\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 155, pp. 1--21.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/155\hfil Existence of positive solutions] {Existence of positive solutions for $p(x)$-Laplacian equations with a singular nonlinear term} \author[J. Liu, Q. Zhang, C. Zhao\hfil EJDE-2014/155\hfilneg] {Jingjing Liu, Qihu Zhang, Chunshan Zhao} % in alphabetical order \address{Jingjing Liu \newline College of Mathematics and Information Science, Zhengzhou University of Light Industry, Zhengzhou, Henan 450002, China} \email{jingjing830306@163.com} \address{Qihu Zhang (corresponding author)\newline College of Mathematics and Information Science, Zhengzhou University of Light Industry, Zhengzhou, Henan 450002, China} \email{zhangqihu@yahoo.com, zhangqh1999@yahoo.com.cn} \address{Chunshan Zhao\newline Department of Mathematical Sciences, Georgia Southern University, Statesboro, GA 30460, USA} \email{czhao@GeorgiaSouthern.edu} \thanks{Submitted July 2, 2013. Published July 7, 2014.} \subjclass[2000]{35J25, 35J65, 35J70} \keywords{$p(x)$-Laplacian; singular nonlinear term; sub-supersolution method} \begin{abstract} In this article, we study the existence of positive solutions for the $p(x)$-Laplacian Dirichlet problem $$ -\Delta _{p(x)}u=\lambda f(x,u) $$ in a bounded domain $\Omega \subset \mathbb{R}^{N}$. The singular nonlinearity term $f$ is allowed to be either $f(x,s)\to +\infty $, or $f(x,s)\to +\infty $ as $s\to 0^{+}$ for each $x\in \Omega $. Our main results generalize the results in \cite{g1} from constant exponents to variable exponents. In particular, we give the asymptotic behavior of solutions of a simpler equation which is useful for finding supersolutions of differential equations with variable exponents, which is of independent interest. \end{abstract} \dedicatory{Dedicated to Professor Xianling Fan on his 70th birthday} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Let $\Omega \subset \mathbb{R}^{N}$ be an open bounded domain with $C^{2}$ boundary. We consider the existence of positive solutions for elliptic problems with variable exponent of the form \begin{equation} \begin{gathered} -\Delta _{p(x)}u=\lambda f(x,u), \quad \text{in }\Omega , \\ u(x)>0, \quad \text{in }\Omega , \\ u(x)=0, \quad \text{on }\partial \Omega , \end{gathered} \label{1.1} \end{equation} where $-\Delta _{p(x)}u=-\operatorname{div}(| \nabla u|^{p(x)-2}\nabla u$) with $\nabla u=(\partial _{x_1}u,\partial_{x_2}u,\dots ,\partial _{x_{N}}u)$ which is so-called $p(x)$-Laplacian, $p(\cdot )$ is a function which satisfies some conditions specified below, $f:\Omega \times (0,\infty )\to [ 0,\infty )$ is a continuous function, and $\lambda >0$ is a real parameter. Throughout this paper, we will denote $d(x)=d(x,\partial \Omega )$. In recent years, the study of differential equations and variational problems with nonstandard $p(x)$-growth condition has been an interesting topic. The $p(x)$-Laplacian arises from the study of nonlinear elasticity, electrorheological fluids and image restoration etc. For example, electrorheological fluids have an extensive applications in robotics, aircraft and aerospace. We refer readers to \cite{a1,c3,h1,r1,r2,z4} for more detailed background of applications. There are many reference papers related to the study of differential equations and variational problems with variable exponent. Far from being complete, we refer readers to \cite{a1,b1,c1,d1,f1,f2,f3,f4,f5,f6,f8,h2,h3,h4,h5,h6,h7,k1,l4,m1,m2,m3,m4,m5,m6,p2,r2,z1,z2,z3,z4} and references cited therein. For example, the regularity of weak solutions for differential equations with variable exponent was studied in \cite{a1,f1}, and existence of solutions for variable exponent problems was studied in a series of papers \cite{c1,f3,f6,h2,h6,l4,m3,m6,p2,z2,z3}. Recently, the applications of variable exponent analysis in image restoration attracted more and more attention \cite{g2,g3,h5,l2}. In this paper, our aim is to study the existence of positive solution for problem \eqref{1.1} with singular nonlinear term $f$. Clearly, if $p(\cdot )\equiv p$, a constant, the operator is the well-known $ p$-Laplacian, and \eqref{1.1} is the usual $p$-Laplacian equation, but for non-constant $p(\cdot )$, $p(x)$-Laplacian problems are more complicated due to the non-homogeneity of $p(x)$-Laplacian. For example, if $\Omega $ is a smooth bounded domain, the Rayleigh quotient \[ \lambda _{p(\cdot )}=\underset{u\in W_0^{1,p(\cdot )}(\Omega )\backslash \{0\}}{\inf }\frac{\int_{\Omega }\frac{1}{p(x)}| \nabla u| ^{p(x)}dx}{\int_{\Omega }\frac{1}{p(x)}| u| ^{p(x)}dx} \] is zero in general, and $\lambda _{p(\cdot )}>0$ only under some special conditions (see \cite{f7}). It is also possible the first eigenvalue and eigenfunction of $p(x)$-Laplacian do not exist, even though the existence of the first eigenvalue and eigenfunction is very important in the study of elliptic problems related to $p$-Laplacian problems. For example, in \cite{g1}, the author use the first eigenfunction and the first eigenvalue to construct subsolutions. Fan \cite{f2} considered the eigenvalue problem of $p(x)$-Laplacian equation with the Neumann boundary condition, the existence of infinite many eigenvalues has been established. Benouhiba \cite{b1} studied the eigenvalue problem \[ -\Delta _{p(x)}u=\lambda V(x)| u| ^{q(x)-2}u, \quad x\in\mathbb{R}^{N}, \] where $10, \quad \text{in }\Omega , \\ u(x)=0, \quad \text{on }\partial \Omega , \end{gather*} where $\Omega $ is a bounded domain in $\mathbb{R}^{N}$ with $C^{1,\omega }$ boundary for some $0<\omega <1$, and singular nonlinearity term $f(x,t)$ could show up when $t\to 0^{+}$. Mohammed make the following two assumptions: \begin{itemize} \item[(1)] For each $\theta \in (0,1)$, there is a constant $C_{\theta }\geq 1 $ such that $g(\theta t)\leq C_{\theta }g(t)$ for all $t>0$; \item[(2)] $f(x,s)\geq a(x)$ for any $(x,s)\in \Omega \times (0,\infty )$. \end{itemize} In \cite{g1}, the authors studied the existence