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\markboth{A.\,Filinovskiy}{Estimates of eigenvalues of a boundary value problem with a parameter}

\title{Estimates of eigenvalues of a boundary value problem with a parameter\thanks{This work was
supported by the RFBR Grant No. 11-01-00989.}}

\author[A.\,Filinovskiy]{Alexey Vladislavovich Filinovskiy\affil{1}\comma\corrauth}

\address{\affilnum{1}\ Department of Fundamental Sciences, N.E. Bauman Moscow State Technical University,
$2^{nd}$ Baumanskaya 5, 105\,005 Moscow, Russia}

\email{{\tt flnv@yandex.ru} (A.\,Filinovskiy)}

\begin{abstract}
We study an eigenvalue problem for the
Laplace operator with a boundary
condition containing a parameter. We
estimate the rate of convergence of the
eigenvalues to the eigenvalues of the
Dirichlet problem for large positive
values of the parameter.
\end{abstract}

\keywords{Laplace operator, boundary value problem, large parameter,
eigenvalues}

\ams{35J05, 35P15, 49R05}

\maketitle



\section{Introduction}

We consider the eigenvalue problem
\begin{align}
 \label{form1.1}
 \Delta u+\lambda u&=0 \quad \rm{in} \quad
\Omega,
 \\
 \label{form1.2}
\frac{\partial u}{\partial\nu}+\alpha
\sigma (x) u&=0 \quad \rm{on} \quad \Gamma,
     \end{align}
where $\Omega \subset R^n$, $n \geq 2$,
is a bounded domain with boundary
$\Gamma=\partial\Omega \in C^2$. By
$\nu$ we denote the outward unit normal
vector to $\Gamma$, $\alpha$ is a real
parameter. The function $\sigma(x)\in
C^1(\Gamma)$ is positive:
\[
0<\sigma_0\le\sigma(x) \le\sigma_1,\,
\sigma_0=\inf\limits_{x\in\Gamma}
\sigma(x) \mbox{ and }
\sigma_1=\sup\limits_{x\in\Gamma}
\sigma(x).
\]

Problem (\ref{form1.1}),
(\ref{form1.2}) with $\sigma(x)=1$ is
known as the Robin (Fourier) problem
for $\alpha>0$ (see \cite[Ch. 7,
Par. 7.2]{hen}), and
 the generalized Robin problem for
all $\alpha$ (\cite{gs}).

There is a sequence of eigenvalues
$\lambda_1(\alpha)<\lambda_2(\alpha)\leq
\dots $ of problem (\ref{form1.1})
- (\ref{form1.2}) enumerated according
to their multiplicities with
\[
\lim\limits_{k \to
\infty}\lambda_k(\alpha)=+\infty.
\]
We also consider the sequence of
eigenvalues $0<\lambda_1^D<\lambda_2^D
\leq \dots$ of the Dirichlet eigenvalue
problem
\begin{align}
 \label{form1.4}
\Delta u+\lambda u&=0 \quad \rm{in}
\quad \Omega,\\
\label{form1.5} u&=0 \quad \rm{on}
\quad \Gamma,
\end{align}
with
\[
\lim\limits_{k \to
\infty}\lambda_k^D=+\infty.
\]
Note that
the eigenvalues $\lambda_1(\alpha)$ and
$\lambda_1^D$ are simple and the
corresponding eigenfunctions
$u_{1,\alpha}(x)$ and $u^D_1(x)$ are
positive.

In this paper, we estimate
$\lambda_k(\alpha)$ for large values of
$\alpha$. We now give some known
results.

It is easy to see that
$\lambda_k(\alpha) \le \lambda_k^D$,
$k=1,2, \dots$. These inequalities give
the upper bound of $\lambda_k(\alpha)$
for all values of $\alpha$. It was
announced in (\cite[Ch. 6, Par. 2,
No. 1]{ch}) that for $n=2$ and a smooth
boundary $\lim\limits_{\alpha \to
+\infty}
\lambda_k(\alpha)=\lambda_k^D$.

Later the properties of the first
eigenvalue $\lambda_1(\alpha)$ were
studied more precisely. Consider the
case $\sigma (x) = 1$. The following
two-sided estimates:
\begin{equation}
\nonumber \lambda_1^D
\left(1+\frac{\lambda_1^D}{\alpha
q_1}\right)^{-1} \le
\lambda_1(\alpha)\le \lambda_1^D
\left(1+\frac{4\pi}{\alpha
|\Gamma|}\right)^{-1}, \qquad \alpha>0,
\end{equation}
were obtained in \cite{sp} for $n=2$.
Here $|\Gamma|$ is the length of
$\Gamma$ and $q_1$ is the first
eigenvalue of the Steklov problem
\begin{align}
\nonumber \Delta ^2 u&=0 \quad {\rm in}
\quad \Omega,
\\
\nonumber u&=0, \quad \Delta
u-q\frac{\partial u}{\partial \nu}=0
\quad {\rm on} \quad \Gamma.
\end{align}
In \cite{f}, for any $n\ge 2$ we
establish the following asymptotic
expansion:
\begin{equation}
\nonumber \lambda_1 (\alpha)
=\lambda_1^D -
\frac{\int_{\Gamma}\left(\frac{\partial
u_1^D}{\partial \nu}\right)^2
ds}{\int_\Omega \bigl(u_1^D\bigr)^2\,
dx}\,{\alpha^{-1}}+o\left({\alpha^{-1}}\right),
\qquad
 \alpha \to +\infty.
 \end{equation}
%where $u_1^D$ is the first
%eigenfunction of the Dirichlet problem
%(\ref{form1.4}), (\ref{form1.5}).

