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 symmetry. Also, t ∈ (0, 2) and p(t) := NN = 2∗ (t) − 1 is the Hardy−2 Sobolev exponent, i.e. for an optimal constant S = St,N,k > 0 we have
Z S
2∗ (t)
2∗2(t)
|u| |y|t
Z ≤
|∇u|2
∀ u ∈ D1,2 (RN )
(1.2)
RN
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 1 |u|p(t)+1 2 J(u) = |∇u| dx − φ(x) dx, u ∈ D1,2 (RN ) (2.1) 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 0 (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 1 1 |u|p(t)+1 2 |∇u| dx − dx (2.2) 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 +
j=1
k2 X
j j wjζn ,Rn
j=1
+
J(un ) = J(u0 ) +
I(vj ) +
j=1
k2 X
j
wˆjηn ,n
j=1
φ(0, η j ) p(t)−1
1
where znj , ζnj , ηnj ∈ Rh with znj →n ∞ for all j, and Rnj →n ∞, jn →n 0. Moreover, k1 X
j
k3 X
ζnj , ηnj → ζ j , η j ∈ Rh ∪ {∞}
2
j
I(wj ) + φ (0, η )
j=1
k3 X
I(wˆj ) + o(1)
j=1
In case φ depends only on y and satisfies (2.4), the vj0 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 , vj0 s and wj0 s are also nonnegative. (ii) If un are cylindrically symmetric or having partial radial symmetry in the y or z variable then u0 , vj0 s and wj0 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 lim inf φ(x) dx = 0 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 lim φ(x) dx ≥ C > 0 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 φ(x) dx = φ(x) dx = δ (2.5) |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
|vn |p(t)+1 φ(y, z) ≤ |y|t
Z φ(y, z + zn )
|vn |p(t)+1 =δ |y|t
(2.6)
−1 ) B(0,Rn
−1 ) B(ζ,Rn
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 = vn ψ + o(1), ∀ψ ∈ D1,2 (RN ) |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 (max{φ(0), φ(∞)})− p(t)+1 if φ satisfies (2.4) 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| (Pφ ) (3.1) 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 U N −2 dx > 2− N −2 S N −1 . (3.2) 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 k1 k2 P P N −2 form β = J(vj ) + I(wj ) where vj0 s are solutions of −∆u = φ(x) |u| |y| u j=0
j=1
and wj0 s are solutions of (1.6) with t = 1. We know from [13] that if wj0 s are S N −1 positive then I(wj ) = 2(N . If wj changes sign, then −1) Z
|∇wj+ |2 =
RN
Z
|wj+ | |y|
2(N −1) N −2
Z and
RN
|∇wj− |2 =
RN
Using this in (1.2) with t = 1 gives Z |∇wj+ |2 ≥ S N −1 and RN
Z
Z
|wj− | |y|
2(N −1) N −2
RN
|∇wj− |2 ≥ S N −1
RN
" and hence I(wj ) = S N −1 . N −1
1 2(N −1)
R
|∇wj |2 =
RN
1 2(N −1)
Similarly since Sφ = S we get J(vj ) ≥
# R
|∇wj+ |2 +
RN S N −1 2(N −1)
R
|∇wj− |2
≥
RN
if vj is positive and
S N −1
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 =
u ∈ D1,2 (RN ) : u 6= 0,
Z
Z
|∇u|2 =
RN
φ(x)
|u|
2(N −1) N −2
|y|
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)
Z
1 |∇u| = 2(N − 1) 2
RN N −1 Sφ 2(N −1)
Also, inf J = Nφ
=
Z φ(x)
|u|
2(N −1) N −2
∀u ∈ Nφ
|y|
RN S N −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 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 t
γ0 (t) := V t , ∀ t. By direct computation one can show that
J(γ0 (λ)) =
1 2(N − 1)
Z
(N −1)2
|∇V λ |2 =
S 2(N − 1)
RN
RN
12
λ
Z φ(x)
2(N −1) N −2
|U | |y|
2−N
Substituting U λ we can see that β ≤ sup J(γ0 (λ)) < λ
S N −1 N −1
holds 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, ( t V if t ≤ ε or t ≥ 1ε γε (t) = 1 V ε if ε < t < 1ε Ctε ε(t−ε) V ε + 1 − 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
t≤ε,t≥ 1ε
Therefore it is enough to show that sup J(Cλε (λV where
Cλε
is chosen so that
Wλε
=
0<λ<1 1 ε Cλ (λV ε
1 2(N − 1)
+ (1 − λ)V ε )) <
S N −1 , N −1
+ (1 − λ)V ε ) ∈ Nφ . Now, !N −1 1
R J(Wλε ) =
1 ε
|∇(λV ε + (1 − λ)V ε )|2
RN
φ(x) (λV
R
2(N −1) 1 ε +(1−λ)V ε ) N −2
!N −2
(3.5)
|y|
RN
Recall that, we have V ε = Cε U ε where R |∇U ε |2 N Cε = R 2(N −1) R ε N −2 φ(x) (U ) |y|
N2−2
RN
Now, Lemma 7.4 gives V ε = (1 + O(εn−1 ))U ε
V
1 ε
1
= (1 + O(εn−1 ))U ε
Using (3.6) we have Z Z Z 1 1 ε 2 2 ε 2 2 |∇(λV ε + (1 − λ)V )| = (1 − λ) |∇U | + λ |∇U ε |2 RN
RN
Z +2λ(1 − λ)
1
∇U ε .∇U ε + O(εN −1 )
RN
13
RN
(3.6)
Since U ε is an extremal of (1.2) and solves (1.6) with t = 1, the above expression simplifies to Z 1 |∇(λV ε + (1 − λ)V ε )|2 = λ2 + (1 − λ)2 S N −1 RN 1
Z
ε
+2λ(1 − λ)
(U )
N N −2
Uε + O(εN −1 ) |y|
RN
Now using Lemma 7.2 we get Z 1 |∇(λV ε + (1 − λ)V ε )|2 = RN
λ2 + (1 − λ)2 S N −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 Z
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 )
RN
Thus combining the above inequalities we get J(Wλε )
Now the quantity
S N −1 ≤ 2(N − 1)
N −1
(λ2 + (1 − λ)2 + o(1)) 2(N −1) N −2
λ
(λ2 +(1−λ)2 )
(3.8)
N −1
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
+ (1 − λ)
N −2 + o(1)
2(N −1) N −2
!N −2
1 2
is less than 2 for λ 6=
1 2
and is equal
we need more refined estimates.
