The Jenkins-Serrin problem for constant mean curvature graphs in the Heisenberg space Ni l 3 (τ)
TTHE JENKINS-SERRIN PROBLEM FOR CONSTANT MEANCURVATURE GRAPHS IN THE HEISENBERG SPACE Nil ( τ ) CARLOS PEÑAFIELA bstract . In this paper we find functions over bounded domains inthe 2-dimensional Euclidean space, whose graphs (in the Heisenbergspace) has constant mean curvature different from zero and taking on(possibly) infinite boundary values over the boundary of the domain.
1. I ntroduction
The classical Dirichlet problem for the constant mean curvature andminimal surfaces (H-surfaces and minimal surfaces from now on) in the3-dimensional Euclidean space R , consists in the determination of a func-tion u = u ( x , y ) satisfying the partial differential equation(1.1) ( + q ) r − pqs + ( + p ) t = H ( + p + q ) in a fixed domain Ω in the ( x , y ) − plane, and taking on assigned continu-ous value on the boundary of Ω .In order to introduce the Jenkins-Sering theory, we divide the equation(1.1) in two cases. • For the case of minimal surfaces in R , the Dirichlet problem wassolved by Radô in 1930, for convex domains, see [16]. Radô basedhis proof on the existence theorem for the parametric problem ofleast area. In [10] Jenkins and Serrin gave an alternate proof en-tirely avoiding reference to the parametric problem.On the other hand, it was well known that in 1934 H. F. Scherkdiscovered his famous minimal surface which is a graph of a func-tion defined over a square and taking on infinite boundary data.More precisely, Scherk find out the surface given by z = log cos x − log cos y , | x | < π /2, | y | < π /2.The graph of this function is a minimal surface in R , the functiontakes on plus infinite and minus infinite boundary data on alter-nates sides of the boundary of the square. We call this surface, theScherk example. This Scherk example can be seem as a solution ofthe Dirichlet problem for the minimal curvature equation, takingon infinite boundary values over the boundary of the square. Mathematics Subject Classification.
Primary 53C42; Secondary 53C30. a r X i v : . [ m a t h . DG ] M a r CARLOS PEÑAFIEL
In [10] Jenkins and Serrin developed an existence and unique-ness theory applicable to situation in which continuity of the data isset aside and which infinite boundary values are allowed on entirearcs of the boundary. Fundamental for they work was the existenceof a solution of the Dirichlet problem in a convex domain with as-signed boundary data, notice that this work generalize the Scherkexample. From now on, we call this kind of surfaces, Jenkins-Serrinsurfaces and the theory development, the Jenkins-Serrin theory.The Jenkins-Serrin theory was extend to H × R (here H label the2-dimensional hyperbolic space) by Nelli and Rosenberg in [13],and to M × R , where M is an arbitrary Riemannian surface, byPinheiro in [15]. And for non-compact domains, the Jenkins-Serrintheory was treated by Mazet L., Rodríguez, M. and Rosenberg, H.in [12].Different of the 3-dimensional product spaces H × R and S × R (here S label the 2-dimensional Euclidean sphere), which are ho-mogeneous and simply connected, having 4-dimensional isometrygroups, we have the Heisenberg space Nil ( τ ) and the universalcover of the space PSL ( R ) (here PSl ( R ) denote the preserving-orientation isometries of the hyperbolic space H ) which we denoteby (cid:103) PSL ( R , τ ) . For more details, see for instance [19], [4].The Jenkins-Serrin theory for compact domains in Nil ( τ ) wastreated by Ana Lucia Pinheiro in [14] and for the (cid:103) PSL ( R , τ ) (in the τ = − • On the other hand, for the case H (cid:54) =
0, the Jenkins-Serrin theoryfor bounded domains was extended to R by Spruck in [18], for H × R and S × R by Hauswirth, Rosenberg and Spruck in [9].For unbounded domains, the Jenkins-Serrin theory was extendedto H × R by Folha and Mello in [6], and in the case M × R (here M label a Hadamard surface) by Folha and Rosenberg in [8].If we consider complete, simply connected, 3-dimensional homogeneousmanifolds having 4-dimensional isometry groups, it result open the case H (cid:54) = ( τ ) and the (cid:103) PSl ( R , τ ) space. As we have remarked, the fundamental fact in theJenkins-Serrin theory, was the existence of a solution of the Dirichlet prob-lem with prescribed continuous boundary data, on a convex domain.Since that in [2] Dajczer and Lira solve this Dirichlet problem for gen-eral 3-manifolds carrying on a non-singular Killing field, we use the ex-istence of such solution to extend the Jenkins-Serrin Theory to Nil ( τ ) in the case H (cid:54) = (cid:103) PSL ( R , τ ) space is treated in [7] and for a Killing Submersion, see [1]. ENKINS-SERRIN PROBLEM 3
The paper is organized as follow, in Section 2 we give the details of theHeisenberg space Nil ( τ ) seem as a Riamannian submersion over the 2-dimensional Euclidean space R . We prove a Maximum principle and wecite the existence theorem of the Dirichlet problem. In Section 3, we dealwith H -sections, and establishes the Jenkins-Serrin problem as well as westudy the properties of the flux of the sequences of solutions. Finally inSection 4, we prove the main theorems.2. P reliminares We denote by Nil ( τ ) the 3-dimensional Lie group endowed with a leftinvariant metric g . For each τ (cid:54) =
0, Nil ( τ ) is a homogeneous simplyconnected Riemannian manifold.In Euclidean coordinates Nil ( τ ) = ( R , g ) , where R label the 3-dimensionalEuclidean space and Nil ( τ ) is endowed with the metric(2.1) g = dx + dy + ( τ ( ydx − xdy ) + dz ) The Lie group Nil ( τ ) is one of the eight Thurston’s geometries (see [19]),and it is well known that there exists a Killing submersion (see [17]) π : Nil ( τ ) −→ R (2.2) ( x , y , z ) (cid:55)−→ ( x , y ) from Nil ( τ ) into the Eucliedan 2-dimensional space R . That is π is aRiemannian submersion, the bundle curvature is τ and the unit vectorfield along the fibers is a Killing vector field. Therefore translations alongthe fibers are isometries. We denote this Killing vector field by ξ .We call a vector field Z ∈ χ ( Nil ( τ )) vertical if it is a non-zero multipleof ξ and horizontal if g ( Z , ξ ) = ( τ ) , we consider thecanonical frame { e = ∂ x , e = ∂ y } of R and consider the horizontal lift E and E of e and e respectively. Then the canonical orthonormal frame { E , E , ξ } of Nil ( τ ) is given by E = ∂ x − τ y ∂ z , E = ∂ y + τ x ∂ z , ξ = ∂ z .Denote by ∇ the Riemannian connection of Nil ( τ ) , so ∇ E E = ∇ E E = τ E ∇ E E = − τ E ∇ E E = − τ E ∇ E E = ∇ E E = τ E ∇ E E = − τ E ∇ E E = τ E ∇ E E = [ E , E ] = τ E , [ E , E ] = = [ E , E ] for more details see [3], [5], [4]. CARLOS PEÑAFIEL
The isometry group of Nil ( τ ) has dimension 4, the isometries of Nil ( τ ) are the translations generated by the Killing vector fields F = ∂ x + τ y ∂ z , F = ∂ y − τ x ∂ z , F = ∂ z and the rotations about the z -axis corresponding to F = − y ∂ x + x ∂ y The translations corresponding to F and F are respectively ( x , y , z ) (cid:55)−→ ( x + t , y , z + τ ty ) and ( x , y , x ) (cid:55)−→ ( x , y + t , z − τ tx ) where t ∈ R .2.1. The mean curvature equation and the maximum principle.
