Entire Spacelike Radial Graphs in the Minkowski Space, Asymptotic to the Light-Cone, With Prescribed Scalar Curvature

www.elsevier.com/locate/anihpc

Entire spacelike radial graphs in the Minkowski space, asymptotic to the light-cone, with prescribed scalar curvature

Pierre Bayard

, Philippe Delanoë

aInstituto de Física y Matemáticas, U.M.S.N.H. Ciudad Universitaria, CP. 58040 Morelia, Michoacán, Mexico bUniversité de Nice-Sophia Antipolis, Laboratoire J.-A. Dieudonné, Parc Valrose, 06108 Nice Cedex 2, France

Received 4 July 2007; accepted 10 March 2008 Available online 27 April 2008

Abstract

We prove the existence and uniqueness inR^n,1of entire spacelike hypersurfaces contained in the future of the originOand asymptotic to the light-cone, with scalar curvature prescribed at their generic pointM as a negative function of the unit vector pointing in the direction of^−−→OM, divided by the square of the norm of^−−→OM (a dilation invariant problem). The solutions are seeked as graphs over the future unit-hyperboloid emanating fromO(the hyperbolic space); radial upper and lower solutions are constructed which, relying on a previous result in the Cartesian setting, imply their existence.

©2008 Elsevier Masson SAS. All rights reserved.

Résumé

On prouve l’existence et l’unicité dansR^n,1d’hypersurfaces entières de genre espace contenues dans le futur de l’origineOet asymptotes au cône de lumière, dont la courbure scalaire est prescrite au point génériqueMcomme fonction négative du vecteur unité pointant en direction de^−−→OM, divisée par le carré de la norme du vecteur^−−→OM (un problème invariant par homothétie).

Les solutions sont cherchées comme graphes sur l’hyperboloïde-unité futur émanant deO(l’espace hyperbolique) ; des solutions supérieure et inférieure radiales sont construites qui, d’après un résultat antérieur en cartésien, impliquent l’existence de telles solutions.

©2008 Elsevier Masson SAS. All rights reserved.

MSC:53C40; 35J65; 34C11

Keywords:Radial graphs; Spacelike; Entire solution; Prescribed scalar curvature; Upper and lower barriers

0. Introduction

The Minkowski spaceR^n,1is the affine Lorentzian manifoldRⁿ×Rendowed with the metric ds²=dX²−dX²_n+1, wheredX²=dX²₁+ · · · +dX_n²,

✩ The first author was supported by the project DGAPA-UNAM IN 101507; the second author is supported by the CNRS.

* Corresponding author.

E-mail addresses:bayard@ifm.umich.mx (P. Bayard), philippe.delanoe@unice.fr (P. Delanoë).

0294-1449/$ – see front matter ©2008 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2008.03.008

settingX=(X, X_n+1)∈Rⁿ×R, and time-oriented bydX_n+1>0. Distinguishing the originOofR^n,1,let H=

x∈R^n,1, |^−−→Ox|²= |x|²−x_n+1²= −1, xn+1>0 ,

be the future unit-hyperboloid, model of the hyperbolic space inR^n,1.Ifϕis a real function defined onH,we define theradial graphofϕby

graph_Hϕ=

X∈R^n,1, ^−−→OX=e^{ϕ(x)−−→}Ox, x∈H . This is a hypersurface contained in the future open solid cone

C⁺=

X∈R^n,1, Xn+1>|X| .

We say that ϕ is spacelike if its graph is a spacelike hypersurface, which means that the metric induced on it is Riemannian. Conversely, a spacelike and connected hypersurface in C⁺ is the radial graph of a uniquely deter- mined functionϕ:H→R.Of course, such a graph may also be considered as the Cartesian graph of some function u:Rⁿ→R

graph_Rnu=

x, u(x)

, x∈ Rⁿ ,

and the correspondence between the two representations is bijective passing from the Cartesian chartX=(X, X_n+1) restricted toC⁺, to the polar chart(x, ρ)∈H×(0,∞)ofC⁺defined by:

ρ=

−|^−−→OX|², ^−−→Ox= 1 ρ

−−→OX.

Recall that the principal curvatures(κ₁, . . . , κ_n)at a point of a spacelike hypersurface are the eigenvalues of its shape endomorphismdN,whereN is the future oriented unit normal field, and themth mean curvature (denoted byH_m) is themth elementary symmetric function of its principal curvatures:H_m=σ_m(κ₁, . . . , κ_n).For each realλ >0, the coneC⁺is globally invariant under the ambient dilationX→λXofR^n,1and the abovemth mean curvature is(−m)- homogeneous; specifically, it transforms likeH_m(λX)=λ^−mH_m(X).It is thus natural to pose, as in [6, Theorem 1], the following inverse problem forH_m: given a positive functionh >0 onHtending to 1 at infinity, find a spacelike hypersurfaceΣinC⁺, asymptotic to∂C⁺at infinity, such that, for each pointX∈Σ, themth mean curvature ofΣ atXis given by:

H_m:= 1 _n

m

H_m(X)= 1 (−|^−−→OX|²)^m2

h(x) ^m, with^−−→Ox=

−−→OX

−|^−−→OX|². (1) By construction, this problem is dilation invariant; moreover, as explained below, the positivity ofhmakes it elliptic.

