On the Yamabe problem and the scalar curvature problems under ...

Math. Ann. 322, 667–699 (2002)

Mathematische Annalen

Digital Object Identifier (DOI) 10.1007/s002080100267

On the Yamabe problem and the scalar curvature problems under boundary conditions Antonio Ambrosetti · YanYan Li · Andrea Malchiodi Received: 18 May 2000 / Published online: 28 February 2002 – © Springer-Verlag 2002

1. Introduction In this paper we prove some existence results concerning a problem arising in conformal differential geometry. Consider a smooth metric g on B = {x ∈ Rn : |x| < 1}, the unit ball on Rn , n ≥ 3, and let ∆g , Rg , νg , hg denote, respectively, the Laplace-Beltrami operator, the scalar curvature of (B, g), the outward unit normal to ∂B = S n−1 with respect to g and the mean curvature of (S n−1 , g). Given two smooth functions R and h , we will be concerned with the existence of positive solutions u ∈ H 1 (B) of  (n − 1) n+2   −4 (n − 2) ∆g u + Rg u = R u n−2 , in B;   

2 n ∂νg u + hg u = h u n−2 , on ∂B = S n−1 . (n − 2)

(1)

It is well known that such a solution is C ∞ provided g, R and h are, see [10]. If u > 0 is a smooth solution of (1) then g = u4/(n−2) g is a metric, conformally equivalent to g, such that R and h are, respectively, the scalar curvature of (B, g ) and the mean curvature of (S n−1 , g ). Up to a stereographic projection, this is equivalent to finding a conformal metric on the upper half sphere S+n = {(x1 , . . . , xn+1 ) ∈ Rn+1 : |x| = 1, xn+1 > 0} such that the scalar curvature of S+n and the mean curvature of ∂S+n = S n−1 are prescribed functions. A. Ambrosetti S.I.S.S.A., via Beirut, 2–4, 34019 Trieste, Italy (e-mail: [email protected]) Y.Y. Li, A. Malchiodi Department of Mathematics, Rutgers University New Brunswick, NJ 08903, USA (e-mail: [email protected]; [email protected]) The first and the third authors have been supported by the italian M.U.R.S.T. under the national project Variational Methods and Nonlinear Differential Equations.

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In the first part of the paper we deal with the the case in which R and h are constant, say R ≡ 1 and h ≡ c, when (1) becomes  (n − 1) n+2   −4 (n − 2) ∆g u + Rg u = u n−2 , in B; (Y )  2 n  n−1  ∂ν u + hg u = cu n−2 , on ∂B = S . (n − 2) g This will be referred as the Yamabe like problem and was first studied in [10– 12]. More recently, the existence of a solution of (1) has been proved in [14,15] under the assumption that (B, g) is of positive type (for a definition see [14]) and satisfies one of the following assumptions: (i) (B, g) is locally conformally flat and ∂B is umbilical; (ii) n ≥ 5 and ∂B is not umbilical. Our main result concerning the Yamabe like problem shows that none of (i) or (ii) is required when g is close to the standard metric g0 on B. Precisely, consider the following class Gε of bilinear forms Gε = {g ∈ C ∞ (B) : g − g0 L∞ (B) ≤ ε, ∇gLn (B) ≤ ε, ∇gLn−1 (S n−1 ) ≤ ε}. (2) Inequalities in (2) hold if for example g−g0 C 1 (B) ≤ ε, or if g−g0 W 2,n (B) ≤ ε. We will show: Theorem 1. Given M > 0 there exists ε0 > 0 such that for every ε with ε ∈ (0, ε0 ), for every c > −M and for every metric g ∈ Gε problem (Y ) possesses a positive solution. In the second part of the paper we will take g = g0 , R = 1 + εK(x), h = c + εh(x) and consider the Scalar Curvature like problem  (n − 1) n+2   −4 (n − 2) ∆u = (1 + εK(x))u n−2 , in B; (Pε )  2 ∂u n  n−1  n−2 + u = (c + εh(x)) u , on S , (n − 2) ∂ν

where ν = νg0 . The Scalar Curvature like problem has been studied in [16] where a non perturbative problem like  (n − 1) n+2   −4 (n − 2) ∆u = R (x)u n−2 , in B;  2 ∂u   + u = 0, on S n−1 , (n − 2) ∂ν

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has been considered. We also mention the paper [9] dealing with the existence of solutions of    ∆u = 0, in B; (3) 2 ∂u n  n−1  n−2 + u = (1 + εh(x)) u , on S ,  (n − 2) ∂ν a problem similar in nature to (Pε ). To give an idea of the existence results we can prove, let us consider the particular cases that either h ≡ 0 or K ≡ 0. In the former, problem (Pε ) becomes  (n − 1) n+2   −4 (n − 2) ∆u = (1 + εK(x))u n−2 , in B; (Pε,K )  2 ∂u n  n−1  n−2 + u = c u , on S , (n − 2) ∂ν Theorem 2. Suppose that K satisfies (K1 ) there exists an absolute maximum (resp. minimum) p of K|S n−1 such that K (p) · p < 0, resp. K (p) · p > 0. Then for |ε| sufficiently small, (Pε,K ) has a positive solution. Another kind of result is the following Theorem 3. Let K|S n−1 be a Morse function and satisfies K (x) · x = 0,  x∈Crit (K|S n−1

∀ x ∈ Crit (K|S n−1 )

(K2 )

(−1)m(x,K) = 1,

(K3 )

):K (x)·x 0.

(5)

Z c = {zµ,ξ : µ2 + |ξ |2 − cκµ − 1 = 0}

(6)

The set

is an n-dimensional manifold, diffeomorphic to a ball in Rn , with boundary ∂Z c corresponding to the parameter values µ = 0, |ξ | = 1.

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We need to study the eigenvalues of I0 (zµ,ξ ), with zµ,ξ ∈ Z c . Recall that, by definition, λ ∈ R is an eigenvalue of I0 (zµ,ξ ) if there exists v ∈ H 1 (B), v = 0 such that I0 (zµ,ξ )[v] = λv and this means that v is solution of the linear problem  (n − 1) 4 n + 2 n−2  −4 − λ) ∆v = z v, (1   (n − 2) n − 2 µ,ξ

2  (n − 1) n  n−2 4 z + λ − 1 v, (1 − λ) ∂ν v = 2(n − 1) c (n − 2) (n − 2) µ,ξ

in B; on S n−1 . (7)

