The role of the spectrum of the Laplace operator on S2 in the H-bubble problem
∗
Roberta Musina Dipartimento di Matematica ed Informatica Universit`a di Udine via delle Scienze, 206 – 33100 Udine, Italy e-mail:
[email protected] Abstract A H-bubble is a regular surface M in R3 parametrized by a conformal map u : S2 → R3 , having mean curvature H(u) at every regular point u ∈ M. We wonder if it is possible to extend the mean curvature function H to a neighbourhood of M, in such a way that M is H-stable, that means that for every smooth K : R3 → R, and for every ε > 0 small, there exists a H + εKbubble Mε diffeomorphic to M and close to M. In this paper we consider the case M = S2 , and we take H to be a radially symmetric extension of the mean curvature function 1 S2 . From the main results it follows in particular that S2 is H-stable if and only if the number 2(1 + H0 (1)) do not belong to the spectrum of the Laplace operator on S2 . We study also the problem of finding | · |−1 + εK-bubbles, where K is any 2 C function on R3 . We prove that if s¯ > 0 is a “good” stationary point for the R Melnikov-type function Γ(s) = − |q|<s K(q) dq, then for |ε| small there exists a | · |−1 + εK-bubble ω ε that converges to the sphere of radius s¯ centered at 0, as ε → 0.
Ref. SISSA 30/2003/M ∗
Work supported by Regione Friuli Venezia Giulia, L.R.3/98 (2002), progetto di ricerca “Equazioni Differenziali in Geometria ed in Fisica Matematica”, and by M.U.R.S.T., progetto di ricerca “Metodi Variazionali ed Equazioni Differenziali Nonlineari”
1
Introduction A H-bubble is a regular surface M ,→ R3 parametrized by a conformal map u : S2 → R3 , having mean curvature H(u) at every regular point u ∈ M. Up to the composition with the stereographic projecton S2 → R2 , every conformal parametrization u of M is a solution to ( ∆u = 2H(u)ux ∧ uy in R2 R (0.1) 2 R2 |∇u| < +∞ . Conversely, if H : R3 → R is a prescribed curvature function, then every sufficientely regular map u satisfying (0.1) parametrizes a H-bubble. The invariance of problem (0.1) with respect to the action of the conformal group of S2 ≈ R2 ∪ {∞} means that the true unknown in this problem is a parametric surface, rather than its parametrization. Problem (0.1) is variational in nature: its solutions are critical points of the functional Z 1 |∇u|2 + VH (u) EH (u) = 2 R2 where VH (u) is the H-volume functional (compare with Section 4.1 and [16]). For a sufficientely regular u it measures the oriented volume enclosed by the surface u with weight H. As regards existence for problem (0.1), only few results are available in the literature. When the H is a nonzero constant H(u) ≡ H0 , Brezis and Coron [5] proved that the only nonconstant solutions to (0.1) are spheres of radius |H0 |−1 anywhere placed in R3 . In [7] some conditions on H are given in order to have the existence of a H-bubble of minimal energy. Let us roughtly state the general problem we are interested in. Let M be a Hbubble, and let H : R3 → R be a smooth extension of the mean curvature function H : M → R. We agree to say that M is “H-stable” if for every K : R3 → R smooth enough and for ε > 0 small there exists a smooth (H + εK)-bubble Mε , such that Mε → M “suitably”, as ε → 0. We wonder how H has to be chosen in order to have H-stability of the surface M. By [8], it results that M is H-stable if H satisfies suitable asymptotic assumptions, and if M can be parametrized by a map u that minimizes the energy EH 2
among all H-bubbles. However, in general, we do not have stability. Let us consider for example the case M = S2 , and let ω : R2 → S2 be a conformal parametrization of S2 (compare with Section 1.1). Then the mean curvature function H is H ≡ 1 on S2 . Let us be more precise in our notion of “stability” in this case. Definition Let H ∈ C 2 (R2 ) with H|S2 ≡ 1. We say that S2 is H-stable if for every K ∈ C 2 (R3 ) and for every ε close to 0 there exists a (H + εK)-bubble ω ε : S2 → R3 such that kω ε − ωkC 1 (S2 ) = O(ε) as ε → 0. It results that S2 is not stable with respect to the extension: H ≡ 1 on R2 . In fact, by Proposition 4.3 in [9], it turns out that if for K ∈ C 1 (R3 ) there exists a sequence ω εj of solutions to ε
ε
∆ω εj = 2(1 + εj K(ω εj ) ) ωxj ∧ ωyj
in R2
with εj → 0 and ω εj → ω in C 1 (S2 , R3 ), then p = 0 is a stationary point for the R Melnikov-type functional R3 → R, p → B1 (p) K(q) dq . In [9] we also give sufficient conditions that guarantee the existence of smooth (1 + εK)-bubbles approaching the unit sphere centered at a point p ∈ R3 , as ε → 0. Other results can be found in [6] and [11]. In this paper we show that the spectral properties of S2 play a crucial role in finding radially symmetric extensions H of 1|S2 that leave S2 stable. More precisely, it turns out that S2 is H-stable if and only if the number 2(1 + H0 (1)) do not belong to the spectrum σ(−∆S2 ) of the Laplace operator on S2 (compare with Section 1.3). The “only if” part is proved in Section 2, where we assume that 2(1 + H0 (1)) = n(n + 1) ∈ σ(S2 ) for an integer n ≥ 0 (notice that this is the case if H0 (1) = 0). We give a necessary condition on K in order to have the existence of a sequence of (H(| · |) + εK)-bubbles, suitably approaching the unit sphere. More precisely, we prove that the restriction of K to S2 has to be orthogonal to the n-eigenspace of −∆S2 . This result is somehow a generalization of Proposition 4.3 in [9]. In the non resonant case, that is, when 2(1 + H0 (1)) ∈ / σ(S2 ), we show that S2 is H-stable. Actually, we prove a slighlty stronger resut. 3
Theorem 0.1 Let H: R \ {0} → R be a given map of class C 2 . We assume that i) H(1) = 1 ii) 2(1 + H0 (1)) 6= n(n + 1) for every n ≥ 0 . Then for every K ∈ C 2 (R3 , R) there exists a unique curve ε 7→ ω ε of class C 1 from a neighborhood I ⊂ R of 0 into C 3 (S2 , R3 ) such that ω 0 = ω and ω ε is a conformally embedded (H(| · |) + εK)-bubble for every ε ∈ I. Let us briefly describe our approach. Let H: R \ {0} → R be a regular function with H(1) = 1. Then ω solves ( ∆ω = 2H(|ω|) ωx ∧ ωy in R2 R (0.2) 2 R2 |∇ω| < +∞ . Now, let K : R3 → R be a any smooth function, and consider the problem of finding for ε small a solution to ( ∆ω ε = 2( H(|ω ε |) + εK(ω ε ) ) ωxε ∧ ωyε in R2 R (0.3) ε 2 R2 |∇ω | < +∞ with ω ε → ω as ε → 0. Solutions to problem (0.3) are (formally) critical points of the energy functional Eε : H 1 (S2 , R3 ) → R, Z 1 |∇u|2 + 2VH (u) + 2εVK (u) = E0 (u) + 2εVK (u) . Eε (u) = 2 R2 The invariance of Eε with respect to the 6-dimensional group G of conformal transformations on R2 causes to Eε some lack of smoothness and lack of compactness that hinders the use of standard tools from the calculus of variations. On the other hand, it quite difficult to avoid lack of compactness by defining Eε on a suitable quoitent space, say on X = H 1 (S2 , R3 )/G. In fact, since no Lp norm is invariant with respect to G, no suitable metric structure can be defined on X. More, the standard implicit function theorem does not apply directly, since the G-invariance of E0 generates a 6-dimensional set T of solutions to the linearized unperturbed problem, that turns out to be ( ∆v = 2 (vx ∧ ωy + ωx ∧ vy ) + 2H0 (1)(ω · v) ωx ∧ ωy in R2 R (0.4) 2 R2 |∇v| < +∞ . 4
Problem (3.2) has been already studied by several authors in case H0 (1) = 0 (see for example [15], [12], [13], [10]), but, far as we know, the first published description of the set of solutions of (3.2) is due to Chanillo and and Malchiodi [10]. In the Appendix we exibit a new and simple proof of their result. In Section 3 we give a complete description of the set of solutions to problem (3.2), for every value of the real parameter H0 (1). In particular we prove that in the “non resonant” case, namely, when 2(1 + H0 (1)) ∈ / σ(−∆S2 ), maps in T are the only solutions to (0.4). Even if there exist notrivial solutions to the linearized problem, this is actually a “non degenerate” case, since the set of solutions to (0.4) is produced by a group G that leaves invartiant also the perturbed functional Eε . In this case, the finite dimensional reduction method leads to the conclusion in Theorem 0.1, with no Melnikov function involved. In case of “resonance”, namely, when 2(1 + H0 (1)) = n(n + 1) ∈ σ(−∆S2 ) for an integer n ≥ 0, then degeneracy do occur. More precisely, we prove that v solves (0.4) if and only if v = τ + Φω, where τ ∈ T and Φ solves −∆S2 Φ = n(n + 1) Φ
on S2 .
