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T-coercivity for the Maxwell problem with sign-changing coefficients Anne-Sophie Bonnet-Ben Dhia, Lucas Chesnel, Patrick Ciarlet

To cite this version: Anne-Sophie Bonnet-Ben Dhia, Lucas Chesnel, Patrick Ciarlet. T-coercivity for the Maxwell problem with sign-changing coefficients. 2012.

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-coercivity for the Maxwell problem with sign-changing coecients

T

1

2

Anne-Sophie Bonnet-Ben Dhia , Lucas Chesnel , Patrick Ciarlet Jr.

Abstract.

December 6, 2012

3

In this paper, we study the time-harmonic Maxwell problem with sign-changing permit-

tivity and/or permeability, set in a domain of

R3 .

We prove, using the

T-coercivity

approach, that the

well-posedness of the two canonically associated scalar problems, with Dirichlet and Neumann boundary conditions, implies the well-posedness of the Maxwell problem. This allows us to give simple and sharp criteria, obtained in the study of the scalar cases, to ensure that the Maxwell transmission problem between a classical dielectric material and a negative metamaterial is well-posed.

Key words.

Maxwell's equations, interface problem, metamaterial, compact imbedding, sign-

changing coecients.

1

Introduction

We investigate the time-harmonic Maxwell problem in a composite material surrounded by a perfect conductor.

A composite material is modeled by non constant electric permittivity

permeability

µ.

ε

and magnetic

It is well-known that some materials, like metals at optical frequencies, are almost dissi-

pationless and have a dielectric permittivity whose real part is negative. More surprising is the possibility of realizing materials, called negative metamaterials, which exhibit both negative real valued permittivity and permeability in some appropriate range of frequencies. The association of classical dielectrics and such negative materials has very exciting potential applications such as plasmonic waveguides, perfect lenses [31, 21, 27], photonic traps, subwavelength cavities [16] ... From a mathematical point of view, the change of sign of the coecients

ε

and/or

µ

in the medium raises a lot of original questions for the

corresponding electromagnetic model, both for the mathematical analysis and the numerical simulation [26, 28, 17]. Indeed, standard theorems proving the well-posedness of the problem and the convergence of conventional numerical methods are no-longer valid in such situations. Consequently, and generally speaking, the questions we have to address are the following.

Can we extend the classical theory to

congurations with sign-changing coecients? And if not, is there a new functional framework in which well-posedness and stability properties can be recovered? For 2D congurations, the corresponding electromagnetic model reduces to a scalar problem involving the operators

−div (σ∇·)

with Dirichlet or Neumann boundary condition,

σ

being equal to

ε−1

or

µ−1 .

Those scalar problems have been thoroughly investigated [5, 35, 7, 24, 2, 12, 3, 10, 13] and sharp results have been recently obtained using the simple variational technique of the

T-coercivity.

The problems

are proved to be of Fredholm type in the classical functional framework if the contrasts (ratios of the values of

σ

across the interface) are outside some interval, which always contains the value

interval reduces to

{−1}

on integral equations). For a contrast equal to in

H1 .

−1.

This

if (and only if ) the interface is smooth (see also [14, 25] for approaches relying

−1,

the problems are severely ill-posed (not Fredholm)

The inuence of corners in the interface, noticed for instance in [30, 32], has been claried in

[15, 8, 29]. When the interface between the dielectric and the negative material has a corner, depending on the value of the contrast in

1 Laboratoire

H1 ,

even for

email: [email protected]

POEMS, UMR 7231 CNRS/ENSTA/INRIA, ENSTA ParisTech, 828, Boulevard des Maréchaux, 91762

Palaiseau, France;

3 Laboratoire

the scalar problems can be ill-posed (not Fredholm) in


Palaiseau, France;

2 Laboratoire

σ,



Palaiseau, France;


1

contrasts dierent from

−1,

because of the onset of strong singularities at the corner. Well-posedness

can be recovered by working in a new functional framework in which a radiation condition at the corner is imposed [4]. For scalar problems, the theory is now quite mature. Now, we wish to obtain such results for Maxwell problems, and rst, the results of well-posedness using variational techniques. These variational methods are interesting because they allow one to consider rather general congurations: Lipschitz-continuous interface between the positive and the negative material and

L∞

coecients

ε, ε−1 , µ, µ−1 .

However,

it appears that the geometric approach followed for studying the scalar problems is dicult to apply because of the nature of functional spaces used for Maxwell problems. dierently. Again, we will use the can construct

T-coercivity

T-coercivity

Therefore, we will proceed

technique but in a dierent form. We will prove that one

operators which allow us to restore some coercivity property as soon as the

associated 3D scalar problems are well-posed. This will provide very simple criteria (those of the scalar problems) to ensure that Maxwell problems are well-posed. When the contrasts in

ε

and/or

µ

lie inside

the critical intervals, the denition of a new functional framework taking into account the gradients of the strong singularities, is still an open question. The outline of the paper is the following. The denition of the problem and the notations are introduced in Section 2. In Section 3, we give an equivalent formulation of the problems, using some classical functional spaces for the study of Maxwell problems:

VN (ε; Ω)

for the electric eld and

VT (µ; Ω)

for

the magnetic eld ; some divergence free condition is included in their denition. Section 4 expresses the main idea of the paper: how to build a

T-coercivity

operator for the Maxwell problems when the associ-

ated scalar problems are well-posed. For the sake of clarity, we present it in the canonical cases the electric eld and

µ = 1 for

ε = 1 for

the magnetic eld. Then, we use these results and a technique due to [19]

to prove some result of compact embedding of

VN (ε; Ω)

and

VT (µ; Ω)

in

L2 (Ω) := L2 (Ω)3 ,

extending

[6]. Again, let us underline that these results are not classical in the literature when the coecients and

µ

change sign.

ε

In Section 6, we state the main theorem of this work, summing up the previous

results: electric and magnetic Maxwell transmission problems are well-posed as soon as the associated 3D scalar problems are well-posed. We illustrate this result on a series of canonical geometries. Finally, we present some generalizations in Section 8. First, we are interested in congurations where the scalar problems are well-posed in the Fredholm sense with a non trivial kernel. Second, we consider the case

4

of a non-simply connected domain whose boundary is not connected . This study covers the case of non simply connected domains with connected boundary and the case of simply connected domains with non connected boundary.

2

Setting of the problem

Ω be a domain in R3 , i.e. an open, connected and bounded subset of R3 with a Lipschitz-continuous boundary ∂Ω. For some ω 6= 0 (ω ∈ R), the time harmonic Maxwell's equations are given by

Let

curl E − iωµH = 0 Above,

E

and

H are respectively the electric J is the current density. We

The source term

curl H + iωεE = J

and

in

Ω.

(1)

and magnetic components of the electromagnetic eld. suppose that the medium

Ω

is surrounded by a perfect

conductor and we impose the boundary conditions

E×n=0 where

and

µH · n = 0

on

∂Ω,

(2)

n denotes the unit outward normal vector to ∂Ω. We assume that the dielectric permittivity ε and µ are real valued functions which belong to L∞ (Ω), with ε−1 , µ−1 ∈ L∞ (Ω).

the magnetic permeability

Let us introduce some classical spaces in the study of Maxwell's equations:

2 3 L2 (Ω) := L (Ω) ;2 H(curl ; Ω) := u ∈ L (Ω) | curl u ∈ L2 (Ω) ; HN (curl ; Ω) := {u ∈ H(curl ; Ω) | u × n = 0 on ∂Ω} ; VN (ξ; Ω) := {u ∈ H(curl ; Ω) | div (ξ u) = 0, u × n = 0 VT (ξ; Ω) := {u ∈ H(curl ; Ω) | div (ξ u) = 0, ξu · n = 0

4 The

Figure 2 at the end of this paper gives an example of such a geometry.

2

on on

∂Ω} , ∂Ω} ,

for for

ξ ∈ L∞ (Ω); ξ ∈ L∞ (Ω).

For simplicity, the current density

J

L2 (Ω)

will be chosen in

with

div J = 0 5 .

2

2

(·, ·) the inner products of L (Ω) and L (Ω) and k · k the associated norms. HN (curl ; Ω), VN (ξ; Ω) and VT (ξ; Ω) are endowed with the inner product

We denote indistinctly

The spaces

H(curl ; Ω),

(·, ·)curl = (·, ·) + (curl ·, curl ·). Classically, we prove that if

(E, H)

satises (1)-(2),

E

and

H

are respectively solutions to the problems

E ∈ H(curl ; Ω) such that: curl µ−1 curl E − ω 2 εE = iωJ E×n = 0 Find

in on

Ω , ∂Ω

H ∈ H(curl ; Ω) such that: curl ε−1 curl H − ω 2 µH = curl ε−1 J µH · n = 0 ε−1 (curl H − J ) × n = 0

(3)

Find

As already announced in the introduction, we want to nd criteria for

Ω . ∂Ω on ∂Ω in

(4)

on

ε

and

µ

to ensure that problems

(3) and (4) are well-posed in the Fredholm sense. Classically, for the study of Maxwell's equations, our strategy will consist in working in the space for the magnetic eld. belongs to the space

VN (ε; Ω) for the E satises (3)

Indeed, for example, if

VN (ε; Ω).

VT (µ; Ω) div (εE) = 0, so E

electric eld and in the space and if

ω 6= 0,

then

Therefore, the Fredholm property for the problem (3) will rely on two

VN (ε; Ω) in L2 (Ω), secondly, the isomorphism property for the principal part curl µ curl · : VN (ε; Ω) → VN (ε; Ω)∗ . In a symmetric way, the Fredholm property 2 for the magnetic eld relies on the compact embedding of VT (µ; Ω) in L (Ω) and on the isomorphism −1 ∗ property for the principal part curl ε curl · : VT (µ; Ω) → VT (µ; Ω) .

arguments: rst the compact embedding of

−1

3

Equivalent formulations

Let us rst give equivalent formulations to problem (1)-(2) in the spaces

VN (ε; Ω)

and

VT (µ; Ω).

3.1 Problem for the electric eld Consider

aε

aε (ϕ, ϕ0 ) = (ε∇ϕ, ∇ϕ0 ) for all ϕ, ϕ0 ∈ H10 (Ω). With the Riesz Aε : H10 (Ω) → H10 (Ω) such that, for all ϕ, ϕ0 ∈ H10 (Ω), H10 (Ω), we dene the norm k·kH1 (Ω) = k∇ · k. Let h·, ·i designate the

the sesquilinear form such that

representation theorem, we dene the operator

(∇(Aε ϕ), ∇ϕ0 ) = aε (ϕ, ϕ0 ). On −1 duality pairing H (Ω) × H10 (Ω). to the electric eld. Here, C > 0

(Hε )

0

Now, let us introduce an assumption for the 3D scalar problem related is a constant.

There exists an isomorphism

|aε (ϕ, Tε ϕ)| ≥ C

2 kϕkH1 (Ω) 0

Tε ,

of

H10 (Ω) such that ∀ϕ ∈ H10 (Ω).

Lemma 3.1

ε 1 1 Assumption (Hε ) holds true if and only if A : H0 (Ω) → H0 (Ω) is an isomorphism. In this −1 case, for all f ∈ H (Ω), there exists a unique solution to the problem

ϕ ∈ H10 (Ω) such a (ϕ, ϕ0 ) = hf, ϕ0 i, Find ε Besides, there exists a constant

Remark 3.2

C>0

independent of

f

that : 0

∀ϕ ∈ H10 (Ω).

such that

(5)

kϕkH1 (Ω) ≤ C kf kH−1 (Ω) . 0

−div (ε∇ϕ) = f in H−1 (Ω). We will see in section 7 ε that, in practice, it is possible to build explicitly such operators T , depending on the values of ε and on the geometry of the domain. Actually, assumption (Hε ) is an assumption on ε and on the geometry of Classically, Eq. (5) can be rewritten

the domain only (for more details see [2]).

5 The

case

div J 6= 0

can be handled similarly with the tools that we propose, see Remark 6.3 below.

3

Proof.

Assume

(Hε )

to be true.

