Martin Costabel IRMAR, Université de Rennes 1 …Corner Singularities Martin Costabel IRMAR,...

Corner Singularities

Martin Costabel

IRMAR, Université de Rennes 1

Analysis and Numerics of Acoustic and Electromagnetic ProblemsLinz, 10–15 October 2016

Martin Costabel (Rennes) Corner Singularities Linz, 11–14/10/2016 1 / 62

PART I : Motivation


Dirichlet problem in 2D: Some general observations

Dirichlet problem: −∆u = f in Ω, u = g on ∂Ω

Known (for g = 0)

(i) If Ω is any bounded open set in Rd , then

∀f ∈ L2(Ω) ∃1u ∈ H10 (Ω) : −∆u = f in Ω

(ii) If Ω is any open set in Rd , then

∀f ∈ L2(Ω) ∃1u ∈ H10 (Ω) : −∆u + u = f in Ω

H10 (Ω): closure of C∞0 (Ω) in H1(Ω). Norm: ‖u‖2

1 =∫

Ω(|u|2 + |∇u|2)dx .Variational formulations:

(i) u ∈ H10 (Ω) :

∫Ω∇u ·∇v dx =

∫Ω

v f dx ∀v ∈ H10 (Ω)

(ii) u ∈ H10 (Ω) :

∫Ω

(∇u ·∇v + uv)dx =

∫Ω

v f dx ∀v ∈ H10 (Ω)

(Counter-)Example in R2 for (i): Ω = x | |x | > R, f = 0, g = 1No solution in H1(Ω), but two “reasonable” solutions in H1

loc(Ω):u ≡ 1 : “harmonic at infinity” andu(x) = log |x |/ log R (except for R = 1) : “electrostatic potential”.


Harmonic functions near infinity

Tool: Polar coordinates and separation of variables (Fourier series).Polar coordinates: r ≥ 0, θ ∈ R/(2πZ), (x1, x2) = (r cos θ, r sin θ)Laplace operator in polar coordinates: ∆ = r−2

((r∂r )2 + ∂2

θ

)Fourier series in θ: u continuous in neighborhood of∞ =⇒

u(x) =∑k∈Z

uk (r)eikθ

∆u = 0⇐⇒ r(r u′k )′ = k2uk ⇐⇒ uk (r) =

αk rk + βk r−k if k 6= 0α0 + β0 log r if k = 0

Dirichlet condition u = 1 for r = R: α0 + β0 log R = 1 and αk Rk + βk R−k = 0.

Two particular solutions: u ≡ 1 and u = log r/ log R, but alsoInfinitely many solutions of the homogeneous problem: uk (r) =

((r/R)k − (r/R)−k

)eikθ

Observations for the exterior Dirichlet problem

1 Existence and uniqueness require a-priori assumptions on the asymptotic behavior ofthe solution at infinity.

2 There may be a choice between different “reasonable” a-priori assumptions.


Dirichlet problem on bounded smooth domains

What everybody knows: Existence, uniqueness, regularity

Assume that Ω is a bounded C∞ domain and f ∈ C∞(Ω).Consider the Dirichlet problem ∆u = f in Ω, u = 0 on ∂Ω.Then

1 The solution exists and is unique.2 The solution is regular: u ∈ C∞(Ω).

Right or wrong?Right for the solution u in H1

0 (Ω), for u ∈ C(Ω), or even for u ∈ L2(Ω).But wrong, in general!Example:Let Ω be the half-disk 0 < r < 1, 0 < θ < π = z ∈ C | |z| < 1, Im z > 0.Define

u =(r k − r−k) sin kθ = Im

(zk + z−k) if z = x1 + ix2 = reiθ

Then∆u = 0 in Ω and u = 0 on ∂Ω .

Observations for the Dirichlet problem in a smooth domain

1 Existence, uniqueness and regularity require a-priori assumptions on the behavior ofthe solution near the boundary.

2 There seems to be a unique choice for “reasonable” a-priori assumptions(equivalent: H1, L2, bounded etc.)


Dirichlet problem on a plane sector

Definition: For 0 < ω < 2π, define

Γω = 0 < r <∞, 0 < θ < ω = z | 0 < arg z < ωAssume f = 0 and g = 0 in a neighborhood of the corner, i.e.∆u = 0 in Γω ∩ B(0,R) and u = 0 on ∂Γω ∩ B(0,R).Fourier series:

u(x) =∞∑

k−1

uk (r) sin kπωθ

Radial differential equation: r(r u′k )′ = ( kπω

)2uk ⇐⇒ uk (r) = αk rkπω + βk r−

kπω

Solutions of the totally homogeneous problem:

sk = rkπω sin kπ

ωθ

s−k = r−kπω sin kπ

ωθ

Regular or singular at the corner?

