Anne-Laure Dalibard

David Gérard-Varet

On shape optimization problems involving the fractional laplacian *

Keywords: fractional laplacian, shape optimization, shape derivative, moving plane method

Our concern is the computation of optimal shapes in problems involving (-∆) 1/2 . We focus on the energy J(Ω) associated to the solution u Ω of the basic Dirichlet problem (-∆) 1/2 u Ω = 1 in Ω, u = 0 in Ω c . We show that regular minimizers Ω of this energy under a volume constraint are disks. Our proof goes through the explicit computation of the shape derivative (that seems to be completely new in the fractional context), and a refined adaptation of the moving plane method.

Introduction

This article is concerned with shape optimization problems involving the fractional laplacian.

The typical example we have in mind comes from the following system:

(-∆) 1/2 u = 1 on Ω, u = 0 on Ω c , (1.1)
set for a bounded open set Ω of R 2 . We wish to find the minimizers of the associated energy

J(Ω) := inf v∈H 1/2 (R 2 ), v| Ω c =0 1 2 (-∆) 1/2 v, v H -1/2 ,H 1/2 - R 2 v . (1.2)
among open sets Ω with prescribed measure (and a smoothness assumption). Beyond this specific example, we wish to develop mathematical tools for shape optimization in the context of fractional operators.

Our original motivation comes from a drag reduction problem in microfluidics. Recent experiments, carried on liquids in microchannels, have suggested that drag is substantially lowered when the wall of the channel is water-repellent and rough [START_REF] Lauga | Microfluidics: The no-slip boundary condition[END_REF][START_REF] Vinogradova | Surface roughness and hydrodynamic boundary conditions[END_REF]. The idea is that the liquid sticks to the bumps of the roughness, but may slip over its humps, allowing for less friction at the boundary. Mathematically, one can consider Stokes equations for the liquid (variable (x, z) = (x 1 , x 2 , z), velocity field u = (v, w) = (v 1 , v 2 , w)):

-∆u + ∇p = 0, div u = 0, z > 0, (1.3) set above a flat surface {z = 0}. This flat surface replaces the rough hydrophobic one, and is composed of areas on which the fluid satisfies alternately perfect slip and no-slip boundary conditions. In the simplest models, these areas of perfect slip and no-slip form a periodic pattern, corresponding to a periodic pattern of humps and bumps. That means that the impermeability condition w = 0 at {z = 0} is completed with mixed Dirichlet/Navier conditions:

∂ z v = 0 in Ω, v = 0 in Ω c (1.4)
where Ω ⊂ R 2 corresponds to the zones of perfect slip and Ω c to the zones of no-slip. The whole issue is to design Ω so that the energy J(Ω) associated with this problem is minimal for a fixed fraction of slip area. Unfortunately, this optimization problem (Stokes operator, periodic pattern) is still out of reach. That is why we start with the simpler equations (1.1) (still difficult, and interesting on their own!). Note that they can be seen as a scalar version of (1.3)- (1.4), replacing the Stokes operator by the Laplacian, and the Navier by the Neumann condition. Using the classical characterization of (-∆) 1/2 as the Dirichlet-to-Neumann operator leads to system (1.1) and energy (1.2). Again, we stress that the methods developed in our paper may be useful in more elaborate contexts.

If the fractional laplacian in (1.1) is replaced by a standard laplacian (which leads to the classical Dirichlet energy), this problem is well-known and described in detail in the book [START_REF] Henrot | Variation et optimisation de formes[END_REF] by Henrot and Pierre (see also [START_REF] Serrin | A symmetry problem in potential theory[END_REF]). In this case, one can show that a smooth domain Ω minimizing the Dirichlet energy under the constraint |Ω| = 1 is a disc. A standard proof of this result has two main steps:

1. One computes the shape derivative associated with the Dirichlet energy for the laplacian.

This leads to the following result: if Ω is a minimizer of the Dirichlet energy and u Ω is the solution of the associated Euler-Lagrange equation, then ∂ n u is constant on ∂Ω.

2. One analyzes an overdetermined problem. More precisely, the idea is to prove that if there exists a function u solving -∆u = 1 in Ω, u = 0 on ∂Ω,

∂ n u = c on ∂Ω, then all the connected components of Ω are discs. This second step is achieved thanks to the moving plane method.

Our goal in this article is to develop the same approach in the context of the fractional laplacian, showing radial symmetry of any smooth minimizer Ω of (1.2). Accordingly, we start with the computation of the shape derivative.

Theorem 1. Let f ∈ C ∞ (R 2 , R), and let Taking into account this extra condition on the fractional normal derivative, we can then determine the minimizing domain:

J f (Ω) := inf v∈H 1/2 (R 2 ), v| Ω c =0 1 2 (-∆) 1/2 v, v H -1/2 ,H 1/2 - R 2 f v . Let ζ ∈ C ∞ 0 (R 2 , R 2
Theorem 3. Let Ω ⊂ R 2 be a C ∞ open set such that the system (-∆) 1/2 u = 1 in Ω, u = 0 in Ω c , ∂ 1/2
n u = c 0 on ∂Ω.

(1.5)

has at least one solution. Assume that Ω is connected. Then Ω is a disc.

Of course, Theorem 1 and Corollary 3 imply immediately the following:

Corollary 4. Let Ω ⊂ R 2 be a C ∞ connected open set such that |Ω| = 1, and
J(Ω) = inf Ω ′ ⊂R 2 , |Ω ′ |=1 J(Ω ′ ).
Then Ω is a disc.

Let us make a few comments on our results. As regards Theorem 1, we stress that, due to the non-locality of the fractional laplacian, the shape derivative of J f is hard to compute. In the case of the classical laplacian, it is obtained through integration by parts, which are completely unavailable in the present context. The idea is to bypass the nonlocality by using an asymptotic expansion of u Ω,f (x) as the distance between x and ∂Ω goes to zero. Such asymptotic expansion follows from general results of [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF], on the solutions of linear elliptic systems in domains with cracks. We insist that the proof of the theorem neither uses scalar arguments, nor Fourier-based calculations. In particular, we believe that its interet goes much beyond Corollary 4 (that is finding the minimizer of (1.2)). It is likely that it can be adapted to vectorial settings, or functionals with non-constant coefficients. Notice also that Theorem 1 and Corollary 2 do not require the connectedness of Ω. For our special case f = 1, they imply that the value of the fractional normal derivative is the same on all the boundaries of the connected components of Ω.

As regards Theorem 3, it is deduced from an adaptation of the moving plane method to our fractional setting. Again, this is not straightforward, as the standard method relies heavily on the maximum principle and Hopf's Lemma, which are essentially local tools. To overcome our non-local problem, we use appropriate three-dimensional extensions of u, to which we can apply maximum principle methods in a classical context. Note however that we need to assume that Ω is connected: this hypothesis is precisely due to the nonlocality of the fractional laplacian.

We conclude this introduction by a brief review of related results. Let us first mention that the condition ∂ 1/2 n u = c 0 appears in other problems related to the fractional laplacian. In [START_REF] Caffarelli | Variational problems for free boundaries for the fractional Laplacian[END_REF], Caffarelli, Roquejoffre and Sire consider a minimization problem for another energy related to the fractional laplacian, and they prove that minimizers satisfy the condition ∂ 1/2 n u = c 0 on ∂{u = 0}. However, we emphasize that the issues of the present paper and those of [START_REF] Caffarelli | Variational problems for free boundaries for the fractional Laplacian[END_REF] are rather different. The problem addressed in [START_REF] Caffarelli | Variational problems for free boundaries for the fractional Laplacian[END_REF] is essentially a free-boundary problem (i.e. Ω is not given, but is defined as {u > 0}), and therefore questions such as the regularity and the non-degeneracy of u, and the regularity of the free boundary, are highly non trivial and are at the core of the paper [START_REF] Caffarelli | Variational problems for free boundaries for the fractional Laplacian[END_REF]. Here, our goal is not to investigate these questions, but rather to derive information on the shape of Ω, assuming a priori regularity. As mentioned before, we rely on article [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF] by Costabel and co-authors, which provides asymptotic expansions for solutions of linear elliptic equations near cracks. As they derive accurate asymptotics, based on pseudo-differential calculus, they need the domain to be C ∞ . In our context, only cruder information is needed (broadly, we need the first term in the expansion, and tangential regularity). Apart from these asymptotic expansions, the proofs of Theorem 1 and Theorem 3 use very little information on the regularity of ∂Ω (existence of tangent and normal vectors, boundedness and regularity of the curvature). Hence, it is likely that our C ∞ regularity requirement can be lowered.

As regards our adaptation of the moving plane method, it relates to other results on the proof of radial symmetry for minimizers of nonlocal functionals: see for instance [START_REF] Lu | An overdetermined problem in riesz potential and fractional laplacian[END_REF] on local Riesz potentials

u(x) = Ω 1 |x -y| N -1
or [START_REF] Cabré | Positive solutions of nonlinear problems involving the square root of the Laplacian[END_REF] on the radial symmetry of solutions of nonlinear equations involving A 1/2 , where A is the Dirichlet laplacian of a ball in R n . Note that in these two papers, "local" maximum principles are still available, which helps. Further references (notably to article [START_REF] Birkner | Comparison results and steady states for the Fujita equation with fractional Laplacian[END_REF]) will be provided in due course.

Let us eventually point out that more direct proofs of the final Corollary 4 might be available. For instance, in the case of the classical laplacian, one way to proceed is to consider the auxiliary problem: minimize

E(u) := 1 2 R 2 |∇u| 2 - R 2 u
under the measure constraint: |{u > 0}| = 1}. Crudely, any minimizer u of E provides a solution Ω u = {u > 0} of the shape optimization problem, and vice versa. We refer once again to [START_REF] Henrot | Variation et optimisation de formes[END_REF] for rigourous statements. In particular, showing that any minimizer of E is radial shows that any minimizing domain is radial (without a priori regularity assumption).

In the case of the fractional laplacian, a close context has been recently investigated by Lopes and Maris in [START_REF] Lopes | Symmetry of minimizers for some nonlocal variational problems[END_REF]. Their result is the following:

Proposition 5. Let s ∈ (0, 1) and assume that F, G : R → R are such that u → F (u), u → G(u) map Ḣs (R d ) into L 1 (R d ). Assume that u ∈ Ḣs (R d ) is a solution of the minimization problem Minimize E(u) := R d |ξ| 2s |û(ξ)| 2 dξ + R d F (u) under the constraint R d G(u) = λ.
Then u is radially symmetric.

Looking closely at their proof, it seems that their arguments can be extended as such to the functionals

F (u) := u, G(u) := 1 u>0
although such F and G do not map Ḣ1/2 (R 2 ) into L 1 (R 2 ). This is likely to yield the radial symmetry of the minimizing domain for our shape functional J (without a priori regularity assumption). See Remark 19 for further discussion.

The plan of our paper is the following: Section 2 collects more or less standard results on the fractional laplacian, which will be used throughout the article. Special attention is paid to regularity properties of solutions of (1.1), that we deduce from regularity results for the Laplace equations in domains with cracks. Section 3 is devoted to the proof of Theorem 1 and Corollary 2. Finally, Section 4 contains the proof of Theorem 3.

Preliminaries

Reminders on the fractional laplacian

We remind here some basic knowledge about (-∆) 1/2 , see for instance [START_REF] Silvestre | Regularity of the obstacle problem for a fractional power of the Laplace operator[END_REF]. We start with Definition 6. For any f ∈ S(R n ), one defines (-∆) 1/2 f through the identity

(-∆) 1/2 f (ξ) = |ξ| f (ξ).
Note that g = (-∆) 1/2 f does not belong to S(R n ) because of the singularity of |ξ| at 0. Nevertheless, it is C ∞ and satisfies for all k ∈ N:

sup x∈R n (1 + |x| n+1 )|g (k) (x)| < +∞
This allows for a definition of (-∆) 1/2 over a large subspace of S ′ (R n ), by duality. We shall only retain

Proposition 7. (-∆) 1/2 extends into a continuous operator from H 1/2 (R n ) to H -1/2 (R n ).
This is clear from the definition.

It is also well-known that (-∆) 1/2 can be identified with the Dirichlet-to-Neumann operator, in the following sense (writing (x, z) ∈ R n × R the elements of R n+1 ):

Theorem 8. Let u ∈ H 1/2 (R n ). One has (-∆) 1/2 u = -∂ z U | z=0 , where U is the unique solution in Ḣ1 (R n+1 + ) := U ∈ L 2 loc (R n+1 + ), ∇U ∈ L 2 (R n+1 + ) of -∆ x,z U = 0 in R n+1 + , U | z=0 = u.
Let us remind that the normal derivative

∂ z U | z=0 ∈ H -1/2 (R n ) has to be understood in a weak sense: for all φ ∈ H 1 (R n+1 + ), < ∂ z U | z=0 , γφ > = - R n+1 + ∇U • ∇φ
where γ is the trace operator (onto H 1/2 (R n )). It coincides with the standard derivative whenever U is smooth.

We end this reminder with a formula for the fractional laplacian:

Theorem 9. Let f satisfying R n |f (x)|
1+|x| n+1 dx < +∞, with regularity C 1,ǫ , ǫ > 0, over an open set Ω. Then, (-∆) 1/2 f is continuous over Ω, and for all x ∈ Ω, one has

(-∆) 1/2 f (x) = C 1 PV R n f (x) -f (y) |x -y| n+1 dy (2.1) = C 1 R n f (x) -f (y) -∇f (x) • (x -y)1 |x-y|≤C |x -y| n+1 dy. (2.2)
for some C 1 = C 1 (n) and any C > 0.

