Juan Casado-Díaz

University of Sevilla

Index

Statemen of the problem

Relaxation via homogenization

Discretization

The gradient method

The case of dimension one

Control problem

CP

min𝑀∇ℳ(Ω)

𝒥(𝑢, 𝑀∆𝑢)

−div 𝑀(𝑥)∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

with Ω ⊂ ℝ𝑁 bounded, open, 𝑓 ∇ 𝐻−1 Ω ,

ℳ Ω a set of measurable matrices functions in

Ω, uniformly bounded and elliptic. It is the set of controls.

Remark. The PDE considered appears for example in electricity

or heat conduction. The matrix depends on the material chosen.

Some interesting choices of ℳ Ω 𝑎𝑟𝑒

ℳ Ω = 𝐴𝑖𝜒𝜔 𝑖 :

𝑚

𝑖=1

𝜔𝑖 ∩ 𝜔𝑗 = ∅, Ω = 𝜔𝑖 ,

m

i=1

|𝜔𝑖 | ≤ 𝜇𝑖

ℳ Ω = 𝑃𝑡 𝐴𝑖𝜒𝜔 𝑖

𝑚

𝑖=1

𝑃: 𝜔𝑖 ∩ 𝜔𝑗 = ∅, Ω = 𝜔𝑖 ,

m

i=1

𝜔𝑖 ≤ 𝜇𝑖 , 𝑃 𝑥 orthogonal matrix

𝐴𝑖 symmetric positive, 𝜇𝑖≥0, 𝜇𝑖

𝑚

𝑖=1

≥|Ω|,

More generally that 𝒥 𝑢, 𝑀∆𝑢 , we take 𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚

Existence of solution: Direct method of the CV

We consider 𝜔𝑛𝑖 (and 𝑃𝑛 ) such that 𝜔𝑛

𝑖 ≤ 𝜇𝑖 , such that

𝑀𝑛 = 𝑃𝑛𝑡 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

𝑃𝑛 or 𝑀𝑛 = 𝐴𝑖𝜒𝜔𝑛𝑖 ,

𝑚

𝑖=1

we have

𝒥(𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 𝜒𝜔𝑛1 , ⋯ , 𝜒𝜔𝑛

𝑚 ) → inf (CP)

with 𝑢𝑛 the solution of

−div 𝑀𝑛(𝑥)∆𝑢𝑛 = 𝑓 in Ω

𝑢𝑛 = 0 on 𝜕Ω,

We know that (for a subsequence)

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω

𝜒𝜔𝑛𝑖 ⇀

∗휃𝑖 in 𝐿∞ Ω ,

0≤휃𝑖 ≤ 1, 휃𝑖

𝑚

𝑖=1

= 1, 휃𝑖𝑑𝑥 ≤ 𝜇𝑖

Ω

𝑃𝑛 ⇀∗

𝑃 in 𝐿∞ Ω 𝑁×𝑁 (𝑛𝑜𝑡 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑖𝑙𝑦 𝑜𝑟𝑡𝑕𝑜𝑔𝑜𝑛𝑎𝑙),

𝑀𝑛∆𝑢𝑛 = 𝑃𝑛𝑡 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

𝑃𝑛∆𝑢𝑛 ⇀ 𝜎 𝑖𝑛 𝐿2 Ω 𝑁

So, −div 𝑀𝑛∆𝑢𝑛 = 𝑓 in Ω ⇒ −div 𝜎 = 𝑓 in Ω

If there exist 𝜔𝑖 ⊂ Ω, measurable, with 휃𝑖 = 𝜒𝜔 𝑖 , P is

orthogonal

𝜎 = lim 𝑃𝑛𝑡 𝐴𝑖

𝑚

𝑖=1

𝜒𝜔 𝑖 𝑃𝑛∆𝑢𝑛 = 𝑃𝑡 𝐴𝑖

𝑚

𝑖=1

휃𝑖 𝑃∆𝑢

and 𝒥 satisfies the semicontinuity property

liminf 𝒥 𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 𝜔𝑛1 , ⋯ , 𝜔𝑛

𝑚 ≥ 𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚 .

Then the sets 𝜔𝑖 the matrix function 𝑃 and the function 𝑢

give a solution of the control problem.

However in general

∄𝜔𝑖 such that 휃𝑖 = 𝜒𝜔 𝑖 , 𝑃 is not orthogonal,

and due to the fact that the product of weak limits is not in

general the weak limit of the product, we have

𝜎 = lim𝑃𝑛𝑡 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

𝑃𝑛∆𝑢𝑛 ≠ 𝑃𝑡 𝐴𝑖휃𝑖

𝑚

𝑖=1

𝑃∆𝑢

F. Murat (1974) proved there is not solution in general

Theorem (S. Spagnolo 1969,

non symmetric case F. Murat, L. Tartar 1974)

Assume 𝐿𝑛 symmetric, 𝛼 ≤ 𝐿𝑛 ≤ 𝛽 a.e. in Ω,

∃𝐿 symmetric 𝛼 ≤ 𝐿 ≤ 𝛽 a.e. in 𝛺 such that (for a subsequence)

If 𝑣𝑛 ⇀ 𝑣 in 𝐻1 Ω

−div 𝐿𝑛(𝑥)∆𝑣𝑛 ⟶ 𝑔 in 𝐻−1(Ω)

Then 𝐿𝑛∆𝑣𝑛 ⇀ 𝐿∆𝑣 𝑖𝑛 𝐿2(Ω)𝑁 .

In particular, 𝑣 satisfies − div 𝐿(𝑥)∆𝑣 = 𝑔 in Ω

We say that 𝐿𝑛 H-converges to 𝐿.

Key ingredient div-curl lemma by F. Murat, L. Tartar

𝑢𝑛 ⇀ 𝑢 in 𝐻1 Ω

𝜎𝑛 ⇀ 𝜎 in 𝐿2 Ω 𝑁

div 𝜎𝑛 ⇀ div 𝜎 in 𝐻−1 Ω

⟹ 𝜎𝑛 ∙ ∆𝑢𝑛 ⇀ 𝜎𝑛 ∙ ∆𝑢 in 𝒟′ Ω .

Returning to the direct method of the calculus of variations

We had

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω

𝑀𝑛∆𝑢𝑛 = 𝑃𝑛𝑡 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

𝑃𝑛∆𝑢𝑛 ⇀ 𝜎 𝑖𝑛 𝐿2 Ω 𝑁

Then 𝜎 = 𝑀∆𝑢 in Ω, for some matrix function 𝑀.

What can we say about 𝑀?

Definition. For 𝑝1 , ⋯ , 𝑝𝑚 ∇ 0,1 , 𝑝𝑖𝑚𝑖=1 = 1, we denote by

𝒦(𝑝1 , ⋯ , 𝑝𝑚 ) the set of constant matrices 𝑀 such that

∃𝜔𝑛1 , ⋯ , 𝜔𝑛

𝑚 , 𝜔𝑛𝑖 ∩ 𝜔𝑛

𝑗= ∅, Ω = 𝜔𝑛

𝑖 ,

𝑚

𝑖=1

(𝑎𝑛𝑑 𝑃𝑛 ∇ 𝐿∞ Ω 𝑁×𝑁)

𝜒𝜔𝑛𝑖 ⇀

∗𝑝𝑖 in 𝐿∞ Ω

𝐴𝑖𝜒𝜔𝑛𝑖

𝑚

𝑖=1

𝐻→ 𝑀 𝑜𝑟 𝑃𝑛

𝑡 𝐴𝑖𝜒𝜔𝑛𝑖

𝑚

𝑖=1

𝑃𝑛

𝐻→ 𝑀

The definition does no depend on Ω.

Theorem. 휃1 , … , 휃𝑚 and 𝑀 are functions such that

∃𝜔𝑛1 , ⋯ , 𝜔𝑛

𝑚 , 𝜔𝑛𝑖 ∩ 𝜔𝑛

𝑗= ∅, Ω = 𝜔𝑛

𝑖 ,

𝑚

𝑖=1

(𝑎𝑛𝑑 𝑃𝑛 ∇ 𝐿∞ Ω 𝑁×𝑁)

𝜒𝜔𝑛𝑖 ⇀

∗휃𝑖 in 𝐿∞ Ω

𝐴𝑖𝜒𝜔𝑛𝑖

𝑚

𝑖=1

𝐻→ 𝑀 𝑜𝑟 𝑃𝑛

𝑡 𝐴𝑖𝜒𝜔𝑛𝑖

𝑚

𝑖=1

𝑃𝑛

𝐻→ 𝑀

If and only if 𝑀(𝑥) ∇ 𝒦(휃1(𝑥), ⋯ , 휃𝑚 (𝑥)) a.e. in Ω.

Theorem (F. Murat, L. Tartar, 1985)

Assume that 𝒥 satisfies

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω

𝜒𝜔𝑛𝑖 ⇀

∗휃𝑖 in 𝐿∞ Ω ∗ -weak,

𝑀𝑛∆𝑢𝑛 = 𝑃𝑛𝑡 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

𝑃𝑛∆𝑢𝑛 ⇀ 𝜎 𝑖𝑛 𝐿2 Ω 𝑁

−div 𝑀𝑛∆𝑢𝑛 → −div𝜎 𝑖𝑛 𝐻−1 Ω .

implies

∃ lim𝑛

𝒥 𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 𝜔𝑛1 , ⋯ , 𝜔𝑛

𝑚 = 𝒥 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 .

Then a relaxation of (CP) is given by

RCP

min 𝒥 𝑢, 𝑀∆𝑢, 휃1 , ⋯ , 휃𝑚

휃1 , ⋯ , 휃𝑚 , 0 ≤ 휃𝑖 ≤ 1, 휃𝑖

𝑚

𝑖=1

= 1, 휃𝑖𝑑𝑥Ω

≤ 𝜇𝑖

𝑀 ∇ 𝒦 휃1 , ⋯ , 휃𝑚 a.e. in Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

Remark. Assume that

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω , 휃𝑛

𝑖 ⇀ 휃𝑖 in 𝐿∞ Ω ∗ -weak,

𝑀𝑛 ∇ 𝒦 휃𝑛1 , . . , 휃𝑛

𝑛 , 𝑀𝑛∆𝑢𝑛 ⇀ 𝜎 𝑖𝑛 𝐿2 Ω 𝑁

−div 𝑀𝑛∆𝑢𝑛 → −div𝜎 𝑖𝑛 𝐻−1 Ω .

implies

liminf𝑛

𝒥 𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 𝜔𝑛1 , ⋯ , 𝜔𝑛

𝑚 ≥ 𝒥 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 .

Then problem

min 𝒥 𝑢, 𝑀∆𝑢, 휃1 , ⋯ , 휃𝑚

휃1 , ⋯ , 휃𝑚 , 0 ≤ 휃𝑖 ≤ 1, 휃𝑖

𝑚

𝑖=1

= 1, 휃𝑖𝑑𝑥Ω

≤ 𝜇𝑖

𝑀 ∇ 𝒦 휃1 , ⋯ , 휃𝑚 a.e. in Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

has a solution but its mínimum can be strictly smaller than the

infimum of (CP)

The assumptions made on the functional 𝒥 hold for example

For

𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚

= 𝑔𝑖 𝑥, 𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

+ 𝑕 𝑥, 𝑢 ∆𝑢𝑑𝑥

Ω

+ 𝑘 𝑥, 𝑢 𝑀∆𝑢𝑑𝑥

Ω

+ 𝑙 𝑥, 𝑢 𝑀∆𝑢 ∙ ∆𝑢𝑑𝑥.

