\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 79, pp. 1--20.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/79\hfil Strongly nonlinear nonhomogeneous problems] {Strongly nonlinear nonhomogeneous elliptic unilateral problems with $L^1$ data and no sign conditions} \author[E. Azroul, H. Redwane, C. Yazough,\hfil EJDE-2012/79\hfilneg] {Elhoussine Azroul, Hicham Redwane, Chihab Yazough} % in alphabetical order \address{Elhoussine Azroul \newline University of Fez, Faculty of Sciences Dhar El Mahraz, Laboratory LAMA, Department of Mathematics, B.P. 1796 Atlas Fez, Morocco} \email{azroul\_elhoussine@yahoo.fr} \address{Hicham Redwane \newline Facult\'e des Sciences Juridiques, Economiques et Sociales, University Hassan 1, B.P. 784, Settat, Morocco} \email{redwane\_hicham@yahoo.fr} \address{Chihab Yazough \newline University of Fez, Faculty of Sciences Dhar El Mahraz, Laboratory LAMA, Department of Mathematics, B.P. 1796 Atlas Fez, Morocco} \email{chihabyazough@gmail.com} \thanks{Submitted September 19, 2011. Published May 15, 2012.} \subjclass[2000]{35J60} \keywords{Entropy solutions; variable exponent; unilateral problem} \begin{abstract} In this article, we prove the existence of solutions to unilateral problems involving nonlinear operators of the form: $$ Au+H(x,u,\nabla u)=f $$ where $A$ is a Leray Lions operator from $W_0^{1,p(x)}(\Omega)$ into its dual $W^{-1,p'(x)}(\Omega)$ and $H(x,s,\xi)$ is the nonlinear term satisfying some growth condition but no sign condition. The right hand side $f$ belong to $L^1(\Omega)$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Partial differential equations with nonlinearities involving non constant exponents have attracted an increasing amount of attention in recent years. The development, mainly by R\.{u}\~{z}icka \cite{R}, of a theory modeling the behavior of electrorheological fluids, an important class of non-Newtonian fluids, seems to have boosted a still far from completed effort to study and understand nonlinear PDE's involving variable exponents. Other applications relate to image processing \cite{CLR}, elasticity \cite{AM}, the flow in porous media \cite{AS} and problems in the calculus of variations involving variational integrals with nonstandard growth \cite{Z}. This in turn, gave rise to a revival of the interest in Lebesgue and Sobolev spaces with variable exponent,where many of the basic properties of these spaces are established by the work of Kov\`{a}cik and Rakosnik \cite{KR}. Many models of the obstacle problem have already been analyzed for constant exponents of nonlinearity. In \cite{AAB} the authors have proved the existence of solution for quasilinear degenerated elliptic unilateral problems associated to the operator $Au+g(x,u,\nabla u)=f$ in which the nonlinear term satisfies the sign condition. The principal part $A$ is a differential elliptic operator of the second order in divergence form, acting from $W_0^{1,p}(\Omega,\omega)$ into its dual $W^{-1,p'}(\Omega,\omega)$ and g having natural growth with respect to $\nabla u$ and $u$ not assuming any growth restrictions, but assuming the sign-condition. Porretta \cite{P} studied the same problem in the classical Sobolev space that is ($p(.)=p$ constant) where the right-hand side is a bounded Radon measure on $\Omega$ and where the sign condition is violated, more precisely the problem treated in \cite{P} is of the form \begin{gather*} Au+g(u)|\nabla u|^{p}=\mu\quad \text{in}\quad\Omega\\ u=0\quad \text{on } \partial\Omega. \end{gather*} The work by Aharouch et al \cite{AA,AAA} can be seen as generalization of \cite{P} in the sense that in \cite{AA} the nonlinearity have taken as $ H(x,u,\nabla u)$ and in \cite{AAA} the degenerated case for the same problem. Recently, Rodriguez et al in \cite{RSU} have proved the existence and uniqueness of an entropy solution to obstacle problem with variable growth and $L^{1}$ data, of the form \begin{gather*} - \Delta_{p(.)} u + \beta (., u) = f \quad \text{in } \Omega \\ u = 0 \quad \text{on } \partial\Omega, \end{gather*} where $\beta$ is some function related to a maximal monotone graph. Besides, while $f(x,u,\nabla u)$, Benboubker, Azroul and Barbara have proved the existence results in Sobolev spaces with variable exponent by using a classical theorem of Lions operators of the calculus of variations (see \cite{BAB}). Recently, while $Au=-\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u),\ H\equiv 0$, Bendahmane and Wittbold \cite{BW} proved the existence and uniqueness of renormalized solution with $L^1$-data, and Wittbold and Zimmermann \cite{WZ} extended the results to the case $Au= -\operatorname{div}(a(x,u))$, (see also Bendahmane and Karlsen \cite{BK}). The objective of our article, is to study the non homogenous obstacle problem with $L^{1}$ data associated to the general nonlinear operator of the form \begin{equation} \label{P} \begin{gathered} Au + H(x,u,\nabla u) = f \quad \text{in } \Omega, \\ u = 0 \quad \text{on } \partial\Omega. \end{gathered} \end{equation} The principal part $Au=-\operatorname{div}(a(x,\nabla u))$ is a differential elliptic operator of the second order in divergence form, acting from $W_0^{1,p(x)}(\Omega)$ into its dual $W^{-1,p'(x)}(\Omega)$ and we suppose that the lower order term satisfies the exact natural growth: $$|H(x,s,\xi)|\leq \gamma(x)+g(s)|\xi|^{p(x)}$$ with $\gamma(x)\in L^{1}(\Omega)$ and $g\in L^{1}(\mathbb{R})$ and $g\geq0$ but not satisfying the sign condition. Under these assumptions the above problem does not admit, in general, a weak solution since the terms $a(u,\nabla u)$ and $H(x,u,\nabla u)$ may not belong to $L^1_{\rm loc}(\Omega)$. In order to overcome this difficulty, we work with the framework of entropy solutions introduced by B\'enilan et al \cite{BBGGPV}. Let us mention that an equivalent notion of solution, called renormalized solution was first introduced by Di-Perna and Lions \cite{DL} for the study of Boltzmann equation. It has been used by many authors to study the elliptic equations (see \cite{BGDM}) and the parabolic equations (see \cite{BM,BMR1,BR}). Note that our paper can be seen as a generalization of \cite{AA} and \cite{RSU}, and as a continuation of \cite{BAB}. The outline of this paper is as follows. In Section 2, we give some preliminaries and notations. In Section 3, the existence of entropy solutions of $\eqref{P}$ is obtained. In Section 4, we give the proof of Proposition \ref{prop1}, Lemma \ref{lem5.2} and Lemma \ref{lem8} (see appendix). \section{Preliminaries} In what follows, we recall some definitions and basic properties of Lebesgue and Sobolev spaces with variable exponents. For each open bounded subset $\Omega$ of $\mathbb{R}^{N}$ $(N \geq 1)$, we denote $$ C^{+}(\overline{\Omega})= \big\{\text{continuous function } p:\overline{\Omega} \to \mathbb{R}^{+} \text{ such that } 1 < p_{-} \leq p_{+} < \infty\big\}, $$ where $p_{-}=\inf_{x\in\overline{\Omega}}p(x)$ and $p_{+}=\sup_{x\in\overline{\Omega}}p(x)$. We define the variable exponent Lebesgue space for $ p\in C^{+}(\overline{\Omega}) $ by: $$ L^{p(x)}(\Omega) = \big\{u : \Omega \to \mathbb{R} \text{ measurable},\, \int_{\Omega} |u(x)|^{p(x)} dx < \infty \big\}. $$ the space $L^{p(x)}(\Omega)$ under the norm $$ \|u\|_{p(x)} = \inf \big\{\lambda > 0,\; \int_{\Omega} |\frac{u(x)}{\lambda}|^{p(x)} \leq 1 \big\} $$ is a uniformly convex Banach space, then reflexive. We denote by $ L^{p'(x)}(\Omega) $ the conjugate space of $L^{p(x)}(\Omega) $ where $\frac{1}{p(x)}+\frac{1}{p'(x)} = 1$. \begin{proposition}[\cite{FZ}] \label{prop1} (i) For any $ u\in L^{p(x)}(\Omega) $ and $ v\in L^{p'(x)}(\Omega)$, we have $$ |\int_{\Omega} u v dx | \leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \|u\|_{p(x)} \|v\|_{p'(x)}. $$ (ii) For all $p_1, p_{2}\in C^{+}(\overline{\Omega}) $ such that $ p_1(x) \leq p_{2}(x) $ and any $x \in \overline{\Omega}$, we have $ L^{p_{2}(x)}(\Omega) \hookrightarrow L^{p_1(x)}(\Omega) $ and the embedding is continuous. \end{proposition} \begin{proposition}[\cite{FZ}] \label{prop2} Let us denote $$ \rho(u) = \int_{\Omega} |u|^{p(x)} dx, \quad \forall u\in L^{p(x)}(\Omega); $$ then the following assertions hold: \begin{itemize} \item[(i)] $ \|u\|_{p(x)} < 1$ (resp. = 1 or $>1$) if and only if $\rho(u) < 1$ (resp. = 1 or $> 1$) \item[(ii)] $ \|u\|_{p(x)} > 1$ implies $ \|u\|_{p(x)}^{p_{-}} \leq \rho(u) \leq \|u\|_{p(x)}^{p_{+}} $, and $ \|u\|_{p(x)} < 1 $ implies $ \|u\|_{p(x)}^{p_{+}} \leq \rho(u) \leq \|u\|_{p(x)}^{p_{-}}$ \item[(iii)] $\|u\|_{p(x)} \to 0$ if and only if $\rho(u) \to 0$, and $\|u\|_{p(x)} \to \infty$ if and only if $\rho(u) \to \infty$. \end{itemize} \end{proposition} We define the variable exponent Sobolev space by $$ W^{1,p(x)}(\Omega) = \{ u\in L^{p(x)}(\Omega) \text{ and } |\nabla u|\in L^{p(x)}(\Omega) \}. $$ where the norm is defined by $$ \|u\|_{1,p(x)} = \|u\|_{p(x)} + \|\nabla u\|_{p(x)} \quad \forall u \in W ^{1,p(x)}(\Omega). $$ We denote by $ W_0^{1,p(x)}(\Omega)$ the closure of $C_0^{\infty}(\Omega)$ in $W^{1,p(x)}(\Omega)$ and $ p*(x) = \frac{N p(x)}{N - p(x)}$ for $p(x) < N$. \begin{proposition}[\cite{FZ}] \label{prop3} (i) Assuming $ 1< p_{-}\leq p_{+} < \infty$, the spaces $ W^{1, p(x)}(\Omega) $ and $ W_0^{1, p(x)}(\Omega) $ are separable and reflexive Banach spaces. (ii) if $q\in C^{+}(\bar{\Omega})$ and $q(x) < p*(x) $ for any $x \in \overline{\Omega}$, then the embedding $ W^{1, p(x)}_0(\Omega) \hookrightarrow\hookrightarrow L^{q(x)}(\Omega)$ is compact and continuous. (iii) There is a constant $C > 0$, such that $$ \|u\|_{p(x)} \leq C \|\nabla u\|_{p(x)} \quad \forall u\in W_0^{1,p(x)}(\Omega). $$ \end{proposition} \begin{remark} \label{rmk2.4} \rm By Proposition \ref{prop3} (iii), we know that $\|\nabla u\|_{p(x)}$ and $\|u\|_{1,p(x)}$ are equivalent norms on $W^{1,p(x)}_0(\Omega)$. \end{remark} \section{Existence of an entropy solutions} In this section, we study the existence of an entropy solution of the obstacle problem. \subsection{Basic assumptions and some Lemmas} Throughout the paper, we assume that the following assumptions hold. Let $\Omega$ be a bounded open set of $\mathbb{R}^{N}$\ $(N\geq 1)$, $ p\in C^+(\overline{\Omega})$ and $(1/ p(x)) + (1/ p'(x))=1$. The function $a:\Omega\times \mathbb{R}^{N}\to \mathbb{R}^{N}$ is a Carath\'{e}odory function satisfying the following conditions: For all $\xi, \eta\in \mathbb{R}^{N}$ and for almost every $x\in\Omega$, \begin{gather} |a(x,\xi)|\leq \beta (k(x)+|\xi|^{p(x)-1}),\label{ass1} \\ [a(x,\xi)-a(x,\eta)](\xi-\eta)>0\quad \forall \xi\neq\eta, \label{ass2} \\ a(x,\xi)\xi\geq \alpha |\xi|^{p(x)}, \label{ass3} \end{gather} where $k(x)$ is a positive function in $L^{p'(x)}(\Omega)$ and $\alpha$ and $\beta$ are a positive constants. Let $H(x,s,\xi):\Omega\times \mathbb{R}\times \mathbb{R}^{N}\to \mathbb{R}$ be a Carath\'{e}odory function such that for a.e. $x\in\Omega$ and for all $s\in \mathbb{R}$, $\xi\in \mathbb{R}^{N}$, the growth condition: \begin{equation} |H(x,s,\xi)|\leq \gamma(x)+g(s)|\xi|^{p(x)}\label{ass4} \end{equation} is satisfied, where $g:\mathbb{R}\to \mathbb{R}^{+}$ is a continuous positive function that belongs to $L^{1}(\mathbb{R})$, while $\gamma(x)$ belongs to $L^{1}(\Omega)$. \begin{equation} f\in L^1(\Omega).