Colette De Coster

Serge Nicaise
email: serge.nicaise@univ-valenciennes.fr

Colette Decoster

Singular behavior of the solution of the Helmholtz equation in weighted L p -Sobolev spaces

Keywords: Helmholtz equation, singular behavior, nonsmooth domains. AMS Subject Classification: 35J05, 35B65

We study the Helmholtz equation (1) -∆u + zu = g in Ω, with Dirichlet boundary conditions in a polygonal domain Ω, where z is a complex number. Here g belongs to L p µ (Ω) = {v ∈ L p loc (Ω) : r µ v ∈ L p (Ω)}, with a real parameter µ and r(x) the distance from x to the set of corners of Ω. We give sufficient conditions on µ, p and Ω that guarantee that problem (1) has a unique solution u ∈ H 1 0 (Ω) that admits a decomposition into a regular part in weighted L p -Sobolev spaces and an explicit singular part. We further obtain some estimates where the explicit dependence on |z| is given.

Introduction

This paper is the first one of a large program of research devoted to the study of (nonlinear) heat equation in nonsmooth domains in weighted L p -Sobolev spaces. Our final goal requires precise informations about the solution of the linear heat equation, in particular its decomposition into a regular part and an explicit singular part. Although this theory is well developed in weighted L 2 -Sobolev spaces [START_REF] Grisvard | Edge behavior of the solution of an elliptic problem[END_REF][START_REF] Kozlov | Singularities of solutions of the first boundary value problem for the heat equation in domains with conical points[END_REF][START_REF] Kozlov | Coefficients in the asymptotic solutions of the Cauchy boundary-value parabolic problems in domains with a conical point[END_REF][START_REF] Anh | Asymptotic formulas for solutions of parameter-depending elliptic boundary-value problems in domains with conical points[END_REF] or in L p -Sobolev spaces [START_REF] Grisvard | Singular behavior of elliptic problems in non hilbertian Sobolev spaces[END_REF], to our best knowledge such a result does not exist in the framework of weighted L p -Sobolev spaces. For maximal regularity type results in weighted L p -Sobolev spaces, we refer to [START_REF] Da Prato | Sommes d'opérateurs linéaires et équations différentielles opérationnelles[END_REF][START_REF] Nazarov | L p -estimates for the solution to the Dirichlet problem and to the Neumann problem for the heat equation in a wedge with edge of arbitrary codimension[END_REF][START_REF] Solonnikov | L p -estimates for solutions of the heat equation in a dihedral angle[END_REF][START_REF] Nazarov | Dirichlet problem for quasilinear parabolic equations in domains with smooth closed edges[END_REF][START_REF] Prüss | H ∞ -Calculus for the sum of noncommuting operators[END_REF].

According to the approach of [START_REF] Grisvard | Singular behavior of elliptic problems in non hilbertian Sobolev spaces[END_REF], the study of the linear heat equation in non-hilbertian Sobolev spaces can be performed with the help of the theory of sums of operators. This theory requires in a first step to obtain uniform estimates of the solution of the Helmholtz equation. Hence the goal of this paper is to make this analysis in L p µ (Ω) for a large range of values of µ and p. Our results extend the ones from [START_REF] Grisvard | Edge behavior of the solution of an elliptic problem[END_REF] to the L p µ (Ω) setting. For the sake of simplicity we have restricted ourselves to the two-dimensional situation. The results of this paper can be easily extended to the case of domains with conical points.

The paper is organized as follows: In section 2 after some recalls, we prove some embeddings and some a priori estimates in H 1 -0 and in L p µ norms. Section 3 is devoted to the proof of the decomposition of the solution of [START_REF] Adeyeye | Generation of analytic semi-group in L p (Ω) by the Laplace operator[END_REF] with uniform estimates with respect to the parameter z.

2 Some preliminary results

Some notations and definitions

In this paper we consider polygonal domains of R 2 in the following sense. Definition 2.1. Let Ω be a bounded domain of R 2 . We say that Ω is a polygonal domain if its boundary is the union of a finite number of line segments Γj , j ∈ {1, . . . , J} (Γ j being supposed to be open). Hence we do not assume that Ω is a Lipschitz domain, that is we include the presence of cracks.

any other vertex of Ω. For j ∈ {1, 2, • • • , J}, let ψ j be the interior angle of Ω at the vertex S j , λ j = π ψ j and (r j , θ j ) the polar coordinates centered at S j such that B(S j , 1) ∩ Ω = {(r j cos θ j , r j sin θ j ) | 0 < r j < 1, 0 < θ j < ψ j } =: D j .

For µ = (µ j ) J j=1 , we define the spaces L p µ (Ω) = {f ∈ L p loc (Ω) | wf ∈ L p (Ω)} with

w = 1 + J j=1 η j (r µ j j -1), (2.1)
where r j (x) is the distance from x to the vertex S j and η j ∈ D(R 2 ) are such that η j ≡ 1 in D j (1/2), η j ≡ 0 on Ω \ D j [START_REF] Adeyeye | Generation of analytic semi-group in L p (Ω) by the Laplace operator[END_REF], where D j (r) is the truncated cone D j (r) = Ω∩B(S j , r). Note that the weight w satisfies w = r µ j j on D j (1/2) and w = 1 on Ω \ ∪ J j=1 D j (1). The space L p µ (Ω) is a Banach space for the norm We denote

f L p µ (Ω) = Ω |f (
|u| V k,p µ (Ω) =   |γ|=k Ω |D γ u(x)| p w p (x) r (|γ|-k)p (x) dx   1/p .
In H 1 0 (Ω) we will denote the norms in the following way For µ and γ, we write µ > γ in case, for all j ∈ {1, . . . , J}, µ j > γ j .

September 2, 2010

Embeddings and consequences

Lemma 2.1. Let Ω be a polygonal domain of R 2 , p ≥ 2 and µ satisfies, ∀j = 1, . . . , J, µ j < 2p-2 p , if p > 2,

µ j ≤ 1, if p = 2. (2.2)
Then the next continuous embeddings hold: 1. L p µ (Ω) ֒→ L 2 1 (Ω), 2. L 2 -1 (Ω) ֒→ (L p µ (Ω)) ′ = L q -µ (Ω), with q = p p-1 , 3. H 1 0 (Ω) ֒→ L q -µ (Ω). Proof. The first assertion follows from the identity r j f = r µ j j f r 1-µ j j , Hölder's inequality and the fact that r

1-µ j j belongs to L q (Ω) if 1 -µ j > -2
q . The second assertion is a consequence of the first one by using duality. For the last assertion, in case p = 2, it is known (see [START_REF] Costabel | Corner Singularities and Analytic Regularity for Linear Elliptic Systems[END_REF]Thm 14.5.5] or Lemma 2.11 below) that H 1 0 (Ω) is continuously embedded in L 2 -1 (Ω). We then conclude observing that, for µ j ≤ 1, we have r

2(-µ j +1) j ∈ L ∞ (Ω).
In the case p > 2, we use the embedding H 1 0 (Ω) ֒→ L 2 -1 (Ω) and the second assertion.

Lemma 2.2. Let Ω be a polygonal domain of R 2 , p ≥ 2 and µ satisfies, for all j = 1, . . . , J,

µ j > -2 p , if p > 2, µ j ≥ -1, if p = 2. Then H 1 0 (Ω) is continuously embedded in L p µ (Ω).
Proof. We have the continuous embedding of H 1 0 (Ω) into L p (Ω) for all p > 1 and r µp j ∈ L s (Ω) for all s > 1 satisfying µps + 2 > 0. The result follows by Hölder inequality as we have by assumption s > 1 such that µps + 2 > 0 and, for f

∈ H 1 0 (Ω), Ω r µp j |f | p ≤ |f | p L s ′ r µp j L s , with 1 s + 1 s ′ = 1. Lemma 2.3. Let Ω be a polygonal domain of R 2 , p ≥ 2, µ satisfy (2.2) and θ A ∈ ]0, π[. Then for all g ∈ L p µ (Ω) and all z ∈ C with | arg z| ≤ θ A , the problem ∀ϕ ∈ H 1 0 (Ω), Ω ∇u • ∇ϕ + z Ω uϕ = Ω gϕ, (2.3)
has a unique solution u ∈ H 1 0 (Ω). Proof. For all g ∈ L p µ (Ω), we know by Lemma 2.1 that the mapping

H : H 1 0 (Ω) → C : u → Ω g ū
is linear and continuous. Observe that, for all z ∈ C with | arg z| ≤ θ A , there exists θ ∈ [0, 2π] such that cos θ > 0 and ℜ(ze iθ ) ≥ 0. Hence the conclusion follows from the Lax-Milgram Lemma as, for all u ∈ H 1 0 (Ω),

ℜ[e iθ Ω (|∇u| 2 + z|u| 2 )] ≥ cos θ Ω |∇u| 2 ≥ α u 2 H 1 0 ,
for some α > 0, due to Poincaré's inequality.

Remark 2.1

The unique solution u ∈ H 1 0 (Ω) of (2.3) is called a weak solution of the problem

-∆u + zu = g, in Ω, u = 0,
on ∂Ω, since -∆u + zu = g in the distributional sense.