We iden-tify the space R with its lift R × { } ⊂ Nil ( τ ) and for a given domain Ω ⊂ R , we also denote by Ω its lift to R × { } .To a function u ∈ C ( Ω ) on Ω , we define the graph of u , denoted by Σ ( u ) , as being the set(2.3) Σ ( u ) = { ( x , y , u ( x , y )) ∈ Nil ( τ ) ; ( x , y ) ∈ Ω } Throughout this paper the surface Σ ( u ) will have mean curvature H > Σ ( u ) .In order to obtain the normal vector N of Σ ( u ) , consider the function u ∗ : Nil ( τ ) −→ R ( x , y , z ) (cid:55)−→ u ∗ ( x , y , z ) = u ( x , y ) and set F ( x , y , z ) = z − u ∗ ( x , y , z ) . Therefore, Σ ( u ) = F − ( ) , that is Σ ( u ) is the level surface of F . It is well known that the function H satisfies(2.4) div Nil ( τ ) (cid:32) ∇ g F |∇ g F | (cid:33) = − H where div Nil ( τ ) and ∇ g denote the divergence and gradient in Nil ( τ ) .Thus, the upward pointing normal N is given by N = ∇ g F |∇ g F | .A straightforward computation shows ∇ g F = − ( u x + τ y ) E − ( u y − τ x ) E + E hence W : = |∇ g F | = + ( u x + τ y ) + ( u y − τ x ) | .Using the Riemannian submersion, the function u satisfies the equation( the mean curvature equation )(2.5) div R (cid:18) α W ∂ x + β W ∂ y (cid:19) = H ENKINS-SERRIN PROBLEM 5 where(2.6) α = τ y + u x , β = − τ x + u y , and W = + α + β .Thus, the surface Σ ( u ) has mean curvature function H if and only if u is asolution of the following PDE(2.7) L H ( u ) : = W [( + β ) u xx + ( + α ) u yy − αβ u xy ] − H = α , β and W from (2.6).Notice that N = W ( − α E − β E + E ) = N h + N v where N h = horizontal part of N = − G ( u ) W = α E + β E W and N v = vertical part of N = E W .Consider two functions u , v : Ω −→ R ,the upwards pointing normal N and N of Σ ( u ) and Σ ( u ) are respec-tively(2.8) N = − G ( u ) W + W E N = − G ( u ) W + W E where W i = W ( u i ) , i =
1, 2.Notice that (cid:104) W i N i , E (cid:105) = (cid:104) G ( u ) − G ( u ) , G ( u ) W − G ( u ) W (cid:105) = ( W + W )( − (cid:104) N , N (cid:105) )= W + W | N − N | ≥ G ( u i ) = − ( u ix + τ y ) E − ( u iy − τ x ) E Setting (cid:101) X i = − π ∗ ( G ( u i )) = α∂ x + β∂ y = ( u ix + τ y ) ∂ x + ( u iy − τ x ) ∂ y and X u i = (cid:101) X i W i we conclude that X u − X u = ∇ u − ∇ u ,where ∇ denotes the gradient in the Euclidean space R .Using the Riemannian submersion (2.9) becomes(2.10) (cid:104)∇ u − ∇ u , X u − X u (cid:105) R ≥ CARLOS PEÑAFIEL
Thus, we have proved the next lemma.
Lemma 2.1.
Let u and u be functions in C ( Ω ) , Ω ⊂ R and set W i = W ( u i ) ,i =
1, 2 . Then (2.11) (cid:104)∇ u − ∇ u , X u − X u (cid:105) ≥ with equality at a point if and only if ∇ u = ∇ u . Following the above notation, the mean curvature function H of thesurface Σ ( u ) satisfy the equation(2.12) div R ( X u ) = H where X u = − π ∗ (cid:18) G ( u ) W (cid:19) .For this situation, we prove the following maximum principle. Theorem 2.2. (Maximum principle) Consider Ω ⊂ R a bounded domain. Letu , v ∈ C ( Ω ) be two functions whose graphs Σ ( u ) and Σ ( v ) have the sameprescribed mean curvature H. Let E ⊂ ∂ Ω be a finite set of points such that ∂ Ω − E consists of smooth arcs, and suppose that u and v extend continuously toeach smooth arc of ∂ Ω − E. If u ≥ v on ∂ Ω , then u ≥ v on Ω .Proof. Consider the set D = { x ∈ Ω , u ( x ) − v ( x ) < } . We can translatethe surfaces Σ ( u ) and Σ ( v ) in the ∂ z direction to assume that u < v on Ω − E , and by contradiction, we are supposing that D is not empty.The boundary of D consists of proper curves in Ω which goes to pointsof E . We can also assume that those curves are regular, ie. ∇ ( u − v ) isnonzero on ∂ D . Denote by (cid:98) D a connected component of D and let (cid:98) D (cid:101) ⊂ (cid:98) D be the domain such that ∂ (cid:98) D (cid:101) is the set of points in ∂ (cid:98) D whose distancefrom, E is greater from (cid:101) >
0, for (cid:101) small enough, together with the union ∪ C (cid:101) of circular arcs contained in (cid:98) D which are part of circles centered atpoints of ∂ (cid:98) D ∩ E , having radius (cid:101) .From the mean curvature equation (2.12), we hace (cid:90) D (cid:101) div R ( X u − X v ) = = (cid:90) (cid:98) D (cid:101) div R ( X u − X v ) = (cid:90) ∂ (cid:98) D (cid:101) (cid:104) X u − X v , η (cid:105) where η is the outward unit conormal to ∂ (cid:98) D (cid:101) .From Lemma 2.1(2.14) (cid:104) X u − X v , ∇ u − ∇ v (cid:105) ≥ ∇ ( u − v ) (cid:54) = u − v ≡ ∂ (cid:98) D (cid:101) − ∪ C (cid:101) , and u − v < (cid:98) D (cid:101) , thevector ∇ ( u − v ) is a positive multiple of η on ∂ (cid:98) D (cid:101) − ∪ C (cid:101) . Therefore from(2.14) (cid:90) ∂ (cid:98) D (cid:101) −∪ C (cid:101) (cid:104) X u − X v , η (cid:105) R ≥ δ > ENKINS-SERRIN PROBLEM 7
On the other hand, on ∪ C (cid:101) , we have (cid:90) ∪ C (cid:101) (cid:104) X u − X v , η (cid:105) ≤ (cid:90) ∪ C (cid:101) |(cid:104) X u − X v , η (cid:105)| ≤ × length ( ∪ C (cid:101) ) thence, making (cid:101) tend to zero, we conclude that (cid:90) (cid:101) (cid:104) X u − X v , η (cid:105) > D is empty and u ≤ v on Ω . (cid:3) The Dirichlet problem.
In [2] Dajczer and Lira have proved the ex-istence and uniqueness of Killing graphs with prescribed mean curvaturein Killing submersions. In our context, the surfaces Σ ( u ) ⊂ Nil ( τ ) whichare the graph of the function u : Ω ⊂ R −→ R are Killing graphs, since they are transverse to the vertical Killing field ξ .From now on, the surfaces Σ ( u ) will be called Killing graphs.For a domain Ω ⊂ R with boundary Γ = ∂ Ω , we denote the corre-spondent Killing cylinders by M = π − ( Ω ) and K = π − ( Γ ) . We denoteby H cyl the inward mean curvature of K and by Ric
Nil ( τ ) the Ricci tensorof Nil ( τ ) . Then it has showed in [2, Theorem 1] the following theorem. Theorem 2.3. (The Dirichelt problem) Let Ω ⊂ R be a domain with compactclosure and C α boundary. Suppose H cyl > and inf Nil ( τ ) Ric
Nil ( τ ) ≥ − Γ H cyl . Let H ∈ C α ( Ω ) and φ ∈ C α ( Γ ) be given functions. If sup Ω | H | ≤ inf Γ H cyl . Then, there exists a unique function u ∈ C α ( Ω ) satisfying u | Γ = φ whoseKilling graph Σ ( u ) has mean curvature H. In this paper the ambient Ricci tensor in the v -direction is defined by Ric
Nil ( τ ) ( v ) = ∑ i = (cid:104) R ( w i , v ) v , w i (cid:105) where R is the curvature tensor and { w , w , v } is an orthonormal basis. Remark . Fix a point p ∈ Nil ( τ ) and take a unit vector v ∈ T p Nil ( τ ) .Let Π be the plane orthogonal to the vector v . After a rotation around thevertical fiber passing by p , we can suppose that Π = [ E , aE + bE ] , a + b = { E , aE + bE } is a orthonormal basis for Π and thence v = − bE + aE . CARLOS PEÑAFIEL
Using the Riemannian connection ∇ , we have − τ ≤ Ric
Nil ( τ ) ( v ) ≤ τ .Denote by Γ ⊂ R a piecewise C smooth simple Jordan curve. Takingthe parametrization c : [ a , b ] −→ Γ s (cid:55)−→ c ( s ) ∈ Γ consider the geodesic curvature k ( s ) of Γ at the point c ( s ) ∈ Γ , it is wellknown that when the fiber of the killing submersion are geodesic, then themean curvature H cyl of the vertical cylinder π − ( Γ ) is given by H cyl = k ( s ) τ must be satisfy(2.15) τ ≤ inf s (cid:18) k ( s ) (cid:19) With this in mind, the Theorem 2.3 can be write in the next form.
Theorem 2.5. (The Dirichelt problem) Let Ω ⊂ R be a domain with compactclosure and C α boundary. Suppose H cyl > and H ≥ | τ | for a constant H. Let φ ∈ C α ( Γ ) be given functions. If H ≤ k ( s ) , where k ( s ) denotes the geodesic curvature of Γ . Then, there exists a unique func-tion u ∈ C α ( Ω ) satisfying u | Γ = φ whose Killing graph Σ ( u ) has mean curva-ture H. Once there, using Theorem 2.5 we can prove an important theorem forsequences of solutions of the mean curvature equation.
