FEM: Domain Decomposition and Homogenization for Maxwell’s ...

FEM: Domain Decomposition and Homogenization for Maxwell’s Equations of Large Scale Problems

FEM: Domain Decomposition and Homogenizationfor Maxwell’s Equations of Large Scale Problems

Karl Hollaus

Vienna University of Technology, Austria

Institute for Analysis and Scientific Computing

February 13, 2012


Outline

1 Domain DecompositionIntroductionPoisson Equation

Nitsche-type mortar methodHybride Nitsche-type mortar methodNumerical Examples

Maxwell’s EquationsSteady StateHybride Nitsche-type mortar methodNovel hybride Nitsche-type mortar methodNumerical Examples

2 HomogenizationIntroductionElectrostatic Problem

Two-scale approach

Eddy Current ProblemTwo-scale approach


Domain Decomposition

Introduction

Motivation

Domain Decomposition can be used for:

Moving parts: Electrical machines, etc.Complicated structure: Printed circuit boards, human body,Parallel computingEtc.



Introduction

Problem

Ω ... Domain∂Ω ... Surface of Ω

Ωi ... SubdomainΩ = Ω1 ∪ Ω2

∂Ωi ... Surface of Ωi

Γ ... InterfaceΓ = ∂Ω1 ∩ ∂Ω2



Introduction

Problem

Aim: Non-matching finite element grids and continuoussolutions!

2D:

3D:



Introduction

Electrostatics

Consider the corresponding Maxwell’s equations

curl E = 0

div D = ρ,

the material relation D = ε E and

appropriate boundary conditions.

Then, an electric scalar potential V can be introduced as

E = −∇V

and the Poisson equation

∆V = −ρε

is obtained.



Poisson Equation

Standard problem

Classical formulation:

−∇(λ(x)∇u(x)) = f in Ω

u = uD on ΓD

λ∂u

∂n= g on ΓN

Weak form:Multiplication of PDE by a test function v and integration over Ω yields∫

Ω

∇(λ∇(u))v dΩ =

∫Ω

fv dΩ,

and integration by parts leads to∫Ω

λ∇u · ∇v dΩ−∫

Γ

λgvn dΓ =

∫Ω

fv dΩ.



Poisson Equation

Standard problem


−∇(λ(x)∇u(x)) = f in Ω

u = uD on ΓD

λ∂u

∂n= 0 on ΓN

Weak form:The boundary conditions together with v = 0 on ΓD leads to the weakform for the finite element method:Find uh ∈ VD := uh ∈ Vh : uh = uD on ΓD, such that∫

Ω

λ∇uh · ∇vh dΩ =

∫Ω

fvh dΩ ∀vh ∈ V0

with V0 := vh ∈ Vh : vh = 0 on ΓD, where Vh is a finite elementsubspace of H1(Ω).



Poisson Equation

Equivalent transmission problem


−∆ui = f in Ωi , i = 1, 2ui = 0 on ∂Ω, i = 1, 2

u1 − u2 = 0 on Γ

∂u1

∂n1+∂u2

∂n2= 0 on Γ,

where ui = u|Ωi .

Definitions:

Jump at the interface: [[u]] := u1n1 + u2n2

Mean value: u := 12 (u1 + u2)



Poisson Equation

Nitsche-type mortar method


−∆ui = f in Ωi , i = 1, 2ui = 0 on ∂Ω, i = 1, 2

u1 − u2 = 0 on Γ

∂u1

∂n1+∂u2

∂n2= 0 on Γ,

where ui = u|Ωi .

Idea of Nitsche-type mortar method for domain decomposition:

2∑i=1

∫Ωi

∇u · ∇v dΩ−∫∂Ωi

∇u · [[v ]] ds =

∫Ωi

fv dΩ



Poisson Equation


Weak form:

Find uh ∈ VD := uh ∈ Vh : uh = 0 on ∂Ω, such that

2∑i=1

∫Ωi

∇uh · ∇vh dΩ−∫

Γ

∇uh · [[vh]] ds−∫Γ

∇vh · [[uh]] ds +αp2

h

∫Γ

[[uh]] · [[vh]] ds =

∫Ωi


with V0 := vh ∈ Vh : vh = 0 on ∂Ω, where Vh is a finite elementsubspace of H1(Ω1)× H1(Ω2).