1 For ω = π, we saw (k ∈ N):sk = Im zk : regular (polynomial, Taylor expansion)s−k = Im z−k : singular (not even in L2)

2 For general ω, derivatives of sk are singular: sk 6∈ C∞ if kπω6∈ N. More precisely:

sk ∈ Hs(Γω ∩ B(0,R)) ⇐⇒ s < kπω

+ 1


Singular functions, kπω 6∈ N

sk = Im zkπω ∈ Hs(Γω ∩ B(0,R)) ⇐⇒ s < kπ

ω+ 1

Observations

1 The condition u ∈ H10 (“finite potential energy”) excludes all k < 0.

Same as H1/2; same as boundedness.2 The condition u ∈ L2 does not exclude all s−k :

If ω > π (non-convex corner), then s−1 ∈ L2 near the corner.(⇒ Non-uniqueness, u = s−1− s1 solves the homogeneous problem in Γω ∩B(0, 1)

)3 If ω > π, then s1 has unbounded derivatives, ∇s1 6∈ H1, hence s1 6∈ H2 near the

corner.

Questions

1 Are the singular functions we have seen sufficient to describe the asymptotic behaviornear a corner also for the inhomogeneous Dirichlet problem?

2 Are they always there?3 What is the best way to describe the a-priori regularity assumptions?4 How to treat more general elliptic boundary value problems?

And finally:5 Why should we care??


Exercise: Singular functions for the 2D Neumann problem and mixed D/N

Neumann problem: ∆u = 0 in Γω , ∂nu = 0 on ∂Γω ∂n = −+ 1

r ∂θ on ∂Γω

We look for homogeneous (of degree λ > 0) solutions of ∆u = 0.We find zλ and zλ.Neumann condition in θ = 0 −→ Re zλ = 1

2 (zλ + zλ) = rλ cosλθ.Neumann condition in θ = ω: sinλω = 0 −→ λ = kπ

ω.

Solution: sk = rkπω cos kπ

ωθ (k ∈ Z).

Mixed Dirichlet/Neumann problem:∆u = 0 in Γω , u = 0 for θ = 0, ∂nu = 0 for θ = ω

Homogeneous solution with Dirichlet condition in θ = 0 −→ rλ sinλθ.Neumann condition in θ = ω: cosλω = 0 −→ λ = (k + 1

2 ) πω

.

Solution: sk = r (k+12 )πω cos(k + 1

2 ) πωθ (k ∈ Z).


Exercise: Singular functions for a 2D transmission problem

Transmission or interface problem:

div a∇u = 0 in R2, a(x) = µ for |θ| < ω, a(x) = 1 for ω < |θ| ≤ π Angle 2ω

Let Ω1 = −ω < θ < ω, Ω2 = ω < |θ| ≤ π and u1,2 = u∣∣Ω1,2

.

Then this problem, understood in the distributional sense, is equivalent to∆uj = 0 in Ωj , j = 1, 2

u1 = u2 for θ = −+ω

µ∂nu1 = ∂nu2 for θ = −+ω

Algorithm

We write u1,2 = α1,2a1,2(x) + β1,2b1.2(x),where aj , bj are a basis of harmonic functions in Ωj , homogeneous of degree λ.Then the 4 interface conditions give a 4× 4 linear system for the 4 coefficients α1, . . . , β2:

Mω,µ(λ)

(α1α2β1β2

)= 0

The characteristic equation det Mω,µ(λ) = 0 gives the possible singularity exponents λ,depending on ω and µ.


Exercise: Singular functions for a 2D transmission problem

Transmission or interface problem:

div a∇u = 0 in R2, a(x) = µ for |θ| < ω, a(x) = 1 for ω < |θ| ≤ π Angle 2ω

Let Ω1 = −ω < θ < ω, Ω2 = ω < |θ| ≤ π and u1,2 = u∣∣Ω1,2

.

Then this problem, understood in the distributional sense, is equivalent to∆uj = 0 in Ωj , j = 1, 2

u1 = u2 for θ = −+ω

µ∂nu1 = ∂nu2 for θ = −+ω

Computation: We can split the problem into odd and even problems with respect to θ.For the odd problem, we have a Dirichlet condition on the line θ ∈ 0, π, and we canchoose u1 = rλ sin θλ, u2 = α sin(π − θ)λ.

Two interface conditions at θ = ω:

sinωλ = α sin(π − ω)λ

µ cosωλ = −α cos(π − ω)λ

Elimination of α gives one characteristic equation

tanωλ+ µ tan(π − ω)λ = 0.

For the even problem, we have a Neumann condition on the line θ ∈ 0, π, and we canchoose u1 = rλ cos θλ, u2 = β cos(π − θ)λ.We end up with the second characteristic equation

tanωλ+ 1µ

tan(π − ω)λ = 0.


Re: Questions

Questions

1 Are the singular functions we have seen sufficient to describe the asymptotic behaviornear a corner also for the inhomogeneous Dirichlet problem?

2 Are they always there?3 What is the best way to describe the a-priori regularity assumptions?4 How to treat more general elliptic boundary value problems?

And finally:5 Why should we care??

Questions 1 to 4 will find systematic answers in the lectures by Monique Dauge.

Here follow some observations to questions 2 and 4 , followed by some storiesconcerning the last question 5 .


Dual singular function

Ω ⊂ R2 bounded domain, coincides with Γω in a neighborhood of the origin, otherwisesmooth.Dirichlet problem u ∈ H1

0 (Ω), ∆u = f in Ω f ∈ L2(Ω).

