A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
G IOVANNI M ANCINI
R OBERTA M USINA
Surfaces of minimal area enclosing a given body in R
3Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 16, n
o3 (1989), p. 331-354
<http://www.numdam.org/item?id=ASNSP_1989_4_16_3_331_0>
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Enclosing
aGiven Body
in R3GIOVANNI MANCINI - ROBERTA MUSINA
Given a
body
in1ft 3,
weconsider
a class ofsurfaces, parametrized by 5~,
whichenclose,
in a weak sense,IT.
To "enclose" means, under someregularity assumption
on the surface underconsideration,
that such a surface isnot contractible
in 1ft 3B11.
The first
problem
we dealwith,
is concerned with the existence of surfaces which minimize the areaintegral
in such a class. In case an is of classC2,
thiswill lead to
finding
a C1,a surfaceparametrized by
a map UOO : 1ft 2 -+which satisfies
and which is not contractible in
(i.e. "encloses" ?i
in a strongsense).
Here b is the second fundamental form of ac
(see
Section1), v
is the innernormal at
a C,
and x A is the characteristic function of the set A C 1ft 2 .This
problem,
which at ourknowledge
was notpreviously
considered in the framework ofparametric surfaces,
is somehow related to theproblem
ofminimal boundaries with obstacles
(see
forexample [12]).
We attack our
problem by
means of a Dirichlet’sPrinciple,
i.e. we lookfor extremals of the Dirichlet
integral
over a suitable class of maps from S2 into In a moreregular setting,
thisproblem
amounts tofinding
not
homotopically
trivial minimalspheres
in a Riemannian manifold N withboundary.
In case lV has emptyboundary, striking
results have been obtained ina celebrated paper
by
Sacks and Uhlenbeck([17],
see also[10], [19]).
Here weperform
ablow-up technique
introduced in this contextby
Sacks and Uhlenbeck.But,
in order to avoidestimates
on the solutions ofEuler-Lagrange equations
related to
approximated problems,
we follow a more directapproach
based ona lemma
by Brezis,
Coron and Lieb[4].
Partially supported by Ministero della Pubblica Istruzione.
Pervenuto alla Redazione il 24 Aprile 1988.
We also consider the case
of disk-type
minimal surfacesspanned by
agiven
wire r over theobstacle n. The
existence andregularity
of an areaminimizing
surface up ,spanned
over0,
wasproved by
Tomi[20] (see
also[9]).
We answer here the rather naturalquestion whether
it exists a second minimal surface urwhich, jointly
with ur, "encloses"fi.
While this is not thecase in
general,
we prove that this occursprovided
Here
Xr
is a suitable class of surfaceswhich, jointly
with ur, "enclose"n.
In the first section of this paper we present
preliminary
remarkson
thefunctional
setting
and we defineprecisely
the class of surfacesenclosing fl.
In Section 2 we
describe
a Dirichlet’sprinciple
for minimal surfacesenclosing
agiven body 1Z
and we prove theexistence
of a closedregular S2-type
minimal surfacespanned
over the obstacleIT.
In Section 3 we
give
an existenceresult
forpairs
of minimal surfacesspanned by
the same wire r over anobstacle a
andenclosing
it.In an
Appendix
we present a resultconcerning
continuousdependence,
upon
boundary data,
of minimizers for the Dirichletintegral
in presence of obstacles. Thisresult,
which we did not find in theliterature,
turns out to bea
key
tool inproving
the basicinequality (see Proposition 3.4)
on which our existence resultsrely.
Notations. denotes the open disk of radius r and center z in
I1~ 2 , ~ ’ ~ ,
’denote the norm and the scalar
product in 1ft 3, ---"
denotes weak convergence in variousspaces, ) . . 100 and I - 2
denote L°° and L2 normsrespectively.
1. -
Preliminary
remarks and statement of the ProblemLet C be the closure of the unbounded connected component of
1ft 3B0,
were 11 is a
given
bounded open connected set in R~. We will assumethroughout
the paper thatthere is an open
neighbourhood
0 of C(1.1)
and a
Lipschitz
retraction 0 --+ C.We shall denote
Using
asmoothing - by averaging -
method(see [18],
and[ 1 ], Appendix),
one
easily
obtains adensity
result which will be useful in thesequel.
