[PPT] - Eigenvalues of the curl operator: variational formulation and PowerPoint Presentation

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Introduction The mathematical problem The variational problem Spectral analysis Finite element approximation Numerical results

Eigenvalues of the curl operator: variational formulation and numerical approximation

Alberto Valli

Dipartimento di Matematica, Universit` a di Trento, Italy

A. Valli

Eigenvalus of the curl

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Joint paper with: Ana Alonso Rodr´ ıguez Dipartimento di Matematica, Universit` a di Trento, Italy Jessika Cama˜ no Departamento de Matem´ atica y F´ ısica Aplicadas, Universidad Cat´

lica de la Sant´

ısima Concepci´

n, Chile

Rodolfo Rodr´ ıguez Departamento de Ingenier´ ıa Matem´ atica, Universidad de Concepci´

n, Chile

Pablo Venegas Departamento de Matem´ atica, Universidad del B´ ıo B´ ıo, Chile

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Eigenvalus of the curl

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Outline

1

Introduction

2

The mathematical problem

3

The variational problem

4

Spectral analysis

5

Finite element approximation

6

Numerical results

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Eigenvalus of the curl

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Introduction The mathematical problem The variational problem Spectral analysis Finite element approximation Numerical results

Introduction

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Eigenvalus of the curl

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Physical framework

By the Lorentz law the density of the magnetic force is given by F = J × B, where J is the current density and B is the magnetic induction. Linear isotropic media: B = µH (the scalar function µ being the magnetic permeability). Eddy current or static approximation: J = curl H. If curl H = λH (λ a scalar function) the magnetic force vanishes: F = curl H × µH = λH × µH = 0 .

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Eigenvalus of the curl

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Physical framework (cont’d)

Fields satisfying curl H = λH are called force-free fields. If λ is a constant are called linear force-free fields. [Clearly, the most interesting case is for λ not identically vanishing.] [In fluid dynamics, force-free fields are called Beltrami fields, and a Beltrami field u that is divergence-free and tangential to the boundary is a steady solution of the Euler equations for incompressible inviscid flows (with pressure given by p = −|u|2/2).]

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Physical framework (cont’d)

A couple of interesting physical remarks: a field which is divergence-free and tangential to the boundary (e.g., the magnetic field), and which minimizes the magnetic energy with fixed helicity is a linear force-free field [Woltjer (1958)]; linear force-free fields are time-asymptotic configurations (they remain force-free as time changes) [Jette (1970)]. [Helicity of a vector field v in a domain Ω, i.e., H(v) = 1 4π

Ω
Ω

v(x) × v(y) · x − y |x − y|3 dxdy , is a “measure of the extent to which the field lines wrap and coil around one another” (Cantarella et al. (2000:a); other physical remarks can be found there and in Cantarella et al. (2000:b)).]

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Eigenvalus of the curl

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The mathematical problem

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Eigenvalus of the curl

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Mathematical framework

Let us focus now on the mathematical aspects of this problem. A linear force-free field is an eigenfunction of the curl operator: curl u = λu . It is thus interesting to see when it is possible to define self-adjoint realizations of the curl operator. The starting point is clearly the Green’s formula (here and in the sequel we write Γ = ∂Ω)

Ω

(v · curl w − curl v · w) =

Γ

v × n · w , and the analysis is driven by the need of obtaining

Γ v × n · w = 0.
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Eigenvalus of the curl

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The spectral problem: Ω simply-connected

It is clear that

Γ v × n · w = 0 when v × n = 0 on Γ. However,

this boundary condition is too strong for the spectral problem. Weaker boundary conditions can be devised (see, e.g., Kress (1972, 1986); Picard (1976, 1998); Yoshida and Giga (1990)). When Ω is a simply-connected domain, it is sufficient to assume that curl v · n = 0 on Γ. Since for an eigenfunction u of the curl

perator the condition u · n = 0 implies curl u · n = 0, it is thus

natural to consider the spectral problem curl u = λu in Ω div u = 0 in Ω u · n = 0

n Γ .

(1) The numerical approximation of (1) has been analyzed in Rodr´ ıguez and Venegas (2014).