of solutions of the nonlinear elliptic problem with constant exponent, \begin{gather*} -\Delta _{p}u=\lambda f(x,u), \quad \text{in }\Omega , \\ u(x)>0, \quad \text{in }\Omega , \\ u(x)=0, \quad \text{on }\partial \Omega , \end{gather*} where $\Omega $ is a bounded domain in $\mathbb{R}^{N}$, $10$ is a real parameter. The nonlinearity term $f$ is allowed to be either $f(x,s)\to +\infty $ or $f(x,s)\to +\infty $ as $s\to 0^{+}$ for each $x\in \Omega $, and the assumptions (1) and (2) are not assumed. Results on elliptic problems with singular nonlinearity are rare (see \cite{l3,z2}). In \cite{z2}, by using the sub-supersolution method, we studied the existence and the boundary asymptotic behavior of solutions of the elliptic problem with variable exponent, \begin{gather*} -\Delta _{p(x)}u=\frac{\lambda }{u^{\gamma (x)}}, \quad \text{in }\Omega , \\ u(x)>0, \quad \text{in }\Omega , \\ u(x)=0, \quad \text{on }\partial \Omega , \end{gather*} where $\Omega \subset \mathbb{R}^{N}$ is a domain with $C^{2}$ boundary, $\lambda $ is a positive parameter which is large enough. Liu \cite{l3} generalized the results of \cite{m7} to $p(x)$-Laplacian by making the similar assumptions. The condition (1) implies that $g(t)\leq Ct^{-a}$ when $t\leq 1$ for some $a>0$, which is invalid for $g(t)=e^{1/t}$, and the condition (2) is a bit strong in some sense. Motivated by \cite{g1}, in this rticle we partly generalized the results to $p(x)$-Laplacian. Before stating our main results, we make the following assumptions throughout this paper: \begin{itemize} \item[(H0)] $p(\cdot )\in C^{1}( \overline{\Omega })$, $10$ such that $p( x) \equiv p$ (a constant) for any $x\in \Omega $ with $d(x)\leq \delta $; \item[(ii)] $\underset{s\to +\infty }{\lim }\frac{g( s) }{ s^{p^{-}-1-\varepsilon }}:=g_{\infty }\in [ 0,+\infty ) $, where $ \varepsilon >0$ is small enough; \item[(iii)] $\alpha ( \cdot ) >N$ on $\overline{\Omega }$. \end{itemize} Then problem \eqref{1.1} has a solution for any positive $\lambda $. \end{theorem} \begin{theorem} \label{thm1.3} Assume that {\rm (H0), (H1), (H2)} hold. Also assume that \begin{itemize} \item[(i)] $\frac{\partial p( \cdot ) }{\partial \nu }<0$ on $\partial \Omega $, where $\nu $ is the inward unit normal vector of $\partial \Omega $; \item[(ii)] $\lim_{s\to +\infty } \frac{g( s) }{ s^{p^{-}-1-\varepsilon }}:=g_{\infty }\in [ 0,+\infty ) $, where $\varepsilon >0$ is small enough; \item[(iii)] $\alpha ( \cdot ) >N$ on $\overline{\Omega }$. \end{itemize} Then problem \eqref{1.1} has a solution for any positive constant $\lambda $. \end{theorem} \begin{theorem} \label{thm1.4} Assume that {\rm (H0), (H1), (H2)} hold. Also assume that \begin{itemize} \item[(i)] Equation \eqref{1.1} is radial; \item[(ii)] $\lim_{s\to +\infty } \frac{g( s) }{ s^{p^{-}-1-\varepsilon }}:=g_{\infty }\in [ 0,+\infty ) $, where $ \varepsilon >0$ is small enough. \end{itemize} Then problem \eqref{1.1} has a solution for any positive $\lambda $. \end{theorem} This paper is organized as follows. In section 2, we will recall some basic facts about the variable exponent Lebesgue and Sobolev spaces which we will use later, and we will also give a general principle of sub-supersolution method. Proofs of our results will be presented in section 3. \section{Preliminaries} Throughout this paper, the letters $c,c_{i},C,C_{i}$ $(i=1,2,\dots )$, denote positive constants which may vary from line to line, but they are independent of the terms which will take part in any limit process. To deal with the $p(x)$-Laplacian problem, we need introduce some functional spaces $L^{p(\cdot )}( \Omega ) $, $W^{1,p(\cdot )}( \Omega )$, $W_0^{1,p(\cdot )}( \Omega ) $ and properties of the $p(x)$-Laplacian which we will use later. Denote by $S(\Omega ) $ be the set of all measurable real-valued functions defined in $\Omega $. Note that two measurable functions are considered as the same element of $S( \Omega ) $ when they are equal almost everywhere. Let \[ L^{p(\cdot )}( \Omega ) =\big\{ u\in S( \Omega ) : \int_{\Omega }| u(x)| ^{p(x)}dx<\infty \big\} , \] with the norm \[ | u| _{p(\cdot )}=| u| _{L^{p(\cdot )}(\Omega )}=\inf \big\{ \lambda >0 : \int_{\Omega }| \frac{u(x)}{\lambda }| ^{p(x)}dx\leq 1 \big\} . \] The space $( L^{p(\cdot )}( \Omega ) ,|\cdot | _{p(\cdot )}) $ becomes a Banach space. We call it variable exponent Lebesgue space. Moreover, this space is a separable, reflexive and uniform convex Banach space; see \cite[Theorems 1.6, 1.10, 1.14]{f8}. The variable exponent Sobolev space \[ W^{1,p(\cdot )}( \Omega ) =\big\{ u\in L^{p(\cdot )}( \Omega ) : | \nabla u| \in L^{p(\cdot)}(\Omega )\big\} , \] can be equipped with the norm \[ \| u\| =| u|_{p(\cdot )}+| \nabla u| _{p(\cdot )},\quad \forall u\in W^{1,p(\cdot )}( \Omega ) . \] Note that $W_0^{1,p(\cdot )}( \Omega ) $ is the closure of $C_0^{\infty}( \Omega ) $ in $W^{1,p(\cdot )}( \Omega ) $. The spaces $ W^{1,p(\cdot )}( \Omega ) $ and $W_0^{1,p(\cdot )}( \Omega ) $ are separable, reflexive and uniform convex Banach spaces (see \cite[Theorem 2.1]{f8}. For $u,v \in S( \Omega ) $, we write $u\leq v$ if $u(x)\leq v(x)$ for a.e. $x\in \Omega $. Let $\rho (x,s)$ be a Carath\'eodory function on $\Omega \times\mathbb{R}$ with property that for any $s_0>0$ there exists a constant $A$ such that \begin{equation} \label{e2b} | \rho (x,s)| \leq A\quad \text{for a.e. $x\in \Omega$ and all }s\in [ -s_0,s_0] . \end{equation} \begin{definition} \label{def2.1}\rm (i) Let $\underline{u},\overline{u}\in W_{\rm loc}^{1,p(\cdot )}( \Omega ) \cap C_0(\overline{\Omega })$ satisfy $\underline{u},\overline{u}>0$ in $\Omega $. We say $\underline{u}$ and $\overline{u}$ are a subsolution and a supersolution of \eqref{1.1} respectively, if \begin{gather*} \int_{\Omega }| \nabla \underline{u}| ^{p(x)-2}\nabla \underline{u}\nabla \phi\,dx \leq \int_{\Omega }\lambda f(x,\underline{u} )\phi\,dx, \\ \int_{\Omega }| \nabla \overline{u}| ^{p(x)-2}\nabla \overline{u}\nabla \phi\,dx \geq \int_{\Omega }\lambda