LEMMA 1.l . For every U E
X(C)
there exists a sequenceUn
E C°° n Xsuch that ’
Furthermore, for
each n,Un
can be taken constantfar
away.In order to
give
our notion of"mappings enclosing f2",
we define the Volume Functional(see
forexample [21]):
which is well defined
since, by
Holderinequality,
Notice that if
Sup
and in L2 for somei ", ,
Now, assuming
forsimplicity, 0 E
C, we define the mapSince p is
Lipschitz
continuous far away from 0, we havep U
E X if U EX ( C) .
Moreover,
if a sequence CX(C)
is bounded in L °° VUin
L’, U,, --+ U
a.e., then -V(pU)
in L2 andV(pU).
Werecall
that,
if U EX(C)
n C’(R 1, R 1)
and U isregular
atinfinity,
that isthen
and
gives
thedegree
ofp U o II E ,S 2 ) ,
where II denotes thestereographic projection
ofS2
onto R 2(see [15],
and[ 1 ],
Lemma1).
We noticethat,
because of thedensity Lemma,
andcontinuity properties
of the volumefunctional,
wehave
This
integer
still denotes thedegree
ofpU,
so thatXe(C)
is the set of mapswhich have a non-zero
degree
with respect to thesphere | £| =
1. Inparticular,
if U E
X(C)
n isregular
atinfinity,
and 0, thenU,
as amap from S2 into
C,
is not contractible.Accordingly,
we setREMARK 1.2. If p : C 2013~
S2
induces anisomorphism
between the secondhomotopy
groups of C andS’ 2 ,
then{U
EXe (C) I
U is continuous andregular
atinfinity,
can be identified with the closure of the set of
smooth,
non-contractible maps from 82 into C,by Hopf’s
Theorem.In the
following section,
we willstudy
PROBLEM 1. Find
U 00 E
continuous andregular
atinfinity,
in thesense
of (1.3),
which has minimal area among all thesurfaces
inXI(C).
In view of the
previous remarks, U 00
will be a closed non-contractible surface in C.Before
ending
thissection,
we wish to state aproblem concerning disk-type
minimal surfaces
spanned by
a wire r over the obstacleS~.
We first recall a well known result
(see [20], [9]).
Let r C C be a closed Jordan curve, and suppose that the classXr ( C)
of maps u EH 1 ( D, C),
whosetrace on a D is a
continuous, weakly
monotoneparametrization
of the curveF,
is not empty. Then, if is of class
C2 ,
there iswhich has minimal area among all surfaces in
Xr (C),
and which satisfies theconformality
conditionsi.e. its area is
given exactly by
We wish to find a second surface
satisfying
theconformality conditions,
which is harmonicwhere
it does nottouch and which
"encloses, jointly
withu",
the obstacleH.
To make moreprecise
the last statement, let us writeand set
where
In order to describe the
geometric
property of surfaces inXr (C),
let usfirst recall
(see [1])
thatfor every with
Actually,
holds for everyFurthermore,
theinteger
in(1.5) gives
thedegree of ;
where
Thus,
if u, v EXr (C)
nCO (D’ R 3),
u = v on a D, the conditionVD (pv)
isequivalent
to thenon-contractibility
of p o U on,
Ugiven by (1.6).
In
addition, if u E C~(D) (which
occurs, e.g., if T n 8C =0,
see[8]),
one can
build,
for every u EXr (C) (see
LemmaB.3),
achange
of variablesgu e such that
and
In this case, if u E then
pencil
is notcontractible,
whereU
corresponds
here to thepair
u,After this
preliminaries
we areready
to statePROBLEM 2. Find u E
Xr (C) of
classC1,a(D)nCO(D)
which has minimalarea in the class
Xr (C).
2. - Closed minimal surfaces
spanned
over obstaclesAs a standard
procedure,
we aregoing
toreplace
the minimization Problem1,
with the
simpler problem:
Find
U 00
EXe(C)
such thatFirst we prove the
following
THEOREM 2.1. Let C be as in the above Section, and assume in addition that aC E C2.