2.2

The Dirichlet problem for (-∆) 1/2 . Regularity properties (2d case).

In view of system (1.1), a key point in our analysis is to know the behavior near the boundary of solutions to the following fractional Dirichlet problem:

(-∆) 1/2 u = f on Ω, u = 0 on Ω c , (2.3)
where Ω is a smooth open set of R 2 and f ∈ C ∞ (R 2 ). Note that in system (2.3), we prescribe the value of u not only at ∂Ω, but in the whole Ω c . This is reminiscent of the non-local character of (-∆) 1/2 : remind that u = U | z=0 , where U satisfies the (3d) local problem

-∆U = 0 in z > 0, ∂ z U = -f over Ω × {0}, U = 0 over Ω c × {0}, (2.4)
whose mixed Robin/Dirichlet boundary condition must be specified over the whole plane {z = 0}.

We were not able to find direct references for regularity properties of problem (2.3), although the C 0,1/2 regularity of u and the existence of ∂ 1/2 n u are evoked in [START_REF] Krzysztof | The boundary Harnack principle for the fractional Laplacian[END_REF][START_REF] Caffarelli | Variational problems for free boundaries for the fractional Laplacian[END_REF][START_REF] Desilva | Regularity in a one-phase free boundary problem for the fractional laplacian[END_REF]. In particular, we could not collect information on transverse and tangential regularity of the solution u near ∂Ω×{0}. We shall use results for the Laplace equation in domains with cracks, in the following way.

Let η ∈ C ∞ c (R 3 ), odd in z, with θ(x, z) = z for x in a neighborhood of Ω and |z| ≤ 1. Then, V (x, z) := U (x, z) + η(x, z)f (x) satisfies -∆V = F in z > 0, ∂ z V = 0 over Ω × {0}, U = 0 over Ω c × {0}, (2.5)
where F := -∆ x,z (ηf ) is smooth, odd in z, and compactly supported in R 3 . We then extend V to {z < 0} by the formula V (x, z) := -V (x, -z), z < 0. In this way, we obtain the system

-∆V = F in R 3 \ (Ω × {0}) ∂ z V = 0 at (Ω × {0}) ± (2.6)
which corresponds to a Laplace equation outside a "crack" Ω × {0} with Neumann boundary condition on each side of the crack. We can now use regularity results for the laplacian in singular domains, such as those of [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF]. First, note that V is C ∞ away from Ω × {0}, by standard elliptic regularity. Let now Γ be a connected component of ∂Ω, and ϕ a truncation function such that ϕ = 1 in a neighborhood of Γ, with Supp ϕ ∩ (∂Ω \ Γ) = ∅. Then, ϕV still satisfies a system of type (2.6), with Ω replaced by Supp ϕ, and F by F -[∆, ϕ]V (which is still smooth). We can then apply [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF]Theorem A.4.3], which leads to the following Theorem 10. Let Γ be a connected component of ∂Ω. We denote by (r, θ) polar coordinates in the planes normal to Γ and centered on Γ, and by s the arc-length on Γ, so that

R 3 \ ((Supp ϕ ∩ Ω) × {0}) = {(s, r, θ), s ∈ (0, L), r > 0, θ ∈ (-π, π)}.
Then, the solution V of (2.6) has the following asymptotic expansion, as r → 0: for any integer K ≥ 0,

V = K k=0 r 1/2+k Ψ k (s, θ) + U reg,K + U rem,K (2.7)
where:

• the coefficients Ψ k are regular over [0, L] × [-π, π],
• U reg,K is regular over R 3 ,

• the remainder

U rem,K satisfies ∂ β U rem,K = o(r K-|β|+1/2 ) for all β ∈ N 3 .
Setting

ψ k (s) := Ψ k (s, π), u rem,K (x) := U rem,K (x, 0 + ), u reg,K (x) := U reg,K (x, 0 + ),
we can get back to the solution u of (2.3) and obtain the asymptotic expansion

u = K k=0 r 1/2+k ψ k (s) + u reg,K + u rem,K . (2.8)
Such formulas will be at the core of the next sections.

3 Shape derivative of the energy J(Ω)

This section is devoted to the proof of Theorem 1 and Corollary 2. The proof of Theorem 1 is rather technical, although it follows a simple intuition. Therefore let us first explain how Corollary 2 is derived. Assume that Theorem 1 holds. Consider the application

J : ζ ∈ C 1 b (R 2 , R 2 ) → J((I + ζ)Ω),
where

C 1 b (R 2 , R 2 ) = C 1 ∩ W 1,∞ (R 2 , R 2
). We recall that C 1 b equipped with the norm • W 1,∞ is a Banach space. We first claim that J is differentiable at ζ = 0. This follows from the differentiability of the application

ζ ∈ C 1 b (R 2 , R 2 ) → v ζ ∈ H 1/2 (R 2 ), where v ζ = u ζ • (I + ζ) and u ζ = u (I+ζ)Ω
is the solution of the Euler-Lagrange equation associated with (I + ζ)Ω. The proof goes along the same lines as the one of Lemma 11 below. Furthermore, the variational formulation of the Euler-Lagrange equation implies (see formula (3.9))

J((I + ζ)Ω) = - 1 2 R 2 v ζ det(I + ∇ζ).
The differentiability of J follows.

Using Theorem 1, we then identify the differential of

J at ζ = 0. Indeed, if (φ t ) t∈R is the flow associated with ζ ∈ C ∞ 0 (R 2 ), d dt J(φ t (Ω)) |t=0
is the Gâteaux derivative of J at point 0 in the direction ζ. We infer that

dJ (0)ζ = C 0 ∂Ω ζ • n(∂ 1/2 n u Ω ) 2 dσ for all ζ ∈ C ∞ 0 (R 2 , R 2
), and by density for all ζ ∈ C 1 b (R 2 , R 2 ). Now, assume that Ω is a bounded domain with C ∞ boundary, which minimizes J under the constraint |Ω| = 1. In other words, ζ = 0 is a minimizer of J (ζ) in the Banach space

C 1 b under the constraint V (ζ) := |(I + ζ)(Ω)| = 1.
According to the theorem of Lagrange multipliers, there exists λ ∈ R such that

∀ζ ∈ C 1 b (R 2 , R 2 ), (dJ (0) + λdV (0)) ζ = 0. (3.1)
It is proved in [START_REF] Henrot | Variation et optimisation de formes[END_REF] that

dV (0)ζ = ∂Ω ζ • n dσ. Thus (3.1) becomes ∃λ ∈ R, ∀ζ ∈ C 1 b (R 2 , R 2 ), ∂Ω (ζ • n) (∂ 1/2 n u Ω ) 2 + λ C 0 dσ = 0.
Since ζ is arbitrary, we infer that ∂ 1/2 n u Ω is constant on Ω. Moreover, since u Ω ≥ 0 on Ω by the maximum principle, the constant is positive. This completes the proof of Corollary 2.

We now turn to the proof of Theorem 1. In the case of the classical laplacian, the shape derivative of the Dirichlet energy is well-known and is proved in the book by Henrot et Pierre [START_REF] Henrot | Variation et optimisation de formes[END_REF]. Let us recall the main steps of the derivation, which will be useful in the case of the fractional laplacian. Let

I f (Ω) := inf u∈H 1 0 (Ω) 1 2 Ω |∇u| 2 - Ω f u , For all ζ ∈ C 1 ∩ W 1,∞ (R 2 )
, consider the flow φ t associated with ζ. Then for all t ∈ R, φ t is a diffeomorphism of R 2 . We recall the following properties, which hold for all t ∈ R:

φ0 = ζ, |det(∇φ t (x))| =: j(t, x) = exp t 0 (div ζ)(φ s (x)) ds dφ -1 t dt |t=0 = -ζ, det(∇φ -1 t ) = j(-t, x). (3.2)
The last line merely expresses the fact that φ -1 t = φ -t , for all t ∈ R. For t ≥ 0, let Ω t = φ t (Ω), and let w t be the solution of the associated Euler-Lagrange equation, namely

-∆w t = f in Ω t , w t = 0 on ∂Ω t .
Eventually, we define z t := w t • φ t . It is proved in [START_REF] Henrot | Variation et optimisation de formes[END_REF] that

d dt I f (Ω t ) |t=0 = - 1 2 ∂Ω (ζ • n)(∂ n w 0 ) 2 dσ.
Indeed, since w t solves the Euler-Lagrange equation, we have The Euler-Lagrange equation yields

I f (Ω t ) = - 1 2 Ωt f w t = - 1 2 Ω z t (y)f • φ t (y)j(t,
-∆ ẇ0 = 0 in Ω.
Gathering all the terms and using the fact that w 0|∂Ω = 0, we deduce that

d dt I f (Ω t ) |t=0 = - 1 2 Ω (f ( ẇ0 + ζ • ∇w 0 ) + (ζ • ∇f )w 0 + f div ζw 0 ) = - 1 2 Ω (-ẇ0 ∆w 0 + div (ζf w 0 )) = 1 2 Ω w 0 ∆ ẇ0 + ∂Ω ẇ0 ∂ n w 0 dσ = - 1 2 ∂Ω (ζ • n)(∂ n w 0 ) 2 dσ.
Therefore the shape derivative of I f is similar to the one of J f , the fractional derivative being merely replaced by a classical derivative.

Unfortunately, the proof in the case of the classical laplacian can only partially be transposed to the fractional laplacian. Indeed, several integration by parts play a crucial role in the computation, and cannot be used in the framework of the fractional laplacian.

We use therefore a different method to estimate dJ f (Ω t )/dt. The main steps of the proof are as follows:

1. As above, we introduce u t = u Ωt,f and v t = u t • φ t . We derive regularity properties and asymptotic expansions for v t , from which we deduce a decomposition of u 0 and u0 .

2. In order to avoid the singularities of u0 near ∂Ω, we introduce a truncation function χ k supported in Ω, and vanishing in the vicinity of the boundary. Using the integral form of the fractional laplacian, we then derive an integral formula for an approximation of dJ f (Ω t )/dt involving u 0 , u0 and χ k .

3. Keeping only the leading order terms in the decomposition of u 0 and u0 , we obtain an expression of dJ f (Ω t )/dt in terms of ∂ 1/2 n u 0 , and we prove that this expression is independent of the choice of the truncation function χ k .

4. We then evaluate the contributions of the remainder terms to the integral formula, and we prove that they all vanish as k → ∞.

Most of the technicalities are contained in steps 3 and 4. However, some more or less formal calculations -performed at the end of step 2 -lead relatively easily to the desired result. Before tackling the core of the proof, let us introduce some notation:

• We denote by Γ 1 , • • • , Γ N the
connected components of ∂Ω, and we parametrize each Γ i by its arc-length s. We denote by L i the length of Γ i .

• The number r ∈ R stands for the (signed) distance to the boundary of Ω. More precisely, |r| is the distance to the boundary, and r > 0 inside Ω, r < 0 outside Ω;

• We denote by U t the three dimensional extension of u t in the half-space, i.e. the function such that

-∆U t = 0 on R 3 + , U t | z=0 = u t . The three-dimensional function V t is then defined by V t (x, z) = U t (φ t (x), z) x ∈ R 2 , z ∈ R, so that V t | z=0 = v t .
• Derivatives with respect to t are denoted with a dot.

Regularity of u 0 and u0 and expansions

We start with the following lemma Lemma 11. For all t in a neighbourhood of zero,

v t ∈ H 1/2 (R 2 ), vt ∈ H 1/2 (R 2 ), u t ∈ H 1/2 (R 2 ), ut ∈ H -1/2 (R 2 ). (3.
3

)
The proof is rather close to the one of Theorem 5.3.2 in [START_REF] Henrot | Variation et optimisation de formes[END_REF]. In order to keep the reading as fluent as possible, the details are postponed to the end of the section. The idea is to prove that V t solves a three-dimensional elliptic equation with smooth coefficients. The implicit function theorem then implies that t

→ V t ∈ H 1 is C 1 in a neighbourhood of t = 0. Since v t is the trace of V t , t → v t ∈ H 1/2 is also a C 1 application. Eventually, the chain-rule formula entails that ut ∈ H -1/2 .
We also derive asymptotic formulas for V t and v t in terms of r: we rely on the results in the paper by Costabel, Dauge and Duduchava [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF], and we use the notations of paragraph 2.2 (see also Theorem 10 of the present paper). We claim that there exists

ψ 0 , ψ 1 , ψ 2 ∈ C 1 (∂Ω), u 1 , u 2 ∈ W 1,∞ (R 2 ) such that u 0 = √ r + ψ 0 (s) + u 1 (x), (3.4
) u0 = 1 r>0 1 √ r ψ 1 (s) + √ r + ψ 2 (s) + u 2 (x), (3.5)
where

ψ 0 (s) = ∂ 1/2 n u 0 (s), (3.6
)

ψ 1 (s) = 1 2 ζ • n(s)ψ 0 (s). (3.7)
Moreover, there exists δ > 0 such that

ut ∈ L ∞ ((-δ, δ), L 1 (R 2 )). (3.8)
These decompositions and the regularity result (3.8) will be proved at the end of the section, after the proof of Lemma 11.

