In this article, we solve the Stokes problem in an exterior domain ofR3, with non-standard boundary conditions

Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 196, pp. 1–28.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

STOKES PROBLEM WITH SEVERAL TYPES OF BOUNDARY CONDITIONS IN AN EXTERIOR DOMAIN

CH ´ERIF AMROUCHE, MOHAMED MESLAMENI

Abstract. In this article, we solve the Stokes problem in an exterior domain ofR³, with non-standard boundary conditions. Our approach uses weighted Sobolev spaces to prove the existence, uniqueness of weak and strong solutions.

This work is based on the vector potentials studied in [7] for exterior domains, and in [1] for bounded domains. This problem is well known in the classical Sobolev spacesW^m,2(Ω) when Ω is bounded; see [3, 4].

1. Introduction and functional setting

Let Ω⁰ denotes a bounded open inR³ of class C^1,1, simply-connected and with a connected boundary∂Ω⁰ = Γ, representing an obstacle and Ω is its complement;

i.e. Ω =R³\Ω⁰. Then a unit exterior normal vector to the boundary can be defined almost everywhere on Γ; it is denoted byn. The purpose of this paper is to solve the Stokes equation in Ω, with two types of non standard boundary conditions on Γ:

−∆u+∇π=f and divu=χ in Ω,

u·n=g and curl u×n=h×n on Γ, (1.1) and

−∆u+∇π=f and divu=χ in Ω, π=π₀, u×n=g×n on Γ and

Z

Γ

u·ndσ= 0. (1.2) Since this problem is posed in an exterior domain, our approach is to use weighted Sobolev spaces. Let us begin by introducing these spaces. A point in Ω will be denoted byx= (x₁, x₂, x₃) and its distance to the origin byr=|x|= (x²₁+x²₂+ x²₃)^1/2. We will use the weights

ρ=ρ(r) = (1 +r²)^1/2. For allminNand allkinZ, we define the weighted space

W_k^m,2(Ω) ={u∈ D⁰(Ω) :∀λ∈N³: 0≤ |λ| ≤m, ρ(r)^k−m+|λ|D^λu∈L²(Ω)},

2000Mathematics Subject Classification. 35J25, 35J50, 76M30.

Key words and phrases. Stokes equations; exterior domain; weighted Sobolev spaces;

vector potentials; inf-sup conditions.

c

2013 Texas State University - San Marcos.

Submitted July 28, 2013. Published September 3, 2013.

1

which is a Hilbert space with the norm kuk_Wm,2

k (Ω)= X^m

|λ|=0

kρ^k−m+|λ|D^λuk²_L2(Ω)

1/2

,

wherek · k_L2(Ω)denotes the standard norm of L²(Ω). We shall sometimes use the seminorm

|u|_Wm,2

k (Ω)= X

|λ|=m

kρ^kD^λuk²_L2(Ω)

1/2

.

In addition, it is established by Hanouzet in [11], for domains with a Lipschitz- continuous boundary, that D(Ω) is dense in W_k^m,2(Ω). We set ˚W_k^m,2(Ω) as the adherence of D(Ω) for the norm k · k_W^m,2

k (Ω). Then, the dual space of ˚W_k^m,2(Ω), denoting by W_−k^−m,2(Ω), is a space of distributions. Furthermore, as in bounded domain, we have form= 1 orm= 2,

W˚_k^1,2(Ω) ={v∈W_k^1,2(Ω), v= 0 on Γ}, W˚₀^2,2(Ω) ={v∈W₀^2,2(Ω), v= ∂v

∂n= 0 on Γ},

where _∂n^∂v is the normal derivative ofv. As a consequence of Hardy’s inequality, the following Poincar´e inequality holds: for m= 0 or m = 1 and for all k in Z there exists a constantC such that

∀v∈W˚_k^m,2(Ω), kvk_W^m,2

k (Ω)≤C|v|_W^m,2

k (Ω); (1.3)

i.e., the seminorm | · |_Wm,2

k (Ω) is a norm on ˚W_k^m,2(Ω) equivalent to the norm k · k_Wm,2

k (Ω).

In the sequel, we shall use the following properties. For all integers mandk in Z, we have

∀n∈Z with n≤m−k−2, Pn⊂W_k^m,2(Ω), (1.4) whereP_ndenotes the space of all polynomials (of three variables) of degree at most n, with the convention that the space is reduced to zero whennis negative. Thus the differencem−k is an important parameter of the spaceW_k^m,2(Ω). We denote byP_n^∆ the subspace of all harmonic polynomials ofPn.

Using the derivation in the distribution sense, we can define the operatorscurl and div onL²(Ω). Indeed, let h·,·i denote the duality pairing betweenD(Ω) and its dual spaceD⁰(Ω). For any function v = (v1, v2, v3)∈L²(Ω), we have for any ϕ= (ϕ1, ϕ2, ϕ3)∈D(Ω),

hcurl v,ϕi= Z

Ω

v·curlϕdx

= Z

Ω

v1(∂ϕ3

∂x₂ −∂ϕ2

∂x₃) +v2(∂ϕ1

∂x₃ −∂ϕ3

∂x₁) +v3(∂ϕ2

∂x₁ −∂ϕ1

∂x₂) dx, and for anyϕ∈ D(Ω),

hdivv, ϕi=− Z

Ω

v·gradϕ dx=− Z

Ω

v₁∂ϕ

∂x₁+v₂∂ϕ

∂x₂ +v₃ ∂ϕ

∂x₃

dx.

We note that the vector-valued Laplace operator of a vector field v= (v1, v2, v3) is equivalently defined by

∆v=grad(divv)−curl curl v. (1.5)

This leads to the following definitions:

Definition 1.1. For all integersk∈Z, we define the space H²_k(curl,Ω) ={v∈W^0,2_k (Ω);curl v∈W_k+1^0,2 (Ω)}, with the norm

kvk_H2

k(curl,Ω)=

kvk²_W0,2

k (Ω)+kcurl vk²_W0,2 k+1(Ω)

1/2

. Also we define the space

H²_k(div,Ω) ={v∈W^0,2_k (Ω); divv∈W_k+1^0,2(Ω)}, with the norm

kvk_H2

k(div,Ω)=

kvk²_W0,2

k (Ω)+kdiv vk²_W0,2 k+1(Ω)

^1/2 . Finally, we set

X²_k(Ω) =H²_k(curl,Ω)∩H²_k(div,Ω).

with the norm X²_k(Ω) =

kvk²_W0,2

k (Ω)+kdivvk²_W0,2

k+1(Ω)+kcurl vk²_W0,2 k+1(Ω)

1/2

. These definitions will also be used with Ω replaced byR³.