The following lemma is well known. Lemma 1. (a) λ = 0 is an eigenvalue of (7) and the corresponding eigenspace is n dimensional and coincides with the tangent space to Z c at zµ,ξ , namely is spanned by Dzµ,ξ . (b) (7) has precisely one negative eigenvalue λ1 (c); all the remaining eigenvalues are positive. Item (a) is proved in [14]. Item (b) easily follows from the fact that zµ,ξ is a Mountain Pass critical point of I0c . Let λ2 (c) denote the smallest positive eigenvalue of I0 (zµ,ξ ). The main result of this section is the following one: Lemma 2. For all M > 0 there exists a positive constant CM such that 1 ≤ |λi (c)| ≤ CM , ∀ c ≥ −M, i = 1, 2. CM Remark. There is a numerical evidence that λ2 (c) ↓ 0 as c ↓ −∞. Proof. We will prove separately that |λi (c)| ≤ CM and that C1M ≤ |λi (c)|. For symmetry reasons it is sufficient to take zµ,ξ = zµ , namely to take ξ = 0. In such a case µ depends only on ξ and (5) yields

1 µ(c) = κc + κ 2 c2 + 4 . 2 Case 1. |λi (c)| ≤ CM . By contradiction suppose there exists a sequence cj → +∞ such that |λi (cj )| → +∞, i = 1, 2. Let vj denote an eigenfunction of (7) with λ = λi (cj ). Then vj solves the problem  ∆vj = aj (x)vj ,

in B;

∂ v = b (x)v , ν j j j

on S n−1 ,

(8)

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where 4 n + 2 n−2 1 zµ(cj ) (x), x ∈ B (λi (cj ) − 1) 4(n − 1)

2 n n−2 n−2 cj z (x) + λi (cj ) − 1 , x ∈ S n−1 . bj (x) = 2(1 − λi (cj )) (n − 2) µ(cj )

aj (x) =

Above, it is worth pointing out that bj is constant on S n−1 . Actually, there results 2 n−2

zµ (x) = κ µ

−1



1 1+ 2 µ

−1

, ∀ x ∈ S n−1 ,

and hence

 

−1 n n−2 1 −1 bj ≡ cj · κ µ (cj ) 1 + 2 + λi (cj ) − 1 , 2(1 − λi (cj )) (n − 2) µ (cj ) ∀ x ∈ S n−1 .

Moreover, since µ ∼ κc as c → +∞, it turns out that bj → −

(n − 2) . 2

(9)

Now, integrating by parts we deduce from (8)    |∇vj |2 dx + aj vj2 dx = bj B

S n−1

B

vj2 dσ.

(10)

Using (9) and a Poincar´e-like inequality, we find there exists C > 0 1   2 − aj vj dx ≥ C vj2 dx. B

B

This leads to a contradiction because aj (x) → 0 in C 0 (B) and vj ≡ 0. Case 2. C1M ≤ |λi (c)|. Arguing again by contradiction, let cj → +∞ and suppose that |λi (cj )| → 0. As before, the corresponding eigenfunctions vj satisfy (10), where now bj → 1, because µ ∼ κc and |λi (cj )| → 0. Choosing vj is such a way that supB |vj | = 1, then (10) yields that vj is bounded in H 1 (B) and hence vj < v0 weakly in H 1 (B). Passing to the limit in    ∇vj · ∇w + aj vj w − bj vj w = 0, ∀w ∈ H 1 (B), B

B

S n−1

1 in the sequel we will use the same symbol C to denote possibly different positive constants.

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it immedately follows that v0 satisfies  ∆v0 = 0, ∂ v = v , ν 0 0

in B; on S n−1 .

(P3 )

The solutions of problem (P3 ) are explicitly known, namely they are the linear functions an B. We denote by X the vector space of these solutions, which is n-dimensional. To complete the proof we will show that v0 ∈ X leads to a contradiction. We know that λ = 0 is an eigenvalue with multiplicity n, and the eigenvectors corresponding to λ = 0 are precisely the elements of Tzµ Z c . Let uj ∈ Tzµ (cj ) Z c with supB |uj | = 1. Then, by using simple computations, one can prove that, up to a subsequence, uj → v strongly in H 1 (B) for some function v ∈ X. We can assume w.l.o.g. that v = v0 (the weak limit of vj ), so it follows that (uj , vj ) → v0 2 = 0. But this is not possible, since vj are eigenvectors corresponding to λ1 < 0, while uj are eigenvectors corresponding to λ = 0 and hence they are orthogonal.   In conclusion, taking into account of Lemma 2, we can state: Lemma 3. The unperturbed functional I0c possesses an n-dimensional manifold Z c of critical points, diffeomorphic to a ball of Rn . Moreover I0c satisfies the following properties (i) I0 (z) = I − K, where K is a compact operator for every z ∈ Z c ; (ii) Tz Z c = KerD 2 I0c (z) for all z ∈ Z c . From (i)-(ii) it follows that the restriction of D 2 I0c to (Tz Z c )⊥ is invertible. Moreover, denoting by Lc (z) its inverse, for every M > 0 there exists C > 0 such that Lc (z) ≤ C for all z ∈ Z c

and for all c > −M.

(11)

3. The Yamabe like problem 3.1. Preliminaries Solution s of problem (1) can be found as critical points of the functional I c : H 1 (B) → R defined in (4). We recall some formulas from [3] which will be useful for our computations. We denote with gij the coefficients of the metric g in some local co-ordinates and with g ij the elements of the inverse matrix (g −1 )ij .

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The volume element dVg of the metric g ∈ Gε , taking into account (2) is 1

dVg = |g| 2 · dx = (1 + O(ε)) · dx 2 .

(12)

The Christoffel symbols are given by Γijl = 21 [Di gkj + Dj gki − Dk gij ]g kl . The components of the Riemann tensor, the Ricci tensor and the scalar curvature are, respectively l l l = Di Γjlk − Dj Γikl + Γim Γjmk − Γjlm Γikm ; Rkj = Rklj ; R = Rg = Rkj g kj . Rkij (13)

For a smooth function u the components of ∇g u are (∇g u)i = g ij Dj u, so (∇g u)i = ∇u · (1 + O(ε)).

(14)

From the preceding formulas and from the fact that g ∈ Gε it readily follows that I c (u) = I0c (u) + O(ε). More precisely, the following lemma holds. The proof is rather technical and is postponed to the Appendix. Lemma 4. Given M > 0 there exists C > 0 such that for c > −M and g ∈ Gε there holds ∇I c (z) ≤ C · ε · (1 + |c|)−

n−2 2

, ∀z ∈ Z c ;

 2 c  D I (z) − D 2 I c (z) ≤ C · ε, ∀z ∈ Z c 0  c  I (z + w) − I c (z + w ) ≤ C · (1 + |c|) · (ε + ρ



2 n−2

(15)

(16)

(17)

) · w − w , ∀z ∈ Z c , w, w ∈ H 1 (B), ∀w, w  ≤ ρ;

4

4

1 + u n−2 + w n−2

 c  ∇I (u + w) − ∇I c (u) ≤ C · w · (18)

2 2 + |c| · u n−2 + |c| · w n−2 , ∀u, w ∈ H 1 (B).

