\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 154, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/154\hfil Optimization of eigenvalues] {Optimization of the principal eigenvalue under mixed boundary conditions} \author[L. Cadeddu, M. A. Farina, G. Porru \hfil EJDE-2014/154\hfilneg] {Lucio Cadeddu, Maria Antonietta Farina, Giovanni Porru} % in alphabetical order \address{Lucio Cadeddu \newline Dipartimento di Matematica e Informatica, Univ. di Cagliari, Via Ospedale 72, 09124 Cagliari, Italy} \email{cadeddu@unica.it} \address{Maria Antonietta Farina \newline Dipartimento di Matematica e Informatica, Univ. di Cagliari, Via Ospedale 72, 09124 Cagliari, Italy} \email{mafarina@unica.it} \address{Giovanni Porru \newline Dipartimento di Matematica e Informatica, Univ. di Cagliari, Via Ospedale 72, 09124 Cagliari, Italy} \email{porru@unica.it} \thanks{Submitted November 26, 2013. Published July 5, 2014.} \subjclass[2000]{47A75, 35J25, 35Q92, 49J20, 49K20} \keywords{Principal eigenvalue; rearrangements; minimization; \hfill\break\indent maximization, symmetry breaking; population dynamics} \begin{abstract} We investigate minimization and maximization of the principal eigenvalue of the Laplacian under mixed boundary conditions in case the weight has indefinite sign and varies in a class of rearrangements. Biologically, these optimization problems are motivated by the question of determining the most convenient spatial arrangement of favorable and unfavorable resources for a species to survive or to decline. We prove existence and uniqueness results, and present some features of the optimizers. In special cases, we prove results of symmetry and results of symmetry breaking for the minimizer. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Suppose that $\Omega \subset \mathbb{R}^2 $ is a smooth bounded domain representing a region occupied by a population that diffuses at rate $D$ and grows or declines locally at a rate $g(x)$ (so that $g(x)>0$ corresponds to local growth and $g(x)<0$ to local decline). Suppose the boundary $\partial\Omega$ is divided in two parts, $\Gamma$ and $\partial\Omega\setminus\Gamma$ so that the $1$-Lebesgue measure of $\Gamma$ is positive. Suppose there is an hostile population outside $\Gamma$ (we have Dirichlet boundary conditions on $\Gamma$), and suppose there is not flux of individuals across $\partial\Omega\setminus\Gamma$ (we have Neumann boundary conditions there). If $\phi(x,t)$ is the population density, the behavior of such a population is described by the logistic equation \begin{gather*} \frac {\partial \phi}{\partial t} =D \Delta \phi+ (g(x)-\kappa \phi)\phi \quad \mbox{in } \Omega\times\mathbb{R}^+,\\ \phi=0 \quad \mbox{on } \Gamma\times\mathbb{R}^+,\quad \frac{\partial \phi}{\partial\nu}=0\quad \mbox{on } (\partial\Omega\setminus\Gamma)\times\mathbb{R}^+, \end{gather*} where $\Delta \phi$ denotes the spatial Laplacian of $\phi(x,t)$, $\kappa$ is the carrying capacity and $\nu$ is the exterior normal to $\partial\Omega$. It is known (see \cite{CC1, CC2}) that the logistic equation predicts persistence if and only if $\lambda_g<1/D$, where $\lambda_g$ is the (positive) principal eigenvalue in $$\Delta u + \lambda g(x) u=0 \quad \mbox{in} \quad \Omega,\quad u=0 \quad \mbox{on} \quad \Gamma, \quad \frac{\partial u}{\partial\nu}=0 \quad \mbox{on} \quad \partial\Omega\setminus\Gamma. $$ Many results and applications related to such eigenvalue problems are discussed in \cite{Be, CCP, Li, MM}. In the present paper we consider the following question: \it for weights $g(x)$ within the set of rearrangements of a given weight function $g_0(x)$, which, if any, minimizes or maximizes $\lambda_g$? \rm The corresponding problem with Dirichlet boundary conditions has been investigated by many authors, see \cite{Co, CM1, CM2, CP1} and references therein. For the case of the $p$-Laplacian see \cite{CEP, Po}. For the case of Neumann boundary conditions see \cite{JP}. Eigenvalue problems for nonlinear elliptic equations are discussed in \cite{FPV}. The problem of competition of more species has been treated in \cite{CL, Lo}. In what follows, $\Omega$ is a bounded smooth domain in $\mathbb R^N$. In applications to population dynamics, we have $1\le N\le 3$, but most of our results hold for general $N$. If $E\subset\mathbb R^N$ is a measurable set we denote with $|E|$ its Lebesgue measure. We say that two measurable functions $f(x)$ and $g(x)$ have the same rearrangement in $\Omega$ if $$|\{x\in\Omega:f(x)\ge\beta\}|=|\{x\in\Omega:g(x)\ge\beta\}|\quad \forall\beta\in \mathbb R.