Linear Force-Free Magnetic Field or Self-adjoint Extension of the

Linear Force-Free Magnetic Fieldor

Self-adjoint Extension of the curlOperator

Hiptmair1, Kotiuga2, Tordeux1

1. Seminar of Applied Mathematics, ETH-Zurich

2. College of BU Engineering Boston University

Linear Force-Free Magnetic FieldorSelf-adjoint Extension of the curl Operator – p. 1/35

Beltrami Fields

Beltrami fields = force-free magnetic fields (Lorentz)

−→Fρ =

−→j ×

−→B = 1

µ

−−→curl(

−→B ) ×

−→B ,

−→Fρ = 0 =⇒

−−→curl(

−→B ) = λ(x)

−→B .

Description of the quasi-static states leading to thesolar eruptions


The problematic

Which boundary condition (BC.) have to be considered atthe boundary?

Our point of view: the mathematical modelizationFor which BC. is the operator curl self-adjoint?Existence of linear force-free fields (opérator withcompact resolvent)

Which BC. have a physical sense? They arenon classical (surfacic differentials)?non local (non trivial topologies)

Numerical computationsmetric free methodsthree dimensional


Some references- Mathematical study of the force-free fields

Boulmezaoud, Maday, Amari (1999)Yoshida, Giga (1990) Picard (1998)

- Functional analysis

Paquet (1982) Buffa, Ciarlet, Costabel, Dauge, · · ·

- Symplectic geometry

Markus, Everitt (2005)

- Numerical methods

Arnold, Falk, Winther (2006) Kotiuga, Gross (2004)Hiptmair (2002) Hiptmair, Ostrowski (2002)

- differential forms and algebraic topology

Bott, Tu (1982)


The formal curl operator

In all the continuation, the pivot space is

(L2(Ω)

)3=

−→u Ω → IR3

∣∣∣∫

Ω

∥∥−→u∥∥2

< +∞.

The formal curl operator

curl : (C∞0 (Ω))3 ⊂

(L2(Ω)

)3−→ (L2(Ω))3

is symmetric∫

Ωcurl(−→u )·−→v =

∫

Ω

−→u ·curl(−→v ) ∀−→u ∈ C∞0 (Ω), ∀−→v ∈ C∞

0 (Ω).

However, this operator is not closed.


Its minimal closure

an unbounded operator T : D(T) ⊂ H −→ H is closed iff(un ∈ D(T), un → u, and, Tun → v

)=⇒

(u ∈ D(T), Tu = v

).

The minimal closure T of T

D(T) ⊂ D(T) ⊂ D(Text) (for any other closed extensions)

is the smallest closed extension of T.

The minimal closure of the formal curl operator is a curloperator with domain

H0(curl,Ω) =−→u ∈

(L2(Ω)

)3∣∣∣ curl(−→u ) ∈

(L2(Ω)

)3,

γt−→u = 0

.


Symmetric and self-adjoint

Why the self-adjoint property is important?

Any self-adjoint operator T (with compact resolvent)admits eigenvectors:

Tu = λu

Self-adjoint curl operators: existence of force-free fields

The curl : H0(curl,Ω) −→(L2(Ω)

)3 operator is symmetric

For bounded operators: symmetric =⇒ self-adjoint

For unbounded operators: symmetric 6=⇒ self-adjoint

One additional condition

D(T) = D(T∗).


The notion of adjoint and self-adjoint

The adjoint T∗ : D(T∗) ⊂ H −→ H of an operator is

D(T∗) =

u ∈ H∣∣∣ (u,Tv)H 6 Cu ‖v‖H ∀v ∈ D(T)

,

(T∗u, v)H = (u,Tv)H.

For the curl-operators, the adjoint of

curl : H0(curl,Ω) ⊂(L2(Ω)

)3−→

(L2(Ω)

)3

iscurl : H(curl,Ω) ⊂

(L2(Ω)

)3−→

(L2(Ω)

)3

with H(curl,Ω) =−→u ∈

(L2(Ω)

)3∣∣∣ curl(−→u ) ∈

(L2(Ω)

)3.


The GKN theorem

Symplectic geometry and self-adjoint operators

Glazman-Krein-Naimark Theorem (symplectic version)

Let T be an unbounded symmetric operator and let T be itsclosure.

