\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 177, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/177\hfil Existence of positive solutions] {Existence of positive solutions for $p(x)$-Laplacian problems} \author[G. A. Afrouzi and H. Ghorbani\hfil EJDE-2007/177\hfilneg] {Ghasem A. Afrouzi, Horieh Ghorbani} % in alphabetical order \address{Ghasem A. Afrouzi \newline Department of Mathematics\\ Faculty of Basic Sciences \\ Mazandaran University, Babolsar, Iran} \email{afrouzi@umz.ac.ir} \address{Horieh Ghorbani \newline Department of Mathematics\\ Faculty of Basic Sciences \\ Mazandaran University, Babolsar, Iran} \email{seyed86@yahoo.com} \thanks{Submitted July 18, 2007. Published December 17, 2007} \subjclass[2000]{35J60, 35B30, 35B40} \keywords{Positive radial solutions; $p(x)$-Laplacian problems; \hfill\break\indent boundary value problems} \begin{abstract} We consider the system of differential equations \begin{gather*} -\Delta_{p(x)} u=\lambda [g(x)a(u) + f(v)] \quad\text{in }\Omega\\ -\Delta_{q(x)} v=\lambda [g(x)b(v) + h(u)] \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega \end{gather*} where $p(x) \in C^1(\mathbb{R}^N)$ is a radial symmetric function such that $\sup|\nabla p(x)| < \infty$, $1 < \inf p(x) \leq \sup p(x) < \infty$, and where $-\Delta_{p(x)} u = -\mathop{\rm div}|\nabla u|^{p(x)-2}\nabla u$ which is called the $p(x)$-Laplacian. We discuss the existence of positive solution via sub-super-solutions without assuming sign conditions on $f(0),h(0)$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} The study of differential equations and variational problems with nonstandard $p(x)$-growth conditions has been a new and interesting topic. Many results have been obtained on this kind of problems; see for example \cite{f1,f2,f3,f4,f5,f6,z3}. In \cite{f3,f4} Fan and Zhao give the regularity of weak solutions for differential equations with nonstandard $p(x)$-growth conditions. Zhang \cite{z1} investigated the existence of positive solutions of the system \begin{equation}\label{eI} \begin{gathered} -\Delta_{p(x)} u= f(v) \quad\text{in }\Omega\\ -\Delta_{p(x)} v= g(u) \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega \end{gathered} \end{equation} where $p(x) \in C^1(\mathbb{R}^N)$ is a function, $\Omega \subset \mathbb{R}^N$ is a bounded domain. The operator $-\Delta_{p(x)} u= -\mathop{\rm div}|\nabla u|^{p(x)-2}\nabla u)$ is called $p(x)$-Laplacian. Especially, if $p(x)$ is a constant $p$, System \eqref{eI} is the well-known $p$-Laplacian system. There are many papers on the existence of solutions for $p$-Laplacian elliptic systems, for example \cite{a1,f1,f2,f3,f4,f5,f6,h1}. In \cite{h1} the authors consider the existence of positive weak solutions for the $p$-Laplacian problem \begin{equation} \label{eII} \begin{gathered} -\Delta_p u= f(v) \quad\text{in }\Omega\\ -\Delta_p v= g(u) \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega\,. \end{gathered} \end{equation} There the first eigenfunctions is used for constructing the subsolution of $p$-Laplacian problems. Under the condition $\lim_{u\to+\infty}f(M(g(u))^{1/(p-1)}/u^{p-1} = 0$, for all $M > 0$, the authors show the existence of positive solutions for problem \eqref{eII}. In this paper, at first, we consider the existence of positive solutions of the system \begin{equation} \label{eP1} \begin{gathered} -\Delta_{p(x)} u= F(x, u, v) \quad\text{in }\Omega\\ -\Delta_{p(x)} v= G(x, u, v) \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega \end{gathered} \end{equation} where $p(x) \in C^1(\mathbb{R}^N)$ is a function, $F(x, u, v) = [g(x)a(u) + f(v)]$, $G(x, u, v) = [g(x)b(v) + h(u)]$, and $\Omega \subset \mathbb{R}^N$ is a bounded domain. Then we consider the system \begin{equation} \label{eP2} \begin{gathered} -\Delta_{p(x)} u= \lambda F(x, u, v) \quad\text{in }\Omega\\ -\Delta_{p(x)} v= \lambda G(x, u, v) \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega \end{gathered} \end{equation} where $p(x) \in C^1(\mathbb{R}^N)$ is a function, $F(x, u, v) = [g(x)a(u) + f(v)]$, $G(x, u, v) = [g(x)b(v) + h(u)]$, $\lambda $ is a positive parameter and $\Omega \subset \mathbb{R}^N$ is a bounded domain. To study $p(x)$-Laplacian problems, we need some theory on the spaces $L^{p(x)}(\Omega)$, $W^{1,p(x)}(\Omega)$ and properties of $p(x)$-Laplacian which we will use later (see \cite{f2}). If $\Omega \subset \mathbb{R}^N$ is an open domain, write \[ C_+(\Omega) = \{h : h \in C(\Omega), h(x)> 1 \text{ for }x \in \Omega \} \] $h^+ = \sup_{x \in \Omega} h(x)$, $h^- = \inf_{x \in \Omega} h(x)$, for any $h \in C(\Omega)$, $L^{p(x)}(\Omega) = \{u | u$ is a measurable real-valued function, $\int_\Omega |u|^{p(x)}dx < \infty \}$. Throughout the paper, we will assume that $p \in C_+(\Omega)$ and $1 < \inf_{x \in \mathbb{R}^N}p(x) \leq \sup_{x \in \mathbb{R}^N}p(x) < N$. We introduce the norm on $L^{p(x)}(\Omega)$by \[ |u|_{p(x)} = \inf\{\lambda > 0 : \int_\Omega |\frac{u(x)}{\lambda}|^{p(x)} dx \leq 1\}, \] and $(L^{p(x)}(\Omega) , |\cdot|_{p(x)})$ becomes a Banach space, we call it generalized Lebesgue space. The space $(L^{p(x)}(\Omega) , |\cdot|_{p(x)})$ is a separable, reflexive and uniform convex Banach space (see \cite[Theorem 1.10, 1.14]{f2}). The space $W^{1,p(x)}(\Omega)$ is defined by $W^{1,p(x)}(\Omega) = \{u \in L^{p(x)}(\Omega) : |\nabla u| \in L^{p(x)}(\Omega)\}$, and it is equipped with the norm \[ \|u\| = |u|_{p(x)} + |\nabla u|_{p(x)}, \quad \forall u \in W^{1,p(x)}(\Omega). \] We denote by $W_0^{1,p(x)}(\Omega)$ the closure of $C_0^\infty(\Omega)$ in $W^{1,p(x)}(\Omega)$. $W^{1,p(x)}(\Omega)$ and $W_0^{1,p(x)}(\Omega)$ are separable, reflexive and uniform convex Banach space (see \cite[Theorem 