Slah Chaabi
email: slah.chaabi@univ-amu.fr

Stephane Rigat
email: stephane.rigat@univ-amu.fr

DECOMPOSITION THEOREM AND RIESZ BASIS FOR AXISYMMETRIC POTENTIALS IN THE RIGHT HALF-PLANE

Keywords: . 2010 Mathematics Subject Classification. 35B07, 33E05, 35J15, 35E05, 42C15 Axially symmetric solutions, Fundamental solutions, Riesz Basis, Elliptic functions and integrals

The Weinstein equation with complex coefficients is the equation governing generalized axisymmetric potentials (GASP) which can be written as Lm[u] = ∆u + (m/x) ∂xu = 0, where m ∈ C. We generalize results known for m ∈ R to m ∈ C. We give explicit expressions of fundamental solutions for Weinstein operators and their estimates near singularities, then we prove a Green's formula for GASP in the right half-plane H + for Re m < 1. We establish a new decomposition theorem for the GASP in any annular domains for m ∈ C, which is in fact a generalization of the Bôcher's decomposition theorem. In particular, using bipolar coordinates, we prove for annuli that a family of solutions for GASP equation in terms of associated Legendre functions of first and second kind is complete. For m ∈ C, we show that this family is even a Riesz basis in some non-concentric circular annuli.

Introduction

In this article, we study the Weinstein differential operator

L m = x -m div (x m ∇•) = ∆ + m x ∂ ∂x
with m ∈ C, well-defined on the right half-plane H + = {(x, y) ∈ R 2 , x > 0} = {z ∈ C, Re z > 0} with the convention x m = exp(m ln x). This class of operators is also called operators governing axisymmetric potentials, they have been studied quite intensively in cases m ∈ N or m ∈ R in [START_REF] Weinstein | On the torsion of shafts of revolution[END_REF][START_REF] Weinstein | On generalized potential theory and on the torsion of shafts. Studies and essays Presented to R[END_REF][START_REF] Weinstein | Discontinuous integrals and generalized potential theory[END_REF][START_REF] Weinstein | Transonic flow and generalized axially symmetric potential theory[END_REF][START_REF] Weinstein | On Tricomi's equation and generalized axiallly symmetric potential theory[END_REF][START_REF] Weinstein | Generalized axially symmetric potential theory[END_REF][START_REF] Weinstein | The singular solutions and the Cauchy problem for generalized tricomi equations[END_REF][START_REF] Weinstein | The method of axial symmetry in partial differential equations[END_REF][START_REF] Weinstein | Elliptic and hyperbolic axially symmetric problems[END_REF][START_REF] Weinstein | On a class of partial differential equations of even order[END_REF][START_REF] Weinstein | The generalized radiation problem and the Euler-Poisson-Darboux equation[END_REF][START_REF] Weinstein | Sur une classe d'équations aux dérivées partielles singulières. La théorie des équations aux dérivées partielles[END_REF][START_REF] Weinstein | On a singular differential operator[END_REF][START_REF] Weinstein | Singular partial differential equations and their applications[END_REF][START_REF] Weinstein | Some applications of theory of generalized axially symmetric potential theory to continuum mechanics[END_REF][START_REF] Weinacht | Fundamental solutions for a class of singular equations[END_REF][START_REF] Weinacht | A mean value theorem in generalized axially symmetric potential theory[END_REF][START_REF] Weinacht | Fundamental solutions for a class of equations with several singular coefficients[END_REF][START_REF] Weinacht | Poisson integral formulas in generalized bi-axially symmetric potential theory[END_REF][START_REF] Copson | On sound waves of finite amplitude[END_REF][START_REF] Copson | On hadamard's elementary solution[END_REF][START_REF] Copson | Partial differential equations[END_REF][START_REF] Huber | A theorem of Phragmén-Lindelöf type[END_REF][START_REF] Huber | On the uniqueness of generalized axially symmetric potentials[END_REF][START_REF] Huber | Some results on generalized axially symmetric potentials[END_REF][START_REF] Erdélyi | Singularities of generalized axially symmetric potentials[END_REF][START_REF] Erdélyi | Symmetric potentials and fractional integration[END_REF][START_REF] Erdélyi | An application of fractional integrals[END_REF]11,12,13,[START_REF] Brelot | Équation de Weinstein et potentiels de Marcel Riesz[END_REF][START_REF] Gilbert | Some properties of generalized axiallly symmetric potentials[END_REF][START_REF] Gilbert | On generalized axiallly symmetric potentials[END_REF][START_REF] Gilbert | Poisson's equation and generalized axiallly potential theory[END_REF][START_REF] Gilbert | Bergman's integral operator method in generalized axiallly symmetric potential theory[END_REF][START_REF] Liu | The Cauchy problem for an axially symmetric equation and the Schwarz potential conjecture for the torus[END_REF]. In this paper, we focus exclusively on case m ∈ C and some results for integer values of m is recalled. The Weinstein equation is written as follows

L m u = 0. (1.1)
The main reason for which we consider the case m ∈ C is that, if we complexify the coordinates by writing z = x + iy, (1.1) can be rewritten

∂ 2 u ∂z∂z + m/2 z + z ∂u ∂z + ∂u ∂z = 0,
which is a particular case of the equation

∂ 2 u ∂z∂z + α z + z ∂u ∂z + β z + z ∂u ∂z = 0
considered with α, β ∈ C in [START_REF] Vekua | New methods for solving elliptic equations[END_REF] (equation (5.7), page 20). Equation (1.1) also appears in physics in the study of the behavior of plasma in a tokamak. The role of tokamak, which has a toroidal geometry, is to control location of the plasma in its chamber by applying magnetic fields on its boundary. It is possible to assume that plasma is axially symmetric what reduces this problem to a plane section in H + , where the magnetic flux in the vacuum between the plasma and the circular boundary of the chamber satisfies a second-order elliptic nonlinear partial differential equation, the so-called Grad-Shafranov equation, which reduces to the homogeneous equation (1.1) with m = -1.

O x y u, ∂ n u boundary of the plasma

L -1 (u) = 0
Note that in this instance, (1.1) takes place in an annular domain rather than in a simply connected domain (see [START_REF] Blum | Numerical Simulation and Optimal Control in Plasma Physics : With applications to Tokamaks[END_REF][START_REF] Shafranov | On magnetohydrodynamical equilibrium configurations[END_REF][START_REF] Blum | Reconstruction of the equilibrium of the plasma in a Tokamak and identification of the current density profile in real time[END_REF]). And this motivates the Decomposition Theorem 5.8. In the sequel, the sense in which the solutions are studied will be specified. We will also look at solutions to the equation in the sense of distributions L m u = δ (x,y) , where δ (x,y) denotes the Dirac mass at (x, y) ∈ H + . A. Weinstein was the first to introduce this class of operators in 1948 in [START_REF] Weinstein | Discontinuous integrals and generalized potential theory[END_REF], he studied the case where m ∈ N * .

He also established the link between the axisymmetric potentials for m ∈ N * and the harmonic functions on R m+2 , that we will recall in the proposition 2. [START_REF] Axler | Harmonic function theory[END_REF]. In [START_REF] Weinstein | On Tricomi's equation and generalized axiallly symmetric potential theory[END_REF][START_REF] Weinstein | Generalized axially symmetric potential theory[END_REF][START_REF] Diaz | On the fundamental solutions of a singular Beltrami operator[END_REF], Weinstein and Diaz-Weinstein established the correspondence principle that we will recall between axisymmetric potentials corresponding to m and those corresponding to 2 -m (proposition 2.3). They deduced an expression of a fundamental solution (where the singular point is taken on the y-axis) for m ∈ R and they made a link between the Weinstein equation and Tricomi equations and their fundamental solutions. Moreover, still in [START_REF] Vekua | New methods for solving elliptic equations[END_REF], Vekua gave means to express fundamental solutions of elliptic equations with analytic coefficients by using the Riemann functions, introduced in the past (see eg [START_REF] Garabedian | Partial Differential Equations[END_REF]) in the real hyperbolic context, he generalized to elliptic equations through the complex operators ∂ z et ∂ z in [START_REF] Vekua | New methods for solving elliptic equations[END_REF]. In heuristic words, in the same way that we can say a harmonic function is the real part of a holomorphic function, or the sum of a holomorphic and an anti-holomorphic function, Vekua expressed the fact that solutions of elliptic equations, and therefore especially GASP, are written as a sum of two functionals, one applied to an arbitrary holomorphic function and the other one applied to an anti-holomorphic functiun also arbitrary. These functionals can be writen explicitly in terms of Riemann function, which are obtained by using the hypergeometric functions ( [START_REF] Vekua | New methods for solving elliptic equations[END_REF]) or using fractional derivations ( [START_REF] Copson | On sound waves of finite amplitude[END_REF]). In [START_REF] Henrici | On the domain of regularity of generalized axiallly symmetric potentials[END_REF], Henrici gave a very interesting introduction to the work of Vekua. More recently, by using the work of Vekua in [START_REF] Savina | On splitting up singularities of fundamental solutions to elliptic equations in C 2[END_REF], Savina gave a series representation of fundamental solutions for the operator Lu = ∆u + a∂ x u + b∂ y u + cu and she studied the convergence of thess series. She gave an application to Helmholtz equation. In [START_REF] Gilbert | Poisson's equation and generalized axiallly potential theory[END_REF], Gilbert studied the non-homogeneous Weinstein equation m ≥ 0, he gave an integral formula for this class of equations, in particular, an explicit solution is given when the second member depends only of one variable. Some Dirichlet problems can be found in [START_REF] Liu | The Cauchy problem for an axially symmetric equation and the Schwarz potential conjecture for the torus[END_REF] in [START_REF] Fokas | Boundary-value problems for the stationary axisymmetric einstein equations: a rotating disc[END_REF] in special geometry ("geometry with separable variable"). Even if some results presented in this paper are known for real values of m, we make a totally self-contained presentation with elementary technics not usually used in the previous quoted papers. For instance, usual arguments involving estimates of hypergeometric integrals are replaced by arguments using Lebesgue dominated convergence theorem. The main result is a decomposition theorem for axisymmetric potentials which is new also for real values of m. We obtain a Liouville-type result for the solutions of Weinstein equation on H + , the interesting side of this result is the fact that there is a lost of strict ellipticity of the Weinstein operator on the boundary of H + . An application of the decomposition theorem is given by showing that an explicit family of axisymmetric potentials constructed with the introduction of bipolar coordinates is a Riesz basis in some annuli.

The plan of the paper will be the following. It should be useful for the reader to keep this plan in mind while reading the paper.

First, in section 2, we recall the notion and interest of fundamental solutions to linear partial differential operators with non constant coefficients.

There is connection between fundamental solutions to L m and fundamental solutions to L * m , where L * m denotes the formal adjoint of L m (defined in the first section). The connection is a consequence of proposition 2.1.

Weinstein Principle, stated in [START_REF] Weinstein | Generalized axially symmetric potential theory[END_REF] (Proposition 2.3) is valid whatever m is real and complex. This is just a straightforward computation, and it makes connection between L m and L 2-m for every m ∈ C. Proposition 2.4, valid only if m ∈ N, is fundamental in the sense that we can compute a fundamental solution of L m just by knowing the usual fundamental solution of the Laplacian in R m+2 .

These computations are done for m ∈ N first, and for m ∈ Z in section 3.

The extension to the case where m ∈ C becomes natural in section 4 (because the same formulas remain still valid, whatever m ∈ Z or m ∈ C). Moreover, the behavior of these fundamental solutions near their singularities are given in proposition 4.2 in a very elementary way, and theorem 4.4 uses these estimates to show the extension of this formula from m ∈ Z to m ∈ C. Section 5 is a preparation section to the Decomposition Theorem. For this, we modify the fundamental solutions built above in order to get fundamental solutions which are vanishing on the boundary of H + . Proposition 5.2 shows that if u is solution to L m u = 0 which is vanishing on the boundary of H + , then u ≡ 0 on H + . We emphasize the fact that, even if this proposition looks obvious, it is not because of the lost of ellipticity of L m on the boundary of H + . We can also note that Proposition 5.2 is a consequence of the maximum principle for pseudo-analytic functions given in a very recent paper by Chalendar-Partington ([14]) for more general σ than x m . But in their situation, there is an assumption on σ, which corresponds in our case by asumming |m| ≥ 1. The proof of proposition 5.2, is then quite long, but not difficult : this is just done by careful estimates of fundamental solutions in some parts of H + . Then everything is in place to prove the Decomposition Theorem 5.8 in the same way than Bôcher's decomposition theorem in [START_REF] Axler | Harmonic function theory[END_REF]. We end this section 5 by giving a Poisson formula in H + (proposition 5.9).

In section 6, we consider the case where the annular domain is a kind of annulus. We introduce the very classical (in physics) bipolar coordinates (cf. [START_REF] Lebedev | Special functions and their applications[END_REF]). In these coordinates, the GASP Equation has a different form (theorem 6.1) and the method of separation of variables gives a basis of solutions in disks and complements of disks in H + (theorem 6.2).

It is moreover shown in section 7 that this forms a Riesz Basis.

