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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 75, pp. 1--27.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/75\hfil Timoshenko-type system of thermoelasticity]
{Energy decay in a Timoshenko-type system for thermoelasticity
of type III with distributed delay and past history}

\author[J. Hao, F. Wang \hfil EJDE-2018/75\hfilneg]
{Jianghao Hao, Fei Wang}

\address{Jianghao Hao (corresponding author)\newline
School of Mathematical Sciences,
Shanxi University,
Taiyuan, Shanxi 030006, China}
\email{hjhao@sxu.edu.cn}

 \address{Fei Wang \newline
School of Mathematical Sciences,
 Shanxi University,
Taiyuan, Shanxi 030006, China}
 \email{2930904441@qq.com}

\dedicatory{Communicated by Anthony Bloch}

\thanks{Submitted May 11, 2017. Published March 17, 2018.}
\subjclass[2010]{35L70, 35L75, 93D20}
\keywords{Timoshenko system; past history; relaxation function;
\hfill\break\indent distributed delay; energy decay}

\begin{abstract}
 In this work, we consider a one-dimensional Timoshenko system of
 thermoelasticity of type III with past history and distributive delay.
 It is known that an arbitrarily small delay may be the source of instability.
 We establish the well-posedness and the stability of the system for the
 cases of equal and nonequal speeds of wave propagation respectively.
 Our results show that the damping effect is strong enough to uniformly
 stabilize the system even in the presence of time delay under suitable
 conditions and improve the related results.
\end{abstract}

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

\section{Introduction}

In this article, we study the following Timoshenko-type system for thermoelasticity
of type III with distributive delay and past history,
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x+\beta\theta_{tx}=0,\quad
  (x,t)\in (0,1)\times(0,\infty) ,\\
\begin{aligned}
&\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)+\int_0^{\infty}g(s)\psi_{xx}(x,t-s)ds\\
&-\beta\theta_{t}+f(\psi)=0,\quad  (x,t)\in(0,1)\times(0,\infty) ,
\end{aligned}\\
\begin{aligned}
&\rho_3\theta_{tt}-\delta\theta_{xx}-\ell\theta_{txx}
 +\gamma\varphi_{tx}+\gamma\psi_t-\int_{\tau_1}^{\tau_2}\mu(\varsigma)\theta_{txx}
  (x,t-\varsigma)d\varsigma\\
&=0,\quad  (x,t)\in(0,1)\times(0,\infty),
\end{aligned}\\
\varphi(x,0)=\varphi_0(x), \quad \varphi_t(x,0)=\varphi_1(x), \quad
 \theta(x,0)=\theta_0(x),\quad  x\in(0,1),\\
\theta_t(x,0)=\theta_1(x),\quad \psi_t(x,0)=\psi_1(x),\quad  x\in(0,1),\\
\psi(x,-t)=\psi_0(x,t),\quad (x,t)\in (0,1)\times(0,\infty),\\
\varphi(0,t)=\varphi(1,t)=\psi(0,t)=\psi(1,t)=\theta(0,t)=\theta(1,t)=0,
\quad t\in(0,\infty),\\
\theta_{tx}(x,-t)=f_0(x,t),\quad (x,t)\in(0,1)\times (0,\tau_2),
\end{gathered}\label{e1.1}
\end{equation}
where $\varphi$ is the longitudinal displacement, $\psi$ is the volume fraction,
$\theta$ is the difference in temperature, the coefficients $\rho_1$,
$\rho_2$, $\rho_3$, $k$, $b$, $\ell$, $\beta$, $\delta$, $\gamma$ are
 positive constants, $\tau_1<\tau_2$ are non-negative constants such that
$\mu:[\tau_1,\tau_2]\to R$ represents distributive time delay,
 $f$ is a forcing term.

In 1921, Timoshenko \cite{4} gave, as a model for a thick beam, the following
system of coupled hyperbolic equations
\begin{equation}
\begin{gathered}
\rho u_{tt}=(K(u_x+\varphi))_x,\\
I_\rho\varphi_{tt}=(EI\varphi_x)_x+K(u_x-\varphi),
\end{gathered} \label{e1.2}
\end{equation}
where $t$ denotes the time variable and $x$ is the space variable along
the beam of length $L$, in its equilibrium configuration, $u$ is the
transverse displacement of the beam and $\varphi$ is the rotation angle
of the filament of the beam. The coefficients $\rho$, $I_\rho$, $E$, $I$
 and $K$ are respectively the density (the mass per unit length), the polar
 moment of inertia of a cross section, Young's modulus of elasticity,
the moment of inertia of a cross section, and the shear modulus.

Since then, the issue of existence and stability of Timoshenko system has
attracted a great deal of attention in the last decades
(e.g. \cite{7,13,6,3,1,8,14,5,2}).
Messaoudi and Said \cite{2} considered the following Timoshenko-type system
 with past history
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x=0,\\
\rho_2\psi_{tt}-b\psi_{xx}+\int_0^\infty g(s)\psi_{xx}(x,t-s)ds
+k(\varphi_x+\psi)=0,
\end{gathered} \label{e1.3}
\end{equation}
where $\rho_1$, $\rho_2$, $k$, $b$ are positive
constants and $g$ is a differentiable function satisfying, for some positive
constant $k_0$ and $1\leq p<3/2$, the  conditions
$$
g(t)>0,\quad \widehat{b}=b-\int_0^\infty g(s)ds>0, \quad g'(t)\leq k_0g^p(t).
$$
They proved that, for the case of equal-speed propagation
$\frac{\rho_1}{k}=\frac{\rho_2}{b}$, the first energy decays exponentially
if $p=1$ and polynomially if $p>1$. When in the opposite
case $\frac{\rho_1}{k}\neq\frac{\rho_2}{b}$, the decay is in the rate
of $1/t^p$. Guesmia and Messaoudi \cite{3} also considered \eqref{e1.3}
and established some general decay results for the equal and nonequal speed
propagation cases where the relaxation function satisfies a relation of the form
$$
g'(t)\leq-\xi(t)g(t).
$$
Guesmia and Messaoudi \cite{1} concerned with the long-time behavior of
the solution of the  Timoshenko system
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x+\gamma h(\varphi_t)
 +\int_0^{\infty}g(s)(a(x)\varphi_x(t-s))_xds=0,\\
\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)=0.
\end{gathered} \label{e1.4}
\end{equation}
They showed that the dissipation given by this complementary controls
guarantees the stability of the system in case of the equal-speed
propagation as well as in the opposite case.

Time delays arise in many applications because most phenomena naturally
depend not only on the present state but also on some past occurrences.
In recent years, the stability of evolution systems with time delay
effects has become an active area of research
(e.g. \cite{9, 7, 8, 14, 10,12, 11}). Apalara \cite{9} considered the
following thermoelasic system of Timoshenko type with a linear frictional
damping and an internal distributed delay acting on transverse displacement,
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-\ell(\varphi_x+\psi)_x+\mu_1\varphi_t
+\int_{\tau_1}^{\tau_2}\mu_2(s)\varphi_t(x,t-s)ds=0,\\
\rho_2\psi_{tt}-b\psi_{xx}+\ell(\varphi_x+\psi)+\delta\theta_{x}=0,\\
\rho_3\theta_{t}+q_x+\delta\psi_{tx}=0,\\
\tau q_t+\beta q+\theta_{x}=0.
\end{gathered} \label{e1.5}
\end{equation}
Under suitable assumptions on the weight of the delay and that of
frictional damping, the author established the well-posedness result
 and proved that the system is exponentially stable regardless of the
speeds of wave propagation.

 Feng and Pelicer \cite{7} were concerned with a Timoshenko system with time delay
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x=0,\\
\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)
 +\mu_1\psi_t+\mu_2\psi_t(x,t-\tau)+f(\psi)=0,
\end{gathered} \label{e1.6}
\end{equation}
where $\mu_2\psi_t(x,t-\tau)$ is time delay. They established the
 well-posedness of the problem with respect to weak solutions  under
suitable assumptions and the exponential stability of the system under
the usual equal wave speed assumption.

Kafini et al.\ \cite{8} considered the following Timoshenko-type system
of thermoelasticity of type III with distributive delay
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x=0,\\
\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)+\beta\theta_{tx}=0,\\
\rho_3\theta_{tt}-\delta\theta_{xx}-k\theta_{txx}
-\int_{\tau_1}^{\tau_2}g(s)\theta_{txx} (x,t-s)ds+\gamma\psi_{tx}=0,
\end{gathered} \label{e1.7}
\end{equation}
where $\tau_1<\tau_2$ are non-negative constants such that
$g:[\tau_1,\tau_2]\to R^+$ represents distributive time delay.
They proved an exponential decay in the case of equal wave speeds and
a polynomial decay result  in the case of nonequal wave speeds with smooth
initial data.

In this present work we consider \eqref{e1.1}, prove the well-posedness
and establish the energy decay rate in case of the equal-speed propagation
as well as in the opposite case.

 The article is organized as follows.
 In Section 2, we introduce some transformations and assumptions needed in our work.
In Section 3, we use the semigroup method to prove the well-posedness of
 problem \eqref{e1.1}. In Section 4, we state and prove our stability results.

 \section{Preliminaries}

In this section, we present some materials needed in the proof of our results.
Throughout this paper, $c$ is used to denote a generic positive constant and
is different in various occurrences.

Firstly, to deal with the delay term, we introduce the new variable
 $$
z(x,\rho,\varsigma,t)=\theta_{tx}(x,t-\varsigma\rho),\quad
 x\in(0,1),\; \rho\in(0,1), \; \varsigma\in(\tau_1,\tau_2),\; t>0.
$$
Then we obtain
$$
\varsigma z_t(x,\rho,\varsigma,t)+z_\rho(x,\rho,\varsigma,t)=0,\quad x\in(0,1),\;
 \rho\in(0,1), \; \varsigma\in(\tau_1,\tau_2),\; t>0.
$$
Then problem \eqref{e1.1} is equivalent to
\begin{equation}
\begin{gathered}
\rho_1\varphi_{tt}-k(\varphi_x+\psi)_x+\beta\theta_{tx}=0,\quad
 (x,t)\in (0,1)\times(0,\infty) ,\\
\begin{aligned}
&\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)
 +\int_0^{\infty}g(s)\psi_{xx}(x,t-s)ds-\beta\theta_{t} +f(\psi)\\
&=0,\quad   (x,t)\in (0,1)\times(0,\infty) ,
\end{aligned}\\
\begin{aligned}
&\rho_3\theta_{tt}-\delta\theta_{xx}-\ell\theta_{txx}+\gamma\varphi_{tx}
 +\gamma\psi_t-\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x (x,1,\varsigma,t)d\varsigma\\
&=0,  \quad  (x,t)\in (0,1)\times(0,\infty) ,
\end{aligned}\\
\begin{aligned}
&\varsigma z_t(x,\rho,\varsigma,t)+z_\rho(x,\rho,\varsigma,t)\\
&=0,\quad   (x,\rho,\varsigma,t) \in (0,1)\times(0,1)
\times(\tau_1,\tau_2)\times(0,\infty), 
\end{aligned}\\
\varphi(x,0)=\varphi_0(x),\quad  \varphi_t(x,0)=\varphi_1(x), \quad
 \theta(x,0)=\theta_0(x), \quad  x\in(0,1),\\
\theta_t(x,0)=\theta_1(x),\quad \psi_t(x,0)=\psi_1(x),\quad x\in(0,1),\\
\psi(x,-t)=\psi_0(x,t),\quad  (x,t)\in (0,1)\times(0,\infty) ,\\
\varphi(0,t)=\varphi(1,t)=\psi(0,t)=\psi(1,t)=\theta(0,t)=\theta(1,t)=0,\quad
 t \in (0,\infty) ,\\
z(x,\rho,\varsigma,0)=f_0(x,\rho \varsigma),\quad  (x,\rho,\varsigma)
 \in (0,1)\times(0,1)\times(\tau_1,\tau_2).
\end{gathered} \label{e2.1}
\end{equation}

We shall use the folloing hypotheses.
\begin{itemize}
\item[(H1)] $\mu:[\tau_1,\tau_2]\to R$ is a bounded function and
\begin{equation}
{\ell-\int_{\tau_1}^{\tau_2}|\mu(\varsigma)|d\varsigma>0.} \label{e2.2}
\end{equation}

\item[(H2)] $g:R_+\to R_+$ is a $C^1$ function satisfying
\begin{equation}
g(0)>0,\quad  b-\int_0^{\infty} g(s)ds=l>0,\quad \int_0^{\infty} g(s)ds=g_0 .
\label{e2.3}
\end{equation}

\item[(H3)] There exists a positive nonincreasing differentiable function
$\xi:R_+\to R_+$ satisfying
\begin{equation}
{g'(t)\leq-\xi(t)g(t),\ t\geq 0.} \label{e2.4}
\end{equation}

\item[(H4)] $f:R\to R$ satisfies
\begin{equation}
{|f(\psi^2)-f(\psi^1)|\leq k_0(|\psi^1|^\varrho+|\psi^2|^\varrho)|\psi^1-\psi^2|,
\quad \psi^1,\psi^2\in R, } \label{e2.5}
\end{equation}
where $k_0>0$, $\varrho>0$. In addition we assume that
\begin{equation}
{0\leq \widehat{f}(\psi)\leq f(\psi)\psi, \quad  \psi\in R, }
\label{e2.6}
\end{equation}
with $ \widehat{f}(\psi)=\int_0^\psi f(s)ds$.
\end{itemize}

