Existence and Non-Existence for a Mean Curvature Equation in Hyperbolic Space

COMMUNICATIONS ON Website: http://AIMsciences.org PURE AND APPLIED ANALYSIS Volume 4, Number 3, September 2005 pp. 549–568

EXISTENCE AND NON-EXISTENCE FOR A MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE

Elias M. Guio Departamento de Matemática Universidade Federal do Espirito Santo, Vitória29060-900 ES, Brazil Ricardo Sa Earp Departamento de Matemática Pontif´ıciaUniversidade Católicado Rio de Janeiro Rio de Janeiro 22453-900 RJ, Brazil (Communicated by Gigliola Staffilani )

Abstract. There exists a well-known criterion for the solvability of the Dirich- let Problem for the constant mean curvature equation in bounded smooth domains in Euclidean space. This classical result was established by Serrin in 1969. Focusing the Dirichlet Problem for radial vertical graphs P.-A. Nitsche has established an existence and non-existence results on account of a criterion based on the notion of a hyperbolic cylinder. In this work we carry out a similar but distinct result in hyperbolic space considering a diﬀerent Dirichlet Problem based on another system of coordinates. We consider a non standard cylinder generated by horocycles cutting orthogonally a geodesic plane P along the boundary of a domain Ω ⊂ P. We prove that a non strict inequality between 0 the mean curvature HC (y) of this cylinder along ∂Ω and the prescribed mean 0 curvature H(y), i.e HC (y) > |H(y)|, ∀ y ∈ ∂Ω yields existence of our Dirichlet Problem. Thus we obtain existence of surfaces whose graphs have prescribed mean curvature H(x) in hyperbolic space taking a smooth prescribed boundary data ϕ. This result is sharp because if our condition fails at a point y a non-existence result can be inferred.

1. Introduction. A classical result in Partial Diﬀerential Equations and Diﬀer- ential Geometry due to Serrin (1969) is the following: Given a constant a, there exists a condition on the boundary of the domain Ω such that the Dirichlet Prob- lem for the mean equation H = a is solvable. In fact, if the mean curvature of the Euclidean cylinder over ∂Ω is bigger than a, then some a priori gradient estimates ensure a posteriori existence for the Dirichlet Problem. This inequality is sharp in the following sense: if it does not hold at a point then it can be inferred non- existence for a certain smooth boundary data. In this article we carry out a similar program in hyperbolic space, taking into account a certain coordinates system and a certain geometric boundary condition in hyperbolic space based on a suitable but not standard notion of “cylinder”. A novelty value in our paper is the deduction of a solution of the related Dirichlet Problem for prescribed (not necessarily constant) mean curvature H(x). Moreover, the sharpness of the result is assured by a non-

2000 Mathematics Subject Classiﬁcation. 35J25, 53A10. Key words and phrases. Dirichlet problem, mean curvature equation, hyperbolic space.

549 550 ELIAS M. GUIO AND RICARDO SA EARP existence result if our criterion in hyperbolic space “fails at a point”. In fact, let P be a n-dimensional totally geodesic plane in hyperbolic space Hn+1 and let Ω be a domain in P with C2 boundary ∂Ω. We shall focus the upper halfspace model of hyperbolic space, i.e Hn+1 := {(x0, x1, . . . , xn); xn > 0}equipped with the metric 2 1 2 2 2 ds = 2 (dx0 + dx1 + ··· + dxn). For sake of simplicity, we fix a hyperplane xn n+1 n+1 P := {x0 = 0}. Notice that the asymptotic boundary of H denoted ∂∞H is n+1 defined by ∂∞H := {xn = 0} ∪ {∞}. We represent by ∂Ω a (n − 1)-dimensional closed connected embedded smooth submanifold of P. We then define C by the n- dimensional cylinder generated by horocycles with asymptotic boundary ∞ cutting orthogonally P. That is,

C := {(t, x1, ··· , xn); t ∈ R, (0, x1, ··· , xn) ∈ ∂Ω}. Notice that in this model P is a vertical hyperplane and our cylinder C is an “Euclidean horizontal cylinder”. We shall consider throughout this paper the following deﬁnition of horizontal graph of a function in hyperbolic space: Let Ω ⊂ P be the domain whose boundary is ∂Ω. Given a function u : Ω → R, the graph of u is deﬁned as the set

G(u) = {(u(x1, . . . , xn), x1, . . . , xn); (0, x1, . . . , xn) ∈ Ω}. Indeed, this is the natural notion of graph in our system of coordinates given by horocycles cutting orthogonally P. We shall commence to describe the equation on which we are going to focus: Given a C1 function H : Ω¯ ⊂ P → R, and a C2 function ϕ : Ω → R, we shall consider the following Dirichlet Problem for the prescribed horizontal mean curvature equation, say Problem (P ): µ ¶ µ ¶ Du n D u div = H + n in Ω (1.1) W (u) xn W (u) u = ϕ on ∂Ω p where Du means the Euclidean gradient of u in P, W (u) = 1 + |Du|2 ∂u n Diu = , i = 1, . . . , n and the symbol “div” denotes the divergence in R = ∂xi {(x1, . . . , xn)}. We shall explain with details some historical backgrounds and mo- tivations. A main uniqueness result obtained by Lucas Barbosa and the second author (see [5], [6]) gives a strong motivation for the study of the above Dirichlet Problem. In fact, under some assumptions about the relation of the size of H with the geometry of ∂Ω , uniqueness of the above Dirichlet Problem (horizontal graphs) can be inferred in the class of immersed compact hypersurfaces with same mean curvature and same boundary. Particularly, if H is constant with H2 6 1, and if the principal curvatures λi of the boundary ∂Ω satisfy an inequality λi > h0 > 1, then any smooth compact connected immersed n-dimensional manifold with boundary ∂Ω is an horizontal graph. Lucas Barbosa and the second author have focused attention exclusively on the case of zero boundary conditions. Horizontal graphs satisfying the assumptions above were constructed to be used as barriers to obtain the cited main uniqueness theorem. Before proceeding with the discussion of our problem in hyperbolic space, we pause to make precise the criterion for the solvability of the Dirichlet Problem in Euclidean space, and for the equation of constant mean curvature equation in a MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 551

