A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
D ONG Z HANG
The existence of nonminimal regular harmonic maps from B
3to S
2Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 16, n
o3 (1989), p. 355-365
<http://www.numdam.org/item?id=ASNSP_1989_4_16_3_355_0>
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Harmonic Maps from
B3 toS2
DONG ZHANG
1. - Introduction
There has been a
great
deal of interest in the harmonic mapproblem
for maps from domains inR3
to,52,
the two-dimensionalsphere (see [BCL], [HLI,2], [ABL], [AL]).
Thelarge
solutions for harmonic maps in two dimensions have been studiedby
H. Brezis and J. Coron[BC].
In addition to thegeometric significance
of these maps,they
also arise as animportant special
case in thetheory
of nematicliquid crystals (see [deG], [E], [HKL1,2], [HLI,2]).
Ageneral
existence and
regularity
theorem was provenby
R. Schoen and K. Uhlenbeck(see [SU1,2,3])
for energyminimizing
maps. In the case of three dimensionaldomains,
their result asserts that minimizers areregular
except for isolatedpoints
in the interior. There are severalimportant
earlier works whichrequire
some restriction on the target manifold. These include the works of J. Eells and J. H.
Sampson [ES]
fortargets
ofnonpositive
curvature, and the works of S.Hildebrandt,
H.Kaul,
and K. O. Widman[HKW1,2]
for maps whoseimages
lie in a convex coordinate
neighbourhood.
If one considers a map p :
aB3
--~,52
which has nonzerodegree,
thenany minimizer which agrees
with p
onaB3 necessarily
possesses at least oneinterior
singular point.
A naturalquestion
arises as to whether thesingular points
are determined
by
theboundary data,
or one canspecify singular
behavior aswell as
boundary
data. Thisquestion
is unresolved.The first
special
case of theproblem
ofprescribing singularities
is thequestion
of whether aboundary
map p of zerodegree
has aregular
harmonicextension to B3. On the
positive side,
the theorem of S.Hildebrandt,
H. Kaul and K. O. Widman[HKW2]
asserts that a map p, whoseimage
lies in an openhemisphere
ofS2,
has aregular
harmonic extension(also having
animage
in this
hemisphere).
On the otherhand,
R. Hardt and F. H. Lin[HL2]
haveconstructed zero
degree boundary
maps pfor
which any minimizer must possess interiorsingular points. Thus,
anyregular
harmonicextension,
in such a case,Pervenuto alla Redazione il 4 Ottobre 1988.
would have
larger
energy than the infimum over all W1.2 extensions of p.In this paper, we show
that,
for ageneral
class ofaxially symmetric
maps p of zerodegree,
there is aregular axially symmetric
harmonic extensionof p
toB3.
. Inparticular,
in many cases these harmonic maps havelarger
energy thana minimizer. In the last section of this paper, we will present some
examples
of
axially symmetric boundary
maps p whose smooth harmonic extensions are not energy minimizers.Our
proof
reduces to theanalysis
of a scalarpartial
differentialequation
intwo variables. This
equation
isdegenerate
on theboundary corresponding
tothe axis of symmetry. Our main theorem can now be stated.
’
THEOREM.
Let p : --+ 82
be aregular non-surjective axially symmetric
map. There exists a
regular axially symmetric
harmonic extensionof
cp toB3.
To make
precise
the symmetry condition werequire:
let B3 c R 3 be theunit ball and
S2
cR3
be the unitsphere;
i.e.and
We are
going
to use thefollowing
coordinates inB3,
where
The Euclidean metric is then
given by
Choose coordinates
(0, ~)
in S2 asfollows,
The metric on S’2 then takes the form
The
axially symmetric
harmonic maps we consider have the formUsing
thesecoordinates,
ourproblem
is now to solve thefollowing equation:
with the
boundary condition 0 = Q on a D,
where Q satisfieswhere e is a
positive
number.We will solve the above
equation by finding
a smooth criticalpoint
of theenergy functional:
which is defined in the Hilbert space
2. - Several Lemmas
LEMMA 1.
Any
criticalpoint 0 of
E is a local minimum in the sense thatprovided 0
ECo (D)
andwhere C is a
fixed
constant.PROOF.
Suppose 0
E H is a criticalpoint
of Eand 0
is a map inCo (D).
Using integration by
parts,equation (1),
andTaylor’s theorem,
we haveSince the first
eigenvalue
of -0 on a bounded domain n c Rn is bounded from belowby {diameter (n)} - 2,
aslong
as therequired
condition is satisfiedby
supp(0), inequality (4)
holds. DThe next lemma is a consequence of Lemma 1 and standard
elliptic regularity theory (see [M]).
LEMMA 2. The critical
points of
theequation (1)
in H areregular
in theinterior
of
D.Using
Lemma1,
we now derive thefollowing
existence result.LEMMA 3.
Suppose ¢1
and are criticalpoints of
E and 0 onD. Let cp; be the
boundary
valuesof Oi,
=1, 2. Suppose
p is afunction
on 8Dlying
between ~pl and y~2, i.e.VI :5 V:5
(P2- Thenis a critical value
of
E and is achievedby
a criticalpoint 0,
i.e. there exists~
E H such thatMoreover,
~1 :5 0:5 tP2
on D and = cp.’
°
PROOF. Consider a
minimizing
sequence(§n )
which has a weak limit0.
Itis easy to see that
E(¢)
= c; infact, 0
is the strong limit ofTherefore,
we may assume
In order to show
that 0
is a criticalpoint
of E, we need to prove that(5)
holdsfor
all 0
ECo .
