14

SIGURD ANGENENT AND JOOST HULSHOF

Lemma 6.2 with

r

=

R'P(t4

+

t)

asserts that

~

=

cp(R'P(t4

+

t), t4

+

t)

2

cp+(BR'P(t4

+

t), t2

+ B2t),

whence

BR'P(t4

+

t)

:S:

R(t2

+

B2t)

=

e-(2+o(l))v't2H2t

=

e-(211+o(l))y't (t/'oo).

Division by

B

and replacement of

t4

+

t

by

t

leads to additional terms all of which

can be absorbed in the o( 1) in the exponential. We therefore find that

R'P ( t) :::;

e-(211+o(l))y't

for all

e

1, and thus

(6.3)

It remains to prove the opposite asymptotic inequality.

Let

B

1 be given, and choose a

t

5

0 so that

r _(t) B

for all

t 2 t

5

.

LEMMA 6.3.

For sufficiently large t

6

t

5

one has

and hence

for all t 2

0, 0

:S:

r

:S:

1.

As we pointed out in the introduction, this implies global existence of our

solution.

CoROLLARY 6.4.

Any solution cp(r, t) of

(1.2}

whose initial data satisfies (1.4),

( 1. 5) exists for all t

0.

Indeed, on any finite time interval [0, T] we have

cp(r, t) :::; cp_(Br, t

6

+

B2t),

so

that on some small interval 0 :::;

r :::; 8

one has

cp(r, t) :::; Cr

for some

C

oo. Thus

the maps

Ftk

defined in (1. 7) cannot converge to the stereographic projection.

PROOF OF LEMMA 6.3. Again,

cp_(Br, t)

converges uniformly to

1r

on any in-

terval [8, 1] with

8

0, in fact, the convergence is in C

1

([8, 1]). It follows from

r_(t) B

that

cp_(B,t)

1r.

On the other hand we may assume w.l.o.g. that

cp(1, 0)

=

1r,

and that

cp(r,

0) :::;

1r-

8(1 -

r)

for some small 8 0. (As before,

even if the initial

cp

fails to satisfy this condition, cp( ยท,

t)

will do so for any small

t

0.) So, for any given

8

0 we can find at 0 such that

cp(r,

0) :::;

cp_(Br, t)

holds for

8 :::; r :::;

1. On the short interval [0, 8] the initial data are bounded by

cp(r,

0) :::;

Cr

for some

C

oo. Clearly, for sufficiently large

t

0 one will have

cp(r,O):::; cp_(Br,t)

for

r

E [0,8].

Let

t

6

be such a large

t.

Then, since

rp(r, t)

=

cp_(Br, t

6

+

B2t)

is a subsolution,

and since

rp(1, t)

=

cp_(B, t6

+

B2t)

2

cp_(r _(t6

+

B2t),

t

6

+

B2t)

=

1r,

it follows from

the Maximum Principle that

rp(r, t)

2

cp(r, t)

for all

r

E [0, 1],

t

2

0, as claimed. D

This implies that

R'P(t)

2

R(t6

+

B2t) =

e-(

211+o(l))v't

(t /'

oo).

Once again, this holds for all

B

1, so that we have

R'P(t) 2

exp(

-(2

+

o(1))y't)

fort/' oo. combined with (6.3) we get

R'P(t)

=

exp(

-(2

+

o(1))y't).