8
where
SIGURD ANGENENT AND JOOST HULSHOF
T1
=
-R3 Rtt'lPl
T2
=
(RRt)
2
(y'l/J~(y)- 1/Jl(y))
1 f"(U; v)
2
T3 =-
v
2 y2
4.3. Estimation of the remainder terms. Assuming that
R
satisfies (2.9),
i.e.
(4.2)
dR
clef
2 arctan
R R
dt=-a(R),
wherea(R) = R'l/J
1
(
1/R) =(2+o(1)) 10g 1/R'
we now estimate the terms
Tj,
beginning with the time derivatives of
R.
LEMMA
4.1.
If R satisfies (
4.2)
then for large t one has
R
-Rt
= (2+o(1)) log 1/R
Ru
=
-a'(R)Rt
=
(4
+
o(1))
R
2
. (logR)
R2
(RRt)t
= (8 + o(1)) (log R)
2
PROOF.
This follows immediately from (4.2) and the fact that
d~V(l
= (2 +
o(1))(-logR)-
1
as
R "\, 0. D
It will be convenient to use
y::S:1
clef
{1
L(y)
= 1 + log+
y
=
1 +logy y
~
1
Then for all y
~
0 we have
cyL(y)
::S: '1/JI(Y) ::S:
CyL(y)
and
1'1/J~(y)l:::;
CL(y)
for certain
constants2
0 c
C
oo.
PROPOSITION
4.2.
Fort
---
oo
one has
R4
ITll + IT21:::;
c
2yL(y)
(log R)
PROOF.
From
T1
= -
R
3
Ru'l/J1
we have
IT1I:::; CR4 (logR)-
2'1/Jl(Y):::;
CR4 (logR)-
2
yL(y).
For T2 we have
D
2Here and elsewhere c and C stand for generic constants, whose value may change from line
to line.
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