Harmonic Maxwell equations and their Finite Element discretization

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discretization

Monique Dauge, Martin Costabel

To cite this version:

Monique Dauge, Martin Costabel. Harmonic Maxwell equations and their Finite Element

discretiza-tion. Doctoral. France. 2019. �cel-02059645�

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discretization

Les ´equations de Maxwell harmoniques et leurs

discr étisations par él éments finis

Monique Dauge & Martin Costabel

IRMAR, Universit ´e de Rennes 1, FRANCE

Cours Doctoral IRMAR 2019

25 & 26 f ´evrier 2019

This document includes the contributions by MD and MC and is available at: http://perso.univ-rennes1.fr/monique.dauge/publis/MaxwellMD.pdf

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Part I. Introduction to Maxwell equations[MD]

1 Notations

2 _{Maxwell equations}

3 _{Variational formulation for cavity problem}

Part II.Trapsin Finite element discretization /Quelques pi `eges[MD]

4 Toy problem – Bench test

5 _{Numerical test / Rien ne va plus}

Part III. Elliptic regularization:badandgoodmethods[MD]

6 _{Standard regularization} 7 Non-convex corners

8 _{Weighted regularization}

Part IV. Edge element techniques[MC]

9 _Goals

10 _{Nedelec’s edge elements} Part V. References

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Part I

Introduction to Maxwell equations

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Outline

1 Notations

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Before stating equations, agree on notations and conventions

Slides written in English / Transparents en anglais, avec quelques traductions Colors

Direction that we will follow Direction that we will leave Important expressions EmphasizeorDanger General notation t ∈ R, time variable ∂t := _∂t∂, time derivative x, space variable In 3 dimensionsx = (x1,x2,x3) In 2 dimensionsx = (x1,x2) For j ∈ {1, 2, 3}, ∂j:= _∂x∂

j partial space derivative

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Operators of order 1 and 2 in 3 dimensions of space

∇ is the gradient operator. For scalar distribution ϕ

∇ϕ =   ∂1ϕ ∂2ϕ ∂3ϕ  

div is the divergence operator: For vector distributionsu = (u1,u2,u3) divu = ∇ · u = ∂1u1+ ∂2u2+ ∂3u3

curl is the curl operator / rotationnel: For vector distributionsu = (u1,u2,u3)

curl u = ∇ × u =   ∂2u3− ∂3u2 ∂3u1− ∂1u3 ∂1u2− ∂2u1  

∆is the Laplace operator (aka Laplacian). For scalar distribution ϕ ∆ϕ = ∂1ϕ + ∂2 22ϕ + ∂32ϕ

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Important relations

div ∇ϕ = ∆ϕ divcurl u = 0 curl ∇ϕ = 0

curl curl u − ∇ div u = −∆u where the vector Laplacian is

∆u =   ∆u1 ∆u2 ∆u3   4/72

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Outline

1 Notations

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Time dependent Maxwell equations

Unknowns are 4 vector functions (fields / champ) with 3 components each E electric field

H magnetic field D electric displacement B magnetic induction

Maxwell equations consist of the 4 relations

∂tB + curl E=0 (1a)

divD = ρ (1b)

∂tD − curl H= −J (1c)

divB = 0 (1d)

(1a) Faraday’s law

(1b) Gauss’s law with ρ the scalar charge density

(1c) Amp `ere’s circuital law, modified by Maxwell, with current densityJ (1d) tells thatB is solenoidal

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Time harmonic Maxwell equations

By partial in time Fourier transformation, or because the dataJ and ρ are time harmonic, we assume thatE, H, D, and B are time harmonic, i.e. that there exists ω ∈ Rsuch that

E(t, x) = e−iωt

E(x), H(t, x) = e−iωtH(x), B(t, x) = e−iωt

B(x), D(t, x) = e−iωtD(x) Then the 4-equation system becomes

curl E − iωB=0 (2a)

divD = ρ (2b)

curl H + iωD=J (2c)

divB = 0 (2d)

Divergence constraints

Applydivto (2a) =⇒ iω divB = 0. Hence (2d) implied if ω 6= 0 Applydivto (2c) =⇒ iω divD = div J. Hence the relation iωρ = div J The 4-equation system is not closed.

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Constitutive equations for linear media

ThenD is proportional to E and B is proportional to H

D = εE and B = µH

with coefficients ε = ε(x) (electric permittivity) and µ = µ(x) (magnetic permeability) depending on the material property atx.

Material coefficients ε and µ can be matrix valued (anisotropic materials). We consider here isotropic materials for which ε and µ are scalar. Particular materials

Vacuum(or free space): ε = ε0and µ = µ0 1

Dielectricmaterial: ε and µ real, ε ≥ ε0and µ ≥ µ0for classical materials, possiblynegative for metamaterials.

Conductingmaterial: µ ≥ µ0real and ε complex valued, with Im ε = σω−1where σ is the conductivity.

Globally in R3, ε and µ arepiecewise constantdepending on which material occupies the space at each point.

1_ε

0=8.854 × 10−12Fm−1and µ0=4π × 10−7Hm−1. Speed of light c = (ε0µ0)−1/2.

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Time harmonic Maxwell equations with constitutive laws

Putting all together we obtain

curl E − iωµH =0 (3a)

div εE = ρ (3b)

curl H + iωεE =J (3c)

div µH = 0 (3d)

Leaving aside the source problemwe take ρ = 0 andJ = 0:

curl E − iωµH=0 (4a)

div εE = 0 (4b)

curl H + iωεE=0 (4c)

div µH = 0 (4d)

The problem is to find triples (ω,E, H) with ω ∈ C, and (E, H) 6= (0, 0) in admissible function spaces

In R3, this is the problem of findingscattering resonances.Suitable radiation conditions at infinity have to be imposed. In general Im ω < 0. In bounded domains, combined with suitable boundary conditions, this is the problem of findingcavity resonances. In general ω ∈ R.

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The cavity problem

Anelectromagnetic cavity Ωis a bounded region of R3_{that is isolated from} an electromagnetic point of view from the outside region R3\ Ω.

This is an idealization of aFaraday cagefor which we consider that Ω is surrounded by alayer of infinite conductivity σ. Then the electric fieldE is zero outside Ω and this causes the boundary condition

E × n = 0 on ∂Ω (the tangential component ofE is 0 (5) Heren is the unitary outward normal field to ∂Ω.

This can be rigorously proved by setting Maxwell equation in a region containing Ω and its surrounding conductive medium and let σ tend to infinity. Going to this limit exhibits the skin effect / effet de peau / in conductive media.

