\documentclass[reqno]{amsart} %\pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \usepackage{hyperref} \AtBeginDocument{{\noindent\small 2004-Fez conference on Differential Equations and Mechanics \newline {\em Electronic Journal of Differential Equations}, Conference 11, 2004, pp. 1--10.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE/Conf/11 \hfil Existence and regularity of entropy solutions] {Existence and regularity of entropy solutions for some nonlinear elliptic equations} \author[L. Aharouch, E. Azroul\hfil EJDE/Conf/11 \hfilneg] {Lahsen Aharouch, Elhoussine Azroul} % in alphabetical order \address{Lahsen Aharouch\hfill\break D\'epartement de Math\'ematiques et Informatique\\ Facult\'e des Sciences Dhar-Mahraz\\ B.P. 1796 Atlas F\es, Maroc} \email{lahrouche@caramail.com} \address{Elhoussine Azroul\hfill\break D\'epartement de Math\'ematiques et Informatique\\ Facult\'e des Sciences Dhar-Mahraz\\ B.P. 1796 Atlas F\es, Maroc} \email{elazroul@caramail.com} \date{} \thanks{Published October 15, 2004.} \subjclass[2000]{35J60} \keywords{Orlicz Sobolev spaces; boundary value problems; Entropy solution} \begin{abstract} This paper concerns the existence and regularity of entropy solutions to the Dirichlet problem \begin{gather*} Au = -\mathop{\rm div} (a(x, u, \nabla u)) = f - \mathop{\rm div}\phi(u) \quad \mbox{in }\Omega \\ u = 0 \quad \mbox{on } \partial \Omega. \end{gather*} In particular, we show the $L^{{\bar q}}$-regularity of the solution to this boundary-value problem. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{thm}{Theorem}[section] \newtheorem{lemma}[thm]{Lemma} \newtheorem{remark}[thm]{Remark} \allowdisplaybreaks \section{Introduction} Let $\Omega$ be a bounded open subset of $\mathbb{R}^N$ $(N\geq 2)$, and let $p$ be a real number such that $2-\frac{1}{N} 0.\label{e1.3} \end{gather} Here$\alpha >0$,$\beta\geq 0$and$c(x)\in L^{p'}(\Omega)$. In the present paper, we study the boundary-value problem $$\begin{gathered} Au := -\mathop{\rm div} a(x, u, \nabla u) = f - \mathop{\rm div}\phi(u) \quad \mbox{in } \Omega \\ u = 0 \quad \mbox{on } \partial \Omega , \end{gathered} \label{e1.4}$$ where the right hand side is assumed to satisfy \begin{gather} f\in L^1 (\Omega),\label{e1.5} \\ \phi\in C^0 (\mathbb{R} , \mathbb{R}^N).\label{e1.6} \end{gather} Recall that, since no growth hypothesis is assumed on the function$\phi$, the term$\mathop{\rm div} \phi(u)$may be meaningless, even as a distribution for a function$v\in W_{0}^{1,r}(\Omega)$,$r>1$(see \cite{b} and \cite{bgdm}). \noindent \textbf{Definition}\; A function$u$is called an entropy solution of the Dirichlet problem \eqref{e1.4} if, \begin{gather*} u\in W_{0}^{1,q}(\Omega),\quad 10,\\ {\int_{\Omega}a(x, u, \nabla u)\nabla T_k (u -\varphi )\,dx \leq \int_{\Omega}f T_k (u -\varphi )\,dx + \int_{\Omega}\phi(u) \nabla T_k (u -\varphi )\,dx},\\ \forall \varphi\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega), \end{gather*} %\label{e1.7} where$T_k (s)$is the truncation operator at height$k>0$defined on$\mathbb{R}$. \smallskip When$\phi = 0$and$f$is a bounded Radon measure, it is known that \eqref{e1.4} admits a weak solution$u$in$W_{0}^{1,q}(\Omega)$with$ 1 0 \label{e1.9} \\ a(x,s,\zeta)\zeta \geq \alpha M(\frac{|\zeta|}{\delta}), \label{e1.10} \end{gather} for a.e. $x\in \Omega$, for all $s\in \mathbb{R}$, and all $\xi\in \mathbb{R}^N$ \section{Preliminaries} Let $M: \mathbb{R}^{+}\to \mathbb{R}^{+}$ be an $N$-function, i.e. $M$ is continuous, convex, with $M(t)>0$ for $t>0$, $\frac{M(t)}{t}\to 0$ as $t\to 0$, and $\frac{M(t)}{t}\to \infty$ as $t\to \infty$. Equivalently, $M$ admits the representation: $$M(t)={\int_{0}^{t}a(s)\,ds}$$ where $a:\mathbb{R}^{+}\to \mathbb{R}^{+}$ is nondecreasing, right continuous, with $a(0)=0$, $a(t)>0$ for $t>0$ and $a(t)$ tends to $\infty$ as $t\to \infty$. The conjugate of $M$ is also an $N$-function and it is defined by $\overline {M}={\int_{0}^{t}\bar{a}(s)\,ds}$, where $\bar a:\mathbb{R}^{+}\to \mathbb{R}^{+}$ is the function $\bar{a}(t)=\sup\{s: a(s)\leq t \}$. An $N$-function $M$ is said to satisfy the $\Delta_2$-condition if, for some $k$, $$M(2t)\leq kM(t) \quad \forall t\geq 0. \label{e2.1}$$ When \eqref{e2.1} holds only for $t\geq t_0 >0$ then $M$ is said to satisfy the $\Delta_2$ condition near infinity. We will extend these $N$-functions into even functions on all $\mathbb{R}$. Moreover, we have the following Young's inequality $$st \leq M(t) + \overline{ M}(s), \quad \forall s,t \geq 0.$$ Given two $N$-functions, we write $P<0$, $\frac{P(t)}{Q(\epsilon t)}\to 0$ as $t\to \infty$. This is the case if and only if $$\lim_{t\to\infty}\frac{Q^{-1}(t)}{P^{-1}(t)}=0.$$ %\par {\bf 2-2} 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 classes of) real valued measurable functions $u$ on $\Omega$ such that $$\int_\Omega M(u(x))\,dx < +\infty \quad (\mbox{resp. } \int_\Omega M(\frac{u(x)}{\lambda})\,dx < +\infty \mbox{ for some } \lambda>0 ).$$ The set $L_M(\Omega)$ is Banach space under the norm $$\|u\|_{M,\Omega}=\inf\big\{\lambda>0: \int_\Omega M(\frac{u(x)}{\lambda})\,dx\leq 1\big\}$$ 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 dual of $E_M(\Omega)$ can be identified with $L_{\overline{M}}(\Omega)$ by means of the pairing $\int_\Omega u v \,dx$, and the dual norm of $L_{\overline{M}}(\Omega)$ is equivalent to $\|.\|_{\overline{M},\Omega}$. %{\bf 2-3} We now turn to the Orlicz-Sobolev space, $W^{1}L_M(\Omega)$ [resp. $W^{1}E_M(\Omega)$] is the space of all functions $u$ such that $u$ and its distributional derivatives up to order 1 lie in $L_M(\Omega)$ [resp. $E_M(\Omega)$]. It is a banach space under the norm $$\|u\|_{1,M}= \sum_{|\alpha|\leq 1}\|D^{\alpha}u\|_M.