Some inequalities

For R > 0 and θ A ∈ ] π 2 , π[ fixed, we define

π + = {z ∈ C | ℜ(z) ≥ 0}, S A = {z ∈ C | |z| ≥ R and | arg z| ≤ θ A }. (2.4)
For a solution u ∈ H 1 0 (Ω) of (2.3) we will use the following notation: u g means the existence of a positive constant C, which is independent of the quantities u , g and z under consideration such that u ≤ C g and u ∼ g means u g and g u .

Lemma 2.4. Let R > 0 and θ A ∈ ] π 2 , π[ be fixed. Let Ω be a polygonal domain of R 2 , p ≥ 2, µ satisfy (2.2), z ∈ π + ∪ S A , and u ∈ H 1 0 (Ω) be the solution of (2.3). Then u satisfies the inequality

|u| H 1 0 (Ω) g L p µ (Ω) . (2.5)
Proof. Applying (2.3) with w = u we have

|u| 2 H 1 0 (Ω) + z Ω |u| 2 = Ω gū. (2.6)
By Lemma 2.1, taking the real and the imaginary parts of (2.6), we obtain

|u| 2 H 1 0 (Ω) + ℜ(z) Ω |u| 2 g L p µ (Ω) |u| H 1 0 (Ω) . (2.7)
and

|ℑ(z)| Ω |u| 2 g L p µ (Ω) |u| H 1 0 (Ω) . (2.8)
Case 1: ℜ(z) ≥ 0. In that case the result can be directly deduced from (2.7).

Case 2:

ℜ(z) < 0. As z ∈ S A ∪ π + , we have ℜ(z) = ρ cos θ, ℑ(z) = ρ sin θ with ρ > R and |ℑ(z)| > ρ sin θ A . Hence we deduce from (2.8) that Ω |u| 2 1 ρ g L p µ (Ω) |u| H 1 0 (Ω) .
(2.9)

Then (2.7) together with ℜ(z) < 0 gives

|u| 2 H 1 0 (Ω) g L p µ (Ω) |u| H 1 0 (Ω) -ℜ(z) Ω |u| 2 g L p µ (Ω) |u| H 1 0 (Ω) -ℜ(z) 1 ρ g L p µ (Ω) |u| H 1 0 (Ω) .
We conclude using the inequality -ℜ(z) ≤ ρ.

Corollary 2.5. Let R > 0 and θ A ∈ ] π 2 , π[ be fixed. Let Ω be a polygonal domain of R 2 , p ≥ 2, µ satisfy (2.2), g ∈ L p µ (Ω), z ∈ π + ∪S A ,
and u ∈ H 1 0 (Ω) be the solution of (2.3). Then we have the inequalities

|u| 2 L 2 (Ω) 1 |z| g L p µ (Ω) |u| H 1 0 (Ω) 1 |z| g 2 L p µ (Ω)
(2.10)

and (1 + |z| 1/2 )|u| L 2 (Ω) g L p µ (Ω) . (2.11)
Proof. By (2.6), we have

z Ω |u| 2 = Ω gū -|u| 2 H 1 0 (Ω) ,
from which we deduce, using Lemmas 2.1 and 2.4,

|z| Ω |u| 2 g L p µ (Ω) |u| H 1 0 (Ω) + |u| 2 H 1 0 (Ω) g L p µ (Ω) |u| H 1 0 (Ω) .
We Let Ω be a bounded domain of R 2 , g ∈ L 2 (Ω), z ∈ π + ∪ S A , and u ∈ H 1 0 (Ω) be the solution of (2.3). Then we have the inequality

(1 + |z|) u L 2 (Ω) g L 2 (Ω) .
Proof. As in Lemma 2.4, we obtain (2.6).

Case 1: ℜ(z) ≥ 0. In that case, we also have

|u| 2 H 1 0 (Ω) ≤ ℜ( Ω gu) ≤ g L 2 (Ω) u L 2 (Ω) , (2.12)
and z

Ω |u| 2 = Ω gu - Ω |∇u| 2 .
Using (2.12), this implies

|z| Ω |u| 2 ≤ | Ω gu| + |u| 2 H 1 0 (Ω) g L 2 (Ω) u L 2 (Ω) ,
from which we deduce

|z| u L 2 (Ω) g L 2 (Ω) .
By Poincaré inequality and (2.12), we have,

u 2 L 2 (Ω) |u| 2 H 1 0 (Ω) ≤ g L 2 (Ω) u L 2 (Ω)
, and the result follows.

Case 2: ℜ(z) < 0. In that case z ∈ S A . As in the proof of Lemma 2.4, we have

ρ u L 2 (Ω) g L 2 (Ω) ,
where ρ = |z| > R. Hence we deduce

(1 + |z|) u L 2 (Ω) |z| u L 2 (Ω) g L 2 (Ω) ,
which concludes the proof.

Lemma 2.7. Let Ω be a polygonal domain of R 2 , p ≥ 2, µ satisfy (2.2), f ∈ L p µ (Ω) and u ∈ H 1 0 (Ω) be the solution of

∀ϕ ∈ H 1 0 (Ω), Ω ∇u • ∇ϕ = Ω f ϕ. (2.13)
Then, for all γ ≥ µ with γ > 2 -2 p -λ and λ = (λ j

) J j=1 = ( π λ j ) J j=1 , u ∈ V 2,p γ (Ω),
and, in particular,

r γ j -2 j u ∈ L p (D j ), r γ j -1 j ∇u ∈ (L p (D j )) 2 , r γ j j ∆u ∈ L p (D j ).
If moreover µ > -λ then we have in particular

u ∈ V 2,p µ+2-2 p (Ω),
and hence,

r µ j -2 p j u ∈ L p (D j ), r µ j +1-2 p j ∇u ∈ (L p (D j )) 2 , r µ j +2-2 p j ∆u ∈ L p (D j ).
Proof. Using regularity results far from the corners of Ω for the Laplace equation with Dirichlet boundary conditions (see [START_REF] Gilbarg | Elliptic partial differential equations of second order[END_REF][START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]), u ∈ W 2,p ( Ω), where, for some δ > 0, Ω is a subdomain of Ω with a smooth boundary, its boundary being the same as Ω except in

J j=1
B(S j , δ), with the estimate

u W 2,p ( Ω) f L p µ (Ω) .
It then remains to look at the behaviour of u near the corners. For any j = 1, • • • , J, consider the cut-off function η j ∈ D(R 2 ) used in (2.1). For shortness we write D = D j (1) and drop the index j. Let us set ũ = ηu.

This function satisfies

-∆ũ = h, in D, ũ = 0, on ∂D, (2.14)
where h = ηh-2∇η•∇u-u∆η. Moreover due to the above results (regularity far from the corners), h belongs to L p µ (D) and

h L p µ (D) h L p µ (Ω) .
For γ ≥ µ we have L p µ (D) ⊂ L p γ (D). As, by [START_REF] Kufner | Weighted Sobolev spaces[END_REF]Remark 9.11],

L p γ (D) = V 0,p γ (D), applying [13, Lemma 11.2 (ii)] (as in [13, Example 11.3]) we prove that u ∈ V 2,2 1 (D). Let γ 1 be such that µ ≤ γ 1 ≤ γ, γ 1 > 2 -2 p -λ and γ 1 ≤ 2 -2 p . As f ∈ L p γ 1 (D), by [13, Thm 9.3] we have u ∈ V 2,p γ 1 (D) if 0 ≤ 2 - 2 p -γ 1 < λ, which is equivalent to 2 - 2 p -λ < γ 1 ≤ 2 - 2 p .
This gives the first part of the result observing that

V 2,p γ 1 (D) ⊂ V 2,p γ (D) if γ 1 ≤ γ.
To prove the second part of the result, we just have to observe that, if

µ > -λ then µ -2 p + 2 ≥ µ and µ + 2 -2 p > -λ + 2 -2 p . Corollary 2.8. Let Ω be a polygonal domain, p ≥ 2, µ > -λ satisfy (2.2), f ∈ L p µ (Ω), u ∈ H 1 0 (Ω)
be the solution of (2.13) and define v = w u with w defined by (2.1). Then we have

v ∈ V 2,p 2-2 p (Ω), r -2 p v ∈ L p (Ω), r 1-2 p ∇v ∈ (L p (Ω)) 2 , r 2-2 p ∆v ∈ L p (Ω).
Proof. This can be easily deduced from Lemma 2.7.

Definition 2.2. For Ω a polygonal domain, p ≥ 2 and µ ∈ R J , we define

D(∆ p, µ ) = {u ∈ H 1 0 (Ω) | ∆u ∈ L p µ (Ω)}.
Before going on let us show that Lemma 2.7 furnishes an existence result in D(∆ p, µ ) for problem (2.3). Lemma 2.9.

