SINGULARITIES AT
t
=
oo
IN EQUIVARIANT HARMONIC MAP FLOW
9
PROPOSITION 4.3.
If
lv2l ::::; v1
for all sufficiently large t, then one has
IT3
I
CR4 L( )
::::;
(logR)2Y Y.
PROOF. The hypothesis lv2l v1implies that 0 ::::;
v
2v1. Since v1
RRt'l/h(y)
we find that
v(y, t)
is bounded by
CR2
lv(y,
t)l ::::;
C IRRtl yL(y) ::::;
I
log Rl
yL(y).
Using
f"(u;v)
=
f"(u)
+ ~vf( 3 l(u;v) we split
T3
into two terms,
T3
=
T4
+
T5
= ~
f"(U(y))
v
2 
~
f(
3l(U;
v) v3.
2 y2 6
y2
Since
f"(U(y))
=
8y(1
y2)/(1
+
y2)2,
we have
IT,
I I
f"(U)
v21
C (RR
)2
y
IY2 11
(yL(y))2
4 
2y2 
t
(1 + y2)2 y2
C R4
y
L( )2
 (logR)
2 1+y2
Y
C R4 L( )
 (log
R)
2
Y Y ·
To estimate
T5
we note that
f(3l(U;v)
=
f(3l(U
+ Ov)
=
4cos2(U + Ov) by the
Mean Value Theorem, and hence
ITo5
I= ~~f(3)(U;v)
31
~
lv31
C(RR
)3
L(
)3
6 y2
v 
3
y2  t y y
0
Using Lemma 4.1 we then get
CR6
3
CR6
R4
IT51::::;
(logR)3yL(y) ::::;
llogRiyL(y)::::;
C (logR)2yL(y).
0
4.4. Choice of v2.
We let
'l/J2
be a solution of M['lf;2]
=
yL(y),
e.g. we could
choose
'l/J2
=
X[yL(y)].
According to Proposition 3.1 we have
1
(4.3)
'l/J2(y)
=
(S +o(1))y3
logy as
y /
oo.
We set
R4
v2
=
k (log R)2
'l/J2(y),
where k E
JR.
is a constant to be specified below. The function
u
=
U(y)
+ v1
(y, t)
+
v2 (y, t)
will be a subsolution if
2at 8v2
J
[v1 + v2]
=
M
[v2] +
R

RRtYV2,y
+ T1 +
T2
+
T3 ::::;
0.
The opposite inequality will generate a supersolution.
LEMMA 4.4.
For any
k E
JR. there is a tk such that
lv2l ::::;
~v1
for t
?:
tk.