F Caubet

M Dambrine
email: marc.dambrine@univ-pau.fr

D Kateb

C Z Timimoun
email: timimoun.chahnaz@univ-oran.dz

A Kohn-Vogelius formulation to detect an obstacle immersed in a fluid

Keywords: geometric inverse problem, order two shape sensitivity, shape calculus, stationary Stokes problem. AMS Subject Classification: 49Q10, 34A55, 49Q12 1

The aim of our work is to reconstruct an inclusion ω immersed in a fluid flowing in a larger bounded domain Ω via a boundary measurement on ∂Ω. Here the fluid motion is assumed to be governed by the Stokes equations. We study the inverse problem of reconstructing ω thanks to the tools of shape optimization by minimizing a Kohn-Vogelius type cost functional. We first characterize the gradient of this cost functional in order to make a numerical resolution. Then, in order to study the stability of this problem, we give the expression of the shape Hessian. We show the compactness of the Riesz operator corresponding to this shape Hessian at a critical point which explains why the inverse problem is ill-posed. Therefore we need some regularization methods to solve numerically this problem. We illustrate those general results by some explicit calculus of the shape Hessian in some particular geometries. In particular, we solve explicitly the Stokes equations in a concentric annulus. Finally, we present some numerical simulations using a parametric method.

of its center of mass can be identified from a single boundary measurement. In [START_REF] Conca | Detection of a moving rigid solid in a perfect fluid[END_REF], using complex analysis tools, Conca et al. proved that this result cannot be generalized to any solid. However, they extended the result to moving ellipses: they proved that a solid with some symmetry properties can be partially detected. As expected for an inverse problem, the numerical experiments which were conducted show some difficulties to reconstruct the object. Those numerical difficulties are explained in a recent paper of Badra et al. (see [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF]). Using a least-squares approach, the authors prove that the problem is severely ill-posed.

Here the fluid motion is assumed to be governed by the classical Stokes equations with non-homogeneous Dirichlet boundary condition on the exterior boundary and homogeneous Dirichlet boundary condition on the interior boundary. In order to simplify the expression, we assume that the exterior forces are null but the same results hold if we add a second member. This problem is classical and was studied by many authors (see for example [START_REF] Alvarez | Identification of immersed obstacles via boundary measurements[END_REF], [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF], [START_REF] Simon | Domain variation for drag in Stokes flow[END_REF], [START_REF] Abdelwahed | Topological optimization method for a geometric control problem in Stokes flow[END_REF], [START_REF] Ballerini | Stable determination of an immersed body in a stationary Stokes fluid[END_REF][START_REF] Ballerini | Stable determination of a body immersed in a fluid: the nonlinear stationary case[END_REF]) due to its importance in many applications involving fluid related technology and receive considerable attention by engineers and mathematicians. It is known to be ill-posed and our goal is to determine what sort of informations on the shape to be detected can be recovered by the tools of shape optimization.

Following previous works on electrical impedance tomography by Afraites et al. in [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF] or [START_REF] Afraites | On second order shape optimization methods for electrical impedance tomography[END_REF] and Badra et al. in [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF], we solve our inverse problem by minimizing a cost functional. In our paper, we consider an other approach than the least-squares cost functional for the same problem than the one studied in [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF]. Indeed, we show how to solve the inverse problem by defining a Kohn-Vogelius type cost functional. Here we make the measurement only on a part of the exterior boundary and not on the whole exterior boundary as in the classical Kohn-Vogelius approach (see [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF] for example). Thus, we consider here the Stokes equations with Dirichlet and mixed boundary conditions. Then, we follow the classical recipe: first we give an explicit formula for the gradient of this functional and compute after the associated shape Hessian to study the stability. We show the compactness of the Riesz operator corresponding to this shape Hessian at a critical point; this explains why the inverse problem is ill-posed. We then illustrate this result by computing explicitly this shape Hessian matrix for some particular geometries; we point out the consequence of the high frequencies on the value of the smallest eigenvalue. Finally, we present some numerical simulations with a regularization method: we use here a parametric model.

The paper is organized as follows. Firstly, we introduce the notations, the overdetermined problem that we consider and the Kohn-Vogelius cost functional J KV . Secondly, in Section 2, we state the main results of this work. We recall an important identifiability result that we quote from Alvarez et al. (see [START_REF] Alvarez | Identification of immersed obstacles via boundary measurements[END_REF]Theorem 1.2]) which ensures that the functional J KV has a unique global minimum. The first order derivatives of the state and of the cost functional are characterized and the shape Hessian is explicitly computed. We then claim that the Riesz operator associated to the shape Hessian is compact. This result is crucial in the sense that it explains all the stability difficulties that one has to face. Finally, in order to illustrate the exponentially ill-posedness of this problem, we present an explicit calculus of the shape Hessian in a concentric annulus and compare it with the electrical impedance tomography case (i.e. the Laplacian case). In Section 3, we prove those results. In the last part of this paper, we present some numerical attempts to effectively reconstruct the inclusion ω using shape derivative informations. Let us point out that we use a parametric model of shape variation in order to highlight the bad conditioning of the Hessian matrix. To remove the oscillations due to the high frequencies, we propose an adaptive method which seems to be efficient. The needed results on Stokes equations with mixed boundary conditions (a theorem of existence and uniqueness of the solution and a local regularity result) are recalled in Appendix A. Moreover, the explicit computations of the solution of the Stokes equations in an annulus are detailed in Appendix B.

Introduction of the general notations. Let us introduce the notations that we adopt in this paper. For a bounded Lipschitz open set Ω ⊂ R N (N = 2 or 3), we denote by L p (Ω), W m,p (Ω) and H s (Ω) the usual Lebesgue and Sobolev spaces. We note in bold the vectorial functions and spaces: L p (Ω), W m,p (Ω), H s (Ω), etc. We denote by |Ω| the measure of Ω. Moreover, n represents the external unit normal to ∂Ω, and for a smooth enough function u, we note ∂ n u the normal derivative of u. Finally, we define the space

L 2 0 (Ω) := p ∈ L 2 (Ω), Ω p = 0 .
The problem setting. Let Ω be a bounded, connected and Lipschitz open subset of R N (N = 2 or 3). Let d 0 > 0 fixed (small). We define O d 0 the set of all open subsets ω of Ω with a C 2,1 boundary such that d(x, ∂Ω) > d 0 for all x ∈ ω and such that Ω\ω is connected. The set O d 0 is referred as the set of admissible geometries. Notice that we make the assumption that the inclusion is far from the boundary ∂Ω. We also define Ω d 0 an open set with a C ∞ boundary such that

{x ∈ Ω ; d(x, ∂Ω) > d 0 /2} ⊂ Ω d 0 ⊂ {x ∈ Ω ; d(x, ∂Ω) > d 0 /3} . Let f ∈ H 1/2 (∂Ω) such that f = 0 satisfying the compatibility condition ∂Ω f • n = 0. (1.1)
Let O be a non-empty subset of ∂Ω and g ∈ H -1/2 (O) be an admissible boundary measurement. For ω ∈ O d 0 , let us consider the following overdetermined Stokes boundary value problem

           -ν∆u + ∇p = 0 in Ω\ω div u = 0 in Ω\ω u = f on ∂Ω u = 0 on ∂ω -ν∂ n u + pn = g on O. (1.2)
Here the constant ν > 0 represents the kinematic viscosity of the fluid, the vectorial function u represents the velocity of the fluid and the scalar function p represents the pressure.

We assume here that there exists ω * ∈ O d 0 such that (1.2) has a solution. Thus, we consider the following geometric inverse problem:

find ω ∈ O d 0 and a pair (u, p) which satisfy the overdetermined system (1.2).

(1.3)

We will tackle the inverse problem of reconstructing ω thanks to the tools of shape optimization. In order to recover the shape of the inclusion ω, a usual strategy is then to minimize a cost functional. In this work, we consider the following Kohn-Vogelius cost functional

J KV (ω) := 1 2 Ω\ω ν |∇(u D -u N )| 2 , where (u D , p D ) ∈ H 1 (Ω\ω) × L 2 0 (Ω\ω) is the unique solution of the Stokes problem with Dirichlet boundary conditions        -ν∆u D + ∇p D = 0 in Ω\ω div u D = 0 in Ω\ω u D = f on ∂Ω u D = 0 on ∂ω (1.4) and (u N , p N ) ∈ H 1 (Ω\ω) × L 2 (Ω\ω)
is the unique solution of the Stokes problem with mixed boundary conditions

           -ν∆u N + ∇p N = 0 in Ω\ω div u N = 0 in Ω\ω -ν∂ n u N + p N n = g on O u N = f on ∂Ω\O u N = 0 on ∂ω.
(1.5)

We refer to [START_REF] Boyer | Éléments d'analyse pour l'étude de quelques modèles d'écoulements de fluides visqueux incompressibles[END_REF] or [START_REF] Galdi | An introduction to the mathematical theory of the Navier-Stokes equations[END_REF] for the results of existence, uniqueness and regularity of the solutions of the Stokes problem with Dirichlet boundary conditions. Notice that we assume the compatibility condition (1.1) associated to problem (1.4) is satisfied. Concerning the mixed boundary conditions, we recall the main results used in this paper in Appendix A. Hence, the existence and the uniqueness of the couple (u N , p N ) is guaranteed by Theorem 11.

Then, we try to minimize the Kohn-Vogelius cost functional J KV :

ω * = argmin ω∈O d 0 J KV (ω). (1.6)
Introduction of the needed functional tools. To define the shape derivatives, we will use the velocity method introduced by Murat and Simon in 1976 in [START_REF] Murat | Sur le contrôle par un domaine géométrique[END_REF]. To this end, we need to introduce the space of admissible deformations

U := {θ ∈ W 3,∞ (R N ); Supp θ ⊂ Ω d 0 }.
Then for V ∈ U and t ∈ [0, T ) (where T > 0 is a fixed real number sufficiently small), we define ω t := (I + t V )(ω). In this paper, V is referred as the perturbation direction and we denote

V n := V • n.
For details concerning the differentiation with respect to the domain, we refer to the papers of Simon [START_REF] Simon | Differentiation with respect to the domain in boundary value problems[END_REF][START_REF] Simon | Second variations for domain optimization problems[END_REF] and the books of Henrot and Pierre [START_REF] Henrot | Variation et optimisation de formes[END_REF] and of Soko lowski and Zolsio [START_REF] Soko | Introduction to shape optimization[END_REF].

Statement of main results

Identifiability result. We quote an identifiability result in the Dirichlet case proved by Alvarez et al. [START_REF] Alvarez | Identification of immersed obstacles via boundary measurements[END_REF]Theorem 1.2]. In order to state it, let us precise a notation: we denote by σ the stress tensor defined by σ(u, p) := ν(∇u + t ∇u) -pI.

Here t ∇u is the transposed matrix of ∇u. Notice that if div u = 0 in Ω\ω, one has -div (σ(u, p)) = -ν∆u + ∇p in Ω\ω.

