Gazzini M., Musina R. - On the Maz'ja inequalities existence and multiplicity results for an...

7/25/2019 Gazzini M., Musina R. - On the Maz'ja inequalities existence and multiplicity results for an elliptic problem involvin

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On the Mazja inequalities: existence and

multiplicity results for an elliptic problem

involving cylindrical weights

Marita Gazzini and Roberta Musina

Abstract.In this paper we discuss existence and multiplicity of solutions to the following class of superlinear

problems:

(1)

div(|x|au) =|x|ba up1 in RN , x= 0

u >0 .

Here u = u(x, y) : Rk RNk R, 1 k < N, ba = N pN2+a

2 and a >(2 N) k

N. The main tools are

some Sobolev-type inequalities with cylindrical weights that were proved by Mazja in 1980.

We also get new existence results for the singular superlinear problem

(2)

v = |x|2 v+ |x|b0 vp1 in RN , x= 0

v >0 ,

by compounding a simple functional change with our results for (1). Here k22

2and b0= Np

N22

.

In particular, we give an alternative proof of a result by Tertikas and Tintarev [16] in case = k222

. For

(2 N)kN

, max

2, 2(N k)N 2 + a

< p 2 , ba= N p N 2 + a

2 , (0.4)

then there exists ca,p> 0 such that

ca,p

RN

|x|ba |u|p d

2/p

RN

|x|a|u|2 d , u Cc (RN). (0.5)

Thanks to inequality (0.5), we can define the Hilbert space D10(RN; |x|ad) by completing

Cc (RN) with respect to the norm u2 =

RN

|x|a|u|2. Then, every minimizer for

SDa,p:= inf D10(RN;|x|ad)

RN|x|a|u|2 d

RN|x|

ba

|u|p

d2/p (0.6)

is, up to a Lagrange multiplier, a solution to (0.1). In case a > 2 k, the existence of

solutions to (0.6) follows from Theorems B1 and B2 in [15]. In the present paper we focus

our attention on the case a 2 k.

While studying (0.6) one has to take into account the action of the groups of dilations in

RN and of translations in RNk. Indeed, it is quite easy to exhibit non compact minimizing

sequences. To overcome this difficulty we use a suitable Rellich type theorem and a rescaling

argument. Our existence result is stated in Theorem 2.1. IfN 3 and p= 2, the group

of translations in the x-variable produces some extra lack of compactness phenomena. For

this reason, to construct a compact minimizing sequence we need the additional assumption

SDa,2

< S, whereS is the Sobolev constant. Some sufficient conditions for SDa,2

< Scan be

found in Section 3.2. Our main existence result in the critical case is stated in Theorem 3.9.

In Section 3 we take a < 2k and we compare the solution uD to the minimization

problem (0.6) with the solution uX D10(RN; |x|ad) L2(RN; |x|a2d) to (0.1), whose

existence was proved in [15]. Under suitable assumptions on the exponents a and p, we are

able to prove that uX do not solve (0.6). Following this idea we prove several multiplicity

results for (0.1). This will be done in Section 3.1 forp 0 ,(0.7)

where,p,b0 R satisfy

k 2

2

2, p (2, 2] , b0 = N p

N 2

2 .

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First of all we use a functional change and Theorem 2.1 to prove the existence of a cylindri-

cally symmetric solution for = k222 (see Theorem 4.1). Existence was already proved

in [16] in case p = 2.

In Subsection 4.2 we state some existence results of singular solutions to (0.7) under the

assumption 0(0.8)

has a solution in D10(RN) which is unique up to translations and up to changes of scale

in RN. Moreover, its L2

- norm is SN/2, where S is the Sobolev constant. In Corollary

4.4 we prove that, for N 7, problem (0.8) has a smooth cylindrically symmetric solution

v : (R3

\ {0}) RN3

R, such thatRN

|v|2 d= +,

RN

|v|2 d < SN/2 .

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 Rj .

It will be often convenient to set

RN0 := (R

k \ {0}) RNk .

