The Dirichlet problem for the constant mean curvature equation in Sol_3
TThe Dirichlet problem for the constant meancurvature equation in Sol ∗ and Ana MenezesApril 17, 2019 Abstract
We prove a version of the Jenkins-Serrin theorem for the existenceof CMC graphs over bounded domains with infinite boundary data inSol . Moreover, we construct examples of admissible domains where theresults may be applied. Keywords:
Constant mean curvature surfaces, Dirichlet problem.
In 1966, Jenkins and Serrin [9] studied the problem of finding necessary andsufficient conditions in a bounded domain in R in order to solve the Dirich-let problem for the minimal surface equation with certain infinite boundarydata. More precisely, they considered domains Ω ⊂ R bounded by arcs A , A , · · · , A j , B , B , · · · , B k of curvature zero and arcs C , C , ..., C m of nonnegative curvature (normal pointing inwards), and wanted to find a function u : Ω → R that satisfies the minimal surface equation and assumes + ∞ on thearcs A i ’s, −∞ on B i ’s and a bounded prescribed data on the arcs C i ’s. Theygave necessary and sufficient conditions for the existence of a solution on Ω interms of the length of the boundary arcs of Ω and of inscribed polygons.Later, Spruck [18] considered the same problem for mean curvature H > . In this case it is natural to expect that the arcs A i and B i be of curvature2 H and − H, respectively, since a solution would be asymptotic to a verticalcylinder of mean curvature H. The existence result for infinite boundary datacould be thought to be proved by taking a limit of n → ∞ in a sequence of ∗ The author was partially supported by CNPq-Projeto Universal 482113/2013-8 andthanks Princeton University for the hospitality during part of the time the research andpreparation of this article were conducted. a r X i v : . [ m a t h . DG ] A p r solutions that take the boundary data n on A i and − n on B i . Nevertheless,the existence of bounded solutions is only guaranteed for convex domains. Animportant idea introduced in [18] was to consider reflected arcs of the family { B i } in order to get a convex domain.After those results many people have worked on these problems in other am-bient spaces, specially in other homogeneous spaces. For instance, in H × R the minimal case ( H = 0) was considered by Nelli and Rosenberg [12] (forbounded domains) and by Collin and Rosenberg [1] (for unbounded domains);and the mean curvature H > . Recently, Nguyen [13] considered theminimal case for both bounded and unbounded domains in Sol and presenteda warped product model of the space where vertical graphs can be considered.Recall that Sol can be viewed as R endowed with the Riemannian metric ds = e x dx + e − x dx + dx . (1.1)As described in [10], Sol admits exactly two foliations by totally geodesicsubmanifolds, the two being similar. In both foliations, each leaf is isometricto the hyperbolic plane H . The approach in [13] was to consider one of thesefoliations by applying the change of coordinates x := x , y := e x , t := x , that turned Sol into the model of a warped product, in fact, Sol = H × y R = { ( x, y, t ) ∈ R : y ≥ } , where the half-plane model is used for H and theRiemannian metric is given by ds = dx + dy y + y dt . (1.2)Seeing Sol as a warped product gives us a natural way to define functions in H and consider their graphs in Sol . There are two interesting properties about this model for Sol : the verticallines are not geodesics and the mean curvature vector of a vertical plane gen-erated by a curve γ is related to its Euclidean geodesic curvature. In fact, if γ is a curve in H , then the mean curvature vector (cid:126)H γ × R of the vertical plane γ × R in Sol is (cid:126)H γ × R = 12 y (cid:126)κ euc , where (cid:126)κ euc denotes the Euclidean geodesic curvature vector of the curve. Hencethe conditions on the boundary of a given domain can be formulated in termsof the Euclidean geodesic curvature of the arcs of the boundary.Recall that Sol is the Thurston geometry of smallest isometry group whichhas dimension three and has no positive isometry with fixed points. A con-sequence of the lack of isometries is that the reflected arcs idea from [18]described above does not make sense in Sol and therefore a definition of ad-missible domain for the case H > C open arcs { A i } , { B i } and { C i } and their endpoints, with Euclidean curvature κ euc ( A i ) = 2 H/y,κ euc ( B i ) = − H/y and κ euc ( C i ) ≥ H/y, respectively, and if the family { B i } isnon empty, a special domain Ω ∗ must be well defined. We require Ω and Ω ∗ tobe simply connected and admit a bounded subsolution to the mean curvatureP.D.E.. Besides, a polygon P ⊂
Ω is said to be an admissible polygon if itis a curvilinear polygon whose sides are curves of Euclidean curvature ± H/y and vertices are chosen from among the endpoints of arcs of the families { A i } and { B i } . For an admissible polygon P , we denote by α and β the totalEuclidean lengths of the arcs in the boundary ∂ P that belong to { A i } and { B i } , respectively, and by (cid:96) the Euclidean perimeter of P . We consider thequantity I ( P ) := (cid:90) P y d a, where d a is the Euclidean area element.Our main results about existence of solutions to the Dirichlet Problem forCMC surfaces with infinite boundary data as described before are the following. Theorem 1.
Let Ω be an admissible domain such that the family { B i } isempty. Assume that the assigned boundary data on the arcs { C i } is boundedbelow. Then the Dirichlet Problem has a solution in Ω if and only if α < (cid:96) + 2 H I ( P ) (1.3) for all admissible polygons P . If the family { A i } is empty, an analogous result holds. Theorem 2.
Let Ω be an admissible domain such that the family { A i } isempty. Assume that the assigned boundary data on the arcs { C i } is boundedabove. Then the Dirichlet Problem has a solution in Ω if and only if β < (cid:96) − H I ( P ) , (1.4) for all admissible polygons P . The next result combines the two above for the case of the family { C i } being empty. Theorem 3.
Let Ω be an admissible domain such that the family { C i } isempty. Then the Dirichlet Problem has a solution in Ω if and only if α = β + 2 H I (Ω) , (1.5) and for all admissible polygons properly contained in Ω2 α < (cid:96) + 2 H I ( P ) and β < (cid:96) − H I ( P ) . (1.6)After studying these results some natural questions arise: Are there do-mains for which they apply? What can be said about them? In Section 11 weconstruct admissible domains where Theorems 1 and 2 apply and in Section13 we exhibit a method to find domains with no { C i } -type arcs that admit asolution to the Dirichlet Problem.One of the main difficulties in this problem was to find the right conditionson the boundary of the domains to be considered and understand the propertiesof horizontal curves whose product by the vertical line has constant meancurvature.We remark that the arguments presented here also can be adapted forproving existence of solutions of the Dirichlet problem with infinite boundarydata in other warped products. In this section we will describe the model that we use for Sol and will statesome results. For more details and proofs see, for instance, Section 2 in [13].The homogeneous Riemannian 3 − manifold Sol is a Lie group which canbe viewed as R with the metric (1.1).A change of coordinates allows us to treat Sol as a warped product, contextin which much is known about constant mean curvature surfaces that aregraphs. The diffeomorphism ϕ : Sol → H × f R given by ϕ ( x , x , x ) = ( x , e x , x ) = ( x, y, t )is an isometry if H is considered with the half-plane model H = { ( x, y ) ∈ R | y > } with metric dx + dy y and f ( x, y ) = y, so that the warped metricis ds = dx + dy y + y dt . (2.1)Notice that (cid:107) ∂ t (cid:107) = y and vertical translations are isometries. In particular, ∂ t is a Killing vector field that is not unitary and vertical lines are not geodesics.In this model, Euclidean properties of curves γ in the half-plane are trans-mitted to the hypersurface γ × R . For example:
Proposition 1.
Let γ be a curve in H . Then the mean curvature vector of γ × R in Sol is (cid:126)H γ × R = 12 y (cid:126)κ euc , where (cid:126)κ euc is the Euclidean curvature vector of γ. Let Ω be a domain in H and denote by G u the graph of a function u overΩ . The upward unit normal vector to G u is given by N = − y ∇ u + y ∂ t (cid:112) y (cid:107)∇ u (cid:107) , (2.2)where ∇ is the hyperbolic gradient operator and (cid:107) . (cid:107) is the hyperbolic norm.The graph G u of u has constant mean curvature H with respect to thenormal pointing up if u satisfies the equationdiv (cid:18) y ∇ uW (cid:19) = 2 yH, (2.3)where W = (cid:112) y (cid:107)∇ u (cid:107) , and div denotes the hyperbolic divergence opera-tor. If u satisfies (2.3), u is called a solution in Ω . Definition 1.
Let Ω be a domain in H and h be a C − function over Ω .
1. The function h is a subsolution in Ω of (2.3) ifdiv (cid:18) y ∇ hW (cid:19) ≥ yH.
2. The function h is a supersolution in Ω of (2.3) ifdiv (cid:18) y ∇ hW (cid:19) ≤ yH. Hence the classical (bounded) Dirichlet problem in a domain Ω for theconstant mean curvature equation is given by div (cid:18) y ∇ uW (cid:19) = 2 yH in Ω ,u = ϕ on ∂ Ω . (2.4) Remark 1.
Throughout this paper we consider a bounded domain Ω ⊂ H that is away from the asymptotic boundary of H , i.e., Ω is contained in somehalfspace { ( x, y ) ∈ H ; 0 < y ≤ y } . Hence, if we need to use curves thatsatisfy the condition κ euc ≥ H/y, we can take the ones that satisfy the clearercondition κ euc ≥ H/y , and if we need curves with κ euc ≤ − H/y, we cantake the ones that satisfy κ euc ≤ − H/y . We start this section by stating a general maximum principle for sub andsuper solutions of the mean curvature equation for boundary data with a finitenumber of discontinuities (whose proof is analogous to the proof of Theorem2.2 in [8]).
Lemma 1 (General Maximum Principle) . Let u be a subsolution and u be asupersolution of (2.3) in a bounded domain Ω ⊂ H . Suppose that lim inf (u − u ) ≥ for any approach to ∂ Ω with the possible exception of a finite numberof points of ∂ Ω . Then u ≥ u in Ω with strict inequality unless u ≡ u . The next result gives an upper bound to the norm of the gradient of asolution u to (2.3) at a point p depending on its value u ( p ) and on the distancefrom p to the boundary of the domain. It has two important consequences forthis work: The Harnack inequality (Theorem 5) and the Compactness Theorem(Theorem 7). It can be found in [3], Theorem 1. Although the statement thereis weaker, their proof yields the next result as stated. Theorem 4 (Interior gradient estimate, [3]) . Let u be a non negative solutionto (2.3) on B R ( p ) ⊂ H . There is a constant C = C ( p, R ) such that (cid:107)∇ u ( p ) (cid:107) ≤ f (cid:18) u ( p ) R (cid:19) for f ( t ) = e C ( t +1) . As a consequence of the interior gradient estimate, we have the Harnackinequality. The proof follows the same steps as in R , which was presented bySerrin in [17], Theorem 5. Theorem 5 (The Harnack Inequality) . Let u be a non negative solution to (2.3) in B R ( p ) . Then there is a function Φ( t, r ) such that u ( q ) ≤ Φ( m, r ) and Φ( t,
0) = t, where m = u ( p ) and r is the distance from q to p. For each t fixed, Φ( t, r ) is a continuous strictly increasing function definedon an interval [0 , ρ ( t )) , for ρ a continuous strictly decreasing function tendingto zero as t tends to infinity and lim r → ρ ( t ) Φ( t, r ) = + ∞ . The main theorem of this section is about the existence of solutions of (2.4) inbounded piecewise C domains for bounded boundary data that are continuousexcept in a finite subset of the boundary. This result is essential in the proofof our main results.Given a piecewise C domain Ω, the outer curvature ˆ κ ( P ) of a point P ∈ ∂ Ωis defined as the supremum of the curvatures of C curves through P that donot intercept Ω with normal vectors pointing to Ω . If there is not such a curve,ˆ κ ( P ) is −∞ . Theorem 6 (Existence Theorem) . Let Ω ⊂ H be a piecewise C domain.Suppose that the outer Euclidean curvature of ∂ Ω satisfies ˆ κ euc ( x, y ) ≥ H/y with possible exception in a finite set E. If the equation (2.3) admits a boundedsubsolution in Ω , then the Dirichlet problem (2.4) for constant mean curvature H is solvable for any bounded ϕ ∈ C ( ∂D \ E ) . Besides, from Lemma 1, thesolution is unique.
