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` e IRMAR ` e R´ egularit´ e et singularit´ es en optimisation de forme et fronti` eres libres
GDR “Applications nouvelles de l’optimisation de forme" 21-23 octobre 2004, Ker Lann
Influence of singularities in rounded corners (Les coins ronds)
Monique DAUGE Adaptation libre d’un article avec Gabriel CALOZ, Martin COSTABEL et Gr´ egory VIAL .
Institut de Recherche MAth´ ematique de Rennes http://perso.univ-rennes1.fr/Monique.Dauge
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♣ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eLaplace operator in polygonal domains
1 Let Ω be a polygonal domain in R2 . Let u be solution of the Dirichlet problem
∆u = f in Ω
and
u = 0 on ∂Ω.
At each of the corners c of Ω , the solution u has singular parts:
- If f is smooth and flat enough, and if the sides of Ω are straight
u =
K
ak rkπ/ω
c
sin kπ ω θc + O(rKπ/ω
c
), rc → 0.
Here: (rc, θc) polar coord. centered in c and ω = ωc opening of Ω at c . The coefficients ak depend on f .
- If the sides of Ω are curved around c and exponents kπ
ω are not integers:
u =
K
ak
Lk
rℓ+kπ/ω
c
ϕk,ℓ(θc) + O(rKπ/ω
c
), rc → 0.
Here the angular functions ϕk,ℓ(θc) depend on the curved sides.
kπ ω
are integers logarithmic terms may appear, except in the situation of a curved crack, despite the fact that kπ
ω = k 2 can be integer!
The above splittings can be realized between Sobolev spaces...
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` e IRMAR
♥ Polygons ♣ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eRounded corners
2 The same Dirichlet problem has no singularities if Ω is smooth... Thus if the corners of the polygon are “rounded" by small arcs of circles. Let us define the radius of these circles as a small parameter ε , and index the domain accordingly: Ωε . The limiting polygon is denoted by Ω0 . Example of “the" corner of a circular sector of opening 270 ◦ , rounded with two values of ε .
Ωε, ε =
1 10
Ωε, ε =
1 20
Let uε be the solution of the Dirichlet pb in Ωε and u0 the solution in Ω0 . In principle, we expect that uε → u0 in energy as ε → 0 . How are the singularities of u0 hiding inside uε ?
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` e IRMAR
♥ Polygons ♥ Rounded corn. ♣ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eRounded corners... from inside or outside?
3 Example of “the" corner of a circular sector of opening 270 ◦ , rounded by two different procedures. From inside ( C1 Bezier curve).
Ωε ⊂ Ω0 .
In blue Ω0 \ Ωε . From outside (exterior arc of circle).
Ω0 ⊂ Ωε .
In red Ωε \ Ω0 .
Ωε Ωε
For convex angles (the exterior of the sector) the situation is reversed. What is the meaning of a convergence of uε towards u0 ? We try to give answers with the help of multi-scale expansions.
SLIDE 5 ` e IRMAR
♥ Polygons ♥ Rounded corn. ♥ Convergence? ♣ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eStarting the expansion
4 Fix the rhs f . Decompose (“deconstruct") u0 :
u0 = χ(r)
ω S π ω + a 2π ω S 2π ω + · · ·
u0,N = O(rN).
Here χ ≡ 1 near the corner c and χ ≡ 0 outside a region where Ω0 coincides with a plane sector of opening ω . The function S
kπ ω
is the k -th singularity. Near c , the domains Ωε are self-similar. By a dilatation of ratio ε−1 and making ε → 0 (“blow-up") we obtain an infinite domain Q which coincides with a infinite sector of opening ω when R → ∞ and reproduces the pattern
Ωε, ε =
1 10
Q R → ∞
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♣ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eThe fundamental result on profiles
5 Let R be the distance to the origin in Q . Relations between slow ( r ) and rapid ( R ) variables through the singularities
R = r ε
and
Sλ(r) = ελSλr ε
The singularities Sλ(r) can be viewed as profiles at infinity Sλ(R) , with the singularity exponents λ ∈ { π
ω, 2 π ω, 3 π ω, · · · }
Theorem 1. For λ ∈ { π
ω, 2 π ω, 3 π ω, · · · } , there exists a solution Kλ ∈ H1 loc(Q) to:
∆Kλ = 0
in
Q, Kλ = 0
∂Q, Kλ − Sλ = O(Rλ) R → ∞.
Moreover, there exist homogeneous functions Kλ,−µ of degree −µ ,
µ ∈ { π
ω, 2 π ω, 3 π ω, · · · } such that for all M > 0
Kλ − Sλ =
ω , 2 π ω , ··· , j π ω ≤ M
Kλ,−µ + O(R−M), R → ∞.
Proof: (i) cut-off by ψ , (ii) variational formulation on Q , (iii) Mellin transform.
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♣ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eThe substitution trick
6 Recall: χ ≡ 1 near the corner, with support where Ω0 coincides with a sector
u0 = χ(r)
aλ Sλ + u0,N, u0,N = O(rN).
Here ΛN =
π
ω, 2 π ω, · · · , j π ω ≤ N
Cut-off ψ : ψ ≡ 1 near infinity, with support where Q coincides with a sector. By convention, ψ
x
ε
- is defined on Ωε and is ≡ 1 outside a ball B(c, κε) .
