\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 83, pp. 1--19.\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-2006/83\hfil Singular perturbation problem]
{Singular perturbation problem for the incompressible Reynolds equation}
\author[I. Ciuperca, I. Hafidi, M. Jai\hfil EJDE-2006/83\hfilneg]
{Ionel Ciuperca, Imad Hafidi, Mohammed Jai}  % in alphabetical order

\address{Ionel  Ciuperca \newline
CNRS-UMR 5208 Universit\'e Lyon1, MAPLY,  101,
69622 Villeurbanne Cedex, France}
\email{ciuperca@maply.univ-lyon1.fr}

\address{Imad Hafidi \newline
CNRS-UMR 5208,  Insa de Lyon, Bat L\'eonard de Vinci, 69621, France}
\email{Imad.Hafidi@insa-lyon.fr}

\address{Mohammed Jai \newline
CNRS-UMR 5208,  Insa de Lyon, Bat L\'eonard de Vinci, 69621, France}
\email{Mohammed.Jai@insa-lyon.fr}

\date{}
\thanks{Submitted January 5, 2006. Published July 21, 2006.}
\subjclass[2000]{76D08, 35B25, 35B40}
\keywords{Reynolds lubrication equation; weighted Sobolev Spaces;
\hfill\break\indent variational inequality; singular perturbation analysis;
maximum principle}

\begin{abstract}
 We study the asymptotic behavior of the solution of a Reynolds
 equation which describe the behavior of the fluid between two
 closes surfaces as the distance between the two surfaces locally
 tends to zero.
\end{abstract}

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


\section{Introduction}

The field of lubricated contact deals with dynamical systems
which consist of two (or more) bodies in relative motion. The
contact between the bodies  is mediated by a lubricant fluid,
which in this work is assumed incompressible. The simplest
such contact is the wedge (or plane slider), used in thrust bearings.
It consists of two planar, rigid surfaces which are not
mutually parallel. It is sketched in Fig. \ref{patin}, in which
the bottom surface is assumed to be moving horizontally towards
the right. This movement entrains lubricant towards the right into
the convergent gap between the surfaces. In turn, this generates
a pressure field and consequently a thrust force, which allows
to equilibrate a load applied to the top of the device.

Under the thin-film hypothesis (the gap thickness $h$ much smaller
than the in-plane dimensions of the contact, with the variations
in $h$ also assumed small) the fluid pressure does not depend
on the vertical coordinate, which is taken across the gap. Upon
normalization and assuming that the system is in a time-independent
state, the pressure satisfies the normalized Reynolds equation \cite{frene}
\begin{gather}
  \nabla \cdot [  h(x)^3  \nabla p  ] = \frac{\partial
h}{\partial x_{1}}
 \quad x=(x_{1},\dots,x_{n}) \in \Omega \label{lubrif1}\\
p= 0 \quad  x \in \partial \Omega \label{lubrif2}
\end{gather}
where $\Omega\subseteq \mathbb{R}^{n}(n=1$ or $2)$ is the domain in
which the two surfaces are in proximity, $p$ is the normalized
pressure, $h(x)$ is the normalized gap thickness and the relative
motion is assumed along the $x_1$-direction.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\caption{Sketch of a slider bearing}
\label{patin}
\end{center}
\end{figure}

Assume, as in Fig. \ref{patin}, that a vertical force $F$ is
applied to the upper surface of the bearing at a point
$x^0=(x_1^0,x_2^0,\dots,x_n^0)$.
To equilibrate this load the upper surface changes its position,
intuitively getting closer to the lower surface as the
applied load increases. Let us define the {\em reference shape} of
the upper surface through a non negative function $h_0$.\\
Into this function we incorporate the so-called {\em attitude}
of the slider (pitch and roll angles), so that the gap
thickness becomes, simply,
\begin{equation}
h(x)=h_0(x) + \epsilon
\end{equation}
where $\epsilon$ represents the minimal distance between the
surfaces. With $h_0$ fixed the pressure becomes a function of
$\epsilon$ satisfying the problem
\begin{equation} \label{problem:P1}
\begin{gathered}
\nabla\cdot\big[\big(h_{0}(x)+\epsilon\big)^{3}\nabla p\big]
=\frac{\partial h_{0}}
{\partial x_{1}} \quad \text{on } \Omega\\
 p=0 \quad \text{on } \partial \Omega
\end{gathered}
\end{equation}
If for any $\epsilon>0$, we denote
\begin{equation}
\label{char}
 g(\epsilon)=\int_{\Omega} p\,dx
\end{equation}
then for equilibrium to hold in a system in which the only degree
of freedom is the vertical position,  the upper
surface must be placed so as to satisfy
\begin{equation}
g(\epsilon) = F
\end{equation}
It is easy to show both that $\lim_{\epsilon\to
+\infty} g(\epsilon) = 0$ and that $g$ is a continuous function, so
that an equilibrium position exists for any positive load $F$
smaller than $\max_{\epsilon} g(\epsilon)$. It is thus
extremely important to analyze the behavior of $g$ in the vicinity of
zero, in particular the conditions under which
$\lim_{\epsilon\to 0} g(\epsilon) = +\infty$. This
guarantees the existence of an equilibrium position for any positive
$F$. A finite limit, on the other hand, guarantees the existence of
an equilibrium position for any $0 < F \leq g(0)$.

It is also important to study the moments of the force exerted by
the pressure for each minimal gap thickness $\epsilon$ defined with
respect to the point $x^0$,
\begin{equation} \label{mom}
\begin{array}{ll}
{ m_{i}(\epsilon)=\int_{\Omega}{p(x_{i}-x_{i}^{0})dx }}
& { i=1,\dots,n}
\end{array}
\end{equation}
because equilibrium also requires that
\begin{equation}
m_{i}(\epsilon)=0 \quad i=1,\dots,n
\end{equation}
and systems with the pitch and roll angles as additional degrees
of freedom have their attitudes defined
by these conditions.

In this article we will assume $n=2$,
$\Omega=]a_1,b_1[ \times ]a_2,b_2[$,
$h_{0}\in C^{0}(\bar{\Omega})$ with $h_0(x) > 0$ a.e $ x \in \Omega$, and
$$
\min_{x\in \overline{\Omega}} h_0(x) = 0.
$$
The goal is to find the limits of $g(\epsilon)$ and
$m_{i}(\epsilon)$, $i=1,2$ as $\epsilon\to 0^+$. Beside its
intrinsic importance, this study provides crucial tools for a
forthcoming analysis on the existence of equilibria for the
dynamical equations of slider bearings.

As we will see later, the results strongly depend on the shape
function $h_0$. We will consider two situations:
\begin{itemize}
\item[(i)] when $h_0$ vanishes on a segment of type $\{x_1= d_1\}$ only, with
$d_1 \in [a_1,b_1]$, which will be called ``line contact case''.
In this case we will assume $h_0 \sim |x_1-d_1|^\alpha$ in a neighborhood
of $\{x_1= d_1\}$, with $\alpha > 0$.
\item[(ii)]  when $h_0$ vanishes only at a single point $d=(d_1,d_2)$ of
$\bar{\Omega}$,  which will be called ``point contact case''.
We will assume  $h_0 \sim |x-d|^\alpha$ in a neighborhood of $\{x= d\}$,
 with $\alpha > 0$.
\end{itemize}

In both the ``line-contact case'' and the ``point-contact case''
we obtain two types of results: (i) convergence of load and momenta
to some finite limits which will be made precise in Section 3 and (ii)
divergence to $+\infty$ of load and momenta.
Problem \eqref{problem:P1} can be seen as a singular perturbation
of the corresponding problem ($\epsilon =0$ )
with small parameter $\epsilon $. This kind of problem has been
studied in \cite{sanchez92, caillerie96, lions73, huet60}.

We can apply here singular perturbation results to obtain the convergence
of $p$ to the solution of the limit problem denoted $p_0$ in a weighted
Sobolev space of type $H^1_0(\Omega,h_0^{2\delta_1},h_0^{2\delta_2} )$
(see Section 2 for the definition). This is not sufficient for the
convergence of load and momenta; we also need a continuous embedding
of $H^1_0(\Omega,h_0^{2\delta_1},h_0^{2\delta_2} )$ into $L^1(\Omega)$
which is obvious if $\delta_1$ is not large.

This singular perturbation approach works only for $\alpha < 1$ in the
``line-contact case'' and for  $\alpha < \frac{3}{2}$ in the
``point-contact case''.
To have a well-posed limit problem we need a Poicar\'e-like inequality
for weighted Sobolev spaces. This subject is well studied in the
literature (see \cite{edmundsopic93,opickufner90,franchi91}).
We prefer  to give here a new elementary result
(Lemma \ref{lem:principal}) well-adapted to our problem.

In cases $\alpha \geq 1$ for ``line-contact'' and
$\alpha \geq \frac{3}{2}$ for ``point-contact'' (divergent cases)
the singular perturbations results cited above are no longer applicable.

This part is more difficult and we use extensively the maximum principle
in order to find an appropriate lower bound for $p$ whose integral tends
to infinity. This proves the divergence to infinity of the load and we
also prove that this divergence, in the ``line-contact case'', is of
order greater than $ \epsilon^{2/\alpha-2}$ for $\alpha > 1$ and greater
than $\log(1/\epsilon) $  for $\alpha = 1$. In the ``point-contact case''
the same result holds with $\alpha$ replaced by $\frac{2}{3} \alpha$.

In order to prove the divergence of momenta we also need to prove that
$p$ is bounded far from the annulation points of $h_0$. This result is
again proved using the maximum principle, assuming $h_0$ to be a tensor
product.
We remark  that in the ``point-contact case'' for
$\alpha \in [\frac{3}{2},2[$ we have divergence of load and momenta while
the limit problem of \eqref{problem:P1} exists in a weighted Sobolev space.
This is because the continuous embedding of this space in $L^1(\Omega)$
does not hold.