An integral formula for an approximation of dJ f (Ω t )/dt

We first use the Euler-Lagrange equation (1.1) in order to transform the expression defining J f (Ω t ). Classically, we prove that the unique minimizer u of the energy

1 2 (-∆) 1/2 v, v H -1/2 ,H 1/2 - R 2 f v in the class {v ∈ H 1/2 (R 2 ), v |Ω c t = 0} satisfies (-∆) 1/2 u, v H -1/2 ,H 1/2 - R 2 f v = 0 ∀v ∈ H 1/2 (R 2 ) s.t. v |Ω c t = 0. Choosing v ∈ C ∞ 0 (Ω t ), we infer that (-∆) 1/2 u = f in Ω. Since u |Ω c t = 0, we infer that u = u Ωt,f = u t , and in particular (-∆) 1/2 u t , u t = Ωt f u t .
The identity above yields

J f (Ω t ) = - 1 2 R 2 f u t = - 1 2 Ωt f u t .
Changing variables in the integral, we obtain

J f (Ω t ) = - 1 2 Ω v t (y)f • φ t (y)j(t, y) dy. (3.9) Since t → v t ∈ H 1/2 is differentiable, t → J f (Ω t ) is C 1 for t close to zero, and
d dt J f (Ω t ) |t=0 = - 1 2 Ω ( v0 f + ζ • ∇f v 0 + v 0 f div ζ). Now, for k ∈ N large enough, we define χ k ∈ C ∞ 0 (R 2 ) by χ k (x) = χ(kr), where χ ∈ C ∞ (R) and χ(ρ) = 0 for ρ ≤ 1, χ(ρ) = 1 for ρ ≥ 2. Then, since u 0 = v 0 , d dt J f (Ω t ) |t=0 = - 1 2 lim k→∞ R 2 χ k ( v0 f + v 0 div (ζf )) = - 1 2 lim k→∞ R 2 χ k u 0 div (ζf ) + d dt R 2 χ k f v t |t=0 = - 1 2 lim k→∞ R 2 χ k u 0 div (ζf ) + d dt R 2 u t (f χ k ) • φ -1 t j(-t, •) |t=0 .
For fixed k and for t in a neighbourhood of zero, there exists a compact set K k such that

K k ⋐ Ω and Supp χ k • φ -1 t ⊂ K k . Since ut ∈ L ∞ t (L 1
x ) according to (3.8), we can use the chain rule and write

d dt R 2 u t (f χ k ) • φ -1 t j(-t, •) |t=0 = R 2 ( u0 f χ k -u 0 f ζ • ∇χ k -u 0 χ k ζ • ∇f -u 0 χ k f div ζ).
Using the decomposition (3.4) together with the definition of χ k , we deduce that

R 2 u 0 f ζ • ∇χ k ≤ C R √ r + k|χ ′ (kr)| dr ≤ C √ k .
Notice also that

R 2 (-u 0 χ k ζ • ∇f -u 0 χ k f div ζ) = - R 2 u 0 χ k div (f ζ).
We now focus on the term involving u0 ; since (-∆) 1/2 u 0 = f on the support of χ k , we have

R 2 u0 f χ k = R 2 u0 χ k (-∆) 1/2 u 0 = R 2 u 0 (-∆) 1/2 ( u0 χ k ).
Notice also that χ k (-∆) 1/2 ( u0 ) = 0. Indeed, ut is smooth on K k for t small enough (see for instance (3.28) below). Hence for x ∈ K k , the integral formula (2.2) makes sense and we have, using (3.8),

(-∆) 1/2 u0 (x) = C 1 R 2 u0 (x) -u0 (y) -∇ u0 (x) • (x -y)1 |x-y|≤C |x -y| 3 dy = C 1 d dt R 2 u t (x) -u t (y) -∇u t (x) • (x -y)1 |x-y|≤C |x -y| 3 dy |t=0 = d dt (f (x)) = 0.
Eventually, we obtain

R 2 u0 f χ k = R 2 u 0 (-∆) 1/2 , χ k u0 .
Gathering all the terms, we infer eventually

d dt J f (Ω t ) |t=0 = - 1 2 lim k→∞ R 2 u 0 (-∆) 1/2 , χ k u0 .
Let us now express the right-hand side in terms of the kernel of the fractional laplacian. Using the integral formula (2.2) together with the expansion (3.5), we infer that

(-∆) 1/2 , χ k u0 (x) = C 1 R 2 u0 (y)(χ k (x) -χ k (y)) -u0 (x)∇χ k (x) • (x -y)1 |x-y|≤C |x -y| 3 dy.
The value of the integral above is independent of the constant C. Therefore the shape derivative of the energy J f is given by

d dt J f (Ω t ) |t=0 = - C 1 2 lim k→∞ I k ,
where

I k := R 2 ×R 2 u 0 (x) |x -y| 3 u0 (y)(χ k (x) -χ k (y)) -u0 (x)∇χ k (x) • (x -y))1 |x-y|≤C dx dy.
(3.10) The next step is to compute the asymptotic value of I k as k → ∞. This part is rather technical, and involves several error estimates. However, the intuition leading the calculations is simple: first, the main order is obtained when u 0 and u0 are replaced by the leading terms in their respective developments. Moreover, because of the truncation χ k , the integral is concentrated on the boundary ∂Ω. All these claims will be fully justified in the next paragraph.

If we follow these guidelines, we end up with

I k ≈ PV T δ ×T δ u 0 (x) u0 (y) |x -y| 3 (χ k (x) -χ k (y)) dx dy,
where T δ is a tubular neighbourhood of ∂Ω of width δ ≪ 1 (see (3.11)). If we change cartesian coordinates for local ones, and if we neglect the curvature of ∂Ω -which is legitimate if δ is small -we are led to

I k ≈ PV [0,δ] 2 (0,L) 2 ψ 0 (s)ψ 1 (s) ((s -s ′ ) 2 + (r -r ′ ) 2 ) 3/2 r r ′ (χ(kr) -χ(kr ′ ) ds ds ′ dr dr ′ .
For simplicity, we have assumed that ∂Ω only has one connected component, of length L. Integrating first with respect to s ′ , and changing variables by setting ρ = kr, ρ ′ = kr ′ , we obtain eventually

I k ≈ 2C L 0 ψ 0 ψ 1 PV ∞ 0 ∞ 0 ρ ρ ′ χ(ρ) -χ(ρ ′ ) |ρ -ρ ′ | 2 dρ dρ ′ ≈ C L 0 ψ 0 ψ 1 ∞ 0 ∞ 0 χ(ρ) -χ(ρ ′ ) √ ρρ ′ (ρ -ρ ′ ) dρ dρ ′ , where C = ∞ 0 dz (1 + z 2 ) 3/2 .
There remains to prove that the value of the integral in the right-hand side does not depend on χ, which we do at the end of the next paragraph, and the formula of Theorem 1 is proved.

Of course, the above calculation is very sketchy, and careful justification must be given at every step. But the general direction follows these formal arguments.

Asymptotic value of I k

We now evaluate the integral I k defined in (3.10). There are two main ideas:

• We prove that the domain of integration can be restricted to a tubular neighbourhood of ∂Ω (see lemma 12).

• Since the integral I k is bilinear in u 0 , u0 , we replace u 0 and u0 by their expansions in (3.4), (3.5). The leading term is obtained when u 0 and u0 are replaced by the first terms in their respective developments. We prove in the next subsection that all other terms vanish as k → ∞.

We begin with the following Lemma (of which we postpone the proof):

Lemma 12. For δ > 0, let

T δ := {x ∈ R 2 , d(x, ∂Ω) < δ}. (3.11)
Choose C = δ 2 in the definition of I k (3.10). Then there exists a constant C δ such that for k > 5/δ,

(T δ ×T δ ) c u 0 (x) |x -y| 3 u0 (y)(χ k (x) -χ k (y)) -u0 (x)∇χ k (x) • (x -y))1 |x-y|≤δ/2 dx dy ≤ C δ √ k .
We henceforth focus our attention on the value of the integral on T δ × T δ . Replacing u 0 and u0 by the first terms in the expansions (3.4), (3.5), we define

I δ k := T δ ×T δ ψ 0 (s) √ r + |x -y| 3 ψ 1 (s ′ ) 1 r ′ >0 √ r ′ (χ(kr) -χ(kr ′ )) -ψ 1 (s) 1 r>0 √ r kχ ′ (kr)n(s) • (x -y)1 |x-y|≤δ/2 dx dy. (3.12)
There is a slight abuse of notation in the integral above, since we use simultaneously cartesian and local coordinates. In order to be fully rigorous, r, r ′ , s, s ′ should be replaced by r(x), r(y), s(x), s(y) respectively. However, the first step of the proof will be to express the integral I δ k in local coordinates, and therefore we will avoid these heavy notations. In fact, the computation of the limit of I δ k is much more technical than the estimation of all other quadratic remainder terms, which we will achieve in the next subsection. We now prove the following: Lemma 13. There exists an explicit constant C 2 , independent of χ and of Ω, such that

lim δ→0 lim k→∞ I δ k = C 2 N i=1 L i 0 ψ 0 (s)ψ 1 (s) ds.
Proof. Throughout the proof, different types of error terms will appear, which we have gathered in Lemma 14 below. We will therefore refer to ( • We use local coordinates instead of cartesian ones, i.e. we change x into (s, r) and y into (s ′ , r ′ ). Since |r|, |r ′ | ≤ δ, the jacobian of this change of coordinates is, for δ > 0 small enough,

|1 + rκ(s)| 1 + r ′ κ(s ′ ) = (1 + rκ(s)) 1 + r ′ κ(s ′ ) ,
where κ is the algebraic curvature of ∂Ω. We refer to the Appendix for a simple proof. Notice that this jacobian is always bounded.

• We write T δ = ∪ N i=1 T i δ , where

T i δ = {x ∈ R 2 , d(x, Γ i ) < δ}.
If δ is small enough, T i δ ∩T j δ = ∅ for i = j, and |x-y| is bounded from below for x ∈ T i δ , y ∈ T j δ by a constant independent of δ. Hence, for i = j,

T i δ ×T j δ ψ 0 (s) √ r + |x -y| 3 ψ 1 (s ′ ) 1 r ′ >0 √ r ′ (χ(kr) -χ(kr ′ )) -ψ 1 (s) 1 r>0 √ r kχ ′ (kr)n(s) • (x -y)1 |x-y|≤δ/2 dx dy ≤ C T i δ ×T j δ 1 r,r ′ >0 √ r √ r ′ dx dy ≤ C (0,δ) 2 √ r √ r ′ dr dr ′ ≤ Cδ 2 .
Therefore, in the integral defining I δ k , we replace the domain of integration T δ ×T δ by ∪ N i=1 T i δ × T i δ , and this introduces an error term of order δ 2 .

• In local coordinates, we write T i δ as (0, L i ) × (-δ, δ). We replace the jacobian

(1 + rκ(s)) 1 + r ′ κ(s ′ ) by (1 + rκ(s)) 2 . Since s, s ′ ∈ (0, L i ) (i.e.
x and y belong to the neighbourhood of the same connected component of ∂Ω), this change introduces error terms bounded by (3.19), (3.21), which vanish as k → ∞ and δ → 0. More details will be given in the fourth step for similar error terms; we refer to Lemma 14.

• We evaluate |x -y| in local coordinates for x, y ∈ T i δ . We have

x = p(s) -rn(s), y = p(s ′ ) -r ′ n(s ′ ), where p(s) ∈ R 2 is the point of ∂Ω with arc-length s. Since the boundary Γ i is C ∞ , p(s) -p(s ′ ) = (s -s ′ )τ (s) + O(|s -s ′ | 2 ),
where τ (s) is the unit tangent vector at p(s) ∈ ∂Ω. Using the Frenet-Serret formulas, we also have

n(s) -n(s ′ ) = -(s -s ′ )κ(s)τ (s) + O(|s -s ′ | 2 ),
where κ is the curvature of Γ i . Gathering all the terms, we infer that

x -y = (s -s ′ )(1 + κ(s)r)τ (s) + (r -r ′ )n(s) + O(|s -s ′ | 2 + |r -r ′ | 2 ), (3.13)
so that |x -y| 2 = (s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 + O |s -s ′ | 3 + |r -r ′ | 3 (3.14)
and

|x -y| -3 = (s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 -3/2 1 + O(|s -s ′ | + |r -r ′ |) .
In particular, there exists a constant C such that • We replace the factor ψ 1 (s ′ ) by ψ 1 (s); this also leads to an error term of the type (3.19).

1 |x -y| 3 ≤ C ((s -s ′ ) 2 + (r -r ′ ) 2 ) 3/2 , and replacing |x -y| -3 by (s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 -3/
• Using (3.13), we infer that

n(s) • (x -y) = r -r ′ + O(|s -s ′ | 2 + |r -r ′ | 2 ).
The second term in the right-hand side of the above equality gives rise to an error term of the type (3.21).

