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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 04, pp. 1--20.\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-2018/04\hfil Nonlinear pseudoparabolic equations]
{Existence and asymptotic behavior of solutions of the
Dirichlet problem for a nonlinear pseudoparabolic equation}


\author[L. T. P. Ngoc, D. T. H. Yen, N. T. Long \hfil EJDE-2018/??\hfilneg]
{Le Thi Phuong Ngoc, Dao Thi Hai Yen, Nguyen Thanh Long}

\address{Le Thi Phuong Ngoc \newline
University of Khanh Hoa,
01 Nguyen Chanh Str., Nha Trang City, Vietnam}
\email{ngoc1966@gmail.com}

\address{Dao Thi Hai Yen \newline
Department of Natural Science,
Phu Yen University,
18 Tran Phu Str., Ward 7, Tuy Hoa City, Vietnam.\newline
Department of Mathematics and Computer Science,
VNUHCM - University of Science,
227 Nguyen Van Cu Str., Dist. 5, HoChiMinh City, Vietnam}
\email{haiyennbh@yahoo.com}

\address{Nguyen Thanh Long \newline
Department of Mathematics and Computer Science,
VNUHCM - University of Science,
227 Nguyen Van Cu Str., Dist. 5, HoChiMinh City, Vietnam}
\email{longnt2@gmail.com}

\dedicatory{Communicated by Dung Le}

\thanks{Submitted January 27, 2017. Published January 4, 2018.}
\subjclass[2010]{34B60, 35K55, 35Q72, 80A30}
\keywords{Nonlinear pseudoparabolic equation; asymptotic behavior;
\hfill\break\indent exponential decay; periodic weak solution;
 Faedo-Galerkin method}

\begin{abstract}
 This article concerns the initial-boundary value problem for nonlinear
 pseudo-parabolic equation
 \begin{gather*}
 u_{t}-u_{xxt}-(1+\mu (u_{x}))u_{xx}+(1+\sigma (u_{x}))u=f(x,t),\quad 0<x<1,\;
 0<t<T, \\
 u(0,t)=u(1,t)=0, \\
 u(x,0)=\tilde{u}_0(x),
 \end{gather*}
 where $f$, $\tilde{u}_0$, $\mu $, $\sigma $ are given functions. Using the
 Faedo-Galerkin method and the compactness method, we prove that there exists
 a unique weak solution $u$  such that
 $u\in L^{\infty }(0,T;H_0^1\cap H^2)$,
 $u'\in L^2(0,T;H_0^1)$ and
 $\| u\| _{L^{\infty }(Q_{T})}\leq \max \{\|\tilde{u}_0\| _{L^{\infty }(\Omega )}$,
 $\| f\|_{L^{\infty }(Q_{T})}\}$. Also we prove that the problem has a unique
 global solution with $H^1$-norm decaying exponentially as $t\to +\infty $.
 Finally, we establish the existence and uniqueness of a weak solution of
 the problem  associated with a periodic condition.
\end{abstract}

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


\section{Introduction}

Consider the following initial-boundary value problem for the
pseudo-parabolic equation arising in third-grade fluid flows
\begin{equation}
u_{t}-(1+\mu (u_{x}))u_{xx}-\alpha u_{xxt}+\big( \gamma +\beta \sigma
(u_{x})\big) u=f(x,t),\quad 0<x<1,\; 0<t<T,  \label{1}
\end{equation}
with the boundary conditions
\begin{equation}
u(0,t)=u(1,t)=0,  \label{2}
\end{equation}
and with the initial condition
\begin{equation}
u(x,0)=\tilde{u}_0(x),  \label{3}
\end{equation}
or the $T$-periodic condition
\begin{equation}
u(x,0)=u(x,T),  \label{3a}
\end{equation}
where $\alpha >0$, $\beta >0$, $\gamma >0$\ are given constants and $f$,
$\tilde{u}_0$, $\mu $, $\sigma $ are given functions satisfying conditions
specified later.

The pseudo-parabolic equation
\begin{equation}
u_{t}-u_{xxt}=F(x,t,u_{x},u_{xx}),\quad 0<x<1,\; t>0  \label{4}
\end{equation}
with the initial condition $u(x,0)=\tilde{u}_0(x)$ and with the difrerent
boundary conditions, has been extensively studied by many authors, see for
example \cite{Amick}, \cite{Arn}, \cite{Bona}, \cite{Medei}, \cite{Zhang}
among others and the references given therein. In these works, many results
about existence, regularity, asymptotic behavior, and decay of solutions
were obtained.

Equations of type \eqref{4} with a one time derivative appearing in the
highest order term are called pseudo-parabolic or Sobolev equations, and
arise in many areas of mathematics and physics. We refer to the monographs
of Al'shin \cite{Al'shin}, and of Carroll \cite{Carroll} for references and
results on pseudoparabolic or Sobolev type equations. Mathematical study of
pseudo-parabolic equations goes back to works of Showalter in the seventies,
since then, numerous of interesting results about linear and nonlinear
pseudo-parabolic equations have been obtained. We also refer to \cite{Sho1}
for asymptotic behavior and to \cite{Sho2} for nonlinear problems.

An important special case of the model is the
Benjamin-Bona-Mahony-Burgers (BBMB) equation
\begin{equation}
u_{t}+u_{x}+uu_{x}-\nu u_{xx}-\alpha ^2u_{xxt}=0,  \label{5}
\end{equation}
it was studied by Amick et al in \cite{Amick}, where $\nu >0$, $\alpha =1$,
$x\in\mathbb{R}$, $t\geq 0$. The authors proved that solution of \eqref{5}
with initial data in $L^1\cap H^2$ decays to zero in $L^2$ norm as
$t\to +\infty $. With $\nu >0$, $\alpha >0$, $x\in [ 0,1]$, $t\geq 0$,
 the model has the form \eqref{5} was also investigated earlier by
Bona and Dougalis in \cite{Bona}, where uniqueness, global existence and
continuous dependence of solutions on initial and boundary data were
established and the solutions were shown to depend continuously on
$\nu \geq 0$ and on $\alpha >0$.

The Benjamin-Bona-Mahony (BBM) equation is introduced in \cite{BenJamin}, in
1972, as a model for$\ $describing long - wave behavior. Since then, the
periodic boundary value problems, the initial value problems and the initial
boundary value problems, for various generalized BBM equations have been
studied. On the other hand, many people have studied the large time
behaviors of solutions to the initial value problems for various generalized
BBM equations, such as BBMB equations with a Burgers-type dissipative term
appended, see \cite{Zhang}. Medeiros and Miranda \cite{Medei} studied
another special case, namely
\begin{equation}
u_{t}+f(u)_{x}-u_{xxt}=g(x,t),  \label{6}
\end{equation}
where $u=u(x,t)$, $0<x<1$, and $t\geq 0$ is the time. They proved existence,
uniqueness of solutions for $f$ in $C^1$ and regularity in the case
$f(s)=s^2/2$.  Arnold et al.\cite{Arn} considered the following
equation from the point of view of periodic solutions
\begin{equation}
-(au_{xt})_{x}+cu_{t}=-(\alpha u_{x})_{x}+\beta u_{x}+\gamma ,\quad
x\in\mathbb{R},\; t\in [ 0,T].  \label{7}
\end{equation}
Here, the authors proved the existence, uniqueness and regularity of
solutions under the hypothesis that $\alpha $, $\beta $ and $\gamma $ are
$C^1$-functions of $x$, $t$ and $u$, and that they are bounded together
with their first derivatives.

It is well known that  equation \eqref{1} arises within frameworks of
mathematical models in engineering and physical sciences on third-grade
fluid flows, see \cite{Aziz,Ha1,Ha2} and references therein.
For example, the following equation of motion for the unsteady flow of
third-grade fluid over the rigid plate with porous medium is investigated
\begin{equation}
\rho \frac{\partial u}{\partial t}
=\mu \frac{\partial ^2u}{\partial y^2}
+\alpha _1\frac{\partial ^{3}u}{\partial y^2\partial t}
+6\beta_3\Big( \frac{\partial u}{\partial y}\Big) ^2
\frac{\partial ^2u}{\partial y^2}
-\frac{\phi }{k}\Big[ \mu
+\alpha _1\frac{\partial }{\partial t}+2\beta _3
 \Big( \frac{\partial u}{\partial y}\Big) ^2
\Big] u,  \label{8}
\end{equation}
for $y>0$, $t>0$,
where $u$ is the velocity component, $\rho $ is the density, $\mu $ the
coefficient of viscosity, $\alpha _1$ and $\beta _3$ are the material
constants, see \cite{Aziz}.

Motivated by the above mentioned works, because of mathematical context,
we study of the existence, uniqueness and exponential decay of solutions for
Dirichlet problem \eqref{1}-\eqref{3} or \eqref{3a}.
This article is organized as follows.
In section 2, under appropriate conditions of $\alpha $, $\beta $, $\gamma $,
$f$, $\tilde{u}_0$, $\mu $, $\sigma $ we prove the existence of a unique
solution on $(0,T)$, for every $T>0$ and the boundedness of the solution.
In section 3, we study exponential decay of solutions.
In section 4, we prove the existence and uniqueness of a $T$-periodic weak
solution.

\section{Preliminaries}

Without loss of generality, we consider Problem \eqref{1} -- \eqref{3}
with $\alpha =\beta =\gamma =1$.

We put $\Omega =(0,1)\ $and denote the usual function spaces used in this
paper by the notations $L^{p}=\,L^{p}(\Omega )$,
$H^{m}=H^{m}( \Omega) $. Let $\langle \cdot ,\cdot \rangle $ be either the
scalar product in $L^2$ or the dual pairing of a continuous linear
functional and an element of a function space. The notation $\|\cdot \|$
stands for the norm in $L^2$ and we denote by $\|\cdot \|_{X}$ the norm
in the Banach space $X $. We call $X'$ the dual space of $X$.