In case of resonance we are not able to handle problem (0.3), exept when “nondegeneracy” can be recovered by using some further invariances of the unperturbed problem. In this case the main tool that can be used is the finite-dimensional reduction method, as formalized by Ambrosetti and Badiale in [1]. This has been done in [9], where in case H ≡ 1 on R2 (and hence n = 1), the invariance of the unperturbed problem with respect to translations on the images is used to recover nondegeneracy. In Section 6 we consider the case H(r) = r−1 , and we study the problem ( ∆u = 2 ( |u|−1 + εK(u) ) ux ∧ uy in R2 R (0.5) 2 R2 |∇u| < +∞ . Notice that 2(1 + H0 (1)) = 0 ∈ σ(−∆S2 ), that is, we have “resonance” and “degeneracy”. Moreover, for every s > 0 the map sω is a solution of the unperturbed problem. Therefore, we may wonder if for some s¯ > 0 there exists a sequence of solutions to (0.5) suitably approaching the sphere of radius s¯, as ε → 0. By the results in Section 2, a necessary condition on K at s¯ is that Z Z d K(q) dq =0 (0.6) K(q) dσ(q) = − ds s=¯ s Bs (0) ∂Bs¯(0) 5
(compare with Proposition 6.1). In Section 6 we use the finite dimensional reduction method to give a partial answer to this question. In order to state our result let us introduce the Melnikov-type function Γ: [0, +∞) → R defined by Z Γ(s) = − K(q) dq . Bs (0)
Theorem 0.2 Let K ∈ C 2 (R3 , R) be given. Assume that there exists an interval [s1 , s2 ] ⊂ (0, +∞) such that Γ0 (s1 )Γ0 (s2 ) < 0. Then for |ε| small enough there exists a conformally embedded smooth solution ω ε to (0.5) such that kω ε − s¯ωkC 3 (S2 ,R3 ) = O(ε) as ε → 0, where Γ(¯ s) = min[s1 ,s2 ] Γ if Γ0 (s1 ) < 0, or Γ(¯ s) = max[s1 ,s2 ] Γ if Γ0 (s1 ) > 0. Notice that our assumption on Γ is equivalent to the existence of a local minimum or maximum s¯ for Γ that is “topologically stable” in the sense of [1]. In particular, the necessary condition (0.6) is fullfilled at s¯. However, up to now we do not know if (0.6) is sufficient to solve (0.5) near s¯ω, for ε small. A more general open problem is stated in Remark 2.2. The paper is organized as follows. 1 Notation and preliminaries A conformal parametrization of the unit sphere The variational space On the spectrum of −∆S2 2 A necessary condition in case of resonance 3 The linearized problem 4 The variational approach The H-volume and the energy functional On the Morse index of ω 5 Proof of Theorem 0.1 6 The case H(r) = 1r A Appendix
6
1 1.1
Notation and preliminaries A conformal parametrization of the unit sphere
By conformal invariance of problem (0.3), it is not restrictive to fix a conformal parametrization ω of the unit sphere. We take ω to be the inverse of the standard stereographic projection from the north pole, that is, µx 2 ω(z) = µy , µ = µ(z) = , (1.1) 1 + |z|2 1−µ being z = (x, y) and |z|2 = x2 + y 2 . Notice that ω solves problem (0.2), since we assume H(1) = 1. From now on, the letter “ω” will denote both the map defined in (1.1) and points in S2 . In the following Lemma we collect some useful properties of the map ω. The proofs are just simple computations. Lemma 1.1 The map ω: R2 → R3 is conformal and it satisfies: i) ii) iii) iv)
∆ω = 2ωx ∧ ωy = −2µ2 ω; 1 |∇ω|2 = |ωx |2 = |ωy |2 = |ωx ∧ ωy | = µ2 ; 2 ωx ∧ ω = ωy , ω ∧ ωy = ωx Z Z 2 |∇ω| = 8π , ω =0. R2
1.2
S2
The variational space
Let s ∈ [1, +∞). Then to every (scalar-valued) map Φ ∈ Ls (S2 ) one can associate ¯ = Φ ◦ ω: R2 → R, that satisfies the map Φ Z Z ¯ s µ2 . kΦkss := |Φ|s = |Φ| S2
R2
In the following, we shall always use the same notation for a map Φ on S2 and for its composition with the inverse of the stereographic projection ω, which is a map on R2 . Using the above identification, the norm of a map Φ ∈ W 1,s (S2 ) is given by Z 1/s Z 1/s s 2 s 2−s kΦkW 1,s = |∇Φ| µ + |Φ| µ R2
R2
7
(see for example [2]). In the same spirit, the duality product between W 1,s (S2 ) and 0 W 1,s (S2 ) can be written as Z Z (Φ|Ψ) = ∇v · ∇Ψ + v · Ψ µ2 . R2
R2
0
for every Φ ∈ W 1,s (S2 ), Ψ ∈ W 1,s (S2 ). For every subspace N of H 1 (S2 ) we denote by N ⊥ its orthogonal in the scalar product (·|·). Identifying R ≡constant maps on S2 , in this notation we have that H∗1 (S2 ) = R⊥ , where Z Z 1 2 1 2 H∗ (S ) := {Φ ∈ H (S ) | Φ = Φ µ2 = 0 } . S2
R2
Following [9], the spaces Ls (S2 , R3 ), W 1,s (S2 , R3 ), H 1 (S2 , R3 ) and H∗1 (S2 , R3 ) will be denoted simply by Ls , W 1,s , H 1 and H∗1 respectively, and the same identification v ≡ v ◦ ω will be used.
1.3
On the spectrum of −∆S2
Consider the following linear problem for scalar-valued functions on R2 : (
−∆Φ = λµ2 Φ R 2 R2 |∇Φ| < +∞
in R2 ,
(1.2)
where λ ∈ R and the weight µ = µ(z) is defined in (1.1). In our notations, a weak solution to problem (1.2) is a map Φ ∈ H 1 (S2 ) such that Z Z ∇Φ · ∇Ψ = λ ΦΨ ∀Ψ ∈ H 1 (S2 ) . R2
S2
Every weak solution to (1.2) is in fact of class C ∞ (S2 ). For a complete description of the eigenvalues and eigenspaces of problem (1.2) we refer for example to [3], Ch. III. The spectrum σ(−∆S 2 ) is the set of all λ ∈ R such that there exists a nontrivial solution Φ ∈ C ∞ (S2 ) to (1.2). It results that σ(−∆S 2 ) = {n(n + 1)| n ∈ N ∪ {0} }. Let us set λn = n(n + 1) and let Λn ⊂ H 1 (S2 ) be the space of solutions to (1.2) with λ = λn . It turns out that dim Λn = 2n + 1 and Λ0 = R ,
Λ1 = {a · ω | a ∈ R3 } ,
8
Λn ⊂ H∗1 (S2 ) ∀n ≥ 1 .