The case $\alpha <0$ has recently attracted
attention (see, for instance,
\cite{los}). It was shown in \cite{los}
that for a piecewise-$C^1$ boundary
\[
\liminf\limits_{\alpha\to -\infty}
\lambda_1(\alpha)/(-\alpha^2)\ge 1.
\]
For $C^1$ boundaries it was proved
(\cite{lz}) that
\[
\lim\limits_{\alpha\to -\infty}
\lambda_1(\alpha)/(-\alpha^2)= 1.
\]
The
$C^1$-condition is optimal. In
\cite{los}, the authors constructed
plane triangle domains for which
\[
\lim\limits_{\alpha\to -\infty}
\lambda_1(\alpha)/(-\alpha^2)> 1.
\]
In \cite{dk}, the authors proved that for $C^1$ boundaries
\begin{equation}
\label{form1.5a} \lim\limits_{\alpha\to
-\infty}
\frac{\lambda_k(\alpha)}{-\alpha^2}= 1
\end{equation}
 for all $k=1,2,\dots$.



\section{Main results}

The main result of this paper reads as
follows.
\begin{theorem}
The eigenvalues $\lambda_k(\alpha)$,
$k=1,2,\dots$, satisfy the estimates
\begin{equation}
\label{form1.8}
0\le\lambda_k^D-\lambda_k(\alpha)\leq
C_1
\alpha^{-1/2}\left(\lambda_k^D\right)^2,
\qquad \alpha>0,
\end{equation}
where the constant $C_1$ does not
depend on $k$.
 \label{theorem1}
\end{theorem}
In the following theorem we gather the
qualitative properties of eigenvalues
of problem (\ref{form1.1}) -
(\ref{form1.2}) (see also \cite[Ch. 6]{ch} for i) and \cite{los} for ii) and
iii) for $\sigma (x)=1$)

\begin{theorem} \label{theorem2}
The eigenvalues have the following
properties:

\begin{itemize}
\item[i$)$] $\lambda_k(\alpha)$, $k=1,2,\dots$,
are continuous functions of $\alpha$
and
\begin{equation}
\label{form1.6aa}
\lambda_k(\alpha_1)\le\lambda_k(\alpha_2),
\qquad \alpha_1<\alpha_2;
\end{equation}

\item[ii$)$] $\lambda_1(\alpha)$ is a concave
function of $\alpha$:
\begin{equation}
\label{form1.7}
\lambda_1(\beta\alpha_1+(1-\beta)\alpha_2)
\geq \beta\lambda_1(\alpha_1)+
(1-\beta)\lambda_1(\alpha_2), \qquad
0<\beta<1;
\end{equation}

\item[iii$)$] $\lambda_1(\alpha)$ is
differentiable and
\begin{equation}
\label{form1.6a}
\lambda_1'(\alpha)=\frac{\int_\Gamma
\sigma u_{1,\alpha}^2\,ds} {\int_\Omega
u_{1,\alpha}^2\,dx}>0; \end{equation}

\item[iv$)$] the following estimate
\begin{equation}
 \label{form1.6c}
 \liminf\limits_{\alpha\to
-\infty} \frac{\lambda_1
'(\alpha)}{-\alpha
  }\ge \sigma_1^2
 \end{equation}
holds.
\end{itemize}
\end{theorem}





\section{Operator treatment}

In this section, we introduce two linear
operators associated with problems
(\ref{form1.1}) - (\ref{form1.2}) and
(\ref{form1.4}) - (\ref{form1.5}) to
derive the eigenvalue estimates
(\ref{form1.8}).

Consider problem (\ref{form1.1}) -
(\ref{form1.2}) in the space
$H^1(\Omega)$ (\cite{cas,mih}). We define an eigenvalue of
problem (\ref{form1.1}),
(\ref{form1.2}) as a value $\lambda$
for which there exists the non-zero function
$u\in H^1(\Omega)$ satisfying the integral
identity
\begin{equation}
\label{eqn1} \int_{\Omega} (\nabla u,
\nabla v)\,dx+\alpha\int_{\Gamma}
\sigma uv \, ds=\lambda \int_{\Omega}
uv \,dx
\end{equation}
for any $v\in H^1(\Omega)$. Relation (\ref{eqn1}) can be rewritten
as
\begin{equation}
 \int_{\Omega}((\nabla u, \nabla v)
+ Muv) \,dx+\alpha\int_{\Gamma} \sigma
uv \, ds=
 \label{eqn2}
 (\lambda+M) \int_{\Omega} uv \,dx, \quad M>0.
\end{equation}
Let us define an equivalent scalar
product in the space $H^1(\Omega)$ by
the formula
\begin{equation} \label{eqn2a}
 [u,v]_M=
\int_{\Omega}((\nabla u, \nabla v)+Muv)
\,dx,\quad \|u\|^2_M=[u,u]_M.
\end{equation}
 Now (\ref{eqn2})
 transforms to
$$ [u,v]_M+\alpha
[Tu,v]_M=(\lambda+M)[Bu,v]_M, $$ where
the linear self-adjoint non-negative
operators $T:H^1(\Omega)\rightarrow
H^1(\Omega)$ and $B:H^1(\Omega)
\rightarrow H^1(\Omega)$ were defined
by the bilinear forms
\begin{equation} \label{eqn3}
[Tu,v]_M=\int_{\Gamma} \sigma uv \, ds,
\quad [Bu,v]_M=\int_{\Omega} uv \,dx,
\quad u,v \in H^1(\Omega).
\end{equation}
Hence we have an equation in the space
$H^1(\Omega)$ with the norm
$\|\cdot\|_M$:
\begin{equation}
\label{eqn4} (I+\alpha
T)u=(\lambda+M)Bu.
\end{equation}
Now we use the inequality (\cite[Ch. 3, Par. 5, Formula 19]{mih})
\begin{equation}
\label{form2.6b} \|v\|_{L_2(\Gamma)}^2
\le \varepsilon \|\nabla
v\|_{L_2(\Omega)}^2+C_\varepsilon
\|v\|_{L_2(\Omega)}^2,
\end{equation}
which is valid for $v \in H^1 (\Omega)$
with an arbitrary $\varepsilon >0$.
Using (\ref{eqn3}), (\ref{form2.6b}),
we obtain
\begin{eqnarray}
\nonumber \|Tu\|_M^2&=&[Tu,Tu]_M=
\int_{\Gamma} \sigma uTu\,ds\le
\sigma_1
\|u\|_{L_2(\Gamma)}\|Tu\|_{L_2(\Gamma)}
\\
\nonumber &\le& \sigma_1
\varepsilon\left(\int_\Omega\left(|\nabla
Tu|^2+\frac{C_\varepsilon}{\varepsilon}
 (Tu)^2\right) \,dx \right) ^{1/2}\\
&&\times
\left(\int_\Omega\left(|\nabla
u|^2+\frac{C_\varepsilon}{\varepsilon}
u^2\right)\,dx\right)^{1/2}\label{est2}\le
C_2 \varepsilon \|Tu\|_M \|u\|_M ,
 \end{eqnarray}
where $\varepsilon>0$,
$M=M_\varepsilon$. It follows from
(\ref{est2}) that
\[
\|Tu\|_{M_\varepsilon} \le C_2
\varepsilon \|u\|_{M_\varepsilon},
\]
and for any arbitrary small
$\varepsilon$ we have $\|\alpha
T\|_{H^1(\Omega)\rightarrow
H^1(\Omega)}<1$ for $|\alpha|<1/C_2
\varepsilon$. Therefore, the inverse
operator $(I+\alpha T)^{-1}$ is bounded
and
\[
\|(I+\alpha
T)^{-1}\| \le
(1-|\alpha|\|T\|)^{-1}.
\]
Hence, equation (\ref{eqn4}) is
equivalent to
\[
\left(I-(\lambda+M)(I+\alpha
T)^{-1}B\right)u=0.
\]
 The operator $B$
is compact (\cite[Ch. 3, Par. 5, Th. 3]{mih}) and the operator $(I+\alpha
T)^{-1} B: H^1(\Omega) \rightarrow
H^1(\Omega)$ is also compact. Hence the
spectrum of problem (\ref{eqn4})
consists of real eigenvalues
$\lambda_j(\alpha)$, $j=1,2,\dots$, of
finite multiplicity with the only limit
point at the infinity. From
(\ref{eqn3}), (\ref{eqn4}) we obtain
the inequality
\begin{equation}
\nonumber \lambda_j (\alpha)\ge
-M_\varepsilon+\left(1-|\alpha|\|T\|\right)
\frac{\|u_{j,\alpha}\|_{M_\varepsilon}^2}
{\|u_{j,\alpha}\|^2_{L_2(\Omega)}}  \ge
-M_\varepsilon
\end{equation}
with the corresponding eigenfunction
$u_{j,\alpha}$.
 Thus, $\lambda_j(\alpha) \to +\infty$, $j \to
\infty$.

