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\begin{document}
\title[Qualitative analysis and stabilization]{Quasilinear parabolic problem with variable exponent: qualitative analysis and stabilization}

\author[J. Giacomoni, V. R{\u a}dulescu and G. Warnault]{Jacques Giacomoni, Vicen\c tiu R{\u a}dulescu and Guillaume Warnault}

\address[J. Giacomoni]{Laboratoire de Math\'ematiques et leurs Applications Pau, UMR CNRS 5142,
Universit\'e de Pau et Pays de l'Adour, Avenue de l'Universit\'e, BP 1155, 64013 Pau Cedex,
 France}
\email{jacques.giacomoni@univ-pau.fr}

\address[V. R{\u a}dulescu]{Department of Mathematics, Faculty of Sciences,
King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia  \& Department of Mathematics, University of Craiova, Street A.I. Cuza 13,
          200585 Craiova, Romania}
\email{vicentiu.radulescu@imar.ro}

\address[G. Warnault]{Laboratoire de Math\'ematiques et leurs Applications Pau, UMR CNRS 5142, Universit\'e de Pau et Pays de l'Adour, Avenue de l'Universit\'e, BP 1155, 64013 Pau Cedex, France}
\email{guillaume.warnault@univ-pau.fr}

%----------classification, keywords, date
\subjclass{Primary 35K55, 35J62; Secondary 35B65}

\keywords{Leray-Lions operator with variable exponents, parabolic equation, local and global in time existence, stabilization}

\date{May 5, 2017}

\begin{abstract}
We discuss the existence and uniqueness of the weak solution
of the following nonlinear parabolic problem
\begin{equation*}
\left\{
\begin{array}{ll}
u_t-{\nabla \cdot\mathbf{a}(x,\nabla u)} = f(x,u)&\text{in } Q_T\eqdef (0,T)\times\Om, \\
u  = 0 &\text{on  }  \Sigma_T\eqdef (0,T)\times\pa\Om,\\
u(0,x)=u_0(x)&\text{in }\Om ,
\end{array}\right. \tag{$P_T$} \end{equation*}
which involves a quasilinear elliptic operator of Leray-Lions type with variable exponents. Next, we discuss the global behaviour of solutions and in particular the convergence to a stationary solution as $t\to\infty$.
\end{abstract}

%%% ----------------------------------------------------------------------
\maketitle
%%% ----------------------------------------------------------------------
\tableofcontents
\section{Introduction and main results}
\noindent

%{\it\footnotesize 2000 Mathematics Subject Classification.}
%{\scriptsize }\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent 
Let $\Om\subset\R^d$ ($d\geq 2$) be a bounded domain with smooth boundary (at least $C^2$).
Our main goal in this paper is to study the existence, uniqueness and global behaviour of the weak solutions to the following problem involving a quasilinear elliptic operator of Leray-Lions type with variable exponents:
 \begin{equation*}\label{Pt}
\left\{
\begin{array}{ll}
u_t-{\nabla\cdot \mathbf{a}(x,\nabla u)}= f(x,u)&\text{in } Q_T:=(0,T)\times\Om, \\
u  = 0 &\text{on  }  \Sigma_T:=(0,T)\times\pa\Om,\\
u(x,0)=u_0(x)&\text{in }\Om\,.
\end{array}\right. \tag{$P_T$}
\end{equation*}
We assume that  $f: \Omega\times \R\mapsto \R$ satisfies
\begin{enumerate}[label=({$ f_\arabic*$})]
\item$0\not\equiv\, f\colon (x,s)\to f(x,s)$ is a Carath\'eodory function and $\ t\mapsto f(x,t)$ is locally Lipschitz uniformly in $x\in\Omega$;
\setcounter{saveenum}{\value{enumi}}
\end{enumerate}
and $\mathbf{a}(x,\xi)=(a_j(x,\xi))_j$ with $a_j(x,\xi)=\phi(x,\vert\xi\vert)\xi_j$, $j=1,\cdot\cdot\cdot,\, d$ be defined  for all $\xi\in\R^d$, such that $\phi$ is differentiable on $\Omega\times (0,\infty)$ and $\phi(x,s)>0$ for $(x,s)\in\Omega\times (0,\infty)$. We assume that $\mathbf{a}$ satisfies the following structural conditions:
\begin{enumerate}[label=({\bf A\arabic*})]
\item $a_j(x,\mathbf{0})=0, \quad\mbox{for all a.e. }x\in\Omega$,
\item $a_j\in C^1\left(\Omega\times\left(\R^d\backslash\{\mathbf{0}\}\right)\right)\cap C^0\left(\Omega\times\R^d\right)$,
\item $\displaystyle\sum_{i,j=1}^d\frac{\partial a_j(x,\xi)}{\partial\xi_i}\eta_i\eta_j\geq\gamma\vert\xi\vert^{p(x)-2}\cdot\vert\eta\vert^2,\quad\forall\, x\in\Omega,\,%\xi\in\R^d\backslash\{\mathbf{0}\}
,\, \forall \xi\in\R^d\backslash\{\mathbf{0}\}$,  $\forall\eta\in\R^d$,
\item $\displaystyle\sum_{i,j=1}^d\left\vert\frac{\partial a_j(x,\xi)}{\partial\xi_i}\right\vert\leq\Gamma\vert\xi\vert^{p(x)-2}$, $\quad\forall x\in\Omega,\;\xi\in\R^d\backslash\{\mathbf{0}\}$,
\end{enumerate}
where $p:\Om \mapsto ]1,+\infty[$ is a Lebesgue measurable function satisfying $1<p^-:=\inf_{\bar \Om} p(x)\leq p(x)\leq p^+:=\sup_{\bar \Om} p(x)<\infty$.
%\item There exists a positive constant $c_1>0$ such that $\vert \mathbf{a}(x,\xi)\vert\leq c_1(1+\vert\xi\vert^{p(x)-1})$.
%\item There exists $C>0$ such that for any $\xi \in \R^d\backslash\{0\}$
%$$\textcolor{red}{|a(x,\xi)|\leq C(1+|\xi|^{p(x)-2})}$$
%where .
%\item For any $\xi, \eta\in \R^d$,
%$$(a(x,\xi)-a(x,\eta)).(\xi-\eta)\geq 0.$$
%\item There exists $k>0$ such that for any $\xi, \eta\in \R^d$,
%$$A(x,\frac{\xi+\eta}{2}) \leq \frac12(A(x,\xi)+A(x,\eta))-k|\xi-\eta|^{p(x)}.$$
%\item For any $\xi\in \R^d$,
%$$|\xi|^{p(x)}\leq a(x,\xi).\xi\leq p(x)A(x,\xi).$$
{More precisely, throughout the paper, we assume  that
$$p\in C(\overline \Om)\cap\mathcal P^{log}(\Om) \mbox{ such that
 } 1<p^{-}\leq p^+< d,$$ % to have suitable embeddings (see Theorem \ref{SE1}),
  where
$$\mathcal P^{log}(\Om)\eqdef \left\{q\in\mathcal P(\Om)\,:\, \frac{1}{q} \mbox{ globally log-H\"older continuous}\right\}.$$
In particular, for any $p\in\mathcal P^{\log}(\Om)$, there exists a function $w$ such that for all $(x,y)\in \Omega^2$
$$ |p(x)-p(y)|\leq w(|x-y|) \ \mbox{and} \ \limsup_{t\ra0^+} \left(-w(t)\ln t\right) <+\infty.$$

 The natural {function} spaces to study problem \eqref{Pt} are the variable exponent  Lebesgue space $L^{p(x)}(\Omega)$ and { the} corresponding Sobolev space $W^{1,p(x)}(\Omega)$  defined by
$$L^{p(x)}(\Om)=\left\{u\,|\,u\mbox{ is measurable on $\Om$ and }\ \displaystyle\int_\Omega |u|^{p(x)}dx<\infty\right\}$$
and
$$W^{1,p(x)}(\Om)\eqdef \left\{u\in L^{p(x)}(\Om):|\nabla u|\in L^{p(x)}(\Om)\right\}.$$
We recall that $L^{p(x)}(\Omega)$ and $W^{1,p(x)}(\Omega)$ are normed linear spaces equipped respectively with the {(Luxemburg)} norms
$$\|u\|_{L^{p(x)}(\Om)}= \inf\left\{\la>0:\int_\Omega \left|\frac{u}{\lambda}\right|^{p(x)}dx\leq 1\right\}$$
and
$$\|u\|_{W^{1,p(x)}}=\|u\|_{L^{p(x)}(\Om)}+\|\nabla u\|_{L^{p(x)}(\Om)}.$$

Under the {above} assumptions on $p$, we define  $\X\eqdef W^{1,p(x)}_0(\Omega)$ the closure of $C^\infty_0(\Om)$ in $W^{1,p(x)}(\Om)$. {Since $\Om$ is a bounded domain,} the Poincar\'e inequality holds and {a natural norm of $\X$ is} $\|u\|_\X=\|\nabla u\|_{L^{p(x)}(\Om}$.}

Now, we define the even function $\Phi:\Omega\times\R\to\R$  by $\Phi(x,t)=\int_0^t\phi(x,\vert s\vert)s\,\mathrm{d}s$, {which is} increasing on $\R^+$, and for a.e. $x\in \Omega$, $\xi\to A(x,\xi)\eqdef \Phi(x,\vert\xi\vert)$.

Furthermore, from ({\bf A1}), ({\bf A3}) and ({\bf A4}), $A$ is strictly convex and satisfies for any fixed $x\in\Omega$
%
\begin{eqnarray}
& \frac{\gamma}{p^+-1}\, |\xi|^{p(x)}
    \leq A(x,\xi) \leq
  \frac{\Gamma}{p^--1}\, |\xi|^{p(x)}
  \quad\mbox{ for all }\ \xi\in \R^d .
\label{growth.A}
\end{eqnarray}
%
These inequalities are a direct consequence of Taylor's formula
combined with ({\bf A3}) and~({\bf A4}),
which yield
\[
  \frac{\gamma}{p^+-1}\, |\xi|^{p(x)}
    \leq A(x,\xi) - A(x,\mathbf{0})
  - \langle \partial_\xi A(x,\mathbf{0}) , \xi\rangle
    \leq
  \frac{\Gamma}{p^--1}\, |\xi|^{p(x)}
\]
for all $(x,\xi)\in \Omega\times \R^d$. Similarly, using ({\bf A4}), we deduce that there exists a positive constant $c_1$ such that
\begin{eqnarray}\label{control-a}
\vert\mathbf{a}(x,\xi)\vert\leq c_1\vert\xi\vert^{p(x)-1},\; \phi(x,\vert\xi\vert)\leq c_1\vert\xi\vert^{p(x)-2}\quad\mbox{for all }x\in\Omega\mbox{ and }\xi\in\R^d.
\end{eqnarray}

We assume that
\begin{enumerate}
\item[({\bf A5})] $(0,\infty)\ni t\to\Phi(x,\sqrt{t})\mbox{ is convex for a.e. }x\in\Omega^+\eqdef\{z\in\Omega; \, p(z)\geq 2\}.$
\end{enumerate}

From the above assumption together with $\Phi(x,0)=0$ and $\Phi(x,\cdot)$ increasing, we obtain a Clarkson-type inequality for the function $\Phi$ (see Lemma 2.1 in \cite{La} for the proof). More precisely, we have
\begin{equation}\label{clarkson}
\begin{split}
\frac{1}{2}\left[\int_{\Omega^+}\Phi(x,\vert\nabla u\vert)\,\mathrm{d}x\right.+\left.\int_{\Omega^+}\Phi(x,\vert\nabla v\vert)\,\mathrm{d}x\right]\geq
\int_{\Omega^+}&\Phi\left(x,\frac{|\nabla (u+ v)|}{2}\right)\,\mathrm{d}x\\
&+\int_{\Omega^+}\Phi\left(x,\frac{|\nabla (u- v)|}{2}\right)\,\mathrm{d}x
\end{split}
\end{equation}
for all $u$, {$v\in \X$}.

From the pointwise version of \eqref{clarkson} {(see Lemma 2.1  in \cite[p. 459]{La})} and from \eqref{growth.A}, we deduce that for $x\in\Omega^+$, $A(x,\cdot)$ is $p(x)$-uniformly convex (see Definition 2.2 in \cite{NaMa} and \cite{FaGu}), that is,
\begin{eqnarray}\label{uniform-convexity}
\mbox{for any }\xi, \eta\in \R^d, \quad A(x,\frac{\xi+\eta}{2}) \leq \frac12(A(x,\xi)+A(x,\eta))-c_0|\xi-\eta|^{p(x)}
\end{eqnarray}
with $c_0=\frac{\gamma}{2^{p^+}(p^+-1)}$.
\begin{rem}
From the Clarkson inequality (see \cite[p. 96]{Br}), relation \eqref{uniform-convexity} is satisfied by $A(x,\xi)=\vert\xi\vert^{p(x)}$ with $p$ satisfying $p(x)\geq 2$. In the case $1<p(x)<2$, a similar inequality (namely, the Morawetz inequality) can be derived (see Lemma 4.1 in \cite{GV}).
\end{rem}

\begin{rem}
\begin{enumerate}[label=\arabic*)]
\item Prototype examples satisfying conditions {({\bf A1})-({\bf A5})} below are: a) $\phi(x,t)=\phi_1(x,t)\eqdef  t^{p(x)-2}$, b) $\phi(x,t)=\phi_2(x,t)\eqdef (1+ t^2)^{\frac{p(x)-2}{2}}$.
\item Assumption ({\bf A4}) can be replaced by {\bf(A4')} {for the validity of the }results in Section 1: there exists $c>0$ such that $\vert\phi(x,s)s\vert\leq c (1+\vert s\vert^{p(x)-1})$ for all $x \in\Omega$ and $s\in\R$.
\end{enumerate}
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

There is an abundant literature devoted to questions on existence and uniqueness of solutions to \eqref{Pt} for $\phi(x,\xi)=\vert\xi\vert^{p(x)-2}$ and $p(x)\equiv p$ (see for instance \cite{BBG} and references
therein).
More recently, parabolic and elliptic problems with variable exponents have been studied
quite extensively, see for example \cite{AMS,AS1,AS2,AS3,CLR,GTW,MR,R}.
The importance of investigating these problems lies in their occurrence in modeling
various physical problems involving strong anisotropic phenomena related to
electrorheological fluids (an important class of non-Newtonian fluids)
\cite{AMS,RR,R}, image processing \cite{CLR}, elasticity \cite{Zhikov}, the processes of filtration in complex media \cite{AntShm}, stratigraphy problems \cite{GV} and also mathematical biology \cite{Frag}.

 Regarding the existing literature on quasilinear parabolic equations with variable exponent, we consider in the present paper a more general class of operators of Leray-Lions type than the $p(x)$-Laplace operator and prove new local existence and regularity results for $(P_T)$ (see Theorems~\ref{PT1} and \ref{continuity u with f}). By constructing new barrier functions, we also provide global existence results (see Theorems~\ref{PT3} and \ref{PT3b}) for the general class of quasilinear parabolic problems considered in the paper that are not known in the literature.

A natural issue is to consider the asymptotic behaviour of the obtained global solutions. For this purpose, introducing an homogeneity condition, say ({\bf A7}), we prove new uniqueness results for the stationary equation associated to $(P_T)$ (see Theorem~\ref{uniqueness-stat}). The proof of the uniqueness results uses  the extension of  convexity properties of the associated energy functional, in the spirit of works \cite{Brezis-Osw} and \cite{Diaz-Saa}, proved in \cite{GiTa}. Combining Theorem~\ref{uniqueness-stat} and Theorem~\ref{continuity u with f}, we are able to prove under suitable assumptions the asymptotic convergence to the stationary solution for global solutions to $(P_T)$ (see Theorem~\ref{asymptotic-behaviour}), extending significantly former results proved in \cite{GTW}. More precisely, regarding the main results established in \cite{GTW}, we first extend by using monotone methods the local existence results for the general class of operators satisfying
  ({\bf A1})-({\bf A5}) (see Theorems~\ref{PT1} and \ref{sol for (S_t)} below). Next, in respect to \cite{GTW}, using new barrier functions and new uniqueness results for the stationary equation (see Theorem~\ref{uniqueness-stat} below), we improve essentially global existence results and asymptotic convergence to a stationary solution (see below Theorems~\ref{PT3}, \ref{PT3b} and \ref{asymptotic-behaviour} respectively). We point out that even if restricting to the $p(x)$-operator, these results are new.

We now state the main results that we will prove in the next sections.

We first  consider the following problem:
\begin{equation*}\label{Lt}
\left\{\begin{array}{ll}
u_t-\nabla\cdot \mathbf{a}(x,\nabla u) = h(x,t) &\text{in } Q_T,  \\
 u  = 0 &\text{on  }
 \Sigma_T,\\
u(x,0)=u_0(x)&\text{in }\Om,
\end{array}\right.\tag{$L_T$}
 \end{equation*}
where $T>0$, $h\in L^2(Q_T)\cap L^q(Q_T)$, $q>\displaystyle \frac{d}{p^-}$. 

Considering the initial data  $u_0\in\X\cap L^\infty(\Om)$, we study the weak solutions of problem \eqref{Lt} defined as follows.
\begin{defi}\label{defi1}
A weak solution to \eqref{Lt} is any function $u\in L^\infty(0,T;\X)$ such that $u_t\in L^2(Q_T)$ and satisfying for any $\varphi\in
C^\infty_0(Q_T)$
\begin{equation*}
\ii u_t\varphi\,\mathrm{d}x\mathrm{d}t+\ii \mathbf{a}(x,\nabla u)\cdot\nabla\varphi\,\mathrm{d}x\mathrm{d}t=\ii
h(x,t)\varphi\,\mathrm{d}x\mathrm{d}t
\end{equation*}
and $u(0,.)=u_0$ a.e. in $\Omega$.
\end{defi}

 We define in {a similar} way {the notion of} weak solutions to the problem \eqref{Pt} as follows:
\begin{defi}\label{ws2}
A solution to \eqref{Pt} is a function $u\in L^\infty(0,T;\X)$ such that $u_t\in L^2(Q_T)$, $f(.,u)\in
L^\infty(0,T,L^2(\Omega))$ and for any $\varphi\in C^\infty_0(Q_T)$
\begin{equation*}
\ii u_t\varphi\,\mathrm{d}x\mathrm{d}t+\ii \mathbf{a}(x,\nabla u)\cdot\nabla\varphi\,\mathrm{d}x\mathrm{d}t=\ii
f(x,u)\varphi\,\mathrm{d}x\mathrm{d}t
\end{equation*}
and $u(0,.)=u_0$ a.e. in $\Omega$.
\end{defi}

 {According to} the above definitions, we establish the following local existence result proved in Section~\ref{local-exist} {and which extends Theorem~2.3 in \cite{GTW}}.
\begin{thm}\label{sol for (S_t)}
Assume that conditions ({\bf A1})-({\bf A5}) are satisfied and $p^->\frac{2 d}{d+2}$. Let $T>0$, $u_0\in \X\cap L^\infty(\Om)$ and $h\in L^2(Q_T)\cap L^q(Q_T)$, $q>\frac{d}{p^-}$. Then problem \eqref{Lt} admits a
unique solution $u$ in the sense of Definition \ref{defi1}. Moreover, $u\in C([0,T],\X).$
\end{thm}

{ Next we deal with the local existence of solutions of problem \eqref{Pt} proved in Section~\ref{local-exist-u} (Subsection~\ref{local}) {which improves Theorem~2.4 in \cite{GTW}}.
\begin{thm}\label{PT1}
Assume that conditions ({\bf A1})-({\bf A5}) are fulfilled. Let $f: \Omega\times \R\mapsto \R$ satisfying ($f_1$) and:
\begin{enumerate}[label=({$ f_\arabic*$})]\setcounter{enumi}{\value{saveenum}}
%\item[$\bf{(f_1')}$] $\ t\ra f(x,t)$ is locally Lipschitz uniformly in $x\in\Omega$;\medskip
\item there exists  $s_0\in \R$  such that $x\mapsto f(x,s_0)\in L^2(\Om)\cap L^q(\Om)$, $q>\frac{d}{p^-}$.
\setcounter{saveenum}{\value{enumi}}
\end{enumerate}
Assume in addition that one of the following hypotheses holds:
\begin{itemize}
\item[(H1)] there exists a nondecreasing locally Lipschitz function $L_0$ such that
\begin{equation*}%\label{condf}
  |f(x,v)|\leq  L_0(v), \quad a.e.\  (x,v)\in \Omega\times \R;
\end{equation*}
\item[(H2)] there exist two
nondecreasing locally Lipschitz functions $L_1$ and $L_2$ such that
\begin{equation*}%\label{condf2}
 L_1(v)\leq f(x,v)\leq  L_2(v), \quad a.e.\ (x,v)\in \Omega\times \R.
\end{equation*}
\end{itemize}
Then, for any $u_0\in \X\cap L^\infty(\Omega)$,  there exists $\tilde T\in (0,+\infty]$ such that for any $T\in [0, \tilde T)$, problem \eqref{Pt} admits a unique solution $u$ in the sense of Definition \ref{ws2}. Moreover, for
any
$r> 1$, we have $u\in C([0,T];L^r(\Om))\cap C([0,T];\X)$.
\end{thm}
In  Appendix~\ref{mild-sol}, using the theory of m-accretive operators, we also provide additional regularity results for weak solutions to $(P_T)$ (see in particular Theorem~\ref{continuity u with f}).