Tε : H10 (Ω) → H10 (Ω) is an isomorphism, ϕ satises (5) if ε 0 ε 0 1 ε problem nd ϕ ∈ H0 (Ω) such that a (ϕ, T ϕ ) = hf, T ϕ i, for all Since

and only if ϕ is a solution of the ϕ0 ∈ H10 (Ω). Lax-Milgram theorem then allows us to conclude that problem (5) is well-posed because (ϕ, ϕ0 ) 7→ aε (ϕ, Tε ϕ0 ) is coercive on H10 (Ω) × H10 (Ω) by assumption. This approach also proves that Aε : H10 (Ω) → H10 (Ω) is an isomorphism. Reciprocally, if Aε : H10 (Ω) → H10 (Ω) is an isomorphism, taking Tε = Aε , one obtains, for all ϕ ∈ H10 (Ω), |aε (ϕ, Tε ϕ)| = kAε ϕk2H1 (Ω) ≥ C kϕk2H1 (Ω) . 0

Theorem 3.3 1) If

(E, H)

Assume that

ω 6= 0. E

satises (1)-(2) then

is a solution of the problem

E ∈ VN (ε; Ω)

Find Z

0

such that for all

µ−1 curl E · curl E 0 − ω 2 εE · E 0

E 0 ∈ VZN (ε; Ω):

2) Assume

(Hε )

Ω

to be true. If

E

satises (6) then

(E, (i ωµ)−1 curl E)

Proof.

1) If E satises (1)-(2), then E is a solution of (3). div (εE) = 0. This allows us to show that E satises (6). 2) Now, let us prove that if

problem

(6)

J · E0.

= iω

Ω

satises (1)-(2).

On the other hand, since

E ∈ VN (ε; Ω) ⊂ HN (curl ; Ω)

satises (6) then

E

ω 6= 0,

there holds

is a solution to the

0 ZE ∈ HN (curl ; Ω): µ−1 curl E · curl E 0 − ω 2 εE · E 0 = iω J · E0.

Find

Z

E ∈ HN (curl ; Ω)

such that for all

Ω

(7)

Ω

E 0 in HN (curl ; Ω), Lemma 3.1 indicates that we can build ϕ ∈ H10 (Ω) such that div (ε∇ϕ) = div (εE ). The element E 0 −∇ϕ belongs to VN (ε; Ω). Taking E 0 −∇ϕ as a test-eld in (6) and observing that (εE, ∇ϕ) = 0 and (J , ∇ϕ) = 0 (recall that div J = 0), one obtains Z Z µ−1 curl E · curl E 0 − ω 2 εE · E 0 = iω J · E0. For all

0

Ω

Ω

But (3) and (7) are equivalents.

E satises (6) then E is (E, (i ωµ)−1 curl E) satises (1)-(2).

Therefore, if

remains to notice that in this case,

a solution of (3).

With the Riesz representation theorem, let us introduce the bounded operator

VN (ε; Ω)

such that for all

E, E 0 ∈ VN (ε; Ω),

There just

AN (ω) : VN (ε; Ω) →

(AN (ω)E, E 0 )curl = (µ−1 curl E, curl E 0 ) − ω 2 (εE, E 0 ).

(8)

3.2 Problem for the magnetic eld For the study of the magnetic eld, we introduce the space

H1# (Ω)

1

Z

ϕ ∈ H (Ω) |

:=

ϕ=0 .

Ω 1 Notice that the only constant in H# (Ω) is 0. Using the Rellich's theorem which indicates that the 1 2 embedding of H (Ω) in L (Ω) is compact and using that Ω is connected, we prove by contradiction the

Lemma 3.4 is denoted

The map

(ϕ, ϕ0 ) 7→ (∇ϕ, ∇ϕ0 )

denes an inner product on

H1# (Ω).

The associated norm

k·kH1 (Ω) . #

the Riesz representation theorem, we dene

aµ (ϕ, ϕ0 ) = (µ∇ϕ, ∇ϕ0 ) for all ϕ, ϕ0 ∈ H1# (Ω). With µ the operator A : H1# (Ω) → H1# (Ω) such that, for all

ϕ, ϕ0 ∈ H1# (Ω), (∇(Aµ ϕ), ∇ϕ0 ) = aµ (ϕ, ϕ0 ).

As for the study of the electric eld, let us introduce a

Consider

µ

a

the sesquilinear form such that

property associated with this 3D scalar problem for the magnetic eld. Again,

(Hµ )


|aµ (ϕ, Tµ ϕ)| ≥ C

Tµ ,

2 kϕkH1 (Ω) #

4

of

C>0

H1# (Ω) such that ∀ϕ ∈ H1# (Ω).

is a constant.

Lemma 3.5

µ 1 1 Assumption (Hµ ) holds true if and only if A : H# (Ω) → H# (Ω) is an isomorphism. In 1 0 this case, for all ` ∈ (H# (Ω)) , there exists a unique solution to the problem

ϕ ∈ H1# (Ω) such that: a (ϕ, ϕ0 ) = `(ϕ0 ), ∀ϕ0 ∈ H1# (Ω). Find µ

Besides, there exists a constant

Remark 3.6 ϕ0 ∈ H1# (Ω),

independent of

Eq. (9) can

`

such that

kϕkH1 (Ω) ≤ C k`k(H1 (Ω))0 . #

#

R f ∈ L2 (Ω) such that Ω f = 0, if one denes ` by `(ϕ0 ) = Ω f ϕ0 for all 2 −1/2 be rewritten −div (µ∇ϕ) = f in L (Ω) and µ∂n ϕ = 0 in H (∂Ω). As in the (Hµ ) is an assumption on µ and on the geometry of the domain only. R

Classically, for

electric case, assumption

Proof.

C>0

(9)

Tµ : H1# (Ω) → H1# (Ω) is an isomorphism, ϕ satises (9) if µ 0 µ 0 1 µ and only if ϕ is a solution to the problem nd ϕ ∈ H# (Ω) such that a (ϕ, T ϕ ) = `(T ϕ ), for all ϕ0 ∈ H1# (Ω). But Lemma 3.4 ensures that (ϕ, ϕ0 ) 7→ aµ (ϕ, Tε ϕ0 ) is coercive on H1# (Ω) × H1# (Ω). Then, Assume

(Hµ )

to be true.

Since

one can conclude that problem (9) is well-posed with Lax-Milgram theorem. In the same way, we prove

Aµ : H1# (Ω) → H1# (Ω) is an isomorphism. Reciprocally, if Aµ is an isomorphism T = Aµ , one nds, for ϕ ∈ H1# (Ω), |aµ (ϕ, Tµ ϕ)| = kAµ ϕk2H1 (Ω) ≥ C kϕk2H1 (Ω) . that

of

H1# (Ω),

taking

µ

#

#

Adapting the proof of Theorem 3.3, one obtains the

Theorem 3.7 1) If

(E, H)

Assume that

ω 6= 0. H

satises (1)-(2) then Find Z

is a solution of the problem

H ∈ VT (µ; Ω)

such that for all

H 0Z∈ VT (µ; Ω):

ε−1 curl H · curl H 0 − ω 2 µH · H 0 =

Ω

2) Assume

(Hµ )

Ω

to be true. If

H

satises (10) then

(i (ωε)−1 (curl H − J ), H)

satises (1)-(2).

With the Riesz representation theorem, let us introduce the bounded operator

VT (µ; Ω)

such that for all

(10)

ε−1 J · curl H 0 .

H, H 0 ∈ VT (µ; Ω),

AT (ω) : VT (µ; Ω) →

(AT (ω)H, H 0 )curl = (ε−1 curl H, curl H 0 ) − ω 2 (µH, H 0 ). 4

The

T-coercivity

For the sake of clarity, we present rst the pler cases:

ε=1

and

µ

VN (1; Ω)

approach in

and in

(11)

VT (1; Ω)

T-coercivity approach for the Maxwell's equations on the simµ = 1 and ε possibly sign-changing

possibly sign-changing for the electric eld;

for the magnetic eld. This study will be useful for two reasons. On the one hand, the results obtained will serve to prove the compact embeddings of

VN (ε; Ω)

this will give us an insight of how to proceed to study the

VT (µ; Ω) in L2 (Ω). On the operators AN (0) and AT (0).

and

other hand,

VN (1; Ω) and VT (1; Ω) (cf. [33, 1]). VN (1; Ω) in L2 (Ω) and of VT (1; Ω) in L2 (Ω) are compact. Furthermore, when ∂Ω is connected (respectively when Ω is simply connected), the map (u, v) 7→ (curl u, curl v) denes an inner product on VN (1; Ω) (resp. on VT (1; Ω)) and the associated norm is 1/2 equivalent to the canonical norm u 7→ (u, u)curl .

Let us recall some classical properties for the spaces The embeddings of

4.1 Study for the electric eld with a permittivity ε = 1 Let us rst study the problem for the electric eld.

Lemma 4.1

Let

Ω

be a simply connected domain such that

Then, there exists an isomorphism

T

of

VN (1; Ω)

∂Ω

is connected. Assume

such that for all

u, v ∈ VN (1; Ω),

(µ−1 curl u, curl Tv) = (µ−1 curl Tu, curl v) = (curl u, curl v).

5

(Hµ )

to be true.

Remark 4.2 such that

Proof.

One can prove a reciprocal assertion. Namely, there exists an isomorphism

(µ−1 curl u, curl Tu) ≥ kcurl uk2 ,

for all

u ∈ VN (1; Ω)

? Definition of T. Consider v ∈ VN (1; Ω). ϕ the unique element of H1# (Ω) such that Z Z 0 µ∇ϕ · ∇ϕ = µ curl v · ∇ϕ0 ,

T of VN (1; Ω)

implies (Hµ ) is true.

i) Dene

Ω

ϕ

∀ϕ0 ∈ H1# (Ω).

Ω

(Hµ ) to be true. µ(curl v−∇ϕ) is a divergence free element of L2 (Ω) such that µ(curl v−∇ϕ)·n = 0 on ∂Ω. Since Ω is simply connected and since ∂Ω is connected, according to theorem 3.17 in [1] (see also theorem 3.6 in [18]), there exists a unique potential Tv ∈ VN (1; Ω) such that curl Tv = µ(curl v − ∇ϕ). This denes a bounded operator T : VN (1; Ω) → VN (1; Ω). ? Positiveness property. For all u, v ∈ VN (1; Ω), we then compute

The function

is well-dened since we have assumed

ii) Observe next that

(µ−1 curl u, curl Tv) = (µ−1 curl u, µ(curl v − ∇ϕ)) = (curl u, curl v), u × n = 0 on ∂Ω. Notice we have also (µ−1 curl Tu, curl v) = (curl u, curl v). ? T is an isomorphism of VN (1; Ω). Let us consider the operator for the electric VN (1; Ω) → VN (1; Ω) dened in (8). The previous identity allows us to write because

eld

AN (ω) :

(AN (0)(Tu), v)curl = (µ−1 curl Tu, curl v) = (curl u, curl v), ∀u, v ∈ VN (1; Ω). Since

∂Ω

that

(u, v) 7→ (curl u, curl v) denes an inner product on VN (1; Ω). AN (0) ◦ T is an isomorphism of VN (1; Ω). Since AN (0) is selfadjoint, we

is connected, the map

quently, the operator

AN (0)

and

Consededuce

are isomorphisms.

T

We deduce the

Proposition 4.3 be true and

Ω be a simply connected domain such that ∂Ω is connected. Assume (Hµ ) to Then, the problem for the electric eld (6) has one and only one solution which

Let

ε = 1.

continuously depends on the data

J,

for all

ω ∈ C\S ,

where

S

is a discrete set.

Proof.

According to Lemma 4.1, the operator AN (0), with ε = 1, is an isomorphism when assumption (Hµ ) holds true. Since VN (1; Ω) is compactly embedded in L2 (Ω), AN (ω)−AN (0) is a compact operator of VN (1; Ω) for all ω ∈ C. The analytic Fredholm theorem allows us to conclude.

Remark 4.4

The analysis presented in this paragraph also holds in the case where the assumption

≥ C > 0. To deal with the case of a sign changing coecient (Hε ) holds true to be able to build an operator T with values 2 embedding of VN (ε; Ω) in L (Ω) is compact.