By the variational principle
(\cite[Ch. 4, Par. 1, No. 4]{mih}) we
have
\be
\label{form1.3}
 \lambda_k
(\alpha)&=&\sup_{v_1,\dots,\,v_{k-1}\in
L_2(\Omega)}\inf_{\scriptsize
\begin{array}{c}v\in H^1(\Omega)\\
 (v,v_j)_{L_2(\Omega)}=0 \\
j=1,\dots,
k-1\end{array}}\frac{\int_\Omega|\nabla
v|^2 dx+\alpha\int_\Gamma \sigma v^2
ds}{ \int_\Omega v^2 dx}\,,
\\
\label{form1.6}
\lambda_k^D&=&\sup_{v_1,\dots,\,v_{k-1}\in
L_2(\Omega)}\inf_{\scriptsize
\begin{array}{c}v\in \stackrel {o}{H}\!\!{}^1(\Omega)\\
 (v,v_j)_{L_2(\Omega)}=0 \\
j=1,\dots,
k-1\end{array}}\frac{\int_\Omega|\nabla
v|^2 dx}{ \int_\Omega v^2 dx}\, , \quad
 k=1,2,\dots.
\ee

To prove inequalities
(\ref{form1.8}) we apply the following
statement (see \cite[Ch.~2,\linebreak  Th.~2.3.1]{hen}).

\begin{theorem} \label{theorem3}
Let $T_1$ and $T_2$ be
two linear self-adjoint, compact and
positive operators on a separable
Hilbert space H. Assume also that
$\mu_k(T_1)$ and $\mu_k(T_2)$ are their
$k$-th respective eigenvalues. Then
\begin{equation}\label{form3.23}
\left|\mu_k(T_1)-\mu_k(T_2)\right| \le
\left\|T_1-T_2\right\|.
\end{equation}
\end{theorem}
Now we give the proof of Theorem 1.