Let θ A ∈ ]0, 2π[, p ≥ 2, µ satisfies (2.2) as well as µ > -λ -2 p , z ∈ C with | arg z| ≤ θ A , and u ∈ H 1 0 (Ω) be the solution of (2.3) with g ∈ L p µ (Ω). Then u ∈ D(∆ p, µ ). Proof. We only need to show that -∆u ∈ L p µ (Ω). (2.15)
Using regularity results far from the corners of Ω for the Laplace equation with Dirichlet boundary conditions (see [START_REF] Gilbarg | Elliptic partial differential equations of second order[END_REF][START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]), u ∈ W 2,p ( Ω), where Ω was introduced in the proof of Lemma 2.7. We directly deduce that u ∈ W 2,p ( Ω). Hence it remains to show (2.15) near each corner S j . For a fixed j ∈ {1, 2, • • • , J}, we then set u j = η j u ∈ H 1 0 (D j ) with η j the cut-off function used in (2.1). The function u j is a weak solution of

-∆u j + z u j = g j , in D j , u j = 0, on ∂D j , (2.16)
where g j = η j g-2∇η j •∇u-u ∆η j . By the regularity u ∈ W 2,p ( Ω) mentioned before, we have

∇u ∈ W 1,p ( Ω) 2 , u ∈ L p ( Ω),
and consequently g j ∈ L p µ j (D j ) with

g j L p µ j (D j ) g L p µ (Ω) .
(2.17)

Hence we can concentrate on (2.16) and prove that

-∆u j ∈ L p µ j (D j ). (2.18)
For the rest of the proof we drop the index j and write D for D j , µ for µ j , ...

As Lemma 2.2 guarantees that H 1 0 (D) ֒→ L p µ ′ (D) for all µ ′ > -2 p , we distinguish different cases: Case 1. µ > -2 p . In that case -∆u = g -zu ∈ L p µ (D). Case 2. -2 -2 p < µ ≤ -2 p .
In that case, we take

µ 1 = µ + 2 if µ < -2 p and µ 1 = 2 -2 p -ǫ, with ǫ ∈ ]0, 2[ if µ = -2 p . Since µ 1 > -2 p , u ∈ H 1 0 (Ω) is solution of -∆u = g -zu ∈ L p µ 1 (D). This implies by Lemma 2.7 that u ∈ V 2,p µ 1 (D). Accordingly r µ 1 -2 u ∈ L p (D), which implies that u ∈ L p µ (D), (2.19) due to µ 1 -2 ≤ µ. This again guarantees (2.18) because -∆u = g -zu. Case 3. -4 -2 p < µ ≤ -2 -2 p . then µ 2 = µ + 2
u ∈ V 2,p µ 2 (Ω). Accordingly r µ 2 -2 u ∈ L p (Ω),
and we conclude that (2.19) holds. The general case follows by induction.

Corollary 2.10.

Let θ A ∈ ]0, 2π[, D = {(r cos θ, r sin θ) | 0 < r < 1, 0 < θ < ψ}, λ = π ψ , p ≥ 2, µ ∈ R, f ∈ L p µ (D), z ∈ C with | arg z| ≤ θ A and u ∈ D(∆ p,µ ) be the solution of -∆u + zu = f, in D, u = 0, on ∂D. (2.20) Assume v = r µ u ∈ V 2,p 2-2 p (D). Then v satisfies (a) ℜ( D r -1 ∂v ∂r |v| p-2 v) = 0; (b) p 2 D |∇v| 2 |v| p-2 + p -2 2 D |v| p-4 v 2 (∇v) 2 -µ 2 D r -2 |v| p + 2µ D r -1 ∂v ∂r |v| p-2 v + z D |v| p = D r µ f |v| p-2 v. Remark 2.2 Recall that, by Corollary 2.8, if µ > -λ satisfies (2.2) then v ∈ V 2,p 2-2 p (D) as u ∈ D(∆ p,µ ) is a solution of (2.20). Proof. Recall that V 2,p 2-2 p (D) = C ∞ S (D) with C ∞ S (D) = {v ∈ C ∞ (D) | S ∈ supp v}.
Step 1: Proof of (a).

For v ∈ C ∞ S (D), we have D r -1 ∂v ∂r |v| p-2 v rdrdθ = - D v ∂ ∂r (|v| p-2 v) drdθ + ψ 0 |v(1, θ)| p dθ and hence D r -1 ∂v ∂r |v| p-2 v rdrdθ = ψ 0 |v(1, θ)| p dθ - D v[ ∂ ∂r ((v v) p-2 2 )v + |v| p-2 ∂v ∂r ] drdθ = ψ 0 |v(1, θ)| p dθ - D v( p -2 2 |v| p-4 (v) 2 ∂v ∂r + p -2 2 |v| p-2 ∂v ∂r + |v| p-2 ∂v ∂r ) drdθ = ψ 0 |v(1, θ)| p dθ - D p -2 2 |v| p-2 v ∂v ∂r drdθ - D p 2 |v| p-2 v ∂v ∂r drdθ
.

We obtain then

p 2 D r -1 |v| p-2 v ∂v ∂r rdrdθ + D r -1 |v| p-2 v ∂v ∂r rdrdθ = ψ 0 |v(1, θ)| p dθ
By density and [17, Thm 1.31, p.27 and Def 1.9 p.15], we can pass to the limit and we see that the previous equality is also valid for v ∈ V 2,p 2-2/p (D). As v = r µ u with u ∈ H 1 0 (D) we have γ 0 v = 0 on ∂D which gives the equality

p 2 D r -1 |v| p-2 v ∂v ∂r + D r -1 |v| p-2 v ∂v ∂r = 0. (2.21)
Then (a) follows by taking the real part of (2.21).

Step 2: Proof of (b). Observe that v satisfies

-∆v -µ 2 r -2 v + 2µr -1 ∂v ∂r + zv = g, in D, v = 0, on ∂D, (2.22) with g = r µ f ∈ L p (D). As v ∈ V 2,p 2-2 p (D), it is meaningful to multiply (2.22)
by |v| p-2 v and integrate. We then obtain

- D ∆v |v| p-2 v -µ 2 D r -2 |v| p + 2µ D r -1 ∂v ∂r |v| p-2 v + z D |v| p = D g |v| p-2 v. (2.23) For v ∈ C ∞ S (D) we have - D ∆v |v| p-2 v = D ∇v • ∇(|v| p-2 v) + ∂D γ 0 (|v| p-2 v) γ 0 (∇v) • ν = D |∇v| 2 |v| p-2 + p -2 2 ( D |∇v| 2 |v| p-2 + D |v| p-4 v2 (∇v) 2 ) + ∂D γ 0 (|v| p-2 v) γ 0 (∇v) • ν.
As in Step 1, we show that we can pass to the limit and the previous equality is also valid for v ∈ V 2,p 2-2/p (D). Again the boundary term is equal to zero and we obtain the equality

- D ∆v |v| p-2 v = p 2 D |∇v| 2 |v| p-2 + p -2 2 D |v| p-4 v(∇v) 2 .
(2.24)

The result follows then from (2.23) and (2.24).

Lemma 2.11.

Let D = {(r cos θ, r sin θ) | 0 < r < 1, 0 < θ < ψ}. For all w ∈ H 1 0 (D), we have D |∇w| 2 ≥ π 2 ψ 2 D 1 r 2 w 2 .
Proof. First observe that

D |∇w| 2 = D (( ∂w ∂r ) 2 + 1 r 2 ( ∂w ∂θ ) 2 ).
Moreover, for all r ∈ ]0, 1[ we have w(r, •) ∈ H 1 0 (0, ψ) and hence, by applying Poincaré inequality in (0, ψ),

ψ 0 ( ∂w ∂θ ) 2 (r, θ) dθ ≥ π 2 ψ 2 ψ 0 w 2 (r, θ) dθ.
We deduce from this inequality that

1 0 ψ 0 (( ∂w ∂r ) 2 + 1 r 2 ( ∂w ∂θ ) 2 ) rdθdr ≥ 1 0 ψ 0 ( ∂w ∂θ ) 2 dθ 1 r dr ≥ π 2 ψ 2 1 0 ψ 0 w 2 (r, θ) dθ 1 r dr = π 2 ψ 2 D w 2 1 r 2 . 2
Remark 2.3 Observe that for w(r, θ) = r β (1 -r) α sin(λθ) with β > 0 and α > 1/2 we have w ∈ H 1 0 (D) and

D |∇w| 2 = ( π 2 ψ 2 + β(α + β) 2α -1 ) D 1 r 2 w 2 ,
which proves the optimality of the previous inequality.

Corollary 2.12.

Let D = {(r cos θ, r sin θ) | 0 < r < 1, 0 < θ < ψ}, λ = π ψ , p ≥ 2, µ ∈ R satisfy 4(p -1)λ 2 -µ 2 p 2 > 0. (2.25) Let f ∈ L p µ (D), z ∈ C with ℜ(z) ≥ 0 and u ∈ D(∆ p,µ
) be the solution of

-∆u + zu = f, in D, u = 0, on ∂D. (2.26) Assume v = r µ u ∈ V 2,p 2-2 p (D).
Then we have the inequalities

ℜ(z) u L p µ (D) ≤ f L p µ (D)
and

|ℑ(z)| u L p µ (D) f L p µ (D) .
Proof. By Corollary 2.10, v = r µ u satisfies

p 2 D |∇v| 2 |v| p-2 + p -2 2 D |v| p-4 v 2 (∇v) 2 -µ 2 D r -2 |v| p + 2µ D r -1 ∂v ∂r |v| p-2 v + z D |v| p = D g |v| p-2 v, (2.27)
with g = r µ f and

ℜ( D r -1 ∂v ∂r |v| p-2 v) = 0.
(2.28)