In some cases the solution of \eqref{problem:P1} is negative, which
does not correspond to the actual fluid behavior since cavitation
takes place for $p<0$. To account for cavitation, problem (P1) is
replaced by the variational inequality \cite{baycham86}
\begin{equation} \label{problem:P2}
\begin{gathered}
\textrm{Find } p \in \mathcal{K}=\{ v \in H^1_0(\Omega): v \geq 0\} \\
 \int_\Omega (h_0(x)+\epsilon)^3 \nabla  p \nabla
(\varphi -p)dx \ge  \int_\Omega h_0 \frac{\partial}{\partial
x_1}  (\varphi -p) \quad \forall \varphi \in \mathcal{K}
\end{gathered}
\end{equation}
The goal is also to find limits of $g$ and $m_i, i=1,2$ (defined as
in \eqref{char} and \eqref{mom} with $p$ now the solution
of \eqref{problem:P2}) when $\epsilon$ goes to 0. We obtain the same kind
of results as in the equation case.

The paper is organized as follows.
In Section 2 we present some preliminary results concerning
weighted Sobolev spaces, in particular  the Poincar\'e-like inequality.
In Section 3 we study the limits  of load and momenta in the equation
case (problem \eqref{problem:P1}) for the different cases cited above.
Finally, Section 4 presents the same study in the inequality case
(problem \eqref{problem:P2}).

\section{Preliminaries}

Let us consider $f_0, f_1 \in C^{0}(\overline{\Omega})$ with $f_k > 0$
a.e. $x\in \Omega, k=0,1$.
We introduce the weighted Sobolev space
\begin{equation*}
H^{1}(\Omega,f_0,f_1)
\end{equation*}
as the set of all measurable functions $\varphi = \varphi(x)$ defined on
$\Omega$ with (generalized) derivatives $D^\alpha
\varphi$ for $\alpha=(\alpha_1,\alpha_2) \in \mathbb{N}^2$ with $
\alpha_{1}+\alpha_{2}\le 1 $ such that
\begin{equation}\label{normder}
   \int_\Omega f_0(x) \varphi^2(x)dx +  \int_\Omega f_1(x)|
   \nabla \varphi |^2(x)dx <\infty
\end{equation}
$H^{1}(\Omega,f_0,f_1)$ is a pre-Hilbert space equipped with the scalar
product
$$
\big( \varphi_1,\varphi_2\big)_{H^{1}(\Omega,f_0,f_1)}=
\int_\Omega f_0(x) \varphi_1(x) \varphi_2(x)dx
+  \int_\Omega f_1(x) \nabla \varphi_1 (x) \cdot  \nabla \varphi_2 (x)dx\,.
$$
Let $ H_{0}^{1}(\Omega,f_0,f_1) $ be the closure of $\mathcal{D}(\Omega) $
with respect to the norm of $H^{1}(\Omega,f_0,f_1)$,
which is a Hilbert space endowed with the same scalar product
as $H^{1}(\Omega,f_0,f_1)$.

\begin{remark} \label{rmk2.1} \rm
 If $1/f_k\in L_{\rm loc}^{1}(\Omega), k=0,1$ then
 $H^{1}(\Omega,f_0,f_1)$ is a
Hilbert space \cite{kufneropic86}.
\end{remark}

We have the following general Poincar\'e-like inequality.

\begin{lemma}\label{lem:principal}
Let $ f \in C^{0}(\overline{\Omega})$  with $f > 0$ a.e.
$x\in \Omega $. Assume that a real $d_1 \in [a_1,b_1]$ exists such that
$f$ is  non-increasing in $x_1$ on
$[a_1,d_1] \times [a_2,b_2]$ and  non-decreasing in $x_1$ on
$[d_1,b_1] \times [a_2,b_2]$, with the obvious convention that for
$d_1=a_1$ (resp. $d_1=b_1$) the function $f$ is only non-decreasing (
resp. non-increasing).
Then for any $\delta_1, \delta_2 \in \mathbb{R}^{+}$ such that
\begin{align*}
K&=\sup_{x_2\in[a_2,b_2]}\int_{a_1}^{d_1}
 {\int_{a_1}^{x_1}{f^{2(\delta_1-\delta_2)}(s,x_2)ds}dx_1} \\
&\quad + \sup_{x_2\in[a_2,b_2]}\int_{d_1}^{b_1}
 {\int_{x_1}^{b_1}{f^{2(\delta_1-\delta_2)}(s,x_2)ds}dx_1}<\infty
\end{align*}
we have
$$
\int_{\Omega}{f^{2\delta_{1}}u^{2}}\le
K\int_{\Omega}{f^{2\delta_{2}}\big|\nabla u\big|^{2}}, \quad \forall u \in  H^1_0(\Omega,f^{2\delta_1},f^{2\delta_2})
$$
\end{lemma}

\begin{proof}
For any ${ u \in \mathcal{D}(\Omega)}$ we have for $x_1 < d_1$:
\begin{align*}
f^{\delta_{1}}(x_{1},x_{2})|u(x_1,x_2)|
& \leq f^{\delta_1}(x_{1},x_{2})
\int_{a_1}^{x_{1}}\big|\frac{\partial u}{\partial x_{1}}(s,x_{2})\big|ds \\
& \leq \int_{a_1}^{x_{1}}f^{\delta_1}(s,x_{2})
\big|\frac{\partial u}{\partial x_{1}}(s,x_{2})\big|ds,
\end{align*}
since $f$ is $x_1$-non-increasing.
We then have
\begin{align*}
&f^{2\delta_{1}}(x_{1},x_{2})u^2(x_{1},x_{2}) \\
&\le
\Big(\int_{a_1}^{x_{1}}{f^{2(\delta_{1}-\delta_{2})}(s,x_{2})ds}\Big)
\Big(\int_{a_1}^{b_1}{f^{2\delta_{2}}(x_1,x_{2})\Big(\frac{\partial
u}{\partial x_{1}}(x_1,x_{2})\Big)^{2}dx_1}\Big)
\quad \forall x_1 \leq d_1
\end{align*}
By integrating  in $x_1$ on $[a_1,d_1]$ first and then in $x_2$ we obtain
\begin{equation} \label{ineg:d1}
\begin{aligned}
&\int_{a_1}^{d_1}\int_{a_2}^{b_2}f^{2\delta_1}u^2dx\\
&\leq
\Big(\sup_{x_2\in [a_2,b_2]}\int_{a_1}^{d_1} & \int_{a_1}^{x_1}
f^{2(\delta_1-\delta_2})(s,x_{2})\,ds\,dx_1 \Big)
 \Big(\int_{\Omega}f^{2\delta_{2}}\big(\frac{\partial u}{\partial
x_1}\big)^2dx\Big)
\end{aligned}
\end{equation}
In the same manner, using the fact that $f$ is $x_1$-non-decreasing
on $[d_1,b_1] \times [a_2,b_2]$, we obtain
\begin{equation} \label{ineg:d2}
    \begin{split}
&\int_{d_1}^{b_1}\int_{a_2}^{b_2}f^{2\delta_1}u^2dx\\
&\leq \Big(\sup_{x_2\in [a_2,b_2]}\int_{d_1}^{b_1}
\int_{x_1}^{b_1}f^{2(\delta_1-\delta_2})(s,x_{2})\,ds\,dx_1 \Big)
\Big(\int_{\Omega}f^{2\delta_{2}}\big(\frac{\partial u}{\partial
x_1}\big)^2dx\Big)
\end{split}
\end{equation}
Adding \eqref{ineg:d1}  and \eqref{ineg:d2} we obtain the desired
inequality and by a density argument we obtain the result.
\end{proof}

\begin{corollary} \label{cosemi1}
For $f, \delta_1, \delta_2$ satisfying assumptions of
Lemma \ref{lem:principal} the semi-norm
$\| f^{\delta_2}\nabla \cdot\|_{L^2(\Omega)}$ is a norm on
$H_{0}^{1}(\Omega,f^{2\delta_1},f^{2\delta_2})$ and is
equivalent to the norm of $H^{1}(\Omega, f^{2\delta_1},f^{2\delta_2})$.
\end{corollary}


\section{Asymptotic behavior of the equation case (problem \eqref{problem:P1})}

In this section we set, for the sake of simplicity, $\Omega= ]-1,0[^2$ and
we suppose that
\begin{equation} \label{h0}
h_0 \geq 0, \quad h_0 \in C^0(\bar{\Omega})
\text{ and $h_0$ is non increasing  in $x_1$}.
\end{equation}
 We study the asymptotic behavior, as $\epsilon$ tends to 0, of the
 solution $p$ of \eqref{problem:P1} which is non negative by the maximum
principle.
We introduce the following limit problem for any $\delta_1 \in \mathbb{R}^{+}$:
\begin{equation}
\label{problem:P0}
\begin{gathered}
 \text{Find } p_{0} \in H_0^1 (\Omega,h_0^{2\delta_1},h_{0}^{3})
 \text{ such that}\\
 \int_{\Omega}{h_{0}^{3}\nabla
p_{0}\nabla\varphi}=\int_{\Omega}{h_{0}\frac{\partial\varphi}{\partial
x_{1}}} \quad \forall \varphi \in  H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3)
\end{gathered}
\end{equation}

\begin{proposition} \label{prop:exist1} Suppose that
$\int_\Omega\frac{dx}{h_0}< +\infty$ and $\delta_1 \in \mathbb{R}^{+}$ is such that
$$
\sup_{x_2 \in [-1,0]} \int_{-1}^0 \int_{-1}^{x_1}
h_0^{2\delta_1-3}(s,x_2)\,ds\,dx_1< +\infty
$$
 Then problem \eqref{problem:P0} admits a unique solution
$p_0 \in H^1_0(\Omega;h_0^{2\delta_1},h_0^3)$ which is independent
on $\delta_1$.
\end{proposition}

\begin{proof} From Lemma \ref{lem:principal} with $d_1=0$ and
$\delta_2 = \frac{3}{2}$ we deduce that the seminorm
$\|h_0^3 \nabla \cdot  \|_{L^2(\Omega)} $ is a norm in
$H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3) $ equivalent to the norm in
$H^{1}(\Omega,h_0^{2\delta_1},h_0^3) $.
On the other hand the application
$$
\varphi \in H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3) \to
 \int_\Omega h_0 \frac{\partial \varphi }{\partial x_1}dx \in \mathbb{R}
$$
is in the dual of $H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3) $ since
we have
$$
{ \Big\vert\int_{\Omega}{h_{0}\frac{\partial \varphi }{\partial x_{1}}}
\Big\vert}  \le \Big(\int_{\Omega}{\frac{dx}{h_{0}}}\Big)^{1/2}\Big[
\int_{\Omega}h_{0}^{3}\big(\frac{\partial \varphi}{\partial x_{1}}
\big)^{2}\Big]^{1/2}\,.
$$
By Lax-Milgram theorem   we have classically
the existence and the uniqueness for any fixed $\delta_1 \in \mathbb{R}^{+}$.
To prove the independence of $p_0$ with respect to $\delta_1$ we
consider $p_0^1$ and $p_0^2$ two solutions corresponding respectively
to $\delta_1^1$ and $\delta_1^2$ with $\delta_1^2<  \delta_1^1$.
We remark that $H^1_0(\Omega,h_0^{2\delta_1^2},h_0^3)
\subset H^1_0(\Omega,h_0^{2\delta_1^1},h_0^3)$ with dense and
continuous embedding. Then $ p_0^2$ satisfies by density the same
problem as $p_0^1$ which by uniqueness gives the result.
\end{proof}