f(x,\overline{u} )\phi\,dx, \end{gather*} for all $\phi \in C_0^{\infty }( \Omega ) $ with $\phi \geq 0$ and $\operatorname{supp}\phi \subset \subset \Omega $. We say $u$ is a solution of \eqref{1.1}, if it is both a subsolution and a supersolution of \eqref{1.1}. (ii) A function $u\in W^{1,p(\cdot )}( \Omega ) \cap C(\overline{ \Omega })$ is called a weak solution of the problem \begin{equation} \begin{gathered} -\Delta _{p(x)}u=\rho (x,u), \quad \text{in }\Omega , \\ u(x)=\varphi (x), \quad \text{on }\Omega , \end{gathered}\label{1.5} \end{equation} where $\varphi (\cdot )\in C(\overline{\Omega })$, if \[ \int_{\Omega }| \nabla u| ^{p(x)-2}\nabla u\nabla \phi\,dx =\int_{\Omega }\rho (x,u)\phi\,dx,\forall \phi \in C_0^{\infty }(\Omega ); \] (iii) $\underline{u},\overline{u}\in W^{1,p(\cdot )}( \Omega ) \cap C(\overline{\Omega })$ are called a weak subsolution and a weak supersolution of the problem \eqref{1.5} respectively if $\underline{u}\leq \varphi $ and $\overline{u}\geq \varphi $ on $\partial \Omega $ and for all $\phi \in C_0^{\infty }( \Omega )$, $\phi \geq 0$, \begin{gather*} \int_{\Omega }| \nabla \underline{u}| ^{p(x)-2}\nabla \underline{u}\nabla \phi\,dx \leq \int_{\Omega }\rho (x,\underline{u})\phi \,dx, \\ \int_{\Omega }| \nabla \overline{u}| ^{p(x)-2}\nabla \overline{u}\nabla \phi\,dx \geq \int_{\Omega }\rho (x,\overline{u})\phi\,dx. \end{gather*} \end{definition} \begin{lemma}[{\cite[Proposition 2.1]{f6}}] \label{lem2.2} The space $(L^{p(\cdot)}(\Omega ),| \cdot | _{p(\cdot )})$ is a separable, uniform convex Banach space, and its conjugate space is $L^{p^{0}(\cdot )}(\Omega )$, where $p^{0}(\cdot )$ is the conjugate function of $p(\cdot )$ satisfying $\frac{1}{p(\cdot )}+\frac{1}{p^{0}(\cdot )}\equiv 1$. For any $u\in L^{p(\cdot )}(\Omega )$ and $v\in L^{p^{0}(\cdot )}(\Omega )$, we have the following H\"{o}lder inequality \[ | \int_{\Omega }uvdx| \leq \int_{\Omega }|uv|\,dx \leq (\frac{1}{p^{-}}+\frac{1}{(p^{0})^{-}})|u| _{p(\cdot )}| v| _{p^{0}(\cdot )} \leq 2| u| _{p(\cdot )}| v| _{p^{0}(\cdot)}. \] \end{lemma} \begin{definition} \label{def2.3}\rm Let $u,v$ $\in W^{1,p(\cdot )}( \Omega) \cap L^{\infty }(\Omega )$. We say that $-\Delta _{p(x)}u+\rho(x,u)\leq -\Delta _{p(x)}v+\rho (x,v)$ in $\Omega $ if \[ \int_{\Omega }| \nabla u| ^{p(x)-2}\nabla u\nabla \phi \,dx+\int_{\Omega }\rho (x,u)\phi\,dx\leq \int_{\Omega }| \nabla v| ^{p(x)-2}\nabla v\nabla \phi\,dx+\int_{\Omega }\rho (x,v)\phi \,dx \] for all $\phi \in C_0^{\infty }( \Omega )$, $\phi \geq 0$. \end{definition} Next we give a comparison principle as follows. \begin{lemma}[{\cite[Lemma 2.3]{z1}}] \label{lem2.4} Let $\rho (x,t)$ be a function satisfying \eqref{e2b} and nondecreasing in $t$. Let $u,v\in W^{1,p(\cdot )}(\Omega )$ satisfy \[ -\Delta _{p(x)}u+\rho (x,u)\leq -\Delta _{p(x)}v+\rho (x,v),\quad ( x\in \Omega ) , \] if $u\leq v$ on $\partial \Omega $, then $u\leq v$ in $\Omega $. \end{lemma} \begin{lemma}[{\cite[Theorem 8.3.1]{d1}}] \label{lem2.5} For every $u\in W_0^{1,p(\cdot )}(\Omega )$, the inequality \[ | u| _{p^{\ast }(\cdot )}\leq C| \nabla u| _{p(\cdot )} \] holds with a constant $C$ depending only on the dimension $N$ and $p^{+}$ and independent of $\Omega $. \end{lemma} \begin{lemma} \label{lem2.6} Suppose the domain $\Omega $ has finite measure, i.e. $| \Omega | <+\infty $, $p(\cdot ),q(\cdot )\in C(\overline{\Omega })$, and $11/n\}$. Let \[ \widetilde{f}(x,u)=\begin{cases} f(x,\overline{u}), & u\geq \overline{u}, \\ f(x,u), & \underline{u}0, \quad \text{in }\Omega _{n}, \\ u(x)=\underline{u}(x), \quad \text{on }\partial \Omega _{n}. \end{gathered} \end{equation} Since $| \widetilde{f}(x,u)| $ is bounded on $\overline{ \Omega _{n}}$ and $\underline{u}\in C_0(\overline{\Omega })$, it is easy to see that \eqref{eIn} has a solution $u_{n}$, satisfy $\underline{u}\leq u_{n}\leq \overline{u}$. By \cite[Theorem 1.2]{f5}, we can see that $\{u_{n}\}_{n\geq n_0+1}$ has uniformly bounded $C^{1,\alpha }$ norm on $\overline{\Omega _{n_0}}$. By the diagonal method, we can choose a subsequence $\{u_{n_{k}}\}$ of $\{u_{n}\}$ such that \[ u_{n_{k}}(x)\to u(x),\quad \nabla u_{n_{k}}(x)\to \nabla u(x),\quad \forall x\in \Omega , \] where $u\in C_0( \overline{\Omega }) \cap C^{1}( \Omega) $. Thus $u$ is a solution of \eqref{1.1} and satisfies $\underline{u}\leq u\leq \overline{u}$. \end{proof} \section{Proofs of main results} To study the existence of solutions of \eqref{1.1}, we need to do some preparation work. Note that by \cite[Theorem 4.2]{f4}, the following problem has a weak solution $\omega _{b}\in W_0^{1,p(\cdot )}( \Omega ) $, \begin{equation} \begin{gathered} -\Delta _{p(x)}\omega =b( x) , \quad \text{in }\Omega , \\ \omega ( x) =0, \quad \text{on }\partial \Omega . \end{gathered} \label{1.6} \end{equation} Since $b( \cdot ) $ is nonnegative, by the comparison principle it follows that $\omega _{b}$ is nonnegative (see \cite[Lemma 2.3]{z1}) and it is positive in $\Omega $ (see \cite[Theorem 1.1]{z1}]). From \cite[Theorem 4.1]{f4}, we see that $\omega _{b}$ is bounded. Then we have $\omega _{b}\in C^{1,\alpha }( \overline{\Omega }) $ and $\frac{ \partial \omega _{b}}{\partial \nu }>0$ on $\partial \Omega $ from the following Lemma. \begin{lemma} \label{lem3.1} (i) \cite[Theorem 1.2]{f1} Let $\omega_{b}$ be a bounded solution of \eqref{1.6}, then $\omega _{b}\in C^{1,\alpha }( \overline{\Omega }) $; (ii) \cite[Theorem 1.2]{z1} Let $\omega _{b}$ be a solution of \eqref{1.6}, $x_1\in \partial \Omega $, $\omega _{b}\in C^{1}(\Omega \cup \{x_1\})$, $\omega _{b}(x_1)=0$. If $\Omega $ satisfies the inward-ball condition at $x_1$, then $\frac{\partial \omega _{b}}{\partial\nu }(x_1)>0$, where $\nu $ is the inward unit normal vector of $\partial \Omega $ on $x_1$. \end{lemma} We will prove the Theorems \ref{thm1.1}--\ref{thm1.4} stated in section 1 by using Lemma \ref{lem2.7}. Next we will construct a supersolution of \eqref{1.1} when $\lambda $ is small enough. Before we begin the proof of Theorem \ref{thm1.1}, we need