Numerical analysis of a finite-element method for the axisymmetric eddy current model of an induction furnace

Numerical analysis of a finite element method for the axisymmetric eddycurrent model of an induction furnace

ALFREDO BERMUDEZ∗Departamento de Matematica Aplicada, Universidade de Santiago de Compostela, 15706,

Santiago de Compostela, Spain

CARLOS REALES†, RODOLFO RODRIGUEZ‡

Departamento de Ingenierıa Matematica, Universidad de Concepcion, Casilla 160-C,Concepcion, Chile.

AND

PILAR SALGADO§

Departamento de Matematica Aplicada, Escola Politecnica Superior, Universidade deSantiago de Compostela, 27002, Lugo, Spain

The aim of this paper is to analyze a finite element method to solve an eddy current problem arisingfrom the modeling of an induction furnace. By taking advantage of the cylindrical symmetry, the three-dimensional problem reduces to a two-dimensional one on a meridional section, provided the currentdensity, written in cylindrical coordinates, has only azimuthal component.A mixed formulation in ap-propriate weighted Sobolev spaces is given. Existence and uniquenessof solution is proved by analyzingan equivalent weak formulation. Moreover, additional regularity is proved under suitable assumptionson the physical coefficients. The problem is discretized by standard finiteelements anda priori errorestimates are proved. Finally, some numerical experiments which allow assessing the performance of themethod are reported.

Keywords: low-frequency harmonic Maxwell equations, eddy current problems, finite element computa-tional electromagnetism, axisymmetric problems.

1. Introduction

An induction heating system consists basically of one or several inductors and metallic workpiecesto be heated. The inductors are supplied with alternating current which induces eddy currents insidethe component being heated due to Faraday’s law. This technique is widely used in the metallurgicalindustry in an important number of applications such as metal smelting, preheating for operations ofwelding, purification systems and, in general, processes needing a high speed of heating in particularzones of a piece of a conductive material. The overall process is highly complex and involves differentphysical phenomena: electromagnetics, heat transfer withphase change and hydrodynamics in the liquidmetal.

∗Email: [email protected]†Email: [email protected]. Permanent address: Departamento de Matematica y Estadıstica, Universidad de Cordoba,

Monterıa, Colombia.‡Email: [email protected]§Email: [email protected]

2 of 22 A. BERMUDEZ, C. REALES, R. RODRIGUEZ & P. SALGADO

Cylindrical symmetry allows reducing very often the original three-dimensional problem to a two-dimensional one. This approach has been followed in some recent papers (Bermudezet al. (2007a,b,2008)), where numerical tools for solving this kind of problems have been proposed and tested. Theaim of this paper is to provide a rigorous mathematical analysis of the finite element method used tosolve the underlying electromagnetic model: an eddy current problem in a two-dimensional meridionaldomain.

There exist several references dealing with the mathematical and numerical analysis of axisymmetricproblems. For instance, the strategy of reducing the dimension in finite element methods was usedfor the axisymmetric Laplace and Stokes equations in Mercier & Raugel (1982) and Belhachmietal. (2002), respectively. The time-dependent and static Maxwell equations in axisymmetric singulardomains were studied in Assouset al. (2002, 2003) by introducing a method based on a splitting ofthe space of solutions into a regular subspace and a singularone. In Lacoste (2000), a method wasintroduced to solve a time-harmonic Maxwell equation in an axisymmetric domain using a Fourierdecomposition. Fourier decomposition in axisymmetric problems was used in Mercier & Raugel (1982)for the Laplace equation, too.

We consider a formulation of the eddy current problem arising from the modeling of an inductionfurnace, which is based on introducing a vector potential for the magnetic field. This vector potentialis shown to have only azimuthal component in meridional coordinates. We introduce suitable weightedSobolev spaces in this two-dimensional setting and consider a mixed formulation, whose solution is themagnetic vector potential and Lagrange multipliers which are constant on each connected component ofthe two-dimensional section of the inductor. To prove well-posedness, we also introduce an equivalentdirect formulation. Then, the existence and uniqueness of the solution of this problem follows from theLax-Milgram lemma.

We discretize the mixed formulation by using piecewise linear finite elements on triangular meshes.We study the convergence of the method by introducing an equivalent direct discrete problem, too.Due to Cea’s lemma, this study reduces to have a suitable interpolation operator. A Clement operatorintroduced in Belhachmiet al. (2002) is used in the general case. A regularity result of thesolution isproved when the magnetic permeability is constant in the whole domain. This allows using a Lagrangeinterpolant and to prove optimal order error estimates in such a case. Moreover, a duality argumentallows improving the order of convergence for the current density, which is typically the variable ofmain interest.

The outline of this paper is as follows: In Section 2, we introduce the eddy current problem ininduction furnaces and the geometric assumptions. Then, wederive a vector potential formulationunder axisymmetric assumptions and introduce adequate boundary conditions. In Section 3, we recallthe definitions of some weighted Sobolev spaces and some of their properties. This allows us to obtain,in Section 4, equivalent variational formulations in mixedand direct forms of the problem. We provethat the problem has a unique solution. At the end of the section, we prove an additional regularityresult. In Section 5, we introduce the finite element method and prove the error estimates. Finally, inSection 6, we report some numerical tests which allow us to asses the performance of the proposedmethod.

2. Statement of the problem

We consider an induction furnace consisting of an inductioncoil surrounding a workpiece, as sketchedin Figure 1. The workpiece consists of a crucible containingthe metal to be heated. The current flowingthrough the coil produces an electromagnetic field. This, inturn, induces eddy currents in the workpiece

FEM for an axisymmetric eddy current model 3 of 22

which, due to the Joule effect, produce heat that melts the metal. The domain of the problem is inprinciple the whole space; however, for computational purposes, we will take an artificial boundeddomainΩ “sufficiently large” and suitable conditions on its boundary.

FIG. 1. Sketch of the induction furnace.

To take advantage of the symmetry of the problem, we will use acylindrical coordinate system(r,θ ,z). Accordingly, the artificial domainΩ will be chosen as a cylinder of radiusR and heightL. Wedenote byeeer , eeeθ andeeez the unit vectors of the local orthonormal basis corresponding to this coordinatesystem. We assume cylindrical symmetry, i.e., we suppose that no field depends on the angular variableθ . We denote byΩ := (0,R)× (0,L) a meridional section (θ = constant) ofΩ . The boundary ofΩconsists of the union ofΓN , ΓR andΓD , as shown in Figure 2:ΓD lies on the revolution axis,ΓR is parallelto this axis andΓN is perpendicular. We denote byΩ0 the section of the workpiece to be heated and byΩ1, . . . ,Ωm the sections of the turns of the coil (see Figure 2). We assumeΩ 0, . . . ,Ωm are connectedand mutually disjoint and denoteΩc := Ω0∪Ω1∪ ·· · ∪Ωm the section of the domain occupied by allthe conductors and byΩa := Ω \Ω c that of the air around the conductors.