Ifp [1, +), R, and if is a domain in RN, then Lp(; |x|d) is the space ofmeasurable mapsu on with

|x||u|p d


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1 Preliminaries

One of the main features in problem (0.1) is its invariance with respect to transforms

u(x, y)(T(, )u) (x, y) := N2+a

2 u(x, y+ ), (1.1)

where(0, +), RNk. In addition, it holds also thatRN

|x|a|(T(, )u)|2 d=

RN

|x|a|u|2 d ,RN

|x|ba |T(, )u|p d=

RN

|x|ba |u|p d (1.2)

for any u Cc (RN), whenever the weights |x|a and |x|ba are locally integrable on RN.

Indeed, these invariances are responsible for the integral inequalities that are the starting

point of the present paper and that were proved by Mazja [14] in 1980.

1.1 Mazja inequality. The space D10(RN; |x|ad)

Assume thata,p,ba in R satisfy (0.4). Then the weights|x|a and|x|b are locally integrable

on RN. The Mazja inequalitystates that there exists ca,p> 0 such that

ca,p

RN

|x|ba |u|p d

2/p

RN

|x|a|u|2 d , u Cc (RN). (1.3)

Inequality (1.3) was proved by Mazja in [14], Section 2.1.6, Corollary 2, under the assump-

tionsN 3 and p 2. As observed in [5], Section 4, ifN= 2 and k = 1, then the same

inequality holds true for any p (2, +).Thanks to (1.3), for any a >(2 N) kNwe can define the Hilbert space D

10(R

N; |x|ad)

by completingCc (RN) with respect to the scalar product

u, v=

RN

|x|a u v d .

Notice that for N3, D10(RN; |x|0d)D10(R

N). From (1.3) it follows that the infimum

SDa,p:= inf uD10(R

N,|x|ad)

RN

|x|a|u|2 dRN

|x|ba |u|p d2/p

is positive, provided that a,p,ba are as in (0.4). It is clear that if SDa,p is achieved by

u D10(RN; |x|ad) thenu is a classical solution to (0.1), up to multiplicative constant. Bystandard elliptic regularity theory and by the maximum principle, u is smooth and positive

on{x= 0}. In addition, u is aweak solutionto (0.1) on RN, that is,RN

|x|au d=

RN

|x|baup1 d , Cc (RN), (1.4)

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and it is entire, namely,

RN

|x|a|u|2 d=RN

|x|baup d 2 k, and for any u Cc (R

N0 ) ifa 2 k (see for example

[14], Section 2.1.6, or [7], Theorem 3.5). The constant

1(a) :=

k 2 + a

2

2(1.6)

is sharp, and it is not achieved. Notice that Hardys inequality implies that D10(RN; |x|ad)

is continuously embedded into L2(RN; |x|a2d), provided that a >2 k.

The Hardy-Sobolev-Mazja inequalitywas proved in [14] (Section 2.1.6, Corollary 3) in

caseN 3, and in Section 4 of [5] in case N = 2, k = 1. It states that for any a R and

for every real exponent p (2, 2] there exists a constant ca,p> 0 such that

ca,p

RN

|x|ba |u|p d

2/p

RN

|x|a|u|2 d 1(a)

RN

|x|a2|u|2 d (1.7)

for any u Cc (RN) ifa >2 k, and for any u Cc (RN0 ) ifa 2 k. We remark that(1.7) fails ifk = N. We refer to [9] for a discussion on this subject.

Inequality (1.7) provides the starting point for applying variational methods to the sin-

gular problem div(|x|au) = |x|a2u + |x|baup1 in RN , x= 0

u >0 ,(1.8)

where is a real parameter. For future convenience we recall here the approach used in [15]

to study (1.8) in case < 1(a). For any a R define the Hilbert space

X10 (RN; |x|ad) := D10(R

N; |x|ad) L2(RN; |x|a2d) .

