%------------------------------------------------------------------------------ % Which version of paper: Here please write first, second, .... % Here please write the corresponding author and his/her e-mail. % Here please write the date of submission of paper or its revisions. %------------------------------------------------------------------------------ % \documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, plainpages]{hyperref} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} %\usepackage{showkeys} \newcommand{\tx}[1]{\mbox{\quad{#1}\quad}} \numberwithin{equation}{section} \begin{document} \setcounter{page}{68} \noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}} \noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent Banach J. Math. Anal. 2 (2008), no. 2, 68--75\\ $\frac{\rule{4.55in}{0.05in}}{{}}$\\ {\footnotesize \textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach \textbf{\large{J}}ournal of \textbf{\large{M}}athematical \textbf{\large{A}}nalysis}}\\ ISSN: 1735-8787 (electronic)\\ \textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\ $\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]} \title[Eigenvalue estimates]{An eigenvalue problem with mixed boundary conditions and trace theorems} \author[C.Bandle]{Catherine Bandle$^1$} \address{$^{1}$ Department of Mathematics, University of Basel, Rheinsprung 21, CH-4051 Basel, Switzerland.} \email{\textcolor[rgb]{0.00,0.00,0.84}{catherine.bandle@unibas.ch}} \dedicatory{This paper is dedicated to Professor J. E. Pecaric\\ \vspace{.5cm} {\rm Submitted by P. K. Sahoo}} \subjclass[2000]{Primary 35P15, 47A75 ; Secondary 49R50, 51M16.} \keywords{Estimates of eigenvalues, trace inequality, comparison theorems for eigenvalues.} \date{Received: 21 April 2008; Accepted 20 May 2008.} \begin{abstract} An eigenvalue problem is considered where the eigenvalue appears in the domain and on the boundary. This eigenvalue problem has a spectrum of discrete positive and negative eigenvalues. The smallest positive and the largest negative eigenvalue $\lambda_{\pm 1}$ can be characterized by a variational principle. We are mainly interested in obtaining non trivial upper bounds for $\lambda_{-1}$. We prove some domain monotonicity for certain special shapes using a kind of maximum principle derived by C. Bandle, J.v. Bellow and W. Reichel in [J. Eur. Math. Soc., 10 (2007), 73--104]. We then apply these bounds to the trace inequality. \end{abstract} \maketitle \section{Introduction and preliminaries} \noindent Let $D\subset \mathbb{R}^N$ be a bounded domain with a Lipschitz boundary and denote by $n$ its outer normal. The eigenvalue problem we are interested in is \begin{align}\label{eigenvalue} \triangle \varphi + \lambda \varphi=0 \tx{in} D, \quad \frac{\partial \varphi}{\partial n}=\lambda \sigma \varphi \tx{on} \partial D. \end{align} Here $\sigma$ is a real number. Notice that $\lambda_0=0$ and $\varphi_0=$const. is always a solution. It was proved in \cite{BBR}, cf. also \cite{ViVa} that problem \eqref{eigenvalue} has