Under additional hypothesis about the growth of $f$ and regularity of the initial data, we are able to prove the existence of global solutions. Precisely, we have the following results showed in Section~\ref{local-exist-u} (Subsection~\ref{global}) {which gives sharper conditions on $f$ than those in Theorem~2.5 in \cite{GTW}} :
\begin{thm}\label{PT3}
Assume that conditions ({\bf A1})-({\bf A5}) are fulfilled. Let $f: \Omega\times \R\mapsto \R$ satisfying ($f_1$) and:
\begin{enumerate}[label=({$ f_\arabic*$})]\setcounter{enumi}{\value{saveenum}}
\item there exists $s_1\in \R$ such that $x\mapsto f(x,s_1)\in L^\infty(\Omega)$;
\item {uniformly with respect to} $x\in \Omega$ we have
$$\limsup_{\vert s\vert\to \infty}\frac{\vert f(x,s)\vert}{\vert s\vert^{p^--1}}<\gamma  \Lambda^{p^-}(p_c)^-$$
where $\Lambda:=\left(\sup_{\|u\|_\X=1} \|u\|_{L^{p^-}(\Om)}\right)^{-1}$ and {$(p_c)^-=\frac{p^-}{p^+-1}$};
\item  {uniformly with respect to} $x\in \Omega$ we have
$$\liminf_{s\to 0}\frac{|f(x,s)|}{|s|^{p^--1}} > \Gamma \Lambda^{p^-}(p^-)_c$$
{where $(p^-)_c=\frac{p^-}{p^--1}$}.
\setcounter{saveenum}{\value{enumi}}
\end{enumerate}
Assume in addition that:
\begin{itemize}
\item[(C1)] $u_0\in\X$ such that $\nabla\cdot \mathbf{a}(x,\nabla u_0)\in L^q(\Om)$ where $q> \frac{d}{p^-}$;
\end{itemize}
Then, for any $T>0$, problem \eqref{Pt} admits a unique  weak solution in the sense of Definition \ref{ws2}. Moreover $u\in C([0,T];\X)$.
\end{thm}
\noindent Furthermore, under the following new hypothesis
\begin{itemize}
\item[({\bf A6})] $\displaystyle\sum_{i=1}^d\left\vert\frac{\partial \phi(x,s)}{\partial x_i}\right\vert\leq C_1(1+\vert\ln(s)\vert)\phi(x,s)\;\mbox{ for all }x\in\Omega,\,s\in (0,\infty)$
\end{itemize}
{ and constructing a suitable supersolution, we prove the existence of bounded} global solutions of \eqref{Pt} with a regular initial data and releasing hypothesis on $f$ :
\begin{thm}\label{PT3b}
Assume conditions ({\bf A1})-({\bf A6}) and $p\in C^\beta(\overline \Om)$, $\beta\in(0,1)$ . Let $f: \Omega\times \R\mapsto \R$ satisfying ($f_1$), ($f_3$) and ($f_4$). Assume in addition
\begin{itemize}
\item[(C2)] $u_0\in C^1_0(\overline \Omega)$.
\end{itemize}
Then, for any $T>0$, problem \eqref{Pt} admits a unique {bounded} weak solution in the sense of Definition \ref{ws2}. Moreover, $u\in C([0,T];\X)$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exa}
A prototype example for $f$ satisfying all conditions ($f_1$)-($f_5$) is $f(x,t)=h(x)t^{r(x)-1}$ with $0\not\equiv\, h\in L^\infty(\Omega)$ nonnegative, $r\in C(\overline{\Omega})$ such that $1<r(x)< p^-$ for all $x\in \overline{\Omega}$.
\end{exa}

\begin{rem}
In the case $\phi(x,t)=\vert t\vert^{p(x)-2}$, $p$ belonging to $C^1(\overline \Omega)$ is a sufficient condition to have ({\bf A6}).
\end{rem}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%
%
% Apr�s le th�or�me de stabilisation


In section~\ref{asympt-conv}, we discuss the question of the asymptotic behaviour of solutions to \eqref{Pt}. {For this purpose, under suitable conditions,  we prove the existence and the uniqueness of the positive stationary solution. Precisely, assuming in addition }
%
\begin{enumerate}[label=({$ f_\arabic*$})]\setcounter{enumi}{\value{saveenum}}
\item the function $\displaystyle s\mapsto \frac{f(x,s)}{s^{p^--1}}$ is nonincreasing for a.e. $x\in \Om$,
\setcounter{saveenum}{\value{enumi}}
\end{enumerate}
%
we have the following {result, even new for the $p(x)$-operator (i.e. $A(x,\xi)=\vert \xi\vert^{p(x)}$):}
%
\begin{thm}\label{uniqueness-stat}
%Let ({\bf A5}) be satisfied.
Assume that  $p\in C^\beta(\overline{\Omega})$, $\beta\in(0,1)$ such that $p(.)\not\equiv p^-$. Let $f:\Omega\times[0,+\infty)\mapsto [0,+\infty)$ be a nonnegative function satisfying ($f_1$), ($f_4$)-($f_6$) and $f(x,0)=0$. Assume conditions ({\bf A1})-({\bf A4}) {and ({\bf A6})}. We suppose in addition that:
\begin{enumerate}
\item[({\bf A7})]  $\xi\mapsto A(x,\xi)$ is $p(x)$-homogeneous, that is,
\begin{equation*}
 A(x,\xi) =
  A\left( x,\, \frac{\xi}{|\xi|} \right)\cdot |\xi|^{p(x)} \,,
    \quad\mbox{ for all }\, \xi\in \R^N\setminus \{\mathbf{0}\};
\end{equation*}
%\item[({\bf A8})] the function $\xi\mapsto A(x,\xi)^{p^-/p(x)}:\R^d\mapsto \R^+$ is strictly convex for every $x\in %\Omega$.
\end{enumerate}
Then, {the stationary problem associated with $({\rm P}_T)$ possesses
a unique nonnegative and nontrivial weak solution}
$u\in \X\cap L^\infty(\Omega)$.
This solution belongs to the class
$C^{1+\alpha}(\overline{\Omega})$, for some $\alpha\in (0,1)$,
and satisfies  the Hopf maximum principle, namely
%
\begin{equation*}%\label{Hopf:problem_1}
  u(x) > 0 \;\mbox{ for all }\, x\in \Omega
  \quad\mbox{ and }\quad
  \frac{\partial u}{\partial\boldsymbol{\nu}} (x) < 0
    \;\mbox{ for all }\, x\in \partial\Omega \,.
\end{equation*}
\end{thm}

The above uniqueness result implies {the following result which improves significantly Theorem~2.10 in \cite{GTW}}.
\begin{thm}\label{asymptotic-behaviour}
Assume hypothesis in Theorem~\ref{uniqueness-stat} and ({\bf A5}) be satisfied.
Let $u_0\in C^{1,+}_0(\overline{\Omega})$ where $C^{1,+}_0(\overline{\Omega})$ denotes the interior of the positive cone of $C^1_0(\overline{\Omega})$. Then, for any $T>0$, there exists a unique weak solution, $u\in C([0,T],\X)\cap L^\infty(Q_T)$, to $({\rm P}_T)$ with initial data $u_0$ and such that $\frac{\partial u}{\partial t}\in L^2(Q_T)$ and $u>0$ in $Q_T$. Furthermore, $u$ verifies
\begin{eqnarray*}%\label{convergence}
u(t)\to u_\infty\quad\mbox{in }L^\infty(\Omega)\quad \mbox{as } t\to\infty
\end{eqnarray*}
where $u_\infty$ is the unique positive stationary solution to $({\rm P}_T)$ given in Theorem \ref{uniqueness-stat}.
\end{thm}
%
\begin{rem}
Alternatively to the hypothesis $p(.)\not\equiv p^-$ in Theorems~\ref{uniqueness-stat} and \ref{asymptotic-behaviour}, we can assume instead of ($f_6$), ($f_6'$): the function $\displaystyle s\mapsto \frac{f(x,s)}{s^{p^--1}}$ is strictly decreasing for a.e. $x\in \Om$.
\end{rem}
%
\begin{rem}
Theorems~\ref{uniqueness-stat} and  \ref{asymptotic-behaviour} hold for other kind of nonlinearities. Indeed, let $f:\Omega\times (0,\infty)\to\mathbb R$ be defined as $f(x,t)= h(x)t^{r(x)-1}-g(x)t^{s(x)-1}$ with $r,s\in C(\overline{\Omega})$ satisfying $1<r(x)\leq p^-\leq s(x)$ with $r^+\neq s^-$ in $\Omega$. Suppose that $h,g\in L^\infty(\Omega)$, are nonnegative and such that $h\not\equiv 0$ and $\frac{h(x)}{g(x)}$ bounded in $\Omega$. Then, statements of Theorems~\ref{uniqueness-stat} and \ref{asymptotic-behaviour} hold and their proofs are similar. We point out that any weak solution $u$ satisfies the uniform bound $u(t,x)\leq \left(\left\Vert\frac{h}{g}\right\Vert_{L^\infty(\Omega)}+1\right)^{\frac{1}{s^--r^+}}$ in $\Omega$.
\end{rem}
%
\begin{rem}
Under an additional asymptotic super homogeneous growth assumption on $f$ and for initial data large enough, blow up in finite time of solutions can also occur. For instance, let $f(x,v)=v^q$ with $q>p^+$ and define the energy functional
\begin{equation*}
E(u)\eqdef\int_\Omega A(x,\nabla u)\,\mathrm{d}x-\int_\Omega\frac{u^{q+1}}{q+1}\,\mathrm{d}x.
\end{equation*}
Then, using a well-known energy method and for any initial data $u_0$ satisfying $E(u_0)<0$, the weak solution to \eqref{Pt} blows up in finite time. For further discussions of global behaviour of solutions (blow up, localization of solutions, extinction of solutions) to quasilinear anisotropic  parabolic  equations involving variable exponents, we refer to \cite{AS2} and \cite{AS3}.
\end{rem}
%
{
\begin{rem}\label{r1.7}
Condition ({\bf A7}) implies that $A(x,\xi)=\frac{1}{p(x)}\langle a(x,\xi),\xi\rangle$. Examples satisfying ({\bf A1})-({\bf A5}) and ({\bf A7}) are given by functions of the form $\phi(x,t)=h(x)t^{p(x)-2}$ where $h\in L^\infty(\Omega)$ such that $h\geq c>0$.
\end{rem}
}
}

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%                      PROOFS                        %
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\section{Existence of solution to \eqref{Lt}}\label{local-exist}
\noindent For the proof of Theorem \ref{sol for (S_t)}, we first consider the following quasilinear elliptic problem:
\begin{equation*}\label{(P)}
\Biggl\{\begin{array}{ll}
u-\la \nabla\cdot \mathbf{a}(x,\nabla u) = g&\text{ in } ~\Om  \\
%u  > 0 ~&\text{  in  }~ \Omega\\
u  = 0 ~&\text{  on   }~
 \pa\Om
\end{array}\tag{P}  \end{equation*}
with $\la>0$ and $g$ a measurable function.  So we have:
\begin{lem}\label{sol of (P)}
Assume conditions ({\bf A1})-({\bf A4}). Let $g\in L^q(\Omega)$, $q>\frac{d}{p^-}$. Then for any $\lambda>0$, problem $(\rm P)$ admits a { unique weak solution $u\in \X$ satisfying}
$$\int_\Om u\varphi\,\mathrm{d}x+\la\int_\Om \mathbf{a}(x,\nabla u).\nabla \varphi\,\mathrm{d}x=\int_\Om g\varphi\,\mathrm{d}x, \quad \forall
\varphi \in \X.$$
{Furthermore}, $u\in L^\infty(\Omega)$.
\end{lem}
\begin{proof}
We define the energy functional $J_\la$ { associated} to \eqref{(P)} given by
\[J_\la(u)=\frac{1}{2}\int_\Om u^2\,\mathrm{d}x+\la\int_\Om A(x,\nabla u)\,\mathrm{d}x-\int_\Om gu\,\mathrm{d}x.\]
From \eqref{growth.A}, $J_\la$ is well-defined and continuously differentiable on $\X$. Indeed, $q>\frac{d}{p^-}$ and $1<p(.)<d$ a.e. in $\Omega$ imply that $L^q\subset (L^{p^*(x)})'$.
%Indeed, \eqref{condq} implies $\X\subset L^{q_c(x)}$.
By \eqref{growth.A}, for {$\|u\|_{\X}\geq 1$}:
\begin{align*}
 J_\la(u)&=\frac{1}{2}\int_\Om u^2\,\mathrm{d}x+\la\int_\Om A(x,\nabla u)\mathrm{d}x-\int_\Om gu\,\mathrm{d}x
\geq\frac{\lambda\gamma}{p^+(p^+-1)}\|u\|_{\X}^{p^{-}}-C\|u\|_{\X}.
\end{align*}
Thus $J_\la$ is coercive. Hence $J_\lambda$ admits a global minimizer $u\in \X$ which is a weak solution to \eqref{(P)}. By ({\bf A3}), $J_\lambda$ is strictly convex on $\X$, which guarantees the uniqueness of the critical point and the uniqueness of the solution to \eqref{(P)}.

To conclude, Corollary \ref{reg2} in Appendix \ref{ApC} implies $u\in L^\infty(\Omega)$.
\end{proof}
\noindent\textbf{Proof of Theorem  \ref{sol for (S_t)}:} {This proof follows the proof of Theorem 2.3 in \cite{GTW}. However the steps 3 and 4 are different due to the more general operator $\mathbf{a}$. For the reader's convenience we have included the complete proof.}

We perform the proof along four steps.

{\textbf{Step 1.} { Time-discretization} of \eqref{Lt}.\\
Let $N\in\mathbb N^*$, $T>0$ and set $\Delta_t=\frac{T}{N}$. For $0\leq n\leq N$, we define $t_n=n\Delta_t$ and for $n\in\{1,\cdots,N\}$, for $(t,x)\in[t_{n-1}, t_n) \times \Omega$
$$h_{\Delta_t}(t,x)=h^n(x):=\frac{1}{\Delta_t}\int_{t_{n-1}}^{t_n} h(s,x)\,\mathrm{d}s.$$
{The} Jensen's inequality implies that $\|h_{\Delta_{t}}\|_{L^q(Q_T)} \leq \|h\|_{L^q(Q_T)}$ and we have $h_{\TT}\ra h$ in  $L^q(Q_T)$.}

Now we define the iterative scheme
\begin{equation}\label{eq1}
u^0=u_0\ \mbox{and for $n\geq 1$, $u^n$ is solution of}\
\left\{\begin{array}{ll}
\DD\frac{u^{n}-u^{n-1}}{\Delta_{t}}-\nabla\cdot \mathbf{a}(x,\nabla u^n) = h^n~&\text{ in }\Om,  \\
u^n = 0 ~&\text{  on   }~\pa\Om.
\end{array} \right.
\end{equation}
The sequence $(u^n)_{n\in\{1,\cdots,N\}}$ is well-defined because existence and uniqueness of $u^1\in\X\cap L^\infty(\Om)$ follow from Lemma \ref{sol of (P)} with $g=\Delta_t h^1+u^0\in L^q(\Omega)$ and by induction we obtain in the same way the existence of $(u^n)$, for any $n=2,\cdots,N$.

For $n=1,\cdots,N$ and $t\in[t_{n-1},t_n)$, we define the functions
\begin{equation}\label{uu}
u_{\Delta_t}(t)=u^n \quad \mbox{and}\quad  \tilde u_{\Delta_t}(t)=\DD\frac{(t-t_{n-1})}{\Delta_t}(u^n-u^{n-1})+u^{n-1},
\end{equation}
which satisfy
\begin{equation}\label{eq2}
 \frac{\pa \tilde u_{\TT}}{\pa t} -\nabla\cdot \mathbf{a}(x, \nabla u_{\TT})=h_{\TT} \quad\mbox{in } Q_T.
\end{equation}
%
\textbf{Step 2.} A priori estimates for $u_{\TT}$ and $\tilde u_{\TT}$.

Multiplying the equation in \eqref{eq1} by $(u^n-u^{n-1})$ and summing from $n=1$ to $N'\leq N$, we get
\begin{equation}\label{eq3}
\sum_{n=1}^{N'}\TT\int_\Om\left(\frac{u^n-u^{n-1}}{\TT}\right)^2\,\mathrm{d}x +\sum_{n=1}^{N'}\int_\Om \mathbf{a}(x,\nabla
u^n)\cdot\nabla(u^n-u^{n-1})\,\mathrm{d}x =\DD\sum_{n=1}^{N'}\int_\Om h^n(u^n-u^{n-1})\,\mathrm{d}x,
\end{equation}
hence by Young's inequality, we obtain:
\begin{equation*}\label{eq6}
\begin{split}
 \frac{1}{2}\sum_{n=1}^{N'}\TT\int_\Om\left(\frac{u^n-u^{n-1}}{\TT}\right)^2\,\mathrm{d}x
+\sum_{n=1}^{N'}\int_\Om \mathbf{a}(x,\nabla u^n)\cdot\nabla(u^n-u^{n-1})\,\mathrm{d}x
&\leq \frac{1}{2}\|h\|^2_{L^2(Q_T)}.
\end{split}
\end{equation*}
Thus we obtain
\begin{equation}\label{b1}
\DD\left(\frac{\pa\tilde u_{\Delta t}}{\pa t}\right)_{\Delta t} \mbox{ is bounded in } L^2(Q_T)  \mbox{ uniformly in  }
\TT.
\end{equation}
Since $A$ is strictly convex and from \eqref{growth.A},
%\begin{equation*}
%\begin{split}
%\int_\Om \mathbf{a}(x,\nabla u^n)\cdot\nabla(u^n-u^{n-1})dx\geq \int_\Om A(x,\nabla u^n)-A(x,\nabla u^{n-1})dx
%0\\&-
%\int_\Om \left(\mathbf{a}(x,\theta_n\nabla u^n+(1-\theta_n)\nabla u^{n-1})-\mathbf{a}(x,\nabla u^n)\right)\cdot\nabla(u^n-u^{n-1})dx.
%\end{split}
%\end{equation*}
%Set $\xi=\theta_n\nabla u^n+(1-\theta_n)\nabla u^{n-1}$ and $\eta=\nabla u^n$, then by (3):
%\begin{equation*}
%\begin{split}
%\int_\Om \left(\mathbf{a}(x,\theta_n\nabla u^n+(1-\theta_n)\nabla u^{n-1})-\mathbf{a}(x,\nabla u^n)\right)\cdot\nabla(u^n-u^{n-1})dx&=-\int_\Omega (1-\theta_n)^{-1}(\mathbf{a}(x,\xi)-\mathbf{a}(x,\eta))\cdot (\xi-\eta)dx\\
%&\leq 0.
%\end{split}
%\end{equation*}
we obtain for any $N'$
\begin{equation*}
\begin{split}
\frac{1}{2}\|h\|^2_{L^2(Q_T)}&\geq \sum_{n=1}^{N'}\int_\Om \mathbf{a}(x,\nabla u^n)\cdot\nabla(u^n-u^{n-1})\,\mathrm{d}x \geq \sum_{n=1}^{N'} \int_\Om A(x,\nabla u^n)-A(x,\nabla u^{n-1})\,\mathrm{d}x\\
&=  \int_\Om A(x,\nabla u^{N'})\,\mathrm{d}x- \int_\Om A(x,\nabla u^0)\,\mathrm{d}x \geq \frac{\gamma}{p^+-1}\int_\Om |\nabla u^{N'}|^{p(x)}\,\mathrm{d}x- \int_\Om A(x,\nabla u^0)\,\mathrm{d}x.\
\end{split}
\end{equation*}
We conclude that
\begin{equation}\label{b2}
\mbox{$(u_{\TT}) \mbox{ and }(\tilde u_{\TT})$ are bounded in $L^\infty(0,T,\X)$ uniformly in
$\TT$.}
\end{equation}
Furthermore, using \eqref{b1} we have
\begin{equation}\label{eq7}
\sup_{[0,T]}\|u_{\TT}-\tilde u_{\TT}\|_{L^2(\Om)}
\leq \DD\max_{n=1,...,N}\|u^n-u^{n-1}\|_{L^2(\Om)}
\leq C\TT^{1/2}.
\end{equation}
Therefore for $\TT\ra 0$, we deduce that there exist $u,v\in L^\infty(0,T,\X)$ such that (up to a subsequence)
\begin{equation}\label{fe}
\tilde u_{\TT}\stackrel{*}{\rightharpoonup}  u \text{ in }L^\infty(0,T,\X),\quad  u_{\TT}\stackrel{*}{\rightharpoonup} v
\text{ in }L^\infty(0,T,\X),
\end{equation}
and
\begin{equation}\label{convL2}
\frac{\pa\tilde u_{\TT}}{\pa t}\rightharpoonup \frac{\pa u}{\pa t}\text { in }L^2(Q_T).
\end{equation}
Inequality \eqref{eq7} implies that $u\equiv v$. By \eqref{fe}, for any $r\geq1$
\begin{equation}\label{convergeLp}
\tilde u_{\TT},\ u_{\TT} \rightharpoonup  u \text{ in }L^r(0,T,\X).
\end{equation}
\textbf{Step 3.} $u$ satisfies \eqref{Lt}.