1
(λV ε + (1 − λ)V ε ) φ(x) |y|
2(N −1) N −2
=
RN
= 2−
2(N −1) N −2
Z
φ(x)
1 ε
(V + V ε ) |y|
2(N −1) N −2
Z +
|x|≤1
|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
(V ε + V ε ) φ |y|
Z
2(N −1) N −2
2(N −1)
(V ε ) N −2 2(N − 1) φ + |y| N −2
Z ≥ |x|≤1
|x|≤1 2(N −1)
=
1 ε
|x|≤1
(U ε ) N −2 2(N − 1) φ + |y| N −2
Z
N
(V ε ) N −2 φV |y|
Z
|x|≤1
Z
N
(U ε ) N −2 + O(εN −1 ) φU |y| 1 ε
|x|≤1
=S
N −1
2(N − 1) + N −2
Z
N
(U ε ) N −2 + o(εN −2 ) U |y| 1 ε
RN
Similarly, Z
1
(V ε + V ε ) φ |y|
2(N −1) N −2
≥S
N −1
2(N − 1) + N −2
Z U
ε (U
1 ε
N
) N −2 + o(εN −2 ) |y|
Rn
|x|>1
Combining all the above inequalities we get Z
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
RN
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 ε + o(ε ) 1 + N −2 CN,k S
S N −1 1 − (N − 1)CN,k S −(N −1) εN −2 + o(εN −2 ) N −1 N −1
if λ = 21 . Thus J(Wλε ) < SN −1 when λ = 12 and hence in a small interval ( 21 − 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<λ<1
Proof of Theorem 3.2 and Theorem 3.3. Let us assume, by contradiction, that (Pφ ) has no solution. In view of Lemma 3.2 and of Lemma 3.1, we 15
infer that β =
S N −1 2(N −1)
and hence there exists a sequence γm ∈ Γ such that
S N −1 S N −1 1 ≤ J(γm (t)) ≤ + 2(N − 1) 2(N − 1) m
∀t ∈ (0, ∞)
We claim that, for m large enough, γm (t) ∈ Ω,λ ∪ Ω,λ
(3.9)
for every t , where
Ω,λ = {U ζ,λ + v :
ζ ∈ Rh , 0 < λ < λ, kvk < }
Ω,λ = {U ζ,λ + v :
ζ ∈ Rh , λ > λ > 0, kvk < }
We notice that Ω,λ ∩ Ω,λ = ∅ for small, λ small and λ large. In fact, Z Z N dx N −1 ζ2 ,λ2 2 ζ1 ,λ1 )| = 2S − 2 U ζ2 ,λ2 [U ζ1 ,λ1 ] N −2 −U |∇(U ≥ 2S N −1 − |y| RN
RN
−S
N 2
S
B
N −2 2(N −1)
Z 1 λ2
(0,
ζ1 −ζ2 |) λ2
N −1 U 2 N −2 |y|
2(NN−1)
+
N −1
U 2 N −2 |y|
Z
|x|≥ λ1
≥ S N −1
1
if λ1 << 1 << λ2 , uniformly in ζ1 , ζ2 . Since by (3.6) γm (t) ∈ Ω,λ for t small while γm (t) ∈ Ω,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 ) ∈ / Ω,λ ∪ Ω,λ . By (3.9), γm (tm ) is a minimizing sequence for J|Nφ . Now using Ekeland’s principle we can choose vm sufficiently close to γm (tm ), still a minimizing sequence for J|Nφ , which is also a S N −1 P-S sequence for J|Nφ and hence for J at level 2(N . By the P-S charac−1) ζm ,λm terisation, we have vm = U + ◦(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 symmetry 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 N −2 u˜ := r 2 u(r, z). Then u˜ solves (see Appendix 7.3) − ∆H u := −(r2 ∆u − (h − 1)r ur ) = c u + up 2
in R+ × Rh
(4.1)
2
+2−2t , p = NN and ∆H denotes the Laplace-Beltrami operwhere c = h −(k−2) 4 −2 ator on the n := h + 1 dimensional hyperbolic space H (half space model) dr2 + |dz|2 h+1 + h 2 H := H = R × R , ds := r2
with associated volume form dVH = rdrdz h+1 and, accordingly, Z Z Z Z drdz 2 2 drdz 2 v dVH = v h+1 , |∇H v| dVH = r2 |∇v|2 h+1 r r R+ ×Rh
H
R+ ×Rh
H
We notice that, if k ≥ 3, u˜ ∈ H 1 (H), in view of the identity (easy to check) Z Z h2 u2 k−2 2 2 2 2 |∇u| + dydz = ωk ) u˜ dVH |∇H u˜| + ( 4 |y|2 2 RN
H
and of the Hardy inequality. Remark 4. The same change of variable sends cylindrically symmetric solutions of the more general equation involving inverse square potentials (see [19]) u up(t) in RN (4.2) − ∆u = λ 2 + |y| |y|t 2
2
into solutions of (4.1) with c = λ+ h −(k−2) . Here, as above, N = k+h, k ≥ 2 4 −2t+2 k h and x = (y, z) ∈ R × R , t ∈ (0, 2) and p(t) := NN , −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 " 2 # k−2 Q+2 1 2 h − , p= < 2∗ − 1 c= 4 α+1 Q−2 2n 2∗ := n−2 if n > 2 and 2∗ = +∞ if n = 2. Again, if k 6= 2, uˆ ∈ H 1 (H) provided u is an extremal of the weighted Sobolev inequality (see [22]) Q−2 Q Z Z 2Q |∇y u|2 + (α + 1)2 |y|2α |∇z u|2 dydz := k|uk|2 ≤ S˜ |u| Q−2 dydz RN
RN
Again, this follows by the easy-to-check identity 2 2 Z Z k−2 h(α + 1) u2 2 2 ωk (α + 1) |∇H uˆ| + uˆ dVH = k|uk| + 2(α + 1) 2 |y|2 RN
H
and the Hardy inequality k|uk|2 ≥ ( k−2 )2 2 Q−2
R RN
u2 |y|2
(see [11]). If u is smooth, Q−2
decay estimates [C(1 + kxk2 )]− 2 ≤ u ≤ C(1 + kxk2 )− 2 , k(y, z)k := 1 2(α+1) |y| + |z|2 2(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) if n = 2 and (p+3)2 (4.1) has at most one positive solution in H 1 (H).