$$ If $g_0(x)$ is a bounded function in $\Omega$ we denote by $\mathcal{G}$ the class of its rearrangements. We make use of the following results proved in \cite{Bu1} and \cite{Bu2}. For short, throughout the paper we shall write increasing instead of non-decreasing, and decreasing instead of non-increasing. Denote with $\overline {\mathcal G}$ the weak closure of ${\mathcal G}$ in $L^p(\Omega)$. It is well known that $\overline {\mathcal G}$ is convex and weakly sequentially compact (see for example \cite[Lemma 2.2]{Bu2}). \begin{lemma}\label{le1} Let $\mathcal G$ be the set of rearrangements of a fixed function $g_0\in L^\infty(\Omega)$, and let $u\in L^p(\Omega)$, $p\ge 1$. There exists $ \hat g\in {\mathcal G}$ such that $$\int_\Omega g\, u\, dx\le \int_\Omega \hat g\, u\, dx \quad \quad \forall g\in \overline{\mathcal G}.$$ \end{lemma} The above lemma follows from \cite[Lemma 2.4]{Bu2}. \begin{lemma}\label{le2} Let $g:\Omega\mapsto\mathbb R$ and $w:\Omega\mapsto\mathbb R$ be measurable functions, and suppose that every level set of $w$ has measure zero. Then there exists an increasing function $\phi$ such that $\phi(w)$ is a rearrangement of $g$. Furthermore, there exists a decreasing function $\psi$ such that $\psi(w)$ is a rearrangement of $g$. \end{lemma} The assertions of the above lemma follow from \cite[Lemma 2.9]{Bu2}. \begin{lemma}\label{le3} Let $\mathcal G$ be the set of rearrangements of a fixed function $g_0\in L^p(\Omega)$, $p\ge 1$, and let $w\in L^q(\Omega)$, $q=p/(p-1)$. If there is an increasing function $\phi$ such that $\phi(w)\in {\mathcal G}$ then $$ \int_\Omega g\, w\, dx\le\int_\Omega \phi(w)\, w\, dx\quad \forall g\in \overline{\mathcal G}, $$ and the function $\phi(w)$ is the unique maximizer relative to $\overline{\mathcal G}$. Furthermore, if there is a decreasing function $\psi$ such that $\psi(w)\in {\mathcal G}$ then $$ \int_\Omega g\, w\, dx\ge\int_\Omega \psi(w)\, w\, dx\quad \quad \forall g\in \overline{\mathcal G}, $$ and the function $\psi(w)$ is the unique minimizer relative to $\overline{\mathcal G}$. \end{lemma} The assertions of the above lemma follow from \cite[Lemma 2.4]{Bu2}. We recall that the $L^q(\Omega)$ topology on $L^p(\Omega)$ is the weak topology if $1\le p<\infty$, and the weak* topology if $p=\infty$ \cite{Bu1}. \section{Optimization of the principal eigenvalue} Let $\Omega$ be a bounded smooth domain in $\mathbb R^N,$ and let $g_0(x)$ be a bounded measurable function in $\Omega$ which takes positive values in a set of positive measure. Suppose $\Gamma$ is a portion of $\partial\Omega$ with a positive $(N-1)$-Lebesgue measure. Let $\mathcal G$ be the class of rearrangements generated by $g_0$. For $g\in\mathcal G$, we consider the eigenvalue problem \begin{equation}\label{e2.1} \Delta u + \lambda g(x) u=0 \quad \mbox{in } \Omega,\quad u=0 \quad \mbox{on }\Gamma, \quad \frac{\partial u}{\partial\nu}=0 \quad \mbox{on } \partial\Omega\setminus\Gamma. \end{equation} We are interested in the principal eigenvalue, that is, a positive eigenvalue to which corresponds a positive eigenfunction. If $$ W^+_\Gamma=\Bigl\{w\in H^1(\Omega):\, w=0 \text{ on } \Gamma,\; \int_\Omega g\, w^2dx>0\Bigr\}, $$ we have \begin{equation}\label{e2.2} \lambda_g=\inf_{w\in W^+_\Gamma}\frac{\int_\Omega |\nabla w|^2dx}{\int_\Omega g\; w^2dx}=\frac{\int_\Omega |\nabla u_g|^2dx}{\int_\Omega g u_g^2dx}, \end{equation} where $u_g$ is positive in $\Omega$ and unique up to a positive constant. Note that, if $u_g$ is a minimizer, so is $|u_g|$, hence $|u_g|$ satisfies equation \eqref{e2.1}. By Harnack's inequality (see, for example, \cite[Theorem 1.1]{Tr}) we have $|u_g|>0$ in $\Omega$. By continuity, we have either $u_g>0$ or $u_g<0$. We also note that if there are a positive number $\Lambda$ and a positive function $v$ such that $$ \Delta v + \Lambda g(x) v=0 \quad \mbox{in } \Omega,\quad v=0 \quad \mbox{on }\Gamma, \quad \frac{\partial v}{\partial\nu}=0 \quad \mbox{on } \partial\Omega\setminus\Gamma, $$ then $\Lambda=\lambda_g$ and $v=cu_g$ for some positive constant $c$ (see \cite[Corollary 5.6]{KLP}). Actually, in \cite{KLP} the authors consider the case of Dirichlet boundary conditions, however, the same proof works in our situation. We investigate the problem of finding $$ \inf_{g\in \mathcal G}\lambda_g,\quad \sup_{g\in \mathcal G}\lambda_g. $$ Let $\overline {\mathcal G}$ be the closure of ${\mathcal G}$ with respect to the weak* topology of $L^\infty(\Omega)$. Recall that $\overline {\mathcal G}$ is convex and weakly sequentially compact. \begin{theorem}\label{teo1} Let $\lambda_g$ be defined as in \eqref{e2.2}. \noindent (i) The problem of finding $$ \min_{g\in\mathcal G}\lambda_g $$ has (at least) a solution. \noindent (ii) If $\hat g$ is a minimizer then $\hat g=\phi\bigl( u_{\hat g}\bigr)$ for some increasing function $\phi(t)$. \end{theorem} \begin{proof} If $g_n$ is a minimizing sequence for $\inf_{g\in\mathcal G}\lambda_g$, we have \begin{equation}\label{e2.3} I=\inf_{g\in\mathcal G}\lambda_g =\lim_{n\to\infty}\lambda_{g_n} =\lim_{n\to\infty}\frac{\int_\Omega |\nabla u_{g_n}|^2dx}{\int_\Omega g_n u_{g_n}^2dx}. \end{equation} We can suppose the sequence $\lambda_{g_n}$ is decreasing, therefore, \begin{equation}\label{e2.4} \int_\Omega |\nabla u_{g_n}|^2dx\le C_1\int_\Omega g_n u_{g_n}^2dx\le C_2\int_\Omega u_{g_n}^2dx, \end{equation} for suitable constants $C_1, C_2$. Let us normalize $u_{g_n}$ so that \begin{equation}\label{e2.5} \int_\Omega u_{g_n}^2dx=1. \end{equation} By \eqref{e2.4} and \eqref{e2.5} we infer that the norm $\| u_{g_n}\|_{H^1(\Omega)}$ is bounded by a constant independent of $n$. Therefore (see \cite{GT}), a sub-sequence of $u_{g_n}$ (denoted again by $u_{g_n}$) converges weakly in $H^{1}(\Omega)$ and strongly in $L^2(\Omega)$ to some function $z\in H^1(\Omega)$ with $z(x)\ge 0$ in $\Omega$, $z=0$ on $\Gamma$ and $$\int_\Omega z^2dx=1.$$ Furthermore, since the sequence $g_n$ is bounded in $L^\infty(\Omega)$, there is a subsequence (denoted again by $g_n$) which converges to some $\eta\in \overline{\mathcal G}$ in the weak* topology of $L^\infty(\Omega)$. We have $$ \int_\Omega g_n u_{g_n}^2dx-\int_\Omega \eta\; z^2dx=\int_\Omega (g_n-\eta)\; z^2dx +\int_\Omega g_n(u_{g_n}^2-z^2)dx. $$ Since $$ \lim_{n\to\infty}\int_\Omega (g_n-\eta)\; z^2dx=0 $$ and since $$ \big|\int_\Omega g_n(u_{g_n}^2-z^2)dx\big|\le C_3\| u_{g_n}+z\|_{L^2(\Omega)}\| u_{g_n}-z\|_{L^2(\Omega)}, $$ we find \begin{equation}\label{e2.6} \lim_{n\to\infty}\int_\Omega g_n u_{g_n}^2dx=\int_\Omega \eta\; z^2dx\ge 0. \end{equation} Furthermore, since $\Bigl(\int_\Omega |\nabla u|^2dx\Bigr)^\frac{1}{2}$ is a norm equivalent to the usual norm in $H^1(\Omega)$ with $u=0$ on $\Gamma$, we have \begin{equation}\label{e2.7} \liminf_{n\to\infty}\int_\Omega |\nabla u_{g_n}|^2dx\ge \int_\Omega |\nabla z|^2dx. \end{equation} We claim that $\int_\Omega \eta\, z^2dx>0$. Indeed, passing to the limit as $n\to\infty$ in $$ \int_\Omega \nabla u_{g_n}\cdot\nabla \psi\, dx=\lambda_{g_n}\int_\Omega g_n\, u_{g_n} \psi\, dx $$ we obtain $$ \int_\Omega \nabla z\cdot\nabla \psi\, dx=I\int_\Omega \eta\, z\, \psi\, dx,\quad \forall\psi\in H^1(\Omega), $$ and $$ \int_\Omega |\nabla z|^2\, dx=I\int_\Omega \eta\, z^2dx. $$ If we had $\int_\Omega |\nabla z|^2\, dx=0$, we would have $z=0$, contradicting the condition $\int_\Omega z^2dx=1$. The claim follows. Now, by Lemma \ref{le1}, we find some $\hat g\in\mathcal G$ such that $$ \int_\Omega \eta\, z^2dx\le \int_\Omega \hat g\, z^2dx. $$ Using this estimate and recalling the variational characterization of $\lambda_{\hat g}$ we find $$ I= \frac{\int_\Omega |\nabla z|^2dx}{\int_\Omega \eta\, z^2dx}\ge \frac{\int_\Omega |\nabla z|^2dx}{\int_\Omega \hat g\, z^2dx}\ge \lambda_{\hat g}\ge I. $$ Therefore, $$ \inf_{g\in\mathcal G}\lambda_g=\lambda_{\hat g}. $$ Part (i) of the theorem is proved. Let us prove that $\hat g=\phi(u_{\hat g})$ for some increasing function $\phi$. By $$ \frac{\int_\Omega |\nabla w|^2dx}{\int_\Omega g\, w^2dx}\ge \frac{\int_\Omega |\nabla u_{\hat g}|^2dx}{\int_\Omega \hat g\, u_{\hat g}^2dx}\quad \forall g\in {\mathcal G}, \forall w\in W^+_\Gamma, $$ with $w=u_{\hat g}$ we obtain \begin{equation}\label{e2.9} \int_\Omega g u_{\hat g}^2dx\le \int_\Omega \hat g\, u_{\hat g}^2dx\quad \forall