Hardy-Sobolev inequalities, hyperbolic symmetry and the Webster scalar curvature problem

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Hardy-Sobolev inequalities, hyperbolic symmetry and the Webster scalar curvature problem D. Castorina, I. Fabbri, G. Mancini∗and K. Sandeep†

Abstract p(t)

We discuss the problem −∆u = φ(x) u|y|t , u ∈ D1,2 (RN ) where −2t+2 x = (y, z) ∈ Rk × Rh = RN , 2 ≤ k < N, t ∈ (0, 2), p(t) := NN −2 in connection with Grushin-type equations and the Webster scalar curvature problem, providing various existence results. We highlight the role of hyperbolic symmetry in nondegeneracy and uniqueness questions and present a uniqueness result for semilinear elliptic equations in Hyperbolic space which applies to the above equation when N > 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|t

u ∈ 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, Universit`a degli Studi ”Roma Tre”, Largo S. Leonardo Murialdo, 1 - 00146 Roma, Italy. E-mail [email protected], [email protected], [email protected]. †

1

−2t+2 = 2∗ (t) − 1 is the Hardysymmetry. Also, t ∈ (0, 2) and p(t) := NN −2 Sobolev exponent, i.e. for an optimal constant S = St,N,k > 0 we have



S

Z

RN

2∗ (t)

 2∗2(t)

|u|  |y|t

≤

Z

|∇u|2

∀ u ∈ D1,2 (RN )

(1.2)

RN

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 C 1 (D1,2 (RN )) variational integral J (see (2.1)). When φ ≡ 1, (1.1) has a special interest, because its solutions are the extremals 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 ∆H v +

h2 − (k − 2)2 v + φv p = 0 4

(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 Q+2

Lv := −∆y v − (α + 1)2 |y|2α ∆z v = v Q−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 Q+2

− ∆H u = φ(|y|2 , z) u Q−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)−1 u |y|t

in 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 

(N − 2)(k − 1) U (y, z) = (1 + |y|)2 + |z|2

 N2−2

(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 solutions 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 operators, 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 cylindrically 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 , ˆ j = 1, . . . , m of index m(ψ; ζj ). Let φ(x) := φ( |x|x2 ) and, for ζ ∈ Rh , set ∆∗ (ζ) := (k − 1)∆y ϕ(0, ζ) + (2k + h − 3)∆z ϕ(0, ζ) Assume the non degeneracy conditions (i) ∆∗ (ζj ) 6= 0 ∀j ˆ ∇φˆ have a limit as x goes to zero and (ii) φ,

Then (1.1) has a solution for |ε| small provided X (−1)m(ψ;ζj ) 6= 1

(1.8)

lim ∇z φˆ 6= 0

x→0

(1.9)

{j: ∆∗ (ζj )>0}

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 Theorem A. We also establish a relation with the Webster scalar curvature equation 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 Z Z 1 |u|p(t)+1 1 2 dx, u ∈ D1,2 (RN ) (2.1) |∇u| dx − φ(x) J(u) = 2 p(t) + 1 |y|t RN

RN

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, lim J(un ) = β, we say un is a n→∞

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, 2−N

−1

un := (φ(0, z0 )) p(t)−1 εn 2 u(ε−1 n (y, z − z0 )) is a P-S sequence for J with no subsequence converging in D1,2 (RN ). Moreover if I denotes the energy associated with (1.6), Z Z |u|p(t)+1 1 1 2 dx (2.2) |∇u| dx − I(u) = 2 p(t) + 1 |y|t RN

RN

−2

then, J(un ) = (φ(0, z0 )) p(t)−1 I(u) + o(1). Lack of compactness due to concentration 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,

lim φ(x) = 1

x→∞

(2.3)

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

φ > 0,

lim φ(y) = 1

y→∞

(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) = λ 5

2−N 2

u(λ−1 (x − (0, z0 )))

