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large coupling, singular perturbation, eigenvalues, supra-conductor
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\documentclass[11pt]{article}


\textheight9in
\textwidth6.5in
\baselineskip14pt
\hoffset-2cm
\voffset-2cm


\parindent=0pt


\def\dsp{\displaystyle}
\def\nl2#1{\left\| #1\right\|_{L^2(\O)}}
\def\nli#1{\left\| #1\right\|_{L^{\infty}(\O)}}

%%%%%%%%%%%Les letres grecques%%%%%%%%%%%%5


\def\l{\lambda}

\def\s{\sigma}

\def\rs{\sqrt{\s}}
\def\pt{\forall\;\;}
\def\dis{\varphi}
\def\ds{\longrightarrow}
\def\D{\Delta}
\def\G{\Gamma}
\def\L{{\cal L}}
\def\O{\Omega}
\def\C{{\mathord{C\mkern-16mu{\phantom t\vrule}{\phantom o}}}}
\def\U{ {\cal U}}
\def\lap{\Delta}
\def\OO{{\cal O}}
\def\li{\lambda^{\infty}}

\def\dd#1#2{\frac{\partial #1}{\partial #2}}
\def\ddn#1{\dsp\frac{\partial #1}{\partial\nu}}

%%%%%%%%%%%%%%%%%%%les ensembles%%%%%%%%%%%%%%%%

\def\R{{\mathord{I\!\! R}}}
\def\I{{\mbox{\rm I$\!$I}}}
\def\N{{\mbox{\rm I$\!$N}}}
\def\CC{{\cal C}}
\def\H{{\cal H}}




\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{remark}{Remark}[section]

%numerotation des equations par section
\def\thesection{\arabic{section}}
\def\theequation{\arabic{section}.\arabic{equation}}


\catcode`\@=12

\def\un{{\bf  1}} 

\begin{document}

%\proofmodefalse

\begin{center}
{\Large \bf Spectral Asymptotic in the Large Coupling Limit}
\end{center}
\begin{center}

Vincent Bruneau, Gilles Carbou
\end{center}

\medskip

\noindent
 Math\'ematiques Appliqu\'ees de Bordeaux,
Universit\'e Bordeaux 1, 351 cours de la Lib\'eration, 33405 Talence cedex,
France. \\


\smallskip

{\small
{\bf Abstract : } in this paper, we study  a singular perturbation of an eigenvalues
problem related to   supra-conductor wave guides.  Using boundary
layer tools we
perform a complete  asymptotic expansion of the eigenvalues as the conductivity
tends to $+\infty$.}

\section{Introduction}



Let $\O$ and  $\OO$ two bounded connected smooth domains of $\R^2$ such that $\O \subset \subset \OO$. We denote
$\U = \OO \setminus \O$.

 We consider the self adjoint operator $\H^\s$ on $L^2(\OO)$ with domain $H^2(\OO) \cap H_0^1(\OO)$ defined by:
\begin{equation}\label{defhs}
\H^\sigma = - \Delta + \s \un_\O,
\end{equation}
where $\s$ is a positive real number. In this paper we study the
asymptotic behavior of the eigenvalues of $\H^\s$ as $\s$ tends to
$+\infty$. The limit operator is $\H^\infty=-\lap$ defined on $H^2(\U)\cap
H^1_0(\U)$, that is $u=0$ on $\partial \U=\partial \OO\cup \partial \O$.  

\begin{remark}
 Such a problem appears in the study of an electromagnetic
wave guide $\OO\times \R$ with section $\OO$, where $\U\times \R$ is a 
dielectric body and $\O\times \R$ is a supra-conductor material with
very large conductivity. For mathematical studies concerning
electromagnetic wave guides, see \cite{ASBND} and \cite{Joly}.
\end{remark}

\vspace{3mm}


%LAIUS BIBLIOGRAPHIQUE 1 %%%%%%%%%%%%%%%%%%%%%%%
According to general monotone convergence  theorems (see \cite{Si78}),
it is well known that   $\H^\s$ tends to $\H^\infty$ in strong resolvent sense,
as $\s \rightarrow + \infty$.  In our case, where the operators are
self-adjoint, non negative with compact  resolvents, this implies that
the eigenvalues of  $\H^\s$ tend to ones of $\H^\infty$  (see also
\cite{thCa}). To our knowledge,  there is no result concerning the
rate of the convergence (see below for more references). 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The goal of this paper is to build an asymptotic expansion of the
eigenvalues of $\H^\s$ when $\s$ tends to $+\infty$. 

We denote  $\dsp R^\s(\xi)=(\H^\s-\xi I)^{-1}$  the resolvent of $\H^\s$,
defined on $L^2(\OO)$ with values in $H^2(\OO)\cap H^1_0(\OO)$ for
$\xi\notin \mbox{spec }\H^\s$.


We first prove that the resolvent of $\H^\s$ admits a first order  asymptotic
expansion~: 

\begin{theorem}
\label{thm0}
We define the operator $R^\infty(\xi)$ by~:~
if $f\in H^1(\OO)$, $R^\infty(\xi)(f)$ is defined by
$R^\infty(\xi)(f)(x)=U_0(x)$ if $x\in \U$ and $R^\infty(\xi)(f)(x)=0$ if
$x\in \O$, where
\[\left\{\begin{array}{l}
U^0=0 \mbox{ on } \partial \O\cup \partial \OO,\\
\\
(-\lap -\xi)U^0=f \mbox{ on }\U.
\end{array}\right.\] 

On the other hand we define $R^\s_1(\xi)$ by~:~if $f\in H^1(\OO)$,
\[
R^\s_1(\xi)(f)(x)=\left\{
\begin{array}{l}
 \psi(x)\alpha_1(x)e^{-\rs\dis(x)} \mbox{ if }x\in\O,\\
\\
U^1(x)\mbox{ if }x\in \U,
\end{array}
\right.
\]
where
\[
\left\{\begin{array}{l}
U^1=0\mbox{ on }\partial \OO,\\
\\
\dsp U^1=\ddn{U^0} \mbox{ on }\partial \O,\\
\\
(-\lap -\xi)U^1=0 \mbox{ on }\U,
\end{array}\right.
\]
and   
\begin{itemize}
\item $\alpha_1$ is an extension of $\dsp U^1|_{\partial \O}$ on $\O$,

\item for $x\in \O$, $\dis(x)=\mbox{dist}(x, \partial \O)$, 

\item $\psi$ is a cut-off function equal
to 1 in a neighborhood of $\partial \O$.
\end{itemize}


\vspace{2mm}
Then when $\s$ tends to $+\infty$, $R^\s(\xi)$ admits
an asymptotic expansion of the type~:
\[R^\s(\xi)=R^\infty(\xi)+\frac{1}{\rs}R^\s_1(\xi)+\frac{1}{\rs}K^\s_2(\xi),\]
and there exists $C$ independent on $\s$ such that  the remainder term  $K^\s_2(\xi)$ satisfies :
\[\forall\; f\in H^1(\OO),\;\;\;
\left\{\begin{array}{l}
\dsp \| K^s_2(\xi)(f)\|_{H^1(\OO)}\leq C\|f\|_{H^1(\OO)},
\\
\\
\dsp \|K^s_2(\xi)(f)\|_{L^2(\OO)}\leq \frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^1(\OO)}.
\end{array}\right.
\]
\end{theorem}


\begin{remark} We denote  $\tilde{R}^\infty(\xi)=(\H^\infty-\xi  I)^{-1}$  the resolvent of
$\H^\infty$, defined on $L^2(\U)$ with values in $H^2(\U)\cap
H^1_0(\U)$. Then the first term of the asymptotic expansion of $R^\s(\xi)$
is  $R^\infty(\xi)=e\tilde{R}^\infty(\xi)r$, where $e:L^2(\U)\ds
L^2(\OO)$ is the extension by zero in $\O$ and  $r:L^2(\OO)\ds L^2(\U)$ is the restriction operator.
\end{remark}


Theorem \ref{thm0} prove the existence of a first  order asymptotic
expansion of the resolvent and we deduce the following theorem
concerning eigenvalues~:

\begin{theorem}
\label{thm1}
Let $\l^{\infty} $ be an eigenvalue of $\H^\infty$ with multiplicity
$m$. We fix $\eta>0$ such that $B(\li,\eta)\cap \mbox{spec
}\H^{\infty}=\{\li\}$, where $B(\li, \eta)=\left\{\xi\in \C, \frac{}{}
\; |\li-\xi|\leq \eta\right\}$.

Then for sufficiently large $\s$, $\H^\s$ has exactly $m$ eigenvalues
counted according to their multiplicities 
in $B(\li,\eta)$, and these eigenvalues $\l^\sigma_i$ admit an
asymptotic expansion of the type~:
\begin{equation}
\label{wiz}
\l^\s_i=\li+\frac{1}{\sqrt{\s}}\mu_i+o(\frac{1}{\rs}).
\end{equation}
Furthermore if we denote $(f_1,\ldots,f_m)$ an $L^2(\U)$
orthonormal basis of the eigenspace ker($\H^\infty-\li I)$, then 
$(\mu_i)_{1\leq i\leq m}$ are the eigenvalues of the matrix $A$ with coefficients~:
\[a_{ij}=\int_{\partial \O}\ddn{f_i}\ddn{f_j}.\]
\end{theorem}


\begin{remark}
The computation of the eigenvalues and eigenvectors of the limit
operator $\H^\infty$ give easily a first order approximation of the eigenvalues of 
$\H^\s$, since $A$ is deduced from the eigenvectors of $\H^\infty$.
\end{remark}

\begin{remark}
In the proof of Theorem \ref{thm1}, we show in fact that the remainder term in
(\ref{wiz}) is $\dsp {\cal
O}\left(\frac{1}{\s^{\frac{3}{4}}}\right)$. This estimate is not
optimal as we will see in Theorem \ref{thm2}.

