The Dirichlet problem of the constant mean curvature equation in Lorentz-Minkowski space and in Euclidean space
TThe Dirichlet problem of the constant mean curvatureequation in Lorentz-Minkowski space and in Euclideanspace
Rafael L´opez ∗ Departamento de Geometr´ıa y Topolog´ıaInstituto de Matem´aticas (IEMath-GR)Universidad de Granada18071 Granada, Spain
Abstract
We investigate the differences and similarities of the Dirichlet problem of themean curvature equation in the Euclidean space and in the Lorentz-Minkowskispace. Although the solvability of the Dirichlet problem follows standards tech-niques of elliptic equations, we focus in showing how the spacelike condition in theLorentz-Minkowski space allows to drop the hypothesis on the mean convexity whichis required in the Euclidean case.Keywords: Euclidean space; Lorentz-Minkowski space; Dirichlet problem; meancurvature, maximum principleMSC: 58G20, 53A10, 53C50 ∗ This research has been partially supported by the grant no. MTM2017-89677-P,MINECO/AEI/FEDER, UE. a r X i v : . [ m a t h . DG ] D ec Introduction
In this paper we investigate the differences and similarities in the study of thesolvability of the Dirichlet problem for the constant mean curvature equation inthe Euclidean space and in the Lorentz-Minkowski space. Firstly we introducethe following notation. Let (cid:15) ∈ {− , } . Denote by R n +1 (cid:15) the vector space R n +1 equipped with the metric (cid:104) , (cid:105) = ( dx ) + ( dx ) + . . . + ( dx n ) + (cid:15) ( dx n +1 ) , where ( x , . . . , x n +1 ) are the canonical coordinates of R n +1 . If (cid:15) = 1 (resp. (cid:15) = − E n +1 (resp. the Lorentz-Minkowski space l n +1 ).We consider the Dirichlet problem for the constant mean curvature equation in R n +1 (cid:15) . Let Ω ⊂ R n be a bounded domain with smooth boundary ∂ Ω and let H be areal number. The Dirichlet problem asks for existence and uniqueness of a function u ∈ C (Ω) ∩ C ( ∂ Ω) such that (1 + (cid:15) | Du | )∆ u + (cid:15)D i uD j uD ij u = 2 H (1 + (cid:15) | Du | ) / in Ω (1) u = 0 on ∂ Ω (2) | Du | < (cid:15) = − D is the gradient operator, D i is the derivative with respect to the variable x i and the summation convention is used. A solution of (1)-(2) describes a hypersur-face with constant mean curvature H in R n +1 (cid:15) whose boundary is contained in thehyperplane x n +1 = 0. If (cid:15) = −
1, the extra condition | Du | < E n +1 (resp. in l n +1 ) with zero meancurvature ( H = 0) is called a minimal (resp. maximal) hypersurface.The example that shows the differences of the theory of constant mean curva-ture hypersurfaces in both ambient spaces is the Bernstein problem which we nowformulate. Suppose that the domain Ω is R n . A graph on R n is called an entiregraph. Let H = 0. The Bernstein problem asks if, besides linear functions, thereare other entire solutions of (1) with zero mean curvature. In the case n = 2,Bernstein proved that planes are the only entire minimal surfaces ([1]). In arbitrarydimension, this result holds if n ≤
7. A famous theorem of Bombieri, De Giorgiand Giusti asserts that there are other entire minimal graphs if n ≥ n -dimensional Lorentz-Minkowski space, Cheng and Yau proved, extendingprevious works of Calabi, that spacelike hyperplanes are the only entire maximalhypersurfaces ([3]).The interest of the study of constant mean curvature (cmc in short) hypersur-faces has its origin in physics. In the Euclidean space E , cmc surfaces are math-ematical models of the shape of a liquid in capillarity problems and of a interface hat separates two medium of different physical properties. In Lorentz-Minkowski l n +1 , cmc spacelike hypersurfaces have been used in General Relativity to provethe positive mass theorem or analyze the space of solutions of Einstein equations([4, 5]).We review briefly the state of the art of the Dirichlet problem for the constantmean curvature equation in both spaces. Assume that u takes arbitrary continuousboundary values u = ϕ on ∂ Ω. In the Euclidean space and for the minimal case H = 0, the Dirichlet problem (1) was solved for n = 2 by Finn [6] and in arbitrarydimension by Jenkins and Serrin [7] proving that the mean convexity of the domain Ωyields a necessary and sufficient condition of the solvability of the Dirichlet problemfor all boundary values ϕ : a domain Ω is said to be mean convex if the meancurvature κ ∂ Ω of ∂ Ω with respect to the inner normal is non-negative. If H (cid:54) = 0, astronger assumption is needed on Ω relating H and κ ∂ Ω and the answer appears inthe seminal paper [8], where Serrin proved the following result. theorem 1.1. The Dirichlet problem (1) in the Euclidean space has a unique solu-tion for any boundary values ϕ if and only if κ ∂ Ω ≥ n | H | n − on ∂ Ω . (4)It is expected that if we assume ϕ = 0 on ∂ Ω, the assumption (4) may be relaxed.Indeed, if ϕ = 0 and n = 2, the Dirichlet problem (1)-(2) has a unique solution if κ ∂ Ω ≥ | H | ([9]): see other results in the Euclidean case. If we drop the convexityassumption of ∂ Ω, it is possible to derive existence results if one assumes smallnesson the domain Ω and certain uniform exterior sphere conditions: see [10, 11, 12]The theory in l n +1 is shorter. The solvability of (1)-(3) with arbitrary boundaryvalues was initially investigated in the maximal case H = 0 assuming the meanconvexity of ∂ Ω ([13, 14]). However, the groundbreaking result is due to Bartnikand Simon in 1982 where the counterpart to Theorem 1.2 in l n +1 is surprisinglysimple because there is not any assumption on ∂ Ω ([15]). theorem 1.2.
The Dirichlet problem (1)- (3) in the Lorentz-Minkowski space has aunique solution for any spacelike boundary values ϕ if and only if ϕ has a spacelikeextension to Ω . This result was later generalized in other Lorentzian manifolds: [16, 17, 18,19, 20]. The method employed in the proof of Theorems 1.1 and 1.2 follows theLeray-Schauder fixed point theorem for elliptic equations because equation (1) is aquasilinear elliptic differential equation: if (cid:15) = −
1, this is assured by the spacelikecondition (3). In order to apply standard methods in the solvability of the Dirichlet roblem, we need to ensure a priori estimates of the height and the gradient forthe prospective solutions. Throughout this paper, we refer to the reader [11] as ageneral guide.The purpose of this work is twofold. Firstly, give an approach to the results inLorentz-Minkowski space comparing with the ones of Euclidean space and showinghow the spacelike condition | Du | < a priori estimates. The second objective is to provide geometricproofs to derive these estimates. For example, Serrin used the distance functionto ∂ Ω as a barrier for the desirable estimates ([8]), and similarly Flaherty in thesolvability in the Lorentzian case when H = 0 ([14]). This distance function isdefined in Ω but loses its geometric sense if we look the graph of u in E or l . Inour case, the a priori estimates will be obtained by a comparison argument betweenthe solutions of (1) and known cmc surfaces, such as, rotational surfaces. In orderto simplify the notation and arguments, we will consider the Dirichlet problem forthe 2-dimensional case, so we will work with surfaces in E and spacelike surfacesin l . In such a case, the mean convexity of the curve ∂ Ω is merely the convexity of ∂ Ω.This paper is organized as follows. After the Preliminaries section devoted tofix some definitions and notations, we derive the constant mean curvature equationin Section 3 obtaining some properties of the solutions showing differences in bothambient spaces. Section 4 describes the method of continuity to solve the Dirich-let problem (1). In Section 5 we obtain the height estimates for solutions of (1)and we prove that the boundary gradient estimates imply global (interior) gradientestimates. In Section 6, we analyze the solvability of the Dirichlet problem in theEuclidean case showing that a strong convexity hypothesis is necessary to solve theproblem. Finally, in Section 7 we solve the Dirichlet problem in Lorentz-Minkowskispace for arbitrary domains and we show the role of the cmc rotational surfaces inthe solvability of the problem.
