A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
R OBERT F INN
E NRICO G IUSTI
On nonparametric surfaces of constant mean curvature
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 4, n
o1 (1977), p. 13-31
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Nonparametric
ROBERT
FINN (**) -
ENRICOGIUSTI (***)
dedicated to Jean
Leray
Let
u(x)
define a surface of mean curvature H _--_ 1 over an n-dimen- sional domainD,
thatis,
letu(x)
be a solution in D ofIt was observed
by
S. Bernstein[1], by
E. Heinz[2]
andby
S. S. Chern[3]
that 5) cannot
strictly
contain a closed ballBR
of radius infact,
anintegration
of(1)
overB R yields
where v is the exterior directed unit normal.
Since 1,
we find from(2)
where 2 and ware volume and surface of the unit
n-ball,
and thus R 1.Finn
[4]
showed that if Ð contains the open ballBi
then 0 coincides with this ball and describes a lowerhemisphere,
= Uo -that
is,
the open ballBl admits, essentially, only
asingle
solution of(1).
This result leads
naturally
to theconjecture
that the manifold of such surfaces defined overBR
isprogressively
more restricted as 1.Speci- (*)
This work wassupported
in partby
N.S.F.grant
MPS72-04967A02 and in partby
the C.N.R. Part of the work wascompleted
while the former authorwas
visiting
at Università di Genova in theSpring
of 1974, andpart
while bothwere at Universitat Bonn in the Summer of 1975.
(**)
StanfordUniversity.
(***)
Universith di Trento.Pervenuto alla Redazione 1’8 Marzo 1976.
fically,
Finnconjectured
that if asolution u(x)
of(1)
is defined overB~
and if then all derivatives ofu(x)
are bounded at the center ofBR,
the bound
depending only
on R and in no other way on the functionu(x).
Further,
as R - 1 all solutions inBR
should tend in any fixedBR, uniformly
with all
derivatives,
y to the lowerhemisphere.
We note the
hypothesis
is necessary; the functionwhich defines a
cylinder
inclined withangle
to theplane u = 0,
covers the ball
B1/n
for any a, but-~
oo asIn the
present
paper we establish theconjecture
in the case n = 2 for all Rexceeding
a critical value 0.565 406 233 2 ..., and we show theconjec-
ture fails if
R Ro .
Our
procedure
is to compare agiven
solutionu(x)
ofwith a
particular
solutionv(x),
which will be chosen tomajorize
thegradient
of
u (x )
at the center x = 0. We obtain such av (x )
as solution of a «capil- lary problem »
in a « moon domain » D as indicated inFigure
1. This solu-Figure: 1
tion has also an
independent
interest as a solution of(3)
in a characteristic domaincorresponding
tosingular data, analogous
to the surfaces describedby Spruck
in[5].
The
comparison
method was usedpreviously by
Finn[6]
to obtain agradient
bound for minimal surfacesz(x, y) depending
on a boundfor
In that case the
comparison
function wasaccordingly
determined as the solution of aparticular
Dirichletproblem, depending
on the solution z.In the
present
work werequire
a « universal »comparison function,
suitablefor any conceivable solution of
(3).
The naturaldefining
condition appears in the «capillary problem »,
in which the solution surface isrequired
to makea
prescribed angle
with thecylinder projecting
onto theboundary
of thedomain of definition. We
prescribe
therespective angles
0 and a on thetwo arcs of
D,
and we show that thecorresponding
solution serves as auniversal
majorant
for thegradient.
We prove the existence of the
majorizing
solutionin §
III.In §
I weapply
thesolution,
in a canonicalconfiguration,
to obtain thegradient bound,
for any solution of(3)
defined in a diskBR
of radiusR > Ro.
Ourestimates are, at least in
principle, explicit. In §
IV we showby
a directconstruction that the choice
I~o
cannot beimproved.
