\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 54, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2007/54\hfil Existence of bounded solutions] {Existence of bounded solutions for nonlinear degenerate elliptic equations in Orlicz spaces} \author[A. Youssfi\hfil EJDE-2007/54\hfilneg] {Ahmed Youssfi} \address{Ahmed Youssfi \newline Department of Mathematics and Informatics\\ Faculty of Sciences Dhar El Mahraz\\ University Sidi Mohammed Ben Abdallah\\ PB 1796 Fez-Atlas, Fez, Morocco} \email{Ahmed.youssfi@caramail.com} \thanks{Submitted December 11, 2006. Published April 10, 2007.} \subjclass[2000]{46E30, 35J70, 35J60} \keywords{Orlicz-Sobolev spaces; degenerate coercivity; \hfill\break\indent $L^\infty$-estimates; rearrangements} \begin{abstract} We prove the existence of bounded solutions for the nonlinear elliptic problem $$ -\mathop{\rm div}a(x,u,{\nabla}u)=f \quad\text{in }{\Omega}, $$ with $u\in W^1_0L_M({\Omega})\cap L^{\infty}(\Omega)$, where $$ a(x,s,\xi)\cdot\xi\geq {\overline M}^{-1}M(h(|s|))M(|\xi|), $$ and $h:{\mathbb{R}^+}{\to }{]0,1]}$ is a continuous monotone decreasing function with unbounded primitive. As regards the $N$-function $M$, no $\Delta_2$-condition is needed. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \section{Introduction} Let $\Omega$ be a bounded open set of $\mathbb{R}^N$, $N\geq2$. We consider the equation \begin{equation} \begin{gathered} -\mathop{\rm div}(a(x,u){\overline{M}}^{-1}(M(|\nabla u|)){\frac{\nabla u} {|\nabla u|}})=f \quad \text{in }{\Omega}, \\ u=0\quad\text{on }{\partial}{\Omega}, \end{gathered}\label{e1.1} \end{equation} where \begin{equation} {\overline{M}}^{-1}(M(\frac{1}{(1+|s|)^{\theta}}))\leq a(x,s)\leq\beta, \label{e1.2} \end{equation} with $0\leq \theta \leq 1$, and $\beta$ is a positive constant. For $M(t)=t^2$, existence of bounded solutions of \eqref{e1.1} was proved under \eqref{e1.2} in \cite{AFT} and in \cite{BDO} when $f\in L^m(\Omega)$ with $m>\frac{N}{2}$. This result was then extended in \cite{ABFOT}, to the study of the problem \begin{equation} \begin{gathered} -\mathop{\rm div} a(x,u,{\nabla}u)=f \quad\text{in }{\Omega},\\ u=0\quad\text{on }{\partial}{\Omega}, \end{gathered} \label{e1.3} \end{equation} in the Sobolev space $W^{1,p}_0(\Omega)$, under the condition \begin{equation} a(x,s,\xi)\cdot\xi\geq{\frac{\alpha}{(1+|s|)^{{\theta}(p-1)}}}|\xi|^p, \label{e1.4} \end{equation} when $f\in L^m(\Omega)$ with $m>\frac{N}{p}$. In this paper, we prove the existence of bounded solutions of \eqref{e1.3} in the setting of Orlicz spaces under a more general condition than \eqref{e1.4} adapted to this situation. The main tools used to get a priori estimates in our proof are symmetrization techniques. such techniques are widely used in the literature for linear and nonlinear equations (see \cite{ABFOT} and the references quoted therein). We remark that our result is in some sense complementary to one contained in \cite{Tal1}. As examples of equations to which our result can be applied, we list: \begin{gather*} -\mathop{\rm div}(\frac{\alpha}{({e+|u|)}^{\gamma}\log(e+|u|)} \frac{e^{|{{\nabla}u}|^p}-1} {|{\nabla}u|^2}{{\nabla}u})=f \quad\text{in }{\Omega}, \\ u=0\quad\text{on }{\partial}{\Omega}, \end{gather*} where ${\alpha}>0$, ${\gamma}<1$ and $M(t)={e^{t^p}}-1$ with $1
0$ and $0\leq{\gamma}\leq1$, here $M(t)={t^p}\log^q{(e+t)}$ with $1
0$ for $t>0$,
$\frac{M(t)}{t}\to 0$ as $t\to 0$ and
$\frac{M(t)}{t}\to \infty$ as $t\to \infty$. The
$N$-function conjugate to $M$ is defined as
${{\overline{M}}(t)}={\rm sup}\{{st-M(t), {s\geq0}}\}$. We will
extend these $N$-functions into even functions on all $\mathbb{R}$.
We recall that (see \cite{Ad})
\begin{equation}
M(t)\leq t{\overline M}^{-1}(M(t))\leq 2M(t)\quad \text{for all }
t\geq 0\label{e2.1}
\end{equation}
and the Young's inequality: for all $s, t\geq0$,
$st\leq{\overline{M}(s)+M(t)}$. If for some $k>0$,
\begin{equation}
M(2t)\leq{kM(t)}\quad \text{for all }t{\geq0},\label{e2.2}
\end{equation}
we said that $M$ satisfies the
${\Delta}_2$-condition, and if \eqref{e2.2} holds only for
$t$ greater than or equal to $t_0$, then $M$ is said to satisfy the
${\Delta}_2$-condition near infinity.
