numerical simulation and optimization of industrial problems€¦ · mathematical modeling in...

Mathematical modeling in electromagnetism

Numerical Simulation and Optimization of Industrial Problems

Alfredo Bermúdez de Castro

Departamento de Matemática Aplicada, Universidade de Santiago de Compostela. Spain

Doctoral Intensive Week in PDEs and ApplicationsUniversidad Complutense de Madrid

October 21–25, 2013

Alfredo Bermúdez de Castro Numerical Simulation and Optimization of Industrial Problems


Contents

1 Mathematical modeling in electromagnetismMaxwell equationsThe eddy currents model

2 Numerical simulation of induction furnaces for silicon metallurgyProblem descriptionMathematical models. CouplingsNumerical methods

3 Electromagnetic and thermal analysis of electric motorsElectromagnetic modelHeat transfer: Galerkin lumped parameter modelsNumerical results

4 Mathematical models for gas transport networksStatement of the problem. Gas transportation networksMathematical modelling: from PDE to algebraic models. estacionario



Contenidos (cont.)

5 Optimization of gas networksOptimization problemsNumerical algorithmsApplications to the the Spanish network



Contents

1 Introduction. Some industrial applications

2 Maxwell equations in the empty space

3 Maxwell equations in material media

4 Electrostatics

5 Direct current

6 Magnetostatics

7 Eddy currents (Foucault currents)


Silicon: production and applications

Silicon is produced in electric arc furnaces by reduction of silicondioxide with carbon :

Si O2 + C = Si + C O2

CIRM. February 9, 2009. – p. 4/68

Silicon: production and applications

Silicon is produced in electric arc furnaces by reduction of silicondioxide with carbon :

Si O2 + C = Si + C O2

Applications of silicon:

Ferrosilicon (silicon steels, can contain more than 2% of other materials)

Metallurgical silicon (e.g. silicon-aluminum alloys, contains about 1% ofother elements)

Chemical silicon (silicones)

Solar silicon (solar cells)

Electronic silicon (semiconductors, the purest silicon, "9N” = 99.9999999of purity)

CIRM. February 9, 2009. – p. 4/68

The reduction furnace

CIRM. February 9, 2009. – p. 9/68

The electrodes

Main components of the reduction furnace.

Electric current enters each electrode through contact clamps.

The current flows down through the column generating heat by Jouleeffect.

Electric arc at the tip of each electrode.

CIRM. February 9, 2009. – p. 11/68

The electrodes

Main components of the reduction furnace.

Electric current enters each electrode through contact clamps.

The current flows down through the column generating heat by Jouleeffect.

Electric arc at the tip of each electrode.

Classical electrodes:

Pure graphite: steel production.

Prebaked: silicon metal production.

Søderberg(“self-baked” electrodes): ferrosilicon production.

CIRM. February 9, 2009. – p. 11/68

The ELSA electrode

Graphite coreMotion system

Casing

Clamps

Pre-baked paste

Liquid paste

Solid paste

Nipple

Supportsystem

CIRM. February 9, 2009. – p. 12/68

The ELSA electrode

Compound electrode: graphite/paste.

Suitable for metallurgical silicon production.

Cheaper than prebaked electrodes.

Paste baking is a crucial point: the liquid-solid interphase has to becontrolled in order to prevent the liquid paste to fall down in the mixture

CIRM. February 9, 2009. – p. 13/68

Numerical results

Modulus the current density, |Jh|, in the conductors.

CIRM. February 9, 2009. – p. 32/68

Numerical results

Magnetic potential Φ̃h in the dielectric.

CIRM. February 9, 2009. – p. 33/68

Numerical results

|Jh|: Horizontal section of one of the electrodes.

CIRM. February 9, 2009. – p. 34/68

Numerical results

|Jh|: Vertical section of one of the electrodes.

CIRM. February 9, 2009. – p. 35/68

Numerical simulation of induction furnaces

Photographs taken from http://www.ameritherm.com

CIRM. February 9, 2009. – p. 36/68

An industrial induction furnace

Silicon for melting and purification

Graphite crucible

Refractory layers

Water-cooled coil

CIRM. February 9, 2009. – p. 37/68

Mathematical model

Coupled problems

Thermal modelElectromagnetic model

Hydrodynamic Model


Material properties

Ohm’s law

Hydrodynamic Model

CIRM. February 9, 2009. – p. 38/68

Mathematical model

Electromagnetic model


Material properties

Ohm’s law

Hydrodynamic Model

curlH = J,

iωB + curlE = 0,

div B = 0,

B = µ(T )H,

J = σ(T )(E + v × B).

CIRM. February 9, 2009. – p. 38/68

Mathematical model

Thermal model


Joule effect

Material properties

Ohm’s law

Hydrodynamic Model

Convection

∂e

∂t+ v · grad e− div (k gradT ) =

|J|2

2σ

CIRM. February 9, 2009. – p. 39/68

Mathematical model

Hydrodynamic model


Joule effect

Material properties

Ohm’s law

Lorentz forceLiquid regionMaterial propertiesBuoyancy forces

Hydrodynamic Model

Convection

ρ(T )v̇ − div (η(T )(gradv + (gradv)t)) + grad p = f(T,H)

div v = 0

CIRM. February 9, 2009. – p. 40/68

Temperature, enthalpy and current density evolution

CIRM. February 9, 2009. – p. 66/68

Enthalpy and velocity field in silicon

CIRM. February 9, 2009. – p. 67/68

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Maxwell equations in free space

1. Charges and currents

Electromagnetic field theory is a discipline concerned with the study of

charges, at rest and in motion, producing currents and electric and

magnetic fields.

Charges are represented by density scalar fields. These densities can be

volume, surface or line densities. From a mathematical point of view

they will be distributions supported on a volume, a surface or a line,

respectively. We will also consider point charges that will be

represented by Dirac measures.

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Suppose there are electric charges of density ρ(x, t) in a region and the

charge at point x moves with velocity v(x, t) at time t. Then we define

the current density field J by

J(x, t) = ρ(x, t)v(x, t). (1)

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Let us consider a surface S. The surface integral

IS =

∫

S

J · ndA (2)

represents the charge flux through the surface S which is also called

the intensity through S.

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In the MKS system charge is measured in Coulomb (C) and current

density in C/sm2 = A/m2, where Coulomb over second is Ampère (A).

Hence, the MKS unit for the current flux or intensity is Ampère.

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2. Electric and magnetic fields

Electric and magnetic fields are fundamentally fields of force that

originate from electric charges.

Electric charges at rest relative to an observation point give rise to an

electric field there, which is static (time independent).

