\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small 2005-Oujda International
Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 21--33.\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}}
\setcounter{page}{21}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil Limit behavior of an oscillating
thin layer] {Limit behavior of an oscillating thin layer}

\author[A. Ait Moussa, J. Messaho \hfil EJDE/Conf/14 \hfilneg]
{Abdelaziz Ait Moussa, Jamal Messaho}  % in alphabetical order

\address{Abdelaziz Ait Moussa and Jamal Messaho\newline
D\'epartement de math\'ematiques et informatique, Facult\'e des
sciences, Universit\'e Mohammed 1er, 60000 Oujda, Morocco.}
\email{moussa@sciences.univ-oujda.ac.ma}
\email{j\_messaho@yahoo.fr}

\date{}
\thanks{Published September 20, 2006.}
\subjclass[2000]{35B40, 82B24, 76M50} 
\keywords{Limit behavior; epi-convergence method; limit problems}

\begin{abstract}
 We study the limit behavior of a thermal problem, of a containing
 structure, an oscillating thin layer of thickness and conductivity
 depending of $\varepsilon$. We use the the epi-convergence method
 to find the limit problems with interface conditions. The obtained
 results are tested numerically.
\end{abstract}

\maketitle \numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}

\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
%\renewcommand{\qedsymbol}{$\boxtimes$}
\section{Introduction}

 In mathematical physics, one meets several
kinds of boundary problems, the heat conduction, electrostatic,
electromagnetic, mechanical of the continuous mediums, where the
unknown $u $ satisfies the transmission conditions on the surface of
separation between two domains $\Omega_1 $ and $\Omega_2$:
\begin{gather}
u_{|_{\Omega_1}}=u_{|_{\Omega_2}} \label{1}\\
\sigma_1|\nabla u |^{p-2}{\frac{\partial u}{\partial
n}}\big|_{\Omega_1} = \sigma_2|\nabla u |^{p-2}{\frac{\partial
u}{\partial n}}|_{\Omega_2} \label{2}
\end{gather}
where $p>1$ and $n $ represents the outward normal vector to the
surface of separation,
 $\sigma_1 $ and $\sigma_2 $ are the associated constants to each domain $\Omega_1$ and $\Omega_2$ respectively. The boundary conditions of
 type \eqref{1} and \eqref{2} are met in
 thermal conductivity problems, where $\sigma_1$ and $\sigma_2$
designate the
 conductivities of two bodies. In the
 electrostatic or magnetostatic problems $\sigma_1$ and
$\sigma_2$
 are the dielectric or permeability constants respectively. A
 transmission problem with the conditions of type (\ref{1}) and
(\ref{2}) and
 $p=2 $, was studied by   Sanchez-Palencia in
 \cite{sanch}.

Our aim in this work is to study the limit behavior of solutions of
a thermal conductivity problem, this last is in a structure
containing an oscillating layer of thickness and conductivity
depending of $\varepsilon $, $\varepsilon$ being a parameter
intended to tend towards $0$.


A similar problems are found in Brillard and al in \cite{brd1}. The
vectorial case one finds it in Ait Moussa and al, and Brillard and
al in \cite{aitmoussa, brd2}.

This paper is organized in the following way. In section
\ref{sec:2}, one expresses the problem to study, and one defines
functional spaces for this
 study in the section \ref{sec:3}. In the section
 \ref{sec:4}, one studies the problem \eqref{Pe}.
 The section \ref{cal:epii} is
 reserved to the determination of the limits problems. Finally in the
 section \ref{sim}, one will give a numerical test illustrating the
 obtained theoretical results.

\section{Position of the problem}\label{sec:2}

One considers a problem of nonlinear thermal conduction in a body
which occupies a bonded domain $\Omega\subset\mathbb{R}^3 $, with a
Lipschitz border $\partial\Omega $, composed of a layer
$B_\varepsilon $, with oscillating border $\Sigma^{^\pm}_\varepsilon
$, of average interface $\sigma$, of very high conductivity, and a
remaining region $\Omega_\varepsilon $ with a constant conductivity
( see figure \ref{fig:1}). The body occupying the domain $\Omega $,
is subject to an outside temperature $f $, $f\colon
\Omega\to\mathbb{R} $, and cooled at the boundary $\partial\Omega $.
 The equations of the problem are:
\begin{equation} \label{Pe}
 \begin{gathered}
 \mathop{\rm div}(|\nabla u^\varepsilon|^{p-2}\nabla u^\varepsilon)+f
 =0 \quad \mbox{in }\Omega_\varepsilon ,\\
 \frac{1}{\varepsilon^{\alpha}}\mathop{\rm div}(|\nabla
 u^\varepsilon|^{p-2}\nabla u^\varepsilon)+f=0 \quad\mbox{in
}B_\varepsilon,
 \\
 [ u^{\varepsilon}] =0
 \quad \mbox{on }\Sigma^{^\pm}_\varepsilon,\\
 |\nabla u^\varepsilon|^{p-2}\frac{\partial u^{\varepsilon}}
 {\partial n}\big|_{\Omega_\varepsilon}=\frac{1}{\varepsilon^{\alpha}}
 |\nabla u^\varepsilon|^{p-2}\frac{\partial u^{\varepsilon}}{\partial
 n}\big|_{{B_\varepsilon}} \quad \mbox{on }\Sigma^{^\pm}_\varepsilon,\\
 u^\varepsilon = 0 \quad\mbox{on }\partial\Omega,
\end{gathered}
\end{equation}
where $n$ the outward normal to $\partial \Omega$,
 $p>1$ and $\alpha \geq 0$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.6\textwidth,height=0.6\textheight,angle=-90]{fig1}
 \caption{Domain $\Omega$.} \label{fig:1}
\end{center}
\end{figure}

Where $\varepsilon $ being a positive parameter intended to tend
towards zero and $\varphi_\varepsilon$
 is a bounded real function, $]0,\varepsilon [^ {2}$-periodic.

\section{Notation and functional setting}\label{sec:3}

Here is the notation that will be used in the sequel:\\
 $x=(x',x_3)$ where $x'=(x_1,x_2)$, $\lambda=1,2$,
$\nabla'=(\frac{\partial}{\partial x_1}$,
$\frac{\partial}{\partial x_2}), Y=]0,1[\times]0,1[$,\\
$\varphi\colon \mathbb{R}^{2}\to [a_1,a_2]$ where $\varphi$ is
$Y$-periodic and $a_2\geq a_1>0$,
$\varphi_{_\varepsilon}(x')=\varphi(\frac{x'}{\varepsilon})$,\\
$\frac{\partial \varphi}{\partial x_{\lambda}}\in
\mathcal{C}(\Sigma) \cap L^{^{\infty}}(\Sigma)$, $
 m(\varphi)= (\frac{1}{\int_{Y}dx'})\int_{Y}\varphi(x')dx',
$ $\displaystyle\eta(t)=\lim_{\varepsilon\to 0}\varepsilon^{1-t}$
with $t\geq 0$.

In the following $C $ will denote any constant with respect to
$\varepsilon $. Also we use the convention $0.+\infty=0$.