The argument used by Hanouzet [11] to prove the denseness ofD(Ω) inW_k^m,2(Ω) can be easily adapted to establish thatD(Ω) is dense in the space H²_k(div,Ω) and in the space H²_k(curl,Ω) and so inX²_k(Ω). Therefore, denoting bynthe exterior unit normal to the boundary Γ, the normal trace v·nand the tangential trace v×ncan be defined respectively in H^−1/2(Γ) for the functions ofH²_k(div,Ω) and in H^−1/2(Γ) for functions ofH²_k(curl,Ω), where H^−1/2(Γ) denotes the dual space of H^1/2(Γ). They satisfy the trace theorems; i.e, there exists a constantC such that

∀v∈H²_k(div,Ω), kv·nk_H−1/2(Γ)≤Ckvk_H2

k(div,Ω), (1.6)

∀v∈H²_k(curl,Ω), kv×nk_H−1/2(Γ)≤Ckvk_H2

k(curl,Ω) (1.7) and the following Green’s formulas holds: For anyv∈H²_k(div,Ω) andϕ∈W_−k^1,2(Ω)

hv·n, ϕiΓ = Z

Ω

v· ∇ϕ dx+ Z

Ω

ϕdivvdx, (1.8)

where h,iΓ denotes the duality pairing between H^−1/2(Γ) and H^1/2(Γ). For any v∈H²_k(curl,Ω) andϕ∈W^1,2_−k(Ω)

hv×n,ϕiΓ= Z

Ω

v·curlϕdx− Z

Ω

curl v·ϕdx, (1.9) whereh,iΓ denotes the duality pairing betweenH^−1/2(Γ) andH^1/2(Γ).

Remark 1.2. If v belongs to H²_k(div,Ω) for some integer k≥1, then divv is in L¹(Ω) and Green’s formula (1.8) yields

hv·n,1iΓ= Z

Ω

divvdx (1.10)

But whenk≤0, then divvis not necessarily inL¹(Ω) and (1.10) is generally not valid. Note also that whenk≤0,W_−k−1^0,2 (Ω) does not contain the constants.

The closures ofD(Ω) inH²_k(div,Ω) and inH²_k(curl,Ω) are denoted respectively by ˚H²_k(curl,Ω) and ˚H²_k(div,Ω) and can be characterized respectively by

H˚²_k(curl,Ω) ={v∈H²_k(curl,Ω) :v×n=0on Γ}, H˚²_k(div,Ω) ={v∈H²_k(div,Ω) :v·n= 0 on Γ}.

Their dual spaces are characterized by the following propositions:

Proposition 1.3. A distributionf belongs to[˚H²_k(div,Ω)]⁰if and only if there exist ψ∈W^0,2_−k(Ω)andχ∈W_−k−1^0,2 (Ω), such thatf =ψ+gradχ. Moreover

kfk_[˚H2

k(div,Ω)]⁰ = max{kψk_W0,2

−k(Ω),kχk_W0,2

−k−1(Ω)}. (1.11) Proof. Letψ∈W^0,2_−k(Ω) and χ∈W_−k−1^0,2 (Ω), we have

∀v∈D(Ω), hψ+gradχ,vi_D⁰_{(Ω)×D(Ω)}= Z

Ω

(ψ·v−χdivv) dx.

Therefore, the linear mapping `: v 7−→ R

Ω(ψ·v−χdivv)dx defined on D(Ω) is continuous for the norm of ˚H²_k(div,Ω). Since D(Ω) is dense in ˚H²_k(div, Ω), ` can be extended by continuity to a mapping still called ` ∈ [˚H²_k(div,Ω)]⁰. Thus ψ+gradχ is an element of [˚H²_k(div,Ω)]⁰.

Conversely, LetE=W^0,2_k (Ω)×W_k+1^0,2(Ω) equipped by the following norm kvkE= (kvk²_W0,2

k (Ω)+kdivvk²_W0,2

k+1(Ω))^1/2.

The mapping T :v∈H˚²_k(div,Ω)→(v,divv)∈E is an isometry from ˚H²_k(div,Ω) in E. Suppose G = T(˚H²_k(div,Ω)) with the E-topology. Let S = T⁻¹ : G → H˚²_k(div,Ω). Thus, we can define the following mapping:

v∈G7→ hf, Svi_[˚H2

k(div,Ω)]⁰×H˚²_k(div,Ω) forf ∈[˚H²_k(div,Ω)]⁰

which is a linear continuous form onG. Thanks to Hahn-Banach’s Theorem, such form can be extended to a linear continuous form onE, denoted by Υ such that

kΥk_E⁰ =kfk_[˚H2

k(div,Ω)]⁰. (1.12)

From the Riesz’s Representation Lemma, there exist functions ψ ∈W^0,2_−k(Ω) and χ∈W_−k−1^0,2 (Ω), such that for anyv= (v1, v2)∈E,

hΥ,viE⁰×E= Z

Ω

v1·ψdx+ Z

Ω

v2χ dx, withkΥkE⁰ = max{kψk_W0,2

−k(Ω),kχk_W0,2

−k−1(Ω)}. In particular, ifv=Tϕ∈G, where ϕ∈D(Ω), we have

hf,ϕi_[˚H2

k(div,Ω)]⁰×H˚²_k(div,Ω) =hψ− ∇χ,ϕi_[˚H2

k(div,Ω)]⁰×H˚²_k(div,Ω),

and (1.11) follows imeddiatly from (1.12).

We skip the proof of the following result as it is similar to that of Proposition 1.3.

Proposition 1.4. A distribution f belongs to [˚H²_k(curl,Ω)]⁰ if and only if there exist functions ψ ∈ W^0,2_−k(Ω) and ξ ∈ W^0,2_−k−1(Ω), such that f = ψ+curlξ.