a discrete spectrum consisting of non negative eigenvalues \begin{align*} \lambda_0=0 <\lambda_1<\lambda_2\leq\lambda_3\leq \dots \tx{if} \sigma \geq 0 \end{align*} and of positive and negative eigenvalues if $\sigma$ is negative. Taking into account their multiplicity they can be ordered as follows: $$ \ldots \leq \lambda_{-n} \leq \ldots \leq \lambda_{-2}\leq \lambda_{-1}<0=\lambda_0<\lambda_1\leq\lambda_2\leq \ldots\leq \lambda_n\leq \ldots $$ Except for $N=1$ both the positive and the negative parts of the spectrum contains infinitely many eigenvalues tending to $\pm\infty.$ As for the classical eigenvalue problems the eigenvalues and eigenfunctions can be obtained by a minimum maximum principle \cite{BBR}. We shall describe it for $\lambda_{\pm1}$ and $\sigma<0$ which will be the main topic of this paper. For this purpose we introduce the following notation. \newline For $u,v\in W^{1,2}(D)$ set \begin{align*} a(u,v):= \int_D uv\:dx+\sigma \int_{\partial D} uv\:ds,\\ :=\int_D (\nabla u,\nabla v)\:dx. \end{align*} The number $$ \sigma_0(D):= - \frac{|D|}{|\partial D|} $$ will play an essential role in our considerations. It turns out cf. \cite{Ba07} that \begin{align}\label{variation} \frac{1}{\lambda_1(D)}= \sup_{\mathcal{K}} a(v,v), \tx{if} \sigma<\sigma_0<0,\\ \nonumber \frac{1}{\lambda_{-1}(D)}=\inf_{\mathcal{K}}a(v,v) \tx{if} \sigma_0<\sigma<0,\\ \nonumber {\mathcal{K}}:=\{v\in W^{1,2}(D),=1\}. \end{align} From here it follows \cite{BBR} that \newline {\sl (i) If $\sigma <\sigma_0$ then $\varphi_1$ is of constant sign and $\lambda_1$ is simple, whereas $\varphi_{-1}$ changes sign. \newline (ii) If $0>\sigma >\sigma_0$ then $\varphi_{-1}$ is of constant sign and $\lambda_{-1}$ is simple, whereas $\varphi_1$ changes sign. \newline (iii) If $\sigma=\sigma_0$ both $\varphi_1$ and $\varphi_{-1}$ change sign.} From \eqref{variation} we get trace inequalities of the type \begin{align} |\sigma| \oint_{\partial D} v^2 ds \geq \int_Dv^2\:dx -\frac{1}{\lambda_1(D)}\int_D|\nabla v|^2\:dx \tx{if} \sigma<\sigma_0<0 \nonumber\\ |\sigma | \oint_{\partial D} v^2 ds \leq \int_Dv^2\:dx +\frac{1}{|\lambda_{-1}(D)|}\int_D|\nabla v|^2\:dx \tx{if} \sigma_0<\sigma<0.\label{trace2} \end{align} In order to exhibit these inequalities we need a lower bound for $\lambda_1(D)$ and an upper bound for $\lambda_{-1}(D)$. Such bounds were given in \cite{Ba07} for$\lambda_1(D)$. In particular it was shown that among all domains of fixed volume $\lambda_1$ is smallest for the ball. \smallskip Notice that unlike the case with Dirichlet boundary conditions the eigenvalues are in general not monotone with respect to the domain. An exception is when $D$ is contained in a ball $B$, $D\subset B$. Then \cite{Ba07}, $\lambda_1(D)\geq \lambda_1(B)$ and $\lambda_{-1}(D)\geq \lambda_{-1}(B)$. \smallskip In this note we give some estimates from above for $\lambda_{-1}$ and compare the corresponding trace inequality \eqref{trace2} with a result of Horgan \cite{Ho79}. %__________________________________________________SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS% \section{Main results} Let us start with two examples. \smallskip \subsection {Ball} Consider the the ball $B_R$ centered at the origin and of radius $R$ and let $(r,\theta)$, $r\in [0,R]$, $\theta\in \mathbb{S}^{N-1}$ be its polar coordinates. The critical value is $$ \sigma_0=-\frac{R}{N}. $$ Separation of variables $\varphi(x)=w(r)\alpha(\theta)$ yields $$ w''+\frac{N-1}{r}w'+\Big(\lambda-\frac{\nu}{r^2}\Big)w = 0, \quad \Delta_\theta \alpha + \nu \alpha=0, $$ where $\Delta_\theta$ is the Laplace-Beltrami operator on $\mathbb{S}^{N-1}$ with eigenfunction $\alpha$ and eigenvalue $\nu$. Hence $\alpha$ must be a spherical harmonic and $\nu=\nu_k=k(k+N-2)$, $k=0,1,2,\ldots$. We are interested only in the case where the eigenfunctions do not change sign. Therefore $\nu=0$ and $\alpha=1$. The equation for $w$ then becomes \begin{eqnarray*} w''+\frac{N-1}{r}w'+\lambda w & = & 0 \mbox{ in } (0,R), \\ w'(R) & = & \sigma \lambda w(R). \end{eqnarray*} By the usual transformation $z(r)=r^{\frac{N-2}{2}}w(r)$ one finds \begin{eqnarray*} z''+\frac{z'}{r} +\Big(\lambda-\frac{((N-2)/2)^2}{r^2}\Big)z=&0 \mbox{ in } (0,R), \\ z'(R)=\Big(\sigma \lambda +\frac{N-2}{2R}\Big)z(R). \end{eqnarray*} Solutions are of the form \begin{equation*} z(r)=\left\{\begin{array}{ll} J_{(N-2)/2}(\sqrt{\lambda}r) & \mbox{ if } \lambda>0, \vspace{\jot}\\ I_{(N-2)/2}(\sqrt{-\lambda}r) & \mbox{ if } \lambda<0, \vspace{\jot}\\ r^{(N-2)/2} & \mbox{ if } \lambda=0, \end{array}\right. \label{form} \end{equation*} where $J_\nu$ is the regular Bessel function of index $\nu$ and $I_\nu$ is the regular modified Bessel function of index $\nu$. The eigenvalues $\lambda_{\pm1}$ are determined by the equations \begin{eqnarray*} \frac{J_{(N-2)/2}'(\sqrt{\lambda}R)}{J_{(N-2)/2}(\sqrt{\lambda}R)} &=& \frac{\sigma\lambda+\frac{N-2}{2R}}{\sqrt{\lambda}} \mbox{ if } \lambda>0,\\ \frac{I_{(N-2)/2}'(\sqrt{-\lambda}R)}{I_{(N-2)/2}(\sqrt{-\lambda}R)} &=& \frac{\sigma\lambda+\frac{N-2}{2R}}{\sqrt{-\lambda}} \mbox{ if } \lambda<0. \end{eqnarray*} If follows from the remark at the end of the Introduction that $\lambda_{\pm1}$ is a decreasing function of $R$. %%%%%%%%%%%%%%%%%%%%%%%%%%Subsection \subsection{N-dimensional rectangle} Consider the rectangle ${\bf R}:=\{x\in \mathbb{R}^N: |x_i|\sigma_0$. \smallskip {\sc Monotonicity} Let $a_1\leq a_2\leq \dots a_N$. Since $x\to x\tan(a x)$ and $x\to \tanh(ax)$ are increasing functions of $a$ for fixed $x>0$, we deduce that $\omega_1\geq \omega_2\geq \dots\geq \omega_N$ and $\nu_1\geq \nu_2\geq \dots \geq\nu_N$. %%%%%%%%%%%%%%%%%%LLLLLLLLLLLLLLLLLLLL \begin{lemma}(i) If $\sigma <\sigma_0$ then $$ \lambda_1({\mathcal{C}}_{a_N})\leq \lambda_1({\bf R})\leq \lambda_1({\mathcal{C}}_{a_1}) . $$ (ii) Let $\sigma>\sigma_0$. Then $$ \lambda_{-1}({\mathcal{C}}_{a_N})\leq \lambda_{-1}({\bf R})\leq \lambda_{-1}({\mathcal{C}}_{a_1}). $$ \end{lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%llllllllllllllllllllllllllllllllllllllllllll \begin{proof} (i) The