\documentclass[reqno]{amsart} \usepackage{amssymb} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 87, pp. 1--8.\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-2005/87\hfil Steklov problem with an indefinite weight] {Steklov problem with an indefinite weight for the $p$-Laplacian} \author[O. Torn\'{e}\hfil EJDE-2005/87\hfilneg] {Olaf Torn\'{e}} \address{Olaf Torn\'{e}\hfill\break Universit\'{e} Libre de Bruxelles, Campus de la Plaine, ULB CP214, Boulevard du Triomphe, 1050 Bruxelles, Belgium} \email{otorne@ulb.ac.be} \date{} \thanks{Submitted August 10, 2004. Published August 14, 2005.} \subjclass[2000]{35J70, 35P30} \keywords{Nonlinear eigenvalue problem; Steklov problem; $p$-Laplacian; \hfill\break\indent nonlinear boundary condition; indefinite weight} \begin{abstract} Let $\Omega\subset\mathbb{R}^{N}$, with $N\geq2$, be a Lipschitz domain and let $1
0$ is the first positive Steklov eigenvalue. Moreover $\lambda_{1}^{+}$ is simple and isolated and it is the only nonzero Steklov eigenvalue associated to an eigenfunction of definite sign. Also, $\lambda^{+}_{2}$ is the second positive Steklov eigenvalue and $\lambda^{-}_{2}$ is the first negative Steklov eigenvalue. \item Assume that $m^{-}\not\equiv0$ and that $\int_{\partial\Omega}m\,d\sigma>0$. Then $\lambda^{+}_{1}=0$ and $\lambda^{-}_{1}<0$ is the first negative Steklov eigenvalue. Moreover $\lambda_{1}^{-}$ is simple and isolated and it is the only nonzero Steklov eigenvalue associated to an eigenfunction of definite sign. Also, $\lambda^{+}_{2}$ is the first positive Steklov eigenvalue and $\lambda^{-}_{2}$ is the second negative Steklov eigenvalue. \item Assume that $\int_{\partial\Omega}m\,d\sigma=0$. Then $\lambda^{+}_{1}=\lambda^{-}_{1}=0$ and there does not exist a nonzero eigenvalue associated to an eigenfunction of definite sign. Also, $\lambda^{+}_{2}$ is the first positive Steklov eigenvalue and $\lambda^{-}_{2}$ is the first negative Steklov eigenvalue. \end{enumerate} In each of the above cases any eigenfunction associated to $\lambda^{+}_{2}$ or $\lambda^{-}_{2}$ has exactly two nodal domains. Moreover, if in case \emph{(1)} there holds $m^{+}\equiv 0$ then there does not exist a positive eigenvalue, however the assertions concerning the negative eigenvalues remain true. A similar statement holds when $m^{-}\equiv 0$ in case \emph{(2)}. \end{theorem} Note that $\lambda=0$ is always a Steklov eigenvalue and that the associated eigenfunctions are just the constant functions. In \cite{Veron}, L. V\'{e}ron discovered an interesting formula for the first nonzero eigenvalue in the usual $p$-Laplacian spectrum on a Riemannian manifold without boundary. There holds a similar formula for the Steklov problem and it will be used to deduce, as in \cite{Veron}, some of the assertions in the above theorem. Let us now state the formula. If $\omega\subset\Omega$ is an open subset, let $W^{1,p}_{*}(\omega)$ denote the subset of $W^{1,p}(\Omega)$ consisting of functions which are zero almost everywhere in $\Omega\setminus\bar{\omega}$. Let \begin{equation} \label{intro_eq_mu_of_omega} \mu^{\pm}(\omega)=\inf \big\{ \frac{1}{p}\int_\Omega |\nabla u|^{p}\,dx;\,u\in W^{1,p}_{*} (\omega),\, \frac{1}{p}\int_{\partial\Omega} m|u|^{p}\,d\sigma=\pm 1 \big\} \end{equation} if this quantity is well defined and $\mu^{\pm}(\omega)=+\infty$ if not. Lastly let $\mathcal{A}$ denote the set of pairs $(\omega,\tilde{\omega})$ where $\omega$ and $\tilde{\omega}$ are disjoint nonempty opens subsets of $\Omega$. \begin{theorem} \label{intro_theo_veron_formula} Let $1
0$ is a positive constant. Now let $\alpha_{k}=\frac{1}{|\Omega|}\int_{\Omega}u_{k}\,dx$ and let $\tilde{u}_{k}=u_{k}-\alpha_{k}$. Since $J(\tilde{u}_{k})=J(u_{k})$ is bounded it follows from the Poincar\'{e}-Wirtinger inequality that the sequence $\tilde{u}_{k}$ is bounded in $W^{1,p}(\Omega)$. We may write \begin{equation} \label{existence_eq_intermediate_palaissmale_proof} \int_{\partial\Omega}m|u_{k}|^{p-2}u_{k}\varphi\,d\sigma =|\alpha_{k}|^{p-2}\alpha_{k} \int_{\partial\Omega}m\big|\frac{\tilde{u}_{k}}{\alpha_{k}}+1\big|^{p-2} \big(\frac{\tilde{u}_{k}}{\alpha_{k}}+1\big)\varphi\,d\sigma \end{equation} If the sequence $|\alpha_{k}|$ is not bounded we may assume $|\alpha_{k}|\to\infty$ so that the integral on the right hand side of (\ref{existence_eq_intermediate_palaissmale_proof}) goes to $\int_{\partial\Omega}m\varphi\,d\sigma\neq0$ and we have a contradiction. Since $\alpha_{k}$ is bounded it follows that the sequence $u_{k}$ is bounded. Using the compactness property of the trace mapping it follows that $u_{k}$ contains a subsequence, again noted $u_{k}$, such that $B'(u_{k})\to B'(u)$ for some $u\in W^{1,p}(\Omega)$. Since $-\Delta_{p}u_{k}=Du_{k}+J(u_{k})B'(u_{k}) \to\lambda B'(u)$ in $W^{1,p}(\Omega)'$ it follows from lemma \ref{existence_lemma_inverse_plap} that $u_{k}-\alpha_{k}\to (-\Delta_{p})^{-1}(\lambda B'(u))$ in $W^{1,p}(\Omega)$. Since $\alpha_{k}$ is bounded we have that $u_{k}$ converges in the $W^{1,p}(\Omega)$ sense. \end{proof} \end{section} \begin{section}{Qualitative properties of the first and second eigenvalues} In this section, we prove theorems \ref{intro_theo_simple_eigenvalues} and \ref{intro_theo_veron_formula}. Let us note that all Steklov eigenfunctions are of class $C^{1,\alpha}(\Omega)$ since they are $p$-harmonic. Moreover, following the procedure outlined in \cite{Lindqvist} one may show that $u\in L^{\infty}(\Omega)$. The proof of this fact is carried out in detail in \cite{OlafDEA}. The following lemma derives from Picone's identity (see \cite{AllegrettoHuang}) and is a standard tool in this context. A little extra care is required in the proof in the Steklov setting. \begin{lemma} Let $1
