# Interaction of a free boundary with a diffusion on a plane: analogy with the
obstacle problem

Luis A. Caffarelli
The University of Texas at Austin
Mathematics Department RLM 8.100, 2515 Speedway Stop C1200
Austin, Texas 78712-1202, U.S.A.
<EMAIL_ADDRESS>
Jean-Michel Roquejoffre
Institut de Mathématiques de Toulouse (UMR CNRS 5219)
Université Toulouse III-Paul Sabatier, 118 route de Narbonne
31062 Toulouse cedex, France
<EMAIL_ADDRESS>
Ignacio Tomasetti
The University of Texas at Austin
Mathematics Department RLM 8.100, 2515 Speedway Stop C1200
Austin, Texas 78712-1202, U.S.A.
<EMAIL_ADDRESS>
###### Abstract

A horizontal $N$-dimensional plane, having a diffusion of its own, exchanges
with the lower half space. There, a reaction-diffusion process, modelled by a
free boundary problem, takes place. We wish to understand whether, and how,
the free boundary meets the plane.

The origin of this problem is a two-dimensional reaction diffusion model
proposed some time ago by the second author, in collaboration with H.
Berestycki and L. Rossi, to model how biological invasions can be enhanced by
a line of fast diffusion. Some counter-intuitive numerical simulations of this
model, due to A.-C. Coulon, have been explained by the first two authors by
transforming the model into a free boundary interacting with a line, and a
careful study of the free boundary. At this occasion, it was noticed that the
free boundary very much like that of the obstacle problem.

The goal of the paper is to explain how this analogy with the obstacle problem
can be pushed further in higher space dimensions.

##  1 Introduction

###  1.1 Model and question

Let $\varphi$ be a smooth concave function of $\mathbb{R}^{N}$, attaining a
positive maximum at $x=0$. We look for a function $u(X)$, defined for
$X=(x,y)\in\mathbb{R}^{N+1}_{-}$ with
$\mathbb{R}^{N+1}_{-}:=\mathbb{R}^{N}\times\mathbb{R}_{-}$, and a surface
$\Gamma\subset\mathbb{R}^{N+1}_{-}$ such that

(1.1) $\left\\{\begin{array}[]{rll}\Delta u=&0\quad(x,y)\in\\{u>0\\}\\\
|\nabla u|=&1\quad((x,y)\in\Gamma:=\partial\\{u>0\\})\\\ \\\
-\Delta_{x}u+u_{y}=&0\quad(x\in\mathbb{R}^{N},y=0)\\\
u(x,0)\geq&\varphi(x).\end{array}\right.$

We ask whether, and how, the free surface $\Gamma$ meets the plane
$\\{y=0\\}$.

###  1.2 Motivation of the model, possible extensions

The study of Model (1.1) is motivated by previous investigations on a two-
dimensional version that modelled the influence of a line of fast diffusion on
the propagation of reaction-diffusion waves. More precisely, in [14] and [15],
the first two authors study o a system of the form

(1.2) $\left\\{\begin{array}[]{rll}-\Delta
u+c\partial_{x}u=&0\quad(x,y)\in\\{u>0\\}\\\ |\nabla
u|=&1\quad((x,y)\in\Gamma:=\partial\\{u>0\\}\\\ \\\
-u_{xx}+c\partial_{x}u+u_{y}=&0\quad\hbox{for $x\in\mathbb{R}$, $y=0$}\\\
u(-\infty,y)=&1,\quad\hbox{uniformly in $y\in\mathbb{R}_{-}$}\\\
u(+\infty,y)=&0\quad\hbox{pointwise in
$y\in\mathbb{R}_{-}$}.\end{array}\right.$

Studying (1.2) was itself motivated by a rather a counter-intuitive numerical
simulation in [17], of a system initially proposed in [6] to model the
influence of transportation networks on biological invasions. See also [4].

Among other things, it is proved in [14] that the free boundary $\Gamma$ hits
the line $\\{y=0\\}$ as an inverted parabola: assuming that $(0,0)$ is the
intersection point, then, in its vicinity, $\Gamma$ may be described as

$y=-\frac{x^{2}}{2},\ \ \ \ x\leq 0,$

and $u(x,0)$ grows in a quadratic fashion for negative $x$. This is typical of
the behaviour of the solutions to the obstacle problem, and the present study
of Model (1.1) is an attempt to understand whether the situation discovered in
[14] is general. Theorems 1.1 and 1.2 below show that this is indeed the case.

We point out that, while Model shows the genericity of the two-dimensional
situation in several space dimensions, our study depends on the global
geometry of the problem, and that properties such as monotonicity in every
direction, help simplifying the analysis. Thus, a general local study of the
interaction of the free boundary of the one phase problem with the plane is
still to be carried out. Whether, in particular, Theorem 1.2 would persist in
a general setting, is something we do not know. Another interesting question
is to replace the equation for $u$ on the plane by the family of equations

$-D\Delta u+u_{y}=0,$

and investigate what happens as $D\o+\infty$. This question, by the way, is
studied in the forthcoming work [16].

Other geometries would be relevant. In particular, the obstacle condition

$u(x,0)\geq\varphi(x)$

could certainly be replaced by a Dirichlet type condition on a vertical
cylinder inside the lower half space. The ideas displayed here would probably
work in this setting without so many changes. The diffusion equation on the
plane $\\{y=0\\}$ could also be replaced by

$-\sum_{1\leq i,j\leq N}\partial_{x_{i}}\bigl{(}a_{ij}(\nabla
u)\bigl{)}+u_{y}=0,$

where the matrix $\bigl{(}a_{ij}(p)\bigl{)}$ satisfies the usual ellipticity
conditions, such as, for instance, in [7]. This last extension could open the
door to the study of interesting directional effects.

###  1.3 Main results

Let us first introduce some notations. As we will have to deal with several
choices of balls, either in the lower half space $\mathbb{R}^{N+1}_{-}$ or the
plane $\\{y=0\\}$ it is convenient to denote them with distinctive signs now.
So, let $B^{N+1}_{R}(X)$ be the ball of $\mathbb{R}^{N+1}$ with centre $X$ and
radius $R$; when $X=0$ we will simply denote it by $B^{N+1}_{R}$. In the same
spirit, we denote by $B_{R}^{N}(x)$ the ball of the plane $\\{y=0\\}$ with
centre $x$ and radius $R$, again denoted by $B_{R}^{N}$ if $x=0$.

Let us now make a set of slightly more stringent assumptions on $\varphi$.
They are probably not indispensable, however they will ensure, in a relatively
painless way, the existence of a solution to (1.1) whose support on the plane
is larger than that of $\varphi$.

  1. 1.

We have $\varphi(x)=1$ if $x\in B_{1}^{N}$.

  2. 2.

There is $\rho_{0}>0$ such that

(1.3) $\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}\subset
B^{N}_{1+\rho_{0}}.$

Moreover $\varphi$ is strictly concave outside $B^{N}_{1}$.

With these assumptions we can prove the

###### Theorem 1.1

If $\rho_{0}>0$ is small enough, Problem (1.1) has a solution $(u,\Gamma)$
which has the following properties:

  1. 1.

there is $\rho_{1}>\rho_{0}$ such that

(1.4) $B^{N}_{1+\rho_{1}}\subset\mathrm{supp}\leavevmode\nobreak\ u(.,0).$

  2. 2.

There is $\theta_{0}>0$ such that, for every $x\in\mathbb{R}^{N}$ outside
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, we have
$\partial_{e}u(x,0)\leq 0$ for every direction $e$ making an angle less or
equal $\theta_{0}$ with $\displaystyle\frac{x}{|x|}$.

  3. 3.

The free boundary $\Gamma$ meets the plane $\\{y=0\\}$ along a Lipschitz
surface, and encloses a domain $\Omega$ of the plane $\\{y=0\\}$.

This result is in the spirit of the work [8] of the first author, where some a
priori monotonicity results of the solution entail the Lipschitz property of
the free boundary in an obstacle type problem.

###### Theorem 1.2

There is $\delta>0$ such that have

$u_{y}(x,0)\geq\delta\quad\hbox{if $x\in\Omega$}.$

Moreover, $u_{y}$ is Hölder-continuous in $\bar{\Omega}$.

