\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 25, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/25\hfil $p(x)$-Laplacian problems with critical exponents] {Existence of solutions for discontinuous $p(x)$-Laplacian problems with critical exponents} \author[X. Shang, Z. Wang \hfil EJDE-2012/25\hfilneg] {Xudong Shang, Zhigang Wang} % in alphabetical order \address{ School of Mathematics, Nanjing Normal University Taizhou College, 225300, Jiangsu, China} \email[Xudong Shang ]{xudong-shang@163.com} \email[Zhigang Wang]{wzg19.scut@163.com} \thanks{Submitted November 7, 2011. Published February 7, 2012.} \subjclass[2000]{35J92, 35J70, 35R70} \keywords{$p(x)$-Laplacian problem; critical Sobolev exponents; \hfill\break\indent discontinuous nonlinearities} \begin{abstract} In this article, we study the existence of solutions to the problem \begin{gather*} -\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u) =\lambda |u|^{p^*(x)-2}u + f(u)\quad x \in \Omega ,\\ u = 0 \quad x \in \partial\Omega, \end{gather*} where $\Omega$ is a smooth bounded domain in ${\mathbb{R}}^{N}$, $p(x)$ is a continuous function with $1 0$. The function $f(u)$ can have discontinuities, so that functionals associated with \eqref{e1.1} may not be differentiable, and standard variational techniques can not be applied. There are many publications for the case when $p(x)$ is a constant function; see for example \cite{AB,ABG,Ba,BT,S}. For the existence of solutions for $p(x)$-Laplacian problems we refer the reader to \cite{Bo,CF,D,FH,FZ,Mi}. The existence of solutions for $p(x)$-Laplacian problems with critical growth is relatively new. In 2012, Bonder and Silva \cite{BS} extended the concentration-compactness principle of Lions to the variable exponent spaces and proved the existence of solutions to the problem \begin{gather*} -\Delta_{p(x)}u = |u|^{q(x)-2}u + \lambda(x)|u|^{r(x)-2}u\quad x \in \Omega ,\\ u = 0 \quad x \in \partial\Omega. \end{gather*} Where $\Omega$ is a smooth bounded domain in ${\mathbb{R}}^{N}$, with $q(x)\leq p^*(x)$ and the set $\{q(x)= p^*(x)\}\neq\emptyset$, we can find a similar result in \cite{F}. Fu \cite{FZh} studied the existence of solutions for $p(x)$-Laplacian equationd involving the critical exponent and obtained a sequence of radially symmetric solutions. In the present paper, we study the discontinuous $p(x)$-Laplacian problems with critical growth for \eqref{e1.1}. To handle the gaps at the discontinuity points, our approach uses nonsmooth critical point theory for locally Lipschitz functionals, we obtain some general results for the simple case when $f$ has only one point of discontinuity. Because $f$ is discontinuous, we say that a function $u \in W_0^{1,p(x)}(\Omega)$ is a solution of the multivalued problem associated to \eqref{e1.1} if $u$ satisfies $-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in \widehat{f}(u) \quad \text{a.e. in } \Omega ,$ where $\widehat{f}(u)$ is the multivalued function $\widehat{f}(u) = [\underline{f}(u),\overline{f}(u)]$ with $\underline{f}(t) =\liminf_{s\to t}f(s),\quad \overline{f}(t) = \limsup_{s\to t}f(s).$ In this article, we assume $f: {{\mathbb{R}}} \to {\mathbb{R}}$ is a measurable function satisfying: \begin{itemize} \item[(F1)] $f(t) = 0$ if $t\leq0$ and for all $t \in {\mathbb{R}}$, there exist the limits: $f(t+0)=\lim_{\delta\to 0^{+}}f(t+\delta);\quad f(t-0)=\lim_{\delta\to 0^{+}}f(t-\delta).$ \item[(F2)] there exist $C_1,C_2> 0$ such that $|f(t)|\leq C_1 + C_2|t|^{q(x)-1}$, where $q(x)\in C(\overline{\Omega})$ such that $p(x) < q(x) < p^*(x)$. \item[(F3)] $f(t) = o({|t|^{p(x)-1}})$ as $t\to 0$. \item[(F4)] $\underline{f}(t)t \geq q_{-}F(t)> 0$, for all $t\in {\mathbb{R}}\backslash \{0\}$, where $F(t)=\int_0^tf(s)ds$. \end{itemize} Note that by hypothesis (F1), $\overline{f}(u) = \max\{f(u-0), f(u+0) \},\quad \underline{f}(u) = \min\{ f(u-0), f(u+0)\}.