Infinite boundary value problems for constant mean curvature graphs

Infinite boundary value problems for constant mean curvature graphs in H2 × R and S2 × R Laurent Hauswirth ∗, Harold Rosenberg †and Joel Spruck

1

‡

Introduction

In this paper we study constant mean curvature graphs in M × R where M = H2 or S2 the hyperbolic plane of curvature -1 or the 2-sphere of curvature 1. Let D be a bounded domain in M with piecewise smooth boundary. We prescribe continuous data on the smooth arcs of boundary D, which may be +∞ on some arcs and −∞ on others. Our purpose is to solve the Dirichlet problem for the constant mean curvature equation in D, assuming the prescribed boundary data.. More precisely, we seek a smooth function u defined in D, whose graph S = {(x, u(x)) : x ∈ D} in M × R (with the standard product metric) has mean curvature H and which takes on the prescribed data on ∂D. The domains D we consider and the geometric properties that guarantee the existence of a solution to this Dirichlet problem, are given in section 7. When M = R2 and H=0, this is the classical theory of Jenkins-Serrin [3]. This theory was extended to H 6= 0 by Spruck [9], again for M = R2 . ∗

Universit´e Marne-La-Valle, Cit´e Descartes, 5 Boulevard Descartes, 77454 Champssur-Marne, Marne-La-Valle, France. † Institut de Math`ematique, 2 Place Jussieu, 75005 Paris, France. ‡ Department of Mathematics, Johns Hopkins University Baltimore, MD 21218, USA. Research supported in part by NSF grant DMS0603707 .

1

For H = 0, the Jenkins-Serrin theory was extended to M = H2 by Nelli and Rosenberg [5] and to arbitrary Riemannian surfaces M by Pinheiro [4]. Our methods also work in this case. We shall always assume that H > 0 with respect to the upward unit normal of S. Then if u tends to +∞ for any approach to a boundary arc γ, then necessarily its curvature κ(γ) = 2H is constant, while if u tends to −∞ on γ then κ(γ) = −2H (see Theorem 4.8). Thus we must deal with non-convex domains D with ∂D piecewise C 2 and consisting of three set of open arcs {Ai } , {Bi } and {Ci } satisfying κ(Ai ) = 2H , κ(Bi ) = −2H and κ(Ci ) ≥ 2H respectively. The generalized Dirichlet problem is to find a solution of (2.1) in D taking on the boundary values +∞ on the Ai , −∞ on the Bi and arbitrary continuous boundary data on the Ci . A precise definition of the admissible domains D is given in Definition 7.1 of section 7. When the family {Ci } is non-empty, we give in Theorem 7.11 necessary and sufficient conditons that there exist a unique solution. These are the generalized Jenkins-Serrin flux conditions (as in [9]) formulated in terms of admissible polygons (see Definiton 7.4 of section 7). When the family {Ci } is empty we give in Theorem 7.12 necessary and sufficent conditions for “Scherk type” solutions to exist; these are unique up to a vertical translation (see Theorem 7.13). Examples of this type are constructed for H2 for H ≤

1 2

in section 8. The proof of these results is long and involved and essentially follows that of [9] using the existence theory and interior gradient estimates of [10] as basic tools. As in all previous proofs, the key idea is to study the flux of monotone increasing and decreasing sequences of solutions along arcs where they diverge. This is carried out in sections 5 and 6. Section 2 contains an important general maximum principle which in conjunction with the special barriers of section 4 allows us to analyse the flux. Section 3 formulates a general Perron existence theorem which is a basic tool for the existence 2

theory of section 7. In some sense, the Scherk type examples with H =

1 2

in

H2 × R are the most interesting as they can be constructed over arbitrarily large domains. We hope in future work to use these to construct interesting complete graphs.

2

The mean curvature equation and a general maximum principle

Let ds2 =

P2

i,j=1

σij dxi dxj is a local Riemannian metric on M and denote

by (σ ij ) the inverse matrix of (σij ). Then M × R is given the product metric ds2 + dt2 where t is a coordinate for R and (see [10]) if S = {graph u} has constant mean curvature H, the height function u(x) ∈ C 2 (Ω) satisfies the divergence form equation (2.1)

M u := div

∇u = 2H , W = (1 + |∇u|2 )1/2 W

where the divergence and gradient ∇u are taken with respect to the metric on M. Equivalently, equation (2.1) can be written in non-divergence form (2.2)

Mu =

2 1 X ij g Di Dj u = 2H , W i,j=1

where D denotes covariant differentiation on M and 2

g ij = σ ij −

X ui uj i , u = σ ij uj . W2 j=1

The main result of this section is a general maximum principle for sub and super solutions of the mean curvature operator for boundary data with a finite number of discontinuities. We follow the argument of [9] which holds in the general setting with little change. For the convenience of the reader we sketch the argument which uses the divergence structure. 3

Lemma 2.1. Let u1 and u2 be functions in C 2 (Ω), Ω ⊂ M and set Wk = W (∇uk ), k = 1, 2. Then (2.3)

< ∇u1 − ∇u2 ,

∇u1 ∇u2 − > ≥0 W1 W2

with equality at a point if and only if ∇u1 = ∇u2 Proof. The downward unit normal Ni to Si = graph ui is given by Ni =

∇ui 1 ∂ . − e, e= Wi Wi ∂t

Hence using < Wi Ni , e >= −1 , ∇u1 ∇u2 1 1 − >=< W1 N1 − W2 N2 , N1 − N2 + ( − )e > W1 W2 W1 W2 = < W1 N1 − W2 N2 , N1 − N2 >= (W1 + W2 )(1− < N1 , N2 >) W1 + W2 ||N1 − N2 ||2 ≥ 0 . = 2

< ∇u1 − ∇u2 ,

(Here, the inner products are in the metric of M × R.) Theorem 2.2.(General maximum principle) Let u1 and u2 satisfy M u1 ≥ nH ≥ M u2 in a bounded domain D ⊂ M. Suppose that lim inf(u2 − u1 ) ≥ 0 for any approach to ∂D with the possible exception of a finite number of points of ∂D. Then u2 ≥ u1 with strict inequality unless u2 ≡ u1 . Proof. Let N and  be positive constants with N large and  small. Define  if u1 − u2 ≥ N  N − 1 2 u − u −  if  < u1 − u2 < N ϕ=  0 if u1 − u2 ≤  Then ϕ is Lipschitz and 0 ≤ ϕ < N and ∇ϕ = ∇u1 − ∇u2 in the set where  < u1 − u2 < N and ∇ϕ = 0 almost everywhere in the complement of this set. About each exceptional boundary point Pi , i = 1, . . . , k we construct a ball of radius  and obtain a domain D ⊂ D bounded by spherical boundaries 4

Ci , i = 1, . . . , k and a subset of ∂D which we simply denote by Γ. Evidently, ϕ ≡ 0 in a neighborhood of Γ. Now let Z ∇u1 ∇u2 J = P ϕ< − , ν > dS W1 W2 Ci where ν is the exterior normal to ∂D . Then J ≤ 2ωn−1 n−1 kN ,

(2.4)

where ωn−1 is the volume of the unit (n − 1) sphere. Since div (ϕ(

∇u1 ∇u2 ∇u1 ∇u2 ∇u1 ∇u2 − )) =< ∇ϕ, − > +ϕ(div − div ) W1 W2 W1 W2 W1 W2

almost everywhere in D, we can apply the divergence theorem and obtain Z ∇u1 ∇u2 ∇u1 ∇u2 (< ∇ϕ, − > +ϕ(div − div )) dV (2.5) J= W1 W2 W1 W2 D By our assumptions, the last term of the integrand in (2.5) is non-negative so by (2.3),(2.4),(2.5) we have Z ∇u1 ∇u2 (2.6) 0≤ < ∇ϕ, − > dV ≤ 2ωn−1 n−1 kN. W1 W2 D As  decreases to zero, the sets D are expanding and thus ∇u1 ≡ ∇u2 in the set {0 < u1 − u2 < N }. Since N is arbitrary, ∇u1 ≡ ∇u2 whenever u1 > u2 . It follows that u1 ≡ u2 + positive constant in any nontrivial component of the set {u1 > u2 }. Suppose there exists one such component. Then the maximum principle ensures u1 ≡ u2 + positive constant in D and by the assumptions of the theorem, the constant must be non-positive, a contradiction. Remark 2.3. With a little more effort using a covering argument as in [6], we can allow the exceptional set E of points on ∂D to have one dimensional Hausdorff measure H1 (E) = 0. 5