\begin{proof}
Consider the boundary value problem
\be
 \label{form3.1}
- \Delta u+u&=&h \quad \rm{in} \quad
\Omega,
 \\
 \label{form3.2}
\frac{\partial u}{\partial\nu}+\alpha
\sigma (x) u&=&0 \quad \rm{on} \quad
\Gamma, \quad \alpha>0,
\ee
with $h \in L_2(\Omega)$. A weak
solution $u \in H^1(\Omega)$ of
problem (\ref{form3.1}),
(\ref{form3.2}) satisfy the integral
identity
\begin{equation}
\label{form3.3} \int_\Omega ((\nabla u,
\nabla v)+uv)dx+\alpha \int_{\Gamma}
\sigma uv\, ds=\int_{\Omega} hv\, dx
\end{equation}
for all $v\in H^1(\Omega)$. Let us
define the scalar product in the space
$H^1(\Omega)$ as
\begin{equation}
\label{form3.4}
(u,v)_{H^1(\Omega),\alpha}=\int_\Omega
((\nabla u, \nabla v)+uv)dx+\alpha
\int_{\Gamma} \sigma uv\, ds
\end{equation}
and the corresponding norm by
\begin{equation}
\nonumber \|u\|_{H^1(\Omega),
\alpha}^2= (u,u)_{H^1(\Omega), \alpha}.
\end{equation}
Due to (\ref{form2.6b}), scalar
product (\ref{form3.4}) is equivalent
to the standard one
\begin{equation}
\label{form3.4a}
(u,v)_{H^1(\Omega)}=\int_\Omega
((\nabla u, \nabla v)+uv)dx.
\end{equation}
 Using
(\ref{form3.3}), (\ref{form3.4}), we
obtain the relation
\begin{equation}
\label{form3.5} (u,v)_{H^1(\Omega),
\alpha}=(h,v)_{L_2(\Omega)}.
\end{equation}
Hence, consider the linear functional
\begin{equation}
\nonumber
 l_h (v)=(h,v)_{L_2(\Omega)}.
\end{equation}
This functional is bounded on the space $H^1(\Omega)$:
\begin{equation}
\label{form3.5a} \left| l_h(v)
\right|\le \|h\|_{L_2(\Omega)}
\|v\|_{L_2(\Omega)}.
\end{equation}
Now, by the Riesz lemma there exists
the unique function $u\in H^1(\Omega)$
satisfying integral identity
(\ref{form3.3}). Applying
(\ref{form3.5}) with $v=u$, we obtain
\begin{equation}
\nonumber \|u\|_{H^1(\Omega), \alpha}^2
\le\|h\|_{L_2(\Omega)}
\|u\|_{H^1(\Omega), \alpha}.
\end{equation}
 Therefore,
\begin{equation}
\label{form3.6} \|u\|_{L_2(\Omega)}\le
\|u\|_{H^1(\Omega),
\alpha}\le\|h\|_{L_2(\Omega)},
\end{equation}
and we can define the bounded linear
operator $A_\alpha
:L_2(\Omega)\rightarrow L_2(\Omega)$
such that $u=A_\alpha h$ and
$\|A_\alpha\| \le 1$. Moreover, if the
domain $\Omega$ with $C^2$ boundary is
bounded, then the space $H^1(\Omega)$
embeds compactly into the space
$L_2(\Omega)$ (\cite[Ch.~1, Th.~1.1.1]{hen}). It means that the operator
$A_\alpha$ is compact. Note that
\be
\nonumber (h, A_\alpha g)_{L_2(\Omega)}
&=&\int_\Omega h A_\alpha g\,
dx=\int_\Omega h v\, dx\\
\nonumber &=&\int_\Omega ((\nabla u,
\nabla v)+uv)dx+\alpha
\int_{\Gamma} \sigma uv\, ds\\
\label{form3.7}&=&\int_\Omega u g\,
dx=(A_\alpha h,g)_{L_2(\Omega)}, \qquad
f,g \in L_2(\Omega),
\ee
with $u=A_\alpha h$, $v=A_\alpha g$,
$u,v \in H^1(\Omega)$. Relation
(\ref{form3.7}) means that $A_\alpha$
is a self-adjoint operator. Now, by
relation (\ref{form3.7}) we have
\be
\nonumber (h, A_\alpha h)_{L_2(\Omega)}
&=& \int_\Omega u h\, dx\\
\nonumber &=&\int_\Omega (|\nabla
u|^2+u^2)dx+\alpha \int_{\Gamma} \sigma
u^2\, ds=\|u\|_{H^1(\Omega), \alpha
}^2>0, \qquad h\ne 0.
\ee
Hence,  the operator $A_\alpha$ is
positive. Finally, $A_\alpha$ is a
self-adjoint positive compact operator
in the Hilbert space $H=L_2(\Omega)$.
By the well-known theorem (\cite[Ch.~1, Th.~1.2.1]{hen}), $A_\alpha$ has a
sequence of eigenvalues $\{\mu_k
(\alpha)\}$, $k=1,2, \dots$ with finite
multiplicities such that $\mu_k
(\alpha)>0$, $\mu_k (\alpha)\searrow
0$, $k \to \infty$. Let us denote by
$u_{k,\alpha}(x)\in L_2(\Omega)$ the
eigenfunction satisfying
 $A_\alpha u_{k, \alpha} =
 \mu_k (\alpha) u_{k, \alpha}$.
 Thus,
\[
 \mu_k (\alpha)
 \left(u_{k, \alpha},v\right)_{H^1(\Omega),\alpha}
  = \left(u_{k,\alpha},v\right)_{L_2(\Omega)}
\]
and
\begin{equation}
\nonumber \mu_k
(\alpha)\left(\int_\Omega ((\nabla
u_{k,\alpha}, \nabla
v)+u_{k,\alpha}v)dx+\alpha
\int_{\Gamma} \sigma u_{k,\alpha}v\,
ds\right) =\int_\Omega u_{k,\alpha}v \,
dx.
\end{equation}
It can be seen that
\[
\mu_k(\alpha)=\frac{1}
{\lambda_k(\alpha)+1}.
\]
Let us note
that for $\alpha > 0$ we have
\[
\mu_k(\alpha)\le
\frac{1}{\lambda_1(\alpha)+1}<1,
\]
so
$\|A_\alpha \|<1$.

