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J. Math. Pures Appl. 81 (2002) 311341

Implicit partial differential equations and the constraintsof nonlinear elasticity

Bernard Dacorogna , Chiara Tanteri

Dpartement de Mathmatiques, EPFL, 1015 Lausanne, Switzerland

Manuscript received 2 July 2001

Abstract

We study a Dirichlet problem associated to some nonlinear partial differential equations under additional constraints that arerelevant in nonlinear elasticity. We also give several examples related to the complex eikonal equation, optimal design, potentialwells or nematic elastomers. 2002 ditions scientifiques et mdicales Elsevier SAS. All rights reserved.

Keywords: Rank one convex hull; Implicit partial differential equations

1. Introduction

We consider here the following Dirichlet problem (as well as some higher-order versions of the problem)

Fi

x,u(x),Du(x) = 0, i = 1, 2, . . . , I , a.e. x ,

u = , on , (1)

where Rn

is a bounded open set, u : Rn

and therefore Du Rn

n

, Fi :Rn

n

R, i = 1, . . . , I , are quasiconvexfunctions and , the boundary datum, is given.This problem has been intensively studied and we refer to Dacorogna and Marcellini [6] for a discussion of these implicit

equations. We will be interested here in considering the case where we require that the solutions satisfy some constraints that arenatural in nonlinear elasticity. The first one is the noninterpenetration of matterthat is expressed mathematically by det Du > 0and the second one is the incompressibility which reads as det Du = 1. These two questions were raised in [6] and are discussedby Dacorogna, Marcellini and Tanteri [7] for the first one and by Mller and Sverak [14] for the second one, using a differentapproach based on the method of convex integration of Gromov.

We will discuss here some theoretical results related to the first (Section 2) and second (Sections 3, 4) cases and deal withseveral relevant examples (Sections 5 to 10). We will also make some general considerations concerning polyconvex hulls(Section 11) and we will conclude in Appendix A with some well known properties ofsingular values of matrices.

We now describe six examples that we will investigate here, but we first recall that we respectively denote by co E, Pco E,Rco E, the convex, polyconvex, rank one convex hull of a given set E Rmn.

The first example has already been considered when n

=2 by Dacorogna and Marcellini [6].

Example 1.1 (Complex eikonal equation). The problem has been introduced by Magnanini and Talenti [12] motivated byproblems of geometrical optics with diffraction. Given Rn a bounded open set, f : R R R, f = f(x,u,v),a continuous function, we wish to find a complex function w W1,(;C),

w(x) = u(x) + iv(x)

* Correspondence and reprints.E-mail addresses: [email protected] (B. Dacorogna), [email protected] (C. Tanteri).

0021-7824/02/$ see front matter 2002 ditions scientifiques et mdicales Elsevier SAS. All rights reserved.PII: S0021-7824( 01)012 35-1

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312 B. Dacorogna, C. Tanteri / J. Math. Pures Appl. 81 (2002) 311341

such thatn

i=1w2xi + f2 = 0, a.e. in ,

w = , on ,(2)

where wxi = w/xi . The problem is then equivalent to

|Dv|2 = |Du|2 + f2, a.e. in ,Dv; Du = 0, a.e. in ,w = , on .

We will solve, in fact, a more restrictive problem, namely:

|Dv|2 = r2 + f2, |Du|2 = r2, a.e. in ,Dv; Du = 0, a.e. in ,w = , on ,

for an appropriate r > 0. In algebraic terms (at this point we can consider f to be constant) we have, letting s =

r2 + f2,

E= = ab R2n: |a| = r, |b| = s and a; b = 0.

Letting

A =

1/r 00 1/s

R22

we will prove that

Pco E = Rco E = R2n: 1(A),2(A) 1,where 1(A),2(A) are the singular values of the matrix AR2n (cf. Appendix A for more details).

The second example is important for optimal design and is related to the method of confocal ellipses of Murat and Tartar [17]and of the results of Dacorogna and Marcellini [4,6]. However the existence part will be obtained without the use of the confocalellipses method, contrary to the one in [4,6].

Example 1.2 (Optimal design). Let R2 be a bounded open set and consider the DirichletNeumann problem:

w(x) {0, 1}, a.e. x ,det D2w(x) 0, a.e. x ,w(x) = (x), Dw(x) = D (x ), x .

The associated algebraic problem is (denoting the set of 2 2 symmetric matrices by R22s ) when we letE = R22s : trace {0, 1}, det 0,

to find that

Rco E = co E =

R22s : 0 trace 1, det 0

.

We will next consider two more academic examples but that exhibit some interesting features. The first one shows how wecan handle some problems depending on singular values under a constraint on the positivity of the determinant.

Example 1.3. Let R2 be an open set and consider

1(Du) + 2(Du) = 1, a.e. in ,det Du > 0, a.e. in ,

u(x) = (x ), x .

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B. Dacorogna, C. Tanteri / J. Math. Pures Appl. 81 (2002) 311341 313

The associated algebraic problem is: given

E =

R22: 1( ) + 2( ) = 1, det 0

,

to prove that

Pco E = Rco E =

R22: 1( ) + 2( ) 1, det 0

.

The fourth problem that we want to discuss is the following second-order problem.

Example 1.4. Let R2 be a bounded open set and consider the DirichletNeumann problem: 2uxi xj= 1, a.e. x , i,j= 1, 2,

u(x) = (x ), Du(x ) = D (x ), x .The algebraic problem is then

E = = (ij) R22s : |ij| = 1, i,j= 1, 2and we will find that

Pco E=

Rco E= = (ij) R22s : |ij| 1, i,j = 1, 2, |11 22|det .

The last two examples concern the incompressibility constraint. The first one is the problem of two potential wells in twodimensions that was resolved by Mller and Sverak in [13] and by Dacorogna and Marcellini in [5,6] for the case det A = det Band by Mller and Sverak in [14] for the case det A = det B. In Section 9 we will show that our results also apply to thisproblem.

Example 1.5 (Potential wells). Let Rn be open andE = SO(2)A SO(2)B

with det A = det B > 0. Let intRco E,

where intRco E stands for the interior (relative to the manifold det =

det A=

det B) of the rank one convex hull of E (cf.

Section 9 for the characterization of Rco E). Then there exists u W1,(;R2) such thatDu(x) E, a.e. in ,u(x) = x, on .

The last example is related to some recent work of DeSimone and Dolzmann [9] on nematic elastomers; we refer to thisarticle for the description of the physical model.

Example 1.6 (Nematic elastomers). Let r < 1 (this is called the oblate case while r > 1 is called the prolate case and it can behandled similarly), let 0 1(A) n(A) denote the singular values of a matrix A Rnn and

E =

A: (A) = r1/(2n), 1 n 1, n(A) = r (1n)/(2n), det A = 1

.

We will prove that

Rco E =

A:n

i=i (A) r

(1)/2n, 2 n, det A = 1

= A: (A) r1/2n, r(1n)/2n, 1 n, det A = 1(this representation formula, under the second form, has been established in [9] when n = 2, 3; actually we will considerbelow a slightly more general case). Our analytical result is then: given intRco E and Rn an open set, there existsu + W1,0 (;Rn) ((x) = x) such that

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1

Du(x) = = n1Du(x) = r1/2n, a.e. x ,

n

Du(x) = r(1n)/2n, a.e. x ,

det Du(x) = 1, a.e. x .

2. Inequality constraints

The results of this section are inspired by those of Dacorogna, Marcellini and Tanteri [7]. We recall first some notations anddefinitions introduced in [6].

