\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 78, 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 (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2008/78\hfil Boundary eigencurve problems] {Boundary eigencurve problems involving the p-Laplacian operator} \author[A. El Khalil, M. Ouanan\hfil EJDE-2008/78\hfilneg] {Abdelouahed El Khalil, Mohammed Ouanan} % in alphabetical order \address{Abdelouahed El Khalil \newline Department of mathematics, College of Sciences\\ Al-Imam Muhammad Ibn Saud Islamic University\\ P. Box 90950, Riyadh 11623 Saudi Arabia} \email{alakhalil@imamu.edu.sa \quad lkhlil@hotmail.com} \address{Mohammed Ouanan \newline University Moulay Ismail\\ Faculty of Sciences et Technology\\ Department of informatics\\ P.O. Box 509, Boutalamine\\ 52000 Errachidia, Morocco} \email{m\_ouanan@hotmail.com} \thanks{Submitted September 16, 2007. Published May 27, 2008.} \subjclass[2000]{35P30, 35J20, 35J60} \keywords{p-Laplacian operator; nonlinear boundary conditions; \hfill\break\indent principal eigencurve; Sobolev trace embedding} \begin{abstract} In this paper, we show that for each $\lambda \in \mathbb{R}$, there is an increasing sequence of eigenvalues for the nonlinear boundary-value problem \begin{gather*} \Delta_pu=|u|^{p-2}u \quad \text{in } \Omega\\ |\nabla u|^{p-2}\frac{\partial u}{\partial \nu}=\lambda \rho(x)|u|^{p-2}u+\mu|u|^{p-2}u \quad \text{on } \partial \Omega\,; \end{gather*} also we show that the first eigenvalue is simple and isolated. Some results about their variation, density, and continuous dependence on the parameter $\lambda$ are obtained. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \textbf{Editor's note:} After publication, we learned that this article is an unauthorized copy of ``On the principal eigencurve of the p-Laplacian related to the Sobolev trace embedding'', Applicationes Mathematicae, 32, 1 (2005), 1-16. The authors alone are responsible for this action which may be in violation of the Copyright Laws. \section{Introduction and Notation} Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^N$, with $N\geq 1$. Let $\rho$ be a function in $L^{\infty}(\partial\Omega)$ with $\rho\not\equiv 0$ and that can change sign. Let $\lambda, p, \mu$ be real numbers, with $1-\infty, \label{e3.1} \end{equation} where $\lambda_{1}=\mu_{1}(0)$ is the reciprocal of the optimal constant in the Sobolev trace embedding $W^{1,p}(\Omega)\hookrightarrow L^p(\partial \Omega)$. By employing the Sobolev trace embedding, we deduce that \begin{itemize} \item $\Psi$ and $\Phi$ are weakly continuous \item $\Psi'$ and $\Phi'$ are compact. \end{itemize} The following lemma is the key to show the existence of eigenvalues. \begin{lemma} \label{lem3.1} For each $\lambda\in\mathbb{R}$, we have \begin{enumerate} \item[(i)] $(\Phi_{\lambda})'$ maps the bounded sets in the bounded sets; \item[(ii)] if $u_{n}\rightharpoonup u$ (weakly) in $W^{1,p}(\Omega)$ and $(\Phi_{\lambda})'(u_{n})$ converges strongly in the space $(W^{1,p}(\Omega))'$, then $u_n\to u$ (strongly) in $W^{1,p}(\Omega)$; \item[(iii)] the functional $\Phi_{\lambda}$ satisfies the Palais-Smale condition on $\mathcal{M}$; i.e., for each sequence $(u_{n})_{n}\subset\mathcal{M}$, if $\Phi_{\lambda}(u_{n})$ is bounded and \begin{equation} (\Phi_{\lambda})'(u_{n})-c_{n}\Psi'(u_n)\to 0, \label{e3.2} \end{equation} with $c_n=\frac{\langle (\Phi_{\lambda})'(u_n),u_n\rangle}{\langle \Psi'(u_n),u_n\rangle}$. Then, $(u_n)_n$ has a convergent subsequence in $W^{1,p}(\Omega)$. \end{enumerate} \end{lemma} \begin{proof} (i) Let $u,v \in W^{1,p}(\Omega).