We denote by $L^{p}(0,T;X)$, $1\leq p\leq \infty $ for the Banach space of
real functions $u:(0,T)\to X$ measurable, such that
\begin{equation*}
\| u\| _{L^{p}(0,T;X)}=\Big(\int_0^{T}\| u(t)\| _{X}^{p}dt\Big)
^{1/p}<\infty \quad \text{for }1\leq p<\infty ,
\end{equation*}
and
\begin{equation*}
\| u\| _{L^{\infty }(0,T;X)}=\operatorname{ess\,sup}_{0<t<T}
\| u(t)\| _{X}\quad \text{for }p=\infty .
\end{equation*}
On $H^1$, we shall use the norm
\begin{equation*}
\|v\|_{H^1}=( \|v\|^2+\|v_{x}\|^2) ^{1/2}.
\end{equation*}

The following lemma is well known.

\begin{lemma} \label{lem2.1}
The imbedding $H^1\hookrightarrow C^{0}(\overline{\Omega })$
is compact and
\begin{equation*}
\|v\|_{C^{0}(\overline{\Omega })}\leq \sqrt{2}\|v\|_{H^1}\quad
\text{for all }v\in H^1.
\end{equation*}
\end{lemma}

\begin{remark} \label{rmk2.1} \rm
 On $H_0^1$, $\|v\|_{H^1}$ and $\|v_{x}\|$ are equivalent norms.
 Furthermore,
\begin{equation*}
\|v\|_{C^{0}(\overline{\Omega })}\leq \|v_{x}\|\quad \text{for all }v\in
H_0^1.
\end{equation*}
\end{remark}

\section{Existence and uniqueness theorem}

Without losing of generality, we consider  problem
\eqref{1}-\eqref{3} with $\alpha =\beta =\gamma =1$.
\begin{equation}
\begin{gathered}
u_{t}-u_{xxt}-\frac{\partial }{\partial x}( u_{x}+\bar{\mu}
(u_{x})) +( 1+\sigma (u_{x})) u=f(x,t), \quad 0<x<1, \; 0<t<T, \\
u(0,t)=u(1,t)=0, \\
u(x,0)=\tilde{u}_0(x),
\end{gathered} \label{c1}
\end{equation}
where $\bar{\mu}(y)=\int_0^{y}\mu (z)dz$, $y\in\mathbb{R}$.

The weak formulation of  \eqref{c1} can be given in the following
manner: Find $u(t)$ defined in the open set $(0,T)$ such that $u(t)$
satisfies the  variational problem
\begin{equation}
\begin{aligned}
&\langle u_{t}(t),w\rangle +\langle u_{xt}(t),w_{x}\rangle +\langle u_{x}(t)
+\bar{\mu}(u_{x}(t)),w_{x}\rangle \\
&+\langle ( 1+\sigma (u_{x}(t)))
u(t),w\rangle =\langle f(t),w\rangle ,
\end{aligned} \label{c2}
\end{equation}
for all $w\in H_0^1$ and the initial condition
\begin{equation}
u(0)=\tilde{u}_0.  \label{c3}
\end{equation}
We make the following assumptions:
\begin{itemize}
\item[(H1)]  $\tilde{u}_0\in H_0^1\cap H^2$; \\
\item[(H2)]  $f\in L^2(0,T;H_0^1)$; \\
\item[(H3)]  $\mu \in C^{0}(\mathbb{R};\mathbb{R})$ such that
$\mu (0)=0$, $\mu (z)>0$, for all $z\in\mathbb{R}$, $z\neq 0$;
\item[(H4)]  $\sigma \in C^1(\mathbb{R};\mathbb{R})$ such that
\begin{itemize}
\item[(i)] $\sigma (0)=0$, $\sigma (z)>0$, $z\sigma '(z)>0$, for all
$z\in \mathbb{R}$, $z\neq 0$,
\item[(ii)]  $y(\int_0^{y}z\sigma '(z)dz) \leq
y^2\sigma (y)$ for all $y\in\mathbb{R}$.
\end{itemize}
\end{itemize}

 An example of the function $\sigma $ satisfying (H4)  is
\begin{equation*}
\sigma (z)=| z| ^q,
\end{equation*}
where $q>1$ is a constant. It is obvious that (H4) holds, because
\begin{gather*}
\sigma (z)=| z| ^q, \quad \sigma '(z)=q|z| ^{q-2}z,  \\
\sigma (0)=0, \quad \sigma (z)>0, z\sigma '(z)=q|z| ^q>0, \quad \forall z\in
\mathbb{R}, \quad z\neq 0,\\
\begin{aligned}
y\Big( \int_0^{y}z\sigma '(z)dz\Big)
&=qy\Big(\int_0^{y}| z| ^qdz\Big)
 =qy\frac{ | y| ^qy}{q+1} \\
&=\frac{q}{q+1}| y| ^{q+2}
 =\frac{q}{q+1}y^2\sigma (y)\leq y^2\sigma (y).
\end{aligned}
\end{gather*}

\begin{theorem} \label{thm3.1}
Let $T>0$ and {\rm (H1)--(H4)} hold. Then,  problem \eqref{c1}
has a unique weak solution $u$ satisfying
\begin{equation}
u\in L^{\infty }(0,T;H_0^1\cap H^2),\quad u'\in L^2(0,T;H_0^1).  \label{c4}
\end{equation}
Furthermore, we  have the estimate
\begin{equation}
\| u\| _{L^{\infty }(Q_{T})}\leq \max \{\| \tilde{u}
_0\| _{L^{\infty }(\Omega )},\| f\| _{L^{\infty
}(Q_{T})}\}.  \label{c5}
\end{equation}
\end{theorem}

Estimate \eqref{c5} appears naturally, both
physical and mathematical context, from the maximum principle in the study
of partial differential equation of the kind of \eqref{c1}.

\begin{proof} The proof consists of several steps.
\smallskip

\noindent\textbf{Step 1:} The Faedo-Galerkin approximation (introduced by
Lions \cite{Lions}). Consider a special orthonormal basis $\{w_{j}\}$ on
$H_0^1:w_{j}(x)=\sqrt{2}\sin (j\pi x)$,
$j\in \mathbb{N}$, formed by the eigenfunctions of the Laplacian
$-\Delta =-\frac{\partial ^2}{\partial x^2}$:
\begin{equation*}
-\triangle w_{j}=\lambda _{j}w_{j},w_{j}\in C^{\infty }([0,1]),\quad
\lambda _{j}=(j\pi )^2,\quad j=1,2,\dots
\end{equation*}
Put
\begin{equation}
u_{m}( t) =\sum_{j=1}^{m}c_{mj}( t) w_{j},  \label{c6}
\end{equation}
where the coefficients $c_{mj}( t) $ satisfy a system of
nonlinear differential equations
\begin{equation}
\begin{gathered}
\begin{aligned}
&\langle u_{m}'( t) ,w_{j}\rangle +\langle u_{mx}'( t) ,w_{jx}\rangle
+\langle u_{mx}( t) +\bar{\mu}(u_{mx}(t)),w_{jx}\rangle   \\
& +\langle ( 1+\sigma (u_{mx}(t)))u_{m}(t),w_{j}\rangle
=\langle f(t),w_{j}\rangle , \quad 1\leq j\leq m,
\end{aligned} \\
u_{m}(0)=u_{0m},
\end{gathered}  \label{c7}
\end{equation}
in which
\begin{equation}
u_{0m}=\sum_{j=1}^{m}\beta _{mj}w_{j}\to \tilde{u}_0\quad
\text{strongly in }H_0^1\cap H^2.  \label{c8}
\end{equation}
System \eqref{c7} can be rewritten in the form
\begin{equation}
\begin{gathered}
\begin{aligned}
&c_{mi}'( t) +c_{mi}( t) +\frac{1}{1+\lambda
_{i}}\big[ \langle \bar{\mu}(u_{mx}(t)),w_{ix}\rangle +\langle \sigma
(u_{mx}(t))u_{m}(t),w_{i}\rangle \big]\\
& =\frac{1}{1+\lambda _{i}}\langle f(t),w_{i}\rangle ,
\end{aligned} \\
c_{mi}( 0) =\beta _{mi}, \quad 1\leq i\leq m.
\end{gathered}  \label{c9}
\end{equation}

It is clear that for each $m$ there exists a solution $u_{m}(t)$ in form
\eqref{c6} which satisfies \eqref{c7} almost everywhere on $0\leq t\leq T_{m}$
for some $T_{m}$, $0<T_{m}\leq T$. The following estimates allow us to take
$T_{m}=T$ for all $m$.
\smallskip

\noindent\textbf{Step 2: A priori estimates}.

(a) First estimate. Multiplying the $j^{th}$ equation of \eqref{c7}$_1$
 by $c_{mj}(t)$\ and summing up with respect to $j$, afterwards,
integrating with respect to the time variable from $0$ to $t$, we obtain after
some rearrangements
\begin{equation}
S_{m}(t)=S_{m}(0)+2\int_0^{t}\langle f(s),u_{m}(s)\rangle ds,
\label{c10}
\end{equation}
where
\begin{equation}
\begin{aligned}
S_{m}(t)
&=\| u_{m}(t)\|_{H^1}^2+2\int_0^{t}\| u_{m}(s)\|_{H^1}^2ds \\
&\quad +2\int_0^{t}\langle \bar{\mu}
(u_{mx}(s)),u_{mx}(s)\rangle ds+2\int_0^{t}\langle \sigma
(u_{mx}(s)),u_{m}^2(s)\rangle ds.
\end{aligned}\label{c11}
\end{equation}

By $u_{0m}\to \tilde{u}_0$ strongly in $H_0^1\cap H^2$, we deduce
\begin{equation}
S_{m}(0)=\| u_{0m}\| _{H^1}^2\leq \bar{S}_0\quad \forall m\in\mathbb{N},  \label{c12}
\end{equation}
where $\bar{S}_0$ always indicates a constant depending on $\tilde{u}_0$.

