Hardy-Sobolev inequalities,

hyperbolic symmetry and the

Webster scalar curvature problem

D. Castorina, I. Fabbri, G. Mancini∗and K. Sandeep†

Abstract

We discuss the problem −∆u = φ(x)up(t)

|y|t , u ∈ D1,2(RN ) wherex = (y, z) ∈ Rk × Rh = RN , 2 ≤ k < N, t ∈ (0, 2), p(t) := N−2t+2

N−2

in connection with Grushin-type equations and the Webster scalarcurvature problem, providing various existence results. We highlightthe role of hyperbolic symmetry in nondegeneracy and uniquenessquestions and present a uniqueness result for semilinear elliptic equa-tions in Hyperbolic space which applies to the above equation whenN > 3, φ = 1 and to semilinear Grushin-type equations as well.

1 Introduction

The main purpose of this paper is to find positive solutions of

−∆u = φ(x)up(t)

|y|tu ∈ D1,2(RN) (1.1)

where x = (y, z) ∈ Rk × Rh, k ≥ 2 and the potential φ : RN → [0,∞)

is assumed to be continuous, bounded and eventually enjoying some kind of

∗Dipartimento di Matematica, Universita degli Studi ”Roma Tre”, Largo S. LeonardoMurialdo, 1 - 00146 Roma, Italy. E-mail castorin@mat.uniroma3.it, fabbri@mat.uniroma3.it,mancini@mat.uniroma3.it.†

1

symmetry. Also, t ∈ (0, 2) and p(t) := N−2t+2N−2

= 2∗(t) − 1 is the Hardy-

Sobolev exponent, i.e. for an optimal constant S = St,N,k > 0 we have

S

∫RN

|u|2∗(t)

|y|t

22∗(t)

≤∫

RN

|∇u|2 ∀ u ∈ D1,2(RN) (1.2)

We recall that t = 2 (Hardy-type inequality) is allowed in (1.2) if k ≥ 3.

Equation (1.1), with a suitable φ, was proposed by two astro-physicists

G.Bertin and L.Ciotti (see [7]) as a model to describe the dynamics of axially

symmetric galaxies. Thanks to (1.2), its solutions turn out to be the critical

points of some C1(D1,2(RN)) variational integral J (see (2.1)).

When φ ≡ 1, (1.1) has a special interest, because its solutions are the ex-

tremals of the Hardy-Sobolev inequality (1.2) (see also [19] for more general

Hardy-Sobolev-Maz’jia inequalities).

Another relevant feature of (1.1) is that, for φ radially symmetric in y, it is

related to an elliptic equation in the n dimensional hyperbolic space H

∆Hv +h2 − (k − 2)2

4v + φvp = 0 (1.3)

where n = h+ 1 and ∆H is the Laplace-Beltrami operator in H.

As a consequence, (1.1) is also related (see Appendix 7.2 ), in the cylindrically

symmetric case, to critical Grushin-type equations (see [22], and, in a more

general context, [16]) in Rk × Rh

Lv := −∆yv − (α + 1)2|y|2α∆zv = vQ+2Q−2 (1.4)

(y, z) ∈ Rk × Rh, where α > 0, k, h ≥ 1, Q = k + h(1 + α).

In the particular case α = 1, k = 2m and h = 1, L acts like the Kohn

laplacian ∆H on the Heisenberg group and (1.1) turns out to be equivalent

to the Webster scalar curvature equation (see Section 3) : if v = v(|y|, z)solves (1.1) (with k = m+ 1, h = 1 = t, then u(y, z) = v(|y|2, z) solves

−∆Hu = φ(|y|2, z) uQ+2Q−2 , Q = 2m+ 2 (1.5)

Going back to (1.1), we recall that scale and translation invariance of (1.2)

induce lack of compactness and non existence phenomena appear: in [6]

2

it is shown that J has no nontrivial critical points if φ = φ(|y|) is strictly

monotone. However, existence of ground state solutions has been established,

by means of concentration-compactness techniques, in some cases ([6], [25]).

We address here the problem of existence of unstable solutions .

As usual, knowledge of the behaviour of Palais-Smale sequences and/or of the

limiting problem (i.e. φ ≡ constant), are crucial tools in detecting unstable

(or even stable) solutions. A first step is to identify positive entire solutions

(i.e. in D1,2(RN)) of the limiting equation

−∆u =|u|p(t)−1u

|y|tin RN (1.6)

This has been done in [13] (see also [30], [1], and [16]) in case t = 1: up to

dilations and translations in z, they are given by

U(y, z) =

[(N − 2)(k − 1)

(1 + |y|)2 + |z|2

]N−22

(1.7)

Here we perform an asymptotic analysis of P-S sequences of J , given in (2.1),

for rather general φ. In case t = 1, thanks to the classification result, we get

enough informations to apply succesfully min-max techniques. In particular,

we obtain a new kind of global existence result:

Theorem A. Let t = 1 and φ = 1 + ρ with ρ ∈ Cc(RN \ y = 0

). Then

(1.1) has a positive solution.

As a matter of fact, due to severe technical difficulties, we limited our global

variational analysis to the case of potentials φ ≡ const. on y = 0, the set

of all possible concentration points. In Section 6 we drop this restriction, but

only in a perturbative setting.

A crucial property to handle perturbed problems is nondegeneracy of so-

lutions of the limiting problem: in case t = 1, we prove nondegeneracy of

solutions (1.7) exploiting hyperbolic symmetry of positive solutions of (1.3).

Hyperbolic symmetry, which was used in [5] in the context of Grushin oper-

ators, turns out to be also a key tool to prove uniqueness for (1.6) as well as

for (1.4) in general cases. As a byproduct, we get a direct ODE proof of the

classification result in [13].

3

As for (1.4), uniqueness was established in case k = h = 1 and, for cylindri-

cally symmetric solutions, in case k ≥ 3, h = 1 in [22]. Here we will present

a uniqueness result (of cylindrically symmetric solutions) for h ≥ 2.

Going back to the existence problem for (1.1) in case t = 1 and general φ, a

tipical result in this direction, based on an index-counting type condition (a

la Bahri-Coron, [8]), is the following

Theorem B. Let t = 1 and N ≥ 4. Let φ = 1 + εϕ, ψ(z) := ϕ(0, z)

and suppose ψ has a finite number of non degenerate critical points ζj,

j = 1, . . . ,m of index m(ψ; ζj). Let φ(x) := φ( x|x|2 ) and, for ζ ∈ Rh, set

∆∗(ζ) := (k − 1)∆yϕ(0, ζ) + (2k + h− 3)∆zϕ(0, ζ) (1.8)

Assume the non degeneracy conditions

(i) ∆∗(ζj) 6= 0 ∀j(ii) φ, ∇φ have a limit as x goes to zero and lim

x→0∇zφ 6= 0

Then (1.1) has a solution for |ε| small provided∑j: ∆∗(ζj)>0

(−1)m(ψ;ζj) 6= 1 (1.9)

The plan of the paper is as follows.

In Section 2, we give a description of the behaviour of P-S sequences of J .

In Section 3, we prove some global existence results for (1.1), including The-

orem A. We also establish a relation with the Webster scalar curvature equa-

tion in the cylindrically symmetric case, giving a striking improvement of

results in [14]. This new result seems not to have any counterpart in the

analogous ’scalar curvature problem’ in RN .

In Section 4 we discuss hyperbolic symmetry for related elliptic equations on

the hyperbolic space, and present uniqueness results for (1.3), with φ ≡ 1,

which apply both to (1.6) and to critical Grushin type equations (1.4).

In section 5-6 we develop the basic tools for an exhaustive analysis of (1.1),

when t = 1, in the perturbative case: we show nondegeneracy and develop a

finite dimensional reduction (similar to the one in [3]) to prove Theorem B

and some other existence/multiplicity results.

4

2 Palais-Smale Characterisation

The main difficulty in the study of (1.1) is its lack of compactness or failure

of Palais-Smale condition. More precisely if J denotes the energy functional

J(u) =1

2

∫RN

|∇u|2 dx− 1

p(t) + 1

∫RN

φ(x)|u|p(t)+1

|y|tdx, u ∈ D1,2(RN) (2.1)

we say that un is a P-S sequence of J if J(un) is bounded and J ′(un)→ 0 in

the dual space of D1,2(RN). If, in addition, limn→∞

J(un) = β, we say un is a

P-S sequence at level β. Scale invariance of (1.2) implies that P-S sequences

do not have, in general, any subsequence converging strongly in D1,2(RN).

For example, if u solves (1.6), then for any z0 ∈ Rh and εn → 0,

un := (φ(0, z0))−1

p(t)−1 ε2−N

2n u(ε−1

n (y, z − z0))

is a P-S sequence for J with no subsequence converging in D1,2(RN). More-

over if I denotes the energy associated with (1.6),

I(u) =1

2

∫RN

|∇u|2 dx− 1

p(t) + 1

∫RN

|u|p(t)+1

|y|tdx (2.2)

then, J(un) = (φ(0, z0))−2

p(t)−1 I(u) +o(1). Lack of compactness due to concen-

tration appears in several interesting problems in geometry and physics (e.g.

the scalar curvature problem) and in many cases an analysis of P-S sequence

has been carried out ([24] and [28] to quote a few). Our aim is to obtain

similar classification results for P-S sequences of J.

Here and in the next section we will assume the (normalized) condition :

φ ∈ C(RN), φ > 0, limx→∞

φ(x) = 1 (2.3)

However, in case φ depends only on y (as in [7]) we will simply assume

φ ∈ C(Rk), φ > 0, limy→∞

φ(y) = 1 (2.4)

Before stating the theorems let us fix some notations. Given z0 ∈ Rh, λ > 0

and a function u, we denote by uz0 , uz0,λ the translated and dilated functions:

uz0(y, z) = u(y, z − z0), uz0,λ(x) = λ2−N

2 u(λ−1(x− (0, z0)))

5

Theorem 2.1. Let φ satisfy (2.3) and un be a P-S sequence for J . Then,

up to a subsequence , un = u0 + u1n + o(1) where o(1) → 0 in D1,2(RN) , u0

is a solution of (1.1) and, for some vj, wj, wj solutions of (1.6),

u1n =

k1∑j=1

vzjnj +

k2∑j=1

wζjn,R

jn

j +

k3∑j=1

wηjn,ε

jn

j

φ(0, ηj)1

p(t)−1

where zjn, ζjn, η

jn ∈ Rh with zjn →n ∞ for all j, ζjn, η

jn → ζj, ηj ∈ Rh∪∞

and Rjn →n ∞, εjn →n 0. Moreover,

J(un) = J(u0) +

k1∑j=1

I(vj) +

k2∑j=1

I(wj) + φ2(0, ηj)

k3∑j=1

I(wj) + o(1)

In case φ depends only on y and satisfies (2.4), the v′js above are solutions

of (1.1) and I(vj) has to be replaced by J(vj) in the energy expansion.

Remark 1. (i) Using standard arguments one can show that if un ≥ 0 in

the above two theorems then u0, v′js and w′js are also nonnegative.