0 bounded C2 domain Ω. Let H³ be the´ mean curvature of ∂Ω. Then the constant Du mean curvature equation div W (u) = nH in Ω , u = ϕ on ∂Ω is solvable for 0 constant H and arbitrary boundary continuous data ϕ if and only if (n − 1)H > n |H| everywhere on ∂Ω. This theorem for smooth boundary data was established by J. Serrin in [26]. Notice that convexity of Ω is the sharp criterion of solvability for arbitrary continuous boundary data of the Dirichlet Problem on a bounded domain for the minimal surface equation in two variables; this was proved by R. Finn [10]. For more than two variables H. Jenkins and J. Serrin in [13] showed that the minimal hypersurface equation on a bounded C2,α domain, say Ω, 0 < α < 1, 0 is solvable for arbitrary ϕ ∈ C2,α(Ω)¯ if and only if the mean curvature H of ∂Ω is non-negative at every point of ∂Ω. Notice that the mean curvature equation in Euclidean space is now classical in the literature (see, for instance [7], [3], [9], [12]). Notice that there exist some recent interesting papers on minimal graphs over unbounded domains (see [29], [15]). We should mention the remarkable work of R. Schoen on the mean curvature equation in Euclidean space and its application to Geometry (see [27]). L. Bers showed the classical minimal surface equation cannot have an isolated singularity (see [2]). The same result for the constant mean curvature equation was established by R. Finn (see [11]). The removable singularity theorem for prescribed mean curvature equations in Rn+1 is due to the H. Rosenberg in a joint work with the second author ( see [22]). Returning to hyperbolic space, we observe that a removable singularity type theorem for the horizontal mean curvature equation in hyperbolic space was done by B. Nelli and the second author (see [20]). Now we shall write down the geometric condition (say, condition (∗)) that plays a crucial role in the whole study of our Dirichlet Problem (P ) for the prescribed mean curvature H(x): 0 HC (y) > |H(y)|, ∀ y ∈ ∂Ω(∗) 0 where HC (y) is the mean curvature of the cylinder generated by horocycles at a boundary point y. We can express the above inequality in the following equivalent analytic form 0 (n − 1)H (y) + xn(y)Dnd(y) > n|H(y)|, ∀ y ∈ ∂Ω(∗) 0 where H (y), is the mean curvature of the boundary of Ω and xn(y)Dnd(y) is the nth vertical component of the Euclidean unit interior normal to ∂Ω at a boundary point y (d(x) is the hyperbolic distance from x ∈ Ω to the boundary ∂Ω). We feel that it is interesting to now give a geometric insight about the second term that appears on the left of the above inequality : We recall that Serrin’s condition in Euclidean space ensures a priori boundary gradient estimates that are essential for the related Dirichlet Problem. On the other hand, looking to the simplest situation in hyperbolic space, it is not hard to see that there exist “spherical caps” of geodesic (minimal) planes cutting orthogonally the plan of the domain Ω at some point y, whose projections are convex curves (Euclidean ellipse) in the hyperbolic plane P. Thus, Serrin inequality holds but we have no control of the gradient at y. However, condition (∗) allows us to infer a priori boundary gradient estimates for the smooth solutions of Dirichlet Problem (P ). We infer these estimates in Section 1. Then in Section 2 we prove existence for Dirichlet Problem (P ) for the prescribed mean curvature H(x), given smooth boundary value data ϕ, if condition (∗) holds. 552 ELIAS M. GUIO AND RICARDO SA EARP

Finally, in Section 3, we obtain non-existence for the Dirichlet Problem (P ), if condition (∗) fails at a point. We would like to mention that in a preliminary version we have derived our existence result for |H(x)| 6 1. However Harold Rosenberg in a private conver- sation recalled to us the height estimates for compact embedded constant mean curvature hypersurfaces in hyperbolic space with constant mean curvature H > 1 (see [21], [14]). In view of this, if condition (∗) holds, our construction provide existence of the Dirichlet Problem (P ) for prescribed H(x)-hypersurfaces (taking prescribed smooth value data). We remark that P.- A. Nitsche (see [19]) studied a different Dirichlet Problem on account of a different notion of a graph, precisely radial vertical graphs. His condition is philosophically similar to ours, but geometrically distinct: Instead of cylinders generated by horocycles the author deals with usual hyperbolic cylinders. He proved that if the strict inequality between the mean curvature of the hyperbolic cylinder along the boundary of the domain (in a totally geodesic plane) and the prescribed mean curvature holds, then existence of his Dirichlet Problem for mean curvature less than 1 can be deduced. If the reverse strict inequality holds then he proved non existence. At last, we would like to mention that there also exists a notion of vertical graphs in hyperbolic space. For instance, the reader is referred to [20] and [23]. We also point out that F. Lin (see [16]) studied regularity for vertical minimal equation in hyperbolic space. E. Toubiana and the second author (see [25]) established the Perron process for these vertical minimal equations in two independent variables. The program accomplished in this paper was motivated by a question raised by the second author (see [23]), then investigated by the first author in his Doctoral Thesis under the supervision of the second author (see [8]).

2. A Priori Boundary Gradient Estimates. Next, we begin our investigation of the Dirichlet Problem (P ) by ﬁrst deriving an estimate given by the following theorem: Theorem 2.1 (a priori boundary gradient estimate). Let Ω be a bounded domain in Hn with C3 boundary ∂Ω, ϕ ∈ C2(Ω) and H(x) ∈ C1(Ω). If u ∈ C2(Ω) ∩ C1(Ω) is a solution of Dirichlet Problem (P ) and if H(x) satisﬁes condition (∗) then sup |Du| 6 c ∂Ω where