$$ Thus, $W^{1}L_M(\Omega)$ and $W^{1}E_M(\Omega)$ can be identified with subspaces of product of $N+1$ copies of $L_M(\Omega)$. Denoting this product by $\prod L_M$, we will use the weak topologies $\sigma(\prod L_M, \prod E_{\overline{M}})$ and $\sigma(\prod L_M, \prod L_{\overline{M}})$. The space $W_{0}^{1}E_M(\Omega)$ is defined as the (norm) closure of the Schwartz space $D(\Omega)$ in $W^{1}E_M(\Omega)$ and the space $W_{0}^{1}L_M(\Omega)$ as the $\sigma(\prod L_M, \prod E_{\overline{M}})$ closure of $D(\Omega)$ in $W^{1}L_M(\Omega)$. %{\bf 2-4} Let $W^{-1}L_{\overline{M}}(\Omega)$ [resp. $W^{-1}E_{\overline{M}}(\Omega)$] denote the space of distributions on $\Omega$ which can be written as sums of derivatives of order $\leq 1$ of functions in $L_{\overline{M}}(\Omega)$ [resp. $E_{\overline{M}}(\Omega)$]. It is a Banach space under the usual quotient norm.(for more details see \cite{a}). We recall some lemmas introduced in \cite{be} which will be used later. \begin{lemma} \label{lem2.1.0} A domain $\Omega$ has the segment property if for every $x\in \partial \Omega$ there exists an open set $G_x$ and a nonzero vector $y_x$ such that $x\in G_x$ and if $z\in {\overline \Omega}\cap G_x$, then $z+ty_x\in \Omega$ for all $00\\ \int_{\Omega}a(x, u, \nabla u)\nabla T_k (u -\varphi )\,dx\leq \int_{\Omega}f T_k (u -\varphi )\,dx + \int_{\Omega}\phi(u) \nabla T_k (u -\varphi )\,dx,\\ \forall\; \varphi\in W_{0}^{1}L_{M}(\Omega)\cap L^{\infty}(\Omega) \end{gathered} \label{e3.1} admits at least one solution$u\in W_{0}^{1,\overline{q}}(\Omega)$. \end{thm} When$p = N$we assume, in addition, that There exists an$N$-function$H$such that$H(t^N)$is equivalent to$M(t)$. \begin{thm} \label{thm3.2} Assume that for$p = N$the above hypothesis hold, \eqref{e1.8}--\eqref{e1.10} hold,$f\in L^1 (\Omega)$,$\phi \in C^0 (\mathbb{R}, \mathbb{R}^N)$,$\int_{.}^{\infty}\frac{t^{N-1}}{M(t)}\,dt <\infty$and$\frac{t^N}{{\overline{H}}^{-1}(e^{t^{N'}})}$remains bounded near infinity. Then \eqref{e3.1} admits at least one solution in$W_{0}^{1,N}(\Omega)$. \end{thm} \begin{proof}[Proof of Theorems \ref{thm3.1} and \ref{thm3.2}] \quad\\ {\bf Step 1 The approximate problem and a priori estimate.} Let$f_n$be a sequence in$W^{-1}E_{\overline{M}}(\Omega)\cap L^1 (\Omega)$such that$f_n \to f$in$L^1 (\Omega)$, and$\|f_n\|_1 \leq \|f\|_1$. Consider the approximate problem $$\begin{gathered} Au_n = f_n - \mathop{\rm div} \phi_n(u_n) \\ u_n \in W_{0}^{1}L_M (\Omega) \end{gathered}\label{e3.3}$$ where$\phi_n (x) = \phi(T_n (x))$. From the work \cite{gm}, there exists at least one solution$u_n$of the approximate problem \eqref{e3.3}. Moreover, as in \cite{bk}, there exists a constant$C = C(p, \alpha, \|f\|_1)$such that $$\|\nabla u_n\|_{L_{\overline{q}}(\Omega)}\leq C,$$ which implies that$u_n$is bounded in$W_{0}^{1,\overline{q}}(\Omega)$. Then there exists$u\in W_{0}^{1,\overline{q}}(\Omega)$and a