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Eigenvalus of the curl

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The spectral problem: Ω not simply-connected

The condition curl v · n = 0 is not enough if the physical domain is not simply-connected. A possible additional condition is the following (see, e.g., Kress (1972, 1986); Picard (1976, 1998); Yoshida and Giga (1990)): the curl operator is self-adjoint if curl v · n = 0 on Γ (which is equivalent to curl v⊥∇H1(Ω)) and curl v⊥KT, where KT is the space of the so-called harmonic Neumann fields h (those fields satisfying curl h = 0 in Ω, div h = 0 in Ω and h · n = 0 on Γ). This space KT is finite dimensional, its dimension being the first Betti number of Ω; it is trivial for a simply-connected domain Ω. It can be proved that H0(curl; Ω) = ∇H1(Ω)

⊥

⊕ KT. The numerical approximation of the eigenvalues and eigenfunctions

f this problem has been studied in Lara et al. (2016).
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The spectral problem: Ω not simply-connected (cont’d)

It is worth noting that the condition curl v⊥KT is not essential, but only sufficient for the proof that the curl operator is self-adjoint. In this respect, an in-depth analysis has been recently presented in Hiptmair et al. (2012). The authors, by incorporating in problem (1) additional conditions related to the first homology group of Γ, devise suitable self-adjoint realizations of the curl operator.

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Homological tools

Let us show how this family of eigenvalue problems can be described. We first need to recall some geometrical results. Let g be the first Betti number Ω; then the first Betti number of Γ is equal to 2g. From algebraic topology we know that:

n Γ there are 2g non-bounding cycles {γj}g

j=1 ∪ {γ′ j}g j=1, that

are the generators of the first homology group of Γ {γj}g

j=1 are the generators of the first homology group of Ω′,

with Ω′ = B \ Ω, B being an open ball containing Ω (tangent vector on γj denoted by tj); {γ′

j}g j=1 are the generators of the first homology group of Ω

(tangent vector on γ′

j denoted by t′ j);

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Homological tools (cont’d)

in Ω there exist g ‘cutting’ surfaces {Σj}g

j=1, that are

connected orientable Lipschitz surfaces satisfying Σj ⊂ Ω and ∂Σj ⊂ Γ, such that every curl-free vector in Ω has a global potential in the ‘cut’ domain Ω0 := Ω \ g

j=1 Σj; each surface

Σj satisfies ∂Σj = γj, ‘cuts’ the corresponding cycle γ′

j and

does not intersect the other cycles γ′

i for i = j;

in Ω′ there exist g ‘cutting’ surfaces {Σ′

j}g j=1, that are

connected orientable Lipschitz surfaces satisfying Σ′

j ⊂ Ω′ and

∂Σ′

j ⊂ Γ, such that every curl-free vector in Ω′ has a global

potential in the ‘cut’ domain (Ω′)0 := Ω′ \ g

j=1 Σ′ j; each

surface Σ′

j satisfies ∂Σ′ j = γ′ j, ‘cuts’ the corresponding cycle

γj, and does not intersect the other cycles γi for i = j.

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Homological tools (cont’d)

[Some misunderstanding appears when looking back at the literature on this topic; it is thus interesting to make clear that: the statement concerning the ‘cutting’ surfaces Σj does not mean that the ‘cut’ domain Ω0 is simply-connected nor that it is homologically trivial: an example in this sense is furnished by Ω = Q \ K, where Q is a cube and K is the trefoil knot.]

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A typical geometrical situation Figure: Toroidal domain. Σ1 and Σ′

1 represent the ’cutting’ surfaces of Ω

and Ω′, respectively.

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New homological conditions

Hiptmair et al. (2012), have shown that the curl operator is self-adjoint in the space of vector fields v with curl v · n = 0 on Γ and such that

γi v · ti = 0 for 1 ≤ i ≤ g1 and
γ′

j v · t′

j = 0 for

g1 + 1 ≤ j ≤ g, where g1 is a fixed number satisfying 0 ≤ g1 ≤ g. If curl v · n = 0 on Γ the choice g1 = g, namely,

γi v · ti = 0

for 1 ≤ i ≤ g1 = g, is equivalent to the previous condition curl v⊥KT [we wll return on this point in the sequel].

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New homological conditions (cont’d)

However, the most interesting physical case is the one given by the choice g1 = 0, i.e., the additional conditions are given by

γ′

j v · t′

j = 0 for 1 ≤ j ≤ g.

In fact, in this case the eigenfunction associated to the eigenvalue

f minimum absolute value realizes the minimum of the magnetic

energy with fixed helicity.

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The eigenvalue problem

Summing up, we consider the following eigenvalue problem: Problem 1. Find λ ∈ C and u ∈ L2(Ω)3, u = 0, such that curl u = λu in Ω div u = 0 in Ω u · n = 0

n Γ
γi u · ti = 0

1 ≤ i ≤ g1

γ′

j u · t′

j = 0

g1 + 1 ≤ j ≤ g , (2) where 0 ≤ g1 ≤ g.

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A basis for KT

A basis of the space KT is given by

ρj

g

j=1, where ρj =

∇φj (the L2(Ω)3-extension of ∇φj), and φj is the unique solution of ∆φj = 0 in Ω \ Σj , ∂nφj = 0

n

∂Ω , [ [ ∂nφj ] ]Σj = 0 , [ [ φj ] ]Σj = 1 . In a similar way we construct a basis

ρ′

j

g

j=1 of the space of

harmonic Neumann vector fields K′

T associated to the domain

Ω′ = B \ Ω.