Ω

Remark: The set 𝒦 휃1, ⋯ , 휃𝑚 is only known in very few

cases

𝑚 = 2 and 𝐴1 , 𝐴2 are scalar matrices:

-Tartar,

-Lurie, Cherkaev

or in the case of rotations, when 𝑚 = 1,2, 𝑁 = 2

-Francfort ,Tartar, Murat

Some interesting examples in H-convergence:

One-dimensional case: 𝑀𝑛 ∇ 𝐿∞ 𝑎, 𝑏 , 0 < 𝛼 ≤ 𝑀𝑛 ≤ 𝛽

Define 𝑀 by 𝑀𝑛−1 ⇀

∗𝑀−1 𝑖𝑛 𝐿∞ 𝑎, 𝑏

Then 𝑀𝑛 ⇀ 𝐻

𝑀

Assume 𝑢𝑛 ⇀ 𝑢 in 𝐻01 𝑎, 𝑏 , 𝑓𝑛 ⟶ 𝑓 in 𝐻−1 𝑎, 𝑏

−𝑑

𝑑𝑥 𝑀𝑛

𝑑𝑢𝑛

𝑑𝑥 = 𝑓𝑛 in 𝑎, 𝑏 .

Then −𝑑

𝑑𝑥 𝑀

𝑑𝑢

𝑑𝑥 = 𝑓 in 𝑎, 𝑏 .

and (corrector result)

𝑑𝑢𝑛

𝑑𝑥−

𝑀

𝑀𝑛

𝑑𝑢

𝑑𝑥→ 0 𝑖𝑛 𝐿2 𝑎, 𝑏

Corollary: In the one-dimensional case

𝒦 𝑝1 , …𝑝𝑚 = 𝑝𝑖

𝐴𝑖

𝑚

𝑖=1

−1

.

It is the harmonic mean of 𝐴1 , … , 𝐴𝑚 , with proportions 𝑝1 , … , 𝑝𝑚 .

Example: We define 𝑢 as the solution of

−4

3

𝑑2𝑢

𝑑𝑥2= 1 in 0,1

𝑢 0 = 𝑢 1 = 0

i.e.

𝑢(𝑥) =3(𝑥 − 𝑥2)

8.

We consider the control problem (CP)

inf |𝑢 − 𝑢 |2𝑑𝑥1

0

−

𝑑

𝑑𝑥 𝜒𝜔 + 2𝜒(0,1)\𝜔

𝑑𝑢

𝑑𝑥 = 1 in (0,1)

𝑢 0 = 𝑢(1) = 0

𝜔 ≤1

2.

Taking 𝜔𝑛 = 2𝑘

2𝑛,2𝑘 + 1

2𝑛 ,

𝑛−1

𝑘=0

the corresponding solution 𝑢𝑛 converges weakly in 𝐻01(0,1)

and then strongly in 𝐿2(0,1). Therefore

inf CP = 0

Therefore, if problem (CP) has a solution, then ∃𝜔 ⊂ (0,1) with

𝜔 ≤ 1/2 such that 𝑢 satisfies

−𝑑

𝑑𝑥 𝜒𝜔 + 2𝜒 0,1 \𝜔

𝑑𝑢

𝑑𝑥 = 1 in 0,1 .

Thus, ∃𝐶 ∇ ℝ, with

𝜒𝜔 + 2𝜒 0,1 \𝜔 𝑑𝑢

𝑑𝑥=

4

3

𝑑𝑢

𝑑𝑥+ C

⇒𝑑𝑢

𝑑𝑥 𝑥 =

−3𝐶 if 𝑥 ∇ 𝜔3

2𝐶 if 𝑥 ∈ 𝜔,

which is contradiction with the expression of 𝑢.

Periodic homogenization

Consider N

s

NYLMY M; ,1,0 #

.

We denote 𝑀휀 𝑥 = 𝑀 𝑥

휀 . It defines a material constructed

by repeating periodically a small structure. It is very frequent in

engineering.

Theorem. We have

Y

NH

N

s

HH

dywMM

Mx

M

with , where

M

−

𝑗

−𝑑𝑖𝑣 𝑀 𝜉 + ∆𝑤𝜉 = 0 in ℝ𝑁 , 𝑤𝜉 ∇ 𝐻#1 𝑌 .

Remark. 𝑤𝜉 = 𝜉𝑖𝑤𝑒𝑖

𝑁𝑖=1

Corrector result. −div 𝑀 𝑥

휀 ∆𝑢𝑛 = 𝑓𝑛 in Ω.

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω , 𝑓𝑛 ⟶ 𝑓 in 𝐻−1 Ω , 𝑢 smooth

∆𝑢𝑛 𝑥 − ∆𝑢 𝑥 − 𝜕𝑖𝑢 𝑥 ∆𝑦𝑤𝑒𝑖 𝑥

휀

𝑁

𝑖=1→ 0 in 𝐿2 Ω 𝑁

A formal proof.

We look for an expansion of 𝑢휀 of the type

𝑢휀 𝑥 = 𝑢0 𝑥,𝑥

휀 + 휀𝑢1 𝑥,

𝑥

휀 + 휀2𝑢2 𝑥,

𝑥

휀 + ⋯

where the functions 𝑢𝑖(𝑥, 𝑦) are periodic in 𝑦.

Replacing this expression in

−div 𝑀 𝑥

휀 ∆𝑢휀 = 𝑓 in Ω, 𝑢휀 = 0 on 𝜕Ω,

we obtain the equation for the limit 𝑢0 𝑥,𝑥

휀 = 𝑢0 𝑥 of 𝑢휀 and the

equation for the corrector 𝑢1.

It can be justified using two-scale convergence (Allaire, Nguetseng)

Other proof: Use the general proof by Tartar.

Particular case: 𝑀 = 𝑀(𝑦1). Then, 𝑤𝜉=𝑤𝜉 (𝑦1) solution of

−𝑑

𝑑𝑦1 𝑀𝑒1 ∙ 𝜉 + 𝑀𝑒1 ∙ 𝑒1

𝑑𝑤𝜉

𝑑𝑦1 = 0

This contains the laminations

Theorem (Dal Maso, Kohn) The matrices constructed via

periodic homogenization are dense in the matrices

constructed using general homogenization. Namely

Take 𝑀𝑛 ⇀𝐻

𝑀. Define 𝑀𝑛𝛿 by

𝑀𝑛𝛿 (𝑥)𝜉 = 𝑀𝑛(𝑥 + 𝛿𝑦) 𝜉 + ∆y𝑤𝜉

𝑛(𝑥, 𝑦) 𝑑𝑦𝑌

with 𝑤𝜉 (𝑥, 𝑦) periodic in y

−𝑑𝑖𝑣y 𝑀𝑛(𝑥 + 𝛿𝑦) 𝜉 + ∆y𝑤𝜉𝑛 = 0 in ℝ𝑁 , a.e. 𝑥 ∇ Ω

Then

𝑀𝑛𝛿 → 𝑀 𝑎. 𝑒.

when 𝑛 → ∞ and then 𝛿 → 0.

Corrector result. (C-D, Couce-Calvo, Martín-Gómez)

−div 𝑀𝑛∆𝑢𝑛 = 𝑓𝑛 in Ω.

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω , 𝑓𝑛 ⟶ 𝑓 in 𝐻−1 Ω ,

limδ

lim𝑛

∆𝑢𝑛 𝑥 + 𝛿𝑦 − ∆𝑢 𝑥

− 𝜕𝑖𝑢 𝑥 ∆𝑦𝑤𝑒𝑖

𝛿 ,𝑛 𝑥

휀

𝑁

𝑖=1

𝐿2 𝛺𝜌 ×𝑌 𝑁

= 0, ∀𝜌 > 0,

with 𝛺𝜌 = 𝑥 ∇ 𝛺: dist 𝑥, 𝜕𝛺 > 𝜌 .

Corollary: The matrices 𝑀 defined as

𝑀𝜉 = 𝑃𝑡(𝑦) 𝐴𝑖

𝑚

𝑖=1

𝜒𝑌𝑖 𝑃(𝑦) 𝜉 + ∆y𝑤𝜉 (𝑦) 𝑑𝑦𝑌

With 𝑤𝜉 periodic

−divy 𝑀(𝑥 + 𝛿𝑦) 𝜉 + ∆y𝑤𝜉 = 0 in ℝ𝑁 , a.e. 𝑥 ∇ Ω

for 𝑌𝑖 ⊂ 𝑌, 𝑌𝑖 = 𝑝𝑖 , 𝑖 = 1, . . 𝑚,

𝑃 a rotation matrix a.e. in 𝑌,

are dense in 𝒦(𝑝1 , … , 𝑝𝑚 ).

A similar result holds in the case without rotations

Remark: In order to deal with (RCP) we only need to know

the set

𝒦 휃1 , ⋯ , 휃𝑚 𝜉, ∀𝜉 ∇ ℝ𝑁 .

Indeed (RCP) can be written as

RCP2

min 𝒥 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚

0 ≤ 휃𝑖 , 휃𝑖

𝑚

𝑖=1

= 1, 휃𝑖𝑑𝑥Ω

≤ 𝜇𝑖

−div𝜎 = 𝑓 in Ω 𝑢 = 0 on 𝜕Ω,

𝜎 ∇ 𝒦 휃1 , ⋯ , 휃𝑚 ∆𝑢 a.e. in Ω.

Thus, the control problem does not provide an optimal

matrix 𝑀, but only 𝑀∆𝑢.

The set 𝒦 𝑝1 , ⋯ , 𝑝𝑚 𝜉, with 𝜉 ∇ ℝ𝑁 , is known in the following

cases:

If 𝑁 = 1, 𝒦 𝑝1 , ⋯ , 𝑝𝑚 = 𝑝𝑖

𝐴𝑖

𝑚

𝑖=1

−1

.

Tartar: Case with rotations, 𝑁 ≥ 2, if 𝜆𝑖 and Λi are the smaller

and bigger eigenvalues of 𝐴𝑖 and denote

𝜆 𝑝 = 𝑝𝑖

𝜆𝑖

𝑚

𝑖=1

−1

, Λ 𝑝 = 𝑝𝑖Λi

𝑚

𝑖=1

,

then

𝒦 𝑝1 , ⋯ , 𝑝𝑚 𝜉 = 𝐵 Λ 𝑝 + 𝜆 𝑝

2𝜉,

Λ 𝑝 − 𝜆 𝑝

2 𝜉 .

C-D, Couce-Calvo, Martín-Gómez

In the case without rotations, 𝑁 ≥ 2, 𝑚 = 2

𝒦 𝑝1 , 𝑝2 𝜉 is the set of 휂 ∇ ℝ𝑁 satisfying

𝐴2 − 𝐴1 −1 𝐴1

𝑝1+

𝐴2

𝑝2 𝐴2 − 𝐴1 −1 𝐴𝑝𝜉 − 휂 ∙ 𝐴𝑝𝜉 − 휂

≤ 𝐴𝑝𝜉 − 휂 ∙ 𝜉,

where

𝐴𝑝 = 𝑝1𝐴1 + 𝑝2𝐴2

In these two cases where 𝒦 𝑝1 , 𝑝2 𝜉 the elements of 𝜕𝒦 𝑝1 , 𝑝2 𝜉 are

obtained using a lamination and it is essentially the unique way to obtain

them.

Let us now deal with more general functionals, they are of the

form

𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚 = 𝑔𝑖 𝑥, 𝑢, ∆𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

(in the case without rotations) or

𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚 = 𝑔𝑖 𝑥, 𝑢, ∆𝑢, 𝑃𝑡𝐴𝑖𝑃∆𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

in the case with rotations

where 𝑔𝑖 𝑥, 𝑠, ξ, 휂 ≤ 𝐶 1 + 𝑠 2 + ξ 2 + 휂 2

Previous relaxation results.

