\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 53, pp. 1--10.\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-2007/53\hfil Ground state solutions] {A remark on ground state solutions for Lane-Emden-Fowler equations with \\ a convection term} \author[H. Xue, Z. Zhang\hfil EJDE-2007/53\hfilneg] {Hongtao Xue, Zhijun Zhang} % in alphabetical order \address{Hongtao Xue \newline School of Mathematics and Informational Science, Yantai University, Yantai, Shandong, 264005, China} \email{ytxiaoxue@yahoo.com.cn} \address{Zhijun Zhang \newline School of Mathematics and Informational Science, Yantai University, Yantai, Shandong, 264005, China} \email{zhangzj@ytu.edu.cn} \thanks{Submitted January 6, 2007. Published April 10, 2007.} \thanks{Z. Zhang is supported by grant 10671169 from NNSFC} \subjclass[2000]{35J60, 35B25, 35B50, 35R05} \keywords{Ground state solution; Lane-Emden-Fowler equation; \hfill\break\indent convection term; maximum principle; existence; sub-solution; super-solution} \begin{abstract} Via a sub-supersolution method and a perturbation argument, we study the Lane-Emden-Fowler equation $$-\Delta u =p(x)[g(u)+f(u)+|\nabla u|^q]$$ in $\mathbb{R} ^N$ ($N\geq3$), where $00, \quad\text{in }\mathbb{R}^N,\\ u(x)\to 0, \quad \text{ as } |x|\to \infty, \end{gathered} \label{e1.1} where$N\geq3$,$00, \quad\text{in }\mathbb{R}^N,\\ u(x)\to 0, \quad\text{as } |x|\to \infty, \end{gathered} \label{e1.3} has at least one solution $u\in C^{2+\alpha} _ {\rm loc} (\mathbb{R}^N)$. For problem \eqref{e1.3}, we see that no monotonicity conditions are imposed on the functions $g(u)$ and $g(u)/u$ . On the other hand, condition \eqref{e1.2} is necessary to prove the existence (see also Lair and Shaker \cite{l2}). Recently, in Ghergu and R\v{a}dulescu \cite{g2}, the same problem \eqref{e1.1} is considered, where $g\in C^1(0,\infty)$ is a positive decreasing function such that $$\lim_{u\to 0^+}g(u)=+\infty,$$ and $f:[0,\infty)\to [0,\infty)$ is a H\"{o}lder continuous function of exponent $0<\alpha<1$ which is non-decreasing such that $f>0$ on $(0,\infty)$ and satisfies (F1)-(F2) and \begin{itemize} \item [(F3)] the mapping $(0,\infty)\ni u\longmapsto\frac{f(u)}{u}$ is non-increasing. \end{itemize} Finally, they showed that in addition to condition \eqref{e1.2} , if the above assumptions are fulfilled, then problem \eqref{e1.1} has at least one ground state solution. In the present paper, we consider the existence of ground state solutions for problem \eqref{e1.1} under more general conditions. Our main result is summarized in the following theorem. \begin{theorem}\label{thm1.1} Assume (G1)--(G3) and (F1)--(F2). Then problem \eqref{e1.1} has at least one solution provided that condition \eqref{e1.2} is fulfilled. \end{theorem} \begin{remark} \label{rmk1.2} \rm Some basic examples of the function $g$ satisfying (G1)--(G3) are: \begin{itemize} \item [(i)] $u^{-\gamma}+u^p+ \sin {\psi(u)}+1$, where $\gamma >0$, $p< 1$ and $\psi\in C^2(\mathbb{R})$; \item [(ii)] $e^{1/{u^\gamma}}+u^p+\cos {\psi(u)}+1$, where $\gamma >0$, $p< 1$ and $\psi\in C^2(\mathbb{R})$; \item [(iii)] $u^{-\gamma}\ln^{-q_1}(1+u)+\ln^{q_2}(1+u)+u^p+\sin {\psi(u)}+2$ with $\psi\in C^2(\mathbb{R})$ , $\gamma >0$, $p<1$, $q_2>0$ and $q_1>0$; \item [(iv)] $u^{-\gamma}+\arctan {\psi(u)}+\pi$ with $\psi\in C^2(\mathbb{R})$ and $\gamma >0$. \end{itemize} \end{remark} \begin{remark} \label{rmk1.3} \rm Some basic examples of the function $f$ satisfying (F1)--(F2) are: \begin{itemize} \item [(i)] $c_1(1+u)^{-\alpha}+c_2u^\gamma+c_3$, where $c_1, c_2, c_3\geq0$, $\alpha>0$, $0<\gamma<1$; \item [(ii)] $e^{1/{(u+1)}}+u^\gamma+\sin{\psi(u)}+1$, where $0<\gamma<1$ and $\psi\in C^2(\mathbb{R})$; \item [(iii)] $\ln^q(u+1)+(1+u)^{-\alpha}$, where $q>0$, $\alpha>0$; \item [(iv)] $u^p\ln^q(u+1)$, where $p, q\in (0, 1)$ and $p+q<1$. \end{itemize} \end{remark} \section{Proof of Theorem \ref{thm1.1}} In this section, we first show the existence of positive solutions for problem \eqref{e1.1} in smooth bounded domains by a sub-supersolution method. Then, via the perturbation argument, we prove Theorem \ref{thm1.1}. First we recall the following auxiliary results. \begin{lemma}[{\cite[Lemma 3]{c3}}] \label{lem2.1} Let $\Omega\subset\mathbb{R}^N$ be a smooth bounded domain. Suppose the boundary-value problem $$-\Delta u =p(x)[g(u)+f(u)+|\nabla u|^q], \quad u>0,\; x\in\Omega, \; u|_{\partial\Omega}=0, \label{e2.1}$$ has a super-solution $\bar{u}$ and a sub-solution $\underline{u}$ such that $\underline{u}\leq\bar{u}$ in $\Omega$, then problem \eqref{e2.1} has at least one solution $u\in C(\bar{\Omega})\cap C^2(\Omega)$ in the ordered interval $[\underline{u}, \bar{u}]$. \end{lemma} \begin{lemma}[{\cite [Lemma 2.3]{z5}}] \label{lem2.2} Suppose (G1)--(G3) are satisfied. Then there exists a function $\overline{g}_0$ such that \begin{itemize} \item [(i)] $\overline{g}_0 \in C^1((0,\infty),(0,\infty))$; \item [(ii)] $\frac{g(s)}{s}\leq\overline{g}_0, \forall\ s>0$; \item [(iii)] $\overline{g}_0(s)$ is non-increasing on $(0,\infty)$; \item [(iv)]$\lim_{s\to 0+}\overline{g}_0(s)=\infty$ and $\lim_{s\to \infty}\overline{g}_0(s)=0$. \end{itemize} \end{lemma} Note that if $g\in C((0,\infty),(0,\infty))$, the function $\overline{g}_0$ still exists in Lemma \ref{lem2.2}. \begin{lemma} \label{lem2.3} Let $\Omega\subset\mathbb{R}^N$ be a smooth bounded domain. Assume (G1)-(G3) and (F1)-(F2) are fulfilled. Then problem \eqref{e2.1} has at least one solution $u\in C(\bar{\Omega})\cap C^2(\Omega)$. \end{lemma} \begin{proof} Let $\underline{u}$ be a solution of $$-\Delta \underline{u} =p(x)g(\underline{u}), \quad \underline{u}>0, \; x\in\Omega, \quad \underline{u}|_{\partial\Omega}=0. \label{e2.2}$$ The existence of $\underline{u}$ follows from the results in Zhang \cite{z5}. Obviously, $\underline{u}$ is a sub-solution of \eqref{e2.1}. The main point is to find a super-solution $\overline{u}$ of \eqref{e2.1} such that $\underline{u}\leq\overline{u}$ in $\Omega$. Then, by Lemma \ref{lem2.1} we deduce that problem \eqref{e2.1} has at least one solution. Denote $\sigma(u):=g(u)+f(u)$. Then $\sigma$ satisfies \begin{itemize} \item %[($\sigma1$)] $\sigma\in C((0,\infty),(0,\infty))$; \item %[($\sigma2$)] $\lim_{u\to 0^+}\frac{\sigma(u)}{u}=+\infty$; \item %[($\sigma3$)] $\lim_{u\to \infty}\frac{\sigma(u)}{u}=0$. \end{itemize} By Lemma \ref{lem2.2}, corresponding to $\sigma$, there exists a function $\overline{\sigma}_0$ satisfying \begin{itemize} \item [(i)] $\overline{\sigma}_0 \in C^1((0,\infty),(0,\infty))$; \item [(ii)] $\frac{\sigma(u)}{u}\leq\overline{\sigma}_0$, for all $u>0$; \item [(iii)] $\overline{\sigma}_0(u)$ is non-increasing on $(0,\infty)$; \item [(iv)] $\lim_{u\to 0+}\overline{\sigma}_0(u)=+\infty$ and $\lim_{u\to \infty}\overline{\sigma}_0(u)=0$, \end{itemize} such that $G(u):=u(\overline{\sigma}_0(u)+\frac{1}{u})$ satisfying \begin{itemize} \item[(G1)] $G\in C^1((0,\infty),(0,\infty))$; \item[(G2)] $\frac{G(u)}{u}$ is decreasing on $(0,\infty)$; \item[(G3)]$\lim_{u\to 0^+} \frac{G(u)}{u}=\infty$; \item[(G4)] $\lim_{u\to \infty}\frac{G(u)}{u}=0$. \end{itemize} Then, we consider the problem $$-\Delta u =p(x)[G(u)+|\nabla u|^q],\quad u>0, \quad x\in\Omega, \quad u|_{\partial\Omega}=0. \label{e2.3}$$ We claim that this problem has at least one classical solution, which is a super-solution of \eqref{e2.1}. Indeed, let $h:[0,\eta]\to [0,\infty)$ be such that \begin{gathered} -h''(t) =\frac{G(h(t))}{h(t)}, \quad 00, \quad 00$small enough. Multiplying by$h'(t)$in \eqref{e2.4} and integrating on$[t,\eta], we combine (G2) and get $$(h')^2(t)\leq 2h(\eta)\frac{G(h(t))}{h(t)}+(h')^2(\eta), \quad 00. Let \phi_1 be the normalized positive eigenfunction corresponding to the first eigenvalue \lambda_1 of -\Delta in H_0^1(\Omega). By H\"{o}pf's maximum principle, there exist \delta>0 and \omega\Subset\Omega such that $$|\nabla\phi_1|>\delta,\quad \text{in } \Omega\setminus\omega.\label{e2.6}$$ For the rest of this paper we denote \begin{gather*} |\phi_1|_\infty:=\max_{x\in\bar{\Omega}}\phi_1(x),\quad |\phi_1|_0:=\min_{x\in\bar{\omega}}\phi_1(x), \\ |p|_{\infty}:=\max_{x\in\bar{\Omega}}p(x),\quad |\nabla\phi_1|_\infty:=\max_{x\in\bar{\Omega}}|\nabla\phi_1(x)|. \end{gather*} And we fix c>0 such that c|\phi_1|_\infty<\eta. Using the monotonicity of h and h', it follows that \begin{gather} 01 be such that \begin{gather} \lambda_1(Mch'(\eta))^{1-q}|\phi|_0>2|p|_{\infty}|{\nabla\phi_1|^q_\infty}, \label{e2.9}\\ M^{1-q}C^{-1}(c\delta)^{2-q}>2|p|_{\infty}.\label{e2.10} \end{gather} By (G4), we can choose M>1 large enough such that $$\frac{G(Mh(c|\phi_1|_0))}{Mh(c|\phi_1|_0)} \leq\frac{\lambda_1c|\phi_1|_0h'(\eta)}{2|p|_{\infty}h(\eta)}.