On the interior Bernoulli free boundary problem for the fractional Laplacian on an interval

We study the structure of solutions of the interior Bernoulli free boundary problem for $(-\Delta)^{\alpha/2}$ on an interval $D$ with parameter $\lambda>0$. In particular, we show that there exists a constant $\lambda_{\alpha,D}>0$ (called the Bernoulli constant) such that the problem has no solution for $\lambda \in (0,\lambda_{\alpha,D})$, at least one solution for $\lambda = \lambda_{\alpha,D}$ and at least two solutions for $\lambda>\lambda_{\alpha,D}$. We also study the interior Bernoulli problem for the fractional Laplacian for an interval with one free boundary point. We discuss the connection of the Bernoulli problem with the corresponding variational problem and present some conjectures. In particular, we show for $\alpha = 1$ that there exist solutions of the interior Bernoulli free boundary problem for $(-\Delta)^{\alpha/2}$ on an interval which are not minimizers of the corresponding variational problem.


Introduction
The interior Bernoulli free boundary problem for the fractional Laplacian is formulated as follows.Here (−∆) α/2 denotes the fractional Laplacian given by denotes the generalized normal derivative given by where n(x) is the outward unit normal vector to K at x.As usual, by a domain we understand a nonempty, connected open set, by a domain of class C 1 we understand a domain, which boundary is locally a graph of a C 1 function.
When α = 2, that is if (−∆) α/2 is replaced by (−∆) and D α/2 n is replaced by the normal derivative Problem 1 is just the classical interior Bernoulli problem, which have been intensively studied, see e.g.[2,1,18,19,20,10,3].It arises in various nonlinear flow laws and several physical situation e.g.electrochemical machining and potential flow in fluid mechanics.
In the classical case (α = 2) it is well known that Problem 1 does not have a solution for any positive level λ.For example, when D is convex, it is proved in [20] that there is some positive constant λ D such that this problem has a solution for level λ if and only if λ ≥ λ D .This constant λ D is called a Bernoulli constant.It is also known that even if there are solutions to the problem for some λ there is no uniqueness in general.For example, if D is a ball in R d (d ≥ 2) there are exactly 2 solutions to the problem for any λ > λ D , while for λ = λ D the solution is unique (see e.g.[18,Section 3]).A very interesting and open question is whether for general convex bounded domains D the structure of the solutions to the Bernoulli problem enjoys similar features.See [10] for some discussion on this question in the classical case.
The main aim of this paper is to study the structure of solutions of Problem 1 in the simplest geometric case i.e. when D is an interval.The main results of our paper are the following theorems.
Bernoulli problems for fractional Laplacians have been investigated for the first time by Caffarelli, Roquejoffre and Sire in [8].Such problems are relevant in classical physical models in mediums where long range interactions are present.Bernoulli problems for (−∆) α/2 have been intensively studied in recent years see e.g.[12,14,13,15,16,17].In these papers mainly regularity of free boundary of solutions of the related variational problem was studied.This variational problem is formulated in Section 5.
One dimensional Bernoulli problems for the fractional Laplacian for α = 1 are related to some reaction-diffusion equations, which can model the combustion of an oil slick on the ground with the temperature diffucing above ground (see [7]) Problem 1 for α = 1 have been already studied in [21].In that paper (in the part devoted to the inner Bernoulli problem) only existence of the related variational problem was studied, without investigating the number of solutions.In our paper we study all solutions to Problem 1 and not only solutions to the variational problem.The relation between Problem 1 and the variational problem is discussed in Section 5.
The main difficulty in proving Theorem 1.1 is caused by nonlocality of the fractional Laplacian.From technical point of view this is manifested by the fact that there is no explicit formula of the Poisson kernel corresponding to (−∆) α/2 for an open set, which is the sum of 2 disjoint intervals, although the explicit formula of the Possion kernel corresponding to (−∆) α/2 for an interval is known.The nonlocal Problem 1 for an interval can be transformed to the local one in one dimension higher (see e.g.[9]) but this does not seem to allow to find the explicit formula for the Poisson kernel corresponding to (−∆) α/2 for the sum of 2 disjoint intervals.
In our paper we study also a simplified version of Problem 1.This is the interior Bernoulli problem for the fractional Laplacian for an interval with one free boundary point.It is formulated as follows.
Clearly, the point a is the unique free boundary point for this solution.
The main result concerning Problem 2 is the following theorem.
For α = 1 we are able to obtain explicit formulas for µ α,D and solutions of Problem 2.
we have 1 solution of Problem 2 given by   For λ > µ 1,D we have 2 solutions (K 1 , u 1 ) and (K 2 , u 2 ) of Problem 2 given by 1 shows graphs of solutions given in Theorem 1.4.Note, that although there is an extensive literature devoted to free boundary problems for fractional Laplacians, explicit solutions of these problems are quite rare.
The paper is organized as follows.In Section 2 we collect well known facts which will be used in the sequel.In Section 3 we study the interior Bernoulli problem for the fractional Laplacian for an interval with one free boundary point.Section 4 contains proofs of main results of our paper.In Section 5 we present some conjectures and discuss the connection of the Bernoulli problem with the corresponding variational problem.

