the possible additional damping that can ensure the exponential
stability and other kinds of stability of the system? We suggest
investigating the case of an additional viscoelastic damping that acts on
the effective rotation angle without resorting to any boundary control.
Viscoelastic material is very efficient in case there is no considerable
change of frequency or temperature in the structure \cite{b1}. The viscoelastic
damping is (according to the Boltzmann Principle) represented by a memory
term in the form of a convolution which arises in the constitutive equation
between the stress and the strain
\begin{equation*}
\int_0^{t}h(t-r)(3s-\psi )_{xx}(r)dr.
\end{equation*}

There are basically three main papers  in this subject \cite{c1,h2,w1}.
In \cite{h2}, the problem has been derived in details. The authors assumed that the
adhesive layer is of negligible thickness and mass and that the restoring
force created by this layer is proportional to the amount of slip at the
interface.

In \cite{w1}, the system is studied assuming that $\sqrt{G/\rho }$ and $\sqrt{
D/I_{\rho }}$ are two different wave speeds. Putting $\xi =3s-\psi $, they
transformed the original system into
\begin{gather*}
\rho w_{tt}+G(3s-\xi -w_x)_x=0, \\
I_{\rho }\xi _{tt}-G(3s-\xi -w_x)-D\xi _{xx}=0, \\
3I_{\rho }s_{tt}+3G(3s-\xi -w_x)+4\gamma s+4\beta
s_t-3Ds_{xx}=0
\end{gather*}
where $0<x<1$ and $t>0$. In addition to the well-posedness, the authors
pointed out that the frictional damping is enough to asymptotically
stabilize the system. However, it is not possible to have exponential
stability. They justified their claim by the fact that the eigenvalues of
two branches are very close to the imaginary axis as their moduli go to
infinity. To achieve exponential decay of solutions they implemented an
additional boundary control
\begin{gather*}
w(0,t)=\xi (0,t)=s(0,t)=0, \\
\xi _x(1,t)=u_1(t):=-k_1\xi _t(1,t),\quad  s_x(1,t)=0, \\
3s(1,t)-\xi (1,t)-w_x(1,t)=u_2(t):=k_2w_t(1,t)
\end{gather*}
where $t>0$. The same system but with the boundary control
\begin{gather*}
\psi (0,t)-w_x(0,t)=u_1(t):=-k_1w_t(0,t)-w(0,t), \\
3s_x(1,t)-\psi _x(1,t)=u_2(t):=-k_2\xi _t(1,t)-\xi (1,t),
\end{gather*}
has been studied in \cite{c1}. The authors proved an exponential stabilization
result in case $k_1\neq \sqrt{\rho /G}$, $k_2\neq \sqrt{I_{\rho }/D}$
and the dominant part of the system is itself exponentially stable.

For the case of a single viscoelastic Timoshenko beam (therefore without
interfacial slip) there exist many papers in the literature. We can cite a few
of them \cite{a1,d1,l1,m4,m5,r1,r2,r3,s1,s2,s3,t4,t5,t6,x1,y1}.

Here, we shall consider the system
\begin{equation}
\begin{gathered}
\rho w_{tt}+G(\psi -w_x)_x=0, \\
I_{\rho }(3s_{tt}-\psi _{tt})-G(\psi -w_x)
-(3s-\psi )_{xx}+\int_0^{t}h(t-r)(3s-\psi )
_{xx}(r)dr=0, \\
I_{\rho }s_{tt}+G(\psi -w_x)+\frac{4}{3}\gamma s+\frac{4}{3}
\alpha s_t-s_{xx}=0,
\end{gathered}  \label{e1}
\end{equation}
where $0<x<1$ and $t>0$, with the boundary conditions
\begin{equation}
\begin{gathered}
\psi (0,t)=s(0,t)=0, \\
s_x(1,t)=\psi _x(1,t)=0, \\
w_x(0,t)=0,\quad w(1,t)=0.
\end{gathered}  \label{e2}
\end{equation}

The well-posedness of the system has been addressed in \cite{c1,w1}
(see \cite{c2,c3,c5,h1,m6} for the viscoelastic term).
 We have weak solutions in $(V_{\ast }^1\times L^2)^{3}$
and strong solutions in $(V_{\ast }^2\times H^1)^{3}$ where
\begin{equation*}
V_{\ast }^{k}=\big\{ v:v\in H^{k}(0,1):v(0)=0\big\} ,\quad k=1,2.
\end{equation*}
We shall discuss the case where the relaxation function
 $h:\mathbb{R} _{+}\to \mathbb{R}_{+}$ is a bounded differentiable function
satisfying the standard conditions (as we shall not be concerned about
finding the largest class of admissible kernels, see 
\cite{c4,m1,m2,m3,p1,t1,t2,t3,t4,t5,t6} for this matter)
\begin{equation}
-\beta _0h\leq h'\leq -\beta _1h,  \label{e3}
\end{equation}
for some positive constants $\beta _0$ and $\beta _1$. Moreover we
assume that
\begin{equation}
\varsigma :=1-\int_0^{\infty }h(r)dr>0.  \label{e4}
\end{equation}

For $G$ we shall  use  the following assumption
\begin{itemize}
\item[(H1)] If $\varsigma \rho <\frac{\gamma }{12}$, then
$G<\min \{\varsigma \rho ,\frac{3\varsigma }{2},\frac{2\gamma -2\sqrt{\gamma
^2-9\gamma \varsigma \rho }}{9}\} $, and if
 $\frac{\gamma }{12} <\varsigma \rho <\frac{\gamma }{9}$ then assume
$G<\min \big\{ \varsigma \rho ,\frac{3\varsigma }{2}\big\} $.
\end{itemize}

