Keywords: perturbation, développement asymptotique, problème de Stokes perturbation, asymptotic expansion, Stokes problem. iii

La premième partie a pour but l'établissement d'un développement asymptotique pour la solution du problème de Stokes avec une petite perturbation du domaine. Dans ce travail, nous avons appliqué la théorie du potentiel. On a écrit les solutions du problème non-perturbé et du problème perturbé sous forme d'opérateurs intégraux. En calculant la différence, et en utilisant des propriétés liées aux noyaux des opérateurs on a établi un développement asymptotique de la solution.

L'objectif principal de la deuxième partie de ce rapport est de déterminer les termes d'ordre élevé du développement asymptotique des valeurs propres et fonctions propres pour l'opérateur de Stokes dus aux changements d'interface de l'inclusion. Dans la troisième partie, nous proposons une méthode pour l'évaluation des integrales singulières provenant de la mise en oeuvre de la méthode des éléments finis de frontière en électromagnetisme. La méthode que nous adoptons consiste en une réduction récursive de la dimension du domaine d'intégration et aboutit à une représentation de l'intégrale sous la forme d'une combinaison linéaire d'intégrales à une dimension dont l'intégrand est régulier et qui peuvent s'évaluer numériquement mais aussi explicitement. Pour la discrétisation du domaine, des triangles plans sont utilisés ; par conséquent, nous évaluons des intégrales sur le produit de deux triangles. La technique que nous avons développée nécessite de distinguer entre diverses configurations géométriques

Introduction Introduction

This thesis has three main objectives:

• The first goal is to derive an asymptotic expansion for the solution of a transmission Stokes problem with a small perturbation on the interface.

• The second goal is to present a schematic way to derive high-order asymptotic expansions for both eigenvalues and eigenfunctions for the Stokes operator caused by small perturbations of the boundary • The third one is to provide and develop a new method of precise evaluation of singular integrals derived from integral equation methods.

Perturbation problem

Perturbation theory is the study of the effects of small disturbances. It's basic idea is to obtain an approximate solution of a mathematical problem by exploiting the presence of a small parameter. An introduction to perturbation theory can be found in [START_REF] Hinch | Perturbation methods[END_REF], [START_REF] Kato | Perturbation Theory for Linear Operators[END_REF] or [START_REF] Van Dyke | Perturbation methods in fluids mechanics[END_REF].

A theoretical approach of this problem remains of interest for two main reasons: to study the behavior of the solution of perturbation problem, and to derive the perturbation approximation. The derived approximation formula can be applied in the theory of inverse problem as well. It allows to recover mechanical properties by inverting the displacement data, for example to recover the geometric features of the domain of perturbation, or the reconstruction of the location of the anomalies.

INTRODUCTION

Stokes transmission problem

The Stokes resolvent system can be obtained by applying the Laplace transform to the system of the continuity and Navier-Stokes equations which describes the moderate Reynolds number flow of a viscous incompressible fluid (for details see [START_REF] Kohr | The interior Neumann problem for the Stokes resolvent system in a bounded domain in R n[END_REF], [START_REF] Kohr | An indirect boundary integral method for an oscillatory Stokes flow problem[END_REF], [START_REF] Varnhorn | Boundary value problems and integral equations for the Stokes resolvent in bounded and exterior domains of R 3 , Theory of the Navier-Stokes Equations[END_REF]). So the study of this problem has a key role in the understanding of the biomechanics of blood flow, the Brownian motion observed by Brown, the motion of swimming microorganisms, and other biological or physico-chemical phenomena.

On the other hand, the potential theory for the Stokes resolvent system was developed by Varnhorn in [START_REF] Varnhorn | The Stokes equations[END_REF] and [START_REF] Varnhorn | Boundary value problems and integral equations for the Stokes resolvent in bounded and exterior domains of R 3 , Theory of the Navier-Stokes Equations[END_REF]. In addition, the fundamental solution for the system of equations in R 3 was obtained by McCraken in [START_REF] Mccracken | The resolvent problem for the Stokes equations on halfspaces in Lp[END_REF]. Also the Dirichlet and Neumann problems for the Stokes resolvent equations on bounded and exterior domains in R n have been studied recently in [START_REF] Borchers | On the boundedness of the Stokes semigroup in twodimensional exterior domains[END_REF] and [START_REF] Varhorn | An explicit potential theory for the Stokes resolvent boundary value problems in three dimensions[END_REF]. A mixed boundary value problem for the same equations has been treated in [START_REF] Kohr | A mixed boundary value problem for the unsteady Stokes system in a bounded domain in Rn[END_REF]. These autors, construct a solution of these problems in the form of appropriate potentials (it can be represented as single layer potentials, double layer potentials, or their linear combination), with the unknown source densities defined via integral equations on the boundary of the domain. In order to show the existence of a solution (u, p) of the boundary value problem, the method of boundary inregral equations based on layer potentials is used.

Potential theory, can also be used to study the transmission problem of the Stokes resolvent equation, which occurs in the case of contact between two materials with different physical characteristics. In a recent work [START_REF] Mitrea | Boundary Value Problems for the Stokes system in arbitrary Lipschitz domains[END_REF], Mitrea and Wright have used layer potential methods to develop a powerful analysis of a transmission boundary value problems in arbitrary Lipschitz domains. In other works [START_REF] Kohr | Stokes-Brinkman transmission problems on Lipschitz and C 1 domains in Riemannian manifolds[END_REF], Kohr

                               (-∆ + κ 2 )u + ∇q = 0 in Ω \ D (-∆ + κ2 )u + ∇q = 0 in D ∇ • u = 0 in Ω u| + -u| -= 0 on ∂D µ ∂ ∂n (u, q)| + -μ ∂ ∂n (u, q)| -= 0 on ∂D u = g on ∂Ω Ω q = 0, (1)
where µ and μ are positive constants related to the physical properties of Ω and D. Here we take κ 2 and κ2 in C \ {z ∈ R, z ≤ 0}. Let κ = √ κ 2 and κ = √ κ2 be particular square roots, which have positive real parts (i.e. Re κ > 0, Re κ > 0).

Let ν be the outward unit normal to ∂D and ν Ω be the outward unit normal to ∂Ω. Then, we denote by ∂ ∂n the conormal derivative. In this thesis, we propose to show the existence and the uniqueness of the solution of (1) and to express it in the form of linear combinations of potential operators.

Asymptotic expansion

We consider ∂D δ a δ-perturbation of ∂D defined by

∂D δ = {x = x + δν(x); x ∈ ∂D, 0 < δ << 1} (2)
and we denote by D δ the domain bounded by ∂D δ . Define (u δ , q δ ) the solution of the transmission problem for the Stokes system with a small boundary perturbation D δ :

                               (-∆ + κ 2 )u δ + ∇q δ = 0 in Ω \ D δ (-∆ + κ2 )u δ + ∇q δ = 0 in D δ ∇ • u δ = 0 in Ω u δ | + -u δ | -= 0 on ∂D δ µ ∂ ∂n (u δ , q δ )| + -μ ∂ ∂n (u δ , q δ )| -= 0 on ∂D δ u δ = g on ∂Ω Ω q δ = 0. (3)

INTRODUCTION

We aim to derive an asymptotic expansion of (u δ -u)| Ω 0 and (u δ -u)| D 0 as δ tends to zero, where D 0 is any close subset of D ∩ D δ and Ω 0 is any subset of Ω \ (D ∪ D δ ). The procedure is based on potential theoretical considerations in connection with a boundary integral equations method, and geometric properties of the perturbed boundary. This work is submitted for publication [START_REF] Balloumi | Asymptotic expansion of the solution of the transmission Stokes system with a small boundary perturbation for an inclusion[END_REF].

Asymptotic behaviors for eigenvalues and eigenfunctions associated to the Stokes operator in the presence of a small boundary perturbations

The field of eigenvalue problems under shape perturbation has been an active research area for several decades. The main objective of the second part in this report is to present a schematic way to derive high-order asymptotic expansions for both eigenvalues and eigenfunctions for the Stokes operator caused by small perturbations of the boundary. Also, we rigorously derive an asymptotic formula which is in some sense dual to the leading-order term in the asymptotic expansion of the perturbations in the Stokes eigenvalues due to interface changes of the inclusion. This is inspired from the fact that the Stokes system can be viewed as the incompressible limit [START_REF] Ammari | Effective viscosity properties of dilute suspensions of arbitrarily shaped particles[END_REF]. Thus we rigourously extended some results elaborated in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF] to the Stokes case. The properties of eigenvalue problems under shape deformations have been a

subject of comprehensive studies such as [START_REF] Albert | Genericity of simple eigenvalues for elliptic pde's[END_REF][START_REF] Albert | Topology of the nodal and critical point sets for eigenfunctions of elliptic operators[END_REF][START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF][START_REF] Babuska | Eigenvalue Problems[END_REF][START_REF] Kato | Perturbation Theory for Linear Operators[END_REF][START_REF] Ortega | Generic simplicity of the eigenvalues of the stokes system in two space dimensions[END_REF] and the area continues to carry great importance [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF][START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Kelliher | Eigenvalues of the Stokes operator versus the Dirichlet Laplacian in the plane[END_REF][START_REF] Khelifi | Asymptotic property and convergence estimation for the eigenelements of the Laplace operator[END_REF][START_REF] Medkova | Boundary value problems for the Stokes equations with jumps in open sets[END_REF][START_REF] Jia | Approximation and eigenvalue extrapolation of Stokes eigenvalue problem by nonconforming finite element methods[END_REF]. A substantial portion of these investigations discusses the properties of smoothness and analyticity of eigenvalues and eigenfunctions with respect to perturbations.

Let us consider the following eigenvalue problem for the Stokes system with homogeneous boundary conditions:

         -∆v + ∇p = λv in Ω ∇.v = 0 in Ω v = 0 in ∂Ω (4)
with Ω ⊂ R 3 a bounded open domain with boundary of class C 2 . Here v = (v 1 , v 2 , v 3 )

denotes the velocity field while the scalar function p is the pressure.

We assume that the boundary ∂Ω is subjected to a small, smooth deformation and that the INTRODUCTION boundary of the deformed domain Ω δ is given by: ∂Ω δ := {x = x + δh(x)ν(x), x ∈ ∂Ω} [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF] where ν(x) is the outward normal vector on ∂Ω and h(x) is a real function in C 2 (∂Ω) that satisfies h(x) C 2 (∂Ω) < 1.

Obviously, the domain Ω δ is of class C 2 and the Dirichlet eigenvalue problem for the Stokes system can be defined in Ω δ as well.

In this work, we derive the asymptotic expansion of eigenvalues, eigenfunctions and the eigenpressures solutions to the Stokes system:

         -∆v δ + ∇p δ = λ δ v δ in Ω δ ∇.v δ = 0 in Ω δ v δ = 0 in ∂Ω δ . (7)
Here we suppose that the eigenvalue λ 0 is simple. Then the eigenvalue λ δ is simple and is near to λ 0 associated to the normalized eigenfunction v δ .

To the best of our knowledge, this is the first work to rigorously investigate the Stokes eigenvalue problem in the presence of the perturbation and derive (formally) high-order terms in the asymptotic expansion of λ δ -λ 0 and v δ -v 0 when δ → 0. However, by the same method, one can derive asymptotic formula for the Neumann problem as well.

Zuazua and Ortega have proved in [START_REF] Ortega | Generic simplicity of the eigenvalues of the stokes system in two space dimensions[END_REF] the regularity of the eigenvalues and eigenfunctions of the Stokes system with respect to the perturbation parameter, by using the Lyapunov-Schmidt method. Their proofs are essentially inspired in the work of J. Albert [START_REF] Albert | Genericity of simple eigenvalues for elliptic pde's[END_REF][START_REF] Albert | Topology of the nodal and critical point sets for eigenfunctions of elliptic operators[END_REF] for the Laplace operator. Our analysis and uniform asymptotic formulas of eigenvalues and eigenfunctions, which are represented by the single-layer potential involving the Green function, are considerably different from those in [START_REF] Kelliher | Eigenvalues of the Stokes operator versus the Dirichlet Laplacian in the plane[END_REF][START_REF] Ortega | Generic simplicity of the eigenvalues of the stokes system in two space dimensions[END_REF].

These results were published in [START_REF] Daveau | Asymptotic Behaviors for Eigenvalues and Eigenfunctions Associated to Stokes Operator in the Presence of Small Boundary Perturbations[END_REF].

Singular integrals in Electromagnetism

Integral equation methods are an alternative for solving linear partial differential equations.

It consists in transforming the partial differential equation posed in a domain, into an integral INTRODUCTION equation posed on the edge. It is in particular a method adapted for diffraction problems that are posed in an unbounded domain (see [START_REF] Givoli | Numerical methods for problems in infinite domains[END_REF]) and for which the consideration of the radiation condition is essential. We are interested in solving time-harmonic scattering problems by the integral aquation method. In order to ensure a unique solution to the boundary value problem we will take the constant impedance operator, known as standard or Leontovich impedence boundary condition. This approximation does not depend on the incident angle at all.

Once the boundary problem has been converted into a boundary integral equation, it is necessary to discretize the edge of the domain using plane polygons. Here we deal with plane triangles and Rao-Wilton-Glisson basis functions f k . In this thesis, we are particularly interested in the singular integrals resulting from the discretization of the variational problem of Maxwell 3-D equation, namely:

Γ h f i (x) • f j (y) 1 |x -y| dxdy (8)
which comes from the potential of the single layer. This integral has been treated by Marc

Lenoire and Nicolas Salles in [START_REF] Lenoir | Evaluation of 3-d singular and nearly singular integrals in Galerkin BEM for thin layers[END_REF] with a purely analytic method.

The second one comes from a double layer potential can be written as follows:

Γ h [f i (x) × f j (y)] •
x -y |x -y| 1+ξ dxdy; ξ = 0 or 2.

(

) 9
This singular character of the integrand makes the precise evaluation of these integrals difficult. It has been the subject of a large number of publications. The method presented in this thesis allows us to evaluate analytically and with great precision, these singular integrals in the variational case. Note that if ξ = 0 the integral is not singular but the method we propose is still valid.

To better understand this singular integrand problem, in the variational case, it should be noted that the domain of integration for these integrals with "problem", is the product of two elements of the mesh, ie in 3-D, the product of two triangles S and T. When the two triangles are distant, it is possible to evaluate the integral with standard numerical integration methods.

But the two problematic cases are the singular case, which occurs when the intersection of the two triangles is not empty(adjacent triangles and triangles with a common vertex). These two cases require separate treatment and this is a field of research still very active.

INTRODUCTION

Reduction process

The presented method makes it possible to reduce a 4-D integral to a linear combination of mono-dimensional integrals whose integrand is regular. It is then possible to numerically evaluate these 1-D integrals, but also explicitly.

This method of reduction relies on several formulas that make it possible to reduce the dimension of the integration domain. So, repeating three times using these formulas we obtain regular 1-D integrals. These formulas are based on the homogeneity of the integrand; therefore, there are two conditions to check:

1) -The integrand must be homogeneous. We must therefore break the integrand into a part homogeneous, which is treated with our method, and a regular part, which can be treated with usual numerical integration methods.

2) -Flat polygons must be used to discretize the boundary of the domain. This in order to maintain the homogeneity of the integrand during the successive stages of reduction.

Organization of this thesis

This thesis is divided into three parts, and composed of eight chapters, corresponding to the goals introduced above.

The first part includes three chapters (Chapter 1, 2, 3) with the following contents:

Chapter 1 is devoted to introduce the notations and some preliminary results of layerpotential theory for Stokes problem and perturbed geometry.

Chapter 2 show the existence and uniqueness of the solutions of a Dirichlet transmission

Stokes problem and presents it in form of potential operators.

In chapter 3 we first, verify the continuity of the solution with respect to the small perturbation δ via the stability of the density function. Secondly, we derive the asymptotic expansion of the solution, after deriving the expansion of the density function.

INTRODUCTION

The second part states the results that have been published in the journal Math Phys Anal Geom, see [START_REF] Daveau | Asymptotic Behaviors for Eigenvalues and Eigenfunctions Associated to Stokes Operator in the Presence of Small Boundary Perturbations[END_REF].

In chapter 4, we derive asymptotic expansions for both eigenvalues and eigenfunctions for the Stokes operator caused by small perturbations of the boundary. Also, we rigorously derive an asymptotic formula.

In the third part we develop a method for the evaluation of integrals with a singular integrand derived from the application of a boundary element method on Maxwell's equation.

In chapter 5, the time-harmonic scattering problem of electromagnetic waves is described as a system of equations with boundary conditions. The uniqueness of the solution of this problem is shown thanks to a variational formulation. It is also concerned with the discretization of the problem with help of Rao-Wilton-Glisson basis functions for mesh triangulation.

The next chapter, is dedicated to the presentation of the method. The integration of homogeneous functions leads to formulas for reducing the size of the integration domain.

In chapter 7, we presents the evaluation of singular integrals when the triangles are Adjacent triangles, and the evaluation of singular integrals when the triangles have common vertex in chapter 8.

We present numerical results in the last chapter.

We have put the proofs for some inequalities and auxiliary results concerning the first part in Appendix A, the final expressions of the mono-dimensional integrals in Appendix B and some preliminary results about Fredholm alternative and integral formulation Maxwell equation in Appendix C.

Part I

Asymptotic expansion for the solution of a Stokes transmission problem with a small perturbation of the domain

CHAPTER 1
Preliminary results

Introduction

In this chapter, we present geometric definitions, notations and basic results which will be used in the next chapters as well as the main properties of the layer potential operators associated to the Stokes resolvent equation on a bounded Lipschitz domains in R 3 .

The outline of this chapter is as follows. The first section contains the setting for basic notations, definitions and related funtional spaces. Section 1.2 is devoted to present some geometric properties of a boundary perturbation. In section 1.3, we describe layer potential techniques for solving the Stokes resolvent problems on a bounded Lipschitz domains in R 3 .

We then present in 1.3.1 the fundamental solution for Stokes resolvent systems and we define in 1.3.2 the properties of the associated layer potential operators.

Euclidean spaces

This section contains some notations and definitions concerning Euclidean spaces and functional spaces, wich plays a segnificant role all along this work.

For x, y ∈ R 3 , x = (x 1 , x 2 , x 3 ), y = (y 1 , y 2 , y 3 ) let x • y := 3 i=1 x i y i be the scalar product of x, y and |x| := √ x • x be the Euclidean norm of x.

Denote by |x -y| the distance between two points x and y of R 3 given by the formula

|x -y| = [ 3 i=1 (x i -y i ) 2 ] 1/2 .
More generally, |u| stands for the modulus of the vector u and we have |u| = ( i u 2 i ) 1/2 . Similarly, the distance between two subsets A and B of R 3 is denoted by dist(A, B) and we have

dist(A, B) = inf x∈A,y∈B |x -y|. Definition 1.1.1. (Derivatives)
1. The gradient is denoted by ∇ = (∂ 1 , ∂ 2 , ∂ 3 ), where ∂ j = ∂ ∂ j denotes the partial derivative with respect to the j-th coordinate (x j ) j=1,2,3 .

2. For all multi-indices α = (α 1 , α 2 , α 3 ), we denote the α-derivative of f by

D α f = ∂ α 1 +α 2 +α 3 f ∂x α 1 1 x α 2 2 x α 3 3 = ∂ |α| f ∂x α 1 1 x α 2 2 x α 3 3 ,
where the order of this derivative is |α| = α 1 + α 2 + α 3 . Let L 2 (D) be the Lebesgue space of square integrable functions on D and let L 2 (∂D) 3 be the space of vector fields φ = (φ 1 , φ 2 , φ 3 ) : R 3 → R 3 whose components φ i belong to L 2 (∂D).

For vector-valued function

v = (v 1 , v 2 , v 3 ), ∇ • v = ∂ 1 v 1 + ∂ 2 v 2 + ∂ 3 v 3 defines the divergence of v.
We have

φ (L 2 (∂D)) d = ∂D (|φ 1 (x)| 2 + |φ 2 (x)| 2 + |φ 3 (x)| 2 )dσ(x).
We recall now the definition of curvature of a curve and definitions of mean and Gauss curvature of a surface that will be appeared in the expansion terms of the area element in chapter 3. For a complete description, we refer to [START_REF] Ammari | Conductivity interface problems. Part I: Small perturbation of an interface[END_REF] and [START_REF] Banchoff | Differential geometry of curves and surfaces[END_REF].

∂D := {x = X(t), t ∈ [a, b]}.
Then the outward unit normal to ∂D, ν(x), is given by ν

(x) = R -π 2 X (t), where R -π 2 is
the rotation by -π/2; the tangent vector at x, T (x) = X (t), and X (t)⊥X (t). Set the curvature τ (x) to be

X (t) = τ (x)ν(x).
Definition 1.1.5. (Gaussian and mean curvature)

Let k 1 and k 2 be the principal curvatures of a regular surface S at a point P . Define 1. The Gaussian curvature of S at P as the product K = k 1 k 2 ;

2. The mean curvature of S at P is the average

H = κ 1 +κ 2 2
of the principal curvatures.

We also remark that for a bounded domain D with boundary of class C 2,α , the outward unit normal ν is a C There exists a positive constant L such that

|ν(x) • {x -y}| ≤ L|x -y| 2 ,
and

|ν(x) -ν(y)| ≤ L|x -y|,
for all x, y ∈ ∂D.

Small perturbation of an interface

Let D ⊂ R 3 be a bounded domain with connected boundary ∂D of class C 2 . For a fixed constant δ ∈ R, we consider ∂D δ as a δ-perturbation of ∂D defined by

∂D δ = {x = x + δν(x); x ∈ ∂D, 0 < δ << 1}. (1.1)
We denote by D δ the domain bounded by ∂D δ and dσ δ ( x) the surface element of ∂D δ at x.

The result presented below is given in [START_REF] Khelifi | Boundary voltage perturbations resulting from small surface changes of a conductivity inclusion[END_REF] with more details.

Lemma 1.2.1. The surface elements dσ on ∂D and dσ δ on ∂D δ are related by the formula

dσ δ ( x) = 2 n=0 δ n σ n (x)dσ(x), (1.2)
where σ n are bounded functions regardless of n. In particular,

σ 0 (x) = 1, σ 1 (x) = -2H(x), σ 2 = K(x), x ∈ ∂D, (1.3)
with H and K denote the mean and the Gaussian curvature of ∂D respectively, as introduced in Definition 1.1.5.

Since ∂D is parallel to ∂D δ , we can conclude that

ν( x) = ν(x), x ∈ ∂D, x ∈ ∂D δ . (1.4)
We then have for x, y ∈ ∂D δ :

x -y = x -y + δ(ν(x) -ν(y)),

and

| x -y| 2 = |x -y| 2 (1 + 2δ x -y, ν(x) -ν(y) |x -y| 2 + δ 2 |ν(x) -ν(y)| 2 |x -y| 2 ).
We also introduce two functions F and G for x = y ∈ ∂D defined by:

E(x, y) := x -y, ν(x) -ν(y) |x -y| 2 , G(x, y) := |ν(x) -ν(y)| 2 |x -y| 2 .
Since ∂D is of class C 2 , there exists a constant C depending only on ∂D such that:

|E(x, y)| + |G(x, y)| 1 2 ≤ C, ∀x, y ∈ ∂D.
Hence, we obtain:

| x -y| = |x -y| 1 + 2δE(x, y) + δ 2 G(x, y) := |x -y| ∞ n=0 δ n L n (x, y),
where the serie converges absolutely and uniformly and the first two terms are:

L 0 (x, y) = 1, L 1 (x, y) = E(x, y).
CHAPTER 1. PRELIMINARY RESULTS

Potential theory

This section presents layer potential theory for the Stokes resolvent system. The main sources used in the preparation of this section are [START_REF] Varhorn | An explicit potential theory for the Stokes resolvent boundary value problems in three dimensions[END_REF] [58] [START_REF] Varnhorn | The boundary value problems of the Stokes resolvent equations in n dimensions[END_REF] and [START_REF] Varnhorn | The Stokes equations[END_REF] 1.3.1 Stokes resolvent system

* = C \ {z ∈ R, z ≤ 0}. Let κ = √ κ 2 be particular
square root, which have positive real parts (i.e. Re κ > 0). We denote by Γ κ = (Γ κ ij ) i,j=1,2,3 and F κ = (F κ ) i=1,2,3 the fundamental tensor and vector of the following Stokes resolvent system:

(-∆ + κ 2 )Γ κ ij (x, y) + ∂ j F κ i (x, y) = δ ij δ y (x) on D, ∂ i Γ κ ij (x, y) = 0 on D, (1.5)
where δ y is the Dirac distribution with mass at y and δ ij is the Kronecker symbol. The components of (Γ κ , F κ ) are obtained in these forms (see [START_REF] Varhorn | An explicit potential theory for the Stokes resolvent boundary value problems in three dimensions[END_REF] [58]):

   Γ κ ij (x, y) = 1 4π δ ij |x-y| e 1 (κ|x -y|) + (x-y) i (x-y) j |x-y| 3 e 2 (κ|x -y|) , F κ i (x, y) = 1 4π (x-y) i |x-y| 3 , (1.6)
where

e 1 ( ) = ∞ n=0 (n + 1) 2 (n + 2)! (-) n = exp(-)(1 + -1 + -2 ) --2 , e 2 ( ) = ∞ n=0 1 -n 2 (n + 2)! (-) n = exp(-)(-1 -3 -1 -3 -2 ) + 3 -2 .
Here and throughout this report we use the Einstein convention for the summation notation omitting the summation sign for the indices appearing twice.

Stress and pressure tensors

Now, let introduce for x, y ∈ ∂D, the stress tensor S associated to the fundamental tensors (Γ κ , F κ ) defined by:

S κ ijk (x, y) = -F κ j (x)δ ik + ∂Γ κ ij (x) ∂ xk + ∂Γ κ ik (x) ∂ xj , i, j, k = 1, 2, 3. (1.7)
where x = x -y = (x 1 , x2 , x3 ) and r = |x|. Combining (1.6) and (1.7), one can obtain:

S κ ijk (x, y) = - 1 ω d δ ik xj r d D 1 (κr) + (δ kj xi r d + δ ij xk r d )D 2 (κr) + xi xj xk r d+2 D 3 (κr) (1.8)
where

D 1 (t) = 8 ( t 2 ) m+1 K m+1 (t) Γ(m)t 2 - 6 t 2 + 1, D 2 (t) = 8 ( t 2 ) m+1 K m+2 (t) Γ(m)t 2 - 6 t 2 + 2 ( t 2 ) m K m (t) Γ(m) , D 3 (t) = -16 ( t 2 ) m+2 K m+2 (t) Γ(m)t 2 + 30 t 2 , with m = 3 2 , t ∈ R, Γ(m)
is the Gamma function and K n for n ≥ 0 is the modified Bessel function of order n. For more details on these functions, see Appendix A.

The pressure tensor associated with the stress tensor S has the following components:

Λ κ ik (x, y) = - 1 4π (2 δ ik r 3 - δ ik κ 2 r -6 xi xk r 5 ), i, k = 1, 2, 3.
(1.9)

Decay behavior of the fundamental and the stress tensor

In order to study the decay behavior of Γ κ ij (x, y) and S κ ijk (x, y) as r = |x -y| → 0, we can apply the following decomposition:

Γ κ ij (x, y) = Γ 0 ij (x, y) + Γ ij (x, y), (1.10)
S κ ijk (x, y) = S 0 ijk (x, y) + S ijk (x, y), (1.11)
where S 0 and Γ 0 are weakly singular Stokes tensors (with respect to κ=0) and the remaining part S and Γ are continuous kernels, so that the continuity behavior of the corresponding surface potential is determined only by the Stokes tensors S 0 and Γ 0 . A detailed proof of the above relations can be found in [START_REF] Varnhorn | The boundary value problems of the Stokes resolvent equations in n dimensions[END_REF] or [START_REF] Varnhorn | Boundary value problems and integral equations for the Stokes resolvent in bounded and exterior domains of R 3 , Theory of the Navier-Stokes Equations[END_REF]. Besides, we note that Γ 0 and S 0 are fundamental tensors for the standard Stokes system [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF]. This property will be very useful in the treatment of some boundary integral representation associated to a transmission system that will be analyzed in the next chapter.