ε

= 1

is replaced by the assumption ε

ε,

we will have to assume in addition that

in

VN (ε; Ω)

and to prove that the

4.2 Study for the magnetic eld with a permeability µ = 1 Now, let us study the problem for the magnetic eld.

Lemma 4.5

Let

Ω


Then, there exists an isomorphism

T

of

VT (1; Ω)

∂Ω


such that for all

(Hε )

to be true.

u, v ∈ VT (1; Ω),

(ε−1 curl u, curl Tv) = (ε−1 curl Tu, curl v) = (curl u, curl v).

Remark 4.6 VT (1; Ω)

Proof.

One can again prove a reciprocal assertion. Namely, there exists an isomorphism T of (ε−1 curl u, curl Tu) ≥ kcurl uk2 , for all u ∈ VT (1; Ω) implies (Hε ) is true.

such that

The proof is very similar to the proof of Lemma 4.1, the boundary conditions being the only

changing elements.

? Definition of T. Consider v ∈ VT (1; Ω). 1 i) First dene ϕ the unique element of H0 (Ω) such that Z Z ε∇ϕ · ∇ϕ0 = ε curl v · ∇ϕ0 , Ω

Ω 6

∀ϕ0 ∈ H10 (Ω).

The function

ϕ


ε(curl v − ∇ϕ)

(Hε )

to be true.

L2 (Ω).

Since Ω is simply connected ∂Ω is connected, according to theorem 3.12 in [1], there exists a unique potential Tv ∈ VT (1; Ω) such that curl Tv = ε(curl v − ∇ϕ). This denes an operator T from VT (1; Ω) to VT (1; Ω). ? Positiveness property. For all u, v ∈ VT (1; Ω), we then compute

ii) Then, notice that

is a divergence free element of

and since

(ε−1 curl u, curl Tv) = (ε−1 curl u, µ(curl v − ∇ϕ)) = (curl u, curl v), ϕ = 0 on ∂Ω. We have also (ε−1 curl Tu, curl v) = (curl u, curl v). ? T is an isomorphism of VT (1; Ω). Let us consider the operator for the VT (1; Ω) → VT (1; Ω) dened in (11). The previous identity allows us to write

because

magnetic eld

AT (ω) :

(AT (0)(Tu), v)curl = (ε−1 curl Tu, curl v) = (curl u, curl v), ∀u, v ∈ VT (1; Ω). Since

Ω

is simply connected, the map

Thus, the operator

AT (0)

and

T

AT (0) ◦ T

(u, v) 7→ (curl u, curl v) denes an inner product on VT (1; Ω). VT (1; Ω). Since AT (0) is selfadjoint, we deduce that

is an isomorphism of

are isomorphisms.

Proceeding as in Proposition 4.3, one proves the

Proposition 4.7 true and

µ = 1.

Let Ω be a simply connected domain such that ∂Ω is connected. Assume (Hε ) to be Then, the problem for the magnetic eld (10) has one and only one solution which

continuously depends on the data

Remark 4.8

J,

for all

ω ∈ C\S ,

where

S

is a discrete set.

Like in Remark 4.4, let us indicate that the analysis of this paragraph also holds in the

= 1 is replaced by the assumption µ ≥ C > 0. To deal with the case µ, we will have furthermore to assume that (Hµ ) holds true to be able to 2 values in VT (µ; Ω) and to prove that the embedding of VT (µ; Ω) in L (Ω) is

case where the assumption µ of a sign changing coecient build an operator

T

with

compact.

5

Compactness results

ξ ∈ L∞ (Ω), XN (ξ; Ω) := u ∈ H(curl ; Ω) | div (ξ u) ∈ L2 (Ω), u × n = 0 XT (ξ; Ω) := u ∈ H(curl ; Ω) | div (ξ u) ∈ L2 (Ω), ξu · n = 0

Dene the spaces, for

2

XN (ε; Ω)

and

on

∂Ω ; ∂Ω .

2

u 7→ (kuk + kdiv (ξ u)k + kcurl uk )1/2 . In XT (µ; Ω) are compactly embedded in L2 (Ω) when (Hε )

Theses spaces are equipped with the norm we prove that

2

on

this paragraph, and

(Hµ )

hold

true, extending the classical theorems of [33, 19, 23] (for another approach, based on the study of the regularity of elds, in 2D, when

ε, µ

change sign, see [11]). This constitutes a more general result than

the one we actually need for our study, namely the compact embedding of

L2 (Ω).

VN (ε; Ω)

and

VT (µ; Ω)

in

5.1 Compact embedding of XN (ε; Ω) in L2 (Ω) Let us start by studying the space of electric elds.

Theorem 5.1

Let

Ω

be a simply connected domain such that XN (ε; Ω) in L2 (Ω) is compact.

∂Ω


(Hε )

to be true.

Then the embedding of

Proof.

(un ) be a bounded sequence of XN (ε; Ω). Dene fn := div (εun ) and F n := curl un . The (fn ) and (F n ) are respectively bounded in L2 (Ω) and in L2 (Ω). Since (Hε ) is true, according 1 to Lemma 3.1, there exists, for all n ∈ N, ϕn ∈ H0 (Ω) such that div (ε∇ϕn ) = div (εun ). Then, we 2 notice that ε(un − ∇ϕn ) is a divergence free element of L (Ω). Since ∂Ω is connected, there exists (see [1], theorem 3.12) w n ∈ VT (1; Ω) such that curl w n = ε(un − ∇ϕn ). Thus, for all n ∈ N, one has un = ∇ϕn + ε−1 curl wn . Let us show now we can extract sequences from (∇ϕn ) and (curl wn ) which 2 converge in L (Ω). Let

sequences

7

(ϕn ) remains bounded in H10 (Ω). The operator Tε of hypothesis (Hε ) is ε 1 1 1 1 continuous from H0 (Ω) to H0 (Ω), (T ϕn ) is also bounded in H0 (Ω). But H0 (Ω) is compactly em2 bedded in L (Ω). Therefore, we can extract a subsequence from (ϕn ) (still denoted (ϕn )) such that (Tε ϕn ) converges in L2 (Ω). Introduce ϕnm = ϕn − ϕm and fnm = fn − fm . By linearity, there holds: −(ε∇ϕnm , ∇ϕ0 ) = (fnm , ϕ0 ), for all ϕ0 ∈ H10 (Ω). Taking ϕ0 = Tε ϕnm , one obtains Lemma 3.1 ensures that

C kϕnm kH1 (Ω) ≤ |(ε∇ϕnm , ∇(Tε ϕnm ))| = |(fnm , Tε ϕnm )| . 2

0

This shows that

(∇ϕn )

is a Cauchy sequence in

L2 (Ω),

and so, that it converges.

w 7→ kcurl wk denes a norm on VT (1; Ω). VT (1; Ω). By the compact imbedding of VT (1; Ω) in L2 (Ω), we can extract a subsequence, still denoted (wn ), which converges in L2 (Ω). According to Lemma 4.5, there exists an isomorphism T of VT (1; Ω) such that −1 (ε curl w, curl Tw) = kcurl wk2 , ∀w ∈ VT (1; Ω).

Now, let us work on the sequence Consequently, the sequence

(wn )

(curl wn ).

We know that

is bounded in

T is continuous, the sequence (Twn ) is bounded in VT (1; Ω). So, we can extract a subsequence (wn ), still denoted (wn ), such that (Twn ) converges in L2 (Ω). Since curl ε−1 curl wn = F n in Ω −1 and (ε curl wn ) × n = 0 on ∂Ω, one has (ε−1 curl wnm , curl w0 ) = (F nm , w0 ), for all w0 ∈ VT (1; Ω) 0 with w nm = w n − w m and F nm = F n − F m . Testing with w = Tw nm leads to:

Since from

2

kcurl wnm k = |(F nm , Twnm )| . This estimate proves that

Corollary 5.2

Let

Ω

(curl wn )

is a Cauchy sequence in


Then, there exists a constant

C

∂Ω

Consequently, it converges.


(Hε )

to be true.

such that 2

2

kuk ≤ C kcurl uk , Thus, the map

L2 (Ω).

∀u ∈ VN (ε; Ω).

(u, v) 7→ (curl u, curl v) denes an inner 1/2 u 7→ (u, u)curl .

product on

(12)

VN (ε; Ω)

and the associated norm

is equivalent to the canonical norm

Proof.

To prove this corollary, it is sucient to show (12). Let us proceed by contradiction assuming

there exists a sequence

(un )

of elements of

VN (ε; Ω)

∀n ∈ N, kun k = 1

such that

and

lim kcurl un k n→∞

= 0.

(un ) (still denoted (un )) which converges u in L2 (Ω). By construction, we have kuk = 1. Then, we check easily that (un ) converges to u in HN (curl ; Ω) with curl u = 0 a.e. in Ω. Since ∂Ω is connected, one deduces (see [9], chapter 2, 1 theorem 8) that there exists a scalar potential ϕ ∈ H0 (Ω) such that u = ∇ϕ in Ω. Finally, we notice that div (εu) = 0 and so div (ε∇ϕ) = 0. Lemma 3.1 implies ϕ = 0 so u = 0. This leads to a contradiction because we must have kuk = 1.

According to Theorem 5.1, we can extract a sequence from to

5.2 Compact embedding of XT (µ; Ω) in L2 (Ω) Now, let us work on the space of magnetic elds.

Theorem 5.3

Let

Ω

be a simply connected domain such that XT (µ; Ω) in L2 (Ω) is compact.

∂Ω


(Hµ )

to be true.

Then, the embedding of

Proof.

Let (un ) be a bounded sequence of elements of XT (µ; Ω). Dene fn := div (µun ) and F n := curl unR . The sequenceR (fn ) is bounded in L2 (Ω) whereas (F n ) is bounded in L2 (Ω). For all n ∈ N, since Ω div (µun ) = ∂Ω µun · n = 0, Lemma 3.5 allows us to build ϕn ∈ H1# (Ω) such that div (µ∇ϕn ) = div (µun ). Observe next that µ(un − ∇ϕn ) is a divergence free element of L2 (Ω) such that µ(un − ∇ϕn ) · n = 0 on ∂Ω. Since Ω is simply connected, according to theorem 3.17 in [1], there exists a potential w n ∈ VN (1; Ω) such that curl w n = µ(un − ∇ϕn ). Thus, for all n ∈ N, we have un = ∇ϕn + µ−1 curl wn . Then, we show as in the proof of Theorem 5.1, using the assumption quences from

(∇ϕn )

and

(curl wn )

which converge in

8

L2 (Ω).

(Hµ ),

that we can extract subse-

As previously, this theorem allows us to obtain the

Corollary 5.4

Let

Ω

C


(Hµ )

to be true.

such that 2

2

kuk ≤ C kcurl uk , Thus, the map

∂Ω


Then, there exists a constant

∀u ∈ VT (µ; Ω).

(u, v) 7→ (curl u, curl v) denes an inner 1/2 u 7→ (u, u)curl .

product on

VT (µ; Ω)

and the associated norm

is equivalent to the canonical norm

6

Well-posedness of Maxwell's equations

We extend the results of Lemmas 4.1 and 4.5 to address the problem for the electric eld (resp. magnetic eld) when

ε

Lemma 6.1

(resp.

Let

Ω

µ)

changes sign. Thus, we precise Remark 4.4 (resp. Remark 4.8).


∂Ω


(Hε )

and

(Hµ )

to be true. Then:

•


Tε

of

VN (ε; Ω)

such that, for all

u, v ∈ VN (ε; Ω),

(µ−1 curl u, curl Tε v) = (µ−1 curl Tε u, curl v) = (curl u, curl v). •


Tµ

of

VT (µ; Ω)

such that, for all

(13)

u, v ∈ VT (µ; Ω),

(ε−1 curl u, curl Tµ v) = (ε−1 curl Tµ u, curl v) = (curl u, curl v).

Proof.

The rst steps in the construction of the operators

Tε

and

Tµ

(14)

will be the same as those of

Lemmas 4.1 and 4.5. However, we recall them for the sake of clarity. Of course, to obtain elds which

VN (ε; Ω) and VT (µ; Ω) (instead of VN (1; Ω) ? Definition of Tε . Consider v ∈ VN (ε; Ω). 1 i) Introduce ϕ the unique element of H# (Ω) such that belong to

Z

µ∇ϕ · ∇ϕ0 =

Ω

µ curl v · ∇ϕ0 ,

VT (1; Ω)),

we shall add one step.

∀ϕ0 ∈ H1# (Ω).