(ii) If un are cylindrically symmetric or having partial radial symmetry in the

y or z variable then u0, v′js and w′js also enjoy the corresponding symmetry.

Proof of Theorem 2.1. We just outline the main steps. Using standard

arguments (see [28]) one can see that any P-S sequence is bounded and hence,

for a subsequence, un n u0 for some u0 ∈ D1,2(RN). Furthermore, un − u0

is again a P-S sequence and J(un) = J(u0)+J(un−u0)+o(1). Still denoting

un − u0 as un, we now consider the case un n 0. If

lim infn→∞

∫RN

φ(x)|un|p(t)+1

|y|tdx = 0

then, along a subsequence, un →n 0 in D1,2(RN) and the proof is complete.

If not, we have, again for a subsequence,

limn→∞

∫RN

φ(x)|un|p(t)+1

|y|tdx ≥ C > 0

Given S as in (1.2), 0 < δ < SN−22−t ||φ||

N−2t−N∞ , let zn ∈ Rh, Rn > 0 be such that

supζ∈Rh

∫B(ζ,R−1

n )

φ(x)|un|p(t)+1

|y|tdx =

∫B(zn,R

−1n )

φ(x)|un|p(t)+1

|y|tdx = δ (2.5)

6

We also have, eventually for a subsequence, Rn →n R0 ∈ [0,∞].

Case 1 . Let R0 ∈ (0,∞). We claim that, for a subsequence, zn → ∞ and

vn := uznn n v for some v ∈ D1,2(RN), a non trivial solution of (1.6) (if

(2.3) holds true) or of (1.1) (if (2.4) holds true). Furthermore, wn := vzn and

un − wn are P-S sequences for J and J(un) = J(wn) + J(un − wn) + o(1).

To prove the claim, notice that vn is bounded in D1,2(RN) and hence, for a

subsequence, vn n v for some v ∈ D1,2(RN). Due to (2.5) ,∫B(ζ,R−1

n )

φ(y, z)|vn|p(t)+1

|y|t≤

∫B(0,R−1

n )

φ(y, z + zn)|vn|p(t)+1

|y|t= δ (2.6)

for every ζ ∈ Rh. Standard arguments (see e.g. [28]) imply that if v = 0

then vn →n 0 in D1,2loc , thanks to the choice of δ. This would contradict (2.6),

and hence v 6= 0. In turn, this implies that zn →∞, because un n 0. Since

un is a P-S sequence we have∫RN

∇vn.∇ψ =

∫RN

φ(y, z + zn)|vn|p(t)−1

|y|tvnψ + o(1), ∀ψ ∈ D1,2(RN)

Taking the limit we see that v solves (1.6) in case (2.3) and (1.1) in case

(2.4). Finally, wn(y, z) := v(y, z− zn) is clearly a P-S sequence and standard

arguments (see [28]) show that also un − wn is a P-S sequence converging

weakly to zero and satisfying J(un) = J(wn) + J(un − wn) + o(1).

Case 2 : Let R0 = 0 or∞. Define vn := R2−N

2n un(R−1

n x+ (0, zn)). As above,

up to a subsequence, vn v 6= 0 in D1,2(RN) where

−∆v = K|v|p(t)−1v

|y|t(2.7)

and K = φ(0) or K = φ(∞) depending on R0 =∞ or 0 respectively. Hence

v = K−1

p(t)−1w where w is a solution of (1.6). If wn := RN−2

2n v(Rn(y, z+ zn)),

then, as above, wn and un − wn are P-S sequences for J converging weakly

to zero and satisfying J(un) = J(wn) + J(un − wn) + o(1).

Next observe that if un is a P-S sequence, then lim infn→∞

J(un) ≥ 0. Also if u is

a solution of (1.1) or (2.7) then J(u) ≥ C > 0. This together with the above

arguments show that if we are given a P-S sequence with limn→∞

J(un) > 0,

7

then we can find a P-S sequence wn such that un−wn is again a P-S sequence

but with energy reduced at least by C. If limn→∞

J(un − wn) > 0 then we can

again take out a P-S sequence from un − wn so that the new P-S sequence

has energy reduced at least by C. Since the energy of any P-S sequence is

nonnegative and bounded, this process stops in finitely many steps. i.e, in

finitely many steps we get a P-S sequence which converges to 0 in D1,2(RN).

3 Global existence results and geometric im-

plications

Using the P-S characterisation obtained in the previous section, we prove

several global existence results for (1.1). These existence results apply to

the Webster scalar curvature problem on the Heisenberg group under the

assumption of cylindrical symmetry.

3.1 Global existence results for (1.1).

We first look for ground state solutions. For u ∈ D1,2(RN), u 6= 0, let

Qtφ(u) =

∫RN

φ(x)|u|p(t)+1

|y|t

− 2p(t)+1 ∫

RN

|∇u|2 and set Stφ := inf Qtφ(u)

where φ satisfies (2.3) or (2.4). Clearly, if u is a minimizer, so it is |u|.When φ depends only on |y|, the question of whether Stφ is achieved or not has

been studied in [25] and [6]. Now, as a consequence of P-S characterisation,

we see that the results in [25] and [6] extend to more general potentials. Let

S be as in (1.2) and let us denote

Cφ =

(

supz∈Rh

φ(0, z)

)− 2p(t)+1

if φ satisfies (2.3)

(maxφ(0), φ(∞))−2

p(t)+1 if φ satisfies (2.4)

8

Theorem 3.1. Stφ ≤ CφS and it is achieved if Stφ < CφS.

Proof. For any u solution of (1.6).

Stφ = infu6=0

Qtφ(u) ≤ inf

z0∈Rh,λ>0Qtφ(uz0,λ) ≤ CφS

Now assume that Stφ < CφS. Using Ekeland’s variational principle we can

choose a minimising sequence un ≥ 0 for Stφ which is also a P-S sequence for

J. Since Stφ < CφS, it follows from Theorem 2.1 that this is possible only if

(1.1) has a nontrivial solution.

However, one cannot expect in general Stφ to be achieved (actually, if φ ≤ 1

then Stφ = S and is achieved iff φ ≡ 1). On the other hand, from the P-S

characterisation one may suspect sublevels J ≤ c to be disconnected if

Stφ = S and c is close to S. A related situation was first observed by J.M.

Coron [10], and then exploited in [9] and [27], to get existence results for

some other problems . To pursue this idea, we have to take t = 1, because

we need a complete knowledge of positive solutions of the limiting equation

(1.6). That is, we consider

(Pφ)

−∆u = φ(x)uNN−2

|y| u ∈ D1,2(RN)

u > 0 in RN(3.1)

We also assume φ to be constant on the set of all possible concentration

points, i.e. φ(0, z) = 1 ∀ z if it satisfies (2.3) or φ(0) = φ(∞) if it satisfies

(2.4) (the case when φ is not constant on y = 0 will be considered in

Section 6). Recalling that U given in (1.7) is extremal for (1.2), and hence∫RN|∇U |2 = SN−1, we have:

Theorem 3.2. Let φ satisfy (2.4) or (2.3) and φ ≡ φ(∞) = 1 on y = 0.Then (Pφ) has a solution provided

infλ∈(0,∞)

∫RN

φ(λx)

|y|U

2(N−1)N−2 dx > 2−

1N−2SN−1. (3.2)

If φ is cylindrically symmetric or radially symmetric in y or z and satisfy

(3.2), then (Pφ) has a solution enjoying the same symmetry.

9

Remark 2. The above condition is satisfied if inf φ > 2−1

N−2 .

We end this Section with a suitable slight reformulation of Theorem A:

Theorem 3.3. Let φ be of the form φ = 1 + ρ where ρ ∈ Cc(RN \ y = 0

).

Then (Pφ) has a solution. In addition if φ is radially symmetric in y or z

then (Pφ) will have a solution with the same symmetry.

3.2 Webster scalar curvature problem.

Consider the Webster scalar curvature equation

−∆Hnu(ξ) = R(ξ)u(ξ)Q+2Q−2 , u(ξ) > 0, ξ ∈ Hn (3.3)

where Hn = Cn×R = R2n+1 is the Heisenberg group, ∆Hn is the Kohn lapla-

cian and Q = 2n + 2 is the homogeneous dimension. This problem arises in

conformal geometry when one looks, on the unit sphere in Cn+1, for a con-

tact form conformally equivalent to the standard contact form and having

prescribed Webster scalar curvature.

Equation (3.3) with R(ξ) = 1 + εK(ξ) has been studied in [20] and an ex-

istence result has been obtained for ε small enough. In the cylindrically

symmetric case, i.e. R(ξ) = R(|Z|, t), ξ = (Z, t) ∈ Cn × R, (3.3) has been

studied in [14]. However the results therein are mainly of ground state type

or for small perturbations of φ ≡ 1.

We will establish in Lemma 7.7 of Appendix, the connection between (3.3)

and (Pφ) when the Webster scalar curvature has cylindrical symmetry. Thus,

Theorem 3.2 and Theorem 3.3 give existence results for (3.3) which are some-

how global in nature if compared with [14] and [20]: Theorem 3.2 can be

viewed as a non perturbative version of Theorem 2.8 in [14]. We now restate

Theorem 3.3 in terms of the Webster scalar curvature equation.

Theorem 3.4. Let R(ξ) be of the form R(Z, t) = 1 + ρ(Z, t) where ρ ∈Cc (IHn \ Z = 0) and is radially symmetric in the complex variable Z.

Then the Webster scalar curvature problem (3.3) has a solution.

10

3.3 Proofs of Theorems 3.1-3.3.

Let Sφ := S1φ. From our assumptions, we get Sφ ≤ S and, by Theorem 3.1,

(Pφ) has a solution if Sφ < S . Therefore we assume that Sφ = S.

From the characterisation of P-S sequences one can see that there are levels

at which a P-S sequence will give rise to a solution. More precisely

Lemma 3.1. Let un be a P-S sequence at a level β ∈(

SN−1

2(N−1), S

N−1

N−1

). Then

(Pφ) admits a positive solution u with J(u) ≤ β.

Remark 3. Observe that in the above Lemma we are not assuming un ≥ 0.

Proof of Lemma 3.1. From Theorem 2.1 we know that β must be of the

form β =k1∑j=0

J(vj) +k2∑j=1

I(wj) where v′js are solutions of −∆u = φ(x) |u|2

N−2 u|y|

and w′js are solutions of (1.6) with t = 1. We know from [13] that if w′js are

positive then I(wj) = SN−1

2(N−1). If wj changes sign, then

∫RN

|∇w+j |2 =

∫RN

|w+j ||y|

2(N−1)N−2

and

∫RN

|∇w−j |2 =

∫RN

|w−j ||y|

2(N−1)N−2

Using this in (1.2) with t = 1 gives∫RN

|∇w+j |2 ≥ SN−1 and

∫RN

|∇w−j |2 ≥ SN−1

and hence I(wj) = 12(N−1)

∫RN|∇wj|2 = 1

2(N−1)

[ ∫RN|∇w+

j |2 +∫

RN|∇w−j |2

]≥

SN−1

N−1. Similarly since Sφ = S we get J(vj) ≥ SN−1

2(N−1)if vj is positive and

J(vj) ≥ SN−1

N−1if vj changes sign. These estimates on I(wj) and J(vj) show

that β must be of the form β = J(v0) where v0 is a solution of Pφ.