Problem \eqref{problem:P1} can be seen as a singular perturbation of
problem \eqref{problem:P0}
with the small parameter $\epsilon $. This kind of problem has been
studied in \cite{sanchez92, caillerie96, lions73, huet60}. Let us
recall a simplified version of a result given in \cite{huet60} for a
more general problem, which allows us to obtain the convergence
of $p$ to $p_0$.

\begin{lemma} \label{lem:huet}
Let $V,W,H$ be three Hilbert spaces, with continuous embeddings
$V\subset W \subseteq H$ and $V$ dense in $W$ and in $H$.
Let $b(\epsilon;u,v), 0 < \epsilon \leq \epsilon_0$, be a sequence
of continuous bilinear forms on $V$, $b(u,v)$ a continuous
bilinear form on $W$ and $f \in H$.
Under the hypotheses
\begin{enumerate}
    \item \label{i1} $\epsilon \to b(\epsilon;u,v)$ is continuous and
    $ \lim_{\epsilon \to 0} b(\epsilon;u,v) = b(u,v)$
    $\forall (u,v) \in V\times V$
    \item \label{i2} $\exists \alpha(\epsilon) > 0$ with $\alpha(\epsilon)
    \to 0$ and $\beta > 0$ such that $ b(\epsilon;u,u)
    \geq \alpha(\epsilon) \|u\|_V^2+\beta\|u\|_W^2, \; \forall u \in
    V$
    \item \label{i3} $\exists \gamma > 0$ such that $ b(u,u) \geq \gamma \|u\|_W^2, \forall u \in W$
    \item \label{i5} For any sequence $w_\epsilon \in V$ for which
    $|b(\epsilon;w_\epsilon,w_\epsilon)|$ is bounded, $b(\epsilon;w_\epsilon,v)-b(w_\epsilon,v)
    \to 0, \forall v \in V$
    \item \label{i6} $\exists  \delta(\epsilon)$ with
    $ \delta(\epsilon)\to 0$ such that $ b(\epsilon;
    v,v)-b(v,v)+ \delta(\epsilon)  b(v,v) \geq 0$
\end{enumerate}
the solution  $u_\epsilon$ of the problem
$$
b(\epsilon;u_\epsilon,v)=(f,v)\quad \forall v \in V, \epsilon
\leq \epsilon_0
$$
 converges strongly in $W$ to the solution $u$ of the
problem
$$
b(u,v) = (f,v) \quad \forall v \in W\,.
$$
\end{lemma}

\begin{proposition} \label{lem:cvgptop0}
Under the hypotheses of Proposition \ref{prop:exist1}, the  solution $p$
of \eqref{problem:P1} converges,  as $\epsilon$ tends to  $0$, to
the solution $p_{0}$ of \eqref{problem:P0}, strongly in
$H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3)$.
\end{proposition}

\begin{proof}
It suffices to apply Lemma \ref{lem:huet} with
$V=H_{0}^{1}(\Omega)$, $W=H=H_{0}^{1}(\Omega,h_0^{2\delta_1},h_0^3)$
endowed with the norm $\|h_0^3 \nabla \cdot  \|_{L^2(\Omega)}$,
$u_\epsilon=p$, and
$$
b(\epsilon;u,v) = \int_\Omega(h_{0}+\epsilon)^{3}\nabla u\nabla vdx, \;
b(u,v) = \int_\Omega h_{0}^{3}\nabla u\nabla v\,dx\,.
$$
 We choose $f$ in the following manner: Since the application
$v \in W \to  \int_\Omega h_0\frac{\partial v}{\partial x_1}$
is an element of $W'$, from Riesz's theorem there exists $f$ in $W$
such that
$$
\int_\Omega h_0\frac{\partial v}{\partial x_1} = (f,v)_W, \quad
 \forall v \in W .
$$
Assumptions  \eqref{i1}-\eqref{i3} and \eqref{i6} are immediately
verified with $ \alpha(\epsilon)=\epsilon^3$,
${ \beta=1}$, ${ \gamma=1}$ and $ \delta(\epsilon) = 0$.

Let us now verify assumption \eqref{i5}.  Let $w_{\epsilon}$ be a sequence in
$H_{0}^{1}(\Omega)$ for which
${\int_{\Omega}{(h_{0}+\epsilon)^{3}\nabla (w_{\epsilon})^{2}}}$ is
bounded, so $ \| w_{\epsilon}\|_W $ is bounded.
For all ${ v \in H_{0}^{1}(\Omega)}$, we have
\begin{align*}
&{ \int_{\Omega}{(h_{0}+\epsilon)^{3}\nabla
w_{\epsilon}\nabla v}-\int_{\Omega}{h_{0}^{3}\nabla
w_{\epsilon}\nabla v}} \\
 &=\int_{\Omega}{\Big[(h_{0}+\epsilon)^{3/2}-h_{0}^{3/2}\Big]
\big(h_{0}+\epsilon\big)^{3/2}\nabla w_{\epsilon}\nabla v}
+\int_{\Omega}{\Big[(h_{0}+\epsilon)^{3/2}-h_{0}^{3/2}\Big]h_{0}^{3/2}\nabla
w_{\epsilon}\nabla v}
\end{align*}
Since $(h_{0}+\epsilon)^{3/2}-h_{0}^{3/2}\to
0\textrm{ in  }L^{\infty}(\Omega)$ we easily obtain the result.
\end{proof}

For simplicity we shall distinguish here two different situations:
(i) the function $h_0$ vanishes on the entire segment $x_1=0$ namely
the line-contact case, (ii) the function $h_0$ vanishes on a unique
interior point, supposed to be  (0,0), namely point-contact case.

\subsection{Line-contact case} \label{equalinecont}

We assume in this section that, in addition to hypothesis \eqref{h0}
\begin{gather*}
 h_0(x) = 0 \quad \text{for } x_1 = 0 \\
 h_0(x) > 0 \quad \text{for } x_1 \neq 0
\end{gather*}
We shall prove, under some supplementary hypotheses, that:
\\
For $  \int_\Omega \frac{dx}{h_0(x)}  < +\infty$ the load and momenta
have finite limits (paragraph \ref{sec:finlimc1}). \\
For $  \int_\Omega \frac{dx}{h_0(x)}  = +\infty$  the load and momenta
have infinite limits (paragraph \ref{sec:infinlimc1}).


\begin{remark} \label{rmk3.34} \rm
If $h_0$ behaves as $(-x_1)^\alpha $ with $ \alpha > 0$ in a
neighbourhood of $x_1 =0$, then $  \int_\Omega \frac{dx}{h_0(x)}$
is finite for $\alpha < 1$ and infinite for $\alpha \geq 1$.
\end{remark}

\subsubsection{ Finite limit case  } \label{sec:finlimc1}


\begin{theorem}\label{thm:limfinic1}
If there exists $ \alpha \in ]0,1[$  and $m_0 > 0$ such that
$$
h_0(x) \geq m_0 (-x_1)^\alpha \quad \forall x \in \Omega
$$
then, for any $ (x_{1}^{0},x_{2}^{0})\in \Omega$ and $\epsilon \to 0$,
we have
\begin{equation*}
\int_\Omega p\,dx \to \int_ \Omega p_0dx, \quad
\int_\Omega(x_{k}-x_{k}^{0})p \to \int_\Omega(x_{k}-x_{k}^{0})p_{0}, \quad
k=1,2
\end{equation*}
where $p$ is the solution of \eqref{problem:P1} and $p_0$ the
solution of problem \eqref{problem:P0}
\end{theorem}

\begin{proof}
Since $\alpha \in ]0,1[$, the hypotheses in Propositions
\ref{prop:exist1}  and \ref{lem:cvgptop0} are satisfied with
$ \delta_1= \frac{1}{2}$.
Then there exists $ p_0  \in H_0^1(\Omega,h_0,h_0^3)$ unique solution
of \eqref{problem:P0}  with $ \delta_1= \frac{1}{2}$ such that
$p \to p_0 $ strongly in $H_0^1(\Omega,h_0,h_0^3)$.
On the other hand, all $u \in H_0^1(\Omega,h_0,h_0^3)$,
we have
$$
\int_\Omega |u|dx\leq \Big( \int_\Omega
\frac{dx}{h_0}\Big)^{1/2}\Big(\int_\Omega h_0 u^2dx \Big)^{1/2}
$$
that is, the continuous embedding of
$ H_0^1(\Omega,h_0,h_0^3)$ in $L^1(\Omega)$ holds.
This implies that $p$ converges to $p_0$ strongly in $L^1(\Omega)$
which ends the proof.
\end{proof}

