The Dirichlet problem of the constant mean curvature equation in Lorentz-Minkowski space and in Euclidean space

We investigate the differences and similarities of the Dirichlet problem of the mean curvature equation in the Euclidean space and in the Lorentz-Minkowski space. Although the solvability of the Dirichlet problem follows standards techniques of elliptic equations, we focus in showing how the spacelike condition in the Lorentz-Minkowski space allows to drop the hypothesis on the mean convexity which is required in the Euclidean case.


Introduction
In this paper we investigate the differences and similarities in the study of the solvability of the Dirichlet problem for the constant mean curvature equation in the Euclidean space and in the Lorentz-Minkowski space. Firstly we introduce the following notation. Let ∈ {−1, 1}. Denote by R n+1 the vector space R n+1 equipped with the metric where (x 1 , . . . , x n+1 ) are the canonical coordinates of R n+1 . If = 1 (resp. = −1), the space is the Euclidean space 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 . Let Ω ⊂ R n be a bounded domain with smooth boundary ∂Ω and let H be a real number. The Dirichlet problem asks for existence and uniqueness of a function u ∈ C 2 (Ω) ∩ C 0 (∂Ω) such that in Ω (1) Here 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 hypersurface with constant mean curvature H in R n+1 whose boundary is contained in the hyperplane x n+1 = 0. If = −1, the extra condition |Du| < 1 in Ω means that the hypersurface is spacelike. A hypersurface in E n+1 (resp. in l n+1 ) with zero mean curvature (H = 0) is called a minimal (resp. maximal) hypersurface. The example that shows the differences of the theory of constant mean curvature hypersurfaces in both ambient spaces is the Bernstein problem which we now formulate. Suppose that the domain Ω is R n . A graph on R n is called an entire graph. Let H = 0. The Bernstein problem asks if, besides linear functions, there are 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 arbitrary dimension, this result holds if n ≤ 7. A famous theorem of Bombieri, De Giorgi and Giusti asserts that there are other entire minimal graphs if n ≥ 8 ( [2]). In contrast, in n-dimensional Lorentz-Minkowski space, Cheng and Yau proved, extending previous works of Calabi, that spacelike hyperplanes are the only entire maximal hypersurfaces ( [3]).
The interest of the study of constant mean curvature (cmc in short) hypersurfaces has its origin in physics. In the Euclidean space E 3 , cmc surfaces are mathematical models of the shape of a liquid in capillarity problems and of a interface that separates two medium of different physical properties. In Lorentz-Minkowski l n+1 , cmc spacelike hypersurfaces have been used in General Relativity to prove the 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 constant mean curvature equation in both spaces. Assume that u takes arbitrary continuous boundary 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 arbitrary dimension by Jenkins and Serrin [7] proving that the mean convexity of the domain Ω yields a necessary and sufficient condition of the solvability of the Dirichlet problem for all boundary values ϕ: a domain Ω is said to be mean convex if the mean curvature κ ∂Ω of ∂Ω with respect to the inner normal is non-negative. If H = 0, a stronger assumption is needed on Ω relating H and κ ∂Ω and the answer appears in the seminal paper [8], where Serrin proved the following result.
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 convexity assumption of ∂Ω, it is possible to derive existence results if one assumes smallness on 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 boundary values was initially investigated in the maximal case H = 0 assuming the mean convexity of ∂Ω ( [13,14]). However, the groundbreaking result is due to Bartnik and Simon in 1982 where the counterpart to Theorem 1.2 in l n+1 is surprisingly simple because there is not any assumption on ∂Ω ( [15]). problem, we need to ensure a priori estimates of the height and the gradient for the prospective solutions. Throughout this paper, we refer to the reader [11] as a general guide.
