Bubbles with constant mean curvature, and almost constant mean curvature, in the hyperbolic space
aa r X i v : . [ m a t h . DG ] A ug Bubbles with constant mean curvature,and almost constant mean curvature, in the hyperbolic space
Gabriele Cora Roberta Musina
Abstract.
Given a constant k > , let Z be the family of round spheres of radius artanh( k − ) in thehyperbolic space H , so that any sphere in Z has mean curvature k . We prove a crucial nondegeneracyresult involving the manifold Z . As an application, we provide sufficient conditions on a prescribed function φ on H , which ensure the existence of a C -curve, parametrized by ε ≈ , of embedded spheres in H havingmean curvature k + εφ at each point. Keywords:
Hyperbolic geometry, prescribed mean curvature.
Let K be a given function on the hyperbolic space H . The K -bubble problem consists infinding a K -bubble , which is an immersed surface u : S → H having mean curvature K at each point. Besides its independent interest, the significance of the K -bubble problemis due to its connection with the Plateau problem for disk-type parametric surfaces havingprescribed mean curvature K and contour Γ, see for instance [1, 12]. In the Euclideancase, the impact of K -bubbles on nonexistence and lack of compacteness phenomena in thePlateau problem has been investigated in [5, 8, 9].To look for K -bubbles in the hyperbolic setting one can model H via the upper half-space ( R , p − δ hj ) and consider the elliptic system∆ u − u − G ( ∇ u ) = 2 u − K ( u ) ∂ x u ∧ ∂ y u (1.1)for functions u = ( u , u , u ) ∈ C ( S , H ). Here we used the stereographic projection to1ntroduce local coordinates on S ≡ R ∪ {∞} and put G ℓ ( ∇ u ) := ∇ u · ∇ u ℓ − |∇ u | δ ℓ = − u X h,j =1 Γ ℓhj ( u ) ∇ u h · ∇ u j , ℓ = 1 , , , (1.2)where Γ ℓhj are the Christoffel symbols. Any nonconstant solution u to (1.1) is a generalized K -bubble in H (see Lemma A.2 in the Appendix and [13, Chapter 2]), that is, u is aconformal parametrization of a surface having mean curvature K ( u ), apart from a finitenumber of branch points. Once found a solution to (1.1), the next step should concern thestudy of the geometric regularity of the surface u , which might have self-intersections andbranch points.A remarkable feature of (1.1) is its variational structure, which means that its solutionsare critical points of a certain energy functional E , see the Appendix for details. Becauseof their underlying geometrical meaning, both (1.1) and E are invariant with respect to theaction of M¨obius transformations. This produces some lack of compactness phenomena,similar to those observed in the largely studied K -bubble problem, raised by S.T. Yau[19], for surfaces in R (see for instance [7, 10, 11, 17] and references therein; see also[3, 6, 18] for related problems). However, the hyperbolic K -bubble problem is definitivelymore challenging, due to the homogeneity properties that characterize the hyperbolic-areaand the hyperbolic-volume functionals.The main differences between the Euclidean and the hyperbolic case are already evidentwhen the prescribed curvature is a constant k > k < /k in R can be parameterized by an embedded k -bubble, which minimizes the energy functional E Eucl ( u ) = 12 ˆ R |∇ u | dz + 2 k ˆ R u · ∂ x u ∧ ∂ y u dz on the Nehari manifold { u = const. | E ′ Eucl ( u ) u = 0 } , see [7, Remark 2.6]. In contrast, noimmersed hyperbolic k -bubble exists if k ∈ (0 , k >
1, then any sphere in H of radius ρ k := artanh 1 k = 12 ln k + 1 k − U : S → H , which solves∆ u − u − G ( ∇ u ) = 2 u − k ∂ x u ∧ ∂ y u on R , ( P )2nd which is a critical point of the energy functional E hyp ( u ) = 12 ˆ R u − |∇ u | dz − k ˆ R u − e · ∂ x u ∧ ∂ y u dz . (1.3)As in the Euclidean case, the functional E hyp is unbounded from below (see Remark A.3).Thus U does not minimize the energy E hyp on the Nehari manifold, which in fact fills { u = const. } .Besides their invariance with respect to M¨obius transformations, the system ( P ) andthe related energy E hyp are invariant with respect to the 3-dimensional group of hyperbolictranslations as well. Thus, any k -bubble generates a smooth 9-dimensional manifold Z ofsolutions to ( P ). We explicitly describe the tangent space T U Z at U ∈ Z in formula (3.4).As a further consequence of the invariances of problem ( P ), any tangent direction ϕ ∈ T U Z solves the elliptic system∆ ϕ − U − G ′ ( ∇ U ) ∇ ϕ = − U − ϕ ∆ U + 2 U − k (cid:0) ∂ x ϕ ∧ ∂ y U + ∂ x U ∧ ∂ y ϕ (cid:1) , (1.4)which is obtained by linearizing ( P ) at U .The next one is the main result of the present paper. Theorem 1.1 (Nondegeneracy)
Let U ∈ Z . If ϕ ∈ C ( S , R ) solves the linear system(1.4), then ϕ ∈ T U Z . In the Euclidean case the nondegeneracy of k -bubbles has been proved in [15, Propo-sition 3.1]. The proof of Theorem 1.1 (see Section 3), is considerably more involved. Itrequires the choice of a suitable orthogonal frame for functions in C ( S , R ) and the com-plete classification of solutions of two systems of linear elliptic differential equations, whichmight have an independent geometrical interest (see Lemmata 3.4, 3.5).As an application of Theorem 1.1, we provide sufficient conditions on a prescribedsmooth function φ : H → R that ensure the existence of embedded surfaces S → H having nonconstant mean curvature k + εφ . Our existence results involve the notion of stable critical point already used in [16] and inspired from [2, Chapter 2] (see Subsection2.2). The main tool is a Lyapunov-Schmidt reduction technique combined with variationalarguments, in the spirit of [2]. 3 heorem 1.2 Let k > and φ ∈ C ( H ) be given. Assume that the function F φk ( q ) := ˆ B H ρk ( q ) φ ( p ) d H p , F φk : H → R (1.5) has a stable critical point in an open set A ⋐ H . For every ε ∈ R close enough to thereexist a point q ε ∈ A , a conformal parametrization U q ε of a sphere of radius ρ k about q ε ,and a conformally embedded ( k + εφ ) -bubble u ε , such that k u ε − U q ε k C = O ( ε ) as ε → .Moreover, any sequence ε h → has a subsequence ε h j such that q ε hj converges to acritical point for F φk . In particular, if ˆ q ∈ Ω is the unique critical point for F φk in Ω , then u ε → U ˆ q in C ( R , H ) . Theorem 1.3
Assume that φ ∈ C ( H ) has a stable critical point in an open set A ⋐ H .Then there exists k > such that for any k > k and for every ε close enough to , thereexists a conformally embedded ( k + εφ ) -bubble. In Section 4 we first show that the existence of a critical point for F φk ( q ) is a necessarycondition in Theorem 1.2. Then we perform the dimension reduction and prove Theorems1.2, 1.3. With respect to correspondent Euclidean results in [7], a different choice of thefunctional setting allows us to weaker the regularity assumption on φ (from C to C ).We conclude the paper with an Appendix in which we collect some partially knownresults about the variational approach to (1.1) and prove a nonexistence result for (1.1)which, in particular, justifies the assumption on the existence of a critical point for φ inTheorem 1.3 . The vector space R n is endowed with the Euclidean scalar product ξ · ξ ′ and norm | ξ | . Wedenote by { e , e , e } the canonical basis and by ∧ the exterior product in R .We will often identify the complex number z = x + iy with the vector z = ( x, y ) ∈ R .Thus, iz ≡ ( − y, x ), z ≡ ( x − y , xy ) and z − ≡ | z | − ( x, − y ) if z = 0.In local coordinates induced by the stereographic projections from the north and thesouth poles, the round metric on the sphere S is given by g hj = µ δ hj , d S = µ dz , where µ ( z ) = 21 + | z | .