Moreover, if u is uniformly bounded and if w ≤ 1 there results   2 c 2 D I (u + w) − D 2 I c (u) ≤ C · (1 + |c|) · w n−2 .

(19)

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3.2. A finite dimensional reduction The aim of this section is to perform a finite dimensional reduction, using Lemma 3. Arguments of this kind has been emploied, e.g. in [1]. The first step is to construct, for g ∈ Gε , a perturbed manifold Zgc  Z c which is a natural constraint for I c , namely: if u ∈ Zgc and ∇I c |Zgc (u) = 0 then ∇I c (u) = 0. For brevity, we denote by z˙ ∈H 1 (B))n an orthonormal n-tuple in Tz Z c . Moreover, if α ∈ Rn we set α z˙ = αi z˙ i . Proposition 1. Given M > 0, there exist ε0 , C > 0, such that ∀ c > −M, ∀ z ∈ Zc ∀ ε ≤ ε0 and ∀ g ∈ Gε there are C 1 functions w = w(z, g, c) ∈ H 1 (B) and α = α(z, g, c) ∈ Rn such that the following properties hold (i) w is orthogonal to Tz Z c ∀z ∈ Z c , i.e. (w, z˙ ) = 0; (ii) ∇I c (z + w) = α z˙ ∀z ∈ Z c ; n−2 (iii) (w, α) ≤ C · ε · (1 + |c|)− 2 ∀z ∈ Z c . Furthermore, from (i)-(ii) it follows that (iv) the manifold Zgc = {z + w(z, g, c) | z ∈ Z c } is a natural constraint for I c . Proof. Let us define 3 Hg : Z c × H 1 (B) × Rn → H 1 (B) × Rn by setting

c ∇I (z + w) − α z˙ . Hg (z, w, α) = (w, z˙ ) With this notation, the unknown (w, α) can be implicitly defined by the equation Hg (z, w, α) = (0, 0). Setting Rg (z, w, α) = Hg (z, w, α) − ∂(w,α) Hg (z, 0, 0) [(w, α)] we have that Hg (z, w, α) = 0

⇔

∂(w,α) Hg (z, 0, 0)[(w, α)] + Rg (z, w, α) = 0.

Let H0 = Hg0 . From (11) it follows easily that ∂(w,α) H0 (z, 0, 0) is invertible uniformly w.r.t. z ∈ Z c and c > −M. Moreover using (16) it turns out that for ε0 sufficiently small and for ε ≤ ε0 also the operator ∂(w,α) Hg (z, 0, 0) is invertible and has uniformly bounded inverse, provided g ∈ Gε . Hence, for such g there results Hg (z, w, α) = 0 ⇔ (w, α) = Fz,g (w, α) −1  Rg (z, w, α). := − ∂(w,α) Hg (z, 0, 0) We prove the Proposition by showing that the map Fz,g is a contraction in some ball Bρ = {(w, α) ∈ H 1 (B) × Rn : w + |α| ≤ ρ}, with ρ of order 3 H depends also on c, but such a dependence will be understood.

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ρ ∼ ε · (1 + |c|)− 2 . We first show that there exists C > 0 such that for all (w, α), (w , α ) ∈ Bρ , all z ∈ Z c and all g ∈ Gε , there holds  

n  Fz,g (w, α) ≤ C · ε · (1 + |c|)− n−2 2 + (1 + |c|) · ρ n−2 , 2  F (w , α ) − F (w, α) ≤ C · (1 + |c|) · ρ n−2 · (w, α) − (w , α ). z,g

z,g

(20) Condition (20) is equivalent to the following two inequalities

 n−2 2 ∇I c (z + w) − D 2 I c (z)[w] ≤ C · ε · (1 + |c|)− 2 + (1 + |c|) · ρ n−2 ; (21) (∇I c (z + w) − D 2 I c (z)[w]) − (∇I c (z + w ) − D 2 I c (z)[w ]) ≤ (22) C · (1 + |c|) · ρ n−2 · (w, α) − (w , α ). 2

Let us first prove (21). There holds ∇I c (z + w) − D 2 I c (z)[w] = ∇I c (z + w) − ∇I c (z) + ∇I c (z) − D 2 I c (z)[w]  1   2 c c D I (z + sw) − D 2 I c (z) [w]ds. = ∇I (z) + 0

Hence it turns out that ∇I c (z+w)−D 2 I c (z)[w] ≤ ∇I c (z)+w· sup D 2 I c (z+sw)−D 2 I c (z). s∈[0,1]

Using (19) we have n

∇I c (z + w) − D 2 I c (z)[w] ≤ ∇I c (z) + C · (1 + |c|) · ρ n−2 . Hence from (15) we deduce that



n−2 n ∇I c (z + w) − D 2 I c (z)[w] ≤ C · ε · (1 + |c|)− 2 + (1 + |c|) · ρ n−2 ,

and (21) follows. We turn now to (22). There holds ∇I c (z + w) − ∇I c (z + w ) − D 2 I c (z)[w − w ]   1 

  2 c 2 c D I (z + w + s(w − w)) − D I (z) [w − w]ds  =   0

≤ sup D 2 I c (z + w + s(w − w)) − D 2 I c (z) · w − w. s∈[0,1]

Using again (19), and taking w, w  ≤ ρ we have that ||D 2 I c (z + w + s(w − w )) − D 2 I c (z)|| ≤ C · (1 + |c|) · ρ n−2 , 2

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proving (22). Taking ρ = 2C · ε · (1 + |c|)− 2 and ε ≤ ε0 , with ε0 sufficiently small, there results  

n C · ε · (1 + |c|)− n−2 2 + (1 + |c|) · ρ n−2 < ρ, 2 C · (1 + |c|) · ρ n−2 < 1.

Then Fz,g is a contraction in Bρ and hence Hg = 0 has a unique solution w = n−2   w(z, g, c), α = α(z, g, c) with (w, α) ≤ 2C · ε · (1 + |c|)− 2 . Remark 1. In general, the preceding arguments give rise to the following result, see [1]. Let Iε (u) = I0 (u) + O(ε) denote a C 2 functional and suppose that I0 has an n-dimensional manifold Z of critical points satisfying (i) − (ii) of Lemma 3. Then for |ε| small there exists a unique w = wε (z) satisfying (i) − (ii) − (iii) of Proposition 1. Furthermore, the manifold Zε = {z + wε (z) : z ∈ Z} is a natural constraint for Iε . Hence any critical point of Iε (z + wε (z)), z ∈ Z is a critical point of Iε . 3.3. Proof of Theorem 1 Throughout this section we will take ε and c is such a way that Proposition 1 applies. The main tool to prove Theorem 1 is the following Proposition Proposition 2. There results lim I c (zµ,ξ + wg (zµ,ξ )) = bc ,

µ→0

uniformly for ξ satisfying (5).