Define s∗−1 = 1π

s−1 + s−1, where s−1(x) = r−πω sin π

ωθ

and s−1 ∈ H10 (Ω) satisfies

∆s−1 = 0 in Ω

s−1 = −s−1 on ∂Ω. (Note: s−1 is smooth on ∂Ω)

Example: If Ω = Γω ∩ B(0, 1), we can take s∗−1 = 1π

(s−1 − s1) = 1π

(r−πω − r

πω ) sin π

ωθ.

We consider ω > π (non-convex corner).Then s∗−1 ∈ L2(Ω) and s∗−1 solves the totally homogeneous Dirichlet problem.

Assumption: u(x) = c s1(x) + o(|x |πω ) as |x | → 0.

Let us compute the scalar product (s∗−1, f ) =∫

Ω s∗−1(x)f (x)dx .

δ > 0, Green’s formula in Ωδ = Ω \ B(0, δ):∫Ωδ

s∗−1(x)f (x)dx =

∫Ωδ

s∗−1(x)∆u(x)dx =

∫∂Ωδ\∂Ω

(s∗−1∂nu − ∂ns∗−1 u

)ds

= 1π

∫ ω

0,r=δ

(r−

πω c π

ωrπω−1 − (− π

ω)r−

πω−1 c r

πω + o(r−1)

)sin2 π

ωθ r dθ

= c + o(1) as δ → 0 .

Hence (s∗−1, f ) = c (“dual singular function”) .


Why care?


Stability, Consistency, and Convergence:A 21st Century Viewpoint

Douglas N. Arnold

School of Mathematics, University of MinnesotaSociety for Industrial and Applied Mathematics

Feng Kang Distinguished Lecture

Institute of Computational Mathematics and Scientific/Engineering ComputingChinese Academy of Sciences

April 7, 2009

1920–1993Feng’s significance for the scientific development of China cannot beexaggerated. He not only put China on the map of applied andcomputational mathematics, through his own research and that of hisstudents, but he also saw to it that the needed resources were madeavailable. . . .Many remember his small figure at international conferences, his eyesand mobile face radiating energy and intelligence. He will be greatlymissed by the mathematical sciences and by his numerous friends.

– Peter Lax, writing in SIAM News, 1993

The failure of the Sleipner A offshore platform

6m

$700,000,000 Richter magnitude 3

3 / 44

Convergence, consistency, and stability of discretizations

L : X ! Y bounded linear operator on Banach spaces.Continuous problem: Given f " Y find u " X such that Lu = f .

Assume it is well-posed: #f $! u s.t. Lu = f , f %! u is continuous

Discrete problem: Lh : Xh ! Yh operator on finite dimensional spaces,fh " Yh. Find uh " Xh such that Lhuh = fh.

The discretization is convergent if uh is sufficiently near u.

The discretization is consistent if Lh and fh are sufficiently nearL and f .

The discretization is stable if the discrete problem is well-posed.

“Fundamental metatheorem of numerical analysis”A discretization which is consistent and stable is convergent.

4 / 44

23 August 1991 Source: D. N. Arnoldhttps://www.ima.umn.edu/~arnold/disasters/sleipner.html


https://www.ima.umn.edu/~arnold/disasters/sleipner.html

Standard finite elements can give bad results

Official cause:

Computations with NASTRAN...

Shear stresses underestimatedby 47%

Loss: ∼ USD 700 million


Fractures emerge near corners


PART II : Maxwell singularities


Some simple numerical tests


Laplace operator: Dirichlet problem in a square (“no corners”)

−∆u = f in Ω = (0, 1)2 ; u = g on ∂Ω

Meshes:

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Solution (f = 1, g = 0):


Dirichlet problem in a square (“no corners”)

Error asymptotics (Rel. L2 error vs. d.o.f.)

101 102 103 10410−6

10−5

10−4

10−3

10−2

10−1

P1P2N−2


Dirichlet problem in an L (“one corner”)

−∆u = f in Ω = (−1, 1)2 \ (0, 1)× (−1, 0) ; u = g on ∂Ω

Meshes:

!1 !0.5 0 0.5 1!1

!0.8

!0.6

!0.4

!0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Solution:


Dirichlet problem in an L, uniform meshes


101 102 103 104 10510−4

10−3

10−2

10−1

P1P2N−1


Dirichlet problem in an L, refined meshes

−∆u = f in Ω = (−1, 1)2 \ (0, 1)× (−1, 0) ; u = g auf ∂Ω

Meshes:

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1


Dirichlet problem in an L, refined meshes


101 102 103 104 10510−5

10−4

10−3

10−2

10−1

P1P2N−1


Dirichlet problem in an L, hp version

−∆u = f in Ω = (−1, 1)2 \ (0, 1)× (−1, 0) ; u = g auf ∂Ω

Meshes: −1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1


Dirichlet problem in an L, hp version


102 103 10410−6

10−5

10−4

10−3

hpN−2

101 102 103 104 10510−6

10−5

10−4

10−3

10−2

10−1

P1 rafP2 rafhpN−2


Worse than slow approximation: Good approximation of the wrong object

Time-harmonic Maxwell equations

curl E = iωµH

curl H = −iωεE + J

In this section: Domain Ω ⊂ R3, ε = µ = 1, J = 0.The condition div E = div H = 0 follows if ω 6= 0.

