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Mathematical Models and Methods in Applied Sciences c World Scientific Publishing Company

A NEW COMPACTNESS RESULT FOR ELECTROMAGNETIC WAVES. APPLICATION TO THE TRANSMISSION PROBLEM BETWEEN DIELECTRICS AND METAMATERIALS

A.-S. Bonnet-Bendhia, P. Ciarlet, Jr., C.M. Zwölf Laboratoire POEMS, UMR 2706 CNRS/ENSTA/INRIA, ´ Ecole Nationale Supérieure de Techniques Avancées, 32, boulevard Victor, 75739 Paris Cedex 15, France Received (Day Month Year) Revised (Day Month Year) Communicated by (xxxxxxxxxx) We consider the time-harmonic Maxwell equations, involving wave transmission between media with opposite sign dielectric and/or magnetic coefficients. We prove that, in the case of sign-shifting dielectric coefficients, the space of electric fields is compactly embedded in L2 . We build a three-field variational formulation equivalent to Maxwell system for sign-shifting magnetic coefficients and show that, under some suitable conditions, the formulation fits into the coercive plus compact framework. Keywords: electromagnetic wave transmission problem; sign-shifting dielectric coefficient; signshifting magnetic coefficient; left-hand materials. AMS Subject Classification: 22E46, 53C35, 57S20

Introduction In recent years, a growing interest for new materials has arisen. At particular frequencies, they behave like materials with negative electric permittivity or/and negative magnetic permeability µ. They include superconductors, left-handed materials 16 , etc. As a consequence, most mathematical approaches fail to resolve the corresponding electromagnetic models. Accordingly, these ”negative” materials raise many challenging questions, from both mathematical and numerical points of view. We consider here the particular case of an interface between ”positive” (dielectric) and ”negative” materials. Our main objective is to study those interface problems, and to provide variational settings, which can be easily discretized, for instance via finite element methods. In 2d configurations, they reduce to scalar problems involving terms like −div (∇·). Those scalar problems have been thoroughly investigated: we refer the reader to Refs. 9, 15, 5. It is now well understood that well-posedness depends crucially on the ratio of the values of taken from both sides of the interface. On the one hand, when its value is precisely equal to −1, the interface problem is ill-posed 9 . On the other hand, well-posedness in the Fredholm sense has been obtained when its absolute value is small enough (or large enough). This 1


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2 A.-S. Bonnet-Bendhia, P. Ciarlet, Jr., C.M. Zwölf

result has been achieved under very weak assumptions (Lipschitz interface, L∞ coefficient ) in Ref. 3 where a variational formulation with an additional vector unknown is used (see also Ref. 4 for an alternate proof). The main objective of this paper is to extend these results to the case of the 3d Maxwell equations, especially when both and µ exhibit a sign-shift at the interface. With no loss of generality, we shall focus on the electric field formulation. Here, the sign-shift of µ in 1 the term curl curl · raises similar difficulties as in the scalar case. In addition, a µ new difficulty appears in the constraint on the field div (e) = 0, coming now from the sign-shift of . The outline of the paper will be as follows. In section 1, we introduce the ad hoc mathematical framework of our study. In section 2, we specifically target the constraint involving a sign-shifting electric permittivity : we prove a new, Weber-like, compactness embedding result for the space of electric fields. Then, in the next section, we focus instead on Maxwell’s equations with a sign-shifting magnetic permeability µ. We build a three-field variational formulation, thus generalizing the approach advocated in Ref. 3 for the scalar problem. Assembling the previous results allows us to prove the well-posedness of this variational formulation, in the general case of sign-shifting electric permittivity and magnetic permeability. Finally we give some concluding remarks, before recalling some elementary results in the Annex. 1. Derivation of the model and mathematical framework Let Ω be an open, bounded and connected set of R3 with a Lipschitz polyhedrala ∂Ω; let n be the unit outward normal to ∂Ω. It is assumed that the domain Ω can be partitioned into two simply connected sub-domains Ω1 and Ω2 with Lipschitz polyhedral boundaries: Ω = Ω1 ∪ Ω2 , Ω1 ∩ Ω2 = ∅; let ni be the unit outward normal to ∂Ωi , i = 1, 2. Then, define the interface Σ = ∂Ω1 ∩ ∂Ω2 . Finally, we introduce Γi = ∂Ωi \ Σ; it is assumed that Γ1 and Γ2 are connected. Both assumptions on the geometry (Ω1 and Ω2 simply connected, Γ1 and Γ2 connected) can be removed. We introduce them for the ease of exposition. In the sequel, we shall introduce functional spaces with elements defined on O, or on (a part of) its boundary ∂O, where O stands for an open, bounded and connected set with a Lipschitz polyhedral boundary. Typically, O ∈ {Ω, Ω1 , Ω2 }. Hereafter we adopt the same notations as in Ref. 3: for all quantities v defined on Ω, vi := v|Ωi (for i = 1, 2) and  max = sup vi (x), vimin = inf vi (x).  If vi > 0 a. e. in Ωi : vi x∈Ωi

x∈Ωi

a Results

x∈Ωi

 If vi < 0 a. e. in Ωi : vi+ = sup |vi (x)|, vi− = inf |vi (x)|. x∈Ωi

can be generalized to the case of an open, bounded and connected set of R3 with a Lipschitz curvilinear polyhedral boundary. For short, we simply write that the boundaries are Lipschitz polyhedral boundaries.


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Transmission problems for electromagnetic waves between dielectrics and metamaterials

3

Let and µ be respectively the dielectric permittivity and the magnetic permeability: we assume that , −1 , µ and µ−1 all belong to L∞ (Ω). The time-harmonic Maxwell equations (ω 6= 0), with perfect conductor boundary condition on ∂Ω, read:  ıωE − curlH = −J      ıωµH + curlE = 0 , div (E) = ρ    div (µH) = 0   E × n|∂Ω = 0

(1.1)

where (E, H) is the electromagnetic field. Quantities ρ and J , respectively the charge and current densities, satisfy the relation iωρ + div J = 0. We assume that ρ ∈ L2 (Ω). Provided E, H and J belong to L2 (Ω) component by component (we shall write E ∈ L2 (Ω) hereafter), one finds that E and H are both in H(curl; Ω). Recall that

H(curl; O) := {p ∈ L2 (O) | curlp ∈ L2 (O)}, H(div ; O) := {p ∈ L2 (O) | div p ∈ L2 (O)}.

The norm on H(op; O) (for op ∈ {curl, div }) is equal to the graph norm. We also introduce H 0 (curl; O) := {p ∈ H(curl; O) | p × n|∂O = 0}. Indeed, thanks to the boundary condition on E, the electric field belongs to H 0 (curl; Ω). One can eliminate one of the two fields (below, H), to find an equivalent, second order system of equations:  1  2  curlE = ıωJ in Ω  ω E − curl µ . (1.2) div (E) = ρ in Ω    E × n|∂Ω = 0 Let us consider the ”electrostatic-like problem”: Find φe ∈ H01 (Ω) such that

div (∇φe ) = ρ .

(1.3)

In the case when is a constant-sign element of L∞ (Ω), solving the problem (1.3) is classical. When exhibits a sign-shift over Ω, (1.3) may be solved using the three-field variational formulation proposed in Refs. 3 (see also Refs. 4, 18 for another solution). More precisely, suppose for instance that 1 > 0. Then, it is proven that the electrostatic probmax lem is well-posed under the assumption that one of the two contrasts R1 := − or 2 /1 min + R2 := 1 /2 is large enough. Note that this type of condition on contrasts (for and/or µ) will systematically appear throughout the paper as a sufficient condition.