c<
h2 4
if n ≥ 3. Then
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) . Uniqueness in this case has been (p+3)2 proven, actually without any symmetry assumption, in [22]. 2 Case k = 2 (where c = h4 both for (1.6) and (1.4)) has to be dealt separately. The starting point to prove Proposition 4.1 is hyperbolic symmetry of positive H 1 (H) solutions of (4.1). Proposition 4.2. Let u ∈ H 1 (H) be a positive solution of (4.1), with p+1 ≤ 2 2∗ and c < h4 . 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) 2 2 where symmetry is proved in case c < h 4−1 . To get the result up to h4 we just use, instead of the ’sharp’ Sobolev inequality ((13)-(14) in [2]), the Poincar´ e-Sobolev inequality 2n For every n ≥ 3 and every p ∈ (2, n−2 ] there is c = c(n, p) > 0 such that p2
Z
Z
p
|u| dVH ≤
c
(n − 1)2 2 |∇H u| − u dVH 4 2
∀u ∈ C0∞ (H)
(4.3)
H
H
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) n−t Let n = h + 1 ≥ 3. Let t ∈ [0, 2), and set p = 2 n−2 . Then there is c = c(n, p) such that p2 Z Z 2 1 u dx p 2 u ≤c |∇u| − dydz ∀u ∈ C0∞ (R+ × Rh ) (4.4) t 2 |y| 4 |y| Rn
Rn
If h = 1, 2.1.6 Corollary 1 in [18] gives, for any α > 0 , q ∈ (1, 2), u ∈ C ∞ (R2 ), 1q Z Z q(α+1)−2 q |y| |u(y, z)| dydz ≤ c(q, α) |y|α |∇u(y, z)|dydz R2
R2
α = p−2 , it easily gives, when applied 4 ! p2 R p−4 p R r 2 |u| ≤ c r|∇u|2 ∀u ∈ C0∞ (R2 ),
Given p > 2 and taking q = to |u|
p+2 2
(4.5)
, the inequality
2p , 2+p
R2
which in turn, when applied to
R2
√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 ∈ H 1 (B) of ∆B u + cu + up = 0 h i2 h i 1−|ξ|2 1−|ξ|2 where, ∆B u = ∆u + (n − 2) < ∇u , ξ >, 2 2 19
(4.6) |ξ| < 1.
We rewrite (4.6) in hyperbolic polar coordinates |ξ| := tanh 2t , v := u(tanh 2t ), to get h v 00 + v 0 + c v + v p = 0, v 0 (0) = 0 (4.7) tanh t 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 2 of (4.1), with c = h4 , which are not in H 1 (H). Actually, they belong to the broader class H, the closure of Cc∞ (H) with respect to the norm Z ||u|| =
|∇H u|2 −
n−1 2
2
!
21
u2 dVH , u ∈ Cc∞ (H)
H
(see also [29]). Functions in H have in fact the following characterization Lemma 4.3. (Classification of H) n−1 u ∈ H iff v(y, z) := r− 2 u(|y|, z), (y, z) ∈ R2 × Rn−1 is in D1,2 (Rn+1 ). Moreover Z 2 ||u||H = |∇v|2 dx. Rn+1
For solutions in H we prove hyperbolic symmetry via a suitable Poincar´e2n Sobolev inequality: ∀n ≥ 3, ∀p ∈ (2, n−2 ], ∃Sn,p > 0 such that if u ∈ H, p(n−1) v is as in Lemma 4.3, t = n + 1 − 2 , then Z Sn,p
p
p2
|v| dy dz |y|t
RN
Z ≤
|∇v|2 dx
(4.8)
RN 2
Finally, we prove that a solution v of (4.7) (with c = h4 ), 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 2 ). So Theorem 4.1 holds true also in case c = h4 , n ≥ 3 for c = 41 > 2(p+1) (p+3)2 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 N u N −2 − ∆u = in RN (5.1) |y| If U is as in (1.7), they are given by Uλ, ζ (y, z) := λ
N −2 2
U (λy, λz + ζ),
ζ ∈ Rh
λ > 0,
Taking derivatives with respect to the parametrs λ and ζ , we see that V0 :=
1 − |y|2 − |z|2 [(1 + |y|)2 + |z|2 ]
N 2
, Vj :=
zj N
,
j = 1, . . . , h (5.2)
[(1 + |y|)2 + |z|2 ] 2
are solutions to the linearized equation (at U ) 2 −∆v = N U N −2 v (N −2) |y| (HSL) v ∈ D1,2 (RN )
in 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 X
cj Vj
j=0
To prove Theorem 5.1, we need first some qualitative properties of solutions of (5.3). We will denote by v ∗ (x) = |x|2−N u( |x|x2 ) the Kelvin transform of v. Theorem 5.2. Let v be a solution of (5.3). Then v is smooth in {y 6= 0} and Holder continuous up to {y = 0} and |v(x)| ≤ 1+|x|cN −2 for some c > 0. Furthermore, if v ∗ (0) = 0, for any α ∈ (0, 1) there exist cα > 0 such that |v(x)| ≤ 1+|x|cNα−2+α . 1 Proof: We have seen in [13] that if u ∈ Hloc (RN ) satisfies −∆u = g(x) |y|t
Lploc (RN )
f (x)u |y|t
+
N where t ∈ (0, 2), f and g are in for some p > 2−t then u is N N locally bounded in R . So v is locally bounded in R 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)| ≤ 1+|x|CN −2 . Finally, a simple bootstrap argument as in [13], Lemma 3.3, shows that v ∈ C 0,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)| ≤ 1+|x|cNα−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 2 N −2
w(y, z) := v(y1 , .., yk ; z) − v(−y1 , .., yk ; z). We have −∆w = NN−2 U |y| w = 0 N in RN + = {x = (y, z) : y1 > 0} and w = 0 on R+ . Hence, multiplying by w and integrating by parts we get 2 Z N U N −2 2 2 0= |∇w| − w (5.4) N − 2 |y| RN + On the other hand, denoted N (1 + |y|)y1 (N − 2)(k − 1) 2 y1 (1 + |y|) NN−2 ∂U (y, z) = U V1 (y, z) := − = 2 2 ∂y1 (k − 1)|y| (1 + |y|) + |z| (k − 1)|y| we see, taking the y1 derivative in (5.1), that 2