\end{remark}



We can not build an asymptotic expansion of the resolvent at any order
in $\L(H^1;H^1)$ as it is done at order 1 in Theorem
\ref{thm0}. Nevertheless it is possible to prove that the eigenvalues
of $H^\s$ admit an asymptotic expansion at any order~:

\begin{theorem}
\label{thm2}
Let $\li$ be an eigenvalue of $\H^\infty$. With the notations of
Theorem \ref{thm1}, for any $i$, $1\leq
i\leq m$, $\l_i^\s$ admits an asymptotic expansion at any order, that
is there exists a sequence $(\mu_i^j)_{j\in \N}$ such that for all
$N$,
\[\lambda^\s_i=\li+\sum_{j=1}^N \frac{1}{\s^{\frac{j}{2}}} \mu_i^j + {\cal
O}\left(\frac{1}{\s^{\frac{N+1}{2}}}\right).\]
\end{theorem}

\vspace{2mm}
\begin{remark} Theorem \ref{thm0} remains valid in all dimension, but
in Theorem \ref{thm1}, we use a Sobolev embedding which prevents the
generalization of our proof at any dimension. Nevertheless the proof
of Theorem \ref{thm2} does not depend on the dimension, so the
eigenvalues of $\H^\s$ admit an asymptotic expansion at any order in
all dimension.
\end{remark}

\vspace{3mm}

%LAIUS BIBLIOGRAPHIQUE%%%%%%%%%%%%%%%%%%%%%%%


Without any assumption on the dimension and on the multiplicity of the
eigenvalues, it  seems to be the first asymptotic result (at all
order) in Large-coupling Limit.  Large coupling limit are essentially
discussed for operators with  continuous spectrum. For example, for
periodic problems let us quote works  of Hempel-Lineau-Herbst (see
\cite{HeLi00} and their references).  They use a Floquet decomposition
(or direct fiber-integral decomposition  \cite{ReSi4}) and a monotonic
convergence theorem in each fiber.  Concerning perturbations of the
Laplacian on $\R^n$ of the form $H_\lambda= -  \Delta + V + \lambda
{\bf 1}_{\Omega}$, with $V \in L^2(\R^n)$, Demuth  \cite{De91} use a
Feynman-Kac formula to prove trace norm convergence  of the resolvent
$(H_\lambda-z)^{-r}$, $r \geq 1+ n/2$, to the  resolvent of associated
Dirichlet problem. More general perturbations of  the type $H_\lambda=
- \Delta + V + \lambda W$, with $V \in L^\infty(\R^n)$  and $ 0 \leq W
\in  L^\infty(\R^n)$ are studied by Gesztesy and al.
\cite{Ge&al88}. Exploiting monotonic convergence theorems they prove
that  the discrete and essential spectrums of $H_\lambda$  tend to the
ones of the  associated Dirichlet operator. Furthermore, in
\cite{Ge&al88}, using WKB machinery,  1-order asymptotic expansions
are given in 1-dimension and for multiplicity  $m\leq 2$. Further
order asymptotic expansion is proved by Ashbaugh-Harrel \cite{AsHa82}
in 1-dimension (on $\R^+$) for $H_\lambda= - \Delta + V + \lambda {\bf
1}_{\Omega}$, with $V$ continuous supported in $[0,1]$ and $W= (x-1)^p
{\bf 1}_{[1, + \infty[}$. By an analytic implicit function theorem they
prove that the eigenvalues of $H_\lambda$ are analytic with respect to
$\lambda^{-\frac{1}{p+2}}$. 




\vspace{5mm}


In the following, we fix an eigenvalue $\li$ of the operator
$\H^\infty$. The resolvent of $\H^\infty$ is compact, so there exists
$\eta>0$ such that spec $\H^\infty\cap B(\li,\eta)=\{\li\}$. 


We will use the following notations~:

\begin{itemize}

\item $E^\s$ is the sum of the eigenspaces associated with the
eigenvalues of $\H^\s$ contained in $B(\li,\eta)$. 

\item $P^\s$ is the spectral projection onto $E^\s$. It is given by~:
\begin{equation}
\label{fandepute}
P^\s=\frac{-1}{2i\pi}\int_{{\cal C}(\li,\eta)}R^\s(\xi)d\xi ,
\end{equation}
where $\CC(\li,\eta)$ is the circle of center $\li$ and of radius
$\eta$.

\item $\tilde{E}^\infty$ is the eigenspace of $\H^\infty$ associated to 
$\li$. We denote $\tilde{P}^\infty$ the spectral projection onto
$\tilde{E}^{\infty}$.

\item $E^{\infty}=e(\tilde{E}^{\infty})$ and
$P^\infty=e\tilde{P}^\infty r$ which satisfies~:
\begin{equation}
\label{defpi}
P^\infty=\frac{-1}{2i\pi}\int_{\CC(\li,
\eta)}R^{\infty}(\xi)d\xi.
\end{equation}


\end{itemize}


In order to perform the asymptotic expansion of the resolvent  we use boundary layers machinery inside $\O$, that is we seek $R^\s(\xi)(f)$ on the form~:
\[R^\s(\xi)(f)(x)=V^0(x,\sqrt{\s}\dis(x))+\frac{1}{\rs}V^1(x,\rs\dis(x))+\ldots,\]
where $\dis(x)$ denotes the distance from $x$ to $\partial \O$. The
term $\psi(x)\alpha_1(x)e^{-\rs\dis(x)}$ represents the formation of
the thin layer near the boundary of $\O$.

\begin{remark} Boundary layers should appear in the case of viscous
perturbation of hyperbolic  and parabolic  systems
(see \cite{GG}, \cite{G} and  \cite{CFG}).
\end{remark}

In the proof of Theorem \ref{thm1}, following Kato \cite{Kato}, we introduce the invertible operator
${\cal A}^\s=1-P^\infty+P^\s P^\infty$. This operator maps $E^\infty$ onto
$E^\s$. The first order asymptotic expansion of the resolvent gives a
first order asymptotic expansion of ${\cal A}^\s$ in the space $\L(H^1;H^1)$.

\vspace{3mm}

We denote  $Q^\s=P^\infty [{\cal A}^{\s}]^{-1}\H^\s {\cal A}^\s P^\infty$. This
operator belongs to $\L(E^\infty)$ and has the same eigenvalues than
$\H^\s P^\s$. Using asymptotic expansions of the resolvent we obtain a
first order asymptotic expansion of $Q^\s$. Applying classical finite 
dimensional results due to Kato (see \cite{Kato}) we deduce Theorem \ref{thm1}.

\vspace{3mm}
In order to prove Theorem \ref{thm2}, we introduce  the unitary
operator ${\cal B}^\s$ defined by~:
\[{\cal B}^\s=(1- W^\s)^{-\frac12} \Big( (1 - P^\s) (1-
P^\infty) + P^\s 
P^\infty \Big),\]
with $W^\s=(P^\s- P^\infty )^2$, and we remark that
$\tilde{Q}^\s=P^\infty [{\cal B}^{\s}]^{-1}\H^\s {\cal B}^\s P^\infty$
has the same eigenvalues than $\H^\s$. On the other hand, with
algebraic arguments, we see that it suffices to perform the asymptotic
expansion of the resolvent for $f\in E^\infty$ that is $f=0 $ on $\O$
and $f\in \CC^\infty(\U)\cap H^1_0(\U)$. This is possible at any order
in an algebra and we can compose the asymptotic expansions to obtain
that $\tilde{Q}^\s$ admit an asymptotic expansion at any order. Since
$\tilde{Q}^\s$ is self-adjoint, the eigenvalues of $\tilde{Q}^\s$
admit then a complete asymptotic expansion.







\vspace{5mm}
The article is organized as follows. In Section 2, we prove technical
estimates and we introduce the boundary layers tools. The asymptotic
expansions of the resolvent are given in Section 3. We conclude the
proof of Theorem \ref{thm1} in Section 4. The last section is devoted to the
proof of Theorem \ref{thm2}.



\section{Preliminaries}

\subsection{Boundary layers tools}

We denote $\dis:\O\ds\R^+$ the distance from $x$ to $\partial \O$. The 
open set $\O$ being smooth, there exists $\O_1\subset \O$ a
neighborhood of the boundary $\partial \O$ such that $\dis$ is smooth
on $\O_1$. Then we have $|\nabla \dis|=1$ on $\O_1$ and $\nabla
\dis=-\nu$ on $\partial \O$, where $\nu$ is the outward unitary normal
on $\partial \O$. 

We have the following proposition~:
\begin{proposition}
\label{proptrans}
Let $\gamma\in H^\frac{1}{2}(\partial \O)$ and $f\in H^{\frac{1}{2}}(\O_1)$. Then there exists an unique
$\alpha\in H^\frac{1}{2}(\O_1) $ such that 
\begin{equation}
\label{trans}
\left\{
\begin{array}{l}
2(\nabla \dis, \nabla \alpha)+\lap \dis \alpha=f\mbox{ in }\O_1\\
\\
\alpha=\gamma\mbox{  on }\partial \O.
\end{array}\right.
\end{equation}
Furthermore there exists a constant $C$ such that~:
\[\|\alpha\|_{H^{\frac{1}{2}}( \O_1)}\leq C\left(\|\gamma\|_{H^{\frac{1}{2}}(\partial \O)}+\|f\|_{H^{\frac{1}{2}}(\O_1)}\right).\]
\end{proposition}

{\bf Proof.} Even if it means reducing $\O_1$, we give a 
parameterization of $\O_1$ of the form~:
\[\begin{array}{rcl}
\Lambda :& \partial \O\times [0,\delta]&\ds  \O_1\\
&(x,s)&\longmapsto  x-s \nu(x)
\end{array}\]

We denote $\tilde{\alpha}=\alpha\circ \Lambda$. Equation (\ref{trans}) 
is equivalent to~:
\begin{equation}
\label{transequ}
\left\{\begin{array}{l}
\dsp 2\dd{\tilde{\alpha}}{s}+\lap
\varphi(\Lambda(x,s))\tilde{\alpha}=(f\circ\Lambda)(x,s),\\
\\
\tilde{\alpha}(x,0)=\gamma(x).
\end{array}\right.
\end{equation}

This is a linear differential equation with regular
coefficients. Cauchy-Lipschitz Theorem shows that there exists a
unique $\tilde{\alpha}\in
\CC^{\infty}(0,\delta;H^{\frac{1}{2}}(\partial \O))$
satisfying (\ref{transequ}) and there exists a constant $C$
independent on $\gamma$ such that~:
\[\|\tilde{\alpha}\|_{\CC^1(0,\delta;H^{\frac{1}{2}})}\leq
C\left(\|\gamma\|_{H^{\frac{1}{2}}(\partial \O)}+\|f\|_{H^{\frac{1}{2}}(\O_1)}\right).\]
Since $\alpha=\tilde{\alpha}\circ
\Lambda^{-1}$, we obtain that $\alpha\in H^{\frac{1}{2}}(
\O_1)$ and that there exists a constant $C$ such that
\[\|\alpha\|_{H^{\frac{1}{2}}(\O_1)}\leq C\left(\|\gamma\|_{H^{\frac{1}{2}}(\partial \O)}+\|f\|_{H^{\frac{1}{2}}(\O_1)}\right).\]



\subsection{Estimates}

First we recall in Lemma \ref{lemme1} and Lemma \ref{lemme2} two
classical {\it a priori} estimates.