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 , wˆj solutions of (1.6), u1n =

k1 X

j zn

vj +

k2 X

j j wjζn ,Rn

+

j=1

j=1

j=1

k3 X

where znj , ζnj , ηnj ∈ Rh with znj →n ∞ for all j, and Rnj →n ∞, ǫjn →n 0. Moreover, J(un ) = J(u0 ) +

k1 X

I(vj ) +

k2 X

j

1

φ(0, η j ) p(t)−1 ζnj , ηnj → ζ j , η j ∈ Rh ∪ {∞} 2

j

I(wj ) + φ (0, η )

j=1

j=1

j

wˆjηn ,ǫn

k3 X

I(wˆj ) + o(1)

j=1

In case φ depends only on y and satisfies (2.4), the vj′ s 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 , vj′ s and wj′ s are also nonnegative. (ii) If un are cylindrically symmetric or having partial radial symmetry in the y or z variable then u0 , vj′ s and wj′ s 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 Z |un |p(t)+1 dx = 0 lim inf φ(x) n→∞ |y|t RN

then, along a subsequence, un →n 0 in D1,2 (RN ) and the proof is complete. If not, we have, again for a subsequence, Z |un |p(t)+1 dx ≥ C > 0 lim φ(x) n→∞ |y|t RN

N −2

N −2

t−N Given S as in (1.2), 0 < δ < S 2−t ||φ||∞ , let zn ∈ Rh , Rn > 0 be such that Z Z |un |p(t)+1 |un |p(t)+1 sup dx = φ(x) dx = δ (2.5) φ(x) |y|t |y|t ζ∈Rh −1 B(ζ,Rn )

−1 B(zn ,Rn )

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 := v zn 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) , Z

−1 B(ζ,Rn )

|vn |p(t)+1 ≤ φ(y, z) |y|t

Z

φ(y, z + zn )

−1 B(0,Rn )

|vn |p(t)+1 =δ |y|t

(2.6)

for every ζ ∈ Rh . Standard arguments (see e.g. [28]) imply that if v = 0 1,2 then vn →n 0 in Dloc , 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 Z Z |vn |p(t)−1 φ(y, z + zn ) vn ψ + o(1), ∀ψ ∈ D1,2 (RN ) ∇vn .∇ψ = |y|t RN

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). 2−N

Case 2 : Let R0 = 0 or ∞. Define vn := Rn 2 un (Rn−1 x + (0, zn )). As above, up to a subsequence, vn ⇀ v 6= 0 in D1,2 (RN ) where − ∆v = K

|v|p(t)−1 v |y|t

(2.7)

and K = φ(0) or K = φ(∞) depending on R0 = ∞ or 0 respectively. Hence N −2

1

v = K − p(t)−1 w where w is a solution of (1.6). If wn := Rn 2 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 inf J(un ) ≥ 0. Also if u is n→∞

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 lim J(un ) > 0, n→∞

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 lim J(un − wn ) > 0 then we can n→∞ 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 implications

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)



=

Z

RN

p(t)+1

|u| φ(x) |y|t

2 − p(t)+1



Z

|∇u|2

and set

Sφt := inf Qtφ (u)

RN

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 Sφt 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  2 − p(t)+1   sup φ(0, z) if φ satisfies (2.3) Cφ = z∈Rh  2  if φ satisfies (2.4) (max{φ(0), φ(∞)})− p(t)+1 8

Theorem 3.1. Sφt ≤ Cφ S and it is achieved if Sφt < Cφ S. Proof. For any u solution of (1.6). Sφt = inf Qtφ (u) ≤ u6=0

inf

z0 ∈Rh ,λ>0

Qtφ (uz0 ,λ ) ≤ Cφ S

Now assume that Sφt < Cφ S. Using Ekeland’s variational principle we can choose a minimising sequence un ≥ 0 for Sφt which is also a P-S sequence for J. Since Sφt < Cφ S, it follows from Theorem 2.1 that this is possible only if (1.1) has a nontrivial solution. However, one cannot expect in general Sφt to be achieved (actually, if φ ≤ 1 then Sφt = 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 Sφt = 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  N −∆u = φ(x) u N −2 u ∈ D1,2 (RN ) |y| (3.1) (Pφ )  u>0 in RN 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 R |∇U |2 = S N −1 , we have:

