A FINITE ELEMENT METHOD WITH LAGRANGE MULTIPLIERS FOR LOW-FREQUENCY HARMONIC MAXWELL EQUATIONS

Transcription

1 A FINITE ELEMENT METHOD WITH LAGRANGE MULTIPLIERS FOR LOW-FREQUENCY HARMONIC MAXWELL EQUATIONS ALFREDO BERMÚDEZ RODOLFO RODRÍGUEZ AND PILAR SALGADO Abstract The aim of this paper is to anayze a finite eement method to sove the ow-frequency harmonic Maxwe equations in a bounded domain containing conductors and dieectrics This system of partia differentia equations is a mode for the so-caed eddy currents probem After writing this probem in terms of the magnetic fied it is discretized by Nédéec edge finite eements on a tetrahedra mesh Error estimates are easiy obtained if the cur-free condition is imposed on the eements in the dieectric domain Then the cur-free condition is imposed at a discrete eve by introducing a piecewise inear mutivaued potentia The resuting probem is shown to be a discrete version of other continuous formuation in which the magnetic fied in the dieectric part of the domain has been repaced by a magnetic potentia Moreover this approach eads to an important saving in computationa effort Probems reated to the topoogy are aso considered in that the possibiity of having a non simpy connected dieectric domain is taken into account Impementation issues are discussed incuding an amenabe procedure to impose the boundary conditions by means of a Lagrange mutipier Finay the method is appied to sove a threedimensiona mode probem: a cyindrica eectrode surrounded by dieectric Key words ow-frequency harmonic Maxwe equations eddy currents probems finite eement computationa eectromagnetism AMS subject cassifications 78M10 65N30 1 Introduction In this paper we anayze a finite eement method with Lagrange mutipiers to sove the eddy currents mode in a bounded domain incuding conductors and dieectrics This mode can be obtained from Maxwe equations by assuming that a fieds are harmonic and the current frequency is ow enough so that the term invoving the dispacement current in Ampère s Law can be negected Such a situation happens for instance in probems reated with machines working at power frequencies In particuar this paper is motivated by the need of a three-dimensiona numerica simuation of a metaurgica furnace (see Bermúdez et a [6 7] for reated works concerning axisymmetric modes) Because of many interesting appications in eectrica engineering numerica soution of eddy currents probems became an important research area eading to a great number of pubications in recent years (see for instance [ ] The books by Bossavit [11] and Sivester and Ferrari [29] aso contain vauabe materia on this subject and incude arge reference ists Whie severa papers dea with the mathematica and numerica anaysis of the fu harmonic Maxwe equations (see for instance the papers by Monk [ ] and Fernandes and Giardi [17]) the number of papers concerning anaysis of the eddy This work was partiay supported by Programa de Cooperación Científica con Iberoamérica Ministerio de Educación y Ciencia Spain Departamento de Matemática Apicada Universidade de Santiago de Compostea Santiago de Compostea Spain (mabermud@usces) Partiay supported by Xunta de Gaicia research project PGIDT00PXI20701PR (Spain) GI 2 MA Departamento de Ingeniería Matemática Universidad de Concepción Casia 160-C Concepción Chie (rodofo@ing-matudecc) Partiay supported by FONDAP in Appied Mathematics (Chie) Departamento de Matemática Apicada Universidade de Santiago de Compostea Santiago de Compostea Spain (mpiar@usces) Partiay supported by Xunta de Gaicia research project PGIDT00PXI20701PR (Spain) 1

2 current mode is much smaer Significant mathematica and numerica resuts have been obtained by MacCamy and coauthors [ ] for a two-dimensiona eddy current probem In the three-dimensiona case et us mention the artice by Ammari et a [4] where a thorough justification of the eddy current mode is given The above mentioned papers dea with the eddy currents probem in the whoe space the infinity being usuay taken into account by means of integra equations A usefu aternative approach is considered by Aonso and Vai [2 3] where the probem in a bounded domain is considered incuding appropriate boundary conditions In these papers a formuation invoving ony the eectric fied is given and then numericay soved by using a domain decomposition technique and Nédéec edge finite eements In the present paper we aso consider the eddy currents probem in a bounded domain which incudes conductors and dieectrics The conductors are not assumed to be totay incuded in the probem domain We consider a formuation in terms of magnetic fied with mixed Neumann and Dirichet boundary conditions The former are the natura conditions for the conducting part of the boundary The atter are imposed on the dieectric part and aow taking into account a the eectromagnetics effects outside the domain Then foowing Bossavit and Verité [13] we introduce a scaar magnetic potentia in the domain occupied by the dieectric This hybrid formuation is discretized by using Nédéec edge eements for the magnetic fied and standard piecewise inear continuous eements for the magnetic potentia The outine of the paper is as foows: In Section 2 we reca the eddy currents mode and obtain a weak formuation invoving the magnetic fied ony Section 3 concerns existence and uniqueness of soution which are proved by using cassica toos Then in Section 4 we introduce a scaar magnetic potentia in the dieectric domain and show that the resuting probem is competey equivaent to the previous one The numerica discretization is introduced in Section 5 where error estimates are obtained under mid reguarity assumption on the soution In order to sove the discretized probem a Lagrange mutipier is proposed in Section 6 to impose the Dirichet boundary conditions The resuting mixed probem is shown to attain a unique soution and to be equivaent to the origina discrete one Finay in Section 7 we report numerica resuts for a test with known anaytica soution; these resuts confirm the predicted order of convergence of the method 2 The eddy currents probem Eddy currents are usuay modeed by the ow-frequency harmonic Maxwe equations Let us reca first the governing equations of eectromagnetism; namey Maxwe equations: (21) D cur H = J t (22) B + cur E = 0 t (23) div B = 0 (24) constitutive aws: div D = ρ (25) (26) B = µh D = ǫe 2

3 and Ohm s aw in conductors: (27) J = σe We have used notations which are standard in eectromagnetism: D is the eectric dispacement E is the eectric fied B is the magnetic induction H is the magnetic fied J is the current density ρ is the eectric charge density µ is the magnetic permeabiity ǫ is the eectric permittivity σ is the eectric conductivity We use bodface etters to denote vector fieds and variabes as we as vector-vaued operators throughout the paper When aternating currents are considered a the fieds have the foowing steadystate form: F(xt) = Re [ e iωt F(x) ] where ω is the anguar frequency Moreover in the ow-frequency harmonic regime the term in (21) incuding the eectric dispacement can be negected Under these assumptions equations (21) (27) reduce to the so-caed eddy currents mode: (28) (29) (210) (211) curh = J iωµh + cure = 0 divb = 0 divd = ρ with (212) (213) (214) B = µh D = ǫe J = σe We are interested in soving these equations in a bounded domain Ω which consists of two parts Ω C and Ω D occupied by conductors and dieectrics respectivey The eectric conductivity σ vanishes in the dieectric domain The boundary of the domain Ω aso spits into two parts: Γ C := Ω C Ω and Γ D := Ω D Ω Finay we denote Γ I := Ω C Ω D the interface between dieectric and conductors Boundary conditions must be added to sove the eddy currents mode in the bounded domain Ω We consider: (215) (216) E n = 0 on Γ C H n = f on Γ D with f being a given tangentia vector fied (ie satisfying f n = 0 on Γ D ) In the equations above n denotes the outer unit norma vector to Ω Throughout the paper n wi denote a unit vector norma to a given surface not necessariy the 3