Note that
\begin{equation*}
y\bar{\mu}(y)=y\int_0^{y}\mu (z)dz\geq 0,\quad \forall y\in\mathbb{R}.
\end{equation*}

On the other hand, we have
\begin{equation}
\begin{aligned}
2\int_0^{t}\langle f(s),u_{m}(s)\rangle ds
&\leq \int_0^{t}\| f(s)\| ^2ds+\int_0^{t}\| u_{m}(s)\| ^2ds \\
&\leq \int_0^{T}\| f(s)\| ^2ds+\frac{1}{2}S_{m}(t).
\end{aligned}\label{c13}
\end{equation}
Therefore,
\begin{equation}
S_{m}(t)\leq 2\bar{S}_0+2\int_0^{T}\| f(s)\|
^2ds\leq C_{T}^{(1)}.  \label{c14}
\end{equation}

(b) Second estimate. Next, by replacing $w_{j}$ in \eqref{c7}$
_1\ $by $-w_{jxx}$, we obtain that
\begin{equation}
\begin{aligned}
&\langle u_{mx}'( t) ,w_{jx}\rangle +\langle \Delta
u_{m}'( t) ,\Delta w_{j}\rangle +\langle \Delta
u_{m}( t) ,\Delta w_{j}\rangle \\
&+\langle u_{mx}(t),w_{jx}\rangle +\langle \mu (u_{mx}(t))\Delta u_{m}(
t) ,\Delta w_{j}\rangle \\
&+\langle \sigma '(u_{mx}(t))u_{m}(t)\Delta u_{m}( t)
+\sigma (u_{mx}(t))u_{mx}(t),w_{jx}\rangle \\
&=\langle f_{x}(t),w_{jx}\rangle ,\text{ }1\leq j\leq m.
\end{aligned} \label{c15}
\end{equation}

Similar to \eqref{c7}$_1$, we have
\begin{equation}
\begin{aligned}
P_{m}(t)&=P_{m}(0)-2\int_0^{t}\Big[ \langle \sigma '(u_{mx}(s))u_{m}(s)
 \Delta u_{m}( s) ,u_{mx}(s)\rangle \\
&\quad +\langle \sigma (u_{mx}(s)),| u_{mx}(s)| ^2\rangle \Big] ds
+2\int_0^{t}\langle f_{x}(s),u_{mx}( s) \rangle ds \\
&=P_{m}(0)+I_1+I_2,
\end{aligned} \label{c16}
\end{equation}
where
\begin{equation}
\begin{aligned}
P_{m}(t)&=\| u_{mx}(t)\| ^2+\| \Delta u_{m}(t) \| ^2+2\int_0^{t}( \|
u_{mx}(s)\| ^2+\| \Delta u_{m}(s)\| ^2)
ds \\
&\quad +2\int_0^{t}\langle \mu (u_{mx}(s)),| \Delta u_{m}( s) | ^2\rangle ds.
\end{aligned}\label{c17}
\end{equation}

From $u_{0m}\to \tilde{u}_0$ strongly in $H_0^1\cap H^2$, we deduce
\begin{equation}
P_{m}(0)=\| u_{mx}(0)\| ^2+\| \triangle u_{m}( 0) \| ^2
=\| u_{0mx}\| ^2+\| \triangle u_{0m}\| ^2\leq \bar{P}_0\quad
\forall m\in  \mathbb{N},   \label{c18}
\end{equation}
where $\bar{P}_0$ always indicates a constant depending on $\tilde{u}_0$.
\smallskip

\noindent\emph{Estimating $I_1$}. Note that
\begin{equation}
\begin{aligned}
&-2\langle \sigma '(u_{mx}(s))u_{mx}(s)\Delta u_{m}( s)
,u_{m}(s)\rangle \\
&=-2\int_0^1\sigma '(u_{mx}(x,s))u_{mx}(x,s)\Delta
u_{m}( x,s) u_{m}(x,s)\,dx \\
&=-2\int_0^1u_{m}(x,s)\frac{\partial }{\partial x}(
\int_0^{u_{mx}(x,s)}z\sigma '(z)dz) \,dx \\
&=-2\Big[  u_{m}(x,s)\big( \int_0^{u_{mx}(x,s)}z\sigma
'(z)dz\Big) \Big|_0^1\\
&\quad -\int_0^1u_{mx}(x,s)\Big(
\int_0^{u_{mx}(x,s)}z\sigma '(z)dz\Big) \,dx\Big] \\
&=2\int_0^1u_{mx}(x,s)\Big(
\int_0^{u_{mx}(x,s)}z\sigma '(z)dz\Big) \,dx \\
&\leq 2\int_0^1u_{mx}^2(x,s)\sigma (u_{mx}(x,s))\,dx \\
&=2\langle \sigma (u_{mx}(s)),| u_{mx}(s)| ^2\rangle ,
\end{aligned} \label{c19}
\end{equation}
since $y\big( \int_0^{y}z\sigma '(z)dz\big) \leq y^2\sigma (y)$ for all
$y\in\mathbb{R}$.
Hence
\begin{equation}
\begin{aligned}
I_1&=-2\int_0^{t}\Big[ \langle \sigma '(u_{mx}(s))u_{m}(s)\Delta u_{m}( s) ,
u_{mx}(s)\rangle  \\
&\quad +\langle\sigma (u_{mx}(s)),| u_{mx}(s)| ^2\rangle \Big]
ds\leq 0.
\end{aligned}\label{c20}
\end{equation}
\smallskip

\noindent \emph{Estimating $I_2$.}
\begin{equation}
\begin{aligned}
I_2&=2\int_0^{t}\langle f_{x}(s),u_{mx}( s) \rangle ds
 \leq \int_0^{T}\| f_{x}(s)\| \|u_{mx}(s)\| ds \\
&\leq \int_0^{T}\| f_{x}(s)\| \sqrt{S_{m}(s)} ds
 \leq \sqrt{C_{T}^{(1)}}\int_0^{T}\|f_{x}(s)\| ds.
\end{aligned}\label{c21}
\end{equation}

It follows from \eqref{c16}, \eqref{c18}, \eqref{c20}, \eqref{c21} that
\begin{equation}
P_{m}(t)\leq \bar{P}_0+\sqrt{C_{T}^{(1)}}\int_0^{T}\|
f_{x}(s)\| ds\leq C_{T}^{(2)}.  \label{c22}
\end{equation}
\smallskip

\noindent (c) Third estimate. Multiplying the $j^{th}$\ equation of
\eqref{c7}$_1$ by $c_{mj}'(t)$\ and summing up with respect to $j$,
afterwards, integrating with respect to the time variable from $0$ to $t$,
we obtain after some rearrangements
\begin{equation}
\begin{aligned}
Q_{m}(t)&=Q_{m}(0)-2\int_0^{t}\langle \sigma
(u_{mx}(s))u_{m}(s),u_{m}'(s)\rangle
ds+2\int_0^{t}\langle f(s),u_{m}'(s)\rangle ds \\
&=Q_{m}(0)+J_1+J_2,
\end{aligned}\label{c23}
\end{equation}
where
\begin{gather}
Q_{m}(t)=\| u_{m}(t)\|_{H^1}^2+2\int_0^{t}\| u_{m}'(s)\|_{H^1}^2ds
+2\int_0^1\tilde{\mu}(u_{mx}(x,t))\,dx, \label{c24} \\
\tilde{\mu}(z)=\int_0^{z}\bar{\mu}(y)dy\geq 0\quad \forall z\in\mathbb{R}.
\nonumber
\end{gather}
\smallskip

\noindent\emph{Estimating $Q_{m}(0)$.}
From  $u_{0m}\to \tilde{u}_0$ strongly in $H_0^1\cap H^2$, we can deduce the
existence of a constant $\bar{Q}_0>0$ independent of $m$ such that
\begin{equation}
Q_{m}(0)=\| u_{0m}\| _{H^1}^2+2\int_0^1
\tilde{\mu}(u_{0mx}(x))\,dx\leq \bar{Q}_0\quad \forall m\in\mathbb{N}.  \label{c25}
\end{equation}
\smallskip

\noindent\emph{Estimating $J_1$.} By \eqref{c22}, we have
\begin{align*}
| u_{mx}(x,s)|
&\leq \| u_{mx}(s)\|_{C^{0}([0,1])}\leq \sqrt{2}\| u_{mx}(s)\| _{H^1} \\
&\leq \sqrt{2}\sqrt{\| u_{mx}(s)\| ^2+\| \Delta
u_{m}(s)\| ^2}\leq \sqrt{2}\sqrt{2\| \Delta
u_{m}(s)\| ^2} \\
&\leq 2\| \Delta u_{m}(s)\| \leq 2\sqrt{P_{m}(s)}\leq 2
\sqrt{C_{T}^{(2)}}.
\end{align*}
Hence
\begin{equation}
\begin{aligned}
J_1
&=-2\int_0^{t}\langle \sigma (u_{mx}(s))u_{m}(s),u_{m}'(s)\rangle ds \\
&\leq 2\sup_{| z| \leq 2\sqrt{C_{T}^{(2)}}}
\sigma (z)\int_0^{t}\| u_{m}(s)\| \| u_{m}'(s)\| ds \\
&\leq 2\sup_{| z| \leq 2\sqrt{C_{T}^{(2)}}}
\sigma (z)\int_0^{t}\sqrt{S_{m}(s)}\| u_{m}'(s)\| ds \\
&\leq 2\sqrt{C_{T}^{(1)}}\sup_{| z| \leq 2\sqrt{C_{T}^{(2)}}}
 \sigma (z)\int_0^{t}\| u_{m}'(s)\| ds \\
&\leq 2TC_{T}^{(1)}\sup_{| z| \leq 2\sqrt{C_{T}^{(2)}}}
 \sigma ^2(z)+\frac{1}{2}\int_0^{t}\|u_{m}'(s)\| ^2ds \\
&\leq 2TC_{T}^{(1)}\sup_{| z| \leq 2\sqrt{C_{T}^{(2)}}}
 \sigma ^2(z)+\frac{1}{4}Q_{m}(t).
\end{aligned}\label{c26}
\end{equation}
\smallskip

\noindent\emph{Estimating $J_2$.}
\begin{equation}
\begin{aligned}
J_2&=2\int_0^{t}\langle f(s),u_{m}'(s)\rangle ds \\
&\leq 2\int_0^{T}\| f(s)\| ^2ds+\frac{1}{2}
\int_0^{t}\| u_{m}'(s)\| ^2ds \\
&\leq 2\int_0^{T}\| f(s)\| ^2ds+\frac{1}{4} Q_{m}(t).
\end{aligned} \label{c27}
\end{equation}