The purpose of this work is twofold. Firstly, give an approach to the results in Lorentz-Minkowski space comparing with the ones of Euclidean space and showing how the spacelike condition |Du| < 1 makes completely different the method of obtaining the a priori estimates. The second objective is to provide geometric proofs to derive these estimates. For example, Serrin used the distance function to ∂Ω as a barrier for the desirable estimates ( [8]), and similarly Flaherty in the solvability in the Lorentzian case when H = 0 ( [14]). This distance function is defined in Ω but loses its geometric sense if we look the graph of u in E 3 or l 3 . In our case, the a priori estimates will be obtained by a comparison argument between the solutions of (1) and known cmc surfaces, such as, rotational surfaces. In order to simplify the notation and arguments, we will consider the Dirichlet problem for the 2-dimensional case, so we will work with surfaces in E 3 and spacelike surfaces in l 3 . 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 to fix some definitions and notations, we derive the constant mean curvature equation in Section 3 obtaining some properties of the solutions showing differences in both ambient spaces. Section 4 describes the method of continuity to solve the Dirichlet 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) gradient estimates. In Section 6, we analyze the solvability of the Dirichlet problem in the Euclidean case showing that a strong convexity hypothesis is necessary to solve the problem. Finally, in Section 7 we solve the Dirichlet problem in Lorentz-Minkowski space for arbitrary domains and we show the role of the cmc rotational surfaces in the solvability of the problem.

Preliminaries
We need to recall some definitions in Lorentz-Minkowski space. In l 3 , the metric , is non-degenerate of index 1 and classifies the vectors of R 3 in three types: a vector v ∈ l 3 is said to be spacelike (resp. timelike, is called spacelike (resp. timelike, lightlike) if the induced metric on U is positive definite (resp. non-degenerate of index 1, degenerate and U = {0}). Any vector subspace belongs to one of the above three types. For 2-dimensional subspaces, U is spacelike (resp. timelike, lightlike) if its orthogonal subspace U ⊥ is timelike (resp. spacelike, lightlike). A curve or a surface immersed in l 3 is said to be 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 and in l 3 , the scalar product of any two timelike vectors is not zero. Thus, if we fix e 3 = (0, 0, 1), which is a timelike vector, it is possible to define a unit orthogonal vector field N on M so N, e 3 is negative (or positive) on M , determining a global orientation. Another consequence is that there do not exist closed spacelike surfaces in l 3 , 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 3 . In case = −1, we also assume that the immersion is spacelike. Let ∇ 0 and ∇ be the Levi-Civita connections in R 3 and M respectively. The Gauss formula is ∇ 0 X Y = ∇ X Y + σ(X, Y ) for any two tangent vector fields X and Y on M , where σ is the second fundamental form. The mean curvature H of M is defined as Let us choose N a unit normal vector field on M with N, N = . Let A = ∇ 0 N stand for the Weingarten endomorphism with respect to N . Then the Gauss formula is ∇ 0 X Y = ∇ X Y + A(X), Y N and A is a diagonalizable map. If κ 1 and κ 2 are the principal curvatures, we have Remark 2.1. In case of timelike surfaces of l 3 , the mean curvature is defined as in (5). However, although A is self-adjoint with respect to the induced metric , , this metric is Lorentzian and it may occur that A is not real diagonalizable.
Example 2.2. 1. Planes of E 3 and spacelike planes of l 3 have zero mean curvature.
2. Round spheres S 2 (r) in E 3 and hyperbolic planes H 2 (r) in l 3 of radius r > 0 can be described up to a rigid motion as {p ∈ l 3 : p, p = r 2 }.

3.
Right circular cylinders of R 3 have constant mean curvature. To be precise, let a ∈ R 3 be a unit vector with a, a = 1 (in l 3 , the vector a is spacelike). Up to a rigid motion, the circular cylinder of axis a and radius r > 0 is For the orientation N (p) = (p− p, a a)/r, the mean curvature is H = − /(2r).
4. Let u = u(x 1 , x 2 ) be a smooth function defined in a open domain Ω ⊂ R 2 and let M be the graph of u. Suppose that M is endowed with the induced metric from R 3 . If = −1, we also assume that M is spacelike, that is, |Du| < 1 in Ω. The mean curvature H of M satisfies Let us notice that (6) coincides with the equation (1).