4 same at each occurrence In genera it wi not be expicity mentioned which this surface is provided this is sufficienty cear from the context To obtain a weak formuation of the boundary vaue probem (28) (216) consider a test function G such that G n = 0 on Γ D From (29) we have (217) iω µh Ḡ + cure Ḡ = 0 Ω Now we can transform the second term above by using a Green s formua: (218) cure Ḡ = E curḡ E n Ḡ dγ Ω Ω Γ C = E curḡ E n Ḡ dγ = E curḡ Ω Γ C Ω where we have used the boundary condition (215) to obtain the ast equaity We observe that equations (28) and (214) and the fact that σ is nu in the dieectric domain ead to curh = 0 in Ω D Because of this we ony need to take test functions G satisfying curg = 0 in Ω D By doing so equations (217) and (218) yied iω µh Ḡ + E curḡ = 0 Ω Ω C Instead in the conductors equations (28) and (214) ead to E = 1 σ curh which aows us to eiminate E in the equation above Thus we finay obtain 1 iω µh Ḡ + curh curḡ = 0 Ω Ω C σ 3 Anaysis of the magnetic fied formuation of the eddy currents probem Let us assume that Ω is simpy connected with a Lipschitz-continuous connected boundary The subdomains Ω C and Ω D are aso assumed to have Lipschitzcontinuous boundaries athough not necessariy connected Finay the boundaries of Γ C Γ D and Γ I are assumed to be Lipschitz-continuous too We use standard notation for Soboev spaces and norms Moreover we reca the definition of some functiona spaces Let endowed with the norm H(curΩ) := { G L 2 (Ω) 3 : curg L 2 (Ω) 3} G H(curΩ) := and for each positive rea number r et endowed with the norm [ G 2 L 2 (Ω) 3 + curg 2 L 2 (Ω) 3 ] 1/2 H r (curω) := { G H r (Ω) 3 : curg H r (Ω) 3} G Hr (curω) := [ G 2 H r (Ω) 3 + curg 2 H r (Ω) 3 ] 1/2 4 Ω

5 Consider the foowing cosed subspaces of H(curΩ): V = {G H(curΩ) : curg = 0 in Ω D } { V 0 = G V : G n = 0 in H 1/2 00 (Γ D ) 3} where H 1/2 00 (Γ D ) 3 denotes the dua space of H 1/2 00 (Γ D )3 which in its turn is the space of functions defined on Γ D that extended by 0 on Ω \ Γ D beong to H 1/2 ( Ω) 3 We assume that µǫσ L (Ω) and that there exist constants µ ǫ and σ such that µ(x) µ > 0 ae in Ω ǫ(x) ǫ > 0 ae in Ω σ(x) σ > 0 ae in Ω C σ(x) = 0 in Ω D Concerning the boundary data f we suppose there exists a fied H f V such that (31) H f n = f in H 1/2 00 (Γ D ) 3 Remark 31 We refer to [1] for necessary and sufficient conditions on f to ensure that there exists H f H(curΩ) such that H f n = f on Γ D in a weak sense in the case Γ C = Γ D = Ω (ie when the conductors Ω C are fuy contained in Ω) We aso refer to [14 15] for simiar conditions in the case that Ω is a Lipschitz poyhedron and Γ C and Γ D are poyhedra surfaces with piecewise smooth boundaries Equation (31) impies an additiona constraint on the data f since H f has to be cur-free in Ω D A necessary condition for the existence of such H f is that div Γ f = 0 on Γ D where div Γ stands for the tangentia divergence operator (see [1] for the resut and a precise definition of div Γ ) In the case Γ C = Γ D = Ω then div Γ f = 0 on Γ D is aso a sufficient condition when Ω has a smooth boundary (see Theorem 41 of [1]) Now we can state a variationa formuation of our probem in terms of the magnetic fied H: Probem MP: To find H V such that (32) (33) H n = f in H 1/2 00 (Γ D ) 3 1 iω µh Ḡ + Ω Ω C σ curh curḡ = 0 G V0 by Let a: H(curΩ) H(curΩ) C be the sesquiinear continuous form defined 1 a(hg) := iω µh Ḡ + curh curḡ Ω Ω C σ This form ceary satisfies (34) a(gg) α G 2 H(curΩ) G V Hence the foowing existence resut is immediatey derived: 5

6 Theorem 31 If there exists H f V such that H f n = f in H 1/2 00 (Γ D ) 3 then probem MP attains a unique soution Proof Consider the transation Ĥ = H H f Then probem MP is equivaent to find Ĥ V 0 such that a(ĥg) = a(h f G) G V 0 and this probem has a unique soution because of inequaity (34) and Lax-Migram Lemma Once the magnetic fied H is known the current density J and the eectric fied E can be readiy computed in the conductors by means of equations (28) and (214) respectivey These are the magnitudes actuay needed in most appications In the foowing theorem we show that the soution of probem MP satisfies some of Maxwe equations (28) (211) and the boundary conditions (215) (216) in a weak sense Theorem 32 Let H V be the soution of probem MP Let B = µh L 2 (Ω) J = curh L 2 (Ω) and E = ( 1 σ J) Ω C L 2 (Ω C ) Then the foowing properties hod true: (35) (36) (37) (38) (39) divb = 0 in Ω iωµh + cure = 0 in Ω C E n = 0 in H 1/2 00 (Γ C ) 3 H n = f in H 1/2 00 (Γ D ) 3 J = 0 in Ω D Proof Given Ψ D(Ω) := {Ψ C (Ω) : suppψ Ω} et G = gradψ V 0 Then (33) yieds µh grad Ψ = 0 Ω Consequenty B = µh H(divΩ) and (35) hods true Now et G D(Ω) 3 be such that suppg Ω C Then G V 0 and (33) yieds 1 iω µh Ḡ + curh curḡ = 0 Ω C Ω C σ Hence E = 1 σ J = 1 σ curh H(curΩ ) and (36) hods true C To prove (37) given ϕ H 1/2 00 (Γ C) 3 we wi show that E n ϕ ΩC = 0 where ΩC denotes the duaity pairing in H 1/2 ( Ω C ) 3 H 1/2 ( Ω C ) 3 and ϕ H 1/2 ( Ω C ) 3 is the natura extension of ϕ by 0 on Ω C \ Γ C To this aim et G H 1 (Ω C ) 3 be such that G ΩC = ϕ and G be the extension by 0 of G to Ω \Ω C Then G V 0 and (33) yieds 0 = iω µh G + E cur Ω G C Ω = iω µh Ḡ + cure Ḡ + E ng ΩC = E n ϕ ΩC Ω C Ω Ω C C 6