\vspace{3mm}

\begin{lemma}
\label{lemme1}
There exists a constant $C$ such that for any $\xi \in \CC(\li,\eta)$,
for any $u\in H^1(\U)$ such that $(-\lap-\xi )u\in L^2(\U)$, we have~: 
\[\|u\|_{L^2(\U)}\leq C
\left(\|u\|_{L^2(\partial \U)}+\|(-\lap-\xi )u\|_{L^2(\U)}\right).\]
\end{lemma}

{\bf Proof.} Let $f\in L^2(\U)$. There exists $w\in H^2(\U)\cap
H^1_0(\U)$ such that $-\lap w-\xi  w=f$ in $\U$. Moreover,
$\|w\|_{H^2(\U)}\leq C\|f\|_{L^2(\U)}$, where $C$ does not depend on
$\xi \in\CC(\li,\eta)$ and $f$.

Then we have~:
\[\begin{array}{rl}
\dsp\int_U uf & =\dsp \int_{U}u(-\lap w-\xi  w)\\
\\
& = -\dsp\int_\U\lap u w -\xi  \int_{\U}uw +\int_{\partial \U}u\ddn{w}.
\end{array}\]


Hence,
\[\begin{array}{rl}
\dsp\left|\int_\U uf\right| & \leq \dsp\|-\lap u
-\xi u\|_{L^2(\U)}\|w\|_{L^2(\U)} +\|u\|_{L^2(\partial
\U)}\|\ddn{w}\|_{L^2(\U)}\\
\\
& \leq \dsp C\left( \|-\lap u-\xi u\|_{L^2(\U)}+\|u\|_{L^2(\partial
\U)}\right)\|f\|_{L^2(\U)}.
\end{array}\]
These estimates are true for all $f\in L^2(\U)$, so by a duality
argument, we obtain~:
\[\|u\|_{L^2(\U)}\leq C \left( \|-\lap u -\xi u\|_{L^2(\U)}+\|u\|_{L^2(\partial
\U)}\right).\]

\vspace{3mm}
\begin{remark}
More general estimates of this type are proved in \cite{LM}.
\end{remark}


\begin{lemma}
\label{lemme2}
There exists a constant $C$ such that for all $v\in H^1(\O)$, 
\[\|v\|^2_{L^2(\partial \O)}\leq
C\left(\|v\|^2_{L^2(\O)}+\|v\|_{L^2(\O)}\|\nabla
v\|_{L^2(\O)}\right).\]
\end{lemma}

{\bf Sketch of the Proof. } We prove the estimate for regular maps and we conclude
the proof by a density argument (see \cite{A}).

\vspace{5mm}
In the asymptotic expansion of the resolvent, the following {\it a
priori} estimate will be used to estimates remainder terms.

\vspace{3mm}

\begin{proposition}
\label{estimation}
Let $g\in L^2(\O)$, $\gamma \in L^2(\partial \O)$ and $\xi \in \CC(\li,\eta)$. There exist
$\s_0>0$ and a constant $C$, independent on $g$, $\gamma$ and $\xi $, such
that for all $\s\geq \s_0$, if $a$ and $b$ are the solution of the
following system, 
\[
\left\{
\begin{array}{lc}
(-\lap -\xi )b+\s b =g \mbox{ on }\O\;\;\;\;& (i)\\
\\
(-\lap -\xi )a =0 \mbox{ on }\U\;\;\;\;& (ii)\\
\\
a=b \mbox { and } \dsp \ddn{b}=\ddn{a}+\gamma \mbox{ on }\partial \O\;\;\;\;& (iii)\\
\\
a=0\mbox{ on }\partial \OO\;\;\;\;& (iv)
\end{array}\right.
\]
then we have the following estimate~:
\begin{equation}
\label{estiprop}
\int_\O|\nabla b|^2+\int_{\U}|\nabla a|^2+\s\int_{\O}|b|^2\leq
\frac{C}{\s}\|g\|^2_{L^2(\O)}+\frac{C}{\sqrt{\s}}\|\gamma\|_{L^2(\partial 
\O)}^2.
\end{equation}
\end{proposition}

\vspace{2mm}

{\bf Proof.} We multiply the first equation by $\bar{b}$ and the
second by $\bar{a}$. We add up the two equalities and we obtain~:
\[\int_\O|\nabla b|^2+\int_{\U}|\nabla
a|^2+\s\int_{\O}|b|^2=\xi \int_{\O}|b|^2+\xi \int_{\U}|a|^2+\int_{\O}g\bar{b}+\int_{\partial
\O}\ddn{b}\bar b-\int_{\partial \O}\ddn{a}\bar{a}.\]
Using $(iii)$, we have~:
\[\int_\O|\nabla b|^2+\int_{\U}|\nabla
a|^2+\s\int_{\O}|b|^2\leq| \xi |\int_{\O}|b|^2+|\xi |\int_{\U}|a|^2+\|g\|_{L^2(\O)}\|b\|_{L^2(\O)}+\int_{\partial
\O}|\gamma||b|.\]

Now, we have the following estimates~:

\begin{itemize}
\item For $\s$ sufficiently large, $\dsp |\xi |\leq \frac{\s}{8}$.

\item $\|g\|_{L^2(\O)}\|b\|_{L^2(\O)}\leq
\dsp\frac{\s}{8}\|b\|^2_{L^2(\O)}+\frac{2}{\s}\|g\|^2_{L^2(\O)}$.

\item According to Lemma \ref{lemme1}, 
\[\int_{\U}|a^2|\leq C\int_{\partial \O}|a^2|,\]
and since $a=b$ on $\partial \O$, using Lemma \ref{lemme2}, since
$|\xi |\leq \li+\eta$, 
\[
\dsp |\xi |\int_{\U}|a|^2  \dsp \leq K\|b\|^2_{L^2(\O)}+K\|b\|_{L^2(\O)}\|\nabla
b\|_{L^2(\O)},\]
and for $\s$ sufficiently large,
\[
|\xi |\int_{\U}|a|^2
 \leq \dsp 
\frac{\sigma}{8}\|b\|^2_{L^2(\O)}+\frac{1}{2}\|\nabla b\|_{L^2(\O)}^2.
\]

\item Finally, for $\s\geq \s_0$, with $\s_0$ sufficiently large,

\[\begin{array}{rl}
\dsp \left|\int_{\partial \O}\gamma \bar{b}\right| & \leq
\|\gamma\|_{L^2(\partial \O)}\|b\|_{L^2(\partial \O)}\\
\\
 & \dsp \leq C\|\gamma\|_{L^2(\partial
 \O)}\|b\|_{L^2(\O)}+C\|\gamma\|_{L^2(\partial
 \O)}\|b\|^{\frac{1}{2}}_{L^2(\O)}\|\nabla b
 \|^{\frac{1}{2}}_{L^2(\O)}\\
\\
& \leq\dsp  \frac{\s}{8}\|b\|^2_{L^2(\O)}+\frac{1}{4}\|\nabla
b\|^2_{L^2(\O)}+\frac{K}{\sqrt{\s}}\|\gamma \|^2_{L^2(\partial \O)}.
\end{array}\]

\end{itemize}

Using these estimates, we obtain the claimed result.


\section{Asymptotic expansions of the resolvent}


We fix $f\in L^2(\OO)$ and we denote $u^\s$ (resp $v^\s$) the
restriction of $R^\s(\xi)(f)$ on $\U$ (resp $\O$). The couple
$(u^\s,v^\s)$ satisfies the following system~:

\begin{equation}
\label{uetv}
\left\{
\begin{array}{lc}
-\lap u^\s-\xi  u^\s=f\;\;\;\;\;\mbox{ on }\U & (1)\\
\\
-\lap v^\s+(\s-\xi )v^\s=f\;\;\;\;\;\mbox{ on }\O\;\;\; & (2)\\
\\
u^\s=v^\s \;\;\;\;\;\mbox{ on }\partial \O & (3)\\
\\
\ddn{u^\s}=\ddn{v^\s}\;\;\;\;\;\mbox{ on } \partial \O & (4)\\
\\
u^\s=0\;\;\;\;\;\mbox{ on } \partial \OO & (5)
\end{array}
\right.
\end{equation}

We will seek asymptotic expansion of $u^\s$ and $v^\s$ of the form~:
\[\begin{array}{l}
\dsp u^\s(x)=U^0(x)+\frac{1}{\rs}U^1(x)+\frac{1}{\s}U^2(x)+\ldots\\
\\
\dsp v^\s(x)=V^0(x,\rs\dis(x))+\frac{1}{\rs} V^1(x,\rs
\dis(x))+\frac{1}{\s} V^2(x,\rs \dis(x)) +\ldots
\end{array}\]
where the profiles $V^i(x,z)$ can be decomposed on the following way~:
\[V^i(x,z)=\widetilde{V}^i(x,z)+\overline{V}^i(x),\]
where $\tilde{V}^i$  and their derivatives tend to zero when $z\ds +\infty$.


We formally replace $u^\s$ and $v^\s$ in (\ref{uetv}) by their profile 
and we say that each order of the asymptotic expansion is zero.

 
\subsection{Order 0}

\begin{proposition}
\label{ordre0}
Let $\li$ and $\eta$ as in Theorem \ref{thm1}. Then for all
$\xi\in\CC(\li,\eta)$, $R^\s(\xi)$ satisfies the asymptotic
expansion~:
\[R^\s(\xi)=R^\infty(\xi)+K_1^\s(\xi),\]
where $K_1^\s(\xi)$ is an operator defined on $L^2(\OO)$. Furthermore,
there exists a constant $C$ such that~:
\[\forall\; \xi \in\CC(\li, \eta), \;\forall\; \s>\s_0,\;\forall\;
f\in L^2(\OO), \;\;
\left\{\begin{array}{l}
\dsp \|K^\s_1(\xi)(f)\|_{H^1(\OO)}\leq
\frac{C}{\s^{\frac{1}{4}}}\|f\|_{L^2(\OO)}
\\
\\
\dsp \|K^\s_1(\xi)(f)\|_{L^2(\OO)}\leq
\frac{C}{\rs}\|f\|_{L^2(\OO)}.
\end{array}\right.
\]

\end{proposition}


\vspace{2mm}
{\bf Proof. } Let $f\in L^2(\OO)$.  We denote $u^\s$ (resp. $v^\s$) the
restriction of $R^\s(\xi)(f)$ on $\U$ (resp. $\O$). We know that $u^\s$ 
and $v^\s$ satisfy (\ref{uetv}). We write 
\[\left\{\begin{array}{l}
u^\s(x)=U^0(x)+a^\s(x),\\
\\
v^\s(x)=b^\s(x),
\end{array}\right.\]
where $U_0$ satisfies~:

\[\left\{\begin{array}{l}
U^0=0 \mbox{ on } \partial \O,\\
\\
U^0=0 \mbox{ on } \partial \OO,\\
\\
(-\lap -\xi)U^0=f \mbox{ on }\U,
\end{array}\right.\] 
that is 
\[U^0=(\widetilde{R}^\infty(\xi)\circ r)(f).\]