α ... Stabilization factorp ... Polynomial orderh ... Mesh size



Poisson Equation


Weak form:

Find uh ∈ VD := uh ∈ Vh : uh = 0 on ∂Ω, such that

2∑i=1

∫Ωi

∇uh · ∇vh dΩ−∫

Γ

∇uh · [[vh]] ds−∫Γ

∇vh · [[uh]] ds +αp2

h

∫Γ

[[uh]] · [[vh]] ds =

∫Ωi


with V0 := vh ∈ Vh : vh = 0 on ∂Ω, where Vh is a finite elementsubspace of H1(Ω1)× H1(Ω2).

The formulation is consistent, stable and bounded w. r. t. a meshdependant energy norm 1.

1Becker, R., P. Hansbo and R. Stenberg, “A finite element method for domaindecomposition with non-matching grids”, 2003



Poisson Equation


Algebraic equation system:

Surface integrals introduce a coupling across the interface Γ:

Ω1︷︸︸︷ Ω2︷︸︸︷∗ ∗

∗ ∗

u1

u2

=

f1

f2

The corresponding entries ∗ are almost all zero.

But the integrals of non-zero entries can be difficult to compute!



Poisson Equation


Coupling terms in the weak form:

· · ·∫

Γ∆

∇ϕ1 · ϕ2n2 dΓ · · ·

Fig.: Partly adjacent finite elements Ti with basis functions ϕi

The determination of the intersection Γ∆ of the two elements T1 and T2

can be expensive. Practically, a remeshing is required.



Poisson Equation

Hybride Nitsche-type mortar method

Definition:

Additional independent variable λ, which is the restriction u1 = u2

on Γ

Function space M := µ ∈ L2(∂Ω1 ∪ ∂Ω2) : µ = 0 on ∂Ω



Poisson Equation


Weak form2:Find (uh, λh) ∈ VD := (uh, λh) ∈ Vh ×Mh : uh = 0 on ∂Ω, such that

2∑i=1

∫Ωi

∇uh·∇vh dΩ−∫∂Ωi

∂uh

∂n(vh − µh) dΓ−

∫∂Ωi

∂vh

∂n(uh − λh) dΓ+

αp2

h

∫∂Ωi

(uh − λh)(vh − µh) dΓ

=

∫Ω

fvh dΩ ∀(vh, µh) ∈ V0

with (uh, λh) ∈ V0 := (vh, µh) ∈ Vh ×Mh : vh = 0 on ∂Ω, where Vh isa finite element subspace of H1(Ω1)× H1(Ω2) and Mh is a subspace ofL2(Γ).

2Egger, H., “A class of hybrid mortar finite element methods for interface problemswith non-matching meshes,” preprint AICES-2009-2, Aachen Institute for AdvancedStudy in Computational Engineering Science, 2009.



Poisson Equation


Matrix of the hybrid Nitsche-type mortar method:

∗ 0

0 ∗

∗

∗ ∗

u1

u2

λ

=

f1

f2

0

Now, coupling of u1 with u2 exists only via the interface variable λ!



Poisson Equation

Poisson Equation

We propose a B-spline basis for Mh3:

Mh = S∆ =λ ∈ R | λ|[ti ,ti+1] ∈ Pn , λ ∈ C n−1 ([0, 1])

Gaussian quadrature benefits from smoothness of the splines.

DeBoor recursion

Bi,r (x) =x − ti

ti+r−1 − tiBi,r−1 (x) +

ti+r − x

ti+r − ti+1Bi+1,r−1 (x)

3K. Hollaus, D. Feldengut, J. Schoberl, M. Wabro, D. Omeragic: “Nitsche-typeMortaring for Maxwell’s Equations”’, PIERS Proceedings, 397 - 402, July 5-8,Cambridge, USA 2010.