\label{e2.11}$$ Next, we show that \overline{u}_0=Mh(c\phi_1) is a super-solution of \eqref{e2.3} provided that M satisfies \eqref{e2.9}-\eqref{e2.11}. We have$$ -\Delta \overline{u}_0=\lambda_1Mc\phi_1 h'(c\phi_1) +Mc^2|\nabla\phi_1|^2\frac{G(h(c\phi_1))}{h(c\phi_1)}. $$\label{e2.12} By (G2), \eqref{e2.7}-\eqref{e2.8} and \eqref{e2.11}, we have$$ \frac{G(Mh(c\phi_1))}{Mh(c\phi_1)} \leq\frac{G(Mh(c|\phi_1|_0))}{Mh(c|\phi_1|_0)} \leq\frac{\lambda_1c|\phi_1|_0h'(\eta)}{2|p|_{\infty}h(\eta)} \leq\frac{\lambda_1c\phi_1h'(c\phi_1)}{2p(x)h(c\phi_1)},\quad\text{in }\omega. $$%\label{e2.13} It follows that $$\lambda_1Mc\phi_1 h'(c\phi_1)\geq2p(x)G(Mh(c\phi_1)), \quad\text{in } \omega. \label{e2.14}$$ From \eqref{e2.8}-\eqref{e2.9}, we have $$\lambda_1Mc\phi_1 h'(c\phi_1)\geq2p(x)|Mch'(c\phi_1)\nabla\phi_1|^q =2p(x)|\nabla\overline{u}_0|^q,\ {\rm in} \quad \omega.\label{e2.15}$$ Since h(0)=0, we get$$ \lim_{x\to \partial\Omega}\Big(\frac{(c\delta)^2}{h(c\phi_1)} -2|p|_{\infty}\Big)=+\infty, $$namely, $$\frac{(c\delta)^2}{h(c\phi_1)}>2|p|_{\infty}>2p(x), \quad\text{in } \Omega\setminus\omega.\label{e2.16}$$ From \eqref{e2.6}, (G2) and \eqref{e2.16}, we have $$Mc^2|\nabla\phi_1|^2\frac{G(h(c\phi_1))}{h(c\phi_1)} \geq Mc^2\delta^2\frac{G(Mh(c\phi_1))}{Mh(c\phi_1)} \geq2p(x)G(Mh(c\phi_1)), \quad\text{in } \Omega\setminus\omega. \label{e2.17}$$ From \eqref{e2.5}-\eqref{e2.6} and \eqref{e2.10}, we have $$Mc^2|\nabla\phi_1|^2\frac{G(h(c\phi_1))}{h(c\phi_1)} \geq 2p(x)|Mch'(c\phi_1) \nabla\phi_1|^q=2p(x)|\nabla\bar{u}_0|^q,\quad \text{in }\Omega\setminus\omega.\label{e2.18}$$ From \eqref{e2.14}-\eqref{e2.15} and \eqref{e2.17}-\eqref{e2.18}, we deduce that$$ -\Delta \overline{u}_0\geq p(x)[G(\overline{u}_0)+|\nabla\overline{u}_0|^q], \quad\text{in } \Omega, $$namely, \overline{u}_0=Mh(c\phi_1) is a super-solution of \eqref{e2.3}. On the other hand, the unique solution \underline{u}_0 of the boundary-value problem$$ -\Delta \underline{u}_0 =p(x)G(\underline{u}_0),\quad \underline{u}_0>0,\; x\in\Omega,\; \underline{u}_0|_{\partial\Omega}=0, % \label{e2.19} $$is a sub-solution of problem \eqref{e2.3}. Here the existence of \underline{u}_0 follows from the results in Goncalves and Santos \cite{g4}. Next, we prove that$$ \underline{u}_0\leq\bar{u}_0 \quad\text{in } \Omega. Assume the contrary; i.e., there exists x_0\in \Omega such that \bar{u}_0(x_0)<\underline{u}_0(x_0). Then, \sup_{x\in \Omega} \left (\ln(\underline{u}_0(x))-\ln (\bar{u}_0(x))\right ) exists and is positive in \Omega. At the point, we have \begin{gather*} \nabla \left (\ln(\underline{u}_0(x_0))-\ln ( \bar{u}_0(x_0))\right)=0,\\ \Delta \left (\ln(\underline{u}_0(x_0))-\ln(\bar{u}_0(x_0))\right)\leq 0. \end{gather*} By (G2), we see that \begin{align*} &\Delta \left (\ln(\underline{u}_0(x_0))-\ln (\bar{u}_0(x_0))\right)\\ & = \frac {\Delta \underline{u}_0(x_0)} {\underline{u}_0(x_0)}- \frac {\Delta \bar{u}_0(x_0)}{\bar{u}_0(x_0)} -\frac {|\nabla \underline{u}_0(x_0)|^2}{(\underline{u}_0(x_0))^2}+\frac {|\nabla \bar{u}_0(x_0)|^2}{(\bar{u}_0(x_0))^2}\\ &=\frac {\Delta \underline{u}_0(x_0)}{\underline{u}_0(x_0)}- \frac {\Delta \bar{u}_0(x_0)}{\bar{u}_0(x_0)}\\ &\geq p(x_0)\Big(\big[\frac{G(\bar{u}_0(x_0))}{\bar{u}_0(x_0)} -\frac{G(\underline{u}_0(x_0))}{\underline{u}_0(x_0)}\big] +\frac{|\nabla \bar{u}_0(x_0)|^q}{\bar{u}_0(x_0)}\Big) > 0, \end{align*} which is a contradiction. Therefore, \overline{u}_0\geq\underline{u}_0 in \Omega. Then by Lemma \ref{lem2.1}, problem \eqref{e2.3} has at least one classical solution denoted by \bar{u}, which is a super-solution of problem \eqref{e2.1}. Finally, we show that \underline{u}\leq\bar{u} in \Omega. Assume the contrary, i.e., there exists x_1\in \Omega such that \bar{u}(x_1)<\underline{u}(x_1). Then, \sup_{x\in \Omega} \left (\ln(\underline{u}(x))-\ln (\bar{u}(x))\right ) exists and is positive in \Omega. At the point, we have \nabla \left (\ln(\underline{u}(x_1))-\ln ( \bar{u}(x_1))\right)=0 \quad\text{and}\quad \Delta \left (\ln(\underline{u}(x_1))-\ln(\bar{u}(x_1))\right)\leq 0. $$Since \overline \sigma_0 is non-increasing on (0,\infty), we have$$ \overline \sigma_0(\bar{u}(x_1))\geq\overline \sigma_0(\underline{u}(x_1)) \geq\frac{g(\underline{u}(x_1))+f(\underline{u}(x_1))}{\underline{u}(x_1)}. Then we obtain \begin{align*} &\Delta \left (\ln(\underline{u}(x_1))-\ln (\bar{u}(x_1))\right)\\ & = \frac {\Delta \underline{u}(x_1)} {\underline{u}(x_1)}- \frac {\Delta \bar{u}(x_1)}{\bar{u}(x_1)} -\frac {|\nabla \underline{u}(x_1)|^2}{(\underline{u}(x_1))^2}+\frac {|\nabla \bar{u}(x_1)|^2}{(\bar{u}(x_1))^2}\\ &= \frac {\Delta \underline{u}(x_1)}{\underline{u}(x_1)}- \frac {\Delta \bar{u}(x_1)}{\bar{u}(x_1)}\\ &= p(x_1)\Big(\big[\frac{G(\bar{u}(x_1))}{\bar{u}(x_1)} -\frac{g(\underline{u}(x_1))}{\underline{u}(x_1)}\big] +\frac{|\nabla \bar{u}(x_1)|^q}{\bar{u}(x_1)}\Big)\\ &= p(x_1)\Big(\big[\overline \sigma_0(\bar{u}(x_1)) -\frac{g(\underline{u}(x_1))}{\underline{u}(x_1)}\big] +\frac{1}{\bar{u}(x_1)}+\frac{|\nabla \bar{u}(x_1)|^q}{\bar{u}(x_1)}\Big)> 0, \end{align*} which is a contradiction. Therefore, \overline{u}\geq\underline{u} in \Omega. By Lemma \ref{lem2.1}, the proof is complete. \end{proof} \begin{lemma}\label{lem2.4} Assume condition \eqref{e1.2} is fulfilled. Then there is a function w such that $$\begin{gathered} -\Delta w \geq p(x)[g(w)+f(w)+|\nabla w|^q], \quad w>0, \; x\in \mathbb{R}^N,\\ w(x)\to 0,\quad\text{as } |x|\to \infty. \end{gathered}\label{e2.20}$$ \end{lemma} \begin{proof} Denote \Psi(r):=r^{1-N}\int_0^rt^{N-1}\varphi(t)dt,\quad \forall r>0. $$By condition \eqref{e1.2} and the L'H\^{o}pital's rule, we have \lim_{r\to 0}\Psi(r)=\lim_{r\to \infty}\Psi(r)=0. Thus, \Psi is bounded on (0,\infty) and it can be extended in the origin by taking \Psi(0)=0. On the other hand, by integration by parts and the L'H\^{o}pital's rule (see details in \cite{g2}) , we get$$ \int_0^\infty\Psi(r)dr=\lim_{r\to \infty}\int_0^r\Psi(t)dt =\frac{1}{N-2}\int_0^\infty r\varphi(r)dr<\infty. $$Let \mu>2 be such that $$\mu^{1-q}\geq2\max_{r\geq0}\Psi^q(r).