(23)

Hence I c |Zgc can be continuously extended to ∂Zgc by setting I c |∂Zgc = bc .

(24)

Postponing the proof of Proposition 2, it is immediate to deduce Theorem 1. Proof of Theorem 1. The extended functional I c has a critical point on the compact manifold Zgc ∪ ∂Zgc . From (24) it follows that either I c is identically constant or it achieves the maximum or the minimum in Zgc . In any case I c has a critical point on Zgc . According to Proposition 1, such a critical point gives rise to a solution of (Y ).   In order to prove Proposition 2 we prefer to reformulate (Y ) in a more convenient form using the stereographic projection σp , trough an appropriate point p ∈ ∂S+n , see Remark 3. In this way the problem reduces to study an elliptic equation in Rn+ , where calculation are easier. More precisely, let g˜ ij : Rn+ → R be the components of the metric g in σp -stereographic co-ordinates, and let

2 1 + |x|2 g ij = g˜ ij . (g) 2

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Then problem (Y ) is equivalent to find solutions of  (n − 1) n+2   −4 ∆g u + Rg u = u n−2 , in Rn+ ;    (n − 2)   2 n on ∂Rn+ = Rn−1 , ∂νg u + hg u = cu n−2 ,   (n − 2)     u > 0, u ∈ D1,2 (Rn ),

(Y )

+

where the symbols have obvious meaning. Solutions of problem (Y ) can be found as critical points of the functional fg : D1,2 (Rn+ ) → R defined in the following way    (n − 1) 1 1 ∗ 2 2 fg (u) = 2 |∇g u| dVg + Rg u dVg − ∗ u2 dVg n n n (n − 2) R+ 2 R+ 2 R+   n−1 + (n − 1) hg u2 dσg − c(n − 2) |u|2 n−2 dσg . ∂ Rn+

∂ Rn+

In general the transformation (g) induces an isometry between H 1 (B) and D1,2 (Rn+ ) given by

n−2 2 2 u(x) #→ u(x) := (x )2 + (xn + 1)2

2x (x )2 + xn2 − 1 × u , , (x )2 + (xn + 1)2 (x )2 + (xn + 1)2 where x = (x1 , . . . , xn−1 ). It turns out that fg (u) = I c (u) as well as

(25)

∇fg (u) = ∇I c (u).

In particular this implies that u solves (Y ) if and only if u is a solution of (Y ). Furthermore, there results – g0 corresponds to the trivial metric δij on Rn+ ; – z0 corresponds to z0 ∈ D1,2 (Rn+ ) given by z0 (x) = z0 (x − (0, a0 c)), c

x ∈ Rn+ ;

a0 =

κ ; 2

– Z c corresponds to Z given by 

 x − (ξ , a0 cµ) c − n−2 n−1 2 Z = zµ,ξ := µ z0 , µ > 0, ξ ∈ R . µ

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Let us point out that the manifold Z is nothing but τp ◦ τS−1 Z c (see Notations). From the preceding items it follows that the equation c

∇fg (z + w) ∈ Tz Z , c

have a unique solution w ⊥ Tz Z and there results w g (z) = wg (z). From this and (25) it follows I c (z + wg (z)) = fg (z + wg (z)).

(26)

Let us now introduce the metric g δ (x) := g(δx), δ > 0 and let fgδ : D1,2 (Rn+ ) → R be the corresponding Euler functional. For all u ∈ D1,2 (Rn+ ) there results

 2−n fgδ (u) = fg δ 2 u(δ −1 x) . Introducing the linear isometry Tδ : D1,2 (Rn+ ) → D1,2 (Rn+ ) defined by Tδ (u) := n−2 δ − 2 u(x/δ) this becomes fgδ (u) = fg (Tδ u) ,

(27)

Furthermore, for all u ∈ D1,2 (Rn+ ) one has ∇fg (u) = Tδ ∇fgδ (Tδ−1 u) D fg (u)[v, w] = D 2

2

(28)

fgδ (Tδ−1 u)[Tδ−1 v, Tδ−1 w].

(29)

c

Arguing as above, there exists w gδ (z0 ) ∈ (Tz0 Z )⊥ such that c

∇fgδ (z0 + wgδ ) ∈ Tz0 Z . and there results wgδ (z0 )(x) = δ

n−2 2

wg (zδ )(δx),

namely wg (zδ ) = Tδ wgδ (z0 ).

(30)

Remark 2. From (27), (28), (29) and using the relations between fg and I c discussed above, it is easy to check that the estimates listed in Lemma 4 hold true, substituting I c with fgδ and z with z. A similar remark holds for Proposition 1.

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We are interested to the behaviour of fgδ as δ → 0. To this purpose, we set     1 ∗ 2 (n − 1) g ij (0)Di uDj u − ∗ |u|2  dVg(0) fg(0) (u) = (n − 2) i,j 2 Rn+  n−1 −c(n − 2) |u|2 n−2 dσg(0) , ∂ Rn+

which is the Euler functional corresponding to the constant metric g(0). Remark 3. Unlike the g δ , the metric g(0) does not come from a smooth metric on B. This is the main reason why it is easier to deal with (Y ) instead of (Y ). Lemma 5. For all u ∈ D1,2 (Rn+ ) there results lim ||∇fgδ (u) − ∇fg(0) (u)|| = 0;

(31)

lim fgδ (u) = fg(0) (u).

(32)

δ→0

δ→0

Proof. For any v ∈ D1,2 (Rn+ ) there holds   ∇fgδ (u) − ∇fg(0) (u), v = θ1 + θ2 + θ3 + θ4 + θ5 , where

   n−1 θ1 = 4 ∇gδ u · ∇gδ v dVgδ − ∇g(0) u · ∇g(0) v dVg(0) ; n−2 Rn+ Rn+  Rgδ u v dVgδ ; θ2 = Rn+   4 |u| n−2 u v (dVgδ − dVg(0) ); θ4 = 2(n − 1) hgδ u v dσgδ ; θ3 = Rn+

θ5 = 2c(n − 1)

 ∂ Rn+

|u|

2 n−2

∂ Rn−1

 u v dσgδ −

∂ Rn+

|u|

2 n−2



u v dσg(0) .