Preliminaries
In this section we present notation and gather some well known facts, which we need in the paper.In particular, we introduce Poisson kernel and Green function corresponding to the fractional Laplacian (−∆) α/2 (for α ∈ (0, 2)).We concentrate only on one-dimensional case and only on facts which are needed in this paper.For the detailed exposition of the potential theory corresponding to the fractional Laplacian we refer the reader to [5] or [11].
We denote Let G be a class of nonempty open sets G in R, G = R which has the representation where n ∈ N and all G k are intervals (bounded or unbounded) and dist(G i , G j ) > 0 for any i, j ∈ {1, . . ., n}, i = j.The class G plays a role of smooth sets in R. Let D ∈ G be bounded and let f : D c → R be bounded and measurable.Let us consider the following outer Dirichlet problem for the fractional Laplacian (−∆) α/2 .We look for a bounded measurable function u : R → R, continuous on D, satisfying Such Dirichlet problem has a unique solution, which is given by the formula (see e.g.[6,Lemma 13]).When f is continuous on D c , it is well known that u is continuous on R.
On the other hand, if measurable, bounded function u : R → R, continuous on D satisfies (−∆) α/2 u(x) = 0, for x ∈ D, then the following mean value property is satisfied (see e.g.[6, (61), (62)]).For any bounded B ⊂ D, B ∈ G and any x ∈ B we have There are known explicit formulas of the Poisson kernels for intervals.For any a, b ∈ R, a < b we have (see e.g.[5, (1.57)]) where C α is defined as in Theorem 1.2.
For any D ∈ G we denote by G D the Green function for D corresponding to (−∆) α/2 .It has the following properties: The function h D,y satisfies the following mean value property (see e.g.[5, (1.46)]).
For any fixed y ∈ D, any bounded For any B ⊂ D, B, D ∈ G we have As in the classical case, there is known representation of the Poisson kernel in terms of the Green function.For any bounded D ∈ G we have (see e.g.[5, (1.49)]) where This is the same constant, which appear in the definition of (−∆) α/2 for dimension d = 1.
In the paper we need some well known facts concerning hypergeometric functions.For the convenience of the reader, we briefly present them below.Let |z| < 1, p, q, r ∈ R and −r / ∈ N. The (Gaussian) hypergeometric function is defined as where (•) n is a Pochhammer symbol.For p ∈ R, r > q > 0 and |z| < 1 we have where B(•, •) is the beta function.Note, that we can rewrite the above as We will also need the following easy result concerning beta function.