Notations. (1) Let N,n,m 1 be integers. For u :Rn Rm we write

DNu =

Nui

xj1 xjN

1im1j1,...,jNn

RmnNs .

(The index s stands here for all the natural symmetries implied by the interchange of the order of differentiation.) When N= 1we have

Rmns =Rmn

while ifm

= 1 andN

= 2 we obtainRn

2

s =Rnns ,i.e., the usual set of symmetric matrices.

(2) For u : Rm we letD[N]u = u ,D u ,...,DNu

stand for the matrix of all partial derivatives ofu up to the order N. Note that

D[N]u RmMs =Rm Rmn Rmn2

s Rmn(N1)

s ,

where

M= 1 + n + + n(N1) = nN 1n

1

.

Hence

D[N]u = D[N1]u, DNu RmMs RmnNs .We now define the main property, called the relaxation property (cf. [6]), in order to get existence of solution.

Definition 2.1 (Relaxation property). Let E, K Rn RmMs RmnN

s . We say that K has the relaxation property withrespect to E if for every bounded open set Rn, for every u, a polynomial of degree N with DNu(x) = , satisfying

x, D[N1]u(x),DNu(x) int K,

there exists a sequence u CNpiec( ;Rm) such that

u u + WN,

0 ;Rm,u

u in W

N,,x, D[N1]u (x),DNu (x)

E int K, a.e. in ,

dist

x, D[N1]u (x),DNu (x); Edx 0 as .

The following theorem is the main abstract existence theorem.

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Theorem 2.2. Let Rn be open. LetFi : RmMs RmnN

s R, Fi = Fi (x,s,), i = 1, 2, . . . , I , be continuous withrespect to all variables and quasiconvex with respect to the variable . LetE, K Rn RmMs Rmn

N

s be such that

E = {(x,s,): Fi (x,s,) = 0, 1 i I1, Fi (x,s,) 0, I1 + 1 i I} ,K (x,s,): Fi (x,s,) 0, 1 i I.

Assume that the set in the right-hand side of the inclusion is bounded uniformly forx

and whenever s vary on a bounded

set ofRmMs and that K has the relaxation property with respect to E. Let CNpiec( ;Rm) such thatx, D[N1](x),DN(x)

E int K, a.e. in ;then there exists (a dense set of) u + WN,0 (;Rm) such that

Fi

x, D[N1]u(x), DNu(x) = 0, i = 1, 2, . . . , I 1, a.e. x ,

Fi

x, D[N1]u(x), DNu(x) 0, i = I1 + 1, . . . , I , a.e. x .

Remark 2.3. (1) This result has been proved by Dacorogna and Marcellini [6] (Theorem 6.3) when I = I1. With substantiallythe same proof it was found by Dacorogna, Marcellini and Tanteri [7].

(2) An interesting case of constraints is when I = I1 + 1 andFI1+1( ) = det

(i.e., det 0). It is actually this constraint that will be used in two of the examples below.

If the set E is given by only one equation the theorem takes a simpler form.

Theorem 2.4. Let Rn be open. Let F: Rn Rnn R be continuous and quasiconvex. Assume that { Rnn: F(x,s,) 0, det > 0} is bounded in Rnn uniformly with respect to x and s in a bounded set ofRn. If C1piec( ;Rn) is such that

F

x,(x),D(x) 0, a.e. x ,

det D(x) > 0, a.e. x ,then there exists (a dense set of) u + W1,0 (;Rn) such that

Fx,u(x),Du(x) = 0, a.e. x ,det Du(x) > 0, a.e. x .

Remark 2.5. The fact that we can treat strict inequalities follows from the observation that, by hypothesis, we can find > 0such that det D > since C1piec. The remaining part of the proof follows from the next theorem.

We have a generalization of the above theorem.

Theorem 2.6. Let Rn be open. Let F, : RmMs RmnN

s R be continuous and respectively quasiconvex andquasiaffine. Assume that

RmnNs : F(x ,s, ) 0, (x,s,) 0

is bounded in Rm

nN

s

uniformly with respect to x

ands in a bounded set ofRm

M

s

. If

CN

piec(

;Rm) is such that

F

x, D[N1](x),DN(x) 0, a.e. x ,

x, D[N1](x),DN(x)

< 0, a.e. x ,then there exists (a dense set of) u + WN,0 (;Rm) such that

F

x, D[N1]u(x),DNu(x) = 0, a.e. x ,

x, D[N1]u(x), DNu(x) 0, a.e. x .

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Proof. We will do the proof when is affine and when there is no dependence on lower-order terms, i.e.,

E = RmnNs : F() = 0, () 0.The general case follows as in [6].

Step 1: We first prove that

Rco E= RmnNs : F() 0, () 0.

Indeed call X the right-hand side. It is clear that E X and that X is rank one convex; we therefore have Rco E X. We nowshow the reverse inclusion. Let X be fixed and assume that F() < 0, otherwise E and the result is trivial. Since isrank one affine, we have for every , a matrix of rank one, that for every tR

( + t ) = () + tD();

.

We therefore choose a matrix of rank one so thatD(); = 0

(in the preceding theorem () = det + and D() = adjn1) and this leads to the desired identity( + t ) = (), tR.

By compactness ofE we deduce that we can find t1 < 0 < t2 so that

F (+ t ) < 0, t (t1, t2), F ( + ti ) = 0, i = 1, 2.We can therefore rewrite

= t2t2 t1

(+ t1) +t1

t2 t1(+ t2)

which leads to Rco E. Note for further reference that we easily obtain

intRco E RmnNs : F() 0, ( )< 0.Step 2: We wish now to show that Rco E has the relaxation property with respect to E, i.e., that for every bounded open set

Rn, for every u, an affine function with DNu(x) = , satisfying

DNu(x)

intRco E,

there exists a sequence u CNpiec( ;Rm) such that

u u + WN,0

;Rm,u

u in W

N,,

DNu (x) intRco E, a.e. in ,

F

DNu (x)

dx 0 as .

IfF() = 0, we choose u = u. So from now on we can assume that F ( ) < 0 and ( ) < 0. Using then the compactnessassumption we can find as in Step 1, a matrix of rank one, t1 < 0 < t2 such that (we let t = + t )

( + t ) = ( ) < 0,F (+ t ) = F (+ t ) < 0, t (t1, t2),

F (+ t1) = F (+ t2) = 0.The approximation lemma (cf. Lemma 6.8 of Dacorogna and Marcellini [6]) with A = t1+ and B = t2 for small enoughand

= t2 t2 t1 2

A + (t1 + )t2 t1 2

B

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leads immediately to the result. Note that in this lemma since rank [A B] = 1(A) = (B) = ( )< 0

and the constructed function satisfies

dist

DNu (x), co{A, B}

a.e. in

we deduce that by choosing sufficiently small we have (Du ) < 0 and F(Du ) < 0 which implies thatDNu (x) intRco E

as wished. We may then apply Theorem 2.2 to get the result.

The more difficult question is to know when the relaxation property holds if the set E is given by more than one equation.One such case is the following theorem that uses the notion of approximation property (cf. Theorem 6.14 in [6]).

Definition 2.7 (Approximation property). Let E K(E) Rn RmMs RmnN

s . The sets E and K(E) are said to have theapproximation property if there exists a family of closed sets E and K(E ), > 0, such that

(1) E K(E) int K(E) for every > 0;(2) for every > 0 there exists 0 = 0() > 0 such that dist(; E) for every E and [0, 0];(3) if int K(E) then K(E ) for every > 0 sufficiently small.