Using the Dominated Convergence Theorem and the integrability of |∇u|2 and ∗ of |u|2 , it is easy to show that θ1 , θ3 and θ5 converge to zero.  As far as θ2 is concerned, we first note that the bilinear form (u, v) → Rn Rg u v dVg is +

uniformly bounded for g ∈ G ε , so it turns out that given η > 0 there exists n uη ∈ Cc∞ (R+ ) such that        Rgδ u v dVgδ − Rgδ uη v dVgδ  ≤ η · v; ∀v ∈ D1,2 (Rn+ ). (33)  n  Rn+  R+

Yamabe and scalar curvature

683

Hence, since it is Rgδ (δ −1 x) = δ 2 Rg (x) (see (13)), it follows that for δ sufficiently small        2 R δ u v dVgδ  ≤ δ Rg L∞ (B) uη ∞ |v| = o(1) · v.   Rn+ g η  supp(uη ) So, using (33) and the arbitrarity of η, one deduces that θ2 = o(1) · v. Similar computations hold for the term θ4 . In the same way one can prove also (32).   We need a more complete description of w 0 (z). For this, according to Remark 3, we shall study the functional fg(0) in a direct fashion. If g ∈ Gε then the constant metric g(0) on Rn+ satisfies g(0) − I d∞ = O(ε) and thus fg(0) can be seen as a perturbation of the functional    (n − 1) 1 n−1 2 2∗ f0 (u) = 2 |∇u| dV0 − ∗ u dV0 − c(n − 2) |u|2 n−2 dσ0 , (n − 2) Rn+ 2 Rn+ ∂ Rn+ corresponding to the trivial metric δij . Then the procedure used in Sect. 3.2 yields to find w0 (z) such that c

(j) w 0 (z) is orthogonal to Tz Z ; c (jj) ∇fg(0) (z + w0 (z)) ∈ Tz Z ; n−2 c (jjj) w 0 (z) ≤ C · ε · (1 + |c|)− 2 ∀z ∈ Z . The following Lemma proves that a property stronger than (jj ) holds. c

Lemma 6. For all z ∈ Z there results ∇fg(0) (z + wg(0) (z)) = 0.

(34)

Hence z + wg(0) (z) solves 

(n−1) −4 (n−2) 2 ∂u (n−2) ∂ν

n

n+2

i,j =1

= cu

g ij (0)Dij2 u = u n−2

n n−2

in Rn+ ; on ∂Rn+ .

(35)

Here ν is the unit normal vector to ∂Rn+ with respect to g(0), namely g(0)(ν, ν) = 1;

g(0)(ν, v) = 0, ∀v ∈ ∂Rn+ .

Proof. The Lemma is a simple consequence of the invariance of the functional under the transformation Tµ,ξ : D1,2 (Rn+ ) → D1,2 (Rn+ ) defined in the following way

x − (ξ , 0) − n−2 2 Tµ,ξ (u) = µ . u µ This can be achieved with an elementary computation. It then follows that wg(0) (zµ,ξ ) = Tµ,ξ (wg(0) (z0 )),

for all µ, ξ .

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Hence, from the invariance of fg(0) , it turns out that fg(0) (zµ,ξ + w g(0) (zµ,ξ )) = fg(0) (Tµ,ξ (z0 + wg(0) (z0 ))) = fg(0) (z0 + wg(0) (z0 )). Since fg(0) (zµ,ξ + wg(0) (zµ,ξ )) is a constant function then, according to (j ) − (jj ), any z + wg(0) (z) is a critical point of fg(0) , proving the lemma.   Let us introduce some further notation: G denotes the matrix g ij (0), νg(0) is the outward unit normal to ∂Rn+ with respect to g ij (0), and e1 , . . . , en is the standard basis of Rn . Lemma 7. The solutions u of problem (35) are, up to dilations and translations, of the form u = z0 (Ax), where A is a matrix which satisfies −1

AG AT = I,

νg(0) =

 (A−1 )j n ej .

(36)

j

In particular, up to dilations, one has that z0 + w g(0) (z0 ) = z0 (A ·). Proof. First of all we prove the existence of a matrix A satisfying (36). The first −1 equality simply means that the bilinear form represented by the matrix G can be diagonalized, and this is standard. The matrix A which satisfies the first equation in (36) is defined uniquely up to multiplication on the left by an orthogonal matrix. Let (x1 , . . . , xn ) be the co-ordinates with respect to the standard basis (e1 , . . . , en ) of Rn , let (f1 , . . . , fn ) be the basis given by f = (A−1 )T e, and let (y1 , . . . , yn ) be the co-ordinates with respect to this new basis. This implies the relation between the co-ordinates x = Ay and the first of (36) implies that the bilinear form g ij (0) is diagonal with respect to y1 , . . . yn . Moreover, by the transitive action of O(n) over S n−1 we can ask that fn = ν; this is exactly the second equation in (36). In this way the matrix A is determined up to multiplication on the left by O(n − 1). We now prove that the function z˜ 0 = z0 (Ax) = z0 (y) is a solution of (35). First of all, since νg(0) is g(0)-orthogonal to ∂Rn−1 , the domain xn > 0 coincides with yn > 0 and the equation in the interior is, by formula (36) n n+2 (n − 1)  2 (n − 1)  ij −4 Dxi xj z˜ 0 (x) = −4 g Ali Akj Dy2k yl z0 (Ay) = z˜ 0n−2 (x). (n − 2) i,j =1 (n − 2) i,j

Yamabe and scalar curvature

685

Moreover, since ν = fn = ∂Rn+



j (A

−1 T )nj ej

=



j (A

−1

)j n ej , it turns out that on

 ∂ z˜ 0 (x) = (A−1 )j n Dxj z0 (Ay) ∂ν j  n (A−1 )j n Akj Dyk z0 (Ay) = Dyn z0 (Ay) = c˜z0n−2 (x). = j,k

Hence also the boundary condition is satisfied. Moreover, the function z0 ∈ D1,2 (Rn+ ) is the unique solution up to dilation and translation of problem (Y ) with g ij = I d, see [14]. As pointed out before, if A and A are two matrices satisfying (36), they differ up to O(n − 1). Then it is easy to check that z0 (Ax) = z0 (A x) and hence z˜ 0 is unique up to dilation and translation. This concludes the proof.   Corollary 1. The quantity fg(0) (z0 + w 0 (z0 )) is independent of g(0). Precisely one has: fg(0) (z0 + w 0 (z0 )) = bc . Proof. There holds fg(0) (z0 + wg(0) (z0 ))   (n − 1) g ij (0)Aki Alj Dk z0 (Ay)Dl z0 (Ay)dVg(0) (y) =2 (n − 2) Rn+ i,j,k,l   1 n−1 2∗ − ∗ |z0 (Ay)| dVg(0) (y) − c(n − 2) |z0 (Ay)|2 n−2 dσg(0) (y). n n 2 R+ ∂ R+ Using the change of variables x = Ay, and taking into account equations (12) and (36) we obtain the claim. This concludes the proof.   Lemma 8. There holds wgδ (z0 ) → wg(0) δ

as δ → 0.

(37)

c

Proof. Define H : D1,2 (Rn+ ) × Rn × Z → D1,2 (Rn+ ) × Rn by setting

δ ∇fgδ (z + w g(0) + w) − α z˙ H (w, α, z) = . ˙ (w, z) One has that ∇fgδ (z + wg(0) + w) = ∇fgδ (z + wg(0) ) + D 2 fgδ (z + wg(0) )[w] + ϑ(w) where



1

ϑ(w) := 0

 D 2 fgδ (z + wg(0) + sw) − D 2 fgδ (z + wg(0) ) [w]ds.