Ω

(Hµ ) to be true. µ(curl v−∇ϕ) is a divergence free element of L2 (Ω) such that µ(curl v−∇ϕ)·n = 0 on ∂Ω. Under the assumptions on the geometry, there exists a unique potential ψ ∈ VN (1; Ω) such that curl ψ = µ(curl v − ∇ϕ). 1 iii) Consider ζ the unique element of H0 (Ω) such that Z Z 0 ε∇ζ · ∇ζ = εψ · ∇ζ 0 , ∀ζ 0 ∈ H10 (Ω). The function

ϕ

Z

and


ii) Remark next that

Ω The function

ζ

Ω

(Hε ) to be true. Tε : VN (ε; Ω) → VN (ε; Ω) such that Tε v = ψ − ∇ζ


iv) Finally, dene the operator

? Definition of Tµ . Consider v ∈ VT (µ; Ω). 1 i) Introduce ϕ the unique element of H0 (Ω) such that Z Z ε∇ϕ · ∇ϕ0 = ε curl v · ∇ϕ0 , Ω The function

ϕ

ε(curl v − ∇ϕ)

ζ

∀ϕ0 ∈ H10 (Ω).

the unique element of

Z Ω

(Hε )

to be true.

L2 (Ω). Under the assumptions that curl ψ = ε(curl v − ∇ϕ).

is a divergence free element of

the geometry, there exists a unique potential iii) Consider

v ∈ VN (ε; Ω).

Ω


ii) Then, notice that

for

H1# (Ω)

µ∇ζ · ∇ζ 0 =

ψ ∈ VT (1; Ω)

such

such that

Z

µψ · ∇ζ 0 ,

Ω

9

∀ζ 0 ∈ H1# (Ω).

on

ζ

The function

(Hµ ) to be true. Tµ : VT (µ; Ω) → VT (µ; Ω) such that Tµ v = ψ − ∇ζ



for

v ∈ VT (µ; Ω).

A simple computation leads to (13) and (14). Proceeding as in Lemmas 4.1 and 4.5, and using Corollaries 5.2 and 5.4 which prove that

VT (µ; Ω), VT (µ; Ω).

(u, v) 7→ (curl u, curl v) denes an Tε and Tµ are respectively

inner product on

one shows a posteriori that

on

isomorphisms of

VN (ε; Ω) VN (ε; Ω)

and and

We now have all the tools to prove the main result of this paper.

Theorem 6.2

Let

Ω


∂Ω

is connected. Assume also that

(Hε )

ε There exists an isomorphism T of 2 ε |(ε∇ϕ, ∇(T ϕ))| ≥ C kϕkH1 (Ω) , 0

H10 (Ω) such that ∀ϕ ∈ H10 (Ω).

(Hµ )

µ There exists an isomorphism T of 2 |(µ∇ϕ, ∇(Tµ ϕ))| ≥ C kϕkH1 (Ω) , #

H1# (Ω) such that ∀ϕ ∈ H1# (Ω).

Then, the following results hold.

•

There exists a unique solution to the problem for the electric eld 0 Find E ∈ VN (ε; Ω) such that for all E ∈ VZ N (ε; Ω): Z 0 0 −1 2 µ curl E · curl E − ω εE · E = iω J · E0, Ω

Ω

which continuously depends on the data

•

J,

for all

ω ∈ C\S

where

S

is a discrete set.

There exists a unique solution to the problem for the magnetic eld Find Z

H ∈ VT (µ; Ω)

such that for all

H 0Z∈ VT (µ; Ω):

ε−1 curl H · curl H 0 − ω 2 µH · H 0 =

Ω

ε−1 J · curl H 0 ,

Ω

which continuously depends on the data

•

(15)

J,

for all

ω ∈ C\S . ω ∈ C∗ \S .

Maxwell's equations (1)-(2) are uniquely solvable for all

Proof.

ε Let us begin with the rst point. Lemma 6.1 ensures the existence of an isomorphism T : VN (ε; Ω) → VN (ε; Ω) such that (u, v) 7→ (µ−1 curl u, curl Tε v) is coercive on VN (ε; Ω) × VN (ε; Ω). ε Now, since T is an isomorphism, E satises (15) if and only if E satises the problem Find

Z

E ∈ VN (ε; Ω)

such that for all

E 0 ∈ VN (ε;ZΩ):

µ−1 curl E · curl (Tε E 0 ) − ω 2 εE · (Tε E 0 ) = iω

Ω

J · (Tε E 0 ).

Ω

It only remains to observe that the Fredholm alternative holds for this problem because the embedding of

VN (ε; Ω)

in

L2 (Ω)

is compact by Theorem 5.1. The second point can be proved in the same way

whereas the third statement can be obtained thanks to Theorems 3.3 and 3.7.

Remark 6.3 0.

However, if one assumes that

Then, one can proceed exactly as

7

div (εE) = (iω)−1 div J 6= (Hε ) is true, one can solve (5) with right-hand side f = (iω)−1 div J . before with (J − iωε∇ϕ, E − ∇ϕ) replacing (J , E) in Eq. (1).

If in Eq. (1) one considers

J

such that

div J 6= 0,

it follows that

Illustrations

We apply Theorem 6.2 in a few simple congurations. We focus on situations where the medium consists

Ω is divided into two sub-domains Ω1 Ω = Ω1 ∪ Ω2 et Ω1 ∩ Ω2 = ∅. We denote Σ := ∂Ω1 \ ∂Ω = ∂Ω2 \ ∂Ω. Let us introduce ε1 and µ1 (resp. ε2 and µ2 ) two elements of L∞ (Ω1 ) (resp. L∞ (Ω2 )). Dene the functions ε and µ such that ε|Ωk = εk and µ|Ωk = µk for k = 1, 2. We assume that Ω1 is lled with a positive material and that Ω2 is lled with a possibly negative material (for ε and/or µ). For that, we make the following assumptions: of two dierent materials. To model this problem, we assume that and

Ω2

with

10

Figure 1: Canonical geometries: symmetric domain, prismatic edge, Fichera's corner, non symmetric cavity.

there exists a constant there exists a constant there exists a constant

ε

σ1+ := sup σ1 ,

s.t. s.t.

ε1 ≥ C > 0 and µ1 ≥ C > 0 a.e. in Ω1 ; ε2 ≥ C > 0 a.e. in Ω2 or ε2 ≤ −C < 0 a.e. in Ω2 ; µ2 ≥ C > 0 a.e. in Ω2 or µ2 ≤ −C < 0 a.e. in Ω2 .

∈ L∞ (Ω)

and

σ2+ := sup |σ2 |,

Ω1

v

s.t.

−1

In particular, notice that

Generally speaking, if

C C C

Ω2

µ−1 ∈ L∞ (Ω). σ1− := inf σ1

and

Ω1

is a measurable function on

Ω,

Then, we dene

σ2− := inf |σ2 |, Ω2

we use the notation

for

σ = ε, µ.

vk := v|Ωk , k = 1, 2.

For

more details concerning the study of the scalar problems, we refer the reader to [2].

7.1 Symmetric domain Let

Ω

be a symmetric domain, in the sense that

Ω1

and

Ω2

can be mapped from one to the other with

Σ is included z = 0 (see Figure 1, left, for an example). Consider the operators R1 and R2 respectively 1 dened by (R1 ϕ1 )(x, y, z) = ϕ1 (x, y, −z) and (R2 ϕ2 )(x, y, z) = ϕ2 (x, y, −z) for ϕ ∈ H (Ω). Dene the operators T1 and T2 such that: ϕ1 in Ω1 ϕ1 − 2R2 ϕ2 in Ω1 T1 ϕ = ; T2 ϕ = . −ϕ2 + 2R1 ϕ1 in Ω2 −ϕ2 in Ω2

the help of a reection symmetry. Without loss of generality, we assume that the interface in the plane

T1 ϕ and T2 ϕ belong to H1 (Ω). As T1 ◦ T1 = T2 ◦ T2 = Id, we deduce that T1 and T2 are ε ε 1 1 1 isomorphisms of H (Ω). The restrictions T1 and T2 of T1 and T2 to H0 (Ω) are isomorphisms of H0 (Ω). R R 1 Let us introduce the linear form γ : H (Ω) → R such that γ(ϕ) = ϕ/ Ω 1. Note that ker(γ) = H1# (Ω). Ω µ µ µ 1 Then, we dene the operators T1 and T2 such that, for all ϕ ∈ H# (Ω), T1 ϕ = T1 ϕ − γ(T1 ϕ) and µ µ µ 1 T2 ϕ = T2 ϕ − γ(T2 ϕ). Notice that T1 ϕ and T2 ϕ are elements of H# (Ω). Moreover, we have

By construction,

Tµ1 (Tµ1 ϕ) = Tµ1 (T1 ϕ − γ(T1 ϕ)) = T1 (T1 ϕ − γ(T1 ϕ)) − γ(T1 (T1 ϕ − γ(T1 ϕ))) = ϕ − T1 (γ(T1 ϕ)) − γ(ϕ − T1 (γ(T1 ϕ))) = ϕ − T1 (γ(T1 ϕ)) + γ(T1 (γ(T1 ϕ))) = ϕ. Thus,

Tµ1 ◦ Tµ1 = Id.

Proposition 7.1

In the same way, we nd

Tµ2 ◦ Tµ2 = Id.

Hence

Tµ1

and

Tµ2

are isomorphisms of

H1# (Ω).

(Symmetric domain)

+ − + ε satises ε ≥ C > 0 a.e. in Ω or max(ε− 1 /ε2 , ε2 /ε1 ) > 1. − + − + Assume that µ satises µ ≥ C > 0 a.e. in Ω or max(µ1 /µ2 , µ2 /µ1 ) > 1. ∗ Then, Maxwell's equations (1)-(2) are uniquely solvable for all ω ∈ C \S where S

Assume that

Proof.

Apply Theorem 6.2. To check that For Take

Tε

equal to

For Take

Tµ

equal to

(Hε )

and

(Hµ )

is a discrete set.

hold true, use the following table [2].

ε≥C>0

+ ε− 1 /ε2 > 1

+ ε− 2 /ε1 > 1

Id

Tε1

Tε2

µ≥C>0

+ µ− 1 /µ2 > 1

+ µ− 2 /µ1 > 1

Id

Tµ1

Tµ2

11

In the case where

ε

and

µ

are constant on each side of the interface, the statement of Proposition 7.1

can be simplied.

Proposition 7.2

(Symmetric domain: piecewise constant coefficients)

are constant numbers. Then, if ε2 /ε1 , µ2 /µ1 ∈ ∗ equations (1)-(2) are uniquely solvable for all ω ∈ C \S where S is a discrete set. Assume that

ε1 , ε2 , µ1

and

µ2

R∗ \{−1},

Maxwell's

7.2 Prismatic edge (r, θ, z) centered (x, y, z) = (r cos θ, r sin θ, z). Let H > 0 0 < α < 2π , dene

Consider the geometry of Figure 1, middle-left. Introduce the cylindrical coordinates on the edge, so that the cartesian coordinates are mapped as denote the height of the cylinder,

R>0

its radius. Given

Ω1 := {(r cos θ, r sin θ, z) | 0 < r < R, 0 < θ < α, 0 < z < H} ; Ω2 := {(r cos θ, r sin θ, z) | 0 < r < R, α < θ < 2π, 0 < z < H} . α (θ−2π), z) and (R2 ϕ2 )(r, θ, z) = R1 and R2 such that (R1 ϕ1 )(r, θ, z) = ϕ1 (r, α−2π 1 ϕ ∈ H (Ω).

Introduce the two operators

ϕ2 (r, α−2π α θ + 2π, z)

for

Proceeding as for the case of the symmetric domain, one obtains the

Proposition 7.3

(Prismatic edge)

α Iα := max( 2π−α , 2π−α α ). − + − + Assume that ε satises ε ≥ C > 0 a.e. in Ω or max(ε1 /ε2 , ε2 /ε1 ) > Iα . − + − + Assume that µ satises µ ≥ C > 0 a.e. in Ω or max(µ1 /µ2 , µ2 /µ1 ) > Iα . ∗ Then, Maxwell's equations (1)-(2) are uniquely solvable for all ω ∈ C \S where S

Dene

Proposition 7.4

is a discrete set.