Theorem 2.8. LetE Rn RmMs RmnN

s be closed and bounded uniformly with respect to x Rn and whenevers vary ona bounded set ofRmMs andRco E has the approximation property with K(E) = Rco E, then it has the relaxation propertywith respect to E.

As a corollary we obtain (cf. Corollary 6.18 in [6] or [7]).

Corollary 2.9. Let Rn be open. LetFi :RmnN

s R, i = 1, 2, . . . , I , be quasiconvex and let

E = RmnNs : Fi ( ) = 0, i = 1, 2, . . . , I 1, Fi ( ) 0, i = I1 + 1, . . . , I .Assume that Rco E is compact andRco E = co E. Let CNpiec( ;Rm) verify

DN(x) E intRco E, a.e. x ,or WN,(;Rm) satisfy

DN(x) compactly contained in intRco E, a.e. x .

Then there exists (a dense set of) u + WN,0 (;Rm) such that

Fi

DNu(x) = 0, a.e. x , i = 1, . . . , I 1,

Fi

DNu(x) 0, a.e. x , i = I1 + 1, . . . , I .

3. The incompressible case

Comparable results to those of the present section are obtained by Mller and Sverak [14] using the ideas of Gromov onconvex integration; here we show how the method of Dacorogna and Marcellini in [6] can also be applied.

In the present section and in Section 9 and 10 we will consider subsets E or K of the manifold det = 1, so when we willwrite int K we will mean the interior relative to the manifold.

We now adapt the definitions of the relaxation and the approximation properties to the present context. Here we give the firstone under a slightly more restrictive form in order to avoid some technicalities. We first let for > 0 and Rn an open set,W be the set of functions u C1piec( ;Rn) such that there exists an open set so that meas( ) < and u ispiecewise affine in . We could consider a more general set but the proof is then more involved.

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Definition 3.1 (Relaxation property). Let E, K Rn Rn Rnn. We say that K has the relaxation property with respect toE if for every bounded open set Rn, for every u, an affine function with Du(x) = , satisfying

x, u(x),Du(x) int K,

there exists a sequence u W1/ such that

u

u+

W1,0 ;Rn,

u

u in W1,,

x, u (x),Du (x) E int K, a.e. in ,

dist

x, u (x),Du (x); Edx 0 as .

The following theorem is the main abstract existence theorem.

Theorem 3.2. Let Rn be open. LetFi : Rn Rnn R, Fi = Fi (x,s,), i = 1, 2, . . . , I , be continuous with respectto all variables and quasiconvex with respect to the variable . LetE Rn Rn Rnn be such that

E

= (x,s,) Rn

Rn

Rnn: Fi (x,s,)

=0, i

=1, 2, . . . , I 1, Fi (x,s,) 0, i

=I1

+1, . . . , I , det

=1.Assume that Rco E has the relaxation property with respect to E and that it is bounded uniformly for x and whenever s

vary on a bounded set ofRnn. Let be an affine function such thatx,(x),D(x)

E intRco E, in ;then there exists (a dense set of) u + W1,0 (;Rn) such that

Fi

x,u(x),Du(x) = 0, i = 1, 2, . . . , I 1, a.e. x ,

Fi

x,u(x),Du(x) 0, i = I1 + 1, . . . , I , a.e. x ,

det Du(x) = 1, a.e. x .

Proof. We will make the proof when there is no dependence on lower order terms, otherwise use the standard procedure in [6].

Step 1: We first observe that Rn

can be assumed bounded, without loss of generality. We then let V be the set offunctions u so that there exists u W1/ , u = on and Du E intRco E a.e. such that u u in L() as .

Note that V and V is a complete metric space when endowed with the C0 norm. Note that by weak lower semicontinuitywe have

V u + W1,0 (;Rn): Fi (Du(x)) 0, i = 1, 2, . . . , I , a.e. x , det Du(x) = 1, a.e. x .Step 2: Let for u V

L(u) =I1

i=1

Fi

Du(x)

dx.

Observe that by quasiconvexity ofFi we have for every u V

liminfus

u,usV

L(us )L(u). (3)

We next immediately see that for every u V (recall that in V we have det Du = 1 and Fi (Du) 0, i = I1 + 1, . . . , I )L(u) = 0 Du(x) E, a.e. in . (4)

We then let

Vk =

u V: L(u) > 1k

.

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We have that Vk is open (cf. (3)). Furthermore it is dense in V. This will be proved in Step 3. If this property has been establishedwe deduce from Baire category theorem that

Vk is dense in V. Thus the result by (4).

Step 3: So it remains to prove that for any u V and any > 0 sufficiently small we can find u Vk so thatu u .

We will prove this property under the further assumption that, for some > 0, small, u W and

Du(x) E intRco E, a.e. in .The general case will follow by definition ofV. By working on each piece where u is affine and by setting u = u on we can assume that u is affine in . The result now follows at once from the relaxation property.

As usual if the set E is defined by only one equation the relaxation property is easier to establish and we therefore have as afirst consequence of the theorem.

Theorem 3.3. Let Rn be open. LetF: Rn Rnn R be continuous and quasiconvex and coercive, with respect tothe last variable , in any direction, uniformly with respect to x ands in a bounded set ofRn. If is affine and is such that

F

x,(x),D(x) 0, a.e. x ,

det D(x) = 1, a.e. x ,then there exists (a dense set of) u + W1,0 (;Rn) such that

F

x,u(x),Du(x) = 0, a.e. x ,

det Du(x) = 1, a.e. x .

Proof. We first let

0 =

x : Fx,(x),D(x) = 0,1 = 0 =

x : Fx,(x),D(x) < 0

and observe that, by continuity, 0 is closed and hence 1 is open; we therefore need only to work on this last set, since in 0we can choose u = .

We may now apply the abstract theorem with

E = (x,s,) Rn Rn Rnn: F(x ,s, ) = 0, det = 1,K = Rco E = (x,s,) Rn Rn Rnn: F(x,s,) 0, det = 1.

The proposition below ensures that all the hypotheses of the abstract theorem are satisfied and therefore the theorem isproved.

Proposition 3.4. Let F:Rn Rn Rnn R be continuous and rank one convex and coercive, with respect to the lastvariable , in any direction, uniformly with respect to x ands in a bounded set ofRn. Let

E = (x,s,) Rn Rn Rnn: F(x,s,) = 0, det = 1.Then

Rco E = (x,s,) Rn Rn Rnn: F(x ,s, ) 0, det = 1.Furthermore Rco E has the relaxation property with respect to E.

Proof. We will do the proof when there is no dependence on lower-order terms, i.e.,

E = Rnn: F() = 0, det = 1.Step 1: We now prove that

Rco E = Rnn: F() 0, det = 1.

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Indeed call X the right-hand side. It is clear that E X and that X is rank one convex; we therefore have Rco E X. We nowshow the reverse inclusion. Let X be fixed and assume that F ( ) < 0, otherwise E and the result is trivial. We can thenfind Rnn a matrix of rank one so that

det(+ t ) = det = 1for every tR. This is easy and follows from the following observation ( being a matrix of rank one)

det(+ t ) = det + tadjn1; ;so choose so that

adjn1; = 0. (5)By compactness ofE we deduce that we can find t1 < 0 < t2 so that

F (+ t ) < 0, t (t1, t2),F (+ ti ) = 0, i = 1, 2.

We can therefore rewrite

= t2t2 t1

(+ t1) +t1

t2 t1(+ t2)

which leads to

Rco E.Step 2: We wish now to show that for every bounded open set Rn, for every u, an affine function with Du(x) = ,

satisfying

Du(x) intRco E,there exists a sequence u W1/ , such that

u u + W1,0 (;Rn),u

u in W

1,,

Du (x) intRco E, a.e. in ,

F

Du (x)

dx 0 as .