Let
Th
EXI (C)
be a solutionof (2.1).
ThenPROOF.
(i)
In view of a resultby
Duzaar[7],
it isenough
to prove such thatAccording
to Duzaar’sresult,
this willimply
for every p~ 11, ool,
and henceUoo
E c1.aby
Sobolevimbedding
Theorem.Given let r > 0 be such that
and let , with Let us consider
Since truncation decreases the Dirichlet
integral,
we can assume so that inparticular,
if T is admissible(i.e.
~ Eand hence
let us consider
Since from
I-- -I- -.
(2.4)
we deduceand hence W E Thus
and
(2.2)
follows from(2.3). Finally, being
again
aminimizer,
it is continuous at z = 0 and we findi.e. is
regular
atinfinity.
(ii) - (iii)
Here werely
on the "EulerEquation"
for the"energy minimizing maps" (i.e.
for minima of(2.1))
establishedby
Duzaar[7]:
Here b is the second fundamental form of is the inner normal at ac, and x A is the characteristic function of the set
A C
JR 2. As a consequence,U 00
is harmonic in the open
set ~ z
E1ft 2 I
AlsoThis
easily implies that, setting (in complex notation),
then i.e. p
and 0
are harmonic. Since(iv)
FromMorrey’s e - conformality
result([13],
see also[16], §226),
wehave that for every U E C°° n
Xe (C)
constant far away, and for every e > 0, there exists aVs
E suchthat 2 1 f )Uz
AUyl +
e.Thus,
forevery C °° n
X (C)
constant far away, we haveand the conclusion follows from the
density
Lemma.The main result in this Section is THEOREM 2.2.
is achieved.
The
proof
is based on a blow uptechnique,
introduced in this class ofproblems by
Sacks und Uhlenbeck[17].
A crucial step in theproof
of Theorem2.2 is the
description
of the behaviour of sequences( Un ) n
Cwhich,
inthe
limit, jump
out of the class. To this extent, we first recall a result in[4]
(see
also[23]).
Let C
X(C) satisfy
Then, eventually passing
to asubsequence,
weakly
in the sense of measures, forsome di e Z ,
ai E Here6a,
denotesthe Dirac measure concentrated at a~ .
We first show
that,
if det "concentrates" at some a, thenUn loses,
in thelimit,
at least as much energy as100.
PROPOSITION 2.3. Let
(Un)n
Csatisfy (2.6).
AssumeU ~
0
for
some index i in(2.7).
Thenfor
every p > 0 smallenough,
PROOF. Fix p > 0 such that
D2p (ai)
We can assume,eventually passing
to asubsequence,
that there existsFor almost every r p, there exists a
subsequence (depending
onr)
suchthat
II
and hence
weakly
inNow,
we denoteby hk
asolution of
Because of the
good
behaviour ofhk
one can prove(see Proposition A. I)
that up tosubsequences
where h minimizes the Dirichlet
integral
with constraint C andboundary
dataU. In
particular
Now,
let us defineWe claim
that,
if r is chosen smallenough,
thenUk
Exe(c), k ( r ) big
enough.
Infact, using (1.2), (2.8), (2.9),
we findfor k >
k (r). Hence, by using again (2.8), (2.9),
we getThus,
for suchgood r’s,
we haveand, letting
r go to zero, we conclude theproof
ofProposition
2.3. ·In
particular,
we getPROPOSITION 2.4. Let
(Un)n
9X~ (C) satisfy (2.6). Assume U g
Then
p ,
PROOF. Let us first remark
that,
in casedi =
0 for every i, thenUn
- U*( Un , U*
defined as in(2.5)) satisfy
the sameassumptions
asUn,
Uand
(compare
with(2.7))
where d
(since U ~ Xe(C) by assumption)
does notdepend
on n,form large. eventually replacing
Un by
we can assumeUn
satisfies theassumptions
inProposition
2.3 andhence,
for some az ,In case
Un is,
inaddition, minimizing,
i.e.more.
we can say PROPOSITION 2.5. Let
(Un)n
Csatisfy (2.6)
and, inaddition, If U 0
xe(C),
then U is a constant, and either(i)
There is(exactly one)
a E 1ft 2 such thatPROOF.