In order to prove it, we need some preliminary results. The first is theCompactness Theorem, which follows from the gradient estimate for solutions(Theorem 4) and the Schauder theory for PDEs and Arzel´a-Ascoli Theorem.
Theorem 7 (Compactness Theorem) . Let ( u n ) be a sequence of solutions of (2.3) uniformly bounded in a bounded domain Ω . Then, up to a subsequence, { u n } converges to a solution u on compact subsets of Ω . The other preliminary result is an application of Theorem 2 of [2] to Sol , which implies existence of constant mean curvature graphs taking C ,α bound-ary data in C ,α domains. Theorem 8 ([2]) . Let Ω ⊂ H be a domain with boundary γ of class C ,α contained in an Euclidean disk B R ( x , y + R ) of radius R centered at ( x , y + R ) , where ( x , y ) ∈ H . If the Euclidean curvature of γ satisfies κ euc ( x, y ) ≥ H/y and either H ≤ √ or R ≤ y (cid:32) H + √ H − √ (cid:33) / √ − , (4.1) then for any ϕ ∈ C ,α ( γ ) , there is a unique solution u ∈ C ,α (Ω) of (2.3) . To see that the above result is a consequence of Theorem 2 of [2], notice thatin Sol , Ric
Sol ≥ − , the mean curvature of the cylinder γ × R is yκ euc / B R ( x , y + R ) ifand only if it is contained in the hyperbolic disk of center ( x , (cid:112) y + 2 Ry ) andradius ln(1+2 R/y ) / . Therefore, Theorem 2 of [2] applies if ln(1+2
R/y ) / ≤ coth − ( H/ √ / √ , which is equivalent to (4.1).In [3] an interior gradient estimate that implies Theorem 8 for only con-tinuous boundary data is obtained. The proof of how the interior gradientestimate implies the generalization to C boundary data is made in Section 3of [3]: The idea is to take sequences of smooth boundary data that approximatethe continuous data ϕ from above and from below and use the interior gradientestimates to guarantee the convergence in compact subsets of the domain ofsequence of solutions. Proof of Theorem 6.
We apply the Perron method as in [17] to obtain a solu-tion. Since (2.3) admits a bounded subsolution in Ω , by translating it down-wards if necessary, the set S ϕ = { v ∈ C (Ω) ∩ C (Ω) | v is a subsolution to (2.3) in Ω and v ≤ ϕ on ∂ Ω } is non empty. If M = sup ϕ, then the constant function v = M is above anyfunction in S ϕ . Let m = inf Ω v , for some v ∈ S ϕ . Using the CompactnessTheorem, the Perron method implies that u = sup v ∈ S ϕ v is a solution to (2.3)in Ω with m ≤ u ≤ M. It remains to show that u extends continuously to ∂ Ω \ E. This is a conse-quence of the existence of local barriers given by Theorem 8. More precisely,for any q ∈ ∂ Ω \ E and for any ε > U ε the subset ofΩ obtained by smoothing the boundary of Ω ∩ B ε ( q ) , for B ε ( q ) the Euclideanball centered at q. We define w + and w − the solutions to (2.3) in U ε withboundary data w + = ϕ + on ∂U ε and w − = ϕ − on ∂U ε , for ϕ + , ϕ − ∈ C ( ∂U ε ) , such that ϕ + ( q ) = ϕ − ( q ) = ϕ ( q ) , ϕ − ≤ ϕ ≤ ϕ + on ∂U ε ∩ ∂ Ω , ϕ + = M on ∂U ε ∩ Ω and ϕ − = m on ∂U ε ∩ Ω . Then w − and w + arelower and upper barriers, w − ≤ u ≤ w + in U ε , which implies the continuity of u up to the boundary.A particular case where a subsolution exists occurs when the domain Ω iscontained in a disk as in Theorem 8. We remark that for any H ≥ , if adomain is taken sufficiently small, the Dirichlet problem is solvable, as we cansee as a consequence of the next corollary. Corollary 9.
Let Ω ⊂ H be a piecewise C domain. Suppose that Ω is con-tained in an Euclidean disk of radius R centered at ( x , y + R ) , where ( x , y ) ∈ H , and that the outer Euclidean curvature of ∂ Ω satisfies ˆ κ euc ( x, y ) ≥ H/y with possible exception in a finite set E. If (4.1) holds, then the Dirichlet prob-lem for constant mean curvature H is solvable for any bounded ϕ ∈ C ( ∂ Ω \ E ) . Proof.
Notice that given a bounded function ϕ ∈ C ( ∂ Ω \ E ) , we can find asolution to (2.3) defined on Ω that is below ϕ on ∂ Ω . For that, take a larger C ,α domain (cid:101) Ω containing Ω , such that (cid:101) Ω is contained in the disk of radius R and, by Theorem 8 applied to the Dirichlet problem on (cid:101) Ω with boundary data u = inf ϕ, there is a solution w, which is (or can be translated downwards tobe) below ϕ on ∂ Ω . Hence, by Theorem 6, the Dirichlet problem has a solutionin Ω . Lemma 2.
Let Ω be a domain and γ be a C arc of ∂ Ω . (i) If κ euc ( x, y ) < H/y, then any point in the interior of γ admits a neigh-borhood U ⊂ Ω in which there is a supersolution u + of (2.3) with exteriornormal derivative ∂u + ∂η = + ∞ along γ. (ii) If γ has κ euc ( x, y ) < − H/y.
Then any point in the interior of γ admitsa neighborhood U ⊂ Ω in which there is a subsolution u − of (2.3) withexterior normal derivative ∂u − ∂η = −∞ along γ. Proof.
Let us prove the first item. Consider w a function of the hyperbolicdistance r to γ. By definition, the function w is a supersolution to (2.3) withnormal derivative ∂w∂η = + ∞ along γ if and only ifdiv (cid:18) y ∇ wW (cid:19) = 1 W (cid:2) (2 yw (cid:48) + y w (cid:48) ) (cid:104)∇ y, ∇ r (cid:105) + y w (cid:48)(cid:48) (cid:3) + y w (cid:48) ∆ rW ≤ yH, (5.1)where W = (cid:112) y w (cid:48) and the gradient and the Laplacian are taken in thehyperbolic metric and also lim r → w (cid:48) ( r ) = −∞ . Since we only have assumptions on the Euclidean curvature of the bound-ary, let us relate the hyperbolic Laplacian with the Euclidean one. Denotingby γ r the curve in Ω parallel to γ with hyperbolic distance r apart from γ, wehave for r > − ∆ r = κ ( γ r ) = yκ euc ( γ r ) + 1 y (cid:104)∇ y, ∇ r (cid:105) . (5.2)We define (cid:102) W = W − yw (cid:48) = (cid:113) y w (cid:48) ) − and we can rewrite (5.1) with the Euclidean curvature1 (cid:102) W (cid:20) − w (cid:48)(cid:48) yw (cid:48) (cid:21) + y κ euc ( γ r ) (cid:102) W − (cid:104)∇ y, ∇ r (cid:105) y w (cid:48) (cid:102) W ≤ yH. (5.3)Take w ( r ) = − r a , for a ∈ (0 , / . Hence when r → , we get w (cid:48) ( r ) → −∞ , (cid:102) W → , (cid:104)∇ y, ∇ r (cid:105) y w (cid:48) (cid:102) W → w (cid:48)(cid:48) w (cid:48) → a ∈ (0 , / . Then given ε > , the first and third terms of (5.3) can be assumed to haveabsolute value less than ε/ , if r is sufficiently small. Moreover, for sufficientlysmall r and restricting to a neigboorhood U where y does not vary much, fromrelation (5.2), the Euclidean curvature of the parallel curves γ r also remainsbounded close to 2 H/y, implying (5.3) in U and then u + = w is a supersolutionin U. To prove the second item, take u − = − w so u − is a subsolution in aneighborhood U. Lemma 3.
Let Ω be a domain bounded by the union of two closed arcs γ and γ , where γ is of class C . Let u ∈ C (Ω) ∩ C ( γ ) and v ∈ C (Ω) ∩ C (Ω) be respectively a solution and a subsolution of (2.3) in Ω and assume that ∂v∂η = −∞ along γ . If lim inf( u − v ) ≥ for any approach to a point of γ , then v ≤ u in Ω . The proof follows from the General Maximum Principle (Lemma 1) exactlyas stated in [18].
Lemma 4.
Let u be a solution of (2 . in a domain Ω , γ ⊂ ∂ Ω be a C arc,and suppose that m ≤ u ≤ M on γ for some constants m, M. Then thereexists a constant c = c (Ω) (only depending on Ω ) such that for any compact C subarc γ (cid:48) ⊂ γ, (i) if κ euc ( γ (cid:48) ) ≥ H/y with strict inequality except for isolated points, thenthere is a neighborhood U of γ (cid:48) in Ω such that u ≥ m − c in U ; (ii) if κ euc ( γ (cid:48) ) > − H/y, then there is a neighborhood U of γ (cid:48) in Ω such that u ≤ M + c in U. Proof. If κ euc ( γ (cid:48) ) ≥ H/y with strict inequality except for isolated points,then, assuming that γ (cid:48) is small enough, there is a curve δ with κ euc ( δ ) > H/y such that γ (cid:48) ∪ δ bounds a domain U. Reverting the orientation of δ so that itencloses U, it has κ euc ( δ ) < − H/y.
By approximating δ to γ (cid:48) if necessary, we can assume that U is containedin the neighborhood of δ given by Lemma 2, where u − is defined. ApplyingLemma 3 for the subsolution v = u − + m − sup U u − , we conclude that u ≥ m − (sup U u − − inf U u − )in U, proving the first part of the lemma for c = sup U u − − inf U u − . We remark that the assumption on the size of γ (cid:48) is not important since wecan divide it into small pieces and obtain U as the union of a finite number ofneighborhoods.An analogous argument with the supersolution u + (by taking v = u + + M − inf U u + ) implies the second assertion of the lemma.1 In this section we will describe some flux formulas that we will need in orderto establish our existence theory for solutions with infinite boundary values.The definition of the flux of a function presented here is very similar to theones presented in previous results in other ambient spaces. The only essentialdifference is that we deal with Killing graphs whose Killing vector field is notunitary ( | ∂ t | = y ), hence the norm of this vector field naturally appears in thedefinition of the flux.Let u ∈ C (Ω) ∩ C ( ¯Ω) be a solution of (2.3) in a domain Ω ⊂ H . Thenintegrating (2.3) over Ω gives (cid:90) Ω yH d σ = (cid:90) ∂ Ω (cid:104) yX u , ν (cid:105) d s, (6.1)where d σ is the area element in H , X u = y ∇ u √ y |∇ u | and ν is the outer normalto ∂ Ω . The right-hand integral is called the flux of u across ∂ Ω . Motivated bythis equality, we define the flux of u across any subarc of ∂ Ω as follows.