Notice that X1

0(R

N

; |x|a

d) = D1

0(R

N

; |x|a

d) if a > 2 k, by (1.5). It turns out thatCc ((R

k \ {0}) RNk) is dense inX10(RN; |x|ad) ([15], Appendix B) and that (1.7) holds

for anyu X10 (RN; |x|ad). The paper [15] deals with the existence of extremals for

infuX1

0(RN;|x|ad)

RN

|x|a|u|2 d RN

|x|a2|u|2 dRN

|x|ba |u|p d2/p

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under the assumption < 1(a). In particular, for = 0 and a = 2k, every solution

uX X10

(RN; |x|ad) to the minimization problem

SXa,p:= inf uX10(R

N;|x|ad)

RN

|x|a|u|2 dRN

|x|ba |u|p d2/p (1.9)

is, up to a Lagrange multiplier, an entire classical solution to (0.1). Notice also thatuX is

a weak entire solution to (0.1) on (Rk \ {0}) RNk. Ifa 2 k, in general, one can not

conclude that uX is indeed a weak solution on the whole of RN. On the other hand, for

a > 2 k one can take advantage of the Hardy inequality (1.5) to show that SDa,p = SXa,p,

and thatuX satisfies (1.4) (see [15], Theorem A.2).

Finally we notice that in generalSDa,p SXa,pfor anya >(2 N)

kN, sinceX

10 (R

N; |x|ad)

is contained inD10(RN; |x|ad). In Section 3 we illustrate some examples in which the strict

inequality SDa,p< SXa,p holds true.

Concerning the existence of minimizers for SXa,p we can state the following result.

Theorem 1.1 Assumea = 2 k, and letp > 2. The infimumSXa,p is achieved by a map

uX X10 (RN; |x|ad) provided that one of the following conditions holds:

a) N= 2 orp (2, 2);

b) N3, p= 2 andSXa,2 < S .

Moreover, ifk= 1 then the support ofuX is a half-plane.

Theorem 1.1 is a direct consequence of Theorems B.1 and B.2 in [15] in case N3; the

same proof can be carried out in case N= 2, k = 1.

We conclude this subsection with a few remarks on the infimum SXa,p. First we state auseful Lemma that was proved in [15], Appendix B.

Lemma 1.2 For anya R the linear operatorLa(u) :=|x|a2 u is a bi-continuous isomor-

phism betweenX10 (RN; |x|ad) andX10(R

N; d) = D10(RN) L2(RN; |x|2d). Moreover,

RN

|x|a|u|2 1(a)

RN

|x|a2|u|2 =

RN

|(Lau)|2 1(0)

RN

|x|2|Lau|2 (1.10)

for anyu X10 (RN; |x|ad).

In the next lemma we point out some remarks on the behavior of the mapa SXa,p.

Lemma 1.3 Assume2 < p < 2. Then the mapa SXa,pis strictly increasing fora 2k,

and it is strictly decreasing fora 2 k.

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Proof. Fora R let1(a) be as in (1.6). Set a:= 2(2k)aand notice that1(a) = 1(a).

Then, by (1.10), it turns out that

SXa,p = inf vX10 (R

N;d)

RN

|v|2 d (1(0) 1(a))RN

|x|2v2 dRN

|x|ba |v|p d2/p

= inf vX1

0(RN;d)

RN

|v|2 d (1(0) 1(a))RN

|x|2v2 dRN

|x|ba |v|p d2/p =SXa,p .

Thus SXa,p= SXa,p. The Lemma is readily proved, since the map a 1(a) is increasing for

a 2 k, and sinceSXa,p is achieved for any a = 2 k.

The case N 3, p = 2 is more difficult. We recall that SXa,2 S for any a R (see

[15], Theorem B.5), and that the map vTdefined in (0.9) achieves the best Sobolev constant

S on D10(RN). Moreover vT L2(RN; |x|2d), that is, vT X10 (RN; d), if and only ifk 3. In this case, a direct computation shows that the map

uT(x, y) := |x|k2vT(x, y)

belongs to X10(RN; |x|2(2k)d) and achieves SX2(2k),2 = S. On the contrary, ifk = 1, 2

thenvT /L2(RN; |x|2d), andSX0,2 =S

X2(2k),2 =Sare not achieved.

In the next Lemma we collect some remarks on the behavior of the map a SXa,2 .

Lemma 1.4 Assume N 3 and let a R. Then the map a SXa,2 is increasing for

a 2 k, and it is decreasing fora 2 k. Moreover:

1. SXa,2 S for anya R.

2. SXa,2 =S andSXa,2 is not achieved in the following cases:

k = 1 andN= 3, ork = 1, N4 anda /(0, 2);

k = 2

k 3, a /[2(2 k), 0].