4e identify the compactified plane R with the sphere S through the inverse of the stere-ographic projection from the north pole, which is explicitly given by ω ( x, y ) = ( µx, µy, − µ ) . (2.1)The identity | ω | ≡ ω · ∂ x ω ≡ ω · ∂ y ω ≡
0. We also notice that ω is aconformal (inward-pointing) parametrization of the unit sphere and satisfies ∆ ω = 2 ∂ x ω ∧ ∂ y ω , − ∆ ω = 2 µ ω∂ x ω · ∂ y ω = 0 | ∂ x ω | = | ∂ y ω | = |∇ ω | = µ .∂ x ω ∧ ω = ∂ y ω , ω ∧ ∂ y ω = ∂ x ω , ∂ x ω ∧ ∂ y ω = − µ ω. (2.2) We adopt as model for the three dimensional hyperbolic space H the upper half-space R = { ( p , p , p ) ∈ R | p > } endowed with the Riemannian metric g hj ( z ) = p − δ hj .The hyperbolic distance d H ( p, q ) in H is related to the Euclidean one bycosh d H ( p, q ) = 1 + | p − q | p q , and the hyperbolic ball B H ρ ( p ) centered at p = ( p , p , p ) is the Euclidean ball of center( p , p , p cosh ρ ) and radius p sinh ρ .If F : H → R is a differentiable function, then ∇ H F ( p ) = p ∇ F ( p ), where ∇ H , ∇ are the hyperbolic and the Euclidean gradients, respectively. In particular, ∇ H F ( p ) = 0 ifand only if ∇ F ( p ) = 0. The hyperbolic volume form is related to the Euclidean one by d H p = p − dp . Let X ∈ C ( H ) and let Ω ⋐ H be open. We say that X has a stable critical point in Ωif there exists r > G ∈ C (Ω) satisfying k G − X k C (Ω) < r has acritical point in Ω.As noticed in [16], conditions to have the existence of a stable critical point p ∈ Ω for X are easily given via elementary calculus. For instance, one can use Browder’s topologicaldegree theory or can assume thatmin ∂ Ω X > min Ω X or max ∂ Ω X < max Ω X. X is of class C and Ω contains a nondegenerate critical point p (i.e. the Hessianmatrix of X at p is invertible), then p is stable. Any function f on R is identified with f ◦ ω − , which is a function on S . If no confusioncan arise, from now on we write f instead of f ◦ ω − .The Hilbertian norm on L ( R , R n ) ≡ L ( S , R n ) is given by k f k = ˆ R | f | µ dz < ∞ . Let m ≥
0. We endow C m ( R , R n ) = { u ∈ C m ( R , R n ) | u ( z − ) ∈ C m ( R , R n ) } ≡ C m ( S , R n )with the standard Banach space structure (we agree that C m ( R , R n ) = C ⌊ m ⌋ ,m −⌊ m ⌋ ( R , R n )if m is not an integer). If m is an integer, a norm in C m ( R , R n ) is given by k u k C m = k u k ∞ + k µ − m ∇ m u k ∞ . (2.3)Since we adopt the upper space model for H , we are allowed to write C m ( R , H ) = C m ( R , R ) = { u ∈ C m ( R , R ) | u > } , so that C m ( R , H ) is an open subset of C m ( R , R ).If ψ, ϕ ∈ C ( R , R ) and τ ∈ R , we put ∇ ψ · ∇ ϕ = ∂ x ψ · ∂ x ϕ + ∂ y ψ · ∂ y ϕ , τ ∇ ϕ = τ ∂ x ϕ + τ ∂ y ϕ (notice that τ ∇ ϕ ( z ) = dϕ ( z ) τ for any z ∈ R ). For instance, we have z h ∇ ϕ = ∂ x ϕ if h = 0 x∂ x ϕ + y∂ y ϕ if h = 1 , iz h ∇ ϕ = ∂ y ϕ if h = 0 − y∂ x ϕ + x∂ y ϕ if h = 1 . For future convenience we notice, without proof, that the next identities hold, ∂ x ω = e − ω ω − e ∧ ωz ∇ ω = e − ω ωz ∇ ω = − ( e − ω ω + e ∧ ω ) ∂ y ω = e − ω ω + e ∧ ωiz ∇ ω = e ∧ ω,iz ∇ ω = e − ω ω − e ∧ ω . (2.4)6he monograph [4] is our reference text for the theory of Sobolev spaces on Riemannianmanifolds. In view of our purposes, it is important to notice that H ( R , R n ) = { u ∈ H ( R , R n ) | |∇ u | + | u | µ ∈ L ( R ) } ≡ H ( S , R n ) . We simply write L ( R ), C m ( R ) and H ( R ) instead of L ( R , R ), C m ( R , R ) and H ( R , R ), respectively. Transformations in
P GL (2 , C ) are obtained by composing translations, dilations, rotationsand complex inversion. Its Lie algebra admits as a basis the transforms z , z i , z z , z iz , z z , z iz . Therefore, for any u ∈ C ( R , H ), the functions z h ∇ u , iz h ∇ u , h = 0 , , , span the tangent space to the manifold { u ◦ g | g ∈ P GL (2 , C ) } at u .Hyperbolic translations are obtained by composing a horizontal (Euclidean) translation p p + ae + be , a, b ∈ R with an Euclidean homothety p tp , t >
0. Therefore, for any u ∈ C m ( R , H ), the functions e , e , u , span the tangent space to the manifold { u q | q ∈ H } at u , where u q := q u + q − ( q · e ) e . (2.5) k -bubbles The proof of Theorem 1.1 needs some preliminary work. We put U = r k ( ω + ke ) , r k := sinh ρ k = 1 k cosh ρ k = 1 √ k − , where ω is given by (2.1). Since U is a conformal parametrization of the Euclidean sphereof radius r k about kr k e , which coincides with the hyperbolic sphere of radius ρ k about e ,then U has curvature k and in fact it solves ( P ). Accordingly with (2.5), we put U q := q U + q − ( q · e ) e (3.1)7notice that U e = U ), and introduce the 9-dimensional manifold Z = (cid:8) U q ◦ g | g ∈ P GL (2 , C ) , q ∈ H (cid:9) . (3.2) Remark 3.1
Any surface U ∈ Z is an embedding and solves ( P ) . Conversely, let U ∈ C ( R , H ) be an embedding. If U solves ( P ) , then it is a k -bubble by Lemma A.2and, thanks to an Alexandrov’ type argument (see for instance [14, Corollary 10.3.2]) itparametrizes a sphere of hyperbolic radius ρ k and Euclidean radius r k . Since U is confor-mal, then ∆ U = 2 r − k ∂ x U ∧ ∂ y U . Therefore U ∈ Z by the uniqueness result in [5]. By the remarks in Subsection 2.4 and since ∇ U q is proportional to ∇ ω , we have that T U q Z = T U Z for any q ∈ H , and T U Z = (cid:10) { z h ∇ ω , iz h ∇ ω | h = 0 , , } (cid:11) ⊕ (cid:10) e , e , U (cid:11) . (3.3)Moreover, any tangent direction τ ∈ T U Z solves (1.4).It is convenient to split C m ( R , R ) in the direct sum of its closed subspaces h ω i ⊥C m := { ϕ ∈ C m ( R , R ) | ϕ · ω ≡ R } , h ω i C m := { ηω | η ∈ C m ( R ) } . Since T U Z = (cid:0) T U Z ∩ h ω i ⊥ C (cid:1) ⊕ (cid:0) T U Z ∩ h ω i C (cid:1) , from (2.4) we infer another useful descriptionof the tangent space, that is T U Z = (cid:8) s − ( s · ω ) ω + t ∧ ω | s, t ∈ R (cid:9) ⊕ (cid:8) ( α · ( kω + e )) ω | α ∈ R (cid:9) . (3.4)We now introduce the differential operator J ( u ) = − div( u − ∇ u ) − u − |∇ u | e + 2 ku − ∂ x u ∧ ∂ y u . Notice that Z ⊂ { J = 0 } . Further, let I ( z ) = z − . Since J ( u ◦ I ) = | z | − J ( u ) ◦ I for any u ∈ C m ( R , R ), m ≥
0, then J is a C map J : C m ( R , H ) → C m ( R , R ) . We denote by J ′ ( u ) : C m ( R , R ) → C m ( R , R ) its differential at u .Finally, J ( U q ◦ g ) = 0 for any g ∈ P GL (2 , C ), q ∈ H , that implies T U Z ⊆ ker J ′ ( U ).In order to prove Theorem 1.1 it suffices to show thatker J ′ ( U ) ⊆ T U Z . ain computations Recall that U = r k ( ω + ke ) solves J ( U ) = 0 to check that J ′ ( U ) ϕ = − div (cid:0) U − ∇ ϕ (cid:1) + 2 U − (cid:2) G ′ ( ∇ U ) ∇ ϕ − ∇ U ∇ ϕ − ϕ ∆ U + k ( ∂ x ϕ ∧ ∂ y U + ∂ x U ∧ ∂ y ϕ ) (cid:3) , where G is given by (1.2). Since ∇ ω = −∇ µ = µ z , thanks to (2.2) we have r k J ′ ( U ) ϕ = − div (cid:0) ( ω + k ) − ∇ ϕ (cid:1) + 2( ω + k ) − (cid:2)(cid:0) G ′ ( ∇ ω ) ∇ ϕ − µ z ∇ ϕ (cid:1) + µ ϕ ω + k ( ∂ x ϕ ∧ ∂ y ω + ∂ x ω ∧ ∂ y ϕ ) (cid:3) , (3.5) G ′ ( ∇ ω ) ∇ ϕ − µ z ∇ ϕ = ∇ ϕ ∇ ω − ( ∇ ϕ · ∇ ω ) e . (3.6)To rewrite (3.5) in a less obscure form, we decompose any ϕ ∈ C m ( R , R ), m ≥
0, as ϕ = P ϕ + ( ϕ · ω ) ω , P ϕ := ϕ − ( ϕ · ω ) ω = µ − (cid:0) ( ϕ · ∂ x ω ) ∂ x ω + ( ϕ · ∂ y ω ) ∂ y ω (cid:1) . (3.7)Accordingly, for ϕ ∈ C ( R , R ) we have J ′ ( U ) ϕ = P (cid:0) J ′ ( U ) ϕ (cid:1) + ( J ′ ( U ) ϕ · ω ) ω , so that we can reconstruct J ′ ( U ) ϕ ∈ C ( R , R ) by providing explicit expressions for P (cid:0) J ′ ( U ) ϕ (cid:1) and J ′ ( U ) ϕ · ω , separately. This will be done in the next Lemma. Lemma 3.2
Let ϕ ∈ C ( R , R ) . Then r k P (cid:0) J ′ ( U ) ϕ (cid:1) = P (cid:16) − div (cid:16) ∇P ϕ ( ω + k ) (cid:17)(cid:17) + 2 µ ( ω + k ) ( iz ∇P ϕ ) ∧ ω − µ ( ω + k ) P ϕ , (3.8) r k ( J ′ ( U ) ϕ ) · ω = − div (cid:16) ∇ ( ϕ · ω )( ω + k ) (cid:17) − k µ ( ω + k ) ( ϕ · ω ) . (3.9) Proof.