But it isenough
to show that(5)
holds forthose 0
ECo satisfying
Let 0
be such aCo
function and considerWe define
Since
01 :5 0 tP2, St
andMt satisfy
thefollowing
conditions:Define
By
the definition of0,
we know thatjE*(~) Using
Lemma1,
wehave
Combining
these twoinequalities,
weget
Therefore
(4)
holds forevery 0.
The lemma isproved.
DWe are
going
to use Lemma 3 to prove our theorem. We will take the trivialsolution,
the zerofunction,
asour Q 1.
The rest of theproof
is to show thatthere exists a
~2,
for anygiven boundary
data p,satisfying
the condition of Lemma 3. Our~2
will be az-independent solution,
i.e. a function of r which is a solution of thefollowing ordinary
differentialequation
3. - The construction
Of 02
Let r =
et , t
E( - oo, 0~ .
Theequation (6)
then becomes~ We are
going
to use a variational method to solve the aboveequation.
Thecorresponding
functional isWe are
going
to solveequation (7)
with thefollowing boundary
condition:This can be done
by minimizing
the functionalEi .
It is easy to see thatSince
the
problem (7), (8)
has a monotoneincreasing positive
solution.Let
0,
be such a solution.Multiply
theequation (7) by 0’
andintegrate
toget
Since
§r
E WI, 2(_ 00, 0),
the constant is zero.Because
we have
Thus we have solved the
following
first orderordinary
differentialequation
Since 0 -=
7r is the solution of(11)
with the initial value T = we see thatuniformly
in any finite interval~-T, 0~. Furthermore,
Take a v E
(0,
and let tv( 1")
be the root of(t)
= v, thenMoreover,
since sin0 (t) 0 (t),
we haveNow let us go back to the
equation (6).
Define~T (r) by
The function
0’
is then a solution of(6)
and will be ourl/J2
with proper choice of T.The
inequality (15),
in terms of r, isSuppose
thefunction
on 8D satisfies thefollowing
conditions:(a) ~ == 0,
forr = 0;
(b) Lipschitz’s
condition at both(0, 1)
and(0, -1);
(c) p
maxaD(p)
7r.From
(b),
we have twopositive
constants k and d such thatFirst,
from(14),
there is a T° such thatFor this
TO,
we havewhen r
60
= min{8, By (13),
we know that(19)
holds for all TTO,
with the same
60.
Secondly,
from(12),
there is a T 1TO
such thatTherefore, 4;2 == 4;1’1
will serve as our upper barrier.Using
Lemma3,
we know that theproblem (1), (2)
has aregular solution 0
whenever theboundary
valuep is
regular
andnon-surjective.
4. - The
proof
of the TheoremLet p
be aregular boundary
functionsatisfying
the conditions of the theoremand 0
be thecorresponding regular
solution of theproblem (1), (2).
Define themap 4) : B 3 -~ S 2 by
~ is an
axially symmetric
map with theboundary
valuesWe are
going
to show that (D is harmonic and smooth onB3.
We know that -0 is harmonic away from the
axis,
i.e. ~ satisfies theequation
or,
equivalently,
We will prove that (D is harmonic in B3
by showing
that(22)
holds for all_ ft - ft ...
Let T be a
Co (B3, R3)
map, and consider thefollowing
cut-off functionThe first
integral
on theright
side is zero. The second and the thirdintegrals
tend to zero as l~ ---~ oo, because the measure of supp goes to zero and is bounded. Hence 4D is harmonic on
B3 .
We know that (D is smooth away from the axis. Since
by
constructionO is continuous on the axis. Therefore (D is smooth
everywhere
in B3(see [HI).
Our main theorem has been
proved.
5. - An
example
In this section we will show that the
regular
harmonic extension we get isnot a energy minimizer among the
Wl,2
extensions. As a matter of fact it is not even a energy minimizer among theaxially symmetric
W l2 extensions. Weare
going
to show thisby constructing
a sequence ofaxially symmetric
maps1,Dk)
cWI, 2 (B3 , 82)
all of them haveboundary values,
sayIV/,}, satisfying
the conditions of our theorem.
Therefore,
pk haveregular axially symmetric
harmonic extensions on
B3
which will be denoted as The difference between(0k)
and145k)
isbut has a
positive
lower bound. Since areaxially symmetric, using
theprevious coordinates,
we can writeFor this reason, we
only
need togive
the definition ofotherwise where k =
4, 5, 6,....
Itsboundary
value iswhere
S1 1/ 2 = ({r, z ) E I r 2 + z 2
=1}. Since Qk only
takes 7r and zero onthe
axis, ~k
is well defined.Recognize that,
in the coordinates we areusing,
the famous harmonic map from
B~
toS 2 namely x 1-+ xl Ixl
can be written asTherefore, Furthermore,
it is easy to see thatIt is obvious that pk is C1 on
SI/2
except twopoints (*,±(1 - 2/k))
andCPk
Hence, by
ourtheorem,
pk has a smoothaxially symmetric
harmonicextension
~k = (B, ~k).
Now,
we aregoing
to estimateFor any is a smooth function of
From Section
3,
we know thatis
positive. Therefore,
ispositive.
REMARK. The same result holds for those
axially symmetric
bounded R3domains
diffeomorphic
to B3 and also foraxially symmetric boundary
maps which have the form ofwhere n is an
integer;
in fact theEuler-Lagrange equation
then becomesAcknowledgements
The author is very
grateful
to his adviser Professor Rick Schoen without whosehelp
this result would not be here.REFERENCES
[ABL]
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Mathematics Department
Stanford University Stanford, California 94305