Notations and preliminaries

Throughout the following, H + = {(x, y) ∈ R 2 , x > 0} will denote the right half-plane. All scalar functions will be complex valued. If Ω is an open set of R n with n ∈ N * , D(Ω) will designate the space of C ∞ functions compactly supported on Ω and the support of an arbitrary function f defined on Ω is supp

f := {x ∈ Ω, f (x) = 0}. Let K be a compact set of Ω, D K (Ω) is the set of functions ϕ ∈ D(Ω) such that supp ϕ ⊂ K.
The partial derivatives of a differentiable function u on an open set Ω ⊂ R n will be denoted ∂u ∂x i or ∂ x i u, or sometimes u x i with i ∈ 1, n (for a < b ∈ N, a, b denotes the set of all integers between a and b).

If α = (α 1 , . . . , α n ) ∈ N n is a multi-index, we will denote ∂ α := ∂ α 1 x 1 • • • ∂ αn xn = ∂ |α| ∂x α 1 1 • • • ∂x αn n with |α| = α 1 + • • • + α n .
It is assumed that the reader is familiar with the terminology of distributions and we refer to [START_REF] Hörmander | The Analysis of Linear Partial Differential Operators I: Distribution Theory And Fourier Analysis[END_REF].

Let L be a linear differential operator on Ω,

L = |α|≤N a α ∂ α
where N ∈ N, the previous summation is performed on the multi-indices α of length |α| ≤ N , a α are C ∞ (Ω) functions. By definition, if T is a distribution, LT will be the distribution : LT = |α|≤N a α ∂ α T . L will designate the adjoint operator of L in the sense of distributions, namely if T is a distribution,

L T = |α|≤N (-1) |α| ∂ α (a α T ). It is noticed that if f, g ∈ D(Ω), we have Lf, g = f, L g . Let a ∈ Ω and L be a differential operator on Ω. A fundamental solution of L on Ω at a ∈ Ω is a distribution T a such that LT a = δ a ,
where the previous equality is taken in the sense of distributions on Ω. This equality can be rewritten

∀ϕ ∈ D(Ω), ϕ(a) = LT a , ϕ = T a , L ϕ .
In particular, if a ∈ Ω and if T a is a fundamental solution of L at a on Ω and if g ∈ D(Ω) is such that g = L(ϕ) with ϕ ∈ D(Ω), then ∀a ∈ Ω, ϕ(a) = T a , g .

Indeed, we have

∀a ∈ Ω, ϕ(a) = δ a , ϕ = L * T a , ϕ = T a , Lϕ = T a , g .
These fundamental solutions is therefore a good tool for solving Lϕ = g on

D(Ω) if g ∈ D(Ω).
If m ∈ N * , the Laplacian in R m will be denoted ∆ m , or ∆ when m = 2. For m ∈ C, L m denotes the Weinstein operator :

∀(x, y) ∈ H + , L m u(x, y) = ∆u(x, y) + m x ∂u ∂x (x, y), where u ∈ C 2 (H + ).
The following notation will be sometimes used : With these notations, the operator L m can be written as follows :

if f (x, y) = (f 1 (x, y), f 2 (x, y)) is a C 1 vector
if u ∈ C 2 (H + ), then L m u(x, y) = x -m div(x m ∇u)(x, y).
It is clear from the Schwarz rule that if u is a function defined on a connected open set of H + such that div(σ∇u) = 0 where σ : H + → C * is a C 1 function, then there is a function v which satisfies the well-known generalized Cauchy-Riemann system of equations :

         ∂v ∂x = -σ ∂u ∂y ∂v ∂y = σ ∂u ∂x
and v satisfies the conjugate equation div( 1 σ ∇v) = 0 (see for exemple [START_REF] Baratchart | Hardy spaces of the conjugated Beltrami equation[END_REF]). This observation justifies the fact that we call L -m with m ∈ C the conjugate operator of L m . L m denotes adjoint operator of L m : for all u ∈ C 2 (H + ) and for all (x, y) ∈ H + ,

L m u(x, y) = ∆u(x, y)- ∂ ∂x m u(x, y) x = ∆u(x, y)- m x ∂u ∂x (x, y)+ m x 2 u(x, y)
This definition is given on H + but it is easily transposed to the case of an open set Ω of H + .

In the case where the functions involved do not depend only of x and y, we will write L m,x,y instead of L m , which means that the partial derivatives are related to the variables x and y, and all other variables are considered to be fixed.

S m L m = L m S m , L -m D = DL m .
Proof. Straightforward computations, whatever m is real or complex.

Remark 2.2.

(1) Let m ∈ C, S m and L m S m are auto-adjoints operators, ie. S m = S m and L m S m = (L m S m ) .

(2) There is a result, which generalizes the first point of this remark about the conjugation of operators L m and L m . Let σ : Ω → C be a C 1 function which does not vanish, the operator defined on C 2 (Ω) by : for u ∈ C 2 (Ω),

P σ u(x, y) = 1 σ(x, y) div (σ(x, y)∇u(x, y)) ,
where

Ω is an open set of R 2 . Then P σ = div σ∇ • σ .
Indeed, if u, v ∈ D(Ω), then we have by using the derivation in the sense of distributions,

P σ u, v = Ω 1 σ(x, y) div (σ(x, y)∇u(x, y)) v(x, y) dxdy = - Ω σ∇u • ∇ v σ dxdy = Ω u div σ∇ v σ = u, P σ v
We define S σ the operator such that for u ∈ C 2 (Ω),

(S σ u)(x, y) = 1 σ(x, y) u(x, y).
Thus, S σ conjugates P σ and P σ , where P σ = div σ∇ • σ because, in an obviously way, we have

S σ P σ = P σ S σ .
If m is a positive integer, we introduce the operator T m : u → v defined as follows : for a function u defined on an open set Ω of H + , the function v is defined on {x ∈ R m+2 , (

x 2 1 + • • • + x 2 m+1 , x m+2 ) ∈ Ω} by v(x 1 , . . . , x m+2 ) = u( x 2 1 + • • • + x 2 m+1 , x m+2
). The two following propositions can be found in Weinstein work ( [START_REF] Weinstein | Generalized axially symmetric potential theory[END_REF]) in the case m ∈ R and they will be useful in the sequel (the (short) proofs are just direct computations, and the proof is the same whatever m is real or complex):

Proposition 2.3. (Weinstein principle [59]) Let Ω be a relatively com- pact open set of H + , if u : Ω → C is C 2 , then for all m ∈ C, L m u = x 1-m L 2-m [x m-1 u]. Proposition 2.4. ([54]) Let Ω be a relatively compact open set of H + . For u ∈ C 2 (Ω)and if m ∈ N, then ∆ m+2 (T m u) = T m (L m u).
The two previous propositions will allow us to calculate fundamental solutions for L m and L m with m ∈ N in a first step, and thereafter, for m ∈ Z. Finally, estimates of these expressions will show that the expressions obtained actually provide fundamental solutions of L m and L m for m ∈ C.

Integral expressions of fundamental solutions for integer values of m.

We recall the definition of the Dirac mass in a point : if (x, y) ∈ R 2 , δ (x,y) is the distribution defined by

∀ϕ ∈ D(R 2 ), δ (x,y) , ϕ = ϕ(x, y).
Let m be a positive integer.

Proposition 3.1. (partially in [START_REF] Diaz | On the fundamental solutions of a singular Beltrami operator[END_REF][START_REF] Weinacht | Fundamental solutions for a class of equations with several singular coefficients[END_REF][START_REF] Weinstein | Discontinuous integrals and generalized potential theory[END_REF]) Let m ∈ N . For (x, y) ∈ H + and (ξ,

η) ∈ H + , E m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 m/2
is a fundamental solution on H + for the operator L m,ξ,η at the fixed point (x, y) ∈ H + , which means that in the sense of distributions, we have H + :

L m,ξ,η E m (x, y, ξ, η) = δ (x,y) (ξ, η).
Moreover, if (ξ, η) ∈ H + is fixed, then in the sense of distributions on H + , we have L m,x,y E m (x, y, ξ, η) = δ (ξ,η) (x, y), which means that E m is a fundamental solution on H + of the operator L m,x,y at the fixed point (ξ, η) ∈ H + .

Proof. Let m ∈ N * . We recall that

E(x) = - 1 m ω m+2 x m , x ∈ R m+2 ,
is a fundamental solution for the Laplacian on R m+2 i. e. that in the sense of distributions, ∆ m+2 E = δ 0 , where ω m+2 is the area of the unit sphere R m+2 . Thus, for all v ∈ D(R m+2 ),

v(t 1 , ..., t m+2 ) = - 1 m ω m+2 τ ∈R m+2 ∆ m+2 v(τ ) dτ 1 dτ 2 ...dτ m+2 ((τ 1 -t 1 ) 2 + • • • + (τ m+2 -t m+2 ) 2 ) m/2
where τ = (τ 1 , ..., τ m+2 ).

Applying this relation to the function v = T m u where u ∈ D(H + ) and due to the proposition 2.4, we have for all (x, y)

∈ H + , u(x, y) = - 1 m ω m+2 R m+2 (L m u)( ξ 2 1 + • • • + xi 2 m+1 , ξ m+2 ) dξ 1 • • • dξ m+2 (ξ 1 -x) 2 + ξ 2 2 + • • • + ξ 2 m+1 + (ξ m+2 -y) 2 m/2
We will simplify this integral expression. For this, we will consider the following hyper-spherical coordinates :

ξ 1 = ξ cos θ 1 ξ 2 = ξ sin θ 1 cos θ 2 . . . = . . . ξ m-1 = ξ sin θ 1 • • • sin θ m-2 cos θ m-1 ξ m = ξ sin θ 1 • • • sin θ m-1 cos θ m ξ m+1 = ξ sin θ 1 • • • sin θ m where ξ = ξ 2 1 + • • • + ξ 2 m+1 ≥ 0, θ m ∈ ]-π, π[ and θ 1 , . . . , θ m-1 ∈ ]0, π[.
The absolute value of the determinant of the Jacobian matrix defined by this system of coordinates is

ξ m sin θ m-1 sin 2 θ m-2 • • • sin m-1 θ 1 Then we have for all (x, y) ∈ H + , u(x, y) = ∞ η=-∞ ∞ ξ=0 L m (u)(ξ, η)E m (x, y, ξ, η)dξdη (3.1) with E m (x, y, ξ, η) = - ξ m m ω m+2 π θm=-π π θ 1 ,...,θ m-1 =0 sin θ m-1 sin 2 θ m-2 • • • sin m-1 θ 1 dθ 1 . . . dθ m (ξ 2 -2xξ cos θ 1 + x 2 + (y -η) 2 ) m/2 Since I := π θm=-π π θ 2 ,...,θ m-1 =0 sin θ m-1 sin 2 θ m-2 • • • sin m-2 θ 2 dθ 2 . . . dθ m-1 dθ m is the area of the unit sphere on R m because ω m = Sm 1 dσ = π θ m-1 =-π π θ 1 ,...,θ m-2 =0 sin θ m-2 sin 2 θ m-3 • • • sin m-2 θ 1 dθ 2 . . . dθ m-1 ,
E m can be written as :

E m (x, y, ξ, η) = - ω m ξ m m ω m+2 π θ=0 sin m-1 θ dθ (ξ 2 -2xξ cos θ + x 2 + (y -η) 2 ) m/2 .
Using the fact that

ω m = 2π m 2 Γ( m 2 )
, we get

E m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η)
E m (x, y, ξ, η) = x ξ -m E m (ξ, η, x, y)
and thanks to the Proposition 2.1, S m conjugates L m and L m , we have in the sense of distributions

L m,x,y E m (x, y, ξ, η) = L m,x,y x ξ -m E m (ξ, η, x, y) = x ξ -m L m,x,y E m (ξ, η, x, y), then L m,x,y E m (x, y, ξ, η) = x ξ -m δ (ξ,η) (x, y) = δ (ξ,η) ,
and this completes the proof.

For m ∈ Z \ N, the previous proposition and the Weinstein principle gives us the following proposition : Proposition 3.2. (partially in [START_REF] Diaz | On the fundamental solutions of a singular Beltrami operator[END_REF][START_REF] Weinacht | Fundamental solutions for a class of equations with several singular coefficients[END_REF][START_REF] Weinstein | Discontinuous integrals and generalized potential theory[END_REF])

Let m ∈ Z \ N * . For (x, y) ∈ H + and (ξ, η) ∈ H + , E m (x, y, ξ, η) = ξ x m-1 E 2-m (x, y, ξ, η) = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η)
= H + ξ 1-m L 2-m (ξ m-1 u)E m (x, y, ξ, η) dξdη.
Denoting v(x, y) = x m-1 u(x, y), we obtain

x 1-m v(x, y) = H + ξ 1-m (L 2-m v)E m (x, y, ξ, η) dξdη, then, for all m ∈ Z \ N * , v ∈ D(H + ) and (x, y) ∈ H + , putting m = 2 -m , we have v(x, y) = H + (L m v) ξ x m -1 E 2-m (x, y, ξ, η) dξdη.
The proof of the second point is totally similar.

Fundamental solutions for the Weinstein equation with complex coefficients

In this section, we will generalize the result obtained in the previous section for m ∈ Z to m ∈ C. More precisely, if Re m ≥ 1, then

E m = - ξ m 2π π θ=0 sin m-1 θ dθ [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2
] m/2 is suitable, and if Re m < 1, then

E m = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 1-m 2 is suitable.
Note that, in the above formulae, by convention, if α = 0 is a real number and µ is a complex number, then

α µ := exp(µ ln α).
Note also that, in preceding equations, integrals are convergent in Lebesgue sense.

