\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 57, 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 (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/57\hfil A multiplicity result] {A multiplicity result for quasilinear problems with convex and concave nonlinearities and nonlinear boundary conditions in unbounded domains} \author[D. A. Kandilakis\hfil EJDE-2005/57\hfilneg] {Dimitrios A. Kandilakis} \address{Dimitrios A. Kandilakis \hfill\break Department of Sciences, Technical University Of Crete, Chania, Crete 73100 Greece} \email{dkan@science.tuc.gr} \date{} \thanks{Submitted September 27, 2004. Published May 31, 2005.} \subjclass[2000]{35J20, 35J60} \keywords{Variational method; fibering method; Palais-Smale condition; genus} \begin{abstract} We study the following quasilinear problem with nonlinear boundary conditions \begin{gather*} -\Delta_{p}u=\lambda a(x)|u|^{p-2}u+k(x)|u|^{q-2}u-h(x)|u|^{s-2}u, \quad \text{in }\Omega,\\ |\nabla u|^{p-2}\nabla u\cdot\eta+b(x)|u|^{p-2}u=0\quad \text{on }\partial\Omega, \end{gather*} where $\Omega$ is an unbounded domain in $\mathbb{R}^{N}$ with a noncompact and smooth boundary $\partial\Omega$, $\eta$ denotes the unit outward normal vector on $\partial\Omega$, $\Delta_{p}u=\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)$ is the $p$-Laplacian, $a$, $k$, $h$ and $b$ are nonnegative essentially bounded functions, $q0\}>0$ and there exist positive constants $K_{1}$ and $\alpha_{2}$, with $\frac{p}{q}<\frac{\alpha_{1}-N} {\alpha_{2}-N}$, such that $k(x)\leq\frac{K_{1}}{\left( 1+|x|\right) ^{\alpha_{2}}}\quad \text{a.e. in }\Omega.$ \item[(H)] $h\in L^{\infty}(\Omega)$, $h\geq0$ a.e. and $m\{x\in\Omega:h(x)>0\}>0$. \item[(B)] $b\in C(\mathbb{R}^{N})$ and $\frac{B_{1}}{\left( 1+|x|\right) ^{p-1}}\leq b(x)\leq\frac{B_{2}}{\left( 1+|x|\right) ^{p-1}},$ where $B_{1},B_{2}>0$. \end{itemize} The growing attention in the study of the p-Laplace operator $\Delta_{p}$ is motivated by the fact that it arises in various applications, e.g. non-Newtonian fluids, reaction-diffusion problems, flow through porus media, glacial sliding, theory of superconductors, biology etc. (see \cite{show98}, \cite{ci-mo-ra}, \cite{pe-re} and the references therein). The existence of nontrivial solutions to equations like (1) with a power like right hand side has received considerable attention since the work of Brezis and Nirenberg \cite{bre-nir}. When $\Omega$ is bounded, $p=2$ and $10$. If $\Omega=\mathbb{R}^{N}$ and $h\geq0$ we refer to \cite{liu-li} where it was shown that \eqref{e1} admits an infinite number of solutions in $D^{1,p} (\mathbb{R}^{N})$. In this paper we study \eqref{e1} in connection with the corresponding eigenvalue problem for the $p$-Laplacian: $-\Delta_{p}u=\lambda a(x)|u|^{p-2}u$ subject to the nonlinear boundary condition in \eqref{e1}. We show that the first eigenvalue $\lambda_{1}$ is positive, simple and isolated, the associated eigenvectors do not change sign and form a vector space of dimension 1. Then we combine the method employed in \cite{liu-li} with the results in \cite{Pfl} in order to show that if $\lambda<\lambda_{1}$ then \eqref{e1} admits an infinite number of solutions, while if $\lambda =\lambda_{1}$ we use the fibering method (which is also applicable in case $\lambda<\lambda_{1}$) to show that it admits at least one nonnegative solution. To be more specific, we establish the following \begin{theorem}\label{T} Suppose that (D), (A), (K), (H) and (B) are satisfied. \begin{itemize} \item[(i)] If $\lambda<\lambda_{1}$ then \eqref{e1} admits infinitely many solutions with negative energy. If