\section{Uniform stabilization}

The `modified' energy of the system \eqref{e1}--\eqref{e2} is given by
\begin{equation}
\begin{aligned}
E(t)&=\frac{1}{2}\Big[ \rho \| w_t\| ^2
+I_{\rho}\| 3s_t-\psi _t\| ^2+3I_{\rho }\|s_t\| ^2+G\| \psi -w_x\| ^2 \\
&\quad  +(1-\int_0^{t}h(r)dr)\| 3s_x-\psi _x\| ^2+3\| s_x\| ^2+4\gamma \|
s\| ^2\\
&\quad +\int_0^1(h \square (3s-\psi )_x)dx\Big] ,
\end{aligned}
\label{e5}
\end{equation}
for $t\geq 0$, where $\| \cdot \| $ denotes the norm in $
L^2(0,1)$ and
\begin{equation*}
(g\square h)(t):=\int_0^{t}g(t-s)|
h(s)-h(t)| ^2ds,\quad t\geq 0.
\end{equation*}
Our result reads as follows.

\begin{theorem} \label{thm1}
For the energy $E(t)$ defined above,
if $\rho =GI_{\rho }$  and {\rm (H1)} holds, then there
exist two positive constants $K$ and $\kappa _0$ such
that
\begin{equation*}
E(t)\leq Ke^{-\kappa _0t},\quad t>0.
\end{equation*}
\end{theorem}

We first give some lemmas that will serve as a support for the proof
of this theorem.

\begin{lemma} \label{lem1}
 If $k$ and $\phi $ are two
differentiable functions then
\begin{align*}
(k\ast \phi )(t)\phi '(t)
&=\frac{1}{2}(k' \square  \phi )(t)+\frac{1}{2}\frac{d}{dt}
\Big[ \Big(\int_0^{t}k(s)ds\Big) \phi ^2(t)-(k\square \phi )(t)\Big] \\
&\quad -\frac{1}{2}k(t)\ \phi ^2(t),\quad t>0
\end{align*}
where $\ast $ stands for the usual convolution.
\end{lemma}

\begin{proof}
The statement of the this follows from the identity
\begin{align*}
\frac{d}{dt}(k\square \phi )(t)
&=(k'\square \phi )(t)+2\Big(\int_0^{t}k(s)ds\Big)\ \phi _t(t)\phi
(t)-2(k\ast \phi )(t)\phi _t(t) \\
&=(k'\square \phi )(t)+\frac{d}{dt}
\Big[ \Big(\int_0^{t}k(s)ds\Big) \phi ^2(t)\Big] -k(t)\ \phi ^2(t)\\
&\quad -2(k\ast \phi )(t)\phi _t(t),\quad t>0.
\end{align*}
\end{proof}

\begin{lemma} \label{lem2}
The energy $E(t)$ given by \eqref{e5} satisfies
\begin{equation*}
\frac{d}{dt}E(t)
=-\dfrac{h(t)}{2}\| (3s-\psi )_x\| ^2-4\alpha \| s_t\| ^2
+\dfrac{1}{2} \int_0^1(h'\square (3s-\psi )_x) dx,\quad t>0.
\end{equation*}
\end{lemma}

\begin{proof}
Multiplying the first equation of \eqref{e1} by $w_t$ and
integrating over $(0,1)$ we obtain
\begin{equation*}
\frac{\rho }{2}\frac{d}{dt}\left[ \| w_t\| ^2\right]
+G\int_0^1(\psi -w_x)_xw_tdx=0
\end{equation*}
or
\begin{equation*}
\frac{\rho }{2}\frac{d}{dt}\left[ \| w_t\| ^2\right]
-G\int_0^1(\psi -w_x)w_{xt}dx+\left[ G(\psi
-w_x)w_t\right] _0^1=0
\end{equation*}
and by our boundary conditions \eqref{e2}
\begin{equation*}
\frac{\rho }{2}\frac{d}{dt}[ \| w_t\| ^2]
-G\int_0^1(\psi -w_x)w_{xt}dx=0,\quad t>0.
\end{equation*}
Note that
\begin{align*}
G\int_0^1(\psi -w_x)w_{xt}dx
&=-G\int_0^1(\psi -w_x)(\psi -w_x-\psi )_tdx \\
&=-\frac{G}{2}\frac{d}{dt}[ \| \psi -w_x\| ^2]
+G\int_0^1(\psi -w_x)\psi _tdx.
\end{align*}
Therefore,
\begin{equation}
\frac{1}{2}\frac{d}{dt}[ \rho \| w_t\|
^2+G\| \psi -w_x\| ^2] -G\int_0^1(
\psi -w_x)\psi _tdx=0,\quad t>0.  \label{e6}
\end{equation}
Similarly multiplying the second equation of \eqref{e1} by $3s_t-\psi _t$ and
integrating over $(0,1)$ we obtain
\begin{align*}
&\frac{I_{\rho }}{2}\frac{d}{dt}[ \| 3s_t-\psi _t\|^2]
 -G\int_0^1(\psi -w_x)(3s_t-\psi_t)dx\\
&-\int_0^1(3s-\psi )_{xx}(3s_t-\psi_t)dx
+\int_0^1(3s_t-\psi _t)\int_0^{t}h(t-r)(3s-\psi
)_{xx}(r)\,dr\,dx=0
\end{align*}
or, using integration by parts and the boundary conditions \eqref{e2}
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\left[ I_{\rho }\| 3s_t-\psi
_t\| ^2+\| 3s_x-\psi _x\| ^2\right]
-G\int_0^1(\psi -w_x)(3s_t-\psi _t)dx \\
&-\int_0^1(3s_t-\psi _t)_x\int_0^{t}h(t-r)(
3s-\psi )_x(r)\,dr\,dx=0,\quad t>0.
\end{aligned} \label{e7}
\end{equation}
By using Lemma \ref{lem2} we see that
\begin{equation}
\begin{aligned}
&\int_0^1(3s_t-\psi _t)_x\int_0^{t}h(t-r)(
3s-\psi )_x(r)\,dr\,dx \\
&=\frac{1}{2}(h'\ \square (3s-\psi )_x)
(t)-\frac{h(t)}{2}\| 3s_x-\psi _x\| ^2 \\
&\quad +\frac{1}{2}\frac{d}{dt}\Big[ \Big(\int_0^{t}h(s)ds\Big)\|3s_x-\psi _x\| ^2
-(h\square (3s-\psi )_x)(t)\Big] ,\quad t>0.
\end{aligned}\label{e8}
\end{equation}
Likewise, multiplying the third equation of \eqref{e1} by $s_t$ and
integrating over $(0,1)$, we obtain
\begin{equation}
\frac{1}{2}\frac{d}{dt}\big[ I_{\rho }\| s_t\| ^2+
\frac{4\gamma }{3}\| s\| ^2+\| s_x\|
^2\big] +G\int_0^1(\psi -w_x)s_tdx+\frac{4\alpha }{3
}\| s_t\| ^2=0,  \label{e9}
\end{equation}
for $t>0$.
Now it is clear from \eqref{e6}--\eqref{e9} that
\begin{equation*}
E'(t)=-4\alpha \| s_t\| ^2-\dfrac{h(t)}{2}
\| (3s-\psi )_x\| ^2+\dfrac{1}{2}
\int_0^1(h'\square (3s-\psi )_x)
dx,\quad t>0.
\end{equation*}
This completes the proof.
\end{proof}

