SINGULARITIES AT t =
oo
IN EQUIVARIANT HARMONIC MAP FLOW 5
We will show in section 3 that 7j;1 is uniquely determined upto a multiple of 7j;0 (y),
and has the following asymptotic behaviour
(2.6)
7/J1
(y)
=
y logy
+
o(y logy), (y /
oo).
Equation (2.5) implies a relation between a and
R,
which in view of the asymptotic
behaviour of 7j;1 implies
(2.7)
We will choose
(2.8)
R2
a=
(2
+
o(1))--
1
.
log R
a(t)
=
-R(t)R1 (t).
The two relations (2.5) and (2.8) imply a differential equation for R, namely
dR arctanR R
(2·9)
dt
= -
2
R7j;
1
(1/ R)
=-
(2
+
o(
1))
log 1/ R
where we have used arctanx
=
x
+
o(x),
and (2.6). Integrating this differential
equation we find
R(t)
=
exp (- (2
+
o(1))
Jt).
(The integration constant can be absorbed in the o(1).) The function
u(y, t)
U(y)
+
a(t)7j;1 (y) is the formal solution found in
[BHK].
3. Inverting
JV(
3.1. A formula for M-
1

We observe that for each ,\ 0 the function
U -. (y)
=
U ( ,\y)
satisfies
d2U-.
1
dU-. f(U-.)
--+-----=0
dy2 y dy y2 '
so that we can differentiate this equation and set ,\
=
1. One finds that
7/Jo(Y)
=
yU1 (Y)
=
1
!yy2
satisfies
M [7/Jo] = 0.
To solve Mu
=
v we apply the standard method of "reduction of order." One puts
u
=
w'lj;0
and computes
1 )
11
1
1 )
7/Jb
I
1
7/Jb 7/J~
-J\1(7/Jow
=
w
+
-w - q(y w
+
2-w
+
--w
+
-w
7/Jo Y 7/Jo Y 7/Jo 7/Jo
11 {
1
27/Jb }
1
J\17/Jo
11 {
1
27/Jb }
1
=w
+ -+-
w +--w=w
+ -+-
w
Y /Jo 7/Jo Y 7/Jo
so that Mu
=
v is equivalent to
wll
+ {
~
+
27/Jb }
wl
=
~.
Y 7/Jo 7/Jo
Multiply with
y'lj;0 (y)
2
and integrate, to get
y'lj;0
(y)
2
w1 (y) =A+
laY
777/Jo(ry)v(ry)dry,
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