Moreover

kfk_[˚H2

k(curl,Ω)]⁰ = max{kψk_W0,2

−k(Ω),kξk_W0,2

−k−1(Ω)}.

Definition 1.5. LetX²_k,N(Ω),X²_k,T(Ω) and ˚X²_k(Ω) be the following subspaces of X²_k(Ω):

X²_k,N(Ω) ={v∈X²_k(Ω);v×n=0on Γ}, X²_k,T(Ω) ={v∈X²_k(Ω);v·n= 0 on Γ},

X˚²_k(Ω) =X²_k,N(Ω)∩X²_k,T(Ω).

2. Preliminary results

Now, we give some results related to the Dirichlet problem and Neumann problem which are essential to ensure the existence and the uniqueness of some vectors potentials and one usually forces either the normal component to vanish or the tangential components to vanish. We start by giving the definition of the kernel of the Laplace operator for any integerk∈Z:

A^∆_k−1={χ∈W_−k^1,2(Ω) : ∆χ= 0 in Ω andχ= 0 on Γ}.

In contrast to a bounded domain, the Dirichlet problem for the Laplace operator with zero data can have nontrivial solutions in an exterior domain; it depends upon the exponent of the weight. The result that we state below is established by Giroire in [10].

Proposition 2.1. For any integer k ≥ 1, the space A^∆_k−1 is a subspace of all functions in W_−k^1,2(Ω) of the form v(p)−p, where p runs over all polynomials of P_k−1^∆ andv(p)is the unique solution in W₀^1,2(Ω) of the Dirichlet problem

∆v(p) = 0 in Ω and v(p) =p onΓ. (2.1) The spaceA^∆_k−1 is a finite-dimentional space of the same dimension asP_k−1^∆ and A^∆_k−1={0} whenk≤0.

Our second proposition is established also by Giroire in [10], it characterizes the kernel of the Laplace operator with Neumann boundary condition. For any integer k∈Z,

N_k−1^∆ ={χ∈W_−k^1,2(Ω) : ∆χ= 0 in Ω and ∂χ

∂n = 0 on Γ}.

Proposition 2.2. For any integer k ≥ 1, N_k−1^∆ the subspace of all functions in W_−k^1,2(Ω) of the formw(p)−p, wherepruns over all polynomials ofP_k−1^∆ andw(p) is the unique solution in W₀^1,2(Ω) of the Neumann problem

∆w(p) = 0 inΩ and ∂w(p)

∂n = ∂p

∂n onΓ. (2.2)

Here also, we set N_k−1^∆ ={0} when k ≤0; N_k−1^∆ is a finite-dimentional space of the same dimension asP_k−1^∆ and in particular, N₀^∆=R.

Next, the uniqueness of the solutions of Problem (1.1) and Problem (1.2) will follow from the characterization of the kernel. For all integerskin Z, we define

Y²_k,N(Ω) ={w∈X²_−k,N(Ω) : divw= 0 and curl w=0in Ω}

Y_k,T² (Ω) ={w∈X²_−k,T(Ω) : divw= 0 and curl w=0in Ω}.

The proof of the following propositions can be easily deduced from [7].

Proposition 2.3. Letk∈Zand suppose thatΩ⁰ is of classC^1,1, simply-connected and with a Lipshitz-continuous and connected boundaryΓ.

• If k <1, thenY_k,N² (Ω) ={0}.

• Ifk≥1, thenY_k,N² (Ω) ={∇(v(p)−p), p∈ P_k−1^∆ }, wherev(p)is the unique solution inW₀^1,2(Ω) of the Dirichlet problem (2.1).

Proof. Letk ∈Z and let w ∈X²_−k,N(Ω) such that divw = 0 and curl w=0 in Ω. Then since Ω⁰ is simply-connected, there existsχ ∈W_−k^1,2(Ω), unique up to an additive constant, such that w =∇χ. But w×n = 0, hence, χ is constant on Γ (Γ is a connected boundary) and we choose the additive constant in χ so that χ = 0 on Γ. Thus χ belongs to A^∆_k−1(Ω) . Due to Proposition 2.1, we deduce that ifk <1,χ is equal to zero and if k≥1,χ =v(p)−p, wherepruns over all polynomials of P_k−1^∆ and v(p) is the unique solution in W₀^1,2(Ω) of problem (2.1) and thusw=∇(v(p)−p). Now, to finish the proof we shall prove that∇(v(p)−p) belongs to Y_k,N² (Ω) and this is a simple consequence of the definition of p and

v(p).

We skip the proof of the following result as it is entirely similar to that of Proposition 2.3.

Proposition 2.4. Let the assumptions of Proposition 2.4 hold.

• If k <1, thenY_k,T² (Ω) ={0}.

• If k ≥ 1, then Y²_k,T(Ω) = {∇(w(p)−p), p ∈ P_k−1^∆ }, where w(p) is the unique solution inW₀^1,2(Ω) of the Neumann problem (2.2)

The imbedding results that we state below are established by Girault in [7]. The first imbedding result is given by the following theorem.

Theorem 2.5. Let k 6 2 and assume that Ω⁰ is of class C^1,1. Then the space X²_k−1,T(Ω)is continuously imbedded inW^1,2_k (Ω). In addition there exists a constant C such that for anyϕ∈X²_k−1,T(Ω),

kϕk_W^1,2

k (Ω)≤C

kϕk_W^0,2

k−1(Ω)+kdivϕk_W^0,2

k (Ω)+kcurlϕk_W^0,2

k (Ω)

. (2.3) If in addition, Ω⁰ is simply-connected, there exists a constant C such that for all ϕ∈X²_k−1,T(Ω) we have

kϕk_W^1,2

k (Ω)≤C(kdivϕk_W^0,2

k (Ω)+kcurlϕk_W^0,2

k (Ω)

+

N(−k)

X

j=2

| Z

Γ

ϕ· ∇w(qj)dσ|), (2.4) where {q_j}^N(−k)_j=2 denotes a basis of {q ∈ P_−k^∆ : q(0) = 0}, N(−k) denotes the dimension of P_−k^∆ and w(q_j) is the corresponding function of N_−k^∆. Thus, the

seminorm in the right-hand side of (2.4)is a norm onX²_k−1,T(Ω)equivalent to the normkϕk_W^1,2

k (Ω).