Then, it follows from \eqref{c23}, \eqref{c25}--\eqref{c27} that
\begin{equation}
Q_{m}(t)\leq 2\Big( \bar{Q}_0+2TC_{T}^{(1)}\underset{|
z| \leq 2\sqrt{C_{T}^{(2)}}}{\sup }\sigma
^2(z)+2\int_0^{T}\| f(s)\| ^2ds\Big) \leq
C_{T}^{(3)}.  \label{c28}
\end{equation}
\smallskip

\noindent\textbf{Step 3: Limiting process.} Thanks to \eqref{c14},
\eqref{c22}, \eqref{c28} there exists a subsequence of $\{u_{m}\}$,
still denoted by $\{u_{m}\}$ such that
\begin{equation}
\begin{gathered}
u_{m}\to u \quad \text{in } L^{\infty }(0,T;H_0^1\cap H^2)
\text{ weakly*,}  \\
u_{m}'\to u' \quad  \text{in } L^2(0,T;H_0^1)\text{ weakly.}
\end{gathered}  \label{c29}
\end{equation}

Using the compactness lemma of Lions \cite[p.57]{Lions}, and applying
Fischer-Riesz theorem, from \eqref{c29}, there exists a subsequence of
$\{u_{m}\}$, denoted by the same symbol satisfying
\begin{equation}
\begin{gathered}
u_{m}\to u  \quad\text{strongly in $L^2(0,T;H_0^1)$
 and a.e. in }Q_{T}, \\
u_{mx}\to u_{x} \quad \text{strongly in $L^2(Q_{T})$
and a.e. in $Q_{T}$}.
\end{gathered}  \label{c30}
\end{equation}

Then, it follows from \eqref{c30}, that
\begin{equation}
\begin{gathered}
\bar{\mu}(u_{mx}(x,t))\to \bar{\mu}(u_{x}(x,t))\quad\text{a.e., $(x,t)$ in }
Q_{T}, \\
\sigma (u_{mx}(x,t))u_{m}(x,t)\to \sigma (u_{x}(x,t))u(x,t) \quad
\text{a.e., $(x,t)$ in } Q_{T}.
\end{gathered} \label{c31}
\end{equation}

On the other hand, by \eqref{c22}, we have
\begin{equation}
\begin{gathered}
\begin{aligned}
| u_{mx}(x,t)|
&\leq \| u_{mx}(t)\|_{C^{0}([0,1])}
 \leq \sqrt{2}\| u_{mx}(t)\| _{H^1} \\
&\leq 2\| \triangle u_{m}( t) \| \leq 2\sqrt{
P_{m}(t)}\leq 2\sqrt{C_{T}^{(2)}};
\end{aligned} \\
| \bar{\mu}(u_{mx}(x,t))|
\leq \sup_{|z| \leq 2\sqrt{C_{T}^{(2)}}} | \bar{\mu}(z)| \leq C_{T}; \\
\begin{aligned}
| \sigma (u_{mx}(x,t))u_{m}(x,t)|
&\leq \|u_{mx}(t)\| | \sigma (u_{mx}(x,t))| \\
&\leq \sqrt{C_{T}^{(2)}}\underset{| z| \leq 2\sqrt{
C_{T}^{(2)}}}{\sup }| \sigma (z)| \leq C_{T}.
\end{aligned}
\end{gathered} \label{c32}
\end{equation}

Applying the dominated convergence theorem,  from \eqref{c31}, \eqref{c32} we obtain
\begin{equation}
\begin{gathered}
\bar{\mu}(u_{mx})\to \bar{\mu}(u_{x}) \quad \text{strongly in }
 L^2(Q_{T}),  \\
\sigma (u_{mx})u_{m}\to \sigma (u_{x})u\quad  \text{strongly in }
 L^2(Q_{T}).
\end{gathered} \label{c33}
\end{equation}

Passing to the limit in \eqref{c7} by \eqref{c8}, \eqref{c29}, \eqref{c30}
and \eqref{c33}, we have $u$ satisfying
\begin{equation}
\begin{gathered}
\begin{aligned}
&\langle u_{t}(t),w\rangle +\langle u_{xt}(t),w_{x}\rangle +\langle u_{x}(t)+
\bar{\mu}(u_{x}(t)),w_{x}\rangle   
+\langle ( 1+\sigma (u_{x}(t))) u(t),w\rangle\\
& =\langle f(t),w\rangle , \quad \forall w\in H_0^1,
\end{aligned} \\
u(0)=\tilde{u}_0.
\end{gathered} \label{c34}
\end{equation}
Furthermore,
\begin{equation*}
u\in L^{\infty }(0,T;H_0^1\cap H^2),\quad
u'\in L^2(0,T;H_0^1).
\end{equation*}
\smallskip

\noindent\textbf{Step 4: Uniqueness of the solution.}
Let $u$ and $v$ be two weak solutions of \eqref{c1} such that
\begin{equation}
u, v\in L^{\infty }(0,T;H_0^1\cap H^2),\quad
u', v'\in L^2(0,T;H_0^1).  \label{c35}
\end{equation}
Then $w=u-v$ satisfies
\begin{equation}
\begin{gathered}
\begin{aligned}
&\langle w_{t}(t),y\rangle +\langle w_{xt}(t),y_{x}\rangle
+\langle w_{x}(t),y_{x}\rangle +\langle w(t),y\rangle \\
&+\langle \bar{\mu}(u_{x}(t))-\bar{\mu}(v_{x}(t)),y_{x}\rangle 
+\langle \sigma (u_{x}(t))u-\sigma (v_{x}(t))v,y\rangle =0,
\quad\forall y\in H_0^1,
\end{aligned} \\
w(0)=0,  \\
u,v,w\in L^{\infty }(0,T;H_0^1\cap H^2), \quad
u_{t},v_{t},w_{t}\in L^2(0,T;H_0^1).
\end{gathered}  \label{c36}
\end{equation}

Take $y=w=u-v$, in \eqref{c36}$_1$ and integrating with respect to $t$,
we obtain
\begin{equation}
\begin{aligned}
\rho (t)
&=-2\int_0^{t}\langle \bar{\mu}(u_{x}(s))-\bar{\mu}
(v_{x}(s)),w_{x}(s)\rangle ds \\
&\quad -2\int_0^{t}\langle \sigma (u_{x}(s))u(s)
 -\sigma (v_{x}(s))v(s),w(s)\rangle ds \\
&=\rho _1(t)+\rho _2(t),
\end{aligned} \label{c37}
\end{equation}
where
\begin{equation}
\rho (t)=\| w(t)\|_{H^1}^2+2\int_0^{t}\| w(s)\|_{H^1}^2ds.  \label{c38}
\end{equation}
\smallskip

\noindent\emph{Estimating $\rho _1(t)$.} Using the monotonicity of
the function $z\mapsto \bar{\mu}(z)$, we obtain
\begin{equation}
\rho _1(t)=-2\int_0^{t}\langle \bar{\mu}(u_{x}(s))-\bar{\mu}
(v_{x}(s)),w_{x}(s)\rangle ds\leq 0.  \label{c39}
\end{equation}
\smallskip

\noindent\emph{Estimating $\rho _2(t)$.} We have
\begin{equation}
\begin{aligned}
[ \sigma (u_{x})u-\sigma (v_{x})v] w
&=[ \sigma (u_{x})w+( \sigma (u_{x})-\sigma (v_{x})) v] w \\
&=\sigma (u_{x})w^2+( \sigma (u_{x})-\sigma (v_{x})) vw \\
&\geq ( \sigma (u_{x})-\sigma (v_{x})) vw.
\end{aligned} \label{c40}
\end{equation}
This implies
\begin{equation}
\begin{aligned}
\rho _2(t)
&=-2\int_0^{t}\langle \sigma (u_{x}(s))u(s)-\sigma
(v_{x}(s))v(s),w(s)\rangle ds \\
&\leq -2\int_0^{t}\langle \left[ \sigma
(u_{x}(s))-\sigma (v_{x}(s))\right] v(s),w(s)\rangle ds \\
&\leq 2\int_0^{t}\| \left[ \sigma
(u_{x}(s))-\sigma (v_{x}(s))\right] v(s)\| \|w(s)\| ds \\
&\leq 2\int_0^{t}\| \sigma
(u_{x}(s))-\sigma (v_{x}(s))\| \| v_{x}(s)\|\| w(s)\| ds.
\end{aligned}\label{c41}
\end{equation}

Put $M=\| u\| _{L^{\infty }(0,T;H_0^1\cap H^2)}
 +\| v\| _{L^{\infty }(0,T;H_0^1\cap H^2)}$ and
$L_{M}=\sup_{| z| \leq M} | \sigma'(z)| $, we have
\begin{equation}
| \sigma (u_{x})-\sigma (v_{x})| \leq L_{M}|w_{x}| .  \label{c42}
\end{equation}
Hence
\begin{equation}
\begin{aligned}
\rho _2(t)
&\leq 2L_{M}\int_0^{t}\| w_{x}(s)\|\| v_{x}(s)\| \| w(s)\| ds \\
&\leq 2ML_{M}\int_0^{t}\| w_{x}(s)\| \| w(s)\| ds \\
&\leq ML_{M}\int_0^{t}\rho (s)ds.
\end{aligned}\label{c43}
\end{equation}
Then,  from \eqref{c37}, \eqref{c39}, \eqref{c43} it follows that
\begin{equation}
\rho (t)\leq ML_{M}\int_0^{t}\rho (s)ds.  \label{c44}
\end{equation}
By Gronwall's lemma, \eqref{c44} leads to $\rho (t)=0$, i.e., $w=u-v=0$.
\smallskip

\noindent\textbf{Step 5: Proof of the estimate}\eqref{c5}.
First, let us assume that
\begin{equation}
u_0(x)\leq M,\quad\text{a. e., $x\in \Omega$, and }
\max \{\| \tilde{u}_0\| _{L^{\infty }},\| f\| _{L^{\infty
}(Q_{T})}\}\leq M.  \label{c45}
\end{equation}
Then $z=u-M$ satisfies the initial and boundary value
\begin{equation}
\begin{gathered}
\begin{aligned}
&z_{t}-z_{xxt}-\frac{\partial }{\partial x}( z_{x}+\bar{\mu}
(z_{x})) +z+(z+M)\sigma (z_{x})  \\
&=f(x,t)-M, \quad 0<x<1, \; 0<t<T,
\end{aligned} \\
z(0,t)=z(1,t)=-M,  \\
z(x,0)=\tilde{u}_0(x)-M.
\end{gathered}  \label{c46}
\end{equation}