The relative motion of charges provides an additional force called

magnetic force. If charges are moving at constant velocities relative to

the observation point, this added magnetic field is static (time

independent).

Accelerated motions produce both time-varying electric and magnetic

fields.

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The symbol for the electric field intensity is the vector E. Its unit is

force per unit charge, in MKS system Newton over Coulomb (N/C).

In its turn, the magnetic field is represented by vector B called the

magnetic flux density.

Its unit is Weber per square meter (Wb/m2) or Tesla (T ).

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The presence of E and B at a point in space may be detected

physically by means of a charge Q moving at velocity v.

The force acting on that charge is given by the Lorentz force law,

F = Q(E+ v ×B). (3)

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In the case of a charge distribution with (volumetric) charge density ρv,

the density of force of the electromagnetic field is

f = ρv(E+ v ×B) = ρvE+ J×B. (4)

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3. Maxwell integral equations in free space

The connection of the electric and magnetic fields to their charge and

current sources is provided by Maxwell equations. Their integral forms

in free space are

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Γ = ∂Ω∫Γ ε0E · ndA = QΩ, (C) (5)

Γ = ∂Ω∫ΓB · ndA = 0, (Wb) (6)

l border of S

∫

l

E · τdl = −d

dt

∫

S

B · ndA, (V ) (7)

l border of S

∫

l

B

µ0· τdl = IS +

d

dt

∫

S

ε0E · ndA, (A), (8)

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where QΩ and IS denotes, respectively, the total charge in volume Ω

and the total current flux through surface S, and V (volts) is Joule

over Coulomb.

Equation (5) and (6) are the Gauss laws for electric charge and

magnetic field, respectively.

Equation (7) is Faraday’s law and equation (8) is Ampére’s law.

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Constant ε0 is called electric permittivity of the free space. Its

value is approximately ε0 ≃10−9

36πin the MKS system, i.e., in C2/Nm2

or Faraday (F ).

Constant µ0 is called magnetic permeability of the free space. Its

value is µ0 = 4π10−7 in the MKS system, i.e. in H/m.

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4. Maxwell’s equations in differential form in free space

In this section we deduce local or differential forms of Maxwell’s

equations from the integral ones given above.

4.1 Obtaining the partial differential equations.

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4.1. Obtaining the partial differential equations.

For the sake of simplicity, let us assume that QΩ and IS are given by

QΩ =

∫

Ω

ρvdV, (9)

IS =

∫

S

J · ndA, (10)

where ρv is the charge density and J is the current density.


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Since the integral forms of Maxwell’s equations must hold for any

volume Ω and any surface S, we can deduce local forms which are

partial differential equations.


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Indeed, by using the Gauss theorem in (5) and (6) we obtain

∫

Ω

(ε0 divE− ρv) dV = 0,

∫

Ω

divBdV = 0,

and then,

ε0 divE = ρv, (11)

divB = 0. (12)


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Now, transforming (7) and (8) by Stokes theorem. We get

∫

S

[curlE · n+∂B

∂t· n]dA = 0,

and hence

∂B

∂t+ curlE = 0. (13)


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Similarly,

∫

S

[1

µ0curlB · n− J · n− ε0

∂E

∂t· n

]dA = 0,

and then

ε0∂E

∂t−

1

µ0curlB = −J. (14)

4.2 Harmonic regime

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4.2. Harmonic regime

Time harmonic fields E and B are generated whenever their charges

and current sources have densities varying sinusoidally in time. More

generally, let us assume that ρv and J are of the form,

ρv(x, t) = Re(eiωtρ̂v(x)),

J(x, t) = Re(eiωtĴ(x)),

where ρ̂v and Ĵ are complex valued fields independent of time.

4.2 Harmonic regime

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Then the solutions of the Maxwell equations also have the form,

E(x, t) = Re(eiωtÊ(x)),

B(x, t) = Re(eiωtB̂(x)),

where, again, Ê and B̂ are complex valued fields independent of time.

4.2 Harmonic regime

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By replacing the above expressions in the differential forms of

Maxwell’s equations we obtain

div(ε0Ê) = ρ̂v, (15)

div B̂ = 0, (16)

curl Ê = −iωB̂, (17)

curl

(B̂

µ0

)= −Ĵ+ iωε0Ê. (18)

4.2 Harmonic regime

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These equations are called Maxwell’s equations for harmonic regime

in the frequency domain. We notice that all fields are complex valued.

However, they do not depend on the time variable.

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Maxwell equations in material regions

5. Conductors and insulators

In terms of their charge conduction properties materials may be

classified as insulators (or dielectrics), which possesses essentially no

free electrons to provide currents under an electric field, and

conductors, in which free, outer orbit electrons are readily available to

produce a conduction current when an electric field is present.

5.1 Electrical conductivity. Ohm’s Law

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5.1. Electrical conductivity. Ohm’s Law

Many conductors exhibit a linear dependence of J on the applied

electric field E, namely,

J = σE. (19)

This is the Ohm’s law. The factor σ is called the electric

conductivity of the material.

5.1 Electrical conductivity. Ohm’s Law

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In the MKS System σ units are (Ωm)−1 or mho/m (0/m). The mho

is also called siemens.

Electric conductivity depends on temperature. It is of the order of

108 (0/m) for the best conductors at room temperature to

10−16 (0/m) for the best insulators.

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6. Electric polarization and electric displacement

Under the action of an electric field, a microscopic distribution of

electric dipoles arises in the dielectric.

The density of these dipoles is denoted by P and called density of

dipole momentum or polarization vector.

Let us recall that the momentum of an electric dipole m has units

C ·m. Hence P has units C/m2.

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The electric field produced by the polarization is equivalent to the

one created by the charge density

ρP = − divP (C/m3).

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Let us assume that a free charge of density ρv exists in the polarized

material. According to the Gauss’s law for electric charge,

div ε0E = ρv + ρP = ρv − divP, (20)

and then,

div(ε0E+P) = ρv.

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The vector field

D = ε0E+P, (21)

is called electric displacement.

In terms of D, the Gauss’s law for electric charge (20) becomes

divD = ρv. (22)

6.1 Electric susceptibility and electric permittivity

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6.1. Electric susceptibility and electric permittivity

Experiments reveal that many dielectric substances are essentially

linear, meaning that P is proportional to the applied electric field, E.

For such materials,

P = χeε0E, (23)

where parameter χe is called electric susceptibility of the dielectric.

The factor ε0 is retained to make χe dimensionless.


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According to the definition of D we have

D = (1 + χe) ε0E. (24)

It is usual to denote 1 + χe by the dimensionless symbol

εr = 1 + χe, (25)

which is called relative (electric) permittivity of the region.