(Prismatic edge: piecewise constant coefficients)

ε1 , ε2 , µ1 and µ2 ε2 /ε1 , µ2 /µ1 ∈ R∗ \[−Iα ; −1/Iα ], where S is a discrete set. Assume that

are constant numbers.

Dene

α , 2π−α Iα := max( 2π−α α ).

Maxwell's equations (1)-(2) are uniquely solvable for all

Then, if

ω ∈ C∗ \S

7.3 Fichera corner Consider the geometry of Figure 1, middle-right. More precisely, dene

Ω := (−1; 1)3 , Ω1 := (0; 1)3

and

Ω2 := Ω\Ω1 . Introduce the operator

Above, for

R1 , R2 ,

(R1 ϕ1 )(x, y, z)

=

(R2 ϕ2 )(x, y, z)

=

` = 1 . . . 7, ϕ`2

ϕ ∈ H1 (Ω),  ϕ1 (−x, y, z) in     ϕ (x, −y, z) in  1    in  ϕ1 (x, y, −z) ϕ1 (−x, −y, z) in   ϕ (−x, y, −z) in  1    ϕ (x, −y, −z) in  1   ϕ1 (−x, −y, −z) in

such that, for

Ω12 Ω22 Ω32 Ω42 Ω52 Ω62 Ω72

:= (−1; 0) × (0; 1)2 := (0; 1) × (−1; 0) × (0; 1) := (0; 1)2 × (−1; 0) := (−1; 0)2 × (0; 1) ; := (−1; 0) × (0; 1) × (−1; 0) := (0; 1) × (−1; 0)2 := (−1; 0)3

ϕ12 (−x, y, z) + ϕ22 (x, −y, z) + ϕ32 (x, y, −z) −ϕ42 (−x, −y, z) − ϕ52 (−x, y, −z) − ϕ62 (x, −y, −z) +ϕ72 (−x, −y, −z).

is the restriction of

ϕ2

to

Ω`2 .

Again, proceeding as for the case of the symmetric domain, one obtains the

Proposition 7.5

(Fichera's corner)

+ − + ε satises ε ≥ C > 0 a.e. in Ω or max(ε− 1 /ε2 , ε2 /ε1 ) > 7. − + − + Assume that µ satises µ ≥ C > 0 a.e. in Ω or max(µ1 /µ2 , µ2 /µ1 ) > 7. ∗ Then, Maxwell's equations (1)-(2) are uniquely solvable for all ω ∈ C \S where S

Assume that

Proposition 7.6 Assume that

is a discrete set.

(Fichera's corner: piecewise constant coefficients)

ε1 , ε2 , µ1

and

µ2

are constant numbers. Then, if ∗ equations (1)-(2) are uniquely solvable for all ω ∈ C \S where

12

ε2 /ε1 , µ2 /µ1 ∈ R∗ \[−7; −1/7], S is a discrete set.

Maxwell's

7.4 Non symmetric cavity Ω := {(x, y, z) ∈ (−a; b) × (0; 1) × (0; 1)}, Ω1 := (−a; 0) × (0; 1) × (0; 1) and Ω2 := (0; b) × (0; 1) × (0; 1) with a > 0 and b > 0. The interface Σ is then equal to {0} × (0; 1) × (0; 1). Assume that ε1 , ε2 , µ1 and µ2 are constant numbers. Let us consider the non symmetric cavity of Figure 1. More precisely, dene

ϕ, ϕ0 ∈ H10 (Ω), (∇(Aε ϕ), ∇ϕ0 ) = (ε∇ϕ, ∇ϕ ) whereas in 3.2, we have introduced the operator A such that, for all ϕ, ϕ0 ∈ H1# (Ω), (∇(Aµ ϕ), ∇ϕ0 ) = (µ∇ϕ, ∇ϕ0 ). For this particular geometry, we know (see [2]) that the operator Aε µ (resp. A ) is Fredholm of index 0 (see Denition 8.1 below) if and only if ε2 /ε1 6= −1 (resp. µ2 /µ1 6= −1). ε µ To apply Theorem 6.2, we need A and A to be isomorphisms. Therefore, we study here the question of ε µ ε 1 ε the injectivity of A and A . Let us start with A . Consider ϕ an element of H0 (Ω) such that A ϕ = 0. The pair (ϕ1 , ϕ2 ) satises the equations Recall that in 3.1, we have dened the operator

Aε

such that, for all

0

µ

∆ϕ1 ∆ϕ2

= 0 = 0

in in

ϕ1 and ϕ2 in Fourier L ((0; 1) × (0; 1))), one obtains

Decomposing

Ω1 ; Ω2 ;

ϕ1 − ϕ2 ε1 ∂x ϕ1 − ε2 ∂x ϕ2

series (the family

2

ϕ1 (x, y, z) = ϕ2 (x, y, z) =

and where

ϕmn 1

and

ϕmn 2

P∞ P∞ n=1 P∞ n=1

on on

Σ; Σ.

{(y, z) 7→ sin(mπy) sin(nπz)}∞ m, n=1

is a basis of

√ 2 2 ϕmn 1 sinh( m + n π(x + a)) sin(mπy) sin(nπz) √ 2 2 ϕmn 2 sinh( m + n π(x − b)) sin(mπy) sin(nπz),

are constant numbers. Besides, the transmission conditions imply,

∗

∗

∀(m, n) ∈ N × N , For each

m=1 P∞ m=1

= 0 = 0

(m, n) ∈ N∗ × N∗ ,

√ √ 2 + n2 πa) 2 2 ϕmn = −ϕmn 1 sinh( m 2 sinh( √m + n πb) √ . 2 + n2 πa) = ϕmn ε cosh( m2 + n2 πb) ϕmn ε cosh( m 1 2 1 2

there exists a non trivial solution to the system (16) (in

(16)

mn (ϕmn 1 , ϕ2 ))

if and

only if

√ √ √ √ = ε2 sinh( m2 + n2 πa) cosh( m2 + n2 πb) + ε1 sinh( m2 + n2 πb) cosh( m2 + n2 πa) √ tanh( m2 + n2 πb) √ = − . tanh( m2 + n2 πa)

0 ⇔

ε2 /ε1

Aε : H10 (Ω) → H10 (Ω) is an isomorphism if and only if ε2 /ε1 is not an element p p Sε := {− tanh( m2 + n2 πb)/ tanh( m2 + n2 πa), (m, n) ∈ N∗ × N∗ } ∪ {−1}.

Consequently,

of (17)

Following the same way, exchanging the sin by cos and the sinh by cosh to satisfy the Neumann

Aµ : H1# (Ω) → H1# (Ω) is an isomorphism if and only if µ2 /µ1 is not p p Sµ := {− tanh( m2 + n2 πa)/ tanh( m2 + n2 πb), (m, n) ∈ N∗ × N∗ } ∪ {−1}.

condition, we prove that of

Remark 7.7

g : z 7→ − tanh(zπb)/ tanh(zπa) is continuous, a < b. Besides, we have limz→+∞ g(z) = −1. both sets Sε and Sµ .

The map

and strictly increasing if accumulation point of

Remark 7.8

an element (18)

a > b −1 is an

strictly decreasing if As a consequence,

For this particular 3D geometry, we obtain a result specic to 2D congurations (see [3]):

the problem with Dirichlet boundary condition for the coecient

ε

is well-posed if and only if the problem µ := ε−1 .

with Neumann boundary condition is well-posed for the coecient We deduce the

Proposition 7.9

(Non symmetric cavity: piecewise constant coefficients)

ε1 , ε2 , µ1 and µ2 are constant numbers. Assume that ε2 /ε1 ∈ R∗ \Sε and µ2 /µ1 ∈ R∗ \Sµ , with Sε and Sµ respectively dened in (17) and (18). Then, Maxwell's equations (1)-(2) are uniquely ∗ solvable for all ω ∈ C \S where S is a discrete set.

Assume that

13

8

Relaxing the main hypotheses

To prove the previous results, we relied extensively on two types of assumptions.

(Hε )

hypotheses domain

Ω

(Hµ )

and

ensure that the scalar problems are well-posed.

On the one hand,

On the other hand, the

is supposed to be simply connected, with a connected boundary. We would like now to relax

these assumptions. Concerning the hypotheses on the geometry, the diculty is well-known (see for instance [9]).

∂Ω

instance, if the boundary

1/2

∇ϕ, so that u 7→ (u, u)curl Ω is not simply connected.

is not connected, the space

is not a norm on

VN (ε; Ω)

VN (ε; Ω)

For

contains non-trivial curl free elds

anymore. The same occurs for

VT (µ; Ω)

when

(Hε ) and (Hµ ) has similar consequences. For instance, suppose that f = 0 has a non-trivial solution ϕ˜: the non-trivial curl free 1/2 VN (ε; Ω). Once more, u 7→ (u, u)curl is not a norm on VN (ε; Ω).

At rst glance, relaxing hypotheses

the homogeneous scalar problem (5) with

∇ϕ˜

eld

belongs to

However, we observe a fundamental dierence between the scalar potentials which are built in the two

ϕ∈ / H10 (Ω),

cases:

whereas

˜ = ∇ϕ˜ ϕ˜ ∈ H10 (Ω). As a consequence, the eld E Z ˜ · E 0 = 0, εE ∀E 0 ∈ VN (ε; Ω),

is such that:

Ω which is not the case for electric eld (6) stated in

˜ is a solution to the homogeneous problem (J = 0) for the E = ∇ϕ. So, E VN (ε; Ω) but not to the homogeneous problem (3) stated in HN (curl ; Ω). In

other words, when the scalar problems have non-trivial kernels, Theorems 3.3 and 3.7 (equivalence with the original Maxwell's problem) are no longer true. Summing up, we see that the diculties which occur when relaxing either hypotheses on the geometry or hypotheses on the scalar problem present some similarities (existence of admissible elds which are both divergence free and curl free) but also some fundamental dierences (formulations in and

VT (µ; Ω)

VN (ε; Ω)

are no longer equivalent to the original problem when the scalar problem have kernels).

Since the non-injectivity of the scalar problems is a diculty which is specic to the presence of signchanging coecients

ε

and/or

µ,

it will be treated rst in Subsection 8.1.

Then, in Subsection 8.2,

we shall check that the usual treatment for non-trivial geometries can be extended to sign-changing coecients.

8.1 Extension to non injective scalar problems 8.1.1 Non injective scalar problems: a new functional framework We have introduced the sesquilinear forms

aε

and

aµ

such that

aε (ϕ, ϕ0 ) = (ε∇ϕ, ∇ϕ0 ), aµ (ϕ, ϕ0 ) = (µ∇ϕ, ∇ϕ0 ),

∀ϕ, ϕ0 ∈ H10 (Ω); ∀ϕ, ϕ0 ∈ H1# (Ω).

With the Riesz representation theorem, we have dened the operators

and

Aε : H10 (Ω) → H10 (Ω) Aµ : H1# (Ω) → H1# (Ω)

s.t. s.t.

(∇(Aε ϕ), ∇ϕ0 ) = aε (ϕ, ϕ0 ), (∇(Aµ ϕ), ∇ϕ0 ) = aµ (ϕ, ϕ0 ),

∀ϕ, ϕ0 ∈ H10 (Ω); ∀ϕ, ϕ0 ∈ H1# (Ω).

Theorem 6.2 indicates that Maxwell's equations (1)-(2) are well-posed in the Fredholm sense when and

A

µ

are isomorphisms. In this section, we want to study the case when

Fredholm operators of index

Denition 8.1 operator i) ii)

L

Let

X

and

0.

Y

A

ε

and

A

µ

are non injective

Let us recall the denition of a Fredholm operator (see [34, 22]).

be two Banach spaces, and let

L:X→Y

be a continuous linear map. The

is said to be a Fredholm operator if and only if the following two conditions are fullled

dim(ker L) < ∞

and

dim(coker L) < ∞

range L

where

Aε

is closed;

coker L :=

Y/range L .

14

L

Besides, the index of a Fredholm operator

is dened by

ind L = dim(ker L) − dim(coker L).

For the example of the non symmetric cavity described in 7.4, this corresponds to consider the situation where where

Aε

κε ∈ Sε \{−1} and κµ ∈ Sµ \{−1}. For the sake of brevity, we will focus on congurations Aµ both have a kernel non reduced to zero. When only one of these two operators is not

and

injective, the study of the Maxwell's equations can be easily inferred from the one we present.