Symmetry axis

ΓN

Ω0

ΓN

ΓDΓR

L

R

z

r

Ωa

Ωm

Ω1

FIG. 2. Sketch of the domainΩ .


Eddy currents are usually modeled by the low-frequency harmonic Maxwell equations. We will usestandard notation in electromagnetism:

• EEE is the electric field,• BBB is the magnetic induction,• HHH is the magnetic field,• JJJ is the current density,• ρ is the electric charge density,• µ is the magnetic permeability,• ε is the electric permittivity,• σ is the electric conductivity.

We use boldface letters to denote vector fields and variables, as well as vector-valued operators, through-out the paper.

In the low-frequency harmonic regime, the electric displacement can be neglected in Ampere’s law,leading to the so-called eddy current model:

curl HHH = JJJ, (2.1)

iωBBB+curl EEE = 000, (2.2)

divBBB = 0, (2.3)

divDDD = ρ. (2.4)

The system (2.1)–(2.4) above needs to be completed by the constitutive relations

BBB = µHHH, (2.5)

DDD = εEEE, (2.6)

and the Ohm’s lawJJJ = σEEE. (2.7)

The electric conductivity satisfies

0 < σ 6 σ 6 σ in conductors, (2.8)

σ ≡ 0 in air, (2.9)

whereas the other physical parameters are bounded above andbelow:

0 < µ 6 µ 6 µ , (2.10)

0 < ε 6 ε 6 ε. (2.11)

We notice that, sinceω 6= 0, equation (2.3) follows from (2.2). As will be shown below,equations (2.1)and (2.2) can be solved independently of (2.4) leading toHHH in the whole domain andJJJ in conductors.

In (Assouset al., 2002, Proposition 2.2), it was shown that the eddy current equations in cylindricalcoordinates lead to two decoupled problems, one for the azimuthal component (eeeθ ) of JJJ and the otherfor the meridional component (eeer , eeez). In our case, the induction furnace has been modeled in Bermudezet al. (2007a) by assuming that all the physical quantities are independent of the angular coordinateθand that the current density field has only azimuthal non-zero component, i.e,

JJJ(r,θ ,z) = Jθ (r,z)eeeθ . (2.12)


Given a vector fieldFFF = Fr(r,θ ,z)eeer + Fθ (r,θ ,z)eeeθ + Fz(r,θ ,z)eeez and a scalar fieldf = f (r,θ ,z),we recall that

curl FFF =

(1r

∂Fz

∂θ− ∂Fθ

∂z

)eeer +

(∂Fr

∂z− ∂Fz

∂ r

)eeeθ +

(1r

∂ (rFθ )

∂ r− 1

r∂Fr

∂θ

)eeez, (2.13)

divFFF =1r

∂ (rFr)

∂ r+

1r

∂Fθ∂θ

+∂Fz

∂z, (2.14)

∇ f =∂ f∂ r

eeer +1r

∂ f∂θ

eeeθ +∂ f∂z

eeez. (2.15)

Notice that from the assumption thatHHH does not depend onθ , (2.1), (2.13) and (2.12) lead to

−∂Hθ∂z

=1r

∂ (rHθ )

∂ r= 0,

which in its turn implies thatrHθ has to be constant inΩ . Now, if HHH ∈ L2(Ω)3, thenHθ eeeθ ∈ L2(Ω)3,too. However,rHθ being constant, this could happen only if this constant is zero. Therefore,Hθ has tovanish and, from (2.5),Bθ will vanish as well. Moreover, from (2.7), (2.8) and (2.12),Er andEz alsovanish in conductors. Therefore, we have

HHH(r,θ ,z) = Hr(r,z)eeer +Hz(r,z)eeez, (2.16)

BBB(r,θ ,z) = Br(r,z)eeer +Bz(r,z)eeez, (2.17)

EEE(r,θ ,z) = Eθ (r,z)eeeθ (in conductors). (2.18)

SinceBBB is divergence-free (cf. (2.3)), there exists a so-called magnetic vector potentialAAA such thatBBB= curl AAA. For the sake of uniqueness, we takeAAA to be divergence-free, too, and satisfyingAAA·nnn= 0 on∂Ω . Thus, we have

curl AAA = BBB in Ω , (2.19)

divAAA = 0 in Ω , (2.20)

AAA·nnn = 0 on∂Ω . (2.21)

From (2.19), (2.17) and (2.13), we obtain

∂Ar

∂z− ∂Az

∂ r= 0 in Ω .

Therefore, sinceΩ is simply connected, there existsϕ ∈ H1(Ω) such thatAr = ∂ϕ∂ r andAz = ∂ϕ

∂z . Onthe other hand, from (2.20), (2.14) and (2.15),

0 = divAAA =1r

∂ (rAr)

∂ r+

∂Az

∂z= ∆ϕ in Ω ,

whereϕ(r,θ ,z) := ϕ(r,z). Thus, we have

∆ϕ = 0 in Ω ,

∂ ϕ∂nnn

= 0 on∂Ω


∂ ϕ∂nnn = ∂ϕ

∂ r = AAAr = AAA·nnn = 0

nnn = eeez

nnn = eeer

∂ ϕ∂nnn = ∂ϕ

∂z = AAAz = AAA ·nnn = 0

FIG. 3. Boundary conditions forϕ in the domainΩ .

(for the deduction of the boundary condition, see Figure 3).Henceϕ is constant and, consequently,Ar = Az = 0. Therefore, we conclude that

AAA(r,θ ,z) = Aθ (r,z)eeeθ in Ω

and, hence, from (2.19) and (2.13),

Br(r,z) = −∂Aθ∂z

and Bz(r,z) =1r

∂ (rAθ )

∂ rin Ω . (2.22)

On the other hand, taking into account again (2.19), we deduce from (2.2) and (2.7) that

curl((

iωAθ +σ−1Jθ)

eeeθ)

= 000 (in conductors),

from which it follows from (2.13) that

∂∂z

(iωAθ +σ−1Jθ

)= 0 in Ωc,

∂∂ r

(r(iωAθ +σ−1Jθ

))= 0 in Ωc.