One can associate to any self-adjoint extension of T acomplete Lagrangian of the symplectic space

S = D(T∗)/D(T)

equipped with the symplectic pairing

[u, v] = (T∗u, v) − (u,T∗v)


Basic definitions of symplectic geometry

A symplectic space is a linear space S equiped with asymplectic pairing [·, ·] −→ IR.

A symplectic pairing isskew-symmetric (u, v ∈ S)

[u, v] = −[v, u].

bilinear (u1, u2, v ∈ S and α1, α2 ∈ IR)

[α1u1 + α2u2, v] = α1[u1, v] + α2[u2, v] ∀u1, u2, v ∈ S.

non-degenerate

If u ∈ S satisfies [u, v] = 0 ∀v ∈ S then u = 0.


Symplectic geometry and orthogonality

What is the (symplectic) orthogonality?

u ⊥S

v ⇐⇒ [u, v] = 0.

A strange world:u ⊥

S

u.

L is a lagrangian if and only if

L ⊥S

L i. e. [u, v] = 0 ∀u, v ∈ L.

L is a complete lagrangian if and only if

L =

u ∈ S

∣∣∣ [u, v] = 0 ∀v ∈ S

= L⊥.


The GKN theorem

Symplectic geometry and self-adjoint operators


Let T be an unbounded symmetric operator and let T be itsclosure.

One can associate to any self-adjoint extension of T acomplete Lagrangian of the symplectic space

S = D(T∗)/D(T)

equipped with the symplectic pairing

[u, v] = (T∗u, v) − (u,T∗v)


Definition of the self-adjoint operator

Self-adjoint extensions Ts areextensions of T

restrictions of T∗

T ⊂ Ts ⊂ T∗

(=⇒ D(T) ⊂ D(Ts) ⊂ D(T∗)

).

=⇒ the domain defines the operator.

∀u ∈ D(Ts), Tsu = T∗u.

L a complete lagrangian of S = D(T∗)/D(T).

The link between the lagrangian subspace and

D(Ts) = ∪x∈L

x = L + D(T)


GKN theorem for curl operators


Let T be the curl with domain C∞0 (Ω).

T curl H0(curl,Ω).

T∗ curl H(curl,Ω)

What are the complete lagrangian of

S = H(curl,Ω)/H0(curl,Ω)

equipped with the symplectic product

[−→u ,−→v ] =

∫

Ωcurl(−→u ) · −→v − −→u · curl(−→v )


A trace theorem

The tangential trace operator

γt : H(curl,Ω) −→ H−1

2 (curl∂Ω, ∂Ω)

−→u 7−→ −→u − (−→u · −→n )−→n .

is continuous and surjective and its kernel is H0(curl,Ω).

Hence,

γt : H(curl,Ω)/H0(curl,Ω) −→ H−1


is bijective.


A trace theorem

We identify the trace space and the symplectic space

H(curl,Ω)/H0(curl∂Ω, ∂Ω) = H−1


The symplectic product becomes

[−→u ,−→v ] =

∫

Ωcurl(−→u ) · −→v − −→u · curl(−→v )

=

∫

∂Ω

(−→u ×−→v)· −→n

=

∫

∂Ω

(γt−→u × γt

−→v)· −→n .

(Stokes theorem)


GKN theorem for the curl operator

Glazman-Krein-Naimark Theorem (symplectique version)


T curl H0(curl,Ω).



S = H(curl,Ω)/H0(curl,Ω)

= H−1/2(curl∂Ω, ∂Ω)


[−→u ,−→v ] =

∫

Ωcurl(−→u )·−→v − −→u ·curl(−→v ) ?

=

∫

∂Ω(γt

−→u × γt−→v ) · −→n ?





T curl H0(curl,Ω).



S = H(curl,Ω)/H0(curl,Ω) = H−1/2(curl∂Ω, ∂Ω)


[−→u ,−→v ] =

∫

Ωcurl(−→u )·−→v − −→u ·curl(−→v ) ?

=

∫

∂Ω(γt

−→u × γt−→v ) · −→n ?





T curl H0(curl,Ω).



S = H(curl,Ω)/H0(curl,Ω) = H−1/2(curl∂Ω, ∂Ω)


[−→u ,−→v ] =

∫

Ωcurl(−→u )·−→v − −→u ·curl(−→v ) =

∫

∂Ω(γt

−→u×γt−→v )·−→n ?