3. SXa,2 is achieved in the following cases:

k = 1, N4, a (0, 2), a = 1;

k 3, a [2(2 k), 0], a= 2 k.

Proof. The monotonicity properties of the map a SXa,2 can be checked as in Lemma

1.3. For the proof of 1.we refer to [15], Theorem B.5. By contradiction, assume thatk = 1,

N = 3 and that SXa,6 is achieved by a map u X10 (R

3; |x|ad), for some a R. Then

a (0, 2) by [15], Theorem B.5. Setv(x, y) =|x|a/2u(x, y). Then, by Proposition B.3 and

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by Lemma 1.2 of [15], we can assume that the support ofv is contained in the half space

(0, +) R2, and that v solves

v= |x|2v+ v5 in (0, +) R2

v >0

v D10((0, +) R2) ,

where

=1

4

a 1

2

2, 0< 2, satisfying also p 2 if

N3. Let ba = N pN2+a

2 as in (0.2). Notice that

ba+ap

2 0 .

Thus, Holder and Mazja inequalities give

|x|a|u|2 d ||p

p2

|x|ap2 |u|p d

2p

c||p

p2

RN

|x|a|u|2 d , (2.1)

where||is the Lebesgue measure of the domain and the constant cdoes not depend onu.

This proves the continuity of the embedding. Now, let uh be a sequence in D10(R

N; |x|ad),

withuh 0 weakly inD10(R

N; |x|ad). For >0 small take a smooth function C(Rk)

such that (x) = 0 if|x| 2, and (x) = 1 if|x| . Notice that for fixed it results

that

|x|a|uh|2 d= o(1)

as h + by the standard Rellich Theorem, since |x| is bounded away from 0 on the

support of. Set := {(x, y) | |x|< }, and notice that

|x|a|uh|2 d =

|x|a|uh+ (1 )uh|2 d

2

|x|a|uh|2 d+ 2

|x|a|1 |2|uh|

2 d

o(1) + c

|x|a|uh|2 d .

Next, fix p >2 as before, and argue as for (2.1) to get

|x|a|uh|2 d o(1) + c||

pp2

RN

|x|a|uh|2 d .

Notice that the sequence uh is bounded in D1

0

(RN; |x|ad) and that || 0 as 0.

Therefore, passing to the limit first as h go to 0, and then as 0, we get that

|x|a|uh|2 d 0,

that proves the compactness of the embedding.

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The next Lemma will allow us to use suitable test functions.

Lemma 2.5 If Cc (RN) thenu D10(RN; |x|ad) for anyu D10(RN; |x|ad).

Proof. We can approximate any fixed u D10(RN; |x|ad) with a sequence uh Cc (R

N).

Using Lemma 2.4 it is easy to prove that uh u in D10(R

N; |x|ad), hence u

D10(RN; |x|ad).

The main step in the proof of Theorem 2.1 is the following Lemma. It is based on a rescaling

argument that is quite close to that used in [15].

Lemma 2.6 Letuh be a bounded sequence inD10(R

N; |x|ad), and letfh 0 be a sequence

in the dual ofD10(RN; |x|ad). Assume that fora, p, ba as in (0.3), (0.2) it holds that

div(|x|auh) = |x|ba |uh|p2uh+ fh .

Then either (up to a subsequence) uh 0 strongly in Lp(RN; |x|bad), or there exist

sequences(th)h (0, +) and(h)h RNk, such that

limh+

K

|x|ba |uh|p d >0 ,

whereuh(x, y) = tN2+a

2

h uh(thx, thy+ h) and

K= {(x, y) Rk RNk | 1

2 0. We argue as follows. Firstly we fix some 0 > 0 in

such a way that

p

p2

0 < limh+

RN

|x|ba |uh|p d , 20 < S

Da,p .