E × n = 0 & H · n = 0 on ∂Ω

Eigenfrequencies of a cavity with perfectly conducting walls.


Worse than slow approximation: Good approximation of the wrong object

Time-harmonic Maxwell equations

curl E = iωµH

curl H = −iωεE + J

In this section: Domain Ω ⊂ R3, ε = µ = 1, J = 0.The condition div E = div H = 0 follows if ω 6= 0.

E × n = 0 & H · n = 0 on ∂Ω

Eigenfrequencies of a cavity with perfectly conducting walls.Second order system for E : curl curl E − ω2E = 0

Simplest variational formulation

Find ω 6= 0, E ∈ H0(curl,Ω) \ 0 such that

∀F ∈ H0(curl,Ω) :

∫Ω

curl E · curl F = ω2∫

ΩE · F

Energy space: H0(curl,Ω) = u ∈ L2(Ω) | curl u ∈ L2(Ω); u × n = 0= closure in H(curl,Ω) of C∞0 (Ω)3


Regularized formulation

Simple variational formulation

E ∈ H0(curl,Ω) \ 0 : ∀F ∈ H0(curl,Ω) :

∫Ω

curl E · curl F = ω2∫

ΩE · F

Galerkin discretization:Restriction to finite-dimensional subspace Vh, h → 0.

Good: Eigenfrequencies are non-negative, discrete.Big Problem: ω = 0 has infinite multiplicityKernel: Electrostatic fields: gradients of all φ ∈ H1

0 (Ω) (+ harmonic forms).

Idea: div E = 0, so we can add a multiple of 0 =∫

Ω div E div F

Regularized formulation: E ∈ X N \ 0 : ∀F ∈ X N :

(RegX )∫

Ωcurl E · curl F + s

∫Ω

div E div F = ω2∫

ΩE · F

Energy space: X N = H0(curl,Ω) ∩ H(div,Ω)Second order system: curl curl E − s∇ div E = ω2E : Strongly elliptic. OK


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

Good approximation: Triangular edge elements (15 nodes per side, P1)

150 160 170 180 190 200 210 2200

2

4

6

8

10

12

14

Eigenvalue ω2k vs. rank k


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

Bad approximation: Nodal triangular elements (15 nodes per side, P1)

0 20 40 60 80 100 120 1400

2

4

6

8

10

12

14

Eigenvalue ω2k vs. rank k


Regularized formulation in the square

0 1 2 3 4 5 6 70

2

4

6

8

10

12

14

ω[s]2 vs. s

Blue circles: computed ω[s]2 with curl-dominant eigenfunctions.Red stars: computed ω[s]2 with div-dominant eigenfunctions.div E satisfies s∆ div E = ω2 div EExtra eigenvalues: s times Dirichlet eigenvalues.


Regularized formulation in the “L”

0 1 2 3 4 50

5

10

15

20

25

30

35

40

45

ω[s]2 vs. s

Gray triangles: computed ω[s]2 with indifferent eigenfunctions.Cyan-Lines: true Maxwell eigenvalues


Regularized formulation in the “L”

Error of the first eigenvalue

101 102 10310-1

100

101

2 3 4 5 6 7 8 9 10

2 3 4 5 6 7 8 9 10

2 3 4 5 6 7 8 9 10

2 3 4 5 6 7 8 9 10

2 3 4 5 6 7 8 9 10

Interp 1

Interp 2

Interp 3

Interp 4

Interp 5

Valeur propre Maxwell n° 1 pénalisée au bord par λ = 101 ; Maillage 1

Error vs. number of d.o.f.

Error remains largerthan 90%.


Solution of the source problem, regularized formulation

Exact solution(2nd component E2 = r−

13 cos θ3 )).

Computation with Q3 elements.

curl curl E −∇ div E = 0 in Ω; E × n = E0 on ∂Ω


Analysis of Maxwell corner singularities is needed!


An integration by parts formula for Maxwell’s equations

Lemma

Let Ω ⊂ R3 be a bounded smooth domain and u, v ∈ C2(Ω). Then∫Ω

(curl u · curl v + div u div v

)+ c(u, v) =

∫Ω∇u ·∇v + b(u, v)

where c(u, v) =

∫∂Ω

(∇τun · vτ − divτ uτ vn

)b(u, v) =

∫∂Ω

((uτ ·∇n) · vτ ) + div n un vn

)

Corollary 1

If either uτ = vτ = 0 or un = vn = 0, then∫Ω


)=

∫Ω∇u ·∇v + b(u, v)

With the help of Corollary 1 one can prove that if Ω is smooth or convex (approximation bysmooth domains with boundaries of positive curvature), then [Saranen 1982, Nedelec 1982]

XN ∪ XT ⊂ H1.