686

A. Ambrosetti et al. c

Recall that D 2 fgδ (z) is invertible on (Tz Z )⊥ . Since wg(0) satisfies (jjj ), then c also D 2 fgδ (z + wg(0) ) is invertible on (Tz Z )⊥ . As a consequence, the equation c c ∇fgδ (z + wg(0) + w) = 0, w ∈ (Tz Z )⊥ is equivalent, on (Tz Z )⊥ , to  −1   w = − D 2 fgδ (z + wg(0) ) ∇fgδ (z + wg(0) ) + ϑ(w) In addition, by Remark 2, we can use the estimates corresponding to (19) of Lemma 4 and to (iii) of Proposition 1, to infer that 

1

ϑ(w) = 0

 D 2 fgδ (z + wg(0) + sw) − D 2 fgδ (z + wg(0) ) [w]ds = o(w).



Then, repeating the arguments used in Sect. 3.2 with small changes, one can δ show that the equation H = 0 has a unique solution w = ω such that ω ≤ C · ∇fgδ (z + wg(0) ). From (34) and (31) it follows that ω → 0 as δ → 0. Since both wg(0) + ω c and wgδ solve (on (Tz Z )⊥ ) the same equation, we infer by uniqueness that wgδ = wg(0) + ω. Finally, since ω → 0 as δ → 0, then (37) follows.   Remark 4. All the preceding discussion has been carried out by taking the stereographic projection σp through an arbitrary p ∈ S n−1 . We are interested to the limit (23). When µ → 0 then ξ → ξ for some ξ ∈ S n−1 and it will be convenient to choose p = −ξ . We are now in position to give: Proof of Proposition 2. As pointed out in Remark 4, we take p = −ξ and use all the preceding results proved so far in this section. With this choice, when (µ, ξ ) → (0, ξ ) with ξ = |ξ | · ξ , zµ,ξ corresponds to zµ := zµ ,0 , for some µ → 0. Next, in view of (26), we will show that lim fg (zµ + wg (zµ )) = bc .

µ →0

By Corollary 1, bc = fg(0) (z0 + w g(0) ) and hence we need to prove that   lim fg (zµ + wg (zµ )) − fg(0) (z0 + w g(0) ) = 0. µ →0

Using (30), we have fg (zµ + wg (zµ )) = fg (zµ + Tµ w gµ (z0 )).

Yamabe and scalar curvature

687

Then we can write fg (zµ + w g (zµu )) − fg(0) (z0 + wg(0) ) = fg (zµ + Tµ w gµ (z0 )) = fg (zµ + Tµ wgµ (z0 )) − fg (zµ + Tµ wg(0) (z0 )) +ïfg (zµ + Tµ wg(0) (z0 )) − fg(0) (z0 + wg(0) ). From (17) with I c substituted by fg , we infer     fg (zµ + Tµ w gµ (z0 )) − fg (zµ + Tµ wg(0) ( ovz0 )) ≤ C · Tµ w gµ (z0 ) − Tµ wg(0) (z0 ) ≤ C · w gµ (z0 ) − wg(0) (z0 ) = o(1) as µ → 0.

Using zµ = Tµ z0 and (27), we deduce

 fg (zµ + Tµ w gµ (z0 )) = fg Tµ (z0 + wgµ (z0 )) = fgµ (z0 + w g(0) ). Finally     fg (zµ + Tµ wgµ (z0 )) − fg(0) (z0 + w g(0) )     = fgµ (z0 + w g(0) ) − fg(0) (z0 + wg(0) ) → 0, according to Lemma 5. Since the above arguments can be carried out uniformly   with respect to ξ ∈ S n−1 , the proof is completed. 4. The scalar curvature problem In this section the value of c is fixed. Therefore its dependence will be omitted. So we will write Iε instead of Iεc , I0 instead of I0c , etc. 4.1. The abstract setting Solutions of problem (Pε ) can be found as critical points of the functional Iε : H 1 (B) → R defined as Iε (u) = I0 (u) − εG(u) where the unperturbed functional I0c (u) is defined by (see Sect. 2)   1 1 n−1 ∗ |u|2 − c(n − 2) |u|2 n−2 I0 (u) = u21 − ∗ 2 2 B S n−1

688

A. Ambrosetti et al.

and the perturbation G has the form   1 n−1 2∗ K(x)|u| dx + (n − 2) h(x)|u|2 n−2 dσ. G(u) = ∗ 2 B S n−1 The existence of critical points of Iε will be faced by means of the perturbation theory studied in [1]. Precisely, let us recall that I0 possesses an n-dimensional manifold Z = Z c , given by (6). Moreover, Z is non-degenerate in the sense that (i) − (ii) of Lemma 3 hold true. Then the results of [1] lead to consider the finite dimensional functional Γ := G|Z and give rise to the following Theorem: Theorem 5. In the preceding setting, let us suppose that either (a) Γ has a strict maximum (minimum) on Z; or (b) there exists an open subset Ω ⊂⊂ Z such that deg(Γ , Ω, 0) = 0. Then Iε has a critical point close to Z, provided ε is small enough. In our specific case, the function Γ (µ, ξ ) = G(zµ,ξ ) has the expression   (n−1) 2 (n−2) 1 2∗ Γ (µ, ξ ) = ∗ K(x)zµ,ξ (x)dx + (n − 2) h(σ )zµ,ξ (σ )dσ, 2 B S n−1

(38)

where µ > 0 and ξ ∈ Rn are related to c by (5), namely by µ2 + |ξ |2 − cκµ − 1 = 0. In order to apply the preceding abstract result we need to study the behaviour of Γ at the boundary of Z, which is given by ∂Z = {zµ,ξ0 : µ = 0, |ξ0 | = 1}. The following lemma will be proved in the Appendix and describes the behaviour of Γ at ∂Z. Below a1 , . . . , a6 denote positive constants defined in the Appendix. Lemma 9. Let |ξ0 | = 1 and let ν denote the outher normal direction to ∂Z at (0, ξ0 ). Γ can be extended to ∂Z and there results: (a) Γ (0, ξ0 ) = a1 K(ξ0 ) + a2 h(ξ0 ); (b) ∂ν Γ (0, ξ0 ) = a3 K (ξ0 ) · ξ0 ; (c) suppose that K (ξ0 ) · ξ0 = 0 and let n > 3. Then   ∂ν2 Γ (0, ξ0 ) = 4 a4 ∆T K(ξ0 ) + a5 D 2 K(ξ0 )[ξ0 , ξ0 ] + a6 ∆T h(ξ0 ) . Furthermore, if n = 3 and ∆T h(ξ0 ) = 0, then  +∞ provided ∆T h(ξ0 ) > 0, 2 ∂ν Γ (0, ξ0 ) = −∞ provided ∆T h(ξ0 ) < 0. The above Lemma is the counterpart of the calculation carried out in [2] for the Scalar Curvature Problem on S n .