3

Existence and compactness theorems

In this section we state the basic existence and compactness theorems we shall use in our study of the Dirichlet problem with infinite boundary data. Theorem 3.1. Compactness theorem Let {un } be a uniformly bounded sequence of solutions of (2.1) in a bounded domain D. Then there exists a subsequence which comverges uniformly on compact subsets (in any topology) to a solution of (2.1) in D. Theorem 3.1 follows from the interior gradient estimate of [10] (see (6.22) at the beginnng of section 6 for a precise statement)and a standard argument (see Corollary 16.7 and Theorem 16.8 of [2]) using Schauder theory and the Arzela-Ascoli theorem. We next turn to existence theorems. For simplicity, we will assume M = H or S and consider only bounded piecewise C 2 domains D. For P ∈ ∂D define the outer curvature κ ˆ (P ) to be the supremum of all (inward) normal curvatures of C 2 curves passing through P and locally supporting D. If no such curve exists we define κ ˆ (P ) to be −∞. Note that κ ˆ (P ) = κ(P ) at all regular points of ∂D. Theorem 3.2. Existence theorem Suppose that κ ˆ (P ) ≥ 2H for all P ∈ ∂D except for a finite set E of exceptional corner points of ∂D. a. If E = ∅ there is a unique solution of (2.1) in D taking on arbitrarily assigned continuous boundary data on ∂D \ E. b. If E 6= ∅ the same result holds if either i. M = H 2 and H ≤ 21 , or ii. there is a bounded subsolution of (2.1) in D. Remark 3.3. Theorem 3.2 is unusual in that existence and uniqueness holds for certain special non-convex domains. Note that D may even be multiply 6

connected. A simple example is provided by an annulus with outer boundary a circle and inner boundary a quadrilateral composed of four (or an even number) of circular arcs (convex toward the annulus). If all the circles have curvature at least 2H, Theorem 3.2 guarantees a unique solution for arbitrary continuous boundary data. Note that the solution is not required to be continous at the corner points and in general cannot be so. By a simple approximation argument we can also allow a finite number of discontinuities in the boundary data on ∂D \ E, that is on the part of the boundary satisfying the generalized curvature condition. Example 3.4. Lens domain A simple but useful example is provided by a lens domain L bounded by an arc γ of curvature 2H and its geodesic reflection γ ∗ . In S 2 for all H > 0 and in H 2 for H >

1 2

we must restict the length

of γ to half that of the geodesic circle of curvature 2H. However in H for 0 0 such that |fn (x) − fn (Q)| < ε and |fn (Q) − f (P )| < ε if d(x, Q) < δ and n large. Now let γ be an arc of constant curvature 2H that supports ∂Dn at Q and let w+ = −w− = w(x) be the upper and lower barriers in Nδ given by Corollary 3.6 with M = 2 sup un . Then by the maximum principle −w(x) − 3ε ≤ un (x) − f (P ) ≤ w(x) + 3ε in Nδ ∩ Dn , Hence there is a uniform modulus of continuity σ(t) so that |un (x) − f (P )| < σ(d(x, P )) for any P ∈ ∂D and x ∈ Dn . It follows that u ∈ C(D) and u = f on ∂D. For part b., we use Perron’s method as in [8]. We note that the existence of a bounded subsolution in case M = H 2 and H ≤

1 2

is automatic since

there are global radial solutions. Otherwise, we assume such a bounded subsolution exists. Note that the constants are always supersolutions. This 8

being the case there is a Perron solution u in D which achieves the assigned boundary data on ∂D − \E because of the existence of local barriers. In all cases, the solution is unique by the General maximum principle, Theorem 2.2.

4

Special barriers and the asymptotic behavior of solutions with infinite boundary values

We begin this section with the construction of special barriers in case M = H and H ≤ 21 . For the half-space model of H = {(x, y) : y ≥ 0} with metric ds2 =

dx2 +dy 2 , y2

(4.7)

note that

  ∂  ux uy ∂ M u = y2 (q (q )+ )  ∂x ∂y 1 + y 2 (u2x + u2y ) 1 + y 2 (u2x + u2y ) 

Lemma 4.1. For u = f (φ(x) − y), (4.8) Mu =

y2 (1 + y 2 f 02 (1 + φ02 ))

 3 2

In particular for φ(x) = 2H +

f 0 φ00 + f 00 (1 + φ02 ) + y 2 f 03 φ00 + yf 03 (1 + φ02 ) √

1 − x2 and f (t) = − a1 log |t| for 0 < |t|
0)   a2 t √ y 2 2 (4.9) M u = 2H 1 + ( 1 − x − 2 ) + O((at) ) 2Hy a √ where t = 2H + 1 − x2 − y. Hence u tends to +∞ on the curve C = √ {(x, y) : y = φ(x)} and in a neighborhood of (0, 1 + 2H), if a > 1 + 2H, u √ is a subsolution for t > 0 and a supersolution for t < 0 while if a < 1 + 2H, u is a supersolution for t > 0 and a subsolution for t < 0 . 9

Proof. Equation (4.8) is a straightforward computation. To derive (4.9) we make a Taylor expansion about t = 0 (using f 0 = − at1 , f 00 = af 02 , φ00 = 3

3

−(1 + φ2 ) 2 = (1 − x2 )− 2 ): 3 1 − x2 2 2 M u = (1 − a t + O(a4 t4 ))(2H − t + 2 2 y √ 1 − x2 = 2H + (a2 − 1)t + O(a2 t2 ) y

√ a2 t2 1 − x2 2 a t+ ) y y

Remark 4.2. For 2H < 1 the curve C = {(x, y) : y = φ(x)} is a so called equidistant circle and has constant curvature 2H. (For 2H=1 the curve C is half a horocycle .) Note that the translates Ct = {(x, y) : y = φ(x)−t)}, t > 0 are also equidistant circles of smaller curvature 2H − t Corollary 4.3. Let D be a domain in H 2 bounded in part by an arc Γ with κ ˆ ≥ 2H (for H ≤ 21 ). Given M > 0 large, let Ct0 be a supporting equidistant circle C at P ∈ Γ of constant curvature 2H − t0 ( 0 < t0 ≤

0 −aM e ) 2a

For C 0 a

translate of C of curvature 2H − eaM t0 , let N (t0 , M ) be the subdomain of D bounded by Γ and C 0 . Then there is a subsolution w < 0 of (2.1) in N with w(P ) = 0 and w = −M on C 0 . Proof. Let P=(0,2H) and {y = φ(x)} be the equidistant circle of Remark 4.2 so that Ct0 is the translate {y = φ(x) − t0 }. Then w = − a1 log φ(x)−y has the t0 required properties. We next construct barriers (with infinite slope) using the (oriented) distance function d(x) to an arc Γ in M. Let w = h(d); then as in [10] (with n=2) from the non-divergence form (2.2) of the equation 1 ij wi wj (σ − )(h0 Di Dj d + h00 Di dDj d) 2 W W 1 h00 = √ (h0 ∆d + ) (1 + h02 ) 1 + h02 h00 h0 = κ(x), 3 − √ 1 + h02 (1 + h02 ) 2

Mw =

(4.10)

10

where κ(x) is the inward curvature of the level set of d(x) passing through x. Lemma 4.4. i. Let Γ be a C 2 arc with κ(Γ) ≤ 2α < 2H with respect to the “interior” unit normal −n . Then in a sufficiently small neighborhood N of any interior point of Γ, there is a supersolution w of (2.1) with exterior normal derivative

∂w ∂n

= +∞ on Γ.

ii. If κ(Γ) ≤ −2α < −2H with respect to the “interior” unit normal −n, then there is a a subsolution w of (2.1) with exterior normal derivative

∂w ∂n

= −∞

on Γ. q Proof. Let h(t) = − 2t . Then

h00 3 (1+h02 ) 2

3

= (1 + 2t)− 2 ,

0 √ h 1+h02

= −(1 +

1

2t)− 2 , so from (4.10) w = h(d) satisfies 3

1

1

M w = (1 + 2d)− 2 + (1 + 2d)− 2 κ(x) ≤ (1 + 2d)− 2 (2α + 2) < 2H for  and d small enough, proving i. For part ii. let w = −h(d). Then 3

1

1

M w = −(1 + 2d)− 2 − (1 + 2d)− 2 κ(x) ≥ (1 + 2d)− 2 (2α − 2) > 2H for  and d small enough, proving ii.

Remark 4.5. For  = 12 |α − H| the barriers of the lemma are defined in an annulus {0 < d(x)
21 ), Delaunay’s onduloids provide explicit such barriers which are radial. For later use we note also that if Γ has constant curvature −2α, then M w ≤ 2α = 2H + 2(α − H). Hence if α is very close to H, w is an approximate supersolution. These special barriers are important because of the following well-known variant of the maximum principle due to Finn.