Let $P$ and $Q$ be two $N$-functions. the notation $P{\ll}Q$ means
that $P$ grows essentially less rapidly than $Q$, i.e.
$$
\forall \epsilon >0,\quad
\frac{P(t)}{Q(\epsilon t)}\to 0\quad \text{as } t\to \infty,
$$
that is the case if and only if
$$
\frac{Q^{-1}(t)}{P^{-1}(t)}\to 0\quad \text{as } t\to \infty.
$$
Let $\Omega$ be an open subset of
$\mathbb{R}^N$. The Orlicz class $K_M({\Omega})$ (resp. the Orlicz
space $L_M({\Omega})$) is defined as the set of (equivalence class
of) real-valued measurable functions $u$ on $\Omega$ such that:
$$
{\int_{\Omega}M(u(x))dx}<{\infty}
\quad \text{(resp.}
{\int_{\Omega}M(\frac{u(x)}{\lambda})dx}<{\infty}
\text{ for some $\lambda>0$)}.
$$
Endowed with the norm
$$
\|u\|_M={\rm inf}\{{\lambda>0}:
{\int_{\Omega}M(\frac{u(x)}{\lambda})dx}<{\infty}\},
$$
$L_M(\Omega)$ is a Banach space and $K_M({\Omega})$ is a convex
subset of $L_M(\Omega)$. the closure in $L_M({\Omega})$ of the set
of bounded measurable functions with compact support in
$\overline{\Omega}$ is denoted by $E_M({\Omega})$.
The Orlicz-Sobolev space $W^1L_M(\Omega)$
(resp. $W^1E_M(\Omega)$) is the space of functions $u$ such that
$u$ and its distributional derivatives up to order $1$ lie in
$L_M(\Omega)$ (resp. $E_M(\Omega)$).
This is a Banach space under the norm
$$
\|u\|_{1,M}=\sum_{|\alpha|\leq1}{\|D^{\alpha}u\|}_M.
$$
Thus, $W^1L_M(\Omega)$ and $W^1E_M(\Omega)$ can be identified with
subspaces of the product of $(N+1)$ copies of $L_M(\Omega)$.
Denoting this product by $\Pi{L_M}$, we will use the weak
topologies $\sigma(\Pi{L_M},\Pi{E_{\overline{M}}})$ and
$\sigma(\Pi{L_M},\Pi{L_{\overline{M}}})$.
The space $W_0^1E_M(\Omega)$ is defined as the norm closure of the
Schwartz space $D(\Omega)$ in $W^1E_M(\Omega)$ and the space
$W_0^1L_M(\Omega)$ as the $\sigma(\Pi{L_M},\Pi{E_{\overline{M}}})$
closure of $D(\Omega)$ in $W^1L_M(\Omega)$.
We say that a sequence $\{u_n\}$ converges to $u$ for the
modular convergence in $W^1L_M(\Omega)$ if, for some $\lambda >0$,
$$
\int_{\Omega}M(\frac{D^{\alpha}u_n-D^{\alpha}u}{\lambda})dx\to 0\quad
\mbox{for all }|\alpha|\leq1;
$$
this implies convergence for
$\sigma(\Pi{L_M},\Pi{L_{\overline{M}}})$.
If $M$ satisfies the $\Delta _2$-condition on $\mathbb{R}^+$
(near infinity only if $\Omega$ has finite measure), then the
modular convergence coincides with norm convergence.
Recall that the norm $\|Du\|_M$ defined on $W_0^1L_M(\Omega)$ is
equivalent to $\|u\|_{1,M}$ (see \cite{Go4}).
Let $W^{-1}L_{\overline{M}}(\Omega)$ (resp.
$W^{-1}E_{\overline{M}}(\Omega)$) denotes the space of
distributions on $\Omega$ which can be written as sums of
derivatives of order $\leq1$ of functions in
$L_{\overline{M}}(\Omega)$ (resp. $E_{\overline{M}}(\Omega)$). It
is a Banach space under the usual quotient norm.
If the open $\Omega$ has the segment property then the space
$D(\Omega)$ is dense in $W_0^1L_M(\Omega)$ for the topology
$\sigma(\Pi{L_M},\Pi{L_{\overline{M}}})$ (see \cite{Go4}).
Consequently, the action of a distribution in
$W^{-1}L_{\overline{M}}(\Omega)$ on an element of
$W_0^1L_M(\Omega)$ is well defined. For an exhaustive treatments
one can see for example \cite{Ad} or \cite{Kr}.
We will use the following lemma,
(see\cite{Mes}), which concerns operators of Nemytskii Type in
Orlicz spaces. It is slightly different from the analogous one
given in \cite{Kr}.
\begin{lemma} \label{lem2.1}
Let $\Omega$ be an open subset of $\mathbb{R}^N$ with finite
measure. let $M$, $P$ and $Q$ be $N$-functions such that $Q{\ll}P$,
and let $f : \Omega\times\mathbb{R}\to \mathbb{R}$ be a
Carath\'eodory function such that, for a.e. $x\in \Omega$ and for
all $s\in \mathbb{R}$,
$$
|f(x,s)|\leq c(x)+k_1 P^{-1}M(k_2|s|),
$$
where $k_1,k_2$ are real constants and $c(x)\in E_Q(\Omega)$. Then
the Nemytskii operator $N_f$, defined by $N_f(u)(x)=f(x,u(x))$, is
strongly continuous from $ P(E_M,\frac{1}{k_2})=\{u\in
L_M(\Omega): d(u,E_M(\Omega))<\frac{1}{k_2}\}$
into $E_Q(\Omega)$.