The method as described above
yields
agradient
bound in a diskbut
provides
no information as to whathappens
outside this disk. We showin § II by
an indirectreasoning
that as .R --~ 1 any solution inB B, toge-
ther with all its
derivatives,
mustapproximate
a lowerhemisphere
in anycompact
subdomain. Here the convergence estimate from above is obtainedsimply
andexplicitly starting
with thegradient
estimate in We found the estimate frombelow, however,
to be more difficult.The results of this paper are
evidently
related to S. Bernstein’s theorem[7]
that a minimal surface
z(x, y)
defined over the(x, y) plane
is itself aplane.
The result of Finn cited above
presents
a naturalanalogue
of that theorem for surfaces ofprescribed
meancurvature;
thepresent
paper in turn extends the result to surfaces defined over a domain « close » to the maximal domain of definition.The
analogous
extension for minimal surfaces couldperhaps
beregarded
as the a
priori
bound on secondderivatives,
dueoriginally
toHeinz,
cf.
[8, 9, 10].
Thistype
of bound has been obtained also for surfaces of pre- scribed mean curvatureby Spruck [11, 12] ;
it doesnot, however,
seem toimply
the moreprecise
information contained in thepresent
results.In the interest of
simplicity
wepresent
all results for the case of con-stant mean curvature H == 1. The case of
general
constantH # 0
is ob- tainedby
asimilarity
transformation. Without essentialchange
a corre-sponding
result can be obtained also for variable.H(x)
bounded from zero, under reasonable smoothnesshypotheses;
we shall not go into this matter here in detail. It seemslikely
that ananalogue
of theequation
class studied in[13, 6]
will also be accessible to the method. We have not however in-vestigated
this direction.Most of the results in this paper have been obtained
independently by
E. Bombieri. His
method, although indirect,
extends to any dimension n andyields
the same value for.Ro
if n = 2. His paper will appear elsewhere.It is a
pleasure
for us to thank E. Bombieri for a number ofhelpful
com-ments. We are indebted also to Giovanni
Giusti,
whogenerously
sacri-ficed his paper
airplane
so that his father could read his co-author’sproposal
for the
proof
of a theorem.I. - The
gradient
estimate.In this section we assume Theorem 4 and use it to prove the central
gradient
bound.LEMMA 1. 1. Let
u(x)
be a solutionof (3)
in a diskBR: Ixl
R. -Let 9) be a domain with theproperties:
a)
theboundary
a9) =1-’1-f- 1’2 +
po+
qo, whereF1, F2.
arerelatively
open and po, I qo arepoints;
b)
there is an arc S andextending
to such that the maximum distanceof
anypoint p E S
to a~ is less thanR;
c)
there exists a bounded solutionv(x) of (3)
inÐ,
such that ’V.for
anyapproach
toF~,
’V. Tv --~-E-
1for
anyapproach
to.r2 .
Then
p ES
We note no condition is
imposed
onv(x)
at po, qo .PROOF OF LEMMA I.1. We may assume 0 as otherwise there is
nothing
to prove. Let andplace 5)
overBR
with p at the center.Then
by b),
D cBR.
We may rotate 0 about p so that thegradients
of uand of v coincide in direction. If
IVvl I
at p we move 5)continuously
so that the
origin
movesalong
~S in the direction1~1 (see Fig. 2).
SinceFigure
2on
hl
there must be apoint
pi E S at which(after
apossible
rota-tion of
Ð)
Vu = Vv.Set w = u - v. We may assume
w(o)
= 0. There cannot hold w - 0 asivvl
-~ oo at We assert that in the indicatedconfiguration
there is aninteger
m ~ 2 and aneighbourhood
of pi that is dividedby (smooth)
levelcurves w = 0
through
pi into 2m distinctregions Ð(1),
...,Ð(2m),
and suchthat w 0 in
Ð(2j),
w > 0 in9)(2J I’).