 Since $p^->\frac{2d}{d+2}$, Theorem \ref{SE1} gives the embedding $\WW\hookrightarrow L^2(\Om)$ is compact. Hence, plugging \eqref{b1}, \eqref{b2} the compactness Aubin-Simon's result (see \cite{AS}) implies that (up to a subsequence),
\begin{equation}\label{new 1}
\tilde u_{\TT}\ra u\in C([0,T],L^2(\Om)).
\end{equation}
Equation \eqref{eq2} multipled by $(u_{\TT}-u)$ yields
\begin{equation*}\label{eq8}
\ii\frac{\pa\tilde u_{\TT}}{\pa t}(u_{\TT}-u)\,\mathrm{d}x\mathrm{d}t+\ii \mathbf{a}(x,\nabla u_{\TT})\cdot \nabla(u_{\TT}-u) \,\mathrm{d}x\mathrm{d}t
 =\ii h_{\TT}(u_{\TT}-u)\mathrm{d}x\mathrm{d}t.
\end{equation*}
Rearranging the terms in the last equations and using \eqref{eq7}-\eqref{convergeLp} we have
\begin{equation*}\label{eq9}
\begin{split}
\ii\left(\frac{\pa\tilde u_{\TT}}{\pa t}-\frac{\pa u}{\pa t}\right)(\tilde u_{\TT}-u)\,\mathrm{d}x\mathrm{d}t
+\ii (\mathbf{a}(x,\nabla u_{\TT})-a(x,\nabla u))\cdot \nabla(u_{\TT}-u)\,\mathrm{d}x\mathrm{d}t
=o_{\TT}(1)
\end{split}
\end{equation*}
{ where $o_{\TT}(1)\to 0$ as $\TT\to 0^+$}. Thus we get
\begin{equation*}\label{eq10}
 \frac{1}{2}\int_\Om|\tilde u_{\TT}(T)-u(T)|^2
 dx+\ii (\mathbf{a}(x,\nabla u_{\TT})-\mathbf{a}(x,\nabla u))\cdot \nabla(u_{\TT}-u)\,\mathrm{d}x\mathrm{d}t=o_{\TT}(1).
\end{equation*}
Using \eqref{new 1}, we obtain
\begin{equation}\label{convcont}
\ii (\mathbf{a}(x,\nabla u_{\TT})-\mathbf{a}(x,\nabla u))\cdot \nabla(u_{\TT}-u) \,\mathrm{d}x\mathrm{d}t\to 0\ \mbox{ as }\Delta_t\to 0^+.
\end{equation}
We now prove that
\begin{eqnarray}\label{convstrong}
\ii\vert\nabla( u_{\Delta_t}-u)\vert^{p(x)}\,\mathrm{d}x\mathrm{d}t\to 0\ \mbox{ as }\Delta_t\to 0^+.
\end{eqnarray}
{To establish \eqref{convstrong}, the general form of $\mathbf{a}$ do not allow to use the algebraic inequalities of \cite{S2} as in the proof of  Theorem 2.3 in \cite{GTW}. The convexity of $\Phi$ and assumption ({\bf A3}) bring the {arguments} to conclude.}

Indeed for this purpose, we distinguish two cases: (i) $p<2$ and (ii) $p\geq 2$. Let us first consider the case $p<2$.
Setting $q(x)=\frac{p(x)(2-p(x))}{2}$ and $\Omega^-=\displaystyle\{x\in\Omega:\, p(x)<2\}$, since $u,\, u_{\Delta_t}\in L^\infty(0,T,\X)$ we have from the H\"older inequality
\begin{equation*}
\begin{split}
\int_{\Omega^-}\vert\nabla (u-u_{\Delta_t})\vert^{p(x)}\,\mathrm{d}x
&\leq C\displaystyle\left\Vert\frac{\vert\nabla(u-u_{\Delta_t})\vert^{p(x)}}{(\vert\nabla u\vert+\vert\nabla u_{\Delta_t}\vert)^{q(x)}}\right\Vert_{L^{\frac{2}{p(x)}}(\Omega^-)}\Vert(\vert\nabla u\vert+\vert\nabla u_{\Delta_t}\vert)^{q(x)}\Vert_{L^{\frac{2}{2-p(x)}}(\Omega^-)}\\
&\leq \tilde C \displaystyle\left\Vert\frac{\vert\nabla(u-u_{\Delta_t})\vert^{p(x)}}{(\vert\nabla u\vert+\vert\nabla u_{\Delta_t}\vert)^{q(x)}}\right\Vert_{L^{\frac{2}{p(x)}}(\Omega^-)}\eqdef \tilde C \mathcal I.
\end{split}
\end{equation*}
Integrating in time the previous inequality and splitting the integral of the right-hand side, we obtain
\begin{equation}\label{pm2}
\begin{split}
\int_0^T\int_{\Om^-} |\nabla(u-u_{\Delta_t})|^{p(x)}\,\mathrm{d}x\mathrm{d}t\leq  C\int_{\mathcal I\leq 1}  &\left(\int_{\Omega^-}\frac{\vert \nabla( u- u_{\Delta_t})\vert^2}{(\vert \nabla u\vert+\vert \nabla u_{\Delta_t}\vert)^{2-p(x)}}\,\mathrm{d}x\right)^{2^{-1}\displaystyle\sup_{\Omega^-} p(x)}\,\mathrm{d}t \\
&+ C\int_{\mathcal I> 1}  \left(\int_{\Omega^-}\frac{\vert \nabla( u- u_{\Delta_t})\vert^2}{(\vert \nabla u\vert+\vert \nabla u_{\Delta_t}\vert)^{2-p(x)}}\,\mathrm{d}x\right)^{\frac{p^-}{2}}\,\mathrm{d}t.
\end{split}
\end{equation}
In the other hand, using {assumption ({\bf A3})}, we deduce that
\begin{equation}\label{pm2b}
\gamma\int_{\Omega^-}\frac{\vert \nabla( u- u_{\Delta_t})\vert^2}{(\vert \nabla u\vert+\vert \nabla u_{\Delta_t}\vert)^{2-p(x)}}\,\mathrm{d}x\leq \int_\Om(\mathbf{a}(x,\nabla u_{\TT})-\mathbf{a}(x,\nabla u))\cdot \nabla(u_{\TT}-u) \,\mathrm{d}x.
\end{equation}
Hence, plugging \eqref{pm2}, \eqref{pm2b} and H\"older's inequality, \eqref{convcont} implies
\begin{equation}\label{case1}
\int_0^T\int_{\Om^-} |\nabla(u-u_{\Delta_t})|^{p(x)}\,\mathrm{d}x\mathrm{d}t\rightarrow 0 \qquad \mbox{as } \quad\Delta_t\rightarrow 0.
\end{equation}
\noindent We now deal with the case $p(x)\geq 2$ and proceed as in \cite{MiRaRe} (see also \cite{RaRe} for other related issues). From the convexity of $\Phi$, we obtain:
\begin{equation*}
\int_{\Omega^+}\Phi(x,\vert\nabla u \vert)\,\mathrm{d}x\leq \int_{\Omega^+}\Phi \left(x,\frac{|\nabla (u+u_{\TT})|}{2}\right)\,\mathrm{d}x+\frac12\int_{\Omega^+}\mathbf{a}(x,\nabla u)\cdot\nabla (u-u_{\TT})\,\mathrm{d}x
\end{equation*}
and similarly
\begin{equation*}
\int_{\Omega^+}\Phi(x,\vert\nabla u_{\TT} \vert)\,\mathrm{d}x\leq \int_{\Omega^+}\Phi \left(x,\frac{|\nabla (u+u_{\TT})|}{2}\right)\,\mathrm{d}x+\frac12\int_{\Omega^+}\mathbf{a}(x,\nabla u_{\TT})\cdot\nabla (u_{\TT}-u)\,\mathrm{d}x.
\end{equation*}
Adding both above relations we get
\begin{equation}\label{unif-conv}
\begin{split}
\frac{1}{2}\int_{\Omega^+}(\mathbf{a}(x,\nabla u)-\mathbf{a}(x,\nabla u_{\TT}))\cdot(\nabla u-\nabla u_{\TT})\,\mathrm{d}x&\geq
\int_{\Omega^+}\Phi(x,\vert\nabla u\vert)dx+\int_{\Omega^+}\Phi(x,\vert\nabla u_{\TT}\vert)\,\mathrm{d}x\\
&-2\int_0^T\int_{\Omega^+}\Phi\left(x,\frac{|\nabla (u+u_{\TT})|}{2}\right)\,\mathrm{d}x.
\end{split}
\end{equation}
Using \eqref{clarkson}, we have
\begin{equation*}
\begin{split}
\int_{\Omega^+}\Phi(x,\vert\nabla u\vert)\,\mathrm{d}x+\int_{\Omega^+}\Phi(x,\vert\nabla u_{\TT}|)\,\mathrm{d}x\geq 2\int_{\Omega^+}&\Phi\left(x,\frac{|\nabla (u+u_{\TT})|}{2}\right)\,\mathrm{d}x\\
&+2\int_{\Omega^+}\Phi\left(x,\frac{|\nabla (u-u_{\TT})|}{2}\right)\,\mathrm{d}x.
\end{split}
\end{equation*}
Therefore, plugging the two last inequalities and \eqref{growth.A}, we deduce that
\begin{equation}\label{estip2}
\begin{split}
\int_{\Omega^+}(\mathbf{a}(x,\nabla u)-\mathbf{a}(x,\nabla u_{\TT}))\cdot\nabla (u-u_{\TT})\,\mathrm{d}x&\geq
 4\int_{\Omega^+}\Phi\left(x,\frac{|\nabla (u-u_{\TT})|}{2}\right)\,\mathrm{d}x\\
&\geq \frac{4\gamma}{2^{p^+}(p^+-1)}\int_{\Omega^+}\vert \nabla(u-u_{\TT})\vert^{p(x)}\,\mathrm{d}x.
\end{split}
\end{equation}
Now from \eqref{convcont}, \eqref{estip2} combining with \eqref{case1}, we obtain \eqref{convstrong}.
This implies that $\nabla u_{\TT}$ converges to $\nabla u$ in $L^{p(x)}(Q_T)$ and $u_{\TT}$ converges to $u$ in $\X$.
Furthermore
\begin{equation}\label{cvc}
\mathbf{a}(x,\nabla u_{\TT})\rightarrow \mathbf{a}(x,\nabla u) \mbox{ in } (L^{p_c(x)}(Q_T))^d
\end{equation}
with $p_c(x)=\frac{p(x)}{p(x)-1}$ the conjugate exponent of $p$. Indeed, we observe that from \eqref{convstrong} we get
\begin{eqnarray*}
\vert\nabla u_{\TT}\vert^{p(x)}\to\vert\nabla u\vert^{p(x)}\mbox{ in }L^1((0,T)\times\Omega)\quad\mbox{ as }\TT\to 0^+.
\end{eqnarray*}
Using Theorem 4.9 in \cite{Br}, we have for a subsequence $\{{\TT}_n\}$,
\begin{eqnarray*}
\nabla u_{{\TT}_n}\to\nabla u\mbox{ a.e. in }(0,T)\times\Omega\quad\mbox{and }\vert\nabla u_{{\TT}_n}\vert^{p(x)}\leq g\in L^1((0,T)\times\Omega).
\end{eqnarray*}
Using the dominated convergence theorem and observing that from \eqref{control-a}:
$$\vert\mathbf{a}(x,\nabla u_{{\TT}_n})\vert\leq c_1\vert\nabla u_{{\TT}_n}\vert^{p(x)-1}\leq g^{\frac{p(x)-1}{p(x)}}\in L^{p_c(x)},$$
we obtain
\begin{equation*}
\mathbf{a}(x,\nabla u_{{\TT}_n})\rightarrow \mathbf{a}(x,\nabla u) \mbox{ in } (L^{p_c(x)}(Q_T))^d
\end{equation*}
from which together with a classical compactness argument we get \eqref{cvc}.

Finally, Step 1, \eqref{convL2} and \eqref{cvc} allow to pass to the limit, in the distribution sense, in equation \eqref{eq2} and we conclude that $u$ is a weak solution of \eqref{Lt}.

Assume that there exist $u$ and $v$  weak solutions of \eqref{Lt}.
Thus,
$$\ii \frac{\partial(u-v)}{\partial t}(u-v)\,\mathrm{d}x\mathrm{d}t-\int_0^T
(\mathbf{a}(x,\nabla u)-\mathbf{a}(x,\nabla v))\cdot\nabla(u-v)\,\mathrm{d}t=0.$$
Since $u(0)=v(0)$, the above equality implies that $u\equiv v$ and we deduce the uniqueness.

\textbf{Step 4.} $u$ belongs to $C([0,T];\X)$.

Since $u\in C([0,T]; L^2(\Om))\cap L^\infty([0,T];\X)$ and $p\in \mathcal P^{\log}(\Omega)$, $u:t\in[0,T]\mapsto\X$ is weakly continuous.

Define $\Psi:\,\X\ni u\mapsto \int_{\Omega}\Phi(x,\vert\nabla u\vert)\,\mathrm{d}x=\int_{\Omega}A(x,\nabla u)\,\mathrm{d}x$. Then $\Psi$ is differentiable and {$\Psi'(u)=-\nabla\cdot \mathbf{a}(x,\nabla u)\in \X'$. {Note that} {$\Psi$ is a semimodular on $\X$} and thus is weakly lower semicontinuous (see Theorem 2.2.8 in \cite{DHHR}). Hence, fixing $t_0\in [0,T]$, we have
\[\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x\leq\liminf_{t\ra t_0}\int_\Om A(x, \nabla u(t))\,\mathrm{d}x.\]
From \eqref{eq3} with $\sum_{n=N''}^{N'}$ for $1\leq N''\leq N'$ {and the convexity of $\Psi$}, it follows that $u$ satisfies for any
$t\in [t_0,T]$:
\begin{equation}\label{eq10b}
\begin{split}
\int_{t_0}^t\int_\Om \left(\frac{\partial u}{\partial t}\right)^2\,\mathrm{d}x\mathrm{d}s+\int_\Om A(x,\nabla u(t))\,\mathrm{d}x\leq
\int_{t_0}^t\int_\Om h&\frac{\partial u}{\partial t}\,\mathrm{d}x\mathrm{d}s\\
&+\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x.
\end{split}
\end{equation}
Passing to the limit, we obtain
\[\limsup_{t\ra t^+_0}\int_\Om A(x,\nabla u(t))\,\mathrm{d}x\leq\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x.\]
Thus we get $\displaystyle\lim_{t\ra t^+_0}\int_\Om A(x,\nabla u(t))\,\mathrm{d}x=\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x$. Hence from \eqref{growth.A} and the dominated convergence theorem, we also obtain $\displaystyle\lim_{t\ra t^+_0}\int_\Om \frac{\vert\nabla u(t)\vert^{p(x)}}{p(x)}\,\mathrm{d}x=\int_\Om \frac{\vert\nabla u(t_0)\vert^{p(x)}}{p(x)}\,\mathrm{d}x$.

Now we prove the left continuity. Let $0<k\leq t-t_0$. Multiplying \eqref{Lt} by $\tau_k(u)(s)=\frac{u(s+k)-u(s)}{k}$ and integrating over $(t_0,t)\times \Omega$, the convexity gives
\begin{equation}\label{eq10t}
\begin{split}
\int_{t_0}^t\int_\Om \tau_k(u)\frac{\partial u}{\partial t}\,\mathrm{d}x\mathrm{d}s&+\int_{t}^{t+k}\int_\Om \frac{A(x,\nabla u(t))}{k}\,\mathrm{d}x\mathrm{d}s\\
&-\int_{t_0}^{t_0+k}\int_\Om \frac{A(x,\nabla u(t_0))}{k}\,\mathrm{d}x\mathrm{d}s\\
&\geq \int_{t_0}^t\int_\Om \tau_k(u)h\,\mathrm{d}x\mathrm{d}t.
\end{split}
\end{equation}
By the dominated convergence theorem as $k\rightarrow 0^+$, we have
\begin{equation*}
\begin{split}
&\int_{t}^{t+k}\int_\Om\frac{ A(x,\nabla u(s))}{k}\,\mathrm{d}x\mathrm{d}s\rightarrow \int_\Om A(x,\nabla u(t)) \,\mathrm{d}x,\\
&\int_{t_0}^{t_0+k}\int_\Om \frac{ A(x,\nabla u(s))}{k}\,\mathrm{d}x\mathrm{d}s\rightarrow\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x.
\end{split}
\end{equation*}
Hence \eqref{eq10t} yields
\begin{equation*}\label{eq10q}
\int_{t_0}^t\int_\Om \left(\frac{\partial u}{\partial t}\right)^2\,\mathrm{d}x\mathrm{d}s+\int_\Om A(x,\nabla u(t)) \,\mathrm{d}x\geq
\int_{t_0}^t\int_\Om h\frac{\partial u}{\partial t}\,\mathrm{d}x\mathrm{d}s+\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x.
\end{equation*}
From the above inequality, we deduce that we have the equality in \eqref{eq10b}. Using the  dominated convergence theorem, we obtain that
$$\displaystyle\lim_{t\ra t_0}\int_\Om A(x,\nabla u(t))\,\mathrm{d}x=\int_\Om A(x,\nabla u(t_0))\,\mathrm{d}x $$
and thus
$$\lim_{t\ra t_0}\int_\Om \vert\nabla u(t)\vert^{p(x)}\,\mathrm{d}x=\int_\Om \vert\nabla u(t_0)\vert^{p(x)}\,\mathrm{d}x.$$
Then from the uniform convexity of $\X$ we deduce that $u\in C([0,T];\X)$.\ $\hfill \square$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Existence of solutions to \eqref{Pt}}\label{local-exist-u}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Existence of a local solution}\label{local}
In the first part, we prove Theorem \ref{PT1}. For this purpose, we proceed as in the proof of Theorem \ref{sol for (S_t)} splitting the proof in several steps. {The proof of Theorem \ref{PT1} is almost similar as the proof of Theorem 2.5 in \cite{GTW}. Several differences appear in Step 3 due the general form of $\mathbf{a}$. For {sake of clarity}, we give the {entire} proof.} 

{\bf Step 1.} Existence of barrier functions.

Consider the equations, for $i\in\{0,1,2\}$
\begin{equation}\label{CL}
 \left\{\begin{array}{ll}
 \DD\frac{dv_i}{dt} = L_i(v_i),& \\
 v_i(0)  =(-1)^i\nu,&
 \end{array}\right.
\end{equation}
where $\nu= \|u_0\|_{\infty}$.