We need to recall some definitions in Lorentz-Minkowski space. In l , the metric (cid:104) , (cid:105) is non-degenerate of index 1 and classifies the vectors of R in three types: a vector v ∈ l is said to be spacelike (resp. timelike, lightlike) if (cid:104) v, v (cid:105) > v = 0 (resp. (cid:104) v, v (cid:105) < (cid:104) v, v (cid:105) = 0 and v (cid:54) = 0). The modulus of v is | v | = (cid:112) |(cid:104) v, v (cid:105)| . A vectorsubspace U ⊂ R is called spacelike (resp. timelike, lightlike) if the induced metricon U is positive definite (resp. non-degenerate of index 1, degenerate and U (cid:54) = { } ).Any vector subspace belongs to one of the above three types. For 2-dimensional ubspaces, U is spacelike (resp. timelike, lightlike) if its orthogonal subspace U ⊥ istimelike (resp. spacelike, lightlike). A curve or a surface immersed in l is said tobe spacelike if the induced metric is positive-definite.The spacelike property is a strong condition. For example, any spacelike surface M is orientable. This is due because a unit vector orthogonal to M is timelike andin l , the scalar product of any two timelike vectors is not zero. Thus, if we fix e = (0 , , N on M so (cid:104) N, e (cid:105) is negative (or positive) on M , determining a globalorientation. Another consequence is that there do not exist closed spacelike sur-faces in l , in particular, any compact spacelike surface has non-empty boundary.Similarly, if a plane contains a closed spacelike curve, the plane must be spacelike.Let M be an orientable surface immersed in R (cid:15) . In case (cid:15) = −
1, we also assumethat the immersion is spacelike. Let ∇ and ∇ be the Levi-Civita connections in R (cid:15) and M respectively. The Gauss formula is ∇ X Y = ∇ X Y + (cid:15)σ ( X, Y ) for any twotangent vector fields X and Y on M , where σ is the second fundamental form. Themean curvature H of M is defined as H = 12 trace( σ ) . (5)Let us choose N a unit normal vector field on M with (cid:104) N, N (cid:105) = (cid:15) . Let A = ∇ N stand for the Weingarten endomorphism with respect to N . Then the Gauss formulais ∇ X Y = ∇ X Y + (cid:15) (cid:104) A ( X ) , Y (cid:105) N and A is a diagonalizable map. If κ and κ arethe principal curvatures, we have H = (cid:15)
12 trace( A ) = (cid:15)
12 ( κ + κ ) . Remark 2.1.
In case of timelike surfaces of l , the mean curvature is defined asin (5). However, although A is self-adjoint with respect to the induced metric (cid:104) , (cid:105) ,this metric is Lorentzian and it may occur that A is not real diagonalizable. Example 2.2.
1. Planes of E and spacelike planes of l have zero mean curva-ture.2. Round spheres S ( r ) in E and hyperbolic planes H ( r ) in l of radius r > can be described up to a rigid motion as { p ∈ l : (cid:104) p, p (cid:105) = (cid:15)r } . If (cid:15) = − , we also assume (cid:104) p, e (cid:105) < , where e = (0 , , . With respect to theGauss map N ( p ) = p/r , the mean curvature is H = − (cid:15)/r . . Right circular cylinders of R (cid:15) have constant mean curvature. To be precise,let a ∈ R (cid:15) be a unit vector with (cid:104) a, a (cid:105) = 1 (in l , the vector a is spacelike). Upto a rigid motion, the circular cylinder of axis a and radius r > is C ( r ) = { p ∈ R (cid:15) : (cid:104) p, p (cid:105) − (cid:104) p, a (cid:105) = (cid:15)r } . For the orientation N ( p ) = ( p −(cid:104) p, a (cid:105) a ) /r , the mean curvature is H = − (cid:15)/ (2 r ) .4. Let u = u ( x , x ) be a smooth function defined in a open domain Ω ⊂ R andlet M be the graph of u . Suppose that M is endowed with the induced metricfrom R (cid:15) . If (cid:15) = − , we also assume that M is spacelike, that is, | Du | < in Ω . The mean curvature H of M satisfies (1 + (cid:15) ( D u ) ) D u − (cid:15)D uD uD u + (1 + (cid:15) ( D u ) ) D u = 2 H (1 + (cid:15) | Du | ) / (6) with respect to the orientation N = ( − (cid:15)D u, − (cid:15)D u, (cid:112) (cid:15) | Du | = ( − (cid:15)Du, (cid:112) (cid:15) | Du | · (7) Let us notice that (6) coincides with the equation (1).
In this section we will derive some properties on the solutions of the cmc equation(1). The mean curvature equation (1) (or (6)) can be expressed in the divergenceform div (cid:16) Du (cid:112) (cid:15) | Du | (cid:17) = 2 H in Ω , (8)with the observation that if (cid:15) = −
1, we assume the spacelike condition | Du | < u ( x , x ) = − (cid:15) (cid:113) r − (cid:15) ( x + x ) , (cid:26) x + x < r (cid:15) = 1( x , x ) ∈ R (cid:15) = − . For (cid:15) = 1, x = u ( x , x ) is defined in a disc and describes a hemisphere in S ( r ),and for (cid:15) = − x = u ( x , x ) is the hyperbolic plane H ( r ). On the other hand, acylinder C ( r ) with axis a = (0 , ,
0) and radius r > u ( x , x ) = − (cid:15) (cid:113) r − (cid:15)x , (cid:26) | x | < r (cid:15) = 1( x , x ) ∈ R (cid:15) = − . quation (8) (with (3) if (cid:15) = −
1) is of quasilinear elliptic type, hence we canapply the machinery for these equations. It is easily seen that the difference of twosolutions of equation (1) satisfies the maximum principle. As a consequence, wegive a statement of the comparison principle in our context. We define the operator Q [ u ] = (1 + (cid:15) | Du | )∆ u − (cid:15)D i uD j uD ij u − H (1 + (cid:15) | Du | ) / . (9)The comparison principle asserts ([11, Th. 10.1]). Proposition 3.1 (Comparison principle) . If u, v ∈ C (Ω) satisfy Q [ u ] ≥ Q [ v ] in Ω and u ≤ v on ∂ Ω , then u ≤ v in Ω . If we replace Q [ u ] ≥ Q [ v ] by Q [ u ] > Q [ v ] ,then u < v in Ω . In particular, the solution of the Dirichlet problem, if it exists, isunique. An immediate consequence is the touching principle.
