\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 $$\begin{gathered} -\mathop{\rm div}(a(x,u){\overline{M}}^{-1}(M(|\nabla u|)){\frac{\nabla u} {|\nabla u|}})=f \quad \text{in }{\Omega}, \\ u=0\quad\text{on }{\partial}{\Omega}, \end{gathered}\label{e1.1}$$ where $${\overline{M}}^{-1}(M(\frac{1}{(1+|s|)^{\theta}}))\leq a(x,s)\leq\beta, \label{e1.2}$$ 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 $$\begin{gathered} -\mathop{\rm div} a(x,u,{\nabla}u)=f \quad\text{in }{\Omega},\\ u=0\quad\text{on }{\partial}{\Omega}, \end{gathered} \label{e1.3}$$ in the Sobolev space $W^{1,p}_0(\Omega)$, under the condition $$a(x,s,\xi)\cdot\xi\geq{\frac{\alpha}{(1+|s|)^{{\theta}(p-1)}}}|\xi|^p, \label{e1.4}$$ 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 $10$ and $0\leq{\gamma}\leq1$, here $M(t)={t^p}\log^q{(e+t)}$ with $10$ 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}) $$M(t)\leq t{\overline M}^{-1}(M(t))\leq 2M(t)\quad \text{for all } t\geq 0\label{e2.1}$$ and the Young's inequality: for all $s, t\geq0$, $st\leq{\overline{M}(s)+M(t)}$. If for some $k>0$, $$M(2t)\leq{kM(t)}\quad \text{for all }t{\geq0},\label{e2.2}$$ 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 $$w^\ast(\mu_w(t))=t\label{e2.3}$$ 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 $\xi$, $\eta$ $\in \mathbb{R}^N$, $\xi \neq \eta$, $$a(x,s,\xi)\cdot\xi\geq {\overline M}^{-1}M(h(|s|))M(|\xi|)\label{e3.1}$$ 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, $$|a(x,s,\xi)|\leq a_0(x)+k_1{\overline P}^{-1}M(k_2|s|)+k_3 {\overline M}^{-1}M(k_4|\xi|)\label{e3.2}$$ where $a_0(x)$ belongs to $E_{\overline M}(\Omega)$ and $k_1, k_2, k_3, k_4$ to $\mathbb{R}_+^\ast$, $$(a(x,s,\xi)-a(x,s,\eta))\cdot(\xi-\eta)>0. \label{e3.3}$$ 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 $$\begin{gathered} A(u):=-\mathop{\rm div}(a(x,u,\nabla u)=f \quad\text{in }{\Omega},\\ u=0\quad\text{on }{\partial}{\Omega}, \end{gathered} \label{e3.4}$$ As regards the data $f$, we assume one of the following two conditions: Either $$f\in L^N(\Omega),\label{e3.5}$$ or $$\begin{gathered} f\in L^m(\Omega)\quad \text{with m=rN/(r+1) for some r>0},\\ \text{and }\int_{.}^{+\infty}(\frac{t}{M(t)})^rdt<+\infty. \end{gathered} \label{e3.6}$$ 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 $$\int_{\Omega}a(x,T_n(u_n),\nabla u_n)\cdot\nabla vdx=\int_{\Omega}f_nvdx\label{e4.1}$$ 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 $t>0$: $$h(t)\leq {\frac{2M(1)}{{{\overline M}^{-1}}(M(1))NC_N^{1/N}}} {\frac{-\mu_n'(t)}{{\mu_n(t)}^{1-\frac{1}{N}}}} B^{-1} \Big(\frac{\int_{\{|u_n|>t\}}|f_n|dx}{{\overline M}^{-1}(M(1))NC_N^{1/N}{{\mu_n(t)}^{1-\frac{1}{N}}}}\Big) \label{e4.2}$$ 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: $$\|u_n\|_{\infty}\leq H^{-1}\Big({\frac{2M(1)}{{\overline M}^{-1}(M(1)){N}C_N^{1/N}}}B^{-1}\Big(\frac{\|f\|_N}{{\overline M}^{-1}(M(1))NC_N^{1/N}}\Big)N|\Omega|^{1/N}\Big). \label{e4.3}$$ 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 $$\|u_n\|_\infty\leq H^{-1}\Big( {\frac{2M(1)\|f\|_m^r}{{({{\overline M}^{-1}(M(1))})^{r+1}}{N^{r}}C_N^{\frac{r+1}{N}}}}\Big(\frac{B^{-1}(c_0)}{c_0^r} +\int_{B^{-1}(c_0)}^{+\infty}\big(\frac{s}{M(s)}\big)^rds\Big)\Big). \label{e4.4}$$ 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 $$\|u_n\|_\infty\leq c. \label{e4.5}$$ \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 $${\int_{\Omega}M(|\nabla u_n|)dx\leq \frac{c\|f\|_{m^\ast} |\Omega|^{1-\frac{1}{m^\ast}}}{{{\overline M}^{-1}(M(h(c)))}}}. \label{e4.6}$$ 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 $$u_n \rightharpoonup u \quad\text{in W_0^{1}L_M(\Omega) for \sigma(\Pi L_M,\Pi E_{\overline M})} \label{e4.7}$$ and $$u_n \to u \quad\text{in E_M(\Omega) strongly and a.e. in \Omega.