The end result of Theorem 1.2 is that $u(x,0)$ behaves, in the vicinity of
$\Gamma\cap\\{y=0\\}$, very much like the solution of the obstacle problem in
its reduced form, that is, according to the notation of [11] (Part I, Point
b)):

$-\Delta_{x}u(x,0)+g(x)H(u)=0\quad\hbox{in $\Omega=\\{u(x,0)>0\\}$},$

where $H$ is the Heaviside function, and the function $g(x)$ is a positive,
Hölder continuous function, a role that the function $u_{y}(x,0)$, can be made
to play thanks to Theorem 1.2. This property is sufficient to study a good
part of the regularity of $\partial\Omega$, see for instance [11], or the
original paper [10], or the review [9]. In particular, Theorem 6 of [11]
implies that $\partial\Omega$ is a $C^{1,\alpha}$ surface.

###  1.4 Organisation of the paper

A first issue is to construct a solution to (1.1) having some nice geometrical
properties, such as the monotonicity in a cone of directions of the plane.
While this should not be indispensable to the study (at least the basic
properties), it greatly helps us in the analysis. And so, the next two
sections are devoted to the construction of such a solution to (1.1), through
an approximation by a singularly perturbed semilinear problem. We prove, in
particular, that the free boundary $\Gamma$ intersects the plane, opening the
way to the analogy with the obstacle problem. In Section 4, we start the
analysis of $u(.,0)$ by proving the basic properties of the solutions to the
obstacle problem. We prove optimal regularity, that is, $u(.,0)$ is $C^{1,1}$
in the vicinity of $\Gamma\cup\\{y=0\\}$, and that $u(x,.)$ grows at least
quadratically as $x$ departs from the free boundary. The last section is
devoted to the Hölder continuity of $u$, thus concluding the analogy.

##  2 Construction of a solution to (1.1): the semilinear approximation

The goal of the next two sections is to construct a solution $(u,\Gamma)$ of
(1.1) that is as classical as possible, that is the free boundary relation is
satisfied in the strong sense away from the place where $\Gamma$ meets the
plane, and the exchange condition $-\Delta_{x}u+u_{y}=0$ is satisfied in the
strong sense, the function $u_{y}$ being $L^{\infty}$ at this stage. For that
we will resort to a smoothing of the free boundary problem by a semilinear
one. This procedure, currently used in mathematical models for premixed
flames, was first used in multidimensional elliptic problems by Berestycki,
Nirenberg and the first author in [3] to approximate one-phase free boundary
problems. See also the work of Lee, Mellet and the first author where this
approximation is used in the homogenisation of this type of free boundary
problems for parabolic equations. The procedure was more recently used in [14]
to construct a travelling wave solution.

As is usual, the coincidence set of $u$ will be the (closed) subset of the
plane $\mathbb{R}^{N}$ where $u(.,0)=\varphi$. We denote it by
$\mathcal{C}_{u}$.

Consider a smooth function $f$, positive on $\mathbb{R}_{+}$, and such that
$\displaystyle\int\beta(u)du=\displaystyle\frac{1}{\sqrt{2}}$, with
$\beta(u)=uf(u)$, and such that $uf^{\prime}(u)+f(u)\geq 0$ for
$u\in\mathbb{R}_{+}$. A suitable multiple of $\displaystyle\frac{1}{1+u}$ will
do the trick.

We will first solve the problem

(2.1) $\left\\{\begin{array}[]{rll}-\Delta
u+\beta(u)=&0\quad(x,y)\in\mathbb{R}^{N+1}_{-}\\\ \\\
-\Delta_{x}u+u_{y}=&0\quad(x\in\mathbb{R}^{N},y=0)\\\
u(x,0)\geq&\varphi(x).\end{array}\right.$

Once a solution $u$ is constructed, we will set, for every $\varepsilon>0$:

(2.2)
$\beta_{\varepsilon}(u)=\displaystyle\frac{u}{\varepsilon^{2}}f(\frac{u}{\varepsilon}),$

and show that, up to a subsequence, the family
$(u_{\varepsilon})_{\varepsilon}$ will converge, as $\varepsilon>0$, to a
solution of (1.1).

Let us therefore start with (2.1). We start the construction in a standard
way: we first look for a solution to the approximate problem

(2.3) $\left\\{\begin{array}[]{rll}-\Delta
u+\beta(u)=&0\quad(x,y)\in\mathbb{R}^{N+1}_{-}\cap B^{N+1}_{R}\\\
-\Delta_{x}u+u_{y}=&0\quad(x\in\mathbb{R}^{N},y=0)\cap B^{N}_{R}\\\ \\\
u(x,0)\geq&\varphi(x)\\\ u(X)=&0\quad(|X|=R).\end{array}\right.$

In what follows, $R$ is assumed to be large enough so that
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$ is inside $B_{R/2}^{N}$ We
have the easy a priori estimate

$0\leq u\leq\varphi(0).$

The energy functional

(2.4) $J_{R}(u)=\int_{B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-}}\biggl{(}|\nabla
u|^{2}+B(u)\biggl{)}dX+\int_{B_{R}^{N}}|\nabla_{x}u(x,0)|^{2}dx,$

with

$\quad B(u)=\int_{0}^{u}\beta(v)dv,$

is easily shown to attain a minimum over all functions $u\in
H^{1}_{0}(B^{N+1}_{R}\cap\mathbb{R}^{N+1}_{-})$, whose trace on the plane
$\\{y=0\\}$ is larger than $\varphi$ and belongs to $H^{1}_{0}(B^{N}_{R})$.
This means that (2.3) has at least a solution. Now, we need to prove enough
smoothness - not such a difficult issue at this stage. The main step is a
uniform Lipschitz bound.

###### Lemma 2.1

There is a constant $M>0$, independent of $\lambda$ and $R$ (but, at this
stage, strongly depending on $\beta$) such that, for all $u\in S_{\lambda,R}$
we have

$\|\nabla u\|_{L^{\infty}(B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-})}\leq M.$

Proof. The only point that is slightly nonclassical here is to handle the
detachment from the coincidence set $\mathcal{C}_{u}$, and the points where
the Wentzell conditions are enforced at the same time. In order to isolate
these two parts, let $R_{0}>0$ be such that $B_{R_{0}}^{N}$ contains the whole
coincidence set, and choose $R\geq 3R_{0}$ once and for all. Outside
$B_{3R_{0}}^{N+1}\cap\mathbb{R}^{N+1}_{-}$, the problem is a standard
semilinear equation with Wentzell boundary conditions, so that $u$ enjoys
$C^{2,\alpha}$ bounds that are uniform in $R$.

To deal with the remaining part, we could use the property
$uf^{\prime}(u)+f(u)\geq 0$ for a quick proof. The proof that we are going to
give, while being slightly longer, shows that this particular property is not
needed here. Let us first recall some basic facts about the first eigenvalue.
If $\mathcal{O}$ is a bounded open subset of $\mathbb{R}^{N+1}_{-}$ with trace
$\mathcal{O}_{0}$ on the plane $\mathbb{R}^{N}$, contained in $B_{2R_{0}}$,
and such that $\mathcal{C}_{u}$ is contained in $\mathcal{O}_{0}$, at nonzero
distance from $\partial\mathcal{O}_{0}$. Consider the eigenvalue problem

(2.5) $\left\\{\begin{array}[]{rll}-\Delta u=&\mu
u\quad((x,y)\in\mathcal{O})\\\ -\Delta_{x}u+u_{y}=&\mu
u\quad(x\in\mathcal{O}_{0},y=0)\\\ \\\
u(x,y)=&0\quad((x,y)\in\mathcal{O},y<0)\\\
u(x,0)=&0\quad(x\in\partial\mathcal{O}_{0}).\end{array}\right.$

From Krein-Rutman’s theorem there is a smallest eigenvalue, called
$\mu_{1}(\mathcal{O})$, associated to a positive eigenfunction
$\Phi_{1}(x,y)$, and characterised as the minimum of the Rayleigh type
quotient

$\frac{\displaystyle{\int_{\mathcal{O}_{0}}|\nabla
u(x,0)|^{2}dx+\int_{\mathcal{O}}\biggl{(}|\nabla
u|^{2}+u^{2}\biggl{)}dxdy}}{\|u\|_{L^{2}(\mathcal{O}_{0})}^{2}+\|u\|_{L^{2}(\mathcal{O})}^{2}},$

taken on all functions $u\in H^{1}(\mathcal{O})$,whose trace on
$\partial\mathcal{O}\cap\\{y<0\\}$ is zero, and such that $u(.,0)\in
H^{1}_{0}(\mathcal{O}_{0})$. Let us define the width of $\mathcal{O}$ as the
smallest $l>0$ such that there is a point $(x,-l)$ in $\mathcal{O}$, one
easily deduces from the Poincaré inequality that $\mu_{1}(\mathcal{O})$ goes
to infinity as the width of $\mathcal{O}$ goes to 0.