The remainder terms satisfy the following system~:
\[\left\{\begin{array}{l}
(-\lap -\xi)a^\s=0\mbox{ in }\U, \\
\\
(-\lap -\xi)b^\s+\s b^\s=f\mbox{ on } \O,\\
\\
a^\s=b^\s\mbox{ on }\partial \O,\\
\\
\dsp \ddn{U^0}+\ddn{a^\s}=\ddn{b^\s}\mbox{ on }\partial \O.
\end{array}
\right.\]

We remark that there exist a constant $C$ independent on $\xi$ and $f$ 
such that~:
\[\|U^0\|_{H^2(\U)}\leq C \|f\|_{L^2(\U)}.\]
We apply proposition \ref{estimation} and we obtain that there exists
a constant $C$ which does not depend on $\xi$ and $f$ such that~:
\[\|\nabla a^\s \|_{L^2(\U)}^2+\|\nabla
b^\s \|_{L^2(\O)}^2+\s\|b^\s \|^2_{L^2(\O)}\leq
C\frac{1}{\s}\|f\|^2_{L^2(\O)}+C\frac{1}{\rs}\|\ddn{U^0}\|_{L^2(\partial\O)}^2 
,\]
thus
\begin{equation}
\label{merde}
\|\nabla a^\s \|_{L^2(\U)}^2+\|\nabla
b^\s \|_{L^2(\O)}^2+\s\|b^\s \|^2_{L^2(\O)}\leq
C\frac{1}{\rs}\|f\|^2_{L^2(\OO)}.
\end{equation}



Now we have~:
\[ K^\s_1(\xi)(f)(x)=\left\{
\begin{array}{l}
a^\s(x)\mbox{ if }x\in \U,\\
b^\s(x)\mbox{ if }x\in\O,
\end{array}\right.\]
and Estimate (\ref{merde}) give that~:
\[\|\nabla K^\s_1(\xi)(f)\|_{L^2(\OO)}\leq
\frac{C}{\s^{\frac{1}{4}}}\|f\|_{L^2(\OO)}.\]
With estimate (\ref{merde}) we have that $\|b^\s\|_{L^2(\O)}\leq
\dsp\frac{C}{\s^{\frac{3}{4}}}\|f\|_{L^2(\OO)}$ and using Lemma
\ref{lemme1} and Lemma
\ref{lemme2}, we can estimate $\|a^\s\|_{L^2(\U)}$ and we obtain~:
\[\begin{array}{rl}
\|a^\s\|_{L^2(\U)}& \leq C \|a^\s\|_{L^2(\partial \O)}\\ \\
 & \leq C\|b^\s\|_{L^2(\partial \O)}\\ \\
 &\leq C \dsp \left( \|b^\s\|_{L^2(
 \O)}+\|b^\s\|_{L^2( \O)}^{\frac{1}{2}}\|\nabla
 b^\s\|_{L^2(]] \O)}^{\frac{1}{2}}\right)\\ \\
 & \leq \dsp \frac{C}{\rs}\|f\|_{L^2(\OO)}.
\end{array}\]

Hence we have~:
\[\| K^\s_1(\xi)(f)\|_{L^2(\OO)}\leq  \dsp \frac{C}{\rs}\|f\|_{L^2(\OO)}.\]


This completes the proof of Proposition \ref{ordre0}.

 
\subsection{First order asymptotic expansion}



{\bf First step. Formal asymptotic expansion}

With the usual notations we seek formal first order asymptotic expansion on the form~:
 
\[\left\{\begin{array}{l}
\dsp u^\s(x)=U^0(x)+\frac{1}{\rs}U^1(x)+\ldots\\
\\
\dsp v^\s(x)=V^0(x,\rs
\dis(x))+\frac{1}{\rs}V^1(x,\rs\dis(x))+\ldots
\end{array}\right.\]

We recall that $u^\s$ and $v^\s $ satisfy (\ref{uetv}). We will replace
$u^\s$ and $v^\s$ by their asymptotic expansion in (\ref{uetv}) and
we will identify the different powers of $\s$.

\vspace{2mm}
We will use the following notations~:
\begin{itemize}
\item $x=(x_1,\ldots, x_n)$ are the coordinates in $\R^n$,

\item $V_z=\dsp \dd{V}{z}$ and $\dsp V_{zz}=\dd{^2V}{z^2}$,

\item $\dsp \nabla V=\dsp \left(\begin{array}{c}
\dsp \dd{V}{x_1}\\
\cdot\\
\cdot\\
\dsp \dd{V}{x_n}\end{array}\right)$ and  $\dsp \nabla V_z=\dsp
\left(\begin{array}{c} 
\dsp \dd{^2V}{x_1\partial z}\\
\cdot\\
\cdot\\
\dsp \dd{^2V}{x_n\partial z}\end{array}\right)$,

\item $\lap V=\dsp \dd{^2V}{x_1^2}+\ldots +\dd{^2V}{x_n^2}$.
\end{itemize}

We remark that if $w(x)=V(x,\rs \dis(x))$, 
\[\nabla w(x)=\nabla V(x, \rs\dis(x))+\rs \nabla \dis(x) V_z(x, \rs
\dis(x)),\]
and
\[\begin{array}{rl}
\lap w(x)=& \s |\nabla \varphi|^2 V_{zz}(x, \rs\dis(x))\\
\\
&\dsp  + \rs\left(2(\nabla \varphi(x), \nabla V_z(x, \rs\dis(x)))+\frac{}{}\lap
\dis(x) V_z(x, \rs \dis(x))\right)\\
\\
& +  \lap V(x, \rs\dis(x)).\end{array}\]

Using that $|\nabla \varphi|=1$ we formally obtain the following
equations~:

\begin{equation}
\label{profilordre1}
\begin{array}{ll}
\mbox{from (\ref{uetv}.1)}, & 
\left\{\begin{array}{ccl}
\mbox{(i)} & \;\;\;\;\;\; & (-\lap -\xi)U^0=f\mbox { in }\U ,\\ \\
\mbox{(ii)} & & (-\lap -\xi)U^1=0 \mbox { in }\U ,
\end{array}\right.
\\ \\
\mbox{from (\ref{uetv}.2)}, & 
\left\{\begin{array}{ccl}
\mbox{(iii)} & \;\;\;\;\;\; & V^0-V^0_{zz}=0 \mbox { in } \O,\\ \\
\mbox{(iv)} &  \;\;\;\;\;\;& \dsp V^1-V^1_{zz}-\left[(2\nabla \dis, \nabla V^0_z)+\lap\dis 
V^0_z\right]=0\mbox { in }\O,
\end{array}\right.
\\ \\
\mbox{from (\ref{uetv}.3)}, & 
\left\{\begin{array}{ccl}
\mbox{(v)} &  \;\;\;\;\;\;& U^0=V^0\mbox { in }\partial \O,\\ \\
\mbox{(vi)} & & U^1=V^1 \mbox { in }\partial \O,
\end{array}\right.
\\ \\
\mbox{from (\ref{uetv}.4)}, & 
\left\{\begin{array}{ccl}
\mbox{(vii)}& \;\;\;\;\; & V^0_z=0 \mbox { in }\partial \O,\\ \\
\mbox{(viii)} & & \dsp \ddn{U^0}=-V^1_z\mbox { in }\partial \O,\\
\end{array}\right.
\\ \\
\mbox{from (\ref{uetv}.5)}, &
\left\{\begin{array}{ccl}
\mbox{(ix)}& \;\;\;\;\;\; & U^0=0\mbox{ in }\partial \OO,\\ \\
\mbox{(x)} & & U^1=0\mbox{ in } \partial \OO.
\end{array}\right.
\end{array}
\end{equation}


Using the equations (\ref{profilordre1}), we obtain~:
\begin{itemize}
\item from $\mbox{(iii)}$, we have $V^0(x,z)=\alpha_0(x)e^{-z}$.
\item from $\mbox{(vii)}$, $\alpha_0(x)=0$ on $\partial \O$.
\item in $\mbox{(iv)}$ we arbitrary cut the equation in two terms and we say
that $V^1-V^1_{zz}=0$, i.e. $V^1(x,z)=\alpha_1(x)e^{-z}$, and we have $
(2\nabla \dis, \nabla \alpha_0)+\lap\dis 
\alpha_0=0$ in $\O$. 

Using Proposition \ref{proptrans} we obtain that~:
\[\alpha_0=0\mbox{ on } \O.\]

\item Using (i) and (ix), since (v) implies that $U^0=0$ on $\partial \O$,
we have~:
\begin{equation}
\label{lkj1}
\left\{
\begin{array}{l}
(-\lap -\xi )U^0=f \mbox{ on }\U,\\
\\
U^0=0 \mbox{ on }\partial \U,
\end{array}
\right.
\end{equation}
that is
\[U^0=(\widetilde{R}^\infty(\xi)\circ r)(f).\]

\item Using (viii) and (vi) we have~:
\[\alpha_1=U^1=\ddn{U^0}\mbox{ on }\partial \O,\]
thus $U^1$ is uniquely determined by~:
\begin{equation}
\label{lkj2}
\left\{
\begin{array}{l}
(-\lap -\xi)U^1=0\mbox{ on }\U,\\ \\
U^1=0\mbox{ on }\partial \OO,\\ \\
U^1=\dsp\ddn{U^0}\mbox{ on }\partial \O.
\end{array}
\right.
\end{equation}

\item Finally, in order to define $\alpha_1$, we will use a classical trace relevement  from $\partial \O$ to $\O$ denoted by 
${\cal E}_\O$ defined on $H^{\frac{3}{2}}(\partial \O)$ with values on $H^2(\O)$, and we
set
\[\alpha_1= {\cal E}_\O( \ddn{U^0}).\]


\end{itemize}


In short, if we fix $f\in H^1(\OO)$, we define $R^\infty (\xi)(f)=e(U^0)$
by (\ref{lkj1}) and we define $R^\s_1(\xi)(f)$ by~:
\begin{equation}
\label{lkj11}
R^\s_1(\xi)(f)(x)=\left\{
\begin{array}{l}
 \psi(x)\alpha_1(x)e^{-\rs\dis(x)} \mbox{ if }x\in\O,\\
\\
U^1(x)\mbox{ if }x\in \U,
\end{array}
\right.
\end{equation}
where $U_1$ satisfies (\ref{lkj2}), $\alpha_1$ is defined by
\begin{equation}
\label{lkj3}
\alpha_1={\cal E}_\O(U^1)\mbox{ on } \O,
\end{equation}
and where $\psi$ is a cut-off function with support in $\overline{\O_1}$ and equal
to 1 in a neighborhood of $\partial \O$. The cut-off function
$\psi$ is used to avoid the problems of non regularity of $\dis$ far
from $\partial \O$. 