Theorem 1 (Alvarez et al., [START_REF] Alvarez | Identification of immersed obstacles via boundary measurements[END_REF]). Let Ω ⊆ R N (N = 2 or 3) be a bounded C

and f ∈ H 3/2 (∂Ω) with f = 0, satisfying the flux condition ∂Ω f • n = 0. For ε * = 0 or ε * = 1, let (u j , p j ) for j = 0, 1, be a solution of        -div (σ(u j , p j )) + ε * div (u j ⊗ u j ) = 0 in Ω\ω j div u j = 0 in Ω\ω j u j = f on ∂Ω u j = 0 on ∂ω j .
Assume that (u j , p j ) are such that

σ(u 0 , p 0 ) n = σ(u 1 , p 1 ) n on O.
Then ω 0 ≡ ω 1 .

We can adapt this result to our problem, i.e. with -ν∂ n u + pn instead of σ(u, p)n, with ε * = 0, with Ω Lipschitz and with f ∈ H 1/2 (∂Ω) (see [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF]Theorem 2.2]). Hence this result states that given a fixed f , two different geometries ω 0 and ω 1 in O d 0 yield two different measures g 1 and g 2 . Thus problem (1.3) admits a unique solution.

Sensitivity with respect to the domain. The following result is based on [8, Propositions 2.1 and 2.5] and ensures that the solutions (u D , p D ) and (u N , p N ) are differentiable with respect to the domain. Moreover, we characterize the shape derivatives of these solutions.

Proposition 2 (First order shape derivatives of the states). Let V ∈ U be an admissible deformation. The solutions (u D , p D ) and (u N , p N ) are differentiable with respect to the domain and the shape derivatives

(u ′ D , p ′ D ) and (u ′ N , p ′ N ) belong to H 2 (Ω d 0 \ω) × H 1 (Ω d 0 \ω). The couples (u ′ D , p ′ D ) ∈ H 1 (Ω\ω) × L 2 0 (Ω\ω) and (u ′ N , p ′ N ) ∈ H 1 (Ω\ω) × L 2 (Ω\ω)
are respectively the only solutions of the following boundary value problems

       -ν∆u ′ D + ∇p ′ D = 0 in Ω\ω div u ′ D = 0 in Ω\ω u ′ D = 0 on ∂Ω u ′ D = -V n ∂ n u D on ∂ω (2.1) and            -ν∆u ′ N + ∇p ′ N = 0 in Ω\ω div u ′ N = 0 in Ω\ω -ν∂ n u ′ N + p ′ N n = 0 on O u ′ N = 0 on ∂Ω\O u ′ N = -V n ∂ n u N on ∂ω. (2.2)
In order to simplify the expressions, we introduce the following notations:

w := u D -u N and q := p D -p N , (2.3)
where (u D , p D ) solves (1.4) and (u N , p N ) solves (1.5).

Proposition 3 (First order shape derivative of the functional). For V ∈ U , the Kohn-Vogelius cost functional J KV is differentiable at ω in the direction V with

DJ KV (ω).V = - ∂ω (ν∂ n w -qn) • ∂ n u D V n + 1 2 ν ∂ω |∇w| 2 V n , (2.4)
where (w, q) is defined by (2.3).

Second order analysis: justification of the instability. We study the stability of the problem. First, we give an explicit formula for the shape Hessian when ω ∈ O d 0 in the following proposition. In order to simplify the expressions, we will use the following notations:

w ′ := u ′ D -u ′ N and q ′ := p ′ D -p ′ N , (2.5)
where (u ′ D , p ′ D ) solves (2.1) and (u ′ N , p ′ N ) solves (2.2).

Proposition 4 (Second order shape derivative of the functional). The solution (u, p) is twice differentiable with respect to the domain. Moreover, for V ∈ U , we have

D 2 J KV (ω).V.V = ∂ω ν∇w ′ : ∇w -ν∂ n w ′ + ν∇w n ′ -q ′ n -q n ′ • ∂ n u D -ν∂ n w -q n • ∂ n u ′ D + ∇u D n ′ V n + ∂ω div 1 2 ν |∇w| 2 -ν∂ n w -q n • ∂ n u D V V n , (2.6)
where (w, q) and (w ′ , q ′ ) are defined respectively by (2.3) and (2.5).

Let ω * be a critical point of the Kohn-Vogelius functional. The following proposition shows that the optimization problem (1.6) is unstable.

Proposition 5 (Compactness at a critical point). Let ω * ∈ O d 0 be a solution of the inverse problem (1.3). We have

D 2 J KV (ω * ).V.V = - ∂ω * (ν∂ n w ′ -q ′ n) • ∂ n u D V n ,
where (w ′ , q ′ ) is defined by (2.5). Moreover, the Riesz operator corresponding to D 2 J KV (ω * ) defined from H 1/2 (∂ω * ) to H -1/2 (∂ω * ) is compact.

The above proposition explains the difficulties encountered to solve numerically this problem. Indeed, the gradient has not a uniform sensitivity with respect to the deformation direction: hence, oscillations of the boundary generated by high frequencies are to be expected (see Section 4 for the numerical simulations).

Explicit computation of the shape Hessian when O = ∂Ω. We illustrate the compact behavior of the shape Hessian in the case O = ∂Ω. We study it in the bidimensional situation for some particular geometries. In order to simplify the expressions, we assume ν = 1. In the following, ( e r , e θ ) := cos θ sin θ , -sin θ cos θ represents the polar coordinates system.

Proposition 6 (Explicit calculus of the shape Hessian in a concentric annulus). Let us assume that the optimum ω * corresponds to the concentric annulus case

Ω ρ = x ∈ R 2 , ρ < |x| < 1 .
Set f = cos(nθ) e r , n > 1 and the associated Neumann boundary condition g (see (B.4)).

Let p > 1 and let us define

ξ n := ρ 2 -n 2 ρ 2n + 2 -1 + n 2 ρ 2+2n -n 2 ρ 4+2n + ρ 2+4n κ n := 2 -1 + ρ 2n + n 2 -1 + ρ 2 -1 + ρ 2n -n -1 + ρ 2 1 + ρ 2n A p := 2 -1 + ρ 2p 2 + p 3 -1 + ρ 2 2 -1 + ρ 4p + 2p 1 + ρ 2 -1 + ρ 4p -p 2 -1 + ρ 2 1 + ρ 2 + 4ρ 2p + ρ 4p + ρ 2+4p B p := ρ 2 -p 2 ρ 2p + 2 -1 + p 2 ρ 2+2p -p 2 ρ 4+2p + ρ 2+4p 3ρ 2 + 8 + p 2 ρ 2p -2 -1 + p 2 ρ 2+2p + p 2 ρ 4+2p + 3ρ 2+4p .
We have =

D 2 J KV (ω * ) [(cos kθ) e r , ( sin
         - 1 2 R n,|n-k| + R n,|n+l| if k -l = 2n - 1 2 R n,|n+k| + R n,|n-l| if l -k = 2n R n,|n+k| + R n,|n-l| if k = l, where R n,p := -8πρ 2+2n ξ 2 n κ 2 n ρ 2p A p B p if p > 1 R n,1 := - πρ 2n ξ 2 n κ 2 n -3 + 12ρ 2 -6ρ 4 -4ρ 6 + ρ 8 + 16ρ 4 ln(ρ) -3 + 3ρ 2 -ρ 4 + ρ 6 1 -ρ 2 + (1 + ρ 2 ) ln(ρ) R n,0 := 0.
Remark 7. In particular, if |n-k| / ∈ {0, 1}, notice that we can write the diagonal elements of the shape Hessian under the form

D 2 J KV (ω * ) [(cos kθ) e r , (cos kθ) e r ] = D 2 J KV (ω * ) [(sin kθ) e r , (sin kθ) e r ] = -8πρ 2+2n ξ 2 n κ 2 n ρ 2|n+k| A |n+k| B |n+k| + ρ 2|n-k| A |n-k| B |n-k|
Let us compare our Stokes case with the electrical impedance tomography (E.I.T.) situation (see [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF]Proposition 3]). We make the computations of the E.I.T. case with Dirichlet boundary conditions (in [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF], they impose Neumann boundary conditions). Proceeding as in [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF] and in this paper, we prove that, in the E.I.T. situation with Dirichlet boundary conditions (and with f = cos(nθ) e r , n > 1), the Hessian matrix is defined similarly as in Proposition 6 with

R n,p := -4πn 2 ρ -2+2n (1 -ρ 2n ) 2 pρ 2p -1 + ρ 4p if p ≥ 1 R n,0 := 0.
Hence our shape Hessian has the same aspect than the Hessian in the E.I.T. case.

Example 8. In order to be complete, we give some examples of the influence of the different parameters on the value of the smallest eigenvalue λ 1 of the shape Hessian D 2 J KV (ω * ) and hence on the reconstruction of the object.

In Table 1, we study the influence of the size of the object ω * . We here assume that f = cos(2θ) e r (i.e. n = 2) and k, l = 1, . . . , 7. The regularizing behavior is emphasized:

Table 1: Influence of the size of the object ρ on the smallest eigenvalue λ 1 of the shape Hessian D 2 J KV (ω * ). ρ 0.9 0.7 0.5 0.3 0.1 0.05 λ 1 8.9401e+05 1.0921e+03 3.9992 0.0019 3.1851e-10 1.9032e-14 the more the object is far of the exterior boundary Ω, the more the functional is degenerated (and the more it is difficult to detect it).

In Table 2, we study the influence of the deformation directions. Here again we assume n = 2 and we fix ρ = 0.7. As we expected taking into account of our compactness result 5, we see that the problem is severely ill-posed for the high frequencies.

In order to highlight this degeneracy and to compare our Stokes case with the E.I.T. case (with Dirichlet boundary conditions), we present in Figure 1 the spectrum of the shape Hessian matrix for 50 Fourier modes. Here again we fix n = 2 and ρ = 0.7. The results are presented in decimal logarithm scale to emphasize the behavior of the spectrum. As a clear consequence, the corresponding mode is poorly distinguished by the functional: the functional is almost flat in the direction of high frequencies. We will have to take into account of this point for the numerical simulations (see Section 4). Figure 1 also suggests that the E.I.T. case is as ill-posed as the Stokes case.

Proof of the main results

In order to simplify the expressions, we only detail the case O = ∂Ω but the proofs in the general case are simple adaptations of the ones detailed here. We recall that we use the notations (w, q) and (w ′ , q ′ ) defined respectively by (2.3) and (2.5). First order shape derivative. The proof of Proposition 2 is directly adapted of the proof of Proposition 2.5 in [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF] and we refer to this paper for the details. The main difficulty is to prove the existence of the shape derivatives: it is obtained through a generalized implicit function theorem proved by Simon (see [START_REF] Simon | Domain variation for drag in Stokes flow[END_REF]Theorem 6]). The characterization of (u ′ D , p ′ D ) and (u ′ N , p ′ N ) is obtained using classical results of shape derivatives calculus (see [START_REF] Henrot | Variation et optimisation de formes[END_REF]Chapter 5]).

