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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 14, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/14\hfil Boundary layers]
{Boundary layers for transmission problems with singularities}
\author[A. Maghnouji and S. Nicaise\hfil EJDE-2006/14\hfilneg]
{Abderrahman Maghnouji, Serge Nicaise}  % in alphabetical order

\address{Abderrahman Maghnouji \hfill\break
Universit\'e de Valenciennes et du Hainaut Cambr\'esis\\
MACS, Institut des Sciences et Techniques de Valenciennes\\
F-59313 - Valenciennes Cedex 9, France}
\email{Abderrahman.Maghnouji@univ-valenciennes.fr}

\address{Serge Nicaise \hfill\break
Universit\'e de Valenciennes et du Hainaut Cambr\'esis\\
MACS, Institut des Sciences et Techniques de Valenciennes\\
F-59313 - Valenciennes Cedex 9, France}
\email{Serge.Nicaise@univ-valenciennes.fr}

\date{}
\thanks{Submitted September 26, 2005. Published January 31, 2006.}
\subjclass[2000]{35J25, 35B30}
\keywords{Transmission  problems; boundary layers; singularities}


\begin{abstract}
 We study two-dimensional transmission  problems for the
 Laplace operator for two diffusion coefficients.
 We describe the boundary layers  of this problem and show 
 that the layers appear only in the  part where the coefficient
 is large. The relationship with the singularities of the 
 limit problem is also described.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{Introduction}

We study  two-dimensional transmission problems (also called
interface problems) for the Laplace operator on polygonal domains  
consisting of different materials connected via an interface line. 
Dirichlet boundary conditions on the exterior boundary and standard 
transmission conditions are imposed. Such problems appear in 
diffusion problems where the conductivity of the materials are 
different on some parts of the domain. 
It is well known that the solutions of such problems have
corner singularities due the jump of the coefficients
\cite{kellogg:71,kellogg:75,leguillon:87,lemrabet:77,nicaise:88,nicaise:93,nicaise:94a}.
On the other hand, for a homogeneous medium having a large
diffusion coefficient, the solution exhibits boundary layers added
to corner singularities. Their relationship and description  are
 well understood nowadays 
\cite{butuzov:73,han:90,ilin:92,kellogg:94,melenk:00}. But to our
knowledge, the description of such a phenomenon is not known for
transmission problems where only one of the  diffusion
coefficients is large. Therefore in this paper we study a
relatively simple example of  a transmission problem that has
corner singularities and  boundary layers.

For a standard problem
\begin{equation}\label{singpertstandard} -\varepsilon\Delta
u_\varepsilon+u_\varepsilon=f \quad \text{in } \Omega,
\end{equation}
when $\Omega$ is a
polygonal domain of the plane,  $f$ is smooth and $\varepsilon>0$ is a
fixed (but small) parameter. An asymptotic expansion of $u_\varepsilon$
is well known \cite{butuzov:73,han:90,kellogg:94,melenk:00} and
may be written as
$$
u_\varepsilon=w_\varepsilon+w^{BL}+w^{CL}+r_\varepsilon,
$$
where $w_\varepsilon$ is the outer expansion, 
$w^{BL}$ describes the boundary layer,
$w^{CL}$ describes the corner layer,
and $r_\varepsilon$ is a remainder that is estimated as a function 
of $\varepsilon$ in some appropriate norms.
Usually the terms $w_\varepsilon, w^{BL}$ and $w^{CL}$
are explicit,
which means that, for numerical purposes for instance,
the behaviour of $u_\varepsilon$
is fully understood by the behaviour of
the terms $w_\varepsilon, w^{BL}$ and $w^{CL}$.

The goal of the present paper is to reproduce
a similar but simpler expansion for a transmission problem
where on a part $\Omega_+$ of the domain  we consider the problem
$$
-\varepsilon\Delta u_\varepsilon+u_\varepsilon=f \quad\text{in } \Omega_+,
$$
and on the other part $\Omega_-$, the problem
$$
-\Delta u_\varepsilon+u_\varepsilon=f \quad\text{in } \Omega_-,
$$
with, of course, transmission conditions on the interface.
By a simpler expansion, we mean that
$w_\varepsilon$, $w^{BL}$, $w^{CL}$ will be reduced to one term.
As we shall see
the situation is more complicated than in the standard case of
problem (\ref{singpertstandard}).
The main reason is that the
solution of the limit problem has singularities in the domain $\Omega_-$.
Let us further notice that surprisingly the solution of our problem has
only layers in the domain $\Omega_+$.

In this paper, the spaces $H^s(\Omega)$, with   $s\geq 0$,
are the standard Sobolev spaces in $\Omega$
with norm $\|\cdot\|_{s,\Omega}$
and semi-norm $|\cdot|_{s,\Omega}$.
The space $H^1_0(\Omega)$ is defined, as usual, by
$H^1_0(\Omega):=\{v\in H^1(\Omega)/ v=0$ on ${\Gamma}\}$.
$L^p(\Omega)$, $p>1$,
are the usual Lebesgue spaces with norm
$\|\cdot\|_{0,p,\Omega}$ (as usual we drop the index $p$ for $p=2$).
Finally, the notation
$a\lesssim b$  means the existence of a positive constant
$C$, which is independent   of the quantities $a$ and $b$
under consideration  and of the parameter $\varepsilon$,   such that $a\le C b$.


This  paper is organized as follows:
In section \ref{sect1} we start with  a one-dimensional problem
in order to describe and understand the typical phenomena.
Section \ref{sect2d} is devoted to the introduction of the
two-dimensional problem and to the (weak) convergence of the solution
 to the solution of  the limit problem.
We go on with the description of
the boundary and corner layers  in section \ref{sectbl},
paying a  particular attention   to the interface layers due to the
singularities. Finally in section  \ref{sectfullexp} we give the expansion
of the solution of our problem.

\section{The one-dimensional case\label{sect1}}

Let $\varepsilon\in ]0,1]$ be  a fixed   parameter.
Consider the following transmission  problem in $]-1,1[$:
\begin{equation} \label{e01}
\begin{gathered}
-\varepsilon^2u_{\varepsilon}''+u_{\varepsilon} =   1\quad  \text{in } ]-1,0[, \\
-w_{\varepsilon}''+w_{\varepsilon}=0          \quad \text{in } ]0,1[,\\
u_{\varepsilon}(-1)= w_{\varepsilon}(1)=0,  \\
u_{\varepsilon}(0)-w_{\varepsilon}(0)=0,   \\
\varepsilon^2 u'_{\varepsilon}(0)- w'_{\varepsilon}(0)=0.
\end{gathered}
\end{equation}
We remark that in  this problem the small parameter $\varepsilon$ appears 
only on $]-1,0[$. Consequently the formal limit problem is the non standard
transmission problem
\begin{equation} \label{e01limit}
\begin{gathered}
u_{0} =   1 \quad  \text{in } ]-1,0[, \\
-w_{0}''+w_{0}=0   \quad \text{in } ]0,1[, \\
u_{0}(-1)= w_{0}(1)=0,  \\
u_{0}(0)-w_{0}(0)=0,  \\
 w'_{0}(0)=0.
\end{gathered}
\end{equation}
This limit problem has a solution $w_0\equiv0$, but has  no solution $u_0$
since $u_0=1$
does not satisfy the boundary condition $u_{0}(-1)=0$
and the transmission condition $u_{0}(0)-w_{0}(0)=0$.
Therefore, we may expect that $u_\varepsilon$ will develop boundary layers at
 0 (transmission layer) and at $-1$ (standard boundary  layer).
We now justify this formal argument.