Theproof
of acorresponding
assertionfor minimal surfaces
given
in[6] applies
withoutchange
to thepresent
situa-tion and we shall not
repeat
it here.We use the same
symbols 0(i)
to denote the connectedcomponents
con-taining
theseregions,
in which w does notchange sign.
Denote
by r(i)
thepoints
of that lie on eitherr1
orr2. Suppose r(1) 0, h~2~
n0, r(3) 0.
Thenclearly r(4) (see Figure 3).
We conclude either there is aregion Ð(2i)
withF(2i)
r~or there is a
region 1)(2i + 1)
with1~~2’ + 1)
r~r1 == 0.
Figure 3
Both cases
respond
to the samereasoning;
it suffices for illustration to consider the situation0. Setting
we find
immediately
since
q > 0
and(We
note thesingularities
ofv(z) at po,
qocause no
difficulty
in(4),
as for any functionf).
Theintegrand
2 - Annali delta Scuola Norm. Sup. di Pisa
on the left side of
(4)
can however beexpressed
as anintegral
ofpositive quadratic
forms in thecomponents
ofVq,
and hence(4) implies
- 0in D. This contradiction
completes
theproof
of the lemma.To
apply
Lemma I.1 we need to know conditions under which domains 9) andcorresponding
solutions v can exist. From Theorem 1 of[14]
we seethat
h1, T2
must have curvaturesm, > 2,
"22,
and thus we are ledby symmetry
considerations to a moon domain D as discussedin §
III. There must hold the necessarycondition,
obtainedby integrating (3)
overD,
Heuristic considerations indicate that in an extremal
situation, r2
willbe concentric with
BR
and will passthrough
the center ofBR.
In thissituation, (5) implies
for the radiiRi, R~
of.r1, .r2
with
One verifies after some calculation that if
R2(Rl)
is determined from(6), (7),
then
.R2 (.R1 )
0.Thus,
the most favorablegeometry
occurs if.Rl
is aslarge
as
possible.
IfB, = -1,
then(6), (7) become, setting
which has the solution ,
(9) Ro= 0.5654062332 ....
We refer to the
corresponding
moonregion
as Do-The
configuration
Do is notpermitted
in Lemma1.1; by
Theorem 1of
[14]
anycomparison
solution v would be unbounded. We may,however,
choose
Ri = § - 8
for anyE > 0,
in which caseR2 = Ro -~- r~(E), q (I) - 0
with 8. We show
in §
III that in thecorresponding
moonregion
boundedcomparison
solutionsalways
exist.Granting
thisassertion,
we choose for pa
point
on the line L ofsymmetry
ofD~
and for S thesegment
onL j oining
pto We obtain
THEOREM 1. Let
u(x)
be a solutionof (3)
in a diskBR:
There exists afunction ~O) > 0,
nonincreasing in R,
nondecreasing in
e andfinite for 1~ > Ro,
~O .R -.Ro,
such thatif
.I~ >Ro
there holdsat all
points
in the diskBR-Ro.
The considerations
of §
IIIpermit
anexplicit
calculation of a suitableC(R; (2).
We showin §
IV that the value.Ro
cannot beimproved.
II. - Behavior as R - 1.
The
proof
of Theorem 1yields
information at most in a diskBl-Ro
ofradius 1-
Ro~0.434.
Weproceed
to establish a bound in any fixedBR,
as 1-~ --~ 1.
II .1. - An upper bound can be found
explicitly
if 1~ >( 1-E-
Weobtain it
by comparing
the solution with arotationally symmetric
solutionof the same
equation.
LEMMA 11.1. - There exists a
unique (rotationally symmetric)
solutionof (3), defined
in the annulusA,,:
81- E,
such that 1 onthe
bounding
circlesFe, F1-e of
and such that~~~~ =
0 on In anyf ixed BR, y~~E~(x)
tendsuniformly
to the lowerhemisphere V(x)
= 1-1/1- r2
as E - 0.