Using in a standard way the concentration function

Qh(t) := supRNk

Bkt(0)BNkt ()

|x|ba

|uh|p

d ,

it is possible to select sequences th > 0 and h RNk such that, for a subsequence, the

rescaled sequence

uh(x, y) := tN2+a

2

h uh(thx, thy+ h)

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is bounded in D10(RN; |x|ad) (with the same D10(R

N; |x|ad) and Lp(RN; |x|bad)-norms

asuh), and satisfiesBk1(0)BNk1 (y)

|x|ba |uh|p d (20)

pp2 y RNk , (2.2)

Bk1(0)BNk1 (0)

|x|ba |uh|p d (0)

pp2 >0 , (2.3)

div(|x|auh) = |x|ba |uh|

p2uh+ fh , (2.4)

with fh 0 in the dual ofD10(R

N; |x|ad). As before, if (up to a subsequence) it happens

that uh u = 0 then we are done. If uh 0, we choose a finite number of points

y1,...,ys RNk in such a way that

BNk1 (0)

sj=1

BNk1/2 (yj ) . (2.5)

Let 1, ...s be cut-off functions, such that j = j (y) Cc (BNk1 (yj )), j 1 on

BNk1/2 (yj ), 0 j 1. Also, fix a map = (x) Cc (B

k1 (0)) s.t. 0 1 and

1 onB k1/2(0). Thanks to Lemma 2.5 we can use 22iuh as test function in (2.4) to find

RN

|x|auh (22juh)d=

RN

|x|ba |uh|p2|j uh|

2 d+ o(1).

Notice thatRN

|x|auh (22juh)d =

RN

|x|a|(j uh)|2 d

RN

|x|a|uh|2(|j x|2 + |yj |2) d

=

RN

|x|a|(j uh)|2 d+ o(1)

by Lemma 2.4. Then, we use Holder inequality, (2.2), and the definition ofSDa,pto infer that

SDa,p

RN

|x|ba |j uh|p d

2p

20

RN

|x|ba |j uh|p d

2p

+ o(1).

Since 20 < SDa,p this implies thatRN

|x|ba |j uh|p d= o(1), and therefore, by (2.5),

Bk1/2(0)BNk1 (0)|x|ba |uh|

p ds

j=1 Bk1/2(0)BNk1/2 (yj)|x|ba |uh|

p d= o(1).

Finally, from (2.3) we get

0< p

p2

0 0 such thatSDa,p< SXa,p if

0< p 2(N k)

N 2 + a< .

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Proof. Fix any map w Cc (RN) such that w 1 on {(x, y) | |x| 1 , |y| 1 }. Then

compute

SDa,p

RN

|x|a|w|2 dRN

|x|ba |w|p d2/p C

{|x|


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achieves the best constantSD2k,pk . Now we set, for a 2 k,

Ra :=RN|x|a|uM|2 d

RN|x|ba |uM|pk d

2pk

. (3.4)

We are going to prove by direct computation that Ra< R2k fora


18/26


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By Remark 2.2 we get that uD D10(RN) is cylindrically symmetric whenever uD

achieves the best constant SD

0,p

. On the other hand, by Lemma 1.2 in [15], if uX

X10(RN; |x|0d) achieves SX0,p, then the support of u

X is contained in a half-space, and

hence it cannot achieveSD0,p. This proved the next result.

Theorem 3.5 Assumek = 1, N3, and2< p 0 ,

withb0 = N pN22 , has at least two distinct (moduloG1) entire classical solutions:

uX D10((0, +) RN1), withuX (x, y) 0 forx 0 .(3.9)

Theorem 2.1 provides the existence of a solution to (3.9) whenever

SDa,2 < S . (3.10)

Notice that a first set of sufficient conditions for (3.10) can be easily obtained from Lemma

1.4. By this argument and by the symmetry result in [10] one can prove the following result

(see also [16] for existence).

Theorem 3.6 Assume N 4 and k = 2. Then the infimum SD2k,2 is achieved by an

entire solutionu todiv(|x|2ku) = |x|N

k2N2 u2

1 inRN , x= 0

u > 0 .

Moreoveru is symmetric: u(x, y) = u(|x|, |y|), and decreasing in the |y|-variable.

From now on we take a


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Lemma 3.7 Inequality (3.10) holds true if

(2 N)kN

< a 2 k and a < 0 .