As $h'(t)\leq 0$, we see that $E'(t)\leq 0$ for all $t>0$.
Therefore the energy is non-increasing and uniformly bounded above by $E(0)$.

Next we shall construct a Lyapunov functional $F$ satisfying the
inequalities
\begin{equation*}
\lambda _1E(t)\leq F(t)\leq \lambda _2E(t)\quad\text{and}\quad
\frac{d}{dt} F(t)\leq -\kappa F(t)
\end{equation*}
for some positive constants $\lambda _1$, $\lambda _2$ and $\kappa $.
The first two inequalities show that $E(t)$ and $F(t)$ are equivalent. The
second one gives the exponential decay of $F(t)$ (and therefore the
exponential decay of $E(t)$ as well). To this end, we define
\begin{equation*}
F(t)=E(t)+\sum\nolimits_{i=1}^{5}\delta _{i}G_{i}(t),\quad \delta
_{i}>0,\;i=1,\dots ,5,\;t\geq 0,
\end{equation*}
where
\begin{gather*}
G_1(t)=I_{\rho }(s_t,s),\quad
G_2(t)=-\rho (w_t,w),\quad
G_{3}(t)=I_{\rho }(3s_t-\psi _t,3s-\psi ),\quad t\geq 0,\\
G_{4}(t)=-\frac{4\gamma \rho }{G}(w_t,\Theta )-\frac{3\rho }{G
}(s_x,w_t)+3I_{\rho }(s_t,\psi -w_x),\quad t\geq 0,
\end{gather*}
with $\Theta (x,t)=\int_x^1s(\xi ,t)d\xi $ and
\begin{equation*}
G_5(t)=-I_{\rho }\Big(3s_t-\psi _t,\int_0^{t}h(t-r)\left[ (3s-\psi
)(t)-(3s-\psi )(r)\right] dr\Big),\quad t\geq 0.
\end{equation*}
Using the Cauchy-Schwarz inequality and the Poincar\'{e} inequality, one can
easily see that all the $G_{i}(t)$, $i=1,\dots ,5$ are bounded (above and
below) by an expression containing the existing terms in the energy $E(t)$.
This leads to the equivalence of $F(t)$ and $E(t)$.

We shall now prove several lemmas with the purpose of creating negative
counterparts of the terms that appear in the energy in the estimations of
the derivatives of the above functionals.

\begin{lemma} \label{lem3}
Along the solutions of \eqref{e1}--\eqref{e2}, we have
\begin{equation*}
G_1'(t)\leq -\| s_x\| ^2+\big(\frac{G}{
4\varepsilon _0}+\varepsilon -\frac{4}{3}\gamma \big)\|s\| ^2
+\varepsilon _0G\| \psi -w_x\|^2
+\big(I_{\rho }+\frac{4\alpha ^2}{9\varepsilon }\big)\|
s_t\| ^2,
\end{equation*}
for all $t>0$ and some $\varepsilon _0,\varepsilon >0$.
\end{lemma}

\begin{proof} Clearly,
\begin{equation*}
G_1'(t)=I_{\rho }\| s_t\| ^2+I_{\rho }(s_{tt},s),\quad t>0
\end{equation*}
and by the third equation in \eqref{e1} we obtain that for $t>0$,
\begin{align*}
G_1'(t)
&=I_{\rho }\| s_t\| ^2-\|
s_x\| ^2-\frac{4\gamma }{3}\| s\| ^2-\frac{
4\alpha }{3}(s_t,s)-G(\psi -w_x,s)\\
&\leq I_{\rho }\| s_t\| ^2-\| s_x\|
^2+\big(\frac{G}{4\varepsilon _0}-\frac{4\gamma }{3}\big)\|
s\| ^2+\varepsilon _0G\| \psi -w_x\|
^2+\varepsilon \| s\| ^2+\frac{4\alpha ^2}{
9\varepsilon }\| s_t\| ^2 \\
&\leq -\| s_x\| ^2+\big(\frac{G}{4\varepsilon _0}
+\varepsilon -\frac{4\gamma }{3}\big)\| s\|
^2+\varepsilon _0G\| \psi -w_x\| ^2+\big(I_{\rho
}+\frac{4\alpha ^2}{9\varepsilon }\big)\| s_t\| ^2.
\end{align*}
\end{proof}

