Breaking of symmetry for a minimization problem

.Voni,nearAnolysrs. Printed m Great

Theory. Bnrain.

.Werhods&

BREAKING

Applrcnrrons.

Vol.

13. No.

5. pp.

OF SYMMETRY

579-587,

0362-%6X/89 53.00 + SX C 1989 Pergamon Press plc

1989.

FOR A MINIMIZATION

PROBLEM

JEAN-FRANCOIS LEON Centre d‘Etudes

de Gramat,

F46500 Gramat,

France

and GABRIELLA TARANTELLO Department

of Mathematics,

University

of California,

Berkeley,

(Received 10 March 1988; receivedforpublicurion Key words and phrases: Rotational symmetry.

variance,

minimization

1. STATEMENT

problem,

CA 94720, U.S.A.

20 June 1988) Hartree-Fock

model,

breaking

of

OF RESULTS

ONE OF THE interesting problems in nuclear physics is the study of the structure of the groundstate for a nucleus with N-nucleons. A good approximation model for such a problem is given by the Hartree-Fock model, where theNwave functions are the solutions of a suitable minimization problem. This yields a study of minimization problems invariant under the rotation of coordinates. For a small number of nucleons, the density of matter preserves this symmetry. However we learn from nuclear physics that this symmetry may break for a heavy nucleus. We refer to Gogny and Lions [l] for further discussion and references on the phenomenon of “breaking of symmetry’ ’ . Here we study this phenomenon for a model problem proposed by Lions in [2] and reminiscent of the Hartree minimization problem. Let B be the unit ball in lR3.For N E N let ui E HA(B),i= 1, . . . , N. Set

where F: [R++ R is a given function and p = Cr= 1 uf. Consider the following set of constraints s = ((u,, . ..) UN): uj E Hi(B),

(uj, l(k) = S,,j)

where (- , -) denotes the usual scalar product in L'(B). The minimization problem under consideration is the following: inf

@(u1, . . . . UN).

(1)N

(~I....,UN)ES

It is an easy exercise to show that if lim F-(t)/t5'3 = 0 and lim F+(t)/t3 = 0 (F-,F+ are, t-+00 I-++00 respectively, the positive and negative part of F) then (1)N achieves a minimum. More precisely we assume: (i) F E C'[O, +oo)is concave and m((t E IF?+: F"(t)= 0))= 0 where for A c IR,m(A) denotes the Lebesgue measure of A; (ii) ,,“y) IF'1 c n/4. * m

579

J.-F. LEON

580

and G. TARANTELLO

THEOREM1. If F satisfies (i), (ii), there exists a sequence Nk E fN, Nk -+ +m as k + +a, such that if (u,, . . . . u,.,~) is a minimum for (l),+ then the associated density p = C,3, u,? is not radial. Remark. Hypothesis (ii) is only technical and we believe it can be removed. In a perturbation

framework,

for each E E R, set N

inf c ]VUj12+ & F(P), (1% (UI.....UV)ESs B and denote by A, I AZ5 .-. 5 A,, I a-0 the sequence of eigenvalues for -A in B under Dirichlet boundary condition (each repeated according to the multiplicity). We obtain the following. j=

1

‘9

THEOREM1’. Assume F satisfies (i) and

sup IF’\ < +oo. For each N E N with AN = AN+, < IO.+mP) AN+Zthere exists &N> 0 such that V E E IRand ]&I < &Nthe density associated with a minimum of (1); is not radial.

Remark. An

eStirTMte

for &NiS given in (2.13).

THEOREM2. Let F satisfy (i) and lim F’(t)/ts’3 = 0. If the density associated with a solution I-+= of (l)N is radial then the associated kinetic energy is radial too. To clarify the proof of theorem 1 we shall sketch it below. Sketch of theproof of theorem 1. It is given in three steps with an argument by contradiction. Step 1. If (u,, . . . . UN) is a solution of (1)N then the associated Lagrange multipliers ~j j=l , . .., N are the first N eigenvalues (with corresponding eigenfunctions Uj) for the eigenvalue problem: -Au + F’(p)u = ,uu (1) Ujas = 0. Furthermore,

if pN+i denotes the (N + 1)th eigenvalue for (1) then PN < Pi+, .