Tartar: 𝑚 = 2, 𝐴1 , 𝐴2 scalar matrices

𝒥 𝑢 = ∆ 𝑢 − 𝑢0 2𝑑𝑥

Ω

For 𝑢0 ∇ 𝐷 dense of 𝐻01 Ω a relaxation is obtained

reemplacing M(Ω) by 𝐴1휃 + 𝐴2 1 − 휃 : 0 ≤ 휃 ≤ 1

Grabovski. Bellido, Pedregal. 𝑢0 arbitrary.

Maestre, Munch, Pedregal, wave equation.

Munch, Pedregal, Periago, heat equation.

Allaire, Gutiérrez. 𝐴1 , 𝐴2 non scalar, but 𝐴2 − 𝐴1

small, not rotation, general functional.

Difficulty: To pass to the limit in (similar with rotations)

lim𝑛

𝑔𝑖 𝑥, 𝑢𝑛 , ∆𝑢𝑛 𝑑𝑥

𝜔𝑛𝑖

𝑚

𝑖=1

where

−div 𝐴𝑖𝜒𝜔𝑛

𝑖

𝑚

𝑖=1

∆𝑢𝑛 = 𝑓 in Ω

𝑢𝑛 = 0 on 𝜕Ω,

Remark. If the sets 𝜔𝑛𝑖 = 𝜔휀

𝑖 have a periodic structure, i.e.

𝜔휀𝑖 = 휀 𝑌𝑖 + 𝑘

𝑘∇ℤ𝑛

, 𝑌𝑖 ⊂ 𝑌,

we know

∆𝑢휀 𝑥 ~∆𝑢 𝑥 + 𝜕𝑖𝑢 𝑥 ∆𝑦𝑤𝑒𝑖 𝑥

휀

𝑁

𝑖=1

Then

lim휀

𝑔𝑖 𝑥, 𝑢휀 , ∆𝑢휀 𝑑𝑥

𝜔휀𝑖

𝑚

𝑖=1

= lim휀

𝑔𝑖 𝑥, 𝑢휀 , ∆𝑢 + 𝜕𝑖𝑢∆𝑦𝑤𝑒𝑖 𝑥

휀

𝑁

𝑖=1

𝜒𝑌𝑖 𝑥

휀 𝑑𝑥

Ω

𝑚

𝑖=1

= 𝑔𝑖 𝑥, 𝑢(𝑥), ∆𝑢(𝑥) + 𝜕𝑖𝑢(𝑥)∆𝑦𝑤𝑒𝑖 𝑦

𝑁

𝑖=1

𝑌𝑖

𝑑𝑦𝑑𝑥

Ω

𝑚

𝑖=1

JCD, J. Couce-Calvo, J.D. Martín-Gómez (2008),

𝐴1 , 𝐴2 arbitrary, not rotation, general functional.

Definition: We write

𝑢𝑛 , 𝜎𝑛 , 휃𝑛1 , ⋯ , 휃𝑛

𝑚 →𝜏

𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚

if

휃𝑛𝑖 →

∗휃𝑖 in 𝐿∞ Ω

𝑢𝑛 ⇀ 𝑢 in 𝐻01 Ω

𝜎𝑛 ⇀ 𝜎 in 𝐿2 Ω 𝑁

div 𝜎𝑛 → div 𝜎 in 𝐻−1 Ω .

We define

𝒡 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 = 𝑔𝑖 𝑥, 𝑢, ∆𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

𝒡 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 = 𝑔𝑖 𝑥, 𝑢, ∆𝑢, 𝑃𝑡𝐴𝑖𝑃∆𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

휃𝑖 = 𝜒𝜔 𝑖 , 𝜎 = 𝐴𝑖𝜒𝜔 𝑖

𝑚

𝑖=1

∆𝑢 𝜎 = 𝑃𝑡 𝐴𝑖𝜒𝜔 𝑖

𝑚

𝑖=1

𝑃∆𝑢

𝒡 휃1 , ⋯ , 휃𝑚 , 𝑢, 𝜎 = +∞ in another case.

Theorem (C-D, Couce-Calvo, Martín-Gómez)

∃𝐻: Ω × ℝ × ℝ𝑁 × ℝ𝑁 × ℝ𝑚 ↦ ℝ ∪ +∞

such that the lower semicontinuous envelope 𝒡 of 𝒡 with

respect to the 𝜏-convergence is given by

𝒡 𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 = 𝐻(𝑥, 𝑢,

Ω

∆𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 )𝑑𝑥

The function 𝐻 can be defined as (case without rotations)

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 = lim𝛿

𝐻𝛿 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚

𝐻𝛿 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 = inf 𝑔𝑖 𝑥, 𝑠, ∆𝑤𝜉 + 𝜉

𝑌𝑖

𝑑𝑦

𝑚

𝑖=1

𝑌1 , ⋯ , 𝑌𝑚 ⊂ 𝑌, 𝑌𝑖 ∩ 𝑌𝑗 = ∅, 𝑌𝑖 = 𝑝𝑖

−div 𝐴𝑖𝜒𝑌𝑖

𝑚

𝑖=1

(∆𝑤𝜉 + 𝜉) = 0 in ℝ𝑁

𝑤𝜉 periodic

𝐴𝑖𝜒𝑌𝑖

𝑚

𝑖=1

(∆𝑤𝜉 + 𝜉) 𝑑𝑦 −

𝑌

휂 < 𝛿

Theorem (JCD, J. Couce-Calvo, J.D. Martín-Gómez)

A relaxation of (CP) is given by

RCP

min 𝐻(𝑥, 𝑢,

Ω

∆𝑢, 𝜎, 휃1 , ⋯ , 휃𝑚 )𝑑𝑥

휃1 , ⋯ , 휃𝑚 , 0 ≤ 휃𝑖 ≤ 1, 휃𝑖

𝑚

𝑖=1

= 1, 휃𝑖𝑑𝑥Ω

≤ 𝜇𝑖

𝑀 ∇ 𝒦 휃1, ⋯ , 휃𝑚 a.e. in Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

If 𝑁 = 1

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 = 𝑝𝑖𝑔𝑖 𝑥, 𝑠,휂

𝐴𝑖

𝑚

𝑖=1

𝑖𝑓 휂 = 𝑝𝑖

𝐴𝑖

𝑚

𝑖=1

−1

𝜉,

+∞ in another case

If 𝑁 > 1, the function 𝐻 is not known in general.

We have the following properties

Dom 𝐻 = 𝑥, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 : 휂 ∇ 𝒦 𝑝1 , ⋯ , 𝑝𝑚 𝜉 .

In its domain 𝐻 is measurable with respect to 𝑥 and continuos with

respect to the other variables.

𝐻 𝑥, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 ≤ 𝐶 1 + 𝑠 2 + ξ 2 + 휂 2

𝐻 satisfies the following convexity property

𝐻 𝑥, 𝛾𝜉1 + 1 − 𝛾 𝜉2 , 𝛾휂1 + 1 − 𝛾 휂2 ,

𝛾𝑝11 + 1 − 𝛾 𝑝2

1 , ⋯ , 𝛾𝑝1𝑚 + 1 − 𝛾 𝑝2

𝑚

≤ 𝛾 𝐻 𝑥, 𝜉1 , 휂1 , 𝑝11 , ⋯ , 𝑝1

𝑚 + 1 − 𝛾 𝐻 𝑥, 𝜉2 , 휂2 , 𝑝21 , ⋯ , 𝑝2

𝑚

if 𝛾 ∇ 0,1 , 𝜉2 − 𝜉1 ∙ 휂2 − 휂1 = 0.

If we replace the functional 𝐽 by

𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚 + 𝑔𝑖 𝑥, 𝑢 𝑑𝑥

𝜔 𝑖

𝑚

𝑖=1

+ 𝑕 𝑥, 𝑢 ∆𝑢𝑑𝑥

Ω

+ 𝑘 𝑥, 𝑢 𝑀∆𝑢𝑑𝑥

Ω

+ 𝑙 𝑥, 𝑢 𝑀∆𝑢 ∙ ∆𝑢𝑑𝑥.

Ω

Then the function 𝐻 is replaced by

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 + 𝑝𝑖𝑔𝑖 𝑥, 𝑢

𝑚

𝑖=1

+ 𝑕 𝑥, 𝑠 𝜉

+ 𝑘 𝑥, 𝑠 𝑀𝜉 + 𝑙 𝑥, 𝑠 𝑀𝜉 ∙ 𝜉

In the case without rotations and 𝑚 = 2, we have

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , 𝑝2 =

𝑝𝑖𝑔𝑖 𝑥, 𝑠, 𝐴𝑗 − 𝐴𝑖 −1 𝐴𝑗 𝜉 − 휂

𝑝𝑖 𝑗 ≠ 𝑖

2

𝑖=1

in 𝜕Dom 𝐻 .

In the case with rotations

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , ⋯ , 𝑝𝑚 =

𝑝𝑖𝑔𝑖 𝑥, 𝑠, 1 −

𝜆 𝑝 λ𝑖

휂 + 𝜆 𝑝 Λ 𝑝 λ𝑖

− 1 𝜉

Λ 𝑝 − 𝜆 𝑝 , Λi − 𝜆 𝑝 휂 + 𝜆 𝑝 Λ 𝑝 − Λ𝑖 𝜉

Λ 𝑝 − 𝜆 𝑝

𝑚

𝑖=1

in 𝜕Dom 𝐻 .

𝜆 𝑝 = 𝑝𝑖

𝜆𝑖

𝑚

𝑖=1

−1

, Λ 𝑝 = 𝑝𝑖Λi

𝑚

𝑖=1

,

In both cases the mixture giving the value of H is obtained by one lamination.

Examples where 𝐻 can be explicitly calculated,

case without rotations, 𝑚 = 2. We assume

𝒥 𝑢, 𝑀∆𝑢, 𝜔1, ⋯ , 𝜔𝑚 = 𝑕 𝑥, 𝑢 𝐴1∆𝑢 ∙ ∆𝑢 𝑑𝑥

𝜔1

+ 𝐺 𝑥, 𝑢, ∆𝑢 𝑑𝑥,

𝜔2

𝑄 𝑥, 𝑠, 𝜉 = 𝐺 𝑥, 𝑠, 𝜉 − 𝑕 𝑥, 𝑠 𝐴2𝜉 ∙ 𝜉 convex in 𝜉 .

Then

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1, 𝑝2 = 𝑕 𝑥, 𝑠 𝜉 ∙ 휂 + 𝑝2𝑄 𝑥, 𝑠,1

𝑝2 𝐴2 − 𝐴1 −1 휂 − 𝐴1𝜉

This value of 𝐻 is obtained using two laminations.

Corollary: 𝐴1 = 𝛼𝐼, 𝐴2 = 𝛽𝐼, 𝛼 < 𝛽,

𝒥 𝑢, 𝑀∆𝑢, 𝜔1 , ⋯ , 𝜔𝑚 = 𝑟 ∆𝑢 2

Ω

𝑑𝑥

If 𝑟 > 0, 𝑝 ∇ (0,1)

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , 𝑝2 =𝑟

𝛽휂 ∙ 𝜉 +

𝑟

𝑝1 𝛽 − 𝛼 휂 − 𝛽𝜉 2

If 𝑟 > 0, 𝑝 ∇ (0,1)

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , 𝑝2 =𝑟

𝛼휂 ∙ 𝜉 −

𝑟

𝑝2 𝛽 − 𝛼 휂 − 𝛼𝜉 2

Remark: For 0r , this gives the results of

Bellido - Pedregal, Grabovsky.