Choose $\mathcal{O}$ and $l_{0}>0$ such that

  * •

$\mathcal{O}$ is a smooth subdomain of $B^{N+1}_{2R_{0}}\cap(-l_{0},0)$,

  * •

$\partial\mathcal{O}$ contains $B_{4R_{0}/5}^{N}\times\\{-l_{0}\\}$ and
$\mathcal{O}_{0}$ contains $B_{4R_{0}/5}^{N}$,

  * •

we have $\mu_{1}(\mathcal{O})\geq\|\partial_{u}\beta(u)\|_{\infty}+1$.

Choose, finally, a subdomain $\mathcal{O}^{\prime}$ of $\mathcal{O}$ such that
$\partial\mathcal{O}^{\prime}\cap\\{y<0\\}$ is at positive distance from
$\partial\mathcal{O}\cap\\{y<0\\}$, and such that, if
$\mathcal{O}_{0}^{\prime}$ is the trace of $\mathcal{O}_{0}$ on the plane
$\mathbb{R}^{N}$, then $\partial\mathcal{O}_{0}^{\prime}$ is at a positive
distance from $\partial\mathcal{O}_{0}$.

Pick any direction $e\in\mathbb{R}^{N}$, let us bound $\partial_{e}u$. As
mentionned before, it is sufficient to do it in $\mathcal{O}^{\prime}$. If
$\Phi_{1}$ is the eigenfunction associated to $\mu_{1}(\mathcal{O})$, then
$C\Phi_{1}$ bounds $|\partial_{e}u|$ on $\mathcal{O}^{\prime}\cap\\{y<0\\}$,
as well as on $\mathcal{C}_{u}$, as soon as $C>0$ is large enough, in any case
larger than $\|\nabla\phi\|_{\infty}$. Now, $v=\partial_{e}u$ satisfies

$-\Delta v+\partial_{u}\beta(u)v=0.$

Since
$\mu_{1}(\mathcal{O}^{\prime})>\mu_{1}(\mathcal{O})\geq\|\partial_{u}\beta(u)\|_{\infty}+1$,
the maximum principle holds and we have $|v|\leq C|\Phi_{1}|$. The same
argument applies to $w:=\partial_{ee^{\prime}}u$, where $e^{\prime}$ is
another direction of $\mathbb{R}^{N}$. We have indeed

$-\Delta
w+\partial_{u}\beta(u)w=\partial_{uu}\beta(u)\partial_{e}u\partial_{e^{\prime}}u=O(1).$

It remains to bound $u_{y}$. We denote it by $v$, and notice that $v$ is
bounded on the plane outside $\mathcal{C}_{u}$, while $v_{y}$ is bounded on
$\mathcal{C}_{u}$, as it is equal to $-\Delta\phi$. Consider
$v(x,y)+y\Delta\varphi(x)$ and note that this time it vanishes on
$\mathcal{C}_{u}$, while still being bounded outside. The preceding argument
may be applied to $v$, this time to the Laplacian in $\mathcal{O}^{\prime}$,
still with Dirichlet conditions on $\mathcal{O}^{\prime}\cap\\{y<0\\}$, but
with Neumann conditions on $\mathcal{O}_{0}^{\prime}$. $\Box$

As a consequence of the equation $-\Delta u+u_{y}=0$, for all
$\alpha\in(0,1)$, there is a uniform $C^{1,\alpha}$ on $B^{N}_{R}$ for
$u(.,0)$, for all $u\in S_{\lambda,R}$. This bound is in turn translated into
a uniform $C^{1,\alpha}$ bound in $B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-}$. Once
this is at hand, we go to the main qualitative property.

###### Lemma 2.2

There is $\delta>0$, independent of $\lambda$ and $R$, such that we have, for
all solution $u$ of (2.3), for all $x$ outside
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, and all unit vector $e$:

(2.6) $\hbox{If $\displaystyle{\biggl{|}\frac{x}{|x|}-e\biggl{|}}\leq\delta$,
then}\ \partial_{e}u(x,0)\leq 0.$

Moreover we have

(2.7) $\partial_{y}u\geq\delta.$

Proof. This is really where we need $uf^{\prime}(u)+f(u)\geq 0$. Let us first
check (2.7) in the most standard way, by proving that $v:=u_{y}$ cannot have a
nonpositive minimum. Compactness will once again ensure $v\leq-\delta$, for
some $\delta>0$ independent of $R$. Assume therefore the existence of a
nonpositive minimum for $v$. As $uf^{\prime}(u)+f(u)\geq 0$, the equation for
$v$

$-\Delta v+\bigl{(}uf^{\prime}(u)+f(u)\bigl{)}v=0\quad\hbox{in
$B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-}$}$

and the Hopf Lemma ensures that the minimum can be assumed neither inside
$B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-}$, nor on
$\partial(B_{R}^{N+1}\cap\mathbb{R}^{N+1}_{-})$. So, it has to be taken on the
plane. However there are new obstructions: on $\mathcal{C}_{u}$ we have

$v_{y}(x,0)=u_{yy}(x,0)=-\Delta_{x}u(x,0)=-\Delta\varphi(x)>0,$

so that the Hopf Lemma precludes a minimum of $v$ on that set. Outside we
have, by the same considerations:

$v_{y}(x,0)+v(x,0)=\beta\bigl{(}u(x,0)\bigl{)},$

which forbids a nonpositive minimum once again.

That $u$ decreases along each direction $e$ of the plane $\\{y=0\\}$, outside
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, is a standard consequence of
the moving plane method [18], as one can perform it by reflecting $u$ across
any hyperplane

(2.8) $H_{\lambda}=\\{(x,y):\ x.e=\lambda\\}.$

Indeed, if $u^{r}$ is the reflection of $u$, we have $-\Delta
u^{r}+\beta(u^{r})=0$ in the reflection of $B_{R}^{N+1}\cap\\{y<0\\}$ across
$H_{\lambda}$, while $-\Delta u^{r}(x,0)=-\partial_{y}u^{r}(x,0)\leq 0$ in the
reflection of $B^{N}_{R}$ across the trace of $H_{\lambda}$. As for $u$, it
still solves the PDE below, and its laplacian is negative when it coincides
with $\varphi$. So, this is sufficient to perform the moving plane method
until the moving plane touches $\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$
. Consider now a point $x_{0}$ at distance $\varepsilon_{0}>0$ small from
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, and its projection
$\bar{x}_{0}$ on $\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$ . Set

$e_{0}=\displaystyle\frac{x_{0}-\bar{x}_{0}}{|x_{0}-\bar{x}_{0}|}.$

Let $H_{\lambda}$ be given by (2.8) with $e=e_{0}$. By convexity, there is
$\lambda_{0}$ such that $H_{\lambda_{0}}$ is the supporting hyperplane of
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$ at $x_{0}$, and all
neighbouring points $x^{\prime}$ are reached by the reflection across
$H_{\lambda}$, $\lambda-\lambda_{0}$ small enough. This yields a uniform cone
of monotonicity for $u$ directed by $e_{0}$. $\Box$

###### Remark 2.3

As $\beta^{\prime}\geq 0$, uniqueness holds in (2.3).

###### Remark 2.4

For more general $f$’s, another way to obtain solutions to (2.3) with
$u_{y}\geq 0$ is to start from the Cauchy Problem

(2.9) $\left\\{\begin{array}[]{rll}u_{t}-\Delta_{x}u+u_{y}=&0,\
u(t,x,0)\geq\varphi(x)\quad(t>0,x\in B_{R}^{N})\\\ u_{t}-\Delta
u+\beta(u)=&0\quad(t>0,x\in\mathbb{R}^{N},y<0)\\\ \\\
u(t,x,y)=&0\quad(t>0,|(x,y)|=R,y\leq 0)\\\ u(0,x,0)=&\varphi(x)\\\
u(0,x,y)=&0.\end{array}\right.$

The equation on the plane $\\{y=0\\}$ is a parabolic obstacle problem, however
(2.9) would yield a solution converging, for large times, to a steady
solution. The property $u_{y}\geq 0$ is inherited from the initial datum This
would not prove that all solutions $u$ to (2.3) satisfy $y_{y}\geq 0$, but
would at least have the merit of exhibiting one.