First,
U = const.by Proposition
2.4.Furthermore,
if 0 forsome i in
(2.7),
there isjust one di i-
0,by Proposition 2.3, which,
at the sametime, implies
IIand hence
In case we have
(see (2.10))
Again by Proposition 2.3,
wehave,
as in theprevious
case,for every
PROOF OF THEOREM 2.2. Let C be
minimizing:
Since truncation does not increase the Dirichlet
integral
and 0 isbounded,
wecan assume the
Un
have a common LOO boundand, passing
to asubsequence,
VU in
~2, Un --~
U a.e. for some U EX(C).
If U EX~ (C), U
is aminimizer
by
the lowersemicontinuity
of the Dirichletintegral.
Ifxe ( C) ,
we want to show
that,
afterrescaling
andtranslating Un,
we can construct anew
minimizing
sequenceweakly converging
inXe(C).
Let us introduce theconcentration function
(see
forexample [11], [3]):
This is a
continuous, non-decreasing function,
withQn (0)
= 0 andThus, given
6E]O, 100 [,
there are, for nlarge,
tn >0, zn
E R2 such that. Notice that
and
Sup
=Sup
+oo.Again
we can find asubsequence ( Un ) n
and
Uoo E X(C)
such thatWe claim that
xe(c). Otherwise, by Proposition 2.5,
either for somea E IR. 2
and Vr > 0,
orBut the first alternative cannot occour, since
for r 1.
Finally,
the second alternative cannot occoureither,
becauseREMARK 2.6. It may
happen
that the "coincidence set"8fl)
isthe all
plane I~ 2 .
Sinceprojections
on convex sets reduce the Dirichletintegral,
this is for
example
the case when 11 is a convex set.Moreover,
in this case, itresults that the
image through
the map isexactly
8fl(otherwise
the mapwould be contractible in
C) and, identifying U 00
with itscomposition
withthe
stereographic projection,
our solutionU 00
is in fact a non-constant harmonic map from thesphere
ontoIn order to avoid this
phenomena,
we could use an observationby
Duzaar[7]:
since the minimizerU 00
satisfiesfor almost every z E the obstacle an has to
satisfy
a"concavity
condition"
(when
viewed fromC)
in order to be"essentially
touched"by
theenveloping
surfaceU 00.
In otherwords,
if b ispositive
defined somewhere oncannot lie
entirely
on and as we havepreviously noticed,
it isharmonic outside the coincidence set.
3. - Pairs of solutions of the Plateau Problem for
disk-type
minimalsurfaces with obstructions
Given the obstacle
0,
we assumeas
in theprevious
Sections thatC,
the unbounded connected component of is of classC2
and satisfies(1.1).
Let r C C be a
given
Jordan curve,parametrizable
with adiffeomorphism ,0 :
9D -~1ft 3.
Let us denoteby ,~r
the class of nCO (a D, JR 3) - weakly
monotone
parametrizations
of r which are normalizedby
athree-point
condition.We suppose that the class of "admissible functions":
is not empty.
The "small solution" u, obtained
by
Tomi[20],
isjust
a solution of theminimum
problem:
r -
and its existence is
easily proved using weakly
lowersemicontinuity
of theDirichlet
integral
and theCourant-Lebesgue
Lemma[5].
In order to find asecond solution as an extremal for the Dirichlet
integral,
we will first prove that the set(see
Section1),
is not empty wheneverXr (C) =,4 0.
Then we will consider thefollowing
minimizationproblem:
Find u E
Xr (C)
such thatAs in Section
2,
one can prove thefollowing
Dirichlet’sPrinciple:
THEOREM 3.1. Let C be as
above,
and solutionof (3.1 ).
ThenThe
regularity
result in(i)
follows from a Theoremby
Hildebrandt([9],
see also Tomi
[20],
Satz[6]
and Duzaar[7]),
via arguments similar to those used in theproof
of Theorem 2.1.Propositions (ii), (iii), (iv)
can be obtained ina standard way
(see [5],
pg. 105 for(ii)
and theMorrey’s e-conformality
result-
[13],
Theorem 1.2 - for(iv)), using
the invariance of the volume functional underreparametrizations
of the domain.0
REMARK 3.2. We notice that in case r C C,
the
conformal map _u isharmonic in a
neighbourhood
of a D, and thus u ECI (D, 1ft 3) (see [8]).