IfF() = 0, we choose u = u. So from now on we can assume that F() < 0 (and det = 1). Using then the coercivityassumption we can find as in Step 1, a matrix of rank one, t1 < 0 < t2 such that (we let t = + t )

det(+ t ) = det = 1,F (+ t ) = F (+ t ) < 0, t (t1, t2),

F (+ t1) = F (+ t2) = 0.So let be large enough. We can then find = () > 0 so that

F (t), t [t1 + 1/,t2 1/].Call A = t1+1/ , B = t21/ , (x) = u(x) and observe that

det A = det B = 1, rank{A B} = 1 and D = = t2 1/t2

t1

2/A + (t1 + 1/)

t2

t1

2/B.

By continuity ofF we can find = ( = ()) > 0 so thatdist

, co{A, B} F()/2.

Therefore apply approximation Lemma 4.1 with < min{1/,} and call u the function that is found in the lemma, to get theresult.

The next question we discuss is to know when the relaxation property holds if the set E is defined by more than one equation.The question is more involved and we need, as in Section 2 or as in [6], the so called approximation property.

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Definition 3.5 (Approximation property). Let E K(E) Rn Rn Rnn. The sets E and K(E) are said to have theapproximation property if there exists a family of closed sets E and K(E ), > 0, such that

(1) E K(E) int K(E) for every > 0;(2) for every > 0 there exists 0 = 0() > 0 such that dist(; E) for every E and [0, 0];(3) if int K(E) then K(E ) for every > 0 sufficiently small.

Theorem 3.6. Let E Rn Rn Rnn be closed and bounded uniformly with respect to x Rn and whenever s vary ona bounded set ofRn andRco E has the approximation property with K(E ) = Rco E, then it has the relaxation property withrespect to E.

Proof. We will make the proof when there is no dependence on lower-order terms. We therefore are given Rn, a boundedopen set and u, an affine function with Du(x) = , with intRco E and we wish to show that there exists a sequence u Wsuch that

u u + W1,0

;Rn,u

u in W1,,

Du (x) E intRco E, a.e. in ,

dist

Du(x); E

dx 0 as 0.

(6)

By the approximation property we have, for some , that Rco E and hence that RJ co E for a certain J. We thenproceed by induction on J.

Step 1: We start with J = 1. We can therefore writeDu = = t A + (1 t)B, rank{A B} = 1, with A, B E .

We then use the approximation Lemma 4.1 to find (setting = A B ) u Wu u near ,u u ,

Du(x) =

A in A,B in B ,

det Du (x) = det A = det B, in ,distDu(x); Rco E , in ,

where we have used the fact that

co{A, B} Rco E .The claim (6) follows by choosing and smaller if necessary.

Step 2: We now let for J > 1

RJ co E .Therefore there exist A, B Rnn such that

= tA + (1 t)B, rank {A B} = 1, A, B RJ1 co E .We then apply the approximation Lemma 4.1 and find that there exist a vector-valued function v W/2 and A, B disjointopen sets such that

meas

(A B ) /2 meas ,

v u near ,v u /2,

Dv(x) =

A in A,B in B ,

dist

Dv(x); Rco E , in .

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We now use the hypothesis of induction on A, B and A, B . We then can find A, B , vA W/4 in A, vB W/4 inB satisfying

meas

A A, measB B /4 meas ,vA v near A, vB v near B ,vA v /2 in A, vB v /2 in B ,distDvA(x); E , a.e. in A,dist

DvB (x); E

, a.e. in B ,

dist

DvA(x); Rco E , a.e. in A,

dist

DvB (x); Rco E , a.e. in B .

Letting = A B andu(x) =

v(x) in (A B ),vA(x) in A,vB (x) in B

we have indeed obtained (6) by choosing and smaller if necessary, and thus the result.

4. The approximation lemma

The following result is due to Mller and Sverak [14] and is an extension of a classical lemma (cf. for example, Lemma 6.8in [6]) to handle constraint on the determinants. For the convenience of the reader we will give the proof of Mller and Sverakwith however a slight variation.

Lemma 4.1. Let Rn be an open set with finite measure. Let t [0, 1] and A, B Rnn with rank{A B} = 1 anddet A = det B > 0. Let be such that

D(x) = t A + (1 t)B, x .Then, for every > 0, there existu C( ;Rn) and disjoint open sets A, B , so that

|meas A tmeas |, |meas B (1 t) meas | ,u near ,u ,

Du(x) =

A in A,

B in B,

det Du(x) = det A = det B, in ,dist

Du(x), co{A, B}

, in .

Remark 4.2. By co{A, B} = [A, B] we mean the closed segment joining A to B .

Proof. We divide the proof into two steps.Step 1: We first assume that

D = t A + (1 t )B = Iand hence det A = det B = 1. We also assume (these assumptions will be removed in Step 2) that the matrix has the form

A B = e1where e1 = (1, 0, . . . , 0), = (0, 2, . . . , n) Rn, i.e.,

A B =

0 0 . . . 0

2 0 . . . 0. .. . .. . . . . . .

n 0 . . . 0

Rnn.

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We can express as union of cubes with faces parallel to the coordinate axes and a set of small measure. Then, by posingu on the set of small measure, and by homotheties and translations, we can reduce ourselves to work with equal to theunit cube.

Let be a set compactly contained in and let C0 () and L > 0 be such that

meas( ) /2,0 (x) 1,

x

,

(x) = 1, x ,D(x)L/, x ,D2(x)L/2, x .(7)

Let us define a C function v : [0, 1] R in the following way: given > 0, divide the interval (0, 1) into three finite unionsIA, IB , J of disjoint open subintervals such that

IA IB J = [0, 1],meas IA = t , meas IB = 1 t ,

v(x1) =

(1 t ) if x1 IA,t ifx1 IB ,

v(x1) t, (1 t), x1 (0, 1),v(x1), v(x1) , x1 (0, 1).We then let

A = {x : x1 IA}, B = {x : x1 IB }and observe that by choosing small enough we havemeas A tmeas , meas B (1 t ) meas .We next define V :Rn Rn by

V(x) = v(x1)(x)(0, 2, . . . , n) + v(x1)

n

i=2i

xi, 2

x1, 3

x1, . . . , n

x1

.

Note that V C and has the following properties (where has been chosen sufficiently small)div V 0 in ,V 0 near ,|DV v e1|, |V| 2 in ,

DV =

(1 t ) e1 in A,t e1 in B .

We can now define u as the flow associated to the vector field V (this is the usual procedure to construct a volume preservingmap), i.e.,

d

dsu(s,x) = Vu(s,x), s [0, 1], u(0, x) = x. (8)

The map u(x) = u(1, x) has all the claimed properties, as will now be shown.(1) Indeed since V 0 near , we have by uniqueness of the solution of the differential system thatu(s,x) x, s [0, 1]

and hence the boundary condition for u is satisfied (recall that by hypothesis we are considering the case (x) = x).(2) Since we have |V| 2 we deduce that

u(s,x) x = s

0

V

u(,x)

d

2s, s [0, 1], x . (9)

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(3) Ifx A B then by uniqueness of solutions we find

u(s,x) = x + sv (x1), s [0, 1],

and, hence,

Du(x) = Dx u(1, x) = I+ v(x1) e1 = A in A,B in B .

(4) Since div V 0 in we have automatically (cf., for example [2], p. 28)

det Dx u(s,x) 1, s [0, 1], x .