subsequence still denoted by$u_n$such that $$\begin{gathered} u_n\rightharpoonup u \quad \mbox{weakly in } W_{0}^{1,{\overline{q}}}(\Omega)\\ u_n\to u \quad \mbox{strongly in L^{{\overline{q}}}(\Omega) and a.e. in }\Omega . \end{gathered} \label{e3.4}$$ Moreover, the use of$T_k (u_n)$as test function in \eqref{e3.3} implies that the sequence$T_k (u_n)$is bounded in$W_{0}^{1}L_{M}(\Omega)$, then there exists a subsequence of$T_k (u_n)$still denoted by$T_k (u_n)$such that $$\begin{gathered} T_k (u_n)\rightharpoonup T_k(u) \quad \mbox{weakly in } W_{0}^{1}L_M(\Omega) \mbox{ for } \sigma(\prod L_M, \prod E_{\overline{M}})\\ T_k(u_n)\to T_k(u) \quad \mbox{strongly in } E_M(\Omega)\mbox{ and a.e. in } \Omega . \end{gathered} \label{e3.5}$$ \noindent {\bf Step 2 Convergence of the gradient.} Let$\Omega_r = \{x\in \Omega : |\nabla T_k (u(x))|\leq r\}$and denote by$\chi_r$the characteristic function of$\Omega_r$. Clearly,$\Omega_r\subset\Omega_{r+1}$and$\rm {meas}(\Omega\backslash \Omega_r)\to 0$as$r \to \infty$.\\ Fix$r$and let$s\geq r. We have, \begin{align*} 0&\leq\int_{\Omega_r}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u))][\nabla T_k(u_n) -\nabla T_k(u)]\,dx\\ &\leq \int_{\Omega_s}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u))][\nabla T_k(u_n) -\nabla T_k(u)]\,dx\\ &=\int_{\Omega_s}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u)\chi_s)][\nabla T_k(u_n) -\nabla T_k(u)\chi_s]\,dx\\ &\leq \int_{\Omega}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u)\chi_s)][\nabla T_k(u_n) -\nabla T_k(u)\chi_s]\,dx. \end{align*} On the other hand, leth>k$and$M = 4k + h$. If one takes$w_n = T_{2k}(u_n -T_{h}(u_n)+T_{k}(u_n)-T_{k}(u))$as test function in \eqref{e3.3}, it is easy to see that$\nabla w_n = 0$when$|u_n|>M. We can write $$\int_{\Omega}a(x, T_M(u_n) , \nabla T_M(u_n))\nabla w_n \,dx = \int_{\Omega}f_n w_n \,dx + \int_{\Omega}\phi_n (u_n)\nabla w_n \,dx.$$ We have \begin{align*} &\int_{\Omega}a(x, T_M(u_n) , \nabla T_M(u_n))\nabla T_{2k}(u_n -T_{h}(u_n) +T_{k}(u_n)-T_{k}(u))\,dx \\ &\geq \int_{\Omega}a(x, T_k(u_n) , \nabla T_k(u_n))(\nabla T_{k}(u_n) -\nabla T_{k}(u))\,dx\\ &\quad - \int_{|u_n|>k}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)|\,dx\\ & = \int_{\Omega}a(x, T_k(u_n) , \nabla T_k(u_n))(\nabla T_{k}(u_n) -\nabla T_{k}(u)\chi_s)\,dx\\ &\quad - \int_{\Omega}a(x, T_k(u_n) , \nabla T_k(u_n))(\nabla T_{k}(u) -\nabla T_{k}(u)\chi_s)\,dx\\ &\quad - \int_{|u_n|>k}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)|\chi_s \,dx\\ &\quad - \int_{|u_n|>k}|a(x, T_M(u_n) , \nabla T_M(u_n))|(|\nabla T_{k}(u)| - |\nabla T_{k}(u)|\chi_s) \,dx\,. \end{align*} Then \begin{align*} &\int_{\Omega}a(x, T_M(u_n) , \nabla T_M(u_n))\nabla T_{2k}(u_n -T_{h}(u_n)+T_{k}(u_n)-T_{k}(u))\,dx\\ &\geq \int_{\Omega}a(x, T_k(u_n) , \nabla T_k(u_n))(\nabla T_{k}(u_n) -\nabla T_{k}(u)\chi_s)\,dx\\ &\quad - \int_{\Omega \backslash \Omega_s}a(x, T_k(u_n), \nabla T_k(u_n))\nabla T_{k}(u)\,dx \\ &\quad - \int_{|u_n|>k}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)|\chi_s \,dx \\ &\quad - \int_{\Omega \backslash \Omega_s}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)| \,dx\,. \end{align*} From this inequality, it follows \begin{aligned} &\int_{\Omega}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u)\chi_s)][\nabla T_k(u_n) -\nabla T_k(u)\chi_s]\,dx\\ & \leq \int_{|u_n|>k}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)|\chi_s \,dx\\ &\quad + \int_{\Omega \backslash \Omega_s}a(x, T_k(u_n) , \nabla T_k(u_n))\nabla T_{k}(u)\,dx \\ &\quad + \int_{\Omega \backslash \Omega_s}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)| \,dx \\ &\quad + \int_{\Omega}f_n T_{2k}(u_n -T_{h}(u_n)+T_{k}(u_n)-T_{k}(u))\,dx \\ &\quad + \int_{\Omega}\phi_n(u_n))\nabla T_{2k}(u_n -T_{h}(u_n)+T_{k}(u_n) -T_{k}(u))\,dx \\ &\quad - \int_{\Omega}a(x, T_k(u_n) , \nabla T_k(u)\chi_s)][\nabla T_k(u_n) -\nabla T_k(u)\chi_s]\,dx \end{aligned} \label{e3.6} Now, we study each term of the right hand side of the above inequality. We denote by\varepsilon_i(t)$($i=1,2,3,\dots$) various sequences of real numbers which tends to$0$when$t$tends to infinity. Remark that$a(x, T_{\mu}(u_n), \nabla T_{\mu}(u_n))$is bounded in$L_{ {\overline{M}}}(\Omega)$for all$\mu >0$. Let$\varepsilon >0$, we have $$M(\frac{|\nabla T_{k}(u)|\chi_s \chi_{\{|u_n|>k\}}}{\varepsilon}) \leq M(\frac{s}{\varepsilon})\in L^1 (\Omega)$$ and $$|\nabla T_{k}(u)|\chi_s \chi_{\{|u_n|>k\}} \to 0 \quad \rm {a.e.}$$ Then by the Lebesgue dominated convergence theorem we deduce that $$|\nabla T_{k}(u)|\chi_s \chi_{\{|u_n|>k\}} \to 0 \quad \mbox{in } L_{M}(\Omega),$$ which implies that the first term in the right hand side of \eqref{e3.6} tends to$0$as$n$tends to$\infty$. Concerning the second and third terms on the right hand side of \eqref{e3.6}, since$|a(x, T_M(u_n) , \nabla T_M(u_n))|$and$|a(x, T_k(u_n) , \nabla T_k(u_n))|$are bounded in$L_{\overline{M}}(\Omega)$then there exist two functions$\varphi$and$\psi$in$L_{\overline{M}}(\Omega)$such that $$\begin{gathered} |a(x, T_M(u_n) , \nabla T_M(u_n))|\to \varphi \quad \mbox{for } \sigma (L_{\overline{M}}, E_M)\\ |a(x, T_k(u_n) , \nabla T_k(u_n))|\to \psi \quad \mbox{for } \sigma (L_{\overline{M}}, E_M)\,. \end{gathered} \label{e3.7}$$ This implies $$\int_{\Omega \backslash \Omega_s}|a(x, T_M(u_n) , \nabla T_M(u_n))||\nabla T_{k}(u)|\,dx \to \int_{\Omega \backslash \Omega_s} \varphi |\nabla T_{k}(u)|\,dx \label{e3.8}$$ and $$\int_{\Omega \backslash \Omega_s}|a(x, T_k(u_n) , \nabla T_k(u_n))||\nabla T_{k}(u)| \,dx \to \int_{\Omega \backslash \Omega_s}\psi |\nabla T_{k}(u)|\,dx\,.