accepted 31 August 2006 Abstract We study a free boundary problem for the Laplace operator, where we impose a Bernoulli-type boundary condition

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aDepartment of Mathematics, Royal Institute of Technology, 100 44 Stockholm, Sweden

bInstitut Élie Cartan de Nancy, Université Henri Poincaré Nancy 1, B.P. 239, Vandœuvre-lés-Nancy Cedex, France Received 10 January 2005; accepted 31 August 2006

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 super-solutions 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.

c

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 Rⁿ. Consider a smooth, bounded and convex domain K such that K ⊂ {x1 = 0}. We seek a bounded domain Ω⊂Rⁿ+= {Rⁿ:x1>0}withK ⊂∂Ω, together with a functionu :Ω→Rsuch that











∆u=0 inΩ, u =1 onK, u =0 on∂Ω\K,

|∇u| =1 on(∂Ω\K)∩Rⁿ+.

Fig. 1shows a geometric illustration of this problem in the casen=2. This problem arises from various areas, for instance shape optimization, fluid dynamics, electrochemistry and electromagnetics. See for example [1,3,5]. We also see a possibility to extend the results in [2,8] to be valid in our case.

1.2. The main theorem

The main theorem of the paper is:

∗Corresponding author.

E-mail addresses:eriklin@math.kth.se(E. Lindgren),Yannick.Privat@iecn.u-nancy.fr(Y. Privat).

doi:10.1016/j.na.2006.08.045

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Fig. 1. The geometric situation inR².

Theorem 1. There is a unique solution to the free boundary problem (P) with∂Ωbeing C²⁺^αfor any0< 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. LetC be the class of smooth, bounded and convex domains inRⁿ such that K belongs to the boundary of the domain. LetΩ ∈C; we denote furthermore byu_Ωthe function fulfilling







∆u_Ω = 0,

u_Ω = 0 in∂Ω\K, u_Ω = 1 inK.

Let us introduce the following classes of domains:

A=

Ω ∈C: lim inf

y→x x∈Ω

|∇u_Ω(y)| ≥1,∀x∈∂Ω∩Rⁿ+

, A0=

(

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 )

.

Translated into terms ofAandB, the aim of this project is to prove thatA∩B 6= ∅. To do this, we use Beurling’s technique. In particular, we show that a subclass ofBhas, in some sense, a minimal element (if it is non-empty). This part of the proof relies mainly onLemma 1, the bound on|∇u|(Lemma 2). These results are proved using the same arguments as in [6–8].

In what follows we prove that the minimal element ofBbelongs toAas well. Mainly, we useLemma 4and some barrier arguments together withLemma 1.

The proof is more or less a synthesis of [6,8]. The big difference in this problem is that the free boundary and the fixed boundary do meet.

2. Preliminaries

Before we start treating the classes we need a bit of preparation. The results in this subsection are more or less already 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. LetΩ ∈C. Then the level sets of u_Ω, i.e. the setsL_ε = {x∈Ω :u_Ω > ε}, are convex.

Proof. Let

K_n= {x∈Rⁿ:dist(x,K) <n/2} Ω_n= {x∈Rⁿ:dist(x,Ω) <n}.

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Lemma 1. We denote by x1 the first coordinate inRⁿ. Let u = u_Ω withΩ ∈ 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 >0uniformly bounded by a constant M(r)in{x :dist(K,x) <r}. Then we will have, after suitable rotations and translations,

u(x)=u(x⁰)+α(x1−x₁⁰)++o(rn) for x0∈∂Ω\ {x:dist(K,x) <r},

for any sequence rn→0such that the rescalings u(rnx)/rnexist in the limit. In particular, we will have lim sup

y→x0

|∇u(y)| =lim sup

y→x0

h∇u(y), vi ≥0,

wherevis a normal vector orthogonal the tangent plane at x₀(or to one of them, if there are several).

Remark 1. We remark an immediate consequence of this theorem; letu andvbe non-negative harmonic functions inside a domainΩsuch thatu =v=0 in some neighborhood ofx∈∂Ωandu≥vinΩ. Then

lim sup

y→x

|∇u(y)| ≥lim sup

y→x

|∇v(y)|.

Indeed, applyLemma 1to the functionsu andv and use that they both attain a minimum atx. The result follows immediately.

Now we just observe that|∇u|²is a subharmonic function ifu is harmonic. To finish up the preparations we show that the gradient is almost uniformly bounded on the boundary. The proof is more or less taken from [6].

Lemma 2. LetΩ ∈ C. Then the gradient of u_Ω is uniformly bounded outside and far enough from K , i.e. for each r₀>0there is a constant M(r₀)such that:

|∇u| ≤M(r0),

for all x∈Ω\N(r0), where N(r0)= {x:dist(K,x) <r0}.