Bernoulli problem with one free boundary point
This section is devoted to the study of the interior Bernoulli problem for the fractional Laplacian for an interval with one free boundary point.
For a ∈ (0, 1) let us define Using ( 10) and ( 11), we obtain . For convenience, throughout the rest of this section we denote Recall that in the whole paper we use notation T α = B(α, 1 − α 2 ).Using the above formulas we obtain Now, fix a ∈ (0, 1) and put u a = 1 − w a , K a = (a, 1).Let n(a) be the outward unit normal vector to K a at a. Then we have It follows that for any fixed a ∈ (0, 1) the function u a and the set K a is a solution of Problem 2 for D = (0, 1) and λ = R(a).Now we will study properties of the function (0, 1) a → R(a).These properties will allow to justify Theorem 1.3.Proof.By differentiating (12) twice, we get Recall that Hence Using this and Lemma 2.1, we finally obtain Proposition 3.2.We have The limit for a → 0 + immediately follows from (12), continuity of F α (a) in a = 1 and the fact that F α (0) = 1.By the inequality (α/2 + 1) n > (1) n = n!, we get From this inequality we obtain Proof.For x ∈ K s = sK we have x/s ∈ K, so u s (x) = u(x/s) = 1.Similarly, for x ∈ (D s ) c = sD c we have x/s ∈ D c , so u s (x) = u(x/s) = 0. We also have Note that for x ∈ D s \ K s = sW we have x/s ∈ W .Using this, [5, (1.62)] and ( 13), we obtain for By substitution y = sz, this is equal to P sW (x, y)u s (y) dy.
By the definition of the fractional Laplacian one easily obtains the following result.Proof of Theorem 1.4.We first show the assertion for D = (0, 1).We assume that a ∈ (0, 1).Recall that we denote u a = 1−w a .By (10) and (11), we get for x ∈ (0, a) By (12), we obtain ).Note that we have Hence we obtain We have d da Thus, the minimum of the function (0, 1) a → −D 1/2 n u a (a) is obtained for a = 1/2 and it is equal to 4/π.Therefore, For λ = µ 1,(0,1) the unique solution is given by K = (1/2, 1) and u = u 1/2 (given by ( 14)).For any λ > µ 1,(0,1) we have −D The above can be reduced to the following quadratic equation: 1,(0,1) 4λ 2 = 0. Since we have chosen λ > µ 1,(0,1) , it has exactly two solutions, given by This implies that Problem 2 has exactly two solutions.The first solution is K = (a 1 , 1) and u = u a 1 .The second solution is K = (a 2 , 1) and u = u a 2 .Functions u a 1 , u a 2 are given by ( 14).This gives the assertion of the theorem for D = (0, 1).The assertion for arbitrary D follows from Lemmas 3.3 and 3.4.

Proofs of main results
In this section we present proofs of Theorems 1.1 and 1.2.
Proposition 4.1.Let x 0 ∈ R, r > 0, D = (x 0 − r, x 0 + r) and λ > 0. Assume that (u, K) is a solution of Problem 1 for D and λ.Then K is symmetric with respect to x 0 .
Before we prove this proposition, we need some estimates of the Green function corresponding to the fractional Laplacian.Proof.By ( 5) and ( 4), for any x, y ∈ (b, w) we have , where c depends on α, b, w.
Lemma 4.6.The function (0, 1) a → f a (x) is continuous for any fixed x ∈ R.
Proof.By formulas ( 19), ( 21) and induction for any fixed n ∈ N and x ∈ R, the function (0, 1) a → f Using this and induction, for any fixed ε > 0 and arbitrary n ∈ N we obtain By this and continuity of (0, 1) a → f (n) a (x), we get the assertion of the lemma.For a ∈ (0, 1) put The next two lemmas concern some properties of the function Ψ. (24) For any a ∈ (0, 1) we have Ψ(a) ∈ (0, ∞) and Ψ is continuous on (0, 1).
On the other hand, by Lemma 4.7, for any a ∈ (0, 1) the function f a is a solution of Problem 1 for D and λ = Ψ(a).Now, using Lemma 4.8 and the fact that the function (0, 1) a → Ψ(a) is continuous and positive on (0, 1), we obtain the assertion of the theorem.