In the sequel, E m will always designate the corresponding formula (depending of Re m ≥ 1 or Re m < 1).

Proposition 4.1. For m ∈ C and (ξ, η) ∈ H + fixed, we have

∀(x, y) ∈ H + \ {(ξ, η)} L m,x,y E m (x, y, ξ, η) = 0.
and for (x, y) ∈ H + fixed, we have

∀(ξ, η) ∈ H + \ {(x, y)} L m,ξ,η E m (x, y, ξ, η) = 0.
Proof. For convenience in the calculations, it should be denoted

f m (x, y, ξ, η, θ) = 1 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2
.

To prove the first equality of the proposition, it suffices to show that

π θ=0 L m,x,y f m (x, y, ξ, η, θ) sin m-1 θ dθ = 0.
Let's compute the derivatives of the function f m :

∂ x f m = -m 2 2(x -ξ) + 4ξ sin 2 θ 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 (= -m(x -ξ cos θ)f m+2 )
and

∂ xx f m = -m [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 + + m 2 m 2 + 1 (2(x -ξ) + 4ξ sin 2 θ 2 ) 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +2
and

∂ yy f m = -m [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 + + m 2 m 2 + 1 (2(y -η)) 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +2
We then have

∆f m = -2m [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 + + m 2 m 2 + 1 (2(x -ξ) + 4ξ sin 2 θ 2 ) 2 + (2(y -η)) 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +2 . However 2(x -ξ) + 4ξ sin 2 θ 2 2 +(2(y -η)) 2 = 4 (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 -4ξ 2 sin 2 θ then ∆f m = m 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 - m(m + 2)ξ 2 sin 2 θ [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +2 . Noting that ∂f m+2 ∂θ = -(m + 2) xξ sin θ [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +2 ,
we have

∆f m = m 2 f m+2 + m ξ x
sin θ ∂f m+2 ∂θ and by integration by parts, we have :

π θ=0 ∆f m sin m-1 θ dθ = m 2 π θ=0 f m+2 sin m-1 θ dθ+m ξ x π θ=0 ∂f m+2 ∂θ sin m θ dθ = m x π θ=0 m (x -ξ cos θ) f m+2 sin m-1 θ dθ = - m x π θ=0 ∂ x f m sin m-1 θ dθ,
and the result is deduced in the case Re m ≥ 1. The proof is totally similar if Re m < 1. The second equality of the proposition can be deduced immediately of the fact that S m conjugates L m and L m (see proposition 2.1).

In the sequel, we will denote

d 2 = (x -ξ) 2 + (y -η) 2 and k = 4xξ d 2 .
The following proposition gives the behavior of these functions near their singularity. And it will be useful to show that they are indeed fundamental solutions for m ∈ C and not only for integer values of m. In particular, we show that the behavior of the fundamental solutions is close to the behavior of fundamental solutions for the Laplacian. This fact is well known for elliptic operators. But we emphasize here that in the proof of this proposition, the estimates of elliptic integrals are totally elementary estimates (using the dominated convergence theorem) and here we do not use estimates arising from classical estimates of hypergeometric functions. From those integral expressions, we deduce the following estimations :

Proposition 4.2. Let m ∈ C. For (x, y) ∈ H + fixed, E m (x, y, ξ, η) ∼ (ξ,η)→(x,y) 1 2π ln (x -ξ) 2 + (y -η) 2
Proof. We start with Re m ≥ 1.

In this case, we have :

E m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 m/2 = - 1 2π ξ d m π θ=0 sin m-1 θ dθ (1 + k sin 2 θ 2 ) m/2
.

Note that when d → 0, k → +∞.

We have the following proposition :

Proposition 4.3. When k → +∞ and m ∈ C, π θ=0 sin m-1 θ dθ (1 + k sin 2 θ 2 ) m/2 ∼ k→+∞ 2 m-1 k m/2 ln k. Proof. Putting u = sin θ 2 , this integral is equal to 2 m 1 0 u m-1 (1 -u 2 ) m-2 2 du (1 + ku 2 ) m/2 = 2 m k m/2 1 0 u m-1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2
.

However 1 0 u m-1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2 - 1 0 u m-1 du ( 1 k + u 2 ) m/2 = - 1 0 u m-1 1 k + u 2 m/2 (1-(1-u 2 ) m- 2
2 )du and by monotone convergence, we obtain

-→ k→+∞ - 1 0 u m-1 (u 2 ) m/2 (1 -(1 -u 2 ) m- 2
2 )du = -

1 0 1 -(1 -u 2 ) m-2 2 u du The change of variable u = 1 √ k sh t gives us 1 0 u m-1 du ( 1 k + u 2 ) m/2 = argsh √ k 0 th m-1 t dt
Since th m-1 t tends to 1 when t → +∞, we deduce that when k → +∞ argsh

√ k 0 th m-1 dt ∼ k→+∞ argsh √ k 0 dt = argsh √ k ∼ k→+∞ 1 2 ln k.
The proof is completed.

Due to Proposition 4.3, we have

E m (x, y, ξ, η) ∼ d→0+ - 1 2π x d m 2 m-1 k m/2 ln k ∼ d→0+ 1 2π ln d.
The case Re m < 1 is analogous. Now, we can prove the main result of this section,

Theorem 4.4. Let m ∈ C. For (x, y) ∈ H + and (ξ, η) ∈ H + , E m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 m/2 if Re m ≥ 1 and E m (x, y, ξ, η) = ξ x m-1 E 2-m (x, y, ξ, η) = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 1-m 2 if Re m < 1
is a fundamental solution on H + for L m,ξ,η at the fixed point (x, y) ∈ H + , which means that in the sense of distributions on H + :

L m,ξ,η E m (x, y, ξ, η) = δ (x,y) (ξ, η).
Moreover, if (ξ, η) ∈ H + is fixed, then in the sense of distributions on H + :

L m,x,y E m (x, y, ξ, η) = δ (ξ,η) (x, y), which means that E m is a fundamental solution on H + of L m,x,y at the fixed point (ξ, η) ∈ H + .
Proof. Let m ∈ C and u ∈ D(H + ). Let (x, y) ∈ H + and ε > 0 such that D((x, y), ε) ⊂ H + where D((x, y), ε) is the disk of center (x, y) and of radius ε.

We put

I ε := H + \D((x,y),ε) L m (u)(ξ, η)E m (x, y, ξ, η)dξdη = = H + \D((x,y),ε) (L m (u)(ξ, η)E m (x, y, ξ, η) -u(ξ, η)L m (E m )(x, y, ξ, η)) dξdη because L m (E m ) = 0 on H + \ D((x, y), ε
). An elementary calculation gives us

L m (u)E m -uL m (E) = ∂ ξ (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m +∂ η ((∂ η u)E m -u(∂ η E m )) .
We will recall the Green formula in the framework that will be useful to us here.

Recall.

Let Ω be an open set of R 2 whose boundary is piecewise C 1differentiable. By denoting n the outer unit normal vector to ∂Ω and ds the arc length element on ∂Ω (positively oriented), if

X = (X 1 , X 2 ) : Ω → C 2 is a C 1 vector field, then Ω div X(x, y)dxdy = ∂Ω X(x, y) • n(x, y)ds
With this reminder applied to the open set Ω = U \ D((x, y), ε) where U is a regular open set of H + containing the support of u, we have

I ε = - t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m cos t+ + ((∂ η u)E m -u(∂ η E m )) sin t εdt Proposition 4.2 shows that t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) [(∂ ξ u) + m ξ u] cos t + (∂ η u) sin t E m εdt -→ ε→0+ 0 because lim ε→0 ε ln ε = 0.
Then, if we want to prove that lim ε→0 I ε exists, we have to prove the existence of lim

ε→0 t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u ((∂ ξ E m ) cos t + (∂ η E m ) sin t) ε dt,
and this limit will be equal to the limit of I ε . Now, we assume that Re m ≥ 1. We denote J ε the integral in the previous expression. A computation gives

J ε = - m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u ξ m-1 ε m π 0 sin m-1 θ dθ 1 + k sin 2 θ 2 m/2 ε cos t dt J ε,1 + + m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u ξ m ε m+2 π 0 sin m-1 θ dθ 1 + k sin 2 θ 2 m/2+1 ε 2 dt J ε,2 + + m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u ξ m ε m+2 π 0 2x sin 2 θ 2 sin m-1 θ dθ 1 + k sin 2 θ 2 m/2+1 ε cos t dt J ε,3
where k = 4xξ ε 2 . We have the following propositions :

Proposition 4.5. When k → +∞ and m ∈ C π θ=0 sin 2 θ 2 sin m-1 θ dθ (1 + k sin 2 θ 2 ) m/2+1 ∼ k→+∞ 2 m-1 k m 2 +1 ln k.
Proof. We put u = sin θ 2 , this integral is equal to

2 m 1 0 u m+1 (1 -u 2 ) m-2 2 du (1 + ku 2 ) m/2+1 = 2 m k m/2+1 1 0 u m+1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2+1 . However 1 0 u m+1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2+1 - 1 0 u m+1 du ( 1 k + u 2 ) m/2+1 = - 1 0 u m+1 1 k + u 2 m/2+1 (1-(1-u 2 ) m-2 2 )du -→ k→+∞ - 1 0 u m+1 (u 2 ) m/2+1 (1 -(1 -u 2 ) m- 2
2 )du = -

1 0 1 -(1 -u 2 ) m-2 2 u du.
The change of variable u = 1 √ k sh t gives us

1 0 u m+1 du ( 1 k + u 2 ) m/2+1 = argsh √ k 0 th m+1 t dt
Since th m+1 t tends to 1 when t → +∞, it follows that when k → +∞ argsh

√ k 0 th m+1 dt ∼ k→+∞ argsh √ k 0 dt = argsh √ k ∼ k→+∞ 1 2 ln k.
The proposition is well proven.

Proposition 4.6. When k → +∞ and m ∈ C π θ=0 sin m-1 θ dθ (1 + k sin 2 θ 2 ) m/2+1 ∼ k→+∞ 2 m mk m 2
Proof. Putting as previously u = sin θ 2 , this integral is equal to

2 m 1 0 u m-1 (1 -u 2 ) m-2 2 du (1 + ku 2 ) m/2+1 = 2 m k m/2+1 1 0 u m-1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2+1 . However 1 0 u m-1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2+1 - 1 0 u m-1 du ( 1 k + u 2 ) m/2+1 = - 1 0 u m-1 1 k + u 2 m/2+1 (1-(1-u 2 ) m- 2
2 )du

We first estimate the right hand side of this equality :

1 0 u m-1 1 k + u 2 m/2+1 (1 -(1 -u 2 ) m- 2
2 )du -

1 0 u m-1 1 k + u 2 m/2+1 m -2 2 u 2 du = 1 0 u m-1 1 k + u 2 m/2+1 1 - m -2 2 u 2 -(1 -u 2 ) m-2 2 du -→ k→+∞ 1 0 u m-1 (u 2 ) m/2+1 1 - m -2 2 u 2 -(1 -u 2 ) m-2 2 du = 1 0 1 -m-2 2 u 2 -(1 -u 2 ) m-2 2 u 3 du. ( * )
As seen in the proof of Proposition 4.5, we have

m -2 2 1 0 u m+1 ( 1 k + u 2 ) m 2 +1 du ∼ k→+∞ m -2 4 ln k. ( * * )
Through (*) and (**), one obtains :

1 0 u m-1 1 k + u 2 m/2+1 (1 -(1 -u 2 ) m-2 2 )du ∼ k→+∞ m -2 4 ln k.
The change of variable u = 1 √ k sh t gives us

1 0 u m-1 du ( 1 k + u 2 ) m/2+1 = k argsh √ k 0 th m-1 t ch 2 t dt = k m th m argsh √ k .
It follows that when k → +∞,

1 0 u m-1 du ( 1 k + u 2 ) m/2+1 ∼ k→+∞ k m .
We thus obtain

1 0 u m-1 (1 -u 2 ) m-2 2 du ( 1 k + u 2 ) m/2+1 ∼ k→+∞ k m .
And

π θ=0 sin m-1 θ dθ (1 + k sin 2 θ 2 ) m/2+1 ∼ k→+∞ 2 m mk m 2
and this completes the proof.

Let us return to the proof of Theorem 4.4. The Proposition 4.3 shows that

J ε,1 ∼ ε→0+ - m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u x m-1 ε m 2 m-1 k m/2 (ln k) ε cos t dt ∼ ε→0+ + m 2πx ε ln ε t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u(x + ε cos t, y + ε sin t) cos t dt
which tends to 0. The Proposition 4.5 shows that

J ε,3 ∼ ε→0+ m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u x m ε m+2 (2x) 2 m-1 k m/2+1 (ln k) ε cos t dt ∼ ε→0+ - m 4πx ε ln ε t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u(x + ε cos t, y + ε sin t) cos t dt
which tends to 0. Finally, the proposition 4.6 shows that

J ε,2 ∼ ε→0+ m 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u x m ε m+2 2 m mk m/2 ε 2 dt ∼ ε→0+ 1 2π t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) u(x + ε cos t, y + ε sin t)dt -→ ε→0+ u(x, y).
So we have proved that for all m ∈ C such that Re m > 0, lim

ε→0+ H + \D((x,y),ε) L m (u)(ξ, η)E m (x, y, ξ, η)dξdη = = H + L m (u)(ξ, η)E m (x, y, ξ, η)dξdη = u(x, y)
therefore that E m is indeed a fundamental solution of L m for all m ∈ C with Re m > 0. Proof for m ∈ C with Re m ≤ 1 is completely similar. We also have the dual assertions for fundamental solutions of L m for all m ∈ C thanks to Proposition 2.1.

