\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 68, pp. 1--27.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/68\hfil Existence and regularity of entropy solutions] {Existence and regularity of entropy solutions for strongly nonlinear $p(x)$-elliptic equations} \author[E. Azroul, H. Hjiaj, A. Touzani \hfil EJDE-2013/68\hfilneg] {Elhoussine Azroul, Hassane Hjiaj, Abdelfattah Touzani} % 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{Hassane Hjiaj \newline University of Fez, Faculty of Sciences Dhar El Mahraz, Laboratory LAMA, Department of Mathematics, B.P. 1796 Atlas Fez, Morocco} \email{hjiajhassane@yahoo.fr} \address{Abdelfattah Touzani \newline University of Fez, Faculty of Sciences Dhar El Mahraz, Laboratory LAMA, Department of Mathematics, B.P. 1796 Atlas Fez, Morocco} \email{atouzani07@gmail.com} \thanks{Submitted May 22, 2012. Published March 8, 2013.} \subjclass[2000]{35J20, 35J25, 35J60} \keywords{Sobolev spaces with variable exponents; entropy solutions; \hfill\break\indent strongly nonlinear elliptic equations; boundary value problems} \begin{abstract} This article is devoted to study the existence of solutions for the strongly nonlinear $p(x)$-elliptic problem \begin{gather*} - \operatorname{div} a(x,u,\nabla u) + g(x,u,\nabla u) = f- \operatorname{div} \phi(u) \quad \text{in } \Omega, \\ u = 0 \quad \text{on } \partial\Omega, \end{gather*} with $ f\in L^1(\Omega) $ and $ \phi \in C^{0}(\mathbb{R}^{N})$, also we will give some regularity results for these solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $ \Omega $ be a bounded open subset of $\mathbb{R}^{N}$ with $N\geq 2$. For $ 2 - \frac{1}{N} < p < N$, Boccardo and Gallou\"{e}t \cite{bocc3} studied the problem \begin{gather*} Au = f \quad \text{in } \Omega, \\ u = 0\quad \text{on } \partial\Omega, \end{gather*} where $ Au = - \operatorname{div} a(x,u,\nabla u) $ is a Leray-Lions operator from $ W_0^{1,p}(\Omega) $ into its dual, and $ f $ is a bounded Radon measure on $ \Omega$. They proved the existence of solutions $ u\in W_0^{1,q}(\Omega) $ for all $ 1 < q < \bar{q} = \frac{N(p-1)}{N-1}$. Moreover, they showed the critical regularity $ u\in W_0^{1,\overline{q}}(\Omega) $ under the assumption $ f\log(1 + |f|) \in L^1(\Omega)$. Boccardo \cite{bocc1} studied the existence of entropy solutions for the problem \begin{equation} \begin{gathered} - \operatorname{div} a(x,u,\nabla u) = f- \operatorname{div} \phi(u) \quad \text{in } \Omega, \\ u = 0\quad \text{on } \partial\Omega, \end{gathered}\label{appp} \end{equation} where $ f\in L^1(\Omega) $ and $ \phi\in C^{0}(\mathbb{R}^{N})$, he proved the solutions existence and some regularity results, under the above assumptions. Aharouch and Azroul \cite{LAEA} studied the problem \eqref{appp} in Oricz-sobolev spaces. They proved the existence of entropy solutions $ u \in W_0^{1,q}(\Omega)$. In the case of $ p = N$, they assume in addition that there exists an N-function $H$ such that $H(t^{N})$ is equivalent to $M(t)$. Kbiri Alaoui, Meskine and Souissi \cite{KBMESO} proved the critical regularity $W_0^{1,\overline{q}}(\Omega)$ of solutions for nonlinear elliptic problems with right-hand side in $L\log^\alpha L(\Omega)$ and $\alpha\geq\frac{N-1}{N}$. Also they proved some regularity results when $\alpha<\frac{N-1}{N}$. In this article, we consider the problem \begin{equation} \begin{gathered} - \operatorname{div} a(x,u,\nabla u) + g(x,u,\nabla u) = f - \operatorname{div} \phi(u) \quad \text{in } \Omega, \\ u = 0\quad \text{on } \partial\Omega, \end{gathered} \label{eq1} \end{equation} where the right hand side is assumed to satisfy \begin{equation} f \in L^1(\Omega)\quad \text{and} \quad \phi \in C^{0}(\mathbb{R}^{N}).\label{ass4} \end{equation} We will study the strongly nonlinear boundary-value problem \eqref{eq1} in the framework of variable exponent Sobolev spaces, we will prove the existence of entropy solutions and some $ \bar{q}(x)$-regularity results. Recall that, since no growth hypothesis is assumed on $\phi$, the term $ \operatorname{div} \phi(v) $ may be meaningless, even as a distribution for a function $ v \in W^{1,r(x)}_0(\Omega)$, $r(x)>1 $ (see \cite{bocc1} and \cite{bocc2} for the case of constant exponent). \begin{definition} \label{def}\rm For $ k>0 $ and $ s \in \mathbb{R} $, the truncation function $ T_k(.) $ is defined by $$ T_k(s)=\begin{cases} s &\text{if } |s|\leq k,\\ k\frac{s}{|s|}&\text{if }|s|>k. \end{cases} $$ \end{definition} This article is organized as follows. In the section $2$ we recall some important definitions and results of variable exponent Lebesgue and Sobolev spaces. We introduce in the section $3$ some assumptions on $ a(x, s, \xi) $ and $ g(x, s, \xi) $ for which our problem has a solutions. The section $4$ contains some important lemmas useful to prove our main results. The section $5$ will be devoted to show the existence of entropy solutions for the problem \eqref{eq1}, also we will give some important $ L^{\bar{q}(x)}$-regularity results for these solutions (the case $ p = 2 - 1/N $ and $ p = N $ are excluded). \section{Preliminaries} Let $ \Omega $ be a bounded open subset of $\mathbb{R}^{N}$ ($N\geq 2$), we say that a real-valued continuous function $ p(.) $ is log-H\"{o}lder continuous in $ \Omega $ if $$ |p(x) - p(y)| \leq \frac{C}{|\log|x-y||} \quad\forall x,y\in \overline{\Omega} \text{ such that } |x-y|<\frac{1}{2}, $$ with possible different constant $ C$. We denote $$ C_{+}(\overline{\Omega})= \{\text{log-H\"{o}lder continuous function } p:\overline{\Omega} \to \mathbb{R} \text{ with } 1 < p_{-} \leq p_{+} < N\}, $$ where $$ p_{-} = \min\{p(x) : x\in \overline{\Omega}\}\quad p_{+} = \max\{p(x) : x\in \overline{\Omega}\}. $$ We define the variable exponent Lebesgue space for $ p\in C_{+}(\overline{\Omega}) $ by $$ L^{p(x)}(\Omega) = \{u : \Omega \to \mathbb{R} \text{ measurable}: \int_{\Omega} |u(x)|^{p(x)} dx < \infty\}, $$ the space $L^{p(x)}(\Omega)$ under the norm $$ \|u\|_{p(x)} = \inf \big\{\lambda > 0 : \int_{\Omega} |\frac{u(x)}{\lambda}|^{p(x)} dx \leq 1 \big\} $$ is a uniformly convex Banach space, and therefore reflexive. We denote by $ L^{p'(x)}(\Omega) $ the conjugate space of $L^{p(x)}(\Omega) $ where $\frac{1}{p(x)} + \frac{1}{p'(x)} = 1$ (see \cite{Fan2,zhao}). \begin{proposition}[Generalized H\"{o}lder inequality \cite{Fan2,zhao}] \label{prop1} (i) For any functions $ 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) $ a.e. in $\Omega$, we have $ L^{p_{2}(x)}(\Omega) \hookrightarrow L^{p_1(x)}(\Omega) $ and the embedding is continuous. \end{proposition} \begin{proposition}[\cite{Fan2,zhao}] \label{prop2} If we 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, $= 1$, $> 1$) if and only if $\rho(u) < 1$ (resp, $= 1$, $> 1$); \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} Now, we define the variable exponent Sobolev space by $$ W^{1,p(x)}(\Omega) = \{ u\in L^{p(x)}(\Omega) \text{ and } |\nabla u|\in L^{p(x)}(\Omega) \}, $$ with the norm $$ \|u\|_{1,p(x)} = \|u\|_{p(x)} + \|\nabla u\|_{p(x)} \quad \forall u \in W ^{1,p(x)}(\Omega). $$ We denote by $ W_0^{1,p(x)}(\Omega)$ the closure of $ C_0^{\infty}(\Omega)$ in $W^{1,p(x)}(\Omega)$, and we define the Sobolev exponent by $ p^{*}(x) = \frac{N p(x)}{N - p(x)}$ for $p(x) < N$. \begin{proposition}[\cite{Fan2, Harjhast}] \label{prop3} \begin{itemize} \item[(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. \item[(ii)] If $q\in C_{+}(\bar{\Omega})$ and $q(x) < p^{*}(x) $ for any $x \in \Omega$, then the embedding $ W^{1, p(x)}_0(\Omega) \hookrightarrow\hookrightarrow L^{q(x)}(\Omega)$ is continuous and compact. \item[(iii)] Poincar\'{e} inequality: there exists a constant $ C > 0$, such that $$ \| u\|_{p(x)} \leq C \|\nabla u\|_{p(x)} \quad \forall u\in W_0^{1,p(x)}(\Omega). $$ \item[(vi)] Sobolev-Poincar\'{e} inequality : there exists an other 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{itemize} \end{proposition} \begin{remark} \label{rmk2.1} \rm By (iii) of Proposition \ref{prop3}, we deduce that $\|\nabla u\|_{p(x)}$ and $\|u\|_{1,p(x)}$ are equivalent norms in $W^{1,p(x)}_0(\Omega)$. \end{remark} \begin{definition}[\cite{DHHR}] \label{def2.1}\rm We denote the dual of the Sobolev space $ W_0^{1,p(x)}(\Omega) $ by $ W^{-1,p'(x)}(\Omega)$, and for each $ F\in W^{-1,p'(x)}(\Omega) $ there exists $ f_0, f_1, \ldots, f_{N} \in L^{p'(x)}(\Omega) $ such that $ {F = f_0 + \sum_{i=1}^{N} \frac{\partial f_{i}}{\partial x_{i}} }$. Moreover, for all $ u\in W_0^{1,p(x)}(\Omega) $ we have $$ \langle F, u\rangle = \int_{\Omega} f_0u dx - \sum_{i=1}^{N} \int_{\Omega} f_{i} \frac{\partial u}{\partial x_{i}} dx. $$ and we define a norm on the dual space by $$ \|F\|_{-1,p'(x)} \simeq \sum_{i=0}^{N} \|f_{i}\|_{p'(x)}. $$ \end{definition} Now, we define $$ T_0^{1,p(x)}(\Omega) := \{\text{measurable function $u$ such that } T_k(u)\in W_0^{1,p(x)}(\Omega) \quad \forall k > 0\}. $$ \begin{proposition}\label{prop1b} Let $ u \in T_0^{1,p(x)}(\Omega)$, there exists a unique measurable function $ v: \Omega\to \mathbb{R}^{N} $ such that $$ v.