\label{ass4'} \end{equation} Finally, let the convex set $$ K_{\psi}=\big\{u\in W_0^{1,p(x)}(\Omega),\, u\geq\psi \text{ a.e. in } \Omega\big\} $$ where $\psi$ is a measurable function such that \begin{equation} \psi^{+}\in W_0^{1,p(x)}(\Omega)\cap L^{\infty}(\Omega)\label{ass5} \end{equation} \begin{lemma}[\cite{BAB}] \label{lem1} Let $ g\in L^{r(x)}(\Omega) $ and $ g_n\in L^{r(x)}(\Omega) $ with $ \|g_n\|_{r(x)} \leq C $ for $ 1< r(x)< \infty$. If $ g_n(x)\to g(x) $ a.e. on $ \Omega$, then $ g_n\rightharpoonup g $ in $ L^{r(x)}(\Omega) $. \end{lemma} \begin{lemma}\label{lem2} Assume that \eqref{ass1}--\eqref{ass3}, and let $ (u_n)_n $ be a sequence in $ W_0^{1,p(x)}(\Omega) $ such that $ u_n \rightharpoonup u $ weakly in $ W_0^{1,p(x)}(\Omega) $ and \begin{equation} \int_{\Omega} [a(x,\nabla u_n) - a(x,\nabla u)]\nabla (u_n - u )dx \to 0.\label{eq37} \end{equation} Then $ u_n\to u$ strongly in $W_0^{1,p(x)}(\Omega)$. \end{lemma} The proof of the above Lemma is a slight modification of the analogues one of \cite[Lemma 3.2]{BAB}. \begin{lemma}\label{lem5.2} Let $ F:\mathbb{R} \to \mathbb{R} $ be a uniformly Lipschitz function with $F(0) = 0$ and $ p\in C_{+}(\bar{\Omega}) $. If $u\in W_0^{1,p(x)}(\Omega)$, then $F(u)\in W_0^{1,p(x)}(\Omega)$, moreover, if $ D $ is the set of discontinuity points of $F'$ is finite, then $$ \frac{\partial (F\circ u)}{\partial x_i} = \begin{cases} F'(u)\frac{\partial u}{\partial x_i} & \text{a.e. in } \{x\in \Omega: u(x)\notin D\} \\ 0 \quad & \text{a.e. in } \{x\in \Omega : u(x)\in D\}. \end{cases} $$ \end{lemma} The proof of the above lemma is presented in the appendix. The following Lemma is a direct deduction from Lemma \ref{lem5.2}. \begin{lemma} \label{lem3.4} Let $u\in W_0^{1,p(x)}(\Omega)$ then $u^{+}=\max(u,0)$ and $u^{-}=\max(-u,0)$ lie in $W_0^{1,p(x)}(\Omega)$. Moreover $$ \frac{\partial u^{+}}{\partial x_{i}} =\begin{cases} \frac{\partial u}{\partial x_{i}}& \text{if } u>0 \\ 0 & \text{if } u\leq0, \end{cases} \quad \frac{\partial u^{-}}{\partial x_{i}}=\begin{cases} 0 & \text{if } u\geq0\\ -\frac{\partial u}{\partial x_{i}} & \text{if } u<0. \end{cases} $$ \end{lemma} \subsection{Definition and existence result of an entropy solution} In this article, $T_k$ denotes the truncation function at height $k\geq 0:\ T_k(r) = \min(k,\ \max(r,\ -k))$. Define $$ T_0^{1,p(x)}(\Omega)=\big\{ u\text{ measurable in } \Omega: T_{k}(u)\in W_0^{1,p(x)}(\Omega),\,\forall\ k>0 \big\}. $$ We now give the following definition and existence theorem. \begin{definition}\label{def1} \rm An entropy solution of the obstacle problem for $\{f,\psi\}$ is a measurable function $u\in T_0^{1,p(x)}(\Omega)$ such that $u\geq\psi$ a.e. in $\Omega$, and $$ \int_{\Omega}a(x,\nabla u)\nabla T_{k}(\varphi-u)dx +\int_{\Omega}H(x,u,\nabla u)T_{k}(\varphi-u)dx\geq\int_{\Omega}f T_{k}(\varphi-u)dx $$ for all $k\geq 0$ for all $\varphi\in {K}_{\psi} \cap L^{\infty}(\Omega)$. \end{definition} \begin{theorem}\label{thm1} Under assumptions \eqref{ass1}, \eqref{ass2}, \eqref{ass3}, \eqref{ass4}, \eqref{ass4'} and \eqref{ass5} there exists at least an entropy solution. \end{theorem} \subsection{Approximate problem} Let $\Omega_n$ be a sequence of compact subsets of $\Omega$ such that $\Omega_n$ is increasing to $\Omega$ as $n\to\infty$. We consider the following sequence of approximate problems \begin{equation} \label{Pn} \begin{gathered} u_n\in K_{\psi}\\ \int_{\Omega}a(x,\nabla u_n)\nabla (u_n-v)dx +\int_{\Omega}H_n(x,u_n,\nabla u_n)(u_n-v)dx \leq\int_{\Omega}f_n(u_n-v)dx \end{gathered} \end{equation} for all $v\in K_{\psi}$, where $f_n$ are regular functions such that $f_n\in L^{\infty}(\Omega)$, strongly converge to $f$ in $L^{1}(\Omega)$ and $\|f_n\|_{L^{1}(\Omega)} \leq \|f\|_{L^{1}(\Omega)}$ and $$ H_n(x,s,\xi)=\frac{H(x,s,\xi)}{1+\frac{1}{n}|H(x,s,\xi)|}\chi_{\Omega_n} $$ where $\chi_{\Omega_n}$ is the characteristic function of $\Omega_n$. Note that $|H_n(x,s,\xi)|\leq |H(x,s,\xi)|$ and $|H_n(x,s,\xi)|\leq n$. \begin{theorem}\label{thm2} For fixed n, the approximate problem \eqref{Pn} has at least one solution. \end{theorem} \begin{proof} Let $X=K_\psi$, we define the operator $G_n:X\to X^{*} $ by $$ \langle G_nu,v\rangle =\int_{\Omega}H_n(x,u,\nabla u)v dx $$ Thanks to H\"{o}lder's inequality, for all $u, v\in X$, \begin{align*} \big|\int_{\Omega}H_n(x,u,\nabla u)v dx \big| &\leq (\frac{1}{p_{-}}+\frac{1}{p'_{-}}) (\int_{\Omega}|H_n(x,u,\nabla u)|^{p'(x)}dx)^{\theta}\|v\|_{L^{p(x)}(\Omega)}\\ &\leq (\frac{1}{p_{-}}+\frac{1}{p'_{-}}) n^{\theta p'_{+}} (\operatorname{meas}(\Omega))^{\theta}\|v\|_{L^{p(x)}(\Omega)} \end{align*} with \begin{equation} \theta=\begin{cases} 1/p'_{-} \quad \text{if } \|H_n(x,u,\nabla u)\|_{L^{p'(x)}(\Omega)}\geq 1\\ 1/p'_{+} \quad \text{if } \|H_n(x,u,\nabla u)\|_{L^{p'(x)}(\Omega)}\leq 1 \end{cases} \end{equation} We deduce that the operator $B_n=A+G_n$ is pseudomonotone (see appendix, Lemma \ref{lem8}). On the other hand, we show that $B_n$ is coercive in the following sense: there exists $v_0\in K_{\psi}$ such that $$ \frac{\langle B_nv,v-v_0\rangle}{\|v\|_{1,p(x)}}\to +\infty\quad \text{if } \|v\|_{1,p(x)}\to \infty \text{ and } v\in K_{\psi}. $$ Let $ v_0\in K_{\psi}$, we use H\"{o}lder inequality and the growth condition to have \begin{align*} \langle Av,v_0\rangle &=\int_{\Omega}a(x,\nabla v)\nabla v_0dx\\ &\leq C(\frac{1}{p^{-}}+\frac{1}{p'^{-}}) \Big(\int_{\Omega}|a(x,\nabla v)|^{p'(x)}\Big)^{\theta'}\|v_0\|_{W_0^{1,p(x)} (\Omega)}\\ &\leq C(\frac{1}{p^{-}}+\frac{1}{p'^{-}})\|v_0\|_{W_0^{1,p(x)}(\Omega)} \Big(\int_{\Omega}\beta(K(x)^{p'(x)}+|\nabla v|^{p(x)})\Big)^{\theta'}\\ &\leq C_0(C_1+\rho(\nabla v))^{\theta'} \end{align*} where \begin{equation} \theta'=\begin{cases} \frac{1}{p'^{-}}\quad \text{if } \|a(x,\nabla v)\|_{L^{p'(x)}(\Omega)}\geq1\\ \frac{1}{p'^{+}}\quad \text{if } \|a(x,\nabla v)\|_{L^{p'(x)}(\Omega)}\leq 1 \end{cases} \end{equation} From \eqref{ass3}, we have \begin{equation} \frac{\langle Av,v\rangle }{\|v\|_{1,p(x)}} -\frac{\langle Av,v_0\rangle}{\|v\|_{1,p(x)}}\geq \frac{1}{\|v\|_{1,p(x)}}(\alpha\rho(\nabla v)-C_0(C_1 +\rho(\nabla v))^{\theta'}) \end{equation} hence $\frac{\rho(\nabla v)}{\|v\|_{1,p(x)}}\to \infty$ as $\|v\|_{1,p(x)}\to \infty$. Since $\frac{}{\|v\|_{1,p(x)}}$ and $\frac{}{\|v\|_{1,p(x)}}$ are bounded, then we have $$ \frac{\langle B_nv,v-v_0\rangle }{\|v\|_{1,p(x)}} =\frac{\langle Av,v-v_0\rangle}{\|v\|_{1,p(x)}} +\frac{\langle G_nv,v\rangle }{\|v\|_{1,p(x)}} -\frac{\langle G_nv,v_0\rangle}{\|v\|_{1,p(x)}}\to\infty $$ as $\|v\|_{1,p(x)}\to\infty$. Finally $B_n$ is pseudomonotone and coercive. Hence by virtue of \cite[Theorem 8.2, chapter 2]{L}, the approximate problem \eqref{Pn} has at least one solution. \end{proof} \subsubsection{A priori estimate} \begin{proposition} \label{prop3.8} Assume that \eqref{ass1}--\eqref{ass5} hold, and let $u_n$ is a solution of the approximate problem \eqref{Pn}. Then, there exists a constant $C$ (which does not depend on the $n$ and $k$) such that $$ \int_{\Omega}|\nabla T_{k}(u_n)|^{p(x)}dx\leq Ck\quad \forall\ k>0. $$ \end{proposition} \begin{proof} Let $v=u_n-\eta\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})$ where $G(s)=\int_0^{s}\frac{g(t)}{\alpha}dt$ and $\eta\geq 0$, we have $v\in W_0^{1,p(x)}(\Omega)$, and for $\eta$ small enough we deduce that $ v\geq\psi$, and thus $v$ is an admissible test function in \eqref{Pn}. Then \begin{align*} &\int_{\Omega} a(x,\nabla u_n)\nabla\Big(\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})\Big)dx\\ &+ \int_{\Omega}H_n(x,u_n,\nabla u_n) \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx \\ &\leq \int_{\Omega}f_n \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx \end{align*} which implies \begin{align*} &\int_{\Omega} a(x,\nabla u_n)\nabla u_n\frac{g(u_n)}{\alpha}\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &+\int_{\Omega} a(x,\nabla u_n)\nabla T_{k}(u_n^{+}-\psi^{+})\exp(G(u_n))dx\\ &\leq -\int_{\Omega}H_n(x,u_n,\nabla u_n) \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &\quad +\int_{\Omega}f_n\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &\leq \int_{\Omega}(f_n+\gamma(x)) \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &\quad + \int_{\Omega}g(u_n)|\nabla u_n|^{p(x)}\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx. \end{align*} In view of \eqref{ass3} and since $\|f_n\|_{L^1(\Omega)}\leq \|f\|_{L^1(\Omega)},\ \gamma \in L^1(\Omega)$ we deduce that \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla T_{k}(u_n^{+}-\psi^{+})\exp(G(u_n)dx\\ &\leq\int_{\Omega}f_n\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})\,dx +\int_{\Omega}\gamma(x) \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})\,dx\\ &\leq (\|f\|_{L^{1}(\Omega)}+\|\gamma\|_{L^{1}(\Omega)}) \exp(\frac{\|g\|_{L^{1}(\mathbb{R})}}{\alpha})k \leq C_1 k \end{align*} where $C_1$ is a positive constant. Consequently, \begin{align*} &\int_{\{|u_n^{+}-\psi^{+}|\leq k\}} a(x,\nabla u_n)\nabla u_n^{+}\exp(G(u_n))dx\\ &\leq\int_{\{|u_n^{+}-\psi^{+}|\leq k\}}a(x,\nabla u_n)\nabla \psi^{+}\exp(G(u_n))dx+C_1k\nonumber \end{align*} Thanks to \eqref{ass3} and Young's inequality, we deduce that \begin{equation} \int_{\{|u_n^{+}-\psi^{+}|\leq k\}}|\nabla u_n^{+}|^{p(x)}dx\leq C_{2}k. \label{eq1} \end{equation} Since $\{x\in\Omega,|u_n^{+}|\leq k\}\subset \{x\in\Omega,|u_n^{+}-\psi^{+}|\leq k+\|\psi^{+}\|_{\infty}\}$, it follows that $$ \int_{\Omega}|\nabla T_{k}(u_n^{+})|^{p(x)}dx =\int_{\{|u_n^{+}|\leq k\}}|\nabla u_n^{+}|^{p(x)} \leq\int_{\{|u_n^{+}-\psi^{+}| \leq k+\|\psi^{+}\|_{\infty}\}}|\nabla u_n^{+}|^{p(x)}dx $$ Moreover, \eqref{eq1} implies \begin{equation} \int_{\Omega}|\nabla T_{k}(u_n^{+})|^{p(x)}dx\leq C_{3}k,\quad\forall k>0, \label{eq2} \end{equation} where $C_{3}$ is a positive constant. On the other hand, taking $v=u_n+\exp(-G(u_n)T_{k}(u_n^{-})$ as test function in \eqref{Pn}, we obtain \begin{align*} &-\int_{\Omega} a(x,\nabla u_n)\nabla (\exp(-G(u_n))T_{k}(u_n^{-}))dx\\ & -\int_{\Omega}H_n(x,u_n,\nabla u_n) \exp(-G(u_n))T_{k}(u_n^{-})dx\\ &\leq-\int_{\Omega}f_n \exp(-G(u_n))T_{k}(u_n^{-})dx \end{align*} Using \eqref{ass4}, we have \begin{align*} &\int_{\Omega} a(x,\nabla u_n)\nabla u_n \frac{g(u_n)}{\alpha} \exp(-G(u_n))T_{k}(u_n^{-})dx\\ &-\int_{\Omega} a(x,\nabla u_n)\nabla T_{k}(u_n^{-})\exp(-G(u_n))dx\\ &\leq\int_{\Omega}\gamma(x)\exp(-G(u_n))T_{k}(u_n^{-})dx +\int_{\Omega}g(u_n)|\nabla u_n|^{p(x)}\exp(-G(u_n))T_{k}(u_n^{-})dx\\ &\quad -\int_{\Omega}f_n\exp(-G(u_n))T_{k}(u_n^{-})dx \end{align*} By \eqref{ass3} and since $\gamma\in L^{1}(\Omega),\|f_n\|_{L^{1}(\Omega)}\leq\|f\|_{L^{1}(\Omega)}$ we have \begin{align*} &-\int_{\Omega} a(x,\nabla u_n)\nabla T_{k}(u_n^{-})\exp(-G(u_n))dx\\ &=\int_{\{u_n\leq 0\}}a(x,\nabla u_n)\nabla T_{k}(u_n)\exp(-G(u_n))dx\leq C_{3}k\nonumber \end{align*} By using again \eqref{ass3} we deduce that \begin{equation} \int_{\{u_n\leq 0\}}|\nabla T_{k}(u_n)|^{p(x)}dx\leq C_4k, \label{eq3} \end{equation} where $C_4$ is a constant positive. Combining \eqref{eq2} and \eqref{eq3}, we conclude \begin{gather} \int_{\Omega}|\nabla T_{k}(u_n)|^{p(x)}dx\leq Ck\quad with \quad C>0,\label{eq4} \\ \|\nabla T_{k}(u_n)\|_{L^{p(x)}(\Omega)}\leq (Ck)^{\theta''},\label{eq80} \end{gather} with \begin{equation} \theta''=\begin{cases} 1/ p^{-} & \text{if } \|\nabla T_{k}(u_n)\|_{L^{p(x)}(\Omega)}\geq1\\ 1/ p^{+} & \text{if } \|\nabla T_{k}(u_n)\|_{L^{p(x)}(\Omega)}\leq1. \end{cases} \end{equation} \end{proof} \subsubsection{Strong convergence of truncations} \begin{proposition} \label{prop3.9} There exist a measurable function $u$ and a subsequence of $u_n$ such that $$ T_{k}(u_n)\to T_{k}(u) \quad\text{strongly in }W_0^{1,p(x)}(\Omega). $$ \end{proposition} The proof of the above proposition is done in two steps. \textbf{Step 1.} We will show that $(u_n)_n$ is a Cauchy sequence in measure in $\Omega$. According to the Poincar\'{e} inequality and \eqref{eq80}, \begin{equation} \begin{split} k \ \text{meas}\{|u_n|> k\} & = \int_{\{|u_n|> k\}} |T_{k}(u_n)| dx \leq \int_{\Omega} |T_{k}(u_n)| dx \\ & \leq { \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big)\|1\|_{p'(x)} \|T_{k}(u_n)\|_{p(x)}} \\ & \leq {\big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) (\operatorname{meas}(\Omega)+1)^{1/p'_{-}} \|T_{k}(u_n)\|_{p(x)} } \leq C k^{1/\gamma} \end{split} \end{equation} Thus \begin{equation} \operatorname{meas}\{|u_n|> k\} \leq C \frac{1}{k^{1-\frac{1}{\gamma}}} \to 0 \quad \text{as } k \to\infty.\label{eqsig2} \end{equation} For all $\delta > 0$, we obtain \begin{align*} \operatorname{meas} \{|u_n - u_{m}|>\delta\} &\leq \operatorname{meas} \{|u_n|>k\} + \operatorname{meas} \{|u_{m}|>k\}\\ &\quad + \operatorname{meas} \{|T_{k}(u_n) - T_{k}(u_{m})|>\delta\}. \end{align*} In view of\eqref{eqsig2}, we deduce that for all $\varepsilon >0$, there exists $k_0>0$ such that \begin{equation} \operatorname{meas}\{|u_n|> k\} \leq \frac{\varepsilon}{3} \quad \text{and} \quad \operatorname{meas}\{|u_n|> k\} \leq \frac{\varepsilon}{3}\quad \forall k \geq k_0. \label{eqsig3} \end{equation} and by \eqref{eq4}, we have $(T_{k}(u_n))_n$ bounded in $W_0^{1,p(x)}(\Omega)$, then there exists a subsequence $(T_{k}(u_n))_n$ such that $T_{k}(u_n)$ converges to $\eta_{k}$ a.e. in $\Omega$, strongly in $L^{p(x)}(\Omega)$ and weakly in $W_0^{1,p(x)}(\Omega)$ as $n$ tends to $\infty$. Thus, we can assume that $(T_{k}(u_n))_n$ is a Cauchy sequence in measure in $\Omega$, then there exists $n_0$ which depend on $\delta$ and $\varepsilon$ such that \begin{equation} \operatorname{meas}\{|T_{k}(u_n)-T_{k}(u_{m})|> \delta\} \leq \frac{\varepsilon}{3} \quad \forall m, n \geq n_0 \text{ and } k\geq k_0. \label{eqsig4} \end{equation} by combining\eqref{eqsig3}$ and\eqref{eqsig4}$, we obtain for all $\delta > 0$, there exists $\varepsilon > 0$ such that $$ \operatorname{meas}\{|u_n-u_{m}|> \delta\} \leq \varepsilon\quad \forall n,\ m \geq n_0(k_0,\delta). $$ Then $(u_n)_n$ is a Cauchy sequence in measure in $\Omega$, thus, there exists a subsequence still denoted $u_n$ which converges almost everywhere to some measurable function $u$, and by Lemma \ref{lem1}, we obtain \begin{equation} T_{k}(u_n) \to T_{k}(u) \text{ strongly in } L^{p(x)}(\Omega) \text{ and weakly in } W_0^{1,p(x)}(\Omega).\label{eq5} \end{equation} \textbf{Step 2.} We will use the following function of one real variable, which is defined as follows \begin{equation} h_{j}(s)= \begin{cases} 1 & \text{if } |s|\leq j\\ 0 & \text{if } |s|\geq j+1\\ j+1-|s| & \text{if } j\leq |s|\leq j+1 \end{cases} \label{eq6} \end{equation} where $j$ is a nonnegative real parameter. To prove the strong convergence of truncation $T_{k}(u_n)$, we have to prove the following assertions: \begin{proposition}\label{prop3.10} The subsequence of $u_n$ solution of problem \eqref{Pn} satisfies, for any $k\geq 0$, Assertion (i): \begin{equation} \lim_{j\to\infty}\lim_{n\to\infty}\int_{\{j\leq|u_n|\leq j+1\}}a(x,\nabla u_n)\nabla u_ndx=0.\label{eq7} \end{equation} Assertion(ii): \begin{equation} \lim_{j\to\infty}\lim_{n\to\infty}\int_{\Omega}a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))(\nabla T_{k}(u_n)-\nabla T_{k}(u))h_{j}(u_n)dx=0.\label{eq8} \end{equation} Assertion(iii): \begin{equation} \lim_{j\to\infty}\lim_{n\to\infty}\int_{\Omega}a(x,\nabla T_{k}(u_n))\nabla T_{k}(u_n)(1-h_{j}(u_n))dx=0.\label{eq81} \end{equation} Assertion(iv): \begin{equation} \lim_{j\to\infty}\lim_{n\to\infty}\int_{\Omega}\Big(a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))\Big)(\nabla T_{k}(u_n)-\nabla T_{k}(u))dx=0.