Case without rotations, 𝑚 = 2,

𝒥 𝑢, 𝑀∆𝑢, 𝜔1, ⋯ , 𝜔𝑚 = 𝐹 𝑥, 𝑢, ∆𝑢 𝑑𝑥

Ω

with 𝐹(𝑥, 𝑠, 𝜉) convex in 𝜉 and such that for every 𝑠 ∇ ℝ and a.e. 𝑥 ∇ Ω

there exists 휁 ∇ ℝ𝑁\ 0 , satisfying that

𝜆 → 𝐹 𝑥, 𝑠, 𝜉 + 𝜆휁 is linear, ∀𝜉 ∇ ℝ𝑁 .

Then 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝1 , 𝑝2 = 𝑝1𝐹 𝑥, 𝑠,1

𝑝1 𝐴2 − 𝐴1 −1 𝐴2𝜉 − 휂

+𝑝2𝐹 𝑥, 𝑠,1

𝑝2 𝐴2 − 𝐴1 −1 휂 − 𝐴1𝜉

This value of 𝐻 is obtained using three laminations.

Particular case: 𝐹 𝑥, 𝑠, 𝜉 = 𝐶 𝑥, 𝑠 𝜉 ∙ 𝜉, where the smaller eigenvalue of

𝐶 𝑥, 𝑠 is zero.

G. Allaire. Shape optimization by the homogenization method. Springer-Verlag 2002.

J.C. Bellido, P. Pedregal. Explicit quasiconvexification for some cost functionals

depending on derivatives of the state in optimal designing. Discr. Contin. Dyn. Syst.

8, 4 (2002), 967-982.

J. Casado-Díaz, J. Couce-Calvo, J.D. Martín-Gómez. Relaxation of a control problem

in the coefficients with a functional of quadratic growth in the gradient. SIAM J.

Control and Optim. 47 (2008), 1428-1459.

Y. Grabovsky. Optimal design for two-phase conducting composites with weakly

discontinuous objective functionals. Adv. Appl. Math. 27 (2001), 683-704.

K.A. Lurie. Applied optimal control theory of distributed systems. Plenum Press, New

York, 1993.

K.A. Lurie, A.V. Cherkaev. Exact estimates of the conductivity of a binary mixture of

isotropic materials. Proc. Royal Soc. Edinburgh 104 A (1986), 21-38.

F. Murat. Un contre-example pour le problème du contrôle dans les coefficients.

C.R.A.S Sci. Paris A 273 (1971), 708-711.

F. Murat, L. Tartar. H-convergence. In Topics in the Mathematical Modelling of

Composite Materials, Birkauser Boston, 1998, 21-43.

S. Spagnolo. Sulla convergenza di soluzioni di equazioni paraboliche

ed ellittiche. Ann. Scuola Norm. Sup. Pisa, 22, 3 (1968), 571-597.

F. Murat, L.Tartar. On the control of coefficients in partial differential

equations. In Topics in the Mathematical Modelling of Composite

Materials. Birkauser, 1998, 1-8.

L. Tartar. Estimations fines de coefficients homogénéisés. In Ennio de

Giorgi colloquium. Pitman, 1985, 168-187.

L. Tartar. Remarks on optimal design problems. In Calculus of

variations, homogenization and continuum mechanics,World

Scientific, 1994, 279-296.

L. Tartar. An introduction to the homogenization method in optimal

design. In Optimal shape design, Springer-Verlag, 2000, 47-156.

Numerical Analysis

To simplify, let us consider the problem 𝛼, 𝛽 > 0

inf 𝐹(𝑥, 𝑢, ∆𝑢)𝑑𝑥

Ω

−div 𝛼𝜒𝜔 + 𝛽𝜒Ω\𝜔 ∆𝑢 = 𝑓 in Ω

𝑢 = 0 𝑜𝑛 𝜕 Ω

𝜔 ≤ 𝜇

We want to solve the relaxed problem

min 𝐻(𝑥, 𝑢,

Ω

∆𝑢, 𝑀∆𝑢, 휃)𝑑𝑥

0 ≤ 휃 ≤ 1, 휃𝑑𝑥Ω

≤ 𝜇, 𝑀 ∇ 𝒦 휃 a.e. in Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

Dificulty. We do not explicitely know the function 𝐻.

A discretization using an upper approximation

Let us consider 𝐻 such that

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 lower semicontinuous, with

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 ≥ 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝

𝐻 𝑥, 𝑠, 𝜉, 𝛼𝜉, 1 = 𝐻 𝑥, 𝑠, 𝜉, 𝛽𝜉, 0 = 𝐹 𝑥, 𝑠, 𝜉 .

For h>0, we consider a triangulation 𝒯𝑕 = 𝑇𝑖 ,𝑕 𝑖=1

𝑛𝑕

, if ,0

diam ,0 ,measurable ,

,,

,,,

1

,

jiTT

hTTTT

hjhi

hihihi

hn

i

hi

and a sequence of closed subspaces consider then we,1

0 HVh

Discretized problem

min 𝐻 (𝑥, 𝑢𝑕 ,

Ω

∆𝑢𝑕 , 𝑀𝑕∆𝑢𝑕 , 휃𝑕)𝑑𝑥

0 ≤ 휃𝑕 ≤ 1, 휃𝑕𝑑𝑥Ω

≤ 𝜇𝑖 , 𝑀𝑕 ∇ 𝒦 휃𝑕 a.e. in Ω

𝑢𝑕 ∇ 𝑉𝑕 , 𝑀𝑕∆𝑢𝑕 ∙ ∆𝑣𝑕𝑑𝑥 = 𝑓𝑣𝑕 𝑑𝑥, ∀

ΩΩ

𝑣𝑕 ∇ 𝑉𝑕

𝑀𝑕 , 휃𝑕 constants in the elements 𝑇𝑖 ,𝑕

Remark. If 𝑉𝑕 = 𝐻01 Ω , ∀𝑕, then we are solving the PDE problem

exactly. To have a true discretization 𝑉𝑕 must be

finite-dimensional. For example a finite dimensional space.

So, in general 𝑉𝑕 is associated to a triangulation which is not

necessarily given by 𝒯𝑕 . Usually it will be a refinement of 𝒯𝑕 .

We assume

i)

lim𝑕→0

min𝑣𝑕∇𝑉𝑕

𝑣 − 𝑣𝑕 𝐻01(Ω) = 0, ∀𝑣 ∇ 𝑉𝑕

ii)

lim𝑕→0

min𝑣𝑕∇𝑉𝑕

𝑤𝑕𝜑 − 𝑣𝑕 𝐻01 Ω = 0,

∀𝑤𝑕 ∇ 𝑉𝑕 bounded in 𝐻01 Ω , ∀𝜑 ∇ 𝐶𝑐

∞ Ω

lim𝑕→0

𝐻 𝑥, 𝑢𝑕 , ∆𝑢𝑕 , 𝜎𝑕 , 휃𝑕 𝑑𝑥

Ω

≥ 𝐻(𝑥, 𝑢, ∆𝑢, 𝜎, 휃)𝑑𝑥

Ω

∀𝑢𝑕 ⇀ 𝑢 in 𝐻01 Ω , 𝑢𝑕 ∇ 𝑉𝑕 , ∀𝜎𝑕 ⇀ 𝜎 in 𝐿2 Ω 𝑁

∀휃𝑕 ⇀∗

휃 in 𝐿∞ Ω , 0 ≤ 휃𝑕 ≤ 1 a.e. in Ω

lim𝑕→0

max𝑣𝑕∇𝑉𝑕

1

𝑣𝑕 𝐻01 Ω

𝜎𝑕 − 𝜎 ∙ ∆𝑣𝑕𝑑𝑥

Ω

= 0.

Properties i), ii), iii) are satisfied for .1

0 HVh

If hV is a usual space of finite elements, it satisfies i), ii).

In the examples where we have an explicit representation of H,

every sequence of closed subspaces hV satisfies iii). For example

𝛼 < 𝛽, 𝐹 𝑥, 𝑢, ∆𝑢 = ∆𝑢 2

and then

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 =1

𝛽휂 ∙ 𝜉 +

1

𝑝 𝛽 − 𝛼 휂 − 𝛽𝜉 2

So we need to prove

lim𝑕→0

1

𝛽𝜎𝑕 ∙ ∆𝑢𝑕 +

1

휃𝑕 𝛽 − 𝛼 𝜎𝑕 − 𝛽∆𝑢𝑕 2 𝑑𝑥

Ω

≥ 1

𝛽𝜎 ∙ ∆𝑢 +

1

휃 𝛽 − 𝛼 𝜎 − 𝛽∆𝑢 2 𝑑𝑥

Ω

It is enough to show

lim𝑕→0

𝜎𝑕 ∙ ∆𝑢𝑕𝑑𝑥

Ω

≥ 𝜎 ∙ ∆𝑢𝑑𝑥

Ω

lim𝑕→0

𝜎𝑕 − 𝛽∆𝑢𝑕 2

휃𝑕𝑑𝑥

Ω

≥ 𝜎 − 𝛽∆𝑢 2

휃𝑑𝑥

Ω

The first one immediately follows from

lim𝑕→0

max𝑣𝑕∇𝑉𝑕

1

𝑣𝑕 𝐻01 Ω

𝜎𝑕 − 𝜎 ∙ ∆𝑣𝑕𝑑𝑥

Ω

= 0.

For the second one use that the application 𝑡, 𝑟 → 𝑡 2

𝑟 is

convex in ℝ × ℝ+.

Remark. Assumption

lim𝑕→0

min𝑣𝑕∇𝑉𝑕

𝑤𝑕𝜑 − 𝑣𝑕 𝐻01 Ω = 0,

∀𝑤𝑕 ∇ 𝑉𝑕 bounded in 𝐻01 Ω , ∀𝜑 ∇ 𝐶𝑐

∞ Ω

allows us to prove a discrete version of the div-curl lemma

If 𝑢𝑕 ⇀ 𝑢 in 𝐻01 Ω , 𝑢𝑕 ∇ 𝑉𝑕 , 𝜎𝑕 ⇀ 𝜎 in 𝐿2 Ω 𝑁

lim𝑕→0

min𝑣𝑕∇𝑉𝑕

1

𝑣𝑕 𝐻01 Ω

𝜎𝑕 − 𝜎 ∙ ∆𝑣𝑕𝑑𝑥

Ω

= 0.

Then 𝜎𝑕 ∙ ∆𝑢𝑕 ⇀ 𝜎 ∙ ∆𝑢 in 𝒟′ Ω .

Theorem: The discrete problem has a solution hhh Mu ,,

Up to a subsequence

𝑢𝑕 ⇀ 𝑢 in 𝐻01 Ω

𝑀𝑕∆𝑢𝑕 ⇀ 𝑀∆𝑢 in 𝐿2 Ω 𝑁

휃𝑕 ⇀∗

휃 in 𝐿∞ Ω

.,,,,,,,,lim

, a.e. )( ,10

on 0 ,in div

,,,,inf

ofsolution ) ,(

0

dxuMuuxHdxuMuuxH

dxKM

ufuM

dxuMuuxH

u,M

hhhhhh

Example 1:

otherwise. ,,

),,(0,,1,,

pH

sxFBHAH

In this case, we are solving a discrete version of the original

(unrelaxed) problem, i.e.

inf 𝐹(𝑥, 𝑢𝑕 , ∆𝑢𝑕)𝑑𝑥

Ω

𝑢𝑕 ∇ 𝑉𝑕

𝛼𝜒𝜔𝑕+ 𝛽𝜒Ω\𝜔𝑕

∆𝑢𝑕

Ω

∙ ∆𝑣𝑕𝑑𝑥 = 𝑓𝑣𝑕

Ω

𝑑𝑥, ∀𝑣𝑕 ∇ 𝑉𝑕

𝜔𝑕 a union of elements of 𝒯𝑕 , 𝜔𝑕 ≤ 𝜇

Remark: Contrarily to the continuous problem, it always has a

solution. This is due to the fact that the set of controls is contained

in a finite-dimensional space.