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The meaning of the line integrals

By the Stokes theorem and integration by parts we can see that the meaning of the line integrals is:

γi u · ti =
Σi curl u · ni =
Σi curl u · ni [

[ φi ] ]Σi =

Ω\Σi curl u · ∇φi =
Ω curl u ·

∇φi =

Ω curl u · ρi =
Γ n × u · ρi ,

and similarly

γ′

j u · t′

j = −

Γ n × u · ρ′

j .

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The “orthogonality” relations

In particular, the fundamental “orthogonality” relations

γi ρs · ti =
Γ n × ρs · ρi = 0
γ′

j ρ′

s · t′ j = −

Γ n × ρ′

s · ρ′ j = 0

γ′

i ρs · t′

i = −

Γ n × ρs · ρ′

i = δs,i

γj ρ′

s · tj =

Γ n × ρ′

s · ρj = δs,j

(3) hold true, together with

γi ∇ω · ti =
Γ n × ∇ω · ρi = 0
γ′

j ∇ω · t′

j = −

Γ n × ∇ω · ρ′

j = 0 .

(4)

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The variational problem

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Variational formulation

Some function spaces: X = {v ∈ H(curl; Ω) : curl v · n = 0 on Γ} , X 0 = {v ∈ X :

γi v · ti = 0 for i = 1, . . . , g1} ,

X ⋆ = {v ∈ X :

γ′

j v · t′

j = 0 for j = g1 + 1, . . . , g} ,

Z = X 0 ∩ X ⋆ , Q = ∇H1(Ω) ⊕ span {ρ1, . . . , ρg1} . The following equality holds true Q = X ⋆ ∩ H(curl 0; Ω) = Z ∩ H(curl 0; Ω) . (5) (for the second equality, the key point is that the cycles γi are bounding in Ω!).

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Variational formulation (cont’d)

A saddle-point formulation of Problem 1 is devised by looking for a vector field u in Z and a Lagrange multiplier q (associated to the divergence-free and boundary constraints) in Q. It reads: Problem 2. Find λ ∈ C and (u, q) ∈ Z × Q, u = 0, such that

Ω

curl u · curl v +

Ω

q · v = λ

Ω

u · curl v ∀ v ∈ Z (6a)

Ω

u · p = 0 ∀ p ∈ Q . (6b)

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Variational formulation (cont’d)

It is worth noting that the Lagrange multiplier q in Problem 2 is vanishing. Moreover, an eigenvalue λ must be different from 0. In fact, taking v = q ∈ Q ⊂ Z in (6a) it follows q = 0. Hence, if we suppose λ = 0, we obtain curl u = 0 in Ω, and consequently u ∈ Q by equality (5). Choosing p = u in (6b) it follows u = 0, and we conclude that λ = 0 is not admissible.

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Equivalence of the eigenvalue problems

Lemma If (λ, u), λ = 0, is a solution to Problem 1, then (λ, u, 0) is a solution to Problem 2. If (λ, u, q) is a solution to Problem 2, then q = 0 and (λ, u) is a solution to Problem 1.

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Equivalence of the eigenvalue problems (cont’d)

From Problem 1 to Problem 2 If (λ, u), λ = 0, is a solution to Problem 1, then clearly (6a) is satisfied with q = 0. Since u ∈ H0(div 0; Ω), it follows that u is

rthogonal to the gradients. Therefore, recalling that p ∈ Q can

be written as p = ∇ψ + g1

i=1 αiρi, we have only to prove that

Ω u · ρi = 0 for i = 1, . . . , g1. We have already seen that
γi u · ti =
Ω curl u · ρi ,

hence the condition

γi u · ti = 0 can be interpreted as

λ

Ω u · ρi = 0, and the thesis follows because λ = 0.

✷ [Here we have also seen that

γi u · ti = 0 for each i = 1, . . . , g

means curl u⊥KT...]

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Equivalence of the eigenvalue problems (cont’d)

From Problem 2 to Problem 1 If (λ, u, q) is a solution to Problem 2, we have already seen that q = 0. Moreover, taking in (6b) p ∈ ∇H1(Ω) ⊂ Q it follows div u = 0 in Ω and u · n = 0 on Γ. Therefore, we only need to prove that curl u = λu in Ω. By integrating by parts (6a) we find curl(curl u − λu) = 0 in Ω and

Γ

(curl u − λu) · n × v = 0 (7) for each v ∈ Z. We know that curl u − λu = ∇ϕ + g

s=1 βsρs, as curl u − λu is

curl-free. Moreover, for v ∈ Z we have on Γ n × v = n × (∇ω +

g1

k=1

ζkρk +

g

l=g1+1

ηlρ′

l) .