inequality $\omega_1\geq \omega_i$ implies $$ \sqrt{\omega_1}\tan(\sqrt{\omega_1}a_1)=\lambda_1|\sigma|\leq N\omega_1|\sigma|. $$ Let $\omega^*$ be the solution $$ \sqrt{\omega^*}\tan(\sqrt{\omega^*}a_1)= N\omega^*|\sigma|. $$ Note that for positive $a$, the function $x\to \frac{\tan(xa)}{x}$ is increasing in $(0, \pi/2a)$. Hence $\omega_1\leq \omega^*$ and by \eqref{rectangle} $$ \sqrt{\omega^*}\tan(\sqrt{\omega^*a_1}\geq \sqrt{\omega_1}\tan(\sqrt{\omega_1a_1}=\lambda_1|\sigma|. $$ The first assertion now follows from \eqref{cube}. The proof of the second assertion is similar with the inequality signs reversed. \smallskip (ii) The inequality $\nu_1\geq \nu_i$ implies $$ \sqrt{\nu_1}\tanh(\sqrt{\nu_1}a_1)=\lambda_{-1}|\sigma|\leq N\nu_1|\sigma|. $$ Let $\nu^*$ be the solution $$ \sqrt{\nu^*}\tanh(\sqrt{\nu^*}a_1)= N\nu^*|\sigma|. $$ Note that for positive $a$, the function $x\to \frac{\tanh(xa)}{x}$ is decreasing in $\mathbb{R}^+$. Hence $\nu_1\geq \nu^*$ and by \eqref{rectangle} $$ \sqrt{\nu^*}\tanh(\sqrt{\nu^*a_1}\leq \sqrt{\nu_1}\tanh(\sqrt{\nu_1a_1}=\lambda_{-1}|\sigma|. $$ The first conclusion follows from \eqref{cube}. The second assertion is obtained by the same argument. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%subsection \subsection{General domains} \subsubsection{Scaling} Let $\alpha>0$ be an arbitrary fixed number. Put $y=\alpha x$. If $(\varphi,\lambda)$ is a solution of \eqref{eigenvalue} then $\psi(y)=\varphi(y/\alpha)$ satisfies $$ \triangle \psi + \frac{\lambda}{\alpha^2}\psi = 0 \tx{in} \alpha D,\quad \frac{\partial \psi}{\partial n}=\alpha^{-1}\lambda\sigma \psi \tx{on} \partial (\alpha D). $$ Hence $\lambda \alpha^{-2}$ corresponds to the eigenvalue in $\alpha D$ with $\sigma$ replaced by $\alpha \sigma$. It was shown in \cite{BBR06} that for $\sigma \in \mathbb{R}^-$ the functions $\sigma \to\sigma \lambda_{\pm 1}(\sigma)$ are continuous and strictly increasing. Hence for $\alpha>1$ $$ \frac{\lambda_{\pm 1}(D)}{\alpha}\geq \lambda_{\pm 1}(\alpha D). $$ From this inequality we obtain that for any domain $D$ and any $\alpha >1$ the following monotonicity result $$ \lambda_1(D)\geq \lambda_1(\alpha D). $$ Such a conclusion can not be drawn for $\lambda_{-1}$. \medskip \noindent {\bf Open problem} {\sl Does $\lambda_{-1}(D)\geq \lambda_{-1}(\alpha D)$ hold for all domains?} \medskip \begin{remark} It is not difficult to see that the above inequality holds for the balls and the rectangles. \end{remark} \subsubsection{Positivity principle and upper bounds for $\lambda_{-1}$} Nontrivial upper bounds for $\lambda_{-1}$ can be obtained from the positivity principle derived in \cite{BBR06}. %%%%%%%%%%%%%%%%%%%%%%%%%%%%LLLLLLLLLLLLLLLLLLLLLLLLL \begin{lemma} \label{2.3} Assume $\sigma \in (\sigma_0, 0)$. If there exists a positive solution $u$ of $$ \triangle u+ \lambda u\leq 0 \tx{in} D, \quad \frac{\partial u}{\partial n} \geq \lambda \sigma u \tx{on} \partial D, $$ then $\lambda \geq \lambda_{-1}$. \end{lemma} %%%%%%%%%%%%%%%%%%%%%%%%lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%TTTTTTTTTTTTTTTTTTTTTT \begin{theorem} \label{2.4} Let $D$ be a star-shaped domain and $B_R$ the greatest ball inscribed in $D$. Suppose that $(\frac{x}{|x|},n)\geq \alpha$ on $\partial D$ and that $\sigma':= \sigma \alpha^{-1}\geq \sigma_0(B_R)=-R/N$. Let $\lambda'_{-1}(B_R)$ be the largest negative eigenvalue of \eqref{eigenvalue} corresponding to $\sigma'$. Then $$ \lambda_{-1}(D)\leq \lambda'_{-1}(B_R). $$ \end{theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt \begin{proof} Let $u$ be the eigenfunction of $B_R$ corresponding to $\lambda'_{-1}(B_R)$ with $\sigma $ replaced by $\sigma':=\sigma/\alpha$. This eigenfunction can be extended to $r>R$ (cf. Section 2.1). From the monotonicity of $\lambda'_{-1}(B_R)$ with respect to $R$ (cf. \cite{Ba07}) it follows that the function $r\to u'(r)/u(r)$ is increasing. This can also be seen directly from the following simple argument. Consider the function $v= \frac{u'(r)}{u(r)}$. It satisfies $$ v'+v^2 +\frac{N-1}{r}v+\lambda'_{-1}=0 \tx{in} \mathbb{R}^+ , \quad v(0)=0. $$ Near $r=0$ obviously $v$ increases. Suppose that $v$ is not monotone. Then there exists a point $r'$ where $v$ assumes a local maximum. This is not possible because $v''(r')=\frac{N-1}{r'}v(r')>0$. On $\partial D$ we have $$ \frac{\partial u}{\partial n}= u'(r)(\frac{x}{|x|},n)\Big |_{\partial D}\geq \lambda'_{-1}\sigma' \alpha u(r)\Big |_{\partial D}=\lambda'_{-1}\sigma u(r)\Big |_{\partial D}. $$ The assertion now follows from Lemma \ref{2.3}. \end{proof} A similar result can be obtained if we inscribe instead of a ball a rectangle ${\bf R}$ as in Section 2.2. %%%%%%%%%%%%%%%%TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT \begin{theorem} \label{2.5} Let $D$ be convex and ${\bf R}\subset D$. Assume that for $i=1,\dots N$, sign$(n_i)$=sign$(x_i)$ on $\partial D$. (This can always be achieved in convex domains.) Let $\lambda_{-1}''$ be the first negative eigenfunction of ${\bf R}$ with $\sigma$ replaced by $\sigma'':= \sigma\beta^{-1}$, where $\beta =\min_{\partial D}\sum_{i=1}^N|n_i|$. Suppose that $\sigma ''>\sigma_0({\bf R})$. Then $$ \lambda_{-1}(D)\leq \lambda''_{-1}({\bf R}). $$ \end{theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt \begin{proof} Let $\varphi$ be the positive eigenfunction corresponding to ${\bf R}$ (cf. Section 2.2). Then since ${\bf R}$ is contained in $D$ we have, using the same notation as in Section 2.2. $$ \varphi^{-1}\frac{\partial \varphi}{\partial n}\Big|_{\partial D} = \sum_{j=0}^Nn_j\frac{x_j}{|x_j|}|\nu_j|\tanh(\sqrt{|\nu_j|}x_j)\geq\beta \sigma''\lambda_1'' = \sigma\lambda_{-1}''. $$ The conclusion follows now from Lemma \ref{2.3}. \end{proof} \begin{remark} From the last theorem we see immediately that if ${\bf R'}$ and ${\bf R}$ are two rectangles such that ${\bf R'}\subset {\bf R}$ then $\lambda_{-1}({\bf R})\leq \lambda_{-1}({\bf R'})$. \end{remark} \smallskip {\bf Open problem} Is it possible to find other domains beside of balls and rectangles for which inclusion results as