Poisson Equation

Poisson Equation

Numerical Example in 2D:f = x - 0.7 in the circle, else f = 0Rectangle: a = 2, b = 1Circle: r = 0.3, at: x = 0.7, y = 0.5FE-order: 5Spline-order: 6, no. of splines: 6

Solution u of the Poisson equation



Poisson Equation

Poisson Equation

Numerical Example in 2D:f = x - 0.7 in the circle, else f = 0Rectangle: a = 2, b = 1Circle: r = 0.3, at: x = 0.7, y = 0.5FE-order: 5Spline-order: 6, no. of splines: 6

(∇u)x of the Poisson equation



Poisson Equation

Poisson Equation

Numerical Example in 3D:f = xz in the cylinder, else f = 0Cuboid: a = 3, b = 2, c = 1Cylinder: r = 0.4, at: x = 1.0, y = 1.0FE-order: 5Spline-order: 6, No. splines: 6

Solution u of the Poisson equation



Poisson Equation

Poisson Equation

Numerical Example in 3D:f = xz in the cylinder, else f = 0Cuboid: a = 3, b = 2, c = 1Cylinder: r = 0.4, at: x = 1.0, y = 1.0FE-order: 5Spline-order: 6, No. splines: 6

(∇u)x of the Poisson equation



Maxwell’s Equations

Complex Representation

The time harmonic case, steady state:

Consider the partial differential equations

curlH = J + jωD in Ω

curlE = −jωB

divB = 0

divD = ρ,

boundary conditions

E × n = 0 on ΓE

H × n = K on ΓH

B · n = b on ΓB

and material relations

J = σE

B = µH

D = εH .




Standard Problem

Magnetic vector potential A:

A magnetic vector potential A can be introduced as

B = curlA.

Taking acount of the above relations, the physical quantaties can bewritten as

E = −iωA, H = µ−1 curlA

and after known manpulations

curlµ−1 curlA + κA = J in Ω

with the complex conductivity κ = jωσ − ω2ε is obtained.




Standard Problem


curlµ−1 curlA + κA = J 0 in Ω

µ−1 curlA× n = K on ∂ΩH

A× n = α on ∂ΩB

A× n = 0 on ∂ΩE ,

where ∂Ω = ∂ΩH + ∂ΩB + ∂ΩE .




Standard Problem

Weak form:

Considering A× n = 0 on ∂Ω lead to the finite element approximation:Find Ah ∈ VD := Ah ∈ Vh : Ah × n = αh on Ω, such that∫

Ω

µ−1 curlAh curl vh dΩ + jω

∫Ω

κAhvh dΩ =

∫Ω0

J 0vh dΩ

for all vh ∈ V0 := vh ∈ Vh : vh × n = 0 on ∂Ω, where Vh is a finiteelement subspace of H(curl,Ω).




Equivalent transmission problem


curlµ−1 curlAi + κAi = J 0 in Ωi , i = 1, 2Ai × n = 0 on ∂Ω, i = 1, 2

A1 × n1 + A2 × n2 = 0 on Γ

µ−1 curlA1 × n1 + µ−1 curlA2 × n2 = 0 on Γ,

where Ai = A|Ωi .





Weak form:

Considering homogeneous boundary conditions on ∂Ω and introducingsymmetry, stabilization and hybridization yields the hybrid formulation:

Find (Ah,λ) ∈ VB := Ah ∈ Vh : Ah × n = 0 on ΓB, such that

2∑i=1

∫Ωi

µ−1curlA ·curl v +κA ·v+

∫∂Ωi

µ−1 curlA · [(v − µ)× n]

+

∫∂Ωi

µ−1 curl v · [(A− λ)× n]+αp2

µh

∫∂Ωi

[(A− λ)× n] · [(v − µ)× n]

=

∫Ω

J 0 · v

for all (v ,µ).




Novel hybride Nitsche-type mortar method

Novel hybride Nitsche-type mortar method 4:

Considering symmetry, stability and hybridization also for φ leads to

2∑i=1

∫Ωi

µ−1 curlA·curl v+κAv+∫∂Ωi

1

µcurlA [(v−∇ψ−µ)×n]+∫

∂Ωi

1

µcurl v [(A−∇φ− λ)× n]+

α1p2

h

∫∂Ωi

1

µ[(A−∇φ−λ)×n][(v −∇ψ−µ)×n]−

∫∂Ωi

κAn(ψ−ψΓ)−∫∂Ωi

κvn(φ−φΓ)+α2p2

h

∫∂Ωi

κ(φ−φΓ)(ψ−ψΓ)

=

2∑i=1

∫Ωi

J 0v−∫∂Ωi

J nψ

4K. Hollaus, D. Feldengut, J. Schoberl, M. Wabro, D. Omeragic: “Nitsche-typeMortaring for Maxwell’s Equations”’, PIERS Proceedings, 397 - 402, July 5-8,Cambridge, USA 2010.