Hence, we deduce that there exist constantsVk ∈ C, k = 0, . . . ,m, such that

iωAθ +σ−1Jθ =Vk

rin Ωk (2.23)

(recall thatΩk are the connected components ofΩc).Next, from (2.1), (2.5), (2.17), (2.22) and (2.12),

curl(− 1

µ∂Aθ∂z

eeer +1

µr∂ (rAθ )

∂ reeez

)= Jθ eeeθ .

Thus, taking into account (2.13) and (2.23), we obtain fork = 0, . . . ,m,

−(

∂∂ r

(1

µr∂ (rAθ )

∂ r

)+

∂∂z

(1µ

∂Aθ∂z

))+ iωσAθ =

σr

Vk in Ωk, (2.24)


whereas using thatJθ vanishes outside the conductors (cf. (2.7), (2.9) and (2.12)),

−(

∂∂ r

(1

µr∂ (rAθ )

∂ r

)+

∂∂z

(1µ

∂Aθ∂z

))= 0 in Ωa. (2.25)

In order to solve equations (2.24) and (2.25), we assume thatthe intensities going through eachcylindrical ring are given data. Thus we add to the model the equations

∫

Ωk

Jθ dr dz= Ik, k = 1, . . . ,m,

Ik being the intensity traversingΩk. Hence, from (2.23), we have fork = 1, . . . ,m,

Vk =1dk

(Ik + iω

∫

Ωk

σAθ dr dz

), (2.26)

where

dk :=∫

Ωk

σr

dr dz.

Additional physical considerations (see Chaboudez & Clain(1997)) allow us to impose

V0 = 0. (2.27)

Notice that, as a consequence of (2.23), this condition has to hold true forAAA andEEE to belong to L2(Ω)3,whenever meas(∂Ω0∩ΓD) > 0 (as is the case for the problem sketched in Figure 2).

Equations (2.24)–(2.27) must be completed with suitable boundary conditions. Following Chaboudez& Clain (1997), we impose onΓR the Robin condition

∂ (rAθ )

∂ r+Aθ = 0 onΓR (2.28)

and onΓN the homogeneous Neumann condition

∂Aθ∂z

= 0 onΓN ; (2.29)

the latter stems from the fact that the radial component of the magnetic induction is close to zero on thisboundary. Finally, the natural symmetry condition along the revolution axis leads to

Aθ = 0 onΓD . (2.30)

3. Weighted Sobolev spaces

In this section we define appropriate weighted Sobolev spaces that will be used in the sequel and es-tablish some of their properties; the corresponding proofscan be found in Belhachmiet al. (2002);Gopalakrishnan & Pasciak (2006); Mercier & Raugel (1982); Kufner (1983). More general resultsabout weighted Sobolev spaces can be found in the last reference. To simplify the notation, we willdenote the partial derivatives by∂r and∂z.


Let Ω be a Lipschitz bounded connected two-dimensional open set.Let ΓD be, as in our example,the intersection between∂Ω and the symmetry axis (r = 0). Let L2

r (Ω) denote the weighted Lebesguespace of all measurable functionsu defined inΩ for which

‖u‖L2r (Ω) :=

∫

Ω|u|2 r dr dz< ∞.

The weighted Sobolev space Hkr (Ω) consists of all functions in L2r (Ω) whose derivatives up to orderk

are also in L2r (Ω). We define the norms and semi-norms in the standard way; in particular,

|u|2H1r (Ω) :=

∫

Ω

(|∂ru|2 + |∂zu|2

)r dr dz,

|u|2H2r (Ω) :=

∫

Ω

(|∂rr u|2 + |∂rzu|2 + |∂zzu|2

)r dr dz.

Let H1r (Ω) := H1

r (Ω)∩L21/r(Ω), where L21/r(Ω) denotes the set of all measurable functionsu de-

fined inΩ for which

‖u‖2L2

1/r (Ω) :=∫

Ω

|u|2r

dr dz< ∞.

H1r (Ω) is a Hilbert space with the norm

‖u‖H1r (Ω) :=

(‖u‖2

H1r (Ω) +‖u‖2

L21/r (Ω)

)1/2

.

Let H2r (Ω) :=

u∈ H1

r (Ω) : ‖u‖H2r (Ω) < ∞

, where

‖u‖2H2

r (Ω):= ‖u‖2

H1r (Ω)

+

∣∣∣∣1r

∂r(ru)

∣∣∣∣2

H1r (Ω)

+‖∂zu‖2H1

r (Ω).

The proof of the following three lemmas can be found in (Gopalakrishnan & Pasciak, 2006, Sec-tion 3.1).

LEMMA 3.1 The set ofC ∞(Ω) functions which vanish in a neighborhood ofΓD is dense inH1r (Ω).

LEMMA 3.2 Consider the notation of Figure 4, with 06 r0 < r1 anda ∈ [r0, r1]. For all u ∈ H1r (S),

u|γa ∈ L2(γa) and there holds

‖u‖2L2(γa)

6 ‖∂ru‖2L2

r (S) +2r1− r0

r1− r0‖u‖L2

1/r (S) .

The preceding result implies that the functions inH1r (Ω) have traces onΓD (r = 0). Moreover, since

the set of the functions inC ∞(Ω) vanishing in a neighborhood ofΓD is dense inH1r (Ω) (cf. Lemma 3.1),

the functions inH1r (Ω) have vanishing traces onΓD .

LEMMA 3.3 For allu∈ H1r (Ω), ∂r(ru) ∈ L2

1/r(Ω) and there holds

‖∂ru‖2L2

r (Ω) +‖u‖2L2

1/r (Ω) 6 ‖∂r(ru)‖2L2

1/r (Ω) 6 2‖∂ru‖2L2

r (Ω) +2‖u‖2L2

1/r (Ω) .


z

z1

z0

S

γa

Ω

r1ar0 r

FIG. 4. Sketch ofS.

The following result has been proved in (Mercier & Raugel, 1982, Theorem 4.7).

LEMMA 3.4 The injection H2r (Ω) → C 0(Ω) is continuous.

Finally, the next lemma is a variation of a result from Ern & Guermond (2004).

LEMMA 3.5 LetΩ be a Lipschitz bounded connected open set. Letf be a continuous linear functionalon H1

r (Ω) whose restriction to constant functions is not zero. Then, there existα > 0 such that

α ‖u‖H1r (Ω) 6 ‖∇u‖L2

r (Ω) + | f (u)| ∀u∈ H1r (Ω).