The exact  solution of this problem \eqref{e01} is
\begin{equation}\label{e02}
\begin{gathered}
u_{\varepsilon}(x)=\alpha \cosh\frac{x}{\varepsilon}+\beta \sinh\frac{x}{\varepsilon}+1,\\
w_{\varepsilon}(x)=\gamma \cosh x+\delta \sinh x,
\end{gathered}
\end{equation}
where  $\alpha,\beta,\gamma$ and $\delta $ are constants 
(depending on  $\varepsilon$)
determined in order to check the boundary and transmission conditions.
This yields a $4\times 4$ linear system that gives after resolution:
\begin{equation}\label{e03}
\begin{gathered}
 \alpha +1=\gamma, \quad \beta=\delta/\varepsilon,\\
 \gamma=-\delta\tanh 1, \quad \delta=-\varepsilon \psi(\varepsilon),
\end{gathered}
\end{equation}
where the function $\psi$ is 
$$
\psi(\varepsilon)=\frac{ \cosh \frac{1}{\varepsilon}-1}{\varepsilon\tanh 1\cosh
\frac{1}{\varepsilon}+ \sinh  \frac{1}{\varepsilon}}.
$$
Since one easily sees that $\psi(\varepsilon)$  approaches  1 as
$\varepsilon$ approaches  0, we deduce that
 $\delta=-\varepsilon\psi(\varepsilon)\sim -\varepsilon$  as $\varepsilon\to 0$.
Due to the identities (\ref{e02}) and (\ref{e03}), we can show that,
as $\varepsilon$ approaches $0$, $u_\varepsilon\to 1$ and $w_\varepsilon\to 0$,
as well as
\begin{gather*}
  u'_{\varepsilon}(-1)=-\frac{\alpha}{\varepsilon}\sinh\frac{1}{\varepsilon}+\frac{\beta}{\varepsilon}
\cosh\frac{1}{\varepsilon}\sim  \frac{1}{\varepsilon},\\
 u'_{\varepsilon}(0)=\frac{\beta}{\varepsilon}\sim  \frac{-1}{\varepsilon},\\
 w'_{\varepsilon}(1)=\gamma\sinh 1+\delta\cosh 1\sim\frac{- \varepsilon}{\cosh 1},\\
  w'_{\varepsilon}(0)=\delta\sim -\varepsilon.
\end{gather*}
 From these equivalences, we may say that
$w_\varepsilon$ has no layer, while $u_\varepsilon$
has a standard boundary layer at $-1$
and a transmission layer at $0$.
We also refer to Figure \ref{fig1} for an illustration of this fact.

\begin{figure}[ht]
\begin{center}
\includegraphics*[width=0.48\textwidth]{fig1}
\includegraphics*[width=0.48\textwidth]{fig2}
\end{center}
\caption{Exact solutions for  $\varepsilon=0.1$ (left) and 
$\varepsilon=0.05$ (right). \label{fig1}}
\end{figure}

Let us give a more precise result,
that will also allow us to underline the fact  that the transmission
layer at $0$ may be seen as a (Dirichlet) boundary layer.

\begin{theorem} \label{dim1}
For any $\varepsilon\in ]0,1]$, the unique solution 
$ ( u_{\varepsilon},w_{\varepsilon}) $ of \eqref{e01} satisfies
\begin{equation} \label{e04}
u_{\varepsilon}(x) =1-\chi^b(x)\exp\big(-\frac{\mathop{\rm dist}(x, -1)}{\varepsilon}\big)
-\chi^i(x) \exp\big(-\frac{\mathop{\rm dist}(x,0)}{\varepsilon}\big)
   +r_\varepsilon(x), \quad \forall x\in ]0,1[,
\end{equation}
where $\chi^b $ and $ \chi^i $ are the two following cut-off functions:
\begin{gather*}
\chi^b=1 \quad \text{on }]-1,-1+\eta[, \\
 \chi^i=1 \quad \text{on } ]-\eta,\eta[,\\
\mathop{\rm supp} \chi^b\cap  \mathop{\rm supp}\chi^i=\emptyset.
\end{gather*}
Moreover,
\begin{equation}\label{estdim1}
\varepsilon\Vert r'_\varepsilon\Vert_{0,]-1,0[}+ \Vert r_\varepsilon\Vert_{0, ]-1,0[}
+\Vert w_\varepsilon\Vert_{1,]0,1[}\lesssim (\varepsilon  e^{\frac{-\eta}{\varepsilon}}+\varepsilon).
\end{equation}
\end{theorem}

\begin{proof} Let us define the functions
$  v^b: x\mapsto - \exp\big(-\frac{\mathop{\rm dist}(x, -1)}{\varepsilon}\big) $,
a solution of
\begin{gather*}
-\varepsilon^2 {v^b}''+v^b=0  \quad\text{in } ]-1,0 [, \\
v^b(-1)+1=0, \\
v^b(+\infty)=0,
\end{gather*}
and $v^i : x\mapsto -\exp\big(-\frac{\mathop{\rm dist}(x,0)}{\varepsilon}\big) $,
a solution of
\begin{gather*}
-\varepsilon^2 {v^i}''+v^i=0 , \quad \text{in } ]-1,0 [, \\
v^i(0)+1=0, \\
v^i(-\infty)=0 .
\end{gather*}
Using these two problems and by substitution of (\ref{e04}) in \eqref{e01},
we see that   $(r_\varepsilon, w_{\varepsilon}) $ is solution  of
\begin{equation} \label{e05}
 \begin{gathered}
-\varepsilon^2 r''_\varepsilon+r_\varepsilon=g_\varepsilon   \quad\text{in } ]-1,0 [, \\
w_{\varepsilon}-w_\varepsilon''=0 \quad \text{in } ]0,1[,\\
r_\varepsilon(-1)=0,  \\
w_{\varepsilon}(1)=0, \\
r_\varepsilon(0)=w_{\varepsilon}(0),  \\
\varepsilon^2 r'_\varepsilon(0)- w'_{\varepsilon}(0)=-\varepsilon,
\end{gathered}
\end{equation}
where
$$
g_\varepsilon:=\varepsilon^2\left ([\chi^b; \frac{d^2}{ dx^2}] e^{-\frac{x+1}{\varepsilon}}
+[\chi^i;\frac{d^2}{ dx^2}] e^{\frac{x}{\varepsilon}}\right),
$$
the bracket $[\chi^b; \frac{d^2}{ dx^2}]$
being defined as usual,
$$
[\chi^b; \frac{d^2}{ dx^2}]h=
\frac{d^2}{ dx^2}(\chi^bh)-\chi^b\frac{d^2}{ dx^2}h=
h \frac{d^2}{ dx^2}\chi^b+2 \frac{d}{ dx}\chi^b \frac{d}{ dx} h.
$$
The variational formulation of this problem is
\begin{equation}\label{varperturbe1d}
\begin{aligned}
&\int_{-1}^0 \varepsilon^2 r'_\varepsilon w'\,dx+\int_{0}^1  w'_\varepsilon w'\,dx
+\int_{-1}^0   r_\varepsilon w\,dx+\int_{0}^1  w_\varepsilon w\,dx\\
&=\int_{-1}^0 g_\varepsilon w\,dx-\varepsilon w(0), \forall w\in H^1_0(]-1,1[).
\end{aligned}
\end{equation}
Since this left-hand side is trivially coercive on $H^1_0(]-1,1[)$,
by the  Lax-Milgram lemma, this problem has a unique solution
$r_\varepsilon\in H^1(]-1,0 [)$
and $w_\varepsilon\in H^1(]0,1[)$ such that 
$r_\varepsilon(-1)=w_{\varepsilon}(1)=0, $
and
$
r_\varepsilon(0)=w_{\varepsilon}(0)$ (this means that the function 
$k_\varepsilon$ defined by $r_\varepsilon$ on $]-1,0 [$
and by $w_\varepsilon$ on $]0,1[$ belongs to $H^1_0(]-1,1[)$).
Moreover by taking as test function in (\ref{varperturbe1d})
$w=k_\varepsilon$ we obtain
\begin{equation}\label{estvarperturbe1d}
\begin{aligned}
&\varepsilon^2\Vert r'_\varepsilon\Vert_{0,]0,1[}^2+ \Vert r_\varepsilon\Vert_{0,]-1,0[}^2+
\Vert w'_\varepsilon\Vert_{0,]0,1[}^2 + \Vert w_\varepsilon\Vert_{0,]0,1[}^2\\
&\leq \Vert g_\varepsilon\Vert_{0,]-1,0[} \Vert  r_\varepsilon \Vert_{0,]-1,0[}+
\varepsilon |w_\varepsilon(0)|.
\end{aligned}
\end{equation}
It then remains to estimate the $L^2$-norm of $g_\varepsilon$.
The properties of $ \chi^b $ and $ \chi^i $ imply that
$\mathop{\rm supp} g_\varepsilon \subset [-1+\eta,-\eta]$.
 Since on this  interval
$ e^{-\frac{(x+1)}{\varepsilon}}\leq e^{-\frac{\eta}{\varepsilon}} $ and
$
 e^{\frac{x}{\varepsilon}}\leq e^{-\frac{\eta}{\varepsilon}},
$
we obtain
\[
\Vert g_\varepsilon\Vert_{0,]-1,0[} \lesssim \varepsilon  e^{\frac{-\eta}{\varepsilon}}.
\]
On the other hand, the   identities (\ref{e02}) and (\ref{e03}) imply that
\[
 \vert w_{\varepsilon}(0)\vert \lesssim \varepsilon \;\;  \forall \varepsilon\in ]0,1[.
\]
These two estimates in (\ref{estvarperturbe1d})
yield
$$
\varepsilon^2\Vert r'_\varepsilon\Vert_{0,]-1,0[}^2+ \Vert r_\varepsilon\Vert_{0,]-1,0[}^2+
\Vert w'_\varepsilon\Vert_{0,]0,1[}^2+ \Vert w_\varepsilon\Vert_{0,]0,1[}^2\lesssim
 \varepsilon e^{\frac{-\eta}{\varepsilon}} \Vert  r_\varepsilon \Vert_{0,]-1,0[}+
\varepsilon^2.
$$
The desired estimate (\ref{estdim1}) follows from Young's inequality.
\end{proof}


Note that the estimate (\ref{estdim1}) is optimal: Direct
calculations yield $\Vert w_\varepsilon\Vert_{0,]0,1[}\sim \varepsilon$.