2.1]{f2}). We define \[ (L(u) , v) = \int_{\mathbb{R}^N} |\nabla u|^{p(x)-2}\nabla u\nabla v dx, \quad \forall u, v \in W^{1,p(x)}(\Omega), \] then $L :W^{1,p(x)}(\Omega) \to(W^{1,p(x)}(\Omega))^*$ is a continuous, bounded and is a strictly monotone operator, and it is a homeomorphism \cite[Theorem 3.11]{f5}. Functions $u, v $ in $W_0^{1,p(x)}(\Omega)$, is called a weak solution of \eqref{eP2}; it satisfies \begin{gather*} \int_\Omega |\nabla u|^{p(x)-2}\nabla u \nabla \xi dx = \int_\Omega \lambda F(x, u, v) \xi dx, \quad \forall \xi \in W_0^{1,p(x)}(\Omega), \\ \int_\Omega |\nabla v|^{q(x)-2}\nabla v \nabla \xi dx = \int_\Omega \lambda G(x, u, v)\xi dx, \quad \forall \xi \in W_0^{1,p(x)}(\Omega). \end{gather*} We make the following assumptions \begin{itemize} \item[(H1)] $p(x) \in C^1(\mathbb{R}^N)$ is a radial symmetric and $\sup |\nabla p(x)| < \infty $ \item[(H2)] $\Omega = B(0 , R) = \{x | |x| < R\}$ is a ball, where $R > 0$ is a sufficiently large constant. \item[(H3)] $a, b \in C^1([0 , \infty))$ are nonnegative, nondecreasing functions such that \[ \lim_{u \to+\infty} \frac{a(u)}{u^{P^- - 1}} = 0, \quad \lim_{u \to+\infty} \frac{b(u)}{u^{P^- - 1}} = 0\,. \] \item[(H4)] $f, h \in C^1([0, \infty))$ are nondecreasing functions, $\lim_{u \to+\infty}f(u) = +\infty$,\\ $ \lim_{u \to+\infty}h(u) = +\infty$, and \[ \lim_{u \to +\infty}\frac{f(M(h(u))^\frac{1}{p^--1})}{u^{p^--1}} = 0, \quad \forall M > 0\,. \] \item[(H5)] $g : [0,+\infty) \to(0, \infty)$ is a continuous function such that $ L_1 = \min_{x\in \bar{\Omega}}g(x)$, and $ L_2 = \max_{x\in \bar{\Omega}}g(x)$. \end{itemize} We shall establish the following result. \begin{theorem} \label{thm1} If {\em(H1)--(H5)} hold, then \eqref{eP1} has a positive solution. \end{theorem} \begin{proof} We establish this theorem by constructing a positive subsolution $(\phi_1 , \phi_2)$ and supersolution $(z_1, z_2)$ of \eqref{eP1}, such that $\phi_1 \leq z_1$ and $\phi_2 \leq z_2$. That is $(\phi_1 , \phi_2)$ and $(z_1, z_2)$ satisfy \begin{gather*} \int_\Omega |\nabla \phi_1|^{p(x)-2}\nabla \phi_1 \cdot \nabla \xi dx \leq \int_\Omega g(x) a(\phi_1) \xi dx + \int_\Omega f(\phi_2) \xi dx, \\ \int_\Omega |\nabla \phi_2|^{p(x)-2}\nabla \phi_1 \cdot \nabla \xi dx \leq \int_\Omega g(x)b(\phi_2) \xi dx + \int_\Omega h(\phi_1) \xi dx, \\ \int_\Omega |\nabla z_1|^{p(x)-2}\nabla z_1 \cdot \nabla \xi dx \geq \int_\Omega g(x) a(z_1) \xi dx + \int_\Omega f(z_2) \xi dx, \\ \int_\Omega |\nabla z_2|^{p(x)-2}\nabla z_2 \cdot \nabla \xi dx \geq \int_\Omega g(x)b(z_2) \xi dx + \int_\Omega h(z_1) \xi dx, \end{gather*} for all $\xi \in W_0^{1,p(x)}(\Omega)$ with $\xi \geq 0$. Then \eqref{eP1} has a positive solution. \subsection*{Step 1} We construct a subsolution of \eqref{eP1}. Denote \begin{gather*} \alpha = \frac{\inf p(x)- 1}{4(\sup |\nabla p(x)| +1)},\quad R_0 = \frac{R-\alpha}{2}, \\ b = \min \{a(0)L_1 + f(0), b(0)L_1 + h(0), -1\}, \end{gather*} and let \[ \phi(r) = \begin{cases} e^{-k(r-R)} - 1, & 2R_0 < r \leq R, \\ e^{\alpha k} - 1 + \int_r^{2R_0}(ke^{\alpha k})^{\frac{p(2R_0)-1}{p(r)-1}} \\ \times [\frac{(2R_0)^{N-1}}{r^{N-1}}\sin(\varepsilon(r-2R_0)+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & 2R_0-\frac{\pi}{2\varepsilon} < r \leq 2R_0, \\[3pt] e^{\alpha k} - 1 + \int_{2R_0- \frac{\pi}{2\varepsilon}} ^{2R_0}(ke^{\alpha k})^{\frac{p(2R_0)-1}{p(r)-1}}\\ \times [\frac{(2R_0)^{N-1}}{r^{N-1}}\sin(\varepsilon_0(r-2R_0)+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & r \leq 2R_0-\frac{\pi}{2\varepsilon}, \end{cases} \] where $R_0$ is sufficiently large, $\varepsilon$ is a small positive constant which satisfies $R_0 \leq 2R_0 -\frac{\pi}{2\varepsilon}$, In the following, we will prove that $(\phi , \phi)$ is a subsolution of \eqref{eP1}. Since \[ \phi'(r) =\begin{cases} e^{-k(r-R)} - 1, & 2R_0 < r \leq R, \\ -(ke^{\alpha k})^{\frac{p(2R_0)-1}{p(r)-1}}\\ \times [\frac{(2R_0)^{N-1}}{r^{N-1}}\sin(\varepsilon(r-2R_0)+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & 2R_0-\frac{\pi}{2\varepsilon} < r \leq 2R_0, \\ 0, & 0 \leq r \leq 2R_0-\frac{\pi}{2\varepsilon}, \end{cases} \] it is easy to see that $\phi \geq 0$ is decreasing and $\phi \in C^1([0 , R]), \phi(x) = \phi(|x|) \in C^1(\bar{\Omega})$. Let $r = |x|$. By computation, \[ -\Delta_{p(x)}\phi = -\mathop{\rm div}|\nabla \phi(x)|^{p(x)-2}\nabla \phi(x)) = -(r^{N-1}|\phi'(r)|^{p(r)-2}\phi'(r))'/r^{N-1}. \] Then \[ -\Delta_{p(x)}\phi= \begin{cases} (k e^{-k(r-R)})^{p(r)-1} \big[-k(p(r)-1)+ p'(r)\ln k \\ - kp'(r)(r-R) + \frac{N-1}{r}\big], & 2R_0 < r \leq R, \\[3pt] \varepsilon (\frac{2R_0}{r})^{N-1} (ke^{\alpha k})^({p(2R_0)-1})\\ \times \cos(\varepsilon(r-2R_0)+ \frac{\pi}{2})(L_1 + 1), & 2R_0-\frac{\pi}{2\varepsilon} < r \leq 2R_0, \\ 0, & 0 \leq r \leq 2R_0-\frac{\pi}{2\varepsilon}, \end{cases} \] If $k$ is sufficiently large, when $2R_0 < r \leq R$, then \[ -\Delta_{p(x)}\phi \leq -k[\inf p(x) -1- \sup|\nabla p(x)|(\frac{\ln k}{k} + R - r) + \frac{N-1}{kr}] \leq -k \alpha. \] Since $\alpha$ is a constant dependent only on $p(x)$, if $k$ is a big enough, such that $-k a < b$, and since $\phi(x) \geq 0$ and $a, f$ are monotone, this implies \begin{equation} \label{e1} -\Delta_{p(x)}\phi \leq a(0) L_1 + f(0)\leq g(x) a(\phi) + f(\phi), \quad 2R_0 < |x| \leq R\,. \end{equation} If $k$ is sufficiently large, then \[ a(e^{\alpha k}- 1) \geq 1,\quad f(e^{\alpha k}-1) \geq 1 , \quad b(e^{\alpha k}- 1) \geq 1,\quad h(e^{\alpha k}- 1) \geq 1 \] where $k$ is dependent on $a, f, b, h, p$, and independent on $R$. Since \begin{align*} -\Delta_{p(x)}\phi &= \varepsilon (\frac{2R_0}{r})^{N-1} (ke^{\alpha k})^({p(2R_0)-1})\cos(\varepsilon(r-2R_0)+ \frac{\pi}{2})(L_1 +1)\\ & \leq \varepsilon(L_1 + 1) 2^N k^{p^+}e^{\alpha