Let us define also Nn :=
n \
Λ⊥ j ,
j=0 1 2 which is a subspace of Notice that N0 = Λ⊥ 0 = H∗ (S ). We conclude with the following well known Lemma, that follows from the variational characterization of the eigenvalues of −∆S2 .
H∗1 (S2 ).
Lemma 1.2 If Ψ ∈ Nn for some integer n ≥ 0, then Z Z Ψ2 . |∇Ψ|2 ≥ n(n + 1)
(1.3)
S2
R2
Moreover, equality holds in (1.3) if and only if Ψ ∈ Λn+1 .
2
A necessary condition in case of resonance
In this Section we show that in the radially symmetric case the condition 2(1 + H0 (1)) ∈ / σ(−∆S2 ) is necessary in order to have H-stability of S2 . Proposition 2.1 Let H ∈ C 1 (R \ {0}) and K ∈ C 0 (R3 ) be given. Assume that H(1) = 1 and 2(1 + H0 (1)) = n(n + 1) for an integer n ≥ 0. If there exist a sequence εj → 0 and a sequence ω εj of solution to problem (0.3), with kω εj − ωkW 1,3 = O(εj ) as j → ∞, then K|S2 ∈ Λ⊥ n. Proof. Set ε = εj and write ω ε = ω + εθε , with θε bounded in W 1,3 . We can assume that θε → θ weakly in W 1,3 , for some θ ∈ W 1,3 . From (0.3) and Lemma 1.1 we get ∆θε = −
2 (H(|ω ε |) − 1) 2 ωµ + 2H(ω ε )(ωx ∧ θxε + θxε ∧ ωy ) − 2K(ω ε )ωµ2 + O(ε) (2.1) ε
uniformly in W −1,3/2 . On the other hand, (H(|ω ε |) − 1 ) = εH0 (1) (ω · θε ) + o(ε) and hence, passing to the limit in (2.1), we infer that θ solves ∆θ = −2H0 (1)(ω · θ)ωµ2 + 2(ωx ∧ θy + θx ∧ ωy ) − 2K(ω)ωµ2 .
(2.2)
Now we test (2.2) with Φω for any Φ ∈ Λn . Since 2(1+H0 (1)) = n(n+1), by Lemma R A.3 in the appendix, we infer that S2 KΦn = 0 for every Φ ∈ Λn , where now K denotes the restriction of the map K : R3 → R to the unit sphere. Proposition 2.1 is completely proved. 9
Remark 2.2 It whould be of interest to know whether the condition K|S2 ∈ Λ⊥ n is sufficient for the existence of a sequence of solutions to problem (0.3) approaching the unit sphere, as ε → 0.
3
The linearized problem
The aim of this Section is to study the linear problem on R2 ,
∆v = 2(vx ∧ ωy + ωx ∧ vy ) + 2H1 (ω · v) ωx ∧ ωy
(3.1)
where H1 ∈ R is given, and v ∈ H 1 . A weak solution to (3.1) is a map v ∈ H 1 such that Z Z Z ∇v · ∇ϕ + 2 ϕ · (vx ∧ ωy + ωx ∧ vy ) − 2H1 (ω · v)(ω · ϕ) = 0 R2
R2
S2
for every ϕ ∈ H 1 (compare with Lemma 1.1, i) ). Notice that every weak solution v ∈ H 1 is of class C ∞ (S2 ) by standard elliptic regularity theory. Notice that for H1 = 0, (3.1) reduces to on R2 .
∆v = 2(vx ∧ ωy + ωx ∧ vy )
(3.2)
Let us introduce the continuous and bilinear form BH1 : H 1 × H 1 → R, Z Z Z BH1 (v, ϕ) = ∇v · ∇ϕ + 2 ϕ · (vx ∧ ωy + ωx ∧ vy ) − 2H1 (ω · v)(ω · ϕ) R2
R2
S2
for every v, ϕ ∈ H 1 . It results that v ∈ H 1 solves (3.1) if and only if BH1 (v, · ) = 0. Our aim is to study the kernel and the sign of BH1 . We notice that BH1 is symmetric by the divergence theorem, since vx ∧ ωy + ωx ∧ vy = div(v ∧ ωy , ωx ∧ v ). With the same argument we get also Z Z Z 2 BH1 (v, v) = |∇v| + 4 ω · vx ∧ vy − 2H1 (ω · v)2 . R2
R2
S2
It is convenient to remark that Z BH1 (ω, v) = −2(1 + H1 ) for every v ∈ H 1 , by Lemma 1.1, i). 10
ω· v S2
(3.3)
Let us consider the following 6-dimensional subset of H 1 : T := { (a · ω)ω + b ∧ ω − a | a, b ∈ R3 }. Every map in τ is pointwise orthogonal to ω, and hence also Z Z τ ·ω =0 ∇τ · ∇ω = −2 S2
R2
for any τ ∈ T , for Lemma 1.1. Simple computations show that every τ ∈ T is a solution to (3.2), hence BH1 (τ, · ) = B0 (τ, · ) = 0
(3.4)
for every H1 ∈ R. As a consequence of the following Propositions (see the Appendix for the proofs), maps in T are the only solutions to (3.1) if and only if 2(1 + H1 ) do not belong to the spectrum of −∆S 2 . Proposition 3.1 If 2(1 + H1 ) = n(n + 1) for some n ≥ 0, then v ∈ H 1 solves (3.1) if and only if there exist τ ∈ T and Φn ∈ Λn such that v = τ + Φn ω . Proposition 3.2 If 2(1 + H1 ) 6= n(n + 1) for every n ≥ 0 then v solves (3.1) if and only if v ∈ T . Remark 3.3 Notice that in case H1 = 0, Proposition 3.1 applies with n = 1. In this case every solution v to (3.2) can be written in the form v = (a · ω)ω + b ∧ ω + c for some a, b, c ∈ R3 , accordingly to the statement of Lemma 9.2 in [10]. For, we just notice that (a · ω)ω + b ∧ ω + c = [−(c · ω)ω + b ∧ ω + c] + [(a + c) · ω] ω ∈ T + Λ1 ω . In order to study the sign of BH1 we introduce the following spaces: Z Xω := {η ∈ H 1 | ∇η · ∇τ = 0 ∀τ ∈ T } Xωn
R2
:= {η ∈ Xω | η · ω ∈ Nn } . 11
R for n ≥ 0. Since N0 = H∗1 (S2 ), then every map η ∈ Xω0 satisfies 2 S2 η · ω = R − R2 ∇η · ∇ω = 0 . In order to describe Xω1 let us recall that N1 = Λ⊥ 1 is the set R R 1 2 of all Φ ∈ H∗ (S ) such that S2 Φ(a · ω) = a · S2 Φω = 0 for every a ∈ R3 , and therefore Xω1 = {η ∈ Xω | η · ω ∈ H∗1 (S2 ) , (η · ω)ω ∈ H∗1 } . (3.5) Lemma 3.4 For n ≥ 1, Xωn is a Hilbert space with respect to the scalar product Z ∇η · ∇ψ . hη, ψi = R2
Proof. First take n = 1, and notice that Xω1 is a closed subspace of H 1 . prove that h·, ·i is a scalar product on Xω1 we remark that it do not contain nonzero constant, since if η ∈ R3 ∪ Xω1 is constant, then, by (3.5) and Lemma R 0 = S2 (η · ω)ω = 4π 3 η, hence η = 0. It remains to prove that there exists a δ such that Z Z 2 |∇η| ≥ δ |η|2 R2
To any 1.1, >0
S2
Xω1 .