Step 2. Assume p = I?. 1 u_f is radial. Consider the family of rotational problems: -Au + tF’(p)u = pu i

4s

= 0

invariant eigenvalue (I),

t E [O, I]. Under the given assumptions, there exists an analytic curve p(t) of eigenvalues for (I)( joining any given eigenvalue of (1) with an eigenvalue of -A. Since the representation of SO(J) is irreducible on the eigenspaces of - d, using the rotational invariance of (l), we even obtain that the multiplicity of p(t) can only increase along this curve. Step 3. By a close study of the sequence of eigenvalues for -A and standard comparison arguments we obtain precise information on the multiplicity of p N+l (or gN)_ Hence for a suitable sequence Nk, Nk -+ +cx, as k -+ +co we show how, under the given assumptions, this prevents the existence of the analytic curve of step 2 joining fiN+i (or fiN) with an eigenvalue of -A.

581

Breaking of symmetry for a minimization problem 2. THE PROOF

I

OF THEOREM

Let (u,, . . . . u,) be a minimum for (l),,,. As is well known ui i= Euler-Lagrange equation: -AUi + F’(~)u; = FLU;

1,

. . . , N satisfies the (2.1)

&las = 0 5 puz5 ... I P,,,. Furthermore, @ is positive semidefinite in (u,, . . . . uN) we obtain the following. with pi E I?, and we can always assumep,

since the Hessian of

LEMMA2.1. The sequence pi, i = 1, . . . , N defines the first N eigenvalues for (1). Moreover if pN+, denotes the (N + 1)th eigenvahre for (1) then r(lN< pN+r. Proof.

Let u E Hi(B) satisfy u*=1,

Then v t E IR, vj

UUj = 0

j

jB

vi

=

1, . . . . N.

B

1, . . . . N

=

therefore @tvj,

I)

1

@c”,

t * * * 9 uN)

VtEIR.

Hence (d/dt)@(Vj,31,=, = 0 (i.e. (2.1)) and

(2.2) By straightforward

computation

jB[VU[*

(2.2) gives

+ IBF’(p)U2

+ 21BF”(/?)U2Uf

2 JBIPUj[* + jBF’(p)Uj.

(2.3)

Thus by the concavity of F we obtain i,tVui* V~~f?,‘(B):ll~llLz=

such that

(IuI(L~

=

+ j~F’(~)u’

> P(N

(2.4)

land(u,Uj)=Oj= l,..., N. Furthermore, if (u, P) is an eigenpair of (1) 1 and (u, Uj) = 0 Vi = 1, . . ., N, then (2.4) readily implies p > pN. n

Arguing by contradiction, operators

assume that p is radial. Since F’ is bounded in [0, +co), the linear L, = -A

+ tF’(p)

tEm

define an analytic family of type (A) in the sense of Kato (see Reed and Simon [3]). Hence v to E IR, if ,utOis an eigenvalue for (1),0 with multiplicity m, there exists a neighborhood U of f,, and m analytic curves pi(t) (perhaps not all distinct) defined v t E U of eigenvalues for (1X such that pi(to) = pro. Moreover, Ipi - pi( I f E;wml iF’(Ol b, - hi.

J.-F. LEON and G. TARANTELLO

582

LEMMA 2.2. Let A be an eigenvalue for -A with multiplicity m. There exists a unique analytic curve p(f) of eigenvalues for (l), , t E R starting from h at t = 0. Furthermore each eigenvalue p(t) must have at least multiplicity m.

Proof. The representation of SO(3) on the eigenspace E associated to I is irreducible (see Appendix 1). Assume that there exist two distinct analytic curves p,(t) and ~z(r) with eigenspaces E,(f) and E,(f) starting from A at I = 0. Since p is radial V t > 0, E,(t) and E,(t) define two invariant subspaces for SO(3). Hence if we let I -, 0 we have E,(t) -, E,(O) # (0) and E,(t) + E,(O) # (01. Now E,(O) and E,(O) define two distinct invariant subspaces of SO(3) in E. This clearly contradicts the irreducibility. Thus, for t near 0 there exists a unique analytic curve p(t) such that ~(0) = A. Furthermore p(t) can be uniquely extended to an analytic curve of eigenvalues for (I), V 1. Hence v t, the multiplicity of p(t) is at least m. R From now on, we denote with 1” a simple eigenvalue of (l),=,. We have the following technical result. There exists a sequence Nk E [r\J,Nk --t +oc, as k -+ +a (a) ANkis a multiple eigenvalue for -A;

LEMMA 2.3.

such that

Proof. It follows by well known properties of the zeros of Bessel’s functions, we have collected these properties in Appendix 1. Using standard notation, we denote by J, the Bessel function of order v. As is we11known the eigenvalues of -A in B c R3 are the square of the zeros of the Bessel function J,,*+, with I E h\l. Furthermore, if J,,z+,(v%) = 0 then I has multiplicity 21 + 1. Let 1 < p be two consecutive simple eigenvalues. There exists a multiple eigenvalue p (with multiplicity three) such that

p= -A+fi 2

+ O(1)

as A -+ +co.

