Gazzini M., Serra E. - The Neumann problem for the Henon equation, trace inequalities and Steklov eigenvalues(26).pdf

7/25/2019 Gazzini M., Serra E. - The Neumann problem for the Henon equation, trace inequalities and Steklov eigenvalues(2

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The Neumann problem for the Henon equation,

trace inequalities and Steklov eigenvalues

Le probleme de Neumann pour lequation de Henon,

inegalites de trace et valeurs propres de Steklov

Marita Gazzini1, Enrico Serra2

1SissaIsas, Trieste, Via Beirut 4, 34014 Trieste, Italy2Dipartimento di Matematica, Universita di Milano, Via Saldini 50, 20133 Milano, Italy

Abstract. We consider the Neumann problem for the Henon equation. We obtain existence results and

we analyze the symmetry properties of the ground state solutions. We prove that some symmetry and

variational properties can be expressed in terms of eigenvalues of a Steklov problem. Applications are

also given to extremals of certain trace inequalities.

Resume. On considere le probleme de Neumann pour lequation de Henon. On obtient des resultatsdexistence et on analyse les proprietes de symetrie des solutions a energie minimale. Nous demontrons

que certaines proprietes variationnelles et de symetrie peuvent etre exprimees a laide des valeurs propres

dun probleme de Steklov. Nous donnons en outre des applications aux fonctions minimizantes pour

certaines linegalites de trace.

Mathematics subject classification:35J65, 46E35.

1 Introduction and statement of the main results

The elliptic equation appearing in the Dirichlet problem

u= |x|up1 in u >0 in u= 0 on ,

(1)

where is the unit ball ofRN,p >2 and >0, was introduced in the paper [12] by M. Henonand now bears his name. In [12], problem (1) was proposed as a model for spherically symmetricstellar clusters and was investigated numerically for some definite values ofp and .

This research was supported by MIUR Project Variational Methods and Nonlinear Differential Equations.

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In the last few years it became evident that in spite of (or thanks to) its simple appearance, the

Henon equation exhibits very interesting features concerning existence, multiplicity and, aboveall, symmetry properties of solutions.Research has been directed up to now only on the Dirichlet problem (1) with the intent ofclassifying the range of solvability (in p) and especially of analyzing the symmetry properties ofthe ground statesolutions.In order to provide a clear motivation for the results contained in the present paper, we brieflyrecall some of the main achievements concerning problem (1). We denote, as usual, 2 = 2NN2 .

The first existence result is due to Ni, who in [14] proved that for every p (2, 2 + 2N2),problem (1) admits at least one radialsolution and pointed out that it is the presence of theweight |x| that enlarges the existence range beyond the usual critical exponent.The most important question for our results is that of symmetry of solutions. The starting point

is the fact that since the function r r

is increasing, GidasNiNirenberg type results ([11])do not apply, and therefore nonradial solutions could be expected. This is the content of thepaper [18] by Smets, Su and Willem, who studied the ground state solutions associated to (1).They proved in particular the following symmetry breaking result.

Theorem 1.1 (Smets, Su, Willem) For everyp (2, 2) no ground state for problem (1) isradial provided is large enough.

Further results on the Dirichlet problem can be found in [19], [6], [4], [5] for residual symmetryproperties and asymptotic behavior of ground states (for p 2 or ) and in [16], [3],[15] for existence and multiplicity of nonradial solutions for critical, supercritical and slightlysubcritical growth; see also [17] for symmetry breaking results for TrudingerMoser type non-

linearities.We stress once more that all the above results have been obtained for the homogeneous Dirichletproblem, while it seems that so far the Neumannproblem has never been studied. The purposeof this paper is to fill this gap and to point out a series of new, interesting and unexpectedphenomena that arise in passing from Dirichlet to Neumann boundary conditions.To describe our results we let be the unit ball of RN, with N 3, and we consider theNeumann analogue to (1), namely

u + u= |x|up1 in

u >0 in

u

= 0 on ,

(2)

where again p >2 and >0. We have denoted by the outer normal to .Solutions of (2) arise from critical points of the functional Q: H

1() \ {0} R defined by

Q(u) =

|u|2 dx +

u2 dx

|u|p|x| dx

2/p = ||u||2

|u|p|x| dx

2/p .

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This functional of course is well defined ifp (2, 2), but as we shall see, its restriction to the

space ofH1rad() ofH

1

radial functions is still well defined ifp (2, 2

+ 2N2).

We will callground statesthe functions that minimizeQoverH1(), while we reserve the term

radial minimizer to functions that minimize Q over H1rad().

The purpose of this paper is to investigate the existence of solutions to (2), in the spirit of [14],and especially to carry out the analysis of the symmetry properties of the ground states ofQ.As far as existence of solutions is concerned, our first result matches completely the one for theDirichlet problem obtained by Ni in [14].

Theorem 1.2 For every >0 and everyp (2, 2 + 2N2 ), there existsu H1rad() such that

Q(u) = inf vH1rad()\{0}

Q(v).

A suitable multiple ofu is a classical solution of (2).