Furthermore, consider a Dirichlet
problem
\begin{align}
 \label{form3.8}
- \Delta y+y&=h \quad \rm{in} \quad
\Omega,
 \\
 \label{form3.9}
y&=0 \quad \rm{on} \quad \Gamma.
     \end{align}
For $h\in L_2(\Omega)$ a weak solution
$y\in \stackrel {o}{H}\!\!{}^1(\Omega)$
of problem (\ref{form3.8}),
(\ref{form3.9}) satisfies the integral
identity
\begin{equation}
\label{form3.10} \int_\Omega ((\nabla
y, \nabla v)+yv)dx=\int_{\Omega} hv\,
dx
\end{equation}
for all $v\in \stackrel
{o}{H}\!\!{}^1(\Omega)$. Define the
scalar product in the space $\stackrel
{o}{H}\!\!{}^1(\Omega)$ by
(\ref{form3.4a}).
 Using
(\ref{form3.4a}), (\ref{form3.10}), we
obtain the relation
\begin{equation}
\label{form3.12} (y,v)_{\stackrel
{o}{H}\;\!\!{}^1(\Omega)}=l_h (v).
\end{equation}
 Now, by (\ref{form3.5a}) and the Riesz
lemma there exists the unique function
$y \in \stackrel
{o}{H}\!\!{}^1(\Omega)$ satisfying
integral identity (\ref{form3.10}).
Using (\ref{form3.10}) with $v=y$, we
obtain
\begin{equation}
\|y\|_{\stackrel
{o}{H}\;\!\!{}^1(\Omega)}^2
\le\|h\|_{L_2(\Omega)} \|y\|_{\stackrel
{o}{H}\;\!\!{}^1(\Omega)}.
\end{equation}
 Therefore,
\begin{equation}
\label{form3.13} \|y\|_{L_2(\Omega)}\le
\|y\|_{\stackrel
{o}{H}\;\!\!{}^1(\Omega)}\le\|h\|_{L_2(\Omega)},
\end{equation}
and we can define the bounded linear
operator $A^{D}:L_2(\Omega)\rightarrow
L_2(\Omega)$ such that $u=A^{D} h$ and
$\|A^D\| \le 1$. If the domain $\Omega$
is bounded, then the space $\stackrel
{o}{H}\;\!\!{}^1(\Omega)$ embeds
compactly into the space $L_2(\Omega)$
(\cite[Ch.~1, Th.~1.1.1]{hen}). Hence, the
operator $A^D$ is compact. Note that
\be
\nonumber (h, A^{D}g)_{L_2(\Omega)}
&=&\int_\Omega h A^{D}g\, dx=\int_\Omega
h v\, dx=\int_\Omega ((\nabla y, \nabla
v)+y v)dx\\
\label{form3.14}&=&\int_\Omega y g\,
dx=(A^{D}h,g)_{L_2(\Omega)}, \quad f,g
\in L_2(\Omega),
\ee
with $y=A^{D}h$, $v=A^{D}g$, $y,v \in
\stackrel {o}{H}\!\!{}^1(\Omega)$. Relation (\ref{form3.14}) means that
$A^{D}$ is a self-adjoint operator.
Now, by (\ref{form3.14}) we have
\begin{equation}
\nonumber (h, A^{D}h)_{L_2(\Omega)} =
\int_\Omega y h\, dx=\int_\Omega
(|\nabla y|^2+y^2)dx=\|y\|_{\stackrel
{o}{H}\;\!{}^1(\Omega)}^2>0, \quad h\ne
0.
\end{equation}
Hence,  the operator $A^{D}$ is
positive. Finally, $A^{D}$ is a
self-adjoint positive compact operator
in the Hilbert space $H=L_2(\Omega)$.
By (\cite[Ch.~1, Th.~1.2.1]{hen}), there exists a sequence
of eigenvalues $\{\mu_k ^D\}$, $k=1,2,
\dots$, of the operator $A^D$ with
finite multiplicities such that
$\mu_k^D>0$, $\mu_k^D \searrow 0$, $k
\to \infty$. Denote by $y_k (x)\in
L_2(\Omega)$ the respective
eigenfunction satisfying
 $A^{D}y_k =
 \mu_k ^D y_k$.
 Thus, $\mu_k ^D
 \left(y_k,v\right)_{\stackrel
{o}{H}\;\!{}^1(\Omega)}
  = \left(y_k,v\right)_{L_2(\Omega)}$ and
\begin{equation}
\nonumber \mu_k^D\int_\Omega ((\nabla
y_k, \nabla v)+y_k v)dx =\int_\Omega
y_k v dx.
\end{equation}
Then,
\[
\mu_k^D=
\frac{1}{\lambda_k^D+1}.
\]
Note that
\[
\mu_k^D \le \frac{1}{\lambda_1^D+1}<1,
\]
so $\|A^D \|<1$.

Now we estimate the norm $\|A_\alpha
-A^D\|_{L_2(\Omega)\rightarrow
L_2(\Omega)}$ for large positive
values of $\alpha$.

Let us remind that in domains with
$C^2$-class boundaries and positive
$\sigma (x) \in C^1 (\Gamma)$ the
functions $u=A_\alpha h$ and $y=A^D h$
are strong solutions and belong to
$H^2(\Omega)$ (\cite[Ch. 4, Par.
2, Th. 4]{mih}). Moreover, the following
estimate
\begin{equation}
\label{form3.15} \|y\|_{H^2(\Omega)}
\le C_2 \|h\|_{L_2(\Omega)}
\end{equation}
holds. Now we use estimate
(\ref{form2.6b}) with $\varepsilon=1$:
\begin{equation}
\label{form3.16} \|y\|_{L_2(\Gamma)}
\le C_3 \|y\|_{H^1(\Omega)}.
\end{equation}
Combining (\ref{form3.15}) and
(\ref{form3.16}) we derive the
inequality
\begin{equation}
\label{form3.17} \|\nabla
y\|_{L_2(\Gamma)} \le C_4
\|y\|_{H^2(\Omega)}.
\end{equation}
Since $\left|\frac{\partial
y}{\partial\nu}\right|
\le|\nabla y|$ on $\Gamma$, from (\ref{form3.15}),
(\ref{form3.17}) we obtain the estimate
\begin{equation}
\label{form3.17a} \left\|
\frac{\partial
y}{\partial\nu}\right\|_{L_2
(\Gamma)}\le C_5 \|h\|_{L_2 (\Omega)}.
\end{equation}

