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
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