The
proof
of the lemma is a formal exercise and we suppress details.A
typical y~~8~(x)
is indicated inFigure
4. The mostgeneral y~~~~(x)
can bedetermined
explicitly
in terms ofelliptic integrals.
Figure 4
THEOREM 2. Let
u(x)
be a solutionof (3)
inBz, .R > l(l + .Ro),
and sup- poseu(O)
= 0. There exists a continuousdecreasing function defined
in!(1 + Ro)
I with0’(1)
=0,
such thatu(x) ~ + a(B) throughout Bn.
PROOF.
By
Theorem1,
there holdsC(R; e)
inBe,
for any 8 R-Ro.
Hencein Be.
Choose 8 in the range
1 2013 JR a ~(1 2013 jRo).
Then andon
Fe.
We show first
u(x) ~ ~~8~(x) + e) throughout
For if there were an open set A cAs
in which u >y~~~ + 8C(R; 8)
we setand find
As in the
proof
of LemmaI.1,
we are led to a contradiction because of theboundary
condition for1jJ(e).
Thusin
defining y~~8~ (x )
= 0 in(12)
will then hold a fortiorithroughout Bl-s.
Now let s - 1 -
R ;
we obtain in the limitthroughout Writing
we obtain
the result to be
proved.
II.2. - We obtain a lower bound
by
an indirectreasoning.
Consider asequence of domains and a
corresponding
sequence of solutionsuj(x)
in We may suppose =0,
allj.
Using
Theorem1,
we conclude there is asubsequence-which
weagain
denote
by
converges in to a solution of(3).
Denote
by
d the set of allpoints
inBl
at which somesubsequence
ofthe Ui will converge to a finite limit. Let Then in
particular, ni(x) > - K > - c)o
for somesubsequence.
Butby
Theorem2, uj(x)
isbounded above in From the Harnack
inequality (see
Serrin[15])
weconclude there is a disk
DK(X)
centered at x and of radius notdepending on j,
such that
uj(x) > - K - 1
in Aparticular
consequence is that J is open.Let v cc 4. Then v can be covered
by
a finite number of disksi =1, , .. , m. Thus,
there is asubsequence
such thatI Ky
o0 on v.Let Vk be the set of
points
x E J whose distance from ad exceeds11k.
The
general gradient
estimatesimply
that eachcorresponding sequence uk
is
equicontinuous
in Hence asubsequence uj
can be found that con-verges in all of
d, uniformly
in each vk, I to a solutionU(x)
of(3)
in J.Clearly, U(x) 1p(x)
in J.Let p I be a sequence of
points
such that and and supposeU(p ~) > - .K > -
oo for all 1. Foreach p,
there is an indexjo(l)
such that for
Applying again
the Harnack in-equality,
y we find pz is the center of a diskDK(pl),
I of radius ogdepending only
on K and on p(and
not in whichu’ (x)
> - .K -1for j
Choosing I sufficiently large
that andthen j
>j,(I),
we finduJ(p) > - .K -1,
and hence p E J. We conclude thatif p
EB1
r~ ad thenlim
U(x) = -
oo for anyapproach
to p from within J.We have
U(o) _ (o) = 0.
Thus the set41 c 4
definedby U(x) - - y(x) > -1
is nonempty; by
the aboveremarks,
the functionis a continuous
Lipschitz
function inAs,
andThe second
integral
on theright
is nonpositive,
asq > 0
and v.T1p(s) ~ 12
v - T U 1 on We have
also, by
Theorem 2 andby
theproperties
of1jl(e),
since C 1;
here C isindependent
of E.Letting
E - 0 we findAg - Bl,
andOnce
again,
theintegrand
is nonnegative
and vanishesonly
ifVq
= 0.We conclude == in the
component J(’)
of jcontaining
theorigin.
Since
Z7(x)
- - 00 at anyboundary point
ofJ(O)
interior toB1,
we conclude also d ==Bi.