N U N −2 (k − 1) LV1 := −∆V1 − V1 − 2 V1 = 0 N − 2 |y| |y| (1 + |y|) N Since V1 > 0 and L(V1 ) ≥ 0 in RN + , then λ1 (L, Ω) ≥ 0 ∀ Ω ⊂ Ω ⊂ R+ i.e. 2 Z N U N −2 2 (k − 1) 2 |∇ϕ| − ϕ − 2 ϕ2 ≥ 0 ∀ϕ ∈ C0∞ (RN +) N N − 2 |y| |y| (1 + |y|) R+
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 ! N1−2 Z 2 Z Z Z Z 1 ϕ2 U N −2 2 + ϕ ≤ |∇ϕ|2 + c |∇U |2 |∇ϕ|2 N N N N 4 RN+ y12 |y| R+ R+ R+ R+ 22
for every ϕ ∈ C0∞ (RN + ), so that we can pass to the limit. Thus we obtain, in R 2 N view of (5.4), RN |y|(k−1) 2 (1+|y|) w ≤ 0. So w ≡ 0 in R+ . +
Thanks to the cylindrical symmetry, and writing v as v(r, z), r = |y| and N −2 N −2 v˜(r, z) := r 2 v(r, z), U˜ := r 2 U (r, z) we easily get (see Appendix 7.3) ∆H v˜ +
h2 − (k − 2)2 N ˜ N2−2 v˜ + U v˜ = 0 4 N −2
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 2z x + e0 1 − r2 − |z|2 , −e0 , x = (r, z). =2 M (r, z) := 2 2 2 2 (1 + r) + |z| (1 + r) + |z| |x + e0 |2 M is a bijection of Rh+1 \ {−e0 } onto itself , M 2 = M = M −1 and M (R+ × Rh ) = B h+1 , the (open) unit ball in Rh+1 . Recall the useful facts: 1 − |ξ|2 2r = 2 (1 + r)2 + |z|2
and
|M ξ|2 =
(1 − ξ0 )2 + |η|2 , (5.6) (1 + ξ0 )2 + |η|2
where (r, z) = M ξ and ξ = (ξ0 , η) ∈ Rh+1 . In particular, |M (r, z)| < 1 if r > 0, |M (r, z)| = 1 ⇔ r = 0 and |M ξ| → +∞ iff (ξ0 , η) → (−1, 0). N −2
For a given v ∈ D1,2 , we will write (M v)(ξ) := r(ξ) 2 v(M ξ) . It can be easily seen that, for k ≥ 3, M is a linear isomorphism between 2 N (k−1)r D1,2 (RN ) and H 1 (B). Now, recalling that NN−2 U˜ N −2 = (1+r) 2 +|z|2 , by (5.6) and since M is an hyperbolic isometry v solves (HSL) iff w = Mv solves h2 − (k − 2)2 N (k − 1) w= (1 − |ξ|2 ) w in B h+1 (5.7) 4 4 N2−2 1−|ξ|2 In particular, (5.7) has the solutions MVj (ξ) = ξj , j = 0, .., h. 4 − ∆B w −
Hence, Theorem 5.1 rewrites as follows; Theorem 5.4. Let v be a solution of (HSL). Then there exist c0 , .., ch ∈ R P such that Mv = hj=0 cj MVj
23
Proof: Given ν a unit vector in Rh+1 , let Rν be the reflection with respect to the hyperplane {< ξ, ν >= 0} and let Oν denote an orthogonal transformation satisfying Oν ν = e0 . We split the proof in three steps. • Step 1. There is a constant c such that ∀ ξ ∈ B h+1 we get 2 N −2 ) 2 < ξ, ν > (Mv)(ξ) − (Mv)(Rν ξ) = c (MV0 )(Oν ξ) = c( 1−|ξ| 4 • Step 2. There exist c0 , .., ch ∈ R such that ω := M(v − is radial.
Ph
j=0 cj Vj )
• Step 3. If ω is a radial solution of (5.7), then ω ≡ 0 . Proof of Step 1. θ(ξ) := (Mv)(Oν−1 ξ) − (Mv)(Rν Oν−1 ξ) is, by rotation invariance, again a solution of (5.7) and θ(ξ) = 0 if ξ0 = 0. Hence M−1 θ is a solution of (HSL) vanishing along {|x| = 1}, and hence it gives, as well as V0 , a solution of the Dirichlet eigenvalue problem 2
U N −2 − ∆φ = λ φ |y|
in B N ,
φ=0
in ∂B N
(5.8)
(with λ = NN−2 ). Since D1,2 (RN ) is compactly embedded into the weighted 2 2 N U N −2 space L R , |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−1 θ = 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)− 21 [(Mv)(ξ)+(Mv)(R0 ξ)] = c0 (MV0 )(ξ) for some c0 , and hence M(v − c0 V0 ) is even in ξ0 . The same argument applied to v 1 = v − c0 V0 with R0 replaced by R1 gives (Mv 1 )(ξ) − (Mv 1 )(R1 ξ) = 2c1 (MV0 )(O1 ξ) = 2c1 (MV1 )(ξ) for some c1 and then M(v − c0 V0 − c1 V1 ) is even in ξ1 and, of course, in ξ0 as well. By iterating, we conclude that, for some c0 , . . . , ch P ω := M(v − hj=0 cj Vj ) is even in all the ξj and hence ∇ω(0) = 0.
24
But, as we showed above, ω(ξ) − ω(Rν ξ) = cν MV0 (Oν ξ) for some cν . Since N −2 ∇(MV0 )(Oν ξ)|ξ=0 = ( 41 ) 2 ν, we have cν = 0, i.e. ω(ξ) = ω(Rν ξ) 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 √ 1 − r − N −2 ) 2 , z(r) := ω( r)( r = |ξ| 2 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 h + 2k − 1 k−1 h+1 00 − r z0 + z=0 r ∈ (0, 1) r(1 − r)z + 2 2 2 This is Gauss’ hypergeometric equation r(1 − r)z 00 + γ − α+ + α− + 1 r z 0 − α+ α− z = 0 with γ=
h+1 , 2
α± =
h + 2k − 3 1 p ± (h + 2k − 3)2 + 8(k − 1) 4 4
Its solutions (see [26], [4]) are, at least locally at r = 0, of the form z(r) = c1 F (α± , γ; r) + c2 bF (α± , γ; r) log r + r1−γ G(r) where G is analytic, b is some, eventually zero, constant and F is the hypergeometric function F (α± , γ; r) =
X (α+ )k (α− )k k≥0
(γ)k k!