The first-order energy associated with \eqref{e2.1} is
\begin{equation}
\begin{aligned}
E(t):=&E_1(\varphi, \psi, \theta) \\
 =&\frac{\gamma}{2}\int_0^1\Big(\rho_1\varphi_t^2+k(\varphi_x+\psi)^2
 +\rho_2\psi_t^2+(b-\int_0^{\infty}g(s)ds)\psi_x^2\Big)dx  \\
 & +\frac{\gamma}{2}(g\circ\psi_x)+\gamma\int_0^1\widehat{f}(\psi)dx
 +\frac{\beta}{2}\int_0^1(\rho_3\theta_t^2+\delta\theta_x^2)dx \\
&  +\frac{\beta}{2}\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}
 \varsigma|\mu(\varsigma )|z^2(x,\rho,\varsigma,t)\,d\varsigma\,d\rho\,dx,
\end{aligned}\label{e2.7}
\end{equation}
where
$$
(g\circ \nu)(t)=\int_0^1\int_0^{\infty}g(s)(\nu(x,t)-\nu(x,t-s))^2\,ds\,dx.
$$

\section{Well-posedness of the problem}

In this section, we give a brief idea about the existence and uniqueness of
solution for \eqref{e2.1} using the semigroup theory \cite{16}. Using the notation
\begin{equation}
\eta^t(x,s)=\psi(x,t)-\psi(x,t-s),\quad t\in R_+, \; (x,t,s)\in(0,1)\times R_+
\times R_+, \label{e3.1}
\end{equation}
which was adopted in articles \cite{2, PRQ} and \cite{RS} and $\eta^t$
is the relative history of $\psi$, we have
\begin{equation}
\begin{gathered}
\eta^t_t+\eta^t_s-\psi_t=0, \quad (x,t,s)\in(0,1)\times R_+ \times R_+ ,\\
\eta^t(0,s)=\eta^t(1,s)=0, \quad (t,s)\in R_+ \times R_+ ,\\
 \eta^t(x,0)=0,\quad (x,t)\in(0,1)\times R_+  .
\end{gathered}\label{e3.2}
\end{equation}
Then the second equation of \eqref{e2.1} can be formulated as
$$
\rho_2\psi_{tt}-b\psi_{xx}+k(\varphi_x+\psi)+g_0\psi_{xx}(x,t)
-\int_0^{\infty}g(s)\eta^t_{xx}(x,s)ds-\beta\theta_{t}+f(\psi)=0.
$$
Let
$$
\eta_0(x,s):=\eta^0(x,s)=\psi_0(x,0)-\psi_0(x,s),\quad
 (x,s)\in(0,1)\times R_+.
$$

Before using the semigroup theory, we introduce three new dependent
variables
$$
u=\varphi_t,\quad   v=\psi_t, \quad \omega=\theta_t.
$$
Then problem \eqref{e2.1} becomes the following problem for an abstract
first-order evolutionary equation,
\begin{equation}
\begin{gathered}
\frac{d}{dt} U+\mathcal{A} U= F( U),\\
 U(0)= U_0=(\varphi_0,\varphi_1,\psi_0,\psi_1,\theta_0,\theta_1,\eta_0,f_0)^T ,
\end{gathered} \label{e3.3}
\end{equation}
where $ U=(\varphi,u,\psi,v,\theta,\omega,\eta^t,z)$ and the linear
operator $\mathcal{A}:D(\mathcal{A})\subset\mathcal{H}\to\mathcal{H}$ is defined by
\begin{gather}
\mathcal{A} U=
\begin{pmatrix}
 -u \\
 -\frac{k}{\rho_1}(\varphi_x+\psi)_x+\frac{\beta}{\rho_1}\omega_x \\
 -v\\
 -\frac{b}{\rho_2}\psi_{xx}+\frac{k}{\rho_2}(\varphi_x+\psi)
 +\frac{g_0}{\rho_2}\psi_{xx}-\frac{1}{\rho_2}\int_0^{\infty}g(s)
 \eta^t_{xx}(x,t,s)ds-\frac{\beta}{\rho_2}\omega \\
 -\omega\\
 -\frac{\delta}{\rho_3}\theta_{xx}-\frac{\ell}{\rho_3}\omega_{xx}
 +\frac{\gamma}{\rho_3}u_x+\frac{\gamma}{\rho_3}v-\frac{1}{\rho_3}
 \int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x (x,1,\varsigma,t)d\varsigma\\
\eta^t_s-v\\
 \frac{1}{\varsigma}z_\rho
\end{pmatrix},
\label{e3.4} \\
 F( U)=
\begin{pmatrix}
0\\
0\\
0\\
-\frac{1}{\rho_2}f(\psi)\\
0\\
0\\
0\\
0
\end{pmatrix}.
\label{e3.5}
\end{gather}

Next, we introduce the energy space
\begin{align*}
\mathcal{H}&=H_0^1(0,1)\times L^2(0,1)\times H_0^1(0,1)\times L^2(0,1)
 \times H_0^1(0,1)\times L^2(0,1)\times L_g \\
&\quad \times L^2((0,1)\times(0,1)\times(\tau_1,\tau_2)),
\end{align*}
where
$$
L_g=\big\{\phi:R_+\to H_0^1(0,1),\, \int_0^1\int_0^{\infty}g(s)\phi_x^2\,ds\,dx
 <\infty\big\},
$$
endowed with the inner product
$$
\langle \phi_1,\phi_2\rangle_{L_g}
=\int_0^1\int_0^{\infty}g(s)\phi_{1x}(s)\phi_{2x}(s)\,ds\,dx.
$$
For any $ U=(\varphi,u,\psi,v,\theta,\omega,\eta^t,z)^T\in\mathcal{H}$,
$\widetilde{ U}=(\widetilde{\varphi},\widetilde{u},\widetilde{\psi},
\widetilde{v},\widetilde{\theta},\widetilde{\omega},\widetilde{\eta^t},
\widetilde{z})^T\in\mathcal{H} $
and for $\ell-\int_{\tau_1}^{\tau_2}|\mu(\varsigma)|d\varsigma>0$
we equip $\mathcal{H}$  with the inner product defined by
\begin{align*}
\langle  U,\widetilde{ U}\rangle _\mathcal{H}
&=\gamma\int_0^1(\rho_1u\widetilde{u}+\rho_2v\widetilde{v}
 +k(\varphi_x+\psi)(\widetilde{\varphi_x}+\widetilde{\psi})
 +b\psi_x\widetilde{\psi_x}-g_0\psi_x\widetilde{\psi_x})dx \\
&\quad  +\gamma<\eta^t,\widetilde{\eta^t}>_{L_g}
 +\beta\int_0^1(\rho_3\omega\widetilde{\omega}+\delta\theta_x\widetilde{\theta_x})dx\\
&\quad +\beta\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma|\mu(\varsigma)
 |z\widetilde{z}(x,\rho,\varsigma,\cdot)\,d\varsigma \,d\rho\,dx.
\end{align*}
The domain of $\mathcal{A}$ is
\begin{align*}
D(\mathcal{A})&=\Big\{ U\in \mathcal{H} : \varphi, \psi, \theta \in
 H^2(0,1)\cap H_0^1(0,1),u,v,\omega \in H_0^1(0,1), \\
&\quad \eta^t\in L_g,z,z_\rho\in L^2((0,1)\times(0,1)\times(\tau_1,\tau_2))\Big\},
\end{align*}
which is dense in $\mathcal{H}$.

\begin{theorem} \label{thm3.1}
Assume $U_0\in \mathcal{H}$ and {\rm (H1)--(H4)} hold. Then, there exists
 a unique solution $ U\in (R_+,\mathcal{H})$ of problem \eqref{e2.1}.
 Moreover, if $U_0\in D(\mathcal{A}) $ then
$$
U\in C(R_+, D(\mathcal{A}))\cap C^1(R_+,\mathcal{H}).
$$
\end{theorem}

\begin{proof}
We use the semigroup approach. Sufficiently, we prove that $\mathcal{A}$
is a maximal monotone operator. First, we prove that $\mathcal{A}$ is monotone.
For any $U\in D(\mathcal{A})$, we have
\begin{align*}
(\mathcal{A}U,U)_ {\mathcal{H}}
&={-\frac{\gamma}{2}(g'\circ\psi_x)+\ell\beta\int_0^1\omega_x^2dx
 -\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x(x,1,\varsigma,
 \cdot)d\varsigma\omega\,dx} \\
&\quad  +\frac{\beta}{2}\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma)
 |z^2(x,1,\varsigma,\cdot)d\varsigma\,dx
 -\frac{\beta}{2}\int_{\tau_1}^{\tau_2}|\mu(\varsigma)
 |d\varsigma\int_0^1\omega_x^2dx.
\end{align*}
Using integration by parts and Young's inequality, we obtain
\begin{align*}
&-\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x(x,1,\varsigma,\cdot)
d\varsigma\omega\,dx \\
&=\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma)z(x,1,\varsigma,\cdot)
 d\varsigma\omega_x\,dx \\
&\geq-\frac{\beta}{2}\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma)|z^2(x,1,
 \varsigma,\cdot)d\varsigma\,dx
-\frac{\beta}{2}\int_{\tau_1}^{\tau_2}|\mu(\varsigma)|d\varsigma
 \int_0^1\omega_x^2dx.
\end{align*}
Consequently,
\[
 (\mathcal{A}U,U)_ {\mathcal{H}}\geq-\frac{\gamma}{2}(g'\circ\psi_x)
+\beta(\ell-\int_{\tau_1}^{\tau_2}|\mu(\varsigma)|d\varsigma)
 \int_0^1\omega_x^2dx\geq0.
\]
Thus, $\mathcal{A}$ is monotone. Next, we prove that the operator $I+\mathcal{A}$
is surjective. Given
$\mathcal{F}=(k_1, k_2, k_3, k_4, k_5, k_6, k_7, k_8)^T\in\mathcal{H}$,
we prove that there exists a unique $U\in D(\mathcal{A})$ such that
\begin{equation}
 (I +\mathcal{A}) U=\mathcal{F}.  \label{e3.6}
\end{equation}
That is,
\begin{equation}
\begin{gathered}
\varphi-u=k_1\in H_0^1,\\
\rho_1u-{k}(\varphi_x+\psi)_x+{\beta}\omega_x=\rho_1k_2 \in L^2(0,1),  \\
\psi-v=k_3\in\ H_0^1,\\
\begin{aligned}
&{\rho_2}v-{b}\psi_{xx}+{k}(\varphi_x+\psi)+\int_0^{\infty}g(s)ds\psi_{xx} \\
&-\int_0^{\infty}g(s)\eta^t_{xx}(x,t,s)ds-{\beta}\omega={\rho_2}k_4\in L^2(0,1),
\end{aligned}\\
\theta-\omega=k_5\in H_0^1, \\
{\rho_3}\omega-{\delta}\theta_{xx}-{\ell}\omega_{xx}+{\gamma}u_x+{\gamma}v-\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x (x,1,\varsigma,\cdot)d\varsigma={\rho_3}k_6\in L^2(0,1),\\
\eta^t+\eta^t_s-v=k_7\in L_g,\\
{\varsigma}z+z_\rho=\varsigma k_8\in L^2((0,1)\times(0,1)\times(\tau_1, \tau_2)).
\end{gathered} \label{e3.7}
\end{equation}
Using lines $7$ and $8$ in the above equation, we obtain
\begin{gather}
\eta^t=e^{-s}\int_0^s e^\tau(v+k_7(\tau)) d\tau, \label{e3.8}\\
z(x,\rho,\varsigma,\cdot)=e^{-\varsigma\rho}\omega_x
+\varsigma e^{-\varsigma\rho}\int_0^\rho e^{\varsigma \tau}
k_8(x,\tau,\varsigma )d\tau.\label{e3.9}
\end{gather}
Inserting $u=\varphi-{k_1}$, $v=\psi-{k_3}$, $\omega=\theta-{k_5}$,
\eqref{e3.8} and \eqref{e3.9} in \eqref{e3.7}$_2$, \eqref{e3.7}$_4$ and
\eqref{e3.7}$_6$, we obtain
\begin{equation}
\begin{gathered}
\rho_1\varphi-k(\varphi_x+\psi)_x+\beta\theta_x=h_1\in L^2(0,1),\\
\begin{aligned}
&\rho_2\psi-b\psi_{xx}+k(\varphi_x+\psi)+\int_0^{\infty}g(s)ds\psi_{xx}\\
&-\int_0^{\infty}g(s)e^{-s}\int_0^s\psi_{xx} e^\tau d\tau ds
 -\beta\theta=h_2\in L^2(0,1),
\end{aligned}  \\
\rho_3\theta-\delta\theta_{xx}-\ell\theta_{xx}+\gamma\varphi_x+\gamma\psi
 -\int_{\tau_1}^{\tau_2}\mu(\varsigma)e^{-\varsigma \rho}\theta_{xx}d\varsigma
=h_3\in L^2(0,1),
\end{gathered}\label{e3.10}
\end{equation}
where
\begin{gather*}
{ h_1=k_2\rho_1+k_1\rho_1+\beta k_{5x},}\\
{ h_2=\rho_2k_4+\rho_2k_3-\beta k_5+\int_0^{\infty}g(s)e^{-s}\int_0^s(k_7-k_3)_{xx}e^\tau d\tau ds,}\\
\begin{aligned}
 h_3&=\rho_3k_5-\ell k_{5xx}+\gamma k_{1x}+\gamma k_3
  +\int_{\tau_1}^{\tau_2}\mu(\varsigma )\varsigma e^{-\varsigma \rho}
 \int_0^\rho e^{\varsigma \tau}k_{8x}(x,\tau,\varsigma )d\tau d\varsigma\\
&\quad  +k_6\rho_3-\int_{\tau_1}^{\tau_2}\mu(\varsigma )
 e^{-\varsigma \rho}k_{5xx}d\varsigma.
\end{aligned}
\end{gather*}
To solve \eqref{e3.10}, we consider the following variational formulation
\begin{equation}
 B((\varphi, \psi, \theta), \ (\varphi_1, \psi_1, \theta_1))
 =G(\varphi_1, \psi_1, \theta_1),
\label{e3.11}
\end{equation}
where $B:[H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)]^2\to R$ is the bilinear
form defined by
\begin{align*}
& B((\varphi,\psi,\theta),(\varphi_1,\psi_1,\theta_1))\\
&=\gamma\rho_1\int_0^1\varphi\varphi_1dx
 +k\gamma\int_0^1(\varphi_x+\psi)(\varphi_{1x}+\psi_{1})dx
 +\gamma\beta\int_0^1\theta_x\varphi_1dx \\
&\quad +\gamma\rho_2 \int_0^1\psi\psi_1dx+b\gamma\int_0^1\psi_x\psi_{1x}dx
  -\gamma\int_0^{\infty}g(s)ds\int_0^1\psi_x\psi_{1x}dx \\
&\quad +\gamma\int_0^1\int_0^{\infty}g(s)e^{-s} \int_0^s\psi_x e^\tau d\tau ds
 \psi_{1x}dx-\beta \gamma\int_0^1\theta \psi_1\,dx     \\
&\quad +\rho_3\beta \int_0^1\theta\theta_1dx+\beta\delta\int_0^1\theta_x\theta_{1x}dx
 +\ell\beta\int_0^1\theta_x\theta_{1x}dx +\gamma\beta\int_0^1\varphi_x\theta_1dx   \\
&\quad +\gamma\beta  \int_0^1\psi\theta_1dx
 +\beta \int_0^1\int_{\tau_1}^{\tau_2}e^{-\varsigma\rho}\mu(\varsigma )
 \theta_xd\varsigma \theta_{1x}dx,
\end{align*}
and $G:[H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)]^2\to R$
is the linear functional
$$
 G[(\varphi_1,\psi_1,\theta_1)]=\gamma\int_0^1h_1\varphi_1dx
 +\gamma\int_0^1h_2\psi_1dx+\beta\int_0^1h_3\theta_1dx.
$$
Now, for $V=H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)$ equipped with the norm
$$
\|\varphi, \psi, \theta\|_V^2=\|\varphi_x+\psi\|_2^2
+\|\varphi\|_2^2+\|\psi_x\|_2^2+\|\theta\|_2^2+\|\theta_x\|_2^2,
$$
using integration by parts we have
\begin{align*}
&B((\varphi,\psi,\theta),(\varphi,\psi,\theta))\\
&=\gamma\rho_1\int_0^1\varphi ^2dx+k\gamma\int_0^1(\varphi_x+\psi)^2dx
    +\gamma\rho_2 \int_0^1\psi^2dx \\
&\quad +(b-\int_0^{\infty}g(s)ds)\gamma\int_0^1\psi_x^2dx
    +\gamma\int_0^1\psi_{x}^2dx\int_0^{\infty}g(s)\int_0^s e^\tau d\tau e^{-s } ds\\
&\quad +\rho_3\beta\int_0^1\theta^2dx+\beta\delta\int_0^1\theta_x^2dx
 +\ell\beta\int_0^1\theta_x^2dx
 +\beta \int_0^1\theta_x^2dx\int_{\tau_1}^{\tau_2}e^{-\varsigma \rho}
 \mu(\varsigma )d\varsigma ,   \\
& \geq\alpha_0\|\varphi, \psi, \theta\|_V^2,
\end{align*}
for some $\alpha_0>0$. Thus, $B$ is coercive.

