Chapter 7 Isoperimetric problem

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1 Chapter 7 Isoperimetric problem Recall that the isoperimetric problem (see the itroductio its coectio with ido s proble) is oe of the most classical problem of a shape optimizatio. It ca be formulated i the followig way : fid, amog all admissible domais with a give perimeter, the oe whose Lebesgue measure is as large as possible. Equivaletly, oe could miimize the perimeter of a set amog all admissible domais whose Lebesgue measure is prescribed. I this chapter, we study sets of fiite perimeter ad compactess of their characteristic fuctio. This is very useful i shape optimizatio whe the perimeter appears i the miimizig fuctioal. The we study the existece of a solutio for the isoperimetric problem. 1. Set of bouded perimeter First, we begi by to defie the otio of perimeter. Here we adopt the otio defied by e Giorgi efiitio Let IR N be a ope set. We deote by { (, IR N ) = ϕ = (ϕ 1,..., ϕ N ) ; ϕ i (). IR N is equipped with the orm ( N ) 1 2 ϕ = sup ϕ i (x) 2 x 1 Eio de Giorgi (Lecce, February 8, Pisa, October 25, 1996) was a ifluetial Italia mathematicia of the 20th cetury. He made importat cotributios to mathematical aalysis, i particular to the study of miimal surfaces ad to the regularity of the solutios of elliptic partial differetial equatios. i=1 1

2 Itroductio to Shape optimizatio N. Igbida 2 efiitio 0.1 Let be a measurable set of. The perimeter of with respect to (or simply perimeter of is = IR N ) is the real umber give by { P () = sup div(ϕ) dx ; ϕ (, IR N ), ϕ 1. Si = IR N the the perimeter is simply deoted by P (). Remark 0.1 Let (, IR N ) = () N the set of distributios T = (T 1, T 2,..., T N ) with T i () for ay i = 1,...N. (, IR N ) = () N is the set of distributios T = (T 1, T 2,..., T N ) with T i () for ay i = 1,...N. We may rewrite div(ϕ) dx as follows : div(ϕ) dx = = = ( N χ i=1 N χ, i=1 N i=1 ) ϕ i dx x i ϕ i x i, χ x i, ϕ i = χ, ϕ,. I the case where is a regular, the perimeter may be defie by usig the lateral boudary, this the aim of the followig propositio. Propositio 0.1 If is a bouded C 1 ope domai, the P () = dσ, where σ is is a boudary elemet., Idea of the proof : By Gree formula, it is ot difficult to see that div(ϕ) dx = ϕ ν dσ, so that P () The idea after is to costruct a sequece ϕ of (, IR N ), such that dσ. ϕ ν uiformly i. Aother iterestig iterpretatio of the perimeter is i term of Rado measures o.

3 Itroductio to Shape optimizatio N. Igbida Prelimiaries o vector valued Rado measure Recall that for ay f L 1 (, IR N ), { f 1 = sup f(x) ϕ(x) dx ; ϕ (, IR N ), ϕ 1. The last formula remais to be true for vector valued Rado measure with fiite mass. Recall that vector valued Rado measure o is a liear form o the space C 0 (, IR N ) the set of vector valued cotiuous fuctios compactly supported i. More precisely, we have Propositio 0.2 Set µ = (µ 1, µ 2,..., µ N ) (, IR N ). The, µ is a Rado measure with fiite mass is ad oly if { µ 1 = sup < µ, ϕ > ; ϕ (, IR N ), ϕ 1. We deote by M b () the space of all Rado measure defied i with fiite mass ad the space ( ) N M b (, IR N ) = M b () equipped with the orm. 1. Propositio 0.3 Let µ = (µ 1, µ 2,..., µ N ) (, IR N ). If, there exists C > 0, such that < µ, ϕ > C ϕ, for ay ϕ (, IR N ), the µ is a vector valued Rado measure with fiite mass. Remark 0.2 I geeral if a distributio is cotiuous with respect to a certai orm i (, IR N ), the it ca be idetified to a elemet of the completed dual of (, IR N ) with respect to to this orm. Recall that M b () ca be idetified to the dual space of C 0 () equipped with the uiform orm. More precisely, each µ M b () ca be idetified to the followig liear applicatio : ϕ C 0 () ϕ(x) dµ(x). So, sice C 0 () is a separable set, we deduce that a douded set of M b () is sequetially compact with respect to the weak topology σ(m b (), C 0 ()). That is Propositio 0.4 If µ is a sequece of Rado measure o such that sup µ 1 <, the there exists µ M b () ad a subsequece that we deote agai by µ such that ϕ(x) dµ ϕ(x) dµ for ay ϕ C 0 (). At last, let us prove the followig compactess result that will be very useful for the study of the perimeter fuctio.