The above theorem  gives an explicit expansion of $u_\varepsilon$,
which also shows that $u_\varepsilon$ has  two layers (at 0 and $-1$).
It further
 says that the natural energy norm of
the remainder $r_\varepsilon$ is of order $\varepsilon$.
Finally it says that $w_\varepsilon$
has no layer and that its natural energy norm
  is of order $\varepsilon$.

The goal of the next sections is to show similar results for a
 polygonal domain on the plane.

\section{The two-dimensional  problem \label{sect2d}}

\begin{figure}[ht] 
\begin{center}
\setlength{\unitlength}{.8mm}
\begin{picture}(80,50)(0,0)
\drawline(0,0)(20,20)(80,20)(60,0)(0,0)
\drawline(20,20)(25,50)(55,50)(52.5,35)(82.5,35)(80,20)
\put(39,9){$\Omega_-$}
\put(47,30){$\Omega_+$}
\put(13,18){$A$}
\put(82,18){$B$}
\end{picture}
\end{center}
\caption{\label{fig2} The domains $\Omega_+$ and $\Omega_-$}
\end{figure}


Let $ \Omega_ +$ and $ \Omega_- $ be two polygonal domains  of
 $ \mathbb{R}^2 $  with respective boundary $\partial  \Omega_ + $ and $\partial  \Omega_ -$
 having in  common a  segment  $ \Sigma=[A,B]$, see Figure \ref{fig2}.
Denote by   $A_1,A_2,\dots,A_N$ the vertices of $\partial\Omega_+$
enumerated  clockwise and so that $ A_1=A $ and $ A_2=B. $
Denote further  by $\omega_j $ the interior angle  of $\Omega_+$ at the vertex   $A_j$,
for any $ j\in\{1,2,\dots,N\}$
and let
    $\varphi_j $ the interior angle  of $\Omega_-$ at the vertex  $A_j$, $ j=1,2$.

For further purposes we denote by $\Omega=\Omega_ +\cup  \Omega_- \cup \Sigma$.
Moreover for a function $u$ defined in $\Omega$,
we denote by $u_+$ (resp. $u_-$)
the restriction of $u$ to $\Omega_+$ (resp. $\Omega_-$).


For $ \varepsilon\in]0,1[,  f_\pm\in \mathcal{C}^\infty(\bar \Omega_\pm)  $ 
and $h\in\mathcal{C}^\infty(\bar \Sigma)$, we consider
the transmission problem in $\Omega$: Find $u^\varepsilon$
solution of
\begin{equation}\label{e1}
\begin{gathered}
- \varepsilon^2\Delta u_+^\varepsilon +u_+^\varepsilon
=f_+\quad \text{in } \Omega_+,\\
 -\Delta u_-^\varepsilon +u_-^\varepsilon=f_- \quad \text{ in  } \Omega_-,\\
 u_+^\varepsilon=0  \quad \text{on }\partial \Omega_+\setminus\Sigma,\\
 u_-^\varepsilon=0  \quad \text{on } \partial \Omega_-\setminus\Sigma,\\
 u_+^\varepsilon-u_-^\varepsilon=0  \quad \text{on } \Sigma,\\
 \varepsilon^2 \frac{\partial u_+^\varepsilon}{\partial \nu} -
\frac{\partial u_-^\varepsilon}{\partial \nu}=h  \quad \text{on } \Sigma,
\end{gathered}
\end{equation}
where $ \nu $ denotes the outward normal vector along  $\Sigma $ oriented
outside $ \Omega_+$. The variational formulation of this  problem consists
in finding  a unique   solution $u^\varepsilon\in H^1_0(\Omega)$ of
\begin{equation}\label{e1var}
\begin{aligned}
&\int_{\Omega_+}  (\varepsilon^2\nabla u_+^\varepsilon\cdot\nabla v_+
+u_+^\varepsilon  v_+)+ \int_{\Omega_-}  (\nabla u_-^\varepsilon\cdot
\nabla v_- +u_-^\varepsilon  v_-) \\
&=\int_{\Omega_+} fv +\int_{\Sigma} h v, \forall v\in H^1_0(\Omega).
\end{aligned}
\end{equation}
Since this left-hand side is a  coercive and continuous bilinear form
on $H^1_0(\Omega)$, this problem has a unique solution
thanks to the Lax-Milgram lemma.

We now look at the limit of the problem and  of $u^\varepsilon$ as $\varepsilon$
goes to zero.
As before the formal limit problem is
\begin{equation}\label{e1limit}
\begin{gathered}
u_+^0=f_+\quad \text{in } \Omega_+,\\
 -\Delta u_-^0 +u_-^0=f_-\quad \text{in } \Omega_-,\\
 u_+^0=0  \quad \text{on }\partial \Omega_+\setminus\Sigma,\\
 u_-^0=0  \quad \text{on } \partial \Omega_-\setminus\Sigma,\\
 u_+^0-u_-^0=0  \quad \text{on } \Sigma,\\
 - \frac{\partial u_-^0}{\partial \nu}=h  \quad\text{on  } \Sigma.
\end{gathered}
\end{equation}
 As in dimension 1, in this limit problem
 $u_-^0$ may be seen as the (unique) solution of a mixed Dirichlet-Neumann
problem  in $\Omega_-$, and since $f_+$ does not satisfy the Dirichlet
boundary condition  $f_+=0$ on $\partial \Omega_+\setminus\Sigma$,
 and $f_+=u_-^0$  on $\Sigma$,  the solution $u^\varepsilon_+$ should develop
boundary layers along $\partial \Omega_+$.
 This will be proved in details in the remainder of this paper.
 Let us first state a weak convergence.

\begin{theorem} \label{conv}
There exists a subsequence of $u_\varepsilon$,
still denoted by $u_\varepsilon$, such that
the  pair $ (u_+^\varepsilon,u_-^\varepsilon) $ converges in
$ L^2(\Omega_+)\times H^1(\Omega_-) $ to $ (u_+^0,u_-^0) $
as   $\varepsilon $ goes to 0,
where
$u_+^0=f_+ $ and $ u_-^0 $ is the unique  variational solution  of the
mixed Dirchlet-Neumann problem
\begin{equation}\label{e2}
\begin{gathered}
-\Delta u_-^0 +u_-^0=f_-\quad \text{in  }  \Omega_-,\\
u_-^0=0  \quad \text{on } \partial \Omega_-\setminus\Sigma,\\
 \frac{\partial u_-^0}{\partial \nu}=-h  \quad\text{on }  \Sigma.
\end{gathered}
\end{equation}
\end{theorem}

Before proving this  theorem, let us introduce some notation
and give a density result.
Let us introduce the following  bilinear and linear  forms:
\begin{equation} \label{e3}
\begin{gathered}
a(u,v)=\int_{\Omega_+}  \nabla u_+\cdot\nabla v_+  ,\\
b(u,v)=\int_{\Omega_-}\nabla u_-\cdot\nabla v_-+\int_{\Omega}u v, \\
F(v)=\int_{\Omega} fv+\int_\Sigma h v.
\end{gathered}
\end{equation}
Let us define the space
$$
W=\{ w\in L^2(\Omega):  w_-\in   H^1(\Omega_-)\text{ and }  w_-=0  
\text{ on  } \partial \Omega_-\setminus\Sigma\},
$$
which is a Hilbert space, equipped with the norm
$\Vert w\Vert^2_W= b(w,w)$.