7 where we have used that E = 1 σ curh in Ω and (36) C Finay (38) and (39) arise expicity in probem MP Remark 32 The theorem above shows that probem MP aows us to determine uniquey the eectric fied E in the conductors In its turn E and Maxwe equation (211) determines the charge density ρ in Ω C In particuar in the interior of any homogeneous subdomain Ω of Ω C (ie Ω Ω C such that ǫ Ω and σ Ω are constant) ρ Ω = div ( ǫ σ curh) Ω = 0 Instead the eectric fied E is not uniquey determined in the dieectric Indeed from the eddy currents mode (28) (214) we obtain the foowing equations for E ΩD : (310) (311) (312) cure = iωµh in Ω D div (ǫe) = ρ in Ω D E n = E ΩC n on Γ I The atter arises from the facts that E ΩC is aready known and E is gobay in H(curΩ) A boundary condition on Γ D is needed to determine a unique soution even in the simpest case of a topoogicay trivia Ω D (ie when Ω D is simpy connected with a connected boundary) A natura condition woud be to impose the norma component of the eectric dispacement D on Γ D ; namey (313) ǫe n = ψ on Γ D The data ψ amounts to eventua surface charges on the outer boundary of the dieectric domain Existence of soution of (310) (313) has been proved in Theorem 42 of [1] in the case that Ω D is smooth and Γ I Γ D = (for instance when Ω C Ω) Even in this simper case a number of additiona constraints reated with the topoogy of Ω D must be added to have uniqueness as can be seen in this reference To the best of the authors knowedge a simiar resut has not been proved for the genera case of Ω D being a Lipschitz poyhedron with Γ I Γ D Nevertheess this is not a drawback for the appication of this eddy currents mode since typicay the goa of these probems is to compute the eectric fied ony in the conductors as said above 4 Introducing a magnetic potentia In this section we show how probem MP can be transformed by repacing the magnetic fied in the dieectric domain Ω D by a (scaar) magnetic potentia We reca that Ω is assumed to be simpy connected with connected boundary Ω Let Ω C = j=j j=0 Ωj with Ω0 being the union of a the connected components of Ω C C C such that Ω \ Ω 0 is simpy connected and Ωj j = 1J the remaining connected C C components of Ω C (see Figure 41) We assume that for each Ω j j = 1J there exists an open cut surface C Σ j Ω D such that Σ j Ω D and Ω D := Ω D \ j=j j=0 Σ j is pseudo-lipschitz and simpy connected (see Figure 41) We aso assume that each one of these surfaces Σ j is connected and Σ j Σ k = for j k (see for instance [5]) Let us arrange the conductors Ω j in such a way that the inner ones are numbered C from j = 1 to K and those going through Ω from j = K + 1 to J In Figure 41 Ω 1 is an exampe of a conductor of the first kind and Ω2 of the second one C C 7

8 Fig 41 Sketch of the domain We aso assume that there exist cross sections of Ω j j = 1J; namey open C surfaces S j Ω j with respective boundaries S C j = S j Γ I which are assumed to be cosed simpe curves We denote these curves γ j Moreover for j = K + 1J we take S j Γ C and γ j Γ C Γ D (see again Figure 41) Let Ω j := Ω \ Σ D D j j = 1J We fix a unit norma n j on each Σ j and denote its two faces Σ j and Σ + j with n j being the outer norma to Ω j aong Σ+ D j We choose an orientation for each γ j by taking its initia and end points on Σ j and Σ + j respectivey We denote by t j the unit vector tangent to γ j For any function Ψ H 1 ( Ω D ) we denote by [[ Ψ]] Σj := Ψ Σ j Ψ Σ + j the jump of Ψ through Σ j aong n j The gradient of Ψ in D ( Ω D ) can be extended to L 2 (Ω D ) 3 and wi be denoted by grãd Ψ Let Θ be the inear space of H 1 ( Ω D ) defined by } Θ = { Ψ H 1 ( Ω D ) : [[ Ψ]] = constant j = 1J Σj Then for Ψ H 1 ( Ω D ) we have that grãd Ψ H(curΩ D ) if and ony if Ψ Θ in which case cur (grãd Ψ) = 0 (see Lemma 311 in [5]) Actuay the kerne of the operator cur : H(curΩ D ) L 2 (Ω D ) 3 is given by (41) Ker (cur) = grãd Θ = gradh 1 (Ω D ) C 8

9 where C is the space of the so-caed Neumann harmonic fieds in Ω D defined by C := { G L 2 (Ω D ) 3 } : curg = 0 div (µg) = 0 in Ω D and G n = 0 on Ω D A basis of the space C is given by the set of functions {grãd Φ j j = 1J} where for each j Φ j H 1 ( Ω j ) is the soution of D (42) µgrãd Φ j grad Ψ = 0 Ψ H 1 (Ω D ) eω j D (43) [[ Φ j ]] Σj = 1 By using Lax-Migram Lemma it is straightforward to see that Φ j is uniquey defined in H 1 ( Ω j )/C (See for instance again [5]) D Therefore according to (41) for a G V there exist unique constants c j j = 1J and a unique scaar fied Ψ H 1 (Ω D )/C such that G ΩD = grãd Ψ with Ψ Θ given by Ψ = Ψ+ J j=1 c Φ j j Furthermore because of (43) the constants c j are the jumps of Ψ across the respective cuts Σ j Consequenty given Ψ Θ we have that Ψ H 1 (Ω) if and ony if [[ Ψ]] Σj = 0 for j = 1J Remark 41 These jumps have a precise physica meaning For instance for the soution H of probem MP et us write H ΩD = grãd Φ with Φ Θ If H is sufficienty smooth by using Stokes theorem and equation (28) we have [[ Φ]] Σj = grãd Φ t j dγ = H ΩD t j dγ = H ΩC t j dγ γ j γ j γ j = curh ΩC n dγ = J n dγ =: I j j = 1J S j S j Thus the jump of the magnetic potentia Ψ across each cut surface Σ j is exacty the current intensity I j through the cross section S j of the conductor Ω j (as defined C above) We introduce the foowing notation: for G C L 2 (Ω C ) 3 and G D L 2 (Ω D ) 3 we denote by (G C G D ) the fied G L 2 (Ω) 3 defined ae by { GC (x) if x Ω G(x) := C G D (x) if x Ω D Let us denote by W the inear space given by { W := (G Ψ) H(curΩ C ) (Θ/C) : (G grãd Ψ) } H(curΩ) Ceary the foowing appication is an isomorphism: W V (G Ψ) (G grãd Ψ) Simiary we define the cosed subspace of W { W 0 := (G Ψ) W : grãd Ψ n = 0 in H 1/2 00 (Γ D ) 3} 9