• The last preliminary step is to replace the indicator function 1 |x-y|≤δ/2 in (3.12) by a quantity depending on s, s ′ , r, r ′ . Using the asymptotic development (3.14) above, it can be easily proved that there exists a constant c such that

1 |x-y|≤δ/2 -1 (s-s ′ ) 2 (1+κ(s)r) 2 +(r-r ′ ) 2 ≤δ 2 /4 ≤ 1 (1-cδ) δ 2 ≤|x-y|≤(1+cδ) δ 2 .
Therefore the substitution between the two indicator functions yields an error term bounded by

N i=1 T i δ ×T i δ dx dy |x -y| 2 |∇χ k (x)|1 (1-cδ) δ 2 ≤|x-y|≤(1+cδ) δ 2 ≤ N i=1 T i δ |∇χ k (x)| dx R 2 dz |z| 2 1 (1-cδ) δ 2 ≤|z|≤(1+cδ) δ 2 ≤ C ln 1 + cδ 1 -cδ ≤ Cδ.
As a consequence, at this stage, we have proved that

I δ k = N i=1 δ 0 δ -δ (0,L i ) 2 ψ 0 (s)ψ 1 (s) √ r(1 + rκ(s)) 2 ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 × (3.15) × 1 r ′ >0 √ r ′ (χ(kr) -χ(kr ′ )) - 1 √ r kχ ′ (kr)(r -r ′ )1 (s-s ′ ) 2 (1+κr) 2 +(r-r ′ ) 2 ≤δ 2 /4 ds ds ′ dr ′ dr + O ln k √ k + δ| ln δ| .
We now evaluate the right-hand side of the above identity. We first prove that the term involving χ ′ (kr) does not contribute to the limit, due to symmetry properties of the integral. This was expected, since this term had a vanishing integral in the beginning; its only role was to ensure the convergence of I k . We then focus on the term involving χ(kr)χ(kr ′ ), and we prove that its asymptotic value is independent of χ.

First, since the integral (3.15) is convergent, we have

(3.15) = N i=1 lim ǫ→0 δ 0 δ -δ (0,L i ) 2 1 |r-r ′ |≥ǫ • • •
Therefore, up to the introduction of a truncation, we can separate the two terms of (3.15). We have in particular

δ -δ ψ 0 (s)ψ 1 (s)1 |r-r ′ |≥ǫ (1 + rκ(s)) 2 ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 kχ ′ (kr)(r -r ′ )1 (s-s ′ ) 2 (1+κr) 2 +(r-r ′ ) 2 ≤δ 2 /4 dr ′ = - δ-r -δ-r ψ 0 (s)ψ 1 (s)1 |ξ|≥ǫ (1 + rκ(s)) 2 ((s -s ′ ) 2 (1 + κ(s)r) 2 + ξ 2 ) 3/2 kχ ′ (kr)ξ 1 (s-s ′ ) 2 (1+κr) 2 +ξ 2 ≤δ 2 /4 dξ. (3.16)
Notice that the above integral only bears on the values of ξ such that |ξ| ≤ δ/2. On the other hand, for all r such that χ ′ (kr) = 0, we have

1/k ≤ r ≤ 2/k, so that if k > 4/δ, δ -r ≥ δ - 2 k > δ 2 , -δ -r < -δ < - δ 2 .
Hence the integral (3.16) is in fact equal to

ψ 0 (s)ψ 1 (s)(1 + rκ(s)) 2 kχ ′ (kr) δ/2 -δ/2 ξ1 |ξ|≥ǫ 1 (s-s ′ ) 2 (1+κr) 2 +ξ 2 ≤δ 2 /4 ((s -s ′ ) 2 (1 + κ(s)r) 2 + ξ 2 ) 3/2 dξ.
Since the integrand is odd in ξ, the integral is identically zero for all ǫ > 0 and for all δ, k such that kδ > 4.

There remains to investigate the first term in (3.15), namely

(0,δ) 2 (0,L i ) 2 ψ 0 (s)ψ 1 (s)(1 + rκ(s)) 2 (χ(kr) -χ(kr ′ )) ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 r r ′ 1 |r-r ′ |≥ǫ ds ds ′ dr ′ dr. (3.17)
We first symmetrize the integral by exchanging the roles of r and r ′ . We have

r r ′ (1 + rκ(s)) 2 - r ′ r (1 + r ′ κ(s)) 2 = r -r ′ √ rr ′ (1 + rκ(s)) 2 + O r ′ r |r -r ′ | , 1 ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 - 1 ((s -s ′ ) 2 (1 + κ(s)r ′ ) 2 + (r -r ′ ) 2 ) 3/2 = O |r -r ′ | ((s -s ′ ) 2 + (r -r ′ ) 2 ) 3/2 .
The term

(0,δ) 2 (0,L i ) 2 ψ 0 (s)ψ 1 (s)(χ(kr) -χ(kr ′ )) r ′ r |r -r ′ | ((s -s ′ ) 2 + (r -r ′ ) 2 )
(3.17) = 1 2 (0,δ) 2 (0,L i ) 2 ψ 0 (s)ψ 1 (s)(1 + rκ(s)) 2 (χ(kr) -χ(kr ′ )) ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 r -r ′ √ rr ′ ds ds ′ dr ′ dr (3.18) +O ln k √ k .
We now compute the integral with respect to s ′ ∈ (0, L i ). Setting

φ(ξ) := ξ 0 dz (1 + z 2 ) 3/2 ,
we have

L i 0 ds ′ ((s -s ′ ) 2 (1 + κ(s)r) 2 + (r -r ′ ) 2 ) 3/2 = 1 (r -r ′ ) 2 (1 + κ(s)r) φ (L i -s)(1 + κ(s)r) |r -r ′ | + φ s(1 + κ(s)r) |r -r ′ | .
Inserting this formula into the integral above and changing variables, we obtain

(3.18) = 1 2 (0,kδ) 2 L i 0 ψ 0 (s)ψ 1 (s) χ(ρ) -χ(ρ ′ ) (ρ -ρ ′ ) √ ρρ ′ 1 + κ(s) ρ k × × φ k(L i -s)(1 + κ(s)ρ/k) |ρ -ρ ′ | + φ ks(1 + κ(s)ρ/k) |ρ -ρ ′ | ds dρ dρ ′ . Since ∞ 0 ∞ 0 χ(ρ) -χ(ρ ′ ) (ρ -ρ ′ ) √ ρρ ′ dρ dρ ′ < ∞, and 0 ≤ φ(ξ) ≤ φ(+∞) < ∞ ∀ξ > 0,
using Lebesgue's theorem, we infer that for all δ > 0, lim k→∞ (3.18) = φ(+∞)

L i 0 ψ 0 (s)ψ 1 (s)ds ∞ 0 ∞ 0 χ(ρ) -χ(ρ ′ ) (ρ -ρ ′ ) √ ρρ ′ dρ dρ ′ .
There only remains to prove that the integral involving χ is in fact independent of χ. We have

I 0 := ∞ 0 ∞ 0 χ(ρ) -χ(ρ ′ ) (ρ -ρ ′ ) √ ρρ ′ dρ dρ ′ = ∞ 0 ∞ 0 1 0 χ ′ (τ ρ + (1 -τ )ρ ′ ) dτ dρ dρ ′ √ ρρ ′ = z=ρ ′ /ρ ∞ 0 ∞ 0 1 0 χ ′ (ρ(τ + (1 -τ )z)) dτ dρ dz √ z = ∞ 0 1 0 1 τ + (1 -τ )z dτ dz √ z = ∞ 0 ln z (z -1) √ z dz.
Gathering all the terms, we infer that lim k→∞

I δ k = I 0 φ(+∞) N i=1 L i 0 ψ 0 ψ 1 + O (δ| ln δ|) .
Passing to the limit as δ → 0, we obtain the result announced in Lemma 13.

Evaluation of remainder terms in the integral I k

⊲ We start with the proof of Lemma 12. The idea is to divide R 2 ×R 2 \T δ ×T δ into subdomains and to evaluate the contribution of every subdomain.

• For (x, y) ∈ (T c δ ∩ Ω) 2 , χ k (x) = χ k (y) = 1 and ∇χ k (x) = 0 for k > δ -1 , so that the contribution of T c δ ∩ Ω × T c δ ∩ Ω is zero. • For x ∈ T c δ ∩ Ω c , u 0 (x) = 0: the contribution of T c δ ∩ Ω c × R 2 is zero. • For x ∈ T c δ ∩ Ω, y ∈ T c δ ∩ Ω c , u0 (
y) = 0 and ∇χ k (x) = 0, so that the contribution of

T c δ ∩ Ω × T c δ ∩ Ω c is zero. • For x ∈ T c
δ ∩Ω, y ∈ T δ , χ k (x) = 1 and ∇χ k (x) = 0, so that the contribution of the sub-domain is

T c δ ∩Ω×T δ u 0 (x) |x -y| 3 u0 (y)(χ k (x) -χ k (y)) -u0 (x)∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy ≤ C T c δ ∩Ω×T δ 1 |x -y| 3 | u0 (y)|(1 -χ k (y))dx dy. Since 1 -χ k is supported in Ω c ∪ {d(y, ∂Ω) ≤ 2/k} and | u0 (y)| ≤ C 1 Ω √ d(y,∂Ω)
, the right-hand side is bounded by

C |δ -2 k | 3 2/k 0 dr √ r ≤ C δ √ k .
• For x ∈ T δ , y ∈ T c δ , the integral

T δ ×T c δ u 0 (x) u0 (x) |x -y| 3 ∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy is in fact supported in ({d(x, ∂Ω) ≤ 2/k} × {d(y, ∂Ω) > δ}) ∩ {|x -y| ≤ δ/2}.
It is easily seen that for k large enough (say k > 5/δ) this set is empty, and therefore the integral is zero.

• There only remains

T δ ×T c δ u 0 (x) |x -y| 3 u0 (y)(χ k (x) -χ k (y)) dx dy.
Using the same kind of estimates as for the domain T c δ ∩ Ω × T δ , it can be proved that this term is bounded by C δ k -3/2 .

⊲ We now estimate the remainder terms on T δ × T δ . We use the following lemma: Lemma 14. Let k ≥ 1 and δ > 0 such that kδ > 5.

Then the following estimates hold: for

1 ≤ i ≤ N , T i δ ×T i δ 1 r,r ′ >0 √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| 2 |χ k (x) -χ k (y)| dx dy = O ln k √ k , (3.19)
T i δ ×T i δ √ r + |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy = O √ δ + ln k √ k , (3.20)
T i δ ×T i δ |∇χ k (x)| |x -y| dx dy = O(δ| ln δ|), (3.21)
T i δ ×T i δ 1 r>0,r ′ <0 √ r |x -y| 3 |χ k (x) -χ k (y)| dxdy = O 1 k 1/2 . (3.22)
Proof. Throughout the proof, we use the following facts:

• The jacobian of the change of variables (x, y) → (s, r, s ′ , r ′ ) is

(1 + rκ(s)) 1 + r ′ κ(s ′ ) ≤ C;
• Using the expansion (3.14), we infer that there exists a constant C such that

1 |x -y| ≤ C ((s -s ′ ) 2 + (r -r ′ ) 2 ) 1/2 .
• For all α ≥ 1, there exists a constant C α such that for all s ∈ (0, L i ), for all r, r ′ ∈ (-δ, δ)

with r = r ′ ,

L i 0 ds ′ ((s -s ′ ) 2 + (r -r ′ ) 2 ) α/2 ≤ C α |r -r ′ | 1-α if α > 1, | ln |r -r ′ | | if α = 1.
(3.23) Indeed, for α > 1

L i 0 ds ′ ((s -s ′ ) 2 + (r -r ′ ) 2 ) α/2 ≤ R dz (z 2 + (r -r ′ ) 2 ) α/2 ≤ ξ=z/|r-r ′ | |r -r ′ | 1-α R dξ (1 + ξ 2 ) α/2 , while for α = 1 L i 0 ds ′ ((s -s ′ ) 2 + (r -r ′ ) 2 ) 1/2 = L i -s |r-r ′ | -s |r-r ′ | dξ (1 + ξ 2 ) 1/2 ≤ L i |r-r ′ | - L i |r-r ′ | dξ (1 + ξ 2 ) 1/2 ≤ 2 ln L i |r -r ′ | .
We now tackle the proof of (3.19)-(3.22).

1. Proof of (3.19): we split the domain (T δ × T δ ) ∩ {r, r ′ ≥ 0} into four subdomains:

• 0 ≤ r, r ′ ≤ 3/k; • 0 ≤ r ′ ≤ 3/k and 3/k < r < δ; • 0 ≤ r ≤ 3/k and 3/k < r ′ < δ; • 3/k < r, r ′ < δ.
Since χ k (x) = 1 for r > 2/k, it is easily seen that χ k (x)χ k (y) = 0 on the last subdomain. Moreover, we obviously have

s,s ′ ∈Γ i 0≤r≤3/k, 3/k<r ′ <δ √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| 2 |χ k (x) -χ k (y)| dx dy ≤ s,s ′ ∈Γ i 0≤r ′ ≤3/k, 3/k<r<δ √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| 2 |χ k (x) -χ k (y)| dx dy.
Therefore we focus on the estimates of the integral on the first two subdomains above. First, since χ k is a Lispchitz function with a Lipschitz constant of order k, we have

s,s ′ ∈Γ i 0≤r,r ′ ≤3/k √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| 2 |χ k (x) -χ k (y)| dx dy ≤ Ck s,s ′ ∈Γ i 0≤r,r ′ ≤3/k √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| dx dy ≤ Ck 0≤r,r ′ ≤3/k (0,L i ) 2 √ r √ r ′ ( √ r + √ r ′ ) 1 ((s -s ′ ) 2 + (r -r ′ ) 2 ) 1/2 ds ds ′ dr dr ′ ≤ Ck 0≤r,r ′ ≤3/k √ r √ r ′ ( √ r + √ r ′ ) | ln |r -r ′ | | dr dr ′ ≤ Ck 3/k 0 dr ′ √ r ′ 3/k -3/k | ln |z| | dz ≤ Ck × 1 √ k × ln k k .
As for the second subdomain, since χ k (x) = 1 for r > 2/k, in fact, the domain of integration in r ′ is only r ′ < 2/k. Since χ k is bounded by 1, we have

s,s ′ ∈Γ i 0≤r ′ ≤3/k, 3/k<r<δ √ r √ r ′ ( √ r + √ r ′ ) 1 |x -y| 2 |χ k (x) -χ k (y)| dx dy ≤ C 0≤r ′ ≤2/k, 3/k<r<δ (0,L i ) 2 √ r √ r ′ ( √ r + √ r ′ ) 1 (s -s ′ ) 2 + (r -r ′ ) 2 ds ds ′ dr dr ′ ≤ C 0≤r ′ ≤2/k, 3/k<r<δ √ r √ r ′ ( √ r + √ r ′ ) 1 |r -r ′ | dr dr ′ ≤ C 2/k 0 dr ′ √ r ′ δ 1/k dz z ≤ C ln k √ k .
2. Proof of (3.20): we use the same type of domain decomposition as above; the only difference lies in the fact that r ′ may take negative values.