We introduce the differential operator L = − div (cid:0) ( ω + k ) − ∇ (cid:1) and start to prove(3.9) by noticing that Lϕ · ω = L ( ϕ · ω ) + 2( ω + k ) − (cid:2) ( ω + k ) ∇ ϕ · ∇ ω − µ ϕ · ( z ∇ ω ) − µ ( ω + k )( ϕ · ω ) (cid:3) . (3.10)Recalling that ω is pointwise orthogonal to ∂ x ω, ∂ y ω , from (3.6) we obtain (cid:0) G ′ ( ∇ ω ) ∇ ϕ − µ z ∇ ϕ (cid:1) · ω = − ( ∇ ϕ · ∇ ω ) ω . ∂ x ϕ ∧ ∂ y ω + ∂ x ω ∧ ∂ y ϕ ) · ω = −∇ ϕ · ∇ ω . Finally, we obtain r k ( J ′ ( U ) ϕ ) · ω = L ( ϕ · ω ) − ω + k ) − µ (cid:2) ϕ · ( z ∇ ω ) − ϕ + ( ω + k )( ϕ · ω ) (cid:3) , and (3.9) follows, because e = z ∇ ω + ω ω , see (2.4).Next, using the equivalent formulation U J ′ ( U ) ϕ = − ∆ ϕ + 2( ω + k ) − (cid:2) G ′ ( ∇ ω ) ∇ ϕ + µ ωϕ + k ( ∂ x ϕ ∧ ∂ y ω + ∂ x ω ∧ ∂ y ϕ ) (cid:3) we find that, for ϕ = ηω , η ∈ C ( R ), it holds U J ′ ( U )( ηω ) · ∂ x ω = U J ′ ( U )( ηω ) · ∂ y ω = 0 , whence we infer P (cid:0) J ′ ( U )( ϕ − P ϕ ) (cid:1) = 0 , for every ϕ ∈ C ( R , R ) . (3.11)Thanks to (3.9) and (3.11) we get P (cid:0) J ′ ( U ) ϕ (cid:1) = J ′ ( U )( P ϕ ), thus to conclude the proofwe can assume that ϕ = P ϕ . Since ϕ is pointwise orthogonal to ω , we trivially have ∂ x ϕ · ω = − ϕ · ∂ x ω , ∂ y ϕ · ω = − ϕ · ∂ y ω . We start to handle (3.6). From e = z ∇ ω + ω ω we get( G ′ ( ∇ ω ) ∇ ϕ − µ z ∇ ϕ ) + ω ( ∇ ϕ · ∇ ω ) ω = ∇ ϕ ∇ ω − ( ∇ ϕ · ∇ ω ) z ∇ ω = (cid:0) ∂ x ϕ − x ( ∇ ϕ · ∇ ω ) (cid:1) ∂ x ω + (cid:0) ∂ y ϕ − y ( ∇ ϕ · ∇ ω ) (cid:1) ∂ y ω . Further, ∂ x ϕ − x ( ∇ ϕ · ∇ ω ) = ∂ x ϕ · ( z ∇ ω + ω ω ) − x ( ∇ ϕ · ∇ ω )= (cid:0) ∂ x ϕ · ( z ∇ ω ) − x ( ∇ ϕ · ∇ ω ) (cid:1) − ω ϕ · ∂ x ω = − ( iz ∇ ϕ ) · ∂ y ω − ω ϕ · ∂ x ω . In a similar way one can check that ∂ y ϕ − y ( ∇ ϕ · ∇ ω ) = ( iz ∇ ϕ ) · ∂ x ω − ω ϕ · ∂ y ω , thus G ′ ( ∇ ω ) ∇ ϕ − µ z ∇ ϕ = µ ( iz ∇ ϕ ) ∧ ω − ω ( ∇ ϕ · ∇ ω ) ω − µ ω ϕ. Next, using (2.2) we can compute ∂ x ϕ ∧ ∂ y ω = ∂ x ϕ ∧ ( ∂ x ω ∧ ω ) = − ( ϕ · ∂ x ω ) ∂ x ω − ( ∂ x ϕ · ∂ x ω ) ω x ω ∧ ∂ y ϕ = ( ω ∧ ∂ y ω ) ∧ ∂ y ϕ = − ( ϕ · ∂ y ω ) ∂ y ω − ( ∂ y ϕ · ∂ y ω ) ω, which give the identity ∂ x ϕ ∧ ∂ y ω + ∂ x ω ∧ ∂ y ϕ = − µ ϕ − ( ∇ ϕ · ∇ ω ) ω , (3.12)that holds for any ϕ ∈ h ω i ⊥C m .Putting together the above informations we arrive at r k J ′ ( U ) ϕ = Lϕ + 2 µ ( ω + k ) ( iz ∇ ϕ ) ∧ ω − µ ( ω + k ) ϕ + 2( ω + k ) (cid:2) µ ϕ − ( ω + k ) ∇ ϕ ·∇ ω (cid:3) ω . Using (3.10) and ϕ = ϕ · ( z ∇ ω ), we conclude the proof. (cid:3) Thanks to Lemma 3.2 we can study the system J ′ ( U ) ϕ = 0 separately, on h ω i ⊥C m first,and on h ω i C m later. In fact, ϕ ∈ ker J ′ ( U ) if and only if the pair of functions ψ := P ϕ ∈ h ω i ⊥C ⊂ C ( R , R ) , η := ϕ · ω ∈ C ( R ) , solves P (cid:16) − div (cid:16) ∇ ψ ( ω + k ) (cid:17)(cid:17) + 2 µ ( ω + k ) ( iz ∇ ψ ) ∧ ω = 2 µ ( ω + k ) ψ , (3.13a) − div (cid:16) ∇ η ( ω + k ) (cid:17) = 2 k µ ( ω + k ) η . (3.13b)We begin by facing problem (3.13a). Firstly, we show that the quadratic form associatedto the differential operator J ′ ( U ) is nonnegative on h ω i ⊥C . Lemma 3.3
Let ψ ∈ h ω i ⊥C . Then ˆ R J ′ ( U ) ψ · ψ dz = r − k ˆ R ( ∂ x ψ · ∂ x ω − ∂ y ψ · ∂ y ω ) + ( ∂ x ψ · ∂ y ω + ∂ y ψ · ∂ x ω ) µ ( ω + k ) dz . Proof.