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By solving problem (1.3) and setting e = E − ∇φe , j = iωJ − ω 2 ∇φe , it is straightforward to prove that the system of equations (1.2) may be rewritten as  1  2  curle = j in Ω ω e − curl  µ . (1.4) div (e) = 0 in Ω    e × n|∂Ω = 0 We note that, by contruction, j ∈ L2 (Ω) and div j = 0, and the field e belongs to the functional space X defined by X := {p ∈ H 0 (curl; Ω) | div (p) = 0 in Ω} . On the one hand, the ”natural” variational formulation of (1.4) is: Find e ∈ X such that 1 ∀ v ∈ X, curle, curlv − ω 2 (e, v)0,Ω = −(j, v)0,Ω . µ 0,Ω

(1.5)

(1.6)

This ”natural” formulation highlights the difficulties we have to cope with. The first one, if exhibits a sign-shift, since in this case, there exists no result ensuring that the embedding of the functional space X into L2 (Ω) is compact. The second one, if µ exhibits a signshift, since µ−1 curlv, curlv 0,Ω has no specific sign, so its coercivity does not hold. We note that the respective roles of and µ can be reversed, if one chooses instead to write the ”natural” variational formulation in h. On the other hand, it is easy to prove that the system of equations (1.4) with a solution in X is equivalent to: Find (e1 , e2 ) ∈ H(curl; Ω1 ) × H(curl; Ω2 ) such that  1  2  curle1 = j 1 in Ω1 ω 1 e1 − curl    µ1    1  2  curle2 = j 2 in Ω2 ω 2 e2 − curl   µ2    div (i ei ) = 0 in Ωi .  ei × ni|Γi = 0 i = 1, 2     e1 × n1|Σ = e2 × n1|Σ     1 e1 · n1|Σ = 2 e2 · n1|Σ     1 1   curle1 × n1|Σ = curle2 × n1|Σ µ1 µ2

(1.7)

In particular, ei belongs to

Xi := {p ∈ H(curl; Ωi ) | div (p) = 0 in Ωi , p × n|Γi = 0}.

(1.8)

This setting shall be used hereafter, to build some well-posed variational formulations. In the sequel, we denote by (·, ·)0 and k · k0 respectively the canonical scalar product and norm of L2 (Ω) and L2 (Ω), whereas we denote by (·, ·)0,i and k · k0,i resp. the canonical


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5

scalar product and norm of L2 (Ωi ) (and of L2 (Ωi )), for i = 1, 2. In order to build equivalent variational formulations, we shall need a few results on traces of vector fields of H(curl; O). We follow Refs. 6, 7. First (cf. Thms 3.9 and 3.10, and subsection 4.1 in Ref. 6), the following integration by parts formula holds ∀f ∈ H(curl; O), ∀g ∈ H(curl; O),

(f , curlg)0,O − (curlf , g)0,O = f × n|∂O , g T |∂O ∂O ,

with g T |∂O the trace of the tangential components of g. Above, h·, ·i∂O is a well-defined duality product between two different ad hoc Hilbert spaces of functions with support on the boundary ∂O, and endowed with the ”natural” – quotient – norm. Namely, T L(∂O) := (p × n)|∂O | p ∈ H(curl; O) , T R(∂O) := (p)T |∂O | p ∈ H(curl; O) .

In addition, the trace mappings v 7→ v×n|∂O and v 7→ v T |∂O are onto, from H(curl; O) to the same trace spaces, respectively T L(∂O) and T R(∂O) (cf. Thm 5.4 in Ref. 7). Second (cf. Thms 3.15 and 3.16 and subsection 4.2 in Ref. 6), given γ ⊂ ∂O (also with a Lipschitz boundary ∂γ) and γ 0 = ∂O \ γ¯ , consider H 0,γ (curl; O) := {p ∈ H(curl; O) | p × n|γ = 0} ; this space is endowed with the usual norm of H(curl; O). Then, one can prove the following integration by parts formula ∀f ∈ H 0,γ (curl; O), ∀g ∈ H(curl; O),

(f , curlg)0,O − (curlf , g)0,O = f × n|γ 0 , g T |γ 0 γ 0 ,

The duality product h·, ·iγ 0 is again considered between appropriate Hilbert spaces: T L(γ 0 ) := (p × n)|γ 0 | p ∈ H 0,γ (curl; O) , T R(γ 0 ) := (p)T |γ 0 | p ∈ H(curl; O) .

The trace mapping v 7→ f × n|γ 0 is onto, from H 0,γ (curl; O) to T L(γ 0 ) (cf. Thm 6.6 in Ref. 7). Finally, for scalar fields that belong to H 1 (O), recall that 1/2

H00 (γ) := {p ∈ H 1/2 (γ) | p˜ ∈ H 1/2 (∂O)}, where p˜ is the continuation of p by zero to the whole boundary, is the ”natural” space for traces on γ, whenever the trace vanishes on ∂O \ γ. This Hilbert space is endowed with the pkH 1/2 (∂O) . ”natural” norm kpkH 1/2 (γ) := k˜ 00


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2. A compactness result for a sign-shifting : extension of the Weber embedding Theorem When is sign-constant over the whole domain Ω, according to 17,13 the embedding of X into L2 (Ω) is compact (we call this result the Weber embedding Theorem, as a tribute to the landmark paper of Weber 17 ). However, when exhibits a sign-shift, there exists to our knowledge no result ensuring that the embedding of the functional space X into L2 (Ω) is compact. We suppose in this section that 1 > 0 and 2 < 0. In order to extend further the range of our theory, we shall establish that, provided at least one of the two global contrasts max in R1 := − or R2 := min /+ 1 2 /1 2 is large enough, the embedding of the functional space XY := {p ∈ H 0 (curl; Ω) | div (p) ∈ L2 (Ω)} into L2 (Ω) is compact. In other words, we propose an extension of the Weber embedding Theorem. As a particular case, the embedding of X – which is a subset of XY – will be compact. We follow the skeleton of the proof given by Hazard-Lenoir for the same result, when > 0 a.e. (Appendix B of Ref. 13). For that, we study separately the case of curl-free elements (the space Y defined below) and the case of divergence-free elements (the space X). First, let us consider the embedding of Y := {p ∈ XY | curlp = 0 in Ω} into L2 (Ω). Theorem 2.1. The embedding of the functional space Y into L2 (Ω) is compact if at least one of the global contrasts R1 or R2 is large enough. Proof: We carry out the proof in the case of a large contrast R1 . NB. The proof, in the case of a large contrast R2 , proceeds symmetrically, with the roles of Ω1 andΩ2 reversed. Let U k be a bounded sequence of Y . In particular, we deal with curl-free fields: as k∈N

Ω is simply connected (see page 31 in Ref. 11) and as its boundary is connected, one can replace each U k by ∇ϕk , with ϕk ∈ H01 (Ω). Our aim is to prove that a subsequence of (∇ϕk )k converges in L2 (Ω). Note that, since finding this subsequence is an iterative process (one extracts a subsequence, then a subsubsequence, etc.), we keep the same notation for all subsequences of a given sequence. By construction, ϕk solves Find ϕk ∈ H01 (Ω) such that div (∇ϕk ) = div (U k ) in Ω. (2.1) According to Corollary 4.3 of Ref. 3, this problem is well-posed for a large contrast R1 . Let us consider pki solution to


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7

1 Find pki ∈ H0,Γ (Ωi ) such that i

0 ∂pki 1/2 (2.2) |Σ = 0 in H00 (Σ) . ∂n The sequence (pki )k is bounded in H 1 (Ωi ), thus by the Sobolev embedding Theorem we can extract a subsequence – still called (pki )k – that converges in L2 (Ωi ). Moreover, since there holds i ∇(pki − pli ), ∇(pki − pli ) 0,i ≤ kdiv (i (U ki − U li ))k0,i k(pki − pli )k0,i , div (i ∇pki ) = div (i U ki ) in Ωi ,

i

the subsequence (pki )k actually converges in H 1 (Ωi ). Let us introduce the auxiliary (sub)sequences of term uki := ϕki − pki ; these fields belong to H 1 (Ωi ). Moreover, (uk1 , uk2 ) satisfies the system of equations  div (i ∇uki ) = 0 in Ωi    k ui |Γi = 0 , (2.3)  uk | − uk | = hkΣ   1Σ k2Σ 1 ∂n1 u1 |Σ = −|2 |∂n1 uk2 |Σ

where the jump is equal to hkΣ := −(pk1 − pk2 )|Σ . By construction, the sequence (hkΣ )k 1/2 converges in H00 (Σ). k l k Let us set ukl = uk −ul and hkl Σ = hΣ −hΣ . From the definition of u , we have, integrating by parts,

kl kl kl 2 ∇ukl 2 , ∇u2 0,2 = 2 ∂n2 u2 , u2 Σ = −H 1/2 (Σ)0 h2 00

This leads to the inequality

∂ukl 2 kl , hkl i 1/2 − 1 ∇ukl 1 , ∇u1 0,1 . (2.4) ∂n Σ H00 (Σ)

∂ukl 2 2 + max k∇ukl (2.5) khkl k 1/2 k 1/2 1 1 k0,1 . ∂n H00 (Σ)0 Σ H00 (Σ) To bound the last term of (2.5), we use (implicitly) a Dirichlet-to-Neumann operator in the process: we go from Ω1 to the interface Σ, and then from Σ to Ω2 . In other words, it is possible to consider (2.3) as a problem where the unknown is defined on Ω1 , i. e. uk1 or ul1 . From Proposition A.1, we can then verify that kl kl kl int int , ku k + kh k ku k ≤ C k∇ukl k ≤ C 1/2 1/2 1/2 0,1 2 H Σ H 1 H 1 1 1 (Σ) (Σ) (Σ) kl 2 − 2 k∇u2 k0,2 ≤ k2