\label{eq9} \end{equation} \end{proposition} The proof of the above proposition is shown in the appendix. Thanks to \eqref{eq9} and lemma \ref{lem2}, we have \begin{gather} T_{k}(u_n)\to T_{k}(u) \quad \text{strongly in } W_0^{1,p(x)}(\Omega)\ \text{as n tends to } +\infty,\label{eq14} \\ \nabla u_n\to \nabla u \quad \text{ a.e. in } \Omega.\label{eq14a} \end{gather} \subsubsection{Passing to the limit} \begin{equation} H_n(x,u_n,\nabla u_n)\to H(x,u,\nabla u)\quad \text{strongly in } L^{1}(\Omega).\label{eq10} \end{equation} Let $v=u_n+\exp(-G(u_n))\int_{u_n}^{0}g(s)\chi_{\{s<-h\}}ds$. Since $v\in W_0^{1,p(x)}(\Omega)$ and $v\geq\psi$ is an admissible test function in \eqref{Pn}, \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla\Big(-\exp(-G(u_n))\int_{u_n}^{0}g(s) \chi_{\{s<-h\}}\Big)\,ds\,dx\\ & +\int_{\Omega}H(x,u_n,\nabla u_n)(-\exp(-G(u_n)) \int_{u_n}^{0}g(s)\chi_{\{s<-h\}}ds)dx\\ &\leq \int_{\Omega}f_n(-\exp(-G(u_n))\int_{u_n}^{0}g(s)\chi_{\{s<-h\}}\,ds\,dx. \end{align*} This implies \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla u_n \frac{g(u_n)}{\alpha} \exp(-G(u_n))(\int_{u_n}^{0}g(s)\chi_{\{s<-h\}}ds)dx\\ &+\int_{\Omega}a(x,\nabla u_n)\nabla u_n \exp(-G(u_n))g(u_n) \chi_{\{u_n<-h\}}dx\\ &\leq \int_{\Omega}\gamma(x) \exp(-G(u_n))\int_{u_n}^{0}g(s) \chi_{\{s<-h\}}\,ds\,dx\\ &\quad + \int_{\Omega}g(u_n)|\nabla u_n|^{p(x)} \exp(-G(u_n)) \int_{u_n}^{0}g(s)\chi_{\{s<-h\}}\,ds\,dx\\ &\quad - \int_{\Omega}f_n\exp(-G(u_n))\int_{u_n}^{0}g(s) \chi_{\{s<-h\}}\,ds\,dx, \end{align*} using \eqref{ass3} and since $\int_{u_n}^{0}g(s)\chi_{\{s<-h\}}ds\leq \int_{-\infty}^{-h} g(s)ds$, we obtain \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla u_n \exp(-G(u_n))g(u_n)\chi_{\{u_n<-h\}}dx\\ &\leq \exp(\frac{\|g\|_{L^{1}(\mathbb{R})}}{\alpha}) \int_{-\infty}^{-h} g(s)ds(\|\gamma\|_{L^{1}(\Omega)}+\|f_n\|_{L^{1}(\Omega)})\\ &\leq \exp(\frac{\|g\|_{L^{1}(\mathbb{R})}}{\alpha})\int_{-\infty}^{-h} g(s)ds(\|\gamma\|_{L^{1}(\Omega)}+\|f\|_{L^{1}(\Omega)}) \end{align*} using again \eqref{ass3}, we obtain \begin{equation} \int_{\{u_n<-h\}}g(u_n)|\nabla u_n|^{p(x)}dx\leq c\int_{-\infty}^{-h}g(s)ds\label{eq11} \end{equation} and since $g\in L^{1}(\mathbb{R})$, we deduce that \begin{equation} \lim_{h\to+\infty}\sup_n\int_{\{u_n<-h\}}g(u_n)|\nabla u_n|^{p(x)}dx=0.\label{eq12} \end{equation} On the other hand, let $$ M=\exp(\frac{\|g\|_{L^{1}(R)}}{\alpha})\int_0^{+\infty}g(s)ds $$ and $h\geq M+\|\psi^{+}\|_{L^{\infty}(\Omega)}$. Consider $$ v=u_n-\exp(G(u_n))\int_0^{u_n}g(s)\chi_{\{s>h\}}ds. $$ Since $v\in W_0^{1,p(x)}(\Omega)$ and $v\geq\psi$, $v$ is an admissible test function in \eqref{Pn}. Then, similarly to \eqref{eq12}, we obtain \begin{equation} \lim_{h\to +\infty}\sup_{n\in N}\int_{\{u_n>h\}}g(u_n)|\nabla u_n|^{p(x)}dx=0.\label{eq13} \end{equation} Combining \eqref{eq14}, \eqref{eq12}, \eqref{eq13} and Vitali's theorem, we conclude \eqref{eq10}. Now, let $\varphi\in K_{\psi}\cap L^{\infty}(\Omega)$ and take $v=u_n-T_{k}(u_n-\varphi)$ as a test function in \eqref{Pn}. We obtain \begin{equation} \begin{gathered} u_n\in K_{\psi}\\ \begin{aligned} &\int_{\Omega}a(x,\nabla u_n)\nabla T_{k} (u_n-\varphi)dx+\int_{\Omega}H_n(x,u_n,\nabla u_n)T_{k}(u_n-\varphi)dx\\ &\leq\int_{\Omega}f_nT_{k}(u_n-\varphi)dx \quad \forall \varphi\in K_{\psi}\cap L^{\infty}(\Omega),\; \forall k>0. \end{aligned} \end{gathered} \label{eq15} \end{equation} Finally, from \eqref{eq14} and \eqref{eq10}, we can pass to the limit in \eqref{eq15}. This completes the proof of Theorem \ref{thm1}. \section{Appendix} \begin{proof}[Proof of Proposition \ref{prop1}] \textbf{Assertion (i):} Consider the function $$ v=u_n-\eta \exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+}. $$ For $j$ large enough and $\eta$ small enough, we can deduce that $v\geq\psi$ and since $v\in W_0^{1,p(x)}(\Omega)$, $v$ is a admissible test function in \eqref{Pn}. Then, we obtain \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla \Big(\exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+} \Big)dx\\ &+\int_{\Omega}H_n(x,u_n,\nabla u_n)\exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+}dx\\ &\leq \int_{\Omega}f_n\exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+}dx. \end{align*} From the growth conditions \eqref{ass3} and \eqref{ass4}, we have \begin{equation} \begin{aligned} &\int_{\Omega}a(x,\nabla u_n)\nabla (T_1(u_n-T_{j}(u_n))^{+})\exp(G(u_n))dx\\ &\leq \int_{\Omega} \gamma(x)\exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+}dx\\ &\quad +\int_{\Omega}f_n\exp(G(u_n))T_1(u_n-T_{j}(u_n))^{+}dx. \end{aligned}\label{eq16} \end{equation} Since $f_n$ converges to $f$ strongly in $L^1(\Omega)$ and $\gamma \in L^1(\Omega)$, by Lebesgue's theorem, the right-hand side approaches zero as $n, j\to\infty$. Therefore, passing to the limit first in $n$, then in $j$, we obtain from \eqref{eq16} \begin{equation} \lim_{j\to\infty}\lim_{n\to \infty}\int_{\{j\leq u_n\leq j+1\}} a(x,\nabla u_n)\nabla u_ndx=0.