Proof. We repeat the steps of the proof of Lemma B.63 from Ern & Guermond (2004) withX := H1r (Ω),

Y := L2r (Ω)×C andZ := L2

r (Ω), and use that the injectionX → Z is compact due to Theorem 4.5 fromMercier & Raugel (1982).

4. Variational formulation

In this section we establish a variational formulation of problem (2.24)–(2.30) for which we will provethe existence and uniqueness of the solution. With this aim,we multiply (2.24) and (2.25) by a testfunction inH1

r (Ω), integrate by parts, use that the functions in this space have a vanishing trace onΓD ,the boundary conditions (2.28) and (2.29), (2.27) and rewrite (2.26) in a convenient way, to obtain thefollowing problem:


PROBLEM 4.1 GivenIII := (I1, . . . , Im) ∈ Cm, find (Aθ ,VVV) ∈ H1

r (Ω)×Cm such that

∫

Ω

1µ

(1r

∂ (rAθ )

∂ r1r

∂ (rZ)

∂ r+

∂Aθ∂z

∂ Z∂z

)r dr dz+

∫

ΓR

1µ

Aθ Z dz

+iω∫

Ωc

σAθ Zr dr dz−m

∑k=1

∫

Ωk

σVkZ dr dz= 0 ∀Z ∈ H1r (Ω),

m

∑k=1

∫

Ωk

σWkAθ dr dz+iω

m

∑k=1

∫

Ωk

σWkVk

rdr dz=

iω

m

∑k=1

WkIk ∀WWW ∈ Cm.

Let a be the sesquilinear form defined inH1r (Ω) by

a(Aθ ,Z) :=∫

Ω

1µ

(1r

∂ (rAθ )

∂ r1r

∂ (rZ)

∂ r+

∂Aθ∂z

∂ Z∂z

)r dr dz+

∫

ΓR

1µ

Aθ Z dz

+ iω

(∫

Ωc

σAθ Zr dr dz−m

∑k=1

1dk

∫

Ωk

σAθ dr dz∫

Ωk

σ Z dr dz

).

It is straightforward to show that Problem 4.1 is equivalentto the following one, withVk given by (2.26),k = 1, . . . ,m:

PROBLEM 4.2 GivenIII ∈ Cm, find Aθ ∈ H1

r (Ω) such that

a(Aθ ,Z) =m

∑k=1

Ikdk

∫

Ωk

σ Z dr dz ∀Z ∈ H1r (Ω).

REMARK 4.1 Problem 4.2 will be used to prove the existence and uniqueness of the solution and errorestimates, but not for the actual numerical approximation,because the term

∫Ωk

σAθ dr dz∫

Ωkσ Z dr dz

would lead to a dense matrix. In fact, for the numerical computations we will use a discretization ofProblem 4.1.

In the following lemma and thereafterC will denote a generic constant, not necessarily the same ateach occurrence.

LEMMA 4.1 There holds

‖u‖H1r (Ω) 6 C

(|u|H1

r (Ω) +‖u‖L2(ΓR)

)∀u∈ H1

r (Ω).

Proof. Let f (u) :=∫

ΓR

1µ u dz, u∈ H1

r (Ω). Because of Lemma 3.5,

α ‖u‖H1r (Ω) 6 ‖∇u‖L2

r (Ω) + | f (u)| 6 |u|H1r (Ω) +‖1/µ‖L2(ΓR) ‖u‖L2(ΓR) ∀u∈ H1

r (Ω),

which allows us to conclude the proof.

REMARK 4.2 The lemma above holds true for any Lipschitz bounded connected domainΩ and anysubsetΓR ⊂ ∂Ω \ΓD with positive measure.

LEMMA 4.2 The sesquilinear forma is H1r (Ω)-elliptic and continuous.


Proof. The ellipticity arises from the definition ofa and (2.10) as follows:

Re(a(Aθ ,Aθ )) >1µ

(‖∂r(rAθ )‖2

L21/r (Ω) +‖∂zAθ‖2

L2r (Ω) +‖Aθ‖2

L2(ΓR)

)

>1µ

(‖∂rAθ‖2

L2r (Ω) +‖Aθ‖2

L21/r (Ω) +‖∂zAθ‖2

L2r (Ω) +‖Aθ‖2

L2(ΓR)

)

> C

(‖Aθ‖2

H1r (Ω) +‖Aθ‖2

L21/r (Ω)

)

= C‖Aθ‖2H1

r (Ω),

where we have used Lemma 3.3 for the second inequality and Lemma 4.1 for the third one. The conti-nuity follows directly from Lemmas 3.2 and 3.3.

Now we are in a position to prove that Problem 4.1 is well posed. Here and thereafter‖III‖Cm :=(∑m

k=1 |Ik|2)1/2 denotes the standard Euclidean norm inCm.

THEOREM 4.1 Problem 4.1 has a unique solution which satisfies

‖Aθ‖H1r (Ω) 6 C‖III‖

Cm .

Proof. Since Problems 4.1 and 4.2 are equivalent, it is enough to show that the latter is well posed. Theright-hand side of Problem 4.2 satisfies

∣∣∣∣∣m

∑k=1

Ikdk

∫

Ωk

σ Z dr dz

∣∣∣∣∣6 C‖III‖Cm‖Z‖L2

r (Ωk).

Hence, the theorem follows from Lemma 4.2 and the Lax-Milgram Lemma.

To end this section we will prove a regularity result for the solution of Problem 4.1, valid at leastwhen the magnetic permeability is constant in the whole domain. With this aim, we will consider aslightly more general framework, which will be also used to prove a double order of convergence inL2

r (Ω) of the numerical method proposed in the following section. Consider the following auxiliaryproblem:

PROBLEM 4.3 Giveng∈ L2r (Ω), findYθ ∈ H1

r (Ω) such that

∫

Ω

1µ

(1r

∂ (rYθ )

∂ r1r

∂ (rZ)

∂ r+

∂Yθ∂z

∂ Z∂z

)r dr dz+

∫

ΓR

1µ

Yθ Z dz=∫

ΩgZr dr dz ∀Z ∈ H1

r (Ω),

LEMMA 4.3 If µ is constant inΩ , then the solution of Problem 4.3 satisfiesYθ ∈ H2r (Ω) and

‖Yθ‖H2r (Ω) 6 C‖g‖L2

r (Ω) .

Proof. The arguments in the proof of Lemma 4.2 show that the sesquilinear form on the left-hand sideof Problem 4.3 isH1

r (Ω)-elliptic, as well. Hence the problem is well posed and its solution satisfies‖Yθ‖H1

r (Ω) 6 C‖g‖L2r (Ω).

