Published online by De Gruyter October 6, 2022

# FEM-BEM Coupling for the Maxwell–Landau–Lifshitz–Gilbert Equations via Convolution Quadrature: Weak Form and Numerical Approximation

Jan Bohn , Michael Feischl and Balázs Kovács

## Abstract

The full Maxwell equations in the unbounded three-dimensional space coupled to the Landau–Lifshitz–Gilbert equation serve as a well-tested model for ferromagnetic materials. We propose a weak formulation of the coupled system based on the boundary integral formulation of the exterior Maxwell equations. We show existence and partial uniqueness of a weak solution and propose a new numerical algorithm based on finite elements and boundary elements as spatial discretization with backward Euler and convolution quadrature for the time domain. This is the first numerical algorithm which is able to deal with the coupled system of Landau–Lifshitz–Gilbert equation and full Maxwell’s equations without any simplifications like quasi-static approximations (e.g. eddy current model) and without restrictions on the shape of the domain (e.g. convexity). We show well-posedness and convergence of the numerical algorithm under minimal assumptions on the regularity of the solution. This is particularly important as there are few regularity results available and one generally expects the solution to be non-smooth. Numerical experiments illustrate and expand on the theoretical results.

MSC 2010: 35Q61; 65M12; 65M38; 65M60

Award Identifier / Grant number: 258734477

Award Identifier / Grant number: 446431602

Funding source: Austrian Science Fund

Award Identifier / Grant number: SFB F65

Funding statement: Funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) “Project-ID 258734477” SFB 1173 as well as the Austrian Science Fund (FWF) under the special research program “Taming complexity in PDE systems” (grant SFB F65). The work of Balázs Kovács is additionally funded by the Heisenberg Programme of the DFG – Project-ID 446431602.

Lemma 23

## Lemma 23 (Discrete Integration by Parts)

For N N and sequences ( a j ) j = 0 , , N , ( b j ) j = 0 , , N , it holds

[ ( t τ a ) τ + , b τ - ] [ 0 , T ] = a N b N - a 0 b 0 - [ a τ + , ( t τ b ) τ + ] [ 0 , T ] , [ a τ + , b τ - ] [ 0 , T ] = ( ( t τ ) - 1 ( a k + 1 ) k ) ( t N - 1 ) b N - [ ( ( t τ ) - 1 ( a k + 1 ) k ) τ - , ( t τ b ) τ + ] [ 0 , T ] .

## Proof

It holds

[ ( t τ a ) τ + , b τ - ] [ 0 , T ] + [ a τ + , ( t τ b ) τ + ] [ 0 , T ] = τ j = 0 N - 1 a j + 1 - a j τ b j + τ j = 0 N - 1 a j + 1 b j + 1 - b j τ = j = 0 N - 1 a j + 1 b j + 1 - a j b j = a N b N - a 0 b 0 .

The second assertion can be shown similarly, by setting c 0 := 0 , c j := ( ( t τ ) - 1 ( a k + 1 ) k ) ( t j - 1 ) = τ k = 0 j - 1 a k + 1 for j = 1 , , N and using c j + 1 - c j τ = a j + 1 for j = 0 , , N - 1 :

[ a τ + , b τ - ] [ 0 , T ] + [ ( ( t τ ) - 1 ( a k + 1 ) k ) τ - , ( t τ b ) τ + ] [ 0 , T ] = τ j = 0 N - 1 c j + 1 - c j τ b j + τ j = 0 N - 1 c j + 1 b j + 1 - b j τ = c N b N - c 0 b 0 = ( ( t τ ) - 1 ( a k + 1 ) k ) ( t N - 1 ) b N .

This concludes the proof. ∎

We define for r N the exponentially weighted spaces, for c R ,

L c 2 ( [ 0 , ) , X ) := { u : [ 0 , ) X e - c u ( ) L 2 ( [ 0 , ) , X ) } ,

equipped with the norm u L c 2 ( [ 0 , ) , X ) := e - c u L 2 ( [ 0 , ) , X ) , and the exponentially weighted spaces of 𝑟-times weakly differentiable functions with zero condition at t = 0 ,

H 0 , * r ( [ 0 , ) , X ) := { ϕ : [ 0 , ) X e - c ϕ ( ) H 0 r ( [ 0 , ) , X ) for a c R } .

Using the Laplace transform ℒ and its inverse, the operator B ( t ) f (for a B : { s > σ 0 } L ( X ) and a suitable function 𝑓) is defined by

B ( t ) f := L - 1 ( B ( s ) L ( f ) ( s ) ) .