\subsection{Functional setting}

First, we introduce the Banach space
$V^\varepsilon=W^{1,p}_0(\Omega)$.
 Let
\begin{gather*}
V^{p}(\Sigma)=\Big\{ u\in W^{1,p}_0(\Omega) : u\big|_{{\Sigma}}\in
W^{1,p}
(\Sigma)\Big\},\\
V^{C}(\Sigma)=\Big\{ u\in W^{1,p}_0(\Omega) :
u\big|_{{\Sigma}}=C\Big\}.
\end{gather*}
The set $V^{C}(\Sigma)$ is a Banach space with the norm of
$W^{1,p}_0(\Omega)$. we show easily that $\displaystyle
V^{p}(\Sigma)$ is a Banach space with the norm
$$u\mapsto \norm{\nabla u}_{L^{^p}(\Omega)^3}
+\norm{ \nabla' u_{|_{\Sigma}}}_{L^{^p}(\Sigma)^2}.
$$
Let
\begin{gather*}
\mathbb{G}^{\alpha}=\begin{cases}\Big\{ u\in W^{1,p}_0(\Omega):
\eta(\alpha)u\big|_{{\Sigma}}\in W^{1,p}(\Sigma)\Big\}
 &\text{if } \alpha\leq 1 ,\\
 V^C&\text{if }\alpha>1.\end{cases}
 \\
\mathbb{D}^{\alpha}=\begin{cases}
\mathcal{D}({\Omega})&\text{if }\alpha\leq 1 ,\\
\Big\{u\in \mathcal{D}({\Omega}): u\big|_{{\Sigma}}=C
\Big\}&\text{if } \alpha>1.\end{cases}
\end{gather*}
It is known that $\displaystyle
\overline{\mathbb{D}^{\alpha}}=\mathbb{G}^{\alpha}$.

 Our goal in this work, is to study the problem
\eqref{Pe} and its limit behavior.

\section{Study of the problem \eqref{Pe}}\label{sec:4}

 The problem \eqref{Pe} is equivalent to the minimization problem
\begin{equation} \label{Pie}
\inf_{v\in V^{\varepsilon}}\Big\{\frac{1}{p}
 \int_{\Omega_\varepsilon} \abs{\nabla v}^p
+\frac{1}{p\varepsilon^{\alpha}}\int_{B_\varepsilon} \abs{\nabla
v}^{p} -\int_{\Omega}f.v\Big\}.
\end{equation}

\begin{proposition}\label{existe}
 For $f\in L^{p'}(\Omega)$,
 the problem \eqref{Pie} admits
 an unique solution $u^\varepsilon $ in $V^\varepsilon $.
\end{proposition}

The proof of this proposition is based on classical convexity
arguments see for example \cite{brez}.

\begin{lemma} \label{lemme1}
For every $f\in L^{p'}(\Omega)$, the family
$(u^\varepsilon)_{\varepsilon>0}$ satisfies:
\begin{gather}
\norm{\nabla u^\varepsilon}_{{L^{p}(\Omega_\varepsilon)}}^p
\leq C\label{assert:1},\\
\frac{1}{\varepsilon^{\alpha}}\norm{\nabla
u^\varepsilon}_{{L^{^p}(B_\varepsilon)}}^p \leq C.\label{assert:2}
\end{gather} Moreover $u^\varepsilon$ is bounded in
$W^{1,p}_0(\Omega)$.
\end{lemma}

\begin{proof}
 Since $u^\varepsilon$ is the solution of the problem \eqref{Pie},
 we have
 \[
 \int_{\Omega_{\varepsilon}}
 \abs{\nabla u^\varepsilon}^{p-2}\nabla
 u^\varepsilon\nabla v+\frac{1}{\varepsilon^{\alpha}}
 \int_{B_{\varepsilon}}\abs{\nabla u^\varepsilon}^{p-2}\nabla
 u^\varepsilon\nabla v=\int_{\Omega}f v, \quad \forall v\in
 V^\varepsilon.
 \]
 In particular, by taking $v=u^\varepsilon$, one obtains
\[
\norm{ \nabla u^\varepsilon}^p_{{L^{p}(\Omega_\varepsilon)}}
+\frac{1}{\varepsilon^{\alpha}}\norm{\nabla
u^\varepsilon}^p_{{L^p(B_\varepsilon)}}=\int_{{\Omega}}fu^\varepsilon.
\]
According to the inequalities of H\"older and Young, one has
\begin{align*}
\norm{ \nabla
u^\varepsilon}^p_{{L^{p}(\Omega_\varepsilon)}}+\frac{1}{\varepsilon^{\alpha}}\norm{\nabla
u^\varepsilon}^p _{{L^p(B_\varepsilon)}} &\leq C\norm{ \nabla
u^\varepsilon}_{{L^{p}(\Omega)}}\\
&\leq C(\norm{ \nabla
u^\varepsilon}_{{L^{p}(\Omega_\varepsilon)}}+\norm{\nabla
u^\varepsilon}_{{L^p(B_\varepsilon)}})\\
&\leq C+\frac{1}{p}\norm{ \nabla
u^\varepsilon}^p_{{L^{p}(\Omega_\varepsilon)}}+\frac{1}{p}\norm{\nabla
u^\varepsilon}^p_{{L^p(B_\varepsilon)}}\\
&\leq C+\frac{1}{p}\norm{ \nabla
u^\varepsilon}^p_{{L^{p}(\Omega_\varepsilon)}}+\frac{1}{\varepsilon^{\alpha}}\frac{1}{p}\norm{\nabla
u^\varepsilon}^p_{{L^p(B_\varepsilon)}}.
\end{align*}
So that
\[
\norm{\nabla
u^\varepsilon}^p_{{L^{p}(\Omega_\varepsilon)}}+\frac{1}{\varepsilon^{\alpha}}\norm{\nabla
u^\varepsilon}^p _{{L^p(B_\varepsilon)}} \leq C.
\]
 Therefore, one will have the assertions \eqref{assert:1}
 and \eqref{assert:2}.
It is clair that for a small enough $\varepsilon$, the solution
($u^\varepsilon$) is bounded in $W^{1,p}_0(\Omega)$.
\end{proof}