\label{e3.9}$$ On the other hand, $$\lim_{n\to \infty}\int_{\Omega}f_n T_{2k}(u_n -T_h (u_n) + T_k (u_n) - T_k (u))\,dx = \int_{\Omega}f T_{2k}(u -T_h (u))\,dx =\varepsilon_3 (h)$$ and, for$nlarge enough, one can write. \begin{align*} &\int_{\Omega}\phi_n (u_n) \nabla T_{2k}(u_n -T_h (u_n) + T_k (u_n) - T_k (u))\,dx\\ &= \int_{\Omega}\phi (T_{4k+h}(u_n)) \nabla T_{2k}(u_n -T_h (u_n) + T_k (u_n) - T_k (u))\,dx, \end{align*} which yields, \begin{align*} &\lim_{n\to \infty}\int_{\Omega}\phi_n (u_n) \nabla T_{2k}(u_n -T_h (u_n) + T_k (u_n) - T_k (u))\,dx \\ &= \int_{\Omega}\phi (u) \nabla T_{2k}(u -T_h (u)\,dx = 0. \end{align*} The right-most term in \eqref{e3.6} tends to0$: Since$a(x, T_k(u_n) , \nabla T_k(u)\chi_s)$converges strongly to$a(x, T_k(u) , \nabla T_k(u)\chi_s)$in$(E_{\overline{M}}(\Omega))^N$, using Lemma \ref{lem2.3} while$ \nabla T_k(u_n)$tends weakly to$\nabla T_k(u)by \eqref{e3.4}. We conclude then that \begin{align*} 0&\leq \limsup_{n\to \infty} \int_{\Omega_r}\big[a(x, T_k(u_n) , \nabla T_k(u_n))\\ &\quad - a(x, T_k(u_n) , \nabla T_k(u)][\nabla T_k(u_n) -\nabla T_k(u)\big]\,dx \\ &\leq \int_{\Omega \backslash \Omega_s}\varphi |\nabla T_{k}(u)|\,dx + \int_{\Omega \backslash \Omega_s}\psi |\nabla T_{k}(u)|\,dx + \int_{\Omega}f T_{2k}(u -T_h (u))\,dx\,. \end{align*} %\label{e3.10} Lettings$and$h$approach infinity we get, $$\int_{\Omega_r}[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u)][\nabla T_k(u_n) -\nabla T_k(u)]\,dx \to 0$$ as$n\to \infty$. Passing to a subsequence if necessary, we can assume that $$[a(x, T_k(u_n) , \nabla T_k(u_n)) - a(x, T_k(u_n) , \nabla T_k(u)][\nabla T_k(u_n) -\nabla T_k(u)]\to 0$$ a.e. in$\Omega_r$. As in \cite{be}, we deduce that there exists a subsequence still denoted by$u_n$such that$ \nabla u_n \to \nabla u \quad \mbox{a.e. in } \Omega $. %\label{e3.11} \smallskip \noindent{\bf Step 3 Passage to the limit.} Let$\varphi \in W_{0}^{1}L_M (\Omega)\cap L^{\infty}(\Omega)$, and set$M = k + \|\varphi\|_{\infty} $with$k>0$. We shall prove that $$\liminf_{n\to \infty}{ \int_{\Omega}a(x, u_n , \nabla u_n)\nabla T_k(u_n - \varphi)\,dx \geq \int_{\Omega}a(x, u , \nabla u)\nabla T_k(u - \varphi)\,dx}.$$ We have: If$|u_n|>M$then$|u_n - \varphi|>kwhich implies \begin{align*} &a(x, u_n, \nabla u_n)\nabla T_k(u_n - \varphi)\\ &=a(x, T_M(u_n) , \nabla T_M (u_n))(\nabla u_n - \nabla\varphi)\chi_{\{|u_n - \varphi|\leq k\}}\\ &=a(x, T_M(u_n) , \nabla T_M (u_n))(\nabla T_M(u_n) - \nabla\varphi)\chi_{\{|u_n - \varphi|\leq k\}}. \end{align*} Let\Omega_s = \{x\in \Omega: |\nabla \varphi|\leq s\}$and denote by$\chi_s$the characteristic function of$\Omega_s. Then \begin{align*} &\int_{\Omega}a(x, u_n , \nabla u_n)\nabla T_k(u_n - \varphi)\,dx\\ & = \int_{\Omega} a(x, T_M(u_n) , \nabla T_M (u_n))(\nabla T_M(u_n) - \nabla\varphi)\chi_{\{|u_n - \varphi|\leq k\}}\,dx \\ & = \int_{\Omega} a(x, T_M(u_n) , \nabla T_M (u_n))(\nabla T_M(u_n) - \nabla\varphi \chi_s)\chi_{\{|u_n - \varphi|\leq k\}}\,dx \\ &\quad - \int_{\Omega}a(x, T_M(u_n) , \nabla T_M (u_n))(\nabla \varphi - \nabla\varphi \chi_s)\chi_{\{|u_n - \varphi|\leq k\}}\,dx, \end{align*} and \begin{aligned} &\int_\Omega a(x, u_n , \nabla u_n)\nabla T_k(u_n - \varphi)\,dx\\ & \geq -\int_{\Omega\backslash \Omega_s} |a(x, T_M(u_n) , \nabla T_M (u_n))||\nabla \varphi |\,dx\\ &\quad + \int_{\Omega} \big[a(x, T_M(u_n) , \nabla T_M (u_n)) - a(x, T_M(u_n) , \nabla \varphi \chi_s)\big]\\ &\quad\times [\nabla T_M(u_n) - \nabla\varphi \chi_s]\chi_{\{|u_n - \varphi|\leq k\}}\,dx \\ &\quad + \int_{\Omega}a(x, T_M(u_n) , \nabla \varphi \chi_s)[\nabla T_M(u_n) - \nabla\varphi \chi_s]\chi_{\{|u_n - \varphi|\leq k\}}\,dx. \end{aligned} \label{e3.12} Similarly to the proof of \eqref{e3.8}, the first term in the right hand side of \eqref{e3.12} is greater than a value\varepsilon_6(s), which implies \begin{aligned} &\liminf_{n\to \infty} \int_{\Omega}a(x, u_n , \nabla u_n)\nabla T_k(u_n - \varphi)\,dx\\ & \geq \lim_{n\to \infty} \int_{\Omega}a(x, T_M(u_n) , \nabla \varphi \chi_s) [\nabla T_M(u_n) - \nabla\varphi \chi_s]\chi_{\{|u_n - \varphi|\leq k\}}\,dx + \varepsilon_6(s) \\ &\quad + \int_{\Omega} [a(x, T_M(u) , \nabla T_M (u)) - a(x, T_M(u) , \nabla \varphi \chi_s)]\\ &\quad \times [\nabla T_M(u) - \nabla\varphi \chi_s]\chi_{\{|u - \varphi|\leq k\}}\,dx. \end{aligned}\label{e3.13} From Lemma \ref{lem2.3}, the first term in the right hand side of \eqref{e3.13} is equal to $$\int_{\Omega}a(x, T_M(u) , \nabla \varphi \chi_s)[\nabla T_M(u) - \nabla\varphi \chi_s]\chi_{\{|u - \varphi|\leq k\}}\,dx + \varepsilon_6(s),$$ then \begin{align*} &\liminf_{n\to \infty}\int_{\Omega}a(x, u_n , \nabla u_n)\nabla T_k(u_n - \varphi)\,dx\\ &\geq \int_{\Omega} a(x, T_M(u) , \nabla T_M (u)) [\nabla T_M(u) - \nabla\varphi \chi_s]\chi_{\{|u - \varphi|\leq k\}}\,dx + \varepsilon_6(s). \end{align*} By lettings\to + \infty, we obtain \begin{align*} &\liminf_{n\to \infty}\int_{\Omega}a(x, u_n ,\nabla u_n)\nabla T_k(u_n - \varphi)\,dx\\ &\geq \int_{\Omega} a(x, T_M(u) , \nabla T_M (u)) [\nabla T_M(u) - \nabla\varphi ]\chi_{\{|u - \varphi|\leq k\}}\,dx \\ & = \int_{\Omega} a(x, u , \nabla u)\nabla T_k(u - \varphi)\,dx. \end{align*} Now takingT_k(u_n - \varphi )$as test function in \eqref{e3.8} and passing to the limit we deduce the desired statement. \end{proof} \begin{remark} \label{rmk3.1} \rm If$M$and$\overline {M} $satisfy the$\Delta_2$condition, instead of \eqref{e1.8} we can assume the condition: $$|a(x, s, \xi)|\leq c(x) + k_1{\overline {M}}^{-1}M(k_2|s|) + k_3{\overline {M}}^{-1}M(k_4|\xi|) \label{e1.8'}.$$ Then we prove the same result as in Theorems \ref{thm3.1} and \ref{thm3.2}. \end{remark} \begin{remark} \label{rmk3.2} \rm If w$f$belongs to$W^{-1}L_{\overline {M}}(\Omega)$the statements of Theorems \ref{thm3.1} and \ref{thm3.2} still hold. \end{remark} \subsection*{Example} Let$2 -\frac{1}{N}1$. 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