Definition 2.
Let γ be a subarc of ∂ Ω . Take η a simple smooth curve in Ωso that γ ∪ η bounds a simply connected domain ∆ η . We define the flux of u across γ to be F u ( γ ) = 2 H (cid:90) ∆ η y d σ − (cid:90) η (cid:104) yX u , ν (cid:105) d s. (6.2)Observe that the first integral does not depend on u , it only depends onthe domain. Moreover, the definition does not depend on the choice of η. Infact, let (cid:101) η be another choice of curve and consider the 2-chain C with orientedboundary η − ˜ η. Using (2.3) and the divergence theorem on C we get2 H (cid:90) ∆ ˜ η y d σ − H (cid:90) ∆ η y d σ = (cid:90) ˜ η (cid:104) yX u , ν (cid:105) d s − (cid:90) η (cid:104) yX u , ν (cid:105) d s. Therefore, the definition is well posed. Notice that if u ∈ C (Ω ∪ γ ) , then we canchoose η to be γ with the inverse orientation, and then F u ( γ ) = (cid:82) γ (cid:104) yX u , ν (cid:105) d s. Notice that the definition of flux makes sense for any curve γ contained inΩ . In fact, if γ ⊂ Ω we can consider a subdomain U ⊂ Ω such that γ ⊂ ∂U and use the definition above.The proof of the next three lemmas follows the same steps as the proof ofProposition 4.6 in [13]. Lemma 5.
Let u be a solution of (2.3) in a domain Ω . Then1. (cid:82) Ω yH d σ = (cid:82) ∂ Ω (cid:104) yX u , ν (cid:105) d s ;
2. For every curve γ in Ω with (cid:96) euc ( γ ) < ∞ , we have | F u ( γ ) | < (cid:96) euc ( γ );
3. For every curve γ in Ω with (cid:96) euc ( γ ) < ∞ , we have | F u ( γ ) | ≤ (cid:96) euc ( γ ) . Lemma 6.
Let u be a solution of (2.3) in a domain Ω and γ ⊂ ∂ Ω be apiecewise C arc satisfying κ euc ( γ ) ≥ H/y and so that u is continuous on γ. Then (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) γ (cid:104) yX u , ν (cid:105) d s (cid:12)(cid:12)(cid:12)(cid:12) < (cid:96) euc ( γ ) . (6.3) Lemma 7.
Let u be a solution of (2.3) in a domain Ω and γ ⊂ ∂ Ω be apiecewise C arc.1. If u tends to + ∞ on γ, then κ euc ( γ ) = 2 H/y and (cid:90) γ (cid:104) yX u , ν (cid:105) d s = (cid:96) euc ( γ ) .
2. If u tends to −∞ on γ, then κ euc ( γ ) = − H/y and (cid:90) γ (cid:104) yX u , ν (cid:105) d s = − (cid:96) euc ( γ ) . The following lemma is a simple extension of Lemma 7.
Lemma 8.
Let Ω be a domain and γ ⊂ ∂ Ω be a compact piecewise C arc. Let { u n } be a sequence of solutions of (2 . in Ω such that each u n is continuouson γ.
1. If the sequence diverges to + ∞ uniformly on compact subsets of γ whileremaining uniformly bounded on compact subsets of Ω , then lim n → + ∞ (cid:90) γ (cid:104) yX u n , ν (cid:105) d s = (cid:96) euc ( γ ) .
2. If the sequence diverges to −∞ uniformly on compact subsets of γ whileremaining uniformly bounded on compact subsets of Ω , then lim n → + ∞ (cid:90) γ (cid:104) yX u n , ν (cid:105) d s = − (cid:96) euc ( γ ) . We have one more useful property of the flux given by the next lemma.
Lemma 9.
Let Ω be a domain and γ ⊂ ∂ Ω be a compact piecewise C arc with κ euc ( γ ) = 2 H/y . Let { u n } be a sequence of solutions of (2 . in Ω such thateach u n is continuous on γ. Then if the sequence diverges to −∞ uniformly oncompact sets of Ω and remains uniformly bounded on compact subsets of γ, weget lim n → + ∞ (cid:90) γ (cid:104) yX u n , ν (cid:105) d s = (cid:96) euc ( γ ) . Proof.
The idea to prove this result is basically the same idea as the proof ofLemma 7.Denote by G u n the graph of u n . Since each G u n is a stable constant meancurvature surface, there exists δ > p k is sufficiently far from the boundary (a condition that depends only on δ ), aneighborhood of each point ( p k , u n ( p k )) in G u n is a graph of bounded geometryover a disk of radius δ centered at the origin of the tangent plane T p k G u n . Now let p ∈ γ and consider a sequence of points p k ∈ Ω that converges to p. Since u n ( p k ) → −∞ , then, for each k, there exists n k large such that for n ≥ n k , a neighborhood of ( p k , u n ( p k )) is a graph of bounded geometry. Nowconsider the sequence of graphs, denoted by G p k ( δ ) , at each point ( p k , u n k ( p k ))that have bounded geometry. If we denote by N p k the unit normal vectorto G p k ( δ ) at p k , we can prove exactly as in Lemma 7 that N p k converges toa horizontal vector. Since u n k → −∞ on p k and { u n k ( p ) } is bounded, weknow that (cid:104)∇ u n k , ν (cid:105) ≥ (cid:104) N u nk , ν (cid:105) ≤ . Hence we conclude that N u nk ( p ) = − ν. Since p is arbitrary, we getlim n → + ∞ (cid:90) γ (cid:104) yX u n , ν (cid:105) d s = lim n → + ∞ (cid:90) γ y (cid:104)− N u n , ν (cid:105) d s = (cid:90) γ y d s = (cid:96) euc ( γ ) . As in the Euclidean case, we have the following consequence of Lemma 7 andthe Local Harnack Inequality (Theorem 5).
Theorem 10.
Let { u n } be a monotone increasing or decreasing sequence ofsolutions of (2 . on a domain Ω ⊂ H . If the sequence is bounded at some pointof Ω , then there exists a nonempty open set U ⊂ Ω such that the sequence { u n } converges to a solution of (2 . in U. The convergence is uniform on compactsubsets of U and the divergence is uniform on compact subsets of V = Ω \ U. If V is nonempty, then its boundary ∂V consists of arcs of curvature κ euc = ± H/y and arcs of ∂ Ω . These arcs are convex to U for increasing sequencesand concave to U for decreasing sequences. In particular, no component of V can consist of a single interior arc. Moreover, we have the following property for the divergence set V. Lemma 10.
Let Ω be a domain bounded in part by an arc γ with κ euc ( γ ) ≥ H/y.
Let { u n } be a monotone increasing or decreasing sequence of solutionsof (2 . in Ω with each u n continuous in Ω ∪ γ. Suppose α is an interior arc of Ω with curvature κ euc ( α ) = 2 H/y forming part of the boundary of the divergenceset V. Then α cannot terminate at an interior point of γ if { u n } either divergesto ±∞ on γ or remains uniformly bounded on compact subsets of γ. Proof.
Let α ⊂ ∂V be an interior arc of Ω with κ euc ( α ) = 2 H/y and supposethat it terminates at an interior point p ∈ γ. Up to restricting ourselves to asubarc of γ that contains p, we can assume that γ is a C arc. By Lemma 7,the sequence { u n } cannot diverge to −∞ on γ ; and if the Euclidean curvatureof γ is not identically 2 H/y, then the sequence { u n } cannot diverge to + ∞ on γ. Hence if the Euclidean curvature of γ is not identically 2 H/y, the sequence { u n } remains uniformly bounded on compact subsets of γ, but it follows fromLemma 4 that a neighborhood of γ is contained in U, a contradiction. Thuswe can assume that κ euc ( γ ) = 2 H/y.
Suppose { u n } diverges to + ∞ on γ and there exists only one such α thatterminates at p. It follows that { u n } diverges to + ∞ on α. In fact, since thereis only one such α, then one sub arc γ (cid:48) of γ with p as an endpoint is containednecessarily in ∂V, hence we can choose a point q ∈ α and r ∈ γ (cid:48) such that thetriangle with vertices p, q, r is entirely contained in V ; thus since the divergencein V is uniform on compact subsets and we already know that u n → + ∞ on γ, we conclude that { u n } diverges to + ∞ on α. Now let q be a point of α close to p and choose a point r on γ near p sothat the Euclidean geodesic segment rq lies in U. Let T be the triangle formedby rq and the arcs (cid:95) qp ⊂ α and (cid:95) pr ⊂ γ of curvature 2 H/y.
By (6.1) we have (cid:90) T yH d σ = F u n ( (cid:95) qp ) + F u n ( (cid:95) pr ) + F u n ( rq ) . Since by Lemma 8 we know thatlim n →∞ F u n ( (cid:95) qp ) = (cid:96) euc ( (cid:95) qp ) , lim n →∞ F u n ( (cid:95) pr ) = (cid:96) euc ( (cid:95) pr )and by Lemma 5 (item 2), F u n ( rq ) > − (cid:96) euc ( rq ) , then we get (cid:82) T yH d σ(cid:96) euc ( rq ) ≥ (cid:96) euc ( (cid:95) qp ) + (cid:96) euc ( (cid:95) pr ) (cid:96) euc ( rq ) − . (7.1)Keeping p fixed, we move the point q to ¯ q and r to ¯ r along the same arcsso that (cid:96) euc ( (cid:95) ¯ qp ) = λ(cid:96) euc ( (cid:95) qp ) and (cid:96) euc ( (cid:95) p ¯ r ) = λ(cid:96) euc ( (cid:95) pr ) for λ < . Observe that (cid:82) T yH d σ has a quadratic dependency on λ while the other terms have a lineardependency. Then the left-hand side of (7.1) tends to zero when λ →
0, and theright-hand side remains uniformly positive, a contradiction. So if such α ⊂ ∂V exists, it should not be unique. However, if we assume that there are at leasttwo arcs α , α ⊂ ∂V that terminate at p, then again we could find a triangle T ⊂ U whose edges are two constant curvature 2 H/y arcs (where u n → + ∞ )and an Euclidean geodesic segment as before (perhaps with ∂T ∩ γ = { p } ),and the same argument would give us a contradiction.If the sequence remains uniformly bounded on compact subsets of γ, wechoose r on γ so that T is contained in V. By Lemma 4 the sequence mustdiverge to −∞ in V. We now reach a contradiction as above using Lemma 9.5
Definition 3.