Actually, introducing the positivity cone [9] ofσm: Γ_m=

κ∈Rⁿ, ∀i=1, . . . , m, σi(κ) >0 , and recalling McLaurin’s inequalities (satisfied onΓ_m):

0< (H_m)^m1 (H_m−1)^m1−1 · · ·H₂

1 2 H₁,

we note that, if a hypersurfaceΣ=graph_Rnusolves (1) with the asymptotic condition, then the time-functionumust assume a minimum onΣand, as readily checked (using e.g. [3, p. 245]), the principal curvatures ofΣat such a min- imum point ofumust lie inΓ_m. Now Eq. (1) combined with McLaurin’s inequalities forces the principal curvatures of Σ to stay inΓ_m everywhere. Let us call any spacelike hypersurface ofC⁺ having this property,m-admissible;

accordingly, a function ϕ:H→R(resp.u:Rⁿ→R) is calledm-admissible, provided graph_Hϕ (resp. graph_Rⁿu) is so. The condition ofm-admissibility is local (and open); one may thus speak of a functionϕ:H→Rbeingm- admissibleat a point(hence nearby) whenever graph_Hϕ is so at that point. We will seek the solution hypersurface Σ as the radial graph of somem-admissible functionϕ:H→Rvanishing at infinity (to comply with the asymptotic condition). Eq. (1) then reads

F_m(ϕ)=h, (2)

with the radial operatorF_mdefined by:

F_m(ϕ)=e^ϕH_m(X)

1

m, X∈graph_Hϕ.

For briefness, we will not compute here explicitly the general expression of the operatorF_m(keeping it for a further study)—its restriction to radial functions will suffice (see Section 3.3 below). We will rely instead on the well-known corresponding Cartesian expression (see e.g. [2]) combined with a few basic properties ofF_mrecorded in the next section (and proved with elementary arguments).

Furthermore, we will essentially restrict to the case m=2 (and freely say ‘admissible’, for short, instead of

‘2-admissible’). SinceH₂ is related to the scalar curvatureS byS= −2H2, our present study is really about the prescription of the scalar curvature, at a generic pointX of a radial graph, as a negative function ofx∈H(withx given as in (1)) divided by the square of the norm of^−−→OX.Aside from the originOof the ambient spaceR^n,1, we will distinguish a pointoinHand setr=r(x)for the hyperbolic distance fromotox∈H; accordingly, a function onH will be calledradialwhenever it factors through a function ofronly. Our main result is the following:

Theorem 1.Forα∈(0,1), leth:H→(0,∞)be a function of classC^2,αwith

r(x)lim→+∞h(x)=1.

Assume that the functionsh⁻andh⁺defined onR⁺by h⁻(r)= sup

r(x)=r

h(x) and h⁺(r)= inf

r(x)=rh(x) satisfy

+∞

0

h⁻−1

+dr <+∞,

+∞

0

1−h⁺

+dr <+∞,

where(h⁻−1)₊(resp.(1−h⁺)₊)means the positive part ofh⁻−1 (resp.1−h⁺). Then the equation

F2(ϕ)=h (3)

has a unique admissible solution of classC^4,αsuch thatlimr(x)→+∞ϕ(x)=0.

Remark 1.From Lemma 4 below, anytime the functionhis radial, the integral convergence conditions of Theorem 1 appears necessary for the existence of bounded solutions.

An analogous problem in the Euclidean setting is solved for the Gauss curvature in [6, Théorème 1], and in [13,5]

some related problems are studied. In the Lorentzian setting, the prescription of the mean curvature for entire graphs is studied in [1] and that of the Gauss curvature in [11,8,4]. In [3], the scalar curvature is prescribed in Cartesian coordinatesx_n+1=u(x₁, . . . , x_n).

The outline of the paper is as follows. In Section 1, we prove that there exists at most one solution vanishing at infinity for Eq. (2) withm∈ {1, . . . , n}. In Section 2, relying on [3], we prove the existence of a solution whenm=2, provided upper and lower barriers are known. The latter are constructed, as radial functions, in Section 3.

1. Uniqueness

We first require a few basic properties of the operatorF_m. It is a non-linear second order scalar differential operator defined onm-admissible real functions onH. The dilation invariance of (1) implies the identity:

Fm(ψ+c)≡Fm(ψ ), (4)

for everym-admissible functionψ:H→Rand constantc; linearizing atψyields dFm(ψ )(1)≡0.

