\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 43, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/43\hfil Multiple positive solutions] {Multiple positive solutions for a singular elliptic equation with Neumann boundary condition in two dimensions} \author[B. S. Kaur, K. Sreenadh\hfil EJDE-2009/43\hfilneg] {Bhatia Sumit Kaur, K. Sreenadh} % in alphabetical order \address{Bhatia Sumit Kaur \newline Department of Mathematics, Indian Institute of Technology Delhi Hauz Khaz, New Delhi-16, India} \email{sumit2212@gmail.com} \address{Konijeti Sreenadh \newline % He prefers to be called K. Sreenadh Department of Mathematics, Indian Institute of Technology Delhi Hauz Khaz, New Delhi-16, India} \email{sreenadh@gmail.com} \thanks{Submitted August 26, 2008. Published March 24, 2009.} \subjclass[2000]{34B15, 35J60} \keywords{Multiplicity; nonlinear Neumann boundary condition; \hfill\break\indent Laplace equation} \begin{abstract} Let $\Omega\subset \mathbb{R}^2$ be a bounded domain with $C^2$ boundary. In this paper, we are interested in the problem \begin{gather*} -\Delta u+u = h(x,u) e^{u^2}/|x|^\beta,\quad u>0 \quad \text{in } \Omega, \\ \frac{\partial u}{\partial\nu}= \lambda \psi u^q \quad \text{on }\partial \Omega, \end{gather*} where $0\in \partial \Omega$, $\beta\in [0,2)$, $\lambda>0$, $q\in [0,1)$ and $\psi\ge 0$ is a H\"older continuous function on $\overline{\Omega}$. Here $h(x,u)$ is a $C^{1}(\overline{\Omega}\times \mathbb{R})$ having superlinear growth at infinity. Using variational methods we show that there exists $0<\Lambda <\infty$ such that above problem admits at least two solutions in $H^1(\Omega)$ if $\lambda\in (0,\Lambda)$, no solution if $\lambda>\Lambda$ and at least one solution when $\lambda = \Lambda$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $\Omega\subset \mathbb{R}^2$ be a bounded domain with $C^2$ boundary and $0\in \partial \Omega$. In this work, we study weak solutions $u\in H^1 (\Omega)$ of the problem \begin{equation} \label{Plambda} \begin{gathered} -\Delta u+u = h(x,u) e^{u^2}/|x|^\beta,\quad u>0 \quad \text{in } \Omega, \\ \frac{\partial u}{\partial\nu}= \lambda \psi u^q \quad \text{on }\partial \Omega, \end{gathered} \end{equation} where $\beta\in [0,2)$, $\lambda>0$, $q\in[0,1)$ and $h(x,t)$ satisfies the following conditions: \begin{enumerate} \item[(H1)] $h(x,t)\in C^{1}(\overline{\Omega}\times \mathbb{R}$), $h(x,u)\ge 0$ for all $u\in \mathbb{R}$, $h(x,u)=0$ if $u<0$; \item[(H2)] $ \frac{\partial h}{\partial t}(x,t)\ge 0$ for $t>0$, $h(x,t) \sim t^k$ as $t\to 0$, uniformly in $x$, for some $k>1$, \item[(H3)] $ \liminf_{t\to \infty} \frac{h(x,t)}{t}>0 $ and $ \limsup_{t\to \infty} \frac{h(x,t)}{t^p}=0$ for some $p>1$ uniformly in $x$; \item[(H4)] For any $\epsilon>0$, $\limsup_{t\to \infty} \frac{\partial g}{\partial t}(x,t) e^{-(1+\epsilon)t^2}=0$. \end{enumerate} where $g(x,u)=h(x,u) e^{u^2}/|x|^\beta$ and $G(x,u)=\int_{0}^{u} g(x,s)ds$. Associated to the problem \eqref{Plambda} we have the functional $J_\lambda: H^1 (\Omega)\to \mathbb{R}$ defined by \begin{equation}\label{eq:a1} J_\lambda (u)=\frac{1}{2}\int_{\Omega} (|\nabla u|^2+|u|^2) -\int_{\Omega}G(x,u) -\frac{\lambda}{q+1}\int_{\partial \Omega} \psi |u|^{q+1}. \end{equation} The Fr$\acute{\text{e}}$chet derivative of this functional is given by \begin{equation*}\label{eq:a2} \langle J_{\lambda}'(u),\phi\rangle =\int_{\Omega}\nabla u.