An integration by parts formula for Maxwell’s equations

Lemma

Let Ω ⊂ R3 be a bounded smooth domain and u, v ∈ C2(Ω). Then∫Ω


)+ c(u, v) =

∫Ω∇u ·∇v + b(u, v)

where c(u, v) =

∫∂Ω

(∇τun · vτ − divτ uτ vn

)b(u, v) =

∫∂Ω

((uτ ·∇n) · vτ ) + div n un vn

)

Corollary 2 [Co 1991]

If Ω is a polyhedron and u ∈ HN ∪ HT , then∫Ω

(| curl u|2 + | div u|2

)=

∫Ω|∇u|2

For the proof one has to show that smooth functions that are zero near the edges andcorners are dense in HN and HT .From Corollary 2 follows that the subspaces HN of XN and HT of XT are closed.This implies that approximation of elements of XN \ HN or XT \ HT by conforming finiteelements is impossible.


Form of Maxwell singular functions, 2D

On a sector Γ = Γω , 0 < ω ≤ 2π, we define spaces of homogeneous functions of degreeλ 6∈ N: Sλ(Γ) = u = rλφ(θ) | φ ∈ C inf([0, ω]) , Sλ(Γ) = Sλ(Γ)2

SλDir(Γ) = u ∈ Sλ(Γ) | u∣∣∂Γ

= 0 , SλN (Γ) = u ∈ Sλ(Γ) | u × n∣∣∂Γ

= 0.We consider homogeneous solutions of the principal part of the regularized Maxwell system

curl curl u − s∇ div u = 0 in Γ

u × n = 0 and div u = 0 on ∂Γ

u ∈ Sλ(Γ)

We rewrite the system by introducing ψ = curl u, q = div u as a triangular system∆q= 0 in Γ, q ∈ Sλ−1

Dir (Γ) (1)

curlψ= s∇q in Γ, ψ ∈ Sλ−1Neu (Γ) = Sλ−1(Γ) (2)

curl u = ψ , div u= q in Γ, u ∈ SλN (Γ) (3)

Solutions: Sums of the followingType 1: q = 0, ψ = 0, u general non-zero solution of (3)Type 2: q = 0, ψ general solution of (2), u particular solution of (3)Type 3: q general solution of (1), ψ and u particular solutions of (2) and (3).

In 2D, Type 2 is easy: curlψ = 0 means ψ = const. This doesn’t exist in Sλ−1(Γ).Type 1: curl u = 0⇒ u = ∇φ, ∆φ = 0 and φ ∈ Sλ+1

Dir (Γ). That is, φ is a Laplace/Dirichletsingular function, λ+ 1 = kπ

ωand

φ = c Im zλ+1 =⇒ u = c(λ+ 1)rλ(sinλθ, cosλθ).


Form of Maxwell singular functions, 2D

Type 1: q = 0, ψ = 0, u general non-zero solution of (3)Type 2: q = 0, ψ 6= 0 general solution of (2), u particular solution of (3)Type 3: q 6= 0 general solution of (1), ψ and u particular solutions of (2) and (3).

In 2D, Type 2 is easy: curlψ = 0 means ψ = const. This doesn’t exist in Sλ−1(Γ).Type 1: curl u = 0⇒ u = ∇φ, ∆φ = 0 and φ ∈ Sλ+1

Dir (Γ). That is, φ is a Laplace/Dirichletsingular function, λ+ 1 = kπ

ωand

φ = c Im zλ+1 =⇒ u = c(λ+ 1)rλ(sinλθ, cosλθ).

Type 3: (1) means q is a Laplace/Dirichlet singular function, λ− 1 = kπω

andq = c Im zλ−1. From (2) we see that ψ is conjugate harmonic to q, henceψ = −c Re zλ−1. A particular solution of (3) is then

u = c2λ rλ(sinλθ,− cosλθ) .

Theorem

At a polygonal corner of opening ω, the non-integer singular exponents of the principal partof the regularized Maxwell system are of the form

λ = kπω −

+1 , k ∈ Z

The divergence-free Maxwell singular functions are of the form

u = rλ((sinλθ, cosλθ) , λ = kπω− 1 , k ∈ Z .


Form of Maxwell singular functions, 3D conical point

Cone in spherical coordinates Γ = (ρ, ϑ) | ρ > 0, ϑ ∈ G ⊂ S2Spaces of homogeneous functions (λ ∈ R \ N)

Sλ(Γ) = u = ρλφ(ϑ) | φ ∈ H1(G), Sλ , SλDir , etc

SλN (Γ) = u = ρλφ(ϑ) ∈ L2loc(Γ \ 0) | curl u, div u ∈ L2

loc(Γ \ 0), u × n = 0 on ∂Γ

SλT (Γ) = u = ρλφ(ϑ) ∈ L2loc(Γ \ 0) | curl u, div u ∈ L2

loc(Γ \ 0), u · n = 0 on ∂Γ

Homogeneous solutions of the principal part of the regularized Maxwell system (s > 0)curl curl u − s∇ div u = 0 in Γ

u × n = 0 and div u = 0 on ∂Γ

u ∈ SλN (Γ), div u ∈ Sλ−1Dir (Γ)

We rewrite the system by introducing ψ = curl u, q = div u as a triangular system∆q = 0 in Γ, q ∈ Sλ−1

Dir (Γ) (1)

curlψ = s∇q in Γ, ψ ∈ Sλ−1T (Γ) (2)

curl u = ψ , div u = q in Γ, u ∈ SλN (Γ) (3)



Solutions of the triangular system:Type 1: q = 0, ψ = 0, u ∈ SλN (Γ) general non-zero solution of (3): curl u = ψ, div u = qType 2: q = 0, ψ ∈ Sλ−1

T (Γ) general solution of (2) : curlψ = s∇q, (→ s = 1)u particular solution of (3)

Type 3: q ∈ Sλ−1Dir (Γ) general solution of (1): ∆q = 0,

ψ and u particular solutions of (2) and (3).