$ Thus $$ \langle (\Phi_{\lambda})'(u),v\rangle =\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla v dx+\int_{\Omega}|u|^{p-2}uv dx+\int_{\partial\Omega}\rho(x)|u|^{p-2}uvd\sigma. $$ By H\"older's inequality, we obtain \begin{align*} |\langle(\Phi_{\lambda})'(u),v\rangle| &\leq \Big(\int_{\Omega}|\nabla u|^{(p-1)p'}dx\Big)^{1/p'} \|\nabla v\|_{p} +\Big(\int_{\Omega}| u|^{(p-1)p'}dx\Big)^{1/p'}\|v\|_{p} \\ &\quad + |\lambda|\|\rho\|_{\infty,\partial\Omega} \Big(\int_{\partial\Omega}| u|^{(p-1)p'}d\sigma\Big)^{1/p'}\|v\|_{p,\partial\Omega} \\ &= \|\nabla u\|_{p}^{p-1}\|\nabla v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}+ |\lambda|\|\rho\|_{\infty,\partial\Omega} \|u\|_{p,\partial\Omega}^{p-1}\|v\|_{p,\partial\Omega}. \end{align*} Now, the trace Sobolev embedding $W^{1,p}(\Omega)\hookrightarrow L^{p}(\partial\Omega)$ ensures the existence of a constance $c>0$ such that $$ \|w\|_{p,\partial\Omega}\leq c\|w\|_{1,p}, \quad \text{for each } w\in W^{1,p}(\Omega). $$ Therefore, $$ \|(\Phi_{\lambda})'(u)\|\leq \|\nabla u\|_{p}^{p-1}\|\nabla v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}+ c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1}\|v\|_{1,p}. $$ It is clear that $$ \|\nabla u\|_{p}^{p-1}\|\nabla v\|_{p}+\|u\|_{p}^{p-1}\|v\|_{p}\leq \|u\|_{1,p}^{p-1}\|v\|_{1,p}. $$ Combining the above inequalities, we conclude that $$ |\langle(\Phi_{\lambda})'(u),v\langle|\leq (1+c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1}\|v\|_{1,p}, $$ for any $u,v\in W^{1,p}(\Omega)$. It follows that $$ \|(\Phi_{\lambda})'(u)\|\leq (1+c^{p}|\lambda|\|\rho\|_{\infty,\partial\Omega})\|u\|_{1,p}^{p-1}, $$ where $\|\cdot\|$ denotes the norm of $(W^{1,p}(\Omega))'$. This implies assertion (i). (ii) We use the condition $(S_+)$ as follows. $(\Phi_{\lambda})'(u_n)$ being a convergent sequence strongly to some $f\in (W^{1,p}(\Omega))'$. Thus, we have by calculation \begin{equation} \langle Au_{n},v\rangle =\langle -\Delta_{p}u_{n},v\rangle+\int_{\Omega}|u_{n}|^{p-2}u_{n}vdx+ \int_{\partial\Omega}|\nabla u_{n}|^{p-2}\nabla u_{n}\nu v\,d\sigma,\label{e3.3} \end{equation} for any $v\in W^{1,p}(\Omega)$, where $A$ is an operator defined from $W^{1,p}(\Omega)$ into $(W^{1,p}(\Omega))'$ by $$ \langle Au,v\rangle =\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla v dx+\int_{\Omega}|u|^{p-2}uv\,dx. $$ This operator satisfies the condition $(S_{+})$ because $-\Delta_{p}$ does it \cite{e1}. If we take $v=u_{n}-u$ in \eqref{e3.3} we obtain \begin{align*} &\langle Au_{n},u_{n}-v\rangle\\ &=\langle -\Delta_{p}u_{n},u_{n}-v\rangle+\int_{\Omega} |u_{n}|^{p-2}u_{n}(u_{n}-u)dx+\int_{\partial\Omega}|\nabla u_{n}|^{p-2}\nabla u_{n}\nu (u_{n}-u)\, d\sigma. \end{align*} Introducing $(\Phi_{\lambda})'(u_{n})$, we deduce that $$ \langle Au_{n},u_{n}-u\rangle=\langle (\Phi_{\lambda})'(u_{n})-f,u_{n}-u\rangle +\langle f,u_{n}-u\rangle -\langle(\Phi_{\lambda})'(u_n),u_n-u\rangle. $$ Using the compactness of $\Phi'$, we find that as $n\to \infty$, $$ \limsup_{n\to +\infty}\langle Au_{n},u_{n}-u\rangle \geq 0. $$ Hence $u_{n}\to u$ strongly in $W^{1,p}(\Omega)$, in virtue of the condition $(S_{+})$. (iii) From \eqref{e3.1} we deduce that $(u_{n})_{n}$ is bounded in $W^{1,p}(\Omega)$. Thus, without loss of generality, we can assume that $u_{n}\rightharpoonup u$ (weakly) in $W^{1,p}(\Omega)$ for some function $u\in W^{1,p}(\Omega)$. It follows that $\Psi'(u_{n})\to \Psi'(u)$ in $(W^{1,p}(\Omega))'$ and $p\Psi(u)=1$, because $p\Psi(u_{n})=1, \forall n\in \mathbb{N}^*$. Hence $u\in\mathcal{M}$. Since $(u_{n})_{n}$ is bounded, then $(i)$ ensures that $\{(\Phi_{\lambda})'(u_{n})\}$ is bounded. By a calculation we obtain via $\eqref{e3.2}$ that $\{(\Phi_{\lambda})'(u_{n})\}$ converges strongly in $(W^{1,p}(\Omega))'$. Consequently, from (ii) we conclude that $u_{n}\to u$ (strongly) in $W^{1,p}(\Omega)$. This achieves the proof of Lemma. \end{proof} Set $\Gamma_{k}=\left\{K\subset \mathcal{M}: K\text{ symmetric, compact and }\ \gamma(K)=k \right\}$, where $\gamma(K)=k$ is the genus of $K$; i.e., the smallest integer $k$ such that there is an odd continuous map from $K$ to $\mathbb{R}^k\backslash\{0\}$. Next, we establish our existence result. \begin{theorem} \label{thm3.1} For each $\lambda\in \mathbb{R}$ and each integer $k\in \mathbb{N}^*$, $$ \mu_{k}(\lambda):=\inf_{K\in \Gamma_{k}}\max_{u\in K}\Phi_{\lambda}(u) $$ is a critical value of $\Phi_{\lambda}$ restricted to $\mathcal{M}$. More precisely, there exists $u_{k}(\lambda)\in\mathcal{M}$ such that $$ \mu_{k}(\lambda)=p\Phi_{\lambda}(u_{k}(\lambda))=\max_{u\in K}p\Phi_{\lambda}(u) $$ and $(u_{k}(\lambda),\mu_{k}(\lambda))$ is a solution of \eqref{e1.1}--\eqref{e1.2}. Moreover, $\mu_{k}(\lambda)\to +\infty$, as $k\to +\infty$. \end{theorem} \begin{proof} In view of \cite{s2}, we need only to prove that for any $k\in \mathbb{N}^*, \Gamma_{k}\neq\emptyset$ and the last assertion. Indeed, since $W^{1,p}(\Omega)$ is separable, there exist $(e_{i})_{i\geq 1}$ linearly dense in $W^{1,p}(\Omega)$ such that $\mathop{\rm supp}e_{i}\cap \mathop{\rm supp}e_{j}=\empty$ if $i\neq j$, where $\mathop{\rm supp}e_{i}$ denotes the support of $e_{i}$. We can suppose that $e_{i}\in \mathcal{M}$ (if not we take $e_{i}'=\frac{e_{i}}{p\Psi(e_{i})}$). Let $k\in \mathbb{N}^*$ and $\mathcal{F}_{k}=\mathop{\rm span}\{e_{1},e_{2},\dots,e_{k}\}$. $\mathcal{F}_{k}$ is a vector subspace and $\dim\mathcal{F}_{k}=k$. If $v\in \mathcal{F}_{k}$, then there exist $\alpha_{1},\dots,\alpha_{k} $ in $\mathbb{R}$ such that $v=\sum_{i=1}^{i=k}\alpha_{i}e_{i}$. Thus \[ \Psi(v)=\sum_{i=1}^{i=k}|\alpha_{i}|^{p}\Psi(e_{i}) =\frac{1}{p}\sum_{i=1}^{i=k}|\alpha_{i}|^{p}, \] because $\Psi(e_{i})=1$, for $ i=1,2,\dots,k$. It follows that the map $v\to (p\Psi(v))^{1/p}$ is a norm on $\mathcal{F}_{k}$. Hence, there is a constant $c>0$ so that $$ c\|v\|_{1,p}\leq (p\Psi(v))^{1/p}\leq \frac{1}{c}\|v\|_{1,p}, \quad \forall v\in \mathcal{F}_{k}. $$ That is, $$ c\|v\|_{1,p}\leq \Big(\int_{\partial\Omega} |v|^pd\sigma\Big)^{1/p}\leq \frac{1}{c}\|v\|_{1,p}, \quad \forall v\in \mathcal{F}_{k}. $$ This implies that the set $$ \mathcal{V}=\mathcal{F}_{k}\cap \left\{v\in W^{1,p}(\Omega): \|v\|_{p,\partial\Omega}\leq 1\right\} $$ is bounded. Because $\mathcal{V}\subset B(0,\frac{1}{c})$, where \[ B(0,\frac{1}{c})=\{v\in W^{1,p}: \|v\|_{1,p}\leq \frac{1}{c}\}. \] Moreover $\mathcal{V}$ is a symmetric bounded neighborhood of the origin $0$. Consequently, from \cite[Proposition 2.3]{s2}, we deduce that $\gamma(\mathcal{F}_{k}\cap \mathcal{M})=k$. Then $\mathcal{F}_{k}\cap \mathcal{M}\in \Gamma_{k}$ (because $\mathcal{F}_{k}\cap \mathcal{M}$ is compact, since it exactly equals to the boundary of $\mathcal{V}$). To complete the proof, it suffices to show that for any $\lambda \in \mathbb{R}$, $\mu_{k}(\lambda)\to +\infty$, as $ k\to +\infty$. Indeed, let $(e_{n},e_{j}^{*})_{n,j}$ be a biorthogonal system such that $e_{n}\in W^{1,p}(\Omega)$, $e_{j}^{*}\in (W^{1,p}(\Omega))'$, the $(e_{n})_{n}$ are dense in $W^{1,p}(\Omega)$; and the $(e_{j}^*)_{j}$ are total in $(W^{1,p}(\Omega))'$. Set for any $k\in \mathbb{N}^*$ $$ \mathcal{F}_{k-1}^{\bot}=\overline{\mathop{\rm span} (e_{k+1},e_{k+2},e_{k+3},\dots)}. $$ Observe that for any for any $K\in\Gamma_{k}$, $K\cap \mathcal{F}_{k-1}^{\bot}\neq \emptyset$ (by \cite[(g) of Proposition 2.3]{s2}). Now, we claim that $$ t_{k}:=\inf_{K\in\Gamma_{k}}\sup_{K\cap\mathcal{F}_{k-1}^{\bot}}p \Phi_{\lambda}(u)\to +\infty, \quad \text{as } k\to +\infty. $$ Indeed, to obtain the contradiction, assume for $k$ large enough that there is $u_{k}\in \mathcal{F}_{k-1}^{\bot}$ with $\int_{\partial\Omega}|u_{k}|^pd\sigma=1$ such that $$ t_{k}\leq p\Phi_{\lambda}(u_{k})\leq M, $$ for some $M>0$ independent of $k$. Therefore, $$ \|u_{k}\|_{1,p}^p-\lambda\int_{\partial\Omega} \rho(x)|u_{k}|^pd\sigma\leq M. $$ Hence $$ \|u_{k}\|_{1,p}^p\leq M+\lambda\|\rho\|_{\infty,\partial \Omega}< \infty. $$ This implies that $(u_{k})_{k}$ is bounded in $W^{1,p}(\Omega)$. For a subsequence of $(u_{k})_{k}$ if necessary, we can suppose that $(u_{k})$ converges weakly in $W^{1,p}(\Omega)$ and strongly in $L^p(\partial\Omega)$. By our choice of $\mathcal{F}_{k-1}^{\bot}$, we have $u_{k}\rightharpoonup 0$ in $W^{1,p}(\Omega)$. Because $\langle e_{n}^{*},e_{k}\rangle$, for all $k\geq n$. This contradicts the fact that $\int_{\partial\Omega}|u_{k}|^{p}d\sigma=1$, for all $k$ and the the claim is proved. Finally, since $\mu_{k}(\lambda)\geq t_{k}$ we conclude that $\mu_{k}(\lambda)\to +\infty$, as $k\to+\infty$ and the proof is complete. \end{proof} \section{Simplicity and isolation of $\mu_{1}(\lambda)$} \subsection{Simplicity} First, observe that solutions of \eqref{e1.1}--\eqref{e1.2}, by an well-known advanced regularity, belong to $C^{1,\alpha}(\bar\Omega)$, see \cite{t1}. \begin{lemma} \label{lem4.1} Eigenfunctions associated to $\mu_{1}(\lambda)$ are either positive or negative in $\Omega$. Moreover if $u\in C^{1,\alpha}(\Omega)$ then $u$ has definite sign in $\bar\Omega$. \end{lemma} \begin{proof} Let $u$ be an eigenfunction associated to $\mu_{1}(\lambda)$,. Since $\Phi_{\lambda}(|u|)\leq \Phi_{\lambda}(u)$ and $\Psi(|u|)=\Psi(u)$, it follows from \eqref{e1.3} that $|u|$ is also an eigenfunction associated to $\mu_{1}(\lambda)$. Using Harnack's inequality, cf. \cite{g1}, we deduce that $|u|>0$ in $\Omega$ and by continuity we conclude that has definite sign in $\bar\Omega.$ In fact $|u|>0$ in $\bar\Omega$ because $\frac{\partial u}{\partial \nu}(x_{0})<0$ for any $x_{0}\in\partial\Omega$ with $u(x_{0})=0$, by applying Hopf's Lemma, see \cite{v1}. \end{proof} \begin{theorem}[Uniqueness] \label{thm4.1} For any $\lambda\in\mathbb{R}$, the eigenvalue $\mu_{1}(\lambda)$ defined by $\eqref{e1.3}$ is a simple; i.e., the set of the eigenfunctions associated with $(\lambda,\mu_{1}(\lambda))$ is $\{tu_{1}(\lambda): t\in \mathbb{R}\}$, where $u_{1}(\lambda)$ denotes the principal eigenfunction associated with $(\lambda,\mu_{1}(\lambda))$. \end{theorem} \begin{proof} By Theorem \ref{thm3.1} it is clear that $\mu_{1}(\lambda)$ is an eigenvalue of the problem \eqref{e1.1}--\eqref{e1.2} for any $\lambda\in\mathbb{R}$. Let $u$ and $v$ be two eigenfunctions associated to $(\lambda,\mu_{1}(\lambda))$, such that $u,v\in\mathcal{M}$. Thus in virtue of Lemma \ref{lem4.1} we can assume that $u$ and $v$ are positives. Note that $W^{1,p}(\Omega)\ni w\to\|\nabla w\|_{p}^{p}$; $w\to \int_{\partial \Omega}|w|^pd\sigma$ and $w\to\int_{\partial\Omega}\rho(x)|w|^pd\sigma$ are linear functionals in $w^p$, for $w^p\geq 0$. Hence if we consider $w=\left(\frac{u^p+v^p}{2}\right)^{1/p}$, then it belongs to $W^{1,p}(\Omega)$ and $\int_{\partial \Omega}|w|^pd\sigma=1$. Consequently, $w$ is admissible in the definition of $\mu_{1}(\lambda)$. On the other hand, by the convexity of $\chi\to |\chi|^{p}$ we have by calculation the following inequalities \begin{equation} \begin{aligned} \int_{\Omega}|\nabla w|^pdx &= \frac{1}{2}\int_{\Omega}|u^{p-1}\nabla u+v^{p-1}\nabla v|^{p}(u^{p}+v^{p})^{1-p}dx\\ &= \frac{1}{2}\int_{\Omega}\big|\frac{u^{p}}{u^{p}+v^{p}}\frac{\nabla u}{u}+ \frac{v^{p}}{v^{p}+u^{p}}\frac{\nabla v}{v}\big|^{p}(u^{p}+v^{p})^{1-p}dx \\ &\leq \frac{1}{2}\int_{\Omega}\Big(\frac{u^{p}}{u^{p}+v^{p}} |\frac{\nabla u}{u}|^{p}+ \frac{v^{p}}{v^{p}+u^{p}} |\frac{\nabla v}{v}|^{p}\Big)dx \\ &\leq \frac{1}{2}\int_{\Omega}(|\nabla u|^{p}+|\nabla v|^{p})dx. \end{aligned} \label{e4.1} \end{equation} By the choice of $u$ and $v$, we deduce that \begin{equation} \left| t\frac{\nabla u}{u}+ (1-t)\frac{\nabla v}{v} \right|^{p}=t\left|\frac{\nabla u}{u}\right|^{p}+(1-t)\left|\frac{\nabla v}{v}\right|^{p},\label{e4.2} \end{equation} with $t=u^{p}/(u^{p}+v^{p})$. Now, we claim that Now, we claim that $u=v \text{ a.e. on }{\overline\Omega}.$ Indeed, consider the auxiliary function $$ F({\chi}_{1},{\chi}_2)=\left|t{\chi}_{1}+ (1-t){\chi}_2\right|^{p} -t\left|{\chi}_{1}\right|^{p}+(1-t)\left|{\chi}_2\right|^{p}. $$ Since $t\neq 0$, critical points of $F$ are solutions of the system \begin{gather} \frac{\partial F({\chi}_{1},{\chi}_2)}{\partial {\chi}_{1}}= pt\left(\left|t{\chi}_{1}+ (1-t){\chi}_2\right|^{p-2}(t{\chi}_{1}-\left|{\chi}_{1} \right|^{p-2}{\chi}_{1}\right)=0;\label{e4.3}\\ \frac{\partial F({\chi}_{1},{\chi}_{2})}{\partial {\chi}_{2}}= p(t-1)\left(\left|t{\chi}_{1}+ (1-t){\chi}_{2}\right|^{p-2}(t{\chi}_{1}-\left|{\chi}_{2} \right|^{p-2}{\chi}_{2}\right)=0.