\chi_{\{|u|\leq k\}} = \nabla T_k(u) \quad \text{for a.e. $x\in \Omega$ and for all } k> 0. $$ We will define the gradient of $u$ as the function $v$, and we will denote it by $v=\nabla u$. \end{proposition} \begin{definition} \label{def2.2}\rm A measurable function $ u $ is an entropy solution of the Dirichlet problem \eqref{eq1} if \begin{gather*} T_k(u) \in W^{1,p(x)}_0(\Omega) \quad \forall k>0,\\ \begin{aligned} &\int_{\Omega}a(x,u,\nabla u) \nabla T_k(u-\varphi) dx + \int_{\Omega}g(x,u,\nabla u) T_k(u-\varphi) dx \\ &\leq \int_{\Omega} fT_k(u-\varphi) dx + \int_{\Omega} \phi(u)\nabla T_k(u-\varphi) dx \end{aligned} \end{gather*} for all $\varphi \in W^{1,p(x)}_0(\Omega) \cap L^{\infty}(\Omega)$. \end{definition} \begin{lemma}\label{lem1.1} Let $\lambda \in \mathbb{R}$ and let $u$ and $v$ be two functions which are finite almost everywhere, and which belong to ${\mathcal{T}}_0^{1,p(x)}(\Omega)$, then $$ \nabla(u+\lambda v)=\nabla u+ \lambda \nabla v\quad \text{ a.e. in } \Omega, $$ where $\nabla u$, $\nabla v$ and $\nabla(u+\lambda v)$ are the gradients of $u$, $v$ and $u+\lambda v$ introduced in the Definition \ref{def2.2}. \end{lemma} \begin{proof} Let $E_n=\{|u| \leq n\}\cap \{|v| \leq n\}$. We have $T_n(u)=u$ and $T_n(v)=v$ in $E_n$, then for every $k>0$ $$ T_k(T_n(u)+\lambda T_n(v))=T_k(u+ \lambda v) \quad \text{a.e. in } E_n, $$ and therefore, since both functions belong to $W_0^{1,p(x)}(\Omega)$, \begin{equation}\label{2.13} \nabla T_k(T_n(u)+\lambda T_n(v))=\nabla T_k(u+ \lambda v) \quad \text{a.e. in } E_n. \end{equation} Since $T_n(u)$ and $T_n(v)$ belong to $W_0^{1,p(x)}(\Omega)$, we have by using a classical property of the truncates functions in $W_0^{1,p(x)}(\Omega)$, and the definition of $\nabla u$ and $\nabla v$, \begin{align*} \nabla T_k(T_n(u)+\lambda T_n(v)) & = \chi_{\{|T_n(u)+\lambda T_n(v)|\leq k \}}(\nabla T_n(u) +\lambda \nabla T_n(v)) \\ & = \chi_{\{|T_n(u)+\lambda T_n(v)|\leq k \}}(\nabla u.\chi_{\{|u|\leq n\}} + \lambda \nabla v .\chi_{\{|v|\leq n\}}) \end{align*} a.e. in $\Omega$. Therefore, \begin{equation}\label{2.14} \nabla T_k(T_n(u)+\lambda T_n(v)) =\chi_{\{|u+\lambda v|\leq k\}} (\nabla u+ \lambda\nabla v) \quad \text{a.e. in } E_n. \end{equation} On the other hand, by definition of $\nabla( u+ \lambda v)$, \begin{equation}\label{2.15} \nabla T_k(u+\lambda v)=\chi_{\{|u+\lambda v|\leq k\}} \nabla( u+ \lambda v) \quad \text{a.e. in } E_n. \end{equation} Putting together \eqref{2.13}, \eqref{2.14} and \eqref{2.15}, we obtain \begin{equation}\label{2.16} \chi_{\{|u+\lambda v|\leq k\}} \nabla( u+ \lambda v) =\chi_{\{|u+\lambda v|\leq k\}} (\nabla u+ \lambda\nabla v) \quad \text{a.e. in } E_n. \end{equation} We have $\cup_{n \in \mathbb{N}} E_n$ (resp. $\cup_{k \in \mathbb{N}} \{ |u+\lambda v| \leq k \}$) differs at most from $\Omega$ by a set of zero Lebesgue measure, since $u$ and $v$ are almost everywhere finite, then \eqref{2.16} holds almost everywhere in $\Omega$. which conclude the proved of Lemma \ref{lem1.1}. \section{Essential assumption} Let $ \Omega $ be a bounded open subset of $\mathbb{R}^{N}$ ($N\geq 2$) and $ p\in C_{+}(\bar{\Omega})$, we consider a Leray-Lions operator from $W_0^{1,p(x)}(\Omega)$ into its dual $W^{-1,p'(x)}(\Omega)$, defined by the formula \begin{equation} Au = - \operatorname{div} \ a(x,u,\nabla u) \end{equation} where $ a:\Omega \times \mathbb{R} \times \mathbb{R}^{N} \to \mathbb{R}^{N} $ is a Carath\'{e}odory function (measurable with respect to $x$ in $\Omega$ for every $(s, \xi)$ in $\mathbb{R} \times \mathbb{R}^{N}$, and continuous with respect to $(s, \xi)$ in $\mathbb{R} \times \mathbb{R}^{N}$ for almost every $x$ in $\Omega$) which satisfies the following conditions \begin{gather} |a(x,s,\xi)|\leq \beta (K(x) + |s|^{p(x) - 1} + |\xi|^{p(x) - 1}),\label{aq1}\\ a(x,s,\xi)\xi \geq \alpha |\xi|^{p(x)},\label{aq2}\\ [a(x,s,\xi) - a(x,s,\overline{\xi})](\xi - \overline{\xi}) > 0 \quad \text{for all $\xi \neq \overline{\xi}$ in $\mathbb{R}^{N}$},\label{aq3} \end{gather} for a.e. $x \in \Omega$, all $(s,\xi) \in \mathbb{R} \times \mathbb{R}^{N}$, where $ K(x) $ is a positive function lying in $ L^{p'(x)}(\Omega) $ and $ \alpha, \beta > 0$. The nonlinear term $ g(x,s,\xi) $ is a Carath\'{e}odory function which satisfies \begin{gather} g(x,s,\xi)s \geq 0,\label{aq4} \\ |g(x,s,\xi)|\leq b(|s|)(c(x) + |\xi|^{p(x)}),\label{aq5} \end{gather} where $ b:\mathbb{R}^{+}\to \mathbb{R}^{+} $ is a continuous, nondecreasing function, and $ c: \Omega \to \mathbb{R}^{+} $ with $c\in L^1(\Omega)$. We consider the problem \begin{equation} \begin{gathered} - \operatorname{div} a(x,u,\nabla u) + g(x,u,\nabla u) = f - \operatorname{div} \phi(u) \quad \text{in } \Omega, \\ u = 0\quad \text{on } \partial\Omega, \end{gathered} \label{aq7} \end{equation} with \begin{equation} f \in L^1(\Omega)\quad \text{and} \quad \phi \in C^{0}(\mathbb{R}^{N}).\label{aqq1} \end{equation} The symbol $\rightharpoonup$ will denote the weak convergence, and the constants $ C_{i}$, $i = 1, 2, \dots $ used in each steps of proof are independent. \end{proof} \section{Some technical Lemmas} \begin{lemma}[\cite{EAM}] \label{lemp1} 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{lemp3} Let $u\in W_0^{1,p(x)}(\Omega)$ then $T_k(u)\in W_0^{1,p(x)}(\Omega)$ with $k>0$. Moreover, we have $T_k(u)\to u$ in $W_0^{1,p(x)}(\Omega)$ as $k\to \infty$. \end{lemma} \begin{proof} Let $k>0$ and $T_k: \mathbb{R} \to \mathbb{R}$, $$ T_k(s)= \begin{cases} s & \text{if } |s|\leq k,\\ k.\text{sign}(s) & \text{if } |s|>k, \end{cases} $$ then for all $ u\in W_0^{1,p(x)}(\Omega) $ we have $ T_k(u)\in W_0^{1, p(x)}(\Omega)$, and \begin{align*} & \int_{\Omega}|T_k(u)-u|^{p(x)}dx+\int_{\Omega}|\nabla T_k(u)-\nabla u|^{p(x)}dx\\ & = \int_{\{|u|\leq k\}}|T_k(u)-u|^{p(x)}dx+\int_{\{|u|>k\}}|T_k(u)-u|^{p(x)}dx \\ & \quad+ \int_{\{|u|\leq k\}}|\nabla T_k(u)-\nabla u|^{p(x)} +\int_{\{|u|>k\}}|\nabla T_k(u)-\nabla u|^{p(x)}dx\\ & = \int_{\{|u|>k\}}|T_k(u)-u|^{p(x)}dx +\int_{\{|u|>k\}}|\nabla u|^{p(x)}dx . \end{align*} Since $T_k(u)\to u$ as $k\to\infty$ and by using the dominated convergence theorem, we have $$ \int_{\{|u|>k\}}|T_k(u)-u|^{p(x)}dx +\int_{\{|u|>k\}}|\nabla u|^{p(x)}dx\to 0 \quad \text{ as } k\to \infty. $$ Finally $\|T_k(u)-u\|_{W_0^{1,p(x)}(\Omega)}\to 0$ as $k\to \infty$. \end{proof} \begin{lemma}[\cite{BenWit}] \label{lemp5} Let $ p(\cdot) $ be a continuous function in $ C_{+}(\overline{\Omega}) $ and $ u $ a function in $ W_0^{1,p(x)}(\Omega)$. Suppose $ {2- \frac{1}{N} < p_{-} \leq p_{+} < N} $, and that there exists a constant $ c_1 $ such that $$ \int_{\{k \leq |u| \leq k+1\}} |\nabla u|^{p(x)} dx \leq c_1\quad \forall k > 0. $$ Then there exists a constant $ c_{2} > 0$, depending on $ c_1$, such that $$ \|u\|_{1,q(x)} \leq c_{2}, $$ for all continuous functions $ q(\cdot) $ on $ \overline{\Omega} $ satisfying $$ 1 \leq q(x) < \frac{N(p(x) - 1)}{N - 1}\quad \text{for all } x \in \overline{\Omega}. $$ \end{lemma} \begin{lemma}\label{lemp4} Assume \eqref{aq1}-\eqref{aq3}, and let $ (u_n)_n $ be a sequence in $ W_0^{1,p(x)}(\Omega) $ such that $ u_n \rightharpoonup u $ in $ W_0^{1,p(x)}(\Omega) $ and \begin{equation} \int_{\Omega} [a(x, u_n, \nabla u_n) - a(x, u_n, \nabla u)]\nabla (u_n - u ) dx \to 0,\label{eqq1} \end{equation} then $ u_n\to u \quad \text{in } \quad W_0^{1,p(x)}(\Omega)$ for a subsequence. \end{lemma} \begin{proof} Let $ D_n = [a(x,u_n,\nabla u_n) - a(x,u_n,\nabla u)]\nabla(u_n - u)$, thanks to \eqref{aq3} we have $ D_n $ is a positive function, and by \eqref{eqq1}, $ D_n\to 0 $ in $ L^1(\Omega) $ as $n \to\infty$. Since $ u_n \rightharpoonup u $ in $ W_0^{1,p(x)}(\Omega) $ then $ u_n \to\ u $ a.e. in $ \Omega$, and since $ D_n\to 0 $ a.e. in $ \Omega$, there exists a subset $ B $ in $ \Omega $ with measure zero such that for all $x\in \Omega\backslash B$, $$ |u(x)|<\infty,\quad |\nabla u(x)|<\infty,\quad K(x) < \infty,\quad u_n \to\ u,\quad D_n\to 0. $$ Taking $ \xi_n = \nabla u_n $ and $ \xi = \nabla u$, we have \begin{align*} D_n(x) & = [a(x,u_n,\xi_n) - a(x,u_n,\xi)](\xi_n - \xi) \\ & = a(x,u_n,\xi_n)\xi_n + a(x,u_n,\xi)\xi - a(x,u_n,\xi_n)\xi - a(x,u_n,\xi)\xi_n \\ & \geq \alpha |\xi_n|^{p(x)} + \alpha |\xi|^{p(x)} - \beta (K(x)+ |u_n|^{p(x) -1} + |\xi_n|^{p(x)-1})|\xi| \\ &\quad - \beta (K(x)+ |u_n|^{p(x)-1} + |\xi|^{p(x)-1})|\xi_n| \\ & \geq \alpha |\xi_n|^{p(x)} - C_{x}(1+ |\xi_n|^{p(x)-1} + |\xi_n|), \end{align*} where $ C_{x} $ depending on $ x$, without dependence on $ n$. (since $ u_n(x)\to u(x) $ then $ (u_n)_n $ is bounded), we obtain $$ D_n(x) \geq |\xi_n|^{p(x)} \big(\alpha - \frac{C_{x}}{|\xi_n|^{p(x)}} -\frac{C_{x}}{|\xi_n|} - \frac{C_{x}}{|\xi_n|^{p(x)-1}}\big), $$ by the standard argument $ (\xi_n)_n $ is bounded almost everywhere in $ \Omega$, (Indeed, if $ |\xi_n|\to \infty $ in a measurable subset $ E\in \Omega $ then $$ \lim_{n \to \infty}\int_{\Omega} D_n(x) dx \geq \lim_{n \to \infty}\int_{E} |\xi_n|^{p(x)} \big(\alpha - \frac{C_{x}}{|\xi_n|^{p(x)}} -\frac{C_{x}}{|\xi_n|} - \frac{C_{x}}{|\xi_n|^{p(x)-1}}\big) dx = \infty, $$ which is absurd since $ D_n\to 0 $ in $ L^1(\Omega) $). Let $ \xi^{*} $ an accumulation point of $ (\xi_n)_n$, we have $ |\xi^{*}|<\infty $ and by the continuity of $ a(.,.,.) $ we obtain, $$ [a(x,u(x),\xi^{*}) - a(x,u(x),\xi)](\xi^{*} - \xi) = 0, $$ thanks to \eqref{aq3} we have $\xi^{*} = \xi$, the uniqueness of the accumulation point implies that $\nabla u_n\to \nabla u$ a.e. in $\Omega$. since $ (a(x,u_n,\nabla u_n))_n $ is bounded in $ (L^{p'(x)}(\Omega))^{N} $ and $ a(x,u_n,\nabla u_n) \to a(x,u,\nabla u) $ a.e. in $\Omega$, by the Lemma $\ref{lemp1}$, we can establish that $$ a(x,u_n,\nabla u_n) \rightharpoonup a(x,u,\nabla u)\quad \text{in } (L^{p'(x)}(\Omega))^{N}. $$ Let us taking $ \bar{y}_n = a(x,u_n,\nabla u_n)\nabla u_n$ and $\bar{y} = a(x,u,\nabla u)\nabla u$, then $ \bar{y}_n \to \bar{y} $ in $ L^1(\Omega)$, according to the condition $ \eqref{aq2} $ we have $$ \alpha|\nabla u_n|^{p(x)} \leq a(x,u_n,\nabla u_n)\nabla u_n, $$ Let $ z_n = \nabla u_n , z = \nabla u $ and $ y_n = \frac{\bar{y}_n}{\alpha}$, $y = \frac{\bar{y}}{\alpha}$, in view of the Fatou Lemma, we obtain $$ \int_{\Omega} 2.y dx \leq \liminf_{n\to \infty} \int_{\Omega} (y_n + y - |z_n - z|^{p(x)})dx, $$ then $ 0 \leq -\limsup_{n\to \infty}\int_{\Omega}|z_n - z|^{p(x)} dx $, and since $$ 0 \leq \liminf_{n\to \infty}\int_{\Omega}|z_n - z|^{p(x)} dx \leq \limsup_{n\to \infty}\int_{\Omega}|z_n - z|^{p(x)} dx \leq 0, $$ it follows that $ { \int_{\Omega} |\nabla u_n -\nabla u|^{p(x)} dx \to 0} $ as $ n\to \infty$, and we get $$ \nabla u_n \to\nabla u \quad in \quad (L^{p(x)}(\Omega))^{N} $$ we deduce that $$ u_n \to u \quad \text{in } W_0^{1,p(x)}(\Omega), $$ which completes our proof. \end{proof} Now, we consider $ \phi_n(s) = \phi(T_n(s)) $ with $ \phi \in C^{0}(\mathbb{R}^{N}) $ and \[ {g_n(x,s,\xi) = \frac{g(x,s,\xi)}{1 + \frac{1}{n}|g(x,s,\xi)|}} \] such that $ g(x,s,\xi) $ satisfies $ \eqref{aq4}-\eqref{aq5}$, note that $$ g_n(x,s,\xi) s\geq 0, \quad |g_n(x,s,\xi)|\leq |g(x,s,\xi)|,\quad |g_n(x,s,\xi)|\leq n \quad \forall n\in \mathbb{N}^{*}. $$ We define the operator $ G_n : W_0^{1,p(x)}(\Omega) \to W^{-1,,p'(x)}(\Omega)$, by $$ \langle G_n u,v \rangle = \int_{\Omega} g_n(x,u,\nabla u)v dx \quad \forall v\in W_0^{1,p(x)}(\Omega). $$ Thanks to the H\"{o}lder inequality, we have that for all $ u, v \in W_0^{1,p(x)}(\Omega)$, \begin{equation} \begin{aligned} &\big|\int_{\Omega} g_n(x,u,\nabla u)v dx\big|\\ & {\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \|g_n(x,u,\nabla u)\|_{p'(x)}\|v\|_{p(x)}} \\ &{ \leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \Big(\int_{\Omega} |g_n(x,u,\nabla u)|^{p'(x)} dx + 1\Big)^{\frac{1}{p'_{-}}} \|v\|_{1,p(x)}} \\ & {\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \Big(\int_{\Omega} n^{p'(x)} dx + 1\Big)^{\frac{1}{p'_{-}}}\|v\|_{1,p(x)}} \\ & {\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \big( n^{p'_{+}}.\operatorname{meas}(\Omega) + 1\big)^{\frac{1}{p'_{-}}}\|v\|_{1,p(x)}} \\ & \leq C_0\|v\|_{1,p(x)} , \end{aligned}\label{aq10} \end{equation} and we define the operator $ R_n = \operatorname{div} \phi_n : W_0^{1,p(x)}(\Omega) \to W^{-1,p'(x)}(\Omega)$, such that $$ \langle R_n(u),v \rangle = \langle \operatorname{div} \phi_n(u),v \rangle = - \int_{\Omega} \phi_n(u)\nabla v dx \quad \forall u,v \in W_0^{1,p(x)}(\Omega), $$ we have \begin{equation} \begin{aligned} {\big|\int_{\Omega} \phi_n(u)\nabla v dx \big|} &{\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \|\phi_n(u)\|_{p'(x)}\|\nabla v\|_{p(x)}} \\ & {\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \Big(\int_{\Omega} |\phi_n(u)|^{p'(x)} dx + 1\Big)^{\frac{1}{p'_{-}}} \|v\|_{1,p(x)}} \\ & {\leq \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big) \big(\sup_{|s|\leq n} (|\phi(s)| + 1)^{p'_{+}} \operatorname{meas}(\Omega) + 1 \big)^{1/p'_{-}}\|v\|_{1,p(x)}} \\ & \leq C_1 \|v\|_{1,p(x)}. \end{aligned}\label{aq11} \end{equation} \begin{lemma}\label{lem1} The operator $ B_n = A + G_n + R_n $ is pseudo-monotone from $ W_0^{1,p(x)}(\Omega) $ into $ W^{-1,p'(x)}(\Omega)$. Moreover, $B_n$ is coercive in the following sense $$ \frac{\langle B_n v,v\rangle}{\|v\|_{1,p(x)}} \to + \infty \quad \text{as} \quad \|v\|_{1,p(x)} \to + \infty \quad \text{for}\quad v \in W_0^{1,p(x)}(\Omega). $$ \end{lemma} \begin{proof} Using H\"{o}lder's inequality and the growth condition \eqref{aq1}, we can show that the operator $ A $ is bounded, and by using \eqref{aq10} and \eqref{aq11} we conclude that $ B_n $ bounded. For the coercivity, we have for any $ u\in W_0^{1,p(x)}(\Omega)$, \begin{align*} {\langle B_n u,u\rangle} & = {\langle A u,u\rangle + \langle G_n u,u\rangle + \langle R_n u,u\rangle} \\ & = {\int_{\Omega} a(x,u,\nabla u)\nabla u dx + \int_{\Omega} g_n(x,u,\nabla u)u dx - \int_{\Omega} \phi_n(u)\nabla u dx} \\ & \geq {\alpha\int_{\Omega} |\nabla u|^{p(x)} dx - \big(\frac{1}{p_{-}} + \frac{1}{p'_{-}}\big)\|\phi_n(u)\|_{p'(x)} \|\nabla u\|_{p(x)} } \\ & \geq {\alpha\|\nabla u\|_{p(x)}^{\delta} - C_1.\|u\|_{1,p(x)}} \quad (\text{using} \eqref{aq11} ) \\ & \geq {\alpha'\|u\|_{1,p(x)}^{\delta} - C_1.\|u\|_{1,p(x)} ,} \quad (\text{using the Poincar\'{e} inequality}) \end{align*} with $$ \delta = \begin{cases} p_{-} & \text{if } \|\nabla u\|_{p(x)} > 1, \\ p_{+} & \text{if } \|\nabla u\|_{p(x)} \leq 1, \end{cases} $$ then, we obtain $$ \frac{\langle B_n u,u\rangle}{\|u\|_{1,p(x)}} \to + \infty \quad \text{as } \|u\|_{1,p(x)} \to + \infty. $$ It remains to show that $B_n$ is pseudo-monotone. Let $ (u_k)_k $ a sequence in $ W_0^{1,p(x)}(\Omega) $ such that \begin{equation} \begin{gathered} u_k \rightharpoonup u \quad \text{in } W_0^{1,p(x)}(\Omega), \\ B_nu_k \rightharpoonup \chi \quad \text{in } W^{-1,p'(x)}(\Omega), \\ \limsup_{k\to \infty} \langle B_nu_k,u_k\rangle \leq \langle \chi,u\rangle. \end{gathered}\label{aq12} \end{equation} We will prove that $$ \chi = B_nu \quad \text{and} \quad \langle B_nu_k,u_k\rangle \to \langle \chi,u\rangle \quad \text{as } k \to + \infty. $$ Firstly, since $W_0^{1,p(x)}(\Omega)\hookrightarrow\hookrightarrow L^{p(x)}(\Omega)$, then $ u_k\to u \text{ in } L^{p(x)}(\Omega) $ for a subsequence still denoted $ (u_k)_k $. We have $ (u_k)_k $ is a bounded sequence in $ W_0^{1,p(x)}(\Omega)$, then by the growth condition $ (a(x,u_k,\nabla u_k))_k $ is bounded in $ (L^{p'(x)}(\Omega))^{N}$, therefore, there exists a function $\varphi\in (L^{p'(x)}(\Omega))^{N}$ such that \begin{equation} a(x,u_k,\nabla u_k) \rightharpoonup \varphi \quad \text{in } (L^{p'(x)}(\Omega))^{N} \text{ as } k\to \infty.\label{aq13} \end{equation} Similarly, since $ (g_n(x,u_k,\nabla u_k))_k $ is bounded in $ L^{p'(x)}(\Omega)$, then there exists a function $ \psi_n\in L^{p'(x)}(\Omega) $ such that \begin{equation} g_n(x,u_k,\nabla u_k) \rightharpoonup \psi_n \quad \text{in } L^{p'(x)}(\Omega) \text{ as } k\to \infty,\label{aq14} \end{equation} and since $ \phi_n = \phi\circ T_n $ is a bounded continuous function and $ u_k\to u$ in $L^{p(x)}(\Omega)$, it follows \begin{equation} \phi_n(u_k) \to \phi_n(u) \quad \text{in } (L^{p'(x)}(\Omega))^{N} \text{ as } k \to \infty. \label{aq15} \end{equation} For all $v\in W_0^{1,p(x)}(\Omega)$, we have \begin{equation} \begin{aligned} \langle\chi,v\rangle & = \lim_{k\to \infty} \langle B_nu_k ,v\rangle\\ & = \lim_{k\to \infty} \int_{\Omega} a(x,u_k,\nabla u_k)\nabla v dx + \lim_{k\to \infty}\int_{\Omega} g_n(x,u_k,\nabla u_k)v dx \\ &\quad - \lim_{k\to \infty} \int_{\Omega} \phi_n(u_k) \nabla v dx\\ & = {\int_{\Omega} \varphi\nabla v dx + \int_{\Omega} \psi_n v dx - \int_{\Omega} \phi_n(u) \nabla v dx}. \end{aligned}\label{aq16} \end{equation} Using \eqref{aq12} and \eqref{aq16}, we obtain \begin{equation} \begin{aligned} &\limsup_{k\to \infty} \langle B_n(u_k),u_k\rangle \\ & = \limsup_{k \to\infty} \Big\{\int_{\Omega} a(x,u_k,\nabla u_k)\nabla u_k dx + \int_{\Omega} g_n(x,u_k,\nabla u_k) u_k dx - \int_{\Omega} \phi_n(u_k)\nabla u_k dx \Big\}\\ &\leq \int_{\Omega} \varphi\nabla u dx + \int_{\Omega} \psi_n u dx - \int_{\Omega} \phi_n(u) \nabla u dx, \end{aligned}\label{aq18} \end{equation} thanks to \eqref{aq14} and \eqref{aq15}, we have \begin{equation} \int_{\Omega} g_n(x,u_k,\nabla u_k) u_k dx \to \int_{\Omega} \psi_n u dx, \quad \int_{\Omega} \phi_n(u_k) \nabla u_k dx \to \int_{\Omega} \phi_n(u) \nabla u dx; \label{aq17} \end{equation} therefore, \begin{equation} \limsup_{k\to \infty} \int_{\Omega} a(x,u_k,\nabla u_k)\nabla u_k dx \leq \int_{\Omega} \varphi\nabla u dx.