for every η ∈ This can be easily done by using a contradiction argument. In case n ≥ 2 it sufficies to notice that Xωn is a closed subspace of Xω1 . The dimension of the negative subspace of the bilinear form BH1 depends on the eigenvalues of −∆S2 that are smaller than 2(1 + H1 ), as the following Proposition shows. Proposition 3.5 i) There exists a constant C0 > 0 such that for every η ∈ Xω0 , Z Z Z |∇η|2 + 4 ω · ηx ∧ ηy ≥ C0 |∇η|2 . R2
R2
R2
ii) If H1 < −1, there exists a constant CH1 > 0 such that for every η ∈ Xω , Z Z Z Z |∇η|2 + 4 ω · ηx ∧ ηy − 2H1 (ω · η)2 ≥ CH1 |∇η|2 . R2
R2
S2
R2
iii) If 2(1 + H1 ) < λn+1 = (n + 1)(n + 2) for some n ≥ 0, there exists a constant CH1 > 0 such that for every η ∈ Xωn , Z Z Z Z |∇η|2 + 4 ω · ηx ∧ ηy − 2H1 (ω · η)2 ≥ CH1 |∇η|2 . R2
R2
S2
R2
Statement i) was proved by Isobe in [12], while ii) and iii) will be proved in the Appendix. 12
4 4.1
The variational approach The H-volume and the energy functional
Let H ∈ C 0 (R3 ) be a given curvature, and let Q: R3 → R3 be any smooth vectorfield such that div Q = H. For every u ∈ H 1 ∩ L∞ let us set Z Q(u) · ux ∧ uy . VH (u) = R2
If u is smooth enough, the real number VH (u) measures the algebraic volume enclosed by the surface parametrized by u with respect to the weight H, and it is independent of the choice of the vectorfield Q. For several results and remarks on the volume functional VH we refer to [4] for the case H ≡ const., and [16] for variable H. We also quote [9] for the proofs of the results that follows. It turns out that the volume functional VH is not smooth on H 1 , even not Gateaux differentiable. Thus, as in [9], we are induced to define VH on a smaller Sobolev space, namely, W 1,3 . Since W 1,3 is complaclty embedded into L∞ , the functional VH : W 1,3 → R enjoies the following regularity properties. Lemma 4.1 Assume that H ∈ C 2 (R3 ). Then the functional VH is of class C 2 on W 1,3 . The Fr´echet differential of VH at u ∈ W 1,3 admits a continuous and linear extension on W 1,3/2 that is given by Z 0 (VH (u)|ϕ) = H(u)ϕ · ux ∧ uy for every ϕ ∈ W 1,3/2 . R2
The second derivative of VH at u is given by Z Z 00 (VH (u) · η|ϕ) = H(u)ϕ · (ηx ∧ uy + ux ∧ ηy ) + (∇H(u) · η)ϕ · (ux ∧ uy ) . R2
R2
for every u, η ∈ W 1,3 and ϕ ∈ W 1,3/2 . Now we are in position to define the energy functional relative to a curvature of the form Hε (u) = H(|u|) + εK(u), with H : R \ {0} → R and K : R3 → R of class C 2 . Since we are interested in finding solutions of problem (0.2) whose image is closed to the unit sphere, we can assume that the map u → H(|u|) is of class C 2 on R2 . For ε ∈ R we set Z 1 |∇u|2 + 2VH(|·|) (u) + 2εVK (u) = E0 (u) + 2εVK (u) . Eε (u) = 2 R2 13
By Lemma 4.1, the functional Eε is of class C 2 on W 1,3 . In particular, for all u ∈ W 1,3 one has Eε0 (u) ∈ W 1,3 and Z Z Z 0 K(u)ϕ · ux ∧ uy , (Eε (u)|ϕ) = ∇u · ∇ϕ + 2 H(|u|)ϕ · ux ∧ uy + 2ε R2
R2
R2
for every ϕ ∈ W 1,3/2 . Hence, every critical point u ∈ W 1,3 of Eε is a weak solution to (0.3).
4.2
On the Morse index of ω
In this Section we assume as in Theorem 0.1 that H(1) = 1. Here we make some considerations on the unperturbed functional Z 1 E0 (u) = |∇u|2 + 2VH (u) . 2 R2 Let ω be the inverse of the stereographic projection, as in Section 1.1. Since |ω| = 1, the map ω is a stationary point for E0 . Let us compute the second derivative of the functional E0 at ω: Z Z Z 00 0 (E0 (ω) · v|ϕ) = ∇v · ∇ϕ + 2 ϕ · (vx ∧ ωy + ωx ∧ vy ) − 2H (1) (ω · v)(ω · ϕ) R2
R2
S2
for every u, v ∈ W 1,3 and ϕ ∈ W 1,3/2 . Hence, (E000 (ω) · v|ϕ) = BH0 (1) (v, ϕ) . Therefore, the kernel and the sign of E0 (ω) depend on H0 (1), according to the results in Section 3. Corollary 4.2 i) If 2(1 + H0 (1)) 6= n(n + 1) for every n ≥ 0, then ker E000 (u) = T ; ii) If 2(1 + H0 (1)) = n(n + 1) for some n ≥ 0, then ker E000 (u) = T + Λn ω, where Λn ω = {Φω | Φ ∈ Λn }. Corollary 4.3 i) If H0 (1) < −1 then E000 (ω) is positive definite on Xω ; ii) If 2(1 + H0 (1)) < λn+1 = (n + 1)(n + 2) for some n ≥ 0 then E000 (ω) is positive definite on Xωn .
14
Now, let us notice that the functional E0 is invariant with respect to the conformal group on R2 ∪ {∞}. Therefore, the mapping ω defined by (1.1) generates a 6− dimensional manifold of solutions u of (0.2), which can be written as Z = Rω ◦ Lα,ζ : R ∈ SO(3), α > 0, ζ ∈ R2 where for α > 0 and ζ ∈ R2 we set Lα,ζ z = α(z − ζ) (see [12]). In the next Lemma we show that the tangent space Tω (Z) to Z at ω equals the space T defined in Section 3. Lemma 4.4 Let e1 , e2 , e3 be the canonical basis of R3 . Then Tω Z = span{ωx , ωy , xωx + yωy , ei ∧ ω | i = 1, 2, 3} = T . Proof. The proof is essentially contained for example in [9]. First we compute ∂ ∂ ∂ = ωx , = ωy , = xωx +yωy . ω ◦ L1,ζ ω ◦ L1,ζ ω ◦ Lλ,0 ∂x ∂y ∂λ ζ=0 ζ=0 λ=1 Next, we differentiate with respect to the variable R ∈ SO(3). Since the Lie algebra of SO(3) admits as a basis the set of matrices {ξ1 , ξ2 , ξ3 } defined by:
0 0 0 ξ1 = 0 0 −1 , 0 1 0
0 0 1 ξ2 = 0 0 0 , −1 0 0
0 −1 0 ξ3 = 1 0 0 , 0 0 0
one has ξ1 ω = e1 ∧ ω ,
ξ2 ω = e2 ∧ ω ,
ξ3 ω = e3 ∧ ω .
This proves that Tω Z is spanned by the maps ωx , ωy , xωx + yωy , ei ∧ ω. The second identity in Lemma 4.4 is readily proved by taking into account of the identities ωx = −(e1 · ω)ω − e2 ∧ ω + e1 ωy = −(e2 · ω)ω + e1 ∧ ω + e2 xωx + yωy = −(e3 · ω)ω + e3 .
15
Remark 4.5 Corollaries 4.2 and 4.3 allows us to compute the Morse index of ω as a critical point of the functional E0 . Actually, denoting by W− the negative subspace of E000 (ω) and by iω it dimension, we have that • If H0 (1) ≤ −1 then W− = Ø and iω = 0; • If 2(1 + H0 (1)) ∈ (n(n + 1), (n + 1)(n + 2)] for an integer n ≥ 1, then W− = ⊕ni=0 Λi and iω = (n + 1)2 .