Let YYY(r,θ ,z) := Yθ (r,z)eeeθ . Using (2.13) and Lemma 3.3, it is easy to show thatcurl YYY ∈ L2(Ω)3

and‖YYY‖H(curl ,Ω)

6 C‖Yθ‖H1r (Ω) 6 C‖g‖L2

r (Ω) . (4.1)


To prove additional regularity we test Problem 4.3 withZ ∈ D(Ω). Hence, using (2.13), (2.14) andthe fact thateeeθ is orthogonal tonnn throughout the whole boundary ofΩ , we have that

curl(

1µ

curl YYY)

= geeeθ in Ω , (4.2)

divYYY = 0 in Ω , (4.3)

YYY ·nnn = 0 on∂Ω . (4.4)

Thus, from (4.1), (4.3) and (4.4), we have thatYYY ∈ H(curl ,Ω)∩H0(div,Ω). Hence, sinceΩ is convex,YYY ∈ H1(Ω)3 (cf. (Amroucheet al., 1998, Theorem 2.17)) and

‖YYY‖H1(Ω)3 6 C‖YYY‖H(curl ,Ω)6 C‖g‖L2

r (Ω) . (4.5)

Next, we prove thatcurl YYY is also in H1(Ω)3. In this case the results from Amroucheet al. (1998)cannot be directly applied, because neithercurl YYY ·nnn nor curl YYY×nnn vanish on∂Ω . This is the reasonwhy we will make a translation by using the functionΦ defined below. Let 0< r1 < r2 < R (recall thatΓR lies on the liner = R) and letϕ ∈ C ∞([0,R]) be such thatϕ(r) ≡ 0 in [0, r1] andϕ(r) ≡ 1 in [r2,R].Let

Φ(r,θ ,z) := − 1R

YYY(r,θ ,z)×ϕ(r)eeer =1R

ϕ(r)Yθ (r,z)eeez,

andΨ := curl YYY + Φ . We will show thatΨ ∈ H0(curl ,Ω)∩H(div,Ω). To prove this, we split∂Ωinto two parts,ΓR andΓN , which correspond to the Robin (ΓR) and the Neumann (ΓN) boundaries of thetwo-dimensional domainΩ , respectively. From (2.13), we have

Ψ ×nnn = (curl YYY +Φ)×eeer =1r

∂ (rYθ )

∂ reeeθ +

1R

ϕ(r)Yθ (r,z)eeeθ = 000 onΓR,

where for the last equality we have used the boundary condition

1r

∂ (rYθ )

∂ r+Yθ = 0 onΓR,

which in its turn is obtained by testing Problem 4.3 withZ ∈ C ∞(Ω) such that supp(Z)∩ (ΓD ∪ΓN) = /0.On the other hand,

Ψ ×nnn = (curl YYY +Φ)×eeez =∂Yθ∂z

eeeθ = 000 onΓN ,

where now the last equality follows by testing Problem 4.3 with Z ∈ C ∞(Ω) such that supp(Z)∩ (ΓD ∪ΓR) = /0. Therefore, by using (4.2) and the regularity ofΦ , we conclude thatΨ ∈ H0(curl ,Ω)∩H(div,Ω). HenceΨ ∈ H1(Ω)3 (cf. (Amroucheet al., 1998, Theorem 2.17), again) and

‖Ψ‖H1(Ω)3 6 ‖curl YYY‖H(curl ,Ω)+‖Φ‖H1(Ω)3 6 C‖g‖L2

r (Ω) ,

where we have used (4.1), (4.2) and (4.5) for the last inequality. Consequently,curl YYY = Ψ −Φ ∈H1(Ω)3 and

‖curl YYY‖H1(Ω)3 6 C‖g‖L2r (Ω) . (4.6)


Finally, from (Assouset al., 2003, Proposition 3.17) we have that‖YYY‖2H1(Ω)3 = 2π ‖Yθ‖2

H1r (Ω)

and

‖curl YYY‖2H1(Ω)3 = 2π ‖∂zYθ‖2

H1r (Ω)

+ 2π∥∥1

r ∂r(rYθ )∥∥2

H1r (Ω)

. Consequently, the definition of theH2r (Ω)-

norm, (4.5) and (4.6) lead to

‖Yθ‖2H2

r (Ω)6

12π

(‖curl YYY‖2

H1(Ω)3 +‖YYY‖2H1(Ω)3

)6 C‖g‖2

L2r (Ω) .

Thus, we conclude the proof.

THEOREM 4.2 If µ is constant inΩ , then the solution of Problem 4.3 satisfiesYθ ∈ H2r (Ω) and


r (Ω) .

Proof. Let J1 denote the first-order Bessel function of the first kind. Define

jm(r) :=

√2

|J2(βmR)| J1(βmr), m= 1,2, . . .

whereβm := αm/R, with αm being themth positive zero of the equation

2J1(x)+xJ1′(x) = 0,

andsn(z) :=

√2cos

(nπzL

), n = 0,1,2, . . .

Then by classical completeness results for Bessel functions (see (Gonzalez-Velasco, 1996, Sections 10.7-8) and Hochstadt (1967)), the set of functionsemn(r,z) = jm(r)sn(z), m= 1,2, . . . , n = 0,1,2, . . . , is acomplete orthogonal system of L2

r (Ω). From this fact and Lemma 4.3, the rest of the proof runs essen-tially as those of Proposition 4.1 and Theorem 4.1 from Gopalakrishnan & Pasciak (2006).

COROLLARY 4.1 If µ is constant inΩ , then the solution of Problem 4.1 satisfiesAθ ∈ H2r (Ω) and

‖Aθ‖H2r (Ω) 6 C‖III‖

Cm .

Proof. It follows from the first equation of Problem 4.1 and Theorem 4.2 applied to Problem 4.3 with

g := −iωσAθ +m

∑k=1

σVk

rχΩk,

χΩk being the characteristic function ofΩk, k = 1, . . . ,m. In its turn,‖g‖L2r (Ω) 6 C‖III‖

Cm, by virtue ofTheorem 4.1 and (2.26).

5. Finite element discretization

In this section we introduce a discretization of Problem 4.1and prove error estimates. LetThh>0 bea regular family of triangulations ofΩ with h being the mesh-size (see Ciarlet (2002)). Let us remarkthat there is no need of assuming that the meshes are compatible with the geometry of the conductordomain (i.e., that each element ofTh is contained either inΩc or in Ωa), although, of course, this kindof meshes make easier the implementation of the method. Fromnow on, the generic constantC willalways be independent of the mesh-size.