\end{lemma}
We recall the definition of decreasing
rearrangement of a measurable function $w:\Omega\to \mathbb{R}$.
If one denotes by $|E|$ the Lebesgue
measure of a set $E$, one can define the distribution function
$\mu_w(t)$ of $w$ as:
$$
\mu_w(t)=|\{x\in \Omega: |w(x)|>t\}|, \quad t\geq 0.
$$
The decreasing rearrangement $w^\ast$ of $w$ is defined as the
generalized inverse function of $\mu_w$:
$$
w^\ast(\sigma)={\rm inf}\{t\in \mathbb{R}: \mu_w(t)\leq \sigma\},\quad
\sigma \in (0,\Omega).
$$
It is shown in \cite{Tal3} that $w^\ast$ is everywhere continuous
and
\begin{equation}
w^\ast(\mu_w(t))=t\label{e2.3}
\end{equation}
for every $t$ between 0 and $\mathop{\rm ess\,sup} |w|$.
More details can be found for example in
\cite{BS,Tal1,Tal2}.
\section{Assumptions and main result}
Let $\Omega$ be an open bounded subset of $\mathbb{R}^N$, $N\geq2$,
satisfying the segment property and $M$ is an $N$-function twice continuously
differentiable and strictly increasing, and $P$ is an $N$-function
such that $P{\ll}M$.
Let $a:{\Omega}\times{\mathbb{R}}\times{\mathbb{R}^N}{\to }{\mathbb{R}^N}$
be a Carath\'eodory function satisfying, for a.e. $x\in \Omega$,
and for all $s\in \mathbb{R}$ and all $\xi$, $\eta$ $\in
\mathbb{R}^N$, $\xi \neq \eta$,
\begin{equation}
a(x,s,\xi)\cdot\xi\geq {\overline M}^{-1}M(h(|s|))M(|\xi|)\label{e3.1}
\end{equation}
where $h: {\mathbb{R}^+}{\to }{\mathbb{R}^*_+}$ is a
continuous monotone decreasing function such that $h(0){\leq}1$
and its primitive $H(s)={\int_0^s{h(t)}dt}$ is
unbounded,
\begin{equation}
|a(x,s,\xi)|\leq a_0(x)+k_1{\overline P}^{-1}M(k_2|s|)+k_3
{\overline M}^{-1}M(k_4|\xi|)\label{e3.2}
\end{equation}
where $a_0(x)$ belongs to $E_{\overline M}(\Omega)$ and
$k_1, k_2, k_3, k_4$ to
$\mathbb{R}_+^\ast$,
\begin{equation}
(a(x,s,\xi)-a(x,s,\eta))\cdot(\xi-\eta)>0. \label{e3.3}
\end{equation}
Let $A$: $D(A){\subset}W_0^1L_M(\Omega){\to }W^{-1}L_{\overline
M}(\Omega)$ be a mapping (non-everywhere defined) given by
$$
A(u):=-{\rm div}\,a(x,u,\nabla u),
$$
We are interested, in proving the existence of
bounded solutions to the nonlinear problem
\begin{equation}
\begin{gathered}
A(u):=-\mathop{\rm div}(a(x,u,\nabla u)=f \quad\text{in }{\Omega},\\
u=0\quad\text{on }{\partial}{\Omega},
\end{gathered} \label{e3.4}
\end{equation}
As regards the data $f$, we assume one of
the following two conditions: Either
\begin{equation}
f\in L^N(\Omega),\label{e3.5}
\end{equation}
or
\begin{equation}
\begin{gathered}
f\in L^m(\Omega)\quad \text{with $m=rN/(r+1)$
for some $r>0$},\\
\text{and }\int_{.}^{+\infty}(\frac{t}{M(t)})^rdt<+\infty.
\end{gathered} \label{e3.6}
\end{equation}
We will use the following concept of solutions:
\begin{definition} \label{def3.1} \rm
Let $f{\in {L^1(\Omega)}}$, a function $u{\in{W^1_0{L_M}(\Omega)}}$ is said
to be a weak solution of \eqref{e3.4}, if $a(\cdot,u,\nabla u)\in
(L_{\overline{M}}(\Omega))^N$ and
$$
\int_{\Omega}a(x,u,\nabla u)\cdot\nabla v\,dx=\int_{\Omega}fv\,dx
$$
holds for all $v\in D(\Omega)$.
\end{definition}
Our main result is the following.
\begin{theorem} \label{thm3.1}
Under the assumptions \eqref{e3.1}, \eqref{e3.2}, \eqref{e3.3} and
either \eqref{e3.5} or \eqref{e3.6}, there exists at least one weak
solution of \eqref{e3.4} in
${W^1_0L_M(\Omega)}\cap{L^{\infty}(\Omega)}$.
\end{theorem}
\begin{remark} \label{rmk3.2} \rm
In the case where $M(t)=t^p$, with $p>1$, assumptions \eqref{e3.5} and
\eqref{e3.6} imply that $m>\frac{N}{p}$. Our result extends those in
\cite{BDO} and \cite{AFT} where $M(t)=t^2$ and
\cite{ABFOT} where $M(t)=t^p$, with $p>1$.
\end{remark}
\begin{remark} \label{rmk3.3} \rm
Note that the result of theorem \eqref{e3.1} is independent of the
function $h$ which eliminates the coercivity of the operator $A$.
The result is not surprising, since if we look for bounded
solutions then the operator $A$ becomes coercive.