\nabla \phi +\int_\Omega u\phi -\int_\Omega g(x,u) \phi -\lambda \int_{\partial \Omega} \psi |u|^{q-1}u \phi,\quad \forall \phi \in H^{1}(\Omega). \end{equation*} Clearly, any positive critical point of $J_\lambda$ is a weak solution of \eqref{Plambda}. The exponential nature of the nonlinearity $g(x,u)$ is motivated by the following version of Moser-Trudinger inequality (due to Adimurthi-Yadava\cite{AY}) \begin{equation}\label{eq:a3} \sup_{\|u\|_{H^1(\Omega)}\le 1 }\int_\Omega e^{2\pi u^2} \le C(|\Omega|), \end{equation} where $C$ is a positive constant. The above imbedding immediately implies that the nonlinear map $H^{1}(\Omega)\ni u\mapsto e^{u^\alpha}\in L^1 (\Omega)$ is a continuous map for all $\alpha\in (0,2]$ and is compact if and only if $\alpha\in (0,2)$. The inequality (\ref{eq:a3}) is a $H^1$ version of the following Moser-Trudinger inequality\cite{M},\cite{T}: \[ \sup_{\|u\|_{H^{1}_{0}(\Omega)}\le 1} \int_\Omega e^{4\pi u^2} \le C(|\Omega|). \] In this case non-compactness of the imbedding $H^{1}_{0}(\Omega)\ni u\mapsto e^{u^2}\in L^1 (\Omega)$ can be shown by using a sequence of functions that are suitable truncations and dilations of the fundamental solution for $-\Delta$ in $\mathbb{R}^2$. These functions are commonly referred to as Moser functions in the literature. In case of the imbedding in (\ref{eq:a3}), the non-compactness can be shown by suitably modifying the Moser functions so that they concentrate at a point on the boundary $\partial \Omega$ (see Lemma 2.1 below). The inequality (\ref{eq:a3}) cannot be used in our case due to the presence of the singularity $|x|^{-\beta}$. To overcome this, first we prove the following singular version of (\ref{eq:a3}) using the methods in \cite{AS}: \[ \sup_{\|u\|_{H^1(\Omega)}\le 1 }\int_\Omega \frac{e^{\alpha u^2}}{|x|^\beta} \le C(|\Omega|) \] where $C$ is a positive constant and $ \frac{\alpha}{2\pi}+\frac{\beta}{2}<1$. Moreover, the above inequality does not hold if $\frac{\alpha}{2\pi}+\frac{\beta}{2}>1$. We prove the following existence and multiplicity results: \begin{theorem} \label{thm1.1} There exists $\Lambda\in(0, \infty)$ such that \eqref{Plambda} admits minimal solution $u_\lambda$ for all $\lambda \in (0,\Lambda)$. \end{theorem} \begin{theorem} \label{thm1.2} There exists $0<\Lambda <\infty$ such that \eqref{Plambda} has at least two solutions for all $\lambda\in (0,\Lambda)$, no solutions for $\lambda>\Lambda$ and at least one solution when $\lambda = \Lambda$. \end{theorem} The minimal solution in the above theorem is obtained using sub-super solution arguments as in \cite{PS} and this minimal solution is shown to be a local minimum of $J_\lambda$. The second solution is obtained using the generalized mountain-pass theorem of Ghoussoub-Priess\cite{GP}. At this point we briefly recall related existence and multiplicity results for elliptic equations. The study of semilinear elliptic problems with critical nonlinearities of Sobolev and Hardy-Sobolev type has recieved considerable interest in recent years. In a recent work \cite{AS}, authors considered Dirichlet Problem for \eqref{Plambda} with superlinear type nonlinearity and studied the existence of positive solutions. In \cite{Deng},\cite{Haitao2} authors studied the existence and multiplicity with Hardy-Sobolev critical exponents. Neumann type Problems are studied in \cite{Do}, \cite{GPR},\cite{Deng} and \cite{PS}. The Multiplicity result for Neumann problem with Sobolev critical nonlinearity has been studied in \cite{GPR} where authors considered the problem \begin{equation} \label{Ptlambda} \begin{gathered} -\Delta u+u = u^{p},\quad u>0 \quad \text{in } \Omega, \\ \frac{\partial u}{\partial \nu}= \lambda \psi u^q \quad\text{on } \partial \Omega, \end{gathered} \end{equation} where $\Omega \subset \mathbb{R}^N$, $N\ge 3$ and $0\tilde{\Lambda}$. \end{theorem} Subsequently, the problem in two dimensions was considered in \cite{PS} where authors