Novel hybride Nitsche-type mortar method

Novel hybride Nitsche-type mortar method:

We propose to descretize by

u, v ... H (curl) conforming finite elements of order p in Ωi

φ, ψ ... H1 conforming finite elements of order p + 1 on ∂Ωi ∩ Γ

φΓ, ψΓ ... tensor product scalar B-splines of order m and

λ, µ ... Nedelec-type B-splines satisfying an exact sequence at Γ




Numerical Example

Numerical Example: Permanent MagnetPDE for magnetostatics:

curlµ−1 curl Ai = f (M) in Ωi

A ... Magnetic vector potentialM ... Magnetisation: M 6= 0 in the permanent magnet, else M = 0

Fig.: Two domains (green and blue) with the permanent magnet (red)




Numerical Example

Numerical Example: Permanent Magnet

Fig.: Vector plot of the magnetic field intensity H




Numerical Example

Numerical Example: Permanent Magnet

Fig.: Scalar plot of the magnetic field intensity Hx




Numerical Example

LWD-Tool

LWD ... logging while drilling

Detail of the tool.




Numerical Example

LWD-Tool

Ambient soil with a borehole.




Numerical Example

LWD-ToolLWD-Tool to reconstruct the ambient material parameters

Detail of borehole with exciting antenna (red ring) and receiver (bluering)

The bore tool is eliminated by surface impedance boundary conditions.




Numerical Example

LWD-Tool

Vector plot of the magnetic field intensity H




Numerical Example

LWD-Tool

Field lines of the magnetic field intensity H




Numerical Example

LWD-Tool

Frequency in kHz Standard: Voltage in µ V NVF: Voltage in µ V20 25.44 -i 18.38 25.43− i 18.37

100 71.68 -i 197.5 71.65− i 197.3400 124.9 -i 963.0 124.9− i 962.3

2000 -634.9 -i 5295 −634.8− i 5290

Table: Numerical results for first order finite elements.

Frequency in kHz Standard: Voltage in µ V NVF: Voltage in µ V20 24.99 -i 18.47 24.98− i 18.47

100 70.25 -i 196.7 70.23− i 196.7400 121.7 -i 957.9 121.7− i 957.7

2000 -648.1 -i 5256 −647.9− i 5255

Table: Numerical results for second order finite elements.

Spline order: 5, spline no.: nϕ = 5, nz = 150


Homogenization

Introduction

Motivation

Laminated transformer core:

Large transformer

Principle of a transformer Laminated transformer core


Homogenization

Introduction

Motivation

Fig.: FE-Model with 100laminates.

Fig.: FE-Model with 100 laminates(Detail, lower right corner).

Finite element models of laminated media lead to large equationssystems.

Homogenization overcomes this problem!


Homogenization

Introduction

Motivation

Problem with laminated medium.

λi ... Material parametern ... No. laminationsd = d1 + d2 ... periodic structure !!!ff ... Fill factorff = d1

d1+d2

For example:Electrostatics

λ=ε ... Electric permittivityu=V ... Electric scalar potential


Homogenization

Electrostatic Problem


Boundary value problem:

Ω ... Domain

Ω = Ω0 ∪ Ωm

Γ ... Boundary

Γ = ΓD ∪ ΓN

−∇(λ(x , y)∇u(x , y)) = 0 in Ω (1)

u = uD on ΓD (2)

λ∂u

∂n· n = α on ΓN (3)


Homogenization


Standard problem


−∇(λ(x)∇u(x)) = f in Ω

u = uD on ΓD

λ∂u

∂n= 0 on ΓN

Weak form:The boundary conditions together with v = 0 on ΓD leads to the weakform for the finite element method:Find uh ∈ VD := uh ∈ Vh : uh = uD on ΓD, such that∫

Ω

λ∇uh · ∇vh dΩ = 0 ∀vh ∈ V0

with V0 := vh ∈ Vh : vh = 0 on ΓD, where Vh is a finite elementsubspace of H1(Ω).


Homogenization


Electrostatic Problem in 2D

Numerical Example:

Fig.: Numerical model.

ff ... Fill factorff = d1

d1+d2

ff = 0.9

λ0 = 1

λ1 = 1000

λ2 = 1

ub − ua = 2

No. laminates = 10


Homogenization



Reference solution: Each laminate is modeled individually!

Fig.: Reference solution u of one halfe of the domain.


Homogenization



Reference solution:

Solution u, mean value u0 andenvelope u1 along x at y = 0.