Proposition 3.2 (Touching principle) . Let M and M be two surfaces in R (cid:15) withthe same constant mean curvature and with possibly non-empty boundaries ∂M , ∂M . If M and M have a common tangent interior point and M lies above M around p , then M and M coincide at an open set around p . The same statementis also valid if p is a common boundary point and the tangent lines to ∂M i coincideat p . A first difference of the Dirichlet problem for the constant mean curvature equa-tion (1) is that in the Euclidean space E the value H is not arbitrary and dependson the size of Ω, whereas in l the value H may be arbitrary. Indeed, from equation(8), the divergence theorem yields2 | H | area(Ω) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∂ Ω (cid:104) Du (cid:112) (cid:15) | Du | , (cid:126)n (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , where (cid:126)n is the outward unit normal vector along ∂ Ω. The idea is to estimate theright-hand side from above. If (cid:15) = 1, we have2 | H | area(Ω) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∂ Ω (cid:104) Du (cid:112) | Du | , (cid:126)n (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:90) ∂ Ω | Du | (cid:112) | Du | < (cid:90) ∂ Ω ∂ Ω) , Proposition 3.3.
A necessary condition for the solvability of the Dirichlet problem (1) in E is | H | < length ( ∂ Ω)2 area (Ω) · (10) et us notice that this upper bound for H does not depend on the boundaryvalues ϕ . In fact, there are explicit examples where all values between 0 and theupper bound in (10) are attained. Indeed, let Ω be a disc of radius ρ and ϕ = 0.Then the value of length( ∂ Ω) / (2 area(Ω)) is 1 /ρ . On the other hand, for each0 < H < /ρ , take the spherical cap of radius 1 / | H | u ( x , x ) = − (cid:114) H − x − x , x + x < ρ . Then u is a graph on Ω with constant mean curvature H for every H going from until /ρ . The limit case H = 1 /ρ corresponds with a hemisphere of radius 1 / | H | .The same computations in l do not provide the same conclusion because | Du | / (cid:112) − | Du | may be arbitrarily large. So, for the hyperbolic planes u ( x , x ) = (cid:114) H + x + x (11)the value | Du | (cid:112) − | Du | = | H | (cid:113) x + x is arbitrary large and the function u is defined in any domain of the plane R andfor any H .A second difference is the question of the existence of entire solutions of (1)with non-zero mean curvature H : recall that the case H = 0 (Bernstein problem)was discussed in the Introduction. In l , the hyperbolic planes (11) show thatfor any H , there are solutions (1) defined in the plane R . Also the cylinders u ( x , x ) = (cid:112) /H + x are other examples of entire solutions of (1)-(3). Howeverin the Euclidean space, we have Proposition 3.4.
Let Ω be a domain of R . If u is a solution of (1) with H (cid:54) = 0 in E , then Ω does not contain the closure of a disk of radius / | H | .Proof. We proceed by contradiction. Assume that D is an open disk of radius 1 / | H | such that D ⊂ Ω. Let x be the center of D . Without loss of generality, we supposethat the sign of H is positive: recall that the mean curvature is computed withrespect to the orientation (7). Let r = 1 /H and S ( r ) be a sphere of radius r whose center lies on the straight-line through x and perpendicular to the ( x , x )-plane. Here, and in what follows, S ( r ) denotes a sphere of radius r whose centermay be changing. We orient S ( r ) by the inward orientation. With this choice oforientation, the mean curvature is H and the orthogonal projection of S ( r ) on R is D . et M be the graph of u . Lift S ( r ) vertically upwards until S ( r ) is completelyabove M . Then, let us descend S ( r ) until the first point p of contact with M .Since D ⊂ Ω and M is a graph on Ω, the contact point p must be interior in bothsurfaces. By the touching principle, the surfaces M and S ( r ) agree on an open setaround p , hence M is included in a sphere of radius 1 /H : this is a contradictionbecause the orthogonal projection onto R would give Ω ⊂ D . In this section, we present the method for solving the Dirichlet problem (1)-(2),which holds in the Euclidean and Lorentzian contexts. We establish the solvabilityof the Dirichlet problem by applying the method of continuity ([11, Sec. 17.2]). Thematrix of the coefficients of second order of (1) is (cid:18) (cid:15) ( D u ) − (cid:15)D uD u − (cid:15)D uD u (cid:15) ( D u ) (cid:19) . The minimum and maximum eigenvalues of this matrix are λ = 1 and Λ = 1 + | Du | if (cid:15) = 1 and λ = 1 − | Du | and Λ = 1 if (cid:15) = −
1. Thus if (cid:15) = −
1, the equation (1) isuniformly elliptic provided | Du | < t ∈ [0 , Q t [ u ] = 0 in Ω u = 0 on ∂ Ω , | Du | < (cid:15) = − Q t [ u ] = (1 + (cid:15) | Du | )∆ u − (cid:15)D i uD j uD ij u − tH (1 + (cid:15) | Du | ) / . A solution u of Q t [ u ] = 0 describe a surface with constant mean curvature tH . Asusual, let A = { t ∈ [0 ,
1] : there exists u t ∈ C ,α (Ω) , Q t [ u t ] = 0 , u t | ∂ Ω = 0 } . The existence of solutions of the Dirichlet problem (1)-(2)-(3) is established if 1 ∈ A .For this purpose, we prove that A is a non-empty open and closed subset of [0 , . The set A is not empty . This is because u = 0 solves the Dirichlet problemfor t = 0.2. The set A is open in [0 , t ∈ A , we need to prove that there exists η > t − η, t + η ) ∩ [0 , ⊂ A . Define the map T ( t, u ) = Q t [ u ] for t ∈ R and u ∈ C ,α (Ω). Then t ∈ A if and only if T ( t , u t ) = 0. If we showthat the derivative of Q t with respect to u , say ( DQ t ) u , at the point u t is anisomorphism, the Implicit Function Theorem ensures the existence of an openset V ⊂ C ,α (Ω), with u t ∈ V and a C function ψ : ( t − η, t + η ) → V for some η >
0, such that ψ ( t ) = u t > T ( t, ψ ( t )) = 0 for all t ∈ ( t − η, t + η ):this guarantees that A is an open set of [0 , DQ t ) u is one-to-one if for any f ∈ C α (Ω), there is a unique solution v ∈ C ,α (Ω) of the linear equation L [ v ] := ( DQ t ) u ( v ) = f in Ω and v = 0 on ∂ Ω. The computation of L will be done in Theorem 5.6, obtaining L [ v ] = ( DQ t ) u v = a ij D ij v + b i D i v, where a ij = a ij ( Du ) is symmetric, b i = b i ( Du, D u ) and L is a linear ellipticoperator whose term for the function v is zero. Therefore the existence anduniqueness is assured by standard theory ([11, Th. 6.14]).3. The set A is closed in [0 , { t k } ⊂ A with t k → t ∈ [0 , k ∈ N , there is u k ∈ C ,α (Ω) such that Q t k [ u k ] = 0 in Ω and u k = 0 in ∂ Ω.Define the set S = { u ∈ C ,α (Ω) : there exists t ∈ [0 ,
1] such that Q t [ u ] = 0 in Ω , u | ∂ Ω = 0 } . Then { u k } ⊂ S . If we see that the set S is bounded in C ,β (Ω) for some β ∈ [0 , α ], and since a ij = a ij ( Du ) in (9), the Schauder theory proves that S is bounded in C ,β (Ω), in particular, S is precompact in C (Ω) (Th. 6.6 andLem. 6.36 in [11]). Hence there is a subsequence { u k l } ⊂ { u k } converging tosome u ∈ C (Ω) in C (Ω). Since T : [0 , × C (Ω) → C (Ω) is continuous,we obtain Q t [ u ] = T ( t, u ) = lim l →∞ T ( t k l , u k l ) = 0 in Ω. Moreover, u | ∂ Ω =lim l →∞ u k l | ∂ Ω = 0 on ∂ Ω, so u ∈ C ,α (Ω) and consequently, t ∈ A . The set S is bounded in C ,β (Ω) if it is bounded in C (Ω), where the norm is defined by (cid:107) u t (cid:107) C (Ω) = sup Ω | u t | + sup Ω | Du t | . Usually, the a priori estimates for | u | are called height estimates and gradientestimates for | Du | . efinitively, A is closed in [0 ,
1] provided we find two constants M and C independent on t ∈ A , such thatsup Ω | u t | < M, sup Ω | Du t | < C. (12)Here we make the observation that whereas in the Euclidean space, the con-stant C can take an arbitrary value, the spacelike condition in the Lorentz-Minkowski space implies that C may be chosen to be C = 1. However, duringthe above process of the method of continuity, we require that Q t is uniformlyelliptic, in particular, we have to ensure that | Du | << l , the constant C in (12) has to satisfy the condition C < Remark 4.1.