In order to solve Pφ it is enough to build, in view of the previous Lemma,

a P-S sequence as at a ’good’ level. We will do this using a Mountain pass

type argument. We proceed as follows. Let Nφ be the Nehari Manifold

11

Nφ =

u ∈ D1,2(RN) : u 6= 0,

∫RN

|∇u|2 =

∫RN

φ(x)|u|

2(N−1)N−2

|y|

Nφ is a C2 submanifold of D1,2(RN) and for every u ∈ D1,2(RN) with u 6= 0

there exists a unique constant C(u) > 0 such that C(u)u ∈ Nφ. It results

J(u) =1

2(N − 1)

∫RN

|∇u|2 =1

2(N − 1)

∫RN

φ(x)|u|

2(N−1)N−2

|y|∀u ∈ Nφ

Also, infNφJ =

SN−1φ

2(N−1)= SN−1

2(N−1). Now, if un ∈ Nφ is a P-S sequence for J |Nφ

then un is also a P-S sequence for J in D1,2(RN). To prove our theorem we

build a P-S sequence at level β ∈(

SN−1

2(N−1), S

N−1

N−1

)for J |Nφ . For U given by

(1.7) and λ > 0 define

Uλ(x) = λ2−N

2 U(λ−1x), V λ = CλUλ ∈ Nφ (3.4)

Now, we say that a continuous maps γ : (0,∞) → Nφ is in Γ if for some

0 < t1 < t2 it results γ(t) = V t for t ≤ t1, and t ≥ t2 . Define

β := infγ∈Γ

[suptJ(γ(t))

]We have the following estimates on β:

Lemma 3.2. Let φ satisfy (3.2) or let φ be of the form φ = 1 + ρ, where

ρ ∈ Cc(RN \ y = 0). Then

SN−1

2(N − 1)≤ β <

SN−1

N − 1

Proof. Since limt→0

J(V t) = SN−1

2(N−1), we have SN−1

2(N−1)≤ β. To prove the upper

bound on β in case φ satisfies (3.2), we just compute suptJ(γ0(t)) where

γ0(t) := V t, ∀ t. By direct computation one can show that

J(γ0(λ)) =1

2(N − 1)

∫RN

|∇V λ|2 =S(N−1)2

2(N − 1)

∫RN

φ(x)|Uλ|

2(N−1)N−2

|y|

2−N

12

Substituting Uλ we can see that β ≤ supλJ(γ0(λ)) < SN−1

N−1holds if (3.2) is

satisfied and the lemma is proved in this case.

Estimating β from above in the other case is much more involved. Define,

for ε > 0,

γε(t) =

V t if t ≤ ε or t ≥ 1

ε

Cεt

(ε(t−ε)1−ε2 V

1ε +

(1− ε(t−ε)

1−ε2

)V ε)

if ε < t < 1ε

where Cεt is chosen such that γε(t) ∈ Nφ. We will show that for ε small enough

suptJ(γε(t)) < SN−1

N−1. Clearly for ε small enough sup

t≤ε,t≥ 1ε

J(γε(t)) < SN−1

N−1.

Therefore it is enough to show that sup0<λ<1

J(Cελ(λV

1ε + (1 − λ)V ε)) < SN−1

N−1,

where Cελ is chosen so that W ε

λ = Cελ(λV

1ε + (1− λ)V ε) ∈ Nφ. Now,

J(W ελ) =

1

2(N − 1)

( ∫RN|∇(λV

1ε + (1− λ)V ε)|2

)N−1

( ∫RN

φ(x) (λV1ε+(1−λ)V ε)

2(N−1)N−2

|y|

)N−2(3.5)

Recall that, we have V ε = CεUε where

Cε =

∫

RN|∇U ε|2

∫RN

φ(x) (Uε)2(N−1)N−2

|y|

N−2

2

Now, Lemma 7.4 gives

V ε = (1 +O(εn−1))U ε V1ε = (1 +O(εn−1))U

1ε (3.6)

Using (3.6) we have∫RN

|∇(λV1ε + (1− λ)V ε)|2 = (1− λ)2

∫RN

|∇U ε|2 + λ2

∫RN

|∇U1ε |2

+2λ(1− λ)

∫RN

∇U ε.∇U1ε +O(εN−1)

13

Since U ε is an extremal of (1.2) and solves (1.6) with t = 1, the above

expression simplifies to∫RN

|∇(λV1ε + (1− λ)V ε)|2 =

(λ2 + (1− λ)2

)SN−1

+2λ(1− λ)

∫RN

(U ε)NN−2

U1ε

|y|+O(εN−1)

Now using Lemma 7.2 we get∫RN

|∇(λV1ε + (1− λ)V ε)|2 =

(λ2 + (1− λ)2

)SN−1 + 2λ(1− λ)(CN,k + o(1))εN−2 (3.7)

Similarly using the inequality (a + b)p > ap + bp for a, b positive and p > 1

and (3.6) , we get∫RN

φ(x)(λV

1ε + (1− λ)V ε)

2(N−1)N−2

|y|≥(λ

2(N−1)N−2 + (1− λ)

2(N−1)N−2

)SN−1+O(εN−1)

Thus combining the above inequalities we get

J(W ελ) ≤ SN−1

2(N − 1)

(λ2 + (1− λ)2 + o(1))N−1(

λ2(N−1)N−2 + (1− λ)

2(N−1)N−2 + o(1)

)N−2(3.8)

Now the quantity(λ2+(1−λ)2)

N−1 λ

2(N−1)N−2 +(1−λ)

2(N−1)N−2

!N−2 is less than 2 for λ 6= 12

and is equal

to 2 when λ = 12. Therefore at λ = 1

2we need more refined estimates.

Assume that λ = 12. Then

∫RN

φ(x)(λV

1ε + (1− λ)V ε)

2(N−1)N−2

|y|=

= 2−2(N−1)N−2

∫|x|≤1

φ(x)(V

1ε + V ε)

2(N−1)N−2

|y|+

∫|x|>1

φ(x)(V

1ε + V ε)

2(N−1)N−2

|y|

14

Now using the inequality (a+ b)p ≥ ap+pap−1b, for p > 1, a > 0, b > 0, (3.6),

Lemma 7.1, Lemma 7.3, Lemma 7.4 and Lemma 7.5 we get∫|x|≤1

φ(V

1ε + V ε)

2(N−1)N−2

|y|≥∫|x|≤1

φ(V ε)

2(N−1)N−2

|y|+

2(N − 1)

N − 2

∫|x|≤1

φV1ε(V ε)

NN−2

|y|

=

∫|x|≤1

φ(U ε)

2(N−1)N−2

|y|+

2(N − 1)

N − 2

∫|x|≤1

φU1ε(U ε)

NN−2

|y|+O(εN−1)

= SN−1 +2(N − 1)

N − 2

∫RN

U1ε(U ε)

NN−2

|y|+ o(εN−2)

Similarly,∫|x|>1

φ(V

1ε + V ε)

2(N−1)N−2

|y|≥ SN−1 +

2(N − 1)

N − 2

∫Rn

U ε (U1ε )

NN−2

|y|+ o(εN−2)

Combining all the above inequalities we get∫RN

φ(x)(V

1ε + V ε)

2(N−1)N−2

|y|≥ 2

−2(N−1)N−2

(2SN−1 +

4(N − 1)

N − 2CN,kε

N−2 + o(εN−2)

)

Plugging this inequality and (3.7) in (3.5), we get

J(W ελ) ≤ SN−1

N − 1

(1 + CN,kS

−(N−1)εN−2 + o(εN−2))N−1(

1 + 2(N−1)N−2

CN,kS−(N−1)εN−2 + o(εN−2))N−2

=SN−1

N − 1

(1− (N − 1)CN,kS

−(N−1)εN−2 + o(εN−2))

if λ = 12. Thus J(W ε

λ) < SN−1

N−1when λ = 1

2and hence in a small interval

(12− δ, 1

2+ δ). In the complement of this interval, we already know from (3.8)

that J(W ελ) < SN−1

N−1when ε is small enough. Hence for ε is small enough

sup0<λ<1

J(W ελ) < SN−1

N−1. This proves the lemma.

Proof of Theorem 3.2 and Theorem 3.3. Let us assume, by contradic-

tion, that (Pφ) has no solution. In view of Lemma 3.2 and of Lemma 3.1, we

15

infer that β = SN−1

2(N−1)and hence there exists a sequence γm ∈ Γ such that

SN−1

2(N − 1)≤ J(γm(t)) ≤ SN−1

2(N − 1)+

1

m∀t ∈ (0,∞) (3.9)

We claim that, for m large enough, γm(t) ∈ Ωε,λ∪Ωε,λ for every t , where

Ωε,λ = U ζ,λ + v : ζ ∈ Rh, 0 < λ < λ, ‖v‖ < ε

Ωε,λ = U ζ,λ + v : ζ ∈ Rh, λ > λ > 0, ‖v‖ < ε

We notice that Ωε,λ ∩ Ωε,λ = ∅ for ε small, λ small and λ large. In fact,∫RN

|∇(U ζ1,λ1 − U ζ2,λ2)|2 = 2SN−1 − 2

∫RN

U ζ2,λ2 [U ζ1,λ1 ]NN−2

dx

|y|≥ 2SN−1−

−SN2

S

∫B 1λ2

(0,ζ1−ζ2λ2|)

U2N−1N−2

|y|

N−2

2(N−1)

+

∫|x|≥ 1

λ1

U2N−1N−2

|y|

N

2(N−1)

≥ SN−1

if λ1 << 1 << λ2, uniformly in ζ1, ζ2.

Since by (3.6) γm(t) ∈ Ωε,λ for t small while γm(t) ∈ Ωε,λ for t large, the

above claim gives a contradiction. So, we are left with the proof of the claim.

By contradiction, let γm(tm) /∈ Ωε,λ ∪ Ωε,λ. By (3.9), γm(tm) is a minimizing

sequence for J |Nφ . Now using Ekeland’s principle we can choose vm suffi-

ciently close to γm(tm), still a minimizing sequence for J |Nφ , which is also a

P-S sequence for J |Nφ and hence for J at level SN−1

2(N−1). By the P-S charac-

terisation, we have vm = U ζm,λm + (1) where λm either goes to zero or to

infinity, a contradiction. Thus the claim is proved.

When (Pφ) has cylindrical symmetry or partial radial symmetry as in the

statements of the theorems, one can obtain a solution with the same sym-

metry just working in the subspace of D1,2(Rn) consisting of functions with

the desired symmetry. This completes the proof of both theorems.

16

4 Hyperbolic symmetry and uniqueness

We present here a uniqueness result concerning a semilinear elliptic equation

on the hyperbolic space, which applies both to (1.6) and to (1.4). In order

to mantain this paper into a reasonable size, we just sketch, here, the main

steps while giving detailed proofs in a separate paper.