Proof of Proposition 3. The Hadamard's formula (see [START_REF] Henrot | Variation et optimisation de formes[END_REF]Theorem 5.2.2]) leads:

DJ KV (ω).V = Ω\ω ν∇w : ∇w ′ + 1 2 ν div (|∇w| 2 V ) = Ω\ω ν∇w : ∇(u ′ D -u ′ N ) + 1 2 ∂ω ν |∇w| 2 V n (3.1)
using the fact that V = 0 on ∂Ω. Let us prove that

Ω\ω ν∇w : ∇u ′ N = 0. (3.2)
Applying Green's Formula and using the system (2.2) solved by (u ′ N , p ′ N ), we get

Ω\ω ν∇w : ∇u ′ N = - Ω\ω ν∆u ′ N • w + ∂(Ω\ω) ν∂ n u ′ N • w = - Ω\ω ∇p ′ N • w + ∂(Ω\ω) ν∂ n u ′ N • w = Ω\ω p ′ N div w - ∂(Ω\ω) p ′ N w • n + ∂(Ω\ω) ν∂ n u ′ N • w = 0, since div w = 0 in Ω\ω, -ν∂ n u ′ N + p ′ N n = 0 on ∂Ω and w = 0 on ∂ω.

Then, let us write

Ω\ω ν∇w : ∇u ′ D as an integral on ∂ω. Proceeding as above, we apply Green's Formula on this term and use the system (2.1) solved by (u ′ D , p ′ D ) to obtain

Ω\ω ν∇w : ∇u ′ D = - Ω\ω ν∆w • u ′ D + ∂(Ω\ω) ν∂ n w • u ′ D = - Ω\ω ∇q • u ′ D + ∂(Ω\ω) ν∂ n w • u ′ D = Ω\ω q div u ′ D - ∂(Ω\ω) q u ′ D • n + ∂(Ω\ω) ν∂ n w • u ′ D = ∂(Ω\ω) (ν∂ n w -qn) • u ′ D . Since u ′ D = 0 on ∂Ω and u ′ D = -∂ n u D V n on ∂ω, we obtain Ω\ω ν∇w : ∇u ′ D = - ∂ω (ν∂ n w -qn) • ∂ n u D V n . (3.3)
Gathering equations (3.1), (3.2) and (3.3), we check that

DJ KV (ω).V = - ∂ω (ν∂ n w -qn) • ∂ n u D V n + 1 2 ν ∂ω |∇w| 2 V n .
Characterization of the shape Hessian.

Proof of Proposition 4. The proof of the existence of the second order shape differentiability of the states (u D , p D ) and (u N , p N ) is directly adapted from [8, Proposition 2.3]. Thus we do not detail it here. From Proposition 3, we have

DJ KV (ω).V = ∂ω 1 2 ν |∇w| 2 -(ν∂ n w -qn) • ∂ n u D V n = Ω\ω div 1 2 ν |∇w| 2 -(ν∂ n w -qn) • ∂ n u D V , since V ≡ 0 on ∂Ω. From Hadamard's formula (see [20, Theorem 5.2.2]), this leads D 2 J KV (ω).V.V = Ω\ω div ν∇w ′ : ∇w -ν∂ n w ′ + ν∇wn ′ -q ′ n -qn ′ • ∂ n u D -(ν∂ n w -qn) • (∂ n u ′ D + ∇u D n ′ ) V + ∂ω div 1 2 ν |∇w| 2 -(ν∂ n w -qn) • ∂ n u D V V n .
Since V = 0 on ∂Ω, we then obtain

D 2 J KV (ω).V.V = ∂ω ν∇w ′ : ∇w -ν∂ n w ′ + ν∇wn ′ -q ′ n -qn ′ • ∂ n u D -(ν∂ n w -qn) • (∂ n u ′ D + ∇u D n ′ ) V n + ∂ω div 1 2 ν |∇w| 2 -(ν∂ n w -qn) • ∂ n u D V V n .
Justifying the ill-posedness of the problem. The instability of the inverse problem (1.3) is proved using the same methods as those used in [START_REF] Badra | Detecting an obstacle immersed in a fluid by shape optimization methods[END_REF]. Therefore, we use a local regularity argument (see Theorem 12) in order to prove the compactness of the Riesz operator corresponding to the shape Hessian at a solution of the inverse problem. An alternative proof could be to use the hydrodynamical potential layers as what is done in [START_REF] Afraites | Detecting perfectly insulated obstacles by shape optimization techniques of order two[END_REF] for the Laplacian case.

In order to prove Proposition 5, we investigate the properties of stability of the cost functional J KV . Then we consider an admissible inclusion ω * ∈ O d 0 which is solution of problem (1.3). Then, ω * realizes the absolute minimum of the criterion J KV , J KV (ω * ) = 0 and then u D = u N and p D = p N in Ω\ω. Therefore

D 2 J KV (ω * ).V.V = - ∂ω * (ν∂ n w ′ -q ′ n) • ∂ n u D V n .
Proof of Proposition 5. We first decompose D 2 J KV (ω * ) as a composition of two operators:

for V ∈ U , D 2 J KV (ω * ).V.V = M (V ), T (V ) H -1/2 (∂ω * ),H 1/2 (∂ω * )
where •, • denote the dual product between H -1/2 (∂ω * ) and H 1/2 (∂ω * ). Here, the operator T :

H 1/2 (∂ω * ) → H 1/2 (∂ω * ) is defined by T (V ) := -∂ n u D V n
and the operator M :

H 1/2 (∂ω * ) → H -1/2 (∂ω * ) is defined by M (V ) := ν∂ n w ′ -q ′ n.
From the systems (2.1) and (2.2) and using the fact that u D = u N in Ω\ω * (and then ∂ n u D = ∂ n u N on ∂ω * ), we obtain the following system:

       -ν∆w ′ + ∇q ′ = 0 in Ω\ω * div w ′ = 0 in Ω\ω * w ′ = -u ′ N on ∂Ω w ′ = 0 on ∂ω * .
Since ∂ n u D does not depend on V , notice that the operator T is linear continuous as multiplier by a smooth function (see [START_REF] Maz | Theory of multipliers in spaces of differentiable functions[END_REF]). Now, let us prove the following lemma which states that the operator M is compact. Hence, the Riesz operator corresponding to the shape Hessian is compact as composition of linear continuous operator with a compact one.

Lemma 9. The operator M is compact.

Proof. We decompose the operator M as follows:

M := M 2 • M 1 with M 1 : V ∈ H 1/2 (∂ω * ) → M 1 (V ) := -u ′ N ∈ H 1/2 (∂Ω), and M 2 : Ψ ∈ H 1/2 (∂Ω) → M 2 (Ψ) := ν∂ n Φ -χn ∈ H -1/2 (∂ω * ),
where (u ′ N , p ′ N ) is solution of (2.2) and where (Φ, χ)

∈ H 1 (Ω\ω * ) × L 2 0 (Ω\ω * ) solves        -ν∆Φ + ∇χ = 0 in Ω\ω * div Φ = 0 in Ω\ω * Φ = Ψ on ∂Ω Φ = 0 on ∂ω * . (3.4)
Let us prove that M 1 is linear continuous and M 2 is compact. We check that

M 1 = M 1,2 • M 1,1 with M 1,1 : V ∈ H 1/2 (∂ω * ) → M 1,1 (V ) := -∂ n u N V n ∈ H 1/2 (∂ω * )
and with

M 1,2 : η ∈ H 1/2 (∂ω * ) → M 1,2 (V ) := -z ∈ H 1/2 (∂Ω),
where the couple (z,

q) ∈ H 1 (Ω\ω * ) × L 2 (Ω\ω * ) is solution of        -ν∆z + ∇q = 0 in Ω\ω * div z = 0 in Ω\ω * -ν∂ n z + qn = 0 on ∂Ω z = η on ∂ω * .
Since ∂ n u N does not depend on V , the operator M 1,1 is linear continuous as multiplier by a smooth function (see [START_REF] Maz | Theory of multipliers in spaces of differentiable functions[END_REF]) and the operator M 1,2 is clearly linear continuous. Thus M 1 is linear continuous.

Finally, we decompose M 2 as follows:

M 2 = M 2,3 • M 2,2 • M 2,1
where

M 2,1 : Ψ ∈ H 1/2 (∂Ω) → (Φ, χ) ∈ H 3 (Ω d 0 \ω * ) × H 2 (Ω d 0 \ω * ) with (Φ, χ) solution of (3.4), M 2,2 : (v, ξ) ∈ H 3 (Ω d 0 \ω * ) × H 2 (Ω d 0 \ω * ) → ν∂ n v -ξn ∈ H 3/2 (∂ω * )
and M 2,3 is the compact imbedding of H 3/2 (∂ω * ) into H -1/2 (∂ω * ). From the local regularity theorem 12, the operator M 2,1 is linear continuous and the operator M 2,2 is clearly linear continuous. Then, by composition M 2 is compact, which concludes the proof.

Explicit computation of the Hessian matrix in a concentric annulus. We recall that ( e r , e θ ) := cos θ sin θ , -sin θ cos θ represents the polar coordinates system.

Proof of Proposition 6. Using the computations of u D , u N , u ′ D and u ′ N (see Appendix B), we will compute the elements of the Hessian D 2 J KV ((cos kθ) e r , (cos lθ) e r ) when the Dirichlet data on the exterior boundary is taken as f = f (ext) = cos (nθ) e r . From the computations detailed in Appendix B, we see that only the elements with k = l or | k -l |= 2n have to be computed; the other elements are zero.