The second imbedding result is given by the following theorem.

Theorem 2.6. Let k 6 2 and assume that Ω⁰ is of class C^1,1. Then the space X²_k−1,N(Ω)is continuously imbedded inW^1,2_k (Ω). In addition there exists a constant C such that for anyϕ∈X²_k−1,N(Ω),

kϕk_W^1,2

k (Ω)≤C

kϕk_W^0,2

k−1(Ω)+kdivϕk_W^0,2

k (Ω)+kcurlϕk_W^0,2

k (Ω)

. (2.5) If in addition,Ω⁰ is simply-connected and its boundaryΓ is connected, there exists a constant C such that for allϕ∈X²_k−1,N(Ω)we have

kϕk_W1,2

k (Ω)≤C(kdivϕk_W0,2

k (Ω)+kcurlϕk_W0,2 k (Ω)

+| Z

Γ

(ϕ·n)dσ|+

N(−k)

X

j=1

| Z

Γ

(ϕ·n)qjdσ|), (2.6) where the term|R

Γ(ϕ·n)dσ|can be dropped if k6= 1 and where{qj}^N(−k)_j=1 denotes a basis of P_−k^∆ . In other words, the seminorm in the right-hand side of (2.6) is a norm onX²_k−1,N(Ω) equivalent to the normkϕk_W1,2

k (Ω).

Finally, let us recall the abstract setting of Babuˇska-Brezzi’s Theorem (see Babuˇska [5], Brezzi [6] and Amrouche-Selloula [4]).

Theorem 2.7. Let X and M be two reflexive Banach spaces and X⁰ and M⁰ their dual spaces. Let a be the continuous bilinear form defined on X ×M, let A∈ L(X; M⁰) andA⁰∈ L(M; X⁰)be the operators defined by

∀v∈X, ∀w∈M, a(v, w) =hAv, wi=hv, A⁰wi andV = kerA. The following statements are equivalent:

(i) There exist β >0 such that

w∈M, w6=0inf sup

v∈X, v6=0

a(v, w)

kvk_Xkwk_M ≥β. (2.7)

(ii) The operatorA :X/V 7→M⁰ is an isomophism and 1/β is the continuity constant ofA⁻¹.

(iii) The operatorA⁰ :M 7→X⁰⊥V is an isomophism and1/β is the continuity constant of(A⁰)⁻¹.

Remark 2.8. As consequence, if the inf-sup condition (2.7) is satisfied, then we have the following properties:

(i) IfV ={0}, then for any f ∈X⁰, there exists a uniquew∈M such that

∀v∈X, a(v, w) =hf, vi and kwkM ≤ 1

βkfkX⁰. (2.8) (ii) If V 6= {0}, then for any f ∈ X⁰, satisfying the compatibility condition:

∀v∈V, hf, vi= 0, there exists a uniquew∈M such that (2.8).

(iii) For any g ∈M⁰, there exists v ∈X, unique up an additive element of V, such that:

∀w∈M, a(v, w) =hg, wi and kvk_X/V ≤ 1 βkgk_M⁰.

3. Inequalities and inf-sup conditions

In this sequel, we prove some imbedding results. More precisely, we show that the results of Theorem 2.5 and the result of Theorem 2.6 can be extended to the case where the boundary conditions v·n = 0 or v×n = 0 on Γ are replaced by inhomogeneous one. Next, we study some problems posed in an exterior do- main which are essentials to prove the regularity of solutions for Problem (1.1) and Problem (1.2).

For any integerk inZ, we introduce the following spaces:

Z²_k,T(Ω) ={v∈X²_k(Ω) and v·n∈H^1/2(Γ)}, Z²_k,N(Ω) ={v∈X²_k(Ω) andv×n∈H^1/2(Γ)}

and

M²_k,T(Ω) ={v∈W^1,2_k+1(Ω), divv∈W_k+2^1,2(Ω), curl v∈W^1,2_k+2(Ω) andv·n∈H^3/2(Γ)}.

Proposition 3.1. Let k = −1 or k = 0, then the space Z²_k,T(Ω) is continuously imbedded in W^1,2_k+1(Ω) and we have the following estimate for any vin Z²_k,T(Ω):

kvk_W1,2

k+1(Ω)≤C kvk_W0,2

k (Ω)+kcurl vk_W0,2

k+1(Ω)+kdivvk_W0,2

k+1(Ω)+kv·nk_H1/2(Γ)

. (3.1) Proof. Letk =−1 or k= 0 and let v any function ofZ²_k,T(Ω). Let us study the Neumann problem

∆χ= divvin Ω and ∂_nχ=v·non Γ. (3.2) It is shown in [7, Theorems 3.7 and 3.9], that Problem (3.2) has a unique solution χinW_k+1^2,2(Ω)/Rifk=−1 andχis unique inW_k+1^2,2(Ω) ifk= 0. With the estimate

k∇χk_W1,2

k+1(Ω)≤C kdivvk_W0,2

k+1(Ω)+kv·nk_H1/2(Γ)

. (3.3)

Let w = v−gradχ, then w is a divergence-free function. Since W^1,2_k+1(Ω) ,→ W^0,2_k (Ω), then w ∈ X²_k,T(Ω). Applying Theorem 2.5, we have w belongs to W^1,2_k+1(Ω) and thenvis in W_k+1^1,2 (Ω). According to Inequality (2.3), we obtain

kwk_W1,2

k+1(Ω)≤C kwk_W0,2

k (Ω)+kcurl wk_W0,2 k+1(Ω)

.

Then, inequality (3.1) follows directly from (3.3).

Similarly, we can prove the following imbedding result.

Proposition 3.2. Suppose that Ω⁰ is of class C^2,1. Then the spaceM²_−1,T(Ω) is continuously imbedded inW^2,2₁ (Ω)and we have the following estimate for anyvin M²_−1,T(Ω):

kvk_W^2,2

1 (Ω)≤C kvk_W^1,2

0 (Ω)+kcurl vk_W^1,2

1 (Ω)+kdivvk_W^1,2

1 (Ω)+kv·nk_H3/2(Γ)

. (3.4) Proof. Proceeding as in Proposition 3.1. Let v in M²_−1,T(Ω). Since Ω⁰ is of class C^2,1, then according to [7, Theorem 3.9], there exists a unique solution χ

inW₁^3,2(Ω)/Rof Problem (3.2). Settingw=v−gradχ. SinceW₁^2,2(Ω) is imbed- ded inW₀^1,2(Ω), it follows from [7, Corollary 3.16], thatw belongs toW₁^2,2(Ω) and moreover we have the estimate

kwk_W2,2

1 (Ω)≤C kwk_W1,2

0 (Ω)+kcurl wk_W1,2 1 (Ω)

.

