\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small \emph{Electronic Journal of
Differential Equations}, Vol. 2010(2010), No. 09, pp. 1--12.\newline 
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/09\hfil On the penalized obstacle problem]
{On the penalized obstacle problem in the unit half ball}

\author[E. Lindgren\hfil EJDE-2010/09\hfilneg]
{Erik Lindgren}  

\address{Erik Lindgren \newline
CERMICS - ENPC, 6 et 8 avenue Blaise Pascal,
Cite Descartes Champs sur Marne,
77455 Marne la Vallee Cedex 2 France}
\email{erik.lindgren@cermics.enpc.fr}

\thanks{Submitted September 21, 2009. Published January 16, 2010.}
\subjclass[2000]{35J70, 35J60, 35J85}
\keywords{Obstacle problem; elliptic equation; regularity; penalization}

\begin{abstract}
  We study the penalized obstacle problem in the  unit half ball,
  i.e. an approximation of the obstacle problem in the
  unit half ball. The main result states that when the approximation
  parameter is small enough and when certain level sets are sufficiently
  close to the hyperplane $\{x_1=0\}$, then these level sets are
  uniformly $C^1$ regular graphs. As a by-product, we also recover some
  regularity of the free boundary for the limiting problem, i.e., for
  the obstacle problem.
\end{abstract}

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

\section{Introduction}
\subsection{Problem}

Given some non-negative boundary data
$g\in L^\infty(B_1^+)\cap H^1(B_1^+)$ we study the penalized
obstacle problem in the unit half ball; that is
\begin{gather*}
\Delta u_\varepsilon=\beta_\varepsilon(u_\varepsilon) 
\quad\text{in }B_1^+=B_1\cap \{x_1>0\},\\
u_\varepsilon= g\quad\text{on }\partial B_1\cap \{ x_1>0\},\\
u_\varepsilon=0 \quad\text{on }B_1\cap \{x_1=0\}.
\end{gather*}
Here we assume that $\beta_\varepsilon$ is any function satisfying the same
assumptions as in \cite{Red93a} and \cite{Red93b},  see section
\ref{sec:main} for details. This would then imply by the maximum
principle that $u_\varepsilon\geq 0$. We will refer to $B_1\cap \{x_1=0\}$
as the fixed boundary.

\subsection{Known result}
The penalized obstacle problem has been well studied,  see
\cite{KS80}, \cite{BK73}, \cite{Red93a} and \cite{Red93b}. Brezis
and Kinderlehrer proved the uniform local $C^{1,1}$ regularity.
Redondo proved, using the result of Kinderlehrer and Stampacchia,
that the level sets of $u_\varepsilon$ are locally uniform
$C^{1,\alpha}$-graphs in some direction. However, the behaviour of
the level sets close to the fixed boundary is so far unknown for
the penalized obstacle problem. The present paper contributes in
this direction.

In the case of the obstacle problem (the limiting case when
$\varepsilon\to 0$), the behaviour of the free boundary has been studied in
\cite{SU03}. There the authors prove that, close to the fixed
boundary, the free boundary is a $C^1$-graph in directions close
to $e_1$. In fact, they prove this in a more general setting,
which includes the obstacle problem as a special case.

\subsection{Main result}\label{sec:main}
Before mentioning the main result of this paper we must  carefully
define which class of solutions we will work with.

\begin{definition} \label{def1} \rm
We say $\beta_\varepsilon \in \mathcal{C}_\varepsilon$ if
\begin{enumerate}
\item For any $\varepsilon>0$, $\beta_\varepsilon$ is uniformly Lipschitz 
 on $\mathbb{R}$,
\item $\beta_\varepsilon(t)\leq 0$ for all $t\leq 0$ and all $\varepsilon>0$,
\item $\beta_\varepsilon(0)=0$ and $\beta_\varepsilon(t)\leq 1$,
\item $\beta''_\varepsilon(t)\leq 0$,
\item There holds
$$
0\leq \frac{1-\beta_\varepsilon(t)}{\varepsilon}\leq \beta_\varepsilon'(t)
\leq \frac{1}{\varepsilon},
$$
whenever $t\geq 0$.
\end{enumerate}
\end{definition}

One can deduce that
 $\Psi_\varepsilon\leq \beta_\varepsilon\leq\Phi_\varepsilon$, where
$$
\Psi_\varepsilon(t)=\begin{cases}
 t/\varepsilon & t\leq 0,\\
1-e^{-t/\varepsilon} & t>0,
\end{cases}
$$
and
$$
\Phi_\varepsilon(t)=\begin{cases}
t/\varepsilon & t\leq \varepsilon,\\
1 & t>\varepsilon.
\end{cases}
$$
In addition, one can see that if 
$\beta_\varepsilon(\cdot)\in \mathcal{C}_\varepsilon$
then $\beta_\varepsilon(s\cdot)\in \mathcal{C}_{\varepsilon/s}$. 
This scaling property
will be used frequently further on.  In what follows we use the notation
$$B_r^+(x)=B_r(x)\cap \{x_1>0\}.$$ We define the local class of
solutions as the following:

\begin{definition} \label{def} \rm
 We say $u_\varepsilon\in P_r(M,\varepsilon)$ if
\begin{enumerate}
\item $\Delta u_\varepsilon=\beta_\varepsilon(u_\varepsilon)$ in 
$B_r^+$ where $\beta_\varepsilon\in\mathcal{C_\varepsilon}$.
\item $u_\varepsilon\geq 0$ on $\partial B_r\cap \{x_1>0\}$.
\item $u_\varepsilon=0$ on $B_r\cap \{x_1=0\}$,
\item $\sup_{B_r^+}u_\varepsilon\leq M$.
%\item $0\in\partial\{u_\varepsilon>0\}$.
\end{enumerate}
\end{definition}