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The electric permittivity of the material is defined by

ε = εrε0 = (1 + χe)ε0 (F/m). (26)

Thus D = εE.

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7. Electric Gauss’s law for materials

It is

divD = ρv, (27)

in differential form, and in integral form

∫

Γ

D · ndA = QΩ, (28)

where Γ is the boundary of any regular volume Ω and QΩ is the total

charge enclosed in Ω.

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8. Magnetic polarization for materials (magnetization)

Magnetic materials are those that exhibit magnetic polarization

when they are subjected to an applied magnetic field.

The magnetization phenomenon is represented by the alignment of

the magnetic dipoles of the material with the applied magnetic field.

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A magnetic material has a number of magnetic dipoles and thus

many magnetic moments.

In the absence of an applied magnetic field the magnetic dipoles and

their corresponding electric loops are oriented in a random way so

that, on a macroscopic scale, the vector sum of the magnetic

moments is equal to zero.

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When the magnetic material is subjected to an applied magnetic

field represented by the magnetic flux density B, then a torque is

exerted on the magnetic dipole given by m×B.

As a consequence, they will tend to align in the direction of B.

Thus the resultant magnetic field at every point in the material

would be grater that its corresponding value at the same point when

the material is absent.

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The macroscopic density of magnetic dipoles is represented by a vector

field M called magnetization density. Its units are (Am2

m3= A/m).

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Let us introduce the magnetic intensity field H defined by

H =B

µ0−M (A/m). (29)

8.1 Magnetic susceptibility. Magnetic permeability.

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8.1. Magnetic susceptibility. Magnetic permeability.

For linear magnetic materials M is proportional to H:

M = χmH, (30)

where the dimensionless parameter χm is called magnetic

susceptibility of the material.


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Inserting this equality in the definition of H we have

H =B

µ0− χmH,

from which it follows that

B = (1 + χm)µ0H.

The parameter

µr = 1 + χm (31)

is called the relative (magnetic) permeability of the material.


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Further, the (magnetic) permeability is defined by

µ = µrµ0, (32)

so, finally,

B = µH. (33)

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9. Magnetic Gauss’s law for materials

Recall that the electric Gauss’s law for material was developed by

adding the effect of electric polarization charge density to the

corresponding law for free-space.

The magnetic Gauss’s law for materials can be developed analogously.

However, no additive term is required in this case because no free

magnetic charges exist physically in any known material.

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Thus B remains divergence-free in materials, that is,

divB = 0,

or

∫

Γ

B · ndA = 0,

where Γ = ∂Ω, for all regular domain Ω.

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10. Ampère’s law for materials

In free-space the curl of B/µ0 has been expressed as the sum of a

conduction current density J plus a displacement current density∂

∂t(ε0E) at any point, i.e.,

curl

(B

µ0

)= J+

∂

∂t(ε0E).

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For materials two additional types of currents occur: JP =∂P

∂tand

JM = curlM, arising from dielectric and magnetic polarization effects,

respectively.

Adding these currents together leads to a revision of the Ampère’s law

for a material region, namely,

curl

(B

µ0

)= J+

∂

∂t(ε0E) +

∂P

∂t+ curlM.

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Grouping the curl terms and the time derivative terms together we

obtain

curl

(B

µ0−M

)= J+

∂

∂t(ε0E+P) .

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Recalling that

H :=B

µ0−M and D := ε0E+P,

we finally get∂D

∂t− curlH = −J, (34)

which is the Ampère’s law for materials.

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11. Maxwell’s equations in differential forms for materials

We summarize the Maxwell’s equations describing the electromagnetic

behaviour of materials:

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divD = ρv,

divB = 0,

∂D

∂t− curlH = −J,

∂B

∂t+ curlE = 0,

D = εE,

B = µH,

J = σE.

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12. Properties of materials: homogeneity, isotropy,

linearity.

A material having parameter µ, ε and σ independent of the position is

termed homogeneous.

Conversely, if one or more of these parameters is space-dependent then

the material is said inhomogeneous.

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In some physical materials such as crystalline substances possessing a

well-ordered atomic or molecular lattice , the polarization fields P or M

resulting from the application of electric or magnetics fields may not

necessarily have the same directions as the applied fields.

Such material are called anisotropic.

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This fact can be modelled by taking tensors rather than scalars for the

characteristic parameters ε, µ and σ. As an example, the constitutive

law D = εE should be replaced by (in coordinates),

D1

D2

D3

=

ε11 ε12 ε13

ε21 ε22 ε23

ε31 ε32 ε33

E1

E2

E3

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Finally, a material is termed nonlinear if one or more of the parameters

ε, µ or σ are dependent on the level of the applied fields. In that case

one would write,

D = ε(|E|)E,

B = µ(|H|)H,

J = σ(|E|)E.

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13. Electromagnetic energy

1

2

∫

E

D(x, t1) · E(x, t1) dVx +

∫

E

B(x, t1) ·H(x, t1) dVx

=1

2

∫

E

D(x, t2) · E(x, t2) dVx +

∫

E

B(x, t2) ·H(x, t2) dVx

+

∫ t2

t1

∫

E

J · E dVx dt = 0. (35)

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Terms

EE(t) =1

2

∫

E

D(x, t) · E(x, t) dVx

and

EM(t) =1

2

∫

E

B(x, t) ·H(x, t) dVx

represents the electric and magnetic energy at time t, so the equality

(35) can be rewritten as

E(t1) = E(t2) +

∫ t2

t1

∫

E

J · E dVx dt,

where E(t) = EE(t) + EM(t) is the energy of the electromagnetic field

at time t.

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This equality say that the electromagnetic energy is conserved as far

as no conductors exist in the space.

Otherwise, it says that the electromagnetic energy decreases along

time due to the (increasing with time) dissipation term∫ t2

t1

∫

E

J · E dVx dt

representing the electromagnetic energy transformed in heat by the

conductors existing in the space (Joule effect).

13.1 Energy balance in a bounded domain'

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13.1. Energy balance in a bounded domain

13.1 Energy balance in a bounded domain'

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EΩ(t1) = EΩ(t2) +

∫ t2

t1

∫

E

J · E dVx dt+

∫ t2

t1

∫

Γ

E×H · n dAx dt,

where EΩ(t) denotes the electromagnetic energy in domain Ω at

time t.

Vector field P := E×H is called the Poynting vector.

The last term in this equality represents the outgoing

electromagnetic flux from Ω through boundary Γ along time interval

(t1, t2).

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Electrostatics

14. Maxwell’s equations for electrostatics

In electrostatics charges do not move, so currents do not exist and then

magnetic field is null. In this situation Maxwell’s equations becomes,

divD = ρv,

curlE = 0,

(36)

with D = εE.