\begin{lemma} \label{lem4}
The derivative of $G_2(t)$ along
solutions of \eqref{e1}--\eqref{e2} satisfies
\begin{equation*}
G_2'(t)\leq -\rho \| w_t\| ^2+(G+\varepsilon _1)\| \psi -w_x\| ^2
+\frac{G}{2\varepsilon _1} \| \psi _x-3s_x\| ^2+\frac{9G}{2\varepsilon _1}
\| s_x\| ^2,\;
\end{equation*}
for all $t>0$ and some $\varepsilon _1>0$.
\end{lemma}

\begin{proof}
Using the first equation in \eqref{e1} and the boundary conditions
\eqref{e2}, we have that for $t>0$,
\begin{align*}
G_2'(t)&=-\rho \| w_t\| ^2-\rho (
w_{tt},w)\\
&=-\rho \| w_t\| ^2+G((\psi -w_x)
_x,w)\\
&=-\rho \| w_t\| ^2-G(\psi -w_x,w_x)+G
\left[ (\psi -w_x)w\right] _0^1 \\
&=-\rho \| w_t\| ^2+G(\psi -w_x,\psi
-w_x)-G(\psi -w_x,\psi )\\
&\leq -\rho \| w_t\| ^2+G\| \psi
-w_x\| ^2+\varepsilon _1G\| \psi -w_x\|
^2+\frac{G}{4\varepsilon _1}\| \psi _x\| ^2 \\
&\leq -\rho \| w_t\| ^2+(G+\varepsilon _1)\|
\psi -w_x\| ^2+\frac{G}{2\varepsilon _1}\| \psi
_x-3s_x\| ^2+\frac{9G}{2\varepsilon _1}\|
s_x\| ^2.
\end{align*}
\end{proof}

\begin{lemma} \label{lem5}
The derivative of $G_{3}(t)$ along
solutions of \eqref{e1}--\eqref{e2} satisfies
\begin{align*}
G_{3}'(t)
&\leq I_{\rho }\| 3s_t-\psi _t\|
^2-(\varsigma -\frac{G}{4\varepsilon _2}-\varepsilon )
\| 3s_x-\psi _x\| ^2+\varepsilon _2G\| \psi
-w_x\| ^2 \\
&\quad +\frac{1-\varsigma }{4\varepsilon }\int_0^1(h\square
(3s_x-\psi _x))dx,\quad t>0
\end{align*}
for $\varepsilon _2>0$, $\varepsilon >0$.
\end{lemma}

\begin{proof}  Using the second equation in \eqref{e1} we find that
\begin{align*}
&I_{\rho }\frac{d}{dt}(3s_t-\psi _t,3s-\psi )=I_{\rho
}\| 3s_t-\psi _t\| ^2-\| 3s_x-\psi _x\| ^2\\
&+[ (3s_x-\psi _x)(3s-\psi )] _0^1
+G((\psi -w_x),(3s-\psi ))\\
&+\Big(
\int_0^{t}h(t-r)(3s_x-\psi _x)(r)dr,3s_x-\psi _x\Big),\quad t>0.
\end{align*}
Then
\begin{align*}
G_{3}'(t)
&\leq I_{\rho }\| 3s_t-\psi _t\|
^2-\| 3s_x-\psi _x\| ^2+\varepsilon _2G\|
\psi -w_x\| ^2+\frac{G}{4\varepsilon _2}\|
3s_x-\psi _x\| ^2 \\
&\quad +\Big(\int_0^{t}h(t-r)\left[ (3s_x-\psi _x)(r)-(3s_x-\psi _x)(t)
\right] dr,3s_x-\psi _x\Big)\\
&\quad +\Big(\int_0^{t}h(r)dr\Big)((3s_x-\psi _x,3s_x-\psi_x)
\end{align*}
for $\varepsilon _2>0$, or
\begin{align*}
G_{3}'(t)
&\leq I_{\rho }\| 3s_t-\psi _t\| ^2-\| 3s_x-\psi _x\| ^2
+\varepsilon _2G\| \psi -w_x\| ^2\\
&\quad +\frac{G}{4\varepsilon _2}\| 3s_x-\psi _x\| ^2
+\varepsilon \| 3s_x-\psi _x\| ^2+\frac{1-\varsigma }{
4\varepsilon }\int_0^1(h\square (3s_x-\psi _x))dx\\
&\quad +(1-\varsigma )\| (3s_x-\psi _x)\| ^2,
\end{align*}
for $\varepsilon >0$. Hence
\begin{align*}
G_{3}'(t)&\leq I_{\rho }\| 3s_t-\psi _t\|
^2-(\varsigma -\frac{G}{4\varepsilon _2}-\varepsilon )
\| 3s_x-\psi _x\| ^2+\varepsilon _2G\| \psi -w_x\| ^2 \\
&\quad +\frac{1-\varsigma }{4\varepsilon }\int_0^1(h\square
(3s_x-\psi _x))dx,\;t>0.
\end{align*}
\end{proof}

\begin{lemma} \label{lem6}
The derivative of $G_{4}(t)$ is
estimated as follows
\begin{align*}
G_{4}'(t)&\leq -(3G-\varepsilon _1)\| \psi
-w_x\| ^2+\varepsilon _1(1+\varepsilon )I_{\rho
}\| 3s_t-\psi _t\| ^2+\varepsilon _1\|
w_t\| ^2 \\
&\quad +\big[ \frac{4\gamma ^2\rho ^2}{\varepsilon _1G^2}+\frac{4\alpha
^2}{\varepsilon _1}+(9+\frac{1}{\varepsilon }+\frac{9}{
4\varepsilon _1})I_{\rho }\big] \| s_t\|^2,\quad t>0,
\end{align*}
for $\varepsilon _1,\varepsilon >0$ provided that $I_{\rho }=\frac{\rho }{G}$.
\end{lemma}