• On the subdomain r ≥ 3/k, r ′ ≥ 3/k, the integral is identically zero;

• On the subdomain 3/k < r < δ, -δ < r ′ < 3/k, we have ∇χ k (x) = 0 and as soon as

r ′ ≥ 2/k, χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 = 0. Therefore 3/k<r<δ,-δ<r ′ <3/k, s,s ′ ∈Γ i √ r + |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy ≤ C 3/k<r<δ,-δ<r ′ <2/k (0,L i ) 2 √ r + ((s -s ′ ) 2 + (r -r ′ ) 2 ) 3/2 ds ds ′ dr dr ′ ≤ C 3/k<r<δ,-δ<r ′ <2/k √ r |r -r ′ | 2 dr dr ′ ≤ C δ 3/k √ r r -2 k dr.

Using the inequality

√ r ≤ r - 2 k + 2 k ,
we infer eventually that the integral is bounded by

C √ δ + ln k √ k . • On the subdomain 0 ≤ r ≤ 3/k, |r ′ | ≥ 3/k, we use the bound |χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 | ≤ Ck|x -y|, so that 0≤r≤3/k,3/k<|r ′ |<δ, s,s ′ ∈Γ i √ r |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy ≤ Ck 0≤r≤3/k,3/k<|r ′ |<δ √ r |r -r ′ | dr dr ′ ≤ -Ck 3/k 0 √ r ln 3 k -r dr ≤ Ck × 1 √ k × ln k k .
• On the subdomain 0 ≤ r ≤ 3/k, |r ′ | ≤ 3/k, we use a Taylor-Lagrange expansion for the function χ k , which yields

|χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 | ≤ Ck 2 |x -y| 2 , so that 0≤r≤3/k,|r ′ |≤3/k, s,s ′ ∈Γ i √ r + |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy ≤ Ck 2 0≤r≤3/k,|r ′ |≤3/k √ r ln |r -r ′ | dr dr ′ ≤ Ck 2 ln k k k -3/2 .
3. Proof of (3.21): this term is easier to estimate than (3.19), (3.20). We merely have

T i δ ×T i δ |∇χ k (x)| |x -y| dx dy ≤ C (-δ,δ) 2 k|χ ′ (kr)| ln |r -r ′ | dr dr ′ ≤ C 2δ -2δ |ln |z| | dz ≤ Cδ| ln δ|.
4. Proof of (3.22): since χ k (y) = 0 if r ′ < 0, we have

T i δ ×T i δ 1 r>0,r ′ <0 √ r |x -y| 3 |χ k (x) -χ k (y)| dxdy = T i δ ×T i δ 1 r>0,r ′ <0 √ r |x -y| 3 χ k (x) dxdy ≤ C 0 -δ δ 0 √ r |r -r ′ | 2 χ(kr) dr dr ′ ≤ C 2/k 0 √ r 1 r - 1 r + δ dr ≤ C √ k .
We now address the rest of the proof of Theorem 2. With the help of Lemma 14, the estimation of the remainder terms in I k is immediate. We set

w 0 (x) := 1 r>0 ψ 1 (s) √ r , w 1 (x) := √ r + ψ 2 (s) + u 2 (x),
so that u0 = w 0 + w 1 . We write

u0 (y)(χ k (x) -χ k (y)) -u0 (x)∇χ k (x) • (x -y)1 |x-y|≤δ/2 = u0 (x) χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 +( u0 (y) -u0 (x))(χ k (x) -χ k (y)).
There exists a constant C such that

|u 1 (x) u0 (x)| + | √ r + ψ 0 (s)w 1 (x)| ≤ C √ r + ∀x ∈ R 2 , so that T δ ×T δ u 1 (x) u0 (x) |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy + T δ ×T δ √ r + ψ 0 (s)w 1 (x) |x -y| 3 χ k (x) -χ k (y) -∇χ k (x) • (x -y)1 |x-y|≤δ/2 dx dy ≤ (3.20).
On the other hand, simple calculations show that

| u0 (x) -u0 (y)| ≤ C|x -y| 1 + 1 r,r ′ >0 √ rr ′ ( √ r + √ r ′ ) +C1 rr ′ <0 1 √ r + + r ′ + , |w 1 (x) -w 1 (y)| ≤ C|x -y| 1 + 1 r,r ′ >0 √ r + √ r ′ +C1 rr ′ <0 ( √ r + + r ′ + ).
As a consequence,

|u 1 (x)|| u0 (x) -u0 (y)| ≤ C|x -y| r + + 1 r,r ′ >0 √ r √ r ′ ( √ r + √ r ′ ) + C1 r>0,r ′ <0 √ r, | √ r + ψ 0 (s)| |w 1 (x) -w 1 (y)| ≤ C|x -y| √ r + + 1 r,r ′ >0 √ r √ r + √ r ′ + C1 r>0,r ′ <0 r, so that T δ ×T δ u 1 (x) |x -y| 3 ( u0 (x) -u0 (y))(χ k (x) -χ k (y)) dx dy + T δ ×T δ √ r + ψ 0 (s) |x -y| 3 (|w 1 (x) -w 1 (y))(χ k (x) -χ k (y)) dx dy ≤ (3.19) + (3.22).
Gathering all the terms, we deduce that lim δ→0 lim k→∞

I k = C 2 N i=1 L i 0 ψ 0 ψ 1 ,
where C 2 is an explicit positive constant. Moreover, using formulas (3.6), (3.7), we have

ψ 0 (s)ψ 1 (s) = - 1 2 (∂ 1/2 n u 0 (s)) 2 ζ • n(s),
where ζ(x) = φ0 (x). Eventually, we deduce that

dJ f (Ω t ) dt |t=0 = - C 1 C 2 4 N i=1 L i 0 (∂ 1/2 n u 0 (s)) 2 ζ • n(s)ds = - C 1 C 2 4 ∂Ω (∂ 1/2 n u 0 (s)) 2 ζ • n(s)dσ(s).
Therefore Theorem 1 is proved, with C 0 := -(C 1 C 2 )/4. Notice that C 0 does not depend on Ω.

3.5 Proofs of Lemma 11 and formulas (3.4), (3.5)

• The proof of (3.3) follows closely the one of Theorem 5.3.2 in [START_REF] Henrot | Variation et optimisation de formes[END_REF]. The only differences come from the mixed boundary conditions and the fact that φ t only affects horizontal variables.

The key point is to prove that V t is differentiable with respect to t with values in H 1 (R 3 + ). To that end, observe that V t is the solution of the elliptic problem

-div (A t ∇V t ) = 0 in R 3 + , ∂ z V t = f • φ t on Ω × {0}, V t = 0 on Ω c × {0}, (3.24)
where

A t = det(∇φ t ) (∇φ t ) -1 (∇φ T t ) -1 0 0 1 .
(Notice that ∇φ t is a 2 × 2 matrix.)

Indeed, (3.24) is easily proved by writing the variational formulation associated with the equation on U t and performing changes of variables. Since the latter are strictly identical to the ones of [START_REF] Henrot | Variation et optimisation de formes[END_REF], we skip the proof.

For further reference, we also write the system derived by lifting the Neumann boundary condition. We set

V t (x, z) = Ṽt (x, z) + f (φ t (x))η(φ t (x), z), where η ∈ C ∞ 0 (R 3 ) is such that η(x, z) = z for x in a neighbourhood of Ω and |z| ≤ 1. Then Ṽt solves -div (A t ∇ Ṽt ) = | det(∇φ t )|(∆(f η))(φ t (x), z) on R 3 + , ∂ z Ṽt = 0 on Ω × {0}, Ṽt = 0 on Ω c × {0}. (3.25) Let V := V ∈ L 2 loc (R 3 + ), ∇V ∈ L 2 (R 3 +
), V = 0 on Ω c × {0} . Then according to the Hardy inequality in R 3 , there exists a constant C H such that for all

V ∈ V, V (1 + |x| 2 + |z| 2 ) -1/2 ∈ L 2 (R 3 + ) and R 3 + |V (x, z)| 2 1 + |x| 2 + |z| 2 dx dz ≤ C H R 3 + |∇V | 2 .
This inequality is usually stated in the whole space, but a simple symmetry argument shows that it remains true in the half-space. Therefore ∇V L 2 is a norm on the Hilbert space V. Now, for t in a neighbourhood of zero and Ṽ ∈ V, define the linear form

F (t, Ṽ ) ∈ V ′ by ∀W ∈ V, W, F (t, Ṽ ) = R 3 + (A t ∇ Ṽ ) • ∇W - R 3 + | det(∇φ t )|(∆(f η))(φ t (x, z))W (x, z) dx dz.
Notice that F (t, Ṽ ) = 0 is the variational formulation associated with the equation (3.25). We then claim that the operator

F : (t, Ṽ ) ∈ R × V → F (t, Ṽ ) ∈ V ′ has C 1 regularity for t small enough. Indeed, t → A t ∈ L ∞ (R 2 , M 3 ) is C ∞ for t in a neighbourhood of zero. On the other hand, for (A, Ṽ ) ∈ L ∞ (R 2 , M 3 ) × V, W , let a(A, Ṽ ) : W ∈ V → R 3 + (A∇ Ṽ ) • ∇W - R 3 + | det(∇φ t )|(∆(f η))(φ t (x, z))W (x, z) dx dz. Since vt ∈ C([-δ, δ], H 1/2 (R 2 )) for δ > 0 small enough, the first term is O(t) in L 2 (R 2 ).
As for the second one, if v 0 were smooth, say v 0 ∈ H 1 (R 2 ), then we could write

d dt v 0 • φ -1 t = dφ -1 t dt • ∇v 0 • φ -1 t .
It can be easily checked that the right-hand side is bounded in

H -1/2 by C v 0 H 1/2 , where C is a constant depending only on φ t . Therefore if v 0 ∈ H 1 , t > 0, v 0 • φ -1 t -v 0 H -1/2 (R 2 ) ≤ Ct v 0 H 1/2 (R 2 ) .
By density, this inequality remains true for all v 0 ∈ H 1/2 . We conclude that ut belongs to

C([-δ, δ], H -1/2 (R 2 )).
• We end this section by proving the asymptotic expansions (3.4), (3.5). We apply the results of Chapter C in [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF] to the function V t . We start by extending V t on R 3 \ Ω × {0} by setting

V t (x, z) = -V t (x, -z) x ∈ R 2 , z < 0.
Without any loss of generality, we assume that the function η ∈ C ∞ 0 (R 3 ) defined in the proof of Lemma 11 is odd with respect to z.

Then the extended function V t satisfies

V t = Ṽt + f (φ t (x))η(φ t (x), z), -div (A t ∇ Ṽt ) = | det(∇φ t )|∆(f η)(φ t (x), z) on R 3 \ Ω × {0} ∂ z Ṽt = 0 on Ω × {0}.
By elliptic regularity, Ṽt ∈ C ∞ (K) for any compact set K ⊂ R 3 \Ω×{0} such that K ∩∂Ω = ∅. For every connected component Γ i of ∂Ω, we introduce a truncation function

ϕ i ∈ C ∞ 0 (R 2 ) such that ϕ i ≡ 1 on a neighbourhood of Γ i , Supp ϕ i ∩ (∂Ω \ Γ i ) = ∅.
Since the support of ∇ϕ i is separated from the zones where Ṽt has singularities, we infer that for all t, ϕ i (x) Ṽt (x, z) solves a boundary value problem which is elliptic of order two in the sense of Agmon, Douglis, Nirenberg (see [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF], [START_REF]Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF] and the definitions in chapter 7 of [START_REF] Dauge | Elliptic boundary value problems on corner domains[END_REF]), with homogeneous boundary conditions of order one and with C ∞ right-hand side. Moreover, we have proved in Lemma 11 that

V t ∈ H 1 loc (R 3 \ Ω × {0}
). Therefore we can apply Corollary C.6.5 in [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF]. As in paragraph 2.2, we denote by (r, θ) polar coordinates in the planes normal to Γ i and centered on Γ i , and by s the arc-length on Γ i , so that

R 3 \ (Supp ϕ i ∩ Ω × {0}) = W π = {(s, r, θ), s ∈ (0, L i ), r > 0, θ ∈ (-π, π)}.
We deduce that for all i, there exist functions ψ0 i (t; s, θ), ψ1 i (t; s, θ), which are smooth with respect to s, θ ∈ [-π, π], such that

ϕ i (x) Ṽt (x, z) = ψ0 i (t; s, θ)r
+ η(φ t (x), z), we have V 1 i ∈ C 1 (W π ), V 1 i (t; x, z) = O(r) for all t, and V t (x, z) = ψ0 (t; s, θ)r 1/2 + V 1 i (t; x, z) for x in a neighbourhood of Γ i . Since u 0 (x) = lim z→0,z>0 V 0 (x, z),
we obtain decomposition (3.4) with

ψ 0 (s) := ψ0 i (0, s, π) for s ∈ Γ i , u 1 (x) = lim z→0 + V 1 i (0; x, z) for x ∈ Supp ϕ i .
Differentiating (3.5) requires regularity results with respect to t on the terms of the decomposition (3.27). Using for instance Theorem C.6.2 in [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF], or looking precisely at the details of the proof in section C of [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF], it is easily proved that the functions ψ0 , ψ1 , u reg and u rem are differentiable with respect to t, and that their t-derivatives are smooth with respect to x.