It provides a physical control and not a relaxed one.

If the solution of the relaxed problem is a finite mixture of the

materials 𝛼 and 𝛽 the convergence can be very slowly.

The construction of algorithms is difficult. The controls are

characteristic functions which only take the value one or zero.

We cannot use step methods.

Example 2: (𝑁 ≥ 2) Since we know the values of 𝐻 in the

boundary of its domain, we can take

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝

= 𝑝𝐹 𝑥, 𝑠,𝛽𝜉 − 휂

𝑝 𝛽 − 𝛼 + (1 − 𝑝)𝐹 𝑥, 𝑠,

휂 − 𝛼𝜉

(1 − 𝑝) 𝛽 − 𝛼

if 휂 ∇ 𝜕𝒦(𝑝)𝜉

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 = +∞ elsewhere.

Recall

휂 ∇ 𝜕𝒦 𝑝 𝜉 ⟺ 휂 − 𝜆 𝑝 𝜉 ∙ 휂 − Λ 𝑝 𝜉 = 0

𝜆 𝑝 = 𝑝

𝛼+

1 − 𝑝

𝛽

−1

, Λ 𝑝 = 𝛼𝑝 + 𝛽(1 − 𝑝)

Remark: In this case we are dealing with the discretization of a

partially relaxed control problem.

The controls obtained are not physical controls in general. However,

we know that these controls can be obtained by using one lamination.

Therefore, we can give simple methods to obtain a physical control

from the relaxed one.

We hope that the convergence is quickly because we are able to work

directly with some fine mixtures.

The set of controls is a differentiable manifold and thus we ca use

more effective algorithms to compute a solution of the discrete

problem.

Clearly, when we know the function 𝐻 we can just take

𝐻 = 𝐻

We hope to obtain the faster convergence.

A lower approximation of 𝐻

Here 𝑁 ≥ 2.

We consider 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 lower semicontinuous, with

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 ≤ 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝

For h>0, we consider a triangulation 𝒯𝑕 = 𝑇𝑖 ,𝑕 𝑖=1

𝑛𝑕 as above

and closed subspaces 𝑉𝑕 ⊂ 𝐻01 Ω satisfying properties i) and ii)

as above and

lim𝑕→0

𝐻 𝑥, 𝑢𝑕 , ∆𝑢𝑕 , 𝜎𝑕 , 휃𝑕 𝑑𝑥

Ω

≥ 𝐻(𝑥, 𝑢𝑕 , ∆𝑢𝑕 , 𝜎𝑕 , 휃𝑕)𝑑𝑥

Ω

∀𝑢𝑕 ⇀ 𝑢 in 𝐻01 Ω , 𝑢𝑕 ∇ 𝑉𝑕 , ∀𝜎𝑕 ⇀ 𝜎 in 𝐿2 Ω 𝑁

∀휃𝑕 ⇀∗

휃 in 𝐿∞ Ω , 0 ≤ 휃𝑕 ≤ 1 a.e. in Ω

lim𝑕→0

min𝑣𝑕∇𝑉𝑕

1

𝑣𝑕 𝐻01 Ω

𝜎𝑕 − 𝜎 ∙ ∆𝑣𝑕𝑑𝑥

Ω

= 0.

We denote

ℭ = co 𝑀, 𝑝 : 𝑀 ∇ 𝒦(𝑝)

Discretized problem

min 𝐻(𝑥, 𝑢𝑕 ,

Ω

∆𝑢𝑕 , 𝑀𝑕∆𝑢𝑕 , 휃𝑕)𝑑𝑥

0 ≤ 휃𝑕 ≤ 1, 휃𝑕𝑑𝑥Ω

≤ 𝜇𝑖 , 휃𝑕 , 𝑀𝑕 ∇ ℭ a.e. in Ω

𝑢𝑕 ∇ 𝑉𝑕 , 𝑀𝑕∆𝑢𝑕 ∙ ∆𝑣𝑕𝑑𝑥 = 𝑓𝑣𝑕 𝑑𝑥, ∀

ΩΩ

𝑣𝑕 ∇ 𝑉𝑕

𝑀𝑕 , 휃𝑕 constants in the elements 𝑇𝑖 ,𝑕

Remark: For every 𝜉 ∇ ℝ𝑁 , the set

휂, 𝑝 ∇ ℝ𝑁 × 0,1 : 휂 ∇ 𝒦(𝑝)𝜉 .

Therefore, if 휃, 𝑀 ∇ ℭ a.e. in Ω and 𝑢 ∇ 𝐻01 Ω there

exists 𝑀 ∇ 𝒦(휃) such that 𝑀 ∆𝑢 = 𝑀∆𝑢.

So, in the discretized problem

(𝑥, 𝑢𝑕 , ∆𝑢𝑕 , 𝑀𝑕∆𝑢𝑕 , 휃𝑕)

belongs to the domain of 𝐻.

Recall that the control problem only provides 𝑀∆𝑢.

Theorem: The discrete problem has a solution hhh Mu ,,

Up to a subsequence

𝑢𝑕 ⇀ 𝑢 𝑖𝑛 𝐻01 Ω

𝑀𝑕∆𝑢𝑕 ⇀ 𝑀∆𝑢 𝑖𝑛 𝐿2 Ω 𝑁

휃𝑕 ⇀∗

휃 𝑖𝑛 𝐿∞ Ω

dxuMuuxHdxuMuuxH

dxKM

ufuM

dxuMuuxH

Mu

hhhhhh

,,,,,,,,lim

, a.e. )( ,10

on 0 ,in div

,,,,inf

ofsolution ) ,,(

0

Remark: Since 𝐻 ≤ 𝐻, it is clear that the functions

(𝑢, 𝑀, 휃) given by the theorem, satisfy

𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 𝑑𝑥 ≤

Ω

min(RCP).

If

𝐼 𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 = 𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 a.e.in Ω,

Then (𝑢, 𝑀, 휃) is a solution of (RCP).

Can we give some condition on 𝐻 to assure (I)?

Optimallity conditions

We assume 𝐻 smooth enough and we consider a solution

(𝑢 , 𝑀 , 휃 ) of

min 𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 𝑑𝑥

Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

휃𝑑𝑥Ω

≤ 𝜇, (θ,𝑀) ∇ ℭ

We can assume 𝑀 ∇ 𝒦 휃 a.e. in Ω.

Take 𝑀, 휃 , 𝑀 ∇ 𝒦 휃 a.e. in Ω, ∫ 휃Ω

𝑑𝑥 ≤ 𝜇.

For 0 < 휀 < 1, we define (𝑀𝒠 , 휃𝒠) by

𝑀휀 = 𝑀 + 휀 𝑀 − 𝑀 , 휃𝒠 = 휃 + 휀 휃 − 휃

and 𝑢휀 as the solution of

−div 𝑀휀∆𝑢휀 = 𝑓 in Ω

𝑢휀 = 0 on 𝜕Ω.

Then,

𝐻 𝑥, 𝑢 , ∆𝑢 , 𝑀 ∆𝑢 , 휃 𝑑𝑥

Ω

≤ 𝐻 𝑥, 𝑢휀 , ∆𝑢휀 , 𝑀휀∆𝑢휀 , 휃휀 𝑑𝑥.

Ω

𝐻 𝑥, 𝑢휀 , ∆𝑢휀 , 𝑀𝒠∆𝑢휀 , 휃𝒠 − 𝐻 𝑥, 𝑢 , ∆𝑢 , 𝑀 ∆𝑢 , 휃

휀𝑑𝑥

Ω

≥ 0.

Taking the limit when 휀 tends to zero and denoting

⋀ = 𝑥, 𝑢 , ∆𝑢 , 𝑀 ∆𝑢 , 휃 and

𝑢 ′ = lim휀→0

𝑢휀 − 𝑢

휀

we get

𝜕𝑠𝐻 ⋀ 𝑢 ′𝑑𝑥

Ω

+ 𝜕𝜉𝐻 ⋀ ∙ ∆𝑢 ′𝑑𝑥

Ω

𝜕휂𝐻 ⋀ ∙ (𝑀 − 𝑀 )∆𝑢 + 𝑀 ∆𝑢 ′ 𝑑𝑥

Ω

+ 𝜕𝑝𝐻 ⋀ (휃 − 휃 )𝑑𝑥

Ω

≥ 0.

Derivating in −div 𝑀𝒠∆𝑢휀 = 𝑓 in Ω, 𝑢휀 = 0 on ∂Ω

we deduce that 𝑢 ′ is the solution of

−div (𝑀 − 𝑀 )∆𝑢 + 𝑀 ∆𝑢 ′ = 0 in Ω, 𝑢 ′ = 0 on ∂Ω.

We introduce the adjoin state 𝑝 as the solution of

−div 𝑀 ∆𝑝 = 𝜕𝑠𝐻 ⋀ − div 𝜕𝜉𝐻 ⋀ + 𝑀 𝜕휂𝐻 ⋀ in Ω

𝑝 = 0 on ∂Ω.

Then 𝜕𝑠𝐻 ⋀ 𝑢 ′ + 𝜕𝜉𝐻 ⋀ ∙ ∆𝑢 ′ + 𝜕휂𝐻 ⋀ ∙ 𝑀 ∆𝑢 ′ 𝑑𝑥

Ω

= − 𝑀 − 𝑀 ∆𝑢 ∙ ∆𝑝 𝑑𝑥

Ω

Then, (𝑢 , 𝑀 , 휃 ) satisfies

(𝑀 − 𝑀 )∆𝑢 ∙ 𝜕휂𝐻 ⋀ − ∆𝑝 𝑑𝑥

Ω

+ 𝜕𝑝𝐻 ⋀ (휃 − 휃 )𝑑𝑥

Ω

≥ 0

∀ 𝑀, 휃 ∇ ℭ, 𝑀 ∇ 𝒦 휃 a.e. in Ω, ∫ 휃𝑑𝑥 ≤ 𝜇.Ω

If 휃 = 휃, we can replace 𝑀 by 𝑀𝜒𝜔 + 𝑀 𝜒Ω\𝜔 ∇ 𝒦 휃 a. e. in Ω

with 𝜔 ⊂ Ω measurable. We have

(𝑀 − 𝑀 )∆𝑢 ∙ 𝜕휂𝐻 ⋀ − ∆𝑝 𝑑𝑥

𝜔

≥ 0 ∀𝜔

i.e. (𝑀 − 𝑀 )∆𝑢 ∙ 𝜕휂𝐻 ⋀ − ∆𝑝 ≥ 0 a.e. in 𝜔.

We have then proved

𝑀 ∆𝑢 ∙ 𝜕휂𝐻 ⋀ − ∆𝑝 = min𝑀∇𝒦 휃

𝑀∆𝑢 ∙ 𝜕휂𝐻 ⋀ − ∆𝑝

a.e. in Ω and thus,

𝑀 ∆𝑢 ∇ 𝜕𝒦 휃 ∆𝑢 a.e. in 𝜕휂𝐻 ⋀ − ∆𝑝 ≠ 0 .