A standard compactness argument now shows, by letting $R\to+\infty$, that a
sequence of minimisers of the energy $J_{R}$ given by (2.4) will converge,
locally on every compact set, to a solution $u$ of (2.1). We denote this
sequence $(u_{R_{n}})_{n}$, we have that the sequence
$(J_{R_{n}}(u_{R_{n}}))_{n}$ is nonincreasing, simply because $u_{R_{n}}$ is
an admissible test function for $J_{R_{n+1}}$. This implies that $u$ has
finite energy, hence, in particular, $u(X)$ tends to 0 as $|X|\to\infty$.

##  3 Construction of a solution to (1.1): letting $\varepsilon\to 0$

Before starting the approximation process, let us define an auxiliary solution
that will help us in proving that the free boundary on the plane actually
exists. Let $\rho_{0}>0$ be such that $B_{\rho_{0}}^{N}$ contains
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, and let $\bar{\varphi}$ be a
spherically symmetric function, equal to $\varphi(0)$ on $B_{\rho_{0}}^{N}$,
and strictly concave outside $\overline{B}_{\rho_{0}}^{N}$. Call
$\bar{u}_{\Lambda}(x,y)$ a solution of (2.1) with $\varphi$ replaced by
$\Lambda\bar{\varphi}$, with $\Lambda>1$. We claim that $\bar{u}_{\Lambda}$ is
axially symmetric and, if $\Lambda>1$ is large enough, that
$\bar{u}_{\Lambda}>u$. This is once again done by the usual sliding type
argument, as $u$ may exceed $\bar{u}_{\Lambda}$ only when it is close to 0,
that is, at infinity. However, the usual maximum principle applies to
$\bar{u}_{\Lambda}-u$ in this area. Decreasing back $\Lambda$ to 1, we see
that $\bar{u}_{1}>u$. This function $\bar{u}_{1}$ will be called $\bar{u}$,
until further notice.

We now set $\beta(u):=\beta_{\varepsilon}(u)$ given by (2.2), and we call
$u_{\varepsilon}$ a so constructed solution to (2.1). We want to pass to the
limit $\varepsilon\to 0$, and the main step is a gradient bound.

###### Lemma 3.1

There is $M>0$ independent of $\varepsilon$ such that $\|\nabla
u_{\varepsilon}\|_{\infty}\leq M$.

Proof. The situation that we have inherited is the following: for all
$\lambda>0$, each level set $\Sigma_{\lambda}=\\{u_{\varepsilon}=\lambda\\}$
is a graph in the vertical direction, that meets the plane $\\{y=0\\}$ because
$u_{\varepsilon}(X)$ goes to 0 as $|X|$ goes to infinity.

Let us consider a point of $\Sigma_{\varepsilon}$ of the form
$(x_{\varepsilon},-\varepsilon)$. Do the classical Lipschitz scaling

$v(\xi,\zeta)=\frac{u\biggl{(}(x_{\varepsilon},-\varepsilon)+\varepsilon(\xi,\zeta)\biggl{)}}{\varepsilon},$

the function $v$ solves

(3.1) $\left\\{\begin{array}[]{rll}-\Delta_{\xi}v+\varepsilon
v_{\zeta}=&0\quad(\xi\in\mathbb{R}^{N},\zeta=1)\\\ \\\ -\Delta
v+\beta(v)=&0\quad(\xi\in\mathbb{R}^{N},\zeta\leq 1)\\\
v(0,0)=&1.\end{array}\right.$

We want to prove that $v(0,1)$ is bounded by a universal constant. Let us set
$v(0,1)=A$ with $A>1$ and let us examine what this implies. Notice that a
crude second derivative bound for $u_{\varepsilon}$ is
$|D^{2}u_{\varepsilon}(X)|\leq\displaystyle\frac{C}{\varepsilon}$, so that, by
interpolation,
$v_{\zeta}(\xi,\zeta)=v_{y}(\xi,\zeta)=O(\displaystyle\frac{1}{\sqrt{\varepsilon}})$.
Thus the equation for $v$ on the plane $\\{\zeta=1\\}$ is really $-\Delta
v=O(\sqrt{\varepsilon})$, so that the Harnack inequality yields

$v(\xi,1)\geq\eta_{0}A\quad\hbox{for $|\xi|\leq 1$},$

with $\eta_{0}>0$ universal. Let $\underline{v}(\xi,\zeta)$ solve

(3.2) $\left\\{\begin{array}[]{rll}v(\xi,\zeta)=&\eta_{0}A\quad(\xi\in
B_{1}^{N},\zeta=1)\\\ \\\ -\Delta v+\|f\|_{\infty}v=&0\quad(\xi\in
B_{1}^{N},\zeta\leq 1)\\\ v(\xi,\zeta)=&0\quad(\xi\in\partial
B_{1}^{N},\zeta\leq 1)\end{array}\right.$

System (3.2) implies that $\underline{v}$ is a subsolution to (3.1), and that
$\underline{v}\leq v$. The argument is of the classical sliding type: we have

(3.3) $\gamma\underline{v}\leq v$

for small $\gamma>0$ (one has to examine $v$ and $\underline{v}$ at infinity,
but this is where the standard maximum principle applies) and (3.3) holds on
compact sets; this implies that (3.3) holds up to $\gamma=1$, if not there
would be a touching point between $\gamma\underline{v}$ and $v$ inside the
cylinder $B_{1}^{N}\times\\{\zeta\leq 1\\}$. However, (3.2) and the strong
maximum principle implies that $v(0,0)$ is a universal multiple of $A$, which
implies that $A$ is universally bounded. By elliptic regularity, $v_{\zeta}$
and $\nabla_{\xi}v$ are bounded on $B_{1}^{N}\times(-1,1)$.

Consider $l_{0}<1$ such that the first eigenvalue of the Dirichlet Laplacian
on the cylinder

$\\{\xi_{1}\geq-2,|\xi_{i}|\leq 2\\}\times(1-2l_{0},1)$

is larger than $1+\|\beta_{1}\|_{\infty}$. Because of the boundedness of
$\partial_{\xi_{i}}v$ and $\partial_{\zeta}v$ on
$\\{\xi_{1}=-1,|\xi_{i}|=2\\}\times(1-l_{0},1)$, a large multiple of the first
eigenfunction of this operator bounds $\partial_{\xi_{i}}v$ and
$\partial_{\zeta}v$ in $\\{\xi_{1}\geq-1,|\xi_{i}|\leq 1\\}\times(1-l_{0},1)$.
Because $v$ decreases in the vertical direction, this process actually bounds
$v$ in the whole slab $\\{\xi_{1}\geq-1,|\xi_{i}|\leq 1\\}\times(-\infty,1)$.

Let $\Sigma_{\varepsilon}$ be the projection of $\Sigma_{\varepsilon}$ onto
the plane $\\{y=0\\}$ and let $\Omega_{\varepsilon}$ the open subset enclosed
by $\tilde{\Sigma}_{\varepsilon}\times\mathbb{R}_{-}$. As the above argument
can be repeated at any point where $u=\varepsilon$, we may conclude, due to
the invariance of the gradient norm under Lipschitz scaling, that $|\nabla
u_{e}|$ is universally bounded on $\partial\Omega_{\varepsilon}$.

Once we are here, it is easy to finish. As $\\{u_{e}=\varepsilon\\}$ is a
graph, it is at distance of the order $\varepsilon$ at least from the plane,
so that the scaling argument will bound $\nabla u_{\varepsilon}$ on the
surface, and below also because $u$ is nonincreasing in the vertical
direction. Above the surface, $u_{\varepsilon}$ is harmonic, so that the
maximum principle and elliptic regularity yield the gradient bound. $\Box$

We need a last lemma ensuring that, for the solution $u(x,y)$ to (1.1) that we
will eventually construct, the set

$\mathrm{supp}\leavevmode\nobreak\ u(,.)\backslash\mathcal{C}$

is nontrivial.