10 which is isomorphicay equivaent to V 0 Thus we are ead to define the foowing probem: Probem HP: To find (H Φ) W such that grãd Φ n = f in H 1/2 00 (Γ D ) 3 1 iω µh Ḡ + curh curḡ + iω µgrãd Ω C Ω C σ Φ grãd Ψ = 0 Ω D (G Ψ) W 0 This is the we known magnetic fied/magnetic potentia hybrid formuation of the eddy currents probem introduced by Bossavit and Verité [13] One main advantage with respect to formuation (32) (33) ies in the fact that a vector fied is repaced by a scaar one in the dieectric domain The foowing emma is an immediate consequence of the isomorphisms between W and V and between W 0 and V 0 : Lemma 41 The pair (H Φ) is soution of probem HP if and ony if (H grãd Φ) is soution of probem MP As a consequence of this emma Theorem 31 yieds existence and uniqueness of soution for probem HP: Coroary 42 Under the assumptions of Theorem 31 probem HP has a unique soution (H Φ) with (H grãd Φ) being the unique soution of probem MP 5 Numerica soution In this section we first introduce a discretization of probem MP and prove its convergence Then we prove that the obtained discrete probem is competey equivaent to a convenient discrete version of probem HP 51 Discretizing the magnetic fied We empoy edge finite eements to approximate the magnetic fied; more precisey the owest-order finite eement of the famiy introduced by Nédéec in [26] This eement beongs to the famiy of the so-caed Whitney eements (see [9]) We assume Ω Ω C and Ω D are Lipschitz poyhedra and consider a famiy of reguar tetrahedra meshes {T h } of Ω such that for every mesh T h each eement K T h is contained either in Ω C or in Ω D (h stands as usua for the corresponding mesh-size) The magnetic fied is approximated in each tetrahedron K by a poynomia vector fied in the space N(K) := { G h P 1 (K) 3 : G h (x) = a x + b ab C 3 x K } An expicit computation shows that vector fieds of this type have constant tangentia components aong each straight ine in the Eucidean space Moreover given six compex numbers β n n = 16 there exists a unique G h N(K) (ie unique ab C 3 ) such that its tangentia component aong the n-th edge of K coincide with β n for n = 16 respectivey Thus these tangentia components aong the edges of K can be taken as the degrees of freedom defining the eements in N(K) These eements are H(cur)-conforming in the sense that G h N(K) their tangentia traces on each trianguar face T of K ony depend on the degrees of freedom of G h on the three edges of T So if we set N h (Ω) := {G h H(curΩ) : G h K N(K) K T h } 10

11 the eements in this space are piecewise inear vector fieds with tangentia traces that are continuous through the faces of the mesh This is the owest-order Nédéec finite eement space introduced in [26] See [19] for a detaied mathematica anaysis and [11] for usefu impementation issues If G is smooth enough (vg G H 2 (Ω) 3 ) then its Nédéec interpoant G I is defined by (51) G I N h (Ω) : G I t dγ = G t dγ edge of T h where from now on t denotes a unit vector tangent to the edge The Nédéec interpoation operator (52) H 2 (Ω) 3 N h (Ω) G G I with G I defined by (51) extends uniquey to H r (curω) with r > 1/2 Indeed according to Soboev Imbedding Theorem and a trace theorem for each K T h G K L p (K) 3 curg K L p (K) 3 and G n K L p ( K) 3 with p = 4/(3 2r) > 2 Then the resut foows by appying Lemma 47 of [5] However the soution H of probem MP does not satisfy in genera cur H = J H r (Ω) 3 with r > 1/2 In fact J ΩD = 0 whereas J ΩC n = ( 1 σ E) Ω C n in genera does not vanish on Γ I ; thus J n has a jump across Γ I (see for instance the probem in Section 7) Nevertheess typicay H ΩC H r (curω C ) and H ΩD H r (curω D ) with r > 1/2 This is enough for H I to be we defined as shown in the foowing Lemma which aso provides an error estimate for the Nédéec interpoant under these assumptions (Here and thereafter C denotes a generic constant not necessariy the same at each occurrence but aways independent of the mesh-size h) Lemma 51 Let r ( 1 2 1] The operator defined by (52) (51) extends uniquey to the space {G H(curΩ) : G ΩC H r (curω C ) and G ΩD H r (curω D )} Furthermore for a G in this space [ ] G G I H(curΩ) Ch r G H r (curω C ) + G H r (curω D ) Proof According to the discussion above since G ΩC H r (curω C ) and G ΩD H r (curω D ) with r > 1/2 then the Nédéec interpoants of G ΩC and G ΩD are we defined in N h (Ω C ) and N h (Ω D ) respectivey Moreover since G H(curΩ) a density argument shows that the degrees of freedom corresponding to the edges Γ I coincide for both interpoants Thus the goba interpoant G I N h (Ω) is we defined aso in this case On the other hand the arguments in the proof of Theorem 54 in [19] can be extended to this case to prove the error estimate above In order to use these eements to discretize probem MP we have to use an approximant f I of the boundary data f such that a discrete version of equation (32) can hod true namey such that there exists H h N h (Ω) satisfying H h n = f I To attain this goa we wi use the two-dimensiona Nédéec interpoant of n f on the trianguar mesh induced by T h on the poyhedra surface Γ D To introduce this interpoant et T Γ D h := {T Γ D : T face of K T h } For each triange T T Γ D h 11

12 consider oca orthogona coordinates (ξ η ζ) such that T is contained in the pane ζ = 0 Let N 2 (T) := { ϕ h P 1 (T) 3 : ϕ h (ξη0) = (a cηb + cξ0) abc C (ξη0) T } This is the owest-order two-dimensiona Nédéec finite eement (see [26]) on the pane ζ = 0 The tangentia components of these vector fieds aong the three edges of the triange T can aso be taken as the degrees of freedom defining them Therefore we define { N 2 h(γ D ) := ϕ h L 2 (Γ D ) 3 : ϕ h T N 2 (T) T T Γ D h and ϕ h t is continuous on edge of T Γ D h Let ϕ be a tangentia vector fied on Γ D (ie satisfying ϕ n = 0 on Γ D ) If ϕ is sufficienty smooth (vg ϕ H 1 (Γ D ) 3 ) then its Nédéec interpoant on Γ D which we denote by ϕ I2 is defined by (53) ϕ I2 N 2 h(γ D ) : ϕ I2 t dγ = ϕ t dγ edge of T Γ D h If G is smooth enough in Ω D (vg G H 2 (Ω D ) 3 ) then its tangentia trace on Γ D n (G ΓD n) is smooth too and satisfies (54) [ ( )] I2 ( ) n G ΓD n = n G I ΓD n on Γ D Indeed a straightforward computation shows that the right hand side above aso beongs to N 2 h(γ D ) On the other hand (51) impies ( ) n G I ΓD n t dγ = G I t dγ = G t dγ ( ) = n G ΓD n t dγ edge of T Γ D h Thus the degrees of freedom defining both sides of (54) coincide and consequenty (54) hods true The foowing emma shows that a simiar resut is vaid for G H r (curω D ): Lemma 52 Let r ( 1 2 1] The inear operator } (55) H 2 (Ω D ) 3 N 2 h(γ D ) G [ n (G ΓD n) ] I 2 with ( ) I2 defined by (53) extends uniquey to H r (curω D ) Furthermore equation (54) hods true for a G in this space Proof As said above if G H 2 (Ω D ) 3 then [ n (G ΓD n) ] I 2 N 2 h (Γ D ) is defined by [ ( I2 n G ΓD n)] t dγ = G t dγ edge of T Γ D h Then by repeating the arguments in the proof of Lemma 51 (ie using Soboev Imbedding Theorem and Lemma 47 of [5]) we prove that the operator defined by (55) and (53) extends uniquey to H r (curω D ) for r > 1/2 12