\label{e2.21}$$ Define$$ \rho(x):=\mu\int_{|x|}^{\infty}\Psi(t)dt, \quad \text{for } x\in \mathbb{R}^N. $$Then \rho is bounded and satisfies \begin{gather*} -\Delta \rho=\mu\varphi(|x|),\quad \rho>0,\; x\in\mathbb{R}^N,\\ \rho(x)\to 0, \quad \text{as } |x|\to \infty. \end{gather*} We claim that there are R>0 and a function w\in C^2(\mathbb{R}^N) such that $$\rho(x)=\frac{1}{R}\int_0^{w(x)}\frac{t}{G(t)+1}dt.\label{e2.22}$$ Indeed, since$$ \lim_{r\to +\infty}\frac{\int_0^r\frac{t}{G(t)+1}dt}{r} =\lim_{r\to +\infty}\frac{r}{G(r)+1}=+\infty, $$we notice first for some R>0$$ |\rho|_\infty\leq\frac{1}{R}\int_0^{R}\frac{t}{G(t)+1}dt, $$and in particular $$w(x)\leq R, \ x\in \mathbb{R}^N.\label{e2.23}$$ From \eqref{e2.22} we have$$ |\nabla w|=R\frac{G(w)+1}{w}|\nabla \rho|=\mu R\Psi(|x|)\frac{G(w)+1}{w}, $$and combining with (G2), we get$$ \frac{1}{R}\frac{w}{G(w)+1}\Delta w+\frac{1}{R}\frac{d}{dw} \left(\frac{w}{G(w)+1}\right)|\nabla w|^2=\Delta \rho, $$i.e.,$$ -\Delta w \geq \mu R\varphi(|x|)\frac{G(w)+1}{w}. Then from \eqref{e2.21}, \eqref{e2.23} and (G2), we obtain \begin{align*} -\Delta w & \geq \mu R\varphi(|x|)\frac{G(w)+1}{w}\\ &\geq R p(x)\frac{G(w)+1}{w}+\frac{\mu}{2}R p(x)\frac{G(w)+1}{w}\\ &\geq p(x)(G(w)+1)+p(x)|\mu R\frac{G(w)+1}{w}\Psi(|x|)|^q\\ & = p(x)(G(w)+1)+p(x)|\nabla w|^q. \end{align*} Hence, \begin{gather*} -\Delta w \geq p(x)[G(w)+1+|\nabla w|^q],\ w>0, \ x\in \mathbb{R}^N,\\ w(x)\to 0, \ {\rm as }\quad |x|\to \infty. \end{gather*} % \label{e2.24} Since G(w)>g(w)+f(w) on (0,+\infty), it follows that w satisfies \eqref{e2.20}. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] Consider the perturbed problem $$-\Delta u_n =p(x)[g(u_n)+f(u_n)+|\nabla u_n|^q], \quad u_n>0, \; x\in B_n, \; u_n|_{\partial B_n}=0, \label{e2.25}$$ where B_n:=\{x\in\mathbb{R}^N; |x|n. Letw$be as in Lemma \ref{lem2.4}, with the same proof above, we deduce that $$u_n(x)\leq w(x),\ x\in \mathbb{R}^N, \ n=1, 2, 3, \dots.\label{e2.26}$$ Now, we need to estimate$\{u_n\}$. For any bounded$C^{2+\alpha}$-smooth domain$\Omega'\subset \mathbb{R}^N$, take$\Omega_1$and$\Omega_2$with$C^{2+\alpha}$-smooth boundaries, and$K_1$large enough, such that $$\Omega'\subset\subset \Omega_1\subset\subset \Omega_2\subset\subset B_n, \quad n\geq K_1.$$ Note that $$u_n(x)\geq \underline{u}(x)>0,\quad \forall x\in B_{K_1}, \label{e2.27}$$ where$B_{K_1}$is the substitution for$\Omega$in the proof of Lemma \ref{lem2.3}. Let $$\Psi_n(x)=p(x)[g(u_n)+f(u_n)+|\nabla u_n|^q],\quad x\in \bar{B}_{K_1}.