###### Lemma 3.2

With the notations of assumptions 1 and 2 on $\varphi$, there is $\rho_{1}>0$
and $\delta_{1}>0$ independent of $\varepsilon$ such that, if
$\rho_{0}\leq\rho_{1}$ we have

$u_{\varepsilon}(x,0)\geq\delta_{1}\ \hbox{if}\ |x|=1+\rho_{0}.$

Proof. Pick $\rho_{0}>0$ such that $\mathrm{supp}\leavevmode\nobreak\
\varphi^{+}$ is contained in $B_{1+\rho_{0}/2}^{N}$. Assume the existence of a
sequence $(\varepsilon_{n})_{n}$ going to 0 for which there is a sequence
$(x_{n})_{n}$ of $\partial B_{1+\rho_{0}}^{N}$ and a sequence
$(\delta_{n})_{n}$ going to 0 such that

$u_{\varepsilon_{n}}(x_{n},0)=\delta_{n}.$

Recall that $u_{\varepsilon_{n}}$ is a minimiser of the energy

$J_{\varepsilon_{n}}(u)=\int_{\mathbb{R}^{N}}|\nabla_{x}u(x,0)|^{2}dx+\int_{\mathbb{R}^{N+1}_{-}}\biggl{(}|\nabla
u|^{2}+B_{\varepsilon_{n}}(u))\biggl{)}dxdy,$

where $B_{\varepsilon_{n}}$ is a primitive of $\beta_{\varepsilon_{n}}$. Let
$\Omega_{n}$ be the $\delta_{n}$ level surface of $u_{\varepsilon_{n}}$ We
have

$J_{\varepsilon_{n}}(u_{\varepsilon_{n}})\geq\int_{\partial\Omega_{n}}|\nabla_{x}u_{\varepsilon_{n}}(x,0)|^{2}dx,$

and the RHS is larger than the infimum, over all functions of
$H^{1}(B^{N}_{1+\rho_{0}})$ that are larger than $\varphi$ and equal to
$\delta_{n}$ on $\Omega_{n}$, in other words, the solution $v_{n}$ of the
classical obstacle problem in $\Omega_{n}$ with the Dirichlet data
$\delta_{n}$ on $\partial\Omega_{n}$. Let us bound its Dirichlet energy from
below; we recall that, if $\bar{x}_{n}$ is the projection of $x_{n}$ onto
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, and

$e_{n}=\frac{x_{n}-\bar{x}_{n}}{|x_{n}-\bar{x}_{n}|},$

then, around $x_{n}$, the level set $\partial\Omega_{n}$ is a uniformly
Lispchitz graph in the direction $e_{n}$. In other words, there is a local
coordinate system around $\bar{x}_{n}$, named $(\xi^{\prime},\xi_{N})$, a
radius $r_{0}>0$, independent of $\rho_{0}$, and a function
$\psi_{n}(\xi^{\prime})$ defined in the ball $B^{N-1}_{r_{0}}$ of
$\mathbb{R}^{N-1}$, expressed as $\\{|\xi^{\prime}|\leq r_{0}\\}$, with
$\|\nabla\psi_{n}\|_{L^{\infty}(\\{|\xi^{\prime}|\leq r_{0}\\})}$ uniformly
bounded, such that, in this coordinate system, we have
$x_{n}=(0,\displaystyle\frac{\rho_{0}}{2})$, and

$\Sigma_{n}\cap
B^{N-1}_{r_{0}}=\\{(\xi^{\prime},\frac{\rho_{0}}{2}+\psi_{n}(\xi^{\prime})),|\xi^{\prime}|\leq
r_{0}\\}.$

Let $\Gamma_{n}$ be the cone with vertex $\bar{x}_{n}$, generated by the ball
$B^{N-1}_{r_{0}}$, its aperture is given by

$\theta_{n}=\mathrm{Arctan}\biggl{(}{2r_{0}}{\rho_{0}}\biggl{)},$

thus bounded from below independently of $\rho_{0}$. And so, there is a
universal constant $C_{0}>0$ such that

$|\Gamma_{n}|\geq C_{0}\rho_{0}.$

In $\Gamma_{n}$, the function $v_{n}$ drops from $v_{n}=1$ on $B^{N}_{1}$ to
$O(\varepsilon_{n})$ on $\Sigma_{n}$; by the Harnack inequality we have

$-\partial e_{n}v_{n}(x)\geq\frac{C}{\rho_{0}},$

$C>0$ universal. All in all, we have

$\int_{\partial\Omega_{n}}|\nabla_{x}u_{\varepsilon_{n}}(x,0)|^{2}dx\geq\int_{\Gamma_{n}}|\nabla
v_{n}|^{2}dx\gtrsim\frac{1}{\rho_{0}},$

a quantity that goes to infinity as $\rho_{0}$ goes to 0. On the other hand,
take any function $v(x,y)$ such that

  1. 1.

$v(.,0)$ is smooth and supported in $B^{N}_{2}$,

  2. 2.

the set $\partial\\{v>0\\}$ is a smooth surface that meets the plane
$\\{y=0\\}$ tangentially,

  3. 3.

$v$ is supported in $B^{N+1}\cap\\{y<0\\}$, smooth in $B^{N+1}\cap\\{y<0\\}$
and decays linearly to 0 in the vicinity of $\partial\\{v>0\\}$.

Then, due to the nondegeneracy of $v$ along the boundary of its zero set we
have, uniformly in $n$:

$J_{\varepsilon_{n}}(u_{\varepsilon_{n}})=O(1),$

independently of $\rho_{0}$. This is a contradiction. $\Box$

Proof of Theorem 1.1. The uniform Lipschitz bound implies the convergence of a
subsequence of $(u_{\varepsilon})_{\varepsilon}$ to a limiting Lipschitz
function $u$. From Lemma 3.2, and the fact that $u$ is Lipschitz, the set

$\Omega=\\{x\in\mathbb{R}^{N},\ u(x,0)>0\\}$

is nontrivial, as the maximum of $u$ is $\varphi(0)$. Moreover, if
$\mathcal{C}$ is its coincidence set, then $\Omega\backslash\mathcal{C}$ has
nonempty interior. This implies that the set
$\\{u>0\\}\cap\mathbb{R}^{N+1}_{-}$ has also a nonempty interior.

Let $\bar{u}_{\varepsilon}$ be the axially symmetric barrier constructed for
$u_{\varepsilon}$ with the aid of the function $\bar{\varphi}$. From [14] (an
analysis based on [1] and [3]), a subsequence of
$(\bar{u}_{\varepsilon})_{\varepsilon}$ converges to a minimiser of the
functional

$\bar{J}(u)=\int_{\mathbb{R}^{N}}|\nabla
u(x,0)|^{2}dx+\int_{\mathbb{R}^{N+1}_{-}}|\nabla
u|^{2}dxdy+|\\{u>0\\}\cap\mathbb{R}^{N+1}_{-}|.$

This implies that $\bar{u}$ develops a free boundary $\bar{\Gamma}$ that is,
due to [1], a $C^{\infty}$ surface in $\mathbb{R}^{N+1}_{-}$ which is, due to
axial symmetry, generated by a $C^{\infty}$ curve $\bar{\Gamma}_{0}$. Assume
that $\bar{\Gamma}$ does not meet the plane, integration of the full equation
(1.1) for $\bar{u}$ yields

$+\infty=\int_{\Gamma}d\sigma(x)=\int_{\mathcal{C}}\partial_{y}\bar{u}<+\infty,$

a contradiction. This implies that $\bar{\Gamma}_{0}$ has finite length, in
other words, that $\bar{\Gamma}$ meets the plane $\mathbb{R}^{N}$ along a
sphere of finite radius.

As $\bar{u}\geq u$, we have

$\\{u>0\\}\subset\\{\bar{u}>0\\},$

so that $\Gamma$ lies above $\bar{\Gamma}$ and intersects the plane
$\mathbb{R}^{N}$ along $\partial\Omega$. The positivity set $\Omega$ of $u$ in
$\mathbb{R}^{N}$ is thus bounded. Notice that $\bar{u}$ was useful in ruling
out an infinite surface $\partial\Omega$ with needles at infinity, something
that is not a priori ruled out by the fact that $\\{u>0\\}$ has finite
perimeter.

