A free boundary problem for the Laplacian with a constant Bernoulli-type boundary condition

TL;DR: In this article, a free boundary problem for the Laplace operator is studied, where a Bernoulli-type boundary condition is imposed, and a solution to this problem is shown.

Abstract: We study a free boundary problem for the Laplace operator, where we impose a Bernoulli-type boundary condition. We show that there exists a solution to this problem. We use A. Beurling's technique, by defining two classes of sub- and supersolutions and a Perron argument. We try to generalize here a previous work of A. Henrot and H. Shahgholian. We extend these results in different directions.

Summary

1. Introduction

• This problem arises from various areas, for instance shape-optimization, fluid dy- namics, electrochemistry and electromagnetics.
• In particular, the free boundary (∂Ω \K) meets the fixed boundary (K) tangentially.
• Moreover, the solution has convex level sets.

2. Preliminaries

• The results in this subsection are more or less already known, but not proved in detail for this particular case.
• To finish up the preparations the authors show that the gradient is almost uniformly bounded on the boundary.
• Ω since the function used will be the same.

3. Beurling’s technique

• As mentioned earlier the authors revisit Beurling’s technique to prove their main result.
• The arguments that the authors will use are more or less the same as the proofs of Henrot and Shahgholian in [6] and [8].
• First, the authors give the steps of this technique and then, corresponding theorems.
• The class B is closed under intersection.
• Then there exists a solution of the free boundary problem (P) in a strong sense.

5. Generalizations and comments

• Let un denote the sequence of functions corresponding to the solution with the free boundary condition |∇un| = λn, and let Ωn be the corresponding domains.
• In the case of the arbritary p-Laplacian there are some small adjustments needed; the authors will have to replace the fact that |∇u|2 is subharmonic by [6, Lemma 2.1] and use that the therein defined operator Lu admits a maximum principle and that Lu|∇u|p ≥.
• Here the authors must use the general expression for the n-dimensional radial symmetric p-capacitary potential, i.e.
• The authors also remark that the C2+α-regularity will not be valid in the general case since the theorem used only applies to elliptic operators.
• As mentioned in the introduction, the result in this paper could probably be extended to results similar to those in [8] and [2] for their case.

Figures

Figure 2. The picture when n = 2.

Figure 1. The geometric situation in R2.

Full text

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A free boundary problem for the Laplacian with
constant Bernoulli-type boundary condition
Erik Lindgren, Yannick Privat
To cite this version:
Erik Lindgren, Yannick Privat. A free boundary problem for the Laplacian with constant Bernoulli-
type boundary condition. Nonlinear Analysis: Theory, Methods and Applications, Elsevier, 2007, 67
(8), pp.2497-2505. �10.1016/j.na.2006.08.045�. �hal-00128760�

A FREE BOUNDARY PROBLEM FOR THE LAPLACIAN WITH
CONSTANT BERNOULLI-TYPE BOUNDARY CONDITION
ERIK LINDGREN AND YANNICK PRIVAT
Abstract. We study a free boundary problem f or the Laplace operator,
where we impose a Bernoulli-type boundary condition. We show that there
exists a solution to this problem. We use A. Beurling’s technique, by defin-
ing two classes of sub- and s upersolutions and a Perron argument. We try to
generalize here a previous work of A. Henrot and H. Shahgholian. We extend
these results in different directions.
1. Introduction
1.1. The problem. The aim of this paper is to prove the existence and uniqueness
of a Bernoulli-type free boundary problem in R
n
. Consider a smooth, bounded
and convex domain K such that K ∩ {x
1
= 0}. We seek a bounded domain
Ω ⊂ R
n
+
= {R
n
: x
1
> 0} with K ⊂ ∂Ω, together with a function u : Ω → R such
that







∆u = 0 in Ω ,
u = 1 on K ,
u = 0 on ∂Ω \ K ,
|∇u| = 1 on (∂Ω \ K) ∩ R
n
+
.
This problem arises from various areas, for ins tance shape-o ptimization, fluid dy-
u = 1
u = 0u = 0
|∇u| = 1
u = 0
Ω
∆u = 0
K
Figure 1. The geometric situation in R
2
.
namics, electrochemistry and electromagnetics. See for example [1], [3] and [5]. We
also see a possibility to extend the results in [2] and [8] to be valid in our case.
Date: February 2, 2007.
1

