[PPT] - Transformation of corner singularities in presence of small or large PowerPoint Presentation

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References Example 1 Example 2 Example 3 Example 4 Profiles

Transformation of corner singularities in presence of small or large parameters

Monique Dauge

IRMAR, Universit´ e de Rennes 1, FRANCE

Analysis and Numerics of Acoustic and Electromagnetic Problems

October 17-22, 2016, Linz, Austria RICAM Special Semester on Computational Methods in Science and Engineering

http://perso.univ-rennes1.fr/monique.dauge

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles

Joint works

M. COSTABEL, M. DAUGE,

Example 2 A singularly perturbed mixed boundary value problem,

Comm. Partial Differential Equations 21, 11 & 12 (1996), 1919–1949.
G. CALOZ, M. COSTABEL, M. DAUGE, G. VIAL

Example 3 Asymptotic expansion of the solution of an interface problem in a polygonal domain with thin layer Asymptotic Analysis 50 (1/2) (2006), 121–173.

M. DAUGE, S. TORDEUX, AND G. VIAL

Example 1 Selfsimilar perturbation near a corner: matching versus multiscale expansions for a model problem, Around the research of Vladimir Maz’ya. II, vol. 12 of Int. Math.

Ser. Springer (2010), 95–134.
M. DAUGE P. DULAR, L. KR¨

AHENB ¨ UHL, V. P´ ERON, R. PERRUSSEL, C.

POIGNARD Example 4 Corner asymptotics of the magnetic potential in the eddy-current model Mathematical Methods in the Applied Sciences 37, 13 (2014) 1924–1955

M. COSTABEL, M. DALLA RIVA, M. DAUGE, P. MUSOLINO

Example 1 Converging expansions for Lipschitz self-similar perforations of a plane sector, In preparation, (2016). 2D eddy current formulation for a conductor surrounded by a dielectric medium Recent discussions with

R. HIPTMAIR, R. CASAGRANDE, K. SCHMIDT, A. SEMIN

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 1: Self-similar perturbations of a 2D corner

See references

1

Let Γ be the infinite plane sector Γ = {x = (x1, x2) ∈ R2, r > 0, θ ∈ (0, ω)} with polar coordinates (r, θ) and the opening ω ∈ (0, π) ∪ (π, 2π).

2

Bounded un-perturbed domain Ω ≡ Ω0, with Ω ⊂ Γ. Assume

Ω bounded connected with curvilinear polygonal boundary, ∃r0 > 0 such that B(0, r0) ∩ Ω = B(0, r0) ∩ Γ

3

Bounded perturbation pattern P, with P ⊂ Γ.

P connected with curvilinear polygonal boundary, ∃R0 > 0 such that B(0, R0)∁ ∩ P = B(0, R0)∁ ∩ Γ

4

Family of perturbed domains, with ε ∈ (0, ε0) Ωε = Ω ∩ εP

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References Example 1 Example 2 Example 3 Example 4 Profiles

Dirichlet Problems for Example 1

Simplifying assumption f ∈ L2(Ω) with support outside a ball B(0, r1) General assumption f ∈ L2(Ω) analytic or C ∞ inside a ball B(0, r1) We choose the simplifying assumptions for the talk because A zero Taylor expansion of f at the corners allows to discard polynomials from corner asymptotics, so to avoid log terms and convergence issues. The family of problems under consideration is the family of Dirichlet problems for ε ∈ [0, ε0) (Pε)

−∆uε = f

in Ωε uε = 0

n ∂Ωε.

Unique solution in uε ∈ H1

0(Ωε)

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References Example 1 Example 2 Example 3 Example 4 Profiles

Two particular cases of Example 1

Question Behavior of uε close to 0 (in the region B(0, r0) ∩ Ωε) as ε → 0. Let us consider two examples 1a A rounded corner: The boundary of P is smooth! 1b A cracked corner: The boundary of P has a crack Σ abutting at 0 (P = Γ \ Σ, with a straight segment Σ). Set α = π ω In both examples, Ω = Ω0 has a corner of opening ω at 0. Therefore [Kondrat’ev 1967] u0 has an expansion as u0 = uK

0,reg + K

k=1

γkr kα sin kαθ, uK

0,reg ∈ Hm+1(Ω ∩ B(0, r0)), m = [(K + 1)α]

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References Example 1 Example 2 Example 3 Example 4 Profiles

Paradoxes on singularities for Example 1

On the one hand u0 = γ1r α sin αθ + u0,reg, u0,reg ∈ H2(Ω ∩ B(0, r0)) On the other hand 1a Rounded corner: uε is smooth in Ωε ∩ B(0, r0) 1b A cracked corner: The first singularity of uε behaves like r

1 2 sin θ

2

Thus 1a uε has no singularity but u0 has one 1b uε has a stronger singularity than u0 How can the singularity r α sin αθ associated with ε = 0 1a Disappear as soon as ε > 0 1b Transform into a stronger singularity as soon as ε > 0 ?