$$ Since$-\Delta u_n(x)=\Psi_n(x)$,$x\in B_{K_1}$, by the interior estimate theorem of Ladyzenskaja and Ural'tseva \cite[Theorem 3.1, p. 266]{l1}, we get a positive constant$C_1$independent of$n$such that $$\max_{x\in\bar{\Omega}_2}|\nabla u_n(x)| \leq C_1\max_{x\in \bar{B}_{K_1}}u_n(x)\leq C_1 \max_{x\in \bar{B}_{K_1}}w(x),\quad \forall \ x\in B_{K_1},$$ %\label{e2.28} i.e.,$|\nabla u_n(x)|$is uniformly bounded on$\bar{\Omega}_2$. It follows that$\{\Psi_n\}_{K_1}^\infty$is uniformly bounded on$\bar{\Omega}_2$and hence$\Psi_n\in L^p(\Omega_2)$for any$p>1$. Since$-\Delta u_n(x)=\Psi_n(x), \ x\in \Omega_2, $we see by \cite[Theorem 9.11]{g3}, that there exists a positive constant$C_2$independent of$n$such that $$\|u_n\|_{W^{2, p}(\Omega_1)}\leq C_2\left(\|\Psi_n\|_{L^p(\Omega_2)}+\|u_n\|_{L^p(\Omega_2)} \right),\quad \forall\ n\geq K_1.$$ % \label{e2.29} Taking$p>N$such that$\alpha< 1-N/p$and applying Sobolev's embedding inequality, we see that$\{\|u_n\|_{C^{1+\alpha}(\bar{\Omega}_1)}\}_{K_1}^\infty$is uniformly bounded. Therefore$\Psi_n\in C^\alpha(\bar{\Omega}_1)$and$\{\|\Psi_n\|_{C^\alpha(\bar{\Omega}_1)}\}_{K_1}^\infty$is uniformly bounded. It follows by Schauder's interior estimate theorem (see \cite[Chapter 1, p. 2]{g3}) that there exists a positive constant$C_3$independent of$n$such that $$\|u_n\|_{C^{2+\alpha}(\bar{\Omega}')}\leq C_3\left(\|\Psi_n\|_{C^\alpha(\bar{\Omega}_1)}+\|u_n\|_{C(\bar{\Omega}_1)} \right),\quad \forall \ n\geq K_1;$$ \label{e2.30} i.e.,$\{\|u_n\|_{C^{2+\alpha}(\bar{\Omega}')}\}_{K_1}^\infty$is uniformly bounded. Using Ascoli-Arzela's theorem and the diagonal sequential process, we see that$\{u_n\}_{K_1}^\infty$has a subsequence that converges uniformly in the$C^2(\bar{\Omega}')$norm to a function$u\in C^2(\bar{\Omega}')$and$u$satisfies $$- \Delta u=p(x)[g(u)+f(u)+|\nabla u|^q], \quad x\in \bar{\Omega}'.$$ By \eqref{e2.27}, we obtain that $$u>0, \quad \forall x\in \bar{\Omega}'.$$ Applying Schauder's regularity theorem we see that$u\in C^{2+\alpha}(\bar{\Omega}')$. Since$\Omega'$is arbitrary, we also see that$u\in C^{2+\alpha}_{\rm loc}(\mathbb{R}^N)$. It follows by \eqref{e2.26} that$\lim_{|x|\to \infty}u(x)=0$. Thus, a standard bootstrap argument shows that$u$is a classical solution to problem \eqref{e1.1}. The proof is complete. \end{proof} At last, it is worth pointing out that Ye and Zhou \cite{y1} proved that in many situations condition \eqref{e1.2} can be replaced by the following more general condition \begin{itemize} \item [(P1)]$-\Delta u=p(x)$has a bounded ground state solution. \end{itemize} Obviously, condition \eqref{e1.2} implies (P1) (see \cite{y1} for details about comparison between condition \eqref{e1.2} and (P1)). Therefore, we have an unsolved problem as follows. \begin{remark} We note that the existence of ground state solutions for problem \eqref{e1.1} is left an open problem if$p$satisfies condition (P1) instead of \eqref{e1.2}. \end{remark} \begin{thebibliography}{00} \bibitem{a1} R. Agarwal, D. 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