Remark: There exist some results showing that in some cases

it is always possible to obtain a solution such that

𝑀 ∆𝑢 ∇ 𝜕𝒦 휃 ∆𝑢 a.e. in Ω

without taking into account the value of 𝜕휂𝐻 ⋀ − ∆𝑝 .

A.e. in 𝜕휂𝐻 ⋀ − ∆𝑝 ≠ 0 the optimal matrix is obtained by

a simple lamination.

Taking in account the last result, we will take 𝐻 such that

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 = 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝

= 𝑝𝐹 𝑥, 𝑠,𝛽𝜉 − 휂

𝑝 𝛽 − 𝛼 + (1 − 𝑝)𝐹 𝑥, 𝑠,

휂 − 𝛼𝜉

(1 − 𝑝) 𝛽 − 𝛼

for 휂 ∇ 𝜕𝒦 𝑝 𝜉.

Theorem. We assume that 𝐻 . , 𝑠, 𝜉, 휂, 𝑝 is Fréchet derivable for

𝑝 ∇ (0,1) and

11

,,,,

p

A

p

BC

pHpH

if the solution (𝑢, 𝑝, 휃) of

min 𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 𝑑𝑥

Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω, 휃𝑑𝑥

Ω

≤ 𝜇, (θ,𝑀) ∇ ℭ

given by the convergence theorem satisfies

𝜕휂𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 − ∆𝑝 ≠ 0 a.e. in Ω

then (𝑢, 𝑝, 휃) is a solution of (RCP).

Example of function 𝐻.

Assume 𝐹(𝑥, 𝑠, 𝜉) convex in 𝜉 and 𝐹(. , 𝑠, 𝜉) smooth.

Then, the function

𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝

= 𝑝𝐹 𝑥, 𝑠,𝛽𝜉 − 휂

𝑝 𝛽 − 𝛼 + (1 − 𝑝)𝐹 𝑥, 𝑠,

휂 − 𝛼𝜉

(1 − 𝑝) 𝛽 − 𝛼

for 𝑥, 𝑠, 𝜉, 휂, 𝑝 ∇ Ω × ℝ × ℝ𝑁 × ℝ𝑁 × [0,1], 휂 ∇ 𝜕𝒦 𝑝 𝜉,

is in the conditions of the above results.

REFERENCES

The most of the results contained in this part are contained in

J. Casado-Díaz, J. Couce-Calvo, M. Luna-Laynez, J.D. Martín-Gómez. Optimal

design problems for a non-linear cost in the gradient: numerical results. Applicable

Anal. 87 (2008), 1461-1487.

See also

G. Allaire. Shape optimization by the homogenization method. Springer-Verlag, 2002.

B. Bourdin. Filters in topology optimization. Int. J. Numer. Meth. Enging. 50 (2001),

2143-2158.

J. Casado-Díaz, J. Couce-Calvo, M. Luna-Laynez, J.D. Martín-Gómez. Discretization

of Coefficient Control Problems with a Nonlinear Cost in the Gradient. In Integral

Methods in Science and Engineering, Vol. 2. Birkauser, 2010, 55-63.

D. Chenais, E. Zuazua. Finite Element Approximation of 2D Elliptic Optimal Design.

J. Math. Pures et Appl., 85 (2006), 225-249.

R. Lipton, A.P. Velo. Optimal design of gradient fields with applications to

electrostatics. In Nonlinear partial differential equations and their applications,

Collège de France Seminar, Vol. XIV. North-Holland, Amsterdam, 2002, 509-532.

Gradient algorithm for the resolution

of the discrete problem

Let us use the previous calculus to give an algorithm to solve

the relaxed control problem

min 𝐻 𝑥, 𝑢, ∆𝑢, 𝑀∆𝑢, 휃 𝑑𝑥

Ω

−div 𝑀∆𝑢 = 𝑓 in Ω

𝑢 = 0 on 𝜕Ω,

휃𝑑𝑥Ω

≤ 𝜇, (θ,𝑀) ∇ ℭ

To simplify the exposition, we present the algoritm applied directly to

the continuous problem and to the function 𝐻 which we assume

smooth.

Assuming an approximation 𝑀𝑛 , 휃𝑛 of a solution, let

us try to construct

𝑀𝑛+1 = 𝑀𝑛 + 휀 𝑀 − 𝑀𝑛 , 휃𝑛+1 = 휃𝑛 + 휀 휃 − 휃𝑛 ,

with 휀 ∇ (0,1) small given, and 𝑀, 휃 , with 𝑀 ∇ 𝒦 휃 , ∫ 휃𝑑𝑥 ≤ 𝜇Ω

,

to choose in such way that

𝐻 𝑥, 𝑢𝑛+1, ∆𝑢𝑛+1, 𝑀𝑛+1∆𝑢𝑛+1, 휃𝑛+1 𝑑𝑥

Ω

≤ 𝐻 𝑥, 𝑢𝑛 , ∆𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 휃𝑛 𝑑𝑥

Ω

The functions 𝑢𝑛 , 𝑢𝑛+1 are the solutions of the corresponding

state equations. We use a first order Taylor expansion to write

𝐻𝑛+1𝑑𝑥

Ω

~ 𝐻𝑛𝑑𝑥

Ω

+휀 𝑀 − 𝑀𝑛 ∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

Ω

+ 𝜕𝑝𝐻𝑛(휃 − 휃𝑛)𝑑𝑥

Ω

,

where we used the index 𝑛 in 𝐻 (idem 𝑛 + 1) to mean that the

functions are valuated in 𝑥, 𝑢𝑛 , ∆𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 휃𝑛 .

The function 𝑝𝑛 is defined by

−div 𝑀𝑛∆𝑝𝑛 = 𝜕𝑠𝐻𝑛 − div 𝜕𝜉𝐻𝑛 + 𝑀𝑛𝜕휂𝐻𝑛 in Ω

𝑝𝑛 = 0 on 𝜕Ω

Let us choose (𝑀, 휃) minimizing

𝑀 − 𝑀𝑛 ∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

Ω

+ 𝜕𝑝𝐻𝑛(휃 − 휃𝑛)𝑑𝑥

Ω

or minimizing 𝑀∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

Ω

+ 𝜕𝑝𝐻𝑛휃𝑑𝑥

Ω

when 𝑀 ∇ 𝒦 휃 , a.e.in Ω, ∫ 휃𝑑𝑥𝜓 ≤ 𝜇Ω

.

Fixing 휃 we deduce that

𝑀∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 = min𝑀 ∇𝒦 휃

𝑀 ∆𝑢 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 .

Taking into account that

휂 ∇ 𝒦 𝑝 𝜉 ⟺ 휂 − 𝜆 𝑝 𝜉 ∙ 휂 − Λ 𝑝 𝜉 ≤ 0

𝜆 𝑝 = 𝑝

𝛼+

1 − 𝑝

𝛽

−1

, Λ 𝑝 = 𝛼𝑝 + 𝛽(1 − 𝑝)

We have that

𝑀∆𝑢𝑛 =Λ 휃 + 𝜆 휃

2∆𝑢𝑛

−Λ 휃 − 𝜆 휃

2

∆𝑢𝑛

𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝜕휂𝐻𝑛 − ∆𝑝𝑛

a.e. where ∆𝑢𝑛, 𝜕휂𝐻𝑛 − ∆𝑝𝑛 are not zero.

Inserting this expression in

𝑀∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

Ω

+ 𝜕𝑝𝐻𝑛휃𝑑𝑥

Ω

,

We deduce that 휃 is a minimum of

Λ 휃 + 𝜆 휃

2∆𝑢𝑛

Ω

∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

− Λ 휃 − 𝜆 휃

2 ∆𝑢𝑛

Ω

𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥 + 𝜕𝑝𝐻𝑛휃𝑑𝑥.

Ω

when ∫ 휃𝑑𝑥 ≤ 𝜇Ω

, 0≤ 휃 ≤ 1.

This is a convex problem. The Kuhn Tucker theorem asserts that it

is equivalent to find 𝑟 ≥ 0, with

𝑟 휃𝑑𝑥

Ω

− 𝜇 = 0

such that 휃 is a minimum of

Λ 휃 + 𝜆 휃

2∆𝑢𝑛

Ω

∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥

− Λ 휃 − 𝜆 휃

2 ∆𝑢𝑛

Ω

𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥 + 𝜕𝑝𝐻𝑛 + 𝑟 휃𝑑𝑥.

Ω

when 0≤ 휃 ≤ 1.

Thus, a.e. in Ω, 휃 = 휃𝑟 is a minimum of

Φ𝑟 휃 =Λ 휃 + 𝜆 휃

2∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛

−Λ 휃 − 𝜆 휃

2 ∆𝑢𝑛 𝜕휂𝐻𝑛 − ∆𝑝𝑛 + 𝜕𝑝𝐻𝑛 + 𝑟 휃.

when 0≤ 휃 ≤ 1.

For 𝑟1 ≤ 𝑟2, we have

Φ𝑟1 휃𝑟1

≤ Φ𝑟1 휃𝑟2

= Φ𝑟2 휃𝑟2

− 𝑟2 − 𝑟1 휃𝑟2

≤ Φ𝑟2 휃𝑟1

− 𝑟2 − 𝑟1 휃𝑟2= Φ𝑟1

휃𝑟1 − 𝑟2 − 𝑟1 휃𝑟2

− 휃𝑟1

i.e. 휃𝑟2≤ 휃𝑟1

a.e. in Ω.

Since we are looking for 𝑟 ≥ 0 such that

휃𝑟𝑑𝑥

Ω

≤ 𝜇 and 𝑟 휃𝑟𝑑𝑥

Ω

− 𝜇 = 0.

we can choose

𝑟 = 0 if 휃0𝑑𝑥

Ω

≤ 𝜇

𝑟 = min 𝑠 ≥ 0: 휃𝑠𝑑𝑥

Ω

= 𝜇 if not

The above rule gives us a procedure to compute 휃, 𝑀 .

Then, we take

휃𝑛+1 = 휃𝑛 + 휀 휃 − 휃𝑛 , 𝑀𝑛+1 = 𝑀𝑛 + 휀 𝑀 − 𝑀𝑛

for some 휀 ∇ (0,1]. How can we choose 휀?

Assume 𝒞 a bounded closed convex set of a Banach space and

𝐹: 𝒞 → ℝ such that there exists

𝐹′′ 𝑥 𝑦 − 𝑥, 𝑦 − 𝑥 , ∀𝑥, 𝑦 ∇ 𝒞

with 𝐹′′ 𝑥 𝑦 − 𝑥, 𝑦 − 𝑥 ≤ 𝐾 𝑦 − 𝑥 2.

We want to solve the problem

min𝒞

𝐹

As above, we consider a descent method to compute the minimum.

Given 𝑥𝑛 let us take 𝑥𝑛+1 = 𝑥𝑛 + 휀(𝑥 − 𝑥𝑛) as above,

with 𝑥 minimizing

𝐹 𝑥𝑛 + 휀𝐹′ 𝑥𝑛 (𝑥 − 𝑥𝑛)

when 𝑥 ∇ 𝒞, i.e. (the minimum exists if for example the space is

reflexive)

𝐹′ 𝑥𝑛 𝑥 = min𝑥∇𝒞

𝐹′ 𝑥𝑛 𝑥

A second order Taylor expansion gives for some 𝑥 ∇ (𝑥𝑛 , 𝑥𝑛+1),

𝐹 𝑥𝑛+1 = 𝐹 𝑥𝑛 + 휀𝐹′ 𝑥𝑛 𝑥 − 𝑥𝑛 +휀2

2𝐹′ 𝑥 𝑥 − 𝑥𝑛 , 𝑥 − 𝑥𝑛

≤ 𝐹 𝑥𝑛 + 휀𝐹′ 𝑥𝑛 𝑥 − 𝑥𝑛 +휀2

2𝐾 𝑥 − 𝑥𝑛 2

Thus, we choose 휀 minimizing

휀𝐹′ 𝑥𝑛 𝑥 − 𝑥𝑛 +휀2

2𝐾 𝑥 − 𝑥𝑛 2,

in 0,1 , i.e.