\label{e4.4} \end{gather} Thus \eqref{e4.2}, \eqref{e4.3} and \eqref{e4.4} imply that $({\chi}_{1}=\frac{\nabla u}{u}, ({\chi}_{2}=\frac{\nabla v}{v})$ is a solution of the above system. Therefore, $$ \left|\frac{\nabla u}{u}\right|^{p-2}\frac{\nabla u}{u}=\left|\frac{\nabla v}{v}\right|^{p-2}\frac{\nabla v}{v}. $$ Hence $$ \frac{\nabla u}{u}=\frac{\nabla v}{v}\text{ a.e. in }{\overline \Omega}. $$ This implies easily that $u=cv$ for some positive constant $c$. By normalization we conclude that $c=1$. The proof is completed. \end{proof} \begin{remark} \label{rmk4.1} \rm Various proofs of the uniqueness result were given in Direchlet $p$-Laplacian case by using $C^{1,\alpha}$-regularity and $L^{\infty}$-estimation of the first eigenfunctions and by applying either Picone's identity \cite{a1}; or Diaz-Sa\'{a}'s inequality \cite{a2,c1,d2}, and or an abstract inequality \cite{l1}. \end{remark} \subsection{Isolation} \begin{proposition} \label{prop4.1} For each $\lambda\in\mathbb{R}$, $\mu_{1}(\lambda)$ is the only eigenvalue associated with $\lambda$, having an eigenfunction that does not change sign on the boundary $\partial\Omega$. \end{proposition} \begin{proof} Fix $\lambda\in\mathbb{R}$ and let $u_{1}(\lambda)$ be the principal eigenfunction associated with $(\lambda,\mu_{1}(\lambda))$. Suppose that there exists an eigenfunction $v$ corresponding to a pair $(\lambda,\mu)$ with $v\geq 0$ on $\partial\Omega$ and $v\in \mathcal{M}$. By the Maximum Principle, $v>0$ on ${\overline\Omega}$. For simplify of notation, set $u=u_{1}(\lambda)$. Let $\epsilon>0$ be small enough, and write \begin{gather} u_{\epsilon}=u+\epsilon, \quad v_{\epsilon}=v+\epsilon, \label{e4.5} \\ \phi(u_{\epsilon},v_{\epsilon}) =\frac{u_{\epsilon}^{p}-v_{\epsilon}^{p}}{u_{\epsilon}^{p-1}}.\label{e4.6} \end{gather} It is clear that $\phi(u_{\epsilon},v_{\epsilon})\in W^{1,p}(\Omega)$ and it is an admissible test function in \eqref{e1.1}--\eqref{e1.2}. Thus we obtain \begin{equation} \begin{aligned} &\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla \phi(u_{\epsilon},v_{\epsilon})dx+\int_{\Omega} u^{p-1}\phi(u_{\epsilon},v_{\epsilon})dx\\ &=\int_{\partial\Omega}(\lambda \rho(x)+\mu_{1}(\lambda)) u^{p-1}\phi(u_{\epsilon},v_{\epsilon})d\sigma \label{e4.7} \end{aligned} \end{equation} and \begin{equation} \int_{\Omega}|\nabla v|^{p-2}\nabla v\nabla \phi(u_{\epsilon},v_{\epsilon})dx+ \int_\Omega v^{p-1}\phi(u_{\epsilon},v_{\epsilon})dx=\int_{\partial \Omega}(\lambda \rho(x)+\mu))v^{p-1}\phi(u_{\epsilon},v_{\epsilon})d\sigma. \label{e4.8} \end{equation} From \eqref{e4.7} and \eqref{e4.8}, we deduce by calculations that \begin{equation} \begin{aligned} &\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla \phi(u_{\epsilon},v_{\epsilon})\, dx+ \int_{\Omega}|\nabla v|^{p-2}\nabla v\nabla \phi(u_{\epsilon},v_{\epsilon})dx+ \int_{\Omega}|v|^{p-2}v\phi(u_{\epsilon},v_{\epsilon})\, dx\\ &=\int_{\partial \Omega} \lambda \rho(x)\Big(\big(\frac{u}{u_{\epsilon}}\big)^{p-1}- \big(\frac{v}{v_{\epsilon}}\big)^{p-1}\Big) (u_{\epsilon}^p-v_{\epsilon}^p)d\sigma \\ &\quad +\mu_{1}(\lambda)\int_{\partial \Omega}u^{p-1}\big[u_{\epsilon}-\big(\frac{v_{\epsilon}}{u_{\epsilon}} \big)^{p-1}v_{\epsilon}\big]d \sigma+\mu\int_{\partial \Omega} u^{p-1}\big[v_{\epsilon}-\big(\frac{u_{\epsilon}}{v_{\epsilon}} \big)^{p-1}u_{\epsilon}\big]d\sigma. \end{aligned} \label{e4.9} \end{equation} On the other hand, by a long calculation again, we obtain \begin{equation} \nabla\phi(u_{\epsilon},v_{\epsilon})= \big\{1+(p-1)\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^p\big\}\nabla u_{\epsilon}- p\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1}\nabla v_{\epsilon} \label{e4.10} \end{equation} and \begin{equation} \int_{\Omega} \left[u^{p-1}\phi(u_{\epsilon},v_{\epsilon})+v^{p-1} \phi(u_{\epsilon},v_{\epsilon})\right]dx =\int_{\Omega}\big[\big(\frac{u}{u_{\epsilon}}\big)^{p-1} -\big(\frac {v}{v_{\epsilon}}\big)^{p-1}\big] (u_{\epsilon}^p-v_{\epsilon}^p)dx.