14.1 Electrostatic potential

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14.1. Electrostatic potential

As in empty space, the electric fields is curl-free which implies existence

of a potential fields V such that,,

E = −∇V.

By replacing in the Gauss’s law for electric charge we get,

− div(ε∇V) = ρv, (37)

which is an elliptic partial differential equation similar to Poisson’s

equation.


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The electric permittivity needs not to be constat in the whole space

but if this is the case the equation [37] is equivalent to

−∆V = −ρvε,

which is a Poisson’s equation similar to that obtained for empty space.


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Let us recall that, in general, the solution of the problem

−∆V = S,

V(x) → 0 as r(x) → ∞,

(38)

is the distribution obtained by convolution,

V = T ∗ S,


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where T is the fundamental solution, i.e., T = Tv with

−∆V = δo,

V(x) → 0 as r(x) → ∞.


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If the distribution S is defined through a compact support function

ρv/ε, i.e.,

S(ψ) =

∫

E

ρv(x)

εψ(x)dVx,

then V is defined by a L1loc(E) function given by (see subsection ??)

V(x) =

∫

E

1

4πε

ρv(y)

|x− y|dVy.


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Also in this case the electric field E = −∇V is given by

E(x) =1

4πε

∫

E

ρv(y)x− y

|x− y|3dVy.

14.2 Electric field created by a set of charged conductors. Capacitance matrix

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14.2. Electric field created by a set of charged conductors.

Capacitance matrix


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This is a case where we do not need to know the charge distribution in

order to determine the electric field.

Let us consider N bounded conductors Ω1, . . . ,ΩN having total

charges Q1, . . . ,QN, respectively.

Let us assume that the complementary set Ω = E\ (Ω1 ∪ . . .ΩN) is

filled with a dielectric with electric permittivity, ε(x), which can be

dependent on the position (non-homogenous media).


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According to the local form of the Gauss’ law, we have

− div(ε∇V) = 0 in Ω,

because the dielectric is assumed to be charge-free.


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In order to get a well-posed problem we need to write boundary

conditions on boundaries Γ1, . . . ,ΓN of Ω1, . . . ,ΩN, respectively.

For this we observe that, in electrostatics, the electric field in

conductors must be null because otherwise we would have an electric

current different from zero by Ohm’s law.

Figura 1: The domain


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Furthermore, E = 0 in Ωi implies the potential is constant in each

conductor. Let V = Vi in Ωi, for i = 1, . . . ,N.

If these potentials were known, then they could be Dirichlet data for

the above Poisson’s equation and we were lead to solve

− div(ε∇V) = 0 in Ω,

V = Vi on Γi,

V(x) → 0 as r(x) → ∞.

(39)


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Moreover, since E = 0 in Ωi, then div(εE) = 0 in Ωi so the charge

density must be null in the interior of each conductor. This implies that

the charge must be concentrated on the surface Γi and surface density,

ρΓi, may depend on the particular point on Γi. Furthermore,

ε+E+ · n+ + ε−E− · n− = −ρΓi.

Figura 2: Conductor Ωi


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Let the region ”minus”be the interior of Ωi and the region “plus”be its

complementary set. Then the previous equation yields

ε+E+ · n+ = −ρΓi,

or

ε+∂V+

∂n+= ρΓi.

This is a Neumann boundary condition for the Poisson equation.


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The main difficulty is that the σi are not known (recall that the data are

charges Q1, . . . ,QN). Thus, in order to solve the problem we observe

that the mapping giving the total charges from the potentials is linear:

RN −→ RN

(V1, . . . ,VN) −→

(Q1 =

∫

Γ1

ε∂V

∂ndAx, . . . ,QN =

∫

ΓN

ε∂V

∂ndAx

),

where V is the solution of problem (39).


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Then we can find a matrix C such that

C−→V =

−→Q ,

where

−→V =

V1...

VN

and

−→Q =

Q1...

QN

.


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We notice that the columns of C are simply the charge vectors

corresponding to potentials (1, 0, . . . , 0), (0, 1, 0, . . . , 0), (0, 0 . . . , 0, 1)

so it can be easily computed by solving N Poisson problems with the

same differential operator.


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Matrix C does not depend on the particular electric charges. It only

depends on ε and the geometry of conductors. It is called capacitance

matrix and its entries have units coulomb per volt or farad.


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Moreover, once C is known, the problem of determining the potentials

in the conductors, and thereby the electric field in the dielectrics, is

simply solved by taking the inverse matrix of C. Indeed,

−→V = C−1

−→Q .


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Direct current

In this case, sources, i.e., charges and currents do not depend on

time but, unlike electrostatics, currents are not null.

Since all fields must be independent on time, Maxwell’s equations

becomes,

divD = ρ, divB = 0,

curlE = 0, curlH = J,

with, D = εE, B = µH and J = σE.


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The interesting feature of this case is that we can compute first the

electric field and the current density, and then the magnetic field.

For the former we notice that, by taking the divergence, Ampère’s law

yields

div J = 0, (46)

and using Ohm’s law,

div(σE) = 0. (47)

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15. Electric potential. Boundary conditions

Moreover curlE = 0 implies the existence of an electric potential V

such that E = −∇V. By replacing in (46) we get

− div(σ∇V) = 0,

which is a Laplace-like partial differential equation similar to the one in

electrostatics.

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In order to solve this equation, which is only valid in conductors

(otherwise σ = 0), we consider boundary conditions.

They can be either Dirichlet or Neumann conditions:

Dirichlet: V = Vd on Γd,

Neumann: −σ∂V

∂n= σE · n = J · n = jn on Γn.

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16. Numerical solution

The weak formulation of the above boundary value problem can be

obtained by standard procedures. It reads as follows:

To find V ∈ H1(Ω) with V = Vd on Γd such that

∫

Ω

σ∇V · ∇zdVx = −

∫

Γn

jnzdAx ∀z ∈ H1(Ω) with z|Γd = 0.

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This weak formulation can be discretized by using a finite element

approximation of the Sobolev space H1(Ω).

This approximation can be done from a tetrahedral mesh, Th, h being

the mesh-size. Let us consider the finite-dimensional space

Vh ={z ∈ C0(Ω) : z|K ∈ P1 ∀K ∈ Th

},

where P1 denotes the space of polynomials of degree less or equal than

one.

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The approximated problem is defined by

To find Vh ∈ Vh such that Vh(p) = Vd(p) ∀p vertex in Γd and

∫

Ω

σ∇Vh · ∇zhdVx = −

∫

ΓN

jdzhdAx,

∀zh ∈ Vh such that zh(p) = 0 ∀p vertex in Γd.