 Let us define the operator ``$m^\varepsilon$'' which transforms the
 definite functions $u$ on $B_\varepsilon $ into functions
 definite on $\Sigma$ by
 \begin{equation} \label{moyenne}
 {m^\varepsilon}u(x_1,x_2)=\frac{1}{2\varepsilon\varphi_{\varepsilon}}
\int_{-\varepsilon\varphi_{\varepsilon}}^{\varepsilon\varphi_{\varepsilon}}
 u(x_1,x_2,x_3)dx_3.
 \end{equation}

\begin{lemma}\label{lemme2}
 The operator $m^\varepsilon$ definite by \eqref{moyenne}
 is linear and bounded of $L^p(B_\varepsilon)$
 (respectively $W^{1,p}(B_\varepsilon) $) in
 $L^p(\Sigma)$ (respectively $W^{1,p}(\Sigma)$), with norm
$\leq C {\varepsilon}^{-\frac{1}{p}} $, moreover, for all $u\in
W^{1,p}(B_\varepsilon) $, one has
 \begin{equation} \label{fort0}
 \norm{m^\varepsilon u-u_{|_{\Sigma}}}_{|_{L^{^p}(\Sigma)}}^{^{p}}
 \leq C \varepsilon^{p-1}\int_{B_{{\varepsilon}}}\abs{\nabla
 u}^p.
 \end{equation}
 \end{lemma}

\begin{proof} One has
 \begin{equation} \label{etiq16}
 \int_{\Sigma}\abs{{m^{\varepsilon}}u}^p dx_1
 dx_2=\int_{\Sigma}(\frac{1}{2\varepsilon\varphi_\varepsilon})^p
\abs{\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 u dx_3}^{p}dx_1dx_2,
\end{equation}
 since $0<a_1\leq\varphi_\varepsilon\leq a_2$, and according to the
inequality of H\"{o}lder, (\ref{etiq16}) becomes
\begin{equation} \label{etiqq17}
 \begin{aligned}
 \int_{\Sigma}\abs{{m^{\varepsilon}}u}^p dx_1 dx_2
&\leq\int_{\Sigma}\frac{1}{2\varepsilon\varphi_\varepsilon}
\Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 \abs{u}^p dx_3\Big)dx_1dx_2 \\
&\leq \frac{1}{2\varepsilon a_1}\int_{\Sigma}
 \Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon
\varphi_\varepsilon}
 \abs{u}^p dx_3\Big)dx_1dx_2,
\end{aligned}
\end{equation}
 since $u\in L^p(B_\varepsilon)$ and \eqref{etiqq17}, it follows that
${m^{\varepsilon}}u\in L^p(\Sigma)$.
 Let $u\in{\mathcal{D}}(\overline{B_\varepsilon})$.
 One has
 \begin{align*}
 \frac{\partial }{\partial
 x_{\lambda}}(m^{\varepsilon}u)(x_1,x_2)
&= \frac{1}{2} \frac{\partial }{\partial
 x_{\lambda}}\Big( \int_{-1}^{1}
 u(x_1,x_2,x_3\varepsilon\varphi_\varepsilon)dx_{_3}\Big) \\
&=\frac{1}{2} \Big( \int_{-1}^{1}
 \frac{\partial u}{\partial
 x_{\lambda}}(x_1,x_2,x_3\varepsilon\varphi_\varepsilon) \\
&\quad + \varepsilon x_3\frac{\partial \varphi_\varepsilon}
{\partial x_\lambda} \frac{\partial u}{\partial
 x_3}(x_1,x_2,x_3\varepsilon\varphi_\varepsilon)dx_{_3}\Big) \\
 &= \frac{1}{2\varepsilon\varphi_\varepsilon} \Big(
\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 \frac{\partial u}{\partial
 x_{\lambda}}+
 (\frac{x_3}{\varepsilon\varphi_\varepsilon})(\varepsilon
 \frac{\partial \varphi_\varepsilon}{\partial x_\lambda})
 \frac{\partial u}{\partial
 x_3}dx_{_3}\Big).
 \end{align*}
So that
 \begin{align*}
 \int_{\Sigma}\big|\frac{\partial }{\partial
 x_{\lambda}}(m^{\varepsilon}u)\big|^p
 &=\int_\Sigma\big|\frac{1}{2\varepsilon\varphi_\varepsilon} \Big(
\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 \frac{\partial u}{\partial x_{\lambda}}+
 (\frac{x_3}{\varepsilon\varphi_\varepsilon})
(\varepsilon\frac{\partial \varphi_\varepsilon}{\partial
 x_\lambda})
 \frac{\partial u}{\partial x_3}dx_{_3}\Big)\big|^p\\
&\leq \frac{1}{2\varepsilon a_1}\int_\Sigma\Big(
\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 \big|\frac{\partial u}{\partial x_{\lambda}}+
 (\frac{x_3}{\varepsilon\varphi_\varepsilon})(\varepsilon\frac{\partial
 \varphi_\varepsilon}{\partial x_\lambda})
 \frac{\partial u}{\partial
 x_3}\big|^pdx_{_3}\Big).
\end{align*}
 However, $\frac{\partial \varphi}{\partial
 x_{_\lambda}}\in \mathcal{C}(\Sigma)\cap L^{^{\infty}}(\Sigma)$, then
$\varepsilon\frac{\partial \varphi_{{\varepsilon}}}{\partial
x_\lambda}$ is bounded, and therefore
\[
 \int_{\Sigma}\big|\frac{\partial }{\partial
 x_{\lambda}}(m^{\varepsilon}u)\big|^p
 \leq \frac{C}{\varepsilon }
 \int_{B_\varepsilon}
 \big(\big|\frac{\partial u}{\partial x_{\lambda}}\big|^p+
 \big|\frac{\partial u}{\partial x_3}\big|^p \big)dx_{_3}\\
 \leq \frac{C}{\varepsilon }
 \int_{B_\varepsilon}
 \abs{\nabla u}^p,
\]
 by density arguments, for all
 $u\in W^{1,p}(B_\varepsilon)$, one has
 \begin{eqnarray*}\label{w1ps}
 \int_{\Sigma}\big|\frac{\partial }{\partial
 x_{\lambda}}(m^{\varepsilon}u)\big|^p
 &\leq&\frac{C}{\varepsilon }
 \int_{B_\varepsilon}
 \abs{\nabla u}^p.
 \end{eqnarray*}
 Let $u\in \mathcal{D}(\overline{B_\varepsilon})$, so that
 \begin{equation}\label{etiq024}
\norm{{m^{\varepsilon}}u-{u}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}}^p=\int_{\Sigma}\Big|
\Big(\frac{1}{2\varepsilon\varphi_\varepsilon}
\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
u(x_1,x_2,x_3)dx_3\Big)-u (x_1,x_2,0)\Big|^pdx_1dx_2,
\end{equation}
 using the inequality of H\"{o}lder, \eqref{etiq024} becomes
 \begin{align*}\label{etiq25}
 \norm{{m^{\varepsilon}}u-{u}_{|_{\Sigma}}  }_{{L^p(\Sigma)}}^p
&\leq \frac{1}{2\varepsilon a_1} \int_{\Sigma}
 \Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon
 \varphi_\varepsilon}
 \abs{u(x_1,x_2,x_3)-u(x_1,x_2,0)}^pdx_3\Big) dx_1 dx_2 \\
 & \leq \frac{C}{\varepsilon } \int_{\Sigma}
\Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
 \big|\int_0^{x_3}\frac{\partial u}{\partial x_3}
 (x_1,x_2,t)dt\big|^pdx_3\Big) dx_1 dx_2 \\
 &\leq \frac{C}{\varepsilon} \int_{\Sigma}
 \Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon
 \varphi_\varepsilon}|x_3|^{p-1}
\Big(\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}
\big|\frac{\partial u}{\partial x_3}
 (x_1,x_2,t)\big|^p dt\Big)dx_3\Big) dx_1 dx_2 \\
 &\leq C\varepsilon^{p-1} \int_{\Sigma}
 \Big(
\int_{-\varepsilon\varphi_\varepsilon}^{\varepsilon\varphi_\varepsilon}\abs{\frac{\partial
u}{\partial x_3}
 }^pdx_3\Big) dx_1 dx_2 \\
 &\leq C\varepsilon^{p-1} \int_{B_\varepsilon}
 \abs{\nabla u }^pdx,
 \end{align*}
 by density arguments, one has for all $u\in W^{1,p}(B_\varepsilon)$
 \begin{equation}\label{etiq25}
 \norm{{m^{\varepsilon}}u-{u}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}}^p
 \leq C\varepsilon^{p-1} \int_{B_\varepsilon}
 \abs{\nabla u
 }^pdx.
\end{equation}
Hence the proof of lemma \ref{lemme2} is complete.
 \end{proof}