\label{e4.11} \end{equation} Therefore, \eqref{e4.9}, \eqref{e4.10} and \eqref{e4.11} yield \begin{equation} \begin{aligned} &\int_{\Omega}\Big[ \big\{1+(p-1)\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^p\big\}|\nabla u_{\epsilon}|^p+ \big\{1+(p-1)\big(\frac{u_{\epsilon}}{v_{\epsilon}}\big)^p\big\}|\nabla v_{\epsilon}|^p\Big]dx\\ &\quad +\int_{\Omega} \Big[-p\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1}|\nabla v_{\epsilon}|^{p-2}\nabla u_{\epsilon}\nabla v_{\epsilon}+p\big(\frac{u_{\epsilon}}{v_{\epsilon}}\big)^{p-1}|\nabla u_{\epsilon}|^{p-2}\nabla u_{\epsilon}\nabla v_{\epsilon}\Big]dx\\ &= J_{\epsilon}+K_{\epsilon}-I_{\epsilon}, \end{aligned} \label{e4.12} \end{equation} with \begin{gather} I_{\epsilon}=\int_{\Omega} \Big(\big(\frac{u}{u_{\epsilon}}\big)^{p-1} -\big(\frac{v}{v_{\epsilon}}\big)^{p-1}\Big) \left(u_{\epsilon}^p-v_{\epsilon}^p\right)dx\,,\label{e4.13} \\ J_{\epsilon}=\lambda\int_{\partial \Omega}\rho(x) \Big(\big(\frac{u}{u+\epsilon}\big)^{p-1} -\big(\frac{v}{v+\epsilon}\big)^{p-1}\Big) \left(u_{\epsilon}^p-v_{\epsilon}^p\right)d\sigma\,,\label{e4.14} \\ K_{\epsilon}=\mu_{1}(\lambda)\int_{\partial \Omega} u^{p-1} \Big[u_{\epsilon}-\big(\frac{v_{\epsilon}}{u_{\epsilon}}\big)^{p-1} v_{\epsilon}\Big]d \sigma +\mu\int_{\partial \Omega} u^{p-1}\Big[v_{\epsilon}-\big(\frac{u_{\epsilon}}{v_{\epsilon}} \big)^{p-1}u_{\epsilon}\Big]d \sigma.\label{e4.15} \end{gather} It is clear that $I_{\epsilon}\geq 0$. Now, thanks to the inequalities of Lindqvist \cite{l1}, we can distinguish tow cases according to $p$. First case: $p\geq 2$. From \eqref{e4.12} we have \begin{equation} J_{\epsilon}+K_{\epsilon}\geq \frac{1}{2^{p-2}-1}\int_{\Omega} \left(\frac{1}{(u+1)^p}+\frac{1}{(v+1)^p}\right)|u\nabla v-v\nabla u|^pdx\geq 0.\label{e4.16} \end{equation} Second case: $10$ independent of $u,v,\lambda $ and $\mu_{1}(\lambda)$. The Dominated Convergence Theorem implies $$ \lim_{\epsilon\to 0^{+}}J_{\epsilon}=\lim_{\epsilon\to 0^+}K_{\epsilon}=(\mu_{1}(\lambda)-\mu)\int_{\partial\Omega} (u^p-v^p)d\sigma=0, $$ because \begin{equation} \int_{\partial\Omega}u^pd\sigma=\int_{\partial\Omega}v^pd\sigma=1. \label{e4.18} \end{equation} Now, letting $\epsilon\to 0^{+}$ in \eqref{e4.16} and \eqref{e4.17}, we arrive at $u\nabla v=v\nabla u$ a.e. on $\Omega$. Thus $$ \nabla \left(\frac{u}{v}\right)=0\quad \text{a.e. on }\Omega. $$ Hence, there exists $t>0$ such that $u=tv$ a.e. on $ \Omega$. By continuity $u=v$ a.e. in $\overline\Omega$; and by the normalization $\eqref{e4.18}$ we deduce that $t=1$ and $u=v$ a.e. on $\partial\Omega$. This implies that $u=v$ a.e. on $\overline\Omega$. Finally, we conclude that $\mu=\mu_{1}(\lambda)$. Which completes the prof. \end{proof} \begin{remark} \label{rmk4.2} \rm Proposition \ref{prop4.1} can also be shown by using {\it Picone's identity}. A similar result was given by \cite{b4} in the restrictive case $\lambda=0$. \end{remark} \begin{corollary} \label{coro4.1} For each $\lambda\in\mathbb{R}$, if $u$ is an eigenfunction associated with a pair $(\lambda, \mu)$ and $\mu\neq\mu_{1}(\lambda)$, then $u$ changes sign on the boundary $\partial\Omega$. Moreover, we have the estimate \begin{equation} \min(|\partial\Omega^{-}|,|\partial\Omega^{+}|)\geq c_{p^{*}}^{-N}(|\lambda|\|\rho\|_{\infty,\partial\Omega} +|\mu|)^{-\eta},\label{e4.19} \end{equation} where $$ \eta=\begin{cases} \frac{N}{p} &\text{if }1N; \end{cases} $$ $c_{p^*}$ is the best constant in the Sobolev trace embedding $W^{1,p}(\Omega)$ in $L^{p^*}(\partial\Omega)$; and $|\partial \Omega^{\pm}|$ denotes the $(N-1)$-dimensional measure of $\partial\Omega^{\pm}$. Here $p^{*}=\frac{p(N-1)}{N-p}$ is