Furthermore, we have:

Lemma 1.For eachm-admissible functionψ, the linear differential operatordF_m(ψ )is elliptic everywhere onH, with positive-definite symbol.

Summarizing for later use, the expression ofdF_m(ψ ), in the chartx∈RⁿofH, at a fixedm-admissible functionψ reads like:

δψ→dF_m(ψ )(δψ )=

1i,jn

B_ij ∂²

∂x_i∂x_j(δψ )+ n

i=1

B_i ∂

∂x_i(δψ ), (5)

with then×nmatrix(Bij)symmetric positive definite (depending onψ, of course, like theBi’s). We now proceed to proving Lemma 1.

Proof. We require the Cartesian operatorv→G_m(v):=F_m(ψ )defined onm-admissible functionsv:Rⁿ→Rby:

graph_Rnv=graph_Hψ. (6)

The ellipticity ofdG_m(v)and the positive-definiteness of its symbol are well-known [10,14,2]. Its expression thus starts out like

dG_m(v)(δv)=

1i,jn

A_ij ∂²

∂Xi∂Xj

(δv)+ lower order terms,

with the matrix(A_ij)symmetric positive definite. Them-admissible functionψ onHsuch that (6) holds, is related tov, in the chartx=(x₁, . . . , x_n)∈Rⁿ, by:

v(X)=

1+ |x|²exp

ψ (x) , with^−−−→OX=e^{ψ(x)−−→}Ox.

Varyingψbyδψthus yields for the corresponding variationδvofvthe following expression:δv(X)=w(X)δψ (x), with

w(X)=

v− n

i=1

X_i ∂v

∂X_i

(X).

Since the graph lies inC⁺and it is spacelike, we havev(X) >|X|and (using Schwarz inequality) n

i=1

X_i ∂v

∂X_i <|X|,

thereforew >0. Moreover, up to lower order terms, we have:

∂²

∂X_i∂X_j(δv)(X)=w(X)

1i,jn

∂²

∂x_k∂x_l(δψ )(x)∂x_k

∂X_i

∂x_l

∂X_j withx_k =√ ^Xk

v²(X)−|X|². We thus find in (5):

B_kl=w(X)

1i,jn

A_ij∂x_k

∂X_i

∂x_l

∂X_j

and the ellipticity ofδψ→dFm(ψ )(δψ )follows. 2

We need also a more specific (ellipticity) property of the operatorFm, namely:

Lemma 2.For each couple(ϕ₀, ϕ₁)ofm-admissible real functions onHand each pointx₀∈Hwhereϕ=ϕ₁−ϕ₀ assumes a local extremum, the whole segmentt ∈ [0,1] →ϕ_t =ϕ₀+t ϕ consists ofm-admissible functions at the pointx₀.

Proof. The analogue of Lemma 2 is fairly standard in the Cartesian setting, using the expression of the operatorG_m introduced in the proof of Lemma 1 (see [2]) together with the well-known fact:∀κ∈Γ_m,∀i∈ {1, . . . , n},^∂σ_∂κ^m_i(κ) >0.

Here, we will simply reduce the proof to that setting (and let the reader complete the argument). Let us first normalize

the situation at an extremum pointx₀∈Hofϕ. From (4), we may assumeϕ(x₀)=0. Moreover, we may assume that ϕ has a local minimum atx₀(if not, exchangeϕ₀andϕ₁). Finally, setting graph_Hϕ_a=graph_Rnu_afora=0,1, and performing if necessary a suitable Lorentz transform (hyperbolic rotation), we may takex0=(0,1)∈Rⁿ×Rthus withua(0)=1. Fort∈ [0,1]and nearx0, setΣt =graph_Rⁿut for the hypersurface graph_Hϕt. We must prove that Σ_t ism-admissible atx₀. ForX_t ∈R^n,1lying inΣ_t, we have:^−−−→OX_t=e^{t ϕ(x)−−−→}OX₀with^−−→Ox=^−−−→OX₀/

−|^−−−→OX₀|². In the Cartesian setting, we thus have (sticking to theRⁿ-valued charts used in the preceding proof):

u_t X_t

=e^{t ϕ(x)}u₀

e^{−t ϕ(x)}X_t , here withx=X₀/

u²₀(X₀)− |X₀|², X_t=e^{t ϕ(x)}X₀, and(X₀, u0(X₀))∈graph_Rⁿu0; moreover, the lemma boils down to proving thatu_tism-admissible atX_t=0. A routine calculation yields atX_t=0 the equalities:

∂u_t

∂X_{t i} (0)= ∂u₀

∂X_0i(0), ∂²u_t

∂X_{t i}∂X_tj (0)= ∂²u₀

∂X_0i∂X_0j(0)+t ∂²ϕ

∂x_i∂x_j(0),

where, in the second one, the matrix [∂²ϕ/∂x_i∂x_j(0)]1i,jn is non-negative. We readily infer [2] that, for each t∈ [0,1], the principal curvaturesκ1t· · ·κnt of the hypersurfaceΣt atx0(each repeated according to its multi- plicity) satisfy:∀i∈ {1, . . . , n}, κ_it κ_i0. The latter implies that then-tuple(κ_1t, . . . , κ_nt)lies in the coneΓ_m, since (κ₁₀, . . . , κ_n0)∈Γ_m. 2

Theorem 2.The operatorF_mis one-to-one onm-admissible functions of classC²vanishing at infinity.