Since J ′ ( U ) ψ · ψ = P (cid:0) J ′ ( U ) ψ (cid:1) · ψ and P ψ = ψ , formula (3.8) gives r k ˆ R J ′ ( U ) ψ · ψ dz = ˆ R |∇ ψ | ( ω + k ) dz + 2 ˆ R ψ · ( iz ∇ ψ ) ∧ ω ( ω + k ) µ dz − ˆ R | ψ | ( ω + k ) µ dz. Now we prove the identity B ψ := 2 ˆ R ψ · ( iz ∇ ψ ) ∧ ω ( ω + k ) µ dz = 2 ˆ R ω · ∂ x ψ ∧ ∂ y ψ ( ω + k ) dz + ˆ R | ψ | ( ω + k ) µ dz . (3.14)11e use polar coordinates ρ, θ on R and notice that ∂ θ ψ = iz ∇ ψ . From ρµ = ∂ ρ ω we get B ψ = − π ˆ dθ ∞ ˆ ( ψ · ∂ θ ψ ∧ ω ) ∂ ρ ( ω + k ) − dρ = ∞ ˆ dρ π ˆ ω · ∂ ρ ψ ∧ ∂ θ ψ − ψ · ∂ ρ ω ∧ ∂ θ ψ ( ω + k ) dθ + ∞ ˆ dρ π ˆ ∂ ρθ ψ · ω ∧ ψ ( ω + k ) dθ = ∞ ˆ dρ π ˆ ω · ∂ ρ ψ ∧ ∂ θ ψ − ψ · ∂ ρ ω ∧ ∂ θ ψ ( ω + k ) dθ + ∞ ˆ dρ π ˆ ω · ∂ ρ ψ ∧ ∂ θ ψ − ψ · ∂ ρ ψ ∧ ∂ θ ω ( ω + k ) dθ . Using the elementary identity ∂ ρ α ∧ ∂ θ β = ρ ( ∂ x α ∧ ∂ y β ), we see that B ψ = 2 ˆ R ω · ∂ x ψ ∧ ∂ y ψ ( ω + k ) dz − ˆ R ψ · ( ∂ x ω ∧ ∂ y ψ + ∂ x ψ ∧ ∂ y ω )( ω + k ) dz , and (3.14) follows from (3.12) (with ϕ replaced by ψ ).Thanks to (3.14), we have the identity r k ˆ R J ′ ( U ) ψ · ψ dz = ˆ R |∇ ψ | + 2 ω · ∂ x ψ ∧ ∂ y ψ − µ | ψ | ( ω + k ) dz, so that we only need to handle the function b ψ := |∇ ψ | + 2 ω · ∂ x ψ ∧ ∂ y ψ − µ | ψ | . We decompose ∂ x ψ and ∂ y ψ accordingly with (3.7), to obtain µ ∂ x ψ = ( ∂ x ψ · ∂ x ω ) ω x + ( ∂ x ψ · ∂ y ω ) ω y − µ ( ψ · ∂ x ω ) ω ,µ ∂ y ψ = ( ∂ y ψ · ∂ x ω ) ω x + ( ∂ y ψ · ∂ y ω ) ω y − µ ( ψ · ∂ y ω ) ω , respectively. Since |∇ ψ | = | ∂ x ψ | + | ∂ y ψ | , we infer µ (cid:0) |∇ ψ | − µ | ψ | ) = ( ∂ x ψ · ∂ x ω ) + ( ∂ x ψ · ∂ y ω ) + ( ∂ y ψ · ∂ x ω ) + ( ∂ y ψ · ∂ y ω ) . Writing µ ω = − ∂ x ω ∧ ∂ y ω , see (2.2), we get µ ω · ( ∂ x ψ ∧ ∂ y ψ ) = − ( ∂ x ψ · ∂ x ω )( ∂ y ψ · ∂ y ω ) + ( ∂ x ψ · ∂ y ω )( ∂ y ψ · ∂ x ω ) , from which it readily follows that µ b ψ = ( ∂ x ψ · ∂ x ω − ∂ y ψ · ∂ y ω ) + ( ∂ x ψ · ∂ y ω + ∂ y ψ · ∂ x ω ) .The proof is complete. (cid:3) emma 3.4 Let ψ ∈ C ( R , R ) be a solution to (3.13a). There exist s, t ∈ R such that ψ = s − ( s · ω ) ω + t ∧ ω , and thus ψ ∈ T U Z ∩ h ω i ⊥ C = { s − ( s · ω ) ω + t ∧ ω | s, t ∈ R } . Proof.
From (3.13a) it immediately follows that ψ is pointwise orthogonal to ω , whichimplies ψ ∈ h ω i ⊥C . Since P ψ = ψ , then J ′ ( U ) ψ = 0 by (3.8) and (3.9), hence ∂ x ψ · ∂ x ω − ∂ y ψ · ∂ y ω = 0 ∂ x ψ · ∂ y ω + ∂ y ψ · ∂ x ω = 0 (3.15)by Lemma 3.3. Since ψ ∈ h ∂ x ω, ∂ y ω i pointwise on R , we can write ψ = f ∇ ω , where f := µ − ( ψ · ∂ x ω, ψ · ∂ y ω ) ∈ C ( R , R ) . We identify f with a complex valued function. A direct computation based on (2.2) showsthat ψ solves (3.15) if and only if f solves ∂ x f + i∂ y f = 0 on R . In polar coordinates wehave that ρ∂ ρ f + i∂ θ f = 0 . (3.16)For every ρ > f ( ρ, · ) in Fourier series, f ( ρ, θ ) = X h ∈ Z γ h ( ρ ) e ihθ , γ h ( ρ ) = 12 π π ˆ f ( ρ, θ ) e − ihθ dθ . The coefficients γ h are complex-valued functions on the half-line R + that solve γ ′ h − hγ h = 0 , because of (3.16). Thus for every h ∈ Z there exists a h ∈ C such that γ h ( ρ ) = a h ρ h . Nowrecall that µψ ∈ L ( R , R ). Since ˆ R µ | ψ | dz = ˆ R µ | f | dz ≥ π ˆ ∞ µ ρ | γ h | dρ = a h ˆ R µ | z | h dz , ∀ h ∈ Z , we infer that γ h = 0 for every h = 0 , ,
2. Thus f ( z ) = P h =0 a h z h , that is ψ = P h =0 a h z h ∇ ω ,and in particular the space of solutions of (3.13a) has (real) dimension 6. The conclusionof the proof follows from the relations (2.4). (cid:3) emma 3.5 Let η ∈ C ( R ) be a solution to (3.13b). There exists α ∈ R such that η = α · ( kω + e ) and thus ηω ∈ T U Z ∩ h ω i C = { ( α · ( kω + e )) ω | α ∈ R } . Proof.
