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.