ε, µ and the geometry are Aε : H10 (Ω) → H10 (Ω) and Aµ : H1# (Ω) → H1# (Ω) are non injective Fredholm operators of index ε µ Nµ ε ε ε µ 0. Let us introduce {λεi }N i=1 a basis of ker A such that (∇λi , ∇λj ) = δij and {λi }i=1 a basis of ker A µ µ ε µ such that (∇λi , ∇λj ) = δij . Dene the spaces S and S such that

Up to the end of this section, we assume that the physical parameters such that

⊥

⊥

H10 (Ω) = ker Aε ⊕ Sε Consider the sesquilinear forms

a ˜ε

and

a ˜µ

and

H1# (Ω) = ker Aµ ⊕ Sµ .

such that

a ˜ε (ϕ, ϕ0 ) = (ε∇ϕ, ∇ϕ0 ), a ˜µ (ϕ, ϕ0 ) = (µ∇ϕ, ∇ϕ0 ),

∀ϕ, ϕ0 ∈ Sε ; ∀ϕ, ϕ0 ∈ Sµ .

With the Riesz representation theorem, dene the operators

and

A˜ε : Sε → Sε A˜µ : Sµ → Sµ

s.t. s.t.

(∇(A˜ε ϕ), ∇ϕ0 ) = a ˜ε (ϕ, ϕ0 ), µ 0 (∇(A˜ ϕ), ∇ϕ ) = a ˜µ (ϕ, ϕ0 ),

∀ϕ, ϕ0 ∈ Sε ; ∀ϕ, ϕ0 ∈ Sµ .

Classically (see [22]), one has the

Proposition 8.2

The operators

A˜ε : Sε → Sε

and

A˜µ : Sµ → Sµ

are isomorphisms.

As mentioned in the introduction of Section 8, Theorems 3.3 and 3.7 do not hold anymore. Indeed,

span(∇λε1 , . . . , ∇λεN ε )

is included in the kernel of problem (6), stated in

of the original problem (3), stated in

HN (curl ; Ω).

VN (ε; Ω),

but not in the kernel

Our objective is therefore to write variational

formulations of Maxwell's problems in some spaces dierent from

VN (ε; Ω)

and

VT (µ; Ω)

in order to

eliminate these articial kernels. A way to achieve that is to enrich the usual spaces by setting

˜ N (ε; Ω) := {u ∈ HN (curl ; Ω) | (εu, ∇ϕ) = 0, ∀ϕ ∈ Sε } ; V ˜ T (µ; Ω) := {u ∈ H(curl ; Ω) | (µu, ∇ϕ) = 0, ∀ϕ ∈ Sµ } . V Notice that we have

˜ N (ε; Ω) VN (ε; Ω) ⊂ V

and

˜ T (µ; Ω). VT (µ; Ω) ⊂ V

(19)

Let us precise the relation

between these spaces.

Lemma 8.3

For

i = 1 . . . N ε,

We deduce

Proof.

there exists

˜ N (ε; Ω) Λεi ∈ V

such that

(εΛεi , ∇λεj ) = δij ,

for

j = 1 . . . N ε.

˜ N (ε; Ω) = VN (ε; Ω) ⊕ span(Λε )N ε . V i i=1 j = 1 . . . N ε,

˜ N (ε; Ω) V

`j (v) = (εv, ∇λεj ). PN ε ε Let us prove that the family `1 , . . . , `N ε is free. Let α1 , . . . , αN ε be N constants such that j=1 αj `j = 0. PN ε PN ε ε ε ˜ For all v ∈ VN (ε; Ω), we have (εv, j=1 αj ∇λj ) = 0. This implies (εw, j=1 αj ∇λj ) = 0 for all w ∈ HN (curl ; Ω). Indeed, for w ∈ HN (curl ; Ω), let us introduce ϕ ∈ Sε the function such that (ε∇ϕ, ∇ϕ0 ) = (εw, ∇ϕ0 ) for all ϕ0 ∈ Sε ; ϕ is well-dened according to Proposition 8.2. We have ε ˜ N (ε; Ω). Since (ε∇ϕ, ∇λε ) = 0, j = 1 . . . N ε , we obtain (εw, PN αj ∇λε ) = 0. w − ∇ϕ ∈ V j j j=1 For

let us introduce the linear form

8.4 (see [20, lemma 4.12])

v−

PN ε

ε j=1 `j (v)Λj belongs

Theorem 8.4 V

Let

on

dened by

PN ε L2 (Ω), we deduce j=1 αj ∇λεj = 0. The family λε1 , . . . , λεN ε we have α1 = · · · = αN ε = 0: `1 , . . . , `N ε is free. Then Theorem ε ˜ N (ε; Ω), hereafter ensures the existence of the (Λi ). Finally, for all v ∈ V to VN (ε; Ω), which ends the proof.

C0∞ (Ω)2 ⊂ HN (curl ; Ω) is ε constitutes a basis of ker A , so

Since

`j

dense in

V be a normed vector space.

For all free family

`1 , . . . , `n ∈ V∗ , there exist x1 , . . . , xn ∈

such that

ì (xj ) = δ ij

with

δ ij = 1

if

15

i=j

and

δ ij = 0

if

i 6= j .

Proceeding in the same way, one can prove the

Lemma 8.5

For

i = 1 . . . N µ,

there exists

We deduce

˜ T (µ; Ω) such that (µΛµ , ∇λµ ) = δij , Λµi ∈ V j i

for

j = 1 . . . N µ.

˜ T (µ; Ω) = VT (µ; Ω) ⊕ span(Λµ )N µ . V i i=1

8.1.2 Non injective scalar problems: equivalent formulations Now, we can give equivalent formulations to problem (1)-(2) in the spaces

Theorem 8.6

˜ N (ε; Ω) V

ω 6= 0 and that the operator Aε : H10 (Ω) → H10 (Ω) ε Nε ˜ N (ε; Ω) be dened as in (19). with a non trivial kernel span{λi }i=1 . Let V 1) If (E, H) satises (1)-(2) then E is a solution of the problem Assume that

Find Z

˜ N (ε; Ω) E∈V

such that for all

Proof.

satises (20) then

is Fredholm of index

0

(20)

J · E0.

Ω

E

˜ T (µ; Ω). V

˜ N (ε; Ω): E 0 Z∈ V

µ−1 curl E · curl E 0 − ω 2 εE · E 0 = iω

2) If

and

Ω


satises (1)-(2).

˜ N (ε; Ω) ⊂ HN (curl ; Ω) E∈V HN (curl ; Ω). For all E 0 in HN (curl ; Ω), ε Proposition 8.2 indicates that we can build ϕ ∈ S such that a ˜ε (ϕ, ϕ0 ) = (εE 0 , ∇ϕ0 ), for all ϕ0 ∈ Sε . 0 ˜ N (ε; Ω). Observing that (εE, ∇ϕ) = 0 and (J , ∇ϕ) = 0 (recall that The element E − ∇ϕ belongs to V div J = 0), one obtains Z Z µ−1 curl E · curl E 0 − ω 2 εE · E 0 = iω J · E0. Let us focus on the proof

satises (20) then

E

2) ⇒ 1).

For that, let us show that if

is a solution to the problem (7) set in

Ω Therefore, if

E

satises (20) then

Ω


satises (1)-(2).

Adapting the proof of Theorem 8.6, one obtains the

Theorem 8.7

ω 6= 0 and that the operator Aµ : H1# (Ω) → H1# (Ω) µ Nµ ˜ T (µ; Ω) be dened as in (19). with a non trivial kernel span{λi }i=1 . Let V 1) If (E, H) satises (1)-(2) then H is a solution of the problem Assume that


˜ T (µ; Ω): HZ0 ∈ V ε−1 curl H · curl H 0 − ω 2 µH · H 0 = ε−1 J · curl H 0 .

Find Z

˜ T (µ; Ω) H∈V

such that for all

Ω

2) If

H

satises (21) then

0

(21)

Ω

(i (ωε)−1 (curl H − J ), H)

satises (1)-(2).

Now, we wish to study formulations (20) and (21). For that, we need some new compactness results.

8.1.3 Non injective scalar problems: compactness results Let us start by proving the compactness result needed to study the problem for the electric eld.

Theorem 8.8 ˜ N (ε; Ω) V

Proof.

in

Let

L2 (Ω)

Ω


∂Ω

is connected. Then the embedding of

is compact.

˜ N (ε; Ω). For all n ∈ N, using Lemma 8.3, dene (un ) be a bounded sequence of V PN ε ε Nε v n ∈ VN (ε; Ω) and (αin ) ∈ C (αin = ì (un )) such that un = v n + i=1 αin Λi . To prove The2 orem 8.8, it is sucient to show that we can extract a subsequence from (v n ) which converges in L (Ω). 2 Dene F n = curl v n . The sequence (F n ) is bounded in L (Ω). For all n ∈ N, we have div (εv n ) = 0. Since ∂Ω is connected, there exists (see [1], theorem 3.12) w n ∈ VT (1; Ω) such that curl w n = εv n . −1 Thus, for all n ∈ N, we can write v n = ε curl wn . Let us show we can extract a sequence from 2 (curl wn ) which converges in L (Ω). Let

We know that

w 7→ kcurl wk

denes a norm on

16

VT (1; Ω).

Consequently, the sequence

(wn )

is

VT (1; Ω). Since curl ε−1 curl wn = F n in Ω and (ε−1 curl wn ) × n = 0 on ∂Ω, one has (ε curl wn , curl w0 ) = (F n , w0 ), for all w0 ∈ VT (1; Ω). Now, let us build a continuous operator T from VT (1; Ω) to itself, to restore some property of positivity. Consider w ∈ VT (1; Ω). ε i) First, dene ϕ the unique element of S such that

bounded in

−1

ε

Z

0

Z

ε∇ϕ · ∇ϕ = Ω

ε(curl w − Ω

N X

βi Λεi ) · ∇ϕ0 ,

∀ϕ0 ∈ Sε ,

i=1

A˜ε : Sε → Sε is an isomorphism. PN ε ε ii) Thanks to the denition of the coecients (βi ), notice that ε(curl w− i=1 βi Λi −∇ϕ) is a divergence 2 free element of L (Ω). Since Ω is simply connected and since ∂Ω is connected, according to theorem 3.12 PN ε ε in [1], there exists a unique potential Tw ∈ VT (1; Ω) such that curl Tw = ε(curl w − i=1 βi Λi − ∇ϕ). This denes a bounded operator T : VT (1; Ω) → VT (1; Ω). Since T is continuous, the sequence (Tw n ) is bounded in VT (1; Ω). So, we can extract a subsequence 2 ε from (w n ), still denoted (w n ), such that (Tw n ) converges in L (Ω). Since for i = 1 . . . N , the sequence ε (βin ), with βin = (ε curl wn , ∇λi ), is bounded in C, we can extract a subsequence from (wn ), still denoted (w n ), such that (βin ) converges. Introducing w nm := w n − w m , F nm := F n − F m and βinm := βin − βim , one nds |(F nm , Twnm )| = (ε−1 curl wnm , curl Twnm ) PN ε 2 ≥ kcurl wnm k − i=1 |βinm ||(curl wnm , Λεi )|. where

βi = (ε curl w, ∇λεi ).

The function

ϕ

is well-dened since

(curl wn ) is a Cauchy sequence in L2 (Ω), so it converges. (v n ) = (ε−1 curl wn ) which converges in L2 (Ω).

This estimate proves that a subsequence from

Thus, we can extract

Proceeding in the same way, we can prove the

Theorem 8.9 ˜ T (µ; Ω) V

in

Let

Ω

L2 (Ω)


∂Ω

is connected. Then the embedding of

is compact.

8.1.4 Non injective scalar problems: well-posedness of Maxwell's equations ˜ N (ε; Ω) → AÑ (ω) : V ˜ N (ε; Ω) and A˜T (ω) : V ˜ T (µ; Ω) → V ˜ T (µ; Ω) such that for all E, E 0 ∈ V ˜ N (ε; Ω) and for all H, H 0 ∈ V ˜ T (µ; Ω), V (AÑ (ω)E, E 0 )curl = (µ−1 curl E, curl E 0 ) − ω 2 (εE, E 0 ), (A˜T (ω)H, H 0 )curl = (ε−1 curl H, curl H 0 ) − ω 2 (µH, H 0 ).