First of all, we notice that α · ( kω + e ) solves (3.13b) for any α ∈ R .By the Hilbert–Schmidt theorem, the eigenvalue problem − div (cid:16) ∇ η ( ω + k ) (cid:17) = λ µ ( ω + k ) η on R , η ∈ C ( R ) , (3.17)has a non decreasing, divergent sequence ( λ h ) h ≥ of eigenvalues which correspond to criticallevels of the quotient R ( η ) := ˆ R |∇ η | ( ω + k ) dz ˆ R | η | ( ω + k ) µ dz , η ∈ H ( R ) \ { } . Clearly, λ = 0 is simple, and its eigenfunctions are constant functions. We claim that thenext eigenvalue is 2 k , and that its eigenspace has dimension 3, that concludes the proof.To this goal, we use the functional change η ( z ) = µ ( z ) µ ( c k z ) Φ( c k z ) , c k := e ρ k = r k + 1 k − . By a direct computation involving the identity ( ω ( z ) + k ) µ ( c k z ) = ( k − µ ( z ) and inte-gration by parts, one gets λ = inf η ∈C R \{ } ´ R η µ dz ( ω k )3 =0 R ( η ) = 2 k + inf Φ ∈C R \{ } ´ R Φ µ dz =0 ˆ R |∇ Φ | dz − ˆ R | Φ | µ dz ˆ R | Φ | ( k − ω ) µ dz . On the other hand, it is well known thatmin Φ ∈C R \{ } ´ R µ dz =0 ˆ R |∇ Φ | dz ˆ R | Φ | µ dz = 2is the first nontrivial eigenvalue for the Laplace-Beltrami operator on the sphere and thatits eigenspace has dimension 3, see for instance [4]. This concludes the proof. (cid:3) emark 3.6 The third eigenvalue λ of (3.17) verifies λ > k by Lemma 3.5, and λ = min n R ( η ) (cid:12)(cid:12)(cid:12) ˆ R η ( ω + k ) µ dz = ˆ R η ( kω j + δ j )( ω + k ) µ dz = 0 , j = 1 , , o . Proof of Theorem 1.1
In fact, we only have to sum up the argument. Let U ∈ Z .Thanks to (3.2), U = U q ◦ g for some q ∈ H , g ∈ P GL (2 , C ). Since T U q ◦ g Z = T U Z ◦ g , ker J ′ ( U q ◦ g ) = ker J ′ ( U ) ◦ g , for every q ∈ H , g ∈ P GL (2 , C ) , it suffices to consider the case U = U .If ϕ ∈ C ( R , R ) solves (1.4) then J ′ ( U ) ϕ = 0, which means P ( J ′ ( U ) ϕ ) = 0 and( J ′ ( U ) ϕ ) · ω = 0. From Lemma 3.2 we infer that P ϕ solves (3.13a) and that ϕ · ω solves(3.13b). Therefore, Lemmata 3.4, 3.5 give the existence of s, t, α ∈ R such that P ϕ = s − ( s · ω ) ω + t ∧ ω , ϕ · ω = α · ( kω + e ) . Thus ϕ = P ϕ + ( ϕ · ω ) ω ∈ T U Z by (3.4), which concludes the proof. (cid:3) J ′ ( U ) To shorten notation we put H = H ( R , R ) . Since integration by parts gives ˆ R − div (cid:16) ∇ ϕ ( ω + k ) (cid:17) · ψ dz = ˆ R ∇ ϕ · ∇ ψ ( ω + k ) dz , ϕ, ψ ∈ C ( R , R ) , the quadratic form ( ϕ, ψ ) ˆ R J ′ ( U ) ϕ · ψ dz (3.18)can be extended to a continuous bilinear form H × H → R via a density argument. Itcan be checked by direct computations (see also Remark 4.2), that the quadratic form in(3.18) is self-adjoint on H , that is, ˆ R J ′ ( U ) ϕ · ψ dz = ˆ R J ′ ( U ) ψ · ϕ dz for any ϕ, ψ ∈ H . (3.19)15ince T U Z is a subspace of L ( R , R ) ≡ L ( S , R ), we are allowed to put T U Z ⊥ := n f ∈ L ( R , R ) (cid:12)(cid:12)(cid:12) ˆ R f · τ µ dz = 0 , ∀ τ ∈ T U Z o . To shorten notation we introduce on L ( R , R ) the equivalent scalar product( f, ψ ) ∗ = ˆ R P f · P ψ ( ω + k ) µ dz + ˆ R ( f · ω )( ψ · ω )( ω + k ) µ dz and the subspaces T U Z ⊥∗ := (cid:8) f ∈ L ( R , R ) | ( f, τ ) ∗ = 0 , ∀ τ ∈ T U Z (cid:9) ,N ∗ := h ω i ⊥∗ = (cid:8) f ∈ L ( R , R ) | ( f, ω ) ∗ = 0 (cid:9) . We are in position to state the main result of this section.
Lemma 3.7
Let q ∈ H . For any v ∈ T U Z ⊥ , there exists ϕ v ∈ H ∩ T U Z ⊥∗ ∩ N ∗ such that J ′ ( U q ) ϕ v = v µ on R . (3.20) If in addition v ∈ C m ( R , R ) for some m ∈ (0 , , then ϕ v ∈ C m ( R , R ) . In view of Lemma 3.2, we split the proof of Lemma 3.7 in few steps.
Lemma 3.8
Let v ∈ T U Z ⊥ be such that v · ω ≡ on R . There exists ϕ ∈ H ∩ T U Z ⊥∗ suchthat ϕ · ω ≡ on R and J ′ ( U ) ϕ = v µ on R . (3.21) Proof.
We introduce X := (cid:8) ψ ∈ H (cid:12)(cid:12) ψ · ω ≡ R (cid:9) ∩ T U Z ⊥∗ , which is a closed subspace of H . Notice that ψ = P ψ for any ψ ∈ X and moreover ˆ R J ′ ( U ) ψ · ψ dz = ˆ R |∇ ψ | ( ω + k ) dz + 2 ˆ R (cid:16) ( ψ · iz ∇ ψ ) ∧ ω ( ω + k ) − | ψ | ( ω + k ) (cid:17) µ dz , use (3.8) and a density argument. Next we put λ := inf ψ ∈ Xψ =0 ˆ R J ′ ( U ) ψ · ψ dz ˆ R ( ω + k ) − | ψ | µ dz , λ ≥ λ is achieved by Rellich theorem.Thus λ >
0, because of Lemma 3.4. It follows that the energy functional I : X → R , I ( ψ ) = 12 ˆ R J ′ ( U ) ψ · ψ dz − ˆ R v · ψ µ dz , is weakly lower semicontinuous and coercive. Thus its infimum is achieved by a function ϕ ∈ X which satisfies ˆ R J ′ ( U ) ϕ · ψ dz = ˆ R v · ψ µ dz , ∀ ψ ∈ X . (3.22)If ψ ∈ H we write ψ = ( P ψ ⊤ + P ψ ⊥ ) + ηω , where η = ψ · ω , P ψ ⊤ ∈ T U Z = ker J ′ ( U ) is the orthogonal projection of P ψ = ψ − ηω onto T U Z in the scalar product ( · , · ) ∗ and P ψ ⊥ := ψ − P ψ ⊤ − ηω ∈ X . We use (3.19) and(3.9) to compute ˆ R J ′ ( U ) ϕ · P ψ ⊤ dz = ˆ R J ′ ( U ) P ψ ⊤ · ψ dz = 0 , ˆ R J ′ ( U ) ϕ · ( ηω ) dz = ˆ R ∇ ( ϕ · ω ) · ∇ η ( ω + k ) dz − k ˆ R ( ϕ · ω ) η ( ω + k ) µ dz = 0 , because ϕ · ω ≡
0. Therefore, (3.22) gives ˆ R J ′ ( U ) ϕ · ψ dz = ˆ R J ′ ( U ) ϕ · P ψ ⊥ dz = ˆ R v · P ψ ⊥ µ dz = ˆ R v · ψ µ dz , as v is orthogonal to T U Z ∋ P ψ ⊤ and to ηω in L ( R , R ). We showed that ϕ solves (3.21),and thus the proof is complete. (cid:3) Lemma 3.9
Let f ∈ H ( R ) be such that f ω ∈ T U Z ⊥ . There exists η ∈ H ( R ) such that ηω ∈ H ∩ T U Z ⊥∗ ∩ N ∗ and J ′ ( U )( ηω ) = f ω µ on R . (3.23)17 roof. We introduce the space Y := n η ∈ H ( R ) (cid:12)(cid:12)(cid:12) ˆ R η ( ω + k ) µ dz = ˆ R η ( τ · ω )( ω + k ) µ dz = 0 , ∀ τ ∈ T U Z o , so that ηω ∈ H ∩ T U Z ⊥∗ ∩ N ∗ for any η ∈ Y , and the energy functional I : Y → R , I ( ϕ ) = 12 ˆ R J ′ ( U )( ηω ) · ( ηω ) dz − ˆ R f η µ dz = 12 ˆ R |∇ η | ( ω + k ) dz − k ˆ R | η | ( ω + k ) µ dz − ˆ R ηf µ dz , compare with (3.9). The functional I is weakly lower semicontinuous with respect to the H ( R ) topology and coercive by Remark 3.6. Thus its infimum is achieved by a function η ∈ Y . To conclude, argue as in the proof of Lemma 3.8 to show that η solves (3.23). (cid:3) Proof of Lemma 3.7.
