Diameter constraint in Shape Optimization

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Diameter constraint in Shape Optimization

Antoine Henrot,Jimmy Lamboley, YannickPrivat

University Paris Dauphine,CEREMADE

September 2013

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Outline

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Introduction and Examples

Outline

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Main question

We consider the Shape Optimization problem minn

J(Ω), Ω∈ S_ad, Diam(Ω) =αo

whereSad is a class of admissibleconvexdomains inR², andJ is shape functional (rotation and translation invariant).

Existence of a solution ?

Geometrical descriptionof the solution.

Numerical computation of the solution.

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Examples

Isodiametrical inequality :

|Ω|

Diam(Ω)^N ≤|B1| 2^N . Spectral Gap theorem :

λ2(Ω)−λ1(Ω)≥ 3π² Diam(Ω)²

B. Andrews, J. Clutterbuck,Proof of the fundamental gap conjecture, J. Amer. Math. Soc.24(2011), 899–916.

Another problem involving eigenvalues : minnp

λ1(∆,Ω)−λ1(∆_∞,Ω), Ωconvex, Diam(Ω) =αo

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Trivia

minn

J(Ω), Ωconvex, Diam(Ω) =αo

IfJ is monotone increasing for inclusion, then solutions are segments.

Examples : J=| · |,J= Per,J =−λk.

IfJ is monotone decreasing for inclusion, then solutions are of constant width αand we have

minn

J(Ω), Diam(Ω) =αo

= minn

J(Ω), Ωconvex of constant widthαo Examples : J=−| · |, J=−Per,J =λk.

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Trivia

minn

IfJ is monotone increasing for inclusion, then solutions are segments.

Examples : J=| · |,J= Per,J =−λk.

IfJ is monotone decreasing for inclusion, then solutions are of constant width αand we have

minn

J(Ω), Diam(Ω) =αo

= minn

J(Ω), Ωconvex of constant widthαo Examples :J =−| · |, J=−Per,J=λk.

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Examples

Generalization of theGap functional: what shape realizes the minimum in : infn

γ[−λ1(Ω)] +λ2(Ω), Ωopen and convex,Diam(Ω) =αo whereγ >0.

Somereverse isoperimetricinequality : what shape realizes the minimum in : minn

γ|Ω| −Per(Ω), Ωconvex,Diam(Ω) =αo .

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General results

Outline

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General results

Parametrization of convex domain with its support function

Figure:a convex domainC

Definition : hC :θ∈T7→max

c∈C

c,

cosθ sinθ

. Geometrical interpretation : dis- tance between the support line or- thogonal tod = (cosθ,sinθ) and the origin.

Examples Segment Σ ={0} ×[−1,1] :hΣ(θ) =|sinθ|

RectangleR with corners (±a,±b) :h_R(θ) =a|cosθ|+b|sinθ|

Ellipse centered at 0 and semi-axesa,b:h_E(θ) =p

a²cos²θ+b²sin²θ.

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General results

Second order arguments

Reformulation

The problem is now of the form min

j(h), h⁰⁰+h≥0, max

θ∈T

{h(θ) +h(θ+π)}=α

.

wherej(hΩ) :=J(Ω).

Example γ|Ω| −P(Ω) = γ

2 Z

T

(h²_Ω−h⁰²_Ω)− Z

T

h_Ω.

Remark: there is some concavity forj.

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General results

Second order arguments

Early work on convexity constraint : Theorem

minn

j(h), h⁰⁰+h≥0o Theorem (L., Novruzi, Pierre, 2011)

Let h be a minimizer. We assume j smooth andlocally concavein the sense that, j⁰⁰(h)(v,v)<0,for any v having a small enough support.

Then h⁰⁰+h is a sum of Dirac massesinside the constraint.

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General results

Second order arguments

Early work on convexity constraint : Examples

Reverse isoperimetry in an annulus : minn

γ|Ω| −P(Ω), Ωconvex, Da⊂Ω⊂Db

o .

J. Lamboley, A. NovruziPolygon as optimal shapes with convexity constraint, SIAM Control Optim.48, no. 5, (2009), 3003–3025

C. Bianchini, A. Henrot,Optimal sets for a class of minimization problems with convex constraints, J. Convex Anal.19(2012), no. 2.