Let us begin by the case k = l. We have

D 2 J KV ((cos kθ) e r , (cos kθ) e r ) = -∂ n (u ′ N -u ′ D ) + (p ′ N -p ′ D )n(ρ, θ), -V n ∂ n u D = ∂ω * (E + F ), with E(θ) = K |n+k|,r cos (n+k)θ + K |n-k|,r cos (n-k)θ K D n,r cos (n+k)θ + K D n,r cos (n-k)θ
and with

F (θ) = L |n+k|,r sin (n+k)θ + L |n-k|,r sin (n-k)θ L D n,r sin (n+k)θ + L D n,r sin (n-k)θ ,
where K D n,r and L D n,r are defined by (B.5) and K n,r and L n,r by (B.6). Since K D n,r = 0 by computations, we have E(θ) = 0 and then we have only to compute

D 2 J KV ((cos kθ) e r , (cos kθ) e r ) = ρ 2π 0 (E(θ) + F (θ))dθ = πρ L |n+k|,r L D n,r + L |n-k|,r L D n,r
Set p 1 = |n + k| and p 2 = |n -k|. Defining ξ n , κ n , A p i , B p i and R n,p i as in the statement of Proposition 6, we obtain after some computations

D 2 J KV ((cos kθ) e r , (cos kθ) e r ) = R n,p 1 (ρ) + R n,p 2 (ρ).
Similarly, we prove that

D 2 J KV ((sin kθ) e r , (sin kθ) e r ) = R n,p 1 (ρ) + R n,p 2 (ρ).
Let us now study the case |k -l| = 2n. Set V := (cos kθ) e r and W := (cos lθ) e r . We denote (w ′ W , q ′ W ) the shape derivative in the direction W (the pair (w ′ , q ′ ) is defined in (2.5)). Then, using the linearity of the Stokes problem and the polarization formula for the quadratic form, we prove that D 2 J KV ((cos kθ) e r , (cos lθ) e r ) =

1 2 ∂ω * (-∂w ′ W + q ′ W n) • ∂ n u D V n + ∂ω * (-∂w ′ V + q ′ V n) • ∂ n u D W n .
Let us focus on the first integral: the case of the second integral is obtained by reversing the roles of k and l. Defining

G(θ) = K |n+k|,r cos (n+k)θ + K |n-k|,r cos (n-k)θ K D n,r cos (n+l)θ + K D n,r cos (n-l)θ
and

H(θ) = L |n+k|,r sin (n+k)θ + L |n-k|,r sin (n-k)θ L D n,r sin (n+l)θ + L D n,r sin (n-l)θ , we have ∂ω * (-∂w ′ W + q ′ W n) • ∂ n u D V n = ∂ω * (G + H) .
Using the fact that K D n,r = 0, we then obtain

• if k -l = 2n ∂ω * (G + H) = ρ L |n-k|,r L D n,r 2π 0 (sin(n -k)θ) sin((n + l)θ) = -ρπ L |n-k|,r L D n,r , • if l -k = 2n ∂ω * (G + H) = ρ L |n+k|,r L D n,r 2π 0 (sin(n + k)θ) sin((n -l)θ) = -ρπ L |n+k|,r L D n,r .
Thus,

D 2 J KV (ω * ) [(cos kθ) e r , (cos lθ) e r ] =            - ρπ 2 L D n,r L n,|n-k| + L n,|n+l| if k -l = 2n - ρπ 2 L D n,r L n,|n+k| + L n,|n-l| if l -k = 2n ρπL D n,r L n,|n+k| + L n,|n-l| if k = l.
We proceed in the same way for D 2 J KV (ω * ) [(sin kθ) e r , (sin lθ) e r ].

Numerical experiments

Since the problem is severely ill-posed, we need some regularization methods to solve it numerically, for example by adding to the functional a penalization in terms of the perimeter. Indeed, this term leads to well posed problems (see [START_REF] Bucur | Variational methods in shape optimization problems[END_REF] or [START_REF] Dambrine | On variations of the shape Hessian and sufficient conditions for the stability of critical shapes[END_REF]). Here we choose to make a parametric regularization using a parametric model of shape variations in order to first highlight the bad conditioning of the shape Hessian matrix.

Framework for the numerical simulations

The numerical simulations presented are made in dimension two using the finite elements library Mélina (see [START_REF] Martin | Finite Element Library Mélina[END_REF]) and the mesh generator Triangle (see [START_REF] Shewchuk | Mesh Generator Triangle[END_REF]). We use a P 3-P 2 finite elements discretization to solve the Stokes equations (1.4) and (1.5). The framework is the following: we assume the kinematic viscosity ν is equal to 1, the exterior boundary is assumed to be the unit circle centered at the origin and we consider the exterior Dirichlet boundary condition

f := n 2 -n 1 = sin θ -cos θ , θ ∈ [0, 2π) ,
where n = (n 1 , n 2 ) is the exterior unit normal. Notice that f is such that the compatibility condition (1.1) is satisfied. In order to have a suitable pair (measure g, domain ω * ), we use a synthetic data: we fix a shape ω * , solve the Stokes problem (1.4) in Ω\ω * using another finite elements method (here a P 4-P 3 finite elements discretization) and extract the measurement g by computing -ν∂ n u + p n on ∂Ω. Except when mentioned, the simulations are performed in the case where O = ∂Ω.

Here, we restrict ourselves to star-shaped domains and use polar coordinates for parametrization: the boundary ∂ω of the object can be then parametrized by

∂ω = x 0 y 0 + r(θ) cos θ sin θ , θ ∈ [0, 2π) ,
where x 0 , y 0 ∈ R and where r is a C 2,1 function, 2π-periodic and without double point.

Taking into account of our main compactness result (Proposition 5), we approximate the polar radius r by its truncated Fourier series

r N (θ) := a N 0 + N k=1 a N k cos(kθ) + b N k sin(kθ),
for the numerical simulations. Indeed this regularization by projection permits to remove high frequencies generated by cos(kθ) and sin(kθ) for k >> 1, for which the functional is degenerated.

Then, the unknown shape is entirely defined by the coefficients (a i , b i ). Hence, for k = 1, . . . , N , the corresponding deformation directions are respectively,

V 1 := V x 0 := 1 0 , V 2 := V y 0 := 0 1 , V 3 (θ) := V a 0 (θ) := cos θ sin θ , V 2k+2 (θ) := V a k (θ) := cos(kθ) cos θ sin θ , V 2k+3 (θ) := V b k (θ) := sin(kθ) cos θ sin θ , θ ∈ [0, 2π
). The gradient is then compute component by component using its characterization (see Proposition 3, formula (2.4)):

∇J KV (ω) k = DJ KV (ω) • V k , k = 1, . . . , 2N + 3.
This equality is simply that

lim t→0 J KV (I + tV k )(ω) -J KV (ω) t = DJ KV (ω) • V k .
Notice that we have to solve only two Stokes problems (problems (1.4) and (1.5)) to compute the gradient. Indeed, the directional perturbations V dependance is explicit in formula (2.4).

The optimization method used for the numerical simulations is here the classical gradient algorithm with a line search (using the Wolfe conditions: see for example [24, eq. (3.6) page 34]). Moreover, we here use N ext := 100 discretization points for the exterior boundary and N int := 75 for the interior boundary. In order to be completely explicit, we detail this algorithm: Algorithm 1 1. fix a number of iterations M and an initial shape ω 0 , 2. mesh Ω\ω i using Triangle (where ω i denotes the i th iterate of the approximate shape), 3. solve problems (1.4) and (1.5) with ω = ω i using Mélina, 4. extract ∇u D , ∇u N , p D and p N on ∂ω i and compute ∇J KV (ω i ) using formula (2.4), 5. use the Wolfe conditions to compute a satisfying step length α i , 6. move the coefficients associated to the shape: ω i+1 = ω i -α i ∇J KV (ω i ), 7. get back to the step 2. while i < M . Remark 10. Here, we do not use an optimization method of order two due to the expression of the shape Hessian (2.6). Indeed, notice that (u ′ D , p ′ D ) and (u ′ N , p ′ N ) (and so (w ′ , q ′ )) depend on the perturbation direction V . Therefore, if we want to compute this shape Hessian (for a shape parametrized by k parameters), then we will have to solve 2 + 2 * k Stokes problems, which would be too costly. that is to say with five parameters. In order to reconstruct this object, we work with shapes parametrized respectively by five and fifteen parameters and we stop each experiment when we obtain a residual value of the cost function J KV : ε ≈ 10 -3 . Figure 2 shows that when Here again, the conclusion is the same (see Figure 3): high frequencies lead oscillations of Here again, more iterations are needed to obtain the residual ε. In order to compare the errors, we represent on the second graph of Figure 3 the logarithm of the residual obtained with 17 parameters (noted log(J 17 )) versus the logarithm of the residual obtained with 21 parameters (noted log(J 21 )). We then obtain a regression line whose equation is given by y = 1.091x -0.064. Hence J 17 J 21 = 0.938 J 0.091 21 and this points out that when the number of iterations is big, the gap between J 17 and J 21 is big too (J 17 is much smaller than J 21 ).

Highlight of the degeneration of the functional

The problem is: how to know the number of parameters with which we must work to well approximate the shape? Indeed, if we work with too few parameters we cannot detect a non-trivial shape and if we work with too many parameters, degeneracy of the functional leads problems. From now, we present some numerical illustrations.

An adaptive method

A solution which seems to be efficient is to use an adaptive method. It consists in increasing gradually the number of parameters during the algorithm to a fixed final number of parameters. For example, if we want to work with twenty-one parameters, we begin by working with two parameters during five iterations, then with three parameters (we add the radius) during five more iterations, and then we add two search parameters every fifteen iterations.

We also adapt the number of discretization points for the interior boundary (we conserve the same for the exterior boundary). We fix a minimum number of discretization points (here 40). Then, when we consider the perturbation directions associated to cos(kθ) and sin(kθ), we discretize the interior boundary using N int = 9 * k points. This method permits to discretize half of a period using five points, which seems suitable.

The algorithm is the same than Algorithm 1 described above. However, the step 6. is replaced by

ω i+1 (1 : m) = ω i (1 : m) -α i ∇J KV (ω i )(1 : m),
where ω i (1 : m) represents the m first coefficients parametrizing the shape ω i (the same notation holds for ∇J KV (ω i )(1 : m)). The number m grows to the fixed final number of parameters following the procedure described previously. This process permits to avoid the oscillations of the boundary as what is shown in Figure 4. Indeed, we saw in Figure 3 Figure 4: Reconstruction of ω * 2 using the adaptive method with 21 parameters that if we work with twenty-one parameters, oscillations appear but we see in Figure 4 that it is not the case using this adaptive method. We can notice that this adaptive method leads to steps in the evolution of the residual. Hence this adaptive method seems to be efficient and permits to reconstruct the obstacle removing the oscillations due to the high frequencies.

Detecting objects with corners

In order to test the performances of our algorithm and of our adaptive method, we want now to detect more complicated obstacles and particularly, we wonder if it is effective to reconstruct objects containing straight lines and corners. The first objective is then to detect the square ω * 3 whose vertices are the points (-0.55, -0.55), (0.55, -0.55), (0.55, 0.55) and (-0.55, 0.55). We see in Figure 5 that this reconstruction is quite efficient. However, if we want to reconstruct a smaller object, the results are significantly worse. Indeed, in Figure 6, we reconstruct the square ω * 4 whose Figure 6: Reconstruction of ω * 4 using the adaptive method with 27 parameters vertices are the points (-0.2, -0.2), (0.2, -0.2), (0.2, 0.2) and (-0.2, 0.2). We see that the detection is not good. This phenomenon concerning the size of the object was underlined using the explicit calculus of the shape Hessian in Table 1 (in Example 8). Thus, the parametrization method coupling with our adaptive method seems to be efficient to reconstruct obstacles, even if the shapes are not trivial. However, the regularizing behavior complicate the detection of small objects or of distant parts of the measurement domain.