Yamabe and scalar curvature

689

4.2. A general existence result Let us consider the auxiliary function ψ : S n−1 → R defined by ψ(x) = a1 K(x) + a2 h(x),

x ∈ S n−1 .

If x ∈ Crit (ψ) we denote by m(x, ψ) its Morse index. Theorem 6. Suppose that either (a) there exists an absolute maximum (resp. minimum) p ∈ S n−1 of ψ such that K (p) · p < 0 (resp. K (p) · p > 0); or (b) ψ is a Morse function satisfying K (x) · x = 0, 

∀ x ∈ Crit (ψ);

(39)

(−1)m(x,ψ) = 1.

(40)

x∈Crit (ψ), K (x)·x 0, there exists C > 0 such that for all c > −M there holds z ≤ C · (1 + |c|)−

n−2 2

for all z ∈ Z c .

(47)

Proof. By symmetry it suffices to take By symmetry it suffices to take ξ = 0 and consider z = zµ . As c → +∞ one has that µ ∼ κc and zµ ∼ µ(n−2)/2 in B. Then the lemma follows by a straight calculation.   Now we start by proving Eq. (15). Since it is clearly ∇I0c (z) = 0, it is sufficient to estimate the quantity ∇I c (z) − ∇I0c (z). Given v ∈ H 1 (B) and setting   (n − 1) (n − 1) ∇g z · ∇g v dVg − 4 ∇z · ∇v dV0 ; α1 = 4 (n − 2) B (n − 2) B  α2 = Rg z v dVg ; B   n+2 n+2 n−2 n−2 z v dV0 − z v dVg ; α4 = 2(n − 1) hg z v dσg ; α3 = B B ∂B   n n α5 = 2(n − 1) c z n−2 v dσg − 2(n − 1) c z n−2 v dσ0 , ∂B

there holds

∂B

 c  ∇I (z) − ∇I0c (z), v = α1 + α2 + α3 + α4 + α5 .

(48)

As far as α1 is concerned, taking into account of equations (12), (14) and the fact n−2 that z ≤ C · (1 + |c|)− 2 (see Lemma 10) one deduces that       ∇g z · ∇g v − ∇z · ∇v dx + C |∇z · ∇v| |dVg − dV0 | |α1 | ≤ C B

≤ C · ε · (1 + |c|)

− n−2 2

B

· v.

(49)

Turning to α2 we recall that the expression of Rg as a function of g, is of the type Rg = DΓ + G2 ;

Γ = Dg, ⇒ Rg = D 2 g + (Dg)2 .  We by estimating the quantity B Rg z v dV0 . Integrating by parts, the term  start 2 B D g z v dV0 transforms into    D 2 g z v dV0 = Dg z v dσ0 + Dg D(zv)dV0 . B

∂B

B

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A. Ambrosetti et al.

Hence, if g ∈ Gε (see expression (2)), from the H¨older inequality it follows that   Rg zv dV0  (D 2 g + (Dg)2 )zv dV0 ≤ C · ε · z · v, B

B

and hence     n−2   Rg zv dV0 + |Rg zv||dVg − dV0 | ≤ C · ε · (1 + |c|)− 2 · v. |α2 | ≤ B

B

(50)

With simple estimates one can also prove that |α3 | ≤ C · ε · (1 + |c|)−

n+2 2

· v.

(51)

The function hg is of the form hg = Dg so, taking into account (2) one finds |α4 | ≤ C · ε · (1 + |c|)−

n−2 2

· v.

(52)

In order to estimate the last term α5 , using the continuous embedding H 1 (B) P→ n−1 L2 n−2 (S n−1 ) and the H¨older inequality one deduces that n

|α5 | ≤ C · ε · (1 + |c|) · z n−2n

L n−2 (S n−1 )

n

· v ≤ C · ε · (1 + |c|) · (1 + |c|)− 2 · v.

Putting together equations (49)-(52) one deduces (15). Turning to equation (19) and given v1 , v2 ∈ H 1 (B), there holds (D 2 I c (z + w) − D 2 I c (z))[v1 , v2 ] = δ1 + δ2 , where



 (n + 2) 4 4 u n−2 v1 v2 dVg − (u + w) n−2 v1 v2 dVg δ1 = (n − 2) B B 

 (n − 1) 2 2 n−2 n−2 c u v1 v2 dσg − (u + w) v1 v2 dσg . δ2 = 2n (n − 2) ∂B ∂B

Using standard inequalities one finds that  4 C · w n−2 for n ≥ 6,

 |δ1 | ≤ 6−n 6−n C · w · u n−2 + w n−2 for n < 6;  4 C · (1 + |c|) · w n−2 for n ≥ 4,

 |δ2 | ≤ 4−n 4−n C · (1 + |c|) · w · u n−2 + w n−2 for n < 4, D 2 I c (z + w) − D 2 I c (z) ≤ C · |c| · w n−2 . 2

(53)

Yamabe and scalar curvature

695

so we obtain the estimate. We now prove inequality (16). Given v1 , v2 ∈ H 1 (B) and setting   (n − 1) (n − 1) ∇g v1 · ∇g v2 dVg − 4 ∇v1 · ∇v2 dV0 ; β1 = 4 (n − 2) B (n − 2) B  β2 = Rg v1 v2 dVg ; B   (n + 2) (n + 2) 4 4 n−2 z v1 v2 dV0 − z n−2 v1 v2 dVg ; β3 = (n − 2) B (n − 2) B  hg v1 v2 dσg ; β4 = 2(n − 1) ∂B   (n − 1) (n − 1) 2 2 β5 = 2n c c z n−2 v1 v2 dσg − 2n z n−2 v1 v2 dσ0 , (n − 2) (n − 2) ∂B ∂B there holds (D 2 I c (z) − D 2 I0c (z))[v1 , v2 ] = β1 + β2 + β3 + β4 + β5 .

(54)

For β1 , taking into account equation (14) one finds         ∇g v1 · ∇g v2 − ∇v1 · ∇v2 dV0 + C |∇v1 · ∇v2 | · dVg − dV0  |β1 | ≤ C B

B

≤ C · ε · v1  · v2 .

(55)

Turning to β2 reasoning as for the above term α2 one deduces that    Rg z v  dVg ≤ C · ε · v1  · v2 . |β2 | ≤

(56)

B

In the same way one can prove that |β3 | ≤ C · ε · z n−2 · v1  · v2  ≤ C · ε · (1 + |c|)−2 · v1  · v2 . 4

(57)

For the term β4 , similarly to the expression α4 above there holds |β4 | ≤ C · ε · v1  · v2 .