c = c(n, Ω, inf xn, sup xn, sup |u|, |ϕ|2, |H|1). Ω Ω Ω Proof. Before getting into details we are going to outline the proof. Recall that d(x) = dist(x, ∂Ω) be the hyperbolic distance from x ∈ Ω to the boundary ∂Ω. 2 Notice that d ∈ C (Γ) where Γ = {x ∈ Ω; d(x) < d0} for some d0 > 0 small enough. To obtain the a priori estimate we shall make use of barriers techniques. We shall construct a supersolution ω+ and a subsolution ω− satisfying (Q is an elliptic operator) : ±Q(ω±) 6 0 in N ∩ Ω (2.1) ±ω± > ±u in ∂(N ∩ Ω), MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 553 where N is a neighborhood of ∂Ω lying in Γ. More precisely, we shall infer that suitable barriers can be chosen of the form ω± = ±ψ(d) + ϕ (2.2) where ψ : [0, ∞) → R, is a C2 function such that ψ(0) = 0, ψ0 > 0 and ψ00 < 0. Now notice that maximum principle yields u(x) 6 ψ(d(x)) + ϕ(x), ∀ x ∈ N ∩ Ω (2.3) −ψ(d(x)) + ϕ(x) 6 u(x), ∀ x ∈ N ∩ Ω Let us recall how to proceed to obtain the a priori estimate by the method of barriers. Now let x0 ∈ ∂Ω be an arbitrary point of the boundary of Ω and let ~v be an unit Euclidean . Let us assume that ~v · n−→ > 0, where · is the inner product and −→ n is an Euclidean unit normal interior to Ω at x0. Then from the above inequalities we get −ψ ◦ d(x + ε~v) ϕ(x + ε~v) − ϕ(x ) 0 + 0 0 ε ε u(x + ε~v) − u(x ) ψ ◦ d(x + ε~v) ϕ(x + ε~v) − ϕ(x ) 6 0 0 6 0 + 0 0 ε ε ε for ε > 0 small enough. Now let ε → 0+. We obtain 0 0 −ψ (0)Dd(x0) · ~v + Dϕ(x0) · ~v 6 Du(x0) · ~v 6 ψ (0)Dd(x0) · ~v + Dϕ(x0) · ~v 1 Notice that |Dd(x0)| = , where |.| stands for the Euclidean norm. Now xn(x0) taking ~v = Du(x0) , by applying Schwarz inequality we deduce therefore |Du(x0)| ψ0(0) |Du(x0)| 6 + sup |Dϕ|, ∀ x0 ∈ ∂Ω. inf(xn) Ω Ω

Henceforth |Du|∂Ω 6 c, moreover, for our future choice of ψ

c = c(n, Ω, inf xn, sup xn, sup |u|, |ϕ|2, |H|1), Ω Ω Ω where 2 |ϕ|2 = max{sup |ϕ|, sup |Dϕ|, sup |D ϕ|} and Ω Ω Ω

|H|1 = max{sup |H|, sup |DH|}. Ω Ω This will prove the desired a priori estimate. We are now going to carry out the details. First, notice that the mean equation (1.1) can be re-written in the form Xn ij Q(u) = a (x, Du)Diju + b(x, Du) = 0 i,j=1 where (using summation convention)

ij 2 a (x, Du) = (1 + |Du| )δij − DiuDju nD u(1 + |Du|2) nH(1 + |Du|2)3/2 b(x, Du) = − n − · xn xn 554 ELIAS M. GUIO AND RICARDO SA EARP

Let us assume momentarily that the barriers ω± are chosen such that ψ satisfies Property (i) ψ(a) > M = sup |ϕ| + sup |u| Ω Ω Property (ii) ψ0(d) d 6 1 Property (iii) ψ0|Dd| > µ where µ > 3|Dϕ| + 8 is a fixed constant. We shall first work with ω+. For sake of simplicity we shall adopt the following notations and conventions: Xn + ij ω = ω and ξ = a (x, Dω)(Diω − Diϕ)(Djω − Djϕ). i,j=1 Recalling that the eigenvalues of the matrix aij(x, Dω) are 1 and (1 + |Dω|2), we Xn 2 ij observe that |Dω − Dϕ| 6 a (x, Dω)(Diω − Diϕ)(Djω − Djϕ). i,j=1 In view of our choice of barriers, see (2.2), we have |Dω − Dϕ| > µ. Employing the inequality above together with Cauchy’ s inequality we infer that |Dω|2 ξ > |Dω − Dϕ|2 > − |Dϕ|2. We also can deduce that µξ > 1 + |Dω|2, with 2 the aid of Schwarz’s inequality. We begin now to estimate Q(ω) in Γ to seek a suitable upper bound of the form Q(ω) 6 (cst)ξ :

Xn ij Q(ω) = a (x, Dω)Dijω + b(x, Dω) i,j=1 Xn ij 00 0 = a (x, Dω)(ψ DidDjd + ψ Dijd + Dijϕ) + b(x, Dω) i,j=1 Xn Xn ψ00 = ψ0 aij(x, Dω)D d + aij(x, Dω)D ϕ + b(x, Dω) + ξ ij ij ψ02 i,j=1 i,j=1 : = A + B + C + D

1. Estimate of A Xn Xn 0 ij 0 2 A = ψ a (x, Dω)Dijd = ψ ((1 + |Dω| )δij − DiωDjω)Dijd i,j=1 i,j=1 Xn Xn 0 2 03 = ψ (1 + |Dω| ) Diid − ψ DidDjdDijd− i=1 i,j=1 Xn Xn 02 0 − 2ψ DidDjϕDijd − ψ DiϕDjϕDijd i,j=1 i,j=1

Now in view of the hyperbolic metric we record that

2 1 |Dd| = 2 (2.4) xn MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 555

Owing to (2.4) we obtain

Xn ψ03D d − ψ03 D dD dD d = n i j ij x3 i,j=1 n (2.5) ψ0(1 + |Dω|2)D d ψ0D d = n − n (2ψ0Dd · Dϕ + |Dϕ|2 + 1) xn xn

Now the last term above can be estimated as follows

0 0 ψ Dnd 0 2 ψ 2 | (2ψ Dd · Dϕ + |Dϕ| + 1)| 6 2 |2(Dω − Dϕ) · Dϕ + |Dϕ| + 1| xn xn |Dω − Dϕ| 6 (2|Dω − Dϕ||Dϕ| + |Dϕ|2 + 1) xn |Dω − Dϕ|2 6 (3|Dϕ| + 1) (2.6) xn

n 02 P Notice that the expression ψ DidDjϕ satisﬁes the inequality i,j=1

Xn Xn 02 0 2 0 |ψ DidDjϕ| =|ψ (Diω − Diϕ)Djϕ| 6 n ψ |Dω − Dϕ||Dϕ| i,j=1 i,j=1 (2.7) 2 2 =n xn|Dω − Dϕ| |Dϕ|

Consequently

Xn 0 ij A = ψ a (x, Dω)Dijd i,j=1 Ã ! Xn 0 2 Dnd 6 ψ (1 + |Dω| ) Diid + xn · i=1 ¸ 2 ¡ 2 2 2¢ 1 + 3|Dϕ| +|Dω − Dϕ| 2n xn|Dϕ| + n xn|Dϕ| sup |Dijd|+ Γ xn