00

00

00

where the local contrast Cint is equal to the ratio max /min . Next, the trace operator, 1 1 1 1/2 1 from H0,Γ2 (Ω2 ) to H00 (Σ) is linear and continuous. Let C1/2 be its norm (see the Annex kl after Proposition A.1 for a discussion): one has kukl 2 kH 1/2 (Σ) ≤ C1/2 k∇u2 k0,2 . Putting 00 everything back together, we obtain int 2 max 2 − k∇ukl 2 k0,2 2 − (C1 C1/2 ) 1 n i 2 h kl max kl ≤ khkl Cint + 2ku k kh k 1/2 1/2 Σ kH 1/2 (Σ) 1 2 Σ 1 H00 (Σ) H00 (Σ) 00 o ∂ukl +k2 2 kH 1/2 (Σ)0 . (2.6) 00 ∂n


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Recall that the subsequence (uk2 )k is bounded in H 1 (Ω2 ), so (uk2 |Σ )k and (2 ∂n uk2 |Σ )k 1/2 1/2 are bounded sequences of respectively H00 (Σ) and H00 (Σ)0 (cf. Proposition A.1). As a consequence, the right-hand side of (2.6) goes to zero when k, l → ∞. From the definition of R1 , we deduce that, provided R1 > (Cint C1/2 )2 , 1

(2.7)

holds, the subsequence (uk2 )k actually is a Cauchy sequence in H 1 (Ω2 ), so it converges. The same is true for (pk2 )k . Therefore, in the sub-domain Ω2 , we conclude that, since ∇ϕk2 = ∇(uk2 + pk2 ), the subsequence (∇ϕk2 )k converges in L2 (Ω2 ). In order to end the proof, we must show that some subsequence of (∇ϕk1 ) also converges in L2 (Ω1 ). To this aim, let us recall the ”natural” variational formulation of (2.1): Find ϕk ∈ H01 (Ω) such that (2.8) ∇ϕk , ∇v 0 = U k , ∇v , ∀v ∈ H01 (Ω). 0

kl

k

l

kl

k

l

Let us set U = U − U , ϕ = ϕ − ϕ and choose in (2.8) the test field v = ϕkl . We have, after integrating by parts, kl kl kl kl kl (1 ∇ϕkl 1 , ∇ϕ1 )0,1 − (|2 |∇ϕ2 , ∇ϕ2 )0,2 = −(div (U ), ϕ )0 .

As (ϕk )k is bounded in H 1 (Ω), we can extract a subsequence that converges in L2 (Ω). Since a (sub)sequence (ϕk2 )k converges in H 1 (Ω2 ) (provided (2.7) holds), the convergence of (∇ϕk1 )k in L2 (Ω1 ) follows. Going back to U k = ∇ϕk , we conclude that we can extract a subsequence of (U k )k that converges in L2 (Ω). Second, let us study the embedding of X into L2 (Ω). In order to achieve a compactness result similar to Theorem 2.1, we add two geometry-related assumptions. For that, let W T (O) := {w ∈ H(curl; O) | div w ∈ L2 (O), w · n|∂O = 0} . We recall that, according to the Weber embedding Theorem, W T (O) is compactly embedded in L2 (O). The first assumption writes χ = 1 in a neighborhood of Σ ∞ ¯ ∃χ ∈ C (Ω) such that w 7→ χw is continuous, from W T (Ω) to H 1 (Ω).

(2.9)

The second assumption writes χ = 1 in a neighborhood of Σ ∞ ¯ ∃χ ∈ C (Ω) s. t. wi 7→ χwi is continuous, from W T (Ωi ) to H 1 (Ωi ), i = 1, 2. (2.10) These two assumptions are independent of the coefficients and µ. On the one hand, (2.9) is verified if, and only if, the domain Ω is locally convex or if the boundary ∂Ω is smooth


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in a neighborhood of the intersection of the interface Σ with ∂Ω (note that Σ ∩ ∂Ω is equal to ∂Γ1 ∩ ∂Γ2 ). One the other hand, (2.10) is verified if, and only if, both Ωi are locally convex in a neighborhood ∂Γ1 ∩ ∂Γ2 and if Σ is smooth. Finally, we consider that the two functions χ can be merged into a single one. Figure 1 pictures an example of an admissible

W1

S

W2

Fig. 1. An admissible configuration.

geometry: both assumptions (2.9) and (2.10) hold true. Figure 2 pictures two unadmissible geometries: on the left, assumption (2.9) is violated; on the right; assumption (2.10) is violated.

W1 S

W2

W1

W2

S Fig. 2. Two unadmissible configurations.

Proposition 2.1. Assume that (2.9) and (2.10) hold. Then, the embedding of the functional space X into L2 (Ω) is compact if at least one of the global contrasts R1 or R2 is large enough. Remark 2.1. We shall explain how the first assumption (2.9) can be removed later on. In this way, a configuration like the one depicted on the left of Figure 2 is now admissible. We proceed in two steps, since its removal adds another layer of technicalities. The final result is established at Theorem 2.2. Proof: We again carry out the proof in the case of a large contrast R1 . NB.Onceagain, the case of a large contrast R2 is handled similarly. Let W k be a bounded sequence of X. Let us introduce and focus on the problem k∈N

below: Find φk ∈ L2 (Ω) such that

 k k  curlφ = W in Ω k . div φ = 0 in Ω  k φ · n|∂Ω = 0

(2.11)


10

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It is well-known that this “magnetic”-like problem is well-posed in the simply-connected domain Ω (see for instance Ref. 8). In particular, φk belongs to W T (Ω), and (φk )k is bounded in W T (Ω) (and in L2 (Ω)). Our aim is to prove that a (sub)sequence of (curlφk )k converges in L2 (Ω). Note that since W k belongs to X, one has −1 curlφk × n|∂Ω = 0, so φk actually solves Find φk ∈ H(curl; Ω) such that  1 k   curlφ = curlW k in Ω curl     div φk = 0 in Ω . (2.12)  φk · n|∂Ω = 0      1 curlφk × n| = 0 ∂Ω

Following a procedure analogous to the proof of the previous Theorem, we are going to isolate the trace of φk on the interface Σ, φk|Σ , which belongs to H 1/2 (Σ) (cf. assumption (2.9)). In a first step, for i = 1, 2, let us consider pki solution to the regularized problem Find pki ∈ H(curl; Ωi ) such that  1 k   curl curlpi − sg(i )∇(div pki ) = curlW ki in Ωi   i . (2.13) pki · ni |∂Ωi = 0   1   curlpk × ni |∂Ω = 0 i i i

Above, sg(i ) is equal to the sign of i . To begin with, let us prove a few results on the two sequences (pki )k , with the help of the Annex. Set i to 1 or 2. According to Proposition A.2, the regularized problem defines a unique pki , the W T (Ωi )-norm of which is bounded by kcurlW ki k0,i . It follows from Corollary A.1 that there exists a subsequence, still denoted by (pki )k , that converges in W T (Ωi ). Furthermore, from assumption (2.10), we deduce that (pki |Σ )k converges in H 1/2 (Σ). Finally (cf. Proposition A.2) (div pki )k is bounded in H 1 (Ωi ). Now, let g kΣ := (pk2 − pk1 )|Σ : by construction, (g kΣ )k converges in H 1/2 (Σ) and one has (g kΣ )T = g kΣ (since g kΣ · n1 = 0). Next, we define the vector field uki := φki − pki . This field, which belongs to H(curl; Ωi ), satisfies the system of equations  1  k   curl curlui − sg(i )∇(div uki ) = 0 in Ωi   i     div uki = −div pki in Ωi     uki · ni|Γi = 0 , (2.14) uk1 |Σ − uk2 |Σ = g kΣ    1   curluki × ni|Γi = 0    i   1 1   curluk2 × n1|Σ  curluk1 × n1|Σ = − 1 |2 |


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k l kl k l kl k l Let us set ukl i := ui − ui , pi := pi − pi and g Σ := g Σ − g Σ . Our aim is to show that (ukl i )kl converges to zero in H(curl; Ωi ) when k, l → ∞.

Integrating by parts the first line of (2.14) for indices k and l with the test field ukl i , we find 1 1 kl kl kl curlukl , curlu curlu × n , (u ) = i i i i i T i i Σ 0,i kl −sg(i )(∇(div pkl i ), ui )0,i .