\label{eq18} \end{equation} On the other hand, consider the test function $v=u_n+\exp(-G(u_n))T_1(u_n-T_{j}(u_n))^{-}$ in \eqref{Pn}. Similarly to \eqref{eq18}, it is easy to see that \begin{equation} \lim_{j\to\infty}\lim_{n\to \infty}\int_{\{-j-1\leq u_n\leq -j\}} a(x,\nabla u_n)\nabla u_ndx=0\label{eq19} \end{equation} Finally, by \eqref{eq18} and \eqref{eq19} we obtain assertion (i). \textbf{Assertion (ii):} On one hand, let $v=u_n-\eta \exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)$ with $h_{j}$ is defined in \eqref{eq6} and $\eta$ small enough such that $v\in K_{\psi}$, then we take $v$ as test function in \eqref{Pn}, we obtain \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla \Big(\eta \exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)\Big)dx\\ &+\int_{\Omega}H_n(x,u_n,\nabla u_n)\Big(\eta \exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)\Big)dx\\ &\leq\int_{\Omega}f_n \eta \exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)dx. \end{align*} Similarly, using \eqref{ass3}$) and \eqref{ass4}$), we deduce \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla(T_{k}(u_n)-T_{k}(u))^{+} \exp(G(u_n))h_{j}(u_n)dx\\ &\leq\int_{\Omega}\gamma(x)\exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)dx\\ &+\int_{\{j\leq u_n\leq j+1\}}a(x,\nabla u_n)\nabla u_n \exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}dx\\ &+\int_{\Omega}f_n\exp(G(u_n))(T_{k}(u_n)-T_{k}(u))^{+}h_{j}(u_n)dx \end{align*} In view of \eqref{eq18}, the convergence $f_n$ to $f$ in $L^1(\Omega)$ and $\gamma \in L^1(\Omega)$, it is easy to see that \begin{equation} \begin{aligned} &\lim_{j\to +\infty} \lim_{n\to +\infty} \int_{\{T_{k}(u_n)-T_{k}(u)\geq 0\}}a(x,\nabla u_n)\nabla(T_{k}(u_n)-T_{k}(u))^{+}\\ &\quad\times \exp(G(u_n))h_{j}(u_n)dx\leq 0. \end{aligned}\label{eq20} \end{equation} Moreover, \eqref{eq20} becomes \begin{align*} & \lim_{j\to +\infty} \lim_{n\to +\infty} \int_{\{T_{k}(u_n)-T_{k}(u)\geq 0,\ |u_n|\leq k\}}a(x,\nabla u_n) \nabla(T_{k}(u_n)-T_{k}(u))\\ &\quad\times \exp(G(u_n))h_{j}(u_n)dx\\ &- \lim_{j\to +\infty} \lim_{n\to +\infty} \int_{\{T_{k}(u_n)-T_{k}(u)\geq 0,\ |u_n|>k\}}a(x,\nabla u_n) \nabla T_{k}(u)\\ &\quad\times \exp(G(u_n))h_{j}(u_n)dx \leq 0 \end{align*} Since $h_{j}(u_n)=0$ if $|u_n|>j+1$, we obtain \begin{align*} & \lim_{j\to +\infty} \lim_{n\to +\infty}\int_{\{T_{k}(u_n)-T_{k}(u) \geq 0,\ |u_n|>k\}}a(x,\nabla u_n)\nabla T_{k}(u)\exp(G(u_n))h_{j}(u_n)dx\\ &= \lim_{j\to +\infty} \lim_{n\to +\infty}\int_{\{T_{k}(u_n)-T_{k}(u)\geq 0,\ |u_n|>k\}}a(x,\nabla T_{j+1}(u_n))\nabla T_{k}(u)\\ &\quad\times \exp(G(u_n))h_{j}(u_n)dx\\ & = \lim_{j\to +\infty} \int_{\{ |u|>k\}}X_j \nabla T_{k}(u) \exp(G(u))h_{j}(u)dx=0, \end{align*} where $X_j$ is the limit of $a(x,\nabla T_{j+1}(u_n))$ in $(L^{p'(x)}(\Omega))^N$ as $n$ goes to infinity and $\nabla T_{k}(u) \chi_{\{|u|>k\}}=0$ a.e. in $\Omega$. Consequently, \begin{equation} \begin{split} &\lim_{j,n\to\infty}\int_{\{T_{k}(u_n)-T_{k}(u)\geq 0\}}\Big(a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))\Big)\\ &\times (\nabla T_{k}(u_n)-\nabla T_{k}(u))h_{j}(u_n)=0. \end{split}\label{eq21} \end{equation} On the other hand, taking $v=u_n+ \exp(-G(u_n))(T_{k}(u_n)-T_{k}(u))^{-}h_{j}(u_n)$ as test function in \eqref{Pn} and reasoning as in \eqref{eq21} we have \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla(-\exp(-G(u_n))(T_{k}(u_n) -T_{k}(u))^{-}h_{j}(u_n))dx\\ &+\int_{\Omega}H_n(x,u_n,\nabla u_n)(- \exp(-G(u_n))(T_{k}(u_n) -T_{k}(u))^{-}h_{j}(u_n))dx\\ &\leq-\int_{\Omega}f_n(\exp(-G(u_n))(T_{k}(u_n)-T_{k}(u))^{-} h_{j}(u_n))dx \end{align*} Similarly to \eqref{eq21}, it is easy to see that \begin{equation} \lim_{j,n\to\infty}\int_{\{T_{k}(u_n)-T_{k}(u)\leq 0\}}a(x,\nabla u_n)\nabla(T_{k}(u_n)-T_{k}(u))\exp(-G(u_n))h_{j}(u_n)dx=0.\label{eq22} \end{equation} Combing \eqref{eq21} and \eqref{eq22} we obtain the desired assertion (ii). \textbf{Assertion (iii):} Let $v=u_n+\exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))$ as test function in\eqref{Pn}. Then we have \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla \Big(-\exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))\Big)dx\\ &+\int_{\Omega}H_n(x,u_n,\nabla u_n) \Big(- \exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))\Big)dx\\ &\leq-\int_{\Omega}f_n\exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))dx \end{align*} Using\eqref{ass4} and \eqref{ass3}, we deduce that \begin{align*} &\int_{\{u_n\leq0\}}a(x,\nabla u_n)\nabla T_{k}(u_n) \exp(-G(u_n))(1-h_{j}(u_n))dx\\ &\leq-\int_{\{-1-j\leq u_n\leq -j\}}a(x,\nabla u_n) \nabla u_n\exp(-G(u_n))T_{k}(u_n)^{-}dx\\ &\quad +\int_{\Omega}\gamma(x)\exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))dx\\ &\quad -\int_{\Omega}f_n\exp(-G(u_n))T_{k}(u_n)^{-}(1-h_{j}(u_n))dx \end{align*} In view of \eqref{eq7}, the second integral tends to zero as $n$ and $j$ approach infinity. By Lebesgue's theorem, it is possible to conclude that the third and the fourth integrals converge to zero as $n$ and $j$ approach infinity. Then \begin{equation} \lim_{j,n\to\infty}\int_{\{u_n\leq0\}}a(x,\nabla T_{k}(u_n))\nabla T_{k}(u_n)(1-h_{j}(u_n))dx=0.