2 ERIK LINDGREN AND YANNICK PRIVAT
1.2. The main theorem. The main theorem of the paper is:
Theorem 1. There is a unique solution to the free boundary problem (P) with ∂Ω
being C
2+α
for any 0 < α < 1. In particular, the free boundary (∂Ω \ K) meets the
fixed boundary (K) tangentially. Moreover, the solution has convex level sets.
1.3. Outline of the proof. The method used is as follows. Let C be the class of
smooth, bounded and convex domains in R
n
such that K belongs to the boundary
of the domain. Let Ω ∈ C, we denote furthermore by u
Ω
the function fulfilling



∆u
Ω
= 0 ,
u
Ω
= 0 in ∂Ω \ K ,
u
Ω
= 1 in K .
Let us introduce the following classes of domains
A =

Ω ∈ C : lim inf
y→x x∈ Ω
|∇u
Ω
(y)| ≥ 1, ∀x ∈ ∂Ω ∩ R
n
+

,
A
0
=

Ω ∈ C : lim sup
y→x x∈ Ω
|∇u
Ω
(y)| > 1, ∀x ∈ ∂Ω \ K

,
B =

Ω ∈ C : lim sup
y→x x∈ Ω
|∇u
Ω
(y)| ≤ 1, ∀x ∈ ∂Ω \ K

.
Transla ted into terms of A and B, the aim of this project is to prove that A∩B 6= ∅.
To do this, we use Beur ling’s technique. In particular, we show that a subclass of
B has, in some sense, a minimal element (if it is non-empty). This part of the proof
relies mainly on Lemma 1, the bound on |∇u| (Lemma 2). These results are proved
using the same arguments as in [6], [7] and [8].
In what follows we prove that the minimal element of B belongs to A as well.
Mainly, we use Lemma 4 and some barrier arguments together with Lemma 1.
The proof is more or less a synthesis of [6] and [8]. The big difference in this problem
is that the free boundary and the fixed boundary do meet.

A fbp for the Laplacian with Bernoulli-type boundary condition 3
2. Preliminaries
Before we start tre ating the classes we need a bit of preparations. The results in
this subsection are more or le ss alr eady known, but not proved in detail for this
particular case.
The first thing we prove is that the level sets of a Dirichlet solution are convex.
Theorem 2. Let Ω ∈ C. Then the level sets of u
Ω
, i.e. the sets L
ε
= {x ∈ Ω :
u
Ω
> ε}, are convex.
Proof. Let
K
n
= {x ∈ Ω : dist(x, K) < n/2}
Ω
n
= {x ∈ Ω : dist(x, Ω) < n} .
Then D
n
= Ω
n
\ K
n
is a convex ring. Therefore, u
n
= u
D
n
has convex level sets
(cf [9]). By standard arguments, u
n
converges to a harmonic function in every
compact subset of Ω. Furthermore we will have C
0,α
-convergence in R
n
. Clearly,
u = 1 on I and u = 0 on R
n
\ (Ω ∪ K). Hence, u
n
→ u
Ω
. Moreover, L
ε
= ∩L
n
ε
,
where L
n
ε
= {x ∈ Ω : u
n
> ε}, which implies that the sets L
ε
are convex. 
Now we prove a fundamental (but not trivial) lemma (based on the same idea than
in [8]), which has some very important consequences when comparing the gradient
on the boundary.
Lemma 1. We denote by x
1
the first coordinate in R
n
. Let u = u
Ω
with Ω ∈ C,
such that u is Lipschitz on the boundary. Suppose that the gradient of u exists on
the boundary and that it is for every r > 0 uniformly bounded by a constant M (r)
in {x : dist(K, x) < r}. Then we will have, after suitable rotations and translations,
u(x) = u(x
0
) + α(x
1
− x
0
1
)
+
+ o(r
n
) for x
0
∈ ∂Ω \ {x : dist(K, x) < r},
for some sequences r
n
. In particular, we will have
lim sup
y→x
0
|∇u(y)| = lim sup
y→x
0
h∇u(y), vi ≥ 0 ,
where v is a normal vector orthogonal the t angent plane at x
0
(or to one of them,
if there are several).
Remark 1. We remark an immediate consequence of this theorem; let u and v be
non-negative harmonic functions inside a domain Ω such that u = v = 0 in some
neighborhood of x ∈ ∂Ω and u ≥ v in Ω. Then
lim sup
y→x
|∇u(y)| ≥ lim sup
y→x
|∇v(y)| .
Indeed, apply Lemma 1 to the functions u and v and use that they both attain a
minimum at x. The result follows immediately.
Now we just o bserve that |∇u|
2
is a subharmonic function if u is harmonic. To
finish up the prepar ations we show that the gradient is almost uniformly bounded
on the boundary. The proof is more or less taken from [6].
Lemma 2. Let Ω ∈ C. Then the gradient of u
Ω
is uniformly bounded outside and
far enough from K, i.e. for each r
0
> 0 there is a constant M (r
0
) such that:
|∇u| ≤ M(r
0
) ,
for all x ∈ Ω \ N (r
0
), where N (r
0
) = {x : dist(K, x) < r
0
}.
Remark 2. In [2], the authors prove this lemma in a more simple case, in the
sense that the convex domain K doesn’t belong to the boundary of Ω. Lemma 3
prove that the gradient of u
Ω
is bounded even if K ∩
Ω 6= ∅.