Proof. Let us argue by contradiction. Letϕ0, ϕ1be twom-admissibleC²functions vanishing at infinity and having the same image byF_m. Fort∈ [0,1], setϕ_t =ϕ₀+t ϕwithϕ=ϕ₁−ϕ₀. Sinceϕ vanishes at infinity, ifϕ≡0, it assumes a non-zero local extremum (a maximum, say, with no loss of generality) at some pointx₀∈H. By Lemma 2, the whole segment t∈ [0,1] →ϕ_t ism-admissible in a neighborhood Ω of x₀ whereϕ thus satisfies the second order linear equationLϕ=0 with L1=0 and the operatorL given byL=₁

0 dF_m(ϕ_t) dt. Combining Lemma 1 above with Hopf’s strong Maximum Principle (see [7]), we getϕ≡ϕ(x₀)throughoutΩ. By connectedness, we infer ϕ≡ϕ(x0)=0 on the whole ofH, contradicting limr(x)→+∞ϕ=0. So, indeed, we must haveϕ≡0, in other words Fmis one-to-one. 2

2. Existence of a solution reduced to that of upper and lower solutions

Theorem 3.Leth:H→Rbe a function of classC^2,α,for someα∈(0,1), such that there existsϕ⁻∈C^4,α(H)with graph_Hϕ⁻strictly convex and spacelike, andϕ⁺∈C²(H)withgraph_Hϕ⁺spacelike, satisfying

F₂(ϕ⁻)h, F₂(ϕ⁺)h and lim

r(x)→+∞ϕ^±=0.

Then the equation F₂(ϕ)=h

has a unique admissible solution of classC^4,αsuch thatlimr(x)→+∞ϕ(x)=0.Moreoverϕsatisfies the pinching:

ϕ⁻ϕϕ⁺.

Remark 2.Sinceϕ is a bounded function, the hypersurfaceM=graph_H(ϕ)is entire. More precisely, denoting by ϕ_minandϕ_maxtwo constants such thatϕ_minϕϕ_max,the functionu:Rⁿ→Rsuch that graph_Rn(u)=graph_H(ϕ) satisfiesuminuumaxwhereumin(resp.umax) is such that graph_Rn(umin)=graph_H(ϕmin)(resp. graph_Rn(umax)= graph_H(ϕmax)).Noting that the graphs ofuminandumaxare hyperboloids, we see that the inequalityuuminimplies thatMis entire, and the inequalityuu_maximplies thatMis asymptotic to the lightcone.

Proof. The asserted uniqueness follows from Theorem 2; so let us focus on the existence part. A straightfor- ward comparison principle, using (5) and Lemma 2, implies ϕ⁻ϕ⁺ on H.Let u⁻,u⁺:Rⁿ→Rbe such that graph_Rn(u^±)=graph_H(ϕ^±).SetH for the function onR^n,1defined by:

H (X)= (ⁿ₂)

|Xn+1|²− |X|²

h

X |X_n+1|²− |X|²

2

. (7)

The spacelike functionsu⁻andu⁺satisfy:

H2[u⁻]H (·, u⁻), H2[u⁺]H (·, u⁺), u⁻u⁺ and lim

|x|→∞

u^±(x)− |x| =0,

whereH₂[u^±]stands for the second mean curvature of the graph ofu^±.Theorem 1.1 in [3] asserts the existence of a functionu:Rⁿ→R,belonging toC^4,α,spacelike, such thatH₂[u] =H (·, u)inRⁿ,lim_|x|→+∞u(x)− |x| =0, andu⁻uu⁺. The functionϕ:H→Rsuch that graph_H(ϕ)=graph_Rn(u) is a solution of our original prob- lem. 2

3. Construction of radial upper and lower solutions

In the sequel of the paper, we first solve the Dirichlet problem on a bounded set inH(Section 3.1) then proceed to proving the existence and uniqueness of an entire solution in the radial case and study its properties (Sections 3.2 and 3.3); finally, we construct the required radial barriers (Section 3.4).

3.1. The Dirichlet problem

Theorem 4.Givenα∈(0,1), letΩ be a uniformly convex bounded open subset ofHwithC^2,αboundary,h:Ω→R be a positive function of class C^2,α,and ϕ₀:Ω →Rbe a spacelike function of classC^2,α whose radial graph is strictly convex. Then the Dirichlet problem

F₂(ϕ)=h inΩ, ϕ=ϕ₀ on∂Ω, (8)

has a unique admissible solution of classC^4,α.