5
Proof of Theorem 0.1
In this Section we assume that 2(1 + H0 (1)) 6= λn = n(n + 1) for every n ≥ 0. Then T = Tω Z = ker E000 (ω) by Corollary 4.2, i). We apply here the arguments of the finite dimensional reductional method to construct a smooth path ε → ω ε such that Eε0 (ω ε ) is tangent to Z and ω ε is orthogonal to T = Tω Z. Then, taking advantage of the fact that the perturbed problem still has the same invariances that generate Tω Z, we show that for |ε| small, ω ε is a critical point of Eε . Let us fix a basis {τ1 , . . . , τ6 } of T with Z ∇τi · ∇τj = δij , R2
where i, j = 1, . . . , 6 and δij is the Kronecker symbol. Our goal is to construct C 1 maps ε → η ε and ε → αε ∈ R6 such that Eε0 (ω
ε
+η ) = −
6 X
αiε ∆τi
(5.1)
i=1
Z
∇η ε · ∇τi = 0
∀ i = 1, . . . , 6.
(5.2)
R2
This will be done later. Let us first prove that from (5.1), (5.2) it follows that αiε = 0 for every i = 1, . . . , 6 if |ε| is small enough. The argument is the same alredy used in [9].
16
Let g1 (t), . . . , g6 (t) be smooth paths of conformal transforms such that gi (0) = Id and (ω ◦ gi )0 (0) = τi (i = 1, . . . , 6). Since for gi = gi (t) it results Z Z d d ε ∇(τj ◦ gi ) · ∇(η ◦ gi ) = ∇τj · ∇η ε = 0 , dt R2 dt R2 then Aεi,j
Z
ε
Z
0
∇((τj ◦ gi )0 (0)) · ∇η ε
∇τj · ∇((η ◦ gi ) (0)) =
:= −
R2
R2
for every i, j = 1, . . . , 6, and consequently kAε k ≤ Ck∇η ε k2 = o(1), where Aε := (Aεi,j )i,j=1,...,6 is a 6 × 6 matrix. Now we use the invariance of Eε with respect to the conformal group, that is, Eε (u) = Eε (u◦g) for every u ∈ W 1,3 and for every conformal mapping g. Writing ω ε = ω + η ε we have that (ω ε ◦ gi )0 (0) = τi + (η ε ◦ gi )0 (0) and, by (5.1), d ε 0 = Eε ((ω ◦ gi )(t)) = (Eε0 (ω ε )|τi ) + (Eε0 (ω ε )|(η ε ◦ gi )0 (0)) dt t=0 Z 6 X = αiε + αjε ∇τj · ∇((η ε ◦ gi )0 (0)) . j=1
R2
Then, αε solves the system Aε αε = αε . If for a sequence ε → 0 it were αε 6= 0, then kAε k ≥ 1, a contradiction with kAε k = o(1). Hence, one has αε = 0 if |ε| is small enough, and therefore Eε0 (ω ε ) = 0, that is, ω ε is a stationary point for Eε . Finally, to prove that ω ε is a smooth Hε bubble, and that kω ε − (pε + ω)kC 3 → 0 as ε → 0, one can follow a boot-strap argument, as in the proof of Theorem 0.1 in [9]. Construction of the map η ε . Our goal is to apply the implicit function theorem to the map F = (F1 , F2 ): R × (W 1,3 × R6 ) → W 1,3 × R6 defined by F1 (ε; η, α) = Eε0 (ω + η) +
6 X
αi ∆τi
i=1
Z F2 (ε; η, α) =
Z ∇η · ∇τ1 , . . . ,
R2
∇η · ∇τ6
.
R2
Notice that F(0; 0, 0) = 0 since E00 (ω) = 0. Moreover, F is of class C 1 . Therefore, we only have to prove that the linear, continuous operator L: W 1,3 × R6 → W 1,3 × R6 defined as L(v, γ) = ∂(η,α) F(0; 0, 0) · (v, γ) is an isomorphism. 17
Notice that L = (L1 , L2 ), with L1 (v, γ) = E000 (ω) · v +
6 X
γi ∆τi
i=1
Z
Z
L2 (v, γ) =
∇v · ∇τ6
∇v · ∇τ1 , . . . , R2
R2
for every v ∈ W 1,3 and γ ∈ R6 . Injectivity If L(v, γ) = 0 then (v, γ) ∈ W 1,3 × R6 satisfies (E000 (ω) · v|ϕ) =
6 X
Z ∇ϕ · ∇τi
γi
(ϕ ∈ W 1,3/2 )
(5.3)
R2
i=1
Z ∇v · ∇τi = 0
(i = 1, . . . , 6).
(5.4)
R2
Taking ϕ = τj in (5.3), we use (E000 (ω) · v|τj ) = (E000 (ω) · τj |v) = 0 since τj ∈ T = ker E000 (ω), and we get γj = 0 for each j = 1, . . . , 6. Hence, (5.3) becomes v ∈ ker E000 (ω). Then, by Corollary 4.2, v ∈ T which, together with (5.4), implies that v = 0. Injectivity is proved. Surjectivity. Fix (u, β) ∈ W 1,3 × R6 . We have to show that there exists (v, γ) ∈ W 1,3 × R6 such that: (E000 (ω) · v|ϕ) =
6 X i=1
Z ∇ϕ · ∇τi + (u|ϕ)
γi
(ϕ ∈ W 1,3/2 )
(5.5)
R2
Z ∇v · ∇τi = βi
(i = 1, . . . , 6).
(5.6)
R2
First, we take ϕ = τj in (5.5) to see that γ has to satisfy γj = −(u|τj )
(j = 1, . . . , 6).
Fix an integer n ≥ 2 such that H0 (1) < λn+1 . Taking Lemma A.5 into account, we are going to choose τv ∈ T , p ∈ R3 , t ∈ R, Φ ∈ ⊕ns=2 Λs and η n ∈ Xωn ∩ W 1,3 in such a way that the map v = τv + p + tω + Φω + η n (5.7) satisfies (5.5) and (5.6).
18
We look for the unique τv ∈ T such that (5.6) is satisfied. In fact, notice that if a map v is given by (5.7) then Z Z ∇τv · ∇τi , ∇v · ∇τi = R2
R2
for every i = 1, . . . , 6, since Φ ∈ Λ⊥ 1 (use Lemma A.1). Therefore we choose τv =
6 X
βi τi .
i=1
Taking ϕ = ej ≡ const. in (5.5), using the previous decomposition of v, and (E000 (ω) · R v|ej ) = (E000 (ω) · ej |v) = −2H0 (1) S2 (v · ω)ωj , we infer that v has to satisfy the identity Z Z − 2H0 (1)
(v · ω)ωj = (u|ej ) = S2
uj .
(5.8)
S2
On the other hand, using the above decomposition of v we get Z Z (v · ω)ωj = p · ωj ω , S2
S2
n ⊥ since ωj ∈ Λ1 ⊂ H∗1 (S2 ), Φ ∈ Λ⊥ 1 , η ·ω ∈ Nn ⊂ Λ1 . Now computing and comparing with (5.8), we choose Z 3 p=− u. 8πH0 (1) S2
Now we set t=−
R
S2
ωj ω =
4π 3 ej ,
(u|ω) . 8π(1 + H0 (1))
This choice of t is made by taking ω as test function in (5.5). In fact, using (3.3) and τv · ω ≡ 0, we have Z 00 0 (E0 (ω) · v|ω) = −2(1 + H (1)) (p · ω + t + Φ + η n · ω) = −8π(1 + H0 (1))t , S2
since ω ∈ H∗1 and Φ, η n · ω ∈ H∗1 (S2 ). Now we look for Φ. Testing (5.5) with ϕ = Ψs ω, where Ψs is any eigenfunction in Λs and s ∈ {2, . . . , n}, and using Lemma A.3, we see that we have to choose P Φ = ns=2 Φs , where Φ ∈ Λs is uniquely determined by: Z 0 t [λs − 2(1 + H (1))] Ψs Φs = (u|Ψs ω) ∀Ψs ∈ Λs , s ∈ {2, . . . , n} . S2
19
Finally, we have to determinate η n ∈ Xωn . Notice that taking ϕ ∈ Xωn in (5.5), we obtain the following condition for η n : (E000 (ω) · η n |ϕ) = (u|ϕ)
for every ϕ ∈ Xωn .