proved the Theorem 1.3. In these works authors obtained the minimal solution using sub-super solution method and the second solution by Mountain-pass arguments. The main ingredient is the local minimum nature of the minimal solution which was obtained using $H^1$ verses $C^1$ local minimizer arguments introduced in \cite{BN}. The special features of this class of problems, considered in this paper, are they involve critical singular growth. In this case main difficulty arise due to the fact that the solutions are not $C^1$ near origin. So the arguments like $C^1$ verses $H^1$ local minimizers cannot be carried out for a singular equation. So to establish the local minimum nature of the minimal of solution we use Perron's method as in \cite{Haitao}. To obtain the mountain-pass type solution, one studies the critical levels and the convergence of Palaise-Smale sequences. The critical levels in our case are different than in \cite{PS} where $\beta=0$ case was studied. We recall the following Hardy-Sobolev inequality, which is used in later sections. \begin{lemma}\label{lem0.1} Let $\Omega\subset \mathbb{R}^2$, then there exists a constant $C>0$ such that for any $u\in H^{1}(\Omega)$ \begin{equation}\label{eq:a4} \Big(\int_{\Omega}\frac{|u|^p}{|x|^\beta} dx\Big)^{1/p}\le C \Big(\int_{\Omega} (|\nabla u|^2+u^2)dx\Big)^{1/2} \end{equation} for any $p<\infty$, and $\beta<2$. \end{lemma} \subsection*{Notation} In this paper we make use of the following notation: If $p\in(0,\infty), p'$ denotes the number $p/(p-1)$ so that $p' \in (1,\infty)$ and $1/p+1/p'=1$; $L^{p}(\Omega, |x|^{-\beta})$ denotes the Lebesgue spaces with measure $|x|^{-\beta}dx$ and norm $\|.\|_{L^{p}(\Omega, |x|^{-\beta})}$; $H^1(\Omega)$ denotes the Sobolev space with norm $\|.\|$; $|A|_n$ denotes the Lebesgue measure of the set $A\subset \mathbb{R}^n$. \section{A singular Moser-Trudinger inequality in $H^1(\Omega)$} In this section we show a singular version of Moser-Trudinger inequality in $H^1(\Omega)$. We recall the following lemma from \cite[Lemma 3.3]{AY}. \begin{lemma}\label{lem1} Let $\partial \Omega$ is a smooth manifold. For every $x_0 \in \partial \Omega$, we can find a $L>0$ such that for each $00$ and $\beta\in[0,2)$, we have \begin{equation} \label{eq:b0} \int_{\Omega} \frac{e^{\alpha u^2}}{|x|^{\beta}}dx <\infty. \end{equation} Moreover for any $\beta\in[0,2)$, \begin{equation}\label{eq:b1} \sup \big\{\alpha: \sup_{\|u\|\le 1} \int_{\Omega} \frac{e^{\alpha u^2}}{|x|^\beta}dx <\infty\big\}= \pi(2-\beta) \end{equation} \end{theorem} \begin{proof} Let $t>1$ be such that $\beta t<2$, then by Cauchy-Schwartz inequality, \[ \int_{\Omega} \frac{e^{\alpha u^2}}{|x|^\beta}dx \le \Big(\int_{\Omega} e^{\frac{\alpha t}{t-1}u^2}dx\Big)^{(t-1)/t} \Big(\int_{\Omega} \frac{1}{|x|^{t\beta}}\Big)^{1/t} \] The above inequality along with (\ref{eq:a1}) implies (\ref{eq:b0}). Now Suppose $\sup \{\alpha: \sup_{\|u\|\le 1} \int_{\Omega} \frac{e^{\alpha u^2}}{|x|^\beta}dx <\infty\}< \pi(2-\beta)$. Then there exists $\alpha$ such that $ \frac{\alpha}{2\pi}+\frac{\beta}{2}<1$. Again choose $t>1$ such that $ \frac{\alpha}{2\pi}+\frac{\beta t}{2}=1$, and using Cauchy-Schwartz inequality we obtain, \begin{equation} \label{eq:b2} \sup_{\|u\|\le 1} \int_{\Omega} \frac{e^{\alpha u^2}}{|x|^\beta}dx \le \sup_{\|u\|\le 1} \Big(\int_{\Omega} e^{2\pi u^2}\Big)^{\frac{\alpha}{2\pi}} \Big(\int_{\Omega}\frac{1}{|x|^{\frac{2}{ t}}}dx\Big)^{\frac{t\beta}{2}}<\infty \end{equation} as $1-\frac{\alpha}{2\pi}=1-(1-\frac{t\beta}{2})=\frac{\beta t}{2}<1$ and $\frac{2}{t}<1$. Next, we show that (\ref{eq:b2}) does not hold if $\frac{\alpha}{2\pi}+\frac{\beta}{2}>1$. Let $w_l$ be the