13 Furthermore we have aso shown above that for G H 2 (Ω D ) [ ( I2 n G ΓD n)] t dγ = = G t dγ ( ) n G I ΓD n t dγ edge of T Γ D h Then a density argument and the fact that n (G I ΓD n) N 2 h(γ D ) aow us to concude that (54) hods true for a G H r (curω D ) If the data f of probem MP is sufficienty smooth we define (56) f I := (n f) I2 n; that is f I is such that n f I = (n f) I2 which means n f I N 2 h(γ D ) : n f I t dγ = n f t dγ edge of T Γ D h The foowing emma shows that this definition aso works under weak smoothness assumptions as those in the previous emmas: Lemma 53 Let G H r (curω D ) with r > 1/2 and et g = G ΓD n Then g I := (n g) I2 n is we defined and satisfies n g I = n (G I ΓD n) on Γ D Proof As a consequence of Lemma 52 (n g) I2 = [ n (G ΓD n) ] I 2 is we defined Hence g I := (n g) I2 n is we defined too Moreover since according to this emma (54) hods true for G H r (curω D ) then n g I = (n g) I2 = [ ] I2 n (G ΓD n) = n (G I ΓD n) on Γ D Now we are in a position to discretize probem MP We introduce the foowing finite-dimensiona spaces: V h := {G h N h (Ω) : curg h = 0 in Ω D } V 0 h := {G h V h : G h n = 0 on Γ D } Finay we define the discrete magnetic probem as foows: Probem DMP: Find H h V h such that H h n = f I on Γ D 1 iω µh h Ḡ h + Ω Ω C σ curh h curḡ h = 0 G h V 0 h It is straightforward to prove existence and uniqueness of soution for this probem under mid smoothness assumptions on the soution of probem MP Moreover an error estimate can be deduced from the standard finite eement approximation theory: 13

14 Theorem 54 Let us assume that the soution H of probem MP satisfies H ΩC H r (curω C ) and H ΩD H r (Ω D ) 3 with r ( 1 2 1] Then f is we defined by (56) I probem DMP attains a unique soution H h and [ ] H H h H(curΩ) Ch r H Hr (curω C ) + H H r (Ω D ) 3 Proof Since H V then curh = 0 in Ω D Hence H ΩD H r (curω D ) Therefore according to Lemma 51 its Nédéec interpoant H I N h (Ω) is we defined and satisfies [ ] (57) H H I H(curΩ) Ch r H H r (curω C ) + H H r (Ω D ) 3 Moreover the arguments of Remark 56 in [19] can be extended to this case to prove that curh ΩD = 0 impies curh I ΩD = 0 Therefore H I V h On the other hand because of Lemma 53 f I is we defined by (56) and satisfies f I = H I ΓD n on Γ D Thus we have proved that there exists H I V h such that H I n = f I on Γ D Hence since V 0 h V 0 the arguments in the proof of Theorem 31 aso appy to probem DMP aowing us to prove existence and uniqueness of a soution H h of this probem Finay to prove the error estimate notice that since V 0 h V 0 then a(h H h G h ) = 0 G h V 0 h Hence since H h n = f I = H I n on Γ D then H h H I V 0 h Therefore because of this and (34) α H H h 2 H(curΩ) a(h H hh H h ) = a(h Hh H H I ) C H H h H(curΩ) H H I H(curΩ) which together with estimate (57) aow us to concude the proof 52 Discretizing the magnetic potentia Probem DMP is actuay just a theoretica method in that its soution requires to impose somehow the cur-free condition in the definition of V h to tria and test functions In what foows we show how to dea efficienty with this cur-free condition by introducing a discrete mutivaued magnetic potentia in the dieectric domain We assume that the cut surfaces Σ j are poyhedra and that the meshes are compatibe with them in the sense that each Σ j is union of faces of tetrahedra K T h for each mesh T h Therefore T Ω D h := {K T h : K Ω D } can aso be seen as a mesh of Ω D Firsty we introduce an approximation of the space Θ Let us denote L h ( Ω D ) := { Ψh H 1 ( Ω D ) : Ψh K P 1 (K) K T Ω D h Then we consider the famiy of finite dimensiona subspaces of Θ given by } Θ h := { Ψ h L h ( Ω D ) : [[ Ψ h ]] Σj = constant j = 1J} The foowing emma shows that the cur-free vector fieds in N h (Ω D ) admit a mutivaued potentia in Θ h : 14

15 Lemma 55 Let G h L 2 (Ω D ) 3 Then G h N h (Ω D ) with curg h = 0 in Ω D if and ony if there exists Ψ h Θ h such that G h = grãd Ψ h in Ω D Such Ψ h is unique up to an additive constant Proof According to (41) curg h = 0 in Ω D if and ony if there exists Ψ h Θ such that G h = grãd Ψ h in Ω D Moreover since Ω D is connected then Ψ h is unique up to an additive constant Now et K T Ω D h be a tetrahedron of the mesh A direct cacuation shows that G h N(K) with curg h K = 0 if and ony if G h K P 0 (K) 3 or equivaenty if and ony if Ψ h K P 1 (K) 3 Thus the emma foows from the definition of Θ h Let us introduce the foowing famiies of finite-dimensiona approximations of W and W 0 respectivey: { W h := (G h Ψ h ) N h (Ω C ) (Θ h /C) : (G h grãd Ψ } h ) H(curΩ) W 0 h := {(G h Ψ h ) W h : grãd Ψ } h n = 0 on Γ D By virtue of Lemma 55 W h and W 0 h are isomorphicay equivaent to V h and V 0 h respectivey Thus we define the foowing discrete probem which turns out to be equivaent to probem DMP: Probem DHP: To find (H h Φ h ) W h such that (58) (59) grãd Φ h n = f I on Γ D 1 iω µh h Ḡ h + Ω C Ω C σ curh h curḡ h + iω µgrãd Φ h grãd Ψh = 0 (G h Ψ h ) W 0 h Ω D Ceary the foowing discrete anaogue of Lemma 41 hods true: Lemma 56 The pair (H h Φ h ) is soution of probem DHP if and ony if (H h grãd Φ h ) is soution of probem DMP As an immediate consequence of these two emmas Theorem 54 yieds an error estimate for the approximation obtained from probem DHP: Coroary 57 Let us assume that the soution (H Φ) of probem HP satisfies H H r (curω C ) and grãd Φ H r (Ω D ) 3 with r ( 1 2 1] Then probem DHP is we posed it attains a unique soution (H h Φ h ) and H H h H(curΩC ) + grãd Φ grãd Φ h L2 (Ω D ) 3 Ch r [ H Hr (curω C ) + grãd Φ Hr (Ω D ) 3 ] 6 Computer impementation For probem DHP to be usefu for computationa purposes we have to introduce effective procedures to impose the foowing constraints: 1 (G h grãd Ψ h ) H(curΩ) which arise in the definition of W h ; 2 [[ Ψ h ]] Σj = constant which arise in the definition of Θ h ; 3 the boundary condition grãd Φ h n = f I on Γ D 15