Since J ′ ( U q ) = q − J ′ ( U ), we can assume that q = e , that is, U q = U . We take any v ∈ T U Z ⊥ , and write v = P v + ( v · ω ) ω , where P v = v − ( v · ω ) ω , as before. Since P v ∈ T U Z ⊥ , by Lemma 3.8 there exists a uniqueˆ ϕ ∈ H ∩ T U Z ⊥∗ such that ˆ ϕ · ω ≡ R and ˆ R J ′ ( U ) ˆ ϕ · ψ dz = ˆ R P v · ψ µ dz , for any ψ ∈ H .Next, notice that ( v · ω ) ω ∈ T U Z ⊥ , so we can use Lemma 3.9 to find η ∈ H ( R ) such that ηω ∈ H ∩ T U Z ⊥∗ ∩ N ∗ solves ˆ R J ′ ( U )( ηω ) · ψ dz = ˆ R ( v · ω )( ψ · ω ) µ dz, for any ψ ∈ H .The function ϕ v = ˆ ϕ + ηω solves (3.20).To conclude the proof we have to show that if v ∈ C m ( R , R ) then ϕ v ∈ C m ( R , R ).Since ω ∈ C ∞ ( R , R ) and ω + k is bounded and bounded away from zero, ϕ v solves alinear system of the form − ∆ ϕ v = A ( z ) ϕ v + B ( z ) ∇ ϕ v + µ ( ω + k ) v , R . A standard bootstrap argument and Schauder regularitytheory plainly imply that ϕ v ∈ C mloc ( R , R ). The function z ϕ v ( z − ) satisfies a linearsystem of the same kind, hence ϕ v ∈ C m ( R , R ), as desired. (cid:3) In this Section we perform the finite dimensional reduction and prove Theorems 1.2, 1.3.By the results in the Appendix, any critical point of the C -functional E ε : C ( R , H ) → R , E ε ( u ) := 12 ˆ R u − |∇ u | dz − k ˆ R u − e · ∂ x u ∧ ∂ y u dz + 2 ε V φ ( u ) = E ( u ) + 2 ε V φ ( u )(notice that E = E hyp , compare with (1.3)), solves∆ u − u − G ( ∇ u ) = 2 u − ( k + εφ ( u )) ∂ x u ∧ ∂ y u on R ( P ε )and has mean curvature ( k + εφ ), apart from a finite set of branch points.Due to the action of the M¨obius transformations and of the hyperbolic translations, forany u ∈ C ( R , R ) we have the identities E ′ ε ( u )( z h ∇ u ) = 0 , E ′ ε ( u )( iz h ∇ u ) = 0 , for h = 0 , , ε ∈ R , (4.1) E ′ ( u ) e = 0 , E ′ ( u ) e = 0 , E ′ ( u ) u = 0 . (4.2)Now we prove that E ε ( U q ) = E ( U ) − εF φk ( q ) , (4.3)where F φk is the Melnikov type function in (1.5). The above mentioned invariances give E ( U q ) = E ( U ). Since the hyperbolic ball B H ρ k ( q ) coincides with the Euclidean ball ofradius q r k about the point q k := ( q , q , kr k q ), the divergence theorem gives F φk ( q ) = ˆ B H ρk ( q ) φ ( p ) d H p = ˆ B q rk ( q k ) p − φ ( p ) dp = ˆ ∂B q rk ( q k ) Q φ ( p ) · ν p . Here Q φ ∈ C ( R , R ) is any vectorfield such that div Q φ ( p ) = p − φ ( p ) and ν p is the outernormal to ∂B q r k ( q k ) at p . The function U q in (3.1) parameterizes the Euclidean sphere ∂Bq r k ( q k ). Since ∂ x U q ∧ ∂ y U q is inward-pointing, we have F φk ( q ) = − ˆ R Q φ ( p ) · ∂ x U q ∧ ∂ y U q dz = − V φ ( U q ) , (4.4)19nd (4.3) is proved. Before going further, let us show that the existence of critical pointsfor F φk is a necessary condition for the conclusion in Theorem 1.2. Theorem 4.1
Let k > , φ ∈ C ( H ) . Assume that there exist sequences ε h ⊂ R \ { } , ε h → , u h ∈ C ( R , H ) and a point q ∈ H such that u h solves ( P ε h ) , and u h → U q in C ( R , H ) . Then q is a stationary point for F φk . Proof.
The function u h is a stationary point for the energy functional E ε h = E + 2 ε h V φ .From (4.2) we have V ′ φ ( u h ) e j = 0 for j = 1 , V ′ φ ( u h ) u h = 0. We can plainly pass tothe limit to obtain V ′ φ ( U q ) e j = 0 for j = 1 , V ′ φ ( U q ) U q = 0. To conclude, use (4.4) andrecall that ∂ q j U q = e j for j = 1 ,
2, and ∂ q U q = U = q − ( U q − q e − q e ). (cid:3) Now we fix m ∈ (0 , J ε : C m ( R , H ) → C m ( R , R ) defined by J ε ( u ) = − div( u − ∇ u ) − u − |∇ u | e + 2( k + εφ ) u − ∂ x u ∧ ∂ y u , is related to the differential of E ε via the identity E ′ ε ( u ) ϕ = ˆ R J ε ( u ) · ϕ dz , u ∈ C m ( R , H ) , ϕ ∈ C m ( R , R ) . (4.5) Remark 4.2
Since E ε is of class C and E ′′ ε ( u )[ ϕ, ψ ] = ˆ R J ′ ε ( u ) ψ · ϕ dz , then the quadratic form in the right hand side is a self-adjoint form on H . We are in position to state and proof the next lemma, which is the main step towardsthe proofs of Theorems 1.2, 1.3.
Lemma 4.3 (Dimension reduction)
Let Ω ⋐ H be an open set. There exists ˆ ε > and a unique C -map [ − ˆ ε, ˆ ε ] × Ω → C m ( R , H ) , ( ε, q ) u εq , such that the following facts hold: i ) u εq parameterizes an embedded S -type surface, and u q = U q ; i ) u εq − U q ∈ T U Z ⊥ ∩ C m ( R , R ) and E ′ ε ( u εq ) ϕ = 0 for any ϕ ∈ T U Z ⊥ ∩ C ( R , R ) ; iii ) for any ε ∈ [ − ˆ ε, ˆ ε ] , the manifold { u εq | q ∈ Ω } is a natural constraint for E ε , that is,if ∇ q E ε ( u εq ε ) = 0 for some q ε ∈ Ω , then u εq ε is a ( k + εφ ) -bubble ; iv ) k E ε ( u εq ) − E ε ( U q ) k C ( Ω ) = o ( ε ) as ε → , uniformly on Ω . Proof.
To shorten the notation, we put C m := C m ( R , R ). For s ≥ δ > s := { p ∈ H | dist( p, Ω) < s } , and U δ := { ν ∈ C m | | ν ( z ) | < δ for every z ∈ R } . We fix s > δ = δ ( s ) > s ⊂ H and ( U q + ν ) · e > q ∈ Ω s , ν ∈ U δ .We define τ := c ∂ x ω , τ := c √ z ∇ ω , τ := c z ∇ ω ,τ := c ∂ y ω , τ := c √ iz ∇ ω , τ := c iz ∇ ω , γ := 2 c ( kω + e ) , (4.6)where c := q π is a normalization constant. Thanks to (3.3), (3.4), we have T U Z = h τ , . . . τ i ⊕ { ( α · γ ) ω | α ∈ R } . Trivially, τ j · ω ≡ R . Elementary computations give ˆ R τ i · τ j µ dz = δ ij , ˆ R γ h γ ℓ µ dz = 0 if h = ℓ , for i, j ∈ { , . . . , } , h, ℓ ∈ { , , } , and moreover ˆ R γ µ dz = ˆ R γ µ dz = k , ˆ R γ µ dz = k + 3 . Construction of u ε q satisfying i ), ii ). By our choices of s and δ , the functions F ( ε, q ; ν, ξ, α ) := µ − J ε ( U q + ν ) − X j =1 ξ j τ j − ( α · γ ) ω ∈ C m , F ( ε, q ; ν, ξ, α ) := (cid:16) ˆ R ν · τ µ dz , . . . , ˆ R ν · τ µ dz ; ˆ R γ ( ν · ω ) µ dz (cid:17) ∈ R × R , are well defined and continuously differentiable on R × Ω s × U δ × ( R × R ). Thus F := ( F , F ) : R × Ω s × U δ × ( R × R ) → C m × ( R × R )21s of class C on its domain. Notice that F (0 , q ; 0 , ,
0) = 0 for every q ∈ Ω s because J ( U q ) = 0. Now we solve the equation F ( ε, q ; ν, ξ, α ) = 0 in a neighborhood of (0 , q ; 0 , , , L ) : C m × ( R × R ) → C m × ( R × R )given by L ( ϕ ; ζ, β ) := µ − J ′ ( U q ) ϕ − X j =1 ζ j τ j − ( β · γ ) ω , L ( ϕ ; ζ, β ) := L ( ϕ ) = (cid:16) ˆ R ϕ · τ µ dz , . . . , ˆ R ϕ · τ µ dz ; ˆ R γ ( ϕ · ω ) µ dz (cid:17) , so that L = ( L , L ) is the differential of F (0 , q ; · , · , · ) evaluated in ( ν, ξ, α ) = (0 , , L is injective we assume that L( ϕ, ζ, β ) = 0 and put v = µ − J ′ ( U q ) ϕ ∈ T U Z .
