Gazzini M., Musina R. - Hardy-Sobolev-Mazya inequalities symmetry and breaking symmetry of extremal functions(14).pdf

7/25/2019 Gazzini M., Musina R. - Hardy-Sobolev-Mazya inequalities symmetry and breaking symmetry of extremal function

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Hardy-Sobolev-Mazya inequalities: symmetry and

breaking symmetry of extremal functions

Marita Gazzini

and Roberta Musina

Abstract. Denote points in Rk RNk as pairs = (x, y), and assume 2 k < N. In this paper we study

problem

v = |x|2 v+ |x|b vp1 in RN , x= 0v >0 ,

where p >2, b = N pN22

and k22

2, the Hardy constant. Our results are the following:

Let p 0 ,

namely, for the Euler-Lagrange equations of the Mazya inequality with cylindrical weights.

Key Words: Variational methods, critical growth, weighted Hardy-Sobolev inequalities.

2000 Mathematics Subject Classification: 35J20, 35J70, 35B33

Ref. SISSA 06/2008/M

Work supported by Cofin. M.I.U.R. Progetto di Ricerca Metodi Variazionali ed Equazioni Differenziali

NonlineariSISSA, Via Beirut 4 34010 Trieste, Italy, e-mail: [email protected] di Matematica ed Informatica, Universita di Udine, via delle Scienze, 206 33100 Udine,

Italy, e-mail: [email protected]

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Introduction

Letk, Nbe positive integers with k < N. We put RN = Rk RNk, and we denote points

in RN as pairs (x, y) Rk RNk. In this paper we study the elliptic problemv= |x|2 v+ |x|b vp1 in RN , x= 0

v >0 ,(0.1)

wherep >2, k22

2and

b= N pN 2

2 . (0.2)

A large number of bibliographical references for (0.1) is available in case k = N: we

quote for example [1], [3], [6], [8], [18] and references there-in. In particular, in [6] and [8]

one can find a careful analysis on breaking symmetry of ground state solutions.

Concerning existence in case k < N we cite [2], [5], [13], [15] and [17]. Existence andmultiplicity results can be found in [9]. Symmetry properties of weak entire solutions to

(0.1) were proved in [11], under the assumptions k 2, = 0 andp (2, 2N).

As noticed in [5] and in [13], solutions that are radially symmetric in the x-variable

receive importance with regard to certain elliptic equations on the n = Nk +1-dimensional

hyperbolic space Hn. More precisely, ifv(x, y) = v(|x|, y) solves (0.1), then the transform

u(r, y) := rN22 v(r, y) gives a solution of

Hnu= u + up1 . (0.3)

Here, Hn is the Laplace-Beltrami operator on Hn and the parameter is given by

= +(N k)2 (k 2)24

.

We refer to [5] and to [13] for a discussion on the relevances between equation (0.3) and

some significant problems in hyperbolic geometry, Yamabe-type equations of Heisenberg

type, Grushing-type equations.

Motivated by these considerations, in the present paper we address our attention towards

cylindrically symmetric solutions (see Definition 1.1 and [12]).

Our first result concerns the existence of entire solutions (see Definition 1.2 and [13]), in

case that p is smaller than the critical Sobolev exponent in RNk+1.

Theorem 0.1 Assume2 k < N, k22

2and2< p


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proved in Section 2, where we state also a non-existence/existence result in the limiting case

p= 2Nk+1.

In Section 3 we investigate the symmetry properties of solutions to (0.1). In case k 2andp (2, 2N), Mancini, Fabbri and Sandeep adopted in [11] the moving plane method to

show that nonnegative entire solutions to

v= |x|b vp1 in RN (0.4)

are cylindrically symmetric. As a matter of fact, their arguments work as well in case k = 1.

We omit the proof of the next result.

Theorem 0.2 Assumek = 1,N3, and 2(N1)N2 < p 0 .

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As corollaries of our theorems for (0.1) we prove an existence result of symmetric solu-

tions, symmetry properties of solutions and breaking symmetry.

Acknowledgments

The authors wish to thank Prof. Gianni Mancini for having proposed them the study of

the breaking symmetry phenomenon for problem (0.1).

Notation For any integer j 1, we denote by BjR(z) the j-dimensional ball of radius R centered

at z Rj . We denote by j the surface measure of the unit sphere Sj1 in Rj .

Ifp [1, +), R, and if is a domain in RN, thenLp(; |x|d) is the space of measurable

mapsu on withR|x||u|p d


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as the closure of maps v Cc ((Rk \ {0}) RNk) with respect to the norm

v =

RN

|v|2 |x|2v2

d

1/2.