With a standardreasoning
we now obtainTHEOREM 3. There exists a
function defined in
andsatisfying 1,
and af unction or*(R; defined
inq;(R)
.R C 1and
satisfying
lima*(R; R)
=0,
such thatif u(x)
is a solutionof (3)
inB.
R--->1
and
u(O) == 0,
thenlu(x) P)
inBit.
That
is,
if .R is close to1,
a solution of(3)
inBR
cannot differsignificantly
from a lower
hemisphere
incompact
subdomains.We note Theorem 3 does not include Theorem
2,
as the boundgiven
inTheorem 2 is uniform
throughout B .
It seems dubious that a bound of thattype
would hold from below.The methods of this section extend without essential
change
to any number n of dimensions. Theresults, however,
ydepend
on Theorem1,
theproof
of which does not seem to extend asgiven.
III. - Existence of the
comparison
surfaces.We establish here the existence of the surfaces
w(z)
in domainsD~,
onwhich the discussion
in § I
is based.THEOREM 4. Let D be a moon domain
(see Figure 1), satisfying
the neces-sary condition
and
for
which therespective
radiiR1, .R2 satisfy Rl 2 , .R2 > ~ .
Then theproblem
has a
(bounded)
solutionsv(x), unique
up to an additive constant.We shall reduce the
proof
of Theorem 4 to thegeneral
existence results of[20].
We introduce first some notation.For any function in an open set A we define
If
flDul
we say that u EB V(A),
the space of functions of boundedA
variation over A.
B V(A)
is a Banach space with the normIf 99B is the characteristic function of a Borel set
E,
we call Af ID99_,l
theperimeter of
E inA;
if A = Rn we refer to theperimeter peE) of
E. Setswith finite
perimeter
will be calledCaccioppoli
sets.For detailed
background
information we refer the reader to[22, 21].
We remark that if the
boundary
aE of E is apiecewise
smooth curve, then theperimeter
of E is thelength
of thepart
of E that lies in A.We shall obtain the solution to
(18)
as a minimum for the functionalWe wish to
apply
the results of[20], Chapter
4. To do so, we must show there is apositive
constant ao such that for everyCaccioppoli
set E c Dwe have
It suffices to prove
we obtain the
remaining inequality by replacing
Eby
D - E andusing (17 ) .
Since every
Caccioppoli
set can beapproximated by piecewise regular
sets
(see [22]),
it suffices to prove(21)
for sets E withpiecewise
smoothboundary.
For such sets(21)
readswhere
We observe also that we can
always
suppose otherwise wecould add to E an
s-neighbourhood
ofTi:
and then pass to the limit as e -> 0+.
We may now note that for fixed the
quantity
attains its minimum whena2E
is an arc of a circlepassing through
theendpoints
of and it will suffice to prove
(22)
for thatconfiguration.
Consider the situation in
Figure
5. Here d is a fixednumber,
0 Cd C 2 ,
Figure
5and a circle
passing through
thepoints
pand q
is determinedby
theposi-
tion of its center on the t axis
(note
that inFigure
5 we have t0).
Wedenote
by I(t)
thelength
of the arc r andby A (t )
the area of theregion
,~.We have
and
If we set
we find
Thus
f’= 0 only
att = to = -~/1 - d2.
The functionf(t)
is illustrated inFigure
6.Figure
6Let tl
and t2
be the valuescorresponding
to the radii.Rl
andR2
of D.The necessary condition
(17)
becomesTo
satisfy
the conditions of Theorem4~we
must haveFor tl we have
Further
The
proof
of(22)
now reduces toshowing
for every
t, t1 c t c t2 .
..We obtain this
inequality
fromLEMMA III.L. Let
f(t)
be aC(l) f unction
in the interval[t,. satisfying (25)
and
(26),
and letg(t)
be aLipschitz- continuous function,
withg(tl)
=g(t2)
= 0.Then there exists a
positive
ao such thatfor
everyt,
PROOF. Since
f’(t1)
>0,
there exist 81> 0 andtl >
t1 such thatand hence
Similarly,
we haveLet m >
f (tl)
be the minimum off (t)
in the interval[t1, t2],
let M and Lbe the maximum of
and
theLipschitz
constantof g, respectively.