\vspace{5mm}



The main properties of the operator $R^\s_1(\xi)$ are listed in the
following lemma.


\begin{lemma}
\label{flop}
The operator $R^\s_1(\xi)$ satisfies~:
\begin{enumerate}
\item $R^\s_1(\xi)(f)$ only depends on $rf$, the restriction of $f$ to
$\U$.

\item $\dsp rR^\s_1(\xi)$ does not depend on $\s$, and there exists a
constant $C$ such that for all $\s$,
\begin{equation}
\label{blofp}
\pt f \in H^1(\OO),\;\;\;\left\|(r
R^\s_1(\xi))(f)\right\|_{H^2(\U)}\leq C\|rf\|_{H^1(\OO)}.
\end{equation}


\item There exists a constant $C$ such that~:
\[\pt f\in H^1(\OO),\;\;\;\dsp\|R^\s_1(\xi)(f)\|_{L^2(\O)}\leq
\frac{C}{\s^{\frac{1}{4}}}\|rf\|_{H^1(\U)},\]
\[\pt f\in H^1(\OO),\;\;\;\dsp\|R^\s_1(\xi)(f)\|_{H^1(\O)}\leq
C\s^{\frac{1}{4}}\|rf\|_{H^1(\U)},\]

\item $P^{\infty}R^\s_1(\xi)P^{\infty}=\dsp \frac{1}{(\li-\xi)^2}A P^\infty$,
where $A$ is a linear map in $E^\infty$. The coefficients of the
matrix  $A$ in an
orthonormal basis $(f_1,\ldots,f_m)$ of $E^{\infty}$ are given by the
formula~:
\[a_{ij}=\int_{\partial \O}\ddn{f_i}\ddn{f_j}.\]
\end{enumerate}
\end{lemma}

{\bf Proof.} Let $f\in H^1(\OO)$. We define $U^0\in H^3(\U)$ by
(\ref{lkj1}). So $U_0$ only depends on $rf$. Hence $\dsp
\ddn{U^0}|_{\partial \O}\in H^{\frac{3}{2}}(\partial \O)$ only depends
on $rf$ and by classical results on elliptic equations we obtain
inequality (\ref{blofp}).

In addition, by property of ${\cal E}_\O$, 
\[\begin{array}{rl}
\|\alpha_1\|_{H^2(\O)}\leq &
C\dsp\|\ddn{U^0}\|_{H^{\frac{3}{2}}(\partial \O)}\\ \\
\leq & C\dsp \|rf\|_{H^1(\U)},
\end{array}
\]
and we have~:
\[\begin{array}{rl}
\left\|R^\s_1(\xi)\frac{}{}(f)\right\|_{L^2(\O)}\leq &
\|\psi\|_{L^{\infty}}\|\alpha_1\|_{L^{\infty}}\|\dsp
e^{-\rs\varphi}\|_{L^2}\\
\\
 \leq & \dsp \frac{C}{\s^{\frac{1}{4}}}\|rf\|_{H^1(\U)}.
\end{array}
\]

Furthermore
\[\begin{array}{rl}
\left\|\nabla (R^\s_1(\xi)\frac{}{}(f))\right\|_{L^2(\O)}\leq 
& \|\psi\|_{L^{\infty}}\|\alpha_1\|_{L^{\infty}}\|\nabla (\dsp
e^{-\rs\varphi})\|_{L^2}+\|\nabla \psi\|_{L^{\infty}}\|\alpha_1\|_{L^{\infty}}\|\dsp
e^{-\rs\varphi}\|_{L^2}\\
& +\|\psi\|_{L^{\infty}}\|\nabla \alpha_1\|_{L^2}\| (\dsp
e^{-\rs\varphi}\|_{L^\infty}\\
\\
\leq & C\s^{\frac{1}{4}}\|rf\|_{H^1(\U)}
\end{array}\]

\begin{remark}
We use here the Sobolev embedding $H^2(\O)\subset L^\infty(\O)$ which
is valid in dimension two. This is only here that there is a
restriction on the dimension.
\end{remark}

\vspace{3mm}

Let us prove now the last assertion.

Let $(f_1, \ldots, f_m)$  be an orthonormal basis of $E^{\infty}$. Let us
compute $P^{\infty}R^\s_1(\xi)P^{\infty}(f_j)$.

We note $U^0\in H^1_0(\U)$ the solution of~:
\[(-\lap-\xi)U^0=f_j\mbox{ in }\U.\]
We remark that $\dsp U^0=\frac{1}{\li-\xi}f_j$.

\vspace{2mm}

We consider $U^1\in H^1(\U)$ the solution of~:
\[\left\{\begin{array}{l}
\dsp (-\lap -\xi)U^1=0\mbox{ in } \U,\\
\\
\dsp U^1=\ddn{U^0}\mbox{ on }\partial \O\\
\\
U^1=0\mbox{ on }\partial \OO.
\end{array}\right.
\]
We have~:
\[P^{\infty}R^\s_1(\xi)P^{\infty}(f_j)=\sum_{i=1}^m
\left(\int_{\U}U^1\cdot f_i\right)f_i.\]

Now we have~:
\[
\begin{array}{rl}
\dsp \int_{\U}U^1 f_i=&\dsp -\frac{1}{\li}\int_{\U}U^1\lap f_i\\
\\
=& \dsp -\frac{1}{\li}\int_{\U}\lap U^1f_i+\frac{1}{\li}\int_{\partial \O}U^1\ddn{f_i}\\
\\
=& \dsp \frac{\xi}{\li}\int_{\U}U^1
f_i+\frac{1}{\li}\int_{\partial\O}\ddn{U^0}\ddn{f_i}\end{array}\]

Hence we obtain that~:
\[\int_{\U}U^1 f_i=\frac{1}{(\li-\xi)^2}\int_{\partial
\O}\ddn{f_j}\cdot \ddn{f_i},\]
thus
\[P^{\infty}R^\s_1(\xi)P^{\infty}(f_j)=\frac{1}{(\li-\xi)^2}\sum_{i=1}^m\left(\int_{\partial
\O}\ddn{f_j}\cdot \ddn{f_i}\right)f_i.\]





We are now able to prove the following proposition which contains the
results of Theorem \ref{thm0}~:
\begin{proposition}
\label{ordre1}
The resolvent $R^\s(\xi)$ admits an asymptotic expansion of the form~:
\begin{equation}
\label{lkj11}
R^\s(\xi)=R^\infty(\xi)+\frac{1}{\rs}R^\s_1(\xi)+\frac{1}{\rs}K^\s_2(\xi),
\end{equation}
where $R^\s_1(\xi)$ is defined by  (\ref{lkj1}), (\ref{lkj2}) and (\ref{lkj3}). 

Furthermore there exists $\s_0>0$ and there exists a constant $C$ independent on $\xi\in \CC(\li,
\eta)$, such that~:
\[\forall\; \s>\s_0, \;\;\;\forall\;\xi \in \CC(\li, \eta), \;\forall\; f\in H^1(\OO),\;\;\;
\left\{\begin{array}{l}
\dsp \| K^\s_2(\xi)(f)\|_{H^1(\OO)}\leq C\|f\|_{H^1(\OO)},
\\
\\
\dsp \|K^\s_2(\xi)(f)\|_{L^2(\OO)}\leq \frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^1(\OO)}.
\end{array}\right.
\]
\end{proposition}




\vspace{2mm}
{\bf Proof.} We fix $f\in H^1(\OO)$, we define $U^0$, $U^1$ and $\alpha_1$ like in (\ref{lkj1}), (\ref{lkj2}), (\ref{lkj3}), 
and  we set~:
\[\left\{ 
\begin{array}{l}
u^\s(x)=U^0(x)+\dsp\frac{1}{\rs}U^1(x)+\frac{1}{\rs}a^\s(x)\mbox{
if }x\in \U,\\
\\
v^\s(x)=\dsp \frac{1}{\rs}\psi(x)\alpha_1(x) e^{-\rs
\dis(x)}+\frac{1}{\rs}b^\s(x)\mbox{ if } x\in \O.
\end{array}\right.
\]

The remainder terms $a^\s$ and $b^\s$ satisfy the following system~:


\begin{equation}
\left\{
\begin{array}{ccl}
\mbox{(i)} & \;\;\;\;\; & (-\lap - \xi )a^\s=0 \mbox { in }\U ,\\ \\
\mbox{(ii)} & & a^\s=b^\s \mbox { in }\partial \O,  \\ \\
\mbox{(iii)} & & \dsp \ddn{a^\s}+\ddn{U^1}=\ddn{b^\s}+\ddn{\alpha_1} \mbox { in
}\partial \O,  \\ \\
\mbox{(iv)} & &\dsp  (-\lap -\xi )b^\s+\s b^\s=g^\s\mbox{ in }\O,\\ \\
\mbox{(v)} & & \dsp a^\s=0\mbox{ on } \partial \OO,
\end{array}\right.
\end{equation}

where
\[\begin{array}{rl}
g^\s= &\dsp  \rs f + \rs\psi 
\left(2(\nabla \dis, \nabla \alpha^1) \frac{}{}+\lap \dis \alpha_1\right)e^{-\rs\dis}\\ \\
& \dsp +\left(\psi\lap \alpha_1+\xi \psi\alpha_1+\lap \psi\alpha_1+ 2(\nabla \psi, \nabla
\alpha_1)\frac{}{}+2\alpha_1(\nabla \psi,\nabla \dis)\right)e^{-\rs\dis}.
\end{array}\]

We have the following estimates~:
\begin{itemize}
\item $\dsp \|\ddn{\alpha_1}\|_{L^2(\partial \O)}\leq
C\|\alpha_1\|_{H^2(\O)}
\leq K\|\alpha_1\|_{H^{\frac{3}{2}}(\partial \O)}\leq
K\|U_0\|_{H^3(\U)}$, and we obtain~:
\[\|\ddn{\alpha_1}\|_{L^2(\partial \O)}\leq K\|f\|_{H^1(\U)}.\]

\item $\dsp \|\ddn{U^1}\|_{L^2(\partial\O)}\leq C\|U^1\|_{H^2(\U)}\leq
C\|U^1\|_{H^{\frac{3}{2}}(\partial \O)}\leq
C\|\ddn{U^0}\|_{H^{\frac{3}{2}}(\partial \O)}$ hence
\[\|\ddn{U^1}\|_{L^2(\partial\O)}\leq K\|f\|_{H^1(\U)}.\]

\item $\dsp\|g^\s\|_{L^2(\O)}\leq
\rs\|f\|_{L^2(\O)}+K\rs\|2(\nabla \varphi, \nabla \alpha_1)+\lap
\varphi \alpha_1)\|_{L^2(\O)}\|e^{-\rs\varphi}\|_{L^{\infty}(\O)}
+K\|\alpha_1\|_{H^2(\O)}$ so
\[\|g^\s\|_{L^2(\O)}\leq
K\rs\|f\|_{L^2(\O)}+K\rs \|f\|_{H^1(\U)}.\]
\end{itemize}

We apply Proposition \ref{estimation} and we obtain that~:
\[\|\nabla a^\s\|_{L^2(\U)}^2+\|\nabla
b^\s\|_{L^2(\O)}^2+\s\|b^\s\|_{L^2(\O)}^2\leq
K\left(\|f\|_{L^2(\O)}^2+\frac{}{}\|f\|_{H^1(\U)}^2\right).\]

Furthermore, applying Lemma \ref{lemme1} and Lemma \ref{lemme2}, since
$a^\s=b^\s$ on $\partial \O$, we obtain that~:
\[\|a^\s\|_{L^2(\U)}\leq
\frac{K}{\s^{\frac{1}{4}}}\left(\|f\|_{L^2(\O)}^2+\|f\|_{H^1(\U)}^2\right).\]

So the proof of Proposition \ref{ordre1} is complete.