We recall that if u ∈ D1,2 is a positive solution of (1.6) then u ∈ C∞ for

y 6= 0 and, as shown in [13] (see also [30]), u is Holder continuous up to

the singular set y = 0. In addition, it was proved in [13] that u is radial

in y (and in z). We want to prove that u has additional symmetries. Set

u := rN−2

2 u(r, z). Then u solves (see Appendix 7.3)

−∆Hu := −(r2∆u− (h− 1)r ur) = c u+ up in R+ × Rh (4.1)

where c = h2−(k−2)2

4, p = N+2−2t

N−2and ∆H denotes the Laplace-Beltrami oper-

ator on the n := h+ 1 dimensional hyperbolic space H (half space model)

H := Hh+1 =

(R+ × Rh, ds2 :=

dr2 + |dz|2

r2

)with associated volume form dVH = drdz

rh+1 and, accordingly,∫H

v2dVH =

∫R+×Rh

v2drdz

rh+1,

∫H

|∇Hv|2dVH =

∫R+×Rh

r2|∇v|2drdzrh+1

We notice that, if k ≥ 3, u ∈ H1(H), in view of the identity (easy to check)∫RN

|∇u|2 +h2

4

u2

|y|2dydz = ωk

∫H

[|∇Hu|2 + (

k − 2

2)2u2

]dVH

and of the Hardy inequality.

Remark 4. The same change of variable sends cylindrically symmetric so-

lutions of the more general equation involving inverse square potentials (see

[19])

−∆u = λu

|y|2+up(t)

|y|tin RN (4.2)

into solutions of (4.1) with c = λ+h2−(k−2)2

4. Here, as above, N = k+h, k ≥ 2

and x = (y, z) ∈ Rk × Rh, t ∈ (0, 2) and p(t) := N−2t+2N−2

,

17

Also , a cylindrically symmetric solution u of critical Grushin type equations

(1.4) gives, via a suitable change of variables (see Appendix 7.3), a solution

of (4.1) with

c =1

4

[h2 −

(k − 2

α + 1

)2], p =

Q+ 2

Q− 2< 2∗ − 1

2∗ := 2nn−2

if n > 2 and 2∗ = +∞ if n = 2. Again, if k 6= 2, u ∈ H1(H)

provided u is an extremal of the weighted Sobolev inequality (see [22])

S

∫RN

|u|2QQ−2dydz

Q−2Q

≤∫

RN

(|∇yu|2 + (α + 1)2|y|2α|∇zu|2

)dydz := ‖|u‖|2

Again, this follows by the easy-to-check identity

ωk(α + 1)

∫H

|∇Hu|2 +

(k − 2

2(α + 1)

)2

u2dVH = ‖|u‖|+(h(α + 1)

2

)2 ∫RN

u2

|y|2

and the Hardy inequality ‖|u‖|2 ≥ (k−22

)2∫

RN

u2

|y|2 (see [11]). If u is smooth,

decay estimates [C(1 + ‖x‖2)]−Q−2

2 ≤ u ≤ C(1 + ‖x‖2)−Q−2

2 , ‖(y, z)‖ :=[|y|2(α+1) + |z|2

] 12(α+1) , which follow by invariance of u with respect to an

appropriate Kelvin tranform (see [22]), imply as well energy bounds.

Theorem 4.1. Let c < 2(p+1)(p+3)2

if n = 2 and c < h2

4if n ≥ 3. Then

(4.1) has at most one positive solution in H1(H).

Remark: Theorem 4.1 clearly yields uniqueness for (1.6) in case k ≥ 3 and

for (1.4) in case k 6= 2, h ≥ 2 and h = 1, k ≥ 3 .

Uniqueness for (1.4) when k = h = 1 is not covered by Proposition 4.1

because, in this case, it results c = 2(p+1)(p+3)2

. Uniqueness in this case has been

proven, actually without any symmetry assumption, in [22].

Case k = 2 (where c = h2

4both for (1.6) and (1.4)) has to be dealt separately.

The starting point to prove Proposition 4.1 is hyperbolic symmetry of positive

H1(H) solutions of (4.1).

Proposition 4.2. Let u ∈ H1(H) be a positive solution of (4.1), with p+1 ≤2∗ and c < h2

4. Then u has hyperbolic symmetry, i.e. there exists a point

x0 ∈ H such that u(x) depends only on the distance between x0 and x in H.

18

The proof of Proposition (4.2) is the same as in [2], Theorem 11 (and 12)

where symmetry is proved in case c < h2−14

. To get the result up to h2

4we

just use, instead of the ’sharp’ Sobolev inequality ((13)-(14) in [2]), the

Poincare-Sobolev inequality

For every n ≥ 3 and every p ∈ (2, 2nn−2

] there is c = c(n, p) > 0 such that

c

∫H

|u|pdVH

2p

≤∫H

[|∇Hu|2 −

(n− 1)2

4u2

]dVH ∀u ∈ C∞0 (H) (4.3)

If n = 2 any p > 2 is allowed. In turn, (4.3) follows easily from the

Hardy-Sobolev-Maz’ja inequalities ([18], 2.1.6 Corollary 3 and 1)

Let n = h+ 1 ≥ 3. Let t ∈ [0, 2), and set p = 2 n−tn−2

. Then there is c = c(n, p)

such that∫Rn

updx

|y|t

2p

≤ c

∫Rn

[|∇u|2 − 1

4

u2

|y|2

]dydz ∀u ∈ C∞0 (R+ × Rh) (4.4)

If h = 1, 2.1.6 Corollary 1 in [18] gives, for any α > 0 , q ∈ (1, 2), u ∈ C∞(R2),∫R2

|y|q(α+1)−2|u(y, z)|qdydz

1q

≤ c(q, α)

∫R2

|y|α|∇u(y, z)|dydz (4.5)

Given p > 2 and taking q = 2p2+p

, α = p−24

, it easily gives, when applied

to |u| p+22 , the inequality

(∫R2

rp−42 |u|p

) 2p

≤ c∫R2

r|∇u|2 ∀u ∈ C∞0 (R2),

which in turn, when applied to u√|y|

, gives (4.3).

In view of Proposition 4.2 , we are led to prove uniqueness for solutions of

(4.1) possessing hyperbolic symmetry. We may take for H the ball model

and prove uniqueness of positive radial solutions u ∈ H1(B) of

∆Bu+ cu+ up = 0 (4.6)

where, ∆Bu =[

1−|ξ|22

]2

∆u+ (n− 2)[

1−|ξ|22

]< ∇u , ξ >, |ξ| < 1.

19

We rewrite (4.6) in hyperbolic polar coordinates |ξ| := tanh t2, v := u(tanh t

2),

to get

v′′ +h

tanh tv′ + c v + vp = 0, v′(0) = 0 (4.7)

and then we apply the energy method introduced in [17]. The most delicate

case corresponds to c positive, because in this case all positive solutions of

(4.7) decay at infinity, but only one has finite energy.

When k = 2, positive entire solutions of (1.6) and (1.4) give rise to solutions

of (4.1), with c = h2

4, which are not in H1(H). Actually, they belong to the

broader class H, the closure of C∞c (H) with respect to the norm

||u|| =

∫H

(|∇Hu|2 −

(n− 1

2

)2

u2

)dVH

12

, u ∈ C∞c (H)

(see also [29]). Functions in H have in fact the following characterization

Lemma 4.3. (Classification of H)

u ∈ H iff v(y, z) := r−n−1

2 u(|y|, z), (y, z) ∈ R2 × Rn−1 is in D1,2(Rn+1).

Moreover

||u||2H =

∫Rn+1

|∇v|2 dx.

For solutions in H we prove hyperbolic symmetry via a suitable Poincare-

Sobolev inequality: ∀n ≥ 3, ∀p ∈ (2, 2nn−2

], ∃Sn,p > 0 such that if u ∈ H,

v is as in Lemma 4.3, t = n+ 1− p(n−1)2

, then

Sn,p

∫RN

|v|p

|y|tdy dz

2p

≤∫

RN

|∇v|2 dx (4.8)

Finally, we prove that a solution v of (4.7) (with c = h2

4), corresponding

to some u ∈ H, has ’fast decay’ and the ’energy method’ in [17] can be

put at work again, if n ≥ 3 (case n = 2 escapes this analysis, because

c = 14> 2(p+1)

(p+3)2). So Theorem 4.1 holds true also in case c = h2

4, n ≥ 3 for

u ∈ H. Details will be presented in a separate paper.

20

5 Hyperbolic symmetry and nondegeneracy

We prove here nondegeneracy of positive entire of solutions of (1.6) when

t = 1, namely of

−∆u =u

NN−2

|y|in RN (5.1)

If U is as in (1.7), they are given by

Uλ, ζ(y, z) := λN−2

2 U(λy, λz + ζ), λ > 0, ζ ∈ Rh

Taking derivatives with respect to the parametrs λ and ζ , we see that

V0 :=1− |y|2 − |z|2

[(1 + |y|)2 + |z|2]N2

, Vj :=zj

[(1 + |y|)2 + |z|2]N2

, j = 1, . . . , h (5.2)

are solutions to the linearized equation (at U)

(HSL)

−∆v = N(N−2)

U2

N−2

|y| v in RN

v ∈ D1,2(RN)(5.3)

The nondegeneration result is the following:

Theorem 5.1. Let v be a solution of (HSL) and Vj be as in (5.2). Then

∃ c0, .., ch ∈ R : v =h∑j=0

cjVj

To prove Theorem 5.1, we need first some qualitative properties of solutions

of (5.3). We will denote by v∗(x) = |x|2−Nu( x|x|2 ) the Kelvin transform of v.

Theorem 5.2. Let v be a solution of (5.3). Then v is smooth in y 6= 0and Holder continuous up to y = 0 and |v(x)| ≤ c

1+|x|N−2 for some c > 0.

Furthermore, if v∗(0) = 0, for any α ∈ (0, 1) there exist cα > 0 such that

|v(x)| ≤ cα1+|x|N−2+α .

Proof: We have seen in [13] that if u ∈ H1loc(RN) satisfies −∆u = f(x)u

|y|t +g(x)|y|t where t ∈ (0, 2), f and g are in Lploc(RN) for some p > N

2−t then u is

locally bounded in RN . So v is locally bounded in RN because U is bounded.

21

Moreover, since U = U∗, it is easily seen that v∗ solves (5.3) as well and

hence it is locally bounded and hence |v(x)| ≤ C1+|x|N−2 .

Finally, a simple bootstrap argument as in [13], Lemma 3.3, shows that

v ∈ C0,α ∀α ∈ (0, 1) and the same holds for v∗. Thus, if v∗(0) = 0, then

|v∗(x)| ≤ cα|x|α , and hence |v(x)| ≤ cα1+|x|N−2+α .

Theorem 5.3. Solutions v(y, z) of (HSL) are radially symmetric in y.