in addition $k>0$ a.e., then it also admits a nonnegative solution. \item[(ii)] If $\lambda=\lambda_{1}$and $k>0$ a.e., then \eqref{e1} admits at least one nonnegative solution with negative energy. \end{itemize} \end{theorem} The proof of Theorem \ref{T} will be given in Sections 4 and 5. \section{Preliminaries} Let $C_{\delta}^{\infty}(\Omega)$ be the space of $C_{0}^{\infty }(\mathbb{R}^{N})-$functions restricted on $\Omega$. Then the weighted Sobolev space $E_{p}$ is the completion of $C_{\delta}^{\infty}(\Omega)$ in the norm $|||u|||_{p}=\Big( \int_{\Omega}\left\vert \nabla u\right\vert ^{p} dx+\int_{\Omega}\frac{1}{(1+|x|)^{p}}\left\vert u\right\vert ^{p}dx\Big)^{1/p}\,.$ By \cite[Lemma 2]{Pfl} we see that if $b(\cdot)$ satisfies (B), then the norm $$\|u\|_{1,p}=\Big(\int_{\Omega}\left\vert \nabla u\right\vert ^{p} dx+\int_{\partial\Omega}b(x)\left\vert u\right\vert ^{p}d\sigma(x)\Big)^{1/p} \label{8}$$ is equivalent to $|||\cdot|||_{p}$ ($\sigma(\cdot)$ being the surface measure on $\partial\Omega$). Let $w_{\alpha}(x):=\frac{1}{(1+|x|)^{\alpha}}$where $\alpha \in\mathbb{R}$. If $\Sigma$ is a measurable subset of $\mathbb{R}^{N}$, we assume that the weighted Lebesgue space $L^{r}(w_{\alpha},\Sigma) :=\{u:\int_{\Sigma}w_{\alpha}(x)|u(x)|^{r}dx<+\infty\},$ $r\in(1,+\infty)$, is supplied with the norm $\|u\|_{w_{\alpha},r}=\Big( \int_{\Sigma}w_{\alpha}(x)|u(x)|^{r}dx\Big) ^{1/r}.$ For a nonnegative measurable function $h:\Sigma\to \mathbb{R}$, the space $L^{s}(h,\Sigma)$ is similarly defined. We associate with it the seminorm $|u|_{h,s}=\big( \int_{\Sigma}h(x)|u(x)|^{s}dx\big) ^{1/s}$. Let $E=E_{p}\cap L^{s}(h,\Omega)$. Then $E$ endowed with the norm $\|\cdot\|_{E}=\|\cdot\|_{1,p}+|\cdot|_{h,s}$ becomes a separable Banach space. \begin{lemma}\label{B} \begin{itemize} \item[(i)] If $p\leq r\leq\frac{pN}{N-p}\quad\text{and }\quad N>\alpha\geq N-r\frac{N-p}{p},$ then the embedding $E\subseteq L^{r}(w_{\alpha},\Omega)$ is continuous. If the upper bound for $r$ in the first inequality and the lower bound for $\alpha$ in the second are strict, then the embedding is compact. \item[(ii)] If $p\leq m\leq\frac{p(N-1)}{N-p}\quad\text{and}\quad N>\beta\geq N-1-m\frac{N-p}{p},$ then the embedding $E\subseteq L^{m}(w_{\beta},\partial\Omega)$ is continuous. If the upper bound for $m$ in the first inequality and the lower bound for $\beta$ are strict, then the embedding is compact. \item[(iii)] If $1\frac{p}{q},$ then the embedding $L^{p}(w_{\alpha_{1}},\Omega)\subseteq L^{q}(w_{\alpha_{2} },\Omega)$ is continuous. \end{itemize} \end{lemma} \begin{proof} The first and second part of the lemma corresponds to \cite[Theorem 1]{Pfl}, while the third is a consequence of the following inequality $\int_{\Omega}\frac{1}{\left( 1+|x|\right) ^{\alpha_{2}}}|u|^{q}dx \leq\Big( \int_{\Omega}\frac{1}{\left( 1+|x|\right) ^{d}}dx\Big) ^{\frac{p-q}{p} } \Big(\int_{\Omega}\frac{1}{\left( 1+|x|\right) ^{\alpha_{1}}} |u|^{p}dx\Big) ^{q/p},$ where $d=(\alpha_{2}p-\alpha_{1}q)/(p-q)$. Note that the integral $\int_{\Omega}\frac{1}{\left( 1+|x|\right) ^{d}}dx$ converges since $d>N$. \end{proof} The energy functional $\Phi_{\lambda}:E\to \mathbb{R}$ corresponding to our problem is \begin{aligned} \Phi_{\lambda}(u)&=\frac{1}{p}\int_{\Omega}|\nabla u|^{p}dx-\frac{\lambda} {p}\int_{\Omega}a|u|^{p}dx-\frac{1}{q}\int_{\Omega}k|u|^{q}dx \\ &\quad +\frac{1}{s}\int_{\Omega}h|u|^{s}dx+\frac{1}{p}\int_{\partial\Omega} b|u|^{p}d\sigma(x). \end{aligned}\label{b} It is clear that if (D), (A), (K), (H) and (B) are satisfied, then $\Phi_{\lambda}(.)