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Outline

1 Notations

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Elimination of magnetic field

Recall equations

curl E − iωµH=0 in Ω (6a)

div εE = 0 in Ω (6b)

curl H + iωεE=0 in Ω (6c)

div µH = 0 in Ω (6d)

E × n=0 on ∂Ω (6e)

Using (6a) it istempting to eliminateHby writing: iωH = _µ1curl E which yields,formallywith (6c)

curl1

µcurl E − ω 2

εE = 0 (7)

Most frequently, one finds (7) in the literature, followed by an integration by parts to find a variational formulation.

We will rather start from the system (6) to find directly the variational formulation, which allows to find variational spaces without doubt.

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The space H(curl; Ω)

Assume thatE ∈ L2(Ω)3andH ∈ L2(Ω)3. Then (6c) and (6a) yields curl E ∈ L2(Ω)3 and curl H ∈ L2(Ω)3

This leads to introduce the space

H(curl; Ω) = {U ∈ L2(Ω)3, curl U ∈ L2(Ω)3}

Lemma [Girault-Raviart, 86]

Let Ω be a bounded Lipschitz domaina_{. Then} _C∞

(Ω)3_{is dense in H(}_{curl; Ω).} a_{A Lipschitz domain is a domain that is (after possible rotations) the epigraph of a Lipschitz} function in the neighborhood of each of its boundary points.

Consequence: ifU ∈ H(curl; Ω), thetangential traceU × nmakes sense in H−1/2(∂Ω)3thanks to the identity, valid for any Φ ∈ H1(Ω)3:

U × n, Φ_H−1/2_(∂Ω)3_{| H}1/2_(∂Ω)3= Z Ω U · curl Φ dx − Z Ω curl U · Φ dx 11/72

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The space H

0

(

curl; Ω)

Then we can introduce the H-curl space with zero tangential traces

H0(curl; Ω) = {U ∈ H(curl; Ω), u × n ∂Ω=0} Then Lemma [Girault-Raviart, 86]

Let Ω be a bounded Lipschitz domain. Then C0∞(Ω) 3

is dense in H0(curl; Ω). And an important consequence

Lemma

Let Ω be a bounded Lipschitz domain. Then

Z

Ω

U · curl V dx =

Z

Ω

curl U · V dx ∀U ∈ H0(curl; Ω), ∀V ∈ H(curl; Ω).

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Towards variational formulation of cavity problem

Recall

curl E − iωµH=0 in Ω (6a)

curl H + iωεE=0 in Ω (6c)

E × n=0 on ∂Ω (6e)

IfE ∈ L2_(Ω)3_and_{H ∈ L}2_(Ω)3_{, then} _{E ∈ H}

0(curl; Ω) and H ∈ H(curl; Ω).

Pick a test functionE0∈ H0(curl; Ω).Multiply (6a) by µ−1on the left, take the

· product with curl E0

on the right, integrate over Ω

Z

Ω

µ−1curl E · curl E0− iωH · curl E0dx = 0 (6a’) Multiply (6c) by iω, take the · product withE0

on the right, integrate over Ω

Z

Ω

iωcurl H · E0− ω2εE · E0dx = 0 (6c’) Add (6a’) and (6c’), use the Lemma on previous slide and obtain

Z Ω µ−1curl E · curl E0− ω2 εE · E0dx = 0 13/72

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Electric spectral problem

Definition

Let Ω be a bounded Lipschitz domain. Theelectric spectral problemis to find pairs (ω,E) with non-zero E ∈ H0(curl; Ω), such that

Z

Ω

µ−1curl E · curl E0− ω2εE · E0dx = 0 ∀E0∈ H0(curl; Ω) (8)

Many questions arise 1 _{Can we find solutions?}

2 _{Do solutions correspond to solutions of the cavity problem?}

3 _{Can we discretize (8) by Finite Element Method (Galerkin projection)}

We address these questions on a simplified two-dimensional problem which 1 _{Encounters the same difficulties as the original 3D problem}

2 _{Has solutions that can be alternatively deduced by solving a scalar} equation.

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Part II

Traps in Finite element discretization

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Outline

4 _{Toy problem – Bench test}

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From 3 to 2 dimensions

Take ε and µ constant equal to 1.

Take as domain Ω a 2-dim. polygon (straight sides).

To find the Maxwell cavity problem in Ω in its TE (Transverse Electric) formulation we go back to the 3-dim. formulation, considered in Ω × R:

curl E − iωH=0 in Ω × R (6a)

divE = 0 in _{Ω × R} (6b)

curl H + iωE=0 in Ω × R (6c)

divH = 0 in Ω × R (6d)

E × n=0 on ∂Ω × R (6e)

and assume that

E and H are function of (x1,x2)only (no dependence in x3) E3=0, H1=H2=0, i.e. E =   E1 E2 0   and H =   0 0 H3  

Note that (6d) is already satisfied. We obtain · · · / · · ·

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The TE cavity problem

∂1E2− ∂2E1− iωH3=0 in Ω (9a)

∂1E1+ ∂2E2=0 in Ω (9b)

∂2H3+iωE1=0 and −∂1H3+iωE2=0 in Ω (9c)

E1n2− E2n1=0 on ∂Ω (9d)

Define the scalar curl (denoted rot) in 2 dimensions as

rotU = ∂1U2− ∂2U1 for U = (U1,U2)

and the spacesH(rot; Ω)andH0(rot; Ω)accordingly.

By the same method as in 3-dim. we find thatU = (E1,E2)is solution of the

Electric Maxwell spectral problem in 2-dim.

Find pairs (ω,U) with non-zero U ∈ H0(rot; Ω), such that Z

Ω

rotU rot U0− ω2_{U · U}0

dx = 0 ∀U0∈ H0(rot; Ω) (10)

Observe that (9c) implies ∂1H3and ∂2H3are in L2(Ω). Hence H3∈ H1(Ω). We find:

Neumann spectral problem

Find pairs (ω, H3)with non-zero H3∈ H1(Ω), such that Z Ω ∇H3· ∇H0− ω2H3H0 dx = 0 ∀H0_{∈ H}1_(Ω) ₍₁₁₎ 17/72

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The electric Maxwell spectral problem (rot-rot eigenmodes)

Proposition 1

Let Ω be a 2-dim. simply connected Lipschitz domain. Let (ω,U) ∈ C × H0(rot; Ω) be a solution of

(∗) Z Ω rotU rot U0− ω2 U · U0dx = 0 ∀U0∈ H0(rot; Ω)

1 _{If ω = 0, then exists a scalar potential ϕ such that} ϕ ∈H01(Ω) and ∇ϕ = U

Conversely, if ϕ ∈ H1

0(Ω), then (0, ∇ϕ) solves (∗).