The main result of this paper is the $C^{1,1}$-estimates and the
uniform (in $\varepsilon$) $C^1$ regularity of the 
$\varepsilon^\gamma$-level  sets
for any $0<\gamma<1$, as stated below in Theorem 1 and 2. A simple
consequence of Theorem \ref{thm:C1} is Corollary \ref{cor:limit}.

\begin{theorem}\label{thm:C11}
 Let $u_\varepsilon\in P_1(M,\varepsilon)$. Then there is a constant 
 $C=C(n,M)$ such that
$$
\|u_\varepsilon\|_{C^{1,1}(B_{1/2}^+)}\leq C.
$$
\end{theorem}


\begin{theorem}\label{thm:C1}
Let $u_\varepsilon\in P_1(M,\varepsilon)$. Assume moreover that
$y\in \partial\{u_\varepsilon>\varepsilon^\gamma\}\cap B_{1/2}^+$ for some
$0<\gamma<1$. Then there are constants $\varepsilon_0,\rho>0$ depending on $M$,
$\gamma$ and the dimension such that
$$
\max(\varepsilon,y_1)<\varepsilon_0
$$
implies that $B_\rho^+(y)\cap \partial\{u_\varepsilon>\varepsilon^\gamma\}$ 
is a $C^1$-graph with the $C^1$-norm bounded independently of $\varepsilon$
and $y_1$. Moreover, the normal of 
$\partial\{u_\varepsilon>\varepsilon^\gamma\}$ at $y$
converges to $e_1$ as $y_1\to 0$.
\end{theorem}

Sending $\varepsilon\to 0$ we obtain the following result for the obstacle
problem, which is a slightly stronger result than the one in
\cite{SU03} since it only requires the free boundary to  be near
the fixed one.

\begin{corollary} \label{cor:limit}
Let $u_0\in P_1(M,0)$, that is $u_0$ is a solution to the
obstacle problem, and suppose $y\in \partial\{ u_0>0\}$. Then there is
a constant $\varepsilon_0$ depending on $M$ such that $y_1<\varepsilon_0$ 
implies that $B_\rho^+(y)\cap \partial\{u_0>0\}$ is a $C^1$-graph with the
$C^1$-norm bounded independently $y_1$.
\end{corollary}

\section{Non-degeneracy}
Here we state the result that the solutions do not grow too slow.
This is important when considering sequences as we will do later,
since it implies the stability of the level sets under
uniform convergence. For the sake of completeness,
we give the standard proof.

\begin{proposition}\label{prop:nondeg}
Let $u_\varepsilon\in P_1(M,\varepsilon)$ and assume $u_\varepsilon(y)\geq \varepsilon$.
Then there is $C=C(n)>0$ such that
$$
\sup_{B_r(y)\cap B_1^+}(u_\varepsilon-u_\varepsilon(y))\geq Cr^2
$$
for $r<\mathop{\rm dist}(y,\partial B_1^+)$.
\end{proposition}

\begin{proof}
Take $y\in \{u_\varepsilon=\delta\}$  for some $\delta\geq \varepsilon$ and let
$$
v=u_\varepsilon-\delta-\frac{|y-z|^2}{8n}.
$$
Then $\Delta v\geq \beta_\varepsilon(u_\varepsilon)-1/4\geq 0$ in 
$\{u_\varepsilon> \delta/2\}\cap
B_r^+(y)$ for any $r$ such that $B_r^+(y)\subset B_1^+$.
Therefore, $v$ attains its maximum on the boundary of $\{u_\varepsilon>
\delta/2\}\cap B_r^+(x_0)$. Furthermore, $v(y)=0$, $v\leq 0$ on
$\{u_\varepsilon=\delta\}$ and $v\leq 0$  on $\{x_1=0\}$. Hence $v$ attains
its positive maximum at some $z\in \partial B_r(y)\cap \{x_1>0\}$. This
yields the desired result.
\end{proof}

\section{$C^{1,1}$-estimates in $B_{1/2}^+$}

In this section we prove that the functions $u_\varepsilon$ are uniformly
bounded in the space $C^{1,1}(B_{1/2}^+)$. The first step is to show that we
have quadratic growth up to the fixed boundary.

\begin{lemma}\label{lem:quadgrowth}
Let $u_\varepsilon\in P_1(M,\varepsilon)$. Then there are constants $C,r_0>0$
depending on $M$ and the dimension such that
$$
\sup_{B_r^+}|u_\varepsilon(x)-x\cdot \nabla u_\varepsilon(0)|\leq Cr^2
$$
for $r<r_0$.
\end{lemma}