 \begin{lemma}\label{lemme3}
 Let $ (u^\varepsilon)_{{\varepsilon >0}}\subset
 V^\varepsilon$ which satisfies
 \eqref{assert:1} and \eqref{assert:2}. Then
\begin{equation} \label{etiq21'}
\norm{\nabla^{'}({m^{\varepsilon}}u^{\varepsilon})}_{{({L^p(\Sigma))^2}}}^p
\leq  C\varepsilon^{\alpha-1}.
\end{equation}
 Moreover $m^\varepsilon u^\varepsilon$ possess a bounded subsequence in
 $L^p(\Sigma)$.
 \end{lemma}

 \begin{proof}
 Thanks to lemma \ref{lemme2}, one has
\[
 \big\|\frac{\partial ({m^{\varepsilon}}u^{\varepsilon})}
 {\partial x_{\lambda}}\big\|_{{L^p(\Sigma)^2}}^p
 \leq C\varepsilon^{-1} \int_{B_\varepsilon}\abs{\nabla
 u^{\varepsilon}}^pdx\,.
\]
According to \eqref{assert:2}, one has
\[
\big\|\frac{\partial ({m^{\varepsilon}}u^{\varepsilon})}{\partial
x_{\lambda}}
 \big\|_{{L^p(\Sigma)^2}}^p
\leq C\varepsilon^{\alpha-1}.
\]

 Let us show that (${m^{\varepsilon}}u^\varepsilon$) is
 a bounded sequence in $L^p(\Sigma)$.
 From (\ref{fort0}), (see, lemma \ref{lemme2}), one has
\[
 \big\|{m^{\varepsilon}}u^\varepsilon-{u^\varepsilon}_{|_{\Sigma}}
 \big\|_{{L^p(\Sigma)}}^p
 \leq C\varepsilon^{p-1} \int_{B_\varepsilon}
 \abs{\nabla u^{\varepsilon} }^pdx.
\]
 According to \eqref{assert:2}, one obtains
\[
 \norm{{m^{\varepsilon}}u^\varepsilon-{u^\varepsilon}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}}^p \leq C\varepsilon^{\alpha+p-1}.
\]
 As $u^\varepsilon$ satisfies \eqref{assert:1} and \eqref{assert:2},
so $u^\varepsilon$ is
 bounded in $W^{1,p}_0(\Omega)$, it follows that there exists $u^*\in W^{1,p}_0(\Omega)$
 and a subsequence of $u^\varepsilon$, still denoted by
  $u^\varepsilon$,
 such that $u^\varepsilon \rightharpoonup
 u^*$ in $W^{1,p}_0(\Omega)$, so ${u^\varepsilon}_{|_{\Sigma}}$ is
 a bounded sequence in $L^p(\Sigma)$.
 Since
\begin{equation} \label{etiq29}
 \norm{{m^{\varepsilon}}u^\varepsilon}_{{L^p(\Sigma)}}
\leq
\norm{{m^{\varepsilon}}u^\varepsilon-{u^\varepsilon}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}}+\norm{{u^\varepsilon}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}},
\end{equation}
from (\ref{etiq29}), there exists a constant $C>0$ such that
$\norm{{m^{\varepsilon}}u^\varepsilon
 }_{{L^p(\Sigma)}}^p\leq C $.
 \end{proof}

 \begin{proposition} \label{prop1}
 The solution of the problem \eqref{Pie},
$(u^\varepsilon)_\varepsilon $, possess a subsequence
 weakly convergent toward an element $u^*$ in $W^{1,p}_0(\Omega) $
satisfying
\begin{enumerate}
 \item If $\alpha=1$:
 ${u^*}\big|_{{\Sigma}}\in W^{1,p}(\Sigma)$.
 \item If $\alpha>1$:
 $ {u^*}\big|_{{\Sigma}}=C$.
 \end{enumerate}
\end{proposition}

\begin{proof} According to lemma \ref{lemme1},
 the sequence $u^{\varepsilon}$ is bounded in
 $W^{1,p}_0(\Omega)$, it follows that there exists an element
 $u^*\in W^{1,p}_0(\Omega)$ and a subsequence of $u^{\varepsilon}$,
still denoted by $u^{\varepsilon}$ such that
 $u^{\varepsilon}\rightharpoonup u^*$ in $W^{1,p}_0(\Omega)$.
 One has
 \[
 \norm{{m^{\varepsilon}}u^\varepsilon-{u^\varepsilon}_{|_{\Sigma}}
 }_{{L^p(\Sigma)}}
 \leq  C\varepsilon^{\frac{\alpha+p-1}{p}}\quad\text{and }
 {u^\varepsilon}_{|_{\Sigma}}\rightharpoonup{u^*}_{|_{\Sigma}}\text{
 in }L^p(\Sigma).
\]
For $\alpha=1$, according to the evaluation (\ref{etiq21'}), the
sequence $\nabla' m^\varepsilon u^{\varepsilon}$
 possess a subsequence, still denoted by $\nabla' m^\varepsilon
u^{\varepsilon}$
 weakly convergent to an element $u^2$ in
 $L^{p}(\Sigma)^2$, as $m^\varepsilon u^{\varepsilon}\rightharpoonup
{u^*}_{|_{\Sigma}}\text{
 in }L^{p}(\Sigma)$, so one concludes that $m^\varepsilon
u^{\varepsilon}\rightharpoonup {u^*}_{|_{\Sigma}}\text{
 in }W^{1,p}(\Sigma)$ and
 $\nabla' {u^*}_{|_{\Sigma}}=u^2$.
 Hence ${u^*}_{|_{\Sigma}}\in W^{1,p}(\Sigma)$.

For $\alpha>1$, one shows, as in the case $\alpha=1$ and taking
$u^2=0$, that
 ${u^*}_{|_{\Sigma}}=C$.
Hence the proof of proposition \ref{prop1} is complete.
 \end{proof}

 The limit behavior of the problem \eqref{Pie}, will be derived with
the epi-convergence method,  (see definition \ref{def:epi}).