In the Euclidean case, the smoothness of the solution on ∂ Ω isguaranteed if the graph close to the boundary point does not blow-up at infinity,that is, | Du | (cid:54)→ ∞ . In the Lorentzian case, we have to prevent the possibilitythat | Du | → as we go to ∂ Ω . The existence of the constant C shows that thesurface cannot ‘go null’ in the terminology of Marsden and Tipler [5, p. 124]. Consider the Dirichlet problem for the cmc equation and arbitrary boundary values div (cid:16) Du (cid:112) (cid:15) | Du | (cid:17) = 2 H in Ω u = ϕ on ∂ Ω , (13)where, in addition, if (cid:15) = −
1, we suppose | Du | < | u | and | Du | for a solution u of (13)in terms of the initial conditions. In Theorems 5.2, 5.4 and 5.5 we will derive theestimates for | u | . For the gradient estimates, we will prove that the supremum of | Du | in Ω is attained at some boundary point (Theorem 5.6).We begin with the height estimates. The main difference between both ambientspaces is that in E there exist estimates of sup Ω | u | depending only on H and ϕ ,whereas in l the size of the domain Ω appears in these estimates, such as showsthe hyperbolic planes (11).The height estimates for cmc graphs in the Euclidean space are obtained withthe functions f ( p ) = (cid:104) p, a (cid:105) , g ( p ) = (cid:104) N ( p ) , a (cid:105) , p ∈ M, where a is a fixed unit vector of R and N is the Gauss map of M . Firstly we needto compute the Beltrami-Laplacian ∆ M of the functions f and g . The following esult holds for cmc surfaces in E and in l without to be necessarily graphs: werefer the reader to [21] for a proof. Lemma 5.1.
Let M be an immersed surface in R (cid:15) . Then ∆ M (cid:104) p, a (cid:105) = 2 H (cid:104) N, a (cid:105) . (14) If, in addition, the immersion has constant mean curvature, then ∆ M (cid:104) N, a (cid:105) + (cid:15) | σ | (cid:104) N, a (cid:105) = 0 , (15) where | σ | is the norm of the second fundamental form. Consider u be a solution of (13) and let M = graph( u ). If we take a = e =(0 , , (cid:104) p, e (cid:105) and (cid:104) N, e (cid:105) inform about u and Du because (cid:104) p, e (cid:105) = (cid:15)u, (cid:104) N, e (cid:105) = (cid:15) (cid:112) (cid:15) | Du | . (16)In particular, sign( g ) = sign( (cid:15) ). Suppose H ≥
0. Then ∆ M f ≥ ≤
0) in E (resp. l ) and the maximum principle implies (cid:104) p, e (cid:105) ≤ max ∂ Ω (cid:104) p, e (cid:105) in E (resp. (cid:104) p, e (cid:105) ≥ min ∂ Ω (cid:104) p, e (cid:105) in l ). Thus u ≤ max ∂ Ω u in both ambient spaces. On theother hand ∆ M ( Hf + (cid:15)g ) = (2 H − | σ | ) g (cid:26) ≤ (cid:15) = 1 ≥ (cid:15) = − . Since | σ | = κ + κ ≥ H , the maximum principle yields Hf + (cid:15)g (cid:26) ≥ min ∂ Ω Hf + g (cid:15) = 1 ≤ max ∂ Ω Hf − g (cid:15) = − . In case (cid:15) = 1, we have Hu + (cid:104) N, e (cid:105) ≥ H min ∂ Ω u + min ∂ Ω (cid:104) N, e (cid:105) ≥ H min ∂ Ω u because (cid:104) N, e (cid:105) ≥
0. Since (cid:104)
N, e (cid:105) ≤
1, we deduce u ≥ − /H + min ∂ Ω ϕ . theorem 5.2. A solution u of (13) in the Euclidean space satisfies min ∂ Ω ϕ − H ≤ u ≤ max ∂ Ω ϕ, if H > ∂ Ω ϕ ≤ u ≤ max ∂ Ω ϕ − H , if H < . e analyze the same argument in l . The reverse Cauchy-Schwarz inequalityfor timelike vectors yields (cid:104) N, e (cid:105) ≤ − − Hu + (cid:104) N, e (cid:105) ≤ H max ∂ Ω ( − u ) + max ∂ Ω (cid:104) N, e (cid:105) ≤ − H min ∂ Ω u − , but it is not possible to bound from below because of the function (cid:104) N, e (cid:105) . Thismakes a key difference with the Euclidean case and concludes that the argumentdone in the Euclidean space is not valid in l . If H = 0, from (14) we deduce: Corollary 5.3.