\begin{proof}
We prove that for any $r$ we have either
$$
\sup_{B_r^+} |u_\varepsilon(x)-x\cdot \nabla u_\varepsilon(0)|\leq Cr^2,
$$
or there is a $k\in \mathbb{N}$ such that
$$
\sup_{B_r^+} |u_\varepsilon(x)-x\cdot \nabla u_\varepsilon(0)|
\leq 4^{-k}\sup_{B_{2^kr}^+}|u_\varepsilon(x)-x\cdot \nabla u_\varepsilon(0)|.
$$
Assume that this statement fails, then there are sequences of 
$\varepsilon_j,r_j\to 0$ and $u_j=u_{\varepsilon_j}$ such that
$$
S_j\geq Cjr_j^2
$$
and
$$
S_j\geq 4^{-k}S_{2^kr_j},
$$
for all $k\in \mathbb{N}$. Here
$$
S_j=\sup_{B_{r_j}^+}|u_j(x)-x\cdot\nabla u_j(0)|.
$$
Let
$$
v_j(x)=\frac{u_j(r_jx)-r_jx\cdot \nabla u_j(0)}{S_j}.
$$
Then
\begin{enumerate}
\item $v_j(0)=|\nabla v_j(0)|=0$,
\item $\sup_{B_1^+}|v_j|=1$,
\item $\sup_{B_{2^k}^+}|v_j|\leq 4^k$ for all $k\in \mathbb{N}$,
\item $\sup_{B_R^+}|\partial_ev_j|\leq C/j$ for any
$e\cdot e_1=0$ and any $R>0$, since by the maximum principle
$|\partial_eu_j(x)|\leq C|x|$,
\item $|\Delta v_j|=C/j$ in $B_{1/r_j}^+$.
\end{enumerate}
Therefore, there is a subsequence, again labelled $v_j$,
converging in $C^{1,\alpha}(\overline{B_R^+})$ for any
$R$ to a function $v_0$ satisfying
\begin{enumerate}
\item $v_0(0)=|\nabla v_0(0)|=0$,
\item $\sup_{B_1^+}|v_0|=1$,
\item $\sup_{B_{2^k}^+}|v_0|\leq 4^k$ for all $k\in \mathbb{N}$,
\item $D_e v_0=0$ for any $e\cdot e_1=0$ in ${\mathbb{R}^n_+}$,
\item $\Delta v_0=0$ in ${\mathbb{R}^n_+}$.
\end{enumerate}
Clearly, (4) implies that $v_0$ is one-dimensional.
Then (1) and (5) implies that $v_0=0$ which contradicts (2).
\end{proof}

This estimate implies, as we will now see, that $u_\varepsilon$
is uniformly $C^{1,1}$. The proof is standard.

\begin{proof}[Proof of Theorem \ref{thm:C11}]
Since interior estimates are already known (see Theorem 2 in
\cite{BK73} even though this is stated in a slightly different form,
this can also be proved with methods similar to the ones in the
proof of Lemma \ref{lem:quadgrowth}), it suffices to obtain
estimates near the fixed boundary.

Take $y\in B_{1/2}^+\cap \{x_1<r_0/2\}$ where $r_0$ is
as in Lemma \ref{lem:quadgrowth}. 
Consider the function
$$
v(x)=\frac{u_\varepsilon(|y|x+y)-(|y|x+y)\cdot \nabla u_\varepsilon(0)}{|y|^2}.
$$
Then, by Lemma \ref{lem:quadgrowth}
\begin{align*}
\sup_{x\in B_1}|v(x)|=\sup_{x\in B_{|y|}}\left|\frac{u_\varepsilon(x+y)-(x+y)\cdot
  \nabla u_\varepsilon(0)}{|y|^2}\right|\\\leq \sup_{x\in B_{2|y|}^+}\left
|\frac{u_\varepsilon(x)-x\cdot
  \nabla u_\varepsilon(0)}{|y|^2}\right|\leq C.
\end{align*}
Moreover, 
$$
|\Delta v (x)|=|\beta_\varepsilon(y^2v+(|y|x+y)\cdot\nabla
u_\varepsilon(0))|\leq 1.
$$
Therefore $\|u\|_{C^{1,\alpha}(B_{1/2})}\leq C$ by elliptic
estimates. In particular, for any $x\in B_\frac12$ there holds
$$
C\geq |\nabla v(x)|=\left|\frac{\nabla u_\varepsilon(y)
-\nabla u_\varepsilon(0)}{|y|}\right|,
$$
and 
$$
C\geq |\nabla v(x)|=\left|\frac{\nabla u_\varepsilon(|y|x+y)
-\nabla u_\varepsilon(0)}{|y|}\right|.
$$
Combining the last to estimates gives
$$
|\nabla u_\varepsilon(|y|x+y)-\nabla u_\varepsilon(y)|\leq C|y|\leq C|yx|.
$$
Hence $D^2u_\varepsilon(y)$ is bounded
independently of $y$.
\end{proof}

\section{Classification of certain global solutions}

For the proofs in this section we will need the following theorem,
which is proved in \cite[Lemma 5.1]{ACF84} (even though slightly
different form) and in \cite[Theorem 9]{Caf98}.

\begin{theorem}\label{thm:acf}
Let $u$ and $v$ be two subharmonic functions such that $u\cdot v=0$.
Then with
$$
\phi(r)=\frac{1}{r^4}\int_{B_r}\frac{|\nabla u|^2}{|x|^{n-2}}\,\textrm{d} x
 \int_{B_r}\frac{|\nabla v|^2}{|x|^{n-2}}\,\textrm{d} x
$$
we have that $\phi$ is monotonically increasing. Moreover, if $\phi$
vanishes for all $r$ then one of the functions vanishes.
\end{theorem}

In the forthcoming sections, we will need the following
characterization of global solutions.