Lemma Explicit solutions, [Co-Dauge ARMA2000]

λ 6= −1, u ∈ SλN (Γ) =⇒ φ = 1λ+1 u · x ∈ Sλ+1

Dir (Γ) ; curl u = 0 =⇒∇φ = u

λ 6= −1, ψ ∈ Sλ−1T (Γ) =⇒ u = 1

λ+1ψ × x ∈ SλN (Γ) ; divψ = 0 =⇒ curl u = ψ

λ 6= − 12 , q ∈ Sλ−1

Dir (Γ)⇒ u = 2qx+ρ2∇q4λ+2 ∈ SλN (Γ) ; ∆q = 0⇒ curl u = 0, div u = q








λ 6= −1, u ∈ SλN (Γ) =⇒ φ = 1λ+1 u · x ∈ Sλ+1

Dir (Γ) ; curl u = 0 =⇒∇φ = u

λ 6= −1, ψ ∈ Sλ−1T (Γ) =⇒ u = 1


λ 6= − 12 , q ∈ Sλ−1


Corollary, Type 1

If λ 6= −1, thenu solution of Type 1 ⇐⇒ u = ∇φ, φ ∈ Sλ+1

Dir (Γ), ∆φ = 0.








λ 6= −1, u ∈ SλN (Γ) =⇒ φ = 1λ+1 u · x ∈ Sλ+1

Dir (Γ) ; curl u = 0 =⇒∇φ = u

λ 6= −1, ψ ∈ Sλ−1T (Γ) =⇒ u = 1


λ 6= − 12 , q ∈ Sλ−1


Corollary, Type 2

If λ 6∈ 0,−1, thenu solution of Type 2 ⇐⇒ curl u = ∇φ, φ ∈ SλNeu(Γ), ∆φ = 0 ; u = 1

λ+1∇φ× x








λ 6= −1, u ∈ SλN (Γ) =⇒ φ = 1λ+1 u · x ∈ Sλ+1

Dir (Γ) ; curl u = 0 =⇒∇φ = u

λ 6= −1, ψ ∈ Sλ−1T (Γ) =⇒ u = 1


λ 6= − 12 , q ∈ Sλ−1


Corollary, Type 3

If λ 6∈ − 12 , 0, then

u solution of Type 3 ⇐⇒ div u = q, q ∈ Sλ−1Dir (Γ), ∆q = 0.

ψ = 1λ∇φ× x , u = 1

λ(2λ+1)

((2λ− 1)qx − ρ2∇q

)


Some formulas from vector analysis

From [Co-Da ARMA 2000]:

260 Martin Costabel & Monique Dauge

1. q = 0,! = 0 andU is a general non-zero solution of (6.3c), respectively (6.4c).2. q = 0, ! is a general non-zero solution of (6.3b), respectively (6.4b) and U aparticular solution of (6.3c), respectively (6.4c).

3. q is a general non-zero solution of (6.3a), respectively (6.4a), ! a particularsolution of (6.3b), respectively (6.4b) and U a particular solution of (6.3c),respectively (6.4c).

6.2. Explicit solutions of first order problems

The Laplace singularities on polyhedral cones were described in Lemma 2.4.They contain Laplace-Beltrami eigenfunctions and have therefore, in contrast tothe two-dimensional case, no analytically known form, in general. But once theseLaplace singularities are known,we are able to provide completely explicit formulasfor the three types of Maxwell singularities.

This section is devoted to the description of solution formulas for the first orderproblems (6.3) and (6.4). All these formulas are based on the scalar product or thevector productwith the vectorx, withxdenoting the vector ofCartesian coordinates(x, y, z), and ρ = |x|.

We begin with three series of formulas. First we give product laws: a and bdenoting vector fields and γ being a scalar function on R3, we have

grad(a · b) = (a · grad) b + (b · grad) a + a × curl b + b × curla, (6.5a)

curl(a × b) = (b · grad) a − (a · grad) b + a div b − b diva, (6.5b)

div(a × b) = b · curla − a · curl b, (6.5c)

curl(γa) = γ curla + grad γ × a, (6.5d)

div(γa) = γ diva + grad γ · a. (6.5e)

Now, using the above formulas for the field x which satisfies

divx = 3, curlx = 0, x · grad = ρ∂ρ and gradx = I,

we obtain for any field a and scalar q

grad(a · x) = (ρ∂ρ + 1)a + x × curla, (6.6a)

curl(a × x) = (ρ∂ρ + 2)a − x diva, (6.6b)

div(a × x) = x · curla, (6.6c)

curl(qx) = grad q × x, (6.6d)

div(qx) = (ρ∂ρ + 3)q. (6.6e)

Finally, with γ = ρ2 and a = grad q, (6.5d) and (6.5e) yield

curl(ρ2 grad q) = −2 grad q × x, (6.6f)

div(ρ2 grad q) = 2ρ∂ρq + ρ2$q. (6.6g)


Form of Maxwell singular functions, 3D conical point, Electric b.c.