\section{Limit behavior}\label{cal:epii}
\setcounter{equation}{0}
 Let
 \begin{gather}\label{F0}
 F^{\varepsilon}(u)=
 \frac{1}{p}  \int_{\Omega_\varepsilon}
 \abs{\nabla u}^p
+\frac{1}{p\varepsilon^{\alpha}}\int_{B_\varepsilon}\abs{\nabla
u}^{p},\quad \forall u\in W^{1,p}_0(\Omega),\\
 G(u)=-\int_{{\Omega}}f u,\quad \forall u\in W^{1,p}_0(\Omega).
 \end{gather}
One denotes by $\tau_f$ the weak topology on $W^{1,p}_0(\Omega)$.

\begin{theorem}\label{cal:epi}
 According to the values of $\alpha$, there exists a functional
 $F^{\alpha}$ defined on $W^{1,p}_0(\Omega)$ with value in
$\mathbb{R}\cup\{+\infty\}$
 such that
$
 \tau_{_f}-\lim_{e}F^{\varepsilon}= F^{\alpha}\quad \mbox{in
}W^{1,p}_0(\Omega),
$
where the functional $F^{\alpha}$ is given by
\begin{enumerate}
\item If $0\leq\alpha<1$:
\[
 F^{\alpha}(u)= \frac{1}{p}
 \int_{\Omega}
 \abs{\nabla u}^p, \quad\forall u\in W^{1,p}_0(\Omega).
\]
 \item If $\alpha\geq 1$:
\[
 F^{\alpha}(u)=  \begin{cases}\displaystyle
 \frac{1}{p}  \int_{\Omega} \abs{\nabla u}^p
+\frac{2m(\varphi)\;\eta(\alpha)}{p}\int_{\Sigma}\abs{\nabla'
u_{|_{\Sigma}}}^{p}
 &\mbox{if } u\in \mathbb{G}^{\alpha},\\
 +\infty &\mbox{if }u\in W^{1,p}_0(\Omega)\setminus
\mathbb{G}^{\alpha}.
 \end{cases}
\]
 \end{enumerate}
\end{theorem}