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Equivalence of the eigenvalue problems (cont’d)

Thus, using the “orthogonality” relations (3) and (4) for ρi and ρ′

j, it follows that curl u − λu = ∇ϕ + g1 i=1 βiρi ∈ Q.

Therefore by (6b) u is orthogonal to λu − curl u and we have 0 =

Ω λu · (λu − curl u)

=

Ω(λu − curl u) · (λu − curl u)

+

Ω curl u · (λu − curl u) .

The last integral vanishes due to (6a), thus it follows curl u = λu in Ω. ✷

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Spectral analysis

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The solution operator

In order to obtain a spectral characterization of Problem 2 we introduce the following solution operator: T : Z − → Z , f − → Tf := w , where (w, q) ∈ Z × Q is the solution of

Ω

curl w · curl v +

Ω

q · v =

Ω

f · curl v ∀ v ∈ Z (8a)

Ω

w · p = 0 ∀ p ∈ Q . (8b)

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The solution operator: well-posedness

The well-posedness of T follows by these three lemmas. Lemma (Poincar´ e) The seminorm |w| =

curl w2

0,Ω + div w2 0,Ω + g1 i=1

Ω w · ρi
2

+ g

j=g1+1

γ′

j w · t′

j

2 1/2

is equivalent to the norm in X ∩ H0(div ; Ω).

Proof. By contradiction, in a somehow standard way. The main

point is showing that a harmonic Neumann field w with g1

i=1

Ω w · ρi
2 = 0 and g

j=g1+1

γ′

j w · t′

j

2

= 0 is null. ✷

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The solution operator: well-posedness (cont’d)

Lemma (ellipticity in the kernel) There exists α > 0 such that

Ω

|curl v|2 ≥ αv2

curl,Ω

∀v ∈ V, where V =

v ∈ Z :
Ω

v · p = 0 ∀p ∈ Q

.
Proof. By the Poincar´

e lemma, as an element v ∈ V satisfies div v = 0 in Ω, v · n = 0 on Γ,

Ω w · ρi = 0 for i = 1, . . . , g1 (as it

is orthogonal to Q) and

γ′

j w · t′

j = 0 for j = g1 + 1, . . . , g (as it

belongs to Z). ✷

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The solution operator: well-posedness (cont’d)

Lemma (inf–sup condition) There exists β > 0 such that sup

v∈Z,v=0

Ω v · p
vcurl,Ω

≥ β p0,Ω ∀ p ∈ Q .

Proof. Just take v = p ∈ Q ⊂ Z, and note that curl v = 0.

✷ Thus problem (8) is well-posed (Babuˇ ska–Brezzi theory for saddle-point problems).

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Looking for the eigenvalues of T

We have that Tu = µu, with µ = 0, if and only if (λ, u, 0) is a solution of Problem 2, with λ = 1/µ. We thus focus on the spectrum of T. The following result is easily proved: Lemma The Lagrange multiplier q in (8a) is null Tf ∈ H(curl; Ω) ∩ H0(div 0; Ω) and curl Tf ∈ H(curl; Ω) ∩ H0(div 0; Ω) T is compact (curl Tf − f) ∈ Q.

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Symmetry of the curl operator

For the analysis of the spectral problem the fundamental theorem is: Theorem (Symmetry of the curl) For all v, w ∈ Z,

Ω

(curl w · v − w · curl v) = 0.

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Symmetry of the curl operator (cont’d)

Proof. Since
Ω

curl w · v −

Ω

w · curl v =

Γ

(n × w) · v , the point is to show that

Γ(n × w) · v = 0. Recall that w ∈ Z can

be written on Γ as n × w = n × (∇ω +

g1

k=1

ζkρk +

g

l=g1+1

ηlρ′

l) ,

and analogously v ∈ Z. Moreover it is easily proved that

Γ

(n × ∇ω) · ∇θ =

Ω

(curl ∇ω) · ∇θ = 0 . Hence, using the orthogonality relations (3) and (4), the result follows at once. ✷

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Self-adjointness of the operator T

The symmetry of the curl operator has this easy but pivotal consequence: Theorem The operator T : Z → Z is self-adjoint.