Proof. We are going to estimate the infimum SDa,p by using the map

vT(x, y) =

1 + |x|2 + |y|2N2

2

already defined in (0.9). In order to simplify notations we set r = |x|,s =|y|, := N+k2 and

:=

+0

tNk1

(1 + t2)N1 dt , a:=

+0

ra+k1

(1 + r2) dr , a :=

+0

r NaN2+k1

(1 + r2) dr .

We start with two identities, that can be proved via simple computations: +

0

tNk1

(1 + t2)N dt=

N+ k 22(N 1)

(3.12)

+0

ra+k1

(1 + r2)1 dr =

N+ k 2

N 2 a a (3.13)

It turns out that

|vT|2 = (N 2)2

1

(1 + r2 + s2)N1

1

(1 + r2 + s2)N

.

First we compute, for every r ,

+0

sNk1

(1 + r2 + s2)N ds =

1

(1 + r2)N +0

sNk11 +

s(1 + r2)

12

2N ds

= 1

(1 + r2)

+0

tNk1dt

(1 + t2)N =

N+ k 2

2(N 1)

1

(1 + r2)

by (3.12). In the same way one gets +0

sNk1

(1 + r2 + s2)N1 ds=

1

(1 + r2)1 .

SetcN,k := kNk, and compute

RN

|x|a

|vT|2

d = cN,k(N 2)2 +

0 ra+k1

dr +0 |v

T|2

sNk1

ds

= cN,k(N 2)2

+0

ra+k1 dr

(1 + r2)1

N+ k 2

2(N 1)

+0

ra+k1 dr

(1 + r2)

= cN,k(N 2)2 N+ k 2

2(N 1)

N+ a

N 2 a a

20


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by (3.13). We also compute

RN

|x| NaN2 |vT|2

d = cN,k +0

r NaN2+k1 dr +0

sNk1(1 + r2 + s2)

N ds

= cN,k N+ k 2

2(N 1)a .

Next, we use Holder inequality (with conjugate exponents NN2 and N2) to estimate

a =

+0

ra+k1 dr

(1 + r2)

N2N

a 2N0 . (3.14)

and we set

Ra :=

RN

|x|a|vT|2 d

RN|x| NaN2 |vT|2 dN2N

,

in such a way that S= R0 andSDa,2 Ra. From the above equality we get

S=

cN,k

N+ k 2

2(N 1)0

2N

N(N 2) (3.15)

since 0 = 0. Therefore, using also (3.14) and (3.15), we can estimate

SDa,2 Ra

cN,k

N+ k 2

2(N 1)0

2N

(N 2)2 N+ a

N 2 a=S

N 2

N

N+ a

N 2 a .

The conclusion readily follows.

We can get new sufficient conditions for (3.11) in the special case N= 2(k 1). Indeed,

in this case the exponent pk defined in (3.2) coincides with the critical exponent 2. The

arguments and the computations of Lemma 3.2, together with the existence Theorem 3.6

and the uniqueness result in [13] lead to the following result.

Lemma 3.8 Inequality (3.11) holds true if2 < k < N,N= 2(k1)and N+22 < a


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k= N+22 ;

k= 3, N7 anda=2;

k= 3, N= 5, 6, ork 4 anda is close enough to (2 N)k/N.

a

(2N) 0 1 21

SX

N

XS=

DS

DS ?

S

a

SD

= SX

SX

S

N(2N)2 0

SD

Fig.2 k = 1 andN4. Fig.3 k = 2.

a0

SD= S

XS

XS

2(2k)(2N)

k

N

2k

SD

a0

SD

= SX

SX

S

SD

(2N)3

N

2 1

?

Fig.4 k = N+22 . Fig.5 k = 3 and N7.

Remark 3.11 We do not know whether SDa,2 < SXa,2 =Sholds true ifk = 1, N= 3 and

a (0, 1). Ifk = 1,N4 anda (0, 1) we know thatSDa,2 , SXa,2 are both achieved, but we

do not know ifSDa,2 < SXa,2 . The same question is still open ifk 3 anda (2(2k), 2k),

unlessN= 2(k+ 1).