Example 24

(a) For the operator B ( s ) = s , f H 0 , * 1 ( [ 0 , ) , X ) , it holds s ( L f ) ( s ) H , and we have

B ( t ) f = t f .

Thus the Laplace differential operator t coincides with the weak derivative t if 𝑓 is weakly differentiable and f ( 0 ) = 0 .

(b) For the operator B ( s ) = s - 1 and for a function f L * 2 ( [ 0 , ) , X ) with Laplace transform s - 1 ( L f ) ( s ) H , we have

B ( t ) = t - 1 f := 0 t f ( τ ) d τ .

Thus the Laplace differential operator t - 1 coincides with the integration over time 0 t d τ .

## Proof

(b) Let f L σ 2 ( [ 0 , , X ) ) ; then we have s - 1 L f ( s ) H ( max ( σ , ε ) ) for ε > 0 . Furthermore, it holds for s > max ( σ , ε ) that r 1 s e - s r f ( r ) L 1 ( [ 0 , ) , X ) , and therefore we have, by Fubini’s theorem,

L ( t - 1 f ) ( s ) = 0 e - s t 0 t f ( r ) d r d t = 0 0 1 r t e - s t f ( r ) d r d t = 0 r e - s t d t f ( r ) d r = 0 1 s e - s r f ( r ) d r = 1 s L f ( s ) .

As s - 1 L f ( s ) H , it holds t - 1 f = L - 1 s - 1 L f ( s ) .

(a) Let f H 0 , σ 1 ( [ 0 , ) , X ) for σ R . It is t - 1 t f = f , and therefore, by (b) for s max ( σ , ε ) > 0 ,

1 s L ( t f ) ( s ) = L f ( s ) .

As L ( t f ) H , it holds t f = L - 1 ( s L f ) . ∎

We want to apply the (inverse) Laplace transform to operators, B ( s ) : X X and convolute with functions f ( t ) X . The difference comparing to the scalar case is now, with the induced norm, L ( X ) is no Hilbert space, but only a Banach space. Plancherel’s formula does not hold in general.

Lemma 25

For B L 1 ( σ 0 + i R , L ( X ) ) H ( σ 0 ) the convolution with the inverse Laplace transform gives for every δ > 0 a well-defined and continuous operator

L - 1 B * : L σ 0 2 ( [ 0 , ) , X ) L σ 0 + δ 2 ( [ 0 , ) , X ) ,

and it holds

L - 1 B * u L σ 0 + δ 2 ( [ 0 , ) , X ) C ( δ ) B L 1 ( σ 0 + i R , L ( X ) ) u L σ 0 2 ( [ 0 , ) , X ) .

## Proof

The proof works by combining Hölder’s inequality

L - 1 B * u L σ 0 + δ 2 ( [ 0 , ) , X ) C ( δ ) e - ( σ 0 + δ / 2 ) ( ) L - 1 B * u L ( [ 0 , ) , X )

and Young’s inequality for convolution

e - ( σ 0 + δ / 2 ) ( ) L - 1 B * u L ( [ 0 , ) , X ) e - ( σ 0 + δ / 2 ) ( ) L - 1 B L ( [ 0 , ) , L ( X ) ) e - ( σ 0 + δ / 2 ) ( ) u L 1 ( [ 0 , ) , X ) .

Then the estimates for the inverse Laplace transform which follow from the equivalence with the Fourier transform,

e - ( σ 0 + δ / 2 ) ( ) L - 1 B L ( [ 0 , ) , L ( X ) ) 1 2 π B L 1 ( σ 0 + i R , L ( X ) ) ,

and again Hölder’s inequality

e - ( σ 0 + δ / 2 ) ( ) u L 1 ( [ 0 , ) , X ) C ( δ ) u L σ 0 2 ( [ 0 , ) , X ) .

conclude the proof. ∎

Lemma 26

## Lemma 26 (cf. [39, Lemma 2.1])

Let r N 0 . For

B H r := H r ( σ 0 ) := { B : { s > σ 0 } L ( X ) holomorphic B ( s ) L ( X ) C | s | r for all s > σ 0 } ,

B ( t ) f exists for every f H 0 , * r ( [ 0 , T ] , X ) and it holds L - 1 ( B ( s ) s - r L f ) H 0 , * r ( [ 0 , T ] , X ) and

B ( t ) f = 1 [ 0 , T ] L - 1 ( B ( s ) s - r L ( t r f ) )

We can define B ( t ) as a continuous operator

B ( t ) : H 0 , * r ( [ 0 , T ] , X ) L 2 ( [ 0 , T ] , X ) .