\begin{proof}
Using the first and third equations in \eqref{e1},
\begin{align*}
G_{4}'(t)&=-\frac{4\gamma \rho }{G}(w_{tt},\Theta )-
\frac{4\gamma \rho }{G}(w_t,\Theta _t)-\frac{3\rho }{G}
(s_{xt},w_t)-\frac{3\rho }{G}(s_x,w_{tt})\\
&\quad +3I_{\rho }(s_{tt},\psi -w_x)+3I_{\rho }(s_t,\psi
_t-w_{xt})\,.
\end{align*}
Then we find that
\begin{align*}
G_{4}'(t)&=4\gamma ((\psi -w_x)_x,\Theta
)-\frac{4\gamma \rho }{G}(w_t,\Theta _t)-\frac{3\rho
}{G}(s_{xt},w_t)+3(s_x,(\psi -w_x) _x)\\
&\quad +3(-G(\psi -w_x)-\frac{4\gamma }{3}s-\frac{4\alpha }{3}
s_t+s_{xx},\psi -w_x)+3I_{\rho }(s_t,\psi
_t-w_{xt}),
\end{align*}
for $t>0$.
Next, by the definition of $\Theta $ and the assumption $I_{\rho }=
\frac{\rho }{G}$, we obtain
\begin{equation*}
G_{4}'(t)=-\frac{4\gamma \rho }{G}(w_t,\Theta _t)
-3G\| \psi -w_x\| ^2-4\alpha (s_t,\psi
-w_x)+3I_{\rho }(s_t,\psi _t),
\end{equation*}
for $t>0$.
Now, clearly
\begin{gather*}
\frac{4\gamma \rho }{G}(w_t,\Theta _t)\leq \varepsilon
_1\| w_t\| ^2+\frac{4\gamma ^2\rho ^2}{
\varepsilon _1G^2}\| s_t\| ^2,
\\
4\alpha (s_t,\psi -w_x)\leq \varepsilon _1\| \psi
-w_x\| ^2+\frac{4\alpha ^2}{\varepsilon _1}\|
s_t\| ^2,
\\
\begin{aligned}
3(s_t,\psi _t)
&\leq \varepsilon _1\| \psi _t\| ^2+\frac{9}{4\varepsilon _1}\| s_t\|
^2 \\
&\leq \varepsilon _1(1+\varepsilon )\| 3s_t-\psi
_t\| ^2+(9+\frac{1}{\varepsilon }+\frac{9}{4\varepsilon
_1})\| s_t\| ^2
\end{aligned}
\end{gather*}
lead to
\begin{align*}
G_{4}'(t)
&\leq \varepsilon _1\| w_t\| ^2+
\frac{4\gamma ^2\rho ^2}{\varepsilon _1G^2}\|
s_t\| ^2-3G\| \psi -w_x\| ^2+\varepsilon
_1\| \psi -w_x\| ^2+\frac{4\alpha ^2}{\varepsilon
_1}\| s_t\| ^2 \\
&\quad +\varepsilon _1(1+\varepsilon )I_{\rho }\|
3s_t-\psi _t\| ^2+(9+\frac{1}{\varepsilon }+\frac{9}{
4\varepsilon _1})I_{\rho }\| s_t\| ^2
\end{align*}
or
\begin{align*}
G_{4}'(t)
&\leq -(3G-\varepsilon _1)\| \psi-w_x\| ^2
 +\varepsilon _1(1+\varepsilon )I_{\rho}\| 3s_t-\psi _t\| ^2+\varepsilon _1\|
w_t\| ^2 \\
&\quad +\big[ \frac{4\gamma ^2\rho ^2}{\varepsilon _1G^2}+\frac{4\alpha
^2}{\varepsilon _1}+\big(9+\frac{1}{\varepsilon }+\frac{9}{
4\varepsilon _1}\big)I_{\rho }\big] \| s_t\|^2,\quad t\geq 0.
\end{align*}
\end{proof}

For the next lemma we need to get away from zero to ensure strict positivity
of $\int_0^{t}h(r)dr$. So for that $t\geq t_0>0$ we have
$\int_0^{t}h(r)dr\geq \int_0^{t_0}h(r)dr=h_0>0$.

\begin{lemma} \label{lem7}
For the functional $G_5(t)$ we have
\begin{align*}
G_5'(t)&\leq G\varepsilon \| \psi -w_x\|
^2+(G+4\varepsilon +2-\varsigma )\frac{1-\varsigma }{
4\varepsilon }\int_0^1(h\ \square (3s-\psi )_x)dx \\
&\quad +(2-\varsigma )\varepsilon \| 3s_x-\psi
_x\| ^2+I_{\rho }(\varepsilon -h_0)\|
3s_t-\psi _t\| ^2 \\
&\quad +\frac{I_{\rho }h(0)}{4\varepsilon }
\int_0^1(| h'| \square (3s-\psi )_x)dx,\quad
t\geq t_0>0
\end{align*}
for $\varepsilon >0$.
\end{lemma}