\begin{proposition}\label{prop:class}
Let $u:\mathbb{R}^n_+\to \mathbb{R}$ be a function with the 
 following properties:
\begin{enumerate}
\item $u\geq 0$,
\item $e_1\in \partial\{u>t\}$ for some $t\geq 0$,
\item $u=0$ on $x_1=0$,
\item $\sup_{{B_r}^+}u\leq Cr^2$ for $r>1$,
\item $\Delta u=\beta_k(u)$ for some $k\geq 0$, with
$\beta_0(t)=\chi_{\{t>0\}}$ and $\beta_k\in \mathcal{C}_k$.
\end{enumerate}
Then $u$ is one-dimensional and monotone.
\end{proposition}

\begin{proof}
First we extend $u$ to $x_1<0$ by odd reflection, that will give us
$$
\Delta u=\beta_k(u)\chi_{\{x_1>0\}}-\beta_k(-u)\chi_{\{x_1<0\}}.
$$
Then we observe that
$$
\Delta (\partial_e u)=(\beta'_k(u)\chi_{\{x_1>0\}}
+\beta_k'(-u)\chi_{\{x_1<0\}})\partial_e u+\beta_k(u)\partial_e(\chi_{\{x_1>0\}})
-\beta_k(-u)\partial_e(\chi_{\{x_1<0\}}).%=(\beta'_k(u)+\beta_k'(-u))\partial_e u
$$
Since $\beta(0)=0$ and $\nabla \chi_{\{x_1>0\}}$ is a finite
measure the last two terms vanish. This implies that $(\partial_e u)^\pm$
are both subharmonic and therefore Theorem \ref{thm:acf} applies. Let
$$
u_R(x)=\frac{u(Rx)}{R^2}.
$$
Then
\begin{enumerate}
\item $\Delta u_R=\beta_k(R^2u_R)\chi_{\{x_1>0\}}
 -\beta_k(-R^2u_R)\chi_{\{x_1<0\}}$ where 
 $\beta_k(R^2t)\in \mathcal{C}_{k/R^2}$,
\item $|u_R(x)|\leq C|x|^2$ for $|x|>1/R$,
\item $e_1/R\in \partial\{u_R>t/R^2\}$,
\item $-e_1/R\in \partial\{u_R<-t/R^2\}$,
\item $|\nabla u_R(0)|\leq C/R$.
\end{enumerate}
We see that $\Delta u_R$ will be uniformly bounded from above
and from below. Moreover, $\sup_{B_1}|u_R|$ is uniformly bounded.
Hence, there is by standard elliptic estimates, a subsequence
$u_j=u_{R_j}$ with $R_j\to \infty$, converging in
$W_\textup{loc}^{2,p}\cap C^{1,\alpha}_\textup{loc}$.
Invoking Proposition \ref{prop:nondeg} we can conclude
that $u_j\to u_\infty$ where $u_\infty$ satisfies
\begin{enumerate}
\item $\Delta u_\infty=\chi_{\{u_\infty>0\}}\chi_{\{x_1>0\}}
 -\chi_{\{u_\infty<0\}}\chi_{\{x_1<0\}}
 =\chi_{\{u_\infty>0\}}-\chi_{\{u_\infty<0\}}$,
\item $|u_\infty(x)|\leq C|x|^2$ for all $x$,
\item $0\in \partial\{u_\infty>0\}\cap \partial\{u_\infty<0\}\cap
\{|\nabla u_\infty|=0\}$.
\end{enumerate}
By applying the classification of global solutions in \cite{SUW04},
we obtain that up to rotations
$$
u_\infty=\frac{(x_1^+)^2}{2}-\frac{(x_1^-)^2}{2}.
$$
Again, Theorem \ref{thm:acf} applies to $\partial_e u_j^\pm$ for any
direction $e$. Thus,
$$
0=\phi(\partial_e u_\infty,1)=\lim \phi(\partial_e u_j,1)
=\lim \phi(\partial_e u,R_j)
=\phi(\partial_e u,\infty).
$$
Since $\phi$ is non-decreasing this implies that $\phi(\partial_e
u,r)=0$ for all $r$. Hence, either $\partial_e u\geq 0$ or $\partial_e u\leq
0$,  for any direction $e$. This implies that $u$ must be
one-dimensional.
\end{proof}

\begin{remark} \label{rmk1} \rm
It seems plausible that a proof of this can be given without the
use of the ACF monotonicity formula. However, such a proof would
be much longer and more technically involved.
\end{remark}

\section{The $C^1$ regularity of certain level sets close to
the fixed boundary}

In this section we will prove that if the $\varepsilon^\gamma$-level set is
close to the fixed boundary then, due to the results in
\cite{Red93a} and \cite{Red93b}, the $\varepsilon^\gamma$-level set is a
uniform (with respect to $\varepsilon$) $C^1$-graph.

The first step is to prove that the set 
$\{u_\varepsilon\leq \varepsilon^\gamma\}$
is not going to be too small in this case, see Figure \ref{fig:md}.
This is needed in order to apply the results in \cite{Red93a}
and \cite{Red93b}, which are of the type ''exterior flatness
implies regularity''. In the rest of this paper $\gamma\in (0,1)$
is fixed, so all estimates and results will depend on $\gamma$.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)
\includegraphics[width=0.8\textwidth]{fig1} %\input{by1.pdf_t}
\end{picture}
\begin{picture}(100,48)(0,0)
\put(45,41){$\partial\{u_\varepsilon>\varepsilon^\gamma\}$}
\put(90,35){$B_1^+$}
\put(38,20){$y$}
\put(58,15){$B_{\sigma y_1}(y)$}
\put(53,3){$B_{\sigma y_1}(y)\cap \{x_1\leq y_1-\sigma y_1/2\}$}
\end{picture}
\end{center}
\caption{The situation when $\partial\{u_\varepsilon>\varepsilon^\gamma\}$ 
is close to the fixed boundary.}
\label{fig:md}
\end{figure}