For $i\in\{0,1,2\}$, the Cauchy-Lipschitz theorem gives the existence of $T_i^{max}\in(0,+\infty]$ and a unique maximal solution $v_i$ to \eqref{CL} on $[0,T_i^{max})$.

If (H1) holds, we take $T\in (0,T_0^{max})$ otherwise, if  (H2) holds, we take $T\in (0,\min(T_1^{max};T_2^{max}))$.

Let $N\in \N^*$. Set $\Delta_t=\frac{T}{N}$ and consider the family $(v_i^n)$ defined by
$v_i^n=v_i(t_n)=v_i(n\Delta_t)$ for $n\in \{1,...,N\}$. Hence for any $i\in\{0,1,2\}$
\begin{equation*}\label{eq 12}
v_i^{n+1}=v_i^{n}+\int_{t_n}^{t_{n+1}}  L_i(v_i(s))\,\mathrm{d}s,\quad \forall n\in \{0,...,N-1\}.
\end{equation*}
Replacing $L_1$ (resp. $L_2$) by $\min(L_1,0)$ (resp. $\max(L_2,0)$) in  (H2), we can assume that $L_1\leq0$ and $L_2\geq0$. We get for $n\in \{0,...,N\}$, $v_1(T) \leq v_1^{n}\leq -\nu$ and for $i=0$ or $i=2$, $\nu \leq v_i^{n}\leq v_i(T)$.

\textbf{Step 2.} Semi-discretization in time of \eqref{Pt}.

Introduce the following iterative scheme $(u^n)$ defined as
\begin{equation}\label{eq 13}
u^0=u_0\quad \mbox{ and }\quad
\Biggl\{\begin{array}{ll}
u^{n}-\Delta_t\nabla\cdot \mathbf{a}(x,\nabla u^n) = u^{n-1}+\Delta_t f(x, u^{n-1})~&\text{ in }\Om,  \\
u^n = 0 ~&\text{  on   }~\pa\Om.
\end{array}
\end{equation}
We just prove the existence of $u^1$. The conditions ($f_1$) and ($f_2$) insure that $f(.,u^0)\in L^q(\Omega)$ with $q>\frac{d}{p^-}$. Thus, Lemma \ref{sol of (P)} applied with $g=u^0+\Delta_tf(x,u^0)\in L^q(\Omega)$ gives the existence of $u^1\in
\X\cap L^\infty(\Om)$.

Let $u_{\TT}$ and $\tilde u_{\TT}$ be defined as in \eqref{uu} and for $t<0$, $u_{\TT}(t)=u_0$. Thus \eqref{eq2} is satisfied with $h_{\TT}(t)\eqdef f(x,u_{\TT}(t-\TT))$.

\textbf{Step 3.} $(u^n)$ is bounded in $L^\infty(\Om)$ uniformly in $\Delta_t$.

We first consider the case where  (H1) is valid. We claim that for all $n$, $|u^n|\leq v_0^n$ in $\Omega$. We just prove it in case of $n=1$. Since $L_0$ and $v_0$ are nondecreasing, we get
\begin{equation*}
u^1-v_0^1- \Delta_t\nabla\cdot \mathbf{a}(x,\nabla u^1)=\int_0^{\Delta_t} (f(x,u_0)-L_0(v_0(s)))\,\mathrm{d}s+u_0-v_0^0\leq 0.
\end{equation*}
Multiplying the previous inequality by $(u^1-v_0^1)^+=\max(u^1-v_0^1,0)$ and integrating on $\omega=\{x\in \Omega \ |\ u^1(x)>v_0^1\}$, we get
$$\int_\omega (u^1-v_0^1)^2\,\mathrm{d}x +C\Delta_t\int_\omega |\nabla u^1|^{p(x)}\,\mathrm{d}x\leq \int_\omega (u^1-v_0^1)^2\,\mathrm{d}x+ \Delta_t\int_{\omega}\mathbf{a}(x,\nabla u^1)\nabla u^1\,\mathrm{d}x\leq 0.$$
Hence, $u^1\leq v_0^1$ and by the same method we have $-v_0^1\leq u^1$.

For (H2) we claim that for all $n$, $v_1^n\leq u^n \leq v_2^n \mbox{ in } \Om.$ Let $n=1$. Since $L_1$, $L_2$, $-v_1$  and $v_2$ are nondecreasing:
\begin{equation*}
\begin{split}
&u^1-v_1^1- \Delta_t\nabla\cdot \mathbf{a}(x,\nabla u^1)=\int_0^{\Delta_t} (f(x,u_0)-L_1(v_1(s)))\,\mathrm{d}s+u_0-v_1^0\geq 0,\\
&u^1-v_2^1- \Delta_t\nabla\cdot \mathbf{a}(x,\nabla u^1)=\int_0^{\Delta_t} (f(x,u_0)-L_2(v_2(s)))\,\mathrm{d}s+u_0-v_2^0\leq 0.
\end{split}
\end{equation*}
Multiplying the first inequality by $(v^1_1-u^1)^+$ and the second inequality by $(u^1-v^1_2)^+$ and integrating respectively
on
$\omega_1=\{x\in \Omega \ |\  v_1^1>u^1(x)\}$ and $\omega_2=\{x\in \Omega \ |\  v_2^1<u^1(x)\}$ and using \eqref{control-a} we get
\begin{equation*}
\begin{split}
-&\int_{\omega_1} (u^1-v_1^1)^2\,\mathrm{d}x -C\Delta_t\int_{\omega_1} |\nabla u^1|^{p(x)}\,\mathrm{d}x\geq -\int_{\omega_1} (u^1-v_1^1)^2\,\mathrm{d}x-\Delta_t\int_{\omega_1}\mathbf{a}(x,\nabla u^1)\nabla u^1\,\mathrm{d}x\geq 0,\\
&\int_{\omega_2} (u^1-v_2^1)^2\,\mathrm{d}x +C\Delta_t\int_{\omega_2} |\nabla u^1|^{p(x)}\,\mathrm{d}x\leq \int_{\omega_2} (u^1-v_2^1)^2\,\mathrm{d}x +\Delta_t\int_{\omega_2} \mathbf{a}(x,\nabla u^1)\nabla u^1\,\mathrm{d}x\leq 0.
\end{split}
\end{equation*}
Then $v_1^1 \leq u^1\leq v_2^1 $. By induction, we deduce that for $n\in \{0,...,N\}$, $v^n_1\leq u^n \leq v_2^n$ in $\Om$.

Therefore
\begin{equation}\label{eq20}
(u_{\TT}),\ (\tilde u_{\TT})  \mbox{ are bounded in $L^\infty(Q_T)$ uniformly in $\TT$}
\end{equation}
and
\begin{equation*}\label{eq21}
(h_{\TT}) \mbox{ is bounded in $L^2(Q_T)$ uniformly in $\TT$.}
\end{equation*}
Indeed, either (H1) holds which implies $$|f(x,u^{n})|\leq L_0(u^n)\leq L_0(v_0(T))$$ or (H2) holds, we have $$|f(x,u^{n})|\leq \max(-L_1(u^n),L_2(u^n))\leq \max(-L_1(v_1(T)),L_2(v_2(T))).$$ %\leq
Hence
\begin{equation*}
\|h_{\Delta_{t}}\|_{L^2(Q_T)}^2=\Delta_t\DD\sum_{n=1}^N\|f(x,u^{n-1})\|^2_{L^{2}(\Om)}\leq C.
\end{equation*}
{\bf Step 4.} End of the proof.

By the same computations of Step 2 of the proof of Theorem \ref{sol for (S_t)}, we obtain estimates and we prove that there exists $u\in L^\infty(0,T,\X)$ such
that
\begin{equation*}
\tilde u_{\TT},\, u_{\TT}\stackrel{*}{\rightharpoonup}  u  \text{ in }L^\infty(0,T,\X)\
\mbox{ and }\ \frac{\pa\tilde u_{\TT}}{\pa t}\rightharpoonup \frac{\pa u}{\pa t}\text { in }L^2(Q_T).
\end{equation*}
Relation \eqref{b1} implies that $(\tilde u_{\TT})$ is equicontinuous in $C([0,T];L^r(\Om))$ for $1\leq r\leq 2$. By the
interpolation inequality  and \eqref{eq20} we obtain that $(\tilde u_{\TT})$ is equicontinuous in $C([0,T];L^r(\Om))$ for
any $r>1$.

 By \eqref{b2} and Step 3, we deduce applying the Ascoli-Arzela theorem that (up to a subsequence)
for any $r>1$
$$\quad \tilde u_{\TT}\ra u \mbox{ in } C([0,T];L^r(\Om)).$$
Since $(u_{\TT})$ is uniformly bounded in $L^\infty(Q_T)$, ($f_1$) implies
\begin{equation*}
\|h_{\TT}(t)-f(.,u(t))\|_{L^2(\Om)}\leq C\|u_{\TT}(t-\TT)-u(t)\|_{L^2(\Om)}.
\end{equation*}
Hence we deduce that $h_{\TT}\ra f(.,u)$ in $L^\infty(0,T,L^2(\Om))$. Next we follow Step 4 of Theorem \ref{sol for (S_t)} and obtain that $u$ is a weak solution to \eqref{Pt}.

Now, we prove the uniqueness of the solution to \eqref{Pt}. Let $w$ be another weak solution of \eqref{Pt}. By ($f_1$), for $t\in [0,T]$:
\begin{equation*}
\begin{split}
\frac12& \|u(t)-w(t)\|_{L^2(\Omega)}^2-\int_0^t\langle \mathbf{a}(x,\nabla u)-\mathbf{a}(x,\nabla w),u-w\rangle \,\mathrm{d}s\\
&=\ii
(f(x,u)-f(x,w))(u-w)\,\mathrm{d}x\mathrm{d}s \leq C\int_0^t \|u(s)-w(s)\|_{L^2(\Omega)}^2\,\mathrm{d}s.
\end{split}
\end{equation*}
Since {$u\ra -\nabla\cdot \mathbf{a}(x,\nabla u)$} is a monotone operator from $\X$ to $\X'$, the second term in the left-hand side is nonnegative. Then, by Gronwall's lemma, we deduce that $u\equiv w$.

Step 5 of the proof of Theorem \ref{sol for (S_t)} again goes through and completes the proof.
\hfill$\square$

%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%
\subsection{Existence of global solution of \eqref{Pt}}\label{global}
 Now we prove Theorem \ref{PT3} and \ref{PT3b}.
To ensure the existence of barrier functions, we will need the following lemmata:{\begin{lem}\label{exist}
Assume conditions ({\bf A1})-({\bf A4}). Let $f:\Omega\times \R\mapsto \R$ be a function satisfying ($f_1$), ($f_3$)-($f_5$) and $h\in L^q(\Omega)$ with $q> \frac{d}{p^-}$. Assume that $f$ and $h$ {are nonnegative} functions. Then the stationary problem
\begin{equation*}\label{E}
 \left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla u)=f(x,u)+h& \mbox{ in } \Omega,\\
u=0& \mbox{ on } \partial \Om
\end{array}\right.\tag{$S$}
\end{equation*}admits a nontrivial weak solution $u\in\X$. Furthermore $u\in L^\infty(\Omega)$.
\end{lem}
We define
the notion of weak solution to \eqref{E} as follows:
\begin{defi}
Any function $w\in \X$ is called a weak solution to \eqref{E} if for all $\varphi\in \X$,
\begin{eqnarray*}
\int_{\Omega}\mathbf{a}(x,\nabla w)\cdot\nabla\varphi\,\mathrm{d}x=\int_\Omega (f(x,w)+h)\varphi\,\mathrm{d}x.
\end{eqnarray*}
%
\end{defi}
\begin{proof}
Consider the energy functional $E$ associated to \eqref{E} given by
\[E(u)=\int_\Om A(x,\nabla u)\,\mathrm{d}x-\int_\Om F(x,u)\,\mathrm{d}x-\int_\Om hu\,\mathrm{d}x\]
where $F(x,t)=\int_0^t f(x,s)\,\mathrm{d}s.$

Define
$$\alpha_\infty :={\displaystyle\sup_{x\in\Omega}}\limsup_{\vert s\vert\to \infty}\frac{\vert f(x,s)\vert}{\vert s\vert^{p^--1}}\  \mbox{ and }\    \beta_0:={\displaystyle\inf_{x\in\Omega}}\liminf_{s\to 0}\frac{|f(x,s)|}{|s|^{p^--1}}.$$
By ($f_3$) and ($f_4)$, for $\varepsilon>0$, there exists a constant $M=M(\varepsilon)$ large enough such that for any $(x,t)\in \Omega\times \R$, $|F(x,t)|\leq M|t|+\frac{\alpha_\infty+\varepsilon}{p^-} |t|^{p^-}$ and such that $\alpha_\infty+\varepsilon <\gamma\Lambda^{p^-}(p_c)^-$. Hence, $E$ is well-defined  and continuous on $\X$. Moreover, by \eqref{growth.A}, for $\|u\|_\X \geq 1$, {there exists $C>0$ such that}
\begin{equation*}
\begin{split}
E(u)&\geq \frac{\gamma}{p^+-1}\int_\Om |\nabla u|^{p(x)}\,\mathrm{d}x-M\int_\Om |u|\,\mathrm{d}x-\frac{\alpha_\infty+\varepsilon}{p^-}\int_\Om |u|^{p^-}\,\mathrm{d}x-C\|h\|_{L^q(\Om)}\|u\|_\X\\
&\geq \frac{\gamma}{p^+-1}\|u\|_\X^{p^-}-\frac{\alpha_\infty+\varepsilon}{p^-}\|u\|_{L^{p^-}(\Om)}^{p^-}-\tilde M\|u\|_\X\\
&\geq \left(\frac{\gamma}{p^+-1}-\frac{\alpha_\infty+\varepsilon }{{\Lambda^{p^-}}p^-}\right)\|u\|_\X^{p^-}-\tilde M\|u\|_\X.
\end{split}
\end{equation*}
Condition ($f_4)$ implies that $E$ is coercive. Thus $E$ admits a global minimizer $u\in \X$ which is a weak solution of \eqref{E}.

We claim that  $u\not\equiv 0$. We need to consider two cases $f(.,0)+h\not\equiv 0$ and $h\equiv 0$.

In the first case, this is obvious. For the second case, we establish that there exists $v\in \X$ such that $E(v)<0$. Consider $\varepsilon>0$ and $v_\varepsilon\in C^1_0(\Omega)$ such that
\begin{equation}\label{beta}
\beta_0>\Gamma(\Lambda+\varepsilon)^{p^-}(p^-)_c,\quad \|v_\varepsilon\|_\X=1\  \mbox{ and  }\  \|v_\varepsilon\|_{L^{p^-}(\Omega)}>\frac{1}{\Lambda+\varepsilon}.
\end{equation}
By ($f_5$), {for $\eta>0$, there exists $s_\eta>0$ small enough such that  for any $x\in \Om$ and $0<s\leq s_\eta$
$$|f(x,s)| \geq (\beta_0-\eta) |s|^{p^--1}.$$}
Hence {we obtain using \eqref{growth.A} and \eqref{beta}, $\theta=\theta(\eta)>0$ small enough such that}
\begin{equation*}
\begin{split}
E(\theta v_\varepsilon)&\leq \theta^{p^-}\left(\frac{\Gamma}{p^--1}\int_\Om |\nabla v_\varepsilon|^{p(x)}\,\mathrm{d}x-\frac{\beta_0-\eta}{p^-}\int_\Om |v_\varepsilon|^{p^-}\,\mathrm{d}x\right)\\
&\leq  \theta^{p^-}\left(\frac{\Gamma}{p^--1}-\frac{\beta_0-\eta}{p^-}\left(\frac{1}{\Lambda+\varepsilon}\right)^{p^-}\right).
\end{split}
\end{equation*}
Choosing $\eta$ small enough and using the first inequality in \eqref{beta}, we conclude that $E(\theta v_\varepsilon)<0$ and we deduce that $u\not\equiv 0$. Finally, by Corollary \ref{reg2}, $u\in L^\infty(\Om)$.
\end{proof}
}

%which ensures the existence of an appropriate supersolution and extends Lemma 2.1 in \cite{Fa} and Lemma 2.2 in \cite{YY}:
\begin{lem}\label{31YY}
Let $p\in C^\beta(\overline\Omega)$, $\beta\in(0,1)$. Assume conditions ({\bf A1})-({\bf A4}) and ({\bf A6}). Let $\lambda\in \R^+$ such that
$$\lambda\geq \lambda^*:=\frac{\gamma (p_c)^-}{2\vert\Omega\vert^{1/d}C_0}$$ where $C_0$ is the best embedding constant of $W^{1,1}_0(\Omega)\subset L^{\frac{d}{d-1}}(\Omega)$. Let $w_\lambda\in \X\cap L^\infty(\Om)$ be the unique solution of
\begin{equation}\label{El}
 \left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla w_\lambda)=\lambda& \mbox{ in } \Omega,\\
w_\lambda=0& \mbox{ on } \partial \Om.
\end{array}\right.\tag{$E_\lambda$}
\end{equation}
Then, there exist two constants $C_1$ and $C_2$ which do not depend to $\lambda$ such that
\begin{equation}\label{lower-upper}
 \|w_\lambda\|_{L^{\infty}}\leq C_1\lambda^{\frac{1}{(p^--1)}} \quad \mbox{and} \quad
w_\lambda(x) \geq C_2 \lambda^{\frac{1}{p^+-1+\mu}}\rho(x)
\end{equation}
where $\mu\in (0,1)$ and $\rho(x)=d(x,\partial\Omega)$ denotes the distance of $x\in\Omega$ to the boundary of $\Omega$.
\end{lem}
\begin{proof}
We first prove the upper bound of \eqref{lower-upper}. For that we follow closely the proof of Lemma 2.1 in \cite{Fa}:
Let $u$ be the solution to \eqref{El} for a fixed $\lambda$ satifying assumptions of the lemma. By the the maximum principle, $u\geq 0$. Using classical regularity results (see \cite{Fa0} and \cite{FZ}), $u\in C^{1,\alpha}(\overline{\Omega})$ and from  \cite{Zh} $u>0$ in $\Omega$ and satisfies the Hopf maximum principle. Now, for $k\geq 0$, set $A_k=\{x\in\Omega\,:\, u(x)>k\}$.  Using $(u-k)^+$ as a  testing function together with \eqref{growth.A} and the Young inequality, we obtain for any $\epsilon>0$:
\begin{equation}\label{estimatenorm}
\begin{split}
\frac{\gamma}{p^+-1}\int_{A_k}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x&\leq\int_{A_k}\mathbf{a}(x,\nabla u).\nabla u\,\mathrm{d}x\\
&=\lambda\int_{A_k}(u-k)\,\mathrm{d}x\\
&\leq \lambda\vert A_k\vert^{\frac{1}{d}}\vert(u-k)^+\vert_{L^{\frac{d}{d-1}}(\Omega)}\\
&\leq \lambda\vert A_k\vert^{1/d}C_0\int_{A_k}\vert\nabla u\vert\,\mathrm{d}x\\
&\leq\frac{\lambda\vert A_k\vert^{1/d}C_0}{p^-}\int_{A_k}\epsilon^{p(x)}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x+\frac{\lambda\vert A_k\vert^{1/d}C_0}{(p^+)_c}\int_{A_k}\epsilon^{-p_c(x)}\,\mathrm{d}x.
\end{split}
\end{equation}
Taking $\epsilon=\left(\frac{\lambda^*}{\lambda}\right)^{1/p^-}$, we have
\begin{eqnarray*}
\frac{\lambda\vert A_k\vert^{1/d}C_0}{p^-}\int_{A_k}\epsilon^{p(x)}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x\leq \frac{\gamma}{2(p^+-1)}\int_{A_k}\vert \nabla u\vert^{p(x)}\,\mathrm{d}x
\end{eqnarray*}
which implies together with \eqref{estimatenorm}
\begin{equation}\label{estimate2}
\begin{split}
\int_{A_k}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x&\leq \frac{2 \lambda C_0(p^+-1)\vert A_k\vert^{1/d}}{\gamma (p^+)_c}\int_{A_k}\epsilon^{-p_c(x)}\,\mathrm{d}x\\
&\leq \frac{2 \lambda C_0(p^+-1)}{\gamma (p^+)_c\epsilon^{(p^-)_c}} \vert A_k\vert^{1+\frac{1}{d}}.
\end{split}
\end{equation}
From \eqref{control-a}, \eqref{estimatenorm} and \eqref{estimate2}, we get
\begin{eqnarray}\label{u-k}
\int_{A_k}(u-k)\,\mathrm{d}x=\frac{1}{\lambda}\int_{A_k}\mathbf{a}(x,\nabla u).\nabla u\,\mathrm{d}x\leq \frac{c_1}{\lambda}\int_{A_k}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x\leq \tilde K\vert A_k\vert^{1+\frac{1}{d}},
\end{eqnarray}
where $\displaystyle \tilde K=\frac{2c_1 C_0(p^+-1)}{\gamma (p^+)_c\epsilon^{(p^-)_c}}$. By Lemma 5.1 in \cite[Chapter 2]{LU} and \eqref{u-k}, we deduce that
\begin{eqnarray*}
\Vert u\Vert_{L^\infty(\Omega)}\leq\tilde K(d+1)^{\frac{d+1}{d}}\vert\Omega\vert^{1/d}
\end{eqnarray*}
from which we easily obtain that $\Vert u\Vert_{L^\infty(\Omega)}\leq C_1\lambda^{\frac{1}{p^--1}}$ where $C_1$ does not depends on $\lambda$.
%with
%\begin{eqnarray*}
%C_1=\displaystyle\frac{(d+1)^{\frac{d+1}{d}}c_1(2 %C_0)^{(p^-)'}}{(p^-)^{1/(p^--1)}c^{(p^-)'}(p^+)'}\vert\Omega\vert^{(p^-)'/d}.
%\end{eqnarray*}
Next, we show the lower bound estimate.  Since $\partial\Omega$ is $C^2$, there exists $\ell\in (0,1)$ small enough such that $\rho$ is  $C^2$ and
\begin{equation}\label{regrho}
\vert\nabla \rho\vert\equiv\, 1 \mbox{ in } \{x\in\Omega\,:\, \rho(x)<3\ell\}
\end{equation}
(see \cite{GiTr} Lemma 14.16 page 355). As in \cite{Zh} we introduce the following function: let $\kappa>0$,
\begin{equation*}%\label{subsolutionM}
v_1(x)=\displaystyle\left\{\begin{array}{ll}
\kappa \rho(x), \quad &\rho(x)<\ell\\
\kappa \ell+\kappa\int_\ell^{\rho(x)}m(t)\,\mathrm{d}t,\quad &\ell\leq \rho(x)\leq 2\ell,\\
\kappa\ell+\kappa\int_\ell^{2\ell}m(t)\,\mathrm{d}t,\quad &2\ell\leq \rho(x)
\end{array}\right.
\end{equation*}
where $\displaystyle m(t)=\left(\frac{2\ell-t}{\ell}\right)^{\frac{2}{p^--1}}.$