Thenv=w+gradχ belongs toW^2,2₁ (Ω) and we have the estimate (3.4).

Although we are under the Hilbertian case but the Lax-Milgram lemma is not always valid to ensure the existence of solutions. Thus, we shall establish two “inf- sup” conditions in order to apply Theorem 2.7. First recall the following spaces for all integersk∈Z:

V²_k,T(Ω) =n

z∈X²_k,T(Ω) : divz= 0 in Ω and Z

Γ

z· ∇(w(q)−q)dσ= 0,∀(w(q)−q)∈ N_−k−1^∆ o and

V²_k,N(Ω) ={z∈X²_k,N(Ω) : divz= 0 in Ω and Z

Γ

(z·n)q dσ= 0,∀q∈ P_−k−1^∆ }.

The first “inf-sup” condition is given by the following lemma.

Lemma 3.3. The following inf-sup Condition holds: there exists a constantβ >0, such that

inf

ϕ∈V_0,T² (Ω),ϕ6=0

sup

ψ∈V²_−2,T(Ω),ψ6=0

R

Ωcurlψ·curlϕdx kψk_X²

−2,T(Ω)kϕk_X²

0,T(Ω)

≥β. (3.5)

Proof. Letg∈W₋₁^0,2(Ω) and let us introduce the Dirichlet problem

−∆χ= divg in Ω, χ= 0 on Γ.

It is shown in [7, Theorem 3.5], that this problem has a solution χ ∈ W˚₋₁^1,2(Ω) unique up to an element ofA^∆₀ and we can chooseχsuch that

k∇χk_W0,2

−1(Ω)≤Ckgk_W0,2

−1(Ω).

Setz=g− ∇χ. Then we havez∈W^0,2₋₁(Ω), divz= 0 and we have kzk_W^0,2

−1(Ω)≤Ckgk_W^0,2

−1(Ω). (3.6)

Letϕany function ofV²_0,T(Ω), by Theorem 2.5 we haveϕ∈X²_0,T(Ω),→W^1,2₁ (Ω).

Then due to (2.4) we can write kϕk_X2

0,T(Ω)≤Ckcurlϕk_W0,2

1 (Ω)=C sup

g∈W^0,2₋₁(Ω),g6=0

R

Ωcurlϕ·gdx kgk_W^0,2

−1(Ω)

. (3.7) Using the fact thatcurlϕ∈H²₁(div,Ω) and applying (1.8), we obtain

Z

Ω

curlϕ· ∇χdx= 0. (3.8)

Now, letλ∈W₀^1,2(Ω) the unique solution of the problem

∆λ= 0 in Ω and λ= 1 on Γ.

It follows from [7, Lemma 3.11] that Z

Γ

∂λ

∂ndσ=C1>0.

Now, setting

ez=z− 1 C1

hz·n,1iΓ∇λ.

It is clear thatez∈W^0,2₋₁(Ω), divez= 0 in Ω and that hez ·n,1iΓ = 0. Due to [7, Theorem 3.15], there exists a potential vectorψ∈W^1,2₋₁(Ω) such that

ez=curlψ, divψ= 0 in Ω and ψ·n= 0 on Γ. (3.9) and we have

∀v(q)∈ N₁^∆, Z

Γ

ψ· ∇v(q)dσ= 0. (3.10) In addition, we have the estimate

kψk_W^1,2

−1(Ω)≤Ckezk_W^0,2

−1(Ω)≤Ckzk_W^0,2

−1(Ω). (3.11)

Using (3.10), we obtain that ψ belongs to V_−2,T² (Ω). Sinceϕ is H¹ in a neigh- borhood of Γ, thenϕhas anH¹extension in Ω⁰ denoted byϕ. Applying Green’se formula in Ω⁰, we obtain

0 = Z

Ω⁰

div(curlϕ) dxe =hcurlϕe·n,1iΓ=hcurlϕ·n,1iΓ.

Using the fact thatcurlϕinH²₁(div,Ω) andλin W₋₁^1,2(Ω) and applying (1.8), we obtain

0 =hcurlϕ·n,1iΓ=hcurlϕ·n, λiΓ= Z

Ω

curlϕ· ∇λdx. (3.12) Using (3.8) and (3.12), we deduce that

Z

Ω

curlϕ·gdx= Z

Ω

curlϕ·zdx= Z

Ω

curlϕ·ezdx. (3.13) From (3.11), (3.6) and (3.13), we deduce that

R

Ωcurlϕ·gdx kgk_W0,2

−1(Ω)

≤C R

Ωcurlϕ·ezdx kezk_W0,2

−1(Ω)

=C R

Ωcurlϕ·curlψdx kcurlψk_W0,2

−1(Ω)

.

Applying again (2.4) and using (3.10), we obtain

R

Ωcurlϕ·gdx kgk_W0,2

−1(Ω)

≤C R

Ωcurlϕ·curlψdx kψk_X2

−2,T(Ω)

,

and the inf-sup Condition (3.5) follows immediately from (3.7).

The second ”inf sup” condition is given by the following lemma:

Lemma 3.4. The following inf-sup Condition holds: there exists a constantβ >0, such that

inf

ϕ∈V²_−2,N(Ω),ϕ6=0

sup

ψ∈V_0,N² (Ω),ψ6=0

R

Ωcurlψ·curlϕdx kψk_X²

0,N(Ω)kϕk_X²

−2,N(Ω)

≥β. (3.14)

Proof. The proof is similar to that of Lemma 3.3. Let g ∈ W^0,2₁ (Ω) and let us introduce the generalized Neumann problem

div(∇χ−g) = 0 in Ω and (∇χ−g)·n= 0 on Γ. (3.15) It follows from [10] that Problem (3.15) has a solutionχ∈W₁^1,2(Ω) and we have

k∇χk_W0,2

1 (Ω)≤Ckgk_W0,2 1 (Ω).