Theorem 2.6. (Compactness theorem) Let { u n } be a uniformly bounded se-quences of solutions of the constant mean curvature equation (2.12) in a boundeddomain Ω . Then there exists a subsequence which converges uniformly on com-pact subsets to a solution of (2.12) in Ω .Proof. From Theorem 2.5, we obtain an interior estimate for the first andsecond derivative as well as to the solutions of the equation (2.12). There-fore on compact subdomains we have the equicontinuity of the secondderivatives. Consequently by Arzela-Ascoli’s Theorem, we obtain a sub-sequence which converges uniformly on compact subsets to a solution of(2.12) in Ω . (cid:3) ENKINS-SERRIN PROBLEM 9 H - sections A section is the image of a map u : Ω ⊂ R → Nil ( τ ) ,such that π ◦ u = Id | Ω , where Id label the identity map restricted to thedomain Ω and π is the canonical projection from Nil ( τ ) over R . Thus,identifying the domain Ω with its lift to R × { } , we identify the section u with the Killing graph Σ ( u ) and by simplicity we denote this surface by Σ u .In this section we are going to consider sections u : Ω ⊂ R → Nil ( τ ) whose Killing graph Σ u has constant mean curvature (CMC-graphs) H .We call such Killing graphs, H -sections or H -surfaces. For a piece of curve γ ⊂ B , we denote by k ( γ ) its geodesic curvature. It has showed in [17,Theorem 3.3] the following theorem. Theorem 3.1.
Let π : Nil ( τ ) → R the Killing submersion and letu : Ω → Nil ( τ ) be an H-section over a domain Ω ⊂ R . Let U be a neighbourhood of an arc γ ⊂ ∂ Ω and ι : U → Nil ( τ ) a section.Assume that for any sequence ( x n ) of Ω which converge to a point x ∈ γ , theheight of u ( x n ) goes to + ∞ , that is, u ( x n ) − ι ( x n ) → + ∞ , then γ is a smoothcurve with k ( γ ) = H. If H > , then γ is convex with respect to Ω if andonly if, the mean curvature vector −→ H of Σ u points up along Σ u . Moreover, Σ u converges to the vertical H-cylinder π − ( γ ) with respect to the C k -topology, forany k ∈ N . As a consequence of this theorem, we prove the following lemma.
Lemma 3.2.
Let u be a solution of (2.12) in a domain Ω bounded in part by anarc γ and suppose that m ≤ u ≤ M on γ . Then, there is a constant c = c ( Ω ) such that for any compact C sub-arc γ (cid:48) ⊂ γ , (i) If k ( γ (cid:48) ) ≥ H, with strict inequality except for isolated points, there is aneighbourhood V of γ (cid:48) in Ω such thatu ≥ m − cin V. (ii) If k ( γ (cid:48) ) > − H, there is a neighbourhood V of γ (cid:48) in Ω such thatu ≤ M + cin V.Proof. First suppose that k > H on some sub-arc γ (cid:48) ⊂ γ , let p be themiddle point of γ (cid:48) and consider the curve γ (cid:48) tangent to γ (cid:48) at the point p having 2 H < k ( γ (cid:48) ) < k (with respect to the interior normal to Ω ), noticethat γ (cid:48) is outside of Ω . Let Γ be the curve joining the endpoints o γ (cid:48) having curvature k ( γ (cid:48) ) with respect to the outward pointing unit normal vector to Ω . Finallyconsider the sub-arc Γ (cid:48) ⊂ Γ which lies in Ω . The sub-arc Γ (cid:48) intersects thearc γ (cid:48) in two points, these two points determine a segment of γ (cid:48) , which wewill call γ (cid:48) again. Now consider the domain ∆ ⊂ Ω bounded by Γ (cid:48) and γ (cid:48) ,where k ( γ (cid:48) ) > H and k ( Γ (cid:48) ) > H wit respect to the interior unit normalof ∆ . Claim 3.3.
The function u = u | Γ (cid:48) satisfies | u | ≤ C, for some constant C.
Proof of the Claim.
Denote by Σ u the graph of u over ∆ . As Σ u is trans-verse to the Killing field ξ , then Σ u is stable, therefore we have curvatureestimates.Now suppose that there is a sequence of points ( p n ) in ∆ such that u ( p n ) goes to − ∞ . Then the tangent planes T u ( p n ) Σ u becomes vertical when n goes to + ∞ . Using the curvature estimates, we obtain a small fixed δ >
0, such that Σ u is locally the graph of a function over a ball of radius δ , centered at the origin of T u ( p n ) Σ u . From Theorem 3.1, Γ (cid:48) must havecurvature k ( Γ (cid:48) ) = − H which contradicts our assumption, analogously inthe case u ( p n ) goes to + ∞ , we get a contradiction, proving the claim.It was shown in [2], that the Dirichlet problem (cid:26) L ( u ) = H , in ∆ u | Γ (cid:48) = C , u | γ (cid:48) = m has a sub-solution, so there is a constant c = c ( Ω ) such that u ≥ m − c .On the other hand, for the case ( ii ) , suppose there exists a decreasingsequence { V n } of neighbourhood of γ , that is V n ⊃ V n + , such that foreach n , there exists a point p n ∈ V n with u ( p n ) = u n > n . Notice that themean curvature vector is pointing upwards and the graph of the function u has constant mean curvature H . Using curvature estimates, we obtaina graph Σ u n over the tangent plane T u ( p n ) Σ ( u ) , so from Theorem 3.1, Σ u n converges to the vertical cylinder π − ( γ ) , thus γ must have constant geo-desic curvature − H , which is a contradiction. This completes the proofof the lemma. (cid:3) The Jenkins-Serrin problem.
We are going assume that H > Σ u . Then if u tends to + ∞ for any approach to a boundary arc γ , necessarily the cur-vature k ( γ ) = H is constant, while if u tends to − ∞ on γ , k ( γ ) = − H .Thus we must deal with non-convex domains Ω with ∂ Ω piecewise C and consists of three set of open arcs { A i } , { B i } and { C i } satisfying k ( A i ) = H , k ( B i ) = − H and k ( C i ) ≥ H respectively. The Jenkins-Serrinproblem consist in to find a solution of (2.12) in Ω taking on the boundary ENKINS-SERRIN PROBLEM 11 values + ∞ on A i , − ∞ on the B i and arbitrary continuous boundary dataon the C i .In order to give a precise announcement of the Jenkins-Serrin problem,we consider the following definitions. Definition 3.4. (Admissible domain) We say that a bounded domain Ω isadmissible if it is simply connected and its ∂ Ω is piecewise C and consistsof three set of C open arcs { A i } , { B i } and { C i } satisfying k ( A i ) = H , k ( B i ) = − H and k ( C i ) ≥ H respectively (with respect to the interior of Ω ). We suppose that no two of the arcs A i and no two of the arcs B i havea common endpoint.The Jenkins-Serrin problem is defined in the next definition. Definition 3.5. (The Jenkins-Serrin problem) Given an admissible domain Ω , the Jenkins-Serrin problem is to find a solution of (2.12) in Ω whichassumes the value + ∞ on each A i , − ∞ on each B i and assigned continuousdata on each of the open arcs C i . Note that the continuous data are allowedto become unbounded at the endpoints. Remark . A solution of the Jenkins-Serrin problem is called a Jenkins-Serrin solution.
Definition 3.7.
Let Ω be an admissible domain. We say that P is an admis-sible polygon if P is a simply domain contained in Ω with ∂ P piecewisesmooth consisting of arcs of constant curvature k = ± H with verticeschosen from among the endpoints of the families { A i } , { B i } and { C i } .For an admissible P , let α ( P ) and β ( P ) be the total length of the arcs ∂ P belonging to { A i } and { B i } respectively. Finally, let l ( P ) be the perimeterof P and A ( P ) be the area of P .3.2. Flux formulas.
In this section, we deal with flux formulas for H -sections which are the crucial tool to obtain the Jenkins-Serrin solutionof the Jenkins-Serrin problem.We have denoted by X u the negative of the projection via π of the hori-zontal part of the normal vector N .Let u ∈ C ( Ω ) ∩ C ( Ω ) be a solution of (2.12) in a domain Ω . Then, inte-grating (2.12) over Ω gives(3.1) 2 H A ( Ω ) = (cid:90) ∂ Ω (cid:104) X u , ν (cid:105) ds where A ( Ω ) denotes the area of Ω and ν is the outer conormal to ∂ Ω . Theright hand integral in (3.1) is called the f lux of u across Ω . Let γ be asubarc of ∂ Ω (homeomorphic to [
0, 1 ] ). Even if u is not differentiable on γ we can define the flux of u across γ as follows. Definition 3.8.