11

Lemma 4.6. Let D be a domain whose boundary is the union of two closed sets γ1 and γ2 with γ2 ∈ C 1 . Let u1 ∈ C 2 (D)∩C 1 (γ2 ) and u2 ∈ C 2 (D)∩C 0 (D) satisfy M u1 ≥ M u2 in D and suppose

∂u2 ∂ν

= +∞ (ν is the outer normal) at

every point of γ2 . Then if lim inf (u2 − u1 ) ≥ 0 for any approach to γ1 , then u2 ≥ u1 in D. Corollary 4.7. Let u be a solution of (2.1) in a domain D and let P be a regular point of ∂D. If κ(P ) < 2H then u cannot tend to +∞ at P. If κ(P ) < −2H then u cannot tend to −∞ at P. Proof. Suppose κ(P ) < 2H. Then there is an arc C0 of curvature < 2H internally tangent to ∂D at P. Let Ct , 0 < t < δ be inward parallel arcs to C, with δ so small that all have curvature < 2H. We can then choose a domain Nt (contained in D) bounded by Ct , Cδ and two geodesics joining them that are strictly interior to D (independent of t). We can also arrange that Nt is sufficiently small that the supersolution w of Lemma 4.4 is welldefined for Γ = Ct . Set M = sup∂Nt \Ct u + supNt w. Then u ≤ w + M in Nt by Lemma 4.6. Letting t → 0 implies u(P ) < ∞. An analogous argument in case κ(P ) < −2H shows u(P ) > −∞. We can now characterize those arcs on which solutions can tend to ±∞. Theorem 4.8. Let D be a domain bounded in part by a C 2 arc γ and let u be a solution of (2.1). If u tends to +∞ for any approach to interior points of γ, then κ(γ) = 2H, while if u tends to −∞ for any approach to interior points of γ, then κ(γ) = −2H. Proof. Suppose u → +∞ on γ. Then by Corollary 4.7 κ(γ) ≥ 2H. Now suppose κ > 2H on some subarc γ 0 ⊂ γ. Then by shrinking γ 0 if necessary, we can find a subdomain ∆ ⊂ D bounded by γ 0 and an arc Γ with curvature (with respect to the interior of ∆) satisfying κ(Γ) < −2H . The maximum 12

principle Lemma 4.6 and part ii of Lemma 4.4 implies u > w + M in ∆ for any constant M, a contradiction. Hence κ(γ) ≡ 2H. Now suppose u → −∞ for any approach to interior points of γ. Then by Corollary 4.7 κ(γ) ≥ −2H. If κ > −2H on some subarc γ 0 ⊂ γ. Then by shrinking γ 0 if necessary, we can find a subdomain ∆ ⊂ D bounded by γ 0 and an arc Γ with curvature (with respect to the interior of ∆) satisfying κ(Γ) < 2H . Then Lemma 4.6 and part i of Lemma 4.4 implies u < w − M in ∆ for any constant M, a contradiction. Thus κ(γ) ≡ −2H. The proof of Theorem 4.8 contains the following useful result. Lemma 4.9. Let u be a solution of (2.1) in a domain D bounded in part by an arc γ and suppose m ≤ u ≤ M on γ. Then there is a constant c=c(D) such that for any compact C 2 subarc γ 0 ⊂ γ, (i) if κ ≥ 2H on γ 0 with strict inequality except for isolated points , there is a neighborhood ∆ of γ 0 in D such that u ≥ m − c in ∆. (ii) if κ > −2H on γ 0 , there is a neighborhood ∆ of γ 0 in D such that u ≤ M + c in ∆.

5

Flux formula on boundary arcs

In this section we collect some flux formulas we will need to develop our existence theory for solutions with infinite boundary values. These formulas are exactly the same as in section 4 of [9] with almost identical proof. The only essential difference is that we use the barrier in part ii of Lemma 4.4 instead of Delaunay solutions. Therefore we only briefly indicate the idea of the proofs which are already contained in the proof of the general maximum principle Theorem 2.2. Let u ∈ C 2 (D) ∩ C 1 (D) be a solution of (2.1) in a domain D . Then

13

integrating (2.1) over D gives ∇u , ν > ds W

Z (5.11)

2HA(D) =

< ∂D

where A(D) is the area of D and ν is the outer normal to ∂D. The right hand integral is called the flux of u across ∂D. Let γ be a subarc of ∂D (homeomorphic to [0, 1]). Even if u is not differentiable on γ we can define the flux of u across γ as follows. Definition 5.1. Choose Γ to be a simple smooth curve in D so that γ ∪ Γ bounds a simply connected domain ∆Γ . We then define the flux of u across γ to be Z (5.12)

Fu (γ) = 2HA(∆Γ ) −

< Γ

∇u , ν > ds W

The last integral in (5.12) is well-defined as an improper integral. To see that this definition is independent of Γ, let Γ0 be another choice of curve and consider the 2-chain R with oriented boundary Γ0 − Γ. Using (2.1) and the divergence theorem on R gives Z 2HA(∆Γ0 ) − 2HA(∆Γ ) = Γ0

∇u < , ν > ds − W

Z < Γ

∇u , ν > ds . W

Therefore the definition makes sense. If u ∈ C 1 (D ∪ γ), then Fu (γ) = ∇u ,ν W

R γ


ds.

The following lemma is now obvious.

Lemma 5.2. Let u be a solution of (2.1) in a domain D and let Γ be a piecewise C 1 curve in D. Then Z Z ∇u ∇u 2HA(D) = < , ν > ds and < , ν > ds ≤ |Γ| . W W ∂D Γ

14

Lemma 5.3. Let D be a domain bounded in part by a piecewise C 2 arc Γ satisfying κ ˆ (Γ) ≥ 2H. Let u be a solution of (2.1) in D which is continuous on Γ. Then Z ∇u < < |Γ| . , ν > ds W Γ

(5.13)

Proof. It suffices to prove (5.13) for a small subarc γ of Γ. To this end let P ∈ Γ and let D = D ∩ B (P ). Then by Theorem 3.2 there is a solution v of (2.1) in D with v = u + 1 on γ and v = u on the remainder of the boundary. Set w = v − u. Then as in the proof of the general maximum principle, Z Z ∇v ∇u ∇v ∇u 0< < ∇w, − > dV = < − , ν > ds . Wv Wu Wv Wu D γ Hence Fu (γ) < Fv (γ) ≤ |γ| .

Lemma 5.4. Let D be a domain bounded in part by an arc γ and let u be a solution of (2.1) in D. Then (i) if u tends to +∞ on γ, we have (ii) if u tends to −∞ on γ, we have

∇u , ν > ds = |γ|. W R < ∇u , ν > ds = −|γ|. W γ

R

γ


0 and let δ be arbitrarily small. Let γ 0 be a subarc of γ of length a small multiple of √  . Choose an arc Λ of constant curvature −2α = −(2H + 4) (with respect to its normal pointing exterior of D) so that the distance between Λ and γ 0 satisfies δ < d(Λ, γ 0 ) ≤ δ + C|γ 0 |2 . Then by by Remark 4.5, the subsolution w of Lemma 4.4 part (ii) (with Γ replaced by Λ) is well-defined in an annulus of size proportional to  ( with γ 0 strictly interior) and w is an approximate supersolution with flux (5.14)

Fw (γ 0 ) ≥ (1 − (δ + C|γ 0 |2 ))|γ 0 | . 15

Now choose a simple arc Γ joining the endpoints of γ 0 so that γ 0 ∪ Γ bounds a simply connected domain ∆ which is contained in the annulus. Set ϕ = u1 − u2 where u1 = u and u2 = w . We may assume ϕ > 0. Let M0 be chosen so large that the the linear measure of Γ ∩ {ϕ > M0 } is less than  . 8

Let Γ0 be the subarc of Γ with endpoints at arc length distance

 16

from

the endpoints of Γ. For M > supΓ0 ϕ, let ∆M be the component of the set ∆ ∩ {ϕ < M } containing Γ0 in its closure. Then ∆M will be bounded by a curve γM with endpoints on Γ close to those of γ 0 (on which ϕ = M ), by subarcs ΓM ⊂ Γ and possibly by other curves γ˜M (with endpoints on Γ) on which ϕ = M . Further we may always choose M so that ∂∆M is piecewise smooth. Now let