\begin{lemma} $H^1_0(\Omega)$ is dense in $ W $. \end{lemma}

\begin{proof} Let $w\in W$.
Since $w_-\in \tilde H^{1/2}(\Sigma)$, by
\cite[Theorem 1.5.2.3]{grisvard:85a}
(trace  Theorem), there  exists $ \tilde w_+\in H^1(\Omega_+) $ such that
\begin{gather*}
 \tilde w_+=w_-\quad \text{on } \Sigma,\\
\tilde w_+=0  \quad \text{on } \partial \Omega_+\setminus\Sigma.
\end{gather*}
Since $ w_+-\tilde w_+$ belongs to $L^2(\Omega_+) $ and   since
$H_0^1(\Omega_+) $  is dense in $ L^2(\Omega_+) $, there exists
a sequence of functions  $ w_+^n\in H_0^1(\Omega_+) $, $n\in \mathbb{N}$ such that
\begin{equation}\label{dense1}
\Vert w_+^n-(w_+-\tilde w_+)\Vert_{0,\Omega_+}\to 0 \quad
\text{as } n\to \infty.
\end{equation}
For all positive integer $ n $, we introduce    the function $\tilde w^n$
  defined in $\Omega$ as follows
\begin{gather*}
 \tilde w_+^n= w_+^n+ \tilde w_+, \\
\tilde w_-^n=w_-.
\end{gather*}
 From the boundary condition satisfied by $\tilde w_+$,
$\tilde w^n$  belongs to $H^1_0(\Omega)$.
Moreover from the definition of $\tilde w^n$
and owing to  (\ref{dense1}), we have
 $$
\Vert \tilde w^n-w\Vert_W=\Vert w_+^n-(w_+-\tilde w_+)\Vert_{0,\Omega_+}
\to 0.
$$
\end{proof}


\begin{proof}[Proof   of Theorem  \ref{conv}]
 From (\ref{e1var}) and the definition of $u^0$, we see that
$u^\varepsilon\in H^1_0(\Omega) $  and  $ u^0\in W $
are the respective solution of
\begin{gather}
\label{conv1}
\varepsilon^2a(u^\varepsilon,v)+b(u^\varepsilon,v) =F(v) ,
\forall v\in H^1_0(\Omega),\\
\label{conv2}
b(u^0,w) =F(w) , \forall w\in W.
\end{gather}

\noindent{\bf Step 1.} $u^\varepsilon $ is weakly convergent to  $ u^0 $
in $ W$. We first  remark that
\[
\Vert u^\varepsilon\Vert_W^2=b(u^\varepsilon,u^\varepsilon)
\leq b(u^\varepsilon,u^\varepsilon)
+\varepsilon^2 a(u^\varepsilon,u^\varepsilon).
\]
Now taking  $ v=u^\varepsilon$ in (\ref{conv1})
and $ w=u^\varepsilon$ in (\ref{conv2})  we obtain
\begin{equation}\label{nic1}
\varepsilon^2a(u^\varepsilon,u^\varepsilon)+b(u^\varepsilon,u^\varepsilon)
=b(u^0,u^\varepsilon).
\end{equation}
Using Cauchy-Schwarz's inequality, we directly have
\[
\vert b(u^0,u^\varepsilon)\vert \leq  \Vert u^0\Vert_W \Vert u^\varepsilon\Vert_W.
\]
These three properties imply that
\begin{equation}\label{nic2}
\Vert u^\varepsilon\Vert_W\leq \Vert u^0\Vert_W.
\end{equation}
Therefore, there exists  $ w \in W  $ and a subsequence of  $u^\varepsilon  $,
still denoted by  $u^\varepsilon$,
weakly   convergent to  $w $ in $ W $.

Now for any fixed  $ v\in H^1_0(\Omega)$, using successively (\ref{nic1}) 
and (\ref{nic2}) we may write
\begin{align*}
\vert a(u^\varepsilon,v)\vert
&\leq  \Vert \nabla u_+^\varepsilon\Vert_{0,\Omega_+}\Vert \nabla v\Vert_{0,\Omega_+} \\
&\leq  \varepsilon^{-1}( \varepsilon^2\Vert \nabla u_+^\varepsilon\Vert^2_{0,\Omega_+}
 +b(u^\varepsilon,u^\varepsilon))^{\frac{1}{2}} \Vert \nabla v\Vert_{0,\Omega_+} \\
&=\varepsilon^{-1}b(u^0,u^\varepsilon)^{\frac{1}{2}} \Vert \nabla v\Vert_{0,\Omega_+} \\
&\leq  \varepsilon^{-1}\Vert u^0\Vert_W^{\frac{1}{2}}\Vert u^\varepsilon
 \Vert_W^{\frac{1}{2}}\Vert \nabla v\Vert_{0,\Omega_+} \\
&\leq  \varepsilon^{-1}\Vert u^0\Vert_W\Vert \nabla v\Vert_{0,\Omega_+}.
\end{align*}
This last estimate implies that
$$
 \lim_{\varepsilon\to 0}\varepsilon^2 a(u^\varepsilon, v)=0,\quad
 \forall v\in H^1_0(\Omega).
$$
Therefore, passing to the limit in (\ref{conv1}), we obtain
$$
 \lim_{\varepsilon\to 0} b(u^\varepsilon, v)= F(v)=b(u^0,v), \quad
\forall v\in H^1_0(\Omega).
$$
Since  $H^1_0(\Omega)$ is dense in $ W, $ we conclude that
$$
b(u^0,v)=b(w,v), \quad  \forall v\in W.
$$
%
Since $b(\cdot,\cdot)$ is the inner product of $W$, we deduce that $ u^0=w$.

\noindent{\bf Step 2.}   $ u^\varepsilon $ is strongly convergent to $u^0$
in $ W$.
\begin{align*}
\Vert  u^\varepsilon -u^0\Vert_W^2
&=b( u^\varepsilon -u^0, u^\varepsilon -u^0)\\
&=b(u^\varepsilon,u^\varepsilon)-b(u^0, u^\varepsilon)
-b(u^\varepsilon-u^0,u^0).
\end{align*}
Taking into account (\ref{nic1}), we obtain
\[
\Vert  u^\varepsilon -u^0\Vert_W^2\leq -b(u^\varepsilon-u^0,u^0).
\]
Then we have the conclusion, by the weak convergence in $W$ of
$u^\varepsilon$ to $u^0$.
\end{proof}


 From this Theorem we may see $u^0$
as the first term of the  outer expansion of $u^\varepsilon$.
Let us now pass to the description of the boundary layers.

\section{Boundary layers\label{sectbl}}

In the sequel let $\mathcal{L}_\varepsilon $ denote the operator
$\mathcal{L}_\varepsilon=I-\varepsilon^2\Delta$.
In this section, we define in   $\Omega_+$, the boundary layer $ v_j^b $
along  $ \Gamma_j=[A_{j-1}, A_j]$, $j=2,3,\dots,N $ and
the interface  layer $ v^i$  along  $\Sigma$, such that
if $ \mathcal{V}_j $ denote a small neighbourhood of $ \Gamma_j $, we have
\begin{equation} \label{b1}
\begin{gathered}
\mathcal{L}_\varepsilon( u_+^\varepsilon-f_+-v_j^b)=\varepsilon^2O(\varepsilon)
 \quad \text{in } \mathcal{V}_j\cap\Omega_+, \\
 f_++v_j^b=0    \quad \text{on } \Gamma_j
\end{gathered}
\end{equation}
and
 \begin{equation} \label{b2}
 \begin{gathered}
 \mathcal{L}_\varepsilon( u_+^\varepsilon-f_+-v^i)=\varepsilon^2O(\varepsilon)
  \quad \text{in } \mathcal{V}_1\cap\Omega_+, \\
 f_++v^i=u_-^0 \quad \text{on } \Sigma,
  \end{gathered}
  \end{equation}
when    $O (\varepsilon) $ denote as usual a   function of $\varepsilon$ bounded
in a neighbourhood of $\varepsilon=0$.
  Note that the situation is not the same along $\Sigma$
  due to the lack of regularity of $u_-^0$ (see below).

\subsection{Some notation  and definitions}

We denote by $ (x,y) $ the Cartesian coordinates of the plane with origin at
$A_1$ and such that $ \Gamma_1\subset \{(x,0), x>0\}$.
Similarly we   denote by  $(x_j,y_j) $  the Cartesian coordinates of
the plane  with origin at $A_j$ and such that
$ \Gamma_j \subset \{(x_j,0), x_j>0\}$.

We now fix two cut-off functions
$\chi_j^1,\chi_j^2\in \mathcal{C}_0^\infty(\mathbb{R}) $ satisfying
$\mathop{\rm supp} \chi_j^1\subset [-a_j,a_j]$, and
$$
\chi_j^1(x)=1 \quad\text{on }]0,l_j[,
$$
where $ l_j $ is the length of $ \Gamma_j $,
and $\mathop{\rm supp} \chi_j^2\subset [-b,b]$, as well as
\[
 \chi_j^2(y)=1 \quad \text{on }]-\frac{b}{2},\frac{b}{2}[,
\]
for a sufficiently small fixed $ b>0 $.