Influence of the size of the domain where measurements are made

Let us consider the influence of the size of O, the part of the boundary where the measurements are assumed to be made. We now restrict the domain O to some part of ∂Ω. We precise that we use here exactly the same algorithm and the same parameters and data that the ones used in Section 4.3. In particular we use the previously described adaptive method and we want to reconstruct the object ω * 2 which is given by (4.1). In Figure 7, we make the measurement only on a semicircle. On the left of Figure 7, O is the upper semicircle. As we could expect, the reconstruction of the object ω * 2 is less efficient than in the case O = ∂Ω (see Figure 4). In particular, the bottom of the obstacle (i.e. the part exposed to the part of the exterior boundary where we do not make the measurement) is poorly detected. We precise that here, we do not obtain a residual ε ≈ 10 -3 as in the case O = ∂Ω but only ε ≈ 6.8 • 10 -3 . On the right of Figure 7, O is the lower semicircle. We obtain similar results. As we expected, we see that in this case only the bottom of the obstacle is well detected.

In Figure 8, O is the right superior quart-circle. Here again, the reconstruction is less Figure 8: Reconstruction of ω * 2 using the adaptive method with 21 parameters restricting the measurement domain (O is a quarter of ∂Ω) efficient than in the case O = ∂Ω and even in the previous case O = 1 2 ∂Ω. We can notice again that only the part exposed to the measurement domain O is well determined. Here, we only obtain a residual ε ≈ 8.4 • 10 -3 . These two simulations emphasize an intuitive idea: the more the measurement domain O ⊂ ∂Ω is small, the more it is difficult to well reconstruct the object ω.

Detecting more than one object

Theoretically, besides the regularity assumptions, the main assumption is that Ω\ω is connected. This assumption does not exclude the case of two or more inclusions in Ω. Thus, we want now to detect numerically two objects: a square ω * 5 whose vertices are the points (-0.6, 0.1), (-0.1, 0.1), (-0.1, 0.6) and (-0.6, 0.6) and a circle ω * 6 centered at (0.35, -0.35) with radius 0.35. Here again we use the adaptive method described in Section 4.3. Since the computation is significantly longer than the one with only one object, we stop the experiment when the residual is ε ≈ 1.5e-02. We then obtain Figure 9. This result is satisfying and we can hope that it would be more efficient if we increase the accuracy of the computation (using a better finite elements discretization for example) and/or if we increase the computation time. Notice that the non-smooth square is not as-well reconstructed as the circle: this is caused by the polar representation we used.

Notice an important remark. The parametric method that we adopted here to detect the objects does not permit to modify the topology of the object ω. Thus, in order to detect two or more inclusions, we have to know how many objects are included in Ω. A solution could be to initialize the algorithm using the notion of topological gradient (see for example [START_REF] Ben Abda | Topological sensitivity analysis for the location of small cavities in Stokes flow[END_REF], [START_REF] Garreau | The topological asymptotic for PDE systems: the elasticity case[END_REF] or [START_REF] Soko | On the topological derivative in shape optimization[END_REF]) which could give us the number of inclusions and their rough location, providing initial shapes for our optimization method.

Conclusion

We have partially reconstructed an obstacle immersed in a fluid by minimizing the Kohn-Vogelius functional. The Kohn-Vogelius approach proposed here is not the classical one and permits to make the measurement only on a part of the exterior boundary and not on the whole exterior boundary. We have used shape optimization methods to compute the gradient and to motivate the use of a parametric model to solve numerically the inverse problem. Indeed, this problem is severely ill-posed. In order to highlight the exponentially ill-posedness of our problem, we have computed explicitly the shape Hessian matrix in a concentric annulus. We have then compared our Stokes case with the Laplacian situation and have concluded that the two cases are as ill-posed. The numerical simulations made in the bi-dimensional case were effective but oscillations appeared when we worked with high frequencies. In order to remove these oscillations, the adaptive method that we proposed seems to be efficient.

There is still some room for improvement principally concerning the numerical issues. Our algorithm permits to reconstruct more than one object if we know how many objects are included in the domain. The use of the notion of topological gradient (see for example [START_REF] Ben Abda | Topological sensitivity analysis for the location of small cavities in Stokes flow[END_REF], [START_REF] Garreau | The topological asymptotic for PDE systems: the elasticity case[END_REF] or [START_REF] Soko | On the topological derivative in shape optimization[END_REF]) could permit to give us the number of inclusions and their rough location, providing initial shapes for our optimization method.

Proof. Step 1: existence and uniqueness. Let us begin by studying the case of null divergence. According to [START_REF] Amrouche | Decomposition of vector spaces and application to the Stokes problem in arbitrary dimension[END_REF]Lemma 3.3], consider H ∈ H 1 (Ω\ω) such that div H = 0, H = h int on ∂ω, H = h ext on ∂Ω\O such that ∂Ω∪∂ω H • n = 0 and satisfying

H H 1 (Ω\ω) ≤ C h int H 1/2 (∂ω) + h ext H 1/2 (∂Ω\O) . (A.2)
Then the couple

(U := u -H, p) ∈ H 1 (Ω\ω) × L 2 (Ω\ω) satisfies            -ν∆U + ∇p = f + ν∆H in Ω\ω div U = 0 in Ω\ω -ν∂ n U + pn = h ext + ν∂ n H on O U = 0 on ∂Ω\O U = 0 on ∂ω.
According to Lax-Milgram's theorem, there exists a unique

U ∈ V O (Ω\ω) such that for all v ∈ V O (Ω\ω) ν Ω\ω ∇U : ∇v = f , v Ω\ω -ν Ω\ω ∇H : ∇v -h O + ν∂ n H, v O (A.3)
and we have, using (A.2),

U H 1 (Ω\ω) ≤ C f [H 1 (Ω\ω)] ′ + h int H 1/2 (∂ω) + h ext H 1/2 (∂Ω\O) + h O H -1/2 (O) .
(A.4)

In particular (A.3) is true for all v ∈ V O (Ω\ω) ∩ H 1 0 (Ω\ω). Then using De Rham's theorem (see for example [6, Lemma 2.7]), there exists p ∈ L 2 (Ω\ω), up to an additive constant, such that for all v ∈ H 1 0 (Ω\ω)

ν Ω\ω ∇U : ∇v - Ω\ω p div v = f H 1 0 (Ω\ω) , v H -1 (Ω\ω),H 1 0 (Ω\ω) -ν Ω\ω ∇H : ∇v. (A.5)
According to [START_REF] Amrouche | Decomposition of vector spaces and application to the Stokes problem in arbitrary dimension[END_REF]Lemma 3.3] or [18, Theorem 3.2], we define ϕ N ∈ H 1 (Ω\ω) such that div ϕ N = 1 in Ω\ω, ϕ N = 0 on ∂Ω\O and ϕ N = 0 on ∂ω with

O ϕ N • n = 0. Let v ∈ H 1 (Ω\ω) such that v = 0 on ∂Ω\O, v = 0 on ∂ω and define c b (v) = 1 ∂(Ω\ω) ϕ N • n ∂(Ω\ω) v • n.
Using again [START_REF] Amrouche | Decomposition of vector spaces and application to the Stokes problem in arbitrary dimension[END_REF]Lemma 3.3] or [18, Theorem 3.2], we define

v 2 ∈ V O (Ω\ω) in such a way that v = v 1 + v 2 + c b (v)ϕ N , where v 1 ∈ H 1 0 (Ω\ω) satisfies the following equality: div v 1 = div (v -c b (v)ϕ N ).
Then, using (A.3) and (A.5), we obtain

Ω\ω ν∇U : ∇v - Ω\ω p div v = f , v Ω\ω -ν Ω\ω ∇H : ∇v -h O + ν∂ n H, v O + Ω\ω ν∇U : ∇(c b (v)ϕ N ) - Ω\ω p div (c b (v)ϕ N ) -f , c b (v)ϕ N Ω\ω + ν Ω\ω ∇H : ∇(c b (v)ϕ N ) + h O + ν∂ n H, c b (v)ϕ N O .
Therefore, choosing the additive constant for p such that

Ω\ω p = ν Ω\ω ∇U : ∇ϕ N -f , c b (v)ϕ N Ω\ω + ν Ω\ω ∇H : ∇ϕ N + h O + ν∂ n H, c b (v)ϕ N O ,
we prove that there exists a unique pair (U , p) ∈ V O (Ω\ω) × L 2 (Ω\ω) such that for all v ∈ H 1 (Ω\ω) with v = 0 on ∂Ω\O and v = 0 on ∂ω,

Ω\ω ν∇U : ∇v - Ω\ω p div v = f , v Ω\ω -ν Ω\ω ∇H : ∇v -h O + ν∂ n H, v O . (A.6)
Step 2: estimate. [START_REF] Amrouche | Decomposition of vector spaces and application to the Stokes problem in arbitrary dimension[END_REF]Lemma 3.3]). Using v in (A.6), and according to (A.4), we obtain

Let v := v + c(p)ϕ N , where c(p) := 1 |Ω\ω| Ω\ω p and v ∈ H 1 0 (Ω\ω) is such that div v = p -c(p) and v H 1 0 (Ω\ω) ≤ C p L 2 (Ω\ω) (see
U H 1 (Ω\ω) + p L 2 (Ω\ω) ≤ C f [H 1 (Ω\ω)] ′ + h int H 1/2 (∂ω) + h ext H 1/2 (∂Ω\O) + h O H -1/2 (O)
and hence

u H 1 (Ω\ω) + p L 2 (Ω\ω) ≤ C f [H 1 (Ω\ω)] ′ + h int H 1/2 (∂ω) + h ext H 1/2 (∂Ω\O) + h O H -1/2 (O)
Step 3: case g = 0. The first part of the theorem is proved for g = 0. The case g = 0 is obtained by a lifting argument. Let us define

u g := u g + c(g)ϕ N - 1 |Ω\ω| Ω\ω ( u g + c(g)ϕ N ) ∈ H 1 (Ω\ω),
where c(g) := 1 |Ω\ω| Ω\ω g and where, according to [18, Theorem 3

.2] or [7, Lemma 3.3], u g ∈ H 1 0 (Ω\ω) is such that div u g = g -c(g) and u g H 1 0 (Ω\ω) ≤ C g L 2 (Ω\ω) . Thus u g is such that Ω\ω u g = 0, div u g = g in Ω\ω and u g H 1 (Ω\ω) ≤ C g L 2 (Ω\ω) .
Thus, defining u 0 := uu g and

f : v ∈ H 1 (Ω\ω) → f , v Ω\ω -ν Ω\ω ∇H : ∇v -ν Ω\ω ∇u g : ∇v, the problem (A.1) is equivalent to        Find (u 0 , p) ∈ V O (Ω\ω) × L 2 (Ω\ω) such that Ω\ω ν∇u 0 : ∇v - Ω\ω p div v = f , v Ω\ω -h O + ν∂ n H , v O ∀v ∈ H 1 (Ω\ω), v = 0 on ∂ω, v = 0 on ∂Ω\O.
Then we proceed in the same manner as the case g = 0.