From (3.19) we find ˆ R | v | µ dz = ˆ R (cid:0) µ − J ′ ( U q ) ϕ ) · v µ dz = ˆ R J ′ ( U q ) ϕ · v dz = ˆ R J ′ ( U q ) v · ϕ dz = 0 , which implies J ′ ( U q ) ϕ = 0, that is, ϕ ∈ T U Z . On the other hand, ϕ ∈ T U Z ⊥ because L ( ϕ ) = 0. Thus ϕ = 0 and therefore also β = ζ = 0.To prove that L is surjective fix v ∈ C m and ( θ, b ) ∈ R × R . We have to find ϕ ∈ C m and ( ζ, β ) ∈ R × R such that L ( ϕ ; ζ, β ) = v and L ( ϕ ) = ( θ, b ). To this goal we introducethe minimal distance projection P ⊤ : L ( R , R ) → T U Z , w P ⊤ w , so that L ( w ) is uniquely determined by P ⊤ w , and vice-versa. We find ζ j and β so that X j =1 ζ j τ j + ( β · γ ) ω = − P ⊤ v . Then, we use Lemma 3.7 to find b ϕ ∈ C m ∩ T U Z ⊥∗ ∩ N ∗ such that J ′ ( U q ) b ϕ = ( v − P ⊤ v ) µ . ϕ ⊤ ∈ T U Z such that L ( ϕ ⊤ ) = ( θ, b ) − L ( b ϕ ).The triple ( ϕ ⊤ + b ϕ, ζ, β ) satisfies L( ϕ ⊤ + b ϕ, ζ, β ) = ( v ; θ, b ) and surjectivity is proved. Weare in the position to apply the implicit function theorem to F , for any fixed q ∈ Ω s .In fact, thanks to a standard compactness argument, we get that there exist ε ′ > C functions ν : ( − ε ′ , ε ′ ) × Ω s → U δ α : ( − ε ′ , ε ′ ) × Ω s → R ξ : ( − ε ′ , ε ′ ) × Ω s → R ν : ( ε, q ) ν εq α : ( ε, q ) α ε ( q ) ξ : ( ε, q ) ξ ε ( q )such that ν q ≡ , α ( q ) = 0 , ξ ( q ) = 0 , F ( ε, q ; ν εq , ξ ε ( q ) , α ε ( q )) = 0 . (4.7)By (4.7), the C function ( − ε ′ , ε ′ ) × Ω s → C m ( R , H ),( ε, q ) u εq := U q + ν εq = (cid:0) q U + q e + q e (cid:1) + ν εq , satisfies i ), if ε ′ is small enough. Further, using (4.5) (see also Lemma A.1) we rewrite thelast identity in (4.7) as E ′ ε ( u εq ) ϕ = ˆ R J ′ ε ( U q + ν εq ) · ϕ dz = X j =1 ξ εj ( q ) ˆ R τ j · ϕ µ dz + ˆ R ( α ε ( q ) · γ )( ω · ϕ ) µ dz ∀ ϕ ∈ C , ˆ R ν εq · τ j µ dz = 0 , ∀ j ∈ { , . . . , } , ˆ R γ ℓ ( ν εq · ω ) µ dz = 0 , ∀ ℓ ∈ { , , } . (4.8)In particular, claim ii ) holds true. Proof of iii ) . As a straightforward consequence of (4.8) we have that ˆ R ∂ q i ν εq · τ j µ dz = 0 , ˆ R γ ℓ ( ∂ q i ν εq · ω ) µ dz = 0 , hence E ′ ε ( u εq ) ∂ q i ν εq = 0 for any i = 1 , ,
3. We infer the identities ∂ q i E ε ( u εq ) = E ′ ε ( u εq )( e i + ∂ q i ν εq ) = E ′ ε ( u εq ) e i , i = 1 , ,∂ q E ε ( u εq ) = E ′ ε ( u εq )( U + ∂ q ν εq ) = E ′ ε ( u εq ) U . (4.9)23ow, from (2.4), (4.6) and (4.8) we find2 c e = τ − τ + k − γ ω , c e = τ + τ + k − γ ω , c U = kr k ( √ τ + k − γ ω ) ,E ′ ε ( u εq ) τ j = ξ εj ( q ) , E ′ ε ( u εq )( γ ℓ ω ) = ( k + 3 δ ℓ ) α εℓ ( q ) , for any j = 1 , . . . , ℓ = 1 , ,
3. Thus by (4.9) we get2 c ∇ q E ε ( u εq ) = M k ξ ε ( q ) + Θ k α ε ( q ) , (4.10)where M k and Θ k are constant matrixes, namely M k = − −
10 0 √ kr k , Θ k = k k
00 0 ( k + 3) r k . On the other hand, from (4.1) and using ∇ U q = r k q ∇ ω we obtain − q r k ξ εj ( q ) = E ′ ε ( u εq )( τ εj ( q )) , (4.11)where, in the spirit of (4.6), we have putted τ ε ( q ) := c ∂ x ν εq , τ ε ( q ) := c √ z ∇ ν εq , τ ε ( q ) := c z ∇ ν εq ,τ ε ( q ) := c ∂ y ν εq , τ ε ( q ) := c √ iz ∇ ν εq , τ ε ( q ) := c iz ∇ ν εq . Notice that ˆ R | τ εj ( q ) | µ dz ≤ ˆ R |∇ z ν εq | µ dz ≤ k ν εq k C ˆ R µ dz = o (1) , (4.12)as ε →
0, uniformly on Ω, see (2.3).For sake of clarity, we make now some explicit computation. We denote by σ ℓh theentries of the 3 × − k M k , and introduce the 6 × A ε ( q ) =( a εjh ( q )) j,h =1 ,..., , whose entries are given by a εjh ( q ) = ˆ R τ h · τ εj ( q ) µ dz − X ℓ =1 σ ℓh ˆ R γ ℓ ( ω · τ εj ( q )) µ dz . Since τ εj µ → L ( R , R ) by (4.12), then A ε → − ε ′ , ε ′ ) × Ω s . In particular, if ˆ ε ∈ (0 , ε ′ ) is small enough, then the determinant of the 6 × A ε ( q ) + q r k Id) is uniformly bounded away from 0 on [ − ˆ ε, ˆ ε ] × Ω.24ssume that ∇ q E ε ( u εq ε ) = 0 for some ε ∈ [ − ˆ ε, ˆ ε ], q ε ∈ Ω. From (4.10) we obtain α ε ( q ε ) = − Θ − k M k ξ ε ( q ε ). Thus (4.8) and (4.11) give − q ε r k ξ ε ( q ε ) = A ε ( q ε ) ξ ε ( q ε ) , and hence ξ ε ( q ε ) = 0, because the matrix ( A ε ( q ε ) + q ε r k Id) is invertible. But then (4.10)and ∇ q E ε ( u εq ε ) = 0 imply that α ε ( q ε ) = 0 as well, hence E ′ ( u εq ε ) = 0 by (4.8). Proof of iv ) . The function ( ε, q ) ν εq is of class C , and in particular ∂ ε ν εq is uniformlybounded in C for ( ε, q ) ∈ [ − ˆ ε, ˆ ε ] × Ω. Thus Taylor expansion formula for ε E ε ( u εq ) − E ε ( U q ) = E ( u εq ) − E ( U q ) + 2 ε (cid:0) V φ ( u εq ) − V φ ( U q ))gives E ε ( u εq ) − E ε ( U q ) = o ( ε ) as ε →
0, uniformly on Ω.Now we estimate ∇ q ( E ε ( u εq ) − E ε ( U q )). We use (4.2), (4.9) to obtain, for j = 1 , ∂ q j ( E ε ( u εq ) − E ε ( U q )) = (cid:0) E ′ ( u εq ) e j − E ′ ( U q ) e j (cid:1) + 2 ε (cid:0) V ′ φ ( u εq ) e j − V ′ φ ( U q ) e j (cid:1) = 2 ε (cid:0) V ′ φ ( u εq ) e j − V ′ φ ( U q ) e j (cid:1) = o ( ε ) , because k u εq − U q k C m = o (1) and V φ is a C -functional.To handle the derivative with respect to q we first argue as before to get ∂ q ( E ε ( u εq ) − E ε ( U q )) = (cid:0) E ′ ( u εq ) U − E ′ ( U q ) U (cid:1) + 2 ε (cid:0) V ′ φ ( u εq ) U − V ′ φ ( U q ) U (cid:1) = E ′ ( u εq ) U + o ( ε ) , uniformly on Ω. Next, from q U = u εq − ( q e + q e ) − ν εq and (4.2) we obtain q E ′ ( u εq ) U = E ′ ( u εq )( u εq − ( q e + q e ) − ν εq )= − E ′ ( u εq ) ν εq = − E ′ ε ( u εq ) ν εq + 2 εV ′ φ ( u εq ) ν εq = 2 εV ′ φ ( u εq ) ν εq because of (4.8). Since ν εq → C m we infer that E ′ ( u εq ) u εq = o ( ε ) uniformly on Ω as ε →
0, which concludes the proof. (cid:3)
Proof of Theorem 1.2.