휀 = min 𝐹′ 𝑥𝑛 𝑥𝑛 − 𝑥

𝐾 𝑥 − 𝑥𝑛 2, 1 ,

which gives

𝐹 𝑥𝑛+1 ≤ 𝐹 𝑥𝑛 − 𝐹′ 𝑥𝑛 𝑥𝑛 − 𝑥 2

2𝐾 𝑥 − 𝑥𝑛 2 if 휀 < 1 .

𝐹 𝑥𝑛+1 ≤ 𝐹 𝑥𝑛 −𝐾 𝑥 − 𝑥𝑛 2

2 if 휀 = 1

Remark. If 𝐹′ 𝑥𝑛 𝑥𝑛 − 𝑥 = 0 ⟺ 𝐹′ 𝑥𝑛 𝑥𝑛 = min𝑥∇𝒞 𝐹′ 𝑥𝑛 𝑥,

then, we cannot decrease and the algorithm stops. In another case

𝐹 𝑥𝑛+1 < 𝐹 𝑥𝑛 .

Since 𝐹 𝑥𝑛 is decreasing (assuming inf𝒞 𝐹 > −∞), we have that

𝐹 𝑥𝑛 converges.

If 휀 < 1, the inequality

𝐹 𝑥𝑛+1 ≤ 𝐹 𝑥𝑛 − 𝐹′ 𝑥𝑛 𝑥𝑛 − 𝑥 2

2𝐾 𝑥 − 𝑥𝑛 2

and 𝒞 bounded shows that

𝐹′ 𝑥𝑛 𝑥𝑛 − 𝑥 = 𝐹′ 𝑥𝑛 𝑥𝑛 − min𝑥∇𝒞

𝐹′ 𝑥𝑛 𝑥 → 0.

Analogously, if 휀 = 1we have that 𝑥 − 𝑥𝑛 tends to zero which also

implies the above statement.

If we assume the space finite dimensional and 𝐹′continuous, we

have (recall that the algorittm will be applied to a discretization of

the Control Problem)

Theorem. There exists a subsequence 𝑥𝑛𝑘of 𝑥𝑛 and 𝑥 ∇ 𝒞, such that

𝑥𝑛𝑘→ 𝑥 and 𝑥 satisfies

𝐹′ 𝑥 𝑥 − 𝑥 ≥ 0, ∀𝑥 ∇ 𝒞,

i.e. 𝑥 satisfies the usual optimality conditions for the problem

min𝒞

𝐹

Proof. Since 𝒞 is compact we can assume the existence of 𝑥𝑛𝑘 and

𝑥 ∇ 𝒞, such that 𝑥𝑛𝑘→ 𝑥 . Using that

𝐹′ 𝑥𝑛 𝑥𝑛 − min𝑥∇𝒞

𝐹′ 𝑥𝑛 𝑥 → 0,

we get the result.

Summarizing

We fix 𝐾 > 0 large enough.

For 휃𝑛 , 𝑀𝑛 , with 𝑀𝑛 ∇ 𝒦 휃𝑛 a.e. in Ω, ∫ 휃𝑛Ω𝑑𝑥 ≤ 𝜇, we define

𝑢𝑛 , 𝑝𝑛 as the solutions of

−div 𝑀𝑛∆𝑢𝑛 = 𝑓 in Ω

𝑢𝑛 = 0 on 𝜕Ω

−div 𝑀𝑛∆𝑝𝑛 = 𝜕𝑠𝐻𝑛 − div 𝜕𝜉𝐻𝑛 + 𝑀𝑛𝜕휂𝐻𝑛 in Ω

𝑝𝑛 = 0 on 𝜕Ω

The index 𝑛 in 𝐻 means that the functions are valuated in

𝑥, 𝑢𝑛 , ∆𝑢𝑛 , 𝑀𝑛∆𝑢𝑛 , 휃𝑛 .

We denote 𝛾 = ∆𝑢𝑛 , 휁 = 𝜕휂𝐻𝑛 − ∆𝑝𝑛

For 𝑟 ≥ 0, we take 휃𝑟 as

If −𝛽 − 𝛼

2𝛼 𝛼 + 𝛽 휁 ∙ 𝛾 + 𝛽 − 𝛼 휁 ∙ 𝛾 + 𝜕𝑝𝐻𝑛 + 𝑟 ≥ 0,

then 휃𝑟 = 0.

If −𝛽 − 𝛼

2𝛽 𝛼 + 𝛽 휁 ∙ 𝛾 − 𝛽 − 𝛼 휁 ∙ 𝛾 + 𝜕𝑝𝐻𝑛 + 𝑟 ≤ 0,

then 휃𝑟 = 1.

In another case

휃𝑟 =1

𝛽 − 𝛼

𝛼𝛽 𝛽 − 𝛼 휁 𝛾 + 휁 ∙ 𝛾

𝛽 − 𝛼 휁 𝛾 − 휁 ∙ 𝛾 − 2 𝜕𝑝𝐻𝑛 + 𝑟 − 𝛼

Then, we take 𝑟 = 0 if 휃0𝑑𝑥

Ω

≤ 𝜇

𝑟 = min 𝑠 ≥ 0: 휃𝑠𝑑𝑥

Ω

= 𝜇 if not

and define

If 휃(𝑥) = 0 ⇒ 𝑀(𝑥) = 𝛽𝐼 , If 휃(𝑥) = 1 ⇒ 𝑀(𝑥) = 𝛼𝐼

If 휃 𝑥 ∇ 0,1 , we distinguish several cases:

If 𝛾(𝑥) and 휁 𝑥 = 0, then 𝑀(휃 𝑥 ) ∇ 𝒦(𝑥) can be arbitrarily

chosen.

If 𝛾 𝑥 ∙ 휁 𝑥 = 𝛾 𝑥 휁 𝑥 , then we take

𝑀 𝑥 = 𝜆 휃(𝑥) 𝛾 𝑥 ⨂𝛾 𝑥

𝛾 𝑥 2 + Λ 휃(𝑥) 𝐼 −

𝛾 𝑥 ⨂𝛾 𝑥

𝛾 𝑥 2

If 𝛾(𝑥) ∙ 휁 𝑥 = − 𝛾 𝑥 휁 𝑥 , then we consider a unitary vector

𝜐 𝑥 orthogonal to 𝛾 𝑥 and we take

𝑀 𝑥 = 𝜆 휃 𝑥 𝜐 𝑥 ⨂𝜐 𝑥 + Λ 휃 𝑥 𝐼 − 𝜐 𝑥 ⨂𝜐 𝑥

In another case, we take

𝜐 𝑥 =𝛾 𝑥

𝛾 𝑥 +

휁 𝑥

휁 𝑥

𝑎𝑛𝑑 𝑀 𝑥 = 𝜆 휃 𝑥 𝜐 𝑥 ⨂𝜐 𝑥

𝑣 𝑥 2 + Λ 휃 𝑥 𝐼 −

𝜐 𝑥 ⨂𝜐 𝑥

𝑣 𝑥 2 .

Finally, for 휀 equals to the minimum of

∫ 𝑀𝑛 − 𝑀 ∆𝑢𝑛 ∙ 𝜕휂𝐻𝑛 − ∆𝑝𝑛 𝑑𝑥Ω

+ ∫ 𝜕𝑝𝐻𝑛 휃𝑛 − 휃 𝑑𝑥Ω

𝐾 ∫ 𝑀𝑛 − 𝑀 ∆𝑢𝑛 2𝑑𝑥 + ∫ 𝑀𝑛 − 𝑀 ∆𝑝𝑛 2𝑑𝑥ΩΩ

and 1, we define

𝑀𝑛+1 = 𝑀𝑛 + 휀 𝑀 − 𝑀𝑛 , 휃𝑛+1 = 휃𝑛 + 휀 휃 − 휃𝑛

Numerical experiments: .0

.\0,1)( ,0,1)(on 0

0,1)(in 1

min

(CP)

22

2

\20,1)(

20,1)(

2

u

udiv

dxu

α=1, β=2, μ=0.1

a=1, b=20, k=0.9

α=1, β=2, µ=0.4

a=1, b=20, k=0.4

α=1, β=20, =0.1α=1, β=20, μ=0.4

.3.0\0,1)( ,0,1)(on 0

0,1)(in 10

min

(CP)

22

2

16/1|)5.0,5.0(|\20,1)(

20,1)(

u

udiv

udxuC

x

C=I,

100

01C

Remark. In these examples the optimal control seems to satisfy

𝛼휃 + 𝛽(1 − 휃) ∆𝑢 = 𝑀∆𝑢.

Recall (Tartar). For 𝑓 ∇ 𝐻−1 Ω , problem

, ,on 0

in )1(

)(min

(CP)

2

dxu

fudiv

dxuu

has a unique solution for every 𝑢 ∇ 𝐻01(Ω) except a set of first

category.

Pedregal have also obtained some existence results for this problem.

In this sense we can also remark that if 𝑓(𝑥, 𝑠, . ) is convex, then

min 𝐻 𝑥, 𝑠, 𝜉, 휂, 𝑝 : 휂 ∇ 𝒦(𝑝)𝜉 = 𝑝𝛼 + 𝛽(1 − 𝑝) 𝜉.

Examples: Ω = 0,1 2

min 𝑢 𝑥 − 1 2𝑑𝑥

Ω

−𝑑𝑖𝑣 𝜒𝜔 + 3 1 − 𝜒𝜔 ∆𝑢 = 𝑥1 in Ω

𝑢 = 0 on ∂Ω

We start with 𝜔 = Ω,

Solution

977.0)( dxx

1.0

1.0

3.0

5.0

7.0

9.0

1

Consider now, 𝑢𝑤 the solution of

−𝑑𝑖𝑣 5𝜒𝜔𝑤

+ 1 − 𝜒𝜔𝑤 ∆𝑢𝑤 = 1 in Ω

𝑢 = 0 on ∂Ω.

Let us solve the problema

min 𝑢 𝑥 − 𝑢𝑤 (𝑥) 2𝑑𝑥

Ω

−𝑑𝑖𝑣 5𝜒𝜔 + 1 − 𝜒𝜔 ∆𝑢 = 1 in Ω

𝑢 = 0 on ∂Ω,

starting from 𝜔 = ∅.

Iteration=0

Iteration=10

Iteration=50

Iteration=100

Iteration=250

Iteration=500

Iteration=1000

Iteration=2000

Iteration=3000

Iteration=4000

Iteration=5000

Iteration=6000

Iteration=7000

Iteration=8000

Iteration=9000

Iteration=10000

Iteration=15000

Iteration=20000

Iteration=30000

Iteration=35000

Iteration=40000

Iteration=50000

Iteration=80000

Iteration=100000

The method converges very slowly.

The main problems hold where ∆𝑢𝑤 is small.

May be it would be better to consider several equations

with different second members.

The one-dimensional case

We have shown that we can solve numerically the control

problem discretizing both the unrelaxed and the relaxed problem.

What is better?

Let us answer this question in the simple case where 𝑁 = 1.

The results of this part are contained in

J. Casado-Díaz, C. Castro, M. Luna-Laynez, E. Zuazua. Numerical

approximation of a one-dimensional elliptic optimal design problem.