2. Estimate of B

Xn Xn ij 2 B = a (x, Dω)Dijϕ 6 (1 + |Dω| ) sup |Dijϕ| i,j=1 i,j=1 Ω

3. Estimate of C

−n 2 n 2 3/2 C = b(x, Dω) = Dnω(1 + |Dω| ) − (1 + |Dω| ) H xn xn

Now write (Big-oh notation)

(1 + |Dω|2)3/2 = (1 + |Dω|2)(ψ0|Dd| + |Dϕ|) + (1 + |Dω|2) O(1), (as |Dw| → ∞) 556 ELIAS M. GUIO AND RICARDO SA EARP

Then 0 −nψ 2 n 2 b(x, Dω) 6 2 (1 + |Dω| )H + (1 + |Dω| )(|H| + |H||Dϕ|) xn xn nψ0D d n − n (1 + |Dω|2) + |Dϕ|(1 + |Dω|2) xn xn ψ0(1 + |Dω|2) = 2 (−nH − nxnDnd) xn (1 + |Dω|2) + (n O(1)|H| + n|H||Dϕ| + n|Dϕ|) xn where O(1) will always represent a term which is bounded by a constant under control. From now on we will omit it. Combining these estimates we find Ã ! 0 2 Xn ψ (1 + |Dω| ) 2 Q(ω) 6 2 xn Diid − (n − 1)xnDnd − nH xn µ i=1 ¶ 1 + 3|Dϕ| ¡ ¢ + |Dω − Dϕ|2 + 2n2x |Dϕ|n2x |Dϕ|2 sup |D d| x n n ij  n Γ  n |H| Xn n|H||Dϕ| n ψ00 + (1 + |Dω|2)  + sup |D ϕ| + + sup |Dϕ|+ ξ x ij x x ψ02 n i,j=1 Ω n n Ω Taking into account the two inequalities µξ > (1 + |Dω|2) and |Dω − Dϕ|2 6 ξ, we infer Ã ! 0 2 Xn ψ (1 + |Dω| ) 2 Q(ω) 6 2 xn Diid − (n − 1)xnDnd − nH xn · i=1 ¡ ¢ 1 + 3|Dϕ| + 2n2x |Dϕ| + n2x |Dϕ|2 sup |D d| + n n ij x  Γ n   n |H| Xn n|H||Dϕ| n ψ00 +  + sup |D ϕ| + + sup |Dϕ| µ ξ + ξ x ij x x ψ02 n i,j=1 Ω n n Ω We pause now to say a few words about basic hyperbolic geometry. We first note n that fixing any two points Q = (x1, ...., xn) and P = (y1, ...., yn) in H , a straightforward computation shows that hyperbolic distance dist(P,Q) has an explicit formula given by s s  nP−1 Pn  (x − y )2 + (x + y )2 + (x − y )2   i i n n i i   i=1 i=1  dist(Q, P ) = ln s s   nP−1 Pn   2 2 2  (xi − yi) + (xn + yn) − (xi − yi) i=1 i=1 We shall need now fundamental formulas in hyperbolic space. Let t = d(x), 0 6 t 6 d0, where d0 is chosen sufficiently small. From a standard computation it follows that (see, for instance [17], [28], [8])

Xn nX−1 tanh t − k x2 (x) D d(x) − (n − 2)x (x)D d(x) = i (2.8) n ii n n 1 − k tanh t i=1 i=1 i MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 557 where k1, k2, ··· , kn−1 the principal curvatures of ∂Ω at y. Here y = y(x) is the closest point in ∂Ω to x. On account of (2.8), we are now going to obtain a bound for the expression Xn 2 xn(x) Diid(x) − (n − 1)xn(x)Dnd(x) − nH(x) i=1 −k As i , is a non-increasing function in the t variable we ﬁnd 1 − ki tanh t Xn nX−1 tanh t x2 (x) D d(x) − (n − 2)x (x)D d(x) 6 −(n − 1)H0(y) + (2.9) n ii n n 1 − k tanh t i=1 i=1 i P where H0 is the mean curvature of ∂Ω at y. As n−1 tanh t = t O(1), in view i=1 1−ki tanh t of our condition (∗) we derive Xn 2 xn(x) Diid(x) − (n − 1)xn(x)Dnd(x) − nH(x) i=1 6 nH(y) + xn(y)Dnd(y) + t O(1) − xn(x)Dnd(x) − nH(x)

= n(H(y) − H(x)) + xn(y)Dnd(y) − xn(x)Dnd(x) + t O(1) Now utilizing the relation between Euclidean and hyperbolic metrics we get Xn 2 xn(x) Diid(x) − (n − 1)xn(x)Dnd(x) − nH(x) 6 (1 + K)t O(1) i=1 n|H(y) − H(x)| |x (y)D d(y) − x (x)D d(x)| where K = sup + sup n n n n x∈Γ |x − y| x∈Γ |x − y| Estimate of Q(ω) · (1 + K)t O(1)ψ0µ 1 + 3|Dϕ| Q(ω) 6 + (2n2x |Dϕ| + n2x |Dϕ|2) sup |D d| + + x2 n n ij x  n Γ  n n |H| Xn n|H||Dϕ| n ψ00 +  + sup |D ϕ| + + sup |Dϕ| µ +  ξ x ij x x ψ02 n i,j=1 Ω n n Ω

Thus µ ¶ ψ00 Q(ω) 6 ν + ξ 0 ψ02 where   n sup |H||Dϕ| + n sup |Dϕ| + n|H| Xn 1 + K Ω Ω  ν0 = 2 + µ + sup |Dijϕ|µ+ inf xn inf xn Ω Ω Ω i,j=1 µ ¶ 2 2 2 1 + 3|Dϕ| + 2n sup |xnDϕ| + n sup xn|Dϕ| sup |Dijd| + + sup Ω Ω Γ Ω xn We turn now to our primary task to build the barriers. Choose then ν = max{ν0, 1}. We have ψ00 Q(ω) 6 (ν + )ξ ψ02 558 ELIAS M. GUIO AND RICARDO SA EARP