\section{Proof of Theorem \ref{thm1}}


\begin{proposition}
\label{vlouf}
The projector $P^\s$ admits an asymptotic expansion of the form~:
\[P^\s=P^\infty+\frac{1}{\rs}P^{1,\s}+\frac{1}{\rs}M^\s,\]
where $P^{1,\s}$ is an linear operator defined on
$H^1(\OO)$. Furthermore, 
there exist a constant $C$ and $\s_0$ such that for all $\s>\s_0$ and
for all $f\in H^1(\OO)$,
\[\left\{\begin{array}{l}
\dsp \|rP^{1,\s}(f)\|_{H^1(\U)}\leq C\|rf\|_{H^1(\U)}\\
\\
\dsp\|P^{1,\s}(f)\|_{H^1(\O)}\leq C\s^{\frac{1}{4}}\|rf\|_{H^1(\U)}\\
\\
\dsp \|P^{1,\s}(f)\|_{L^2(\O)}\leq \frac{C}{\s^{\frac{1}{4}}}\|rf\|_{H^1(\U)}
\end{array}\right.\]
and
\[
\left\{\begin{array}{l}
\|M^\s(f)\|_{H^1(\OO)}\leq C\|f\|_{H^1(\OO)}\\
\\
\|M^\s(f)\|_{L^2(\OO)}\leq\dsp  \frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^1(\U)}.
\end{array}\right.
\]
Furthermore, $rP^{1,\s}$ is independent on $\s$ and
$P^\infty P^{1,\s}P^{\infty}=0$.
\end{proposition}

{\bf Proof. } The operators $\H^\s$ and $\H^\infty$ are self-adjoint
thus the projections $P^\sigma$ and $P^\infty$ are given by the
relations (\ref{fandepute}) and (\ref{defpi}).

Therefore, according to Lemma \ref{flop} and Proposition \ref{ordre1},
if we set
\[P^{1,\s}=\frac{-1}{2i\pi}\int_{\CC(\li,\eta)}R^{1,\s}(\xi)d\xi\]
and 
\[M^\s=\frac{-1}{2i\pi}\int_{\CC(\li,\eta)}K^{2,\s}(\xi)d\xi,\]
the first assertion and the estimates are straightforward.

The second assertion is a consequence of Lemma \ref{flop} since
\[\int_{\CC(\li,\eta)}(\li-\xi)^{-2}d\xi=0.\]


\vspace{3mm}
{\bf Proof of Theorem \ref{thm1}.}

Using Kato's method (see \cite{Kato}), we introduce the operator
${\cal A}^\s$~:
\[{\cal A}^\s=I-P^\infty+P^{\sigma}P^{\infty}.\]
This operator maps $E^\infty$ into $ E^\sigma$ and leaves the orthogonal
of $E^\infty$ invariant.
















Using Proposition \ref{vlouf}, we obtain an asymptotic expansion of
${\cal A}^\s$~:
\begin{equation}
\label{dasus}
{\cal A}^\s=I+\frac{1}{\rs}P^{1,\s}P^\infty +\frac{1}{\rs}M^\s P^\infty,
\end{equation}

We remark that there exists a constant $C$ such that~:
\begin{equation}
\label{putain1}
\pt \s>\s_0, \;\;\pt f\in H^1(\OO),\;\;
\left\{
\begin{array}{l}
\dsp \left\|\frac{1}{\rs}P^{1,\s}P^{\infty}(f)\right\|_{H^1(\OO)}\leq
\frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^1(\OO)}\\
\\
\dsp \left\|\frac{1}{\rs}M^{\s}P^{\infty}(f)\right\|_{H^1(\OO)}\leq
\frac{C}{\rs}\|f\|_{H^1(\OO)}.
\end{array}\right.
\end{equation}
So ${\cal A}^\s$ is invertible in $\L(H^1(\O);H^1(\O))$ and
\[ [{\cal A}^\s]^{-1}=Id-\frac{1}{\rs}P^{1,\s}P^{\infty}-\frac{1}{\rs} M^\s
P^{\infty}+\sum_{n=2}^{+\infty}\frac{(-1)^n}{\s^{\frac{n}{2}}}\left[P^{1,\s}P^{\infty}+\frac{}{}M^{\s}P^{\infty}\right]^n.\]

Using assertion (\ref{putain1}) we remark that~:
\[\left\|\sum_{n=3}^{+\infty}\frac{(-1)^n}{\s^{\frac{n}{2}}}\left[P^{1,\s}P^{\infty}+\frac{}{}M^{\s}P^{\infty}\right]^n\right\|_{\L(H^1(\OO))}\leq
\frac{K}{\s^{\frac{3}{4}}}.\]
Furthermore, using that $P^\infty P^{1,\s} P^\infty=0$, we obtain that
\[\begin{array}{rl}
P^\infty [{\cal A}^\s]^{-1} = & P^\infty -\frac{1}{\rs}P^\infty M^\s P^\infty
+\frac{1}{\s}P^{\infty} M^\s P^\infty M^\s P^\infty \\ \\ 
 & + \dsp
 \sum_{n=3}^{+\infty}\frac{(-1)^n}{\s^{\frac{n}{2}}}\left[P^{1,\s}P^{\infty}+\frac{}{}M^{\s}P^{\infty}\right]^n.
\end{array}
\]
Now since $P^\infty$ is a regularizing operator,
\[\left\|P^\infty M^\s P^\infty\right\|_{\L(H^1;H^1)}\leq C\|M^\s
P^\infty\|_{\L(H^1;L^2)}\leq \frac{K}{\s^{\frac{1}{4}}}.\]

So  we
have~:
\begin{equation}
\label{dasusinv}
P^\infty ({\cal A}^\s)^{-1}=P^\infty+\frac{1}{\rs}N^\s,
\end{equation}
with 
\[\left\|N^\s\right\|_{\L(H^1)}\leq \frac{K}{\s^{\frac{1}{4}}}.\]

As a consequence, the eigenvalues of $\H^\s$ contained in
$B(\li,\eta)$ are the eigenvalues of the operator $Q^\s$ defined on
$E^\infty$ by~:
\[Q^\s=P^\infty ({\cal A}^\s)^{-1} \H^\s P^\s {\cal A}^\s P^\infty.\]

Since we have the relation~:
\begin{equation}
\label{hsps}
H^\s P^\s=-\frac{1}{2i\pi}\int_{\CC(\li, \eta)}\xi R^\s(\xi)d\xi,
\end{equation}
we are lead to study the asymptotic expansion of $P^\infty ({\cal A}^\s)^{-1}
R^\s P^\s {\cal A}^\s P^\infty$.

\vspace{2mm}
Using Proposition \ref{ordre1}, and equations (\ref{dasus}) and (\ref{dasusinv}) we
have~:
\[\begin{array}{rl}
P^\infty ({\cal A}^\s)^{-1}
R^\s P^\s {\cal A}^\s P^\infty=&  \dsp \left(P^\infty
+\frac{1}{\rs}N^\s\right)\left(R^\infty(\xi)+\frac{1}{\rs}R^\s_1(\xi)+\frac{1}{\rs}K^\s_2\right)\\
& \times \left(P^\infty+\frac{1}{\rs}P^{1,\s}P^\infty+\frac{1}{\rs}M^\s
P^\infty\right),
\end{array}
\]
{\it i.e.}
\[P^\infty ({\cal A}^\s)^{-1}
R^\s P^\s {\cal A}^\s P^\infty= T_1+\ldots T_6\]
with
\[\begin{array}{l}
\dsp T_1= P^\infty R^\infty(\xi) P^\infty   +   \frac{1}{\rs} P^\infty R^\infty(\xi) P^{1,\s}P^{\infty}   +   \frac{1}{\rs} P^\infty
R^\infty(\xi) M^\s P^\infty   ,  \\
\\
\dsp T_2=  \frac{1}{\rs}  P^\infty R^\s_1(\xi) P^\infty  + \frac{1}{\s}   P^\infty R^\s_1(\xi) P^{1,\s}P^{\infty}   +  \frac{1}{\s}  P^\infty
R^\s_1(\xi)   M^\s P^\infty   ,  \\
\\
\dsp T_3=  \frac{1}{\rs}  P^\infty K^{\s}_2(\xi) P^\infty     +    \frac{1}{\s}   P^\infty K^{\s}_2(\xi)  P^{1,\s}P^{\infty}    +    \frac{1}{\s}  P^\infty
K^{\s}_2 (\xi) M^\s P^\infty  ,    \\
\\
\dsp T_4=  \frac{1}{\rs}  N^\s  R^\infty(\xi) P^\infty   +  \frac{1}{\s}    N^\s  R^\infty(\xi)  P^{1,\s}P^{\infty}   +   \frac{1}{\s}     N^\s
R^\infty(\xi)   M^\s P^\infty  ,  \\
\\
\dsp T_5 = \frac{1}{\s}  N^\s R^\s_1(\xi) P^\infty  + \frac{1}{\s^{\frac{3}{2}}} N^\s R^\s_1(\xi)  P^{1,\s}P^{\infty}  + \frac{1}{\s^{\frac{3}{2}}}  N^\s R^\s_1(\xi)M^\s P^\infty   ,
\\
\\
\dsp T_6 =  \frac{1}{\s} N^\s K^{\s}_2(\xi)  P^\infty   + \frac{1}{\s^{\frac{3}{2}}} N^\s K^{\s}_2(\xi)  P^{1,\s}P^{\infty}    + \frac{1}{\s^{\frac{3}{2}}}N^\s K^{\s}_2 (\xi)  M^\s P^\infty  .   
\end{array}\]















