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\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)\\
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$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.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,\\
<u,v>:=\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),<v,v>=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 ${\bf R}:=\{x\in \mathbb{R}^N: |x_i|<a_i\}$,  $i=1,2\dots N$. Separation of variables yields for the eigenfunctions corresponding to $\lambda_{\pm1}$
\begin{eqnarray*}
\varphi_1= \prod_{i=1}^N\cos(\sqrt{\omega_1}x_i),
\quad \lambda_1= \sum_{i=1}^N\omega_i,\\
\varphi_{-1}= \prod_{i=1}^N\cosh(\sqrt{\nu_1}x_i, \lambda_{-1} =-\sum_{i=1}^N\nu_i
\end{eqnarray*}
where
\begin{equation}\label{rectangle}
-\sqrt{\omega_i}\tan(\sqrt{\omega_i}a_i)= \lambda_1\sigma, \quad \sqrt{\nu_i}\tanh(\sqrt{\nu_i}a_i)= \lambda_{-1}\sigma.
\end{equation}
For a cube $\mathcal{C}_a$ where $a_i=a$, we get $\omega_i=\omega$ and $\nu_i=\nu$  for all $i$ where
\begin{equation}\label{cube}
-\tan(\sqrt{\omega}a)=N\sqrt{\omega}\sigma, \quad \tanh\sqrt{\nu}a=N\sqrt{\nu}\sigma.
\end{equation}
The critical value for ${\bf R}$ is
$$
\sigma_0=-\left(\sum_{j=1}^N\frac{1}{a_j}\right)^{-1}.
$$
It follows from the previous remark that $\omega_i\neq 0$, $i=1.\dots.N$  exist only if $\sigma<\sigma_0$ and $\nu_i\neq 0$, $i=1,\dots,N$
exist only if $\sigma>\sigma_0$.
\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 $|\sigma|<A/N$. Namely, choosing
$\epsilon= \frac{T^2}{A^2(|\sigma|^{-1}-N/A)}$, we obtain
$$
\frac{1}{\lambda_{-1�}(D)} = \int_D \varphi_{-1}^2\:dx+\sigma \oint_{\partial D} \varphi_{-1}^2\:ds
\geq \frac{\sigma T^2}{A^2(|\sigma|^{-1}-N/A)}.
$$
Observe that this is a rather crude bound because the inequality
\eqref{ho} is not optimal in contrast to \eqref{trace2} which is
attained for $v=\varphi_{-1}$. Another disadvantage is that this
estimate does not yield an upper bound if $ -\sigma_0<\sigma <-A/N$.
From \eqref{trace2} we get we obtain
\begin{eqnarray*}
\oint_{\partial D} v^2 ds \leq\frac{1}{|\sigma \lambda_{-1}(D)|}\int_D|\nabla v|^2\:dx+ |\sigma | ^{-1} \int_Dv^2\:dx \tx{if} \sigma_0<\sigma<0.
\end{eqnarray*}
It turns out \cite{BBR06} that
$$
\sigma \lambda_{-1}(D;\sigma) \nearrow,\quad \lim_{\sigma\to \sigma_0^+} \sigma \lambda_{-1}(D;\sigma)=0 \quad \lim_{\sigma \to 0^-}\sigma \lambda_{-1}(D;\sigma)=\infty.
$$
Consequently for any given $\epsilon >0$ 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}

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