k p^+} , 2R_0-\frac{\pi}{2\varepsilon} < |x| < 2R_0\,. \end{align*} Let $\varepsilon = 2^{-N} k^{-p^+}e^{-\alpha k p^+}$. Then \begin{equation} \label{e2} -\Delta_{p(x)}\phi \leq L_1 + 1 \leq g(x) a(\phi) + f(\phi), 2R_0-\frac{\pi}{2\varepsilon} < |x| < 2R_0. \end{equation} Obviously, \begin{equation} \label{e3} -\Delta_{p(x)}\phi = 0 \leq L_1 + 1 \leq g(x)a(\phi) + f(\phi), |x| <2R_0-\frac{\pi}{2\varepsilon}. \end{equation} Since $\phi(x) \in C^1(\Omega)$, combining \eqref{e1}, \eqref{e2}, \eqref{e3}, we have \[ -\Delta_{p(x)}\phi \leq g(x)a(\phi) + f(\phi) \] for a.e. $x \in \Omega$. Similarly we have \[ -\Delta_{p(x)}\phi \leq g(x) b(\phi) +h(\phi), \] for a.e. $x \in \Omega$. Let $(\phi_1, \phi_2) = (\phi, \phi)$, since $\phi(x) \in C^1(\bar{\Omega})$, it is easy to see that $(\phi_1 , \phi_2)$ is a subsolution of \eqref{eP1}. \subsection*{Step 2} We construct a supersolution of \eqref{eP1} Let $z_1$ be a radial solution of \begin{gather*} -\Delta_{p(x)}z_1(x) = (L_2 + 1) \mu,\quad\text{in } \Omega, \\ z_1 = 0 \quad \text{on } \partial\Omega\,. \end{gather*} We denote $z_1 = z_1(r) = z_1(|x|)$, then $z_1$ satisfies \[ -(r^{N-1}|z_1'|^{p(r)-2}z_1')' = r^{N-1}(L_2 + 1)\mu, z_1(R) = 0, z_1'(0) = 0\,. \] Then \begin{equation} \label{e4} z_1' = -|\frac{r(L_2 + 1 )\mu}{N}|^{\frac{1}{p(r)-1}}, \end{equation} and \[ z_1 = \int_r^R |\frac{r(L_2 + 1 )\mu}{N}|^{\frac{1}{p(r)-1}}dr. \] We denote $\beta = \beta((L_2 + 1)\mu) = \max_{0 \leq r \leq R}z_1(r)$, then \[ \beta((L_2 + 1)\mu) = \int_0^R |\frac{r(L_2 + 1)\mu}{N}|^{\frac{1}{p(r)-1}}dr = ((L_2 + 1 )\mu)^{\frac{1}{p(q)-1}}\int_0^R |\frac{r}{N}|^{\frac{1}{p(r)-1}}dr, \] where $q \in [0 , 1]$. Since $\int_0^R |\frac{r}{N}|^{\frac{1}{p(r)-1}}dr$ is a constant, then there exists a positive constant $C \geq 1$ such that \begin{equation} \label{e5} \frac{1}{C}((L_2 +1)\mu)^{\frac{1}{p^+ -1}} \leq \beta((L_2 + 1)\mu) = \max_{0 \leq r \leq R} z_1(r) \leq C((L_2 + 1)\mu) ^{\frac{1}{p^- - 1}}\,. \end{equation} We consider \begin{gather*} -\Delta_{p(x)} z_1 =(L_2 + 1)\mu \quad\text{in }\Omega\\ -\Delta_{p(x)} z_2=(L_2 + 1)h(\beta((L_2 + 1)\mu)) \quad\text{in }\Omega\\ z_1=z_2= 0 \quad\text{on } \partial \Omega\,. \end{gather*} Then we shall prove that $(z_1 , z_2)$ is a supersolution for \eqref{eP1}. For $\xi \in W^{1,p(x)}(\Omega)$ with $\xi \geq 0$, it is easy to see that \begin{align*} \int_{\Omega}|\nabla z_2|^{p(x)-2}\nabla z_2 \cdot \nabla \xi dx &= \int_{\Omega}(L_2 + 1)h(\beta((L_2 + 1 )\mu))\xi dx\\ &\geq \int_{\Omega}L_2 h(\beta((L_2 + 1 )\mu))\xi dx + \int_{\Omega} h(z_1)\xi dx. \end{align*} Similar to \eqref{e5}, we have \[ \max_{0 \leq r \leq R}z_2(r) \leq C[(L_2 + 1) h(\beta((L_2 + 1)\mu))]^{\frac{1}{(p^- - 1)}}. \] By (H3), for $\mu$ large enough we have \[ h(\beta((L_2 + 1)\mu)) \geq b(C[(L_2 + 1) h(\beta((L_2 + 1 )\mu))]^{\frac{1}{p^- - 1}}) \geq