Proof: Given a solution v, it is enough to show that y → v(y, z) is symmetric

in any direction for every z ∈ Rh, i.e. given any ν ∈ Rk, we have that v(y +

tν, z) = v(y− tν, z) for all z ∈ Rh, t ∈ R and y ∈ Rk orthogonal to ν. Thanks

to rotation invariance, it is enough to show that all solutions v of (HSL) are

symmetric in some given direction, e.g. v(y1, .., yk; z) ≡ v(−y1, .., yk; z). Set

w(y, z) := v(y1, .., yk; z) − v(−y1, .., yk; z). We have −∆w = NN−2

U2

N−2

|y| w = 0

in RN+ = x = (y, z) : y1 > 0 and w = 0 on RN

+ . Hence, multiplying by w

and integrating by parts we get

0 =

∫RN+|∇w|2 − N

N − 2

U2

N−2

|y|w2 (5.4)

On the other hand, denoted

V1(y, z) := −∂U∂y1

(y, z) =(1 + |y|)y1

(k − 1)|y|

[(N − 2)(k − 1)

(1 + |y|)2 + |z|2

]N2

=y1(1 + |y|)(k − 1)|y|

UNN−2

we see, taking the y1 derivative in (5.1), that

LV1 := −∆V1 −N

N − 2

U2

N−2

|y|V1 −

(k − 1)

|y|2(1 + |y|)V1 = 0

Since V1 > 0 and L(V1) ≥ 0 in RN+ , then λ1(L,Ω) ≥ 0 ∀ Ω ⊂ Ω ⊂ RN

+ i.e.∫RN+|∇ϕ|2 − N

N − 2

U2

N−2

|y|ϕ2 − (k − 1)

|y|2(1 + |y|)ϕ2 ≥ 0 ∀ϕ ∈ C∞0 (RN

+ )

and for w as well by density, due to Theorem 5.2 and to the fact that, thanks

to the Hardy-Sobolev-Maz’ja inequality (4.4), we have

1

4

∫RN+

ϕ2

y21

+

∫RN+

U2

N−2

|y|ϕ2 ≤

∫RN+|∇ϕ|2 + c

(∫RN+|∇U |2

) 1N−2 ∫

RN+|∇ϕ|2

22

for every ϕ ∈ C∞0 (RN+ ), so that we can pass to the limit. Thus we obtain, in

view of (5.4),∫

RN+(k−1)

|y|2(1+|y|)w2 ≤ 0. So w ≡ 0 in RN

+ .

Thanks to the cylindrical symmetry, and writing v as v(r, z), r = |y| and

v(r, z) := rN−2

2 v(r, z), U := rN−2

2 U(r, z) we easily get (see Appendix 7.3)

∆Hv +h2 − (k − 2)2

4v +

N

N − 2U

2N−2 v = 0 in H (5.5)

Now, let us rewrite this equation on B, the ball model for Hn, by means of

the standard hyperbolic isometry between the two models, i.e. the Moebius

map M (see [23]). If e0, ej, j = 1, . . . , h is the standard basis in R×Rh, then

M(r, z) :=

(1− r2 − |z|2

(1 + r)2 + |z|2,

2z

(1 + r)2 + |z|2

)= 2

x+ e0

|x+ e0|2−e0, x = (r, z).

M is a bijection of Rh+1 \ −e0 onto itself , M2 = M = M−1 and

M(R+ × Rh) = Bh+1, the (open) unit ball in Rh+1. Recall the useful facts:

1− |ξ|2

2=

2r

(1 + r)2 + |z|2and |Mξ|2 =

(1− ξ0)2 + |η|2

(1 + ξ0)2 + |η|2, (5.6)

where (r, z) = Mξ and ξ = (ξ0, η) ∈ Rh+1. In particular, |M(r, z)| < 1 if

r > 0, |M(r, z)| = 1⇔ r = 0 and |Mξ| → +∞ iff (ξ0, η)→ (−1, 0).

For a given v ∈ D1,2, we will write (M v)(ξ) := r(ξ)N−2

2 v(Mξ) .

It can be easily seen that, for k ≥ 3, M is a linear isomorphism between

D1,2(RN) and H1(B). Now, recalling that NN−2

U2

N−2 = N(k−1)r(1+r)2+|z|2 , by (5.6)

and since M is an hyperbolic isometry v solves (HSL) iff w =Mv solves

−∆Bw −h2 − (k − 2)2

4w =

N(k − 1)

4(1− |ξ|2)w in Bh+1 (5.7)

In particular, (5.7) has the solutions MVj(ξ) =(

1−|ξ|24

)N−22ξj, j = 0, .., h.

Hence, Theorem 5.1 rewrites as follows;

Theorem 5.4. Let v be a solution of (HSL). Then there exist c0, .., ch ∈ Rsuch that Mv =

∑hj=0 cjMVj

23

Proof: Given ν a unit vector in Rh+1, let Rν be the reflection with re-

spect to the hyperplane < ξ, ν >= 0 and let Oν denote an orthogonal

transformation satisfying Oνν = e0. We split the proof in three steps.

• Step 1. There is a constant c such that ∀ ξ ∈ Bh+1 we get

(Mv)(ξ)− (Mv)(Rνξ) = c (MV0)(Oνξ) = c(1−|ξ|24

)N−2

2 < ξ, ν >

• Step 2. There exist c0, .., ch ∈ R such that ω :=M(v −∑h

j=0 cjVj)

is radial.

• Step 3. If ω is a radial solution of (5.7), then ω ≡ 0 .

Proof of Step 1. θ(ξ) := (Mv)(O−1ν ξ) − (Mv)(RνO

−1ν ξ) is, by rotation

invariance, again a solution of (5.7) and θ(ξ) = 0 if ξ0 = 0. Hence M−1θ is

a solution of (HSL) vanishing along |x| = 1, and hence it gives, as well as

V0, a solution of the Dirichlet eigenvalue problem

−∆φ = λU

2N−2

|y|φ in BN , φ = 0 in ∂BN (5.8)

(with λ = NN−2

). Since D1,2(RN) is compactly embedded into the weighted

space L2

(RN , U

2N−2

|y| dx

)(see Lemma 6.3 below), standard compact operator

theory and maximum principle for the Laplacian tell us that (5.8) has a

simple first eigenvalue with a corresponding positive eigenfunction and then

M−1θ = cV0 for some c, i.e. θ = cMV0.

Proof of Step 2. Let, with notations as above, Rj := Rej , Oj := Oej .

The very same arguments of Step 1 gives (Mv)(ξ)−12[(Mv)(ξ)+(Mv)(R0ξ)] =

c0(MV0)(ξ) for some c0, and hence M(v − c0V0) is even in ξ0.

The same argument applied to v1 = v − c0V0 with R0 replaced by R1 gives

(Mv1)(ξ)− (Mv1)(R1ξ) = 2c1 (MV0)(O1ξ) = 2c1 (MV1)(ξ)

for some c1 and then M(v − c0V0 − c1V1) is even in ξ1 and, of course, in ξ0

as well. By iterating, we conclude that, for some c0, . . . , ch

ω :=M(v −∑h

j=0 cjVj) is even in all the ξj and hence ∇ω(0) = 0.

24

But, as we showed above, ω(ξ)− ω(Rνξ) = cνMV0(Oνξ) for some cν . Since

∇(MV0)(Oνξ)|ξ=0= (1

4)N−2

2 ν, we have cν = 0, i.e. ω(ξ) = ω(Rνξ) for all ξ

and all ν and hence ω is radial .

Proof of Step 3. Let v be a solution of (HSL), ω = Mv, ω(ξ) = ω(|ξ|).Set

z(r) := ω(√r)(

1− r2

)−N−2

2 , r = |ξ|

Notice that, by decay estimates in Proposition 5.2 and (M1), z is bounded

in (0, 1). Also, straightforward computations (see Appendix 7.3) show that

z satisfies the ODE

r(1− r)z′′ +[(

h+ 1

2

)−(h+ 2k − 1

2

)r

]z′ +

k − 1

2z = 0 r ∈ (0, 1)

This is Gauss’ hypergeometric equation

r(1− r)z′′ +[γ −

(α+ + α− + 1

)r]z′ − α+α−z = 0

with

γ =h+ 1

2, α± =

h+ 2k − 3

4± 1

4

√(h+ 2k − 3)2 + 8(k − 1)

Its solutions (see [26], [4]) are, at least locally at r = 0, of the form

z(r) = c1F (α±, γ; r) + c2

[bF (α±, γ; r) log r + r1−γG(r)

]where G is analytic, b is some, eventually zero, constant and F is the hyper-

geometric function

F (α±, γ; r) =∑k≥0

(α+)k(α−)k

(γ)kk!rk (x)k := x(x+ 1)..(x+ k − 1)

We just have to check that this equation has no nontrivial bounded solution

in (0, 1).

Clearly, in order a solution z to be bounded it has to be of the form cF (α±, γ; r).

However, it is known ([12], page 61) that the hypergeometric series, which is

convergent for 0 ≤ r < 1, has infinite sum at r = 1 if α+ + α− − γ ≥ 0, as it

is the case here, because α+ + α− − γ = k − 2. The proof is complete.

25

6 Small Perturbation : Existence and Multi-

plicity

The main purpose of this section is to study problem (Pφ) when φ need not

be constant on the set of all possible concentration points.

We know from Theorem 3.2 that if φ = φ(|y|) and φ(0) = φ(∞), then (Pφ)

admits a solution if ||φ− φ(∞)||∞ is small enough. However no such general

existence results is possible if φ(0) 6= φ(∞). In fact it follows from Pohozaev

identity (see [6]) that if φ is monotonically increasing or decreasing then (Pφ)

has a solution if and only if φ is a constant.

In this section we look for conditions which, breaking monotonicity, generate

solutions. For example, we find that if φ′ has opposite signs near 0 and ∞then we have solution for Pφ if ‖φ− φ(∞)‖∞ is small enough.

We will also obtain conditions which yield non-uniqueness of solutions. To

make things more precise, let us consider the problem

(P εφ)

−∆u = (1+εφ(x))

|y| uNN−2 u ∈ D1,2(RN)

u > 0 in RN

where ε > 0 is small and φ ∈ L∞(RN) ∩ C(RN).

Motivated by the potentials in [7], we first consider the case where φ just

depends on |y|, i.e., φ = φ(|y|) and is differentiable. Since φ is bounded, we

have to require to ψ′(r) to decay at infinity. Accordingly, we assume

∃α ∈ (0, 2), ∃ψ′(∞) 6= 0 : rα ψ′(r)→r→∞ ψ′(∞) (6.1)

Theorem 6.1. Let φ = ψ(|y|) or φ = ψ(|z|). Let us assume ψ ∈ C1([0,∞))

and satisfies (6.1). Then

(i) (P εφ) has a solution for ε small enough if

ψ′(0) · ψ′(∞) < 0 (6.2)

(ii) (P εφ) has at least two solutions for ε small enough if

ψ′(0) > 0, ψ′(∞) > 0 and ψ(0) ≥ ψ(∞)

or

ψ′(0) < 0, ψ′(∞) < 0 and ψ(0) ≤ ψ(∞)

(6.3)

26

Remark 5. (i) The above theorem can be generalised to the case ψ′(0) = 0

by imposing conditions on ψ′′(0) or on the decay of ψ′(r) as r → 0

(ii) One can have a similar theorem when the potential is of the form φ(x) =

ψ(|y|, |z|), by imposing conditions on the decay of ∇ψ at infinity.