Take an open set Ω ⋐ R containing the closure of A , let u εq bethe function given by Lemma 4.3 and notice that, by (4.4), E ε ( U q ) = E ( U q ) − εF φk ( q ).Thus for ε ∈ [ − ˆ ε, ˆ ε ] , ε = 0 we can estimate (cid:13)(cid:13)(cid:13) ε (cid:0) E ε ( u εq ) − E ( U q ) (cid:1) + F φk ( q ) (cid:13)(cid:13)(cid:13) C ( A ) = 12 | ε | k E ε ( u εq ) − E ε ( U q ) k C ( A ) = o (1) , iv ) in Lemma 4.3. Recalling the definition of stable critical pointpresented in Subsection 2.2, we infer that for any ε ≈ ε (cid:0) E ε ( u εq ) − E ( U q ) (cid:1) has a critical point q ε ∈ A , to which corresponds the embedded ( k + εφ )-bubble u ε := u εq ε by iii ) in Lemma 4.3. The continuity of ( ε, q ) u εq gives the continuity of ε u ε .The last conclusion in Theorem 1.2 follows via a simple compactness argument andthanks to Theorem 4.1. (cid:3) Proof of Theorem 1.3.
Recalling that q k := ( q , q , kr k q ), we write F φk ( q ) = ˆ B rk (0) ( p + kr k ) − φ ( q p + q k ) dp . Since r k → kr k = k ( k − − / → k → ∞ , we infer that q k → q uniformly oncompact sets of R and 34 πr k F φk → φ as k → ∞ , uniformly on Ω. Next, we easily compute ∂ q j F φk ( q ) = ˆ B rk (0) ( p + kr k ) − ∂ q j φ ( q p + q k ) dp , j = 1 , ,∂ q F φk ( q ) = ˆ B rk (0) ( p + kr k ) − ∇ φ ( q p + q k ) · ( p + kr k e ) dp , and thus we obtain, by the same argument,34 πr k ∇ F φk → ∇ φ as k → ∞ ,uniformly on Ω. It follows that for k large enough, F φk has a stable critical point in Ω ⋐ H ,since having a stable critical point is a C -open condition. Thus Theorem 1.1 applies andgives the conclusion of the proof. (cid:3) Appendix
Let K ∈ C ( H ). Take any vectorfield Q K ∈ C ( R , R ) such that div Q K ( p ) = p − K ( p )for any p ∈ R (here div = P j ∂ j is the Euclidean divergence). The functional V K ( u ) := ˆ R Q K ( u ) · ∂ x u ∧ ∂ y u dz , u ∈ C ( R , H ) , u , with respect to theweight K . In fact, if u parameterizes the boundary of a smooth open set Ω ⋐ R and if ∂ x u ∧ ∂ y u is inward-pointing, then the divergence theorem gives V K ( u ) = − ˆ ∂ Ω Q K ( u ) · ν du = − ˆ Ω p − K dp = − ˆ Ω K d H . Clearly, the functional V K does not depend on the choice of the vectorfield Q . Notice thatif K ≡ k is constant, then V k ( u ) = − k ˆ R u − e · ∂ x u ∧ ∂ y u dz , u ∈ C ( R , H ) . In the next Lemma we collect few simple remarks about the energy functional E ( u ) = 12 ˆ R u − |∇ u | dz + 2 V K ( u ) . (A.1) Lemma A.1
Let K ∈ C ( H ) . i ) The functional E : C ( R , H ) → R is of class C , and its differential is given by E ′ ( u ) ϕ = ˆ R ( u − ∇ u · ∇ ϕ − u − |∇ u | e · ϕ ) dz + 2 ˆ R u − K ( u ) ϕ · ∂ x u ∧ ∂ y u dz ; ii ) If u ∈ C ( R , H ) , then E ′ ( u ) extends to a continuous form on C ( R , R ) , namely E ′ ( u ) ϕ = ˆ R ( − div( u − ∇ u ) − u − |∇ u | e + 2 u − K ( u ) ∂ x u ∧ ∂ y u ) · ϕ dz ; iii ) If K ∈ C ( H ) , then E is of class C on C ( R , H ) . In the next Lemma we show that critical points for E are in fact hyperbolic K -bubbles. Lemma A.2
Let K ∈ C ( H ) and let u ∈ C ( R , H ) be a nonconstant critical point for E . Then u is conformal, that is, | ∂ x u | = | ∂ y u | , ∂ x u · ∂ y u = 0 , hence it parameterizes an S type surface in H , having mean curvature K , apart from afinite number of branch points. roof. Put α = u − ( | ∂ x u | − | ∂ y u | ), β = − u − ∂ x u · ∂ y u , ϕ = α + iβ and notice that | ϕ | ≤ c u |∇ u | ∈ L ∞ ( R ). By direct computation we find( ∂ x α − ∂ y β ) u = u ∂ x u · ∆ u − ( | ∂ x u | − | ∂ y u | ) ∂ x u − ∂ x u · ∂ y u ) ∂ y u , ( ∂ y α + ∂ x β ) u = − u ∂ y u · ∆ u − ( | ∂ x u | − | ∂ y u | ) ∂ y u + 2( ∂ x u · ∂ y u ) ∂ x u . (A.2)Since u solves (1.1), it holds that u ∂ x u · ∆ u = 2 G ( ∇ u ) · ∂ x u = 2( ∂ x u · ∂ y u ) ∂ y u + ( | ∂ x u | − | ∂ y u | ) ∂ x u ,u ∂ y u · ∆ u = 2 G ( ∇ u ) · ∂ y u = 2( ∂ x u · ∂ y u ) ∂ x u − ( | ∂ x u | − | ∂ y u | ) ∂ y u . (A.3)Putting together (A.2) and (A.3) we obtain ∂ x α − ∂ y β = ∂ y α + ∂ x β = 0, namely, ϕ is anholomorphic function. Since ϕ is bounded and vanishes at infinity then ϕ ≡ R , hence u is conformal.The last conclusion follows from Proposition 2.4 and Example 2.5(4) in [13]. (cid:3) Remark A.3
Here we take K ≡ k constant and point out two simple facts about theenergy functional E hyp in (1.3).By (4.2), the Nehari manifold contains any nonconstant function. Secondly, E hyp isunbounded from below. In fact, for t > we have E hyp ( ω + te ) = 12 ˆ R ( ω + t ) − µ dz + k ˆ R ( ω + t ) − ω µ dz = 4 π (cid:0) − kt − t − k t + 1 t − (cid:1) . Notice that ω + te approaches a horosphere as t → , and that lim t → E hyp ( ω + te ) = −∞ . Remark A.4
Differently from the Euclidean case, see for instance [5], the geometric andcompactness properties of the energy functional E are far from being understood (also inthe case of a constant curvature), and would deserve a careful analysis. We conclude the paper by pointing out a necessary condition for the existence of em-bedded K bubbles.Let K ∈ C ( H ) be given, and let u ∈ C ( R , H ) be an embedded solution to (1.1). ByLemma A.2, u is a conformal parametrization of the open set Ω ⊂ R , which is the boundedconnected component of R \ u ( S ). We can assume that the nowhere vanishing normalvector ∂ x u ∧ ∂ y u is inward pointing. Since u is a critical point of the energy functional in(A.1), then for j = 1 , E ′ ( u ) e j = V ′ K ( u ) e j = ˆ R u − K ( u ) e j · ∂ x u ∧ ∂ y u dz = − ˆ Ω div( p − K ( p ) e j ) dp
28y the divergence theorem. Thus ˆ Ω p − ∂ p j K ( p ) dp = 0 . In a similar way, from E ′ ( u ) u = 0 and since div( p − K ( p ) p ) = p − ∇ K ( p ) · p , one gets ˆ Ω p − ∇ K ( p ) · p dp = 0 . In particular, ∂ p K, ∂ p K and the radial derivative of K can not have constant sign in Ω.We infer the next nonexistence result (see [7, Proposition 4.1] for the Euclidean case). Theorem A.5
Assume that K ∈ C ( H ) satisfies one of the following conditions, i ) K ( p ) = f ( ν · p ) for some direction ν orthogonal e , where f is strictly monotone; ii ) K ( p ) = f ( | p | ) , where f is strictly monotone.Then (1.1) has no embedded solution u ∈ C ( R , H ) . Acknowledgements . This work is partially supported by PRID-DMIF Projects PRIDENand VAPROGE, Universit`a di Udine.
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Gabriele Cora , Dipartimento di Scienze Matematiche, Informatiche e Fisiche, Universit`a di UdineEmail: [email protected].
Roberta Musina , Dipartimento di Scienze Matematiche, Informatiche e Fisiche, Universit`a di UdineEmail: [email protected]., Dipartimento di Scienze Matematiche, Informatiche e Fisiche, Universit`a di UdineEmail: [email protected].