The following proposition is roughly a consequence of the previous theorem. It is of course a very classical proposition : we juste recall very shortly the proof.

Proposition 4.7. Let m ∈ C and let Ω be a relatively compact open set of H + whose boundary is piecewise C 1 -differentiable. Then, for (x, y) ∈ Ω and u ∈ C 2 (Ω), by denoting n the outer unit normal vector to ∂Ω and ds the arc length element on ∂Ω (positively oriented), we have

u(x, y) = Ω L m (u)E m dξdη - ∂Ω (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m , (∂ η u)E m -u(∂ η E m ) • n ds
where u := u(ξ, η) and E m := E m (x, y, ξ, η).

Proof. Indeed, if u ∈ C 2 (Ω), we have for (x, y) ∈ Ω and ε > 0 such that D((x, y), ε) ⊂ Ω :

Ω\D((x,y),ε) L m (u)E m dξ dη = Ω\D((x,y),ε) (L m (u)E m -L m (E m )u)dξ dη.
Thanks to the Green formula previously recalled, this last integral is equal to

∂Ω (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m , (∂ η u)E m -u(∂ η E m ) • n ds - t∈[0,2π] (ξ,η)=(x,y)+ε(cos t,sin t) (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m cos t+ + ((∂ η u)E m -u(∂ η E m )) sin t εdt,
and from what we saw in the previous proof, this last expression tends to

∂Ω (∂ ξ u)E m -u(∂ ξ E m ) + m ξ uE m , (∂ η u)E m -u(∂ η E m ) • n ds + u(x, y), when ε → 0. Due to integrability E m near (x, y), we get lim ε→0 Ω\D((x,y),ε) L m (u)E m dξ dη = Ω L m (u)E m dξ dη,
and the proof of the proposition is complete.

Liouville-type result and decomposition theorem for the axisymmetric potentials

In the previous section, we have just seen that there are two different expressions of the fundamental solutions depending on the values of m. For the rest, each of the expressions have different behaviors according to the value of m. We will look at the two cases separately : Re m < 1 and Re m ≥ 1.

More specifically, we need fundamental solutions which vanish at the boundary of H + , which means that it tend to zero on the y-axis and to zero at infinity. For Re m < 1, the formula

E m (x, y, ξ, η) = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 1-m 2
shows that E m satisfies this property (E m (x, y, •, •) tends to 0 when x → 0+ and (x, y) → +∞).

For Re m ≥ 1, the expression

E m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ dθ [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m/2 no longer satisfies this property. Contrariwise, E m (x, y, ξ, η) -E m (-x, y, ξ, η)
is also a fundamental solution on H + , and satisfies this property. Then, we will put -For Re m < 1 :

F m (x, y, ξ, η) = E m (x, y, ξ, η) -For Re m ≥ 1 : F m (x, y, ξ, η) = E m (x, y, ξ, η) -E m (-x, y, ξ, η).
We will need the following definition of convergence to the boundary of H + .

Definition. Let u : H + → R be a function defined on H + . We write lim

∂H + u = 0 if and only if ∀ε > 0, ∃N ∈ N, ∀n ≥ N, ∀(x, y) ∈ H + , x ≤ 1 n or (x, y) ≥ n =⇒ |u(x, y)| ≤ ε.
In other words, this amounts to considering that the boundary ∂H + of H + consists of y-axis points and points at infinity and to say that the concept of punctual convergence to the boundary of H + is a uniform convergence.

Indeed, we do not need the uniform convergence. More precisely, we have the following proposition : Let ε > 0. There is A > 0 such that for all (ξ, η) ∈ H + ,

ξ 2 + η 2 ≥ A ⇒ |u(ξ, η)| ≤ ε.
Similarly, for all y ∈ R, there is α y ∈]0, 1[ such that for all (ξ, η)

∈ H + ξ 2 + (η -y) 2 < α y ⇒ |u(ξ, η)| ≤ ε.
The interval [-A, A] is compact. By the Lebesgue covering lemma, there is α > 0 such that for all y ∈ [-A, A], the ball B(y , α) is included in one of the balls B(y, α y ) with y ∈ [-A, A].

In particular, if (ξ, η) ∈ H + is such that 0 < ξ < α, then |u(ξ, η)| ≤ ε. This completes the proof.

The following proposition is a Liouville-type result for the axisymmetric potentials in the right half-plane and this result is not immediate because there is the loss of strict ellipticity of the Weinstein operator on the y-axis. In [START_REF] Baratchart | Pseudo-holomorphic functions at the critical exponent[END_REF] (see Theorem 7.1), we can found an interesting result on the description of a class of non-strictly elliptic equations with unbounded coefficients.

Proposition 5.2. Let u ∈ C 2 (H + ) such that L m u = 0 and lim

∂H + u = 0. Then u ≡ 0 on H + .
Proof. For (ξ, η) ∈ H + and N ∈ N * , we define

φ N (ξ, η) = θ 1 (N ξ)θ 2 ξ N θ 2 η N
where θ 1 and θ 2 are smooth functions on R, valued on [0, 1] and such that θ

1 (t) = 1 for t ≥ 1, θ 1 (t) = 0 for t ≤ 1 2 , θ 2 (t) = 1 for t ∈ -1 2 , 1 2
and θ 2 (t) = 0 for t ∈ R \ ]-1, 1[. We assume that all derivatives of θ 1 and θ 2 vanish at -1, - 1 2 , 1 2 , 1 .

1

1 1 2 θ 1 1 1 1 2 θ 2 If u ∈ C 2 (H + ) satisfies L m u = 0, then uφ N ∈ C 2 (H +
) and is compactly supported on H + . Throughout the following, we fix (x, y) ∈ H + . For N sufficiently large, thanks to Proposition 4.7 (true if E m is replaced by F m ), we have

u(x, y) = u(x, y)φ N (x, y) = H + L m (uφ N )F m dξdη (because the function L m (uφ N ) is identically zero in a neighborhood of the singularity of F m ), thus u(x, y) = H + [L m (u)φ N + uL m (φ N ) + 2∇u • ∇φ N ]F m dξdη = H + u[L m (φ N )F m -2 div (F m ∇φ N )]dξdη = D 1 ∪•••∪D 8 u[L m (φ N )F m -2 div (F m ∇φ N )]dξdη = - D 1 ∪•••∪D 8 u[L -m (φ N )F m + 2∇F m • ∇φ N ]dξdη
where D 1 , . . . , D 8 are the following domains (which depend of N) :

D 1 = 1 2N , 1 N × - N 2 , N 2
, D 2 = 1 N , N 2 × N 2 , N , D 3 = N 2 , N × - N 2 , N 2
, D 4 = 1 N , N 2 × -N, - N 2 , , D 5 = 1 2N , 1 N × N 2 , N , D 6 = N 2 , N × N 2 , N , D 7 = N 2 , N × -N, - N 2 and D 8 = 1 2N , 1 N × -N, - N 2 .
Figure :

Domains D i D 2 D 4 D 6 D 7 D 3 D 1 D 5 D 8 1 2N 1 N N N/2 φ N ≡ 1 -N -N/2 N N/2 ξ η Since lim ∂H + u = 0, then u N := sup (ξ,η)∈D 1 ∪•••∪D 8 |u(ξ, η)| -→ N →+∞ 0.
We will estimate each integrals supported on D 1 , . . . , D 8 . For this, we need the following lemmas which will give us estimates of each terms when N tends to infinity. We recall that, if (u N ) N and (v N ) N are complex sequences, On D 3 , we have

u N = O(v N )
sup ∂φ N ∂ξ = O 1 N and sup ∂φ N ∂η = 0.
On D 5 ∪ D 8 , we have

sup ∂φ N ∂ξ = O(N ) and sup ∂φ N ∂η = O 1 N .
On D 6 ∪ D 7 , we have

sup ∂φ N ∂ξ = O 1 N and sup ∂φ N ∂η = O 1 N . On D 1 ∪ D 5 ∪ D 8 , we have sup |L -m (φ N )| = O(N 2 ). On D 2 ∪ D 3 ∪ D 4 ∪ D 6 ∪ D 7 , we have sup |L -m (φ N )| = O 1 N 2 . Proof. *For (ξ, η) ∈ D 1 , φ N (ξ, η) = θ 1 (N ξ) and thus ∂φ N ∂ξ (ξ, η) = N θ 1 (N ξ) , ∂φ N ∂η (ξ, η) = 0, L -m φ N (ξ, η) = N 2 θ 1 (N ξ) - m N ξ θ 1 (N ξ),
which give us sup

D 1 ∂φ N ∂ξ = O(N ), sup D 1 ∂φ N ∂η = 0, sup D 1 |L -m (φ N )| = O(N 2 )
since the derivatives of θ 1 are bounded and for (ξ, η)

∈ D 1 , one gets ξ ≥ 1 2N . *For (ξ, η) ∈ D 2 , φ N (ξ, η) = θ 2 η N and thus ∂φ N ∂ξ (ξ, η) = 0 , ∂φ N ∂η (ξ, η) = 1 N θ 2 η N , L -m φ N (ξ, η) = 1 N 2 θ 2 η N , which give us sup D 2 ∂φ N ∂ξ = 0, sup D 2 ∂φ N ∂η = O 1 N , sup D 2 |L -m (φ N )| = O 1 N 2 *So does same with D 4 . * For (ξ, η) ∈ D 3 , φ N (ξ, η) = θ 2 ξ N
and thus

∂φ N ∂ξ (ξ, η) = 1 N θ 2 ξ N , ∂φ N ∂η (ξ, η) = 0, L -m φ N (ξ, η) = 1 N 2 θ 2 ξ N - 1 N m ξ θ 2 ξ N ,
which give us sup

D 3 ∂φ N ∂ξ = O 1 N , sup D 3 ∂φ N ∂η = 0, sup D 3 |L -m (φ N )| = O 1 N 2 * For (ξ, η) ∈ D 5 , φ N (ξ, η) = θ 1 (N ξ)θ 2 η N and thus ∂φ N ∂ξ (ξ, η) = N θ 1 (N ξ)θ 2 η N , ∂φ N ∂η (ξ, η) = 1 N θ 1 (N ξ)θ 2 η N , L -m φ N (ξ, η) = N 2 θ 1 (N ξ)θ 2 η N + 1 N 2 θ 1 (N ξ)θ 2 η N - m ξ N θ 1 (N ξ)θ 2 η N which give us sup D 5 ∂φ N ∂ξ = O(N ), sup D 5 ∂φ N ∂η = O 1 N , sup D 5 |L -m (φ N )| = O(N 2 ). * So does same with D 8 . *For (ξ, η) ∈ D 6 , φ N (ξ, η) = θ 2 ξ N θ 2 η N and thus ∂φ N ∂ξ (ξ, η) = 1 N θ 2 ξ N θ 2 η N , ∂φ N ∂η (ξ, η) = 1 N θ 2 ξ N θ 2 η N , L -m φ N (ξ, η) = 1 N 2 θ 2 ξ N θ 2 η N + 1 N 2 θ 2 ξ N θ 2 η N - m N ξ θ 2 ξ N θ 2 η N which give us sup D 6 ∂φ N ∂ξ = O 1 N , sup D 6 ∂φ N ∂η = O 1 N , sup D 6 |L -m (φ N )| = O 1 N 2 .
* So does same with D 7 . Hence the lemma resulting.

We now estimate the following quantities for i ∈ {1, . . . , 8} :

D i |F m | dξdη, D i |∂ ξ F m | dξdη et D i |∂ η F m | dξdη.
Lemma 5.4. For Re m < 1, we have :

-for i = 1 :

D i |F m |dξ dη = O 1 N 2 , D i ∂F m ∂ξ dξdη = O 1 N .
-for i = 2, 4 :

D i |F m |dξ dη = O N 2 , D i ∂F m ∂η dξdη = O (N ) .
-for i = 3 :

D i |F m |dξ dη = O N 2 , D i ∂F m ∂ξ dξdη = O (N ) .
-for i = 5, 8 :

D i |F m |dξ dη = O 1 N 2 , D i ∂F m ∂ξ dξdη = O 1 N , D i ∂F m ∂η dξdη = O 1 N 2 .
-for i = 6, 7 :

D i |F m |dξ dη = O N 2 , D i ∂F m ∂ξ dξdη = O (N ) , D i ∂F m ∂η dξdη = O (N ) .
Proof. For Re m < 1, we have

F m (ξ, η) = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 1-m 2 .
therefore there is a constant C 1 such that for all (ξ, η) ∈ H + , we have

|F m (ξ, η)| ≤ C 1 ξ [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 .
(5.1)

Similarly, we have

∂F m ∂ξ = F m ξ - ξx 1-m 2π (m -2) π θ=0 [(ξ -x) + 2x sin 2 θ 2 ] sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 2-m 2 ,
and as before, as

∀θ ∈ [0, π], [(ξ -x) + 2x sin 2 θ 2 ] sin 1-m θ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 2-m 2 ≤ |(ξ -x) + 2x sin 2 θ 2 | (x -ξ) 2 + (η -y) 2 2-Re m 2 = |ξ -x cos θ| (x -ξ) 2 + (η -y) 2 2-Re m 2 ≤ ξ + x (x -ξ) 2 + (η -y) 2 2-Re m 2 ,
there exists a constant C 2 such that for all N large enough and for all (ξ, η) ∈ H + , we have

∂F m ∂ξ ≤ C 2    1 (x -ξ) 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) (x -ξ) 2 + (η -y) 2 2-Re m 2    .
(5.2) Finally, as

∂F m ∂η = (2 -m)(η -y) ξx 1-m 2π π θ=0 sin 1-m θ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 2-m 2 ,
there exists a constant C 3 such that for all N large enough and for all (ξ, η) ∈ H + , we have

∂F m ∂η ≤ C 3 ξ |η -y| 3-Re m .
(5.3)

Using these inequalities, we estimate integrals of these functions on the domains D i .