The main point of interest in the previous theorem is the fact that the presence of the weight|x| allows one to obtain existence of solutions beyond the usual critical threshold.Whenp (2, 2), the functionalQ can be minimized directly in H

1(), without the symmetryconstraint; of course in this case the infimum is attained because p is subcritical. The naturalquestion that arises is to ascertain whether this minimizer, the ground state, is still a radialfunction. This is the question addressed in [18] for the Dirichlet problem and answered in thenegative for all p, provided is large enough.In the Neumann problem the situation is much more complex, and we point out from thebeginning that we cannot give a complete solution in the whole interval (2 , 2).

The main breakthrough for the study of the symmetry properties of the ground states is thepossibility to describe the precise asymptotic behavior, as , of radial minimizers. This isobtained by the construction of a limit functional in terms of a classical eigenvalue problem, theSteklov problem (see Definition 2.6). A further important point in all the symmetry questions isplayed by the number 2=

2N2N2, the critical exponent for the embedding ofH

1() intoLp().Indeed, in contrast to the Dirichlet problem, as , the denominator in Q behaves like atrace norm (in a sense to be made precise) resulting in worse growth properties of the levels ofradial minimizers.The consequence is the following result; Theorem 1.1 should be kept in mind for comparison.

Theorem 1.3 For everyp (2, 2), no ground state ofQ is radial provided is large enough.

The new limitation p >2 does not come from a weakness of the arguments, but is a structuralfact, peculiar of the Neumann problem. Indeed, we will prove in Section 6 the following symmetryresult, in opposition to the Dirichlet case.

Theorem 1.4 For everyp close enough to 2, the ground state ofQ is radial for every large.

The two above theorem are our main results concerning symmetry of ground states. They arecompleted by a further analysis of the variational properties of radial minimizers, which unveilsa rather unexpected phenomenon. Indeed it is quite natural to expect that symmetry of ground

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states breaks down for large because the the second derivative Q(u) at a radial minimizer

u becomes indefinite on H1

() when exceeds a certain threshold. This is what is describedin [18].In the Neumann problem, the situation is completely different: radial minimizers continue tobe local minima for Q over H

1() no matter how large is. This means, for example forp (2, 2

), that the formation of nonsymmetric ground states does not manifest locally aroundradial minimizers, but can be justified only by globalproperties of the functional. We think thatthis fact is rather interesting, also in connection with an analogous result for minimizers of thetrace inequality (see below). The precise result is the following.

Theorem 1.5 For all p (2, 2) if N 4 (resp. for all p (2, p), for some p (2, 2) if

N= 3) and all large enough, the minimizers ofQ overH1rad() are local minima ofQ over

the whole spaceH1(). The limitation forN= 3 is not removable.

As an application of the analysis of the sign of the second derivative of Q we obtain twouniqueness results, for radial minimizers in essentially the whole interval (2, 2), and for groundstates when p is close to 2.

Theorem 1.6 i) In the same range ofps as in the previous theorem, Q has a unique positiveradial minimizer (of unitary norm) for all large enough.ii) There exists p (2, 2) such that for allp (2,p), Q has a unique positive ground state (ofunitary norm) for all large enough.

Throughout the paper we use more or less explicitly a link between the functional Q and the

functionalSp: H1

() \ H10 () R defined, for p [2, 2] by

Sp(u) = ||u||2

|u|p d

2/p ,

whose infimum is the best constant in the Sobolev trace inequality. Existence of extremals forthis functional and their symmetry properties have been studied in [7], [9], [10] and [13] (seealso the references therein). As we have anticipated earlier, the functionalSp plays the role of alimiting functional forQ when . It is therefore clear that many properties of minimizersofQ for large and ofSp should coincide. In particular, we obtain the following analogue ofTheorem 1.5 for Sp.

Theorem 1.7 For all p (2, 2) the minimizers of Sp over H1rad() are local minima of Sp

over the whole spaceH1().

We point out that it is not known whether the ground states for Sp are radial for a genericp (2, 2), but only for p close to 2. The contribution of the previous theorem and of thediscussion that surrounds it in Section 5 is the fact that nonradial ground states for Sp, if theyexist, cannotbifurcatefrom the branchp upof radial minimizers, and should therefore appearas separated objects, whose existence springs out only after a certain p not too close to 2. In

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view of these considerations, and also of the arguments presented in Section 5, we propose the

followingConjecture. For all N3 and for all p (2, 2), the best constant Sp for the trace inequalityon the unitball ofRN is attained by a radialfunction.

The paper is structured as follows. In Section 2 we establish the main existence result (Theorem1.2), we carry out the asymptotic analysis of radial minimizers, and we prove Theorem 1.3.Section 3 is devoted to the variational properties of radial minimizers, and contains the proof ofTheorem 1.5. In Section 4 we prove the first part of the uniqueness result Theorem 1.7. Section5 is entirely devoted to the applications of our results to the trace inequality. Finally, in Section6, we prove Theorem 1.4 and the second part of the uniqueness result.

Notation. The space H1() is endowed with the norm ||u||2 = |u|2 dx+ u2 dx. AnyLp() norm is denoted by || ||p. The expression , is the scalar product in H1(). Ifu is aradial function, we will write freelyu(x) asu(|x|) oru(), for = |x|. The symbol dstands forthe standardN 1 dimensional measure on the unit sphere. When we deal with functionals Qdefined onH1() \ {0}, we will frequently write infH1() Qinstead of the heavier infH1()\{0} Q,and likewise for similar expressions. The letterCdenotes positive constants that may changefrom line to line.