LetVh :=

uh ∈ H1

r (Ω) : uh|T ∈ P1 ∀T ∈ Th

,

with P1 being the complex-valued linear functions in the coordinatesr andz:

P1 := p(r,z) = c0 +c1r +c2z : c0,c1,c2 ∈ C .

The finite element approximation of Problem 4.1 is defined as the solution(Ahθ ,VVVh) of the following

problem:

PROBLEM 5.1 GivenIII := (I1, . . . , Im) ∈ Cm, find (Ah

θ ,VVVh) ∈ Vh×Cm such that

∫

Ω

1µ

(1r

∂ (rAhθ )

∂ r1r

∂ (rZh)

∂ r+

∂Ahθ

∂z∂ Zh

∂z

)r dr dz+

∫

ΓR

1µ

Ahθ Zh dz

+iω∫

Ωc

σAhθ Zhr dr dz−

m

∑k=1

∫

Ωk

σVhk Zh dr dz= 0 ∀Zh ∈ Vh,

m

∑k=1

∫

Ωk

σWhk Ah

θ dr dz+iω

m

∑k=1

∫

Ωk

σWhk Vh

k

rdr dz=

iω

m

∑k=1

Whk Ik ∀WWWh ∈ C

m.

It is straightforward to see that Problem 5.1 is equivalent to the following one, withVVVh :=(Vh1 , . . . ,Vh

m)given by

Vhk =

1dk

(Ik + iω

∫

Ωk

σAhθ dr dz

), k = 1, . . . ,m. (5.1)

PROBLEM 5.2 GivenIII ∈ Cm, find Ah

θ ∈ Vh such that

a(Ahθ ,Zh) =

m

∑k=1

Ikdk

∫

Ωk

σ Zh dr dz ∀Zh ∈ Vh.

We will use Problem 5.2 to prove well-posedness and error estimates. However, as stated in Re-mark 4.1, for the computer implementation of this approach we will use Problem 5.1 to avoid densematrices.

THEOREM 5.1 Problem 5.1 has a unique solution(Ahθ ,VVVh). Moreover, there exists a constantC > 0,

independent ofh, such that if(Aθ ,VVV) is the solution of Problem 4.1, then

‖Aθ −Ahθ‖H1

r (Ω) +m

∑k=1

|Vk−Vhk | 6 C inf

Zh∈Vh

‖Aθ −Zh‖H1r (Ω).

Proof. Since Problems 5.1 and 5.2 are equivalent, we use the latter for the estimate forAθ , whichfollows immediately from Cea’s lemma (see for instance Ciarlet (2002)). The estimate forVk, k =1, . . . ,m, follows from the latter, (2.26) and (5.1).

According to the theorem above, there only remains to prove thatAθ can be conveniently approxi-mated by a function inVh. With this purpose, in the most general case, we resort to a Clement operatorstable for functions inH1

r (Ω) (which, recall, vanish onΓD). Such operators have been studied forweighted Sobolev spaces in Belhachmiet al. (2002) and Mercier & Raugel (1982).

In particular, we consider the Clement operatorΠh : H1r (Ω)→Vh defined in (Belhachmiet al., 2002,

Eq. (36)). The proof of the following lemma can be found in (Belhachmiet al., 2002, Theorem 2).


LEMMA 5.1 If 1 6 l 6 2, then there exists a constantC > 0, independent ofh, such that for allu ∈Hl

r(Ω)∩ H1r (Ω),

‖u− Πhu‖H1r (Ω) 6 Chl−1‖u‖Hl

r (Ω)∩H1r (Ω) .

On the other hand, when the solutionAθ is sufficiently smooth, we are able to use the Lagrangeinterpolation operatorΠh. In fact, according to Lemma 3.4, such interpolant is well defined for functionsin H2

r (Ω). Moreover, for functions in H2r (Ω), there holds the following error estimate, whose proof canbe found in (Mercier & Raugel, 1982, Lemma 6.3).

LEMMA 5.2 There exists constantsC > 0, independent ofh, such that for allu∈ H2r (Ω),

‖Πhu‖H1r (Ω) 6 C‖u‖H2

r (Ω) and ‖u−Πhu‖H1r (Ω) 6 Ch‖u‖H2

r (Ω) .

Now we are in a position to establish the main result of this paper.

THEOREM 5.2 Let (Aθ ,VVV) be the solution of Problem 4.1 and(Ahθ ,VVVh) the solution of Problem 5.1.

There exists a constantC > 0, independent ofh, such that ifAθ ∈ H2r (Ω), then

‖Aθ −Ahθ‖H1

r (Ω) +m

∑k=1

|Vk−Vhk | 6 Ch‖Aθ‖H2

r (Ω) .

Proof. It follows from Theorem 5.1 and Lemma 5.2.

The solution(Ahθ ,VVVh) of Problem 5.1 allows us to compute the three-dimensional electromagnetic

quantities. In fact, recalling (2.17) and (2.22), we define the computed magnetic induction by

BBBh := −∂Ahθ

∂zeeer +

1r

∂ (rAhθ )

∂ reeez.

Analogously, from (2.12) and (2.23), the computed current density is given by

JJJh := Jhθ eeeθ ,

with Jhθ vanishing in the dielectric and defined in the conductors as follows:

Jhθ

∣∣∣Ωk

:= σ

(Vh

k

r− iωAh

θ

), k = 0,1, . . . ,m,

whereVh0 := V0 = 0 (cf. (2.27)). Notice that, in particular, the current density in the workpiece to be

heated, which is typically the quantity of main interest, isgiven byJJJh = −iσωAhθ eeeθ .

In what follows we obtain error estimates for these three-dimensional quantities.

COROLLARY 5.1 Under the same hypotheses as in Theorem 5.2, we have

‖BBB−BBBh‖L2(Ω)3 6 Ch‖Aθ‖H2r (Ω) .

Proof. Recalling (2.17) and (2.22), we have

‖BBB−BBBh‖2L2(Ω)3 = ‖∂z(Aθ −Ah

θ )‖2L2

r (Ω) +‖∂r(r (Aθ −Ahθ ))‖2

L21/r (Ω)

6 ‖∂z(Aθ −Ahθ )‖2

L2r (Ω) +2‖∂r(Aθ −Ah

θ )‖2L2

r (Ω) +2‖Aθ −Ahθ‖2

L21/r (Ω)

6 C‖Aθ −Ahθ‖2

H1r (Ω)

6 Ch‖Aθ‖2H2

r (Ω) ,

where we have used Lemma 3.3 for the first inequality and Theorem 5.2 for the last one.