\subsubsection{Infinite-limit case } \label{sec:infinlimc1}

For any $\delta > 0$, we shall define
$$
\Omega_\delta= ]-\delta,0[ \times ]-1,0[
$$
In this paragraph we need the following supplementary hypothesis:
There exist $\delta_0 \in ]0,\frac{1}{2}[$, $\alpha \geq 1$
and $0 < m_1 \leq M_1$ such that
\begin{equation} \label{h1}
h_0 \in W^{1,\infty}(\Omega)\quad\text{and}\quad
\alpha m_1 (-x_1)^{\alpha-1} \leq -\frac{\partial
h_0}{\partial
    x_1} \leq \alpha M_1 (-x_1)^{\alpha-1}, \quad \forall
    x \in \Omega_{2\delta_0}
\end{equation}


\begin{lemma} \label{lem:pmineq}
For any  $\phi \in C^{2}([-2\delta_{0},0])$ with
$\phi(-2\delta_{0})=0$ and $q_{2}\in C^{2}([-1,0])$ with
$q_{2}(-1)=q_{2}(0)=0 $, there is $c>0$ small enough such that:
$$
p(x)\ge c q_{1}(x_1)\phi(x_{1})q_2(x_2) \quad
 \forall x=(x_{1},x_{2})\in \Omega_{2\delta_0}
$$
with
$$
{ q_{1}(x_{1})=\frac{(-x_{1})^{\alpha+1}}
{\big(M_1(-x_{1})^{\alpha}+\epsilon)^{3}}}
$$
\end{lemma}

\begin{proof}
We apply the maximum principle. Since the function
$q_1(x_1)\phi(x_1) q_2(x_2)$, vanishes on $\partial
\Omega_{2\delta_0}$, it suffices to prove
$$
-c\nabla\cdot\Big[(a+h_{0})^{3}\nabla(q_{1}(x_{1})\phi(x_{1}) q_{2}(x_{2}))
\Big]
\le - \frac{\partial h_{0}}{\partial x_{1}}\quad
\forall x \in \Omega_{2\delta_0}
$$
Dividing by  $ (-x_1)^{\alpha-1}$ and using
\eqref{h1} it suffices to prove the existence of a positive
constant $K_1$ independent of $\epsilon$ such that
$$
-\frac{\nabla\cdot\big[(a+h_{0})^{3}\nabla(q_{1}(x_{1})\phi(x_{1}) q_{2}(x_{2}))\big]}
{(-x_1)^{\alpha-1} } \le K_{1}
$$
which is equivalent to
\begin{equation} \label{eq:prmax}
\begin{split}
&3(\epsilon+h_0)^{2}q_2\phi q_1^{'}\big[\frac{-\frac{\partial
h_0}{\partial x_1}}{(-x_1)^{\alpha-1}}\big]
 -(\epsilon+h_0)^3q_2\phi \frac{q_1''}{(-x_1)^{\alpha-1}}
 -2(\epsilon+h_0)^3q_2\phi^{'}\frac{q_1^{'}}{(-x_1)^{\alpha-1}}\\
&+3(\epsilon +h_0)^2q_1q_2\phi^{'}\big[\frac{-\frac{\partial
h_0}{\partial x_1}}{(-x_1)^{\alpha-1}}\big]
-(\epsilon+h_0)^3q_2\phi''\frac{q_1}{(-x_1)^{\alpha-1}}\\
&-(\epsilon+h_0)^3\phi q_2''\frac{q_1}{(-x_1)^{\alpha-1}}
-3(\epsilon+h_0)^2q_{1}\phi q_2^{'}\big[\frac{\frac{\partial
h_0}{\partial x_2}}{(-x_1)^{\alpha-1}}\big] \le K_1
\end{split}
\end{equation}
On the other hand, it is easy to see that a constant
$K_2$ independent of  $\epsilon$ exists such that:
\begin{gather*}
\vert q_{1}^{'}(x_{1})\vert\le
K_{2}\frac{(-x_{1})^{\alpha}}{\big(M_1(-x_{1})^{\alpha}+\epsilon\big)^{3}}
\quad \forall x_{1}\in[-2\delta_0,0], \\
 \vert q_{1}''(x_{1})\vert\le K_{2}\frac{(-x_{1})^{\alpha-1}}
{\big(M_1(-x_{1})^{\alpha}+\epsilon\big)^{3}}
 \quad  \forall x_{1}\in[-2\delta_0,0].
\end{gather*}
Integrating \eqref{h1} on $[x_1,0]$ we obtain
\begin{equation} \label{eq:hh0}
m_1 (-x_1)^\alpha \leq h_0(x) \leq M_1 (-x_1)^\alpha\quad
\forall x \in \Omega_{2\delta_0}
\end{equation}
Using  the above inequalities we obtain
\eqref{eq:prmax} which concludes the proof.
\end{proof}

Now we are able to give a first result in the case $\alpha\ge1$.

\begin{theorem} \label{thm:liminfinic1}
For $ \alpha\ge 1$ we have
$\int_\Omega p\,dx \to +\infty$ as $\epsilon \to 0$.
Moreover there exists $K > 0$ such that for $\epsilon  $
small enough we have
\begin{gather*}
\int_\Omega p\,dx \geq K \epsilon ^{\frac{2}{\alpha}-2}\quad
 \text{for } \alpha > 1,\\
\int_\Omega p\,dx \geq K \log (\frac{1}{\epsilon}) \quad
\text{for } \alpha = 1.
\end{gather*}
\end{theorem}

\begin{proof}
We apply Lemma \ref{lem:pmineq} with $\phi = 1$ on $[-\delta_0,0] $
and $q_2 \geq 0 $ with
$$
q_2 \in C^2([-1,0])\cap H^1_0(]-1,0[)
$$
and $\int_{-1}^0q_2(x_2) > 0 $. We deduce that there exists a constant
$c> 0$ independent of $\epsilon$ such that
$$
p(x_1,x_2) \geq cq_2(x_2)\phi(x_1)q_1(x_1),\quad \forall x \in
\Omega_{2\delta_0}
$$
with $q_1$ given in Lemma \ref{lem:pmineq}.
Taking into account the fact that $p$ is non negative on all of
$\Omega $ we obtain
\begin{equation}\label{eq:mincharge}
\int_\Omega p\,dx \geq c \int_{\Omega_{\delta_0}} q_2(x_2)q_1(x_1)dx =
c\int_{-1}^0 q_2(x_2)dx_2\int_{-\delta_0}^0 q_1(x_1)dx_1\,.
\end{equation}
On the other hand, for $ \alpha > 1$ and $\epsilon$
small enough we have
$$
\int_{-\delta_0}^0 q_1(x_1)dx_1 \geq \int_{-\epsilon^{1/\alpha}}^0
q_1(x_1)dx_1  \geq
\frac{\epsilon^{1+\frac2\alpha}}{(M_1+1)^3\epsilon^3(\alpha+2)}
$$
which, with \eqref{eq:mincharge}, gives the result for $\alpha > 1$.

For $\alpha =1$ an elementary calculation gives
\begin{equation*}
\begin{split}
\int_{-\delta_0}^0q_1(x_1)dx_1
=& \frac{-1}{3M_1^3}\int_{-\delta_0}^0
\frac{\frac{d}{dx_1}\left((-M_1x_1+\epsilon)^3
\right)}{(-M_1x_1+\epsilon)^3}dx_1\\
&-\frac{2\epsilon}{M_1^2}\int_{-\delta_0}^0
\frac{-M_1x_1}{(-M_1x_1+\epsilon)^3}dx_1
-\frac{\epsilon^2}{M_1^2}\int_{-\delta_0}^0
\frac{dx_1}{(-M_1x_1+\epsilon)^3}
\end{split}
\end{equation*}
We easily prove that the last two terms of the right-hand side are
bounded by a constant independent of $\epsilon$.
With the help of \eqref{eq:mincharge} we obtain the result.
\end{proof}

In the particular case when $h_0$ is symmetric in the $x_2$
direction we have the following asymptotic behavior of the
$x_2$-momentum.

\begin{theorem}\label{thm:liminfinim2c1}
Suppose that $h_0$ is
symmetric in $x_2$ with respect to $x_2=-1/2$ and $\alpha \geq 1$.
Then  for $\epsilon \to 0$, we have
\begin{gather*}
   \int_{\Omega} (x_2-x_2^0)p\,dx \to +\infty \quad\text{if }
  x_2^0 \in ]-1, -\frac{1}{2}[, \\
   \int_{\Omega} (x_2-x_2^0)p\,dx \to -\infty \quad\text{if }
  x_2^0 \in ]-\frac{1}{2},0[ ,\\
   \int_{\Omega} (x_2-x_2^0)p\,dx = 0 \quad\text{if }
  x_2^0  = -\frac{1}{2}\,.
\end{gather*}
\end{theorem}

\begin{proof}
By symmetry of $h_0$ the function $\bar{p}(x_1,x_2) = p(x_1,-1-x_2)$
is also a solution of problem \eqref{problem:P1}, so that by uniqueness
$p=\bar{p}$. We then have
$$
\int_\Omega (x_2-x_2^0)p\,dx = \int_\Omega (x_2+\frac{1}{2})p\,dx
-(x_2^0+\frac{1}{2})\int_\Omega p\,dx\,.
$$
The first integral of the right-hand side is equal to 0 by symmetry.
Then we have the result by Theorem \ref{thm:liminfinic1}.
\end{proof}

Now we shall give the behavior of the $x_1$-moment in the particular
case when $h_0$ is a tensor product. This is often the case in
practice for the contact line.
We begin by the following lemma which means that $p$ is bounded
uniformly in $\epsilon$ far from the line contact.

\begin{lemma} \label{lem:bornsup}
Suppose that $h_0(x_1,x_2) = a_1(x_1)a_2(x_2)$ with
$a_2 \in C^0([-1,0])$, $a_2 > 0$,
$a_1 \in H^1(]-1,0[)$, $a_1 \geq 0 $,
$ \frac{\partial a_1}{ \partial x_1} \leq 0  $.
Then there exists $C > 0 $ such that
$$
p(x) \leq C \int_{-1}^{x_1} \frac{ds}{(\bar{h}_0(s)+\epsilon)^2},
\quad \forall x \in \Omega
$$
where $\bar{h}_0(x_1) = a_{2m}a_1(x_1) $ with
$ a_{2m} = \min_{x_2 \in [-1,0]} a_2(x_2)$
\end{lemma}