Now we can   introduce the cut-off function along $ \Gamma_j $ by
\begin{equation}\label{b3}
\chi_j^b(x,y)=\chi_j^1(x)\; \chi_j^2(y) .
\end{equation}
We finally take  $ \chi^i=\chi_1^b$.

Now we assume that $ f_+ $ is the restriction to $\Omega_+$
of a smooth function
$\tilde f_+\in \mathcal{C}^\infty (\mathbb{R}^2) $ and that
$\Omega_+$ is convex, i.e.,  $ 0<\omega_j <\pi$,
for all $j=1,\dots, N$.
This last assumption is simply made to simplify the construction of
corner layers. Using the method of   \cite{melenk:00}, we probably
can treat the non convex case.

\subsection{Construction of  $ v_j^b$}
They are standard, see for instance \cite{han:90,ilin:92}.
For $j=2,\dots,N$, $ v_j^b  $ is
the  unique solution    of the problem
\begin{gather*}
v_j^b -\varepsilon^2{v^b_j}''=0 \quad \text{in }  y_j>0 , \\
v_j^b=-\tilde f_+(x_j,.)\quad  \text{at }  y_j=0 , \\
v_j^b=0  \quad \text{at } y_j=+\infty .
\end{gather*}
It is explicitly given by
\begin{equation}\label{b4}
v_j^b(x_j,y_j)=-\tilde f_+(x_j,0)\; e^{-y_j/\varepsilon}.
\end{equation}
Since  $ \omega_j < \pi $, the function $\chi_j^b v^b_j $ is well
defined in  $\Omega_+$ and  satisfies the conditions (\ref{b1}).
Moreover it has the regularity $\mathcal{C}^\infty(\bar\Omega_+) $
and for any $(x_j,y_j)\in\Omega_+$,
\begin{align*}
\mathcal{L}_\varepsilon(\chi_j^bv_j^b)(x_j,y_j)
&=(I-\varepsilon^2\Delta_{(x_j,y_j)})(\chi_j^b v^b_j)(x_j,y_j)\\
&=\varepsilon^2 \chi_j^b(x_j,y_j)e^{-\frac{ y_j}{\varepsilon}}({\partial}_{x_j})^2
\tilde f_+(x_j,0)  +\varepsilon^2[ \chi_j^b; \Delta_{(x_j,y_j)}] v_j^b,
\end{align*}
where we recall that
$[ \chi;  \Delta_{(x_j,y_j)}] v:= \chi \Delta_{(x_j,y_j)} v-
\Delta_{(x_j,y_j)}(\chi v)$.
Since
\begin{gather*}
\vert   \chi_j^be^{-\frac{ y_j }{\varepsilon}}({\partial}_{x_j})^2 f_+(x_j,0)\vert
 \lesssim  1,\\
\vert [ \chi_j^b \; ; \; \Delta_{(x_j,y_j)}] v_j^b\vert
\lesssim  \frac{1}{\varepsilon}e^{-\frac{b}{2\varepsilon}},
\end{gather*}
we  deduce that
\begin{equation}\label{b5}
 \Vert \mathcal{L}_\varepsilon(\chi_j^b v_j^b)\Vert_{0,\Omega_+}
\lesssim \varepsilon^2+\varepsilon e^{-\frac{b}{2\varepsilon}}.
\end{equation}

\subsection{Construction of   $ v^i$}
In  general the solution  $ u_-^0 $  of problem (\ref{e2}) has only the
regularity $ u_-^0\in H^1(\Omega_-)$.
Consequently if we proceed as in the previous subsection,
namely if we take
\[
v^i(x_1,y_1)= (\tilde u_-^0-\tilde f_+ )(x_1,0)\;  e^{-y_1/\varepsilon},
\]
the  regularity of $  v^i $ is not  sufficient to obtain an  estimate
similar to  (\ref{b5}). To overcome this difficulty, we shall use the
decomposition of $  u_-^0$ into a  regular part and singular one.

For $ j=1,2$, we recall that the  singular exponents associated with
the mixed  Dirichlet-Neumann problem near $A_j$ are given by
(see \cite{grisvard:85a,dauge:88})
$$
\Lambda_j=\{\lambda_k= \frac{\pi}{2\varphi_j} +\frac{k\pi}{\varphi_j},
k\in \mathbb{Z}\}.
$$
Let  $( r_j,\theta_j) $ be the polar     coordinates  centred at $ A_j $
and such that  $\theta_j =0 $ on $ \Sigma $,
and $\theta_j =-\omega_j$ on the other edge of $ \Omega_-$ having $A_j$
as extremity.
For  $ \lambda_k\in \Lambda_j$, we denote
\begin{equation}\label{e6}
S_{j,\lambda_k}( r_j,\theta_j) =r_j^{\lambda_k}\sin \lambda_k
(\varphi_j+\theta_j), \quad -\varphi_j<\theta_j<\omega_j.
\end{equation}
Recall that this function satisfies
\begin{equation}
\label{singulariteDN}
\begin{gathered}
\Delta S_{j,\lambda_k}=0, \\
S_{j,\lambda_k}(r_j,-\varphi_j) =0,\\
\frac{\partial}{\partial \theta} S_{j,\lambda_k}(r_j,0) =0.
\end{gathered}
\end{equation}
According to  \cite[Corollary 4.4.3.8]{grisvard:85a},
the solution   $u_-^0\in H^1(\Omega_-) $ of (\ref{e2}) admits the
decomposition
\begin{equation}\label{e7}
u_-^0=u_{-,r}^0+\sum_{j=1,2}\eta_j
\sum_{\lambda_k\in \Lambda_j,0<\lambda_k<2}
 C_{j,\lambda_k}S_{j,\lambda_k},
\end{equation}
where   $ u_{-,r}^0\in H^3(\Omega_-\cap \mathcal{V}_1) $,
 $ C_{j,\lambda_k} $ are real   constants and
   $\eta_j $  is a (radial) cut-off function equal to 1 in neighbourhood of
$A_j$ and    equal to zero outside another neighbourhood of $A_j, j=1,2$.
Using this expansion, we can define
\begin{equation}\label{e71}
 v^i=v_r^i+v_s^i
\end{equation}
where
\begin{equation}\label{e61}
\begin{gathered}
v_r^i(x,y)= (\tilde u_{-,r}^0-\tilde f_+ )(x,0)e^{-y/\varepsilon},\\
v_s^i(x,y)=\sum_{j=1,2}\eta_j
\sum_{\lambda_k\in \Lambda_j,0<\lambda_k<2}
 C_{j,\lambda_k}S_{j,\lambda_k}e^{-y/\varepsilon},
\end{gathered}
\end{equation}
where $\tilde u_{-,r}^0(\cdot)$ is an extension to the real line of
$u_{-,r}^0(\cdot,0)$.
Since $u_{-,r}^0(\cdot,0)$ belongs to $H^{\frac52}(\Sigma)$,
this extension may be chosen in $H^{\frac52}(\mathbb{R})$ and by  the Sobolev
embedding Theorem, $v^i_r\in \mathcal{C}^2(\bar \Omega_+)$.
Therefore, as in the previous subsection,
we have
\begin{equation}\label{i3}
\Vert \mathcal{L}_\varepsilon (\chi^iv_r^i )\Vert_{0,\Omega_+}
\lesssim \varepsilon^2+\varepsilon e^{-\frac{b}{2\varepsilon}}.
\end{equation}
On the other hand, Leibniz's rule yields
\begin{align*}
&\Delta (\chi^i\eta_j S_{j,\lambda_k} e^{-y/\varepsilon}) \\
&= \chi^i\eta_j S_{j,\lambda_k}\Delta e^{-y/\varepsilon}
 +2\nabla(\chi^i\eta_j S_{j,\lambda_k})\cdot\nabla e^{-y/\varepsilon}
 +\Delta(\chi^i\eta_j S_{j,\lambda_k} )e^{-y/\varepsilon}\\
&=  e^{-y/\varepsilon} \{\frac{1}{\varepsilon^2}   \chi^i\eta_j S_{j,\lambda_k}
 -\frac{2}{\varepsilon} \frac{\partial (\chi^i\eta_j S_{j,\lambda_k})}{\partial y}
+\chi^i\eta_j \Delta S_{j,\lambda_k} -[\chi^i\eta_j;\Delta]S_{j,\lambda_k} \},
\end{align*}
and therefore (reminding  $\Delta S_{j,\lambda_k}=0$)
\[
\mathcal{L}_\varepsilon(\chi^i\eta_j S_{j,\lambda_k} e^{-y/\varepsilon})
=\varepsilon^2e^{-y/\varepsilon}
 \Big(
\frac{2}{\varepsilon}\frac{\partial  }{\partial y}(\chi^i\eta_j S_{j,\lambda_k})
+[\chi^i\eta_j; \Delta] S_{j,\lambda_k}
\Big).
\]
 From this identity, we deduce that
\begin{equation}\label{i4}
 \Vert \mathcal{L}_\varepsilon(\chi^i v_s^i )\Vert_{0,\Omega_+}
\lesssim \varepsilon^{1+\lambda}+\varepsilon e^{-b/(2\varepsilon)},
\end{equation}
where $\lambda=\min_{k=1,2}\min\{\lambda_k: \lambda_k\in \Lambda_k\}$.
As $ v^i=v_r^i+v_s^i, $ the estimates  (\ref{i3}) and (\ref{i4}) lead to
\begin{equation}\label{i5}
 \Vert \mathcal{L}_\varepsilon (\chi^iv^i)\Vert_{0,\Omega_+}
\lesssim \varepsilon^{1+\lambda}+\varepsilon e^{-\frac{b}{2\varepsilon}}.
\end{equation}
At this stage if we set
\begin{equation}
\label{i6}
U_+:= f_++\sum_{j=2}^{N}\chi_j^bv_j^b+\chi^iv^i \quad
\text{in } \Omega_+,
\end{equation}
then we may write (since $\mathcal{L}_\varepsilon u^\varepsilon_+=f_+$)
\[
\mathcal{L}_\varepsilon (u^\varepsilon_+-U_+)=
\varepsilon^2\Delta f_+-\mathcal{L}_\varepsilon (\sum_{j=2}^{N^+}\chi_j^bv_j^b)
-\mathcal{L}_\varepsilon (\chi^iv^i).
\]
And by (\ref{b5}) and (\ref{i5}), we arrive at
\begin{equation}\label{i7}
 \Vert\mathcal{L}_\varepsilon (u^\varepsilon_+-U_+)\Vert_{0,\Omega_+}
\lesssim \varepsilon^{1+\lambda}+\varepsilon e^{-\frac{b}{2\varepsilon}}.
\end{equation}
At this stage we can say that    $U_+$ approaches $u_\varepsilon^+$ in the 
interior of  $\Omega_+$,  satisfies the Dirichlet boundary condition in 
the interior  of $\Gamma_j$, $j=2,\dots, N$
  and the correct interface condition in the interior  of $\Sigma$.
But the correct boundary/interface conditions are not satisfied near the 
corners $A_j$. Therefore, corner correctors have to be introduced.