g\in\mathcal G. \end{equation} The function $u_{\hat g}$ satisfies the equation \begin{equation}\label{e2.10} -\Delta u_{\hat g}=\lambda_{\hat g}\hat g u_{\hat g}. \end{equation} Recall that $u_{\hat g}>0$ in $\Omega$. By equation \eqref{e2.10}, the function $u_{\hat g}$ cannot have flat zones neither in the set $$ F_1=\{x\in\Omega: \hat g(x)<0\} $$ nor in the set $$ F_2=\{x\in\Omega: \hat g(x)>0\}. $$ By Lemma \ref{le2}, there is an increasing function $\phi_1(t)$ such that $\phi_1(u^2_{\hat g})$ is a rearrangement of $\hat g(x)$ on $F_1\cup F_2$. Define $$ \alpha=\inf_{x\in \Omega\setminus F_1} u^2_{\hat g}(x). $$ Using \eqref{e2.9}, one proves that $u^2_{\hat g}(x)\le \alpha$ in $F_1$ (see \cite[Lemma 2.6]{BM} for details). Now define $$ \beta=\sup_{x\in \Omega\setminus F_2} u^2_{\hat g}(x). $$ Using \eqref{e2.9} again one shows that $u^2_{\hat g}(x)\ge \beta$ in $F_2$. Since $$ \sup_{F_1}\phi_1( u^2_{\hat g})=\sup_{F_1}\hat g(x)\le 0 $$ we have $\phi_1(t)\le 0$ for $t<\alpha$. Similarly, since $$ \inf_{F_2}\phi_1( u^2_{\hat g})=\inf_{F_2}\hat g(x)\ge 0 $$ we have $\phi_1(t)\ge 0$ for $t>\beta$. We put \[ \tilde\phi(t)=\begin{cases} \phi_1(t) & \text{if } 0\le t< \alpha\\ 0 & \text{if } \alpha\le t \le\beta\\ \phi_1(t) & \text{if } t > \beta. \end{cases} \] The function $\tilde\phi(t)$ is increasing. Furthermore, $\tilde\phi(u^2_{\hat g})$ is a rearrangement of $\hat g(x)$ in $\Omega$ (the functions $\hat g$ and $\tilde\phi(u^2_{\hat g})$ have the same rearrangement on $F_1\cup F_2$, and both vanish on $\Omega\setminus (F_1\cup F_2)$). By \eqref{e2.9} and Lemma \ref{le3} we must have $\hat g=\tilde\phi(u^2_{\hat g})$. Part (ii) of the theorem follows with $\phi(t)=\tilde\phi\bigl(t^2\bigr)$. \end{proof} Let us prove a continuity result. \begin{proposition}\label{prop1} Let $\lambda_g$ be defined as in \eqref{e2.2}. Suppose $g_n\in \mathcal G$, $g\in\overline{\mathcal G}$ and $g_n\rightharpoonup g$ as $n\to\infty$ with respect to the weak* convergence in $L^\infty(\Omega)$. \noindent (i) If $g(x)>0$ in a subset of positive measure then $$\lim_{n\to\infty}\lambda_{g_n}=\lambda_g.$$ \noindent (ii) If $g(x)\le 0$ in $\Omega$ then $$\lim_{n\to\infty}\lambda_{g_n}=+\infty.$$ \end{proposition} \begin{proof} To prove Part (i), we follow an argument similar to that used in \cite[Lemma 4.2]{CEP} in the case of Dirichlet boundary conditions and $g(x)\ge 0$. Let $u_{g_n}$ be the eigenfunction corresponding to $g_n$ normalized so that $$ \int_\Omega u_{g_n}^2dx=1. $$ We have $$ \lambda_{g_n}=\frac{\int_\Omega|\nabla u_{g_n}|^2dx}{\int_\Omega g_n u_{g_n}^2dx}\le \frac{\int_\Omega|\nabla u_{g}|^2dx}{\int_\Omega g_n u_{g}^2dx}, $$ where $u_g$ is the principal eigenfunction corresponding to $g$ normalized so that $$ \int_\Omega u_{g}^2dx=1. $$ Since $$ \lambda_{g}=\frac{\int_\Omega|\nabla u_{g}|^2dx}{\int_\Omega g\, u_{g}^2dx}, $$ we have $$ \lambda_{g_n}\le\frac{\int_\Omega |\nabla u_{g}|^2dx}{\int_\Omega g_n u_{g}^2dx}= \lambda_{g}\frac{\int_\Omega g u_{g}^2dx}{\int_\Omega g_n u_{g}^2dx}. $$ The assumption $g_n\rightharpoonup g$ with respect to the weak* convergence in $L^\infty(\Omega)$ yields $$ \lim_{n\to\infty}\int_\Omega g_nu_{g}^2dx=\int_\Omega gu_{g}^2dx. $$ Therefore, for $\epsilon>0$ we find $\nu_\epsilon$ such that, for $n>\nu_\epsilon$ we have $\lambda_{g_n}<\lambda_{g}+\epsilon$. It follows that $$ \limsup_{n\to\infty}\lambda_{g_n}\le \lambda_{g}. $$ To find the complementary inequality we use the equation $$ -u_{g_n}\Delta u_{g_n}=\lambda_{g_n}g_nu_{g_n}^2. $$ Integrating over $\Omega$, recalling that $u_{g_n}=0$ on $\Gamma$, that the normal derivative of $u_{g_n}$ on $\partial\Omega\setminus \Gamma$ vanishes, and using the inequality $\lambda_{g_n}<\lambda_{g}+\epsilon$ (for $n$ large), we find a constant $C$ such that $$ \int_\Omega |\nabla u_{g_n}|^2dx\le (\lambda_{g}+\epsilon)\int_\Omega g_n u_{g_n}^2dx\le C, $$ where the boundedness of $g_n$ and the normalization of $u_{g_n}$ have been used. We infer that the norm $\| u_{g_n}\|_{H^1(\Omega)}$ is bounded by a constant independent of $n$. A sub-sequence of $u_{g_n}$ (denoted again by $u_{g_n}$) converges weakly in $H^{1}(\Omega)$ and strongly in $L^2(\Omega)$ to some function $z\in H^1(\Omega)$ with $z\ge 0$, $z=0$ on $\Gamma$, and $$ \int_\Omega z^2dx=1. $$ As a consequence, $$ \liminf_{n\to\infty}\int_\Omega |\nabla u_{g_n}|^2dx\ge \int_\Omega |\nabla z|^2dx. $$ Moreover, arguing as in the proof of Theorem \ref{teo1} we obtain $$ \lim_{n\to\infty}\int_\Omega g_n u_{g_n}^2dx=\int_\Omega g\, z^2dx. $$ An argument similar to that used in the proof of Theorem \ref{teo1} shows that we cannot have $$ \int_\Omega |\nabla z|^2\, dx=\int_\Omega g z^2dx=0. $$ Therefore, it follows that $$ \liminf_{n\to\infty}\lambda_{g_n} =\liminf_{n\to\infty}\frac{\int_\Omega|\nabla u_{g_n}|^2dx} {\int_\Omega g_n u_{g_n}^2dx}\ge \frac{\int_\Omega|\nabla z|^2dx} {\int_\Omega g\, z^2dx}\ge \lambda_g. $$ Part (i) of the proposition follows. To prove Part (ii), we argue by contradiction. Suppose there is a sub-sequence of $\lambda_{g_n}$, still denoted $\lambda_{g_n}$, and a real number $M$ such that $$ \lambda_{g_n}=\frac{\int_\Omega |\nabla u_{g_n}|^2dx} {\int_\Omega g_n u_{g_n}^2dx}\le M $$ and $$ \int_\Omega u_{g_n}^2dx=1. $$ It follows that $$ \int_\Omega |\nabla u_{g_n}|^2dx\le M \int_\Omega g_n u_{g_n}^2dx\le \tilde M. $$ Therefore, there is a sub-sequence of $u_{g_n}$ (denoted again by $u_{g_n}$) which converges weakly in $H^{1}(\Omega)$ and strongly in $L^2(\Omega)$ to some function $z\in H^1(\Omega)$, $z(x)\ge 0$ $z=0$ on $\Gamma$, and such that $$ \int_\Omega z^2dx=1. $$ Furthermore, up to a subsequence, we may suppose that $$ \lim_{n\to\infty}\lambda_{g_n}=\tilde\lambda. $$ For $n=1, 2, \dots $ we have $$ \int_\Omega \nabla u_{g_n}\cdot \nabla \psi dx=\lambda_{g_n}\int_\Omega g_n u_{g_n}\psi dx\quad \ \forall \psi\in H^1(\Omega). $$ Letting $n\to\infty$ we find $$ \int_\Omega \nabla z\cdot \nabla \psi dx=\tilde\lambda\int_\Omega g z\psi dx\quad \ \forall \psi\in H^1(\Omega). $$ By the latter equation we find $z\in C^1(\Omega)$. Furthermore, putting $\psi=z$ we find $$ \int_\Omega |\nabla z|^2dx= \tilde\lambda \int_\Omega g z^2dx\le 0, $$ where the assumption $g(x)\le 0$ has been used. It follows that $|\nabla z|=0$ in $\Omega$. Therefore, $z=0$, contradicting the condition $\int_\Omega z^2dx=1$. The proof is complete. \end{proof} \begin{proposition}\label{prop2} Let $\lambda_g$ be defined as in \eqref{e2.2}, and let $J(g)=1/\lambda_g$. The map $g\mapsto J(g)$ is Gateaux differentiable with derivative $$J'(g;h)=\frac{\int_\Omega h\, u_g^2dx}{\int_\Omega |\nabla u_g|^2dx}.$$ Furthermore, if $g$ satisfies $\int_\Omega g(x)dx\ge 0$, the map $g\mapsto \lambda_g$ is strictly concave. \end{proposition} In case we have Dirichlet boundary conditions, the proof of this proposition is well known (see, for example, \cite[Proposition 1]{CCP}). The same proof also works under our boundary conditions. \begin{theorem}\label{teo12} Let $\lambda_g$ be defined as in \eqref{e2.2}. The problem of finding $$ \max_{g\in\mathcal{\overline G}}\lambda_g $$ has a solution; if $\int_\Omega g_0(x)dx\ge 0$, the maximizer $\check g$ is unique; if $\int_\Omega g_0(x)dx> 0$, we have $\check g=\psi\bigl( u_{\check g}\bigr)$ for some decreasing function $\psi(t)$; finally, if $g_0(x)\ge 0$ then the maximizer $\check g$ belongs to $\mathcal G$. \end{theorem} \begin{proof} Since the functional $g\mapsto \lambda_g$ is continuous with respect to the weak* topology of $L^\infty(\Omega)$ (by Proposition \ref{prop1}), and since $\mathcal{\overline G}$ is weakly compact, a maximizer $\check g$ exists in $\mathcal{\overline G}$. Assuming $\int_\Omega g_0(x)dx \ge 0$, the uniqueness of the maximizer follows from the strict concavity of $\lambda_g$ (see Proposition \ref{prop2}). If $\int_\Omega g_0(x)dx> 0$, the maximizer $\check g$ is positive in a subset of positive measure, therefore, $\lambda_{\check g}$ is finite and $u_{\check g}(x)>0$ a.e. in $\Omega$. If $00\}$ nor in the set $F_4=\{x\in\Omega: \check g(x)<0\}$. By Lemma \ref{le2}, there is a decreasing function $\psi_1(t)$ such that $\psi_1(u^2_{\check g})$ is a rearrangement of $\check g(x)$ on $F_3\cup F_4$. Following the proof of \cite[Theorem 2.1]{BM}, we introduce the class $\mathcal W$ of rearrangements of our maximizer $\check g$. Of course, $\mathcal W\subset {\mathcal{\overline G}}$. Define $$ \gamma=\inf_{x\in \Omega\setminus F_3} u^2_{\check g}(x). $$ Using \eqref{e21}, one proves that $u^2_{\check g}(x)\le \gamma$ in $F_3$. Define $$ \delta=\sup_{x\in \Omega\setminus F_4} u^2_{\check g}(x). $$ Using \eqref{e21} again one shows that $u^2_{\check g}(x)\ge \delta$ in $F_4$. Now we put \[ \tilde\psi(t)=\begin{cases} \psi_1(t) & \text{if } 0\le t< \gamma\\ 0 & \text{if } \gamma\le t \le\delta\\ \psi_1(t) & \text{if } t > \delta. \end{cases} \] The function $\tilde\psi(t)$ is decreasing and $\tilde\psi(u^2_{\check g})$ is a rearrangement of $\check g(x)$ in $\Omega$. Indeed, the functions $\check g$ and $\tilde\psi(u^2_{\check g})$ have the same rearrangement on $F_3\cup F_4$, and both vanish on $\Omega\setminus (F_3\cup F_4)$. By \eqref{e21} and Lemma \ref{le3} we must have $\check g=\tilde\psi(u^2_{\check g})\in \mathcal W$. Note that, in general, the maximizer $\check g$ does not belong to $\mathcal G$ (see next Theorem \ref{teo2.5}). Assuming $g_0(x)\ge 0$, we can prove that $\check g\in\mathcal G$. Indeed, by \eqref{e22}, the function $u_{\check g}$ cannot have flat zones in the set $F=\{x\in\Omega: \check g(x)>0\}$. If $|F|<|\Omega|$, since $\check g\in\mathcal{\overline G}$, by \cite[Lemma 2.14]{Bu2} we have $|F|\ge |\{x\in\Omega: g_0(x)>0\}|$. Therefore there is $g_1\in \mathcal G$ such that its support is contained in $F$. By Lemma \ref{le3}, there is a decreasing function $\psi_1(t)$ such that $\psi_1(u^2_{\check g})$ is a rearrangement of $g_1(x)$ on $F$. Define $$ \gamma=\inf_{x\in \Omega\setminus F} u^2_{\check g}(x). $$ Using \eqref{e21}, one proves that $u^2_{\check g}(x)\le \gamma$ in $F$. By using equation \eqref{e21} once more we find that $u^2_{\check g}(x)< \gamma$ a.e. in $F$. Now define \[ \tilde\psi(t)=\begin{cases} \psi_1(t) & \text{if } 0\le t< \gamma\\ 0 & \text{if } t \ge\gamma. \end{cases} \] The function $\tilde\psi(t)$ is decreasing and $\tilde\psi(u^2_{\check g})$ is a rearrangement of $g_1\in\mathcal G$ on $\Omega$. Indeed, the functions $g_1$ and $\tilde\psi(u^2_{\check g})$ have the same rearrangement on $F$, and both vanish on $\Omega\setminus F$. By \eqref{e21} and Lemma \ref{le3} we must have $\check g=\tilde\psi(u^2_{\check g})\in \mathcal G$. Hence, in case of $|F|<|\Omega|$, the conclusion follows with $\psi(t)=\tilde\psi(t^2)$. If $|F|=|\Omega|$, the proof is easier and we do not need the introduction of the function $g_1$. The statement of the theorem follows. \end{proof} \begin{theorem} \label{teo2.5} Suppose $u\in H^2(\Omega)\cap C^0(\Omega)$ with $u=0$ on $\Gamma$ and $\frac{\partial u}{\partial\nu}=0$ on $\partial\Omega\setminus\Gamma$. Here $\Gamma\subset \partial\Omega$ is supposed to be smooth and to have a $(N-1)$-Lebesgue positive measure. Let $u(x)>0$ in $\Omega$ and $$ -\Delta u=\Lambda \psi(u)u \quad \text{a.e. in }\ \Omega $$ for some $\Lambda>0$ and some decreasing bounded function $\psi$. Then, either $\Delta u\le 0$ or $\Delta u\ge 0$ a.e. in $\Omega$. \end{theorem} \begin{proof} By contradiction, suppose that the essential range of $\Delta u$ contains positive and negative values. Since $u>0$ and $-\Delta u=\Lambda \psi(u)u$, $\psi(t)$ takes positive and negative values for $t>0$. Let \[ \beta=\sup\{t:\psi(t)\ge 0\},\quad \Omega_\beta=\{x\in\Omega:u(x)>\beta\}. \] By our assumptions, the open set $\Omega_\beta$ is not empty. On the other hand, since $\psi$ is decreasing and $u>0$ we have $$ -\Delta u<0\quad \text{in } \Omega_\beta,\quad u=\beta\quad \text{on } \partial\Omega_\beta\setminus\Gamma_\beta\quad \text{and } \frac{\partial u}{\partial\nu}=0\quad \text{on } \Gamma_{\beta}, $$ where $\Gamma_{\beta}$ is a suitable subset of $\partial\Omega\setminus\Gamma$. By second Hopf's boundary Lemma, $u$ cannot have its maximum value on $\Gamma_{\beta}$. Therefore, the maximum principle for subharmonic functions yields $u(x)\le\beta$ in $\Omega_\beta$. This contradicts the definition of $\Omega_\beta$, and the theorem follows \end{proof} \section{Symmetry} \subsection{The one-dimensional case} Let $N=1$ and $\Omega=(0,L)$. Given a measurable function $f:\Omega\to \mathbb R$, we denote by $f^*$ the decreasing rearrangement of $f$ ($f^*$ is non increasing on $(0,L)$). Similarly, we denote by $f_*$ the increasing rearrangement of $f$. The following results are well known. \begin{lemma}\label{le4} Let $N=1$ and $\Omega=(0,L)$. \noindent (i) If $f(x)$ and $g(x)$ belong to $L^\infty(\Omega)$ then \begin{equation}\label{e2.11} \int_\Omega