 Suppose that $w=\left(A^D-A_\alpha\right) h$. By
 (\ref{form3.1}), (\ref{form3.2}), (\ref{form3.8}),
(\ref{form3.9}) the function $w$ is a
solution of the boundary value problem
\begin{align}
 \label{form3.18}
- \Delta w+w&=0 \quad \rm{in} \quad
\Omega,
 \\
 \label{form3.19}
\frac{\partial w}{\partial\nu}+\alpha
\sigma w&=\frac{\partial y}{\partial
\nu} \quad \rm{on} \quad \Gamma.
     \end{align}
Multiplying equation
(\ref{form3.18}) by $w$ and integrating
it over $\Omega$ with respect to
boundary condition (\ref{form3.19}), we
get the relation
\begin{equation}
\label{form3.20}
 \int_\Omega \left(|\nabla
w|^2+w^2
\right)dx+\frac{1}{\alpha}\int_\Gamma
\left(\frac{\partial
w}{\partial\nu}\right)^2
\frac{ds}{\sigma}=\frac{1}{\alpha}\int_\Gamma
\frac{\partial
w}{\partial\nu}\frac{\partial
y}{\partial\nu}\frac{ds}{\sigma},
\qquad \alpha >0.
\end{equation}
Then we obtain the inequality
\begin{equation}
\nonumber \|w\|^2_{L_2
(\Omega)}+\frac{1}{\alpha} \left\|
\frac{\partial
w}{\partial\nu}\right\|^2_{L_2
(\Gamma)} \le \frac{C_6}{\alpha}\left\|
\frac{\partial
w}{\partial\nu}\right\|_{L_2
(\Gamma)}\left\| \frac{\partial
y}{\partial\nu}\right\|_{L_2 (\Gamma)}
\end{equation}
and, consequently,
\begin{equation}
\nonumber
 \|w\|^2_{L_2
(\Omega)}+\frac{1}{\alpha} \left\|
\frac{\partial
w}{\partial\nu}\right\|^2_{L_2
(\Gamma)} \le\frac{1}{2\alpha}\left\|
\frac{\partial
w}{\partial\nu}\right\|^2_{L_2
(\Gamma)}+\frac{C_6^2}{2\alpha}\left\|
\frac{\partial
y}{\partial\nu}\right\|^2_{L_2
(\Gamma)}.
\end{equation}
Therefore, we have the estimate
\begin{equation}
\label{form3.21} \|w\|_{L_2 (\Omega)}
\le \frac{C_6}{\sqrt{2\alpha}}\left\|
\frac{\partial
y}{\partial\nu}\right\|_{L_2 (\Gamma)},
\quad \alpha>0.
\end{equation}
Combining (\ref{form3.21}) with
(\ref{form3.17a}), we get
\begin{equation}
\nonumber
 \|w\|_{L_2 (\Omega)} \le
C_7\alpha^{-1/2}\|h\|_{L_2 (\Omega)},
\quad \alpha>0,
\end{equation}
with the constant $C_6$ independent of
$\alpha$. Thus, for all $h\in L_2
(\Omega)$ we have the estimate
\begin{equation}
\nonumber
\left\|\left(A^D-A_\alpha\right)
h\right\|_{L_2 (\Omega)} \le
C_7\alpha^{-1/2}\|h\|_{L_2 (\Omega)}
\end{equation}
and
\begin{equation}
\label{form3.22} \left\|A^D-A_\alpha
\right\| \le C_7\alpha^{-1/2}, \quad
\alpha>0.
\end{equation}

Now we apply (\ref{form3.23}) to the
operators $T_1=A_\alpha$, $T_2=A^D$.
Then, by the relations
\begin{equation}
\nonumber \mu_k
(\alpha)=\frac{1}{\lambda_k(\alpha)+1},
\quad \mu_k^D=\frac{1}{\lambda_k^D+1},
\end{equation}
and inequalities (\ref{form3.23}),
(\ref{form3.22})
 we get the estimate
\begin{equation}
\label{form3.24}
\left|\frac{1}{\lambda_k(\alpha)+1}-
\frac{1}{\lambda_k^D+1}\right| \le
C_7\alpha^{-1/2}.
\end{equation}
Therefore,
\begin{equation}
\label{form3.25}
\left|\lambda_k^D-\lambda_k(\alpha)\right|
\le
C_7\alpha^{-1/2}\left(\lambda_k^D+1\right)
\left(\lambda_k(\alpha)+1\right).
\end{equation}
and taking into account
inequalities (\ref{form2.01}) (see
Section 4), we obtain the estimate
\begin{equation}
\label{form3.26}
0\le\lambda_k^D-\lambda_k(\alpha) \le
C_7\alpha^{-1/2}\left(\lambda_k^D+1\right)^2
\le
C_1\alpha^{-1/2}\left(\lambda_k^D\right)^2.
\end{equation}

The proof of Theorem 1 is completed.
\end{proof}

\section{General properties of eigenvalues}

In this Section, we give the proof of
Theorem 2.

\begin{proof}
Due to (\ref{form1.3}), $\lambda_k(\cdot)$ is an increasing
function. Using (\ref{form1.6}) and the
inclusion $\stackrel
{o}{H}\!\!{}^1(\Omega)\subset
H^1(\Omega)$, we have
\be
\nonumber
\lambda_k(\alpha)&=&\sup_{v_1,\dots,\,v_{k-1}\in
L_2(\Omega)}\inf_{\scriptsize
\begin{array}{c}v\in H^1(\Omega)\\
 (v,v_j)_{L_2(\Omega)}=0 \\
j=1,\dots,
k-1\end{array}}\frac{\int_\Omega|\nabla
v|^2 dx+\alpha\int_\Gamma \sigma v^2
ds}{
\int_\Omega v^2 dx}\\
& \le& \nonumber
\sup_{v_1,\dots,\,v_{k-1}\in
L_2(\Omega)}\inf_{\scriptsize
\begin{array}{c}v\in \stackrel {o}{H}\!\!{}^1(\Omega)\\
 (v,v_j)_{L_2(\Omega)}=0 \\
j=1,\dots,
k-1\end{array}}\frac{\int_\Omega|\nabla
v|^2 dx+\alpha\int_\Gamma \sigma v^2
ds}{ \int_\Omega v^2
dx}\\
\label{form2.01}
 &=&\sup_{v_1,\dots,\,v_{k-1}\in
L_2(\Omega)}\inf_{\scriptsize
\begin{array}{c}v\in \stackrel {o}{H}\!\!{}^1(\Omega)\\
 (v,v_j)_{L_2(\Omega)}=0 \\
j=1,\dots,
k-1\end{array}}\frac{\int_\Omega|\nabla
v|^2 dx}{ \int_\Omega v^2
dx}=\lambda_k^D.
\ee
The continuity of $\lambda_k(\alpha)$
was proved in (\cite[Ch. 6, Par.
2, No. 6]{ch}).