\subsection{Corner correctors}
For all $j=1,\dots, N$ consider polar coordinates   $( r_j,\theta_j) $
centered at $ A_j $  and such that  $\Gamma_j\subset \{(r_j,0), r_j>0\}$
 and therefore
$$
\Gamma_{j-1}\subset \{(r_j\cos \omega_j,r_j\sin \omega_j), r_j>0\}
$$
 (here and below the index are considered modulo $N$,
 i.e. ${ }_0={ }_N$).
Denote
\begin{gather*}
S_j=\{(r_j,\theta_j), \; r_j>0, 0<\theta_j<\omega_j\},\\
\tilde\Gamma_{j-1}=\{(r_j,\omega_j), \; r_j>0\},\\
\tilde\Gamma_{j}=\{(r_j\cos \omega_j,r_j\sin \omega_j), \; r_j>0\},
\end{gather*}
 and let $ R_j>0 $ be fixed sufficiently small so that
\begin{gather*}
\mathop{\rm supp}\chi_{j-1}^b\cap \mathop{\rm supp}\chi_j^b\cap S_j\subset
B(A_j,\frac{R_j}{2}), \\
B(A_j,R_j)\cap B(A_k,R_k)=\emptyset \quad \text{if } k\neq j.
\end{gather*}
To each vertex $ A_j $ we associate a radial cut-off function $\chi_j^c$
such that
$$
 \chi_j^c(r)=\begin{cases}
1 &\text{if } r<\frac{R_j}{2}, \\ 0 &\text{if } r>R_j.
\end{cases}
$$
In the sector $S_j, $ according to the definition of  the function $  U_+ $
we may write
\begin{equation}\label{e13}
U_+(x,y)=f_+(x,y)+ \chi_{j-1}^b(x_{j-1},y_{j-1})v^b_{j-1}(x_{j-1},y_{j-1})
+\chi_j^b(x_j,y_j)v^b_j(x_j,y_j),
\end{equation}
where for shortness we write $v^b_1=v^i$, $\chi_1^b=\chi^i$.
By construction of the boundary layers $ v_j^b $, we then have
\[
U_+\big|_{\partial S_j}=\begin{cases}
\chi_j^bv^b_j &\text{on }  \tilde  \Gamma_{j-1},
\\
\chi_{j-1}^bv^b_{j-1} &\text{on }  \tilde \Gamma_j.
\end{cases}
\]
Now we introduce   the   changes   of coordinates
\begin{align*}
\Psi_j &: (r_j,\theta_j)\longmapsto (x_j,y_j)=(r_j\cos\theta_j,r_j\sin\theta_j),
\\
\Phi_j &: (x_j,y_j)\longmapsto (x_{j-1},y_{j-1}).
\end{align*}
Using the fact that $\tilde \Gamma_{j-1}$ (resp.
$\tilde  \Gamma_j$) is parametrized by $(x_j,y_j)=(r_j\cos\omega_j$,
$r_j\sin\omega_j)$ (resp.   $ (x_j,y_j)=(r_j, 0) $)
and using the definition of $v_j^b$
and $v^i$, we see that
\[
U_+\mid_{\partial S_j}=\begin{cases}
g_j^1(r_j) \exp\big(-\frac{r_j\sin{\omega_j}}{\varepsilon}\big))
  &\text{on }  \tilde  \Gamma_{j-1}, \\
g_j^2(r_j) \exp\big(-\frac{r_j\sin{\omega_j}}{\varepsilon}\big)
  &\text{on }  \tilde \Gamma_j,
\end{cases}
\]
where, except in the case $j=k=1$ and $j=k=2$, the functions $ g_j^k $
are smooth, while in the exceptional case,
due to (\ref{e71}) and (\ref{e61}),  we have
\begin{align}
 \label{defg11}
g^1_1(r_1)&= g^1_{1,r}(r_1)+g^1_{1,s}(r_1), \\
 \label{defg22}
g^2_2(r_2)&= g^2_{2,r}(r_2)+g^2_{2,s}(r_2),\\
g^1_{1,r}(r_1)&=
\chi^i\circ \Psi_1(r_1,\omega_1)   v_r^i(r_1\cos\omega_1,0),
\\
g^1_{1,s}(r_1)&= \chi^i\circ \Psi_1(r_1,\omega_1) \eta_1(r_1)
\sum_{\lambda_k\in \Lambda_1, 0<\lambda_k<2}C_{1,\lambda_k}
S_{1,\lambda_k}(r_1,\omega_1),
\nonumber \\
g^2_{2,r}(r_2)&=\chi^i\circ\Phi_2^{-1}\circ \Psi_2(r_2,\omega_2)
 v_r^i (-r_2\cos\omega_2+l_1, 0),
\nonumber\\
g^2_{2,s}(r_2)&= \chi^i\circ\Phi_2^{-1}\circ \Psi_2(r_2,\omega_2)
\eta_2(r_2)\sum_{\lambda_k\in \Lambda_2, 0<\lambda_k<2}C_{2,\lambda_k}
 S_{2,\lambda_k}(r_2,\omega_2).
\nonumber
\end{align}