Every B H r is causal, and for every sufficiently smooth extension f ~ of 𝑓 on [ 0 , ) , it holds

B ( t ) f = 1 [ 0 , T ] L - 1 ( B ( s ) L f ~ ) .

Theorem 27

## Theorem 27 (Herglotz Theorem on [ 0 , T ] , cf. [8, Lemma 2.2])

Let B , R H r ( σ 0 ) for σ 0 R , and suppose that a ( , ) : X × X C is sesquilinear and continuous. If there exists c > 0 such that, for all w X , all s > σ 0 ,

a ( w , B ( s ) w ) c R ( s ) w X 2 ,

then it holds, for all w H 0 , * r ( [ 0 , T ] , X ) , for all σ σ 0 ,

0 T e - 2 σ t a ( w ( t ) , B ( t ) w ( t ) ) d t c e - 2 σ T R ( t ) w L 2 ( [ 0 , T ] , X ) 2 .

## Proof

The assertion can be shown as in the scalar case by a discrete Herglotz theorem (cf. [31, Lemma 2.1]) and the convergence of CQ.∎

Theorem 28

## Theorem 28 (Discrete Herglotz Theorem on [ 0 , T ] , cf. [8, Lemma 2.1 and 2.3])

Let B H m ( σ 0 ) for σ 0 R + . For N N sufficiently large and a sequence ( w n ) n = 0 , , N X , it holds

τ j = 0 N ( B ( t τ ) w ) ( t j ) 2 C τ j = 0 N ( ( t τ ) r w ) ( t j ) 2 .

The constant 𝐶 depends on σ 0 , 𝑇 and 𝐵, but not on 𝜏.

## Proof

We extend 𝑤 to a sequence ( w n ) n N such that ( ( t τ ) r w ) ( t j ) = 0 for all j > N . This is always possible by an iterative procedure, as we can write ( ( t τ ) r w ) ( t k + 1 ) = w k + 1 / τ r - f ( ( w n ) n k ) , where f ( ( w n ) n k ) does not depend on w k + 1 . Now we compute iteratively w N + 1 such that ( ( t τ ) r w ) ( t N + 1 ) = 0 , w N + 2 such that ( ( t τ ) r w ) ( t N + 2 ) = 0 , .

Now we define the finite sequence w M j := w j for j = 0 , , M and w M j = 0 , j > M . As in Lemma 15, we have, for ρ = e - 2 σ 0 τ , | ζ | < ρ and sufficiently small 𝜏,

( δ ( ζ ) τ ) 1 - e - 2 σ 0 τ τ = 0 2 σ 0 e - τ r d r 2 σ 0 e - 2 τ σ 0 > σ 0 .

With arguments similar to [8, Lemma 2.1, Lemma 2.3], we obtain

τ j = 0 e - 4 σ 0 t j ( B ( t τ ) w ) ( t j ) 2 C τ j = 0 e - 4 σ 0 t j ( ( t τ ) r w M ) ( t j ) 2 .

For j M , it is w j C t j r (this can be shown by discrete integration) and therefore

τ j = 0 e - 4 σ 0 t j ( ( t τ ) r w M ) ( t j ) 2 τ j = 0 N e - 4 σ 0 t j ( ( t τ ) r w M ) ( t j ) 2 + C ( τ , r ) e - 4 σ 0 t M t M r

and the limit M exists on the right-hand side. We obtain by discrete causality (i.e., B ( t τ ) w ( t j ) is independent of w n , n > j ), for M > N ,

τ j = 0 N e - 4 σ 0 t j ( B ( t τ ) w ) ( t j ) 2 = τ j = 0 N e - 4 σ 0 t j ( B ( t τ ) w M ) ( t j ) 2 τ j = 0 e - 4 σ 0 t j ( B ( t τ ) w M ) ( t j ) 2 .

Combining the previous estimates for the limit M gives

τ j = 0 N e - 4 σ 0 t j ( B ( t τ ) w ) ( t j ) 2 C τ j = 0 N e - 4 σ 0 t j ( ( t τ ) m w ) ( t j ) 2 .

Now the bounds e - 4 σ 0 T e - 4 σ 0 t j 1 yield the assertion. ∎

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