\label{aq19} \end{equation} On the other hand, using \eqref{aq3}, we have \begin{equation} \int_{\Omega} (a(x,u_k,\nabla u_k) - a(x,u_k,\nabla u))(\nabla u_k - \nabla u) dx \,. \geq 0,\label{aq21} \end{equation} Then \begin{align*} &\int_{\Omega} a(x,u_k,\nabla u_k)\nabla u_k dx \\ &\geq - \int_{\Omega} a(x,u_k,\nabla u)\nabla u dx +\int_{\Omega} a(x,u_k,\nabla u_k)\nabla u dx + \int_{\Omega} a(x,u_k,\nabla u)\nabla u_k dx, \end{align*} and by \eqref{aq13}, we get $$ \liminf_{k\to \infty} \int_{\Omega} a(x,u_k,\nabla u_k)\nabla u_k dx \geq \int_{\Omega} \varphi \nabla u dx, $$ this implies, thanks to \eqref{aq19}, that \begin{equation} \lim_{k\to \infty} \int_{\Omega} a(x,u_k,\nabla u_k)\nabla u_k dx = \int_{\Omega} \varphi \nabla u dx.\label{aq22} \end{equation} By combining of \eqref{aq16}, \eqref{aq17} and \eqref{aq22}, we deduce that $$ \langle B_nu_k , u_k\rangle \to \langle \chi , u\rangle \quad \text{as } k \to +\infty. $$ Now, by \eqref{aq22} we can obtain $$ \lim_{k \to +\infty} \int_{\Omega} (a(x,u_k,\nabla u_k) - a(x,u_k,\nabla u))(\nabla u_k - \nabla u) dx = 0, $$ in view of the Lemma \ref{lemp4}, we obtain $$ u_k \to u , \quad W_0^{1,p(x)}(\Omega), \quad \nabla u_k \to \nabla u\quad \text{a.e. in } \Omega, $$ then $$ a(x,u_k, \nabla u_k) \rightharpoonup a(x,u, \nabla u),\quad \phi_n(u_k) \to \phi_n(u) \quad \text{in } (L^{p'(x)}(\Omega))^{N}, $$ and $$ g_n(x,u_k, \nabla u_k) \rightharpoonup g_n(x,u, \nabla u) \quad \text{in } L^{p'(x)}(\Omega), $$ we deduce that $ \chi = B_n u$, which completes the proof. \end{proof} \section{Main results} In the sequel we assume that $\Omega$ is an open bounded subset of $\mathbb{R}^{N}$ $(N\geq 2)$, and let $ p(.)\in C_{+}(\overline{\Omega})$. We will prove the following existence results \begin{theorem}\label{thm1} Assuming that \eqref{aq1}-\eqref{aq5} hold, $ p(.)\in C_{+}(\overline{\Omega})$, $f\in L^1(\Omega) $ and $ \phi \in C^{0}(\mathbb{R}^{N})$, then the problem \begin{equation} \begin{gathered} T_k(u) \in W_0^{1,p(x)}(\Omega) \quad \forall k > 0,\\ \begin{aligned} &\int_{\Omega} a(x,u,\nabla u)\nabla T_k(u-\varphi) dx + \int_{\Omega} g(x,u,\nabla u) T_k(u-\varphi) dx \\ & \leq \int_{\Omega} fT_k(u-\varphi) dx + \int_{\Omega} \phi(u)\nabla T_k(u-\varphi) dx, \quad \forall \varphi \in W_0^{1,p(x)}(\Omega) \cap L^{\infty}(\Omega), \end{aligned} \end{gathered}\label{aq8} \end{equation} has at least one solution. \end{theorem} The above theorem is prove in the following 5 steps. \subsection*{Step 1: Approximate problems} Let $ (f_n)_n $ be a sequence in $ W^{-1,p'(x)}(\Omega)\cap L^1(\Omega) $ such that $ f_n \to f $ in $ L^1(\Omega) $ with $ \|f_n\|_1 \leq \|f\|_1 $ and we consider the approximate problem \begin{equation} \begin{gathered} Au_n + g_n(x,u_n,\nabla u_n) = f_n - \operatorname{div} \phi_n(u_n)\\ u_n\in W_0^{1,p(x)}(\Omega), \end{gathered}\label{aq9} \end{equation} with $ \phi_n(s) = \phi(T_n(s)) $ and $ {g_n(x,s,\xi) = \frac{g(x,s,\xi)}{1 + \frac{1}{n}|g(x,s,\xi)|}}$. In view of the Lemma \ref{lem1}, there exists at least one weak solution $ u_n \in W_0^{1,p(x)}(\Omega) $ of the problem \eqref{aq9}, (cf. \cite{lions}). \subsection*{Step 2: A priori estimates} Taking $ T_k(u_n) $ as a test function in \eqref{aq9}, we obtain \begin{equation} \begin{aligned} &\int_{\Omega} a(x,u_n, \nabla u_n)\nabla T_k(u_n) dx + \int_{\Omega} g_n(x,u_n, \nabla u_n) T_k(u_n) dx \\ &= \int_{\Omega} f_n T_k(u_n) dx + \int_{\Omega} \phi_n(u_n)\nabla T_k(u_n) dx. \end{aligned}\label{aq23} \end{equation} Thanks to \eqref{aq2} and Young's inequality, we obtain \begin{equation} \begin{aligned} &{\alpha\int_{\Omega} |\nabla T_k(u_n)|^{p(x)} dx} \\ & {\leq \int_{\Omega} a(x,T_k(u_n), \nabla T_k(u_n))\nabla T_k(u_n) dx + \int_{\Omega} g_n(x,u_n, \nabla u_n) T_k(u_n) dx} \\ & {= \int_{\Omega} f_n T_k(u_n) dx + \int_{\Omega} \phi_n(T_k(u_n))\nabla T_k(u_n) dx} \\ & {\leq k\int_{\Omega} |f_n| dx + \int_{\Omega} \frac{|\phi_n(T_k(u_n))|}{(\frac{\alpha}{2} p(x))^{\frac{1}{p(x)}}}(\frac{\alpha}{2} p(x))^{\frac{1}{p(x)}}|\nabla T_k(u_n)| dx} \\ & {\leq k\|f_n\|_1 + \int_{\Omega} \frac{|\phi_n(T_k(u_n))|^{p'(x)}}{p'(x) (\frac{\alpha}{2} p(x))^{\frac{p'(x)}{p(x)}}} dx +\int_{\Omega}\frac{\frac{\alpha}{2} p(x)|\nabla T_k(u_n)|^{p(x)}}{p(x)} dx} \\ & {\leq k\|f\|_1 + C_{2}\int_{\Omega} |\phi_n(T_k(u_n))|^{p'(x)} dx + \frac{\alpha}{2}\int_{\Omega}|\nabla T_k(u_n)|^{p(x)} dx,} \end{aligned}\label{aq24} \end{equation} and since \begin{align*} {\int_{\Omega} |\phi_n(T_k(u_n))|^{p'(x)} dx} & {\leq \int_{\Omega} \sup_{|s|\leq k}|\phi_n(s)|^{p'(x)} dx} \\ & {\leq \int_{\Omega} \sup_{|s|\leq n}|\phi(s)|^{p'(x)} dx} \\ & {\leq \big(\sup_{|s|\leq n}|\phi(s)| +1\big)^{p'_{+}}.\operatorname{meas}(\Omega),} \end{align*} by \eqref{aq24}, we obtain $$ \frac{\alpha}{2}\|\nabla T_k(u_n)\|_{p(x)}^{\gamma} \leq \frac{\alpha}{2}\int_{\Omega} |\nabla T_k(u_n)|^{p(x)} dx \leq k\|f\|_1 + C_{3}, $$ with $$ \gamma = \begin{cases} p_{+} & \text{if } \|\nabla T_k(u_n)\|_{p(x)} \leq 1, \\ p_{-} & \text{if } \|\nabla T_k(u_n)\|_{p(x)} > 1, \end{cases} $$ we deduce that \begin{equation} \|\nabla T_k(u_n)\|_{p(x)} \leq C_{4}k^{\frac{1}{\gamma}}\quad \text{for all } k\geq 1,\label{aq25} \end{equation} where $C_{4}$ is a constant that does not depend on $k$. Now, we show that $(u_n)_n$ is a Cauchy sequence in measure. Indeed, we have \begin{align*} k \operatorname{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)^{\frac{1}{p'_{-}}}\|T_k(u_n)\|_{p(x)} }\\ & \leq C_{5} k^{\frac{1}{\gamma}}, \end{align*} according to the Poincar\'{e} inequality and \eqref{aq25}. Therefore, \begin{equation} \operatorname{meas}\{|u_n|> k\} \leq C_{5}\frac{1}{k^{1-\frac{1}{\gamma}}} \to 0 \quad \text{as } k \to \infty.\label{aq26} \end{equation} Since for all $\delta > 0 $, \begin{align*} &\operatorname{meas} \{|u_n - u_{m}|>\delta\} \\ &\leq \operatorname{meas} \{|u_n|>k\} + \operatorname{meas} \{|u_{m}|>k\} +\operatorname{meas} \{|T_k(u_n) - T_k(u_{m})|>\delta\}, \end{align*} using \eqref{aq26}, we get that for all $ \varepsilon > 0$, there exists $ k_0> 0 $ such that \begin{equation} \operatorname{meas}\{|u_n|> k\} \leq \frac{\varepsilon}{3}, \quad \operatorname{meas}\{|u_{m}|> k\} \leq \frac{\varepsilon}{3}\quad \forall k \geq k_0(\varepsilon), \label{aq27} \end{equation} On the other hand, by \eqref{aq25}, the sequence $(T_k(u_n))_n$ is bounded in $W_0^{1,p(x)}(\Omega)$, then there exists a subsequence still denoted $(T_k(u_n))_n$ such that $$ T_k(u_n) \rightharpoonup \eta_k \quad \text{in } W_0^{1,p(x)}(\Omega) \quad \text{as } n\to \infty. $$ and by the compact embedding, we obtain $$ T_k(u_n) \to \eta_k \quad \text{in } L^{p(x)}(\Omega) \text{ and a.e. in } \Omega. $$ Therefore, we can assume that $(T_k(u_n))_n$ is a Cauchy sequence in measure in $\Omega$, then for all $ k> 0 $ and $ \delta, \varepsilon > 0 $ there exists $n_0 = n_0(k, \delta,\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. \label{aq281} \end{equation} Combining \eqref{aq27} and \eqref{aq281}, we obtain that for all $\delta, \varepsilon > 0$, there exists $n_0 = n_0(\delta,\varepsilon)$ such that $$ \operatorname{meas}\{|u_n-u_{m}|> \delta\} \leq \varepsilon\quad \forall n, m \geq n_0, $$ it follows that $(u_n)_n$ is a Cauchy sequence in measure, then there exists a subsequence still denoted $ (u_n)_n $ such that $$ u_n \to u \quad \text{a.e. in } \Omega. $$ We obtain \begin{equation} \begin{gathered} T_k(u_n) \rightharpoonup T_k(u) \quad \text{in } W_0^{1,p(x)}(\Omega) \\ T_k(u_n) \to T_k(u) \quad \text{in $L^{p(x)}(\Omega)$ and a.e. in } \Omega. \end{gathered}\label{aq28} \end{equation} \subsection*{Step 3: Convergence of the gradient} In the sequel, we denote by $ \varepsilon_{i}(n)$ $i = 1, 2, \ldots $ various functions of real numbers which converge to $ 0 $ as $ n $ tends to infinity. Let $ \varphi_k(s) = s\exp(\gamma s^{2}) $ where $ \gamma = \big(\frac{b(k)}{2\alpha}\big)^{2}$, it is obvious that $$ \varphi'_k(s) - \frac{b(k)}{\alpha}|\varphi_k(s)| \geq \frac{1}{2}\quad \quad \forall s\in \mathbb{R}, $$ we consider $ h> k > 0 $ and $ M = 4k + h$, we set $$ \omega_n = T_{2k}(u_n - T_{h}(u_n) + T_k(u_n) - T_k(u)). $$ Taking $ \varphi_k(\omega_n) $ as a test function in \eqref{aq9}, we obtain \begin{align*} &\int_{\Omega} a(x,u_n, \nabla u_n) \varphi'_k(\omega_n)\nabla \omega_n dx + \int_{\Omega} g_n(x,u_n, \nabla