4 Existence results for (0.7)

In this section we deal with classical solutions to problem (0.7) under the assumption

1(0) =

k 2

2

2.

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In particular, solutions u we are looking for satisfy

RN|x|

b0 |u|p d < +, even if they

might be singular, in the sense that one or both of the integralsRN

|u|2 d ,

RN

|x|2|u|2 d

might be unbounded. We distinguish the cases = 1(0) and < 1(0).

4.1 Existence for = 1(0)

In this Subsection we study problemv=

k22

2|x|2 v+ |x|b0 vp1 in RN , x= 0

v >0,(4.1)

where p > 2, p 2 ifN 3, and b0 = NpN22 . We are going to give an alternative

proof of a result by Tertikas and Tintarev [16], that works also for p (2, 2) andN2.

Theorem 4.1 Problem (4.1) has a cylindrically symmetric classical solutionv such thatRN

|x|b0 |v|p d =

RN

|v|2 1(0)|x|

2|v|2

d


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4.2 Existence of singular solutions for < 1(0)

Here we study problem (0.7) in case < 1(0) =

k222

. A first solution v X D10(RN) L2(RN; |x|2d) can be find by studying the minimization problem for the the infimum SXa,pin (1.9). This was done in [15], Theorems 1 and 2. More precisely, SXa,pis achieved, provided

thatp 0, N4 and k = 2.

Our aim is to use here the results in Sections 2 and 3 to find new classical solutions

v Lp(RN; |x|b0d) to (0.7) that are singular in the sense thatRN

|x|2|v|2 ddiverges.

Notice that Theorem 3.5 already provides the existence of a solution u / L2(RN; |x|2d)

whenk = 1, = 0 and p (2, 2).

To handle the case k 1 we use a simple trick: assume

1(0) N kN

2

< < 1(0), 2(N k)

N k 2

1(0)

< p 2 , (4.2)

and define

a= ak,:= 2 k 2

1(0) .

Notice that with this choice, assumptions (0.3) on a and p are satisfied by (4.2). Moreover,

it turns out that ba = N pN2+a

2 = b pa2 , accordingly with (0.2). Assume that u is a

solution to (0.1) with respect to this choice of the parameters a, b and with respect to the

samep. Then the map

v= Lau:=|x|a2 u

is aC(RN0 )- solution to (0.7). In addition, if for example u D10(R

N; |x|ad), then

RN

|x|b0 |v|p d=RN

|x|ba |u|p d 2 andp 2 ifN 3. Then problem (0.7) has a

solutionv Lp(RN; |x|b0d) that satisfies

RN

|x|2v2 d= + ,

provided that one of the following conditions is satisfied:

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i) p2 andp

N k 2

1(0)

is close enough to 2(N k);

ii) p= pk = 2(Nk+1)

Nk , 2 k N+22 and1(0) p

2k < < 1(0);

iii) p= 2, k= 1, 2 and1(0)

NkN

2<


26/26

[10] M. Gazzini - R. Musina, On the Hardy-Sobolev-Mazja inequalities: symmetry and breaking

symmetry of extremal functions, preprint 2007.

[11] G. Mancini - I. Fabbri - K. Sandeep,Classification of solutions of a critical Hardy Sobolev

operator, J. Differential Equations 224 (2006), 258-276.

[12] G. Mancini - K. Sandeep,Cylindrical symmetry of extremals of a Hardy-Sobolev inequality,

Ann. Mat. Pura Appl. (4) 183 (2004), 165-172.

[13] G. Mancini - K. Sandeep,On a semilinear elliptic equation inHN, preprint 2007.

[14] V.G. Mazja, Sobolev Spaces, Springer-Verlag, Berlin, 1980.

[15] R. Musina,Ground state solutions of a critical problem involving cylindrical weights, Nonlinear

Analysis TMA, to appear.

[16] A. Tertikas - K. Tintarev, On existence of minimizers for the Hardy-Sobolev-Mazya in-

equality, Ann. Mat. Pura e Appl. 186 (2007), 645-662.

[17] S. Terracini, On positive entire solutions to a class of equations with a singular coefficient

and critical exponent, Adv. Diff. Eq. 2 (1996), 241-264.

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