Translation in terms of differential forms


Let T be the exterior derivative ∗d with domain C∞0 (Λ1(Ω)).

T ∗d H0(d,Λ1(Ω)).

T∗ ∗d H(d,Λ1(Ω))

with d the exterior derivative and ∗ the Hodge operator.


S = H(d,Λ1(Ω))/H0(d,Λ1(Ω)) = H−1/2(d,Λ1(∂Ω))


[ω, η] =

∫

Ωdω ∧ η − ω ∧ dη =

∫

∂Ωi∗inω ∧ i∗inη ?


Differential forms and vector proxies

For a three dimensional manifold

div

curl

grad

grad

curl

div

scalar

vector

flux

scalar density

d

d

d

∗d∗

∗d∗

∗d∗

Λ0(Ω)

Λ1(Ω)

Λ2(Ω)

Λ3(Ω)


Differential forms and vector proxies

For a two dimensional oriented manifold

curl∂Ω

grad∂Ω

curl∂Ω

div∂Ω

scalar

vector

flux

d

d

∗d∗

∗d∗

Λ0(Ω)

Λ1(Ω)

Λ2(Ω)


Hodge decomposition and forms

All ω in H−1

2 (d,Λ1(∂Ω)) can be decomposed into

ω = dϕω + ∗d ∗ ψω + hω

= dϕω + ∗dψω + hω

with

ϕω ∈ H1

2 (Λ0(∂Ω)) and hϕω= 0,

ψω ∈ H3

2 (Λ0(∂Ω)) and hψω= 0,

hω ∈ C∞(Λ1(∂Ω)) and dhω = 0, ∗d ∗ hω = 0,



All ω in H−1

2 (d,Λ1(∂Ω)) can be decomposed into

ω = dϕω + ∗d ∗ ψω + hω

= dϕω + ∗dψω + hω

The symplectic pairing can be computed as follows

[ω, η] =

∫

∂Ωω ∧ η

=

∫

∂Ωdϕω ∧ ∗dψη −

∫

∂Ωdϕη ∧ ∗dψω +

∫

∂Ωhω ∧ hη.



All −→u in H−1

2 (curl∂Ω, ∂Ω) can be decomposed into

−→u = grad∂Ω(ϕ−→u ) + curl∂Ω(ψ−→

u ) +−→h −→

u

with

ϕ−→u ∈ H

1

2 (∂Ω) and hϕ−→u

= 0,

ψ−→u ∈ H

3

2 (∂Ω) and hψ−→u

= 0,

−→h −→

u ∈(C∞(∂Ω)

)3 and curl∂Ω(−→h −→

u ) = 0, div∂Ω(−→h −→

u ) = 0,



All −→u in H−1

2 (curl∂Ω, ∂Ω) can be decomposed into

−→u = grad∂Ω(ϕ−→u ) + curl∂Ω(ψ−→

u ) +−→h −→

u

The symplectic pairing can be computed as follows

[−→u ,−→v ] =

∫

∂Ω(−→u ×−→v ) · −→n

=

∫

∂Ωgrad∂Ω(ϕ−→

u ) · grad∂Ω(ψ−→v )

−

∫

∂Ωgrad∂Ω(ψ−→

u ) · grad∂Ω(ϕ−→v )

+

∫

∂Ω

(−→h −→

u ×−→h −→

v

)· −→n .


Complete lagrangian of closed forms

If ω is closed, then ω can be written as

dω = 0 =⇒ ω = dϕω + hω.

and the symplectic pairing can be computed by

[ω, η] =

∫

∂Ωhω ∧ hη for dω = 0 and dη = 0.

Moreover, the symplectic orthogonal of closed forms arethe coboundaries

ω = dϕω + hω

⊥

=ω = dϕω


Complete lagrangian of closed forms

Theorem. To each complete lagrangian L of the space ofclosed forms corresponds a complete lagrangian LH of thespace of harmonic forms

L =

dϕω + hω

∣∣∣ hω ∈ LH

.

What are the complete lagrangian LH of the space ofharmonic forms?

For a trivial topology, there is no non-vanishing harmonicforms. Hence the set of closed forms is a completeLagrangian.