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 Multiplicity
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 φ0 has opposite signs near 0 and ∞ then we have solution for Pφ if kφ − φ(∞)k∞ is small enough. We will also obtain conditions which yield non-uniqueness of solutions. To make things more precise, let us consider the problem ( (Pφε )
−∆u =
(1+εφ(x)) NN u −2 |y|
u ∈ D1,2 (RN ) in RN
u > 0
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 0 (r) to decay at infinity. Accordingly, we assume ∃α ∈ (0, 2), ∃ψ 0 (∞) 6= 0 :
rα ψ 0 (r) →r→∞ ψ 0 (∞)
(6.1)
Theorem 6.1. Let φ = ψ(|y|) or φ = ψ(|z|). Let us assume ψ ∈ C 1 ([0, ∞)) and satisfies (6.1). Then (i) (Pφε ) has a solution for ε small enough if ψ 0 (0) · ψ 0 (∞) < 0
(6.2)
(ii) (Pφε ) has at least two solutions for ε small enough if ψ 0 (0) > 0, or ψ 0 (0) < 0,
ψ 0 (∞) > 0 ψ 0 (∞) < 0 26
and
ψ(0) ≥ ψ(∞)
and
ψ(0) ≤ ψ(∞)
(6.3)
Remark 5. (i) The above theorem can be generalised to the case ψ 0 (0) = 0 by imposing conditions on ψ 00 (0) or on the decay of ψ 0 (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 lim φ(x) = φ(∞) ∈ R. Then (Pφε ) has x→∞
a solution for ε small enough if there exist (λ0 , ζ0 ) ∈ (0, ∞) × Rh+1 satisfying 2(N −1) R N −2 φ(λ0 y, λ0 z + ζ0 ) U |y| > sup φ(0, ζ)S N −1 ζ∈Rh RN (6.4) or 2(N −1) R N −2 φ(λ0 y, λ0 z + ζ0 ) U |y| < inf φ(0, ζ)S N −1 h ζ∈R RN
If both conditions in (6.4) are satisfied for two distinct points (λ0 , ζ0 ) and 0 0 (λ0 , ζ0 ) of (0, ∞)×Rh then (Pφε ) 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 (Pφε ), building a natural constraint for the energy functional u ∈ D1,2 (RN )
Eε (u) = E0 (u) − εG(u) where
E0 (u) =
1 2
R RN
2
|∇u| −
N −2 2(N −1)
N −2 G(u) = 2(N − 1)
R RN
Z RN
|u|
2(N −1) N −2
|y|
and
−1) φ(x) 2(N |u| N −2 |y|
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 N2−2 2−N (N − 2)(k − 1) Uλ,ζ (y, z) = λ 2 U (λ−1 y, λ−1 (z − ζ)) and U = (1 + |y|)2 + |z|2 27
Clearly, Tλ0 ,ζ0 , the tangent space of Z at Uλ0 ,ζ0 , is spanned by ∂Uλ,ζ ∂Uλ,ζ V0 = for j = 1, .., h and Vj = ∂λ (λ0 ,ζ0 ) ∂ζj (λ0 ,ζ0 ) We start stating the crucial property of the linearized equation. 00
Lemma 6.3. For any (λ, ζ) ∈ R+ ×Rh , the operator E0 (Uλ,ζ ) is a self adjoint Fredholm operator of index zero. Furthermore 00 Ker E0 (Uλ0 ,ζ0 ) = Tλ0 ,ζ0 00
Proof: Since E0 (Uλ,ζ ) = I −Gλ,ζ , where
hGλ,ζ (u), vi =
N N −2
2
R RN
1 U N −2 uv, |y| λ,ζ
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 < N2N . Now, by Holder and Hardy-Sobolev inequalities: −2 ! N1−1 −1 R U N2−2 |un | |v| R U 2 NN −2 dx ≤ c kun k kvk ≤ for R large, |y| |y|
|x|≥R
|x|≥R
2
while, if
0<δ<
1 , N −2
! N2N−1 R
c
|un |
2N N −1
×
|x|≤R
R h RN
|x|≤R
R
then
v |y|(1−δ)
i N2N −2δ
U N −2 |un | |v| dx |y| −2δ ! N2N
×
≤
R h |x|≤R
1 |y|δ
2N i 1+2δ
! 1+2δ 2N
! N2N−1 ≤ c
R
2N
|un | N −1
≤ kvk
for
n large, because if t :=
|x|≤R 2N is N −2δ
2N (1−δ) , N −2δ
2N δ the corresponding Hardy-Sobolev exponent. Moreover, 1+2δ <2 then 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 Tλ,ζ the orthogonal complement of Tλ,ζ in D1,2 (RN ), Lemma 00 ⊥ into itself. Moreover, 6.3 tells that E0 (Uλ,ζ ) induces an isomorphism from Tλ,ζ 00
00
−1 k(E0 (Uλ,ζ ))−1 kL(Tλ,ζ kL(T1,0 ⊥ ) = k(E0 (U1,0 )) ⊥ )
∀(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) ∈ Tλ,ζ ,
0
Eε (Uλ,ζ + w(λ, ζ, ε)) ∈ Tλ,ζ ,
kw(λ, ζ, ε)k ≤ 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 N −2 Γ(λ, ζ) := 2(N − 1)
Z RN
−1) φ(y, z) 2(N Uλ,ζN −2 |y|
Then we have the following expansion: Φε (λ, ζ) =
S N −1 − εΓ(λ, ζ) + O(ε2 ) 2(N − 1)
Consequently, if Γ(λ, ζ) has a stable critical point (λ0 , ζ0 ), Φε will have a critical point uε = Uλε0 ,ζ0ε + w(λε0 , ζ0ε , ε) for any |ε| 1 and (λε0 , ζ0ε ) → (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 are indeed Cyl
1,2 critical points of Eε , where DCyl is the completion of {u ∈ C0∞ : u(x) = R v(|y|, |z|)} with respect to the norm ||u||2 = |∇u|2 . Therefore in this Rn
1,2 case we can do the finite dimensional reduction of Eε : DCyl → R to get the same results with the only difference that there wouldn’t be any dependence 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 2−N Uλ (x) = λ 2 U (λ−1 x).