Z < ∇ϕ,

J= ∆M

∇u1 ∇u2 − > dV W1 W2

Since div ϕ(

∇u1 ∇u2 ∇u1 ∇u2 ∇u1 ∇u2 − ) =< ∇ϕ, − > +ϕ(div − div ) W1 W2 W1 W2 W1 W2

in D, we can apply the divergence theorem and obtain Z Z ∇u1 ∇u2 ∇u1 ∇u2 < ϕ< J = M − , ν > ds + − , ν > ds W1 W2 W1 W2 γM ΓM Z ∇u1 ∇u2 (5.15) + M < − , ν > ds + O(M |∆|) W1 W2 γ˜M = J1 + J2 + J3 + O(M |∆|). We can estimate J2 as follow: (5.16)

 J2 ≤ M + 2|Γ|M0 4

˜ M of ΓM ; then To estimate J3 we complete γ˜M to closed curves by subarcs Γ Z Z ∇u1 ∇u2 ∇u1 ∇u2 J3 = M < − , ν > ds − M < − , ν > ds W1 W2 W1 W2 ˜M γ˜M +Γ Γ˜M 16

Since u2 is a subsolution and u1 is a solution of (2.1),  J3 ≤ M . 4

(5.17)

From (5.15), (5.16), (5.17) and the non-negativity of J, we find Z  M0 ∇u1 ∇u2 (5.18) 0≤ − , ν > ds + + 2|Γ| + O(|∆|) . < W1 W2 2 M γM Consider now the domain ∆0M bounded by γ 0 , γM and parts of Γ. Then Z Z ∇u1 ∇u2 ∇u1 ∇u2 < < − , ν > ds ≥ − − , ν > ds W1 W2 W1 W2 γ0 −γM  − (5.19) − |∆0M | . 2 Combining (5.18) and (5.19) (remembering the difference in orientation of the common term) gives Z M0 ∇u1 ∇u2 − , ν > ds ≥ − − 2|Γ| + O(|∆|) . (5.20) < W1 W2 M γ0 We can now let M → ∞ and |∆| → 0 in (5.20) to obtain Fu (γ 0 ) ≥ Fw (γ 0 ) −  Recalling (5.14) and letting δ → 0 gives (5.21)

[Fu (γ 0 ) ≥ |γ 0 | − K|γ 0 |3 − 

p Note that we cannot let  tend to zero in (5.21) since |γ 0 | = K |γ| . Instead q we divide our original arc γ into N = K small pieces γ 0 . Then summing (5.21) gives Fu (γ) ≥ |γ| − 2 |γ|3 −

√ K .

Since  is arbitrary, Fu (γ) = |γ|. The following lemma is a simple extension of Lemma 5.4. 17

Lemma 5.5. Let D be a domain bounded in part by an arc γ and let {un } be a sequence of solutions of (2.1) in D with each un continuous on γ. Then (i) if the sequence tends to +∞ uniformly on compact subsets of γ while remaining uniformly bounded on compact subsets of D, we have Z ∇un lim < , ν > ds = |γ| . n→∞ γ Wn (ii) if the sequence tends to −∞ uniformly on compact subsets of γ while remaining uniformly bounded on compact subsets of D, we have Z ∇un lim < , ν > ds = −|γ| . n→∞ γ Wn . We also will need the following variant of Lemma 5.5. Lemma 5.6. Let D be a domain bounded in part by an arc γ with κ(γ) = 2H and let {un } be a sequence of solutions of (2.1) in D with each un continuous on γ. Then if the sequence diverges to −∞ uniformly on compact subsets of D while remaining uniformly bounded on compact subsets of γ, we have Z ∇un lim < , ν > ds = |γ| . n→∞ γ Wn We shall omit the proof of Lemmas 5.5 and 5.6 which uses the technique of Lemma 5.4. See the proof of Lemma 4.5 of [9] for more details.

6

The local Harnack inequality and monotone convergence theorem

Let u be a non-negative or non-positive solution of (2.1) in Bρ (P ) for ρ < R(P ), the injectivity radius at P. Then it follows from Theorem 1.1 of [10] that (6.22)

|∇u(P )| ≤ f (

|u(P )| 2 ) , f (t) = eC(1+t) ρ 18

where C depends on ρ and the local geometry of M but is independent of u. Following Serrin [8] we prove Theorem 6.1. Local Harnack inequality There exists a function Φ(t, r) such that (6.23)

|u(Q)| ≤ Φ(m, r)

where m = |u(P )| and r = d(P, Q) the distance from P to Q. For each fixed t, Φ(t, r) is a continuous strictly increasing function defined on an interval 0 ≤ r < ρ(t) with Φ(t, 0) = t , Φ(t, r) → ∞ as r → ρ(t) , where ρ(t) is a continuous strictly decreasing function tending to zero as t tends to infinity. Proof. Suppose u ≥ 0 (otherwise let v = −u). Consider points Q lying on a fixed geodesic emanating from P and let u(r) denote the values of u along this ray. Then from (6.22), du u(r) ≤ f( ). dr ρ−r Define Φ(t, r) by the conditions (6.24)

dΦ Φ = f( ) , Φ(t, 0) = t . dr ρ−r

Then u(r) ≤ Φ(m, r) whenever Φ is well-defined. The solution of (6.24) is implicitly defined by Φ = (ρ − r)v and Z v dv 1 t = log , v0 = . ρ−r ρ v0 f (v) + v Since the integral converges as v tends to infinity, Φ is defined only on an interval 0 ≤ r < ρ(t) where clearly ρ(t) → 0 as t → ∞ . This completes the proof. 19

The existence theorems of the following sections depend upon the limiting behavior of monotone increasing and monotone decreasing sequences of solutions in a fixed domain. As in the Euclidean case, we have the following consequence of Theorems 4.8 and 6.1. Theorem 6.2. Monotone convergence theorem Let {un } be a monotonically increasing or decreasing sequence of solutions of (2.1) in a fixed domain D. If the sequence is bounded at a single point of D, there exists a non-empty open set U ⊂ D such that {un } converges to a solution in U. The convergence is uniform on compact subsets of U and the divergence is uniform on compact subsets of V = D \ U . If V is nonempty, ∂V consists of arcs of curvature ±2H and parts of ∂D. These arcs are convex to U for increasing sequences and concave to U for decreasing sequences. In particular, no component of V can consist of a single interior arc. We have more information about the set V if we have more knowledge about the boundary behavior of the sequence {un } and the curvature of ∂D. Lemma 6.3. Let D be a domain bounded in part by an arc C with κ ˆ (C) ≥ 2H. Let {un } be an increasing or decreasing sequence of solutions of (2.2) in D with each un continuous in D ∪ C. Suppose γ is an interior arc of D of curvature 2H forming part of the boundary of V. Then γ cannot terminate at an interior point of C if {un } either diverges on C to ±∞ or remains uniformly bounded on compact subsets of C.

Proof. Let γ be an arc in ∂V as in Lemma 6.3 that terminates at an interior point P of C. By considering only a small neighborhood of P, we may assume that C is C 2 . By Theorem 4.8 the sequence {un } cannot diverge to −∞ on C. Moreover, if the curvature of C is not identically 2H and {un } remains uniformly bounded on compact subsets of C, Lemma 4.9 insures that 20

a neighborhood of C is either contained in U or V, a contradiction. Hence assume κ(C) ≡ 2H. Suppose {un } diverges to +∞ on C and there exists exactly one such γ terminates at P. Let Q be a point of γ close to P and choose R on C close to P so that the geodesic segment RQ lies in U. Let T _

be the triangle formed by RQ and the constant curvature 2H arcs QP and _

P R. Then by (5.11) _

(6.25)

_

2HA(T ) = Fun (QP ) + Fun (P R) + Fun (RQ)

while by Lemma 5.5 _

_

_

_

lim Fun (QP ) = | QP | , lim Fun (P R) = | P R | .

