6 SIGURD ANGENENT AND JOOST HULSHOF

and, integrating again,

u(y)

=

A¢"o(Y)

+ {

B +loy

TJ'¢o~TJ) 2 l''

ry''l/Jo(TJ')v(ry')dry'dry} 'l/Jo(y).

Here

- J

dy

'l/Jo(Y)

=

Y'¢o(y)2

is a solution of the homogeneous equation M¢"0

=

0 which is singular at

y

=

0.

Since we shall always require solutions of Mu

=

v

to be regular at y

=

0,

we set

the coefficient A

=

0, and choose B so that the solution we find vanishes at

y

=

1.

This leads to

(3.1) u(y)

=

Xv(y)

~f

'l/Jo(y)

1Y

TJ'¢o~TJ)

2

lo

11

ry''l/Jo(TJ')v(ry')dry'dry.

LEMMA

3.1. If v(y)

=

(C

+

o(1))ya.for y /'

oo,

then, assuming a

=1-

-1, -3,

X()- 1+o(1) a.+

2

v y-

(a+1)(a+3)y

as y /'

oo.

If a=

-1,

then

Xv(y)

=

(~C+o(1))

ylogy

More generally, ifv(y)

=

(1 + o(1))ya. (logy)

13

,

with

f3

-1, then for a

=1-

-1, -3,

1 + o(1)

f3

Xv(y)

=

ya.+2

(logy)

(a+1)(a+3) '

while for a=

-1 ,

f3

-1,

one has

1 + o(1)

f3+1

Xv(y)

=

(a+ 3) (/3 + 1) y (logy) .

PROOF.

For large y one has '¢0 (y)

=

(2 + o(1))y-

1

,

so, assuming a

=1-

-1, -3,

loy ry'¢0 (ry)v(ry)dry= loy

(2+o(1))rya.(logry)13

dry

2+o(1) a.+l(l ){3

=

a+

1

y ogy '

and thus

2

= -

rya.+

1

(logry)13

dry

/,Y----:-:--=-

1

1ry''l/Jo(TJ')v(ry')dry'dry

11

/,Y

ry2+o(1)

1

TJ'l/Jo(TJ) o

1

4 a+1

1 + o(1) a.+3 (

f3

2(a + 1)(a +

3)y

logy) ·

Multiply with '¢0 (y) ""'2/y, and the proposition follows. The case a= -1 follows

by a similar computation. 0