u_n) \varphi_k(\omega_n) dx\\ & = \int_{\Omega} f_n\varphi_k(\omega_n) dx + \int_{\Omega} \phi_n(u_n)\varphi'_k(\omega_n)\nabla \omega_n dx, \end{align*} it is easy to see that $ \nabla \omega_n = 0 $ on $ \{|u_n|> M\}$, and since $ g_n(x,u_n, \nabla u_n) \varphi_k(\omega_n) \geq 0 $ on $ \{|u_n| > k\}$, we have \begin{equation} \begin{aligned} &{\int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n) \nabla \omega_n dx + \int_{\{|u_n| \leq k\}} g_n(x,u_n, \nabla u_n) \varphi_k(\omega_n) dx} \\\ & \leq \int_{\Omega} f_n\varphi_k(\omega_n) dx + \int_{\{|u_n| \leq M\}} \phi_n(T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx. \end{aligned}\label{aq29} \end{equation} We have \begin{equation} \begin{aligned} &\int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n) \nabla \omega_n dx \\ & = \int_{\{|u_n|\leq k\}} a(x,T_k(u_n), \nabla T_k(u_n))\varphi'_k(\omega_n) \nabla T_{2k}(u_n - T_k(u)) dx \\ &\quad + \int_{\{|u_n| > k\}} a(x,T_{M}(u_n), \nabla T_{M}(u_n)) \varphi'_k(\omega_n)\nabla T_{2k}(u_n-T_{h}(u_n)\\ &\quad +T_k(u_n) - T_k(u)) dx. \end{aligned}\label{aq30} \end{equation} On the one hand, since $ |u_n - T_k(u)|\leq 2k $ on $ \{|u_n|\leq k\}$, we have \begin{equation} \begin{aligned} &{\int_{\{|u_n|\leq k\}} a(x,T_k(u_n), \nabla T_k(u_n))\varphi'_k(\omega_n) \nabla T_{2k}(u_n - T_k(u)) dx} \\ & = \int_{\{|u_n|\leq k\}} a(x,T_k(u_n), \nabla T_k(u_n)) \varphi'_k(\omega_n )\nabla (T_k(u_n) - T_k(u)) dx \\ & = \int_{\Omega} a(x,T_k(u_n), \nabla T_k(u_n))\varphi'_k (\omega_n)\nabla (T_k(u_n) - T_k(u)) dx \\ &\quad - \int_{\{|u_n| > k\}} a(x,T_k(u_n), \nabla T_k(u_n))\varphi'_k(\omega_n)\nabla (T_k(u_n) - T_k(u)) dx. \end{aligned}\label{aq31} \end{equation} Since $ 1 \leq \varphi'_k(\omega_n) \leq \varphi'_k(2k)$, it follows that \begin{align*} &{-\int_{\{|u_n| > k\}} a(x,T_k(u_n), \nabla T_k(u_n)) \varphi'_k(\omega_n)\nabla (T_k(u_n) - T_k(u)) dx } \\ & = \int_{\{|u_n| > k\}} a(x,T_k(u_n), \nabla T_k(u_n)) \varphi'_k(\omega_n)\nabla T_k(u) dx \\ &\leq \varphi'_k(2k)\int_{\{|u_n| > k\}} |a(x,T_k(u_n), \nabla T_k(u_n))| |\nabla T_k(u)| dx, \end{align*} and since $ (|a(x,T_k(u_n), \nabla T_k(u_n))| )_n $ is bounded in $ L^{p'(x)}(\Omega)$, then there exists $ \vartheta \in L^{p'(x)}(\Omega) $ such that $$ |a(x,T_k(u_n), \nabla T_k(u_n))| \rightharpoonup \vartheta \quad \text{in }\quad L^{p'(x)}(\Omega), $$ then $$ \int_{\{|u_n| > k\}} |a(x,T_k(u_n), \nabla T_k(u_n))| |\nabla T_k(u)| dx \to \int_{\{|u| > k\}} \vartheta |\nabla T_k(u)| dx = 0, $$ and we obtain \begin{equation} \int_{\{|u_n| > k\}} a(x,T_k(u_n), \nabla T_k(u_n)) \varphi'_k(\omega_n)\nabla (T_k(u_n) - T_k(u)) dx = \varepsilon_0(n),\label{aqp} \end{equation} with $ \varepsilon_0(n) $ tend to $0$ as $ n\to\infty$. On the other hand, for the second term on the right hand side of \eqref{aq30}, taking $ z_n = u_n - T_{h}(u_n) + T_k(u_n) - T_k(u)$, \begin{equation} \begin{aligned} &\int_{\{|u_n| > k\}} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n) \nabla T_{2k}(u_n-T_{h}(u_n)+T_k(u_n) - T_k(u)) dx \\ &= \int_{\{|u_n| > k\}\cap \{|z_n|\leq 2k\}} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n)\nabla (u_n-T_{h}(u_n)+T_k(u_n)\\ &\quad - T_k(u)) dx \\ &= \int_{\{|u_n| > k\}\cap \{|z_n|\leq 2k\}} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n)\nabla (u_n - T_k(u)).\chi_{\{|u_n|> h\}} dx \\ &\quad - \int_{\{|u_n| > k\}\cap \{|z_n|\leq 2k\}} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n)\nabla T_k(u).\chi_{\{|u_n|\leq h\}} dx \\ &\geq - \int_{\{|u_n| > k\}} |a(x,T_{M}(u_n), \nabla T_{M}(u_n))| |\nabla T_k(u)| \varphi'_k(\omega_n) dx. \end{aligned}\label{aq32} \end{equation} By combining \eqref{aq30}-\eqref{aqp} and \eqref{aq32}, we obtain \begin{align*} & \int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n) \nabla \omega_n dx \\ &\geq \int_{\Omega} a(x,T_k(u_n), \nabla T_k(u_n))\varphi'_k(\omega_n)\nabla (T_k(u_n) - T_k(u)) dx \\ &\quad - \int_{\{|u_n| > k\}} |a(x,T_{M}(u_n), \nabla T_{M}(u_n))| |\nabla T_k(u)| \varphi'_k(\omega_n) dx - \varepsilon_0(n), \end{align*} %5.16 which is equivalent to \begin{align*} &{\int_{\Omega} ( 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)) \varphi'_k(\omega_n) dx} \\ &\leq \int_{\{|u_n| > k\}} |a(x,T_{M}(u_n), \nabla T_{M}(u_n))| |\nabla T_k(u)| \varphi'_k(\omega_n) dx \\ & \quad + \int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n)) \varphi'_k(\omega_n)\nabla \omega_n dx \\ & \quad - \int_{\Omega} a(x,T_k(u_n), \nabla T_k(u))(\nabla T_k(u_n) - \nabla T_k(u)) \varphi'_k(\omega_n) dx + \varepsilon_0(n). \end{align*} We obtain \begin{equation} \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))) \varphi'_k(\omega_n) (\nabla T_k(u_n) - \nabla T_k(u)) dx} \\ & \leq \varphi'_k(2k)\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_{M}(u_n), \nabla T_{M}(u_n)) \varphi'_k(\omega_n) \nabla \omega_n dx \\ & + \varphi'_k(2k)\int_{\Omega} |a(x,T_k(u_n), \nabla T_k(u))| |\nabla T_k(u_n) - \nabla T_k(u)| dx + \varepsilon_0(n). \end{aligned}\label{aq33} \end{equation} Now, we study each terms on the right hand side of the above inequality. For the first term, we have $ (|a(x,T_{M}(u_n),\nabla T_{M}(u_n))|)_n $ is bounded in $ L^{p'(x)}(\Omega)$, and since $$ |\nabla T_k(u)|^{p(x)}\chi_{\{|u_n|>k\}} \leq |\nabla T_k(u)|^{p(x)}, $$ and $$ |\nabla T_k(u)|^{p(x)}\chi_{\{|u_n|>k\}} \to 0, \quad \text{a.e. in $\Omega$ as } n\to \infty, $$ by the Lebesgue dominated convergence theorem, we deduce that $$ |\nabla T_k(u)|\chi_{\{|u_n|>k\}} \to 0, \quad \text{in $L^{p(x)}(\Omega)$ as } n\to \infty, $$ which implies that the first term in the right hand side of \eqref{aq33} tends to $0$ as $n$ tends to $\infty$, and we can write \begin{equation} \varphi'_k(2k)\int_{\{|u_n| > k\}} |a(x,T_{M}(u_n),\nabla T_{M}(u_n))| |\nabla T_k(u)| dx = \varepsilon_1(n).\label{aq34} \end{equation} For the third term on the right-hand side of \eqref{aq33}, we have $$ |a(x,T_k(u_n),\nabla T_k(u))| \to |a(x,T_k(u),\nabla T_k(u))| \quad \text{in $L^{p'(x)}(\Omega)$ as } n\to\infty, $$ and since $\nabla T_k(u_n)$ tends weakly to $\nabla T_k(u)$ in $(L^{p(x)}(\Omega))^{N}$, we obtain $$ \varphi'_k(2k)\int_{\Omega} |a(x,T_k(u_n),\nabla T_k(u))| |\nabla T_k(u_n) - \nabla T_k(u)| dx \to 0 \quad \text{as } n\to \infty, $$ then \begin{equation} \varphi'_k(2k)\int_{\Omega} |a(x,T_k(u_n),\nabla T_k(u))| |\nabla T_k(u_n) - \nabla T_k(u)| dx = \varepsilon_{2}(n).\label{aq38} \end{equation} By \eqref{aq33} we conclude that \begin{equation} \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)))\varphi'_k(\omega_n) (\nabla T_k(u_n) - \nabla T_k(u)) dx \\ &\leq \int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx + \varepsilon_{3}(n). \end{aligned}\label{aq39} \end{equation} Now, we turn to the second term on the left-hand side of \eqref{aq29}; by \eqref{aq5} we have \begin{align*} &\big|\int_{\{|u_n|\leq k\}} g_n(x,u_n,\nabla u_n)\varphi_k(\omega_n) dx\big| \\ & \leq \int_{\{|u_n|\leq k\}} b(|u_n|)(c(x) + |\nabla T_k(u_n)|^{p(x)})|\varphi_k(\omega_n)| dx \\ &\leq b(k)\int_{\{|u_n|\leq k\}} c(x)|\varphi_k(\omega_n)| dx\\ &\quad + \frac{b(k)}{\alpha}\int_{\Omega} a(x,T_k(u_n),\nabla T_k(u_n))\nabla T_k(u_n)|\varphi_k(\omega_n)| dx \\ & \leq b(k)\int_{\{|u_n|\leq k\}} c(x)|\varphi_k(\omega_n)| dx + \frac{b(k)}{\alpha}\int_{\Omega} (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))|\varphi_k(\omega_n)| dx \\ &\quad + \frac{b(k)}{\alpha}\int_{\Omega} a(x,T_k(u_n), \nabla T_k(u))(\nabla T_k(u_n) - \nabla T_k(u)) |\varphi_k(\omega_n)| dx \\ &\quad + \frac{b(k)}{\alpha}\int_{\Omega} a(x,T_k(u_n), \nabla T_k(u_n)) \nabla T_k(u) |\varphi_k(\omega_n)| dx. \end{align*} Then \begin{equation} \begin{aligned} &\frac{b(k)}{\alpha} \int_{\Omega} ( 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)) |\varphi_k(\omega_n)| dx \\ & \geq \big|\int_{\{|u_n|\leq k\}} g_n(x,u_n,\nabla u_n) \varphi_k(\omega_n) dx\big| - b(k) \int_{\{|u_n|\leq k\}} c(x)|\varphi_k(\omega_n)| dx \\ &\quad - \frac{b(k)}{\alpha}\int_{\Omega} a(x,T_k(u_n), \nabla T_k(u))(\nabla T_k(u_n) - \nabla T_k(u)) |\varphi_k(\omega_n)| dx \\ &\quad - \frac{b(k)}{\alpha}\int_{\Omega} a(x,T_k(u_n),\nabla T_k(u_n)) \nabla T_k(u) |\varphi_k(\omega_n)| dx. \end{aligned}\label{aq40} \end{equation} We have \begin{equation} \int_{\{|u_n|\leq k\}} c(x)|\varphi_k(\omega_n)| dx \to \int_{\{|u|\leq k\}} c(x)|\varphi_k(T_{2k}(u - T_{h}(u)))| dx = 0\quad\text{as } n\to \infty. \end{equation} Concerning the third term on the right hand side of \eqref{aq40}, we have \begin{align*} &\int_{\Omega} a(x,T_k(u_n),\nabla T_k(u))(\nabla T_k(u_n) - \nabla T_k(u))|\varphi_k(\omega_n)| dx \\ &\leq \varphi_k(2k)\int_{\Omega} |a(x,T_k(u_n),\nabla T_k(u))| |\nabla T_k(u_n) - \nabla T_k(u)| dx, \end{align*} and by \eqref{aq38}, we deduce that \begin{equation} \int_{\Omega} a(x,T_k(u_n),\nabla T_k(u))(\nabla T_k(u_n) - \nabla T_k(u)) |\varphi_k(\omega_n)| dx \to 0\quad \text{as } n \to \infty.