Let us now state and prove a local regularity result on the solution of the Stokes problem with mixed boundary conditions. We first introduce the following notations: for k, m ∈ N, k < m and for two open sets Ω 1 et Ω 2 such that Ω 2 ⊂ Ω 1 , we denote by X k,m (Ω 1 , Ω 2 ) the space of functions in H k (Ω 1 ) such that their restriction to Ω 2 belongs to H m (Ω 2 ). Similarly, we note X * ,m (Ω 1 , Ω 2 ) the space of functions in H 1 (Ω 1 )

′ such that their restriction to Ω 2 belongs to H m (Ω 2 ).

⊂ ∂C ′ , C \∂ω ⊂ C ′ and C ′ ⊂ Ω. Let (f , g, h O , h ext , h int ) ∈ X * ,k (Ω\ω, C ′ ) × X 0,k+1 (Ω\ω, C ′ ) × H -1/2 (O) × H 1/2 (∂Ω\O) × H k+ 1 2 (∂ω).
We consider (u, p) ∈ H 1 (Ω\ω) × L 2 (Ω\ω) the solution of the following Stokes problem

           -ν∆u + ∇p = f in Ω\ω div u = g in Ω\ω -ν∂ n u + pn = h O on O u = h ext on ∂Ω\O u = h int on ∂ω. (A.7)
Then (u, p) belongs to H k+2 (C) × H k+1 (C) and the following estimate holds:

u H k+2 (C) + p H k+1 (C) ≤ C f X * ,k (Ω\ω,C ′ ) + g X 0,k+1 (Ω\ω,C ′ ) + h O H -1/2 (O) + h ext H 1/2 (∂Ω\O) + h int H k+ 1 2 (∂ω) .
Proof. First, let us consider the case k = 0. We define

V = C ∪ ω, V ′ = C ′ ∪ ω and ϕ ∈ C ∞ c (Ω) such that 0 ≤ ϕ ≤ 1, ϕ ≡ 1 in V and ϕ = 0 in Ω\V ′ . Let (u, p) ∈ H 1 (Ω\ω) × L 2 (Ω\ω)
the solution of problem (A.7) given by Theorem 11. Using (A.7) we check

       -ν∆(ϕu) + ∇(ϕp) = f in C ′ div (ϕu) = g in C ′ ϕu = 0 on ∂V ′ ϕu = h int on ∂ω,
where f := ϕf -νu∆ϕ -2ν∇u∇ϕ + p∇ϕ belongs to L 2 (C ′ ) and g := ϕg + u • ∇ϕ belongs to H 1 (C ′ ). From the regularity of the solutions of the Stokes equations with Dirichlet boundary conditions (see for example [START_REF] Boyer | Éléments d'analyse pour l'étude de quelques modèles d'écoulements de fluides visqueux incompressibles[END_REF] or [START_REF] Galdi | An introduction to the mathematical theory of the Navier-Stokes equations[END_REF]), ϕu ∈ H 2 (C ′ ), ϕp ∈ H 1 (C ′ ) and from the expression of f and g we obtain

ϕu H 2 (C ′ ) + ϕp H 1 (C ′ ) ≤ C ϕf L 2 (C ′ ) + ϕg H 1 (C ′ ) + u H 1 (Ω\ω) + p L 2 (Ω\ω) + h int H 1 2 (∂ω)
.

Using this inequality and the estimate on u H 1 (Ω\ω) and p L 2 (Ω\ω) given by Theorem 11, we obtain the announced estimate for k = 0. We then proceed by induction for the cases k ≥ 1.

B Resolution of the Stokes equations in a concentric annulus

In this section, we solve explicitly some Stokes boundary value problems set in the concentric annulus

Ω ρ = {x ∈ R 2 , ρ < |x| < 1}.
The pair ( e r , e θ ) := cos θ sin θ , -sin θ cos θ represents the polar coordinates system.

In order to simplify the expressions, we here assume ν = 1 and O = ∂Ω. We first use the partial differential equations to derive the special form of the solution, then reduce the resolution of the boundary value problem to some linear system.

B.1 Using the PDE

Noticing that the incompressibility condition div u = 0 implies that the pressure p is harmonic, we pass in polar coordinates (r, θ) in which p can be expanded in Laurent's series. Hence we seek the velocity u and the pressure under the form: We distinguish the case n = 0 of the other Fourier modes.

u(r, θ) = u r (r, θ) u θ (r, θ) =     n∈Z u n,r (r)e inθ n∈Z u n,θ (r)e inθ
Case n 0. We are led to solve:

u ′′ 0,r + 1 r u ′ 0,r - 1 r 2 u 0,r = β 0 r and u ′′ 0,θ + 1 r u ′ 0,θ - 1 r 2 u 0,θ = 0, whose solutions are u 0,r (r) = A 0,r r + B 0,r r + β 0 2 r ln r and u 0,θ (r) = A 0,θ r + B 0,θ r .
The incompressibility condition imposes:

u 0,r (r) = B 0,r r and u 0,θ (r) = A 0,θ r + B 0,θ r .
Case n = 0. We are led to solve the system:

u ′′ n,r + 3 r u ′ n,r -n 2 -1 r 2 u n,r = |n| α n r |n|-1 -β n r -|n|-1 u ′′ n,θ + 1 r u ′ n,θ -n 2 +1 r 2 u n,θ = in α n r |n|-1 + β n r -|n|-1 -2r -2 u n,r .
As we have a triangular system, we first determine the radial component u n,r . We get after some calculations

u n,r (r) = A n,r r n-1 + B n,r r -n-1 + αn 8 r 2 -βn 2 ln r if n = ±1 u n,r (r) = A n,r r n-1 + B n,r r -n-1 + |n| 4 αn |n|+1 r |n|+1 + βn |n|-1 r -|n|+1
else.

Then we solve the equation in the angular component and impose the incompressibility condition to get

                     u n,θ (r) = A n,r i r n-1 -B n,r i r -n-1 + α n i sg(n) 2 - n 8 r 2 - i 2 β n (sg(n) + n ln r) if n = ±1, u n,θ (r) = A n,r i r n-1 -B n,r i r -n-1 + α n i sg(n) 2 - n 4(|n| + 1) r |n|+1 +β n i - sg(n) 2 + n 4(|n| -1) r -|n|+1 else,
where sg(n) denotes the sign of n. Since we deal with states taking real values, we have obviously

A -n,r = B n,r α -n = α n β -n = β n .
The solution u. Noticing u n = u n,r u n,θ , we have for n > 1

u n (r) = A n,r r n-1 1 i + B n,r r n+1 1 -i + α n r n+1 n 4(1+n) n+2 4(1+n) i + β n r 1-n n 4(n-1) 2-n 4(n-1) i with p = n∈Z *
α n r |n| + β n r -|n| + α 0 . We characterize in the same way the solution for n ≤ 1.

The normal derivative ∂ n u. We easily obtain for n > 1

∂ n u n (r) = (n -1)A n,r r n-2 1 i + (n + 1) B n,r r n+2 -1 i + α n r n n 4 n+2 4 i + β n r -n -n 4 n-2 4 i (B.1)
and we characterize ∂ n u n in the same way for n ≤ 1.

The Neumann boundary condition -∂ n u + pn. We easily obtain for n > 1

-∂ n u n (r) + p n n = A n,r r n-2 1 -n i(1 -n) + B n,r r n+2 n + 1 -i(n + 1) + α n r n 1 -n 4 -n+2 4 i + β n r -n 1 + n 4 2-n 4 i (B.2)
and we characterize -∂ n u n + p n n in the same way for n ≤ 1.

B.2 Solving the boundary value problems

In order to compute the shape hessian at Ω ρ , we have to solve two types of boundary value problem: first a Dirichlet's one then a mixed one with Dirichlet boundary condition on the inner boundary and a Neumann boundary condition on the outer one. We recall that we assume for these computations that O = ∂Ω. Thus Problem (1.5)

is now        -ν∆u N + ∇p N = 0 in Ω\ω div u N = 0 in Ω\ω -ν∂ n u N + p N n = g on ∂Ω u N = 0 on ∂ω.
From now, we only focus on the cases n > 1. We point out that we will choose boundary data such that all the coefficient will be real valued. Thus

A -n,r = B n,r = B n,r α -n = α n = α n β -n = β n = β n .
The solution u D of the Dirichlet Stokes system. Let us consider f (ext) ∈ L 2 (S 1 ) and f (int) ∈ L 2 (ρS 1 ). One seeks u and p such that

u = f (ext) (θ) = f (ext) r (θ) f (ext) θ (θ) on ∂Ω = S 1 u = f (int) (θ) = f (int) r (θ) f (int) θ (θ) on ∂ω = ρS 1 .
The boundary conditions are expanded in Fourier series:

f (ext) r (θ) = n∈Z f (ext)
n,r e inθ and f

(ext) θ (θ) = n∈Z f (ext) n,θ e inθ , f (int) r (θ) = n∈Z f (int)
n,r e inθ and f

(int) θ (θ) = n∈Z f (int) n,θ e inθ .
We have to seek the (unique) solution of the following linear system

                       A D n,r + B D n,r + n 4n + 4 α D n + n 4n -4 β D n = f (ext) n,r A D n,r - B D n,r + n + 2 4n + 4 α D n + 2 -n 4n -4 β D n = f (ext) n,θ i ρ n-1 A D n,r + ρ -n-1 B D n,r + n 4n + 4 ρ n+1 α D n + n 4n -4 ρ 1-n β D n = f (int) n,r ρ n-1 A D n,r -ρ -n-1 B D n,r + n + 2 4n + 4 ρ n+1 α D n + 2 -n 4n -4 ρ 1-n β D n = f (int) n,θ i .
We point out that we will choose boundary data such that all the coefficients A D n,r , B D n,r , α D n and β D n will be real valued. We denote

M n = (ρ n -ρ -n ) 2 -n 2 (ρ -ρ -1 ) 2 .
Using the fact that

ρ 2 -n 2 ρ 2n + 2 -1 + n 2 ρ 2+2n -n 2 ρ 4+2n + ρ 2+4n = ρ 2 ρ 2n M n ,
we then get the solution of the linear system:

A D n,r = n 2 (1 -ρ 2 ) + (n + 2)(ρ -2n -1) 2M n f (ext) n,r + n((1 -ρ -2n ) + n 2 (1 -ρ 2 ) 2M n f (ext) n,θ i + nρ -n (ρ -ρ -1 ) -ρ(ρ n -ρ -n ) 2M n f (int) n,r + nρ -n (ρ -ρ -1 ) + (n + 2)ρ(ρ n -ρ -n ) 2M n f (int) n,θ i B D n,r = n 2 (1 -ρ 2 ) + (n -2)(1 -ρ 2n ) 2M n f (ext) n,r + (1 -ρ 2n ) + n(ρ 2 -1) 2M n f (ext) n,θ i + (1 -n)ρ -n (ρ -1 -ρ) + ρ n-1 -ρ 1-n ) M n f (int) n,r + n 2 ρ n (ρ -ρ -1 ) + nρ(ρ n -ρ -n ) 2M n f (int) n,θ i α D n = (1 -ρ -2n ) + (n -2)(1 -ρ -2 ) M n f (ext) n,r + 2(1 + n) n(1 -ρ -2 ) -(1 -ρ -2n ) M n f (ext) n,θ i + (n -2)ρ -n (ρ -ρ -1 ) + ρ -1 (ρ n -ρ n-1 ) 2M n f (int) n,r + (1 + n)ρ n (ρ -ρ -1 ) + (ρ n+1 -1 ρ n+1 ) 2M n f (int) n,θ i β D n = (n -2)(1 -ρ -2 ) + (1 -ρ 2n ) M n f (ext) n,r + n(1 -ρ -2 ) + (1 -ρ 2n ) M n f (ext) n,θ i +2(1 -n) (1 + n)ρ n (ρ -ρ -1 ) + (ρ n+1 -1 ρ n+1 ) 2M n f (int) n,r +2(n -1) nρ n (ρ -ρ -1 ) + ρ -1 (ρ n -ρ -n ) M n f (int) n,θ i .
Remark 13. We quote from Engliš and Peetre that M n is called an Almansi determinant (see [START_REF] Engliš | A Green's function for the annulus[END_REF]). This arise in connection with the following interpolation problem: try to reconstruct a function f of the type f (x) = P (x) + e µx Q(x) where P and Q are poynomials of respective degree m, and n, given its value at m + n points x 1 , x 2 , . . . , x m+n .