On the other hand, using H\"older and Poincar\'e inequalities,
we obtain
$$
{|B((\varphi,\psi,\theta),(\varphi_1,\psi_1,\theta_1))|
 \leq c\|\varphi, \psi, \theta\|_V\|\varphi_1, \psi_1, \theta_1\|_V.}
$$
Similarly
$$
{|G(\varphi_1,\psi_1,\theta_1)|\leq c\|\varphi_1, \psi_1, \theta_1\|_V.}
$$
Consequently, by the Lax-Milgram Lemma, system \eqref{e3.10} has a unique solution
$$
(\varphi, \psi, \theta)\in H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1)
$$
satisfying
$$
B((\varphi, \psi, \theta),(\varphi_1, \psi_1, \theta_1))
=G(\varphi_1, \psi_1, \theta_1), \quad (\varphi_1, \psi_1, \theta_1)\in V.
$$
The substitution of $\varphi$, $\psi$ and $\theta$ into \eqref{e3.7}$_1$,
\eqref{e3.7}$_3$ and \eqref{e3.7}$_5$ yields
$$
(u, v, \omega)\in H_0^1(0,1)\times H_0^1(0,1)\times H_0^1(0,1).
$$
Similarly, inserting $v$ in \eqref{e3.8} and bearing in mind \eqref{e3.7}$_7$,
we obtain
$\eta^t\in L_g$.
At the same time, inserting $\omega$ in \eqref{e3.9} and using \eqref{e3.7}$_8$,
we obtain
$$
z, z_\rho\in L^2((0,1)\times(0,1)\times(\tau_1, \tau_2)).
$$
Moreover, if we take $(\varphi_1, \theta_1)\equiv(0,0)\in H_0^1(0,1)
\times H_0^1(0,1)$ in \eqref{e3.11}, we obtain
\begin{align*}
&k\int_0^1(\varphi_x+\psi)\psi_1dx+\rho_2\int_0^1\psi\psi_1dx
+b\int_0^1\psi_x\psi_{1x}dx \\
&-\int_0^{\infty}g(s)ds\int_0^1\psi_x\psi_{1x}dx
 -\beta\int_0^1\theta\psi_1dx+\int_0^{\infty}g(s)(1-e^{-s})ds\int_0^1\psi_x\psi_{1x}dx \\
&=\int_0^1h_2\psi_1dx.
\end{align*}
Hence we obtain
\begin{align*}
& b\int_0^1\psi_x\psi_{1x}dx-\int_0^{\infty}g(s)ds\int_0^1\psi_x\psi_{1x}dx
 +\int_0^{\infty}g(s)(1-e^{-s})ds\int_0^1\psi_x\psi_{1x}dx\\
&=\int_0^1[- k(\varphi_x+\psi)-\rho_2\psi+\beta\theta+h_2]\psi_1dx,\quad
  \psi_1\in H_0^1(0,1).
\end{align*}
By noting that
$-k(\varphi_x+\psi)-\rho_2\psi+\beta\theta+h_2\in L^2(0,1)$,
we obtain
$\psi\in H^2(0,1)\cap H_0^1(0,1)$.
Consequently using integration by parts we have
\begin{align*}
&\int_0^1[-b\psi_{xx}+\int_0^{\infty}g(s)ds\psi_{xx}dx
-\int_0^{\infty}g(s)(1-e^{-s})ds\psi_{xx} \\
& +k(\varphi_x+\psi)+\rho_2\psi-\beta\theta-h_2]\psi_{1}dx =0,
\quad \psi_1\in H_0^1(0,1).
\end{align*}
Therefore,
\[
-b\psi_{xx}+\int_0^{\infty}g(s)ds\psi_{xx}dx
-\int_0^{\infty}g(s)(1-e^{-s})ds\psi_{xx}+k(\varphi_x+\psi)
 +\rho_2\psi-\beta\theta=h_2 .
\]
This gives \eqref{e3.10}$_2$.
Similarly, if we take $(\varphi_1,\psi_1)\equiv(0,0)\in H_0^1(0,1)\times H_0^1(0,1)$
 in \eqref{e3.11}, we can show that
$$
\theta\in H^2(0,1)\cap H_0^1(0,1),
$$
and \eqref{e3.10}$_3$ are satisfied.

If we take $(\psi_1,\theta_1)\equiv(0,0)\in H_0^1(0,1)\times H_0^1(0,1)$
in \eqref{e3.11}, we can show that
$$
\varphi\in H^2(0,1)\cap H_0^1(0,1),
$$
and \eqref{e3.10}$_1$ are satisfied.

Finally, from \eqref{e3.8} we can get $\eta^t\in L_g$. From \eqref{e3.9},
we know $z, z_\rho\in L^2((0,1)\times(0,1)\times(\tau_1, \tau_2))$.
Hence, there exists a unique $ U\in D(\mathcal{A})$ such that \eqref{e3.6}
is satisfied. Therefore, $\mathcal{A}$ is a maximal monotone operator.

Now, we prove that the operator $F$ defined in \eqref{e3.3} is locally
 Lipschitz in $\mathcal{H}$.
Let $ U=(\varphi, u, \psi, v, \theta, \omega, \eta^t, z)^T$ and
$ U_1=(\varphi_1, u_1, \psi_1, v_1, \theta_1, \omega_1, \eta^t_1, z_1)^T$,
then we have
$$
\|F( U)-F( U_1)\|_\mathcal{H}\leq\|f(\psi)-f(\psi_1)\|_{L^2}.
$$
By using \eqref{e2.5}, H\"older and Poincar\'e inequalities, we can get
$$
\|f(\psi^2)-f(\psi^1)\|_{L^2}\leq c(\|\psi^1\|_{2\varrho}^\varrho
+\|\psi^2\|_{2\varrho}^\varrho)\|\psi^1-\psi^2\|\leq c\|\psi_x^1-\psi_x^2\|,
$$
which gives us
$$
\|F( U)-F( U_1)\|_\mathcal{H}\leq c\| U- U_1\|_\mathcal{H}.
$$
Then the operator $F$ is locally Lipschitz in $\mathcal{H}$.
The proof complete.
\end{proof}

\section{Proof of stability results}

In this section, we state and prove our stability results for the
 energy of system \eqref{e2.1} by using the multiplier technique.
To achieve our goal, we need the following lemmas.

\begin{lemma} \label{lem4.1}
Let $(\varphi,\ \psi,\ \theta)$ be the solution of \eqref{e2.1}, then we have
\begin{equation}
E'(t)\leq\frac{\gamma}{2}(g'\circ\psi_x)(t)
-\beta\Big(\ell-\int_{\tau_1}^{\tau_2}|\mu(\varsigma )|d\varsigma \Big)
\int_0^1\theta_{tx}^2dx.
\label{e4.1}
\end{equation}
\end{lemma}

\begin{proof}
Multiplying \eqref{e2.1}$_1$ by $\gamma\varphi_t$, \eqref{e2.1}$_2$ by
$\gamma\psi_t$, \eqref{e2.1}$_3$ by $\beta\theta_t$, integrating over $(0,1)$
with respect to $x$, multiplying equation \eqref{e2.1}$_4$ by
$\beta|\mu(\varsigma)|z$ and integrating over
$(0,1)\times(0,1)\times(\tau_1,\tau_2)$ with respect to $\rho, x$ and $\varsigma$,
summing them up, we obtain
\begin{equation}
\begin{aligned}
E'(t)
&=\frac{\gamma}{2}(g'\circ\psi_x)-\ell\beta\int_0^1\theta_{tx}^2dx
 +\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x(x,1,\varsigma ,t)d\varsigma
 \theta_t(x,t)dx \\
&-\frac{\beta}{2}\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma )
 |z^2(x,1,\varsigma ,t)d\varsigma\,dx +\frac{\beta}{2}\int_{\tau_1}^{\tau_2}
 |\mu(\varsigma )|d\varsigma \int_0^1\theta_{tx}^2(x,t)dx.
\end{aligned} \label{e4.2}
\end{equation}
At the same time, using integration by parts and Young's inequality, we have
\begin{equation}
\begin{aligned}
&\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma)z_x(x,1,\varsigma ,t)d\varsigma
 \theta_t(x,t)dx \\
&=-\beta\int_0^1\int_{\tau_1}^{\tau_2}\mu(\varsigma )z(x,1,\varsigma ,t)d\varsigma
  \theta_{tx}(x,t)dx \\
&\leq\frac{\beta}{2}\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma )
 |z^2(x,1,\varsigma ,t)d\varsigma\,dx
 +\frac{\beta}{2}\int_{\tau_1}^{\tau_2}|\mu(\varsigma )|d\varsigma
\int_0^1\theta_{tx}^2(x,t)dx.
\end{aligned}\label{e4.3}
\end{equation}
A combination of \eqref{e4.2} and \eqref{e4.3} gives
$$
E'(t)\leq\frac{\gamma}{2}(g'\circ\psi_x)
 -\beta\Big(\ell-\int_{\tau_1}^{\tau_2}|\mu(\varsigma )|d\varsigma \Big)
 \int_0^1\theta_{tx}dx.
$$
Thus \eqref{e4.1} follows.
\end{proof}

We note here that if
$E(t)=E(t, \varphi, \psi, \theta, z)=E_1(t)$,
denotes the energy defined in \eqref{e2.7} then
$$
E_2(t)=E(t, \varphi_t, \psi_t, \theta_t, z_t),
$$
denotes the second order energy and one can easily obtain that
$$
E_2'(t)\leq\frac{\gamma}{2}(g'\circ\psi_{xt})-c\int_0^1\theta_{ttx}^2dx,
$$
in which $c$ is some positive constant.