Then(28)
will hold if ao is chosen so thatwhich proves the
lemma,
and hence also(20).
We can now
apply
the results of[20], Chapter 4,
whichyield
the existenceof a minimum for the functional
(19).
Thesingularities
ofaD
cause noessential
difficulty,
as one seesby
asimple
modification of thereasoning in §
4E. Theminimizing
functionu(x)
isunique
up to an additive constant and is bounded andregular
in D. It remainsonly
to show that theboundary
data are achieved
strictly.
Let us start with
1-’1.
Sincewe can suppose in D. On the other
hand,
forpositive
functionsvex)
we have
and thus the function
u ( x )
minimizes the functionalin the class
where
B2
denotes the ball of radiusR2.
It follows as in
[23]
that the setminimizes the functional
in the sense that for every set V which coincides with U outside some
compact
set and which contains we haveThis means that U minimizes the functional
(33)
inB2 X R+,
with obstacle From a result of Miranda[24]
there follows that a Uis a
C~1~
surface in aneighbourhood
of L. On the otherhand,
sinceu > 1
inD,
2 U contains at least the vertical surfaceT1
X(0, 1).
Thisimplies
and establishes the first of the
boundary
conditions in(18).
Toobtain the other one, we note that since
u(x)
is bounded(say u M)
the function u minimizes also the functionalamong all functions
vEBV(D), v c .lVl.
’Ve can thenrepeat
theprevious reasoning, obtaining
theregularity
atr2.
IV. -
Ro
is bestpossible.
LEMMA IV.1. Let
B,
be a ballof
radius .R centered at0,
and let B beIf .R C Ro,
the set Eis the
unique
minimumfor
thefunctional
where
Figure
7PROOF. It is
equivalent
to show that B is theunique
minimum ofin the class
As in the
proof
of Theorem4,
it suffices to compare B with domains E that coincide with B outsideBR and
are boundedby
a circular arcpassing through
theendpoints
ofr1.
If 1~ is the circular arc
bounding E
one has to show thatunless E = B.
We consider
separately
two cases.(I) [B[.
In this case the arc r lies in B and one verifiesreadily
(using
the function(23) f(t))
that(36)
holds.(II)
LetBo
be a ball concentric withBR
and with radiusRo,
and let
p’, q’
be the intersections ofaBo
with aB. LetLet E’ be the domain
coinciding
with B outsideBR,
boundedby
acircular arc T’
passing through p’
andq’,
and such thatIEl = JEJ.
Wethen have
On the other
hand,
since we have(cf.
Figure 7)
On the other hand
and hence
(37), (38) imply (36).
This proves the lemma.To show
.Ro
is bestpossible,
we setand,
for~=1,2,...,
letLet uk
minimize the functionalin the class
The sequence u,, is
obviously nondecreasing; proceeding
as in[25],
weconclude that the set
minimizes Hence from Lemma IV. 1 we have
P = E,
and therefore o E ap.Suppose
there exist B > 0 and 1~ oo such thatThen if
0 E P,
the ball-I
would lie inP ;
if we would haveP r~
B.
=0.
In either case0 ~ aP,
a contradiction.Thus,
the sequence Uk cannot have boundedgradient
in anyneighborhood
of 0.REFERENCES
[1] S. BERNSTEIN, Sur les
surfaces definies
au moyen de leur courbure moyenne et totale, Ann. Sci. Ecole Norm.Sup.,
27 (1910), pp. 233-256.[2]
E. HEINZ,Über
Flächen miteineindeutiger Projektion auf
eine Ebene, derenKrümmung
durchUngleichungen eingeschränkt
sind, Math. Ann., 129(1955),
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