$ is continuously differentiable and its critical points correspond to solutions of \eqref{e1}. \section{The principal eigenvalue} In this section we examine the properties of the first eigenvalue $\lambda_{1}$ and the associated eigenvectors of the following problem $$\begin{gathered} -\Delta_{p}u=\lambda a(x)|u|^{p-2}u \quad\text{in }\Omega\\ |\nabla u|^{p-2}\nabla u\cdot\eta+b(x)|u|^{p-2}u=0 \quad\text{on }\partial\Omega. \end{gathered} \label{e7}$$ \begin{proposition}\label{K} Suppose that $10$ such that the interval $(0,\lambda_{1}+\xi)$ does not contain any eigenvalue other than $\lambda_{1}$. \end{itemize} \end{proposition} \begin{proof} (i) Let $I$, $J:E_{p}\to \mathbb{R}$ be defined by $I(u)=\int_{\Omega}|\nabla u|^{p}dx+\int_{\partial\Omega}b(x)|u|^{p} d\sigma(x), \quad J(u)=\int_{\Omega}a(x)|u|^{p}dx.$ Then the operators $I$, $J$ are continuously Fr\'{e}chet differentiable, $I(.)$ is coercive, $J'$ is compact and $J'(u)=0$ implies that $u=0$. Theorem 6.3.2 in \cite{Ber} implies the existence of a principal eigenvalue satisfying $$\lambda_{1}=\inf\limits_{J(u)=1}I(u). \label{12}$$ The positivity of $\lambda_{1}$follows by a standard argument.\smallskip \noindent (ii) Let $u_{1}$ be an eigenfunction corresponding to $\lambda_{1}$. Since $|u_{1}|$ is also a minimizer in (\ref{12}), we may assume that $u_{1}\geq0$. We will show first that $w_{\alpha_{1}}u_{1}$is essentially bounded in $\Omega$. To that purpose for $M>0$ define $u_{M}(x):=\min \{u_{1}(x),M\}$. Multiplying \eqref{e7} by $u_{M}^{kp+1}$, $k>0,$ and integrating over $\Omega$, we obtain $$\int_{\Omega}\,|\nabla u_{1}|^{p-2}\,\nabla u_{1}\cdot\nabla(u_{M} ^{kp+1})\,dx+\int_{\partial\Omega}b(x)\,u_{M}^{(k+1)p}\,d\sigma(x) \leq\lambda_{1}\int_{\Omega}a(x)\,u_{1}^{(k+1)p}\,dx\,. \label{o}$$ Note that \begin{align*} \int_{\Omega}\,|\nabla u_{1}|^{p-2}\,\nabla u_{1}\cdot\nabla(u_{M}^{kp+1})\,dx & =(kp+1)\int_{\Omega}\,|\nabla u_{M}|^{p}u_{M}^{kp}dx \\ & =\frac{kp+1}{(k+1)^{p}}\int_{\Omega}\,\,|\nabla u_{M}^{k+1}|^{p}\,dx,\,. \end{align*} So since $\frac{kp+1}{(k+1)^{p}}\leq1$, it follows that \begin{aligned} & \int_{\Omega}\,|\nabla u_{1}|^{p-2}\,\nabla u_{1}\cdot\nabla(u_{M} ^{kp+1})\,dx+\int_{\partial\Omega}b(x)\,u_{M}^{(k+1)p}\,d\sigma(x)\\ & \geq c_{1}\dfrac{kp+1}{(k+1)^{p}}\Big( \int_{\Omega}\frac{1} {(1+|x|)^{\alpha_{1}}}u_{M}^{(k+1)p^{\ast}}\,dx\Big)^{p/p^{\ast}}, \end{aligned} \label{6} due to the embedding $E_{p}\subseteq L^{p^{\ast}}(w_{\alpha_{1}},\Omega)$. By hypothesis (A), (\ref{o}) and (\ref{6}) we get that \begin{align*} & \Big( \int_{\Omega}\frac{1}{(1+|x|)^{\alpha_{1}}}u_{M}^{(k+1)p^{\ast} }\,dx\Big) ^{1/p^{\ast}}\\ & \leq\left( \frac{\lambda_{1}A_{2}(k+1)^{p}}{c_{3}(kp+1)}\right) ^{1/p}\Big( \int_{\Omega}\frac{1}{\left( 1+|x|\right) ^{\alpha _{1}}}\,u_{1}^{(k+1)p}dx\Big) ^{1/p}\,, \end{align*} so $\|u_{M}\|_{w_{\alpha_{1}},(k+1)p^{\ast}}\leq\Big( \frac{\lambda_{1} A_{2}(k+1)^{p}}{c_{3}(kp+1)}\Big) ^{1/((k+1)p)} \|u_{1}\|_{w_{\alpha _{1}},(k+1)p}.$ A bootstrap argument, as in the proof of \cite[Lemma 3.2]{dra-ku-ni}, shows that $w_{\alpha_{1}}u_{1}$ is essentially bounded. Theorems 1.9 and 1.11 in \cite{dra-ku-ni} imply that $u_{1}\in C_{\rm loc}^{1,\delta}(\Omega)$ and $u_{1}>0$ in $\Omega$. We show next that $E_{1}$ is one dimensional by employing a technique similar to the one exposed in \cite{al-hu}. Namely, we shall prove that if for $\lambda>0$, $w_{1}$ is a solution of $$-\Delta_{p}u\leq\lambda a(x)|u|^{p-2}u\text{\qquad in }\Omega, \label{33}$$ and $z_{1}$ is a solution of $$-\Delta_{p}u\geq\lambda a(x)|u|^{p-2}u\text{\qquad in }\Omega, \label{34}$$ $w_{1}$, $z_{1}>0$ on $\Omega$ and satisfying the boundary condition in \eqref{e1}, then $z_{1}=cw_{1}$ for some constant $c>0$. For $\varepsilon>0$ let $z_{1\varepsilon}=z_{1}+\varepsilon$. If $\varphi\in C_{\delta}^{\infty}(\Omega)$, $\varphi\geq0$, then $\frac{\varphi^{p}}{(z_{1\varepsilon})^{p-1}}\in E_{p}$. By Picone's identity \cite{al-hu}, we get \begin{align*} 0&\leq\int_{\Omega}|\nabla\varphi|^{p}dx-\int_{\Omega}\nabla\Big( \frac{\varphi^{p}}{z_{1\varepsilon}^{p-1}}\Big) \cdot|\nabla z_{1} |^{p-2}\nabla z_{1}dx \\ &=\int_{\Omega}|\nabla\varphi|^{p}dx+\int_{\Omega}\frac{\varphi^{p} }{z_{1\varepsilon}^{p-1}}\Delta_{p}z_{1}dx-\int_{\partial\Omega}\frac {\varphi^{p}}{z_{1\varepsilon}^{p-1}}|\nabla z_{1}|^{p-2}\nabla z_{1}\cdot\eta d\sigma(x) \\ & \leq\int_{\Omega}|\nabla\varphi|^{p}dx-\lambda\int_{\Omega}\frac {\varphi^{p}}{z_{1\varepsilon}^{p-1}}a(x)z_{1}^{p-1}dx -\int_{\partial\Omega}\frac{\varphi^{p}}{z_{1\varepsilon}^{p-1}}|\nabla z_{1}|^{p-2}\nabla z_{1}\cdot\eta d\sigma(x)\,, \end{align*} while the boundary condition implies that $0\leq\int_{\Omega}|\nabla\varphi|^{p}dx-\lambda\int_{\Omega}a(x)\frac {\varphi^{p}}{z_{1\varepsilon}^{p-1}}z_{1}^{p-1}dx+\int_{\partial\Omega }b(x)\frac{\varphi^{p}}{z_{1\varepsilon}^{p-1}}z_{1}^{p-1}d\sigma(x).$ If we let $\varepsilon\to 0$ and $\varphi\to w_{1}$ in $E_{p}$, we get $$0\leq\int_{\Omega}|\nabla w_{1}|^{p}dx-\lambda\int_{\Omega}a(x)w_{1} ^{p}dx+\int_{\partial\Omega}b(x)w_{1}^{p}d\sigma(x). \label{cf}$$ We can now work as in Theorem 2.1 in \cite{al-hu} to conclude that $E_{1}$ is a vector space of dimension 1. The same technique can be used to demonstrate that positive solutions in $\Omega$ correspond only to the first eigenvalue. Assume for instance, that there exists an eigenpair $(\lambda ^{\ast},u_{2})$ such that $\lambda^{\ast}>\lambda_{1}$ and $u_{2}\geq 0$ a.e. in $\Omega$. Then $u_{1}$ is a solution of (\ref{33}) with $\lambda=\lambda_{1}$ and $u_{2}$ is a solution of (\ref{34}) with $\lambda=\lambda^{\ast}$. But then $u_{2}=cu_{1}$ for some $c>0$, a contradiction.\smallskip \noindent (iii) Assume that there exists a sequence of eigenpairs ${(\lambda }_{n}{,}u_{n}{)}$ with $\lambda_{n}\to \lambda_{1}$ and $\lambda_{n}\in(\lambda_{1},\lambda_{1}+\delta)$, $\delta>0$, for every $n\in \mathbb{N}$. Without loss of generality, we may also assume that $\|u_{n} \|_{1,p}=1$ for all $n\in\mathbb{N}$. Hence, there exists $\tilde{u}\in E_{p}$ such that $u_{n}\to \tilde{u}$ weakly in $E_{p}$. The simplicity of $\lambda_{1}$ implies that $\tilde{u}=u_{1}$ or $\tilde {u}=-u_{1}$. Let us suppose that $u_{n}\to u_{1}$ weakly in $E_{p}$. Multiplying \eqref{e7} by $u_{n}-u_{m}$ and integrating by parts we get \begin{align*} & \int_{\Omega}(|\nabla u_{n}|^{p-2}\nabla u_{n}-|\nabla u_{m}|^{p-2}\nabla u_{m})(\nabla u_{n}-\nabla u_{m})\,dx\\ & +\int_{\partial\Omega}b(x)(|u_{n}|^{p-2}u_{n}-|u_{m}|^{p-2}u_{m} )(u_{n}-u_{m})\,d\sigma(x)\\ &=\lambda_{n}\int_{\Omega}a(x)\left( |u_{n}|^{p-2}u_{n}-|u_{m}|^{p-2} u_{m}\right) (u_{n}-u_{m})\,dx \\ &\quad +(\lambda_{n}-\lambda_{m})\int_{\Omega}a(x)|u_{m}|^{p-2}u_{m}(u_{n} -u_{m})\,dx\,. \end{align*} Exploiting the compactness of the operator $J$ and the monotonicity of the $p$-Laplacian operator, we obtain $\int_{\Omega}\,|\nabla u_{n}|^{p}\,dx\to \int_{\Omega}\,|\nabla u_{1}|^{p}\,dx.$ The strict convexity of $L^{p}(\Omega)$ implies that $u_{n}\to u_{1}$ in $E_{p}$. For a fixed $n\in\mathbb{N}$ and for every $\phi\in E_{p}$ we have $\int_{\Omega}\,|\nabla u_{n}|^{p-2}\nabla u_{n}\nabla\phi\,dx+\int _{\partial\Omega}b(x)|u_{n}|^{p-2}u_{n}\phi\,d\sigma(x) =\lambda_{n}\int_{\Omega}a(x)|u_{n}|^{p-2}u_{n}\phi\,dx\,.