RN

Theorem 3.2. Let φ satisfy (2.4) or (2.3) and φ ≡ φ(∞) = 1 on {y = 0}. Then (Pφ ) has a solution provided Z −1) 1 φ(λx) 2(N inf (3.2) U N −2 dx > 2− N −2 S N −1 . λ∈(0,∞) |y| RN

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

1

inf φ > 2− 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 Q+2

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

(3.3)

where Hn = Cn × R = R2n+1 is the Heisenberg group, ∆Hn is the Kohn laplacian 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 contact 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 existence 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 somehow 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 (IH n \ {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 := Sφ1 . 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 ∈ (Pφ ) admits a positive solution u with J(u) ≤ β.



S N −1 S N −1 , 2(N −1) N −1



. Then

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 2 k2 k1 P P N −2 I(wj ) where vj′ s are solutions of −∆u = φ(x) |u| |y| u J(vj ) + form β = j=1

j=0

and wj′ s are solutions of (1.6) with t = 1. We know from [13] that if wj′ s are S N −1 . If wj changes sign, then positive then I(wj ) = 2(N −1) Z

|∇wj+ |2 =

RN

Z

RN

|wj+ | |y|

2(N −1) N −2

and

RN

S N −1 . N −1

1 2(N −1)

|∇wj− |2 =

RN

Using this in (1.2) with t = 1 gives Z |∇wj+ |2 ≥ S N −1 and and hence I(wj ) =

Z

R

RN

|∇wj |2 =

Z

RN

|wj− | |y|

2(N −1) N −2

|∇wj− |2 ≥ S N −1

RN

1 2(N −1)

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

Z

"

R

|∇wj+ |2 +

RN S N −1 2(N −1)

R

RN

|∇wj− |2

#

≥

if vj is positive and

J(vj ) ≥ N −1 if 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

  2(N −1) Z Z  |u| N −2  φ(x) |∇u|2 = = u ∈ D1,2 (RN ) : u 6= 0,  |y|  RN

RN

Nφ is a C 2 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 1 J(u) = 2(N − 1) N −1 Sφ 2(N −1)

Also, inf J = Nφ

1 |∇u| = 2(N − 1)

Z

2

RN

=

S N −1 . 2(N −1)

Z

φ(x)

|u|

RN

2(N −1) N −2

∀u ∈ Nφ

|y|

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 S N −1 S N −1 build a P-S sequence at level β ∈ 2(N −1) , N −1 for J|Nφ . For U given by (1.7) and λ > 0 define U λ (x) = λ

2−N 2

U (λ−1 x),

V λ = Cλ U λ ∈ 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 := inf sup J(γ(t)) γ∈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 S N −1 S N −1 ≤β< 2(N − 1) N −1 Proof. Since lim J(V t ) = t→0

S N −1 , 2(N −1)

we have

S N −1 2(N −1)

≤ β. To prove the upper

bound on β in case φ satisfies (3.2), we just compute sup J(γ0 (t)) where γ0 (t) := V t , ∀ t. By direct computation one can show that J(γ0 (λ)) =

1 2(N − 1)

Z

RN

|∇V λ |2 =

(N −1)2



S  2(N − 1)

Z

RN

12

t

λ

φ(x)

2(N −1) N −2

|U | |y|

2−N 

Substituting U λ we can see that β ≤ sup J(γ0 (λ)) < λ

satisfied and the lemma is proved in this case.