(5) Finally, it remains to show that

dist

Du(x), co{A, B} a.e. in .We first set

L(x) = v(x1)(x) e1.

Returning to (8) we get

d

ds Dx u(s,x) = DVu(s,x)Dx u(s , x), s [0, 1], Dx u(0, x) = Iand we compare the solution of this system with the one of

d

dsF(s,x) = L(x)F(s,x), s [0, 1], F (0, x) = I.

Using the properties ofV and (9) we get thatF(s,x) Dx u(s,x) , s [0, 1], x .The conclusion then follows from the observation that

F(s,x) = esL(x) = I+ sv (x1)(x) e1, s [0, 1],

and the facts that [

0, 1], v

[t, (1

t )

].

Step 2: We now consider the general case. Since A B is a matrix of rank one we can find a, b Rn (replacing a by |b|awe can assume that |b| = 1) such that

C1A C1B = a b,

where C = D = t A + (1 t )B. We can then find R = (rij) SO(n) Rnn (i.e., a rotation) so that b = e1R and hencee1 = bRt. We then set = R andA = RC1ARt and B = RC1BR t.We observe that by construction, setting = Ra, we have

t

A + (1 t )

B = I,

A

B = e1.

Note that this implies in particular that 1 = 0 (since det A = det B = 1) and hence = (0, 2, . . . , n). We may thereforeapply Step 1 to and to (y) = RC1(Rty) and find A, B andu C( ;Rn) with the claimed properties. By settingu(x) = CR tu(R x), x ,A = RtA, B = RtB

we get the result by recalling that

Du(x) = CR tDu(Rx)R.

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5. The complex eikonal equation

We now discuss Example 1.1. We will first derive a theorem on the rank one convex hull and then go back to the differentialequation. First recall the notation of Appendix A. For a matrix Rmn,

=

11 1n...

.

.

.

m1 mn

=

1

...

m

= (1, . . . , n)

we let 0 1( ) mn( ) be its singular values.We then have the following

Theorem 5.1. (1) Letm n, r1, . . . , r m > 0 and

E = Rmn: i; j= ri rjij,where ij is the Kronecker symbol. Let

A = diag

1

r1, . . . ,

1

rm

Rmm

then

E = Rmn: i (A) = 1, i = 1, . . . , m, Rco E = co E = Rmn: m(A) 1.(2) Letm n, r1, . . . , rn > 0 and

E = Rmn: i ; j = ri rjij,where ij is the Kronecker symbol. Let

A = diag

1

r1, . . . ,

1

rn

Rnn

then

E = Rmn: i (A) = 1, i = 1, . . . , n, Rco E = co E = Rmn: n(A) 1.Remark 5.2. (1) The first case with m

=2 will apply to the complex eikonal equation.

(2) If we consider the second case with m = 3, n = 2, r1 = r2 and

= Du(x,y) =u1x u1yu2x u2y

u3x u3y

then Du E means, in geometrical terms, that the surface has been parametrized globally by isothermal coordinates.

(3) The case m = n = 2 has been established in [6]. Recently Bousselsal and Le Dret [1] (still when m = n = 2), in thecontext of nonlinear elasticity, found that (cf. their Theorem 3.11 with = 0) ifr1 = r2 = 1 then

F = R22: i 2 + 1; 2 1, i = 1, 2 Rco E.This is of course compatible with the result of [6] and of the above theorem. Note however that F = Rco E since, for example,

=13

0

23

0 Rco E, / F.Proof. Obviously the two cases are transposed one from each other and we therefore will only deal with the second one. Westart by letting

X = Rmn: n(A) 1.Step 1: We first prove that Rco E co E X. The first inclusion Rco E co E always holds and the second one follows

from the following two observations.

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First we note that the set X is convex since the function n(A) is convex (cf., for example, Lemma 7.10 in [6]). Nextobserve that the inclusion E X holds. Indeed if E (note that A = (1/r1, . . . , n/rn)) then

E (A)i; (A)j= ij A O(m,n) (A) = 1, 1 n.Step 2: We now discuss the reverse inclusions X Rco E co E. Let X. Replacing by A we can assume, without

loss of generality, that A = Inn. Applying Theorem A.4, we can find R O(n) such that

R = = 1, . . . ,n, with i ;j= i jij, i ( ) = i .Since the sets E and X are invariant under the (right) action of O(n) we will assume, without loss of generality, that

= (1, . . . , n), with i; j = 0, i = j, i ( ) = |i | 1, 1 i n.It is therefore sufficient to show that such belongs to Rco E.

Assume first that |i | > 0, i = 1, . . . , n and write

= (1, . . . , n) =1 + |1|

2+ + 1 |1|

2 = 1 + |1|

2

1

|1|, 2, . . . , n

+ 1 |1|

2

1|1|

, 2, . . . , n

.

Note that rank{+ } = 1 and that if = (1 , 2 , . . . ,

n ) then

i

; j = 0, i = j,

1

= 1,

i 1, i = 2, . . . , n .

Iterating the procedure with the second component and then with the other ones we conclude that can be written as a rank oneconvex combination of elements of E, i.e., Rco E.

The case where some of the |i | = 0 is handled similarly. For example if1 = 0 we write

= (1, . . . , n) =1

2+ + 1

2 = 1

2(e1, 2, . . . , n) +

1

2(e1, 2, . . . , n),

where e1 is any vector ofRm such that

|e1| = 1, i; e1 = 0, i = 2, . . . , n .Iterate again the procedure to deduce that Rco E, as claimed. This concludes the proof.

We can finally apply the results in Section 2 to obtain the following existence theorem for the complex eikonal equation (thecase n

=2 is already in [6]).

Corollary 5.3. Let Rn be a bounded open set, f : R R R, f = f(x,u,v), a continuous function and W1,(;C). Then there exists w W1,(;C) satisfying

ni=1

w2xi + f2 = 0, a.e. in ,

w = , on ,(10)

where wxi = w/xi . Or, in other words, there exists (u,v) W1,(;R2) such that

|Dv|2 = |Du|2 + f2, a.e. in ,Dv; Du = 0, a.e. in ,(u,v)

=(

1,

2), on .

Proof. In fact we solve a more restrictive problem of the type of the above theorem, i.e.,

|Du|2 = r2, a.e. in ,|Dv|2 = r2 + f2, a.e. in ,Dv; Du = 0, a.e. in ,(u,v) = (1, 2), on ,

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where r > 0 is chosen so large that

2(AD) 1 for > 0, i.e.,

D(x) compactly contained in intRco E, a.e. x .

We may then apply Corollary 2.9 with F1 = 1 + 2 2, F2 = 2 1 (in case f is constant, otherwise proceed as in [6]).

6. A problem of optimal design

We will denote, in this section, the set of 2 2 symmetric matrices by R22s . Our algebraic result is as follows.

Theorem 6.1. Let

E = R22s : trace {0, 1}, det 0,then

Rco E = co E =

R22s : 0 trace 1, det 0

, intRco E =

R22s : 0 < trace < 1, det > 0

.

Remark 6.2. Note that it is slightly surprising that the rank one convex hull is in fact convex since the function det is notconvex.

Proof. We call

X = R22s : 0 trace 1, det 0, Y = R22s : 0 < trace < 1, det > 0.Step 1: We first prove that

Rco E co E X.The first inclusion always holds and the second one follows from the fact that E X and that X is convex. Indeed let , X,0 t 1 we wish to show that t + (1 t ) X. It is clear that the first inequality in the definition of X holds since

trace is linear. We now show the second one. Observe first that since det

=1122

212, det

=1122

212 0 and

trace , trace 0 then 11, 22, 11, 22 0 and we therefore have (assume below that 11, 11 > 0 otherwise the inequalitybelow is trivial)

; 1122 + 1122 21212 11 21211

+ 1121211

21212 =(1112 1112)2

1111 0.