Choose ζ to be a simple smooth curve in Ω so that γ ∪ ζ bounds a simple connected domain ∆ ζ . We then define the flux of u across γ to be(3.2) F u ( γ ) = H A ( ∆ ζ ) − (cid:90) ζ (cid:104) X u , ν (cid:105) ds .Notice that the integral in (3.2) is well defined as an improper integral.To see that, this definition is independent of ζ , let ζ (cid:48) be another choice ofcurve and consider the 2-chain R with oriented boundary ζ (cid:48) − ζ . By theDivergence Theorem and equation (2.12) we have2 H A ( ∆ ζ (cid:48) ) − H A ( ∆ ζ ) = (cid:90) ζ (cid:48) (cid:104) X u , ν (cid:105) ds − (cid:90) ζ (cid:104) X u , ν (cid:105) ds .Therefore, the definition makes sense. Thus, if u ∈ C ( Ω ∪ γ ) , then F u ( γ ) = (cid:90) γ (cid:104) X u , ν (cid:105) ds .Thus, we obtain the following lemma. Lemma 3.9.
Let u be a solution of (2.12) in a domain Ω and let ζ be a piecewiseC curve in Ω . Then H A ( Ω ) = (cid:90) ∂ Ω (cid:104) X u , ν (cid:105) ds and | (cid:90) ζ (cid:104) X u , ν (cid:105) ds | ≤ | ζ | .Now we will prove some interesting lemmas. Lemma 3.10.
Let Ω be a domain bounded in part by a piecewise C arc ζ satis-fying k ( ζ ) ≥ H. Let u be a solution of (2.12) in Ω which is continuous in ζ .Then (3.3) | (cid:90) ζ (cid:104) X u , ν (cid:105) ds | < | ζ | . Proof.
It is suffices to prove (3.3) for a small subarc γ of ζ . To this end let p ∈ ζ and let Ω (cid:101) = Ω ∩ B (cid:101) ( p ) . Then by the Theorem 2.5, there is a solution v of (2.12) in Ω (cid:101) with v = u + γ and v = u on the remainder of theboundary. Set w = v − u , then by Lemma 2.10 < (cid:90) Ω (cid:101) (cid:104)∇ w , X v − X u (cid:105) dv = (cid:90) γ (cid:104) X v − X u , ν (cid:105) ds .Thence F u ( γ ) < F v ( γ ) ≤ | γ | . (cid:3) Lemma 3.11.
Let Ω be a domain bounded in part by an arc γ and let u be asolution of (2.12) in Ω . Then, (i) if u tends to + ∞ on γ , we have (cid:82) γ (cid:104) X u , ν (cid:105) ds = | γ | , (ii) if u tends to − ∞ on γ , we have (cid:82) γ (cid:104) X u , ν (cid:105) ds = −| γ | .Proof. Suppose u → + ∞ on γ . Notice that the upwards normal vector N on the surface Σ u is becoming horizontal when we approach of γ . Then atpoints sufficiently near to γ , we have (cid:104) X u , ν (cid:105) ≥ − (cid:101) , ENKINS-SERRIN PROBLEM 13 where ν is the outer conormal to γ ⊂ ∂ Ω , and (cid:101) > (cid:90) γ (cid:104) X u , ν (cid:105) ds ≥ (cid:90) γ ( − (cid:101) ) ds ,which implies (cid:90) γ (cid:104) X u , ν (cid:105) ds ≥ | γ | .On the other hand, if u → − ∞ on γ , we have(3.4) (cid:104) X u , ν (cid:105) ≤ − + (cid:101) ,at points sufficiently near γ , and for (cid:101) > (cid:104) X u , ν (cid:105) <
0, Lemma (3.9) implies(3.5) F u ( γ ) ≥ −| γ | .Now from (3.4) (cid:90) γ (cid:104) X u , ν (cid:105) ds ≤ (cid:90) γ ( − + (cid:101) ) ds for all (cid:101) >
0, small enough. Thence(3.6) F u ( γ ) ≤ −| γ | .We conclude from (3.5) and (3.6) that F u ( γ ) = −| γ | . (cid:3) The following lemma is a simple extension of Lemma 3.11.
Lemma 3.12.
Let Ω be a domain bounded in part by an arc γ and let { u n } be asequence of solutions of (2.12) in Ω with each u n continuous on γ . Then (i) if the sequence tends to + ∞ con compact subsets of γ while remaininguniformly bounded on compact subsets of Ω , we have lim n −→ + ∞ (cid:90) γ (cid:104) X u n , ν (cid:105) ds = | γ | (ii) if the sequence tends to − ∞ con compact subsets of γ while remaininguniformly bounded on compact subsets of Ω , we have lim n −→ + ∞ (cid:90) γ (cid:104) X u n , ν (cid:105) ds = −| γ | We also need the next lemma.
Lemma 3.13.
Let Ω be a domain bounded in part by an arc γ with k ( γ ) = Hand let { u n } be a sequence of solutions of (2.12) in Ω with each u n continuous on γ . Then if the sequence diverges to − ∞ uniformly on compact subsets of Ω whileremaining uniformly bounded on compact subsets of γ , we have lim n −→ + ∞ (cid:90) γ (cid:104) X u , ν (cid:105) ds = | γ | . Proof.
Note that the sequence { X u n = − π ∗ ( N hu n ) } converges uniformly tothe outer normal ν on compact subsets of Ω . This implies(3.7) lim n −→ + ∞ (cid:104) X u n , ν (cid:105) = (cid:3) Divergence lines.
Now we focus our attention in the study of con-vergence of H -sections { u n } over a domain Ω ⊂ R .Remember that, we have denoted by Gu m the gradient in Nil ( τ ) of thefunction u n . We beginning this section with the next lemma. Lemma 3.14.
Let p ∈ Ω and suppose that | Gu n ( p ) | is uniformly bounded.Then, there exists a subsequence of { v n = u n − u n ( p ) } converging uniformly toa solution of (2.12) in a neighbourhood of p in Ω .Proof. Notice that the surfaces Σ v n are simply the translation of the surfaces Σ u n along the Killing field ξ . So N v n ( q ) = N u n ( q ) and Gv n ( q ) = Gu n ( q ) for all q ∈ Ω .As Σ v n is stable, since the sections are transverse to the Killing field ξ ,curvature estimates guarantee the existence of disks D δ n ( p ) ⊂ T p Σ v n withsmall positive radius δ , independent of n (it depends only on the distanceof p to ∂ Ω ), where each Σ v n is a local graph (denoted by Σ ( v n , δ ) ) over D δ n ( p ) , having bounded geometry.As | Gv n ( p ) | is bounded and taking into account N = − Gv n W n + ξ ,there exists a subsequence of N v n ( p ) which converges to a non-horizontalvector and thus, the tangent planes associated to this subsequence con-verge to a non-vertical plane Π ( p ) .Notice that the sequence of graphs { Σ ( v n , p ) } have height and slopeuniformly bounded, thence there is a subsequence which converges uni-formly to a H -graph over the disk D δ ( p ) ⊂ Π ( p ) . Moreover, since Π ( p ) isnot vertical, there exists a geodesic ball B ( p , δ ) ⊂ Ω centered at p of radius δ with 0 < δ ≤ δ , such that, the H -graph is a H -section. We conclude thatthere exists a neighbourhood of p in Ω where a subsequence of the { v n } converges to a solution of equation (2.12). (cid:3) The Lemma 3.14 motives the following definition.
Definition 3.15.
We say that the set C = { p ∈ Ω ; | Gu n ( p ) | is bounded } is the convergence set of the sequence { u n } , and D = Ω − C is the diver-gence set of { u n } .Notice that, from Lemma 3.14, the convergence set C is an open subsetof Ω . ENKINS-SERRIN PROBLEM 15
Lemma 3.16.
Let C (cid:48) be a connected bounded component of C . Then for any p ∈ C (cid:48) , there exists a subsequence { v n (cid:48) = u n (cid:48) − u n (cid:48) ( p ) } which converges uniformlyon compact subsets of C (cid:48) to a solution of (2.12) over C (cid:48) .Proof. Taking a countable dense set { p n } in C (cid:48) . According to the Lemma3.14, each point p m has a neighbourhood where a subsequence of { v n } converges to an H -section. This convergence is uniform on compact sets ofthe neighbourhood. By a diagonal process, we constructed a subsequenceof { v n } which converges uniformly on compact subsets of C (cid:48) to an H -section, this happens for any p ∈ C (cid:48) . (cid:3) Now, we study the divergent set D , where the sequence {| Gu n ( p ) |} diverges. If p ∈ D , | Gv n ( p ) | is unbounded and we can consider a subse-quence { v n (cid:48) } of { v n = u n − u n ( p ) } such that | Gv n (cid:48) ( p ) | → + ∞ . Therefore { N v n (cid:48) ( p ) } converges to a horizontal vector N ( p ) .We are going consider the identification. ι ( Ω ) ≈ Ω Lemma 3.17.