Multiplying equation \eqref{c46} by $v\in H_0^1$, then integrating
by parts with respect to variable $x$,  after some rearrangements, one has
\begin{equation}
\begin{aligned}
&\langle z_{t}(t),v\rangle +\langle z_{xt}(t),v_{x}\rangle
+\langle z_{x}(t)+\bar{\mu}(z_{x}(t)),v_{x}\rangle \\
&+\langle z(t)+(z(t)+M)\sigma (z_{x}(t)),v\rangle \\
&=\langle f(t)-M,v\rangle ,\quad \text{for all  }v\in H_0^1.
\end{aligned}\label{c47}
\end{equation}

From assumption (H1)--(H4) we deduce that the solution of
the initial and boundary value problem \eqref{c1} satisfies
$u\in L^{\infty}(0,T;H_0^1\cap H^2)$,
$u'\in L^2(0,T;H_0^1)$, so that
we are allowed to take $v=z^{+}=\frac{1}{2}(| z| +z)$ in
\eqref{c47}). Thus, it follows that
\begin{equation}
\begin{aligned}
&\langle z_{t}(t),z^{+}(t)\rangle +\langle
z_{xt}(t),z_{x}^{+}(t)\rangle +\langle z_{x}(t)+\bar{\mu}
(z_{x}(t)),z_{x}^{+}(t)\rangle \\
&+\langle z(t)+(z(t)+M)\sigma (z_{x}(t)),z^{+}(t)\rangle \\
&=\langle f(t)-M,z^{+}(t)\rangle .
\end{aligned} \label{c48}
\end{equation}
Hence
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}( \| z^{+}(t)\|
^2+\| z_{x}^{+}(t)\| ^2) +\|
z_{x}^{+}(t)\| ^2+\| z^{+}(t)\| ^2 \\
&=-\langle \bar{\mu}(z_{x}^{+}(t)),z_{x}^{+}(t)\rangle -\langle
(z^{+}(t)+M)\sigma (z_{x}^{+}(t)),z^{+}(t)\rangle \\
&\quad +\langle f(t)-M,z^{+}(t)\rangle \leq 0,
\end{aligned}\label{c49}
\end{equation}
since
$M\geq \max \{\| \tilde{u}_0\| _{L^{\infty }},
\|f\| _{L^{\infty }(Q_{T})}\}$ and
\begin{equation}
\begin{aligned}
\langle z_{t}(t),z^{+}(t)\rangle
&=\int_0^1z_{t}(x,t)z^{+}(x,t)\,dx
 =\int_{0,z>0}^1(z^{+}(x,t))_{t\,}z^{+}(x,t)\,dx \\
& =\frac{1}{2}\frac{d}{dt}\int_{0, z>0}^1|z^{+}(x,t)| ^2\,dx
 =\frac{1}{2}\frac{d}{dt}\int_0^1 | z^{+}(x,t)| ^2\,dx\\
& =\frac{1}{2}\frac{d}{dt}\| z^{+}(t)\| ^2,
\end{aligned}\label{c50}
\end{equation}
and on the domain $z>0$ we have $z^{+}=z$, $z_{x}=(z^{+})_{x}$ and
$z_{t}=(z^{+})_{t}$.

Integrating \eqref{c49}, we obtain
\begin{equation}
\| z^{+}(t)\| ^2+\| z_{x}^{+}(t)\|^2
\leq \| z^{+}(0)\| ^2+\|z_{x}^{+}(0)\| ^2.  \label{c51}
\end{equation}

Since $z^{+}(x,0)=(u(x,0)-M)^{+}=(\tilde{u}_0(x)-M)^{+}=0$,
$z_{x}^{+}(x,0)=0$, we obtain
$\| z^{+}(t)\| ^2+\|z_{x}^{+}(t)\| ^2=0$. Thus
 $z^{+}=0$ and $u(x,t)\leq M$, for a.e.  $(x,t)\in Q_{T}$.

The case $-M\leq u_0(x)$, a.e., $x\in \Omega $, and
$M\geq \max\{\| \tilde{u}_0\| _{L^{\infty }}$,
$\|f\| _{L^{\infty }(Q_{T})}\}$ can be dealt with, in the same manner
as above, by considering $z=u+M$ and
$z^{-}=\frac{1}{2}(|z| -z)$, we also obtain $z^{-}=0$ and hence
$u(x,t)\geq -M$, for a.e. $(x,t)\in Q_{T}$.

From the above, one obtains $| u(x,t)| \leq M$, a.e. $(x,t)\in Q_{T}$, i.e.,
\begin{equation}
\| u\| _{L^{\infty }(Q_{T})}\leq M,  \label{c52}
\end{equation}
for all $M\geq \max \{\| \tilde{u}_0\| _{L^{\infty}},\| f\| _{L^{\infty }(Q_{T})}\}$.
This implies \eqref{c5}.
The proof is complete.
\end{proof}

\section{Exponential decay of solutions}

This section investigates the decay of the solution of  \eqref{c1}. For
this purpose, we make the following assumption.
\begin{itemize}
\item[(H5)]  $f\in L^2(\mathbb{R}_{+};H_0^1)$ and there exist two constants
$C_0>0$, $\gamma _0>0$ such that $\| f(t)\| \leq C_0e^{-\gamma _0t}$, for all
$t\geq 0$.
\end{itemize}

\begin{theorem} \label{thm4.1}
Assume that {\rm (H1), (H3)--(H5)} hold. Then,  problem \eqref{c1}
has a unique weak solution $u$ satisfying
\begin{equation}
u\in L^{\infty }(0,T;H_0^1\cap H^2),\quad
u'\in L^2(0,T;H_0^1)\quad \text{for all }T>0,  \label{d1}
\end{equation}
and there exist positive constants $C$, $\gamma $ such that
\begin{equation}
\| u(t)\| _{H^1}\leq C\exp (-\gamma t)\quad \text{for all }t\geq 0.  \label{d2}
\end{equation}
\end{theorem}

\begin{proof}
Multiplying the $j^{th}$ equation of \eqref{c7}$_1$ by $c_{mj}(t)$
 and summing with respect to $j$,  after some
rearrangements, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{m}(t)\|_{H^1}^2+2\| u_{m}(t)\| _{H^1}^2
+2\langle \bar{\mu} (u_{mx}(t)),u_{mx}(t)\rangle
+2\langle \sigma (u_{mx}(t)),u_{m}^2(t)\rangle \\
&=2\langle f(t),u_{m}(t)\rangle .
\end{aligned}\label{d3}
\end{equation}
Note that
\begin{equation}
\begin{aligned}
2\langle f(t),u_{m}(t)\rangle
&\leq 2\| f(t)\| \|u_{m}(t)\| \leq 2\| f(t)\| \| u_{m}(t)\| _{H^1} \\
&\leq \frac{1}{2\delta }\|f(t)\| ^2+2\delta \| u_{m}(t)\| _{H^1}^2,
\end{aligned}\label{d4}
\end{equation}
for all $\delta >0$.

It follows from \eqref{d3}, \eqref{d4} that
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{m}(t)\|
_{H^1}^2+2(1-\delta )\| u_{m}(t)\| _{H^1}^2 \\
&\leq \frac{1}{2\delta }\| f(t)\| ^2\leq
 \frac{1}{2\delta }C_0^2e^{-2\gamma _0t},\quad \text{for all }\delta >0.
\end{aligned}\label{d5}
\end{equation}
Choose $\delta $ and $\gamma $ such that
\begin{equation}
0<\delta <1,\quad 0<\gamma <\min \{1-\delta ,\gamma _0\}.  \label{d6}
\end{equation}
Then  from \eqref{d5}, \eqref{d6} we have
\begin{equation}
\frac{d}{dt}\| u_{m}(t)\| _{H^1}^2+2\gamma \|u_{m}(t)\| _{H^1}^2
\leq \frac{1}{2\delta }C_0^2e^{-2\gamma_0t}.  \label{d7}
\end{equation}
Integrating \eqref{d7}, we obtain
\begin{equation}
\| u_{m}(t)\| _{H^1}^2\leq \Big( \| \tilde{u}
_0\| _{H^1}^2+\frac{C_0^2}{4\delta ( \gamma
_0-\gamma ) }\Big) e^{-2\gamma t}.  \label{d8}
\end{equation}
Letting $m\to +\infty $ in \eqref{d8}, we obtain
\begin{equation}
\begin{aligned}
\| u(t)\| _{H^1}^2
&\leq \liminf_{m\to +\infty } \| u_{m}(t)\| _{H^1}^2\\
&\leq \Big( \| \tilde{u}_0\| _{H^1}^2+\frac{C_0^2
}{4\delta ( \gamma _0-\gamma ) }\Big) e^{-2\gamma t},\quad \text{for all }t\geq 0.
\end{aligned}\label{d9}
\end{equation}
This implies \eqref{d2}, and completes the proof.
\end{proof}