We now choose ψ setting

1 ψ(d) = ln(1 + kd), 0 6 d 6 a . ν

We must require that a and k satisfy

(i) ψ(a) > M = sup |ϕ| + sup |u|, (ii) ψ0(d)d 6 1, (iii) ψ0|Dd| > µ . (2.10) Ω Ω

Notice that ψ(0) = 0, ψ0 > 0 and ψ00 < 0. We must verify now the property (i), (ii) and (iii). eνM −1 Property (i) Let us take a = k , choosing k big enough to guarantee a 6 d0. 1 Hence ln(1 + ka) = M. ν Property (ii) Since ψ0(d)d in increasing, it follows that

ka ψ0(d)d 6 ψ0(a)a = 6 1 ν(1 + ka)

Property (iii) Notice that

0 0 0 0 ψ ψ ψ (a) k ψ |Dd| = > > = νM > µ, xn sup xn sup xn νe sup xn Ω Ω Ω if k is chosen big enough to ensure

νM k > µνe sup xn Ω

In view of the above construction we get a function ψ satisfying the required + ψ00 conditions to obtain the supersolution ω = ω . Note also that ψ02 + ν = 0. We have established therefore that the upper barrier ω satisﬁes (2.1) with the plus sign, namely

Q(ω) 6 0 in N ∩ Ω, ω > u on ∂(N ∩ Ω)

Thus owing to maximum principle we infer the ﬁrst inequality in (2.3), as desired. We now must seek for the lower barrier ω− satisfying the second inequality in (2.3). So we need to ﬁnd ω− satisfying (2.1) (with the minus sign). We will skip the details of this construction. We summarize it as follows. The computation of Q(w) is similar as before. Notice that

Xn ij 00 0 Q(ω) = a (x, Dω)(−ψ DidDjd − ψ Dijd + Dijϕ) + b(x, Dω) i,j=1 Xn Xn ψ00 = − ψ0 aij(x, Dω)D d + aij(x, Dω)D ϕ + b(x, Dω) − ξ ij ij ψ02 i,j=1 i,j=1 MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 559

Now working as in the estimate of the upper barrier we derive that Ã ! 0 2 Xn −ψ (1 + |Dω| ) 2 Q(ω) > 2 xn Diid − (n − 1)xnDnd + nH xn · i=1 1 + 3|Dϕ| − (2n2x |Dϕ| + n2x |Dϕ|2) sup |D d| + n n ij x  Γ n   n|H| Xn n|H||Dϕ| n ψ00 +  + sup |D ϕ| + sup |Dϕ| µ ξ − ξ x ij x x ψ02 n i,j=1 Ω n n Ω

Using once more condition (∗) and fundamental formula (2.8) we can calculate Xn 2 xn(x) Diid(x) − (n − 1)xn(x)Dnd(x) + nH(x) 6 (1 + K)tO(1) i=1 We conclude therefore µ ¶ ψ00 Q(ω) > − ν + ξ ψ02 since ψ0(d)d 6 1 and ψ0|Dd| > µ, where ν is deﬁned as before. Consequently, now 1 choose ψ = ln(1+kd) as before, to conclude that ω = −ψ ◦d+ϕ is a lower barrier ν satisfying (2.1) (with the minus sign), as desired. Thus the proof of Theorem 2.1 is now established. ¡

We shall need a version of Theorem 2.1 for a family t, t ∈ [0, 1] to proceed further and to infer an existence theorem in Section 2, using degree theory. The proof is the same as before with tH in place of H and tϕ in place of ϕ. We next give the precise statement for future reference.

Theorem 2.2 (a priori boundary gradient estimate for a family t). Let Ω be a bounded domain in Hn with C3 boundary ∂Ω, ϕ ∈ C2(Ω) and H(x) ∈ C1(Ω). If t 2 1 u = u ∈ C (Ω) ∩ C (Ω) is a solution of the following Dirichlet Problem (Pt) (using summation convention), then

ij Qt(u) = a (x, Du; t)Diju + b(x, Du; t) = 0 in Ω (2.11) u = tϕ in ∂Ω where ij 2 a (x, Du; t) = (1 + |Du| )δij − DiuDju nD u tnH(x) b(x, Du; t) = − n (1 + |Du|2) − (1 + |Du|2)3/2 xn xn If the satisfying condition (∗) holds then sup |Du| 6 c ∂Ω where t c = c(n, Ω, inf xn, sup xn, sup |u |, |ϕ|2, |H|1). Ω Ω Ω 560 ELIAS M. GUIO AND RICARDO SA EARP

3. Existence Results. In hyperbolic space horospheres play a role of geometric barriers and give rise to a geometric principle. This idea has been useful in many papers. We refer to [24] and to the references there. Let us brieﬂy summarize it now. Consider M a smooth compact immersed surface into hyperbolic space with boundary a curve γ. Assume that the mean curvature of M is not bigger than 1. Now take any horosphere H that involves γ. Then H involves entirely M. In fact, on the contrary, we obtain a copy Ht by hyperbolic translating H involving M and touching M at an interior point. This will give a contradiction with the maximum principle. A similar result holds for hypersurfaces M with mean curvature less than some h, provide M is inside a n- sphere of mean curvature h. We will need the following theorem (see [5]). Theorem 3.1 (height estimate). Let Ω ⊂ Hn be a bounded domain. Assume that 1 t 0 2 H ∈ C (Ω). If u = u ∈ C (Ω) ∩ C (Ω) is a solution of Dirichlet Problem (Pt) (see (2.11))

Qt(u) = 0 in Ω (3.1) u = tϕ, in ∂Ω then t sup |u | 6 c1, Ω where c1 is a constant independent of t. Proof. On account of the previous discussion, we have a priori height estimates if the mean curvature is less than or equal to 1. We check the case H is constant with |H| > 1. In fact, if we prove height estimates for constant mean curvature less or equal a, then a straightforward argument will provide height estimates for |H| 6 a. We will proceed the proof as follows. Let S be a big n− sphere cutting orthogonally the geodesic hyperplane P containing Ω, such that S involves strictly both Ω and the graph of ϕ. Assume that S has mean curvature h ( h is bigger than 1, but h ≈ 1). This obviously can be done by compactness. Now notice that if M is an horizontal graph of mean curvature less than h taking boundary value data ϕ on ∂Ω then necessarily M lies inside S. Indeed, take a half part of S, say S+ far away from M, doing horizontal Euclidean translations (hyperbolic isometries). Then move S+ back towards M. By maximum principle S+ (i.e a translated copy of S+) cannot touch M during this movement, until it reaches the original position. Notice that since the graph of ut has mean curvature tH, where H = a (constant), we infer that for t varying in some interval [t0, 1], such that t0H = h we have priori height estimates independent of t (see [14] or [21]). But we have also a priori height estimates independent of t for 0 6 t 6 t0 according with the previous geometric argument. So then we have the desired estimates. ¡ Combining together a priori boundary gradient estimates (see Theorem 2.1, The- orem 2.2) with height estimates (see Theorem 3.1) we can infer global gradient estimates by applying some elliptic theory, such as Lemma 3.1 in [20], Lemma 5.2 in [5], or Theorem 15.2 in [12] . Then using Ladyzhenskaya and Ural’tseva classical estimates we can infer a priori global H¨olderestimates for the gradient. We have therefore the following result. Theorem 3.2 (global gradient estimate). Let Ω be a bounded domain in Hn with C3 boundary ∂Ω, ϕ ∈ C3(Ω) and H ∈ C2(Ω). If u = ut ∈ C2(Ω) is a solution of MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 561