Let p ∈ Ω and { u n } be a sequence of solutions of (2.12) in thedomain Ω . (i) If p ∈ C , there is a subsequence of { v n } where v n = u n − u n ( p ) , con-verging uniformly in a neighbourhood of p ∈ Ω . (ii) If p ∈ D , there is a compact arc L p ( (cid:101) δ ) of curvature H, containing p suchthat after passing to a subsequence, { N v n ( p ) } converges to a horizontalvector whose projection via π is orthogonal to L p ( (cid:101) δ ) , having the samedirection as the curvature vector to the graph of v n at ι ( p ) ≈ p.Proof. The first part of the lemma was proved in Lemma 3.16. We denoteby Σ v n the graph of the function v n over Ω . Note that N v n ( q ) = N u n ( q ) and the convergence and divergence set are the same for { u n } and { v n } .Since H -sections are stables, they have curvature estimates, see [17].The curvature estimates give us an δ > n ( δ only dependson the distance from p to ∂ Ω ). Such that a neighbourhood of v n ( p ) in Σ v n isa graph in geodesic coordinates, with height and slope uniformly boundedover the disk D δ n ( v n ( p )) of radius δ centered at the origin of T p Σ v n . We callthis graph Σ ( v n ( p ) , δ ) Suppose that p ∈ D . Since | Gv n ( p ) | is unbounded, there is a subse-quence of N v n ( p ) that converges to a horizontal vector N ( p ) . So, for thissubsequence, the tangent planes T p Σ v n converges to a vertical plane Π ( p ) and the graphs Σ ( v n ( p ) , δ ) converge to a constant mean curvature H graph Σ ( p , δ ) over a disk of radius δ (cid:48) ≤ δ , centered at the origin of Π ( p ) . By thechoice of the direction of the normal vector and the choice of H >
0, thelimit of the curvature vectors of Σ ( v n ( p ) , δ ) has the same direction as thenormal limit.Taking the curve L p ⊂ Ω passing through p , orthogonal to π ∗ ( N ( p )) with curvature 2 H and the curvature vector at p having the same directionas π ∗ ( N ( p )) . We want to prove Σ ( p , δ (cid:48) ) ⊂ π − ( L p ) . Since Σ ( p , δ (cid:48) ) is tangent to π − ( L p ) at v n ( p ) , if Σ ( p , δ (cid:48) ) is in one side of π − ( L p ) , by the maximum principle, we have that Σ ( p , δ (cid:48) ) ⊂ π − ( L p ) . Ifthis is not the case, Σ ( p , δ (cid:48) ) ∩ π − ( L p ) is composed by k ≥ v n ( p ) . These curves separate Σ ( p , δ (cid:48) ) in 2 k components and theadjacent components lies in alternative sides of π − ( L p ) . Moreover, thecurvature vector alternates from pointing down to pointing up when onegoes from one component to other. This implies that the normal vector to Σ ( p , δ (cid:48) ) points down and up. So, for n big enough, the normal vectors to Σ ( v n ( p ) , δ ) would point down and up, which does not occur.Let L p ( (cid:101) δ ) ⊂ Ω , δ (cid:48) ≥ (cid:101) δ be the curve contained in Σ ( p , δ (cid:48) ) ∩ ι ( L p ) whichcontains p ≈ ι ( p ) , and has length 2 (cid:101) δ . Since Σ ( p , δ (cid:48) ) ⊂ π − ( L p ) , we havethat, for all q ∈ L p ( (cid:101) δ ) the normal vector to Σ ( p , δ (cid:48) ) at q is a horizontalvector normal to L p ( (cid:101) δ ) at q . (cid:3) Lemma 3.18.
Let { u n } be a sequence of solutions of (2.12) in Ω . Given p ∈ D ,there is a curve L ⊂ Ω of curvature H which passes through p such that, afterpassing to a subsequence, the sequence of normal vectors { N u n | L } converges toa horizontal vector, whose projection via π is normal to L, and having the samedirection as the curvature vector of L. This curve L contains the compact arcL p ( (cid:101) δ ) given in Lemma 3.17.Proof. Let L be the curve of constant curvature 2 H in Ω , which contains L p ( (cid:101) δ ) joining the points of ∂ Ω ( L p ( (cid:101) δ ) is given in the Lemma 3.17). Given p , q ∈ Ω , denote by pq the compact arc in L between p and q . We define Λ = { q ∈ L ; T ( q ) is true } where • T ( q ) = there is a subsequence of { u n } such that { N u n | pq } becomeshorizontal, whose projection via π becomes orthogonal to L , hav-ing the same direction as the curvature vector of L .We want to prove that Λ = L . Since p ∈ Λ , Λ is not empty. We willprove that Λ is open and closed. First, we will prove that Λ is open.Let q be a point in Λ . denote by u Λ ( n ) the subsequence associated to Λ .Since Λ ⊂ D , Lemma 3.17 give us a curve L q ( (cid:101) δ ) through q such that, afterpassing to a subsequence { N u Λ ( n ) | L q ( δ ) } becomes horizontal and has thesame direction as the curvature vector of L q ( δ ) . Note that this subsequenceof { N u Λ ( n ) | L q ( δ ) } converges to a horizontal vector normal to L q ( δ ) and to L simultaneously, so L q ( δ ) ⊂ L , thus Λ is open.Now, we will prove that Λ is closed. We take a convergent sequence { q n } in Λ , q n → q ∈ L . We will show that q ∈ Λ .For each m , there is a subsequence of { u Λ ( n ) } such that { N u Λ ( n ) | pq m } be-comes horizontal with the same direction as the curvature vector of L in pq m . By the diagonal process, we obtain a subsequence of { u Λ ( n ) } suchthat { N u Λ ( n ) | pq m } converges to a horizontal vector having the same direc-tion as the curvature vector of L in pq m for all m . Then by Lemma 3.17, we ENKINS-SERRIN PROBLEM 17 can find a curve L q m ( δ ) having constant curvature 2 H through q m (for m large, δ depends only on the distance from q to ∂ Ω ) such that { N u Λ ( n ) | pq m } converges to a horizontal vector having the same direction as the curva-ture vector to L q m ( δ ) . So, L q m ⊂ L and since q m → q , we have that, for all m large enough, q ∈ L q m ( δ ) . Consequently q ∈ Λ . (cid:3) An important conclusion of the Lemma 3.18 is that the divergence setis given by D = (cid:83) i ∈ I L i , where L i is a curve called a divergence curve,having curvature 2 H . More precisely, we have the following definition. Definition 3.19. (Divergence line) Let L ⊂ Ω be a curve having constantcurvature k ( L ) = H , p ∈ L and { v n = u n − u n ( p ) } . If there exists a subse-quence of { N v n | L } which converges to a horizontal vector whose projectionvia π is orthogonal to L , then we said that L is a divergence line of { u n } . Lemma 3.20.
Let { u n } be a sequence of solutions of (2.12) in Ω . Suppose thatthe divergence set D of { u n } is composed of countable number of divergence lines.Then there exists a subsequence of { u n } , again denoted by { u n } such that • the divergence set of { u n } is composed of a countable number of pairwisedisjoint divergence lines,Proof. Suppose that D (cid:54) = ∅ and let L be a divergence line of { u n } , Lemma3.17 guarantees that, after passing to a subsequence, { N u n ( q ) } convergesto a horizontal vector whose projection via π is orthogonal to L at q for all q ∈ L . The divergence set of this sequence is contained in thedivergence set of the original sequence, so the divergence set associated tothis subsequence has only countable number of lines. This subsequenceis still denoted by { u n } and its divergence set by D . If there is a line L (cid:54) = L in D , we can find a subsequence such that { N u n ( q ) } convergesto a horizontal vector whose projection via π is orthogonal to L at q foreach q ∈ L . This implies that L ∩ L = ∅ . In fact, if this does not occur,we take a point q ∈ L ∩ L , such that the sequence { N u n ( q ) } convergesto a horizontal vector whose projection is orthogonal to L and L at q having the same direction as the curvature vector of L and L . Thenthe uniqueness of a curve through q having constant curvature 2 H witha given tangent vector shows that L = L . We continues this process toget a subsequence of { u n } still denoted by { u n } , whose divergence set iscomposed of a countable number of pairwise disjoint divergence lines. (cid:3) Lemma 3.21.
Let Ω be a domain bounded in part by an arc C having k ( C ) ≥ H.Let { u n } be an increasing or decreasing sequence of solutions of (2.12) in Ω witheach u n continuous in Ω ∪ C. Suppose that γ is an interior arc of Ω of curvature H forming part of the boundary D . Then γ cannot terminate at an interior pointof C if { u n } either diverges to ± ∞ or remains uniformly bounded on compactsubsets of C. Proof.