\begin{proof}
(a) One is going to determine the upper epi-limit:
 Let $u\in \mathbb{G}^{\alpha}\subset W^{1,p}_0(\Omega)$,
 there exists a sequence $(u^n)$
 in
 $\mathbb{D}^{\alpha}$ such that
$$
u^n \to u \text{ in } \mathbb{G}^{\alpha}\mbox{, when } n\to
+\infty.
$$
So that $u^n \to u$ in $W^{1,p}_0(\Omega)$. Let $\theta$ be a smooth
function satisfying
\begin{gather*}
    \theta(x_3)=1 \text{ if }\abs{x_3}\leq 1,\; \theta(x_3)=0\text{ if }\abs{x_3}\geq
    2\; \text{ and }\abs{\theta'(x_3)}\leq 2 \;\forall x\in\mathbb{R},
\end{gather*}
and set
\begin{gather*}
    \theta_{\varepsilon}(x)=\theta(\frac{x_3}{\varepsilon\varphi_{\varepsilon}});
\end{gather*}
we define
\begin{gather*}
    u^{\varepsilon,n}=\theta_{\varepsilon}(x){u^n}_{|_{\Sigma}}+(1-\theta_{\varepsilon}(x))u^n,
\end{gather*}
One shows easily that $u^{\varepsilon,n}\in V^\varepsilon$ and
$u^{\varepsilon,n}\to
 u^n$ in
$\mathbb{G}^{\alpha}$, when $\varepsilon\to 0$.
 Since
\[
 F^\varepsilon(u^{\varepsilon,n})=
 \frac{1}{p} \int_{\Omega_\varepsilon}
 \abs{\nabla u^{\varepsilon,n}}^p
+\frac{1}{p\varepsilon^{\alpha}}\int_{B_\varepsilon}\abs{\nabla
u^{\varepsilon,n}}^{p},
\]
so that
\begin{align}
 F^\varepsilon(u^{\varepsilon,n})
&= \frac{1}{p} \int_{\abs{x_3}>2\varepsilon\varphi_{\varepsilon}}
 \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\abs{\nabla u^{\varepsilon,n}}^p +\frac{1}{p}
 \int_{\varepsilon\varphi_{\varepsilon}<\abs{x_3}<2\varepsilon\varphi_{\varepsilon}}
 \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\abs{\nabla u^{\varepsilon,n}}^p
+\frac{1}{p\varepsilon^{\alpha}}\int_{B_\varepsilon}\abs{\nabla
u^{\varepsilon,n}}^{p}\nonumber\\
 &= \frac{1}{p}  \int_{\abs{x_3}>2\varepsilon\varphi_{\varepsilon}}
 \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\abs{\nabla u^{n}}^p
 +\frac{1}{p}
 \int_{\varepsilon\varphi_{\varepsilon}<\abs{x_3}<2\varepsilon\varphi_{\varepsilon}}
 \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\abs{\nabla u^{\varepsilon,n}}^p
+\frac{2\varepsilon^{1-\alpha}}{p}\int_{\Sigma}\varphi_{{\varepsilon}}
\abs{\nabla' {u^{n}}_{|_{\Sigma}}}^{p}.\label{sup1}
 \end{align}
Since $\varphi_{\varepsilon}$ is bounded, one  verifies easily that
\begin{align}
 \lim_{\varepsilon\to 0}\Big\{ \frac{1}{p}
 \int_{\varepsilon\varphi_{\varepsilon}<\abs{x_3}<2\varepsilon\varphi_{\varepsilon}}
 \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\abs{\nabla
 u^{\varepsilon,n}}^p\Big\}=0.\label{sup2}
\end{align}
\begin{enumerate}
\item If $\alpha\leq 1$:
 Since  $\varphi_{_\varepsilon}\rightharpoonup^* m(\varphi)$
in $L^{\infty}(\Sigma)$ and  $\varepsilon^{1-\alpha}\to
\eta(\alpha)$, it follows that
\[
 \lim_{\varepsilon\to
0}\frac{2\varepsilon^{1-\alpha}}{p}\int_{\Sigma}\varphi_{{\varepsilon}}\abs{\nabla'
{u^{n}}_{|_{\Sigma}}}^{p}
 =\frac{2m(\varphi) \eta(\alpha)}{p}\int_{\Sigma}\abs{\nabla'
{u^{n}}_{|_{\Sigma}}}^{p}.
\]
 By passage to the upper limit, one has
 \begin{align*}
 \limsup_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon,n})
&= \limsup_{\varepsilon\to 0}\Big(\frac{1}{p}
 \int_{\abs{x_3}>2\varepsilon\varphi_{\varepsilon}}\!\!\!\!\!\!\!\!\!\!\!
 \abs{\nabla u^{n}}^p
+\frac{2\varepsilon^{1-\alpha}}{p}\int_{\Sigma}\varphi_{{\varepsilon}}\abs{\nabla'
{u^{n}}_{|_{\Sigma}}}^{p}\Big)\\
 &= \frac{1}{p} \int_{\Omega}
 \abs{\nabla
u^{n}}^p+\frac{2m(\varphi)\;\eta(\alpha)}{p}\int_{\Sigma}\abs{\nabla'
{u^{n}}_{|_{\Sigma}}}^{p}.
 \end{align*}
\item If $\alpha>1$: By passage to the upper limit, one has
 \begin{align*}
 \limsup_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon,n})
&= \limsup_{\varepsilon\to 0}\Big(\frac{1}{p}
 \int_{\abs{x_3}>2\varepsilon\varphi_{\varepsilon}}\!\!\!\!\!\!\!\!\!\!\!
 \abs{\nabla u^{n}}^p
\Big)\\
 &= \frac{1}{p} \int_{\Omega}
 \abs{\nabla
u^{n}}^p.
 \end{align*}
\end{enumerate}
Since $u^{n}\to u$ in $\mathbb{G}^{\alpha}$, when $n\to+\infty$.
According to the classical result, diagonalization's lemma
\cite[Lemma 1.15]{attouch}, there exists a function
$n(\varepsilon):\mathbb{R}^+\to\mathbb{N} $ increasing to $+\infty$
when $\varepsilon\to 0$, such that
$u^{\varepsilon,n(\varepsilon)}\to u$ in $\mathbb{G}^{\alpha}$ when
$\varepsilon\to 0$.
 While $n$ approaches $+\infty$, one will have
\begin{enumerate}
 \item If $\alpha\neq1$:
\begin{align*}
 \limsup_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon,n(\varepsilon)})
 &\leq \limsup_{n\to +\infty}\limsup_{\varepsilon\to 0}
 F^\varepsilon(u^{\varepsilon,n})\\
 &\leq\frac{1}{p}  \int_{\Omega} \abs{\nabla u}^p.
\end{align*}
 \item If $\alpha=1$:
 \begin{align*}
 \limsup_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon,n(\varepsilon)})
&\leq  \limsup_{n\to +\infty}\limsup_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon,n})\\
&\leq \frac{1}{p} \int_{\Omega}
 \abs{\nabla u}^p+\frac{2m(\varphi)
 \eta(\alpha)}{p}\int_{\Sigma}\abs{\nabla'{u}_{|_{\Sigma}}}^{p}.
\end{align*}
\end{enumerate}
If $u\in W^{1,p}_0(\Omega)\setminus\mathbb{G}^{\alpha}$, it is clear
that, for every  $u^\varepsilon\in W^{1,p}_0(\Omega)$,
$u^\varepsilon\rightharpoonup u$ in $W^{1,p}_0(\Omega)$, one has
\[
 \limsup_{\varepsilon\to  0}F^\varepsilon(u^{\varepsilon}) \leq
+\infty.
\]
(b) One is going to determine the lower epi-limit.
 Let $u\in \mathbb{G}^{\alpha}$ and $(u^\varepsilon)$ be a sequence in
$W^{1,p}_0(\Omega)$ such that
 $u^\varepsilon{\rightharpoonup}u$ in $W^{1,p}_0(\Omega)$, so that
\begin{equation}
 \chi_{{\Omega_{\varepsilon}}}\nabla u^\varepsilon
\rightharpoonup \nabla u\quad \text{ in
}L^{^p}(\Omega)^3.\label{eq:liminf2}
\end{equation}
\begin{enumerate}
\item If $\alpha\neq 1$:
 Since
\[
 F^\varepsilon(u^{\varepsilon}) \geq
 \frac{1}{p}  \int_{\Omega_\varepsilon}
 \abs{\nabla u^{\varepsilon}}^p.
\]
 According to \eqref{eq:liminf2} and by passage to the lower limit, one obtains
\[
 \liminf_{\varepsilon\to 0}F^\varepsilon(u^{\varepsilon})
\geq \frac{1}{p} \int_{\Omega} \abs{\nabla u}^p.
\]
\item If $\alpha=1$:
 If $\displaystyle\liminf_{\varepsilon\to
 0}F^\varepsilon(u^\varepsilon)=+\infty$, there is nothing to prove,
because
 $$
\frac{1}{p} \int_{\Omega} \abs{\nabla u}^p+\frac{2m(\varphi)
 \eta(\alpha)}{p}\int_{\Sigma}\abs{\nabla' {u}_{|_{\Sigma}}}^{p}\leq
+\infty.
$$
 Otherwise, $\displaystyle\liminf_{\varepsilon\to
 0}F^\varepsilon(u^\varepsilon)<+\infty$, there exists a
 subsequence of $F^\varepsilon(u^\varepsilon)$ still denoted by
 $F^\varepsilon(u^\varepsilon)$ and a constant
 $C>0$, such that $F^\varepsilon(u^\varepsilon)\leq C$, which implies that
\begin{equation}
\begin{gathered}
 \frac{1}{p\varepsilon^{\alpha}}\int_{B_{\varepsilon}}
 \abs{\nabla u^\varepsilon}^p \leq  C.
\end{gathered}\label{eqliminf33}
\end{equation}
 So $u^\varepsilon$ satisfies the hypothesis of the lemma
 \ref{lemme3}, and according to this last, $\nabla' m^\varepsilon
 u^\varepsilon$ is bounded in $L^{^p}(\Sigma)^2$, so there exists
 an  element $u_1\in L^{^p}(\Sigma)^2$ and a subsequence
of $\nabla' m^\varepsilon u^\varepsilon$, still denoted by
 $\nabla' m^\varepsilon u^\varepsilon$, such that $\nabla'
 m^\varepsilon u^\varepsilon\rightharpoonup  u_1$ in
$L^{^p}(\Sigma)^2$, since
${u^\varepsilon}_{|_{\Sigma}}\rightharpoonup
 u_{|_{\Sigma}}$ in $ L^{^p}(\Sigma)$, and thanks to (\ref{fort0}) and
(\ref{eqliminf33}),
 one
 has $m^\varepsilon{u^\varepsilon}\rightharpoonup
 u_{|_{\Sigma}}$ in $ L^{^p}(\Sigma)$, then
$m^\varepsilon{u^\varepsilon}\rightharpoonup
 u_{|_{\Sigma}}$ in $ W^{1,p}(\Sigma)$, therefore
 $u_1=\nabla' u_{|_{\Sigma}}$, so that $\nabla' m^\varepsilon
u^\varepsilon\rightharpoonup
 \nabla' u_{|_{\Sigma}}$ in $L^{^p}(\Sigma)^2$.
 One has
 \begin{align*}
 F^\varepsilon(u^{\varepsilon})
&\geq  \frac{1}{p}  \int_{\Omega_\varepsilon}
 \abs{\nabla u^{\varepsilon}}^p
 +\frac{1}{p\varepsilon^{\alpha}}
 \int_{B_\varepsilon} \abs{\nabla u^{\varepsilon}}^p\\
 &\geq
 \frac{1}{p}  \int_{\Omega_\varepsilon}
 \abs{\nabla u^{\varepsilon}}^p +\frac{2\varepsilon^{1-\alpha}}{p}
 \int_{\Sigma}\varphi_{{\varepsilon}}
 \abs{\nabla' m^\varepsilon u^{\varepsilon}}^p.
 \end{align*}
 Using the sub-differential inequality of
$$
v\to \frac{2\varepsilon^{1-\alpha}}{p}
 \int_{\Sigma}\varphi_{{\varepsilon}}
 \abs{v}^p, \forall v\in L^{^p}(\Sigma)^2,
$$
 one has
 \begin{align*}
 F^\varepsilon(u^{\varepsilon})
&\geq  \frac{1}{p} \int_{\Omega_\varepsilon}
 \abs{\nabla u^{\varepsilon}}^p
 +\frac{2\varepsilon^{1-\alpha}}{p}
 \int_{\Sigma}\varphi_{{\varepsilon}}
 \abs{ \nabla' u_{|_{\Sigma}}}^p\\
&\quad +\frac{2\varepsilon^{1-\alpha}}{p}
 \int_{\Sigma}\varphi_{{\varepsilon}}
 \abs{ \nabla' u_{|_{\Sigma}}}^{p-2}\nabla' u_{|_{\Sigma}}
(\nabla' m^\varepsilon u^{\varepsilon}-\nabla' u_{|_{\Sigma}}).
 \end{align*}
 Thanks to lemma \ref{period}, one has $\varphi_{{\varepsilon}}\to
m(\varphi)$
 in $L^{^{p'}}(\Sigma)$, so according to (\ref{eq:liminf2}) and by passage to the lower limit, one obtains
\[
 \liminf_{\varepsilon\to 0}F^\varepsilon(u^{\varepsilon})
\geq  \frac{1}{p} \int_{\Omega} \abs{\nabla u}^p
+\frac{2m(\varphi)\;\eta(\alpha)}{p}
 \int_{\Sigma} \abs{ \nabla' u_{|_{\Sigma}}}^p.
\]
\end{enumerate}
 If $u\in W^{1,p}_0(\Omega)\setminus\mathbb{G}^{\alpha}$ and
$u^\varepsilon\in W^{1,p}_0(\Omega)$, such that
$u^\varepsilon\rightharpoonup u$ in $W^{1,p}_0(\Omega)$. \\Assume
that
\[
 \liminf_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon})<+\infty.
\]
So there exists a constant $C>0$ and a subsequence of
$F^\varepsilon(u^{\varepsilon})$, still denoted by
$F^\varepsilon(u^{\varepsilon})$, such that
\begin{align}\label{eqliminf333}
 F^\varepsilon(u^{\varepsilon})<C.
\end{align}
For $0\leq \alpha<1$, there is nothing to prove.\\
Otherwise,
 One takes the same way as in the case $u\in\mathbb{G}^{\alpha=1}$, one has $\nabla' m^\varepsilon
 u^\varepsilon$ is bounded in $L^{^p}(\Sigma)^2$, so there exists
 an  element $u_1\in L^{^p}(\Sigma)^2$ and a subsequence
of $\nabla' m^\varepsilon u^\varepsilon$, still denoted by
 $\nabla' m^\varepsilon u^\varepsilon$, such that $\nabla'
 m^\varepsilon u^\varepsilon\rightharpoonup  u_1$ in
$L^{^p}(\Sigma)^2$, since
${u^\varepsilon}_{|_{\Sigma}}\rightharpoonup
 u_{|_{\Sigma}}$ in $ L^{^p}(\Sigma)$, and thanks to (\ref{fort0}) and
(\ref{eqliminf333}),
 one
 has $m^\varepsilon{u^\varepsilon}\rightharpoonup
 u_{|_{\Sigma}}$ in $ L^{^p}(\Sigma)$, then
$m^\varepsilon{u^\varepsilon}\rightharpoonup
 u_{|_{\Sigma}}$ in $ W^{1,p}(\Sigma)$, so that $u\in\mathbb{G}^{\alpha}$ what contradicts the fact
that $u\not\in
 \mathbb{G}^{\alpha}$, so that
\[
 \liminf_{\varepsilon\to
 0}F^\varepsilon(u^{\varepsilon})=+\infty.
\]
 Hence the proof of theorem \ref{cal:epi} is complete.
 \end{proof}