(2.15)

−1

Then, we use two identities on Σ, namely 1 −1 curlukl curlukl 1 × n1 = |2 | 2 × n2 , kl kl kl and (u1 )T := g Σ + (u2 )T , to reach: 1 1 kl 2 kl kl kcurlu k ≤ curlu × n , g 1 1 0,1 1 Σ max 1 1 Σ X 1 kl kl 2 k∇(div pi )k0,i kukl + i k0,i + − kcurlu2 k0,2 . (2.16) 2 i=1,2

The next step consists in evaluating the terms in the right-hand-side. As far as the first term on the right-hand side is concerned, note that (1 −1 curluk1 ×n1 |Σ )k is bounded in T L(Σ), since (1 −1 curluk1 )k is itself bounded in H(curl; Ω1 ). Moreover, as (g kΣ )k converges in T R(Σ) this term goes to zero when k, l → ∞. About the second term, we recall that (∇(div pki ))k is a bounded sequence in L2 (Ωi ). Then, one can extract a subsequence of (uki )k which converges in L2 (Ωi ). Indeed, one has uki = φki − pki by construction, and • since (pki )k is bounded in W T (Ωi ), the Weber embedding Theorem tells us that there exists a subsequence that converges in L2 (Ωi ) ; • similarly, since (φk )k is bounded in W T (Ω), one can extract a subsequence that converges in L2 (Ω). Its restriction to Ωi converges in L2 (Ωi ). The third term in the right-hand side cannot be handled as straightforwardly. Let us proceed as follows. Thanks to the system of equations (2.14) governing (uk2 )k , we infer first from Proposition A.3 that there exists Creg > 0 independent of (uk2 )k such that kl kl 2 kl k ku k + kg k ≤ C ku k ≤ C kcurlukl 1/2 1/2 1/2 reg reg 1 H Σ H 2 H 2 0,2 (Σ) (Σ) . (Σ) Next, we infer from assumption (2.10) that there exists a constant c > 0 such that 2 kl 2 kl 2 kl kcurlukl 2 k0,2 ≤ c kcurlu1 k0,1 + kdiv p1 k0,1 + kg Σ kH 1/2 (Σ) .

From the above, we upgrade the estimate (2.16) to 1 1 c kl 2 kl kl curlu1 × n1 , g Σ − − kcurlu1 k0,1 ≤ max 1 2 1 Σ X kl + k∇(div pi )k0,i kukl i k0,i i=1,2

+

c 2 kl kdiv pkl 1 k0,1 + kg Σ kH 1/2 (Σ) . − 2

(2.17)


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Now, the last term in the right-hand side converges to zero when k, l → ∞. Indeed, • since (div pki )k is bounded in H 1 (Ωi ), we know from the Sobolev embedding Theorem that there exists a subsequence that converges in L2 (Ωi ) ; • g kΣ converges in H 1/2 (Σ). Therefore, we see that R1 > c

(2.18)

ensures that the subsequence (curluk1 )k actually is a Cauchy sequence in L2 (Ω1 ), so it converges. But we already proved that a subsequence (curlpk1 )k converges in L2 (Ω1 ). As a consequence, (curlφk1 )k does converge too. Therefore, in the sub-domain Ω1 , we conclude that, since W k1 = 1 −1 curlφk1 , the subsequence (W k1 )k converges in L2 (Ω1 ). To conclude the proof, we go back to (2.12) and we introduce the related ”natural” variational formulation (cf. Ref. 8). Find φk ∈ W T (Ω) such that 1 k curlφ , curlw = curlW k , w , ∀w ∈ W T (Ω). 0 0

Set φkl := φk − φl and W kl := W k − W l and choose w = φkl in the above (for indices k and l). This yields 1 1 kl kl kl kl − = curlW kl , φkl . curlφ1 , curlφ1 curlφ2 , curlφ2 1 |2 | 0 0,1 0,2

We already noted that there exists a subsequence (φk )k that converges in L2 (Ω). Since (curlφk1 )k converges in L2 (Ω1 ), we infer that (curlφk2 )k converges in L2 (Ω2 ). Thus (W k2 )k converges in L2 (Ω2 ), and so does (W k )k in L2 (Ω), which ends the proof. Theorem 2.2. Assume that (2.10) holds. Then, the embedding of the functional space X into L2 (Ω) is compact if at least one of the global contrasts R1 or R2 is large enough. Proof: We follow step by step the proof of Proposition 2.1, bearing in mind that, on the one hand, since assumption (2.10) still holds, all the H 1 (Ωi )-regularity results on (pki )k remain valid. On the other hand, without assumption (2.9), φk|Σ does not automatically belong to H 1/2 (Σ). To address this difficulty, we rely on the continuous decomposition of elements of W T (Ω) into a regular part and a gradient part, first obtained by Birman and Solomyak2. Consider 1 1 2 W reg T (Ω) := W T (Ω) ∩ H (Ω), Ψ := ψ ∈ H (Ω)/R | ∆ψ ∈ L (Ω), ∂n ψ|∂Ω = 0 . The space of potentials Ψ can be endowed with the equivalent norm kψkΨ := k∆ψk0 . According to Ref. 2, one can introduce the continuous splitting W T (Ω) = W reg T (Ω) + ∇Ψ, in the following sense: ∃CBS > 0, ∀w ∈ W T (Ω), ∃(w R , ψ) ∈ W reg T (Ω) × Ψ, . w = wR + ψ, kwR kW T (Ω) + kψkΨ ≤ CBS kwkW T (Ω)


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13

In other words, ∇Ψ contains the singular parts of elements of W T (Ω) (if Ω is not convex). Thus, one can write the continuous splittings, for all k, φk = φkR + ∇ψ k , (φkR , ψ k ) ∈ W reg T (Ω) × Ψ. The regular parts are handled as before, whereas the gradient parts have to be tackled separately. This plays a role only in the estimate of the third term in (2.16), since the other two terms can be estimated as before. To that aim, one has to consider a variant of Proposition A.3. More precisely, one considers a regularized problem with boundary data made up of two parts: a regular part, which belongs to the same trace space as in the original Proposition A.3, plus a singular part, equal to the trace on the interface of an element of ∇Ψ. Since one has k∆ψk0 = k∇ψkW T (Ω) , one reaches the same conclusion as before, id est (2.17), for an ad hoc constant c. Then, provided (2.18) holds, there exists a subsequence (curluk1 )k which converges in L2 (Ω1 ). The end of the proof is unchanged. Remark 2.2. Probably, one should be able to handle the case of piecewise smooth interfaces, thus also removing assumption (2.10). The idea in this general case can be outlined as follows. Consider each edge and/or corner of the interface. One field, for instance pk1 , is locally regular: χ? pk1 ∈ H 1 (Ω1 ), for an ad hoc truncation function χ? . Whereas the other field, pk2 can be locally singular: χ? pk2 6∈ H 1 (Ω2 ) is possible. Then, this behavior is inherited by uk1 and uk2 . Therefore, in order to bound the third term as in (2.17), one should probably isolate the singular and regular parts of ukl 2 , and proceed by controlling its kl kl singular part by g Σ , resp. its regular part by u1 . Theorem 2.3. Assume that (2.10) holds. The embedding of the functional space XY into L2 (Ω) is compact if at least one of the global contrasts R1 or R2 is large enough. Proof: It is based on the standard Helmholtz decomposition of vector fields, here on elements of XY . Given xy ∈ XY , solve Find φ ∈ H01 (Ω) such that div (∇φ) = div (xy) in Ω. This problem is well-posed 3,4 : its solution φ exists (and it is unique) and moreover kφkH 1 (Ω) ≤ C() kdiv (xy)k0 (with C() > 0 independent of xy). Let y := ∇φ and x := xy − y. By construction, y ∈ Y , with div (y) = div (xy) in Ω . x ∈ X, with curlx = curlxy in Ω We note that kyk0 = k∇φk0 ≤ C() kdiv (xy)k0 , so it follows that kxk0 ≤ kxyk0 + C() kdiv (xy)k0 . Combining the above, we deduce that, from any bounded sequence (xy k )k in XY , one can build two bounded sequences (y k )k (in Y ) and (xk )k (in X). From each sequence, we extract a subsequence that converges in L2 (Ω), according to Theorems 2.1 and 2.2. Aggregating the two, we obtain a subsequence (xy k )k that converges in L2 (Ω).