Next we show that for a suitable value of $\kappa$, $v_1\in C^1(\overline{\Omega})$ is a subsolution to \eqref{El}. Precisely,
{\begin{eqnarray*}%\label{v1p(x)}
\nabla\cdot \mathbf{a}(x,\nabla v_1)=\sum_{i=1}^d \frac{\partial a_i}{\partial x_i}(x,\nabla v_1)+\sum_{i,j=1}^d \frac{\partial a_i}{\partial \xi_j}(x,\nabla v_1)\frac{\partial^2 v_1}{\partial x_i\partial x_j}(x).
\end{eqnarray*}}
Using the definition of $v_1$, \eqref{regrho} and {noting that for any $i,\,j\in\{1,...,d\}$
$$\frac{\partial a_i}{\partial \xi_j}(x,\xi)=\frac{1}{|\xi|}\frac{\partial \phi}{\partial \xi_j}(x,|\xi|)\xi_i\xi_j+\phi(x,|\xi|)\delta_{ij},$$}
where $\delta_{ij}$ is the Kronecker symbol, we get
$$
\displaystyle{\nabla\cdot \mathbf{a}(x,\nabla v_1(x))=\left\{\begin{array}{ll}
\kappa\sum_{i=1}^d \frac{\partial \phi}{\partial x_i}(x,|\kappa|)\frac{\partial \rho}{\partial x_i}(x) +\kappa\phi(x,|\kappa|)\Delta \rho \quad&\mbox{if } \rho(x)<\ell\\
\ & \\
 0\quad &\mbox{if }2\ell\leq \rho(x)
\end{array}\right.}
$$
and in $\{x\in\Omega\,:\, \ell\leq \rho(x)\leq 2\ell\}$,
\begin{equation*}%\label{Dv1bis}
\begin{split}
\nabla\cdot \mathbf{a}(x,\nabla v_1)&=\kappa m(\rho)\sum_{i=1}^d \frac{\partial \phi}{\partial x_i}(x,|\kappa|m(\rho))\frac{\partial \rho}{\partial x_i}(x)\\
&+\kappa\sum_{i,j=1}^d \frac{\partial a_i}{\partial \xi_j}(x,\kappa m(\rho)\nabla \rho)\left(m'(\rho)\frac{\partial \rho}{\partial x_i}(x)\frac{\partial \rho}{\partial x_j}(x)+m(\rho)\frac{\partial^2 \rho}{\partial x_i\partial x_j}(x)\right).
\end{split}
\end{equation*}
Then, using hypotheses ({\bf A4}), ({\bf A6}) and relations \eqref{control-a} and \eqref{regrho}, we obtain that $\vert\nabla\mathbf{a}(x,\nabla v_1(x))\vert\leq C\kappa^{p(x)-1+\mu}$ a.e. on $\Omega$, for any $\mu\in (0,1)$, where $C=C(\ell,\mu,p,\Omega)$ and independent of $\kappa$. Choosing $\kappa$ such that $2C\kappa^{p^+-1+\mu}=\lambda$, we have that $v_1$ is  a subsolution to \eqref{El} and since $w_\lambda\geq v_1$, the lower bound in \eqref{lower-upper} is proved.
\end{proof}
\begin{rem}
About the $C^{1,\alpha}(\overline \Omega)$-regularity of the solution of \eqref{El}, we apply Theorem 1.2 in \cite{Fa0}.  More precisely, we need the condition on $\mathbf{a}$: for $\delta\in (0,1)$, there exists  $\tilde c>0$ such that for any $x,\ y\in \overline \Omega$, $\eta\in \R^d$:
$$|\mathbf{a}(x,\eta)-\mathbf{a}(y,\eta)|\leq \tilde c|x-y|^\beta(1+|\eta|^{p^+-1+\delta})$$
where $\beta$ is the H\"older exponent of $p$.
Obviously, condition ({\bf A6}) implies the above inequality.
\end{rem}
\noi Now we give the proof of the existence of global solutions.

\smallskip
\noindent \textbf{Proof of Theorem \ref{PT3} and \ref{PT3b}:} We consider the stationary quasilinear elliptic problem associated to \eqref{Pt}:
\begin{equation*}\label{S0}
 \left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla u)=f(x,u)& \mbox{ in } \Omega,\\
u=0& \mbox{ on } \partial \Om
\end{array}\right.\tag{$P_\infty$}
\end{equation*}
Thus we claim that if (C1) or (C2) hold then there exist $\underline u,\overline u\in \X\cap L^\infty(\Om)$, a sub- and a supersolution of \eqref{S0} such that $\underline u\leq u_0\leq\overline u$.

First, consider that (C1) holds. For $(x,s)\in \Omega\times\R$, define
$$G(x,s)=|\nabla\cdot \mathbf{a}(x,\nabla u_0(x))|+|f(x,s)|.$$
Consider the following problems:
\begin{equation*}
\left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla\underline u)=-G(x,\underline u)& \mbox{ in } \Omega,\\
u=0& \mbox{ on } \partial \Om
\end{array}\right.
\quad\mbox{and}\quad \left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla\overline u)=G(x,\overline u)& \mbox{ in } \Omega,\\
u=0& \mbox{ on } \partial \Om.
\end{array}\right.
\end{equation*}
{Lemma \ref{exist} implies the existence of $\underline u$ and $\overline u\in \X\cap L^\infty(\Om)$.} Moreover
$$-\nabla\cdot \mathbf{a}(x,\nabla\underline u)=-G(.,\underline u)\leq -\nabla\cdot \mathbf{a}(x,\nabla u_0) \ \mbox{ and }\  -\nabla\cdot \mathbf{a}(x,\nabla\overline u)=G(.,\overline u)\geq -\nabla\cdot \mathbf{a}(x,\nabla u_0) \ \mbox{ a.e in }\ \Omega.$$
Hence the weak comparison principle implies $\underline u\leq u_0$ and $\underline u$ is a subsolution of \eqref{S0}.
Similarly we have that $\overline u\geq u_0$ and $\overline u$ is a supersolution of \eqref{S0}.

Now, if (C2) holds, we use Lemma \ref{31YY} above. Precisely, from assumptions ($f_3$) and ($f_4$), for $\varepsilon >0$ there exists $M_0=M_0(\alpha_\infty,\varepsilon)>0$ such that for any $M\geq M_0$
\begin{eqnarray*}%\label{control-f-parab}
\vert f(x,s)\vert\leq M+(\alpha_\infty+\varepsilon)\vert s\vert^{p^--1}\quad\mbox{for any }(x,s)\in\Omega\times\R.
\end{eqnarray*}
From Lemma \ref{exist}, there exists a positive solution, $\tilde w_{M,\varepsilon}\in \X\cap L^\infty(\Om)$  to
\begin{eqnarray}\label{sursolutionc2}
\left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla w)=M+ (\alpha_\infty+\varepsilon) |w|^{p^--1}& \mbox{ in } \Omega,\\
\tilde w=0& \mbox{ on } \partial \Om.
\end{array}\right.
\end{eqnarray}
Moreover, we have $\tilde w_{M,\varepsilon}\in C^1(\overline \Om)$ (see \cite[Theorem 1.2]{Fa0} and \cite[Theorem 4.4]{FZ}).

Fix $0<\lambda<M$, let $w_\lambda$ be the solution of \eqref{El}. By Lemma \ref{31YY} we deduce that $\Vert w_\lambda\Vert_{L^\infty(\Om)}\to\infty$ as $\lambda\to\infty$. From the weak comparison principle, $w_\lambda\leq \tilde w_{M,\varepsilon}$. Therefore,
\begin{equation*}%\label{asymptotic-beh}
\Vert \tilde w_{M,\varepsilon}\Vert_{L^\infty({\Omega})}\to\infty \mbox{ as }M\to\infty.
\end{equation*}
%\begin{equation}\label{sur}
%-\nabla\cdot \mathbf{a}(x,\nabla w_\lambda)=\lambda \geq C\left(1+ C_1^\beta\lambda^{\frac{\beta}{p^--1}}\right)\geq C(1+w_\lambda^\beta)\geq |%f(x,w_\lambda)|.
%\end{equation}
Moreover, since $u_0\in C^1_0(\overline \Omega)$, there exists $K>0$ such that for any $x\in \Omega$, $|u_0(x)|\leq K\mbox{dist}(x,\partial \Omega)$. Hence choosing $\lambda$ and $M$ large enough, we have by Lemma \ref{31YY}: $\tilde w_{M,\varepsilon}\geq w_\lambda\geq |u_0|$ in $\overline \Omega$.

Set $\overline u=\tilde w_{M}$ and $\underline u=- \tilde w_{M,\varepsilon}$. We deduce for $M$ large enough$,  \overline u$ and $\underline u$ are respectively a super- and a subsolution of \eqref{S0} such that $\underline u\leq u_0\leq\overline u$.

Now we proceed as in the proof of Theorem \ref{PT1}. We define the sequence $(u^n)$ as follows:
\begin{equation*}\label{iteration}
\left\{
\begin{array}{l l}
u^{n}-\Delta_t\nabla\cdot \mathbf{a}(x,\nabla u^{n})=u^{n-1}+\Delta_t f(x,u^{n-1})& \mbox{ in } \Omega,\\
u^n=0& \mbox{ on } \partial \Om
\end{array}\right.
\end{equation*}
for $n=1,2,...,N$ with $u^0=u_0$. we prove for $n\geq 1$, $\underline u\leq u^n\leq \overline u$ in $\Om$.
Indeed for $n=1$, we have
\[\underline u-u^1-\Delta_t(\nabla\cdot \mathbf{a}(x,\nabla\underline u)-\nabla\cdot \mathbf{a}(x,\nabla u^1))\leq \underline
u-u^0+\Delta_t(f(x,u^0)-f(x,\underline u)).\]
Since $s\mapsto f(x,s)$ is Lipschitz on $[-M_1,M_1]$ uniformly in $x\in \Om$, where $M_1$ is the maximum of $\|\underline u\|_{L^\infty}$ and $\|\overline u\|_{L^\infty}$ thus, for $\Delta_t$ small enough, the function $Id-\Delta_t f$ is nondecreasing. Then we have
\[\underline u-u^1-\Delta_t(\nabla\cdot \mathbf{a}(x,\nabla\underline u)-\nabla\cdot \mathbf{a}(x,\nabla u^1))\leq (Id-\Delta_tf)(\underline u-u^0).\]
Hence the right-hand side of the above inequality is nonpositive and thus by the weak comparison principle
we have $\underline u\leq u^1$. Similarly we prove $u^1\leq \overline u$.

By induction, for $n\geq 1$, $\underline u\leq u^n\leq\overline u$ in $\Om$. Thus $(u^n)$ is uniformly
bounded in $L^\infty(\Om)$.
The rest of the proof follows from steps 3 and 4 of the proof of Theorem \ref{PT1}.\hfill$\square$




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\section{Stabilization}\label{asympt-conv}
%

\subsection{Existence and uniqueness of the solution of the stationary problem}
In the {purpose} of investigating the behaviour of the global solution to \eqref{Pt} as $t\to\infty$, we consider the stationary problem:
%
\begin{equation}\label{sta}
\left\{\begin{array}{ll}
-\nabla\cdot \mathbf{a}(x,\nabla u)=f(x,u)\quad&\mbox{in }\Omega,\\
u>0&\mbox{in }\Omega,\\
u=0\;&\mbox{on }\partial\Omega.
\end{array}\right.\tag{$P_+$}
\end{equation}
We define the notion of a weak solution as follows:
\begin{defi}\label{def-weak-sol}
Any positive function $w\in \X\cap L^\infty(\Omega)$ is called a weak solution to \eqref{sta} if for any $\varphi\in\X$,
%
\begin{eqnarray*}
\int_{\Omega}\mathbf{a}(x,\nabla w)\cdot\nabla\varphi\,\mathrm{d}x=\int_\Omega f(x,w)\varphi\,\mathrm{d}x.
\end{eqnarray*}
%
\end{defi}
%
\noindent We first discuss the existence and the uniqueness of the weak solution to $({\rm P})$.
In the proof of Theorem \ref{uniqueness-stat}, we will use the following ray-strict convexity result on the energy functional proved in \cite{GiTa}. We start by a definition:
%
\begin{defi}\label{def-ray-convex}\nopagebreak
\begingroup\rm
Let $X$ be vector space. A functional\/ $\mathcal{W}\colon \dotV\eqdef \{ v\colon \Omega\to (0,\infty)\colon v\in X\}\to \RR$
will be called {\em ray\--strictly convex\/}
({\em strictly convex\/}, respectively)
if it satisfies for all $v_1, v_2\in \dotV$ and for all $\theta\in (0,1)$
\begin{equation}\label{e:ray-strictly}
  \mathcal{W}\left( (1-\theta) v_1 + \theta v_2 \right)
  \leq (1-\theta)\cdot \mathcal{W}(v_1)+{ \theta\cdot \mathcal W(v_2)}
\end{equation}
where the inequality is strict unless
$v_2 / v_1$ is a constant
(always strict if\/ $v_1\neq v_2$, respectively).
%
\endgroup
\end{defi}
%
With the above definition, we have:
{\begin{thm}\label{thm-ray-convex}
%
Let\/ $r\in [1,\infty)$ and\/
$p\colon \Omega\to (1,\infty)$ satisfy $1 < p^{-}\leq
      p^{+} < \infty
  \quad\mbox{ and }\quad r\leq p^{-}$.
Assume that \begin{math}
  A\colon \overline{\Omega}\times \RR^d\to \RR_+
\end{math}
%
is continuous satisfying ({\bf A7}) and %
\begin{enumerate}%\label{e:A-conv}
\item[({\bf A8})] $\xi\mapsto N(x,\xi)\eqdef A(x,\xi)^{r/p(x)}
  \colon \RR^d\to \RR_+$ is strictly convex for every $x\in \Omega$.
\end{enumerate}
Then (the restriction of) the functional\/ $\mathcal{W}_A\colon X\eqdef\{v\in L^{p(x)/r}(\Omega)\,:\,|v|^{1/r}\in \X\}\to \RR_+$ defined by
%
\begin{equation*}%\label{def:W(v)}
  \mathcal{W}_A(v)\eqdef
  \int_{\Omega}
  A\left( x,\, \nabla (|v(x)|^{1/r}) \right) \,\mathrm{d}x
\end{equation*}
to the convex cone $\dotV$ is ray\--strictly convex on~$\dotV$.

Furthermore, if\/ $p(x)\not\equiv r$ in $\Omega$, i.e., if\/
$r = p^{-}\equiv p(x)\equiv p^{+}$ does not hold in~$\Omega$,
then $\mathcal{W}_A$ is even strictly convex on~$\dotV$.
\end{thm}}
%
\begin{rem}\label{rem-ray-convex}\nopagebreak
\begingroup\rm
We note that the function
%
\begin{math}
  \xi\mapsto A(x,\xi) = N(x,\xi)^{p(x)/r} \colon \R^d\to \R_+
\end{math}
%
is strictly convex for each fixed $x\in \Omega$,
thanks to the power function
$t\mapsto t^{p(x)/r}\colon \R_+\to \R_+$
being strictly monotone increasing and convex.
Consequently,
$A(x,\xi) > A(x,\mathbf{0}) = 0$ for all $x\in \Omega$ and
$\xi\in \R^d\setminus \{\mathbf{0}\}$, and
$A\colon \overline{\Omega}\times \mathbb S^{d-1}\to \R_+$
is bounded below and above on the compact set
%
\begin{math}
  \overline{\Omega}\times \mathbb S^{d-1}\subset \R^d\times \R^d
\end{math}
%
by some positive constants; hence, the ``coefficient''
%
\begin{math}
  A\left( x,\, \frac{\xi}{|\xi|} \right)
\end{math}
%
, if
$\xi = \nabla (|v|^{1/r})\neq \mathbf{0}$,
is bounded from below and above by some positive constants.
Consequently, we recover that the ratio of the functionals in
$\int_{\Omega}A(x,\nabla u)\,\mathrm{d}x$ and $\int_{\Omega}\vert\nabla u\vert^{p(x)}\,\mathrm{d}x$
is bounded from below and above by the same positive constants as in \eqref{growth.A}.