Suppose that γ is an arc in ∂ D that terminates at an interior point p of C . By considering only a small neighbourhood of p , we may assumethat C is C . By Theorem 3.1, the sequence { u n } cannot diverge to − ∞ on C . Moreover, id the curvature of C is not identically 2 H and { u n } remains uniformly bounded on compact subsets of C , Lemma 3.2 insuresthat a neighbourhood of C is contained in C a contradiction. Hence assume k ( C ) ≡ H . Suppose { u n } diverges to + ∞ on C and there exists exactlyone such γ terminates at p . Let q be a point of γ close to p and choose s on C close to p so that the geodesic segment sq lies in C . Let T be thetriangle formed by rq and the constant curvature 2 H arcs qp and pr . Thenby Lemma 3.9(3.8) 2 H A ( T ) = F u n ( pq ) + F u n ( pr ) + F u n ( rq ) while by Lemma 3.12(3.9) lim n −→ + ∞ F u n ( qp ) = | qp | , lim n −→ + ∞ F u n ( pr ) = | pr | .From (3.8), (3.9) and Lemma 3.10 we conclude(3.10) 2 H A ( T ) | rq | ≥ | qp | + | pr || rq | − p fixed, we move q to q (cid:48) and r to r (cid:48) along the same arcs so that | q (cid:48) p | = λ | qp | and | pr (cid:48) | = λ | pr | and form the triangle T (cid:48) by joining q (cid:48) to r (cid:48) by a geodesic. Then the left hand side of (3.10) tends to zero as λ → γ and γ terminates at p .Then again we can find a triangle T ⊂ C whose edges are two constant cur-vature 2 H arcs and a geodesic segment as before (perhaps ∂ T ∩ C = { p } ).The same arguments gives a contradiction.In the case the sequence remains uniformly bounded on compact sub-sets of C and there is exactly on γ , we choose r on C so that T is containedin D . By Lemma 3.2 (ii) the sequence must be divergent to − ∞ on D . Wenow reach a contradiction as above by using Lemma 3.12. If there are twoarcs terminates at p , then D is necessarily the convex lens domain formedby γ and γ (that is γ = γ ∗ , see Remark 4.3). Choose the point q on γ and r on γ and form T in D . Then (3.8), (3.9) and (3.10) still hold and wereach a contradiction as before. (cid:3) We ended this section with the following theorem.
Theorem 3.22. (Monotone convergence theorem) Let { u n } be a monotonicallyincreasing or decreasing sequence of solutions of (2.12) in a fixed domain Ω . Ifthe sequence is bounded in a single point of Ω , there is a non-empty open set C ⊂ Ω such that { u n } converges to a solution in C . The convergence is uniformon compact subsets of C and the divergence is uniform on compact subsets of D = Ω − C . If D is non-empty, ∂ D consists of arcs of curvature ± H and parts
ENKINS-SERRIN PROBLEM 19 of ∂ Ω . These arcs are convex to C for increasing sequences and concave to C fordecreasing sequences.
4. T he main theorems
In this section we deal with the solution for the Jenkins-Serrin problem,see Section 3.1 for notations and definitions. More precisely, we will provethe following two theorems.
Theorem 4.1. (Main Theorem 1) Consider the Jenkins-Serrin problem in an ad-missible domain Ω and suppose the family { C i } is empty. Then, there exists asolution for the Jenkins-Serrin problem if and only if (4.1) α ( ∂ Ω ) = β ( ∂ Ω ) + H A ( Ω ) and for all admissible polygons P (4.2) 2 α ( P ) < l ( P ) + H A ( P ) and 2 β ( P ) < l ( P ) − H A ( P ) where α ( P ) and β ( P ) are the total length of the arcs in ∂ P belonging to { A i } and { B i } respectively, finally l ( P ) denotes the perimeter of P and A ( Ω ) denotesthe area of P . Following the notations of Theorem 4.1, we have the next theorem.
Theorem 4.2. (Main Theorem 2) Consider the Jenkins-Serrin problem in an ad-missible domain Ω and suppose the family { C i } is non-empty. Then, there existsa Jenkins-Serrin solution of the Jenkins-Serrin problem if and only if (4.3) 2 α ( P ) < l ( P ) + H A ( P ) and 2 β ( P ) < l ( P ) − H A ( P ) for all admissible polygon P . Before we prove the theorems, we will use the flux formulas to see thateach Jenkins-Serrin solution u satisfies the equations (4.1),(4.2) and (4.3)for an admissible polygon P .Notice that an admissible polygon P can be write in the form P = {∪ i A P i } ∪ {∪ j B P i } ∪ {∪ k η P k } where A P i , B P j are arcs in P ∩ ∂ Ω and { η P k } is composed by 2 H -curves in Ω and possible arcs C k in ∂ Ω .For an admissible polygon P , the flux formulas yields F u ( ∂ P ) = H A ( P ) .Where, we conclude that F u ( {∪ i A P i } ) = − F u ( P − {∪ i A P i } ) + H A ( P ) F u ( {∪ j B P i } ) = − F u ( P − {∪ i B P i } ) + H A ( P ) Taking the firs equality for instance, we have α ( P ) = Σ i | A P i | = F u ( {∪ i A P i } )= − F u ( P − {∪ i A P i } ) + H A ( P ) ≤ | F u ( P − {∪ i A P i } ) | + H A ( P )= | F u ( {∪ j B P j } ∪ { η P k } ) | + H A ( P ) ≤ | F u ( {∪ j B P j } ) | + | F u ( { η P k } ) | + H A ( P )= β ( P ) + | F u ( { η P k } ) | + H A ( P )= l ( P ) − α ( P ) − |{ η P k }| + | F u ( { η P k } ) | + H A ( P ) < l ( P ) − α ( P ) + H A ( P ) where we used that F u ( B j ) = −| B j | and F u ( η k ) < | η k | . Thus2 α ( P ) < l ( P ) + H A ( P ) .Analogously, we can see2 β ( P ) < l ( P ) − H A ( P ) .On the other hand, when {∪ k C k } = ∅ and ∂ P = ∂ Ω , from flux formulaswe have F u ( {∪ i A P i } ∪ {∪ j B P i } ) = H A ( Ω ) that is F u ( {∪ i A P i } ) + F u ( {∪ j B P i } ) = H A ( Ω ) which implies α ( P ) − β ( P ) = H A ( Ω ) . Remark . (Lens domain) We call a domain D , a lens domain, if D isbounded by an arc γ of curvature 2 H and its geodesic reflection γ ∗ .In order to prove these two last theorems, we are going prove someresults. Proposition 4.4.
Consider the Jenkins-Serrin problem in an admissible domain Ω and suppose the family { B j } is empty, k ( C k ) > H and the assigned data fon the arcs C k are bounded below. Then, there exists a Jenkins-Serrin solution forthis Jenkins-Serrin problem if and only if (4.4) 2 α ( P ) < l ( P ) + H A ( P ) for all admissible polygon P .Proof. Let u n be the solution of (2.12) in Ω such that u n = (cid:26) n on ∪ A i min ( n , f ) on ∪ C i ,such solution exits and is unique by the Dirichlet Theorem (Theorem 2.5).Moreover by the Maximum principle (Theorem 2.2), the sequence { u n } ismonotone increasing, so the Monotone Convergence Theorem (Theorem ENKINS-SERRIN PROBLEM 21 p ∈ Ω such that u n ( p ) → + ∞ ,then there is a divergence line L ( p ) passing through p , thus the divergenceset D is not empty. Supposing that the divergence set D is composed bya countable divergence lines, so from Lemma 3.20, we have that D is theunion of pairwise disjoint divergence lines.By Lemma 3.18 an interior arc of Ω which bounds D must be of curva-ture 2 H and by Lemma 3.21, it can terminates only among the endpointsof the arcs A i or C i . Moreover by Lemma 3.2 a neighbourhood of each C i is contained in C . Therefore, the boundary of each component of D is anadmissible polygon P with vertices among those of the A i and C i . Alsothe curvature 2 H arcs forming the boundary which are not among the A i are concave to D by Theorem 3.1.By Lemma 3.9 applied to each u n in P (4.5) 2 H A ( P ) = F u n ( ∂ P − ∪ (cid:48) A i ) + F u n ( ∪ (cid:48) A i ) where ∪ (cid:48) A i is the union of the arcs A i which are part of P . Then byLemma 3.12(4.6) lim n −→ + ∞ F u n ( ∂ P − ∪ (cid:48) A i ) = α ( P ) − l ( P ) = − ( l ( P ) − α ( P )) .But | F u n ( ∪ (cid:48) A i ) | ≤ α ( P ) , hence from (4.5)2 H A ( P ) ≤ ( α ( P ) − l ( P )) + α ( P ) that is 2 H A ( P ) + l ( P ) ≤ α ( P ) ,contradicting our assumption (4.4). Thus D is empty and the sequenceconverges uniformly on compact subsets of Ω to a solution u . Since each u n is uniformly bounded in a neighbourhood of each C i by Lemma 3.2, astandard barrier argument shows that u = f on ∪ C i .Since the necessity of (4.4) is clear, this completes the proof. (cid:3) Similarly, we have the next theorem.