At ourknowledge,
it is not known aC 1-regularity
result up to theboundary
in thegeneral
case.In order to solve
(3 .1 ),
we first proveLEMMA 3.3. The set
Xr (C)
is not empty andActually,
we want to prove a moregeneral
result:-
PROOF OF STATEMENT
(3.2).
Denotedby
U a solution of Problem1,
noticethat,
under ourregularity assumptions
onaC,
U is continuous andregular
atinfinity,
that isWe set
Let A :
~0,1~
-~ C be aLipschitz
map with map v, isgiven by
Our
where ~r is the
Lipschitz
retraction in( 1.1 )
and ife
1,
~2 r e. For esmall, u(z) - u(0)
issmall, if I z
e,since
u is con-tinuous,
and hencebelongs to
a smallneighbourhood
of C.
Similarly,
if is close toU(oo)
andhence
belongs
to aneighbourhood
of C as well. Thus Vs is well defined andon A direct
computation
shows thatwhile
and thus
To end the
proof,
it isenough
to observe thatby (3.3),
andhence,
if e is smallenough,
sinceREMARK 3.4.
Equality
in Lemma 3.3 cannot beexcluded,
ingeneral.
Moreover,
the sequence(vc)c
in theproof
of(3.2)
showsthat,
wheneverequality
occurs, there existminimizing
sequences for Problem(3.1)
whichweakly
converge to the small solution u and hence do not havestrongly
convergent
subsequences.
Equality
occours, forexample,
in the"degenerate case",
i.e. when rreduces to a
point zO.
In this case the set of "admissible functions" isXr (C) - (u
E= z°
onaD, 0)
and7~. Actually, Ir
=100’
since in this caseXf (C)
is embedded in a natural way inX (C) .
It is
quite likely
thatIr
is not achieved wheneverequality
holds. This isthe case if n is the unit ball and r reduces to a
point,
e.g. inan = S2. In facta minimizer for the Dirichlet
integral
would be a non-constant harmonic map from the disk into ,S2 with constantboundary data;
but this cannot occour inview of a
uniqueness
result due to Lemaire[10].
Notice also that such a minimizer would also be an extremal for the Bononcini-Wente
isoperimetric inequality:
for every v E constant on
a D,
which is known not to exist
([22]).
The main result in this Section is
THEOREM 3.5. Let
C,
r be as above. ThenIr
isachieved, provided
PROOF. We
split
theproof
into two steps.STEP 1. There is
vo
EXr (C)
such thatSTEP 2.
1vo
:= Inf is achieved inPROOF OF STEP 1. Here we do not make use of
assumption (3.4).
Let(u,,),, 9 Xr (C) be
suchthat f IVunl2
->Ir.
We can assumeSup
oo,and
since
isequicontinuous on a D, by Courant-Lebesgue Lemma,
wecan also assume un -
vo weakly
in~I1
and v°uniformly
on a D forsome v°
EXr (C). Thus,
if4ihn = 0, hn =
Un - v° on 0uniformly
and
weakly
in As a consequence wn :=hn )
is well defined forlarge n (here
~r is the retraction of someneighbourhood
of C ontoC) and,
with easy
computations
Since E
Hol,
it isenough
to prove, in view of(3.5),
thatVD (pg).
But thisreadily follows, because const.
--; 0)
that, by
LemmaSince
(see Corollary B.4),
theproof
iscomplete. ·
PROOF OF STEP 2. The argument we present here
applies
to thesolvability
of Dirichlet
problems,
and hence wegive
it in this moregeneral
form. Letv E
Xr (C)
and letIf
w g Xrl (C) then, applying Proposition
2.4 to the sequencewe
immediately
getFrom
(3.6),
it follows that the infimumis achieved
provided (compare
with(3.2)):
This ends the
proof
ofStep
2 since(3.7) holds,
with v = vogiven by Step 1,
in view of
assumption (3.4). ·
REMARK 3.6. It is
interesting
to reformulate Theorem 3.5 from thepoint
ofview of relaxation. If we define the energy associated to the minimum
problem
as
then the relaxed
functional,
in theweak H1-topology,
is definedby
Slight
modifications in ourarguments
show thatNow,
let u be a minimumpoint
for the functional sc-E,
that isIf
(3.4) holds,
thennecessarily
u EXr (C),
and hence u is also a solution ofour minimization Problem 2.