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Self-adjointness of the operator T (cont’d)

Proof. We know that (curl Tf − f) and (curl Tg − g) belong to
Q. Since Tf, Tg satisfy (8b) we have that
Ω Tf · g

=

Ω Tf · g +
Ω Tf · (curl Tg − g)

=

Ω Tf · curl Tg =
Ω curl Tf · Tg

=

Ω(f − curl Tf) · Tg +
Ω curl Tf · Tg

=

Ω f · Tg ,

having used the symmetry of the curl operator in Z. On the other hand, from (8a) and the symmetry again, we obtain

Ω curl Tf · curl g =
Ω f · curl g =
Ω curl f · g

=

Ω curl f · curl Tg ,

which ends the proof. ✷

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Spectral analysis of the operator T

We are now in a position to obtain a spectral characterization of T. Theorem The spectrum of T is given by sp(T)={0} ∪ {µn}∞

n=1, where

µ0 = 0 is an infinite-multiplicity eigenvalue (and its associated eigenspace is Q) {µn}∞

n=1 is a sequence of finite-multiplicity eigenvalues which

converges to 0 and the associated eigenfunctions {un}∞

n=1 are

a Hilbertian basis of Z.

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Spectral analysis of the operator T (cont’d)

Proof. The spectral result is classical; we only need to prove that

KerT=Q. We easily have KerT =

f ∈ Z :
Ω

f · curl v = 0 ∀ v ∈ Z

⊂ Q .

Conversely, f ∈ Q ⊂ Z satisfies curl f = 0 in Ω, thus by the symmetry of the curl operator for all v ∈ Z it holds

Ω f · curl v =
Ω curl f · v = 0.

✷

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Finite element approximation

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N´ ed´ elec finite elements

We consider a family of triangulations Th of the polyhedral domain Ω. For k ≥ 1 and T ∈ Th Pk is the set of polynomials of degree not greater than k

Pk is the subset of homogeneous polynomials of degree k

N k(T) = Pk−1(T)3 ⊕ {p ∈ Pk(T)3 : p(x) · x = 0}. The corresponding global space to approximate H(curl; Ω) is the well-known N´ ed´ elec finite element space: N k

h = {vh ∈ H(curl; Ω) : vh ∈ N k(T) ∀T ∈ Th}.

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Discrete spaces

Whence, the natural approximation space for Z is Zh = Z ∩ N k

h ,

namely, Zh = {vh ∈ N k

h : curl vh · n = 0 on Γ,

γi vh · ti = 0 for i = 1, . . . , g1
γ′

j vh · t′

j = 0 for j = g1 + 1, . . . , g} .

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Discrete spaces (cont’d)

To discretize the Lagrange multiplier q ∈ Q we use the finite element space Qh = Zh ∩ H(curl 0; Ω) . Note that Qh = {vh ∈ N k

h : curl vh = 0 in Ω,

γ′

j vh · t′

j = 0 , for j = g1 + 1, . . . , g} ,

since the cycles γi are bounding in Ω and therefore a curl-free vector field has vanishing line integral on them.

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The discrete problem

We are now in position to introduce a finite element discretization

f Problem 2.

Problem 3. Find λh ∈ C and (uh, qh) ∈ Zh × Qh, uh = 0, such that

Ω

curl uh · curl vh +

Ω

qh · vh = λh

Ω

uh · curl vh ∀ vh ∈ Zh

Ω

uh · ph = 0 ∀ ph ∈ Qh . As for the continuous problem we have that the Lagrange multiplier qh is null and that the eigenvalues satisfy λh = 0.

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The discrete solution operator

We also consider the corresponding discrete solution operator: Th : Z − → Z , f − → Thf := wh , where (wh, qh) ∈ Zh × Qh is the solution of

Ω

curl wh · curl vh +

Ω

qh · vh =

Ω

f · curl vh ∀ vh ∈ Zh

Ω

wh · ph = 0 ∀ ph ∈ Qh .

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The discrete solution operator: well-posedness

We need to satisfy the Babuˇ ska–Brezzi conditions. Lemma (discrete ellipticity in the kernel) There exists α > 0, independent of h, such that

Ω

|curl vh|2 ≥ αvh2

curl,Ω

∀vh ∈ Vh, where Vh =

vh ∈ Zh :
Ω

vh · ph = 0 ∀ph ∈ Qh

.
Proof. More technical than in the continuous case, but along a

well-established path. ✷

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The discrete solution operator: well-posedness (cont’d)

Lemma (discrete inf-sup condition) There exists β > 0, independent of h, such that sup

vh∈Zh,vh=0

Ω vh · ph
vhcurl;Ω

≥ β ph0,Ω , ∀ ph ∈ Qh.

Proof. As in the continuous case, the inf–sup condition is easily

checked by taking vh = ph ∈ Qh ⊂ Zh. ✷

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Convergence of the solution operators

The classical theory for compact operators can be used to prove that the eigenvalues and eigenfunctions of Problem 2 are well-approximated by those of Problem 3. To this aim, a fundamental result is: Lemma (Convergence in norm) There exists C > 0, independent of h, such that for all f ∈ Z (T − Th)fcurl;Ω ≤ Chmin{s,k}fcurl;Ω .