2 _{If ω 6= 0, then div}_{U = 0 and exists a scalar potential}a_{ψ ∈}_H1_(Ω)_{s. t.} ψ ∈H1(Ω) and −rot ψ = U→

and (ω2_{, ψ)}_{is an eigenpair of the Neumann problem} (∗∗)

Z

Ω

∇ψ · ∇ψ0− ω2ψ ψ0dx = 0 ∀ψ0∈ H1(Ω) Conversely, if (ω2, ψ)is an eigenpair (∗∗), then (ω,−rot ψ) solves (∗).→ a−_{rot ψ is the vector curl in 2-dim. :}→ −_{rot ψ = (∂}→

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Proof

1 _{If ω = 0, then rot}_{U = 0.}

As Ω is simply connected, there exists a potential ϕ such that ∇ϕ =U. SinceU × n = 0 on ∂Ω, then ϕ is constant on ∂Ω.

The simple connectedness implies that ∂Ω has one component, so ϕ can be chosen in H1

0(Ω).

2 _{If ω 6= 0, choose as test function}_U0_{= ∇ϕ}0_{, with ϕ}0_{∈ H}1

0(Ω). Then (∗) ⇒ Z

Ω

U · ∇ϕ0dx = 0 ∀ϕ0∈ H01(Ω)

Therefore, in the sense of duality

div U , ϕ0

H−1_{(Ω) |}_H1 0(Ω)=

0 ∀ϕ0∈ H1 0(Ω)

Hence divU = 0. This implies the existence of a scalar potential ψ s.t.−rot ψ = U. As→ rot−rot ψ = −∆ψ→ and −rot ψ ·→ −rot ψ→ 0= ∇ψ · ∇ψ0

U ∈ H0(rot; Ω) ⇐⇒ ψ ∈D(∆Neu; Ω) :=v ∈ H1(Ω), ∆v ∈ L2(Ω) & ∂nv

∂Ω=0

With the test functionsU0=−rot ψ→ 0for any ψ0∈ D(∆Neu_{; Ω), (∗) implies that ψ satisfies}

(∗ ∗ ∗) Z Ω ∆ψ ∆ψ0− ω2_{∇ψ · ∇ψ}0 dx = 0 ∀ψ0∈ D(∆Neu_{; Ω)} 19/72

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End of proof

Integrating by parts (∗ ∗ ∗) implies

Z Ω ∆ψ ∆ψ0+ ω2ψ ∆ψ0 dx = 0 ∀ψ0∈ D(∆Neu_{; Ω)} i.e. Z Ω ∆ψ + ω2ψ∆ψ0dx = 0 ∀ψ0∈ D(∆Neu; Ω) Denote by L2

◦(Ω)the space of functions L2(Ω)orthogonal to constants on Ω

L2◦(Ω) = v ∈ L2(Ω), Z Ω v dx = 0

Now, we can choose ψ ∈ L2

◦(Ω), and still have

−→

rot ψ = U. The operator ∆Neu

∆Neu: D(∆Neu_{; Ω) −→}_L2 ◦(Ω) is onto / surjectif Hence Z Ω ∆ψ + ω2ψv dx = 0 ∀v ∈ L2◦(Ω)

and, since ∆ψ + ω2_ψ_{belongs to L}2 ◦(Ω)

∆ψ + ω2ψ =0

Finishing the proof is now easy.

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The TE cavity problem versus the rot-rot spectral problem

Corollary

Let Ω be a 2-dim. simply connected Lipschitz domain. The solutions ω, (E1,E2,H3))of the TE cavity problem (9) are

1 _{ω =}_{0 with E}₁₌_E₂₌_{0 and H}₃_{non-zero constant.}

2 _{ω 6=}_{0 such that ω}2_{is an eigenvalue of}_∆Neu_{, the positive Laplace operator with}

Neumann conditions: ∆Neu_{= −∆}_{with operator domain D(∆}Neu_{; Ω). Then}

(E1,E2,H3) = (

−→

rot ψ, −iωψ)

with ψ eigenvector of ∆Neu_{associated with ω}2_. Remarks on 3-dim. domains

If Ω is a 3-dim. simply connected Lipschitz domain, the solutions of (∗)

Z

Ω

curl U · curl U0− ω2U · U0dx = 0 ∀U0∈ H0(curl; Ω) are related to the cavity problem in a similar way:

ω =0 =⇒ divU 6= 0 and ω 6=0 =⇒ divU = 0

and the solutions of the cavity problem can be deduced from those of (∗) when ω 6= 0.

But, in 3-dim. there is no scalar potential in general. The 2-dim. serves as a bench test / banc d’essai / for 3-dim.

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Outline

4 _{Toy problem – Bench test}

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The square

Consider Ω = (0, π) × (0, π).

By separation of variables, we find that the eigenpairs of ∆Neuare (

ω2=j2 1+j22

ψ(x1,x2) =cos(j1x1)cos(j2x2) for any integers j1, j2∈ {0, 1, 2, . . .} Using Proposition 1, this implies that the solutions of the electric Maxwell spectral problem (∗) Z Ω rotU rot U0dx = ω2 Z Ω U · U0dx ∀U0∈ H0(rot; Ω)

correspond to eigenvalues ω2_{equal to} 1 _{0 (with infinite multiplicity)}

2 _{1, 1, 2, 4, 4, 5, 5, 8, 9, 9, 10, 10, 13, 13, . . .} (with repetition according to multiplicity)

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Finite element method

a _{Let a be bilinear (or sesquilinear) form well defined on a product space} V × V a(u, v ) =X i X j X |α|≤1 X |β|≤1 Z Ω aαβ∂αui∂βvjdx

Spectral problem associated with a: Find pairs (λ,u), with 0 6= u ∈ V s. t. (†) a(u, v ) = λhu, v iL2_(Ω)|L2_(Ω) ∀v ∈ V

b _{Let e}_{V be a finite dimensional subspace of V .}

Galerkin projection of problem (†): Find pairs (˜λ,u), with 0 6=e u ∈ ee V s. t. (‡) a(u,e v ) = ˜e λheu,v ie L2_(Ω)|L2_(Ω) ∀ev ∈ eV

c _The_{Finite Element Method [FEM]}_{consists in constructing and}

implementing suitable spaces eV . In general, they are based on amesh of Ω (subdivision into triangular or quadrilateral elements in 2-dim.) and piecewise (mapped-)polynomialsin each element of the mesh.

Analysis of FEM: proving (ordisproving) convergence whendim eV → ∞.