\begin{proof}
We recall that
\begin{equation*}
G_5(t)=-I_{\rho }\Big(3s_t-\psi _t,\int_0^{t}h(t-r)[ (3s-\psi
)(t)-(3s-\psi )(r)] dr\Big),\quad t>0
\end{equation*}
and therefore
\begin{align*}
G_5'(t)
&=-I_{\rho }(3s_{tt}-\psi _{tt},\int_0^{t}h(t-r)
 [ (3s-\psi )(t)-(3s-\psi )(r)] dr)\\
&\quad -I_{\rho }\Big(3s_t-\psi _t,\int_0^{t}h'(t-r)[
(3s-\psi )(t)-(3s-\psi )(r)] dr\Big)\\
&\quad -I_{\rho }\Big(\int_0^{t}h(r)dr\Big)\| 3s_t-\psi
_t\| ^2,\quad t>0.
\end{align*}
In view of the second equation in \eqref{e1} and the boundary conditions
\eqref{e2} we write
\begin{equation}
\begin{aligned}
G_5'(t)&=-\Big(G(\psi -w_x)+(3s-\psi
)_{xx},\int_0^{t}h(t-r)[ (3s-\psi )(t)-(3s-\psi )(r)]
dr\Big)\\
&\quad +\Big(\int_0^{t}h(t-r)(3s-\psi )_{xx}(r)dr,\int_0^{t}h(t-r)[
(3s-\psi )(t)-(3s-\psi )(r)] dr\Big)\\
&\quad -I_{\rho }\Big(3s_t-\psi _t,\int_0^{t}h'(t-r)[
(3s-\psi )(t)-(3s-\psi )(r)] dr\Big)\\
&\quad -I_{\rho }\Big(\int_0^{t}h(r)dr\Big)\| 3s_t-\psi _t\| ^2,\quad
t>0.
\end{aligned}\label{e10}
\end{equation}
It is easy to see that for $t>0$,
\begin{equation}
\begin{aligned}
&-G\Big(\psi -w_x,\int_0^{t}h(t-r)\left[ (3s-\psi )(t)-(3s-\psi )(r)
\right] dr\Big)\\
&\leq G\varepsilon \| \psi -w_x\| ^2+\frac{G(1-\varsigma
)}{4\varepsilon }\int_0^1(h\ \square (3s-\psi )_x)dx,
\end{aligned} \label{e11}
\end{equation}
\begin{equation}
\begin{aligned}
&\Big((3s-\psi )_{xx},\int_0^{t}h(t-r)\left[ (3s-\psi
)(t)-(3s-\psi )(r)\right] dr\Big)\\
&=-\Big((3s-\psi )_x,\int_0^{t}h(t-r)\left[ (3s-\psi
)_x(t)-(3s-\psi )_x(r)\right] dr\Big)\\
&\leq \varepsilon \| 3s_x-\psi _x\| ^2+\frac{
1-\varsigma }{4\varepsilon }\int_0^1(h\square (3s-\psi)_x)dx,
\end{aligned} \label{12}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\Big(\int_0^{t}h(t-r)(3s-\psi )_{xx}(r)dr,\int_0^{t}h(t-r)[
(3s-\psi )(t)-(3s-\psi )(r)] dr\Big)\\
&=\| \int_0^{t}h(t-r)[ (3s-\psi )_x(t)-(3s-\psi )_x(r)] dr\| ^2 \\
&\quad -\Big(\int_0^{t}h(r)dr\Big)\Big((3s-\psi )_x,\int_0^{t}h(t-r)
 [ (3s-\psi )_x(t)-(3s-\psi )_x(r)] dr\Big)\\
&\leq \| \int_0^{t}h(t-r)[ (3s-\psi )_x(t)-(3s-\psi )_x(r)] dr\| ^2
+(1-\varsigma ) \Big\{ \varepsilon \| 3s_x-\psi_x\| ^2\\
&\quad +\frac{1}{4\varepsilon }\| \int_0^{t}h(t-r)
 [ (3s-\psi )_x(t)-(3s-\psi )_x(r)] dr\| ^2\Big\}
\\
&\leq (1+\frac{1-\varsigma }{4\varepsilon })(1-\varsigma
)\int_0^1(h\ \square (3s-\psi )_x)dx+\varepsilon
(1-\varsigma )\| 3s_x-\psi _x\| ^2,
\end{aligned} \label{e13}
\end{equation}
for $t>0$.
Further
\begin{equation}
\begin{aligned}
&I_{\rho }(3s_t-\psi _t,\int_0^{t}h'(t-r)[ (3s-\psi
)(t)-(3s-\psi )(r)] dr)\\
&\leq \varepsilon I_{\rho }\| 3s_t-\psi _t\| ^2
+\frac{ I_{\rho }h(0)}{4\varepsilon }\int_0^1(| h'| \square (3s-\psi )_x)dx,\quad
t>0.
\end{aligned} \label{e14}
\end{equation}
Taking into account estimates  \eqref{e11}--\eqref{e14},
 in \eqref{e10} and considering $ t\geq t_0>0$, we obtain
\begin{align*}
G_5'(t)
&\leq G\varepsilon \| \psi -w_x\| ^2+
 \frac{G(1-\varsigma )}{4\varepsilon }\int_0^1(h\ \square (3s-\psi
 )_x)dx+\varepsilon \| 3s_x-\psi _x\| ^2 \\
&\quad +\frac{1-\varsigma }{4\varepsilon }\int_0^1(h\square (3s-\psi)_x)dx
 +(1+\frac{1-\varsigma }{4\varepsilon })(
1-\varsigma )\int_0^1(h\ \square (3s-\psi )_x)dx
\\
&\quad +\varepsilon (1-\varsigma )\| 3s_x-\psi
_x\| ^2+\varepsilon I_{\rho }\| 3s_t-\psi_t\| ^2\\
&\quad +\frac{I_{\rho }h(0)}{4\varepsilon }\int_0^1(
| h'| \square (3s-\psi )_x)dx
-I_{\rho }h_0\| 3s_t-\psi _t\| ^2
\end{align*}
or, for $t\geq t_0>0$
\begin{align*}
G_5'(t)&\leq G\varepsilon \| \psi -w_x\|
^2+(G+4\varepsilon +2-\varsigma )\frac{1-\varsigma }{
4\varepsilon }\int_0^1(h\ \square (3s-\psi )_x)dx \\
&\quad +(2-\varsigma )\varepsilon \| 3s_x-\psi
_x\| ^2+I_{\rho }(\varepsilon -h_0)\|
3s_t-\psi _t\| ^2 \\
&\quad +\frac{I_{\rho }h(0)}{4\varepsilon }\int_0^1(| h'| \square (3s-\psi )_x)dx.
\end{align*}
The proof is complete.
\end{proof}