Proposition 4.5.
Consider the Jenkins-Serrin problem in an admissible domain Ω and suppose that the family { A i } is empty, k ( C i ) > H and the assigned dataf on the arcs C i are bounded above. Then, there exists a Jenkins-Serrin solution ifand only if (4.7) 2 β ( P ) < l ( P ) − H A ( P ) for all admissible polygon P . Now we use Proposition 4.4 and Proposition 4.5 to construct some use-full barriers in order to remove the assumption k ( C i ) > H . Example . Let B = B δ ( P ) be a ball of small radius δ centered at P , andlet Q and R be the "antipodal" points on ∂ B . choose points Q and Q on ∂ B and symmetric with respect to the geodesic through QPR . Nowlet B be an arc of curvature − H (as seem from P ) joining Q and Q
22 CARLOS PEÑAFIEL and set A = B ∗ i (the geodesic reflection of B across its endpoints). Let R and R on ∂ B be reflections on ∂ B of Q and Q (with respect to thegeodesic orthogonal to QPR through P ) and define B and A . Then for δ small compared with H , the domain B + bounded by A , A and parts of ∂ B satisfies the conditions of Proposition 4.4 and similarly, the domain B − bounded by B , B and parts of ∂ B satisfies the condition of Proposition4.5.Let u + be the solution of (2.12) in B + with boundary values + ∞ on A ∪ A and the constant value M on the remainder of the boundary.Similarly, let u − be the solution of (2.12) in B − with boundary values − ∞ on B ∪ B and the constant value − M on the remainder of the boundary.With this example we obtain the following proposition. Proposition 4.7.
Let Ω be a domain bounded in part by an arc γ and let { u n } be a sequence of solutions of (2.12) in Ω which converges uniformly on compactsubsets of Ω to a solution u. Suppose each u n is continuous on Ω ∪ γ , then (i) Suppose the boundary values of u n converges uniformly on compact sub-sets of γ to a bounded limit f . If k ( γ ) ≥ H, then u is continuous on Ω ∪ γ and u = f on γ . (ii) If k ( γ ) = H and the boundary values of u n diverges uniformly to + ∞ on compact subsets of γ , then u takes on the boundary values + ∞ on γ . (iii) If k ( γ ) = − H and the boundary values of u n diverges to − ∞ on com-pact subsets of γ , then u takes on the boundary values − ∞ on γ .Proof. ( i ) It suffices to prove that the sequence { u n } is uniformly boundedin the intersection of Ω with a neighbourhood of any interior point P of γ .Orient the ball of Example 4.6, so that the geodesic joining QPR is tangentto ∂ Ω . We may choose the points Q i i =
1, 2 and δ small enough so thatthe arc joining Q and R lies in a compact subset of Ω . Then if M is largeenough u n ≤ u + in Ω ∩ B + and u n ≥ u − in Ω ∩ B − .Therefore, the sequence is uniformly bounded in a neighbourhood of P . ( ii ) Let P be an interior point of γ . Similarly as in ( i ) , we obtain thatthere exists M large enough so that u n ≥ − M in W = B (cid:101) ( P ) ∩ Ω .Let v m be the solution of (2.12) in W with boundary values m on γ ∩ B (cid:101) ( P ) and − M on the remaining boundary. By the Maximum Principle u n ≥ v m for n sufficiently large so u ≥ v m in W . In particular, u ( P ) > m for every m and u must take on the value + ∞ at P . ( iii ) Again for P interior to γ , u n ≤ M in W = B (cid:101) ( P ) ∩ Ω . Let v m be the solution of (2.12) in W with boundary values − m on γ ∩ B (cid:101) ( P ) and M on the remaining boundary. By the Maximum Principle u n ≤ v m for n sufficiently large, so u ≤ v m in W . Since the v m are monotonically ENKINS-SERRIN PROBLEM 23 decreasing (and converges to a solution with infinite boundary values on γ ∩ B (cid:101) ( P ) ), u must takes on the value − ∞ at P . (cid:3) We can now extend Proposition 4.4 and Proposition 4.5 to allow thearcs C i to satisfy k ( C i ) ≥ H . The only changed needed in the proofof Proposition 4.4 is to use part ( i ) of Proposition 4.7 to show that thesolution takes on the required boundary data on the arcs C i . The extensionof Proposition 4.5 is more delicate. Since if k ( C i ) = H for some i , we donot know that the sequence is bounded bellow in a neighbourhood of C i .However by Lemma 3.21 a neighbourhood of C i is either contained in C or in D . We have already handled the former case. In the latter case,consider a component of D whose boundary is an admissible polygon P (with vertices among those of the B i and C i ) containing a subset ∪ (cid:48) C i ofthe arcs C i of curvature 2 H and a subset ∪ (cid:48) B i of the arc B i . The interiorarcs of Ω which are in P are convex to D . By Lemma 3.9 applied to each u n in P (4.8) 2 H A ( P ) = F u n ( ∪ (cid:48) B i ) + F u n ( ∪ (cid:48) C i ) + F u n ( ∂ P − ∪ (cid:48) B i − ∪ (cid:48) C i ) Then by Lemma 3.12(4.9) lim n −→ + ∞ F u n ( ∂ P − ∪ (cid:48) B i − ∪ (cid:48) C i ) = l ( P ) − β ( P ) − Σ (cid:48) | C i | and by Lemma 3.13(4.10) lim n −→ + ∞ F u n ( ∪ (cid:48) C i ) = Σ (cid:48) | C i | But | F u n ( ∪ (cid:48) B i ) | ≤ β ( P ) , hence2 H A ( P ) ≥ − β ( P ) + Σ (cid:48) | C i | + ( l ( P ) − Σ (cid:48) | C i | − β ( P )) = l ( P ) − β ( P ) contradicting our assumption (4.7).Thus D is empty and the sequence converges uniformly on compactsubsets of Ω to a solution u . Finally we use parts ( i ) and ( ii ) of Proposition4.7 to show that our solution achieves the boundary values.We state these result as Theorem 4.8.
Consider the Jenkins-Serrin problem in an admissible domain Ω and suppose the family { B i } is empty and the assigned data f on the arcs C i arebounded bellow. Then, there exists a solution if and only if (4.11) 2 α ( P ) < l ( P ) + H A ( P ) for all admissible polygons P . Theorem 4.9.
Consider the Jenkins-Serrin problem in an admissible domain Ω and suppose the family { A i } is empty and the assigned data f on the arcs C i arebounded above. Then, there exists a solution if and only if (4.12) 2 β ( P ) < l ( P ) − H A ( P ) for all admissible domain. Now we prove the Main Theorem 2 (Theorem 4.2). That is, we allowboth families { A i } and { B i } to occur and allow the data f on the { C i } tobe unbounded both above and bellow as we approach the endpoints. Proof. (Main Theorem 2) By Theorem 4.8 the first condition of (4.3) guar-antees the existence of a solution u + of (2.12) in Ω ∗ such that u + = + ∞ on ∪ A i ∪ B ∗ i max ( f , 0 ) on ∪ C i .Similarly, by Theorem 4.9, the second condition of (4.3) guarantees theexistence of a solution u − of (2.12) in Ω ∗ such that u − = − ∞ on ∪ B i ∪ A i min ( f , 0 ) on ∪ C i .Now let u n be the solution of (2.12) in Ω ∗ such that u n = n on ∪ A i − n on ∪ B ∗ i f n on ∪ C i ,where f n is the truncation of f above by n and bellow by − n .By the Maximum Principle u n ≤ u + in Ω ∗ and u − ≤ u n in Ω .Therefore the sequence { u n } is uniformly bounded on compact subsetsof Ω and a subsequence converges uniformly on compact subsets to asolution u in Ω . By Proposition 4.7, u takes on the boundary assigneddata. The necessity of the condition (4.3) follows essentially as in theTheorem 4.8 and Theorem 4.9. (cid:3) Finally, we focus our attention in the proof of the Main Theorem 1.