In both ambient spaces, if u is a solution of (13) for H = 0 then min ∂ Ω ϕ ≤ u ≤ max ∂ Ω ϕ. As expected, in the Lorentz-Minkowski space there does not exist height esti-mates depending only on H and ϕ . An example is the following. For r > m > r , let u m ( x , x ) = (cid:112) r + x + x − m defined in the round disc Ω √ m − r = { ( x , x ) ∈ R : x + x < m − r } . The graph of u m is a piece of the hyperbolicplane H ( r ) which has been displaced vertically downwards a distance equal to m .Then u m is a solution of (13) in Ω √ m − r with ϕ = 0 and the height on u m , namely | u m | = m − r , goes to ∞ as m (cid:37) ∞ .Motivated by these examples, we will deduce height estimates for a solution of(13) in terms of the size of Ω (see [23] for a height estimate in terms of the areaof the surface). The estimates that we will deduce are of two types: the first onesare given in terms of the diameter of Ω and second ones depend on the width ofnarrowest strip containing Ω. theorem 5.4. If u be a solution of (13) in l , then min ∂ Ω ϕ − | H | (cid:32)(cid:114) diam (Ω) H − (cid:33) ≤ u ≤ max ∂ Ω ϕ + 1 | H | (cid:32)(cid:114) diam (Ω) H − (cid:33) (17) and equality holds if and only if the graph of u describes a hyperbolic cap. In theparticular case ϕ = 0 , we have sup Ω | u | ≤ | H | (cid:32)(cid:114) diam (Ω) H − (cid:33) . Proof.
The inequalities are obtained by comparing M = graph( u ) with hyperboliccaps with mean curvature | H | coming from below and from above. There is noloss of generality in assuming that Ω is included in the closed disk D ρ of center the rigin and radius ρ = diam(Ω) /
2. Consider the hyperbolic plane H ( r ) defined bythe function u ( x , x ) = (cid:112) r + x + x , where r = 1 / | H | .Let us take H ( r ; s ) the compact part obtained when we intersect H ( r ) withthe horizontal plane of equation x = s . Then ∂ H ( r ; s ) is a circle of radius ρ , with s = (cid:112) ρ + r and H ( r ; s ) = { ( x , x , x ) ∈ H ( r ) : x ≤ s } . Move vertically down H ( r ; s ) until to be disjoint from M . Next move upwards H ( r ; s ) until that H ( r ; s ) touches M the first time. If the contact between bothsurfaces occurs at some common interior point, the comparison principle and thenthe touching principle implies that u describes part of the hyperbolic plane H ( r ; s ).In such a case, the left inequality of (17) holds trivially.In case that the first contact occurs between a point of H ( r ; s ) with a boundarypoint of M , we can arrive until the value s = min ∂ Ω ϕ , hencemin ∂ Ω ϕ − (cid:112) r + ρ + (cid:113) r + x + x ≤ u in Ω.Evaluating at the origin,min ∂ Ω ϕ − | H | − (cid:114) H + ρ ≤ u in Ω,which coincides with the left inequality in (17) because ρ = diam(Ω) / u ( x , x ) = − (cid:112) r + x + x .A second height estimate can be deduced by comparing u with spacelike cylin-ders. We need to introduce the following notation. Given a bounded domain A ⊂ R , consider the set L of all pairs of parallel straight-lines ( L , L ) in R such that A is included in the planar strip determined by L and L . SetΘ( A ) = min { dist( L , L ) : ( L , L ) ∈ L} . Observe that the domain A is included in a strip of width Θ(Ω) and this strip isthe narrowest one among all strips containing A in its interior. Notice also thatΘ( A ) ≤ δ ( A ). theorem 5.5. If u is a solution of (13) in l , then min ∂ Ω ϕ − | H | (cid:16)(cid:112) H − (cid:17) ≤ u ≤ max ∂ Ω ϕ + 12 | H | (cid:16)(cid:112) H − (cid:17) . (18) n the particular case ϕ = 0 , we have sup Ω | u | ≤ | H | (cid:16)(cid:112) H − (cid:17) . Notice that the estimates (18) and (17) are not comparable.
Proof.
The argument is similar to the proof of Theorem 5.4 by replacing the role ofthe hyperbolic planes by cylinders. After a rigid motion if necessary, assume thatΩ is included in the strip | x | < Θ(Ω) /
2. Consider the cylinder C ( r ) u ( x , x ) = (cid:113) r + x , where r = 1 / (2 | H | ). Consider the value s such that the intersection of C ( r ) withthe plane of equation x = s is formed by two parallel straight-lines separated adistance equal to Θ(Ω): this occurs when the value s is s = (cid:114) r + Θ(Ω) . Denote by C ( r ; s ) the part of C ( r ) below the plane of equation x = s , which isa graph on a strip of width Θ(Ω). Let us move down the cylinders C ( r ; s ) untilthat do not intersect M = graph( u ). After, we move upwards C ( r ; s ) until the firsttouching point with M . If this point is a common interior point, then M is includedin the cylinder C ( r ) and the left inequality in (18) is trivially satisfied. If the pointis not interior, we can arrive until the height x = s where s = min ∂ Ω ϕ . Thenmin ∂ Ω ϕ − (cid:114) r + Θ(Ω) (cid:113) r + x ≤ u in Ω . At the points x = 0, we deducemin ∂ Ω ϕ + r − (cid:114) r + Θ(Ω) ≤ u in Ω . This inequality is just the left inequality in (18). The right inequality in (18) isproved by comparing with the cylinders u ( x , x ) = − (cid:112) r + x .We finish this section investigating how to derive the a priori estimates (12) of | Du | in Ω. Recall that we have to find a constant C depending only on the initialdata such that | Du | ≤ C in Ω, with the observation that if (cid:15) = −
1, we require that
C <
1. We will prove that it suffices to find this estimate only in boundary points.We present two proofs of this result which hold in both ambient spaces. heorem 5.6. If u is a solution of (13), then sup Ω | Du | = max ∂ Ω | Du | . (19) Proof 1.
For each i = 1 ,
2, define the functions v i = D i u . Differentiate (9) withrespect to the variable x k , k ∈ { , } . After some computations, we obtain (cid:0) (1 + (cid:15) | Du | ) δ ij − (cid:15)D i uD j u (cid:1) D ij v k +2 (cid:0) (cid:15)D i u ∆ u + 3 H (1 − | Du | ) D i u − (cid:15)D j uD ij u (cid:1) D i v k = 0 . (20)Hence v k satisfies a linear elliptic equation of type a ij D ij v k + b i D i v k = 0 , where a ij = a ij ( Du ) and b i = b i ( Du, D u ). By the maximum principle, | v k | hasnot a maximum at some interior point. Consequently, the maximum of | Du | on thecompact set Ω is attained at some boundary point. Proof 2 .