{Conversely, if we assume that $ \xi\mapsto A(x,\xi)$ is strictly convex for each fixed $x\in \Omega$, then for any $1< r\leq p^-$,  $\xi\mapsto N(x,\xi)$ is strictly convex if $r\neq 1$. {Indeed applying Lemma~2.1 in \cite{BrFr} to the function $F(\xi)=A(x,\xi)^{1/p(x)}$, we deduce that $F$ is convex and thus, for any $r>1$, $\xi\mapsto N(x,\xi)=(F(\xi))^r$ is strictly convex}.}
%
\endgroup
\end{rem}
{For the reader's convenience, we give the proof:}

\noindent{\it Proof of\/} Theorem~\ref{thm-ray-convex}.
Recalling {\rm Definition~\ref{def-ray-convex}},
let us consider any $v_1, v_2\in \dotV$ and $\theta\in (0,1)$.
Let us denote
$v = (1-\theta) v_1 + \theta v_2$; hence, $v\in \dotV$.
We obtain easily
%
\begin{equation*}
    \nabla (v_i(x)^{1/r}) = \frac{v_i^{1/r}}{r}\, \frac{\nabla v_i}{v_i}
    \quad\mbox{ for }\, i=1,2
\end{equation*}
and
\begin{equation}\label{grad}
\begin{split}
\nabla (v(x)^{1/r}) &= \frac{1}{r}\,     \frac{ (1-\theta) \nabla v_1 + \theta\nabla v_2 }%
         { ( (1-\theta) v_1 + \theta v_2 )^{1 - (1/r)} }\\
 &= \frac{v^{1/r}}{r}\,    \frac{ (1-\theta) \nabla v_1 + \theta\nabla v_2 }{v}\\
&= \frac{v^{1/r}}{r}\,     \left( (1-\theta)\, \frac{v_1}{v}\cdot \frac{ \nabla v_1 }{v_1}
             + \theta\, \frac{v_2}{v}\cdot \frac{ \nabla v_2 }{v_2}
   \right)
\end{split}
\end{equation}
%
with the convex combination of coefficients
$(1-\theta)\, \frac{v_1}{v}$ and $\theta\, \frac{v_2}{v}$,
%
\begin{equation*}
  (1-\theta)\, \frac{v_1}{v} + \theta\, \frac{v_2}{v} = 1 \,.
\end{equation*}
%
Now let $x\in \Omega$ be fixed.
Since $\xi\mapsto N(x,\xi)$ is strictly convex, by our hypothesis,
we may apply the identities from above to conclude that
%
\begin{equation}\label{e:v<v_1+v_2}
N\left(x, (1-\theta)\, \frac{v_1}{v}\cdot \frac{ \nabla v_1 }{v_1}
             + \theta\,  \frac{v_2}{v}\cdot \frac{ \nabla v_2 }{v_2}
   \right)\leq (1-\theta)\, \frac{v_1}{v}\cdot
  N\left(x, \frac{ \nabla v_1 }{v_1} \right)
         + \theta\, \frac{v_2}{v}\cdot
  N\left(x, \frac{ \nabla v_2 }{v_2} \right).
\end{equation}
%
The equality holds if and only if
%
\begin{equation}
\label{e:v_1=v_2}
    \frac{ \nabla v_1(x) }{v_1(x)}
  = \frac{ \nabla v_2(x) }{v_2(x)} \,,
    \quad\mbox{ which is equivalent to }\quad
  \nabla\genfrac{(}{)}{}0{v_2(x)}{v_1(x)} = 0 \,.
\end{equation}
%
Notice that the homogeneity conditions ({\bf A7})
 and  ({\bf A8}) yield
%
\begin{equation}
\label{p-hom:N}
  N(x,t\xi) = |t|^r\, N(x,\xi) \quad\mbox{ for all }\,
    t\in \RR \,,\ \xi\in\RR^d \,.
\end{equation}
%
Consequently, plugging \eqref{grad}, \eqref{e:v<v_1+v_2} and \eqref{p-hom:N} we obtain
%
\begin{equation}\label{eq:v<v_1+v_2}
\begin{split}
   N\left( x,\nabla (v(x)^{1/r})\right)
  &= \frac{v}{r^r}\cdot
  N\left( x, (1-\theta)\, \frac{v_1}{v}\cdot \frac{ \nabla v_1 }{v_1}
               + \theta\, \frac{v_2}{v}\cdot \frac{ \nabla v_2 }{v_2}
   \right)\\
&\leq (1-\theta)\, \frac{v_1}{r^r}\cdot
  N\left( x, \frac{ \nabla v_1 }{v_1} \right)
       + \theta\,   \frac{v_2}{r^r}\cdot
  N\left( x, \frac{ \nabla v_2 }{v_2} \right)\\
& = (1-\theta)\cdot N\left( x, \nabla (v_1(x)^{1/r})\right)
      + \theta\cdot N\left( x, \nabla (v_2(x)^{1/r})\right) \,.
\end{split}
\end{equation}
Finally, by {\rm Remark~\ref{rem-ray-convex}},
we conclude that inequality~\eqref{eq:v<v_1+v_2} entails
%
\begin{equation}
\label{eq:A:v<v_1+v_2}
\begin{aligned}
&   A\left( x, \nabla (v(x)^{1/r})\right)
  \leq
    (1-\theta)\cdot A\left( x, \nabla (v_1(x)^{1/r})\right)
      + \theta\cdot A\left( x, \nabla (v_2(x)^{1/r})\right) \,.
\end{aligned}
\end{equation}
We integrate the last inequality \eqref{eq:A:v<v_1+v_2} over $\Omega$ to derive
the convexity of the restriction of
the functional $\mathcal{W}_A$ to the convex cone $\dotV\subset X$.

To derive that $\mathcal{W}_A$
is even ray\--strictly convex on $\dotV$, let us consider any pair
$v_1, v_2\in \dotV$ with $v_1\not\equiv v_2$ in~$\Omega$.
We observe that the equality in the convexity inequality
\eqref{e:ray-strictly} forces both conditions,
\eqref{e:v_1=v_2} and $p(x)/r = 1$, to hold simultaneously
at almost every point $x\in \Omega$.
These conditions are then equivalent with
$v_2/v_1\equiv \mathrm{const}$ ($\neq 1$) in $\Omega$ and
$p(x)\equiv r$ in $\Omega$.
Thus, if $p(x)\not\equiv r$ in $\Omega$,
then $\mathcal{W}_A$ is even strictly convex on~$\dotV$.
%\null\hfill\qed
\qed
%%%%%%%%%%%%%%%%%

A consequence of Theorem \ref{thm-ray-convex} is the following extension of D\'iaz-Saa inequality also proved in \cite{GiTa} {in a weaker form}:
{\begin{thm}\label{thm-Diaz-Saa}
Let\/ $r\in [1,\infty)$ and\/
$p\colon \Omega\to (1,\infty)$ satisfy conditions in Theorem \ref{thm-ray-convex}.
Assume that\/
%
\begin{math}
  A\colon \overline{\Omega}\times \RR^d\to \RR_+
\end{math}
%
satisfies ({\bf A1})-({\bf A4}), ({\bf A7}). Then the following inequality\/
%
\begin{equation}
\label{in_A:Diaz-Saa}
  \int_{\Omega}
  \left(
{}- \frac{ \mathrm{div}\, \mathbf{a}(x, \nabla v_1(x)) }{ v_1(x)^{r-1} }
  + \frac{ \mathrm{div}\, \mathbf{a}(x, \nabla v_2(x)) }{ v_2(x)^{r-1} }
  \right)
  (v_1^r - v_2^r) \,\mathrm{d}x\geq 0
\end{equation}
%
holds (in the sense of distributions)
for all pairs $v_1, v_2\in \X$, such that
$v_1 > 0$, $v_2 > 0$ a.e.\ in $\Omega$ and both
$v_1/v_2$, $v_2/v_1\in L^{\infty}(\Omega)$.
Moreover, if the equality in \eqref{in_A:Diaz-Saa} occurs,
then we have the following two statements:
%
\begin{itemize}
%
\item[{\rm (a)}]
$\;$
$v_2/v_1\equiv \mathrm{const} > 0$ in $\Omega$.
%
\item[{\rm (b)}]
$\;$
If also $p(x)\not\equiv r$ in $\Omega$, then even
$v_1\equiv v_2$ holds in $\Omega$.
%
\end{itemize}
%

\end{thm}}
%
\proof
Recalling {\rm Definition~\ref{def-ray-convex}},
let us consider any pair
$w_1, w_2\in \X$, such that
$w_1 > 0$, $w_2 > 0$ a.e.\ in $\Omega$ and both
$w_1/w_2$, $w_2/w_1\in L^{\infty}(\Omega)$.
Consequently, there is a sufficiently small number
$\delta\in (0,1)$ such that
%
\begin{equation*}
  v\eqdef (1-\theta) w_1^r + \theta w_2^r\in \dotV
    \quad\mbox{ and }\quad
  v^{1/r}\in \X
    \;\mbox{ for all }\, \theta\in (-\delta, 1+\delta) \,.
\end{equation*}
%
The function
%
\begin{equation*}
  \theta\mapsto W(\theta)\eqdef \mathcal{W}(v)
  = \mathcal{W}_A\left( (1-\theta) w_1^r + \theta w_2^r\right)
    \colon (-\delta, 1+\delta)\to \RR_+
\end{equation*}
%
is convex and differentiable with the derivative
%
\begin{equation*}%\label{e:dW_A/d_theta}
  W'(\theta) =
  \int_{\Omega} \mathbf{a}(x, \nabla (v(x)^{1/r}))\cdot
  \nabla \genfrac{(}{)}{}0{ w_2^r - w_1^r }{ v^{ 1 - \frac{1}{r} } }
  \,\mathrm{d}x \,.
\end{equation*}
%
The monotonicity of the derivative
%
\begin{math}
  \theta\mapsto W'(\theta)\colon (-\delta, 1+\delta)\mapsto \RR
\end{math}
%
yields $W'(0)\leq W'(1)$, which is equivalent with
%
\begin{equation}\label{in_A:Diaz-Saa:int}
\int_{\Omega} \mathbf{a}(x, \nabla w_1(x))\cdot \nabla
  \left( w_1 - \frac{w_2^r}{w_1^{r-1}} \right) \,\mathrm{d}x
\geq
  \int_{\Omega} \mathbf{a}(x, \nabla w_2(x))\cdot \nabla
  \left( \frac{w_1^r}{w_2^{r-1}} - w_2 \right) \,\mathrm{d}x,
\end{equation}
%
thanks to
$v =  w_1^r$ if $\theta = 0$, and
$v =  w_2^r$ if $\theta = 1$.

It is now easy to see that inequality~\eqref{in_A:Diaz-Saa} is a distributional interpretation of \eqref{in_A:Diaz-Saa:int} after integration by parts.

Finally, let us assume that the equality in \eqref{in_A:Diaz-Saa} is valid.
This forces $W'(0) = W'(1)$ above; hence, $W'(\theta) = W'(0)$ for all $\theta\in [0,1]$, by the monotonicity of $W'\colon [0,1]\mapsto \RR$.
It follows that $W\colon [0,1]\to \RR$ must be linear, i.e.,
$W(\theta) = (1-\theta) W(0) + \theta W(1)\in \RR$
for all $\theta\in [0,1]$.
Recalling our definition of $W$ , {Remark~\ref{rem-ray-convex}, assumption ({\bf A3}) (which implies the strict convexity of $\xi\mapsto A(x,\xi)$ for each fixed $x\in \Omega$) and Theorem~\ref{thm-ray-convex}},
we conclude that $w_2/w_1\equiv \mathrm{const} > 0$ in $\Omega$.
This proves statement {\rm (a)}.

To verify statement {\rm (b)}, suppose that the constant above
$w_2/w_1\equiv \mathrm{const} \neq 1$ in $\Omega$.
Then the equality in both inequalities,
\eqref{eq:v<v_1+v_2} and \eqref{eq:A:v<v_1+v_2},
is possible only if $p(x)\equiv r$ in $\Omega$.
Statement {\rm (b)} follows.
%\null\hfill\qed
\qed
%
\begin{rem}\label{rem-ray-convexb}\nopagebreak
\begingroup\rm
{An alternative proof of Theorem~\ref{thm-Diaz-Saa} is to show a Picone inequality in a similar fashion as in \cite{BrFr} (see Proposition~2.9 and Remark~2.10). Precisely, we can derive the following Picone inequality:
%
\begin{equation}\label{picone-id}
\begin{split}
A(x,\nabla v_2)^{p(x)-r/p(x)}A(x,\nabla v_1)^{r/p(x)}&\geq \left\langle \frac{1}{p(x)}\partial_\xi A(x,\nabla v_2),\nabla(v_1^r/v_2^{r-1})\right\rangle\\
&= \left\langle {\frac{1}{p(x)}}\mathbf{a}(x,\nabla v_2),\nabla(v_1^r/v_2^{r-1})\right\rangle
\end{split}
\end{equation}
for any $v_1$, $v_2$ satisfying assumptions in Theorem~\ref{thm-Diaz-Saa} and each fixed $x\in \Omega$. Combining \eqref{picone-id}, Remark \ref{r1.7} and the Young inequality, we then easily derive \eqref{in_A:Diaz-Saa:int} and \eqref{in_A:Diaz-Saa}. Note that inequality \eqref{picone-id} being pointwise is stronger than the Diaz-Saa inequality \eqref{in_A:Diaz-Saa} and is of independent interest.}
%
\endgroup
\end{rem}
{
\noindent{\bf Proof of Theorem \ref{uniqueness-stat}:} Let $u$ be a weak solution to \eqref{sta}. From ($f_1$), ($f_3$) and ($f_6$), there exists $M>0$ large enough such that:
%
\begin{equation}\label{growth-f}
0\leq f(x,s)\leq M+|s|^{p^--1} \quad\forall s\geq 0\mbox{ and }x\in\Omega.
\end{equation}
%
Hence, from Theorem~4.1 in \cite{FZ} page~312, it follows that $u$ is bounded and from Theorem~1.2 in \cite{Fa0}, belongs to $C^{1,\alpha}(\overline{\Omega})$. From the Hopf boundary point lemma in \cite{Zh} (see Theorems~1.2), $u$ satisfies $\partial u/\partial\nu<0$ on $\partial\Omega$. Therefore, for any pair $u$, $v$ of weak solutions to \eqref{sta}, $u/v$ and $v/u$ belongs to $L^\infty(\Omega)$. Then let ${\mathcal J}_{p^-}$ defined in $\dotV$ by
%
\begin{equation*}%\label{convexfunct}
{\mathcal J}_{p^-}(w)\eqdef\int_{\Omega} A(x,\nabla (w^{1/{p^-}}))\,\mathrm{d}x-\int_{\Omega}F(x, w^{1/{p^-}})\,\mathrm{d}x
\end{equation*}
%
where $F(x,t)=\int_0^tf(x,s)\,\mathrm{d}s$.

From Theorem~\ref{thm-ray-convex} and ($f_6$), ${\mathcal J}$ is strictly convex on $\dotV$. Let $w_1=u^{p^-}$ and $w_2=v^{p^-}$. Hence, the function $J_{p^-}\colon t\to {\mathcal J}_{p^-}(w_2+t(w_1-w_2)^+)$ is well defined, convex and differentiable on $[0,1]$. Since $u$, $v$ are weak solutions to \eqref{sta}, $J_{p^-}'(0)=J_{p^-}'(1)=0$.

According to Theorem~\ref{thm-Diaz-Saa} with $v_1=w_2+t(w_1-w_2)^+$ and $v_2=w_2$, assertion (b) implies  $(w_1-w_2)^+\equiv 0$, that is $u\leq v$. Interchanging the role of $u$ and $v$, we get $u=v$. This proves the uniqueness of the weak solution to \eqref{sta}.

Now, supposing that ($f_4$), ($f_5$) and $f(x,0)=0$ are satisfied.  Extending $f$ by $f(x,s)=0$ for $s\leq 0$, we apply Lemma \ref{exist}, with $h=0$ to obtain the existence of a solution $u\in \X\cap L^\infty(\Om)$.
%
%\begin{eqnarray*}
%{\mathcal J}(v)\eqdef \int_{\Omega}A(x,\nabla v)\,\mathrm{d}x-\int_{\Omega}F(x,v)\,\mathrm{d}x
%\end{eqnarray*}
%
%for any non negative $v\in W^{1,p(x)}_0(\Omega)$. Extending $f$ by $f(x,s)=0$ for $s\leq 0$, ${\mathcal J}$ is well defined and $C^1$ in $W^{1,p(x)}_0(\Omega)$. From (f4), ${\mathcal J}$ is coercive and $-\infty<\displaystyle\inf_{w\in W^{1,p(x)}_0(\Omega)}{\mathcal J}(w)<0$. Since $W^{1,p(x)}_0(\Omega)$ is reflexive and from the compact embedding $W^{1,p(x)}_0(\Omega)$ in $L^{p-}(\Omega)$, ${\mathcal J}$ admits a non negative global minimizer $u\in W^{1,p(x)}_0(\Omega)$ and $u\not\equiv 0$.
From the strong maximum principle given by Theorem~1.1 in \cite{Zh}, $u$ is positive in $\Omega$ and then a weak solution to \eqref{sta}. This completes the proof of Theorem~\ref{uniqueness-stat}.\hfill\qed}
%%%%%%%%%%%%%%%%

\subsection{Proof of Theorem \ref{asymptotic-behaviour}}
Let $T>0$. Note that from assumptions ($f_1$), ($f_6$), $f(x,0)=0$ and since $f$ is locally Lipschitz in respect to the second variable uniformly in $x\in\Omega$, there exists $M>0$ large enough such that \eqref{growth-f} is still valid.
%
%\begin{equation}\label{control-f}
%0\leq f(x,s)\leq M(1+s)^q\quad\forall\,(x,s)\in\Omega\times\RR^+.
%\end{equation}
%
 The existence of $u\in W^{1,2}(0,T;L^2(\Omega))\cap C([0,T], \X)\cap L^\infty(Q_T)$, unique weak solution to $({\rm P}_T)$ with initial data $u_0\in \X\cap L^\infty(\Omega)$ and $T$ small enough follows from Theorem \ref{PT1}. The $L^\infty$ bound is provided by the barrier function $v$ solution to:
%
\begin{equation*}%\label{barrier-sup}
 \left\{\begin{array}{ll}
 \displaystyle\frac{dv}{dt} = M+v^{p^--1},\quad t\in(0,T), & \\
 v(0)  =\|u_0\|_{\infty}.&
 \end{array}\right.
\end{equation*}
%
Note that the uniqueness of the weak solution and the statement $u\leq v$ in $Q_T$ follow from the local Lipschitz property of $f$ and the monotonicity of $\nabla\cdot \mathbf{a}(x,\nabla\cdot)$.
To get the existence of global weak positive solutions to $({\rm P}_T)$, we need to construct a subsolution $\underline{u}$ and a supersolution $\bar{u}$ independent of $t$:
%
\begin{equation*}%\label{subsolution}
 \left\{
\begin{array}{l l}
-\nabla\cdot \mathbf{a}(x,\nabla\underline{u})=\lambda \underline{u}^{p^--1}& \mbox{ in } \Omega,\\
\underline{u}=0& \mbox{ on } \partial \Omega
\end{array}\right.
\end{equation*}
and $\bar{u}=\tilde w_{M,\varepsilon}$ defined in \eqref{sursolutionc2}
where $\lambda>0$ is small enough and $M>0$ large enough. From Theorem 4.1 in \cite{FZ}, $\underline{u}$ and $\bar{u}$ are bounded. From Theorem 1.2 in \cite{Fa0} and Theorems 1.1 and 1.2 in \cite{Zh}, they belong to $ C^{1,+}_0(\overline{\Omega})\cap C^{1,\alpha}(\overline\Omega)$ for some $\alpha\in (0,1)$. Hence, it is easy to prove that $\Vert \underline{u}\Vert_{C^1(\overline{\Omega})}\to 0$ as $\lambda\to 0^+$. From Lemma \ref{31YY}, we have that there exist $C_1, C_2>0$ independent of $M$ such that $C_1M^{\frac{1}{p^+-1+\nu}}\rho(x)\leq \bar{u}(x)$ for some $0<\nu<1$. Therefore, from ($f_4$) and ($f_5$) for $\lambda$ small enough and $M$ large enough we have that $\underline{u}$ (resp. $\bar{u}$) is a subsolution (resp. a supersolution) to \eqref{sta} and $\underline{u}\leq u_0\leq\bar{u}$. Thus, $u$ is a global weak solution to $({\rm P}_T)$ and using the weak comparison principle on the discrete time-approximated scheme \eqref{eq 13} we obtain $\underline{u}\leq u(t)
 \leq \bar{u}$ for any $0\leq t<\infty$. Using Theorem \ref{continuity u with f}, we obtain that $u_1$ (resp. $u_2$) the weak solution to $({\rm P}_T)$ with initial data $\underline{u}$ (resp. $\bar{u}$) is a mild solution. Then, $u_1$ (resp. $u_2$) belongs to $C([0,\infty), C_0(\overline{\Omega}))$ and since $\underline{u}$ (resp. $\bar{u}$) is a subsolution (resp. a supersolution) $[0,\infty)\ni t\to u_1(t)$ (resp. $[0,\infty)\ni t\to u_2(t)$) is nondecreasing (resp. non increasing). Hence, $u_1$ and $u_2$ converge (pointwise) to a positive steady state as $t\to\infty$. From Theorem \ref{uniqueness-stat}, $({\rm P}_T)$ admits a unique positive and continuous stationary (weak) solution $u_\infty$ and hence by Dini's theorem we infer that
\begin{eqnarray*}
\Vert u_1(t)-u_\infty\Vert_{L^\infty(\Omega)}\to 0,\; \Vert u_2(t)-u_\infty\Vert_{L^\infty(\Omega)}\to 0\quad \mbox{as }t\to\infty
\end{eqnarray*}
which reveals $\Vert u(t)-u_\infty\Vert_{L^\infty(\Omega)}\to 0$ as $t\to\infty$ since $u_1(t)\leq u(t)\leq u_2(t)$.
\hfill\qed