Settingz=g− ∇χ, then we havez∈H˚²₁(div,Ω) and divz= 0 with the estimate kzk_W0,2

1 (Ω)6Ckgk_W0,2

1 (Ω). (3.16)

Letϕ be any function of V²_−2,N(Ω). Due to Theorem 2.6, we have X²_−2,N(Ω),→ W^1,2₋₁(Ω) and by (2.6) we can write

kϕk_X²

−2,N(Ω)≤Ckcurlϕk_W^0,2

−1(Ω)=C sup

g∈W₁^0,2(Ω),g6=0

R

Ωcurlϕ·gdx kgk_W0,2

1 (Ω)

. (3.17) Observe thatcurlϕbelongs toH²₋₁(div,Ω) withϕ×n=0on Γ andχ∈W₁^1,2(Ω).

Then using (1.8), we obtain Z

Ω

curlϕ· ∇χ dx=hcurlϕ·n, χiΓ = 0. (3.18) Due to [7, Proposition 3.12], there exists a potential vectorψ∈W^1,2₁ (Ω) such that z=curlψ, divψ= 0 in Ω and ψ×n=0 on Γ, (3.19)

Z

Γ

ψ·ndσ= 0. (3.20)

In addition, we have

kψk_W1,2

1 (Ω)≤Ckzk_W0,2

1 (Ω). (3.21)

Then, we deduce that ψ belongs to V²_0,N(Ω). Using (3.16), (3.18) and (3.19), we deduce that

R

Ωcurlϕ·gdx kgk_W0,2

1 (Ω)

≤C R

Ωcurlϕ·zdx kzk_W0,2

1 (Ω)

=C R

Ωcurlϕ·curlψdx kcurlψk_W0,2

1 (Ω)

.

Applying again (2.6) and using (3.20), we obtain

R

Ωcurlϕ·gdx kgk_W0,2

1 (Ω)

≤C R

Ωcurlϕ·curlψdx kψk_X2

0,N(Ω)

,

and the inf-sup condition (3.14) follows immediately from (3.17).

4. Elliptic problems with different boundary conditions Next, we study the problem

−∆ξ=f and divξ= 0 in Ω, ξ×n=g×n on Γ and

Z

Γ

(ξ·n)q dσ= 0, ∀q∈ P_k^∆. (4.1)

Proposition 4.1. Let k = −1 or k = 0 and suppose that g×n = 0 and let f ∈[˚H²_k−1(curl,Ω)]⁰ with divf = 0in Ωand satisfying the compatibility condition

∀v∈Y_1−k,N² (Ω), hf,vi_[˚H2

k−1(curlΩ)]⁰×H˚²_k−1(curlΩ)= 0. (4.2) Then, Problem (4.1)has a unique solution in W^1,2_−k(Ω) and we have

kξk_W1,2

−k(Ω)6Ckfk_[˚H2

k−1(curl,Ω)]⁰. (4.3)

Moreover, if f in W_−k+1^0,2 (Ω) and Ω⁰ is of class C^2,1, then the solution ξ is in W^2,2_−k+1(Ω)and satisfies the estimate

kξk_W2,2

−k+1(Ω)6Ckfk_W0,2

−k+1(Ω). (4.4)

Proof. (i) On the one hand, observe that Problem (4.1) is reduced to the variational problem: Findξ∈V²_−k−1,N(Ω) such that

∀ϕ∈X²_k−1,N(Ω), Z

Ω

curlξ·curlϕdx=hf,ϕiΩ, (4.5) where the duality on Ω is

h·,·iΩ=h·, ·i_[˚H2

k−1(curl,Ω)]⁰×H˚²_k−1(curl,Ω).

On the other hand, (4.5) is equivalent to the problem: Findξ∈V²_−k−1,N(Ω) such that

∀ϕ∈V²_k−1,N(Ω), Z

Ω

curlξ·curlϕdx=hf,ϕiΩ. (4.6) Indeed, every solution of (4.5) also solves (4.6). Conversely, assume that (4.6) holds, and letϕ∈X²_k−1,N(Ω). Let us solve the exterior Dirichlet problem

−∆χ= divϕ in Ω and χ= 0 on Γ. (4.7) It is shown in [7, Theorem 3.5] that problem (4.7) has a unique solution χ ∈ W_k^2,2(Ω)/A^∆_−k.

First case. ifk= 0, we set

ϕe =ϕ− ∇χ− 1 C1

hϕ− ∇χ,1iΓ∇(v(1)−1),

wherev(1) is the unique solution inW₀^1,2(Ω) of the Dirichlet problem (2.1) and C1=

Z

Γ

∂v(1)

∂n dσ.

It follows from [7, Lemma 3.11] that C1 > 0 and since ∇(v(1)−1) belongs to Y²_1,N(Ω), we deduce thatϕe belongs toV²_−1,N(Ω).

Second case. ifk=−1, for each polynomialpinP₁^∆, we takeϕe of the form ϕe =ϕ− ∇χ− ∇(v(p)−p)

where v(p) is the unique solution in W₀^1,2(Ω) of the Dirichlet problem (2.1). The polynomialpis chosen to satisfy the condition

Z

Γ

(ϕe·n)q dσ= 0 ∀q∈ P₁^∆. (4.8)

To show that this is possible, letT be a linear form defined byT :P₁^∆→R⁴, T(p) =Z

Γ

∂(v(p)−p)

∂n dσ, Z

Γ

∂(v(p)−p)

∂n x1dσ, Z

Γ

∂(v(p)−p)

∂n x2dσ, Z

Γ

∂(v(p)−p)

∂n x3dσ ,

where{1, x1, x2, x3}denotes a basis ofP₁^∆. It is shown in the proof of [8, Theorem 7], that if

Z

Γ

∂(v(p)−p)

∂n q dσ= 0 ∀q∈ P₁^∆,

thenp= 0. This implies thatT is injective and so bijective. And so, there exists a uniquepin P₁^∆ so that condition (4.8) is satisfied and since∇(v(p)−p) belongs toY²_2,N(Ω), we prove thatϕe ∈V²_−2,N(Ω).