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 2: Neumann-Robin boundary conditions

See references

Take a polygon Ω with straight sides Choose a side, denoted by ΣR. Denote ∂Ω \ ΣR by ΣN Assume for simplicity that the openings of Ω at the two ends of ΣR are the same, say ω Assume for simplicity that π

ω is not a rational number

Assume for simplicity that f ∈ L2(Ω) with support outside neighborhoods

f the ends of ΣR

Then the Neumann-Robin problem is for ε ∈ [0, ε0) (Pε)      −∆uε = f in Ω ∂nuε = 0

n ΣN

ε∂nuε + uε = 0

n ΣR

When ε = 0, we obtain a mixed Neumann-Dirichlet problem (P0)      −∆u0 = f in Ω ∂nu0 = 0

n ΣN

u0 = 0

n ΣR

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 2: Variational formulations and singularities

Variational spaces: Vε = H1(Ω), ε > 0, and V0 = {u ∈ H1(Ω), u

ΣR = 0}.

Variational formulations (Pε) uε ∈ H1(Ω), ∀v ∈ H1(Ω),

Ω

∇uε · ∇v dx + 1 ε

ΣR

uε v dσ =

Ω

f v dx (P0) u0 ∈ V0, ∀v ∈ V0,

Ω

∇u0 · ∇v dx =

Ω

f v dx Singularities of u0 (for a Neumann-Dirichlet corner) recall that α = π

ω

r α/2 cos 1

2αθ,

r 3α/2 cos 3

2αθ,

r 5α/2 cos 5

2αθ, . . .

Singularities of uε (for a Neumann-Robin corner): r α cos αθ + ε−1γ1r α+1 cos(α + 1)θ + ε−2γ2r α+2 cos(α + 2)θ + . . . r 2α cos 2αθ + ε−1γ′

1r 2α+1 cos(2α + 1)θ + ε−2γ′ 2r 2α+2 cos(2α + 2)θ + . . .

i.e. Neumann singularity plus successive shadows. Negative powers of ε

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 3: Thin layer

See references

Take a polygon Ω with straight sides Consider the union Σ of two consecutive sides σ1 and σ2. Let O = σ1 ∩ σ2 be the corner inside Σ. Assume for simplicity that the opening ω at O is such that π

ω ∈ Q

Assume for simplicity that the openings of Ω at the two exterior corners c1 and c2 in Σ equal π

2

Define the layer Lε of width ε as the polygon formed by Σ, the segments σℓ,ε, ℓ = 1, 2, parallel to σℓ at the distance ε outside Ω, and two perpendicular segments of length ε issued from c1 and c2. Assume for simplicity that f ∈ L2(Ω) with support outside a neighb. of O Set Ωε = Ω ∪ Σ ∪ Lε and consider the transmission problem, with a positive number β = 1 uε ∈ H1

0(Ωε), ∀v ∈ H1 0(Ωε),

Ω

∇uε · ∇v dx + β

Lε

∇uε · ∇v dx =

Ω

f v dx. Thus u0 is solution of the Dirichlet pb on Ω = Ω0.

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 3: Singularities

Singularities of u0 at 0 (Dirichlet) recall that α = π

ω

r α sin αθ, r 2α sin 2αθ, r 3α sin 3αθ, . . . Singularities of uε at 0: Those of the transmission problem          −∆UΩ = F in ΓΩ −∆U L = 0 in ΓL [U] = 0

n ∂ΓΩ ∩ ∂ΓL

∂nUΩ + β ∂nU L = 0

n ∂ΓΩ ∩ ∂ΓL

Here ΓΩ is the sector of opening ω and ΓL is (the interior of) its complement UΩ and UL are the restrictions of U to ΓΩ and ΓL, respectively. n exterior derivative. The singularities of the transmission problems are different from the Dirichlet singularities.