Writing v = v 1 + i v 2 with v i : D → R, (2.27) becomes p 2 D [( ∂v 1 ∂r ) 2 + ( ∂v 2 ∂r ) 2 + 1 r 2 ( ∂v 1 ∂θ ) 2 + 1 r 2 ( ∂v 2 ∂θ ) 2 ](v 2 1 + v 2 2 )|v| p-4 + p -2 2 D |v| p-4 (v 2 1 -v 2 2 -2iv 1 v 2 ) [( ∂v 1 ∂r ) 2 -( ∂v 2 ∂r ) 2 + 2i ∂v 1 ∂r ∂v 2 ∂r + 1 r 2 ( ∂v 1 ∂θ ) 2 - 1 r 2 ( ∂v 2 ∂θ ) 2 + 2i r 2 ∂v 1 ∂θ ∂v 2 ∂θ ] -µ 2 D r -2 |v| p + 2µ D r -1 ∂v ∂r |v| p-2 v + z D |v| p = D g |v| p-2 v.
(2.29) By taking the real part of (2.29) and using (2.28) we obtain

D |v| p-4 {(p -1)[(v 1 ∂v 1 ∂r + v 2 ∂v 2 ∂r ) 2 + (v 1 1 r ∂v 1 ∂θ + v 2 1 r ∂v 2 ∂θ ) 2 ] + [(v 2 ∂v 1 ∂r -v 1 ∂v 2 ∂r ) 2 + (v 2 1 r ∂v 1 ∂θ -v 1 1 r ∂v 2 ∂θ ) 2 ]} -µ 2 D r -2 |v| p + ℜ(z) D |v| p = ℜ( D g |v| p-2 v). (2.30) Denoting w = (v 2 1 + v 2 2 ) p/4 = |v| p/2 , then (2.30) gives 4(p -1) p 2 D [( ∂w ∂r ) 2 + ( 1 r ∂w ∂θ ) 2 ] -µ 2 D r -2 w 2 + ℜ(z) D w 2 ≤ ℜ( D g |v| p-2 v).
(2.31) By Lemma 2.11, we have

4(p -1) p 2 D |∇w| 2 -µ 2 D r -2 w 2 ≥ ( 4(p -1)λ 2 p 2 -µ 2 ) D r -2 w 2 .
Hence, we deduce from (2.31) that

ℜ(z) v p L p ≤ g L p v p-1 L p , in case 4(p -1)λ 2 p 2 -µ 2 ≥ 0.
This gives the first inequality.

As

4(p-1)λ 2 p 2
-µ 2 > 0 and ℜ(z) ≥ 0, we also have

D w 2 1 r 2 = D |v| p 1 r 2 g L p v p-1 L p , (2.32) D |∇w| 2 = p 2 4 D |v| p-4 [(v 1 ∂v 1 ∂r + v 2 ∂v 2 ∂r ) 2 + (v 1 1 r ∂v 1 ∂θ + v 2 1 r ∂v 2 ∂θ ) 2 ] g L p v p-1 L p (2.33) and D |v| p-4 [(v 2 ∂v 1 ∂r -v 1 ∂v 2 ∂r ) 2 + (v 2 1 r ∂v 1 ∂θ -v 1 1 r ∂v 2 ∂θ ) 2 ] g L p v p-1 L p . (2.34)
By taking the imaginary part of (2.29) we obtain (p -2)

D |v| p-4 [v 2 1 ∂v 1 ∂r ∂v 2 ∂r -v 2 2 ∂v 1 ∂r ∂v 2 ∂r -v 1 v 2 ( ∂v 1 ∂r ) 2 + v 1 v 2 ( ∂v 2 ∂r ) 2 + v 2 1 1 r ∂v 1 ∂θ 1 r ∂v 2 ∂θ -v 2 2 1 r ∂v 1 ∂θ 1 r ∂v 2 ∂θ -v 1 v 2 ( 1 r ∂v 1 ∂θ ) 2 + v 1 v 2 ( 1 r ∂v 2 ∂θ ) 2 ] + 2µ D r -1 |v| p-2 (v 1 ∂v 2 ∂r -v 2 ∂v 1 ∂r ) + ℑ(z) D |v| p = ℑ( D g|v| p-2 v).
Hence, we deduce from (2.32)-(2.33)-(2.34) that

|ℑ(z)| D |v| p g L p v p-1 L p and hence |ℑ(z)| v L p g L p .
The conclusion follows observing that v L p = u L p µ and g L p = f L p µ .

Before going on let us show that this result is mainly optimal.

Lemma 2.13. Let D and λ = π ψ as in Corollary 2.12, and let p ≥ 2 and µ ∈ R. If

λ 2 < 2(1 - 1 p ) 2 , (2.35)
then, for all µ ∈ ] 2(1

-1 p ) 2 + λ 2 , 2 -2 p [, there exist f ∈ L p µ (D), z ∈ R + and u ∈ D(∆ p,µ ) solution of (2.26) such that r µ u ∈ V 2,p 2-2 p (D) and z u L p µ (D) > f L p µ (D) .
(2.36)

Proof. Take p ≥ 2, z ∈ R + , and u ∈ D(∆ p,µ ) such that r µ u ∈ V 2,p 2-2 p (D),
u > 0 on D and u L p µ (D) = 1. Let us consider the function q : R + 0 → R + defined by

q(X) = -X∆u + u p L p µ (D) u p L p µ (D)
,

in order that -∆u + zu p L p µ (D) z p u p L p µ (D)
= q(1/z).

Clearly we have q(0) = 1 and therefore if q satisfies q ′ (0) < 0, then there exists a positive real number X close enough to 0 (and then a positive real number z large enough) such that q(X) < 1 which will show (2.36).

In view of the definition of q, we directly see that

q ′ (0) = p D u p-1 (-∆u)r µp dx.
Hence we are reduced to find a non zero positive function u ∈ D(∆ p,µ ) such that D u p-1 (-∆u)r µp dx < 0.

(2.37)

For that purpose, according to the proof of the previous Corollary and to Remark 2.3 we take u(r, θ) = ϕ(r) sin(λθ), with ϕ(r) = r -µ+β (1 -r) α . Then we see that u ∈ H

I := -[(β -µ) 2 -λ 2 ] 1 0 r βp-2 (1 -r) αp r dr +[2(β -µ) + 1]α 1 0 r βp-1 (1 -r) αp-1 r dr -[α(α -1)] 1 0 r βp (1 -r) αp-2 r dr < 0 (2.40)
By using the definition of the Beta function, we get that

I = -B(pβ, pα + 1)[(β -µ) 2 -λ 2 ] + B(pβ + 1, pα)[2(β -µ) + 1]α -B(pβ + 2, pα -1)[α(α -1)].
Finally the relation B(x, y) = Γ(x) Γ(y) Γ(x + y) and the recurrence relation Γ(z + 1) = z Γ(z) lead to

I = αp Γ(pβ)Γ(pα -1) Γ(pα + pβ + 1) {-[(β -µ) 2 -λ 2 ] + [2(β -µ) + 1]β}(pα -1) -(α -1)(pβ + 1)β .
In conclusion we are looking for parameter α, β and µ satisfying

β > max(µ, 2 - 2 p ), α > 2 - 1 p and [(p -1)β + p(λ 2 -µ 2 )]α + (p -1)β 2 -(λ 2 -µ 2 ) < 0. Hence, if 2(p -1) 2 + p 2 (λ 2 -µ 2 ) < 0, then for β > 2 -2 p close enough to 2 -2 p , µ ≤ 2 -2
p and α large, the above conditions are satisfied. Remark 2.4 Observe that, in the notations of the previous proof, if [START_REF] Aibeche | L p regularity of transmission problems in dihedral domains[END_REF]

(p - 1) 2 + p 2 (λ 2 -µ 2 ) ≥ 0, β > 2 -2
p and α > 2 -1 p then we have

[(p -1)β + p(λ 2 -µ 2 )]α + (p -1)β 2 -(λ 2 -µ 2 ) ≥ 0.
In order to go further, we need estimates like in Corollary 2.12 for z in a larger part of the complex plane.

Corollary 2.14. First Lemma 2.9 guarantees that, for all z > 0, the range of A + zI is L p µ (D). Second by Corollary 2.12 we have, for all z > 0,

Let D = {(r cos θ, r sin θ) | 0 < r < 1, 0 < θ < ψ}, λ = π ψ , p ≥ 2, µ > -λ satisfy (2.
z (A + zI) -1 ≤ 1.
By [18, Thm I-4.2], this implies that -A is dissipative. As L p µ (D) is reflexive, we have by [START_REF] Pazy | Semigroups of linear operators and applications to partial differential equations[END_REF]] that D(∆ p,µ ) is dense in L p µ (D). Hence by Hille-Yosida Theorem [18, Thm I-3.1], -A is the infinitesimal generator of a C 0 semigroup of contractions T (t) for t ≥ 0.

By Corollary 2.12, we have also a constant C > 0 such that, for all σ ≥ 0 and all This proves the required inequality.

τ ∈ R |τ | (A + (σ + iτ )I) -1 ≤ C,
Lemma 2.15. Let Ω be a bounded domain with a smooth boundary and p ≥ 2, Then there exists θA ∈ ] π 2 , π[ such that, for all h ∈ L p ( Ω), all z ∈ C with | arg z| ≤ θA , and v ∈ H 1 0 ( Ω) weak solution of

-∆v + z v = h, in Ω, v = 0, on ∂ Ω, we have (1 + |z|) v L p ( Ω) h L p ( Ω) . Proof. Recall that v satisfies ∀ϕ ∈ H 1 0 ( Ω), Ω ∇v • ∇ϕ + z Ω vϕ = Ω hϕ.
Step 1: ℜz ≥ 0. By [1, Theorem 1.6], we have

|z| v L p ( Ω) h L p ( Ω) .
As moreover

v L p ( Ω) v H 1 0 ( Ω) h L p ( Ω) ,
we obtain the result.