Dirichlet Problem (Pt) (see (2.11)) and if H(x) satisfies condition (∗) then t sup |u |1,β 6 c, Ω where c is a constant independent of t. Remark 3.1. By virtue of classical elliptic regularity, it suffices to assume that H ∈ C1,α(Ω), 0 < α < 1. Indeed, with this assumption if u is a C2 solution of 3 Dirichlet Problem Pt then u is of classe C and the argument developed in [20] or [5] can be applied. Moreover, a more refined derivation can be done to infer the 1 same result for H ∈ C (Ω), working with a weak form of (Pt). This remark can also be extended to the existence theorem in next section. But we will not go into such analysis in this article (see [12]).

Of course we intent to solve Dirichlet Problem (Pt) for t = 1, i.e the horizontal mean equation in hyperbolic space. We will be able to carry out this plan by applying degree theory in the light of the a priori estimates. We now state our main existence result. Theorem 3.3 (existence theorem). Let Ω be a bounded domain with in Hn with C3 boundary ∂Ω, ϕ ∈ C3(Ω) and H ∈ C2(Ω). Assume condition (∗) holds. Then Dirichlet Problem (P ) has a unique solution u ∈ C2(Ω). Proof. In view of our a priori estimates we will be able to use degree theory methods to obtain existence of a fixed-point of certain map associated to the solutions of our Dirichlet Problem (P ). We will proceed as follows. 1,β 1,β 1,β Given v ∈ C (Ω) we then define a map Tt : C (Ω) → C (Ω) by t Tt : v 7→ u , where ut the unique solution of the second order elliptic linear problem ¡ ¢ 2 2 3/2 Pn 2 t n(1+|Dv| ) t tn(1+|Dv| ) (1+|Dv| )δij −DivDjv Diju − Dnu − H=0 in Ω i,j=1 xn xn ut = tϕ in ∂Ω t That is u = Ttv. Existence and uniqueness of ut is assured by linear elliptic theory. Moreover, ut ∈ C2,αβ(Ω), by regularity theory. With the aid of Arzela-Ascoli theorem and global Schauder estimates we deduce Tt is compact. A standard argument shows continuity of Tt (see [12]). From our estimates we can apply degree theory (see, [1], [4] or [18]) to conclude. We will proceed the final step of the proof as follows. We define the set 1,β C = {v ∈ C (Ω); |v|1,β < c + 1} where c is a constant given by Theorem 3.2. Notice that for our choice, the equation t t u − Ttu = 0 (3.2) has no solutions in ∂C. Thus deg (I − Tt, C, 0) = cst is well defined and independent of t ∈ [0, 1] , owing to homotopy invariance. But according to maximum principle, for t = 0 the trivial solution u0 ≡ 0 is the unique 1,β solution for every v ∈ C (Ω). It follows that T0 ≡ 0 and we deduce

deg (I − T0, C, 0) = deg (I, C, 0) = 1 562 ELIAS M. GUIO AND RICARDO SA EARP

We have henceforth that Dirichlet Problem (P ) has a solution, as desired. ¡ The following Corollaries are immediate consequences of Theorem 3.3. Corollary 3.1. Let Ω be a bounded domain in Hn with C3 boundary ∂Ω, ϕ ∈ C3(Ω) and H ∈ C2(Ω). Assume |H(x)| 6 1, ∀ x ∈ Ω. Assume further H0(y) > n + 1 , ∀ y ∈ ∂Ω. Then there exists a C2 extension u of ϕ whose graph has pre- n − 1 scribed mean curvature H(x). Corollary 3.2. Let Ω be a ball entirely contained in n-dimensional hyperbolic space, i.e Ω ⊂ Hn, and let ϕ be a smooth function deﬁned on ∂Ω. Then there exists a minimal solution u in Ω of the horizontal minimal equation taking the prescribed boundary value data ϕ on ∂Ω.

4. Non-existence Results. Our construction to infer non-existence is based on classical maximum principle. We shall need the following issue stated here for future references. Theorem 4.1 (see [12]). Let Ω be a bounded domain in Rn and let Γ a C1 open subset of ∂Ω. Then if Q is an elliptic operator and if u ∈ C0(Ω) ∩ C2(Ω ∪ Γ), v ∈ C0(Ω) ∩ C2(Ω) satisﬁes Q(u) > Q(v) in Ω, u 6 v in ∂Ω \ Γ, ∂v = −∞ in Γ, ∂ν it follows that u 6 v em Ω. We will now established our non-existence result. Denote diam (Ω), the hyperbolic diameter of Ω. Theorem 4.2. Let Ω be a bounded domain in Hn with C2 boundary ∂Ω. Let us assume that condition (∗) fails at a point y , that is 0 (n − 1)H (y) + xn(y)Dnd(y) < n|H(y)|, for some y ∈ ∂Ω(†) where H ∈ C0(Ω) satisﬁes one of the following conditions: (i) It does not change sign in Ω and n > 3 or ln 3 (ii) It does not change sign in Ω, n = 2 and diam (Ω) < or 2 (iii) |H| > 1 in Ω. Under one of the above hypothesis it follows that there exists ϕ ∈ C∞(Ω), such that Dirichlet Problem (P ) is not solvable. That is, there exists no solution u of the horizontal prescribed mean curvature equation in Ω, taking the prescribed boundary value data ϕ on ∂Ω. Proof. We will proceed the proof in two steps. Firstly, we will work in a neighborhood of y to obtain an upper bound of a solution of Dirichlet Problem (P ) in this neighborhood. Secondly, we will focus the region outside of this neighborhood to get a similar majoration. Then, by putting together these two estimates we will provide an upper bound of any solution of our Dirichlet Problem (P ) satisfying condition (†). As we did to obtain a priori gradient estimates, we will again seek the desired majorations applying fair estimates in the spirit of the calculations of the Section 1 by computing Q(ω). MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 563