Layer potential operators for the Stokes system

In this section, we present layer potential operators associated to the Stokes equation in the bounded Lipschitz domain D ⊂ R 3 . A special attention is devoted to the invertibility of the potential operators associated with the standard Stokes equation and the compactness of the complementary layer potential operators. 3 , we define the single-layer potential S κ D φ : R 3 \∂D → R 3 and the associated pressure potential, V D φ : R 3 \∂D → R as follows:

Definition 1.3.1. For a density φ = (φ 1 , φ 2 , φ 3 ) ∈ L 2 (∂D)
S κ D,i φ(x) := ∂D Γ κ ij (x, y)φ j (y)dσ(y); i, j = 1, 2, 3, (1.12
)

V κ D φ(x) := ∂D F κ j (x, y)φ j (y)dσ(y); j = 1, 2, 3. (1.13)
Likewise for a given density ψ = (ψ 1 , ψ 2 , ψ 3 ) ∈ L 2 (∂D) 3 , we define the double-layer poten-

tial D κ D ψ : R 3 \∂D → R 3 , by D κ D,i ψ(x) := ∂D -S κ ijk (x, y)ν k (y)ψ j (y)dσ(y); i, j, k = 1, 2, 3, (1.14)
where ν is the outward unit normal to ∂D. Moreover, the associated pressure potential

W κ D ψ : R 3 \∂D → R is defined by W κ D ψ(x) := ∂D -Λ jk (x, y)ν k (y)ψ j (y)dσ(y); j, k = 1, 2, 3. (1.15)
We can easily check that (S κ D , V κ D ) and (D κ D , W κ D ) are solutions of the Stokes system in R 3 \ ∂D,

(-∆ + κ 2 )S κ D φ(x) + ∇V κ D φ(x) = 0; ∇ • S κ D (x) = 0, (1.16
)

(-∆ + κ 2 )D κ D ψ(x) + ∇W κ D ψ(x) = 0; ∇ • D κ D (x) = 0. (1.17)
We need also to introduce the principal value of the double-layer potential (or the boundary version of the double layer potential).

Definition 1.3.2. For x ∈ ∂D and a density φ = (φ 1 , φ 2 , φ 3 ) ∈ L 2 (∂D) 3 , the principal value of the double-layer potential is given by:

K κ D,i (φ)(x) := p.v. ∂D -S κ ijk (x, y)ν k (y)φ j (y)dσ(y); i, j, k = 1, 2, 3, (1.18)
and (K κ D ) * the adjoint operator of K κ D :

(K κ D,i ) * (φ)(x) := p.v. ∂D -S κ ijk (x, y)ν k (x)φ j (y)dσ(y); i, j, k = 1, 2, 3. (1.19)
where p.v. means the principal value of a singular integral.

Trace and jump relations

Denote by ∂ ∂n the conormal derivative defined by

∂ ∂n (u, q) = (∇u + (∇u) T )ν -qν on ∂D, (1.20)
where (∇u) T denotes the matrix transposed of ∇u = (∂ i u k ) i,k=1,2,3 (see [START_REF] Ammari | A method of biological tissues elasticity reconstruction using magnetic resonance elastography measurements[END_REF], [START_REF] Boyer | Mathematical tools for the study of the incompressible Navier-Stokes equations and related models[END_REF], [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF] and [START_REF] Temam | Navier-Stokes equations theory and numerical analysis[END_REF]). In the lemma below we give trace relations for the single and double layer potential.

(φ, ψ) ∈ L 2 (∂D) 3 × L 2 (∂D) 3 one has. D κ D (ψ)| ± = (∓ 1 2 I + K κ D )(ψ), (1.21)
∂ ∂n (D κ D (ψ), W κ D (ψ))| + = ∂ ∂n (D κ D (ψ), W κ D (ψ))| -, (1.22)
∂ ∂n (S κ D (φ), V κ D (φ))| ± = (± 1 2 I + (K κ D ) * )(φ), (1.23)
S κ D (ψ)| + = S κ D (ψ)| -= S κ D (ψ). (1.24)
Here and throughout this report, we use the subscripts | + (respectively | -) for the limiting value of a field evaluated from the external side of D (or the internal side of D, respectively).

v| ± (x) := lim

D ± y→x v(y).
The formulas (1.21), (1.22), (1.23) and (1.24) have been proved in [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF] when κ = 0.

Since D κ D -D 0 D and S κ D -S 0 D are smooth operators and according to (1.10) and (1.11), we can extend the result to the case κ = 0. Furthermore, using (1.21) and (1.23) we can easily show that the single and double layer potentials satisfy the following jump ralations on ∂D:

D κ D (ψ)| + -D κ D (ψ)| -= -φ on ∂D, (1.25)
∂ ∂n S κ D (φ)| + - ∂ ∂n S κ D (φ)| -= ϕ on ∂D.
(1.26)

Decay behavior of layer potentials

Here we give another result about the decay behavior at infinity of the layer potentials proved by Varnhorn in [START_REF] Varhorn | An explicit potential theory for the Stokes resolvent boundary value problems in three dimensions[END_REF] p. 15 and [60].

Lemma 1.3.1. For the single and double layer potentials

S κ D [φ], V κ D [φ], D κ D [ψ] and W κ Ω [ψ],
we have the following decay behavior as |x| → ∞:

i) S κ D [φ](x) = O(|x| -1 ), as |x| → ∞.
ii)

V κ D [φ](x) = O(|x| -2 ), as |x| → ∞.
iii)

D κ D [ψ](x) = O(|x| -2 ), as |x| → ∞. iv) W κ D [ψ](x) = O(|x| -3 ), as |x| → ∞.

Compactness of layer potential operators

According to theorem 2.27 in [START_REF] Kress | Linear Integral Equations[END_REF], the following lemma holds about the compactness of operators with kernels defined by (1.10) and (1.11).

Lemma 1.3.2. For x ∈ ∂D, we define the operators :

K D,i [φ](x) := p.v. ∂D -S ijk (x, y)ν k (y)φ j (y)dσ(y); i, j, k = 1, 2, 3, (1.27)
K * D,i [φ](x) := p.v. ∂D -S ijk (x, y)ν k (x 0 )φ j (y)dσ(y); i, j, k = 1, 2, 3, (1.28)
S D,i [φ](x) := ∂D Γ ij (x, y)φ j (y)dσ(y); i, j, k = 1, 2, 3, (1.29)
S D (φ)| + = S D (φ)| -= S D (φ).
(1.30)

These operators have continuous kernel. Moreover, S D , K D and K * D are compact operators on C(∂D) 3 and on L 2 (∂D) 3 .

Invertibility of layer potentials

We recall that the single layer potential associated with κ = 0 is defined as follows

S 0 D,i φ(x) := ∂D Γ 0 ij (x, y)φ j (y)dσ(y), i = 1, • • • , 3.
where D ⊂ R 3 is a bounded domain with boundary ∂D. It then follows from [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF]

: L 2 0 (∂D) → H 1 0 (∂D) is invertible.
ii) The operator λI + K 0 D and λI + (K 0 D ) * are invertible on L 2 (∂D) 3 for |λ| > 1/2.

Lemma 1.3.4. [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF] Let D ⊂ R 3 be a bounded domain with boundary ∂D. Then, we have:

(1/2)I + K κ D : L 2 0 (∂D) → L 2 0 (∂D) is invertible and so is (1/2)I + (K 0 D ) * : L 2 0 (∂D) → L 2 0 (∂D).

Conclusion

By referring to previous works, we recalled theorems and definitions on which our work in the next chapters, will be based.

Introduction

This chapter is devoted to resolving the Dirichlet transmission Stokes problem and presenting its solution in form of a linear combination of potential operators. In the next section 2.1, we construct the solution of this problem, in the form of appropriate potentials and we reduce the transmission problem to a system of boundary integral equations. Then, in section 2.2, we prove unicity of the unknown source densities which implies unicity of the solution.

Boundary integral representation

Let Ω ⊂ R 3 be a bounded domain with a connected Lipschitz boundary ∂Ω and suppose

that Ω contains an inhomogeneity D with a C 2 -boundary ∂D.We assume that there exists a constant c 0 > 0 such that inf x∈D dist(x, ∂Ω) > c 0 which means that D is away from the boundary ∂Ω.

Using arguments given in the previous chapter, we are looking for a solution (u, q) to the transmission Stokes system with Dirichlet boundary condition:

                           (-∆ + κ 2 )u + ∇q = 0 in Ω \ D (-∆ + κ2 )u + ∇q = 0 in D ∇ • u = 0 in Ω u| + -u| -= 0 on ∂D µ ∂ ∂n (u, q)| + -μ ∂ ∂n (u, q)| -= 0 on ∂D u = g on ∂Ω Ω q = 0, (2.1)
where µ and μ are positive constants related to the physical properties of Ω and D. Note that κ 2 and κ2 are complex numbers in C * such that

C * = C \ {z ∈ R, z ≤ 0}.
Let κ = √ κ 2 and κ = √ κ2 be particular square roots, which have positive real parts (i.e. Re κ > 0, Re κ > 0).

From Gauss's law theorem, we have the compatibility condition:

∂Ω g • ν Ω = 0,
where ν Ω is the outward unit normal to ∂Ω. Let ν be the outward unit normal to ∂D.

In order to solve (2.1), We adopte a method of boundary integral equations, wich consist on transforming the transmission problem posed in Ω \ D and in D, into integral equations defined on the boundarys ∂D and ∂Ω. This transition from a volume problem to a surface problem is carried out using layer potentials. Indeed, according to (1.16) and (1.17) in chapter [START_REF] Abramowitz | Handbook of mathematical functions[END_REF], for some triplet (φ, ψ, θ) ∈ L 2 (∂D) 3 × L 2 (∂D) 3 × L 2 (∂Ω) 3 , the boundary integral representation:

u = S κ D [φ] in D, u = S κ D [ψ] + D κ Ω [θ] in Ω \ D, (2.2)
q = V κ D [φ] in D, q = V κ D [ψ] + W κ Ω [θ] in Ω \ D, (2.3)
solves        (-∆ + κ 2 )u + ∇q = 0 in Ω \ D (-∆ + κ2 )u + ∇q = 0 in D ∇ • u = 0 in Ω.
So, we will look for a solution (u, q) to the transmission problem (2.1) in the form of (2.2)-(2.3). Otherwise, we will look for (φ, ψ, θ) ∈ L 2 (∂D) 3 × L 2 (∂D) 3 × L 2 (∂Ω) 3 satisfying the boundary conditions:

       u| + -u| -= 0 on ∂D µ ∂ ∂n (u, q)| + -μ ∂ ∂n (u, q)| -= 0 on ∂D u = g on ∂Ω Remark 2.1.1.
For the same problem, several boundary integral representations can be obtained.

Existence and unicity of the solution

           S κ D | + μ(-1 2 I + (K κ D ) * ) 0 -S κ D | - -µ( 1 2 I + (K κ D ) * ) S κ D -D κ Ω -µ ∂ ∂n D κ Ω 1 2 I + K κ Ω                φ ψ θ     =     0 0 g     .
(2.4) Denote by A κ the matrix corresponding to the left hand side of (2.4).

In order to study the existence and unicity of the solution of the system (2.4), we introduce the space H(∂D) defined by:

H(∂D) := {(φ, ψ) ∈ L 2 (∂D) 3 × L 2 (∂D) 3 ; φ -ψ ∈ L 2 0 (∂D)}
We annonce the theorem.

Theorem 2.2.1. Let (φ, ψ, θ) ∈ H(∂D)×L 2 0 (∂Ω) be the unique solution of (2.4). Then (u, q) represented by (2.2)-(2.3) is the unique solution of (2.1). This is equivalent to the invertibility of A κ . We represent A κ by a sum of two matrices A inv and A com defined by

A com =            S D μ(K D ) * 0 -S D -µ(K D ) * S κ D -D κ Ω -µ ∂ ∂n D κ Ω K Ω            , (2.5)
and

A inv =            S 0 D μ(-1 2 I + (K 0 D ) * ) 0 -S 0 D -µ( 1 2 I + (K 0 D ) * ) 0 0 0 1 2 I + K 0 Ω            . (2.6)
We know that S κ D is compact on L 2 (∂Ω) and that D κ Ω and ∂ ∂n D κ Ω which appear in the third column of A com are also compact on L 2 (∂D) since ∂D and ∂Ω do not intersect. According to lemma 1.3.2, A com is compact. Hence, it sufficies to prove that A inv is invertible and that A κ is injective to show the invertibility of A κ according to the Fredholm alternative.

Invertibility of A inv

We prove this first result.

Lemma 2.2.1. The operator

A inv : H(∂D) × L 2 0 (∂Ω) → H 1 0 (∂D) × L 2 (∂D) 3 × L 2 0 (∂Ω) is invertible. Proof. Let (f, h, g) ∈ H 1 0 (∂D) × L 2 (∂D) 3 × L 2 0 (∂Ω), the solution (φ, ψ, θ) of the system A inv     φ ψ θ     =     f h g     ,
is given by

φ = (S 0 D ) -1 (f ) + ψ ψ = 1 μ-µ (-1 2 1+λ 1-λ I + (K 0 D ) * ) -1 [h -μ(-1 2 I + (K 0 D ) * )(S 0 D ) -1 (f )] θ = ( 1 2 I + K 0 Ω ) -1 (g) (2.7)
where 0 < λ = µ μ . As stated in Lemmas 1.3.3 and 1.3.4, the operators S 0 D , -1 2 1+λ 1-λ I + (K 0 D ) * and 1 2 I + K 0 Ω are invertible and the proof is then completed.

Injectivity of A inj

Now, we focus on to the injectivity of A κ and we give theorem.

Theorem 2.2.2. The operator

A κ : H(∂D) × L 2 0 (∂Ω) → H 1 0 (∂D) × L 2 (∂D) 3 × L 2 0 (∂Ω) is injective .
Proof. Suppose that there exists (φ 0 , ψ 0 , θ 0 ) ∈ H(∂D)×L 2 0 (∂Ω) such that A κ (φ 0 , ψ 0 , θ 0 ) T = (0, 0, 0) T and consider (u, q) defined by

u = S κ D [φ 0 ] , u = S κ D [ψ 0 ] + D κ Ω [θ 0 ] , q = V κ D [φ 0 ] in D, q = V κ D [ψ 0 ] + W κ Ω [θ 0 ] in Ω \ D.
We can see that the couple (u, q) is a solution of (2.1) with g = 0. Then, using integration by parts [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF] on D -= D and D + = Ω \ D, we obtain

2 D - Eu : Eu + κ2 D - u.u = ∂D ∂(u, q) ∂n | -.u| -, and
2 D + Eu : Eu + κ 2 D + u.u = - ∂D ∂(u, q) ∂n | + .u| + + ∂D ∂(u, q) ∂n | -.u| -,
where E(u) is the deformation tensor define by

E(u) := 1 2 (∇u + (∇u) T ).
The system (2.1) garantees that u| -= 0 on ∂D, u| -= u| + and

∂(u,q) ∂n | + = μ µ ∂(u,q)
∂n | -on ∂D. Thus, we have

2 D + Eu : Eu + κ2 D + u.u = - ∂D μ µ ∂(u, q) ∂n | -.u| - = - μ µ [ D - 2Eu : Eu + κ2 D - u.u].
Therefore, we get:

u = 0, q = c 1 , E(u) = 0, on D + , u = 0, q = c 2 , E(u) = 0, on D -, (2.8)
where c 1 and c 2 are real constants. Using the argument that Ω q = 0, we have

Ω q = c 1 D - +c 2 D + = 0. (2.9)
On the other hand, the fifth line of system (2.1) gives c 1 = c 2 µ μ . Taking account of (2.9), we obtain c 1 = c 2 = 0.

On the other hand, we define the solution to the system:

                 (-∆ + κ2 )v + ∇p = 0 in R 3 \ D ∇ • v = 0 in R 3 \ D v = 0 on ∂D |v||∇v| = o(|x| -2 ) as |x| → +∞ |v||p| = o(|x| -2 )
as |x| → +∞.

(2.10)

We know that

S κ D (φ 0 )| + = S κ D (φ 0 )| -= u| -on ∂D.
By (2.8), we have

S κ D (φ 0 )| + = 0. Then, (v, p) = (S κ D (φ 0 ), V κ D (φ 0 )
) is solution of exterior Dirichlet problem (2.10) which admits a unique solution [START_REF] Varnhorn | The boundary value problems of the Stokes resolvent equations in n dimensions[END_REF]. Thus, we have (v, p) = (0, 0) which implies

(S κ D (φ 0 ), V κ D (φ 0 )) = (0, 0) on R 3 \ D.
The jump formula on ∂D yields:

∂ ∂n (S κ D (φ 0 ), V κ D (φ 0 ))| -- ∂ ∂n (S κ D (φ 0 ), V κ D (φ 0 ))| + = φ 0 = 0; x ∈ ∂D.
Consider now (v, p) the solution to the following interior Dirichlet problem:

       (-∆ + κ 2 )v + ∇p = 0 in D ∇ • v = 0 in D v = 0 on ∂D (2.11)
We have

S κ D (ψ 0 )| -+ D κ Ω (θ 0 )| -= S κ D (ψ 0 )| + + D κ Ω (θ 0 )| + = u| + on ∂D.
Then, we get

S κ D (ψ 0 )| -+ D κ Ω (θ 0 )| -= 0 on ∂D. Finally we obtain (v, p) = (S κ D (ψ 0 ) + D κ Ω (θ 0 ), V κ D (ψ 0 ) + W κ Ω (θ 0 ))
a solution of (2.11).

However, interior Dirichlet problem (2.11) admits a unique solution (v, p) where the unknown p is determined up to an additive constant [START_REF] Varnhorn | The boundary value problems of the Stokes resolvent equations in n dimensions[END_REF]. Thus, we get

(S κ D (ψ 0 ) + D κ Ω (θ 0 ), V κ D (ψ 0 ) + W κ Ω (θ 0 )) = (0, c).
On the other hand, the trace formula on ∂D give:

∂ ∂n (v, p)| + - ∂ ∂n (v, p)| -= ∂ ∂n (S κ D (ψ 0 ), V κ (ψ 0 ))| + + ∂ ∂n (D κ Ω (θ 0 ), W κ Ω (θ 0 ))| + - ∂ ∂n (S κ D (ψ 0 ), V κ (ψ 0 ))| -- ∂ ∂n (D κ Ω (θ 0 ), W κ Ω (θ 0 ))| - = ∂ ∂n (S κ D (ψ 0 ), V κ (ψ 0 ))| + - ∂ ∂n (S κ D (ψ 0 ), V κ (ψ 0 ))| - = ψ 0 = cν.
We deduce that ψ 0 = 0 on ∂D since ψ 0 -φ 0 ∈ L 2 0 (∂D). Therefore, we have

(S κ D (ψ 0 ), V κ D (ψ 0 )) = (0, 0) on R 3 \ ∂D which implies that (S κ D (ψ 0 ), V κ D (ψ 0 )) = (0, 0) on D + .
Then, we get (u, q) = (D κ Ω (θ 0 ), W κ Ω (θ 0 )) = (0, 0) on D + . Now, we consider the solution to the following exterior problem :

                 (-∆ + κ2 )v + ∇p = 0 in R 3 \ Ω ∇ • v = 0 in R 3 \ Ω ∂ ∂n (v, p) = 0 on ∂Ω |v||∇v| = o(|x| -2 ) as |x| → +∞ |v||p| = o(|x| -2 )
as |x| → +∞.

(2.12)

Then, (D κ Ω (θ 0 ), W κ Ω (θ 0 )) is solution of (2.12) which is an exterior Neumann problem which has a unique solution. We deduce that (v, q) = (0, 0). We know that (D κ Ω (θ 0 ))| --(D κ Ω (θ 0 ))| + = θ 0 = 0 on ∂Ω which implies that (D κ Ω (θ 0 ), W κ Ω (θ 0 )) = (0, 0) in R 3 \ ∂Ω and in particularly in D + . Then, we get

(u, q) = (S κ D [ψ 0 ], V κ D [ψ 0 ]) on D + .
We showed that (φ 0 , ψ 0 , θ 0 ) = (0, 0, 0), which is equivalent to the injectivity of A κ .

Conclusion

Based on the properties shown in the first chapter, we have shown the existence and uniqueness of a solution for the Stokes transmission problem and we gave an explicit expression in the form of single and double layer operators.

CHAPTER 3

Asymptotic expansion Introduction

The aim of this chapter is to give the expression of the unique solution of the perturbed problem and to derive high-order terms in asymptotic expansion of the displacement field resulting from a small perturbation of the interface.

Asymptotic expansion of the density functions

In the next section, we consider the perturbed transmission Stokes problem.

Let (u δ , q δ ) be the solution to the perturbed problem:

                               (-∆ + κ 2 )u δ + ∇q δ = 0 in Ω \ D δ (-∆ + κ2 )u δ + ∇q δ = 0 in D δ ∇ • u δ = 0 in Ω u δ | + -u δ | -= 0 on ∂D δ µ ∂ ∂n (u δ , q δ )| + -μ ∂ ∂n (u δ , q δ )| -= 0 on ∂D δ u δ = g on ∂Ω Ω q = 0. (3.1)
with ∂D δ a δ-perturbation of ∂D defined by

∂D δ = {x = x + δν(x); x ∈ ∂D, 0 < δ << 1} (3.2) CHAPTER 3. ASYMPTOTIC EXPANSION
Using theorem 2.2.1, (u δ , q δ ) can be represented by:

u δ = S κ D δ [φ δ ] , u δ = S κ D δ [ψ δ ] + D κ Ω [θ δ ] , q δ = V κ D δ [φ δ ] in D δ , q δ = V κ D δ [ψ δ ] + W κ Ω [θ δ ] in Ω \ D δ , where (φ δ , ψ δ , θ δ ) ∈ H(∂D δ ) × L 2 0 (∂Ω) is the unique solution of the following system        S κ D δ [φ δ ] -S κ D δ [ψ δ ] -D κ Ω [θ δ ] = 0 on ∂D δ μ(-1 2 I + (K κ D δ ) * )[φ δ ] -µ( 1 2 I + (K κ D δ ) * )[ψ δ ] -µ ∂ ∂n D κ Ω [θ δ ] = 0 on ∂D δ S κ D δ [ψ δ ] + ( 1 2 I + K κ Ω )[θ δ ] = g on ∂Ω. (3.3)
Let Ψ δ (x) = x + δν(x) be the diffeomorphism from ∂D to ∂D δ . The following estimates hold.

Lemma 3.1.1. There exists a constant C depending only on D such that for any functions

(φ δ , ψ δ , θ δ ) ∈ H(∂D δ ) × L 2 0 (∂Ω), we have: i) S κ D δ [φ δ ] • Ψ δ -S κ D [φ δ • Ψ δ ] L 2 (∂D) 3 ≤ Cδ φ δ L 2 (∂D δ ) 3 .
ii)

S κ D δ [ψ δ ] • Ψ δ -S κ D [ψ δ • Ψ δ ] L 2 (∂D) 3 ≤ Cδ ψ δ L 2 (∂D δ ) 3 .
iii)

(K κ D δ ) * [φ δ ] • Ψ δ -(K κ D ) * [φ δ • Ψ δ ] L 2 (∂D) 3 ≤ Cδ φ δ L 2 (∂D δ ) 3 . iv) (K κ D δ ) * [φ δ ] • Ψ δ -(K κ D ) * [φ δ • Ψ δ ] L 2 (∂D) 3 ≤ Cδ φ δ L 2 (∂D δ ) 3 .
Proof. The two last inequalities have been proven in Lemma 3.1 in [?]. Following the same steps, we can easily obtain the first and the second ones.

Let us consider the asymptotic behavior of

S κ D δ [φ δ ], S κ D δ [ψ δ ], D κ Ω [θ δ ], (K κ D δ ) * [φ δ ] and (K κ D δ ) * [ψ δ ] on ∂D δ , and S κ D δ [ψ δ ] on ∂Ω as δ → 0. Denote by φ = φ δ • Ψ δ the vectorial function with components φj(j=1,2,3) .
Besides, for all x ∈ ∂D δ and r = x -y = 0, we define the integral operators:

S κ D δ,i [φ δ ]( x) = ∂D δ Γ κ ij ( x, y)φ δ,j ( y)dσ δ ( y), S κ D δ,i [ψ δ ]( x) = ∂D δ Γ κ ij ( x, y)ψ δ,j ( y)dσ δ ( y), (K κ D δ,i ) * [φ δ ]( x) = p.v. ∂D δ -S κ ijk ( x, y) ν k ( x)φ δ,j ( y)dσ δ ( y), (K κ D δ,i ) * [ψ δ ]( x) = p.v. ∂D δ -S κ ijk ( x, y) ν k ( x)ψ δ,j ( y)dσ δ ( y), D κ Ω,i [θ δ ]( x) = ∂Ω -S κ ijk ( x, y)ν k (y)θ δ,j (y)dσ(y),
and for x ∈ ∂Ω S κ D δ,i [ψ δ ](x) = ∂D δ Γ κ ij (x, y)ψ δ,j ( y)dσ δ ( y).
By Taylor expansions of

Γ κ ij ( x -y), Γ κ ij (x -y), S ijk ( x, y) ν k ( x)
and S ijk ( x, y)ν k (y), we obtain:

Γ κ ij ( x, y) = Γ κ ij (x, y) + ∞ n=1 δ n |α|=n (ν(x) -ν(y)) α α! ∇ α Γ κ ij (x, y) :=Γ κ,n ij (x,y) , Γ κ ij (x, ỹ) = Γ κ ij (x, y) + ∞ n=1 δ n |α|=n (-ν(y)) α α! ∇ α y Γ κ ij (x, y) :=Γ κ,n ij (x,y) , -S κ ijk ( x, y)ν k (y) = -S κ ijk (x, y)ν k (y) :=H ij (x,y) - ∞ n=1 δ n |α|=n (-ν(x)) α α! ∇ α x S κ ijk (x, y)ν k (y) :=H n ij (x,y) , -S κ ijk ( x, y) ν k ( x) = -S κ ijk (x, y)ν k (x) :=H ij (x,y) - ∞ n=1 δ n |α|=n (ν(x) -ν(y)) α α! ∇ α x S κ ijk (x, y)ν k (y) :=H n ij (x,y)
.

We then introduce the operators with (1.2):

S κ,n D,i [φ δ ]( x) = m+q=n ∂D Γ κ,m ij (x, y)σ q (x) ψj (y)dσ(y), n 0, S κ,n D,i [ψ δ ]( x) = m+q=n ∂D Γ κ,m ij (x, y)σ q (x) ψj (y)dσ(y), n 0, D κ,n Ω,i [θ δ ]( x) = ∂Ω H κ,n ij (x, y)θ δ,j (y)dσ(y), n 0, K κ,n D,i [ψ δ ]( x) = p.v. m+q=n ∂D H κ,m ij (x, y)σ q (x) ψj (x)dσ(x), n 0.
where

φ(x) = φ δ • Ψ(x) and ψ(x) = ψ δ • Ψ(x) . Note that S κ,0 D = S κ,0 D = S κ D , D κ,0 D = D κ D , K κ,0 D = (K κ D )
* and that σ q = 0 for all q ≥ 3. We can obtain the following theorem whose demonstration is similar to that of lemma 3.1.1.