\begin{proof}
We apply again the maximum principle. Let
$ q(x_1)=C \int_{-1}^{x_1} 1/(\bar{h}_0(s)+\epsilon)^2\,ds$.
Since $q \geq 0$ and $p=0$ on $\partial\Omega$ it suffices to show
that for $C > 0$ large enough we have
$$
-\frac{\partial}{\partial x_1}\big[(h_0+\epsilon)^3
\frac{\partial q}{\partial x_1} \big]
\geq -\frac{\partial h_0}{\partial x_1} \quad \forall x \in \Omega,
$$
that is
\begin{equation} \label{eq:ineq}
-C\frac{\partial h_0}{\partial x_1}
\frac{(h_0+\epsilon)^3}{(\bar{h}_0+\epsilon)^3} E(x)
 \geq -\frac{\partial h_0}{\partial x_1}
\end{equation}
with
$$
E(x) = 3 \frac{\bar{h}_0+\epsilon}{h_0+\epsilon}-2 \frac{a_{2m}}{a_2(x_1)}
$$
Now we have
$$
E(x) = \frac{a_{2m}}{a_2(x_2)}
\big[ 3 \frac{a_1+\frac{\epsilon}{a_{2m}}}{a_1+\frac{\epsilon}{a_{2}}}-2\big]
=  \frac{a_{2m}}{a_2(x_2)}  \frac{a_1+\epsilon(\frac{3}{a_{2m}}
- \frac{2}{a_{2}})  }{a_1+\frac{\epsilon}{a_{2}}}\,.
$$
We easily obtain
$$
E(x) \geq  \frac{a_{2m}}{a_{2M}}\quad \text{with} \quad
a_{2M}= \max_{x_2\in [-1,0]} a_2(x_2)
$$
which proves \eqref{eq:ineq} by taking
$C \geq a_{2M}/a_{2m}$
since $h_0 + \epsilon \geq  \bar{h}_0 + \epsilon$ and
$\frac{\partial h_0}{\partial x_1} \leq 0 $.
\end{proof}

\begin{theorem} \label{thm:liminfinim1c1}
Assuming \eqref{h1} to hold, and assuming further that  the function
$h_0$ is of the form  $h_0(x)=g_1(x_1)g_2(x_2)$ with  $g_1, g_2 \in
W^{1,\infty}(]-1,0[)$, then
$$
\int_\Omega (x_1-x_1^0)p\,dx \to +\infty \quad \text{as }
 \epsilon \to 0 \quad \text{ for any } x_1^0 \in ]-1,0[\,.
$$
\end{theorem}

\begin{proof}
We choose $\delta > 0$ such that $- \delta > \max (x_1^0,-\delta_0)$.
Then we have
\begin{equation*}
\int_\Omega (x_1-x_1^0)p\,dx \geq (-\delta-x_1^0)\int_{\Omega_\delta}p\,dx
+ \int_{-1}^{-\delta}  \int_{-1}^0 (x_1-x_1^0)p\,dx
\end{equation*}
We prove as in Theorem  \ref{thm:liminfinic1}  that the first integral
of the right-hand side tends to $+\infty$  since $-\delta-x_1^0 > 0 $.
Applying  Lemma \ref{lem:bornsup} with $a_1 = (-x_1)^\alpha g_1$ and
$a_2=g_2$ we easily prove that the second integral is bounded by a
constant independent of $\epsilon$. We then have the result.
\end{proof}

\begin{remark} \label{rmk3.11} \rm
An  interesting open question is to obtain an upper bound for
$p$ which allows to say that $p$ is bounded uniformly in $\epsilon$
far from $x_1=0$, without the global hypotheses that $h_0$ is a tensor product.
\end{remark}

\subsection{Point-contact case} \label{sec:ptct}

We now assume that $h_0$ is, in a neighbourhood of $x=0$, equivalent
to $|x|^\alpha $ with $\alpha > 0$, where $|\cdot|$ denotes the Euclidean norm.
 For simplicity we make here the following (non-essential) hypothesis
$$
h_0(x) = |x|^\alpha h_1(x)
$$
with $h_1 \in W^{1,\infty}(\Omega)$  and $ h_1 > 0 $. We denote
$$
m= \inf_{x \in \bar{\Omega}} h_1(x), \quad M= \sup_{x \in \bar{\Omega}} h_1(x)
$$
We shall prove that for $0 < \alpha < \frac{3}{2}$
(paragraph \ref{sec:finlimc2}) we have finite limits  of load and momenta
while for $\alpha \geq \frac{3}{2}$ (paragraph \ref{sec:infinlimc2})
they tend to $+\infty$.
We begin by the following existence, uniqueness and convergence results.

\begin{proposition}\label{prop:limcont}\quad
\begin{itemize}
\item If $  0 < \alpha < \frac{4}{3}$ then for any $0<\delta_1 < \frac{3}{4}$
there is a unique solution $p_0 \in H^1_0(\Omega,h_0^{2\delta_1},h_0^3) $
of \eqref{problem:P0} and $ p\to p_0$ in $H^1_0(\Omega,h_0^{2\delta_1},h_0^3) $.

\item If $  \frac{4}{3}\le \alpha < 2$ then for any
$\delta_1 > \frac{3}{2}- \frac{1}{\alpha}$ there is a unique solution
$p_0 \in H^1_0(\Omega,h_0^{2\delta_1},h_0^3) $ of \eqref{problem:P0} and
$ p\to p_0$ in $H^1_0(\Omega,h_0^{2\delta_1},h_0^3) $.
\end{itemize}
\end{proposition}

\begin{proof}
The first hypothesis of Proposition \ref{prop:exist1} is obvious.
The second one is evident for $  \frac{4}{3}\le \alpha < 2$ and
$\delta_1 \geq \frac{3}{2}- \frac{1}{\alpha}$.
For $  0< \alpha < \frac{4}{3}$ and
$ 0 < \delta_1 < \frac{3}{4}$ we use the inequality
$\sqrt{s^2+x_2^2}  \geq |s|$ and the result is immediate.
\end{proof}

\subsubsection{ Finite-limit case
($\alpha < \frac{3}{2}$) } \label{sec:finlimc2}
In the following we prove that the limits of load and momenta are
finite  for $ \alpha<\frac{4}{3}$ without supplementary hypotheses.
For $\frac{4}{3}\leq\alpha<\frac{3}{2}  $ we prove the same result
but adding a restrictive supplementary assumption on $h_0$.

\begin{theorem} \label{thm:limfinic2}
For $ 0<\alpha<\frac{4}{3}$ we have for any $ (x_1^0,x_2^0)\in\Omega$:
$$
\int_\Omega p \to \int_\Omega p_0, \quad \int_\Omega (x_k-x_k^0)p
\to  \int_\Omega (x_k-x_k^0)p_0; \quad k=1,2
$$
with $p_0$ solution of the limit problem \eqref{problem:P0}.
\end{theorem}

\begin{proof}
 From Proposition \ref{prop:limcont} we have $p\to p_0$ in
$H^1_0(\Omega,h_0^{2\delta_1},h_0^3) $-strongly for any $\delta_1$ such that
\begin{equation}\label{eq:delta1m}
0<\delta_1 <  \frac{3}{4}
\end{equation}
On the other hand we remark that if
\begin{equation} \label{eq:delta1M}
0 < \delta_1 < \frac{1}{\alpha}
\end{equation}
then $ \int_\Omega h_0^{-2\delta_1}$ is finite, which by Cauchy-Schwartz
inequality gives the continuous embedding of
$ H^1_0(\Omega,h_0^{2\delta_1},h_0^3)$ in $L^1(\Omega)$.
This will prove the three desired convergence.
Now the existence of at least a $\delta_1$ satisfying \eqref{eq:delta1m}
and \eqref{eq:delta1M} is assured if
$ 3/4< 1/\alpha$ which is equivalent to $\alpha <  4/3 $.
\end{proof}

\begin{theorem}\label{thm:limfinic3}
For  $ 4/3 \leq \alpha < 3/2$, under the supplementary  hypothesis
$\triangle h_0 \geq 0 $ on $\Omega$ we have the same convergence as in
Theorem \ref{thm:limfinic2}.
\end{theorem}

\begin{proof}
We need here an estimation of $p$ in a stronger norm than
$\|h_0^{3/2} \nabla \cdot \|_{L^2(\Omega)} $ in order to obtain
$\|h_0^{\delta_1}  p \|_{L^2(\Omega)} $ bounded with a better parameter
$ \delta_1$ than in Theorem \ref{thm:limfinic2}. We prove in the following that $\|h_0^{(3-\delta)/2} \nabla p \|_{L^2(\Omega)}  $ is bounded for an appropriate $\delta > 0$.
Taking  $ \varphi=\big(h_{0}+\epsilon\big)^{-\delta}p$ with
$0<\delta<\frac{2}{\alpha}-1\leq \frac{1}{2}$ as a test function in the
 variational formulation  of $\eqref{problem:P1}$ we obtain
\begin{equation}
\label{alp}
\int_{\Omega}{\big(h_{0}+\epsilon\big)^{3-\delta}\big\vert\nabla
p\big\vert^{2}}
= \delta\int_{\Omega}{\big(h_{0}+\epsilon\big)
^{2-\delta}\nabla h_{0}p\nabla p}
-\int_{\Omega}{\frac{\partial h_{0}}{\partial
x_{1}}\big(h_{0}+\epsilon\big)^{-\delta}p}
=:I_{1}+I_{2}
\end{equation}
Using Green's formula we deduce
\begin{align*}
 I_{1} & =-\frac{\delta}{2}\int_{\Omega}{\nabla \cdot\Big[\big(h_{0}
+\epsilon\big)^{2-\delta}\nabla h_{0}\Big]p^{2}}\\
 &=-\frac{\delta}{2}\int_{\Omega}{\Big[(2-\delta)
\big(h_{0}+\epsilon\big)^{1-\delta}\big\vert\nabla
h_{0}\big\vert^{2}+\big(h_{0}+\epsilon\big)^{2-\delta}\Delta
h_{0}\Big]p^{2}}
\end{align*}
which is negative, thanks to the additional hypothesis $\Delta h_{0}\ge0$.
On the other hand
\begin{equation*}
\begin{split}
|I_2|& \leq \frac{1}{1-\delta} \int_\Omega
\big| (h_0+\epsilon)^{1-\delta}  \frac{\partial p}{\partial x_1}\big|\\
&= \frac{1}{1-\delta} \int_\Omega (h_0+\epsilon)^{(3-\delta)/2} \left | \frac{\partial p}{\partial x_1}\right| (h_0+\epsilon)^{-(1+\delta)/2}\\
& \leq \frac{1}{1-\delta} \Big( \int_\Omega(h_0+\epsilon)^{3-\delta}
\big| \frac{\partial p}{\partial x_1}\big|^2\Big)^{1/2}
 \Big( \int_\Omega(h_0+\epsilon)^{-1-\delta} \Big)^{1/2}\\
& \leq \frac{1}{2} \int_\Omega(h_0+\epsilon)^{3-\delta}
\big| \frac{\partial p}{\partial x_1}\big|^2dx
+  \frac{1}{2} \frac{1}{(1-\delta)^2}
\int_\Omega \frac{dx}{(h_0+\epsilon)^{1+\delta}}dx
\end{split}
\end{equation*}
The last integral of the above inequality is bounded uniformly in $\epsilon$
 due to the hypotheses on $\delta$.
We deduce from \eqref{alp} that
$$
\Big(\int_{\Omega}{h_{0}^{3-\delta}\big\vert\nabla p\big\vert^{2}dx}
\Big)^{1/2}\le C
$$
Applying Lemma \ref{lem:principal} with
$f=h_0, \delta_2=\frac{3}{2}-\frac{\delta}{2}$ and
$\delta_1 > \frac{3}{2}-\frac{\delta}{2}-\frac{1}{\alpha}$ we deduce that
$p$ is bounded in $H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$.
We then infer the existence of
$\xi \in H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$ and of a
subsequence of $\epsilon$ such that $ p \to \xi$ weakly in
$ H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$.
From the continuous embedding of $H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$
 in $H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3})$ and by identification and
uniqueness of $p_0$ we deduce that $p \to p_0$ weakly in
$H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$  for the entire sequence.