$ \begin{theorem} \label{thm1.1} Suppose $f$ satisfies {\rm (F1)-(F4)}. Then there exists $\lambda_0 > 0$ such that \eqref{e1.1} has a nontrivial solution for every $\lambda\in(0,\lambda_0)$. \end{theorem} One of the main motivations is to consider the particular case associated with \eqref{e1.1}, $$\label{e1.2} \begin{gathered} -\Delta_{p(x)}u= \lambda |u|^{p^*(x)-2}u + bh(u-a)|u|^{q(x)-2}u\quad x \in \Omega , \\ u = 0 \quad x \in \partial\Omega, \end{gathered}$$ where $h(t)=0$ if $t \leq 0$ and $h(t)=1$ if $t > 0$, $a$ and $b$ are positive real parameters, $p(x) 0$, there exists $\lambda_0 > 0$ such that for every $\lambda\in(0,\lambda_0)$, Equation \eqref{e1.2} has a nontrivial solution satisfying $\operatorname{meas}\{x\in\Omega: u(x)>a\}>0$. \end{theorem} The rest of this article is organized as follows: In section 2 we introduce some necessary preliminary knowledge; in section 3 contains the proof of our main results. \section{Preliminaries} First, we recall some definitions and properties of generalized gradient of locally Lipschitz functionals, which will be used later. Let $X$ be a Banach space, $X^*$ be its topological dual and $\langle\cdot,\cdot\rangle$ be the duality. A functional $I: X\to {\mathbb{R}}$ is said to be locally Lipschitz if for every $u\in X$ there exists a neighborhood $U$ of $u$ and a constant $K>0$ depending on $U$ such that $|I(u)-I(v)|\leq K\|u-v\|,\quad \forall u,v\in U.$ For a locally Lipschitz functional $I$, we define the generalized directional derivative at $u\in X$ in the direction $v\in X$ by $I^0(u;v)=\limsup_{h\to 0,\, \delta\downarrow 0} \frac{I(u+h+\delta v)-I(u+h)}{\delta}.$ It is easy to show that $I^0(u;v)$ is subadditive and positively homogeneus. The generalized gradient of $I$ at $u$ is the set $\partial I(u)=\{w\in X^*:I^0(u;v) \geq \langle w,v\rangle, \forall v\in X \}.$ Then, for each $v \in X$, $I^0(u;v) = \max \{ \langle \omega, v \rangle: \omega \in \partial I(u) \}$. A point $u \in X$ is a critical point of $I$ if $0 \in \partial I(u)$. It is easy to see that if $u \in X$ is a local minimum or maximum, then $0 \in \partial I(u)$. Next, we recall some definitions and basic properties of the generalized Lebesgue-Sobolev spaces $L^{p(x)}(\Omega)$ and $W_0^{1,p(x)}(\Omega)$, where $\Omega \subset {\mathbb{R}}^{N}$ is an arbitrary domain with smooth boundary. Set \begin{gather*} C_{+}(\overline{\Omega})= \{p(x)\in C(\overline{\Omega}): p(x)> 1, \forall x\in\overline{\Omega}\},\\ p_{+} = \max_{x\in\Omega}p(x),\quad p_{-} = \min_{x\in\Omega}p(x). \end{gather*} For any $p(x)\in C_{+}(\overline{\Omega})$, we define the variable exponent Lebesgue space $L^{p(x)}(\Omega) = \{u\in M(\Omega):\int_{\Omega}|u(x)|^{p(x)}dx< \infty\},$ with the norm $|u|_{p(x)}=\inf\{\mu > 0 : \int_{\Omega}|\frac{u}{\mu}|^{p(x)}dx\leq 1\},$ where $M(\Omega)$ is the set of all measurable real functions defined on $\Omega$. Define the space $W_0^{1,p(x)}(\Omega)=\{u\in L^{p(x)}(\Omega): |\nabla u|\in L^{p(x)}(\Omega)\},$ with the norm $\|u\|= |u|_{p(x)}+|\nabla u|_{p(x)}.$ \begin{proposition}[\cite{FZH,KR}] \label{prop2.1} There is a constant $C > 0$ such that for all $u\in W_0^{1,p(x)}(\Omega)$, $|u|_{p(x)}\leq C|\nabla u|_{p(x)}.$ So $|\nabla u|_{p(x)}$ and $\|u\|$ are equivalent norms in $W_0^{1,p(x)}(\Omega)$. Hence we will use the norm $\|u\| = |\nabla u|_{p(x)}$ for all $u\in W_0^{1,p(x)}(\Omega)$. \end{proposition} \begin{proposition}[\cite{FZH,KR}] \label{prop2.2} Set $\rho(u)=\int_{\Omega}|u|^{p(x)}dx$. For $u,u_n\in L^{p(x)}(\Omega)$, we have: \begin{itemize} \item[(1)] $|u|_{p(x)}<1$ $(=1;>1)$ $\Leftrightarrow \rho(u)<1$ $(=1;>1)$. \item[(2)] If $|u|_{p(x)} > 1$, then $|u|^{p_{-}}_{p(x)}\leq \rho(u) \leq |u|^{p_{+}}_{p(x)}$. \item[(3)] If $|u|_{p(x)}< 1$, then $|u|^{p_{+}}_{p(x)}\leq \rho(u) \leq |u|^{p_{-}}_{p(x)}$. \item[(4)] $\lim_{n\to\infty} u_n=u \Leftrightarrow \lim_{n\to\infty} \rho (u_n-u) = 0$. \item[(5)] $\lim_{n\to\infty} |u_n|_{p(x)} =\infty \Leftrightarrow \lim_{n\to\infty} \rho (u_n) = \infty$. \end{itemize} \end{proposition} \begin{proposition}[\cite{FZH}] \label{prop2.3} If $q(x)\in C_{+}(\Omega)$ and $q(x) < p^*(x)$ for any $x\in\Omega$, the imbedding $W^{1,p(x)}(\Omega)\to L^{q(x)}(\Omega)$ is compact. \end{proposition} \begin{proposition}[\cite{KR}] \label{prop2.4} The conjugate space of $L^{p(x)}(\Omega)$ is $L^{q(x)}(\Omega)$, where $\frac{1}{p(x)}+\frac{1}{q(x)} =1$. For any $u \in L^{p(x)}(\Omega)$ and $v \in L^{q(x)}(\Omega)$, $\int_{\Omega}|uv|dx\leq (\frac{1}{p_{-}}+ \frac{1}{q_{-}})|u|_{p(x)}|v|_{q(x)}.