On D 1 : Inequality (5.1) give us

D 1 |F m |dξdη = O(1) ξ= 1 N ξ= 1 2N η= N 2 η=-N 2 ξ dξ dη [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 = O(1/N 2 ) η= N 2 η=-N 2 dη (x -1 N ) 2 + (η -y) 2 1-Re m 2 = O(1/N 2 ).
Then, thanks to (5.2), we have

D 1 ∂F m ∂ξ dξdη = O(1) ξ= 1 N ξ= 1 2N η= N 2 η=-N 2    1 (x -ξ) 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) (x -ξ) 2 + (η -y) 2 2-Re m 2    dξdη = O(1/N ) η= N 2 η=-N 2 dη (x -1 N ) 2 + (η -y) 2 1-Re m 2 + O(1/N 2 ) = O(1/N ).
On D 2 : due to inequality (5.1), we have

D 2 |F m |dξdη = O(1) N 2 ξ= 1 N N η= N 2 ξ dξdη [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 = O(1) N 2 ξ= 1 N N η= N 2 ξ |η -y| 2-Re m dξdη = O(N 2 ) N η= N 2 dη |η -y| 2-Re m = O(N 2 ) 1 (N -y) 1-Re m - 1 ( N 2 -y) 1-Re m = O(N Re m+1
). Then, thanks to (5.3), we have

D 2 ∂F m ∂η dξdη = O(1) N 2 ξ= 1 N N η= N 2 ξ |η -y| 3-Re m dξ dη = O(N 2 ) N η= N 2 dη |η -y| 3-Re m = O(N Re m )
On D 3 : due to inequality (5.1), we have

D 3 |F m |dξdη = O(1) N ξ= N 2 N 2 η=-N 2 ξ dξdη [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 = O(1) N ξ= N 2 N 2 η=-N 2 ξ dξdη (x -N 2 ) 2 + (η -y) 2 1-Re m = O(N 2 ) N 2 η=-N 2 dη (x -N 2 ) 2 + (η -y) 2 1-Re m 2 = O(N 2 ) N 2 η=-N 2 dη [1 + (η -y) 2 ] 1-Re m 2 = O(N 2 ).
Then, thanks to (5.2), we have

D 3 ∂F m ∂ξ dξdη = O(1) N ξ= N 2 N 2 η=-N 2    1 (x -ξ) 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) (x -ξ) 2 + (η -y) 2 2-Re m 2    dξdη = O(N ) N 2 η=-N 2 dη (x -N 2 ) 2 + (η -y) 2 1-Re m 2 +O(N 3 ) N 2 η=-N 2 dη (x -N 2 ) 2 + (η -y) 2 2-Re m 2 = O(N ) + O(N 3 ) N 2 η=-N 2 dη (x -N 2 ) 4-Re m = O(N ) + O(N Re m ) = O(N ).
On D 4 : this case is analogous to the case D 2 .

On D 5 : due to inequality (5.1), we have

D 5 |F m |dξdη = O(1) 1 N ξ= 1 2N N η= N 2 ξ dξdη [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 = O(1/N 2 ) N η= N 2 dη (η -y) 2-Re m = O(1/N 2 ) 1 (N -y) 1-Re m - 1 ( N 2 -y) 1-Re m = O(1/N 3-Re m
). Then, thanks to (5.2), we have

D 5 ∂F m ∂ξ dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2    1 (x -ξ) 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) (x -ξ) 2 + (η -y) 2 2-Re m 2    dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2    1 x -1 N 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) x -1 N 2 + (η -y) 2 2-Re m 2    dξdη = O 1 N The estimate (5.3) gives D 5 ∂F m ∂η dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2 ξ dξ dη |η -y| 3-Re m = O 1 N 2
On D 6 : due to (5.1), we have

D 6 |F m |dξdη = O(1) N ξ= N 2 N η= N 2 ξ dξdη [(x -ξ) 2 + (η -y) 2 ] 1-Re m 2 = O(N 2 ) N η= N 2 dη (η -y) 2-Re m = O(N 2 ) 1 (N -y) 1-Re m - 1 ( N 2 -y) 1-Re m = O(N 1+Re m
). Then, thanks to (5.2), we have

D 6 ∂F m ∂ξ dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2    1 (x -ξ) 2 + (η -y) 2 1-Re m 2 + ξ(x + ξ) (x -ξ) 2 + (η -y) 2 2-Re m 2    dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2 1 (η -y) 2-Re m + ξ(x + ξ) (η -y) 4-Re m dξdη = O(N ) + O(N 3 ) N η= N 2 dη (η -y) 4-Re m = O(N ) + O(N Re m ) = O(N ).
The estimate (5.3) gives

D 6 ∂F m ∂η dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2 ξ dξ dη |η -y| 3-Re m = O(N 2 ) η=N η= N 2 dη |η -y| 3-Re m = O(N Re m ).
On D 7 : this case is analogous to the case D 6 .

On D 8 : this case is analogous to the case D 5 .

Lemma 5.5. For Re m ≥ 1, all estimations obtained on the Lemma 5.4 are true.

Proof. For Re m ≥ 1, we have

F m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ 1 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m/2 - 1 [(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 ] m/2
dθ.

Since for all (ξ, η) ∈ H + , we have

(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 m/2 = x 2 + ξ 2 + 2xξ cos θ + (y -η) 2 m/2 , then for all (ξ, η) ∈ H + , (x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 m/2 ≥ (x -ξ) 2 + (y -η) 2 Re m 2 (5.4)
and there is a constant C 1 such that for all (ξ, η) ∈ H + , we have

|F m | ≤ C 1 ξ Re m ((x -ξ) 2 + (y -η) 2 )
Re m 2 .

(5.5)

This inequality does not suffice to estimate integrals supported on D 1 . We can improve inequality (5.5) as follows :

We rewrite F m as

F m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θK m (x, y, ξ, η, θ)dθ
where

K m (x, y, ξ, η, θ) = 1 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m/2 - 1 [(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 ] m/2
.

For (x, y) ∈ H + fixed, θ ∈ [0, π] fixed and η ∈ R fixed, we define the function g m on [-1/N, 1/N ] with 1/N < x by

g m (ξ) = 1 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m/2
. This function is well defined because

(x-ξ) 2 +4xξ sin 2 θ 2 +(y-η) 2 = x 2 +ξ 2 -2xξ cos θ+(y-η) 2 ≥ (x-|ξ|) 2 +(y-η) 2
and this last term is larger than (x -1/N ) 2 > 0.

We have

K m (x, y, ξ, η, θ) = g m (ξ) -g m (-ξ) thus |K m (x, y, ξ, η, θ)| ≤ 2ξ sup [-ξ,ξ] |g m | ≤ 2|m|ξ |ξ -x| + 2x [(x -ξ) 2 + (y -η) 2 ] 1+ 1 2 Re m
, which implies that there exists a constant c 1 such that

∀(ξ, η) ∈ D 1 , |F m | ≤ c 1 ξ Re m+1 (x -ξ) 2 + (η -y) 2 1+ 1 2 Re m . (5.6)
Similarly, we have

∂F m ∂ξ = m F m ξ + mξ m 2π π θ=0 sin m-1 θ (ξ -x) + 2x sin 2 θ 2 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 - (ξ + x) -2x sin 2 θ 2 [(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1
dθ.

(5.7) and as before,

∀θ ∈ [0, π], [(ξ -x) + 2x sin 2 θ 2 ] sin m-1 θ [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 ≤ (ξ -x) + 2x sin 2 θ 2 [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 = |ξ -x cos θ| [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 ≤ ξ + x [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1
and thanks to (5.4) :

∀θ ∈ [0, π], [(ξ + x) -2x sin 2 θ 2 ] sin m-1 θ [(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 ≤ (ξ + x) -2x sin 2 θ 2 [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 = |ξ + x cos θ| [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 ≤ ξ + x [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1
.

Those estimations, the formula (5.7) and the inequality (5.5) show that there is a constant C 2 such that large enough N and for all (ξ, η) ∈ H + , we have

∂F m ∂ξ ≤ C 2 ξ Re m-1 [(x -ξ) 2 + (y -η) 2 ] Re m 2 + ξ Re m (ξ + x) [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1
.

(5.8) We can improve this inequality on D 1 , for this, we need to use the inequality (5.6) instead of (5.5) and we obtain that there is two constants C 2 and C 2 (which do not depend of N ) such that for all (ξ, η)

∈ D 1 ∂F m ∂ξ ≤ C 2 ξ Re m [(x -ξ) 2 + (y -η) 2 ] 1+ Re m 2 + ξ Re m (ξ + x) [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 ≤ C 2 ξ Re m [(x -ξ) 2 + (y -η) 2 ] 1+ Re m 2 (5.9)
Finally,

∂F m ∂η = m(η -y)ξ m 2π π θ=0 sin m-1 θ 1 [(x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1 - 1 [(x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 ] m 2 +1
dθ.

Similarly, there is a constant C 3 such that for all large enough N and for all (ξ, η) ∈ H + , we have

∂F m ∂η ≤ C 3 |η -y|ξ Re m ((x -ξ) 2 + (y -η) 2 ) Re m 2 +1
.

(5.10)

Thanks to those inequalities, we will estimate the integrals of those functions on each domain D i .

On D 1 : due to (5.6), we have

D 1 |F m |dξdη = O(1) ξ= 1 N ξ= 1 2N η= N 2 η=-N 2 ξ Re m+1 dξdη (x -ξ) 2 + (η -y) 2 1+ 1 2 Re m = O(N ) ξ= 1 N ξ= 1 2N ξ Re m+1 dξ = O(N ) 1 N Re m+2 - 1 2N
Re m+2

= O(1/N Re m+1 ). Then thanks to (5.9), we have

D 1 ∂F m ∂ξ dξdη = O(1) ξ= 1 N ξ= 1 2N η= N 2 η=-N 2 ξ Re m dξdη (x -ξ) 2 + (η -y) 2 1+ 1 2 Re m . = O(N ) ξ= 1 N ξ= 1 2N ξ Re m dξ = O(1/N Re m ).
On D 2 : due to (5.5), we have

D 2 |F m |dξdη = O(1) N 2 ξ= 1 N N η= N 2 ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 = O(1) N 2 ξ= 1 N N η= N 2 ξ Re m dξdη |y -N 2 | Re m = O(N 2
), because we integrate a bounded function (independently of N ) on a domain with measure controlled by O(N 2 ).

Then, the inequality (5.10) implies

D 2 ∂F m ∂η dξdη = O(1) N 2 ξ= 1 N N η= N 2 |η -y|ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 +1 = O(1) N 2 ξ= 1 N N η= N 2 ξ Re m dξdη |y -η| Re m+1 = O(1) N 2 ξ= 1 N N η= N 2 N Re m dξdη | N 2 -y| Re m+1 = O(N ).
On D 3 : due to (5.5), we have

D 3 |F m |dξdη = O(1) N ξ= N 2 N 2 η=-N 2 ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 = O(1) N ξ= N 2 N 2 η=-N 2 ξ Re m dξdη (x -N 2 ) 2 + (y -η) 2 Re m 2 = O(N Re m+1 ) N 2 η=-N 2 dη (x -N 2 ) 2 + (y -η) 2 Re m 2 = O(N 2 ).
Then, thanks to (5.8), we have

D 3 ∂F m ∂ξ dξdη = O(1) N ξ= N 2 N 2 η=-N 2 ξ Re m-1 [(x -ξ) 2 + (y -η) 2 ] Re m 2 + ξ Re m (ξ + x) [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 dξdη = O(N Re m ) N 2 η=-N 2 dη [(x -N 2 ) 2 + (y -η) 2 ] Re m 2 +O(N Re m+2 ) N 2 η=-N 2 dη [(x -N 2 ) 2 + (y -η) 2 ] Re m 2 +1 = O(N ) + O(N ) = O(N ).
On D 4 : this case is analogous to the case D 2 .