b(z_2). \] Hence \begin{equation} \label{e6} \int_{\Omega}|\nabla z_2|^{p(x)-2}\nabla z_2 \cdot \nabla \xi dx \geq \int_{\Omega}g(x) b(z_2)\xi dx + \int_{\Omega}h(z_1) \xi dx, \end{equation} Also \[ \int_{\Omega}|\nabla z_1|^{p(x)-2}\nabla z_1 \cdot \nabla \xi dx = \int_{\Omega}(L_2 + 1)\mu \xi dx. \] By (H3), (H4), when $\mu$ is sufficiently large, according to \eqref{e5}, we have \begin{align*} (L_2 + 1)\mu &\geq [\frac{1}{C}\beta((L_2 +1 )\mu)]^{p^- - 1}\\ &\geq L_2 a(\beta ((L_2 + 1)\mu)) + f[C[(L_2 + 1)^{\frac{1}{(p^- - 1)}}(h(\beta((L_2 + 1)\mu)))^{\frac{1}{(p^- - 1)}}] \\ &\geq g(x) a(z_1) + f(z_2), \end{align*} then \begin{equation} \label{e7} \int_{\Omega}|\nabla z_1|^{p(x)-2}\nabla z_1 \cdot \nabla \xi dx \geq \int_{\Omega}g(x) a(z_1) \xi dx + \int_{\Omega} f(z_2)\xi dx. \end{equation} According to \eqref{e6} and \eqref{e7}, we can conclude that $(z_1 , z_2)$ is a supersolution of \eqref{eP1}. Let $\mu$ be sufficiently large, then from \eqref{e4} and the definition of $(\phi_1 , \phi_2)$, it is easy to see that $\phi_1 \leq z_1$ and $\phi_2 \leq z_2$. This completes the proof. \end{proof} Now we consider the problem \begin{equation} \label{eP2'} \begin{gathered} -\Delta_{p(x)} u=\lambda F(x, u, v) \quad\text{in }\Omega\\ -\Delta_{p(x)} v=\lambda G(x, u, v) \quad\text{in }\Omega\\ u=v= 0 \quad\text{on } \partial \Omega. \end{gathered} \end{equation} If $p(x) \equiv p$ (a constant), because of the homogenity of $p$-Laplacian, \eqref{eP1} and \eqref{eP2} can be transformed into each other; but, if $p(x)$ is a general function, since $p(x)$-Laplacian is nonhomogeneous, they cannot be transformed into each other. So we can see that $p(x)$-Laplacian problem is more complicated than than that of $p$-Laplacian, and it is necessary to discuss the problem \eqref{eP2} separately. \begin{theorem} \label{thm2} If $p(x)\in C^1(\bar{\Omega})$, $\Omega = B(0, R)$, and {\rm (H3)--(H5)} hold, then there exists a $\lambda^*$ which is sufficiently large, such that \eqref{eP2} possesses a positive solution for any $\lambda \geq \lambda^*$. \end{theorem} \begin{proof} We construct a subsolution of \eqref{eP2}. Let $\beta \leq \frac{R}{4}$ satisfy \begin{equation} \label{e8} |p(r_1) - p(r_2)| \leq \frac{1}{2}, \forall r_1, r_2 \in [R-2\beta, R]. \end{equation} In the following we denote \begin{equation} \label{e9} \begin{gathered} \delta = \min \{\frac{\inf p(x)-1}{4(\sup |\nabla p(x)| + 1)}\},\quad p_*^+ = \sup_{R-2\beta \leq |x| \leq R}p(x), \quad p_*^- = \inf_{R-2\beta \leq |x| \leq R}p(x), \\ b = \min \{a(0) L_1 + f(0), b(0)L_1 + h(0), -1\}. \end{gathered} \end{equation} Let $\alpha \in (0, \beta]$, and set \[ \phi(r) = \begin{cases} e^{-k(r-R)} - 1, & R- \alpha < r \leq R,\\ e^{\alpha k} - 1 + \int_r^{R-\alpha} (ke^{\alpha k})^{\frac{p(R-\alpha)-1}{p(r)-1}} [\frac{(R-\alpha)^{N-1}}{r^{N-1}}\\ \times \sin(\varepsilon(r-(R-\alpha))+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & R-2\beta < r \leq R - \alpha, \\[3pt] e^{\alpha k} - 1 + \int_{R - \alpha - \frac{\pi}{2\varepsilon}} ^{R - \alpha }(ke^{\alpha k})^{\frac{p(R - \alpha )-1}{p(r)-1}}[\frac{(R - \alpha)^{N-1}}{r^{N-1}}\\ \times \sin(\varepsilon(r- (R - \alpha))+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & r \leq R - 2\beta, \end{cases} \] where $\varepsilon = \frac{\pi}{2(2\beta - \alpha) }$ which satisfies $\varepsilon(R - 2\beta - (R - \alpha)) + \frac{\pi}{2} = 0 $. In the following, we will prove that $(\phi , \phi)$ is a subsolution of \eqref{eP2}. Since \[ \phi'(r) = \begin{cases} e^{-k(r-R)} - 1, & R - \alpha < r \leq R, \\ -(ke^{\alpha k})^{\frac{p(R - \alpha )-1}{p(r)-1}}\\ [\frac{(R -\alpha )^{N-1}}{r^{N-1}}\\ \times \sin(\varepsilon(r- (R - \alpha))+ \frac{\pi}{2})(L_1 + 1)]^{\frac{1}{p(r)-1}}dr, & R - 2\beta < r \leq R - \alpha, \\ 0, & r \leq R - 2\beta. \end{cases} \] It is easy to see that $\phi \geq 0$ is decreasing and $\phi \in C^1([0 , R]), \phi(x) = \phi(|x|) \in C^1(\Omega)$. Let $r = |x|$. By computation, \[ -\Delta_{p(x)}\phi(x) = \begin{cases} (k e^{-k(r-R)})^{p(r)-1} [-k(p(r)-1)\\ + p'(r)\ln k - kp'(r)(r-R) + \frac{N-1}{r}], & R - \alpha < r \leq R, \\[3pt] \varepsilon (\frac{R - \alpha}{r})^{N-1} (ke^{\alpha k})^{({p(R - \alpha )-1})}\\ \times \cos(\varepsilon(r- (R - \alpha))+ \frac{\pi}{2})(L_1 + 1 ), & R - 2\beta < r \leq R - \alpha, \\ 0, & r \leq R - 2\beta. \end{cases} \] If $k$ is sufficiently large, when $R - \alpha < r \leq R$, then we have \[ -\Delta_{p(x)}\phi \leq -k^{p(r)}[\inf p(x) -1- \sup|\nabla p(x)|(\frac{\ln k}{k} + R - r) + \frac{N-1}{kr}] \leq -k^{p(r)} \delta. \] If $k$ satisfies \begin{equation} \label{e10} k^{p_*^-}\delta = -\lambda b , \end{equation} and since $\phi(x) \geq 0$ and $a, f$ is monotone, it means that \begin{equation} \label{e11} -\Delta_{p(x)}\phi \leq \lambda(a(0) L_1 + f(0))\leq \lambda (g(x) a(\phi) + f(\phi)), R - \alpha < |x| \leq R. \end{equation} From (H3), (H4) there exists a positive constant $M$ such that $a(M- 1) \geq 1$, $f(M-1) \geq 1$, $b(M- 1) \geq 1$, $h(M- 1) \geq 1 $. Let \begin{equation} \label{e12} \alpha k = \ln M. \end{equation} Since \begin{align*} -\Delta_{p(x)}\phi(x) &= \varepsilon (\frac{R - \alpha}{r})^{N-1} (ke^{\alpha k})^({p(R - \alpha )-1})\cos(\varepsilon(r-(R - \alpha))+ \frac{\pi}{2})(L_1 + 1)\\ &\leq \varepsilon(L_1 + 1) 2^N (ke^{\alpha k})^{ p_*^+ - 1} , R - 2\beta < |x| < R - \alpha , \end{align*} if \begin{equation} \label{e13} \varepsilon 2^N (ke^{\alpha k})^{ p_*^+ - 1} \leq \lambda, \end{equation} then \begin{equation} \label{e14} -\Delta_{p(x)}\phi(x) \leq \lambda( L_1 + 1) \leq \lambda(g(x) a(\phi) + f(\phi)), \quad R - 2\beta < |x| < R - \alpha . \end{equation} Obviously \begin{equation} \label{e15} -\Delta_{p(x)}\phi(x) = 0 \leq \lambda