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,