\end{remark}
\begin{remark} \label{rmk3.4}\rm
The principal difficulty in dealing with the problem \eqref{e3.4} is
the non coerciveness of the operator $A$, this is due to the
hypothesis \eqref{e3.1}, so the classical methods used to prove the
existence of a solution for \eqref{e3.4} can not be applied (see
\cite{GM} and also \cite{Go3}). To get rid of this difficulty, we
will consider an approximation method in which we introduce a
truncation. The main tool of the proof will be $L^\infty$ a priori
estimates, obtained by mean of a comparison result, which then
imply the $W^1_0L_{M}(\Omega)$ estimate, since if $u$ is bounded
the operator $A$ becomes uniformly coercive.
\end{remark}
\section{Proof of theorem \ref{thm3.1}}
For $s\in \mathbb{R}$ and $k>0$ set: $T_k(s)=\max (-k,\min (k,s))$ and
$G_k(s)=s-T_k(s)$.
Let $\{f_n\}\subset W^{-1}E_{\overline M}(\Omega)$ be a
sequence of smooth functions such that
$$
f_n\to f\quad \text{strongly in } L^{m^*}(\Omega)
$$
and
$$
\|f_n\|_{m^*}\leq\|f\|_{m^*},
$$
where $m^*$ denotes either $N$ or $m$, according as we assume
\eqref{e3.5} or \eqref{e3.6}, and consider the
operator:
$$
A_n(u)=-\mathop{\rm div}a(x,T_n(u),\nabla u).
$$
By assumption \eqref{e3.1}, we have
\begin{align*}
\langle A_n(u),u\rangle
&=\int_{\Omega}a(x,T_n(u),\nabla u)\cdot\nabla u\,dx\\
&\geq {\overline M}^{-1}(M(h(n)))\int_{\Omega}M(|\nabla u|)dx.
\end{align*}
Thus, $A_n$ satisfies the classical conditions from which derives,
thanks to the fact that
$f_n{\in}{W^{-1}E_{\overline{M}}(\Omega)}$, the existence of a
solution $u_n{\in}{W^1_0{L_M}(\Omega)}$, (see \cite{GM} and also
\cite{Go3}), such that
\begin{equation}
\int_{\Omega}a(x,T_n(u_n),\nabla
u_n)\cdot\nabla vdx=\int_{\Omega}f_nvdx\label{e4.1}
\end{equation}
holds for all $v{\in {W^1_0L_M(\Omega)}}$.
To prove the $L^\infty$ a priori estimates, we will need the
following comparison lemma, whose proof will be given in the
appendix.
\begin{lemma} \label{lem4.1}
Let $B(t)={\frac{M(t)}{t}}$ and
$\mu_n(t)=|\{x\in \Omega: |u_n(x)|>t\}|$, for all $t>0$.
We have for almost every $t>0$:
\begin{equation}
h(t)\leq {\frac{2M(1)}{{{\overline M}^{-1}}(M(1))NC_N^{1/N}}}
{\frac{-\mu_n'(t)}{{\mu_n(t)}^{1-\frac{1}{N}}}}
B^{-1}
\Big(\frac{\int_{\{|u_n|>t\}}|f_n|dx}{{\overline
M}^{-1}(M(1))NC_N^{1/N}{{\mu_n(t)}^{1-\frac{1}{N}}}}\Big)
\label{e4.2}
\end{equation}
where $C_N$ is the measure of the unit ball in
$\mathbb{R}^N$.
\end{lemma}
\noindent\textbf{step 1: $L^{\infty}$-bound.}
If we assume \eqref{e3.5}, using the inequality
$\int_{\{|u_n|>t\}}|f_n|dx\leq\|f\|_N{\mu_n(t)^{1-\frac{1}{N}}}$,
\eqref{e4.2} becomes
$$
h(t)\leq
{\frac{2M(1)(-\mu_n'(t))}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}{\mu_n(t)}^{1-\frac{1}{N}}}}
B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big).
$$
Then we integrate between $0$ and $s$, we get
$$
H(s)\leq
{\frac{2M(1)}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}}}B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big)\int_{0}^{s}
{\frac{-\mu_n'(t)}{\mu_n(t)^{1-\frac{1}{N}}}}dt;
$$
hence, a change of variables yields
$$
H(s)\leq {\frac{2M(1)}{{\overline M}^{-1}(M(1)){N}C_N^{1/N}}}
B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big)
\int_{\mu_n(s)}^{|\Omega|}{\frac{dt}{t^{1-\frac{1}{N}}}}.
$$
By \eqref{e2.3} we get
$$
H(u_n^\ast(\sigma))\leq
{\frac{2M(1)}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}}}B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big)\int_{\sigma}^{|\Omega|}
{\frac{dt}{t^{1-\frac{1}{N}}}}.
$$
So that
$$
H(u_n^\ast(0))\leq {\frac{2M(1)}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}}}B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big)N|\Omega|^{1/N}.
$$
Since $u_n^\ast(0)= \|u_n\|_{\infty}$, the assumption made on $H$
(i.e., $\lim_{s\to {+\infty}}H(s)=+\infty$) shows that the
sequence $\{u_n\}$ is uniformly bounded in $L^{\infty}(\Omega)$.
Moreover if we denote by $H^{-1}$ the inverse function of $H$, one
has:
\begin{equation}
\|u_n\|_{\infty}\leq H^{-1}\Big({\frac{2M(1)}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}}}B^{-1}\Big(\frac{\|f\|_N}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big)N|\Omega|^{1/N}\Big).