Proof. We first prove uniqueness, by contradiction: letϕ₀, ϕ₁be two admissible solutions of (8), and, fort∈ [0,1],set ϕ_t=ϕ₀+t ϕwithϕ=ϕ₁−ϕ₀.Sinceϕvanishes on∂Ω,ifϕ≡0,it assumes a non-zero local extremum. Following the arguments of the proof of Theorem 2 we obtain a contradiction with the Hopf’s strong Maximum Principle. Let us focus now on the existence part. Settingx=(x,

1+ |x|²)∈Rⁿ×R,and Ω=

e^ϕ0(x)x, x∈Ω

, u0

e^ϕ0(x)x

=e^ϕ0(x)

1+ |x|², problem (8) is equivalent to the Dirichlet problem:

H₂[u] =H (·, u) inΩ, u=u₀ on∂Ω, (9)

whereH₂is the scalar curvature operator acting on spacelike graphs defined onΩ⊂Rⁿ,andHis defined onΩ×R by (7).

Let us consider the Banach space E=

v∈C^2,α(Ω), v=0 on∂Ω , and the open convex subset ofE

U=

v∈E, sup

Ω

D(v+u₀)<1

.

We first note that for everyv∈U, graph_Rn(v+u₀)belongs to the dependence setKof graph_Rnu₀.Here, by definition, X∈R^n,1belongs toKif for everyξ∈R^n,1withξ, ξ0 andξ=0,the rayX+R.ξ meets graph_Rnu₀.The setK is a compact subset of the open coneC⁺.

For each(v, t )∈U× [0,1], we know from [2,15] that the Dirichlet problem

H2[u] =t H (·, v+u0)+(1−t )H2[u0] inΩ, u=u0 on∂Ω (10) has a unique admissible solution (belonging toC^4,α). We define the map

T : [0,1] ×U→E, (t, v)→u

whereuis such thatu=u+u₀is the admissible solution of (10).

For eacht ∈ [0,1]the fixed points of T (t,·)are under control: indeed, suppose T (t, u)=u,then the function u=u+u₀solves the Dirichlet problem

H2[u] = ˜H (·, u) inΩ, u=u0 on∂Ω (11) where

˜

H (·, u)=t H (·, u)+(1−t )H₂[u₀]. (12)

The followinga prioriestimates are carried out in [3, p.251]: there existϑ∈(0,1)andC >0 such that sup

Ω

|Du|<1−ϑ and u_2,α,Ω< C. (13)

The constantsϑ, Conly depend on diam(Ω),infKH ,˜ ˜H2,K,u_04,Ω,and on a positive lower bound on the min- imum eigenvalue ofD²u₀onΩ.The expression ofH˜ implies that they are independent of the parametert∈ [0,1].

In order to prove that T (1,·)has a fixed point, we now consider the (nonempty) convex subset of the Banach spaceE:

Uϑ,C=

v∈U, D(v+u0)<1−ϑandv+u0_2,α,Ω< C , and the mapT : [0,1] ×U_ϑ,C→E.Then the following properties hold:

(i) T is continuous with compact image due to the above estimates on the solutions of the Dirichlet problem (10);

(ii) T (0,·)≡0 by definition;

(iii) for everyt∈ [0,1], T (t,·)does not have any fixed point on∂Uϑ,C,since each fixed point ofT (t,·)belongs to Uϑ,Cby the definitions ofϑandC.

An elementary version of the Leray–Schauder theorem (due to Browder and Potter [12]) implies thatT (1,·)has a fixed point, which proves that (8) has a solution. 2

3.2. Existence and uniqueness of entire radial solutions

The aim of this section is to prove the following result:

Theorem 5.Forα∈(0,1), leth:R⁺→Rbe a positive function of classC^2,α constant on some neighborhood of0 and letϕ0be a real number. Recallr=r(x)denotes the hyperbolic distance ofx∈Hfrom a fixed origino∈H. The problem:

F₂(ϕ)(x)=h(r) for allx∈H, ϕ(o)=ϕ₀, (14)

admits a unique admissible radial solutionϕ:H→Rof classC^4,α.

Proof. Existence: letBi denote the ball inHwith centeroand radiusi∈N^∗, andϕibe the admissible solution of the Dirichlet problem:

F₂(ϕ)=h, ϕ_|∂Bi=0, (15)

given by Theorem 4. By radial symmetry and uniqueness,ϕ_i is a radial function:ϕ_i(x)=f_i(r)for some function f_i: [0, i] →R. By uniqueness again, forj > i, the functionϕ_j−ϕ_imust be constant onB_i. Thereforef_j(r)≡f_i(r) forr∈ [0, i].We may thus definegonR⁺byg=f_ion each[0, i].Now the functionϕdefined by

ϕ(x)=ϕ₀+ r

0

g(u) du

is a radial solution of (14).