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 4: Conducting material in 2D eddy current formulation

Take a conductor body M (conductivity σ) surrounded by a dielectric body N. Harmonic Maxwell equations curl E = −iκµH and curl H = (iκǫ + σ1M)E + j0 The eddy current formulation consists in neglecting κǫ with respect to σ: curl E = −iκµH and curl H = σ1ME + j0 and add the gauge condition div E = 0 in N. Assume that the domains and the fields do not depend on the variable x3 and that H = (0, 0, u)⊤. With, now, Ω ⊂ R2, M ⊂ Ω, N = Ω \ M the 2D eddy current formulation for a magnetic potential u is (Pδ) uδ ∈ H1

0(Ω), ∀v ∈ H1 0(Ω),

Ω

∇uδ · ∇v dx + i δ2

M

uδ v dx =

N

f v dx with δ−2 = κµσ

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 4: Singularities

The parameter δ is (proportional to) the skin depth. It tends to 0 when σ tends to infinity. Assume that Ω is smooth. Assume for simplicity that the conductor M has only one corner in O. Let 2π − ω be its opening, i.e. ω is the opening of N at O. Assume for simplicity that the opening ω is such that π

ω ∈ Q

Assume for simplicity that f ∈ L2(N) with support outside a neighb. of O In strong form (Pδ) is −∆uδ + i δ2 1M uδ = f in Ω, with uδ ∈ H1

0(Ω)

and (P0) is −∆u0 = f in N, with u0 ∈ H1

0(N)

Singularities of u0 at 0 (Dirichlet) recall that α = π

ω

r α sin αθ, r 2α sin 2αθ, r 3α sin 3αθ, . . . Singularities of uδ at 0 ?

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References Example 1 Example 2 Example 3 Example 4 Profiles

Example 4: Singularities, continued

See references

Singularities at a corner are determined by the principal part of the

perator.

The principal part of −∆ +

i δ2 1M is −∆. The corner O is inside Ω.

Hence no principal singularity! By the way the solution u belongs to H2(Ω). But, still, singularities appear as shadows of harmonic polynomials. Indeed (here * = wildcard) 1 + δ−2r 2 log2r ϕ∗

2(θ) + log r ϕ∗ 1(θ) + ϕ∗

+ δ−4r 4 4

ℓ=0 logℓr ϕ∗ ℓ(θ) + . . .

r cos θ + δ−2r 3 3

ℓ=0 logℓr ϕ∗ ℓ(θ) + δ−4r 5 5 ℓ=0 logℓr ϕ∗ ℓ(θ) + . . .

r sin θ + δ−2r 3 3

ℓ=0 logℓr ϕ∗ ℓ(θ) + δ−4r 5 5 ℓ=0 logℓr ϕ∗ ℓ(θ) + . . .

The logarithmic terms of highest degree are present! The strongest singularity of uδ is r 2 log2 r, Versus the strongest singularity of u0 that is r

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References Example 1 Example 2 Example 3 Example 4 Profiles

Outline

1 Joint works and discussions

2 Example 1: Self-similar perturbations of a 2D corner

3 Example 2: Neumann-Robin boundary conditions

4 Example 3: Thin layers

5 Example 4: Conducting material in 2D eddy current formulation

6 Profiles and corner layers Two strategies Example 1: Self-similar perturbations of a 2D corner Example 2: Neumann-Robin Conclusion

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References Example 1 Example 2 Example 3 Example 4 Profiles Two strategies

Two strategies

We have a family of singularly perturbed problems (Pε) with solutions uε and want to understand How uε tends to u0 How singularities transform as ε → 0 2 strategies, in the framework of multiscale analysis: A Convert the singularities sλ (homogeneous of degree λ) of u0 into profiles K λ solutions of a model problem with ε = 1. Roughly A0sλ = 0, A1K λ = 0 and K λ − sλ = O(r λ) B Leave the singular part as is, but introduce a corrector Y λ (a corner layer) to reduce the distance to uε.

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 1: Self-similar perturbations of a 2D corner

Example 1: Profiles and corner layers

u0 ≃

λ∈Λ cλsλ, with Λ = N∗ π ω and sλ = r λ sin λθ. We have, on the sector Γ:

∆sλ = 0

in Γ sλ = 0

n ∂Γ.