 In the sequel, one is interested to limit problem determination
partner to the problem \eqref{Pie}, when $\varepsilon$ approaches
zero. Thanks to the epi-convergence results, (see  theorem
\ref{th:epi}, Proposition \ref{stab:epi}) and Theorem \ref{cal:epi},
and according to $\tau_f$-continuity of $G$ in $W^{1,p}_0(\Omega)$,
one has $F^\varepsilon +G\;\tau_f$-epi-converges toward $F^\alpha
+G$ in $W^{1,p}_0(\Omega)$.

\begin{proposition} \label{prop5.2}
For every $f\in L^{p'}(\Omega)$ and according to the parameter
values of $\alpha$, there exists $u^*\in W^{1,p}_0(\Omega)$
satisfying
\begin{gather*}
 u^\varepsilon \rightharpoonup u^* \text{in }W^{1,p}_0(\Omega),\\
 F^{\alpha}(u^*)+G(u^*)=\inf_{v\in \mathbb{G}^{\alpha}}
 \Big\{F^{\alpha}(v)+G(v) \Big\}.
\end{gather*}
\end{proposition}

\begin{proof}
Thanks to lemma \ref{lemme1}, the family ($u^\varepsilon$) is
bounded in $W^{1,p}_0(\Omega)$, therefore it possess a $\tau_f-$
cluster point $u^*$ in $W^{1,p}_0(\Omega)$. And thanks to a
classical epi-convergence result (see  theorem \ref{th:epi}), one
has $u^*$ is a solution of the problem Find
\begin{equation} \label{Pia}
 \inf_{v\in W^{1,p}_0(\Omega)}\Big\{F^{\alpha}(v)+G(v)\Big\}.
\end{equation}
Since $F^{\alpha}$ equals $+\infty$ on $W^{1,p}_0(\Omega)\setminus
\mathbb{G}^{\alpha}$, \eqref{Pia} becomes
\begin{equation}\label{Piaa}
 \inf_{v\in \mathbb{G}^{\alpha}}\Big\{F^{\alpha}(v)+G(v)\Big\}.
\end{equation}
According to the uniqueness of solutions of the problem \eqref{Pia},
so that $u^\varepsilon$ admits
 an unique $\tau_f$-cluster point $u^*$, and therefore
 $u^\varepsilon\rightharpoonup u^*$ in $W^{1,p}_0(\Omega)$.
 \end{proof}
\begin{remark}\rm
One shows that the limit behavior of a constituted structure of two
mediums of constant conductivity united by an oscillating non linear
thin layer of thickness $\varepsilon$, which the conductivity
depends on the negative powers of $\varepsilon$, is describes by a
problem with interface $\Sigma$, ( $\Sigma$ the middle interface of
the thin layer). Following the powers of $\varepsilon$, to the
interface $\Sigma$, one has, on the interface $\Sigma$, the heat
continuity, a bidimensional problem or the constant heat.
\end{remark}