$ \end{proposition} \begin{proposition}[\cite{FH}] \label{prop2.5} If $|u|^{q(x)}\in L^{\frac{s(x)}{q(x)}}(\Omega)$, where $q(x),s(x) \in L^{\infty}_{+}(\Omega)$, $q(x)\leq s(x)$, then $u\in L^{s(x)}(\Omega)$ and there is a number $\overline{q}\in [q_{-},q_{+}]$ such that $||u|^{q(x)}|_{\frac{s(x)}{q(x)}} = (|u|_{s(x)})^{\overline{q}}$. \end{proposition} Let $I_{\lambda}(u): W_0^{1,p(x)}(\Omega) \to {\mathbb{R}}$ be the energy functional defined as $$I_{\lambda}(u) = \int_{\Omega}\frac{1}{p(x)}|\nabla u|^{p(x)}dx - \lambda\int_{\Omega}\frac{1}{p^*(x)}|u|^{p^*(x)}dx - \int_{\Omega}F(u)dx,$$ denote $\Phi(u)= \int_{\Omega}F(u)dx$. We say that $I_{\lambda}(u)$ satisfies the nonsmooth $(PS)_{c}$ condition, if any sequence $\{u_n\}\subseteq X$ such that $I_{\lambda}(u_n)\to c$ and $m(u_n) = \min \{ \|w\|_{X^*}: w\in\partial I_{\lambda}(u_n)\} \to 0$, as $n\to\infty$, possesses a convergent subsequence. To prove our main results, we use the generalizations of the mountain pass theorem \cite{Ch}. \begin{theorem} \label{thm2.1} Let $X$ be a reflexive Banach space, $I: X\to {\mathbb{R}}$ is locally Lipschitz functional which satisfies the nonsmooth $(PS)_{c}$ condition, $I(0)=0$ and there are $\rho, r > 0$ and $e\in X$ with $\|e\| > r$, such that $I(u)\geq \beta \quad \text{if } \|u\| = r \quad \text{and} \quad I(e)\leq 0.$ Then there exists $u \in X$ such that $0 \in \partial I(u)$ and $I(u)=c$. Where \begin{gather*} c = \inf_{\gamma\in \Gamma}\max_{0\leq t\leq 1}I(\gamma(t)),\\ \Gamma = \{\gamma \in C([0,1], X)|\gamma(0)=0, \gamma(1) = e \}. \end{gather*} \end{theorem} Recall the concentration-compactness principle for variable exponent spaces. This will be the keystone that enable us to verify that $I_{\lambda}$ satisfies the nonsmooth $(PS)_{c}$ condition. \begin{proposition}[\cite{BS}] \label{prop2.6} Let $\{u_n\}$ converge weakly to $u$ in $W_0^{1,p(x)}(\Omega)$ such that $|u_n|^{p^*(x)}$ and $|\nabla u_n|^{p(x)}$ converge weakly to nonnegative measures $\nu$ and $\mu$ on ${\mathbb{R}}^{N}$ respectively. Then, for some countable set $J$, we have: \begin{itemize} \item[(i)] $\nu = |u|^{p^*(x)} + \sum_{j\in J}\nu_{j}\delta_{x_{j}}$, \item[(ii)] $\mu \geq |\nabla u|^{p(x)} + \sum_{j\in J}\mu_{j}\delta_{x_{j}}$, \item[(iii)] $S\nu_{j}^{\frac{1}{p^*(x_{j})}} \leq \mu_{j}^{\frac{1}{p(x_{j})}}$, \end{itemize} where $x_{j}\in \Omega$, $\delta_{x_{j}}$ is the Dirac measure at $x_{j}$, $\nu_{j}$ and $\mu_{j}$ are constants and $S$ is the best constant in the Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely $$S = \inf \big\{\frac{\|u\|_{1,p(x)}}{|u|_{p^*(x)}}: u\in W_0^{1,p(x)}(\Omega), u\not=0\big\}.$$ \end{proposition} \section{Proof of main results} In this section, we denote by $u_n \rightharpoonup u$ the weak convergence of a sequence $u_n$ to $u$ in $W_0^{1,p(x)}(\Omega)$, and $o(1)$ denote a real vanishing sequence, $C$ and $C_{i}, i= 1, 2, \dots$ are positive constants, $|A|$ is the Lebesgue measure of $A$ and $p'(x)$ as the conjugate function of $p(x)$. $u \in W_0^{1,p(x)}(\Omega)$ is called a solution of \eqref{e1.1} if $u$ is a critical point of $I_{\lambda}(u)$ and satisfies $-\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u) - \lambda |u|^{p^*(x)-2}u \in [\underline{f}(u),\overline{f}(u)]\quad \text{a.e. } {x\in\Omega }.