REMARK 3.7. A
simple
variant of Problem 1 arises if wedrop
theconnectivity assumption
on the obstacle ~; related results arepresented
in[14],
where are also considered extensions tohigher
dimensions.It would be of interest to describe the limit
problem
as the connectedcomponents
of 11 become infinite while their size go to zero. This could also be a way to deal with a muchdeeper
variant of Problem1, namely
the case ofthin obstacles. Problems of this kind have been considered in the framework of minimal boundaries
(see [6]).
Appendix
AWe
present
here a resultconcerning
continuousdependence
of minimizers for the Dirichletintegral, subjected
to obstacleconditions,
withrespect
toH1
weak convergence of
boundary
values.Let C C R 3 be a closed set
satisfying
There is 6 > 0 and a
Lipschitz
retractionLet us denote
Let
hn
Esatisfy
and
and
PROPOSITION A.1. Let
hn satisfy (A.2), (A.3).
Thenif hn -
h, we havePROOF.
Since f |Ah| 12
lim it isenough
to proveTo prove
(A.4),
let usconsider,
for r(in polar coordinates).
Since27Vn is well defined on where 7r is the retraction
given by (A.1).
Moreoverif L denotes the
Lipschitz
constant for 7r.Now, let h
EH I (D, C)
be such thatand we define
Since wn E
C)
and wn= hn
on aD, we haveAn easy
computation gives, using (A.3),
and hence lim sup
(A.4).
COROLLARY A.2. Let
hn satisfy (A.2). If hn -1
h, thenPROOF. For a.e. r 1, we have
Sup
. Sinceclearly
Proposition
A.1applies
to obtainhn - h
inH 1 ( Dr)
andfor every v E
H 1 ( D,. , C ) ,
with v - h EThus,
"
w - h E
setting
we see
that,
for a.e. r 1:because
wr - h e HÓ(Dr).
Butand
hence, sending
r to 1 in(A.6),
we get(i). -
REMARK A.3. To
complete
these continuousdependence results,
it wouldbe of interest to prove
that,
if in addition to theassumptions
inCorollary A.2,
one also assumes
uniformly
onaD,
thenhn -; h uniformly
on D.Since we do not need this
result,
we do not go into details.Appendix
BFor convenience of the
reader,
we list here a fewsimple properties
of thevolume functional
(see [21]
and[2], Appendix).
PROOF. From
(1.2),
one sees thatbecause k n -~ 0 in L°° and
weakly
in H1. Now, the secondintegral
in theright
hand side goes to zeroby
Lemma A.7 in[2],
while the third one goes tozero
by
Lemma A.6 in[2]. ·
LEMMA B.2. Let g E
C°(D, D)
be an orientationpreserving bilipschitz homeomorphism.
ThenThis follows from the chain rule:
be a
nondecreasing function,
with. Then
PROOF. Since , After
properly extending
we can
regularize
it to getuniformly,
Then we set
By
theprevious
Lemma we getVD (u
ogn)
=VD (u) .
Butbecause an - a
uniformly
and u ECI (D) imply
uniformly,
whileCOROLLARY B.4. Assume r is
parametrizable
with adiffeomorphism 7° :
8D -~ 1ft3. ThenPROOF. Given with
(assuming
for
simplicity 0 V C).
It isenough
to proveSince,
for agiven
u EXr (C),
UI,9D is aweakly
monotonereparametrization
of
r,
there is a map a~ :~0, 2~r~
-~[0, 21rl,
continuous andnondecreasing,
with=
0,
= 2Jr, such thatBy
LemmaB.3, setting
and hence
because
u ( z )
= for every z Ea D,
so that Lemma 1 in[ 1 ] applies..
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