On D 5 : due to (5.5), we have

D 5 |F m |dξdη = O(1) 1 N ξ= 1 2N N η= N 2 ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 = O(1/N Re m+1 ) N η= N 2 dη |y -N 2 | Re m = O(1/N 2Re m
). Then, thanks to (5.8), we have

D 5 ∂F m ∂ξ dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2 ξ Re m-1 [(x -ξ) 2 + (y -η) 2 ] Re m 2 + ξ Re m (ξ + x) [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2 ξ Re m-1 |y -η| Re m + ξ Re m (ξ + x) |y -η| Re m+2 dξdη = O(1/N 2Re m-1 ).
With the inequality (5.10), we have

D 5 ∂F m ∂η dξdη = O(1) ξ= 1 N ξ= 1 2N η=N η= N 2 |η -y|ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 +1
= O( 1)

ξ= 1 N ξ= 1 2N η=N η= N 2 ξ Re m dξdη |y -η| Re m+1 = O(1/N 2Re m+1 )
On D 6 : due to (5.5), we have

D 6 |F m |dξdη = O(1) N ξ= N 2 N η= N 2 ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 = O(N Re m+1 ) N η= N 2 dη ( N 2 -y) Re m = O(N 2
). Then, thanks to (5.8), we obtain

D 6 ∂F m ∂ξ dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2 ξ Re m-1 [(x -ξ) 2 + (y -η) 2 ] Re m 2 + ξ Re m (ξ + x) [(x -ξ) 2 + (y -η) 2 ] Re m 2 +1 dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2 ξ Re m-1 |y -η| Re m + ξ Re m (ξ + x) |y -η| Re m+2 dξdη = O(N ) + O(N Re m+2 ) η=N η= N 2 dη |y -η| Re m+2 = O(N ).
Finally, the inequality (5.10) implies

D 6 ∂F m ∂η dξdη = O(1) ξ=N ξ= N 2 η=N η= N 2 |η -y|ξ Re m dξdη ((x -ξ) 2 + (y -η) 2 ) Re m 2 +1 = O(N Re m+1 ) η=N η= N 2 dη |y -η| Re m+1 = O(N ).
On D 7 : this case is analogous to the case D 6 .

On D 8 : this case is analogous to the case D 5 .

In the following table, we summarize the results obtained on the previous lemmas :

i sup D i |L -m φ N | D i |F m |dξdη (|∂ ξ φ N |, |∂ η φ N |) D i |∂ ξ F m | D i |∂ η F m | 1 O(N 2 ) O(1/N 2 ) (O(N ), 0) O( 1 N ) × 2 O(1/N 2 ) O(N 2 ) (0, O( 1 N )) × O(N ) 3 O(1/N 2 ) O(N 2 ) (O( 1 N ), 0) O(N ) × 4 O(1/N 2 ) O(N 2 ) (0, O( 1 N )) × O(N ) 5 O(N 2 ) O(1/N 2 ) (O(N ), O( 1 N )) O( 1 N ) O( 1 N 2 ) 6 O(1/N 2 ) O(N 2 ) (O( 1 N ), O( 1 N )) O(N ) O(N ) 7 O(1/N 2 ) O(N 2 ) (O( 1 N ), O( 1 N )) O(N ) O(N ) 8 O(N 2 ) O(1/N 2 ) (O(N ), O( 1 N )) O( 1 N ) O( 1 N 2 )
We can easily check that for each i ∈ {1, . . . , 8}, the quantities sup

D i |L -m φ N | D i |F m |, sup D i |∂ ξ φ N | D i |∂ ξ F m | and sup D i |∂ η φ N | D i |∂ η F m | stay bounded. Therefore, u (x, y) = o(1)
when N → +∞. Thus u ≡ 0 and this completes the proof of the Proposition 5.2.

Lemma 5.6. Let u ∈ D(H + ) and let (x, y) ∈ H + , we define

U (x, y) = H + u(ξ, η)F m (x, y, ξ, η)dξ dη, then lim (x,y)
→+∞ U = 0, and for all y ∈ R, lim

U = 0. (0,y)
Moreover, U ∈ C ∞ (H + \ supp u) and for all (x, y) ∈ supp u, we have L m,x,y U (x, y) = 0.

Proof. When (ξ, η) is fixed, and since

F m (x, y, ξ, η) = - ξx 1-m 2π π θ=0 sin 1-m θ dθ (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 1-m 2 for Re m < 1, then F m (x, y, ξ, η) -→ (x,
F m (x, y, ξ, η) = - ξ m 2π π θ=0 sin m-1 θ 1 (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 m 2 - - 1 (x + ξ) 2 -4xξ sin 2 θ 2 + (y -η) 2 m 2 dθ then F m (x, y, ξ, η) -→ (x,y) →+∞
0 and the first result of the lemma is shown.

For the second result, it suffices to see, for Re m < 1, that

F m (x, y, ξ, η) ∼ (x,y)→(0,y ) - ξx 1-m 2π[ξ 2 + (y -η) 2 ] 1-m/2 π 0 sin 1-m θ dθ
which implies the desired result. Now, we assume that Re m ≥ 1. Let (ξ, η) be fixed in the support of u, which is a compact set of H + . In particular, there exist M > 0 and α > 0 which do not depend of u such that (ξ, η) ≤ M et ξ ≥ 2α. Let y be in R.

By denotting for

x ∈ [-α, α], f m (x) = 1 (x -ξ) 2 + 4xξ sin 2 θ 2 + (y -η) 2 m 2
, By the mean value inequality, for x > 0 near 0, we have

|f m (x) -f m (0)| ≤ x sup [0,α] |f m | and |f m (-x) -f m (0)| ≤ x sup [-α,0] |f m |. then |f m (x) -f m (-x)| ≤ 2x sup [-α,α] |f m | ≤ 2x|m| 3M + α α Re m+2 .
In particular, sup

(ξ,η)∈supp u y∈R |F m (x, y)| = O(x)
when x → 0+. The second result is proved.

The last result can be deduced of the fact that if (x, y) = (ξ, η) are both in

H + , then L m,x,y F m (x, y, ξ, η) = 0. Remark 5.7. If U ∈ D(H + ), then L m,x,y U = u, but this identity is not necessary true if U ∈ D(H + ).
In particular, we can not say that in the Lemma 5.6, we have L m U = u. Now, we will prove a decomposition theorem for axisymmetric potentials, it is interesting to compare the following theorem to known result in [6, Theorem 2 section 4] (the fundamental difference is that in this work, the conductivity is not extended in all domain by reflection through the boundary ∂Ω).

Note that, due to our construction of the fundamental solutions, the proof of this theorem is more and less the same than the proof of the decomposition theorem, chapter 9, in [START_REF] Axler | Harmonic function theory[END_REF]. Note that in our situation, the domain of definition of our functions is H + , and not C.

u = v + w,
where v ∈ C 2 (Ω) satisfies L m v = 0 in Ω and w ∈ C 2 (H + \ K) satisfies L m w = 0 in H + \ K with lim ∂H + w = 0.
Proof. For E ⊂ C and ρ > 0, we define

E ρ = {x ∈ C, d(x, E) < ρ} (E ρ is a neighborhood of E).
At first, we assume that Ω is a relatively compact open set of H + . We choose ρ as small as K ρ and (∂Ω) ρ are disjoint. There is a function

ϕ ρ ∈ D(H + ) compactly supported on Ω \ K such that ϕ ρ ≡ 1 in a neighborhood of Ω \ (K ρ ∪ (∂Ω) ρ ). Figure : ϕ ρ ≡ 1 on the gray domain x y K K ρ Ω \ {K ρ ∪ (∂Ω) ρ } ∂Ω (∂Ω) ρ ρ ρ For z = x + iy ∈ Ω \ (K ρ ∪ (∂Ω) ρ ), we denote F z (ζ) := F m (x, y, ξ, η) and L ζ := L m,ξ,η for ζ = ξ + iη,
Thanks to Proposition 4.7, we have

u(z) = uϕ ρ (z) = Ωρ F z (ζ)L ζ (uϕ ρ )(ζ)dξdη = (∂Ω)ρ F z (ζ)L ζ (uϕ ρ )(ζ)dξdη + Kρ F z (ζ)L ζ (uϕ ρ )(ζ)dξdη = v ρ (z) + w ρ (z).
Then, the last result of Lemma 5.6 shows us that v ρ satisfies L m v ρ = 0 on Ω\(∂Ω) ρ and w ρ satisfies L m w ρ = 0 on H + \K ρ . We also have lim ∂H + w ρ = 0. Now, we assume that σ < ρ. As previously, we obtain the decomposition

u = v σ + w σ on Ω \ (K σ ∪ (∂Ω) σ ). We claim that v ρ = v σ on Ω \ (∂Ω) ρ and w ρ = w σ on H + \ K ρ . To see this, note that if z ∈ Ω \ (K ρ ∪ (∂Ω) ρ ), then v ρ (z) + w ρ (z) = v σ (z) + w σ (z).
We will designate by (1) the Weinstein equation L m u = 0. Thus w ρ -w σ satisfies (1) on

H + \ K ρ , which is equal to v σ -v ρ on Ω \ (K ρ ∪ (∂Ω) ρ ), therefore v σ -v ρ extends to a solution of (1) on Ω \ (∂Ω) ρ .
Finally, w ρ -w σ extends to a solution of (1) on H + , and lim

∂H + w ρ -w σ = 0.
Due to Proposition 5.2, we have

w ρ = w σ ,
and then v ρ = v σ . For z ∈ Ω, we can define v(z) = v ρ (z) for ρ as small as z ∈ Ω \ (∂Ω) ρ .

Similarly, for z ∈ H + \ K, we put w(z) = w ρ (z) for small ρ. We have proved the desired decomposition u = v + w. Now, assume that Ω is an arbitrary domain of H + and let u be a solution of L m u = 0 on Ω \ K. We choose a ∈ H + and R large enough so that K ⊂ D(a, R) and D(a, R) be a relatively compact set of H + . Let ω = Ω∩D(a, R).

Note that K is a compact set of ω which is a relatively compact open set of H + and u satisfies (1) on ω \ K. Applying the results demonstrated for relatively compact open sets, we obtain u(z) = ṽ(z) + w(z)

for z ∈ ω \ K, where ṽ satisfies (1) on ω and w satisfies (1) on H + \ K with lim ∂H + w = 0. Note that V = u -w satisfies (1) on Ω \ K and V can be extended into a solution of (1) in a neighborhood of K because V = ṽ on ω.

The sum u = V + w provides us a desired decomposition of u.

As before, if we have another decomposition

u = v + w with v ∈ C 2 (Ω), L m v = 0 and with w ∈ C 2 (H + \ K), L m w = 0 and lim ∂H + w = 0, then we have V -v = w -w on Ω \ K.
The function w -w can be extended on H + into a solution of L m (w -w) = 0 on H + with lim ∂H + (w -w) = 0. Thanks to Proposition 5.2, we obtain w = w, then V = v, which completes the proof of the decomposition theorem.

The following proposition is a Poisson formula for the axisymmetric potentials in H + : Proposition 5.9. Let m ∈ C be such that Re m < 1 and u : R → R be a continuous and bounded function. Then there is a unique axisymmetric potential U ∈ C 2 (H + ) such that lim (x,y) →+∞ U (x, y) = 0 and for all y ∈ R, lim

U = u(y).

Moreover, we have for all (x, y) ∈ H + ,

U (x, y) = C m x 1-m ∞ η=-∞ u(η) dη (x 2 + (y -η) 2 ) 1-m 2 where C m = 1-m 2π π θ=0 sin 1-m θ dθ = 1 2 m π Γ 2 (1-m 2 ) Γ(1-m) . Proof. We define f (x, y) = x 1-m (x 2 +(y-η) 2 ) 1-m 2 .
To show that U is a solution of L m U = 0, it suffices to prove that L m f = 0 by differentiation under the integral sign. We have

∂ x f = (1 -m)x -m (x 2 + (y -η) 2 ) 1-m 2 - (2 -m)x 2-m (x 2 + (y -η) 2 ) 2-m 2 and ∂ xx f = - m(1 -m)x -m-1 (x 2 + (y -η) 2 ) 1-m 2 - (2 -m)(3 -2m)x 1-m (x 2 + (y -η) 2 ) 2-m 2 + (2 -m)(4 -m)x 3-m (x 2 + (y -η) 2 ) 3-m 2 and ∂ yy f = - (2 -m)x 1-m (x 2 + (y -η) 2 ) 2-m 2 + (2 -m)(4 -m)(y -η) 2 x 1-m (x 2 + (y -η) 2 ) 3-m 2 .
Then,

∆f = m(2 -m)x 1-m (x 2 + (y -η) 2 ) 2-m 2 - m(1 -m)x -m-1 (x 2 + (y -η) 2 ) 1-m 2
and we deduce that L m f (x, y) = 0. We have

U (x, y) = C m x 1-m ∞ η=-∞ u(η) dη (x 2 + (y -η) 2 ) 1-m 2 = C m x ∞ η=-∞ u(η) dη (1 + ( y-η x ) 2 ) 1-m 2
By a change of variable t = y-η x , we obtain

U (x, y) = C m ∞ t=-∞ u(y -tx) dt (1 + t 2 ) 1-m 2
Thanks to the dominated convergence theorem, it suffices to show that

C m ∞ t=-∞ dt (1 + t 2 ) 1-m 2 = 1 -m 2π π θ=0 sin 1-m θ dθ ∞ t=-∞ dt (1 + t 2 ) 1-m 2 = 1.
To see this, according [START_REF] Ablowitz | Complex Variables : Introduction and Applications[END_REF] (page 258), note that

∞ t=-∞ dt (1 + t 2 ) 1-m 2 = B 1 2 , 1 -m 2 = Γ(1/2)Γ 1-m 2 Γ 1 -m 2
where B is the Euler beta function and

1 -m 2π π θ=0 sin 1-m θ dθ = 1 -m 2π 2 1-m B 1 - m 2 , 1 - m 2 = 1 -m 2π 2 1-m Γ 2 1 -m 2 Γ (2 -m) .
Then, using the duplication formula for the Γ function,

Γ(2z) = π -1/2 2 2z-1 Γ(z)Γ z + 1 2
and the recurrence formula Γ(z + 1) = zΓ(z), we obtain the desired result,

Γ(1/2)Γ 1-m 2 Γ 1 -m 2 1 -m 2π 2 1-m Γ 2 1 -m 2 Γ (2 -m) = 1.
The uniqueness follows from the proposition 5.2. So, we proved the proposition.