(φ δ , ψ δ , θ δ ) ∈ H(∂D δ ) × L 2 0 (∂Ω): (K κ D δ ) * [φ δ ] • Ψ δ -(K κ D ) * [ φ] - N n=1 δ n K D κ,n [ φ] (L 2 (∂D)) 3 ≤ Cδ N +1 φ (L 2 (∂D)) 3 S κ D δ [φ δ ] • Ψ δ -S κ D [ φ] - N n=1 δ n S κ,n D [ φ] (L 2 (∂D)) 3 ≤ Cδ N +1 φ (L 2 (∂D)) 3 S κ D δ [φ δ ] -S κ D [ φ] - N n=1 δ n S κ,n D [ φ] (L 2 (∂D)) 3 ≤ Cδ N +1 φ (L 2 (∂D)) 3 D κ D δ [θ δ ] -D κ D [θ δ ] - N n=1 δ n D κ,n D [θ δ ] (L 2 (∂Ω)) 3 ≤ Cδ N +1 θ δ (L 2 (∂Ω)) 3
where

φ := φ δ • Ψ δ , ψ := ψ δ • Ψ δ . Let (φ (n) , ψ (n) , θ (n) )
, n ≥ 1 be the solution of the following system:

                                 S κ D [φ (n) ] -S κ D [ψ (n) ] -D κ Ω [θ (n) ] = n-1 m=0 -S κ,n-m D [φ (m) ] + S κ,n-m D [ψ (m) ] + D κ,n-m Ω [θ (m) ] on ∂D μ(-1 2 I + (K κ D ) * )[φ (n) ] -µ( 1 2 I + (K κ D ) * )[ψ (n) ] -µ ∂ ∂n D κ Ω [θ (n) ] = n-1 m=0 -μ((K κ,n-m D ) * )[φ (m) ] + µ((K κ,n-m D ) * )[ψ (m) ] + µ ∂ ∂n D κ,n-m Ω [θ (m) ] on ∂D S κ D [ψ n ] + ( 1 2 I + K κ Ω )[θ (n) ] = δ 0n g + n-1 m=0 -S κ,n-m D [ψ (m) ] -K κ,n-m Ω [θ (m) ] on ∂Ω. (3.4)
We recursively construct the sentence (φ

(n) , ψ (n) , θ (n) ) ∀n ∈ N * .
Let define

φ N = N n=0 δ n φ (n) , ψ N = N n=0 δ n ψ (n) , θ N = N n=0 δ n θ (n) , (3.5)
where (φ (0) , ψ (0) , θ (0) ) = (φ, ψ, θ) the unique solution of (2.4). Taking into account Theorem 3.1.1, one deduces from system (3.4) that:

                                     N n=0 δ n S κ,n D [ φ -φ N ] - N n=0 δ n S κ,n D [ ψ -ψ N ] - N n=0 δ n D κ,n Ω [θ δ -θ N ] = O(δ N +1 ) on ∂D N n=0 μ(- δ 0n 2 I + δ n (K κ,n D ) * )[ φ -φ N ] - N n=0 µ( δ 0n 2 I + δ n (K κ,n D ) * )[ ψ -ψ N ] -N n=0 µ ∂ ∂n D κ,n Ω [θ δ -θ N ] = O(δ N +1 ) on ∂D N n=0 δ n S κ,n D [ ψ -ψ N ] + D κ Ω [θ δ -θ N ] = O(δ N +1 ) on ∂Ω.
Then we get the following lemma.

Lemma 3.1.2. Let N ∈ N, there exists C depending only on N and Ω such that

φ - N n=0 δ n φ (n) (L 2 (∂D)) 3 ≤ Cδ N +1 , ψ - N n=0 δ n ψ (n) (L 2 (∂D)) 3 ≤ Cδ N +1 , θ δ - N n=0 δ n θ (n) (L 2 (∂Ω)) 3 ≤ Cδ N +1
where(φ (n) , ψ (n) , θ (n) ) are defined by the recursive relations (3.4).

Asymptotic expansion for the solution of the perturbed problem

In this section, we develop the asymptotic behavior of u δ -u as δ → 0. According to the boundary integral representation of the solution, given in section 3.1, we have:

u δ -u = S κ D δ [φ δ ] -S κ D [φ] in D 0 , S κ D δ [ψ δ ] -S κ D [ψ] + D κ Ω [θ δ ] -D κ Ω [θ] in Ω 0 , (3.6)
where

D 0 is any close subset of D ∩ D δ and Ω 0 is any subset of Ω \ (D ∪ D δ ).
We obtain with Taylor expansion of Γ κ ij (x, ỹ) and (1.2):

S κ D δ [φ δ ](x) = ∂D δ Γ κ ij (x, ỹ) φj (y)dσ δ (ỹ) (3.7)
= ∂D Γ κ ij (x, y) + N n=1 δ n Γ κ,n ij (x, y) φ j (y) + N n=1 δ n φ (n) j (y) 1 + N n=1 δ n σ n (x) dσ(y)+O(δ N +1 ) = ∂D Γ κ ij (x, y)φ(y)dσ(y) + N n=1 δ n m+q+k=n ∂D Γ κ,m ij (x, y)σ q (y)φ (k) (y)dσ(y) + O(δ N +1 ) = S κ D [φ](x) + N n=1 δ n m+q+k=n ∂D Γ κ,m ij (x, y)σ q (y)φ (k) (y)dσ(y) + O(δ N +1 ) ∀x ∈ ∂D.
By the same way, S κ D δ [ψ δ ](x) for x ∈ ∂D can be rewritten as follows:

S κ D δ [ψ δ ](x) = ∂D δ Γ κ ij (x, ỹ) ψj (y)dσ δ (ỹ) (3.8) = S κ D [ψ](x) + N n=1 δ n m+q+k=n ∂D Γ κ,m ij (x, y)σ q (y)ψ (k) (y)dσ(y).
Similarly, we have

D κ Ω [θ δ ](x) = ∂Ω -S κ ijk (x, y)ν k (y)θ δ (y)dσ(y) (3.9) = ∂Ω -S κ ijk (x, y)ν k (y) θ j (y) + N n=1 δ n θ (n) j (y) dσ(y) + O(δ N +1 ), x ∈ Ω 0 .
Then, we get the following result.

Theorem 3.2.1. Define for n ∈ N the vector-valued functions u n = (u n i ) i=1,2,3 as follows:

u n i (x) =                      m+q+k=n ∂D Γ κ,m ij (x, y)σ q (y)φ (k) j (y)dσ(y) ∀x ∈ D 0 ; m+q+k=n ∂D Γ κ,m ij (x, y)σ q (y)ψ (k) j (y)dσ(y) + ∂Ω -S κ ijl (x, y)ν l (y)θ (n) j (y)dσ(y) ∀x ∈ Ω 0 .
Then the following formula holds uniformly for x ∈ Ω 0 ∪ D 0 :

u δ (x) -u(x) = N n=1 δ n u n (x) + O(δ N +1 ). (3.10)
The remainder O(δ N +1 ) depends on N , Ω and Ω 0 .

Computation of the first order approximation

Let us compute the first order approximation of

u δ explicitly. If we put (φ (0) , ψ (0) , θ (0) ) = (φ, ψ, θ) defined by (2.2)-(2.
3), (φ (1) , ψ (1) , θ (1) ) is then defined by the following system:

                                 S κ D [φ (1) ] -S κ D [ψ (1) ] -D κ Ω [θ (1) ] = -S κ,1 D [φ (0) ] + S κ,1 D [ψ (0) ] + D κ,1 Ω [θ (0) ] on ∂D μ(-1 2 I + (K κ D ) * )[φ (1) ] -µ( 1 2 I + (K κ D ) * )[ψ (1) ] -µ ∂ ∂n D κ Ω [θ (1) ] = -μ((K κ,1 D ) * )[φ (0) ] + µ((K κ,1 D ) * )[ψ (0) ] + µ ∂ ∂n D κ,1 Ω [θ (0) ] on ∂D S κ D [ψ (1) ] + ( 1 2 I + K κ Ω )[θ (1) ] = -S κ,1 D [ψ (0) ] -K κ,1 Ω [θ (0) ] on ∂Ω.
Consider the vector-valued function u 1 = (u 1 i ) i=1,2 with:

u 1 i (x) =                      m+q+k=1 ∂D Γ κ,m ij (x, y)σ q (y)φ (k) j (y)dσ(y) in D 0 ; m+q+k=1 ∂D Γ κ,m ij (x, y)σ q (y)ψ (k) j (y)dσ(y) + ∂Ω -S κ ijl (x, y)ν l (y)θ (1)
j (y)dσ(y) in Ω 0 .
Therefore, u 1 i takes the form:

u 1 i (x) = ∂D Γ κ ij (x, y)τ (y)φ (0) j (y)dσ(y) + ∂D Γ κ ij (x, y)φ (1)
j (y)dσ(y)

+ ∂D Γ κ,1 ij (x, y)φ 0 j (y)dσ(y); x ∈ D 0
and more explicitly:

u 1 i (x) = ∂D Γ κ ij (x, y)τ (y)φ (0) j (y)dσ(y) + ∂D Γ κ ij (x, y)φ (1) j (y)dσ(y)
+ ∂D Γ κ,1 ij (x, y)φ (0) j (y)dσ(y) + ∂Ω -S κ ijk (x, y)ν k (y)θ (1)
j (y)dσ(y); x ∈ Ω 0 .

Thus, we get the following formula for

x ∈ Ω 0 ∪ D 0 u δ (x) -u(x) = δu 1 (x) + O(δ 2 ).

Conclusion

We present a result about the construction and justification of the asymptotic expansion of the solution of the transmission Stokes system with a boundary perturbation of an inclusion.

We derive high-order terms in the asymptotic expansion of the solution with layer potential techniques. A reconstruction procedure to determine the localization of the inhomogeneity and the shape of the inclusion based on boundary measurements may be based on our results here.

The method, which is explicitly carried out here for the interior Dirichlet problem of the Stokes operator in three dimensions, can also be used for many other boundary value problems, whenever a suitable potential theory is available, for example in the case of harmonic, elastic, or hydrodynamic boundary value problem.

Part II Asymptotic expansion for Eigenvalues and Eigenfunctions in the Presence of Small Boundary Perturbations Introduction

The main objective of the second part on this report, is to present a schematic way to derive high-order asymptotic expansions for both eigenvalues and eigenfunctions for the Stokes operator caused by small perturbations of the boundary. Also, we rigorously derive an asymptotic formula which is in some sense dual to the leading-order term in the asymptotic expansion of the perturbations in the Stokes eigenvalues due to interface changes of the inclusion.

The following is an outline of this chapter. In Section 4.1, we describe the main problem in this part. In Section 4.2, we develop a boundary integral formulation for solving the eigenvalue problem (4.4), and we present some preliminary results. Finally, Section 4. chapter by developing asymptotic expansion of the perturbations in the eigenfunctions and in eigenpressures using the same problem configurations.

Model problem

Let Ω ⊂ R 3 be a bounded open domain with boundary of class C 2 . We consider the following eigenvalue problem for the Stokes system with homogeneous boundary conditions:

         -µ∆v + ∇p = λv in Ω ∇.v = 0 in Ω v = 0 in ∂Ω.
Here v = (v 1 , v 2 , v 3 ) denotes the velocity field, µ the viscosity while the scalar function p is the pressure. Now, if we suppose that µ is a positive constant, then the first equation in the above system is equivalent to:

-∆v + ∇(p/µ) = ( λ/µ)v in Ω.
Then, if µ is constant we have the following eigenvalue problem

         -∆v + ∇p = λv inΩ ∇.v = 0 inΩ v = 0 in∂Ω (4.1)
with p := p/µ and λ := λ/µ.

It is well known that this eigenvalue problem admits a sequence of a no decreasing positive

eigenvalues 0 < λ 1 ≤ λ 2 ≤ • • • ≤ λ n ≤ • • • tending to infinity as n → +∞.
The eigenfunctions {v n } n≥1 ⊂ (H 1 0 (Ω)) 3 and the eigenpressures {p n } n≥1 ⊂ L 2 (Ω) may be taken so that {v n } n≥1 constitutes an orthonormal basis of

H(Ω) := {v ∈ (H 1 0 (Ω)) 3 : ∇ • v = 0 in Ω}.
The pressure p is determined up to an additive constant.

We assume that the boundary ∂Ω is subjected to a small, smooth deformation and that the boundary of the deformed domain Ω δ is given by: Obviously, the domain Ω δ is of class C 2 and the Dirichlet eigenvalue problem for the Stokes system can be defined in Ω δ as well.

∂Ω δ := {x = x + δh(x)ν(x), x ∈ ∂Ω} (4.
In this chapter, we derive the asymptotic expansion of eigenvalues, eigenfunctions and the eigenpressures solutions to the Stokes system:

         -∆v δ + ∇p δ = λ δ v δ in Ω δ ∇.v δ = 0 in Ω δ v δ = 0 in ∂Ω δ . (4.4)
Here we suppose that the eigenvalue λ 0 is simple. Then the eigenvalue λ δ is simple and is near to λ 0 associated to the normalized eigenfunction v δ .

Integral equations method

We now develop a boundary integral formulation for solving the perturbed eigenvalue problem (4.4). The components of the fundamental Stokes tensor Γ = (Γ ij ) 3 i,j=1 and those of the associated pressure vector P = (P ij ) 3 i,j=1 , which determine the fundamental solution (Γ, P ) of the Stokes system in R 3 , are given by (see for instance [START_REF] Ammari | A method of biological tissues elasticity reconstruction using magnetic resonance elastography measurements[END_REF], [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF])

   Γ ij (λ, x) = -1 4π δ ij e i √ λ|x| |x| -1 4πλ ∂ x i ∂ x j ( e i √ λ|x| -1 |x| ) P i (x) = -1 4π x i |x| 3 , (4.5)
where ∂x i denotes ∂/∂x i for i = 1, 2, 3 and i 2 = -1. We recall that the i th row

Γ i of Γ satisfies    -∆Γ i + ∇P i (x) -λΓ i = e i δ(x) in R 3 ∇.Γ i = 0 in R 3 ; (4.6)
in the sense of distributions and where (e i ; i = 1, 2, 3) is the orthonormal basis of R 3 . Note that we used the Einstein convention for the summation notation omitting the summation sign for the indices appearing twice.

The potential theory for the Stokes system

Let us denote by ϕ = (ϕ 1 , ϕ 2 , ϕ 3 ) a complex vector-valued function with class C 0 (∂Ω).

The hydrodynamic single-layer potential with density ϕ ∈ C 0 (∂Ω) 3 is the vector function

S(λ)ϕ(x) defined by S(λ)ϕ(x) := ∂Ω Γ(λ, |x -y|)ϕ(y) dσ(y), x ∈ R 3 \∂Ω. (4.7)
The pressure term Q corresponding to the single layer potential is the function given by

Qϕ(x) := ∂Ω P (x, y)ϕ(y) dσ(y), x ∈ R 3 \∂Ω.
For a careful study of these potentials, one can refer to [START_REF] Kohr | The interior Neumann problem for the Stokes resolvent system in a bounded domain in R n[END_REF], [START_REF] Ladyzhenskaya | Mathematical theory of the viscous incompressible[END_REF], [START_REF] Varnhorn | The Stokes equations[END_REF].

Taking into account the well known properties of Green function Γ, one obtains the result that the pair (Sϕ, Qϕ) are smooth functions in each of the domains Ω and R 3 \Ω respectively. Also these functions are classical solutions to the Stokes system (4.1).

The continuity and jump relations of the Stokes surface potentials on the boundary ∂Ω are described in the following proposition (see [START_REF] Ammari | A method of biological tissues elasticity reconstruction using magnetic resonance elastography measurements[END_REF], [START_REF] Depauw | Solutions des équations de Navier Stokes incompressibles dans un domaine exterieur[END_REF] pp. 41-42 or [START_REF] Varnhorn | The Stokes equations[END_REF] p. 66 ):

Proposition 4.2.1. Let ϕ ∈ C 0 (∂Ω) 3 and let S denotes the surface potential defined in (4.7).

Then on the boundary ∂Ω the following continuity and jump relations are satisfied:

(S(λ)ϕ) + = (S(λ)ϕ)| -= S(λ)ϕ ∂S(λ)(ϕ) ∂ν (x) ± = ± ϕ(x) 2 + ∂Ω ∂Γ(λ,|x-y|) ∂ν(x)
ϕ(y) dσ(y).

Boundary integral formulation

In this section, we give a boundary integral formulation in order to solve the eigenvalue problems (4.1) and (4.4).

Proposition 4.2.2. Suppose h satisfies (4.3). Then, there exists δ 0 > 0 such that the map

Ψ δ (x) defined by Ψ δ (x) = x + δh(x)ν(x) (4.8)
is a C 2 -diffeomorphism from ∂Ω to ∂Ω δ for δ < δ 0 . In addition, the following equality holds

det (∇Ψ δ ) = 1 + tr∇(hν)δ + 1 2 tr∇(hν) 2 -tr ∇(hν) 2 δ 2 (4.9) + det (∇(hν)) δ 3 ,
where tr means the trace of a matrix. Moreover, we have tr∇(hν) = div(hν).

proof Recall that the function h(x) is C 2 on ∂Ω, then the map Ψ δ (x) is also C 2 . A simple calculation yields the equality (4.9). Consequently, for δ small enough the map

Ψ δ (x) is a C 2 -diffeomorphism from ∂Ω to ∂Ω δ .
We further denote Ψ -1 δ the reciproque function of Ψ δ (x). Thanks to Ψ -1 δ , we can define the operator A δ (λ) as follows:

A δ (λ)ϕ(x) = S Ω δ (λ)ϕ(Ψ -1 δ ) (Ψ δ (x)), ϕ ∈ (L 2 (∂Ω)) 3 (4.10)
where S Ω δ (λ) is the hydrodynamic single-layer potential given by (4.7) when we have replaced the boundary ∂Ω by ∂Ω δ .

For i, j ∈ {1, 2, 3}, we can define the j th -component of the vector-valued function A δ (λ)

as follows:

(A δ (λ)ϕ) j (x) = S Ω δ (λ)ϕ j (Ψ -1 δ ) (Ψ δ (x)), ϕ ∈ (L 2 (∂Ω)) 3 . (4.11)
The j th -component of the single-layer potential S Ω δ (λ) is given by

S Ω δ (λ)ϕ j (x) := ∂Ω δ Γ ji (λ, |x -ỹ|)ϕ i (ỹ) dσ δ (ỹ), x ∈ R 3 \∂Ω δ , j = 1, 2, 3, (4.12)
where ϕ i is the i th -component of the vector-valued function ϕ. Using Proposition 4.2.2, relations (4.11)-(4.12) and the continuity relations given by Proposition 4.2.1, we obtain for

x ∈ ∂Ω that A δ (λ)ϕ j (x) = ∂Ω Γ ji (λ, |Ψ δ (x) -Ψ δ (y)|) det (∇Ψ δ (y)) ϕ i (y) dσ(y), j = 1, 2, 3. (4.13)
Let A 0 the operator defined as in (4.10) by

A 0 φ = S(λ)φ,
where φ ∈ (L 2 (∂Ω)) 3 . Then, we have the following result, which is a slight variation of the Lemma 6.1 due to Ammari et al. [START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF] for the scalar eigenvalue problem.

Proposition 4.2.3. The operator-valued function A 0 (λ) :

H -1/2 (∂Ω) 3 → H 1/2 (∂Ω) 3 is
Fredholm of index zero in C\iR -. In addition the Dirichlet eigenvalues of the Stokes system (4.1) are exactly its real zeros.

From Proposition 4.2.3 we know that if λ 0 is an eigenvalue of (4.1) then λ 0 is a real zero of A 0 (λ). Moreover, for 0 small enough, the function A -1 0 (λ) is meromorphic in D 0 (λ 0 ), where D 0 (λ 0 ) means the disc of center λ 0 and radius 0 , and λ 0 is its unique pole in D 0 .

Furthermore we have the following Laurent expansion:

A -1 0 (λ) = (λ -λ 0 ) -1 0 + R 0 (λ), (4.14)
where 0 : KerA 0 (λ 0 ) → KerA 0 (λ 0 ), and R 0 (λ) is a holomorphic function.

Our main results in this section are summarized in the following theorem.

Theorem 4.2.1. Suppose that the eigenvalue λ 0 of (4.1) is with multiplicity 1. Then, there exist a positive constant δ 0 ( 0 ) such that for |δ| < δ 0 , the operator-valued function λ → A δ (λ) has a real zero λ(δ) in D 0 (λ 0 ). This zero is exactly the eigenvalue of the perturbed eigenvalue problem (4.4), and is an analytic function with respect to δ in ]-δ 0 , δ 0 [. It satisfies λ(0) = λ 0 . Moreover, the following assertions hold:

A -1 δ (λ) = (λ -λ(δ)) -1 (δ) + R δ (λ), (δ) : Ker(A δ (λ(δ)) → Ker(A δ (λ(δ)), (4.15)
where R δ (λ) is a holomorphic function with respect to (δ, λ) ∈] -δ 0 , δ 0 [×D 0 (λ 0 ).

Asymptotic formula 4.3.1 High-order terms in the expansion of A δ

We now present some basic results related to shape perturbation. The methods to be used here differ from those in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF][START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF], but the expressions deduced are the same. To begin, let (τ 1 (x), τ 2 (x)) be the orthornormal basis of the tangent plan to the surface ∂Ω at a regular point x. Their cross product is then orthogonal to ∂Ω at the point x. By changing their order, we can assume that τ 1 × τ 2 is a vector pointing towards the exterior of the surface ∂Ω. Then dividing it by its length yields the unit normal vector ν(x), that is: Set

ν 0 (x) = τ 1 (x) × τ 2 (x) |τ 1 (x) × τ 2 (x)| , (4.16)
τ δ 1 = gradΨ δ • τ 1 , andτ δ 2 = gradΨ δ • τ 2 .
Using Proposition 4.2.2, we find that:

τ δ 1 = τ 1 + δM τ 1 , and τ δ 2 = τ 2 + δM τ 2 , (4.17)
where the (3 × 3)matrix M is given by:

M =     ∂ 1 (hν 1 ) ∂ 2 (hν 1 ) ∂ 3 (hν 1 ) ∂ 1 (hν 2 ) ∂ 2 (hν 2 ) ∂ 3 (hν 2 ) ∂ 1 (hν 3 ) ∂ 2 (hν 3 ) ∂ 3 (hν 3 )    
with ν i means the i-th (i = 1, 2, 3) component of the vector ν.

For δ sufficiently small, one can see that the outward unit normal vector to ∂Ω δ is given by

ν δ (x) = τ δ 1 (x) × τ δ 2 (x) |τ δ 1 (x) × τ δ 2 (x)| , (4.18)
for x ∈ ∂Ω. Then, the following asymptotic expansion holds.

Proposition 4.3.1. Let ν 0 be given by (4.16). Then, the outward unit normal ν δ (x) to ∂Ω δ at

x, can be expanded uniformly as

ν δ (x) = ν (0) (x) + ∞ n=1 δ n ν (n) (x), x ∈ ∂Ω,
where the vector-valued functions ν (n) are uniformly bounded. In particular, for x ∈ ∂Ω:

ν (0) (x) = ν(x), ν (1) = 1 |τ 1 × τ 2 | τ 1 × M τ 2 + M τ 1 × τ 2 -ν 0 • (τ 1 × M τ 2 + M τ 1 × τ 2 ) ν 0 .
Proof. Considering the expansions (4.17) for δ sufficiently small, the relation (4.18) becomes:

ν δ = a + δb + δ 2 c |a + δb + δ 2 c| , (4.19)
where a, b, and c are vector-valued functions given by:

a = τ 1 × τ 2 , b = M τ 1 × τ 2 + τ 1 × M τ 2 , and c = M τ 1 × M τ 2 .
So that, by expanding the quotient (4.19) as δ tends to zero, we get the desired results.

Next, one can use Proposition 4.2.2 to get the uniformly convergent expansion for the surface element as follows:

Proposition 4.3.2. Let ỹ = Ψ δ (y)
where Ψ δ (y) is given by (4.8) for y ∈ ∂Ω. Then, the following expansion for the surface element dσ δ (ỹ) holds uniformly for y ∈ ∂Ω:

dσ δ (ỹ) = det (∇Ψ δ ) dσ(y) = σ 0 (y) + σ 1 (y)δ + σ 2 (y)δ 2 + σ 3 (y)δ 3 dσ(y), (4.20)
where

σ 0 ≡ 1, σ 1 (y) = ∇ • (hν), σ 2 (y) = 1 2 tr∇(hν) 2 -tr ∇(hν) 2 , and σ 3 (y) = det (∇(hν)). Set x = Ψ δ (x), x ∈ ∂Ω (4.21) ỹ = Ψ δ (y), y ∈ ∂Ω, (4.22)
and define

Θ(x, y) := 1 δ Ψ δ (x) -Ψ δ (y) -(x -y) for δ = 0. (4.23)
Recall that Ψ δ is a C 2 vector-valued function on ∂Ω, then Θ is also a C 2 function. Moreover, the following holds.

Proposition 4.3.3. The vector-valued function Θ(x, y) is C 2 on ∂Ω × ∂Ω and there exists a constant C > 0 that only depends on Ω and h such that:

|Θ(x, y)| ≤ C|x -y|, |Θ(x, y) • (x -y)| ≤ C|x -y| 2 ,
for all x, y ∈ ∂Ω.

Proof. Expression (4.23) shows that:

Θ(x, y) = h(x)ν(x) -h(y)ν(y).
Since ∂Ω is a C 2 surface, there exists a constant C > 0 such that:

|ν(x) -ν(y)| ≤ C |x -y|, and |ν(x) • (x -y)| ≤ C |x -y| 2 ,
for all x, y ∈ ∂Ω.

The last inequalities and the C 2 regularity of h yield the results of the proposition. Now, by using (4.23) we obtain:

x -ỹ = x -y + δΘ(x, y), (x, y) ∈ ∂Ω × ∂Ω, (4.24)
and the following results hold.

Lemma 4.3.1. Let r > 0 be a fixed real and m ≥ 2 be a fixed integer. The following asymptotic expansions

e i √ λ|x-ỹ| |x -ỹ| m = e i √ λ|x-y| |x -y| m 1 + δT (m) 1 (x, y) + n≥2 δ n T (m) n (λ; x, y) , (4.25)
∂ xi ∂ xj e i √ λ|x-ỹ| -1 |x -ỹ| = R (ij) 0 (λ; x, y) + δR (ij) 1 (λ; x, y) + n≥2 δ n R (ij) n (λ; x, y) (4.26)
hold uniformly for (λ, x, y) ∈ B r (0)×∂Ω×∂Ω, where B r (0) is a ball in the complex plane of center zero and radius r. In addition the functions T

R (ij) 0 (λ; x, y) := ∂ x i ∂ x j e i √ λ|x-y| -1 |x -y| , and
R (ij) 1 (λ; x, y) = (x-y)•Θ(x, y) -λ 2 δ ij -3i √ λ δ ij r -iλ 3/2 r i r j r +7λ 2 r i r j r 2 +15i √ λ r i r j r 3 e i √ λr r 5 +(x-y)•Θ(x, y) 4i √ λ δ ij r -15 δ ij r 2 -λ 2 [-5 r i r j r 2 + (r i θ j + r j θ i ) (x -y) • Θ(x, y) -7i √ λ[-6 r i r j r 3 + 1 r (r i θ j + r j θ i ) (x -y) • Θ(x, y) ] +15[-7 r i r j r 4 + 1 r 2 (r i θ j + r j θ i ) (x -y) • Θ(x, y) e i √ λr r 5 +15 (x -y) • Θ(x, y) r 7 δ ij -[-7 r i r j r 2 + (r i θ j + r j θ i ) (x -y) • Θ(x, y) ] ,
where r = |x -y|, r j = x j -y j , θ j means the j th component of Θ, and δ ij means the Kronecker index.