Uniqueness: assume thatϕ1andϕ2are admissible radial solutions of (14):ϕ1(x)=f1(r),ϕ2(x)=f2(r)where f1, f2are functionsR⁺→R.For each realR >0, set

ϕ_1,R(x)= − R

r

f₁(u) du and ϕ_2,R(x)= − R

r

f₂(u) du.

The functionsϕ_1,R andϕ_2,Rare both admissible solutions of the Dirichlet problem (15) onB_R.As such, they must coincide onB_R,hencef₁=f₂on[0, R],which implies the desired result. 2

3.3. Properties of the radial solutions

The following lemma describes the monotonicity of a solutionϕof Eq. (14) depending on the sign ofh−1:

Lemma 3.Leth:R⁺→Randϕ:H→Rbe as in Theorem5, and letf :R⁺→Rbe such that ϕ(x)=f[r(x)],

∀x∈H.

(i) Ifh1,thenf is non-increasing;in particular, ifϕ0=0, the functionϕis non-positive.

(ii) Ifh1,thenf is non-decreasing;in particular, ifϕ0=0, the functionϕis non-negative.

Proof. Here, we need to calculate explicitly the expression of Eq. (14) in the radial case. Sete₁, . . . , e_n+1,for the standard orthonormal basis of the vector spaceR^n,1.Fixx∈Hand take, with no loss of generality,

o=en+1=(0, . . . ,0,1), x=(sinhr,0, . . . ,0,coshr)

withr, the hyperbolic distance betweenoandx.Consider the orthonormal basis ofTxHdefined by:

∂_r=coshre1+sinhre_n+1, and ∂_ϑ=e_ϑ, ϑ=2, . . . , n,

and the vectors, tangent toM=graph_Hϕate^ϕ(x)x, induced by the embeddingx∈H→e^ϕ(x)x∈M, given by:

u_r=e^f(fx+∂_r), u_ϑ=e^f∂_ϑ, ϑ=2, . . . , n.

The future oriented unit normal toMate^ϕ(x)xis the vector:

N (r)= f

1−f²

∂_r+ 1

1−f²

x. (16)

LetSbe the shape endomorphism ofMate^ϕ(x)x,with respect to the future unit normalN (r).Using the formulas D_∂r∂_r(x)=x, D_∂ϑ∂_r(x)= 1

tanhr∂_ϑ

whereDdenotes the canonical flat connection ofR^n,1and∂_r the unit radial vector field ofHwith respect to the point o,we readily get:

S(ur)=dN (∂r)= e^−f

1−f² f

1−f²+1

ur,

and, forϑ=2, . . . , n,

S(uϑ)=dN (∂ϑ)= e^−f

1−f² f

tanhr+1

uϑ.

The principal curvatures ofMatr >0 are thus equal to:

e^−f

1−f² f

1−f²+1

(simple), e^−f

1−f² f

tanhr +1

(multiplicityn−1).

Settings=s(r)for the hyperbolic distance fromotoN (r),we infer from (16):

s(r)=r+Argth(f). (17)

In terms of the new radial unknowns(r), forr >0, the principal curvatures read

e^−fcosh(r−s)s, e^−fsinhs

sinhr, . . . , e^−fsinhs sinhr

, (18)

and the equationF2(ϕ)=hreads

2scosh(r−s)sinhrsinhs=nh²sinh²r−(n−2)sinh²s. (19)

We now prove the first statement of the lemma. Sincef=tanh(s−r),we must prove:sron[0,+∞).Suppose firsth <1.Sinces(0)=0 ands(0)=h(0) <1 (from (19)), there existsr0>0 such thatsron[0, r0].Moreover, we get from (19):

s 1

2 cosh(r−s)

nsinhr

sinhs −(n−2)sinhs sinhr

.

We observe that the functions(r)=ris a solution of the ODE:

s= 1

2 cosh(r−s)

nsinhr

sinhs −(n−2)sinhs sinhr

on[r0,+∞).So the comparison theorem for solutions of ordinary differential equations impliessron[r0,+∞).

Suppose onlyh1,fixA >0 and considerh_δ=h−δ,whereδ is some small positive constant such thath_δ>0 on[0, A]. Denoting byϕδandsδ the corresponding solutions of (14) and (19) on the ball of radiusA,the function sδ−ris non-positive; we now prove thatsδ−rconverges uniformly tos−rasδtends to zero, which will yield the desired result. SetBAfor the ball of radiusAinHcentered atoandU= {ψ∈ C^2,α(BA), ψ+ϕis admissible inBA, ψ_|∂B_A=0}; consider the auxiliary map:

Φ:ψ∈U→Φ(ψ ):=F₂(ψ+ϕ)∈C^α(B_A).