4 Itroductio to Shape optimizatio N. Igbida 4 Propositio 0.5 Let IR N be a ope set ad f a sequece of L 1 () such that f is sequece of vector valued Rado measure. If f 1 + f 1 C, the there exists f L 1 () such that f M b () N ad a subsequece that we deote agai by f such that f f i L 1 loc() ad f f i M b () N weak. Remark Let us deote by BV () = { f L 1 () ; f M b () N. This is the set of fuctios of bouded variatio i. Propositio?? implies that the ijectio of BV () equipped with the orm ito L 1 loc() is compact. 2. I geeral we do t have the covergece i L 1 (). Take, f (IR N ) ad f (x) = f(x + ). The, the assumptios of Propositio?? are fulfilled, ad f f i L 1 loc() ad ot i L 1 (). For bouded domai, the covergece holds to be true i L 1 () if has Lipschitz boudary. Proof of Propositio?? : Thaks to Propositio??, there exists f k, f M b () ad µ M b () N such that f k f i M b () weak ad f k µ i M b () N weak Sice () C 0 (), the the covergeces are true i the sese of distributio ad we have µ = f. SIce takig f (x + h) f (x) = ε = 1 0 f (x + th) h dt, { x ; d(x, ) > ε ad h < ε, we deduce that ε f (x + h) f (x) dx 1 h f (x + th) dt dx 0 ε f 1 h.

5 Itroductio to Shape optimizatio N. Igbida 5 This implies that { lim sup f (x + h) f (x) dx = 0, h 0 ε so that, by usig Riesz-Frechet-Kolmogorof Theorem 2 we deduce that f f i L 1 loc(). 1.3 Properties of the perimeter First, we have the followig characterizatio the perimeter ad sets of fiite perimeter. Propositio 0.6 Let be a measurable set. We have P () < χ is a Vector valued Rado measure with fiite mass ad P () = χ 1 (total variatio of χ computed i the sese of distributio). Proof : This as a immediate cosequece of Propositio??. Remark 0.4 For ay ope domai IR N, A ad B measurable sets, we have Ideed, formally we have so that P (A B) P (A) + P (B). χ A B = (1 χ B ) χ A + (1 χ A ) χ B, χ A B (1 χ B ) χ A + (1 χ A ) χ B χ A + χ B χ A B. The mai difficulty here is to approximate χ A ad χ B by regular fuctios f A ad f B such that f A 1 ad f B 1 coverges to χ A 1 ad χ A 1, respectively. Before to give some compactess results cocerig the perimeter fuctio of domais, let us give some cotiuity depedece results of the volume ad the perimeter with respect to the covergece of the characteristic fuctios. 2 Let IR N, ω ad f be a bouded sequece of L p () with 1 p <. If for ay ε > 0 there exists δ > 0 such that δ < dist(ω, c ) ad h δ implies that sup f (. + h) f (.) L 1 (ω) < ε, the the sequece ( ) f /ω is relatively compact i L 1 (ω). If moreover, for ay ε > 0, there exists ω s.t. sup the the covergece hlods to be true i L 1 (). f < ε, \ω

6 Itroductio to Shape optimizatio N. Igbida 6 Propositio 0.7 Let ad be bouded measurable sets of IR N. If χ χ i L 1 loc(), the P () lim if P ( ) I other words, the perimeter is l.s.c for the strog covergece of the characteristics fuctios. Proof : Let ϕ (, IR N ). We have div(ϕ) dx = lim lim if lim if div(ϕ) dx ( { sup ) div(ϕ) dx ; ϕ 1 P ( ). The, by takig the sup over ϕ (, IR N ) such that ϕ 1, we deduce the result. Exercise 0.1 Let ad be bouded measurable sets of IR N. Prove that 1. If χ χ i L 1 loc(), the 2. If χ χ i L 1 (), the lim if. = lim. Solutio : Use Fatou Lemma. Ideed, thaks to the assumptio, there exists a subsequece that we deote agai by, such that sice χ χ a.e. i. So = IR χ N lim if IR χ N =. The secod part of the Theorem for the volume follows by = IR χ N = lim IR χ N =. Remark The coclusio of the first part of the propositio remais true if χ χ i L 1 loc() weak. Sice, this implies the strog covergece i L 1 loc (IRN ). 2. If χ χ i L 1 loc(), the the iequality may be strict for the volume (take = B(x, 1) with x.) At last, by usig Propositio??, we have the followig compactess result. Theorem 0.1 Let be a sequece of measurable sets of a ope domai of IR N. Assume that + P ( ) C (idepedet of ).