Proof. For m = 2 and x = y we have:

|x -ỹ| 2 = |x -y + δΘ(x, y)| 2 = |x -y| 2 1 + δc (2) 1 (x, y) + δ 2 c (2) 2 (x, y) ,
where

c (2) 1 (x, y) = 2Θ(x, y) • (x -y) |x -y| 2 , c (2)
2 (x, y) = Θ(x,y)•Θ(x,y)

|x-y| 2 . Proposition 4.3.1 shows that c

(2)

1 (x, y) and c

(2) 2 (x, y) are bounded uniformly on ∂Ω × ∂Ω.

For m > 2, we have:

|x -ỹ| m = |x -y + δΘ(x, y)| m = |x -y| m 1 + δ Θ(x, y) |x -y| m ,
where x = y.

Using (4.23), Proposition 4.2.2 and Proposition 4.3.3, we see that the regular vector-valued function (x, y) → Θ(x,y) |x-y| is well defined on ∂Ω × ∂Ω, and it is independent of δ. Therefore, we can expand:

|x -ỹ| m = c (m) 0 (x, y) + ∞ n=1 δ n c (m) n (x, y) uniformly on ∂Ω × ∂Ω, (4.27)
where the first coefficients c

(m) 0 (x, y) = |x -y| m , c (m) 1 (x, y) = m < x -y, Θ(x, y) > |x-y| m-2 and c (m) 2 (x, y) = m 2 |x-y| m | Θ(x,y) |x-y| | 2 +(m-2) < x-y |x-y| 2 , Θ(x,y) |x-y| > 2 . Moreover, if m is even, then c (m)
n (x, y) = 0 for n ≥ m + 1. Now combining (4.27) for m = 1 with the well known asymptotic expansion of the exponential function, we immediately get

e i √ λ|x-ỹ| = ∞ n=0 δ n K n (λ; x, y) uniformly on B r (0) × ∂Ω × ∂Ω, (4.28)
where the first coefficients K 0 (λ; x, y) = e i √ λc

(1) 0 = e i √ λ|x-y| , and the coefficients K n (λ; x, y) can be deduced recursively from c

n . Thanks to relations (4.24), (4.27) and (4.28), we can obtain the desired result given by (4.25) where the coefficients T To prove relation (4.26), we firstly expand

∂ xi ∂ xj e i √ λ|x-ỹ| -1 |x -ỹ| = 1 |x -ỹ| ∂ xi ∂ xj e i √ λ|x-ỹ| -1 +∂ xi 1 |x -ỹ| ∂ xj e i √ λ|x-ỹ| -1 (4.29) +∂ xj 1 |x -ỹ| ∂ xi e i √ λ|x-ỹ| -1 + ∂ xi ∂ xj 1 |x -ỹ| e i √ λ|x-ỹ| -1 .
To simplify, let us denote r = |x -y|, r i = x i -y i , r = |x -ỹ|, and rj defined by the relation Now, by using the following result

∂ xj r = rj r . ( 4
∂ xj e i √ λ|x-ỹ| -1 = ∂ xj (i √ λr e i √ λr = i √ λ rj r e i √ λr , (4.32)
we get that

∂ xi ∂ xj e i √ λ|x-ỹ| -1 = i √ λ ∂ xi ( rj r )e i √ λr + rj r ∂ xi e i √ λr = i √ λ δ ij r - ri rj r3 + i √ λ ri rj r2 e i √ λr . ( 4
∂ xi ∂ xj e i √ λ|x-ỹ| -1 |x -ỹ| 3 = i √ λ δ ij r4 -3 δ ij r5 -λ 2 ri rj r5 -7i √ λ ri rj r6 + 15 ri rj r7 e i √ λr + 3 δ ij r5 -15 ri rj r7 . (4.34)
On the other hand, the components of the vectorial relation (4.24) can be given as follows

xi -ỹi = x i -y i + δθ i (x, y). i = 1, 2, 3 (4.35)
where θ i (x, y) means the i th component of the vector-valued function Θ(x, y).

Then, by relations (4.30) and (4.35) we deduce that

rj = xi -ỹi = x i -y i + δθ i (x, y) = r i + δθ i (x, y), i = 1, 2, 3. (4.36)
Using both relations (4.24) and (4.36), we get the following expansion

ri rj = α (ij) 0 + α (ij) 1 δ + α (ij) 2 δ 2 , (4.37)
where the first coefficients: α

(ij) 0 = r i r j , α (ij) 1
= r i θ j + r j θ i and α

(ij) 2 = θ i • θ j .
Now regarding (4.27) and using the fact that c

ri rj rm = β (ij) 0,m + δβ (ij) 1,m + ∞ n=2 δ n β (ij) n,m , (4.39)
where the first coefficients:

β (ij) 0,m (x, y) = α (ij) 0 (c (m) 0 ) -1 (x, y), and
β (ij) 1,m (x, y) = -α (ij) 0 c (m) 1 (c (m) 0 ) -2 (x, y) + α (ij) 1 (c (m) 0 ) -1 (x, y).
To achieve the proof, we may insert all expansions (4.28), (4.38) (for m ∈ {4, 5}), and (4.39)(for m ∈ {5, 6, 7} into (4.34). We get that

∂ xi ∂ xj e i √ λ|x-ỹ| -1 |x -ỹ| 3 = R (ij) 0 (x, y) + δR (ij) 1 (x, y) + ∞ n=2 δ n R (ij) n (x, y)
where the first coefficient:

R (ij) 0 (λ; x, y) = i √ λδ ij κ (4) 0 -3δ ij κ (5) 0 -λ 2 β (ij) 0,5 -7i √ λβ (ij) 0,6 +15β (ij) 0,7 K 0 +3δ ij κ (5) 0 -15β (ij) 0,7 = i √ λδ ij (c (4) 0 ) -1 -3δ ij (c (5) 0 ) -1 -λ 2 r i r j (c (5) 0 ) -1 -7i √ λr i r j (c (6) 0 ) -1 +15r i r j (c (7) 0 ) -1 e i √ λc (1) 0 +3δ ij (c (5)
0 ) -1 -15r i r j (c

(7) 0 ) -1 .
Using the fact that c (m) 0

= |x -y| m , we find that

R (ij) 0 (λ; x, y) := ∂ x i ∂ x j e i √ λ|x-y| -1 |x -y| 3 .
Based on (4.34), we find that

R (ij) 1 (λ; x, y) = i √ λδ ij κ (4) 0 -3δ ij κ (5) 0 -λ 2 β (ij) 0,5 -7i √ λβ (ij) 0,6 + 15β (ij) 0,7 K 1 + i √ λδ ij κ (4) 1 -3δ ij κ (5) 1 -λ 2 β (ij) 1,5 -7i √ λβ (ij) 1,6 + 15β (ij) 1,7 K 0 + 3δ ij κ (5) 1 -15β (ij) 1,7 .
So that, by using the fact that

K 1 = i √ λΘ • (x-y) r e i √ λc (1)
0 , we get

R (ij) 1 (λ; x, y) = i √ λΘ• (x -y) r i √ λδ ij (c (4) 0 ) -1 -3δ ij (c (5) 0 ) -1 -λ 2 r i r j (c (5) 0 ) -1 -7i √ λr i r j (c (6) 0 ) -1 +15r i r j (c (7) 0 ) -1 e i √ λc (1) 0 + i √ λδ ij (c (4) 0 ) -2 c (4) 1 -3δ ij (c (5) 0 ) -2 c (5) 1 -λ 2 [-r i r j (c (5) 0 ) -2 c (5) 1 + (r i θ j + r j θ i )(c (5) 0 ) -1 ] -7i √ λ[-r i r j (c (6) 0 ) -2 c (6) 1 +(r i θ j +r j θ i )(c (6) 0 ) -1 ]+15[-r i r j (c (7) 0 ) -2 c (7)
1 +(r i θ j +r j θ i )(c

(7) 0 ) -1 ] e i √ λc (1) 0 +3δ ij (c (5) 0 ) -2 c (5) 1 -15[-r i r j (c (7) 0 ) -2 c (7) 1 + (r i θ j + r j θ i )(c (7) 0 ) -1 ].
Using the explicit forms of the coefficients c given above, we immediately get the desired expression of R (ij) 1 (λ; x, y). The other coefficients can be deduced easily by the same manner in terms of c (m) n . Now using (4.35), we obtain for i, j = 1, 2, 3 that:

(x i -ỹi )(x j -ỹj ) = ĝ0 (x, y) + δĝ 1 (x, y) + δ 2 ĝ2 (x, y), (4.40)
where

ĝ0 (x, y) = (x i -y i )(x j -y j ), ĝ1 (x, y) = θ i (x, y)(x j -y j ) + θ j (x, y)(x i -y i ), and
ĝ2 (x, y) = θ i (x, y) • θ j (x, y).
Now, by using (4.20), (4.27) and (4.40), we immediately get

(x i -ỹi )(x j -ỹj ) |x -ỹ| m dσ δ (ỹ) = T0 (x, y) + δ T1 (x, y) + n≥2 δ n Tn (x, y) dσ(y), (4.41)
where the Taylor coefficients Tn can be given explicitly with the aid of ĝ0 , ĝ1 , and ĝ2 .

Next, the following result holds.

Lemma 4.3.2. The following uniform expansion holds on R × ∂Ω × ∂Ω:

(x i -ỹi )(x j -ỹj )|x -ỹ|e i √ λ|x-ỹ| dσ δ (ỹ) = E δ (λ; x, y) dσ(y) = ∞ n=0
δ n E n (λ; x, y) dσ(y)

(4.42) with E 0 (λ; x, y) = ĝ0 (x, y)|x -y|e i √ λ|x-y| ,
and the other coefficients E n are deduced from those σ n , c

n , K n and ĝn .

Proof. From the proof of Lemma 4.3.1 (for m = 1) and from relations (4.20) and (4.40), one can get that

(x i -ỹi )(x j -ỹj )|x -ỹ|e i √ λ|x-ỹ| dσ δ (ỹ) = ĝ0 (x, y) + δĝ 1 (x, y)+ δ 2 ĝ2 (x, y) ∞ n=0 δ n c (1) n ∞ n=0 K n (λ; x, y) σ 0 (y) + δσ 1 (y)+ δ 2 σ 2 (y) + δ 3 σ 3 (y) dσ(y).
By collecting terms of equal powers in the above relation, one can deduce easily the uniform expansion (4.42) with

E 0 (λ; x, y) = ĝ0 (x, y)|x -y|e i √ λ|x-y| .
Now, from (4.5) we have

Γ ij (λ, |x -y|) = e i √ λr 4π δ ij r - xi xj r 3 + e i √ λr 4πλ i δ ij r 4 -4i xi xj r 4 -3irx i xj + 1 4πλ -3 δ ij r 5 + 15 xi xj r 7 + e iλr 4πλ - 3
Γ ij (λ, |x -ỹ|) dσ δ (ỹ) = Γ (0) ij (λ, |x -y|) + δΓ (1) ij (λ, |x -y|) (4.43) + n≥2 δ n Γ (n) ij (λ, |x -y|) dσ(y), y ∈ ∂Ω
where the first coefficients:

Γ (0) ij (λ, |x -y|) := δ ij 4π T (1) 0 + i λ T (4) 0 -3 r 0 λ 2 -3 T (5) 0 λ 2 + 1 4π ( 15 λ 2 -1)T (3) 0 ĝ0 - 4i 4πλ T (4) 0 ĝ0 - 3i 4πλ E 0 + 15 4πλ 2 T0 , and
Γ (1) ij (λ, |x -y|) := δ ij 4π T (1) 1 + i λ T (4) 1 -3 r 1 λ 2 -3 T (5) 1 λ 2 + 1 4π ( 15 λ 2 -1) T (3) 0 ĝ1 + T (3) 1 ĝ0 - 4i 4πλ T (4) 0 ĝ1 + T (4) 1 ĝ0 - 3i 4πλ E 1 + 15 4πλ 2 T1 .
In Proposition 4.3.4, the coefficients r 0 and r 1 are deduced from (4.27) for m = 5.

Now, introduce a sequence of components of integral operators (A (n)

i ) n≥0 , defined for any ϕ ∈ L 2 (∂Ω) 3 by:

A (n) ϕ i (x) = ∂Ω Γ (n) ij (x,
y)ϕ j (y) dσ(y), for i, j ∈ {1, 2, 3} and n ≥ 0.

Using previous results, il is clear that we know explicitly the first terms

A (0) i , A (1)
i . For any positive integer N , we can by recursive method get the term A (N ) i . Then, the following theorem holds. Theorem 4.3.1. Let A δ be the operator defined by (4.10). Let Ψ δ (x) be the diffeomorphism defined as in Lemma 4.2.2. Let N be a positive integer. There exists a positive constant C depending only on N , and h C 2 such that for any φ ∈ L 2 (∂Ω δ ) 3 and i, j ∈ {1, 2, 3}, the i th -component A δ i defined by (4.13) satisfies:

A δ φ i oΨ δ -A (0) ϕ) i - N n=1 δ n A (n) ϕ i L 2 (∂Ω) ≤ Cδ N +1 ϕ L 2 (∂Ω) 3 ,
where ϕ = φoΨ δ .

Asymptotic expansion of the eigenvalues

To develop asymptotic behaviors of eigenvalues and eigenfunctions with respect to the parameter of perturbation δ, we may use the results of Theorem 4.3.1. Then, the following asymptotic expansion related to the operator A δ appears clearly.

Proposition 4.3.5. Suppose that we have all hypothesis of Theorem 4.3.1. Then, the operator A δ (λ) defined by (4.10) can be expanded uniformly for x ∈ ∂Ω as follows:

A δ (λ)ϕ = A (0) (λ)ϕ + δA (1) (λ)ϕ + δ 2 A (2) (λ)ϕ + • • • ; as δ → 0, (4.44)
where ϕ ∈ L 2 (∂Ω) 3 , the i th -component of the first term is given by

A (0) i (λ)ϕ i (x) = ∂Ω Γ (0)
ij (x, y)ϕ j (y) dσ(y), for i, j ∈ {1, 2, 3}, and more generally, the i th -component of the term with order n is given by

A (n) (λ)ϕ i (x) = ∂Ω Γ (n)
ij (x, y)ϕ j (y) dσ(y), for i, j ∈ {1, 2, 3} and n ≥ 1.

The coefficients Γ

(n) ij are given by (4.43).

Now, based on the well known works [START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF], we tray to give (formally) an asymptotic expansion related to the eigenvalues λ δ if the eigenvalue λ 0 is simple. But, To get an explicit asymptotic formula for λ δ we may develop our analysis in some Hölder space. This will be given in Theorem 4.3.2. Before that, let a j (δ) denotes:

a j (δ) := 1 2iπ tr ∂D 0 (λ -λ 0 ) j (A δ ) -1 (λ)∂ λ A δ (λ)dλ. (4.45)
The functions a j (δ) is analytic in a complex neighborhood of 0 and satisfies: a j (δ) = a j (δ).

The following main result holds.

Proposition 4.3.6. Let λ 0 be a simple eigenvalue of the problem (4.1). Let the operator A δ be defined by (4.10). Then, there exists a small positive number δ 0 such that the eigenvalue λ(δ) is analytic in ] -δ 0 , δ 0 [ and satisfies:

λ(δ) = λ 0 + δλ 1 + n≥2 λ n δ n , (4.46)
where the first coefficients are (formally) given by:

       λ 1 = 1 2iπ tr ∂D 0 (λ -λ 0 ) (A (0) ) -1 (λ)A (1) (λ)(A (0) ) -1 (λ)∂ λ A (0) (λ) dλ, λ 2 = 1 2iπ tr ∂D 0 (λ -λ 0 ) (A (0) ) -1 A (1) (A (0) ) -1 ∂ λ A (1) + (A (0) ) -1 A (2) (A (0) ) -1 ∂ λ A (0) +(A (0) ) -1 A (1) (A (0) ) -1 2 ∂ λ A (0) dλ.
Proof. Let λ δ be the eigenvalue of the problem (4.4). It is well known that λ δ is also simple such that λ δ → λ 0 as δ → 0.

On the other hand, if we take the curl of the first equation in (4.4), we see that there exists a function w = w(v δ ) called vorticity associated to v δ such that:

   ∆w + λ δ w = 0 in Ω v δ = 0 in ∂Ω.
That is, w is an eigenfunction of the negative Laplacian, but with homogeneous boundary conditions on the velocity v δ .

Moreover, let u δ be the stream function for v δ given as in Lemma 2.10 of [START_REF] Kelliher | Eigenvalues of the Stokes operator versus the Dirichlet Laplacian in the plane[END_REF]. Then, w = ∆u δ and ∇u δ = 0 on ∂Ω. Since u δ is determined only up to a constant, we can then assume So,

λ δ -λ 0 = 1 2iπ tr ∂D 0 (λ -λ 0 )(A (0) ) -1 (λ)∂ λ A δ (λ)ϕ dλ+ 1 2iπ tr ∂D 0 (λ -λ 0 ) ∞ k=1 (A (0) ) -1 (λ) A (0) (λ) -A δ (λ) (A (0) ) -1 (λ) k ∂ λ A δ (λ) dλ
By using (4.14), we find that 1 2iπ tr

∂D 0 (λ -λ 0 )(A (0) ) -1 (λ)∂ λ A δ (λ) dλ = 0.
This result is a direct consequence of the fact that R 0 (λ) and ∂ λ A δ (λ) are holomorphic in the variable λ.

Now we have:

λ δ -λ 0 = 1 2iπ tr ∂D 0 (λ -λ 0 )(A (0) ) -1 (λ) A (0) (λ) -A δ (λ) (A (0) ) -1 (λ) ∂ λ A δ (λ) dλ+ 1 2iπ tr ∂D 0 (λ -λ 0 )(A (0) ) -1 (λ) A (0) (λ) -A δ (λ) (A (0) ) -1 (λ) 2 ∂ λ A δ (λ) dλ+ 1 2iπ tr ∂D 0 (λ -λ 0 ) k≥3 (A (0) ) -1 (λ) A (0) (λ) -A δ (λ) (A (0) ) -1 (λ) k ∂ λ A δ (λ) dλ.
Inserting expression (4.44) into above relation, we may get:

λ δ -λ 0 = 1 2iπ tr ∂D 0 (λ-λ 0 )(A (0) ) -1 δA (1) (A (0) ) -1 +δ 2 A (2) (A (0) ) -1 +• • • ∂ λ A (0) +δ∂ λ A (1) +δ 2 ∂ λ A (2) + • • • ϕ dλ + 1 2iπ tr ∂D 0 (λ -λ 0 )(A (0) ) -1 δA (1) (A (0) ) -1 + δ 2 A (2) (A (0) ) -1 + • • • 2 ∂ λ A (0) + δ∂ λ A (1) + δ 2 ∂ λ A (2) + • • • dλ + • • •
If we collect the same powers of δ, then we get the desired results.

Asymptotic formula in Hölder space

In section we suppose that any shape deformation like (4.2) occurs inside a bounded domain Ω in a Hölder space. Based on the works [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF][START_REF] Ammari | A method of biological tissues elasticity reconstruction using magnetic resonance elastography measurements[END_REF][START_REF] Ammari | Effective viscosity properties of dilute suspensions of arbitrarily shaped particles[END_REF][START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Boyer | Mathematical tools for the study of the incompressible Navier-Stokes equations and related models[END_REF], we may advance the asymptotic expansions for simple and/or multiple eigenvalues of the Stokes operator and we may give explicitly the fist term correction. Since the Stokes system can be viewed as the incompressible limit and as done in [START_REF] Ammari | Effective viscosity properties of dilute suspensions of arbitrarily shaped particles[END_REF], our method in this section may be deeply based on the ones developed in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF] for the elastic case.

Suppose that in this section that Ω ⊂ R 2 is a bounded domain with C and σ 1 may be given by

(σ s ) ijlk = µ s (δ ki δ lj + δ kj δ li ) for i, j, k, l = 1, 2 and s = 0, 1, (4.48)
where µ 0 and µ 1 are the viscosity constants of the flow in Ω\D and D, respectively. Given two (2 × 2) matrices A and B we denote by A : B the contraction, i.e., A :

B = ij a ij b ij .
Now, it is useful to introduce the strain rate tensor D for the flow as follows:

D(v) := 1 2 (∇v + (∇v) T ) = 1 2 ( ∂v i ∂x j + ∂v j ∂x i ) 1≤i,j≤3
.

Let σ D := σ 0 χ Ω\D + σ 1 χ D . Then according to [START_REF] Boyer | Mathematical tools for the study of the incompressible Navier-Stokes equations and related models[END_REF], one can rewrite the eigenvalue problem (4.1) as the generalized one:

               -div(σ D D(v 0 ) -p 0 Id) = λ 0 v 0 in Ω ∇.v 0 = 0 in Ω v 0 = 0 in ∂Ω v 0 = 1 (4.49)
where (v 0 , p 0 , λ 0 ) ∈ (H 2 (Ω)) 2 × H 1 (Ω) × R * + and Id means the identity. Here we refer the reader to ( [14], Theorem IV.5.8) for more details about the regularity properties. On the other hand, if S mens the Stokes operator, it is well know that (see for example, [START_REF] Ammari | Effective viscosity properties of dilute suspensions of arbitrarily shaped particles[END_REF][START_REF] Boyer | Mathematical tools for the study of the incompressible Navier-Stokes equations and related models[END_REF]) there exist an orthogonal projection (the Leray projection) P such that:

S(v) = P (-∆v), ∀v ∈ (H 2 (Ω)) 2 ∩ H(Ω).
In consequence, the Stokes operator enters the general framework about spectral properties of -∆.

The δ-perturbation, denoted by D δ , of the domain D is given by

∂D δ := {x = x + δh(x)ν(x), x ∈ ∂D} (4.50)
where here h(x) is assumed to be real function in C 1,1 (∂D) that satisfies h C 1,1 (∂D) < c 0 for some positive constantc 0 . (4.51)

Let σ D δ := σ 0 χ Ω\D δ + σ 1 χ D δ . Therefore, (4.4) can be generalized to the following perturbed eigenvalue problem

               -div(σ D δ D(v δ ) -p δ Id) = λ δ v δ in Ω ∇.v δ = 0 in Ω v δ = 0 in ∂Ω v δ = 1 (4.52)
where (v δ , p δ , λ δ ) ∈ (H 2 (Ω)) 2 × H 1 (Ω) × R * + . Our purpose in this section is to develop a rigourously asymptotic behavior of the eigenvalue of (4.52) assumed to be more precisely. The main result is the following. Theorem 4.3.2. Suppose that Ω ⊂ R 2 with class C 1,1 . Let λ 0 be a simple eigenvalue of the problem (4.49). Assume that we have (4.51), and the pressure still the same if we have (4.50). Then, the first term correction in the asymptotic expansion of the eigenvalue λ δ is given through the following formula:

λ δ -λ 0 = δ ∂D h(x)V[D(v e 0 )](x) : D(v e 0 (x))ds(x) + O(δ 1+β ),
where β some positif constant and V means the viscous moment tensor (VMT) given by

V[D(v e 0 )] = [ 1 σ 1 (V 1 ∇v e 0 (x)τ ) ⊗ τ + ( σ 0 σ 1 ∇v e 0 (x)ν) ⊗ ν].
Here, we denote by ν, τ respectively the outward normal vector and the tangent vector to ∂D and V 1 a given operator expresses the transmission conditions for v 0 . Proof. To derive the corresponding formulae for the eigenvalues we will use an idea close to Osborn's theorem [START_REF] Osborn | Spectral approximation for compact operators[END_REF] which gives estimates for the convergence of the eigenvalues of a sequence of compact operators. For any f ∈ L 2 (Ω), define the linear operator T f = u , where u is the solution to the problem

-div(σ D δ D(u δ ) -p δ Id) = f in Ω, u δ = 0 on ∂Ω. (4.53)
For the same given function f , we define the linear operator T f = u 0 , where u 0 is the solution

to the problem -div(σ D D(u 0 ) -p 0 Id) = f in Ω, u 0 = 0 on ∂Ω. (4.54)
Clearly T and T are are compact self-adjoint operators. One may use standard energy estimates based on Korn and Poincaré inequalities to get these facts. Moreover, it is easy to see that {T δ ; δ ≥ 0} with T 0 = T are collectively compact and that T δ → T pointwise as δ → 0. To get the last result, one may prove that T δ f converges to T f in L 2 (Ω) for every

f ∈ L 2 (Ω).
By assumption, in the presence of (4.50), there exist a function p such that p δ = p 0 = p. Then, let u δ = T δ f and u 0 = T δ f . For any w ∈ H 1 (Ω), we have:

Ω ∇ • (σ D δ D(u δ ) -pId).w = Ω f.w, so, Ω (σ D δ D(u δ ) -pId) : D(w) = - Ω f.w.
Similarly,

Ω (σ D D(u 0 ) -pId) : D(w) = - Ω f.w.
Consequently, choosing w = u δ -u 0 and subtracting these two equations we get

Ω (σ D δ D(u δ ) -σ D D(u 0 )) : D(u δ -u 0 ) = 0 which gives Ω σ D δ D(u δ -u 0 ) : D(u δ -u 0 ) = - Ω (σ D δ -σ D )D(u 0 ) : D(u δ -u 0 ).
Hence, by using successively Korn's inequality and Hölder's one, one can get

∇(u δ -u 0 ) L 2 (Ω) ≤ C ∇u 0 L 2 (D δ D) ,
where denotes the symmetric difference. It then follows by the Poincaré inequality that

u δ -u 0 H 1 (Ω) ≤ C ∇u 0 L 2 (D δ D) .
Consequently, using the last inequality and the fact that |D δ D| → 0 as δ → 0 and that

∇u 0 ∈ L 2 (Ω) we obtain that T δ → T pointwise as δ → 0 in L 2 (Ω).
Hence all hypotheses hold for the theorem of Osborn. Now if we set,

µ 0 = 1 λ 0 and µ δ = 1 λ δ ,
then according to the problem (4.53)(resp. (4.54)) we can see that ( µ δ , v δ ) (resp.( µ 0 , v 0 )) is eigenpair of T δ (resp. of T 0 ). So, a theorem of Osborn [START_REF] Osborn | Spectral approximation for compact operators[END_REF] yields

µ 0 -µ δ -≺ (T δ -T )v 0 , v 0 ≤ (T -T δ )v 0 2 L 2 (Ω) , (4.55)
where C is independent of δ and v 0 is the solution of (4.49). Furthermore, if v δ is the solution to (4.52), then

v 0 -v δ L 2 (Ω) ≤ (T -T δ )v 0 L 2 (Ω) .
According to [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF][START_REF] Boyer | Mathematical tools for the study of the incompressible Navier-Stokes equations and related models[END_REF] we can extend the regularity results obtained by De Giorgi and Nash in the scalar case [START_REF] Gilbarg | Elliptic Partial Differential Equations of Second Order[END_REF] to get

v δ C 1,α (D δ ) + v δ C 1,α (Ω d 0 /2 \D δ ) ≤ C, for some α > 0, (4.56)
where C is a positif constant and for d 0 > 0 we have Ω d 0 /2 := {x ∈ Ω : dist(x, ∂Ω) > d 0 /2}.