Denoting by s(x), r(x) the local coordinates of a point x ∈ R 2 , we have

u t (x) = ψ0 (t, s(φ -1 t (x)), π)(r(φ -1 t (x))) 1/2 + V 1 (t, φ -1 t (x), 0 + ) so that ut = ∂ t ψ0 + ∂ t (φ -1 t (x)) • ∇s(φ -1 t (x))∂ s ψ0 (t, s(φ -1 t (x)), π)(r(φ -1 t (x))) 1/2 + 1 2 ∂ t (φ -1 t (x)) • ∇r(φ -1 t (x)) 1 r(φ -1 t (x)) 1/2 (3.28) +∂ t V 1 (t, φ -1 t (x), 0 + ) + ∂ t (φ -1 t (x)) • ∇V 1 (t, φ -1 t (x), 0 + ). Therefore ut ∈ L ∞ t (L 1
x ), which proves (3.8). We recall that φ 0 = φ -1 0 = Id, and that ζ = φ0 . Notice also that since n is the outward pointing normal, ∇r = -n. We infer

u0 = ζ • n 2 √ r + ψ 2 (s) √ r + u 2 (x),
where

ψ 2 (s) = ∂ t ψ0 + ζ • τ ∂ s ψ0 (0, s, π), u 2 (x) = ∂ t V 1 (0, x, 0 + ) + ζ • ∇V 1 (0, x, 0 + ).
Thus (3.5) is proved.

Radial symmetry by the moving plane method

The aim of this section is to prove Theorem 3. This theorem extends the well-known theorem of Serrin [START_REF] Serrin | A symmetry problem in potential theory[END_REF] on the classical laplacian, for which (1.5) is replaced by

     -∆u = 1, x ∈ Ω, u = 0, x ∈ ∂Ω, ∂ n u = c 0 , x ∈ ∂Ω (4.1)
The proof of Serrin uses the celebrated moving plane method, and we will adapt it to our fractional setting.

Reminders on the moving plane method

We remind here the main arguments of Serrin's proof. The starting point is the introduction of a family of hyperplanes (lines in our 2d case), say H λ := {x ∈ R 2 , x 1 = λ}, parametrized by λ ∈ R. We also define, for any function u defined on R 2 , R λ u(x 1 , x 2 ) := u(2λx 1 , x 2 ) the reflection of u with respect to H λ . For λ small enough, H λ does not intersect the domain Ω. Increasing λ, that is moving H λ from left to right, one reaches a first contact position, corresponding to

λ 0 := inf{λ, H λ ∩ Ω = ∅}.
Up to a translation, we can always assume that λ 0 = 0. Thus, for λ > 0, we can consider the cap Σ λ := Ω ∩ {x 1 < λ} and its reflection Σ ′ λ with respect to H λ . Note that, for λ > 0 small enough, Σ ′ λ is non-empty and included in Ω. As λ increases, it remains included in Ω at least until one of the following two geometric configurations is reached (see Figure 1):

1. Σ ′ λ is internally tangent to the boundary of Ω at some point P not on H λ .

2. H λ is orthogonal to the boundary of Ω at some point Q.

Let Λ be the first value of λ for which configuration 1 or 2 holds. Note that we may have Σ ′ λ ⊂ Ω for some λ > Λ, but this will be irrelevant in the proof. The main point in Serrin's proof is to show that one has R Λ u = u inside Σ Λ , where u is the solution of (4.1). This fact yields easily the symmetry of Ω with respect to H Λ . As the direction of H Λ was chosen arbitrarily, it will follow that for any direction, there is an axis of symmetry for Ω with that direction. This property implies that Ω is a disk.

To show the identity R Λ u = u inside Σ Λ , one argues by contradiction through three main steps. Assume that the equality does not hold. Then

• Step 1. One shows that the function w := u -R Λ u is positive inside Σ ′ Λ .
• Step 2. One obtains an upper bound for w near the tangency point P (configuration 1) or Q (configuration 2). As the normal derivative of u is constant along ∂Ω, one has w = ∂ n w = 0 at P or Q. For configuration 2, one can show furthermore that the second derivatives cancel at Q: ∂ ij w = 0 for all i, j = 1, 2. Denoting by r the distance at the tangency point, these properties imply that

w = O(r 2 ) at P (configuration 1), or w = O(r 3 ) at Q (configuration 2). (4.2)
• Step 3. One shows a lower bound which contradicts (4.2). For configuration 1, this lower bound is obtained easily. Indeed, one knows from Step 1 that w > 0 inside Σ ′ Λ . Hence, Hopf's lemma implies ∂ n w > 0 at P (where n is the inward normal), which shows that w should grow at least linearly with r inside the domain. For configuration 2, Σ ′ Λ is not regular enough at Q to apply Hopf's lemma. However, one can still show that Lemma 1]. It follows that w should grow at least like r 2 inside the domain. In both cases, we reach the targeted contradiction. We refer to [START_REF] Serrin | A symmetry problem in potential theory[END_REF] for all details.

∂ s w > 0 or ∂ 2 s w > 0 at Q for any direction s entering Σ ′ Λ non-tangentially: see [14,
Our ambition is to transpose this scheme of proof to problem (1.5). Due to the nonlocal character of (-∆) 1/2 , it requires many modifications:

• In the case of the Laplacian, Step 1 follows from a simple application of the strong maximum principle, as w ≥ 0 on ∂Σ ′ Λ . In our fractional (and therefore nonlocal) setting, such use of the maximum principle is impossible: it would require that w ≥ 0 on the whole R 2 \ Σ ′ Λ , which is not true (w is an odd function). The appropriate treatment of Step 1 will be addressed in paragraph 4.2.

• In the case of the Laplacian, Steps 2 and 3 rely on the regularity of u up to the boundary. In the case of (-∆) 1/2 , we only have C 0,1/2 regularity of u, which implies substantial changes. Loosely:

the upper bounds will follow from the asymptotic expansion (2.8) of u near the boundary.

the lower bounds will follow from the construction of refined subsolutions.

Additionally, we emphasize that the constant c 0 in (1.5) is necessarily strictly positive. Indeed, c 0 ≥ 0 by the maximum principle. The fact that c 0 > 0 is a consequence of the Hopf Lemma for the fractional laplacian. In the present context, we may present a self-contained proof: if c 0 = 0, we jump to paragraph 4.3. The arguments developed there (for the function w instead of u) allow us to conclude that u ≡ 0. This is excluded by the equation (-∆) 1/2 u = 1. Therefore we restrict our analysis to c 0 > 0.

Positivity of w

We shall first prove that if non-identically 0, w := R Λ uu is positive inside Σ Λ (which amounts to achieving Step 1). We shall rely on ideas developed by M. Birkner, J. Lopez-Mimbela and A. Wakolbinger in article [START_REF] Birkner | Comparison results and steady states for the Fujita equation with fractional Laplacian[END_REF]. They show there the radial symmetry of solutions of some semilinear fractional problems

(-∆) α u + F (u) = 0, x ∈ B(0, 1), u = 0 in R n \ B(0, 1)
set in the unit ball B(0, 1) of R n . Their proof is based on an adaptation of the moving plane method, and its baseline can be used to show the positivity of w. Nevertheless, several changes are needed, and simplifications of the original arguments can be made, as we now describe.

First of all, we introduce the set

L := {λ ∈]0, Λ[, s.t. ∀0 < γ ≤ λ, R γ u -u ≥ 0 on Σ γ , and ∂ 1 u > 0 on H γ ∩ Ω}.
The main point is to show that L =]0, Λ[. We proceed in several steps:

• Statement 1: ]0, ǫ] ⊂ L for ǫ > 0 small enough. To prove that ]0, ǫ[⊂ L, it is enough to show that ∂ 1 u > 0 for all x ∈ Ω with 0 < x 1 < ǫ. From the condition ∂ 1/2 n u = c 0 > 0,
and the expansion (2.7), we know that

∂ r u = 1 2 c 0 r -1/2 + O(1), ∂ s u = O(1)
as the distance r to the boundary (measured inside the domain) goes to zero. It follows that

∂ 1 u = ∂ 1 r ∂ r u + ∂ 1 s ∂ s u = 1 2 c 0 (∂ 1 r) r -1/2 + O(1), r → 0.
Moreover, for 0 < λ < Λ, for all inward normal vectors n along ∂Ω ∩ {x 1 < λ}, one has e 1 • n > 0, where e 1 = (1, 0). It follows that ∂ 1 r has a positive lower bound for 0 < x 1 < ǫ small enough. The statement follows.

• Statement 2: L is an open subset of ]0, Λ[.
The point is to show that if λ ∈ L, [λ, λ + ǫ[⊂ L for ǫ > 0 small enough. This statement will be deduced from the following:

Lemma 15. Let λ < Λ. Then i) For all x ∈ ∂Ω ∩ {x 1 < λ}, R λ u(x) -u(x) > 0. ii) Assume that R λ u -u ≥ 0 on Σ λ . Then, R λ u -u > 0 on Σ λ .
In particular, one has R λ uu > 0 on Σ λ for any λ ∈ L.

Proof. We remind that u > 0 in Ω by the maximum principle. i) As λ < Λ, for all x = (x 1 , x 2 ) ∈ ∂Ω ∩ {x 1 < λ}, x λ := (2λx 1 , x 2 ) ∈ Ω. In particular:

R λ u(x) -u(x) = u(x λ ) > 0.
ii) We assume that w := R λ uu ≥ 0 on Σ λ . Note that on {x 1 < λ} \ Σ λ , one has w = R λ u ≥ 0, so that w ≥ 0 on the whole half-space {x 1 < λ}. Note also that by i) and the continuity of u, the fonction w is not identically zero in Σ λ . We want to show that w > 0 in Σ λ . We introduce the harmonic extension W of w, defined on R 3 + . We remind that

∆ W = 0 on R 3 + , W | z=0 = w on R 2 .
Moreover, as (-∆) 1/2 u = 1 in Ω, ∂ z W | z=0 = 0 in Σ λ . We can express W with the Poisson kernel:

W (x 1 , x 2 , z) = z 2π R 2 dt 1 dt 2 ((x 1 -t 1 ) 2 + (x 2 -t 2 ) 2 + z 2 ) 1/2 w(t 1 , t 2 ).
As R λ w =w, this integral formula can be written

W (x 1 , x 2 , z) = z 2π {t 1 <λ} dt 1 dt 2 1 ((x 1 -t 1 ) 2 + (x 2 -t 2 ) 2 + z 2 ) 1/2 - 1 (x 1 + (t 1 -2λ)) 2 + (x 2 -t 2 ) 2 + z 2 ) 1/2 w(t 1 , t 2 ).
As w ≥ 0 on {t 1 < λ} and not identically zero, we deduce from this formula that W > 0 for x 1 < λ, z > 0.

To conclude on the positivity of w, we assume a contrario that w(x * ) = 0 for some x * = (x 1 , x 2 ) ∈ Σ λ . The function W is smooth up to the boundary in the vicinity of (x * 1 , x * 2 , 0), and 0 = W (x * 1 , x * 2 , 0) < W (x, z) for all (x, z) ∈ R 3 + . By Hopf's lemma, we should have ∂ z W (x * 1 , x * 2 , 0) > 0 hence reaching a contradiction. This concludes the proof.

Back to Statement 2: we argue by contradiction, as in [3, paragraph 4.2]. We assume that there is λ ∈ L and a decreasing sequence (λ n ) ⊂]0, Λ[\L converging to λ. Up to the extraction of a subsequence, we can assume that there exists a sequence (γ n ) such that λ < γ n < λ n and 1. either there exists a sequence (x n ), x n ∈ Σ γn for all n, such that x n → x * and u(x n ) > R γn u(x n ) for all n.

2. or there exists a sequence (x n ), x n ∈ H γn ∩ Ω for all n, such that x n → x * and ∂ 1 u(x n ) ≤ 0 for all n.

We first consider case 1: since λ ∈ L, we have R λ uu ≥ 0 on Σ λ . As u is continuous, we get u(x * ) -R λ u(x * ) ≥ 0. By Lemma 15, we cannot have

x * ∈ Σ λ ∪ (∂Ω ∩ {x 1 < λ}). Hence, x * ∈ H λ ∩ Ω. If x * ∈ H λ ∩ ∂Ω, one has ∂ 1/2 n u(x * ) = c 0 > 0.
Moreover, as λ < Λ, e 1 • n * > 0 for n * the inward normal vector at x * : indeed, since λ < Λ, configuration 2 has not been met, and therefore e 1 • n(y) does not vanish for all y ∈ ∂Ω such that y 1 ≤ λ. Since e 1 • n(y) = 1 when y = (0, y 2 ) ∈ ∂Ω, we infer e 1 • n * > 0.