Inequality (\ref{form1.7}) can be
proved by the following:
\be
\nonumber \lambda_1(\beta\alpha_1+
(1-\beta)\alpha_2)
&=&\inf\limits_{v\in
\,H^1(\Omega)}\frac{\int_\Omega|\nabla
v|^2\, dx+(\beta\alpha_1+
(1-\beta)\alpha_2)\int_\Gamma \sigma
v^2\, ds}{\int_\Omega v^2\, dx}
\\[1ex]
\nonumber
 &\ge& \beta\inf\limits_{v\in
\,H^1(\Omega)}\frac{\int_\Omega|\nabla
v|^2\, dx+\alpha_1\int_\Gamma \sigma
v^2\, ds}{\int_\Omega v^2\, dx}
\\[1ex]
\nonumber &&+
(1-\beta)\inf\limits_{v\in
\,H^1(\Omega)}\frac{\int_\Omega|\nabla
v|^2\, dx+\alpha_2\int_\Gamma \sigma
v^2\, ds}{\int_\Omega v^2\, dx}
\\[1ex]
\nonumber
&=&\beta\lambda_1(\alpha_1)+(1-\beta)\lambda_1(\alpha_2),
\qquad 0<\beta<1.
\ee
The eigenvalue $\lambda_1(\alpha)$ is
simple for all $-\infty<\alpha<\infty$.
The family of self-adjoint operators
$(I+\alpha T)^{-1} B$ in the space $H^1
(\Omega)$ with norm (\ref{eqn2a})
satisfies the conditions of the
asymptotic perturbation theorem
(\cite[Ch. 8, Par. 4, Th. 2.9]{k}). It
means that the eigenvalue
$\lambda_1(\alpha)$ is a differentiable
function of $\alpha$. So
\begin{equation}
\label{form2.1} \lim\limits_{j\to
\infty}
\frac{\lambda_1(\alpha_j)-\lambda_1(\alpha)}
{\alpha_j-\alpha}=\lambda_1'(\alpha)
\end{equation}
for an arbitrary sequence $\alpha_j \to
\alpha$, $j \to \infty$, $\alpha_j \ne
\alpha$. Let $\alpha_j \to \alpha$, $j
\to \infty$, and
$\|u_{1,\alpha_j}\|_{L_2(\Omega)}=1$,
$u_{1,\alpha_j}\ge 0$. Therefore,
$\|u_{1,\alpha_j}\|_{H^1(\Omega)}\le
C_8$. By (\ref{eqn1}), the functions
$u_{1,\alpha_j}$ satisfy
\begin{equation}
\label{form2.10} \int_{\Omega} (\nabla
u_{1,\alpha_j}, \nabla
v)\,dx+\alpha_j\int_{\Gamma} \sigma
u_{1,\alpha_j}v \,
ds=\lambda_1(\alpha_j) \int_{\Omega}
u_{1,\alpha_j}v \,dx.
\end{equation}
Now, we can choose a subsequence
$u_{1,\alpha_j} \rightharpoonup u$
weakly in $H^1(\Omega)$ and
$\|u_{1,\alpha_j} -u\|_{L_2(\Omega)}\to
0$, $\|u_{1,\alpha_j}
-u\|_{L_2(\Gamma)}\to 0$. It means that
$u\ge 0$ and $\|u\|_{L_2(\Omega)}=1$.
Due to (\ref{form2.10}), $u$ satisfies
the integral identity
\begin{equation}
\label{form2.11} \int_{\Omega} (\nabla
u, \nabla v)\,dx+\alpha\int_{\Gamma}
\sigma u v \, ds=\lambda_1(\alpha)
\int_{\Omega} u v \,dx.
\end{equation}
Hence, by the uniqueness of the first
positive normalized eigenfunction
$u=u_{1,\alpha}$ and
\begin{equation}
\label{form2.12} \|u_{1,\alpha_j}
-u_{1,\alpha}\|_{L_2(\Omega)}\to 0,
\quad j \to \infty.
\end{equation}
Now, we have
\begin{align}
\nonumber
 \int_\Omega |\nabla
(u_{1,\alpha_j}
&-u_{1,\alpha})|^2dx+\alpha \int_\Gamma
\sigma (u_{1,\alpha_j} -u_{1,\alpha})^2
ds \\
\nonumber
&=\lambda_1(\alpha)\int_\Omega
(u_{1,\alpha_j} -u_{1,\alpha})^2dx \\
\nonumber&\quad+
(\lambda_1(\alpha_j)-\lambda_1(\alpha))\int_\Omega
u_{1,\alpha_j}(u_{1,\alpha_j}
-u_{1,\alpha}) dx\\&\quad-
(\alpha_j-\alpha)\int_\Gamma \sigma
u_{1,\alpha_j}(u_{1,\alpha_j}
-u_{1,\alpha}) ds.
 \label{form2.13}
\end{align}
It follows from (\ref{form2.13}) that
\begin{align}
\nonumber \|u_{1,\alpha_j}
-u_{1,\alpha}\|^2_{H^1(\Omega)} \le&\,
C_9 \Bigl( |\alpha| \|u_{1,\alpha_j}
-u_{1,\alpha} \|^2
_{L_2(\Gamma)}\\
\nonumber&+(|\lambda_1(\alpha)|+1)
\|u_{1,\alpha_j} -u_{1,\alpha} \|^2
_{L_2(\Omega)}
\\
\nonumber&
+|\lambda_1(\alpha_j)-\lambda_1(\alpha)|\,
\|u_{1,\alpha_j} -u_{1,\alpha} \|
_{L_2(\Omega)}\|u_{1,\alpha_j} \|
_{L_2(\Omega)}\\
& + |\alpha_j-\alpha|\,
\|u_{1,\alpha_j} -u_{1,\alpha} \|
_{L_2(\Gamma)}\|u_{1,\alpha_j} \|
_{L_2(\Gamma)} \Bigr). \label{form2.14}
\end{align}
Applying (50) and (16) with sufficiently
small $\varepsilon$ we
obtain
\begin{equation}
 \|u_{1,\alpha_j}
-u_{1,\alpha}\|^2_{H^1(\Omega)}
\label{form2.14a} \le
C_{10}\left(\|u_{1,\alpha_j}