The constant mean curvature equation
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 divergence form div Du with the observation that if = −1, we assume the spacelike condition |Du| < 1 in Ω. For instance, spheres and hyperbolic planes of Example 2.2 are graphs of the functions is defined in a disc and describes a hemisphere in S 2 (r), and for = −1, x 3 = u(x 1 , x 2 ) is the hyperbolic plane H 2 (r). On the other hand, a cylinder C(r) with axis a = (0, 1, 0) and radius r > 0 is the graph of the function is of quasilinear elliptic type, hence we can apply the machinery for these equations. It is easily seen that the difference of two solutions of equation (1) satisfies the maximum principle. As a consequence, we give a statement of the comparison principle in our context. We define the operator The comparison principle asserts ([11, Th. 10.1]).
, then u < v in Ω. In particular, the solution of the Dirichlet problem, if it exists, is unique.
An immediate consequence is the touching principle. A first difference of the Dirichlet problem for the constant mean curvature equation (1) is that in the Euclidean space E 3 the value H is not arbitrary and depends on the size of Ω, whereas in l 3 the value H may be arbitrary. Indeed, from equation (8), the divergence theorem yields where n is the outward unit normal vector along ∂Ω. The idea is to estimate the right-hand side from above. If = 1, we have

Proposition 3.3. A necessary condition for the solvability of the Dirichlet problem
Let us notice that this upper bound for H does not depend on the boundary values ϕ. In fact, there are explicit examples where all values between 0 and the upper 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 each 0 < H < 1/ρ, take the spherical cap of radius 1/|H| Then u is a graph on Ω with constant mean curvature H for every H going from 0 until 1/ρ. The limit case H = 1/ρ corresponds with a hemisphere of radius 1/|H|. The same computations in l 3 do not provide the same conclusion because |Du|/ 1 − |Du| 2 may be arbitrarily large. So, for the hyperbolic planes the value |Du| is arbitrary large and the function u is defined in any domain of the plane R 2 and for 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 3 , the hyperbolic planes (11) show that for any H, there are solutions (1) defined in the plane R 2 . Also the cylinders u(x 1 , are other examples of entire solutions of (1)-(3). However in the Euclidean space, we have Proposition 3.4. Let Ω be a domain of R 2 . If u is a solution of (1) with H = 0 in E 3 , then Ω does not contain the closure of a disk of radius 1/|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 suppose that the sign of H is positive: recall that the mean curvature is computed with respect to the orientation (7). Let r = 1/H and S 2 (r) be a sphere of radius r whose center lies on the straight-line through x and perpendicular to the (x 1 , x 2 )plane. Here, and in what follows, S 2 (r) denotes a sphere of radius r whose center may be changing. We orient S 2 (r) by the inward orientation. With this choice of orientation, the mean curvature is H and the orthogonal projection of S 2 (r) on R 2 is D.
Let M be the graph of u. Lift S 2 (r) vertically upwards until S 2 (r) is completely above M . Then, let us descend S 2 (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 both surfaces. By the touching principle, the surfaces M and S 2 (r) agree on an open set around p, hence M is included in a sphere of radius 1/H: this is a contradiction because the orthogonal projection onto R 2 would give Ω ⊂ D.

The solvability techniques of the Dirichlet problem
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 solvability of the Dirichlet problem by applying the method of continuity ( [11,Sec. 17.2]). The matrix of the coefficients of second order of (1) is The minimum and maximum eigenvalues of this matrix are λ = 1 and Λ = 1+|Du| 2 if = 1 and λ = 1 − |Du| 2 and Λ = 1 if = −1. Thus if = −1, the equation (1) is uniformly elliptic provided |Du| < 1 uniformly in Ω.