Reasoning as in the proof of Statement 1, we obtain that u is strictly increasing with x 1 in the vicinity of x * (inside the domain). Hence, for n large enough, we get

R γn u(x n ) -u(x n ) > 0 in contradiction with the assumption u(x n ) > R γn u(x n ) for all n. If x * ∈ H λ ∩ Ω, then ∂ 1 u(x * ) > 0 (because λ ∈ L)
. This means again that u is strictly increasing with x 1 in the vicinity of x * , which yields the same contradiction as before.

It remains to consider case 2. With the same reasoning as in case 1, we get that x * ∈ H λ ∩ Ω, and that u is strictly increasing with x 1 in the vicinity of x * . This contradicts the assumption ∂ 1 u(x n ) ≤ 0 for all n.

• Statement 3: L =]0, Λ[. From the previous statements on L, we know that L =]0, Λ max [ for some Λ max ≤ Λ. Again, we shall argue by contradiction and assume that Λ max < Λ. For all λ < Λ max , one has R λ uu > 0 in Σ λ , so that by continuity of u, R Λmax uu ≥ 0 in Σ Λmax . Lemma 15 implies in turn that R Λmax uu > 0 in Σ Λmax .

We want to show that ∂ 1 u > 0 on H Λmax ∩ Ω. This would imply that Λ max ∈ L, so the contradiction.

As w := R Λmax uu is odd with respect to H Λmax , it is enough to prove that ∂ 1 w < 0 on H Λmax ∩ Ω. We introduce its harmonic extension, as in the proof of Lemma 15: it satisfies

∆ W = 0 on R 3 + , W | z=0 = w on R 2 , ∂ z W | z=0 = 0 on Σ Λmax .
We extend W to the lower half-space by setting

W (x 1 , x 2 , z) := W (x 1 , x 2 , -z) for z < 0. As ∂ z W | z=0 = 0 on Σ Λmax × {0}, this extension is harmonic through Σ Λmax , that is outside (R 2 \ Σ Λmax ) × {0}.
Notice also that as in the proof of Lemma 15, w ≥ 0 on {x 1 < Λ max }, and therefore W ≥ 0 on {x 1 < Λ max }. Let x * ∈ H Λmax ∩ Ω. W is smooth and harmonic near (x * 1 , x * 2 , 0). Moreover, one has 0 = w(x * ) = W (x * 1 , x * 2 , 0) < W (x, z), say for all (x, z) ∈ O := Σ Λmax × R Applying Hopf's lemma in O (for which (x * 1 , x * 2 , 0) is a boundary point), we obtain that

∂ 1 W (x * 1 , x * 2 , 0) < 0 that is ∂ 1 w(x * ) < 0, as expected.
As a conclusion of our analysis, we obtain that R λ uu > 0 on Σ λ for all λ < Λ. By continuity of u, we obtain that w := R Λ uu ≥ 0 on Σ Λ . From there, there are two possibilities:

• either w = 0 on Σ Λ . It follows easily that Ω is symmetric with respect to H Λ . As explained in paragraph 4.1, the direction of H Λ being arbitrary, it follows that Ω is a disk.

• or w(x) > 0 for some point x in Σ Λ . But then, following exactly the proof of point ii) in Lemma 15, we obtain that w > 0 on Σ Λ (or w := u -R Λ u > 0 on Σ ′ Λ ). The rest of the section aims at excluding the second possibility. From now on, we assume that w > 0 on Σ ′ Λ , and will establish contradictory lower and upper bounds on w. First of all, we state a lemma, to be used in both configurations 1 and 2.

Lemma 16. Let W be the harmonic extension of w to R 3 + . For all R > 0 sufficiently large there exists c R > 0 such that

W (x, z) ≥ c R (x 1 -Λ) z, for all Λ ≤ x 1 ≤ R, -R ≤ x 2 ≤ R, 0 ≤ z ≤ R.
Proof. Let R > 0 large, and

Q R :=]Λ, R[×] -R, R[×]0, R[. As W and W ′ := x → (x 1 -Λ) z
are both harmonic functions on Q R , we can use the maximum principle to compare them: we just have to show that W ≥ c R W ′ on ∂Q R for c R > 0. We take R large enough so that the vertical sides of the cube (other than the one supported by {x 1 = Λ}) do not intersect Ω × {0}. Then, we distinguish between the different sides:

• On ∂Q R ∩ {z = 0} or ∂Q R ∩ {x 1 = Λ}, one has W ≥ 0 (remind that w ≥ 0 on {x 1 ≥ Λ}), and W ′ = 0. The inequality is clear.

• Let Γ 1,R := ∂Q R ∩ {x 1 = R}. Let ǫ > 0.
As W is continous and > 0 on the compact set Γ 1,R ∩ {z ≥ ǫ}, one can find c = c ǫ > 0 such that W ≥ c W ′ . To have the same inequality on the whole Γ 1,R , it is enough to show that lim inf

(x,z)∈Γ 1,R , z→0 + W (x, z) W ′ (x, z) = 1 R -Λ lim inf (x,z)∈Γ 1,R , z→0 + W (x, z) z > 0. (4.3)
This is a consequence of Hopf's lemma: W is positive harmonic on Q R , satisfies W = 0 on Γ 1,R ∩ {z = 0}. It follows that ∂ z W > 0 on Γ 1,R ∩ {z = 0}. Note that ∂ z W exists thanks to our choice of R, away from Ω. It is furthermore continuous, so that it is bounded from below by a positive constant. Inequality (4.3) follows.

• On Γ 3,R := ∂Q R ∩ {z = R}, one can proceed exactly as in the case of Γ 1,R : use the positivity of W away from {x 1 = Λ}, and use Hopf's lemma to get

∂ 1 W ≥ c > 0 at Γ 3,R ∩ {x 1 = Λ}.
• We now turn to Γ 2,R = ∂Q R ∩ {x 2 = ±R}. Away from the edge x 1 = Λ, z = 0, we can as before use the positivity of W and Hopf's lemma to obtain W ≥ cW ′ . It then remains to handle the vicinity of X = (Λ, ±R, 0). The main point is to show that lim inf

(x,z)∈Γ 2,R (x,z)→X W (x, z) (x 1 -Λ) z > 0. ( 4.4)
We can not apply Hopf's lemma: W is harmonic in the dihedra {x 1 > Λ, z > 0}, but it does not satisfy the interior sphere condition at X. As W (x, z) = 0 both for x 1 = Λ and z = 0, one has

∂ k z W (x, z) = 0 for x 1 = Λ, ∀k ∈ N ∂ k 1 W (x, z) = 0 for z = 0, ∀k ∈ N. (4.5)
((x, z) in the vicinity of X). Hence, to prove (4.4), it is enough to show that

∂ 1 ∂ z W (x, z) ≥ c > 0, (x, z) ∈ B(X, δ) ∩ {x 1 ≥ Λ, z ≥ 0}, for small enough δ. By continuity of ∂ 1 ∂ z W , it is enough to show that ∂ 1 ∂ z W (X) > 0.
This positivity condition follows straightforwardly from a famous lemma of Serrin, see [14, Lemma 1, p308]. In our context, it reads: for any vector s entering the region {x 1 > Λ, z > 0}, we have ∂ s W (X) > 0 or ∂ 2 s W (X) > 0. Taking s = e 1 + e 3 , we obtain that

(∂ 1 + ∂ z )W (X) > 0, or (∂ 1 + ∂ z ) 2 W (X) > 0.
Combining this statement with (4.5), we see that the first condition is not realized, and that the second one amounts to ∂ 1 ∂ z W (X) > 0 as expected.

Contradictory bounds on w: configuration 1

Lower bound

We start with configuration 1, that is when Σ ′ Λ is internally tangent to ∂Ω at a point P not in H Λ . We introduce a small open disk D R ⊂ Ω of radius R, also tangent to ∂Ω at P . We take R small enough so that ∂D R ∩ ∂Ω = {P }. Let x R be the center of D R , and let φ ∈ [0, 2π] parametrizing ∂D R . Finally, for any x in D R , denote by ρ = ρ(x) the distance between x and the circle ∂D R . One has x = x R + ((Rρ) cos φ, (Rρ) sin φ). The aim of this paragraph is to prove the following

Proposition 17. Let β > 1/2. There exists c β > 0, ρ β > 0 such that for all x ∈ D R with 0 < ρ(x) < ρ β , w(x) ≥ c β ρ β .
First, we shall extend the 2d coordinate system (ρ, φ) to a 3d coordinate system (ρ, φ, θ) in the R/2-neighborhood of C R := ∂D R × {0} (see Figure 2). Any (x, z) = (x 1 , x 2 , z) in this neighborhood has a unique projection p(x, z) on C R , which reads:

p(x, z) = (x R , 0) + (R cos φ, R sin φ, 0).
We introduce ρ = |(x, z)p(x, z)| the distance between (x, z) and C R , and θ ∈ [-π, π[ the oriented angle between the vector (x R , 0)p(x, z) and (x, z)p(x, z). Then, one can write (x, z) = p(x, z) + (-ρ cos θ cos φ, -ρ sin θ sin φ, ρ sin θ) = (x R , 0) + ((Rρ cos θ) cos φ, (Rρ sin θ) sin φ, ρ sin θ)

The triplet (ρ, θ, φ) defines a system of orthogonal curvilinear coordinates in the R/2 neighborhood of C R . For θ = 0, it matches the 2d coordinates (ρ, φ) introduced above.

To prove Proposition 17, we shall rely again on the harmonic extension W of w. It satisfies

∆W = 0 on R 3 + , W | z=0 = w on R 2 , ∂ z W | z=0 = 0 on Σ ′ Λ .
We can extend W to R 3 into an even function of z (still denoted W ). Thanks to the last condition, it is harmonic through Σ ′ Λ . We will show that for β > 1 2 ,

W (x, z) ≥ c β ρ β cos(θ/2), ∀ρ = ρ(x) < δ small enough. ( 4.6)
This will yield Proposition 17 for θ = 0. This lower bound on W will follow from the maximum principle. We introduce U δ := {x, ρ(x) < δ} \ {θ = π}. The open set U δ is obtained by removing from the δ-neighborhood of C R the part of the plane {z = 0} outside D R . Let W β given in (ρ, θ, φ) coordinates by W β (x, z) = ρ β cos(θ/2). We will show that for δ >0 small enough, one has:

i) ∆W β ≥ 0 (= ∆W ) in U δ . ii) W ≥ c β W β on ∂U δ , for some c β > 0.
The lower bound (4.6) will follow. Proof of i). One can check that the Laplacian in coordinates (ρ, θ, φ)

reads ∆f = 1 ρ(R -ρ cos θ) ∂ ∂ρ ρ(R -ρ cos θ) ∂ ∂ρ f + ∂ ∂θ R -ρ cos θ ρ ∂ ∂θ f + ∂ ∂φ ρ R -ρ cos θ ∂ ∂φ f . (4.7)
This implies that

∆W β = 1 ρ(R -ρ cos θ) (β 2 -1/4)ρ β-1 (R -ρ cos θ) -βρ β cos θ cos(θ/2) - 1 2 ρ β sin θ sin(θ/2) = 1 ρ(R -ρ cos θ) (β 2 -1/4)ρ β-1 (R -ρ cos θ) -βρ β cos θ -ρ β sin 2 (θ/2) cos(θ/2) = 1 ρ(R -ρ cos θ) (β 2 -1/4) R ρ β-1 + O(ρ β ) cos(θ/2) ≥ 0
for δ (and ρ < δ) small enough. Note that we used the identity sin θ = 2 sin(θ/2) cos(θ/2) in the second line.

Proof of ii). We must distinguish between different zones.

• On ∂U δ ∩ {θ = ±π}, one has W ≥ 0, W β = 0, so that the inequality is clear.

• The compact set ∂U δ ∩{θ = 0} is included in Σ ′ Λ , so that W > 0 there, and as W is continuous, it is even bounded from below by a positive constant. In particular, one can find c > 0 such that W ≥ cW β .

• For any ǫ > 0, we also know that W > 0 on the compact set ∂U δ ∩ {|z| ≥ ǫ}. We obtain again that W ≥ cW β for some c = c ǫ > 0.

• It remains to show that for ǫ > 0 small enough, one has

W ≥ cW β on ∂U δ ∩ {0 < |z| ≤ ǫ}.
Due to the fact that W is non-negative in R 3 and even in the variable z, it is enough to show that lim inf

(x,z)∈∂U δ , z→0 + |W (x, z)/W β (x, z)| > 0.
To show such a property, we argue by contradiction. Assume that we can find a sequence (x k , z k ) satisfying:

(x k , z k ) ⊂ R 3 + ∩ ∂U δ , z k → 0 + , W (x k , z k )/W β (x k , z k ) → 0.
Up to the extraction of a subsequence, we can assume that x k , z k → X for some X ∈ ∂U δ with X = (X 1 , X 2 , 0). There are two possibilities:

either W (X) > 0. As W β is bounded, this yields easily a contradiction.

or W (X) = 0. In particular, X is outside C R , which implies in turn W β (X) = 0. More generally, W (x 1 , x 2 , 0) = 0 for all (x 1 , x 2 ) close to (X 1 , X 2 ). Moreover, as W β is smooth near X, we obtain

|W β (x, z)| = O(z), as z → 0 + , x in the vicinity of X.
From there, we get that lim sup

k→∞ |W (x k , z k )|/z k = 0 (4.8)
As x k 1 remains away from Λ, this limit contradicts Lemma 16.