When φ does not have any symmetry we have the following :

Theorem 6.2. Let φ ∈ C(RN) with limx→∞

φ(x) = φ(∞) ∈ R. Then (P εφ) has

a solution for ε small enough if there exist (λ0, ζ0) ∈ (0,∞)×Rh+1 satisfying∫RN

φ(λ0y, λ0z + ζ0)U2(N−1)N−2

|y| > supζ∈Rh

φ(0, ζ)SN−1

or∫RN

φ(λ0y, λ0z + ζ0)U2(N−1)N−2

|y| < infζ∈Rh

φ(0, ζ)SN−1

(6.4)

If both conditions in (6.4) are satisfied for two distinct points (λ0, ζ0) and

(λ′0, ζ

′0) of (0,∞)×Rh then (P ε

φ) has at least two solutions for ε small enough.

Finally, by imposing conditions only on the critical points of φ|y=0 , we can

obtain an existence result, Theorem B in the introduction, similar to the

ones in [3], [8], [20], under a Bahri-Coron type condition.

We will prove our theorems by developing a finite dimensional reduction of

the problem (P εφ), building a natural constraint for the energy functional

Eε(u) = E0(u)− εG(u) u ∈ D1,2(RN)

where E0(u) = 12

∫RN|∇u|2 − N−2

2(N−1)

∫RN

|u|2(N−1)N−2

|y| and

G(u) =N − 2

2(N − 1)

∫RN

φ(x)

|y||u|

2(N−1)N−2

This procedure is rather standard, so we will skip most proofs (see e.g. [14]

for the details).

We recall (see [13] and (1.7)) that E0 has an h + 1-dimensional manifold of

critical points given by Z = Uλ,ζ(y, z) : λ > 0, where

Uλ,ζ(y, z) = λ2−N

2 U(λ−1y, λ−1(z − ζ)) and U =

[(N − 2)(k − 1)

(1 + |y|)2 + |z|2

]N−22

27

Clearly, Tλ0,ζ0 , the tangent space of Z at Uλ0,ζ0 , is spanned by

V0 =∂Uλ,ζ∂λ

∣∣∣∣(λ0,ζ0)

and Vj =∂Uλ,ζ∂ζj

∣∣∣∣(λ0,ζ0)

for j = 1, .., h

We start stating the crucial property of the linearized equation.

Lemma 6.3. For any (λ, ζ) ∈ R+×Rh, the operator E′′0 (Uλ,ζ) is a self adjoint

Fredholm operator of index zero. Furthermore

Ker(E′′

0 (Uλ0,ζ0))

= Tλ0,ζ0

Proof: Since E′′0 (Uλ,ζ) = I−Gλ,ζ , where 〈Gλ,ζ(u), v〉 = N

N−2

∫RN

1|y|U

2N−2

λ,ζ uv,

it is enough to prove that Gλ,ζ is compact. So, let un 0 in D1,2(RN).

By Rellich compactness theorem, we can assume that un → 0 in Lploc(RN)

∀p < 2NN−2

. Now, by Holder and Hardy-Sobolev inequalities:∫|x|≥R

U2

N−2 |un| |v||y| dx ≤ c

( ∫|x|≥R

U2N−1N−2

|y|

) 1N−1

‖un‖ ‖v‖ ≤ ε for R large,

while, if 0 < δ < 1N−2

, then∫|x|≤R

U2

N−2 |un| |v||y| dx ≤

c

( ∫|x|≤R

|un|2NN−1

)N−12N

×

( ∫RN

[v

|y|(1−δ)

] 2NN−2δ

)N−2δ2N

×

( ∫|x|≤R

[1|y|δ

] 2N1+2δ

) 1+2δ2N

≤ c

( ∫|x|≤R

|un|2NN−1

)N−12N

≤ ε‖v‖ for n large, because if t := 2N(1−δ)N−2δ

,

then 2NN−2δ

is the corresponding Hardy-Sobolev exponent. Moreover, 2Nδ1+2δ

< 2

and hence the third integral is finite. Thus Gλ,ζ(un)→n 0.

As for the last statement, this is exactly the content of Theorem 5.1

Now, denoted by T⊥λ,ζ the orthogonal complement of Tλ,ζ in D1,2(RN), Lemma

6.3 tells that E′′0 (Uλ,ζ) induces an isomorphism from T⊥λ,ζ into itself. Moreover,

‖(E ′′0 (Uλ,ζ))−1‖L(T⊥λ,ζ) = ‖(E ′′0 (U1,0))−1‖L(T⊥1,0) ∀(λ, ζ)

In turn, this implies, using the contraction lemma, the following

28

Lemma 6.4. There exist positive constants C and ε0 and a smooth function

w = w(λ, ζ, ε) : R+ × Rh × (−ε0, ε0)→ D1,2(RN) such that

w(λ, ζ, ε) ∈ T⊥λ,ζ , E′

ε(Uλ,ζ + w(λ, ζ, ε)) ∈ Tλ,ζ , ‖w(λ, ζ, ε)‖ ≤ C|ε|

for any λ > 0, ζ ∈ Rh and ε ∈ (−ε0, ε0).

Lemma 6.5. Let ε0 and w(λ, ζ, ε) be as in Lemma 6.4. Define

Zε := Uλ,ζ + w(λ, ζ, ε) : (λ, ζ) ∈ R+ × Rh, |ε| ≤ ε0

Then every critical point of Φε := Eε|Zε is also a critical point of Eε.

We are then left with the study of Φε := Eε|Zε . In the following Lemma,

S will denote, as usual, the best constant in (1.2) when t = 1.

Lemma 6.6. Let

Γ(λ, ζ) :=N − 2

2(N − 1)

∫RN

φ(y, z)

|y|U

2(N−1)N−2

λ,ζ

Then we have the following expansion:

Φε(λ, ζ) =SN−1

2(N − 1)− εΓ(λ, ζ) +O(ε2)

Consequently, if Γ(λ, ζ) has a stable critical point (λ0, ζ0), Φε will have a

critical point uε = Uλε0,ζε0 +w(λε0, ζε0 , ε) for any |ε| 1 and (λε0, ζ

ε0)→ (λ0, ζ0)

as ε→ 0.

Remark 6. If the potential φ is cylindrically symmetric, i.e., φ(x) = ψ(|y|, |z|)then it is well known that the positive critical points of Eε|D1,2

Cylare indeed

critical points of Eε, where D1,2Cyl is the completion of u ∈ C∞0 : u(x) =

v(|y|, |z|) with respect to the norm ||u||2 =∫

Rn|∇u|2. Therefore in this

case we can do the finite dimensional reduction of Eε : D1,2Cyl → R to get

the same results with the only difference that there wouldn’t be any depen-

dence on ζ. Hence Z,Zε will be one dimensional manifolds parametrised by

λ and the Melnikov function will be Γ(λ) = G(Uλ) where U is as before and

Uλ(x) = λ2−N

2 U(λ−1x).

29

Proof of Theorem 6.1. We know from Lemma 6.6 that if u ∈ Zε then

Eε(u) = SN−1

2(N−1)+ O(ε). In addition if u ∈ Zε is also a critical point of

Eε and changes sign then as in the proof of Lemms 3.1 we can show that

Eε(u) ≥ SN−1

N−1+O(ε). Hence the critical points of Eε on Zε does not change

sign when ε is small enough. Therefore from Lemma 6.5 , Lemma 6.6 and

the above remark it follows that the theorem will be proved once we establish

that Φε(λ) attains its maximum or minimum at λ0 ∈ (0,∞).

We assume that φ(x) = ψ(|y|), when φ(x) = ψ(|z|) the proof is similar.

Recall

Γ(λ) =N − 2

2(N − 1)

∫RN

ψ(λ|y|)U2(N−1)N−2

|y|

Differentiating with respect to λ,

∂Γ

∂λ(λ0) =

N − 2

2(N − 1)

∫RN

(ψ′(λ0|y|)|y|)U

2(N−1)N−2

|y|dx

Now by dominated convergence theorem

limλ0→0

∂Γ

∂λ(λ0) =

N − 2

2(N − 1)ψ′(0)

∫RN

U2(N−1)N−2 dx

and

limλ0→∞

λα0∂Γ

∂λ(λ0) =

N − 2

2(N − 1)ψ′(∞)

∫RN

U2(N−1)N−2

|y|αdx

where ψ′(∞) is as defined in (6.1) If ψ satisfies (6.2) then Γ(λ) is either

increasing near 0 and decreasing near ∞ or decreasing near 0 and increasing

near ∞. Hence either maximum or minimum of Γ is attained at λ0 ∈ (0,∞)

but not at 0 or ∞. Therefore for ε small enough Φε attains its maximum or

minimum at λε and λε → λ0 as ε→ 0. Hence Uλε + w(λε, ε) is a solution of

(P εφ) for ε small enough bifurcating from the critical Manifold Z at Uλ0 .

If ψ satisfies (6.3) then Γ(λ) attains its maximum and minimum at λ1 ∈(0,∞) and λ2 ∈ (0,∞) respectively and not at 0 or ∞. Hence as before we

obtain two branches of solutions for (P εφ) for ε small enough bifurcating from

the critical Manifold Z at Uλ1 and Uλ2 .

30

Proof of Theorem 6.2. Proof follows exactly as before since the condition

(6.4) assures that Φε has a maximum or minimum and hence a critical point

converging to (λ0, ζ0) ∈ (0,∞) × Rh as ε → 0. If φ satisfies both conditions

of (6.4) then as in the proof of Theorem 6.1 we get multiplicity.

Proof of Theorem B. We follow closely [20], where the Authors prove a

similar result for the Webster scalar curvature problem on the CR sphere.

The theorem can be established through the above finite dimensional reduc-

tion and a topological degree argument, namely proving that there exist an

open set Ω ⊂ R+ × Rh such that deg(∇Γ,Ω, 0) 6= 0.

We start proving that the critical points of the Melnikov function

Γ(λ, ζ) =N − 2

2(N − 1)

∫RN

ϕ(λy, λz + ζ)

|y|U

2(N−1)N−2

lay in a bounded region λ2 + |ζ|2 ≤ R. To this extent, notice first that Γ ∈C2((0,∞)× Rh), it can be extended continously up to λ = 0 by setting

Γ(0, ζ) := limλ→0+

Γ(λ, ζ) =SN−1(N − 2)

2(N − 1)ψ(ζ)

and it results

∇ζΓ(0, ζ) = limλ→0+

∇ζΓ(λ, ζ) =SN−1(N − 2)

2(N − 1)∇ψ(ζ) (6.5)

D2ζΓ(0, ζ) = lim

λ→0+D2ζΓ(λ, ζ) =

SN−1(N − 2)

2(N − 1)D2ψ(ζ) (6.6)

where the limits are uniform for ζ on compact sets.