To compute the term ≺ (T δ -T )v 0 , v 0 appearing in (4.55), we may firstly recall that T v 0 = µ 0 v 0 and T δ v 0 = ũδ where ũδ is the solution to

-∇ • (σ D δ D(ũ δ ) + pId) = v 0 , inΩ ũδ = 0, on∂Ω. (4.57)
Putting ṽ0 = μ0 v 0 , then

-∇ • (σ D D(ṽ 0 ) + pId) = v 0 , inΩ ṽ0 = 0, on∂Ω. (4.58)
From (4.57) and (4.58), we have

≺ (T δ -T )v 0 , v 0 = ≺ ũδ -ṽ0 , v 0 = µ 0 Ω v 2 0 + Ω v 0 ũδ dx = µ 0 Ω (σ D -σ D δ )D(ũ δ ) : D(v 0 )dx = µ 0 D δ \D (σ 0 -σ 1 )D(ũ i δ ) : D(v e 0 )dx -µ 0 D\D δ (σ 0 -σ 1 )D(ũ e δ ) : D(v i 0 )dx.
Now, as done in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF] we put x t := x + th(x)ν(x) for x ∈ ∂D and t ∈ [0, δ]. It is clearly that the Jacobian determinant of the change of variables (x, t)

∈ ∂D × [0, δ] → x t ∈ D δ D is |h(x)| + O(δ)
for δ small enough, and hence we get

I 1 := µ 0 D δ \D (σ 0 -σ 1 )D(ũ i δ ) : D(v e 0 )dx = µ 0 δ 0 ∂D∩{h>0} h(x)(σ 0 -σ 1 )D(ũ i δ )(x t ) : D(v e 0 )(x t )ds(x)dt + O(δ 2 ).
Moreover,

I 2 := -µ 0 D\D δ (σ 0 -σ 1 )D(ũ e δ ) : D(v i 0 )dx = -µ 0 δ 0 ∂D∩{h<0} h(x)(σ 0 -σ 1 )D(ũ e δ )(x t ) : D(v i 0 )(x t )ds(x)dt + O(δ 2 ).
On the other hand, based on the result of Li and Nirenberg proved in [START_REF] Li | Estimates for elliptic systems from composite material[END_REF], the following gradient estimate holds:

ũδ C 1,α (D δ ) + ũδ C 1,α (Ω d 0 /2 \D δ ) ≤ C[ ũδ L 2 (Ω) + v 0 L ∞ (Ω d 0 /2 ) ], (4.59)
where ũδ is the solution of (4.57), and the positive constant

C is independent of δ. Remark that ũδ H 1 (Ω) ≤ C v 0 L 2 (Ω) ≤ C, it follows from (4.56) that ũδ C 1,α (D δ ) + ũδ C 1,α (Ω d 0 /2 \D δ ) ≤ C. (4.60)
Now using Lemma 2.2 in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF], and the relations (4.58)-(4.61), one may obtain:

(T δ -T )v 0 L 2 (Ω) = ũδ -ṽ0 L 2 (Ω) ≤ Cδ η+1/2 . (4.61)
On the other hand, we use the gradient estimates (4.57) and (4.58) for ũδ and v 0 , to approximate

D(ũ i δ )(x t ) : D(v e 0 )(x t ) = D(ũ i δ )(x δ ) : D(v e 0 )(x δ ) + O(δ α )
for δ sufficiently small. To proceed with our proof we may investigate the transmission conditions of v 0 along the interface ∂D. One can easily see from the equation in (4.49) that

v 0 satisfies: v e 0 = v i 0 and (σ 0 D(v e 0 ))ν = (σ 1 D(v i 0 ))ν where v e 0 = v 0 | Ω\D and v i 0 = v 0 | D .
Let τ be the unit tangential vector field to ∂D, and following the approach used in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF], one can find that

(σ 1 D(v i 0 ))τ = (V 1 D(v e 0 ))τ on ∂D, (4.62)
where

V 1 := 2µ 0 I 4 + 2(µ 1 -µ 0 )I 2 ⊗ (τ ⊗ τ ).
Here I 2 is the (2 × 2) identity matrix, I 4 is the identity 4-tensor and the viscosity constants µ 0 , µ 1 are given in (4.48).

Applying the transmission conditions (4.62) for the function ũδ , one may get

D(ũ i δ )(x δ ) = σ -1 1 [(σ 1 D(ũ i δ )(x δ )τ ) ⊗ τ + (σ 1 D(ũ i δ )(x δ )ν) ⊗ ν] = σ -1 1 [(V 1 D(ũ e δ )(x δ )τ ) ⊗ τ +(σ 0 D(ũ e δ )(x δ )ν) ⊗ ν].
Now, regarding Lemma 3.2 in [START_REF] Ammari | Reconstruction of small interface changes of an inclusion from modal measurements II: The elastic case[END_REF], which still applied to Stokes case, on may approximate

D(ũ i δ )(x δ ) = μ0 σ 1 [(V 1 D(v e 0 )(x)τ ) ⊗ τ + (σ 0 D(v e 0 )(x)ν) ⊗ ν] + O(δ α 2(α+2) ). (4.63)
Hence,

I 1 = δ µ 2 0 ∂D∩{h>0} h(x)V[D(v e 0 )](x) : D(v e 0 )(x)ds(x) + O(δ 1+ α 2(α+2) )
and

I 2 = δ µ 2 0 ∂D∩{h<0} h(x)V[D(v e 0 )](x) : D(v e 0 )(x)ds(x) + O(δ 1+ α 2(α+2) ),
where α > 0 and V[D(v e 0 )] deduced from (4.63). Therefore,

≺ (T δ -T )v 0 , v 0 = I 1 + I 2 = δ µ 2 0 ∂D h(x)V[D(v e 0 )](x) : D(v e 0 )(x)ds(x) + O(δ 1+ α 2(α+2) ),
and by considering the above relation, (4.55) and (4.61) we obtain

μ0 -μδ = δ μ2 0 ∂D h(x)V[D(v e 0 )](x) : D(v e 0 )(x)ds(x) + O(δ 1+β )
with β = inf(η, α 2(α+2) ). To achieve the proof, we recall that

λ δ -λ 0 = λ 0 λ δ (1/λ 0 -1/λ δ ) = λ 0 λ δ (μ 0 -μδ )
and we take into consideration that λ 0 λ δ = λ 2 0 + δ(λ 0 λ 1 ) + δ 2 (λ 0 λ 2 ) + • • • .

Now we consider the case of a multiple eigenvalue. Let the operators T and T δ be defined as in the proof of Theorem 4.3.2. We suppose that μ0 be a nonzero eigenvalue of T with multiplicity m ≥ 1. Then, for δ small, T δ has a set of m eigenvalues μj δ (counted according to their multiplicity) such that μj δ → μ0 for each j = 1,

• • • , m as δ → 0. Let μδ = 1 m m j=1
μj δ , and let {u 0,1 , u 0,2 , • • • , u 0,m } is an orthonormal basis for Ker(T -μ0 I), then as written in (4.55) there exists a constant C (independent of δ) such that

µ 0 -μδ - 1 m m j=1 ≺ (T δ -T )u 0,j , u 0,j ≤ C (T -T δ )u 0,j 2 L 2 (Ω) , (4.64)
Moreover, for each j = 1, • • • , m, there is an eigenfunction u δ,j corresponding to μj δ such that u δ,j L 2 (Ω) = 1, and

u δ,j -u 0,j L 2 (Ω) ≤ C (T -T δ )| Ker(T -μ 0 I) . (4.65)
For more details, we can refer to [START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Khelifi | Asymptotic property and convergence estimation for the eigenelements of the Laplace operator[END_REF].

Suppose now that the eigenvalue λ 0 of the problem (4.49) is with geometric multiplicity m and let v 0,j ; j = 1, • • • , m be L 2 -orthonormal eigenfunctions corresponding to λ 0 . Let λ j δ be the eigenvalues of problem (4.52) generated by splitting from λ 0 and let v j δ be the associated eigenfunction (normalized with respect to L 2 ) such that v j δ → v 0,j as δ → 0. Based on (4.64) and (4.65), one can use similar approaches as in the proof of Theorem 4.3.2 to get the following result.

Corollary 4.3.1. Suppose that Ω ⊂ R 2 with class C 1,1 . Let λ 0 be an eigenvalue, of the problem (4.49), with (geometric) multiplicity m ≥ 1. Assume that we have (4.51), and the pressure still the same if we have (4.50). Then, there exist m eigenvalues λ j δ , j = 1, • • • , m (repeated according to their multiplicities) such that λ j δ → λ 0 as δ → 0 and the following asymptotic expansion holds:

1 λ 0 - 1 m m j=1 1 λ j δ = δ m m j=1 ∂D h(x)V[D(v e 0,j )](x) : D(v e 0,j )(x)ds(x) + O(δ 1+β ),
where β some positif constant and V given by Theorem 4.3.2.

Asymptotic formula for eigenfunctions and eigenpressures

This section is dedicated to develop asymptotic expansions for both eigenfunctions and eigenpressures associated to (4.4). Here we suppose that we have all hypothesis and conditions of the Section 1 and Section 2. Define the operator:

B δ (λ)ϕ(x) = (W(λ)ϕ)(Ψ -1 δ )(Ψ δ (x)),
where ϕ ∈ L 2 (∂Ω), Ψ δ given by Section 2.2, and W(λ) is the operator associated to hydrodynamic double layer potential [?, [START_REF] Kohr | The interior Neumann problem for the Stokes resolvent system in a bounded domain in R n[END_REF][START_REF] Varnhorn | The Stokes equations[END_REF]. Then, the following main result holds.

Theorem 4.3.3. Suppose that we have (4.2) and let λ 0 be a simple eigenvalue of (4.1). Let A δ be the operator defined by (4.10), and

M δ = A δ + B δ . Let O 0 be a bounded neighborhood of Ω in R 3 .
Then there exists a constant δ 1 > 0 smaller than δ 0 such that the eigenfunction v δ corresponding to the eigenvalue, λ δ , in (H 1 (Ω δ )) 3 ∩ H(Ω δ ) can be chosen to depend holomorphically in (x, δ) ∈ O 0 ×] -δ 1 , δ 1 [. Moreover this eigenfunction satisfies the following asymptotic formulae

v δ (x) = v 0 (x) + n≥1 v n (x)δ n , (4.66)
where the function v 0 is the eigenfunction solution of (4.1) associated to λ 0 . The terms v n are computed from the Taylor coefficients of the operator valued function M δ and of those of the function a(δ) = (a ij (δ)) 1≤i,j≤3 .

Proof. From [START_REF] Kohr | The interior Neumann problem for the Stokes resolvent system in a bounded domain in R n[END_REF][START_REF] Medkova | Boundary value problems for the Stokes equations with jumps in open sets[END_REF] we deduce that there exist a continuous function ϕ(t, δ), which is ana-

lytic in R 2 ×] -δ 0 , δ 0 [ and such that v δ (x) = S(λ δ )ϕ + W(λ δ )ϕ, x ∈ Ω (4.67)
solves the eigenvalue problem (4.4). Moreover, the function given by

U (δ)(x) = M(λ δ )ϕ(Ψ -1 , δ)
satisfies the eigenvalue problem (4.4) in Ω δ with the boundary conditions:

U (δ)| ∂Ω δ = 0.
Here, M(λ δ )ϕ(Ψ -1 , δ) = M δ (λ)ϕ. Now, by (4.67), we deduce that v δ (x) = U (δ)(x) = M(λ δ )ϕ(Ψ -1 , δ) is jointly analytic with respect to (x, δ) in { x -Ψ δ=0 (y) ≤ z 0 }×] -δ 0 , δ 0 [, where z 0 is a positive constant. The function v δ (x) is jointly analytic in the variables (x, δ) ∈ O 0 ×] -δ 0 , δ 0 [.

CHAPTER 4. ASYMPTOTIC BEHAVIORS FOR EIGENVALUES AND EIGENFUNCTIONS ASSOCIATED TO THE STOKES OPERATOR WITH A SMALL BOUNDARY PERTURBATION

We shall now give the asymptotic expansion of the function v δ (x) when δ tends to 0. Integral equation (4.7) gives us

v δ (x) = ∂Ω M (λ δ , |x -Ψ δ (y)|)ϕ(y, δ)|∇Ψ δ (y)|dσ(y),
where M is the kernel of the operator M δ . The perturbed eigenvalue λ δ is in a small neighborhood of λ 0 for small values of δ. Then we have the following Taylor expansion

M (λ δ ), |x -Ψ δ (y)|)|∇Ψ δ (y)| = M (λ 0 , |x -Ψ(y)|)|∇Ψ(y)| + k≥1 δ k M k (x, y),
which holds uniformly in x ∈ O 0 and y ∈ ∂Ω. The analyticity of the function ϕ(y, δ) with respect to δ immediately gives

ϕ(t, δ) = ϕ 0 (y) + k≥1 δ k ϕ k (y),
uniformly in y ∈ ∂Ω. Substituting the last two asymptotic into (4.68) we find

v δ (x) = v δ=0 (x) + k≥1 δ k [ k n=1 ∂Ω ϕ k-n (y)M n (x, y)dσ(y)].
The next result provide us with the asymptotic expansion of the eigenpressures.

Corollary 4.3.2. Suppose that we have all hypothesis of Theorem (4.3.3). Then the eigenpressures p δ solution of (4.4) have the following uniform asymptotic expansion:

p δ (x) = p 0 (x) + n≥1 p n (x)δ n , (4.68)
where the function p 0 is the eigenpressure solution of (4.1) associated to λ 0 . The terms p n are computed from the Taylor coefficients λ n and

v n = (v 1 n , v 2 n , v 3
n ) as follows:

p n (.; x i ; .) = ∆v i n dx i + n k=0 λ k v i n-k dx i , where i = 1, 2, 3.
Proof. From system (4.4) we have

∇p δ = ∆v δ + λ δ v δ .

Mathematical model

We consider the scattering problem of electromagnetic waves (E, H) by a perfect conducting body with a complex coating. Scattering waves take place when incident waves bounce off an obstacle in a variety of directions as depicted in figure 5.1. The amount of scattering waves that take place depends on the wavelength k 0 of the incident waves and structure of the object. We refer by Ω -to the region of space embodying the scatterer, with a Lipschitz- Electromagnetic waves propagate in Ω + = R n \Ω -. We define the incident electromagnetic field by (E inc , H inc ), the diffracted or scattered field (E sc , H sc ) and the total electromagnetic fields (E, H) in Ω + as:

   E = E inc + E sc inΩ + , H = H inc + H sc inΩ + .
(5.1)

Waves propagation medium is characterized by two physical quantities (electrical permittivity) and µ (magnetic permeability), where = 0 and µ = µ 0 for free space. The fields inside the coating are governed by a set of equations that take into account the detailed electromagnetic properties of the coating r and µ r .

The determination of the total electromagnetic field (E, H) induced in Ω + by an incident field on the impedance scatterer is governed by the following Maxwell equations:

   rotE + iωµH = 0 rotH -iω E = 0 (5.2)
where ω is the pulsation associated to the time-harmonic electromagnetic fields defined as

follows    E(x, t) = (E(x)e iωt ) H(x, t) = (H(x)e iωt ) (5.3)
Let Z be the impedance operator. It depends on incident angle, medium thickness d and the physical characteristics r and µ r of the coating material. We consider boundary condition that binds the tangent electric and magnetic fields. The medium characteristics give an impedance at each point of the surface Γ.

E tg -Z(n × H) = 0 on Γ (5.4)
where subscript tg denotes tangent component on the surface Γ defined as:

E tg = n × (E × n)
and n is the exterior normal vector to the surface. Find (E,H) such that

               rotE + ik 0 µH = 0 in Ω + rotH -ik 0 E = 0 in Ω + E tg -Z(n × H) = 0 on Γ lim r→∞ r(E × n r + H) = 0
(5.6)

Integral formulation

The boundary integral method places their unknowns on Γ the boundary of Ω. It consists in expressing the electromagnetic field as a function of potentials defined on the boundary.

We will use the boundary integral representation to solve problem 5.1.1. We express the electromagnetic field as a function of potentials defined on Γ. The Stratton-Chu formulation (Appendix C eq. (C.30)) helps us to characterize the electromagnetic fields in terms of surface current densities. These current densities are uknowns in the integral formulation of the problem. We introduce current densities J and M on the boundary Γ as follows

M = [E × n] + -; J = [n × H] + -,
where [ ] + -denotes difference between upper (+) and lower (-) values of interface. We introduce the variational form of the operators (B -S), (P + Q) and I as follows:

(B -S)A, ψ = i Γ kGA • ψ - 1 k G∇ y • A∇ x • ψdydx (5.7) (P + Q)A, ψ = 1 2 Γ ψ • (n × A)dx + Γ (ψ × A) • ∇ x Gdydx (5.8) IA, ψ = Γ A • ψdx (5.9)
where G is the Green kernel G(x, y) = e -ιk|x-y| 4π|x-y| . Using the Stratton-Chu formulae, we can give a variationnal formulation of (5.6): Find (J, M) such that:

< Z 0 (B -S)J, Ψ J > + < (P + Q)M, Ψ J >=< E inc , Ψ J >, ∀Ψ J ,
(5.10)

-< (P + Q)J, Ψ M > + < 1 Z 0 (B -S)M, Ψ M >=< H inc , Ψ M >, ∀Ψ M . (5.11)
For more details see Appendix C. Note however that these two equations are completely equivalent.

Variational formulation

In order to insure a unique solution to the boundary value problem it is necessary to applay boundary condition. We recall impedance boundary condition (IBC) that relates current densities J and M n × M = ZJ.

(5.12)

The simplest IBC is Leontovich IBC, Z = const. Usually, it is taken for incident wave perpendicular to plane

a 0 = z 0 µ r r tan ( √ µ r r k 0 d) (LIBC)
The impedance boundary condition (5.12) can be written

n × M = a 0 J. (5.13)
We multiply three dimensional LIBC (5.13) by test functions and integrate on the surface Γ, that gives us a weak form of the boundary condition. So, we take Ψ J and n × Ψ M as test functions to get the weak forms for (5.10) and (5.11), respectively

Γ (n × M) • Ψ J ds = Γ a 0 J • Ψ J ds, (5.14) Γ (n × M) • (n × Ψ M )ds = Γ a 0 J • (n × Ψ M )ds (5.15)
We insert the LIBC into (5.10) through (5.14), by replacing the operator P that was defined in (5.8) then we obtain:

< P M, Ψ J >= 1 2 Γ Ψ J • (n × M)ds = a 0 2 Γ J • Ψ J ds
Using the formula of vector analysis

Ψ M • (n × J) = -J • (n × Ψ M ),
we put weak form of LIBC (5.15) into 5.11 by replacing < P J, Ψ M >. We get:

< P J, Ψ M >= 1 2 Γ Ψ M • (n × J)ds = - 1 2 Γ (n × Ψ M ) • Jds = -1 2a 0 Γ (n × M) • (n × Ψ M )ds = -1 2a 0 Γ M • Ψ M ds (5.16)
Now we define bilinear operator

A(U, Ψ) =< Z 0 (B -S)J, Ψ J > + 1 Z 0 < (B -S)M, Ψ M > + < QM, Ψ J > -< QJ, Ψ M > + a 0 2 < J, Ψ J > + 1 2a 0 < M, Ψ M >
that combines integral equations (5.10)- (5.11), where operator P parts are replaced.

Finally, we introduce the following problem:

Problem 5.1.2. Find U = (J, M) such that A(U, Ψ) = F (Ψ) (5.17)
for all Ψ = (Ψ J , Ψ M ). Where

F (Ψ) = Γ E inc • Ψ J ds + Γ H inc • Ψ M ds.

Existence and uniqueness theorem

According to theorem C.1.4 from Appendix, there exists a unique solution of the problem (5.1.2), if the bilinear operator A(U, Ψ) verifies the continuity and coercivity conditions.

For the sake of simplicity, we consider the operator A(U, Ψ) as a sum of three bilinear operator

A(U, Ψ) =< Z 0 (B -S)J, Ψ J > + 1 Z 0 < (B -S)M, Ψ M > + < QM, Ψ J > -< QJ, Ψ M > + a 0 2 < J, Ψ J > + 1 2a 0 < M, Ψ M >
Let us prove the continuity of A(U, Ψ).

Continuity of the operator A

Lemma 5.2.1. The bilinear operator A(U, Ψ) is continuous on V for all Ψ ∈ V .

Proof. : We have to show that there exists C > 0 such that for all

Ψ ∈ V |A(U, Ψ)| ≤ C U V Ψ V
According to theorems (C.1.1), (C.1.2) and triangle inequality (property of a norm) we have

|A(U, Ψ)| ≤ | < Z 0 (B -S)J, Ψ J > | + |Z -1 0 | | < (B -S)M, Ψ M > | + | < QM, Ψ J > | +| < QJ, Ψ M > | + |a 0 | 2 | < J, Ψ J > | + 1 2|a 0 | | < M, Ψ M > | ≤ ≤ Z 0 (B -S)J -1/2,rot Γ Ψ J -1/2,div Γ +|Z -1 0 | (B -S)M -1/2,rot Γ Ψ M -1/2,div Γ + QM -1/2,rot Γ Ψ J -1/2,div Γ + QJ -1/2,rot Γ Ψ M -1/2,div Γ + a 0 2 J L 2 (Γ) Ψ J L 2 (Γ) + 1 2a 0 M L 2 (Γ) Ψ M L 2 (Γ) ≤ C 1 U V Ψ V

Coercivity of the operator

A Lemma 5.2.2. Bilinear form A(U, Ψ) verifies coercivity inequality for all U ∈ V = [H -1/2 (div, Γ)∩ L 2 (Γ)] 4 .
Proof. : We have to show that there exist α > 0 such that

[A(U, U * )] ≥ α U 2 V -C U 2 H , ∀U ∈ V.
From [START_REF] Lange | Equations intégrales espace-temps pour les équations de Maxwell : calcul du champ diffracté par un obstacle dissipatif[END_REF], we know that there exists α 1 such that

(A) = (< Z 0 (B -S)J, J * >) + (< Z -1 0 (B -S)M, M * >) + (< QM, J * >) -(< QJ, M * >) + ( a 0 2 Γ J • J * ds) + ( 1 2a 0 Γ M • M * ds) ≥ ≥ α J 2 -1/2,div Γ + M 2 -1/2,div Γ + (a 0 ) 2 J 2 L 2 (Γ) + (a 0 ) 2|a 0 | 2 M 2 L 2 (Γ)

Discretization

We recall bilinear form that was formulated in the previous chapter. So we solve the problem

A(U, Ψ) =< E inc , Ψ J > + < H inc , Ψ M > (5.18)
where

A(U, Ψ) =< Z 0 (B -S)J, Ψ J > +Z -1 0 < (B -S)M, Ψ M > + < QM, Ψ J > -< QJ, Ψ M > + a 0 2 < J, Ψ J > + 1 2a 0 < M, Ψ M >
We assume that the initial boundary problem is defined in a 3-D domain. The boundary Γ of the domain is therefore a surface and can be discretized using triangles.

Triangular mesh element

The first step is to approach the surface of the obstacle by a surface Γ h composed of finite number of two dimensional elements. These elements are triangular facets denoted by T i for i = 1 to N T :

Γ h = N T i=1 T i .
We will call it an initial mesh (or original mesh).We denote by N e the total number of edges of the mesh component Γ h . The discretization of unknowns J and M should verify a conditionflow conservation of these currents. One way to ensure this is to use the basis functions of Rao-Wilton-Glisson introduced below.

Rao-Wilton-Glisson basis functions

We introduce local numbering of a triangle T . The vertices (a T j ) j=1,3 are arranged in clockwise order. Triangle edges are numbered so that the edge T j connects vertices a T j and a T j+1 . Moreover, we give an orientation ν n to each edge n. Consider the two triangles sharing this edge. We note T + n the triangle so that the direction of the edge n coincides with the forward direction (locally defined) of this triangle. For the other triangle, which we will denote T - n , the direction of the edge coincides with the indirect sense.

Each RWG basis function is associated with an edge and ensures the conservation of flux through this edge. If we denote |T | the area of a triangle T , the n th basis function is defined as follows:

f n (x) =          ln 2|T + n | (x -a + i-1 ) if x ∈ T + n ln 2|T - n | (a - j-1 -x) if x ∈ T - n 0 if x / ∈ T + n ∪ T - n (5.19)
The density is proportional to div Γ f n , where

div Γ f n (x) =          + ln |T + n | on T + n -ln |T - n | on T - n 0 elsewhere.
(5.20)

Discrete problem

We decompose the electric and magnetic currents in the following form:

J(y) = Ne i=1 J i f i (y), M(y) = Ne i=1 M i f i (y).
Considering to these assumptions we are looking for approximated solution of the problem (5.18). We introduce the following discrete system where

A h (U h , Ψ h ) = Ne i=1 < E inc , f i > + Ne i=1 < H inc , f i >, ( 5
A h (U h , Ψ h ) = Ne i,j=1 < Z 0 (B -S)f j , f i > J j + Z -1 0 Ne i,j=1 < (B -S)f j , f i > M j + Ne i,j=1 < Qf j , f i > M j - Ne i,j=1 < Qf j , f i > J j + a 0 2 Ne i,j=1 < f j , f i > J j + 1 2a 0 Ne i,j=1 < f j , f i > M j .
Since the basis function f i is defined on two triangles, namely adjacent to the i th edge ,

T + i and T - i .
The same for the basis function f j , they are defined on two triangles, T + j and T - j . We define [(B -S)] matrix elements

(B -S) i,j = i T + i ∪T - i T + j ∪T - j kG(x, y)f j (y) • f i (x) - 1 k G(x, y)(div Γ f i (x))(div Γ f j (y))dxdy.
Analogically

Q i,j = -i T + i ∪T - i T + j ∪T - j [f i (x) × f j (y)] • ∇ Γ G(x, y)dydx.
For lonely integrals we have

I i,j = a 0 2 Ntr t=1 J j Tt f j • f i dx = T + i ∪T - i f j • f i dx.
In the following, we will detail the evaluation of the double layer operator Q i,j . The odher terms was evaluated in previous works.

Double layer operator

We denote by S the triangle T + i and by T the triangle T + j . We detail the evaluation of Q i,j on S × T and it will be the same for other combinations of triangles. On each triangle, the current is written as a linear combination of three basis functions associated to three edges of a triangle.

K i,j = S×T f i (x) • ∇ y G(x, y) × f j (y) dx dy; 1 ≤ i, j ≤ 3 = 1 4|S||T | S×T (x -a i ) • ∇ y G(x, y) × (y -b j ) dx dy (5.22)
where

G(x, y) = - e ιk||x-y|| 4π||x -y|| ,
and

∇ y G(x, y) = - (x -y) 4πr 3 - k 2 (x -y) 8πr - ιk 3 (x -y) 8π - (x -y)(1 -ιkr) 4π Σ ∞ n=0 (ιk) n+3 r n (n + 3)! (5.23) Let define the integral R i,j R i,j = S×T (x -a i ) • R(x, y) × (y -b j ) dx dy; i, j = 1, 2, 3 (5.24)
where

R(x, y) = - ιk 3 (x -y) 8π - (x -y)(1 -ιkr) 4π Σ ∞ n=0 (ιk) n+3 r n (n + 3)! (5.25)
and

S i,j = S×T (x -a i ) • (x -y) ||x -y|| ξ+1 × (y -b j )dxdy; i, j = 1, 2, 3; ξ = 0, 2 (5.26)
The regular integral R i,j can be calculated numerically with the usual known methods such as the Gauss method. In the following chapters, we propose a method for the analytical evaluation of the singular term.

Presentation of the reduction method Introduction

As we saw in the previous chapter, we must evaluate singular integrals derived from a double layer operator, arising from the boundary element method applied to resolve a scattering problem. There was some attempt to evaluate it with purely numerical methods but the result was not very accurate. The aim is to present a new approach for the explicit and precise calculation. We propose a method that does not require any numerical integration for the evaluation of singular integrals. It is based on a recursive reduction of an m-dimensional integral into a linear combination of (m-1)-dimensional integrals. It leads to a linear combination of 1-dimensional regular integrals with factors depending only on geometric quantities. These integrals can be evaluated either numerically or explicitly. This method makes it possible to obtain explicit formulae to evaluate singular integrals with a very high precision. It was introduced by M. Lenoir in 2006, see [START_REF] Lenoir | Influence coefficients for variational integral equations[END_REF] to calculate a singular integral in the form of a single layer operator.

6.1 Presentation of the method

Integration of homogeneous functions

The reduction process is based on several simplification formulae presented in the next section. These formulae reduce the dimension of the integration domain. It can be used when the integrand is homogeneous, or positively homogeneous. Definition 6.1.1. A function is said to be homogeneous of degree q or q-homogeneous, with respect to the variable z ∈ R n , if

f (λz) = λ q f (z); ∀λ ∈ R; λ = 0. (6.1)
Definition 6.1.2. A function is said to be positively homogeneous of degree q, with respect to

the variable z ∈ R n , if f (λz) = λ q f (z); ∀λ ∈ R; λ > 0. (6.2)
Definition 6.1.3. A function is said to be q-homogeneous with parameter if it is an homogeneous function with respect to a pair of variables (z, h), where z is the integration variable and h is a real parameter

f (λz, λh) = λ q f (z, h) (6.3)
Theorem 6.1.1.

[39] Let f : Ω ⊂ R n → R be an homogeneous function of degree q with q + n = 0. Then,

(q + n) Ω f (z)dz = ∂Ω (z| ν)f (z)∂z (6.4)
where ν is the exterior normal to Ω, ∂z is the surface element on ∂Ω, and (z| ν) is the inner scalar product.

Theorem 6.1.2.

[39] Let g : R n ×R + → R be an homogeneous function of degree q referred to as an homogeneous function with parameter h.

Ω f (z, h)dz = h q+n ∂Ω (z| ν) +∞ h g(z, u) u q+n+1 du∂z (6.5) Theorem 6.1.3. [39] Let f : Ω ⊂ R n → R
a positively homogeneous function of degree q with q = -n. We introduce an arbitrary homogeneous positive function θ : Ω → R of degree p > 0. Then we have

Ω f (z)dz = 1 p ∂Ω ( - → z | - → ν )f (z) ln(θ(z))ds z (6.6)
These formulae will allow us to reduce the dimension of the integration domain.

Parametrization and flat polygons

It is important to note that the homogeneity of the integrand depends on the parameterization and the origin chosen in the computation of the integral. We will start by presenting a small calculation to detail the importance of the choice of the origin.

Let α and β be two segments of the plane (see Figure 6.4), we calculate the integral α×β x -y ||x -y|| ds x ds y (6.7) with x ∈ α and y ∈ β.

An infinity of parametrizations is possible. But, in practice, only the use of the intersection Figure 6.4: Influence of the origin on the homogeneity of the supports of the two segments as origin ensures the homogeneity of the integer in (6.7).

• If we chose o 1 as origin in Figure (6.4), we obtain the following parametrization in terms of (s, t):

           x 1 (s) = s , x 2 (s) = 0 , y 1 (t) = t cos(θ) , y 2 (t) = t sin(θ).
which gives

s t s -t cos(θ) + t sin(θ) (s -t cos(θ)) 2 + (t sin(θ)) 2 ds x ds y (6.8)
where θ is a fixed angle. The integrand is homogeneous with respect to the pair of variables (s, t).

• If we chose o 2 as an origin, we notice that there is a constant that appears in the parameterization and we obtain

           x 1 (s) = s , x 2 (s) = 1 , y 1 (t) = t cos(θ), y 2 (t) = t sin(θ). Then s t s -t cos(θ) -1 + t sin(θ) (s -t cos(θ)) 2 + (1 -t sin(θ)) 2 ds x ds y (6.9)
The integrand is no longer homogeneous.

We understand that it will therefore be very important during the process of reducing to carefully choose the origin in order to maintain the homogeneous character of the integrand.

We will also see that a judicious choice of the origin makes it possible to simplify the calculations.

The reducing process of the integration's domain dimension, requires the use of flat polygons for discretization. In the contrary case, after a first step of reducing the dimension, the integrand is no more homogeneous. therefore, it is no longer possible to use formulae (6.4), (6.5) or (6.6) to continue reducing the dimension of the integration domain until onedimensional integrals are obtained.

Indeed, the use of flat polygons ensures that the product of the two flat polygons is a polyhedron. So the boundary ∂S × T will be composed of hyperplanes on which the outgoing normal is constant.

Therefore, the scalar product present in formulae (6.4), (6.5) and (6.6) is piecewise constant and it will be possible to get the scalar product out of the integral, to reduce to an homogeneous integrand and thus repeat the process.

Geometric tools and Notation

We present some geometric notations needed for the calculations.

The mesh elements

Let the triangles S and T be defined (respectively) by their vertices a i and b j , i, j = 1, 2, 3

and its sides α i respectively β j , i, j = 1, 2, 3 such as α i and β j are opposed to a i respectively b j (see Figure 6.5). We introduce |α i | and |β j | the length of the side α i respectively β j . Let α i and β j be the vectors Figure 6.6). We denote by λ i and ϑ j the outside normal vector to α i respectively β j .

(a i+2 -a i+1 ) = (a + i -a - i ) respectively (b i+2 -b i+1 ) = (b + i -b - i ) (see

Calculation of the projection

Note p i (x) the orthogonal projection (see Figure 6.7) of a point x on the side α i of the triangle S and g i (x) = ||x -p i (x)|| the distance from x to the support of α i . When x = a i , we note that p i = p i (a i ) and that g i = g i (a i ) is the length of the height (see Figure 6.9). We will also use the signed distance from a point to the support of α i which involves the outgoing normal to the triangle on α i . This distance is defined by γ i = (p i (x) -x| λ i ) (see Figure 6.9). We projection of y on the support of β j , similar to p i (x), is denoted by q j (y) and the distance from y to the support of β j is denoted by d j (y) = ||y -q j (y)||. We define the signed distance δ j (y) from y to the support of β j by δ j (y) = (q j (y) -y| ϑ j ). Finally, d j = d j (b j ) is the equivalent of g i , namely the length of the height from b j to β j . (see Figure 6.9). Figure 6.9: Distances signees

Calculation of abscissas

On side α (resp β) the abscissa s (resp t) is defined with respect to an origin o α (resp o β ) and a unit vector α |α| (resp β |β| ). The abscissas of the ends a ± and b ± are respectively denoted by s ± and t ± .

s k = (a k -o α | α |α| ) t l = (b l -o β | β |β| )

Application of the reduction method

We propose to evaluate the integral

I(x, y) = S×β j x -y ||x -y|| 3 dx dy, (6.10)
where S is a triangle, x a point in the triangle S, β j is a segment in the same plane as S and y is a point in β j . Let b + j = b j+1 and b - j = b j+2 be the two extremities of β. The reduction process can be applied, provided that one takes some common point to β j and S as origin, so that the function f (z) = x-y ||x-y|| 3 is homogeneous on S × β j with q = 2, n = 3. The origin can be taken anywhere on the support of β j , but a wise choice is to choose one of the ends of β j as origin. Indeed, some of the distances being canceled, one will have a simplification.

We choose for origin the point b -. Formula (6.4) provides

I(x, y) = 1 5 ∂(S×β j ) ((x, y)| - → ν ) x -y ||x -y|| 3 ∂(x, y), (6.11)
where ν is the outward normal at the border of the edge of S × β j and ∂(x, y) is the surface

element along ∂(S × β j ). We have ∂(S × β j ) = (∂S × β j ) ∪ (S × ∂β j ). ((x, y)| - → ν ) b + j ×S = |β j |; ((x, y)| - → ν ) |α i ×β j = γ i (b - j )
where |β j | denotes the length of the segment β j . Then, we get

I(x, y) = |β j | 5 S x -b + j ||x -b + j || 3 ds x + 3 i=1 γ i (b - j ) 5 α i ×β j x -y ||x -y|| 3 ds x ds y . (6.12)
So we get a linear combinaison of 2-D integrals.

Conclusion

This chapter allowed to present the reduction formulae of the dimension of the integration domain, to explain the reduction process and also to emphasize the importance of the parametrization and the use of flat polygons on which our method is based. An important point in the following is the choice of the origin during the stages of reduction dimension of the integration domain, in order to maintain the homogeneity of the integrand

Evaluation of a singular integral-Adjacent triangles Introduction

In this chapter, we will apply the reduction process to evaluate singular integrals appearing in chapter 5, on the product of two triangular mesh element S and T, specially the singular integral

S i,j = S×T (x -a i ) • (x -y) ||x -y|| ξ+1 × (y -b j )dxdy; i, j = 1, 2, 3; ξ = 0, 2
Note that if ξ = 0 the integral S i,j is not singular but the method we propose is still valid.

Here we will deal with the case where triangles have an edge in common.

Barycentric basis functions

In order to simplify the expression of the integrand and to facilitate the application of the reduction formulae we will express the Rao-Wilton-Glisson basis functions as a linear combination of barycentric basis functions.

We begin by presenting the basis functions we will use and some of their properties. Definition 7.1.1. Let φ i be a basis functions associated with the vertices of the triangle S:

φ i (x) = 1 + ( ---→ x -a i | - → e i ); i = 1, 2, 3 (7.1)
where

e i = µ 1 --→ α i+1 + µ 2 --→ α i+2 , with µ 1 = || --→ α i+2 || 2 + ( --→ α i+1 | --→ α i+2 ) || --→ α i+2 || 2 || --→ α i+1 || 2 -( --→ α i+1 | --→ α i+2 ) 2 , µ 2 = - || --→ α i+1 || 2 + ( --→ α i+1 | --→ α i+2 ) || --→ α i+2 || 2 || --→ α i+1 || 2 -( --→ α i+1 | --→ α i+2 ) 2 .
The basis functions φ i verify φ i (a i ) = 1, φ i (a i+1 ) = φ i (a i+2 ) = 0, besides

φ i |α i = 0, φ i (x) |α i+1 = 1 - x -a i |α i+1 | , φ i (x) |α i+2 = 1 - x -a i |α i+2 | . (7.2)
The same, if the basis functions ψ are associated with the top of the triangle T :

ψ j (y) = 1 + ( ---→ y -b j | - → l j ); j = 1, 2, 3 (7.3)
We rewrite the Rao-Wiston-Glisson basis functions as a combination of barycentric basis functions and we use properties above to determine θ, κ, ζ, ρ, η and µ verifying:

(x -a i ) = θφ i (x) + κφ i+1 (x) + ζφ i+2 (x), (y -b j ) = ρψ j (y) + ηψ j+1 (y) + µψ j+2 (y). (7.4)
We obtain

(x -a i ) = 0 + (a i+1 -a i )φ i+1 (x) + (a i+2 -a i )φ i+2 (x),
(y -b j ) = 0 + (a j+1 -a j )ψ j+1 (y) + (a j+2 -a j )ψ j+2 (y). (7.5)

Decomposition into homogeneous integrands

We are interested in calculating the singular integral

S i,j = S×T (x -a i ) • (x -y) ||x -y|| ξ+1 × (y -b j )dxdy; i, j = 1, 2, 3; ξ = 0, 2 (7.6)
Using (7.5), the integral (7.6) becomes

S i,j = S×T (κ × η) • (φ i+1 (x)ψ j+1 (y)) (x,y) ||x-y|| ξ+1 dxdy + S×T (κ × µ) • (φ i+1 (x)ψ j+2 (y)) (x-y) ||x-y|| ξ+1 dxdy + S×T (ζ × η) • (φ i+2 (x)ψ j+1 (y)) (x-y) ||x-y|| ξ+1 dxdy + S×T (ζ × µ) • (φ i+2 (x)ψ j+2 (y)) (x-y)
||x-y|| ξ+1 dxdy. So we have to evaluate four integrals of the same type:

I i,j = S×T (φ i (x)ψ j (y)) (x -y) ||x -y|| ξ+1 dxdy; ξ = 0, 2; i, j = 1, 2, 3 (7.8)
Since the basis function is affine, the integrand is no longer homogeneous, so we are led to decompose it into a linear part and a constant part by making a change of origin:

φ i (x) = 1 + ( ---→ o -a i | - → e i ) + ( ---→ x -o| - → e i ) = φ i (o) + (x| - → e i ); i = 1, 2, 3, ψ j (y) = 1 + ( ---→ o -b j | - → l j ) + ( ---→ y -o| - → l j ) = ψ j (o) + (y| - → l j ); j = 1, 2, 3. (7.9)
where o is the new origin, x = x -o and y = y -o. More generally, we note α and S the segment α and the triangle S after this change of origin. According to (7.8) and (7.9), we constate that the integral I i,j must be decomposed into four integrals whose integrands are homogeneous.

I i,j = S×T (x| - → e i )(y| - → l j )(x -y) ||x -y|| 1+ξ dxdy + φ i (o) S×T (x| - → e i )(y| - → l j )(x -y) ||x -y|| 1+ξ dxdy +ψ j (o) S×T (x| - → e i )(x -y) ||x -y|| 1+ξ dxdy + ψ j (o)φ i (o) S×T (x -y) ||x -y|| 1+ξ dxdy. (7.10)
Once this decomposition has taken place, we obtain homogeneous integrands we can then apply the reduction method intermediate of the formulae (6.4) or (6.5). During the calculation, after each step of reduction, we will focus on expressing the results in terms of φ i and ψ j in order to benefit the best of the properties (7.2) of the basic functions. We distinguish two specific case where a singularity occurs. The two triangles S and T have a common stop or a common vertex.

Three-dimensional reduction

We are dealing with the case of triangles with an edge in common(see The equation (7.11) becomes

I 1,1 = S×T (y| - → l 1 )(x -y) ||x -y|| 1+ξ dxdy + S×T (x| - → e 1 )(y| - → l 1 )(x -y) ||x -y|| 1+ξ dxdy + S×T (x| - → e 1 )(x -y) ||x -y|| 1+ξ dxdy + S×T x -y ||x -y|| 1+ξ dxdy. (7.13)
The four integrands are homogeneous of degrees q = 1 -ξ, q = 2 -ξ, q = 1 -ξ and q = -ξ with n = 4. We apply the formula (6.4), it gives the following result

I 1,1 = 3 i=1 γ i (a 1 ) 5 -ξ α i ×T (y| - → l 1 )(x -y) ||x -y|| 1+ξ ds x dy + 3 j=1 δ j (b 1 ) 5 -ξ S×β j (y| - → l 1 )(x -y) ||x -y|| 1+ξ dxds y + 3 i=1 γ i (a 1 ) 6 -ξ α i ×T (x| - → e 1 )(y| - → l 1 )(x -y) ||x -y|| 1+ξ ds x dy + 3 j=1 δ j (b 1 ) 6 -ξ S×β j (x| - → e 1 )(y| - → l 1 )(x -y) ||x -y|| 1+ξ dxds y + 3 i=1 γ i (a 1 ) 5 -ξ α i ×T (x| - → e 1 )(x -y) ||x -y|| 1+ξ ds x dy + 3 j=1 δ j (b 1 ) 5 -ξ S×β j (x| - → e 1 )(x -y) ||x -y|| 1+ξ dxds y + 3 i=1 γ i (a 1 ) 4 -ξ α i ×T x -y ||x -y|| 1+ξ ds x dy + 3 j=1 δ j (b 1 ) 4 -ξ S×β j x -y ||x -y|| 1+ξ dxds y , (7.14)
where γ i (x) and δ j (y) are the signed distances respectively of x to the support of α i and of y to the support of β j (see Figure 6.9).

In particular γ i (x) = 0 for all x belonging to the support of α i . γ i (a i ) = g i and δ j (b j ) = d j are the lengths of the height from the vertex a i or b j to the support of α i (respectively β j ).

Thus, the formula will be simplified and we get We express the results in terms of φ 1 and ψ 1 in order to benefit the best of the properties (7.2) of the basic functions.

I 1,1 = g 1 5-ξ α 1 ×T (y| -→ l
(x| - → e 1 ) = φ 1 (x) -1; (y| - → l 1 ) = ψ 1 (y) -1
we obtain

I 1,1 = g 1 5-ξ α 1 ×T ψ 1 (y)(x-y)
||x-y|| 1+ξ ds x dy - x-y ||x-y|| 1+ξ dxds y . Note that there are terms that vanish because of the fact that φ 1 |α i = 0, ψ i |β i = 0.

We have integrals in three dimensions we reapply the procedure to reduce dimention to two.

Two-dimensional reduction

The integrands are no more homogeneous. It is possible to continue the reduction by taking the second common vertex as the origin a 2 = b 2 = o. We reexpress the basis function as in (7.12) and we obtain

φ 1 (x) = φ 1 (a 2 ) + (x| - → e 1 ) = (x| - → e 1 ); ψ 1 (x) = ψ 1 (b 2 ) + (y| - → l 1 ) = (y| - → l 1 )
91 CHAPTER 7. EVALUATION OF A SINGULAR INTEGRAL-ADJACENT TRIANGLES we get

I 1,1 = ( g 1 5-ξ -g 1 6-ξ ) α 1 ×T (y| -→ l 1 )(x-y) ||x-y|| 1+ξ ds x dy + ( g 1 6-ξ -2 g 1 5-ξ + g 1 4-ξ ) α 1 ×T (x-y) ||x-y|| 1+ξ ds x dy +( d 1 5-ξ -d 1 6-ξ ) S×β 1 (x| -→ e 1 )(x-y) ||x-y|| 1+ξ dxds y + ( d 1 4-ξ + d 1 6-ξ -2d 1 5-ξ ) S×β 1 (x-y)
||x-y|| 1+ξ dxds y .

We apply the formula (6.4) a second time then, we get

I 1,1 = ( g 1 5-ξ -g 1 6-ξ ) d 2 4-ξ T (y| -→ l 1 )(a 3 -y) ||a 3 -y|| 1+ξ dy + ( g 1 5-ξ -g 1 6-ξ ) d 2 4-ξ α 1 ×β 2 (y| -→ l 1 )(x-y) ||x-y|| 1+ξ ds x ds y +( g 1 6-ξ -2 g 1 5-ξ + g 1 4-ξ ) |α 1 | 3-ξ T (a 3 -y) ||a 3 -y|| 1+ξ ds x dy + ( g 1 6-ξ -2 g 1 5-ξ + g 1 4-ξ ) d 2 3-ξ α 1 ×β 2 (x-y) ||x-y|| 1+ξ ds x ds y +( d 1 5-ξ -d 1 6-ξ ) |β 1 | 4-ξ S (x| -→ e 1 )(x-b 3 ) ||x-b 3 || 1+ξ dxds y + ( d 1 5-ξ -d 1 6-ξ ) g 2 4-ξ α 2 ×β 1 (x| -→ e 1 )(x-y) ||x-y|| 1+ξ dxds y +( d 1 6-ξ -d 1 5-ξ ) |β 1 | 4-ξ S (x-b 3 ) ||x-b 3 || 1+ξ ds x + ( d 1 4-ξ + d 1 6-ξ -d 1 5-ξ ) g 2 4-ξ α 2 ×β 1 (x-y)
||x-y|| 1+ξ ds x ds y .

Recall that φ 1 (x) = (x| -→ e 1 ); ψ 1 (x) = (y| -→ l 1 ), thanks to what we obtain

I 1,1 = ( |S| 5-ξ -|S| 6-ξ ) 2 4-ξ P ψ 1 (a 3 , T, ξ) + ( g 1 5-ξ -g 1 6-ξ ) d 2 4-ξ Q(α 1 , β 2 , ξ) +( |S| 6-ξ -2 |S| 5-ξ + |S| 4-ξ ) 2 3-ξ P (a 3 , T, ξ) + ( g 1 6-ξ -2 g 1 5-ξ + g 1 4-ξ ) d 2 3-ξ Q(α 1 , β 2 , ξ) -( |T | 5-ξ -|T | 6-ξ ) 2 4-ξ P φ 1 (b 3 , S, ξ) -( d 1 5-ξ -d 1 6-ξ ) g 2 4-ξ Q 1,φ 1 (β 1 , α 2 , ξ) -( |S| 6-ξ -|S| 5-ξ ) 2 4-ξ P (b 3 , S, ξ) + ( d 1 4-ξ + d 1 6-ξ -d 1 5-ξ ) g 2 4-ξ Q(α 2 , β 1 , ξ),
where

P (a, T, ξ) = T (a -y) ||a -y|| 1+ξ ds x ds y , Q(α i , β j , ξ) = α i ×β j (x -y) ||x -y|| 1+ξ ds x ds y , P ψq (a, T, ξ) = T ψ q (y)(a -y) ||a -y|| 1+ξ dy, Q 1,ψq (α i , β j , ξ) = α i ×β j ψ q (y)(x -y) ||x -y|| 1+ξ ds x ds y . I 3,2 = P (a 3 , T, ξ)( 2|S| (6-ξ)(4-ξ) - 2|S| (6-ξ)(5-ξ) ) + P φ 3 (b 3 , S, ξ)( 2|T | (6-ξ)(5-ξ) -2|T | (6-ξ)(4-ξ) ) +P ψ 2 (a 3 , T, ξ)( 2|S| (6-ξ)(5-ξ) ) + Q 1,φ 3 (β 2 , α 1 , ξ)( d 2 g 1 (6-ξ)(5-ξ) -d 2 g 1 (6-ξ)(4-ξ) ) +Q 1,φ 3 (β 1 , α 2 , ξ)(-d 1 g 2 (6-ξ)(4-ξ) + d 1 g 2 (6-ξ)(5-ξ) ) + Q φ 3 ,ψ 2 (α 2 , β 1 , ξ)( d 1 g 2 (6-ξ)(5-ξ) ) (7.23)
where

Q φp,ψq (α i , β j , ξ) = α i ×β j φ p (x)ψ q (y)(y -x)
||x -y|| 1+ξ ds x ds y In this case the reduction formulae are not usable. In order to operate the reduction, it is necessary to break P into two parts.

P (a, T, ξ) = T (a -â) (h 2 + ||â -y|| 2 ) 1+ξ 2 ds x ds y + T (â -y) (h 2 + ||â -y|| 2 ) 1+ξ 2
ds x ds y (7.25) where â be the projection of a into the support of T and h the distance h = |a -â| as in Figure (7.3). Taking â as a new origin we obtain homogeneous integrands of degrees -1 -ξ and -ξ with parameter h. We can then apply the formula (6.5) which gives

P (a, T, ξ) = (a -â) 3 j=1 δ j (â)S 1 (â, β j , h, ξ) + 3 j=1 δ j (â)R 2 (â, β j , h, ξ) (7.26)
where S 1 and R 2 are defined by

S 1 (a, β, h, ξ) = β h 1-ξ +∞ h du u 2-ξ (u 2 +||a-y|| 2 ) 1+ξ 2 ds y = β f 1 (h, ||a -y||, ξ). (7.27) and R 2 (a, β, h, ξ) = β (a -y)h 2-ξ +∞ h du u 3-ξ (u 2 +||a-y|| 2 ) 1+ξ 2 ds y = β (a -y)f 2 (h, ||a -y||, ξ) (7.28)
where Since a belongs to the support of T we can consider a as an origin and get homogeneous integrand with q = -ξ and n = 2 but we must distinguish two different cases in order to apply the suitable reduction formula.

f n (h, s, ξ) = h n-ξ +∞ h du u n+1-ξ (u 2 + s 2 )
• If n + q = 0, we apply (6.6) and we get First of all, we must find a new origin to ensure homogeneity of the integrand. Let α be the projection of α on the plane parallel to α going through β and let o be the intersection of α with β as in Figure (7.4). We will consider o as a new origin. Define z = ǎ -a and h = ||z||. We express Q as a sum of two integrals whose integrands are homogeneous of degrees q = -1 -ξ and -ξ.

P (a, T, ξ) = 3 j=1 δ j 2 -ξ L(a, β j ) (7.
Q(α, β, ξ) = α×β (x -x) (h 2 + ||x -y|| 2 ) 1+ξ 2 ds x ds y + α×β (x -y) (h 2 + ||x -y|| 2 ) 1+ξ 2
ds x ds y (7.35) where x is a point α. This decomposition allows us to apply the formula (6.5) and gives

Q(α, β, ξ) = (x -x)[ k=± ks k S 1 (ǎ k , β, h, ξ) + l=± lt l S 1 (b l , α, h, ξ)] +[ k=± ks k R 2 (ǎ k , β, h, ξ) - l=± lt l R 2 (a, β, h, ξ)] (7.36)
where the abscissae are evaluated by s k = (ȃ k -o| α |α| ) and t l = (b l -o| β |β| ). the integrand is homogeneous of degree q = -ξ and n = 2. We must distinguish two cases:

• If q + n = 0 the formula (6. a -and a + , we get two homogeneous integrands with parameter d and degrees q = -1 -ξ respectively q = -ξ.

Q(α, β, ξ) = z[s + β d 1-ξ +∞ d du u 2-ξ (u 2 +||a + -ỹ|| 2 ) 1 2 ds ỹ + l=± lt l α d 1-ξ +∞ d du u 2-ξ (u 2 +||x-bl || 2 ) 1 2 ds x ] +s + β d 2-ξ (a + -y) +∞ d du u 3-ξ (u 2 +||a + -y|| 2 ) 1 2 ds ỹ +[ l=± lt l α d 2-ξ (x -bl ) +∞ d du u 3-ξ (u 2 +||x-bl || 2 ) 1 2 ds x ] (7.40)
which is equivalent to

Q(α, β, ξ) = l=± lt l [ zS 1 ( bl , β, d, ξ) -R 2 ( bl , β, h, ξ)] +s + [S 1 (a + , β, d, ξ) + R 2 (a + , β, d, ξ)] (7.41)
7.5.3 Calculation of P ψ q (a, T, ξ)

To calculate the singular integral

P ψq (a, T, ξ) = T ψ q (y)(a -y) ||a -y|| 1+ξ dy, (7.42)
we are inspired by the section (7.5.1). There are two different geometrical cases.

7.5.3.1 Case where a does not belong to the support of T

Note that a = a 3 is a vertex of S. To maintain the homogeneity of the integrands the origin must belong to the support of the triangle T . So we consider â the projection of a on the support of T (see Figure 7.3) and we break P ψq (a, T, ξ) into two parts

P ψq (a, T, ξ) = (a -â) T ψ q (y) (h 2 + ||â -y|| 2 ) 1+ξ 2 dy + T ψ q (y)(â -y) (h 2 + ||â -y|| 2 ) 1+ξ , (7.43)
where h = |a -â| and ξ = 0, 2. Next, we express ψ q (y) with the new origin o = â, ψ q (y) = ψ q (â) + (y|l q ), which gives

P ψq (a, T, ξ) = (a -â)[ψ q (â) T 1 (h 2 + ||â -y|| 2 ) 1+ξ 2 dy + T (y|l q ) (h 2 + ||â -y|| 2 ) 1+ξ 2 dy] +ψ q (â) T (â -y) (h 2 + ||â -y|| 2 ) 1+ξ 2 dy + T (y|l q )(â -y) (h 2 + ||â -y|| 2 ) 1+ξ 2
dy. (7.44) We notice that the four intergrands are homogeneous with parameter h and degrees (respectively) q = -1 -ξ, q = -ξ, q = -ξ and q = 1 -ξ. We can then apply (6.5) and we 99 CHAPTER 7. EVALUATION OF A SINGULAR INTEGRAL-ADJACENT TRIANGLES obtain

P ψq (a, T, ξ) = (a -â)ψ q (â) 3 j=1 δ j (â) β j h 1-ξ +∞ h du u 2-ξ (u 2 +||â-y|| 2 ) 1+ξ 2 dy +(a -â) 3 j=1 δ j (â) β j h 2-ξ (y|l q ) du u 3-ξ (u 2 +||â-y|| 2 ) 1+ξ 2 dy +ψ q (â) 3 j=1 δ j (â) β j h 2-ξ (â -y) +∞ h du u 3-ξ (u 2 +||â-y|| 2 ) 1+ξ 2 dy + 3 j=1 δ j (â) β j h 3-ξ (y|l q )(â -y) +∞ h du u 4-ξ (u 2 +||â-y|| 2 ) 1+ξ 2
dy.