(5.9)
This is equivalent to say that η n is a critical point of the functional J: Xωn → R defined by 1 J(ψ) = (E000 (ω) · ψ|ψ) − (u|ψ) (ψ ∈ Xωn ) . 2 In addition, J is weakly lower semicontinuous on Xωn (use Lemma 3.4 and for example [9], Lemma A.4), and it is bounded from below and coercive by Corollary, 4.3, since 2(1 + H0 (1)) < λn+1 . Therefore, its minimum is attained, and in particular there exists η n ∈ Xωn solving (5.9). In order to show that η n ∈ W 1,3 one can argue as in [9], proof of Lemma 2.1. Finally, it remains to check that the function v = τv + p + tω + Φω + η n (with τv , p, t, Φ and η n defined as above) satisfies (5.5). This can be done using the decomposition in Lemma A.5. We omit details. Surjectivity is proved.
6
The case H(r) =
1 r
In this section we study problem (0.5), where K ∈ C 2 is a given perturbation of the curvature | · |−1 . Let us introduce the function Γ : [0, +∞) → R, Z Γ(s) = − K(q) dq . Bs (0)
In order to compute the derivative of Γ we notice that Γ(s) = −s3 and therefore Z 0 2 Γ (s) = −s (3K(sq) + ∇K(sq) · (sq)) dq . B1 (0)
Thus, 1 Γ (s) = − s 0
Z
Z K
div(K(q)q) dq = Bs (0)
∂Bs (0)
by the divergence Theorem. We start with a corollary of the result in Section 2.
20
R
B1 (0) K(sq)
dq ,
Proposition 6.1 Let K ∈ C 2 (R3 , R), s > 0. Assume that there exist a sequence εj → 0 and a sequence ω εj of |·|−1 +εj K(·)-bubbles, with kω εj −sωkC 1 (S2 ,R3 ) = O(εj ) as j → ∞. Then Γ0 (s) = 0. Proof. Set ε = εj , ω ˜ ε := s−1 ω ε , and notice that ω ˜ ε solves ∆˜ ω ε = 2(
1 ˜ ωε) ) ω + ε K(˜ ˜ xε ∧ ω ˜ yε , |˜ ωε|
˜ where K(q) = sK(sq). Moreover, k˜ ω ε − ωkW 1,3 = O(ε) as ε → 0. Therefore, Proposition 2.1 applies with n = 0, being H(r) = r−1 and H0 (1) = −1. Since R ˜ On the other Λ0 = R, Proposition 2.1 implies that K ∈ H∗1 (S2 ), that is, 0 = S2 K. hand, Z Z −1 ˜ K=s K = s−1 Γ0 (s) , S2
∂Bs (0)
and the Proposition is completely proved. In order to prove the existence Theorem 0.2, we fix a small radius s0 > 0 and any map H of class C 2 (R2 ) such that H(u) = |u|−1 for |u| > s0 . Let us consider the functionals Eε relative to the mean curvature functions H + εK. Then every critical point u for Eε with |u| > s0 solves problem (0.5). Now we look at the unpertubed problem: ( ∆u = 2 |u|−1 ux ∧ uy in R2 R (6.1) 2 R2 |∇u| < +∞ . Notice that for every s > 0 the surface parametrized by sω is a solution to problem (6.1) and hence, for s > s0 , it is a critical point of the unperturbed functional E0 . Therefore, E0 has a 7−dimensional manifold of critical points, namely: Z−1 = {su|u ∈ Z , s > s0 } , where Z is the 6-dimensional manifold of solutions that has been introduced in Section 4.2. The tangent space Tω Z−1 now is given by Tω Z−1 = T ⊕ < ω >= ker E000 (ω) by Corollary 4.2. For future convenience we notice also that all maps in Z−1 have the same energy. In fact, for conformal invariance, E0 (su) = E0 (sω) and d E0 (sω) = E00 (sω) · ω = 0 . ds 21
In addition, E000 (sω) = E000 (ω) for every s > s0 , by direct computation. Let us recall that a smooth manifold Z ⊂ W 1,3 is a natural constraint for Eε if every constrained critical point for Eε |Z is a critical point for Eε on W 1,3 . This is equivalent to say that the following implication holds: u∈Z ,
Eε0 (u) ∈ Tu Z =⇒ Eε0 (u) = 0 .
Now we are in position to state the main Lemma. Lemma 6.2 Let s¯0 , S be fixed real numbers, with 0 < s¯0 < S. Then there exist ε¯ > 0, and (unique) C 1 -maps (ε, s) → η ε (s) ∈ Xω0 ∩ W 1,3 , and (ε, s) → α0ε (s) ∈ R defined on a neighbourhood of [−¯ ε, ε¯] × [¯ s0 , S], such that η 0 (s) = 0, αs0 = α0 (s) = 0 and Eε0 (sω + η ε (s)) = −α0ε (s) ∆ω . Moreover, for every fixed ε ∈ [−¯ ε, ε¯] the set Zε := {sω + η ε (s)| s¯0 < s < S} is a natural constraint for Eε . R Proof. Take as in Section 5 a basis {τ1 , . . . , τ6 } of T with R2 ∇τi · ∇τj = δij . In the first step we construct maps η ε (s), α0ε (s) as in the statement of the Lemma, and a map (ε, s) → (α1ε (s), . . . , α6ε (s)) ∈ R6 such that Eε0 (sω + η ε ) = − α0ε ∆ω −
6 X
αiε ∆τi .
i=1
This can be done following the arguments in Section 5. In particular, one has to apply the implicit function theorem to the map F = (F1 , F2 ): R × R × (W 1,3 × R × R6 ) → W 1,3 × R × R6 defined by F1 (ε, s; η, α) = Eε0 (sω + η) + α0 ∆ω +
6 X
αiε ∆τi
i=1
Z F2 (ε, s; η, α) =
Z ∇η · ∇ω;
R2
R2
Z ∇η · ∇τ1 , . . . ,
∇η · ∇τ6
.
R2
Notice that we have used the notation: α = (α0 ; α1 , . . . , α6 ) for a point in R × R6 . Notice also that condition F2 (ε, s; η, α) = 0 is equivalent to η ∈ Xω0 . In checking 22
injectivity one has to use T ⊕ < ω >= ker E000 (ω) = ker E000 (sω) for s > s0 , because of Corollary 4.2, ii). When proving surjectivity, one has to notice that 2(1 + H0 (1)) = 0 < λ1 , and hence Corollary 4.3 ii), applies with n = 0. The decomposition Lemma that can be used in this case is Lemma A.4. The second step consists in proving that αiε = 0 for i = 1, . . . , 6, if ε is small enough. Fix as in the proof of Theorem 0.1 smooth paths of conformal transforms gi = gi (t) such that gi (0) = Id and (ω ◦ gi )0 (0) = τi for i = 1, . . . , 6. Notice as before that kAε k = o(1), where Aε := (Aεi,j )i,j=1,...,6 and Aεi,j
Z := −
∇τj · ∇((η ε ◦ gi )0 (0)) .