In general,H(Rk RNk) contains the spaceD10(RN) L2(RN; |x|2d). More precisely,

H(Rk RNk) = D10(R

N) L2(RN; |x|2d)

for 0 small enough. More precisely, we choose 0 such that

2(p 1)p2p

0 Sp , (3.2)

where the Sp is the infimum in (1.3). Since |s|< || in s, it turns out that

ws |x|2ws |x|

bA()ws , (3.3)

pointwise on s\ 2s, where

A() :=vp1s v

p1

vs v

satisfies

0 A()(p 1)vp2 on{ws 0}. (3.4)

As in [11], the idea is to use ws := min{ws, 0} 0 as test function for (3.3), but,

differently from [11], the maps ws and ws are not smooth enough. Thus we have to use

suitable cut-off functions. For >0 small set

(x) =

0 if|x| 2log |x|/2

| log | if2


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For any large integer h choose a cut off function h Cc (R

N), with

h() = 1 if || hh() = 0 if || 2h , 0 1 , ||h

4

h .

We are allowed to use 22hw

s as test function for (3.3) on s. Set

h:= hws .

After integration by parts and simple computations one getss

|h|

2 |x|2|h|2

d

s

|x|bA()|h|2 d +

RN

|(h)|2|ws|

2 d . (3.5)

The left hand side in (3.5) is bounded from below by

Sp

s

|x|b|h|p d

2/p.

To estimate the right hand side we notice thats

|x|bA()|h|2 d (p 1)

s

|x|bA()|v|p2|h|2 d

(p 1)p2p

0

s

|x|b|h|p d

2/p

by (3.4), Holder inequality and (3.1). Comparing with (3.5) and with (3.2) we infer

1

2 Sp

,hs|x|

b

|ws|p

d2/p

RN

|(h)|2

|ws|

2

d , (3.6)

where

,hs ={(x, y) s | |xs|> , ||< h} .

In order to handle the right hand side in (3.6) we computeRN

|(h)|2|ws|

2 d 2(I,h1 + I,h2 ),

where

I,h1 =

RN

||22h|w

s|

2 d , I ,h2 =

RN

|h|22|w

s|

2 d .

To estimate the first integral we notice that v is smooth on {x= 0} and that |ws | v ons. Therefore, since k 2,

I,h1 ch

Rk

||2 d

ch| log |2

2

rk3 dr ch| log |

,

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with constants ch that depend only on the measure of BN2h and on the L

-norm of v on

BN2h s. Thus, I,h1 0 for h fixed, as 0. Concerning the second integral, we use

Holder inequality, b 0 and |ws| v on s to get

I,h2

RN

|h|2|v|2 d

BN2h\B

Nh

|x|b|v|p d

2p

RN

|| 2bp2 |h|

2pp2 d

p2p

c

BN2h\B

Nh

|x|b|v|p d

2p

.

Thus I,h2 0 as h + uniformly in , since v Lp(RN; |x|bd). In conclusion, we

have proved that

1

2Sp ,hs |x|

b|ws|p d

2/p

ch| log |

+ c

BN2h\BNh |x|

b|v|p d

2p

.

Since|ws| v Lp(RN; |x|bd), then passing to the limit in (3.6), first as 0 and then

as h +, we get that ws 0 a.e. on s. More precisely, ws > 0 on s\ 2s by (3.3)

and by the maximum principle. The proof now can be carried out as in [11].

4 Breaking symmetry

The reason for the phenomenon described in Theorem 0.4 is that cylindrically symmetric

solutions become highly unstable as , that is, their Morse index becomes too large

(see Remark 4.2). This is a consequence of the next crucial Theorem. Its proof was inspired

by the papers by Kawohl [10] and by Smets, Su and Willem [16].

Theorem 4.1 Assumek 2 and letv = 0 be a local minimum for

R(v) =

RN

|v|2 d RN

|x|2v2 dRN

|x|b|v|p d2/p

onD10(RN) L2(R2; |x|2d), such thatv(x, y) = v(|x|, y) for a.e. y RNk. Then

RN

|v|2 d

RN

|x|2v2 dk 1

p 2

RN

|x|2v2 d .

Proof. Take any h D10(RN) L2(RN; |x|2 d), and set

z(t) = RN|(v+ th)|2 d RN|x|

2(v+ th)2 d ,

n(t) =

RN

|x|b|v+ th|p d

2/p,

g(t) = z(t)

n(t)=R(v+ th).

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Since 0 is a local minima for g, then g(0) = 0 and g(0) 0. To simplify notations we

assumen(0) = 1. Thus we get z (0) z(0)

n(0) , that is,RN

|v|2

RN

|x|2v2

RN

|h|2 RN

|x|2h2

(p 1)RN

|x|b|v|p2h2 (p 2)(RN

|x|b|v|p2vh)2

. (4.1)

Now, let f1 H1(Sk1) be an eigenfunction of the Laplace operator on Sk1 with respect

to the eigenvaluek 1. Thus, f1 solves the minimization problem

inffH1(Sk1)RSk1

f =0

Sk1

|f|2 d

Sk1|f|2 d

=k 1.