Estimates of | Du | are obtained by means of the function (cid:104) N, e (cid:105) because(16). From equation (15)∆ M (cid:104) N, e (cid:105) = − (cid:15) | σ | (cid:104) N, e (cid:105) = | σ | (cid:112) (cid:15) | Du | ≤ , and the maximum principle impliesinf Ω (cid:104) N, e (cid:105) = min ∂ Ω (cid:104) N, e (cid:105) . Thus inf Ω (cid:15) (cid:112) (cid:15) | Du | = min ∂ Ω (cid:15) (cid:112) (cid:15) | Du | , which is equivalent to (19).To summarize, the problem of finding gradient estimates of | Du | in Ω is pass-ing to a problem of estimates along the boundary, exactly, finding a constant C depending only on the initial data such thatmax ∂ Ω | Du | < C. (21)In the proofs of the existence results in the following sections, the method toobtain the constant C in (21) is by an argument of super and subsolutions and thenwe apply the next result. emma 5.7. Let x ∈ ∂ Ω be a boundary point. Suppose that there is a neighborhood U of x and two functions w + , w − ∈ C (Ω ∩ U ) such that Q [ w + ] ≤ ≤ Q [ w − ] in Ω ∩ Uw − ≤ u ≤ w + in ∂ (Ω ∩ U ) w − ( x ) = u ( x ) = w + ( x ) | Dw − | , | Dw + | ≤ C. Then | Du | ≤ C .Proof. The comparison principle yields w − ≤ u ≤ w + in Ω ∩ U , concluding that | Du | ≤ {| Dw − | , | Dw + |} . In this section we address the Dirichlet problem (1) in the Euclidean space. By The-orem 5.2, we know that the value H is not arbitrary. Without to assume convexityon ∂ Ω, there are results of existence assuming some smallness on the value H andon the size of Ω ([10, 11]. Thanks to this smallness on initial data, it is possible toobtain height and boundary gradient estimate of the solution. If we assume convex-ity, there are different hypothesis that ensure the solvability of the Dirichlet problemand relate the size or the convexity of Ω with the value H ([9, 24, 25, 12, 26, 27, 28]).Theorem 1.1 solves the Dirichlet problem in the Euclidean space for arbitraryboundary values. If we now suppose that u = 0 on ∂ Ω, the hypothesis (4) can beweakened assuming κ ∂ Ω ≥ | H | . We give two proofs of this result. The first onewill be proved in arbitrary dimension and, although the idea appears generalized inother ambient spaces ([29, 30, 31, 32]), as far as we know, in the literature there isnot specifically a statement in the Euclidean space. Here we follow [32]. theorem 6.1. Let H (cid:54) = 0 . If the mean curvature of ∂ Ω satisfies κ ∂ Ω > | H | , thenthe Dirichlet problem div (cid:16) Du (cid:112) | Du | (cid:17) = nH in Ω u = 0 on ∂ Ω (22) in arbitrary dimension has a unique solution. roof. Firstly, we observe that the solutions u t of the method of continuity (Section4) are ordered in decreasing sense according the parameter t . Indeed, if t < t ,then Q t [ u t ] = 0 and Q t [ u t ] = ( t − t )(1 + | Du t | ) > Q t [ u t ] . Since u t = 0 = u t on ∂ Ω, the comparison principle yields u t < u t in Ω. Thus, u ≤ u t < t , where for the value t = 1, u is the solution u of (1). Byusing Lemma 5.7, this implies that it suffices to find a priori height and gradientestimates for the prospective solution u of (1).If u is a solution of (22), then − u is a solution of (22) for the value − H . Thus,and without loss of generality, we suppose H >
0. Let M be the graph of u . Bythe height estimates of Theorem 5.2, we know − /H < u < a priori height estimates. According to Theorem 5.6, we need to find a priori boundary gradient estimates. However, we will be able to find the gradient estimateson the domain Ω.We use again the function Hf + g as in Theorem 5.2. Since ∆ M ( Hf + g ) ≤ u = 0 on ∂ Ω, the maximum principle ensures the existence of a boundary point q ∈ ∂ Ω where Hf + g attains its minimum, so H (cid:104) p, e (cid:105) + (cid:104) N, e (cid:105) ≥ min ∂ Ω (cid:104) N, e (cid:105) = (cid:104) N ( q ) , e (cid:105) . (23)Furthermore, the maximum principle on the boundary implies H (cid:104) ν ( q ) , e (cid:105) + (cid:104) dN q ν, e (cid:105) ≥ , where ν is the inward unit conormal vector along ∂ Ω. If σ is the second fundamentalform, this inequality can be written as( H − σ ( ν ( q ) , ν ( q ))) (cid:104) ν ( q ) , e (cid:105) ≥ . Since u < u = 0 on ∂ Ω yields (cid:104) ν ( q ) , e (cid:105) <
0, hence H − σ ( ν ( q ) , ν ( q )) ≤
0. If { v , . . . , v n − } is a orthonormal basis of the tangent spaceto ∂ Ω at the point q , the above inequality implies n − (cid:88) i =1 σ ( v i , v i ) = nH − σ ( ν ( q ) , ν ( q )) ≤ ( n − H. (24)Denote by ∇ ∂ Ω and σ ∂ Ω the Levi-Civita connection and second fundamental formof ∂ Ω as submanifold of Ω, respectively. Let η be the unit normal vector field of ∂ Ωin Ω. The Gauss formula gives ∇ v i v i = ∇ v i v i + σ ( v i , v i ) N ( q ) = ∇ ∂ Ω v i v i − σ ∂ Ω ( v i , v i ) η ( q ) + σ ( v i , v i ) N ( q ) . hen σ ( v i , v i ) = σ ∂ Ω ( v i , v i ) (cid:104) N ( q ) , η ( q ) (cid:105) . From (24), (cid:104) N ( q ) , η ( q ) (cid:105) n − (cid:88) i =1 σ ∂ Ω ( v i , v i ) ≤ ( n − H. Since (cid:80) n − i =1 σ ∂ Ω ( v i , v i ) = ( n − κ ∂ Ω , we have (cid:104) N ( q ) , η ( q ) (cid:105) κ ∂ Ω ( q ) ≤ H, so (cid:104) N ( q ) , η ( q ) (cid:105) κ ∂ Ω ( q ) ≤ H . Since (cid:104) N ( q ) , e (cid:105) + (cid:104) N ( q ) , η ( q ) (cid:105) = 1, we deduce (cid:104) N ( q ) , e (cid:105) ≥ (cid:115) − H κ ∂ Ω ( q ) = (cid:113) κ ∂ Ω ( q ) − H κ ∂ Ω ( q ) := C. From (23) and because H (cid:104) p, e (cid:105) ≤ M , we find (cid:104) N, e (cid:105) ≥ C in Ω . Finally, we conclude from (16) | Du | ≤ √ − C C in Ωobtaining the desired gradient estimates in Ω.The second proof is done in the two-dimensional case, where the mean convexityis now the convexity in the Euclidean plane. The proof uses spherical caps to findthe boundary gradient estimates (21). theorem 6.2. Let H (cid:54) = 0 . If the curvature of ∂ Ω satisfies κ ∂ Ω ≥ | H | , then theDirichlet problem (22) has a unique solution.Proof. We start as in the proof of Theorem 6.1 and we follow the same notation.We only need to find the a priori boundary gradient estimates. Set κ = min q ∈ ∂ Ω κ ∂ Ω ( q ) > r = 1 /κ . irstly, we prove Theorem 5.4 in case of strict inequality κ ∂ Ω > H . Let x ∈ ∂ Ωbe a fixed but arbitrary boundary point. Consider D r a disc of radius r such that x ∈ C r ∩ ∂ Ω and Ω ⊂ D r where C r is the boundary of D r . This is possible because κ > H . Consider C /H a circle of radius 1 /H and concentric to C r . Notice that r < /H . After a translation we suppose that the center of D r is the origin ofcoordinates.Let S (1 /H ) be the hemisphere of radius 1 /H whose boundary is C /H andbelow the plane Π of equation x = 0. Let us lift up