\begin{proposition}\label{prop:md}
Let $u_\varepsilon\in P_1(M,\varepsilon)$ with 
$|\nabla u_\varepsilon(y)|<y_1$ for some
$y\in B_1^+$. Then for any $\sigma>0$ there is
$\tau =\tau(\sigma,M,\gamma,n)$ such that
$y\in \{u_\varepsilon>\varepsilon^\gamma\}$, $y\cdot e_1=y_1<\tau$ and
$\varepsilon<\tau$ imply that
\begin{equation}\label{eq:cone}
B_{y_1\sigma}(y)\cap \{x_1\leq y_1-\frac{y_1\sigma}{2}\}\subset
\{u_\varepsilon\leq \varepsilon^\gamma\}.
\end{equation}
\end{proposition}

\begin{proof}
If this is not true then there exist
$\{u_{\varepsilon_j}>\varepsilon_j^\gamma\}\ni y_1^j\to 0$,
$\varepsilon_j\to 0$, $u_{\varepsilon_j}\in P_1(M,\varepsilon_j)$ and
$$
x^j\in B_{y_1^j\sigma}(y)\cap \{x_1\leq y_1^j
-\frac{y_1^j\sigma}{2}\}\cap \{u_{\varepsilon_j}> \varepsilon_j^\gamma\},
\quad\text{for $j=1,2,\ldots$}
$$
For
$$
v_j(x)=\frac{u_{\varepsilon_j}(y_1^jx+y^j)}{(y_1^j)^2}
$$
we have
\begin{enumerate}
\item $\Delta v_j=\beta_{\varepsilon_j}(v_j (y_1^j)^2)\in
 \mathcal{C}_{\varepsilon_j/(y_1^j)^2}$ for $x_1>-1$,
\item $v_j\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-1\}}v_j\leq C(M)\rho^2
 +C|\nabla u_{\varepsilon_j}(y^j)|\rho/y_1^j
 +\varepsilon_j^\gamma/(y_1^j)^2+v_j(0)$
 for $\rho<r_0/y^j_1$,
\item $0\in \partial\{v_j>\varepsilon_j^\gamma/(y_1^j)^2\}$,
\item $|\nabla v_j(0)|=|\nabla u_{\varepsilon_j}(y^j)|/|y^j|\leq 1$,
\item $v_j=0$ on $x_1=-1$,
\item $z_j=(x_j-y_j)/y_1^j\in B_{\sigma}\cap \{x_1\leq
 -\frac{\sigma}{2}\}\cap \partial\{v_{\varepsilon_j}
 >\varepsilon_j^\gamma/(y_1^j)^2\}$.
\end{enumerate}
All these properties together with Proposition \ref{prop:nondeg}
allow us to pass to the limit for a subsequence and obtain a
limit function $v_0$ and a limit point $z=\lim_j z_j$.
We observe that the $C^{1,1}$-estimates give
$$
0=u_{\varepsilon_j}(y^j-y_1^je_1)\geq u_{\varepsilon_j}(y^j)
-\nabla u_{\varepsilon_j}(y^j)\cdot e_1y_1^j-C(y_1^j)^2,
$$
which implies $\varepsilon_j^\gamma<C(y_1^j)^2$. Clearly, this gives
$\varepsilon_j/(y_1^j)^2\to 0$. We split this into two cases, depending
on the limit of $\varepsilon_j^\gamma/(y_1^j)^2$.


\noindent\textbf{Case 1:} $\varepsilon_j^\gamma/(y_1^j)^2\to 0$. Then
\begin{enumerate}
\item $\Delta v_0=\chi_{\{v_0>0\}}$ for $x_1>-1$,
\item $v_0\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-1\}}v_0\leq C(M)\rho^2$ for $\rho>1$,
\item $0\in \partial\{v_0>0\}$,
\item $|\nabla v_0(0)|=0$,
\item $v_0=0$ on $x_1=-1$,
\item $z\in B_\sigma\cap \{x_1\leq - \sigma/2\}\cap \left(\{v_0>0\}
\cup \partial\{v_0>0\}\right)$.
\end{enumerate}
Note that(1) follows from the fact that $\Delta v_j\in
\mathcal{C}_{\varepsilon_j/(y_1^j)^2}$ where $\varepsilon_j/(y_1^j)^2\to 0$.
Moreover, (5) is a simple consequence of (4). Theorem B in
\cite{SU03} now implies that $v_0=(x_1^+)^2/2$. This contradicts
(7) whenever $1>\sigma>0$.


\noindent\textbf{Case 2:} $\varepsilon_j^\gamma/(y_1^j)^2\to t>0$. This is
similar to case 1. The only difference is that the origin will be
in the $t$-level set instead of in the zero level set which also
means that we do not know that the gradient vanishes at the
origin. Instead of using the result from \cite{SU03} we use
Proposition \ref{prop:class} to say that $v_0$ must be
one-dimensional and monotone. This yields a contradiction for $1>\sigma >0$.
\end{proof}

\begin{corollary}\label{cor:c1}
Let $u_\varepsilon\in P_1(M,\varepsilon)$. Then for any $0<\gamma<1$ there
is an $\varepsilon_0(M,\gamma,n)$ such that 
$y\in\partial\{u_\varepsilon>\varepsilon^\gamma\}$
 and $y_1<\varepsilon_0$ imply that  
$\partial \{u_\varepsilon>\varepsilon^\gamma\}\cap B_{ry_1}(y)$
is a uniform (with respect to $\varepsilon$) $C^1$-graph for some $r>0$.
\end{corollary}

\begin{proof} If $|\nabla u(y)|\geq y_1$ then, by the implicit function theorem,
$\partial \{u_\varepsilon>\varepsilon^\gamma\}\cap B_{ry_1}(y)$ is a 
$C^{1,\alpha}$-graph
for some $r>0$.