For t ∈ [0, 1], define the family of Dirichlet problems The existence of solutions of the Dirichlet problem (1)- For this purpose, we prove that A is a non-empty open and closed subset of [0, 1]. We analyze these three issues.
1. The set A is not empty. This is because u = 0 solves the Dirichlet problem for t = 0.

The set
A is open in [0, 1]. Given t 0 ∈ A, we need to prove that there exists The computation of L will be done in Theorem 5.6, obtaining and L is a linear elliptic operator whose term for the function v is zero. Therefore the existence and uniqueness is assured by standard theory ([11, Th. 6.14]).

The set
Then {u k } ⊂ S. If we see that the set S is bounded in C 1,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 2,β (Ω), in particular, S is precompact in C 2 (Ω) (Th. 6.6 and Lem. 6.36 in [11]). Hence there is a subsequence where the norm is defined by Usually, the a priori estimates for |u| are called height estimates and gradient estimates for |Du|.
Definitively, A is closed in [0, 1] provided we find two constants M and C independent on t ∈ A, such that Here we make the observation that whereas in the Euclidean space, the constant 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, during the above process of the method of continuity, we require that Q t is uniformly elliptic, in particular, we have to ensure that |Du| << 1 in Ω. Definitively, in l 3 , the constant C in (12) has to satisfy the condition C < 1.

Height and gradient estimates
Consider the Dirichlet problem for the cmc equation and arbitrary boundary values where, in addition, if = −1, we suppose |Du| < 1 in Ω. In this section we investigate the problem of finding estimates of |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 the estimates 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 ambient spaces is that in E 3 there exist estimates of sup Ω |u| depending only on H and ϕ, whereas in l 3 the size of the domain Ω appears in these estimates, such as shows the hyperbolic planes (11).
The height estimates for cmc graphs in the Euclidean space are obtained with the functions where a is a fixed unit vector of R 3 and N is the Gauss map of M . Firstly we need to compute the Beltrami-Laplacian ∆ M of the functions f and g. The following result holds for cmc surfaces in E 3 and in l 3 without to be necessarily graphs: we refer the reader to [21] for a proof.
Lemma 5.1. Let M be an immersed surface in R 3 . Then If, in addition, the immersion has constant mean curvature, then where |σ| is the norm of the second fundamental form.
In particular, sign(g) = sign( ). Suppose H ≥ 0. Then ∆ M f ≥ 0 (resp. ≤ 0) in E 3 (resp. l 3 ) and the maximum principle implies p, e 3 ≤ max ∂Ω p, e 3 in E 3 (resp. p, e 3 ≥ min ∂Ω p, e 3 in l 3 ). Thus u ≤ max ∂Ω u in both ambient spaces. On the other hand Since |σ| 2 = κ 2 1 + κ 2 2 ≥ 2H 2 , the maximum principle yields We analyze the same argument in l 3 . The reverse Cauchy-Schwarz inequality for timelike vectors yields N, e 3 ≤ −1 ( [22]). Then the same computation gives but it is not possible to bound from below because of the function N, e 3 . This makes a key difference with the Euclidean case and concludes that the argument done in the Euclidean space is not valid in l 3 . If H = 0, from (14) we deduce: In both ambient spaces, if u is a solution of (13) for H = 0 then As expected, in the Lorentz-Minkowski space there does not exist height estimates depending only on H and ϕ. An example is the following. For r > 0 and m > r, let u m (x 1 , The graph of u m is a piece of the hyperbolic plane H 2 (r) which has been displaced vertically downwards a distance equal to m. Then u m is a solution of (13) in Ω √ m 2 −r 2 with ϕ = 0 and the height on u m , namely |u m | = m − r, goes to ∞ as m ∞. 