Remark 14. To compare the factorization in the case of the Laplacian and the case of the Stokes system, we recall that in the case of the Laplacian, we find a denominator equal to

ρ |n| -ρ -|n| = ρ -|n| (ρ 2|n| -1) = ρ -|n| (ρ |n| -1)(ρ |n| + 1).
In the case of Stokes system, we rewrite the denominator as

M n = (ρ |n| -ρ -|n| ) + |n|(ρ 2 -ρ -2 ) (ρ |n| -ρ -|n| ) -|n|(ρ 2 -ρ -2 ) .
Application to our case for u D . We get

     A D n,r B D n,r α D n β D n      =        n 2 (1-ρ 2 )+(n+2)(ρ -2n -1) 2Mn n((1-ρ -2n )+n 2 (1-ρ 2 ) 2Mn n 2 (1-ρ 2 )+(n-2)(1-ρ 2n ) 2Mn (1-ρ 2n )+n(ρ 2 -1) 2Mn (1-ρ -2n )+(n-2)(1-ρ -2 ) Mn n(1-ρ -2 )-(1-ρ -2n ) Mn (n-2)(1-ρ -2 )+(1-ρ 2n ) Mn 2(1 -n) n(1-ρ -2 )+(1-ρ 2n ) Mn              f n,r f n,θ i       . We choose f (ext) (θ) = f (θ) = cos (nθ) e r , n > 1 and f (int) = 0. Hence f n,r = 1 2 , f n,θ = 0 and      A D n,r B D n,r α D n β D n      = 1 2        n 2 (1-ρ 2 )+(n+2)(ρ -2n -1) 2Mn n 2 (1-ρ 2 )+(n-2)(1-ρ 2n ) 2Mn (1-ρ -2n )+(n-2)(1-ρ -2 ) Mn (n-2)(1-ρ -2 )+(1-ρ 2n ) Mn        .
The solution u N of the mixed Stokes system. The coefficients are denoted A N n,r , B N n,r and α N n and β N n . To fulfill the boundary conditions

-∂ n u N + p N e r = g (ext) if |x| = 1 and u N = g (int) if |x| = ρ,
we have to seek the (unique) solution of the following linear system

                       (1 -n)A N n,r + (1 + n)B N n,r + 4 -n 4 α N n + 4 + n 4 β N n = g (ext) n,r (1 -n)A N n,r -(1 + n)B N n,r - 2 + n 4 α N n + 2 -n 4 β N n = g (ext) n,θ i ρ n-1 A N n,r + ρ -n-1 B N n,r + ρ n+1 n 4n + 4 α N n + ρ 1-n n 4n -4 β N n = g (int) n,r ρ n-1 A N n,r -B N n,r ρ -n-1 + ρ n+1 n + 2 4n + 4 α N n + ρ 1-n 2 -n 4n -4 β N n = g (int) n,θ i . (B.3) We denote N n = 3ρ 2 + (8 + n 2 )ρ 2n -2(-1 + n 2 )ρ 2+2n + n 2 ρ 4+2n + 3ρ 2+4n
which can be written under the form

N n = 3(ρ n -ρ -n ) 2 + n 2 (ρ -ρ -1 ) 2 + 8(1 + ρ -2 ).
The solution of the system corresponding to a general Neumann problem with Dirichlet conditions on the interface is obtained after expanding the inverse of the matrix corresponding to (B.3). We get

A N n,r = - (n + 2)(ρ -2n -1) + n 2 (ρ 2 -1) + 4n 2(n -1)N n g (ext) n,r + n((1 -ρ -2n ) + n 2 (1 -ρ 2 ) 2N n g (ext) n,θ i + 3(n + 2)ρ(ρ n -ρ -n ) + 4(n + 2)ρ -n (ρ -ρ -1 ) + 4(n + 4)ρ -1-n 2N n g (int) n,r + n(n + 4)ρ -n (ρ -ρ -1 ) + 3nρ(ρ n -ρ -n ) + 4(n + 2)ρ -1-n 2N n g (int) n,θ i B N n,r = - n 2 (1 -ρ 2 ) + (n -2)(ρ 2n -1) + 4n 2(1 + n)N n g (ext) n,r - (4 + n)(ρ 2n -1) + n 2 (ρ 2 -1) + 4(n + 2) 2(1 + n)N n g (ext) n,θ i - (3(n -2)ρ(ρ -n -ρ n ) -(n 2 + 8)ρ n (ρ -1 -ρ) -2(n + 2)ρ1 + n 2N n g (int) n,r + n 2 ρ n (ρ -ρ -1 ) -nρ(ρ n -ρ -n ) -4n(ρ 1-n -ρ n-1 ) -4(n + 2)ρ n-1 2N n g (int) n,θ i α N n = 2 (2 -n)(ρ -2 -1) + (ρ -2n -1) + 4 N n g (ext) n,r +2 (4 + n)(1 -ρ -2 ) + (1 -ρ -2n ) -4 N n g (ext) n,θ i -2(1 + n) (n -1)ρ(ρ -n -ρ n ) + 3ρ(ρ n -ρ -n ) + (4 -n)ρ 1-n N n g (int) n,r +2(1 + n) (n -1)ρ -n (ρ -ρ -1 ) + 3(ρ n-1 -ρ 1-n ) + 4ρ 1-n N n g (int) n,θ i β N n = (n + 2)(ρ -2 -1) + (ρ 2n -1) + 4 N n g (ext) n,r + (4 -n)(ρ -2 -1) + (ρ 2n -1) + 2(n -2) N n g (ext) n,θ i +2(n -1) (1 + n)ρ n (ρ -ρ -1 ) + 3ρ -1 (ρ n -ρ -n ) + (n + 4)ρ n-1 N n g (int) n,r +2(n -1) nρ n (ρ -ρ -1 ) -ρ -1 (ρ n -ρ -n ) -4ρ-1 -n N n g (int) n,θ i .
The Dirichlet-to-Neumnn map. Here, we have to take the Neumann data

g (ext) = g (ext) r e r + g (ext) θ e θ
corresponding to the choice of the Dirichlet data f (ext) . In order to simplify the expression and to get closer to our case, we assume f (int) = 0. Omitting the exponent (ext) , a straightforward calculation gives

g n,r = A D n,r (1 -n) + B D n,r (1 + n) + 4 -n 4 α D n + 4 + n 4 β D n g n,θ = i A D n,r (1 -n) -B D n,r (1 + n) - 2 + n 4 α D n + 2 -n 4 β D n ,
where Fourier coefficients A D n,r , B D n,r , α D n and β D n are given above. The expression with respect to the Fourier coefficients f n,r and f n,θ is then given by

g n,r = - 4(2 -n 2 )(1 -ρ -2 ) + 2n(ρ -2n -ρ 2n ) + (ρ n -ρ -n ) 2 + n 2 (ρ -ρ -1 ) 2 M n f n,r + 4n(1 -ρ -2 ) + 2(ρ -2n -ρ -2n ) + n 3 (ρ -ρ -1 ) 2 + n(ρ n -ρ -n ) 2 M n f n,θ i and g n,θ = i 4n(1 -ρ -2 ) + 2(ρ -2n -ρ 2n ) + n(ρ n -ρ -n ) 2 + n 3 (ρ -ρ -1 ) 2 M n f n,r - 4n 2 (1 -ρ -2 ) + 2n(ρ -2n -ρ 2n ) + (ρ -n -ρ n ) 2 + n 2 (ρ -1 -ρ) 2 M n f n,θ .
We illustrate the mapping (f n,r , f n,θ ) → (g n,r , g n,θ ) via the linear transformation

g n,r g n,θ = ∆ (n) 1,1 ∆ (n) 1,2 ∆ (n) 2,1 ∆ (n) 2,2 f n,r f n,θ where
∆ (n) 1,1 = - 4(2 -n 2 )(1 -ρ -2 ) + 2n(ρ -2n -ρ 2n ) + (ρ n -ρ -n ) 2 + n 2 (ρ -ρ -1 ) 2 M n ∆ (n) 1,2 = -i 4n(1 -ρ -2 ) + 2(ρ -2n -ρ -2n ) + n 3 (ρ -ρ -1 ) 2 + n(ρ n -ρ -n ) 2 M n ∆ (n) 2,1 = ∆ (n) 1,2 ∆ (n) 2,2 = - 4n 2 (1 -ρ -2 ) + 2n(ρ -2n -ρ 2n ) + (ρ -n -ρ n ) 2 + n 2 (ρ -1 -ρ) 2 M n .
We can deduce the Dirichlet-to-Neumann map from this representation: in the trigonometric basis it is represented by a block diagonal matrix; its n-th diagonal is given by the matrix ∆

(n) = (∆ (n) i,j ) i,j=1,2 .
Application to our case for u N . In our case, with f (ext) (θ) = f (θ) = cos (nθ) e r , n > 1 and f (int) = 0, we then obtain

       g n,r = - 4(2 -n 2 )(1 -ρ -2 ) + 2n(ρ -2n -ρ 2n ) + (ρ n -ρ -n ) 2 + n 2 (ρ -ρ -1 ) 2 M n f n,r g n,θ = i 4n(1 -ρ -2 ) + 2(ρ -2n -ρ 2n ) + n(ρ n -ρ -n ) 2 + n 3 (ρ -ρ -1 ) 2 M n f n,r . (B.4) Hence      A N n,r B N n,r α N n β N n      =         -(n+2)(ρ -2n -1)+n 2 (ρ 2 -1)+4n 2(n-1)Nn n((1-ρ -2n )+n 2 (1-ρ 2 ) 2Nn
n 2 (1-ρ 2 )+(n-2)(ρ 2n -1)+4n

2(1+n)Nn -(4+n)(ρ 2n -1)+n 2 (ρ 2 -1)+4(n+2) 2(1+n)Nn 2 (2-n)(ρ -2 -1)+(ρ -2n -1)+4 Nn 2 (4+n)(1-ρ -2 )+(1-ρ -2n )-4 Nn (n+2)(ρ -2 -1)+(ρ 2n -1)+4 Nn

B.3

The formulae for the shape derivatives u ′ D and u ′

N

In this section, we deform the domain via the vector field V = cos kθ e r . We then have V n = cos kθ. Here again, we only focus on the cases n > 1.