It is easy to obtain the following inequalities, we omit their proofs.

\begin{lemma} \label{lem4.2}
 The following inequalities hold,
\begin{gather}
\int_0^1\Big(\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)^2dx
 \leq 2g_0(g\circ\psi_x)(t)+2g_0\int_0^1 \psi_x^2dx,  \label{e4.4} \\
\int_0^1\Big(\int_0^{\infty}g(s)(\psi_x(t)-\psi_x(t-s))ds\Big)^2dx
 \leq g_0(g\circ\psi_x)(t),  \label{e4.5} \\
\int_0^1\Big(\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))ds\Big)^2dx
 \leq d_1(g\circ\psi_x)(t),  \label{e4.6} \\
\int_0^1\Big(\int_0^{\infty}g'(s)(\psi(t)-\psi(t-s))ds\Big)^2dx
 \leq -d_2(g\circ\psi_x)(t),  \label{e4.7} \\
\int_0^1\Big(\int_0^{\infty}g'(s)(\psi_x(t)-\psi_x(t-s))ds\Big)^2dx
 \leq -g(0)(g'\circ\psi_x)(t), \label{e4.8}
\end{gather}
in which $d_1$ and $d_2$ are positive constants.
\end{lemma}

\begin{lemma} \label{lem4.3}
 Let $(\varphi, \psi,\theta)$ be the solution of \eqref{e2.1}.
Then for any positive constant $\varepsilon_1$, the functional
$$
J_1(t):=-\rho_1\int_0^1\varphi_t\varphi\,dx-\rho_2\int_0^1\psi_t\psi\,dx
$$
satisfies
\begin{equation}
\begin{aligned}
J_1'(t)
&\leq -\rho_1\int_0^1\varphi_t^2dx+(k+\beta\varepsilon_1)\int_0^1(\varphi_x+\psi)^2dx
+\frac{\beta}{\varepsilon_1}\int_0^1\theta_t^2dx \\
&\quad -\rho_2\int_0^1\psi_t^2dx
 +c(1+\varepsilon_1)\int_0^1\psi_x^2dx+c(g\circ\psi_x)(t).
\end{aligned} \label{e4.9}
\end{equation}
\end{lemma}

\begin{proof}
By computations, using \eqref{e2.1}, we obtain
\begin{align*}
J_1'(t)
&=-\rho_1\int_0^1\varphi_t^2dx+k\int_0^1(\varphi_x+\psi)^2dx
 -\beta\int_0^1\theta_t\varphi_x\,dx-\rho_2\int_0^1\psi_t^2dx \\
&\quad +b\int_0^1\psi_x^2dx-\beta\int_0^1\theta_t\psi\,dx
 -\int_0^1\psi_x\int_0^{\infty}g(s)\psi_x(x,t-s)\,ds\,dx \\
&\quad +\int_0^1f(\psi)\psi\,dx.
\end{align*}
By using Young's inequality and Poincar\'e inequality, we obtain for
$\varepsilon_1>0$,
\begin{equation}
\begin{aligned}
J_1'(t)
&\leq -\rho_1\int_0^1\varphi_t^2dx+k\int_0^1(\varphi_x+\psi)^2dx
 +\frac{\beta\varepsilon_1}{2}\int_0^1\varphi_x^2dx
 -\rho_2\int_0^1\psi_t^2dx \\
&\quad +\frac{\beta}{\varepsilon_1}\int_0^1\theta_t^2dx
  +(2b+\frac{\beta\varepsilon_1}{2})\int_0^1\psi_x^2dx \\
&\quad +\frac{1}{4b}\int_0^1\Big(\int_0^{\infty}g(s)\psi_x(x,t-s)ds\Big)^2dx
 +\int_0^1f(\psi)\psi\,dx.
\end{aligned} \label{e4.10}
\end{equation}
By using the Cauchy-Schwarz inequality and Poincar\'e inequality, we have
\begin{equation}
\int_0^1\varphi_x^2dx\leq2\int_0^1(\varphi_x+\psi)^2dx
+2\int_0^1\psi^2dx\leq2\int_0^1(\varphi_x+\psi)^2dx
 +2\int_0^1\psi_x^2dx,
\label{e4.11}
\end{equation}
and
\begin{equation}
\int_0^1|f(\psi)\psi|dx
\leq \int_0^1|\psi|^{\varrho}|\psi\|\psi|
\leq\|\psi\|_{2({\varrho}+1)}^{\varrho}\|\psi\|_{2({\varrho}+1)}\|\psi\|dx
\leq c\int_0^1\psi_x^2dx.
\label{e4.12}
\end{equation}
The substitution of \eqref{e4.4}, \eqref{e4.11} and \eqref{e4.12} into
 \eqref{e4.10} gives \eqref{e4.9}.
\end{proof}

\begin{lemma} \label{lem4.4}
 Let $(\varphi, \psi, \theta)$ be the solution of \eqref{e2.1}.
Then for any positive constant $\varepsilon_2$, the functional
$$
J_2(t):=\rho_3\int_0^1\theta_t\theta\,dx+{\ell\over 2}\int_0^1\theta_x^2dx
+\gamma\int_0^1\varphi_x\theta\,dx
$$
satisfies
\begin{equation}
\begin{aligned}
J'_2(t)
&\leq -{\delta\over 2} \int_0^1\theta_x^2dx
 + \Big(\rho_3+{\gamma^2\over{2\varepsilon_2}}\Big) \int_0^1\theta_t^2dx
 +\varepsilon_2\int_0^1\psi_x^2dx   \\
&\quad + {\gamma^2\over \delta }\int_0^1\psi_t^2dx
 +\varepsilon_2\int_0^1(\varphi_x+\psi)^2dx \\
&\quad +\frac{c}{\delta}\int_0^1\int_{\tau_1}^{\tau_2}z^2(x,1,\varsigma,t)
 d\varsigma\,dx.
\end{aligned}\label{e4.13}
\end{equation}
\end{lemma}

\begin{proof}
By differentiating $J_2(t)$ and using \eqref{e2.1}, we obtain
\begin{align*}
J_2'(t)&= \rho_3\int_0^1\theta_t^2dx
 +\gamma\int_0^1\varphi_x\theta_tdx
 -\delta\int_0^1\theta_x^2dx-\gamma\int_0^1\psi_t\theta\,dx   \\
&\quad -\int_0^1 \int_{\tau_1}^{\tau_2}\mu(\varsigma )z(x,1,\varsigma ,t)
 d\varsigma \theta_x\,dx .
\end{align*}
By using Young's and Poincar\'e inequalities, we obtain for any
$\varepsilon_2>0$
\begin{gather}
 \gamma\int_0^1\varphi_x\theta_tdx\leq \varepsilon_2
 \int_0^1(\varphi_x+\psi)^2dx+\varepsilon_2\int_0^1\psi_x^2dx
 +{\gamma^2\over{2\varepsilon_2}}\int_0^1\theta_t^2dx,  \label{e4.14}\\
 \gamma\int_0^1\psi_t\theta\,dx\leq {\delta\over 4}
 \int_0^1\theta_x^2dx+{\gamma^2\over{\delta}}\int_0^1\psi_t^2dx. \label{e4.15}
 \end{gather}
By using (H1), we have
\begin{equation}
\begin{aligned}
 &\int_0^1 \int_{\tau_1}^{\tau_2}\mu(\varsigma )z(x,1,\varsigma ,t)d\varsigma
\theta_x\,dx \\
&\leq\frac{\delta}{4} \int_0^1\theta_x^2dx+\frac{1}{\delta}
 \int_0^1\Big( \int_{\tau_1}^{\tau_2}\mu(\varsigma )z(x,1,\varsigma ,t)d\varsigma
  \Big)^2dx \\
&\leq\frac{\delta}{4} \int_0^1\theta_x^2dx
 +\frac{c}{\delta}\int_0^1\int_{\tau_1}^{\tau_2}z^{2}(x,1,\varsigma ,t)d\varsigma\,dx.
\end{aligned} \label{e4.16}
\end{equation}
Thus, \eqref{e4.13} is established.
\end{proof}

\begin{lemma} \label{lem4.5}
Let $(\varphi,\ \psi,\ \theta)$ be the solution of \eqref{e2.1}.
Then for any positive constant $\varepsilon_3$ the functional
$$
{J_3(t):=-\rho_2\int_0^1\psi_t\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx}
$$
satisfies
\begin{equation}
\begin{aligned}
J'_3(t)
&\leq \varepsilon_3c\int_0^1\psi_x^2dx-(\rho_2g_0-\varepsilon_3\rho_2)
 \int_0^1\psi_t^2dx+c(\varepsilon_3+\frac{1}{\varepsilon_3})(g\circ\psi_x)(t)\\
&\quad +\varepsilon_3k\int_0^1(\varphi_x+\psi)^2dx+\varepsilon_3\beta^2
 \int_0^1\theta_t^2dx-\frac{c}{4\varepsilon_3}(g'\circ\psi_x)(t).
\end{aligned} \label{e4.17}
\end{equation}
\end{lemma}

\begin{proof}
First, we note that
\begin{align*}
&\frac{\partial}{\partial t}\Big(\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))ds\Big)\\
&=\frac{\partial}{\partial t}\Big(\int_{-\infty}^tg(t-s)(\psi(t)-\psi(s))ds\Big)\\
&=\int_{-\infty}^tg'(t-s)(\psi(t)-\psi(s))ds+\int_{-\infty}^tg(t-s)\psi_t(t)ds \\
&=g_0\psi_t(t)+\int_0^{\infty}g'(s)(\psi(t)-\psi(t-s))ds.
\end{align*}
Then, by differentiating $J_3(t)$ and using \eqref{e2.1}, we find
\begin{equation}
\begin{aligned}
J'_3(t)
&=b\int_0^1\psi_x\int_0^{\infty}g(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx
 -g_0\rho_2\int_0^1\psi_t^2dx \\
&\quad -\rho_2\int_0^1\psi_{t}\int_0^{\infty}g'(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\quad +k\int_0^1(\varphi_x+\psi)\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\quad -\beta\int_0^1\theta_t\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\quad +\int_0^1f(\psi)\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\quad -\int_0^1\int_0^{\infty}g(s)\psi_x(t-s)ds\int_0^{\infty}
 g(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx.
\end{aligned} \label{e4.18}
\end{equation}
By using Young's and Poincar\'e inequalities,
\begin{gather}
\begin{aligned}
&b\int_0^1\psi_x\int_0^{\infty}g(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx \\
&\leq\varepsilon_3b\int_0^1\psi_x^2dx+\frac{b}{4\varepsilon_3}
 \int_0^1\Big(\int_0^{\infty}g(s)(\psi_x(t)-\psi_x(t-s))ds\Big)^2dx \\
&\leq\varepsilon_3b\int_0^1\psi_x^2dx+\frac{bg_0}{4\varepsilon_3}(g\circ\psi_x)(t),
\end{aligned} \label{e4.19}\\
\begin{aligned}
&-\rho_2\int_0^1\psi_{t}\int_0^{\infty}g'(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\leq\varepsilon_3\int_0^1\rho_2\psi_t^2dx+\frac{\rho_2}{4\varepsilon_3}
 \int_0^1\Big(\int_0^{\infty}g'(s)(\psi(t)-\psi(t-s))ds\Big)^2dx \\
&\leq\varepsilon_3\int_0^1\rho_2\psi_t^2dx
 -\frac{\rho_2d_2}{4\varepsilon_3}(g'\circ\psi_x)(t),
\end{aligned} \label{e4.20} \\
\begin{aligned}
&k\int_0^1(\varphi_x+\psi)\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\leq\varepsilon_3k\int_0^1(\varphi_x+\psi)^2dx
 +\frac{k}{4\varepsilon_3}\int_0^1
 \Big(\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))ds\Big)^2dx \\
&\leq\varepsilon_3k\int_0^1(\varphi_x+\psi)^2dx+\frac{kd_1}{4\varepsilon_3}
 (g\circ\psi_x)(t),
\end{aligned} \label{e4.21} \\
\begin{aligned}
& \beta\int_0^1\theta_t\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\leq\varepsilon_3\beta^2\int_0^1\theta_t^2dx+\frac{1}{4\varepsilon_3}
 \int_0^1\Big(\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))ds\Big)^2dx \\
&\leq\varepsilon_3\beta^2\int_0^1\theta_t^2dx
 +\frac{d_1}{4\varepsilon_3}(g\circ\psi_x)(t),
\end{aligned} \label{e4.22} \\
\begin{aligned}
&\int_0^1f(\psi)\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\leq k_0\int_0^1|\psi|^\varrho|\psi|\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))\,ds\,dx \\
&\leq k_0 \|\psi\|_{2(\varrho+1)}^\varrho\|\psi\|_{2(\varrho+1)}
 \Big(\int_0^1(\int_0^{\infty}g(s)(\psi(t)-\psi(t-s))ds)^2dx\Big)^{1/2} \\
&\leq\varepsilon_3c\int_0^1\psi_x^2dx+\frac{d_1}{4\varepsilon_3}(g\circ\psi_x)(t),
\end{aligned} \label{e4.23} \\
\begin{aligned}
&\int_0^1\int_0^{\infty}g(s)\psi_x(t-s)ds\int_0^{\infty}
 g(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx \\
&\leq\varepsilon_3\int_0^1\Big(\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)^2dx\\
&\quad +\frac{1}{4\varepsilon_3}\int_0^1
 \Big(\int_0^{\infty}g(s)(\psi_x(t)-\psi_x(t-s))ds\Big)^2dx \\
& \leq\Big(2\varepsilon_3+\frac{1}{4\varepsilon_3}\Big)
 g_0(g\circ\psi_x)+2\varepsilon_2g_0\int_0^1\psi_x^2dx.
\end{aligned} \label{e4.24}
\end{gather}
By substituting \eqref{e4.19}-\eqref{e4.24} into \eqref{e4.18},
 we obtain \eqref{e4.17}.
\end{proof}