2 Radial minimizers and their asymptotic properties

Throughout the paper is the unit ball ofRN, with N3; the numbers

2 = 2NN 2

and 2= 2N 2N 2

are the critical exponents for the embedding ofH1() into Lp() and Lq() respectively.We begin with the study of radial minimizers of the functional Q :H

1() \ {0} R definedby

Q(u) =

|u|2 dx +

u2 dx

|u|p|x| dx

2/p = ||u||2

|u|p|x| dx

2/p .

This functional is well defined and of classC2 overH1() ifp 2. We shall see in a while that

its restriction to H

1

rad() is still well defined and C

2

for a much wider interval ofps.To begin with we first establish some properties that have been first proved by Ni in [14] in thecontext of the Dirichlet problem; we now give the H1 versions.

Lemma 2.1 There exists a positive constantCsuch that for allu H1rad() there results

|u(x)| C ||u||

|x|N22

(3)

for allx \ {0}.

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Proof. By the radial Lemmaof [14], we have that there exists C >0 such that

|u(x)| |u(1)| + C||u||2

|x|N22

for alluradial. Since |u(1)| can be controlled by ||u||, the inequality follows.

Denote by Lp(, |x|dx) the space ofLp functions on with respect to the measure |x|dx.The previous lemma allows us to establish the following essential property.

Proposition 2.2 The spaceH1rad()embeds compactly into Lp(, |x|dx)for everyp [1, 2+

2N2).

Proof. By the growth estimate (3) we see that

|u|p|x| dx Cp||u||p

|x|pN22 dx.

The last integral is finite for every p [1, 2 + 2N2), which shows that for all these ps theembedding is continuous. With a standard interpolation argument one obtains the compactnessof the embedding in the same range.

Notice that the continuity of the embedding makes Q a C2 functional over H1rad(), for every

p [1, 2 + 2N2).We are now ready to give the main existence result. It matches completely the analogous onefor the Dirichlet problem obtained in [14].

Theorem 2.3 For every >0 and everyp (2, 2

+

2

N2 ), there existsu H

1

rad() such thatQ(u) = inf

vH1rad

()Q(v).

Proof. The proof is standard. Notice that for every p (1, 2 + 2N2), by Proposition 2.2,we have infH1

rad() Q > 0. Let un be a minimizing sequence for Q, normalized by ||un||= 1.

Then some subsequence un has a weak limit u in H1rad(). By Proposition 2.2 the limit u

cannot vanish identically, since in that case we would haveu

pn|x|

dx 0. ThenQ(u) liminfQ(un) = infH1

rad() Q.

Corollary 2.4 For every andp as in Theorem 2.3, (a suitable multiple of) the minimizeru

is a classical solution of problem

u + u= |x|up1 in

u

= 0 on.

Moreoveru is strictly positive in.

Proof. This follows from the symmetric criticality principle, standard elliptic regularity andthe maximum principle.

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For every >0 and for a given p (2, 2 + 2N2), let

m,r = minH1rad

()Q. (4)

Then any (positive) minimizeru ofQ overH1rad(), when normalized by ||u||= 1, satisfies

u+ u= m

p/2,r |x|up1 in

u

= 0 on .

We are interested in the behavior ofm,r and u when . It turns out that it is possibleto describe it in terms of a classical eigenvalue problem.

We begin with a fundamental result.

Lemma 2.5 The asymptotic relation

( + N)

|u|p|x| dx=

|u|p d+ o(1) as (5)

holdsi) uniformly on bounded subsets ofH1rad(), ifp (2, 2

),ii) uniformly on bounded subsets ofH1(), ifp (2, 2).

Proof. Notice that (+N)|x| = div(|x|x). For u H1rad() or u H1(), andp accordingly,

we can write, applying the divergence Theorem,

( + N)

|u|p|x| dx=

|u|p div(|x|x) dx=

|u|p|x|x dp

|u|p2uu x|x| dx

=

|u|p dp

|u|p2uu x|x| dx

since on we have =x and |x|= 1. We just have to show that the last integral is o(1) as with the required uniformity.To this aim, we first use the Holder inequality to write

|u|p2uu x|x| dx |u|

2 dx1/2

|u|2p2|x| dx

1/2

||u||

|u|2p2|x| dx

1/2.

Assume now thatp (2, 2) and that u is radial. By Lemma 2.1 we have

|u|2p2|x| dx C||u||2p2

|x|(2p2)N22 dx= ||u||2p2o(1)

as .

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If, on the other hand, u is not radial, but p is strictly less than 2, we notice that 2p 2< 2,

so that, by the Holder and Sobolev inequalities,

|u|2p2|x| dx

|u|2

dx

2p22

|x|

2

22p+2

22p+22

||u||2p2o(1)

as .Thus in both cases

|u|p2uu x|x| dx

||u||po(1) as ,which gives the required uniformity.

In order to state the main result of this section we need to introduce an auxiliary problem. This

is one of the classical eigenvalue problems, and we refer to [2] and [13] for more details.

Definition 2.6 The eigenvalue problem

u + u= 0 in

u

=u on

(6)

is called the Steklov problem.