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Let’s go / On y va

0 20 40 60 80 100 120 140 0 2 4 6 8 10 12 14 Triangular mesh ∼ 450 Lagrange elements of degree 1

Sort computed eigenvalues by increasing order

˜

λ1≤ ˜λ2≤ · · ·

Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 140 Ordinates: value of ˜λn

Horizontal lines = exact values for λj

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Another try / Un autre essai

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14

Mesh with one square element of degree 8 : e u = (˜u1, ˜u2) with e u1, eu2∈ Q8= P8⊗ P8

˜

λ1≤ ˜λ2≤ · · · Il y a encore un probl `eme Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 70

Ordinates: value of ˜λn

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Another try / Un autre essai

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14

Mesh with one square element of degree 8 : e u = (˜u1, ˜u2) with e u1, eu2∈ Q8= P8⊗ P8

˜

λ1≤ ˜λ2≤ · · · Il y a encore un probl `eme Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 70

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Try something else (breaking identity between components)

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14

Mesh with one square element of mixed degrees 7&8 :

e

u = (˜u1, ˜u2) with

e

u1∈ P7⊗ P8 eu2∈ P8⊗ P7 Sort computed eigenvalues by increasing order

˜

λ1≤ ˜λ2≤ · · · Il n’y a plus de probl `eme This is an“edge element” cf lecture by Martin. Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 70

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Part III

Elliptic regularization: bad and good methods

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Outline

6 _{Standard regularization}

7 _{Non-convex corners}

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Rappels on Dirichlet and Neumann scalar Laplace operators

Let Ω be a bounded Lipschitz domain. Denote by a∇the bilinear form a∇(u, v ) :=

Z

Ω

∇u · ∇v dx, for u, v ∈ H1(Ω).

a _{The positive Dirichlet Laplacian ∆}Dir_{is defined from H}1

0(Ω)into its dual space

H−1_(Ω)_by

∆Dir(u) = F with F , v_H−1_{(Ω) |H}1 0(Ω)

:= a∇(u, v )

NB: Since H−1(Ω)is a space of distributions in Ω, we haveF = −∆u.

Since a∇is coercive on H01(Ω), ∆Diris invertible with compact inverse. The domain (in

the sense of domain of unbounded operators) is

D(∆Dir; Ω) = {v ∈ H₀1(Ω), F ∈ L2(Ω)}

The operator ∆Dir_{defines an isomorphism from D(∆}Dir_{; Ω)}_{onto L}2_(Ω).

The spectrum of ∆Diris discreteand formed by asequence of positive eigenvalues λDirn

that tends to infinity as n → +∞.

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Rappels on Dirichlet and Neumann scalar Laplace operators

Let Ω be a bounded Lipschitz domain. Denote by a∇the bilinear form a∇(u, v ) :=

Z

Ω

∇u · ∇v dx, for u, v ∈ H1(Ω).

b _{The non-negative Neumann Laplacian ∆}Neu_{is defined from H}1_(Ω)_{into its dual}

space H1_(Ω)0_by

∆Neu(u) = F with F , v_H1(Ω)0_{| H}1(Ω):= a∇(u, v )

NB: Since H1_(Ω)0_{is not a space of distributions in Ω, it may happen that F 6= −∆u}

Since a∇+Id is coercive on H1(Ω), ∆Neu+Id is invertible with compact inverse.

D(∆Neu; Ω) = {v ∈ H1(Ω), F ∈ L2(Ω)}

F ∈ L2_(Ω)_{means that there exists a function f ∈ L}2_(Ω)_{such that hF , v i =}R

Ωf v dx. We deduce that D(∆Neu; Ω) = {v ∈ H1(Ω), ∆v ∈ L2(Ω) and ∂nv ∂Ω=0}

The spectrum of ∆Neu_{is discrete}_{and formed by a sequence of non-negative} eigenvalues / valeurs propres positives ou nulles /λNeu

n → +∞as n → +∞.

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Blowing up the kernel / Exploser le noyau /of the rot-rot operator

Recall that we want to compute FEM approximations of the eigenpairs (λ,U) with λ = ω2and non-zeroU ∈ H0(rot; Ω), solution of

(∗) Z Ω rotU rot U0dx = λ Z Ω U · U0dx ∀U0∈ H0(rot; Ω)

The “standard” approximation theory [Osborn, 75] [Babuˇska-Osborn, 91] applies if there is acompact embeddingof the space V corresponding to the left hand side of (∗) into the space H corresponding to its right hand side. But in our case

V = H0(rot; Ω) and H = L2(Ω)2

The embeddingH0(rot; Ω) −→ L2(Ω)2is not compact. The symptom is the infinite

dimensional kernel.

Since we are interested by the divergence-free solutions of (∗), a natural idea is to

regularize the rot-rot bilinear form by the div-div form. Notation

For any chosen s > 0 / pour tout s fix ´e, / set

a[s](U, U0) = Z

Ω

rotU rot U0+sdivU div U0 dx well defined on the new space

XN(Ω) = {V ∈ H0(rot; Ω), divV ∈ L2(Ω)}

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The divergence

Lemma

Let Ω be a 2-dim. Lipschitz domain. Choose s > 0. Let (λ,U) ∈ R × XN(Ω)be an

eigenpair of a[s] (∗) a[s](U, U0) = λ Z Ω U · U0dx ∀U0∈ XN(Ω) Then divU ∈ H1 0(Ω)and [ 1 or 2 holds]

1 _div_{U =: Φ is an eigenvector of}_{s ∆}Dir_{with eigenvalue λ:} Φ ∈H1 0(Ω) solves s Z Ω ∇Φ · ∇Φ0dx = λ Z Ω Φ Φ0dx ∀Φ0∈ H1 0(Ω) 2 _div_{U = 0.} Proof

Set Φ := divU. Choose as test function U0= ∇Φ0with Φ0∈ D(∆Dir_{; Ω) =}n_{v ∈ H}1 0(Ω), ∆v ∈ L2(Ω) o . ThenU0= ∇Φ0_{belongs to X} N(Ω)since: Φ0_{∈ H}1_{(Ω) =⇒}_U0 ∈ L2_(Ω) Φ0 ∂Ω=0 =⇒U 0_{× n} ∂Ω=0 ∆Φ0∈ L2_{(Ω) =⇒}_div_U0 ∈ L2_(Ω) 30/72

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The divergence: Proof of Lemma

Set Φ := divU. Choose as test function U0= ∇Φ0with Φ0∈ D(∆Dir_{; Ω). Then (∗) ⇒} s Z Ω Φ div ∇Φ0dx = λ Z Ω U · ∇Φ0dx ∀Φ0_{∈ D(∆}Dir_{; Ω)} Observe that div ∇Φ0= −∆Dir_Φ0 Z Ω U · ∇Φ0dx= −hdivU, Φ0i_H−1_{(Ω) |}_H1 0(Ω)= − Z Ω Φ Φ0dx

Therefore, we have the orthogonality condition

Z

Ω

Φs ∆DirΦ0− λΦ0dx = 0 ∀Φ0∈ D(∆Dir; Ω)

In other words divU = Φ belongs to the orthogonal of the range of the self-adjoint operator s ∆Dir_{− λ Id}_: Φ ∈ range(s ∆Dir− λ Id) ⊥ Then 1 _or 2 _holds

1 _Φ_{is a non-zero element in the kernel of} _{s ∆}Dir_{− λ Id}_{, i.e. is an eigenvector of} s ∆Dir_{with eigenvalue λ.}

2 _{Φ =}_0.