As in \cite{17}, we introduce the multiplier $\omega$ which is the solution of
\begin{equation}
-w_{xx}=\psi_x,\quad  w(0)=w(1)=0.
\label{e4.25}
\end{equation}

\begin{lemma} \label{lem4.6}
The solution of \eqref{e4.25} satisfies
\begin{gather*}
\int_0^1w_x^2dx\leq\int_0^1\psi^2dx\leq\int_0^1\psi_x^2dx, \\
\int_0^1w_t^2dx\leq\int_0^1w_{xt}^2dx\leq\int_0^1\psi_t^2dx.
\end{gather*}
\end{lemma}

\begin{lemma} \label{lem4.7}
Let $(\varphi,\psi, \theta)$ be the solution of \eqref{e2.1}.
Then for any positive constant $\varepsilon_4$ the functional
$$
J_4(t):= \int_0^1(\rho_1\varphi_tw+\rho_2\psi_t\psi)dx,
$$
satisfies
\begin{equation}
\begin{aligned}
J_4'(t)
&\leq -\frac{l}{2}\int_0^1\psi_x^2(t)dx+\frac{3\beta^2}{l}\int_0^1\theta_t^2(t)dx
+\varepsilon_4\int_0^1\varphi_t^2dx \\
&\quad +(\rho_2+\frac{\rho_1}{4\varepsilon_4})\int_0^1\psi_t^2dx
 +\frac{3(b-l)}{2l}(g\circ\psi_x)(t)-\int_0^1\widehat{f}(\psi)dx.
\end{aligned}\label{e4.26}
\end{equation}
\end{lemma}

\begin{proof}
A simple differentiation of $J_4(t)$ and together with \eqref{e2.1} give
\begin{equation}
\begin{aligned}
J_4'(t)&= \beta\int_0^1\theta_tw_x\,dx+k\int_0^1w_x^2dx
 +\rho_1\int_0^1\varphi_tw_tdx+\beta\int_0^1\theta_t\psi\,dx   \\
&\quad -b\int_0^1\psi_x^2dx-k\int_0^1\psi^2dx+\rho_2\int_0^1\psi_t^2dx   \\
&\quad +\int_0^1\int_0^{\infty}g(s)\psi_x(x,t-s)ds\psi_x(t)dx
 -\int_0^1f(\psi)\psi\,dx,
\end{aligned}\label{e4.27}
\end{equation}
where we have used integration by parts, \eqref{e4.25} and the boundary
 conditions in \eqref{e2.1}. By using Young's, Poincar\'e inequalities,
Lemma \ref{lem4.2} and Lemma \ref{lem4.6}, we have
\begin{gather*}
 \beta\int_0^1\theta_t\omega_x\,dx
 \leq\frac{l}{6}\int_0^1\omega^2_x\,dx+\frac{3\beta^2}{2l}\int_0^1\theta^2_tdx
 \leq\frac{l}{6}\int_0^1\psi^2_x\,dx+\frac{3\beta^2}{2l}\int_0^1\theta^2_tdx, \\
\rho_1\int_0^1\varphi_t\omega_tdx\leq\varepsilon_4\int_0^1\varphi^2_tdx
 +\frac{\rho_1^2}{4\varepsilon_4}\int_0^1\omega^2_tdx
 \leq\varepsilon_4\int_0^1\varphi^2_tdx
  +\frac{\rho_1^2}{4\varepsilon_4}\int_0^1\psi^2_tdx, \\
 \beta\int_0^1\theta_t\psi\,dx\leq\frac{l}{6}\int_0^1\psi^2_x\,dx
 +\frac{3\beta^2}{2l}\int_0^1\theta^2_tdx,
\end{gather*}
\begin{align*}
&\int_0^1\int_0^{\infty}g(s)\psi_x(x,t-s)ds\psi_x(t)dx \\
&=\int_0^1\int_0^{\infty}g(s)(\psi_x(x,t-s)-\psi_x(t)+\psi_x(t))ds\psi_x(t)dx \\
&=\int_0^1\int_0^{\infty}g(s)(\psi_x(x,t-s)-\psi_x(t))ds\psi_x(t)dx
 +\int_0^{\infty}g(s)ds\int_0^1\psi_x^2(t)dx \\
&\leq\frac{l}{6}\int_0^1\psi_x^2(t)dx
 +\frac{3}{2l}\int_0^1\Big(\int_0^{\infty}g(s)(\psi_x(x,t-s)-\psi_x(t))ds\Big)^2dx \\
&\quad +\int_0^{\infty}g(s)ds\int_0^1\psi_x^2(t)dx \\
&\leq\frac{l}{6}\int_0^1\psi_x^2(t)dx+\frac{3(b-l)}{2l}(g\circ\psi_x)(t)
 +\int_0^{\infty}g(s)ds\int_0^1\psi_x^2(t)dx.
\end{align*}
Thus, \eqref{e4.26} is established.
\end{proof}

\begin{lemma} \label{lem4.8}
Let $(\varphi, \psi, \theta)$ be the solution of \eqref{e2.1}.
Then for any positive constant $\varepsilon_5$ the functional
\[
J_5(t):=\rho_2\int_0^1\psi_t(\varphi_x+\psi)dx+\frac{b\rho_1}{k}
\int_0^1\varphi_t\psi_x\,dx-\frac{\rho_1}{k}
\int_0^1\varphi_t\int_0^{\infty}g(s)\psi_x(t-s)\,ds\,dx
\]
satisfies
\begin{equation}
\begin{aligned}
J_5'(t)
&\leq  \Big[\varphi_x(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds)\Big]_{x=0}^{x=1}
 +\rho_2\int_0^1\psi_t^2dx \\
&\quad-\frac{k}{2}\int_0^1(\varphi_x+\psi)^2dx
  +c(1+\frac{1}{\varepsilon_5})\int_0^1\theta_{tx}^2dx \\
&\quad +c\big(1+\frac{1}{\varepsilon_5}\big)\int_0^1\psi_{x}^2dx
 +c\varepsilon_5\big(g\circ\psi_x\big)(t)
 -\frac{c}{\varepsilon_5}(g'\circ\psi_x)(t) \\
&\quad +\varepsilon_5\int_0^1\varphi_{t}^2dx
 -\int_0^1\widehat{f}(\psi)dx
  +\big(\frac{b\rho_1}{k}-\rho_2\big)\int_0^1\varphi_t\psi_{xt}dx.
\end{aligned} \label{e4.28}
\end{equation}
\end{lemma}

\begin{proof}
First, we note that
\begin{align*}
&\frac{d}{d t}\Big(\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)\\
&=\frac{d}{d t}\Big(\int_{-\infty}^tg(t-s)\psi_x(s)ds\Big) \\
&=g(0)\psi_x(t)+\int_{-\infty}^tg'(t-s)\psi_x(s)ds \\
&=g(0)\psi_x(t)+\int_0^{\infty}g'(s)\big(\psi_x(t-s)-\psi_x(t)+\psi_x(t)\big)ds \\
&=g(0)\psi_x(t)+\int_0^{\infty}g'(s)\big(\psi_x(t-s)-\psi_x(t)\big)ds
 +\int_0^{\infty}g'(s)ds\psi_x(t) \\
&=\int_0^{\infty}g'(s)\left(\psi_x(t-s)-\psi_x(t)\right)ds.
\end{align*}
By using equation \eqref{e2.1} and integration by parts, we obtain
\begin{align*}
J'_5(t)
&=\Big[\varphi_x(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds)\Big]_{x=0}^{x=1}
 +\rho_2\int_0^1\psi_t^2dx-k\int_0^1(\varphi_x+\psi)^2dx \\
&\quad +\beta\int_0^1\theta_t(\varphi_x+\psi)dx
 +\frac{\beta}{k}\int_0^1\theta_{tx}\int_0^{\infty}g(s)
 \left(\psi_x(t-s)-\psi_x(t)\right)\,ds\,dx \\
&\quad +\frac{\rho_1}{k}\int_0^1\varphi_{t}\int_0^{\infty}g'(s)\left(\psi_x(t)
 -\psi_x(t-s)\right)\,ds\,dx
 +\big(\frac{b\rho_1}{k}-\rho_2\big)\int_0^1\varphi_{t}\psi_{xt}dx \\
&\quad +\frac{\beta}{k}\int_0^{\infty}g(s)ds\int_0^1\theta_{xt}\psi_x\,dx
 -\frac{b\beta}{k}\int_0^1\theta_{xt}\psi_x\,dx\\
&\quad -\int_0^1f(\psi)\varphi_x\,dx 
 -\int_0^1\psi f(\psi)\,dx.
\end{align*}
By using Young's, Poincar\'e inequalities, Lemma 4.2, we know that for any
$\varepsilon_5>0$,
\begin{gather}
\beta\int_0^1\theta_t(\varphi_x+\psi)dx
 \leq\frac{k}{4}\int_0^1(\varphi_x+\psi)^2dx
 +\frac{\beta^2}{k}\int_0^1\theta_{tx}^2dx,\label{e4.29}\\
\begin{aligned}
&\frac{\beta}{k}\int_0^1\theta_{tx}\int_0^{\infty}g(s)(\psi_x(t-s)-\psi_x(t)ds)^2dx \\
&\leq c\varepsilon_5(g\circ\psi_x)(t)+\frac{1}{4\varepsilon_5}\int_0^1\theta_{tx}^2dx,
\end{aligned} \label{e4.30} \\
\begin{aligned}
&\frac{\rho_1}{k}\int_0^1\varphi_{t}\int_0^{\infty}g'(s)
 \left(\psi_x(t)-\psi_x(t-s)\right)\,ds\,dx \\
&\leq-\frac{c}{\varepsilon_5}(g'\circ\psi_x)(t)
 +\varepsilon_5\int_0^1\varphi_{t}^2dx,
\end{aligned} \label{e4.31} \\
\frac{\beta}{k}\int_0^{\infty}g(s)ds\int_0^1\theta_{xt}\psi_x\,dx
 -\frac{b\beta}{k}\int_0^1\theta_{xt}\psi_x\,dx
\leq\varepsilon_5\int_0^1\theta^2_{xt}dx
 +\frac{c}{\varepsilon_5}\int_0^1\psi^2_x\,dx, \label{e4.32} \\
\begin{aligned}
\int_0^1|\varphi_{x}f(\psi)|dx
&\leq \|\varphi_x\|\|\psi\|_{2(\varrho+1)}^\varrho\|\psi\|_{2(\varrho+1)}\\
&\leq \frac{k}{8}\int_0^1\varphi_{x}^2dx+\frac{2}{k}\int_0^1\psi_{x}^2dx \\
&\leq \frac{k}{4}\int_0^1(\varphi_{x}+\psi)^2dx+(\frac{2}{k}+\frac{k}{4})
 \int_0^1\psi_{x}^2dx.
\end{aligned} \label{e4.33}
\end{gather}
Substituting \eqref{e4.29}-\eqref{e4.33} into $J_5'(t)$, gives \eqref{e4.28}.
\end{proof}