29
Proof of Theorem 6.1. We know from Lemma 6.6 that if u ∈ Zε then S N −1 + O(ε). In addition if u ∈ Zε is also a critical point of Eε (u) = 2(N −1) Eε and changes sign then as in the proof of Lemms 3.1 we can show that N −1 Eε (u) ≥ SN −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 2(N −1) Z U N −2 N −2 ψ(λ|y|) Γ(λ) = 2(N − 1) |y| RN
Differentiating with respect to λ, ∂Γ N −2 (λ0 ) = ∂λ 2(N − 1)
Z
0
(ψ (λ0 |y|)|y|)
U
2(N −1) N −2
dx
|y|
RN
Now by dominated convergence theorem ∂Γ N −2 0 lim (λ0 ) = ψ (0) λ0 →0 ∂λ 2(N − 1)
Z U
2(N −1) N −2
dx
RN
and lim
λ0 →∞
∂Γ (λ0 ) λα0 ∂λ
N −2 0 = ψ (∞) 2(N − 1)
Z
U
2(N −1) N −2
|y|α
dx
RN 0
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 (Pφε ) 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 (Pφε ) 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 reduction 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 Z −1) ϕ(λy, λz + ζ) 2(N N −2 U N −2 Γ(λ, ζ) = 2(N − 1) RN |y| lay in a bounded region λ2 + |ζ|2 ≤ R. To this extent, notice first that Γ ∈ C 2 ((0, ∞) × Rh ), it can be extended continously up to {λ = 0} by setting Γ(0, ζ) := lim+ Γ(λ, ζ) = λ→0
S N −1 (N − 2) ψ(ζ) 2(N − 1)
and it results ∇ζ Γ(0, ζ) = lim+ ∇ζ Γ(λ, ζ) = λ→0
S N −1 (N − 2) ∇ψ(ζ) 2(N − 1)
(6.5)
S N −1 (N − 2) 2 D ψ(ζ) (6.6) λ→0 2(N − 1) where the limits are uniform for ζ on compact sets. ˆ the Melnikov function associated to ϕ(x) Denoted by Γ ˆ := ϕ( |x|x2 ), it follows ˆ has no critical point close to λ = 0, ζ = from (6.5) and assumption (ii) that Γ Dζ2 Γ(0, ζ) = lim+ Dζ2 Γ(λ, ζ) =
0. We now claim that ∇Γ(λ, ζ) = 0
⇔
ˆ (∇Γ)
λ ζ , λ2 + |ζ|2 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 and hence, with the change of variables x → |x|x2 , we obtain U λ , ζ λ2 +|ζ|2 λ2 +|ζ|2
2(N − 1) Γ(λ, ζ) = N −2
Z RN
−1) φ(x) 2(N Uλ,ζN −2 = |y|
Z
31
RN
φ( |x|x2 ) |y|
|x|(2−N ) Uλ,ζ
x |x|2
−1) 2(N N −2
Z = RN
ˆ φ(x) U λ , ζ λ2 +|ζ|2 λ2 +|ζ|2 |y|
2(N −1) N −2
=
λ 2(N − 1) ˆ ζ Γ( 2 , 2 ) 2 N −2 λ + |ζ| λ + |ζ|2
Now, besides (6.5), (6.6), we also have , by oddness, Z −1) N −2 h∇ϕ(λy, λz + ζ), (y, z)i 2(N ∂Γ (0, ζ) := lim+ U N −2 = 0 λ→0 2(N − 1) RN ∂λ |y| ∂ 2Γ N −2 [A∆y ϕ(0, ζ) + B∆z ϕ(0, ζ)] (λ, ζ) = 2 λ→0 ∂λ 2(N − 1) lim
(6.7)
(6.8)
where the limit is uniform for ζ on compact sets and 1 A := k
Z |y|U
1 B := h
2(N −1) N −2
RN
Z
−1) |z| 2 2(N U N −2 |y|
(6.9)
RN
Notice that, since we assumed N ≥ 4, A and B are finite. Changing to polar s coordinates and using the change of variable t = 1+r we obtain A = 2hωk ωh I(2k + 1, N + k − 2)I(h − 1, N − 1) B = 2kωk ωh I(2k − 3, N + k − 4)I(h + 1, N − 1) R∞ ρn Γ( n+1 Γ 2m−n−1 ) 2 ) ( 2 , where Γ denotes the Euler where I(m, n) = (1+ρ 2 )m dρ = 2Γ(m) 0
Gamma function. Then, by direct computations, A=
k−1 B 2k + h − 3
and hence, following the notation introduced in (1.8) in Theorem B, (6.8) rewrites ∂ 2Γ B(N − 2) ∆∗ (ζ) (6.10) lim 2 (λ, ζ) = λ→0 ∂λ 2(N − 1)(2k + h − 3) 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, ζ) ∆∗ (ζj ) > 0 ∂λ
32
(6.11)
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 1 + h Bs := (λ, ζ) ∈ R × R : |(λ, ζ) − (s, 0)| < s − 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-Poincar´e characteristic of Bs and ∂− Bs is the set of > 0 if s is large. points in ∂Bs where ∇Γ points inward Bs , i.e. with ∂Γ ∂λ 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 X ind(∇(Γ|∂− Bs )) = (−1)m(ψ;ζj ) {j: ∆∗ (ζj )>0}
and this concludes the proof. Remark 7. (ii) in Theorem B, i.e. ∃ V = V1 + V2 ∈ Rk × (Rh \ {0}) such ˆ ˆ ∇z φ) ˆ →x→0 V1 + V2 amounts to require that that ∇φ(x) = (∇y φ, ∇φ(x) =
1 [P x + ◦(1)] |x|2
and
∇ψ(z) =
1 [Qz + ◦(1)] R2
where P x := 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 V2 , and , as an easy consequence, exactly two nondegenerate zeros , ± √2|V | 2
deg(∇ψ, BR , 0) = 1 + (−1)h 33
(6.12)
This formula has a simple geometric interpretation: (ii) also amounts to say that, denoted by π the stereographic projection of S N onto RN , then Φ := φ◦π extends to a C 1 (S N ) map with the south pole as a regular point. Similarly for Ψ := ψ ◦ π. The degree formula for ∇ψ then follows from the Poincar´e-Hopf index Theorem for the tangent vector field on S h given by ∇Ψ. Remark 8. One can reformulate Theorem B using the alternative Morse formula (see [15]) X deg(∇Γ, Bs , 0) = (−1)h+1 +ind(∇(Γ|∂+ Bs )) = (−1)h+1 + (−1)m(ψ;ζj ) {j: ∆∗ (ζj )<0}
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 S 2n+1 and the ndimensional Heisenberg group (see [14] and [20]). Let K be a function in R2n × R such that K ◦ F is a smooth function on S 2n+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) (ii)
∂iK (0, t) ∂ri
≡ 0, i = 1, 2, 3 and ∃ c 6= 0 : t2 K(0, t) −→ c.