0$ on $\partial\Omega$. Let $\varepsilon>0$. By the maximum principle of Vazquez $v>0$ in $\Omega$ so that $\frac{v}{v+\varepsilon}$ converges in $L^{p}(\Omega)$ to $1_{\Omega}$ as $\varepsilon\to 0$. On the other hand $\nabla\frac{v}{v+\varepsilon}$ converges to $0$ a.e. as $\varepsilon\to0$. Taking $\varphi=\frac{1}{(v+\varepsilon)^{p-1}}$ as testing function in equation (\ref{intro_eq_weak_steklov}) satisfied by $v$ we have $$ (p-1)\int_{\Omega}\frac{|\nabla v|^{p}}{(v+\varepsilon)^{p}}\,dx =\lambda\int_{\partial\Omega}m\big(\frac{v}{v+\varepsilon}\big)^{p-1}\,d\sigma $$ so that $$ |\nabla\frac{v}{v+\varepsilon}|^{p} =\big(\frac{\varepsilon}{v+\varepsilon}\big)^{p} \frac{|\nabla v|^{p}}{(v+\varepsilon)^{p}} \leq\frac{|\nabla v|^{p}}{v^{p}}\in L^{1}(\Omega). $$ By the dominated convergence theorem we have that $\frac{v}{v+\varepsilon}\to 1_{\Omega}$ in $W^{1,p}(\Omega)$. By continuity of the trace mapping we have that $\frac{v}{v+\varepsilon}\to 1_{\partial\Omega}$ in $L^{1}{\partial\Omega})$ as $\varepsilon\to0$ and it follows that $v>0$ on $\partial\Omega$. Now let $\varepsilon>0$. By Picone's identity we have \begin{align*} 0&\leq\int_{\Omega}|\nabla u|^{p}\,dx-\int_{\Omega}|\nabla v|^{p-2}\nabla v\nabla\big(\frac{u^{p}}{(v+\varepsilon)^{p-1}}\big)\,dx \\ &=\lambda\int_{\partial\Omega}m u^{p}\,d\sigma-\tilde{\lambda}\int_{\partial\Omega} m\big(\frac{v}{v+\varepsilon}\big)^{p-1}u^{p}\,d\sigma \end{align*} and equality holds if and only if $v$ is a multiple of $u$. Going to the limit $\varepsilon\to0$ and using the fact that $v>0$ on $\partial\Omega$ we get the desired inequality. \end{proof} \begin{proof}[Proof of theorem \ref{intro_theo_simple_eigenvalues}] We begin by proving the assertions relating to $\lambda_{1}^{\pm}$. It follows immediatly from (\ref{intro_eq_eigenvalue_LS}) that we have \begin{equation} \label{qual_eq_charact_of_la1_as_minimum} \lambda^{\pm}_{1}=\pm\inf\{\int_{\Omega}|\nabla u|^{p}\,dx;\,u\in W^{1,p}(\Omega), \,\int_{\partial\Omega}m|u|^{p}\,d\sigma=\pm1\} \end{equation} Assume first that $\int_{\partial\Omega} m\,d\sigma<0$ and $m^{+}\not\equiv0$. The minimum $\lambda_{1}^{-}$ is achieved by a constant function so that $\lambda_{1}^{-}=0$. Now let $u_{k}$ be a minimizing sequence for $\lambda_{1}^{+}$. Let $\alpha_{k}=\frac{1}{|\Omega|}\int_{\Omega}u_{k}\,dx$. By Poincar\'{e}-Wirtinger's inequality the sequence $\tilde{u}_{k}=u_{k}-\alpha_{k}$ is bounded and, moreover, we may write $$ 1=|\alpha_{k}|^{p}\int_{\partial\Omega}m\big|\frac{\tilde{u}_{k}}{\alpha_{k}}+1 \big|^{p}\,d\sigma $$ If the sequence $|\alpha_{k}|$ is not bounded we may assume that $|\alpha_{k}|\to\infty$ so that the integral goes to $\int_{\partial\Omega}m\,d\sigma<0$ and we have a contradiction. Thus $\alpha_{k}$ is bounded and we have that $u_{k}$ is bounded. It follows that the infimum $\lambda^{+}_{1}$ is achieved. Since the minimiser cannot be constant we have that $\lambda_{1}^{+}>0$ and it is clear that $\lambda^{+}_{1}$ is the first positive eigenvalue. Now let $u$ be an eigenfunction associated to $\lambda_{1}^{+}$ so that $|u|$ is a minimiser for (\ref{qual_eq_charact_of_la1_as_minimum}) and is thus an eigenfunction associated to $\lambda_{1}^{+}$. It follows from the maximum principle of Vazquez that $|u|>0$ in $\Omega$ and we conclude that $u$ has constant sign. Taking $\lambda=\lambda_{1}^{+}$ in (\ref{qual_eq_conclusion_of_picone}) we see that no eigenvalue $\tilde{\lambda}>\lambda^{+}_{1}$ can be associated to a positive eigenfunction. Taking $\lambda=0$ in (\ref{qual_eq_conclusion_of_picone}) and $u\equiv1$ an associated eigenfunction, we see that no eigenvalue $\tilde{\lambda}<0$ can be associated to a positive eigenfunction. Thus $\lambda_{1}^{+}$ is the only nonzero eigenvalue associated to an eigenfunction of definite sign. Taking $\lambda=\tilde{\lambda}=\lambda_{1}^{+}$ in (\ref{qual_eq_conclusion_of_picone}) we see that any eigenfunction $v$ associated to $\lambda^{+}_{1}$ must be a multiple of $u$, so that $\lambda^{+}_{1}$ is simple. The case where $\int_{\partial\Omega} m\,d\sigma>0$ and $m^{-}\not\equiv0$ follows from the previous case by taking $-m$ as weight function. Now assume that $\int_{\partial\Omega} m\,d\sigma=0$ and suppose by contradiction that $\lambda_{1}^{+}>0$. If $v$ is an eigenfunction associated to $\lambda_{1}^{+}$ then $|v|$ is a minimiser in (\ref{qual_eq_charact_of_la1_as_minimum}) and it is thus also an eigenfunction associated to $\lambda^{+}_{1}$. However, taking $\tilde{\lambda}=\lambda_{1}^{+}$, $\lambda=0$ and $u\equiv1$ in (\ref{qual_eq_conclusion_of_picone}) we see that $v$ is constant, since it must be a multiple of $u$. But this implies that $\lambda_{1}^{+}=0$ which is a contradiction. In the same way it can be shown that $\lambda_{1}^{-}=0$. Now we prove the assertions concerning $\lambda_{2}^{\pm}$. Since $\lambda^{+}_{1}$ is simple we have $$ \gamma\big(\{u\in\Sigma^{+};\,u \mbox{ is an eigenfunction associated to }\lambda^{+}_{1}\}\big)=1 $$ Thus by theorem \ref{intro_theo_existence_of_eigenvalues} there holds $\lambda_{1}^{+}<\lambda_{2}^{+}$. Likewise we have $\lambda_{2}^{-}<\lambda_{1}^{-}$. Let $\lambda\neq\lambda_{1}^{\pm}$ be an eigenvalue. Following a similar reasoning to \cite{AnaneTsouli} we now show that either $\lambda\leq\lambda_{2}^{-}$ or $\lambda_{2}^{+}\leq\lambda$. First assume $\lambda>0$ and let $u$ be an eigenfunction associated to $\lambda$. Since $u$ must change sign we may assume that $u$ is normalized in such a way that $\frac{1}{p}\int_{\partial\Omega}m|u^{+}|^{p}\,d\sigma=\frac{1}{p}\int_{\partial\Omega}m|u^{-}|^{p}\,d\sigma=1$ and $\frac{1}{p}\int_{\Omega}|\nabla u^{+}|^{p}\,dx=\frac{1}{p}\int_{\Omega}|\nabla u^{-}|^{p}\,dx=\lambda$. The set $C=\{\alpha u^{+}+\beta u^{-};\,\alpha,\beta\in\mathbb{R}\mbox{ such that } |\alpha|^{p}+|\beta|^{p}=1\}$ is in $\mathcal{C}_{2}^{+}$. Thus $$ \lambda_{2}^{+}\leq\max_{|\alpha|^{p}+|\beta|^{p}=1} \frac{1}{p}\int_{\Omega}|\nabla(\alpha u^{+}+\beta u^{-})|^{p}\,dx=\lambda $$ Similarly if $\lambda<0$ we may show that $\lambda\leq\lambda_{2}^{-}$. It remains only to show that any eigenfunction associated to $\lambda_{2}^{+}$ or $\lambda_{2}^{-}$ has exactly two nodal domains. To prove this we use formula (\ref{intro_eq_veron_formula}) which will be proved below. Let $u$ be an eigenfunction associated to $\lambda_{2}^{\pm}$. Assume that $u$ has at least three nodal domains $\omega_{1}$, $\omega_{2}$ and $\omega_{3}$ with, say, $u>0$ in $\omega_{1}$ and $\omega_{2}$ and $u<0$ in $\omega_{3}$. In a similar situation the authors of \cite{CuestaDefigGossez} show that $\partial\omega_{1}\cap\Omega\nsubseteq\partial\omega_{2}\cap\Omega$ and $\partial\omega_{2}\cap\Omega\nsubseteq\partial\omega_{1}\cap\Omega$. Their proof relies on the maximum principle and carries over to the case considered here. It follows that there exist disjoint open sets $\tilde{\omega}_{1},\tilde{\omega}_{2}\subset\Omega$ such that $\omega_{1}\varsubsetneq\tilde{\omega}_{1}$ and $\omega_{2}\varsubsetneq\tilde{\omega}_{2}$. Now it follows from standard arguments that $\mu^{\pm}(\tilde{\omega}_{1})<\mu^{\pm}(\omega_{1})$ and $\mu^{\pm}(\tilde{\omega}_{2})<\mu^{\pm}(\omega_{2})$ thus contradicting (\ref{intro_eq_veron_formula}). \end{proof} \begin{proof}[Proof of theorem \ref{intro_theo_veron_formula}] We only prove the formula for $\lambda_{2}^{+}$ since the proof for $\lambda_{2}^{-}$ is similar. Denote \begin{equation} \label{qual_eq_mu_in_prf_veron_formula} \mu=\inf_{(\omega,\tilde{\omega})\in\mathcal{A}}\textstyle\max \{\mu^{+}(\omega),\mu^{-}(\tilde{\omega})\} \end{equation} Let $(\omega,\tilde{\omega})\in\mathcal{A}$ and let $\psi$ and $\tilde{\psi}$ be minimizers for $\mu^{+}(\omega)$ and $\mu^{+}(\tilde{\omega})$ respectively, normalized so that $\frac{1}{p}\int_{\partial\Omega}m\psi^{p}\,d\sigma=\frac{1}{p}\int_{\partial\Omega}m\tilde{\psi}^{p}\,d\sigma=1$. The set $C=\{\alpha\psi+\beta\tilde{\psi} ;\,\alpha,\beta\in\mathbb{R}\mbox{ such that }|\alpha|^{p}+|\beta|^{p}=1\}$ is in $\mathcal{C}_{2}^{+}$. Consequently \begin{align*} \lambda_{2}^{+}&\leq \max_{|\alpha|^{p}+|\beta|^{p}=1} \textstyle\frac{1}{p}\int_{\Omega} |\nabla(\alpha\psi+\beta\tilde{\psi})|^{p}\,dx \\ &=\max_{|\alpha|^{p}+|\beta|^{p}=1}\textstyle |\alpha|^{p}\mu^{+}(\omega)+|\beta|^{p}\mu^{+}(\tilde{\omega}) \\ &\leq\max\{\mu^{+}(\omega),\mu^{+}(\tilde{\omega})\} \end{align*} Hence $\lambda^{+}_{2}\leq\mu$. Now let $u$ be an eigenfunction associated to $\lambda^{+}_{2}$ normalized so that $\frac{1}{p}\int_{\partial\Omega}m|u^{+}|^{p}\,d\sigma =\frac{1}{p}\int_{\partial\Omega}m|u^{-}|^{p}\,d\sigma=1$. Denote $\Omega^{\pm}=\{x\in\Omega;\,\pm u(x)>0\}$. We have that $$ \mu^{+}(\Omega^{\pm})\leq\frac{1}{p}\int_{\Omega}|\nabla u^{\pm}|^{p}\,dx =\lambda_{2}^{+} $$ Hence $\mu\leq\lambda_{2}^{+}$. In fact using Picone's identity it can be shown that $\mu^{+}(\Omega^{\pm})=\lambda_{2}^{+}$ so that the infimum (\ref{qual_eq_mu_in_prf_veron_formula}) is achieved by the nodal domains of an eigenfunction associated to $\lambda^{+}_{2}$. It remains to show that this is the only case in which (\ref{qual_eq_mu_in_prf_veron_formula}) is achieved. Let $(\omega,\tilde{\omega})\in\mathcal{A}$ be such that $\max\{\mu^{+}(\omega),\mu^{+}(\tilde{\omega})\} =\lambda_{2}^{+}$. Let $\psi$ and $\tilde{\psi}$ be minimizers for $\mu^{+}(\omega)$ and $\mu^{+}(\tilde{\omega})$ respectively, normalized as usual, and let $C$ be as above. Then the infimum $\lambda^{+}_{2}$ given by (\ref{intro_eq_eigenvalue_LS}) is in fact a minimum which is achieved by $C$. It is then a straightforward consequence of Ljusternik-Schnirelman theory that some element of $C$ must be an eigenfunction associated to $\lambda_{2}^{+}$. 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