Using the previous lemmas we now give the proof of our main result.


\begin{proof}[Proof of Theorem \ref{thm1}]
Gathering the estimates in the previous lemmas we find that
\begin{align*}
F'(t)&=E'(t)+\sum\nolimits_{i=1}^{5}\delta
_{i}G_{i}'(t)\leq -4\alpha \| s_t\| ^2-\dfrac{
h(t)}{2}\| (3s-\psi )_x\| ^2 \\
&\quad +\dfrac{1}{2}\int_0^1(h'\square (3s-\psi
 )_x)dx-\delta _1\| s_x\| ^2+\delta
 _1(\frac{G}{4\varepsilon _0}+\varepsilon -\frac{4}{3}\gamma
 )\| s\| ^2 \\
&\quad +\delta _1\varepsilon _0G\| \psi -w_x\| ^2+\delta
_1(I_{\rho }+\frac{4\alpha ^2}{9\varepsilon })\|
s_t\| ^2-\delta _2\rho \| w_t\| ^2 \\
&\quad +\delta _2(G+\varepsilon _1)\| \psi
-w_x\| ^2+\frac{G\delta _2}{2\varepsilon _1}\|
3s_x-\psi _x\| ^2+\frac{9G\delta _2}{2\varepsilon _1}\| s_x\| ^2\\
&\quad +\delta_3I_{\rho }\| 3s_t-\psi_t\| ^2
 -\delta_3(\varsigma -\frac{G}{4\varepsilon _2}-\varepsilon
 )\| 3s_x-\psi _x\| ^2+\delta_3\varepsilon _2G\| \psi -w_x\| ^2 \\
&\quad +\delta_3\frac{1-\varsigma }{4\varepsilon }
 \int_0^1(h\square (3s_x-\psi _x))dx-\delta _{4}(3G-\varepsilon _1)\|
 \psi -w_x\| ^2 \\
&\quad +\delta _{4}\varepsilon _1(1+\varepsilon )I_{\rho }\|
3s_t-\psi _t\| ^2+\delta _{4}[ \frac{4\gamma ^2\rho
^2}{\varepsilon _2G^2}+\frac{4\alpha ^2}{\varepsilon _1}+(9+
\frac{1}{\varepsilon }+\frac{9}{4\varepsilon _1})I_{\rho }]
\| s_t\| ^2 \\
&\quad +\varepsilon _1\delta _{4}\| w_t\| ^2+\delta
_5G\varepsilon \| \psi -w_x\| ^2+\delta
_5\varepsilon (2-\varsigma )\| 3s_x-\psi _x\| ^2 \\
&\quad +\delta _5(G+4\varepsilon +2-\varsigma )\frac{1-\varsigma }{
4\varepsilon }\int_0^1(h\ \square (3s-\psi )_x)dx\\
&\quad +\delta_5I_{\rho }(\varepsilon -h_0)\| 3s_t-\psi_t\| ^2
+\delta _5\frac{I_{\rho }h(0)}{4\varepsilon }\int_0^1(|
h'| \square (3s-\psi )_x)dx,\quad t\geq t_0>0
\end{align*}
or
\begin{equation}
\begin{aligned}
F'(t)&\leq -\left\{ 4\alpha -\delta _1(I_{\rho }+\frac{
4\alpha ^2}{9\varepsilon })-\delta _{4}\left[ (9+\frac{1}{
\varepsilon }+\frac{9}{4\varepsilon _1})I_{\rho }+\frac{4\alpha ^2
}{\varepsilon _1}+\frac{4\gamma ^2\rho ^2}{\varepsilon _2G^2}
\right] \right\} \| s_t\| ^2 \\
&\quad -(\delta _1-\frac{9G\delta _2}{2\varepsilon _1})
\| s_x\| ^2+\delta _1(\frac{G}{4\varepsilon _0
}+\varepsilon -\frac{4}{3}\gamma )\| s\| ^2 \\
&\quad -\left[ \delta _{4}(3G-\varepsilon _1)-\delta _1\varepsilon _0G-\delta
_2(G+\varepsilon _1)-\delta_3\varepsilon _2G-\delta
_5G\varepsilon \right] \| \psi -w_x\| ^2 \\
&\quad -\Big\{ \delta_3(\varsigma -\frac{G}{4\varepsilon _2}
-\varepsilon )-\frac{G\delta _2}{2\varepsilon _1}-\delta
_5\varepsilon (2-\varsigma )\Big\} \| 3s_x-\psi
_x\| ^2 \\
&\quad -(\delta _2\rho -\varepsilon _2\delta _{4})\|
w_t\| ^2+I_{\rho }\left[ \delta_3+\delta _{4}\varepsilon
_1(1+\varepsilon )+\delta _5(\varepsilon
-h_0)\right] \| 3s_t-\psi _t\| ^2 \\
&\quad-\Big\{ \dfrac{\beta _1}{2}-\delta_3\frac{1-\varsigma }{4\varepsilon }
-\delta _5(G+4\varepsilon +2-\varsigma )\frac{1-\varsigma }{
4\varepsilon }\\
&\quad -\delta _5\frac{\beta _0I_{\rho }h(0)}{4\varepsilon }
\Big\} \int_0^1(h\ \square (3s-\psi )_x)dx.
\end{aligned}
\label{e15}
\end{equation}

Our strategy for selecting the different coefficients and parameters is as
follows: all the $\delta _{i}$, $i=1,\dots 5$ will be determined in terms of
only one of them (here $\delta _1$). This $\delta _1$ will be
accountable in front of $\alpha $ and $\beta _1$ in the coefficients of
the first and the last term in \eqref{e15}. From the beginning, we have managed in
our estimations to balance the largest coefficients (here $1/\varepsilon $)
on the terms that appear in the derivative of the energy. This will allow us
to ignore $\varepsilon $ at the beginning of the process of selection.