Finally, using (4.2), we obtain fork= 0 andk=−1 that hf,∇(v(p)−p)iΩ= 0 and hf,∇(v(1)−1)iΩ= 0 and asD(Ω) is dense in ˚H²_k−1(curl,Ω), we obtain that

hf,∇χi_Ω= 0.

Then we have Z

Ω

curlξ·curlϕdx= Z

Ω

curlξ·curlϕedx=hf,ϕiΩ.

Then Problem (4.5) and Problem (4.6) are equivalent. Now, to solve Problem (4.6), we use Lax-Milgram lemma fork= 0 and the inf-sup condition (3.14) fork=−1.

Let us start byk= 0. We consider the bilinear forma:V²_−1,N(Ω)×V_−1,N² (Ω)→R such that

a(ξ,ϕ) = Z

Ω

curlξ·curlϕdx.

According to Theorem 2.6,ais continuous and coercive onV²_−1,N(Ω). Due to Lax- Milgram lemma, there exists a unique solution ξ ∈ V²_−1,N(Ω) of Problem (4.6).

Using again Theorem 2.6, we prove that this solution ξ belongs to W^1,2₀ (Ω) and the following estimate follows immediately

kξk_W1,2

0 (Ω)6Ckfk_[˚H2

−1(curl,Ω)]⁰. (4.9)

When k =−1, we have that Problem (4.6) satisfies the inf-sup condition (3.14).

Let us consider the mapping ` : V<sup>2</sup><sub>−2,N</sub>(Ω) →R such that `(ϕ) = hf,ϕiΩ. It is clear that ` belongs to (V²_−2,N(Ω))⁰ and according to Remark 2.8, there exists a unique solutionξ∈V²_0,N(Ω) of Problem (4.6). Due to Theorem 2.6, we prove that this solutionξbelongs toW₁^1,2(Ω). It follows from Remark 2.8 i) that

kξk_W1,2

1 (Ω)6Ckfk_[˚H2

−2(curl,Ω)]⁰. (4.10) (ii) We suppose in addition thatf is inW^0,2_−k+1(Ω) fork=−1 ork= 0 and Ω⁰ is of classC^2,1and we setz=curlξ, whereξ∈W^1,2_−k(Ω) is the unique solution of Problem (4.1). Then we have

z∈W_−k^0,2(Ω), curl z=f ∈W^0,2_−k+1(Ω), divz= 0 and z·n= 0 on Γ

and thus z belongs to X²_−k,T(Ω). By Theorem 2.5, we prove that z belongs to W^1,2_−k+1(Ω) and using (2.4), we prove thatzsatisfies

kzk_W1,2

−k+1(Ω)6Ckfk_W0,2

−k+1(Ω). (4.11)

As a consequenceξsatisfies

ξ∈W^1,2_−k(Ω), curlξ∈W^1,2_−k+1(Ω), divξ= 0 and ξ×n=0 on Γ.

Applying [7, Corollary 3.14], we deduce thatξbelongs toW^2,2_−k+1(Ω) and using in addition the boundary condition of (4.1) we prove that

kξk_W2,2

−k+1(Ω)6Ckcurlξk_W1,2

−k+1(Ω). (4.12)

Finally, estimate (4.4) follows from (4.11) and (4.12).

Corollary 4.2. Letk=−1ork= 0and letf ∈[˚H²_k−1(curl,Ω)]⁰ withdivf = 0in Ωandg∈H^1/2(Γ)and satisfying the compatibility condition (4.2). Then, Problem (4.1)has a unique solutionξ inW^1,2_−k(Ω)and we have:

kξk_W^1,2

−k(Ω)6C kfk_[˚H2

k−1(curl,Ω)]⁰+kg×nk_H1/2(Γ)

. (4.13)

Moreover, iff inW^0,2_−k+1(Ω),ginH^3/2(Γ)andΩ⁰is of classC^2,1, then the solution ξis in W^2,2_−k+1(Ω) and satisfies

kξk_W2,2

−k+1(Ω)6C kfk_W0,2

−k+1(Ω)+kg×nk_H3/2(Γ)

. (4.14)

Proof. Letk= 0 ork=−1 and let g∈H^1/2(Γ). We know that there exists ξ₀ in H¹(Ω) with compact support satisfying

ξ₀=gτ on Γ and divξ₀= 0 in Ω,

where gτ is the tangential component of gon Γ. Since support ofξ₀ is compact, we deduce thatξ₀ belongs toW^1,2_−k(Ω) for k=−1 ork= 0 and satisfies

kξ₀k_W1,2

−k(Ω)≤Ckgτk_H1/2(Γ). (4.15) Settingz=ξ−ξ₀, then Problem (4.1) is equivalent to: findz∈W^1,2_−k(Ω) such that

−∆z=f+ ∆ξ₀ and divz= 0 in Ω, z×n=0 on Γ and

Z

Γ

(z·n)q dσ= 0, ∀q∈ P_k^∆. (4.16) Observe thatF =f −curl curlξ₀ belongs to [˚H²_k−1(curl,Ω)]⁰. Since D(Ω) is dense in ˚H²_k−1(curl,Ω), we have for any v∈Y_1−k,N² (Ω):

hcurl curlξ₀,viΩ= Z

Ω

curlξ₀·curl vdx= 0.

Thus F satisfies the compatibility condition (4.2). Due to Proposition 4.1, there exists a uniquez∈W^1,2_−k(Ω) solution of problem (4.16) such that

kzk_W^1,2

−k(Ω)6CkFk_[˚H2

k−1(curl,Ω)]⁰ ≤C kfk_[˚H2

k−1(curl,Ω)]⁰+kcurlξ₀k_W^0,2

−k(Ω)

. (4.17) Then ξ=z+ξ₀ belongs to W_−k^1,2(Ω) is the unique solution of (5.8) and estimate (4.13) follows immediately from (4.15) and (4.17).