(7.45)

To simplify the formula, we express the basic functions in this form (y|l q ) = ψ q (y) -ψ q (â)

and we delete the terms that vanish, thanks to (7.2)

P ψq (a, T, ξ) = (a -â)ψ q (â) 3 j=1 δ j (â) β j f 1 (h, ||â -y||, ξ) -f 2 (h, â -y, ξ)ds y +(a -â)[δ q+1 (â) β q+1 ψ q (y)f 2 (h, â -y, ξ)ds y + δ q+2 (â) β q+2 ψ q (y)f 2 (h, â -y, ξ)ds y ] +ψ q (â) 3 j=1 δ j (â) β j (â -y)(f 2 (h, ||â -y||, ξ) -f 3 (h, â -y, ξ))ds y +δ q+1 (â) β q+1 ψ q (y)(â -y)f 3 (h, ||â -y||, ξ)ds y + δ q+2 (â) β q+2 ψ q (y)(â -y)f 3 (h, â -y, ξ)ds y (7.46)
otherwise we can write

P ψq (a, T, ξ) = (a -â)ψ q (â) 3 j=1 δ j (â)(S1(â, β, h, ξ) -S2(â, β, h, ξ)) +(a -â)[δ q+1 (â)S ψq 2 (â, β q+1 , h, ξ) + δ q+2 (â)S ψq 2 (â, β q+2 , h, ξ)] +ψ q (â) 3 j=1 δ j (â)R 2 (â, β j , h, ξ) -R 3 (â, β j , h, ξ) +δ q+1 (â)D ψq 3 (â, β q+1 , h, ξ) + δ q+2 (â)D ψq 3 (â, β q+2 , h, ξ) (7.
P ψq (a, T, ξ) = T ψ q (y)(a -y) ||a -y|| 1+ξ dy, (7.49)
Since o = a is the new origin, ψ q (y) can be written ψ q (y) = ψ q (a) + (y|l q )

We obtain P ψq (a, T, ξ) = ψ q (a)

T (a -y) ||a -y|| 1+ξ dy + T (y|l q )(a -y) ||a -y|| 1+ξ dy (7.50)

Once we have homogeneous integrands, the reduction formula (6.5) gives Let o = α ∩ β be the new origin. The basis function ψ q (y) can be expressed as following

Q 1,ψq (α, β, ξ) = ψ q (o)[ k=± ks k R 2 (â k , β, h, ||â k -y||, ξ) -R 2 (b l , α, h, ||x -b l ||, ξ)] + k=± ks k [D ψq 3 (â k , β j , ||â k -y||, ξ) -ψ q (o)R 3 (â k , β, ||â k -y||, ξ)] + l=± lt l [(-ψ q (b l ) + ψ q (o))R 3 (b l , α, ||x -b l ||, ξ)] + zψ q (o)[ k=± ks k S 1 (â k , β, h, ξ) + l=± lt l S 1 ( bl , α, h, ξ)] + z k=± ks k [S ψq 2 (â k , β, h, ||â k -y||, ξ) -ψ q (o)S 2 (â k , β, h, ||â k -y||, ξ)] + z l=± lt l [(ψ q (b l ) -ψ q (o))S 2 (b l , α, h, ||x -b l ||, ξ)]
ψ q (y) = ψ q (o) + (y|l q ) which gives Q 1,ψq (α, β, ξ) = ψ q (o) α×β (x -y) ||x -y|| 1+ξ ds x ds y + α×β (y|l q )(x -y) ||x -y|| 1+ξ ds x ds y (7.56)
The integrands are homogeneous of degrees q = -ξ and q = 1 -ξ with n = 2.

• If q + n = 0, formula (6.4) gives

Q 1,ψq (α, β, ξ) = ψ q (o)[ k=± ks k 2 S(a k , β, 0) - l=± lt l 2 S(b l , α, 0)] + k=± ks k 3 S ψq(y) (a k , β, 0) -ψ q (o) k=± ks k 3 S(a k , β, 0) - l=± lt l 3 (ψ q (b l ) -ψ q (o))S(b l , α, 0), (7.57)
where

S ψq(y) (a k , β, 0) = β ψ q (y)(a k -y) ||a k -y|| ds y (7.58)
• If q + n = 0 it will be appropriate to apply formula (6.6), which leads us to this expression

Q 1,ψq (α, β, ξ) = ψ q (o)[ k=± ks k L(a k , β) - l=± lt l L(b l , α)] + k=± ks k L ψq (a k , β) - k=± ks k ψ q (o)L(a k , β) - l=± lt l (ψ q (b l ) -ψ q (o))L(b l , α) (7.59)
where

L ψq (a k , β) = β ψ q (y)(a k -y) ||a k -y|| 3 ds y (7.60)
7.5.5 Calculation of Q ϕ p ,ψ q (α, β, ξ)

Here we deal with the singular integral

Q ϕp,ψq (α, β, ξ) = α×β φ p (x)ψ q (y)(y -x)
||x -y|| 1+ξ ds x ds y (7.61)

Intersecting triangles

We project α on the plane parallel to α and going through β. Let α be the projection of α and

x the projection of x. Then, we provide an equivalent expression of the basis functions:

φ p (x) = 1 + µ (x -a p | α) |α| 2 = 1 + µ (x -âp | α) |α| 2 (7.62) ψ q (y) = 1 + ν (y -b q )| β |β| 2 (7.63)
where µ = ±1(µ = 1 if a p = a + and µ = -

1 if a p = a -). Likewise, ν = ±1(ν = 1 if b p = b + and ν = -1 if b p = b -).
We choose the intersection of the supports of α and β, as origin and we note it o. we obtain :

φ p (x) = 1 + µ (o -âp | α) |α| 2 + µ (x -o| α) |α| 2 = C p + µ (x -o) α (7.64) ψ q (y) = 1 + ν (o -b q | β) |β| 2 + ν (y -o| β) |β| 2 = C q + µ (y -o| β) |β| 2 (7.65)
where

C p = 1 + µ (o-âp)| α |α| 2 and C q = 1 + ν (o-bq)| β |β| 2 . Then, the expression of Q ϕp,ψq becomes Q φp,ψq (α, β, ξ) = C q C q α×β (x -y) (h 2 + ||x -y|| 2 ) 1+ξ 2 ds xds y + C p ν |β| 2 α×β (y -o| β)(x -y) (h 2 + ||x -y|| 2 ) 1+ξ 2 ds xds y + C q µ |α| 2 α×β (x -o| α)(x -y) (h 2 + ||x -y|| 2 ) 1+ξ 2 ds xds y + µµ |α| 2 |β| 2 α×β (y -o| β)(x -o| α)(x -y) (h 2 + ||x -y|| 2 ) 1+ξ 2
ds xds y (7.66)

In the formula below the integrands are homogeneous with variable h and of degrees q = -ξ, q = 1 -ξ, 1 -ξ and 2 -ξ. Formula (6.5) gives

Q φp,ψq (α, β, ξ) = C p C q [ k=± kŝ k β (â k -y)h 2-ξ +∞ h du u 3-ξ (u 2 +||â k -y|| 2 ) 1+ξ 2 ds y + l=± l tl α(x -b l )h 2-ξ +∞ h du u 3-ξ (u 2 +||x-b l || 2 ) 1+ξ 2 ds x ] + Cpν |β| 2 [ k=± kŝ k β (y -o| β)(â k -y)h 3-ξ +∞ h du u 4-ξ (u 2 +||â k -y|| 2 ) 1+ξ 2 ds y + l=± l tl α(b l -o| α)(x -b l )h 3-ξ +∞ h du u 4-ξ (u 2 +||x-b l || 2 ) 1+ξ 2 ds x ] + Cqµ |α| 2 [ k=± kŝ k β (â k -o| α)(â k -y)h 3-ξ +∞ h du u 4-ξ (u 2 +||â k -y|| 2 ) 1+ξ 2 ds y + l=± l tl α(x -o| α)(x -b l )h 3-ξ +∞ h du u 4-ξ (u 2 +||x-b l || 2 ) 1+ξ 2 ds x ] + νµ |α| 2 |β| 2 [ k=± kŝ k β (â k -o| α)(y -o| β)(â k -y)h 4-ξ +∞ h du u 5-ξ (u 2 +||â k -y|| 2 ) 1+ξ 2 ds y + l=± l tl α(x -o| α)( bl -o| α)(x -b l )h 4-ξ +∞ h du u 5-ξ (u 2 +||x-b l || 2 ) 1+ξ 2 ds x ] (7.67)
which is equivalent to

Q φp,ψq (α, β, ξ) = C p C q [ k=± kŝ k R 2 (â k , β, h, ξ)ξ -l=± l tl R 2 (b l , α, h, ξ)] + k=± kŝ k [ Cqµ |α| 2 (â k -o| α)R 3 (â k , β, h, ξ) + Cpν |β| 2 M 3 (â k , o, β, h, ξ)] + l=± l tl [-Cpν |β| 2 (b l -o| β)R 3 (b l , α, h, ξ) -Cqµ |α| 2 M 3 (b l , o, α, h, ξ)] + νµ |α| 2 |β| 2 [ k=± kŝ k (â k -o| α)M 4 (â k , o, β, h, ξ) -l=± l tl (b l -o| β)M 4 (b l , o, α, h, ξ)] (7.68)
where

M 3 (â k , o, β, h, ξ) = β (â k -o| α)(â k -y)f 3 (h, ||â k -y||, ξ)ds y (7.69) M 4 (â k , o, β, h, ξ) = β (a -o| α)(a -y)f 4 (h, ||â k -y||, ξ)ds y (7.70)
are given in Appendix (B).

Coplanar triangles

As we saw in section (7.5.2), we need to distinguish two geometrical configurations for the supports of the sides α and β.

Two intersecting segments

When the supports of α and β are intersecting, we choose the intersection of the supports, denoted o, as the origin. Then we use the same decompositions as (7.64) and (7.65).

φ p (y) = C p + µ (x -o| α) |α| 2 (7.71) ψ q (y) = C q + µ (y -o| β) |β| 2 (7.72)
In case where ξ -2 = 0:

If we combine (7.71, 7.72) with (6.4) we obtain

Q ϕp,ψq (α, β, 0) = CpCq 2 [ k=± ks k S(a k , β, 0) -l=± lt l S(b l , α, 0] + Cpµ 3|β| 2 [ k=± ks k N (a k , o, β, 0) -l=± lt l (b l -o| β)S(b l , α, 0)] + Cqν 3|α| 2 [ k=± ks k (a k -o| α)S(a k , β, 0) -l=± lt l N (b k , o, α, 0)] + νµ 4|α| 2 |β| 2 k=± ks k [(a k -o| α)N (a k , o, β, 0) -l=± lt l (b l -o| β)N (b k , o, α, 0)]. (7.73)
where

N (a k , o, β, ξ) = β (y -o| β)(a k -y) ||a k -y|| 1+ξ ds y (7.74) is given in Appendix (B)

Conclusion

This chapter presents the evaluation of the integral S i,j when using affine basis functions in each element. We began by presenting the basis functions used and some of their properties that will simplify the calculations. We only dealt with the case of adjacent triangles. The case of triangles with common vertices presents no particular difficulty. This will be the subject of the next chapter.

-

g 1 6 -ξ U ψ 3 ,φ 1 (β, S, ξ) (8.3) I 2,1 = ( g 1 5 -ξ - g 1 6 -ξ )U φ 2 ,1 (α 1 , T, ξ) + ( d 1 6 -ξ - d 1 5 -ξ )U 1,ϕ 2 (β 1 , S, ξ) + g 1 6 -ξ U φ 2 ,ψ 1 (α 1 , T, ξ) (8.4) I 2,2 = g 1 6 -ξ U φ 2 ,ψ 2 (α 1 , T, ξ) - d 1 6 -ξ U ψ 1 ,φ 2 (β, S, ξ) (8.5)
I 2,3 = g 1 6 -ξ U φ 2 ,ψ 3 (α 1 , T, ξ) - d 1 6 -ξ U ψ 3 ,ϕ 2 (β 1 , S, ξ) (8.6)
I 3,1 = ( g 1 5 -ξ - g 1 6 -ξ )U ϕ 3 ,1 (α 1 , T, ξ) -( d 1 5 -ξ - d 1 6 -ξ )U 1,ϕ 3 (β 1 , S, ξ) + g 1 6 -ξ U ψ 3 ,φ 1 (α 1 , T, ξ) (8.7) I 3,2 = g 1 6 -ξ U φ 3 ,ψ 2 (α 1 , T, ξ) - d 1 6 -ξ U ψ 2 ,ϕ 3 (β 1 , S, ξ) (8.8)
I 3,3 = g 1 6 -ξ U φ 3 ,ψ 3 (α 1 , T, ξ) - d 1 6 -ξ U ψ 3 ,ϕ 3 (β 1 , S, ξ) (8.9)
where U 1,ψq (α i , T, ξ), U ψq,1 (β j , S, ξ) and U φp,ψq (α i , T, ξ) are defined as following

U 1,ψq (α i , T, ξ) = α i ×T ψ q (y)(x -y) ||x -y|| ξ+1 ds x dy (8.10) U ϕp,1 (α, T, ξ) = α i ×T ϕ p (x)(x -y) ||x -y|| ξ+1 ds x dy (8.11) U ϕp,ψq (α i , T, ξ) = α i ×T ϕ p (x)ψ q (y)(x -y)
||x -y|| ξ+1 ds x dy (8.12)

Reduction to dimension two

Let ω be the intersection of the supports of T and S. Here also we need to distinguish two different cases. Recall that U 1,ψq (α, T, ξ) is defined by U 1,ψq (α, T, ξ) = α×T ψ q (y)(x -y) ||x -y|| ξ+1 ds x dy and ψ q (y) = ψ q (I) + (y| -→ l q )

We get

U 1,ψq (α, T, ξ) = α×T ψ q (I)(x -y) ||x -y|| ξ+1 ds x dy + α×T (y| - → l q )(x -y) ||x -y|| ξ+1 ds x dy
and then the reduction formula gives

U 1,ψq (α, T, ξ) = ψ q (I)(1 - 3 -ξ 4 -ξ )U (α, T, ξ) + 1 4 -ξ Σ k=± ks k P ψq (a k , T, ξ) + 3 j=1 δ j (I) 4 -ξ Q 1,ψq (α, β j , ξ) (8.14) 8.2.1.3 Computation of U ϕp,1
Recall that U ϕp,1 (α, T, ξ) is defined by

U ϕp,1 (α, T, ξ) = α×T ϕ p (x) (x -y) ||x -y|| ξ+1 ds x dy and ϕ p (x) = ϕ p (I) + (x| - → e p )
We get

U ϕp,1 (α, T, ξ) = α×T ϕ p (I)(x -y) ||x -y|| ξ+1 ds x dy + α×T (x| - → e p )(x -y)
||x -y|| ξ+1 ds x dy applying the reduction formula (6.4)

U ϕp,1 (α, T, ξ) = ϕ p (I)(1 - 3 -ξ 4 -ξ )U (α, T, ξ) + 1 4 -ξ k=± ϕ p (a k )ks k P (a k , T, ξ) + 3 j=1 δ j (I) 4 -ξ Q ϕp,1 (α, β j , ξ) (8.15)

Intersection of two triangles w is parallel to α

When the support of α is parallel to w, the homogeneity of the integrand is lost. Then, the projection of α is carried out on the plane of the triangle T in order to decompose the norm ||x -y||. Let α and â± be the respective projections of α and a ± on the T plane . We denote by z = x -x and h = ||x -x|| the distance between the segment α and α which plays the role of the parameter.

8.2.2.1 Computation of U (α, T, ξ) U (α, T, ξ) = α×T (x -x) + (x -y) (h 2 + ||x -y|| 2 ) ξ+1 2 ds x dy (8.16)
Thus, the integrand is homogeneous with respect to the pair formed by the integration variables (x, y) and the parameter h. So we have

U (α, T, ξ) = U 1 (α, T, h, ξ) -zU 2 (α, T, h, ξ)
where

U 1 (α, T, h, ξ) = α×T (x -y) (h 2 + ||x -y|| 2 ) ξ+1 2 ds xdy U 2 (α, T, h, ξ) = α×T 1 (h 2 + ||x -y|| 2 ) ξ+1 2 ds xdy
The next step, we chose the extremity âof α as a new origine, we apply the reduction formula (6.5) and we get

U (α, T, ξ) = |α| T (â + -y)f 3 (h, ||â + -y||)dy + 3 j=1 δ j (â -) α×β f 2 (h, ||x -y||)ds xds y -z[|α| T f 2 (h, ||â + -y||, ξ)dy + 3 j=1 δ j (â -) α×β f 2 (h, ||x -y||)ds xds y ] otherwise U (α, T, ξ) = |α|P 3 (â + , T, h, ξ) + 3 j=1 δ j (â -)Q 2 (α, β, h, ξ) -z[|α|P 2 (â + , T, h, ξ) + 3 j=1 δ j (â -)Q 2 (α, β, h, ξ)] 8.2.2.2 Computation of U 1,ψq (α, T, ξ)
We are inspired by what we did with U (α, T, ξ)

U 1,ψq (α, T, ξ) = α×T ψ q (y)(x -y) ||x -y|| ξ+1 ds x dy = U 1,ψq 2 (α, T, ξ) -zU 1,ψq 1 (α, T, ξ)
where

U 1,ψq 1 (α, T, ξ) = α×T ψ q (y) (h 2 + ||x -y|| 2 ) ξ+1 2 ds x dy U 1,ψq 2 (α, T, ξ) = α×T ψ q (y)(x -y) (h 2 + ||x -y|| 2 ) ξ+1 2 ds x dy
We calculate each term separately. Let's start with evaluating U 1,ψq 2

(α, T, ξ) U 1,ψq 2 (α, T, ξ) = ψ q (â -)U 2 (α, T, h, ξ) + |α|P ψq 4vect (â + , T, h, ξ) 110 CHAPTER 8. TRIANGULAR MESH ELEMENTS WITH COMMON VERTEX +δ q+1 (â -)Q 1,ψq 4vect (α, β q+1 , h, ξ) + δ q+2 (â -)Q 1,ψq 4vect (α, β q+2 , h, ξ) -ψ q (â -)[|α|P 4vect (â + , T, h, ξ) + 3 j=1 δ j (â -)Q 4vect (α, β j , h, ξ)] (8.17)
where (α, T, ξ). We replace the basis functions in their equivalent expression with âas the origin, the integrant is homogeneous with parameter, we apply the formula (6.5) Which gives U φ(p),ψq 1

P ψq 4vect (â + , T, h, ξ) = T (â + -y)ψ q (y)f 4 (h, ||â + -y||, ξ)dy Q 1,ψq 4vect (α, β q+1 , h, ξ) =
(α, T, ξ) = φ p (â -)ψ q (â -) α×T 1 (h

Conclusion

In this chapter, we gave formulae of evaluation of the integral I i,j , on two triangles having a vertex in common.

CHAPTER 9

Numerical results

In the previous chapter we discretized the variational formulation of a three-dimentional scattering problem. Here we present our numerical results.

Radar Cross Section

In radiation theory it's well known that the energy intercepted by an object can be reflected, transmitted or absorbed through the target. We can assume that most of the energy is reflected.

The spatial distribution of this energy depends on the size, shape and composition of the target, and on the frequency and nature of the incident wave. This distribution of energy is called scattering, and the target itself is often referred to as a scatterer. The radar cross section (RCS) of the body is a measure of the energy scattered in a particular direction for a given illumination [START_REF] Rahmat-Samii | Hoppe Impedance Boundary Conditions in Electromagnetics[END_REF].

Bistatic scattering is the name given to the situation when the scattering direction is not back toward the source of the radiation. If E and H represent fields scattered by an object illuminated by incident plane wave E inc traveling in the direction of the unit vector k, the bistatic radar cross section in the observation direction r is

σ(r, k) = lim r→∞ 4πr 2 |E| 2 |E inc | 2 .
This cross section is defined as the area through which an incident plane wave carries sufficient power to produce, by omnidirectional radiation, the same scattered power density as that observed in a given far field direction. The monostatic radar cross section is defined Radar cross section is the measure of a target's ability to reflect radar signals in the direction of the radar receiver, i.e. it is a measure of the ratio of backscatter power per steradian (unit solid angle) in the direction of the radar (from the target) to the power density that is intercepted by the target.

The radar cross section of a target can be viewed as a comparison of the strength of the where µ r and r denote the magnetic permeability and electical permittivity of the coating. To validate the code, we test it on a sphere with different sphere-meshes and we compare it with Mie series. We take the sphere's radius is 1m and incident waves pulsation ω = 0.9GHz.

In figure (9.1), we plot the exact RCS calculated by Mie Series(continuous black line) and we compare it with our code results in different sphere mesh with inner radius r = 0.95 and coating thikness d = 0.05. In 9.1, the azure continuous line is the ucp code calculated on the mesh λ/10, red continuous line in our code calculated on the mesh λ/15 and the blue discontinue line is our code on the mesh λ/20.

In the second test, we decrease the coating thikness to d = 0.125 and we increase the inner sphere radius, for the same frequence as the previous test. In figure where s is the abscissa on β which will be calculated as follows s = y -q(a) and t ± where t ± are abscisses of the extremities of beta with q(a) origin and d = q(a) -a. where t ± are abscissae of the extremities of β with q(a) origin and d = q(a) -a.

if ξ = 0 S 1 (a, β, h, 0) = ------------------------------------------ ------------------------------------------ ds ----------------------------------------- ----------------------------------------- We remind that, on a stop β q , the basis function ψ q vanishes. So, the calculations will be performed only on β = β q+1 and β = β q+2 .The basis function ψ q on β can be written as --------------------------------------------------------

  et Wendlend used layer potential techniques to treat transmission boundary problems for the Stokes and Brinkman operators on C 1 and Lipschitz domains. In this work, we consider a boundary value transmission problem for the Stokes system. Let INTRODUCTION Let us consider a boundary value transmission problem for the Stokes system:

Definition 1 . 1 . 2 .

 112 (The space C k ) Given a non-negative integer k and an open domain D ⊂ R 3 , let C k (D) denote the space of all functions u defined in D which have all their derivatives D α u of order |α| ≤ k continuous in D. To simplify the notation, we set C 0 (D) ≡ C(D). Define C k (D) as the space of all u ∈ C k (D) such that D α u can be extended from D to a continuous function on D, the closure of the set D, for all |α| ≤ k. C k (D) can be equipped with the norm φ C k (D) = max i=0,...,k sup D |D i φ|. Definition 1.1.3. (Lebesgue spaces)

CHAPTER 1 .

 1 PRELIMINARY RESULTS Definition 1.1.4. (Curvature) Consider D a bounded open set in R 2 with a connected boundary ∂D, a, b ∈ R with a < b, and X(t) : [a, b] → R 2 an arclength parametrization of ∂D. Namely, X is a C 2 -function satisfying |X (t)| = 1 for all t ∈ [a, b] and

Theorem 3 . 1 . 1 .

 311 Let N ∈ N. There exists C depending on N and D such as for any

  3 is dedicated to develop asymptotic expansion of the perturbations in the eigenvalues, in eigenfunctions and in eigenpressures. More precisely, in the Section 4.3.1, we present some basic results related to shape perturbation and we develop high-order terms in the expansion of the integral operator. In Section 4.3.2, we (formally) provide by layer potentials an asymptotic expansion for the perturbed eigenvalues if the unperturbed eigenvalue is simple. But in section 4.3.3, we rigourously derive the leading-order term for the perturbed Stokes eigenvalues in Hölder space and for both simple and multiple eigenvalues. Here the problems to be study are more general than the one introduced in Section 4.1. In Section 4.3.4, we end our CHAPTER 4. ASYMPTOTIC BEHAVIORS FOR EIGENVALUES AND EIGENFUNCTIONS ASSOCIATED TO THE STOKES OPERATOR WITH A SMALL BOUNDARY PERTURBATION

CHAPTER 4 .

 4 ASYMPTOTIC BEHAVIORS FOR EIGENVALUES AND EIGENFUNCTIONS ASSOCIATED TO THE STOKES OPERATOR WITH A SMALL BOUNDARY PERTURBATIONfor x ∈ ∂Ω. Evidently ν 0 = ν, where ν was introduced in section 4.1.

n

  (λ; x, y) and R (ij) n (λ; x, y) are smooth and bounded uniformly on B r (0) × ∂Ω × ∂Ω. The first coefficients are given by:

= 0

 0 for each integer m. Then one can expand1 rm = κ (m) 0 (x, y) + ∞ n=1 δ n κ (m) n (x, y) uniformly ,(4.38)where the first coefficients: κ 2 (x, y). By using (4.37)-(4.38), we get that

δ ij r 5 + 15 xi xj r 3 ,

 3 where r = |x -y| and xi = x i -y i . Then, by inserting relations (4.25), (4.40), (4.41) and that given by Lemma 4.3.2 into above identity, we immediately get the following main results Proposition 4.3.4. Let the perturbed boundary ∂Ω δ defined by (4.2). Let x and ỹ given by (4.21)-(4.22), and the surface element dσ δ (ỹ) given by (4.20). Then, the components Γ ij , 1 ≤ i, j ≤ 3 of the fundamental Stokes tensor can be expanded uniformly as:

  ,1 boundary.Let D be an open subset of Ω such that dist(∂Ω, D) ≥ d 0 > 0 representing an inclusion made of a different Newtonian fluid material. Assume that the boundary ∂D is of class C 2,1 . We denote by σ 0 and σ 1 the stress tensor fields in Ω\D and D, respectively. We assume that both Ω\D and D are occupied by isotropic and homogeneous Newtonian fluids. Then, the tensors σ 0

Figure 5 . 1 :

 51 Figure 5.1: Scattering Problem

The boundary condition ( 5 . 4 )

 54 is called impedance boundary condition (IBC). The simplest form of which is known as Leontovich IBC or standard IBC (SIBC), where Z = constant [38, 51, 52]. We should notice that asymptotic behavior of the fields (E, H) is depicted by the Silver-Müller radiation condition: lim r→∞ r(E × n r + H) = 0. (5.5) where r = |x| and n r = x |x| , x ∈ R 3 . So, we have the next problem: Problem 5.1.1.

Figure 5 . 2 :

 52 Figure 5.2: Triangles T + n and T - n adjacent to edge n

. 21 )Figure 5 . 3 :

 2153 Figure 5.3: Rao-Wilton-Glisson basis function on triangle elements associated to the edge n

  vertex of the triangle T , and integrals on α × T . The last reduction allows to reduce to one dimensional integrals. The integral on S × β is reduced to a linear combination of integrals on α × β. The reduction of the integral on α × β depends on the geometric configuration of the supports of the two segments. When their supports are intersecting, we obtain integrals on a × β and on α × b. When the supports of the segments are parallel (see Figure6.2), we note α and a, the respective projections of α and a on the support of β. The integral on α × β then reduces to integrals on α × b and integrals on a × β. All 1-D integrals are regular as we will see in the next chapters.

Figure 6 . 3 :

 63 Figure 6.3: Evolution of the integration domain during the reduction process on the product of two coplanar triangles S and T.

Figure 6 . 5 :

 65 Figure 6.5: Definition of Triangle S Figure 6.6: Orientation of the triangle S

Figure 6 . 7 : Projection Figure 6 . 8 :

 6768 Figure 6.7: Projection Figure 6.8: Projections and distances

CHAPTER 7 .