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The scalar rot

We have a similar statement concerning the scalar rot ofU: Lemma

Let Ω be a 2-dim. Lipschitz domain. Choose s > 0. Let (λ,U) ∈ R × XN(Ω)be an eigenpair of a[s]

(∗) a[s](U, U0) = λ Z

Ω

U · U0dx ∀U0_{∈ X} N(Ω)

Then rotU ∈ H1_(Ω)_{and [} 1 _or 2 _holds] 1 _rot_{U = 0.}

2 _rot_{U =: Ψ is an eigenvector of}_∆Neu_{with eigenvalue λ:} Ψ ∈H1(Ω) solves Z Ω ∇Ψ · ∇Ψ0dx = λ Z Ω Ψ Ψ0dx ∀Ψ0∈ H1_(Ω) Proof

Similar as before. Now the test functions areU0=−rot Ψ→ 0_{with any Φ}0_{∈ D(∆}Neu_{; Ω).}

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Spectrum of the regularized form a[s]

Theorem

Let Ω be a 2-dim. simply connected Lipschitz domain. Choose s > 0. Let λDir

n , ΦDirn

n≥1be a complete system of eigenpairs of ∆ Dir

Let λNeu n , ΨNeun

n≥0be a complete system of eigenpairs of ∆ Neu_,

with λNeu

0 =0 and ΨNeu0 =1

Then a complete system of eigenpairs for a[s] is given by the union of sλDirn ,UDivn n≥1 and λNeun ,UMaxn n≥1 where 1 _rot_UDiv

n =0 and divUDivn = ΦDirn 2 _div_UMax

n =0 and rotUMaxn = ΦNeun

Proof. It suffices to set UDivn = − 1 λDirn ∇ΦDir n and UMaxn = 1 λNeun −→ rot ΦDirn

Since Ω is simply connected, there is no non-zero fieldU ∈ XN(Ω)such that

divU = rot U = 0.

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Spectrum of a[s] and Maxwell spectral problem

The spectrum of a[s]: λ ∈ R, U ∈ XN(Ω)

(∗)

Z

Ω

rotU rot U0+sdivU div U0

dx = λ[s]

Z

Ω

U · U0dx ∀U0∈ XN(Ω)

has clearly two well separated parts:

A part that depends linearly of s and with curl-free eigenvectors

A part independent of s with divergence free eigenvectors. This is the spectrum we are looking for.

How to distinguish them in numerical computations? Two techniques:

Calculate the ratio

τ ( eU) = k rot eUk 2 sk div eUk2

We expectlarge values for approximation of divergence free eigenvectorsandsmall values for the others.

Calculate eigenvalues for several different values of s

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Spectrum of a[s] on the square, s = 0

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14

Mesh with one square element of degree 8 : e u = (˜u1, ˜u2_{) ∈ (Q}8)2 with b.c. eu × n ∂Ω=0 Sort computed eigenvalues by increasing order

˜

λ1≤ ˜λ2≤ · · · Extra multiplicities

Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 70 Ordinates: value of ˜λn

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Spectrum of a[s], s = 0.002 on the square

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14

Mesh with one square element of degree 8 : e u = (˜u1, ˜u2) ∈ (Q8)2 with b.c. eu × n ∂Ω=0 Sort computed eigenvalues by increasing order ˜ λ1≤ ˜λ2≤ · · · τ (u) = k rote uke 2 (sk divuke 2 )−1 ?when τ small◦ when τ large

Le probl `eme semble disparaˆıtre Abscissa: rank of computed eigenvalue 1 ≤ n ≤ 70

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Spectrum of a[s] on the square, dependence in s

0 0.5 1 1.5 2 2.5 3 3.5 4 0 2 4 6 8 10 12 14

Mesh with one square element of degree 8 : e u = (˜u1, ˜u2) ∈ (Q8)2 with b.c. eu × n ∂Ω=0 Sort computed eigenvalues by increasing order ˜ λ1≤ ˜λ2≤ · · · τ (u) = k rote uke 2 (sk divuke 2 )−1 ?when τ small◦ when τ large

Le probl `eme a disparu Abscissa: Value of s

Ordinates: all smallest values of ˜λn≤ 14, n = 1, 2, 3, . . . Horizontal lines = exact values for λj

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Outline

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Spectrum of a[s] on a L-shape domain, dependence in s

0 1 2 3 4 5 0 5 10 15 20 25 30 35 40 45 Ω = (−1, 1)2_{\ [0, 1] × [−1, 0]}

Mesh with 9 quadrilateral elements of degree 10 : e u ∈ (Q10)2 with b.c. u × ne ∂Ω=0

Sort computed eigenvalues by increasing order ˜ λ1≤ ˜λ2≤ · · · τ (eu) = k roteuk 2_{(sk div} e uk2₎−1 ?when τ small◦ when τ large Patatras !

Abscissa: Value of s

Ordinates: all smallest values of ˜λn≤ 45, n = 1, 2, 3, . . .

Horizontal lines = values close to exact for λj (by computing Neumann eigenvalues)

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Spectrum of a[s] on a L-shape domain: Interpretation

We observe

One (large) half of eigenvalues seems to be correctly approximated

The other (smaller) half is completely missedand replaced by something else that does not have a clear behavior in s (neither linear nor constant). The situation does not improveif we increase the polynomial degree or the density of the mesh (or both)

The diagnosis is that

We converge towards something that we don’t expect What? Why?

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Spectrum of a[s] on a L-shape domain: Explanation

Recall that

XN(Ω) = {V ∈ H0(rot; Ω), divV ∈ L2(Ω)} Denote by HN(Ω)the space

HN(Ω) =H1(Ω)2∩ XN(Ω) = {V ∈ H1(Ω)2, V × n

∂Ω=0}

The explanation is the conjunction of three facts:

1 _{In L-shape domain Ω, HN}_(Ω)_{is strictly smaller that XN(Ω). Moreover, a} large part of eigenvectorsUDivn andU

Max

n do not belong to HN(Ω) 2 _{The discrete Finite Element spaces are contained in HN}_(Ω) 3 _H

N(Ω)is closed for the topology of XN(Ω)

Conclusion: A large part of the eigenvectors of a[s] cannot be approximated by a plain Finite Element discretization in XN(Ω).

Let us explain each point in more detail.