Complete lagrangian of coclosed forms

If ω is coclosed, then ω can be written as

d ∗ ω = 0 =⇒ ω = ∗dψω + hω.

and the symplectic pairing can be computed by

[ω, η] =

∫

∂Ωhω ∧ hη for d ∗ ω = 0 and d ∗ η = 0.

Moreover, the symplectic orthogonal of coclosed forms is

ω = ∗dψω + hω

⊥

=ω = ∗dψω


Complete lagrangian of coclosed forms

Theorem. To each complete lagrangian L of the space ofcoclosed forms corresponds a complete lagrangian LH ofthe space of harmonic forms

L =

d ∗ ϕω + hω

∣∣∣ hω ∈ LH

.

What are the complete lagrangian LH of the space ofharmonic forms?

For a trivial topology, there is no non-vanxishing harmonicforms. Hence the set of coclosed forms is a completeLagrangian.


Two self-adjoint curl for trivial topology

The ∗d operator or curl operator

∗d : H(d,Λ1(Ω)) −→ L2(Λ1(Ω))

curl : H(curl,Ω) −→(L2(Ω)

)3

are selfadjoint when equipped with the boundary condition

d(i∗inω) = 0 or d ∗ (i∗inω) = 0 on ∂Ω

curl∂Ω(γt−→u ) = 0 or div∂Ω(γt

−→u ) = 0 on ∂Ω


Property of harmonic forms

The space of harmonic forms

is finite dimensional (existence of a good cover)

is non degenerate when equipped with the symplecticpairing. Indeed,

(hω is harmonic

)⇐⇒

(∗ hω is harmonic

).

[hω, ∗hω] =

∫

∂Ωhω ∧ ∗hω =

∥∥∥hω

∥∥∥2

L2(Λ1(Ω))

Existence of a symplectic basis (Darboux theorem)


Construction of a symplectic basis

Definition of a symplecitic basis:

A basis κini=1 ∪ κ′i

ni=1 of S is a symplectic basis if and

only if

[κi, κj ] = [κ′i, κ′j ] = 0 ∀i, j ∈ [[1, n]],

[κi, κ′j ] = −[κ′i, κj ] = δi,j ∀i, j ∈ [[1, n]],

The symplectic pairing has the following matrixrepresentation [

0n Idn

−Idn 0n

]

A good way to design complete lagrangian.


Examples of complete lagrangians

κini=1 ∪ κ′i

ni=1 a symplectic basis.

Spanκini=1 is a complete lagrangian

Spanκ′ini=1 is also a a complete lagangian

Plenty of other possibilities since the symplectic basis isnot unique?


A basis from the embedding

Since we can compute the relative (to the boundary)homology (H2(Ω, ∂Ω)), we can speak about cuts

A symplectic basis can be found by considering theboundary of the cuts of inside and the boundaries of thecut of outside.



The κi solve these problems

Find κi ∈ Λ1(∂Ω) such that

dκi = 0 and d ∗ κi = 0 on ∂Ω∫

`inj

κi = δi,j and∫

`outj

κi = 0



The κ′i solve these problems

Find κ′i ∈ Λ1(∂Ω) such that

dκ′i = 0 and d ∗ κ′i = 0 on ∂Ω∫

`inj

κ′i = 0 and∫

`outj

κ′i = δi,j


Examples of self-adjoint curl

`

l

`1 `2

The limit conditions:

curl∂Ω(γt−→u ) = 0 sur ∂Ω,


∫

`

−→u · d−→` = 0.


∫

`i

−→u · d−→` = 0, i = 1 ou 2.



`

`1 `2




∫

`

−→u · d−→` = 0.


∫

`i

−→u · d−→` = 0, i = 1 ou 2.



`

`1`2




∫

`

−→u · d−→` = 0.


∫

`i

−→u · d−→` = 0, i = 1 ou 2.


Perspectives

Computation of eigenvectors of the curl

Find u ∈ Λ1(∂D) and λ ∈ R such that∗ du = λu dans Ω,

BC. sur ∂Ω,

(1)

Withney forms

Arnold, Falk, Winther (2006)Kotiuga, Gross (2002)Hiptmair (2002) Hiptmair, Ostrowski (2002)


Linear Force-Free Magnetic Field or Self-adjoint Extension of the

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