(6.26)

n→∞

n→∞

From (6.25), (6.26) and Lemma 5.3 we can conclude _

_

| QP | + | P R | 2HA(T ) ≥ −1 |RQ| |RQ|

(6.27)

Keeping P fixed, we move Q to Q’ and R to R’ along the same arcs so _

_

_

_

that | Q0 P | = λ| QP | and | P R0 | = λ| P R | and form the triangle T 0 by joining Q’ to R’ by a geodesic. Then the left hand side of (6.27) tends to zero as λ → 0 while the right hand side of (6.27) remains uniformly positive, a contradiction. The only other possibility is that two arcs γ1 and γ2 terminate at P. Then again we can find a triangle T ⊂ U whose edges are two constant curvature 2H arcs and a geodesic segment as before (perhaps with ∂T ∩ C = {P }). The same argument gives a contradiction. In case the sequence remains uniformly bounded on compact subsets of C and there is exactly one γ, we choose R on C so that T is contained in V. By Lemma 4.4 the sequence must be divergent to −∞ in V. We now reach a contradiction as above by using Lemma 5.5. If there are two arcs terminating at P, then V is necessarily the convex lens domain formed by γ1 and γ2 . Choose the point Q on γ1 and R on γ2 and form T in V . Then (6.25), (6.26) and (6.27) still hold and we reach a contradiction as before. 21

7

Existence and uniqueness

In this section we prove existence theorems for constant mean curvature graphs in which the boundary values ±∞ are allowed on entire arcs of the boundary. As we have seen in Theorem 4.8, these arcs must be of curvature ±2H with respect to the domain. Definition 7.1. Admissible domain We say a bounded domain D is admissible if it is simply connected and ∂D is piecewise C 2 and consists of three set of C 2 open arcs {Ai } , {Bi } and {Ci } satisfying κ(Ai ) = 2H , κ(Bi ) = −2H and κ(Ci ) ≥ 2H respectively (with respect to the interior of D). We suppose that no two of the arcs Ai and no two of the arcs Bi have a common endpoint. In addition if the family {Bi } is nonempty, we assume that the set D∗ formed by replacing each arc Bi by Bi∗ , the geodesic reflection of Bi across its endpoints, is a well-defined domain. Finally we assume that there is a bounded subolution of (2.1) in D∗ (recall this is automatic if κ ˆ (∂D∗ ) ≥ 2H or if D ⊂ H 2 and H ≤ 12 ) so that by Theorem 3.2, the Dirichlet problem with finite data is well-posed in D∗ ) Remark 7.2. Note that we do not insist that κ ˆ (P ) ≥ 2H for P an endpoint of the Ai or Bi so our definition allows admissible domains to be nonconvex. Definition 7.3. Dirichlet Problem Given an admissible domain D, the generalized Dirichlet problem is to find a solution of (2.1) in D which assumes the value +∞ on each Ai , −∞ on each Bi and assigned continuous data on each of the open arcs Ci . Note that the continuous data is allowed to become unbounded at the endpoints. We shall need one more definition before we can start considering some special cases of the Dirichlet problem.

22

Definition 7.4. Admissible polygon Let D be an admissible domain. We say that P is an admissible polygon if P is a simple domain contained in D with ∂P piecewise smooth consisting of arcs of constant curvature κ = ±2H with vertices chosen from among the endpoints of the families {Ai } , {Bi } and {Ci }. For an admissible P, let α and β be the total length of the arcs in ∂P belonging to {Ai } and {Bi } respectively. Finally, let ` be the perimeter of P and A(P ) be the area of P. Proposition 7.5. Consider the Dirichlet problem in an admissible domain D and suppose the family {Bi } is empty, κ(Ci ) > 2H and the assigned data f on the arcs Ci are bounded below. Then there exists a solution if and only if (7.28)

2α < ` + 2HA(P) for all admissible polygons P.

Proof. Let un be the solution of (2.1) in D such that  n on ∪Ai un = min (n, f ) on ∪Ci Such a solution exists and is unique by Theorem 3.2. Moreover by the general maximum principle, the sequence {un } is monotone increasing so the monotone convergence theorem 6.2 applies. Suppose V is nonempty. By Lemma 6.3 an interior arc of D which bounds V must be of curvature 2H and can terminate only at an endpoint of some Ai or Ci . Moreover by Lemma 4.9, a neighborhood of each Ci is contained in U. Therefore the boundary of each component of V is an admissible polygon P with vertices among those of the Ai and Ci . Also the curvature 2H arcs forming the boundary which are not among the Ai are concave to V. By Lemma 5.2 applied to each un in P, (7.29)

2HA(P) = Fn (∂P − Σ0 Ai ) + Fn (Σ0 Ai )

23

where Σ0 Ai is the union of the arcs Ai which are part of P. Then by Lemma 5.5 (7.30)

lim Fn (∂P − Σ0 Ai ) = −(` − α) .

n→∞

But |Fn (Σ0 Ai )| ≤ α, hence ` − α ≤ α − 2HA(P), contradicting our assumption (7.28). Thus V is empty and the sequence converges uniformly on compact subsets of D to a solution u. Since each un is uniformly bounded in a neighborhood of each Ci by Lemma 4.9, a standard barrier argument shows that u = f on ∪Ci . Using (7.29) and Lemmas 5.3 and 5.4, the necessity of (7.28) is clear. This completes the proof. Similarly, we have Proposition 7.6. Consider the Dirichlet problem in an admissible domain D and suppose the family {Ai } is empty, κ(Ci ) > 2H and the assigned data f on the arcs Ci are bounded above. Then there exists a solution if and only if (7.31)

2β < ` − 2HA(P) for all admissible polygons P.

The proof is completely analogous using monotone decreasing sequences in D∗ , which must diverge in a neighborhood of each Bi∗ by Lemma 4.9. The simple case of one B and one C already illustrates the importance of considering the domain D∗ , which is a key idea. We now use Propositions 7.5 and 7.6 to construct some useful barriers and to remove the assumption κ(Ci ) > 2H. Example 7.7. Let B = Bδ (P ) be a ball of small radius δ and let Q and R be “antipodal” points on ∂B. Choose points Q1 and Q2 on ∂B and symmetric with respect to the geodesic through QPR. Now let B1 be an arc of curvature 24

-2H (as seen from P) joining Q1 and Q2 and set A1 = B1∗ . Let R1 and R2 on ∂B be reflections of Q1 and Q2 (with respect to the geodesic orthogonal to QPR through P)and define B2 and A2 . Then for δ small compared with H, the domain B + bounded by A1 , A2 and parts of ∂B satisfies the conditions of Proposition 7.5 and similarly the domain B − bounded by B1 , B2 and parts of ∂B satisfies the conditions of Proposition 7.6. Let u+ be the solution of (2.1) in D+ with boundary values +∞ on A1 ∪ A2 and the constant value M on the remainder of the boundary. Similarly let u− be the solution of (2.1) in D− with boundary values −∞ on B1 ∪ B2 and the constant value -M on the remainder of the boundary.

Using these examples we prove Proposition 7.8. Let D be a domain bounded in part by an arc γ and let {un } be a sequence of solutions of (2.1) in D which converge uniformly on compact subsets of D to a solution u. Suppose each un is continuous in D ∪γ. Then (i). Suppose the boundary values of un converge uniformly on compact subset of γ to a bounded limit f. If κ(γ) ≥ 2H, then u is continuous in D ∪ γ and u = f on γ. (ii). If κ(γ) = 2H and the boundary values of un diverge uniformly to +∞ on compact subsets of γ, then u takes on the boundary values +∞ on γ. (iii). If κ(γ) = −2H and the boundary values of un diverge uniformly to −∞ on compact subsets of γ, then u takes on the boundary values −∞ on γ. Proof. (i). It suffices to prove that the sequence {un } is uniformly bounded in the intersection of D with a neighborhood of any interior point P of γ. Orient the ball B of Example 7.7 so that geodesic joining QPR is tangent to ∂D. We may choose the points Qi i = 1, 2 and δ small so that the boundary

25

arc joining Q2 and R2 lies is in a compact subset of D. Then if M is large enough, un ≤ u+ in D ∩ B + and un ≥ u− in D ∩ B − . Therefore the sequence is uniformly bounded in a neighborhood of P. (ii). Let P be an interior point of γ, Similarly as in (i), we obtain that there exists M large enough so that un ≥ −M in N = B (P ) ∩ D. Let vm be the solution of (2.1) in N with boundary values m on γ ∩ B (P ) and -M on the remaining boundary. By the general maximum principle un ≥ vm for n sufficiently large so u ≥ vm in N. In particular, u(P ) > m for every m and u must take on the value +∞ at P. (iii). Again for P interior to γ, un ≤ M in N = B (P ) ∩ D. Let vm be the solution of (2.1) in N with boundary values -m on γ ∩ B (P ) and M on the remaining boundary. By the general maximum principle un ≤ vm for n sufficiently large so u ≤ vm in N. Since the vm are monotonically decreasing (and converges to a solution with boundary values + − ∞ on γ ∩ B (P )), u must take on the value −∞ at P. We can now extend Propositions 7.5 and 7.6 to allow the arcs Ci to satisfy κ(Ci ) ≥ 2H. The only change needed in the proof of Proposition 7.5 is to use part (i) of Proposition 7.8 to show that the solution takes on the required boundary data on the arcs Ci . The extension of Proposition 7.6 is more delicate since if κ(Ci ) = 2H for some i, we do not know that the sequence is bounded below in a neighborhood of Ci . However by Lemma 6.3, a neighborhood of Ci is either contained in U or V. We have already handled the former case. In the latter case, consider a component of V whose boundary is an admissible polygon P (with vertices among those of the Bi and Ci ) containing a subset Σ0 Ci of the Ci of curvature 2H and a subset Σ0 Bi of the Bi . The interior arcs of D which are in P are convex to