The eigenvalues k of this problem are known to be (recall that is the unit ball in RN)

k = 1

N

2 +

Ik+N/22(1)

Ik+N/22(1) , k= 1, 2, . . . , (7)

whereIis the modified Bessel function of the first kind of order . The associated eigenfunctionsare also known (see [13]); thefirsteigenfunction, corresponding to1, is radial and never vanishesin . The first eigenvalue is simpleand it is characterized by

1= minvH1()

||u||2

u2 d. (8)

With the aid of the Steklov problem we can now describe the asymptotic behavior of the radialminimizers ofQ. The asymptotics for the solutions of the Dirichlet problem for the Henonequation has been obtained in [4] and [5]; in that case the situation is completely different and

much more complex.In the statement of the next result, 1 and 1, positive in and normalized by ||1||= 1, arethe first eigenvalue and eigenfunction of the Steklov problem (6).

Theorem 2.7 Letp (2, 2) and letu, with ||u||= 1, be a minimizer ofQ overH1rad(),

so thatm,r =Q(u). Then, as ,

m,r ( + N)2/p||12/p1 (9)

u 1 in H1(). (10)

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Proof. Let u be any (nonnegative) function in H1rad(), with ||u|| = 1. By Lemma 2.5, as

,

Q(u)

( + N)2/p =

1

(( + N)u

p|x| dx)2/p =

1

(u

p d+ o(1))2/p =

1

(u

p d)2/p+ o(1),

whereo(1) does not depend on u.Since u is radial,

up d

2/p=||2/pu(1)2 =||2/p1

u2 d,

so that for u = u,

m,r

( + N)2/p =

Q(u)

( + N)2/p =||12/p 1

u2 d + o(1)

||12/p minvH1

rad()

||v||=1

1v

2 d+ o(1) =||12/p1+ o(1)

because 1 is attained by a radial function.On the other hand, for every u H1rad() with||u||= 1,

m,r( + N)2/p

= Q(u)

( + N)2/p

Q(u)

( + N)2/p =||12/p

1u

2 d+ o(1).

Choosing u = 1 we obtain

m,r( + N)2/p

||12/p1+ o(1),

and (9) is proved.To prove (10) notice that since ||u||= 1, there is a subsequence, still denoted u, that convergesto some uweakly in H1() and strongly in Lq() for q


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Theorem 2.8 Assume thatp (2, 2). Then for every large enough (depending onp) we

havemin

uH1()

||u||2

(|u|

p|x| dx)2/p < min

uH1rad()

||u||2

(|u|

p|x| dx)2/p. (11)

Proof. We estimate the growth of the lefthandside of (11) as in [18]. Take a nonnegativefunction v C10 () and extend it to zero outside . Let x = (1 1/, 0, . . . , 0) and setv(x) =v((x x)). Then suppv B1/(x). Therefore, by standard changes of variable,

vp|x|

dx=

B1/(x)

vp|x| dx (1 2/)

B1/(x)

vp dx= N(1 2/)

vp dx

and

Q(v)

2N|v|2 dx + Nv2 dx2N/p(1 2/)2/p (

vp dx)2/p C

2N+2N/p

.

Since by Theorem 2.7 the righthandside of (11) is m,r 2/p, we see that (11) holds for all

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where, denotes the scalar product in H1().

Notice that since u solves

u+ u= mp/2,r |x|up1 in

u

= 0 on ,

(12)

the condition v TuS is equivalent tou

p1 v|x|

dx= 0.

Lemma 3.1 Let p (2, 2) and let u be a minimizer of Q over H1rad(), normalized by

u S. Then for everyv TuS,

Q(u) v

2

= 2m,r

||v||2

(p 1)mp/2

,ru

p2

v2

|x|

dx

. (13)

Proof. SetN(u) =||u||2 andD(u) = (u

p|x| dx)2/p, so that Q(u) =N(u)/D(u). For everycritical pointu H1() ofQ and every v H

1(), we have

Q(u) v2 =

D(u)N(u) v2 N(u)D(u) v2

D(u)2 .

NowN(u) v2 = 2||v||2 and

D(u) v2 = 2(2 p)up|x| dx

2/p2

up1v|x| dx

2

+ 2(p 1)

up|x| dx

2/p1

up2v2|x| dx,

so that

Q(u) v2 = 2

||v||2 ||u||2

(2 p)

up1v|x| dxup|x| dx

2+ (p 1)

up2v2|x| dxup|x| dx

(u

p|x| dx)2/p . (14)

This holds for every critical point u of Q and every v H1(). If u = u S we have

(up|x| dx)2/p =m,r and ifv TuS, then up1v|x| dx= 0. Therefore in this case (14)

reduces toQ(u) v

2 = 2m,r

||v||2 (p 1)mp/2,r

up2 v2|x| dx

. (15)

In order to study the sign ofQ(u) we need some more precise estimates on u.In what follows u is a minimizer ofQ over H

1rad(), normalized by ||u||= 1.

Lemma 3.2 The functionsu are uniformly bounded inC1() as .

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Proof. We first prove a uniform bound in L(). Since ||u|| = 1 and u is radial, there is

C >0 such that||u||L(\B1/2(0)) C||u||= C. (16)

By Lemma 2.1, there is a positive constant C such that

|u(x)| C ||u||

|x|N22

= C

|x|N22

for allx \ {0} and all .