Periodic micro-shapefunction φ(x), one periode isshown.


Homogenization



Two-scale approach:

Thus, the following approach can be made:

u(x , y) = u0(x , y) + φ(x)u1(x , y)

u0 ... mean valueu1 ... envelope of the staggered partφ ... periodic-micro shape function

Inserting this approach into the bilinear form∫Ω

λ∇u · ∇v dΩ

leads to ∫Ω

λ∇(u0 + φu1) · ∇(v0 + φv1) dΩ = 0.


Homogenization



Homogenization of the weak form:

The finite element matrix would be calculated by

∫ΩFE

∂u0

∂x∂u0

∂y

u1∂u1

∂x∂u1

∂y

T

λ 0 λφx λφ 00 λ 0 0 λφλφx 0 λφ2

x λφφx 0λφ 0 λφφx λφ2 00 λφ 0 0 λφ2

∂v0

∂x∂v0

∂y

v1∂v1

∂x∂v1

∂y

dΩ,

with φx = (∇φ)x .

The micro shape function φ is a highly oscillating function.

To homogenize the weak form, the coefficients λ, λ∇φ, λφ, etc. areaveraged over the periode d :

λ =1

d

∫ d

0

λ(x)dx =λ1d1 + λ2d2

d


Homogenization



Homogenization of the weak form:

λφx =1

d

∫ d

0

λ(x)φx (x)dx = 2λ1 − λ2

d

λφ =1

d

∫ d

0

λ(x)φ(x)dx = 0

λφ2x =

1

d

∫ d

0

λ(x)φx (x)φx (x)dx =4

d(λ1

d1+λ2

d2)

λφxφ =1

d

∫ d

0

λ(x)φx (x)φ(x)dx = 0

λφ2 =1

d

∫ d

0

λ(x)φ(x)φ(x)dx =λ1d1 + λ2d2

3d


Homogenization



Numerical experiments have shown that neglecting the derivatives inφ∇u1 and φ∇v1, respectively, yields a more accurate solution.Weak form:Find(uh0, uh1) ∈ VD := (uh0, uh1) : uh0 ∈ Vh , uh1 ∈ Wh , uh0 = uD on ΓD,such that

∫Ω

∂x u0

∂y u0

u1

T λ 0 λφx

0 λ 0

λφx 0 λφ2x

∂x v 0

∂y v 0

v 1

dΩ = 0

for all(v h0, v h1) ∈ V0 := (v h0, v h1) : v h0 ∈ Vh , v h1 ∈ Wh , v h0 = 0 on ΓD,where Vh is a finite element subspace of H1(Ω) and Wh a finite elementsubspace of L2(Ωm), respectively.The micro-shape function φ is in the space of periodic and continuousfunctions Hper (Ωm).


Homogenization


Electrostatic Problem in 2D

Numerical Example:

Fig.: Numerical model.

ff ... Fill factorff = d1

d1+d2

ff = 0.9

λ0 = 1

λ1 = 1000

λ2 = 1

ub − ua = 2

No. laminations = 10


Homogenization



Comparison of the results:

Fig.: Reference solution. Fig.: Homogenized solution.

The agreement between reference solution and homogenized solution isobviously excellent!


Homogenization



Comparison of the results: flux density λ (∇φ)x

Fig.: Reference solution. Fig.: Homogenized solution.


Homogenization


Eddy Current Problem in 2D

Boundary value problem:

Ω ... DomainΩ = Ω0 ∪ Ωm

Γ ... Boundary

Γ = ΓH ∪ ΓB

µ ... Magnetic permeability

σ ... Electric conductivity

J 0 ... Impressed current density in a coil

B ... Magnetic flux density

Assumptions:- Linear material properties

- Time harmonic case, steady

state


Homogenization

Eddy Current Problem


Maxwell’s equations for the eddy current problem:

curlH = J in Ωm

curlE = −jωB

divB = 0

J = σE

B = µH

curlH = J 0 in Ω0

divB = 0

B = µH

H × n = K on ΓH

B · n = b on ΓB


Homogenization


Standard Problem

Weak form:Considering homogeneous boundary conditions on Γ leads to the finiteelement approximation:Find Ah ∈ VB := Ah ∈ Vh : Ah × n = αh on ΓB, such that∫

Ω

µ−1 curlAh curl vh dΩ + jω

∫Ω

σAhvh dΩ =

∫Ω0

J 0vh dΩ

for all vh ∈ V0 := vh ∈ Vh : vh × n = 0 on ∂ΩB, where Vh is a finiteelement subspace of H(curl,Ω).