S (1 /H ) until its intersectionwith Π is C r . Denote by S r the piece of S (1 /H ) below Π at this position. SeeFigure 1. The surface S r is a small spherical cap which is the graph of w − ( x , x ) = − (cid:114) H − x − x , x + x ≤ r . We prove now that M lies in the bounded domain determined by S r ∪ D r . Forthis, we move down S r by vertical translations until S r does not intersect M andthen, move upwards S r until the initial position. Since the mean curvature of S r is H and Ω ⊂ D r , the touching principle implies that there is not a contact before that S r arrives to its original position. Once we have arrived to the original position, ina neighborhood of the point x , the surface M lies sandwiched between S r and Π.Then Q [ w + ] = − H < Q [ w − ] = Q [ u ]and consequently by Lemma 5.7max ∂ Ω | Du | < max ∂S r {| Dw − | , | Dw + |} = max ∂S r | Dw − | = Hr √ − H r , where this constant depends only on r and H .Until here, we have obtained the existence of a solution for each 0 < H < κ .Moreover, and since the gradient is bounded from above in Ω depending only onthe initial data, the solution obtained is smooth in Ω. Now, we proceed by provingthe existence of a solution of (1) in the case H = κ : in case that Ω is a round diskof radius r (and κ = 1 /r ), the solution is u ( x , x ) = r − (cid:112) r − x − x .Let us consider an increasing sequence H n → H and u n the solution of (1)for the value H n for the mean curvature: the solution exists because κ > H n .By the monotonicity of H n and the comparison principle, the sequence { u n } ismonotonically increasing and converges uniformly on compact sets of Ω. Let u =lim u n . Standard compactness results involving Ascoli-Arzel´a theorem guaranteethat u ∈ C (Ω) and Q [ u ] = 0. It remains to check that u ∈ C (Ω) and u = 0 on ∂ Ω.Let x ∈ ∂ Ω and { x m } ⊂ Ω with x m → x . Consider the hemisphere S ( r ) as above Ω MS r C r Figure 1: Proof of Theorem 6.2. and let D r be the open disk of radius r = 1 /H such that S ( r ) = graph( v ), with v ∈ C ∞ ( D r ) ∩ C ( D r ). Place D r such that x ∈ ∂D r . We know that Ω ⊂ D r andby the touching principle, 0 < u n < v on Ω. For each n ∈ n , 0 < u n ( x m ) < v ( x m ).Then 0 ≤ u ( x m ) ≤ v ( x m ). Letting m → ∞ , 0 ≤ u ( x ) ≤
0. This proves thecontinuity of u up to ∂ Ω and that u = 0 on ∂ Ω. In this section we address the Dirichlet problem in l following the ideas of theEuclidean case in the above section. The first result that we present is motivatedby Theorem 6.2, where we assumed a strong convexity of ∂ Ω comparing with thevalue H , namely, κ ∂ Ω ≥ | H | . In contrast, in Lorenz-Minkowski space this convexityassumption changes by merely the convexity κ ∂ Ω ≥ ∂ Ω. theorem 7.1. If κ ∂ Ω ≥ , then the Dirichlet problem div (cid:16) Du (cid:112) − | Du | (cid:17) = 2 H in Ω u = 0 on ∂ Ω (25) has a unique solution.Proof.
With a similar argument as in Theorem 6.2, the solutions u t of the methodof continuity are ordered by u t < u t if t < t , so it suffices to get the a priori estimates for the solution u of (25). Without loss of generality, we suppose H > he height estimates were proved in Theorem 5.4 (or 5.5) and we showed that thereexists K = K (Ω , H ) > − K < u < . (26)In order to find the a priori boundary gradient estimates, consider the cylinder C ( r )determined by v ( x , x ) = (cid:112) r + x , where r = 1 / (2 H ). For each m > r , let C ( r ; m ) = { ( x , x , x ) ∈ C r : x ≤ m } . This surface is a graph on the strip Ω r,m = { ( x , x ) ∈ R : −√ m − r ≤ x ≤√ m − r } . Take m sufficiently large so m fulfills the next two conditions: v ( x = (cid:112) m − r ) − v ( x = 0) = m − r > K (27)diam(Ω) < width(Ω r,m ) = 2 (cid:112) m − r . (28)Let us restrict v in the half-strip U = { ( x , x ) ∈ Ω r,m : 0 < x < (cid:112) m − r } and ˜ C ( r ; m ) denotes the graph of v on U . The boundary of ˜ C ( r ; m ) is formed bytwo parallel straight-lines L ∪ L = { v ( x = 0) } ∪ { v ( x = (cid:112) m − r } , where L is contained in the plane x = r and L in the plane x = m , with r < m .Let x ∈ ∂ Ω be a fixed but arbitrary point of the boundary of Ω. After a rotationabout a vertical axis and a horizontal translation, we suppose x = ( √ m − r , U (this is possible by (28)) and the tangent line L to ∂ Ω at x isparallel to the x -line. By vertical translations, we displace vertically down ˜ C ( r ; m )until it does not intersect M = graph( u ). Then we move vertically upwards until˜ C ( r ; m ) intersects M for the first time.We claim that the first time that ˜ C ( r ; m ) touches M occurs when L arrivesto the plane of equation x = 0 and consequently, L = L . Firstly, the touchingprinciple prohibits an interior tangent point between M and ˜ C ( r ; m ). On the otherhand, it is not possible that a boundary point of of C ( r ; m ), namely, a point of L ∪ L , touches a point of M because (26) and (27). Definitively, we can move˜ C ( r ; m ) until L coincides with L , in particular, x ∈ L ∩ ∂ Ω . t this position, ˜ C ( r ; m ) is the graph of the function w − ( x , x ) = (cid:113) r + x − m. Thus M is contained between w − and w + = 0 in Ω ∩ U with w − ( x ) = w + ( x ) = u ( x ) = 0. We are in position to apply Lemma 5.7 because Q [ w + ] < Q [ u ] = Q [ w − ] and w − ≤ u ≤ w + in ∂ (Ω ∩ U ). We conclude that | Du | ≤ C , where theconstant C in (21) is C = | Dw − | | x = √ m − r = √ m − r m . The key in the above proof is that the pieces of cylinders ˜ C ( r ; m ) of l havearbitrary large height and are graphs on strips of arbitrary width (see (28)). Thisgives a priori height estimates by choosing m sufficiently large in (28). Furthermore,the same cylinders provide us the boundary gradient estimates.With a similar argument, we can derive a priori boundary gradient estimatesby using hyperbolic caps. The only difference is that we have to assume strictlyconvexity κ ∂ Ω > C ( r, m ) by Euclidean circularcylinders. Let H > v ( x , x ) = − (cid:112) r − x , r = 1 / (2 H ) whose mean curvature is H with the orientation given in (7). The onlycaution is to assure that the width of any strip containing the (convex) domain Ωis less than 1 / | H | as well as its height is less than 1 / (2 H ). Again this gives notonly the height estimates but also the boundary gradient estimates. With the sameideas as in Theorem 7.1, we prove ( [12]): theorem 7.2. Let