sequence of Moser functions concentrated around $0\in \partial\Omega$, as in Lemma \ref{lem1}, then \begin{align*} \int_{\Omega} \frac{e^{\alpha w_{l}^{2}}}{|x|^\beta} dx &\ge e^{\frac{\alpha}{\pi}\log \frac{R}{l}}\int_{B(l)} \frac{1}{|x|^\beta}dx \\ &= \big(\frac{L}{l}\big)^{\alpha/\pi} \frac{l^{2-\beta}}{2\pi (2-\beta)} \\ &=\frac{2\pi L^{\alpha/2\pi}}{(2-\beta)} \frac{1}{l^{2(\frac{\alpha}{2\pi}+\frac{\beta}{2}-1)}} \end{align*} Since $\frac{\alpha}{2\pi}+\frac{\beta}{2}>1$, the limit $l\to 0$ of right-hand side of the above inequality is infinity. Therefore, \[ \sup_{\|u\|\le 1} \int_{\Omega} \frac{e^{\alpha w_{l}^{2}}}{|x|^\beta}=\infty. \] Hence the required supremum is $\pi(2-\beta)$. \end{proof} \section{Existence of minimal solution} In this section, we show that there exists a $\Lambda>0$ such that $J_\lambda$ possesses minimal solution for $\lambda \in (0,\Lambda)$. First we show that there exists a solution for $\lambda$ small. We can show the following strong comparison principle using Hopf lemma and weak comparison arguments. \begin{lemma}\label{lem3.0} Let $u\ne 0$ satisfies $-\Delta u+u \ge 0 $ in $\Omega$ and $\frac{\partial u}{\partial \nu}\ge 0$ on $\partial \Omega$ then $u>0$ in $\overline{\Omega}$. \end{lemma} \begin{lemma}\label{lem3.1} There exists $\lambda_0>0$, small such that \eqref{Plambda} admits a solution for all $\lambda\in (0,\lambda_0)$. \end{lemma} \begin{proof} \textbf{Step 1:} There exists $\lambda_0, \delta>0$ and $R_0>0$ such that $J_\lambda (u)\ge \delta$ for all $\|u\|= R_0$ and all $\lambda<\lambda_0$. From assumption (H2) for $h(x,u)$ and H\"older's inequality, we obtain that for some $C_1 >0$, \begin{align*} \int_\Omega G(x,u)&\le C_1 \int_\Omega |u|^{k+1} \frac{e^{u^2}}{|x|^\beta}\\ &=\int_{\Omega}\frac{|u|^{k+1}}{|x|^{\beta/p^{'}}}\frac{e^{u^2}}{|x|^{\beta/p}}\\ &\le C_1 \|u\|_{L^{p'(k+1)}(\Omega, |x|^{-\beta})}^{k+1} \Big(\int_\Omega e^{p \|u\|^{2} (\frac{u}{\|u\|})^2}\Big)^{1/p}. \end{align*} Now since $\beta<2$ it is possible to choose $p>1$ and $R>0$ such that $\frac{pR^2}{2\pi}+\frac{\beta}{2}<1$. Then, by (\ref{eq:b2}) and Hardy-Sobolev inequality(\ref{eq:a4}), the last inequality gives for some $C_2>0$, \begin{equation}\label{eq:c1} \int_{\Omega} G(x,u) \le C_2 \|u\|^{k+1},\quad \forall \; \|u\| \leq R. \end{equation} Also, by H\"{o}lder's inequality and the trace imbedding $H^1(\Omega)\hookrightarrow L^2 (\partial \Omega)$ we get for some $C_3>0$, \begin{equation} \label{eq:c2} \int_{\partial \Omega} \psi |u|^{q+1} \le C_3 \|\psi\|_{L^\infty (\partial \Omega)} \|u\|_{L^{q+1}(\partial \Omega)}^{q+1} \le C_3 \|u\|^{q+1}. \end{equation} Thus, from (\ref{eq:c1}), (\ref{eq:c2}) we have that, for $R_0^2\in (0,2\pi)$ small enough \begin{equation}\label{eq:c3} J_\lambda (u)\ge \frac{1}{2} \|u\|^{2} -C_2 \|u\|^{k+1} -\lambda C_3 \|u\|^{q+1},\quad \forall \; \|u\| =R_0. \end{equation} Since $k>1$, we may choose $\lambda_0 > 0$ small enough so that $J_\lambda (u)>\delta$ for some $\delta>0$ and for all $\lambda \in (0,\lambda_0)$. \smallskip \noindent\textbf{Step 2:} $J_\lambda$ possesses a local minimum close to the origin for all $\lambda\in (0,\lambda_0)$. It is easy to see that $J_\lambda (tu)<0$ for $t>0$ small enough and any $u\in H^{1}(\Omega)$. Indeed, $\min_{\|u\|\le R_0} J_\lambda (u)<0$ and if this minimum is achieved at some $u_\lambda$, then necessarily $\|u_\lambda\|0$ in $\Omega$. \end{proof} \begin{lemma}\label{lem3.2} Let $\Lambda = \sup \{\lambda>0: \text{\eqref{Plambda} has a solution}\}$. Then $0<\Lambda<\infty$. \end{lemma} \begin{proof} Let $u_\lambda$ be a solution of \eqref{Plambda}. Taking $\phi\equiv 1$ in $\Omega$ in $\langle J'(u_\lambda),\phi\rangle=0$, we