In the other case, $|\nabla u_\varepsilon(y)|<y_1$, let
$$
v(x)=\frac{u_\varepsilon(y_1x+y)}{y_1^2}.
$$
Moreover, take $y_1<\tau(\sigma,M,\gamma,n)$ as in  Proposition
\ref{prop:md} with $\sigma$ small enough and so that $\varepsilon/y_1^2$ is
sufficiently small. This is possible since we have $\varepsilon^\gamma\leq
Cy_1^2$.

If $|\nabla u_\varepsilon(y)|<y_1$ we know by Proposition \ref{prop:md}  that
(\ref{eq:cone}) holds. Thus,
$$
\frac{|\{v\leq \varepsilon^\gamma/y_1^2\}\cap B_\sigma|}{|B_\sigma|}\geq c_0.
$$
 From \cite[Proposition 19]{Red93a} and with \cite[Theorem 16]{Red93b}
it follows that, for $\sigma$ sufficiently small,  $\partial
\{v>\varepsilon^\gamma/y_1^2\}\cap B_{r}(y)$ is a $C^{1,\alpha}$-graph for
some $r>0$.

By rescaling and combining the two cases, we
obtain the desired result.
\end{proof}

\begin{remark} \label{rmk2} \rm
Noteworthy is that in the proof of Corollary \ref{cor:c1} we apply
the result in \cite{Red93a,Red93b} for the  rescaled function
$v$. For this we need $\varepsilon/y_1^2$ to be small. This is assured by
the assumption $\gamma<1$. To extend this result to $\gamma=1$ we
need to come up with something different. In addition we note that
in Corollary \ref{cor:c1} we obtain that the level sets are $C^1$
but that the $C^1$-norm might blow up as $y_1\to 0$.
\end{remark}


\section{Tangential touching property of the $\varepsilon^\gamma$-level sets}

Here we prove that the $\varepsilon^\gamma$-level set can be approximated
with planes, provided that $\varepsilon$ is small enough and that the
$\varepsilon^\gamma$-level set is close enough to the fixed boundary,
see  Figure \ref{fig:cones}. In what follows, we will use the notation
$$
K_\delta(z)=\{|x_1-z_1|<\delta\sqrt{(x_2-z_2)^2+\cdots+(x_n-z_n)^2}\}.
$$
To avoid cumbersome notation we will also write
$K_\delta(0)=K_\delta$.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)
\includegraphics[width=0.8\textwidth]{fig2} %\input{cones.pdf_t}
\end{picture}
\begin{picture}(100,50)(0,0)
\put(50,44){$\partial\{u_\varepsilon>\varepsilon^\gamma\}$}
\put(90,35){$B_1^+$}
\put(37,35){$K_\delta(y)$} 
\put(39,20){$y$}
\put(63,16){Angle$=\arctan\delta$}
\put(3,9){$B_\rho(y)$}
\end{picture}
\end{center}
\caption{The level set $\partial\{u_\varepsilon>\varepsilon^\gamma\}$ is inside
$K_\delta(y)$ when $y$ is close to the fixed boundary.}
\label{fig:cones}
\end{figure}

\begin{proposition}\label{prop:dirmono}
Let $u\in P_1(\varepsilon,M)$. Then for any $\delta>0$ there are
$\varepsilon_0(M,\gamma,n,\delta)$ and $\rho(M,\gamma,n,\delta)$  such that
$y \in \partial\{u_\varepsilon>\varepsilon^\gamma\}$ and 
$\max(\varepsilon,y_1,|\nabla
u_\varepsilon(y)|)<\varepsilon_0$ imply
\begin{equation}\label{eq:tang}
\partial\{u_\varepsilon>\varepsilon^\gamma\}\cap B_\rho(y)\cap B_1^+
\subset K_\delta(y)\cap B_\rho(y)\cap B_1^+.
\end{equation}
\end{proposition}

\begin{proof}
If this is not true then for $\delta>0$ there are
$\varepsilon_j\to 0$, 
$y^j\in \partial\{u_{\varepsilon_j}>\varepsilon_j^\gamma\}$
with $\max(|y^j_1|,|\nabla u_{\varepsilon_j}(y^j)|)|\to 0$
and $x^j$ with  $r_j=|x^j-y^j|\to 0$, such that
\begin{equation}\label{eq:notcone}
x^j\in K_\delta(y^j)^c\cap \partial\{u_{\varepsilon_j}>\varepsilon_j^\gamma\}.
\end{equation}
We want to prove that this cannot be true. The proof is divided
into two different cases with some sub-cases depending on the rate
of convergence of $y_1^j$, $\varepsilon_j$, $|\nabla u_{\varepsilon_j}(y^j)|$ 
and $r_j$.

\noindent\textbf{Case A:} $|\nabla u_{\varepsilon_j}(y^j)|<y_1^j$. Then we
have cases A1-A3 as follows. By the $C^{1,1}$-estimates,
$$
0=u(y^j-y^j_1e_1)\geq u(y^j)-|\nabla u(y^j)\cdot e_1y^j_1|-C(y^j_1)^2
\geq \varepsilon_j^\gamma-C|y_1^j|^2.
$$