4 ERIK LINDGREN AND YANNICK PRIVAT
Proof. We observe that, by barrier a rguments and the use of Lemma 1 we have
that away from K, ∇u = 0 where ∂Ω is not C
1
. So we can suppose that ∂Ω is C
1
away from K.
Since |∇u|
2
is subharmonic inside Ω it suffices thus to show that the gradient is
bounded on ∂(Ω \ N(r
0
)).
Let r
0
> 0 and let
b
K = K \ N (r
0
).
(i) First Case: x ∈
b
K.
Take B
r
0
/3
and B
r
0
/2
be c e ntered at (x − r
0
/3x
2
). Consider now ˆu, the
capacitary potential on B
r
0
/2
\ B
r
0
/3
, i.e. the harmonic function being zero
on ∂B
r
0
/2
and one o n ∂B
r
0
/3
. Then we have by the comparison principle
that u ≥ bu inside (B
r
0
/2
\ B
r
0
/3
) ∩ Ω, which by Lemma 1 implies
|∇u(x)| ≤ |∇bu(x)| ≤ M (r
0
) .
We observe that we could repeat this for every point in
b
K without changing
the radius of the balls. Hence, the inequality is valid for all x ∈
b
K.
(ii) Second Case: x ∈ ∂N (r
0
) ∩ Ω.
We use more or less the same arguments as above; let the balls B
r
0
/3
and B
r
0
/2
be centered at the same point such that B
r
0
/3
⊂ {u ≤ u(x)} and
B
r
0
/3
∩{u = u(x)} = x. This is possible since the level sets are by Theorem
2 convex. Le t ˆu denote the harmonic function inside B
r
0
/2
\ B
r
0
/3
such
that ˆu = u(x) on ∂B
r
0
/3
and ˆu = 0 on ∂B
r
0
/2
. The comparison principle
and Lemma 1 together imply
|∇u(x)| ≤ |∇ˆu(x)| ≤ M(r
0
) .
Clearly, this inequality is valid for all x ∈ ∂N(r
0
) ∩ Ω since the function
used will be the same.
x ∈
b
K
x ∈ ∂N (r
0
) ∩ Ω
x ∈ ∂Ω \ (K ∪ N (r
0
))
N (r
0
)
{u = u(x)}
Figure 2. The picture when n = 2.
(iii) Third Case: x ∈ ∂Ω \ (K ∪ N(r
0
)).
Let B
r
0
/3
and B
r
0
/2
be centered at the same point such that B
r
0
/3
∩Ω = x.
We pick again a capa c itary potential ˆu o n B
r
0
/2
\ B
r
0
/3
, such that ˆu = 0
on ∂B
r
0
/3
and ˆu = 1 on ∂B
r
0
/2
. Then, as before, we obtain
|∇u(x)| ≤ |∇ˆu(x)| ≤ M(r
0
) ,
uniformly.
The result follows. 