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3. Recovering coerciveness in the case of a sign-shifting µ In this section we assume that the sign of µ shifts from Ω1 to Ω2 , with µ1 > 0 and µ2 < 0. We make no further assumption on the sign of . Recall that j belongs to L2 (Ω) and that div j = 0. The aim is to build a three-field variational formulation, equivalent to (1.7), and then to show that this formulation is well-posed under suitable conditions. 3.1. A three-field formulation To begin with, keeping both e1 and e2 leads to a reformulated – equivalent – definition of X. According to the definition (1.8) of the spaces X1 and X2 , to which e1 and e2 belong respectively, and adding the compatibility conditions on the interface, we are led to introduce the functional space X := {(v, w) ∈ X1 × X2 | v × n1|Σ = w × n1 |Σ , 1 v · n1|Σ = 2 w · n1|Σ } and the auxiliary unknown e2 :=

1 curle2 . |µ2 |

Now, let us consider the test functions (v 1 , v 2 ) ∈ X and v 2 ∈ H(curl; Ω2 ) and • take the L2 −scalar product of the first Eq. of (1.7) with v 1 : 1 2 curle1 , v 1 ω (1 e1 , v 1 )0,1 − curl = (j 1 , v 1 )0,1 µ1 0,1 and let us integrate by parts: 1 curle1 , curlv 1 µ1 0,1 1 curle1 = (j 1 , v 1 )0,1 . − v 1 × n1 , µ1 T Σ 1 Since, by definition, curle1 = −(e2 )T and v 1 × n1 = v 2 × n1 on Σ, µ1 T we obtain 1 curle1 , curlv 1 − ω 2 (1 e1 , v 1 )0,1 − hv 2 × n1 , (e2 )T iΣ µ1 0,1 ω 2 (1 e1 , v 1 )0,1 −

= −(j 1 , v 1 )0,1 .

(3.1)

• take the L2 −scalar product of the second Eq. of (1.7) with curlv 2 ; multiply the resulting equality by a constant factor ϑ > 0: ϑω 2 (2 e2 , curlv 2 )0,2 + ϑ(curle2 , curlv 2 )0,2 = ϑ(j 2 , curlv 2 ) .

(3.2)


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• consider the identity 1 1 curle2 , curlv 2 curle2 , v 2 = curl |µ2 | |µ2 | 0,2 0,2 1 ; − v 2 × n2 , curle2 |µ2 | T Σ according to the definition of e2 this last equation leads to (n2 = −n1 on Σ) 1 curle2 , curlv 2 − (curle2 , v 2 )0,2 − h(v 2 × n1 , e2 )T iΣ = 0 . (3.3) |µ2 | 0,2 • consider the identity (e2 , curlv 2 )0,2 = (curle2 , v 2 )0,2 + he2 × n2 , (v 2 )T iΣ ; since e2 × n2 |Σ = −e1 × n1 |Σ , we get: (|µ2 |e2 , v 2 )0,2 − (e2 , curlv 2 ) − he1 × n1 , (v 2 )T iΣ = 0 .

(3.4)

Adding up the previous contributions (3.1)-(3.4), we introduce the variational formulation (3.5): Find U = ((e1 , e2 ), e2 ) ∈ X × H(curl; Ω2 ) such that ∀V = ((v 1 , v 2 ), v 2 ) ∈ X × H(curl; Ω2 ) , Aϑ (U, V ) = Lϑ (V ) .

(3.5)

We call (3.5) the three-field formulation. The forms Aϑ and Lϑ are respectively defined by 1 Aϑ (U, V ) := curle1 , curlv 1 − ω 2 (1 e1 , v 1 )0,1 + ϑω 2 (2 e2 , curlv 2 )0,2 µ1 0,1 1 +ϑ(curle2 , curlv 2 )0,2 + curle2 , curlv 2 |µ2 | 0,2 −(curle2 , v 2 )0,2 + (|µ2 |e2 , v 2 )0,2 − (e2 , curlv 2 )0,2 −2hv2 × n1 , (e2 )T iΣ − he1 × n1 , (v 2 )T iΣ (3.6) and Lϑ (V ) := −(j 1 , v 1 )0,1 + ϑ(j 2 , curlv 2 )0,2 .

(3.7)

It is important to note that, in the definition of the bilinear form Aϑ , the two boundary terms hv 2 × n1 , (e2 )T iΣ and he1 × n1 , (v 2 )T iΣ are independent of the coefficients and µ. In addition, we remark that this is true for any choice of the strictly positive factor ϑ (which we will fit to some optimal value when we establish the coercivity of Aϑ ). 3.2. Equivalence with the initial problem Proposition 3.1. The three-field formulation (3.5) is equivalent to problem (1.7). Proof: It is already known that (3.5) follows from (1.7), so let us focus on the reciprocal assertion.


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To begin with, we note that, since ei ∈ Xi (i = 1, 2), one has div ei = 0 in Ωi . Next, one finds that e1 × n1 |Σ = e2 × n1 |Σ , 1 e1 · n1 |Σ = 2 e2 · n1 |Σ and ei × ni |Γi = 0 (i = 1, 2), according to the definition of X . Let us choose in (3.5) test functions v 1 which span (D(Ω1 ))3 , (v 2 , v 2 ) = (0, 0) and differentiate in the sense of distributions in (D0 (Ω1 ))3 : 1 2 curle1 − j 1 , v 1 = 0. ω 1 e1 − curl µ1 Thus the first Eq. of (1.7) is recovered. From there, we shall establish simultaneously that |µ2 |e2 = curle2 and that the second Eq. of (1.7) is recovered. We introduce two elements of L2 (Ω): τ := curle2 −|µ2 |e2 , and η := ω 2 2 e2 +curle2 − j 2 , and prove that both fields vanish over Ω2 . To start with, we know that div η = 0 in the whole of Ω, since (e1 , e2 ) ∈ X and div j = 0. Choose first in (3.5) (v 1 , v 2 ) = (0, 0) and v 2 ∈ (D(Ω2 ))3 : ϑω 2 (2 e2 , curlv 2 )0,2 + ϑ(curle2 , curlv 2 )0,2 − ϑ(j 2 , curlv 2 )0,2 +(|µ2 |e2 , v 2 )0,2 − (e2 , curlv 2 )0,2 = 0 . After differentiating in the sense of distributions, this last equation leads to ϑcurl(ω 2 2 e2 + curle2 − j 2 ) + |µ2 |e2 − curle2 = 0 in (D0 (Ω2 ))3 , which implies ϑcurlη = τ in L2 (Ω2 ).

(3.8)

Then, let us take in (3.5) (v 1 , v 2 ) = (0, 0) and v 2 ∈ (D(Ω2 ))3 : 1 − (curle2 , v 2 )0,2 = 0 . curle2 , curlv 2 |µ2 | 0,2 Let us differentiate again in the sense of distributions to obtain 1 curl curle2 − e2 = 0 in (D0 (Ω2 ))3 |µ2 |

(3.9)

that we may rewrite as curl

1 τ |µ2 |

= 0.

(3.10)

Now, let us prove that the tangential trace of η vanishes over ∂Ω2 . For that, choose in (3.5) (v 1 , v 2 ) = (0, 0) and v 2 ∈ H(curl, Ω2 ): ϑω 2 (2 e2 , curlv 2 )0,2 + ϑ(curle2 , curlv 2 )0,2 + (|µ2 |e2 , v 2 )0,2 −(e2 , curlv 2 )0,2 − he1 × n1 , (v 2 )T iΣ − ϑ(j 2 , curlv 2 )0,2 = 0 . In terms of η, this reads: ϑ(η, curlv 2 )0,2 + (|µ2 |e2 , v 2 )0,2 − (e2 , curlv 2 )0,2 − he1 × n1 , (v 2 )T iΣ = 0 .


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Since η and e2 belong to H(curl; Ω2 ), we can integrate by parts to find

ϑ(curlη, v 2 )0,2 − ϑ η × n2 , (v 2 )T ∂Ω + (|µ2 |e2 , v 2 )0,2 2 −(curle2 , v 2 )0,2 − he2 × n2 , (v 2 )T i∂Ω2 − he1 × n1 , (v 2 )T iΣ = 0 . Then, we can remove a number of terms. • We already proved that 0 = ϑcurlη − τ = ϑcurlη + |µ2 |e2 − curle2 in Ω2 . • In addition, there holds he2 × n2 , (v 2 )T i∂Ω2 = he2 × n2 , (v 2 )T iΣ , since e2 ∈ H 0,Γ2 (curl, Ω2 ). • Finally, e2 × n2|Σ = −e1 × n1|Σ , since (e1 , e2 ) ∈ X . The conclusion is: hη × n2 , (v 2 )T i∂Ω2 = 0, for all v 2 ∈ H(curl, Ω2 ). Since v 2 7→ (v 2 )T |∂Ω2 is onto, this finally leads to η × n2 |∂Ω2 = 0 .