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 area of the surface). The estimates that we will deduce are of two types: the first ones are given in terms of the diameter of Ω and second ones depend on the width of narrowest strip containing Ω. theorem 5.4. If u be a solution of (13) in l 3 , then and equality holds if and only if the graph of u describes a hyperbolic cap. In the particular case ϕ = 0, we have Proof. The inequalities are obtained by comparing M = graph(u) with hyperbolic caps with mean curvature |H| coming from below and from above. There is no loss of generality in assuming that Ω is included in the closed disk D ρ of center the origin and radius ρ = diam(Ω)/2. Consider the hyperbolic plane H 2 (r) defined by the function u(x 1 , x 2 ) = r 2 + x 2 1 + x 2 2 , where r = 1/|H|. Let us take H 2 (r; s) the compact part obtained when we intersect H 2 (r) with the horizontal plane of equation x 3 = s. Then ∂H 2 (r; s) is a circle of radius ρ, with s = ρ 2 + r 2 and Move vertically down H 2 (r; s) until to be disjoint from M . Next move upwards H 2 (r; s) until that H 2 (r; s) touches M the first time. If the contact between both surfaces occurs at some common interior point, the comparison principle and then the touching principle implies that u describes part of the hyperbolic plane H 2 (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 2 (r; s) with a boundary point of M , we can arrive until the value s = min ∂Ω ϕ, hence Evaluating at the origin, which coincides with the left inequality in (17) because ρ = diam(Ω)/2. The right hand inequality in (17) is proved with a similar argument by taking the hyperbolic planes u(x 1 , x 2 ) = − r 2 + x 2 1 + x 2 2 .
A second height estimate can be deduced by comparing u with spacelike cylinders. We need to introduce the following notation. Given a bounded domain A ⊂ R 2 , consider the set L of all pairs of parallel straight-lines (L 1 , L 2 ) in R 2 such that A is included in the planar strip determined by L 1 and L 2 . Set Θ(A) = min{dist(L 1 , L 2 ) : (L 1 , L 2 ) ∈ L}.
Observe that the domain A is included in a strip of width Θ(Ω) and this strip is the narrowest one among all strips containing A in its interior. Notice also that Θ(A) ≤ δ(A). theorem 5.5. If u is a solution of (13) in l 3 , then In the particular case ϕ = 0, we have Notice that the estimates (18) and (17)  At the points x 1 = 0, we deduce This inequality is just the left inequality in (18). The right inequality in (18) is proved by comparing with the cylinders u(x 1 , x 2 ) = − r 2 + x 2 1 .
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 initial data such that |Du| ≤ C in Ω, with the observation that if = −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. theorem 5.6. If u is a solution of (13), then Proof 1. For each i = 1, 2, define the functions v i = D i u. Differentiate (9) with respect to the variable x k , k ∈ {1, 2}. After some computations, we obtain Hence v k satisfies a linear elliptic equation of type where a ij = a ij (Du) and b i = b i (Du, D 2 u). By the maximum principle, |v k | has not a maximum at some interior point. Consequently, the maximum of |Du| on the compact set Ω is attained at some boundary point.

Thus inf
which is equivalent to (19).
To summarize, the problem of finding gradient estimates of |Du| in Ω is passing to a problem of estimates along the boundary, exactly, finding a constant C depending only on the initial data such that In the proofs of the existence results in the following sections, the method to obtain the constant C in (21) is by an argument of super and subsolutions and then we apply the next result.
Lemma 5.7. Let x 0 ∈ ∂Ω be a boundary point. Suppose that there is a neighborhood U of x 0 and two functions w + , w − ∈ C 2 (Ω ∩ U) such that Then |Du| ≤ C.

The Dirichlet problem with zero boundary values: the Euclidean case
In this section we address the Dirichlet problem (1) in the Euclidean space. By Theorem 5.2, we know that the value H is not arbitrary. Without to assume convexity on ∂Ω, there are results of existence assuming some smallness on the value H and on the size of Ω ( [10,11]. Thanks to this smallness on initial data, it is possible to obtain height and boundary gradient estimate of the solution. If we assume convexity, there are different hypothesis that ensure the solvability of the Dirichlet problem and 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 arbitrary boundary values. If we now suppose that u = 0 on ∂Ω, the hypothesis (4) can be weakened assuming κ ∂Ω ≥ |H|. We give two proofs of this result. The first one will be proved in arbitrary dimension and, although the idea appears generalized in other ambient spaces ( [29,30,31,32]), as far as we know, in the literature there is not specifically a statement in the Euclidean space. Here we follow [32].
in arbitrary dimension has a unique solution.