\label{eq82} \end{equation} On the other hand, we take $v=u_n-\eta \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})(1-h_{j}(u_n))$ which is an admissible test function in \eqref{Pn}, we have \begin{align*} &\int_{\Omega}a(x,\nabla u_n)\nabla\Big(\eta \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})(1-h_{j}(u_n))\Big)dx\\ &+\int_{\Omega}H_n(x,u_n,\nabla u_n)\Big(\eta \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})(1-h_{j}(u_n))\Big)dx\\ &\leq\int_{\Omega}f_n\Big(\eta \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})(1-h_{j}(u_n))\Big)dx \end{align*} Which takes, by using \eqref{ass4} and \eqref{ass3}, the from \begin{equation} \begin{aligned} &\int_{\Omega}a(x,\nabla u_n)\nabla T_{k}(u_n^{+}-\psi^{+}) \exp(G(u_n))(1-h_{j}(u_n))dx\\ &\leq -\int_{\{j\leq u_n\leq j+1\}}a(x,\nabla u_n)\nabla u_n \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &\quad +\int_{\{-j-1\leq u_n\leq -j\}}a(x,\nabla u_n)\nabla u_n \exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})dx\\ &\quad +\int_{\Omega}\gamma(x)\exp(G(u_n))T_{k}(u_n^{+} -\psi^{+})(1-h_{j}(u_n))dx\\ &\quad +\int_{\Omega}f_n\exp(G(u_n))T_{k}(u_n^{+}-\psi^{+})(1-h_{j}(u_n))dx =\varepsilon_1(j,n) \label{eq83} \end{aligned} \end{equation} By \eqref{eq7} and Lebesgue's theorem, we conclude that $\varepsilon_1(j,n)$ converges to zero as $n$ and $j$ appraoch infinity. From \eqref{eq83}, we have \begin{align*} &\int_{\{|u_n^{+}-\psi^{+}|\leq k\}}a(x,\nabla u_n)\nabla u_n^{+} \exp(G(u_n))(1-h_{j}(u_n))dx\\ &\leq\int_{\{|u_n^{+}-\psi^{+}|\leq k\}}a(x,\nabla u_n) \nabla \psi^{+}\exp(G(u_n)(1-h_{j}(u_n)))dx+\varepsilon_1(j,n) \end{align*} Thanks to \eqref{ass1} and Young's inequality, it is possible to conclude that $$ \int_{\{|u_n^{+}-\psi^{+}|\leq k\}}a(x,\nabla u_n)\nabla \psi^{+} \exp(G(u_n)(1-h_{j}(u_n)))dx\leq\varepsilon_{2}(j,n), $$ where $\varepsilon_{2}(j,n)$ converges to zero as $n$ and $j$ go to infinity. Since $\exp(G(u_n))$ is bounded, $$ \int_{\{|u_n^{+}-\psi^{+}|\leq k\}}a(x,\nabla u_n) \nabla u_n^{+}(1-h_{j}(u_n)))dx\leq\varepsilon_{3}(j,n). $$ Since $\{x\in\Omega,\quad |u_n^{+}|\leq k\}\subset \{x\in\Omega,\quad |u_n^{+}-\psi^{+}|\leq k+\|\psi^{+}\|_{\infty}\}$, hence \begin{align*} &\int_{\{|u_n^{+}|\leq k\}} a(x,\nabla u_n)\nabla u_n(1-h_{j}(u_n)))dx\\ & \leq\int_{\{|u_n^{+}-\psi^{+}|\leq k+\|\psi^{+}\|_{\infty}\}} a(x,\nabla u_n)\nabla u_n(1-h_{j}(u_n)))dx \leq\varepsilon_{3}(j,n) \end{align*} Which, for all $k\geq 0$, yields \begin{equation} \lim_{j,n\to\infty}\int_{\{u_n\geq0\}}a(x,\nabla T_{k}(u_n))\nabla T_{k}(u_n)(1-h_{j}(u_n))dx=0,\label{eq84} \end{equation} using \eqref{eq82} and \eqref{eq84}, we conclude \eqref{eq81} of assertion (iii). \textbf{Assertion(iv):} First we have \begin{align*} &\int_{\Omega} (a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))) (\nabla T_{k}(u_n)-\nabla T_{k}(u))dx\\ &=\int_{\Omega} (a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))) (\nabla T_{k}(u_n)-\nabla T_{k}(u))h_{j}(u_n)dx\\ &\quad +\int_{\Omega}(a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u)))(\nabla T_{k}(u_n)-\nabla T_{k}(u))(1-h_{j}(u_n))dx \end{align*} Thanks to \eqref{eq8}, the first integral of the right hand side converges to zero as $n$ and $j$ tend to infinity. For the second term, we have \begin{align*} &\int_{\Omega}a(x,\nabla T_{k}(u_n))-a(x,\nabla T_{k}(u))(\nabla T_{k}(u_n)-\nabla T_{k}(u))(1-h_{j}(u_n))dx\\ &=\int_{\Omega}a(x,\nabla T_{k}(u_n))\nabla T_{k}(u_n)(1-h_{j}(u_n))\,dx\\ &\quad-\int_{\Omega}a(x,\nabla T_{k}(u_n))\nabla T_{k}(u)(1-h_{j}(u_n))\,dx\\ &\quad -\int_{\Omega}a(x,\nabla T_{k}(u))(\nabla T_{k}(u_n)-\nabla T_{k}(u))(1-h_{j}(u_n))\,dx \end{align*} By \eqref{eq81}, the first integral of the right-hand side approaches zero as $n$ and $j$ tend to infinity, and since $a(x,\nabla T_{k}(u_n))$ in $(L^{p'(x)}(\Omega))^{N}$ and $\nabla T_{k}(u)(1-h_{j}(u_n))$ converges to zero, hence the second integral converges to zero. For the third integral, it converges to zero because $\nabla T_{k}(u_n)\to \nabla T_{k}(u)$ weakly in $(L^{p(x)}(\Omega))^{N}$. Finally we conclude that, $$ \lim_{n\to\infty}\int_{\Omega}\Big(a(x,\nabla T_{k}(u_n)) -a(x,\nabla T_{k}(u))\Big)(\nabla T_{k}(u_n)-\nabla T_{k}(u))dx=0. $$ The proof of Proposition \ref{prop1} is complete. \end{proof} \begin{proof}[Proof of Lemma \ref{lem5.2}] Take at first the case of $ F\in C^{1}(\mathbb{R}) $ and $ F'\in L^{\infty}(\mathbb{R})$. Let $ u\in W_0^{1,p(x)}(\Omega)$. Since $ \overline{C_0^{\infty}(\Omega)}^{W^{1,p(x)}(\Omega)} = W_0^{1,p(x)}(\Omega)$, there exists $u_n\in C_0^{\infty}(\Omega)$ such that $ u_n \to u $ in $W_0^{1,p(x)}(\Omega)$, then $ u_n\to u $ a.e, in $\Omega$ and $\nabla u_n\to \nabla u$ a.e. in $\Omega$, then $F(u_n)\to F(u)$ a.e. in $\Omega$. In the the other hand, we have $|F(u_n)| = |F(u_n)- F(0)|\leq \|F'\|_{\infty} |u_n|$, then \begin{gather*} |F(u_n)|^{p(x)} \leq (\|F'\|_{\infty} + 1)^{p_{+}}|u_n|^{p(x)},\\ |\frac{\partial F (u_n)}{\partial x_i}|^{p(x)} =|F'(u_n)\frac{\partial u_n}{\partial x_i}|^{p(x)} \leq M |\frac{\partial u_n}{\partial x_i}|^{p(x)}, \end{gather*} where $M = (\|F'\|_{\infty} + 1)^{p_{+}}$. Then $ F(u_n) $ is bounded in $ W_0^{1,p(x)}(\Omega) $ and we obtain $ F(u_n) \rightharpoonup \nu$ in $W_0^{1,p(x)}(\Omega)$, then $F(u_n) \to \nu $ strongly in $L^{q(x)}(\Omega)$ with $10 $$ holds for almost all $x\in \Omega$, for $r, s\geq 0$, and for $\xi\neq \eta$, then we are able to prove the following result. \end{remark} \begin{proposition} \label{prop4.5} Let $u$ and $v$ be two entropy solutions of \eqref{P}, where $f\in L^1(\Omega)$ and $f\geq 0$, then one has $$ \lim_{k\to +\infty} k \int_{\{|u-v|\geq k\}} [ H(x,u,Du)-H(x,v,Dv) ]\operatorname{sign}(u-v)\,dx\leq 0, $$ and the condition $$ \lim_{k\to +\infty} k \int_{\{|u-v|\geq k\}} [ H(x,u,Du)-H(x,v,Dv) ] \operatorname{sign}(u-v)\,dx\geq 0 $$ implies $u=v$. \end{proposition} For a proof of the above propositions, see \cite[Proposition 2.2]{SSL} for $p(.)=p$ constant. 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