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( ·,
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).
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