$ \begin{lemma} \label{lem3.1} The function $\Phi(u)$ is locally Lipschitz on $W_0^{1,p(x)}(\Omega)$. \end{lemma} \begin{proof} By (F2), Proposition \ref{prop2.4} and \ref{prop2.5}, for all $u,v \in W_0^{1,p(x)}(\Omega)$, \begin{align*} |\Phi(u)- \Phi(v)|& \leq \int_{\Omega}\big|\int^{v}_u|f(t)|dt\big|dx \\ &\leq \int_{\Omega}\big|\int^{v}_u|C_1 + C_2|t|^{q(x)-1}|dt\big|dx \\ &\leq (|C_1|_{\frac{q(x)}{q(x)-1}} + C_3|u|^{\overline{q}-1}_{q(x)} + C_3|v|^{\overline{q}-1}_{q(x)})|u-v|_{q(x)}. \end{align*} From Proposition \ref{prop2.3}, we obtain that there is a neighborhood $U \subset W_0^{1,p(x)}(\Omega)$ of $u$ such that $|\Phi(u)- \Phi(v)| \leq K\|u-v\|,$ where $K>0$ depends on $\max \{ \|u\|, \|v\|\}$. So, $\Phi(u)$ is locally Lipschitz in $W_0^{1,p(x)}(\Omega)$. The proof is complete. \end{proof} From Lemma \ref{lem3.1}, by Chang's results we have that $I_{\lambda}(u)$ is locally Lipschitz and $\omega \in \partial I_{\lambda}(u)$ if and only if there is $\overline{\omega}\in W^{-1,p^{'}(x)}(\Omega)$ such that for all $\varphi\in W_0^{1,p(x)}(\Omega)$, $$\label{e3.1} \langle \omega, \varphi\rangle = \int_{\Omega}|\nabla u|^{p(x)-2}\nabla u\nabla \varphi dx -\lambda\int_{\Omega}|u|^{p^*(x)-2}u\varphi dx - \int_{\Omega}\overline{\omega}\varphi dx,$$ and $$\overline{\omega}(x)\in [\underline{f}(u(x)),\overline{f}(u(x))]\quad \text{a.e. } {x\in\Omega }.$$ \begin{lemma} \label{lem3.2} Assume {\rm (F1), (F2)}. Let $\{u_n\}$ be a bounded sequence in $W_0^{1,p(x)}(\Omega)$ such that $I_{\lambda}(u_n) \to c$ and $m(u_n)\to 0$. Then there exists a subsequence (denoted again by $u_n$) and some $u \in W_0^{1,p(x)}(\Omega)$, such that $|\nabla u_n|^{p(x)-2}\nabla u_n \rightharpoonup |\nabla u|^{p(x)-2}\nabla u \quad \text{weakly in } [L^{\frac{p(x)}{p(x)-1}}(\Omega)]^{N}.$ \end{lemma} \begin{proof} The proof is similar to that of \cite[Theorem 1]{Zh}. Because $\{u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$, there exist a subsequence and $u \in W_0^{1,p(x)}(\Omega)$ such that $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$ and $u_n \to u$ in $L^{p(x)}(\Omega)$ as $n\to \infty$. We claim that the set $J$ given by Proposition \ref{prop2.6} is finite. Choose a function $\varphi(x)\in C_0^{\infty}({\mathbb{R}}^{N})$ such that $0 \leq \varphi(x)\leq 1$, $\varphi(x) \equiv 1$ on $B(0,1)$ and $\varphi(x) \equiv 0$ on ${\mathbb{R}}^{N}\setminus B(0,2)$. Let $\varphi_{j,\varepsilon}(x)= \varphi (\frac{x-x_{j}}{\varepsilon})$, for any $x\in {\mathbb{R}}^{N}$, $\epsilon > 0$ and $j\in J$. It is clear that $\{\varphi_{j,\varepsilon}u_n\} \subset W_0^{1,p(x)}(\Omega)$ for any $j\in J$, and is bounded in $W_0^{1,p(x)}(\Omega)$. Take $\varphi = \varphi_{j,\varepsilon}u_n$ in $\langle \omega_n, \varphi\rangle$, we obtain $$\label{e3.3} \begin{split} &\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u_n\nabla \varphi_{j,\varepsilon}dx + \int_{\Omega}|\nabla u_n|^{p(x)}\varphi_{j,\varepsilon}dx \\ & -\lambda\int_{\Omega}| u_n|^{p^*(x)}\varphi_{j,\varepsilon}dx -\int_{\Omega}\overline{\omega}_n\varphi_{j,\varepsilon}u_ndx = o(1). \end{split}$$ Taking into account that $\omega_n \in \partial I_{\lambda}(u_n)$, by (F2) and $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$, we infer that $\overline{\omega}_n$ is bounded in $W^{-1,p^{'}(x)}(\Omega)$, and so there exists $\overline{\omega}_0 \in W^{-1,p^{'}(x)}(\Omega)$ such that $$\label{e3.4} \overline{\omega}_n\rightharpoonup \overline{\omega}_0 \quad \text{in } W^{-1,p^{'}(x)}(\Omega)\quad \text{and}\quad \overline{\omega}_0 \in [\underline{f}(u),\overline{f}(u)].$$ Let $n\to \infty$ in \eqref{e3.3}, by Proposition \ref{prop2.6}, we have $$\label{e3.5} \begin{split} &\lim_{n\to \infty} \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n \cdot u_n\nabla \varphi_{j,\varepsilon}dx \\ &=\lambda\int_{\Omega}\varphi_{j,\varepsilon}d\nu - \int_{\Omega}\varphi_{j,\varepsilon}d\mu + \int_{\Omega}\overline{\omega}_0\varphi_{j,\varepsilon}u\,dx. \end{split}$$ By H\"older inequality it is easy to check that \begin{align*} 0&\leq |\lim_{n\to\infty}\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla \varphi_{j,\varepsilon}\cdot u_n\,dx| \\ &\leq \Big(\int_{\Omega}|\nabla u_n|^{p_{+}}dx\Big)^{\frac{p_{+}-1}{p_{+}}} \Big(\int_{\Omega}| u_n|^{p_{+}}|\nabla\varphi_{j,\varepsilon}|^{p_{+}}dx \Big)^{\frac{1}{p_{+}}}\\ &\quad + \Big(\int_{\Omega}|\nabla u_n|^{p_{-}}dx\Big)^{\frac{p_{-}-1}{p_{-}}} \Big(\int_{\Omega}| u_n|^{p_{-}}|\nabla\varphi_{j,\varepsilon}|^{p_{-}}dx \Big)^{\frac{1}{p_{-}}}\\ &\leq C_4(\int_{\Omega}|\nabla u_n|^{p_{+}}dx)^{\frac{p_{+}-1}{p_{+}}} \Big(\int_{B(x_{j},2\varepsilon)}| u_n|^{(p_{+})^*}dx\Big)^{\frac{1}{(p_{+})^*}}\\ &\quad + C_5\Big(\int_{\Omega}|\nabla u_n|^{p_{-}}dx\Big)^{\frac{p_{-}-1}{p_{-}}} \Big(\int_{B(x_{j},2\varepsilon)}| u_n|^{(p_{-})^*}dx\Big)^{\frac{1}{(p_{-})^*}} &\to 0, \quad \text{as }\epsilon \to 0. \end{align*} From \eqref{e3.5}, as $\epsilon \to 0$, we obtain $\lambda \nu_{j} = \mu_{j}$. From Proposition \ref{prop2.6}, we conclude that $$\label{e3.7} \nu_{j} = 0 \quad \text{or}\quad \nu_{j} \geq S^{N} \max\{\lambda^{-\frac{N}{p_{+}}},\lambda^{-\frac{N}{p_{-}}}\}.