7 Itroductio to Shape optimizatio N. Igbida 7 The, there exists a measurable set ad a subsequece k ad Moreover, if is bouded the χ k χ i L 1 loc() χ k χ i M b () N weak. χ k χ i L 1 (). such that Proof : Thaks to Propositio??, there exists f L 1 (), such that f M b () N, ad χ k f i L 1 loc() χ k f i M b () N weak. Sice the L 1 loc covergece implies the a.e. covergece, we deduce that there exists a measurable set such that f = χ a.e. (ideed, a.e. x we have f(x)(1 f(x)) = lim f k (x)(1 f k (x))). At last, if is bouded, the the covergece follows by Lebesgue k domiated Theorem. 2. Existece of optimal shape for the Isoperimetric problem Let IR N be a ope domai ad V 0 (0, ) a give real umber. Cosider the problems (0.1) P ( ) = mi {P () ; measurable = V 0 (0.2) P ( ) = mi {P () ; measurable = V 0 First, recall that the Isoperimetric Iequality i the plae is a geometric iequality ivolvig the square of the circumferece L of a closed curve i the plae ad the area A of a plae regio it ecloses, as well as its various geeralizatios. More precisely, it is give by 4πA L 2. ozes of proofs of the Isoperimetric Iequality i the plae have bee foud. I 1902, Hurwitz published a short proof usig the Fourier series that applies to arbitrary rectifiable curves (ot assumed to be smooth). A elegat direct proof based o compariso of a smooth simple closed curve with a appropriate circle was give by E. Schmidt i It uses oly the arc legth formula, expressio for the area of a plae regio from Gree s theorem, ad the Cauchy?Schwarz iequality. I geeral, for ay measurable set A, we have (see Federer, Herbert (1969). Geometric measure theory. Spriger-Verlag. ISBN ) : (0.3) P (A) c N A N 1 N with c N = N B N (0, 1) 1 N. The iequality is a equality if ad oly if A is a boule.

8 Itroductio to Shape optimizatio N. Igbida 8 Remark 0.6 The N dimesioal Isoperimetric Iequality is equivalet (for sufficietly smooth domais) to the Sobolev iequality o IR N, with optimal costat: ( IR N u N N 1 ) N 1 N 1 B N (0, 1) 1/N IR N u, for ay u W 1,1 (IR N ). We have Theorem 0.2 Assume that <. The, the problems (??) ad (??) has a solutio. Moreover, if cotais a boule B of measure V 0, the ay solutio of (??) is a boule of measure V 0. Proof : First let us see that the set of admissible sets is oempty. Sice is a ope domai, there exists a sequece of boule B p such that = B p. For large k, we have w k = p k B p V 0 ad ω k is a ope domai with bouded perimeter. Moreover, there exists B(0, r) such that ω k B(0, r) = V 0 ad P (ω k B(0, r)) is fiite. Now, let be a miimizig sequece for the problem (??). Sice P ( ) = χ 1 is bouded, thaks to Theorem??, there exists a measurable set ad a subsequece k such that χ k χ i L 1 (). We have = V 0, ad usig the l.s.c of the perimeter fuctio, we deduce that P ( ) lim P ( ). This implies that solves the problem (??). For the the problem (??), the proof follows i the same way. At last, thaks to (??), we deduce that if cotais a boule B such that B = V 0, the the optimal shape for (??) is B. Remark If is ote of fiite measure the the problem (??) may have o solutio. For istace, take { = (x, y) IR 2 ; y 2π 1 arcta(x)

9 Itroductio to Shape optimizatio N. Igbida 9 ad V 0 = π the surface of a disk of radius equals to 1. Thaks to (??), we have if {P () ; measurable = V 0 2π. However, takig ( ) ) = B ( + 1, 0), 2π 1 arcta() B ((2, 0), ε, ( 2 with πε 2 + π 2π arcta()) 1 = π, we ca verify that lim P ( ) = 2π A extesio The more importat poit i the proof is the compactess of the sequece of characteristic fuctios χ, that follows from the fact that P ( ) is bouded. The same approach remais true for a more geeral situatio. Assume we have some PE, such that Moreover, assume that y() L 2 () is a solutio of the PE stated i. (0.4) χ χ i L 1 () y( ) y() i L 2 (). For a give f L 2 (), K L 1 () ad τ > 0, we deduce that the fuctioal J() = (y() f) 2 dx + K(x)dx + τp (). The, cosider the shape optimizatio problem { (0.5) mi J() ; ad (0.6) mi {J() ;, = V 0, where V 0 is give. Theorem 0.3 Assume that < ad V 0 ]0, [. The, the problems (??) ad (??) have solutios. Proof : The two problems (??) ad (??) have lower boud K 1, ad the set of admissible sets such that J() < is o empty (see the proof of Theorem??). Let be a miimizig sequece ad M a upper boud for J( ). The, < ad τ P () M + K 1.

10 Itroductio to Shape optimizatio N. Igbida 10 The, Theorem?? implies that there exists a ope set ad a subsequece k such that χ k χ i L 1 (). This implies that = V 0 for the problem (??). Moreover, thaks to the assumptio (??), we have y( k ) y( ) i L 2 (). Sice P (.) is l.s.c. we deduce that which implies that is a solutio. J( ) lim if J( k ) = if J,