L_1 + 1 \leq \lambda(g(x)a(\phi) + f(\phi)), \quad |x| < R - 2\beta \,. \end{equation} Combining \eqref{e10}, \eqref{e12} and \eqref{e13}, we only need \[ \varepsilon 2^N |\frac{-b}{\delta}\lambda|^\frac{p_*^+ - 1}{p_*^-}M^{p_*^+ - 1} \leq \lambda, \] and according to \eqref{e8}, \eqref{e9}, we only need \[ (\frac{\pi}{\beta}2^N M^{p_*^+ - 1}|\frac{-b}{\delta}|^\frac{p_*^+ - 1}{p_*^-})2p_*^- \leq \lambda\,. \] Let \[ \lambda^* = (\frac{\pi}{\beta}2^N M^{p_*^+ - 1}|\frac{-b}{\delta}|^\frac{p_*^+ - 1}{p_*^-})^{2p_*^-}\,. \] If $\lambda \geq \lambda^*$ is sufficiently large, then \eqref{e13} is satisfied. Since $\phi(x) = \phi(|x|) \in C^1(\Omega)$, according to \eqref{e11}, \eqref{e14} and \eqref{e15}, it is easy to see that if $\lambda$ is sufficiently large, then $(\phi_1, \phi_2)$ is a subsolution of \eqref{eP2}. \subsection*{Step 2} We construct a supersolution of \eqref{eP2}. Similar to the proof of Theorem \ref{thm1}, we consider \begin{gather*} -\Delta_{p(x)} z_1 =\lambda(L_2 + 1)\mu \quad\text{in }\Omega\\ -\Delta_{p(x)} z_2=\lambda(L_2 + 1)h(\beta(\lambda(L_2 + 1)\mu)) \quad\text{in }\Omega\\ z_1=z_2= 0 \quad\text{on } \partial \Omega\,, \end{gather*} where $\beta = \beta(\lambda(L_2 + 1)\mu) = \max_{0 \leq r \leq R}z_1(r)$. It is easy to see that \begin{align*} \int_{\Omega}|\nabla z_2|^{p(x)-2}\nabla z_2 \cdot \nabla \xi dx &= \int_{\Omega}\lambda(L_2 + 1)h(\beta(\lambda(L_2 + 1 )\mu))\xi dx\\ &\geq \int_{\Omega}\lambda L_2 h(\beta(\lambda(L_2 + 1 )\mu))\xi dx + \int_{\Omega}\lambda h(z_1)\xi dx. \end{align*} Similar to \eqref{e5}, we have \[ \max_{0 \leq r \leq R}z_2(r) \leq C[\lambda(L_2 + 1)h(\beta(\lambda(L_2 + 1)\mu))]^{\frac{1}{(p^- - 1)}}. \] By (H3) for $\mu$ large enough we have \[ h(\beta(\lambda(L_2 + 1)\mu)) \geq b(C[\lambda(L_2 + 1)h(\beta(\lambda(L_2 + 1 )\mu))]^{\frac{1}{p^- - 1}}) \geq b(z_2). \] Hence \begin{equation} \label{e16} \int_{\Omega}|\nabla z_2|^{p(x)-2}\nabla z_2 \cdot \nabla \xi dx \geq \int_{\Omega}\lambda g(x) b(z_2)\xi dx + \int_{\Omega} \lambda h(z_1) \xi dx. \end{equation} Also \[ \int_{\Omega}|\nabla z_1|^{p(x)-2}\nabla z_1 \cdot \nabla \xi dx = \int_{\Omega}\lambda(L_2 + 1)\mu \xi dx. \] By (H3), (H4), when $\mu$ is sufficiently large, according to \eqref{e5}, we have \begin{align*} (L_2 + 1)\mu &\geq \frac{1}{\lambda} [\frac{1}{C}\beta(\lambda(L_2 + 1 )\mu)]^{p^- - 1}\\ &\geq L_2 a(\beta (\lambda(L_2 + 1)\mu)) + f(C[\lambda(L_2 + 1) h(\beta(\lambda(L_2 + 1)\mu))]^{\frac{1}{(p^- - 1)}})\,. \end{align*} Then \begin{equation} \label{e17} \int_{\Omega}|\nabla z_1|^{p(x)-2}\nabla z_1 \cdot \nabla \xi dx \geq \int_{\Omega}\lambda g(x) a(z_1) \xi dx + \int_{\Omega} \lambda f(z_2)\xi dx. \end{equation} According to \eqref{e16} and \eqref{e17}, we can conclude that $(z_1 , z_2)$ is a supersolution of \eqref{eP2}. 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