Discussion
In this section we discuss properties of solutions of the inner Bernoulli problem for the fractional Laplacian on intervals and balls.We also discuss the connection of the Bernoulli problem with the corresponding variational problem.
Let us start with the following conjecture.A standard way to show such result is to study properties of the function Ψ (which is defined by (22)).In Section 4 it is shown that (0, 1) a → Ψ(a) is continuous and positive and lim a→0 + Ψ(a) = lim a→1 − Ψ(a) = ∞.So to justify Conjecture 5.1 it is enough to prove that for each α ∈ (0, 2) functions (0, 1) a → Ψ(a) are unimodal.Graphs of Ψ for α = 1 and α = 1/2 (see Figure 2) suggest that these functions have this property.It seems that it is possible to justify Conjecture 5.1 using a computer assisted proof.Below we present some rough idea of such a proof.Of course, this is only the idea, we are not claiming in any way that this is a formal proof.
By (23), we have Φ(a, −y)g a (y) dy , (28) where g a (y) = 1 − f a (y).Fix α ∈ (0, 2).It is easy to show that d da Ψ(a), d 2 da 2 Ψ(a) are well defined for any a ∈ (0, 1).Note also that for any y ∈ (0, 1) the function (0, y) a → g a (y) is decreasing.Using this and (28), one can obtain that there exists a 0 (depending on α) such that d da Ψ(a) < 0 for all a ∈ (0, a 0 ].Now it is enough to show that Ψ is convex on [a 0 , 1).Note that (cf.21) for a ∈ (0, 1), x ∈ (a, 1) we have where for n ∈ N, n ≥ 2. For some n 0 ∈ N denote Ψ = Ψ (1) + Ψ (2) , where a (y) dy , Now, the strategy of the proof could be the following.One could obtain that for sufficiently large k ∈ N a (y) dy is sufficiently small for a ∈ [a 0 , 1).Then one should be able to show that there exist ε > 0 and n 0 ∈ N so that d 2 da 2 Ψ (1) (a) > ε for a ∈ [a 0 , 1) and d 2 da 2 Ψ (2) (a) ≤ ε.Some numerical experiments suggest that this is possible.This would show that (0, 1) a → Ψ(a) is unimodal.Of course, the above way of proving Conjecture 5.1 demands numerical estimates in many steps and it is beyond the scope of this paper.
Well known results for the classical inner Bernoulli problem for balls (see e.g. Figure 3 in [18]) suggest that the following hypothesis holds.
It seems, however, that this result is much more difficult to prove than Conjecture 5.1.Even the proof of result similar to Theorem 1.1 for balls seems quite challenging.
One of the standard ways to study Bernoulli problems (in the classical or fractional case) is to investigate appropriate variational problems (see e.g.[2,8,17]).Fix d ≥ 1, α ∈ (0, 2) and a bounded domain D ⊂ R d .Let us define the energy functional on the Sobolev space depending on the parameter λ > 0, where and |{x ∈ D : u(x) < 1}| denotes the Lebesgue measure of {x ∈ D : u(x) < 1}.A similar definition appears in Section 1.1 in [17].
Let us consider the following problem of finding minimizers to this energy functional.This problem is connected with the inner Bernoulli problem for the fractional Laplacian.
Problem 3. Given α ∈ (0, 2), λ > 0 and a bounded domain For α = 1 this variational problem was studied in Section 3 in [21].In that paper the problem was investigated for general bounded domains.For the particular case when D is an interval [21] implies the following result.Proposition 5.3.Let x 0 ∈ R, r > 0 and D = (x 0 −r, x 0 +r).There exists a constant Λ 1,D such that for any λ ≥ Λ 1,D there exists a solution u of Problem 3 for α = 1, λ, D, which is symmetric with respect to x 0 , continuous on R and nonincreasing on [x 0 , ∞).The function u and K = {x ∈ R : u(x) = 1} is a solution of Problem 1 for α = 1, λ, D. For any λ ∈ (0, Λ 1,D ) there are no solutions of Problem 3 for α = 1, λ, D.
Note that the above result does not guarantee the uniqueness of solution of Problem 3. The results for the classical Bernoulli problem for balls suggest that uniqueness holds (cf.Conjecture 5.6 below).
Proof.By translation and scaling we may assume that x 0 = 0 and r = 1.By arguments as in the proof of Lemma 2.5 in [21] (by changing in that proof E λ to I λ,(−1,1) ), we obtain that there exists a solution of Problem 3, which is symmetric with respect to 0, continuous on R and nonincreasing on [0, ∞).It follows that {x ∈ R : u(x) = 1} = [−a, a] for some a ∈ (0, 1).By Theorem 1.7 (d) in [21], Using again Theorem 1.7 in [21], we obtain that u and The next result gives an inequality between the variational constant Λ 1,D and the Bernoulli constant λ 1,D for any interval D. The proof of this result is computerassisted.
Proposition 5.4.Let x 0 ∈ R, r > 0 and D = (x 0 − r, x 0 + r).We have It is well known that the analogous result holds for the classical inner Bernoulli problem on a ball (see e.g.Example 11 in [18]).
Fix α ∈ (0, 2).For any D ∈ G we denote by P D the Poisson kernel for D corresponding to (−∆) α/2 .P D : D × (D) c → (0, ∞) has the following properties.For any x ∈ D we have (D) c P D (x, y) dy ≤ 1, and if D ∈ G is additionally bounded, then (D) c P D (x, y) dy = 1.

Corollary 5 . 5 .
For α = 1 and any interval D there exists a solution for the inner Bernoulli problem for the fractional Laplacian on D, which is not a minimizer of the corresponding variational problem on D (Problem 3).