\section{Existence and  uniqueness of a T-periodic weak solution}

In this section, we shall consider  problem \eqref{1}, \eqref{2}, \eqref{3a}
with the constants $\alpha =\beta =\gamma =1$,
\begin{equation}
\begin{gathered}
u_{t}-u_{xxt}-(1+\mu (u_{x}))u_{xx}+( 1+\sigma (u_{x}))
u=f(x,t), \quad  0<x<1, \; 0<t<T,  \\
u(0,t)=u(1,t)=0,  \\
u(x,0)=u(x,T).
\end{gathered}  \label{n1}
\end{equation}
We make the following assumptions:
\begin{itemize}
\item[(H6)]  $f$ is $T$-periodic in $t$, i.e., $f(x,0)=f(x,T)$.
\end{itemize}

\begin{remark} \label{rmk5.1}\rm
The weak formulation of problem \eqref{n1}
 can be given in the following manner:
Find $u\in L^{\infty}(0,T;H_0^1\cap H^2)$ with $u'\in L^2(0,T;H_0^1)$, such
that $u$ satisfies the  variational equation
\begin{equation}
\begin{gathered}
\begin{aligned}
&\int_0^{T}\langle u'(t)+u(t),w(t)\rangle\,dt
 +\int_0^{T}\langle u_{x}'(t)+u_{x}(t),w_{x}(t)\rangle\,dt\\
& +\int_0^{T}\langle \bar{\mu}(u_{x}(t)),w_{x}(t)\rangle\,dt  
 +\int_0^{T}\langle \sigma (u(t),u_{x}(t))u(t),w(t)\rangle\,dt \\
&=\int_0^{T}\langle f(t),w(t)\rangle\,dt, \quad\text{for all }
w\in L^2(0,T;H_0^1),
\end{aligned} \\
u(0)=u(T).
\end{gathered} \label{n2}
\end{equation}
\end{remark}

\begin{theorem} \label{thm5.1}
Let $T>0$ and {\rm (H2), (H3), (H4), (H6)}
hold. Then  problem \eqref{n1}  has a weak solution $u$
 such that
\begin{equation}
u\in L^{\infty }(0,T;H_0^1\cap H^2)\text{ and }
u'\in L^2(0,T;H_0^1).  \label{n3}
\end{equation}
Furthermore, if $\| u\| _{L^{\infty
}(0,T;H_0^1\cap H^2)}\leq R$, with
$R\underset{|z| \leq \sqrt{2}R}{\sup }| \sigma '(z)| <2$,
then the solution is unique.
\end{theorem}

\begin{proof} The proof consists of several steps.
\smallskip

\noindent\textbf{Step 1:}
 Consider the basis $\{w_{j}\}$ as above. Let $W_{m}$ be the
linear space generated by $w_1,w_2,\dots,w_{m}$. We consider the following
problem.

Find a function $u_{m}(t)$ in the form \eqref{c6} satisfying the nonlinear
differential equation system \eqref{c7}$_1$ and the $T$-periodic
condition
\begin{equation}
u_{m}(0)=u_{m}(T).  \label{n4}
\end{equation}

We consider an initial value problem given by \eqref{c7}, where $u_{0m}$ is
given in $W_{m}$.

It is clear that for each $m$, there exists a solution $u_{m}(t)$ in the
form \eqref{c6} which satisfies \eqref{c7} almost everywhere on
$0\leq t\leq T_{m}$ for some $T_{m}$, $0<T_{m}\leq T$.
The following a priori estimates
allow us to take $T_{m}=T$\ for all $m$.
\smallskip

\noindent\textbf{Step 2: A priori estimates.}
Multiplying the $j^{th}$\ equation of \eqref{c7}$_1$ by $c_{mj}(t)$ and
summing with respect to $j$, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{m}(t)\|
_{H^1}^2+2\| u_{m}(t)\| _{H^1}^2+2\langle \bar{\mu}
(u_{mx}(t)),u_{mx}(t)\rangle \\
&+2\| \sqrt{\sigma (u_{mx}(t))}u_{m}(t)\| ^2\\
&=2\langle f(t),u_{m}(t)\rangle .
\end{aligned} \label{n5}
\end{equation}
We estimate without difficulty the  term
$2\langle f(t),u_{m}(t)\rangle $ as follows
\begin{equation}
2\langle f(t),u_{m}(t)\rangle
\leq \frac{1}{2\delta _1}\| f(t)\| ^2+2\delta _1\| u_{m}(t)\| ^2
\leq \frac{1}{2\delta _1}\| f(t)\| ^2+2\delta_1\| u_{m}(t)\| _{H^1}^2,  \label{n6}
\end{equation}
for all $\delta _1$, $0<\delta _1<1$.

Hence,  from \eqref{n5}, \eqref{n6}  it follows that
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{m}(t)\|_{H^1}^2+2(1-\delta _1)\| u_{m}(t)\|
_{H^1}^2+2\langle \bar{\mu}(u_{mx}(t)),u_{mx}(t)\rangle \\
&+2\| \sqrt{\sigma (u_{mx}(t))}u_{m}(t)\| ^2\\
&\leq \frac{1}{2\delta _1}\| f(t)\| ^2.
\end{aligned} \label{n7}
\end{equation}

Next, multiplying the $j^{th}$ equation of \eqref{c14} by $c_{mj}(t)$ and
summing with respect to $j$, we obtain
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{mx}(t)\|
_{H^1}^2+2\| u_{mx}(t)\| _{H^1}^2+2\| \sqrt{
\mu (u_{mx}(t))}\triangle u_{m}(t)\| ^2 \\
&+2\langle \sigma '(u_{mx}(t))u_{m}(t)\Delta
u_{m}( t) +\sigma (u_{mx}(t))u_{mx}(t),u_{mx}(t)\rangle \\
&=2\langle f_{x}(t),u_{mx}(t)\rangle .
\end{aligned}\label{n8}
\end{equation}
Similarly, we have
\begin{equation}
\begin{aligned}
&2\langle \sigma '(u_{mx}(t))u_{m}(t)\Delta
u_{m}( t) +\sigma (u_{mx}(t))u_{mx}(t),u_{mx}(t)\rangle \\
&=2\int_0^1u_{m}(x,t)u_{mx}(x,t)\sigma '(u_{mx}(x,t))\Delta u_{m}( x,t)\,dx \\
&\quad +2\int_0^1u_{mx}^2(x,t)\sigma (u_{mx}(x,t))\,dx \\
&=2\int_0^1u_{m}(x,t)\frac{\partial }{\partial x}\Big(
\int_0^{u_{mx}(x,t)}y\sigma '(y)\Big)\,dx
 +2\int_0^1u_{mx}^2(x,t)\sigma (u_{mx}(x,t))\,dx \\
&=-2\int_0^1u_{mx}(x,t)\Big( \int_0^{u_{mx}(x,t)}y\sigma '(y)\Big) \,dx
 +2\int_0^1u_{mx}^2(x,t)\sigma (u_{mx}(x,t))\,dx \\
&=2\int_0^1\Big[ u_{mx}^2(x,t)\sigma (u_{mx}(x,t))-u_{mx}(x,t)
 \Big( \int_0^{u_{mx}(x,t)}y\sigma
'(y)\Big) \Big] \,dx\geq 0,
\end{aligned} \label{n9}
\end{equation}
and this implies
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| u_{mx}(t)\| _{H^1}^2+2(1-\delta
_1)\| u_{mx}(t)\| _{H^1}^2+2\| \sqrt{\mu
(u_{mx}(t))}\triangle u_{m}(t)\| ^2\\
&\leq \frac{1}{2\delta _1}\| f_{x}(t)\| ^2,
\end{aligned}\label{n10}
\end{equation}
for all $\delta _1$, $0<\delta _1<1$.

It follows from \eqref{n7}, \eqref{n10} that
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\left[ \| u_{m}(t)\| _{H^1}^2+\|
u_{mx}(t)\| _{H^1}^2\right] +2(1-\delta _1)( \|
u_{m}(t)\| _{H^1}^2+\| u_{mx}(t)\|
_{H^1}^2) \\
&\leq \frac{1}{2\delta _1}\| f(t)\|_{H^1}^2.
\end{aligned}  \label{n11}
\end{equation}
Integrating \eqref{n11}, we have
\begin{equation}
\begin{aligned}
&\| u_{m}(t)\| _{H^1}^2+\| u_{mx}(t)\| _{H^1}^2 \\
&\leq \Big( \| u_{0m}\|
_{H^1}^2+\| u_{0mx}\| _{H^1}^2-R^2\Big)e^{-2(1-\delta _1)t} \\
&\quad +\Big( R^2+\frac{1}{2\delta _1}\int
_0^{t}e^{2(1-\delta _1)s}\| f(s)\|
_{H^1}^2ds\Big) e^{-2(1-\delta _1)t} \\
&\leq \Big( \| u_{0m}\| _{H^1}^2+\|
u_{0mx}\| _{H^1}^2-R^2\Big) e^{-2(1-\delta _1)t}+R^2,
\end{aligned}\label{n12}
\end{equation}
where $R^2=\underset{0\leq t\leq T}{\sup }R_1(t)$,
\begin{equation}
R_1(t)=\begin{cases}
\frac{1}{2\delta _1}\frac{1}{e^{2(1-\delta _1)t}-1}
\int_0^{t}e^{2(1-\delta _1)s}\| f(s)\|
_{H^1}^2ds, & 0<t\leq T,  \\
\frac{1}{4\delta _1(1-\delta _1)}\| f(0)\|
_{H^1}^2, & t=0.
\end{cases}  \label{n13}
\end{equation}

Therefore, if we choose $u_{0m}$ such that
$\| u_{0m}\| _{H^1}^2+\| u_{0mx}\| _{H^1}^2\leq R^2$, we
obtain from \eqref{n12} that
\begin{equation}
\| u_{m}(t)\| _{H^1}^2+\| u_{mx}(t)\|
_{H^1}^2\leq R^2,\quad \text{i.e., $T_{m}=T$  for all }m.  \label{n14}
\end{equation}

Let $\bar{B}_{m}(0,R)$ be a closed ball in the space $W_{m}$ of linear
combinations of the functions $w_1$, $w_2,\dots$, $w_{m}$, with the norm
\[
\| u_{0m}\| _{\ast }=\sqrt{\| u_{0m}\| _{H^1}^2+\| u_{0mx}\| _{H^1}^2}.
\]
Let us define
\begin{equation}
\begin{gathered}
\mathcal{F} _{m}:\bar{B}_{m}(0,R)\to \bar{B}_{m}(0,R) \\
u_{0m}\mapsto u_{m}(T).
\end{gathered} \label{n15}
\end{equation}