Proof. For f ∈ Z we have that Tf ∈ H(curl; Ω) ∩ H0(div 0; Ω)

and curl Tf ∈ H(curl; Ω) ∩ H0(div 0; Ω); hence it is enough to use a C´ ea-type estimate and the fact that the Lagrange multipliers vanish. ✷

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Convergence results

The convergence results are: let λ be an eigenvalue of Problem 2 with multiplicity m and E ⊂ Z the corresponding eigenspace; then, there exist exactly m eigenvalues λ(1)

h , . . . , λ(m) h

f Problem 3 which converge to

λ as h → 0 let Eh be the direct sum of the eigenspaces corresponding to λ(1)

h , ..., λ(m) h

; then δ(E, Eh) → 0 as h → 0, where

δ(E, Eh) := max{δ(E, Eh), δ(Eh, E)} ,

with δ(M, N) := sup x∈M

x=1

dist(x, N).

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Error estimates

The following error estimates hold true: Theorem Let r > 0 be such that E ⊂ Hr(curl; Ω). There exist constants C1, C2 > 0, independents of h, such that, for small h,

δ(E, Eh) ≤ C1hmin{r,k} ,

(9) and |λ − λ(i)

h | ≤ C2h2 min{r,k},

i = 1, ..., m . (10)

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Numerical results

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Test 1: Domain with first Betti number g = 1 and g1 = 1. Ω is a toroidal domain of rectangular cross section 0.005 ≤ R ≤ 1 and −1/2 ≤ z ≤ 1/2 the least positive eigenvalue is λ ≈ 1.73457π ≈ 5.449 (of multiplicity 2) [Morse (2007)] we use the lowest-order N´ ed´ elec elements N 1

h on tetrahedra.

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Nh λh,1 λh,2 10286 5.624 5.625 18993 5.562 5.564 38304 5.517 5.518 60758 5.500 5.500 λext 5.452 5.449

rder

2.12 2.12 λ 5.449 5.449 Table: Test 1. Smallest positive eigenvalues computed on different meshes.

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Test 2: Domain with first Betti number g = 1 and g1 = 0. Ω as in the Figure below (r1 = 1 and r2 = 0.5) no analytical solution is available we use the lowest-order N´ ed´ elec elements N 1

h on tetrahedra.

Figure: Test 2. Half of the toroidal domain for the numerical test.

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Nh λh,1 λh,2 λh,3 λh,4 λh,5 15554 5.066 6.633 6.636 6.710 6.716 33901 4.986 6.438 6.441 6.505 6.506 65720 4.958 6.372 6.376 6.432 6.433 129187 4.931 6.311 6.312 6.367 6.368 195745 4.919 6.282 6.282 6.336 6.336 247239 4.915 6.272 6.272 6.326 6.326 λext 4.896 6.231 6.228 6.280 6.281

rder

2.22 2.31 2.25 2.28 2.31 Table: Test 2. Smallest positive eigenvalues computed on different meshes.

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10

4

10

5

10

6

10

2

10

1

10

error N

Error Fitted linear model

Figure: Test 2. Error curve for the smallest positive eigenvalue: loglog plot of the computed error |λh,1 − λext| versus the number of tetrahedra Nh.

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Figure: Test 2. Beltrami field corresponding to the smallest positive eigenvalue.

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A remark: except for the smallest eigenvalue λh,1, the eigenvalues are similar to those computed by Lara et al. (2016) for the case g = 1 and g1 = 1 the smallest eigenvalue is the most interesting one from the physical point of view, as the associated eigenfunction realizes the minimum of the magnetic energy with fixed helicity.

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Test 3: Domain with first Betti number g = 2 and different values

f g1.

Ω as in the Figure below no analytical solution is available we use the lowest–order N´ ed´ elec elements on hexahedra.

Figure: Test 3. Two-fold toroidal domain for the numerical test.

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Nh λh,1 λh,2 λh,3 λh,4 λh,5 λh,6 λh,7 1280 9.345 9.456 11.471 11.559 12.209 12.238 12.414 4320 8.946 9.047 10.784 10.843 11.406 11.423 11.568 10240 8.815 8.912 10.561 10.611 11.149 11.162 11.297 20000 8.756 8.851 10.460 10.506 11.033 11.044 11.175 34560 8.724 8.819 10.406 10.449 10.971 10.981 11.110 λext 8.658 8.754 10.297 10.342 10.850 10.858 10.988

rder

2.13 2.16 2.16 2.19 2.19 2.19 2.22 Table: Test 3. Smallest positive eigenvalues computed on different meshes for the problem corresponding to the case g = 2, g1 = 0.