(3.11)

Recalling first (3.10) and then (3.8) and finally (3.11), we reach 1 1 1 1 curl = curl = . 0= τ,η curlη, η curlη, curlη ϑ |µ2 | |µ2 | |µ2 | 0,2 0,2 0,2 Thus curlη = 0 and using (3.8) once again yields τ = 0. Moreover, since η belongs to L2 (Ω2 ) and satisfies div η = 0 and curlη = 0 with a vanishing tangential trace over Ω2 , η vanishes over Ω2 (see for instance Ref. 8). The second Eq. of (1.7) is thus recovered. In order to conclude the proof, we must also recover the last Eq. of (1.7). To this aim, let us take in (3.5) v 2 = 0 1 curle1 , curlv 1 − ω 2 (1 e1 , v 1 )0,1 − 2hv1 × n1 , (e2 )T iΣ + µ 1 0,1 1 curle2 , curlv 2 − (curle2 , v 2 )0,2 = −(j 1 , v 1 )0,1 , |µ2 | 0,2 and let us integrate by parts to get 1 1 curle1 − ω 2 1 e1 + j 1 , v 1 curle2 − e2 , v 2 curl + curl − µ1 |µ2 | 0,1 0,2 1 1 curle2 + curle1 v 1 × n1 , = 0. |µ2 | µ1 T Σ From this last equation, using the first Eq. of (1.7) together with (3.9), it is straightforward to recover the last Eq. of (1.7). 3.3. Finding a well-posed variational setting for the three-field formulation Below, we build a splitting of the bilinear form Aϑ in a two term sum, so that the first term is coercive over X × H(curl; Ω2 ), and the second is a compact perturbation of the first


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one. Let us write Aϑ = Aϑcoer + Aϑcomp , with

1 1 1 + max (e1 , v 1 )0,1 + curle1 , curlv 1 curle2 , curlv 2 µ1 µ1 |µ2 | 0,1 0,2 +(e2 , v 2 )0,2 + ϑ(curle2 , curlv 2 )0,2 + (|µ2 |e2 , v 2 )0,2 −2hv1 × n1 , (e2 )T iΣ − he1 × n1 , (v 2 )T iΣ (3.12) 1 Aϑcomp (U, V ) := − ω 2 1 + µmax e1 , v 1 − (e2 + curle2 , v 2 )0,2 1 0,1 (3.13) 2 −((1 − ϑω 2 )e2 , curlv 2 )0,2 ,

Aϑcoer (U, V ) :=

where U = (e1 , e2 , e2 ) and V = (v 1 , v 2 , v 2 ) both belong to X × H(curl; Ω2 ). Let us prove that Aϑcoer is coercive under some suitable conditions. ∞ Since µ2 and µ−1 2 both belong to L (Ω2 ), the two norms 1/2

:= [(|µ2 |·, ·)0,2 + (curl ·, curl ·)0,2 ] k · kH(curl;Ω f 2)

k · kH(curl;Ω c 2)

1 := (·, ·)0,2 + ( curl ·, curl ·)0,2 |µ2 |

1/2

are equivalent to the X2 usual norm. Then the term |hv 2 × n1 , (v 2 )T iΣ | can be bounded from above by kv2 kH(curl;Ω . |hv 2 × n1 , (v 2 )T iΣ | ≤ kv2 kH(curl;Ω c f 2) 2)

(3.14)

On the other hand, for interface terms involving fields defined on Ω1 and on Ω2 , we introduce some constant. Let c ∈ R+ ? be defined as ( ) hv 1 × n1 , (v 2 )T iΣ c := sup , v 1 ∈ X1 \ {0}, v 2 ∈ H(curl; Ω2 ) \ {0} . kv 1 kH(curl;Ω1 ) kv 2 kH(curl;Ω2 ) (3.15) Note that we have Ω2 ) , ∀(v 1 , v 2 ) ∈ X1 × H(curl; hv 1 × n1 , (v 2 )T iΣ ≤ ckv1 kH(curl;Ω ) kv2 kH(curl;Ω ) 1 2

(3.16)

with an optimal c. max Let us introduce now the first global contrast in µ R1µ equal to the ratio µ− . 2 /µ1 Proposition 3.2. Assume that R1µ > (5/4)c2

(3.17)

ϑ holds with c defined by (3.16). Then, for any ϑ ≥ max(1, µ− 2 ), Acoer is coercive over {X × H(curl; Ω2 )}.


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Proof: Let us first compute the value of Aϑcoer (V, V ), with V = ((v 1 , v 2 ), v 2 ): 1 1 + max (v 1 , v 1 )0,1 + Aϑcoer (V, V ) = curlv 1 , curlv 1 µ1 0,1 µ1 1 + (v 2 , v 2 )0,2 + curlv 2 , curlv 2 |µ2 | 0,2 ϑ(curlv 2 , curlv 2 )0,2 + (|µ2 |v 2 , v 2 )0,2 − 3hv 2 × n1 , (v 2 )T iΣ . Thus, introducing the real parameter η ∈ [0, 3], Aϑcoer (V, V ) may be bounded from below by 1 1 + max (v 1 , v 1 )0,1 + Aϑcoer (V, V ) ≥ curlv 1 , curlv 1 µ1 µ 1 0,1 1 + (v 2 , v 2 )0,2 + curlv 2 , curlv 2 |µ2 | 0,2 ϑ(curlv 2 , curlv 2 )0,2 + (|µ 2 |v 2 , v 2 )0,2 − (3 − η) hv 1 × n1 , (v 2 )T iΣ − η hv 2 × n1 , (v 2 )T iΣ . Then, the term hv 2 × n1 , (v 2 )T iΣ is bounded by (3.14), whereas hv 1 × n1 , (v 2 )T iΣ is bounded as in (3.16). Let us introduce β1 and β2 , two real and strictly positive parameters such that β1 + β2 = 1; we deduce 1 kv 1 k2H(curl;Ω1 ) + kv2 k2H(curl;Ω c 2) µmax 1 +(β1 + β2 ) (|µ2 |v 2 , v 2 )0,2 + ϑkcurlv 2 k20,2 −(3 − η)ckv 1 kH(curl;Ω1 ) kv 2 kH(curl;Ω2 ) −ηkv2 kH(curl;Ω kv 2 kH(curl;Ω . c f 2) 2)

Aϑ coer (V, V ) ≥

Since ϑ ≥ max(1, µ− 2 ), one actually has 1 2 kv 1 k2H(curl;Ω1 ) + β1 µ− 2 kv 2 kH(curl;Ω2 ) µmax 1 +kv 2 k2H(curl;Ω + β2 kv2 k2H(curl;Ω c f ) )

Aϑ coer (V, V ) ≥

2

2

(3.18)

−(3 − η)ckv 1 kH(curl;Ω1 ) kv2 kH(curl;Ω2 ) −ηkv 2 kH(curl;Ω kv 2 kH(curl;Ω . c f 2) 2) Now, the idea is to control the negative terms with (a fraction of) the positive ones. Let us recall the standard result 2 given m, p ∈ R+ ? such that m > p , + 2 2 ∃λ ∈ R? , ∀x, y ∈ R, mx + y − 2pxy ≥ λ(x2 + y 2 ).

Then, let us set x1 = kv 1 kH(curl;Ω1 ) , x2 = kv 2 kH(curl;Ω , f 2)

y1 = kv 2 kH(curl;Ω2 ) y2 = kv 2 kH(curl;Ω c 2)

(3.19)


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and m1 :=

1 µmax µ− 1 2 β1

m 2 = β2 ,

,

c(3 − η) 2β1 µ− 2 p2 = η/2 .

p1 :=

With this new notations, the inequality (3.18) can be rewritten as 2 2 2 2 Aϑcoer (V, V ) ≥ β1 µ− 2 m1 x1 + y1 − 2p1 x1 y1 + m2 x2 + y2 − 2p2 x2 y2 .