Setf(x) =mp/2,r |x|u(x)

p1; then, recalling thatm,r C 2/p,

||f||L(B1/2(0)) C |x| 1

|x|(p1)N2

2

C

2(p1) 2N

2

=o(1) (17)

as .Therefore, by (16) and (17) we see that u solves

u+ u=f in B1/2(0)u C on B1/2(0)

with ||f||L(B1/2(0)) 0 as . By standard elliptic estimates, we obtain that u is

uniformly bounded in C1,(B1/2(0)), for all (0, 1). In view of (16) we obtain that u isuniformly bounded in L() as .To complete the proof we only have to show that there is a C1 bound also on \ B1/2(0).

Since u is radial we see that it solves

uN 1

u+ u= m

p/2,r

up1

and u(1) = 0. Integrating this equation over [t, 1], with t 12 we obtain

u(t) = (N 1)

1t

1

u d

1t

u+ mp/2,r

1t

up1 d.

Therefore, using the fact thatu is bounded in L() and the growth ofm,r,

|u

(t)| (N 1)u()

1t

+ 1t

u()

2 d + C+ Cmp/2

,r

+1

+ 11t

C+ C

+ 1C

for all t [ 12 , 1]. This, together with the estimate in C1,(B1/2(0)), gives the required bound in

C1().

Remark 3.3 Notice that one cannot hope to obtain uniform C2() estimates, even thougheach u does lie in C

2(). This is due to the fact that the righthandside of the equationbehaves like |x|. Now, while this term goes to zero locally uniformly in , on the boundaryit blows up like . Therefore u cannot be bounded in C

0 up to the boundary of .

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At first sight, a rather confusing consequence of the lack ofC2 bounds is that if one tries to

pass navely to the limit in (12) as , then one can do it in the equation (in the weakform, for instance), but not in the boundary conditions. Thus it may (and does) happen thatlimits of solutions of homogeneous Neumann problems do not satisfy a homogeneous Neumanncondition. We have already observed this fact when we have shown thatu 1, a solution ofthe Steklovproblem.

Remark 3.4 In view of the preceding Lemma, we can assure that the convergence ofu to1takes place also in C0(). Since1 is strictly positive in , then for some C >0 and all large,

minx

u(x) C.

We can now continue the study of the second derivative ofQ.

Lemma 3.5 Let p (2, 2) and let u be a minimizer of Q over H1rad(), normalized byu S. Then, as ,

( + N)

up2 v2|x| dx=

up2 v2 d+ o(1), (18)

uniformly forv in bounded subsets ofH1().

Proof. We apply the divergence Theorem exactly like in Lemma 2.5. We obtain

( + N)

up2 v2|x| dx=

up2 v2 d

(p 2)v2up3 u x|x|

dx 2up2 vv x|x|

dx

and we just have to show that the two last integral vanish as .Now By Lemma 3.2 and Remark 3.4 we have

v2up3 u x|x| dx

C

v2|x| dx C||v||22

|x| 2

22 dx

222

C||v||2o(1)

as , while

up2 vv x|x| dx

C

|v||v||x| dx C||v||

v2|x| dx

1/2C||v||2o(1),

as in the previous computation. Thus (18) is proved.

We can now prove the main result on the sign ofQ

(u).

Proposition 3.6 Letp (2, 2)and letu be a minimizer of Q overH1rad(), normalized by

u S. Then the inequalitymin

vTuS

||v||=1

Q(u) v2 >0 (19)

holdsi) for allp (2, 2) ifN4,ii) for allp (2, p), for somep (2, 2

) ifN= 3provided is large enough (depending onp).

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Proof. It is simple to see that the minimum in (19) is attained, so we do not supply details for

this. We assume that (19) is false for an unbounded sequence ofs (which we denote by A),so that

minvTuS

||v||=1

Q(u) v2 0 for all A. (20)

This means that for all A there existsv TuS, with||v||= 1 such that Q(u) v 0,

namely, by (13),

1 (p 1)mp/2,r

up2 v2|x|

dx 0. (21)

Recalling from Theorem 2.7 the asymptotic behavior ofm,r, we can write the preceding in-equality for in A as

1 (p 1)

||p/21p/21 + o(1)

( + N)

up2 v2|x| dx

= (p 1)

||p/21p/21 + o(1)

up2 v2 d+ o(1)

,

where for the last equality we have used Lemma 3.5.Nowu 1 in C

0() and v, being bounded in H1(), admits a subsequence (still denoted

v) converging to some v weakly in H1() and strongly in L2().

Passing to the limit as in A in the preceding inequality we find

1 (p 1)||p/21p/21

p21 v2 d. (22)

If v is identically zero we have reached a contradiction and the proof is complete. Assumetherefore that v 0.Since 1 is radial and normalized by ||1||= 1, we have

1= 1

21 d

= 1

1(1)2||,

so that 1(1)p2 =||1p/2

1p/21 . Inserting this in (22) we see that

1 (p 1)||p/21p/21 ||

1p/21p/21

v2 d= (p 1)1

v2 d,

that is, 1v

2 d (p 1)1.