get \[ \int_\Omega u_\lambda=\int_\Omega g(u_\lambda) +\lambda \int_{\partial\Omega} \psi u_{\lambda}^{q}. \] Since $\|u_\lambda\|_{L^1(\Omega)}\le C_1 \|u_\lambda\|_{L^p (\Omega)}$ and $\int_\Omega g(x,u_\lambda) \ge C_2 \|u_\lambda\|_{L^p(\Omega)}^{p}$ for some $p>1$, for some constants $C_1, C_2>0$, we immediately obtain from above equation that $\|u_\lambda\|_{L^p(\Omega)}$ is bounded by a constant independent of $\lambda$ for any $p\ge 1$. Now taking $\phi=u_{\lambda}^{-q}$ in $\langle J'(u_\lambda),\phi\rangle=0$ we get, \[ -q\int_\Omega u_{\lambda}^{-1-q} |\nabla u_\lambda|^2 + \int_\Omega u_{\lambda}^{1-q}= \int_{\Omega} u_{\lambda}^{-q} g(x,u_\lambda) +\lambda \int_{\partial\Omega} \psi . \] From the above equation it follows that $\Lambda$ is finite and it is positive by Lemma \ref{lem3.1}. \end{proof} \begin{lemma} \label{lem3.3} There exists a solution for \eqref{Plambda} for all $\lambda\in(0,\Lambda)$. \end{lemma} \begin{proof} Let $\lambda\in (0,\Lambda)$, then choose $\lambda_2 \in (0,\Lambda)$ such that $\lambda<\lambda_2$. Let $u_{\lambda_2}$ be a solutions of $(p_{\lambda_2})$. Let $\mu=\min_{\partial \Omega} u_{\lambda_2}$. Let $v_\lambda$ be the solution of \begin{equation} \label{eq:c4} \begin{gathered} -\Delta u+u=0, \quad u>0 \quad \text{in } \Omega,\\ \frac{\partial u}{\partial \nu}=\lambda \psi \mu^q\quad \text{on } \partial \Omega. \end{gathered} \end{equation} Clearly, $u_{\lambda_2}$ is a super solution of (\ref{eq:c4}) above and hence $u_{\lambda_2}>v_{\lambda}$ in $\overline{\Omega}$ by Lemma \ref{lem3.1}. Let $\tilde{h}_\lambda$ and $\tilde{f}_\lambda$ be the cut-off functions defined as $$ (x\in \Omega, t \in \mathbb{R}) \quad \tilde{g}_\lambda (x,t)=\begin{cases} g(x,v_{\lambda}(x)) & t< v_{\lambda}(x),\\ g(x,t) & v_{\lambda}(x)\le t \le u_{\lambda_2}(x),\\ g(u_{\lambda_2}(x)) & t > u_{\lambda_2}(x), \end{cases} $$ $$ (x \in \partial\Omega,t \in \mathbb{R}) \quad \tilde{f}_\lambda (x,t)= \begin{cases} \lambda \psi (x) \mu^q & t< v_{\lambda}(x),\\ \lambda \psi (x) t^q & v_{\lambda}(x)\le t \le u_{\lambda_2}(x),\\ \lambda \psi (x) u_{\lambda_2}^q(x) & t > u_{\lambda_2}(x). \end{cases} $$ Let $\tilde{G}_\lambda (x,u)=\int_{0}^{u} \tilde{g}_{\lambda}(x,t)dt$ $ (x \in \Omega)$, $\tilde{F}_\lambda (x,u)=\int_{0}^{u} \tilde{f}_{\lambda}(x,t)dt$, $(x \in \partial\Omega)$. Then the functional $\tilde{J}_\lambda :H^1(\Omega) \to \mathbb{R}$ given by \[ \tilde{J}_\lambda (u)=\frac{1}{2}\int_\Omega |\nabla u|^2 +|u|^2 -\int_\Omega \tilde{G}_\lambda (x,u) -\int_{\partial} \tilde{F}_{\lambda}(x,u) \] is coercive and bounded below. Let $u_\lambda$ denote the global minimum of $\tilde{J}_\lambda$ on $H^1 (\Omega)$. Clearly $u_\lambda$ is a solution of \eqref{Plambda}. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] From Lemma \ref{lem3.1} we know that there exists a solution $u_\lambda $ for \eqref{Plambda} for all $\lambda \in (0,\Lambda)$. Let $v_\lambda$ be the unique solution of \begin{equation} \label{eq:c5} \begin{gathered} -\Delta u+u=0,\quad u>0 \quad \text{in } \Omega,\\ \frac{\partial u}{\partial \nu}=\lambda \psi u^q\quad \text{on } \partial \Omega. \end{gathered} \end{equation} Clearly $u_\lambda$ is super solution of (\ref{eq:c5}) and hence by Lemma \ref{lem3.0} we have $v_\lambda \le u_\lambda$ in $\overline{\Omega}$. Now Define the sequence $\{u_n\}$ using the monotone iteration \begin{gather*} u_1=v_\lambda\\ -\Delta u_{n+1}+u_{n+1}=\frac{g(x,u_n)}{|x|^\beta} \quad \text{in } \Omega\\ \frac{\partial u_{n+1}}{\partial \nu} =\lambda \psi u_{n}^{q} \quad \text{on } \partial \Omega, \end{gather*} for $n=1,2,3,\dots $. By the comparison theorem in Lemma \ref{lem3.0}, we