We prove that $\mathcal{F} _{m}$ is continuous. Let $u_{0m}$,
$\bar{u}_{0m}\in \bar{B}_{m}(0,R)$ and let $y_{m}(t)=u_{m}(t)-\bar{u}_{m}(t)$,
 where
$u_{m}(t)$ and $\bar{u}_{m}(t)$ are solutions of the system \eqref{c7}$_1$
on $[0,T]$ satisfying the initial conditions $u_{m}(0)=u_{0m}$ and $\bar{u}
_{m}(0)=\bar{u}_{0m}$, respectively. Then, $y_{m}(t)$ satisfies the
 differential equation
\begin{equation}
\begin{aligned}
&\langle y_{m}'( t) +y_{m}(t),w_{j}\rangle
+\langle y_{mx}'( t) +y_{mx}( t) ,w_{jx}\rangle \\
&+\langle \bar{\mu}(u_{mx}(t))-\bar{\mu}(\bar{u}_{mx}(t)),w_{jx}\rangle \\
&+\langle \sigma (u_{mx}(t))u_{m}(t)-\sigma (\bar{u}_{mx}(t))
\bar{u}_{m}(t),w_{j}\rangle
=0,
\end{aligned}\label{n16}
\end{equation}
$1\leq j\leq m$, with initial condition
\begin{equation}
y_{m}(0)=u_{0m}-\bar{u}_{0m}.  \label{n17}
\end{equation}
Using the same arguments as before, we can show that
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\| y_{m}(t)\| _{H^1}^2
+2\|y_{m}(t)\| _{H^1}^2+2\langle \bar{\mu}(u_{mx}(t))-\bar{\mu}(
\bar{u}_{mx}(t)),y_{mx}(t)\rangle   \\
& +2\langle \sigma (u_{mx}(t))u_{m}(t)
 -\sigma (\bar{u}_{mx}(t))\bar{u}_{m}(t),y_{m}(t)\rangle
=0.
\end{aligned}\label{n18}
\end{equation}

On the other hand, we have
\begin{gather}
\langle \bar{\mu}(u_{mx}(t))-\bar{\mu}(\bar{u}_{mx}(t)),y_{mx}(t)\rangle
\geq 0;  \label{n19} \\
\begin{aligned}
& 2\langle \sigma (u_{mx}(t))u_{m}(t)-\sigma (\bar{u}_{mx}(t))
\bar{u}_{m}(t),y_{m}(t)\rangle \\
&=2\| \sqrt{\sigma (u_{mx}(t))}y_{m}(t)\| ^2
 +2\langle \sigma (u_{mx}(t))-\sigma (\bar{u}_{mx}(t)),\bar{u}_{m}(t)y_{m}(t)\rangle .
\end{aligned}\label{n20}
\end{gather}
Putting $\tilde{K}_{R}=$ $\underset{| z| \leq \sqrt{2}R}{\sup
}| \sigma '(z)| $, we have
\begin{equation}
\begin{aligned}
&2\langle \sigma (u_{mx}(t))-\sigma (\bar{u}_{mx}(t)),\bar{u}
_{m}(t)y_{m}(t)\rangle \\
&\leq 2\| \bar{u}_{mx}(t)\|
\| y_{m}(t)\| \| \sigma (u_{mx}(t))-\sigma (\bar{u}
_{mx}(t))\| \\
&\leq 2\tilde{K}_{R}\| \bar{u}
_{mx}(t)\| \| y_{m}(t)\| \|y_{mx}(t)\| \\
&\leq \tilde{K}_{R}\| \bar{u}
_{mx}(t)\| \| y_{m}(t)\| _{H^1}^2\leq R\tilde{K
}_{R}\| y_{m}(t)\| _{H^1}^2.
\end{aligned}\label{n21}
\end{equation}
It follows from \eqref{n18}-\eqref{n21} that
\begin{equation}
\frac{d}{dt}\| y_{m}(t)\| _{H^1}^2+(2-R\tilde{K}
_{R})\| y_{m}(t)\| _{H^1}^2\leq 0.  \label{n22}
\end{equation}

Integrating  inequality \eqref{n22}, we obtain
\begin{equation*}
\| y_{m}(T)\| _{H^1}^2\leq e^{(R\tilde{K}
_{R}-2)T}\| u_{0m}-\bar{u}_{0m}\| _{H^1}^2,
\end{equation*}
or
\begin{equation}
\| \mathcal{F} _{m}(u_{0m})-\mathcal{F} _{m}(\bar{u}_{0m})\| _{H^1}
\leq \exp \Big( \big( \frac{1}{2}R\tilde{K}_{R}-1\big) T\Big)
\| u_{0m}-\bar{u}_{0m}\| _{H^1}.  \label{n23}
\end{equation}

Note that, on $W_{m}$, $\| v_{0m}\|_{H^1}$ and
$ \| v_{0m}\| _{\ast }=\sqrt{\| v_{0m}\| _{H^1}^2+\| v_{0mx}\|_{H^1}^2}$
are equivalent norms, hence, there exist two constants
 $ D_{1m}>0$, $D_{2m}>0$ such that
\begin{equation}
D_{1m}\| v_{0m}\| _{\ast }\leq \| v_{0m}\|
_{H^1}\leq D_{2m}\| v_{0m}\| _{\ast }\quad \text{for all }
v_{0m}\in W_{m}.  \label{n24}
\end{equation}
It follows from \eqref{n23}, \eqref{n24} that
\begin{equation}
\| \mathcal{F} _{m}(u_{0m})-\mathcal{F} _{m}(\bar{u}
_{0m})\| _{\ast }
\leq \frac{D_{2m}}{D_{1m}}\exp ( ( \frac{1}{2}R
\tilde{K}_{R}-1) T) \| u_{0m}-\bar{u}_{0m}\|
_{\ast }
\label{n25}
\end{equation}
 for all $u_{0m}$, $\bar{u}_{0m}\in W_{m}$.

Hence, $\mathcal{F} _{m}:\bar{B}_{m}(0,R)\to \bar{B}_{m}(0,R)$ is
continuous. Applying$\ $the$\ $fixed point theorem of Brouwer,
we have (for every $m$) a function $u_{0m}\in \bar{B}_{m}(0,R)$
such that the solution of the initial value problem \eqref{c7}
is a $T$-periodic solution of the
system \eqref{c7}$_1$. This solution satisfies the inequality \eqref{n14}
a.e., in $[0,T]$ and consequently, by \eqref{n11} we have
\begin{equation}
\begin{aligned}
&\| u_{m}(t)\| _{H^1}^2+\|
u_{mx}(t)\| _{H^1}^2+2(1-\delta_1)\int_0^{t}( \| u_{m}(s)\|
_{H^1}^2+\| u_{mx}(s)\| _{H^1}^2) ds \\
&\leq R^2+\frac{1}{2\delta _1}\int_0^{T}\| f(s)\| _{H^1}^2ds\leq C_{T}.
\end{aligned} \label{n26}
\end{equation}

On the other hand, we multiplying the $j^{th}$\ equation of
\eqref{c7}$_1$ by $c_{mj}'(t)$\ and summing up with respect to $j$, afterwards,
integrating with respect to the time variable from $0$ to $T$, we obtain after
some rearrangements
\begin{equation}
\begin{aligned}
&2\int_0^{T}\| u_{m}'(t)\| _{H^1}^2dt+\int_0^{T}\frac{d}{dt}\Big[
\| u_{m}(t)\| _{H^1}^2+2\int_0^1\tilde{\mu
}(u_{mx}(x,t))\,dx\Big]\,dt \\
&+2\int_0^{T}\langle \sigma
(u_{mx}(t))u_{m}(t),u_{m}'( t) \rangle\,dt \\
&=2\int_0^{T}\langle f(t),u_{m}'( t) \rangle\,dt,
\end{aligned}\label{n27}
\end{equation}
where $\tilde{\mu}(z)=\int_0^{z}\bar{\mu}(y)dy\geq 0$ for all
$z\in \mathbb{R}$.

From \eqref{n4}, we obtain
\begin{equation}
\begin{aligned}
&\int_0^{T}\frac{d}{dt}\left[ \|
u_{m}(t)\| _{H^1}^2+2\int_0^1\tilde{\mu}
(u_{mx}(x,t))\,dx\right]\,dt \\
&=\| u_{m}(T)\| _{H^1}^2-\| u_{m}(0)\|
_{H^1}^2+2\int_0^1\left[ \tilde{\mu}(u_{mx}(x,T))-\tilde{
\mu}(u_{mx}(x,0))\right] \,dx=0.
\end{aligned}\label{n28}
\end{equation}
Moreover,
\begin{equation}
\begin{aligned}
2\int_0^{T}\langle f(t),u_{m}'( t) \rangle\,dt
&\leq 2\int_0^{T}\| f(t)\| \|
u_{m}'(t)\|\,dt \\
&\leq 2\int_0^{T}\|
f(t)\| ^2dt+\frac{1}{2}\int_0^{T}\|
u_{m}'(t)\| ^2dt.
\end{aligned} \label{n29}
\end{equation}
Putting $\sigma _{R}=$ $\underset{| z| \leq \sqrt{2}R}{\sup }
\sigma (z)$, we have
\begin{equation}
\begin{aligned}
&2\int_0^{T}\langle \sigma (u_{mx}(t))u_{m}(t),u_{m}'( t) \rangle\,dt \\
&\leq 2\sigma _{R}\int_0^{T}\|u_{m}( t) \| \| u_{m}'( t)\|\,dt  \\
& \leq 2R\sigma _{R}\int_0^{T}\| u_{m}'( t)
\|\,dt\leq 2TR^2\sigma _{R}^2+\frac{1}{2}\int_0^{T}
\| u_{m}'(t)\| ^2dt.
\end{aligned}\label{n30}
\end{equation}

It follows from \eqref{n27}, \eqref{n28}, \eqref{n29} and \eqref{n30}, that
\begin{equation}
\int_0^{T}\| u_{m}'(t)\|_{H^1}^2dt
\leq 2TR^2\sigma _{R}^2
+2\int_0^{T}\|f(t)\| ^2dt\leq C_{T},  \label{n31}
\end{equation}
for all $m\in \mathbb{N}$, for all $t\in [ 0,T]$, where $C_{T}$
always indicates a bound depending on $T$.
\smallskip