$ Let $\mathcal{U}_{n}^{-}=:\{x\in\overline{\Omega}:u_{n}(x)<0\}$. By (iii) we must have $m(\mathcal{U}_{n}^{-})>0$. By choosing $\phi\equiv u_{n}^{-} =\min\{0,u_{n}\}$, it follows that $\int_{\mathcal{U}_{n}^{-}}\,|\nabla u_{n}^{-}|^{p}\,dx+\int_{\partial \Omega\cap\mathcal{U}_{n}^{-}}b(x)|u_{n}^{-}|^{p}\,dx =\lambda_{n}\int_{\mathcal{U}_{n}^{-}}a(x)|u_{n}^{-}|^{p}\,dx\,.$ Thus $$\|u_{n}^{-}\|_{1,p}^{p}\,\leq\,A_{2}\,(\lambda_{1}+\delta)\|u_{n}^{-} \|_{L^{p}(w_{\alpha_{1}},\mathcal{U}_{n}^{-})}^{p}, \label{35}$$ by (A). Denote by $B_{r}$ the ball with radius $r>0$ centered at $0\in\mathbb{R}^{n}$. For $\varepsilon\in(0,1)$ there exists $r_{\varepsilon ,n}>0$ such that $$\|u_{n}^{-}\|_{1,p}^{p}\,\leq A_{2}\,\,(\lambda_{1}+\delta)(\|u_{n} ^{-}\|_{L^{p}(w_{\alpha_{1}},\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}} )}^{p}+\varepsilon\|u_{n}^{-}\|_{1,p}^{p})\,. \label{45}$$ Apply once again the H\"{o}lder inequality to derive that \begin{aligned} &\|u_{n}^{-}\|_{L^{p}(w_{\alpha_{1}},\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon ,n}})}^{p} \\ &\leq\Big( \int_{\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}}}\frac {1}{\left( 1+|x|\right) ^{\frac{\alpha_{1}p^{\ast}}{p^{\ast}-p}}}dx\Big) ^{\frac{p^{\ast}-p}{p^{\ast}}}\Big(\int_{\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}}}|u_{n}^{-}|^{p^{\ast}}dx\Big) ^{p/p^{\ast}}. \end{aligned}\label{aa} By Lemma \ref{B} (i), $$\Big(\int_{\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}}}|u_{n} ^{-}|^{p^{\ast}}dx\Big) ^{p/p^{\ast}}\leq c_{2}\|u_{n}^{-} \|_{1,p}^{p} \label{44}$$ for some $c_{2}>0$. On combining (\ref{35})-(\ref{44}) we get $1\,-\varepsilon\leq c_{3}\Big( \int_{\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}}}\frac{1}{\left( 1+|x|\right) ^{\frac{\alpha_{1} p^{\ast}}{p^{\ast}-p}}}dx\Big) ^{\frac{p^{\ast}-p}{p^{\ast}}},$ so $m(\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}})>c_{4}>0$, where the constant $c_{4}$ is independent of $n\in\mathbb{N}$. It is clear that there exists $R>0$ such that $$m(B_{R}\cap(\mathcal{U}_{n}^{-}\cap B_{r_{\varepsilon,n}}))>\frac{c_{4}}{2} \label{38}$$ for every $n\in\mathbb{N}$. Since $u_{n}\to u_{1}$ in $E_{p}$ we have that $u_{n}\to u_{1}$ in $L^{p^{\ast}}(w_{\alpha_{1}} ,B_{R}\cap\Omega)$. By Egorov's Theorem, $u_{n}$ converges uniformly to $u_{1}$ on $B_{R}\cap\Omega$ with the exception of a set with arbitrarily small measure. But this contradicts (\ref{38}) and the conclusion follows. \end{proof} \begin{remark} \label{remark} \rm If $u_{1}$ is continuous at $x_{0}\in\partial\Omega$, then $u_{1}(x_{0})>0$. Indeed, if $u_{1}(x_{0})=0$, then by \cite[Theorem 5]{vaz} we would have $|\nabla u_{1}(x_{0})|^{p-2}\nabla u_{1}(x_{0})\cdot\eta(x_0)<0$, contradicting \eqref{e1}. \end{remark} \section{The case $\lambda<\lambda_{1}$} We need the following lemma in order to show that $\Phi_{\lambda}$ is coercive. \begin{lemma} \label{C}If $\lambda<\lambda_{1}$ then the norm $|||u|||_{1,p}:=\Big( \int_{\Omega}|\nabla u|^{p}dx+\int_{\partial\Omega }b|u|^{p}dx-\lambda\int_{\Omega}a|u|^{p}dx\Big) ^{1/p}$ is equivalent to $\|u\|_{1,p}$. \end{lemma} \begin{proof} Suppose that there exists $u_{n}\in E_{p}$, $n\in\mathbb{N}$, such that $\|u_{n}\|_{1,p}=1$ and $\int_{\Omega}|\nabla u_{n}|^{p}dx+\int_{\partial\Omega}b|u_{n}|^{p} d\sigma(x)-\lambda\int_{\Omega}a|u_{n}|^{p}dx\to 0.$ In view of (\ref{12}), $0\leq(\lambda_{1}-\lambda)\int_{\Omega}a|u_{n}|^{p}dx\leq\int_{\Omega}|\nabla u_{n}|^{p}dx+\int_{\partial\Omega}b|u_{n}|^{p}d\sigma(x)-\lambda\int_{\Omega }a|u_{n}|^{p}dx\to 0.$ Hence, $\int_{\Omega}a|u_{n}|^{p}dx\to 0$, which shows that $\|u_{n}\|_{1,p}\to 0$. This is a contradiction with $\|u_{n} \|_{1,p}=1$. \end{proof} We can now prove our first result concerning \eqref{e1}.