(2.5)

In fact, using property (A2) in Appendix 1 with I = 0 (i.e. v = t) we have ,4 = k2n2 7

/.I = (k + 1)‘~’

for some integer k 2 1. Again by (A2) with I = 1 (i.e. v = t) we obtain an eigenvalue ,ii with multiplicity three such that /i =

k%c2+ kn2 + O(1)

for k large.

This readily implies (2.5). Let N > No be two integers. Obviously, N-l AN

-

1N,,

=

c

K=No

t&+,

-

&)

= s1 + s2 + s,

+ s,

(2.6)

Breaking

of symmetry

for a minimization

583

problem

where Si = CkcCi(Ak+r - &) i = 1, . ...4 and Cr = (k, N, 4 k 5 N - 1: Lk is simple); k,N,,sklN-

C,=

;

l:I,ismultipleand~I”simple:O
I 1:I,ismultipleand~15simple:O
I$=

;

C, = (k, No 5 k 5 N - 1: & is multiple k d C, U C,). We shall estimate each S;, i = 1, . . ., 4. Notice, first of all, that if No is large enough, then any given k E (N,, , . . . , N - 1) belongs to only one of the set C;. In fact, if fl > A are consecutive simple eigenvalues then ,D - 1 = O(a) (see (A2) in Appendix 1). Set 01 = lim sup&+,

- &J.

k E CI

We do not know whether or not (Yis finite, and its seems difficult to get good estimates for cr. However, if CY= +co then theorem 1 follows under the more general assumption that sup IF’(t)1 < + CQ. This is shown in Appendix 2. Hence assume that CY< +co. For N,

t E [O.+m)

large enough and (Al) in Appendix 1 we obtain:

+dxp

s, I

(2.7)

Since we have s3=

c ANO 5 XI

tAk+l

c O
-

lk) I

and the term in the bracket is bounded by n/2; again by (A2) we get s3

I

(2.8)

O(q..y).

To estimate S2, write

where AE is the smallest eigenvalue larger than Af + (n/2). Now in view of (2.5) it is clear that 1: - A; I &l;+,

- Af) + O(1)

where Af,, is the first eigenvalue larger than ,lf. Therefore, s2 I

*AN + O&r).

Finally, let p = lim sup&+, k E Cd

- &).

(2.9)

581

J.-F.

LEON and G. TARANTELLO

There exist at most (l/rc)l, distinct eigenvalues of -A below I, Appendix 1). Thus V E > 0 by choosing N,, = N,,(E) large enough

(see (Al) and (A27 in

(2.10) In conclusion, by (2.6)-(2.10) we have that v E > 0 3 N,, = No(c) such that v N > N,, P+c n

>

&V + WV)

so necessarily

We can now conclude the proof of theorem 1. Set /lF’Ij =

sup

I.2 [O,+-I

lF’(t)l. Let E > 0 be such

that IlF’II < (n/4) - (e/2). Consider J and p two consecutive eigenvalues for -A such that A is multiple

,-A>I_E;

(2.11)

(2.12) By lemma 2.3 there exists an infinite sequence of such A’s. Take N E N such that A = AN = Aj.J+,,

We claim that the density associated to the minimization problem (l)N cannot be radial. To see this, notice first that pN+, must be simple. In fact, by lemma 2.1 and standard comparison arguments we have PN < pN+l 5 1 +

IPII < P - ll~‘lI 5

PN+Z.

Moreover by lemma 2.2 we have that the analytic curve p(t) of eigenvalues for (l), through r(lN+tmust start from a simple eigenvalue of -A, i.e. ~(0) = A’. But since

II40 - /WI 5 IPII It - fl it follows:

On the other hand, by the comparison

principle we have

Breaking

of symmetry

for a minimization

problem

585

and so

which is possible in view of (2.12).

W

Remark. If we assume that F has a definite sign, then a sharper use of the comparison principle allows us to replace the assumption llF’j[ < 77/d with IlF’II < n/2. If E. is an eigenvalue for -A with multiplicity s 2 1 and u, , . . . . u, is a complete set of eigenfunctions for d then the partial density

is radial. See lemma 3.1 of next section for a more general proof of this fact. Furthermore, any eigenspace of -A corresponding to a multiple eigenvalue does not contain radial eigenfunctions. This implies that for N E N such that 4V = &+I < &V+* the density associated to the “free”

minimization problem (i.e. F = 0), inf (ut..... UN) Es 1 B i=l! lvu;l*

cannot be radial. By the arguments of theorem 1, we can then extend this fact to the perturbed minimization problem

hence obtain theorem 1’ with EN = inf(lN+2 - AN, AN - Afrr, Ah+, - A,,,)

(2.13)

where JR and AR+, are two consecutive simple eigenvalues such that Ah c A, < &+i . Remark. Although we treated the minimization problem (l)N in fR3the results given here can be generalized in IR”. 3. THE

PROOF

OF THEOREM

2

Let us start with the following. LEMMA 3.1. Assume p radial. Let p be an eigenvalue of (1) with multiplicity s, and let (U1, . . . . u,) be the corresponding complete set of eigenfunctions. Then the partial density

P, =

c4

i=l

is radial.