\noindent\textbf{Step 3:}
The limiting process. By \eqref{n14} and \eqref{n31}
 we deduce that, there exists a subsequence of $\{u_{m}\}$, still denoted
by $\{u_{m}\}$ such that
\begin{equation}
\begin{gathered}
u_{m}\to u \quad \text{in  $L^{\infty }(0,T;H_0^1\cap H^2)$ weakly*,}  \\
u_{m}'\to u' \quad \text{in $L^2(0,T;H_0^1)$  weakly.}
\end{gathered}  \label{n32}
\end{equation}
From \eqref{n4}, we obtain
\begin{equation}
u(0)=u(T).  \label{n33}
\end{equation}

Using the compactness lemma of Lions \cite[p.57]{Lions} and applying
Fischer-Riesz theorem, from \eqref{n32}, there exists a subsequence of
$\{u_{m}\}$, denoted by the same symbol satisfying
\begin{equation}
\begin{gathered}
u_{m}\to u \quad \text{strongly in  $L^2(0,T;H_0^1)$
and a.e. in $Q_{T}$},  \\
u_{mx}\to u_{x} \quad  \text{strongly in  $L^2(Q_{T})$
and a.e. in $Q_{T}$}.
\end{gathered} \label{n34}
\end{equation}

Applying an argument similar to the one used in the proof of Theorem \ref{thm3.1},
 we have
\begin{equation}
\begin{gathered}
\bar{\mu}(u_{mx})\to \bar{\mu}(u_{x}) \quad \text{strongly in }
L^2(Q_{T}),  \\
\sigma (u_{mx})u_{m}\to \sigma (u_{x})u\quad \text{strongly in }
L^2(Q_{T}).
\end{gathered}  \label{n35}
\end{equation}

Denote by $\{\zeta _{i}$, $i=1,2,\dots\}$ the orthonormal base in the real
Hilbert space $L^2(0,T)$. The set $\{\zeta _{i}w_{j}$, $i$, $j=1,2,\dots\}$
forms an orthonormal base in $L^2(0,T;H_0^1)$. From \eqref{c7}$_1$
we have
\begin{equation}
\begin{aligned}
&\int_0^{T}\langle u_{m}'(t)+u_{m}(t),w_{j}\zeta _{i}(t)\rangle\,dt
+\int_0^{T}\langle u_{mx}'(t)+u_{mx}(t),w_{jx}\zeta _{i}(t)\rangle\,dt \\
&+\int_0^{T}\langle \bar{\mu}(u_{mx}(t)),w_{jx}
\zeta _{i}(t)\rangle\,dt
+\int_0^{T}\langle \sigma
(u_{mx}(t))u_{m}(t),w_{j}\zeta _{i}(t)\rangle\,dt \\
&=\int_0^{T}\langle f(t),w_{j}\zeta _{i}(t)\rangle\,dt,
\end{aligned} \label{n36}
\end{equation}
for all $i,j$, $1\leq j\leq m$, $i\in \mathbb{N}$.

For $i$ and $j$ fixed, we deduce from \eqref{n32} that
\begin{equation}
\begin{gathered}
\int_0^{T}\langle u_{m}'(t)+u_{m}(t),w_{j}\zeta
_{i}(t)\rangle\,dt\to \int_0^{T}\langle u'(t)+u(t),w_{j}\zeta _{i}(t)\rangle\,dt, \\
\int_0^{T}\langle u_{mx}'(t)+u_{mx}(t),w_{jx}\zeta
_{i}(t)\rangle\,dt\to \int_0^{T}\langle u_{x}'(t)+u_{x}(t),w_{jx}\zeta _{i}(t)\rangle\,dt.
\end{gathered} \label{n37}
\end{equation}
Furthermore, by \eqref{n35}, we have
\begin{equation}
\begin{gathered}
\int_0^{T}\langle \bar{\mu}(u_{mx}(t)),w_{jx}\zeta
_{i}(t)\rangle\,dt\to \int_0^{T}\langle \bar{\mu}
(u_{x}(t)),w_{jx}\zeta _{i}(t)\rangle\,dt, \\
\int_0^{T}\langle \sigma (u_{mx}(t))u_{m}(t),w_{j}\zeta
_{i}(t)\rangle\,dt\to \int_0^{T}\langle \sigma
(u_{x}(t))u(t),w_{j}\zeta _{i}(t)\rangle\,dt.
\end{gathered} \label{n38}
\end{equation}

Passing to the limit in \eqref{n36} by \eqref{n37}, \eqref{n38}, we have
\begin{equation}
\begin{aligned}
&\int_0^{T}\langle u'(t)+u(t),w_{j}\zeta
_{i}(t)\rangle\,dt+\int_0^{T}\langle u_{x}'(t)
+u_{x}(t),w_{jx}\zeta _{i}(t)\rangle\,dt \\
&+\int_0^{T}\langle \bar{\mu}(u_{x}(t)),w_{jx}\zeta
_{i}(t)\rangle\,dt+\int_0^{T}\langle \sigma
(u_{x}(t))u(t),w_{j}\zeta _{i}(t)\rangle\,dt \\
&=\int_0^{T}\langle f(t),w_{j}\zeta _{i}(t)\rangle\,dt.
\end{aligned} \label{n39}
\end{equation}
This equation holds for every $i$, $j\in\mathbb{N}$, i.e., the equation
\begin{equation}
\begin{aligned}
&\int_0^{T}\langle u'(t)+u(t),w(t)\rangle
\,dt+\int_0^{T}\langle u_{x}'(t)+u_{x}(t),w_{x}(t)\rangle\,dt \\
&+\int_0^{T}\langle \bar{\mu}(u_{x}(t)),w_{x}(t)
\rangle\,dt+\int_0^{T}\langle \sigma (u_{x}(t))u(t),w(t)\rangle\,dt \\
&=\int_0^{T}\langle f(t),w(t)\rangle\,dt,\quad \text{for all }w\in
L^2(0,T;H_0^1),
\end{aligned} \label{n40}
\end{equation}
is satisfied.
\smallskip

\noindent\textbf{Step 4: Uniqueness of the solutions.}
 Let $u$ and $\bar{u}$ be two
solutions of \eqref{n2} such that 
$\| u\| _{L^{\infty}(0,T;H_0^1\cap H^2)}\leq R$, $\| \bar{u}\|
_{L^{\infty }(0,T;H_0^1\cap H^2)}\leq R$, with 
$R\sup_{|z| \leq \sqrt{2}R} | \sigma '(z)| <2$. 
Then $v=u-\bar{u}$ satisfies 
\begin{equation}
\begin{gathered}
\begin{aligned}
&\int_0^{T}\langle v'(t)+v(t),w(t)\rangle\,dt
+\int_0^{T}\langle v_{x}'(t)+v_{x}(t),w_{x}(t)\rangle
\,dt  \\
& +\int_0^{T}\langle \bar{\mu}(u_{x}(t))-\bar{\mu}
(\bar{u}_{x}(t)),w_{x}(t)\rangle\,dt  \\
& +\int_0^{T}\langle \sigma (u_{x}(t))u(t)-\sigma
(\bar{u}_{x}(t))\bar{u}(t),w(t)\rangle\,dt=0, \quad \forall 
w\in L^2(0,T;H_0^1),
\end{aligned}  \\
v(0)=v(T),  \\
v, u, \bar{u}\in L^{\infty }(0,T;H_0^1\cap H^2), \quad
v', u', \bar{u}'\in L^2(0,T;H_0^1).
\end{gathered} \label{n41}
\end{equation}

Taking $w=v$ in \eqref{n41}$_1$ and using \eqref{n41}$_2$, we obtain
\begin{gather}
\int_0^{T}\langle v'(t),v(t)\rangle\,dt=\frac{1}{2}
\| v(T)\| ^2-\frac{1}{2}\| v(0)\|^2=0;  \label{n42} \\
\int_0^{T}\langle v_{x}'(t),v_{x}(t)\rangle\,dt=\frac{1
}{2}\| v_{x}(T)\| ^2-\frac{1}{2}\|
v_{x}(0)\| ^2=0;  \label{n43} \\
\int_0^{T}\langle \bar{\mu}(u_{x}(t))-\bar{\mu}(\bar{u}
_{x}(t)),v_{x}(t)\rangle\,dt\geq 0;  \label{n44} \\
\begin{aligned}
&\int_0^{T}\langle \sigma (u_{x}(t))u(t)-\sigma (
\bar{u}_{x}(t))\bar{u}(t),v(t)\rangle\,dt \\
&=\int_0^{T}\| \sqrt{\sigma (u_{x}(t))}v(t)\|
^2dt+\int_0^{T}\langle [ \sigma (u_{x}(t))-\sigma (\bar{u}
_{x}(t))] \bar{u}(t),v(t)\rangle\,dt.
\end{aligned}\label{n45}
\end{gather}

As for \eqref{n21}, we have
\begin{equation}
\int_0^{T}\langle [ \sigma (u_{x}(t))-\sigma (\bar{u}
_{x}(t))] \bar{u}(t),v(t)\rangle\,dt\leq \frac{1}{2}R\tilde{K}
_{R}\int_0^{T}\| v(t)\| _{H^1}^2dt,
\label{n46}
\end{equation}
with $\tilde{K}_{R}=\sup_{| z| \leq \sqrt{2}R} | \sigma '(z)| $.
Hence
\begin{equation}
\begin{aligned}
&\int_0^{T}\| v(t)\|_{H^1}^2dt+\int_0^{T}\langle \bar{\mu}(u_{x}(t))-\bar{\mu}(
\bar{u}_{x}(t)),v_{x}(t)\rangle\,dt 
+\int_0^{T}\| \sqrt{\sigma (u_{x}(t))} v(t)\| ^2dt \\
&\leq \frac{1}{2}R\tilde{K}_{R}\int_0^{T}\| v(t)\| _{H^1}^2dt.
\end{aligned} \label{n47}
\end{equation}
By $\frac{1}{2}R\tilde{K}_{R}=\frac{1}{2}R\underset{|
z| \leq \sqrt{2}R}{\sup }| \sigma '(z)| <1$, we deduce 
from \eqref{n47} that $\int_0^{T}
\| v(t)\| _{H^1}^2dt=0$, i.e., $v=u-\bar{u}=0$.
This completes the proof.
\end{proof}

\subsection*{Acknowledgments}
 The authors wish to express their gratitude
to the anonymous referees and the editor for their valuable comments.

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\end{document}