Remark 5.10. One may question the existence of such a reproducing for-

mula if Re m ≥ 1. In fact, if m ∈ N * and if u ∈ C 2 (H + ) satisfies L m (u) = 0 on H + , then the function v defined on R m+2 by v(x 1 , . . . , x m+2 ) = u x 2 1 + • • • + x 2 m+1 , x m+2
is an harmonic function on (R m+1 ) * × R. In particular, if m ≥ 2, the Proposition 18 in [START_REF] Dautray | Analyse mathématique et calcul numérique pour les sciences et techniques[END_REF], page 310 shows that v can be extended to an harmonic function on R m+2 , which tends to 0 at infinity. We then deduce that v ≡ 0, implying u ≡ 0. This shows that solving L m (u) = 0 with u tending to 0 at infinity and with prescribed values of u on the y-axis is a problem which does not make sense. In this case, the fact that there is no solution to this Dirichlet problem is a consequence of the loss of ellipticity of L m on the boundary of H + . Therefore, we do not deal with the case Re m ≥ 1.

Fourier-Legendre decomposition

First, we will introduce a specific system of coordinates named bipolar coordinates (τ, θ) (see [START_REF] Lebedev | Special functions and their applications[END_REF]) and numerical applications on extremal bounded problems using this system of coordinates have been realized in [START_REF] Fischer | Solutions to conjugate Beltrami equations and approximation in generalized Hardy spaces[END_REF][START_REF] Fischer | Bounded extremal problems in Hardy spaces for the conjugate Beltrami equation in simplyconnected domains[END_REF][START_REF] Fischer | Some inverse problems around the Tokamak tore supra[END_REF]. Let α > 0. We suppose that there is a positive charge at A = (-α, 0) and a negative charge at B = (α, 0) (the absolute values of the two charges are identical). The potential generated by this charges at a point M is ln M A M B

(modulo a multiplicative constant). The bipolar coordinates are linked to the Cartesian coordinates by the following formulas :

x = α sh τ ch τ -cos θ , y = α sin θ ch τ -cos θ .
Let R > 0 and a = √ R 2 + α 2 , the disk of center (a, 0) and of radius R is defined in terms of bipolar coordinates by

τ ≥ τ 0 = ln a R + a 2 R 2 -1 = argch a R .
Moreover, the right half-plane is

H + = {(τ, θ) : τ ∈]0 + ∞], θ ∈ [0, 2π[}.
The level lines τ = τ 0 are circles of center(α coth τ 0 , 0) and radii α/sh τ 0 . This implies that for all τ 0 , τ 1 such that 0 < τ 0 < τ 1 , the set {(τ, θ), τ ≥ τ 0 } is a closed disk and the set {(τ, θ), 0 < τ < τ 1 } is the complement on H + of the closed disk {τ ≥ τ 1 }. x y

1 -1 τ = 1/2 τ = 1 τ = 1/3 τ = -1/3 θ = π/6 θ = π/3 θ = 11π/6 θ = 5π/3 τ = -1/2 τ = -1 θ = 0 θ = 0 τ = 0 θ = π
The following theorem is known for m = -1 by physicists ( [START_REF] Alladio | Analysis of mhd equilibria by toroidal multipolar expansions[END_REF][START_REF] Van Milligen | Expansion of vacuum magnetic fields in toiroidal harmonics[END_REF][START_REF] Segura | Evaluation of toroidal harmonics[END_REF][START_REF] Ch | Electrostatic problem for a torus and a disk[END_REF][START_REF] Rau | Developments in determining the gravitational potential using toroidal functions[END_REF][START_REF] Love | The dielectric ring in a uniform, axial, electrostatic field[END_REF]. We extend this result to complex values of m : Theorem 6.1. Let u be a solution of L m u = 0 in an open set of H + and putting

v m (τ, θ) = sh m-1 2 τ (ch τ -cos θ) -m/2 u(τ, θ)
where by definition,

sh m-1 2 τ (ch τ -cos θ) -m/2 = exp m -1 2 ln sh τ - m 2 ln(ch τ -cos θ) then ∂ 2 v m ∂τ 2 + ∂ 2 v m ∂θ 2 + coth τ ∂v m ∂τ + 1 4 - (m -1) 2 4 sh 2 τ v m = 0.
Proof. We have Thus, we obtain

∂u ∂τ = α 1 -ch τ cos θ (ch τ -cos θ) 2 ∂u ∂x - sh τ sin θ (ch τ -cos θ) 2
∂u ∂x = 1 α (1 -ch τ cos θ) ∂u ∂τ -sh τ sin θ ∂u ∂θ ,
and

∂ 2 u ∂τ 2 = α 2 (ch τ -cos θ) 4 (1 -ch τ cos θ) 2 ∂ 2 u ∂x 2 + sh 2 τ sin 2 θ ∂ 2 u ∂y 2 -2(1 -ch τ cos θ)sh τ sin θ ∂ 2 u ∂x∂y + α (ch τ -cos θ) 3 sh τ (cos 2 θ + ch τ cos θ -2) ∂u ∂x + sin θ ch 2 τ -2 + cos θch τ ∂u ∂y and ∂ 2 u ∂θ 2 = α 2 (ch τ -cos θ) 4 sh 2 τ sin 2 θ ∂ 2 u ∂x 2 + (ch τ cos θ -1) ∂ 2 u ∂y 2 +2(1 -ch τ cos θ)sh τ sin θ ∂ 2 u ∂x∂y + α (ch τ -cos θ) 3 sh τ (2 -cos 2 θ -cos θch τ ) ∂u ∂x
+ sin θ(2 -ch 2 τ -ch τ cos θ) ∂u ∂y In particular, we have

∂ 2 u ∂τ 2 + ∂ 2 u ∂θ 2 = α 2 (ch τ -cos θ) 2 ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 .
Therefore, we obtain

L m,x,y u = ch τ -cos θ α 2 ∂ 2 u ∂τ 2 + ∂ 2 u ∂θ 2 + m(1 -ch τ cos θ) sh τ (ch τ -cos θ) ∂u ∂τ - m sin θ ch τ -cos θ ∂u ∂θ .
We put

u(τ, θ) = (ch τ -cos θ) m/2 sh m-1 2 τ v m (τ, θ)
and we calculate L m,x,y u in terms of F (τ, θ). Denoting

r m ∂ 2 v m ∂τ 2 + ∂ 2 v m ∂θ 2 + ∂v m ∂τ 2 ∂r m ∂τ + m sh τ 1 -ch τ cos θ ch τ -cos θ r m + + ∂v m ∂θ 2 ∂r m ∂θ - m sin θ ch τ -cos θ r m +v m ∂ 2 r m ∂τ 2 + ∂ 2 r m ∂θ 2 + m(1 -ch τ cos θ) sh τ (ch τ -cos θ) ∂r m ∂τ - m sin θ ch τ -cos θ ∂r m ∂θ = 0 with 2 ∂r m ∂τ + m sh τ 1 -ch τ cos θ ch τ -cos θ r m = r m coth τ, 2 ∂r m ∂θ - m sin θ ch τ -cos θ r m = 0 and ∂ 2 r m ∂τ 2 + ∂ 2 r m ∂θ 2 + m(1 -ch τ cos θ) sh τ (ch τ -cos θ) ∂r m ∂τ - m sin θ ch τ -cos θ ∂r m ∂θ = 1 4 - (m -1) 2 4sh 2 τ r m .
And this completes the proof.

We seek v m by separation of variables : v m (τ, θ) = A m (τ )B m (θ). From the equation satified by v m (see Theorem 6.1), we obtain

A m A m + coth τ A m A m + 1 4 - (m -1) 2 4 sh 2 τ = - B m B m .
The term on the right depends only of θ and the left one depends only of τ , thus we deduce that both are constant. Let n ∈ C such that this constant is equal to n 2 . We then have

   A m + coth τ A m + 1 4 - (m -1) 2 4sh 2 τ -n 2 A m = 0, B m + n 2 B m = 0.
B m is naturally a 2π-periodic function (because θ represents an angle), therefore n should necessarily be an integer.

To examine the equation satisfied by A m , we carry out the following change of function

A m (τ ) = C m (ch τ ). Then, C m satisfies sh 2 τ C m (ch τ ) + 2 ch τ C m (ch τ ) + 1 4 -n 2 - (m -1) 2 4sh 2 τ C m (ch τ ) = 0
which can be rewritten as

(1-ch 2 τ )C m (ch τ )-2 ch τ C m (ch τ )+ n 2 - 1 4 - ((m -1)/2) 2 1 -ch 2 τ C m (ch τ ) = 0.
(LAH) This equation is named Hyperbolic Associated Legendre equation. Note that if we put z = ch τ and u

(z) = C m (ch τ ), then (1 -z 2 )u -2zu + ν(ν + 1) - µ 2 1 -z 2 u = 0 (LA) where ν = n - 1 2 and µ = m -1 2 .
This equation is named Associated Legendre equation, and it can be reduced to the Legendre equation if µ = 0 :

(1 -z 2 )u -2zu + ν(ν + 1)u = 0. (L)

Two independent solutions of (LA) are given in section 8 and denoted P µ ν (ch τ ) and Q µ ν (ch τ ). Starting from this investigation of solutions in the form of separate variables, we can state the following theorem Theorem 6.2. Let m ∈ C. Let 0 < τ 0 . Let u be a smooth solution of L m u = 0 on the disk τ ≥ τ 0 and let v be a smooth solution of L m v = 0 on H + \ {τ > τ 0 } which is the complement on H + of the disk {τ > τ 0 } and we assume that lim ∂H + v = 0. Then there are two sequences (a n ) n∈Z and (b n ) n∈Z of 2 (Z) (which are even rapidly decreasing) such that :

u = +∞ n=-∞ a n Q m-1 2 n-1 2 (ch τ )sh 1-m 2 τ (ch τ -cos θ) m 2 e inθ and v = +∞ n=-∞ b n P m-1 2 n-1 2 (ch τ )sh 1-m 2 τ (ch τ -cos θ) m 2 e inθ .
The sequence (a n ) is unique. In addition, the convergence of the first series is uniform on every compact set [τ 1 , τ 2 ] of the disk τ > τ 0 with τ 0 ≤ τ 1 < τ 2 . And the convergence of the second one is uniform on every compact set [τ 3 , τ 4 ] of the complement of the disk τ > τ 0 on H + with 0 < τ 3 < τ 4 ≤ τ 0 .

If Re m < 1, then the sequence (b n ) is unique.

Proof. Indeed, decomposing the function

θ → u(τ 0 , θ)(ch τ 0 -cos θ) -m/2 sh m-1 2 τ 0
by Fourier series with respect to the variable θ, to yield the Fourier expansion for u(τ 0 , •)

u(τ 0 , θ) = sh 1-m 2 τ 0 (ch τ 0 -cos θ) m 2 +∞ n=-∞ a n e inθ ,
where a n ∈ 2 (Z) satisfies

a n = 1 2π 2π 0 (ch τ 0 -cos θ) -m/2 sh m-1 2 τ 0 u(τ 0 , s)e -ins ds.
This function is a smooth function of the variable θ, therefore we deduce that the sequence (a n ) n is rapidly decreasing when |n| → +∞. The function

ũ(τ, θ) = sh 1-m 2 τ (ch τ -cos θ) m 2 +∞ n=-∞ a n Q m-1 2 n-1 2 (ch τ ) Q m-1 2 n-1 2 (ch τ 0 )
e inθ coincides with u on the circle τ = τ 0 . Moreover, thanks to the Proposition 8.1, we have when |n| → +∞,

Q m-1 2 n-1 2 (ch τ ) Q m-1 2 n-1 2 (ch τ 0 ) ∼ sh τ 0 sh τ e |n|(τ 0 -τ )
and this equivalence is uniform in all compact set [τ 1 , τ 2 ] with 0 < τ 0 ≤ τ 1 < τ 2 . It follows that the series of functions which defines ũ is normally converging on any compact sets [τ 1 , τ 2 ] of the disk τ ≥ τ 0 . So does same for derivatives with respect to τ and θ (which are expressed also with the Associated Legendre functions as mentioned in the section 8).