the critical Sobolev exponent. \end{corollary} \begin{proof} Set $u^{+}=\max (u,0)$ and $u^{-}=\max(-u,0)$. It follows from $\eqref{e2.1}$, where we put $v=u^{-}$, that $$ \int_{\Omega} |\nabla u^-|^pdx+\int_{\Omega} |u^-|^pdx=\int_{\partial\Omega}(\lambda\rho(x)+\mu)|u^-|^pd\sigma. $$ Thus \begin{align*} \|u^-\|_{1,p}&\leq \left(|\lambda|\|\rho\|_{\infty,\partial \Omega}+|\mu|\right)\int_{\partial \Omega^-}|u^-|^pd\sigma\\ &\leq \left(|\lambda|\|\rho\|_{\infty,\partial \Omega}+|\mu|\right)|\partial\Omega^-|^{p/N} \Big(\int_{\partial \Omega}|u^-|^{p^*}\Big)^{p/p^*}. \end{align*} By the Sobolev embedding $W^{1,p}(\partial\Omega)\hookrightarrow L^{p^*}(\partial \Omega)$, we deduce that $$ |\partial \Omega^-|\geq c_{p^*}^{-N}\left(|\lambda|\|\rho\|_{\infty,\partial \Omega}+|\mu|\right)^{-\eta}. $$ For $\partial\Omega^+$ the same estimate follows by taking $v=u^+$ in \eqref{e2.1}. Hence \eqref{e4.19} follows. \end{proof} \begin{remark} \label{rmk4.1b} \rm (i) The right-hand side of \eqref{e4.19} is positive because $\rho\not\equiv 0$ and if $\lambda=0$ then $\mu$ shall be an eigenvalue of $p$-Laplacian related to trace embedding, so $\mu-\lambda_{1}>0$, with $\lambda_{1}$ is the first eigenvalue of \eqref{e1.1}--\eqref{e1.2} in the case $(\lambda=0)$. (ii) As an easy consequence of Corollary \ref{coro4.1}, we get that the number of the nodal components of each eigenfunction of \eqref{e1.1}--\eqref{e1.2} is finite. \end{remark} Using Proposition \ref{prop4.1} and Corollary \ref{coro4.1}, we can state the following important result. \begin{theorem} \label{thm4.2} For each $\lambda\in\mathbb{R}$, $\mu_{1}(\lambda)$ is isolated. \end{theorem} \section{Variations of the weight} Let $\mu_{1}(\lambda)=\mu_{1}(\rho)$ and $u_{1}(\lambda)=u_{1}(\rho)$ (for indicating the dependance of the weight $\rho$). \begin{theorem} \label{thm5.1} For each $\lambda\in \mathbb{R}$, if $(\rho_{k})_{k}$ is a sequence of functions in $L^{\infty}(\partial \Omega)$ that converges to $\rho$ and $\rho\not\equiv 0$, then \begin{gather} \lim_{k\to \infty}\mu_{1}(\rho_{k})=\mu_{1}(\rho)\,,\label{e5.1} \\ \lim_{k\to \infty}\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^p=0\,.\label{e5.2} \end{gather} \end{theorem} \begin{proof} If $\lambda =0$, the result is evident because $\mu_{1}(\rho_{k})=\mu_{1}(\rho)$, for all $k\in \mathbb{N}^*$. If $\lambda\neq 0$, then for $v\in\mathcal{M}$, $$ |\lambda\int_{\partial\Omega}(\rho_{k}-\rho)|v|^{p}d\sigma| \leq |\lambda|\|\rho_{k}-\rho\|_{\infty,\partial\Omega}. $$ Using the convergence of $\rho_{k}$ to $\rho$ in $L^{\infty}(\partial \Omega)$, for all $\epsilon>0$, there exists $k_{\epsilon}\in \mathbb{N}$ such that for all $k\geq k_{\epsilon}$, $$ |\lambda \int_{\partial\Omega}(|(\rho_{k}-\rho)|v|^{p}d\sigma|\leq |\lambda|\frac{\epsilon}{|\lambda|}=\epsilon. $$ This implies \begin{gather} \lambda\int_{\partial\Omega}\rho|v|^{p}d\sigma \leq \epsilon+\lambda\int_{\partial\Omega} \rho_{k}|v|^pd\sigma\,, \label{e5.3}\\ \lambda\int_{\partial \Omega }\rho_{k}|v|^pd\sigma \leq \epsilon+\lambda \int_{\partial\Omega}\rho|v|^pd\sigma, \label{e5.4} \end{gather} for $v\in\mathcal{M}$, $\epsilon >0$ and $k\geq k_{\epsilon}$. On the other hand, we have $\rho\not\equiv 0$. We take $k_{\epsilon}$ large enough so that $\rho_{k}\not\equiv 0$. Thus $$ \mu_{1}(\rho_{k})\leq \|v\|_{1,p}^p-\lambda \int_{\partial\Omega} \rho_{k}|v|^pd\sigma. $$ Combining \eqref{e5.3} and \eqref{e5.4}, we obtain $$ \mu_{1}(\rho_{k})\leq \|v\|_{1,p}^p-\lambda\int_{\partial\Omega}\rho |v|^pd\sigma +\epsilon. $$ Passing to the infimum over $v\in\mathcal{M}$, we find $$ \mu_{1}(\rho_{k})\leq\mu_{1}(\rho)+\epsilon, \quad \mu_{1}(\rho)\leq \mu_{1}(\rho_{k})+\epsilon,\quad \forall \epsilon>0,\; \forall k>k_{\epsilon}. $$ Hence, we conclude the convergence \eqref{e5.1}. For the strong convergence \eqref{e5.2} we argue as follows. We have for $k$ large enough, $\rho_{k}\not\equiv 0$ and \begin{equation} \mu_{k}(\rho_{k})=\|u_{1}(\rho_{k})\|_{1,p}^p-\lambda\int_{\partial \Omega}\rho_{k}(u_{1}(\rho_{k}))^pd\sigma.