$$ It implies that $J$ is a finite set. Without loss of generality, let $J = \{1,2,\dots,m\}$. For any $\delta > 0$, we denote $\Omega_{\delta}= \{x\in \Omega|\text{dist}(x,x_{j})> \delta\}$. Choose $R$ large enough such that $\overline{\Omega} \subset \{x\in {\mathbb{R}}^{N}| |x|< R \}$, $\psi (x) \in C^{\infty}({\mathbb{R}}^{N})$, $0 \leq \psi (x) \leq 1$, $\psi (x)\equiv 0$ on $B(0,2R)$ and $\psi (x)\equiv 1$ on ${\mathbb{R}}^{N}\setminus B(0,3R)$. Take $\epsilon > 0$ small enough such that $B(x_{i},\epsilon) \cap B(x_{j},\epsilon) =\emptyset$, $\forall i,j \in J$, $i \neq j$ and $\cup_{j=1}^{m}B(x_{j},\epsilon) \subset B(0,2R)$. We take $\psi_{\epsilon}(x) = 1 - \sum_{j=1}^{m}\varphi_{j,\varepsilon} - \psi (x), \quad x\in {\mathbb{R}}^{N}.$ Then $\psi_{\epsilon}(x) \in C^{\infty}({\mathbb{R}}^{N})$, $\operatorname{supp}\psi_{\epsilon} \subset B(0,3R)$ with $\psi_{\epsilon}(x) = 0$ on $\cup_{j=1}^{m}B(x_{j},\epsilon/2)$ and $\psi_{\epsilon}(x) = 1$ on $({\mathbb{R}}^{N}\setminus B(x_{j},\epsilon))\cap B(0,2R)$. As $\{\psi_{\epsilon}u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$, let $\varphi = \psi_{\epsilon}u_n$ in $\langle \omega_n, \varphi\rangle$, we obtain \begin{align*} %3.8 &\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u_n\nabla \psi_{\epsilon}dx + \int_{\Omega}|\nabla u_n|^{p(x)}\psi_{\epsilon}dx \\ &-\lambda\int_{\Omega}| u_n|^{p^*(x)}\psi_{\epsilon}dx -\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} u_ndx = o (1). \end{align*} By \eqref{e3.4} and $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$, we can easily obtain $\lim_{n\to \infty}\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} u_ndx = \int_{\Omega}\overline{\omega}_0\psi_{\epsilon} u dx.$ Since $\psi_{\epsilon}(x) = 0$ on $\cup_{j=1}^{m}B(x_{j},\frac{\epsilon}{2})$ and $\nu = |u|^{p^*(x)} + \sum_{j\in J}\nu_{j}\delta_{x_{j}}$, we obtain $\lim_{n\to \infty}\int_{\Omega}| u_n|^{p^*(x)}\psi_{\epsilon}dx = \int_{\Omega}\psi_{\epsilon}d\nu = \int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx.$ Hence $$\label{e3.9} \begin{split} \lim_{n\to \infty}\int_{\Omega}|\nabla u_n|^{p(x)}\psi_{\epsilon}dx & = \lim_{n\to \infty}(-\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n \cdot u_n\nabla \psi_{\epsilon}dx) \\ &\quad +\lambda\int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx +\int_{\Omega}\overline{\omega}_0\psi_{\epsilon} udx. \end{split}$$ In the same way, taking $\varphi = \psi_{\epsilon}u$ in $\langle \omega_n, \varphi\rangle$, we obtain \begin{align*} &\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u\nabla \psi_{\epsilon}dx + \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla u\cdot\psi_{\epsilon}dx\\ & -\lambda\int_{\Omega}| u_n|^{p^*(x)-2}u_nu\psi_{\epsilon}dx -\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} udx = o(1). \end{align*} Thus $$\label{e3.10} \begin{split} &\lim_{n\to \infty}\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla u \cdot\psi_{\epsilon}dx \\ &= \lim_{n\to \infty}(-\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u\nabla \psi_{\epsilon}dx) +\lambda\int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx +\int_{\Omega}\overline{\omega}_0\psi_{\epsilon} udx. \end{split}$$ So, from \eqref{e3.9} and \eqref{e3.10}, as $n\to \infty$, we have \begin{align*} 0&\leq \int_{\Omega_{\delta}}(|\nabla u_n|^{p(x)-2}\nabla u_n - |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx \\ & \leq \int_{\Omega}\psi_{\epsilon}(|\nabla u_n|^{p(x)-2}\nabla u_n - |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx\\ &= \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla \psi_{\epsilon}\cdot(u-u_n)dx\\ &\quad + \int_{\Omega}\psi_{\epsilon}|\nabla u|^{p(x)-2}\nabla u\cdot(\nabla u- \nabla u_n)dx + o(1) \\ &\leq \|\nabla\psi_{\epsilon}\|_{\infty}\cdot||\nabla u_n|^{p(x)-1}|_{p^{'}(x)} \cdot|u-u_n|_{p(x)} \\ &\quad + \int_{\Omega}\psi_{\epsilon}|\nabla u|^{p(x)-2}\nabla u\cdot(\nabla u_n- \nabla u)dx + o(1). \end{align*} Thus we have $$\label{e3.11} \lim_{n\to \infty}\int_{\Omega_{\delta}}(|\nabla u_n|^{p(x)-2}\nabla u_n - |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx = 0.