\smallskip \subsection*{Proof of Theorem \ref{T}(i)} We will show that $\Phi_{\lambda}$ satisfies the Palais-Smale condition in $E$. So let $\{u_{n}\}_{n\in\mathbb{N}}$ be a sequence in $E$ such that $\Phi_{\lambda }(u_{n})$ is bounded and $\Phi_{\lambda}'(u_{n})\to 0$. By Lemma \ref{C} we get \begin{align*} \Phi_{\lambda}(u) & =\frac{1}{p}\Big( \int_{\Omega}|\nabla u|^{p} dx+\int_{\partial\Omega}b|u|^{p}d\sigma(x)-\lambda\int_{\Omega}a|u|^{p} dx\Big) \\ &\quad -\frac{1}{q}\int_{\Omega}k|u|^{q}dx+\frac{1}{s}\int_{\Omega}h|u|^{s}dx\\ &\geq\frac{1}{p}|||u|||_{1,p}^{p}-c_{5}|||u|||_{1,p}^{q}+\frac{1} {s}|u|_{h,s}^{s}, \end{align*} implying that $\Phi_{\lambda}(.)$ is coercive. Thus $\{u_{n}\}_{n\in \mathbb{N}}$ is bounded in $E$. Without loss of generality, we may assume that $u_{n}\to \overline{u}$ strongly in $L^{p}(w_{\alpha_{1}},\Omega)$ and $L^{q}(w_{\alpha_{2}},\Omega)$ and weakly in $L^{p}(w_{p-1},\partial\Omega)$, $E_{p}$and $L^{s}(h,\Omega)$. Thus \begin{gather} \int_{\Omega}a(x)|u_{n}-\overline{u}|^{p}dx\to 0,\quad\int_{\Omega }k(x)|u_{n}-\overline{u}|^{q}dx\to 0\,, \label{5} \\ \int_{\partial\Omega}b(x)|\overline{u}|^{p-2}\overline{u}(u_{n}-\overline {u})d\sigma(x)\to 0,\quad\int_{\Omega}|\nabla\overline{u}|^{p-2} \nabla\overline{u}\nabla(u_{n}-\overline{u})dx\to 0\,, \label{10} \\ \int_{\Omega}h(x)|\overline{u}|^{s-2}\overline{u}(u_{n}-\overline {u})dx\to 0\,. \label{11} \end{gather} Therefore, by (\ref{5})-(\ref{11}), $\left\langle \Phi_{\lambda}'(\overline{u}),u_{n}-\overline {u}\right\rangle \to 0.$ Since $\Phi_{\lambda}'(u_{n})\to 0$, we also have that $\left\langle \Phi_{\lambda}'(u_{n})-\Phi_{\lambda}' (\overline{u}),u_{n}-\overline{u}\right\rangle \to 0.$ Thus \begin{aligned} &\int_{\Omega}\left( |\nabla u_{n}|^{p-2}\nabla u_{n}-|\nabla u|^{p-2} \nabla\overline{u}\right) (\nabla u_{n}-\nabla\overline{u})dx\\ &-\lambda\int_{\Omega}a(x)\left( |u_{n}|^{p-2}u_{n}-|\overline{u} |^{p-2}\overline{u}\right) (u_{n}-\overline{u})dx\\ & -\int_{\Omega}k(x)\left( |u_{n}|^{q-2}u_{n}-|\overline{u}|^{q-2}\overline {u}\right) (u_{n}-\overline{u})dx\\ &+\int_{\partial\Omega}b(x)\left( |u_{n}|^{p-2}u_{n}-|\overline{u} |^{p-2}\overline{u}\right) (u_{n}-\overline{u})d\sigma(x)\\ &+\int_{\Omega}h(x)\left( |u_{n}|^{s-2}u_{n}-|\overline{u}|^{s-2}\overline {u}\right) (u_{n}-\overline{u})dx\to 0\,. \end{aligned} \label{4} On combining (\ref{5})-(\ref{4}) we get \begin{align*} & \int_{\Omega}\left( |\nabla u_{n}|^{p-2}\nabla u_{n}-|\nabla\overline {u}|^{p-2}\nabla\overline{u}\right) (\nabla u_{n}-\nabla\overline{u})dx\\ & +\int_{\partial\Omega}b(x)\left( |u_{n}|^{p-2}u_{n}-|\overline{u} |^{p-2}\overline{u}\right) (u_{n}-\overline{u})d\sigma(x)\\ &+\int_{\Omega}h(x)\left( |u_{n}|^{s-2}u_{n}-|\overline{u}|^{s-2}\overline {u}\right) (u_{n}-\overline{u})dx\to 0\,. \end{align*} We can now use the inequality \begin{align*} 0 & \leq\Big\{ \Big( \int_{\Omega}|f_{1}|^{r}dx\Big) ^{1/r' }-\Big( \int_{\Omega}|f_{2}|^{r}dx\Big) ^{1/r'}\Big\}\\ &\quad\times \Big\{ \Big( \int_{\Omega}|f_{1}|^{r}dx\Big) ^{1/r}-\Big( \int_{\Omega}|f_{2}|^{r}dx\Big) ^{1/r}\Big\}\\ &\leq\int_{\Omega}\left( |f_{1}|^{r-2}f_{1}-|f_{2}|^{r-2}f_{2}\right) (f_{1}-f_{2})dx, \end{align*} where $f_{1}$, $f_{2}\in L^{r}(\Omega)$, $r>1$, $r'=r/(r-1)$, to obtain $\|\nabla u_{n}\|_{p}\to \|\nabla\overline{u}\|_{p},\quad \|h^{\frac {1}{s}}u_{n}\|_{s}\to \|h^{\frac{1}{s}}\overline{u}\|_{s}\,.