∂4K (0, tj ) ∂r4
= 0 ∀j
|t|→∞
Let R(ξ) := 1 + K. Then (3.3) has a solution for 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 U ε be defined as in (1.7) and (3.4), then 2(N −1)
(U ε ) (N −2) dx = |y|
Z |x|≥1
2(N −1)
1
Z
(U ε ) (N −2) dx = O(εN −1 ). |y|
|x|≤1
Proof. Since U is equal to its Kelvin transform we get 2(N −1)
(U ε ) (N −2) dx = |y|
Z
1
Z
2(N −1)
(U ε ) (N −2) dx. |y|
|x|≤1
|x|≥1
Now by a change of variable, 2(N −1)
(U ε ) (N −2) dx = |y|
Z
1 ε
Z
U
2(N −1) (N −2)
1 2ε
Z
dx.
or |z| ≥
1 2ε .
N −1
1 |y|≥ 2ε
|y| ((1 + |y|)2 + |z|2 )
Z
1
+C
N −1
1 1 ,|z|≥ 2ε |y|< 2ε
−2(N −1)
|x|
|y| ((1 + |y|)2 + |z|2 ) Z
Z
dx + C
1 |x|≥ 2ε
Therefore
1
dx ≤ C
|y|
≤ Cε
2(N −1) (N −2)
|y|
implies either |y| ≥
|x|≥ 1ε
Z
U
|x|≥ 1ε
|x|≥1
Since |x| ≤ |y| + |z|, |x| ≥
Z
1 1 |y|< 2ε |z|≥ 2ε
dx
1 |y| (1 + |z|2 )
dx
N −1
dy dz = O(εN −1 )
This proves the Lemma. Lemma 7.2. Let U and U ε be defined as in (1.7) and (3.4), then Z Z N 1 ε N 1 Uε N −2 U (U ε ) N −2 dx = Uε dx = εN −2 (CN,k + o(1)) |y| |y| RN
RN
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 fact that U is invariant under Kelvin transform. Now by a change of variable Z RN
1
N
(U ε ) N −2
Uε dx = |y|
Z
N Z x NN−2 U (εx) (U (x)) N −2 U( ) dx = εN −2 U (ε2 x) dx ε |εy| |y|
RN
RN
35
N
Let W = U 2(N −1) then by direct calculation one can see that W ∈ D1,2 (RN ). Hence from the Hardy Sobolev inequality (1.2), we get −1 N N −2 2(N −1) N Z Z Z (U (x)) N −2 W N −2 |∇W |2 <∞ dx = dx ≤ S −1 |y| |y| RN
RN
RN
Hence by Dominated convergence theorem, we get N N Z Z (U (x)) N −2 (U (x)) N −2 2 dx → U (0) dx. U (ε x) |y| |y| RN
RN
This proves the Lemma. Lemma 7.3. Let U and U ε be defined as in (1.7) and (3.4), then Z Z N 1 ε ε N 1 N −2 U ε N −2 U Uε (U ) dx = dx = o(εN −2 ) |y| |y| |x|≥1
|x|≤1
Proof. Again the first equality follows by a Kelvin-transform argument. Now as in the proof of the previous Lemma, N Z Z 1 ε N (U (x)) N −2 ε N −2 U N −2 2 (U ) dx = ε U (ε x) dx = o(εN −2 ) |y| |y| |x|> 1ε
|x|≥1 N
as
R RN
(U (x)) N −2 |y|
R
dx < ∞ and hence
|x|> 1ε
N N −2
U (ε2 x) (U (x)) |y|
dx → 0. This proves the
Lemma. Lemma 7.4. Let ρ ∈ Cc (RN \ {0}), then 2(N −1)
Z ρ(x)
(U ε ) (N −2) dx = O(εN −1 ) = |y|
RN
1
Z ρ(x)
2(N −1)
(U ε ) (N −2) dx |y|
RN
Proof. Assume that support of ρ is contained in the annulus {x : 0 < R1 < |x| < R2 }. Then by change of variable, 2(N −1)
Z
(U ε ) (N −2) ρ(x) dx = |y|
RN
Z ρ(εx)
U
2(N −1) (N −2)
Z
U
dx ≤ C
|y|
RN
2(N −1) (N −2)
|y| |x|≥
dx = O(εN −1 ).
R1 ε
The other estimate follows similarly. Similar computations yield Lemma 7.5. Let ρ ∈ Cc (RN \ {0}), then Z Z 1 1 NN−2 U ε N Uε ρ(x) (U ε ) N −2 dx = o(εN −2 ) = ρ(x) U ε dx |y| |y| RN
RN
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 Q+2
L(u) = u Q−2
(7.1)
Q+2 where Q = m1 + 2m2 is the ”appropriate” dimension and Q−2 is the corresponding critical exponent. This problem has been investigated in [21] and [22]. Analogous to the Scalar curvature problem, let us consider Q+2 L(u) = K(y, z)u Q−2 in Rm1 × Rm2 (7.2) u>0 in Rm1 × Rm2
We are going to show that under some symmetry assumptions (7.2) is closely related to our Hardy-Sobolev problem (Pφ ). Let φ be radially symmetric in the y variable and let u(y, z) be a solution of (Pφ ) which is ˜ also 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 N
N −2 k−1 ˜ z) θ(r, z) θr (r, z) − ∆z θ(r, z) = φ(r, −θrr (r, z) − r r
Define ψ(r, z) = θ(r2 , z), (r, z) ∈ [0, ∞) × Rh Then, ψr (r, z) = 2rθr (r2 , z),
ψrr (r, z) = 2θr (r2 , z)+4r2 θrr (r2 , z),
∆z ψ(r, z) = ∆z θ(r2 , z)
Hence ψ solves −ψrr (r, z) −
2k − 3 ˜ 2 , z)ψ(r, z) NN−2 ψr (r, z) − 4r2 ∆z ψ(r, z) = 4φ(r r
Thus, if we define v(y, z) = ψ(|y|, z)
(7.3)
˜ 2 , z). then v solves (7.2) with m1 = 2k − 2, m2 = h and K(y, z) = 4φ(|y| 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 the Grushin problem (7.2) has a solution if (Pφ ) has a solution in RN = Rk × Rh which is p ˜ radially symmetric in y, where k = m12+2 , h = m2 and φ(y, z) = 14 K( |y|, z).