586

Proof.

J.-F. LEON and G. TARANTELLO

Since p is radial, by separation of variables in the equation - Au + F’(p)u = ,UU

(3.1)

we obtain that any eigenfunction can be written asf,(r)YA(O, $), where YAdenotes a standard harmonic function and f,(r) depends only on 1x1 = r. As is well known, if f,(r)Y,f,,(O, 4) is an eigenfunction for (3.1), so it is for any m: --Is

fi(r)YA(R 4)

m 5 1.

Since p, does not depend on the choice of the orthonormal sponding to p we can write: Ps =

basis in the eigenspace corre-

[F, f:(r) ,nE_IVZAe, 4))’ = C C f:(r) I E I&,

P

for some finite set 1, c N and positive constant C. So p is radial.

n

To conclude the proof of theorem 2, observe that if fi = pi, i = 1, . . . , N, then by lemma 2.1 we have a complete set of eigenfunctions for p in (ur , . . . , u,]. Hence p,,, is radial Vi= 1, . .., N. Furthermore for p = pci, i = 1, . .., N, pB satisfies: -AP,,

+ F’Wp,

= w,

- 2

c lvu,? w, = P

which implies that c lV412 fW= P is radial. Therefore

Cj=, I VUjl’ is radial.

W

Acknowledgemenrs-This work was completed while the first author was visiting the University of California at Berkeley. He wishes to thank the Department of Mathematics for their kind hospitality. REFERENCES GOGNY D. & LIONS P. L., Hartree-Fock method in nuclear physics, RAIRO M*AN 20, 571-637 (1986). LIONS P. L., Symmetry breaking in vector valued minimization problems, in Vuriufional Problems. Longman,

London (1988). REED M. & SIMON B., Mefhods of Mathematical Analysis, IV, Analysis of Operators. Academic Press, New York

(1978). ~‘JAT&N, Theory of Bessel Functions. Academic Press, New York (1944). APPENDIX

I

We recall here well known properties for the Bessel functions. See Watson [4] for details. (1) Let v 2 -i fixed. For x large (independent of v) the number of zeros N,(x) for J. smaller than x satisfy: N,(x) 5 ;.

(Al)

This is well known for v = -i

and more generally follows from the fact that the function v -+ N,(x) is nonincreasing. (2) The simple eigenvalues of -A in B C I?'are the square of the zeros of J,,2 (i.e. (k?r’, k-N)). Denote by j... the ordered sequence of zeros for Jr. For v 2 -i fixed we have

642)

Breaking

of symmetry

for a minimization

problem

587

i,. , 2 v. Let I, I’ be two distinct This last property

implies

immediately

that

the representation

We assume

here that (Y = ,I$t, formulated

I,,,

as follows:

VA

- I,

= +a.

zeros.

of SO(3) on each eigenspace

APPENDIX

equivalently

642’)

then J,,2+, and J,,L+l. have no common

integers,

(A3)

of -A

is irreducible.

2

We are unable

to prove or disprove

this assertion.

This can be

> 0 there exists k E N such that the inequality kn < j,+,,r,k,

< kn

+ 2

cannot occur V (I, k’) E N x N. We shall prove that, if so, then theorem I holds under the weaker assumption llF’ll < tco. Let I < ,u be two consecutive eigenvalues for -A such that p - I > 2IjF’II and I is simple. Moreover by (2.5) we can always assume that if I is large enough then L’ - p > 2/IF’ll where I’ is the smallest simple eigenvalue larger than p. Let NE N be such that p = I,. We shall prove that the density associated to a solution of (l),v cannot be radial. To see this, notice that pIN must be simple. In fact, ti,v+t >,4z4vHence the analytic

curve ~(1) of eigenvalues

IIF’II> ~.v_,+ IIF’II2 PN-,.

for (l), through 1fl.v -

As for theorem

1, one sees that this is impossible.

P,~ must start from a simple eigenvalue

1’15 ll~‘ll.

We leave the details

to the reader.

L’ of

- A, and