Remark 2. In [2], the authors prove this lemma in a simpler case, in the sense that the convex domain K does not belong to the boundary ofΩ.Lemma 3proves that the gradient ofu_Ωis bounded even ifK ∩Ω6= ∅.

Proof. We observe that, by barrier arguments and the use ofLemma 1, we have that away from K,∇u = 0 where

∂Ωis notC¹. So we can suppose that∂ΩisC¹away fromK.

Since|∇u|²is subharmonic insideΩit suffices thus to show that the gradient is bounded on∂(Ω\N(r0)).

Letr0>0 and letKb=K \N(r0).Fig. 2is an illustration of this proof in the casen =2. All the cases below are summarized.

(i) First case: x ∈Kb.

TakeBr₀/3andBr₀/2centered at(x−r0/3x₂). Consider nowu, the harmonic function being zero onˆ ∂Br₀/2

and one on∂Br₀/3. Then we have by the comparison principle thatu ≥

bu inside(Br₀/2\Br₀/3)∩Ω, which by Lemma 1implies

|∇u(x)| ≤ |∇

bu(x)| ≤M(r₀).

We observe that we could repeat this for every point inKbwithout changing the radius of the balls. Hence, the inequality is valid for allx∈Kb.

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Fig. 2. The picture whenn=2.

(ii) Second case: x∈∂N(r₀)∩Ω.

We use more or less the same arguments as above; let the ballsB_r₀_/₃andB_r₀_/₂be centered at the same point such thatBr₀/3⊂ {u≤u(x)}andBr₀/3∩ {u =u(x)} =x. This is possible since the level sets are byTheorem 2 convex. Letuˆdenote the harmonic function inside Br₀/2 \ Br₀/3such thatuˆ =u(x)on∂Br₀/3anduˆ =0 on

∂Br₀/2. The comparison principle andLemma 1together imply

|∇u(x)| ≤ |∇ ˆu(x)| ≤M(r₀).

Clearly, this inequality is valid for allx∈∂N(r0)∩Ωsince the function used will be the same.

(iii) Third case: x∈∂Ω\(K∪N(r₀)).

Let B_r₀_/₃and B_r₀_/₂ be centered at the same point such that B_r₀_/₃ ∩ Ω = x. We pick again a capacitary potentialuˆonB_r₀_/₂\B_r₀_/₃, such thatuˆ =0 on∂B_r₀_/₃anduˆ=1 on∂B_r₀_/₂. Then, as before, we obtain

|∇u(x)| ≤ |∇ ˆu(x)| ≤M(r0), uniformly.

The result follows.

In the sequel we will use the following:

Proposition 1. LetΩ∈Cand x∈∂Ω. Then lim inf

y→x |∇u_Ω(x)|>0.

Proof. Using exactly the same arguments as in the proof of Lemma 2 in [9] we obtain that for everyx∈∂Ωthere is avsuch that

lim inf

y→x h∇u_Ω(x), vi>0,

which immediately implies the desired result.

3. Beurling’s technique

As mentioned earlier we revisit Beurling’s technique to prove our main result. The arguments that we will use are more or less the same as the proofs of Henrot and Shahgholian in [6,8]. First, we give the steps of this technique and then, corresponding theorems.

• The classBis closed under intersection.

• Consequence: we consider a decreasing sequence of convex domains inB. Then, the interior of the closure belongs toB.

• We use this argument in order to construct a candidate for being the solution of our problem.

• Then, we check that this candidate is the solution of the free boundary problem.

We now give the theorems used.

Theorem 3. LetΩ₁andΩ₂be inB. ThenΩ₁∩Ω₂∈B.

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ThenΩ ∈B.

Proposition 2.Assume that there exist two domainsΩ₀ ∈A₀andΩ₁∈ B. We denote bySthe class of domains D such thatΩ₀⊂D and D∈B. Then there exists a minimal element in the classSfor the inclusion.

Proof. Although this result is already proved in [6], we recall it in order to clarify the construction.

Denote byIthe intersection of all domains in the classS, and let Ω=

◦

I.

We claim thatΩis the minimal elements inS. It suffices to show thatΩ∈B. We observe that we can find a sequence of domains(U_n)n∈N∈Ssuch that

\

n≥0

U_n=I.

Consider now the sequence(Ω_n)n∈Ndefined byΩ₁=D1and Ω_n+1=Ω_n∩U_n+1 for alln∈N.

ByTheorem 3, eachΩ_nbelongs toBand byTheorem 4, Ω:=

◦

\

n≥0

Ω_n

belongs toB. Whence the result.

Further on, we will denote byΩthe minimal element inS. Moreover, a pointx∈∂Ωis said to be extremal if there is a supporting lineLtoΩsuch thatL∩Ω =x. We denote the set of extremal points on∂Ω\K byE_Ω. We will now prove that theu_Ω will be inAas well, but first we need some lemmas, and again, the proofs are the same as in [6].