SIAM: Multiscale Modeling and Simulation. To appear.

We consider the control problem (CP) 𝐹1, 𝐹2 ∇ 𝑊1,∞

inf 𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

𝜔

+ 𝐹2 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

(0,1)\𝜔

−

𝑑

𝑑𝑥 𝛼𝜒𝜔 + 𝛽𝜒(0,1)\𝜔

𝑑𝑢

𝑑𝑥 = 𝑓 in (0,1)

𝑢 0 = 𝑢(1) = 0

𝜔 ≤ 𝜇

Relaxed formulation (RCP)

min 휃𝐹1 𝑥, 𝑢,𝜆(휃)

𝛼

𝑑𝑢

𝑑𝑥 + (1 − 휃)𝐹2 𝑥, 𝑢,

𝜆(휃)

𝛽

𝑑𝑢

𝑑𝑥 𝑑𝑥

1

0

−

𝑑

𝑑𝑥 𝜆(휃)

𝑑𝑢

𝑑𝑥 = 𝑓 in (0,1)

𝑢 0 = 𝑢(1) = 0

휃1

0

𝑑𝑥 ≤ 𝜇.

Semidiscretization

For a partition 𝒫𝑟 = 𝑡0 = 0 < 𝑡1 < ⋯ < 𝑡𝑚𝑟= 1 of (0,1),

with diam 𝒫𝑟 =𝑟.

We consider the problems

inf 𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

𝜔

+ 𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

(0,1)\𝜔

𝑈𝑆𝐷 −

𝑑

𝑑𝑥 𝛼𝜒𝜔 + 𝛽𝜒(0,1)\𝜔

𝑑𝑢

𝑑𝑥 = 𝑓 in (0,1)

𝑢 0 = 𝑢(1) = 0

𝜔 is a union of intervals of 𝒫𝑟, 𝜔 ≤ 𝜇

inf 휃𝐹1 𝑥, 𝑢,𝜆 휃

𝛼

𝑑𝑢

𝑑𝑥

1

0

+(1 − 휃)𝐹2 𝑥, 𝑢,𝜆(휃)

𝛽

𝑑𝑢

𝑑𝑥 𝑑𝑥

𝑅𝑆𝐷 −

𝑑

𝑑𝑥 𝜆(휃)

𝑑𝑢

𝑑𝑥 = 𝑓 in (0,1)

𝑢 0 = 𝑢(1) = 0

휃 is constant in the intervals of 𝒫𝑟, 휃1

0

𝑑𝑥 ≤ 𝜇.

Remark. We have

min(𝑅𝑆𝐷) ≤ min(𝑈𝑆𝐷)

Lemma. Assume 휃 ∇ 𝐿∞ 0,1; 0,1 , then there exist

휃𝑟𝑈 ∇ 𝐿∞ 0,1; 0,1 , 휃𝑟

𝑅 ∇ 𝐿∞ 0,1; 0,1 , constants in the

intervals of 𝒫𝑟 with

휃𝑟𝑈𝑑𝑥,

1

0

휃𝑟𝑅𝑑𝑥

1

0

≤ 휃𝑑𝑥,1

0

such that for every 𝑓 ∇ 𝐿1 0,1 , the solutions 𝑢, 𝑢𝑟𝑈 , 𝑢𝑟

𝑅 in 0,1 of

−𝑑

𝑑𝑥 𝜆 휃

𝑑𝑢

𝑑𝑥 = 𝑓, 𝑢 0 = 𝑢 1 = 0

−𝑑

𝑑𝑥 𝜆 휃𝑟

𝑈 𝑑𝑢𝑟

𝑈

𝑑𝑥 = 𝑓, 𝑢𝑟

𝑈 0 = 𝑢𝑟𝑈 1 = 0

−𝑑

𝑑𝑥 𝜆 휃𝑟

𝑅 𝑑𝑢𝑟

𝑈

𝑑𝑥 = 𝑓, 𝑢𝑟

𝑅 0 = 𝑢𝑟𝑅 1 = 0

satisfy:

For 휃𝑟𝑅

𝑢 − 𝑢𝑟𝑅 𝐿1 0,1 + 𝜆 휃

𝑑𝑢

𝑑𝑥− 𝜆 휃𝑟

𝑅 𝑑𝑢𝑟

𝑅

𝑑𝑥

𝐿∞ 0,1

= 𝑜 𝑟 .

Moreover, if 𝑓 ∇ 𝐿∞ 0,1 and 휃 ∇ 𝐵𝑉 0,1 , then 𝑜 𝑟 = 𝐶𝑟2

For 휃𝑟𝑈

𝑢 − 𝑢𝑟𝑈 𝐿1(0,1) ≤ 𝐶 𝑟

and, assuming 𝑓 ∇ 𝑊1,𝑙(0,1)

𝜆 휃 𝑑𝑢

𝑑𝑥− 𝜆 휃𝑟

𝑈 𝑑𝑢𝑟

𝑈

𝑑𝑥

𝐿∞ 0,1

≤ 𝐶𝑟𝑙+1𝑙+2.

Theorem. We have

min 𝑅𝐶𝑃 − min(𝑅𝑆𝐷) ≤ 𝑜 𝑟 .

Moreover, if 𝑓 ∇ 𝐿∞ 0,1 and there exists a control 휃 ∇ 𝐵𝑉 0,1 ,

then 𝑜 𝑟 = 𝐶𝑟2.

Theorem. We have

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷) ≤ 𝑟.

Moreover, if 𝑓 ∇ 𝑊1,𝑙 0,1 and the functions 𝐹𝑖(𝑥, 𝑠, 𝜉) do not

depend on 𝑠, then

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷) ≤ 𝑐𝑟𝑙+1𝑙+2 .

Example: Assume 𝐹1 = −𝛼 𝜉 2, 𝐹2 = −𝛽 𝜉 2, 𝛼 < 𝛽,

𝑓 ∇ 𝐿1 0,1 , 𝑓 ≢ 0, 𝑓 𝑡 = 𝑓 1 − 𝑡 . We take

𝒡 𝑡 = 𝑓 𝑠 𝑑𝑠𝑡

12

, 𝜇 ≤ 𝐹 𝑡 ≠ 0 .

Then, the optimal controls are the functions 휃 ∇ 𝐿1 0,1; [0,1]

satisfying

휃 𝑠 𝑑𝑠1

0

= 𝜇, 𝒡 𝑠 휃 𝑠 𝑑𝑠1

0

= 0

휃 𝑡 = 1 if 𝒡 𝑡 > 𝛾0

0 if 𝒡 𝑡 < 𝛾0,

where

𝛾0 = inf 𝛾 > 0: 𝒡 𝑡 > 𝛾 < 𝜇

Remark. In this case the original problem has solution.

For 𝛼 < 𝛽, we consider

inf −α 𝑑𝑢

𝑑𝑥

2

𝑑𝑥𝜔

− β 𝑑𝑢

𝑑𝑥

2

𝑑𝑥(0,1)\𝜔

−

𝑑

𝑑𝑥 𝛼𝜒𝜔 + 𝛽𝜒(0,1)\𝜔

𝑑𝑢

𝑑𝑥 = 1 in (0,1)

𝑢 0 = 𝑢(1) = 0

𝜔 ≤2

3

We take 𝒫𝑛 = 𝑘10−𝑛 : 𝑘 = 0, … , 10𝑛 . Then

lim𝑛

min 𝑅𝐶𝑃 − min(𝑅𝑆𝐷)

10−2𝑛=

2(β − α)

27αβ

lim𝑛

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷)

10−𝑛=

β − α

54αβ

The above example shows:

The estimate

min 𝑅𝐶𝑃 − min(𝑅𝑆𝐷) ≤ 𝑐𝑟2

cannot be improved.

We do not know if (for 𝐹𝑖 𝑥, 𝑠, 𝜉 independent of 𝑠)

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷) ≤ 𝑐𝑟𝑙+1𝑙+2 , for every 𝑙

can be improved to

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷) ≤ 𝑐𝑟,

but the inequality

min 𝑅𝐶𝑃 − min(𝑈𝑆𝐷) = 𝑜(𝑟)

does not hold.

Conclusion: Discretizing the relaxed control problem the

convergence is a lot faster than discretizing the unrelaxed one. This

assertion holds even if the unrelaxed continuous problem has

solution.

Full discretization

We now consider a second partition 𝒬𝑕 of diameter 𝑕 < 𝑟,

which is a refinement of 𝒫𝑟 . We define

𝑉𝑕 = 𝑢 ∇ 𝐶00 0,1 : 𝑢 is affine in each interval of 𝒬𝑕

We consider

inf 𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

𝜔

+ 𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 𝑑𝑥

(0,1)\𝜔

𝑈𝐷

𝑢 ∇ 𝑉𝑕

𝛼𝜒𝜔 + 𝛽𝜒(0,1)\𝜔 𝑑𝑢

𝑑𝑥

1

0

𝑑𝑣

𝑑𝑥𝑑𝑥 = 𝑓 𝑣𝑑𝑥, ∀𝑣 ∇ 𝑉𝑕

1

0

𝜔 is a union of intervals of 𝒫𝑟, 𝜔 ≤ 𝜇

and

inf 휃𝐹1 𝑥, 𝑢,𝑑𝑢

𝑑𝑥 + 1 − 휃 𝐹2 𝑥, 𝑢,

𝑑𝑢

𝑑𝑥 𝑑𝑥

𝜔

𝑅𝐷

𝑢 ∇ 𝑉𝑕

𝜆 휃 𝑑𝑢

𝑑𝑥

1

0

𝑑𝑣

𝑑𝑥𝑑𝑥 = 𝑓 𝑣𝑑𝑥, ∀𝑣 ∇ 𝑉𝑕

1

0

휃 is constant in the intervals of 𝒫𝑟, 휃𝑑𝑥1

0

≤ 𝜇

Remark. We have chosen 𝒬𝑕 a refinement of 𝒫𝑟 . If contrarily, we

choose 𝑟 < 𝑕 and 𝒫𝑟 a refinement of 𝒬𝑕 , we can find examples

showing that the discretizations defined above do not convergence

in general.

Lemma. Assume 휃 ∇ 𝐿∞ 0,1; 0,1 constant in each interval of

𝒬𝑕 . for 𝑓 ∇ 𝐿1 0,1 , we consider 𝑢 and 𝑢𝑕 the solutions of

−

𝑑

𝑑𝑥 𝜆(휃)

𝑑𝑢

𝑑𝑥 = 𝑓 in (0,1)

𝑢 0 = 𝑢(1) = 0

𝑢𝑕 ∇ 𝑉𝑕

𝜆 휃 𝑑𝑢𝑕

𝑑𝑥

1

0

𝑑𝑣

𝑑𝑥𝑑𝑥 = 𝑓 𝑣𝑑𝑥, ∀𝑣 ∇ 𝑉𝑕 .

1

0

Then

𝑢 − 𝑢𝑕 𝑊1,1(0,1) ≤ 𝐶𝑕 𝑓 𝐿1(0,1).

Using this result, we have the following theorem

Theorem. Taking 𝑟 = 𝑕 and 𝑓 ∇ 𝑊1,𝑙(0,1) we have

min 𝑅𝐶𝑃 − min(𝑈𝐷) ≤ 𝑐𝑕𝑙+1𝑙+2.

Taking 𝑓 ∇ 𝐿∞ 0,1 , 𝑟 = 𝑕 and assuming that there exists an

optimal control in 𝐵𝑉 0,1 , we have

min 𝑅𝐶𝑃 − min(𝑅𝐷) ≤ 𝑐𝑕.