We already know that $u(x,0)$ is decreasing in the direction
$\displaystyle\frac{x}{|x|}$, as soon as $x$ is outside
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$ . Let us prove that it
actually decreases in a cone of directions of axis
$\displaystyle\frac{x}{|x|}$. For this, it is enough to prove that the
distance between $\partial\Omega$ and $\mathrm{supp}\leavevmode\nobreak\
\varphi^{+}$ is positive. This, however, is not difficult to see, as $u(.,0)$
is $C^{1,\alpha}$ for all $\alpha\in(0,1)$. Should there be a coincidence
point between $\partial\Omega$ and $\mathrm{supp}\leavevmode\nobreak\
\varphi^{+}$, this would imply, by the convexity of
$\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$, a derivative discontinuity at
that point. This proves that the boundary $\partial\Omega$ is a bounded
Lipschitz surface of $\mathbb{R}^{N}$ with no boundary.

We finally notice that, because $u(x,y)$ is decreasing in a cone of directions
of $\mathbb{R}^{N+1}_{-}$, $\Gamma\cap\\{y<0\\}$ is a Lipschitz surface; due
to [12] is is $C^{1,\alpha}$ and the free boundary condition $|\nabla
u|=-u_{\nu}=1$ ($\nu$ is the exterior normal to $\\{u>0\\}$) is satisfied. As
a consequence of [19], $\Gamma\cap\\{y<0\\}$ is analytic. Inside
$\Omega\backslash\mathcal{C}$, classical ellliptic theory implies that the
equation $-\Delta u+u_{y}=0$ is satisfied in the strong sense in the whole
plane $\mathbb{R}^{N}$, as well as in the classical sense in
$\Omega\backslash\mathcal{C}$. $\Box$

##  4 Optimal regularity and nondegeneracy

Let $u$ be a solution of (1.1). We are now ready to start the analogy with the
obstacle problem in the vicinity of $\partial\Omega$, that is, where the free
boundary $\Gamma$ generated by the one phase problem in the lower half space
$\mathbb{R}^{N+1}_{-}$ meets the plane $\mathbb{R}^{N}$. To have an idea of
what to expect, it is good to recall what happens in the axially symmetric
case, that is, when the function $\varphi$ is spherically symmetric in
$\mathbb{R}^{N}$. In the coordinate system $(r=|x|,y)$, Problem (1.1) reduces
to

(4.1) $\left\\{\begin{array}[]{rll}\Delta
u+\displaystyle\frac{N-1}{r}u_{r}=&0\quad(x,y)\in\\{u>0\\}\\\ |\nabla
u|=&1\quad((x,y)\in\Gamma:=\partial\\{u>0\\})\\\ \\\
-u_{rr}-\displaystyle\frac{N-1}{r}u_{r}+u_{y}=&0\quad(x\in\mathbb{R}^{N},y=0)\\\
u(r,0)\geq&\varphi(r).\end{array}\right.$

The positivity set of $u$ on the line $\\{y=0\\}$ is a segment $[0,R_{0})$. In
the vicinity of the point $(R_{0},0)$ we have [14] that $\Gamma$ is a graph in
the $r$ variable:

$y=\psi(r),\ r<R_{0},\quad\quad\hbox{with}\
\psi(r)=-\frac{(r-R_{0})^{2}}{2}+o_{r\to R_{0}^{-}}(R_{0}-r)^{2}.$

Moreover, we have

(4.2) $\displaystyle\lim_{r\to R_{0}^{-}}u_{y}(r,0)=1,$

so that the ODE for $u(r,0)$, together with the initial data
$u(R_{0},0)=u_{r}(R_{0},0)=0$ yields

$u(r,0)=\frac{(r-R_{0})^{2}}{2}+o_{r\to R_{0}^{-}}(R_{0}-r)^{2}.$

Coming back to the multi-D case, the least one can expect is therefore a
quadratic detachment of $\Gamma$ from the plane, as well as a quadratic
behaviour of $u$ in the vicinity of $\partial\Omega$. This is what we will
endeavour to prove in this section.

The first question is now whether $u(x,0)$ enjoys a better regularity than
Lipschitz. The obstacle problem suggests $u(.,0)\in C^{1,1}(\mathbb{R}^{N})$,
and this is what we will prove. The following theorem is sufficient for our
purpose.

###### Theorem 4.1

There is $C>0$ universal such that

$u(x,0)\leq C|x|^{2}.$

Proof. Let $U^{x}(r)$ be the average of $u(.,0)$ over the $N$-dimensional
sphere of origin $x$ and radius $r$:

$U^{x}(r)=\frac{1}{r^{N-1}}\int_{\partial
B_{r}^{N}(x)}u(x^{\prime},0)d\sigma(x^{\prime}).$

We have - notice that this is also valid if $B_{r}^{N}(x)$ intersects the
coincidence set, although we will not need it:

$U^{x}_{rr}+\frac{N-1}{r}U^{x}_{r}\leq C,\quad U^{x}(0)=0,$

because of the equation $\Delta_{x}u=u_{y}$ and the boundedness of $u_{y}$.
Because $U^{x}(r)$ is nonnegative we have also $U^{x}_{r}(0)=0$. This entails

$U^{x}(r)\leq Cr^{2},$

for a possibly different $C>0$. Translate the picture so that the origin
$(0,0)$ is a free boundary point (that is, $(0,0)\in\partial\Omega)$ and
$x\in\Omega$. As $u_{y}\geq 0$ and $\varphi$ concave we have
$\Delta_{x}u(x,0)\geq 0.$ The mean value formula yields

$\begin{array}[]{rll}u(x,0)\leq&\displaystyle\frac{1}{|x|^{2}}\int_{B^{N}_{|x|}(X)}u(x^{\prime},0)dx^{\prime}\\\
\leq&\displaystyle\frac{1}{|x|^{2}}\int_{B^{N}_{2|x|}(0)}u(x^{\prime},0)dx^{\prime}\\\
=&4U^{0}(|x|)\\\ \leq&C|x|^{2}.\end{array}$

This is the theorem. $\Box$

The next main item to prove is that $u(x,0)$ grows at least quadratically away
from a free boundary point. The geometric setting is the following: $\Gamma$
is a graph over $\Omega$ of the form

(4.3) $y=\psi(x),\quad x\in\Omega,$

where $\psi$ is $C^{\infty}$ in $\Omega$ and $\psi\equiv 0$ on
$\partial\Omega$, that we recall to be a Lipschitz surface without boundary.

###### Theorem 4.2

Assume the origine to be on the free boundary $\partial\Omega$. There is $C>0$
and $\Lambda>0$ universal such that

(4.4) $u(x,0)\geq
C|x|^{2},\quad-\Lambda|x|^{2}\leq\psi(x)\leq-\frac{1}{\Lambda}|x|^{2}.$

if $x\in\Omega$.

The upper estimate for $\psi$ is straightforward once the estimate for
$u(x,0)$ is known, and results from the gradient bound for $u$:

$C|x|^{2}\leq u(x,0)=u(x,0)-u(x,\psi(x))\leq\|u_{y}\|_{\infty}|\psi(x)|.$

As for the estimate for $u(x,0)$, we will need a better understanding of the
behaviour of $u$ and $\Gamma$ near a free boundary point, and the key element
is the following equivalent of (4.2).

###### Theorem 4.3

We have

(4.5) $\lim_{d(x,\partial\Omega)\to 0,x\in\Omega}u_{y}(x,0)=1.$

Theorem 4.3 implies the lower estimate for $\psi$ in (4.4). It will in turn
result from the following

###### Lemma 4.4

Pick $M>0$. Consider a solution $v(x,y)$ in of the free boundary problem in
the whole lower half plane

(4.6) $\left\\{\begin{array}[]{rll}\Delta v=&0\quad(x,y)\in\\{u>0\\}\\\
|\nabla v|=&1\quad((x,y)\in\Gamma_{v}:=\partial\\{v>0\\})\\\ \\\
v(x,0)=&M\quad(x\in\mathbb{R}^{N}).\end{array}\right.$

Assume the existence of $0<d_{1}<d_{2}$ such that

$\Gamma_{v}\subset\\{-d_{2}\leq y\leq-d_{1}\\}.$

Assume the existence of $\theta_{0}>0$ such that
$\partial_{x/|x|+e^{\prime}}v(x,y)\geq 0$, for
$|e^{\prime}|\leq\theta_{0}$.Then we have

$v(x,y)=(y+M)^{+}.$

Proof. Consider the comparison function

$v_{\Lambda}(y)=(y+\Lambda)^{+},$

which is a solution of (4.6). For large $\Lambda>0$, the assumptions on
$\Gamma_{v}$ imply