Theorem Electric Maxwell Singularities

• At a 3D conical point, the singular exponents λ 6= 0, λ > −1 of the principal part of theregularized Maxwell equations are of the form

λDir − 1, λNeu or λDir + 1,where λDir and λNeu are Dirichlet resp. Neumann singular exponents of the Laplacian.• The last type does not appear for the divergence-free Maxwell equations.• The exponents λ = 0 and λ = −1 appear for certain non-Lipschitz topologies.

Similar theorem for Magnetic Boundary Conditions...

Rule of thumb: The main singularity is a gradient.


Birman-Solomyak decomposition theorems (“Helmholtz-like”)

Decomposition Theorem: Electric b. c. [Birman-Solomyak 1987]

Let Ω be a bounded Lipschitz domain (+ cracks, in R2 or R3). Then

∃KN : XN (Ω)→ HN (Ω) bounded such that curl KN = curl .

That is, for u ∈ XN (Ω) we have

u = ∇φ+ w with w = KN u ∈ HN (Ω), φ ∈ H10 (∆,Ω) .

Decomposition Theorem: Magnetic b. c. [Birman-Solomyak 1987 + Filonov 1997]

Let Ω be a bounded Lipschitz domain (+ cracks, in R2 or R3). Then

∃KT : XT (Ω)→ HT (Ω) bounded such that curl KT = curl ,

that is, for u ∈ XT (Ω) we have

u = ∇φ+ w with w = KT u ∈ HT (Ω), φ ∈ H1Neu(∆,Ω) ,

if and only if∀ v ∈ H1(Ω) ∃ψ ∈ H2(Ω) : n · v = ∂nψ on ∂Ω .

This is satisfied if Ω is piecewise C3/2+ε, ε > 0.

There exists a domain Ω ∈ C3/2 for which it is not satisfied.


XN and HN

We have seen: u ∈ XN \ HN ←→∇φ, φ ∈ H10 (∆) \ H2.

Corollary

(i) Let Ω be a Lipschitz polygon in R2. Then HN (Ω) is a closed subspace of finitecodimension of XN . The codimension is equal to the number of non-convex corners of Ω.

(ii) Let Ω be a Lipschitz polyhedron in R3. Then either XN (Ω) = HN (Ω) (when Ω is convex)or HN (Ω) is a closed subspace of infinite codimension of XN (due to non-convex edges).

Consequences of XN 6= HN ?


Two variational formulations

Regularized time-harmonic Maxwell boundary value problem (ε = µ = 1)

(BVP)

curl curl u − s∇ div u − ω2u = f in Ω

u × n = 0 and div u = 0 on ∂Ω

u ∈ XN (Ω)

Weak formulation: Integration by parts against v ∈ C1(Ω) with v × n = 0 gives

(P)∫

Ω

(curl u · curl v + s div u div v − ω2u · v

)=

∫Ω

f · v

We assume now that Ω is a Lipschitz polyhedron in R3.Then C∞(Ω) ∩ XN (Ω) is dense in HN (Ω), and we see that

The boundary value problem (BVP) is equivalent to

Find u ∈ XN (Ω) such that (P) is satisfied ∀v ∈ HN (Ω) .

Non-symmetric ! We have two symmetric versions:

(PX ) Find u ∈ XN (Ω) such that (P) is satisfied ∀v ∈ XN (Ω) Maxwell

(PH ) Find u ∈ HN (Ω) such that (P) is satisfied ∀v ∈ HN (Ω) Lamé


Two variational formulations

Two unique solutions

For f ∈ L2(Ω) with div f = 0 and all ω > 0 except for a discrete set of frequences, bothproblems (PX ) and (PH ) have unique solutions.If Ω is non-convex, these solutions are, in general, different.The eigenfrequencies are different.The solution of (PX ) satisfies div u = 0 (Maxwell solution),the solution of (PH ) has, in general, div u 6∈ H1(Ω) (Lamé or “pseudo-Maxwell” solution).


Different versions of the decomposition

The splitting into a singular and regular part

u = using + ureg

is not unique. Different motivations −→ different splittings.Motivation 1: Numerical approximation by singular function method.The principle:Assume u ∈ V is solution of a variational problem

∀ v ∈ V : a(u, v) =< f, v > .

Consider Galerkin approximation by subspaces Vh ⊂ V : uh ∈ Vh such that

∀ vh ∈ Vh : a(uh, vh) =< f, vh > .

We assume that there holds Céa’s Lemma: ‖u − uh‖ ≤ C inf‖u − vh‖ | vh ∈ Vh.If there is a splitting u = γs + ureg with a known singular function s, one can define theapproximation space as an augmented finite element space

Vh := spans ⊕ V 0h .

Céa’s Lemma now gives

‖u − uh‖ ≤ C inf‖ureg − vh‖ | vh ∈ V 0h .