Regularity of the solution: Suppose in addition that Ω⁰ is of class C^2,1, f in W^0,2_−k+1(Ω) andginH^3/2(Γ). Then the functionξ₀defined above belongs toH²(Ω) with compact support and thusξ₀ belongs toW_−k+1^2,2 (Ω) and we have

kξ₀k_W2,2

−k+1(Ω)≤Ckgτk_H3/2(Γ). (4.18) Using again Proposition 4.1, we prove thatzbelongs toW^2,2_−k+1(Ω) and satisfies

kzk_W^2,2

−k+1(Ω)6CkFk_W^0,2

−k+1(Ω).

Thenξis in W_−k+1^2,2 (Ω) and estimate (4.14) follows from (4.18).

The next theorem solves an other type of exterior problem.

Theorem 4.3. Let k = −1 or k = 0 and let v belongs to W_k^0,2(Ω). Then, the following problem

−∆ξ=curl v and divξ= 0 in Ω, ξ·n= 0 and (curlξ−v)×n=0 on Γ, Z

Γ

ξ· ∇(w(q)−q)dσ= 0, ∀(w(q)−q)∈ N_−k^∆

(4.19)

has a unique solutionξ inW^1,2_k (Ω) and we have kξk_W1,2

k (Ω)≤Ckvk_W0,2

k (Ω). (4.20)

Moreover, ifv∈W^1,2_k+1(Ω)andΩ⁰is of classC^2,1, then the solutionξis inW^2,2_k+1(Ω) and satisfies the estimate

kξk_W2,2

k+1(Ω)≤Ckvk_W1,2

k+1(Ω). (4.21)

Proof. At first observe that if ξ ∈ W^1,2_k (Ω) is a solution of Problem (4.19) for k=−1 ork= 0, thencurlξ−vbelongs toH²_k(curl,Ω) and thus (curlξ−v)×n is well defined in Γ and belongs toH^−1/2(Γ).

On the other hand, note that (4.19) can be reduced to the variational problem:

Findξ∈V²_k−1,T(Ω) such that

∀ϕ∈X²_−k−1,T(Ω) Z

Ω

curlξ·curlϕdx= Z

Ω

v·curlϕdx. (4.22) Indeed, every solution of (4.19) also solves (4.22). Conversely, letξ∈V²_k−1,T(Ω) a solution of the problem (4.22). Then,

∀ϕ∈D(Ω), hcurl curlξ−curl v,ϕi_D⁰_{(Ω)×D(Ω)}= 0.

Then

−∆ξ=curl v in Ω. (4.23) Moreover, by the fact that ξ belongs to the space V²_k−1,T(Ω) we have divξ = 0 in Ω and ξ·n = 0 on Γ. Then, it remains to verify the boundary condition (curlξ−v)×n=0on Γ. Now settingz=curlξ−v, thenzbelongs toH²_k(curl,Ω).

Therefore, (4.23) becomes

curl z=0 in Ω. (4.24)

Letϕ∈X²_−k−1,T(Ω), by Theorem 2.5 we have X²_−k−1,T(Ω),→W^1,2_−k(Ω). Thank’s to (1.9) we obtain

Z

Ω

z·curlϕdx=hz×n,ϕi_H−1/2(Γ)×H^1/2(Γ)+ Z

Ω

curl z·ϕdx. (4.25) Compare (4.25) with (4.22) and using (4.24), we deduce that

∀ϕ∈X²_−k−1,T(Ω), hz×n,ϕiΓ= 0.

Let nowµany element of the spaceH^1/2(Γ). As Ω⁰ is bounded, we can fix once for all a ballB_R, centered at the origin and with radiusR, such that Ω⁰⊂B_R. Setting Ω_R= Ω∩B_R, then we have the existence ofϕinH¹(Ω_R) such thatϕ=0on∂B_R andϕ =µ_ton Γ, where µ_tis the tangential component of µon Γ. The function ϕcan be extended by zero outsideB_Rand the extended function, still denoted by ϕ, belongs toW^1,p_α (Ω), for anyαsince its support is bounded. Thusϕ, belongs to W^1,2_−k(Ω). It is clear thatϕbelongs toX²_−k−1,T(Ω) and

hz×n,µiΓ=hz×n,µ_tiΓ=hz×n,ϕiΓ= 0. (4.26) This implies thatz×n=0on Γ which is the last boundary condition in (4.19).

On the other hand, let us introduce the problem: Findξ∈V²_k−1,T(Ω) such that

∀ϕ∈V_−k−1,T² (Ω) Z

Ω

curlξ·curlϕdx= Z

Ω

v·curlϕdx. (4.27) Problem (4.27) can be solved by Lax-Milgram lemma ifk= 0 and by Lemma 3.3 ifk=−1.

We start by the casek=−1. Observe that Problem (4.27) satisfies the inf-sup condition (3.5). Let consider the mapping ` : V<sup>2</sup><sub>0,T</sub>(Ω) → R such that `(ϕ) = R

Ωv·curlϕdx. It is clear that` belongs to (V_0,T² (Ω))⁰ and according to Remark 2.8, there exists a unique solutionξ∈V²_−2,T(Ω). Applying Theorem 2.5, we deduce that this solutionξbelongs toW^1,2₋₁(Ω). It follows from Remark 2.8 i) and Theorem 2.6 that

kξk_W1,2

−1(Ω)≤Ck`k_(V2

0,T(Ω))⁰ ≤Ckvk_W0,2

−1(Ω). (4.28) For k = 0, let us consider the bilinear formb :V_−1,T² (Ω)×V²_−1,T(Ω) → Rsuch that

b(ξ,ϕ) = Z

Ω

curlξ·curlϕdx.

According to Theorem 2.5,bis continuous and coercive onV_−1,T² (Ω). Due to Lax- Milgram lemma, there exists a unique solution ξ ∈ V²_−1,T(Ω) of Problem (4.27).

Using again Theorem 2.5, we prove that this solution ξ belongs to W^1,2₀ (Ω) and estimate (4.20) follows immediately.

Next, we extend (4.27) to any test function inX²_−k−1,T(Ω). Letϕ∈X²_−k−1,T(Ω) and let us solve the exterior Neumann problem

∆χ= divϕ in Ω and ∂ χ

∂n = 0 on Γ. (4.29)

It is shown in [7, Lemma 3.7 and Theorem 3.9] that this problem has a unique solutionχinW_−k−1^1,2 (Ω) if k=−1 and unique up to a constant if k= 0. Set

ϕe =ϕ− ∇χ. (4.30)