(3.20)

According to (3.19), the form Aϑcoer (V, V ) is coercive if both condition m1 > p21 and m2 > p22 are satisfied. In other words, provided both c2 (3 − η)2 µ− 2 > and β2 > (η/2)2 hold. max µ1 4β1

(3.21)

Since β2 < 1, we consider from now on η ∈ [0, 2[. Moreover, since β1 + β2 = 1, the second condition in (3.21) is equivalent to β1−1 > (1 − η 2 /4)−1 . Then the first condition in (3.21) is satisfied for some suitable β1 (η) (depending here on η) if (3 − η)2 µ− 2 . > c2 max µ1 4 − η2

(3.22)

Now, f : η 7→ (3 − η 2 )/(4 − η 2 ) takes his minimal value at η = 4/3, and f (4/3) = 5/4. For this optimal value, condition (3.22) reduces to (3.17) for some suitable β1 (4/3). Theorem 3.1. The variational formulation (3.5) fits into the coercive plus compact framework, for ϑ ≥ max(1, µ− 2 ), if the following conditions are met: (1) for a constant-sign , if the global contrast R1µ is large enough. (2) for a sign-shifting , if the global contrast R1µ is large enough, if assumption (2.10) holds, and if one of the global contrasts R1 or R2 is large enough. Proof: When is sign-constant over the whole domain Ω, we already noted that the embedding of X into L2 (Ω) is compact. Due to this result, Aϑcomp is a compact perturbation of Aϑcoer , which is coercive provided that the condition (3.17) on R1µ holds. In the case of a sign-shifting , condition (3.17) has to be supplemented with a condition like (2.18) on R1 , to ensure that the embedding of X into L2 (Ω) is compact. Remark 3.1. The constant c depends only on the geometry, so the lower bound in (3.17) is fixed by the geometrical configuration. To derive a similar result in the case of a big value of the second global contrast in µ R2µ := µmin /µ+ 1 2 , one simply builds an alternate three-field formulation by choosing e1 := −1 µ1 curle1 as the auxiliary unknown.


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4. Concluding remarks According to the previous results, one can solve problem  1  2  curle = j in Ω ω e − curl  µ . div (e) = 0 in Ω    e × n|∂Ω = 0

under very weak assumptions on the coefficients, which cover many challenging configurations of practical interest. Recall that when and µ are constant-sign coefficients, the problem is well-posed under the assumption that , −1 , µ, µ−1 all belong to L∞ (Ω). These assumptions are always implicit below. • If only exhibits a sign-shift at an interface: there are two possible ways to achieve well-posedness, under the condition that at least one of the two global contrasts R1 or R2 is large enough: – According to the compactness result of theorem 2.2, the variational formulation (1.6) fits into the coercive plus compact framework, provided the interface is ”smooth” in the sense of assumption (2.10). – Or, one can choose a three-field formulation for the magnetic field (like (3.5)), along the same lines as those of section 3, under the weaker assumption that the interface is Lipschitz. • If only µ exhibits a sign-shift at an interface: one can proceed as in the previous case by reversing the roles of the electric and magnetic fields. • If both and µ exhibit a sign-shift at interfaces that can be different: one can use a three-field formulation ((3.5), or in the magnetic field) together with the compactness result. In this case, it is required that one of the two global contrasts R1 or R2 is large enough, and also that one of the two global contrasts R1µ or R2µ is large enough. In this configuration, one interface must be ”smooth” in the sense of assumption (2.10), while the other one is Lipschitz. µ sign-constants µ sign-shifts

sign-constants NF NF or 3F

sign-shifts NF or 3F 3F

The above table summarizes, for all possible transitions between media, which formulation(s) can be chosen for solving the problem. There, the acronyms ”NF” and ”3F” denote respectively the ”natural” variational formulation and a three-field variational formulation. Acknowledgments The authors wish to thank the anonymous referee for her/his very careful reading of the first version of the manuscript. Her/his remarks led to significant improvements in the final version.


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References 1. C. Amrouche, C. Bernardi, M. Dauge, V. Girault, Vector potentials in three-dimensional nonsmooth domains, Math. Meth. Appl. Sci., 21, 823-864, 1998. 2. M. Sh. Birman, M. Z. Solomyak, Maxwell operator in regions with nonsmooth boundaries, Sib. Math. J., 28, 12-24, 1987. 3. A.-S. Bonnet-Bendhia, P. Ciarlet, Jr., C. M. Zwölf, Two- and three-field formulations for wave transmission between media with opposite sign dielectric constants, J. Comput. Appl. Math., 204, 408-417, 2007. 4. A.-S. Bonnet-Bendhia, P. Ciarlet, Jr., C. M. Zwölf, Time harmonic wave diffraction problems in materials with sign-shifting coefficients, submitted to J. Comput. Appl. Math.. 5. A.-S. Bonnet-Bendhia, M. Dauge, K. Ramdani, Analyse spectrale et singularités d’un problème de transmission non coercif, C.R. Acad. Sci Ser. I, 328, 717-720, 1999. 6. A. Buffa, P. Ciarlet, Jr., On traces for functional spaces related to Maxwell’s equations. Part I: an integration by parts formula in Lipschitz polyhedra, Math. Meth. Appl. Sci., 24, 9-30, 2001. 7. A. Buffa, P. Ciarlet, Jr., On traces for functional spaces related to Maxwell’s equations. Part II: Hodge decompositions on the boundary of Lipschitz polyhedra and applications, Math. Meth. Appl. Sci., 24, 31-48, 2001. 8. P. Ciarlet, Jr., Augmented formulations for solving Maxwell equations, Comput. Methods Appl. Mech. Engrg., 194, 559-586, 2005. 9. M. Costabel, E. Stephan, A direct boundary integral method for transmission problems, J. of Math. Anal. and Appl., 106, 367-413, 1985. 10. P. Fernandes, G. Gilardi, Magnetostatic and electrostatic problems in inhomogeneous anisotropic media with irregular boundary and mixed boundary conditions, Math. Models Methods Appl. Sci., 7, 957-991, 1997. 11. V. Girault, P.-A. Raviart, Finite element methods for Navier-Stokes equations, Springer-Verlag, Berlin, 1986. 12. P. Grisvard, Elliptic problems in nonsmooth domains, Pitman, Boston, 1985. 13. C. Hazard, M. Lenoir, On the solution of time-harmonic scattering problems for Maxwell’s equations, SIAM J. Math. Anal., 27, 1597-1630, 1996. 14. J. G. Ma, I. Wolff, Modeling the microwave properties of superconductors, IEEE Trans. Microwave Theory Tech., 43, 1053-1059, 1995. 15. K. Ramdani, Superconducting lines: mathematical and numerical analysis (in French). PhD Thesis, Université Paris 6, 1999. 16. D. R. Smith, W. J. Padilla, D. C. Vier, S. C. Nemat-Nasser, S. Schutz, Composite medium with simultaneously negative permeability and permittivity, Phys. Rev. Lett., 84, 4184-4187, 2000. 17. C. Weber, A local compactness theorem for Maxwell’s equations, Math. Meth. Appl. Sci., 2, 12-25, 1980. 18. C. M. Zwölf, Variational methods to model electromagnetic wave transmission between dielectrics and metamaterials (in French). PhD Thesis, Université de Versailles-Saint-Quentinen-Yvelines, 2007.

A. Annex We recall or prove here a series of elementary results, set in an open, bounded and connected set O with a Lipschitz polyhedral boundary ∂O. Let γ ⊂ ∂O be connected (also with a Lipschitz boundary ∂γ), and γ 0 := ∂O \ γ. Let α ∈ L∞ (O) be positive, with α−1 ∈ L∞ (O).


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A.1. Results on scalar fields The first result deals with the lifting of some scalar data that is prescribed on a part of a boundary: for the sake of completeness, we report the proof. We define the local contrast Cαint equal to the ratio αmax /αmin . Recall that 1/2

H00 (γ) := {p ∈ H 1/2 (γ) | p˜ ∈ H 1/2 (∂O)}, where p˜ is the continuation of p by zero to the whole boundary. This Hilbert space is pkH 1/2 (∂O) . endowed with the ”natural” norm kpkH 1/2 (γ) := k˜ 00

1/2

Proposition A.1. Let h ∈ H00 (γ) and define u as the solution to Find u ∈ H 1 (O) such that   div (α∇u) = 0 in O . u| = h  γ u|γ 0 = 0

(A)

Then the inequality k∇uk0,O ≤ Cαint khkH 1/2 (γ) holds. 00

Proof: Integrating by parts, one finds: (α∇u, ∇u)0 = hα∇u · n, ui = (H 1/2 (γ))0 hα∇u · n, hiH 1/2 (γ) . 00

00

The normal trace mapping γn : H(div ; O) → (H 1/2 (∂O))0 , v 7→ v · n|∂O , is such that kγn k = 1, according to page 28 of Ref. 11. Moreover, given v ∈ H(div ; O): 1/2

kv · n|γ k(H 1/2 (γ))0 = 00

sup

(H00 (γ))0

hv · n|γ , piH 1/2 (γ) 00

kpkH 1/2 (γ)

1/2 p∈H00 (γ)

00

hv · n|∂O , p˜i = sup ≤ kv · n|∂O k(H 1/2 (∂O))0 . k˜ pkH 1/2 (∂O) 1/2 (γ) p∈H 00

Combining the above yields: αmin k∇uk20 ≤ kα∇ukH(div ;O) khkH 1/2 (γ) ≤ αmax k∇uk0 khkH 1/2 (γ) , 00

00

i. e. the expected result.