Upper bound

We now turn to an upper bound for w. Let x ∈ Σ ′ Λ . For x close to P , we can write in a unique way x = x * + r n, resp. x = x * + r n, r, r > 0, where x * ∈ ∂Ω, resp. x * ∈ ∂Σ ′ Λ , and n = n(x * ), resp. n = n(x * ), refers to the inward normal to ∂Ω at x , resp. to the inward normal to ∂Σ ′ Λ at x. We now use the results of Costabel et al. [START_REF] Costabel | Asymptotics without logarithmic terms for crack problems[END_REF], and the fact that ∂

Contradictory bounds on w: configuration 2

In this paragraph, we investigate configuration 2, in which H Λ is orthogonal to ∂Ω at some point Q.

Lower bound

We shall prove the following Proposition 18. Let ν = (1, ν) be a vector entering Σ ′ Λ at Q. Let β > 1/2. There exists c β > 0 such that w(Q + t ν) ≥ c β t 1+β , for t small enough.

To prove this proposition, we consider again a small disk D R , of radius R, tangent to ∂Ω at Q. Exactly as in the proof of Proposition 17, we introduce the 3d coordinates (ρ, θ, φ) in a R/2-neighborhood of C R = ∂D R × {0}. We also introduce the 3d harmonic extension W of w, even with respect to the variable z. Our aim is to prove a lower bound on W which implies Proposition 18, namely W (x, z) ≥ c β ρ β cos(θ/2) cos φ, ∀π/2 ≤ φ ≤ π/2, ∀θ ∈ [-π, π], ∀ρ = ρ(x) < ρ β (4.10) with ρ β > 0 small enough. Therefore, let W β (x, z) = ρ β cos(θ/2) cos φ. Comparison between W and W β will come from the maximum principle. The main change with respect to configuration 1 is that W = 0 on the hyperplane {x = (x 1 , x 2 , z), x 1 = Λ} (corresponding to φ = ±π/2). We have to restrict to the open set

U + δ = U δ ∩ {x 1 > Λ} = {ρ < δ, x 1 > Λ, θ = ±π}.
We shall prove that a) ∆W β ≥ 0 (= ∆W ) on U + δ .

b) W ≥ c β W β on ∂U + δ , δ > 0 small. Proof of a). We use again formula (4.7). It gives

∆W β = 1 ρ(R -ρ cos θ) (β 2 -1/4) R ρ β-1 + O(ρ β ) cos(θ/2) cos φ - ρ β+1 R -ρ cos θ cos(θ/2) cos φ = 1 ρ(R -ρ cos θ) (β 2 -1/4) R ρ β-1 + O(ρ β ) cos(θ/2) cos φ ≥ 0.
Proof of b). Let ǫ > 0. Away from H Λ , that is over ∂U + δ ∩ {x 1 > Λ + ǫ}, we can perform exactly the same analysis as in the proof of Proposition 17, point ii). Thus, W ≥ c W β , over ∂U + δ ∩ {x 1 > Λ + ǫ} for some c = c ǫ > 0.

It remains to treat the vicinity of H Λ inside ∂U + δ . For x 1 = Λ (that is φ = ±π/2), one has W (x, z) = W β (x, z) = 0. Also, for θ = ±π, one has W (x, z) ≥ 0, W β (x, z) = 0. Everywhere else, W β (x, z) > 0 and W (x, z) > 0. Hence, it remains to show that lim inf (x,z)∈∂U + δ \{|θ|=π}, x 1 →Λ + |W (x, z)/W β (x, z)| > 0.

Again, we argue by contradiction: we assume that there is a sequence

(x k , z k ) ⊂ ∂U + δ \ {θ = π}, x k 1 → Λ + , W (x k , z k )/W β (x k , z k ) → 0.
Up to a subsequence, we can assume that (x k , z k ) → X for some X = (Λ, X 2 , Z) ∈ ∂U + δ . As W and W β are even in z, we can also assume that z k ≥ 0 for all k. There are several cases:

• (X 1 , X 2 ) = (Λ, X 2 ) ∈ Ω. W is harmonic and positive on {(x, z), |(x, z) -X| ≤ ǫ, x 1 > Λ}, ǫ > 0 small enough. Moreover, for all (x, z) in this set, we have W (x, z) > 0 = W (Λ, x 2 , z).

We can apply Hopf's lemma, which yields ∂ 1 W (Λ, x 2 , z) ≥ C > 0. Hence,

lim inf k→∞ W (x k , z k ) |x k 1 -Λ| > 0 which yields in turn lim inf k→∞ W (x k , z k ) |W β (x k , z k )| > 0. Indeed, |W β (x k , z k )| = |W β (x k , z k ) -W β (Λ, x k 2 , z k )| = O(|x k 1 -Λ|)
due to the regularity of W β near X. Thus, we reach a contradiction.

• (X 1 , X 2 ) = (Λ, X 2 ) / ∈ Ω, Z = 0. We are still in a situation where we can apply Hopf's lemma to W near X, which yields the same contradiction as above.

• (X 1 , X 2 ) = (Λ, X 2 ) / ∈ Ω, Z = 0. Although W is harmonic and positive on the dihedra {x 1 > Λ, z > 0}, we can not apply Hopf's lemma because the dihedra does not satisfy the interior sphere condition at X. Nevertheless, thanks to Lemma 16, we get that for some c ′ > 0, for all k large enough

W (x k , z k ) ≥ c (x k 1 -Λ)z k ≥ c ′ π 2 -|φ k | (π -θ k ) ≥ c ′′ W β (x k , z k ).
This ends the proof of (4.10), and so the proof of Proposition 18.

Upper bound

We now look for an upper bound for w along the ray Q + t ν, ν = (1, ν), t > 0 small. We show w(Q + t ν) = O(t 2 ), as t goes to 0 + . (4.11)

Of course, such an upper bound leads to a contradiction with Proposition 18 (for 1/2 < β < 1).

As in configuration 1, this bound follows from the behaviour of u near ∂Ω, as given by (2.8). We remind that for x near Q, u(x) = c 0 r 1/2 + C 1 (x 1 ) r 3/2 + u reg (x) where • r = r(x) is a normal coordinate in the vicinity of ∂Ω (which corresponds to r = 0).

• c 0 is the constant fractional normal derivative at ∂Ω, C 1 is a smooth function of x 1 .

• u reg is C 2 near ∂Ω, and vanishes at ∂Ω.

We can then write w(x) = c 0 w 1/2 (x) + C 1 w 3/2 (x) + w reg (x) + O(|x 1 -Λ|r 3/2 ) with C 1 := C 1 (Λ), w 1/2 (x) := r 1/2 (x)r 1/2 (x), w 3/2 (x) := r 3/2 (x)r 3/2 (x), w reg (x) := u reg (x)u reg (x), where x := (2Λx 1 , x 2 ). Note that w 1/2 , w 3/2 , w reg vanish at Q.

Let us first prove that ∇w reg (Q) = 0. The idea is the same as in the paper of Serrin [START_REF] Serrin | A symmetry problem in potential theory[END_REF]. Clearly, w reg (Λ, •) = 0, so that ∂ 2 w reg (Λ, •) = 0. In particular, ∂ 2 w reg (Q) = 0. We then have to show that ∂ 1 w reg (Q). Near Q, we can write ∂Ω as a graph: x 2 = ψ(x 1 ), with ψ ′ (Λ) = 0. Moreover, u reg (x 1 , ψ(x 1 )) = 0 ⇒ ∂ 1 u reg + ψ ′ ∂ 2 u reg = 0, so that ∂ 1 u reg (Λ, 0) = 0, and from there, ∂ 1 w reg (Q) = 2 ∂ 1 u reg (Q) = 0. It follows that w reg (Q + t ν) = O(t 2 ) for t → 0 + .

We still have to control w α , α = 1/2 or 3/2. For x close enough to ∂Ω, we denote p = p(x) its orthogonal projection on ∂Ω. Near Q, it reads p = (p 1 , ψ(p 1 )), with

  ), and let (φ t ) t∈R be the flow associated with ζ, namely φt = ζ(φ t ), φ 0 = Id.

Figure 1 :

 1 Figure 1: Configuration 1 (on the left) and configuration 2 (on the right).

Figure 2 :

 2 Figure 2: The coordinates (ρ, θ, φ).

1 / 2 n

 12 u = c 0 at the boundary. Following (2.8), we get the expansionsu(x) = c 0 r 1/2 + O(r), R Λ u(x) = c 0 r1/2 + O(r), r, r → 0.Now, we remark that the inward normal vectors n and n coincide at P . The expansions above then lead to the upper boundw(P + r n) = O(r), r > 0 small enough. (4.9)Furthermore, the coordinate r along the normal at P coincides with the coordinate ρ from Proposition 17. Applying this proposition with β = 3/4, we get w(P + r n) ≥ c r 3/4 , r > 0 small enough Comparison between this lower bound and the upper bound (4.9) gives a contradiction.

x 1 -p 1 x 2 -

 2 (x) 2 = (x 1p 1 ) 2 + (x 2ψ(p 1 )) 2 = (1 + ψ ′ (p 1 ) 2 )(x 2ψ(p 1 ))2 , andw α (x) = (1 + ψ ′ (p 1 ) 2 ) α/2 |x 2ψ(p 1 )| α -(1 + ψ ′ (p 1 ) 2 ) α/2 |x 2ψ(p 1 )| αwith p := p(x). One may then Taylor expand w α (Q + t ν) with respect to t (x 1 = Λ + t, x 2 = ψ(Λ) + νt). Indeed, relation (4.12):Λ + tp 1 + (ψ(Λ) + νtψ(p 1 ))ψ ′ (p 1 ) = 0, resp. 2Λ -(Λ + t) -p1 + (ψ(Λ) + νtψ(p 1 ))ψ ′ (p 1 ) = 0 leads to p 1 = Λ + t + O(t 2 ), resp. p1 = Λt + O(t 2). Then, since ψ ′ (Λ) = 0, ψ(p 1 ) = ψ(Λ) +

  y) dy.

	Differentiating z t = w t • φ t with respect to t, we obtain
	żt = ẇt + φt • ∇w t ,
	and thus in particular
	ẇ0 + ζ • n ∂ n w 0 = 0 on ∂Ω.

  2 generates yet another error term bounded by (3.19), (3.21).

  ∞ (R 3 ) for all t and u rem (t; •) ∈ C 2 (W π ) with ∂ β u rem (t; x, z) = o(r 3/2-|β| ) for all t and for any multi-index β. Setting

	V 1 i (t; x, z) := ψ1

1/2 + ψ1 i (t; s, θ)r 3/2 + u reg (t; x, z) + u rem (t; x, z), (3.27) with u reg (t; •) ∈ C i (t; s, θ)r 3/2 + u reg (t; x, z) + u rem (t; x, z)

  ψ ′′ (Λ) 2 t 2 + O(t 3 ), ψ(p 1 ) = ψ(Λ) + ψ ′′ (Λ) 2 t 2 + O(t 3 ), so that |x 2ψ(p 1 )| α = |νt| α 1 -αψ ′′ (Λ) 2ν t + O(t 2 ) , resp. |x 2ψ(p 1 )| α = |νt| α 1 -′ (p 1 ) 2 ) α/2 = 1 + α 2 ψ ′′ (Λ)t 2 + O(t 3 ), resp. (1 + ψ ′ (p 1 ) 2 ) α/2 = 1 + α 2 ψ ′′ (Λ)t 2 + O(t 3 ).

	αψ ′′ (Λ) 2ν	t + O(t 2 )
	and
	(1 + ψ

Combining previous Taylor expansions yields: w α (Q + t ν) = O(t 2+α ) as t → 0 + . Together with the estimate w reg (Q + t ν) = O(t 2 ), it implies the upper bound (4.11).

Acknowledgements

The authors wish to thank Jean-Michel Roquejoffre and Cyril Imbert for helpful discussions, and they acknowledge grant ANR-08-JC-JC-0104.

Then the application

has C 1 regularity since it is the sum of a bilinear and continuous function and a constant term (with respect to A, V ).

Let Ṽ0 ∈ V be the solution of (3.25) for t = 0, i.e.

i.e. the scalar product on V. Therefore the differential

is an isomorphism. The implicit function theorem implies that there exists a C 1 function t → Ṽ (t) in a neighbourhood of zero such that

Uniqueness for equation (3.25) yields Ṽt = Ṽ (t). We infer immediately that t → V t is C 1 in a neighbourhood of zero. We now use the following trace result: for any U ∈ V,

Indeed, it is well known that if U ∈ V, the following estimates hold (with constants which do not depend on Ω)

) . There only remains to prove that the

The result on ut follows almost immediately by differentiating the formula

with respect to t. The only difficulty comes from the fact that v t is a priori not smooth enough with respect to x in order to use the chain rule. However, we can write

Appendix

This appendix is devoted to the calculation of the jacobian of the change of variables x = (x 1 , x 2 ) → (s, r), where x belongs to a tubular neighbourhood T δ of ∂Ω, s is the arc-length, and r = d(x, ∂Ω)(21 x∈Ω -1). Using the same notation as in Section 3, we write

where p(s) ∈ R 2 is the point of ∂Ω with arc-length s, and n(s) is the outward pointing normal.

We deduce that ∂x ∂s = τ (s)r dn ds = (1 + κ(s)r)τ (s),

Let θ be the oriented angle between e 1 and n. Then, up to a change of the orientation of the curve,

We infer that the jacobian matrix of the change of variables is