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Spectrum of a[s] on a L-shape domain. Point

1

There existsone corner singularityS = ∇Φsings. t. [Birman-Solomyak 87] XN(Ω)=HN(Ω) ⊕hS i

where Φsing∈ D(∆Dir; Ω) but Φsing6∈ H2(Ω). In polar coordinates (r , θ) Φsing(x) =χ(r ) r2/3sin(2θ

3)

where χ is a smooth function equal to 1 if r <1₄ and to 0 if r > 1₂. We have

D(∆Dir; Ω)= (H2∩ H1

0)(Ω) ⊕hΦsingi In fact, since we are in 2-dim.

S = ∇Φsing= −→

rot Ψsing

where Ψsing∈ D(∆Neu; Ω) but Ψsing6∈ H2(Ω). Note Ψsing= χ(r ) r2/3cos(2θ₃) Almost all eigenvectors ΨNeu

n of ∆Neuthat areeven with respect to the

diagonalx1+x2=0 “contain” this singularity, i.e.

ΨNeun − cnΨsing ∈ H2(Ω)for some coefficient cn6= 0

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Spectrum of a[s] on a L-shape domain. Points

2

and

3 2 _{The FEM space are made of functions}

e u that are a _{piecewise polynomials} b _{in the space XN}_(Ω). We observe rotu is in Le

2_{(Ω) =⇒}_{no tangential jump for} e

u between two elements. diveu is in L

2

(Ω) =⇒no normal jump foru between two elements.e Finally, both components ofu are continuous over Ω.e

Thereforeeu belongs to HN

3 _{A sequence}_u_m_{∈ HN}_{(Ω), m ≥ 1, that is converging for the topology of} XN(Ω)will never converge to a limit outside HN(Ω)by virtue of

Theorem [Costabel, 91] [Costabel-Dauge, 99] Let Ω be a Lipschitz polygon.

The space HN(Ω)is a closed subspace in XN(Ω).

=⇒In L-shape, instead of the Maxwell spectral problem, we are solving a Lam ´e system with elasticity coefficients depending on s

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Outline

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Introduction of a weight

Let Ω be a polygon with one non-convex cornerc of opening ω > π This applies to the L-shape domain Ω with its non-convex corner at the origin.

Let r = |x − c| be the distance function to the non-convex corner c. Choose a number γ ∈ [0, 1]. This will be the exponent of a weight function.

Notation

For any chosen s > 0 set aγ[s](U, U 0 ) = Z Ω

rotU rot U0+s r2γdivU div U0 dx well defined of the new space

X_Nγ(Ω) = {V ∈ H0(rot; Ω), rγdivV ∈ L2(Ω)}

If γ > 0, the norm in the divergence is relaxed (the norm is smaller than without weight)

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Spectrum of a

γ

[s]

on a L-shape domain, γ = 0

0 1 2 3 4 5 0 5 10 15 20 25 30 35 40 45 Ω = (−1, 1)2_{\ [0, 1] × [−1, 0]}

Sort computed eigenvalues by increasing order ˜ λ1≤ ˜λ2≤ · · · τ (eu) = k roteuk 2_{(sk div} e uk2₎−1 ?when τ small◦ when τ large Abscissa: Value of s

(58)

Spectrum of a

γ

[s]

on a L-shape domain, γ = 0.35

0 2 4 6 8 10 0 5 10 15 20 25 30 35 40 45 Ω = (−1, 1)2_{\ [0, 1] × [−1, 0]}

(59)

Spectrum of a

γ

[s]

on a L-shape domain, γ = 0.5

0 2 4 6 8 10 0 5 10 15 20 25 30 35 40 45 Ω = (−1, 1)2_{\ [0, 1] × [−1, 0]}

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Spectrum of a

γ

[s]

on a L-shape domain, γ = 1

0 5 10 15 20 0 5 10 15 20 25 30 35 40 45 Ω = (−1, 1)2_{\ [0, 1] × [−1, 0]}

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Theoretical result

Theorem Costabel-Dauge, 02

Let Ω be a polygon with one non-convex cornerc of opening ω > π and let r be the distance function toc. Let γ be such that

1 − π

ω< γ <1 Then, for any s > 0

HN(Ω)is dense in X_Nγ(Ω)

The eigenvalues of aγ[s] are correctly approximated by Lagrange finite elements

The eigenvalues of aγ[s] are

the sλDir,γn with the eigenvalues λ Dir,γ

n of the Dirichlet realization of v 7→ rγ_∆(rγ_{v )}

the λNeu(independent of s and γ)

This theorem extends to 3-dim. polyhedra. The weight is the distance to the union of non-convex edges.

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Singularities and weighted spaces

Let Ω be a polygon with one non-convex cornerc of opening ω > π Define for m ∈ N and β ∈ R the weighted Sobolev space

Kβm(Ω) = {v ∈ L 2 loc(Ω), r β+|α| ∂αxv ∈ L2(Ω) ∀α ∈ N2, |α| ≤m} Observe that The Laplacian ∆ = ∂2 1+ ∂ 2 2 is continuous from K 2 γ−2(Ω)into K 0 γ(Ω). IfU ∈ Xγ N(Ω), then divU ∈ K 0 γ(Ω).

The singularity Φsinghas the form, in polar coordinates centered atc Φsing(x) =χ(r ) rπ/ωsin(πθ ω) and Φsing∈ K2 γ−2(Ω)if and only if π ω >1 − γ, i.e. γ >1 − π ω Theorem [Kondrat’ev, 67] Let γ ∈ [0, 1]. Ifγ >1 −π ω, ∆ isomorphism Kγ−22 (Ω) ∩H 1 0(Ω) −→K 0 γ(Ω) If γ < 1 −π ω, ∆ isomorphism Kγ−22 (Ω) ∩H 1 0(Ω)⊕ hΦsingi−→ Kγ0(Ω) 49/72

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Part IV

Edge element techniques

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Outline

9 _Goals

10 _{Nedelec’s edge elements}

(65)

References: Monk 2003, Arnold–Falk–Winther 2006

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Main goal

Construction of finite element spaces VN that allowspectrally correct approximations of the Maxwell cavity problem, using the original variational formulation in H0(curl; Ω).

VN ⊂ H0(curl; Ω), piecewise polynomials on meshTN uN ∈ VN : Z Ω curl uN· curl u 0 − ω2uN· u 0 dx = 0 ∀u0∈ VN

N ∼ dim VN→ ∞ either through

1 _{mesh refinement with fixed degree polynomials:}_{“h-version”, or} 2 _{fixed mesh, increasing degree of polynomials:}_{“p-version”}

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Goal Nedelec’s edge elements

Interface conditions

Remark:Ideally, one would like to have

VN ⊂ H0(curl; Ω) ∩ H(div; Ω)=XN(Ω)(†).