\appendix
\section{Regularity result}\label{ApC}
\noindent We begin by recalling the following compactness embedding:
\begin{thm}\label{SE1}
Let $p\in \mathcal P^{log}(\Omega)$ satisfies $1\leq p^-\leq p^+<d$. Then,
$W^{1,p(x)}(\Om)\hookrightarrow L^{\al(x)}(\Om)$ for any $\al\in L^\infty(\Omega)$  such that for all $x\in \Om$, $\al(x)\leq p^*(x)=\frac{d\,p(x)}{ d-p(x)}.$
Also the previous embedding is compact for $\al(x)<p^*(x)-\varepsilon$ a.e. in $\Om$ for any $\varepsilon>0$.
\end{thm}
\noindent Next we recall the regularity result due to Fan and Zhao \cite{FZ}:
\begin{prop}[Theorem 4.1 in \cite{FZ}]\label{reg0}
{Assume conditions ({\bf A1})-({\bf A4}).} Let $p\in C(\overline\Omega)$ and $u\in \X$ satisfying
\begin{equation*}
\int_\Omega \mathbf{a}(x,\nabla u) .\nabla \varphi\,\mathrm{d}x=\int_\Omega f(x,u) \varphi\,\mathrm{d}x, \quad \forall \varphi \in \X,
\end{equation*}
where $f$ satisfies for all $(x,t)\in \Omega\times\R$, $|f(x,t)|\leq c_1+c_2|t|^{r(x)-1}$ with $r\in C(\overline\Om)$ and $\forall x\in\overline \Om$, \break $1<r(x)<p^*(x)$. Then $u\in L^\infty(\Omega)$.
\end{prop}
\noindent For $f(x,.)=f(x)$, we have the following proposition.
\begin{prop}\label{rg1}
{Assume conditions ({\bf A1})-({\bf A4}).} Let $p\in C(\bar\Omega)$ { with $p^-<d$} and $u\in \X$ satisfying
\begin{equation}\label{ws}
\int_\Omega \mathbf{a}(x,\nabla u) .\nabla \varphi\,\mathrm{d}x=\int_\Omega f \varphi\,\mathrm{d}x, \quad \forall \varphi \in \X,
\end{equation}
where $f\in L^{q}(\Omega)$, $q>\frac{d}{p^-}$. Then $u\in L^\infty(\Omega)$.
\end{prop}
\noi To prove Proposition \ref{rg1}, we have the following regularity lemma (see Fusco and Sbordone \cite{FS} and Giacomoni et al. \cite{GTW})
{\begin{lem}\label{fsb1}
Let $u\in W_0^{1,p}(\Omega)$, $1<p<d$, satisfying for any  $B_R$, $R<R_0$, and for all $\sigma\in (0,1)$, and
any $k\geq k_0>0$
\begin{equation*}\label{equa}
\begin{split}
\int_{A_{k,\sigma R}} |\nabla u|^p\,\mathrm{d}x &\leq C\left[ \int_{A_{k, R}}
\left|\frac{u-k}{R(1-\sigma)}\right|^{p^*}\,\mathrm{d}x+k^\alpha|A_{k,R}|+ |A_{k,R}|^{\frac{p}{p^*}+\varepsilon}\right.\\
&\left.+\left(\int_{A_{k, R}} \left|\frac{u-k}{R(1-\sigma)}\right|^{p^*}\,\mathrm{d}x\right)^{\frac{p}{p^*}}|A_{k,R}|^\delta\right]
\end{split}
\end{equation*}
where $A_{k,R}=\{x\in B_R\cap \Omega\ |\ u(x)>k\}$, $0<\alpha<p^*=\frac{dp}{d-p}$ and $\varepsilon,\,\delta>0$. Then $u\in L^\infty(\Omega)$.
\end{lem}


\noindent \textbf{Proof of Proposition \ref{rg1}:} We follow the idea of the proof of Theorem 4.1 in \cite{FZ}.

Let $x_0\in \overline\Omega$, $B_{R}$ the ball of radius $R$ centered in $x_0$ and $K_R:=\Omega\cap B_R$. We define
$$p^+:=\max_{K_{R}} p(x) \quad \mbox{and} \quad p^-:=\min_{K_{R}} p(x)$$
and we choose $R$ small enough such that $p^+< (p^-)^*:=\frac{dp^-}{d-p^-}$. 
Fix $(s,t)\in (\R_+^*)^2$, $t<s<R$ then  $K_t\subset K_s\subset K_R$. Define $\varphi\in
C^\infty(\Omega)$, $0\leq \varphi\leq 1$ such that
$$\varphi=\left\{\begin{array}{l l}
1 & \mbox{in } B_t\\
0 & \mbox{in }\R^d\backslash B_s
\end{array}\right.$$
satisfying $|\nabla\varphi|\lesssim 1/(s-t)$. Let $k\geq 1$, using the same notations as previously $A_{k,\lambda}=\{y\in K_\lambda \ | \ u(y)>k\}$ and taking $\varphi^{p^+}(u-k)^+\in \X$ in \eqref{ws} as test function, we obtain
{\begin{equation}\label{B3}
\int_{A_{k,s}} \mathbf{a}(x,\nabla u)\cdot\nabla u\varphi^{p^+}\,\mathrm{d}x+p^+\int_{A_{k,s}}\mathbf{a}(x,\nabla u) \cdot\nabla \varphi
\,(u-k)^+\varphi^{p^+-1}\,\mathrm{d}x=\int_{A_{k,s}} f\varphi^{p^+}(u-k)\,\mathrm{d}x.
\end{equation}}
Hence by Young's inequality, for $\epsilon>0$, we have
\begin{equation*}
\begin{split}
p^+\int_{A_{k,s}}\mathbf{a}(x,\nabla u) \cdot\nabla \varphi (u-k)\varphi^{p^+-1}\,\mathrm{d}x\leq& \varepsilon \int_{A_{k,s}} |\mathbf{a}(x,\nabla u)|^{\frac{p(x)}{p(x)-1}}\varphi^{(p^+-1)\frac{p(x)}{p(x)-1}}\,\mathrm{d}x\\
& +c\varepsilon^{-1}\int_{A_{k,s}} (u-k)^{p(x)}|\nabla\varphi|^{p(x)}\,\mathrm{d}x.
\end{split}
\end{equation*}
Since $|\nabla\varphi|\leq c/(s-t)$ and for any $x\in K_R$, $p^+\leq (p^+-1)\frac{p(x)}{p(x)-1}$, we have
$\varphi^{(p^+-1)\frac{p(x)}{p(x)-1}}\leq \varphi^{p^+}$.\\
Hence using \eqref{control-a}, this implies
\begin{equation}\label{est1}
p^+\int_{A_{k,s}} \mathbf{a}(x,\nabla u)\cdot\nabla \varphi\,(u-k) \varphi^{p^+-1}\,\mathrm{d}x\lesssim \varepsilon \int_{A_{k,s}} |\nabla
u|^{p(x)}\varphi^{p^+}\,\mathrm{d}x + \varepsilon^{-1}\int_{A_{k,s}} \left(\frac{u-k}{s-t}\right)^{p(x)}\,\mathrm{d}x.
\end{equation}
Using H\"older inequality we estimate the right-hand side of \eqref{B3} as follows:
\begin{equation*}
\int_{A_{k,s}} f\varphi^{p^+}(u-k)\,\mathrm{d}x\leq \|f\|_{L^q}\left(\int_{A_{k,s}}
(u-k)^{\frac{q}{q-1}}\,\mathrm{d}x\right)^{\frac{q-1}{q}}.
\end{equation*}
Since $q>\frac{d}{p^-}$, we have $\frac{(p^-)^*}{p^-}\frac{q-1}{q}>1$. So, applying once again the H\"older inequality, we
obtain
\begin{equation}\label{est2}
\int_{A_{k,s}} f\varphi^{p^+}(u-k)\,\mathrm{d}x\lesssim  \left(\int_{A_{k,s}}
(u-k)^{\frac{(p^-)^*}{p^-}}\,\mathrm{d}x\right)^{\frac{p^-}{(p^-)^*}}|A_{k,s}|^\delta,
\end{equation}
where $\delta= \frac{q-1}{q}-\frac{p^-}{(p^-)^*}>0$. Set $\{u-k>s-t\}=\{x\in K_R\ |\ u(x)-k>s-t\}$ and
its complement as $\{u-k\leq s-t\}$. Now we split the integral in the right-hand side of \eqref{est2} as follows:
\begin{equation}\label{est3}
\begin{split}
\int_{A_{k,s}\cap\{u-k>s-t\}} \left(\frac{u-k}{s-t}\right)^{\frac{(p^-)^*}{p^-}}(s-t)^{\frac{(p^-)^*}{p^-}}\,\mathrm{d}x &+
\int_{A_{k,s}\cap\{u-k\leq s-t\}} \left(\frac{u-k}{s-t}\right)^{\frac{(p^-)^*}{p^-}}(s-t)^{\frac{(p^-)^*}{p^-}}\,\mathrm{d}x\\
&\lesssim \int_{A_{k,s}} \left(\frac{u-k}{s-t}\right)^{(p^-)^*}\,\mathrm{d}x+|A_{k,s}|:=\mathcal I.
\end{split}
\end{equation}
In the same way, the second term in the right-hand side of \eqref{est1} can be estimated as follows.
\begin{equation}\label{est4}
\int_{A_{k,s}\cap\{u-k>s-t\}} \left(\frac{u-k}{s-t}\right)^{p(x)}\,\mathrm{d}x +\int_{A_{k,s}\cap\{u-k\leq s-t\}}
\left(\frac{u-k}{s-t}\right)^{p(x)}\,\mathrm{d}x \lesssim \mathcal I.
\end{equation}
Moreover, we have
\begin{equation}\label{est5}
\int_{A_{k,s}} \mathbf{a}(x,\nabla u)\cdot\nabla u\varphi^{p^+}\,\mathrm{d}x \gtrsim \int_{A_{k,s}} |\nabla u|^{p(x)} \varphi^{p^+}\,\mathrm{d}x \geq 0.
\end{equation}
Finally, plugging \eqref{est1}-\eqref{est5} and we obtain for $\varepsilon$ small enough
\begin{equation*}
\int_{A_{k,s}} |\nabla u|^{p(x)}\varphi^{p^+}\,\mathrm{d}x\lesssim  \mathcal I + |A_{k,s}|^\delta\mathcal I^{\frac{p^-}{(p^-)^*}}
\end{equation*}
where the constant depends on $p,\,R$ and $\varepsilon$. Moreover we have
$$\mathcal I^{\frac{p^-}{(p^-)^*}}\lesssim \left(\int_{A_{k,s}}
\left(\frac{u-k}{s-t}\right)^{(p^-)^*}\,\mathrm{d}x\right)^{\frac{p^-}{(p^-)^*}}+|A_{k,s}|^{\frac{p^-}{(p^-)^*}}.$$
Hence using the Young's inequality, we obtain the following estimate.
\begin{equation*}
\begin{split}
\int_{A_{k,t}} |\nabla u|^{p^-}\,\mathrm{d}x\leq\int_{A_{k,s}} |\nabla u|^{p(x)}\varphi^{p^+}\,\mathrm{d}x+|A_{k,s}|
&\lesssim \int_{A_{k,s}} \left(\frac{u-k}{s-t}\right)^{(p^-)^*}\,\mathrm{d}x +
|A_{k,s}|+|A_{k,s}|^{\frac{p^-}{(p^-)^*}+\delta}\\
&+{|A_{k,s}|^\delta\left(\int_{A_{k,s}}
\left(\frac{u-k}{s-t}\right)^{(p^-)^*}\,\mathrm{d}x\right)^{\frac{p^-}{(p^-)^*}}}.
\end{split}
\end{equation*}
By Lemma \ref{fsb1}, we deduce that $u$ bounded in $\Omega$.
$\hfill \square$

Combining Propositions \ref{reg0} and \ref{rg1}, we have the following corollary:
\begin{coro}\label{reg2}
Let $p\in C(\bar\Omega)$ { such that $p^-<d$} and $u\in W^{1,p(x)}_0(\Omega)$ satisfying
\begin{equation*}
\int_\Omega \mathbf{a}(x,\nabla u) \cdot\nabla \varphi\,\mathrm{d}x=\int_\Omega(f(x,u)+g) \varphi\,\mathrm{d}x, \quad \forall\, \varphi \in \
\X,
\end{equation*}
where $f$ satisfies $|f(x,t)|\leq c_1+c_2|t|^{r(x)-1}$ with $r\in C(\overline\Om)$ and $\forall x\in\overline \Om$, $1<r(x)<p^*(x)$ and $g\in L^{q}$, $q>\frac{d}{p^-}$. Then $u\in L^\infty(\Omega)$.
\end{coro}

\section{Existence of mild solutions to \eqref{Lt} and \eqref{Pt}}\label{mild-sol}

We use the theory of maximal accretive operators in Banach spaces (see Chapters 3 and 4 in \cite{Ba}), which provides the existence of mild solutions. More precisely, observing that the operator $A_0(\cdot)\eqdef-\nabla\cdot \mathbf{a}(x,\nabla(\cdot))$, with Dirichlet boundary conditions, is m-accretive in $L^\infty(\Omega)$ with
\begin{equation*}
{\mathcal D}(A_0)=\left\{u\in\X\cap L^\infty(\Omega)\,\vert\, A_0u\in L^\infty(\Omega)\right\},
\end{equation*}
we get the following properties, which essentially follow from Theorems \ref{sol for (S_t)} with Theorem 4.2 (page 130) and Theorem 4.4 in \cite[p. 141]{Ba}:
\begin{thm}\label{continuity de u} Assume conditions ({\bf A1})-({\bf A5}). Let $T>0$, $h\in L^\infty(Q_T)$ and let $u_0$ be in $\X\cap\overline{{\mathcal D}(A_0)}^{L^\infty}$. Then,
\begin{enumerate}
\item[(i)]the unique weak solution $u$ to \eqref{Lt} belongs to $\mathcal C([0,T]; \mathcal C_0(\overline{\Omega}))$;
\item[(ii)] if $v$ is another mild solution to \eqref{Lt}  with the initial datum $v_0\in\X\cap\overline{{\mathcal D}(A_0)}^{L^\infty}$ and the right-hand side $k\in L^\infty(Q_T)$, then the following estimate holds:
\begin{equation}
\|u(t)-v(t)\|_{L^\infty(\Omega)}\leq \|u_0-v_0\|_{L^\infty(\Omega)}+\int_0^t\|h(s)-k(s)\|_{L^\infty(\Omega)}\, \mathrm{d}s,\quad 0\leq t\leq T;\label{estimation2}
\end{equation}
\item[(iii)] if $u_0\in {\mathcal D}(A_0)$ and $h\in W^{1,1}(0,T;L^\infty(\Omega))$ then $u\in W^{1,\infty}(0,T;L^\infty(\Omega))$ and $\nabla\cdot \mathbf{a}(x,\nabla u)\in L^\infty(Q_T)$, and the following estimate holds:
\begin{equation}
\left \|\frac{\partial u}{\partial t}(t)\right \|_{L^\infty(\Omega)}\leq \|\nabla\cdot \mathbf{a}(x,\nabla u_0)+h(0)\|_{L^\infty(\Omega)}+\int_0^T\left \|\frac{\partial h}{\partial t}(t)\right \|_{L^\infty(\Omega)}\,\mathrm{d}\tau.\label{estimationbiss2}
\end{equation}
\end{enumerate}
\end{thm}

The m-accretivity of $A_0$ follows from the following proposition.
\begin{prop}\label{maccretive}
Assume conditions ({\bf A1})-({\bf A4}). Let $f:\Om\times \R\mapsto \R$ be a function satisfying ($f_1$) and nonincreasing with respect to the second variable. Assume further that $x\to f(x,0)$ belongs to $L^\infty(\Omega)$. Then, $A_f$ defined by $A_f(u)\eqdef-\nabla\cdot \mathbf{a}(\cdot,\nabla u)-f(\cdot,u)$ is m-accretive in $L^\infty(\Omega)$.
\end{prop}
\begin{proof}
 First, let $h\in L^\infty(\Omega)$ and $\lambda>0$. Then,
\begin{equation*}
\left\{\begin{array}{ll}
 u+\lambda A_f(u)=h &\mbox{in }\Omega,\\
 u =0, &\mbox{on } \partial\Omega
\end{array}\right.
\end{equation*}
admits a unique solution, $u\in\X\cap L^\infty(\Omega)$.
Indeed, for $\mu>0$ large enough $\mu$ and $-\mu$ are respectively supersolution and subsolution to the above equation and then from the weak comparison principle, $u\in [-\mu,\mu]$ and $u$ is obtained by a minimization argument and  a truncation argument. The uniqueness of the solution follows from the strict convexity of the associated energy functional. Next we prove the accretivity of $A_f$. Let $h$ and $g\in L^\infty(\Omega)$ and set $u$ and $v$ the unique solutions to
\begin{equation*}
%\label{accretive}
\begin{aligned}
& u+\lambda A_fu=h\quad\mbox{in }\,\Omega,\\
& v+\lambda A_fv=g\quad\mbox{in }\,\Omega.
\end{aligned}
\end{equation*}
Substracting the two above equations and using the test function $w\eqdef \left(u-v-\|h-g\|_{L^\infty(\Omega)}\right)^+$, we get $u-v\leq\|h-g\|_{L^\infty(\Omega)}$ and reversing the roles of $u$ and $v$, we get that $\|u-v\|_{L^\infty(\Omega)}\leq\|h-g\|_{L^\infty(\Omega)}$. This proves the proposition.
\end{proof}
\noindent {The proof of Theorem~\ref{continuity de u}, similar to the proof of Theorem~2.8 in \cite{GTW}, is given below}.
\noindent{\bf Proof of Theorem \ref{continuity de u}:} We follow the approach developed in the proof of Theorems 4.2 and 4.4 in \cite{Ba}. Let $u_0$, $v_0$ be in $\overline{{\mathcal D}(A_0)}^{L^\infty(\Omega)}$.
For $z\in {\mathcal D}(A_0)$ and $r,k$ in $L^\infty(Q_T)$, set
$$
\vartheta(t,s)=\|r(t)-k(s)\|_{L^\infty(\Om)}\quad \forall \, (t,s)\in [0,T]\times [0,T];
$$
\begin{equation*}
\begin{split}
b(t,r,k)&=\|u_0-z\|_{L^\infty(\Om)}+ \|v_0-z\|_{L^\infty(\Om)}+|t|\|A_0 z \|_{L^\infty(\Om)}\\
&+\int_0^{t^+}\|r(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau+\int_0^{t^-}\|k(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau,\;t\in [-T,T],
\end{split}
\end{equation*}
and
$$
\Upsilon(t,s)=b(t-s,r,k)+
\left\{
\begin{array}{ll}
\displaystyle \int_0^{s}\vartheta(t-s+\tau,\tau)\,\mathrm{d}\tau & \mbox{ if } 0\leq s\leq t\leq T,\\
\displaystyle \int_0^{t}\vartheta(\tau,s-t+\tau)\,\mathrm{d}\tau &  \mbox{ if } 0\leq t\leq s\leq T,
\end{array}
\right.
$$
the solution of
\begin{equation}\label{eqPsi}
\left\{
\begin{array}{rcll}
\displaystyle \frac{\partial \Upsilon}{\partial t}(t,s)+\frac{\partial \Upsilon}{\partial s}(t,s)&=&\displaystyle\vartheta(t,s)\; &(t,s)\in [0,T]\times [0,T],\\
\displaystyle \Upsilon(t,0)&=& \displaystyle b(t,r,k)\; &t\in [0,T],\\
 \displaystyle  \Upsilon(0,s)& =& \displaystyle b(-s,r,k)\; &s\in [0,T].
\end{array}
\right.
\end{equation}
Moreover, {let us denote} by $(u_\epsilon^n)$ the solution of \eqref{eq1} with $\Delta_t=\epsilon$, $h=r$, $r^n=\frac{1}{\epsilon}\int_{(n-1)\epsilon}^{n\epsilon}r(\tau,\cdot)\,\mathrm{d}\tau$ and  $(u_\eta^n)$ the solution of \eqref{eq1} with  $\Delta_t=\eta$, $h=k$, $k^n=\frac{1}{\eta}\int_{(n-1)\eta}^{n\eta}k(\tau,\cdot)\,\mathrm{d}\tau$ respectively. For $(n,m)\in \mathbb{N}^*$ elementary calculations { lead} to
\begin{equation*}
\begin{split}
u_\epsilon^n-u_\eta^m+\frac{\epsilon\eta}{\epsilon+\eta}(A_0 u_\epsilon^n-A_0 u_\eta^m)=&\frac{\eta}{\epsilon+\eta}(u_\epsilon^{n-1}-u_\eta^m)\\
+\frac{\epsilon}{\epsilon+\eta}(u_\epsilon^n-u_\eta^{m-1})+&\frac{\epsilon\eta}{\epsilon+\eta}(r^n-k^m),
\end{split}
\end{equation*}
and since $A_0$ is m-accretive in $L^\infty(\Omega)$ we first verify that $\digamma_{n,m}^{\epsilon,\eta}=\|u_\epsilon^n-u_\eta^m\|_{L^\infty(\Om)}$ obeys
\begin{equation*}
\begin{split}
\digamma_{n,m}^{\epsilon,\eta}&\leq \frac{\eta}{\epsilon+\eta}\digamma_{n-1,m}^{\epsilon,\eta}+\frac{\epsilon}{\epsilon+\eta}\digamma_{n,m-1}^{\epsilon,\eta}+\frac{\epsilon\eta}{\epsilon+\eta}\|r^n-k^m\|_\infty,\\
\digamma_{n,0}^{\epsilon,\eta}&\leq  b(t_n,r_\epsilon,k_\eta)\quad\mbox{ and }\quad  \digamma_{0,m}^{\epsilon,\eta}\leq  b(-s_m ,r_\epsilon,k_\eta),
\end{split}
\end{equation*}
and thus, with an easy inductive argument, that $\digamma_{n,m}^{\epsilon,\eta}\leq \Upsilon_{n,m}^{\epsilon,\eta}$ where $\Upsilon_{n,m}^{\epsilon,\eta}$ satisfies
\begin{equation*}
\begin{split}
\Upsilon_{n,m}^{\epsilon,\eta}&= \frac{\eta}{\epsilon+\eta}\Upsilon_{n-1,m}^{\epsilon,\eta}+\frac{\epsilon}{\epsilon+\eta}\Upsilon_{n,m-1}^{\epsilon,\eta}+\frac{\epsilon\eta}{\epsilon+\eta}\|h_\epsilon^n-h_\eta^m\|_\infty,\\
\Upsilon_{n,0}^{\epsilon,\eta}&=  b(t_n,r_\epsilon,k_\eta)\quad\mbox{ and }\quad  \Upsilon_{0,m}^{\epsilon,\eta}=  b(-s_m ,r_\epsilon,k_\eta).
\end{split}
\end{equation*}
For $(t,s)\in (t_{n-1},t_n)\times (s_{m-1},s_m)$, set
\begin{equation*}
\begin{split}
&\vartheta^{\epsilon,\eta}(t,s)=\|r_\epsilon(t)-k_\eta(s)\|_{\infty},\\
&\Upsilon^{\epsilon,\eta}(t,s)=\Upsilon_{n,m}^{\epsilon,\eta},\\
&b_{\epsilon,\eta}(t,r,k)=b(t_n,r_\epsilon,k_\eta)
\end{split}
\end{equation*}
and $$b_{\epsilon,\eta}(-s,r,k)=b(-s_m,r_\epsilon,k_\eta).$$
 Then $\Upsilon^{\epsilon,\eta}$ satisfies the following discrete version of \eqref{eqPsi}:
\begin{equation*}
\begin{split}
\frac{\Upsilon^{\epsilon,\eta}(t,s)-\Upsilon^{\epsilon,\eta}(t-\epsilon,s)}{\epsilon}+\frac{\Upsilon^{\epsilon,\eta}(t,s)-\Upsilon^{\epsilon,\eta}(t,s-\eta)}{\eta}&= \vartheta^{\epsilon,\eta}(t,s),\\
\Upsilon^{\epsilon,\eta}(t,0)= b_{\epsilon,\eta}(t,r,k)\quad\mbox{ and }\quad  \Upsilon^{\epsilon,\eta}(0,s)&=  b_{\epsilon,\eta}(s,r,k),
\end{split}
\end{equation*}
and from $b_{\epsilon,\eta}(\cdot,r,k)\to b(\cdot,r,k)$ in $L^\infty([0,T])$. Furthermore,
\begin{equation*}
\begin{split}
&\sum_{n=1}^{N_n}\int_{t_{n-1}}^{t_n}\Vert r(s)-r_n\Vert_\infty\,\mathrm{d}s\to 0, \quad\mbox{ as } \epsilon\to 0^+, \\
&\sum_{m=1}^{N_m}\int_{s_{m-1}}^{s_m}\Vert k(s)-k_m\Vert_\infty\,\mathrm{d}\tau\to 0\quad\mbox{ as }\eta\to 0^+.
\end{split}
\end{equation*}
The above statements follow easily from the fact that $r, k\in L^1(0,T; L^\infty(\Omega))$ and a density argument.