First step of the proof Let us fix y ∈ ∂Ω, let δ = diam (Ω) and let a be a positive real number such that a < δ. Define Ω˜ := {x ∈ Ω; a < d(x) < δ}. We then define ω setting ω(x) = sup u + ψ(d), ∂Ω\Ba(y) where d(x) = dist (x, y) and ψ :(a, δ) → R is a C2 function to be chosen such that ψ(δ) = 0, ψ0 6 0, ψ0(a) = −∞ and ψ00 > 0 (4.1) We return now to our elliptic operator Q(ω). We intend to build ψ = ψ(d) satisfying above properties such that ω is a supersolution for our Dirichlet Problem (P ), i.e Q(ω) 6 0 in Ω.˜ Let us start our long and tedious computations to produce the barriers recalling Xn 2 Q(ω) = ((1 + |Dω| )δij − DiωDjω)Dijω− i,j=1 µ ¶ n D ω − n + H (1 + |Dω|2)3/2 2 1/2 xn (1 + |Dω| ) Xn Xn 2 02 =(1 + |Dω| ) Diiω − ψ DidDjdDijω− i=1 i,j=1 µ ¶ n ψ0D d − n + H (1 + |Dω|2)3/2 x (1 + |Dω|2)1/2 n Ã ! Xn Xn 2 00 2 0 =(1 + |Dω| ) ψ (Did) + ψ Diid − i=1 i=1 Xn 02 ¡ 00 2 2 0 ¢ − ψ ψ (Did) (Djd) + ψ DidDjdDijd i,j=1 µ ¶ n ψ0D d − n + H (1 + |Dω|2)3/2 2 1/2 xn (1 + |Dω| ) We will need the following basic formulas in hyperbolic geometry. Xn 1 |Dd| = ( (D d)2)1/2 = i x i=1 n Xn n − 1 x2 D d − (n − 2)x D d = n ii n n tanh d i=1 Substituting (4.2) in the above expression we get ψ00 Xn ψ02ψ00 ψ03D d Q(ω) = (1 + |Dω|2) + (1 + |Dω|2)ψ0 D d − + n x2 ii x4 x3 n i=1 n n 2 n(1 + |Dω| ) 0 n 2 3/2 − ψ Dnd − H(1 + |Dω| ) xn xn Xn ψ00 n(1 + |Dω|2) = (1 + |Dω|2)ψ0 D d + − ψ0D d ii x2 x n i=1 n n

n 2 3/2 03 Dnd − H(1 + |Dω| ) + ψ 3 xn xn 564 ELIAS M. GUIO AND RICARDO SA EARP

Now substituting (4.2) we discover Ã ! (1 + |Dω|2) Xn (1 + |Dω|2) ψ0 x2 D d − (n − 2)x D d − 2 ψ0D d x2 n ii n n x n n i=1 n 00 03 Dnd ψ n 2 3/2 + ψ 3 + 2 − H(1 + |Dω| ) xn xn xn 2 2 (1 + |Dω| ) 0 n − 1 (1 + |Dω| ) 0 = 2 ψ − 2 ψ Dnd xn tanh d xn 00 03 Dnd ψ n 2 3/2 + ψ 3 + 2 − H(1 + |Dω| ) xn xn xn 2 2 (1 + |Dω| ) 0 n − 1 (1 + |Dω| ) 0 = 2 ψ − ψ Dnd xn tanh d xn 00 0 Dnd ψ n 2 3/2 − ψ + 2 − H(1 + |Dω| ) xn xn xn We thereby obtain

2 00 (1 + |Dω| ) 0 n − 1 ψ n 2 3/2 Q(ω) 6 2 ψ ( − 2) + 2 − H(1 + |Dω| ) (4.2) xn tanh d xn xn taking into account (4.2) and (4.1). Let us assume momentarily that H(x) is non- negative. The other case has the same analysis. Now owing to our choice we see that the quantity µ1 = inf(n − 1 − 2 tanh d) is Ω strictly positive, hence

2 µ 00 ¶ (1 + |Dω| ) n − 1 − 2 tanh d 0 2 ψ Q(ω) 6 2 ψ + xn 02 xn tanh d ψ 2 µ 0 00 ¶ 1 + |Dω| µ1ψ µ2ψ 6 2 + 02 xn tanh d ψ

2 where µ2 = sup xn. Ω Of course, if n > 3 (case (i)) then µ1 is always strictly positive. We then deﬁne

Z δµ ¶−1/2 −1/2 sinh t µ1 ψ(d)=µ ln dt, where µ := =µ(diam(Ω), inf xn, sup xn, n). d sinh a µ2 Ω Ω It is straightforward to check that ω is the required barrier. Observe ﬁrst that Q(ω) 6 0 in Ω.˜ Now in view of Theorem 4.1 we have therefore u(x) 6 ω(x), x ∈ Ω.˜ From this it follows that,

sup u 6 ω(a) = sup u + ψ(a). (4.3) ∂Ba(y)∩Ω ∂Ω\Ba(y)

By our assumptions if n = 2 (case (ii)) then µ1 is positive hence we arrive analogously to the same conclusion. Let us turn to the estimates to treat the remaining case. Case(iii) Employing (4.2) we deduce MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 565

Now using H > 1 we infer (1 + |Dω|2)3/2 (1 + |Dω|2)3/2 −n H 6 − n xn xn 2 2 (1 + |Dω| ) 0 (1 + |Dω| ) 6 − n |Dω| = nψ 2 xn xn We conclude henceforth 2 00 (1 + |Dω| ) 0 n − 1 ψ Q(ω) 6 2 ψ ( + n − 2) + 2 xn tanh d xn (1 + |Dω|2) n − 1 ψ00(1 + |Dω|2) 6 ψ0( + n − 2) + x2 tanh d ψ02 n ·µ ¶ ¸ 1 + |Dω|2 n − 1 ψ00 = + n − 2 ψ0 + x2 x2 tanh d n ψ02 n µ ¶ 1 + |Dω|2 µ ψ00 6 1 ψ0 + µ 2 2 02 xn tanh d ψ where 2 µ1 = inf(n − 1 + (n − 2) tanh d) > 0, µ2 = sup xn Ω Ω We may therefore select