Considering the boundary terms that appears in \eqref{e4.28},
we define the function
$$
q(x)=2-4x, \quad x\in [0,1].
$$

\begin{lemma} \label{lem4.9}
Let $(\varphi, \psi, \theta)$ be the solution of \eqref{e2.1}.
Then we have that for a positive constant $\varepsilon_6$,
\begin{equation}
\begin{aligned}
&\Big[\varphi_x\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds \Big)
\Big]_{x=0}^{x=1} \\
& \leq-\frac{\varepsilon_6}{k}\frac{d}{dt}\int_0^1\rho_1q(x)\varphi_t\varphi_x\,dx
 +c(\varepsilon_6+\frac{1}{\varepsilon_6}
 +\frac{1}{\varepsilon_6^3})\int_0^1\psi_x^2dx
 +c\varepsilon_6\int_0^1(\varphi_x+\psi)^2dx \\
&\quad -\frac{\rho_2}{4\varepsilon_6}\frac{d}{dt}\int_0^1q(x)
 \psi_t\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx
 +c\big(\frac{1}{\varepsilon_6} +\frac{1}{\varepsilon_6^3}\big)(g\circ\psi_x)(t)\\
&\quad  +\frac{c}{\varepsilon_6}\int_0^1\psi_t^2dx
 -\frac{c}{\varepsilon_6}(g'\circ\psi_x)(t)+c(\varepsilon_6
 +\frac{1}{\varepsilon_6})\int_0^1\theta_{tx}^2dx
 +\frac{2\rho_1\varepsilon_6}{k}\int_0^1\varphi_t^2dx.
\end{aligned} \label{e4.34}
\end{equation}
\end{lemma}

\begin{proof}
By using Young's and Poincar\'e inequalities, we obtain for any
$\varepsilon_6>0$,
\begin{align*}
&\Big[\varphi_x\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds \Big)
\Big]_{x=0}^{x=1} \\
&=\varphi_x(1)\Big(b\psi_x(1)-\int_0^{\infty}g(s)\psi_x(1,t-s)ds\\
&\quad -\varphi_x(0)\Big(b\psi_x(0)-\int_0^{\infty}g(s)\psi_x(0,t-s)ds \Big)
\Big) \\
&\leq\frac{1}{4\varepsilon_6}
 \Big[(b\psi_x(1)-\int_0^{\infty}g(s)\psi_x(1,t-s)ds)^2 \\
&\quad +\Big(b\psi_x(0)-\int_0^{\infty}g(s)\psi_x(0,t-s)ds\Big)^2\Big] 
 +\varepsilon_6[\varphi_x(1)^2+\varphi_x(0)^2].
\end{align*}
By computation we have
\begin{align*}
&\frac{d}{dt}\int_0^1\rho_2q(x)\psi_t
 \Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx \\
&= -\Big[(b\psi_x(1)-\int_0^{\infty}g(s)\psi_x(1,t-s)ds)^2
+\Big(b\psi_x(0)  -\int_0^{\infty}g(s)\psi_x(0,t-s)ds\Big)^2\Big] \\
&\quad +2\int_0^1(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds)^2dx \\
&\quad +\beta\int_0^1q(x)\theta_t\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds
 \Big)dx \\
&\quad -k\int_0^1q(x)(\varphi_x+\psi)
 \Big(b\psi_x -\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx 
 +2\rho_2b\int_0^1\psi_t^2dx \\
&\quad -\int_0^1q(x)f(\psi)(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds)dx\\
&\quad +\rho_2\int_0^1q(x)\psi_t\int_0^{\infty}g'(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx,
\end{align*}
which gives
\begin{align*}
&\frac{1}{4\varepsilon_6}\Big[(b\psi_x(1)-\int_0^{\infty}g(s)\psi_x(1,t-s)ds)^2
 +(b\psi_x(0)-\int_0^{\infty}g(s)\psi_x(0,t-s)ds)^2\Big] \\
&=-\frac{1}{4\varepsilon_6}\frac{d}{dt}\int_0^1\rho_2q(x)\psi_t
 \Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx \\
&\quad +\frac{1}{2\varepsilon_6}\int_0^1(b\psi_x
 -\int_0^{\infty}g(s)\psi_x(t-s)ds)^2dx \\
&\quad +\frac{\beta}{4\varepsilon_6}\int_0^1q(x)
 \theta_t\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx \\
&\quad  -\frac{k}{4\varepsilon_6}\int_0^1q(x)(\varphi_x+\psi)(b\psi_x
 -\int_0^{\infty}g(s)\psi_x(t-s)ds)dx
 +\frac{\rho_2b}{2\varepsilon_6}\int_0^1\psi_t^2dx  \\
&\quad -\frac{1}{4\varepsilon_6}\int_0^1q(x)f(\psi)
 \Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx \\
&\quad +\frac{\rho_2}{4\varepsilon_6}\int_0^1q(x)\psi_t\int_0^{\infty}g'(s)
 (\psi_x(t)-\psi_x(t-s))\,ds\,dx.
\end{align*}
In what follows, we use Young's and Poincar\'e inequalities,
Lemma 4.2, (H2), and the fact $0\leq q(x)\leq 4$, $x\in [0,1]$, and obtain
\begin{gather*}
\begin{aligned}
&\frac{1}{2\varepsilon_6}\int_0^1(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds)^2dx \\
&\leq\frac{1}{\varepsilon_6}\int_0^1b^2\psi^2_x\,dx
 +\frac{1}{\varepsilon_6}\int_0^1\Big(\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)^2dx \\
&\leq\frac{c}{\varepsilon_6}\int_0^1\psi^2_x\,dx
 +\frac{c}{\varepsilon_6}(g\circ\psi_x)(t),
\end{aligned} \\
\begin{aligned}
&\frac{\beta}{4\varepsilon_6}\int_0^1q(x)\theta_t(b\psi_x
 -\int_0^{\infty}g(s)\psi_x(t-s)ds)dx \\
&\leq\frac{\varepsilon^2_6}{8\varepsilon_6}\int_0^1q^2(x)\theta_t^2dx
+\frac{\beta^2}{8\varepsilon^3_6}\int_0^1(b\psi_x
 -\int_0^{\infty}g(s)\psi_x(t-s)ds)^2dx \\
&\leq c\varepsilon_6\int_0^1\theta_{tx}^2dx
 +\frac{c}{\varepsilon^3_6}(g\circ\psi_x)(t)
 +\frac{c}{\varepsilon^3_6}\int_0^1\psi^2_x\,dx,
\end{aligned} \\
\begin{aligned}
&-\frac{k}{4\varepsilon_6}\int_0^1q(x)(\varphi_x+\psi)(b\psi_x
 -\int_0^{\infty}g(s)\psi_x(t-s)ds)dx \\
&\leq\frac{\varepsilon^2_6}{8\varepsilon_6}\int_0^1q^2(x)(\varphi_x+\psi)^2dx
+\frac{k^2}{8\varepsilon^3_6}\int_0^1(b\psi_x
-\int_0^{\infty}g(s)\psi_x(t-s)ds)^2dx \\
&\leq c\varepsilon_6\int_0^1(\varphi_x+\psi)^2dx
 +\frac{c}{\varepsilon^3_6}\int_0^1\psi^2_x\,dx
 +\frac{c}{\varepsilon^3_6}(g\circ\psi_x)(t),
\end{aligned} \\
-\frac{1}{4\varepsilon_6}\int_0^1q(x)f(\psi)
\Big(b\psi_x-\int_0^{\infty}g(s)\psi_x(t-s)ds\Big)dx
\leq \frac{c}{\varepsilon_6}\int_0^1\psi^2_x\,dx
 +\frac{c}{\varepsilon_6}(g\circ\psi_x)(t),\\
\frac{\rho_2}{4\varepsilon_6}\int_0^1q(x)\psi_t\int_0^{\infty}
g'(s)(\psi_x(t)-\psi_x(t-s))\,ds\,dx
\leq \frac{c}{\varepsilon_6}\int_0^1\psi^2_tdx
-\frac{c}{\varepsilon_6}(g'\circ\psi_x)(t).
\end{gather*}
At the same time we can get
\begin{align*}
\frac{d}{dt}\int_0^1\frac{\rho_1}{k}q(x)\varphi_t\varphi_x\,dx
&= -[\varphi_x^2(1)+\varphi_x^2(0)]+2\int_0^1\varphi_x^2dx
 +\int_0^1q(x)\varphi_x\psi_x\,dx \\
&\quad -\frac{\beta}{k}\int_0^1q(x)\theta_{tx}\varphi_x\,dx
 +\frac{2\rho_1}{k}\int_0^1\varphi_t^2dx,
\end{align*}
which gives
\begin{align*}
\varepsilon_6[\varphi_x^2(1)+\varphi_x^2(0)]
&= -\frac{d}{dt}\frac{\varepsilon_6}{k}\int_0^1\rho_1q(x)\varphi_t\varphi_x\,dx
 +2\varepsilon_6\int_0^1\varphi_x^2dx +\varepsilon_6\int_0^1q(x)\varphi_x\psi_x\,dx\\
& \quad -\frac{\beta\varepsilon_6}{k}\int_0^1q(x)
 \theta_{tx}\varphi_x\,dx+\frac{2\rho_1\varepsilon_6}{k}\int_0^1\varphi_t^2dx.
\end{align*}
By using Young's and Poincar\'e inequalities we have
\begin{gather*}
2\varepsilon_6\int_0^1\varphi_x^2dx
\leq4\varepsilon_6\int_0^1(\varphi_x+\psi)^2dx+4\varepsilon_6\int_0^1\psi_x^2dx,\\
\varepsilon_6\int_0^1q(x)\varphi_x\psi_x\,dx
\leq c \varepsilon_6\int_0^1(\varphi_x+\psi)^2dx+c\varepsilon_6\int_0^1\psi_x^2dx.\\
-\frac{\beta}{k}\int_0^1q(x)\theta_{tx}\varphi_x\,dx
 \leq c \varepsilon_6\int_0^1(\varphi_x+\psi)^2dx+c\varepsilon_6\int_0^1\psi_x^2dx
 +c\varepsilon_6\int_0^1\theta_{tx}^2dx.
\end{gather*}
Thus, we obtain \eqref{e4.34}.
\end{proof}

\begin{lemma} \label{lem4.10}
Let $(\varphi, \psi, \theta)$ be the solution of \eqref{e2.1}.
Then for some positive constant $\alpha_1$ the functional
\[
J_6(t):=\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma
e^{-\varsigma \rho}|\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\,d\varsigma \,d\rho \,dx
\]
satisfies
\begin{equation}
\begin{aligned}
J_6'(t)
&\leq -\alpha_1\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma
 |\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\,d\varsigma\,d\rho\,dx
 +\ell\int_0^1\theta_{tx}^2dx \\
&\quad -\alpha_1\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma )
 |z^2(x,1,\varsigma ,t)d\varsigma\,dx.
\end{aligned} \label{e4.35}
\end{equation}
\end{lemma}

\begin{proof}
 Differentiating $J_6(t)$ and using \eqref{e2.1}, we obtain
\begin{align*}
J_6'(t)
&=2\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma e^{-\varsigma \rho}|
\mu(\varsigma )|z(x,\rho,\varsigma,t)z_t(x,\rho,\varsigma ,t)\,d\varsigma\,d\rho\,dx\\
&=-2\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}e^{-\varsigma \rho}
 |\mu(\varsigma )|z(x,\rho,\varsigma,t)z_\rho(x,\rho,\varsigma ,t)
 \,d\varsigma\,d\rho\,dx \\
&=-\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\frac{\partial}{\partial\rho}
 \left(e^{-\varsigma \rho}|\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\right)
 \,d\varsigma\,d\rho\,dx \\
&\quad -\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma 
 e^{-\varsigma \rho}|\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)
 \,d\varsigma\,d\rho\,dx \\
&=-\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma )
 |\left(e^{-\varsigma }z^2(x,1,\varsigma ,t)-z^2(x,0,\varsigma ,t)
 \right)d\varsigma\,dx \\
& -\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma
  e^{-\varsigma \rho}|\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)
 \,d\varsigma\,d\rho\,dx. 
\end{align*}
Using the fact that $z(x,0,\varsigma ,t)=\theta_{tx}(x,t)$ and 
$e^{-\varsigma }\leq e^{-\varsigma \rho}\leq1$, for all 
$\rho\in[0,1]$, we obtain
\begin{align*}
J_6'(t)
&\leq -\int_0^1\int_{\tau_1}^{\tau_2}|\mu(\varsigma )
 |e^{-\varsigma }z^2(x,1,\varsigma ,t)d\varsigma\,dx
 +\int_{\tau_1}^{\tau_2}|\mu(\varsigma )|d\varsigma \int_0^1\theta_{tx}^2(x,t)dx \\
&\quad -\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma e^{-\varsigma }
 |\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\,d\varsigma\,d\rho\,dx.
\end{align*}
Because $-e^{-\varsigma}$ is an increasing function, we have 
$-e^{-\varsigma}\leq-e^{-\tau_2}$ for all $\varsigma\in [\tau_1, \tau_2]$. 
Finally, setting $\alpha_1=e^{-\tau_2}$ and recall \eqref{e2.2}, we obtain 
\eqref{e4.35}.
\end{proof}

Now we define the  Lyapunov functional $L(t)$ by
\begin{align*}
L(t)&= NE(t)+\frac{1}{8}J_1(t)+J_2(t)+N_1J_3(t)+N_2J_4(t)+J_5(t)+N_3J_6(t) \\
&\quad +\frac{\varepsilon_6\rho_1}{k}\int_0^1q(x)\varphi_t\varphi_x\,dx
+\frac{\rho_2}{4\varepsilon_6}\int_0^1q(x)\psi_t(b\psi_x-\int_0^{\infty}g(s)ds)dx.
\end{align*}
where $N$, $N_1$, $N_2$, $N_3$, $\varepsilon_6$ are positive constants to be 
chosen properly later.