26

V. By Lemma 5.2 applied to each un in P, (7.32)

2HA(P) = Fn (Σ0 Bi ) + Fn (Σ0 Ci ) + Fn (∂P − Σ0 Bi − Σ0 Ci )

Then by Lemma 5.5 (7.33)

lim Fn (∂P − Σ0 Bi − Σ0 Ci ) = ` − β − Σ0 |Ci | .

n→∞

and by Lemma 5.6 lim Fn (Σ0 Ci ) = Σ0 |Ci |

(7.34)

n→∞

But |Fn (Σ0 Bi )| ≤ β, hence 2HA(P) ≥ −β + Σ0 |Ci | + (` − Σ0 |Ci | − β) = ` − 2β , contradicting our assumption (7.31). Thus V is empty and the sequence converges uniformly on compact subsets of D to a solution u. Finally, we use parts (i) and (iii) of Proposition 7.8 to show that our solution achieves the boundary values. We state these results as Theorem 7.9. Consider the Dirichlet problem in an admissible domain D and suppose the family {Bi } is empty and the assigned data f on the arcs Ci are bounded below. Then there exists a solution if and only if (7.35)

2α < ` + 2HA(P) for all admissible polygons P.

Theorem 7.10. Consider the Dirichlet problem in an admissible domain D and suppose the family {Ai } is empty and the assigned data f on the arcs Ci are bounded above. Then there exists a solution if and only if (7.36)

2β < ` − 2HA(P) for all admissible polygons P.

In the following theorem we allow both families {Ai } and {Bi } to occur and allow the data f on the {Ci } to be unbounded both above and below as we approach the endpoints. 27

Theorem 7.11. Consider the Dirichlet problem in an admissible domain D and suppose the family {Ci } is nonempty. Then there exists a solution if and only if (7.37)

2α < ` + 2HA(P) and 2β < ` − 2HA(P)

for all admissible polygons P. Proof. By Theorem 7.9 the first condition of (7.37) guarantees the existence of a solution u+ of (2.1) in D∗ such that  on ∪Ai  +∞ 0 on ∪Bi∗ u+ =  max (f, 0) on ∪Ci Similarly by Theorem 7.10, the second condition of (7.37) guarantees the existence of a solution u− of (2.1) in D∗ such that  on ∪Bi  −∞ − 0 on ∪Ai u =  min (f, 0) on ∪Ci Now let un be the solution of (2.1) in D∗ such that  on ∪Ai  n −n on ∪Bi∗ un =  fn on ∪Ci where fn is the truncation of f above by n and below by -n. By the general maximum principle, un ≤ u+ in D∗ and u− ≤ un in D . Therefore the sequence {un } is uniformly bounded on compact subsets of D and so a subsequence converges uniformly on compact subsets to a solution u in D. By Proposition 7.8, u takes on the assigned boundary data. The necessity of the conditions (7.37) follows essentially as in the Theorems 7.9 28

and 7.10. We turn next to the important case when the family {Ci } is empty. Theorem 7.12. Consider the Dirichlet problem in an admissible domain D and suppose the family {Ci } is empty. Then there exists a solution if and only if (7.38)

α = β + 2HA(D)

and (7.39)

2α < ` + 2HA(P) and 2β < ` − 2HA(P)

for all other admissible polygons P. Proof. Let vn be the solution of (2.1) in D∗ with boundary values n on each Ai and 0 on each Bi∗ . For 0 < c < n we define for n ≥ 1 Ec = {vn − v0 > c} ∩

Fc = {vn − v0 < c};

we suppress the dependence of these sets on n. Let Eci and Fci denote respectively the components of Ec and Fc whose closure contains respectively Ai and Bi∗ . By the general maximum principle, Ec = ∪Eci and Fc = ∪Fci . If c is sufficiently close to n, the sets {Eci } will be distinct and disjoint (to see this, note that we can seperate any two of the Ai by a curve joining two of the Bi∗ on which vn − v0 is bounded away from n). Now define µ(n) to be the infimum of the constants c such that the sets {Eci } are distinct and disjoint. The sets {Eµi } will again be distinct although there must be at least i

j

one pair (i, j) , i 6= j such that E µ ∩ E µ is nonempty. This implies that given any Fµi there is some Fµj distinct from it. Now let u+ i , i = 1, . . . , k be the solution of (2.1) in D∗ taking on the boundary values +∞ on Ai and 0 on the remaining boundary. This solution exist by Theorem 7.9 since the 29

solvability condition 7.35 follows trivially from (7.38), (7.39). Also let u− i be ˜ bounded by ∪Ai , B ∗ , ∪j6=i Bj taking on the solution of (2.1) in the domain D i the boundary values −∞ on ∪j6=i Bj and 0 on the remainder of the boundary. We claim this solution exists by Theorem 7.10. To verify (7.36), we need ˜ which contain the lens domain L only consider admissible polygons P˜ in D formed by Bi and Bi∗ . Let P be the corresponding admissible polygon for D formed by deleting L. By (7.38), (7.39) we have 2β = 2(|Bi | + Σ0j6=i |Bj |) ≤ ` − 2HA(P) , or equivalently ˜ + (2HA(L) − 2|Bi |) . 2β˜ = 2Σ0j6=i |Bj | ≤ `˜ − 2HA(P) However since a solution (for example vn ) exists in L, we have by Lemma 5.3 2HA(L) < 2|Bi | so conditon (7.36) is satisfied. We now set − − ∗ u+ = min u+ i in D and u = min ui in D i

i

∗ We note that if we compare each u+ i to a fixed bounded subsolution in D

(which exists since D is admissible), then by the general maximum principle there is a constant N > 0 such that u+ i > −N, i = 1, . . . , k. Finally we set un = vn − µ(n). We now claim that un ≤ u+ + M in D∗ and un ≥ u− − M in D . where M = N + supD∗ |v0 |. S Suppose un > v0 at some point P. Then vn − v0 > µ(n) at P so that P is in some Eµi . Applying the general maximum principle in the domain Eµi , we obtain + un ≤ u+ i + N + sup |v0 | ≤ u + M at P . i Eµ

30

On the other hand, suppose un < v0 at some point P ∈ D. Then vn − v0 > µ(n) at P so that P is in some Fµi . By what has been shown above, there is a corresponding j = j(i) so that Fµi ∩ Fµj = ∅. Applying the general maximum principle in Fµi , we obtain − un ≥ u− j − sup |v0 | ≥ u − M . Fµi

Therefore the claim is justified and the sequence {un } is uniformly bounded on compact subsets of D. By the compactness principle, a subsequence {un } converges uniformly on compact subsets of D to a solution u. We still must show that u takes on the required boundary values. We observe that a subsequence µ(n) diverges to +∞ otherwise we can extract a subsequence converging to a finite limit µ0 . Each un would then be bounded below in D∗ uniformly in n, and the boundary values of un would tend uniformly on compact subsets of ∪Bi∗ to −µ0 and diverge uniformly to +∞ on ∪Ai . Once again we could find a subsequence converging uniformly on compact subsets to a solution in D∗ . By Propositon 7.8 we would have  +∞ on ∪Ai v= −µ0 on ∪Bi∗ We can now obtain a contradiction to (7.38) by a flux argument. By Lemma 5.2, (7.40)

2HA(D) = Fv (ΣAi ) + Fv (ΣBi ) ,

while by Lemmas 5.3 and 5.4 (7.41)

|Fv (ΣBi )| < β and Fv (ΣAi ) = α .