With the Riesz representation theorem, let us introduce the bounded operators

Now, let us state the main result when the geometry and the physical coecients

ε, µ

are such that the

scalar problems are well-posed in the Fredholm sense with a non trivial kernel.

Theorem 8.10

Let

Ω


such that div J = 0. ε Assume that the operator A : ε Nε span{λi }i=1 . µ Assume that the operator A : µ Nµ span{λi }i=1 . Then, the following results hold.

H10 (Ω) → H10 (Ω)

∂Ω

is connected. Consider

J ∈ L2 (Ω)


0

with a non trivial kernel


0

with a non trivial kernel

H1# (Ω) → H1# (Ω)

˜ N (ε; Ω) → V ˜ N (ε; Ω) is a Fredholm operator of index 0. ω ∈ C, the operator AÑ (ω) : V 0 0 0 ∗ ˜ ˜ ˜ N (ε; Ω), if Moreover, for ω ∈ C , E ∈ VN (ε; Ω) satises (AN (ω)E, E )curl = iω(J , E ), for all E ∈ V −1 and only if the pair (E, (i ωµ) curl E) satises the Maxwell's equations (1)-(2). ˜ T (µ; Ω) → V ˜ T (µ; Ω) is a Fredholm operator of index 0. Moreover, for • For all ω ∈ C, A˜T (ω) : V ∗ ˜ ˜ ˜ T (µ; Ω), if and ω ∈ C , H ∈ VT (µ; Ω) satises (AT (ω)H, H 0 )curl = (ε−1 J , curl H 0 ), for all H 0 ∈ V −1 only if the pair (i (ωε) (curl H − J ), H) satises the Maxwell's equations (1)-(2). •

For all

17

Proof.

AÑ (ω) is a Fredholm operator of index 0. For all ω ∈ C, using Theorem 8.8, ˜ ˜ we can prove that AN (ω) − A˜ N (0) is a compact operator of VN (ε; Ω). Consequently, according to [22, ˜ theorem 2.26], AN (ω) is a Fredholm operator of index 0 if and only if A˜ N (0) is a Fredholm operator of ˜ε ˜ ˜ index 0. In the sequel, we work on A˜ N (0). We build a bounded operator T : VN (ε; Ω) → VN (ε; Ω) to ˜ restore some positivity up to a compact perturbation. Let us consider u ∈ VN (ε; Ω). µ i) First, dene ϕ the unique element of S such that Let us prove that

µ

Z

0

Z

µ∇ϕ · ∇ϕ =

µ (curl u −

Ω

Ω

N X

βi Λµi ) · ∇ϕ0 ,

∀ϕ0 ∈ Sµ ,

i=1

βi = (µ curl u, ∇λµi ). The function ϕ is well-dened since A˜µ : Sµ → Sµ is an isomorphism. PN µ µ 2 ii) Then, notice that µ(curl u − i=1 βi Λi − ∇ϕ) is a divergence free element of L (Ω) such that PN µ µ(curl u − i=1 βi Λµi − ∇ϕ) · n = 0 on ∂Ω. Since Ω is simply connected and since ∂Ω is connected, according to theorem 3.17 in [1], there exists a unique potential ψ ∈ VN (1; Ω) such that curl ψ = PN µ µ(curl u − i=1 βi Λµi − ∇ϕ). ε iii) Consider ζ the unique element of S such that Z Z ε∇ζ · ∇ζ 0 = ε ψ · ∇ζ 0 , ∀ζ 0 ∈ Sε . where

Ω The function

ζ

Ω

A˜ε : Sε → Sε is an isomorphism. ˜ ε u = ψ − ∇ζ ˜ N (ε; Ω) → V ˜ N (ε; Ω) such that T T :V

is well-dened since


˜ N (ε; Ω) → V ˜ N (ε; Ω) ˜ε : V K

˜ε

and the operator

such that µ

˜ε

(K u, v)curl = (u, v) +

N X

(µ curl u, ∇λµi )(Λµi , curl v),

˜ N (ε; Ω). ∀v ∈ V

i=1 According to Theorem 8.8, we know that the embedding of

˜ε

K

˜ N (ε; Ω) in L2 (Ω) is compact. V

Consequently,

is the sum of a compact operator and a nite rank operator. Therefore, it is a compact operator.

Now, for all

˜ N (ε; Ω), u, v ∈ V

we obtain

˜ ε u), v)curl (AÑ (0)(T

=

˜ ε u), curl v) (µ−1 curl (T

=

˜ ε u, v)curl . (curl u, curl v) + (u, v) − (K

˜ε + K ˜ ε is a right parametrix for AÑ (0). Thus, the ˜ ε = Id. This proves that T A˜T (0) ◦ T ˜ selfadjoint operator AN (0) is Fredholm of index 0 (use [22, Lemma 2.23]). In the same way, we prove ˜ ˜ that A˜ T (ω) : VT (µ; Ω) → VT (µ; Ω) is a Fredholm operator of index 0 for all ω ∈ C. Finally, the We deduce

equivalence with Maxwell's equations (1)-(2) comes from Theorems 8.6 and 8.7.

Remark 8.11

To apply the analytic Fredholm theorem to prove that Maxwell's equations (1)-(2) are ω ∈ C∗ \S where S is a discrete set, it remains to show that there exists ω ∈ C

uniquely solvable for all such that

AÑ (ω)

or

A˜T (ω)

is invertible. However, we have not been able to prove this result.

8.2 Extension to a non trivial geometry 8.2.1 Non trivial geometry: a decomposition of the functional spaces Classical congurations for Maxwell's equations include non-topologically trivial domains, and/or domains with a non-connected boundary. We study these congurations here. To avoid multiple sub-cases, we focus our work on the case of a non simply connected domain whose boundary is not connected. In these geometries, the functions u of VN (ε; Ω) VT (µ; Ω)) do not necessarily write under the form u = ε−1 curl ψ (resp. u = µ−1 curl ψ ) where ψ belongs to VT (1; Ω) (resp. VN (1; Ω)). However, imposing more restrictive conditions to the functions of VN (ε; Ω) and VT (µ; Ω), we can recover these results of existence for the potentials. Figure 2 presents an example of such a geometry. (resp.

To introduce the spaces adapted to the study of Maxwell's equations in this kind of domains, we use the notations of [1].

18

Figure 2: An example of domain which is not simply connected and whose boundary is not connected. The domain is made of the torus without the purple inclusions. It is not simply connected because of the toroidal structure. The boundary is not connected because the boundary of the torus and the ones of the spheres are not connected. The green disk represents a cut

Σ1

which is such that

Ω \ Σ1

is simply

connected.

Notations for domains with a non connected boundary. connected components of the boundary

∂Ω.

Since we assume that

∂Ω

We denote

Γi , i = 0 . . . I , the I ≥ 1.

is not connected, we have

Let us introduce

H1Γ (Ω) := ϕ ∈ H1 (Ω) | ϕ|Γ0 = 0, ϕ|Γi = cst, i = 1 . . . I . Let us start by characterizing this space. Using a lifting function, we prove the

Proposition 8.12

Assume

(Hε )

to be true. Then, for

i = 1 . . . I,

there exists a unique solution

pi

to

the problem

pi ∈ H1Γ (Ω) such that: div (ε∇pi ) = 0 in Ω pi = δik on Γk , k = 1 . . . I. Find

Then, we have

H1Γ (Ω) = H10 (Ω) ⊕ span(pi )Ii=1 . Dene

ˆ N (ε; Ω) := u ∈ HN (curl ; Ω) | (εu, ∇ϕ) = 0, ∀ϕ ∈ H1 (Ω) . V Γ Notice that

ˆ N (ε; Ω) ⊂ VN (ε; Ω). V

Lemma 8.13 δik ,

(Hε ) to be true. For We deduce

Assume

k = 1 . . . I.

for

and

Proof.

Let us precise the link between these two spaces.

i = 1 . . . I , there exists P i ∈ VN (ε; Ω) such that (εP i , ∇pk ) =

VN (ε; Ω)

=

HN (curl ; Ω)

=

ˆ N (ε; Ω) ⊕ span(P i )I V i=1 ˆ N (ε; Ω) ⊕ span(P i )I ⊕ ∇H1 (Ω). V 0 i=1

`k on VN (ε; Ω) such that `k (v) = (εv, ∇pk ). Let PI `1 , . . . , Ì is free. Consider I constants α1 , . . . , αI such that k=1 αk `k = 0. PI PI For all v ∈ VN (ε; Ω), we have (εv, k=1 αk ∇pk ) = 0. We deduce (εw, k=1 αk ∇pk ) = 0 for all w ∈ HN (curl ; Ω). Indeed, if w ∈ HN (curl ; Ω), let us dene ϕ ∈ H10 (Ω) the function such that (ε∇ϕ, ∇ϕ0 ) = (εw, ∇ϕ0 ) for all ϕ0 ∈ H10 (Ω). We have w − ∇ϕ ∈ VN (ε; Ω). Since (ε∇ϕ, ∇pk ) = 0, PI k = 1 . . . I , we nd (εw, k=1 αk ∇pk ) = 0. For

k = 1 . . . I,

let us dene the linear form

us prove that the family

PI C0∞ (Ω)2 ⊂ HN (curl ; Ω) is dense in L2 (Ω), we obtain k=1 αk ∇pk = 0. But the family p1 , . . . , pI is free. Consequently, we have α1 = · · · = αI = 0 and the family `1 , . . . , Ì is also free. Theorem 8.4 then Since

allows us to obtain the desired result.

Notations for non simply connected domains. surfaces

i) ii) iii) iv)

Σj , j = 1 . . . J

each surface

Σj

We will assume that there exist connected open

called cuts such that: is an open subset of a smooth variety;

Σj is contained in ∂Ω, j = 1 . . . J ; Σj ∩ Σk is empty for j 6= k ; SJ Ω˙ := Ω \ i=1 Σj is pseudo-lipschitz [1]

the boundary of the intersection the open set

19

and simply connected.

The domain to

2

L (Ω)

Ω

J = 0. The extension through Σj , j = 1 . . . J .

is said topologically trivial when we can take

e·

is denoted

whereas

[·]Σj

denotes the jump

operator from

˙ L2 (Ω)

In this denition of

the jump, we assume that a convention has been established for the sign. We also assume that a unit vector

n

normal to

Σj , j = 1 . . . J ,

is chosen, consistent with the choice of the sign of the jump. Let us

introduce the space of scalar potentials

˙ := Θ(Ω)

˙ | ϕ ∈ H (Ω) 1

Z Ω

ϕ e=0

and

[ϕ]Σj = cst, j = 1 . . . J

.

Let us present a result of decomposition of this space.

Proposition 8.14

(Hµ )

Assume

to be true. Then for

j = 1 . . . J,

there exists a unique solution

qj

to

the problem

˙ qj ∈ Θ(Ω) div (µ∇qj ) = µ∂n qj = [qj ]Σk = [µ∂n qj ]Σk = Find

such that:

Ω˙ on ∂Ω k = 1...J k = 1 . . . J.

0 0 δjk , 0,

in

(22)

We then have

˙ = H1 (Ω) ⊕ span(qj )J . Θ(Ω) # j=1

Proof.

Since we have assumed

˙ rj ∈ Θ(Ω)

1 ≤ j ≤ J , let qj = rj − ϕ where ϕ is

it. For

(Hµ )

to hold true, problem (22) has at most one solution. Let us build

be a function such that


Z

µ∇ϕ · ∇ϕ0 =

Ω One classically checks that of the space

qj

H1# (Ω)

Z

[rj ]Σk = δjk

for

k = 1 . . . J.

Then, let us dene

such that

gj · ∇ϕ0 , µ∇r

∀ϕ0 ∈ H1# (Ω).

Ω satises problem (22). This allows to obtain the result of decomposition

˙ . Θ(Ω)

Let us introduce

n o ˆ T (µ; Ω) := u ∈ H(curl ; Ω) | (εu, ∇ϕ) g = 0, ∀ϕ ∈ Θ(Ω) ˙ . V Observe that we have

ˆ T (µ; Ω) ⊂ VT (µ; Ω). V

Working as in the proof of Lemma 8.13, we can precise

the relation between these two spaces.