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17. Conductance matrix. Resistance matrix

Let us consider a conducting domain Ω with boundary Γ. Let

Γ0, . . . ,ΓN the ports of Ω, i.e. the connected subsets of Γ through

which direct currents enter or leave Ω.

Figura 3: Domain

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We assume that no current flux exists across the rest of the boundary

Γn = Γ− (Γ0 ∪ . . . ∪ ΓN), i.e.

This assumption leads to the boundary condition

σ∂V

∂n= −J · n = 0 on Γn.

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We also suppose that current enters or leaves the domain Ω

perpendicularly to the boundary (i.e., J× n = 0) which yields

E× n = 0 on Γ0 ∪ . . . ∪ ΓN.

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Since E = −∇V, condition E× n = 0 on Γ0 ∪ . . . ∪ ΓN gives the

Dirichlet boundary conditions

V = Vi (constant) on Γi, i = 0, . . . ,N,

because each Γi is connected.

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If we know the values Vi, i = 1, . . . ,N, we can state the well-posed

boundary-value problem

− div(σ∇V) = 0,

σ∂V

∂n= 0 on Γn,

V = Vi on Γi, i = 0, . . . ,N,

(48)

which can be solved by numerical methods.

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However, there are many cases where we know the intensities through

each Γi instead of the potentials.

Firstly, we notice that once problem (48) has been solved, intensities

entering the domain are given by

Ii = −

∫

Γi

J · ndA =

∫

Γi

σ∂V

∂ndA, i = 1, . . . ,N.

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Moreover,

N∑

i=0

−

∫

Γi

J · ndA = −

∫

Γ

J · ndA = −

∫

Ω

div JdV = 0.

Furthermore, since V is only defined up to a constant, we can choose

V0 = 0, i.e.,

V(x) = 0 on Γ0.

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Let us consider the mapping giving intensities from potentials, i.e.,

(V1, . . . ,VN) ∈ RN −→ (I1, . . . , IN) ∈ R

N, (49)

with

Ii =

∫

Γi

σ∂V

∂ndA, i = 1, . . . ,N,

and V being the solution of (48). Let

I0 = −N∑

i=1

Ii.

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Since this mapping is linear, there exists a matrix G, called

conductance matrix, such that

G−→V =

−→I ,

where

−→V =

V1...

VN

and

−→I =

I1...

IN

.

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The inverse of G, denoted by R, is called resistance matrix. Since we

have

−→V = G−1

−→I = R

−→I ,

this matrix allows us to obtain potentials from intensities.

Similar to capacitance matrix in electrostatics, the conductance matrix

can be obtained by solving N problems like (48). Indeed, the k-th

column of G is the vector of intensities corresponding to the potentials

Vi = δik on Γi, i = 1, . . . ,N.

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Magnetostatics

18. Maxwell’s equations for magnetostatics

If we know the static current density J, the problem of determining the

corresponding magnetic field is called the magnetostatic problem.

Equations for magnetostatics are

curlH = J, (50)

divB = 0, (51)

B = µH. (52)

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19. Vector magnetic potential

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Since divB = 0, there exists a vector field A such that

B = curlA.

Actually, there exist many of such vector fields.

Indeed, if curlA = B, then

curl(A+∇ϕ) = B

for all scalar field ϕ.

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In order to uniquely determine A a so-called gauge condition should be

added.

An example is the Coulomb’s gauge:

divA = 0.

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Moreover, by using the constitutive law B = µH, Ampère’s law yields

curl

(1

µcurlA

)= J.

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Now we have to solve this equation together with the gauge condition.

Things become much simpler if µ is constant in the whole space, let

say, µ = µ0. In this case we have,

curl(curlA) = µ0J. (53)

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By adding the term ∇(divA) (which is null by the Coulomb’s gauge)

to the left-hand side, and using the vector equality,

−∆A = curl curlA−∇(divA),

equation (53) yields

−∆A = µ0J.

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In a fixed cartesian system of coordinates, this equation is equivalent to

−∆Ai = µ0Ji, i = 1, 2, 3. (54)

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Then we can use the fundamental solution of the Poisson’s equation to

write the solutions of (54) by convolution with their right-hand sides.

We have

Ai(x) =

∫

Ω

µ0Ji(y)

4π|x− y|dVy, i = 1, 2, 3,

and hence,

A(x) =

∫

Ω

µ0J(y)

4π|x− y|dVy (Wb/m). (55)

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Remark 19.1 In the previous integrals Ω denotes any bounded

domain in the affine space containing the support of J.

Of course, J may also be a distribution supported on a surface S or

on a line l in which case Ω should be replaced in the above formula

by S or l, respectively.

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By taking the curl in (55), we get

B(x) = curlxA(x) = curlx

∫

Ω

µ0J(y)

4π|x− y|dVy

=

∫

Ω

µ04π

curlx

(J(y)

|x− y|

)dVy.

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But curl(φu) = ∇φ× u+ φ curl u. In our case J(y) does not depend

on x, so

curlx

(J(y)

|x− y|

)= ∇x

(1

|x− y|

)× J(y) = −

x− y

|x− y|3× J(y).

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Hence,

B(x) =

∫

Ω

µ04π

J(y)× (x− y)

|x− y|3dVy. (56)

This integral for B, expressed directly in terms of the static current

distribution J in free space, is known as the Biot-Savart law.

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In the general case, when µ is not constant, introducing the gauge

condition and solving the problem is more difficult. A way to do it is

described below.

For the sake of simplicity, let us suppose that Ω is a bounded domain

containing the support of the current density J and such that we may

assume the boundary condition

H× n = 0 in Γ = ∂Ω (57)

is satisfied.

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Firstly, we notice that divB = 0 implies

∫

Γ

B · n dA = 0,

by using Gauss’s theorem.

Let us assume Ω is connected and simply connected. Then Theorem

3.5 in the book by Girault and Raviart affirms that, if B ∈ L2(Ω), there

exists a unique vector field A ∈ H(curl,Ω) ∩H(div,Ω) such that

curlA = B in Ω, (58)

divA = 0 in Ω, (59)

A · n = 0 on Γ. (60)

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Furthermore, if Ω is smooth (in particular C1,1) then A ∈ H1(Ω).

By using Ampère’s law we have

curl(1

µcurlA) = J, (61)

and boundary condition (57) becomes

1

µcurlA× n = 0 on Γ. (62)

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One can show that problem (61), (62) together with (59) and (60) has

a unique solution A ∈ H(curl,Ω) ∩H(div,Ω).