We say that a bounded domain Ω is admissible if:i) it is simply connected;ii) its boundary consists in the union of C open arcs { A i } , { B i } and { C i } and their endpoints, satisfying κ euc ( A i ) = 2 H/y, κ euc ( B i ) = − H/y and κ euc ( C i ) ≥ H/y, respectively (with respect to the interior of Ω) and notwo of the arcs A i and no two of the arcs B i have a common endpoint;iii) if the family { B i } is non empty, we require that for all i, if p i and q i are initial and final points of B i , there is an arc B ∗ i ⊂ H \ Ω of κ euc =2 H/y such that the domain W bounded by B i ∪ B ∗ i ∪ { p i , q i } is thesmallest convex domain bounded by arcs of Euclidean curvature 2 H/y (with respect to the interior of W ) that connects p i and q i .iv) We define Ω ∗ as the domain obtained by replacing the boundary arcs B i of ∂ Ω by B ∗ i . We require Ω ∗ to be simply connected and admit a boundedsubsolution of (2.3). If { B i } is empty, we require Ω to admit a boundedsubsolution of (2.3).Corollary 9 states that, if the domain is small enough (depending on H ),it admits a solution and hence condition iv) is satisfied. Remark 2.
We compare condition iii) to its analogous version in R formu-lated in [18]: “An arc B i of constant Euclidean curvature H is required tohave length less than π/ H, so that it consists in less than half-circle and henceits reflection along the line connecting its endpoints, together with B i boundsthe smallest convex set bounded by two curves of curvature H, oriented in-wards, that connect the endpoints of B i . Thus, B ∗ i can be taken as the reflectionof B i .” Since in our setting we do not have reflections in all directions, thisfact is no longer true, and therefore we need condition iii) in Definition 3.This condition is necessary in the proofs of our results, since we first establisha sequence of solutions in Ω ∗ and then we prove that the sequence diverges in Ω ∗ \ Ω , using our knowledge of the divergence set (Section 7), which can onlybe applied if iii) holds. Definition 4 (Admissible polygon) . Let Ω be an admissible domain. We saythat
P ⊂
Ω is an admissible polygon if it is a polygon whose sides are curves ofEuclidean curvature ± H/y, which we call 2
H/y − curves, and whose verticesare chosen from among the endpoints of arcs of the families { A i } and { B i } . Definition 5 (Dirichlet problem) . Let Ω be an admissible domain and fix
H >
0. The generalized Dirichlet problem is to find a solution of (2.3) in Ωof mean curvature H , which assumes the value + ∞ on each arc A i , −∞ oneach arc B i and prescribed continuous data on each arc C i .6 For an admissible polygon P , we denote by α and β the total Euclideanlength of the arcs in ∂ P which belong to A i and B i , respectively, by (cid:96) theperimeter of P , and I ( P ) = (cid:82) P y d a, where d a is the Euclidean area element.We remark that the quantity I ( P ) has already appeared in Section 6 (seeequation (6.1)). In fact, it corresponds to the hyperbolic integral: I ( P ) = (cid:90) P y d a = (cid:90) P y d σ. In this section we present necessary and sufficient conditions for some admis-sible domains to have a solution to (2.3) that assigns a continuous boundarydata in the arcs C i , + ∞ in the arcs A i ’s and −∞ in the arcs B i ’s. We willprove Theorem 1 here but in order to prove Theorem 2 we first need to showthe result for the case where κ ( C i ) > H/y and construct barriers. The dif-ference between the proofs of these two theorems is that on Theorem 1 theassumption that the assigned boundary data is bounded below allows us touse a bounded subsolution as a barrier from below; however in Theorem 2 theassigned boundary data is bounded from above and the bounded subsolutiondoes not work, so in this case we need to construct a barrier (see Proposition7).
Proof of Theorem 1.
First suppose the inequality 2 α < (cid:96) + 2 H I ( P ) holds forany admissible polygon P . Let u n be the solution of (2.3) in Ω assuming the boundary data u n = (cid:26) n on A i min { f, n } on C i . Each u n exists and is unique by Theorem 6. From the General MaximumPrinciple (Lemma 1), the sequence { u n } is monotone increasing and the seconditem of Lemma 4 implies that { u n } is bounded above in a neighborhood of eacharc C i ; hence the Monotone Convergence Theorem (Theorem 10) applies.Let U be the open set in which the sequence { u n } converges to a solutionof (2 .
3) and let V be its divergence set. From Theorem 10 and Lemma 10,each connected component of V must be bounded by curves of ∂ Ω and interiorcurves of κ euc = − H/y, if oriented to V. Besides, its vertices must be amongthe endpoints of { A i } and, from Lemma 4, item ii) , ∂V cannot contain anyboundary arc of the family { C i } . Hence any connected component of V is anadmissible polygon.If V is not an empty set, let P be a connected component of V. Since each u n is a solution to (2.3) in P , equality (6.1) implies that2 H I ( P ) = (cid:90) P yH d σ = (cid:90) ∪ i A i (cid:104) yX u n , ν (cid:105) d s + (cid:90) ∂ P\∪ i A i (cid:104) yX u n , ν (cid:105) d s, (9.1)7where ∪ i A i takes only the i ’s such that A i is a part of ∂ P . From Lemma 5 and 8, we conclude that (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∪ i A i (cid:104) yX u n , ν (cid:105) d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ α and lim n → + ∞ (cid:90) ∂ P\∪ i A i (cid:104) yX u n , ν (cid:105) d s = − ( (cid:96) − α ) , which implies that 2 H I ( P ) ≤ − ( (cid:96) − α ) + α, a contradiction. We then concludethat V is an empty set and that U = Ω . It remains to see that u takes the required boundary data. To see that u = f i on each C i , we follow the same barrier argument as the proof of Theorem6, noticing that for each p ∈ C i , the sequence u n is uniformly bounded in aneighborhood of p, a consequence of Lemma 4 and the fact that Ω admits aglobal subsolution to the Dirichlet Problem.Let us notice that u indeed goes to + ∞ on the arcs { A i } . Recall that u isa global solution to the Dirichlet Problem with u ≤ u n in Ω . Given p ∈ A i , let U p be a neighborhood of p for which the existence result (Theorem 6) applies.Take v k the solution to the Dirichlet problem in U p ∩ Ω with boundary data m = min U u on ∂U p ∩ Ω and k on ∂ Ω ∩ U p . Then for n ≥ k, u n ≥ v k fromthe Maximum Principle. Therefore, u ≥ v k for all k and u diverges to + ∞ on U p ∩ ∂ Ω . Reciprocally, assume that the Dirichlet problem has a solution in Ω . Weknow the equality (6.1) holds in any admissible polygon P in Ω . By Lemma 7, we have (cid:90) ∪ i A i (cid:104) yX u , ν (cid:105) d s = α. Using Lemma 5 for the curves ∂ P \ ∪ i A i that are not on ∂ Ω and Lemma6 for the ones on ∂ Ω , we get (cid:90) ∂ P\∪ i A i (cid:104) yX u , ν (cid:105) d s > − ( (cid:96) − α ) . Then,2 H I ( P ) = (cid:90) ∪ i A i (cid:104) yX u , ν (cid:105) d s + (cid:90) ∂ P\∪ i A i (cid:104) yX u , ν (cid:105) d s > − ( (cid:96) − α ) + α, and therefore, for any admissible polygon, we have2 α < (cid:96) + 2 H I ( P ) . Proposition 2.
Let Ω be an admissible domain such that the family { A i } isempty. Assume that κ euc ( C i ) > H/y and that the assigned boundary data onthe arcs { C i } is bounded above. Then the Dirichlet Problem has a solution in Ω if and only if β < (cid:96) − H I ( P ) for all admissible P . The proof of this proposition in Sol is analogous to the proof of this resultin other ambient spaces (see for instance Proposition 7.2 in [6]) using the sameadaptations used in the above result.
10 More on admissible domains
In this section we state and prove some properties of the 2
H/y − curves in orderto construct admissible domains around any point of a 2 H/y − curve. Thesedomains will be constructed (in Section 11) so that they satisfy the hypothesisof Theorem 1 and Proposition 2 and therefore will admit solutions to theDirichlet Problem. Later we will use these surfaces as barriers to improveProposition 2 for the case of κ euc ( C i ) ≥ H/y and then prove Theorem 2.The first result about these curves presents their shape and parametriza-tion. These curves were also exhibited in [10] in the Lie Group model of Sol . Proposition 3.
An arc of Euclidean curvature H/y is part of the trace of acurve γ P , P = ( w, z ) ∈ R , parametrized by t (cid:55)→ ( x ( t ) , y ( t )) , t ∈ R , where x ( t ) = w + ze / H H S H ( t ) y ( t ) = ze sin2( t/ H . The function S H is defined by S H ( t ) = (cid:90) − cos t − − ue u/ H √ − u du, t ∈ [ − π, (cid:90) − cos t − ue u/ H √ − u du, t ∈ [0 , π ] , for t ∈ [ − π, π ] , and is extended to R using the relation S (2 nπ + t ) = 2 nS ( π ) + S ( t ) , t ∈ [ − π, π ] , n ∈ Z . Proof.
A curve in R , parameterized by γ ( t ) = ( x ( t ) , y ( t )) has Euclidean cur-vature κ euc = − H/y, H (cid:54) = 0 , if and only if x (cid:48) ( t ) y (cid:48)(cid:48) ( t ) − x (cid:48)(cid:48) ( t ) y (cid:48) ( t ) = − Hy ( t ) ( x (cid:48) ( t ) + y (cid:48) ( t ) ) / , (10.1)which holds for γ P defined above.9 Remark 3.
A clearer parametrization of γ P that sees it as a graph in thehorizontal direction and does not take into account the sign of κ euc is γ P ( y ) =( x ( y ) , y ) , y ∈ (cid:0) z, ze / H (cid:1) , where x ( y ) = w ± ze / H H (cid:90) − H ln( y/z ) − − ue u/ H √ − u du. (10.2) We name the function in the integrand by g H ( u ) = − ue u/ H √ − u . (10.3)The shape of a 2 H/y − curve (see Figure 1) repeats periodically if one moveshorizontally. We define some useful quantities associated to these curves. Definition 6.
For any
H > , we define L H = e / H H (cid:90) − g H ( u ) du (10.4) M H = e / H H (cid:90) − − g H ( u ) du (10.5)and T = T H as the number in the interval (0 ,
1) such that (cid:90) T H − g H ( u ) du = 0 . (10.6)Figure 1: Curve of Euclidean curvature 2 H/y.
We can see from Proposition 3 that if the base point P = P has coordi-nates P = (0 , z ) , the other indicated points in the figure have the followingcoordinates:0 P +2 = γ P ( − π/
2) = (cid:0) zL H , ze / H (cid:1) ,P − = γ P ( π/
2) = (cid:0) − zL H , ze / H (cid:1) ,P +4 = γ P ( π ) = (cid:0) zM H , ze /H (cid:1) ,P − = γ P ( − π ) = (cid:0) − zM H , ze /H (cid:1) ,P = γ P ( T H ) = (cid:16) , ze TH H (cid:17) . Since the coordinates are all proportional to z, we conclude that movingupwards makes the period and height of the 2 H/y − curves grow linearly. Sincean Euclidean dilation centered at the origin is an hyperbolic isometry in thismodel, this shows that in H all the 2 H/y − curves are isometric. H/y − curves joining aligned points To build the admissible domains around any point in a 2
H/y − curve that satisfythe conditions of Theorem 1 and Proposition 2, we must analyse all inscribedpolygons, that is, all 2 H/y − curves contained in the domain that connect thevertices of the boundary of the domain. In order to simplify this analysis weconstruct (Section 11) admissible domains using four vertices of a rectanglewith sides parallel to the x and y axes.To verify the hypotheses of Theorem 1 and Proposition 2 we need thenext results about 2 H/y − curves connecting horizontally or vertically alignedpoints.Given two horizontally aligned points p = ( − w, z ) and q = ( w, z ) we wouldlike to know what are all embedded 2 H/y − curves from p to q. The next propo-sition gives sufficient conditions on the Euclidean distance 2 w between p and q for the existence of at most three embedded 2 H/y − curves connecting them. Proposition 4.