To simplify computations it is convenient to take

Sk1

|f1|2

d = 1 ,Sk1

|f1|2

d= k 1.

Notice that we are allowed to use h(x, y) = v(|x|, y)f1(x/|x|) as test function in (4.1). Indeed,

h D10(RN) L2(RN; |x|2 d) since v D10(R

N) L2(RN; |x|2 d). It turns out thatRN

|h|2 =

RN

|v|2 + (k 1)

RN

|x|2v2 ,

RN

|x|2h2 =

RN

|x|2v2 ,

RN

|x|b|v|p2h2 = 1 ,

RN

|x|b|v|p2vh = 0 .

Thus from (4.1) we infer that

RN

|v|2

RN

|x|2

v2

1

p 1

RN|v|

2

RN

|x|2

v2

+ (k 1)RN

|x|2

v2

.

The conclusion easily follows.

Proof of Theorem 0.4. Assume that v H,cyl(Rk RNk) achieves Sp . Then, by

Theorem 4.1 and by Hardy inequality one hask 2

2

2

RN

|x|2v2 d

RN

|v|2 d

RN

|x|2v2 d

k 1

p 2RN|x|2v2 d .

Thus k22

2 k1p2 . Equality can not hold, since the Hardy constant is not achieved.

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Remark 4.2 Set ={u H(Rk RNk)|

RN

|x|b|v|p d= 1} and

E(v) = RN

|v|2 |x|2v2 d , E : R.Then the Morse index of any cylindrically symmetric solution v to (0.1) diverges to +

as . More precisely, for j 1 let j be the eigenspace ofSk1 relative to the

eigenvalue j = j(k+ j 2). Then E(v) is negative definite on i for any i = 1,...,j,

provided that 0 ,(5.1)

where

p= 2NN 2 + 2( )

.

On the exponents , we assume, as in [9], that

< k N 2

2N , 2.

In the spherically symmetric case k = Nproblem (5.1) is related to the Caffarelli-Kohn-

Nirenberg inequalities [4]. Existence, non existence and breaking symmetry of extremals

functions were discussed in [6]. For k < Nthe counterpart of the Caffarelli-Kohn-Nirenberg

inequalities are the Mazya inequalities ([14], Section 2.1.6). Existence results can be found

in [17], where = = k22 , and in [15] for = k22 . Some multiplicity results for (5.1)were proved in [9].

We say that a classical solution u to (5.1) is -cylindrically symmetricif

i) for any choice of y RNk, u(, y) is radially symmetric in Rk, and the map |x|

|x|u(|x|, y) is decreasing.

ii) there exists y0 RNk such that, for any choice ofx Rk \ {0}, u(x, ) is symmetric

decreasing abouty0 in RNk.

In order to state our results for problem (5.1) we need to introduce a suitable Sobolev

space (see [9] for details). For < k N22N we denote byD10(R

N; |x|2d) the completion of

Cc (R

N

) with respect to the scalar product

u, v=

RN

|x|2 u v d .

As a Corollary to Theorem 0.1 we easily get the following result.

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Corollary 5.1 Assume2 k < N and (5.2). IfNk + 2 assume in addition that

k 1

N k+ 1 < .

Then there exists a-cylindrically symmetric solution

u D10(RN; |x|2d) L2(RN; |x|2(+1)d) .

Proof. Let us define

:=

k 2

2

2

k 2 2

2

2.

By direct computation and by the results in [15], Appendix B, one can prove that a map

u D10(RN; |x|2d) L2(RN; |x|2(+1)d) is a classical solution to (5.1) if and only if

v:= |x|u D10(RN)L2(RN; |x|2d) is a solution to (0.1), with respect to the parameter

. The conclusion easily follows from Theorem 0.1.

With the same trick, from Theorems 0.3 and 0.4 one can prove the following results.

Corollary 5.2 Assume2 k < N, (5.2), ,0 k2and letu Lp(RN; |x|pd)

be a classical solution to (5.1). If = = 0 assume in addition thatu has a nonremovable

singularity on{x= 0}. Thenu is-cylindrically symmetric.

Corollary 5.3 Assume2 k < N, (5.2) and , and letu be a solution to the minimum

problem

infD10(R

N;|x|2d)

RN

|x|2|w|2 d

RN|x|p|w|p d

2/p .

Thenu is not radially symmetric inx if

+ 1 N(k 1)

(k 2 2)2 + 2(k 1) . (5.3)

Remark 5.4 Corollary 5.3 holds also in the spherical casek = N2. However, for k = N

(5.3) Felli and Schneider (see also [6]) proved the stonger estimate:

+ 1 N

2

1

N 2 2(N 2 2)2 + 4(N 1)

.

References

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elliptic equation arising in astrophysics, Arch. Rat. Mech. Anal. 163 (2002), 252-293.

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Gazzini M., Musina R. - Hardy-Sobolev-Mazya inequalities symmetry and breaking symmetry of extremal functions(14).pdf