Proof. Firstly, we observe that the solutions u t of the method of continuity (Section 4) are ordered in decreasing sense according the parameter t. Indeed, if t 1 < t 2 , Since u t 1 = 0 = u t 2 on ∂Ω, the comparison principle yields u t 2 < u t 1 in Ω. Thus, where for the value t = 1, u 1 is the solution u of (1). By using Lemma 5.7, this implies that it suffices to find a priori height and gradient estimates 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. By the height estimates of Theorem 5.2, we know −1/H < u < 0 in Ω. This gives the 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 estimates on the domain Ω.
We use again the function Hf + g as in Theorem 5.2. Since ∆ M (Hf + g) ≤ 0 and u = 0 on ∂Ω, the maximum principle ensures the existence of a boundary point q ∈ ∂Ω where Hf + g attains its minimum, so Furthermore, the maximum principle on the boundary implies H ν(q), e 3 + dN q ν, e 3 ≥ 0, where ν is the inward unit conormal vector along ∂Ω. If σ is the second fundamental form, this inequality can be written as (H − σ(ν(q), ν(q))) ν(q), e 3 ≥ 0.
Finally, we conclude from (16) in Ω obtaining the desired gradient estimates in Ω.
The second proof is done in the two-dimensional case, where the mean convexity is now the convexity in the Euclidean plane. The proof uses spherical caps to find the boundary gradient estimates (21). 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 0 = min q∈∂Ω κ ∂Ω (q) > 0 and r = 1/κ 0 .
Firstly, we prove Theorem 5.4 in case of strict inequality κ ∂Ω > H. Let x ∈ ∂Ω be 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 0 > H. Consider C 1/H a circle of radius 1/H and concentric to C r . Notice that r < 1/H. After a translation we suppose that the center of D r is the origin of coordinates.
Let S 2 (1/H) be the hemisphere of radius 1/H whose boundary is C 1/H and below the plane Π of equation x 3 = 0. Let us lift up S 2 (1/H) until its intersection with Π is C r . Denote by S r the piece of S 2 (1/H) below Π at this position. See Figure 1. The surface S r is a small spherical cap which is the graph of We prove now that M lies in the bounded domain determined by S r ∪ D r . For this, we move down S r by vertical translations until S r does not intersect M and then, 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, in a neighborhood of the point x, the surface M lies sandwiched between S r and Π. Then and consequently by Lemma 5.7 where this constant depends only on r and H. Until here, we have obtained the existence of a solution for each 0 < H < κ 0 . Moreover, and since the gradient is bounded from above in Ω depending only on the initial data, the solution obtained is smooth in Ω. Now, we proceed by proving the existence of a solution of (1) in the case H = κ 0 : in case that Ω is a round disk of radius r (and κ 0 = 1/r), the solution is u( 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 0 > H n . By the monotonicity of H n and the comparison principle, the sequence {u n } is monotonically increasing and converges uniformly on compact sets of Ω. Let u = lim u n . Standard compactness results involving Ascoli-Arzelá theorem guarantee that u ∈ C 2 (Ω) and Q[u] = 0. It remains to check that u ∈ C 0 (Ω) and u = 0 on ∂Ω. Let x ∈ ∂Ω and {x m } ⊂ Ω with x m → x. Consider the hemisphere S 2 (r) as above and let D r be the open disk of radius r = 1/H such that S 2 (r) = graph(v), with v ∈ C ∞ (D r ) ∩ C 0 (D r ). Place D r such that x ∈ ∂D r . We know that Ω ⊂ D r and by the touching principle, 0 < u n < v on Ω. For each n ∈ n, 0 < u n (x m ) < v(x m ).
