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 B3 to S2
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4e série, tome 16,
no 3 (1989), p. 355-365
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The Existence of Nonminimal Regular
Harmonic Maps from B3 to S2
DONG ZHANG
1. - Introduction
There has been a great deal of interest in the harmonic map problem for maps
from domains in R3 to ,52, the two-dimensional sphere (see [BCL], [HLI,2],
[ABL], [AL]). The large solutions for harmonic maps in two dimensions have
been studied by H. Brezis and J. Coron [BC]. In addition to the geometric
significance of these maps, they also arise as an important special case in the
theory of nematic liquid crystals (see [deG], [E], [HKL1,2], [HLI,2]). A general
existence and regularity theorem was proven by R. Schoen and K. Uhlenbeck
(see [SU1,2,3]) for energy minimizing maps. In the case of three dimensional
domains, their result asserts that minimizers are regular except for isolated
points in the interior. There are several important earlier works which require
some restriction on the target manifold. These include the works of J. Eells and
J. H. Sampson [ES] for targets of nonpositive curvature, and the works of S.
Hildebrandt, H. Kaul, and K. O. Widman [HKW1,2] for maps whose images
lie in a convex coordinate neighbourhood.
If one considers a map p : aB3 --~ ,52 which has nonzero degree, then
any minimizer which agrees with p on aB3 necessarily possesses at least one
interior singular point. A natural question arises as to whether the singular points
are determined by the boundary data, or one can specify singular behavior as
well as boundary data. This question is unresolved.
The first special case of the problem of prescribing singularities is the
question of whether a boundary map p of zero degree has a regular harmonic
extension to B3. On the positive side, the theorem of S. Hildebrandt, H. Kaul
and K. O. Widman [HKW2] asserts that a map p, whose image lies in an open
hemisphere of S2, has a regular harmonic extension (also having an image
in this hemisphere). On the other hand, R. Hardt and F. H. Lin [HL2] have
constructed zero degree boundary maps p for which any minimizer must possess
interior singular points. Thus, any regular harmonic extension, in such a case,
Pervenuto alla Redazione il 4 Ottobre 1988.
356
would have larger energy than the infimum over all W1.2 extensions of p.
In this paper, we show that, for a general class of axially symmetric maps p
of zero degree, there is a regular axially symmetric harmonic extension of p to
B3. In particular, in many cases these harmonic maps have larger energy than
a 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 the analysis of a scalar partial differential equation in
two variables. This equation is degenerate on the boundary corresponding to
the axis of symmetry. Our main theorem can now be stated.
.
’
--+ 82 be a regular non-surjective axially symmetric
THEOREM. Let p :
map. There exists a regular axially symmetric harmonic extension of cp to B3.
To make precise the symmetry condition we
unit ball and S2 c R3 be the unit sphere; i.e.
require:
let B3 c R 3 be the
and
We
are
going
to use the
following
coordinates in
B3,
where
The Euclidean metric is then
Choose coordinates
The metric
The
on
(0, ~)
in S2
as
follows,
S’2 then takes the form
axially symmetric
Using
given by
harmonic maps
these coordinates,
our
problem
we
is
consider have the form
now
to solve the
following equation:
357
with the
boundary condition 0 = Q on a D,
where e is
where Q satisfies
positive number.
solve the above equation by finding
a
We will
energy functional:
a
smooth critical
point
of the
which is defined in the Hilbert space
2. - Several Lemmas
LEMMA 1.
provided 0
E
where C is
a
PROOF.
Any critical point 0 of E
Co (D)
fixed
is
a
local minimum in the
sense
that
and
constant.
Suppose 0
E H
is
a
critical
point
of E
and 0 is
a
map in
Co (D).
358
Using integration by parts, equation (1),
and
Taylor’s theorem,
we
have
Since the first eigenvalue of -0 on a bounded domain n c Rn is bounded
from below by {diameter (n)} - 2, as long as the required condition is satisfied
D
by supp (0), inequality (4) holds.
The next lemma is
a
consequence of Lemma 1 and standard
elliptic regularity
theory (see [M]).
LEMMA 2. The critical
interior of D.
Using
Lemma 1,
points of
we now
the
derive the
in H
equation (1)
following
are
regular
existence result.
are critical points of E and 0
LEMMA 3. Suppose ¢1 and
D. Let cp; be the boundary values of Oi, 1, 2. Suppose p is a function
lying between ~pl and y~2, i.e. VI :5 V:5 (P2- Then
=
is
~
critical value
E H such that
a
of E
and is achieved
by
a
in the
critical
point 0,
on
on
8D
i.e. there exists
°
Moreover, ~1 :5 0:5 tP2
on
D and
’
=
cp.
PROOF. Consider a minimizing sequence (§n ) which has a weak limit 0. It
is easy to see that E(¢)
Therefore,
c; in fact, 0 is the strong limit of
we may assume
=
359
In order to show that 0 is a critical point of E, we need to prove that
for all 0 E Co . But it is enough to show that (5) holds for those
(5) holds
0 E Co
satisfying
Let 0
be such
Co
a
function and consider
We define
Since
tP2, St and Mt satisfy the following conditions:
01 :5 0
Define
By
the definition of
0,
we
know that
jE*(~)
Using
Lemma 1,
we
have
Combining
Therefore
We
are
these two
inequalities,
we
get
(4) holds for every 0. The lemma is proved.
going
solution, the
D
to use Lemma 3 to prove our theorem. We will take the trivial
function, as our Q 1. The rest of the proof is to show that
zero
there exists a ~2, for any given boundary data p, satisfying the condition of
Lemma 3. Our ~2 will be a z-independent solution, i.e. a function of r which
is a solution of the following ordinary differential equation
360
3. - The construction
Let
r
=
Of 02
et , t E ( - oo, 0~ .