One has to discretize H0(curl; Ω) and H(div; Ω) separately and differently.

(†)

As defined in Part III, XN(Ω)is not a discrete space.Different meaning of subscript “N”!

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Interface conditions

Not good.

(†)

As defined in Part III, XN(Ω)is not a discrete space.Different meaning of subscript “N”!

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Interface conditions

Not good.

A Lemma

Let Ω = Ω1∪ Γ ∪ Ω2and u,u defined on Ω, uj =u|Ωj. Then 1 _{If uj}_{∈ H}1_(Ωj_), _then _{u ∈ H}1_{(Ω) ⇐⇒ [[u]]}_Γ₌₀ 2 _If_u_j _{∈ H(div; Ωj}_{), then} _{u ∈ H(div; Ω) ⇐⇒ [[u · n]]}_Γ₌₀ 3 _If_u

j ∈ H(curl; Ωj), then u ∈ H(curl; Ω) ⇐⇒ [[u × n]]Γ=0

4 _{If uj}_{∈ L}2_(Ωj), _then _{u ∈ L}2_(Ω)

Proof of 3_{: For}_u0_{∈ (C}∞ 0 (Ω))

3_:

hcurl u, u0i = hu, curl u0i = Z Ω1 u · curl u0+ Z Ω2 u · curl u0 = Z Ω1 curl u · u0+ Z Γ u × n · u0ds + Z Ω2 curl u · u0− Z Γ u × n · u0ds = Z Ω (curl u)|Ω\Γ· u0+ Z Γ [[u × n]]Γ· u0ds (†)_{As defined in Part III, X}

N(Ω)is not a discrete space.Different meaning of subscript “N”!

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Interface conditions

Not good.

Corollary

If u,u are piecewise polynomial on a mesh, then 1 _{u ∈ H}1_{(Ω) ⇐⇒ [[u]] = 0 on all interfaces} 2 _{u ∈ H(div; Ω) ⇐⇒ [[u · n]] = 0 on all interfaces} 3 _{u ∈ H(curl; Ω) ⇐⇒ [[u × n]] = 0 on all interfaces} 4 _if_{u ∈ H(div; Ω) ∩ H(curl; Ω), then u ∈ H}1_(Ω)3

(†)_{As defined in Part III, X}

N(Ω)is not a discrete space.Different meaning of subscript “N”!

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How to describe finite element spaces or “A finite element”

Ingredients

1 _MeshTN_._Elements_{K ∈}TN, triangles, quadrilaterals, tetrahedra, hexahedra (cubes)...

One assumes Ω =S K ∈TNK

2 _{On each K , a local}_{space of polynomials}_PK_{. Usually defined on a} reference element and mapped to K .

3 _{Degrees of freedom}_ΣK_{. Linear forms that describe a basis of the dual} space (PK)0. Adjacent elements may havejoint degrees of freedom.

One can then define

4 _a_{global finite element space}

VN = {u : Ω → C| | u|K∈ PK; joint d.o.f. have a unique value } 5 _an_{interpolation projector}_πN

uN = πNu ⇐⇒ uN∈ VN & ∀ t ∈ ΣK : t(u − uN) =0 The element is calledW -conformingif VN ⊂ W .

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Spectrally correct approximation

according to [Caorsi et al 2000] 1 _{Number the increasing sequence of non-zero eigenfrequencies ω ,}

repeated according to multiplicity

0 < ω(1)≤ ω(2)≤ . . . ≤ ω(i)≤ . . .

2 _Let_{ε ∈ (0, ω}(1)_{). Number the increasing sequence of discrete} eigenfrequencies ωN > ε, repeated according to multiplicity

ε < ω(1)_N ≤ ωN(2)≤ . . . ≤ ω (i) N ≤ . . .

Good spectral approximation

(SCA) Spectrally Correct Approximation

∀i ∈ N, lim N→∞ω

(i) N = ω

(i)

and the eigenspaces converge. (SFA) Spurious-Free Approximation

∃α > 0, ∀N ∈ N, ω(i)_N 6∈ (0, α] for all ω_N(i)

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Necessary and sufficient conditions for convergence

[Caorsi et al 2000] (CAS)

Completeness of the Approximating Subspaces

∀ u ∈ H0(curl; Ω) : lim N→∞ u_Ninf∈VN

ku − uNk

H(curl,Ω)=0

(CDK)

Completeness of the Discrete Kernels

∀ k ∈ ker curl : lim

N→∞ kN∈ker curl ∩Vinf N

kk − kNk

L2(Ω)=0.

(DCP) [KIKUCHI1989]

Discrete Compactness Property

Any sequenceuN N→∞with uN ∈ VN∩(kerVNcurl) ⊥ and kuNk H(curl;Ω)≤ 1 contains a subsequence thatconverges in L2(Ω)

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Spurious-free spectrally correct approximation

The ideal situation of eigenvalue approximation is to have (SFA) + (SCA): With the exact eigenvalues

0 < ω(1)≤ ω(2)_{≤ . . . ≤ ω}(i)_{≤ . . .}

and the discrete eigenvalues

0 < ω_N(1)≤ ω(2)N ≤ . . . ≤ ω (i) N ≤ . . .

ω_N(i)converges toω(i), together with the eigenspaces. Theorem [CAORSI- FERNANDES- RAFFETTO2000] (CAS) + (CDK) + (DCP) =⇒ (SFA) + (SCA)

(CDK) means: Have enough gradients in VN

For (DCP), must control discrete divergence-free functions.(Main difficulty!)

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Outline

9 _Goals

10 _{Nedelec’s edge elements}

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Div-conforming elements on simplices

Notation:_Pk polynomials of total degree k .

e

Pk homogeneous polynomials of degree k . =⇒ Pk= Pk −1⊕ ePk Definition K tetrahedron. PK =Dk := (Pk −1)3⊕ ePk −1x ⊂ (Pk)3 Example k = 1:u ∈ D1⇔u = a + bx,a ∈ C3,b ∈ C. “Raviart-Thomas element” RT1. Degrees of freedom: 1 _{u 7→}R fu · n q, q ∈ Pk −1(f ), f face of K 2 _{u 7→}R Ku · q, q ∈ (Pk −1) 3

Global FE spaceWN, interpolation projectorwN, well defined on

H12+δ_(Ω)3_{, δ > 0.}

Note:u ∈ Dk ⇒ u · n ∈ Pk −1(f ) for all faces f of K . WN ⊂ H(div; Ω)

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Curl-conforming elements on simplices

Definition(K tetrahedron)

PK =Rk := (Pk −1)3⊕ {p ∈ (ePk)3| p · x = 0 ⊂ (Pk)3

Example k = 1:u ∈ R1⇔u = a + b × x,a, b ∈ C3. “Nedelec edge element”.