Combining (7.40) and (7.41) gives α − β < 2HA(D), a contradiction. In the same way, we see n − µ(n) diverges to +∞. Summing up we have shown that the boundary values of un , namely, −µ(n) on ∪Bi∗ and n − µ(n) on ∪Ai 31

diverge to −∞ and +∞ respectively. A now familiar argument shows that un diverges to −∞ on ∪Bi . The necessity of the conditions (7.38) and (7.39) is straigtforward. The theorem is proved. We end the section with a maximum principle that is valid for solutions with infinite boundary values. This result immediately proves uniqueness for the Dirichlet problem in an admissible domain if the family {Ci } is nonempty and uniqueness up to translation if the family {Ci } is empty. Theorem 7.13. Let D be a domain whose boundary contains two families {Ai } and {Bi }of arcs satisfying κ(Ai ) = 2H and κ(Bi ) = −2H. Let E be a finite set of exceptional points on ∂D and let C denote the set of points in ∂D \ E which are not in the closure of the families {Ai } and {Bi }. Let u1 and u2 be solutions of (2.1) in D taking on the values +∞ on each Ai and −∞ on each Bi . If C is nonempty, assume lim sup u1 − u2 ≤ 0 for any approach to points of C. If C is empty, assume u1 ≤ u2 at some point P in D. Then in either case u1 ≤ u2 in D. Proof. For simplicity we give the proof in case E is empty; the general case requires only a slight modification. As in the proof of the general maximum principle, let N and ε be positive constants with N large and ε small. Define  if u1 − u2 ≥ N  N −ε u1 − u2 − ε if ε < u1 − u2 < N ϕ=  0 if u1 − u2 ≤ ε Then ϕ is Lipschitz and vanishes in a neighborhood of any point of C. Moreover, 0 ≤ ϕ < N and ∇ϕ = ∇u1 − ∇u2 in the set where ε < u1 − u2 < N and ∇ϕ = 0 almost everywhere in the complement of this set. Let Dε,δ denote the set of points in D whose distance from ∂D is at least δ and whose distance from the endpoints of {Ai } and {Bi } is at least ε, where δ < ε. For small δ, Γ := ∂Dε,δ consists of arcs A0i parallel to Ai , Bi0 parallel to Bi , C 0 parallel to

32

C (all at distance δ) and small spherical arcs of radius ε with centers at the endpoints of the Ai and Bi . Now let Z ∇u1 ∇u2 J = ϕ< − , ν > ds, W1 W2 Γ where ν is the outer normal to Γ with respect to the domain Dε,δ . We can estimate J from above in two ways: (7.42) Z Z Z ∇u2 ∇u1 ∇u2 J = ϕ(1− < , ν >)ds− ϕ(1− < , ν >)ds ≤ M (1− < , ν >)ds W2 W1 W2 Γ Γ Γ\C 0 or (7.43) Z Z Z ∇u1 ∇u2 ∇u1 J = ϕ(1+ < , ν >)ds− ϕ(1+ < , ν >)ds ≤ M (1+ < , ν >)ds W1 W2 W1 Γ Γ Γ\C 0 since for small enough δ, ϕ ≡ 0 in a neighborhood of C 0 . Now applying Lemmas 5.2 and 5.4 in the domain bounded by Ai , A0i and the two spherical arcs of radius ε gives Z ∇uj (1− < (7.44) , ν >)ds = O(ε) , j = 1, 2 . Wj A0i Similarly considering the domain bounded by Bi , Bi0 and the two spherical arcs of radius ε gives Z ∇uj (7.45) (1+ < , ν >)ds = O(ε) , j = 1, 2 . Wj Bi0 where ν is the outer normal to Bi0 with respect to the domain Dε,δ . Using (7.42, (7.43), (7.44) and (7.45) we find that J = O(N ε). On the other hand by Lemma 2.1 Z (7.46)

< ∇ϕ,

J= Dε,δ

∇u1 ∇u2 − > dV ≥ 0 . W1 W2

Now consider a sequence of values of ε decreasing to zero with corresponding values δ(ε) also decreasing to zero. Then Dε,δ is increasing so that J is 33

C2

C1 dα I

cα b

b α/2

O

Figure 1: Triangle T (α) monotone increasing (recall Lemma 2.1). It follows that ∇u1 ≡ ∇u2 in the set where 0 < u1 − u2 < N . Since N is arbitrary, ∇u1 ≡ ∇u2 in the set where u1 > u2 . It follows that if the set u1 > u2 contains any nontrivial component, then u1 ≡ u2 +positive constant in that component and so by the maximum principle, u1 ≡ u2 + positive constant in D. By the assumptions of the theorem, the constant must be nonpositive, a contradiction. Thus u1 ≤ u2 in D. The same proof holds if C is empty.

8

Examples of Scherk type

Suppose M = H2 and H 6 12 . We now construct polygons with an arbitrary number of sides that are admissible domains satisfying the Jenkins-Serrin conditions. Let Sθ be the sector of angle θ in a Euclidean circle of radius b and center O and denote by Tα the triangle (O, C1 , C2 ), where the angle at O is α and the points C1 , C2 are on the circle. Let I be the mid-point of [C1 , C2 ]

34

and let 2dα := dist(C1 , C2 ) and cα := dist(O, I), see figure 1. We define a lens LH (2d), to be a region bounded by a geodesic of length 2d and a segment of curvature κ = 2H. When the geodesic is fixed, there are two lenses with this boundary. By T (α) + LH (2dα ) we mean T (α) union the lens LH (2dα ) whose intersection with T (α) is the geodesic [C1 , C2 ]. Also T (α)−LH (2dα ) is the complement in T (α) of the lens LH (2dα ) having [C1 , C2 ] in its boundary and intersecting T (α) in a region; (cf. figure 2). We consider the domain D(α, β) ⊂ Sα+β defined as (see figure 3): D(α, β) = (T (α) + LH (2dα )) ∪ (T (β) − LH (2dβ )). The boundary ∂D(α, β) consists of two geodesics of length b, and two arcs A, B of constant curvature κ = 2H, with curvature vectors pointing inside D(α, β) on A and outside D(α, β) on B. Proposition 8.1. Let θ ∈ [0, π]. There exists Sα ∪ Sβ = Sθ , a partition of Sθ such that: |A| − |B| = 2H|D(α, β)| where |D(α, β)| = Area(D(α, β)). The perimeter |∂D(α, β)| = 2b + |A| + |B|. Proof. In this proof we will use several formulas of hyperbolic geometry which we derive in the appendix.

We consider the function F (α, β) =

2H|D(α, β)| + |B| − |A|. By Lemma 9.3 (here we are using H 6

1 ), 2

we

have F (t) = F (tθ, (1 − t)θ)) is a continuous function on t ∈ [0, 1] with F (0) = F (0, θ) > 0 and F (1) = F (θ, 0) < 0. Fix an integer n, and apply Proposition 8.1 to the sector S 2π . We do n n

consecutive reflections through the geodesic sides of S 2π to obtain a simple n

closed curve ∂D(n) composed of n arcs of constant curvature κ = +2H and n arcs with curvature κ = −2H. Since the Euclidean circle centered at O 35

L(2dα)

β

α

T (β) − L(2dβ )

T (α) + L(2dα)

Figure 2: Domain D(α, 0) et D(0, β) of radius b has curvature greater than one, such domains D are inscribed in this circle. Here the sign of κ is with respect to the disk D(n) bounded by ∂D(n). It is easy to see that D(n) is an admissible domain (see definition 7.1) with only boundary components of type Ai and Bi . Thus to solve the Dirichlet problem with boundary data +∞ on the Ai and −∞ on the Bi of ∂D(n), it suffices to show the conditions (7.38) and (7.39) of theorem 7.12 are satisfied. Now we denote D(n) by D and ∂D(n) by ∂D. So we must show that for P an admissible polygon inscribed in D we have (7.38)

n|A| = n|B| + 2HA(D), for P = D,

and 7.39a

2λ|A| < |∂P| + 2HA(P)

7.39b

2µ|B| < |∂P| − 2HA(P) 36

B

kg = 1 kg = −1

A

D(α, β)

0

Figure 3: Domain D(α, β) for P = 6 D and λ, µ the number of Ai and Bi ’s sides in ∂P respectively. The condition (7.38) is satisfied for P = D by construction: the perimeter of P is given by |∂D| = n|A| + n|B| and (by Proposition 8.1) n(|A| − |B|) = 2HA(D). Now let P an admissible polygon in D, P = 6 D. Definition 8.2. An elementary admissible polygon P is an admissible polygon contained in a half-disk such that (D − P) have only one connected component. We first check that the Jenkins-Serrin inequalities (7.39) are satisfied for elementary polygons. The elementary polygons have only one additional side C, the boundary of a lens which can be concave or convex with respect to P. When it is convex (resp. concave) we denote the admissible polygon by Pi (resp. Pe ). We have three cases, depending on (λ, µ) = (k, k), (k + 1, k) 37

or (k, k + 1) with k chosen so that λα + µβ 6 π. The perimeter is given by |∂P| = λ|A| + µ|B| + |C|. We remark that |C| is the length of an equidistant curve which bound a lens associated to a geodesic of length 2dλα+µβ . The relationship between the area is A(Pi (λ, µ)) = A(Pe (λ, µ)) + 2A(LH (2dλα+µβ )). Thus it suffices to prove