COROLLARY 5.2 Under the same hypotheses as in Theorem 5.2, if for allg∈ L2r (Ω) the solutionYθ of

Problem 4.3 satisfies‖Yθ‖H2r (Ω) 6 C‖g‖L2

r (Ω), then

‖JJJ−JJJh‖L2(Ω)3 6 Ch2‖Aθ‖H2r (Ω) .

Proof. SinceJJJ andJJJh vanish in the dielectric, we have

‖JJJ−JJJh‖2L2(Ω)3 = 2π

m

∑k=0

‖Jθ −Jhθ‖2

L2r (Ωk)

.

Now, from (2.23) and the definition ofJhθ , we have

‖Jθ −Jhθ‖L2

r (Ωk)6 C

(|Vk−Vh

k |+‖Aθ −Ahθ‖L2

r (Ωk)

), k = 0,1, . . . ,m

(recallVh0 = V0 = 0).

In what follows, we will use a duality argument to estimate∥∥Aθ −Ah

θ∥∥

L2r (Ω)

. With this end, for each

f ∈ L2r (Ω), letYθ ∈ H1

r (Ω) denote the solution of the problem

a(Z,Yθ ) =

∫

ΩZ f r dr dz ∀Z ∈ H1

r (Ω).

Because of Lemma 4.2 and the Lax-Milgram Lemma, this problemhas a unique solution, which satisfies

‖Yθ‖H1r (Ω) 6 C‖ f‖L2

r (Ω) .

Moreover, proceeding as was done to prove the equivalence ofProblems 4.1 and 4.2, we have thatYθalso solves Problem 4.3 with

g := f − iωσYθ +m

∑k=1

σWk

rχΩk,

where

Wk :=iωdk

∫

Ωk

σYθ dr dz, k = 1, . . . ,m.

Therefore, according to the hypothesis of this corollary, the expression ofg and the estimate forYθabove, there holds


r (Ω) 6 C‖ f‖L2r (Ω) . (5.2)

Now we proceed with the duality argument:

‖Aθ −Ahθ‖L2

r (Ω) = supf∈L2

r (Ω)

∣∣∣∣∫

Ω(Aθ −Ah

θ ) f r dr dz

∣∣∣∣‖ f‖L2

r (Ω)

= supf∈L2

r (Ω)

∣∣a(Aθ −Ahθ ,Yθ )

∣∣‖ f‖L2

r (Ω)

= supf∈L2

r (Ω)

∣∣a(Aθ −Ahθ ,Yθ −ΠhYθ )

∣∣‖ f‖L2

r (Ω)

6 supf∈L2

r (Ω)

Ch‖Aθ‖H2r (Ω) h‖Yθ‖H2

r (Ω)

‖ f‖L2r (Ω)

6 Ch2‖Aθ‖H2r (Ω) ,


where we have used the Galerkin orthogonality, Theorem 5.2,Lemma 5.2 and the estimate (5.2).On the other hand, to estimate

∣∣Vk−Vhk

∣∣, k = 1, . . . ,m, we use (2.26), (5.1) and the estimate above,to write

|Vk−Vhk | =

∣∣∣∣iωdk

∫

Ωk

σ(Aθ −Ahθ ) dr dz

∣∣∣∣6 C‖Aθ −Ahθ‖L2

r (Ωk)6 Ch2‖Aθ‖H2

r (Ω) .

Thus, we conclude the proof.

REMARK 5.1 As shown in the proof above, under the assumptions of thiscorollary, the computedconstantsVh

k also converge quadratically.

COROLLARY 5.3 If µ is constant inΩ , then

‖BBB−BBBh‖L2(Ω)3 6 Ch‖III‖Cm and ‖JJJh−JJJ‖L2(Ω)3 6 Ch2‖III‖

Cm .

Proof. It is a direct consequence of Corollaries 4.1, 5.1 and 5.2 andTheorem 4.2.

6. Numerical experiments

The numerical method analyzed above has been implemented ina FORTRAN code; several numericaltests have been already reported in Bermudezet al.(2007a). In this section, we will apply this code to acouple of problems, to assess the orders of convergence proved for the physical quantitiesBBB andJJJ. First,we will consider a problem with known analytical solution, which does not fit exactly in the theoreticalframework considered in the previous sections. We will alsoapply the code to another problem lying inthis framework: the simulation of an industrial furnace. Since no analytical solution is available in thiscase, we will assess the orders of convergence by comparing the obtained results with those obtainedwith the same method on extremely refined meshes.

6.1 Test with analytical solution

Let us consider an infinite cylinder consisting of a core metal surrounded by a crucible and an extremelythin coil. The multi-turn coil is modeled as a continuous single coil with a uniform surface currentdensity (see Figure 5). The solution of the electromagneticproblem can be obtained in the whole space,even for an axisymmetric crucible composed by different materials, provided the physical properties areconstants in each material. We refer the reader to the appendix from Bermudezet al.(2007a) for furtherdetails.

In particular, for the problem described in Figure 5, the azimuthal component of the vector potentialreads as follows:

Aθ (r,z) =

α1 I1(r√

iωµσ), 0 < r < R1,

α2 I1(r√

iωµσ)+β1K1(r√

iωµσ), R1 < r < R2,

α3µ0r2

+β2

r, R2 < r < R3,

βext

r, r > R3,


FIG. 5. Sketch of the domain for the analytical test.

with I1 and K1 being the first-order modified Bessel functions of the first and second kind, respectively.The coefficientsµ andσ are constant in each material and the constantsα1, α2, α3, β1, β2 andβext arechosen so thatAθ and 1

µr∂ (rAθ )

∂ r are continuous atr = R1, r = R2 andr = R3.We denote byRext andHext the width and height of the rectangular box enclosing the domain for the

finite element computations (see Figure 5, again). We remarkthat, in this case, due to the infinite heightof the domain, the Robin condition (2.28) is not valid. For validation purposes, we will use Dirichletboundary conditions atr = Rext, as well as atr = 0, and homogeneous Neumann conditions on thehorizontal edges.

The geometrical data and physical parameters used for this problem are displayed in Table 1.