S N −1 N −1

holds if (3.2) is

Estimating β from above in the other case is much more involved. Define, for ε > 0, ( t V  if t ≤ ε or t ≥ 1ε    γε (t) = 1 V ε + 1 − ε(t−ε) V ε if ε < t < 1ε Ctε ε(t−ε) 1−ε2 1−ε2

where Ctε is chosen such that γε (t) ∈ Nφ . We will show that for ε small enough N −1 N −1 sup J(γε (t)) < SN −1 . Clearly for ε small enough sup J(γε (t)) < SN −1 . t≤ε,t≥ 1ε

t

Therefore it is enough to show that sup J(Cλε (λV where

Cλε

is chosen so that

J(Wλε ) =

Wλε

=

1 2(N − 1)

0 1 and (3.6) , we get Z

RN

1

(λV ε + (1 − λ)V ε ) φ(x) |y|

2(N −1) N −2



≥ λ

2(N −1) N −2

2(N −1) N −2

+ (1 − λ)



S N −1 +O(εN −1 )

Thus combining the above inequalities we get J(Wλε )

S N −1 ≤  2(N − 1)

N −1

(λ2 + (1 − λ)2 + o(1))

2(N −1) N −2

λ

+ (1 − λ)

N −2 + o(1)

2(N −1) N −2

(3.8)

N −1

Now the quantity

(λ2 +(1−λ)2 )

2(N −1) 2(N −1) λ N −2 +(1−λ) N −2

to 2 when λ = 21 . Therefore at λ = Assume that λ = 12 . Then Z

RN

= 2−

2(N −1) N −2

  

Z

|x|≤1

!N −2

1 2

is less than 2 for λ 6=

φ(x)

1

(V + V ε ) |y|

and is equal

we need more refined estimates.

(λV ε + (1 − λ)V ε ) φ(x) |y| 1 ε

1 2

2(N −1) N −2

+

Z

2(N −1) N −2

|x|>1

14

=

φ(x)

1 ε

(V + V ε ) |y|

2(N −1) N −2

  

Now using the inequality (a + b)p ≥ ap + pap−1 b, for p > 1, a > 0, b > 0, (3.6), Lemma 7.1, Lemma 7.3, Lemma 7.4 and Lemma 7.5 we get 1

Z

|x|≤1

(V ε + V ε ) φ |y|

≥

2(N −1)

Z

|x|≤1

(V ε ) N −2 2(N − 1) φ + |y| N −2

2(N −1)

Z

=

2(N −1) N −2

|x|≤1

2(N − 1) (U ε ) N −2 + φ |y| N −2 =S

N −1

2(N − 1) + N −2

Z

RN

Z

|x|≤1

N

Z

(V ε ) N −2 φV |y| 1 ε

|x|≤1

N

(U ε ) N −2 φU + O(εN −1 ) |y| 1 ε

N

(U ε ) N −2 + o(εN −2 ) U |y| 1 ε

Similarly, Z

|x|>1

1

(V ε + V ε ) φ |y|

2(N −1) N −2

≥S

N −1

2(N − 1) + N −2

Z

Rn

U

ε (U

1 ε

N

) N −2 + o(εN −2 ) |y|

Combining all the above inequalities we get Z

RN

1

(V ε + V ε ) φ(x) |y|

2(N −1) N −2

≥2

−2(N −1) N −2



2S

N −1

 4(N − 1) N −2 N −2 CN,k ε + o(ε ) + N −2

Plugging this inequality and (3.7) in (3.5), we get S N −1 J(Wλε ) ≤  N −1 =


N −1 1 + CN,k S −(N −1) εN −2 + o(εN −2 ) N −2 2(N −1) −(N −1) N −2 N −2 1 + N −2 CN,k S ε + o(ε )


S N −1 1 − (N − 1)CN,k S −(N −1) εN −2 + o(εN −2 ) N −1 N −1

if λ = 12 . Thus J(Wλε ) < SN −1 when λ = 21 and hence in a small interval ( 12 − δ, 12 + δ). In the complement of this interval, we already know from (3.8) N −1 that J(Wλε ) < SN −1 when ε is small enough. Hence for ε is small enough N −1 sup J(Wλε ) < SN −1 . This proves the lemma. 0 0, kvk < ǫ}

We notice that Ωǫ,λ ∩ Ωǫ,λ = ∅ for ǫ small, λ small and λ large. In fact, Z Z N dx ζ1 ,λ1 ζ2 ,λ2 2 N −1 |∇(U −U )| = 2S − 2 U ζ2 ,λ2 [U ζ1 ,λ1 ] N −2 ≥ 2S N −1 − |y| RN