-u_{1,\alpha}\|^2_{L_2(\Omega)}+(\alpha_j-\alpha)^2
\|u_{1,\alpha_j}\|^2
_{H^1(\Omega)}\right).
\end{equation}
Due to (\ref{form2.6b}),
(\ref{form2.12}) and (\ref{form2.14a})
we get
\begin{equation}
\nonumber
 \|u_{1,\alpha_j}
-u_{1,\alpha}\|_{L_2(\Gamma)}\le C_{11}
\|u_{1,\alpha_j}
-u_{1,\alpha}\|_{H^1(\Omega)}\to 0,
\quad j \to \infty. \label{form2.15}
\end{equation}
Therefore,
\begin{equation}
 \int_\Gamma \sigma u^2_{1,\alpha_j} ds \to
\int_\Gamma \sigma u^2_{1,\alpha} ds, \quad j
\to \infty. \label{form2.16}
\end{equation}
Now, to obtain (\ref{form1.6a}) we use
the inequalities
\begin{align}
\nonumber
\lambda_1(\alpha_j)-\lambda_1(\alpha)&=
\lambda_1(\alpha_j)- \inf\limits_{v\in
\,H^1(\Omega)}\frac{\int_\Omega|\nabla
v|^2\, dx+\alpha\int_\Gamma \sigma
v^2\, ds}{\int_\Omega v^2\,
dx}\\
\nonumber
&\ge\lambda_1(\alpha_j)-\frac{\int_\Omega|\nabla
u_{1,\alpha_j}|^2\,
dx+\alpha\int_\Gamma \sigma
u_{1,\alpha_j}^2\, ds}{\int_\Omega
u_{1,\alpha_j}^2\, dx}
=(\alpha_j-\alpha)\frac{\int_\Gamma
\sigma u_{1,\alpha_j}^2\,ds}
{\int_\Omega u_{1,\alpha_j}^2\,dx}
\end{align}
and
\begin{align}
 \nonumber
\lambda_1(\alpha_j)-\lambda_1(\alpha)&=
 \inf\limits_{v\in
\,H^1(\Omega)}\frac{\int_\Omega|\nabla
v|^2\, dx+\alpha_j\int_\Gamma \sigma
v^2\, ds}{\int_\Omega v^2\,
dx} -\lambda_1(\alpha)\\
\nonumber &\le \frac{\int_\Omega|\nabla
u_{1,\alpha}|^2\,
dx+\alpha_j\int_\Gamma \sigma
u_{1,\alpha}^2\, ds}{\int_\Omega
u_{1,\alpha}^2\, dx}-\lambda_1(\alpha)=
(\alpha_j-\alpha)\frac{\int_\Gamma
\sigma u_{1,\alpha}^2\,ds} {\int_\Omega
u_{1,\alpha}^2\,dx}\, .
\end{align}
Therefore, for $\alpha_j>\alpha$
\begin{equation}
\label{form1.9} \frac{\int_\Gamma
\sigma u_{1,\alpha_j}^2\,ds} {\int_\Omega
u_{1,\alpha_j}^2\,dx}\le
\frac{\lambda_1(\alpha_j)-\lambda_1(\alpha)}
{\alpha_j-\alpha}\le\frac{\int_\Gamma
\sigma u_{1,\alpha}^2\,ds} {\int_\Omega
u_{1,\alpha}^2\,dx}\,.
\end{equation}
   Finally, it follows from (\ref{form2.1}),
   (\ref{form2.16}) and
(\ref{form1.9}) that
\[
\lambda_1'(\alpha)= \frac{\int_\Gamma
\sigma u_{1,\alpha}^2\,ds} {\int_\Omega
u_{1,\alpha}^2\,dx}\,.
\]
By (\cite[Ch. 4, Par. 2, Th. 4]{mih}),
$u_{1,\alpha} \in H^2 (\Omega)$ and it
satisfies equation (\ref{form1.1})
almost everywhere and the boundary
condition in the sense of trace (the
so-called strong solution). In the case
$\int_\Gamma \sigma u_{1,\alpha}^2
ds=0$, by (\ref{form1.2}) we have:
\begin{equation}
\nonumber u_{1,\alpha}=\frac{\partial
u_{1,\alpha}} {\partial\nu}=0 \quad
\text{on}\quad \Gamma. \end{equation}
 Applying the uniqueness theorem to the Cauchy problem
for second-order elliptic equations
(\cite[Ch.~1, Par.~3, Th.~1.46]{kl}), we get
$u_{1,\alpha}=0$ in $\Omega$. This
contradiction proves that $\lambda
'_1(\alpha)>0$ for all $\alpha$. Taking
into account (\ref{form1.6a}), we have
the inequality
$\lambda_1(\alpha)<\lambda_1^D$.

By combining the result from \cite{lz} with (\ref{form1.6a}) we
obtain the relations
\be
\nonumber \alpha\lambda '
_1(\alpha)&=&\frac{\alpha\int_\Gamma
\sigma u_{1,\alpha}^2\, ds}{\int_\Omega
u_{1,\alpha}^2\, dx}\le
\frac{\int_\Omega|\nabla
u_{1,\alpha}|^2\, dx+\alpha\int_\Gamma
\sigma u_{1,\alpha}^2\, ds}{\int_\Omega
u_{1,\alpha}^2\,
dx}\\
\nonumber  &=&\lambda_1(\alpha)=
-\alpha^2\sigma_1^2(1+\varrho(\alpha)), \qquad
\varrho(\alpha) \to 0, \quad \alpha \to
-\infty.
\ee
Hence,
\begin{equation}
\nonumber
 \frac{\lambda_1
'(\alpha)}{-\alpha
  }\ge \sigma_1^2( 1+\varrho(\alpha)), \quad \alpha
  <0,
 \end{equation}
 and inequality (\ref{form1.6c}) is
 proved.

This completes the proof of Theorem 2.
\end{proof}


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\end{document}