Page 565 - Advanced_Engineering_Mathematics o'neil

15.3 Bessel Functions 545

Make the change of variables y = ux −1/2 to obtain

1

u = 1 − u. (15.10)
4x 2
2
As x increases, 1 − 1/4x → 1 and this differential equation for u more closely approximates
−x
x
u − u = 0, with solutions e and e . We therefore argue intuitively that, as x increases, any

solution of the differential equation for u which approaches ∞ as x →∞ will be approximated
x
by ce for some c.
x
More generally, this suggests the further transformation u =ve . Substitute this into equation
(15.10) to obtain
1

v + 2v + v = 0. (15.11)
4x 2
We will attempt a solution of this equation by a series of the form
1 1 1
v(x) = 1 + c 1 + c 2 + c 3 + ··· .
x x 2 x 3
Substitute this into equation (15.11) and rearrange terms to obtain
1 1 1 1

−2c 1 + + 2c 1 − 4c 2 + c 1
4 x 2 4 x 3

1 1
+ 6c 2 − 6c 3 + c 2
4 x 4

1 1
+ 12c 3 − 8c 4 + c 3 + ··· = 0.
4 x 5
Each coefficient of a power of 1/x must vanish, so
1
−2c 1 + = 0
4
9
c 1 − 4c 2 = 0
4
25
c 2 − 6c 3 = 0
4
49
c 3 − 8c 4 = 0
4
and so on. Then
1 9 3 2
c 1 = ,c 2 = c 1 = ,
8 16 2 · 8 2
2 2 2
2 2
3 5 3 5 7
c 3 = ,c 4 = ,
3!8 3 4!8 4
and so on. The pattern is clear and we write
2 2
2 2 2
1 1 3 2 1 3 5 1 3 5 7 1
v(x) = 1 + + + + + ··· . (15.12)
3
8 x 2 · 8 x 2 3!8 x 3 4!8 4 x 4
2
This suggests an expansion of the form
2 2 2
2 2
ce x
1 3 2 1 3 5 1 3 5 7 1
I 0 (x) = √ 1 + + + + + ··· , (15.13)
3
2
x 8 x 2 · 8 x 2 3!8 x 3 4!8 4 x 4
in which c is an appropriately chosen positive constant. The series on the right is actually a
divergent series. However, the partial sum of the first N terms approximates I 0 (x) as closely as
we like for x sufficiently large. Such an approximation is called an asymptotic expansion.Byan
√
analysis we will omit, it can be shown that c = 1/ 2π.
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October 14, 2010 15:20 THM/NEIL Page-545 27410_15_ch15_p505-562