Lemma 3. Let x∈ E_Ω ∩Rⁿ+. Then lim sup

y→x

|∇u_Ω(y)| =1. This result is already proved in [6].

Lemma 4. LetΩbe a convex domain, and let u=u_Ω. Then for x∈∂L_t, lim sup

z→x

|∇u(z)| ≥lim sup

z→y

|∇u(z)|, where y∈∂Ωis the point fulfilling

x₂(y)=max

z∈∂Ωx₁(z),

where x₁is the direction perpendicular to the tangent plane (or one of them) of ∂L_t at x₀.

Proof. This is more or less the same argument as in [6]. We adapt and recall the proof. Suppose that∂L_tis notC¹at x. Then by barrier arguments andLemma 1,|∇u(x)| = ∞.

Suppose that∂ΩisC¹aty, and denote byx1the perpendicular direction to∂L_t, such thatx1(x)=0.

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Let v=u+

lim sup

z→y

|∇u(z)| −ε

x1.

Thenvis harmonic inΩ∩ {x1>0}and by the maximum principle,vattains its maximum either atxor aty. Suppose thatvattains its maximum aty. ThenLemma 1would imply

lim sup

z→y

∂v

∂n(z)≥0, and then lim sup

z→y

∂u

∂n(z)+

lim sup

z→y

|∇u(z)| −ε

hx1,ni ≥0

wherenis the outward pointing normal vector aty. As we have seen,∇uis perpendicular to the boundary, and thus lim sup

z→y

|∇u(z)| = −lim sup

z→y

∂u

∂n(z).

−ε≥0. Hence,vattains its maximum atx, and by [7, Lemma 2.7] we must have lim sup

z→x

∂v

∂x₁(z)≤0, i.e. lim sup

z→x

∂u

∂x₁(z)+lim sup

z→y

|∇u(z)| −ε≤0. Moreover,

lim sup

z→x

|∇u(z)| = −lim sup

z→x

∂u

∂x₁(z).

Thus,

−lim sup

z→x

|∇u(z)| +lim sup

z→y

|∇u(z)| −ε≤0, which whenε→0 becomes

lim sup

z→x

|∇u(z)| ≥lim sup

z→y

|∇u(z)|.

Theorem 5. Assume that there exist two domainsΩ₀ ∈ A₀and Ω₁ ∈ B. Then there exists a solution of the free boundary problem(P)in a strong sense.

Proof. The first step of the proof will be to show that|∇u_Ω(x)| ≥1 for allx ∈∂Ω∩Rⁿ+. ByTheorem 2the level sets L_t := {x : u_Ω(x) >t}, with 0<t <1, are convex. As a consequence, we can useLemma 4. The rest of the proof is given by geometric arguments. We denote byT the hyperplane tangent toL_t atx. By the property of the level sets, there exists at least one point yinE_Ω such thatyis at the largest distance toxin a direction orthogonal toT. Then, Lemma 1implies

|∇u_Ω(x)| ≥lim sup

z→y z∈Ω

|∇u_Ω(z)| =1.

The theorem follows.

4. Proof ofTheorem 1

To prove the existence it is byTheorem 5sufficient to prove that the classesA₀andBare non-empty.

Proposition 3. The classesA₀andBare non-empty.

Proof. LetΩ∈C. ByLemma 4, there is a neighborhoodU ofK such that|∇u|> δinsideU\K. Letεbe so small that{u≥1−ε} ⊂U, and let

w=u−(1−ε)

ε .

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δ

for someC > 0, on every ball B(x,r)outside N(r0), since pickingδ0small enough we will haveΩ_δ ⊂U for all δ < δ0. Pickingδeven smaller, we will finally obtain that|∇v_δ|>1 on every compact subset ofΩ_δ. Since|∇v_δ|is subharmonic this impliesΩ_δ ∈A₀forδsmall enough.

In order to construct an element inB, letu_R be the capacitary potential of the ballsB_R/B₁centered at the origin.

That is, except forn=2,u_Ris given explicitly by uR= |x|²⁻ⁿ−R²⁻ⁿ

R₀²⁻ⁿ−R²⁻ⁿ .

Moreover, denote byuthe solution to the Dirichlet problem withΩbeing the upper half-disc ofB_R. By the maximum principle,u_R ≥uinΩ, and byLemma 1

|∇uR| ≥ |∇u|,

on∂BRT{u>0}. Thus, forRbig enough we will have|∇u| ≤1. Hence,B₀is non-empty.