\begin{lemma} \label{lem4.11}  
Let $(\varphi,\ \psi,\ \theta)$ be the solution of \eqref{e2.1}. 
For $N$ large enough, there exist two positive constants $\alpha_2$ and 
$\alpha_3$ satisfies
$$
\alpha_2 E(t)\leq L(t)\leq\alpha_3 E(t).
$$
\end{lemma}

\begin{proof}
By the same arguments as in \cite{7},  using 
$\int_0^1\varphi_x(t)\leq 2\int_0^1(\varphi_x+\psi)^2dx+2\int_0^1\psi_x^2(t)dx$ 
and Lemma 2, we can deduce
\begin{align*}
|L(t)-NE(t)|
&\leq\gamma_1\int_0^1\varphi^2_tdt+\gamma_2\int_0^1\psi^2_t\,dt
+\gamma_3\int_0^1(\varphi_x+\psi)^2dx +\gamma_4\int_0^1\psi^2_x\,dx\\
&\quad +\gamma_5\int_0^1\theta^2_tdx+\gamma_6\int_0^1\theta^2_x\,dx
+\gamma_7(g\circ\psi_x)+\gamma_8\int_0^1\widehat{f}(\psi)dx \\
&\quad +\gamma_9\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}
 \varsigma |\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\,d\varsigma\,d\rho\,dx \\
& \leq c E(t)
\end{align*}
in which $\gamma_i$ ($i=1,\dots, 9$) are positive constants as in \cite{7}.
\end{proof}


\begin{lemma} \label{lem4.12}
Assume that {\rm (H1)--(H4)} hold, then, there exist two positive constants 
$\beta_1$ and $\beta_2$ such that for $t>0$,
\begin{equation}
\begin{aligned}
&\xi (t)L'(t)+\beta_1E'(t) \\
&\leq -\alpha_4\xi (t)E(t) +\beta_2\xi (t)\int_t^{\infty}g(s)ds
 +\xi (t)\big({{\rho_1b\over k}-\rho_2}\big)\int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx,
\end{aligned} \label{e4.36}
\end{equation}
with $\beta_1=\frac{2\alpha_5}{\gamma}$ and
$\beta_2=\alpha_5\left(\frac{8E(0)}{\gamma(b-g_0)}+2c_0\right)$.
\end{lemma}

\begin{proof}
By differentiating $L(t)$, using Lemma 4.3-Lemma 4.10 and letting 
$\varepsilon_3=\frac{1}{4N_1}$, we obtain
\begin{align*}
L'(t)
&\leq [\frac{N\gamma}{2}-c(N_1^2+\frac{1}{\varepsilon_5}
 +\frac{1}{\varepsilon_6})](g'\circ\psi_x)(t)
 -[\frac{\rho_1}{8}-\varepsilon_4N_2-\varepsilon_5
 -\frac{2\rho_1\varepsilon_6}{k}]\int_0^1\varphi_t^2dx \\
&\quad -[\frac{k}{8}-c(\varepsilon_1+\varepsilon_2+\varepsilon_6)]
 \int_0^1(\varphi_x+\psi)^2dx-(N_2+1)\int_0^1\widehat{f}(\psi)dx
 -\frac{\delta}{2}\int_0^1\theta_{x}^2dx\\
&\quad -[N_1\rho_2\int_0^{\infty}g(s)ds-\frac{9}{8}\rho_2
 -c(1+\frac{1}{\varepsilon_6}+(1+\frac{1}{\varepsilon_4})N_2)]
  \int_0^1\psi_t^2dx \\
&\quad -[\frac{N_2l}{2}-c(1+\varepsilon_1+\varepsilon_2+\frac{1}{\varepsilon_5}
 +\varepsilon_6+\frac{1}{\varepsilon_6}+\frac{1}{\varepsilon_6^3})]
 \int_0^1\psi_x^2dx\\
&\quad +(\frac{b\rho_1}{k}-\rho_2)\int_0^1\varphi_t\psi_{xt}dx 
 -\alpha_1N_3\int_0^1\int_0^1\int_{\tau_1}^{\tau_2}\varsigma 
 |\mu(\varsigma )|z^2(x,\rho,\varsigma ,t)\,d\varsigma\,d\rho\,dx \\
&\quad -(\alpha_1N_3-\frac{c}{\delta})\int_0^1\int_{\tau_1}^{\tau_2}
 \varsigma |\mu(\varsigma )|z^2(x,1,\varsigma ,t)d\varsigma\,dx \\
&\quad +c(1+N_1^2+N_2+\varepsilon_5+\frac{1}{\varepsilon_6}
 +\frac{1}{\varepsilon_6^3})(g\circ\psi_x)(t) \\
&\quad -[Nc-c(1+\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}
+N_2+\frac{1}{\varepsilon_5}+\frac{1}{\varepsilon_6}+N_3)]
 \int_0^1\theta_{tx}^2dx.
\end{align*}
We choose $\varepsilon_1, \varepsilon_2, \varepsilon_5, \varepsilon_6$ 
small enough such that
$$
\mu_1=\frac{\rho_1}{8}-\varepsilon_5-\frac{2\rho_1\varepsilon_6}{k}>0,\quad
 \frac{k}{8}-c(\varepsilon_1+\varepsilon_2+\varepsilon_6)>0,
$$
and then we choose $N_2$ large enough such that
$$
\frac{N_2l}{2}-c(1+\varepsilon_1+\varepsilon_2+\frac{1}{\varepsilon_5}
+\varepsilon_6+\frac{1}{\varepsilon_6}+\frac{1}{\varepsilon_6^3})>0.
$$
We then select $\varepsilon_4$ so small that
$$ 
\mu_1-\varepsilon_4N_2>0.
$$
Next, we pick $N_1$, $N_3$ large enough such that
$$
N_1\rho_2\int_0^{\infty}g(s)ds-\frac{9}{8}\rho_2-c(1+\frac{1}{\varepsilon_6}
+(1+\frac{1}{\varepsilon_4})N_2)>0,\quad \alpha_1N_3-\frac{c}{\delta}>0.
$$
Finally, we choose $N$ large enough such Lemma 4.11 remains valid and
$$
\frac{N\gamma}{2}-c(N_1^2+\frac{1}{\varepsilon_5}
+\frac{1}{\varepsilon_6}),\quad Nc-c(1+\frac{1}{\varepsilon_1}
+\frac{1}{\varepsilon_2}+N_2+\frac{1}{\varepsilon_5}+\frac{1}{\varepsilon_6}+N_3)>0.
$$
Consequently, by using Poincar\'e inequality and \eqref{e2.7}, we obtain
\begin{equation}
\begin{aligned}
L'(t)
&\leq- \alpha_4E(t)+\alpha_5\int_0^1\int_0^{\infty}g(s)(\psi_x(x,t)-\psi_x(x,t-s))^2
 \,ds\,dx \\
&\quad +\big({{\rho_1b\over k}-\rho_2}\big)\int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx,
\end{aligned} \label{e4.37}
\end{equation}
where $\alpha_4$ and $\alpha_5$ are positive constants.

Using (H3) and \eqref{e4.1}, we obtain that for all $t\in R_+$,
\begin{align*}
&\xi(t)\int_0^1\int_0^tg(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx \\
&\leq{\int_0^1\int_0^t\xi(s)g(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx }\\
&\leq{-\int_0^1\int_0^tg'(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx}\\
&\leq{-\int_0^1\int_0^{\infty}g'(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx }\\
&\leq{-\frac{2}{\gamma}E'(t). }
 \end{align*}
On the other hand, using the definition of $E(t)$ and the fact that $E(t)$  
is nonincreasing, we ask for $t,s\in R_+$,
\begin{align*}
 \int_0^1(\psi_x(x,t)-\psi_x(x,t-s))^2dx
&{\leq 2\int_0^1 \psi_x^2(x,t)dx+2\int_0^1 \psi_x^2(x,t-s)dx}\\
&{\leq4\sup_{s>0}\int_0^1\psi^2_x(x,s)dx+2\sup_{\tau<0}\int_0^1\psi^2_x(x,\tau)d\tau}\\
&{\leq4\sup_{s>0}\int_0^1\psi^2_x(x,s)dx+2\sup_{\tau>0}\int_0^1\psi^2_{0x}(x,\tau)d\tau}\\
&{\leq{8E(0)\over \gamma(b-g_0)}+2c_0.    }
\end{align*}
Then, we obtain
\begin{align*}
&\xi (t)\int_0^1\int_t^{\infty}g(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx\\
&\leq\left({8E(0)\over \gamma(b-g_0)}+2c_0\right)\xi (t)\int_t^{\infty}g(s)ds. 
\end{align*}
Then, we deduce that, for all $t\in  R_+$,
\begin{align*}
& {\xi(t)\int_0^1\int_0^{\infty}g(s)(\psi_x(x,t)-\psi_x(x,t-s))^2\,ds\,dx}\\ 
&{\leq-\frac{2}{\gamma}E'(t)+\Big( {8E(0)\over \gamma(b-g_0)}+2c_0 \Big)\xi(t)
 \int_t^{\infty}g(s)ds.}
\end{align*}
The proof is complete.
\end{proof}

\begin{theorem} \label{thm4.13}
Assume that {\rm (H1)--(H4)} hold.

 1. If $\frac{k}{\rho_1}=\frac{b}{\rho_2}$ holds, then, for any 
$((\varphi_0, \varphi_1), (\psi_0, \psi_1), (\theta_0, \theta_1))
\in (H_0^1(0,1)\times L^2(0,1))^3$ satisfying, for some $c_0\geq0,$
$$
\int_0^1\psi_{0x}^2(x,s)dx\leq c_0, \ \forall s>0,
$$
there exist constants $\epsilon_1$, $\epsilon_2>0$ such that, for all 
$t\in R_+$ and for all $\epsilon_0\in[0,\epsilon_1]$,
\begin{equation}
{ E(t)\leq\epsilon_2\Big(1+\int_0^t(g(s))^{1-\epsilon_0}ds\Big)
e^{-\epsilon_0\int_0^t\xi(s)ds}+\epsilon_2\int_t^{\infty}g(s)ds.     }
\label{e4.38}
\end{equation}

2. If $\frac{k}{\rho_1}\neq\frac{b}{\rho_2}$ , then, for any 
$((\varphi_0, \varphi_1), (\psi_0, \psi_1), (\theta_0, \theta_1))
\in (H_0^1(0,1)\times L^2(0,1))^3$ satisfying, for some $c_0\geq0$
\[
\max\Big\{\int_0^1\psi_{0x}^2(x,s)dx, \int_0^1\psi_{0xs}^2(x,s)dx\Big\}
\leq c_0, \quad  s>0,
\]
there exists a constant $\epsilon_2>0$ such that, for all $t\in R_+$,
\begin{equation}
 E(t)\leq\frac{\epsilon_2(1+\int_0^t\xi(s)
\int_s^{\infty}g(\tau)d\tau ds)}{\int_0^t\xi(s)ds}.
\label{e4.39}
\end{equation}
\end{theorem}

\begin{proof}
First, we define
$$
L_1(t)=\xi(t) L(t)+\beta_1E(t) , \ \ r(t)=\xi(t)\int_t^{\infty}g(s)ds.
 $$
Clearly, $L_1(t)$ and $E(t)$ are equivalent, that is, exist  positive
constants $\alpha_6$ and $\alpha_7$, such that
\[
 \alpha_6E(t)\leq L_1(t)\leq \alpha_7E(t)
\]
Then  using Lemma 4.12 we have
\begin{equation}
L_1'(t)\leq-\epsilon_1\xi(t)L_1(t)+\beta_2r(t)
+ \Big({{\rho_1b\over k}-\rho_2}\Big)\xi(t)\int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx,
\label{e4.40}
\end{equation}
with $\epsilon_1=\frac{\alpha_4}{\alpha_7}$. This inequality still holds,
for any $\epsilon_0\in[0,\epsilon_1]$, that is
\begin{equation}
 L_1'(t)\leq-\epsilon_0\xi(t)L_1(t)+\beta_2r(t)+\Big({{\rho_1b\over k}
-\rho_2}\Big)\xi(t)\int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx.
\label{e4.41}
\end{equation}
Now, we distinguish two cases.
\smallskip