\section{Numerical solutions}\label{sim}

 We showed that for
a small enough $\varepsilon$, the solution $u^\varepsilon$ of the
problem \eqref{Pie}, in a certain sense, approaches the solution
$u^*$ of the limit problem \eqref{Piaa}. In this section we interest
to the numerical treatment of the problem \eqref{Pie} and
\eqref{Piaa}, to illustrate the obtained theoretical results. Take
the problems \eqref{Pie} and \eqref{Piaa}, with
\begin{gather*}
 \Omega=]-1,1[\times]-1,1[,\\
f(x,y)=0.01\exp(-x^2-y^2),\\
\varphi_\varepsilon(x)=1.2+\sin(\pi\frac{x}{\varepsilon}).
\end{gather*}
We solve numerically the problems \eqref{Pie} and \eqref{Piaa},
using the language FreeFem++ (see,\cite{freefem++}), with the finite
elements method and using Newtons method, with $p=3.5$ and
$\varepsilon=1e-06$, and one will have the results shown in figures.
\newpage
\begin{figure}[!h]
 \begin{center}
\includegraphics[width=0.95\textwidth]{fig2}
 \end{center}
\vspace*{-1cm}
 \caption{Solution to \eqref{Pie} (left) and to the limit problem
 \eqref{Piaa}
 for $\alpha=0.01$ (right) \label{0}}
\end{figure}

\begin{figure}[!h]
 \begin{center}
 \vspace*{-.5cm}
\includegraphics*[width=0.95\textwidth]{fig3}
 \end{center}
 \vspace*{-.5cm}
 \caption{Solution to \eqref{Pie} (left) and to the limit problem \eqref{Piaa}
 for $\alpha=1$ (right) \label{fini}}
\end{figure}

\begin{figure}[!h]
 \begin{center}
\includegraphics[width=0.95\textwidth]{fig4}
 \end{center}
 \vspace*{-.5cm}
 \caption{Solution to \eqref{Pie} (left) and to the limit problem \eqref{Piaa}
 for $\alpha=4.5$ (right)\label{inf}}
\end{figure}
 Figures \ref{0}, \ref{fini}, \ref{inf} show that
the solution of \eqref{Pie} approach the one of the limit problem
\eqref{Piaa}, for $\alpha=0.01$, 1 and 4.5 with an error of order
$1.64e-010$,   $1.1e{-05}$, and $9e-014$, respectively.

\section{Appendix}

\begin{definition}[{\cite[Definition 1.9]{attouch}}] \label{def:epi}
 Let $(\mathbb{X},\tau)$ be a metric space and
$(F^{\varepsilon})_\varepsilon$
 and $F$ be functionals defined on $\mathbb{X}$ and with value in
 $\mathbb{R}\cup\{+\infty\}$. $F^{\varepsilon}$
 epi-converges to $F$ in $(\mathbb{X},\tau)$, noted
$\tau-lim_e F^\varepsilon=F$,
 if the following assertions are satisfied
 \begin{itemize}
 \item For all $x\in \mathbb{X}$, there exists $x_\varepsilon^0$,
$x_\varepsilon^0\stackrel{\tau}  {\to} x$ such that
$\displaystyle\limsup_{\varepsilon\to 0}
F^\varepsilon(x_\varepsilon^0)\leq F(x)$.

 \item For all $x\in \mathbb{X}$ and all
 $x_\varepsilon$ with $x_\varepsilon\stackrel{\tau}{\to} x$,
 $\displaystyle\liminf_{\varepsilon\to 0} F^\varepsilon(x_\varepsilon)\geq F(x)$.
 \end{itemize}
 \end{definition}

 Note the following stability result of the
 epi-convergence.

 \begin{proposition}[{\cite[p. 40]{attouch}}] \label{stab:epi}
 Suppose that $F^{\varepsilon}$ epi-converges to $F$ in
$(\mathbb{X},\tau)$
 and that $G\colon \mathbb{X}\to \mathbb{R}\cup\{+\infty\}$,
is $\tau-continuous$. Then
 $F^{\varepsilon}+G$ epi-converges to $F+G$ in $(\mathbb{X},\tau)$
 \end{proposition}

 This epi-convergence is a special case of the
 $\Gamma-$convergence introduced by De Giorgi (1979) \cite{degiorgi}.
It is well  suited to the asymptotic analysis of sequences of
minimization
 problems since one has the following fundamental result.

\begin{theorem}[{\cite[theorem 1.10]{attouch}}] \label{th:epi}
 Suppose that
 \begin{enumerate}
 \item \label{th:1} $F^{\varepsilon}$ admits a minimizer on
 $\mathbb{X}$,
 \item \label{th:2}The sequence $(\overline u^{\varepsilon})$ is
$\tau$-relatively  compact,
 \item \label{th:3}The sequence $F^\varepsilon$ epi-converges to
 $F$ in this topology $\tau$.
 \end{enumerate}
 Then every cluster point $\overline u$ of the sequence
$(\overline{u}^{\varepsilon})$ minimizes $F$
 on $\mathbb{X}$ and
 $$
 \lim_{\varepsilon'\to  0}F^{\varepsilon'}(\overline{u}^{\varepsilon'})
=F(\overline u),
 $$
 if $(\overline{u}^{\varepsilon'})_{{\varepsilon'}}$ denotes
 the subsequence of
 $(\overline{u}^{\varepsilon})_{{\varepsilon}}$ which
 converges to $\overline u$.
 \end{theorem}

 \begin{lemma}\label{period}
 Let $\varphi \in L^{^\infty}(\Sigma)$, a $Y$-periodic,
 $Y=]0,1[\times]0,1[$. Let
 $$
\varphi_{\varepsilon}(x)=\varphi(\frac{x}{\varepsilon})\mbox{, for a
small enough }
 \varepsilon >0.
$$
So that
 \begin{gather*}
 \varphi_{{\varepsilon}}\to m(\varphi) \quad\mbox{in $L^s(\Sigma)$ for
}
 1\leq s<\infty, \\
 \varphi_{{\varepsilon}}\rightharpoonup^*
 m(\varphi)\quad \mbox{in }L^\infty(\Sigma).
 \end{gather*}
 \end{lemma}

 \begin{proof}
 Since $\varphi_\varepsilon$ is a $\varepsilon Y$-periodic, so one has
\begin{equation}\label{lem:ann}
 \begin{gathered}
 \varphi_{\varepsilon}\rightharpoonup m(\varphi)
\quad \mbox{in } L^s(\Sigma) \mbox{ for } 1\leq s<\infty, \\
 \varphi_{\varepsilon}\rightharpoonup^*
 m(\varphi)\quad \mbox{ in }L^\infty(\Sigma).
 \end{gathered}
\end{equation}
 Since $\varphi$ is bounded a.e. in $\Sigma$, so for every $s\geq 1$,
there exists a constant $C>0$, such that
\begin{equation} \label{lem:annn}
 \begin{aligned}
 \int_{\Sigma}\abs{\varphi_{{\varepsilon}}-m(\varphi)}^s
&\leq C \int_{\Sigma} \abs{ \varphi_{{\varepsilon}}-m(\varphi)} \\
&\leq C \Big(\int_{\varphi \geq m(\varphi)}
 (\varphi_{{\varepsilon}}-m(\varphi))-\int_{\varphi \leq m(\varphi)}
 (\varphi_{{\varepsilon}}-m(\varphi))\Big).
 \end{aligned}
\end{equation}
Passing to the limit in (\ref{lem:annn}), one  has
 $\varphi_{{\varepsilon}}\to m(\varphi)$  in $L^s(\Sigma)$  for
$1\leq s<\infty$.
 \end{proof}

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\end{document}