Ansatz for uε
uε =
aλ ελ χ(r)Kλr ε
r ε
ε,
Theorem 2. The remainder ρ1
ε solves
∆ρ1
ε =
ελ+µf λ,µ(x) + O(εN),
in
Ωε, ρ1
ε = 0,
∂Ωε.
Consequence:
ρ1
ε = O(ε
2π ω ) .
Proof: Develop ∆
ε
ε
- using Theorem 1 and ∆Sλ = 0 .
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♣ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eA multiscale expansion for uε
7 We fix N > 0 . The solution uε can be split into pieces according to
uε = ψ r ε
ελψ r ε
+
ελ aλ +
aλ,µ εµ χ(r)Kλr ε
as
ε → 0.
The aλ,µ are coefficients. The functions uλ,N−λ are O(rN−λ) as r → 0 . Support of ψ
x
ε
SLIDE 9 ` e IRMAR
♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♣ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eEnergy (H1 ) estimates and H2 estimates
8 The three pieces (bleu-blanc-rouge) of the H1 norm of “uε − u0 " The worst term (provided a π
ω = 0 ).
The strength in ε -power.
u01,Ω0\Ωε uε − u01,Ωε∩Ω0 uε1,Ωε\Ω0 S
π ω (x)
ε
π ω χ(r)
π ω
r ε
π ω
r ε
π ω K π ω
r ε
π ω )
O(ε
π ω )
O(ε
π ω )
We have a C1 dependence in ε if and only if the “rounded" corners are convex. The limit u0 belongs to H2(Ω0) iff π
ω ≥ 1 .
The H2 -norm of uε is equivalent (as ε → 0 ) to ε
π ω χ(r)K π ω x
ε
ε
π ω χ(r)K π ω x
ε
π ω −1εmin{0,1− π ω } =
1
if
π ω ≥ 1
ε
π ω −1
if
π ω < 1.
What happens if the self-similar perturbation at a corner contains itself one or more corners?
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♣ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eZig-zag corners
9 The perturbed domain for two values of ε , the limit domain, the profile domain.
Ωε, ε =
1 10
Ωε, ε =
1 20
Ω0 Q R → ∞
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♣ Cracked ♥ Generalizations ♥ Coefficients ♥ Conclusion
` eCracked corners
10 The perturbed domain for two values of ε , the limit domain, the profile domain.
Ωε, ε =
1 10
Ωε, ε =
1 20
Ω0 Q R → ∞
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♣ Generalizations ♥ Coefficients ♥ Conclusion
` eGeneralizations
11 Let Ω0 be a polygonal domain. Generalization of the previous results holds in any situation where
Ωε → Ω0 with self-similar patterns
which means that for each corner c of Ω0 , there exists a domain Qc which coincides with the infinite sector of opening ωc outside a ball B(0, κc) so that
Ωε ∩ B(c, εκc) = εQc
and Ωε coincides with Ω0 outside the union of the balls B(c, εκc) . The regularity of Q has no influence on the convergence in H1 -norm. A perturbation of the boundary of Ω0 can be done around a fake corner (a regular point). The above results hold with ω = π . For smooth non-flat right hand sides, new profiles Lm
α have to be introduced
for any α = (α1, α2) with m = |α| + 2 :
∆Lm
α = xα
in
Q
and
Lm
α = 0
∂Q, (∗) Lm
α − Tm α = O(Rm)
as
R → ∞,
where Tm
α is the solution of degree m of pb (∗) posed on the plane sector.
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♣ Coefficients ♥ Conclusion
` eCoefficients of singularities
12 To each corner d of Q corresponds a corner dε of Ωε . Let ν be the opening of d and (Rd, Θd) the polar coordinates at d in Q . The polar coordinates at dε in Ωε are
ρε := ε Rd
and
ϑ := Θd.
The solution uε has a (first) singularity
bε ρ
π ν
ε sin π
ν ϑ.
What is the behavior of bε as ε → 0 ? The first profile K
π ω has a singularity in d :
K
π ω (R, Θ) = γR π ν
d sin π
ν Θd.
We recall that
uε = a1ε
π ω K π ω
r ε
Therefore
bε = a1γ ε
π ω ε− π ν
Example: ω = 3π
2 , ν = 2π . Hence: bε = O(ε
1 6 ) .
SLIDE 14 ` e IRMAR
♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♣ Conclusion
` eExercises: Find coefficient asymptotics at crack tips
13 EX 1. Trouver l’asymptotique au fond des fissures.
Ωε, ε =
1 10
Ωε, ε =
1 20
ε
ε
ε
ε
EX 2. Consid´
erer successivement les conditions de Dirichlet et de Neumann. Ωε, ε =
1 10
Ωε, ε =
1 20
d−
ε
ε
d−
ε
ε
Drawn with fig4tex
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♥ Polygons ♥ Rounded corn. ♥ Convergence? ♥ Starting ♥ Profiles ♥ Substitution ♥ Multiscale ♥ Estimates ♥ Zig-zag ♥ Cracked ♥ Generalizations ♥ Coefficients ♣ Conclusion
` ePost-conclusion
14 Thin layers and impedance boundary conditions
Ωε Q R → ∞ Ω0 Q R → ∞