Step 2. Extension. The extension to {z ∈ C | | arg z| ≤ θA } can be made as in Corollary 2.14.

Corollary 2.16. Under the assumptions of Corollary 2.14, for all f ∈ L q -µ (Ω), there exists a unique φ ∈ L q -µ (Ω) solution of

∀v ∈ D(∆ p, µ ), Ω φ ((-∆ + z)v) = Ω f v. Moreover φ satisfies |z| φ L q -µ (Ω) f L q -µ (Ω) .
Proof. By Corollary 2.14 and Lemma 2.15, the linear operator

A : D(∆ p, µ ) ⊂ L p µ (Ω) → L p µ (Ω) : u → -∆u satisfies, for z ∈ C with | arg(z)| ≤ θ A , |z| (A + z I) -1 f L p µ (Ω) f L p µ (Ω) . Moreover D(∆ p, µ ) is dense in L p µ (Ω) (because D(Ω) ⊂ D(∆ p, µ
)). Hence we can define the adjoint A * of the operator A The result follows from [18, Lemma I-10.1] observing that A * is such that u ∈ D(A * ) ⊂ L q -µ (Ω) if and only if there exists g ∈ L q -µ (Ω) such that ∀v ∈ D(∆ p, µ ),

A * : D(A * ) ⊂ (L p µ (Ω)) * → (L p µ (Ω)) * where D(A * ) = {x * ∈ (L p µ (Ω)) * | ∃y * ∈ (L p µ (Ω)) * , ∀x ∈ D(A), Ω x * Ax = Ω y * x} A * x * = y * , for x * ∈ D(A * ).
Ω g v = - Ω u∆v.
and, for u ∈ D(A * ), in the above notations, A * (u) = g.

3 Uniform decomposition Theorem 3.1. Let R > 0 be fixed. Let p ≥ 2, ǫ > 0 fixed and Ω be a bounded polygonal domain of R 2 . Denote by S j , j = 1, • • • , J, the vertices of ∂Ω.

Set

Ω ǫ = {x ∈ Ω | ∀j = 1, • • • , J, dist (x, S j ) > ǫ},
and let µ ∈ R J . Then, there exists

θ A ∈ ] π 2 , π[ such that, for all g ∈ L p µ (Ω), all z ∈ π + ∪S A , the solution u ∈ H 1 0 (Ω) of (2.3) satisfies u ∈ W 2,p (Ω ǫ ) and |u| W 2,p (Ωǫ) + (1 + |z| 1/2 ) |u| W 1,p (Ωǫ) + (1 + |z|) |u| L p (Ωǫ)
g L p µ (Ω) . (3.1) Proof. Let θ A be given by Corollary 2.15.

Step 1: Regularity H 2 . Let us fix η a cut-off function such that η = 0 on j=1,...,J B(S j , ǫ/2) and η = 1 on Ω ǫ . Let Ω a regular domain such that Ω ǫ ⊂ Ω ⊂ Ω, the boundary of Ω satisfying ∂ Ω = ∂Ω except near the corner, and v = ηu ∈ H 1 0 ( Ω). In that case v is a weak solution of

-∆v + z v = ηg -2∇η • ∇u -∆η u =: h, in Ω, v = 0, on ∂ Ω. (3.2)
By Lemma 2.4 and Poincaré inequality, we have h ∈ L 2 ( Ω) and satisfies

h L 2 ( Ω) g L p µ (Ω) + ∇u L 2 ( Ω) + u L 2 ( Ω) g L p µ (Ω) . (3.3)
By Lemma 2.4 and Corollary 2.6 we obtain

|v| H 1 0 ( Ω) + (1 + |z|) v L 2 ( Ω) h L 2 ( Ω) . (3.4)
We can also consider v ∈ H 1 0 ( Ω) as a weak solution of

-∆v + v = h + (1 -z) v =: h 1 , in Ω, v = 0, on ∂ Ω,
where h 1 ∈ L 2 ( Ω) and, by (3.4),

h 1 L 2 ( Ω) h L 2 ( Ω) + (1 + |z|) v L 2 ( Ω) h L 2 ( Ω) g L p µ (Ω)
. By the strong ellipticity of -∆ and the fact that the boundary conditions cover -∆ we have (see [START_REF] Costabel | Corner Singularities and Analytic Regularity for Linear Elliptic Systems[END_REF]), v ∈ H 2 ( Ω) and

v H 2 ( Ω) h 1 L 2 ( Ω) h L 2 ( Ω) . (3.5)
As v ∈ H 2 ( Ω) we have ∆v ∈ L 2 ( Ω). Multiplying (3.2) by -∆v and integrating, we obtain

z Ω |∇v| 2 = -z Ω v∆v = - Ω h ∆v - Ω |∆v| 2 .
This implies, using (3.5),

|z| Ω |∇v| 2 ≤ h L 2 ( Ω) ∆v L 2 ( Ω) + ∆v 2 L 2 ( Ω) h 2 L 2 ( Ω) .
Combining (3.3), (3.4) and (3.5) we prove the inequality

|v| H 2 ( Ω) + (1 + |z|) |v| H 1 ( Ω) + (1 + |z|) |v| L 2 ( Ω) h L 2 ( Ω) g L p µ (Ω) . As v = u on Ω ǫ , this implies u ∈ H 2 (Ω ǫ ) and |u| H 2 (Ωǫ) + (1 + |z|) |u| H 1 (Ωǫ) + (1 + |z|) |u| L 2 (Ωǫ) g L p µ (Ω) .
Step 2: Regularity W 2,p (Ω ǫ ). As in Step 1, let η 1 be a cut-off function such that η 1 = 0 near the corners and η 1 = 1 on Ω. Let Ω 1 a regular domain such that Ω ⊂ Ω 1 ⊂ Ω, the boundary of Ω 1 satisfying ∂Ω 1 = ∂Ω except near the corner, and w = η 1 u ∈ H 1 0 (Ω 1 ). In that case w is a weak solution of

-∆w + z w = η 1 g -2∇η 1 • ∇u -∆η 1 u =: h1 , in Ω 1 , w = 0, on ∂Ω 1 .
By

Step 1, we have that w ∈ H 2 (Ω 1 ) and

|w| H 2 (Ω 1 ) + (1 + |z|) |w| H 1 (Ω 1 ) + (1 + |z|) |w| L 2 (Ω 1 ) g L p µ (Ω) .
As w = u on Ω this shows u ∈ H 2 ( Ω) and

|u| H 2 ( Ω) + (1 + |z|) |u| H 1 ( Ω) + (1 + |z|) |u| L 2 ( Ω) g L p µ (Ω) . (3.6)
This implies that

h := ηg -2∇η • ∇u -∆η u ∈ L p ( Ω) with h L p ( Ω) g L p µ (Ω) + |u| W 1,p ( Ω) + |u| L p ( Ω) g L p µ (Ω) , (3.7)
where we have used Sobolev inequality and (3.6).

We can now proceed in the same way as before but with a given function

h ∈ L p ( Ω). Hence v ∈ H 1 0 ( Ω) is a weak solution of -∆v + v = h + (1 -z) v =: h 1 , in Ω, w = 0, on ∂ Ω.
By the elliptic regularity, for all p ∈ [2, p] we have v ∈ W 2,p ( Ω) and

v W 2, p( Ω) h 1 L p( Ω) .
Using Corollary 2.6, we obtain

(1 + |z|) v L 2 ( Ω) h L 2 ( Ω) ,
which allows to prove in particular

v H 2 ( Ω) h 1 L 2 ( Ω) ≤ h L 2 ( Ω) + (1 + |z|) v L 2 ( Ω) h L 2 ( Ω) ≤ h L p ( Ω) .
By Corollary 2.15, we deduce that

(1 + |z|) v L p ( Ω) h L p ( Ω)
and hence

h 1 L p ( Ω) g L p µ (Ω)
. By interpolation (see [START_REF] Grisvard | Elliptic problems in nonsmooth domains[END_REF]Thm 1.4.3.3]) we obtain

v W 1,p ( Ω) ≤ ǫ v W 2,p ( Ω) + Kǫ -1 v L p ( Ω) .

Applying this inequality with

ǫ = 1 1+ √ |z| , this gives |v| W 2,p ( Ω) + (1 + |z|) |v| W 1,p ( Ω) + (1 + |z|) |v| L p ( Ω) g L p µ (Ω) . The result follows as v = u on Ω ǫ . Lemma 3.2. Let R > 0 and θ A ∈ ] π
2 , π[ be fixed. Let C be the cone with interior angle ψ and λ = π ψ . For z ∈ π + ∪ S A and u ∈ V 2,p µ (C) ∩ H 1 0 (C) with u = 0 for r > r 0 , if µ satisfies (2.2) and, for all k ∈ Z * , 2 -2 p -µ = kλ, then we have

|u| V 2,p µ (C) + |u| V 1,p µ-1 (C) + u L p µ-2 (C) + |z| 1/2 |u| V 1,p µ (C) + |z| u L p µ (C) (-∆ + z)u L p µ (C) + |z| u L p µ (C) . Proof. By [14, Thm 6.2], we have, for θ ∈ [-θ A , θ A ] and u ∈ E 2,p µ (C), u E 2,p µ (C) (-∆ + e iθ )u L p µ (C) + u L p (S) , (3.8)
where

S = {x ∈ R 2 | δ 1 < |x| < δ 2 } ∩ C with 0 < δ 1 < δ 2 , E 2,p µ (C) = adh • E 2,p µ (C ∞ 0 (C \ {0})
where

u E 2,p µ =   C r pµ 2 |α|=0 (r p(|α|-2) + 1) |D α y u(y)| p   1/p . As u L p (S) u L p µ (C)
, the result can be deduced from a change of variable.