Denoted by Γ the Melnikov function associated to ϕ(x) := ϕ( x|x|2 ), it follows

from (6.5) and assumption (ii) that Γ has no critical point close to λ = 0, ζ =

0. We now claim that

∇Γ(λ, ζ) = 0 ⇔ (∇Γ)

(λ

λ2 + |ζ|2,

ζ

λ2 + |ζ|2

)= 0

and hence Γ has no critical points of large norm. To prove the claim,

just notice that, by direct computations, the Kelvin transform of Uλ,ζ is

U λλ2+|ζ|2

, ζ

λ2+|ζ|2and hence, with the change of variables x→ x

|x|2 , we obtain

2(N − 1)

N − 2Γ(λ, ζ) =

∫RN

φ(x)

|y|U

2(N−1)N−2

λ,ζ =

∫RN

φ( x|x|2 )

|y|

[|x|(2−N)Uλ,ζ

(x

|x|2

)] 2(N−1)N−2

31

=

∫RN

φ(x)

|y|

[U λλ2+|ζ|2

, ζ

λ2+|ζ|2

] 2(N−1)N−2

=2(N − 1)

N − 2Γ(

λ

λ2 + |ζ|2,

ζ

λ2 + |ζ|2)

Now, besides (6.5), (6.6), we also have , by oddness,

∂Γ

∂λ(0, ζ) := lim

λ→0+

N − 2

2(N − 1)

∫RN

〈∇ϕ(λy, λz + ζ), (y, z)〉|y|

U2(N−1)N−2 = 0 (6.7)

limλ→0

∂2Γ

∂λ2(λ, ζ) =

N − 2

2(N − 1)[A∆yϕ(0, ζ) + B∆zϕ(0, ζ)] (6.8)

where the limit is uniform for ζ on compact sets and

A :=1

k

∫RN

|y|U2(N−1)N−2 B :=

1

h

∫RN

|z||y|

2

U2(N−1)N−2 (6.9)

Notice that, since we assumed N ≥ 4, A and B are finite. Changing to polar

coordinates and using the change of variable t = s1+r

we obtain

A = 2hωkωhI(2k + 1, N + k − 2)I(h− 1, N − 1)

B = 2kωkωhI(2k − 3, N + k − 4)I(h+ 1, N − 1)

where I(m,n) =∞∫0

ρn

(1+ρ2)mdρ =

Γ(n+12 )Γ( 2m−n−1

2 )2Γ(m)

, where Γ denotes the Euler

Gamma function. Then, by direct computations,

A =k − 1

2k + h− 3B

and hence, following the notation introduced in (1.8) in Theorem B, (6.8)

rewrites

limλ→0

∂2Γ

∂λ2(λ, ζ) =

B(N − 2)

2(N − 1)(2k + h− 3)∆∗(ζ) (6.10)

Now, assumption (i), (6.10) and (6.7) imply that if ζj is a critical point of ψ

and (λ, ζ), λ 6= 0 is close to (0, ζj), then

∂Γ

∂λ(λ, ζ) ∆∗(ζj) > 0 (6.11)

32

In particular, it follows that Γ has no critical points, for σ > 0 small and

R > 0 large, in((0, σ]× Rh)

)∩λ2 + |ζ|2 ≤ R2. As a consequence, denoted

Bs :=

(λ, ζ) ∈ R+ × Rh : |(λ, ζ)− (s, 0)| <

(s− 1

s

)Γ has no critical point on ∂Bs for s large and hence deg(∇Γ, Bs, 0) is well

defined.

It remains to prove that deg(∇Γ, Bs, 0) is not zero. To compute the degree,

we will use, as in [20] the Morse formula ( see [15])

deg(∇Γ, Bs, 0) = χ(Bs)− ind(∇(Γ|∂−Bs ))

where χ(Bs) is the Euler-Poincare characteristic of Bs and ∂−Bs is the set of

points in ∂Bs where ∇Γ points inward Bs, i.e. with ∂Γ∂λ> 0 if s is large.

Since χ(Bs) = 1, we have just to compute ind(∇(Γ|∂−Bs )).

As s goes to infinity, zeros (λ(s), ζ(s) of the vector field ∇(Γ|∂Bs ) converge ,

by (6.5)-(6.7), to points (0, ζj) and are in fact in one-to-one correspondence

with the critical points of ψ.

In particular, by (6.6) , (λ(s), ζ(s)) are nondegenerate zeros of ∇(Γ|∂Bs ) ,

exactly like the critical points of ψ, and , if (λ(s), ζ(s))→s→+∞ (0, ζj), they

have index m(ψ, ζj).

By (6.8), zeros of∇(Γ|∂−Bs ) correspond to critical points of ψ with ∆∗(ζj) > 0.

Thus

ind(∇(Γ|∂−Bs )) =∑

j: ∆∗(ζj)>0

(−1)m(ψ;ζj)

and this concludes the proof.

Remark 7. (ii) in Theorem B, i.e. ∃V = V1 + V2 ∈ Rk × (Rh \ 0) such

that ∇φ(x) = (∇yφ,∇zφ)→x→0 V1 + V2 amounts to require that

∇φ(x) =1

|x|2[Px+ (1)] and ∇ψ(z) =

1

R2[Qz + (1)]

where Px := V −2 < V, x > x, x ∈ RN and Qz := V2−2 < V2, z > z, z ∈ Rh

and (1) goes to zero as |x| (respectively |z|) goes to infinity. Now, Q has

exactly two nondegenerate zeros , ± V2√2|V2|

, and , as an easy consequence,

deg(∇ψ,BR, 0) = 1 + (−1)h (6.12)

33

This formula has a simple geometric interpretation: (ii) also amounts to say

that, denoted by π the stereographic projection of SN onto RN , then Φ := φπextends to a C1(SN) map with the south pole as a regular point. Similarly for

Ψ := ψ π. The degree formula for ∇ψ then follows from the Poincare-Hopf

index Theorem for the tangent vector field on Sh given by ∇Ψ.

Remark 8. One can reformulate Theorem B using the alternative Morse

formula (see [15])

deg(∇Γ, Bs, 0) = (−1)h+1+ind(∇(Γ|∂+Bs )) = (−1)h+1+∑

j: ∆∗(ζj)<0

(−1)m(ψ;ζj)

where ∂+Bs is the set of points in ∂Bs where ∇Γ points putward Bs. However

this formula, taking into account (6.12), gives the same as (1.9).

We wish to conclude with an easy Corollary of Theorem B, regarding the

Webster scalar curvature equation in the cylindrically symmetric case. In

the following F will denote the CR equivalence between S2n+1 and the n-

dimensional Heisenberg group (see [14] and [20]).

Let K be a function in R2n × R such that K F is a smooth function on

S2n+1, with the south pole as a regular point.

Also assume that K = K(r, t), r = |y|, y ∈ R2n, t ∈ R, (i.e. K has cylindrical

symmetry) and that K(0, t) has only a finite number of nodegenerate critical

points tj.

Corollary 6.7. Let K be as above. Assume:

(i) ∂iK∂ri

(0, t) ≡ 0, i = 1, 2, 3 and ∂4K∂r4

(0, tj) = 0 ∀ j(ii) ∃ c 6= 0 : t2K(0, t) −→

|t|→∞c.

Let R(ξ) := 1 + εK. Then (3.3) has a solution for ε small, provided K(0, t)

has more than one minima.

Remark 6.1. Notice that, by the above Remarks, K has the same, nonzero,

number of minima and maxima.

34

7 Appendix

7.1 Error estimates.

Lemma 7.1. Let U and Uε be defined as in (1.7) and (3.4), then∫|x|≥1

(Uε)2(N−1)(N−2)

|y|dx =

∫|x|≤1

(U1ε )

2(N−1)(N−2)

|y|dx = O(εN−1).

Proof. Since U is equal to its Kelvin transform we get∫|x|≥1

(Uε)2(N−1)(N−2)

|y|dx =

∫|x|≤1

(U1ε )

2(N−1)(N−2)

|y|dx.

Now by a change of variable,∫|x|≥1

(Uε)2(N−1)(N−2)

|y|dx =

∫|x|≥ 1

ε

U2(N−1)(N−2)

|y|dx.

Since |x| ≤ |y|+ |z|, |x| ≥ 1ε implies either |y| ≥ 1

2ε or |z| ≥ 12ε . Therefore∫

|x|≥ 1ε

U2(N−1)(N−2)

|y|dx ≤ C

∫|y|≥ 1

2ε

1

|y| ((1 + |y|)2 + |z|2)N−1dx

+C∫

|y|< 12ε ,|z|≥

12ε

1

|y| ((1 + |y|)2 + |z|2)N−1dx

≤ Cε∫

|x|≥ 12ε

|x|−2(N−1) dx+ C

∫|y|< 1

2ε

∫|z|≥ 1

2ε

1

|y| (1 + |z|2)N−1dy dz = O(εN−1)

This proves the Lemma.

Lemma 7.2. Let U and Uε be defined as in (1.7) and (3.4), then∫RN

(Uε)NN−2

U1ε

|y|dx =

∫RN

(U

1ε

) NN−2 Uε

|y|dx = εN−2(CN,k + o(1))

where CN,k is a positive constant depending only on N and k.

Proof. As in the previous lemma, the equality of the first two terms follows from the factthat U is invariant under Kelvin transform. Now by a change of variable

∫RN

(Uε)NN−2

U1ε

|y|dx =

∫RN

(U(x

ε)) NN−2 U(εx)

|εy|dx = εN−2

∫RN

U(ε2x)(U(x))

NN−2

|y|dx

35

Let W = UN

2(N−1) then by direct calculation one can see that W ∈ D1,2(RN ). Hence fromthe Hardy Sobolev inequality (1.2), we get

∫RN

(U(x))NN−2

|y|dx =

∫RN

W2(N−1)N−2

|y|dx ≤

S−1

∫RN

|∇W |2

N−1N−2

<∞

Hence by Dominated convergence theorem, we get∫RN

U(ε2x)(U(x))

NN−2

|y|dx→ U(0)

∫RN

(U(x))NN−2

|y|dx.

This proves the Lemma.

Lemma 7.3. Let U and Uε be defined as in (1.7) and (3.4), then∫|x|≥1

(Uε)NN−2

U1ε

|y|dx =

∫|x|≤1

(U

1ε

) NN−2 Uε

|y|dx = o(εN−2)

Proof. Again the first equality follows by a Kelvin-transform argument. Now as in theproof of the previous Lemma,∫

|x|≥1

(Uε)NN−2

U1ε

|y|dx = εN−2

∫|x|> 1

ε

U(ε2x)(U(x))

NN−2

|y|dx = o(εN−2)

as∫

RN

(U(x))NN−2

|y| dx < ∞ and hence∫

|x|> 1ε

U(ε2x) (U(x))NN−2

|y| dx → 0. This proves the

Lemma.