(4.7) $\\{v>0\\}\subset\\{v_{\Lambda}>0\\}$

Lower $\Lambda$ until one reaches the threshold $\Lambda_{0}$ where (4.7)
holds no more. Notice that $\Lambda_{0}\geq M$> In case of a strict inequality
there is a contact point between $v$ and $v_{\Lambda_{0}}$, either at finite
distance or at infinity. The assumptions on $v$ imply that $\Gamma_{v}$ is
Lipschitz, thus smooth. Therefore, in the finite distance case, the standard
strong maximum principle and Hopf Lemma preclude this situation, unless
$\Lambda_{0}=M$ and $v\equiv v_{\Lambda_{0}}$. In the infinite distance case,
the usual compactness argument also yields $\Lambda_{0}=M$ and

(4.8) $v(x,y)\leq(y+M)^{+}.$

Now, for small $\Lambda>0$ we have $v\geq v_{\Lambda}$. Increase $\Lambda>0$
such that this is not possible anymore: we have in the end $v\equiv v_{M}$ or

(4.9) $v(x,y)\geq(y+M)^{+}.$

Putting (4.8) and (4.9) together yields the result. $\Box$

Proof of Theorem 4.3. It suffices to prove that the detachment of $\Gamma$
from the plane is at most quadratic, that is, the existence of a universal
constant $C>0$ such that, for all $x_{0}\in\partial\Omega$ we have

(4.10) $|\psi(x)|\leq C|x-x_{0}|^{2}.$

Assume therefore the existence of $x_{0}\in\partial\Omega$ and a sequence
$(x_{n})_{n}$ in $\Omega$ tending to $x_{0}$ such that

$\lim_{x_{n}\to x_{0}}\frac{|\psi(x_{n})|}{|x_{n}-x_{0}|^{2}}=+\infty.$

Set $h_{n}=|\psi(x_{n})|$ Rescale in a Lipschitz fashion with $h_{n}$:

$v_{n}(\xi,\zeta)=\frac{u(x_{n}+h_{n}\xi,h_{n}\zeta)}{h_{n}}.$

From Theorem 4.1, the sequence $(v_{n}(.,0))_{n}$ converges locally uniformly
to 0. Below, it develops a free boundary $\Gamma_{n}$ of the form

$\zeta=\psi_{n}(\xi),\quad\hbox{with $\psi_{n}(0)=-1$}.$

As $\Gamma_{n}$ is uniformly Lipschitz, it is uniformly smooth [12], so that
it converges to a smooth free boundary. This the function $(v_{n})_{n}$
converges, at least in $C^{1}_{loc}$, to a solution $v_{\infty}$ of the one-
phase free boundary problem in $\mathbb{R}^{N+1}_{-}$, while being 0 on the
plane $\mathbb{R}^{N}$. This implies $\partial_{\zeta}v_{\infty}(\xi,0)<0$, a
contradiction with the fact that $u_{y}(x,0)\geq 0$. $\Box$

Proof of Theorem 4.2. Pick $x_{0}\in\partial\Omega$, and let $\bar{x}_{0}$ be
its projection onto $\mathrm{supp}\leavevmode\nobreak\ \varphi^{+}$; let
$e=\displaystyle\frac{x_{0}-\bar{x}_{0}}{|x_{0}-\bar{x}_{0}|}$. We always may
assume that $e=e_{N}$. The solution $u(x,0)$ is decreasing inside a cone
directed by $e_{N}$, so that all its level sets are Lipschitz graphs in the
direction $(-e_{N})$. Assuming without generality that $x_{0}$ is the origin,
it is enough to prove, for $h\in(0,1)$:

(4.11) $u(-\sqrt{h}e_{N},0)\geq Ch,$

with a universal $C>0$. Rescale in a quadratic way:

$U_{h}(x)=\frac{u(\sqrt{h}e_{N},0)}{h},$

we already know from optimal regularity that $(U_{h})_{h}$ is bounded. We
would like to show that $\delta_{h}=U_{h}(-e_{N})$ is bounded from below as
$h\to 0$. For all $\delta>0$, let $\Lambda_{\delta}$ be the $\delta$-level
surface of $u$, we have $\Lambda_{0}=\partial\Omega$. Let
$\Lambda_{\delta}^{h}$ the $\delta/h$ level surface of $U_{h}$. The dilation
of coordinates preserving the Lipschitz norm, the surfaces
$\Lambda_{\delta}^{h}$ are globally uniformly Lipschitz, at least as long as
$\delta/h$ is less than the supremum of $u$. And so, up to a subsequence, the
surfaces $\Lambda_{0}^{h}$ and $\Lambda_{\delta_{h}/h}^{h}$ converge,
respectively, to Lipschitz graphs along the direction $(-e_{N})$ that we
denote by $\Lambda_{0}^{0}$ and $\Lambda^{+}_{0}$. Let $\delta_{0}$ be the
limit of a convergent subsequence of $(\delta_{h}/h)_{h}$, it may be 0 or a
very small constant, and we have to make sure that it is larger than a
universal constant.

Let $\Omega^{0}$ be enclosed by the surfaces $\Lambda_{0}^{0}$ and
$\Lambda^{+}_{0}$. Still because of optimal regularity, its width is bounded
from below. Up to a subsequence, $(U_{h})_{h}$ converges to a solution $U$ of

$\begin{array}[]{rll}-\Delta U+1=&0\quad(x\in\Omega^{0})\\\
U(x)=&0\quad(x\in\Lambda^{0}_{0})\\\
U(x)=&\delta_{0}\quad(x\in\Lambda^{+}_{0})\end{array}$

If $\delta_{0}=0$ we have $U<0$ in $\Omega^{0}$, a contradiction. And so, by
continuity, the positivity of $U$ cannot be preserved if $\delta_{0}$ is below
a universal constant. This proves the claim. $\Box$

##  5 Hölder continuity of the vertical derivative up to $\partial\Omega$

The task is made easier by the Lipschitz character of $\partial\Omega$, as
well as the fact that the free boundary $\Gamma$ below, that we recall to be a
graph $\\{y=\psi(x)\\}$, is a $C^{\infty}$ surface, until it meets the plane.
This will yield an easy upper bound for $u_{y}(x,0)-1$ as $x$ approaches
$\partial\Omega$. A lower bound will be accessible only in an integral sense,
as $x$ approaches $\partial\Omega$. Scaling around the points that are well-
behaved for $u_{y}$ will entail the Hölder continuity of $u_{y}$.

###### Lemma 5.1

There is $\alpha\in(0,1]$ and $C>0$ universal such that, for all $x\in\Omega$
we have

(5.1) $u_{y}(x,0)\leq 1+C|x|^{\alpha}.$

Proof. This lemma involves the specificity of the problem only in a relatively
weak manner. Set $v=u_{y}$, we use once again the Wentzell condition that
entails $v_{y}(x,0)\leq 0$ in $\Omega$. Let $\Omega_{1}$ be a smooth open
subset of $\mathbb{R}^{N}$ such that

$\mathcal{C}_{u}\subset\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega.$

For $M>0$ to be large, let $v_{M}$ be the unique solution of the linear
problem

(5.2) $\left\\{\begin{array}[]{rll}-\Delta
v_{M}=&0\quad(x\in\Omega\backslash\Omega_{1},y\in\mathbb{R})\\\
v_{M}(x,y)=&1\quad(x\in\partial\Omega,y\in\mathbb{R})\\\
v_{M}(x,y)=&M\quad(x\in\partial\Omega_{1},y\in\mathbb{R}).\end{array}\right.$

We claim that $v_{M}\geq v$ as soon as we have

(5.3)
$M>1+\|u_{y}\|_{L^{\infty}(\Omega_{1}\times\mathbb{R}_{-}\cap\\{u>0\\})}.$

Notice first that $\\{u>0\\}\subset\Omega\times\mathbb{R}_{-}$, so that it
suffices to compare $v$ and $v_{M}$ on $\Gamma$ and
$\Omega\backslash\Omega_{1}$. As $v_{M}$ is even in $y$, we have

$v_{y}(x,0)\leq\partial_{y}v_{M}(x,0)=0\quad\hbox{for
$x\in\Omega\backslash\Omega_{1}$.}$

Then, note that $v_{M}\geq 1$ in its domain of definition, so that

$v(x,y)\leq 1\leq v_{M}(x,y)\quad\hbox{for
$(x,y)\in\Gamma\cap\biggl{(}(\Omega\backslash\Omega_{1})\times\mathbb{R}_{-}\biggl{)}$}.$

Finally, (5.3) implies that $v\leq v_{M}$ on
$\\{u>0\\}\cap\partial\Omega_{1}\times\mathbb{R}_{-}$. Therefore, the claim
$v\leq v_{M}$ follows from the maximum principle, and so does the lemma: as
$\partial\Omega$ is Lipschitz, $v_{M}$ satisfies (5.1) for $C$ and $\alpha$
universal. $\Box$

Let us turn to the lower bound. For $x\in\Omega$ close to $\partial\Omega$ let
us set $h=|x|$.