$$ Denote $g_n(x)=(|\nabla u_n|^{p(x)-2}\nabla u_n- |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u),$ then $g_n(x) \geq 0$, and by \eqref{e3.11}, $g_n(x) \to 0$ a.e. on $\Omega_{\delta}$. Let $E$ be a compact subset of $\Omega_{\delta}$, suppose $g_n(x) \to 0$ a.e. on $E$. If $\nabla u_n$ were not convergence to $\nabla u$ everywhere on $E$, there would at least exist $x_0\in E$ such that $\lim_{n\to \infty} \nabla u_n(x_0) \neq \nabla u(x_0).$ Then \begin{align*} |\nabla u_n(x_0)|^{p(x_0)} &= |\nabla u_n(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0) \\ & + |\nabla u(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0) - |\nabla u(x_0)|^{p(x_0)} + g_n(x_0). \end{align*} By the interpolation inequality, \begin{align*} ||\nabla u_n(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0)| & \leq |\nabla u_n(x_0)|^{p(x_0)-1}\cdot|\nabla u(x_0)| \\ & \leq \epsilon_1|\nabla u_n(x_0)|^{p(x_0)} + C_{\epsilon_1}|\nabla u(x_0)|^{p(x_0)}, \end{align*} and \begin{align*} ||\nabla u(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0)| & \leq |\nabla u(x_0)|^{p(x_0)-1}\cdot|\nabla u_n(x_0)| \\ & \leq \epsilon_2|\nabla u(x_0)|^{p(x_0)} + C_{\epsilon_2}|\nabla u_n(x_0)|^{p(x_0)}. \end{align*} We choose $\epsilon_1,\epsilon_2$ properly, because $g_n(x_0)$ is bounded, then $|\nabla u(x_0)|\leq C$. Let $\nabla u(x_0) = \eta$, so we can assume $\nabla u_n(x_0)\to \overline{\eta} \neq \eta$. Thus $g_n(x_0) \to (|\overline{\eta}|^{p(x_0)-2}\overline{\eta} - |\eta|^{p(x_0)-2}\eta)(\overline{\eta} - \eta) > 0.$ This contradicts $g_n(x_0) \to 0$. Hence, $\nabla u_n(x_0)\to \nabla u(x_0)$ everywhere on $E$. So $\nabla u_n(x_0)\to \nabla u(x_0)$ a.e. on $\Omega_{\delta}$. Since $\delta$ is arbitrary, we obtain $\nabla u_n(x_0)\to \nabla u(x_0)$ a.e. on $\Omega$. Since $\{|\nabla u_n|^{p(x)-2}\nabla u_n\}$ is integrable in $L^{1}(\Omega)$, we obtain that as $n\to \infty$, $|\nabla u_n|^{p(x)-2}\nabla u_n \rightharpoonup |\nabla u|^{p(x)-2}\nabla u \quad \text{weakly in } [L^{\frac{p(x)}{p(x)-1}}(\Omega)]^{N}.$ The proof is complete \end{proof} \begin{lemma} \label{lem3.3} Suppose $f$ satisfies {\rm (F2), (F4)}. Then $I_{\lambda}$ satisfies the nonsmooth $(PS)_{c}$ condition provided $c< (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}\max\{\lambda^{1-\frac{N}{p_{+}}}, \lambda^{1-\frac{N}{p_{-}}}\}$. \end{lemma} \begin{proof} Let $\{u_n\}\subset W_0^{1,p(x)}(\Omega)$ be such that $I_{\lambda}(u_n) \to c$ and $m(u_n)\to 0$ as $n\to \infty$. We must show the existence of a subsequence of $\{u_n\}$ which converges strongly in $W_0^{1,p(x)}(\Omega)$. First, we show that $\{u_n\}$ is bounded. We know that $$\label{e3.12} \begin{split} I_{\lambda}(u_n) &= \int_{\Omega}\frac{1}{p(x)}|\nabla u_n|^{p(x)}dx - \lambda\int_{\Omega}\frac{1}{p^*(x)}|u_n|^{p^*(x)}dx - \int_{\Omega}F(u_n)dx \\ & \geq \frac{1}{p_{+}}\int_{\Omega}|\nabla u_n|^{p(x)}dx - \frac{\lambda}{p^*_{-}}\int_{\Omega}|u_n|^{p^*(x)}dx - \int_{\Omega}F(u_n)dx. \end{split}$$ Let $\omega_n\in \partial I_{\lambda}(u_n)$ such that $\|\omega_n \| = m(u_n)= o (1)$. From \eqref{e3.1} we have $$\label{e3.13} \langle \omega_n, u_n\rangle = \int_{\Omega}|\nabla u_n|^{p(x)}dx -\lambda\int_{\Omega}|u_n|^{p^*(x)}dx - \int_{\Omega}\overline{\omega}_nu_ndx,$$ where $\overline{\omega}_n(x)\in [\underline{f}(u_n),\overline{f}(u_n)]$ for a.e. $x\in\Omega$. By (F4) we obtain $$\label{e3.14} \frac{1}{q_{-}}\overline{\omega}_nu_n \geq \frac{1}{q_{-}}\underline{f}(u_n)u_n \geq F(u_n).