We therefore deduce that

det

t + (1 t )= t2 det + t (1 t ); + (1 t)2 det 0.Step 2: We now prove that

X Rco E.Since X is compact, as usual, it is enough to prove that X Rco E. However it is easy to see that

X = E R22s : 0 trace 1, det = 0and therefore the proof will be completed once we will show that the second set in the right-hand side is in Rco E. Assume that is such that 0 < t = trace < 1 and det = 0. We can then write

=

x

x(t x)x(t x) t x

= t 1 + (1 t)2 = t

(1 )

(1 ) 1

+ (1 t)

0 00 0

,

where x = t . The result follows from the facts that 1, 2 E and det(1 2) = 0.Step 3: The fact that Y = intRco E is easy.

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Combining the above theorem with Corollary 2.9 we get

Theorem 6.3. Let R2 be an open set and C2piec( ) satisfy

0(x) 1, a.e. x ,det D2(x) > 0, a.e. x ,

or W2,() such that(x) 1 , a.e. x ,det D2(x) , a.e. x ,

for some > 0; then there exists w + W2,0 ()

w(x) {0, 1}, a.e. x ,det D2w(x) 0, a.e. x .

Remark 6.4. The above theorem has been proved (except the case with W2, boundary data) in Theorem 3.12 in [6] using themethod of confocal ellipses of Murat and Tartar. However the proof we have here (cf. also [7]) relies on the abstract existenceresult of Section 2 and on the algebraic theorem above. Of course the use of the abstract theorem is more flexible, because we

could imagine, for example, to replace 1 by a function a(x,u,Du), which is out of reach by the explicit method.

7. A first academic example

Inspired by the two preceding examples we look to the problem (recalling that we denote the singular values of a matrix R22 by 0 1( ) 2( )).

Theorem 7.1. Let0 , < 1 and

E = R22: 1( ) + 2( ) = 1 , det then

Rco E= R22: 1( ) + 2( ) 1 , det ,

intRco E = R22: 1( ) + 2( ) < 1 , det > .Remark 7.2. Note that if (1 )2 4 < 0 then E = . It can also be proved that if f( ) = 1( ) + 2( ) and g() =||2 2det then

co E = R22: f( ) 1 ,g() (1 )2 4,which in the case = 0 takes the simpler form

co E = R22: 1( ) + 2( ) 1 .Proof. We let

X =

R22: 1( ) + 2( ) 1 , det .

The fact that Rco E X is elementary since E X and the functions 1( ) + 2( ) and det are polyconvex. Sowe only need to show the converse inclusion. The compactness of X implies that the result will be proved if we can show thatX Rco E. Since

X = E R22: 1( ) + 2( ) 1 , det = ,we only need to show that any R22 with 1( ) + 2() < 1 and det = belongs to Rco E. Choose R22 beany matrix of rank one such that

; 1122 + 1122 1221 2112 = 0.

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Define then for tRt = + t

and observe that by construction det t = det = . Using again the compactness argument we can find t1 < 0 < t2 such thatt1 , t2 E, i.e.,

1(ti ) + 2(ti ) = 1 , i = 1, 2;the result then follows at once (the representation formula for intRco E is easily deduced).

As a corollary we obtain

Corollary 7.3. Let R2 be an open set and C1piec( ) be such that1(D) + 2(D) < 1, det D > 0, a.e. in .

Then there exists u + W1,0 () satisfying1(Du) + 2(Du) = 1, det Du > 0, a.e. in .

Proof. This theorem follows either directly from Theorem 2.4 or via the approximation property as done below. First find0 > 0 so that

det D 0 > 0.

Let 0 and

E =

R22: 1( ) + 2( ) = 1 , det

(E = E0 ) hence according to the above theorem we have

Rco E =

R22: 1( ) + 2( ) 1 , det

.

We may then combine Theorem 2.2 and Theorem 2.8 to get the result.

8. A second academic example

We will now consider Example 1.4 and we will compute its rank one convex hull. Recall that the set of 2 2 symmetricmatrices is denoted by R22s .

Theorem 8.1. Letaij > 0, i,j= 1, 2, with a12 = a21. Let

E = = (ij) R22s : |ij| = aij, i,j = 1, 2,then

co E = = (ij) R22s : |ij| aij, i,j= 1, 2.(1) Ifa11a22 a212 < 0 then

Rco E =

= (ij) R22s : |12| = a12, |11| a11, |22| a22

.

(2) Ifa11a22 a212 = 0 then

Rco E = = (ij) R22s : |ij| aij, i,j= 1, 2, |a2211 a1122|det = 1122 + 212.(3) Ifa11a22 a212 > 0 then

Rco E = (ij) R22s : |ij| aij, i,j= 1, 2, |a2211 a1122| a11a22 a212 det .Remark 8.2. (1) If we consider the 2 2 matrix 0 then it is clear that in Case 1: 0 / Rco E, while in Case 2: 0 Rco E but0 / intRco E. It can be shown, however, that in Case 3: 0 intRco E.

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(2) To apply the above theorem to partial differential equations one needs that int Rco E = ; this does not happen in Case 1,contrary to the two other cases. However we need also the approximation property (cf. Definition 2.7) of the rank one convexhull that we are not able, at the moment, to prove.

(3) In Case 3 we were not able to find a complete characterization of Rco E; the set given in the right-hand side of theinclusion is too large.

Proof. The representation formula for the convex hull is trivial.Case 1: We denote by

X = = (ij) R22s : |12| = a12, |11| a11, |22| a22.(1) It is clear that X Rco E. Indeed write any X (assume without loss of generality that 12 = a12) as

=

11 a12a12 22

= a11 + 11

2a11

a11 a12a12 22

+ a11 11

2a11

a11 a12a12 22

and similarlya11 a12

a12 22

= a22 + 22

2a22

a11 a12a12 a22

+ a22 22

2a22

a11 a12a12 a22

to deduce that Rco E.

(2) We now show the reverse inclusion. Observe first that trivially E

X. Therefore to get the claimed result it is enough

to show that X is a rank one convex set. So let , X with det( ) = 0 and 0 < t < 1. Note that since , X then(12 12)2 is either 0 or 4a212. The second case cannot happen since we would have

0 = det( ) = (11 11)(22 22) (12 12)2 |11| + |11||22| + |22| 4a212

4a11a22 4a212 < 0,(11)

which is absurd. So the only case that can happen is 12 = 12 (with |12| = a12). The claimed result t + (1 t) X is thenimmediate.

Case 2: As before we call

X = = (ij) R22s : |ij| aij, i,j= 1, 2, |a2211 a1122|det = 1122 + 212.(1) We easily see that E X and that X is a rank one convex (in fact even polyconvex) set since all the functions involved

with the inequalities are polyconvex and thus rank one convex. We therefore have Rco E X.(2) We now discuss the inclusion X

Rco E. We start by observing that if we can show (cf. below) that X

Rco E then

the result will follow. Indeed if int X, since X is compact, we can find for every R22s with rank = 1, t1 < 0 < t2,such that

+ t1, + t2 Xand hence since X Rco E we get that Rco E.

We now wish to show that if X then Rco E. Note first that the last inequality in the definition ofX is equivalent,bearing in mind that a11a22 a212 = 0, to

0 a212 212 (a11 11)(a22 + 22), 0 a212 212 (a11 + 11)(a22 22). (12)Observe that if either |11| = a11 or |22| = a22 then by (12) necessarily |12| = a12. However if|12| = a12 then, by the sameargument as in Case 1, we deduce that Rco E.