Proof. (Main Theorem 1) let v n be a solution of (2.12) in Ω ∗ with boundaryvalues n on each A i and 0 on each B ∗ i . For 0 < c < n we define for n ≥ E c = { v n − v > } and F c = { v n − v < c } ;we suppress the dependence of these sets on n . Let E ic and F ic denoterespectively the components of E c and F c whose closure contains respec-tively A i and B ∗ i . By the maximum principle E c = ∪ E ic and F c = ∪ F ic . If c is sufficiently close to n , the sets { E ic } will be distinct and disjoint (to seethis, note that we can separate any two of the A i by a curve joining two ofthe B ∗ i on which v n − v is bounded away from n ). Now we define µ ( n ) tobe the infimum of the constants c such that the sets { E ic } are distinct anddisjoints. The sets { E i µ } will again be distinct although there must be atleast one pair ( i , j ) , i (cid:54) = j such that E i µ ∩ E j µ is nonempty. This implies that ENKINS-SERRIN PROBLEM 25 given any F i µ there is some F j µ distinct from it. Now let u + i , i =
1, ..., k bethe solution of (2.12) in Ω ∗ taking on the boundary values + ∞ on A i and 0on the remaining boundary. This solution exists by Theorem 4.8 since thesolvability condition (4.11) follows trivially from (4.1) and (4.2). Also let u − i be the solution of (2.12) in the domain (cid:101) Ω bounded by ∪ A i , B ∗ i , ∪ j (cid:54) = i B j taking on the boundary values − ∞ on ∪ j (cid:54) = i B j and 0 on the remaining ofthe boundary. In order to know that the solution exists by Theorem 4.9,we need to verify (4.12), thus we need only consider the admissible poly-gon P in (cid:101) Ω which contain the lens domain L formed by B i and B ∗ i . Let P be the corresponding admissible polygon for Ω formed by deleting L . By(4.1) and (4.2) we have2 β = ( | B i | + ◦ ∑ i (cid:54) = j | B j | ) ≤ l ( P ) − H A ( P ) or equivalently2 β = ◦ ∑ i (cid:54) = j | B i | ≤ l ( (cid:101) P )) − H A ( (cid:101) P ) + ( H A ( L ) − | B i | ) .However since a solution (for example v n ) exits in L , we have by Lemma3.10 2 H A ( L ) < | B i | so condition (4.12) is satisfied.We now set u + ( p ) = max i { u + i ( p ) } in Ω ∗ and u − ( p ) = min i { u − i ( p ) } in Ω .We note that if we compare each u + i to a fixed bounded solution in Ω ∗ (which exists since Ω is admissible), then by the maximum principlethere is a constant N > u + i > − N , i =
1, ..., k . Finally we set u n = v n − µ ( n ) .We now claim that u n ≤ u + + M in Ω ∗ and u n ≥ u − − M in Ω .where M = N + sup Ω ∗ | v | . Suppose u n > v at some point p . Then v n − v > µ ( n ) at p , so that p is in some E i µ . Applying the maximum principle in thedomain E i µ , we obtain u n ≤ u + i + N + sup E i µ | v o | ≤ u + + M at p .On the other hand, suppose u n < v at some point p ∈ Ω . Then v n − v > µ ( n ) at p , so that p is in some F i µ . By what has been shown above,there is a corresponding j = j ( i ) so that F i µ ∩ F j µ = ∅ , we obtain u n ≥ u − j − sup F i µ | v o | ≥ u − − M . Therefore the claim is justified and the sequence { u n } is uniformlybounded on compact subsets of Ω . By the compactness principle, a sub-sequence { u n } converges uniformly on compact subsets of Ω to a solution u . We still must show that u takes on the required boundary values. Weobserve that a subsequence µ ( n ) diverges to + ∞ otherwise we can ex-tract a subsequence converging to a finite limit µ . Each u n would thenbe bounded below in Ω ∗ uniformly in n , and the boundary values of u n would tend uniformly on compact subsets of ∪ B ∗ i to − µ and divergeuniformly to + ∞ on ∪ A i . Once again we could find a subsequence con-verging uniformly on compact subsets to a solution in Ω ∗ . By Proposition4.7 we would have v = (cid:26) + ∞ on ∪ A i − µ on ∪ B ∗ i We can now obtain a contradiction to (4.1) by a flux argument. By Lemma3.9(4.13) 2
H A ( Ω ) = F v ( Σ A i ) + F v ( Σ B i ) while by Lemma 3.10 and Lemma 3.11(4.14) | F v ( Σ B i ) | < β and F v ( Σ A i ) = α Combining (4.13) and (4.14) gives α − β < H A ( Ω ) , a contradiction. In thesame way, we see n − µ ( n ) diverges to + ∞ . Summing up we have shownthat the boundary values of u n , namely, − µ ( n ) on ∪ B ∗ i and n − µ ( n ) on ∪ A i diverges to − ∞ and + ∞ respectively. Therefore u n diverges to − ∞ on ∪ B i . Since the necessity conditions (4.1) and (4.2) is straightforward,we conclude the theorem. (cid:3) We ended this paper with the following maximum principle at the in-finity, the prove of this theorem is similar to this one in [14, Theorem 4.1]
Theorem 4.10. (Maximum principle at infinity) Let Ω ⊂ R be an admissibledomain, suppose that the family { C k } is nonempty, and let u, v be solutions ofthe Dirichlet problem in Nil ( τ ) , with the same continuous values on each arcC k ⊂ ∂ Ω . Then u = v on Ω . R eferences [1] Cui, Q. and Penafiel, C. The Jenkin-Serrin Problem for Constant Mean CurvatureGraphs in Killing Submersion. Pre-print.[2] Dajczer, M. and Lira, J. Killing graphs with prescribed mean curvature and Riemanniansubmersions. Ann. Inst. H. Poincaré Anal. Non Linéaire , (3): 763-775, 2009.[3] Dajczer, M., Alias, L. J. and Rosenberg, H. The Dirichlet problem for CMC surfaces inHeisenberg space. pre-print[4] Daniel, Benoit. Isometric Immersions into 3-dimensional Homogeneous Manifolds.Comment. Math. Helv., ISSN 0010-2571. 82-1, 87-131. 2007.[5] Figueroa, C.; Mercuri, F.; Pedrosa, R. Invariant surfaces of the Heisenberg group.. Ann.Mat. Pura Appl , 177 (1999), 173–194.[6] Folha, A. and Melo, S. The Dirichlet problem for constant mean curvature graphs in H × R over unbounded domains. Pac. J. of Math. , : 1, 2011. ENKINS-SERRIN PROBLEM 27 [7] Folha, A. and Penafiel, C. The Generalized Dirichlet Problem for Constant Mean Cur-vature Graphs Over Unbounded Domains in the Space (cid:103)
PSL ( R , τ ) over unboundeddomains. Pre-print.[8] Folha, A. and Rosenberg, H. The Dirichlet problem for constant mean curvature graphsin M × R . Geomtry and Topology , : 1171-1203, 2012.[9] Hauswirth, L. Rosenberg, H. and Spruck, J. Infinite boundary value problems for con-stant mean curvature graphs in H × R and S × R . Amer. J. Math. , (1): 195-226,2009.[10] Jenkins, H. and Serrin, J. Variational problems of minimal surface type II. Boundaryvalue problems for the minimal surface equation. Arch. Rational Mech. Anal. , : 321-342, 1966.[11] Leandro, C. and Rosenberg, H. Removable singularities for sections of Riemanniansubmersions of prescribed mean curvature. Bull. Sci. Math. , (4): 445-452, ISSN 0007-4497, 2009.[12] Mazet, L. Rodríguez, M. and Rosenberg, H. The Dirichlet problem for the minimalsurface equation-with possible infinite boundary data- over domains in a Riemanniansurface. Proc. London Math. Soc. , (102): 985-1023, 2011.[13] Nelli, B. and Rosenberg, H. Minimal surfaces in H × R . Bull. Braz. Math. Soc. , :263-292, 2002.[14] Pinheiro, A. L. Minimal vertical graphs in Heisenberg space. Preprint. , (0): 0-0.[15] Pinheiro, A. L. The theorem of Jenkins-Serrin in M × R . Bull. Braz. Math. Soc. , (1):117-148, 2009.[16] Radö, T. The problem of least area and the problem of Plateau. Math. Z. , : 763-796,1930.[17] Rosenberg H., Souam R. and Toubiana E. General Curvature Estimates for Sta-ble H-Surfaces in 3-Manifolds and Applications J. Differential Geom. , (3): 623-648,MR2669367, 2010.[18] Spruck J. Infinite boundary value problem for surfaces of constant mean curvature. Arch. Rationa Mech. Anal. , : 1-31, 1972.[19] Thurston, William. Three-Dimensional Geometry and Topology. Princeton, 1997.[20] Younes R. Minimal surfaces in (cid:103) PSL ( R ) Illinois J. Math. , (2): 671-712, 2010.(2): 671-712, 2010.