Notice now that v, 1= limv, u= 0 by strong convergence ofu and weak convergenceofv. Thus v is orthogonal to 1 in H

1(); this, together with the fact that 1 is simple yields

2= minwH1()w,1=0

||w||2w

2 d

||v||2v

2 d

1v

2 d (p 1)1.

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Therefore assuming that (19) is false for an unbounded sequence ofs implies the inequality

21

p 1 (23)

on the eigenvalues of the Steklov problem (6). We now complete the proof by showing that thisinequality cannot hold. Recall from (7) that

k = 1 N

2 +

Ik+N/22(1)

Ik+N/22(1), k= 1, 2, . . . ,

whereI is the modified Bessel function of the first kind of order .Since (see [1]) I(x) =I+1(x) +

xI(x) holds for all x and all, we see that

k = k 1 +Ik+N/21(1)

Ik+N/22(1). (24)

We also recall from [1] that I1(x) I+1(x) = 2xI(x), so thatI1(1)/I(1) 2. Therefore

21

=1 + IN/2+1(1)/IN/2(1)

IN/2(1)/IN/21(1) >

1

IN/2(1)/IN/21(1)=

IN/21(1)

IN/2(1) 2

N

2 =N.

Thus, if (23) holds, then by our choice ofp,

N 2, and (p 1)1 < 2 for all p (2, 2

) ifN 4 and all p (2, p)ifN= 3, as have already proved in Proposition 3.6. This is a contradiction, and the proof iscomplete.

5 A detour on the trace inequalities

In this section we analyze a little more closely the relations between the minimization ofQ andsome Sobolev trace inequalities. Although we have already used some more or less evident link

between the two, we have not yet formalized the question.In our context the trace inequalities state that the embedding ofH1() intoLp() is continuousfor p [1, 2]; that is, for all p [1, 2] there exists C >0 such that

|u|p d

2/pC||u||2

for everyu H1(). We set

Sp= inf uH1()

||u||2

(|u|

p d)2/p (31)

and we recall thatSpis attained forp [1, 2) because the corresponding embedding is compact.

Ifp = 2 the embedding is no longer compact and the situation is more complex, see [9] andreferences therein, and only partial results are known. However, combining the condition ofTheorem 1 of [9] with the results of [8], one can say that for the unit ball the constant S2 isattained.The question of the symmetry of minimizer ofSp has been treated in [13], [10], and [7] (see alsothe references in these papers).Roughly speaking it turns out that radial symmetry of minimizers depends on the size of thedomain. Confining ourselves to the context where is the unit ball, the main results aboutsymmetry (deduced from [7], [10] and [13]) take the following form: denoting by the ball

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of radius centered at zero, then the functions that attain Sp in (31) are radial for all small

enough, and nonradial for all large enough.The same kind of phenomenon takes place for a fixed domain, say , but when p varies: it hasbeen proved in [13] (for more general problems) that minimizers of (31) are radial for all p closeenough to 2, and nonradial forp large.For further reference we quote a part of Theorem 2 of [13], specialized to our context. In itsstatement1 denotes, as usual, the first eigenvalue of the Steklov problem (6).

Theorem 5.1 (Lami Dozo, Torne) If

p 1> 1

21(1 (N 1)1), (32)

then no minimizer of (31) is radial.

Below we will give an interpretation of the number appearing in the righthandside of (32).Notice that minimizers ofSp, normalized by ||u||= 1, are solutions of

u + u= 0 in

u

=Sp/2p u

p1 on .(33)

Let us now return to the Henon problem. In the rest of this section we always assume thatp (2, 2). The link with the trace inequalities is given byii) of Lemma 2.5; indeed, denotingbySp : H

1() \ H10 () R the functional

Sp(u) = ||u||2

|u|p d

2/p ,

Lemma 2.5 shows that as ,

Q(u)

( + N)2/p =Sp(u) + o(1)

uniformly on bounded subsets of H1(). This fact, that we have already used, has furtherconsequences that we now examine, especially in connection with the results on Q(u) of

Section 3.

Theorem 5.2 For all p (2, 2) the minimizers of Sp over H1rad() are local minima of Sp

over the whole spaceH1().

Proof. It is a simplified version of the proof of Proposition 3.6. Indeed we will show that ifuis a minimizer ofSp over H

1rad(), then

minvTuS||v||=1

Sp (u) v2 >0, (34)

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whereTuS={v H1() : u, v= 0}.

Notice that, by (33), u, v= 0 is equivalent tou

p1

v d= 0.Since u minimizes Sp among radial functions, we have that u = 1 and Sp(u) = Sp(1) =||12/p1; therefore, with the same arguments as in Lemma 3.1 we see that

Sp (u) v2 =Sp (1) v

2 = 2||12/p1

||v||2 (p 1)||p/21

p/21

p21 v2 d

(35)

for allv TuS.Recalling that1(1)

p2 =||1p/21p/21 , and taking||v||= 1, we arrive at

Sp (u) v2 = 2||12/p1

1 (p 1)1

v2 d

.

Since v TuS, we havev2 d 1/2, so that

Sp (u) v2 2||12/p1

1 (p 1)

12

for all v TuS, with ||v|| = 1. By the results at the end of the proof of Proposition 3.6 thenumber in the righthandside of the preceding inequality is uniformly positive, for all p (2, 2),which shows that Sp (u) is positive definite on TuS. Hence u = 1 is a local minimum for Spover H1().

