Page 20 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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Unit 1: Bessel s Functions

Notes
2s
n
s
( 1) s x ( 1) ( /2) n 2s 1
x
= n x
s !(n s )! 2 (s 1)!(n s )!
s 0 s 1
( 1) s x n 2s 1
x
= n J n ( ) x
s 1 (s 1)!(n s )! 2
In the last sum, let us replace s by r as
s = r + 1, then

( 1) r 1 x n 1 2r
x
x J n ( ) = n J n ( ) x 2
x
r 0 r !(n 1 r )!

( 1) r x n 1 2r
x
x
or x J n ( ) = n J n ( ) x
r 0 r !(n 1 r )! 2
= n J (x) x J (x)
n n+1
As the last sum is equal to J (x). Thus
n+1
x
x J n ( ) = n J (x) x J n+1 (x)
n
II. x J n ( ) + n J (x) = x J n 1 (x)
x
n
Again, we have

s
s
( 1) (n 2 ) x n 2s
x
x J n ( ) = n 2s
s 0 s !(n s )! 2
s
( 1) (2n 2s n ) x n 2s
=
s !(n s )! 2
s 0
s
s
( 1) (2n 2 ) x n 1 2s x
x
= nJ n ( )
s !(n s )! 2 2
s 0
s
( 1) (n s ) x n 1 2s
x
= x nJ n ( )
s !(n s )! 2
s 0
n 1 2s
( 1) x
x
= x n J n ( )
s !(n 1 s )! 2
s 0
{As (n + s)! = (n + s)(n 1 + s)!}
Thus identifying the sum with J (x), we have
n 1
x J (x) = x J (x) n J (x)
n n 1 n

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