16 We fix some notation to dea with these constraints We choose an orientation for each edge of the mesh T h and denote P and P + its initia and end points respectivey and t its unit tangent vector pointing from P to P + Regarding the first constraint we have the foowing resut: Lemma 61 Let (G h Ψ h ) N h (Ω C ) (Θ h /C) Then (G h grãd Ψ h ) H(curΩ) if and ony if G h t dγ = Ψ h (P + ) Ψ h (P ) edge of T h : Γ I Proof Since G h N h (Ω C ) and grãd Ψ h N h (Ω D ) then (G h grãd Ψ h ) H(curΩ) if and ony if the tangentia traces on Γ I of G h and grãd Ψ h coincide; that is if and ony if ( n (G h n) = n grãd Ψ ) h n on Γ I Now the equation above hods true if and ony if the degrees of freedom of G h and grãd Ψ h coincide on a the edges Γ I and this reads G h t dγ = grãd Ψ h t dγ = Ψ h (P + ) Ψ h (P ) This emma shows that the constraint (G h grãd Ψ h ) H(curΩ) can be readiy imposed by eiminating the degrees of freedom of G h associated with the edges Γ I in terms of those of Φ h corresponding to the vertices of the mesh on this interface Regarding the second constraint for Ψ h Θ h et us denote c hj := [[ Ψ h ]] Σj j = 1J In order to hande the mutivaued character of the functions Ψ h Θ h for each cut surface Σ j we in principe distinguish the degrees of freedom of Ψ h on Σ + j from those on Σ j Then the atter can be eiminated by using Ψ h Σ j = Ψ h Σ + j + [[ Ψ h ]] Σj = Ψ h Σ + j + c hj j = 1J This eimination must be carried out for the soution (H h Φ h ) W h of probem DHP as we as for the test functions (G h Ψ h ) W 0 h For the former the arguments in Remark 41 can be repeated at discrete eve to show that each jump [[ Φ h ]] Σj represents the current intensity through the conductor Ω j corresponding to the discrete C soution (H h Φ h ) Because of this we denote these jumps I hj := [[ Φ h ]] Σj j = 1J For j = 1K (ie for inner conductors Ω j) I C hj are additiona unknowns of the discrete probem Instead for j = K + 1J (ie for conductors Ω j going C through Ω) I hj can be computed in advance from the data of the discrete probem Indeed since γ j Γ D and grãd Φ h n = f I on Γ D then I hj = grãd Φ ( h t j dγ = n grãd Φ ) (61) h n t j dγ = n f I t j dγ γ j γ j γ j 16

17 For the test functions (G h Ψ h ) W 0 h by repeating these arguments and using that grãd Ψ h n = 0 on Γ D we have ( c hj = n grãd Ψ ) (62) h n t j dγ = 0 j = K + 1J γ j Hence ony the constants c hj for j = 1K must be taken into account as genuine degrees of freedom in the definition of W 0 h Remark 61 The computed and exact intensities through the conductors Ω j C j = K + 1J coincide Indeed because of (61) Lemma 53 (51) the fact that each γ j is union of edges in T h and Remark 41 we have [ I hj = n f I t j dγ = n (grãd Φ) ] I ΓD n t j dγ γ j γ j = (grãd Φ) I t j dγ = grãd Φ t j dγ = [[ Φ]] Σj = I j γ j γ j Regarding the third constraint we impose the boundary condition by means of a Lagrange mutipier Let Γ D be the pseudo-lipschitz connected poyhedra surface defined by Γ D := Γ D \ J j=k+1 ( Σj Γ D ) Let { } L h ( Γ D ) := ν h H 1 ( Γ D ) : ν h T P 1 (T) T T Γ D h { } L h (Γ D ) := ν h H 1 (Γ D ) : ν h T P 1 (T) T T Γ D h Hence given ν h L h ( Γ D ) we have that ν h L h (Γ D ) if and ony if [[ν h ]] Σj Γ D = 0 for j = K + 1J Let grãd Γ denote the surface gradient operator Since we wi use this operator acting ony on piecewise inear functions we give a definition vaid in this case (for its genera definition on poyhedra surfaces see [14 15]): grãd Γ : L h ( Γ D ) L 2 t(γ D ) 3 := { ϕ L 2 (Γ D ) 3 : ϕ n = 0 on Γ D } is defined on each eement T T Γ D h by (grãd Γ ν h ) T = 2 (ν h T ) where 2 is the usua gradient of a function of two variabes; ie using oca coordinates (ξ η ζ) such that T is in the pane ζ = 0 ( (νh (grãd Γ ν h ) T := T ) (ν ) h T ) 0 ξ η For a Ψ h L h ( Ω D ) we have Ψ h ΓD L h ( Γ D ) and it is straightforward to show that [ grãd Γ ( Ψ h eγd ) = n (grãd Ψ ] (63) h ) ΓD n ae in Γ D 17

18 The foowing emma provides a weak formuation of the boundary condition (58) in probem DHP: Lemma 62 Let Ψ Θ be such that grãd Ψ H r (Ω D ) with r > 1/2 Let g = grãd Ψ ΓD n and g I = (n g) I2 n (we defined because of Lemma 53) Let Ψ h Θ h be such that (64) [[ Ψ h ]] Σj = n g I t j dγ j = K + 1J γ j Then grãd Ψ h n = g I on Γ D if and ony if (65) grãd Γ Ψh grad Γ ν h dγ = Γ D n g I grad Γ ν h dγ Γ D ν h L h (Γ D )/C Proof If grãd Ψ h n = g I on Γ D then because of (63) we have (65) Conversey et us assume that (65) hods true Since Ψ H 1+r ( Ω D )/C with r > 1/2 then its Lagrange interpoant Ψ L L h ( Ω D )/C is we defined Given a cut surface Σ j j = 1 J for a the vertices P of T h such that P Σ j we have [[ Ψ L (P)]] Σj = [[ Ψ(P)]] Σj = constant (the same for a such P) Then [[ Ψ L ]] Σj = constant and hence Ψ L Θ h /C Thus because of Lemma 55 grãd Ψ L N h (Ω D ) On the other hand et (grãd Ψ) I N h (Ω D ) be the Nédéec interpoant of grãd Ψ We have (grãd Ψ) I t dγ = grãd Ψ L t dγ edge of T Ω D h Indeed if grãd Ψ were smooth (vg grãd Ψ H 2 (Ω D ) 3 ) then (grãd Ψ) I t dγ = grãd Ψ t dγ = Ψ(P + ) Ψ(P ) = Ψ L (P + ) Ψ L (P ) = grãd Ψ L t dγ Hence because of a density argument this is aso true for grãd Ψ H r (curω D ) 3 Therefore since we have shown that grãd Ψ L N h (Ω D ) and that its degrees of freedom coincide with those of (grãd Ψ) I for a edges of T Ω D then we have h (66) (grãd Ψ) I = grãd Ψ L in Ω D Consequenty [[ Ψ L ]] Σj = grãd Ψ L t j dγ = (grãd Ψ) I t j dγ γ j γ j [ = n (grãd Ψ) ] I ΓD n t j dγ = n g I t j dγ γ j γ j j = K + 1 J the ast equaity because of Lemma 53 Let ν h := ( Ψ h Ψ L ) eγd L h ( Γ D )/C Because of the equation above and (64) we have [[ν h ]] Σj Γ D = [[ Ψ h ]] Σj [[ Ψ L ]] Σj = 0 18 j = K + 1 J