Lemma 7.4. Let ρ ∈ Cc(RN \ 0), then∫RN

ρ(x)(Uε)

2(N−1)(N−2)

|y|dx = O(εN−1) =

∫RN

ρ(x)(U

1ε )

2(N−1)(N−2)

|y|dx

Proof. Assume that support of ρ is contained in the annulus x : 0 < R1 < |x| < R2.Then by change of variable,∫

RN

ρ(x)(Uε)

2(N−1)(N−2)

|y|dx =

∫RN

ρ(εx)U

2(N−1)(N−2)

|y|dx ≤ C

∫|x|≥R1

ε

U2(N−1)(N−2)

|y|dx = O(εN−1).

The other estimate follows similarly.Similar computations yield

Lemma 7.5. Let ρ ∈ Cc(RN \ 0), then∫RN

ρ(x) (Uε)NN−2

U1ε

|y|dx = o(εN−2) =

∫RN

ρ(x)(U

1ε

) NN−2 Uε

|y|dx

36

7.2 Linking to the Grushin operator

The Grushin operator in Rm1 × Rm2 , m1 ≥ 1,m2 ≥ 1 is the differential operator

L = −∆y − 4|y|2∆z

where x ∈ Rm1 × Rm2 is denoted by (y, z). The critical exponent problem for L is

L(u) = uQ+2Q−2 (7.1)

where Q = m1 +2m2 is the ”appropriate” dimension and Q+2Q−2 is the corresponding critical

exponent. This problem has been investigated in [21] and [22]. Analogous to the Scalarcurvature problem, let us considerL(u) = K(y, z)u

Q+2Q−2 in Rm1 × Rm2

u > 0 in Rm1 × Rm2(7.2)

We are going to show that under some symmetry assumptions (7.2) is closely related toour Hardy-Sobolev problem (Pφ).Let φ be radially symmetric in the y variable and let u(y, z) be a solution of (Pφ) which isalso radially symmetric in the y variable, i.e. we assume that φ(y, z) = φ(|y|, z), u(y, z) =θ(|y|, z) and θ solves

−θrr(r, z)−k − 1r

θr(r, z)−∆zθ(r, z) = φ(r, z)θ(r, z)

NN−2

r

Defineψ(r, z) = θ(r2, z), (r, z) ∈ [0,∞)× Rh

Then,

ψr(r, z) = 2rθr(r2, z), ψrr(r, z) = 2θr(r2, z)+4r2θrr(r2, z), ∆zψ(r, z) = ∆zθ(r2, z)

Hence ψ solves

−ψrr(r, z)−2k − 3r

ψr(r, z)− 4r2∆zψ(r, z) = 4φ(r2, z)ψ(r, z)NN−2

Thus, if we definev(y, z) = ψ(|y|, z) (7.3)

then v solves (7.2) with m1 = 2k − 2,m2 = h and K(y, z) = 4φ(|y|2, z).Thus we have proved the following

Lemma 7.6. Let m1 be even and K(y, z) = K(|y|, z) be radially symmetric in y, then theGrushin problem (7.2) has a solution if (Pφ) has a solution in RN = Rk × Rh which isradially symmetric in y, where k = m1+2

2 , h = m2 and φ(y, z) = 14K(

√|y|, z).

37

The requirement of m1 to be even in the above lemma may look like a restriction, howeverwe see that it is natural when we deal with the Webster scalar curvature equation in theHeisenberg group Hn = Cn × R = R2n × R. Let us denote a point in Hn by ξ = (Z, t) ∈Cn × R, we will also denote Z by Z = x+ iy

Consider the problem

−∆Hu(ξ) = R(ξ)u(ξ)Q+2Q−2 , u > 0 in Hn (7.4)

where ∆H is the Heisenberg sublaplacian and Q = 2n + 2 is the homogeneous dimensionof Hn. Recall that

∆H =n∑i=1

(X2i + Y 2

i

)and

Xi =∂

∂xi+ 2yi

∂

∂t, Yi =

∂

∂yi− 2xi

∂

∂t, i = 1...n.

If u(ξ) is cylindrically symmetric, i.e., u(ξ) = θ(|Z|, t) then by direct calculation one cansee that ∆Hu becomes

∆Hu = ∆Zu+ 4|Z|2utt

where ∆Z is the Euclidean laplacian in R2n. Hence it is a Grushin operator on R2n × R.Thus Lemma 7.6 translates to

Lemma 7.7. Let n ≥ 1 and R(ξ) = R(|Z|, t) be radially symmetric in Z, then the Websterscalar curvature problem (7.4) has a solution if (Pφ) has a solution in RN = Rk×R whichis radially symmetric in y, where k = n+ 1, h = 1 and φ(y, z) = 1

4R(√|y|, z).

7.3 Linking to the Hyperbolic Laplacian

Let ∆Hu := −[r2∆u− (h− 1)rur

]denote the Laplace-Beltrami operator on the h + 1

dimensional hyperbolic space H, written in euclidean coordinates (half space model).1. From Hardy-Sobolev equation to (1.3)Let u, v ∈ C∞(R+ × Rh) satisfy the equations

−urr −k − 1r

ur −∆zu =uN+2−2tN−2

rtin R+ × Rh

−vrr −k − 1r

vr −∆zv =N

N − 2U

2N−2

rv in R+ × Rh

Then, after setting u := rN−2

2 u(r, z), v := rN−2

2 v(r, z), U := rN−2

2 U(r, z), wehave

−∆Hu =h2 − (k − 2)2

4u+ u

N+2−2tN−2

−∆Hv =h2 − (k − 2)2

4u+

N

N − 2U

2N−2

rv

38

In fact,

ur = rN−2

2 ur + N−22 r

N−42 u, urr = r

N−22 urr + (N − 2)r

N−42 ur + (N−2)(N−4)

4 rN−6

2 u

and hence,−[r2∆u− (h− 1)rur

]=

−rN+2

2

[urr + ∆zu+ (k − 1)

urr

]+h2 − (k − 2)2

4rN−2

2 u =

h2 − (k − 2)2

4u+ r

N+22 −t

(r−

N−22 u

)N+2−2tN−2

=h2 − (k − 2)2

4u+ u

N+2−2tN−2

Similarly,

−∆Hv = −rN+2

2 ∆v +h2 − (k − 2)2

4rN−2

2 v =

h2 − (k − 2)2

4v + r

N+22

N

N − 2U

2N−2

rr−

N−22 v =

h2 − (k − 2)2

4v +

N

N − 2(r

N−22 U)

2N−2 v

Remark 7.1. Given u cylindrically symmetric, let, as above, u(y, z) := |y|N−22 u(y, z). In

dimension N ≥ 3 it results

ωk

∫H

[|∇u|2 + (

k − 22

)2u2

]dVH =

∫RN

|∇u|2 +h2

4u2

|y|2dydz

where ωk is the area of the unit sphere in Rk. In particular, thanks to Hardy inequality,

u ∈ D1(Rk × Rh) ⇔ u ∈ H1(Hh+1)

The above identity readily follows from

|∇u|2 = rN−2 |∇u|2 +(N − 2

2

)2

rN−4u2 + (N − 2)rN−3uur

and then integrating:

ωk

∫R+×Rh

[rN |∇u|2 +

(N − 2

2

)2

rN−2u2 + (N − 2)rN−1uur + (k − 2

2)2rN−2u2

]drdz

rh+1

=∫

RN

|∇u|2 + (N − 2

2)2u2

|y|2− (N − 2)(k − 2)

2u2

|y|2+ (

k − 22

)2u2

|y|2dydz

2. From Grushin-type equations to (1.3)Let u(y, z) = φ(|y|, z), (y, z) ∈ Rk × Rh. Let φ ∈ C∞(R+ × Rh) satisfy

Lu := −φrr −(k − 1)r

φr − (1 + α)2r2α∆zφ = φ(r, z)Q+2Q−2 in Rk × Rh (7.5)

where , α > 0, r = |y|, Q = k + h(1 + α). Let

Φ(r, z) := γQ−2

2 rβ φ(rγ , z), γ :=1

α+ 1, β :=

Q− 22(1 + α)

39

Then

−∆HΦ − 14

[h2 −

(k − 2α+ 1

)2]

= ΦQ+2Q−2

In fact, Φr = γQ−2

2[γrβ+γ−1φr(rγ , z) + βrβ−1φ(rγ , z)

],

Φrr = γQ−2

2[γ2rβ+2γ−2φrr(rγ , z) + γ(2β + γ − 1)rβ+γ−2φr(rγ , z) + β(β − 1)rβ−2φ(rγ , z)

],

−∆HΦ := −[r2∆Φ− (h− 1)rΦr

]=

14

[h2 −

(k − 2α+ 1

)2]γQ−2

2 rβφ(rγ , z)−

−γQ+2

2 rβ+2γ

[φrr(rγ , z) +

r2−2γ

γ2φzz(rγ , z) + (k − 1)r−γφr(rγ , z)

]because β(h− β) = 1

4

[h2 −

(k−2α+1

)2], 2β+γ−h

γ = k− 1. Since 2− 2γ = 2αα+1 and

β + 2γ = β Q+2Q−2 , we finally get

−∆HΦ− 14

[h2 −

(k − 2α+ 1

)2]

Φ =[γQ−2

2 rβφ(r1

α+1 , z)]Q+2Q−2

= ΦQ+2Q−2

2. From (1.3) to the Gauss hypergeometric equationLet v be a solution of the linearized equation (5.3), ω =Mv, ω(ξ) = ω(|ξ|), i.e.

(1− r2

2)2(ωrr +

h

rωr)− (h− 1)

1− r2

2rωr +

h2 − (k − 2)2

4ω +

N(k − 1)4

(1− r2)ω = 0

in (0, 1). Then φ(r) := (1−r22 )−

N−22 ω(r) satisfies in (0, 1) the equation

1− r2

2(φrr +

h

rφr)− (k − 1)

1− r2

2rφr + (k − 1)φ(r) = 0 r ∈ (0, 1)

In fact, φr = ( 1−r22 )−

N−22 ωr + N−2

2 ( 1−r22 )−

N2 r ω(r) ,

φrr = (1− r2

2)−

N+22

[(1− r2

2)2ωrr + (N − 2)

1− r2

2r ωr +

N − 24

(1− r2 +Nr2)],

1− r2

2(φrr +

h

rφr)− (k − 1)

1− r2

2rφr =

(1− r2

2)−

N2

[(1− r2

2)2(ωrr +

h

rωr

)+ (h− 1)

1− r2

2rωr +

N − 24

[h+ 1− (k − 1)r2]ω]

=

−(1− r2

2)−

N2

[h2 − (k − 2)2

4+N(k − 1)

4(1− r2)− (N − 2)(h+ 1)

4+

(N − 2)(k − 1)4

r2]

=

= −(1− r2

2)−

N2 (1− r2)

k − 12

ω = −(k − 1)φ

because (N−2)(h−k+2)+N(k−1)−(N−2)(h+1)4 = k−1

2 , (N−2)(k−1)−N(k−1)4 = −k−1

2 .Finally, setting z(r) := φ(

√r), from φ′(r) = 2rz′(r2), φ′′(r) = 2z′(r2) + 4r2z′′(r2)

we get, from the equation for φ and writing t = r2,

0 = 2t(1− t)z′′ + z′[h+ 1− (h+ 2k − 1)t] + (k − 1)z

40

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