\label{e5.5} \end{equation} Thus $$ \|u_{1}(\rho_{k})\|_{1,p}^p\leq |\mu_{1}(\rho_{k})|+|\lambda|\|\rho_{k}\|_{\infty,\partial\Omega}. $$ From \eqref{e5.1} and the convergence of $\rho_{k}$ to $\rho$ in $L^{\infty}(\partial\Omega)$, we deduce that $(u_{1}(\rho_{k}))_{k}$ is a bounded sequence in $W^{1,p}(\Omega)$. Since $W^{1,p}(\Omega)$ is reflexive and compactly embedded in $L^p(\partial\Omega)$ we can extract a subsequence of $(u_{1}(\rho_{k}))_{k}$ again labelled by $k$, such that $u_{1}(\rho_{k})\rightharpoonup u$ (weakly) in $W^{1,p}(\Omega)$ and $u_{1}(\rho_{k})\to u$ (strongly) in $L^p(\partial\Omega)$, as $k\to \infty$. We can also suppose that $u_{1}(\rho_{k})\to u$ in $L^p(\Omega)$. Passing to a subsequence if necessary, we can assume that $u_{1}(\rho_{k})\to u$ a.e. in $\bar\Omega$. Thus $u\geq 0$ a.e. in $\overline\Omega$. We will prove that $u\equiv u_{1}(\rho)$. To do this, using the Dominated Convergence Theorem in $\partial\Omega$, we deduce that $$ \int_{\partial\Omega}\rho_{k}(u_{1}(\rho_{k}))^pd\sigma\to \int_{\partial\Omega}\rho u^p d\sigma, $$ as $k\to \infty$. By \eqref{e5.5}, \eqref{e5.1} and the lower weak semi-continuity of the norm we obtain that \begin{equation} \|u\|_{1,p}^p\leq\mu_{1}(\rho)+\lambda\int_{\partial \Omega}\rho u^{p}d\sigma.\label{e5.6} \end{equation} The normalization $\int_{\partial\Omega} u^pd\sigma=1$ is proved. Moreover, $u\geq 0$ a.e. in $\bar\Omega$, because $u_{1}(\rho_{k})>0$ in $\overline \Omega$ Thus $u$ is an admissible function in the variational definition of $\mu_{1}(\lambda)$. So $$ \mu_{1}(\lambda)\leq \|u\|_{1,p}^p-\lambda \int_{\partial \Omega}\rho u^pd\sigma. $$ This and \eqref{e5.6} yield \begin{equation} \mu_{1}(\rho)=\|u\|_{1,p}^p-\lambda\int_{\partial\Omega}\rho u^pd\sigma.\label{e5.7} \end{equation} By the uniqueness of the principal eigenfunction associated to $\mu_{1}(\lambda)$, we must have $u\equiv u_{1}(\rho)$. Consequently the limit function $u_{1}(\rho)$ is independent of the choice of the (sub)sequence. Hence, $u_{1}(\rho_{k})$ converges to $u_{1}(\rho)$ at least in $L^p(\partial\Omega)$ and in $L^p(\Omega)$. To complete the proof of $\eqref{e5.2}$, it suffices to use the Clarckson's inequalities related to uniform convexity of $W^{1,p}(\Omega)$. For this we distinguish two cases. First case: $p\geq 2$. We have \begin{align*} & \int_{\Omega} \Big|\frac{\nabla u_{1}(\rho_{k})-\nabla u_{1}(\rho)}{2}\Big|^pdx + \int_{\Omega} \left|\frac{\nabla u_{1}(\rho_{k})+\nabla u_{1}(\rho)}{2}\right|^pdx \\ & \leq \frac{1}{2}\int_{\Omega} |\nabla u_{1}(\rho_{k})|^pdx+ \frac{1}{2}\int_{\Omega}|\nabla u_{1}(\rho)|^pdx \end{align*} and \begin{align*} &\mu_{1}(\rho_{k})\int_{\partial \Omega}\Big(\frac{u_{1}(\rho_{k})+ u_{1}(\rho)}{2}\Big)^pd\sigma \\ &\leq \int_{\Omega} \Big|\frac{\nabla u_{1}(\rho_{k}) + \nabla u_{1}(\rho)}{2}\Big|^pdx - \lambda\int_{\partial \Omega}\rho_{k}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma. \end{align*} Moreover, $$ \int_{\Omega} \Big|\frac{u_{1}(\rho_{k})-u_{1}(\rho)}{2}\Big|^pdx\leq \int_{\Omega}\Big|\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big|^pdx+ \frac{1}{2}\|u_{1}(\rho_{k})\|_p^p+\frac{1}{2}\|u_{1}(\rho)\|_{p}^{p}. $$ Hence \begin{align*} &\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^p \\ &\leq -\mu_{1}(\rho_{k})\int_{\partial \Omega}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma- \lambda\int_{\partial \Omega}\rho_{k}\Big(\frac{u_{1}(\rho_{k})+u_{1}(\rho)}{2}\Big)^pd\sigma\\ &\quad + \frac{1}{2}\Big(\mu_{1}(\rho_{k})- \lambda \int_{\partial\Omega}\rho_{k}(x)u_{1}(\rho_{k})d\sigma\Big) + \frac{1}{2}\Big(\mu_{1}(\rho)-\lambda\int_{\partial\Omega}\rho u_{1}^pd\sigma\Big). \end{align*} Then, by using the Dominated Convergence Theorem we deduce that $$ \limsup_{k\to +\infty}\|u_{1}(\rho_{k})-u_{1}(\rho)\|_{1,p}^{p}=0. $$ Second case: $1