f_*(x)g^*(x)dx\le\int_\Omega f(x)g(x)dx\le \int_\Omega f^*(x) g^*(x)dx. \end{equation} (ii) If $u\in H^{1}(\Omega)$, $u(x)\ge 0$ and $u(L)=0$, then $u^*\in H^1(\Omega)$, $u^*(x)\ge 0$, $u^*(L)=0$ and \begin{equation}\label{e2.12} \int_\Omega (u')^2dx\ge \int_\Omega ((u^*)')^2dx. \end{equation} \end{lemma} For a proof of the above lemma, see, for example, \cite{Ba, Ka}. Note that, (i) is often proved for non negative functions. However, replacing $f$ by $f+M$ and $g$ by $g+M$ with a suitable constant $M$, one gets the result for bounded functions. \begin{theorem}\label{teo2} Let $\mathcal G$ be a class of rearrangements generated by a bounded function $g_0$. Let $g\in \mathcal G$, and let $\lambda_g$ be defined as in \eqref{e2.2} with $\Omega=(0,L)$ and $u(L)=0$. Then we have $\lambda_g\ge \lambda_{g^*}$. \end{theorem} \begin{proof} If $g\in \mathcal G$ and if $u_g$ is a corresponding (positive) principal eigenfunction, we have \begin{equation}\label{e2.13} \lambda_g=\frac{\int_\Omega (u_g')^2dx}{\int_\Omega g u_g^2dx}. \end{equation} Since $u_g>0$ we have $(u_g^*)^2=(u_g^2)^*$, and by \eqref{e2.11} we find \begin{equation}\label{e2.14} \int_\Omega g u_g^2\; dx\le \int_\Omega g^*\; (u_g^*)^2 dx. \end{equation} Note that $u_g^*(L)=0$. Using \eqref{e2.11}, \eqref{e2.12}, and recalling the variational characterization of $\lambda_{g^*}$ we find $$ \lambda_g=\frac{\int_\Omega (u_g')^2dx}{\int_\Omega g u_g^2dx} \ge \frac{\int_\Omega ((u_g^*)')^2dx}{\int_\Omega g^* (u_g^*)^2 dx} \ge \frac{\int_\Omega (u_{g^*}')^2dx}{\int_\Omega g^* u_{g^*}^2 dx}= \lambda_{g^*}. $$ The proof is complete. \end{proof} \begin{example} \label{examp1} \rm Let us apply the result of Theorem \ref{teo2} to the following example. For $0<\alpha\le \beta0$ is the corresponding principal eigenvalue and $u$ is a corresponding eigenfunction, we have \[ -u''=\begin{cases} \Lambda u, & 00,\quad z(t)<1+t(\beta-\alpha). $$ It follows that equation \eqref{e2.15} has a unique solution $\Lambda=\Lambda(\alpha)$ such that $$ \frac{1}{\alpha}\arctan\frac{1}{1+\sqrt\Lambda(\beta-\alpha)} <\sqrt{\Lambda}<\frac{\pi}{2\alpha}. $$ It is clear that $\Lambda\to\infty$ as $\alpha\to 0$. \end{example} \begin{theorem}\label{teo3} Let $\mathcal G$ be a class of rearrangements generated by a function $g_0$ defined in $(0,L)$ such that $\int_0^L g_0(x)dx>0$. If $g\in \mathcal G$, let $\rho$ such that $\int_0^\rho g_*(x)dx=0$. Define $\check g=0$ for $00$ is the corresponding principal eigenvalue and $u$ is a corresponding eigenfunction, we have \[ -u''=\begin{cases} 0, & 00$ must satisfy the condition $$ \sqrt{\Lambda}=\frac{\pi}{2(L-\alpha)}. $$ \end{example} \begin{remark} \label{rmk1} \rm The conclusions of Theorems \ref{teo2} and \ref{teo3} continue to hold in the following case. Let $\Omega=(0,L)\times (0,\ell)$, and let \[ \begin{cases} u=0 & \text{on }\{x_1=L\},\\ \frac{\partial u}{\partial\nu}=0 & \text{on } \{x_1=0\}\cup\{x_2=0\}\cup\{x_2=\ell\}. \end{cases} \] Suppose the function $g_0$ depends on $x_1$ only, and the class $\mathcal G$ is the (restricted) family of all rearrangements of $g_0$ in $\Omega$ depending on $x_1$ only. In this situation, the principal eigenfunctions depend on $x_1$ only, and the optimization of the corresponding principal eigenvalue is essentially a one-dimensional problem discussed in Theorems \ref{teo2} and \ref{teo3}. \end{remark} \subsection{$\alpha$-sector} For $0<\alpha\le\pi$, consider the domain (in polar coordinates $(r,\theta)$) \begin{equation}\label{e3.90} D=\{(r,\theta):\ 0\le r0$. If $g\in \mathcal G$, let $D_\rho\subset D$ be the $\alpha$-sector such that $\int_{D_\rho} g_*(x)dx=0$. Define $\check g=0$ for $x\in D_\rho$, and $\check g=g_*$ for $D\setminus D_\rho$. Let $\lambda_g$ be defined as in \eqref{e2.2} where $\Omega=D$ and $\Gamma$ is the portion of $\partial D$ with $r=R$. Then $\lambda_g\le \lambda_{\check g}$. \end{theorem} \begin{proof} Let $u_g$ be a principal eigenfunction corresponding to $g\in \mathcal G$, and let $u_{\check g}$ be a principal eigenfunction corresponding to $\check g$. We have \begin{equation}\label{e3.13} \lambda_g=\frac{\int_D |\nabla u_g|^2dx}{\int_D g u_g^2dx}\le \frac{\int_D |\nabla u_{\check g}|^2dx}{\int_D g u_{\check g}^2dx}. \end{equation} On the other hand, the function $u_{\check g}$ satisfies the problem $$ -\Delta u_{\check g}=\lambda_{\check g} \check g u_{\check g}\quad \text{in }D, $$ with $u=0$ on $\Gamma$ and $u_\theta=0$ on the segments $\theta=0$ and $\theta=\alpha$. The solution $u_{\check g}$ is radial and (since $\check g=0$ for $x\in D_\rho$) its derivative (with respect to $r$) is a constant in $(0,\rho)$. Since $u'(0)=0$, this constant must be zero, and the function $u_{\check g}$ is a positive constant in $D_\rho$. Furthermore, since $$ -ru_{\check g}'(r)=\lambda_{\check g} \int_0^r t\check g u_{\check g}dt\ge 0, $$ $u_{\check g}(r)$ is decreasing on $(0,R)$. It follows that $$ u_{\check g}=u_{\check g}^*,\quad u_{\check g}^2=(u_{\check g}^2)^*. $$ Hence, since $g_*(r)$ is increasing, by the left hand side of \eqref{e3.9} we find \begin{equation}\label{e3.14} \int_D g u_{\check g}^2dx\ge \int_D g_* u_{\check g}^2dx. \end{equation} Furthermore, since $u_{\check g}$ is a constant in $D_\rho$ and since $\int_{D_\rho} g_*dx=0$, we have $$ \int_D g_* u_{\check g}^2dx=c^2\int_{D_\rho} g_*dx +\int_{D\setminus D_\rho} g_* u_{\check g}^2dx =\int_D \check g u_{\check g}^2dx. $$ Therefore, by \eqref{e3.14} we find $$ \int_D g u_{\check g}^2dx\ge \int_D \check g u_{\check g}^2dx. $$ The latter inequality and \eqref{e3.13} yield $$ \lambda_g\le \frac{\int_D |\nabla u_{\check g}|^2dx}{\int_D \check g u_{\check g}^2 dx}= \lambda_{\check g}. $$ The theorem is proved. \end{proof} In case the class $\mathcal G$ is generated by $g_0=\chi_E-\chi_F$ with $E\cap F=\emptyset$, $|E|>|F|$, the maximum of $\lambda_g$ is attained for $\check g=\chi_G$, where $G$ is the set $$ G=\Big\{(r,\theta)\in D:\, r^2\ge R^2-\frac{2(|E|-|F|)}{\alpha}\Big\}. $$ If $|E|\le |F|$, we have $\sup\lambda_g=+\infty$. \section{Symmetry breaking} Concerning the minimum, the symmetry of the data may not be inherited by the solution. \begin{theorem}\label{teo4.1} Let $N=2$ and $\Omega=B_{a,a+2}$, the annulus of radii $a$, $a+2$. Suppose $g_0=\chi_E$, where $E$ is a measurable set contained in $\Omega$ and such that $|E|=\pi\rho^2$, $0<\rho<1$. Let $\mathcal G$ be the family of rearrangements of $g_0$. Consider the eigenvalue problem $(2.1)$ in $\Omega$ with $\Gamma$ being the circle with radius $a+2$. If $a$ is large enough then a minimizer of $\lambda_g$ in $\mathcal G$ cannot be radially symmetric with respect to the center of $B_{a,a+2}$. \end{theorem} \begin{proof} Recall that \begin{equation}\label{2.160} \lambda_g=\inf\Big\{\frac{\int_\Omega|\nabla w|^2dx}{\int_\Omega g\, w^2 dx},\quad w\in H^1(\Omega):\quad w=0\quad \text{on}\quad \Gamma,\quad \int_\Omega g\, w^2dx>0\Big\}. \end{equation} Let $E=B_\rho$ be a disc with radius $\rho$ and such that its center $x_0$ lies on $|x|=a+1$. If $g=\chi_{B_\rho}$, the function $z=(\rho^2-|x-x_0|^2)^+$ vanishes on $\Gamma$, hence, if $|x-x_0|=r$, \begin{equation}\label{e2.16} \lambda_g\le \frac{\int_{B_\rho}|\nabla z|^2dx}{\int_{B_\rho} z^2 dx}= \frac{\int_0^\rho 4 r^3 dr}{\int_0^\rho r(\rho^2-r^2)^2dr}=\frac{6}{\rho^2}. \end{equation} Note that this upper bound is independent of $a$. Now suppose $g=\chi_E$, with $E$ radially symmetric with respect to the center of $B_{a,a+2}$. With $r=|x|$, put $g(x)=h(r)=\chi_{E_1}$, $E_1$ being the intersections of $E$ with a ray of $B_{a+2}$. The corresponding eigenfunction is radially symmetric (by uniqueness), and the inferior in \eqref{e2.16} can be taken over all $v\in H^1_{rad}$ (the class of radially symmetric functions in $H^1(\Omega)$) with $v(a+2)=0$. We have $$ \lambda_g=\inf\Big\{\frac{\int_a^{a+2}r (v')^2dr}{\int_a^{a+2}r h v^2 dr},\quad v\in H^1_{rad}: v(a+2)=0\Big\}. $$ We find $$ \frac{\int_a^{a+2}r (v')^2dr}{\int_a^{a+2}r h v^2dr} \ge \frac{\int_a^{a+2}a (v')^2dr}{\int_a^{a+2}(a+2) h v^2 dr}= \frac{a}{a+2}\frac{\int_a^{a+2} (v')^2dr}{\int_a^{a+2} h v^2dr}. $$ The $1$-measure of $E_1$ depends on the location of $E$, however we have $$ |E_1|\le \sqrt{a^2+\rho^2}-a:=\ell. $$ Note that $\ell\to 0$ as $a\to\infty$. Using classical inequalities for decreasing rearrangements we find $$ \frac{\int_a^{a+2}(v')^2dr}{\int_a^{a+2} h\, v^2 dr}\ge \frac{\int_a^{a+2}((v^*)')^2dr}{\int_a^{a+2} h^*\, (v^*)^2 dr}\ge \frac{\int_{-1}^1(w')^2dt}{\int_{-1}^1 g^*\, w^2 dt}, $$ where $w(t)=v^*(r)$, $t=r-(a+1)$, and \[ g^*=\begin{cases} 1, & -1