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 .
∗
Université Marne-La-Valle, Cité Descartes, 5 Boulevard Descartes, 77454 Champssur-Marne, Marne-La-Valle, France.
†
Institut de Mathèmatique, 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 κ̂(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 κ̂(P ) to be −∞. Note that κ̂(P ) = κ(P ) at all
regular points of ∂D.
Theorem 3.2. Existence theorem Suppose that κ̂(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<H ≤
1
2
there are no restrictions needed. In all cases, a subsolution of
(2.1)exists, so there is existence and uniqueness for the Dirichlet problem in
the lens domain.
Lemma 3.5. Theorem 3.2 holds when ∂D ∈ C 2+α and κ(P ) ≥ 2H for all
P ∈ ∂D.
Proof. The existence theory developed in [10](specifically Theorems 1.4 and
5.4) along with a special local barrier construction for H in case H <
1
2
(see
Corollary 4.3 of the next section) gives the existence of a unique solution
u ∈ C 2+α (D) of (2.1).
Using Lemma 3.5 we construct local barriers needed in the proof of Theorem 3.2.
Corollary 3.6. Let γ be an arc of constant curvature 2H and Q an interior point of γ. Let Nδ ⊂ Bδ (Q) be a smooth (convex) domain obtained by
smoothing the convex domain bounded by γ and ∂Bδ (Q). Consider smooth
7
boundary data f ≤ 0 on ∂Nδ with f ≡ 0 in a neighborhood of Q and f ≡ −M
on a neighborhood of ∂Bδ (Q). Then there is a smooth solution w− of (2.1)
in Nδ with boundary values f.
The family w− and w+ = −w− depending on M, δ are local lower and
upper barriers at Q.
Proof of Theorem 3.2:
For the existence in case a., we approximate D by smooth (convex) domains Dn ⊂ D satisfying κ(∂Dn ) ≥ 2H by rounding each corner point of ∂D.
Extend the boundary data f continuously to all of D and let fn be smooth
functions in Dn converging uniformly to f. Then Theorems 1.4 and 5.4 of [10]
give a unique smooth solution un in Dn with un = f on ∂Dn and each un is
uniformly bounded independent of n (see also [1]). Thus by the compactness
theorem, a subsequence of the un converges uniformly on compact subsets
to a solution u of (2.1) in D. It remains to show that u = f on ∂D. Fix
P ∈ ∂D, Q ∈ ∂Dn with d(Q, P ) < δ. Given ε > 0 choose δ > 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
a
( 0 sufficiently small independent of a > 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
κ̂ ≥ 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) <
}
C
of fixed size For M = R2 , S and H (here we must
restrict H > 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 κ̂(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
<
Proof. We sketch the proof of (i) which is a slight modification of Lemma
4.3 of [9]; part (ii) is similar. By Theorem 4.8, κ(γ) = 2H. Fix > 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
To estimate J3 we complete γ̃M to closed curves by subarcs Γ̃M of ΓM ; then
Z
Z
∇u1 ∇u2
∇u1 ∇u2
J3 = M
<
−
, ν > ds − M
<
−
, ν > ds
W1
W2
W1
W2
γ˜M +Γ̃M
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 κ̂(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 κ̂(∂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 κ̂(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
the solution of (2.1) in the domain D̃ bounded by ∪Ai , Bi∗ , ∪j6=i Bj taking on
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
only consider admissible polygons P̃ in D̃ which contain the lens domain L
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
2β̃ = 2Σ0j6=i |Bj | ≤ `˜ − 2HA(P̃) + (2HA(L) − 2|Bi |) .
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., Gálvez, 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, Über ein verallgemeinertes Dirichletsches Problem für die Minimalflächengleichung und hebbare Unstetigkeiten ihrer
Lösungen. 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.
42
[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.
43