Some comments are in order. It is not known whether the best constant Sp is attained by aradial function for all p (2, 2). If the ground states are not radial for some p, it is quite

natural to expect that they bifurcate from the branch of radial minimizers. Theorem 5.2 showshowever that this is definitely not the case: nondegeneracy of radial minimizers over the wholespace H1() rules out any bifurcation phenomenon. Nonradial ground states, if any exist, arerather separated objects, located far away from the radial minimizers and whose existencebegins only after a certain value ofp.It is also interesting to compare our results with Theorem 5.1, by Lami Dozo and Torne (theactual result from [13] is much more general and applies to a wider class of problems). Theorem5.1 states that if

p 1> 1

21(1 (N 1)1), (36)

then no minimizer of (31) is radial; the argument consists in showing that a suitable (small)

variation of a radial minimizer makes the the functional Sp(u) decrease. Therefore we are in thepresence of a local phenomenon, around radial minimizers.On the other hand, Theorem 5.2 shows that radial minimizers are local minima over the wholespaceH1(), for allp (2, 2); the key argument is the fact already proved that for all these p,

p 1>21

. (37)

A natural question is to compare the conditions (36) and (37); our intent is to give a naturalinterpretation of (36). The next result shows that the two conditions coincide.

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Proposition 5.3 There results

1

21(1 (N 1)1) =

21

. (38)

Proof. We recall from (24) that

k = k 1 +Ik+N/21(1)

Ik+N/22(1)

for allk= 1, 2, . . .. In particular,

1=IN/2(1)

IN/21(1) and 2 =

IN/2+1(1)

IN/2(1) + 1.

To prove (38) we have to show that 11 N+ 1 =2.

The already used recursive relationI1(x) =I+1(x) + 2xI(x), for = N/2 and x = 1 reads

IN/21(1) =IN/2+1(1) + N IN/2(1).

Therefore

1

1 N+ 1 =

IN/21(1)

IN/2(1) N+ 1 =

IN/2+1(1) + NIN/2(1)

IN/2(1) N+ 1 =

IN/2+1(1)

IN/2(1) + 1 =2.

Although (36) or (37) are only sufficient conditions for the existence of nonradial minimizers,the fact that2/1> N/(N 1) = 2 1 for allN3 (proved in Section 3) and the variationalproperties of the radial minimizers described in this section seem to provide some evidencetowards the validity of the following

Conjecture. For all N3 and for all p (2, 2), the best constant Sp for the trace inequalityon the unitball ofRN is attained by a radialfunction.

6 Symmetry for slow growth

In this final section we return to the Neumann problem for the Henon equation. We have seen

that for every p (2, 2) the minimizers ofQ are not radial provided is sufficiently large.In the interval (2, 2) the situation is less clear, since it depends on the symmetry properties ofthe minimizers of the trace inequality, which are not precisely known for the unit ball. We pointout that even if one knows that the minimizers ofSp are radial, it is not clear a priori that alsothe minimizers ofQ should be radial.In this section we investigate the symmetry of minimizers when p is close to 2. It is interestingto keep in mind the behavior of minimizers for the Dirichlet problem described in [18]: in thatcase the authors showed that forp close to 2 minimizers are nonradial only if is very large (thethreshold between radial and nonradial minimizers tends to infinity as p 2). However the

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symmetry breaking phenomenon persists, as for a fixed p close to 2 one has nonradial solutions

for very large .We show in Theorem 6.1 below that this is not the case for minimization in H1(): for p closeto 2 minimizers are radial for al l large enough.Of course we take advantage of some result for the limit problem given by the minimizationof Sp. The precise result we need is contained in Theorem 4 of [13]. There it is proved thatthere exists p (2, 2] such that for every p (2,p ] the problem minuH1() Sp(u) has aunique solution, which is radial. Of course this solution (with norm equal to one) is 1, andSp(1) =||

12/p1.

Theorem 6.1 Letp (2,p ). For every large enough the problem

minuH1()

||u||2

|u|p|x| dx

2/p (39)

has a unique positive solution (normalized by||u||= 1), and it is a radial function.

Proof. The proof of this theorem follows very closely that of Theorem 4.1; the main differencecomes from the fact that in the present case we are not dealing with radial functions, whichtends to complicate things. On the other hand we will profit of the fact that we are now workingwith p


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Therefore we have that ||u||= 1 and that u is a minimizer for Sp. By the above quoted results

and the assumption p < p , it must be u= 1. We conclude that every sequence of minimizersfor Q converges to1 strongly in H

1().We now show that Q has a unique minimizer for large. Suppose this is not true; then forevery in an unbounded set A, there exist two distinct minimizers u and v. As in the proofof Theorem 4.1, if we set = (u v)/||u v||, we see that it solves

+ = (p 1)m

2/p |x|c in

= 0 on ,

(40)

with

c= 10 (v+ t(u v))

p2

dt.

Notice that|c| (u+ v)p2.