\label{aq41} \end{equation} For the last term of right hand side of \eqref{aq40}, we have $ (a(x,T_k(u_n),\nabla T_k(u_n)))_n $ is bounded in $ (L^{p'(x)}(\Omega))^{N}$, then there exists $ \varphi \in (L^{p'(x)}(\Omega))^{N} $ such that \[ a(x,T_k(u_n),\nabla T_k(u_n)) \rightharpoonup \varphi \] in $ (L^{p'(x)}(\Omega))^{N}$, and since $$ \nabla T_k(u) |\varphi_k(\omega_n)| \to \nabla T_k(u) |\varphi_k(T_{2k}(u - T_{h}(u)))| \quad\text{in } (L^{p(x)}(\Omega))^{N} , $$ it follows that \begin{equation} \begin{aligned} &\int_{\Omega} a(x,T_k(u_n),\nabla T_k(u_n))\nabla T_k(u) |\varphi_k(\omega_n)| dx\\ & \to \int_{\Omega} \varphi\nabla T_k(u)|\varphi_k(T_{2k}(u - T_{h}(u)))| dx = 0. \end{aligned}\label{aq42} \end{equation} Combining \eqref{aq40}, \eqref{aq41} and \eqref{aq42}, we obtain \begin{equation} \begin{aligned} &\frac{b(k)}{\alpha}\int_{\Omega} ( 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)) |\varphi_k(\omega_n)| dx \\ & \geq \Big|\int_{\{|u_n|\leq k\}} g_n(x,u_n,\nabla u_n) \varphi_k(\omega_n) dx\Big| + \varepsilon_{4}(n). \end{aligned}\label{aq43} \end{equation} Thanks to \eqref{aq39} and \eqref{aq43}, we obtain \begin{equation} \begin{aligned} &\int_{\Omega} \big(a(x,T_k(u_n), \nabla T_k(u_n)) - a(x,T_k(u_n), \nabla T_k(u))\big)\\ &\quad\times\big(\nabla T_k(u_n) - \nabla T_k(u)\big) \Big(\varphi'_k(\omega_n) - \frac{b(k)}{\alpha}.|\varphi_k(\omega_n)|\Big) dx \\ & \leq \int_{\Omega} a(x,T_{M}(u_n), \nabla T_{M}(u_n)) \varphi'_k(\omega_n)\nabla \omega_n dx \\ &\quad - \big|\int_{\{|u_n|\leq k\}} g_n(x,u_n,\nabla u_n) \varphi_k(\omega_n) dx\big| + \varepsilon_{5}(n) \\ & \leq \int_{\Omega} f_n\varphi_k(\omega_n) dx + \int_{\{|u_n| \leq M\}} \phi_n(T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx + \varepsilon_{5}(n). \end{aligned}\label{aq44} \end{equation} We have $ \omega_n\rightharpoonup T_{2k}(u-T_{h}(u)) $ weak-$*$ in $ L^{\infty}(\Omega) $ then \begin{equation} \int_{\Omega} f_n\varphi_k(\omega_n) dx \to \int_{\Omega} f\varphi_k(T_{2k}(u-T_{h}(u))) dx\quad \text{as } n\to\infty,\label{aq45} \end{equation} and for $n$ large enough (for example $n \geq M$), we can write $$ \int_{\Omega} \phi_n(T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx = \int_{\{|u_n|\leq M\}} \phi(T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx, $$ it follows that \begin{equation} \begin{aligned} &\int_{\Omega} \phi_n(T_{M}(u_n))\varphi'_k(\omega_n)\nabla \omega_n dx \\ &\to \int_{\Omega} \phi(T_{M}(u))\varphi'_k(T_{2k}(u-T_{h}(u)))\nabla T_{2k}(u-T_{h}(u)) dx\quad \text{as } n\to\infty. \end{aligned}\label{ap455} \end{equation} Combining \eqref{aq44} and \eqref{ap455}, we obtain \begin{equation} \begin{aligned} & \frac{1}{2}\int_{\Omega} ( 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} f\varphi_k(T_{2k}(u-T_{h}(u))) dx \\ &\quad + \int_{\Omega} \phi(T_{M}(u))\varphi'_k(T_{2k}(u-T_{h}(u))) \nabla T_{2k}(u-T_{h}(u)) dx + \varepsilon_{6}(n). \end{aligned}\label{aq46} \end{equation} Taking $ \Psi(t) = {\int_0^{t} \phi(\tau)\varphi'_k(\tau- T_{h}(\tau)) d\tau}$, then $ \Psi(0) = 0_{\mathbb{R}^{N}} $ and $ \Psi \in C^1(\mathbb{R}^{N})$. By the Divergence Theorem (see also \cite{bocc2}), we obtain \begin{align*} & \int_{\Omega} \phi(T_{M}(u))\varphi'_k(T_{2k}(u-T_{h}(u))) \nabla T_{2k}(u-T_{h}(u)) dx \\ &= \int_{\{h < |u| \leq 2k + h\}} \phi(u)\varphi'_k(u-T_{h}(u)) \nabla u dx \\ & = \int_{\{|u| \leq 2k + h\}} \phi(T_{2k + h}(u)) \varphi'_k(T_{2k + h}(u) -T_{h}(u))\nabla T_{2k + h}(u) dx \\ & \quad - \int_{\{|u| \leq h\}} \phi(T_{h}(u))\varphi'_k(T_{h}(u) - T_{h}(u)) \nabla T_{h}(u) dx \\ & = \int_{\Omega} \operatorname{div} \Psi(T_{2k + h}(u)) dx - \int_{\Omega} \operatorname{div} \Psi(T_{h}(u)) dx \\ & = \int_{\partial \Omega} \Psi(T_{2k + h}(u)).\overrightarrow{n} dx - \int_{\partial \Omega} \Psi(T_{h}(u)).\overrightarrow{n} dx \\ & = \sum_{i = 1}^{N}\Big(\int_{\partial \Omega} \Psi_{i}(T_{2k + h}(u)).n_{i} dx - \int_{\partial \Omega} \Psi_{i}(T_{h}(u)).n_{i} dx\Big) = 0, \end{align*} since $ u = 0 $ on $ \partial \Omega$, with $ \Psi = (\Psi_1, \ldots ,\Psi_{N}) $ and $ \overrightarrow{n} = (n_1,n_{2}, \ldots ,n_{N}) $ the normal vector on $ \partial \Omega$. Then, by letting $h$ tend to infinity in \eqref{aq46}, we obtain \begin{equation} \int_{\Omega} (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 \label{aq47} \end{equation} as $n\to \infty$. Using Lemma \ref{lemp4}, we deduce that \begin{equation} T_k(u_n) \to T_k(u) \quad \text{in } W_0^{1,p(x)}(\Omega);\label{aq48} \end{equation} then $$ \nabla u_n\to \nabla u \quad \text{a.e. in } \Omega. $$ \subsection*{Step 4: Equi-integrability of $ g_n(x,u_n, \nabla u_n)$} To prove that $$ g_n(x,u_n,\nabla u_n) \to g(x,u,\nabla u) \quad \text{strongly in } L^1(\Omega), $$ using Vitali’s theorem, it is sufficient to prove that $ g_n(x, u_n,\nabla u_n) $ is uniformly equi-integrable. Indeed, taking $ T_1(u_n - T_{h}(u_n)) $ as a test function in \eqref{aq9}, we obtain \begin{equation} \begin{aligned} &\int_{\Omega} a(x,u_n, \nabla u_n)\nabla T_1(u_n - T_{h}(u_n)) dx + \int_{\Omega} g_n(x,u_n,\nabla u_n)T_1(u_n - T_{h}(u_n)) dx \\ & = \int_{\Omega} f_n T_1(u_n - T_{h}(u_n)) dx + \int_{\Omega} \phi_n(u_n) \nabla T_1(u_n - T_{h}(u_n)) dx, \end{aligned}\label{aq49} \end{equation} which is equivalent to \begin{equation} \begin{aligned} & \int_{\{h < |u_n|\leq h+ 1\}} a(x,u_n, \nabla u_n) \nabla u_n dx + \int_{\{h \leq |u_n|\}} g_n(x,u_n,\nabla u_n)T_1(u_n - T_{h}(u_n)) dx \\ & = \int_{\{h \leq |u_n|\}} f_n T_1(u_n - T_{h}(u_n)) dx + \int_{\{h < |u_n|\leq h+ 1\}} \phi_n(u_n) \nabla u_n dx. \end{aligned}\label{aq50} \end{equation} Taking $ \Phi_n(t) = {\int_0^{t} \phi_n(\tau) d\tau}$, we have $ \Phi_n(0) = 0_{\mathbb{R}^{N}} $ and $ \Phi_n \in C^1(\mathbb{R}^{N})$. In view of the Divergence theorem, \begin{align*} &\int_{\{h < |u_n|\leq h+ 1\}} \phi_n(u_n)\nabla u_n dx \\ & = \int_{\{|u_n|\leq h+ 1\}} \phi_n(u_n)\nabla u_n dx - \int_{\{|u_n|\leq h\}} \phi_n(u_n)\nabla u_n dx \\ & = \int_{\Omega} \phi_n(T_{h+1}(u_n))\nabla T_{h+1}(u_n) dx - \int_{\Omega} \phi_n(T_{h}(u_n))\nabla T_{h}(u_n) dx \\ & = \int_{\Omega} \operatorname{div} \Phi_n(T_{h+1}(u_n)) dx - \int_{\Omega} \operatorname{div} \Phi_n(T_{h}(u_n)) dx \\ & = \int_{\partial\Omega} \Phi_n(T_{h+1}(u_n)).\overrightarrow{n} d\sigma - \int_{\partial\Omega} \Phi_n(T_{h}(u_n)).\overrightarrow{n} d\sigma = 0. \end{align*} Since $ u_n = 0 $ on $\partial \Omega$, with $ \Phi_n = (\Phi_{n,1}, \ldots ,\Phi_{n,N})$, and since $$ \int_{\{h < |u_n|\leq h+ 1\}} a(x,u_n, \nabla u_n) \nabla u_n dx \geq 0, $$ it follows that \begin{align*} \int_{\{h + 1\leq |u_n|\}} |g_n(x,u_n,\nabla u_n)| dx & {= \int_{\{h + 1 \leq |u_n|\}} g_n(x,u_n,\nabla u_n)T_1(u_n - T_{h}(u_n)) dx} \\ & {\leq \int_{\{h \leq |u_n|\}} g_n(x,u_n,\nabla u_n)T_1(u_n - T_{h}(u_n)) dx} \\ & {\leq \int_{\{h \leq |u_n|\}} f_nT_1(u_n - T_{h}(u_n)) dx} \\ & {\leq \int_{\{h \leq |u_n|\}} |f_n| dx,} \end{align*} thus, for all $ \eta > 0$, there exists $ h(\eta) > 0 $ such that \begin{equation} \int_{\{h(\eta) \leq |u_n|\}} |g_n(x,u_n,\nabla u_n)| dx \leq \frac{\eta}{2}.\label{aq51} \end{equation} On the other hand, for any measurable subset $ E\subset \Omega$, we have \begin{equation} \begin{aligned} \int_{E} |g_n(x,u_n,\nabla u_n)| dx &\leq \int_{E \cap \{|u_n| < h(\eta)\}} b(h(\eta))(c(x) + |\nabla u_n|^{p(x)}) dx \\ &\quad + \int_{\{|u_n| \geq h(\eta)\}} |g_n(x,u_n,\nabla u_n)| dx, \end{aligned}\label{aq52} \end{equation} thanks to \eqref{aq48}, there exists $ \beta(\eta) > 0 $ such that \begin{equation} \int_{E \cap \{|u_n| < h(\eta)\}} b(h(\eta))(c(x) + |\nabla u_n|^{p(x)}) dx \leq \frac{\eta}{2} \quad \text{for } \operatorname{meas}(E) \leq \beta(\eta).\label{aq53} \end{equation} Finally, by combining \eqref{aq51}, \eqref{aq52} and \eqref{aq53}, we obtain \begin{equation} \int_{E} |g_n(x,u_n,\nabla u_n)| dx \leq \eta, \quad \text{with } \operatorname{meas}(E) \leq \beta(\eta).