Let us consider the variational problem:

Find A ∈ V such that∫

Ω

1

µcurlA · curlφ dV +

∫

Ω

divA divφ dV =

∫

Ω

J · φ dV ∀φ ∈ V,

where V is the functional space,

V = {A ∈ H(curl,Ω) ∩H(div,Ω)/A · n = 0 on Γ} .

(63)

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Since the bilinear form,

a(A,φ) =

∫

Ω

1


∫

Ω

divA divφ dV,

is continuous and coercive in V, the above problem has unique solution

A.

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If Ω is smooth, then V ⊂ H1(Ω) and one can use continuous piecewise

linear finite elements to discretize each component of A.

Otherwise, if the boundary of Ω is only Lipschitz-continuous, then A is

singular at reentrant corners and these standard Lagrangian finite

elements do not converge. This drawback can be overcomed by adding

a singular space of approximation as it it done by A. S. Bonnet, P.

Ciarlet Jr. and co-workers.

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Another alternative consists of using edge Nédélec finite elements for

the mixed formulation:

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Find A ∈ W and ϕ ∈ H such that

∫

Ω

1


∫

Ω

∇ψ · φ dV =

∫

Ω

J · φ dV ∀φ ∈ W,

∫ΩA · ∇ϕ dV = 0 ∀ϕ ∈ H,

where W = H(curl,Ω) and H ={ϕ ∈ H1(Ω) :

∫Ω ϕ dV = 0

}.

(64)

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This formulation can be discretized by using edge Nédélec finite

elements to approximate A and piecewise linear continuos finite

elements to approximate ψ. Moreover, by taking φ = ∇ψ in the first

equation it is easy to see that ψ ≡ 0 in Ω.

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Lowest-order Nédélec tetrahedral edge elements for the magnetic

field Hh:

X h := {Gh ∈ H(rot,Ω) : Gh|T ∈ N (T ) ∀T ∈ Th} ,

Gh|T ∈ N (T ) iff Gh(x) = a× x+ b, a,b ∈ C3.

• Basis functions:

Φe = λm∇λn − λn∇λm.

• curlΦe = 2∇λm ×∇λn.

• Hh · τ e continuous.

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20. Scalar magnetic potential

Another technique to solve magnetostatic problems consists in

introducing a scalar potential. We recall that we have to solve the

system

curlH = J,

divB = 0,

B = µH.

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Let H be a vector field such that curlH = J. Notice that H can be

obtained, for instance, by the Biot and Savart’s law, namely,

H(x) =

∫

Ω

1

4π

J(y)× (x− y)

|x− y|3dVy. (65)

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In general, H(x) 6= H(x) because div(µH) need not to be null.

However, curlH = curlH and then there exists a scalar field ϕR such

that

H = H−∇ϕR,

which is called the reduced scalar magnetic potential.

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Then we seek for ϕR such that,

div(µ(H−∇ϕR)) = 0,

or

− div(µ∇ϕR) = − div(µH) in E ,

and satisfying ϕR → 0 at infinity.

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The inconvenient of this method is that it suffers from the so called

cancellation error because the two terms H and ∇ϕR are of the same

order of magnitude and opposite direction in the magnetic materials.

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Now we present an alternative approach.

Firstly, we split the affine space E into three sub-domains:

the support of the current density J, ΩJ,

the set occupied by air and other non-magnetic materials (i.e., for

which µ = µ0, the permeability of the empty space) to be called ΩA,

and the domain filled with magnetic materials (µ 6= µ0) where

J = 0, to be denoted by ΩM.

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Let us assume that ΩR =: int(ΩA ∪ ΩJ) and ΩM are simply connected.

Since curlH = 0 in ΩM there exists a scalar field ϕ, defined in ΩM,

such that

H|ΩM = −∇ϕ.

Moreover, in ΩR we have as before,

H = −∇ϕR +H.

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Now equations divB = 0 and B = µ0H yield,

− div(µ0∇ϕR) = − div(µ0H) = 0 in ΩR,

− div(µ∇ϕ) = 0 in ΩM

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to which we have to add the interface conditions on ΓI = Ω̄R ∩ Ω̄M

B+ · n+ +B− · n− = 0,

H+ × n+ +H− × n− = 0,

In terms of the scalar potentials they read

µ∂ϕ

∂n= µ0

∂ϕR∂n

− µ0H · n on ΓI ,

−∇ϕ× n = −∇ϕR × n+H× n on ΓI .

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One can show that a weak formulation of this problem is the following:

Find ϕR ∈ W1(ΩR), ϕ ∈ H

1(ΩM) and λ ∈ H1/2(ΓI) such that

∫

ΩR

µ0∇ϕR · ∇zR dV +

∫

ΩM

µ∇ϕ · ∇z dV

+ < λ, ∇z × n−∇zR × n >Γ=

∫

ΩR

µH · ∇zR dV,

< −∇ϕR × n+H× n+∇ϕ× n,β >Γ= 0,

∀(zR, z) ∈ W1(ΩR)× H

1(ΩM) ∀β ∈ H1/2(ΓI).

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20.0.1. Remarks

1. Function λ is a Lagrange multiplier to impose the continuity of the

tangential component of H on the interface.

2. If ΩM is not simply connected we have to introduce cuts across

which the potential ϕ jumps.

3. The above formulation can be discretized by using 3D piecewise

linear finite elements on a tetrahedral mesh of Ω for ϕR and ϕ, and

2D piecewise linear finite elements on boundary Γ for the Lagrange

multiplier λ.

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The eddy currents model

In this chapter we deal with the well known eddy currents model which

is obtained from Maxwell equations by neglecting the electric

displacement in the Ampère’s law.

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21. The time-harmonic eddy currents model in a

bounded domain

The eddy currents model is obtained from Maxwell’s equations when

the electromagnetic field is slowly time-varying; more precisely, when

the current carrying system is small compared with the electromagnetic

wavelength associated with the dominant time scale of the problem.

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In this case fields are propagated instantaneously so we are dealing with

a c→ ∞ limit, c being the propagation velocity of waves. This leads to

neglect field radiation effects. In other words, term ∂D∂t is suppressed in

the Ampère’s law.

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Thus, the eddy current model is

∂B

∂t+ curlE = 0, (66)

curlH = J, (67)

divB = 0, (68)

divD = ρV , (69)

B = µH, (70)

D = εE, (71)

J = σE+ JS, (72)

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where JS ∈ H(div,Ω) is a divergence-free source current. We also

suppose that σ is positive in conductors, null in dielectrics and

supp(JS) ∩ supp(σ) = ∅. (73)

Since we have neglected the term ∂D∂t

in Ampère’s law, we do not keep

the Gauss’ law for electric charge, divD = 0, in the conductors.

Instead, from (67) and (72) we get

div(σE) = 0 in the conductors .