Since S, up to subsequences we can assume that weakly in H1() and strongly

in Lp() and in Lp(). We claim that 0. To see this we multiply the equation in (40) by and we use Lemma 2.5 to obtain

1 = ||||2 = (p 1)m2/p

c2|x|

dx= (p 1)(||2/p1p/21 + o(1))( + N)

c2|x|

dx

C

( + N)

cp

p2 |x|

dx

12/p ( + N)

p|x| dx

2/p

C

( + N)

(u+ v)p|x| dx12/p

p d+ o(1)

2/p

C

(u+ v)p d+ o(1)

12/p

p d+ o(1)

2/pC

p d+ o(1)

2/p.

If is zero, then the strong convergence of in Lp() gives a contradiction.

We now pass to the limit in the weak form of (40), which is

, = (p 1)m2/p

c|x| dx

for all H1().

We write

( + N)

c|x| dx= ( + N)

p21 |x| dx +

(c p21 )|x|

dx

= ( + N)

p21 |x| dx +

p21 ( )|x| dx +

(c p21 )|x|

dx

and we evaluate the three terms in the righthandside separately. For the first one we have

( + N)

p21 |x| dx=

p21 d

(p21 ) |x|xdx,

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and, by Holder inequality,

(p21 ) |x|x dx

C

|||||x| dx + C

(|||| + ||||)|x| dx

C||||2||||2

|x|Ndx

1/N+ C(||||2||||2+ ||||2||||2)

|x|Ndx

1/N=o(1)

as . Therefore

( + N)

p21 |x| dx=

p21 d+ o(1).

For the second term we apply Holder inequality to obtain, by Lemma 2.5,

( + N)

p21 ( )|x| dx

( + N)

p1|x| dx

12/p

( + N)

| |p|x| dx

1/p

( + N)

||p|x| dx

1/p

=

p1 d+ o(1)

12/p

| |p d+ o(1)

1/p

||p d+ o(1)

1/p=o(1)

because strongly in Lp().

Finally, for the third term we write

( + N)

(c p21 )|x|

dx

( + N)

|c p21 |

pp2 |x| dx

p2p

( + N)

||p|x| dx

1/p

( + N)

||p|x| dx1/p

and we readily recognize, as above, that the last two integrals are uniformly bounded as .Recalling the definition ofc it is easy to see that the first integral goes to zero as .Putting together the above estimates we can say that

, = (p 1)(||p/21

p/21 + o(1))

p2d+ o(1)

,

so that, when ,

, = (p 1)1

d, (41)

for all H1() (we have used the fact that p21 ||1p/2

1p/21 on ). Equation (41)

says that is a (nontrivial) weak solution of

+ = 0 in

= (p 1)1 on .

Since we know that (p 1)1 is not an eigenvalue of the Steklov problem (6), we conclude that 0, a contradiction.Therefore u v for all large. In other words, problem (39) has a unique solution forp (2,p ) and large. Since (39) is invariant under rotations, this solution must be radial.

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[2] C. Bandle, Isoperimetric inequalities and applications. Monographs and Studies in Mathe-matics, 7. Pitman, Boston, Mass.-London, 1980.

[3] M. Badiale, E. Serra, Multiplicity results for the supercritical Henon equation.Adv. Non-linear Stud. 4 (2004), no. 4, 453467

[4] J.Byeon, Z.-Q.Wang,On the Henon equation: asymptotic profile of ground states, preprint,2002.

[5] J.Byeon, Z.-Q.Wang, On the Henon equation: asymptotic profile of ground states, II, J.Diff. Eqs. 216(2005), 78108.

[6] D. Cao, S. Peng,The asymptotic behaviour of the ground state solutions for Henon equation.J. Math. Anal. Appl. 278 (2003), no. 1, 117.

[7] M. del Pino, C. Flores, Asymptotic behavior of best constants and extremals for trace em-beddings in expanding domains.Comm. PDE 26 (2001), no. 11, 21892210.

[8] J.F. Escobar, Sharp constant in a Sobolev trace inequality. Indiana Univ. Math. Jour. 37(1988), no. 3, 687698.

[9] J. Fernandez Bonder, J.D. Rossi,On the existence of extremals of the Sobolev trace embed-ding theorem with critical exponent. Bull. London Math. Soc.37(2005), no. 1, 119125.

[10] J. Fernandez Bonder, E. Lami Dozo, J.D. Rossi,Symmetry properties for the extremals ofthe Sobolev trace embedding.Ann. IHP Anal. Non Lineaire 21(2004), no. 6, 795805.

[11] B. Gidas, W.M. Ni, L. Nirenberg, Symmetries and related properties via the maximumprinciple. Comm. Math. Phys. 68 (1979), 209243.

[12] M. Henon, Numerical experiments on the stability of spherical stellar systems. Astronomyand Astrophysics 24(1973), 229238.

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[14] W.M. Ni,A nonlinear Dirichlet problem on the unit ball and its applications. Indiana Univ.Math. Jour. 31 (1982), no. 6, 801807.

[15] A. Pistoia, E. Serra, Multipeak solutions for the Henon equation with slightly subcriticalgrowth. Preprint, 2005.

[16] E. Serra, Non radial positive solutions for the Henon equation with critical growth. Calc.Var. and PDEs 23(2005), no. 3, 301326.

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[17] S. Secchi, E. Serra, Symmetry breaking results for problems with exponential growth in the

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