Lemma 8.15 δjk ,

for

(Hµ ) to be true. For We deduce

Assume

k = 1 . . . J.

and

Remark 8.16

VT (µ; Ω)

=

H(curl ; Ω)

=

gk ) = j = 1 . . . J , there exists Qj ∈ VT (µ; Ω) such that (µQj , ∇q ˆ T (µ; Ω) ⊕ span(Qj )J V j=1 ˆ T (µ; Ω) ⊕ span(Qj )J ⊕ ∇H1 (Ω). V j=1 #

ˆ N (ε; Ω) can be written as u = u of V ˆ ε curl ψ with ψ belonging to VT (1; Ω). Similarly, theorem 3.17 of [1] ensures that for every u ∈ ˆ T (µ; Ω), there exists a unique ψ ∈ V ˆ N (1; Ω) such that u = µ−1 curl ψ . In the sequel, we will adapt V Theorem 3.12 in [1] states that every element

−1

the proofs of the previous sections using these results of existence of vector potentials. Assume

(Hε )

and

(Hµ )

to be true.

Remark that Theorems 3.3, 3.7 which prove the equivalence

between initial Maxwell's equations and formulations in sumption concerning the topology of the domain.

VN (ε; Ω), VT (µ; Ω)

do not require any as-

Therefore, they are true for the geometry we are

considering. In the sequel, we will work with these formulations set in prove the compactness results of Theorems 5.1, 5.3 in the case where

VN (ε; Ω), VT (µ; Ω). We Ω is not simply connected

non connected boundary. This will be the subject of the next Subsection.

20

should with a

Remark 8.17

ˆ N (ε; Ω), V ˆ T (µ; Ω)? A priori, the electric eld V ˆ N (ε; Ω). To see this, we use belong to the space V

Can we work with formulations set in

which satises Maxwell's equations has no reason to Lemma 8.13 and we decompose

E

under the form

ˆ+ E=E

I X

αi P i ,

i=1

with

ˆ N (ε; Ω) ˆ∈V E

and

(α1 , . . . , αI ) ∈ CI .

For

i = 1 . . . I,

testing with

∇pi

in (6), we nd

αi = (εE, ∇pi ) = (iω)−1 (J , ∇pi ) = (iω)−1 hJ · n, 1iΓi , h·, ·iΓi denotes the duality product between H1/2 (Γi ) and its dual space. Above, we have used the properties div J = 0 in Ω and pi = δik on Γk , k = 1 . . . I . Thus, if there exists 0 ≤ i ≤ I such that ˆ N (ε; Ω). But this also proves that to know the eld E , it is hJ · n, 1iΓi 6= 0, then E does not belong to V ˆ sucient to determine E . Following the same reasoning, we can check that the magnetic eld is always ˆ T (µ; Ω), regardless of the source term J . an element of V where

8.2.2 Non trivial geometry: compactness results Let us work rst on the space of electric elds.

Theorem 8.18 Assume

(Hε )

Let

Ω

R3 with a Lipschitz-continuous VN (ε; Ω) in L2 (Ω) is compact.

be a bounded connected open subset of

to be true. Then the embedding of

boundary.

Proof.

Let (un ) be a bounded sequence of VN (ε; Ω). For all n ∈ N, using Lemma 8.13, let us dene ˆ N (ε; Ω) and (α1n , . . . , αIn ) ∈ CI the elements such that un = v n + PI αin P i . To prove vn ∈ V i=1 Theorem 8.18, it is sucient to show that we can extract a subsequence from (v n ) which converges in L2 (Ω). Dene F n = curl v n . The sequence (F n ) is bounded in L2 (Ω). In addition, for all n ∈ N, 1 there holds (εv n , ∇ϕ) = 0 for all ϕ ∈ HΓ (Ω). Therefore, according to theorem 3.12 in [1], there exists ˆ wn ∈ VT (1; Ω) such that curl wn = εv n . Thus, for all n ∈ N, we have v n = ε−1 curl wn . Let us prove 2 that we can extract a subsequence from (curl w n ) which converges in L (Ω).

ˆ T (1; Ω) (cf. [1, corollary 3.16]). Consequently, the V −1 sequence Since curl ε curl wn = F n in Ω and (ε−1 curl wn ) × n = 0 −1 ˆ T (1; Ω). on ∂Ω, we have (ε curl wn , curl w0 ) = (F n , w0 ) for all w0 ∈ V ˆ ˆ ˆ Now, let us build a bounded operator T from VT (1; Ω) to VT (1; Ω) to recover some positivity. Consider ˆ T (1; Ω). w∈V 1 i) First, dene ϕ the unique element of H0 (Ω) such that Z Z 0 ε∇ϕ · ∇ϕ = ε curl w · ∇ϕ0 , ∀ϕ0 ∈ H10 (Ω). w 7→ kcurl wk denes a ˆ T (1; Ω). (wn ) is bounded in V

We know that

Ω

norm on

Ω

ϕ is well-dened because we have assumed (Hε ) to be true. PI βi := (ε curl w, ∇pi ) for i = 1 . . . I , let us then notice that we have (ε(curl w − i=1 βi P i − ∇ϕ), ∇ϕ0 ) = 0 for all ϕ0 ∈ H1Γ (Ω). Consequently, according to theorem 3.12 of [1], there exists a unique ˆ ∈V ˆ = ε(curl w − PI βi P i − ∇ϕ). This denes a bounded ˆ T (1; Ω) such that curl Tw potential Tw i=1 ˆ from V ˆ T (1; Ω) to V ˆ T (1; Ω). operator T ˆ is continuous, the sequence (Tw ˆ n ) is bounded in V ˆ T (1; Ω). So we can extract a subsequence Since T 2 ˆ from (w n ), still denoted (w n ), such that (Tw n ) converges in L (Ω). Since for i = 1 . . . I , the sequence (βin ), with βin = (ε curl wn , ∇pi ), is bounded in C, we can extract a subsequence from (wn ), still denoted (w n ), such that (βin ) converges. Let us introduce w nm := w n − w m , F nm := F n − F m and βinm := βin − βim . We have ˆ nm ) = (ε−1 curl wnm , curl Tw ˆ nm ) (F nm , Tw PI 2 ≥ kcurl wnm k − i=1 |βinm ||(curl wnm , P i )|.

The function ii) Dening

(curl wn ) is a Cauchy sequence of L2 (Ω). Therefore, −1 subsequence from (v n ) = (ε curl wn ) which converges in L2 (Ω).

This estimate proves that can extract a

21

it converges. Thus, we

Proceeding in the same way, we prove the

Theorem 8.19 (Hµ )

Assume

Let

Ω

R3 with a Lipschitz-continuous VT (µ; Ω) in L2 (Ω) is compact.


to be true. Then the embedding of

boundary.

8.2.3 Non trivial geometry: well-posedness of Maxwell's equations Let us state now the main result of this section concerning the well-posedness of Maxwell's equations in non trivial geometries.

Theorem 8.20

R3 with a Lipschitz-continuous boundary. Consider J ∈ L (Ω) such that div J = 0. Assume (Hε ) and (Hµ ) to be true. Then, we have the following Let

Ω


2

results. For all ω ∈ C, the operator for the electric eld AN (ω) : VN (ε; Ω) → VN (ε; Ω) dened in (8) is 0 ∗ a Fredholm operator of index 0. Moreover, for ω ∈ C , E ∈ VN (ε; Ω) satises (AN (ω)E, E )curl = 0 0 iω(J , E ), for all E ∈ VN (ε; Ω), if and only if the pair (E, (i ωµ)−1 curl E) satises the Maxwell's

•

equations (1)-(2).

ω ∈ C, AT (ω) : VT (µ; Ω) → VT (µ; Ω) dened in (11) is a Fredholm operator of index 0. ω ∈ C∗ , H ∈ VT (µ; Ω) satises (AT (ω)H, H 0 )curl = (ε−1 J , curl H 0 ), for all H 0 ∈ VT (µ; Ω), if and only if the pair (i (ωε)−1 (curl H − J ), H) satises the Maxwell's equations (1)-(2).

•

For all

Moreover, for

Proof.

AN (ω) is a Fredholm operator of index 0. For all ω ∈ C, using Theorem 8.18, AN (ω) − AN (0) is a compact operator of VN (ε; Ω). So, it is sucient to show that AN (0) is Fredholm of index 0. Again, we are going to build a right parametrix Tε : VN (ε; Ω) → VN (ε; Ω) for the operator AN (0). Consider u ∈ VN (ε; Ω). 1 i) First, dene ϕ the unique element of H# (Ω) such that Let us prove that

one can prove that

Z

µ∇ϕ · ∇ϕ0 =

Z

Ω

µ curl u · ∇ϕ0 ,

∀ϕ0 ∈ H1# (Ω).

Ω

ϕ is well-dened because we have assumed (Hµ ) to hold true. gj ) for j = 1 . . . J , notice that we have (µ(curl w − PJ βj Qj − ii) Dening βj := (µ curl w, ∇q j=1 0 0 g ˙ ∇ϕ), ∇ϕ ) = 0 for all ϕ ∈ Θ(Ω). Therefore, according to theorem 3.17 in [1], there exists a unique ˆ N (1; Ω) such that curl ψ = µ(curl w − PJ βj Qj − ∇ϕ). potential ψ ∈ V

The function

j=1

ζ

iii) Consider


Z

H10 (Ω)

ε∇ζ · ∇ζ 0 =

such that

Z

Ω The function

ζ

∀ζ 0 ∈ H10 (Ω).

(Hε ) to hold true. Tε : VN (ε; Ω) → VN (ε; Ω) such that Tε u = ψ − ∇ζ

is well-dened because we have assumed

iv) Finally, let us dene the operator operator

ε ψ · ∇ζ 0 ,

Ω

K ε : VN (ε; Ω) → VN (ε; Ω) (K ε u, v)curl = (u, v) +

and the

such that

J X gj )(Qj , curl v), (µ curl u, ∇q

∀v ∈ VN (ε; Ω).

j=1 According to Theorem 8.18, we know that the embedding of quently,

K

ε

VN (ε; Ω)

in

is the sum of a compact operator and a nite rank operator.

operator. Now, for all

v ∈ VN (ε; Ω),

L2 (Ω)

is compact.

Conse-

Therefore, it is a compact

we nd

(AN (0)(Tε u), v)curl

=

(µ−1 curl (Tε u), curl v)

=

(curl u, curl v) + (u, v) − (K ε u, v)curl .

AT (0) ◦ Tε + K ε = Id. We deduce that Tε is indeed a right parametrix for AN (0). This proves that the selfadjoint operator AN (0) is Fredholm of index 0. Similarly, we prove that AT (ω) : VT (µ; Ω) → VT (µ; Ω) is a Fredholm operator of index 0 for all ω ∈ C. Finally, the equivalence with

Thus, we have

Maxwell's equations (1)-(2) comes from Theorems 3.3 and 3.7.

22

Remark 8.21

Again, to apply the analytic Fredholm theorem to prove that Maxwell's equations (1)-(2) ω ∈ C∗ \S where S is a discrete set, it remains to show that there exists

are uniquely solvable for all

such that AN (ω) or AT (ω) is invertible. This results does not seem simple to obtain. However, according to Remark 8.17, we observe that the elements of ker AN (ω) (resp. ker AT (ω)) always belong

ω∈C

ˆ N (ε; Ω) (resp. V ˆ T (µ; Ω)). On the other hand, the map u 7→ kcurl uk denes a norm on V ˆ N (ε; Ω) V ˆ T (µ; Ω). But one ingredient is still missing to achieve the injectivity of AN (ω) or AT (ω). and on V

to

Acknowledgments This work was supported by the ANR project METAMATH, grant ANR-11-MONU-016 of the French Agence Nationale de la Recherche.

References [1] C. Amrouche, C. Bernardi, M. Dauge, and V. Girault. Vector potentials in three-dimensional nonsmooth domains. Math. Meth. Appl. Sci., 21:823864, 1998. [2] A.-S. Bonnet-Ben Dhia, L. Chesnel, and P. Ciarlet Jr.

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24

PhD thesis, Université Versailles - Saint-

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