get that the sequence $\{u_n\}$ is monotone; i.e., $u_1\le u_2\le \dots \le u_n\le u_{n+1}\dots \le u_\lambda$. By standard monotonicity arguments we obtain a solution $u_\lambda$ of \eqref{Plambda} which will be also the minimal solution. \end{proof} \section{Existence of local minimum for $J_\lambda$ with $\lambda\in (0,\Lambda)$} In this section we show that the solution $u_\lambda$ obtained in Theorem 1.1 is a local minimum for $J_\lambda$ in $H^1(\Omega)$. We adopt the approach of \cite{Haitao} to prove the following theorem. \begin{theorem} \label{thm3.1} The solution $u_\lambda$ as in Lemma\ref{lem3.1} is a local minimum for $J_\lambda$ in $H^1(\Omega)$. \end{theorem} \begin{proof} Let $\lambda_2 >\lambda$ and $u_{\lambda_2}$ be the minimal solution of $(P_{\lambda_2})$. Suppose if $u_\lambda$ is not a local minimum for $J_\lambda$, then there exists a sequence $\{u_n\}\subset H^1(\Omega)$ such that $u_n\to u_\lambda$ strongly in $H^{1}(\Omega)$ and $J_\lambda (u_n)0 \quad \text{in } \Omega, \\ \frac{\partial v_\lambda}{\partial\nu} = \tilde{f}_\lambda(x,v_\lambda) \quad\text{on }\partial \Omega. \end{gathered} \end{equation} Hence it follows that $w_\lambda=u_\lambda+v_\lambda$ is a solution of \eqref{Plambda}. Therefore, to show the existence of second solution, it is enough to find a $00$, there exists $s_\epsilon >0$ such that $\tilde{G}_\lambda (x,s)\le \epsilon \tilde{g}_{\lambda}(x,s)s$ for all $s\ge s_\epsilon $. Using (\ref{eq:d3}) together with this relation, we get, \begin{equation} \label{eq:d5} \begin{aligned} \frac{1}{2}\|v_n\|^2 & \le \int_{\Omega \cap \{v_n\le s_\epsilon\}} \tilde{G}_\lambda (x,v_n) + \epsilon \int_{\Omega}\tilde{g}_\lambda (x,v_n)v_n + \int_{\partial\Omega} \tilde{f}_\lambda (x,v_n)v_n +\rho +o_n (1) \\ &\le C_\epsilon +\epsilon \int_{\Omega} \tilde{g}_\lambda (x,v_n)v_n + \int_{\partial \Omega} \tilde{f}_{\lambda} (x,v_n) v_n +\rho +o_n (1). \end{aligned} \end{equation} From (\ref{eq:d4}) with $\phi=v_n$ we obtain, \begin{equation}\label{eq:d6} \int_\Omega \tilde{g}_\lambda (x,v_n) v_n + \int_{\partial\Omega} \tilde{f}_\lambda (x,v_n)v_n \le \|v_n\|^2 +o_n (1)\|v_n\|. \end{equation} From the definition of $\tilde{h}_\lambda$ we have \[ \int_{\partial\Omega}\tilde{f}_{\lambda}(x,v_n)v_n\le C \|v_n\|^{1+q} \le C+\frac{1}{4}\|v_n\|^2. \] Hence, plugging the above inequality into (\ref{eq:d5}) we get \[ \frac{1}{4} \|v_n\|^2 +\epsilon o_n (1) \|v_n\|\le C_\epsilon +\rho +o_n (1). \] This shows that $\sup_n \|v_n\|<\infty$ and hence by (\ref{eq:d6}) the claim in Step 1 follows. Since $\{v_n\}\subset H^1(\Omega)$ is bounded, up to a subsequence, $v_n \rightharpoonup v_0$ in $H^1 (\Omega)$ for some $v_0 \in H^1 (\Omega)$. \smallskip \noindent\textbf{Step 2:} \begin{gather}\label{eq:d7} \lim_{n \to \infty} \int_{\Omega} \tilde{g}_\lambda (x,v_n) =\int_\Omega \tilde{g}_\lambda (x,v_0),\quad \lim_{n\to \infty} \int_{\partial\Omega}\tilde{f}_\lambda (x,v_n) =\int_{\partial\Omega}\tilde{f}_\lambda (x,v_0), \\ \label{eq:d8} \lim_{n \to \infty} \int_{\Omega} \tilde{G}_\lambda (x,v_n) =\int_\Omega \tilde{G}_\lambda (x,v_0),\quad \lim_{n\to \infty}\int_{\partial\Omega}\tilde{F}_\lambda (x,v_n) =\int_{\partial\Omega}\tilde{F}_\lambda (x,v_0). \end{gather} Let $\mu(A)=\int_{A}|x|^{-\beta}dx$ and $|B|_1$ denote the one dimensional Lebesgue measure of a set $B\subset \partial\Omega$. We first show that $\{\tilde{g}_\lambda (.,v_n)\}$ and $\{\tilde{f}_\lambda (.,v_n)\}$ are equi-integrable families in $L^1 (\Omega)$ and $L^1(\partial\Omega)$ respectively, i.e., given $\epsilon>0, $ there exists a $\delta>0$ such that for any $A\subset \Omega, B\subset \partial\Omega$, with $\mu(A) +|B|_1<\delta$, we