###### Lemma 5.2

There is $\lambda\in(0,1]$, $\alpha\in(0,1]$ and $C>0$ universal such that

(5.4) $\frac{1}{|B_{\lambda h}^{N}|}\int_{B_{\lambda
h}^{N}(x)}|u_{y}(x,0)-1|dx\leq Ch^{\alpha}.$

Proof. Once again because $\partial\Omega$ is Lipschitz we may choose
$\lambda>0$ universal such that

$B_{\lambda h}^{N}(x)\subset\\{x^{\prime}\in\Omega:\
d(x^{\prime},\partial\Omega)\geq\frac{h}{2}\\}.$

For $q>0$ let $\Sigma_{q}$ be the piece of cylinder

$\Sigma_{q}=\\{u>0\\}\cap\biggl{(}B_{k}^{N}(x)\times\mathbb{R}_{-}\biggl{)},$

and set

$\Gamma_{q}=\Gamma\cap\Sigma_{q}.$

From Theorem 4.2, the set $\Sigma_{\lambda h}(x)$ is included in a cylinder
with height of the order no less, and no more than $h^{2}$. Integrate the
equation for $u$ in it. We have

$\int_{B_{\lambda h}^{N}(x)}u_{y}(x^{\prime},0)dx^{\prime}=|\Gamma_{\lambda
h}(x)|+O(h^{N+1}).$

As we have

$|\Gamma_{\lambda h}(x)|\geq|B_{\lambda h}^{N}(x)|=|B_{\lambda h}^{N}|,$

we have

$\frac{1}{|B_{\lambda h}^{N}|}\int_{B_{\lambda
h}^{N}(x)}\biggl{(}u_{y}(x^{\prime},0)-1\biggl{)}dx^{\prime}\geq Ch,$

an estimate that, combined to Lemma 5.1 and (5.1), implies our lemma. $\Box$

Proof of Theorem 1.2. The issue is now to translate Lemmas 5.1 and 5.2 into an
inequality for the Hölder quotients of $u$ at scales ranging from $h$ to much
smaller than $h$. Assume again that $u$ is translated and rotated conveniently
so that $0\in\partial\Omega$. Consider $h>0$ small. The preceding lemmas,
together with Markov’s inequality, imply that

(5.5) $\frac{\bigl{|}\\{|u_{y}-1|\geq
h^{\alpha/2}\\}\bigl{|}}{|B_{h}^{N}|}\leq Ch^{\alpha/2}.$

We cover $B_{h}^{N}$ with balls of radius $h^{1+\alpha/N}$, inequality (5.1)
guarantees that, if $B_{h^{1+\alpha/N}}^{N}(x_{i})$ is one of these balls,
then there is $\tilde{x}_{i}\in B_{h^{1+\frac{\alpha}{2N}}}^{N}(x_{i})$ such
that

(5.6) $u_{y}(\tilde{x}_{i},0)=1+O(h^{\alpha/2}).$

Consider such a $\tilde{x}_{i}$ and rescale $u$ around $\tilde{x}_{i}$:

(5.7)
${u\biggl{(}(\tilde{x}_{i},0)+h^{2}(\xi,\zeta)\biggl{)}}={h^{2}}v_{h}(\xi,\zeta),$

where we have dropped the index $i$ for commodity, all the more as the
upcoming considerations will not depend on $i$. The situation is now the
following: the function $v(\xi,0)$ is bounded, from optimal regularity, in
$B_{\alpha/N-1}^{N}$, while $D^{2}_{\xi}v(\xi,0)$ is $O(h^{2})$ uniformly in
$\xi\in B_{\alpha/N-1}^{N}$. By interpolation, $\nabla_{\xi}v(\xi,0)$ is
$O(h)$ uniformly in $\xi\in B_{\alpha/N-1}^{N}$. From Theorem 4.3, we have,
for some $M>0$:

$v_{h}(\xi,\zeta)=(M_{h}+\zeta)^{+}o_{h\to 0}(1).$

the constant $M_{h}$ being universally bounded from above and below. So, we
may write

$v_{h}(\xi,\zeta)=\biggl{(}M_{h}+\zeta+w_{h}(\xi,\zeta)\biggl{)}^{+},$

the (rescaled) free boundary $\Gamma_{h}$ being a graph of the form
$\\{\zeta=-M_{h}+\varphi_{h}(\xi)\\}$, the function $\varphi_{h}$ tending to 0
uniformly in $B_{\alpha/N-1}^{N}$. And so, everything boils down to showing
that these $o_{h\to 0}(1)$ are actually (small) powers of $h$. A first step
towards this is to prove that the quantity

$N_{h}:=\frac{w_{h}(0,0)}{h^{\frac{\alpha}{2N}}}$

is bounded, that is, $v_{h}(0,\zeta)$ is $h^{\frac{\alpha}{2N}}$-close to
$(M_{h}+\zeta)^{+}$. Assume this is not true; because $\nabla_{x}w_{h}=O(h)$
we have $v_{h}(\xi,0)\sim h^{\frac{\alpha}{2N}}N_{h}$ in
$B^{N}_{h^{\alpha/N-1}}$. As $h\to 0$ we have, up to a subsequence,

$\lim_{h\to 0}\frac{w_{h}}{h^{\frac{\alpha}{2N}}N_{h}}=\bar{w}(\xi,\zeta)$

at least in the $C^{1}$ sense, the function $\bar{w}$ being harmonic in the
strip $\mathbb{R}^{N}\times[-M,0]$ ($M$ is a limit of $(M_{h})_{h}$) with
$\bar{w}(\xi,0)\equiv 1$ and $\bar{w}(\xi,-M)\equiv 0$. Thus we have
$\bar{w}_{\zeta}(0,0)>0$, something that contradicts (5.7). So, $(N_{h})_{h}$
is bounded, so that $w_{h}$ is uniformly of order $h^{\frac{\alpha}{2N}}$ in
$B^{N}_{h^{\alpha/N-1}}$.

The second step towards an estimate of $\partial_{y}u$ is to write
$\partial_{y}w_{h}$ in integral form. Let $P_{h}(\xi,\zeta,\xi^{\prime})$ be
the Poisson kernel of the Laplacian in the positivity domain of $u$,
remembering that it decays exponentially fast, uniformly in $h$, as
$|\xi-\xi^{\prime}|$ grows (because the free boundary is bounded and bounded
away from the plane), and that its derivative $\partial_{y}P_{h}$ has a
singularity of the form $\displaystyle\frac{1}{|\xi-\xi^{\prime}|^{N+1}}$
uniformly in $h$ we have

$\partial_{y}w_{h}(\xi,0)\sim\int_{B_{h^{\alpha/N-1}}^{N}}\partial_{y}P_{h}(\xi,0,\xi^{\prime})(w_{h}(\xi^{\prime},0)-w_{h}(\xi,0)d\xi^{\prime},$

which entails that $\partial_{y}w_{h}(\xi,0)$ is of order
$h^{\frac{\alpha}{2N}}$. It is also Hölder continuous, its Hölder quotient of
order $\beta$ ($\beta<1$) being controlled by $h^{\frac{\alpha}{2N}}$. Scaling
back, this proves the Hölder continuity of $u$ with exponent
$\beta<\displaystyle\frac{\alpha}{2N}$. $\Box$

Acknowledgement. L.A. Caffarelli is supported by NSF grant DMS-1160802. The
research of J.-M. Roquejoffre has received funding from the ERC under the
European Union’s Seventh Frame work Programme (FP/2007-2013) / ERC Grant
Agreement 321186 - ReaDi. He also acknowledges a J.T. Oden fellowship for a
long term visit in 2018-19.

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