  The boundary condition imposed at $ v_j^b $  on $\Gamma_j $
implies   $v_j^b(A_j)=v^b_{j-1}(A_j)=-f_+(A_j)$, $j=3,\dots,N$.
On the other hand $ u_-^0\in H^1(\Omega_-)$  and  satisfies the
Dirichlet condition on $\partial\Omega_-\setminus\Sigma $. By the  continuity
of $u_-^0$ (due to the expansion (\ref{e7})) we get
$u_-^0(A_1)=u_-^0(A_2)=0$,
and consequently $v^b_1 (A_j)=-f_+(A_j), j=1,2$.
All together the next compatibility conditions are satisfied
\begin{equation}\label{e20}
g_j^1(0) =g_j^2(0) \quad  \forall  j=1,\dots, N.
\end{equation}
Now we look for
explicit functions $ u_j^c $ defined in the cone $S_j$
and satisfying the boundary conditions
\begin{gather*}
 u_j^c=-g_j^1 \quad\text{on } \tilde \Gamma_{j-1}, \\
 u_j^c=-g_j^2 \quad\text{on } \tilde \Gamma_{j}.
 \end{gather*}
Since the   term $g^1_{1,r}$ and $g^2_{2,r}$ are sufficiently smooth
(namely $H^{5/2}$), they can be treated as the functions $ g_j^k $,
for $j>2$. As a consequence we split
$ u_j^c =u_{j,r}^{c}+u_{j,s}^{c}$, where
   $ u_{j,s}^{c}=0 $ for $ j\neq 1,2$ and
\begin{gather} \label{bcu1sc}
u_{1,s}^{c}(r_1,\theta_1)=\begin{cases}
0 & \text{if }  \theta_1=0, \\
-g^1_{1,s}(r_1) &\text{if } \theta_1=\omega_1,
\end{cases}\\
\label{bcu2sc}
u_2^{1c}(r_2,\theta_2)=\begin{cases}
0 & \text{if }  \theta_2=\omega_2, \\
-g^2_{2,s}(r_2) &\text{if } \theta_2=0,
\end{cases}
\end{gather}
and
\begin{gather} \label{bcujrc1}
u_{j,r}^{c}=-\hat g_j^1 \quad \text{on } \tilde \Gamma_{j-1},  \\
u_{j,r}^{c}=-\hat g_j^2 \quad \text{on } \tilde \Gamma_{j}. \label{bcujrc2}
 \end{gather}
where $\hat g_j^k=g_j^k$ except if $j=k=1$
and $j=k=2$;
in that last cases, we take
$\hat g_1^1=g_{1,r}^1$,  $\hat g_2^2=g_{2,r}^2$.

For our purpose, we introduce the functions
$$
\sigma_{j,\lambda_k}(r_j,\theta_j)=
\begin{cases}
\frac{r_j^{\lambda_k} \sin(\lambda_k(\varphi_j+\omega_j)}{\omega_j}
  \theta_j & \text{if }  \sin(\lambda_k\omega_j)=0, \\[3pt]
\frac{S_{j,\lambda_k}(r_j,\omega_j)}{\sin\lambda_k\omega_j}
\sin(\lambda_k\theta_j) & \text{if }  \sin(\lambda_k\omega_j)\neq 0,
\end{cases}
$$
so that it fulfils $\sigma_{j,\lambda_k}(r_j,0)=0 $ and
$ \sigma_{j,\lambda_k}(r_j,\omega_j)=S_{j,\lambda_k}(r_j,\omega_j)$.
Note that the first choice is also valid  in the (generic) case
$\sin(\lambda_k\omega_j)\neq 0$, but in this case the second choice
gives rise to  a harmonic function.


\begin{lemma} \label{lu1c} Let
\begin{gather} \label{lu1c1}
u_{1,s}^{c}(r_j,\theta_j)=
-\chi^i\circ \Psi_j(r_1,\omega_1) \eta_1(r_1)
\sum_{\lambda_k\in \Lambda_1, 0<\lambda_k<2}C_{1,\lambda_k}
\sigma_{1,\lambda_k},
\\
u_{2,s}^{c}(r_j,\theta_j)=
-\chi^i\circ\Phi_j^{-1}\circ \Psi_j(r_2,\omega_2)\eta_2(r_2)
\sum_{\lambda_k\in \Lambda_2, 0<\lambda_k<2}C_{2,\lambda_k}
\sigma_{2,\lambda_k}.
\end{gather}
Then they respectively satisfy (\ref{bcu1sc})
and (\ref{bcu2sc}) and by setting  $\alpha_j=\sin\omega_j$,
\begin{equation}\label{estusc}
\Vert e^{-\frac{\alpha_j r_j}{\varepsilon}} u_{j,s}^{c}\Vert_{0,S_j}
+\varepsilon \Vert e^{-\frac{\alpha_j r_j}{\varepsilon}}\nabla u_{j,s}^{c}\Vert_{0,S_j}
\lesssim \varepsilon^{1+\lambda}, \quad j=1,2.
\end{equation}
 Moreover
$$
\Delta (\chi_j^c e^{-\frac{\alpha_j r_j}{\varepsilon}}  u_{1,s}^{c}(r_j,\theta_j))
\in L^p(S_j),
$$
  for all $p\in [1,\frac{2}{2-\lambda})$, where
$\lambda=\min_{k=1,2}\min\{\lambda_k: \lambda_k\in \Lambda_k\}$.
\end{lemma}

\begin{proof} For  simplicity, let us set
 \begin{gather*}
 \hat \chi^i(r_1)=
\chi^i\circ \Psi_1(r_1,\omega_1) \eta_1(r_1) \quad \text{if } j=1,
\\
\hat \chi^i(r_2)=\chi^i\circ\Phi_2^{-1}\circ \Psi_2(r_2,\omega_2)
\eta_2(r_2)  \quad \text{if }  j=2.
\end{gather*}
Since the function
 $ e^{-\frac{r}{\varepsilon}\alpha_j}D^{\gamma} \sigma_{j,\lambda_k}$ 
behaves like
  $e^{-\frac{r}{\varepsilon}\alpha_j} r_j^{\lambda_k-\vert\gamma\vert}$
at $ 0 $ and at $\infty$, we have
 \begin{equation} \label{lu1c2}
 \Vert \hat\chi^ie^{-\frac{r}{\varepsilon}\alpha_j} D^{\gamma}\sigma_{j,\lambda_k}
\Vert_{0,S_j}
\lesssim \Vert \hat\chi^ir^{\lambda_k-|\gamma|}e^{-\frac{r_j}{\varepsilon}
\alpha}\Vert_{0, S_j}.
\end{equation}
For   $\vert\gamma\vert \leq 1< \lambda_k+1$, by the scaling
$ \rho_j=\frac{r_j}{\varepsilon}$, we obtain
\begin{equation} \label{lu1c3}
\begin{aligned}
\Vert \hat\chi^ir^{\lambda_k-\gamma}e^{-\frac{r_j}{\varepsilon}\alpha}
\Vert^2_{0, S_j}
&\lesssim \int_0^\infty  r^{2(\lambda_k-|\gamma|)}e^{-2\frac{r_j}{\varepsilon}
\alpha} r\,dr
\\
&=\varepsilon^{2(\lambda_k-|\gamma|+1)}\int_0^\infty
\rho^{2(\lambda_k-\gamma)}e^{-2\rho\alpha} \rho\,d\rho
\\
&\lesssim \varepsilon^{2(\lambda_k-|\gamma|+1)}.
\end{aligned}
\end{equation}
 The estimate  (\ref{estusc})  follows directly
from (\ref{lu1c2}) and (\ref{lu1c3}).
The regularity of
$\Delta (\chi_j^c e^{-\frac{\alpha_j r_j}{\varepsilon}} u_{1,s}^{c}(r_j,\theta_j))
\in L^p(S_j)$ is proved in a similar manner.
\end{proof}


\begin{lemma} \label{lu2c}
There exists $u_{j,r}^{c}\in H^1(S_j) $  satisfying
(\ref{bcujrc1}) and (\ref{bcujrc2})
 and such that
\begin{equation}\label{esturc}
\Vert \chi_j^ce^{-\frac{\alpha_j r_j}{\varepsilon}} u_{j,r}^{c}\Vert_{0,S_j}
+\varepsilon \Vert \chi_j^c e^{-\frac{\alpha_j r_j}{\varepsilon}}\nabla u_{j,r}^{c}
\Vert_{0,S_j}
\lesssim\varepsilon.\end{equation}
Moreover
$$
\Delta (\chi_j^c e^{-\frac{\alpha_j r_j}{\varepsilon}}
u_{1,r}^{c}(r_j,\theta_j))\in L^p(S_j),
$$
 for all $p\in [1,2)$.
\end{lemma}