Lemma There exists a unique solution K λ in the perturbation pattern P of the problem      ∆K λ = 0 in P K λ = 0

n ∂P

K λ = sλ + O(1) as R → ∞ Proof: Set K 1

−1(P) = {u ∈ L2 loc(P), ∇u, r −1u ∈ L2(P)}. Define Y λ as the

variational solution in K 1

−1(P) of the problem

∆Y λ = 0

in P Y λ = −sλ

n ∂P \ ∂Γ

(note that ∂P \ ∂Γ is compact and sλ = 0 on ∂P ∩ ∂Γ). Then set K λ = sλ + Y λ

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 1: Self-similar perturbations of a 2D corner

Example 1: Profiles and corner layers, asymptotics at infinity

Then K λ x

ε

is harmonic and satisfies the Dirichlet condition of εP.

Formally u′

ε :≃ λ∈Λ cλ ελK λ x ε

satisfies the Dirichlet condition of εP.

But may differ from uε on ∂Ω \ ∂Γ. This error is far from O. To analyze it, one needs The complete asymptotic at infinity of K λ as negative powers of r. Lemma There exists a sequence of numbers γλ,λ′ so that K λ(X) ≃ sλ(X) +

λ′∈Λ

γλ,λ′R−λ′ sin λ′θ, R → ∞ Then ελK λx ε

≃ sλ(x) +
λ′∈Λ

ελ+λ′γλ,λ′r −λ′ sin λ′θ, |x| ≥ εR0

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 1: Self-similar perturbations of a 2D corner

Example 1: Finite expansions

A Choose m ∈ N. Using the profiles K λ uε = u[m]

0 (x) +

λ≤m

ελu[m]

λ (x) +

λ+λ′≤m

cλ,λ′ελ+λ′K λx ε

+ O(εm)

The terms u[m]

λ

are more or less regular: u[m] ∈ Hm (in fact in K m

−m). This

is a moving asymptotics. But it allows to capture the transformation of singularities through the profiles K λ B Choose m ∈ N. Using the corner layers Y λ uε = u0(x) +

λ≤m

ελuλ(x) +

λ+λ′≤m

cλ,λ′ελ+λ′Y λx ε

+ O(εm)

In certain configurations (essentially when the perturbation consists of small holes dug in Ω), it is possible to prove that the series converge. Here the assumption that f ≡ 0 close to 0 is important. When this assumption is relaxed to analyticity, we can only prove conditional convergence. The conditions that Ω ⊂ Γ and P ⊂ Γ can be relaxed, with the help of cut-offs: χ0 localizes near 0 and multiply rapid terms, whereas χ∞ localizes near infinity and χ∞ x

ε

multiply slow terms.

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Profiles

Denote sλ = r λ cos λθ, for λ ∈ Λ = (N∗ − 1

2) π ω , the singularities of the mixed

Neumann-Dirichlet problem. They satisfies on the sector Γ      ∆sλ = 0 in Γ ∂nsλ = 0 if θ = 0, sλ = 0 if θ = ω. Denote by S λ(Γ) the space of homogeneous functions of degree λ on Γ. Lemma There exists a unique solution K λ in the sector Γ to the problem          ∆K λ = 0 in Γ ∂nK λ = 0 if θ = 0, ∂nK λ + K λ = 0 if θ = ω, K λ = sλ + O(Rλ) as R → ∞ that has the form K λ = sλ +

L

ℓ=1

sλ

ℓ + O(R

1 2 ),

with sλ

ℓ ∈ S λ−ℓ(Γ),

L = [λ + 1

2].

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Profiles

Proof. Denote sλ

0 := sλ. For ℓ = 1, . . . , there exists unique sλ ℓ ∈ S λ−ℓ(Γ)

such that      ∆sλ

ℓ

= 0 in Γ ∂nsλ

ℓ

= 0 if θ = 0, sλ

ℓ

= −∂nsλ

ℓ−1

if θ = ω, The existence as homogeneous solutions and the uniqueness is due to the assumption π

ω ∈ Q that implies that λ − ℓ does not hit the Mellin spectrum

Λ = (Z∗ − 1

2) π ω of the mixed Neumann-Dirichlet problem. Introduce the sum

X λ = χ∞ L

ℓ=0 sλ ℓ

with a radial smooth cut-off, χ∞ ≡ 1 for |x| ≥ 2 and χ∞ ≡ 0 for |x| ≤ 1 Then,

n the Robin side θ = ω and by construction

∂nX λ + X λ = χ∞ ∂nsλ

L

The trace ∂nsλ

L is homogeneous of degree λ − L − 1 that is < − 1

2. Thus

χ∞ ∂nsλ

L belongs to L2(R+). We enter the variational space V

· · · /· · ·

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Profiles

The variational space for the mixed Neumann-Robin problem on the sector Γ      ∆U = F in Γ ∂nU = 0 if θ = 0, ∂nU + U = G if θ = ω, is V =