some background. Define \[ g_{\#}(s)=\begin{cases} g(s), & \text{when }s<1, \\ g(s), & \text{when }s\geq 1\text{ and } \limsup_{s\to +\infty } \frac{g(s)}{s^{p^{-}-1-\varepsilon }}<+\infty , \\ g(1)s^{p^{-}-1-\varepsilon }, & \text{when }s\geq 1\text{ and } \limsup_{s\to +\infty } \frac{g(s)}{s^{p^{-}-1-\varepsilon }}=+\infty . \end{cases} \] Without loss of generality, we assume that $g_{\#}( s) =C_{\ast }s^{p^{-}-1-\varepsilon }$ for $s\geq 1$. There exists $M_0=M_0(\delta )$ large enough such that \begin{equation} g_{\#}( s) <\delta s^{p^{-}-1},\quad \forall s\geq M_0. \label{1.8} \end{equation} Now we define a continuous function $\hat{g}:( 0,\infty ) \to ( 0,\infty ) $ by \[ \hat{g}( s) :=\sup \big\{ \frac{g_{\#}( t) }{ t^{p^{-}-1}},t>s\big\} ,\quad s>0. \] It follows from \eqref{1.8} and the definition of $\hat{g}$ that \begin{itemize} \item[(i)] $\hat{g}$ is non-increasing; \item[(ii)] $\hat{g}( s) \geq \frac{g_{\#}( s) }{s^{p^{-}-1}},s>0$; \item[(iii)] $\hat{g}( s) <\delta $, for all $s\geq M_0$. \end{itemize} We also define a $C^{1}$-function \[ H( s) :=\Big( \frac{2}{s}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt\Big) ^{\frac{1}{p^{-}-1}},\quad s>0. \] \begin{lemma} \label{lem3.2} The function $H$ satisfies \begin{itemize} \item[(i)] $H$ is strictly decreasing, and $-H'( s) \geq \frac{ 2^{\varepsilon }-1}{p^{-}-1}\frac{H( s) }{s}$; \item[(ii)] $\hat{g}( s) \leq [ H( s) ]^{p^{-}-1}\leq \hat{g}(s/2)$, $s>0$; \item[(iii)] $H( s) \to +\infty $ as $s\to 0^{+}$, $H( s) \to 0^{+}$, when $s\to +\infty $. \end{itemize} \end{lemma} \begin{proof} We only need to prove $-H'( s) \geq \frac{2^{\varepsilon }-1}{p^{-}-1}\frac{H( s) }{s}$, the rest is easy to be verified. By computations \[ -H'( s) =\frac{1}{p^{-}-1}\Big( \frac{2}{s}\int_{\frac{s }{2}}^{s}\hat{g}( t) dt\Big) ^{\frac{1}{p^{-}-1}-1} \Big( \frac{2}{s^{2}}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt+\frac{2}{s}(\frac{1 }{2}\hat{g}(\frac{s}{2})-\hat{g}( s) )\Big) . \] By condition (H1), when $s\leq 1$, $s^{p^{-}}\hat{g}( s) $ is decreasing, then we have $\frac{1}{2}\hat{g}(\frac{s}{2})-\hat{g}(s) \geq 0$, and then \[ -H'( s) \geq \frac{1}{p^{-}-1}\Big( \frac{2}{s}\int_{ \frac{s}{2}}^{s}\hat{g}( t) dt\Big) ^{\frac{1}{p^{-}-1}-1}\frac{ 2}{s^{2}}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt\geq \frac{ 2^{\varepsilon }-1}{p^{-}-1}\frac{H( s) }{s}. \] Here we note that $\hat{g}(s)=C_{\ast }s^{-\varepsilon }$ for $s\geq 1$. When $s\geq 2$, we have \begin{align*} \frac{2}{s}\Big(\frac{1}{2}\hat{g}(\frac{s}{2})-\hat{g}( s)\Big) &= C_{\ast }\frac{2}{s}\Big( \frac{1}{2}(\frac{s}{2})^{-\varepsilon} -s^{-\varepsilon }\Big) \\ &= C_{\ast }\frac{2}{s}(2^{\varepsilon -1}-1)s^{-\varepsilon } \\ &= \frac{2}{s}(2^{\varepsilon -1}-1)\hat{g}( s) \\ &\geq \frac{2}{s}(2^{\varepsilon -1}-1)\frac{2}{s}\int_{\frac{s}{2}}^{s} \hat{g}( t) dt \\ &= (2^{\varepsilon }-2)\frac{2}{s^{2}}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt, \end{align*} and \begin{align*} -H'( s) &= \frac{1}{p^{-}-1}\Big( \frac{2}{s}\int_{ \frac{s}{2}}^{s}\hat{g}( t) dt\Big) ^{\frac{1}{p^{-}-1} -1}\Big( \frac{2}{s^{2}}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt+ \frac{2}{s}(\frac{1}{2}\hat{g}(\frac{s}{2})-\hat{g}( s) )\Big) \\ &\geq \frac{1}{p^{-}-1}\Big( \frac{2}{s}\int_{\frac{s}{2}}^{s}\hat{g} ( t) dt\Big) ^{\frac{1}{p^{-}-1}-1}\Big( \frac{2}{s^{2}}\int_{ \frac{s}{2}}^{s}\hat{g}( t) dt+(2^{\varepsilon }-2)\frac{2}{s^{2}} \int_{\frac{s}{2}}^{s}\hat{g}( t) dt\Big) \\ &= \frac{1}{p^{-}-1}\Big( \frac{2}{s}\int_{\frac{s}{2}}^{s}\hat{g}( t) dt\Big) ^{\frac{1}{p^{-}-1}-1}\Big( \frac{2}{s^{2}} (2^{\varepsilon }-1)\int_{\frac{s}{2}}^{s}\hat{g}( t) dt\Big) \\ &= \frac{2^{\varepsilon }-1}{p^{-}-1}\frac{H( s) }{s}. \end{align*} Note that $s^{p^{-}}\hat{g}( s) $ is decreasing for $s\leq 1$. When $10. \] The proof is complete. \end{proof} As a consequence of Lemma \ref{lem3.2}, we can define the function \begin{equation} \eta ( s) :=\int_0^{s}\frac{1}{H( t) }dt,s\geq 0, \label{1.9} \end{equation} for it is easy to show that $\eta \in C^{2}(0,\infty )$. \begin{lemma} \label{lem3.3} The function $\eta $ satisfies \begin{itemize} \item[(i)] $\eta :( 0,\infty ) \to ( 0,\infty ) $ is strictly increasing; \item[(ii)] let $\psi =\eta ^{-1}$ be the inverse function of $\eta $. Then $\psi '(s)=H(\psi ( s) ),s>0$. \end{itemize} \end{lemma} Denote $\Omega _{\sigma }=\{x\in \Omega : \omega _{b}( x) <\sigma \}$, where $\sigma >0$ is a small positive constant. \begin{lemma} \label{lem3.4} Assume that {\rm (H0)} and {\rm (H1)} hold. Then there is a supersolution $v$ of \eqref{1.1} such that $v\in W_{\rm loc}^{1,p(\cdot)} ( \Omega ) \cap C_0( \overline{\Omega })$ when $\lambda $ is small enough. \end{lemma} \begin{proof} Define \begin{equation} v( x) :=\begin{cases} \psi ( k_1\omega _{b}( x) ) , & x\in \Omega _{\sigma}, \\ \omega _{b}( x) +\psi ( k_1\sigma ) -\sigma , & x\in \Omega \backslash \Omega _{\sigma }, \end{cases} \label{2.1} \end{equation} where $\omega _{b}$ is given by \eqref{1.6}, and $k_1>1$ is a constant. Obviously, $v\in C_0(\overline{\Omega })\cap W_{\rm loc}^{1,p(\cdot )}(\Omega ) $. From the definition of $g$ and $\hat{g}$, Lemma \ref{lem3.2}, and Lemma \ref{lem3.3}, it follows that \begin{gather*} \psi '( k_1\omega _{b}( x) ) =H( \psi( k_1\omega _{b}( x) ) ) =H( v( x) ) ,x\in \Omega _{\sigma }, \\ \psi ^{\prime \prime }( s) \leq 0,\text{ for all }s\geq 0. \end{gather*} We will prove this Lemma in three steps. \smallskip \noindent \textit{Step 1}. We will prove that $v$ is a super-solution of \eqref{1.1} in $\Omega _{\sigma }$; i.e., \[ \int_{\Omega }| \nabla v| ^{p( x) -2}\nabla v\nabla \phi\,dx\geq \int_{\Omega }\lambda bg( v) \phi\,dx \geq \int_{\Omega }\lambda f( x,v) \phi\,dx, \] for any $\phi \in C_0^{\infty }( \Omega _{\sigma }) $ with $ \phi \geq 0$ and $\operatorname{supp}\phi \subset \subset \Omega _{\sigma }$. By computation, we have \begin{align*} &\int_{\Omega }| \nabla v| ^{p( x) -2}\nabla v\nabla \phi\,dx \\ &= \int_{\Omega }[ k_1\psi '(k_1\omega _{b}) ] ^{p( x) -1}| \nabla \omega _{b}| ^{p( x) -2}\nabla \omega _{b}\nabla \phi\,dx \\ &= \int_{\Omega }| \nabla \omega _{b}| ^{p( x) -2}\nabla \omega _{b}\nabla \left\{ \phi [ k_1\psi '( k_1\omega _{b}) ]^{p( x) -1}\right\}\,dx \\ &\quad -\int_{\Omega }( k_1) ^{p( x) }| \nabla \omega _{b}| ^{p( x) }\phi ( p( x) -1) [ \psi '( k_1\omega _{b}) ] ^{p( x) -2}\psi ^{\prime \prime }( k_1\omega _{b})\,dx \\ &\quad -\int_{\Omega }( k_1) ^{p( x) -1}| \nabla \omega _{b}| ^{p( x) -2}\nabla \omega _{b}\nabla p[ \phi \psi '( k_1\omega _{b}) ^{p( x) -1} ] \ln k_1\psi '( k_1\omega _{b})\,dx. \end{align*} By Lemma \ref{lem3.2}, we have $-H'( v) \geq \frac{2^{\varepsilon }-1}{p^{-}-1}\frac{H( v) }{v}$ which implies \[ -\psi ^{\prime \prime }( k_1\omega _{b}) =-H'( v) \psi '( k_1\omega _{b}) =-H'( v) H( v) \geq \frac{2^{\varepsilon }-1}{p^{-}-1}\frac{ H( v) }{v}H( v) . \] Note that $00$ is a parameter. It is easy to see that $\phi \in C_0^{1}(\overline{\Omega })$. By computations it follows that \[ -\Delta _{p(x)}\mu \phi =\begin{cases} -k(k\mu e^{kd(x)})^{p(x)-1}\big[(p(x)-1)+(d(x) +\frac{\ln k\mu }{k})\nabla p(x)\nabla d(x)\\ +\frac{\Delta d(x)}{k}\big], \quad \text{if }d(x)<\sigma , \\[4pt] \Big\{\frac{1}{2\ell -\sigma }\frac{2(p(x)-1)}{p^{-}-1}-(\frac{2\ell -d(x)}{ 2\ell -\sigma })[(\ln k\mu e^{k\sigma }(\frac{2\ell -d(x)}{2\ell -\sigma })^{ \frac{2}{p^{-}-1}})\nabla p(x)\nabla d(x)\\ +\Delta d(x)]\Big\} (k\mu e^{k\sigma })^{p(x)-1}(\frac{2\ell -d(x)}{2\ell -\sigma })^{ \frac{2(p(x)-1)}{p^{-}-1}-1},\\ \quad \text{if }\sigma 0$ and $u$ be the unique solution of the problem \begin{equation} \begin{gathered} -\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u)=Mb(x), \quad \text{in }\Omega , \\ u=0, \quad \text{on }\partial \Omega . \end{gathered} \label{2.12} \end{equation} Set \[ \rho _0=\frac{p^{-}}{2| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha^{-}}}C_0}. \] Then $| u| _{\infty }\leq C^{\ast}M^{1/(p^{-}-1)}$ when $M\geq \rho _0$ and $| u|_{\infty }\leq C_{\ast }M^{1/(p^{+}-1)} $ when $M<\rho _0$, where $C^{\ast }$ and $C_{\ast }$ are positive constants depending only on $p^{+}, p^{-}, N, | b(\cdot )| _{L^{\alpha ^{-}}(\Omega )},| \Omega | $ and $C_0$. \end{lemma} \begin{proof} Let $u$ be the solution of \eqref{2.12}. Then $u\geq 0$. For $k\geq 0$, set $A_{k}=\{x\in \Omega :u(x)>k\}$. By taking $(u-k)^{+}$ as a test function of \eqref{2.12}, it follows from \eqref{2.11} and Young inequality that \begin{equation} \begin{aligned} \int_{A_{k}}| \nabla u| ^{p(x)}dx &= M\int_{A_{k}}b(u-k)dx \\ &\leq M| b(\cdot )| _{L^{N}(A_{k})}|(u-k)^{+}| _{L^{N/(N-1)}(A_{k})} \\ &\leq M| b(\cdot )| _{L^{\alpha ^{-}}(A_{k})}| A_{k}| ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}| (u-k)^{+}| _{L^{N/(N-1)}(A_{k})} \\ &\leq M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| A_{k}| ^{\frac{1}{N}-\frac{1}{\alpha ^{-}} }C_0\int_{A_{k}}\varepsilon | \nabla u| \varepsilon ^{-1}dx \\ &\leq M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| A_{k}| ^{\frac{1}{N}-\frac{1}{\alpha ^{-}} }C_0\int_{A_{k}}\Big( \frac{(\varepsilon | \nabla u| )^{p(x)}}{p(x)}+\frac{(\varepsilon ^{-1})^{p^{0}(x)}}{p^{0}(x)}\Big)\,dx \\ &\leq \frac{M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}C_0}{ p^{-}}\int_{A_{k}}\varepsilon ^{p(x)}| \nabla u| ^{p(x)}dx \\ &\quad +\frac{M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| A_{k}| ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}C_0}{ (p^{+})^{0}}\int_{A_{k}}\varepsilon ^{-p^{0}(x)}dx. \end{aligned} \label{2.12b} \end{equation} When $M\geq \rho _0$ we can take \begin{equation} \varepsilon =\Big( \frac{p^{-}}{2M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{ 1}{\alpha ^{-}}}C_0}\Big) ^{1/p^{-}}=( \frac{\rho _0}{M}) ^{1/p^{-}}, \label{2.14} \end{equation} then $\varepsilon \leq 1$ and \begin{align*} &\frac{M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}C_0}{ p^{-}}\int_{A_{k}}\varepsilon ^{p(x)}| \nabla u| ^{p(x)}dx \\ &\leq \frac{M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}C_0}{ p^{-}}\varepsilon ^{p^{-}}\int_{A_{k}}| \nabla u| ^{p(x)}dx\\ &=\frac{1}{2}\int_{A_{k}}| \nabla u| ^{p(x)}dx. \end{align*} Consequently, from the inequality above and \eqref{2.12b} it follows that \begin{equation} \begin{aligned} \int_{A_{k}}| \nabla u| ^{p(x)}dx &\leq \frac{2M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| A_{k}| ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}C_0}{(p^{+})^{0}} \int_{A_{k}}\varepsilon ^{-p^{0}(x)}dx \\ &\leq \frac{2M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}C_0\varepsilon ^{-(p^{-})^{0}}}{(p^{+})^{^{0}}}| A_{k}| ^{1+\frac{1}{N}-\frac{1}{\alpha ^{-}}}. \end{aligned} \label{2.15} \end{equation} Note that $b(\cdot )\geq 1$. From \eqref{2.12} and \eqref{2.15}, we have \begin{equation} \int_{A_{k}}(u-k)dx\leq \int_{A_{k}}b(x)(u-k)dx=\frac{1}{M} \int_{A_{k}}| \nabla u| ^{p(x)}dx\leq \gamma | A_{k}| ^{1+\frac{1}{N}-\frac{1}{\alpha ^{-}}}, \label{2.16} \end{equation} where \begin{equation} \gamma =\frac{2| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}C_0\varepsilon ^{-(p^{-})^{^{0}}}}{(p^{+})^{0}}. \label{2.17} \end{equation} By the \cite[Lemma 5.1, Chapter 2]{l1}, \eqref{2.16} implies \begin{equation} | u| _{\infty }\leq \gamma (\frac{\alpha ^{-}N}{\alpha ^{-}-N}+1)^{2}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}}. \label{2.18} \end{equation} From \eqref{2.14}, \eqref{2.17} and \eqref{2.18}, we see that \begin{equation} | u| _{\infty }\leq C^{\ast }M^{1/(p^{-}-1)}, \label{2.19} \end{equation} where \begin{equation} C^{\ast }=\frac{(\frac{\alpha ^{-}N}{\alpha ^{-}-N}+1)^{2}(2C_0| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}})^{(p^{-})^{0}}}{ (p^{+})^{0}(p^{-})^{(p^{-})^{0}/p^{-}}}. \label{2.20} \end{equation} When $M<\rho _0$, take \[ \varepsilon =\Big( \frac{p^{-}}{2M| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{ 1}{\alpha ^{-}}}C_0}\Big) ^{1/p^{+}}=( \frac{\rho _0}{M}) ^{1/p^{+}}. \] Note that in this case $\varepsilon >1$. Using similar arguments as above we obtain \[ | u| _{\infty }\leq C_{\ast }M^{1/(p^{+}-1)}, \] where \[ C_{\ast }=\frac{(\frac{\alpha ^{-}N}{\alpha ^{-}-N}+1)^{2}(2C_0| b(\cdot )| _{L^{\alpha ^{-}}(\Omega )}| \Omega | ^{\frac{1}{N}-\frac{1}{\alpha ^{-}}})^{(p^{+})^{0}}}{ (p^{+})^{0}(p^{-})^{(p^{+})^{0}/p^{+}}}. \] The proof is complete. \end{proof} \begin{lemma} \label{lem3.6} Suppose there is a small $\delta >0$ such that $p(x) \equiv p$ (a constant) for any $x\in \Omega $ with $d(x)\leq \delta$ and $N<\alpha ( \cdot ) \in C( \overline{\Omega }) $. Let $M>1$ and $u$ be the unique solution of the problem \begin{equation} \begin{gathered} -\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u)=Mb(x), \quad \text{in }\Omega , \\ u=0, \quad \text{on }\partial \Omega , \end{gathered}\label{3.2} \end{equation} where $0\leq b(\cdot )\in L^{\alpha (\cdot )}(\Omega )$. Then $| \nabla u(\cdot )| \leq CM^{\frac{1}{p^{-}-1}}$ on $\partial \Omega$. \end{lemma} \begin{proof} By Lemma \ref{lem3.5}, we have $u(x)\leq C_{\#}M^{\frac{1}{p^{-}-1}}$ for all $x\in \Omega $. Let $u_2$ be the solution of the following $p$-Laplacian equation (with constant exponent) \begin{gather*} -\operatorname{div}(|\nabla u_2|^{p-2}\nabla u_2)=\varkappa b(x), \quad \text{in }\Omega , \\ u_2=0, \quad \text{on }\partial \Omega , \end{gather*} where $\varkappa $ is a positive parameter. It is easy to see that $u_2\in C^{1,\alpha }(\overline{\Omega })$. Then $\frac{\partial u_2}{\partial \nu }>0$ on $\partial \Omega $, where $\nu $ is the inward unit normal vector. We can also see that $u_2>0$ on $\partial (\Omega \backslash \overline{\Omega _{\varepsilon }})$ when $\varepsilon \in (0,\delta )$ is small enough. Let $\varkappa $ be large enough, we have $u_2\geq 2C_{\#}$ on $\partial (\Omega \backslash \overline{\Omega _{\varepsilon }})$. It means that $u_2M^{\frac{1}{p^{-}-1}}\geq u$ on $\partial \Omega _{\varepsilon }$. Define $u_3=u_2M^{\frac{1}{p^{-}-1}}$. Since $p( x) \equiv p$ for any $x\in \Omega $ with $d(x)\leq \delta $, we have \[ -\operatorname{div}(|\nabla u_3|^{p(x)-2}\nabla u_3)=-\operatorname{div}(|\nabla u_3|^{p-2}\nabla u_3)=M^{\frac{p-1}{p^{-}-1}}\varkappa b(x)\geq Mb(x) \quad\text{in }\Omega _{\varepsilon }. \] Therefore, $u_3=u_2M^{\frac{1}{p^{-}-1}}\geq u$ on $\overline{\Omega _{\varepsilon }}$ and $| \nabla u| \leq | \nabla u_3| \leq CM^{\frac{1}{p^{-}-1}}$ on $\partial \Omega $. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] At first, we construct a supersolution of \eqref{1.1}. Denote $k_2=\psi ( k_1\sigma ) $. Let $\omega _2$ be the solution of the problem \begin{gather*} -\Delta _{p(x)}\omega _2=b( x) k_2^{p^{-}-1-\frac{ \varepsilon }{2}}, \quad \text{in }\Omega \backslash \overline{\Omega _{\sigma }}, \\ \omega _2>0, \quad \text{in }\Omega \backslash \overline{\Omega _{\sigma }},\\ \omega _2=0, \quad \text{on }\partial (\Omega \backslash \overline{\Omega _{\sigma }}). \end{gather*} Define \begin{equation} v_2( x) :=\begin{cases} \psi ( k_1\omega _{b}( x) ) , & x\in \Omega _{\sigma}, \\ \omega _2(x)+\psi ( k_1\sigma ) , & x\in \Omega \backslash \Omega _{\sigma }. \end{cases}\label{aa} \end{equation} For a large enough constant $k_1$, we will prove that $v_2$ is a supersolution of \eqref{1.1} in three steps. \smallskip \noindent\textit{Step 1}. When $k_1$ is large enough, we will check that $v_2$ is a supersolution of \eqref{1.1} in $\Omega _{\sigma }$, namely, \[ \int_{\Omega }| \nabla v_2| ^{p( x) -2}\nabla v_2\nabla \phi\,dx\geq \int_{\Omega }\lambda bg( v_2) \phi\,dx\geq \int_{\Omega }\lambda f( x,v_2) \phi \,dx, \] for any $\phi \in C_0^{\infty }( \Omega _{\sigma }) $ with $ \phi \geq 0$ and $\operatorname{supp}\phi \subset \subset \Omega _{\sigma }$. As in the proof of Lemma \ref{lem3.4}, we only need to prove that \begin{gather} \begin{aligned} &\int_{\Omega }( k_1) ^{p( x) }| \nabla \omega _{b}| ^{p( x) }\phi ( p( x) -1) [ H( v_2) ] ^{p( x) -1}\frac{ 2^{\varepsilon }-1}{p^{-}-1}\frac{H( v_2) }{v_2}dx \\ &\geq \int_{\Omega }( k_1) ^{p( x) -1}| \nabla \omega _{b}| ^{p( x) -1}| \nabla p| [ \phi \psi '( k_1\omega _{b}) ^{p( x) -1}] | \ln k_1H( v_2)|\,dx, \end{aligned} \label{5.2} \\ k_1H(v_2(x))\geq 1,\forall x\in \Omega _{\sigma }, \label{5.3} \\ \frac{k_1^{p^{-}-1}}{[v_2{}(x)]^{p^{-}-1}}\geq 1,\forall x\in \Omega _{\sigma }. \label{5.2b} \end{gather} We can see that \eqref{5.2} is valid, provided \begin{equation} | \nabla \omega _{b}| \frac{2^{\varepsilon }-1}{p^{-}-1} k_1\frac{H( v_2) }{v_2}\geq c_1\frac{2^{\varepsilon }-1}{ p^{-}-1}k_1\frac{H( v_2) }{v_2}\geq | \nabla p\| \ln k_1H( v_2) | \quad \text{in }\Omega _{\sigma }. \label{3.1} \end{equation} According to the assumption on $g$, without loss of generality, we assume that \[ g(s)\geq cs^{-1}\text{ for }s\leq 1, \quad \text{and}\quad g(s)=cs^{\theta }\text{ for }s\geq 1, \] where $\theta =p^{-}-1-\varepsilon$. Thus \begin{gather*} \hat{g}( s) \geq cs^{-p^{-}}\text{ for }s\leq 1, \quad \text{and}\quad \hat{ g}(s)=cs^{\theta +1-p^{-}}\text{ for }s\geq 1, \\ H( s) \geq c_1s^{-\frac{p^{-}}{p^{-}-1}}\text{ for }s\leq 1,\quad \text{and}\quad H(s)=c_2s^{\frac{\theta +1-p^{-}}{p^{-}-1}}\text{ for }s\geq 2, \\ \eta ( s) \leq c_3s^{1+\frac{p^{-}}{p^{-}-1}}\text{ for }s\leq 1,\quad \text{and}\quad c_{4}s^{2-\frac{\theta }{p^{-}-1}}\leq \eta (s)\leq c_{5}s^{2- \frac{\theta }{p^{-}-1}}\text{ for }s\geq 3. \end{gather*} Then $\psi (s)$ satisfies \[ c_{7}s^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\leq \psi (s)\leq c_{8}s^{\frac{1 }{2-\frac{\theta }{p^{-}-1}}}\quad \text{for }s\geq 3. \] Let $s_0\geq 3$ such that $\eta (s_0)\geq 3$. Denote \[ \Omega _{\sigma }^{+}=\{x\in \Omega _{\sigma }: k_1\omega _{b}(x)\geq \eta (s_0)\},\quad \Omega _{\sigma }^{-}=\{x\in \Omega_{\sigma }: k_1\omega _{b}(x)<\eta (s_0)\}. \] Here we note that $v_2=\psi (k_1\omega _{b})$ on $\overline{\Omega_{\sigma }}$. Since $\psi $ is strictly increasing, we have $k_1\omega _{b}\geq \eta (s_0)$ if and only if $v_2=\psi (k_1\omega _{b})\geq \psi (\eta (s_0))=s_0\geq 3$. When $v_2\geq s_0$, we have \begin{gather*} c_{7}(k_1\omega _{b})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\leq v_2\leq c_{8}(k_1\omega _{b})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}} \quad\text{on } \Omega _{\sigma }^{+}, \\ c_{9}(k_1\omega _{b})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{p^{-}-1}}\leq H( v_2) \leq c_{10}(k_1\omega _{b})^{ \frac{1}{2-\frac{\theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{p^{-}-1}} \quad \text{on }\Omega _{\sigma }^{+}, \\ | \nabla \omega _{b}| \frac{2^{\varepsilon }-1}{p^{-}-1} k_1\frac{H( v_2) }{v_2}\geq c_{11}k_1(k_1\omega _{b})^{ \frac{1}{2-\frac{\theta }{p^{-}-1}}(\frac{\theta +1-p^{-}}{p^{-}-1}-1)}= \frac{c_{11}}{\omega _{b}}\geq \frac{c_{11}}{\sigma } \text{ on }\Omega _{\sigma }^{+}, \\ \begin{aligned} | \nabla p\| \ln k_1H( v_2)| &\leq | \nabla p| (\ln k_1+| \ln H( v_2) | )\\ &\leq | \nabla p| (\ln k_1+c_{12}| \ln k_1\omega _{b}| ) \leq c_{13}\ln k_1 \quad \text{on }\Omega _{\sigma }^{+}. \end{aligned} \end{gather*} Denoting $\sigma =\frac{c_{11}}{c_{13}\ln k_1}$, we obtain \[ | \nabla \omega _{b}| \frac{2^{\varepsilon }-1}{p^{-}-1} k_1\frac{H( v_2) }{v_2}\geq c_1\frac{2^{\varepsilon }-1}{ p^{-}-1}k_1\frac{H( v_2) }{v_2}\geq | \nabla p\| \ln k_1H( v_2) | \quad\text{on }\Omega _{\sigma }^{+}. \] Since $\psi $ is strictly increasing, we have $k_1\omega _{b}\leq \eta(s_0)$ if and only if $v_2=\psi (k_1\omega _{b})\leq s_0$. Note that $H( v_2) $ is decreasing. It follows that $H( v_2) \geq H(s_0)$ on $\Omega _{\sigma }^{-}$. Thus \eqref{3.1} is valid when $k_1$ is large enough. Thus \eqref{3.1} is valid, and then \eqref{5.2} is valid. Obviously, \begin{align*} k_1H(v_2(x)) &\geq k_1H(\psi (k_1\sigma )) \\ &\geq k_1c_{9}(k_1\frac{c_{11}}{c_{13}\ln k_1})^{\frac{1}{2-\frac{ \theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{p^{-}-1}} \\ &= c_{9}(k_1)^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}(\frac{c_{11}}{ c_{13}\ln k_1})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{ p^{-}-1}}\to +\infty, \end{align*} for all $x\in \Omega _{\sigma }$ as $k_1\to +\infty$. Thus \eqref{5.3} is valid. Note that $\frac{\theta }{p^{-}-1}<1$. Then by the above computation, \[ v_2\leq c_{8}(k_1\omega _{b})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\leq c_{8}(k_1\sigma )^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\leq k_1 \] as $k_1$ is large enough. Thus \eqref{5.2b} is valid. \smallskip \noindent\textit{Step 2.} We will check that $v_2$ is a supersolution of \eqref{1.1} in $\Omega \backslash \overline{\Omega _{\sigma }}$ when $k_1$ is large enough; i.e., \[ \int_{\Omega }| \nabla v_2| ^{p( x) -2}\nabla v_2\nabla \phi\,dx\geq \int_{\Omega }\lambda bg( v_2) \phi\,dx\geq \int_{\Omega }\lambda f( x,v_2) \phi \,dx, \] for all $\phi \in C_0^{\infty }( \Omega \backslash \overline{\Omega _{\sigma }})$, $\phi \geq 0$. By the definition of $\omega _2$ and Lemmas \ref{lem3.5} and \ref{lem3.6}, we have \[ \omega _2\leq C_1(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}},\quad | \nabla \omega _2| \leq C_2(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p-1}}. \] Since $v_2=\omega _2+\psi ( k_1\sigma ) $ in $\Omega \backslash \overline{\Omega _{\sigma }}$, we have \[ c_{7}(k_1\sigma )^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\leq \psi ( k_1\sigma ) \leq v_2\leq C_1(k_2)^{\frac{p^{-}-1-\frac{ \varepsilon }{2}}{p^{-}-1}}+\psi ( k_1\sigma ) \leq (C_1+1)\psi ( k_1\sigma ) . \] Since $k_1\sigma =\frac{c_{11}k_1}{c_{13}\ln k_1}$ is large enough (as long as $k_1$ is large enough) and $v_2(\cdot )$ is large enough in $\Omega \backslash \overline{\Omega _{\sigma }}$, the assumption (ii) of Theorem \ref{thm1.2} implies that $\frac{g( v_2(x)) }{[v_2(x)]^{p^{-}-1-\frac{ \varepsilon }{2}}}$ is small enough. Therefore, we see that \[ \frac{1}{\lambda }>\frac{g( v_2(x)) }{[v_2(x)]^{p^{-}-1-\frac{ \varepsilon }{2}}}. \] Note that $k_2=\psi ( k_1\sigma ) $. We have \[ v_2(x)^{p^{-}-1-\frac{\varepsilon }{2}}\leq [ C_1(k_2)^{\frac{ p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}}+\psi ( k_1\sigma ) ]^{p^{-}-1-\frac{\varepsilon }{2}}\leq C_3k_2^{p^{-}-1-\frac{\varepsilon }{2}},\quad \forall x\in \Omega \backslash \overline{\Omega _{\sigma }}, \] which implies \[ (k_2)^{p^{-}-1-\frac{\varepsilon }{2}}\geq C_3\lambda (k_2)^{p^{-}-1- \frac{\varepsilon }{2}}\frac{g( v_2(x)) }{[v_2(x)]^{p^{-}-1- \frac{\varepsilon }{2}}}\geq \lambda g( v_2(x)) ,\quad \forall x\in \Omega \backslash \overline{\Omega _{\sigma }}. \] We can see that $v_2$ is a supersolution of \eqref{1.1} in $\Omega \backslash \overline{\Omega _{\sigma }}$; i.e., for any $\phi \in C_0^{\infty }( \Omega \backslash \overline{\Omega _{\sigma }}) $ with $\phi \geq 0$, we have \begin{align*} \int_{\Omega }| \nabla v_2| ^{p( x) -2}\nabla v_2\nabla \phi\,dx &= \int_{\Omega }bk_2^{p^{-}-1-\frac{\varepsilon }{2}}\phi\,dx \\ &\geq \int_{\Omega }\lambda bC_3k_2^{p^{-}-1-\frac{\varepsilon }{2}} \frac{g( v) }{v_2{}^{p^{-}-1-\frac{\varepsilon }{2}}}\phi \,dx\\ &\geq \int_{\Omega }\lambda f( x,v_2) \phi\,dx. \end{align*} \smallskip \textit{Step 3}. When $k_1$ is large enough, we will prove that $v_2$ is a supersolution of \eqref{1.1} in $\Omega $. When $\omega _{b}( x) =\sigma $, it is easy to check that \begin{align*} k_1\psi '( k_1\omega _{b}) &= k_1H(v_2) \\ &\geq k_1c_{9}(k_1\omega _{b})^{\frac{1}{2-\frac{\theta }{p^{-}-1}} \frac{\theta +1-p^{-}}{p^{-}-1}} \\ &= c_{9}(k_1)^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}\sigma ^{\frac{1}{2- \frac{\theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{p^{-}-1}} \\ &= c_{9}(k_1)^{\frac{1}{2-\frac{\theta }{p^{-}-1}}}(\frac{c_{11}}{ c_{13}\ln k_1})^{\frac{1}{2-\frac{\theta }{p^{-}-1}}\frac{\theta +1-p^{-}}{ p^{-}-1}}. \end{align*} Then \[ | \nabla \omega _2| \leq C(k_2)^{\frac{p^{-}-1-\frac{ \varepsilon }{2}}{p^{-}-1}}\leq C(\frac{c_{11}k_1}{c_{13}\ln k_1})^{ \frac{1}{2-\frac{\theta }{p^{-}-1}}\frac{p^{-}-1-\frac{\varepsilon }{2}}{ p^{-}-1}}<| \nabla \psi ( k_1\omega _{b}) | \] as $k_1\to +\infty$. Thus we know that \[ (k_1\psi '( k_1\omega _{b}( x) ) | \nabla \omega _{b}( x) | )^{p(x)-1}-| \nabla \omega _2( x) | ^{p(x)-1}>0, \quad\text{when }\omega _{b}( x) =\sigma . \] Therefore, when $\sigma =c_{11}/(c_{13}\ln k_1)$ and $k_1$ is large enough, similar argument as to the step 3 of the proof of Lemma \ref{lem3.4} implies $v_2$ is a supersolution of \eqref{1.1}. It is easy to see that $\mu \phi $ defined in the proof of Theorem \ref{thm1.1} is a subsolution of \eqref{1.1} and $\mu \phi \leq v_2$ when $\mu $ is small enough. By Lemma \ref{lem2.7}, we can get the existence of a solution to \eqref{1.1}. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.3}] At first, similar to the proof of Theorem \ref{thm1.2}, we will prove that $v_2$ defined by \eqref{aa} is also a supersolution of \eqref{1.1} for a large enough constant $k_1$. Similar to the proof of Theorem \ref{thm1.2}, we consider the solution $\omega _2$ of the problem \begin{equation} \begin{gathered} -\Delta _{p(x)}\omega _2=b( x) k_2^{p^{-}-1-\frac{ \varepsilon }{2}}, \quad \text{in }\Omega \backslash \overline{\Omega _{\sigma }} , \\ \omega _2( x) >0, \quad \text{in }\Omega \backslash \overline{ \Omega _{\sigma }}, \\ \omega _2( x) =0, \quad \text{on }\partial (\Omega \backslash \overline{\Omega _{\sigma }}), \end{gathered} \label{3.11} \end{equation} where $k_2=\psi (k_1\sigma )$. We have \[ \omega _2\leq C_1(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}}. \] Next we only need to prove that \[ | \nabla \omega _2| \leq C_2(k_2)^{\frac{p^{-}-1- \frac{\varepsilon }{2}}{p^{-}-1}}. \] Now we consider $(\gamma k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{ p^{-}-1}}\omega _{b}$, where $\gamma \geq 1$ is a constant. Here we note that $\nabla \omega _{b}\cdot \nu =| \nabla \omega _{b}| $ on $\partial (\Omega \backslash \overline{\Omega _{\sigma }})$ and $\nabla p\cdot \nu <0$ on $\partial (\Omega \backslash \overline{\Omega _{\sigma }})$, where $\nu $ is the inward unit normal vector on $\partial (\Omega \backslash \overline{\Omega _{\sigma }})$. There exists a small enough positive constant $\delta >0$ such that $\nabla \omega _{b}\nabla p<0$ in $(\Omega \backslash \overline{\Omega _{\sigma }})_{\delta }^{\#}:=\{x\in \Omega \backslash \overline{\Omega _{\sigma }}: d(x,\partial (\Omega \backslash \overline{\Omega _{\sigma }}))<\delta \}$. By computations it follows that \begin{align*} &-\Delta _{p(x)}(\gamma k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{ p^{-}-1}}\omega _{b} \\ &= (\gamma k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2} }{p^{-}-1}(p(x)-1)}(-\Delta _{p(x)}\omega _{b}-\frac{p^{-}-1-\frac{ \varepsilon }{2}}{p^{-}-1}| \nabla \omega _{b}| ^{p(x)-2}\nabla \omega _{b}\nabla p\ln \gamma k_2) \\ &\geq (\gamma k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1} (p(x)-1)}(-\Delta _{p(x)}\omega _{b}) \\ &\geq b(\gamma k_2)^{p^{-}-1-\frac{\varepsilon }{2}}\text{ in }(\Omega \backslash \overline{\Omega _{\sigma }})_{\delta }^{\#}. \end{align*} Since $\omega _{b}$ is positive and continuous, there exists a large enough positive $\gamma $ such that $\gamma ^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{ p^{-}-1}}\omega _{b}>2C_1$ for $d(x,\partial (\Omega \backslash \overline{ \Omega _{\sigma }}))=\delta $. Therefore $(\gamma k_2)^{\frac{p^{-}-1- \frac{\varepsilon }{2}}{p^{-}-1}}\omega _{b}$ is a supersolution of \eqref{3.11} in $(\Omega \backslash \overline{\Omega _{\sigma }})_{\delta }^{\#}$. By the comparison principle, we have $(\gamma k_2)^{\frac{p^{-}-1-\frac{ \varepsilon }{2}}{p^{-}-1}}\omega _{b}\geq \omega _2$ in $(\Omega \backslash \overline{\Omega _{\sigma }})_{\delta }^{\#}$, and then \[ | \nabla \omega _2| \leq | (\gamma k_2)^{ \frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}}\nabla \omega _{b}| \leq C_2(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}} \quad \text{on }\partial (\Omega \backslash \overline{\Omega _{\sigma }}). \] Note that \[ \max_{x\in \overline{\Omega \backslash \Omega _{\sigma }}} \omega _2(x)\leq C_3(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{ p^{-}-1}} \] and $k_2=\psi (k_1\sigma )$. Since $v_2=\omega _2+\psi (k_1\sigma )$ in $\Omega \backslash \overline{\Omega _{\sigma }}$, we have \begin{align*} c_{7}(k_1\sigma )^{\frac{1}{2-\frac{\theta }{p^{-}-1}}} &\leq \psi (k_1\sigma ) \leq v_2\\ &\leq \max_{x\in \overline{\Omega \backslash \Omega _{\sigma }}} \omega _2(x)+\psi (k_1\sigma )\\ &\leq C_3(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}}+\psi (k_1\sigma ) \leq C_{4}\psi (k_1\sigma ). \end{align*} As in the proof of Theorem \ref{thm1.2}, we can see that the $v_2$ defined in the proof of Theorem \ref{thm1.2} is a supersolution of \eqref{1.1}. It is easy to see that $\mu \phi $ defined in the proof of Theorem \ref{thm1.1} is a subsolution of \eqref{1.1}, and $\mu \phi \leq v_2$ when $\mu $ is small enough. By Lemma \ref{lem2.7}, we obtain the existence of solution of \eqref{1.1}. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.4}] The proof is similar to that of Theorem \ref{thm1.2}. We will prove that $v_2$ defined in \eqref{aa} is a supersolution of \eqref{1.1} for a large enough constant $k_1$. Since \eqref{1.1} is radial, we may assume the both solutions $\omega _{b}( \cdot ) $ and $\omega _2( \cdot ) $ are radial. We only need to prove that $v_2$ defined in \eqref{aa} is also a supersolution of \eqref{1.1} for a large enough constant $k_1$. Since $\omega _2$ is radial, it is easy to see that \[ \omega _2\leq C_1(k_2)^{\frac{p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}} , | \nabla \omega _2| \leq C_2(k_2)^{\frac{ p^{-}-1-\frac{\varepsilon }{2}}{p^{-}-1}}. \] Similar to the proof of Theorem \ref{thm1.2}, we can see that the $v_2$ defined in \eqref{aa} is a supersolution of \eqref{1.1}. The proof is complete. \end{proof} \subsection*{Acknowledgments} The authors would like to thank Professor Julio G. Dix for his suggestions, and the anonymous referees for their valuable comments. 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