This proves the continuity of u up to ∂Ω and that u = 0 on ∂Ω.

The Dirichlet problem with zero boundary values: the Lorentzian case
In this section we address the Dirichlet problem in l 3 following the ideas of the Euclidean case in the above section. The first result that we present is motivated by Theorem 6.2, where we assumed a strong convexity of ∂Ω comparing with the value H, namely, κ ∂Ω ≥ |H|. In contrast, in Lorenz-Minkowski space this convexity assumption changes by merely the convexity κ ∂Ω ≥ 0 of ∂Ω.
has a unique solution.
Proof. With a similar argument as in Theorem 6.2, the solutions u t of the method of continuity are ordered by u t 1 < u t 2 if t 2 < t 1 , so it suffices to get the a priori estimates for the solution u of (25). Without loss of generality, we suppose H > 0.
The height estimates were proved in Theorem 5.4 (or 5.5) and we showed that there exists K = K(Ω, H) > 0 such that In order to find the a priori boundary gradient estimates, consider the cylinder C(r) determined by v(x 1 , x 2 ) = r 2 + x 2 1 , where r = 1/(2H). For each m > r, let This surface is a graph on the strip Ω r, Take m sufficiently large so m fulfills the next two conditions: Let us restrict v in the half-strip andC(r; m) denotes the graph of v on U. The boundary ofC(r; m) is formed by two parallel straight-lines where L 1 is contained in the plane x 3 = r and L 2 in the plane x 3 = m, with r < m. Let x 0 ∈ ∂Ω be a fixed but arbitrary point of the boundary of Ω. After a rotation about a vertical axis and a horizontal translation, we suppose x 0 = ( √ m 2 − r 2 , 0), Ω is contained in U (this is possible by (28)) and the tangent line L to ∂Ω at x 0 is parallel to the x 2 -line. By vertical translations, we displace vertically downC(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 thatC(r; m) touches M occurs when L 2 arrives to the plane of equation x 3 = 0 and consequently, L = L 2 . Firstly, the touching principle prohibits an interior tangent point between M andC(r; m). On the other hand, it is not possible that a boundary point of of C(r; m), namely, a point of L 1 ∪ L 2 , touches a point of M because (26) and (27). Definitively, we can movẽ C(r; m) until L 2 coincides with L, in particular, At this position,C(r; m) is the graph of the function Thus M is contained between w − and w + = 0 in Ω ∩ U with w − (x 0 ) = w + (x 0 ) = u(x 0 ) = 0. We are in position to apply Lemma 5.
. We conclude that |Du| ≤ C, where the constant C in (21) is The key in the above proof is that the pieces of cylindersC(r; m) of l 3 have arbitrary large height and are graphs on strips of arbitrary width (see (28)). This gives 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 estimates by using hyperbolic caps. The only difference is that we have to assume strictly convexity κ ∂Ω > 0. After Theorem 7.1, we can come back to Euclidean space asking if it is possible a similar argument by replacing the pieces of cylinders C(r, m) by Euclidean circular cylinders. Let H > 0 and consider the circular cylinder v(x 1 , x 2 ) = − r 2 − x 2 1 , r = 1/(2H) whose mean curvature is H with the orientation given in (7). The only caution is to assure that the width of any strip containing the (convex) domain Ω is less than 1/|H| as well as its height is less than 1/(2H). Again this gives not only the height estimates but also the boundary gradient estimates. With the same ideas as in Theorem 7.1, we prove ( [12]): then the Dirichlet problem (22) has a unique solution.
Comparing this result with Theorem 6.2, the domain here is merely convex even can contain segments of straight-lines; in contrast, the domain Ω is small in relation to the value of 1/H.