So we now assume that |ij| < aij and (since X ) one of the inequalities in (12) is an equality and without loss ofgenerality say the first one while the second one is a strict inequality. If we call

V1 =

= (ij) R22s : a212 212 = (a11 11)(a22 + 22)

,

V2 =

= (ij) R22s : a212 212 = (a11 + 11)(a22 22)

,

Y1 = X V1then relint Y1 (the relative interior ofY1). We can then choose

=

21 12

12 22

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with 1, 2 = 0 so that

+ t V1, tR;

this is always possible by choosing

1 = 1, 2 = a12 + 12a11

11

or more generally any nonzero solution of

22(a11 11) + 21212 21(a22 + 22) = 0.

Then since relint Y1 and Y1 is compact we can find t1 < 0 < t2, such that

+ t1, + t2 Y1.

But Y1 means that either |ij| = aij for a certain i, j and this case has already been dealt with or V2. Hence the onlycase that requires still to be analyzed is when |ij| < aij and V1 V2, i.e. when in (12) the two inequalities are actuallyequalities. Note that any V1 V2 is of the form

= 11

1 a22/a11a22/a11 a22/a11

and thus if we denote by

=

1 a22/a11a22/a11 a22/a11

,

which is a matrix of rank one, we find that

+ t V1 V2, tR.

The usual argument then applies, namely if|ij| < aij and V1 V2 we can find t1 < 0 < t2, such that one of the inequalities|ij| < aij becomes an equality; in which case we conclude that Rco E by the previous steps.

Case 3: The claimed inclusion follows for the same reasons as in Case 2.

9. The case of potential wells

We now discuss how to apply Theorem 3.2 to the case of two potential wells under incompressibility constraint (this resulthas recently been obtained by Mller and Sverak [14] and we show here how our method gives the same result). We start, asusual, with some algebraic considerations.

Proposition 9.1. Let0 < < 1,

=

00 1/

and

E = SO(2)I SO(2).

Let

F() =

(11 22)2 + (12 + 21)2 +

1

11 22

2+

12 +1

21

2,

G() = (11 22)2 + (12 + 21)2 +

1

11 22

2+

12 +1

21

2.

Then F andG are convex and invariant under the (left) action ofSO(2). Moreover

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E =

R22: F() = 1

, G() =

1

2, det = 1

,

Rco E =

R22: F() 1

, det = 1

,

intRco E =

R22: F ( ) < 1

, det = 1

.

Remark 9.2. Note that 1 = diag(1 , 1/(1 )) intRco E and 2 = diag(/(1 ),(1 )/) intRco E for > 0sufficiently small.

Proof. The result follows from the representation obtained by Sverak and Corollary 8.3 of [6].(1) The fact that F and G are convex and invariant under the (left) action of SO(2) is easy.(2) Let us show now that if

X =

R22: F() = 1

, G() =

1

2, det = 1

then E = X. The inclusion E X is easy since

F(I) = F() =1

, G(I ) = G() = 1 2

, det I = det = 1and F, G and det are invariant under the (left) action of SO(2). We now discuss the reverse inclusion. Let X then either

(11 22)2 + (12 + 21)2 = 0

which implies that SO(2) or1

11 22

2+

12 +1

21

2= 0

which implies that SO(2). In either cases we find that E.(3) Call

Y

= R22: F()

1

, det

=1.

We now show that Rco E = Y. To prove this we use the representation formula established by Sverak (cf. [6]), i.e.,

Rco E =

=

y1 y2y2 y1

1 00 1

+

z1 z2z2 z1

00 1/

,

y21 + y22 +

z21 + z22 1, det = 1

.

Expressing y1, y2, z1, z2 in terms ofij we find

11 = y1 + z1

12 =

y2 +1

z2

21

=y2

+z2

22 = y1 +1

z1

1

y1 =

1

11 22

1

y2 = 12 +1

21

1

z1 = (11 22)

1

z2 = (12 + 21)

.

Since

F() =

1

y21 + y22 +

z21 + z22

we get immediately the result.

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(4) We now discuss the representation formula for int RcoE. Call Z the right-hand side in the formula. The inclusionZ intRco E is clear and so we show the reverse one. Let intRco E; then up to a rotation we can always assume that12 = 0 and hence since det = 1 we deduce that if

t =

11 021 + t 1/11

,

then =

0

and

det t 1, tR.Since intRco E we find that t Rco E for all t small enough. Observe finally that the function t F (t) is strictly convex(note however that the function F() is not strictly convex) and therefore ift= 0 is small enough we have

F ( ) 0. Let intRco E

then there exists u W1,(;R2) such that

Du(x) E, a.e. in ,u(x) = x, on .

Remark 9.4. If det A = det B this result was already obtained by Mller and Sverak [13] and Dacorogna and Marcellini [5].

Proof. Step 1: We start with some algebraic considerations. Observe that there is no loss of generality if we assume that

A = I and B = 00 1/ .Indeed first diagonalize BA1, i.e., find Ra , Rb SO(2) so that

Ra BA1Rb = =

00 1/

we therefore deduce that

RbEA1Rb = SO(2)I SO(2).Step 2: We define for (0, 1]

I =

1 00 1/(1 )

, =

/(1 ) 0

0 (1 )/

.

Observe that I , intRco E and ifE = SO(2)I SO(2)

and accordingly F , G , we then have

E intRco E and K(E ) = Rco E intRco E.Therefore E and Rco E have the approximation property with K(E) = Rco E . Hence combining Proposition 9.1 withTheorem 3.6 and Theorem 3.2 we get the result.

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10. The case of nematic elastomers

The problem considered here has been introduced by DeSimone and Dolzmann [9]. We begin with the computation of therank one convex hull; this follows from [6,8] (in the case n = 2, 3 cf. [9]).

Theorem 10.1. Let0 1(A) n(A) denote the singular values of a matrix A Rnn and

E = A: i (A) = ai , i = 1, . . . , n , det A = ni=1

ai,

where 0 < a1 an. The following then holds:

Pco E = Rco E =

A:n

i=i (A)

ni=

ai , = 2, . . . , n , det A =n

i=1ai

.

Moreover if0 < a1 < < an and is sufficiently small so that0 < a1 = (1 )1na1 a2 = (1 )a2 an = (1 )an,

then E andRco E have the approximation property with K(E) = Rco E, where

E = A: i (A) = ai , i = 1, . . . , n , det A =

n

i=1 ai.Remark 10.2. (i) Note that when n = 3 and a1 = a2 = r1/6, a3 = r1/3, r < 1, we recover the result of DeSimone andDolzmann, namely:

Pco E = Rco E = A R33: i (A) r1/6, r1/3, det A = 1.(ii) The hypothesis that all the ai are different is too strong and can be weakened; it is enough to assume that the ai are not

all equal. It is clear also that if all the ai are equal, then intRco E = , since then E = Rco E.

Proof. Let us denote by

X =

A Rnn:n

i=i (A)

n

i=ai , = 2, . . . , n , det A =

n

i=1ai

.

