Page 14 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

,
Unit 1: Bessel s Functions

Notes
3
( 1) C
So C = 0
6 6
(n 1)(n 2)(n 3) 3 (2)
( 1) 4 C
C = 0
8 8
(n 1)(n 2)(n 3)(n 4) 4 (2)
    
    
    

Substituting the values of k, C , C , C ,... in equation for y we get
0 1 2
2 4
k x k x
y = x n C 0 0 0 ....
(n 1) 1 2 (n 1)(n 2) 2 2
2 1 x 4
1 x
1
= C x n 1 .... ...(19)
0
(n 1) 1 2 (n 1)(n 2) 2 2
If we now take C to be
0
1
C = ...(20)
0 n
2 (n 1)
Where (n) is a gamma function.
As you know the properties of gamma functions n (n) = (n + 1), for any value of n, so we get
various values of (n). The equation for y becomes

x n 1 x 2 ( 1) 2 x 4
y = 1 .....
2 n (n 1) (n 1) 1 2 (n 1)(n 2)1.2 2

1 x n 1 x n 2 ( 1) 2 x 4
= ...
(n 1) 2 (n 2) 1 2 (n 3) .2 2

( 1) s x n 2s
or y = ...(21)
(n 1 s s 2
)
s 0
Here we have used the fact that
(n + 1) (n + 1) = (n + 2),
(n + 2) (n + 2) = (n + 3) and so on.

Also 1, 2, 3... s = s (s 1)
,
The above solution is called Bessel s function J (x). Thus
n
( 1) s 1 x n 2s
J (x) = ...(22)
n (n 1 s ) (s 1) 2
s 0
For k = n and if n is not an integer then the other solution for k = n is obtained from the
equation of J (x) by replacing n n i.e.
n

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