Given two horizontally aligned points p = ( − w, z ) and q =( w, z ) with Euclidean distance w < zK ( H ) , any embedded H/y − curve from p to q is: i) symmetric about reflection on the vertical line x = 0 , ii) either contained in the region { y ≥ z } or contained in the region { y ≤ z } . The constant K = K ( H ) depends only on H and is given by expression (10.7) below.Moreover, there is exactly one such curve above the line y = z and there areat most two such curves below the line y = z. If w < ze − / H L H , there aretwo curves below the line y = z , one which is shorter, without point of verticaltangency, and another one that has two points of vertical tangency (that is, ithas the two P − and P +2 type of points, see Figure 1); in particular its P -typeof point is in the line y = ze − / H or below it. Proof.
All possible 2
H/y − curves connecting p to q are parts of a curve γ P de-scribed in Proposition 3. Hence their possible shapes are obtained intersecting γ P with horizontal lines.To look at all possible intersections we move downwards a horizontal lineand look at parts of γ P that join two intersection points (horizontally aligned)and are embedded. We first find curves above the horizontal line connectingthe two intersection points which exist if2 w = d euc ( p, q ) ≤ ze − H ( M H + L H ) , where the equality corresponds to the case where p is a P − -point and q is a P +2 -point from the next loop of γ P .If p is a point in the arc (cid:95) P P +2 P of a loop of a curve γ P and q is a point inthe arc (cid:95) P P − P of the next loop of γ P , the part of γ P joining the two pointsis an embedded arc but is not contained in { y ≥ z } nor in { y ≤ z } . We wantto avoid this kind of curves.In order to do that we fix the line y = z and look at the family of all 2 H/y -curves that intersects this line and have the P -type point below it. We define d ( t ) = w ( t ) − w ( t ) , for any t ∈ [0 , T H ] , the Euclidean distance between( w ( t ) , z ) and ( w ( t ) , z ) , the first and second intersections of γ (0 ,ze − t/ H ) withthe half line { ( x, y ) | y = z and x ≥ } . (Observe that since translations alongthe x -direction are isometries, we can restrict ourselves to the family of 2 H/y -curves whose P -type point is of the form (0 , ze − t/ H )).We compute w and w and then look for the minimum value of d. On onehand, w ( t ) is the x -coordinate of γ (0 ,ze − t/ H ) when the y -coordinate is z, then w ( t ) = ze − t/ H e / H H (cid:90) − t − g H ( u ) du. On the other hand, w ( t ) is the distance between (0 , ze − t/ H ) and the next P -point of γ (0 ,ze − t/ H ) , which is 2 M H ze − t/ H , minus w ( t ) , hence w ( t ) = 2 M H ze − t/ H − w ( t )and then d ( t ) = 2 M H ze − t/ H − w ( t )= 2 ze − t/ H (cid:18) M H − e / H H (cid:90) − t − g H ( u ) du (cid:19) = z ¯ d ( t ) . We notice that ¯ d is a continuous function in [0 , T H ] that depends onlyon H and we look for the minimum value of ¯ d. We claim that there is a point2 t ∈ (0 , T H ) such that ¯ d is decreasing in [0 , t ] and increasing in [ t , T H ] . Therefore, ¯ d ( t ) is the minimum value of ¯ d. Differentiating ¯ d, we have¯ d (cid:48) ( t ) = 2 e − t/ H (cid:18) − M H H + e / H (2 H ) (cid:90) − t − g H ( u ) du − e / H H g H ( − t ) (cid:19) , which has the same sign as f ( t ) = − M H H + e / H (2 H ) (cid:90) − t − g H ( u ) du − e / H H g H ( − t ) . Differentiating f we have f (cid:48) ( t ) = e t/ H H (1 − ( − t ) ) / > . Moreover, lim t → + f ( t ) = −∞ and f can be extended continuously to [0 , , and thenlim t → − f ( t ) = − M H H + e / H (2 H ) (cid:90) − g H ( u ) du − lim t → − e / H H g H ( t ) = + ∞ . Hence f takes the value zero in (0 ,
2) and therefore ¯ d (cid:48) too. It is easy to seefrom the geometric definition of ¯ d, which also can be extended to [0 , , that t ∈ (0 , T H ) . Therefore t is the unique solution of f ( t ) = 0 , that is, M H H = e / H (2 H ) (cid:90) − t − g H ( u ) du − e / H H g H ( − t ) . Let us define K ( H ) = ¯ d ( t ) = 2 e − t / H (cid:18) M H − e / H H (cid:90) − t − g H ( u ) du (cid:19) (10.7)for t the solution above.Therefore, if p and q are closer than zK ( H ) , any 2 H/y − curve connectingthem is either above the line y = z as described in the beginning of this proofor below it.As a consequence of the Maximum Principle, we can show that the curveabove is unique. In fact, let us denote by p and q two horizontally aligned pointswhose distance is less than zK ( H ). Then any 2 H/y − curve joining them thatis above the line y = z does not contain any point with vertical tangency, itsinterior is contained in the open vertical slab bounded by the vertical linespassing through p and q , and it is symmetric with respect to the vertical line3passing through the midpoint between p and q . In particular, if there were twosuch curve, say c and c , they would be disjoint apart from the endpoints andone would be above the other (see Figure 2). Hence moving c farway upwardsand translating it back, we would find a first point of contact between c and c , but since both of them satisfy that at each point their Euclidean geodesiccurvature is 2 H/y we would get a contradiction with the Maximum Princeple,because the curve c was below c and therefore its y -point is smaller than thecorresponding y -point of c , making the geodesic curvature of c be less thanthe geodesic curvature of c . Therefore, there is only one curve above.Figure 2: Horizontally aligned points.The case of curves below y = z is different, we might not have uniqueness.By looking at the intersections of γ P with horizontal lines, it seems that for p and q sufficiently close there are usually two curves connecting p to q below y = z. We will show this is always the case.To demonstrate that, we again analyse the distance between two intersec-tions of a line y = z and a family of curves { γ (0 ,ze − t/ H ) } t ∈ (0 , T H ) . The differenceis that now we are interested in the distance l ( t ) between the first intersectionof γ (0 ,ze − t/ H ) with { y = z, x ≥ } and the last intersection of γ (0 ,ze − t/ H ) with { y = z, x ≤ } . As done before, we compute l ( t ) = 2 w ( t ) = 2 ze − t/ H e / H H (cid:90) − t − g H ( u ) du = z ¯ l ( t ) . (10.8)Again differentiating and in this case knowing that l (0) = l (1 + T H ) = 0 , we find a value t ∈ (0 , T H ) such that ¯ l is increasing in [0 , t ] and decreasingin [ t , T H ] , with a maximum value ¯ l ( t ) , and for each distance between 0and z ¯ l ( t ) , there are two 2 H/y − curves joining a pair of points that are thisdistance apart.We remark that ¯ l is e − t/ H multiplied by a function that attains its max-imum at t = 1; hence, the maximum value of ¯ l occurs in t ∈ (0 ,
1) and ¯ l isdecreasing in (1 , T H ) . Therefore, given 2 w ∈ (0 , z ¯ l (1)) , the curves that jointwo points 2 w apart are one with P -point above y = ze − / H and the otherwith P -point below y = ze − / H . Since¯ l (1) = 2 e − / H L H , the proof is concluded.The next result is about a domain Ω with only two horizontally alignedvertices p and q contained in the line y = z. γ + the curve that joins p and q contained above the line y = z and by γ − the curve below joining p and q without point of vertical tangency.If (cid:96) euc ( γ ± ) stands for the Euclidean length of γ ± , observe that (cid:96) euc ( γ + ) < (cid:96) euc ( γ − ) + 2 H I (Ω) , since (cid:96) euc ( γ + ) < (cid:96) euc ( γ − ) , because γ − has greater Euclidean curvature than γ + . Moreover, we have the following lemma.
Lemma 11.
In the context described above, let Ω be the domain bounded belowby γ − and above by γ + . Then, if p = ( − w, z ) and q = ( w, z ) are sufficientlyclose, it holds (cid:96) euc ( γ − ) < (cid:96) euc ( γ + ) + 2 H I (Ω) . Proof.
This inequality (cid:96) euc ( γ − ) < (cid:96) euc ( γ + ) + 2 H I (Ω) is a consequence of themore general one (cid:96) euc ( γ − ) < d euc ( p, q ) + 2 H I (Ω − ) , (10.9)where Ω − is the region bounded below by γ − and above by the straight line pq. We will show that this inequality holds if the base point P = (0 , ze − a/ H )of γ − is such that a + e a/ H < , the reason why p and q are required to besufficiently close.In order to do that, we compute and estimate (cid:96) euc ( γ − ) − d euc ( p, q ) and I (Ω − ) . We have (cid:96) euc ( γ − ) = 2 ze − a/ H e / H H (cid:90) − a − e u/ H √ − u du and d euc ( p, q ) = 2 w = 2 ze − a/ H e / H H (cid:90) − a − g H ( u ) du, so, using that the exponential function is increasing, we get (cid:96) euc ( γ − ) − d euc ( p, q ) = ze − a/ H e / H H (cid:90) − a − e u/ H (cid:114) u − u du ≤ zH (cid:90) − a − (cid:114) u − u du = zH (cid:90) a (cid:114) t − t dt. On the other hand, for x γ − the x -coordinate of γ − ,2 H I (Ω − ) = 4 H (cid:90) zze − a/ H x γ − ( y ) y dy = ze − a/ H e / H H (cid:90) a (cid:90) − t − g H ( u ) dudt ≥ ze − a/ H H (cid:90) a (cid:90) − t − − u √ − u dudt = ze − a/ H H (cid:90) a √ t − t dt, y = ze t − a/ H andused (10.2), and in the inequality we used again that the exponential functionis increasing.Hence (10.9) is implied by ze − a/ H H (cid:90) a √ t − t dt ≥ zH (cid:90) a (cid:114) t − t dt, which is a consequence of √ t − t ≥ e a/ H (cid:114) t − t that holds for t ∈ (0 , a ) , if a satisfies a + e a/ H < . The constructions of domains around points in 2
H/y − curves also needsome properties of 2 H/y − curves that join vertically aligned points. Proposition 5.
Given two vertically aligned points p = (0 , z ) and q = (0 , t ) with Euclidean distance < t − z < z ( e T H / H − , there are two H/y − curves γ I and − γ I joining p and q contained in the slab { z ≤ y ≤ t } ; one of themis contained in { x ≥ } and the other one is the reflection of the first about x = 0 . Moreover, any other H/y − curve with endpoints p and q intersects theline y = ze − / H , intersects both half planes { x < } and { x > } , and haslength greater than the length of γ I . The fact that other curves intersect the line y = ze − / H is importantbecause it allows the construction of domains that do not contain these curvesin its interior and, therefore, we do not need to care about their length. Proof.