Choosing now $ \delta=\frac{2}{\alpha}-1-\eta, \delta_1=\frac{3}{2}-\frac{\delta}{2}-\frac{1}{\alpha}+\frac{\eta}{2}=2-\frac{2}{\alpha}+\eta$ with $0 < \eta < \frac{3}{\alpha} - 2$ we obtain $\int_\Omega h_0^{-2\delta_1} <+\infty$ which implies the continuous embedding of $H^1_0(\Omega,h_0^{2\delta_1} ,h_0^{3-\delta})$ in $L^1(\Omega)$.\\
 We then obtain $p\to p_0$ weakly in $L^1(\Omega)$
which gives the desired convergence.
\end{proof}


\subsubsection{Infinite-limit case$(\alpha\ge 3/2)$} \label{sec:infinlimc2}

In this paragraph we use the polar coordinates $ r,\theta$:
$$
 x_1=r\cos{\theta}, \quad x_2=r\sin{\theta}
$$
and we denote  $\tilde{\Omega}$, the image of  $\Omega$ by this change
of variables.
For simplicity notations we use the same notation as in cartesian
coordinates  (for example $ h_0(r,\theta)$ means
$h_0(r\cos{\theta},r\sin{\theta})$).
We set:
$$
\tilde{\Omega}_{r}=]0,r[ \; \times \; ]-\pi,-\frac{\pi}{2}[ \;
\subset \; \tilde{\Omega}, \quad \forall r \in ]0,1[
$$
 The problem   \eqref{problem:P1} becomes
\begin{equation} \label{eqpolair}
\begin{gathered}
\begin{aligned}
&\frac{\partial}{\partial r}\big[ ( a +  h_{0}(r,\theta) )^{3}r
\frac{\partial p}{dr}\big]+
\frac{\partial}{\partial\theta}\big[ \frac{( a +  h_{0}(r,\theta) )^{3}}{r}
\frac{\partial p}{\partial\theta}\big]\\
&=r\frac{\partial h_{0}}{\partial r} \cos{\theta}-\sin{\theta}
\frac{\partial h_{0}}{\partial \theta} \quad  \text{in }  \tilde{\Omega}
\end{aligned}
\\
{ p=0} \quad  \text{in }  \partial\tilde{\Omega}\,.
\end{gathered}
\end{equation}
Let us remark that from the relation $h_0=r^\alpha h_1(r,\theta)$
there exists a positive constant $K > 0$ and $r_1 \in ]0,1[$ such that
\begin{equation}\label{eq:h0min}
\frac{\partial h_{0}}{\partial r} \geq Kr^{\alpha-1}, \quad
\forall (r,\theta) \in \tilde{\Omega}_{r_1}\,.
\end{equation}
We recall that $h_0$ is non-increasing in  $x_1$ which is equivalent
in polar coordinates to
\begin{equation} \label{eq:h0dec1}
r \frac{\partial h_0}{\partial r} \cos \theta
\leq \sin \theta \frac{\partial h_0}{\partial \theta}
\quad \text{ on } \tilde{\Omega}
 \end{equation}

We need here the following supplementary local condition:
There exists $r_2 > 0$ and $\beta \in ]0,1[$ such that
\begin{equation} \label{eq:h0dec2}
\beta r \frac{\partial h_0}{\partial r} \cos \theta
\leq \sin \theta \frac{\partial h_0}{\partial \theta}
\quad \text{ on } \tilde{\Omega}_{r_2}\,.
 \end{equation}

\begin{remark} \label{rmk3.15} \rm
Hypothesis \eqref{eq:h0dec2} is a little stronger locally
than \eqref{eq:h0dec1} and is  true if for example
$ \frac{\partial h_0}{\partial \theta}  \leq 0 $
locally (in particular if $h_0$ is radial) since
$ \frac{\partial h_0}{\partial r} > 0$ locally and
$\cos \theta \leq 0$
\end{remark}

We now give an analog of Lemma \ref{lem:pmineq}  in the line-contact case.

\begin{lemma} \label{lem:liminfc2}
Suppose that  \eqref{eq:h0dec2} is fulfilled and set
$r_0=\min\{r_1,r_2\} $. Then  for any $\phi \in C^2([0,r_0]) $ with
$\phi(r_0) = 0 $, a constant $c > 0$ exists such that
$$
p(r,\theta) \geq c q_1(r) \phi (r) q_2(\theta) \quad
\forall (r,\theta) \in \tilde{\Omega}_{r_0}
$$
with
$$
q_1(r) = \frac{r^{\alpha+1}}{(Mr^\alpha+\epsilon)^3},
\quad \text{and } q_2(\theta) = \cos^3 \theta \sin \theta
$$
\end{lemma}

\begin{proof}
It suffices to prove the inequality
\begin{equation} \label{eq:inegr}
\frac{\partial}{\partial r}\Big[(\epsilon+h_{0})^{3}r
\frac{\partial}{\partial r}\big(cq_{1}q_{2}\phi\big)\Big]+
\frac{\partial}{\partial \theta}\Big[\frac{(\epsilon+h_{0})^{3}}{r}
\frac{\partial}{\partial \theta}\big(cq_{1}q_{2}\phi\big)\Big]
\ge r\frac{\partial h_0}{\partial r}\cos{\theta}-\sin{\theta}
\frac{\partial h_0}{\partial \theta}
\end{equation}
From \eqref{eq:h0dec2}, we have
$$
 r\frac{\partial h_0}{\partial r}\cos{\theta}
-\sin{\theta}\frac{\partial h_0}{\partial \theta}
\leq (1-\beta) r \frac{\partial h_0}{\partial r}\cos \theta\,.
$$
Carrying out  the differentiations in the left-hand side of \eqref{eq:inegr}
and dividing by $ r  \frac{\partial h_0}{\partial r} \cos \theta$,
we obtain the following inequality in $\tilde{\Omega}_{r_0}$,
\begin{equation} \label{eq:inegr1}
\begin{aligned}
&  \frac{q_{2}(\theta)}{\cos{\theta}}\phi \Big[3(\epsilon+h_{0})^{2}
q_{1}^{'}(r) +(a+h_{0})^{2} q_{1}^{'}(r)\frac{(\epsilon+h_{0})}{r}
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1}\\
&+(\epsilon+h_{0})^{3}q_{1}''(r)\Big(\frac{\partial h_{0}}{\partial r}
\Big)^{-1}\Big]\\
&  +\frac{q_{2}(\theta)}{\cos{\theta}}\phi^{'}(r)
\Big[3\big(\epsilon+h_{0}\big)^{2}q_{1}+
\big(\epsilon+h_{0}\big)^{3}\frac{q_{1}}{r}
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1}+
2\big(\epsilon+h_{0}\big)^{3}q_{1}^{'}(r)
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1}\Big]\\
& +\frac{q_{2}(\theta)}{cos{\theta}}(\epsilon+h_{0})^{3}q_{1}\phi''
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1}
+3\frac{ q_{2}^{'}(\theta)}{\cos{\theta}}\phi (\epsilon+h_{0})^2q_1
\frac{\partial h_{0}}{\partial \theta}\frac{1}{r^2}
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1} \\
&+\frac{ q_2''}{\cos{\theta}}\phi \frac{(\epsilon+h_{0})^{3} q_{1}}{r^2}
\Big(\frac{\partial h_{0}}{\partial r}\Big)^{-1} \\
&\le \frac{1-\beta}{c}.
\end{aligned}
\end{equation}
Remark also that there is a constant $K_1 >0$ such that
\begin{equation}\label{eq:ineqq1}
\vert q_{1}^{'}(r)\vert \le K_{1}
\frac{r^{\alpha}}{\big(Mr^{\alpha}+\epsilon\big)^{3}}, \quad
\vert q_{1}''(r)\vert \le K_{1}
\frac{r^{\alpha-1}}{\big(Mr^{\alpha}+\epsilon\big)^{3}}
\end{equation}
Using now  \eqref{eq:h0min}, \eqref{eq:ineqq1} and the expression of $q_2$
we obtain that the absolute value of the left-hand side of \eqref{eq:inegr1}
is bounded by a constant. Taking $c$ small enough we obtain the result.
\end{proof}

\begin{theorem}\label{thm:liminfinic2}
Under hypothesis \eqref{eq:h0dec2} we have
$\int_\Omega p\,dx \to +\infty$  for $\epsilon \to 0$
Moreover there exists $K > 0$ such that for $\epsilon$ small enough we have
\begin{gather*}
 \int_\Omega p\,dx \geq K \epsilon ^{\frac{3}{\alpha}-2}\quad
   \text{for } \alpha > \frac{3}{2}, \\
 \int_\Omega p\,dx \geq K \log (\frac{1}{\epsilon}) \quad
 \text{for } \alpha = \frac{3}{2}\,.
\end{gather*}
\end{theorem}