Since Φ(0)=h and since, classically [7] (recalling (5)), the linearized map dΦ(0) is an isomorphism from {ξ ∈ C^2,α(B_A), ξ_|∂BA=0} to C^α(B_A), the inverse function theorem implies: ∀ε >0,∃δ₀>0,∀δ∈(0, δ0), the solutionψ_δ∈U ofF₂(ψ_δ+ϕ)=h_δ satisfies|ψ_δ|2,αε.Sinceϕ_δ=ψ_δ+ϕ−ψ_δ(o),we obtain|ϕ_δ−ϕ|2,α2ε, which implies the convergence ofϕ_δtoϕinC¹and thus the uniform convergence ofs_δtos.

The proof of statement (ii) is analogous and thus omitted. 2

Our next lemma provides a simple necessary and sufficient condition for an entire radial solution to be bounded.

Lemma 4.Leth:R⁺→Randϕ:H→Rbe as in Theorem5.

(i) Assumeh1,andlimr→∞h=1.Then

r(x)lim→+∞ϕ(x) >−∞ if and only if +∞

0

(1−h) drconverges.

(ii) Assumeh1,andlimr→∞h=1.Then

r(x)lim→+∞ϕ(x) <+∞ if and only if

+∞

0

(h−1) drconverges.

Proof. Let us prove statement (i), thus assumingh1,with limr→∞h=1.We stick to the notations used in the proof of Lemma 3. From (17), we get at once:

ϕ(x)=ϕ₀− r(x)

0

tanh

u−s(u)

du. (20)

Statement (i) amounts to proving that _+∞

0 tanh(u−s(u)) duconverges if and only if so does_+∞

0 (1−h) dr. We split the proof of this fact into five steps.

Step 1.The solutionsof (19) is an increasing function.

Let us consider in the(r, s)plane the curveCwith equation:

nh²sinh²r=(n−2)sinh²s, r, s0.

The slope of its tangent at(0,0)is

n

n−2h(0).Since the solutionssatisfiess(0)=0 ands(0)=h(0),we infer that the graph of s stays under the curveC near 0. Noting that the following vector field, associated to the differential equation (19):

(r, s)→

2 cosh(r−s)sinhrsinhs, nh²sinh²r−(n−2)sinh²s ,

is horizontal onC, and that the heightsof the curveC is increasing withr, we conclude that the solutionsof (19) remains trapped belowC.In other wordsnh²sinh²r(n−2)sinh²sfor allr,and (19) implies:s0.

Step 2.r−shas a limit at+∞.

By contradiction, assume lim inf(r−s) <lim sup(r−s)=δ.Thus there exists a sequencer_k→ +∞such that rk−s(rk)→δands(rk)=1.Denotings(rk)bysk,we get from Eq. (19):

1= 1

2 cosh(rk−s_k)

nh²(rk)sinhr_k

sinhs_k −(n−2)sinhs_k sinhr_k

. (21)

We distinguish two cases:

First case:δ <+∞.We then haves_k→ +∞, ^sinhr_sinhs^k

k ∼e^rk−sk ∼e^δ and ^sinhs_sinhr^k

k ∼e^sk−rk∼e^−δ asktends to infinity (here and below, the equivalence∼between two quantities means that their quotient has limit 1). So (21) yields

1= 1

2 coshδ

ne^δ−(n−2)e^−δ .

Usinge^δe^−δwe get 1_cosh^eδ_δ,which is absurd.

Second case:δ= +∞.First assuming thats_k is not bounded, and sincesis an increasing function (Step 1), we have:

s_k→ +∞, ^sinh_sinh^r_s^k

k ∼e^rk−sk→ +∞and ^sinhs_sinhr^k

k ∼e^sk−rk→0 asktends to infinity. Eq. (21) yields

1∼ n

2 cosh(rk−s_k)e^rk−sk,

which is absurd since cosh(rk−s_k)∼^erk−sk₂ .If we now assumes_k bounded, sincesis an increasing function with s(0) >0, we get thatskconverges tol >0,and, since^sinh_sinh^s_r^k

k →0,we obtain from (21):

1∼ n

2 cosh(rk−s_k) sinhr_k

sinhl ,

with sinhr_k∼^e₂^rk,cosh(rk−s_k)∼^erk−sk₂ ∼^e−₂^le^rk;so 1=ⁿ_2sinh^el _l,which is absurd.

Step 3.r−stends to 0 at infinity.

Having proved thatr−sconverges, let us setδ=limr→+∞r−sand prove by contradiction thatδ=0.There are two cases:

First case:0< δ <+∞.We gets→ +∞,hence ^sinh_sinh^r_s ∼e^r−s∼e^δ, ^sinhs_sinhr ∼e^s−r ∼e^−δas rtends to infinity, and thus, from (19):

s→ 1 2 coshδ

ne^δ−(n−2)e^−δ .