$ Exploiting the strict convexity of $L^{p}(\Omega)$ and $L^{s}(\Omega)$ we derive that $\nabla u_{n}\to \nabla\overline{u}$ in $\left(L^{p}(\Omega)\right) ^{N}$ and $u_{n}\to \overline{u}$ in $L^{s}(h,\Omega)$. Consequently, $u_{n}\to \overline{u}$ in $E$, proving the claim. Now let $Z=\{x\in\Omega:$ $k(x)=0\}$ and $E_{0}=\{u\in E:u(x)=0$ a.e. in $Z\}$. Define a norm on $E_{0}$ by $\|u\|_{E_{0}}=\|k^{1/q}u\|_{q}$. Consider the family $\Sigma$ of closed and symmetric subsets of $E\backslash \{0\}$. For $A\in\Sigma$ we define the genus $\gamma(A)$ of $A$ as the minimum of the $n\in\mathbb{N}$ such that there exists a continuous function $\varphi:A\to \mathbb{R}^{n}\backslash\{0\}$ with $\varphi (-x)=-\varphi(x)$. If no such $n$ exists, we define $\gamma(A)=+\infty$. We claim that for $n\in\mathbb{N}$ there exists $\varepsilon>0$ such that $\gamma(\{u\in E:\Phi_{\lambda}(u)\leq-\varepsilon\})\geq n$. It will be enough to show that the set $\{u\in E:\Phi_{\lambda}(u)\leq-\varepsilon\}$ contains an $n$-dimensional sphere centered at $0\in\mathbb{R}^N$. So let $E_{0}^{n}$ be an $n$-dimensional subspace of $E_{0}$. Then \begin{align*} \Phi_{\lambda}(u) & =\frac{1}{p}\Big( \int_{\Omega}|\nabla u|^{p} dx+\int_{\partial\Omega}b|u|^{p}d\sigma(x)-\lambda\int_{\Omega}a|u|^{p} dx\Big) \\ &\quad -\frac{1}{q}\int_{\Omega}k|u|^{q}dx+\frac{1}{s}\int_{\Omega}h|u|^{s}dx\\ & \leq\frac{1}{p}|||u|||_{1,p}^{p}-\frac{1}{q}\|u\|_{E_{0}}^{q}+\frac{1} {s}|u|_{h,s}^{s}\,. \end{align*} Since all norms on $E_{0}^{n}$ are equivalent, we have that $\Phi_{\lambda }(u)\leq c_{1}'\|u\|_{E_{0}^{n}}^{p}+c_{2}'\|u\|_{E_{0}^{n} }^{s}-c_{3}'\|u\|_{E_{0}^{n}}^{q}$, so there exists $\varepsilon>0$ and $\delta>0$ such that $\Phi_{\lambda}(u)\leq-\varepsilon$ for $\|u\|_{E_{0}^{n}}=\delta$. Thus $\{u\in E_{0}^{n}:\|u\|_{X}=\delta \}\subseteq\{u\in E:\Phi_{\lambda}(u)\leq-\varepsilon\}$, implying that $\gamma(\{u\in E:\Phi_{\lambda}(u)\leq-\varepsilon\})\geq n$. Let $\Sigma _{n}=\{A\in\Sigma:\gamma(A)\geq n\}$. Then the numbers $c_{n}=\inf_{A\in\Sigma_{n}}\sup_{u\in A}\Phi_{\lambda}(u)$ are critical values of $\Phi_{\lambda}$, providing an infinite sequence of critical points of $\Phi_{\lambda}$. For more details we refer to \cite{az-al}. For the existence of a nonnegative solution, see Remark \ref{R} in the next section. \section{The case $\lambda=\lambda_{1}$} In this section we apply the fibering method introduced by Pohozaev \cite{poh1}, \cite{poh2} in order to show that \eqref{e1} admits at least one nonnegative solution.\smallskip \subsection*{Proof of Theorem \ref{T} (ii)} We decompose the function $u\in E$ as $u(x)=rv(x)$ with $r\in\mathbb{R}$ and $v\in E$. By (\ref{b}) we have that \begin{align*} \Phi_{\lambda_{1}}(rv) &=\frac{|r|^{p}}{p}\Big( \int_{\Omega}|\nabla v|^{p}-\lambda_{1}\int_{\Omega}a|v|^{p}+\int_{\partial\Omega}b|v|^{p} d\sigma(x)\Big) \\ & \quad -\frac{|r|^{q}}{q}\int_{\Omega}k|v|^{q} +\frac{|r|^{s}}{s}\int_{\Omega}h|v|^{s}. \end{align*} If $u$ is a critical point of $\Phi_{\lambda_{1}}$, then $\frac{\partial \Phi_{\lambda_{1}}}{\partial r}=0$, so we will search for the critical points of $\Phi_{\lambda_{1}}$ among the ones which satisfy this equation, that is \begin{aligned} &|r|^{p-q}\Big( \int_{\Omega}|\nabla v|^{p}dx-\lambda_{1}\int_{\Omega }a|v|^{p}dx+\int_{\partial\Omega}b|v|^{p}d\sigma(x)\Big) +|r|^{s-q} \int_{\Omega}h|v|^{s}dx\\ &=\int_{\Omega}k|v|^{q}dx\,. \end{aligned}\label{c} Since $k>0$ a.e., for every $v\in E\backslash\{0\}$ there exists a unique $r=r(v)>0$ satisfying (\ref{c}). By using the implicit function theorem \cite[Thm. 4.B, p.150]{Zeid}, we see that the function $v\to r(v)$ is continuously differentiable for $v\neq0$. Clearly, $$r(\mu v)\mu v=r(v)v\quad \text{for every }\mu> 0\,.