U ∈ L2

loc(Γ),

∇U ∈ L2(Γ), U

θ=ω ∈ L2(R+)
and the variational formulation is

U ∈ V, ∀V ∈ V,

Γ

∇U · ∇V dX +

θ=ω

U V dR =

Γ

F V dx +

θ=ω

G V dR Thus we find Y λ ∈ V solution of      ∆Y λ = 0 in Γ ∂nY λ = 0 if θ = 0, ∂nY λ + Y λ = −X λ if θ = ω, and our profile is finally K λ = X λ + Y λ = χ∞

sλ +

L

ℓ=1 sλ ℓ

+ Y λ,

with sλ

ℓ ∈ S λ−ℓ, Y λ ∈ V

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Full expansion at infinity

The constructive approach can be carried on, but will never account for the whole asymptotics at infinity, because part of it is hiding in a variat. solution. Need for an analytic tool to find this asymptotics: The Mellin transform. Since the mixed Robin-Neumann is not an homogeneous operator, this is a non-classical setting. Let us analyze Y λ. Set u = χ∞ Y λ and g = −χ2

∞ ∂nsλ L, so that

     ∆u = 0 in Γ ∂nu = 0 if θ = 0, ∂nu + u = g if θ = ω, Introduce for µ ∈ C and the Mellin transform of u and g: U[µ] = ∞ r −µ u(r, ·) dr r , G[µ] = ∞ r −µ g(r) dr r U[µ] is holomorphic for Re µ > 0 G[µ] is meromorphic with a unique pole of order 1 at µ0 = λ − [λ + 1

2] − 1

µ0 < − 1

2

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Full expansion at infinity

The mixed Neumann-Robin becomes after Mellin transform (∗)      (∂2

θ + µ2)U[µ] = 0

in (0, ω) ∂θU[µ] = 0 if θ = 0, U[µ] = −∂θU[µ + 1] + G[µ] if θ = ω, The Mellin symbol A[µ] : H2(0, ω) → L2(0, ω) × R × R of the mixed NR is the

perator on the left of (∗). The poles of A[µ]−1 are simple and situated on

Λ = (Z∗ − 1

2) π ω . Re-write (∗) as

U[µ] = A[µ]−1 0, 0, G[µ] − ∂θU[µ + 1](ω)

Since U[µ] is holomorphic for Re µ ≥ 0, the rhs above is meromorphic

for Re µ > −1. Hence meromorphic extension of U[µ] for Re µ > −1. Can be continued step by step up to Re µ → −∞. The poles are µ0 − ℓ (∀ℓ ∈ N) and λ′ − ℓ (∀λ′ ∈ Λ ∩ R−, ∀ℓ ∈ N)

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References Example 1 Example 2 Example 3 Example 4 Profiles Example 2: Neumann-Robin

Example 2: Full expansion at infinity of profiles

Recall that the profile is K λ and the corner layer is Y λ. They are associated with the singularity sλ = r λ cos λθ, for λ ∈ Λ ∩ R+ Lemma There exist homogeneous functions sλ

λ′,ℓ ∈ S λ′−ℓ so that

K λ ≃ sλ +

ℓ≥1

sλ

ℓ +

λ′∈Λ, λ<0
ℓ≥0

sλ

λ′,ℓ

r → ∞ Y λ ≃

ℓ>L

sλ

ℓ +

λ′∈Λ, λ<0
ℓ≥0

sλ

λ′,ℓ

r → ∞ The material is ready for the description of the solutions uε and u0, along the lines of Example 1. The analysis is more difficult due the singular perturbational character of the operator ε∂n + I

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References Example 1 Example 2 Example 3 Example 4 Profiles Conclusion

Conclusion

What about examples 3 and 4? The new difficulty there is the presence of a further anisotropic scale (thin layer for Example 3 and boundary layer – skin effect – for Example 4). This was solved 10 years ago for Example 3, and is still under investigation for Example 4.

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References Example 1 Example 2 Example 3 Example 4 Profiles Conclusion

Conclusion

What about examples 3 and 4? The new difficulty there is the presence of a further anisotropic scale (thin layer for Example 3 and boundary layer – skin effect – for Example 4). This was solved 10 years ago for Example 3, and is still under investigation for Example 4.

Thank you for your attention

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