R2
With the same argument we get also Z Z Z ε 0 0 ε ∇ω · ∇(η ◦ gi ) (0) = − ∇(ω ◦ gi ) (0) · ∇η = − R2
R2
∇τi · ∇η ε = 0
R2
since η ε ∈ Xω . Now we use the invariance of Eε with respect to the conformal group. Writing ω ε = sω + η ε we have that (ω ε ◦ gi )0 (0) = sτi + (η ε ◦ gi )0 (0) and, by R R2 ∇ω · ∇τi = 0, 0=
6 X d Eε ((ω ε ◦ gi )(t)) = Eε0 (ω ε ) · (ω ε ◦ gi )0 (0) = sαiε − Aεij αjε . dt t=0 j=1
Then, as in Theprem 0.1 we get that αiε = 0 if |ε| is small enough, for i = 1, . . . , 6. Finally, let us check that the set Zε is a natural constraint for Eε , that is, every constained critical point for Eε on Zε is a critical point for Eε on W 1,3 . Assume that s¯ω + η ε (¯ s) ∈ Zε is a constrained critical point for Eε on Zε that is, fε0 (¯ s) = 0, where fε (s) = Eε (sω + η ε (s)) . Now, computing the derivative fε0 (s)
=
Eε0 (sω
ε
ε
+ η (s)) · (ω + ∂s η (s)) =
α0ε (s)
Z R2
|∇ω|2 = 8πα0ε (s)
R R since R2 ∇ω · ∇∂s η ε (s) = ∂s R2 ∇ω · ∇η ε (s) = 0, because η ε (s) ∈ Xω0 . Thus, α0ε (¯ s) = 0 that is, Eε0 (¯ sω + η ε (¯ s)) = 0.
23
Proof of Theorem 0.2 concluded. Let us compute for s > s0 fixed the derivative of the map Z 1 fε (s) = |∇(sω + η ε (s))|2 + 2VHε (sω + η ε (s)) 2 R2 R with respect to ε. In the computations one has to use R2 ∇η ε (s) · ∇ω = 0, that implies also Z Z ∇η ε (s) · ∇ω =
0 = ∂ε R2
∇(∂ε η ε (s)) · ∇ω .
R2
0 (sω)·g +2V (sω) . Set g(s) := ∂ε η ε (s)|ε=0 . Then it turns out that ∂ε fε (s)|ε=0 = 2VH K On the other hand, VK (sω) = Γ(s) and Z Z 0 2VH (sω) · g = 2s2 H(sω)g · ωx ∧ ωy = −s ∇g · ∇ω = 0 . R2
R2
Thus, one gets ∂ε fε (s)|ε=0 = 2Γ(s). Since we have already observed that E0 (sω) = E0 (ω) for every s > s0 , we get the following expansion for fε : fε (s) = E0 (ω) + 2Γ(s) ε + O(ε2 ) , that readily leads to the conclusion, following the arguments in [9], proof of Theorem 0.2. Remark 6.3 The case Γ ≡ 0 is completely open. Notice that this happends for example when K is an odd function with respect to one of its variables.
A
Appendix
In this Section we prove Propositions 3.1, 3.2 and 3.5. We start with some technical Lemmata. Lemma A.1
R
R2
∇(Φω) · ∇τ = 0 for every Φ ∈ Λ⊥ 1 and for every τ ∈ T .
∞ 2 Proof. Fix Φ ∈ Λ⊥ 1 ∩ C (S ). Since |ω| ≡ 1 and therefore ω · ωx = ω · ωy = 0, for every a ∈ R2 it results Z Z Z ∇(Φω) · ∇(a · ω)ω = ∇Φ · ∇(a · ω) + 2 Φ(a · ω) = 0 R2
R2
S2
24
because a · ω ∈ Λ1 (compare with Lemma 1.1, i) and Section 1.3). Finally, for every b ∈ R3 use ω ∧ ωy = ωx and ωx ∧ ω = ωy to compute Z Z ∇(Φω) · ∇(b ∧ ω) = b · (Φx ωy − Φy ωx ) = 0 R2
R2
by the Gauss-Green theorem. The Lemma is completely proved, since maps τ in T can be written as τ = (a · ω)ω + b ∧ ω − a with a, b in R3 . Lemma A.2 If η ∈ H 1 is pointwise orthogonal to ω, then BH1 (Φω, η) = 0 for every Φ ∈ H 1 (S2 ). Proof. Fix, let us notice that for every η ∈ C ∞ it results ω · (ηx ∧ ωy + ωx ∧ ηy ) = −∇η · ∇ω .
(A.1)
Here we have used as before ω ∧ ωy = ωx and ωx ∧ ω = ωy . Now, assume in addition that η · ω ≡ 0 on R2 . Then BH1 (Φω, η) = B0 (Φω, η) and ∇η · ω = −η · ∇ω. Therefore, Z Z BH1 (Φω, η) = (Φ · ∇ω · ∇η + ∇Φ · ω · ∇η ) − 2 Φ∇η · ∇ω 2 R2 R Z Z = − Φ∇ω · ∇η − η · ∇ω · ∇Φ 2 R2 Z ZR = − ∇ω · ∇(Φη) = −2 Φ(η · ω) = 0. R2
S2
The next simple Lemma is actually a crucial step. Lemma A.3 Let Φn ∈ Λn for an integer n ≥ 0. Then i) −∆(Φn ω) + 2 ((Φn ω)x ∧ ωy + ωx ∧ (Φn ω)y ) = (λn − 2)µ2 Φn ω R ii) BH1 (Φn ω, ϕ) = (λn − 2(1 + H1 ) ) S2 Φn (ω · ϕ) for every ϕ ∈ H 1 . Proof. Using Lemma 1.1 and (3.3) we immediately get the Lemma in case n = 0, since Λ0 = R. Now we assume n ≥ 1, and we let Φn be as in the statement. Then Φn ∈ H∗1 (S2 ) solves −∆Φn = λn µ2 Φn in R2 . 25
Notice that, by i) and ii) in Lemma 1.1, (Φn ω)x ∧ ωy + ωx ∧ (Φn ω)y = Φn ∆ω + ∇Φn · ∇ω . On the other hand, ∆(Φn ω) = Φn ∆ω + ω∆Φn + 2∇Φn · ∇ω, and i) readily follows from Lemma 1.1. Finally, ii) is an immediate consequence of i). The following decomposition results were used in the proofs of Theorems 0.1 and 0.2. Lemma A.4 For every v ∈ H 1 there exist unique τv ∈ T , p ∈ R3 , t ∈ R, η 1 ∈ Xω1 such that v = τv + p + tω + η 1 . R 1 1 2 ⊥ Proof. Fix any v ∈ H 1 , and set t := 4π S2 v · ω, so that v · ω − t ∈ H∗ (S ) = Λ1 ⊕ Λ1 . Thus, we can write v · ω = t + p · ω + Ψ with p ∈ R3 and Ψ ∈ Λ⊥ 1 . Since T \ {0} do not contain constant maps, then < ·, · > is a scalar product on T , and hence there exists a unique τv ∈ T solving Z Z 2 |∇(τv − v)| = min |∇(τ − v)|2 . τ ∈T
R2
Notice that
R2
Z ∇(τv − v) · ∇τ = 0 ∀ τ ∈ T . R2
Then, define η 1 := v − τv − p − tω . To conclude, we have just to notice that η 1 ∈ Xω1 since for every τ ∈ T it results Z Z Z ∇η 1 · ∇τ = ∇(v − τv ) · ∇τ − t ∇ω · ∇τ = 0 , S2
R2
R2
and η 1 · ω = v · ω − p · ω − t = Ψ ∈ Λ⊥ 1. Corollary A.5 Let n ≥ 2 be an integer. Then for every v ∈ H 1 , there exist unique τv ∈ T , p ∈ R3 , t ∈ R, Φ ∈ ⊕ns=2 Λs ω, η n ∈ Xωn such that v = τv + p + tω + Φω + η n . 26