The latter expression is larger than 1, which contradictsrs.

Second case:δ= +∞.We first note that ^sinh_sinhr^s →0 (ifsis bounded this is trivial; ifsis not bounded,s→ +∞since s is increasing, and we have ^sinhs_sinhr ∼e^s−r →0 sincer−s→ +∞). Moreover we have lim infnh^{2 sinh}_sinhs^r n since rs.We thus infer from Eq. (19):

s∼ n

2 cosh(r−s) sinhr sinhs.

Assumings→ +∞,we get ^sinhr_sinhs ∼e^r−s and cosh(r−s)∼^er₂^−s,hences∼n,which is impossible sincesr.

Finally, assumingsbounded yieldss→l >0;sincer−s→ +∞,we infer cosh(r−s)∼^er₂^−s and sinhr∼^e₂^r, hence from (19),e^−ss∼ⁿ_2sinhl¹ and thuss∼ⁿ_2sinhl^el ,which contradicts the boundedness assumption ons.

Step 4.limr(x)→+∞ϕ(x) >−∞if and only ifε(r):=r−sis integrable on[0,+∞).

This is straightforward from (20) combined with tanh(u−s(u))∼ε(u)which holds asu→ +∞due to Step 3.

Step 5.εis integrable on[0,+∞)if and only ifβ:=1−h²is integrable on[0,+∞).

First observation: limr→∞s=1. Indeed, at infinity, we haver−s→0, sos→ +∞,hence:

sinhr

sinhs ∼e^r−s∼1, sinhs

sinhr ∼e^s−r∼1, and (19) yieldss→1.

Using Step 3, the assumptions onhand the preceding observation, we get ε(r)→0, β(r)→0, and ε(r)=1−s(r)→0

asrtends to infinity. Plugging the definitions ofεandβin (19) and using the expansions coshε=1+o(ε), sinh(r−ε)=sinhr

1−ε+o(ε) , yields

(n−1)ε+ε+o(ε)=n

2β. (22)

Fixing a realδ >0,there readily existsr_δ>0 such that, for allrr_δ, ε+(n−1−δ)εn

2β, (23)

and

ε+(n−1+δ)εn

2β. (24)

Integrating (23), we get, forrr_δ, ε(r)e^{−(n−1−δ)r}

C(rδ)+n 2 r

rδ

β(u)e^{(n−1−δ)u}du

.

Integrating again and using Fubini Theorem yields, withδsuch thatn−1−δ >0,

+∞

r_δ

ε(r) drC(r_δ)+n 2

+∞

r_δ

β(u)e^{(n−1−δ)u} +∞

u

e^{−(n−1−δ)r}dr

du,

C(r_δ)+ n

2(n−1−δ) +∞

r_δ

β(u) du.

We conclude thatεis integrable providedβ=1−h²is integrable.

Analogously, using (24), we get ε(r)e^{−(n−1+δ)r}

C(r_δ)+n 2 r

r_δ

β(u)e^(n−1+δ)udu

, and

+∞

r_δ

ε(r) drC(r_δ)+n 2

+∞

r_δ

β(u)e^(n−1+δ)u +∞

u

e^{−(n−1+δ)r}dr

du,

C(r_δ)+ n

2(n−1+δ) +∞

r_δ

β(u) du.

Takingδ >0 arbitrary, we find thatβ is integrable ifεis integrable.

The proof of statement (ii) is analogous and thus omitted. 2 3.4. Construction of radial barriers

Lemma 5.Leth:H→Rbe a positive and continuous function on the hyperbolic space such that

r(x)lim→+∞h(x)=1

and such that the functionsh⁻andh⁺defined onR⁺by h⁻(r)= sup

r(x)=r

h(x) and h⁺(r)= inf

r(x)=rh(x) satisfy

+∞

0

(h⁻−1)₊dr <+∞,

+∞

0

(1−h⁺)₊dr <+∞,

where(h⁻−1)₊(resp.(1−h⁺)₊)means the positive part ofh⁻−1 (resp.1−h⁺). Then there existϕ⁻, ϕ⁺∈C^∞(H), with strictly convex spacelike graphs, satisfying:

F₂(ϕ⁻)h, F₂(ϕ⁺)h and lim

r→+∞ϕ^±=0.

Proof. First, considering 1+(h⁻−1)₊instead ofh⁻ and 1−(1−h⁺)₊instead ofh⁺,we may suppose without loss of generality thath⁻andh⁺are two continuous functions such that:∀x∈H,withr=r(x),

h⁻(r)h(x)h⁺(r) >0, (25)

h⁻1h⁺, lim

r→+∞h⁻(r)= lim

r→+∞h⁺(r)=1, (26)

and +∞

0

(h⁻−1) dr <+∞,

+∞

0

(1−h⁺) dr <+∞. (27)