Theorem 3.3. Let R > 0, p ≥ 2, Ω be a bounded polygonal domain of R 2 . Denote by S j , j = 1, • • • , J, the vertices of ∂Ω enumerated clockwise and, for j ∈ {1, 2, • • • , J}, let ψ j be the interior angle of Ω at the vertex S j , λ j = π ψ j and λ = (λ j ) 1≤j≤J . Let µ > -λ satisfies (2.2), (2.25) and, for all k ∈ Z * and all j ∈ {1, 2,

• • • , J}, 2 -2 p -µ j = kλ j . Then, there exists θ A ∈ ] π 2 , π[ such that, for all g ∈ L p µ (Ω), all z ∈ π + ∪S A , the unique solution u ∈ D(∆ p, µ ) of -∆u + zu = g, in Ω, u = 0,
on ∂Ω, admits the decomposition

u = u R + J j=1 η j (r) 0<λ ′ j <2-2 p -µ j ∃k∈N,λ ′ j =kλ j c λ ′ j (z) P j,λ ′ j (r √ z)e -r √ z r λ ′ j sin(λ ′ j θ), (3.9
)

with u R ∈ V 2,p µ (Ω), c λ ′ j (z) ∈ C, η j ∈ D(R 2 ) is a cut-off function such that η j ≡ 1 in D j (1/2), η j ≡ 0 on Ω \ D j (1), and P j,λ ′ j (s) = l j,λ ′ j -1 i=0 s i i! with l j,λ ′ j > 2 -µ j -2 p -λ ′ j .
Moreover, the following inequalities are satisfied (a

) |u R | V 2,p µ (Ω) + |u R | V 1,p µ-1 (Ω) + |u R | L p µ-2 (Ω) g L p µ (Ω) ; (b) |u R | V 2,p µ (Ω) + |z| 1/2 |u R | V 1,p µ (Ω) + |z| |u R | L p µ (Ω) g L p µ (Ω) ; (c) J j=1 0<λ ′ j <2-2 p -µ j ∃k∈N,λ ′ j =kλ j |c λ ′ j (z)| (1 + |z| 1-1 p - µ j +λ ′ j 2 ) g L p µ (Ω) .
Proof. Let θ A be the minimum of the ones obtained by Theorem 3.1 and by Corollary 2.14. By Theorem 3.1, we have the required result in the interior of Ω. It remains to prove the regularity near the corners. Let j ∈ {1, 2, • • • , J} be fixed and observe that v = η j u is the solution of

-∆v + z v = g 1 , in D j , v = 0, on ∂D j , (3.10)
September 2, 2010 with D j = {(r j cos θ j , r j sin θ j ) | 0 < r j < 1, 0 < θ j < ψ j } and g 1 = η j g -2∇η j • ∇u -∆η j u. By Theorem 3.1, we have

∇u ∈ W 1,p (Ω ǫ ), u ∈ L p (Ω ǫ ),
and g 1 ∈ L p µ j (D j ) with

g 1 L p µ j (D j ) g L p µ (Ω) . (3.11)
Hence we can concentrate on (3.10). For the rest of the proof we drop the index j and write D for D j , µ for µ j , ...

Step 1: v ∈ V 2,2 1 (D) and |v| V 2,2 1 (D) + |v| V 1,2 0 (D) + |v| V 0,2 -1 (D) g 1 L 2 1 (D) g L p µ (Ω) . (3.12)
Using Lemma 2.4 and Corollary 2.5, we have

g 1 L 2 1 (D) rg L 2 (D) + u H 1 (Ω) ( D |r| 2(1-µ)p p-2 ) p-2 2p r µ g L p (D) + u H 1 (Ω) g L p µ (Ω) .
This proves

g 1 L 2 1 (D) g L p µ (Ω) . (3.13)
Moreover v can be considered as a solution of

-∆v = g 1 -z v =: g 2 , in D, v = 0, on ∂D. (3.14)
By Lemma 2.7, v ∈ L p µ (D) and by Corollary 2.14, we obtain

g 2 L p µ (D) g 1 L p µ (D) + |z| v L p µ (D) g 1 L p µ (D) g L p µ (Ω) . (3.15)
Applying [13, Lemma 11.2(ii)] (as in [START_REF] Kufner | Some applications of weighted Sobolev spaces[END_REF]Example 11.3]), this allows to conclude v ∈ V 2,2 1 (D) and

v V 2,2 1 (D) = |v| V 2,2 1 (D) + |v| V 1,2 0 (D) + |v| V 0,2 -1 (D) g 2 L 2 1 (D) g L p µ (Ω) .
The case p = 2 and µ = 1 is proved.

Step 3: Estimates (c) on |c λ ′ (z)|. By (3.18) we know that, for all z ∈ π + ∪S A , the mapping L p µ (D) → C : g 1 → c λ ′ (z) is linear and continuous. Hence by Riesz theorem, there exists a unique

w λ ′ z ∈ L q -µ (D) = (L p µ (D)) ′ such that c λ ′ (z) = D g 1 wλ ′ z dx. Claim: The function w λ ′ z is characterized by (i) for all v ∈ V 2,p µ (D) ∩ D(∆ p,µ ), D (-∆v + z v) wλ ′ z = 0; (ii) for all ξ ∈ ]0, 2 -2 p -µ[ such that ∃k ∈ N, λ ′ = kλ, we have D (-∆ + z)(η(r) r ξ sin(ξθ)) wλ ′ z = δ ξλ ′ .
This can be easily deduced from (3.16) as this implies

v ∈ V 2,p µ (D) ∩ D(∆ p,µ ) ⇐⇒ ∀λ ′ ∈ ]0, 2 -2 p -µ[ with ∃k ∈ N, λ ′ = kλ, c λ ′ (z) = 0. and D (-∆ + z)(η(r) r ξ sin(ξθ)) wλ ′ z = δ ξλ ′ .
The reversed implications can be deduced by uniqueness of w λ ′ z . To obtain the good estimate on c λ ′ (z), we will decompose w λ ′ z in an appropriate way. To this aim define the function

ψ λ ′ = A λ ′ e -r √ z Q( √ zr) r -λ ′ sin(λ ′ θ), (3.20)
with

A λ ′ = 1 λ ′ ψ , Q(s) = m i=0
α i s i , with m fixed below and α i such that

α 0 = α 1 = 1, ∀i ∈ {1, . . . , m -1}, (2λ ′ -1 -2i) α i = (i + 1) (2λ ′ -i -1) α i+1 .
The positive integer m is fixed as follows: if λ ′ -1 ≥ 1 then we take m > λ ′ -1 ≥ m -1, hence we have m ≥ 1 and for i ∈ {1, . . . , m -1} we have i + 1 ≤ m ≤ λ ′ and hence i + 1 = 2λ ′ . If λ ′ -1 < 1 then we take m = 1.

The above choice leads to

(-∆ + z)(ηψ λ ′ ) = -η A λ ′ (2λ ′ -1 -2m)α m r m-λ ′ -1 z m+1 2 e -r √ z sin(λ ′ θ) +(-2∇ψ λ ′ • ∇η -ψ λ ′ ∆η).
Let t > 1 be such that (m -λ ′ -1)t + 2 > 0, by the above computation, we see that (-∆ + z)(ηψ λ ′ ) ∈ L t (D). Hence we can define

φ λ ′ ∈ H 1 0 (D) such that ∀v ∈ H 1 0 (D), D (∇φ λ ′ • ∇v + zφ λ ′ v) = D (-∆ + z)(ηψ λ ′ ) v. (3.21)
Let us show that

w λ ′ z = ηψ λ ′ -φ λ ′ . (3.22)
To this aim, let us prove that ηψ λ ′ -φ λ ′ satisfies the conditions (i) and (ii) above.