The formula for the shape derivative u ′ D . For n > 1, we set The formula for the shape derivative u ′ N . We adopt the same strategy. We have only to write the corresponding Neumann condition on the exterior boundary (i.e. for |x| = 1). For n > 1, we set

K N n,r = A N n,r (1 -n)ρ n-2 + ρ -n-2 B N n,r (1 + n) -α N n ρ n n 4 + β N n ρ -n n 4 L N n,r = A N n,r (1 -n)ρ n-2 -ρ -n-2 B N n,r (1 + n) -α N n ρ n n + 2 4 + β N n ρ -n 2 -n 4

lθ) e r ] = 0 andD 2 J

 02 KV (ω * ) [(cos kθ) e r , (cos lθ) e r ] = D 2 J KV (ω * ) [(sin kθ) e r , (sin lθ) e r ]

Figure 1 :

 1 Figure 1: The typical spectrum of the shape Hessian in decimal logarithm scale

First

  , we want to detect an obstacle ω * 1 contained in the class of objects with which we work. We want to detect the obstacle which boundary is parametrized by: ∂ω * 1 = 0.15 0.15 + 0.65 -0.25 cos θ -0.15 sin θ cos θ sin θ , θ ∈ [0, 2π) ,

Figure 2 :

 2 Figure 2: Reconstruction of ω * 1 with different numbers of parameterswe work with five parameters, the residual ε corresponds to a very good approximation of the object. But the more we add parameters, the more the functional flattens and the more the approximation corresponding to the residual ε is poor. Indeed, we see oscillations of the final object when we work with high frequencies. We can also notice that it is longer to obtain the residual ε with many parameters.The same experiment was conducted to detect a more complicated obstacle which does not belong to the discretized set of objects. It is parametrized by:

Figure 3 :

 3 Figure 3: Reconstruction of ω * 2 with different numbers of parameters the boundary and then a poorer approximation.Here again, more iterations are needed to obtain the residual ε. In order to compare the errors, we represent on the second graph of Figure3the logarithm of the residual obtained with 17 parameters (noted log(J 17 )) versus the logarithm of the residual obtained with 21 parameters (noted log(J 21 )). We then obtain a regression line whose equation is given by

Figure 5 :

 5 Figure 5: Reconstruction of ω * 3 using the adaptive method with 27 parameters

Figure 7 :

 7 Figure 7: Reconstruction of ω * 2 using the adaptive method with 21 parameters restricting the measurement domain (O is a half of ∂Ω)

Figure 9 :

 9 Figure 9: Reconstruction of ω * 5 and ω * 6 using the adaptive method with 19 parameters

Theorem 12 (

 12 Local regularity result). Let k ∈ N, ν > 0, Ω a bounded Lipschitz open set of R N (N ∈ N * ) and ω an open set with a C k+1,1 boundary such that ω ⊂⊂ Ω and Ω\ω is connected. Let O be an open subset of the exterior boundary ∂Ω and ν > 0. Let C and C ′ two smooth open subsets of Ω\ω such that ∂ω ⊂ ∂C, ∂ω

  p(r, θ) = n∈Z p n (r)e inθ , with p 0 (r) = α 0 + 0 ln r and p n (r) = α n r |n| + β n r -|n| , n ∈ Z * .

( 4 -

 4 n)(ρ -2 -1)+(ρ 2n -1)+2(n

ρ n n 4 4 (B. 5 )=

 445 = A D n,r (1 -n)ρ n-2 + ρ -n-2 B D n,r (1 + n) -α D n (1 -n)ρ n-2 -ρ -n-2 B D n,r (1 + n) -α D n ρ n n + 2 4 + β D n ρ -n 2 -nthe respective Fourier coefficients of -∂ n u D (ρ, θ). As it has been noticed, when n > 1, we haveK D -n,r = K D n,rand L D -n,r = -L D n,r .We will then focus on the coefficientsA D ′ n,r , B D ′ n,r , α D ′ n and β D ′ n for n > 1.Let us point out that the computations give K D n,r = 0. We have (see (B.1))-∂ n u D (ρ, θ) = K D n,r e inθ + K D -n,r e -inθ e r + i L D n,r e inθ + L D -n,r e -inθ e θ = i L D n,r e inθ -L D n,r e -inθ e θand then-∂ n u D V n = i L D n,r e i(n+k)θ + L D -n,r e -i(n+k)θ 2 e θ + i L D n,r e i(n-k)θ + L D -n,r e -i(n-k)θ 2 e θ . Hence -∂ n u D V n = -L D n,r sin ((n + k)θ) + sin ((n -k)θ) e θ .We denoteu ′ D = p∈I n,k u D ′ p,r (r)e ipθ e r + p∈I n,k u D ′ p,θ (r)e ipθ e r ,whereI n,k = {n + k, n -k, -n -k, -n + k}.Let us assume that -1, 0, 1 ∈ I n,k (the other cases can be treated similarly). We know that                     u D ′ p,r (r) = A D ′ p,r r p-1 + B D ′ p,r r p+1 + α D ′ p r |p|+1 |p| 4|p| + 4 + β D ′ p r 1-|p| |p| 4|p| -4 u D ′ p,θ (r) = i A D ′ p,r r p-1 -B D ′ p,r r p+1 + α p r |p|+1 1We use the superposition principles: we use the fact that we have to use a linear system associated to a Dirichlet problem and then the coefficients associated to u D ′ p,r (r) when p = |n + k| and p = |n -k| must satisfy the following linear system                     p-1 A D ′ p,r + ρ -p-1 B D ′ p,r + p 4p + 4 ρ p+1 α D ′ p + p 4p -4 ρ 1-p β D ′ p = K D n,r ρ p-1 A D ′ p,r -ρ -p-1 B D ′ p,r + p + 2 4p + 4 ρ p+1 α D ′ p + 2 -n 4p -4 ρ 1-p β D ′ p = L D n,r .We remark that the second side of the linear system is independent of k. Once the coefficients are computed, we use (B.2) to get(-∂ n u ′ D + p ′ D n)(ρ, θ) = K D ′ |n+k|,r cos (n + k)θ + K D ′ |n-k|,r cos (n -k)θ e r -L D ′ |n+k|,r sin (n + k)θ + L D ′ |n-k|,r sin (n -k)θ e θ ,whereK D ′ p,r = A D ′ p,r (1 -p)ρ p-2 + ρ -p-2 B D ′ p,r (1 + p) + α D ′ p ρ p 1 -A D ′ p,r (1 -p)ρ p-2 -ρ -p-2 B D ′ p,r (1 + p) -α D ′ p ρ p p + 2 4 + β D ′ p ρ -p 2 -p 4 .

- 2 n 2 ( 1 p 2 = 4 .

 22124 the respective Fourier coefficients of -∂ n u N (ρ, θ). The computations give (n + 2)ρ -n (1 -ρ-2 ) + 4ρ -2 (ρ -n -ρ n ) + 2n(n -1)ρ n (ρ -2 -1) (ρ n + ρ -n )(ρ -2 -1) + 2n(ρ n -ρ -n ) + (1 + ρ -2 ) -p)A N ′ p,r + (1 + p)ρ -p-2 B N ′ p+ 4)ρ -p (ρ -ρ -1 ) + 3pρ(ρ p -ρ -p ) + 4(p + 2)ρ -1ρ p (ρ -ρ -1 ) -pρ(ρ p -ρ -p ) -4p(ρ 1-p -ρ p-1 ) -4(p + 2)ρ p-1 2N p L N n,r α N ′ p = 2(1 + p) (p -1)ρ -p (ρ -ρ -1 ) + 3(ρ p-1 -ρ 1-p ) + 4ρ 1-p N p L N n,r β N ′ p = 2(p -1) pρ p (ρ -ρ -1 ) -ρ -1 (ρ p -ρ -p ) -4ρ -1-p N p L N n,r .We have to compute now(-∂ n u ′ N + p ′ N n)(ρ, θ) = K N ′ |n+k|,r cos (n + k)θ + K N ′ |n-k|,r cos (n -k)θ e r -L N ′ |n+k|,r sin (n + k)θ + L N ′ |n-k|,r sin (n -k)θ e θwhereK N ′ p,r = A N ′ p,r (1 -p)ρ p-2 + ρ -p-2 B N ′ p,r (1 + p) + α N ′ p ρ n (1 -A N ′ p,r (1 -p)ρ p-2 -ρ -p-2 B N ′ p,r (1 + p) -α N ′ p ρ p p + 2 4 + β N ′ p ρ -p 2 -p Finally it comes -∂ n (u ′ Nu ′ D ) + (p ′ N -p ′ D )n (ρ, θ) = K |n+k|,r cos (n + k)θ+ K |n-k|,r cos (n -k)θ e r -L |n+k|,r sin (n + k)θ + L |n-k|,r sin (n -k)θ e θ where K p,r = K D ′ p,r -K N ′ p,r and L p,r = L D ′ p,r -L N ′ p,r . (B.6)

Table 2 :

 2 Influence of the high frequencies on the smallest eigenvalue λ 1 of the shape Hessian D 2 J KV (ω * ).

	k, l 1,. . . ,5	1,. . . ,10 1,. . . ,15 1,. . . ,30	1,. . . ,40	1,. . . ,60
	λ 1 6.4483e+03 137.7707 14.5852 6.9862e-04 3.6934e-06 4.7546e-12

Acknowledgments The authors thank Pr. G. Vial for his help on the use of the Mélina library. The work of F. Caubet and M. Dambrine is part of the project ANR-09-BLAN-0037 Geometric analysis of optimal shapes (GAOS) financed by the French Agence Nationale de la Recherche (ANR).

A Some results on the Stokes problem with mixed conditions

We recall classical results about the Stokes problem with mixed boundary conditions: a theorem of existence and uniqueness of the solution and a local regularity result. In all this appendix, we note C a generic positive constant, only depending on the geometry of the domain and on the dimension, which may change from line to line. ′ and H 1 (Ω\ω) and the duality product between H -1/2 (∂Ω) and

Then, the problem

admits a unique solution (u, p) ∈ H 1 (Ω\ω) × L 2 (Ω\ω) and the following estimate holds: