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The variational formulationExistence and uniqueness

The discrete approximations

An E -based mixed-FEM and BEM symmetric

coupling for a time-dependent eddy current

problem

Ramiro Acevedo1, Salim Meddahi2

1Universidad del Cauca, Colombia2Universidad de Oviedo, Spain

WONAPDE 2010

Universidad de Concepcion, January 11 - 15

Acevedo & Meddahi A mixed-FEM and BEM coupling for eddy current problem

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Outline

1 The variational formulationThe model problemAn E -based formulationThe mixed-FEM and BEM symmetric coupling

2 Existence and uniquenessThe reduced problemA stable decomposition of the kernel of bWell-posedness

3 The discrete approximationsAnalysis of the semi-discrete schemeAnalysis of the fully-discrete scheme


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The model problemAn E-based formulationThe mixed-FEM and BEM symmetric coupling

The model problem

J : a given time dependent current density,

Ωc: a passive conductor.

Our aim:

to compute the eddy currents in the conductor.


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The time-dependent eddy current problem

Find an electric field E (x, t) and a magnetic field H (x, t):

∂ t (µH ) + curlE = 0 in R3 × (0, T ),

curlH = J + σE in R3 × [0, T ),

div(εE ) = 0 inR3 \ Ωc

× [0, T ),

Σi

εE · n = 0 in [0, T ), i = 0, · · · , I,

H (x, 0) = H 0(x) in R3,

H (x, t) = O (1/ |x|) as |x| → ∞,E (x, t) = O (1/ |x|) as |x| → ∞.


Th i i l f l i Th d l bl

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The geometrical setting of the domain

c

Γ Ω

Ω

Σ

d

Ω ⊂ R3 : Ωc ∪ suppJ ⊂ Ω. Ωd := Ω \ Ωc, Ω := R3 \ Ω


Th i ti l f l ti Th d l bl

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The modified problem

E DOES NOT necessarily satisfy

E |Ω · n = 0 on Γ.

The shifted electric field

E ∗ := E −G, G :=

0 in Ωc,gradψ in Ωd,gradψext in Ω.

ψ and ψext are chosen such that

E ∗ ∈ H(curl,R3), div(E ∗|Ωd) = 0,

div(E ∗|Ω) = 0, E ∗|Ω · n = 0 on Γ.


The variational formulation The model problem

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The E -based strong problem: Equations in Ω

The unknown

u(x, t) :=

t0

E ∗(x, s) ds

To eliminate H :

∂ t (µH ) + curlE ∗ = 0

curlH = J + σE ∗

=⇒ σ∂ tu + curlµ−1 curlu = f ,

where

f := curlH 0 − J .



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The E -based strong problem: the gauge conditions

The unknown

u(x, t) :=

t0E ∗(x, s) ds

The gauge condition

divu = 0 in Ωd × [0, T ),

Σiε0u · n = 0 in [0, T ), i = 0, · · · , I.

The initial condition

u(x, 0) = 0 in Ω.



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Transmission conditions: some kind of traces

The standard trace

γ : H1(Ω) → H1/2(Γ); γv := v|Γ

The normal trace

γ n : H(div; Ω) → H−1/2(Γ); γ nv := v · n

The tangential trace

γ τ

: H(curl; Ω) → H−1/2 (divΓ

; Γ ) ; γ τ v := v × n

The tangential component trace

πτ : H(curl; Ω) → H−1/2 (curlΓ; Γ ) ; πτ v := n× (v ×n)



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The E -based strong problem: Transmission conditions

The unknown

u(x, t) := t

0E ∗(x, s) ds

γ −n

(u) = 0 on Γ × [0, T ),

π+

τ u = π−

τ u on Γ × [0, T ),

γ −τ

µ−10 curlu

= γ +τ

µ−10 curlu

on Γ × [0, T ).



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The ariational formulationExistence and uniqueness



Equations in Ω := R3 \ Ω and asymptotic conditions

The unknown

u(x, t) :=

t0E ∗(x, s) ds

curl curlu = 0 in Ω × [0, T ),

divu = 0 in Ω × [0, T ),

u(x, t) = O (1/ |x|) as |x| → ∞,

curlu(x, t) = O (1/ |x|) as |x| → ∞.



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Existence and uniquenessThe discrete approximations

pAn E-based formulationThe mixed-FEM and BEM symmetric coupling

A mixed formulation in Ω

σ∂ tu+ curlµ−1 curlu = f in Ω

The main variable u belongs to

W :=

v ∈ L2(0, T ;H(curl; Ω)) :

v|Ωc∈ W 1(0, T,H(curl; Ωc),H(curl; Ωc)) .



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An E-based formulationThe mixed-FEM and BEM symmetric coupling


A Lagrange multiplier p belongs to

M (Ωd) :=

q ∈ H1(Ωd) :

Ωd

q = 0, q|Σi= C i, i = 1, . . . , I

Ωd

εv · grad q = 0 ∀q ∈ M (Ωd)

div v = 0 in Ωd; Σi

ε0v · n = 0, i = 0, · · · , I.



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·, ·τ,Γ: Duality pairing between H−1/2 (divΓ; Γ) and H−1/2 (curlΓ; Γ)with L2

τ (Γ) as pivot space.

Find u ∈ W , p ∈ L2(0, T ; M (Ωd)) such that

d

dt [(σu(t),v)0,Ωc + (v,grad p(t))0,Ωd ] +

µ−1 curlu, curlv0,Ω

−γ −τ

µ−1 curlu(t)

,πτ v

τ,Γ

= (f (t),v)0,Ω ∀v ∈ H(curl; Ω),

(u(t),grad q)0,Ωd= 0 ∀q ∈ M (Ωd),

u|Ωc(0) = 0,

Remark. If (u, p) verify the previous equations,

divΓ γ −τ µ−1 curlu = 0


The variational formulationE i d i

The model problemA E b d f l i

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Non local boundary conditions on Γ

We introduce the space

H1/20 (Γ) :=

η ∈ H1/2(Γ) :

Γ

η = 0

divΓγ −τ

µ−1 curlu

= 0

=⇒ ∃!λ ∈ H1/20 (Γ) : curlΓ λ = γ −τ

µ−1 curlu

The first equation of the mixed formulation in Ω, leads to

d

dt[(σu(t),v)0,Ωc

+ (v,grad p(t))0,Ωd] +

µ−1 curlu, curlv

0,Ω

− curlΓ λ,πτ vτ,Γ = (f (t),v)0,Ω ∀v ∈ H(curl; Ω).


The variational formulationE i t d i

The model problemA E b d f l ti

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Non local boundary conditions on Γ

Homogeneous equations in Ω

+Decay conditions as |x| → ∞

=⇒

for any x ∈ Ω

u has an integral represen-tation formula on Γ

Integral operators on Γ:

Sφ(x) := γ x →

Γ

E (x,y)φ(y) dsy

,

V η(x) := πτ

x →

Γ

E (x,y)η(y) dsy

,

Kη(x) := γ +τ x → curlx Γ E (x,y)η(y) dsy ,

where,

E (x,y) :=1

4π |x− y|.



The model problemAn E based formulation

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The mixed-FEM and BEM symmetric coupling

Find u ∈ W , p ∈ L2(0, T ; M (Ωd)) and λ ∈ L2(0, T ; H1/20 (Γ)):

d

dt[(σu(t),v)0,Ωc

+ (v,grad p(t))0,Ωd] +

µ−1 curlu, curl v

0,Ω

+ µ−10 S (curlΓπτ u), curlΓπτ v1/2,Γ

+ K curlΓ λ(t),πτ vτ,Γ = (f (t), v)0,Ω ,

− curlΓ η,V (curlΓ λ)τ,Γ + µ−10 K (curlΓ η),πτ uτ,Γ = 0,

(u(t),grad q)0,Ωd= 0,

u|Ωc(0) = 0.

for any v ∈ H(curl; Ω), η ∈ H1/20 (Γ) and q ∈ M (Ωd).



The reduced problemThe reduced problem

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The reduced problemWell-posedness

Existence and uniqueness: The reduced problem

Hiptmair (2002)

The operator V is elliptic in H−1/2 (divΓ0;Γ)

We introduce

R : H−1/2(Γ) → H1/20 (Γ)

ξ → θ,:

curlΓ χ,V (curlΓ θ)τ,Γ = ξ, χ1/2,Γ∀χ ∈ H

1/20 (Γ).

Consequently,

λ = µ−10 R(curlΓK

∗πτ u)




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Existence and uniqueness: The reduced problem

Problem RP. Find u ∈ W , p ∈ L2(0, T ; M (Ωd)) such that:

d

dt[(σu(t),v)0,Ωc

+ (v,grad p(t))0,Ωd]

+ µ−1 curlu, curlv0,Ω

+ c(u,v) = (f (t),v)0,Ω

∀v ∈ H(curl; Ω),

(u(t),grad q)0,Ωd= 0 ∀q ∈ M (Ωd),

u|Ωc(0) = 0,

where

c(u,v) := µ−10 (curlΓ S curlΓ + K curlΓ RcurlΓK

∗)πτ u,πτ vτ,Γ .




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Existence and uniqueness: The kernel of b.

The bilinear form:b(v, q) := (v,grad q)0,Ω; v ∈ H(curl; Ω), q ∈ M (Ωd)

The kernel of b

V (Ω) := v ∈ H(curl; Ω) : b(v, q)0,Ω = 0 ∀q ∈ M (Ωd)

u|Ωdbelongs to

V (Ωd) := v ∈ HΣ(curl; Ωd) : b(v, q)0,Ω = 0 ∀q ∈ M (Ωd)

=v ∈ HΣ(curl; Ωd) : div v = 0 in Ωd;

γ nv = 0 on Γ; γ nv, 11/2,Σi = 0

.

(Fernandes & Gilardi, 1997)

There exists a real number s > 1/2 such that V (Ωd) is continuouslyimbedded in Hs(Ωd)3.




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qThe discrete approximations

pWell-posedness

Existence and uniqueness: a stable decomposition of V (Ω).

On the space V (Ωd):

vH(curl;Ωd) ≤ C curlv0,Ωd.

There exists an extension operator E : H(curl; Ωc) → V (Ω)

µ−10 (curl E vc, curlw)0,Ωd

+ c(E vc,w) = 0 ∀w ∈ V (Ωd).

A stable decomposition of V (Ω)

An inner product in V (Ω):

(u,v)V (Ω) := (u, v)σ +

µ−1 curlu, curlv0,Ω

+ c(u,v),

·V (Ω) is equivalent to the H(curl; Ω)-norm in V (Ω),

V (Ω) = V (Ωd) ⊕ E (H(curl; Ωc))




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qThe discrete approximations

pWell-posedness

Existence and uniqueness

Sketch of the proof

We look for u := ud + E uc . . . ,

ud is the solution of:

µ−10 (curlud(t), curlv)0,Ωd

+ c(ud(t),v)

= (f (t),v)0,Ωd ∀v ∈ V (Ωd),




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The discrete approximations Well-posedness


Sketch of the proof

We look for u := ud + E uc . . . ,

uc is the solution of:

d

dt(uc(t),v)σ +

µ−1 curl E uc(t), curl E v

0,Ω

+ c(E uc(t), E v)

= (f (t), E v)0,Ω ∀v ∈ H(curl; Ωc),

with the initial condition

uc(0) = 0.




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The discrete approximations Well-posedness


Sketch of the proof:

Lagrange multiplier p: Inf-Sup condition of b.

Remark

The Lagrange multiplier p of our problem vanishes identically.

γ τ µ−10 curlu = curlΓ λ



Th di i i

Analysis of the semi-discrete schemeAnalysis of the fully-discrete scheme

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The discrete approximationsAnalysis of the fully discrete scheme

Space discretization

We approximate u with edge Nedelec elements:

X h(Ω) := v ∈ H(curl; Ω) : v|K ∈ N m, ∀K ∈ T ,

where

N m := x → a + b × x, ∀x ∈ K : a, b ∈ R3 .



Th di t i ti


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The discrete approximationsy y

We approximate p in

M h(Ωd) :=

q ∈ H1(Ωd) : q|K ∈ P1 ∀K ∈ T h, (q, 1)0,Ωd

= 0,

q|Σi = C i, i = 1, . . . , I

We approximate λ in

Λh(Γ) := ϑ ∈ H1/20 (Γ) : ϑ|F ∈ P1 ∀F ∈ T h(Γ)





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The discrete approximationsy y

Well-posedness

Sketch (as in the continuous case)

define Rh and obtain the reduced semi-discrete problem,

the seminorm in H(curl; Ωd) is a norm in V h(Ωd),define an extension discrete operator E h,

obtain a stable decomposition of the discrete kernel of b,

look for uh = ud,h + E h(uc,h), discrete inf-sup condition of b,

etc.





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Semi-discrete error estimates

If u ∈ H1(0, T ;H(curl;Ω)) and eh(t) := u(t) − uh(t),

Cea’s Lemma

supt∈[0,T ]

eh(t)2H(curl;Ω) + T 0

eh(t)2H(curl;Ω) dt + T 0

∂ teh(t)2σ dt

≤C

T 0

inf

v∈Xh(Ω)∂ tu(t) − v2

H(curl;Ω) + inf χ∈Λh(Γ)

∂ tλ(t) − χ21/2,Γ

dt

+ sup[0,T ]

inf χ∈Λh(Γ)

λ(t) − χ21/2,Γ + supt∈[0,T ]

inf v∈Xh(Ω)

u(t) − v2H(curl;Ω)





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A convergence result

If u and µ−1 curlu belong to

H1(0, T ;Hr(curl;Ω)), (1/2 < r ≤ 1)

supt∈[0,T ]

eh(t)2H(curl;Ω) +

T 0

∂ teh(t)2σ dt ≤ Ch2r,

T 0 λ(t) − λh(t)

2

1/2,Γdt ≤ Ch

2r

.





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The fully-discrete scheme

Time discretization: implicit Euler scheme.

Existence and uniqueness: as in the semi-dicrete case.

Convergence analysis: analogous results to the semi-discretecase.


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