Mahler problem in 2d :

min{|Ω||Ω^◦|, Ωconvex} is achieved for Ω = [−1,1]²

E. Harrell, A. Henrot, J. Lamboley,Analysis of the Mahler volume, Preprint

Faber-Krahn versus reverse isoperimetry : minn

λ1(Ω)−P(Ω), Ωconvex ⊂D, |Ω|=V0

o , the boundary∂Ω^∗∩D is polygonal (difficulty: estimateλ⁰⁰₁(Ω))

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General results

Second order arguments

Application to diameter constraint

minn

, j(h_Ω) =J(Ω) Theorem (Henrot, L., Privat, 2013)

LetΩbe a minimizer. We assume j smooth andlocally concave.

Then any connected component of

∂Ω\ {x∈∂Ω,x belongs to a diameter of Ω}

is made of a finite number of segments.

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General results

First order argument

minn

, j(h_Ω) =J(Ω) Theorem (Henrot, L., Privat, 2013)

LetΩbe a minimizer. We assume j is of integral form j(h) =

Z

T

G(h,h⁰), and

∂1G(α,0)>0 or ∂1G(α,0)> α∂22G(α,0) ThenΩsaturates the diameter constraint at a finite number of points.

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Application toγ| · | −Per(·)

Outline

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Reverse isoperimetric estimate

Forγ >0, we seek to minimize

J_γ(Ω) =γ|Ω| −Per(Ω).

among the sets Ω convex such that Diam(Ω) = 1.

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Results

j(h) =γ Z

T

(h²−h⁰²)− Z

T

h.

Theorem

Letγ >0andΩγ be a solution. Then

every point of∂Ω_γ is either diametrical or belongs to a segment of ∂Ω_γ. Ifγ >1/2, thenΩγ is a polygon with diametrical corners.

The segment is a solution if and only ifγ≥ ^√⁴

3.

Ifγ≤ ¹₂, then the Reuleaux triangle is the unique minimizer.

Proof :The first two statements : application of the previous section. . .

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Large values of γ

We start with an under-statement : forγ >4 then the segment is the unique solution.

We fix a diameter [AB] chosen as an axis, and the upper part of the shape is a graph of a concave functionu

(x,u(x)),x ∈

−1 2,1

2

.

We consider the perturbationTt: (x,y)7→(x,(1−t)y) fort≥0 (affinity).

We write the optimality condition fort 7→J(T_t(Ω_γ)) minimized in [0,1] by t= 0, this leads to

Z ¹₂

−¹₂

f⁰²(x)

p1 +f⁰²(x)dx≥γ Z ¹₂

−¹₂

f(x)dx.

We prove that for everyf ∈H₀¹(−¹₂,¹₂) concave, Z ¹₂ f⁰²(x) Z ¹₂

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Large values of γ

,→ The previous statement is not optimal.

With more efforts, we can actually prove Proposition

Ifγ≥ ^√⁴₃, then :

γ|Ω| −Per(Ω)≥ −2, for every Ω convex of diameter 1.

This is no longer true forγ < ^√⁴

3.

In other words, the segment is solution if and only ifγ≥ ^√⁴

3.

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Proof (optimality of the segment Σ)

Step 1 : shape reformulation

Σ is solution if and only if γ≥γ^∗:= sup

P(Ω)−P(Σ)

|Ω| , Ω convex,Diam(Ω) = 1

Step 2 : 1-sided shape reformulation

We denote Σ = [AB] withA= (−1/2,0),B= (1/2,0), andH=R×R+. Then,

γ^∗= sup

P_H(Ω)−1

|Ω| , Ω convex,Σ⊂Ω⊂T

,

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Proof (optimality of the segment Σ)

Step 3 : Angle reformulation in the class of polygons

We optimize the numberpand the anglesθ₀,θ₁, . . .,θ_p−1at the left and similarly at the right.

,→We end up withp≤1.

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Proof (optimality of the segment Σ)

Step 4 : optimization in the class of quadrilateral

We optimize the position ofLandR.

,→The solutions areL=R=D (equilateral triangle) or

L=A,R=B]

(segment)

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Small values of γ

γ= 0. SoJ(Ω) =−Per(Ω)≥ −πand any set of constant width is optimal.

Ifγ≤ ¹₂,

with a non-local perturbation, we prove that Ωγ has no segment in its boundary.

Using the first point in our Theorem, it implies Ωγ is of constant width.

We conclude with the Blaschke-Lebesgue Theorem.

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Transition value γ ∈ (

¹₂

,

^√⁴

3

), incomplete

Let Ωγ be an optimal shape (which is a polygon).

Let [AB] be a diameter.Then eitherAorB is diametrically opposed to at least two points.

Proof :order two argument.

Situations which remains to be excluded :

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Perspectives

Complete the description forγ∈(¹₂,^√⁴

3).

Replace| · |byλ1. Replace Per byλ1. Dimension 3 !