\begin{proof} Using polar coordinates and the non-negativity of $p$
we have
$$
\int_\Omega p dx \geq \int_{\tilde{\Omega}_\rho} rp(r,\theta)\,dr\,d\theta,
\quad \forall \rho \in ]0,1]
$$
Applying Lemma \ref{lem:liminfc2} with $\phi =1 $ on
$[0,\frac{r_0}{2}]$ we show that there exists a $c > 0$ such that
$$
\int_\Omega p dx \geq c \int_{-\pi}^{-\pi/2} q_2(\theta)d\theta
\cdot \int_0^{r_0/2} rq_1(r)dr\,.
$$
As in the proof of Theorem \ref{thm:liminfinic1} with some elementary
computations we obtain the result.
\end{proof}

\begin{theorem} \label{thm3.18}
Under hypothesis  \eqref{eq:h0dec2} if moreover
$h_1(r,\theta) = g_1(r)g_2(\theta)$ with $g_1 \in C^1[0,\sqrt{2}]$,
$g_2 \in C^1[-\pi,-\frac{\pi}{2}]$, $g_1(r) > 0, g_2(\theta) > 0$ and
$\frac{d}{dr} \big(r^\alpha g_1(r) \big) \geq 0$ we have
$$
\int_\Omega (x_k-x_k^0)p\,dx \to +\infty, \quad \text{as } \epsilon \to 0,
\quad k=1,2\,.
$$
\end{theorem}

\begin{proof}
First we prove that there exists  $K > 0$ large enough such that
\begin{equation} \label{eq:limsupp}
p(r,\theta) \leq K \int_r^{\sqrt{2}} \frac{ds}{(\overline{h}_0(s)
+\epsilon)^2}, \quad \forall (r,\theta) \in \tilde{\Omega}
\end{equation}
with
$$
\overline{h}_0(r) = g_{2m}r^\alpha g_1(r), \quad \text{and}\quad
 g_{2m} = \min_{\theta \in [-\pi,-\frac{\pi}{2}]} g_2(\theta)\,.
$$
We use the maximum principle as in the proof of Lemma \ref{lem:bornsup}.
It suffices to prove the following inequality
\begin{equation}\label{eq:Er}
K \frac{(\epsilon+h_0)^3}{(\epsilon+\overline{h}_0)^2}
+ Kr  \frac{\partial h_0}{\partial r}
\frac{(\epsilon+h_0)^3}{(\epsilon+\overline{h}_0)^3}E(r,\theta)
 \geq -r \frac{\partial h_0}{\partial r} \cos \theta
+ \sin \theta \frac{\partial h_0}{\partial \theta}
\end{equation}
with
$$
E(r,\theta) = 3 \frac{\epsilon+\overline{h}_0}{\epsilon+h_0}
-2 \frac{g_{2m} }{g_2(\theta)}.
$$
We consider two situations

\noindent\textbf{Case 1: $r \leq r_1$ with $r_1$ given in \eqref{eq:h0min}.}
As in the proof of Lemma \ref{lem:bornsup} we have
$E(r,\theta) \geq  \frac{g_{2m}}{g_{2M}}$ with
$ g_{2M} =  \max_{\theta \in [-\pi,-\frac{\pi}{2}]} g_2(\theta)$.
Then it suffices to prove for $r \leq r_1$
\begin{equation} \label{eq:in}
Kr\frac{\partial h_0}{\partial r} \frac{g_{2m}}{g_{2M}}
\geq  -r \frac{\partial h_0}{\partial r} \cos \theta
+ \sin \theta \frac{\partial h_0}{\partial \theta}
\end{equation}
with $K > 0$ large enough.
From \eqref{eq:h0min} the function
$ \big| \frac{\partial h_0}{\partial \theta}/
   \big(r \frac{\partial h_0}{\partial r} \big) \big| $
is bounded for $r \leq r_1 $. Now dividing by
$ r \frac{\partial h_0}{\partial r} $ the inequality \eqref{eq:in}
is obvious, which proves \eqref{eq:Er} for $r \leq r_1$.

\noindent\textbf{Case 2: $r > r_1$.}
We shall prove
\begin{equation} \label{eq:inn2}
K(\epsilon +\bar{h}_0) \geq -r \frac{\partial h_0}{\partial r}
 \cos \theta + \sin \theta \frac{\partial h_0}{\partial \theta}
\quad \text{for } r > r_1
\end{equation}
for $K > 0$ large enough, which implies  \eqref{eq:Er} for $ r > r_1$.
We have, from hypothesis on $h_0$, $\bar{h}_0(r) \geq \bar{h}_0(r_1) $
so $ K(\epsilon +\bar{h}_0(r)) \geq K  \bar{h}_0(r_1)$ for $r \geq r_1 $.
Since the right-hand side of \eqref{eq:inn2} is bounded, the result
is obvious.

 From the tow cases above, the proof of \eqref{eq:limsupp} is complete.
\end{proof}

Now we have
\begin{equation}\label{eq:mom}
\int_\Omega (x_k-x_k^0)p\,dx = \int_{\tilde{\Omega}_\delta}
r(r\cos \theta-x_1^0)p\,dr\,d\theta
+  \int_{\tilde{\Omega}-\tilde{\Omega}_\delta}r(r\cos \theta-x_1^0)
p\,dr\,d\theta\,.
\end{equation}
We choose $0 < \delta < \min(r_0,\frac{|x_1^0|}{2})$ with $r_0$ given
in Lemma \ref{lem:liminfc2}.
We have $r\cos \theta-x_1^0 \geq -r+|x_1^0| \geq \frac{|x_1^0|}{2} $ for
$ r < \delta$, so
$$
 \int_{\tilde{\Omega}_\delta}r(r\cos \theta-x_1^0)p\,dr\,d\theta
 \geq  \frac{|x_1^0|}{2}   \int_{\tilde{\Omega}_\delta}rp\,dr\,d\theta
$$
and this last integral goes to $+\infty$ as in the proof of
Theorem \ref{thm:liminfinic2}. Now using \eqref{eq:limsupp}
we easily prove that the second integral of the right-hand side
of \eqref{eq:mom} is bounded which ends the proof for $k=1$.
The case $k=2$ is similar.

\section{Asymptotic behavior in the inequality case (Problem \eqref{problem:P2})}

In this section we suppose for simplicity that $\Omega=]-1,1[^2 $ and that
$h_0$ is non-increasing in $x_1$ on $ \Omega_1$ and non-decreasing in
$x_1$ on $ \Omega_2$ where we denote
$$
\Omega_1=]-1,0[ \times ]-1,1[ \quad \text{and}\quad
 \Omega_2=]0,1[ \times ]-1,1[\,.
$$
We study the asymptotic behaviour of the solution $p$ of  \eqref{problem:P2}
when $\epsilon \to 0$. In order to introduce the limit problem we define
$\mathcal{K}_{\delta_1}$ as the closure of
$\mathcal{K}= \{ \varphi \in H^1_0(\Omega): \varphi \geq 0\}$ with respect
to the norm of  $H_0^1(\Omega,h_0^{2\delta_1} ,h_0^3)$ .
We remark that $\mathcal{K}_{\delta_1} $ is a closed convex set in
$H_0^1(\Omega,h_0^{2\delta_1} ,h_0^3)$.

We now define the limit problem
\begin{equation} \label{problem:P02}
\begin{gathered}
\text{Find } p_0  \in \mathcal{K}_{\delta_1} \text{ such that } \\
 \int_\Omega h_0^3 \nabla  p_0 \nabla
(\varphi -p_0)dx \ge \int_\Omega h_0 \frac{\partial}{\partial
x_1}  (\varphi -p) \quad \forall \varphi \in  \mathcal{K}_{\delta_1}
\end{gathered}
\end{equation}
We now give the following existence, uniqueness and convergence results.

\begin{proposition} \label{prop:ineq}
Suppose that $ \int_\Omega \frac{dx}{h_0} < +\infty $ and
$\delta_1 \in \mathbb{R}^{+}$ is such that
\begin{align*}
K&=\sup_{x_2\in[-1,1]} \int_{-1}^{0}\int_{-1}^{x_1}h_0^{2\delta_1-3}(s,x_2)
\,ds\,dx_1 \\
&\quad +\sup_{x_2\in[-1,1]} \int_{0}^{1}\int_{0}^{x_1}h_0^{2\delta_1-3}(s,x_2)
\,ds\,dx_1<\infty
\end{align*}
Then Problem \eqref{problem:P02} admits an unique solution
$p_0  \in \mathcal{K}_{\delta_1} $ which is independent of $\delta_1 $.
Also the solution $p$ of problem  \eqref{problem:P2}  converges,
when $\epsilon \to 0$,  to $p_0$ strongly in
$H_0^1(\Omega,h_0^{2\delta_1} ,h_0^3)$.
\end{proposition}