37
The requirement of m1 to be even in the above lemma may look like a restriction, however we see that it is natural when we deal with the Webster scalar curvature equation in the Heisenberg 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 Q+2
− ∆H u(ξ) = R(ξ)u(ξ) Q−2 , u > 0 in Hn
(7.4)
where ∆H is the Heisenberg sublaplacian and Q = 2n + 2 is the homogeneous dimension of Hn . Recall that n X ∆H = Xi2 + Yi2 i=1
and Xi =
∂ ∂ + 2yi , ∂xi ∂t
Yi =
∂ ∂ − 2xi , i = 1...n. ∂yi ∂t
If u(ξ) is cylindrically symmetric, i.e., u(ξ) = θ(|Z|, t) then by direct calculation one can see that ∆H u becomes ∆H u = ∆Z u + 4|Z|2 utt 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 Webster scalar curvature problem (7.4) has a solution if (Pφ ) has a solution in RN = Rk × R which p is radially symmetric in y, where k = n + 1, h = 1 and φ(y, z) = 41 R( |y|, z).
7.3
Linking to the Hyperbolic Laplacian
Let ∆H u := − 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 N +2−2t
k−1 u N −2 −urr − ur − ∆z u = r rt
R+ × Rh
in 2
−vrr − Then, after setting have
k−1 N U N −2 vr − ∆z v = v r N −2 r
u ˜ := r
N −2 2
u(r, z),
v˜ := r
N −2 2
in
v(r, z),
N +2−2t h2 − (k − 2)2 u ˜+u ˜ N −2 4 2 ˜ N2−2 h − (k − 2)2 N U −∆H v˜ = u ˜+ v˜ 4 N −2 r
−∆H u ˜=
38
R+ × Rh ˜ := r N2−2 U (r, z), U
we
In fact, u ˜r = r
N −2 2
ur +
N −2 N 2−4 u, 2 r
u ˜rr = r
N −2 2
urr + (N − 2)r
N −4 2
ur +
(N −2)(N −4) N −6 r 2 u 4
and hence, − r2 ∆˜ u − (h − 1)r˜ ur = h N +2 ur i h2 − (k − 2)2 N −2 + −r 2 urr + ∆z u + (k − 1) r 2 u= r 4 +2−2t N −2 N N N +2−2t N +2 h2 − (k − 2)2 h2 − (k − 2)2 −2 u ˜ + r 2 −t r− 2 u ˜ = u ˜+u ˜ N −2 4 4 Similarly, N +2 h2 − (k − 2)2 N −2 −∆H v˜ = −r 2 ∆v + r 2 v= 4 2
N +2 N −2 2 h2 − (k − 2)2 h2 − (k − 2)2 N U N −2 − N −2 N v˜ + r 2 r 2 v˜ = v˜ + (r 2 U ) N −2 v˜ 4 N −2 r 4 N −2
Remark 7.1. Given u cylindrically symmetric, let, as above, u ˜(y, z) := |y| dimension N ≥ 3 it results Z Z k−2 2 2 h2 u2 ωk |∇˜ u|2 + ( ) u ˜ dVH = dydz |∇u|2 + 2 4 |y|2
N −2 2
u(y, z). In
RN
H
where ωk is the area of the unit sphere in Rk . In particular, thanks to Hardy inequality, u ∈ D1 (Rk × Rh )
⇔
u ˜ ∈ H 1 (Hh+1 )
The above identity readily follows from 2 N −2 |∇˜ u|2 = rN −2 |∇u|2 + rN −4 u2 + (N − 2)rN −3 uur 2 and then integrating: Z " ωk
r
N
2
|∇u| +
N −2 2
2 r
N −2 2
u + (N − 2)r
N −1
# k − 2 2 N −2 2 drdz uur + ( ) r u 2 rh+1
R+ ×Rh
Z =
|∇u|2 + (
N − 2 2 u2 (N − 2)(k − 2) u2 k − 2 2 u2 ) − +( ) dydz 2 2 2 |y| 2 |y| 2 |y|2
RN
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 −
Q+2 (k − 1) φr − (1 + α)2 r2α ∆z φ = φ(r, z) Q−2 r
where , α > 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 k−2 1 2 h − = Φ Q−2 −∆H Φ − 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 k−2 1 2 h − γ 2 rβ φ(rγ , z)− −∆H Φ := − r ∆Φ − (h − 1)rΦr = 4 α+1 2−2γ Q+2 r β+2γ γ γ −γ γ −γ 2 r φrr (r , z) + φzz (r , z) + (k − 1)r φr (r , z) γ2 2 2α k−2 because β(h − β) = 41 h2 − α+1 , 2β+γ−h = k − 1. Since 2 − 2γ = α+1 and γ β + 2γ = β
Q+2 Q−2 ,
we finally get " 2 # Q+2 h Q−2 i Q−2 Q+2 1 k−2 1 2 h − = Φ Q−2 −∆H Φ − Φ = γ 2 rβ φ(r α+1 , z) 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
1 − r2 h 1 − r2 (φrr + φr ) − (k − 1) rφr + (k − 1)φ(r) = 0 2 r 2 N −2
2
N
− 2 −2 φr = ( 1−r ωr + N 2−2 ( 1−r r ω(r) , 2 ) 2 ) 1 − r2 − N +2 1 − r2 2 1 − r2 N −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 h 1 − r2 N −2 2 2 ( ) ( ) ωrr + ωr + (h − 1) rωr + [h + 1 − (k − 1)r ]ω = 2 2 r 2 4 N (k − 1) (N − 2)(h + 1) (N − 2)(k − 1) 2 1 − r2 − N h2 − (k − 2)2 ) 2 + (1 − r2 ) − + r = −( 2 4 4 4 4 = −(
1 − r2 − N k−1 ) 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 . √ 0 0 2 00 0 2 Finally, setting z(r) := φ( r), from φ (r) = 2rz (r ), φ (r) = 2z (r ) + 4r2 z 00 (r2 ) we get, from the equation for φ and writing t = r2 ,
0 = 2t(1 − t)z 00 + z 0 [h + 1 − (h + 2k − 1)t] + (k − 1)z
40
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