2λ|A| − |∂P(λ, µ)| < A(Pe (λ, µ)) < A(Pi (λ, µ)) < |∂P(λ, µ)| − 2µ|B| We have

A(Pi (λ, µ)) = k|D(α, β)| + (λ − k)|D(α, 0)| + (µ − k)|D(0, β)| −|T (λα + µβ)| + |LH (2dλα+µβ )|. By definition of F (α, β) and Lemma 9.3 of the appendix we have 2HA(Pi (λ, µ)) − |∂P(λ, µ)| + 2µ|B| = (λ − k)F (α, 0) + (µ − k)F (0, β) −F (0, λα + µβ) < 0 To check the sign, we remark F (α, 0) < 0, F (0, λα + µβ) > 0 and since dα < dλα+µβ , we use |A| < |C|: F (0, β) − F (0, λα + µβ) = −F (α, 0) − F (0, λα + µβ) < (1 − 4H 2 )(|A| − |C|) +4H arcsin(2H tanh dα ) − 4H arcsin(2H tanh dλα+µβ ) < 0 Similarly we have

2HA(Pe (λ, µ)) + |∂P(λ, µ)| − 2λ|A| = (λ − k)F (α, 0) + (µ − k)F (0, β) −F (λα + µβ, 0) > 0 38

To check the sign, we remark F (0, β) > 0, F (λα + µβ, 0) < 0 and F (α, 0) − F (λα + µβ, 0) > 0 by lemma 9.3. Now let P be an admissible polygon. We can write P = D −

P

Pi where

each Pi is an elementary admissible polygon or its complement in D. To conclude we prove the following lemma: Lemma 8.3. We consider P an admissible polygon with P = D−

P

Pi for a

finite collection of admissible polygons Pi satisfying |∂Pi | = λi |A|+µi |B|+|Ci | 2λi |A| − |∂Pi | < 2HA(Pi ) < |∂Pi | − 2µi |B|. Then 2λ|A| − |∂P| < 2HA(P) < |∂P| − 2µ|B|. Proof. We have |∂P| = (n −

P

λi )|A| + (n −

P

µi )|B| +

P

|Ci |. Using

2HA(P ) = n(|A| − |B|) we have

n(|A| − |B|) +

X

2µi |B| −

X

|∂Pi | < 2HA(P ) −

X

2HA(Pi )

X X 2HA(Pi ) < n(|A| − |B|) − 2λi |A| + |∂Pi | P P which gives the result by observing λ = n − λi and µ = n − µi . 2HA(P ) −

9 9.1

X

Appendix Hyperbolic geometry

Lemma 9.1. In a Euclidean circle of radius R = tanh 2b , centered at the origin O in the disk model of H, we consider a triangle (O, C1 , C2 ) where the angle at the center is α and dist(O, C1 ) = dist(O, C2 ) = b. The point I is the 39

mid-point of [C1 , C2 ]. With b fixed, we note dist(O, I) = cα , dist(C1 , I) = dα and the area |T (α)| = Area(O, C1 , C2 ). 1) (Pythagoras) cosh b = cosh dα cosh cα 2) sinh dα = sinh b sin(α/2) and tanh cα = tanh b cos(α/2) 3) sin

sinh cα sinh dα |T (α)| = 2 1 + cosh b

Lemma 9.2. We consider the half-lens LH (2d) bounded by a geodesic γ of length 2d and A a constant curvature κ = 2H arc joining the end points of γ. Then the area is A(LH (2d)) = |LH (2d)| = 2H|A| − 2 arcsin(2H tanh d). Proof. By Gauss-Bonnet, |LH (2d)| = 2H|A|−2α where π −α is the exterior angle at the edge of the lens. We use the half-space model of the hyperbolic space. The geodesic segment of hyperbolic length 2d is the vertical Euclidean segment between the points p− = (0, 1 − tanh d) and p+ = (0, 1 + tanh d) in R+,3 . We consider an arc of a circle with radius R = 1/2H, passing by p− and p+ . Then the center of the circle is O = (x, 1) with x2 +tanh2 d = 1/4H 2 . The hyperbolic curvature of the circle is κ = 2H (the point O is at heigth 1). The lens LH (2d) is the region bounded by the arc and the vertical segment between p− and p+ . Elementary trigonometry gives that sin α = 2H tanh d.

Lemma 9.3. We consider the function F (α, β) = 2H(|T (α)| + |LH (2dα )| + |T (β)| − |LH (2dβ )|) + |B| − |A|. with (α, β) ∈ [0, π]2 and α + β = θ ∈ [0, π]. Then we have F (0, β) > 0 and for 0 < α < γ 6 π, 0 > F (α, 0) > F (γ, 0). 40

Proof. We consider F (α, β) = 2H(|T (α)|+|LH (2dα )|+|T (β)|−|LH (2dβ )|)+ |B| − |A| for some value α + β = θ ∈ [0, π] and by the two lemmas above we have  F (α, β) = 4H arcsin

sinh cα sinh dα 1 + cosh b



 + 4H arcsin

sinh cβ sinh dβ 1 + cosh b



−4H arcsin(2H tanh dα ) + 4H arcsin(2H tanh dβ ) + (1 − 4H 2 )(|B| − |A|) When α = 0, we have dα = 0 and then F (0, β) > 0 when π > β > 0. When β = 0, we prove for α ∈]0, π[   sinh cα sinh dα F (α, 0) = 4H arcsin −4H arcsin(2H tanh dα )+(4H 2 −1)|A| < 0. 1 + cosh b To see that we remark that F (0, 0) = 0 by construction and we prove that F (γ, 0) < F (α, 0) for 0 < α < γ < π. We rewrite F (α, 0) − F (γ, 0) = (1 − 4H 2 )(|Aγ | − |Aα |) +4H arcsin

sinh b 1+cosh b


tanh dα cos α/2 − 4H arcsin

sinh b 1+cosh b


tanh dα cos γ/2

+4H arcsin

sinh b 1+cosh b


tanh dα cos γ/2 − 4H arcsin

sinh b 1+cosh b


tanh dγ cos γ/2

+4H arcsin (2H tanh dγ ) − 4H arcsin (2H tanh dα ) When α < γ, we have dα < dγ and then |Aα | < |Aγ |. Then the first line is positive. The second line of the equation above is positive (cos γ/2 < cos α/2). For the last two lines, we can see that the derivative of the function g(t) = 2 arcsin(ta) − 2 arcsin(tb) has the sign of a2 − b2 . With a = tanh dα and b = tanh dγ , we have  g

sinh b cos γ/2 1 + cosh b



and F (α, 0) is decreasing as desired. 41

> g(1) > g(2H)

References [1] Aledo, J. A., Espinar, J. H., G´alvez, J. A, Height estimates for surfaces with positive constant mean curvature in M2 × R, to appear in Illinois J. Math. [2] Gilbarg, D. and Trudinger, N., Elliptic Partial Differential Equations of Second Order, Springer-Verlag 1983. [3] Jenkins, H. and Serrin, J., Variational problems of minimal surface type II: Boundary value problems for the minimal surface equation. Arch. Rat. mech. Anal. 21 (1966), 321–342. [4] Pinheiro, Ana Lucia, The theorem of Jenkins-Serrin in M × R, preprint 2005. [5] Nelli, B. and Rosenberg, H., Minimal surfaces in H 2 × R, Bull. Braz. Math. Soc 33 (2002), 263–292. ¨ [6] Nitsche, J. C. C, Uber ein verallgemeinertes Dirichletsches Problem f¨ ur die Minimalfl¨achengleichung und hebbare Unstetigkeiten ihrer L¨osungen. Math. Ann. 158 (1965), 203–214. [7] Serrin, J., The problem of Dirichlet for quasilinear elliptic differential equations with many independent variables, Philos. Trans. Roy. Soc. London Ser. A 264(1969), 413-496. [8] Serrin, J., The Dirichlet problem for surfaces of constant mean curvature, Proc. London Math. Soc. 21(1970), 361-384. [9] Spruck, J., Infinite boundary value problems for surfaces of constant mean curvature, Arch. Rat. Mech. Anal. 49(1972), 1-31.

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[10] Spruck, J., Interior gradient estimates and existence theorems for constant mean curvature graphs in M × R, preprint. [11] Zhang, S., Curvature estimates for CMC surfaces in three dimensional manifolds, Math. Z. 249 (2005), 613–624.

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