Z δ µ ¶−1/2 −1/2 sinh t µ1 ψ(d) = µ ln dt, where = µ = µ(diam(Ω), inf xn, sup xn, n). d sinh a µ2 Ω Ω Clearly, ω satisfies Q(ω) 6 0 in Ω˜ thus we arrive to the same conclusion as before, namely sup u 6 sup u + ψ(a) ∂Ba(y) ∩ Ω ∂Ω\Ba(y) First step of the proof is now completed. Second step of the proof We wish now to examine the estimates in the complemen- tary domain Ωε,a := {x ∈ Ω; ε < d(x) < a}. We observe that until now we have not used condition (†). We will use it now. Let us assume that there exists η > 0 such that 0 (n − 1)H (y) + xn(y)Dnd(y) 6 nH(y) − 5η. (4.4) Let S ⊂ Rn be a smooth quadric surface satisfying 0 (i) HS(y) 6 H (y) + η (ii) S ∩ Ba(y) ⊂ Ω Let us consider now distance function d(x) = dist(x, S), in Ωε,a, choosing a sufficiently small to ensure smoothness of d(x). We want to find ω = sup u + Ψ(d) ∂Ba(y)∩Ω where Ψ : (ε, a) → R, 0 < ε < a such that Ψ(a) = 0, Ψ0 6 0, Ψ0(ε) = −∞ and 00 Ψ > 0. Furthermore, we also want Q(ω) 6 0 in Ωε,a. As before, we will pass in performing the computation of Q(ω) using (4.2) once again. Hence Xn Ψ03 Ψ00 nΨ0 Q(ω) = Ψ0(1 + |Dω|2) D d + D d + − (1 + |Dω|2)D d ii x3 n x2 x n i=1 n n n n − H(1 + |Dω|2)3/2 xn 566 ELIAS M. GUIO AND RICARDO SA EARP

Using the identity

03 0 0 Ψ Ψ 2 Ψ Dnd 3 Dnd = (1 + |Dω| )Dnd − xn xn xn We therefore obtain Ã ! Ψ0 Xn Ψ0D d Ψ00 Q(ω) = (1 + |Dω|2) x2 D d − (n − 1)x D d − n + x2 n ii n n x x2 n i=1 n n n − H(1 + |Dω|2)3/2 xn We observe that

−Ψ0 (1 + |Dω|2)3/2 = (1 + |Dω|2) + Ψ0(1 + |Dω|2)o(1) (as |Dω| → ∞) xn Since −Ψ0 −Ψ0 Dnd 6 2 xn xn and 00 00 2 00 Ψ Ψ (1 + |Dω| ) 2 Ψ 2 6 2 2 = (1 + |Dω| ) 02 xn xn |Dw| Ψ We infer " Ψ0 Xn Q(ω) 6 (1 + |Dω|2) x2 D d − (n − 1)x D d + nH+ x2 n ii n n n i=1 µ ¶ 00 ¸ 2 Ψ +(n sup |H| + 1)o(1) + sup xn 03 . Ω Ω Ψ

Let us now denote by St = {x ∈ Ω; d(x) = t} the parallel hypersurface to S at a ﬁxed distance t. In view of formula (2.8), we can write

Xn 2 xn Diid − (n − 2)xnDnd = −(n − 1)HSt i=1

where HSt is the mean curvature of St. Now applying a standard continuity argument we have η (i) |H(x) − H(y)| < n Xn Xn 2 2 (ii) |xn(x) Diid(x) − xn(y) Diid(y)| < η i=1 i=1 η (iii) |x (x)D d(x) − x (y)D d(y)| < n n n n n − 1 for x ∈ Ωε,a (recall that y ∈ ∂Ω is ﬁxed). MEAN CURVATURE EQUATION IN HYPERBOLIC SPACE 567

Hence " Ψ0(1 + |Dω|2) Xn Q(ω) 6 x2 (y) D d(y) − (n − 1)x (y)D d(y)+ x2 n ii n n n i=1 µ ¶ 00 ¸ 2 Ψ +nH(y) − 3η + o(1) + sup xn 03 Ω Ψ Ψ0(1 + |Dω|2) 6 2 [−(n − 1)Hs(y) − xn(y)Dnd(y) + nH(y) xn µ ¶ 00 ¸ 2 Ψ −3η + o(1) + sup xn 03 Ω Ψ 0 2 Ψ (1 + |Dω| ) 0 6 2 [−(n − 1)H (y) − xn(y)Dnd(y)+ + nH(y) xn µ ¶ 00 ¸ 2 Ψ −4η + o(1) + sup xn 03 Ω Ψ Now using condition (†) in the form given by (4.4) we have 0 2 · µ ¶ 00 ¸ Ψ (1 + |Dω| ) 2 Ψ Q(ω) 6 2 5η − 4η + o(1) + sup xn 03 xn Ω Ψ 0 2 · µ ¶ 00 ¸ Ψ (1 + |Dω| ) 2 Ψ = 2 η + o(1) + sup xn 03 xn Ω Ψ We may now deﬁne Ψ(d) = k((a−ε)1/2 −(d−ε)1/2) , ε < d < a, hence we obtain Q(ω) 6 0 in Ωε,a, taking k suﬃciently big. According to Theorem 4.1 we therefore have u(x) 6 sup u + Ψ(d(x)), ∀ x ∈ Ωε,a. ∂Ba(y)∩Ω Thus 1/2 u(x) 6 sup u + Ψ(ε), ∀ x ∈ Ωε,a = sup u + k(a − ε) , ∀ x ∈ Ωε,a. (4.5) ∂Ba(y)∩Ω ∂Ba(y)∩Ω We lastly combine estimates (4.3) and (4.5) letting ε → 0+. We conclude therefore u(y) 6 sup u + ψ(a) + k(a)1/2. ∂Ω\Ba(y) This means u cannot be arbitrarily prescribed on ∂Ω. That is, there exists a smooth function ϕ in Ω such that our Dirichlet Problem (P ) taking boundary value data ϕ on ∂Ω is not solvable. ¡

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