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10

3

10

4

10

5

10

2

10

1

10 10

1

error N

Fitted linear model Error Fitted linear model Error Fitted linear model Error

Figure: Test 3. Error curve for the three smallest positive eigenvalue for the problem corresponding to the case g = 2, g1 = 0: loglog plot of the computed error |λh,1 − λext| versus the number of hexahedra Nh.

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Nh λh,1 λh,2 λh,3 λh,4 λh,5 λh,6 1280 9.412 11.465 11.559 12.206 12.234 12.414 4320 9.007 10.779 10.843 11.403 11.421 11.568 10240 8.874 10.557 10.611 11.146 11.160 11.297 20000 8.814 10.456 10.506 11.030 11.043 11.175 34560 8.781 10.402 10.449 10.968 10.980 11.110 λext 8.714 10.293 10.342 10.846 10.857 10.988

rder

2.13 2.16 2.19 2.19 2.19 2.22 Table: Test 3. Smallest positive eigenvalues computed on different meshes for the problem corresponding to the case g = 2, g1 = 1.

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Nh λh,1 λh,2 λh,3 λh,4 λh,5 1280 11.462 11.559 12.203 12.230 12.414 4320 10.776 10.843 11.399 11.419 11.568 10240 10.554 10.611 11.141 11.159 11.297 20000 10.454 10.506 11.025 11.042 11.175 34560 10.399 10.449 10.962 10.979 11.110 λext 10.291 10.342 10.841 10.856 10.988

rder

2.16 2.19 2.19 2.19 2.22 Table: Test 3. Smallest positive eigenvalues computed on different meshes for the problem corresponding to the case g = g1 = 2.

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Figure: Test 3. Left to right. Eigenfunctions corresponding to the eigenvalues λh,1 = 8.815 and λh,2 = 8.912 for the case g = 2, g1 = 0 (mesh with Nh = 10240).

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Figure: Test 3. Left to right. Eigenfunctions corresponding to the eigenvalues λh,3 = 10.561 and λh,4 = 10.611 for the case g = 2, g1 = 0 (mesh with Nh = 10240).

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Figure: Test 3. Eigenfunction corresponding to the eigenvalue λh,1 = 8.874 for the case g = 2, g1 = 1 (mesh with Nh = 10240).

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References

J. Cantarella, D. DeTurck, H. Gluck and M. Teytel,

Isoperimetric problems for the helicity of vector fields and the Biot–Savart and curl operators, J. Math. Phys., 41 (2000:a), 5615–5641.

J. Cantarella, D. DeTurck and H. Gluck, The spectrum of the

curl operator on spherically symmetric domains, Phys. Plasmas, 7 (2000:b), 2766–2775.

R. Hiptmair, P.R. Kotiuga and S. Tordeux, Self-adjoint curl
perators, Ann. Mat. Pura Appl. (4), 191 (2012), 431–457.

A.D. Jette, Force-free magnetic fields in resistive magnetohydrostatics, J. Math. Anal. Appl., 29 (1970), 109–122.

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References (cont’d)

R. Kress, Ein Neumannsches Randwertproblem bei kraftfreinen

Feldern, Meth. Verf. Math. Phys., 7 (1972), 81–97.

R. Kress, On constant-alpha force-free fields in a torus, J.
Engrg. Math., 20 (1986), 323–344.
E. Lara, R. Rodr´

ıguez and P. Venegas, Spectral approximation

f the curl operator in multiply connected domains, Discrete
Contin. Dyn. Syst. Ser. S, 9 (2016), 235–253.

E.C. Morse, Eigenfunctions of the curl in annular cylindrical and rectangular geometry, J. Math. Phys., 48 (2007), pp. 11, 083504.

R. Picard, Ein Randwertproblem in der Theorie kraftfreier

Magnetfelder, Z. Angew. Math. Phys., 27 (1976), 169–180.

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References (cont’d)

R. Picard, On a selfadjoint realization of curl and some of its

applications, Ricerche Mat., 47 (1998), 153–180.

R. Rodr´

ıguez and P. Venegas, Numerical approximation of the spectrum of the curl operator, Math. Comp., 83 (2014), 553–577.

L. Woltjer, A theorem on force-free magnetic fields, Proc.
Nat. Acad. Sci. U.S.A., 44 (1958), 489–491.
Z. Yoshida and Y. Giga, Remarks on spectra of operator rot,
Math. Z., 204 (1990), 235–245.
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Proof. We define the following operator:

R : Vh − → H(curl; Ω) , vh − → Rvh = vh − Φvh , with Φvh ∈ Q is such that

Ω

Φvh · p =

Ω

vh · p ∀p ∈ Q . From the inclusions Q ⊂ Z and Zh ⊂ Z we have at once Rvh ∈ Z, and from the definition of Φvh it follows Rvh⊥Q.

