2
SIGURD ANGENENT AND JOOST HULSHOF
Here the right hand side is the component of
!:1Ft
E
JR3 which is tangential to
S
2
.
We will always assume "Dirichlet-type" boundary conditions, i.e.
Ft
I
8D
2
is fixed.
As is well known (see
[S85],[S90])
a classical solution of (1.1) exists for each C
1
initial data
F0
:
D
2
----+
S
2
that satisfies the boundary condition. If {
Ft
I
0
s;
t T}
is a maximal classical solution then each
Ft
with
t
0 is
coo.
If this solution only
exists for finite time, i.e. if
T
oo then
lim sup
IV'
Ftl
=
oo.
t/'T D2
There exist examples of solutions which become singular in finite time
[CDY],[CG].
If a solution exists for all time
t E
[0,
oo ), and if its gradient does not blow up,
i.e. if
sup
IY'Ftl
oo,
D
2
x[O,oo)
then the maps
Ft
must converge to harmonic maps (in the sense of w-limit sets:
any sequence
tj /
oo has a subsequence {
tjk}
for which the maps
Ftjk
converge in
coo
to a harmonic map.) However, even if a solution does not become singular in
finite time its gradient can still become unbounded as
t /
oo. This will certainly
happen if there is no limiting harmonic map
F
00
:
D
2
----+
S
2
for the maps
Ft
to
converge to. In this note we consider an example of such a solution and determine
the precise rate at which the gradient grows as
t
----+
oo.
1.1.
A problem with long time blow-up.
If one makes the ansatz
(
cos
e
sin cp(r,
t) )
Ft ( r,
e)
=
sin
e
sin
cp ( r, t)
cos cp(
r, t)
one finds that harmonic map flow is equivalent to the following PDE for
cp:
8cp 82 cp
1
8cp
f (
cp)
(1.
2)
at
=
8r2
+ -:;.
8r -
---:;.2'
where 0
s;
r
s;
1, and where
f( cp)
=
~sin
2cp.
We consider the equivariant harmonic map flow equation (1.2) with boundary
condition
(1.3) cp(O,
t)
=
0 cp(1,
t)
=
7r.
Let cp(r,
t)
be a solution of (1.2) whose initial data satisfy
(1.4)
{
cp(·,
0)
E
C
1
([0, 1]),
0 cp(r,O) 1 for 0
s;
r
s;
1,
cp(O,
0)
=
0,
cp(1,
0)
=
7r,
Moreover, we will assume that
(1.5)
::J!Ro
E
(0, 1) : cp(Ro, 0)
=
2.7l'
LEMMA
1.1.
For each
t
0
there is a unique R'P(t)
E
(0, 1)
such that
cp(R'P(t),
t)
=
7r
/2.
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