\begin{proof} We simply take
\[
u_{j,r}^{c}(r,\theta)=(\hat g_j^1(r)-\hat g_j^2(r))
\frac{\theta}{\omega_j}+\hat g_j^2(r),
\]
which clearly satisfies  (\ref{bcujrc1}) and (\ref{bcujrc2}).
As  $\hat g_j^1 \in \tilde H^{\frac{5}{2}}(\tilde \Gamma_{j-1}) $,
$\hat g_j^2 \in \tilde H^{\frac{5}{2}}(\tilde \Gamma_{j}) $
 and are equal to zero for $ r>R_j$, we deduce that
 $\chi_j^c u_{j,r}^{c}, \chi_j^c\frac{\partial u_{j,r}^{c}}{\partial r}
 \in L^\infty(S_j)  $  and
$$
\chi_j^c\frac{1}{r}\frac{\partial u_{j,r}^{c}}{\partial \theta}
=\chi_j^c(\frac{\hat g_j^1(r)-\hat g_j^1(0)}{r}\frac{1}{\omega_j}
-\frac{\hat g_j^2(r)-\hat g_j^2(0)}{r}\frac{1}{\omega_j})\in L^\infty(S_j).
$$
Consequently it holds
$$
\Vert\chi_j^c e^{-\frac{\alpha_j r}{\varepsilon}}u_{j,r}^{c}\Vert_{0,S_j}
+\varepsilon\Vert\chi_j^c e^{-\frac{\alpha r}{\varepsilon}}\nabla u_{j,r}^{c}
\Vert_{0,S_j}\lesssim \Vert e^{-\frac{\alpha r}{\varepsilon}}\Vert_{0,S_j}.
$$
By the change of variable $ \rho=\frac{r}{\varepsilon}$, one has
$ \Vert e^{-\frac{\alpha r}{\varepsilon}}\Vert_{0,S_j}\lesssim \varepsilon$
and the estimate  (\ref{esturc})    follows.
The second assertion is proved similarly.
\end{proof}

\section{The full decomposition \label{sectfullexp}}

We are now ready to formulate the main result of this paper.

\begin{theorem}\label{dim2}
Assume that $ f_+ $ is the restriction to $\Omega_+$
of a smooth function
$\tilde f_+\in \mathcal{C}^\infty (\mathbb{R}^2) $ and that $\Omega_+$ is convex.
Write for shortness
$$
U_c=\sum_{k=1}^N\chi_k^c e^{-\sin\omega_k\frac{ r_k}{\varepsilon}}u_k^{c}.
$$
Then the  unique solution
 $u^{\varepsilon}\in H^1_0(\Omega)$ of (\ref{e1})
 admits the splitting
\begin{equation}\label{the1}
\begin{gathered}
u_+^{\varepsilon} = f_++\sum_{j=2}^{N}\chi_j^bv_j^b+\chi^iv^i
+U_c+r_+^\varepsilon \quad \text{in }   \Omega_+,
\\
u_-^{\varepsilon} =  u_-^0  +r_-^\varepsilon \quad \text{in }   \Omega_-,
\end{gathered}
\end{equation}
where $r^{\varepsilon}\in H^1_0(\Omega)$ is the variational solution of
\begin{equation}\label{varreps}
\begin{aligned}
&\int_{\Omega_+}  (\varepsilon^2\nabla r_+^\varepsilon\cdot\nabla v_+
+r_+^\varepsilon  v_+)+ \int_{\Omega_-}  (\nabla r_-^\varepsilon\cdot\nabla v_-
+r_-^\varepsilon  v_-)\\
&=\int_{\Omega_+} f^\varepsilon v-\int_{\Sigma} h^\varepsilon v
 -\int_{\Omega_+}  (\varepsilon^2\nabla U_c\cdot\nabla v_+ +U_c v_+),
\quad \forall v\in H^1_0(\Omega),
\end{aligned}
\end{equation}
where $f^\varepsilon=\mathcal{L}_\varepsilon(u_+^\varepsilon-U_+)$
and $ h^\varepsilon=\varepsilon^2\frac{\partial }{\partial \nu}(f_+-U_+)$.
Moreover,
\begin{equation}\label{the2}
\varepsilon\Vert \nabla r_+^\varepsilon\Vert_{0,\Omega_+}
+ \Vert r_+^\varepsilon\Vert_{0,\Omega_+}
+\Vert r_-^\varepsilon\Vert_{1,\Omega_-}\lesssim \varepsilon .
\end{equation}
\end{theorem}

\begin{proof}
By construction, $r^{\varepsilon}$ clearly belongs to $H^1_0(\Omega)$,
and satisfies $\Delta r_\pm^{\varepsilon}\in L^p(\Omega_\pm)$, for some
$p\in (1,2)$.
Therefore applying \cite[Theorem 1.5.3.11]{grisvard:85a}, we may write
\begin{equation}\label{varreps1}
\begin{aligned}
&\int_{\Omega_+}  (\varepsilon^2\nabla r_+^\varepsilon\cdot\nabla v_+
+r_+^\varepsilon  v_+)+ \int_{\Omega_-}  (\nabla r_-^\varepsilon\cdot\nabla v_-
+r_-^\varepsilon  v_-) \\
&= \int_{\Omega_+} \mathcal{L}_\varepsilon r_+^\varepsilon v_+
 +\langle\varepsilon^2\frac{\partial r_+^\varepsilon}{\partial \nu}
 -\frac{\partial r_-^\varepsilon}{\partial \nu},
v\rangle_{\tilde H^{\frac12}(\Sigma)^\star-\tilde H^{\frac12}(\Sigma)},
\quad  \forall v\in \mathcal{D}(\Omega).
\end{aligned}
\end{equation}
We remark that the splitting (\ref{the1}) means that
$$
r_+^\varepsilon=u_+^\varepsilon-U_+-U_c.
$$
Since Lemmas \ref{lu1c} and \ref{lu2c} guarantees that
$U_c\in H^1(\Omega_+)$
and  $\Delta U_c\in L^p(\Omega_+)$, for some $p\in (1,2)$,
again the application  of \cite[Theorem 1.5.3.11]{grisvard:85a} yields
\[
\int_{\Omega_+} \mathcal{L}_\varepsilon U_c v_+
=\int_{\Omega_+}  (\varepsilon^2\nabla U_c\cdot\nabla v_+ +U_c v_+)
-\langle\varepsilon^2\frac{\partial U_c}{\partial \nu},
v\rangle_{\tilde H^{\frac12}(\Sigma)^\star-\tilde H^{\frac12}(\Sigma)},
\quad \forall v\in \mathcal{D}(\Omega).
\]
Inserting this expression in (\ref{varreps1}), we obtain
(\ref{varreps}) since $\mathcal{D}(\Omega)$ is dense in $H^1_0(\Omega)$.

Now taking $v=r^\varepsilon$ in (\ref{varreps}), applying Cauchy-Schwarz's
inequality and a trace theorem (in $\Omega_-$), we get
\begin{equation}
\label{estpreliminaire}
 \varepsilon\Vert \nabla r_+^\varepsilon\Vert_{0,\Omega_+}+\Vert r_+^\varepsilon 
 \Vert_{0,\Omega_+}
+ \Vert r_-^\varepsilon\Vert_{1,\Omega_-}
\lesssim\Vert f^\varepsilon\Vert_{0,\Omega_+}
+ \Vert h^\varepsilon\Vert_{0,\Sigma}
+\varepsilon\Vert \nabla U_c\Vert_{0,\Omega_+}+\Vert U_c \Vert_{0,\Omega_+}.
\end{equation}
The estimate
 (\ref{the2}) follows from this one  if we can show that
 each  term of this right-hand side
 is bounded by $\varepsilon$.
The first term is estimate with the help of (\ref{i7}).
For the second term, due to (\ref{i6}), we may write
$$
h^\varepsilon=-\varepsilon^2\frac{\partial}{\partial \nu} (f_++ \chi_N^b v_N^b
+ \chi^i v^i).
$$
Now  by (\ref{singulariteDN}) we remark that
\begin{gather*}
| \frac{\partial}{\partial \nu} f_+| \lesssim 1, \\
\vert \frac{\partial}{\partial \nu} (\chi_N^b v_N^b )\vert
=\vert \frac{\partial \chi_N^b}{\partial \nu}  v_N^b 
+\frac{\partial v_N^b}{\partial \nu}
 \chi_N^b\vert\lesssim \frac{1}{\varepsilon},
\\
\vert \frac{\partial}{\partial \nu} (\chi^i v^i)\vert
= \vert \frac{\partial \chi^i}{\partial \nu}  v^i 
 +\frac{\partial v_i}{\partial \nu}
 \chi^i\vert\lesssim \frac{1}{\varepsilon}.
\end{gather*}
These estimates lead to
$\Vert h^\varepsilon\Vert_{0,\Sigma}\lesssim\varepsilon$.
Finally for the last terms of the right-hand side, using (\ref{estusc}),
  (\ref{esturc}) and Leibniz's rule, we get
$$
\varepsilon\Vert \nabla U_c\Vert_{0,\Omega_+}+\Vert U_c \Vert_{0,\Omega_+}
\lesssim\varepsilon.
$$
\end{proof}

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\end{document}