Proof. Compare M = graph(u) with the cylinders C(r) = graph(v). An argument as in Theorem 7.1 proved that the hypothesis (29) ensures that −1/(2H) < u < 0 in Ω: in fact, for this estimate it suffices that (29) holds for one pair of lines (L 1 , L 2 ) ∈ L. The boundary gradient estimates follow comparing with quarter of cylinders C(r) defined in the strip 0 ≤ x 1 ≤ 1/(2H).
The following result solves affirmatively the Dirichlet problem in the Lorentz-Minkowski space (25) for arbitrary domains. For this, we will use cmc rotational spacelike surfaces of l 3 as barriers. We now describe the rotationally symmetric solutions of (1).
Consider a rotational surface about the x 3 -axis obtained by the curve (r, 0, w(r)), 0 ≤ a < r < b. With respect to the orientation (7), the mean curvature H satisfies The spacelike condition is equivalent to w 2 < 1. Multiplying by r, a first integral is for a constant c ∈ R, or equivalently w = ± Hr 2 + c If c = 0, the solution is w(r) = 1/H 2 + r 2 , up to a constant, that corresponds with a hyperbolic plane H 2 (1/H). Let H > 0 and c < 0. Since w 2 < 1, the function w is defined in (0, ∞). By (31), w > 0 and w vanishes at a unique point, namely, r 0 = −c/H. It is also clear that lim r→0 w (r) = −1. Consider w = w(r; c) be the solution of (31) parametrized by the constant c assuming initial condition w(r 0 ) = 0, (so w (r 0 ) = 0).
Let S(c) denote the graph of w(r; c) with r 2 = x 2 1 + x 2 2 . See Fig. 2, left. Let ξ c = lim r→0 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. See Fig. 2, right. At this point, the surface is tangent to the (backward) light-cone from w(0; c), namely, The following result has not a counterpart in the Euclidean space. Proof. If H = 0, the solution is the function u = 0. Let H = 0. By changing u by −u if necessary, without loss of generality we suppose that H > 0. We know by Theorem 5.4 that u < 0 in Ω. As in Theorem 7.1, it suffices to find 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 + ] = −2H < 0 in Ω and w + = u along ∂Ω. In order to find lower barriers for u, we will take pieces of cmc rotational surfaces S(c) for suitable choices of the parameter c depending only on the initial data.
Since Ω is smooth (C 2 is enough), Ω satisfies a uniform exterior circle condition. This means that there exists a small enough ε > 0 depending only on Ω with the following property: for any boundary point 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 > 0 with ε ≤ ε.
By the height estimates of Theorem 5.4, there exists a constant K > 0 depending only on the initial data such that −K < u < 0 in Ω. Let c < 0 be sufficiently small with the next two properties r 0 (c) > diam(Ω), w(ε; c) > K.
Given ε, the last inequality is a consequence of ξ c → ∞ as r 0 → −∞. Let w − = w(r; c). Let x ∈ ∂Ω be a boundary point and let D ε be the disc given by the uniform exterior 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, we denote by the same symbol S(c; ε) any vertical translation of this surface which corresponds 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 of the uniform exterior circle condition is x 2 1 + x 2 2 < ε 2 . We move vertically down the surface S(c; ε) until that it does not intersect M = graph(u). Then we come back by lifting vertically upwards S(c; ε). Claim. It is possible to move upwards S(c; ε) without touching M until that we place S(c; ε) just at the position where the boundary circle C 1 coincides with ∂D ε . See Fig. 3.
This occurs because the touching principle forbids a first contact at some common interior point. The other possibility is that during the vertical displacement, and before to arrive to the final position, some boundary point of S(c; ε), namely, a point of C 2 , touches M : the circle C 1 does not touch M because D ε ∩ Ω = ∅. The other circle C 2 projects onto R 2 in the circle x 2 1 + x 2 2 = r 2 0 which contains Ω inside by the first property of (33). Finally, the circle C 2 does not touch M because the vertical distance between C 1 and C 2 is w − (ε; c) − w − (0; c) = w(ε; c) > K by (33).