Proof. Fix v ∈ H 1 and find p, t, τv , η 1 as in Lemma A.4. Since n η 1 · ω ∈ Λ⊥ 1 = (⊕s=2 Λs ) ⊕ Nn ,
˜ ∈ Nn with η 1 · ω = Φ + Ψ. ˜ Then Φ and we can find unique Φ ∈ ⊕ns=2 Λs and Ψ n 1 η := η − Φω give the right decomposition. Proof of Proposition 3.1. The proof of the “if” part is Lemma A.3 and (3.4). To prove the converse, let v ∈ H 1 be a solution to (3.1). Multiplying (3.1) by ω and using (A.1) we see that −ω · ∆v = 2∇v · ∇ω + 2H1 (v · ω) µ2 Hence, from ∆(v · ω) = v · ∆ω + 2∇v · ∇ω + ω · ∆v and from Lemma 1.1 we infer that v · ω solves −∆(v · ω) = λn µ2 (v · ω) , that is, v · ω ∈ Λn . Set v˜ := v − (v · ω)ω. We only have to prove that v˜ ∈ T . Notice that v˜ is pointwise orthogonal to ω, and since both v and (v · ω)ω solve (3.1) (use the “if” part of the Lemma), then v˜ solves (3.2). Then Proposition 3.1 follows from [10], Appendix. One can obtain a new proof arguing as follows. First, use Lemma A.4 to write v˜ in the form v˜ = τ +p+tω +η with τ ∈ T , p ∈ R3 , t ∈ R and η ∈ Xω1 . Our goal is to prove that v˜ = τ . Use B0 (τ, ·) = B0 (p, ·) = 0 for R every τ ∈ T , p ∈ R3 and (3.3) to prove that 0 = B0 (˜ v , ω) = −2(4πt+ S2 η·ω) = −8πt since η · ω ∈ H∗1 (S2 ). Hence, t = 0. Next, notice that both v˜ and τ are pointwise orthogonal to ω, and hence p · ω = −η · ω. On the other hand, η ∈ N1 ⊂ Λ⊥ 1 , and thus p · ω ∈ Λ1 ∩ Λ⊥ ˜ = τ + η. Therfore, also η · ω ≡ 0 and η 1 , that is, p = 0 and v solves (3.2), that is, ∆η = 2(ηx ∧ ωy + ωx ∧ ηy ) . (A.2) It remains to prove that η = 0. Since η · ω = 0, using the conformality of ω we can write η µ2 = (η · ωx )ωx + (η · ωy )ωy . Then, notice that η ∧ ω µ2 = (η · ωx )ωy − (η · ωy )ωx ηx ∧ ωy + ωx ∧ ηy
= −(∇η · ∇ω) ω − η µ2
ηx ∧ ωx + ηy ∧ ωy = η ∧ ω µ2 + (ηx · ωy − ηy · ωx )ω 27
(A.3) (A.4) (A.5)
Here we have used Lemma 1.1 and the identities: η · ω = ω · ωx = ω · ωy = 0. Now, since η ∈ Xω and η solves (A.2), then for every b ∈ R3 it results Z Z Z 0=− ∇η · ∇(b ∧ ω) = 2 (b ∧ ω) · (ηx ∧ ωy + ωx ∧ ηy ) = 2b · η∧ω R2
R2
S2
R by (A.3), that is, S2 η ∧ ω = 0 . Since ∆(η ∧ ω) = (∆η) ∧ ω + 2(ηx ∧ ωx + ηy ∧ ωy ) + η ∧ (∆ω), one infer from (A.4), (A.5) and Lemma 1.1 that −∆(η ∧ ω) = 2 η ∧ ω µ2 + 2(ηx · ωy − ηy · ωx )ω . Therefore, one gets Z
2
|∇(η ∧ ω)|
Z =2
R2
|η ∧ ω|2
S2
and since η ∧ ω ∈ H∗1 , by Lemma 1.2 we can conclude that each coordinate of the vector-valued map η ∧ ω belongs to Λ1 . Consequently, there exists a 3 × 3 matrix M with η ∧ ω = M ω. Notice that from M ω · ω = 0 on S2 it follows that the bilinear map R3 × R3 → R, (p, q) → M p · q is anti-symmetric, and hence M is of the form 0 −a3 a2 M = a3 0 −a1 , −a2
a1
0
with ai ∈ R for i = 1, 2, 3. Therefore, M ω = a ∧ ω, where a := (a1 , a2 , a3 ) ∈ R3 . Finally, η = ω ∧ (a ∧ ω) = a − (a · ω)ω ∈ T which, together with η ∈ Xω and η · ω = 0, implies η = 0. Proof of Proposition 3.2. We only have to prove that if v solves (3.1), then v ∈ T . Let Φn ∈ H 1 (S2 ) be the projection of the map ω · v on Λn . Testing (3.1) with Φn ω and using Lemma A.3 we get Z Z 0 = BH1 (v, Φn ω) = [λn − 2(1 + H1 )] Φn (ω · v) = [λn − 2(1 + H1 )] Φ2n S2
S2
and thus Φn = 0 for every n ≥ 0, that is, ω · v ≡ 0 on S2 ≡ R2 . Therefore, v solves (3.2), and hence Proposition 3.1 gives the existence of a, b ∈ R3 and Φ ∈ Λ1 such that v = (a · ω)ω + b ∧ ω − a + Φω . On the other hand, 0 = v · ω = Φ, and the Lemma is proved. 28
Proof of Proposition 3.5 Statement i) was proved by Isobe in [12]. Now we R 1 prove ii). Fix any η ∈ Xω , and set α := 8π ˜ := η − αω. Notice that R2 ∇η · ∇ω, η Z Z ∇˜ η · ∇ω = 0 , (A.6) η˜ · ω = 2 S2
S2
that is, η˜ ∈ Xω0 . Therefore, by i), we have that Z BH1 (˜ η , η˜) = B0 (˜ η , η˜) ≥ C0
|∇˜ η |2 .
R2
Now, it results BH1 (ω, η˜) = 0 because of (A.6), (3.3), and therefore BH1 (η, η) = α2 BH1 (ω, ω) + BH1 (˜ η , η˜) ≥ −8π(1 + H1 )α2 + C0
Z
|∇˜ η |2
S2
and ii) is proved. Now we prove iii). We can assume H1 ≥ −1. Let η ∈ Xωn , and write η as η = Φω + η ⊥ , with Φ := η · ω ∈ Nn and η ⊥ pointwise orthogonal to ω. Computing the partial derivatives of Φω and using the conformality of ω together with Lemma 1.1 - ii), we compute |∇(Φω)|2 = |∇Φ|2 + 2µ2 |Φ|2 . On the other hand, by Lemma 1.1 - i), we get Φω · ((Φω)x ∧ ωy + ωx ∧ (Φω)y ) = 2|Φ|2 ω · ωx ∧ ωy = −2µ2 |Φ|2 , and therefore Z BH1 (Φω, Φω) =
Z
2
|∇Φ| − 2(1 + H1 ) R2
2
Φ ≥ S2
2(1 + H1 ) 1− λn+1
Z
|∇Φ|2 ,
R2
since Φ ∈ Nn (compare with (1.3)). In addition, ⊥
⊥
⊥
⊥
BH1 (η , η ) = B0 (η , η ) ≥ C0
Z
|∇η ⊥ |2
R2
⊥ ∈ X 0 ). Finally, use by i) (use also Lemma A.1 and Nn ⊂ Λ⊥ ω 1 to prove that η ⊥ 1 2 Lemma A.2 to check that BH1 (Φω, η ) = 0 since Φ ∈ Nn ⊂ H∗ (S ), and hence
BH1 (η, η) = BH1 (Φω, Φω) + BH1 (η ⊥ , η ⊥ ) Z Z 2(1 + H1 ) 2 |∇Φ| + C0 |∇η ⊥ |2 . ≥ 1− λn+1 R2 R2 The proof is complete. 29
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