\label{e4.3}
\end{equation}
Now, we assume that \eqref{e3.6} is filled. Then, using the inequality
$$
\int_{\{|u_n|>t\}}|f_n|dx\leq\|f\|_m{\mu_n(t)^{1-\frac{1}{m}}}
$$
in \eqref{e4.2}, we obtain
$$
H(s)\leq \frac{2M(1)}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\int_{0}^{s}{\frac{-\mu_n'(t)}{\mu_n(t)^{1-\frac{1}{N}}}}
B^{-1}\Big(\frac{\|f\|_m}{{\overline
M}^{-1}(M(1))NC_N^{1/N}\mu_n(t)^{\frac{1}{m}-\frac{1}{N}}}\Big)dt.
$$
A change of variables gives
$$
H(s)\leq \frac{2M(1)}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\int_{\mu_n(s)}^{|\Omega|}
B^{-1}\Big(\frac{\|f\|_m}{{\overline
M}^{-1}(M(1))NC_N^{1/N}\sigma^{\frac{1}{m}-\frac{1}{N}}}\Big)
{\frac{d\sigma}{\sigma^{1-\frac{1}{N}}}}.
$$
As above, \eqref{e2.3} gives
$$
H(u_n^\ast(\tau))\leq \frac{2M(1)}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\int_{\tau}^{|\Omega|}
B^{-1}\Big(\frac{\|f\|_m}{{\overline
M}^{-1}(M(1))NC_N^{1/N}\sigma^{\frac{1}{m}-\frac{1}{N}}}\Big)
{\frac{d\sigma}{\sigma^{1-\frac{1}{N}}}}.
$$
Then, we have
$$
H(\|u_n\|_\infty) \leq \frac{2M(1)}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\int_{0}^{|\Omega|}
B^{-1}\Big(\frac{\|f\|_m}{{\overline
M}^{-1}(M(1))NC_N^{1/N}\sigma^{\frac{1}{m}-\frac{1}{N}}}\Big)
{\frac{d\sigma}{\sigma^{1-\frac{1}{N}}}}.
$$
A change of variables gives
$$
H(\|u_n\|_\infty) \leq {\frac{2M(1)\|f\|_m^r}{{({{\overline
M}^{-1}(M(1))})^{r+1}}{N^{r}}
C_N^{\frac{r+1}{N}}}}\int_{c_0}^{+\infty}rt^{-r-1}B^{-1}(t)dt,
$$
where $c_0= \frac{\|f\|_m}{{\overline M}^{-1}(M(1))NC_N^{1/N}
|\Omega|^{\frac{1}{rN}}}$.
Then, an integration by parts yields
$$
H(\|u_n\|_\infty)\leq {\frac{2M(1)\|f\|_m^r}{{({{\overline
M}^{-1}(M(1))})^{r+1}}{N^{r}}C_N^{\frac{r+1}{N}}}}\Big(\frac{B^{-1}(c_0)}{c_0^r}
+\int_{B^{-1}(c_0)}^{+\infty}\big(\frac{s}{M(s)}\big)^rds\Big).
$$
The assumption made on $H$ guarantees that the sequence $\{u_n\}$
is uniformly bounded in $L^{\infty}(\Omega)$. Indeed, denoting by
$H^{-1}$ the inverse function of $H$, one has
\begin{equation}
\|u_n\|_\infty\leq H^{-1}\Big(
{\frac{2M(1)\|f\|_m^r}{{({{\overline
M}^{-1}(M(1))})^{r+1}}{N^{r}}C_N^{\frac{r+1}{N}}}}\Big(\frac{B^{-1}(c_0)}{c_0^r}
+\int_{B^{-1}(c_0)}^{+\infty}\big(\frac{s}{M(s)}\big)^rds\Big)\Big).
\label{e4.4}
\end{equation}
Consequently, in both cases the sequence $\{u_n\}$ is uniformly
bounded in $L^{\infty}(\Omega)$, so that in\vspace{2mm} the
sequel, we will denote by $c$ the constant appearing either in
\eqref{e4.3} or in \eqref{e4.4},
that is
\begin{equation}
\|u_n\|_\infty\leq c. \label{e4.5}
\end{equation}
\noindent \textbf{Step 2: Estimation in $W_0^{1}L_M(\Omega)$.}
It is now easy to obtain an estimate in
$W_0^{1}L_M(\Omega)$ under either \eqref{e3.5} or \eqref{e3.6}. Let
$m^\ast$ denotes either $N$ or $m$ according as we assume
\eqref{e3.5} or \eqref{e3.6}. Taking $u_n$ as test function in
\eqref{e4.1}, one has
$$
\int_{\Omega}a(x,T_n(u_n),\nabla u_n)\cdot\nabla u_ndx=
\int_{\Omega}f_nu_ndx.
$$
Then by \eqref{e3.1} and \eqref{e4.5}, we obtain
\begin{equation}
{\int_{\Omega}M(|\nabla u_n|)dx\leq \frac{c\|f\|_{m^\ast}
|\Omega|^{1-\frac{1}{m^\ast}}}{{{\overline M}^{-1}(M(h(c)))}}}.
\label{e4.6}
\end{equation}
Hence, the sequence $\{u_n\}$ is bounded in $W_0^{1}L_M(\Omega)$.
Therefore, there exists a subsequence of $\{u_n\}$, still denoted
by $\{u_n\}$, and a function $u$ in $W_0^{1}L_M(\Omega)$ such that
\begin{equation}
u_n \rightharpoonup u \quad\text{in
$W_0^{1}L_M(\Omega)$ for $\sigma(\Pi L_M,\Pi E_{\overline M})$}
\label{e4.7}
\end{equation}
and
\begin{equation}
u_n \to u \quad\text{in $E_M(\Omega)$ strongly and a.e. in $\Omega$.}
\label{e4.8}
\end{equation}
\noindent \textbf{Step 3: Almost everywhere convergence of the
gradients.}
Let us begin with the following lemma which we will use later.
\begin{lemma} \label{lem4.2}
The sequence $\{a(x,T_n(u_n),\nabla u_n)\}$ is bounded in
$(L_{\overline M}(\Omega))^N$.
\end{lemma}
\begin{proof}
We will use the dual norm of $(L_{\overline M}(\Omega))^N$.
Let $\varphi\in (E_M(\Omega))^N$ such that $\|\varphi\|_M=1$.
By \eqref{e3.3} we have
$$
\Big(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\frac{\varphi}{k_4})\Big)\cdot
\Big(\nabla u_n-\frac{\varphi}{k_4}\Big)\geq0.
$$
Let $\lambda=1+k_1+k_3$, by using \eqref{e3.2}, \eqref{e4.5}, \eqref{e4.6}
and Young's inequality we get
\begin{align*}
&\int_{\Omega}a(x,T_n(u_n),\nabla u_n)\varphi dx\\
&\leq k_4\int_{\Omega}a(x,T_n(u_n),\nabla u_n)\cdot\nabla u_n\,dx
-k_4\int_{\Omega}a(x,T_n(u_n),\frac{\varphi}{k_4})\cdot\nabla u_n\,dx\\
&\quad +\int_{\Omega}a(x,T_n(u_n),\frac{\varphi}{k_4})\cdot\varphi dx\\
&\leq k_4c\|f\|_{m^\ast}|\Omega|^{1-\frac{1}{m^\ast}}+
k_4\lambda\frac{c\|f\|_{m^\ast}|\Omega|^{1-\frac{1}{m^\ast}}}{{\overline
M}^{-1}M(h(c))}\\
&\quad +(1+k_4)\Big(\int_{\Omega}{\overline
M}(a_0(x))dx+k_1{\overline M}{\overline P}^{-1}M(k_2c)|\Omega|\Big)
+k_3(1+k_4)+\lambda,
\end{align*}
which completes the proof.
\end{proof}
From \eqref{e4.5} and \eqref{e4.8} we obtain that
$u\in W^1_0L_M(\Omega)\cap L^{\infty}(\Omega)$, so that by
\cite[Theorem 4]{Go2} there exists a sequence $\{v_j\}$ in
$D(\Omega)$ such that $v_j{\to }{u}$ in
$W_0^1L_M(\Omega)$ as $j{\to }{\infty}$ for the modular
convergence and almost everywhere in $\Omega$, moreover
$\|v_j\|_{\infty}{\leq}(N+1)\|u\|_{\infty}$.
For $s>0$, we denote by $\chi_j^s$ the
characteristic function of the set
$$
\Omega_j^s=\{x\in\Omega:|\nabla v_j(x)|\leq s\}
$$
and by $\chi^s$ the characteristic function of the set
$\Omega^s=\{x\in\Omega:|\nabla u(x)|\leq s\}$.
Testing by $u_n-v_j$ in \eqref{e4.1}, we obtain
\begin{equation}
\int_\Omega a(x,T_n(u_n),\nabla u_n)\cdot(\nabla u_n-\nabla v_j)
dx = \int_\Omega f_n(u_n-v_j)dx\label{e4.9}
\end{equation}
Denote by $\epsilon_i(n,j)$, $(i=0,1,\dots)$, various sequences of real
numbers which tend to $0$ when $n$ and $j\to \infty$, i.e.
$$
\lim_{j\to \infty}\lim_{n\to \infty}\epsilon_i(n,j)=0.
$$
For the right-hand side of \eqref{e4.9}, we have
\begin{equation}
\int_\Omega f_n(u_n-v_j)dx=\epsilon_0(n,j).\label{e4.10}
\end{equation} The left-hand side of \eqref{e4.9} is written as
\begin{equation}
\begin{aligned}
&\int_\Omega a(x,T_n(u_n),\nabla u_n)\cdot(\nabla
u_n-\nabla v_j)\,dx\\
&= \int_\Omega\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla v_j\chi^s_j )\right)\cdot\left(\nabla
u_n-\nabla v_j\chi^s_j\right)\,dx\\
&\quad + \int_\Omega a(x,T_n(u_n),\nabla v_j\chi^s_j
)\cdot(\nabla u_n-\nabla v_j\chi^s_j)dx\\
&\quad - \int_{\Omega\setminus\Omega^s_j}a(x,T_n(u_n),\nabla
u_n)\cdot\nabla v_j\,dx
\end{aligned}\label{e4.11}
\end{equation}
We will pass to the limit over $n$ and $j$, for $s$ fixed, in the
second and the third terms of the right-hand side of
\eqref{e4.11}. By Lemma \ref{lem4.2}, we deduce that there exists
$l \in (L_{\overline M}(\Omega))^N$ and up to a subsequence
$a(x,T_n(u_n),\nabla u_n)\rightharpoonup l$ weakly in
$(L_{\overline M}(\Omega))^N$ for $\sigma(\prod L_{\overline
M},\prod E_M)$.
Since $\nabla v_j\chi_{\Omega\setminus\Omega^s_j}\in (E_M(\Omega))^N$, we
have by letting $n\to \infty$,
$$
-\int_{\Omega\setminus\Omega^s_j} a(x,T_n(u_n),\nabla
u_n)\cdot\nabla v_jdx\to -\int_{\Omega\setminus\Omega^s_j}
l\cdot\nabla v_jdx.
$$
Using the modular convergence of ${v_j}$, we
get as $j\to \infty$
$$
-\int_{\Omega\setminus\Omega^s_j} l\cdot\nabla
v_j\,dx\to -\int_{\Omega\setminus\Omega^s} l\cdot\nabla u\,dx.
$$
Hence, we have proved that the third term
\begin{equation}
-\int_{\Omega\setminus\Omega^s_j} a(x,T_n(u_n),\nabla
u_n)\cdot\nabla v_jdx=-\int_{\Omega\setminus\Omega^s}l\cdot\nabla
udx\,+\,\epsilon_1(n,j).\label{e4.12}
\end{equation}
For the second term, as $n \to \infty$, we have
$$
\int_\Omega a(x,T_n(u_n),\nabla v_j\chi^s_j )\cdot\left(\nabla
u_n-\nabla v_j\chi^s_j\right)dx \to \int_\Omega
a(x,u,\nabla v_j\chi^s_j )\cdot\left(\nabla u-\nabla
v_j\chi^s_j\right)dx,
$$
since $a(x,T_n(u_n),\nabla v_j\chi^s_j)\to a(x,u,\nabla v_j\chi^s_j)$
strongly in $(E_{\overline M}(\Omega))^N$ as
$n \to \infty$ by lemma \ref{lem2.1} and \eqref{e4.8}, while
$\nabla u_n\rightharpoonup\nabla u$ weakly in $(L_M(\Omega))^N$
by \eqref{e4.7}.
And since $\nabla v_j\chi^s_j\to \nabla u\chi^s$ strongly
in $(E_M(\Omega))^N$ as $j\to \infty$, we obtain
$$
\int_\Omega a(x,u,\nabla v_j\chi^s_j)\cdot
\left(\nabla u-\nabla v_j\chi^s_j\right)dx\to 0
$$
as $j\to \infty$. So that
\begin{equation}
\int_\Omega a(x,T_n(u_n),\nabla v_j\chi^s_j )\cdot\left(\nabla
u_n-\nabla v_j\chi^s_j\right)dx=\epsilon_2(n,j).\label{e4.13}
\end{equation}
Consequently, combining \eqref{e4.10}, \eqref{e4.12} and \eqref{e4.13},
we obtain
\begin{equation}
\begin{aligned}
&\int_\Omega\left(a(x,T_n(u_n),\nabla u_n)-a(x,T_n(u_n),\nabla
v_j\chi^s_j)\right)\cdot\left(\nabla u_n-\nabla v_j\chi^s_j\right)dx\\
&= \int_{\Omega\setminus\Omega^s}l\cdot\nabla
u\,dx+\epsilon_3(n,j).
\end{aligned}\label{e4.14}
\end{equation}
On the other hand
\begin{align*}
&\int_\Omega\left(a(x,T_n(u_n),\nabla u_n)-a(x,T_n(u_n),\nabla
u\chi^s)\right)\cdot\left(\nabla u_n-\nabla u\chi^s\right)dx\\
&= \int_\Omega\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla v_j\chi^s_j)\right)\cdot\left(\nabla
u_n-\nabla v_j\chi^s_j\right)dx\\
&\quad + \int_\Omega a(x,T_n(u_n),\nabla
u_n)\cdot\left(\nabla
v_j\chi^s_j- \nabla u\chi^s\right)dx\\
&\quad - \int_\Omega a(x,T_n(u_n),\nabla
u\chi^s)\cdot\left(\nabla
u_n- \nabla u\chi^s\right)dx\\
&\quad + \int_\Omega a(x,T_n(u_n),\nabla
v_j\chi^s_j)\cdot\left(\nabla u_n-\nabla v_j\chi^s_j\right)dx.
\end{align*}
We can argue as above in order to obtain
\begin{gather*}
\int_\Omega a(x,T_n(u_n),\nabla u_n)\cdot\left(\nabla
v_j\chi^s_j- \nabla u\chi^s\right)dx= \epsilon_4(n,j),
\\
\int_\Omega a(x,T_n(u_n),\nabla u\chi^s)\cdot\left(\nabla
u_n- \nabla u\chi^s\right)dx= \epsilon_5(n,j),
\\
\int_\Omega a(x,T_n(u_n),\nabla v_j\chi^s_j)\cdot\left(\nabla
u_n- \nabla v_j\chi^s_j\right)dx= \epsilon_6(n,j).
\end{gather*}
Then, by
\eqref{e4.14} we have
\begin{align*}
&\int_\Omega\left(a(x,T_n(u_n),\nabla u_n)-a(x,T_n(u_n),\nabla
u\chi^s)\right)\cdot\left(\nabla u_n-\nabla u\chi^s\right)dx\\
&= \epsilon_7(n,j)+
\int_{\Omega\setminus\Omega^s}l\cdot\nabla udx.