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We are interested in obtaining the magnetic field everywhere but the

electric field only in the conductors. If we wanted to get this field also

in the dielectrics then equation divD = 0 would be retained in the

model, in the dielectrics.

In what follows we consider the harmonic case. This means that all

fields are of the following form

G(x, t) = Re(eiωtĜ(x)).

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By replacing in (66)-(72) we obtain the electromagnetic Helmholtz

equations for the complex phasors Ĝ (for the sake of simplicity we will

drop the “hat” in the notation of phasors):

iωB+ curlE = 0, (74)

curlH = J, (75)

divB = 0, (76)

B = µH, (77)

J = JS + σE. (78)

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To solve these equations, we restrict them to a simply connected 3D

bounded domain Ω consisting of two parts, ΩCand Ω

D, occupied by

conductors and dielectrics, respectively. The mathematical framework

we are going to analyze covers eddy current problems posed on

different geometrical settings. We sketch a particular case in Figure 4

including several connected components of the conducting domain with

different topological properties.

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Figura 4: Sketch of the domain.

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The domain Ω is assumed to have a Lipschitz-continuous connected

boundary ∂Ω. We denote by ΓC, Γ

Dand Γ

Ithe open surfaces such that

Γ̄C:= ∂Ω

C∩ ∂Ω is the outer boundary of the conducting domain,

Γ̄D:= ∂Ω

D∩ ∂Ω that of the dielectric domain and Γ̄

I:= ∂Ω

C∩ ∂Ω

Dthe

interface between both domains. We also denote by n a unit normal

vector to a given surface.

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As shown in Fig. 4, the connected components of the conducting

domain are of two types: “inductors” which cross the boundary of Ω,

and “workpieces” which have their closure included in Ω. Let us denote

Ω1C, . . . ,ΩL

Cthe former and ΩL+1

C, . . . ,ΩM

Cthe latter.

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We assume that the outer boundary of each inductor, ∂ΩnC∩ ∂Ω

(n = 1, . . . , L), has two disjoint connected components, both being the

closure of open surfaces: the current entrance ΓnJ, where the inductor is

connected to an alternate electric current source, and the current exit

ΓnE. We denote Γ

J:= Γ1

J∪ · · · ∪ ΓL

Jand Γ

E:= Γ1

E∪ · · · ∪ ΓL

E.

Furthermore, we assume that Γ̄nJ∩ Γ̄m

J= ∅, Γ̄n

E∩ Γ̄m

E= ∅,

1 ≤ m,n ≤ L, m 6= n, and Γ̄J∩ Γ̄

E= ∅.

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Let us assume that µ and σ are frequency-independent and there exist

constants µ, µ, σ and σ such that

0 < µ ≤ µ(x) ≤ µ, a.e. x ∈ Ω,

0 < σ ≤ σ(x) ≤ σ, a.e. x ∈ ΩC

and σ ≡ 0 in ΩD.

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We have to complete the model with suitable boundary conditions. For

the moment, let us consider the following ones:

E× n = 0 on ΓE, (79)

E× n = 0 on ΓJ, (80)

µH · n = 0 on ∂Ω. (81)

Conditions (79) and (80) mean that the electric current enters and

exits domain Ω perpendicularly to the boundary whereas condition (81)

implies that the magnetic field is tangential to the boundary.

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Formal calculations allow us to show that boundary condition (81)

implies that the tangential component of the electric field E is a

gradient. Indeed, after integrating iωµH · n on any surface S contained

in ∂Ω, by using (74), (77) and Stokes’ Theorem, we obtain

0 = iω

∫

S

µH · n dS = −

∫

S

curlE · n dS = −

∫

∂S

E · t dl

= −

∫

∂S

n× (E× n) · t dl,

with t being a unit vector tangent to ∂S.

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Therefore, since ∂Ω is simply connected, we can assert that there exists

a sufficiently smooth function V defined in Ω up to a constant, such

that V|∂Ω is a surface potential of the tangential component of E,

namely, E× n = − gradV × n on ∂Ω. On the other hand, (79) and

(80) imply that V must be constant on each connected component of

ΓJand Γ

E. The complex number Vn := V|Γn

E

− V|ΓnJ

is the voltage drop

along conductor ΩnC.

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Many physical applications involve current intensities and voltage drops

as boundary data. Let us suppose that the boundary data consist of the

voltage drops Vn, for n = 1, . . . , L̂, and the input current intensities

through each surface ΓnJ, In, for n = L̂+ 1, . . . , L. We notice that the

latter can be written as∫

ΓnJ

J · n dA = In, n = L̂+ 1, . . . , L.

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Moreover, equation (75) yields div J = 0 and hence, by Gauss’

theorem,

0 =

∫

ΩnC

div J dV =

∫

∂ΩnC

J · n dA =

∫

ΓnJ

J · n dA+

∫

ΓnE

J · n dA,

because J · n = 0 on ∂ΩnC\ (Γn

E∪ Γn

J). Thus,

∫

ΓnJ

J · n dA = −

∫

ΓnE

J · n dA, n = 1, . . . , L. (82)

These equalities simply means that the input current intensity coincides

with the output one for each conductor ΩnC, n = 1, . . . , L.

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We summarize the strong problem defined in Ω to be considered in the

next section:

iωB+ curlE = 0, (83)

curlH = J, (84)

divB = 0, (85)

B = µH, (86)

J = JS + σE, (87)

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E× n = 0 on ΓE, (88)

E× n = 0 on ΓJ, (89)

µH · n = 0 on ∂Ω, (90)

V = VnJ on ΓnJ, n = 1, . . . , L̂, (91)

V = VnE on ΓnE, n = 1, . . . , L̂, (92)

∫

ΓnJ

J · n dA = In, n = L̂+ 1, . . . , L. (93)

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22. Magnetic vector potential/scalar electric potential

formulation

We want to formulate and solve the above problem in terms of a

magnetic vector potential A and an electric scalar potential V.

Firstly, from (85), (90) and Theorem 1.3.6 from Girault and Raviart

book we deduce the existence of a vector field A ∈ H(curl,Ω) such

that

curlA = B in Ω, (94)

divA = 0 in Ω, (95)

A× n = 0 on Γ. (96)

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We notice that the latter equality guarantees boundary condition (90).

Indeed, from Stokes’ Theorem we have∫

S

B·n dS =

∫

∂S

A·t dl =

∫

∂S

n×A×n·t dl = 0 for all surface S ⊂ Γ.

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By replacing (94) in (83) we obtain

curl(iωA+ E) = 0 in Ω

and then, in particular,

iωA+ E = − gradV in Ω, (97)

for some V ∈ H1(Ω).