have $\sup_n \int_A |\tilde{g}_\lambda(x,v_n)|+\int_B \tilde{f}_\lambda (x,v_n) \le \epsilon$. Once this is shown, (\ref{eq:d7}) follows from Vitali's convergence theorem. Relation (\ref{eq:d8}) follows from (\ref{eq:d7}) by (H3) and the fact that the trace imbedding $H^1(\Omega) \hookrightarrow L^2(\partial\Omega)$ is compact. Let $\tilde{C}=\sup_n \Big(\int_\Omega \tilde{g}_\lambda (x,v_n)v_n +\int_{\partial\Omega} \tilde{f}_\lambda (x,v_n)v_n \Big)$. By Step 1, $\tilde{C}<\infty$. Given $\epsilon>0$, define \[ \mu_{\epsilon}^{1} =\max_{x\in \overline{\Omega}, |s| \le \frac{4\tilde{C}}{\epsilon}}|\tilde{g}_\lambda (x,s)||x|^{\beta} ,\; \mu_{\epsilon}^{2}=\max_{x\in \partial \Omega, |s| \le \frac{4\tilde{C}}{\epsilon}} |\tilde{f}_\lambda (x,s)|. \] Then for any $A\subset \Omega, B\subset \partial\Omega$ with $\mu(A)\le \frac{\epsilon}{4\mu_{\epsilon}^{1}}, |B|_1 \le \frac{\epsilon}{4\mu_{\epsilon}^{1}}$ we get, \begin{align*} &\int_A |\tilde{g}_\lambda (x,v_n)| + \int_B |\tilde{f}(x,v_n)|\\ &\le \int_{A\cap \{|v_n|\ge \frac{4\tilde{C}}{\epsilon}\}}\frac{|\tilde{g}_\lambda(x,v_n)v_n| }{|v_n|}+ \int_{A\cap\{|v_n|\le \frac{4\tilde{C}}{\epsilon}\}}|\tilde{g}_\lambda(x,v_n)|\\ &\quad + \int_{B\cap\{|v_n|\ge \frac{4\tilde{C}}{\epsilon}\}}\frac{|\tilde{f}_\lambda(x,v_n)v_n| }{|v_n|}+ \int_{B\cap\{|v_n|\le \frac{4\tilde{C}}{\epsilon}\}}|\tilde{f}_\lambda(x,v_n)| \\ &\le \frac{\epsilon}{2}+\mu(A)\mu_{\epsilon}^{1}+|B|_1 \mu_{\epsilon}^{2}\le \epsilon \end{align*} This completes Step 2 and the proof of the Lemma. \end{proof} Now we note that $\tilde{J}_{\lambda}(0)=0$ and $v=0$ is a local minimum for $\tilde{J}_{\lambda}$. It is also clear that $\lim_{t\to \infty} \tilde{J}_{\lambda}(tv)=-\infty$ for any $v\in H^1 (\Omega)\backslash\{0\}$. Hence, we may fix $e\in H^1 (\Omega)$ such that $\tilde{J}_{\lambda}(e)<0$. Let $\Gamma=\{\gamma:[0,1]\to H^1 (\Omega); \gamma$ is continuous, $\gamma(0)=0, \gamma(1)=e\}$. We define the mountain-pass level \[ \rho_0=\inf_{\gamma\in \Gamma}\sup_{t\in [0,1]} \tilde{J}_{\lambda} (\gamma(t)). \] It follows that $\rho_0\ge 0$. Let $R_0=\|e\|$. If $\rho_0=0$ we obtain that $\inf \{\tilde{J}_{\lambda}| \|v\|=R\}=0$ for all $R\in (0,R_0)$. We now let $F=H^1(\Omega)$ if $\rho_{0}>0$ and $F=\{\|v\|=\frac{R_0}{2}\}$ if $\rho_0=0$. We can now prove the following upper bound for $\rho_0$. \begin{lemma}\label{lem4.4} $\rho_0<\frac{\pi}{2}(2-\beta)$. \end{lemma} \begin{proof} Let $\{w_n\}$ be the sequence as in Lemma \ref{lem1} by taking $n=\frac{1}{l}$. We now suppose $\rho_0\ge \frac{\pi}{2}(2-\beta)$ and derive a contradiction. This means that (thanks to Lemma 3.1 in \cite{NT}) for some $t_n>0$, \[ \tilde{J}_{\lambda}(t_n w_n)=\sup_{t>0}\tilde{J}_{\lambda}(t_n w_n) \ge \frac{\pi}{2}(2-\beta), \quad \forall n. \] Then the above inequality gives, \begin{equation}\label{eq:d10} \frac{t_{n}^{2}}{2}-\int_{\Omega}\tilde{G}_\lambda(x,t_n w_n)- \int_{\partial\Omega}\tilde{F}_\lambda (x,t_n w_n) \ge \frac{\pi}{2}(2-\beta), \quad \forall n. \end{equation} In particular, \begin{equation}\label{eq:d11} t_{n}^{2}\ge \pi(2-\beta), \quad \text{for all large } n. \end{equation} Since $t_n$ yields is a maximum for the map $t \mapsto \tilde{J}_{\lambda}(t w_n)$ on $(0,\infty)$, $\frac{d}{dt}(\tilde{J}_{\lambda}(t w_n))|_{t=t_n}=0$. That is, \begin{equation}\label{eq:d12} t_{n}^{2}=\int_{\Omega}\tilde{g}_\lambda (x,t_n w_n)t_n w_n + \int_{\partial\Omega} \tilde{f}_\lambda (x,t_n w_n)t_n w_n. \end{equation} We note that $\inf_{x\in \overline{\Omega}} \tilde{g}_\lambda (x,s) \ge e^{s^2}$ for $s$ large. Since $t_n w_n\to \infty$ on $\{|x|\le \frac{\delta}{n}\}$ we get from (\ref{eq:d12}), \begin{align*} t_{n}^{2}&\ge \int_{\{|x|\le \frac{\delta}{n}\}\cap\overline{\Omega}}\tilde{g}_\lambda (x,t_n w_n)t_n w_n \\ &\ge \int_{\{|x|\le \frac{\delta}{n}\}} \frac{e^{t_{n}^{2} w_{n}^{2}}}{|x|^\beta} t_n w_n\\ &=\frac{\sqrt{\pi}\delta^2}{\sqrt{2}} e^{t_{n}^{2}\frac{\log n}{2\pi} }(t_n (\log n)^{1/2})\Big(\int_{|x|\le \delta/n \cap \overline{\Omega}} |x|^{-\beta}dx\Big)\\ &=C e^{\frac{t_{n}^{2}}{\pi}\log n} (\frac{\delta}{n})^{(2-\beta)} t_n (\log n)^{1/2}\\ &=C e^{(\frac{t_{n}^{2}}{\pi}-(2-\beta))\log n} t_n (\log n)^{1/2} \end{align*} Clearly the above inequality implies that $\{t_n\}$ is bounded sequence. Using (\ref{eq:d11}) in the above inequality, \[ t_{n}^{2}\ge \sqrt{\frac{\pi}{2} } \delta^2 e^{\epsilon \log n } t_n (\log n)^{1/2},\quad \text{for some } \epsilon>0 \] which implies $t_n \to \infty$ as $n\to \infty$, a contradiction. This contradiction shows that $\rho_0<\frac{\pi}{2}(2-\beta)$. \end{proof} \begin{lemma}\label{lem4.5} $\tilde{J}_{\lambda}$ has a critical point $v_\lambda$ of mountain-pass type with $v_\lambda >0$ in $\Omega$. \end{lemma} \begin{proof} Let $\{v_n\}\subset H^1 (\Omega)$ be a $(P.S)_{F,\rho}$ sequence for $\tilde{J}_{\lambda}$(for the existence of such a sequence, see \cite{GP}). Then, by Lemma \ref{lem4.3}, up to a subsequence, $v_n \rightharpoonup v_\lambda$ in $H^1 (\Omega)$ for some $v_\lambda \in H^1 (\Omega)$. Clearly $v_n \to v_\lambda$ point wise a.e. in $\Omega$. If $v_\lambda\not\equiv 0$,then it is easy to see from (\ref{eq:d2}) using weak maximum principle that $v_\lambda$ has no nonpositive local minimum in $\Omega$. By an application of Hopf maximum principle and the fact that $\frac{\partial v_\lambda}{\partial \nu} \ge 0$ on $\partial \Omega$, we conclude that $\min_{\overline{\Omega}}v_\lambda$ is achieved inside $\Omega$. Hence $v_\lambda >0$ in $\overline{\Omega}$. So it is enough to show that $v_\lambda \not\equiv 0$. We divide the proof into steps: \noindent\textbf{Case 1:} $\rho_0 =0$ In this case, from (\ref{eq:d3}) we get \begin{align*} o_n(1)=J(v_n) &=\frac{1}{2}\int_{\Omega} \left(|\nabla v_n|^2 + v_{n}^{2}\right)-\int_{\Omega}\tilde{G}(x,v_n) dx -\int_{\partial \Omega}\tilde{F}(x,v_n)\\ &=\frac{1}{2}\|v_n\|^{2} +o_n (1) \end{align*} \noindent\textbf{Case 2:} $\rho_0 \in (0,\frac{\pi}{2}(2-\beta))$ Since $J(v_n)\to \rho_0$, we obtain, $\|v_n\| \to 2\rho_0 \; \text{as}\; n\to \infty$. This and Lemma 4.4 immediately imply that $\|v_n\|\le \pi(2-\beta)-\epsilon_0$ for some $\epsilon_0>0$. Let $0<\delta<\frac{\epsilon_0}{\pi(2-\beta)-\epsilon_0}$ and let $q_0=\frac{\pi(2-\beta)}{(1+\delta)(\pi(2-\beta)-\epsilon_0)}$. Then $q_0>1$. Now we may choose $q$ such that $10$. Hence $v_\lambda \not\equiv 0$. \end{proof} \begin{proof}[Proof of Theorem 1.2] From Lemma \ref{lem4.5} and the arguments in section 5, we obtain that apart from $u_\lambda$ we obtain a second solution $\tilde{u}_\lambda$ for all $\lambda\in (0,\Lambda)$, and by definition of $\Lambda$, \eqref{Plambda} has no solution for $\lambda>\Lambda$. When $\lambda=\Lambda$, from Lemma \ref{lem4.4} it is clear that $J_\lambda (u_\lambda)\le J_\lambda (v_\lambda)<0$. Let $\{\lambda_n\}$ be a sequence such that $\lambda_n \to \Lambda$ and $\{u_{\lambda_n}\}$ be the corresponding sequence of solutions to $(P_{\lambda_n})$. Then, \begin{equation}\label{eq:a5} \limsup_{n \to \infty} J_{\lambda_n}(u_{\lambda_n}) \le 0,\quad J'_{\lambda_n}(u_{\lambda_n}) = 0. \end{equation} Now (\ref{eq:a5}) implies that $\{u_{\lambda_n}\}$ is a bounded sequence in $H^{1}(\Omega)$. Hence there exists $u_\Lambda$ such that $u_{\lambda_n} \rightharpoonup u_\Lambda$ in $H^{1}(\Omega)$. 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