Since $P^\infty$ is a regularizing operator, we remark that 
\[\|P^\infty K^\s_2(\xi)\|_{\L(H^1;H^1)}\leq
\|P^{\infty}\|_{\L(L^2;H^1)}\|K^\s_2(\xi)\|_{\L(H^1;L^2)}
\leq \frac{C}{\s^{\frac{1}{4}}}.\]

Using Lemma \ref{flop}, Proposition \ref{ordre1} and Proposition
\ref{vlouf}, we estimate the terms $T_1, \ldots , T_6$. We recall that
$P^\infty P^{1,\s} P^\infty=0$.

\begin{itemize}
\item $P^\infty R^\infty(\xi)=\dsp \frac{1}{\li-\xi}P^\infty$. Thus
$\dsp T_1=\frac{1}{\li-\xi}P^\infty+\frac{1}{\li-\xi}\tau_1^\s$
with $\dsp \tau_1^\s=\frac{1}{\rs}P^\infty M^\s P^\infty$, so
\[\|\tau_1^\s\|_{\L(H^1)}\leq \frac{K}{\s^{\frac{3}{4}}}.\]

\item $\dsp T_2=\frac{1}{\rs}\frac{1}{(\li-\xi)^2}A P^\infty +\tau^\s_2$ where
$\dsp \tau_2^\s=\frac{1}{\s}\left(P^\infty
R^\s_1(\xi)P^{1,\s}P^\infty +P^\infty R^\s_1(\xi)M^\s P^\infty\right),$
 and we have~:
\[\|\tau_2^\s\|_{\L(H^1)}\leq \frac{C}{\s}.\]

\item 
$\|T_3\|_{\L(H^1)}\leq \dsp 
\frac{1}{\rs}\|K^\s_2(\xi)\|_{\L(H^1;L^2)}+ \frac{1}{\s}\|P^{\infty}
K^\s_2(\xi)\|_{\L(H^1)}\left(\|P^{1,\s}P^\infty\|_{\L(H^1)}+\|M^\s
P^\infty\|_{\L(H^1)}\right)$. 
Since $P^\infty$ is a regularizing operator, and using Proposition
\ref{vlouf} we obtain that
\[\|T_3\|_{\L(H^1)}\leq\frac{K}{\s}.\]

\end{itemize}

In the same way we prove that $T_4$, $T_5$ and $T_6$ are bounded in
$\L(H^1)$ by $\dsp \frac{K}{\s^{\frac{3}{4}}}$.






Therefore we obtain that
\[P^{\infty}({\cal A}^\s)^{-1}R^\s(\xi){\cal A}^\s P^{\infty}=(\li-\xi)^{-1}P^\infty
+\frac{1}{\rs} (\li-\xi)^{-2} A P^{\infty} +o(\s^{-\frac{1}{2}}).\]

Hence integrating this asymptotic expansion on $\CC(\li, \eta)$, we obtain that~:
\[P^{\infty}({\cal A}^\s)^{-1}\H^\s P^\s {\cal A}^\s P^{\infty}=\li P^\infty
+\frac{1}{\rs} A P^{\infty} +o(\s^{-\frac{1}{2}}).\]

Theorem \ref{thm1} is then a consequence of the well known results in
finite dimensional spaces (see \cite{Kato}).



















\section{Further order asymptotic}

In this section we seek an asymptotic expansion of the eigenvalues of
$\H^\s$ at any order. The main
difficulty comes from the fact that the asymptotic at any order of the resolvent
operator is not valid in an algebra. The first order approximation is
only in ${\cal L}(H^1, L^2)$ and the further order ones are in ${\cal
L}(H^k, L^2)$ with $k$ sufficiently large. Then we can not compose the
asymptotic expansions in the study of $Q^\s=P^\infty ({\cal A}^\sigma)^{-1}
H^\s {\cal A}^\s P^\infty$.

The second difficulty is that $Q^\s$ is not self-adjoint and the
existence of an asymptotic expansion of the operator at any order does
not imply the existence of the corresponding asymptotic expansion for
the eigenvalues.

We consider $W^\s=(P^\s- P^\infty )^2$. This is a self-adjoint
operator which tends to zero as $\s$ tends to $+\infty$, according to
Proposition \ref{ordre0}. Consequently, for $\s$ sufficiently large, we
can define~:

\begin{equation}
\label{Bsigma}
{\cal B}^\s := (1- W^\s)^{-\frac12} \Big( (1 - P^\s) (1-  P^\infty) + P^\s
P^\infty \Big).
\end{equation}
Since $W^\s$ commute with $P^\s$ and $P^\infty$, ${\cal B}^\s$ maps $E^\infty$
in $E^\s$ and $(E^\infty)^\perp$ in $(E^\s)^\perp$. Moreover using
that $P^\infty P^\infty =P^\infty$ and that $P^\s P^\s=P^\s$, we
remark that
\[{\cal B}^{\s}\;^\tau \!B^\s=(1-W^\s)^{-\frac{1}{2}}
(1-W^\s)(1-W^\s)^{-\frac{1}{2}}=I,\]
that is ${\cal B}^\s$ is unitary and then $\H^\s P^\s$ and
$\tilde{Q}^\s=P^\infty ({\cal B}^\s)^{-1} H^\s {\cal B}^\s P^\infty$
have the same eigenvalues.

The crucial result to obtain the asymptotic expansion at any order is
the following proposition~:

\begin{proposition}
\label{prop5.1}
For ${\cal B}^\s$ defined above, we have:
$$P^\infty ({\cal B}^\s)^{-1} H^\s {\cal B}^\s P^\infty =
 (1- W^\s P^\infty )^{-\frac12} P^\infty H^\s P^\s P^\infty
  (1- W^\s P^\infty )^{-\frac12} P^\infty.$$
\end {proposition}

\vspace{2mm}
{\bf Proof.} By definition of ${\cal B}^\s$ and by using that $W^\s$ commute with $P^\s$ and
$P^\infty$ we have~:
$${\cal B}^\s P^\infty = (1- W^\s )^{-\frac12}P^\s P^\infty =
 P^\s P^\infty (1- W^\s )^{-\frac12}P^\infty.$$
In the same way, $P^\infty ({\cal B}^\s)^{-1} =
 P^\infty (1- W^\s )^{-\frac12}P^\infty P^\s$.
Furthermore since $W^\s P^\infty = P^\infty W^\s$ and $P^\infty P^\infty
= P^\infty$, we have~:
$$ (1- W^\s )^{-\frac12}P^\infty = (1- W^\s P^\infty
)^{-\frac12}P^\infty.$$
This complete the proof of the proposition.


\vspace{3mm}

Therefore using the relation (\ref{hsps}) the asymptotic expansion of
$P^\infty ({\cal B}^\s)^{-1} H^\s {\cal B}^\s P^\infty$
will be a consequence of the  asymptotic expansion of
$ (1- W^\s P^\infty )^{-\frac12} P^\infty R^\s(\xi) P^\infty
  (1- W^\s P^\infty )^{-\frac12} P^\infty$.
Moreover since $W^\s P^\infty = P^\infty W^\s P^\infty = P^\infty -
P^\infty P^\s P^\infty$, according to relation (\ref{fandepute}), we are reduced to discuss the asymptotic
expansion of
$P^\infty  R^\s( \xi) P^\infty$. We remark that if $g
\in H^1(\OO)$ then $P^\infty(g)\in \CC^{\infty}(\U)$ and vanishes on $\O$, thus it suffices
to perform an asymptotic expansion of $R^\s(\xi)(f)$ with $f\in
H^1_0(\U)\cap \CC^{\infty}(\U)$ and $f=0$ in $\O$.

\begin{proposition}
\label{putain}
Let $N\in \N$. Let $f\in H^1(\OO)$ such that $f=0$ in $\O$ and such that the
restriction of $f$ in $\U$ is in $ \CC^\infty (\U)$. Then
$R^\s(\xi)(f)$ admits an asymptotic expansion of the form~:
\[R^\s(\xi)(f)=\left\{
\begin{array}{l}
\dsp \sum_{k=0}^N
\frac{1}{\s^{\frac{k}{2}}}U^k(x)\;\;+\frac{1}{\s^\frac{N}{2}}a^\s(x)
\mbox{ if }x\in \U,\\
\\
\dsp \sum_{k=1}^N\frac{1}{\s^{\frac{k}{2}}} \psi(x)\alpha_k(x)e^{-\rs \dis(x)}+
\frac{1}{\s^\frac{N}{2}}b^\s(x)
\mbox{ if }x\in \O,
\end{array}
\right.
\]
and there exists a constant $C$ independent on $f$ and on $\s>\s_0$
sufficiently large such that
 the remainder terms $a^\s$ and $b^\s$ satisfy~:
\[\left\{
\begin{array}{l}
\dsp \|a^\s\|_{L^2(\U)}\leq 
 \frac{C}{\s}\|f\|_{H^{2N}(\U)},\\
\\
\dsp \|\nabla a^\s\|_{L^2(\U)} \leq  \dsp \frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^{2N}(\U)},\\
\\
\|b^\s\|_{L^2(\O)}\leq \dsp \frac{C}{\s^{\frac{3}{4}}}\|f\|_{H^{2N}(\U)},\\
\\
\dsp \|\nabla b^\s\|_{L^2(\O)}\leq
\frac{C}{\s^{\frac{1}{4}}}\|f\|_{H^{2N}(\U)}.
\end{array}
\right.
\]

\end{proposition}

\vspace{2mm}
{\bf Proof.} 

\vspace{2mm}
{\bf First step. Formal asymptotic expansion.}

As in the Section 3, with the usual notations, we seek $(u^\s, v^\s)$ of
the form:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\[\left\{\begin{array}{l}
\dsp u^\s(x)=U^0(x)+\frac{1}{\rs}U^1(x)+\frac{1}{\s}U^2(x) +...\\
\\
\dsp v^\s(x)=V^0(x,\rs
\dis(x))+\frac{1}{\rs}V^1(x,\rs\dis(x))+\frac{1}{\s}V^2(x,\rs\dis(x))+
...
\end{array}\right.\]

We recall that $u^\s$ and $v^\s $ satisfy (\ref{uetv}). Replacing
$u^\s$ and $v^\s$ by their asymptotic expansion in (\ref{uetv}) and
identifying the different powers of $\rs$, we obtain~:
\begin{itemize}
\item In $\U$, for $k=0$~:
\[\left\{
\begin{array}{l}
(-\lap -\xi )U^0=f \mbox{ on }\U\\
\\
U^0=0 \mbox{ on }\partial \U
\end{array}
\right.
\]
that is
\[U^0=(\widetilde{R}^\infty(\xi)\circ r)(f).\]

\item In $\O$, for $k\geq0$~:
\[\left\{
\begin{array}{l}
(-\lap -\xi )V^{k-1} -\left[(2\nabla \dis, \nabla V^k_z)+\lap\dis
V^k_z\right] + (V^{k+1} - V_{zz}^{k+1}) =0\mbox { in
}\O \\
\\
-V^k_z = \dsp \ddn{U^{k-1}}-\ddn{V^{k-1}} \mbox{ on }\partial \O
\end{array}
\right.
\]
with $V^{-1}=V^0=0$ (see Section 3).