$$ From \eqref{e3.12}, \eqref{e3.13} and \eqref{e3.14}, we obtain $$\label{e3.15} C_{6}(1 + \|u_n\|) \geq I_{\lambda}(u_n) - \frac{1}{q_{-}}\langle \omega_n, u_n\rangle \geq (\frac{1}{p_{+}} -\frac{1}{q_{-}})\int_{\Omega}|\nabla u_n|^{p(x)}dx.$$ If $\|u_n\| > 1$, by Proposition \ref{prop2.2}, we obtain $(\frac{1}{p_{+}}-\frac{1}{q_{-}})\|u_n\|^{p_{-}} \leq C_{6}(1 + \|u_n\|).$ Thus $\{ u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$. Then there exist a subsequence and $u \in W_0^{1,p(x)}(\Omega)$ such that $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$, so we know that $\{ |u_n|^{p^*(x)-2}u_n\varphi\}$ is uniformly integrable in $L^{1}(\Omega)$. By this fact, Lemma \ref{lem3.2} and $m(u_n)\to 0$, taking $n\to \infty$ in $\langle \omega_n, \varphi\rangle$, we have $0 = \int_{\Omega}|\nabla u|^{p(x)-2}\nabla u\nabla \varphi dx -\lambda\int_{\Omega}|u|^{p^*(x)-2} u\varphi dx - \int_{\Omega}\overline{\omega}\varphi dx, \hskip 0.3cm \forall\hskip0.1cm \varphi\in C_0^{\infty}(R^{N}).$ So we derive that $$\label{e3.16} -\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in [\underline{f}(u),\overline{f}(u)].$$ Now we applying Proposition \ref{prop2.6} to prove that $\nu_{j} =0$ in \eqref{e3.7}. Assume $\nu_{j}\neq 0$ for some $j\in J$. From \eqref{e3.15}, we have $I_{\lambda}(u_n)- \frac{1}{q_{-}}\langle \omega_n, u_n\rangle \geq (\frac{1}{p_{+}}-\frac{1}{q_{-}})\int_{\Omega}|\nabla u_n|^{p(x)}dx.$ Since $I_{\lambda}(u_n) \to c$ and $m(u_n)\to 0$, using Proposition \ref{prop2.6}, taking $n\to \infty$, we obtain \begin{align*} c &\geq (\frac{1}{p_{+}}-\frac{1}{q_{-}})\int_{\Omega}|\nabla u|^{p(x)}dx + (\frac{1}{p_{+}}-\frac{1}{q_{-}})\sum_{j\in J}\mu_{j} \\ &\geq(\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N} \max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}. \end{align*} Since $c< (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N} \max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}$, then $\nu_{j} =0$ for all $j\in J$. Hence we have $$\label{e3.17} \int_{\Omega}| u_n|^{p^*(x)}dx \to \int_{\Omega}| u|^{p^*(x)}dx.$$ So we can use \cite[Lemma 2.1]{F}. Set $v_n= u_n - u$ and we have \begin{gather} \int_{\Omega}| u_n|^{p^*(x)}dx = \int_{\Omega}| v_n|^{p^*(x)}dx + \int_{\Omega}| u|^{p^*(x)}dx + o(1), \label{e3.18} \\ \int_{\Omega}|\nabla u_n|^{p(x)}dx = \int_{\Omega}|\nabla v_n|^{p(x)}dx + \int_{\Omega}|\nabla u|^{p(x)}dx + o(1). \label{e3.19} \end{gather} Thus, by \eqref{e3.17} and \eqref{e3.18}, $u_n\to u$ strongly in ${L^{p^*(x)} (\Omega)}$. From \eqref{e3.13}, using \eqref{e3.16} and \eqref{e3.19}, we obtain $\langle \omega_n, u_n\rangle = \int_{\Omega}|\nabla v_n|^{p(x)}dx + \int_{\Omega}|\nabla u|^{p(x)}dx -\lambda\int_{\Omega}|u_n|^{p^*(x)}dx - \int_{\Omega}\overline{\omega}_nu_ndx + o(1).$ By \eqref{e3.4} and \eqref{e3.17}, letting $n \to \infty$, we conclude that $\int_{\Omega}|\nabla v_n|^{p(x)}dx \to 0.$ This fact and Proposition \ref{prop2.2} imply that $u_n\to u$ strongly in $W_0^{1,p(x)}(\Omega)$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.4} Suppose $f$ satisfies {\rm (F2), (F3)}. Then, for every $\lambda > 0$, there are $\alpha, \rho > 0$, such that $I_{\lambda}(u) \geq \alpha$, $\|u\|=\rho$. \end{lemma} \begin{proof} By (F2) and (F3), we have $|f(t)|\leq \epsilon |t|^{p(x)-1}+ C|t|^{q(x)}\leq \epsilon |t|^{p(x)-1} + C(\epsilon)|t|^{p^*(x)-1}.$ Therefore, $$\label{e3.20} \begin{split} I_{\lambda}(u)&= \int_{\Omega}\frac{1}{p(x)}|\nabla u|^{p(x)}dx - \lambda\int_{\Omega}\frac{1}{p^*(x)}|u|^{p^*(x)}dx - \int_{\Omega}F(u)dx \\ &\geq \frac{1}{p_{+}}\int_{\Omega}|\nabla u|^{p(x)}dx - \frac{\epsilon}{p_{-}}\int_{\Omega}| u|^{p(x)}dx - \frac{\lambda + C(\epsilon)}{p^*_{-}}\int_{\Omega}|u|^{p^*(x)}dx. \end{split}$$ we can take $\|u\| < 1$ sufficiently small such that $|u|_{p(x)} < 1$ and $|u|_{p^*(x)} < 1$. From \eqref{e3.20}, Propositions \ref{prop2.1} and \ref{prop2.4}, and the definition of $S$, using the usual arguments, we obtain \begin{align*} I_{\lambda}(u) &\geq \frac{1}{p_{+}}\|u\|^{p_{+}} - \frac{\epsilon}{p_{-}}|u|^{p_{-}}_{p(x)} -\frac{\lambda + C(\epsilon)}{p^*_{-}}|u|^{p^*_{-}}_{p^*(x)} \\ &\geq \frac{1}{2p_{+}}\|u\|^{p_{+}} - \frac{\lambda + C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|u\|^{p^*_{-}} \\ &= (\frac{1}{2p_{+}} - \frac{\lambda + C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|u\|^{p^*_{-} -p_{+}})\|u\|^{p_{+}}. \end{align*} Considering $g(t) = \frac{1}{2p_{+}} - \frac{\lambda + C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|t\|^{p^*_{-}-p_{+}},$ since $p_{+}< p^*_{-}$, we have $g(t) \to \frac{1}{2p_{+}}$ as $t \to 0$. Hence, there exists $\rho > 0$ such that $g(\rho) > 0$. So, we obtain $\alpha$ and $\rho > 0$, such that $I_{\lambda}(u) \geq \alpha, \quad \|u\|=\rho.$ The proof is complete. \end{proof} Next, we choose $\varphi(x) \in W_0^{1,p(x)}(\Omega)$, such that $\|\varphi\| = 1$. \begin{lemma} \label{lem3.5} Suppose $f$ satisfies {\rm (F4)}. Then, there exists $\lambda_0 > 0$, $t_0 > 0$ such that $I_{\lambda}(t_0\varphi) < 0$, and for all $\lambda \in (0, \lambda_0)$, $\sup_{t \geq 0}I_{\lambda}(t\varphi) < (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N} \max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.$ \end{lemma} \begin{proof} By (F4), we have $f(u)u \geq\underline{f}(u)u\geq q_{-}F(u), \quad \forall u\neq0.$ This implies $F(t u) \geq t^{q_{-}}F(u)$, for all $t\geq 1$. Then, for any $t > 1$, $% \label{e3.21} I_{\lambda}(t\varphi) \leq \frac{t^{p_{+}}}{p_{-}} - \frac{\lambda t^{p^*_{+}}}{p^*_{+}}\int_{\Omega}|\varphi|^{p^*(x)}dx - \int_{\Omega}F(t\varphi)dx \leq \frac{t^{p}_{+}}{p_{-}} - t^{q_{-}}\int_{\Omega}F(\varphi)dx = J_1(t\varphi).$ Since $q_{-} > p_{+}$ and $F(\varphi) > 0$, there exists $t_0 > 0$ sufficiently large such that $I_{\lambda}(t_0\varphi) < 0$ and $\|t_0\varphi\| > \rho$ with $\rho$ given by Lemma \ref{lem3.4}. If $0\leq t < 1$, then $I_{\lambda}(t\varphi) \leq \frac{t^{p_{-}}}{p_{-}} - \int_{\Omega}F(t\varphi)dx = J_2(t\varphi).$ Let $J(t\varphi) = \max \{ J_1(t\varphi), J_2(t\varphi)\}$, so we have $\sup_{t \geq 0}I_{\lambda}(t\varphi) \leq \sup_{t \geq 0}J(t\varphi).$ Hence, we can find $\lambda_0 > 0$ such that $\sup_{t \geq 0}J_{\lambda}(t\varphi) < (\frac{1}{p_{+}} -\frac{1}{q_{-}})S^{N}\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.$ So, for all $\lambda\in (0, \lambda_0)$, we have $\sup_{t \geq 0}I_{\lambda}(t\varphi) <(\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N} \max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.$ The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] It is obvious that $I_{\lambda}(0) = 0$. By Lemmas \ref{lem3.1}, \ref{lem3.3}--\ref{lem3.5}, according to Theorem \ref{thm2.1}, there exist $\lambda_0 > 0$, and for all $\lambda \in (0, \lambda_0)$, we can find an $u \in W_0^{1,p(x)}(\Omega)$ such that $I_{\lambda}(u) > 0$ and $0 \in \partial I_{\lambda}(u)$. Hence, $u$ is a nontrivial solution of \eqref{e1.1}. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] In \eqref{e1.2}, $f(u)= bh(u-a)|u|^{q(x)-2}u$ has only one discontinuity point $a$, so by the consequence of Theorem \ref{thm1.1}, we obtain that an $u\in W_0^{1,p(x)}(\Omega)$ is a nontrivial nonnegative solution of \eqref{e1.2}. That is, $$\label{e3.22} -\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in \widehat{f}(u) \quad \text{a.e. in } \Omega$$ where $\widehat{f}(u)$ is the multivalued function given by $$\label{e3.23} \widehat{f}(s) = \begin{cases} \{f(s)\} & s\neq a,\\ [0, b h(x)u^{q(x)-1}] & s = a. \end{cases}$$ If $V = \{ x\in \Omega: u(x)=a\}$ exists, by \eqref{e3.22} and \eqref{e3.23}, we have $-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in [0, b h(x)u^{q(x)-1}]\quad \text{a.e. in } V.$ Using the Morrey-Stampacchia's theorem \cite{Mo}, we have $- \Delta_{p(x)}u =0$ a.e. $x\in V$. So $-\lambda a^{p^*(x)-1} \geq 0 \quad \text{a.e. in } V.$ This is a contradiction. Thus $|V|= 0$. The proof is complete. \end{proof} \subsection*{Acknowledgments} The author wants to thank the anonymous referees for their carefully reading this paper and their useful comments. \begin{thebibliography}{00} \bibitem{AB} C. O. Alves, A. M. 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