19 Then ν h L h (Γ D )/C and because of (63) (66) and Lemma 53 we have [ grad Γ ν h = grãd Γ ( Ψ h eγd ) n (grãd Ψ ] L ) ΓD n = grãd Γ ( Ψ h eγd ) n g I Thus using this ν h in (65) we obtain grãd Γ ( Ψ h eγd ) n g I = 0 Hence by using again (63) we concude the proof Now we are in a position to set a new discrete probem incuding the three constraints as we have just described To this aim we introduce the foowing discrete spaces: { Z h := (G h Ψ h c h ) N h (Ω C ) (Θ h /C) C J : [[ Ψ h ]] Σj = c hj j = 1J } and G h t dγ = Ψ h (P + ) Ψ h (P ) edge T h : Γ I { Z 0 h := (G h Ψ } h c h ) Z h : c hj = 0 j = K + 1J The new discrete probem which wi be shown to be equivaent to probem DHP in the next theorem is the foowing one: Probem DLP: To find (H h Φ h I h ) Z h and λ h L h (Γ D )/C such that (67) I hj = (n f I ) t j dγ j = K + 1J γ j 1 (68) iω µh h Ḡ h + Ω C Ω C σ curh h curḡ h + iω µgrãd Φ h grãd Ψh Ω D + grad Γ λ h grãd Γ Ψh dγ = 0 (G h Ψ h c h ) Z 0 h Γ D (69) grãd Γ Φh grad Γ ν h dγ = (n f I ) grad Γ ν h dγ ν h L h (Γ D )/C Γ D Γ D First we prove that probem DLP is we posed: Theorem 63 Let f I be any vector fied defined on Γ D such that the integras in the right hand sides of equations (67) and (69) are we defined Then probem DLP attains a unique soution Proof Probem DLP reduces to a inear system with the same number of equations than unknowns Then it is enough to prove that for f I = 0 this probem attains ony the nu soution So et (H h Φ h I h ) Z h and λ h L h (Γ D )/C satisfying (67) (69) with f I = 0 Equation (67) impies I hj = 0 for j = K + 1 J; hence (H h Φ h I h ) Z 0 h and [[ Φ h ]] Σj = I hj = 0 too Thus if we define ν h := Φ h eγd L h ( Γ D )/C we have [[ν h ]] Σj Γ = 0 for j = K + 1J and then ν h L h (Γ D )/C D Now by testing (69) with this ν h we obtain grãd Γ ( Φ h ΓD ) = 0 Hence by testing (68) with (H h Φ h I h ) (which was aready shown to beong to Z 0 h) we obtain iω µ H h Ω C Ω C σ curh h 2 + iω µ grãd Φ h 2 = 0 Ω D 19

20 Hence H h = 0 in Ω C and grãd Φ h = 0 in Ω D Consequenty Φ h = 0 in L h ( Ω D )/C and I hj = [[ Φ h ]] Σj = 0 for j = 1 K too Thus it ony remains to prove that λ h = 0 To do this et us show first that there exists (G h Ψ h c h ) Z 0 h satisfying Ψ h ΓD = λ h Indeed et Ψ h L h ( Ω D )/C be the unique function in this space satisfying for each vertex P of T Ω D h Ψ h (P) = λ h (P) if P Γ D Ψ h (P) = 0 if P / Γ D Let c hj = [[ Ψ h ]] Σj j = 1 J Because of the definition of Ψ h ceary c h = 0 Then Ψ h Θ h Finay et G h N h (Ω C ) be the unique function in this space satisfying for each edge of T h such that Ω C G h t dγ = Ψ h (P + ) Ψ h (P ) if Γ I G h t dγ = 0 if Γ I Therefore (G h Ψ h c h ) Z 0 h and Ψ h ΓD = λ h Now by testing (68) with this (G h Ψ h c h ) since we aready know that H h = 0 in Ω C and grãd Φ h = 0 in Ω D we obtain Γ D grãd Γ λ h 2 dγ = 0 Then λ h = 0 in L h (Γ D )/C and we concude the proof Now it is very simpe to show that probems DHP and DLP are equivaent: Theorem 64 Let us assume that the soution (H Φ) of probem HP satisfies H H r (curω C ) and grãd Φ H r (Ω D ) 3 with r ( 1 2 1] If ((H h Φ h I h )λ h ) is soution of probem DLP then (H h Φ h ) is soution of probem DHP Conversey if (H h Φ h ) is soution of probem DHP and I hj = [[ Φ h ]] Σj j = 1J then there exists λ h L h (Γ D )/C such that ((H h Φ h I h )λ h ) is soution of probem DLP Proof Let ((H h Φ h I h )λ h ) be soution of probem DLP Since (H h Φ h I h ) Z h then (H h Φ h ) W h (because of Lemma 61) and I hj = [[ Φ h ]] Σj j = 1J Therefore because of (67) Φ h satisfies assumption (64) in Lemma 62 Then (69) impies (58) On the other hand et (G h Ψ h ) W 0 h and c hj = [[ Ψ h ]] Σj j = 1J Because of Lemma 61 and equation (62) (G h Ψ h c h ) Z 0 h Since grãd Ψ h n = 0 on Γ D then because of (63) we have grãd Γ ( Ψ h ΓD ) = 0 Therefore by testing (68) with such (G h Ψ h c h ) we obtain (59) Thus (H h Φ h ) is soution of probem DHP Conversey et (H h Φ h ) be soution of probem DHP Since the assumptions of Coroary 57 are fufied this soution is unique On the other hand Theorem 63 shows that there exists aso a unique soution ((H h Φ h I h)λ h ) of probem DLP But then we have aready proved that (H h Φ h ) is soution of probem DHP Hence (H h Φ h ) = (H h Φ h ) and we concude the proof Probem DLP is the one we have actuay impemented The degrees of freedom for this probem are the foowing ones: 20

21 for H h N h (Ω C ): H h t dγ edge Ω C \ Γ I ; for Φ h L h ( Ω D )/C: Φ h (P) vertex P Ω D (one of them fixed to zero) for I h C J : I hj j = 1J (I hk+1 I hj are directy computed from f); for λ h L h (Γ D )/C: λ h (P) vertex P Γ D (one of them fixed to zero) Remark 62 We have imposed the boundary condition of probem DHP by means of a Lagrange mutipier However this is not the ony way of doing it An aternative procedure consists of using the fact that for each edge Γ D Φ h (P + ) Φ h (P ) = grãd Φ ( h t dγ = n grãd Φ ) h n t dγ = n f I t dγ = n f t dγ Therefore the vaues of Φ h (P) can be obtained for each vertex P Γ D by the foowing procedure: 1 fix arbitrariy the vaue of Φ h at a given vertex P 0 Γ D : Φ h (P 0 ) = 0 (this can be done because Φ h Θ h /C); 2 for each other vertex P Γ D (those on Γ D Σ j j = K + 1J must be counted twice): (a) find a path Γ P joining P 0 with P which does not cross any Σ j Γ D j = K+1J and which consists of adequatey oriented edges Γ D : Γ P := ± 1 ± NP ; (b) evauate: Φ h (P) = ± n f t 1 dγ ± ± n f t NP 1 NP The main drawback of this procedure is that step 2(a) is rather compicate to impement (see [11]) The strategy we have proposed is more expensive than this one in terms of degrees of freedom (one unknown per vertex on Γ D is added instead of being eiminated) Nevertheess one neat advantage is that its impementation is quite straightforward 7 Numerica experiments In this section we present some numerica resuts obtained with a code deveoped by us which impements in Matab the method described above We have soved a particuar probem with known anaytica soution to vaidate the computer code and to test the performance and convergence properties of the method The geometry of the domain is simiar to that of an eectric furnace with ony one eectrode More precisey we consider a domain Ω containing a conductor Ω C and dieectric Ω D as shown in Figure 71 We assume that Ω C and Ω = Ω C Ω D are coaxia cyinders of radius R C and R D respectivey and height L To obtain the data for a test probem in this domain with known anaytica soution we consider that Ω C and Ω are bounded sections of respective infinite cyinders The eectric conductivity σ is taken as constant in Ω C 21