H > and Ω ⊂ R be a bounded domain with κ ∂ Ω ≥ . Ifdist ( L , L ) < H , for all ( L , L ) ∈ L , (29) then the Dirichlet problem (22) has a unique solution. Comparing this result with Theorem 6.2, the domain here is merely convex evencan contain segments of straight-lines; in contrast, the domain Ω is small in relationto the value of 1 /H . roof. Compare M = graph( u ) with the cylinders C ( r ) = graph( v ). An argumentas in Theorem 7.1 proved that the hypothesis (29) ensures that − / (2 H ) < u < one pair of lines( L , L ) ∈ L . The boundary gradient estimates follow comparing with quarter ofcylinders C ( r ) defined in the strip 0 ≤ x ≤ / (2 H ).The following result solves affirmatively the Dirichlet problem in the Lorentz-Minkowski space (25) for arbitrary domains. For this, we will use cmc rotationalspacelike surfaces of l as barriers. We now describe the rotationally symmetricsolutions of (1).Consider a rotational surface about the x -axis obtained by the curve ( r, , w ( r )),0 ≤ a < r < b . With respect to the orientation (7), the mean curvature H satisfies w (cid:48)(cid:48) (1 − w (cid:48) ) / + w (cid:48) r √ − w (cid:48) = 2 H. (30)The spacelike condition is equivalent to w (cid:48) <
1. Multiplying by r , a first integral is Hr + c = rw (cid:48) √ − w (cid:48) for a constant c ∈ R , or equivalently w (cid:48) = ± Hr + c (cid:112) r + ( Hr + c ) · (31)If c = 0, the solution is w ( r ) = (cid:112) /H + r , up to a constant, that correspondswith a hyperbolic plane H (1 /H ).Let H > c <
0. Since w (cid:48) <
1, the function w is defined in (0 , ∞ ). By (31), w (cid:48)(cid:48) > w (cid:48) vanishes at a unique point, namely, r = (cid:112) − c/H . It is also clearthat lim r → w (cid:48) ( r ) = −
1. Consider w = w ( r ; c ) be the solution of (31) parametrizedby the constant c assuming initial condition w ( r ) = 0 , (so w (cid:48) ( r ) = 0) . (32)Let S ( c ) denote the graph of w ( r ; c ) with r = x + x . See Fig. 2, left. Let ξ c = lim r → w ( r ; c ). The functions w ( r ; c ) have the following properties.1. S ( c ) presents a singularity at the intersection point with the rotation axis. SeeFig. 2, right. At this point, the surface is tangent to the (backward) light-conefrom w (0; c ), namely, x + x = ( x − ξ c ) , x < ξ c . . lim c →−∞ r ( c ) = + ∞ and lim c →−∞ ξ c = + ∞ .3. lim c → r ( c ) = 0 and lim c → ξ c = 0. Figure 2: Left: profiles of generating curves of cmc rotational spacelike surfaces S ( c ) forvalues c = 1, c = 2 and c = 3. Right: a cmc rotational spacelike surface. The following result has not a counterpart in the Euclidean space. theorem 7.3. If Ω is a bounded smooth domain, then the Dirichlet problem (25) has a unique solution.Proof. If H = 0, the solution is the function u = 0. Let H (cid:54) = 0. By changing u by − u if necessary, without loss of generality we suppose that H >
0. We knowby Theorem 5.4 that u < a priori estimates for the solution u of (1) which corresponds with the value t = 1. Moreover,the function w + = 0 is an upper barrier because Q [ w + ] = − H < w + = u along ∂ Ω. In order to find lower barriers for u , we will take pieces of cmc rotationalsurfaces S ( c ) for suitable choices of the parameter c depending only on the initialdata.Since Ω is smooth ( C is enough), Ω satisfies a uniform exterior circle condition.This means that there exists a small enough ε > x ∈ ∂ Ω, there is a disc D ε of radius ε and depending on x such that D ε ∩ Ω = ∅ , D ε ∩ Ω = { x } . As consequence, the same property holds for every ε (cid:48) > ε (cid:48) ≤ ε .Fix the above ε . Let w = w ( r ; c ) be a solution of (31)-(32) defined only in theinterval [ ε, r ] and let S ( c ; ε ) be its graph. Here, and in what follows, we identifythe function w = w ( r ) of one variable with the rotationally symmetric function of wo variable w = w ( x , x ) by setting x + x = r . Then the boundary of S ( c ; ε )are the circles ∂S ( c ; ε ) = C ∪ C := { ( x , x , w ( ε ; c )) : x + x = ε } ∪ { ( x , x ,
0) : x + x = r } . By the height estimates of Theorem 5.4, there exists a constant
K > − K < u < c < r ( c ) > diam(Ω) , w ( ε ; c ) > K. (33)Given ε , the last inequality is a consequence of ξ c → ∞ as r → −∞ . Let w − = w ( r ; c ).Let x ∈ ∂ Ω be a boundary point and let D ε be the disc given by the uniformexterior circle condition. We now prove that it is possible to choose a suitable S ( c ; ε )such that S ( c ; ε ) is a lower barrier for u around the point x . In what follows, wedenote by the same symbol S ( c ; ε ) any vertical translation of this surface whichcorresponds with the functions w ( r ; c ) + k for different choices of the constant k .After a horizontal translation, we suppose x = ( ε,
0) and that the disc D ε ofthe uniform exterior circle condition is x + x < ε . We move vertically down thesurface S ( c ; ε ) until that it does not intersect M = graph( u ). Then we come backby lifting vertically upwards S ( c ; ε ). M Ω S(c; ε )C Figure 3: The surface S ( c ; ε ) is a lower barrier for the graph M . Claim.
It is possible to move upwards S ( c ; ε ) without touching M until that weplace S ( c ; ε ) just at the position where the boundary circle C coincides with ∂D ε .See Fig. 3.This occurs because the touching principle forbids a first contact at some com-mon interior point. The other possibility is that during the vertical displacement, nd before to arrive to the final position, some boundary point of S ( c ; ε ), namely, apoint of C , touches M : the circle C does not touch M because D ε ∩ Ω = ∅ . Theother circle C projects onto R in the circle x + x = r which contains Ω insideby the first property of (33). Finally, the circle C does not touch M because thevertical distance between C and C is w − ( ε ; c ) − w − (0; c ) = w ( ε ; c ) > K by (33).Once we have placed S ( c ; ε ) so that C = ∂D ε , the lower barrier is w − = w ( r ; c ) − w ( ε ; c ) defined in the annulus U = { ( x , x ) : ε < x + x < r } . Wededuce that w − < u in Ω ∩ U . This proves that | Du ( x ) | < | Dw − ( x ) | by Lemma 5.7and this value depends only on the initial data, namely, | Dw − ( x ) | = − ddr (cid:12)(cid:12)(cid:12) r = ε w ( r ; c ) = − Hε + c (cid:112) ε + ( Hε + c ) · . This gives the constant C in (21). References [1] Bernstein, S.N. Sur une th´eor`eme de g´eometrie et ses applications aux´equations d´eriv´ees partielles du type elliptique. Comm. Soc. Math. Kharkov, , , 38–45.[2] Bombieri, E., De Giorgi, E., Giusti, E. Minimal cones and the Bernsteinproblem, Invent. Math. , , 243–268.[3] Cheng, S. Y., Yau, S. T. Maximal space-like hypersurfaces in the Lorentz-Minkowski spaces. Ann. of Math. (2) , , 407–419[4] Choquet-Bruhat, Y., York, J. The Cauchy Problem. In: General Relativityand Gravitation
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