\label{k}$$ Also, in view of (\ref{c}) $$\Phi_{\lambda_{1}}(r(v)v)=\big( \frac{r^{q}}{p}-\frac{r^{q}}{q}\big) \int_{\Omega}k|v|^{q}dx+\big( \frac{r^{s}}{s}-\frac{r^{s}}{p}\big) \int_{\Omega}h|v|^{s}dx\leq 0\,.\label{3}$$ Let $H(v)=\int_{\Omega}|\nabla v|^{p}dx-\lambda_{1}\int_{\Omega}a|v|^{p} dx+\int_{\partial\Omega}b|v|^{p}d\sigma(x)+\int_{\Omega}h|v|^{s}dx.$ The variational characterization of $\lambda_{1}$ and hypothesis (H) imply that $H(v)\geq0$ for every $v\in E$. Let $W=\{v\in E:$ $H(v)=1\}$. By (\ref{12}), $W$ is bounded in $L^{s}(h,\Omega)$. Since $(H'(v),v)=p\Big( \int_{\Omega}|\nabla v|^{p}dx-\lambda_{1} \int_{\Omega}a|v|^{p}dx+\int_{\partial\Omega}b|v|^{p}d\sigma(x)\Big) +s\int_{\Omega}h|v|^{s}dx$ we see that $(H'(v),v)\neq0$ for $v\in W$. In view of \cite[Lemma 3.4]{Dra-Poh}, any conditional critical point of the function $\widehat{\Phi}_{\lambda_{1}}(v):=\Phi_{\lambda_{1}}(r(v)v)$ subject to $H(v)=1$ provides a critical point $r(v)v$ of $\Phi_{\lambda_{1}}$. Consider the problem $M_{1}=\inf\{\Phi_{\lambda_{1}}(r(v)v):v\in W\}.$ Suppose that $\{v_{n}\}_{n\in\mathbb{N}}$ is a minimizing sequence in $W$, that is $\Phi_{\lambda_{1}}(r(v_{n})v_{n})\to M_{1}$ and $H(v_{n})=\Big( \int_{\Omega}|\nabla v_{n}|^{p}dx-\lambda_{1}\int_{\Omega }a|v_{n}|^{p}dx+\int_{\partial\Omega}b|v_{n}|^{p}d\sigma(x)\Big) +\int_{\Omega}h|v_{n}|^{s}dx=1.$ Assume that $\|v_{n}\|_{1,p}\to +\infty$and let $u_{n}=\dfrac{v_{n} }{a_{n}}$ where $a_{n}=\|v_{n}\|_{1,p}$. Then $a_{n}^{p}\Big( \int_{\Omega}|\nabla u_{n}|^{p}dx-\lambda_{1}\int_{\Omega }a|u_{n}|^{p}dx+\int_{\partial\Omega}b|u_{n}|^{p}d\sigma(x)\Big) +a_{n} ^{s}\int_{\Omega}h|u_{n}|^{s}dx=1,$ so, by (\ref{12}), $$0\leq\int_{\Omega}|\nabla u_{n}|^{p}dx-\lambda_{1}\int_{\Omega}a|u_{n} |^{p}dx+\int_{\partial\Omega}b|u_{n}|^{p}d\sigma(x)\leq\dfrac{1}{a_{n}^{p} }\to 0\label{22}$$ and $$0\leq\int_{\Omega}h|u_{n}|^{s}dx\leq\dfrac{1}{a_{n}^{s}}\to 0.\label{25}$$ Thus $$\underset{n\to \infty}{\lim}\lambda_{1}\int_{\Omega}a|u_{n} |^{p}dx=1.\label{32}$$ Since $\|u_{n}\|_{1,p}=1$, by passing to a subsequence if necessary, we may assume that $u_{n}\to u$ weakly in $E_{p}$. In view of (\ref{32}) we get $\lambda_{1}\int_{\Omega}a|u|^{p}dx=1,$ so $u\neq0$. The lower semicontinuity of the norm of $E_{p}$ implies that $\int_{\Omega}|\nabla u|^{p}dx+\int_{\partial\Omega}b|u|^{p}d\sigma(x)\leq1,$ and (\ref{22}) gives $\int_{\Omega}|\nabla u|^{p}dx+\int_{\partial\Omega}b|u|^{p}d\sigma (x)=\lambda_{1}\int_{\Omega}a|u|^{p}dx.$ Thus $u$is an eigenfunction corresponding to $\lambda_{1}$. But then $\int_{\Omega}h|u|^{s}dx\leq\underset{n\to \infty}{\lim\inf}\int _{\Omega}h|u_{n}|^{s}dx=0\,,$ by (\ref{25}), a contradiction. Thus $\{v_{n}\}_{n\in\mathbb{N}}$ is bounded in $E_{p}$. Since $\{v_{n}\}_{n\in\mathbb{N}}$ is also bounded in $L^{s}(h,\Omega)$ we conclude that $\{v_{n}\}_{n\in\mathbb{N}}$ is bounded in $E$. Going back to (\ref{c}) we get that $r(W)$ is also bounded. Consequently, $I=\{\Phi_{\lambda_{1}}(r(v)v):v\in W\}$ is a bounded interval in $\mathbb{R}$ with endpoints $A,B$, $AG(r(v_{0}))=\Phi_{\lambda_{1}}(r(v_{0} )v_{0}).\label{g} Let$\gamma\geq1$be such that $$\Big( \int_{\Omega}|\nabla\gamma v_{0}|^{p}dx+\int_{\partial\Omega}b|\gamma v_{0}|^{p}d\sigma(x)-\lambda_{1}\int_{\Omega}a|\gamma v_{0}|^{p}dx\Big) +\int_{\Omega}h|\gamma v_{0}|^{s}dx=1, \label{n}$$ implying that$\gamma v_{0}\in W$. On combining (\ref{k}), (\ref{g}) and (\ref{n}) we obtain $\Phi_{\lambda_{1}}(r(\gamma v_{0})\gamma v_{0})=\Phi_{\lambda_{1}} (r(v_{0})v_{0})<\Phi_{\lambda_{1}}(dv_{0})\leq\underset{n\to +\infty }{\lim\inf}\Phi_{\lambda_{1}}(r(v_{n})v_{n})=A,$ that is$\Phi_{\lambda_{1}}(r(\gamma v_{0})\gamma v_{0})