Remark A.1. This result typically allows us to study the Dirichlet-to-Neumann operator 1/2 1/2 S : h 7→ α∂n u|γ , defined from H00 (γ) to (H00 (γ))0 , since one finds 1/2

kShk(H 1/2 (γ))0 ≤ αmax Cαint khkH 1/2 (γ) , ∀h ∈ H00 (γ). 00

00

Now, assume that the domain O can be partitioned into O = O1 ∪ O2 , O1 ∩ O2 = ∅, with Lipschitz polyhedral boundaries ∂Oi . We let Σ = ∂O1 ∩ ∂O2 be the interface with a Lipschitz boundary ∂Σ, and Bi = ∂Oi \ Σ. On the interface, let us choose to measure 1/2 elements of H00 (Σ) thanks to the norm 1 kpk := k˜ pkH 1/2 (∂O1 ) , with a continuation by


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zero to ∂O1 . It is well-known (cf. for instance page 19 of Ref. 12) that an equivalent (and 1/2 more intrinsic) norm on H00 (Σ) is 1/2 Z p(σ)2 2 kpk :=: kpk + dσ , 1/2 0 H (Σ) Σ d(σ, ∂Σ) where σ 7→ d(σ, ∂Σ) is the distance to the boundary of Σ. The equivalence constants between the two norms are completely determined by the geometry of ∂O1 near Σ. On the 1/2 1 (O2 ) to H00 (Σ). other hand, the trace mapping v2 7→ v2 |Σ is also continuous from H0,B 2 1/2

According to the above, if we measure elements of H00 (Σ) with 1 k · k, the norm of the trace mapping depends on both geometrical configurations of ∂O1 and of ∂O2 near the interface. So, the continuity modulus should be written as C1/2 : 1 kv2 |Σ k

1 ≤ C1/2 k∇v2 k0,O2 , ∀v2 ∈ H0,B (O2 ). 2

A.2. Elementary result on vector fields The proposed results deal with the well-posedness of regularized problems. We recall a result on norms in W T (O), as proven in Proposition 7.4 of Ref. 10 and Corol1/2 lary 3.16 of Ref. 1. The semi-norm k · kW T (O) : w 7→ kcurluk20,O + kdiv uk20,O defines a norm, which is equivalent to the full norm of W T (O), provided that the domain O is simply-connected. Proposition A.2. Assume that O is simply-connected. Let f ∈ L2 (O). Then, the regularized problem below admits one, and only one, solution: Find u ∈ W T (O) such that   curl (αcurlu) − ∇(div u) = f in O u · n|∂O = 0 . (B)  αcurlu × n|∂O = 0

Moreover, one has div u ∈ H 1 (O). Finally, the norms kukW T (O) and kdiv uk1,O depend continuously on kf k0,O . Proof: Classically, an equivalent variational formulation of (B) is Find u ∈ W T (O) such that

(αcurlu, curlw)0,O + (div u, div w)0,O = (f , w)0,O , ∀w ∈ W T (O).

(C)

Well-posedness follows. In addition, taking w = u in (C) yields αmin kcurluk20,O + kdiv uk20,O ≤ kf k0,O kuk0,O , so kukW T (O) depend continuously on kfk0,O . To prove that one has div u ∈ H 1 (O), let us introduce the scalar Neumann problem Find d ∈ H 1 (O) ∩ L20 (O) such that (∇d, ∇w)0,O = −(f , ∇w)0,O , ∀w ∈ H 1 (O) ∩ L20 (O).

(D)


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NB. We recall that L20 (O) := {v ∈ L2 (O) | (v, 1)0,O = 0}. According to the Poincaré-Wirtinger inequality, this problem is well-posed, and kdk1,O depends continuously on kf k0,O . Now, let us compare d to div u. To that aim, we introduce a second scalar Neumann problem. Let δ ∈ R be such that (δ, 1)0,O = (div u, 1)0,O . According to the above, δ depends continuously on kfk0,O . The second scalar problem reads Find v ∈ H 1 (O) ∩ L20 (O) such that (∇v, ∇w)0,O = (d + δ − div u, w)0,O , ∀w ∈ H 1 (O) ∩ L20 (O).

(E)

Since by construction d + δ − div u is orthogonal to constants, the right-hand side defines a linear form on H 1 (O) ∩ L20 (O). This problem is also well-posed. One finds easily that v ∈ H 1 (O) ∩ L20 (O) can be characterized by the relations ∆v = div u − (d + δ) in O, and ∂n v |∂O = 0, so that ∇v belongs to W T (O). Taking successively w = ∇v in (C), w = v in (D) and integrating by parts, we find (div u, div u − (d + δ))0,O = (f , ∇v)0,O = −(∇d, ∇v)0,O = (d, div u − (d + δ))0,O . In other words, (div u − d, div u − (d + δ))0,O = 0. But div u − (d + δ) is orthogonal to constants: (δ, div u − (d + δ))0,O = 0. It follows kdiv u − (d + δ)k0,O = 0, id est div u = d + δ in O. We conclude that kdiv uk1,O depends continuously on kf k0,O . Corollary A.1. Let (f k )k be a bounded sequence in L2 (O), and let (uk )k be the corresponding sequence of solutions to the regularized problems (B) with f = f k . Then, there exists a subsequence of (uk )k that converges in W T (O). Proof: According to Proposition A.2, the sequence (uk )k is bounded in W T (O). Thanks to the Weber embedding Theorem, there exists a subsequence, still denoted by (uk )k , that converges in L2 (O). Taking the difference of Eq. (C) for two indices k and l with the same test field w = uk − ul , one finds αmin kcurl(uk − ul )k20,O + kdiv (uk − ul )k20,O ≤ kf k − f l k0,O kuk − ul k0,O . In other words, the subsequence (uk )k is a Cauchy subsequence in W T (O), so it converges. Let us carry on with a final result on regularized problems with data on the boundary. Let us introduce first 1/2

H 00n (γ 0 ) := {ϕ ∈ H 1/2 (γ 0 ) | ∃φ ∈ H 1 (O), φ · n|γ = 0, ϕ = φ|γ 0 }. Using standard results, we know that ( 1/2 ∃Clif t > 0, ∀ϕ ∈ H 00n (γ 0 ), ∃φ ∈ H 1 (O), . φ · n|γ = 0, ϕ = φ|γ 0 and kφkH 1 (O) ≤ Clif t kϕkH 1/2 (γ 0 )

(F)

Proposition A.3. Assume that O is simply-connected. 1/2 Let ϕ ∈ H 00n (γ 0 ). Then, the regularized problem below admits one, and only one, solution:


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Find u ∈ H(curl; O) ∩ H(div ; O) such that  curl (αcurlu) − ∇(div u) = 0 in O    u · n|γ = 0 .  αcurlu × n|γ = 0   u|γ 0 = ϕ

(G)

Moreover, the norms kcurluk0,O and kdiv uk0,O depend continuously on kϕkH 1/2 (γ 0 ) . Proof: Let us consider φ satisfying (F). We remark that u − φ belongs to W T − (O) := {w ∈ W T (O) | w |γ 0 = 0}, which is a closed subspace of W T (O). Introduce u0 := u − φ. Then one can reformulate the system of equations in u0 as the equivalent variational formulation Find u0 ∈ W T − (O) such that (αcurlu0 , curlw)0,O + (div u0 , div w)0,O = −(αcurlφ, curlw)0,O − (div φ, div w)0,O , ∀w ∈ W T − (O). Since W T − (O) can be endowed with the norm of W T (O), well-posedness of the variational formulation in u0 follows. Existence of u is achieved. To obtain uniqueness and continuity with respect to the data, we proceed as follows. We know that ku0 kW T (O) depends continuously on kφkW T (O) , which is itself bounded by kφkH 1 (O) , and so by kϕkH 1/2 (γ 0 ) according to (F). Then, ku0 + φkW T (O) depends continuously on kϕkH 1/2 (γ 0 ) . We conclude that u is unique and that the norms kcurluk0,O and kdiv uk0,O depend continuously on kϕkH 1/2 (γ 0 ) .