Particularly, the function ũ is well defined on the disk τ ≥ τ 0 and coincides with u on the circle τ = τ 0 . Due to the fact that the solution of an elliptic equation is uniquely determined by its boundary values (this follows from the maximum principle), we deduce that ũ the unique axisymmetric potential on the disk τ ≥ τ 0 which coincides with u on the circle τ = τ 0 . For v, the proof is completely similar. Indeed, decomposing the function

θ → v(τ 0 , θ)(ch τ 0 -cos θ) -m/2 sh m-1 2 τ 0
by Fourier series with respect to the variable θ, to yield the Fourier expansion for v(τ 0 , •)

v(τ 0 , θ) = sh 1-m 2 τ 0 (ch τ 0 -cos θ) m 2 +∞ n=-∞ b n e inθ , where b n ∈ 2 (Z) satisfies b n = 1 2π 2π 0 (ch τ 0 -cos θ) -m/2 sh m-1 2 τ 0 v(τ 0 , s)e -ins ds.
This function is a smooth function of the variable θ, therefore we deduce that the sequence (b n ) n is rapidly decreasing when |n| → +∞. The function ṽ(τ, θ) = sh

1-m 2 τ (ch τ -cos θ) m 2 +∞ n=-∞ b n P m-1 2 n-1 2 (ch τ ) P m-1 2 n-1 2 (ch τ 0 ) e inθ
coincides with v on the circle τ = τ 0 . Moreover, thanks to the Proposition 8.1, we have when |n| → +∞,

P m-1 2 n-1 2 (ch τ ) P m-1 2 n-1 2 (ch τ 0 ) ∼ sh τ 0 sh τ e |n|(τ -τ 0 )
and this equivalence is uniform in all compact set [τ 1 , τ 2 ] with 0 < τ 1 < τ 2 ≤ τ 0 . It follows that the series of functions which defines ṽ is normally converging on any compact sets [τ 1 , τ 2 ] of the complementary of the disc τ > τ 0 . So does same for derivatives with respect to τ and θ.

Particularly, the function ṽ is well defined on the complementary of the disk τ > τ 0 and coincides with v on the circle τ = τ 0 . We will show that lim τ →0+ ṽ = 0.

If Re m < 1, we have when n ∈ N and thanks to the formula (8.1)

P m-1 2 n-1 2 (ch τ ) = 2 m-1 2 √ πΓ 1 -m 2 sh 1-m 2 τ π 0 (ch τ + sh τ cos θ) n+ m 2 -1 sin 1-m θ dθ then lim τ →0+ P m-1 2 n-1 2 (ch τ ) = 0
and in addition, for n > 1 -Re m 2 , we have

P m-1 2 n-1 2 (ch τ ) ≤ 2 Re m-1 2 sh 1-Re m 2 τ √ π Γ 1 -m 2 π 0 (ch τ + sh τ cos θ)) n+ Re m 2 -1 sin 1-Re m θ dθ ≤ 2 Re m-1 2 sh 1-Re m 2 τ √ π Γ 1 -m 2 π 0 (ch τ + sh τ )) n+ Re m 2 -1 sin 1-Re m θ dθ ≤ C m sh 1-Re m 2 τ e (n+ Re m 2 )τ thus n>1-Re m 2 sup τ ∈[0, τ 0 2 ] b n P m-1 2 n-1 2 (ch τ ) P m-1 2 n-1 2 (ch τ 0 )
e inθ < +∞ by the Proposition 8.1, we obtain

P m-1 2 n-1 2 (ch τ 0 ) ∼ n→+∞ n m 2 -1 √ 2π sh τ 0 e nτ 0 .
So, we can deduce that lim τ →0+ ṽ = 0. It remains to prove the uniqueness of the previous decomposition where Re m < 1. This will result in the next paragraph which will establish the fact that the family

A :=    Q m-1 2 n-1 2 (ch τ ) Q m-1 2 n-1 2 (ch τ 0 ) (ch τ -cos θ) m/2
(ch τ ) Q m-1 2 n-1 2 (ch τ )        
n∈Z is a basis of solutions on the disk τ ≥ τ 1 and the other half part is a basis of solutions on τ ≤ τ 0 , which is the complement on H + of a some disk, with 0 < τ 0 < τ 1 . This fact is known for m = -1, namely for µ = 1. We extend this result for complex values of m.

us recall the definition of a Riesz basis. (x n ) n∈N is a quasi-orthogonal or Riesz sequence of a Hilbert space X if there are two constants c, C > 0 such that for all sequences (a n ) n∈Z in 2 , we have

c 2 n |a n | 2 ≤ n a n x n 2 ≤ C 2 n |a n | 2 .
If the family (x n ) n∈Z is complete, it is a Riesz basis. The matrix of scalar product { x i , x j } i,j is named Gram matrix associated to {x i } i .

To prove that {x i } i is a Riesz basis, a convenient characterization with the Gram matrix is the following property :

Property ([43, p. 170]). A family {x i } i is a Riesz basis for a some Hilbert space if {x i } i is complete on this Hilbert space and if the Gram matrix associated to {x i } i defines a bounded and invertible operator on 2 (N).

Let A and B the two families of solutions of L m [u] = 0, respectively inside the disk τ > τ 0 and outside the other one τ > τ 1 , with 0 < τ 0 < τ 1 The annulus defined in terms of bipolar coordinates {0 < τ 0 < τ < τ 1 } will be denoted A. The space L 2 (∂A) is equipped of the following inner product : for f, g ∈ L 2 (∂A), f, g = 1 2π

A :=    Q m-1 2 n-1 2 (ch τ ) Q m-1 2 n-1 2 (ch τ 0 ) (ch τ -cos θ) m/2
2π 0 f (τ 0 , θ)g(τ 0 , θ) sh Re m-1 τ 0 (ch τ 0 -cos θ) Re m dθ + 1 2π 2π 0 f (τ 1 , θ)g(τ 1 , θ) sh Re m-1 τ 1 (ch τ 1 -cos θ) Re m dθ.

We have the following proposition : Proposition 7.1. C is a Riesz basis in the Hilbert space L 2 (∂A).

Proof. Indeed, in order to build the Gram matrix of C, we first calculate all its scalar products. We obtain for all n ∈ Z, In all other cases, the inner product is zero, the Gram matrix is diagonal by blocks and each blocks is expressed as the 2 × 2 matrix : The Gram matrix G can be written as

M n =              1 + Q m-1 2 n-1 2 (ch τ 1 ) Q m-1 2 n-
G =                  M 0 0 • • • • • • • • • • • • • • • • • • 0 M -1 0 • • • • • • • • • • • • • • • . . . 0 M 1 0 • • • • • • • • • • • • . . . . . . 0 M -2 0 • • • • • • • • • . . . .
(M n ) = 1 - Q m-1 2 n-1 2 (ch τ 1 ) Q m-1 2 n-1 2 (ch τ 0 ) P m-1 2 n-1 2 (ch τ 0 ) P m-1 2 n-1 2 (ch τ 1 ) 2
Let's show that M n is invertible. Suppose the contrary, if M n is not invertible, then det(M n ) = 0, which is equivalent to then it implies that τ 0 = τ 1 , it is not possible. We deduce that the matrix M n is invertible and this completes the proof.

  If u ∈ D(H + ), we define S m u ∈ D(H + ) by (S m u)(x, y) = x -m u(x, y). If u ∈ D(H + ), we define Du ∈ D(H + ) by (Du)(x, y) = ∂u ∂x (x, y). These operators satisfy the following proposition : Proposition 2.1. S m conjugates L m and L m , D conjugates L -m and L m , which means that

Proposition 5 . 1 .

 51 Let u : H + → C. We have lim The direct implication is easy. Conversely, we assume lim (x,y) →+∞ u(x, y) = 0 and ∀y ∈ R, lim (0,y) u = 0 and we have to show lim ∂H + u = 0.

Lemma 5 . 3 .

 53 means that there exists a constant M such that, for every N sufficiently large, |u N | ≤ M |v N | ; u N = o(v N ) means that for every ε > 0, for every N sufficiently large, |u N | ≤ ε|v N |. On D 1 , we have sup ∂φ N ∂ξ = O(N ) and sup ∂φ N ∂η = 0. On D 2 ∪ D 4 , we have

  y) →+∞ 0 and the first result of the lemma is shown. Similarly, if Re m ≥ 1,

Theorem 5 . 8 .

 58 Let m ∈ C. Let Ω be an open set of H + and let K be a compact set of Ω. If u ∈ C 2 (Ω \ K) satisfies L m u = 0 in Ω \ K, then u has a unique decomposition as follows :

Figureθ

  Figure : Bipolar coordinates

Figure :

 : Figure : Level lines (with α = 1)

  sin θ (ch τ -cos θ) 2 ∂u ∂x + ch τ cos θ -1 (ch τ -cos θ) 2 ∂u ∂y .

r 2 shθ ch τ -cos θ r m and ∂ 2 r m ∂θ 2 = m 4 (∂ 2 r m ∂τ 2 = 1 4 (

 244 m (τ, θ) = (ch τ -cos θ) m/ch τ -cos θ) 2 2 cos θch τ + m sin 2 θ -2 r m then ∂r m ∂τ = 1 (ch τ -cos θ)sh τ ch 2 τ + (m -1)ch τ cos θ -m r m and ch τ -cos θ) 2 sh 2 τ ch 4 τ -2ch 3 τ cos θ + (m -1) 2 ch 2 τ cos 2 θ+ +2(m -1)ch 2 τ + (4 -2m 2 )ch τ cos θ + 2(m -1) cos 2 θ + m(m -2) r m .The equationL m,x,y u = 0 can be rewritten as

2 (ch τ 1 )( 2 shCorollary 6 . 3 . 2 (= 1 2π 2π 0 ( 1 2= 1 2π 2π 0 ( 1 2 2 sh

 21263201012 ch τ -cos θ) m/The solution of the Dirichlet problem for L m u = 0 on D((a, 0), R) where u = ϕ on ∂D((a, 0), R) is given byu(τ, θ) = sh 1-m 2 τ (ch τ -cos θ) ch τ 0 ) e inθwhere {τ = τ 0 } corresponds to the circle of center (a, 0) and radius R and wherec n ch τ 0 -cos θ) -m/2 sh mτ 0 ϕ(a + R cos s, R sin s)e -ins ds. of L m v = 0 on H + \ D((a, 0), R), which is equal to ϕ on ∂D((a, 0), R) where c n ch τ 0 -cos θ) -m/2 sh mτ 0 ϕ(a + R cos s, R sin s)e -ins ds.Moreover, if Re m < 1, then v satisfies lim ∂H + v = 0, and the function v constructed above is the unique solution of the Dirichlet problem L m v = 0 on H + \ D((a, 0), R) which vanishes on ∂H + .7. Riesz basisWe will prove that the half part of the following family   (ch τ -cos θ) m/

  b n ) n∈ZLet C the union of the two previous families :C := (c n ) n∈Z := (c 2n = a n et c 2n+1 = b n ) n∈Z

2 (

 2 ch τ 0 ) c 2n+1 , c 2n = c 2n , c 2n+1

2 (ch τ 1 ) 2 (ch τ 1 )

 2121 = 0. Therefore, there is λ ∈ C \ {0} (which depends on m, n, τ 0 and τ 1 ) such that Then, by the asymptotic of Associated Legendre functions (see Proposition 8.1 in the Annex), on the one hand, we have both λ ∼ n→+∞ πe iπ m-1 2 e -2nτ 0 and on the other hand, we have λ ∼ n→+∞ πe iπ m-1 2 e -2nτ 1 ,

  function on an open set of R 2 and C 2 valued, then

	div (f ) :=	∂f 1 ∂x	+	∂f 2 ∂y	.
	∇f :=	∂f ∂x	,	∂f ∂y	.

Similarly, if f : R 2 → C is a C 1 scalar function on an open set of R 2 and C valued, then

Annex : Associated Legendre functions of first and second kind

In this section, we provide the main formulas of integral representation for the Associated Legendre function of the first and the second kind with z = ch τ > 1 (see [START_REF] Abramowitz | Handbook of mathematical functions[END_REF][START_REF] Lebedev | Special functions and their applications[END_REF][START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF]) :

with Re ν > -1 and Re(µ + ν) < 0.

with Re µ < 1 2 .

2 , Re(ν -µ + 1) < 0 and Re(ν + µ + 1) > 0.

with Re ν > -1 and Re(µ + ν + 1) > 0 (see [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] pages 4, 5 and 6).

We have also the following relations satisfied by the Legendre functions (see [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] page 6 and [START_REF] Abramowitz | Handbook of mathematical functions[END_REF], formula 8.2.2)

In addition, we have the Whipple formulas connecting the associated Legendre functions of first and second kind (see [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] page 6)

We also have the recursion formulas (see [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] pages 6 et 7)

All of these formulas are used to explicitly calculate the values of P µ ν (ch τ ) and Q µ ν (ch τ ) for all τ > 0 and (µ, ν) ∈ C 2 . If µ and τ are fixed, the following proposition collects the behavior of Associated Legendre functions of the first and second kind when ν = n -1 2 with n ∈ Z and |n| → +∞. Proposition 8.1. We fix τ > 0 and µ ∈ C. Then if ν = n -1 2 with n ∈ Z, we have :

These equivalences are locally uniform with respect to the variable τ , that is to say uniform on all interval [τ 0 , τ 1 ] with 0 < τ 0 < τ 1 .

Proof. If ν = n -1 2 with n ∈ N (see [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] page 48), we have

A straightforward application of the Stirling formula shows that when ν → +∞

which gives us the first estimate.

The second one is obtained directly thanks to the relation P µ ν = P µ -ν-1 . The third estimate follows directly from the formula (8.3) of [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF] :

and the last estimation arises from the fact that for ν = n -1 2 with n ∈ Z, we have

The locally uniform character of these equivalences come from explicits expressions of P µ ν and Q µ ν in terms of hypergeometric functions ( [START_REF] Erdélyi | Higher Transcendental Functions[END_REF], tables pages 124-138) and estimations of these special functions (always locally uniform with respect to their parameters ( [START_REF] Virchenko | Generalized associated Legendre functions and their applications[END_REF], pages 178-182).