Homogenization



Numerical example:

Fig.: Numerical example,dimensions in mm.

α = 1.0Vs/mff = 0.9

µ0 = 4π10−7Vs/Am

µ = µrµ0

µr1 = 1000

σ1 = 2 · 106S/m

f = 50Hz

δ ... Penetration depth

δ = 1.6mm


Homogenization



Reference solution:

Eddy currents in laminates, 2D problem

Two-scale approach:

A = A0 + φ(0,A1)T +∇(φw)

A0 ... represents the mean value of the solution

φ(0,A1)T considers J y and

w∇(φw) models J x at the end of the laminates.


Homogenization



Homogenized weak form:

Inserting the approach into the bilinear form yields:∫Ω

µ−1curl(A0 + φ(0,A1)T +∇(φw)

)curl

(v0 + φ(0, v1)T +∇(φq)

)dΩ

+jω

∫Ω

σ(A0 + φ(0,A1)T +∇(φw)

) (v0 + φ(0, v1)T +∇(φq)

)dΩ =

∫Ω0

J 0v dΩ

(4)


Homogenization



Finite element stiffness matrix:

Bilinearform: ∫Ω

µ−1curlA · curlv dΩ

Homogenized finite element matrix:∫ΩFE

(curlA0

A1

)T (ν νφx

νφx νφ2x

)(curlv0

v1

)dΩ,

with the reluctivity ν = µ−1.


Homogenization



Finite element mass matrix:

Bilinearform:

jω

∫Ω

σA · v dΩ

Homogenized finite element matrix:

jω

∫ΩFE

(A0)x

(A0)y

A1

w∂w∂x∂w∂y

T

σ 0 0 σφx σφ 00 σ σφ 0 0 σφ

0 σφ σφ2 0 0 σφ2

σφx 0 0 σφ2x σφxφ 0

σφ 0 0 σφxφ σφ2 0

0 σφ σφ2 0 0 σφ2

(v0)x

(v0)y

v1

q∂q∂x∂q∂y

dΩ


Homogenization



Homogenized weak form:Find (A0h,A1h,w h) ∈ VB := (A0h , A1h , w h) : A0h ∈ Uh , A1h ∈Vh , w h ∈ Wh and A0h × n = αh on ΓB, such that

A(A0h,A1h; v0h, v 1h) + B(A0h,A1h,w h; v0h, v 1h, qh) = f (v0h) (5)

for all (v0h, v 1h, qh) ∈ V0 := (v0h , v 1h , qh) : v0h ∈ Uh, v 1h ∈Vh , qh ∈ Wh and v0h × n = 0 on ΓB, where Uh is a finite elementsubspace of H(curl,Ω), Vh a finite element subspace of L2(Ωm) and Wh

a finite element subspace of H1(Ωm), respectively.The micro-shape function φ is in the space of periodic and continuousfunctions Hper (Ωm).

Again, a coarse finite element mesh suffices for the homogenized medium!


Homogenization



Numerical example:

Fig.: Numerical example,dimensions in mm.

α = 1.0Vs/mff = 0.9

µ0 = 4π10−7Vs/Am

µ = µrµ0

µr1 = 1000

σ1 = 2 · 106S/m

f = 50Hz

δ ... Penetration depth

δ = 1.6mm


Homogenization



Comparison of the results: Eddy current density J

Reference solution. Homogenized solution.

... solution in the upper left corner: d1 = 1.8mm, 10 laminations

The agreement between reference solution and homogenized solution isobviously excellent!


Homogenization



Comparison of the results: Losses and computational costs

Table: Eddy current losses

Losses in W/md in mm Reference solution Homogenized solution

0.5 11.59 11.581.0 45.70 45.462.0 177.2 174.16

Table: Computational costs for d=0.5mm and 40 laminations

Model FE NDOFReference solution 104 452 783 555

Homogenized solution 1 286 11 117

FE ... No. finite elementsNDOF ... No. degrees of freedom


Homogenization


Thank you for your attention!

FEM: Domain Decomposition and Homogenization for Maxwell’s ...

Documents

Transcript of FEM: Domain Decomposition and Homogenization for Maxwell’s ...