The proof follows the same initial steps as the one for horizontallyaligned points. Since the reflection about a vertical line is an isometry, anycurve considered has its reflected correspondent, which we omit for shortness.Intercepting a 2
H/y − curve with a vertical line, we realize that there are threepossible types of curves joining points p and q vertically aligned:Type I: Occurs when p is between P and P +2 and q is between P +2 and P and looks like the figure of letter D without the vertical segment. Thisis the only type of curve contained in the horizontal slab determined bythe two horizontal lines passing through p and q. Type II: Occurs when p is between P and P − and q is between P and P − . Type III: Occurs when p is between P − and P and q is between P and P − . Notice that if p and q are endpoints of a curve of Type II, then its P -point, P = ( x , y ) , has the following relations with the coordinates of p and q : y < z < y e / H (because p is between P and P − ) and6 y e (1+ T H ) / H < t < y e / H (because q is between P and P − ) . Hence z < y e / H < te − T H / H , implying ze T H / H < t and t − z > z ( e T H / H − . We may conclude that if t − z < z ( e T H / H −
1) then no curve of Type IIfrom p = (0 , z ) to q = (0 , t ) exists.Notice also that a curve of Type III always intersects the line y = ze − / H because p is above its P +2 -point and, therefore, its P -point, P = ( x , y ) , issuch that y e / H < z. Having this observed we prove that the curves of Type I exist and areunique in each side of the vertical line. The existence is a consequence of t − z < z ( e T H / H − < z ( e (1+ T H ) / H −
1) = d ( P , P ) , for P = (0 , z ) . Theuniqueness is a consequence of the Maximum Principle. In fact, if there weretwo 2
H/y − curves from p to q in the same side of the line x = 0 , by movinghorizontally one of them we would find a last interior contact point, whichwould be a tangency point. From the Maximum Principle, the translation ofone curve would coincide with the other. Since their endpoints are the same,they have to be the same.It remains to show that a curve of Type III, denoted by γ III , from p to q is longer than a 2 H/y − curve of Type I from p to q. Let us assume withoutloss of generality that γ I is contained in { x ≥ } and γ III contains pointswith y -coordinate greater than z in { x ≥ } . Hence γ III is contained in thecomplement of the open region O I bounded by γ I and − γ I , the reflection of γ I about the vertical line x = 0 . In fact, if it were not so, there would bea region in the slab { z ≤ y ≤ t } bounded on the left by an arc of γ III andon the right by an arc of γ I , both with mean curvature vectors pointing tothe left; then moving γ III to the right there would be a last interior contactpoint contradicting the Maximum Principle. Therefore, since γ III is outsidethe convex domain O I , it is longer than its projection on O I , defined by π ( v ) = { u ∈ O I | d ( u, v ) ≤ d ( u, w ) , ∀ w ∈ O I } , which is γ I together with a part of − γ I , in particular, it is longer than γ I . The last result of this subsection is about domains bounded by a 2
H/y − curveof Type I and its reflection. Lemma 12.
Let p = (0 , z ) and q = (0 , t ) be two vertically aligned points. If γ I and − γ I are the H/y − curves joining p and q of Type I and Ω is the regionbetween them, then if (cid:96) denotes the Euclidean length of γ I (the same of − γ I ),it holds (cid:96) > H I (Ω) . Proof.
By prolonging γ I , we find the y − coordinate of its P − point, say P =( x , y ) . Then there are a ∈ (0 ,
1) and b ∈ (1 , T H ) such that z = y e a/ H and t = y e b/ H . Moreover, the P +2 − point of γ I is P = ( x γ ( y e / H ) , y e / H )for x γ such that γ I ( y ) = ( x γ ( y ) , y ) . The bottom part of γ I from p to P has length7 (cid:96) b = (cid:18) y e / H H (cid:19) (cid:90) − a e u/ H √ − u du computed as in the proof of Lemma 11. The upper part has length (cid:96) u = (cid:18) y e / H H (cid:19) (cid:90) − b e u/ H √ − u du. The bottom part of Ω, denoted by Ω b , bounded below by the bottom partsof − γ I and γ I and above by the line y = y e / H satisfies2 H I (Ω b ) = 2 (cid:32) H (cid:90) y e / H y e a/ H x ( y ) − x ( y e a/ H ) y dy (cid:33) ≤ H (cid:90) y e / H y e a/ H x ( y e / H ) − x ( y e a/ H ) y dy, because x γ ( y ) ≤ x ( y e / H ) for all y ∈ (cid:0) y e a/ H , y e / H (cid:1) . Hence2 H I (Ω b ) ≤ H (1 − a )2 H (cid:0) x ( y e / H ) − x ( y e a/ H ) (cid:1) = 2(1 − a ) (cid:18) y e / H H (cid:19) (cid:90) − a − ue u/ H √ − u du, and analogously for the upper part of Ω , denoted by Ω u :2 H I (Ω u ) ≤ b − (cid:18) y e / H H (cid:19) (cid:90) b − ue u/ H √ − u du. Therefore, since (1 − a )( − u ) < u ∈ ( a − , , we get2 (cid:96) b = 2 (cid:18) y e / H H (cid:19) (cid:90) − a e u/ H √ − u du> − a ) (cid:18) y e / H H (cid:19) (cid:90) − a − ue u/ H √ − u du ≥ H I (Ω b ) . (10.10)The same holds for the upper part of γ I because ( b − u ) < u ∈ (0 , b −
1) :2 (cid:96) u = 2 (cid:18) y e / H H (cid:19) (cid:90) − b e u/ H √ − u du> b − (cid:18) y e / H H (cid:19) (cid:90) b − ue u/ H √ − u du ≥ H I (Ω u ) . (10.11)Hence, adding inequalities (10.10) and (10.11), the proof is concluded.8
11 Construction of some admissible domains { B i } = ∅ Proposition 6.
There are domains U where the existence result from Theorem1 holds, that is, the family { B i } is empty and α < (cid:96) + 2 H I ( P ) , for alladmissible polygons P in U. We exhibit some of these domains which are bounded by 2 arcs of circles C and C of κ euc ( C i ) > H/y and two arcs A and A of Euclidean curvature2 H/y, if oriented inwards.
Proof.
Let R be a rectangle in H such that:- the sides of R are parallel to the x and y axes;- Denoting by d the Euclidean length of the diagonal of R and by p =( x ( p ) , y ( p )) the Euclidean center of R , it holds d < y ( p )3 + 2 H ; (11.1)- R is sufficiently small so that a subsolution exists (see Corollary 9).Let q and q be the two endpoints of a diagonal of R with x ( q ) < x ( q )and y ( q ) < y ( q ) . Let s and s be two vertical segments of length ε 1) whose medium points are q and q , respectively, where T H is the number described in Definition 6. For i = 1 , , we replace s i by A i , anarc of Type I (see the proof of Proposition 5) of a 2 H/y − curve that joins theendpoints of s i , is oriented in the direction of p and has length less than d/ . This is possible if ε is taken sufficiently small. These two arcs will be part of ∂U. To build the remaining part of ∂U we connect the lowest endpoint of s tothe lowest endpoint of s by a semicircle C whose diameter is the straight seg-ment that joins these two points. The arc C is built analogously by connectingthe upper endpoints of s and s . Observe that C and C are semicircles withsame diameter.Our domain U is the region bounded by the arcs A , A , C and C . Claim : U is an admissible domain.Notice that κ euc ( C i ) = 2 d > Hy min ≥ Hy , where y min = min { y ( q ); q ∈ ¯ U } > y ( p ) − d − d/ H + 32 < y ( p ) d and therefore H < y ( p ) d − ≤ y min d . U with empty { B i } . Claim : U admits a solution to the Dirichlet problem (Definition 5).We need to check that for any admissible polygon P ⊂ U , the inequality2 α < (cid:96) + 2 H I ( P ) holds. Recall (cid:96) denotes the Euclidean perimeter of the poly-gon and α denotes the sum of the Euclidean lengths of the arcs A i ’s containedin P . First observe that if A and A are not contained in P , then the inequalityholds trivially, since α = 0 . Hence we only consider the cases where at leastone of the arcs A i ’s is in P . If P has as vertices the two endpoints of an s i , then P contains A i andanother arc τ with the same endpoints as A i . Since ε < y ( q )( e T H / H − , Proposition 5 implies that A i is the shortest 2 H/y -arc joining its endpoints.Hence, (cid:96) = α + l ( τ ) > α and the inequality follows.In any other case, P contains at least an endpoint of each s i , therefore theperimeter of P inside U is ( (cid:96) − α ) > d. Since α is at most two times d/ , itholds that α ≤ d < (cid:96) − α implying that 2 α < (cid:96) + 2 H I ( P ) . { A i } = ∅ In the next proposition we not only exhibit domains for which Proposition 2implies the existence of a solution to the Dirichlet Problem attaining −∞ onthe arcs of type B i , but also we prove the existence of such domains aroundany point in a 2 H/y − curve. We will use these solutions as barriers to proveTheorem 2. Proposition 7. Let γ be a H/y − curve. Given any point p ∈ γ, there is adomain U, containing p , such thati) the boundary of U consists of two arcs of circles C and C of κ euc ( C i ) > H/y and two arcs B and B of Euclidean curvature − H/y, if orientedinwards; ii) the intersection ∂U ∩ γ is contained in B ∪ B ; iii) U admits a solution to the Dirichlet problem (Definition 5). Remark 4. The main difference from the construction of the domain in Propo-sition 7 to the previous construction in Proposition 6 is that instead of replacingvertical segments, we will replace horizontal segments by curves of Euclideancurvature − H/y. The reason is because the inequality β < (cid:96) − H I ( P ) hasa minus sign and therefore is more delicate to be satisfied by any admissiblepolygon. If we did the procedure with vertical lines, non convex admissiblepolygons with two vertices could arise and controling their length minus I ( P ) is very complicated.Proof. We make a construction analogous to Proposition 6. Assuming thatthe tangent vector to γ at p = ( x ( p ) , y ( p )) is neither horizontal nor vertical,we take the rectangle R as in the proof of Proposition 6. But here instead ofbeing centered at p, R just contains p and is such that γ ∩ R is a graph on thehorizontal and vertical directions. Besides, we assume that γ ∩ ∂R consists inthe two diagonal endpoints q and q . (See Figure 4).Figure 4: Rectangle R .Denoting by d the length of the diagonal of R, we assume d < min (cid:26) y ( p )3 (1 − e − / H ) , y ( p )8 H + 3 (cid:27) . (11.2)As in the proof of last proposition, let s and s be two horizontal seg-ments whose medium points are q and q , respectively, both of length ε < y ( p ) e − /H L H , where L H is given by (10.4). We replace s i by B i , the short-est arc of κ euc = − H/y (normal pointing away from the segments s i ’s) thatjoins the endpoints of s i , with length less than d/ B i ’s to be as small as we want because their lengths becomesmaller with ε ). Let C (resp. C ) be the semicircle that joins the left (resp.the right) endpoints of the segments s and s , of diameter d. We claim thatthe region U bounded by B , C , B and C is the region that we are lookingfor.It follows from d < y ( p )8 H +3 < y ( p )2 H +3 , as in the proof of Proposition 6, that κ euc ( C i ) > H/y. Since ε < y ( p ) e − /H L H and (11.2) implies that y ( q ) > y ( p ) − d >y ( p ) e − / H , it follows from Proposition 4 that B and B are the only arcs1Figure 5: Construction of the domain U with empty { A i } .of κ euc = − H/y joining their endpoints contained in ¯ U and, moreover, wehave the existence of the arcs B ∗ i . Notice that in order to see there is no other arc joining the endpoints of s (the upper segment) contained in U, it is sufficient to check that y min ≥ y ( q ) e − / H . Since y min = min { y ; ( x, y ) ∈ ¯ U } , it holds y min > y ( p ) − d − d/ y ( p ) − d/ , and then y ( q ) > y ( p ) and (11.2) implies y min ≥ y ( q ) e − / H . Claim : U admits a solution to the Dirichlet problem (Definition 5).Let P be an admissible polygon in ¯ U and let (cid:96) denote its Euclidean perime-ter. Because of Proposition 2, it is sufficient to prove the inequality: β + 2 H I ( P ) < ( (cid:96) − β ) . (11.3)For any polygon we have β ≤ (cid:96) ( B ) + (cid:96) ( B ) ≤ d/ d/ d. If A ( U ) denotes the Euclidean area of U, we have A ( U ) ≤ π ( d/ + εd ≤ d and then 2 H I ( P ) ≤ Hy min A ( U ) ≤ Hd y min ≤ Hd y ( p ) − d/ . If P has 3 or 4 vertices, then ( (cid:96) − β ) is at least 2 d. If P has two vertices,each of them must belong to a different B i , and then ( (cid:96) − β ) is at least 2 d aswell.Hence the inequality (11.3) is a consequence of d + 4 Hd y ( p ) − d/ < d, which holds because of (11.2).If p is a point of horizontal or vertical tangency in γ, the constructionbecomes simpler but it is very similar. If γ (cid:48) ( p ) is vertical, we take a very thin2rectangle centered at p. We assume its height d satisfies d < min (cid:26) y ( p )(1 − e − / H ) , y ( p )8 H + 1 (cid:27) . We call s and s the bottom and top of the rectangle and we assume theirlength is ε < y ( p ) e − /H L H . Exactly as in the previous case, we replace s i by B i , the shortest arc of κ euc = − H/y (normal pointing inwards) that joinsthe endpoints of s i , with length less than d/ 2; and we take C (resp. C ) thesemicircle of diameter d that joins the left (resp. the right) endpoints of thesegments s and s . Now the proof follows the exact same steps as in theprevious case.Let us assume that the tangent vector γ (cid:48) ( p ) is horizontal. Then let R be arectangle centered at p with sides parallel to the axes such that:- γ intersects ∂R in the vertical sides of R, which we assume to have length h < y ( p ) e − T H / H − 11 + e − T H / H = 2 y ( p ) tanh( T H / H ) . (11.4)- The basis has length b ≤ y ( p )4 H + 1 . (11.5)We replace the vertical sides of R by arcs B and B of curvature − H/y if oriented inwards. B and B have the same endpoints as the vertical sidesand exist because of Proposition 5. Notice that we can apply Proposition 5because given any point q in the basis of the rectangle, the hypothesis (11.4)implies that h < y ( q )( e − T H / H − y ( p ) = y ( q ) − h/ . We also assume h is small enough so that the length of B i is less than b/ . We replace the horizontal segments of ∂R by arcs of circle C and C suchthat the segments are diameters of the arcs. The domain U is then boundedby C , B , C and B . We assume U is small enough to admit a subsolution.Observe that U ∗ is obtained by reflecting B i about the vertical line that joinsits endpoints and that U does not contain a 2 H/y − arc joining two verticallyaligned vertices of ∂U. As before, the assumption (11.5) implies that the arcs C i have Euclideancurvature greater than 2 H/y. To see that U admits a solution to the Dirichlet problem, we analyse allpossible admissible polygons P ⊂ U . First notice that for any P , it cannot have as vertices only the endpointsof an arc B i because, from Proposition 5, there is no other 2 H/y − curve con-necting these points contained in U. Therefore ( (cid:96) − β ) is always greater than2 b. Besides, since the Euclidean area of U satisfies A ( U ) ≤ bh + b π/ ≤ b , as for the general case, we have32 H I ( P ) ≤ HA ( U ) y min ≤ Hy ( p ) − b b ≤ Hb y ( p ) − b . Since β ≤ (cid:96) ( B ) + (cid:96) ( B ) < b, inequality (11.2) is a consequence of b + 4 Hy ( p ) − b b < b, which holds from (11.5). 12 Existence Theorems - part II Let us finish by proving the remaining existence results. Proof of Theorem 2: We proceed as in Proposition 2 and consider the sequenceof solutions u n to the Dirichlet Problem with boundary values − n on B ∗ i andmax { f i , − n } on C i . Then Theorem 10 implies that the limit function u existsand is a solution to the Dirichlet Problem in the convergence set U. It remains to prove that the divergence set V satisfies V ∩ Ω = ∅ . Noticethat in this case, if κ euc ( C i ) ≡ H/y, Lemma 4 does not imply that { u n } isbounded from below in a neighborhood of C i . From Lemma 10, a neighborhoodof C i is either contained in U or in V. If it is contained in U, by taking the neighborhood U − from Proposition 7,bounded by B (cid:48) ∪ C (cid:48) ∪ B (cid:48) ∪ C (cid:48) , with C (cid:48) ⊂ Ω and C (cid:48) ∩ Ω = ∅ we construct abarrier. Since { u n } is uniformly bounded in compact subsets of U, inf C (cid:48) u n isfinite. Let m = min { inf C (cid:48) u n , inf U − ∩ ∂ Ω f i } . Let u − : U − → R be the solution of the Dirichlet Problem in U − taking value m on C (cid:48) ∪ C (cid:48) , then the Maximum Principle implies that u n ≥ u − for all n andthe sequence is uniformly bounded. Proceeding as in the proof of Theorem 6,we conclude that u = f i on C i . If the neighborhood of C i is contained in V, let P be an admissible polygonwhich is a connected component of V ∩ Ω with C i on its boundary. From thedefinition of the flux,2 H I ( P ) = F u n (Σ B i ) + F u n (Σ C i ) + F u n (Σ D i ) , for the arcs B i , C i of ∂ Ω that are in P , and D i the arcs of P contained in theinterior of Ω. Lemma 8 applied to the arcs D i and Lemma 9 to the arcs C i imply lim n → + ∞ F u n (Σ D i ) = (cid:96) − β − Σ (cid:96) euc ( C i ) , n → + ∞ F u n (Σ C i ) = Σ (cid:96) euc ( C i ) . Besides, from Lemma 5, | F u n (Σ B i ) | ≤ β. Then,2 H I ( P ) ≥ − β + Σ (cid:96) euc ( C i ) + ( (cid:96) − β − Σ (cid:96) euc ( C i )) = (cid:96) − β, contradicting inequality (1.4). Thus, { u n } is uniformly bounded in a neighbor-hood of C i and the proof may follow the steps of Proposition 2. Therefore, theset V ∩ Ω must be empty and the barrier argument from Theorem 6 impliesthat u takes the boundary values f i on C i . It remains to show that u diverges to −∞ on the arcs B i . Since { u n } isa monotonically decreasing sequence u n ≤ u ≤ M, it is bounded from abovein Ω . As in Theorem 1, given p ∈ B i , let U be a neighborhood of p for whichTheorem 6 (existence result) applies. Take v k the solution to the Dirichletproblem in U ∩ Ω with boundary data M on ∂U ∩ Ω and − k on ∂ Ω ∩ U. Thenfor n ≥ k, u n ≤ v k by the Maximum Principle. Therefore, u ≤ v k for all k and u diverges to −∞ on U ∩ ∂ Ω . Theorem 11. Let Ω be an admissible domain such that the family { C i } isnon empty. Assume that κ euc ( C i ) ≥ H/y. Then the Dirichlet Problem has asolution in Ω if and only if α < (cid:96) + 2 H I ( P ) and β < (cid:96) − H I ( P ) (12.1) for all admissible P . Proof. Let Ω ∗ be the correspondent domain to Ω . Let u n be the solution in Ω ∗ to the Dirichlet problem with boundary data u n = n on the arcs of the family { A i } , u n = − n on the ones from { B ∗ i } and u n = f n on the arcs of the family { C i } . The function f n coincides with f i if | f i | < n, takes value n if f i > n and − n otherwise.Define also u + : Ω → R as the solution that goes to + ∞ on A i , vanisheson B i and coincides with max { f i , } on C i . It exists from Theorem 1. By theMaximum Principle, u n ≤ u + for all n. Analogously, there is u − : Ω → R thatvanishes on A i and diverges to −∞ on B i which, from the Maximum Principle,is below all u n . Therefore, the sequence { u n } is uniformly bounded in compact subsets ofΩ and the already presented arguments lead us to conclude that the solutionexists.It remains to see that u takes the required boundary data. For the boundaryarcs A i , the construction of the function v k from Theorem 1 with u replaced by u − , implies that u goes to infinity on A i . Also for the arcs B i , the argument fromTheorem 2 with u replaced by u + gives the result. If κ euc ( C i ) ≥ H/y, withstrict inequality except for isolated points, then the sequence ( u n ) is uniformly5bounded in a neighborhood of C i by Lemma 4. If κ euc ( C i ) = 2 H/y, Lemma 4only provides the boundedness from above. To obtain the boundedness frombelow, we proceed as in the proof of Theorem 2 using Proposition 7.The necessity of conditions (12.1) follows as in the previous results.Theorem 3 was stated in the Introduction and here we describe the stepsof its proof. Proof of Theorem 3. Differently from the last result, if one takes a sequence u n of solutions, there is no guarantee that it is bounded in compact subsets ofΩ . That is why this proof is more delicate. Nevertheless the idea used in [9] todemonstrate the existence result for surfaces in R also works in this setting.We briefly describe the construction of two functions that will play the rolesof u + and u − . For each i such that A i is an arc of ∂ Ω , let u + i be defined in Ω ∗ as thesolution to the Dirichlet problem with boundary values u + i = + ∞ on A i andzero on the remaining part of the boundary. The existence of u + i follows fromTheorem 1. Besides, since Ω is an admissible domain, it has a subsolution andtherefore, the Maximum Principle implies the existence of N > u + > − N. If i is such that B i is an arc of ∂ Ω , let u − i be defined in Ω i , the domainbounded by ( ∪ j A j ) ∪ B ∗ i ∪ ( ∪ j (cid:54) = i B j ) as the solution to the Dirichlet problem withboundary values u − i = −∞ on ∪ j (cid:54) = i B j and zero on the remaining part of theboundary. The existence of u − i follows from Theorem 2. For that we observethat any admissible polygon P in Ω i satisfies (1.4). If P is contained in Ω , thisis a consequence of the hypothesis (1.5). If P contains the domain W boundedby B i and B ∗ i , then P = P ∪ W for P the admissible polygon in Ω obtainedremoving W from P . Therefore (1.4) implies that 2 β ( P ) < (cid:96) ( P ) − H I ( P ) . Since from Lemma 6, (cid:96) ( ∂W ) ≥ H I ( P ) , we get2 β ( P ) < (cid:96) ( P ) − H I ( P ) + (cid:96) ( ∂W ) − H I ( W ) . Observing that β ( P ) = β ( P ) + β i and (cid:96) ( P ) = (cid:96) ( P ) + β i − β ∗ i , it follows2 β < (cid:96) ( P ) − H I ( P ) . Set u + : Ω ∗ → R as u + = max i u + i and u − : Ω → R as u − = min i u − i . Let v n be the solution to the Dirichlet problem in Ω ∗ with boundary values n on A i and 0 on B ∗ i . For each n > , for c ∈ (0 , n ) , define the sets E nc = { v n − v > c } and F nc = { v n − v < c } . For c sufficiently close to n, E cn is a union of distinct and disjoint subsetsof Ω ∗ , each of them containing one arc A i . We suppress the dependence on n and denote each component of E cn by E ci . Define µ ( n ) as the infimum of6the constant c < n such that E ci are all distinct and disjoint. Then define u n = v n − µ ( n ) . We claim that for M = N + sup Ω ∗ v , it holds u − − M ≤ u n in Ω and u n ≤ u + + M in Ω ∗ . This claim follows from the Maximum Principle as it didin other ambient spaces, see for instance [18].Consequently, { u n } is uniformly bounded on compact subsets of Ω and,therefore, it has a converging subsequence. Define u as the limit of this se-quence. Once again following the ideas of [18], one proves that µ ( n ) and n − µ ( n ) diverge to + ∞ . Since u n = − µ ( n ) on B ∗ i and u n = n − µ ( n ) on A i , the conclusions in the proofs of Theorems 1 and 2 will prove that u is thesolution to the Dirichlet problem.Conversely, the necessity of conditions (1.5) and (1.6) follows as in Theorem1 and Proposition 2. 13 Admissible domains with empty { C i } We devote this section to prove the existence of domains for which Theorem3 applies. We remark that even in R where we have a 6-dimensional isom-etry group, the construction of these domains requires some effort, as it waspresented in [18].In order to search for domains to apply Theorem 3 we start with the domain U bounded by one loop of a 2 H/y − curve γ, which starts at a P − point, goesdown to the P = (0 , y ) − point and then comes back up to P = (0 , y e TH H ) . From (10.2), we may assume γ is parameterized by γ + ( t ) = ( x ( t ) , y e t/ H ) and γ − ( t ) = ( − x ( t ) , y e t/ H ) , t ∈ [0 , T H ] , x ( t ) given by (10.2).We proceed to construct a family of domains Ω s ⊂ U. Given s ∈ [0 , , let D + ( s ) = γ + ( s ) and D − ( s ) = γ − ( s ) . Let E + ( s ) be the second intersection pointof the vertical line through D + ( s ) with γ + ; and E − ( s ) be the fourth vertexof a rectangle with sides parallel to the axes and vertices D + ( s ) , D − ( s ) and E + ( s ). (See Figure 6).From the definition of γ, one can see that E + ( s ) = ( x ( s ) , y e ϕ ( s ) / H ) , where ϕ : [0 , → [1 , T H ] is a bijection given by (cid:90) ϕ ( s ) − s − g H ( u ) du = 0 . (13.1)The domain Ω s is bounded by the subarc of γ + from D + ( s ) to E + ( s ) andof γ − from D − ( s ) to E − ( s ) , which we name A + and A − , respectively. Theremaining part of the boundary of Ω s consists in two arcs B D and B E . B D isthe unique 2 H/y − arc (oriented downwards) that joins D − ( s ) and D + ( s ) and B E is the unique 2 H/y − arc (oriented upwards) that joins E − ( s ) and E + ( s )and it is not contained in γ. (See Figure 6)For s = 0 , the B − type arcs are empty and Ω = U. For small values of s ,Ω s is a well defined domain and then there is s for which B D ( s ) intercepts B E ( s ) at one point and for s > s , Ω s is no longer a domain.7Figure 6: Construction of Ω s .Let E be the P − point of the arc B E and let (0 , y e e ( s ) / H ) be its coordi-nates. Let also D be the P − point of B D , D = (0 , y e d ( s ) / H ) . Some computa-tions imply that for s ∈ (0 , 1) the functions e and d are given by the followingexpressions: e e ( s ) / H (cid:90) − ϕ ( s ) − e ( s ) − g H ( u ) du = (cid:90) − s − g H ( u ) du (13.2)and e ( d ( s ) − / H (cid:90) − ( d ( s ) − s ) − g H ( u ) du = (cid:90) − s − g H ( u ) du. (13.3)For s ∈ [0 , s ] , define α ( s ) and β ( s ) as the length of the A − type arcs and B − type arcs, respectively, that bound Ω s . As a first step to find a domain for Theorem 3 we have the following fact. Theorem 12. There is s (cid:63) ∈ (0 , s ) such that α ( s (cid:63) ) = β ( s (cid:63) ) + 2 H I (Ω s (cid:63) ) . Proof. Since F ( s ) := α ( s ) − β ( s ) − H I (Ω s ) is a continuous function on s ∈ [0 , s ] , it is sufficient to see that F (0) > F ( s ) < . For s = 0, we are in the same situation as in Lemma 12, which asserts that F (0) > . For s = s , some computations must be done. First notice that s is the solution of d ( s ) = e ( s ) . We denote d ( s ) by d and ϕ ( s ) by ϕ . Besides, as it was already computed in the proof of Lemma 11, the lengthof an arc in a 2 H/y − curve with P = ( z, − point and that goes from height ze t / H to height ze t / H is L ( t , t ) = ze / H H (cid:90) t t g H ( u ) − u du. Therefore α ( s ) < β ( s ) is equivalent to (cid:90) ϕ − s − g H ( u ) − u du < e ( d − H (cid:90) − ( d − s ) g H ( u ) − u du + e d H (cid:90) − ϕ − d − g H ( u ) − u du (cid:90) ϕ − s − e u/ H √ − u du < (cid:90) ϕ − s − e u/ H (cid:112) − (1 − | u − c | ) du for c = d − . A straightforward computation splitting the interval ( s − , ϕ − 1) as( s − , c ] ∪ [ c, ϕ − 1) shows that the above inequality for the integrands holdson ( s − , c ] if d < s and holds on [ c, ϕ − 1) if 2( ϕ − < d .Hence it is enough to show that 2( ϕ − < d < s . Claim 1: The function e satisfies e ( s ) ≥ ϕ ( s ) − 1) for all s ∈ (0 , . Fix s . Consider Q s = (0 , y e ϕ ( s ) − / H ) and take γ Q s the 2 H/y − curvewith Q s as its P − point. The intersections of γ Q s and the horizontal line { y = y e ϕ ( s ) / H } are at a distance 2 x Q s given by2 x Q s = y e / H H e ϕ ( s ) − H (cid:90) − ϕ ( s ) − g H ( u ) du, which is greater than 2 x ( s ) if and only if (cid:90) s − − g H ( u ) du < e ϕ ( s ) − H (cid:90) − ϕ ( s ) − g H ( u ) du, which is a consequence of the positivity of the integrand in the interval ( − , − ϕ ( s )) that contains ( − , s − 1) (from the definition of ϕ (13.1)) and of the factthat e ϕ ( s ) − H > . Consider the function l that associates to each point P on the segment V = { (0 , y ); y ∈ ( y e ( ϕ ( s ) − − T H ) / H , y e ϕ ( s ) / H ) } the distance between the twointersections of γ P with { y = y e ϕ ( s ) / H } . As described in the proof of Proposi-tion 4 (see (10.8)), l is continuous, takes value 0 on the extrema of the segment V and has a unique critical point which is a maximum. Since l ( E ) = l ((0 , y ))and l ( Q s ) > l ( E ) , we have that the point Q s is necessarily between E and(0 , y ). In particular, Q s is below E and the claim follows.Since d = d ( s ) = e ( s ) , the first inequality is proved. Claim 2: The function d satisfies d ( s ) ≤ s for all s ∈ (0 , s ] . For fixed s ∈ (0 , s ] , define F s ( t ) as the left-hand side of expression (13.3),that is, F s ( t ) = e ( t − / H (cid:90) − ( t − s ) − g H ( u ) du, and consider X ( s ) = (cid:82) − s − g H ( u ) du (the right-hand side of (13.3)). It is easyto see that F s ( s ) = 0 < X ( s ) and that F s (2 s ) = e s − / H (cid:90) − s − − ue − u/ H √ − u du > X ( s ) , s ∈ [0 , . Since d is the solution of F s ( d ) = X ( s ) and F s is continuous,then s < d ( s ) < s. Therefore, the second inequality is true and the theorem is proved.We conclude by applying Theorem 3 to Ω s (cid:63) . Theorem 13. The domain Ω s (cid:63) for y small enough is an admissible domainwith { C i } empty that admits a solution to the Dirichlet problem.Proof. Let us denote Ω s (cid:63) by Ω . From the last result, α = β + 2 H I (Ω) . (13.4)We assume y is small enough so that a subsolution exists in Ω . It remains to see that for any admissible polygon P properly contained inΩ , it holds that 2 α < (cid:96) + 2 H I ( P ) and 2 β < (cid:96) − H I ( P ) . We analyse each possibility depending on the number of vertices of P :1. If P has two vertices, three configurations are possible:1.1 Two vertices horizontally aligned: Actually this configuration is notpossible because B and B are the only 2 H/y − arcs in Ω connectingthese points;1.2 Two vertices vertically aligned: The unique possibility inside Ω isto have ∂ P as an A − type arc (either A − or A + ) and its reflectionabout the vertical line that connects its endpoints. For this case, (cid:96) = 2 α < (cid:96) + 2 H I ( P ) and 2 β = 0 < (cid:96) − H I ( P ) from Lemma 12.1.3 The vertices are the ends of a diagonal of R ; say D − and E + . Thefirst inequality is trivial. Let us check the second one.Let us denote by α − and α + the length of the arcs A − and A + , respectively. We know α − = α + . And let us denote by β D and β E the length of the arcs B D and B E , respectively.Let δ be a 2 H/y − arc from D − to E + . If the curvature vector of δ points to the left (notice it points in only one direction becausesince δ is contained in ¯Ω it cannot have neither a P − point nora P − point), the Maximum Principle implies that δ is outside thedomain U − bounded by A − and its reflection about the vertical linethrough its endpoints. Consider the curve obtained joining δ to thehorizontal segment h from E + to E − . This curve connects D − to E − and is outside the convex domain U − . Therefore its length isgreater than the length of its projection on U − , which is exactly thereflection of A − . Therefore, (cid:96) ( δ )+ (cid:96) ( h ) > α − and then (cid:96) ( δ )+ β i > α − i = D and E. The same argument applies if δ has curvaturevector pointing to the right, with A + replacing A − . Since β = 0 and2 H I ( P ) < H I (Ω) = α + + α − − β D − β E < δ + δ = (cid:96), the second inequality follows.2. If P has 3 vertices, two configurations are possible:2.1 P consists in one A − type arc, say A − , one B − type arc, say B E , and one interior arc δ as described above.Here α = α − and, as shown above, 2 α < ( α − + β E + δ ) ≤ (cid:96) +2 H I ( P ) . Besides, β = β E and the inequality 2 β < (cid:96) − H I ( P ) is equivalentto β E + 2 H I ( P ) < α − + δ, which from (13.4) is equivalent to2 α − − β D − H I (Ω) + 2 H I ( P ) < α − + δ, which is implied by the facts that α − − β D < δ and 2 H I ( P ) < H I (Ω) . 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Departamento de Matem´aticaUniversidade Federal de Santa MariaAv. Roraima 1000, Santa Maria RS, 97105-900, Brazil Email: [email protected] Mathematics DepartmentPrinceton UniversityFine Hall, Washington Road, Princeton NJ, 08544, USA