We deduce that $\rho_{\epsilon,\eta}=\|\Upsilon^{\epsilon,\eta}-\Upsilon\|_{L^\infty([0,T]\times [0,T])} \to 0$ as $(\epsilon,\eta)\to 0$ (see for instance  \cite[Chap.4, Lemma~4.3, p.~136]{Ba} and \cite[Chap.4, proof of Theorem~4.1, p.~138]{Ba}). Then from
\begin{equation}
\|u_\epsilon(t)-u_\eta(s)\|_{\infty}=\digamma^{\epsilon, \eta}(t,s)\leq \Upsilon^{\epsilon, \eta}(t,s)\leq \Upsilon(t,s)+\rho_{\epsilon,\eta},\label{eqconv}
\end{equation}
we obtain with $t=s$, $r=k=h$, $v_0=u_0$:
$$
\|u_\epsilon(t)-u_\eta(t)\|_{L^\infty(\Om)}\leq 2\|u_0-z\|_{L^\infty(\Om)}+\rho_{\epsilon, \eta},
$$
and since $z$ can be chosen in $\mathcal{D}(A_0)$ arbitrary close to $u_0$, we deduce that $u_\epsilon$ is a Cauchy sequence in $L^\infty(Q_T)$ and then that $u_\epsilon \to u$ in $L^\infty(Q_T)$. Thus, passing to the limit in \eqref{eqconv} with $r=k=h$, $v_0=u_0$ we obtain
\begin{equation*}
\begin{split}
\|u(t)-u(s)\|_{L^\infty(\Om)}&  \leq2\|u_0-z\|_{L^\infty(\Om)}+|t-s|\|A_0 z \|_{L^\infty(\Om)}\\
&+\int_0^{|t-s|}\|h(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau\\ &
+\int_{0}^{\max(t,s)}\|h(|t-s|+\tau)-h(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau,
\end{split}
\end{equation*}
which, together with the density ${\mathcal D}(A_0)$ in $L^\infty(\Omega)$ and  $h\in L^1(0,T;L^\infty(\Omega))$, yields $u\in C([0,T];L^\infty(\Omega))$. Analogously, from \eqref{eqconv} with $\varepsilon=\eta=\Delta_t$, $r=k=h$, $v_0=u_0$ and $t=s+\Delta_t$ we deduce that
\begin{equation*}
\begin{split}
\|u_{\Delta_t}(t)-\Tilde u_{\Delta_t}(t)\|_{L^\infty(\Om)}&\leq 2\| u_{\Delta_t}(t)- u_{\Delta_t}(t-\Delta_t)\|_{L^\infty(\Om)}\\
& \leq 4\|u_0-z\|_{L^\infty(\Om)}+2\Delta_t\|A_0 z \|_{L^\infty(\Om)}\\
&\ \ \ +2\int_0^{\Delta_t } \|h(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau\\
&\ \ \ +2\int_{0}^{t}\|h(\Delta_t+\tau)-h(\tau)\|_{\infty}\,\mathrm{d}\tau,
\end{split}
\end{equation*}
which gives the limit $\Tilde u_{\Delta_t}\to u$ in $C([0,T];L^\infty(\Omega))$ as $\Delta_t\to 0^+$. Note that since $\Tilde u_{\Delta_t}\in C([0,T];C_0(\overline{\Omega}))$, the uniform limit $u$ belongs to $C([0,T];C_0(\overline{\Omega}))$.
Moreover, passing to the limit in \eqref{eqconv} with $t=s$ we obtain
$$
\|u(t)-v(t)\|_{L^\infty(\Om)}\leq \|u_0-z\|_{L^\infty(\Om)}+\|v_0-z\|_{L^\infty(\Om)}+\int_0^t\|r(\tau)-k(\tau)\|_{\infty}\,\mathrm{d}\tau,
$$
and \eqref{estimation2} follows since we can choose $z$ arbitrary close to $v_0$.
Finally, if $A_0 u_0\in L^\infty(\Omega)$ and $h\in W^{1,1}(0,T;L^\infty(\Omega))$ and if we assume (without loss of generality) that $t>s$ then with $z=v_0=u(t-s)$ and $(r,k)=(h,h(\cdot+t-s))$ in the last above inequality we obtain
\begin{equation}\label{mes}
\begin{split}
\|u(t)-u(s)\|_{L^\infty(\Om)}\leq  \|u_0&-u(t-s)\|_{L^\infty(\Om)}\\
&+\int_{0}^{s}\|h(\tau)-h(\tau+t-s)\|_{L^\infty(\Om)}\,\mathrm{d}\tau.
\end{split}
\end{equation}
From  \eqref{estimation2} with $v=u_0$, $k=A_0 u_0$:
\begin{equation}\label{mest}
 \|u_0-u(t-s)\|_{L^\infty(\Om)}\leq \int_0^{t-s}\|A_0 u_0-h(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau.
\end{equation}
Using \eqref{mest} and  gathering Fubini's theorem and
$$h(\tau)-h(\tau+t-s)=\int_\tau^{\tau+t-s}\frac{{\rm d}h}{{\rm d}t}(\sigma )\,\mathrm{d}\sigma,$$
the right-hand side of \eqref{mes} is smaller than
\begin{equation*}
\begin{split}
(t-s)\|A_0 u_0-h(0)\|_{L^\infty(\Om)}+&\int_{0}^{t-s}\|h(0)-h(\tau)\|_{L^\infty(\Om)}\,\mathrm{d}\tau\\
&+ \int_{0}^{s}\|h(\tau)-h(\tau+t-s)\|_{L^\infty(\Om)}\,\mathrm{d}\tau.
\end{split}
\end{equation*}
Thus
\begin{equation}\label{mainestimate}
\begin{split}
\|u(t)-u(s)\|_{L^\infty(\Om)}\leq  (t-s)\|A_0 u_0&-h(0)\|_{L^\infty(\Om)}\\
&+(t-s)\int_0^{T}\left \|\frac{\mathrm{d}h}{{\rm d}t}(\tau)\right\|_{L^\infty(\Om)}\,\mathrm{d}\tau.
\end{split}
\end{equation}
Dividing the expression \eqref{mainestimate} by $|t-s|$,
we get that $u$ is a Lipschitz function  and since $\frac{\partial u}{\partial t}\in L^2(Q_T)$, passing to the limit $|t-s|\to 0$ we obtain that
$\frac{u(t)-u(s)}{t-s}\to \frac{\partial u}{\partial t}$ as $s\to t$  weakly in $L^2(Q_T)$ and *-weakly in $L^\infty(Q_T)$. Furthermore,
\begin{equation*}
\left\|\frac{\partial u}{\partial t}\right\|_{L^\infty(\Om)}\leq \displaystyle\liminf_{s\to t}\frac{\|u(t)-u(s)\|_{L^\infty(\Om)}}{|t-s|}.
\end{equation*}
\medskip Therefore, we get $u\in W^{1,\infty}(0,T;L^\infty(\Omega))$ as well as inequality \eqref{estimationbiss2}.\qed

\noindent Concerning problem \eqref{Pt}, we deduce the following similar result:
\begin{thm}\label{continuity u with f}
Assume that conditions in Theorem \ref{continuity de u} and hypothesis on $f$ in Theorem \ref{PT1} are satisfied. Let $u_0\in \X\cap\overline{{\mathcal D}(A_0)}^{L^\infty}$. Then, the unique weak solution to \eqref{Pt}
belongs to $C([0,T];C_0(\overline{\Omega}))$ and
\begin{enumerate}
\item[(i)] there exists $\omega>0$ such that if $v$ is another weak solution to \eqref{Pt} with the initial datum $v_0\in \X\cap\overline{{\mathcal D}(A_0)}^{L^\infty}$ then the following estimate holds for $T<\tilde{T}$:
$$
\|u(t)-v(t)\|_{L^\infty(\Omega)}\leq e^{\omega t} \|u_0-v_0\|_{L^\infty(\Omega)},\quad 0\leq t\leq T.
$$
\item[(ii)]  If $u_0\in {\mathcal D}(A_0)$ then $u\in W^{1,\infty}(0,T;L^\infty(\Omega))$ and $\nabla\cdot \mathbf{a}(x,\nabla u)\in L^\infty(Q_T)$, and the following estimate holds:
$$
\left \|\frac{\partial u}{\partial t}(t)\right \|_{L^\infty(\Omega)}\leq e^{\omega t} \|\nabla\cdot \mathbf{a}(x,\nabla u_0)+f(x,u_0)\|_{L^\infty(\Omega)}.
$$
\end{enumerate}
\end{thm}




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\end{document}













%%%%%%%%%%
%\begin{proof}
%Fusco and Sbordone have already proved in \cite{FS} the local boundedness of $u$ in the case $u\in W^{1,p}(\Omega)$ and the inequality is satisfied for any $B_R\subset \subset \Omega$. We claim { that the result is still valid in our situation}. { For that we will prove  the boundednes of $u$ in a neighborhood of the boundary $\partial\Omega$}.\\
%Let $x_0\in \partial \Omega$, $B_R$ the ball centred in $x_0$. We define $K_R:=B_R\cap \Omega$ and we set
%$$r_h=\frac{R}{2}+\frac{R}{2^{h+1}}, \quad \tilde r_h=\frac{r_h+r_{h+1}}{2} \ \mbox{ and } \ k_h=k\left(1-\frac{1}{2^{h+1}} \right) \quad \mbox{for any $h\in \N$}.$$
%Define also
%$$I_h=\int_{A_{k_h,r_h}} |u(x)-k_h|^{p^*}dx \quad \mbox{and} \quad \varphi(t)=\left\{
%\begin{array}{l l}
%1 & \mbox{if}\ 0\leq t\leq \frac12\\
0 & \mbox{if}\ t\geq \frac34
\end{array}
\right.
$$
{ satisfying} $\varphi\in C^1([0,+\infty))$ and $0\leq \varphi \leq 1$. We set {$\varphi_h(x)=\varphi\left(\frac{2^{h+1}}{R}(|x|-\frac{R}{2})\right)$}. Hence $\varphi_h=1$ {\color{blue}on $B_{r_{h+1}}$} and $\varphi_h=0$ { on $\R^d\backslash B_{\tilde r_{h+1}}$}.

We have
\begin{equation*}
\begin{split}
I_{h+1}&=\int_{A_{k_{h+1},r_{h+1}}} |u(x)-k_{h+1}|^{p^*}dx=\int_{A_{k_{h+1},r_{h+1}}} (|u(x)-k_{h+1}|\varphi_h(x))^{p^*}dx\\
&\leq \int_{K_R} ((u(x)-k_{h+1})^+\varphi_h(x))^{p^*}dx.
\end{split}
\end{equation*}
Since $u\in W^{1,p}_0(\Omega)$, $(u-k_{h+1})^+\varphi_h\in W^{1,p}_0(K_R)$. Thus
\begin{equation*}
\begin{split}
I_{h+1}&{\lesssim} \left(\int_{K_R} |\nabla((u-k_{h+1})^+\varphi_h)|^{p}dx\right)^{\frac{p^*}{p}}\\
&\lesssim \left(\int_{A_{k_{h+1},\tilde r_{h}}} |\nabla u|^{\color{blue}p}dx+\int_{A_{k_{h+1},\tilde r_{h}}}(u-k_{h+1})^{\color{blue}p}dx\right)^{\frac{p^*}{p}}
\end{split}
\end{equation*}
where we use the notation $f\lesssim g$ in the sense there exists a constant $c>0$ such that $f\leq c g$. Since $\tilde r_h<r_h$, we have
\begin{equation*}
\begin{split}
I_{h+1}&\lesssim \left(2^{hp^*}\int_{A_{k_{h+1},r_{h}}}|u-k_{h+1}|^{p^*}dx +k_{h+1}^\alpha|A_{k_{h+1},r_{h}}|+|A_{k_{h+1},r_{h}}|^{\frac{p}{p^*}+\varepsilon}\right.\\
&\left.+2^{hp}\left(\int_{A_{k_{h+1},r_{h}}}|u-k_{h+1}|^{p^*}dx\right)^{\frac{p}{p^*}}|A_{k_{h+1},r_{h}}|^\delta+{\int_{A_{k_{h+1},r_{h}}}|u-k_{h+1}|^{p^*}dx}\right)^{\frac{p^*}{p}}.
\end{split}
\end{equation*}
Moreover, for any $h$, $k_h\leq k_{h+1}$, this implies
\begin{equation}\label{Jh1}
\begin{split}
I_h&=\int_{A_{k_{h},r_{h}}}|u-k_{h}|^{p^*}dx \geq \int_{A_{k_{h+1},r_{h}}}|u-k_{h}|^{p^*}dx\\
&\geq \int_{A_{k_{h+1},r_{h}}}|k_h-k_{h+1}|^{p^*}dx=|A_{k_{h+1},r_{h}}||k_{h+1}-k_h|^{p^*}.
\end{split}
\end{equation}
Then, for any $k>k_0$ and $h\in \N$
$$  |A_{k_{h+1},r_{h}}|+k_{h+1}^\alpha|A_{k_{h+1},r_{h}}|\lesssim 2^{hp^*}I_h$$
where the constant in the notation depends only on $k_0$, $p$ and $\alpha$. Replacing in \eqref{Jh1}, we obtain
\begin{equation}\label{Jh2}
I_{h+1} \lesssim \left(2^{hp^*}I_h+ 2^{h(p+\varepsilon p^*)}I_h^{\frac{p}{p^*}+\varepsilon}+2^{h(p+\delta p^*)}I_h^{\frac{p}{p^*}+\delta}\right)^{\frac{p^*}{p}}.
\end{equation}
Setting $M=\frac{p}{p^*}\max(p^*,\, p+\varepsilon p^*,\, p+\delta p^*)$ and $\theta=\min(1-\frac{p}{p^*},\,\varepsilon,\,\delta)$ and noting
$$I_h\leq \int_{K_R} (|u-k_h|^+)^{p^*}dx\leq \int_{K_R}|u|^{p*}\leq \|u\|_{W^{1,p}_0}^{p*}.$$
Hence \eqref{Jh2} becomes
\begin{equation}\label{Jh3}
I_{h+1}\lesssim 2^{hM}I_h^{1+\frac{\theta p^*}{p}}
\end{equation}
where the constant depends on $||u||_{W^{1,p}_0}$, $k_0$, $\alpha$ and $p$. We need the following to conclude.
\begin{lem}[Lemma 4.7, Chapter 2, \cite{LU}]
Let $(x_n)$ be a sequence such that $x_0\leq \lambda^{-\frac1\eta}\mu^{-\frac{1}{\eta^2}}$ and $x_{n+1}\leq \lambda\mu^nx_n^{1+\eta}$, for any $n\in \N^*$ with $\lambda$, $\eta$ and $\mu$ are positive constants and $\mu>1$. Then $(x_n)$ converges to $0$ as $n\rightarrow +\infty$.
\end{lem}
\noindent It suffices to prove that $I_0$ is small enough. Indeed $u\in L^{p^*}(\Omega)$ implies
$$I_0=\int_{A_{\frac{k}{2},R}} |u-\frac{k}{2}|^{p^*}dx \rightarrow 0 \quad\mbox{ as }\quad k\rightarrow \infty.$$
Hence for $k$ large enough, $I_0\leq C^{-\frac1\eta}(2^M)^{-\frac{1}{\eta^2}}$ with $\eta=\frac{\theta p^*}{p}$. Thus $I_h$ converges to $0$ as $h\rightarrow +\infty$ and
$$\int_{A_{k,\frac{R}{2}}} |u-k|^{p^*}dx=0.$$
We deduce that $u\leq k$ on $K_{\frac{R}{2}}$. In the same way, we prove that $-u\leq k$ on $K_{\frac{R}{2}}$. \\
Since $\overline \Omega$ is compact, we conclude that $u\in L^\infty(\Omega)$.
\end{proof}}
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