Degrees of freedom: 1 _{e edge:}_{u 7→}R equ · t, q ∈ Pk −1(e) 2 _{f face:}_{u 7→} 1 |f | R fu · q, q ∈ (Pk −2(f )) 3 3 _{u 7→}R Ku · q, q ∈ (Pk −3) 3

Global FE spaceVN, interpolation projectorrN, well defined on

{u ∈ H12+δ_(Ω)3_{| curl u ∈ L}p_{(Ω)}, δ >}_{0, p > 2}

Note:For k = 1, only 1 _{appear (“edge element”).} VN ⊂ H(curl; Ω)

A decomposition

(Pk)3=Rk⊕ ∇ePk +1

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A commuting diagram

These spaces and theirdensely definedprojection operators satisfy the commuting diagram H1 0(Ω) grad −→ H0(curl; Ω) curl −→ H0(div; Ω) div −→ L2_(Ω)   yπ 0 N   yπ 1 N =rN   yπ 2 N=wN   yπ 3 N VN0= Pk grad −→ VN1=VN curl −→ VN2=WN div −→ VN3= Pk −1

Lagrange Nedelec Raviart-Thomas discont.

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Edge elements on hexhedra

(

K

cube)

Notation:_Qk ,l,m = Pk ⊗ Pl⊗ Pm 1 _H1_{conforming: PK} _{= (Qk ,k ,k}₎3 2 _H(_{curl) conforming: P} K = Qk −1,k ,k× Qk ,k −1,k× Qk ,k ,k −1 3 _{H(div) conforming: PK} _{= Qk ,k −1,k −1}_{× Qk −1,k ,k −1}_{× Qk −1,k −1,k} 4 _L2_{conforming: PK} _{= (Qk −1,k −1,k −1)}3

One has once again the De Rham complex and the commuting diagram (“cochain projections”)

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Recall: Approximation on the square Ω = (0, π) × (0, π)

One square element ((Q8)2_{∩ H0(}_{curl; Ω))}

0 10 20 30 40 50 60 70 0 2 4 6 8 10 12 14 Wrong multiplicities

Too many discrete divergence free functions

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Approximation on the square Ω = (0, π) × (0, π)

Spurious eigenfunctions in (Q8)2∩ H0(curl; Ω):Graph of x -component

0 1 2 3 4 0 1 2 3 4 0 0.2 0.4 0.6 0.8 1 0 1 2 3 4 0 1 2 3 4 !0.4 !0.2 0 0.2 0.4 0.6 0.8 1

True first eigenfunction Spurious eigenfunction

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Spurious eigenfunctions in V

_N

_{= (Q8,8)}

2

_{∩ H0(}

_{curl; Ω)}

NotationP08= P8(0, π) ∩ H01(0, π), Ω = (0, π) 2_. VN= (Q8,8)2∩ H0(rot; Ω) = P8⊗ P08 × P 0 8⊗ P8 u =f (x )g(y ) 0 ,f ∈ P8,g ∈ P08,v = v₁(x , y ) v2(x , y )

(rotu, rot v ) − ω2(u, v ) = ZZ Ω f (x )g0(y ) ∂2v1− ∂1v2dxdy − ω2 ZZ Ω f (x )g(y )v1(x , y )dx dy = Zπ 0 g0(y )∂y Z π 0 f (x )v1(x , y )dxdy − ω2 Z π 0 g(y ) Zπ 0 f (x )v1(x , y )dxdy − ZZ Ω f (x )g0(y )∂1v2(x , y )dx dy =0 ∀ v ⇐⇒ 1 Rπ 0g 0_{(y )φ}0_{(y )dy − ω}2Rπ

0g(y )φ(y )dy = 0 ∀ φ(y ) = Rπ 0f (x )v1(x , y )dx⇔ ∀ φ ∈ P 0 8 2 Rπ 0 f (x )ϕ 0_{(x )dx} Rπ 0 g 0_{(y )ψ(y )dy = 0} _{∀ϕ ∈ P}0 8, ψ ∈ P8 64/72

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Goal Nedelec’s edge elements

Spurious eigenfunctions in V

_N

_{= (Q8,8)}

2

_{∩ H0(}

_{curl; Ω)}

1 Rπ 0g

0_{(y )φ}0_{(y )dy − ω}2Rπ

0g(y )φ(y )dy = 0 ∀ φ ∈ P 0 8

This is the1D Laplace–Dirichlet(–P8-Galerkin) eigenvalue problem

⇐⇒ (f , ϕ) =0 ∀ϕ or (g, ψ) =0 ∀ψ.

(g0_{, ψ) =}_{0 ∀ψ ∈ P}

8⇒ g0=0 ⇒ g = 0 ⇒u = 0impossible

f ∈ P8, (f , ϕ0) =0 ∀ϕ ∈ P08 has 2 independent solutions:

dim P0

0=9 − 2 = 7 and dim P8=9

1st solutionf = 1, 2nd solutionf (x ) = L8(π2x − 1)(Legendre polynomial).

The 2nd solution gives spurious eigenfunctions that are discrete divergence-free, but not divergence-free. In fact, their divergence goes to infinity with the polynomial degree, in violation of condition (DCP).

Remedy: Eliminate 2nd solution by replacing P8with P7.

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Spurious eigenfunctions in V

_N

_{= (Q8,8)}

2

_{∩ H0(}

_{curl; Ω)}

1 Rπ 0g

0_{(y )φ}0_{(y )dy − ω}2Rπ

0g(y )φ(y )dy = 0 ∀ φ ∈ P 0 8

This is the1D Laplace–Dirichlet(–P8-Galerkin) eigenvalue problem 2 Rπ 0 f (x )ϕ 0_{(x )dx} Rπ 0 g 0_{(y )ψ(y )dy = 0} _{∀ϕ ∈ P}0 8, ψ ∈ P8 ⇐⇒ (f , ϕ0_{) =}_{0 ∀ϕ} _or _(g0_{, ψ) =}_{0 ∀ψ.} (g0_{, ψ) =}_{0 ∀ψ ∈ P} 8⇒ g0=0 ⇒ g = 0 ⇒u = 0impossible

f ∈ P8, (f , ϕ0) =0 ∀ϕ ∈ P08 has 2 independent solutions:

dim P0

0=9 − 2 = 7 and dim P8=9

1st solutionf = 1, 2nd solutionf (x ) = L8(_π2x − 1)(Legendre polynomial).

The 2nd solution gives spurious eigenfunctions that are discrete divergence-free, but not divergence-free. In fact, their divergence goes to infinity with the polynomial degree, in violation of condition (DCP).

Remedy: Eliminate 2nd solution by replacing P8with P7.