22 Fig 71 Sketch of the domain Coordinate system and the magnetic permeabiity µ constant in the whoe Ω We consider that an aternating current J goes through the conductor Ω C in the direction of its axis; this current is assumed to be axiay symmetric with an intensity I(t) = I 0 cos(ωt) We anayze this probem using a cyindrica coordinate system (r θ z) with the z-axis coinciding with the common axis of both cyinders (see Figure 71) We denote e r e θ and e z the unit vectors in the corresponding coordinate directions Because of the assumed conditions on J ony the z-component of the eectric fied E = 1 σj does not vanish in the conductor Moreover it depends on the radia coordinate r but is independent of the other two coordinates z and θ Consequenty ony the θ-component of the magnetic fied H = i ωµ cure does not vanish and it aso depends ony on the coordinate r Then taking into account the expression of the cur operator in cyindrica coordinates we have H(rθz) = H θ (r)e θ with H θ satisfying the equation iωµh θ (r) d { } 1 d dr σr dr [rh θ(r)] = 0 r (0R C ) and the boundary conditions H θ (0) < H θ (R C ) = I 0 2πR C To sove this probem we perform the change of variabe x = γr where γ = iωµσ C Then we obtain the equation x 2 d2 dx 2 H θ (x) + x d dx H θ (x) (x 2 + 1) H θ (x) = 0 x (0γR C ) where H θ (x) = H θ (x/γ) 22

23 This is a Besse equation the soution of which is given by H θ (x) = αi 1 (x) with I 1 being the modified Besse function of the first kind and α a constant to be obtained from the boundary condition at x = γr C Thus the magnetic fied in the conductor is given by H(rθz) = I 0 2πR C I 1 (γr) I 1 (γr C ) e θ r (0R C ) θ [02π] z R On the other hand the magnetic fied created by an infinite circuar cyindrica conductor of radius R C carrying an axiay aigned and symmetric current of intensity I 0 is computed using the Ampère s circuita aw (see for instance [28]) In cyindrica coordinates it is given by H(rθz) = H θ (r)e θ with H θ (r) = I 0 2πr r R C Once more the magnitude of H θ depends ony on the radia coordinate r Moreover from this expression it is aso possibe to know the mutivaued magnetic potentia Φ which corresponds to the magnetic fied in the dieectric domain Indeed taking into account the expression of the gradient operator in cyindrica coordinates we obtain Φ(rθz) = I 0 2π θ r > R θ [02π] z R C Notice that the scaar potentia depends ony on the variabe θ and experiments a jump of magnitude I 0 across the cut surface Σ paced at θ = 0 Now we consider again the bounded cyinder of Figure 71 The boundary conditions added to define propery this probem are the foowing: on the exterior boundary of the dieectric domain (ie the atera surface of the cyinder Ω and the outer part of its top and bottom surfaces) we consider the condition H n = f with f being obtained from the anaytica soution: f(r D θz) = I 0 e z θ [02π] z [ L 2πR 2 ] L 2 D f ( ) rθ ± L I 0 I 1 (γr) 2 = ± 2πR C I 1 (γr C ) e r r [R C R D ] θ [02π]; on the top and bottom surfaces of the conducting cyinder Ω C we impose E ( rθ ± L 2 ) n = 0 r (0RD ) θ [02π]; which is true in this case because the eectric fied has vanishing r- and θ-components and thus it aigns with the norma vector n on these surfaces Finay we have used the foowing geometrica and physica data: R C = 1m; R D = 2m; L = 1m; σ = (Ωm) 1 ; µ = µ 0 = 4π 10 7 Hm 1 (magnetic permeabiity of free space); I 0 = 62000A; ω = 50Hz 23

24 To determine the order of convergence the numerica method has been used on severa successivey refined meshes and we have compared the obtained numerica soutions with the anaytica one Figures 72 and 73 show the coarsest meshes used for conductor and dieectric domains respectivey Z X O Y Fig 72 Coarsest mesh on the conductor domain Z X O Y Fig 73 Coarsest mesh on the dieectric domain Tabe 71 shows the H(curΩ)-norms of the approximate soutions H h computed on severa meshes and their corresponding errors The tota number of degrees of freedom (dof) for each mesh are aso incuded Tabe 71 H(curΩ)-norm of errors and exact soution Mesh-size Number dof Computed soution Error h h/ h/ h/ h/ Figure 74 shows a og-og pot of the errors measured in H(curΩ)-norm versus the number of dof for the same meshes A inear dependence on the mesh-size is obtained by cacuating the sope of the ine These O(h) errors agree with the 24

25 theoretica resuts since the soution is smooth and hence the hypotheses of Theorem 54 are fufied for r = 1 Fig 74 Error versus number of dof (og-og scae) Finay Figures 75 and 76 show the intensity of the computed magnetic fied H h in the conductor domain Ω C and the computed magnetic potentia Φ h in the dieectric domain Ω D The former is presented in a section of Ω C to show its behavior in the interior of this domain Acknowedgments The authors thank Professor Aain Bossavit for many vauabe discussions REFERENCES [1] A Aonso and A Vai Some remarks on the characterization of the space of tangentia traces of H(cur Ω) and the construction of an extension operator Manuscripta Math 89 (1996) pp [2] A domain decomposition approach for heterogeneous time-harmonic Maxwe equations Comput Methods App Mech Engrg 143 (1997) pp [3] An optima domain decomposition preconditioner for ow-frequency time-harmonic Maxwe equations Math Comp 68 (1999) pp [4] H Ammari A Buffa and J C Nédéec A justification of eddy currents mode for the Maxwe equations SIAM J App Math 60 (2000) pp [5] C Amrouche C Bernardi M Dauge and V Giraut Vector potentias in threedimensiona non-smooth domains Math Meth App Sci 21 (1998) pp [6] A Bermúdez J Buón and F Pena A finite eement method for the thermoeectrica modeing of eectrodes Comm Numer Methods Engrg 14 (1998) pp [7] A Bermúdez M C Muñiz F Pena and J Buón Numerica computation of the eectromagnetic fied in the eectrodes of a three-phase arc furnace Internat J Numer Methods Engrg 46 (1999) pp [8] A Bossavit Magnetostatic probems in mutipy connected regions: some properties of the cur operator IEE Proc 135 Pt A (1988) pp [9] Whitney forms: a cass of finite eements for three-dimensiona computations in eectromagnetism IEE Proc 135 Pt A (1988) pp