\begin{proof}
 We apply Lemma \ref{lem:principal} with $d_1=0$ and $\delta_2=\frac{3}{2} $
and we obtain classically the first result.
 Taking $\varphi=0$ in \eqref{problem:P2} we obtain
$$
\int_\Omega (h_0+\epsilon)^3|\nabla p|^2 dx
\leq \int_\Omega h_0 \frac{\partial p}{\partial x_1}
=  \int_\Omega (h_0+\epsilon) \frac{\partial p}{\partial x_1}
$$
which leads to
\begin{equation} \label{eq:in1}
\left \| (h_0+\epsilon)^{3/2}\nabla p\right\|_{L^2(\Omega)}
\leq \Big( \int_\Omega \frac{dx}{h_0(x)}\Big)^{1/2}
\end{equation}
which implies
\begin{equation} \label{eq:in2}
\left \| h_0^{3/2}\nabla p\right\|_{L^2(\Omega)} \leq
\Big( \int_\Omega \frac{dx}{h_0(x)}\Big)^{1/2}.
\end{equation}
 From Lemma \ref{lem:principal} with $d_1=0$ and
$\delta_2=3/2 $ we deduce that $p$ is bounded in
$H_0^1(\Omega,h_0^{2\delta_1},h_0^3)$. Then an element
$\xi \in  \mathcal{K}_{\delta_1} $ exists such that, up to a subsequence,
$ p \to \xi $ weakly in $H_0^1(\Omega,h_0^{2\delta_1} ,h_0^3)$.
We now pass to the limit in all terms in the inequality
\begin{equation} \label{eq:iv}
\int_\Omega  (h_0+\epsilon)^3 \nabla p \cdot \nabla \varphi
 \geq \int_\Omega  (h_0+\epsilon)^3 |\nabla p|^2 + \int_\Omega h_0
\frac{\partial}{\partial x_1}  (\varphi -p) \quad
\forall \varphi \in  \mathcal{K}.
\end{equation}
Writing for any $ \varphi \in  \mathcal{K}$,
\begin{align*}
\int_\Omega  (h_0+\epsilon)^3 \nabla p \cdot \nabla \varphi
&= \int_\Omega \big((h_0+\epsilon)^{3/2}-h_0^{3/2} \big)
(h_0+\epsilon)^{3/2} \nabla p \cdot \nabla \varphi \\
&\quad + \int_\Omega \big((h_0+\epsilon)^{3/2}-h_0^{3/2} \big)
 h_0^{3/2} \nabla p \cdot \nabla \varphi
 +  \int_\Omega h_0^3 \nabla p \cdot \nabla \varphi
\end{align*}
and using \eqref{eq:in1} and \eqref{eq:in2} we deduce
\begin{equation}\label{eq:lim1}
\int_\Omega  (h_0+\epsilon)^3 \nabla p \cdot \nabla \varphi \to
 \int_\Omega h_0^3 \nabla \xi \cdot \nabla \varphi \quad \forall
 \varphi \in \mathcal{K}\,.
\end{equation}
Writing also
$$
\int_\Omega h_0  \frac{\partial}{\partial x_1}  (\varphi -p)
= \int_\Omega h_0^{-1/2}h_0^{3/2}  \frac{\partial}{\partial x_1}  (\varphi -p),
$$
we obtain
\begin{equation}\label{eq:lim2}
\int_\Omega   h_0  \frac{\partial}{\partial x_1}  (\varphi -p)
\to \int_\Omega   h_0  \frac{\partial}{\partial x_1}  (\varphi -\xi)
 \quad \forall  \varphi \in \mathcal{K}.
\end{equation}
Finally we have
$$
 \int_\Omega  (h_0+\epsilon)^3 |\nabla p|^2
\geq  \int_\Omega  h_0^3 |\nabla p|^2
$$
which gives
\begin{equation}\label{eq:lim3}
\lim \inf  \int_\Omega  (h_0+\epsilon)^3 |\nabla p|^2
\geq \int_\Omega  h_0^3 |\nabla \xi|^2\,.
\end{equation}
 From \eqref{eq:iv}-\eqref{eq:lim3} we deduce
$$
 \int_\Omega h_0^3 \nabla \xi \cdot \nabla \varphi
\geq \int_\Omega  h_0^3 |\nabla \xi|^2
+ \int_\Omega   h_0  \frac{\partial}{\partial x_1}  (\varphi -\xi)\quad
\forall  \varphi \in \mathcal{K}\,.
$$
By denseness  and uniqueness we deduce that $\xi = p_0 $ and that the
entire sequence $p$ converges to $p_0$.
It remains to prove the strong convergence. We have
$$
 \int_\Omega h_0^3  |\nabla (p-p_0)|^2 \leq  \int_\Omega (h_0+\epsilon)^3   |\nabla p|^2 +
 \int_\Omega   h_0^3  |\nabla p_0|^2 -2 \int_\Omega   h_0^3
\nabla p \cdot \nabla p_0\,.
$$
Taking $ \varphi = 0 $ in \eqref{problem:P2} and  \eqref{problem:P02} and passing to the limit we deduce
$$
\lim_{\epsilon \to 0}  \int_\Omega  h_0^3  |\nabla (p-p_0)|^2 \leq 2  \int_\Omega h_0 \frac{\partial p_0}{\partial x_1}
-2 \int_\Omega   h_0^3  |\nabla p_0|^2.
$$
The right hand-side of the above inequality is 0
(take  $ \varphi = 0 $ and  $ \varphi = 2p_0 $ in \eqref{problem:P02})
which proves the result.
\end{proof}

In the following we shall use the classical notation
\begin{gather*}
\Omega_\epsilon^0 = \{ x \in \Omega: p(x) =0 \} \quad \text{(cavitation zone)}\\
\Omega_\epsilon^+ = \{ x \in \Omega: p(x) > 0\} \quad \text{(active zone)}
\end{gather*}
It is well known that if $x \in \Omega_\epsilon^0 $ then
$ \frac{\partial h_0}{\partial x_1} \leq 0$ which implies the
inclusion
$\Omega_1 \subset \Omega_\epsilon^+$
so that in $\Omega_1 $, $p$ satisfies
\begin{equation}\label{eq:equap}
\nabla \cdot \left[ (h_0+\epsilon)^3\nabla p \right]
=  \frac{\partial h_0}{\partial x_1}.
\end{equation}
The next lemma will be useful for the proofs in the infinite-limit cases.

\begin{lemma} \label{lem:min}
Let  $\Omega^*$ be an open subset of $ \Omega_1 $ with Lipschitz
boundary and $p^*$ the solution of \eqref{eq:equap} with $ p^* =0$
on $\partial  \Omega^*$. Then $p \geq p^* $ on $ \Omega^*$.
\end{lemma}

\begin{proof}
Since $p$ and $p^*$ satisfy  \eqref{eq:equap} on $ \Omega^*$, we have the
result by the maximum principle since $p\geq 0 $ on  $\partial  \Omega^*$
and $p^*=0 $ on  $\partial  \Omega^*$.
\end{proof}

\subsection{Line-contact case}

We suppose for simplicity
$h_0(x) = (-x_1)^\alpha h_1(x), \forall x \in \Omega $ with $\alpha > 0$,
$h_1 \in W^{1,\infty}(\Omega)$ and $h_1 > 0$.
We have the following result.

\begin{theorem} \label{thm4.3}
For any $\alpha \in ]0,1[ $ and $(x_1^0,x_2^0) \in \Omega$ we have
\begin{gather*}
\int_\Omega p\,dx \to \int_\Omega p_0 dx \\
\int_\Omega (x_k-x_k^0)p\,dx \to \int_\Omega (x_k-x_k^0)p_0dx , \quad k=1,2
\end{gather*}
\end{theorem}

\begin{proof}
The hypotheses of Proposition \ref{prop:ineq} are satisfied with
$\delta_1 = 1/2$. Then the proof is exactly as the proof of
Theorem \ref{thm:limfinic1}.
\end{proof}

Now for the infinite-limit case we use  Lemma \ref{lem:min} with
$\Omega^* = \Omega_1 $. Performing for $p^*$ the same kind of estimates
as for $p$ in paragraph \ref{sec:infinlimc1} we easily obtain the
following result.

\begin{theorem}
For $ \alpha \geq 1$ we have
\begin{enumerate}
\item $ \int_\Omega p\,dx \to +\infty$
\item If $h_0$ is symmetric in $x_2$ with respect to $x_2=0 $ then
\begin{itemize}
\item $\int_\Omega (x_2-x_2^0)p\,dx \to +\infty $ for $x_2^0 < 0 $
\item $\int_\Omega (x_2-x_2^0)p\,dx \to -\infty $ for $x_2^0 > 0 $
\item  $\int_\Omega (x_2-x_2^0)p\,dx =0$  for $x_2^0 = 0 $
\end{itemize}
\item We assume that $h_1$ is of the form
$h_1(x) = g_1(x_1)g_2(x_2) $, $g_2 \in C^0([-1,1]) $,
$g_1  \in H^1(]-1,1[)$,  $g_2 > 0$, $g_1 > 0$ and
$\frac{d}{dx_1}((-x_1)^\alpha g_1) \leq 0$.
Then for any $x_1^0 \in ]-1,0[ $ we have
$$
\int_\Omega (x_1-x_1^0)p\,dx \to +\infty \quad \text{as } \epsilon \to 0
$$
\end{enumerate}
\end{theorem}

\begin{remark} \label{rmk4.5} \rm
In the above theorem we obtained the behaviour of the $x_1$-moment
for $x^0 \in \Omega_1$ only. The problem is open when  $ x^0$ is such
that $x_1^0 \geq 0$.
\end{remark}

\subsection{Point-contact case}

We suppose $h_0(x) = |x|^\alpha h_1(x) $ with $h_1$ as in Section
 \ref{sec:ptct}.
The analogous of Theorem \ref{thm:limfinic2}  for the inequality
problem is the following.

\begin{theorem} \label{thm4.6}
For $0 < \alpha < 4/3$ we have for any $(x_1^0,x_2^0) \in \Omega$
\begin{gather*}
 \int_\Omega p\,dx \to \int_\Omega p_0\,dx \,,\\
\int_\Omega (x_k-x_k^0)p\,dx \to \int_\Omega (x_k-x_k^0)p_0dx, \quad k=1,2\,.
\end{gather*}
\end{theorem}

The proof the above theorem uses Proposition \ref{prop:ineq} and is
 exactly as the proof of Theorem  \ref{thm:limfinic2}.

For the infinite-limit case we pass again to polar coordinates.
We have the following result which is immediate applying
Lemma \ref{lem:min} with $\Omega^* = ]-1,0[^2 $ which reduces
the problem to the equation case.

\begin{theorem} \label{thm4.7}
Suppose that  $r_2 > 0$ and $\beta \in ]0,1[$ exist such that
$$
\beta r \frac{\partial h_0}{\partial r} \cos \theta \leq
\sin \theta \frac{\partial h_0}{\partial \theta}, \quad
\forall (r,\theta) \in [0,r_2]\times ]-\pi, -\frac{\pi}{2} [\,.
$$
Then  for $\alpha \geq 3/2$, we have
$$
\int_\Omega p\,dx \to + \infty\,.
$$
Also fir $h_1$ of the form $h_1(r,\theta) = g_1(r)g_2(\theta) $ with
$g_1 \in C^1[0,\sqrt{2}]$, $g_2 \in C^1[-\pi,-\frac{\pi}{2}]$,
$g_1 > 0$, $g_2 > 0$ and $\frac{d}{dr}(r^\alpha g_1(r)) \geq 0$, we have
$$
\int_\Omega (x_k-x_k^0)p\,dx \to + \infty \quad \text{as } \epsilon \to 0
$$
for all $x_k^0 \in ]-1,0[$, $k=1,2$\,.
\end{theorem}

We remark that for $ \alpha \in [4/3, 3/2[$,
we are not able to obtain a result as in Theorem \ref{thm:limfinic3},
since we can not take a test function $\varphi$ in \eqref{problem:P2}
such that $ \varphi -p = -c(h_0+\epsilon)^{-\delta}p$ with $c$
independent of $\epsilon$.

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\end{document}