The condition (i) can be deduced from the Green formula, see Lemma 3.5 below. Consider then the condition (ii). By the Green formula (apply [START_REF] Nicaise | Polygonal interface problems, Methoden und Verfahren der mathematischen Physik[END_REF]Corollary 1.42] and remind that H In case ξ = λ ′ we have

Hence we have

D (-∆ + z)(η(r)r ξ sin(ξθ)) (η ψλ ′ -φλ ′ ) = D (-∆ + z)(η(r)r ξ sin(ξθ)) η ψλ ′ - D (-∆ + z)(ηψ λ ′ ) η(r)r ξ sin(ξθ) = - D ∆(η(r)r ξ sin(ξθ)) η ψλ ′ + D ∆(η ψλ ′ ) η(r)r ξ sin(ξθ).
D (-∆ + z)(η(r)r λ ′ sin(λ ′ θ)) (η ψλ ′ -φλ ′ ) = A λ ′ ψ 2 lim ǫ→0 2λ ′ e -ǫ √ z Q( √ zǫ) + √ ze -ǫ √ z ǫ (Q( √ zǫ) -Q ′ ( √ zǫ)) = A λ ′ ψλ ′ = 1.
This proves that w λ ′ z = ηψ λ ′ -φ λ ′ . Estimate on w λ ′ z L q -µ (D) . Observe that, as z ∈ π + ∪ S A , we can write z = |z|e iθ with θ ∈ [-θ A , θ A ] and we have |e -qr √ z | = e -qr|z| 1/2 cos(θ/2) e -γqr|z| 1/2 with γ = cos(θ A /2) > 0. Hence we obtain

ψ λ ′ q L q -µ (D) 1 0 e -γqr|z| 1/2 |Q( √ zr)| q r (-λ ′ -µ)q+1 dr.
Making the change of variable s = r|z| 1/2 we obtain

ψ λ ′ q L q -µ (D) |z| -1 2 ((-µ-λ ′ )q+2) +∞ 0 e -γqs s (-λ ′ -µ)q+1 |Q(s)| q ds.
where the integral +∞ 0 e -γqs s (-λ ′ -µ)q+1 |Q(s)| q ds < +∞ as (-λ ′ -µ)q + 2 > 0. Hence we have

ψ λ ′ L q -µ (D) |z| -1 q + µ+λ ′ 2 . (3.23) Recall that (-∆ + z)(ηψ λ ′ ) = -η A λ ′ (2λ ′ -1 -2m)α m r m-λ ′ -1 z m+1 2 e -r √ z sin(λ ′ θ) +(-2∇ψ λ ′ • ∇η -ψ λ ′ ∆η).
Hence, we see that (-∆ + z)(ηψ λ ′ ) ∈ L q -µ (D) as λ ′ < -µ + 2 -2 p and m ≥ 1. Making the change of variables s = r|z| 1/2 we obtain

e -r √ z r m-λ ′ -1 sin(λ ′ θ) q L q -µ (D) 1 0 |e -qr √ z |r (m-µ-λ ′ -1)q+1 dr ( +∞ 0 e -γqs s (m-µ-λ ′ -1)q+1 ds)|z| -q 2 (m-µ-λ ′ -1)-1 ,
where the integral +∞ 0 e -γqs s (m-µ-λ ′ -1)q+1 ds < +∞. Hence we obtain

e -r √ z r m-λ ′ -1 z m+1 2 sin(λ ′ θ) L q -µ (D) |z| -1 2 ((-µ-λ ′ -2)+ 2 q ) . (3.24)
Let us denote by r λ ′ := (-2∇ψ λ ′ • ∇η -ψ λ ′ ∆η). For |z| ≥ 1, we have

r λ ′ q L q -µ (D) 1 r 0 |e -rq √ z | r (-λ ′ -µ)q+1 |Q(r √ z)| q dr + 1 r 0 |e -rq √ z | r (-λ ′ -µ-1)q+1 |Q(r √ z)| q dr + 1 r 0 |e -rq √ z | r (-λ ′ -µ)q+1 |Q(r √ z)| q |z| q 2 dr + 1 r 0 |e -rq √ z | r (-λ ′ -µ-2)q+1 |Q(r √ z)| q dr + 1 r 0 |e -rq √ z | r (-λ ′ -µ)q+1 |Q ′ (r √ z)| q |z| q 2 dr 1 r 0 e -γrq|z| 1 2 |z| q 2 (|Q(r √ z)| q + |Q ′ (r √ z)| q ) dr + 1 r 0 e -γrq|z| 1 2 |Q(r √ z)| q dr e -γr 0 q|z| 1 2 (|z| (l+1)q 2 + 1).
September 2, 2010

Hence we obtain

r λ ′ L q -µ (D) e -|z| 1/2 γ , (3.25)
for some γ > 0.

We then deduce from (3.24) and (3.25) that

(-∆ + z)(ηψ λ ′ ) L q -µ (D) |z| 1-1 q + µ+λ ′ 2 . As (-∆ + z)(ηψ λ ′ ) ∈ L q -µ (D), by Lemma 3.4 below, φ λ ′ ∈ H 1 0 (D) ֒→ L q
-µ (D) satisfies the assumptions of Corollary 2.16, hence it satisfies

|z| φ λ ′ L q -µ (D) (-∆ + z)(ηψ λ ′ ) L q -µ (D) |z| 1-1 q + µ+λ ′ 2 i.e. φ λ ′ L q -µ (D) |z| -1 q + µ+λ ′ 2 . ( 3
w λ ′ L q -µ (D) |z| -1 q + µ+λ ′ 2
and by Hölder inequality

|c λ ′ | = | D g 1 wλ ′ | ≤ g 1 L p µ (D) w λ ′ L q -µ (D) |z| -1 q + µ+λ ′ 2 g 1 L p µ (D) .
Hence, using (3.11), we conclude that, for |z| ≥ 1,

|c λ ′ | |z| -1 q + µ+λ ′ 2 g L p µ (Ω) . ( 3
-∆+z)v R L p µ (C) +|z| v R L p µ (C) . We have (-∆ + z)v R L p µ (C) + |z| v R L p µ (C) (-∆ + z)v L p µ (D) + |z| v L p µ (D) + 0<λ ′ <2-2 p -µ ∃k∈N,λ ′ =kλ |c λ ′ (z)| (-∆ + z)(η(r) ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ)) L p µ (D) + |z| 0<λ ′ <2-2 p -µ ∃k∈N,λ ′ =kλ |c λ ′ (z)| η(r) ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ) L p µ (D) .
Recall that v is a solution of (3.10) and hence, by Corollary 2.14 and (3.11),

(-∆ + z)v L p µ (D) + |z| v L p µ (D) g 1 L p µ (D)
g L p µ (Ω) . It remains to estimate the two last terms.

Since λ ′ p + µp + 2 > 0, we have

η(r) ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ) p L p µ (D) 1 0 |P λ ′ (r √ z) e -γr|z| 1 2 r λ ′ sin(λ ′ θ)| p r µp+1 dr ∞ 0 |P λ ′ (s)| p e -γsp s λ ′ p+µp+1 ds |z| -1 2 (λ ′ p+µp+2) |z| -1 2 (λ ′ p+µp+2) .
We then deduce from (3.27) that

|z| 0<λ ′ <2-2 p -µ |c λ ′ (z)| η(r) ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ) L p µ (D) 0<λ ′ <2-2 p -µ |z| 1-1 q + λ ′ +µ 2 g L p µ (Ω) |z| -1 2 (λ ′ +µ)-1 p g L p µ (Ω) . Moreover if l λ ′ ≥ 2, we have (-∆ + z)(ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ)) p L p µ (D) = z e -r √ z r λ ′ sin(λ ′ θ) P (r √ z) p L p µ (D)
with P a polynomial function of degree l λ ′ -2. As before we obtain

(-∆ + z)(ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ)) p L p µ (D) |z| -1 2 (λ ′ p+µp+2-2p) since λ ′ p + µp + 2 > 0. Hence we have (-∆ + z)(ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ)) L p µ (D) |z| 1-1 p -λ ′ +µ 2 . (3.28) If l λ ′ = 1, then we have (-∆ + z)(ψ λ ′ (r √ z) r λ ′ sin(λ ′ θ)) = (2λ ′ + 1) √ z e -r √ z r λ ′ -1 sin(λ ′ θ),
and the same arguments than before yield (3.28) since here λ ′ -1 + µ + 2 p > 0 (reminding that if l λ ′ = 1, then 2 -µ -2 p -λ ′ < 1). As the function η does not interfere the result follows from (3.27). | φ| H 1 ( Â) , September 2, 2010 and since r is equivalent to 1 on Â, the previous estimate leads to (3.31). Now for all j ∈ N, we set A j = {(r cos θ, r sin θ) | ǫ2 -(j+1) < r < ǫ2 -j , 0 < θ < ψ}.

We then see that the linear mapping

x → x = ǫ2 -j x, is a bijection from  into A j . Now we perform the change of variables x = ǫ2 -j x and using (3.31), we obtain r=ǫ2 -j ∂ r v φrdθ |φ| H 1 ( Âj ) |v| H 1 ( Âj ) + φ L q -µ (A j ) ∆v L p µ (A j ) .

(3.32)

Since the measure of A j tends to zero as j goes to infinity, we deduce from this estimate that

I j (ǫ) := r=ǫ2 -j ∂ r v φrdθ → 0 as j → ∞.
Moreover Green's formula on B j = {(r cos θ, r sin θ) | ǫ2 -j < r < ǫ, 0 < θ < ψ}, yields I 0 (ǫ) = B j (∆v φ + ∇v • ∇ φ) + I j (ǫ).

Accordingly passing to the limit in j, we obtain that

I 0 (ǫ) = r=ǫ ∂ r v φrdθ = D(ǫ) (∆v φ + ∇v • ∇ φ),
where we recall that D(ǫ) = {(r cos θ, r sin θ) | 0 < r < ǫ, 0 < θ < ψ}. Passing to the limit as ǫ goes to zero, by Lebesgue's bounded convergence theorem, we find that I 0 (ǫ) → 0 as ǫ → 0.