\noindent\textbf{Case A1:} $|y_1^j|=o(r_j)$. Let
$$
v_j(x)=\frac{u_{\varepsilon_j}(r_jx+y^j)}{r_j^2}.
$$
Then
\begin{enumerate}
\item $\Delta v_j=\beta_{\varepsilon_j}(v_jr_j^2)\in 
\mathcal{C}_{\varepsilon_j/r_j^2}$
 for $x_1>-y_1^j/r_j$,
\item $v_j\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-y_1^j/r_j\}}v_j
 \leq C(M)\rho^2+C|\nabla u_{\varepsilon_j}(y^j)|\rho/r_j
 +\varepsilon_j^\gamma/r_j^2+v_j(0)$
 for $\rho<r_0/r_j$ for some $r_0>0$,
\item $0\in \partial\{v_j>\varepsilon_j^\gamma/r_j^2\}$ with
 $\varepsilon_j^\gamma/r_j^2\leq C(M)|y^j|^2/r_j^2\to 0$,
\item $|\nabla v_j(0)|=|\nabla u_{\varepsilon_j}(y^j)/r_j|\leq |y^j_1|/r_j\to 0$,
\item $v_j=0$ on $x_1=- y_1^j/r_j$,
\item $z_j=(x^j-y^j)/r_j\in \partial B_1 \cap K^c_\delta\cap\partial 
 \{v_j>\varepsilon_j^\gamma/r_j^2\}$.
\end{enumerate}
We can, by standard compactness arguments, pass to the limit for
a subsequence $v_{j_k}\to v_0$. Using Proposition \ref{prop:nondeg},
we see that $v_0$ satisfies
\begin{enumerate}
\item $\Delta v_0=\chi_{\{v_0>0\}}$ in $x_1>0$,
\item $v_0\geq 0$,
\item $\sup_{B_\rho^+}v_0\leq C(M)\rho^2$ for all $\rho>1$,
\item $0\in \partial\{v_0>0\}$,
\item $|\nabla v_0(0)|=0$,
\item $v_0=0$ for $x_1=0$,
\item $z_0\in \partial B_1\cap K^c_\delta\cap \partial \{v_0>0\}$.
\end{enumerate}
\cite[Theorem B]{SU03} implies together with (1)-(6) that
$$
v_0=\frac{(x_1^+)^2}{2},
$$
which contradicts (7).

\noindent\textbf{Case A2:}
$0<1/A<|y_1^j|/r_j<A<\infty$. This case is very similar to the
ones in the proof of of Proposition \ref{prop:md}.
We obtain that the limit will be one-dimensional via
either \cite{SU03} or Proposition \ref{prop:class}.
This leads to a contradiction.

\noindent\textbf{Case A3:}  $y_1^j/r_j\to \infty$. We recall that,
as in the proof of Proposition \ref{prop:md}, we have
$\varepsilon_j^\gamma<C(y^j_1)^2$. This implies
$(y^j_1)^2/\varepsilon_j^\gamma\not\to 0$ and 
$(y^j_1)^2/\varepsilon_j\to \infty$.
Let
$$
v_j(x)=\frac{u_{\varepsilon_j}(y_1^jx+y^j)}{(y^j_1)^2}.
$$
Then
\begin{enumerate}
\item $\Delta v_j=\beta_{\varepsilon_j}(v_j|y_1^j|^2)
 \in \mathcal{C}_{\varepsilon_j/(y_1^j)^2}$ on $v_j>0$ for $x_1>-1$,
\item $v_j\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-1\}}v_j
 \leq C(M)\rho^2+C|\nabla u_{\varepsilon_j}(y^j)|\rho/y_1^j
 +\varepsilon_j^\gamma/(y_1^j)^2+v_j(0)$ for $\rho<r_0/y^j_1$,
\item $0\in \partial\{v_j>\varepsilon_j^\gamma/(y_1^j)^2\}$,
\item $|\nabla v_j(0)|=\nabla u_{\varepsilon_j}(y^j)/|y^j|\leq 1$,
\item $v_j=0$ on $x_1=-1$,
\item $z_j=(x^j-y^j)/y_1^j\in K^c_\delta\cap
 \{v_j>\varepsilon_j^\gamma/(y_1^j)^2\}$.
\end{enumerate}
When $y_1^j$ is small enough, Corollary \ref{cor:c1} assures that
$\partial\{v_j>\varepsilon_j^\gamma/(y_1^j)^2\}\cap B_{c_0}$ is a uniform
$C^1$-graph for $c_0$ small. Since $|z_j|\to 0$, (7) implies
that $\partial\{v_j>\varepsilon_j^\gamma/(y_1^j)^2\}$ 
has a tangential direction in $K_\delta^c$.
All these properties allow us, together with
Proposition \ref{prop:nondeg}, to pass to the limit for
a subsequence and obtain a function $v_0$ satisfying
\begin{enumerate}
\item $\Delta v_0=\chi_{\{v_0>0\}}$ for $x_1>-1$,
\item $v_0\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-1\}}v_0\leq C(M)\rho^2+t$ for all
 $\rho >1$,
\item $0\in \partial\{v_0>t\}$ for some bounded $t$,
\item $|\nabla v_0(0)|\leq 1$,
\item $v_0=0$ on $x_1=-1$,
\item The set $\partial\{v_0>t\}\cap B_{c_0}$ has a tangential
 direction in $K_\delta^c$, for some small $c_0$.
\end{enumerate}
As in cases 1 and 2 in the proof of Proposition \ref{prop:md},
(1)-(7) imply that $v_0$ is one-dimensional.
 Therefore it can only depend on $x_1$, which contradicts (7),
since (7) says  that the level set of $v_0$ has a tangent
lying in $K_\delta^c$.