 7 EVALUATION OF A SINGULAR INTEGRAL-ADJACENT TRIANGLES

Figure 7 . 1 )Figure 7 . 1 :

 7171 Figure 7.1: Adjacent triangles

1 5||x-y|| 1+ξ dxds y + g 1 6-ξ α 1 1 6-ξ S×β 1 ( 1 5-ξ α 1 1 5-ξ S×β 1 (

 111111111 1 )(x-y) ||x-y|| 1+ξ ds x dy + d ×T (x| -→ e 1 )(y| -→ l 1 )(x-y) ||x-y|| 1+ξ ds x dy + d x| -→ e 1 )(y| -→ l 1 )(x-y) ||x-y|| 1+ξ dxds y + g ×T (x| -→ e 1 )(x-y) ||x-y|| 1+ξ ds x dy + d x| -→ e 1 )(x-y) ||x-y|| 1+ξ dxds y + g 1 4-ξ α 1 ×T x-y ||x-y|| 1+ξ ds x dy + d 1 4-ξ S×β 1x-y ||x-y|| 1+ξ dxds y .(7.15)

g 1 5-ξ α 1 1 5-ξ S×β 1 ψ 1 1 6-ξ α 1 ×T φ 1 1 6-ξ α 1 ×T ψ 1 1 6-ξ α 1 1 6-ξ S×β 1 φ 1 1 6-ξ S×β 1 φ 1 1 6-ξ S×β 1 ψ 1 1 5-ξ S×β 1 φ 1

 1111111111111111111111111 ×T (x-y) ||x-y|| 1+ξ ds x dy + d (y)(x-y) ||x-y|| 1+ξ dxds y -d 1 5-ξ S×β 1 (x-y) ||x-y|| 1+ξ dxds y + g 1 6-ξ α 1 ×T φ 1 (x)ψ(y)(x-y) ||x-y|| 1+ξ ds x dy g (x)(x-y) ||x-y|| 1+ξ ds x dy -g (y)(x-y) ||x-y|| 1+ξ ds x dy + g ×T (x-y) ||x-y|| 1+ξ ds x dy + d (x)ψ(y)(x-y) ||x-y|| 1+ξ dxds y -d (x)(x-y) ||x-y|| 1+ξ dxds y -d (y)(x-y)||x-y|| 1+ξ dxds y+ d 1 6-ξ S×β 1 (x-y) ||x-y|| 1+ξ dxds y + g 1 5-ξ α 1 ×T φ 1 (x)(x-y) ||x-y|| 1+ξ ds x dy -g 1 5-ξ α 1 ×T (x-y) ||x-y|| 1+ξ ds x dy + d (x)(x-y) ||x-y|| 1+ξ dxds y -d 1 5-ξ S×β 1 (x-y) ||x-y|| 1+ξ dxds y + g 1 4-ξ α 1 ×Tx-y ||x-y|| 1+ξ ds x dy + d 1 4-ξ S×β 1

7. 5 3 7. 5 . 1 . 1 Figure 7 . 2 :

 5351172 Figure 7.2: Triangles in intersecting planes

S 1

 1 (a, β, h, ξ) and R 2 (a, β, h, ξ) are calculated in (B.2) and (B.24). 7.5.1.2 Case where a belongs to the support of T

Figure 7 . 3 : 2

 732 Figure 7.3: Projection of a on the support of T

Figure 7 . 4 :

 74 Figure 7.4: Segments neither parallel nor secants

k 2 S• 1 (d 2 +Figure 7 . 5 :

 21275 Figure 7.5: Segments are parallel

  47) where D ψq n (a, β p , h, ξ) = δ q+1 (a) βp ψ κ (y)(a -y)f n (h, a -y, ξ)ds y (7.48) 7.5.3.2 Case where a belongs to the support of T It suffices to consider a as origin to have homogeneity.

7. 5 . 4 . 2

 542 Case where α and β are secants

8. 2 . 1 3 j=1 3 j=1Q

 2133 Intersection of two triangles w is secant with α 8.2.1.1 Computation of U (α, T, ξ) We want to evaluate the integral U (α, T, ξ) = α×T (x -y) ||x -y|| ξ+1 ds x dy; ξ = 0, 2. (8.13) We consider α ∩ w = I as a new origin = ||x -y|| 1+ξ ds x ds y + k=± ks k 3 -ξ T a k -y ||a k -y|| 1+ξ dy = (α, β j , ξ) + k=± ks k 3 -ξ P (a k , T, ξ) P (a k , T, ξ) and Q(α, β j , ξ) were already computated in B 8.2.1.2 Computation of U 1,ψq

1 ,ψq 1 ( 3 j=1δ

 113 α×β q+1 ψ q (y)f 4 (h, ||x -y||, ξ)ds xds y P 4vect (â + , T, h, ξ) = T (â + -y)f 4 (h, ||â + -y||, ξ)dy Q 4vect (α, β q+1 , h, ξ) = α×β q+1 f 4 (h, ||x -y||, ξ)ds xds yare performed below. On the other hand we haveU α, T, h, ξ) = ψ q (â -)U 1 (α, T, h, ξ) + |α|P ψq 3scal (â + , T, h, ξ) +δ q+1 (â -)Q 1,ψq 3scal (α, β q+1 , h, ξ) + δ q+2 (â -)Q 1,ψq 3scal (α, β q+2 , h, ξ) -ψ q (â -)[|α|P 3scal (â + , T, h, ξ) + j (â -)Q 3scal (α, β, h, ξ)](8.18)whereP ψq 3scal (â + , T, h, ξ) = T (â + -y)ψ q (y)f 3 (h, ||â + -y||, ξ)dy Q 1,ψq 3scal (α, β q+1 , h, ξ) = α×β q+1 ψ q (y)f 3 (h, ||x -y||, ξ)ds xds y P 3scal (â + , T, h, ξ) = T (â + -y)f 3 (h, ||â + -y||, ξ)dy 8.2.2.3 Computation of U φ(p),ψq (α, T, ξ)We want to evaluate the singular integralU ϕp,ψq (α i , T, ξ) = α i ×T ϕ p (x)ψ q (y)(x -y) ||x -y|| ξ+1 ds x dy = α×T ϕ p (x)ψ q (y)(x -x) (h 2 + ||x -y|| 2 ) ξ+1 2 ds xdy α×T ϕ p (x)ψ q (y)(x -y) (h 2 + ||x -y|| 2 )We calculate each term separately. Let's start with calculating U φ(p),ψq 1

Figure 9 . 1 :

 91 Figure 9.1: Bistattic RCS for a coated unit sphere, when d = 0.05, r = 0.95, µ r = 1.0, r = 2 and ω = 0.9GHz

Figure 9 . 2 :Figure 9 . 3 :

 9293 Figure 9.2: Bistattic RCS for a coated unit sphere, when d = 0.0125, µ r = 1.0, r = 2 and ω = 0.9GHz

42 .

 42 (9.2), we plot the exact RCS calculated by Mie Series(continuous black line) the red continuous line is our code calculated on the mesh λ/10, azure continuous line represent our code calculated on the mesh λ/15 and the blue discontinuous line represent our code on the mesh λ/20In figure9.3, we consider a coating thikness d = 0.125, a frequence f = 014313GHz, the same constant for the magnetic permiability µ r = 1.0 and electical permittivity r = 2073.In figure(9.3), we plot the exact RCS calculated by Mie Series(continuous red line) and the RCS calculated with our code CHAPTER B One dimensional integrales B.1 Calculation of S(a, β, ξ) S(a, β, ξ) = β a -y ||a -y|| 1+ξ ds y (B.1) Let q(a) be the projection of a on the support of β and d = ||q(a) -a||. Then S(a, β, ξ) = (a -q(a))

B. 2 3 • S 1 (u 2 + d 2 + s 2 )

 231222 Calculation of S 1 , S 2 and S (a, β, h, ξ) S 1 (a, β, h, ξ) = β f 1 (h, ||a -y||, ξ)ds y ξ (u 2 + ||a -y|| 2 ) f n (h,s, ξ) were be defined in 7.29. Let q(a) be the projection of a into the support of β, then S 1 (a, β, h, ξ) =

(B. 21 )B. 4 3 • R 1 R 2 (u 2 + d 2 + s 2 )(u 2 + d 2 + s 2 )R 3 2 • S ψq 2

 214312222222322 Calculation of R 1 , R 2 and R (a, β, h, ξ) R 1 (a, β, h, ξ) = β (a -y)f 1 (h, ||a -y||, ξ) = β (a -y)h 1-ξ +∞ h du u 2-ξ (u 2 +||a-y|| 2 ) 1+ξ 2 ds y (B.22) --------------------------------------------------------• R 2 (a, β, h, ξ) R 2 (a, β, h, ξ) = β (a -y)f 2 (h, ||a -y||, ξ)ds y = (a -q(a)) β f 2 (h, ||a -y||, ξ)ds y + β (q(a) -y)f 2 (h, ||a -y||, ξ)ds y = (a-q(a)) ξ (u 2 + d 2 + s 2 ) ξ (u 2 + d 2 + s 2 ) (a, β, h, ξ) = (a -q(a))S 2 (a, β, h, ξ) -β |β| T 2 (a, β, h, ξ) (B.24) --------------------------------------------------------• R 2 (a, β, h, ξ) R 3 (a, β, h, ξ) = β (a -y)f 3 (h, ||a -y||, ξ)ds y = (a -q(a)) β f 3 (h, ||a -y||, ξ)ds y + β (q(a) -y)f 3 (h, ||a -y||, ξ)ds y = (a-q(a)) (a, β, h, ξ) = (a -q(a))S 3 (a, β, h, ξ) -β |β| T 3 (a, β, h, ξ) (B.26)B.5 Calculation of S ψ q (a, β, h, ξ) S ψq 2 (a, β, h, ξ) = β ψ q (y)f 2 (h, ||a -y||, ξ)ds y (B.27)

2

 2 (a, β, h, ξ) = β f 2 (h, ||y -a||, ξ)ds y -1 |β| β (y -b q )f 2 (h, ||y -a||, ξ)ds y s)f 2 (h, √ d 2 + s 2 , ξ)ds y (B.29) where ε = -1 if b q = b -, ε = 1 if b q = b + andz is the abscissa of b q Finally, we obtain S ψq 2 (a, β, h, ξ) = (1 -εz |β| )S 2 (a, β, h, ξ) + ε |β| T 2 (a, β, h, ξ) (B.30)

  2 + ||x -y|| 2 )

						ξ+1 2	ds xdy+ α×T	ϕ p (â -)(y| -→ f q ) (h 2 + ||x -y|| 2 )	2 ξ+1	ds xdy
	+	α×T	(x| -→ e p )ψ q (y)(y) (h 2 + ||x -y|| 2 ) ξ+1 2	ds xdy +	α×T	(x| -→ e p )(y| -→ f q ) (h 2 + ||x -y|| 2 )	2 ξ+1	ds xdy

  (u 2 + d 2 + s 2 )

									t +			+∞
								t -	h 1	h	du u 2 1 2	ds y
				=		t +	√	h 2 + d 2 + s 2 -h d 2 + s 2	ds
				t -
						t +
				=				√	1 h 2 + d 2 + s 2 -h	ds	(B.5)
				t -
	S 1 (a, β, h, 0) = [	h s	-	√	h 2 + s 2 s	+ lg(s +	√	h 2 + s 2 )] t + t -	(B.6)
	for ξ = 2
									t +			+∞
	S 1 (a, β, h, 2) =	t -	h -1	h	du (u 2 + d 2 + s 2 )	2 3	ds y
			t +
		=	1 h(d 2 + s 2 )	(1 -	√	h h 2 + d 2 + s 2	)ds
			t -
	= [	1 dh	arctan(	s d	) -	1 dh	arctan(	d	√	hs h 2 + d 2 + s 2	)] t + t -	(B.7)

  (u 2 + ||a -y|| 2 ) (u 2 + d 2 + s 2 )

						+∞
			=	β	h 2-ξ	h	du u 3-ξ 1+ξ 2	ds y	(B.8)
						t +			+∞
			S 2 (a, β, h, ξ) =	t -	h 2-ξ	h	du u 3-ξ (u 2 + d 2 + s 2 )	2 1+ξ	ds	(B.9)
	if ξ = 0
							t +	+∞
	[	S 2 (a, β, h, 0) = 2d 2 arcsinh( h 2 + d 2 s √ h 2 + d 2 ) -t -	h 2 2d 2	du u 3 1 2 sh 2 h √ s 2 + d 2 arcsinh( √ d 2 + s 2 ds h	)] t + t -	(B.10)
	for ξ = 2
			S 3	ds
									2
	= [	= d 2 ( t -t + √ d 2 + s 2 ) -(d 2 + s 2 ) √ s arcsinh( 1 h 2 + d 2 + s 2 √ d 2 + s 2 h ) -+	1 (d 2 + s 2 ) 1 d 2 log(s + 3 2 arcsinh √ d 2 + h 2 + s 2 )] t + ( √ d 2 + s 2 ) h t -	ds	(B.11)

• S 2 (a, β, h, ξ)

S 2 (a, β, h, ξ) = β f 2 (h, ||a -y||, ξ)ds y 2 (a, β, h, 2) = t + t - +∞ h du u(u 2 + d 2 + s 2 )

  + h 2 + s 2 )(d 2 + s 2 -2h 2 ) 3(d 2 + s 2 ) 2 ds 2 + s 2 + 2h 2 (h -√ d 2 + h 2 + s 2 ) (d 2 + s 2 ) 2 √ d 2 + h 2 + s 2 ds = [ s(-h + s √ d 2 + h 2 + s 2 ) d 2 (d 2 + s 2 )

	if ξ = 0 if ξ = 2 if ξ = 0 if ξ = 2	t + t -2h 3 + ( h 3-ξ √ +∞ h h 3 s -h 2 s S 3 (a, β, h, 0) = t + 1 3 [ √ d 2 + h 2 s 2 u 4-ξ (u 2 + d 2 + s 2 ) du d 2 (d 2 + s 2 ) + h 3 arctan( s d ) 1+ξ 2 d 3 -h 3 d 3 arctan( hs d √ d 2 + h 2 + s 2 ) + lg(s + √ d 2 + h 2 + s 2 )] ds S 3 (a, β, h, 2) = t + t --h d 3 (arctan( s d ) -arctan( hs d √ d 2 + h 2 + s 2 t + t -sf 1 (h, √ d 2 + s 2 , ξ)ds T 1 (a, β, h, 0) = t + t -sh +∞ h du u 2 (u 2 + ||a -y|| 2 ) 1 2 ds t + +∞ d 2 = T 1 (a, β, h, 2) = t -du sh -1 h 3 (u 2 + ||a -y|| 2 ) 2	))]	(B.12) (B.13) (B.14) (B.15)
		t -

• S 3 (a, β, h, ξ)

S 3 (a, β, h, ξ) = β f 3 (h, ||a -y||, ξ)ds y S 3 (a, β, h, ξ) = d B.3 Calculation of T 1 , T 2 and T 3 • T 1 (a, β, h, ξ) T 1 (a, β, h, ξ) =

  + s 2 ) 2 ( √ d 2 + h 2 + s 2 (d 2 + s 2 -2h 2 ) + 2h 3 )ds [ -h 3 + (d 2 + h 2 + s 2 ) 3 sd 2 + s 3 + 2sh(h -√ d 2 + h 2 + s 2 ) √ d 2 + h 2 + s 2 (d 2 + s 2 ) 2

	if ξ = 2
			t +
	T 3 (a, β, h, 2) =
			t -[ h -	√ d 2 + h 2 + s 2 d 2 + s 2	] t + t -
				t +	sf 3 (h,	√	d 2 + s 2 , ξ)ds	(B.19)
				t -
	if ξ = 0
		t +
	T 3 (a, β, h, 0) =	t -	s 3(d 2 2 3(d 2 + s 2 )	] t + t -	(B.20)

• T 3 (a, β, h, ξ) T 3 (a, β, h, ξ) =

Remerciements

that u δ = 0 on ∂Ω.

Thus, u δ satisfies the following eigenvalue problem for the Dirichlet biharmonic operator:

Note that Temam [START_REF] Temam | Theory and numerical analysis[END_REF] exploits the similar correspondence between the Stokes problem and the biharmonic problem in the proof of the regularity of solutions to the Stokes system and to justify several results. Moreover, as pointed out by Ashbaugh in [START_REF] Ashbaugh | On universal inequalities for the low eigenvalues of the buckling problem[END_REF], there is a similar correspondence between the eigenvalue problems for the Dirichlet Laplacian and system (4. [START_REF] Mitrea | Boundary Value Problems for the Stokes system in arbitrary Lipschitz domains[END_REF] with the boundary condition replaced by ∆u δ = 0.

Then, one can exploit this correspondence to use the approach used in [START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF] in order to develop an asymptotic expansion for the eigenvalue.

On the other hand, it is well known [START_REF] Ammari | Layer Potential Techniques in Spectral Analysis[END_REF][START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF] that there exits a polynomial-valued function δ → Q δ (λ) of degree 1, analytic in ] -δ 0 , δ 0 [ and of the form:

such that the perturbation λ δ -λ 0 is precisely its zero. For the existence of Q δ one can follow the general approach used, for example, in [START_REF] Ammari | Splitting of resonant and scattering frequencies under shape deformation[END_REF] for the case of Laplace operator.

Writing:

Then we have

Therefore, by (4.45) we have

On the other hand, for δ in a small neighborhood of 0, the following Neumann series converges uniformly with respect to λ in ∂D 0 :

Hence, we can expand the function p δ in powers of δ as we have done for λ δ and v δ . Moreover, we have: Boundary element method -Application to electromagnetism

Introduction

Here, we are interested in solving scattering problem of electromagnetic waves, by the boundary element method. We present the boundary integral method, for the Maxwell's equations.

It allows reducing the external problem to a system of integral equations defined on the surface of an object and it takes into account the boundary conditions.

In the next section 5.1, we present mathematical model of scattering problem which includes Boundary conditions on the surface and radiation condition far from an object. We use the Stratton-Chu formulation to give a boundary integral formulation in section 5.1.2, then we write it under the variational formulation with a Leontovich Boundary Condition. The Section 5.2 will be dedicated to recall existence and uniqueness of a solution. In the last section, we discretize the border of the domain using plane triangles and Rao-Wilton-Glisson basis functions.

Scattering problem

The scattering problem is represented as a system of equations including boundary conditions on the surface of an object and radiation conditions far from an object. Here, we recall the uniqueness theorem and we intoduce an integral formulation of the problem.

Conclusion

We have established a system of equations (5.6) describing the scattering problem of a coated object with the boundary condition on Γ. We have introduced the Leontovich impedance boundary conditions (LIBC). Radiation conditions helped us to establish the uniqueness of solution. The Stratton-Chu formulation indicates how the volume problem reduced to the problem on a surface of an object. This constitutes the main concept of integral method. We set 3D LIBC (5.13) into EFIE and MFIE (5.10 -5.11) and we introduce the problem (5.1.2).

We discretized the surface of a three dimensional object by triangulation. We approximated the unknowns in terms of RWG basis functions. The singular integrals which intervene there, will be evaluated in the following parts.

Reduction process

Consider the integral

where f is an homogeneous function of degree q, z = (x, y) with x ∈ S and y ∈ T (see We give also the other terms calculated with the same procedure:

)

(7.17)

the symmetrical terms

Integrands are homogeneous of degrees q = -ξ and q = 1 -ξ. We apply (6.4) and we get

We replace (y|l q ) by (y|l q ) = ψ q (y) -ψ q (a), then we get

which is equivalent to

We want to evaluate the singular integral on two edges

where β = α = α 3 , β = β 3 = β q . We are going to study two different cases. Let α be the projection of α on the plane containing β and parallel to α, x a point in α, h = |a -ǎ| and z = x -x (see Figure 7.4). We decompose Q 1,ψq in two parts

ds xds y (7.55)

We will follow the same approach as previously. We consider o = α ∩ β the new origin and we replace in the equation (7.55) the basic function ψ q (y) by the equivalent expression

Triangular mesh elements with common vertex

In this chapter, we continue to apply the reduction process to evaluate singular integrals appearing in chapter 5, on the product of two triangular mesh element S and T. We deal with the case where triangles have a common vertex.

Reduction to dimension three

Here we deal with the case where the triangles have a vertex in common. Let a 1 = b 1 be the common vertex. We keep the same local orientation in S, then, a 2 = a + 1 and a 3 = a - 1 . But, we take the opposite direction for T , so, b 2 = b - 1 and b 3 = b - 1 . we want to evaluate 9 different singular integrals in the form 7.8

As already mentioned before the integral I i,j is not singular in the case ξ = 0 but the method still valid and it will not cost us extra calculations. It will be wise to choose the common vertex a 1 = b 1 , as origin. In this case the basis functions are written

Applying Theorem 6.4 we get

Physical constant

• π = 3.1415926535897932384...;

• c = 2.99792458.10 8 ms -1 speed of light;

• µ = 4π.10 -7 free space permeability in (henry/m);

• 0 = 1/(µ 0 c 2 ) free space permittivity in (farad/m);

Modified Bessel functions I and K

The differential equation

where ν is a real constant, is called the modified Bessel's equation and its solutions I ν (z) and K ν (z) are known as modified Bessel functions.

To have the formulas for I ν (z) and K ν (z) we need the following preliminary definitions ( [START_REF] Abramowitz | Handbook of mathematical functions[END_REF]).

Definition A.1.1. The Euler constant (sometimes called Gamma) is defined as the limit of the expression

The abbreviation ln stands for the natural logarithm, also called the base e logarithm.

Definition A.1.2. (Euler's formula). For all complex numbers z, except the non-positive integers (z = 0, -1, -2, . . .),

If n is a positive integer,the function is defined by Γ(n) = (n -1)!.

.

We also have the following decay behaviour for K ν (z) as ν is fixed (Reν > 0) and z → 0

A.2 Approximation of unit normal vector

The following result can be found in [START_REF] Ammari | Conductivity interface problems. Part I: Small perturbation of an interface[END_REF].

Lemma A.2.1. Let ν( x) be the outward unit normal vector to ∂Ω δ at x. Then the following formula holds

Proof. Under the notations as in Chapter 5, it follows that x = X(t) = X(t)

is a parametrization of ∂Ω δ (where R stands for the rotation operator),

Using the simplified notations ρ(t) for ρ(X(t)) and ρ (t) for the tangential derivative of ρ(t), then we have (A.1).

A.3 Fundamental result of Riesz theory

Theorem A.3.1. (Fredholm alternative). Let T : X → X be a linear compact operator and let ker(I -T ) = {0}.

Then I -T has the bounded inverse (I -T ) -1 : X → X, where I is the identity operator.

Theorem A.3.2. Let A : X → X be a compact linear operator on a normed space X. Then

Theorem A.3.3. Let X be a normed space, A : X → X compact. If φ -Aφ = 0 has only trivial solution φ = 0 then φ -Aφ = f has a unique solution φ ∈ X for all f ∈ X, and φ depends continuously on f . Theorem A.3.4. (see [START_REF] Kress | Linear Integral Equations[END_REF], Theorem 3.4). Let A : X → X be a compact linear operator on a normed space X. Then I -A is injective if and only if it is surjective. If I -A is injective (and therefore also bijective), then the inverse operator (I -A) -1 : X → X is bounded, i.e., I -A is an isomorphism.

A.4 Singular kernel integrals

Denote by D ⊂ R d a bounded domain with the boundary ∂D of class C 1 . We consider the integral operator P defined as follows

defines a bounded operator on L 2 (R n ).

CHAPTER C

Integral formulation of Maxwell equation

C.1 Properties of integral operators

Some solutions of the problems of diffraction of electromagnetic waves can be expressed with the help of potential of single and dual layer defined on the surface of the obstacle. We present here the properties of the harmonic potential.

Definition C.1.1. We introduce the potentials (B -S) and Q, that are defined by

and G(x, y) is the Green kernel giving the outgoing solutions to the scalar Helmholtz equation:

According to theorem 4.6 in [START_REF] Lange | Equations intégrales espace-temps pour les équations de Maxwell : calcul du champ diffracté par un obstacle dissipatif[END_REF](Chapter I, p.43) we have:

(rot, Γ) and we have that:

According to theorem 2.2 in [START_REF] Lange | Equations intégrales espace-temps pour les équations de Maxwell : calcul du champ diffracté par un obstacle dissipatif[END_REF](Chapter II, p.61) we have:

and it verifies the inequality:

and the coercivity relation ∀φ ∈ H -1/2 (div, Γ):

C.1.1 Fredholm alternative

Let us mention the abstract theorems known as Fredholm alternative [START_REF] Nédélec | Acoustic and electromagnetic equations: Integral representation for harmonic problems[END_REF].

Theorem C.1.3. Let V be a Hilbert space. Let H be a Hilbert space which contains V . Let a(u, v) be continuous bilinear form on V × V which satisfies:

We consider the variational problem

Suppose that the injection of V into H is compact. Then the variational problem (VP) satisfies the Fredholm alternative i.e.

-either it admits a unique solution in V ,

-or it has a finite dimensional kernel and a unique solution up to any element in this kernel, when the duality product of the right-hand side g vanishes on every element in this kernel.

We take the existence and uniqueness theorem from [START_REF] Nédélec | Acoustic and electromagnetic equations: Integral representation for harmonic problems[END_REF] (p.245, Theorem 5.6.1).

Theorem C.1.4. (Existence and uniqueness): Let V and W be Hilbert spaces. Let A(., .) be a bilinear form continuous on V × V which satisfies

where H is Hilbert space containing V . Let B(q, v) be a bilinear form continuous on W × V which satisfies:

where L is a Hilbert space containing W .

Consider the following variational problem, with g 1 ∈ V * and g 2 ∈ W * :

Denote by V 0 the kernel of the bilinear form B in V , i.e.

V 0 = {u ∈ V, B(q, u) = 0, ∀q ∈ W } Suppose that the injection form V 0 into H is compact and the injection from W into L is compact. Suppose that there exists an element u g 2 ∈ V such that:

B(q, u g 2 ) = (g 2 , q), ∀q ∈ V.

Then the variational problem (C.10) satisfies the Fredholm alternative, i.e.

-either it admits a unique solution in V × W , -or it admits a finite dimensional kernel, and a solution defined up to any element in this kernel, when the right-hand side (g 1 , g 2 ) vanishes on any element in this kernel.

C.2 Integral formulation of the problem

We are looking for a presentation of the fields (E, H) outside the object in terms of electromagnetic currents (J, M) on the surface Γ [START_REF] Nédélec | Acoustic and electromagnetic equations: Integral representation for harmonic problems[END_REF]. We have Maxwell's equations on the surface

The superposition theorem lets us consider two cases. In the first case we suppose that M = 0, in the second J = 0. In both cases we want to get E and M expressions in terms of J and M.

After we combine these cases to get the general form of fields E and M. In case M = 0, we are looking for E in the form

with Lorentz gauge condition divA = k 2 V . The divergence of the second equation of (C.11)

with support of gauge condition, shows that V satisfies scalar Helmholtz equation

It can be expressed as a potential simple layer density of -ik -1 Z 0 divJ

In this expression G is the Green kernel, fundamental solution of Helmholtz equation, i.e. verifies Maxwell's equations and radiation condition.

We have that A verifies the next vector Helmholtz equation

According to earlier condition the right side of the first term is equal to k 2 ∇V . When (C.12) allows us to express (C.17) as

The last equivalence derived from Maxwell's equations (C.11).

Then A is a potential of simple layer density -ikZ 0 J. So we can write

We write them via the operators We recall that on a boundary, we have

where the subscript t designates the tangential component E tg := n×(E×n) of the respective vector field on Γ. Later, to write variational formulation, we will multiply it by a test function and integrate over the surface Γ.