\end{align*}
For $r\leq s$, we write
\begin{align*}
0&\leq \int_{{\Omega}^r}\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n -\nabla u\right)dx\\
&\leq \int_{\Omega^s}\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n-\nabla u\right)dx\\
&= \int_{\Omega^s}\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u\chi^s)\right)\cdot\left(\nabla
u_n-\nabla u\chi^s\right)dx\\
&\leq \int_\Omega\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u\chi^s)\right)\cdot\left(\nabla
u_n-\nabla u\chi^s\right)dx\\
&\leq \epsilon_7(n,j)+
\int_{\Omega\setminus\Omega^s}l\cdot\nabla udx.
\end{align*}
Which implies by passing at first to the limit superior
over $n$ and then over $j$,
\begin{align*}
0&\leq \limsup_{n\to \infty}\int_{{\Omega}^r}\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n-\nabla
u\right)dx\\
&\leq \int_{\Omega\setminus\Omega^s}l\cdot\nabla udx.
\end{align*}
Letting $s\to +\infty$ in the previous
inequality, we conclude that as $n\to \infty$,
\begin{equation}
\int_{{\Omega}^r}\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n-\nabla
u\right)dx \to 0.\label{e4.15}
\end{equation}
Let $B_n$ be defined by
$$
B_n=\left(a(x,T_n(u_n),\nabla
u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n-\nabla u\right).
$$
As a consequence of \eqref{e4.15}, one has $B_n\to 0$
strongly in $L^1({\Omega}^r)$, extracting a subsequence, still
denoted by $\{u_n\}$, we get
$B_n\to 0$\quad a.e in ${\Omega}^r$.
Then, there exists a subset $Z$ of ${\Omega}^r$, of zero measure,
such that: $B_n(x)\to 0$ for all $x\in
{{\Omega}^r}\setminus Z$.
Using \eqref{e3.2}, we obtain for
all $x\in {{\Omega}^r}\setminus Z$,
$$
B_n(x)\geq {\overline M}^{-1}M(h(c))M(|\nabla u_n(x)|)-
c_1(x)\left(1+{\overline M}^{-1}M(k_4|\nabla u_n(x)|)+ |\nabla
u_n(x)|\right),
$$
where $c$ is the constant appearing in \eqref{e4.5} and
$c_1(x)$ is a
constant which does not depend on $n$. Thus, the sequence
$\{\nabla u_n(x)\}$ is bounded in $\mathbb{R}^N$, then for a
subsequence
$\{u_{n'}(x)\}$, we have
\begin{gather*}
\nabla u_{n'}(x)\to \xi\quad\text{in }\mathbb{R}^N, \\
\left(a(x,u(x),\xi)-a(x,u(x),\nabla
u(x))\right)\cdot\left(\xi-\nabla u(x)\right)=0.
\end{gather*}
Since $a(x,s,\xi)$ is strictly monotone, we have $\xi=\nabla
u(x)$, and so $\nabla u_n(x)\to \nabla u(x)$ for the whole
sequence. It follows that
$$
\nabla u_n\to \nabla u \quad \text{a.e. in }\Omega^r.
$$
Consequently, as $r$ is arbitrary, one can deduce that
\begin{equation}
\nabla u_n\to \nabla u\quad \text{a.e. in } \Omega. \label{e4.16}
\end{equation}
\noindent\textbf{Step 4: Passage to the limit.}
Let $v$ be a function in $D(\Omega)$. Taking $v$ as test function
in \eqref{e4.1}, one has
$$
\int_{\Omega}a(x,T_n(u_n),\nabla u_n)\cdot\nabla v dx= \int_{\Omega}f_nv\,dx.
$$
Lemma \ref{lem4.2}, \eqref{e4.8} and \eqref{e4.16} imply that
$$
a(x,T_n(u_n),\nabla u_n)\rightharpoonup a(x,u,\nabla u)\quad
\text{weakly in }(L_{\overline M}(\Omega))^N \text{ for } \sigma(\Pi
L_{\overline M},\Pi E_M ),
$$
so that one can pass to the limit in
the previous equality to obtain
$$
\int_{\Omega}a(x,u,\nabla u)\cdot\nabla v dx= \int_{\Omega}fv\,dx.
$$
Moreover, from \eqref{e4.5} and \eqref{e4.8} we have $u\in
W^1_0L_M(\Omega)\cap L^{\infty}(\Omega)$. This completes the proof
of theorem \ref{thm3.1}.
\begin{remark} \label{rmk4.3} \rm
Note that the $L^{\infty}$-bound in step 1 can be proven
under the weaker assumption
$$
\|f\|_{m,\infty}= \sup_{s>0}s^{\frac{1}{m}-1}\int_{0}^{s}f^\ast(t)dt
< \infty,
$$
which is equivalent to say that $f$ belongs to the Lorentz space
$L(m,\infty)$. Indeed, one can use the inequality
$$
\int_{\{|u_n|>t\}}|f_n|dx\leq\int_{0}^{\mu_n(t)}f^\ast(t)dt
$$
(see \cite{Tal1, Tal2}) in \eqref{e4.1} to obtain: If $f$ belongs
to $L(N,\infty)$, then
$$
h(t)\leq {\frac{2M(1)(-\mu_n'(t))}{{\overline
M}^{-1}(M(1)){N}C_N^{1/N}{\mu_n(t)}^{1-\frac{1}{N}}}}
B^{-1}\Big(\frac{\|f\|_{N,\infty}}{{\overline
M}^{-1}(M(1))NC_N^{1/N}}\Big),
$$
and if we assume that $f$ belongs to $L(m,\infty)$ with $m