\noindent\textbf{Case 1:}
 If $\frac{k}{\rho_1}=\frac{b}{\rho_2}$ holds. Because the last term in 
\eqref{e4.41} vanishes, then \eqref{e4.41} implies that, for all $t\in  R_+$,
$$
\frac{d}{dt}\left( e^{\epsilon_0\int_0^t\xi(s)ds}L_1(t)\right)
\leq\beta_2e^{\epsilon_0\int_0^t\xi(s)ds}r(t).
$$
Therefore, by integrating over $[0,T]$ with  $T\geq 0$, we have
$$
L_1(T)\leq e^{-\epsilon_0\int_0^T\xi(s)ds}
\Big(L_1(0)+\beta_2\int_0^Te^{\epsilon_0\int_0^t\xi(s)ds}r(t)dt\Big),
$$
which implies
\begin{equation}
E(T)\leq {1\over \alpha_6}e^{-\epsilon_0\int_0^T\xi(s)ds}\Big(L_1(0)
+\beta_2\int_0^Te^{\epsilon_0\int_0^t\xi(s)ds}r(t)dt\Big).  \label{e4.42}
\end{equation}
Because
\begin{align*}
e^{\epsilon_0\int_0^t\xi(s)ds} r(t)dt
&={1\over \epsilon_0}\frac{d}{dt} \Big( e^{\epsilon_0\int_0^t\xi(s)ds}\Big)
\int_t^{\infty}g(s)ds \\
&={1\over \epsilon_0}\frac{d}{dt}\Big( e^{\epsilon_0\int_0^t\xi(s)ds}
\int_t^{\infty}g(s)ds\Big)+g(t)e^{\epsilon_0\int_0^t\xi(s)ds},
\end{align*}
 by integration we obtain
\begin{align*}
&\int_0^Te^{\epsilon_0\int_0^t\xi(s)ds}r(t)dt \\
&={1\over\epsilon_0}\Big( e^{\epsilon_0\int_0^T\xi(s)ds}
 \int_T^{\infty}g(s)ds-\int_0^{\infty}g(s)ds
 +\int_0^T e^{\epsilon_0\int_0^t\xi(s)ds}g(t)dt \Big).
\end{align*}
Consequently
\begin{equation}
\begin{aligned}
E(T)
&\leq {1\over \alpha_6}\Big(L_1(0)e^{-\epsilon_0\int_0^T\xi(s)ds}
 +{\beta_2\over \epsilon_0} \int_T^{\infty}g(s)ds \Big)     \\
&\quad +{\beta_2\over \alpha_6\epsilon_0}e^{-\epsilon_0\int_0^T\xi(s)ds}
 \int_0^Te^{\epsilon_0\int_0^t\xi(s)ds}g(t)dt.
\end{aligned} \label{e4.43}
\end{equation}
On the other hand, for all $t\in R_+$,
\begin{align*}
&\frac{d}{dt}\Big( e^{\epsilon_0\int_0^t\xi(s)ds}(g(t))^{\epsilon_0}\Big) \\
&=  \epsilon_0e^{\epsilon_0\int_0^t\xi(s)ds}\xi(t)(g(t))^{\epsilon_0}
 -\epsilon_0e^{\epsilon_0\int_0^t\xi(s)ds}g^{\epsilon_0-1}(t)g'(t) \\
&=\epsilon_0e^{\epsilon_0\int_0^t\xi(s)ds}(g(t))^{\epsilon_0}(\xi(t)+g^{-1}(t)g'(t))\\
\leq0,
\end{align*}
and then $e^{\epsilon_0\int_0^t\xi(s)ds}(g(t))^{\delta_0}\leq (g(0))^{\epsilon_0}$.
Therefore
\begin{equation}
  \int_0^T e^{\epsilon_0\int_0^t\xi(s)ds} g(t)dt\leq (g(0))^{\epsilon_0}
\int_0^T(g(t))^{1-\epsilon_0}dt.
\label{e4.44}
\end{equation}
Finally, \eqref{e4.43} and \eqref{e4.44} give \eqref{e4.38} for any
classical solution of \eqref{e2.1} with
\[
 \epsilon_1=\frac{1}{\alpha_6}\max\Big\{L_1(0),\frac{\beta_2}{\epsilon_0},
 \frac{\beta_2}{\epsilon_0}(g(0))^{\epsilon_0}\Big\}.
\]
By denseness arguments, \eqref{e4.38} remains valid for any weak solution
of \eqref{e2.1}.
\smallskip

\noindent\textbf{Case 2:}
 $\frac{k}{\rho_1}\neq\frac{b}{\rho_2}$. We estimate the last term of \eqref{e4.41} 
as follows: for any $\varepsilon > 0$,
there exists a positive constant $c_\varepsilon$ (depending on $\varepsilon$) 
such that
\begin{align*}
&\Big({{\rho_1b\over k}-\rho_2}\Big)\int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx \\
&=\frac{\rho_1b-\rho_2k}{g_0 k}\int_0^1\varphi_t(x,t)
 \int_0^{\infty}g(s)(\psi_{xt}(x,t)-\psi_{xt}(x,t-s))\,ds\,dx \\ 
&\quad +\frac{\rho_1b-\rho_2k}{g_0 k}\int_0^1\varphi_t(x,t)
 \int_0^{\infty} g(s)\psi_{xt} (x,t-s)\,ds\,dx.
\end{align*}
By using Young's inequality we obtain
\begin{align*}
& \frac{\rho_1b-\rho_2k}{g_0 k}\int_0^1\varphi_t(x,t)\int_0^{\infty}g(s)
 (\psi_{xt}(x,t)-\psi_{xt}(x,t-s))\,ds\,dx \\
&\leq c\int_0^1|\varphi_t(x,t)|\int_0^{\infty}g(s)|\psi_{xt}(x,t)
 -\psi_{xt}(x,t-s)|\,ds\,dx \\
&\leq\frac{\varepsilon}{2}E(t)+c_\varepsilon\int_0^1
 \int_0^{\infty}g(s)(\psi_{xt}(x,t)-\psi_{xt}(x,t-s))^2\,ds\,dx.
\end{align*}
At the same time we have
\begin{gather*}
\int_0^{\infty}g(s)\psi_{xt}(x,t-s)ds=\int_0^{\infty}g'(s)
\left(\psi_{xt}(t-s)-\psi_x(t)\right)ds, \\
\begin{aligned}
&  \frac{\rho_1b-\rho_2k}{g_0 k}\int_0^1\varphi_t(x,t)\int_0^{\infty}g(s)
 \psi_{xt}(x,t-s)\,ds\,dx \\
&=\frac{\rho_1b-\rho_2k}{g_0 k}\int_0^1\varphi_t(x,t)
 \int_0^{\infty}g'(s)\left(\psi_{xt}(t-s)-\psi_x(t)\right)\,ds\,dx \\
&\leq\frac{\varepsilon}{2}E(t)-c_\varepsilon(g'\circ\psi_{xt})(t) \\
&\leq\frac{\varepsilon}{2}E(t)-\frac{2c_\varepsilon}{\gamma}E'(t),
\end{aligned}\\
\begin{aligned}
& \Big({{\rho_1b\over k}-\rho_2}\Big) \int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx \\
&\leq\varepsilon E(t)+c_\varepsilon\int_0^1\int_0^{\infty}g(s)(\psi_{xt}(x,t)
 -\psi_{xt}(x,t-s))^2\,ds\,dx-\frac{2c_\varepsilon}{\gamma}E'(t).
\end{aligned}
\end{gather*}
Thus, we have
\begin{align*}
& \xi(t)\Big({{\rho_1b\over k}-\rho_2}\Big)
 \int_0^1\varphi_t(x,t)\psi_{xt}(x,t)dx \\
&\leq\varepsilon \xi(t) E(t)+c_\varepsilon\xi(t)
 \int_0^1\int_0^{\infty}g(s)(\psi_{xt}(x,t)-\psi_{xt}(x,t-s))^2\,ds\,dx
 -\frac{2c_\varepsilon}{\gamma}\xi(t)E'(t).
\end{align*}
Consequently,
\begin{equation}
\begin{aligned}
L_1'(t)
&\leq -\alpha_8\xi(t)E(t)+\beta_2r(t)-\frac{2c_\epsilon}{\gamma}\xi(t)E'(t) \\
&\quad +c_\epsilon\xi(t)\int_0^1\int_0^{\infty} g(s)(\psi_{xt}(x,t)
 -\psi_{xt}(x,t-s))^2\,ds\,dx,
\end{aligned}\label{e4.45}
\end{equation}
where $\alpha_8=\epsilon_0\alpha_6-\varepsilon$. By using the definition of
 $E_2(t)$ and $E'_2(t)$, we have
\begin{gather}
  \xi(t)\int_0^1\int_0^{t} g(s)(\psi_{xt}(x,t)-\psi_{xt}(x,t-s))^2\,ds\,dx
\leq-\frac{2}{\gamma} E_2'(t), \label{e4.46} \\
\begin{aligned}
&\xi(t)\int_0^1\int_t^{\infty} g(s)(\psi_{xt}(x,t)-\psi_{xt}(x,t-s))^2\,ds\,dx \\
&\leq \Big( {8E_2(0)\over \gamma(b-g_0)}+2c_0 \Big)r(t).
\end{aligned}\label{e4.47}
\end{gather}
Hence, combining \eqref{e4.45}, \eqref{e4.46} and \eqref{e4.47} we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\Big(L_1(t)+\frac{2c_\varepsilon}{\gamma}E_2(t)
 +\frac{2c_\varepsilon}{\gamma}\xi(t)E(t)\Big) \\
&\leq-\alpha_8\xi(t)E(t)+\beta_3r(t)+ \frac{2c_\varepsilon}{\gamma}\xi'(t)E(t),
\end{aligned}\label{e4.48}
\end{equation}
where $\beta_3=\beta_2+({8E_2(0)\over \gamma(b-g_0)}+2c_0)c_\varepsilon$.
 Because $\xi$ is nonincreasing, the last term of \eqref{e4.48} is nonpositive,
therefore, by integration on
[0,T] and using the fact $E(t)$ is nonincreasing, we obtain
$$
\alpha_8E(T)\int_0^T\xi(t)dt\leq L_1(0)+\frac{2c_\varepsilon}{\gamma}E_2(0)
+\frac{2c_\varepsilon}{\xi(0)}E(0)+\beta_3\int_0^Tr(t)dt,
$$
which gives \eqref{e4.39} with
\[
\epsilon_2=\frac{1}{\alpha_8}\max\{L_1(0)+\frac{2c_\varepsilon}{\gamma}E_2(0)
+\frac{2c_\varepsilon}\xi(0)E(0),\beta_3 \}.
\]
This completes the proof.
\end{proof}


\subsection*{Acknowledgements}
The authors want to thank the anonymous referee for valuable comments
and suggestions which lead to the improvement of this paper.
 This work was partially supported by NNSF of China (61374089).

\begin{thebibliography}{20}
\bibitem{9} T. Apalara; 
\emph{Well-posedness and exponential stability for a linear damped Timoshenko 
system with second sound and internal distributed delay}, 
Electronic Journal of Differential Equations, 254 (2014), 1-15.

\bibitem{7} B. W. Feng, M. Pelicer; 
\emph{Global existence and exponential stability for a nonlinear Timoshenko 
system with delay}, Boundary Value Problems, 206 (2015), 1-13.

\bibitem{13} A. Guesmia, S. Messaoudi, A. Wehbe;
 \emph{Uniform decay in mildly damped  Timoshenko systems with non-equal 
wave speed propagation}, Dynamic Systems and Applications, 21 (2012), 133-146.

\bibitem{6} A. Guesmia; 
\emph{On the stabilization for Timoshenko system with past history and 
frictional damping controls}, Palestine Journal of  Mathematics, 2 (2013), 187-214.

\bibitem{3} A. Guesmia, S. Messaoudi; 
\emph{A general stability result in a  Timoshenko system with infinite memory:
 A new approach}, Mathematical Methods in the Applied Sciences, 37 (2014), 384-392.

\bibitem{1} A. Guesmia, S. Messaoudi; 
\emph{Some stability results for Timoshenko systems with cooperative frictional
 and infinite-memory dampings in the displacement}, Acta Mathematica Scientia,
 36 (2016), 1-33.

\bibitem{8} M. Kafini, S. Messaoudi, M. I. Mustafa; 
\emph{Energy decay result in a Timoshenko-type system of thermoelasticity 
of type \textrm{III} with distributive delay}, 
Journal of Mathematical Physics, 54 (2013), 1-14.

\bibitem{14} M. Kafinia, S. Messaoudi, M., I. Mustafa, T. Apalara; 
\emph{Well-posedness and stability results in a Timoshenko-type system of 
thermoelasticity of type \textrm{III} with delay}, 
Zeitschrift f\"ur angewandte Mathematik und Physik ZAMP, 66 (2015), 1499-1517.

\bibitem{17} S. Messaoudi, M. Al-Gharabli; 
\emph{A general decay result of a nonlinear system of wave equations with 
infinite memories}, Applied Mathematics and Computation, 259 (2015), 540-551.

\bibitem{5} S. Messaoudi, T. Apalara;
 \emph{General stability result in a memory-type porous thermoelasticity 
system of type \textrm{III}}, Arab Journal of Mathematical Sciences,
 20 (2014), 213-232.

\bibitem{2} S. Messaoudi, B. Said-Houari;
 \emph{ Uniform decay in a Timoshenko-type system with past history}, 
Journal of Mathematical Analysis and Applications, 360 (2009), 459-475.

\bibitem{10} M. Mustafa; 
\emph{A uniform stability result for thermoelasticity of type \textrm{III} 
with boundary distributed delay}, Journal of Mathematical Analysis and Applications,
415 (2014), 148-158.

\bibitem{12} S. Nicaise, C. Pignotti; 
\emph{Stabilization of the wave equation with boundary or internal distributed delay}, 
Differential Integral Equations, 21 (2008), 935-958.

\bibitem{PRQ} P.X. Pamplona, J. E. M. Rivera, R. Quintanilla;
 \emph{On the decay of solutions for porous-elastic systems with history}, 
Journal of Mathematical Analysis and Applications, 379 (2011), 682-705.

\bibitem{11} S. H. Park; 
 \emph{Decay rate estimates for a weak viscoelastic beam equation with 
time-varing delay}, Applied Mathematics Letters, 31 (2014), 46-51.

\bibitem{16} A. Pazy;
 Semigroups of Linear Operator and Applications to Partial Differential Equations. 
Springer, New York (1983).

\bibitem{RS} J. E. M. Rivera, H. D. F. Sare; 
\emph{Stability of Timoshenko systems with past history}, 
Journal of Mathematical Analysis and Applications, 339(2008), 482-502.

\bibitem{4} S. Timoshenko; 
\emph{On the correction for shear of the differential equation for transverse 
vibrations of prismatic bars}, Philosophical Magazine, 41 (1921), 744-746.

\end{thebibliography}

\end{document}