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From (88), (89) and (96) we deduce

gradΓV := n× gradV × n = −n× E× n = 0 on ΓJ∪ ΓE,

which implies that V must be constant on each connected component

of ΓJand Γ

E.We notice that the complex number Vn := V

nE −V

nJ is the

voltage drop along conductor ΩnC.

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From (83), (86), (87) and (97) we deduce

σ(iωA+ gradV) + curl(1

µcurlA) = JS in Ω. (98)

We notice that, since σ = 0 in ΩD, we only need to compute V in Ω

C.

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Finally, from (87) and (97), boundary conditions (93) become∫

ΓnJ

σ(iωA+ gradV) · n dS = −In, n = L̂+ 1, . . . , L. (99)

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Summarizing, the problem to be solved reads as follows:

Given a solenoidal field JS with support ΩS included in ΩD, and

complex numbers VnJ , VnE, n = 1, . . . , L̂ and In, n = L̂+ 1, . . . , L, find

a vector field A defined in Ω, and a scalar field V defined in ΩCand

null on ΓnJ , n = L̂+ 1, . . . , L, such that

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σ(iωA+ gradV) + curl(1

µcurlA) = JS in Ω, (100)

divA = 0 in Ω, (101)

A× n = 0 on Γ, (102)

σ(iωA+ gradV) · n = 0 on ∂ΩnC\ (Γn

E∪ Γn

J), (103)

n = 1, . . . L,

V = VnJ on ΓnJ, n = 1, . . . , L̂,

(104)

V = VnE on ΓnE, n = 1, . . . , L̂,

(105)∫

ΓnJ

σ(iωA+ gradV) · n dS = −In, n = L̂+ 1, . . . , L.

(106)

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In what follows we write a weak formulation of this problem. Firstly, let

us multiply (100) by the conjugate of a test function G ∈ H(rot,Ω),

integrate in Ω and use a Green’s formula. We get∫

ΩC

σ(iωA+gradV)·Ḡ dV +

∫

Ω

1

µcurlA·curl Ḡ dV =

∫

ΩS

Js ·Ḡ dV.

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Next, we notice that, in principle, the traces appearing in (102) and

(106) are not well defined for A ∈ H(rot,Ω) and V ∈ H1(ΩC) so we

write a weak formulation of them. For this purpose, we first notice

that, by taking the divergence operator of (100), we deduce

div(σ(iωA+ gradV)

)= 0 in Ω

C. (107)

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Now, given two vectors−→V J,

−→V E ∈ R

L̂ let us define the affine space

L(−→V J,

−→VE) = {V ∈ H

1(ΩC) : V = VnJ on Γ

nJ, V = VnE on Γ

nE,

n = 1, . . . , L̂,

V = constant on ΓnJand on Γn

E, n = L̂+ 1, . . . , L}.

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We denote L0 the above space corresponding to null vectors−→V J and

−→VE. By multiplying equation (107) by the conjugate of a test function

Z ∈ L0, integrating in ΩC, using a Green’s formula, (103) and (106) we

get∫

ΩC

σ(iωA+ gradV) · grad Z̄ dV =L∑

n=L̂+1

InZ̄n, (108)

with Zn = Z|ΓnE

− Z|ΓnJ

, n = L̂+ 1, . . . , L. Finally, by multiplying the

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gauge condition (101) by the conjugate of a test function Ψ ∈ H10(Ω)

and using a Green’s formula we obtain∫

Ω

A · grad Ψ̄ dV = 0.

Let W0 be the function space

W0 = {G ∈ H(rot,Ω) : G× n = 0 on Γ}.

Next, we summarize the whole weak problem:

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Find A ∈ W0, V ∈ L(−→V J,

−→VE) and ϕ ∈ H

10(Ω) such that∫

ΩC

σ(iωA+ gradV) · Ḡ dV +

∫

Ω

1

µcurlA · curl Ḡ dV

+

∫

Ω

gradϕ · Ḡ dV =

∫

ΩS

Js · Ḡ dV ∀G ∈ W0,

∫

ΩC


n=L̂+1

InZ̄n ∀Z ∈ L0,

∫

Ω

A · grad Ψ̄ dV = 0 ∀Ψ ∈ H10(Ω).

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Remark 22.1 We notice that in this formulation the scalar

potential V only appears under the gradient operator. Since the

gradient of V does not change if V is translated by a constant in

each connect component ΩnC, we can replace the above problem by

the following one:

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Find A ∈ W0, V ∈ M(−→V) and ϕ ∈ H10(Ω) such that∫

ΩC

σ(iωA+ gradV) · Ḡ dV +

∫

Ω

1

µcurlA · curl Ḡ dV

∫

Ω

gradϕ · Ḡ dV =

∫

ΩS

Js · Ḡ dV ∀G ∈ W0,

∫

ΩC


n=L̂+1

InZ̄n ∀Z ∈ M0,

∫

Ω

A · grad Ψ̄ dV = 0 ∀Ψ ∈ H10(Ω),

where M(−→V) is the affine space

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M(−→V) := {V ∈ H1(Ω

C) : V = 0 on Γn

J, n = 1, . . . , L, V = Vn on Γ

nE,

n = 1, . . . , L̂,V = constant on ΓnE, n = L̂+ 1, . . . , L}

and M0 is the above space for the null vector−→V .

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Remark 22.2 Existence and uniqueness of a solution to this

problem can be shown by using the Babuska-Brezzi theory of mixed

formulations.

Remark 22.3 Numerical discretization can be done by using

Nédélec edge finite elements to approximate the vector potential and

continuous piecewise linear nodal elements to approximate ϕ and V.

Alfredo BermúdezDolores Gómez Pilar Salgado

Bermúdez • Góm

ezSalgado

The book represents a basic support for a master course in electromagnetism oriented to numerical simulation. The main goal of the book is that the reader knows the boundary-value problems of partial differential equations that should be solved in order to perform computer simulation of electromagnetic processes. Moreover it includes a part devoted to electric circuit theory based on ordinary differential equations. The book is mainly oriented to electric engineering applications, going from the general to the specific, namely, from the full Maxwell’s equations to the particular cases of electrostatics, direct current, magnetostatics and eddy currents models. Apart from standard exercises related to analytical calculus, the book includes some others oriented to real-life applications solved with MaxFEM free simulation software.

Bermúdez • Gómez • SalgadoMathematical Models and Numerical Simulation in Electromagnetism

74

AB

UN

ITEX

TU

NIT

EXT

springer.com

ISBN 978-3-319-02948-1

Mathematical Models and Numerical Simulation in Electromagnetism

Mathem

atical Models and Num

erical Simulation in

Electromagnetism

ABC

Mathematics

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