in Theorems \ref{2.4} and \ref{2.5} hold? %%%%%%%%%%%%%%%%%%%%%%%%%SSSSSSSSSSSSSSSSSSSSSSSSSSSS \section{Trace inequalities} It is well-known that for Lipschitz domains a function $v\in W^{1,2}(D)$ has a trace which satisfies the following inequality: for any given $\epsilon >0$ there exists a positive number $c(\epsilon)$ independent of $v$ such that \begin{equation}\label{ho} \oint_{\partial D} v^2\: ds \leq \epsilon \int_D|\nabla v|^2\:dx +c(\epsilon) \int_Dv^2\:dx . \end{equation} The smallest such constant will be denoted by $c^*(\epsilon)$. For star-shaped domains Horgan \cite{Ho79} has shown that \begin{equation}\label{ho2} c^*(\epsilon)\leq \frac{N}{A} + \frac{T^2}{A^2\epsilon} \tx{where $A=\min_{\partial D} (x,n)$ and}T=\max_{\partial D}|x|. \end{equation} Equations \eqref{ho} and \eqref{ho2} can be used to estimate $\lambda_{-1}(D)$ from above provided $|\sigma|0$ we can find $\sigma_0<\sigma_\epsilon <0$ such that \newline \noindent $\sigma_\epsilon\lambda_{-1}(D;\sigma_\epsilon)=\epsilon.$ %%%%%%%%%%%%%%%%%%%%%%%%%CCCCCCCCCCCCCCCCCCCCCCCCCCC \begin{corollary} Let $D$ be an arbitrary (not necessarily star-shaped) Lipschitz domain. Let $\sigma_\epsilon$ be the unique solution of $\sigma \lambda_{-1}(D)=\epsilon^{-1}$. Then the optimal constant $c^*(\epsilon)$ in \eqref{ho} is $|\sigma^{-1}_\epsilon|$. The equality sign holds for the corresponding eigenfunction $\varphi_{-1}$. \end{corollary} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%cccccccccccccccccccccccccccccc Under the conditions of Theorem \ref{2.4} we can replace $\lambda_{-1}(D)$ by the upper bound $\lambda_{-1}'(B_R)$. If $\tilde\sigma_\epsilon$ is the solution of $\sigma \lambda'_{-1}(B_R)=\epsilon^{-1}$ then the optimal constant satisfies $$ c^*(\epsilon)\leq\frac{1}{ |\tilde\sigma_\epsilon|}. $$ \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{BBR} C. Bandle, J.v. Below and W. Reichel, \textit{Parabolic problems with dynamical boundary conditions: eigenvalue expansions and blow up}, Rend. Lincei Mat. Appl., \textbf{17} (2006), 35--67. \bibitem {BBR06} C. Bandle, J.v. Below and W. Reichel, \textit{Positivity and anti-maximum principles for elliptic operators with mixed boundary conditions}, J. Eur. Math. Soc., \textbf{10} (2007), 73--104. \bibitem{Ba07} C. Bandle, \textit{A Rayleigh-Faber-Krahn inequality and some monotonicity properties for eigenvalue problems with mixed boundary conditions}, to appear in Proceedings of the Conference on Inequalities and Applications '07. \bibitem{Ho79} C.O. Horgan, \textit{Eigenvalue estimates and the trace theorem}, J. Math. Anal. Appl., \textbf{69} (1979), 231--242. \bibitem{ViVa} J. L. Vazquez and E. Vitillaro, \textit{Heat equation with dynamical boundary conditions of locally reactive type}, Semigroup Forum, \textbf{74} (2007), no. 1, 1--40. \end{thebibliography} \end{document} %------------------------------------------------------------------------------ % End of journal.tex %------------------------------------------------------------------------------ .