Step 1: The fact that Rco E Pco E X is easy, since E X and the functions

A n

i=i (A)

ni=

ai , = 2, . . . , n ,

are polyconvex and A det A ni=1 ai is quasiaffine.Step 2: As usual by compactness ofX it is enough to prove that X Rco E. We show the result by induction.(1) n = 1. This is trivial.(2) n 2. Any A X can, without loss of generality, be assumed of the form

A =

x1

. . .

xn

with 0 x1 x2 xn, ni= xi ni= ai , = 2, . . . , n , and ni=1 xi = ni=1 ai . Since A X we deduce thatn

i= xi =n

i= ai , for a certain {2, . . . , n}. We can then apply the hypothesis of induction to{x1, . . . , x1} and {a1, . . . , a1}

and to

{x , . . . , xn} and {a , . . . , an}.Indeed for the second one this follows from the hypotheses

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ni=

xi

ni=

ai , = + 1, . . . , n ,n

i=xi =

ni=

ai

while (note that if = 2 then necessarily x1 = a1 and this part is trivial , so we will assume that 3) for the first one we have1

i=xi =

n

i=xi

n

i=xi

1=

n

i=xi

n

i=ai

1

n

i=ai

n

i=ai

1=

1

i=ai , = 2, . . . , 1,

and1i=1

xi =1i=1

ai .

We can therefore deduce, by hypothesis of induction, that A Rco E.Step 3: We now observe that the approximation property follows from the fact that (if 0 < a1 < < an)

intRco E =

A:n

i=i (A) 0 withI

i=1 ti = 1, Ai K

Ai = X, i = 1, . . . , I .The set of extreme points in the polyconvex sense ofK is denoted by Kpext.

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Theorem 11.6. LetK Rmn be a nonempty compact polyconvex set and Kpext be its extreme points in the polyconvex sense.Then there exists :Rmn R=R{+} a polyconvex function so that

Kpext =

x K: (x) = 0, (x) 0 x K.

Proof. We first define

f(x) = |x|2, ifx K,+, otherwise and (x) = Pf (x) f(x), ifx K ,+, otherwise.In the convex case it is the function that is the Choquet function. Observe that :Rmn R=R{+} is polyconvex andthat

(x) 0, ifx K, (x) = 0, ifx Kpext.Indeed the inequality is clear since in K the function f is finite and, by definition, Pf is always not larger than f. We nowshow that

(x) = 0 x Kpext.Note that ifx K then

(x)

= |x

|2

+inf

xiK(m,n)+1

i=1 ti |xi |2: T(x) =+1

i=1 ti T (xi ), ti 0 with+1

i=1 ti = 1.Therefore if x Kpext, we deduce, by definition, that in the infimum the only admissible xi are xi = x; and hence we have(x) = 0. We now show the converse implication, i.e., (x) = 0 x Kpext. From the above representation formula we obtain,since (x) = 0 and x K , that

|x|2 = supxiK

(m,n)+1

i=1ti |xi |2: T(x) =

+1i=1

ti T (xi )

.

Combining the above with the convexity of the function x |x|2 we get that

|x|2 (m,n)+1

i=1ti |xi |2

(m,n)+1i=1

ti xi

2

= |x|2;

the strict convexity ofx |x|2 implies then that xi = x. Thus x Kpext.

We now have the following version of Minkowski theorem.

Theorem 11.7. Let E Rmn be a non empty compact set. Let Epext be the extreme points in the polyconvex sense of Pco E,then

Pco E = Pco Epext.

Proof. We adapt here an idea of Zhang [18].Step 1: We first prove that if K is a compact and polyconvex set then it has at least one extreme point in the polyconvex

sense, i.e., Kpext = . Let co K be the convex hull of K , which is a compact and convex set. It is a well established fact in

convex analysis that co K has then at least one extreme point (in the convex sense). Since, by definition, any extreme point (in

the convex sense) is an extreme point in the polyconvex sense, we deduce the result.Step 2: We next let

K = Pco E; L = Pco Epext.The only nontrivial inclusion is K L. We then define

g(X) =

dist(X; L), ifX K ,+, otherwise, f(X) = Pg(X) 0. (14)

We could also have taken, if we want a function that is finite everywhere,

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g(X) = dist(X; L)(mn)+1, f (X) = Pg(X) 0.Observe that

L = X K: f(X) = 0 (15)(this follows from the polyconvexity ofL). Let

a = maxf(X): X K 0. (16)We will show that a = 0 which by (15) implies K L as claimed. Let

Ea =

X K: f(X) = a = , K = Pco Ea K. (17)Since f is polyconvex and nonnegative we getK X K: 0 f(X) a. (18)It follows from Step 1 that

= Kpext Ea. (19)Assume for a moment that we can show

= Kpext Epext (20)then combining (19), (20) and the fact that

Epext L =

X K: f(X) = 0

we would then deduce that

= Kpext Ea L.This implies at once that a = 0 and hence K L as claimed.

So it only remains to show (20). We thus let Kpext be such thatT ( ) =

Ii

=1

ti T (i ), ti > 0 withI

i

=1

ti = 1, i K,

and we wish to show that

i = , i = 1, . . . , I , (21)which implies that Epext and hence (20). Ordering differently the i K , if necessary, we have

i Ea f (i ) = a i = 1, . . . , I 1,

i KEa f (i ) < a i = I1 + 1, . . . , I .

We first show that I1 = I. If this were not the case we would have from (19) and the polyconvexity off that

a = f( )I

i

=1

ti f (i ) = aI1

i

=1

ti +I1+1i

=1

ti f (i ) < a

which is absurd, thus I1 = I. However since Kpext (where K = Pco Ea ) we therefore deduce that (21) holds and hence Epext which is the claimed inclusion (20).

Acknowledgements

This research has been partially financed by Fonds National Suisse (21-50472.97). We would like to thank A. DeSimone,G. Dolzmann and P. Marcellini for interesting discussions.

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Appendix A. Singular values

We recall here the definition and some properties ofsingular values of matrices (cf. work of Horn and Johnson [11, p. 152]and [6, p. 171]). First we write any A Rmn as

A

=

a11 a1n...

.

.

.

am1 amn

=

a1

.

.

.

am = (a1, . . . , an).We start with the following

Definition A.1. We denote by O(m,n) the set of orthogonal matrices R Rmn, i.e.,RtR = Inn

where Inn denotes the identity matrix in Rnn. When m = n, we write O(n) = O(n,n).

Remark A.2. Ifm = n, then in generalRtR = Inn RRt = Imm

(i.e., R O(m,n)Rt O(n,m)); while ifm = n then these two properties are equivalent (i.e., R O(n) Rt O(n)).

We now give the definition of the singular values.

Definition A.3. (1) Let m n and A Rmn. The singular values ofA, denoted by 0 1(A) m(A), are defined tobe the square root of the eigenvalues of the symmetric and positive semidefinite matrix AAt Rmm.

(2) Let m n and A Rmn. The singular values ofA, denoted by 0 1(A) n(A), are defined to be the squareroot of the eigenvalues of the symmetric and positive semidefinite matrix AtA Rnn.

The following theorem is the standard decomposition theorem (cf. Horn and Johnson [10, Lemma 7.3.1]).

Theorem A.4. (1) Letm n, A Rmn and0 1(A) m(A) be its singular values then there exists R O(m) suchthat

RA = A =a1

...am , with ai ;aj= ai ajij, i (A) = ai .Furthermore there exists Q O(n,m) Rnm (i.e., QtQ = Imm) such that

RAQ =

1(A) 0... . . . ...0 m(A)

.(2) Letm n, A Rmn and0 1(A) n(A) be its singular values then there exists R O(n) such that

AR = A = (a1, . . . ,an), with ai ;aj = |ai ||aj|ij, i (A) = |ai |.Furthermore there exists Q Rnm with QQt = Inn (i.e., Qt O(m,n)) such that

QAR =1(A) 0... . . . ...0 n(A)

.

References

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