\label{aq54} \end{equation} Then $ (g_n(x,u_n,\nabla u_n))_n $ is equi-integrable, and by the Vitali's Theorem we deduce that \begin{equation} g_n(x,u_n,\nabla u_n) \to g(x,u,\nabla u) \quad\text{in } L^1(\Omega). \end{equation} \subsubsection*{Step 5: Passage to the limit} Let $ \varphi\in W_0^{1,p(x)}(\Omega)\cap L^{\infty}(\Omega) $ and $ M = k + \|\varphi\|_{\infty} $ with $ k> 0$, we will show that $$ \liminf_{n \to \infty} \int_{\Omega} a(x,u_n,\nabla u_n) \nabla T_k(u_n - \varphi) dx \geq \int_{\Omega} a(x,u,\nabla u) \nabla T_k(u - \varphi) dx. $$ If $|u_n|>M$ then $ |u_n- \varphi|\geq |u_n|-\|\varphi\|_{\infty} > k$; therefore $\{|u_n- \varphi|\leq k\}\subseteq \{|u_n|\leq M\}$, which implies that \begin{equation} \begin{aligned} & a(x,u_n,\nabla u_n)\nabla T_k(u_n - \varphi) \\ & = a(x,u_n,\nabla u_n)\nabla (u_n - \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{aligned}\label{aq55} \end{equation} Then \begin{equation} \begin{aligned} &\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)) - a(x,T_{M}(u_n),\nabla \varphi))\\ &\quad\times (\nabla T_{M}(u_n) - \nabla \varphi)\chi_{\{|u_n- \varphi|\leq k\}} dx \\ & \quad + \int_{\Omega} a(x,T_{M}(u_n),\nabla \varphi)(\nabla T_{M}(u_n) - \nabla \varphi)\chi_{\{|u_n- \varphi|\leq k\}} dx, \end{aligned}\label{aq56} \end{equation} we obtain \begin{equation} \begin{aligned} &\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)) - a(x,T_{M}(u),\nabla \varphi))(\nabla T_{M}(u) - \nabla \varphi)\chi_{\{|u- \varphi|\leq k\}} dx\\ & \quad + \lim_{n\to +\infty} \int_{\Omega} a(x,T_{M}(u_n), \nabla \varphi)(\nabla T_{M}(u_n) - \nabla \varphi)\chi_{\{|u_n- \varphi|\leq k\}} dx. \end{aligned}\label{aq57} \end{equation} Note that the second term in the right hand side of \eqref{aq57} is equal to $$ \int_{\Omega} a(x,T_{M}(u),\nabla \varphi)(\nabla T_{M}(u) - \nabla \varphi)\chi_{\{|u- \varphi|\leq k\}}dx. $$ Finally, we have \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 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, taking $T_k(u_n-\varphi)$ as a test function in \eqref{aq9} and passing to the limit, we conclude the desired statement. This completes the 5 steps for the proof of Theorem \ref{thm1}. \begin{theorem}\label{thm2} Assume that \eqref{aq1}-\eqref{aq5} and \eqref{aqq1} hold, $ p(.)\in C_{+}(\bar{\Omega}) $ such that $ 2 - \frac{1}{N} < p_{-}\leq p_{+} < N$. Then problem \eqref{aq8} has at least one solution $ u \in W_0^{1,q(x)}(\Omega)$ for all continuous functions $ q(.)\in C_{+}(\bar{\Omega}) $ such that $ 1< q(x) < \bar{q}(x) = \frac{N(p(x) - 1)}{N-1}$. \end{theorem} \begin{proof} Let $ (f_n)_n $ be a sequence in $ W^{-1,p'(x)}(\Omega)\cap L^1(\Omega) $ such that $ f_n\to f $ in $ L^1(\Omega) $ and $ \|f_n\|_1 \leq \|f\|_1$. we consider the approximate problem \begin{equation} \begin{gathered} Au_n + g_n(x,u_n,\nabla u_n) = f_n - \operatorname{div} \phi_n(u_n)\\ u_n\in W_0^{1,p(x)}(\Omega), \end{gathered}\label{aq58} \end{equation} where $\phi_n(s)= \phi(T_n(s))$ and $ { g_n(x,s,\xi) = \frac{g(x,s,\xi)}{1+ \frac{1}{n}|g(x,s,\xi)|}}$. Thanks to the first step in the proof of Theorem \ref{thm1}, there exists at least one weak solution $ u_n\in W_0^{1,p(x)}(\Omega) $ for this approximate problem. Let $ \psi_k(t) $ be a real valued function \begin{equation} \psi_k(t) = \begin{cases} 0 &\text{if } 0 \leq t \leq k, \\ t - k &\text{if } k < t \leq k +1, \\ 1 &\text{if } k+1 < t , \\ -\psi_k(-t) &\text{otherwise }, \end{cases}\label{aq59} \end{equation} and we define the sets $$ B_0 = \{x\in \Omega: |u_n| \leq 1\},\quad B_k = \{x\in \Omega: k < |u_n| \leq k +1\} \quad \text{for }k \in \mathbb{N}^{*}. $$ Taking $ \psi_k(u_n) $ as a test function in the approximate problem \eqref{aq58}, we obtain \begin{align*} & \int_{\Omega} a(x, u_n, \nabla u_n)\nabla \psi_k(u_n) dx + \int_{\Omega} g_n(x, u_n, \nabla u_n)\psi_k(u_n) dx \\ & = \int_{\Omega} f_n\psi_k(u_n) dx + \int_{\Omega} \phi_n(u_n)\nabla \psi_k(u_n) dx. \end{align*} Then \begin{align*} & \int_{B_k} a(x, u_n, \nabla u_n)\nabla u_n dx + \int_{\{|u_n|> k \}} g_n(x, u_n, \nabla u_n)\psi_k(u_n) dx \\ & = \int_{\{|u_n|> k \}} f_n\psi_k(u_n) dx + \int_{B_k} \phi_n(u_n)\nabla u_n dx\,. \end{align*} By the Divergence theorem, \begin{equation} \begin{aligned} \int_{B_k} \phi_n(u_n)\nabla u_n dx & = {\int_{\{|u_n|\leq k+1\}} \phi_n(u_n)\nabla u_n dx - \int_{\{|u_n|\leq k\}} \phi_n(u_n)\nabla u_n dx} \\ & = {\int_{\Omega} \phi_n(T_{k+1}(u_n))\nabla T_{k+1}(u_n) dx - \int_{\Omega} \phi_n(T_k(u_n))\nabla T_k(u_n) dx} \\ & = \int_{\Omega} \operatorname{div} \Phi_n(T_{k+1}(u_n)) dx - \int_{\Omega} \operatorname{div} \Phi_n(T_k(u_n)) dx = 0\,. \end{aligned}\label{aq60} \end{equation} Since $ \psi_k(u_n) $ has the same sign as $ u_n$, $ g_n(x, u_n, \nabla u_n)\psi_k(u_n) \geq 0 $ and we obtain $$ \int_{B_k} a(x, u_n, \nabla u_n)\nabla u_n dx \leq \int_{\{|u_n| > k \}} f_n\psi_k(u_n) dx \leq \int_{\Omega} |f_n| dx, $$ using \eqref{aq2}, we deduce that \begin{equation} \alpha\int_{B_k} |\nabla u_n|^{p(x)} dx \leq \|f\|_1\quad \text{for all } k\geq 0. \end{equation} In view of the Lemma \ref{lemp5}, there exists a constant $ C $ that does not depend on $ n $ such that $$ \big\|u_n\big\|_{1,q(x)} \leq C, $$ for any continuous exponent $ q(\cdot)\in C_{+}(\overline{\Omega}) $ with $ {1< q(x) < \bar{q}(x) = \frac{N(p(x) - 1)}{N-1}}$. By using the same steps in the proof of Theorem \ref{thm1}, we can show that there exists a subsequence still denoted $ (u_n)_n $ which converge to $ u$, then $$ \big\|u\big\|_{1,q(x)} \leq C, $$ where $ u $ is solution of \ref{aq8}. \end{proof} \begin{theorem}\label{thm3} Assume that \eqref{aq1}--\eqref{aq5} and \eqref{aqq1} hold, $p(.)\in C_{+}(\overline{\Omega}) $ such that $ 2 - \frac{1}{N} < p_{-}\leq p_{+} < N$. If $ f\log(1+ |f|)\in L^1(\Omega)$ then \eqref{aq8} has at least one solution $ u \in W_0^{1,\bar{q}(x)}(\Omega) $ with $ {\bar{q}(x) = \frac{N(p(x) - 1)}{N-1}}$. \end{theorem} \begin{proof} Let $ (f_n)_n $ be a sequence in $ W^{-1,p'(x)}(\Omega)\cap L^1(\Omega) $ such that $ f_n\to f $ in $ L^1(\Omega)$, with $ \|f_n\|_1 \leq \|f\|_1 $ and $ \|f_n\log(1+|f_n|)\|_1 \leq \|f\log(1+|f|)\|_1 $ (for example $ f_n = T_n(f) $). We consider the approximate problem \begin{equation} \begin{gathered} A u_n + g_n(x,u_n, \nabla u_n) = f_n - \operatorname{div} \phi_n(u_n)\\ u_n\in W_0^{1,p(x)}(\Omega), \end{gathered} \label{aq70} \end{equation} where $\phi_n(s)= \phi(T_n(s))$ and $ {g_n(x,s,\xi) = \frac{g(x,s,\xi)}{1+ \frac{1}{n}|g(x,s,\xi)|}}$, there exists at least one weak solution $ u_n\in W_0^{1,p(x)}(\Omega) $ for this approximate problem. Let $ \psi_k(t) $ be defined by \eqref{aq59}, and $$ S_k = \Big\{x\in \Omega, \quad k < |u_n|\Big\} = \cup_{r=k}^{\infty} B_{r} \quad \quad \forall k\in \mathbb{N}. $$ By using $ \psi_k(u_n) $ as a test function in the approximate problem \eqref{aq70}, we obtain \begin{equation} \alpha \int_{B_k} |\nabla u_n|^{p(x)} dx \leq \int_{S_k} |f_n| dx\quad \text{for all } k\in \mathbb{N}. \end{equation} Let $ {\bar{q}(x) = \frac{N(p(x) - 1)}{N-1}}$, we have \begin{align*} \int_{\Omega} \frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)} dx & = {\sum_{k=0}^{\infty} \int_{B_k} \frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)} dx} \\ & \leq {\sum_{k=0}^{\infty} \frac{1}{k+1} \int_{B_k} |\nabla u_n|^{p(x)} dx} \\ & \leq {\frac{1}{\alpha}\sum_{k=0}^{\infty} \frac{1}{k+1} \int_{S_k} |f_n| dx} \\ & \leq {\frac{1}{\alpha}\sum_{k=0}^{\infty} \frac{1}{k+1} \sum_{s=k}^{\infty} \int_{B_{s}} |f_n| dx} \\ & = {\frac{1}{\alpha}\sum_{k=0}^{\infty} \sum_{s=k}^{\infty} \int_{B_{s}} |f_n|\frac{1}{k+1} dx} \\ & = {\frac{1}{\alpha}\sum_{s=0}^{\infty} \sum_{k=0}^{s} \int_{B_{s}} |f_n|\frac{1}{k+1} dx}\,. \end{align*} Since $\sum_{k=0}^{\infty} \sum_{s=k}^{\infty} v_{s,k} = \sum_{s=0}^{\infty} \sum_{k=0}^{s} v_{s,k}$, the above expression equals \begin{align*} {\frac{1}{\alpha}\sum_{s=0}^{\infty} \int_{B_{s}} |f_n|(\sum_{k=0}^{s} \frac{1}{k+1} ) dx} & \leq \frac{1}{\alpha}\sum_{s=0}^{\infty} \int_{B_{s}} |f_n|[1 + \log(1 + s)] dx \\ & {\leq \frac{1}{\alpha}\sum_{s=0}^{\infty} \int_{B_{s}} |f_n|[1 + \log(1 + |u_n|)] dx} \\ & {\leq \frac{1}{\alpha} \int_{\Omega} |f_n|[1 + \log(1 + |u_n|)] dx,} \end{align*} and since $ a b \leq a \log(1+a) + e^{b} $ for all $ a, b \geq 0$, we obtain \begin{align*} &\frac{1}{\alpha} \int_{\Omega} |f_n| [1 + \log(1 + |u_n|)] dx\\ & = {\frac{1}{\alpha} \int_{\Omega} |f_n| dx + \frac{1}{\alpha} \int_{\Omega} |f_n| \log(1 + |u_n|) dx} \\ & \leq { \frac{1}{\alpha} \int_{\Omega} |f_n| dx + \frac{1}{\alpha} \int_{\Omega} |f_n| \log(1 + |f_n|) dx + \frac{1}{\alpha} \int_{\Omega} (1 + |u_n|) dx} \\ & \leq \frac{1}{\alpha}\|f\|_1 + \frac{1}{\alpha}\|f\log(1 + |f|)\|_1 + \frac{1}{\alpha} \int_{\Omega} (1 + |u_n|) dx\,. \end{align*} In view of the Theorem \ref{thm2} we have $ u_n\in W_0^{1,q(x)}(\Omega)$; then $ {\int_{\Omega}|u_n| dx} $ is bounded. It follows that \begin{equation} \int_{\Omega} \frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)} dx \leq C_1,\label{aq722} \end{equation} with $ C_1 $ is a constant that does not depend on $ n$. Now, observe that $ \overline{\Omega} $ is compact, therefore, we can cover it with a finite number of balls $ (B_{i})_{i=1,\ldots,m}$, with $ B_{i} = B(x_{i},\delta)$. we denote $$ p_{i-} = \min \{ p(x):x\in \overline{B_{i}\cap\Omega}\} \quad \text{and}\quad p_{i+} = \max \{ p(x) : x\in \overline{B_{i}\cap\Omega}\}, $$ since $ p(\cdot) $ is a real-valued continuous function on $ \overline{\Omega}$, then, by taking $ \delta > 0 $ small enough such that \begin{equation} \frac{(N - p_{i-})(p_{i-} - 1)^{2}}{N + p_{i-}^{2} - 2p_{i-}} + p_{i-} > p_{i+} \quad \text{in $B_{i}\cap\Omega$ for } i = 1,\ldots,m, \end{equation} and there exists a constant $ a > 0 $ such that $$ \operatorname{meas}(B_{i} \cap \Omega) > a \quad \text{ for } i = 1,\ldots, m. $$ By the Generalized H\"{o}lder inequality, we have \begin{equation} \begin{aligned} &{\int_{B_{i}\cap\Omega} |\nabla u_n|^{\bar{q}(x)} dx}\\ & = {\int_{B_{i}\cap\Omega} \frac{|\nabla u_n|^{\bar{q}(x)}}{(1+ |u_n|) ^{\frac{\bar{q}(x)}{p(x)}}}(1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}} dx} \\ & = \int_{B_{i}\cap\Omega} \big(\frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)}\big) ^{\frac{\bar{q}(x)}{p(x)}} (1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}} dx \\ & \leq \big(\frac{N(p_{i+} - 1)}{(N-1)p_{i-}} + \frac{N - p_{i-}}{(N-1)p_{i-}}\big) \big\| \big(\frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)}\big) ^{\frac{\bar{q}(x)}{p(x)}} \big\|_{L^{\frac{p(x)}{\overline{q}(x)}} (B_{i}\cap\Omega)} \\ &\quad\times \big\| (1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}} \big\|_{L^{\frac{p(x)}{p(x) - \overline{q}(x)}}(B_{i}\cap\Omega)} \end{aligned}\label{aq71} \end{equation} On the one hand, using \eqref{aq722} we have \begin{equation} \begin{aligned} \big\| \big(\frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)}\big) ^{\frac{\bar{q}(x)}{p(x)}} \big\|_{L^{\frac{p(x)}{\overline{q}(x)}} (B_{i}\cap\Omega)} & { \leq \Big(\int_{B_{i}\cap\Omega} \frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)} dx + 1 \Big)^{\frac{(N-1)p_{i+}}{N(p_{i-} - 1)}}} \\ & { \leq \Big(\int_{\Omega} \frac{|\nabla u_n|^{p(x)}}{(1+ |u_n|)} dx + 1\Big)^{\frac{(N-1)p_{+}}{N(p_{-} - 1)}}} \\ & \leq (C_1 + 1)^{\frac{(N-1)p_{+}}{N(p_{-} - 1)}} \end{aligned} \label{aq72} \end{equation} On the other hand, thanks to the Sobolev-Poincar\'{e} inequality, we have \begin{align*} {\|u_n\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)}} & {\leq \|u_n - \overline{u}_{n,i}\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)} + \|\overline{u}_{n,i}\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)}} \\ & {\leq c \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)} + \|\overline{u}_{n,i}\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)}} \end{align*} with $ {\overline{u}_{n,i} = \frac{1}{|B_{i}\cap\Omega|} \int_{B_{i}\cap\Omega} u_n dx}$, and since $$ {\bar{q}^{*}(x) = \frac{\bar{q}(x)}{p(x) - \bar{q}(x)} = \frac{N(p(x) - 1)}{N-p(x)}}, $$ we obtain \begin{align*} &\int_{B_{i}\cap\Omega} (1+ |u_n|)^{\frac{\bar{q}(x)}{p(x) - \bar{q}(x)}} dx \\ & \leq {C_{2}\int_{B_{i}\cap\Omega} ( 1 + |u_n|^{\frac{\bar{q}(x)}{p(x) - \bar{q}(x)}} ) dx} \\ & = {C_{2}\big( \operatorname{meas}(B_{i}\cap\Omega) + \int_{B_{i}\cap\Omega} |u_n|^{\bar{q}^{*}(x)} dx\big)} \\ & \leq {C_{2}\big( \operatorname{meas}(B_{i}\cap\Omega) + \|u_n\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)}^{\sigma_1}\big)} \\ & \leq C_{3}\big( \operatorname{meas}(B_{i}\cap\Omega) + \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)}^{\sigma_1} + \|\overline{u}_{n,i}\|^{\sigma_1}_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)}\big), \end{align*} with $$ \sigma_1 = \begin{cases} \frac{N(p_{i+} - 1)}{N-p_{i+}} & \text{if } \|u_n\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)} > 1, \\ \frac{N(p_{i-} - 1)}{N-p_{i-}} & \text{if } \|u_n\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)} \leq 1, \end{cases} $$ since $ |\overline{u}_{n,i}| \leq \frac{1}{a}\int_{\Omega} |u_n| dx$, it follows that $ \|\overline{u}_{n,i}\|_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)} $ is bounded and \begin{equation} \begin{aligned} &{\big\|(1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}}\big\|_{L^{\frac{p(x)}{p(x) - \bar{q}(x)}}(B_{i}\cap\Omega)}} \\ & \leq {\big(\int_{B_{i}\cap\Omega} (1 + |u_n|)^{\frac{\bar{q}(x)}{p(x) - \bar{q}(x)}} dx\big)^{\sigma_{2}}} \\ & \leq {\big(C_{3}\big( \operatorname{meas}(B_{i}\cap\Omega) + \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)}^{\sigma_1} + \|\overline{u}_{n,i}\|^{\sigma_1}_{L^{\bar{q}^{*}(x)}(B_{i}\cap\Omega)} \big)\big)^{\sigma_{2}}} \\ & \leq {C_{4} (1 + \|\nabla u_n\|_{L^{\bar{q}(x)} (B_{i}\cap\Omega)}^{\sigma_1\sigma_{2}}),} \end{aligned} \label{aq73} \end{equation} and $$ \sigma_{2} = \begin{cases} \frac{N - p_{i-}}{(N - 1)p_{i-}}& \text{if } \big\|(1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}}\big\|_{L^{\frac{p(x)}{p(x) - \bar{q}(x)}}(B_{i}\cap\Omega)} > 1, \\ \frac{N - p_{i+}}{(N - 1)p_{i+}}& \text{if } \big\|(1+ |u_n|)^{\frac{\bar{q}(x)}{p(x)}}\big\|_{L^{\frac{p(x)}{p(x) - \bar{q}(x)}}(B_{i}\cap\Omega)} \leq 1. \end{cases} $$ By combining \eqref{aq71}, \eqref{aq72} and \eqref{aq73}, we obtain \[ \int_{B_{i}\cap\Omega} |\nabla u_n|^{\bar{q}(x)} dx \leq C_{5} + C_{5} \|\nabla u_n\|_{L^{\bar{q}(x)} (B_{i}\cap\Omega)}^{\sigma_1 \sigma_{2}}\,. \] Then \begin{equation} \begin{aligned} &\|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)}^{\pi} - C_{5} \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)}^{\sigma_1 \sigma_{2}} \\ &\leq \int_{B_{i}\cap\Omega} |\nabla u_n|^{\bar{q}(x)} dx - C_{5} \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)}^{\sigma_1 \sigma_{2}} \leq C_{5}, \end{aligned}\label{aq74} \end{equation} with $$ \pi = \begin{cases} \bar{q}_{i-} & \text{if } \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)} > 1, \\ \bar{q}_{i+} & \text{if } \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)} \leq 1, \end{cases} $$ and since $ \sigma_{1 }\sigma_{2} < \bar{q}_{i-} \leq \pi $ in $ B_{i}\cap\Omega$, it follows that $ \|\nabla u_n\|_{L^{\bar{q}(x)}(B_{i}\cap\Omega)} $ is bounded. Indeed, we have \begin{align*} &\frac{(N - p_{i-})(p_{i-} - 1)^{2}}{N + p_{i-}^{2} - 2p_{i-}} + p_{i-} > p_{i+} \\ & \Longleftrightarrow \frac{Np_{i-}^{2} - Np_{i-} + N - p_{i-}}{N + p_{i-}^{2} - 2p_{i-}} > p_{i+} \\ & \Longleftrightarrow \quad (p_{i-} - 1)(N- p_{i+})p_{i-} - (N - p_{i-})(p_{i+} - 1) > 0 \\ & \Longleftrightarrow \quad \bar{q}_{i-} = \frac{N(p_{i-} - 1)}{N - 1} > \frac{(N - p_{i-})}{(N - 1)p_{i-}} \frac{N(p_{i+} - 1)}{N- p_{i+}} \geq \sigma_{1 }\sigma_{2}. \end{align*} We conclude that there exists some constants $ r_{i} > 0 $ such that $ \int_{B_{i}\cap\Omega} |\nabla u_n|^{\bar{q}(x)} dx \leq r_{i} $ for all $ i = 1,\ldots,m$, it follows that \begin{equation} \int_{\Omega} |\nabla u_n|^{\bar{q}(x)} dx = \sum_{i = 1}^{m} \int_{B_{i}\cap\Omega} |\nabla u_n|^{\bar{q}(x)} dx \leq C_{6}, \end{equation} and by the Poincar\'{e} inequality, we obtain $$ \big\|u_n\big\|_{1,\bar{q}(x)} \leq C_{7}, $$ with $ C_{7} $ is a constant that does not depend on $ n$, we deduce that $$ \big\|u\big\|_{1,\bar{q}(x)} \leq C_{7}, $$ where $ u $ is solution of \eqref{aq8}. \end{proof} \begin{theorem}\label{thm4} Let $ p(.)\in C_{+}(\bar{\Omega})$. Assume \eqref{aq1}-\eqref{aq5} hold with $ f\in W^{-1,p'(x)}(\Omega)$ and $ \phi\in C^{0}(\mathbb{R}^{N})$. Then $ \eqref{aq8} $ has at least one solution $ u \in W_0^{1,p(x)}(\Omega)$. \end{theorem} \begin{proof} Let $ u_n\in W_0^{1,p(x)}(\Omega) $ a weak solution of the approximate problem \begin{equation} \begin{gathered} A u_n + g_n(x,u_n, \nabla u_n) = f - \operatorname{div} \phi_n(u_n)\\ u_n\in W_0^{1,p(x)}(\Omega), \end{gathered} \label{aq75} \end{equation} where $\phi_n(s)= \phi(T_n(s))$ and $ {g_n(x,s,\xi) = \frac{g(x,s,\xi)}{1+ \frac{1}{n}|g(x,s,\xi)|}}$. By taking $ u_n $ as a test function in \eqref{aq75}, we obtain $$ \int_{\Omega} a(x,u_n,\nabla u_n) \nabla u_n dx + \int_{\Omega} g_n(x,u_n,\nabla u_n) u_n dx = \int_{\Omega} f u_n dx + \int_{\Omega}\phi_n(u_n) \nabla u_n dx. $$ By the Divergence theorem, $ \int_{\Omega}\phi_n(u_n)\nabla u_n dx = 0$, and since $ g_n(x,u_n,\nabla u_n) u_n \geq 0$, we obtain \begin{align*} \alpha\int_{\Omega} |\nabla u_n|^{p(x)} dx & \leq \int_{\Omega} a(x,u_n,\nabla u_n)\nabla u_n dx \\ & \leq (\frac{1}{p_{-}} + \frac{1}{p'_{-}})\|f\|_{-1,p'(x)}\|u_n\|_{1,p(x)}, \end{align*} it follows that $$ \|\nabla u_n\|_{p(x)}^{\gamma} \leq C_1\|f\|_{-1,p'(x)} \|u_n\|_{1,p(x)} \quad \text{with } \gamma = \begin{cases} p_{-} & \text{if } \|\nabla u_n\|_{p(x)} > 1, \\ p_{+} & \text{if } \|\nabla u_n\|_{p(x)} \leq 1, \end{cases} $$ by using the Poincar\'{e} inequality, we obtain $$ \|u_n\|_{1,p(x)}^{\gamma} \leq C_{2}\|u_n\|_{1,p(x)}. $$ Then $\|u_n\|_{1,p(x)} \leq C_{3}$, with $C_{3}$ independent of $ n$, and $$ \|u\|_{1,p(x)} \leq C_{3}, $$ where $ u $ is solution of the problem \eqref{aq8}. \end{proof} \begin{thebibliography}{00} \bibitem{LAEA} L. 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