\item In $\U$, for $k\geq1$~:
\[\left\{
\begin{array}{l}
(-\lap -\xi )U^k=0 \mbox{ on }\U\\
\\
U^k = V^k \mbox{ on }\partial \O \\
\\
U^k = 0 \mbox{ on }\partial \OO
\end{array}
\right.
\]
\end{itemize}

\vspace{2mm}


Then we construct functions $V^k$ and $U^k$, $k \geq 1$, by
induction. We set $V^k(x,z) = \alpha_k(x) e^{-z}$
with $\alpha_k$ solution of
\begin{equation}\label{reca}
\left\{
\begin{array}{l}
 (2\nabla \dis, \nabla \alpha_k)+\lap\dis
\alpha_k  =(\lap +\xi )\alpha_{k-1} \mbox { in }\O \\
\\
\alpha_k = \dsp \ddn{U^{k-1}}-\ddn{\alpha_{k-1}} \mbox{ on }\partial \O
\end{array}
\right.
\end{equation}

and $U^k$,  solution of
\begin{equation}\label{recu}
 \left\{
\begin{array}{l}
(-\lap -\xi )U^k=0 \mbox{ on }\U\\
\\
U^k = \alpha_k \mbox{ on }\partial \O \\
\\
U^k = 0 \mbox{ on }\partial \OO
\end{array}
\right.
\end{equation}
where $U^0= r R^\infty(\xi)(f)$ and $\alpha_0=0$.



\vspace{3mm}

{\bf Second step. Properties of the profiles.}

Using (\ref{reca}) we remark that if $\alpha_{k-1}\in H^p(\O)$ and
$U^{k-1}\in H^p(\U)$ then, $\dsp \alpha_k|_{\partial \O}\in
H^{p-\frac{3}{2}}(\partial \O)$ and $\dsp (\lap +\xi )\alpha_{k-1}\in
H^{p-2}(\O)$. Hence $\alpha_k\in H^{p-2}(\O)$ with Proposition
\ref{proptrans}.

Using classical results on elliptic equations, we obtain now that
$U^{k}\in H^{p-1}(\U)$. 

A  straightforward induction allows us to claim that there exists a
constant $C$ independent on $f$ and $\s$  such that~:
\begin{equation}
\label{con}
\pt k\leq N, \;\;\;\;\|\alpha_k\|_{H^{2(N-k)+2}(\O)} + \|U^k\|_{H^{2(N-k)+2}(\U)}\leq
C\|f\|_{H^{2N}(\U)}.
\end{equation}


\vspace{2mm}
{\bf Third step. Estimation of the remainder terms.}

\vspace{2mm}

For any $N\in \N$ we decompose $u^\s$ and $v^\s$ on the
form~:
\[\left\{
\begin{array}{l}
u^\s(x)=\dsp \sum_{k=0}^N {\s}^{-\frac{k}2} U^k (x)+{\s}^{-\frac{N}2}
a^\s(x)\mbox{
if }x\in \U,\\
\\
v^\s(x)=\dsp \sum_{k=0}^N {\s}^{-\frac{k}2}\psi(x)\alpha_k(x) e^{-\rs
\dis(x)}+{\s}^{-\frac{N}2}b^\s(x)\mbox{ if } x\in \O.
\end{array}\right.
\]

The remainder terms $a^\s$ and $b^\s$ satisfy the following system~:


\begin{equation}
\label{restesordre1}
\left\{
\begin{array}{ccl}
\mbox{(i)} & \;\;\;\;\; & (-\lap - \xi )a^\s=0 \mbox { in }\U, \\ \\
\mbox{(ii)} & & a^\s=b^\s \mbox { in }\partial \O,\\ \\
\mbox{(iii)} & & \dsp \ddn{a^\s}+\ddn{U^N}=\ddn{b^\s}+\ddn{\alpha^N} \mbox {
in }\partial \O,\\ \\
\mbox{(iv)} & &\dsp  (-\lap -\xi )b^\s+\s b^\s=g^\s\mbox{ in }\O,\\ \\
\mbox{(v)} & & a^\s=0\mbox{ on }\partial \O,
\end{array}\right.
\end{equation}

where
\[\begin{array}{rl}
g^\s= 
& \dsp \sum_{k=0}^{N-1}\s^{\frac{N-k}{2}}\left(2(\nabla \psi, \nabla
\varphi)\alpha_{k+1}+\lap \psi \alpha_k +2 (\nabla \psi, \nabla
\alpha_k)\right)e^{-\rs \dis}\\
\\
&\dsp + \left( 2(\nabla \psi, \alpha_N)+\lap \psi
\alpha_N\right)e^{-\rs\dis}+\psi (\lap +\xi )\alpha_{N}e^{-\rs\dis},
\end{array}\]
which is bounded in $L^2(\O)$ by $C\|f\|_{H^{2N}(\U)}$ because $\dis(x)>\varepsilon >0$ on the
support of the derivative of $\psi$, and using estimate (\ref{con}).


We apply Proposition \ref{estimation} and we obtain that~:
\[\|\nabla a^\s\|_{L^2(\U)}^2+\|\nabla
b^\s\|_{L^2(\O)}^2+\s\|b^\s\|_{L^2(\O)}^2\leq
\frac{K}{\rs}\|f\|_{H^{2N}(\U)}.\]

Furthermore, applying  Lemma \ref{lemme1}, we obtain that~:
\[\|b^\s\|_{L^2(\partial \O)}\leq
\frac{K}{\s}\|f\|_{H^{2N}}(\U).\]
and applying Lemma \ref{lemme1} it concludes the proof of Proposition
\ref{putain}. 


\vspace{5mm}
We consider $f\in E^\infty$. We remark that since $P^\infty f=f$ and
since $P^\infty$ is a regularizing operator, there exists a constant
$C$ such that~:
\begin{equation}
\label{macarel}
\| f\|_{H^{2N}(\U)}=\|P^\infty f\|_{H^{2N}(\U)}\leq C\|f\|_{L^2(\U)}.
\end{equation}

We introduce now the profiles $U^0, \ldots, U^N$ defined by
(\ref{recu}) and we set
\[A^k(\xi)(f)=P^\infty(U^k).\]
Using (\ref{con}) and (\ref{macarel}), since $P^\infty$ is a
regularizing operator, we obtain that there exists a constant $C$
independent on $f$, $\xi$ and $k\in \{1, \ldots, N\}$ such that~:
\[\|A^k(\xi)(f)\|_{L^2(\U)}\leq C\|f\|_{L^2(\U)}.\]

Furthermore if we denote $K^\s_N(\xi)(f)=P^\infty(a^\s)$ where $a^\s$
is defined by (\ref{restesordre1}), we obtain the proposition~:


\begin{proposition}
\label{prop5.2}
For any $N \in \N$
$$P^\infty R^\s(\xi) P^\infty = (\lambda^\infty - \xi)^{-1} P^\infty
+ \sum_{k=1}^N \frac{1}{\s^{\frac{k}{2}}} A^k(\xi) P^\infty +\frac{1}{\s^{\frac{N}{2}}} K^\s_N,$$
with $\| K^\s_N \|_{L^2(\U)} \leq \frac{C}{\rs} $.
\end{proposition}



\begin{corollary}
There exist $(P^k)_{k \geq 2} \in {\cal L}(E^\infty)^\N$ such that for
any $N \geq 2$:
$$P^\infty P^\s P^\infty = P^\infty + \sum_{k=2}^N \frac{1}{\s^{\frac{k}{2}}} P^k
P^\infty + \frac{1}{\s^{\frac{N}{2}}}P^\s_N,$$
with $\| P^\s_N \|_{L^2(\U)} \leq \frac{C}{\rs} $.
\end{corollary}


\vspace{2mm}
{\bf Proof.} It is a direct consequence of the above Proposition using the relation
(\ref{fandepute}). The coefficient of $\s^{-\frac12}$ vanishes because
$\dsp \int_{C} (\lambda^\infty - \xi)^{-2} d \xi =0$. For $k \geq 2$,
$\dsp P^k =\frac{-1}{2i\pi}
\int_{\CC(\li, \eta)} A^k(\xi) d\xi$.

\vspace{5mm}

{\bf Proof of Theorem \ref{thm2}.}
Let ${\cal B}^\s$ defined by (\ref{Bsigma}). Since ${\cal B}^\s$ maps $E^\infty$ on
$E^\s$, the eigenvalues of $H^\s P^\s$ are the ones of $\tilde{Q}^\s= P^\infty
({\cal B}^\s)^{-1} H^\s {\cal B}^\s P^\infty$. This operator is self-adjoint then
exploiting the perturbation theory in finite dimension we will deduce
the asymptotic expansion of the eigenvalues of $\tilde{Q}^\s$ from the one of
$\tilde{Q}^\s$ itself. According to the proposition \ref{prop5.1},
$$\tilde{Q}^\s =
 (1- W^\s P^\infty )^{-\frac12} P^\infty H^\s P^\s P^\infty
  (1- W^\s P^\infty )^{-\frac12} P^\infty.$$
Using the relations $W^\s P^\infty = P^\infty - P^\infty P^\s P^\infty$
and (\ref{hsps}), from  Proposition \ref{prop5.2} and its
Corollary we deduce that $\tilde{Q}^\s$ admits an asymptotic expansion at any
order with respect to $1/\sqrt{\s}$. Since $\tilde{Q}^\s$ is a self
adjoint operator, using finite dimensional results (see \cite{Kato}),
we obtain that the eigenvalues of $\H^\s$ admit an asymptotic expansion
at any order. This complete the proof of Theorem \ref{thm2}.


\vspace{3mm}
{\bf Aknowledgements.} The authors would like to thank A. S. Bonnet
Bendhia for many stimulating discussions about this subject.


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