If V 0h is a regular finite element space, the convergence order is thus determined by the

regularity of ureg ∈ H1+ε, which should therefore be as high as possible.Martin Costabel (Rennes) Corner Singularities Linz, 11–14/10/2016 49 / 62

Different splittings in 2D

Assumption:• Ω is a bounded domain, coincides near the origin with Γω , ω > π, and is smoothelsewhere.• u ∈ XN is solution of the regularized Maxwell system (PX ) with ω = 0 and smooth righthand side f, div f = 0.Decomposition 1. “Natural decomposition”. s = χ∇rλ1 sinλ1θ with λ1 = π

ω.

Limit 1 + ε∗ of regularity determined by second singular function: ureg ∈ Hs, s < 1 + ε∗ for1 + ε∗ = λ2 = 2π

ω.

Decomposition 2. Divergence-free singular function: s = χ curl rλ1 cosλ1θ. This is thesame as 1.Decomposition 3. Projection on HN . Let φ ∈ H1

0 (Ω) solve ∆φ = s∗−1 with the dualsingular function s∗−1 of the Dirichlet problem, and define s = ∇φ.Here ureg is computable: It is the “pseudo-Maxwell” or “Lamé” solution of (PH ).

Then s, and hence ureg, contains a term in S−πω

+2−1, hence ε∗ = min 2πω− 1, 1− π

ω.

For π < ω < 3π/2, this is worse than Decomposition 1, worst for ω close to π.Decomposition 4. Orthogonal decomposition in XN , s ∈ H⊥

N . Here ε∗ is the same as inDecomposition 3.Note: In Decomposition 3 & 4, additional (spurious) singular functions appear in both s andin ureg that are absent in u.


“Making convergence possible again”: The WRM

Idee: The problem comes from the fact that the regularized Maxwell bilinear form

a(u, v) =

∫Ω

(curl u · curl v + s div u div v

)defines a space XN = H0(curl) ∩ H(div) in which smooth functions (and piecewisepolynomials) are not dense.The term

∫Ω div u div v was rather arbitrary, because div u = 0 for Maxwell solutions. It can

be replaced by a different bilinear form, that is, the norm ‖ div u‖L2(Ω) can be replaced by anorm ‖ div u‖Y with a space Y satisfying the two requirements:

1 The corresponding space

X YN = H0(curl) ∩ u ∈ L2 | div u ∈ Y

is still compactly embedded in L2(Ω).2 C∞(Ω) ∩ X Y

N is dense in X YN .

This was first proposed in [Co-Dauge NuMa 2002] “Weighted regularization of Maxwellequations in polyhedral domains. A rehabilitation of nodal finite elements”.There, Y is a weighted L2 space:

‖q‖2Y =

∫Ωρ2γ |q|2 , ρ distance to the non-convex edges and corners


Weighted regularization

Theorem [Co-Da 2002]

Let Ω be a polygon in R2 or a polyhedron in R3.Define DY (∆) = q ∈ H1

0 (Ω) | ∆q ∈ Y. Then

(i) Decomposition X YN = HN + ∇DY (∆).

(ii) If H2 ∩ H10 (Ω) is dense in DY (∆), then condition 2 is satisfied.

(iii) There exists 0 < γ∗ < 1 such that for weight exponents γ∗ < γ < 1, conditions 1

and 2 are satisfied.γ∗ = max1− π

ωe, 1

2 − λDirc .

As a consequence, for this choice of weights, any conforming finite element Galerkinmethod for the regularized Maxwell system (source problem or eigenvalue problem)converges in X Y

N .

More recently, results with other choices of Y have appeared: e.g.Y = H−s for 0 < s < 1, or a discretized version thereof, ‖ · ‖Y ,h = hs‖ · ‖L2

[Bonito-Guermond MathComp 2011]


Weighted Regularization in the “L”

Computations with Q10 elements on refined mesh.

Legend:

Blue circles: computed ω[s]2 with curl-dominant eigenfunctions.Red stars: computed ω[s]2 with div-dominant eigenfunctions.Gray triangles: computed ω[s]2 with indifferent eigenfunctions.Cyan Lines: true Maxwell eigenvalues (calculated from scalar Neumann problem)


The ideal: Computations in the square

0 1 2 3 4 5 6 70

2

4

6

8

10

12

14

ω[s]2 vs. s


The unweighted reality: Computations in the L (γ = 0)

0 1 2 3 4 50

5

10

15

20

25

30

35

40

45

ω[s]2 vs. sThis is Lamé, not Maxwell !


Towards the ideal: WRM in the L (γ = 0.35)

0 2 4 6 8 100

5

10

15

20

25

30

35

40

45

ω[s]2 vs. s


Towards the ideal: WRM in the L (γ = 0.5)

0 2 4 6 8 100

5

10

15

20

25

30

35

40

45

ω[s]2 vs. s


The ideal recovered: WRM in the L (γ = 1)

0 5 10 15 200

5

10

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35

40

45

ω[s]2 vs. s


WRM on Fichera’s corner



1st Maxwelleigenmode onFichera’s corner

Q4 elements

WeightedRegularization


Another singularity


Martin Costabel IRMAR, Université de Rennes 1 …Corner Singularities Martin Costabel IRMAR,...

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