} \label{e4.8}$$ \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 $$\int_\Omega a(x,T_n(u_n),\nabla u_n)\cdot(\nabla u_n-\nabla v_j) dx = \int_\Omega f_n(u_n-v_j)dx\label{e4.9}$$ 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 $$\int_\Omega f_n(u_n-v_j)dx=\epsilon_0(n,j).\label{e4.10}$$ The left-hand side of \eqref{e4.9} is written as \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} 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 $$-\int_{\Omega\setminus\Omega^s_j} a(x,T_n(u_n),\nabla u_n)\cdot\nabla v_jdx=-\int_{\Omega\setminus\Omega^s}l\cdot\nabla udx\,+\,\epsilon_1(n,j).\label{e4.12}$$ 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 $$\int_\Omega a(x,T_n(u_n),\nabla v_j\chi^s_j )\cdot\left(\nabla u_n-\nabla v_j\chi^s_j\right)dx=\epsilon_2(n,j).\label{e4.13}$$ Consequently, combining \eqref{e4.10}, \eqref{e4.12} and \eqref{e4.13}, we obtain \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} 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 $n\to \infty$, $$\int_{{\Omega}^r}\left(a(x,T_n(u_n),\nabla u_n)-a(x,T_n(u_n),\nabla u)\right)\cdot\left(\nabla u_n-\nabla u\right)dx \to 0.\label{e4.15}$$ 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 $$\nabla u_n\to \nabla u\quad \text{a.e. in } \Omega. \label{e4.16}$$ \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 $mt\}}|f_n|dx. $$Then \eqref{e3.1} yields$$ {\frac{1}{k}}\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}M(h(|u_n|)) M(|\nabla u_n|)dx\leq \int_{\{|u_n|>t\}}|f_n|dx.$$Letting$k\to 0^+$we obtain $$-\frac{d}{dt}\int_{\{|u_n|>t\}}{{\overline M}^{-1}}M(h(|u_n|)) M(|\nabla u_n|)dx\leq \int_{\{|u_n|>t\}}|f_n|dx,\label{eA.2}$$ for almost every$t>0$. The hypotheses made on the$N$-function$M$, which are not a restriction, allow to affirm that the function$C(t)={\frac{1}{B^{-1}(t)}}is decreasing and convex (see \cite{Tal1}). Hence, Jensen's inequality yields \begin{align*} &C\Big(\frac{\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}(M(h(|u_n|))) M(|\nabla u_n|)dx}{\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}(M(h(|u_n|)))|\nabla u_n|dx}\Big)\\ &=C\Big(\frac{\int_{\{t<|u_n|\leq t+k\}}B(|\nabla u_n|){{\overline M}^{-1}}(M(h(|u_n|)))|\nabla u_n|dx}{\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}(M(h(|u_n|))) |\nabla u_n|dx}\Big)\\ &\leq{\frac{\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}(M(h(|u_n|)))dx}{\int_{\{t<|u_n|\leq t+k\}}{{\overline M}^{-1}}(M(h(|u_n|)))|\nabla u_n|dx}}\\ & \leq \frac{{\overline M}^{-1}(M(h(t)))(-\mu_n(t+k)+\mu_n(t))}{{\overline M}^{-1}(M(h(t+k)))\int_{\{t<|u_n|\leq t+k\}}|\nabla u_n|dx}. \end{align*} Taking into account that{\overline M}^{-1}(M(h(t)))\leq {\overline M}^{-1}(M(1))$, using the convexity of$C$and then letting$k\to 0^+$, we obtain for almost every$t>0, \begin{align*} &\frac{{\overline M}^{-1}(M(1))}{{\overline M}^{-1}(M(h(t)))}\; C\Big(\frac{{-\frac{d}{dt}}\int_{\{|u_n|>t\}}{\overline M}^{-1}(M(h(|u_n|)))M(|\nabla u_n|)dx}{{\overline M}^{-1}(M(1))({-\frac{d}{dt}}\int_{\{|u_n|>t\}}|\nabla u_n|dx)}\Big)\\ &\leq \frac{-\mu_n'(t)}{{-\frac{d}{dt}}\int_{\{|u_n|>t\}}|\nabla u_n|dx}. \end{align*} Now we recall the following inequality from \cite{Tal1}: $${-\frac{d}{dt}}\int_{\{|u_n|>t\}}|\nabla u_n|dx\geq NC_N^{1/N}{{\mu_n(t)}^{1-\frac{1}{N}}}\quad \text{for almost every }t>0. \label{eA.3}$$ The monotonicity of the functionC, \eqref{eA.2} and \eqref{eA.3} yield \begin{align*} &{\frac{1}{{\overline M}^{-1}(M(h(t)))}}\\ &\leq {\frac{-\mu_n'(t)}{{\overline M}^{-1}(M(1)) NC_N^{1/N}{{\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). \end{align*} Using \eqref{e2.1} and the fact that0