RN

−S

N 2

      S    

B

1 λ2

Z

(0,

ζ1 −ζ2 |) λ2

N −2  2(N −1)

N −1 U 2 N −2    |y| 



 + 

N −1

Z

|x|≥ λ1

1

 2(NN−1)

U 2 N −2   |y| 



   ≥ S N −1  

if λ1 0, r = |y|, Q = k + h(1 + α). Φ(r, z) := γ

Q−2 2

rβ φ(rγ , z),

in Rk × Rh

Let γ :=

39

1 , α+1

β :=

Q−2 2(1 + α)

(7.5)

Then

Q−2

In fact,

"  2 # Q+2 1 2 k−2 −∆H Φ − h − = Φ Q−2 4 α+1  β+γ−1  γr φr (rγ , z) + βrβ−1 φ(rγ , z) ,

Φr = γ 2  Q−2  Φrr = γ 2 γ 2 rβ+2γ−2 φrr (rγ , z) + γ(2β + γ − 1)rβ+γ−2 φr (rγ , z) + β(β − 1)rβ−2 φ(rγ , z) , "  2 #  2  Q−2 1 k−2 2 γ 2 rβ φ(rγ , z)− h − −∆H Φ := − r ∆Φ − (h − 1)rΦr = 4 α+1   2−2γ Q+2 r γ −γ γ β+2γ γ φrr (r , z) + −γ 2 r φzz (r , z) + (k − 1)r φr (r , z) γ2   2  2α k−2 = k − 1. Since 2 − 2γ = α+1 and , 2β+γ−h because β(h − β) = 41 h2 − α+1 γ β + 2γ = β

Q+2 Q−2 ,

we finally get " 2 #  Q+2 i Q−2 h Q−2 Q+2 1 1 k−2 2 Φ = γ 2 rβ φ(r α+1 , z) = Φ Q−2 −∆H Φ − h − 4 α+1

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

1 − r2 2 h 1 − r2 h2 − (k − 2)2 N (k − 1) ) (ωrr + ωr ) − (h − 1) rωr + ω+ (1 − r2 )ω = 0 2 r 2 4 4 2

− in (0, 1). Then φ(r) := ( 1−r 2 )

N −2 2

ω(r) satisfies in (0, 1) the equation

h 1 − r2 1 − r2 (φrr + φr ) − (k − 1) rφr + (k − 1)φ(r) = 0 2 r 2 2

N −2

N

− 2 −2 φr = ( 1−r ωr + N 2−2 ( 1−r r ω(r) , 2 ) 2 )   1 − r2 N −2 1 − r2 − N +2 1 − r2 2 2 2 2 ) ) ωrr + (N − 2) r ωr + (1 − r + N r ) , =( ( 2 2 2 4

In fact, φrr

2

r ∈ (0, 1)

1 − r2 h 1 − r2 (φrr + φr ) − (k − 1) rφr = 2 r 2     1 − r2 − N 1 − r2 2 1 − r2 h N −2 2 2 ( ( ) ) ωrr + ωr + (h − 1) rωr + [h + 1 − (k − 1)r ]ω = 2 2 r 2 4   1 − r2 − N h2 − (k − 2)2 N (k − 1) (N − 2)(h + 1) (N − 2)(k − 1) 2 −( ) 2 + (1 − r2 ) − + r = 2 4 4 4 4 = −(

k−1 1 − r2 − N ) 2 (1 − r2 ) ω = −(k − 1)φ 2 2

(N −2)(k−1)−N (k−1) because (N −2)(h−k+2)+N 4(k−1)−(N −2)(h+1) = k−1 = − k−1 2 , 4 2 . √ ′ ′ 2 ′′ ′ 2 Finally, setting z(r) := φ( r), from φ (r) = 2rz (r ), φ (r) = 2z (r ) + 4r2 z ′′ (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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