(58)

Turning to β5 , using the H¨older inequality one deduces that 2

|β5 | ≤ C · c · ε · (1 + |c|) · z n−22

L n−2 (S n−1 )

· v1  · v2  ≤ C · ε · v1  · v2 . (59)

Putting together equations (55)-(59) (59) one deduces inequality (16). Equation (17) follows from similar computations.

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A.2. Proof of Lemma 9 Given ξ0 | = 1, we introduce a reference frame in Rn such that en = −ξ0 . Let α = α(µ) be such that the pair (µ, ξ ), with ξ = αξ0 , satisfies (5). Setting γ (µ) = Γ (µ, −α(µ)en ), one has that Γ (0, ξ0 ) = γ (0), ∂ν Γ (0, ξ0 ) = −γ (0), ∂ν2 Γ (0, ξ0 ) = γ (0). In order to evaluate the above quantities, it is convenient to make a change of variables. This will considerably simplify the calculation when we deal with γ (0) and γ (0). Let ψ : Rn+ → B be the map given by (y , yn ) ∈ Rn+ → (x , xn ) ∈ B; x =

2y (y )2 + yn2 − 1 , x = . n (y )2 + (yn + 1)2 (y )2 + (yn + 1)2

Here and in the sequel,if x ∈ Rn we will set x = (x1 , . . . , xn−1 ) so that x = (x , xn ). By using simple computations it turns out that γ (µ) = γ˜ (µ), ˜ where 1 γ˜ (µ) ˜ = ∗ 2

 Rn+

c 2∗ ˜ K(y)(z µ,0 ˜ ) (y)dy + (n − 2)

and µ˜ =

2µ ; 1+ + α(µ) µ2

 ∂ Rn+

n−1 c 2 n−2 ˜ h(ω)(z (ω)dω, µ,0 ˜ )

˜ K(y) = K(ψ(y)).

Let us point out that the derivatives of K and K˜ satisfy the following relations: ˜ 0) = 2Dxn K(ξ0 ); Dyn K(0, ˜ 0) = 2Dx (ξ0 ); Dy K(0,   2 ˜ Dyn K(0, 0) = 4 Dx2n K − Dxn K (ξ0 );   ˜ 0) = 4 Dx2 K − Dxn K (ξ0 ); Dy2 K(0,   ˜ 0) = 4 Dx2 ,xn K − Dx K (ξ0 ). Dy2 ,yn K(0, The change of variables y = µq, ˜ ω = µσ ˜ yields   1 n−1 c 2∗ c 2 n−2 ˜ ˜ µσ ) (σ )dσ. K(µq)(z ˜ h( ˜ )(z1,0 γ˜ (µ) ˜ = ∗ 1,0 ) (q)dq + (n − 2) 2 Rn+ ∂ Rn+ (60)

Yamabe and scalar curvature

697

Hence, passing to the limit for µ˜ → 0, it follows that ˜ ˜ + a2 h(0) = a1 K(ξ0 ) + a2 h(ξ0 ), γ (0) = γ˜ (0) = a1 K(0) with 1 a1 = ∗ 2

 Rn+



∗ z02 (q , qn

− κc/2)dq,

a2 = (n − 2)

2 n−1

∂ Rn+

z0 n−2 (σ, κc/2)dσ.

Let us now evaluate the first derivative. There holds d µ˜ d γ˜ (0) · (0) = 2γ˜ (0). d µ˜ dµ

γ (0) =

Moreover from formula (60) we deduce  1 ∗ c ˜ µq) ˜ = ∗ ∇ K( ˜ · q |z1,0 (q)|2 dq + (n − 2) γ˜ (µ) + 2  Rn n−1 c ˜ µσ × ∇ h( ˜ ) · σ |z1,0 (σ )|2 n−2 (σ )dσ. ∂ Rn+

(61)

For symmetry reasons when µ˜ → 0, the parallel component to ∂Rn+ in the first integral and the second integral vanishes, hence it follows that  2 ∗ c ˜ γ (0) = 2γ˜ (0) = ∗ Dn K(0) qn |z1,0 (q)|2 dq = −a3 K (ξ0 ) · ξ0 , (62) n 2 R+ where 4 a3 = ∗ 2



∗

Rn+

qn z02 (q , qn − κc/2)dq.

We are interested in the study of the second derivative only in the case in which the first derivative vanishes, namely when K (ξ0 ) · ξ0 = 0. As for the second derivative, there holds: γ˜ (µ) ˜ =

1 2∗

 R+ n

n 

∗

c 2 ˜ µq)q Dij2 K( ˜ i qj |z1,0 (q)| dq

i,j =1



+(n − 2)

n−1  ∂ Rn+ i,j =1

:= δ(µ) ˜ + ρ(µ). ˜

(n−1)

c ˜ µσ Dij2 h( (σ )|2 (n−2) dσ ˜ )σi σj |z1,0

(63)

Now we have to distinguish the case n = 3 and the case n > 3. In fact the boundary integral ρ(µ) ˜ in (63) is uniformly dominated by a function in L1 (∂Rn+ )

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A. Ambrosetti et al.

if and only if n > 3. However it is possible to determine the sign of this integral also for n = 3: it turns out that  1 ∗ c ˜ lim δ(µ) ˜ = ∗ |q |2 |z1,0 (q)|2 dq · ∆T K(0) µ→0 ˜ 2 (n − 1) Rn+  1 ∗ c 2 ˜ qn2 |z1,0 (q)|2 dq · Dnn K(0); + ∗ 2 Rn+ and  ˜  ˜ = (+∞) · ∆T h(0), lim ρ(µu)  µ→0 ˜  (n−1) (n − 2)  c ˜  ˜ = |σ |2 |z1,0 (σ )|2 (n−2) dσ · ∆T h(0),  lim ρ(µ) µ→0 ˜ (n − 1) ∂ Rn+

for n = 3; for n > 3.

Hence we have that  for n = 3; (+∞) · ∆T h(ξ0 ) γ˜ (0) = (64)  2 a4 ∆T K(ξ0 ) + a5 D K(ξ0 )[ξ0 , ξ0 ] + a6 ∆T h(ξ0 ) for n > 3, where

 4 ∗ |q |2 z02 (q , qn − κc/2)dq, (n − 1)2∗ Rn+  4 ∗ qn2 z02 (q , qn − κc/2)dq, a5 = ∗ 2 Rn+  (n − 2) 2 n−1 |σ |2 z0 n−2 (σ, κc/2)dσ. a6 = 4 (n − 1) ∂ Rn+

a4 =

Finally, since γ (0) = 4γ˜ (0), the lemma follows. References 1. A. Ambrosetti, M. Badiale, Homoclinics: Poincar´e-Melnikov type results via a variational approach, Ann. Inst. Henri. Poincar´e Analyse Non Lin´eaire 15 (1998), 233–252. 323, (N +2)

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