This proves (3.29) thanks to (3.30). On Â, we notice that v belongs to W 1,p ( Â) and by a standard trace theorem, we deduce that These two estimates in (3.36) lead to

|J ǫ | ǫ 2-2 p -µ-λ ′ v V 2,p µ (D) .
Since 2 -2 p -µ -λ ′ is positive, we obtain (3.35).

  x)| p w p (x) dx 1/p . V k,p µ (Ω) is now defined as the closure of C ∞ S (Ω) = {v ∈ C ∞ (Ω) | S j ∈ supp v} with respect to the norm u V k,p µ (Ω) =   |γ|≤k Ω |D γ u(x)| p w p (x) r (|γ|-k)p (x) dx

(|∇u| 2 +

 2 |u| 2 ).

  2) and (2.25). Then there exists θ A ∈ ] π 2 , π[ such that, for all f ∈ L p µ (D), all z ∈ C with | arg z| ≤ θ A and u ∈ D(∆ p,µ ) solution of (2.26), we have the estimate |z| u L p µ (D) f L p µ (D) . Proof. Observe that by Lemma 2.7, we have that D(∆ p,µ ) ⊂ L p µ (D). Hence we can define the operator A : D(∆ p,µ ) ⊂ L p µ (D) → L p µ (D) : u → -∆u.

hence, by [ 18 , 2 +

 182 Thm II-5.2], there exists δ ∈ ]0, π 2 [ and M > 0 such that ρ(-A) ⊃ Σ := {z ∈ C | | arg z| < π δ} ∪ {0}, and, for all z ∈ Σ, |z| (A + zI) -1 ≤ M.

For

  z ∈ C with | arg(z)| ≤ θ A , we have z ∈ ρ(A) and by [18, Lemma I-10.2], z ∈ ρ(A * ) and (z I + A * ) -1 = ((z I + A) -1 ) * .

  + z)(η(r)r ξ sin(ξθ)) φλ ′ = D ∇(η(r)r ξ sin(ξθ)) • ∇ φλ ′ + z D η(r)r ξ sin(ξθ) φλ ′ By definition of φ λ ′ we obtain D ∇(η(r)r ξ sin(ξθ)) • ∇ φλ ′ + z D η(r)r ξ sin(ξθ) φλ ′ = D (-∆ + z)(ηψ λ ′ ) η(r)r ξ sin(ξθ).

September 2 ,

 2 2010Let us denote by D ǫ = D \ B(0, ǫ), then we can write by the Green formula-D ∆(η(r)r ξ sin(ξθ)) η ψλ ′ + D ∆(η ψλ ′ ) η(r)r ξ sin(ξθ) = lim ǫ→0 -Dǫ ∆(η(r)r ξ sin(ξθ)) η ψλ ′ + Dǫ ∆(η ψλ ′ ) η(r)r ξ sin(ξθ) r)r ξ sin(ξθ)) η ψλ ′ + ∂Dǫ ∂ ∂ν (η ψλ ′ ) η(r)r ξ sin(ξθ) r)r ξ sin(ξθ)) η ψλ ′ -∂ ∂r (η ψλ ′ ) η(r)r ξ sin(ξθ) | r=ǫ dθ.By definition of ψ λ ′ we deduce that, for ξ = λ ′ , D (-∆ + z)(η(r)r ξ sin(ξθ)) wλ ′ z = 0, as ψ 0 sin(ξθ) sin(λ ′ θ)dθ = 0 if ξ = λ ′ .

Step 4 :

 4 Estimates (a) and (b) on v R . By (3.19) we know that v R ∈ V 2,p (C) ∩ H 1 0 (C) with C the cone with interior angle ψ. By Lemma 3.2, to obtain the estimates (a) and (b), it remains to estimate (

Lemma 3 . 4 .∂ r v φd θ φ H 1 (

 341 Let p ≥ 2, D = {(r cos θ, r sin θ) | 0 < r < 1, 0 < θ < ψ} and µ satisfy (2.2). Let φ ∈ H 1 0 (D) and v ∈ D(∆ p,µ ) then we haveD (∆v φ + ∇v • ∇ φ) = 0. (3.29)Proof. For all ǫ ∈ (0, 1), we set D ǫ = {x ∈ D | r(x) > ǫ}. Since v is regular far from the origin, we can apply Green's formula on D ǫ and findDǫ (∆v φ + ∇v • ∇ φ) = -ǫ r=ǫ ∂ r v φdθ. (3.30)Since Hölder's inequality and Lemma 2.1 guarantee that the integrant of left-hand side of (3.30) is integrable on the whole domain D, Lebesgue's convergence theorem allows to conclude that the left-hand side of (3.30) tends to the left-hand side of (3.29) as ǫ → 0. Hence it remains to show that the right-hand side of (3.30) tends to 0. Now denote by  the annulus defined as follows = {(r cos θ, r sin θ) | 1/2 < r < 1, 0 < θ < ψ}.For any v ∈ H 1 ( Â) such that ∆u ∈ L p ( Â) and any φ ∈ H 1 ( Â) such that φ = 0 on θ = 0 and θ = ψ, let us show thatr=1 ∂ r v φd θ | φ| H 1 ( Â) |v| H 1 ( Â) + φ L q -µ ( Â) ∆v L p µ ( Â) . (3.31) Indeed by taking an arbitrary cut-off function η ∈ D(R) such that η(1) = 1 and η(1/2) = 0, Green's formula yields r=1 ∂ r v φd θ = Â(∆v(η φ) + ∇v • ∇(η φ)). By Hölder's inequality we obtain r=1 Â) |v| H 1 ( Â) + φ L q ( Â) ∆v L p ( Â) . Since Poincaré's inequality guarantees that φ H 1 ( Â)

Lemma 3 . 5 .

 35 With the notations of the proof of Theorem 3.3, the function ηψ λ ′ -φ λ ′ (with ψ λ ′ and φ λ ′ defined by(3.20) and (3.21)) satisfies, for all v ∈ V 2,p µ (D) ∩ D(∆ p,µ ), D (-∆v + z v) (η ψλ ′ -φλ ′ ) = 0. (3.33)Proof. For shortness we now skip the indices z and λ ′ and we use the notations of the Proof of Lemma 3.4. Thanks to (3.29), we may writeD (-∆v + z v) φ = D (∇v • ∇ φ + zv φ),and using (3.21), we obtainD (-∆v + z v) φ = D v(-∆ + z)(η ψ).Consequently if we prove thatD (-∆v + z v)η ψ = D v(-∆ + z)(η ψ),(3.34)then the difference between these two last identity yields (3.33).It remains to prove(3.34). For that purpose, we again apply Green's formula in D ǫ = {x ∈ D | r(x) > ǫ}, and findDǫ (-∆v + z v)η ψ = Dǫ v(-∆ + z)(η ψ) + J ǫ , where J ǫ = ǫ r=ǫ (v∂ r ψ -∂ r v ψ) dθ.Hence as before it suffices to show thatJ ǫ → 0 as ǫ → 0.(3.35)For that last purpose we remark that∂ r ψ = A[ √ zr) -λ ′ Q( √ zr)] e - √ zr r -λ ′ -1 sin(λ ′ θ), consequently |J ǫ | ǫ -λ ′ r=ǫ |v(ǫ, θ)| dθ + ǫ -λ ′ +1 r=ǫ |∂ r v(ǫ, θ)| dθ. (3.36) Now we estimate each term of this right-hand side separately. For the first one, we perform the change of variables x = ǫx that maps  into A 0 , and by setting v(x) = v(x), we obtain r=ǫ |v(ǫ, θ)| dθ = r=1 |v(1, θ)| d θ.

r=1 |v( 1 ,

 1 θ)| d θ ( r=1 |v(1, θ)| p d θ) 1 p v W 1,p ( Â) . Since v = 0 on a part of the boundary of Â, v W 1,p ( Â)|v| W 1,p ( Â) , and therefore we obtain r=1 |v(1, θ)| d θ |v| W 1,p ( Â) .

Finally. 2 p -µ A 0 r (µ- 1 )p |∇v| p 1 p ǫ 2 - 2 p 2 p -µ A 0 r µp |∇∂ r v| p 1 p ǫ 1 - 2 p

 201222012 as r is equivalent to 1 on Â, we arrive atr=1 |v(1, θ)| d θ Â r(µ-1)p |∇v| p 1 pGoing back to A 0 , we have proved thatr=ǫ |v(ǫ, θ)| dθ ǫ 2--µ v V 2,p µ (D) .In a similar way we show that r=ǫ |∂ r v(ǫ, θ)| dθ ǫ 1--µ v V 2,p µ (D) .

[START_REF] Maz | L p -estimates of solutions of elliptic boundary value problems in domains with edged[END_REF]

Step 2: L p -decomposition in case p > 2 or in case p = 2 and µ < 1. As µ < 2p-2 p and, by Step 1, v ∈ V 2,2 1 (D), applying [START_REF] Kufner | Some applications of weighted Sobolev spaces[END_REF]Cor (iv) of Thm 10.2 and Thm 10.3], we have

As µ < 2p-2 p , by [13, section 0.10], we have

1 (D). Hence, using (3.12), (3.15) and (3.17), we obtain

The space

Now in order to have uniform estimates with respect to |z|, we choose another decomposition. We rewrite (3.16) in the following way

where

way that (1 -ψ λ ′ (r √ z)) r λ ′ sin(λ ′ θ) ∈ V 2,p µ (D).