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Consequently, Rvh ∈ V ⊂ H(curl; Ω) ∩ H0(div 0; Ω) ⊂ Hs(Ω)3 for s > 1/2. In addition, curl(Rvh) = curl vh ∈ curl(N k

h ). Hence we

have that IN

h (Rvh) is well-defined, and

Rvh − IN

h (Rvh)0,Ω ≤ C {hsRvhs,Ω + hcurl(Rvh)0,Ω} .

(11) Since Rvh ∈ V we obtain Rvhs,Ω ≤ CRvhY ≤ Ccurl(Rvh)0,Ω. (12) Employing the previous result and (11), we have Rvh − IN

h (Rvh)0,Ω ≤ C(hs + h)curl(Rvh)0,Ω.

(13)

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The N´ ed´ elec interpolant IN

h Φvh is defined by

IN

h Φvh = IN h vh − IN h (Rvh) = vh − IN h (Rvh) and, since

curl IN

h (Rvh) = curl vh, it follows that curl(IN h Φvh) = 0 in Ω.

Furthermore,

γ′

j IN

h Φvh · t′ j =

γ′

j Φvh · t′

j = 0 for j = g1 + 1, . . . , g,

as Φvh ∈ Q. In conclusion, IN

h Φvh ∈ Qh.

Employing this result and the fact that IN

h vh = vh, we obtain

vh2

0,Ω

=

Ω

vh · IN

h vh =

Ω

vh ·

IN

h (Rvh) + IN h Φvh

=
Ω

vh · IN

h (Rvh) ,

as vh ∈ Vh. Using inequalities (13) and (12), and the fact that curl(Rvh) = curl vh in Ω, we obtain

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vh0,Ω ≤ IN

h (Rvh)0,Ω ≤ Rvh − IN h (Rvh)0,Ω + Rvh0,Ω

≤ C(hs + h)curl vh0,Ω + Ccurl vh0,Ω ≤

Ccurl vh0,Ω,

which concludes the proof. ✷

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Proof. First note that, by proceeding as in the proof of

Lemma 11, we have that, for all f ∈ E, (T − Th)fcurl;Ω ≤ Chmin{r,k} (Tfr,Ω + curl Tfr,Ω) ≤ Chmin{r,k} sup

g∈E

Tgr,Ω + curl Tgr,Ω gcurl;Ω fcurl;Ω ≤ C ′hmin{r,k}fcurl;Ω , (14) where we have used the fact that E is finite dimensional for the last inequality. Therefore (9) follows from (14) and Babuska and Osborn [Chap. II, Theor. 7.1].

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To prove (10), let f, g ∈ E ⊂ V be two eigenfunctions. Then, in particular, g satisfies curl g = λg in Ω . (15) If we define u := Tf and uh := Thf, then the following identities are satisfied

Ω

curl u · curl v =

Ω

f · curl v ∀v ∈ Z

Ω

curl uh · curl vh =

Ω

f · curl vh ∀vh ∈ Zh . Subtracting both identities, we obtain

Ω

curl(u − uh) · curl vh = 0 ∀vh ∈ Zh. (16)

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In addition, let w := Tg and wh := Thg. Then we have analogous identities for g and, in particular,

Ω

curl w · curl v =

Ω

g · curl v ∀v ∈ Z. (17) Thus, thanks to (15), the fact that u, uh, g ∈ Z, Lemma 6, (16) and (17), we obtain

Ω(T − Th)f · g

= λ−1

Ω(u − uh) · curl g

= λ−1

Ω curl(u − uh) · g

= λ−1

Ω curl(u − uh) · curl(w − wh)

= λ−1

Ω curl((T − Th)f) · curl((T − Th)g) .

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Using Cauchy–Schwarz inequality and (14) we get

Ω(T − Th)f · g
≤

λ−1(T − Th)fcurl;Ω(T − Th)gcurl;Ω ≤ Cλ−1h2 min{r,k}fcurl;Ωgcurl;Ω. (18) On the other hand, thanks to (16), Cauchy–Schwarz inequality, (14), Lemma 5 and Monk [Theorem 5.41], we obtain

Ω curl(T − Th)f · curl g
=
Ω curl(u − uh) · curl g
=
Ω curl(u − uh) · curl(g − IN

h g)

≤ Chmin{r,k} fcurl;Ω hmin{r,k}(gr;Ω + curl gr,Ω)

≤ C ′h2 min{r,k} fcurl;Ω gcurl;Ω . (19)

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Thus, from (18) and (19) we conclude that sup

f,g ∈ E

Ω(T − Th)f · g +
Ω curl(T − Th)f · curl g
fcurl;Ωgcurl;Ω

≤ Ch2 min{r,k} . Estimate (10) follows from Babuska and Osborn [Chap. II, Theor. 7.3] and the fact that T is self adjoint. ✷

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