Let us ignore for the moment the first and the last terms in \eqref{e15}.
We shall, at the same time, ignore the terms having
coefficients in $\varepsilon $. The focus will be on
\begin{gather*}
\delta _1-\frac{9G}{2\varepsilon _1}\delta _2>0, \quad
\frac{G}{4\varepsilon _0}-\frac{4}{3}\gamma <0, \\
\delta _{4}(3G-\varepsilon _1)-\delta _1\varepsilon _0G-\delta
_2(G+\varepsilon _1)-\delta_3\varepsilon _2G>0, \\
\delta_3(\varsigma -\frac{G}{4\varepsilon _2})-\frac{G}{
2\varepsilon _1}\delta _2>0, \\
\delta _2\rho -\varepsilon _2\delta _{4}>0, \quad
\delta_3+\delta _{4}\varepsilon _1-\delta _5h_0<0,
\end{gather*}
or
\begin{equation}
\begin{gathered}
\frac{9G}{2\varepsilon _1}\delta _2<\delta _1, \quad
\frac{G}{4\varepsilon _0}<\frac{4}{3}\gamma , \\
\delta _1\varepsilon _0G+\delta _2(G+\varepsilon _1)
+\delta_3\varepsilon _2G<\delta _{4}(3G-\varepsilon _1), \\
\frac{G}{2\varepsilon _1}\delta _2<\delta_3(\varsigma -\frac{G}{
4\varepsilon _2}), \\
\varepsilon _2\delta _{4}<\delta _2\rho , \quad
\delta_3+\delta _{4}\varepsilon _1<\delta _5h_0.
\end{gathered} \label{e16}
\end{equation}
Let $\varepsilon _0=\frac{G}{4\gamma }$ so that the second inequality in
\eqref{e16} is satisfied. Put $\varepsilon _2=\frac{G}{2\varsigma }$,
$\varepsilon _1=G$ and ignore the last inequality (we will take $\delta _5$
large enough as it does not appear elsewhere), we will be left with
\begin{equation}
\begin{gathered}
\frac{9}{2}\delta _2<\delta _1, \\
\delta _1\frac{G}{4\gamma }+2\delta _2+\delta_3\frac{G}{2\varsigma }
<2\delta _{4}, \\
\delta _2<\varsigma \delta_3, \quad
\frac{G}{2\varsigma }\delta _{4}<\delta _2\rho .
\end{gathered} \label{e17}
\end{equation}
Note that $2\delta _2<\delta _{4}<\frac{2\varsigma }{G}\delta _2\rho $
is valid if $G<\varsigma \rho $ and $\delta _{4}=\frac{G+\varsigma \rho }{G}
\delta _2$. Therefore \eqref{e17} reduces to
\begin{gather*}
\frac{9}{2}\delta _2<\delta _1, \\
\delta _1\frac{G}{4\gamma }+\delta_3\frac{G}{2\varsigma }<\frac{
G+\varsigma \rho }{G}\delta _2, \\
\delta _2<\varsigma \delta_3.
\end{gather*}
By  assumption (H1) we may have
\[
\delta _1\frac{G}{4\gamma }<\frac{G+\varsigma \rho }{2G}\delta _2
<\frac{G+\varsigma \rho }{9G}\delta_1,\quad
\delta_3\frac{G}{2\varsigma }<\frac{G+\varsigma \rho }{2G}
\delta _2<\frac{G+\varsigma \rho }{2G}\varsigma \delta_3.
\]
These inequalities ensure the possibility of selecting (for instance) $\delta _2$
and $\delta_3$ in terms of $\delta _1$. It is now possible to select $
\delta _5$ (satisfying the last relation in \eqref{e16}) in terms of $\delta _1$
and then $\varepsilon $. Finally, $\delta _1$ is chosen so small that the
coefficients of the first and the last terms in \eqref{e15} are satisfied. We end
up with an inequality of the form
\begin{equation*}
F'(t)\leq -CF(t),\quad t\geq t_0>0.
\end{equation*}
This gives the exponential decay of $F(t)$ on $[t_0,\infty )$. The
exponential decay of the energy follows from the equivalence with $F(t)$ and
the statement of the theorem for $t\geq 0$ is clear. The proof is complete.
\end{proof}

\subsection*{Remark}
 It would be nice to remove the conditions on $G$ although $\rho =GI_{\rho }$
(equal wave speeds) seems natural as we have a similar one
in the theory of Timoshenko beams. The assumption (H1) looks
technical and we believe that it may be improved considerably through a
better choice of the functionals and adequate estimations. Investigations on
other boundary conditions would also be of great importance.

\subsection*{Acknowledgments}
 The authors would like to acknowledge the support
provided by King Abdulaziz City for Science and Technology (KACST) through
the Science and Technology Unit at King Fahd University of Petroleum and
Minerals (KFUPM) for funding this work through project No. AC -32- 49. The
authors would like to thank also the anonymous referee for pointing out an
error in the original paper and suggesting a way to fix it.

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