\noindent\textbf{Case B:} $|\nabla u_{\varepsilon_j} (y^j)|>y_1^j$. This
implies by the $C^{1,1}$-estimates that
$$
0=u(y^j-y^j_1e_1)\geq u(y^j)-|\nabla u(y^j)\cdot e_1y^j_1|-C(y^j_1)^2
\geq \varepsilon_j^\gamma-C|\nabla u_{\varepsilon_j}(y^j)|^2,
$$
which gives $\varepsilon_j^\gamma\leq C|\nabla u(y^j)|^2$. Let $s_j=|\nabla
u(y^j)|$. For the first two cases B1 and B2, let us define
$$
v_j(x)=\frac{u_{\varepsilon_j}(s^jx+y^j)}{s_j^2}.
$$
Then
\begin{enumerate}
\item $\Delta v_j=\beta_{\varepsilon_j}(v_js_j^2)\in
 \mathcal{C}_{\varepsilon_j/s_j^2}$ on $v_j>0$ for $x_1>-y_1^j/s_j$,
\item $v_j\geq 0$,
\item $\sup_{B_\rho\cap \{x_1>-y_1^j/s_j\}}|v_j|\leq C(M)\rho^2
 +\varepsilon_j^\gamma/(s_j)^2+v_j(0)$ for $\rho<r_0/s_j$,
\item $0\in \partial\{v_j>\varepsilon_j^\gamma/(s_j)^2\}$,
\item $|\nabla v_j(0)|=1$,
\item $v_j=0$ on $x_1=-y_1/s_j$,
\item With $z_j=(x^j-y^j)/s_j\in K^c_\delta\cap
 \partial \{v_j>\varepsilon_j^\gamma/(s_j)^2\}$.
\end{enumerate}
For this, we consider two different cases.

\textbf{Case B1:} $|z_j|=r_j/s_j\to 0$. This yields a situation
similar to case A3. Via Corollary \ref{cor:c1} and Proposition
\ref{prop:class} we obtain a contradiction.

\textbf{Case B2:} $A<|z_j|<1/A$. Then the situation is similar to
the one in case A2. Thus, we obtain, either by Proposition
\ref{prop:class} or Theorem B in \cite{SU03}, that the limit is
one dimensional. This leads to a contradiction.

\textbf{Case B3:} $|z_j|\to \infty$, i.e. $s_j=o(r_j)$. Then
$\varepsilon_j^\gamma\leq Cs_j^2=o(r_j^2)$. By the assumptions in this
case, $y_1^j=o(r_j)$. Let
$$
v_j(x)=\frac{u_{\varepsilon_j}(r_jx+y^j)}{r_j^2}.
$$
Then similar arguments as in case A1 leads to contradiction. Since
all cases lead to a contradiction, the theorem is proved.
\end{proof}

\subsection{The $C^1$ regularity of the $\varepsilon^\gamma$-level set}

Now we are ready to prove the uniform $C^1$ regularity of
the $\varepsilon^\gamma$-level sets.

\begin{proof}[Proof of Theorem \ref{thm:C1}]
 We apply Corollary \ref{cor:c1} and obtain that for $y_1<\varepsilon_0$,
the level set $\partial\{u_\varepsilon>\varepsilon^\gamma\}\cap B_{ry_1}(y)$ is
differentiable. In addition, we know that the normal at a point
$y$,  denoted by $n_y$, is continuous with modulus of continuity
$\sigma(\cdot/y_1)$, where $\sigma$ is a modulus of continuity.
Hence, it might blow up when $y_1\to 0$. We need to prove that
$n_y$ is uniformly continuous.

Take two points 
$y,z\in \partial\{u_\varepsilon>\varepsilon^\gamma\}\cap \{x_1<\varepsilon_0\}$.
 From Proposition \ref{prop:dirmono} it follows that for any
$\tau>0$ there is $\delta_\tau$ such that $y_1<\delta_\tau$
implies $\|n_y-e_1\|<\tau/2$. We split the proof into three cases:

\noindent\textbf{Case 1:} $y_1,z_1<\delta_\tau/2$. Then obviously
$\|n_y-n_z\|\leq \tau$.

\noindent\textbf{Case 2:} $y_1<\delta_\tau/2$ and
$z_1>\delta_\tau/2$. Then $\|z-y\|<\delta_\tau/2$ implies
$\|n_y-n_z\|\leq \tau$.

\noindent\textbf{Case 3:} $y_1,z_1>\delta_\tau/2$. From the
arguments above,
$$
\|n_y-n_z\|\leq \sigma(2\|y-z\|/\delta_\tau),
$$
which implies that $\|n_y-n_z\|\leq \tau$ if $\|y-z\|$ is small
enough.
\end{proof}

Finally we give the proof of Corollary \ref{cor:limit}.

\begin{proof}[Proof of Corollary \ref{cor:limit}]
By Theorem \ref{thm:C11} we have uniform $C^{1,1}$-estimates and
by Theorem \ref{thm:C1} we have uniform $C^1$-estimates of the
sets $\partial \{u_\varepsilon>\varepsilon^\gamma\}$ when $y_1$ and 
$\varepsilon$ are smaller than
$\varepsilon_0$. Letting $\varepsilon\to 0$ we have $u_\varepsilon\to u_0$. 
Observing that Proposition \ref{prop:nondeg} implies that the limiting set is
indeed the free boundary, $\partial\{u_0>0\}$, the result follows.
\end{proof}

\subsection*{Acknowledgements}
I would like to take the opportunity
to thank my supervisor Henrik Shahgholian for introducing me to
the problem and for all his pertinent comments and suggestions.
Furthermore, I am grateful to R\'egis Monneau for carefully
reading the paper and coming with many valuable comments.

This research project is part of the ESF program GLOBAL. The
author has been supported by grant KAW 2005.0098 from the Knut and
Alice Wallenberg Foundation. Part of this work was carried out
during a one-month stay at the Petroleum Institute, Abu Dhabi,
partially supported by STINT foundation. Moreover, I thank the
Petroleum Institute for the hospitality during my visit.

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\end{document}