To prove the regularity we argue by contradiction. Suppose that∂Ω\K is notC¹atx. Then, sinceΩis convex,

∂Ωhas a corner atx. But then, by barrier arguments andLemma 1we know that lim sup

y→x

|∇(y)| =0,

which is a contradiction. Hence,∂Ω∩Rⁿ+must beC¹. Hence, we have that the free boundary (∂Ω\K) meets the fixed boundary (K) tangentially if the set(∂Ω\K)∩ {x₁=0}is non-empty. But as we have constructed the element inA₀ with(∂Ω\K)∩ {x₁ =0} 6= ∅this is clearly true for the minimal element in the classS as well. Now, applying [4, Thm. 1.4] we obtain that∂ΩisC²⁺^αfor any 0< α <1.

The uniqueness follows from the following arguments. Letu1,Ω₁andu2,Ω₂be two different solutions to (P), such thatΩ₂\Ω₁6= ∅. Thenut =u2(t x)is a solution in

Ω_t = {x:t x∈Ω}. Moreover, we haveu_t =0 and

lim sup

y→x

|∇u_t(y)| =t

on∂Ω_t\Kt. Now pickt >0 such that there is at least onex ∈∂Ω_t\Kt ∩∂Ω₁\K and such thatΩ_t ⊂Ω. By the maximum principle we have thatu1≥ut inΩ_t. UsingLemma 1we obtain

1=lim sup

y→x

|∇u1(y)| ≥lim sup

y→x

|∇u_t(y)| =t >1, which is a contradiction. Whence the uniqueness.

5. Generalizations and comments

5.1. Investigation of a limit in the case n=2

Obviously, the result is valid if we replace the Bernoulli-type boundary condition with|∇u| =λ >0 as well. Now we will determine the limit of the functions corresponding toλasλ→0.

Let λn be a decreasing sequence such that λ0 = 1 and λn → 0. Let un denote the sequence of functions corresponding to the solution with the free boundary condition|∇un| =λn, and letΩ_nbe the corresponding domains.

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Obviously, being a sequence of bounded functions,u_nhas a subsequential limit. We first observe that asn → ∞,Ω_n becomes unbounded in every direction.

Indeed, suppose thatΩ = limΩ_n is bounded in some direction. Then we can construct a domain D which is bounded in the same direction such thatΩ⊂Dproperly. ByLemma 1,|∇u_D|> δinD\K for someδ >0. Now, let nbe big enough such thatλn < δ. Then, by takingD∈A₀(λn)we can construct the minimal element inB(λn)as in Proposition 2. By the uniqueness the constructed set is nothing else butΩ_n. Hence,Ω_n⊂D, which is a contradiction.

Hence,Ω_nbecomes unbounded in every direction asn → ∞, which implies that limΩ_n=R²+. We now wish to find the limit function, which will certainly fulfill

u =1 inK,

u =0 in{x2=0} \K.

We claim that the functionu(z)=arg(z+1)−arg(z−1)is the limit ofu_n. Indeed,ufulfills the boundary conditions on{x₂ =0}, and it is harmonic. So, by the properties of harmonic functions in the upper half-plane,uis the unique function having the properties above.

5.2. The p-Laplacian

In the case of the arbitraryp-Laplacian there are some small adjustments needed; we will have to replace the fact that|∇u|²is subharmonic by [6, Lemma 2.1] and use that the therein defined operatorL_uadmits a maximum principle and thatL_u|∇u|^p≥0. We also have to use that also the p-Laplacian admits a maximum and a comparison principle, but in this case this will not make any difference.

Another difference is that theC^∞-convergence inside compact subsets has to be replaced byC¹^,β-convergence.

In the construction of an element inA₀we use Cauchy’s estimate which does not hold for the arbitraryp-Laplacian.

However, this does not cause any difficulties since what we need is justC¹-convergence inside compact subsets ofΩ_δ which we have in the case of the p-Laplacian as well.

Some slight modifications are also needed in the construction of an element inB. Here we must use the general expression for then-dimensional radial symmetricp-capacitary potential, i.e.











|x|

p−n

p−1 −R

p−n p−1

R

p−n p−1

0 −R

p−n p−1

for p6=n, log|x| −logR

logR₀−logR for p=n.

We also remark that theC²⁺^α-regularity will not be valid in the general case since the theorem used only applies to elliptic operators. However, theC¹-regularity will still hold.

5.3. The case of unbounded and irregular K

While the irregularity causes some small problems in [6] concerning the lemma corresponding to Lemma 1, irregularities on ∂K would in our case not cause any trouble at all, since we stay away from ∂K in our estimate of|∇u|.

The case of an unbounded K can be handled in the same way as in [6], and similarly, the uniqueness will not be true in general.

5.4. Usefulness

As mentioned in the introduction, the result in this paper could probably be extended to results similar to those in [8,2] for our case. This together with results similar to those in [7] might yield a proof of the existence of solutions to the following problem:







∆_pu = f inΩ, u=1 onK, u=0 on∂Ω\K,

for certain class of functions f. HereΩ⊂Rⁿ+andK ⊂∂Ω.

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