Table 1. Analytical test. Geometrical data and physical parameters

Inner radius of crucible (R1): 0.05 mOuter radius of crucible (R2): 0.07 mRadius of the induction coil (R3): 0.09 mRext: 0.2 mHext: 0.1 mFrequency: 3700 HzRMS intensity/unit of length: 30460 Am−1

Electrical conductivity of metal: 1234568 (Ohm m)−1

Electrical conductivity of crucible: 240000 (Ohm m)−1

Magnetic permeability of all materials: 4π10−7 Hm−1

The numerical method has been used on several successively refined meshes and the obtained nu-merical approximations have been compared with the analytical solution. Figure 6 shows log-log plots


of the errors in L2(Ω)3-norm for the computed current densityJJJ and the magnetic inductionBBB versusthe number of degrees of freedom (d.o.f.). We can observe a quadratic dependence on the mesh-sizehfor JJJ and a linear dependence forBBB, which agree with the theoretically predicted orders of convergence.

102

103

104

105

10−2

10−1

100

101

102

Rel

ativ

e er

ror

(%)

Number of d.o.f.

Relative error (%)

y=Ch2

102

103

104

105

10−2

10−1

100

101

102

Rel

ativ

e er

ror

(%)

Number of d.o.f.

Relative error (%)y=Ch

FIG. 6. Analytical test. Error in L2(Ω)3-norm forJJJ (left) andBBB (right) versus number of d.o.f. (log-log scale).

6.2 Simulation of an industrial furnace

Our next goal is to study the convergence behavior of the method applied to a problem lying in theframework of the theoretical results. With this aim, we haveconsidered the simulation of an industrialproblem: a small furnace composed by a graphite crucible containing silicon in its interior and a 4-turnscoil (see Figure 7). The geometrical and physical data are displayed in Table 2.

Since in this case there is no analytical solution to comparewith, we have used as a reference so-lution one computed with the same finite element method over an extremely fine mesh. The numericalapproximationsJJJh andBBBh obtained with several successively refined meshes have beencompared withthe reference one. Figure 8 shows the coarsest used mesh. Figure 9 shows log-log plots of the cor-responding errors measured in L2(Ω)3-norm versus the number of degrees of freedom for differentmeshes. Notice that a quadratic dependence forJJJ and a linear dependence forBBB can be observed again.

Finally, we have also compared the computed constantsVh1 , . . . ,Vh

4 with those corresponding to thereference solution. Figure 10 shows log-log plots of the errors for each constant versus the number ofdegrees of freedom for the different meshes. In this case, a quadratic order of convergence can be clearlyappreciated, in agreement with Remark 5.1.

Funding

PGIDIT06PXIB207052PR, Xunta de Galicia (Spain) to A.B., R.R. and P.S, Consolider MATHEMAT-ICA. CSD2006-00032, Ministerio de Educacion y Ciencia (Spain) to A.B., MECESUP UCO0406


FIG. 7. Sketch of the geometry of the industrial furnace.

(Chile) to C.R., FONDAP in Applied Mathematics (Chile) to R.R.

REFERENCES

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ASSOUS, F., CIARLET JR., P. & LABRUNIE, S. (2002) Theoretical tools to solve the axisymmetric Maxwellequations,Math. Methods Appl. Sci.25, 49–78.

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BERMUDEZ, A., GOMEZ, D., MUNIZ , M.C. & SALGADO , P. (2007b) A FEM/BEM for axisymmetric electro-magnetic and thermal modelling of induction furnaces,Internat. J. Numer. Methods Engrg.71, 856–878.

BERMUDEZ, A., GOMEZ, D., MUNIZ , M.C., SALGADO , P. & VAZQUEZ, R. (2008) Numerical simulation of athermo-electromagneto-hydrodynamic problem in an induction heating furnace (submitted).

CIARLET, P. (2002)The Finite Element Method for Elliptic Problems.New York: SIAM.CHABOUDEZ, C. & CLAIN , S. (1997) Numerical modeling in induction heating for axisymmetric geometries,

IEEE Trans. Magn.33, 739–745.


Table 2. Industrial furnace. Geometrical data and physical parameters

Height of silicon (A): 0.046 mInner radius of crucible (B): 0.021 mOuter radius of crucible (C): 0.025 mCrucible height (D): 0.08 mCrucible width (E): 0.004 mCoil diameter (F): 0.005 mCoil height (G): 0.02 mDistance coil-crucible (K): 0.035 mDistance between coil turns (L): 0.006 mVertical distance from crucible to the bottom (V): 0.5 mVertical distance from silicon to the top (W): 0.45 mWidth of the rectangular box (R): 0.5 mNumber of coil turns: 4Frequency: 3700 HzRMS coil current (in each turn): 3000 AElectrical conductivity of silicon: 1234568 (Ohm m)−1

Electrical conductivity of crucible (graphite): 240000 (Ohm m)−1

Electrical conductivity of coil (copper): 2×107 (Ohm m)−1

Magnetic permeability of all materials: 4π10−7 Hm−1

FIG. 8. Initial mesh of the industrial furnace: global view (left) and detail near the workpiece (right).

ERN, A. & GUERMOND, J. (2004)Theory and Practice of Finite Elements.New York: Springer.


102

103

104

105

10−2

10−1

100

101

102

Rel

ativ

e er

ror

(%)

Number of d.o.f.

Relative error (%)

y=Ch2

102

103

104

105

100

101

102

Rel

ativ

e er

ror

(%)

Number of d.o.f.

Relative error (%)y=Ch

FIG. 9. Industrial furnace. Error in L2(Ω)3-norm forJJJ (left) andBBB (right) versus number of d.o.f. (log-log scale).

102

103

104

105

10−2

10−1

100

101

102

Rel

ativ

e er

ror

(%)

Number of d.o.f.

Relative error (%)−V1




y=Ch2

FIG. 10. Industrial furnace. Error versus number of d.o.f. (log-log scale) forV1, V2, V3 andV4.

GONZALEZ-VELASCO, E.A. (1996)Fourier Analysis and Boundary Value Problems.San Diego, CA: AcademicPress.

GOPALAKRISHNAN, J. & PASCIAK , J. (2006) The convergence of V-cycle multigrid algorithms for axisymmetricLaplace and Maxwell equations,Math. Comp.75, 1697–1719.

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by use ofH(rot) conforming finite element,Numer. Math.84, 577–609.MERCIER, B. & RAUGEL, G. (1982) Resolution d’un probleme aux limites dans un ouvert axisymetrique par

elements finis enr, zet series de Fourier enθ , RAIRO, Anal. Numer. 16, 405–461.

Numerical analysis of a finite-element method for the axisymmetric eddy current model of an induction furnace

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