The
equation (6)
~ We
then becomes
are going to use a variational
corresponding functional is
method to solve the above
equation.
We
with the
condition:
going
are
This
can
to solve
be done
equation (7)
by minimizing
following boundary
the functional Ei . It is easy to
see
The
that
Since
the
problem (7), (8)
0, be such
Let
a
has a monotone increasing positive solution.
solution. Multiply the equation (7) by 0’ and
integrate
get
Since
§r
E
WI, 2 (_
00, 0),
the constant is
zero.
Because
we
have
Thus
we
have solved the
Since 0 -=
7r
following
is the solution of
first order
ordinary
differential
(11) with the initial value
T
=
equation
we see
that
to
361
uniformly
Take
in any finite interval
a v E
and let tv ( 1") be the root of
(0,
Moreover, since sin
Now let
us
~-T, 0~. Furthermore,
0 (t)
0 (t),
go back to the
we
the
function
equation (6).
Define
on 8D satisfies the
(a) ~ == 0, for r = 0;
(b) Lipschitz’s condition
(c) p maxaD (p) 7r.
From
(b),
we
have two
First, from (14), there is
For this
TO,
we
v,
then
~T (r) by
our
following
at both
(0, 1)
l/J2 with proper choice
conditions:
and
(0, -1);
constants k and d such that
T° such that
have
when r
60 = min
with the same 60.
Secondly,
a
positive
=
have
The function 0’ is then a solution of (6) and will be
of T.
The inequality (15), in terms of r, is
Suppose
(t)
from
{8,
By (13),
(12), there is
a
T1
we
know that
TO such that
(19) holds for all
T
TO,
362
Therefore, 4;2 ==
4;1’1
4. - The
of the Theorem
will serve as our upper barrier. Using Lemma 3, we know
that the problem (1), (2) has a regular solution 0 whenever the boundary value
p is regular and non-surjective.
proof
Let p be a regular boundary function satisfying the conditions of the theorem
and 0 be the corresponding regular solution of the problem (1), (2). Define the
map 4) : B 3 -~ S 2 by
~ is
We
an
are
axially symmetric
map with the
boundary
values
to show that (D is harmonic and smooth on
going
B3.
We know that -0 is harmonic away from the axis, i.e. ~ satisfies the
or,
equivalently,
We will prove that (D is harmonic in B3
_
equation
ft
Let T be
-
ft
a
by showing
that
(22) holds for all
...
Co (B3, R3)
map, and consider the
following
cut-off function
363
The first integral
tend to zero as l~
on
the
side is zero. The second and the third
because the measure of supp
goes to
is bounded. Hence 4D is harmonic on B3 .
---~
right
oo,
We know that (D is smooth away from the axis. Since
O is continuous on the axis. Therefore (D is smooth
Our main theorem has been proved.
5. - An
by
integrals
zero
and
construction
everywhere
in B3
(see [HI).
example
In this section
we
will show that the
regular
harmonic extension
we
get is
not a energy minimizer among the Wl,2 extensions. As a matter of fact it is
not even a energy minimizer among the axially symmetric W l2 extensions. We
by constructing a sequence of axially symmetric maps
all
of them have boundary values, say IV/,}, satisfying
1,Dk) WI, 2 (B3 , 82)
the conditions of our theorem. Therefore, pk have regular axially symmetric
harmonic extensions on B3 which will be denoted as
The difference between
and
is
(0k)
145k)
are
to show this
going
c
but
using
has
the
previous
For this reason,
lower bound. Since
coordinates, we can write
a
we
positive
only
need to
give
are
axially symmetric,
the definition of
otherwise
where k
=
4, 5, 6,.... Its boundary value is
where
I r 2 + z 2 1}. Since Qk only takes 7r and zero on
= ({r, z ) E
the axis, ~k is well defined. Recognize that, in the coordinates we are using,
the famous harmonic map from B~ to S 2 namely x 1-+ xlIxl can be written as
S1 1/ 2
=
364
Furthermore, it is easy
Therefore,
It is obvious that pk is C1 on
Hence, by
CPk
extension
Now,
our
theorem,
except two points (*,±(1 - 2/k)) and
SI/2
has a smooth
harmonic
axially symmetric
pk
~k = (B, ~k).
we are
going
to estimate
is
For any
From Section 3,
is
to see that
we
a
smooth function of
know that
is
positive. Therefore,
positive.
REMARK. The same result holds for those axially symmetric bounded R3
domains diffeomorphic to B3 and also for axially symmetric boundary maps
which have the form of
where n is
an
integer;
in fact the
Euler-Lagrange equation
then becomes
Acknowledgements
The author is very grateful to his adviser Professor Rick Schoen without
whose help this result would not be here.
365
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Stanford University
Stanford, California 94305