Page 32 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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Unit 1: Bessel s Functions

Notes
Example 6: Show that

x 1
n
n
(a) x J n 1 ( )dx = x J n , n 1.
x
2 n (n 1)
0
1 1
n
x
(b) x J n 1 ( )dx = n , n .
2 (n 1) 2
0
Solution:
(a) From recurrence formula II, we have
d n
n
x
x J n ( ) = x J ( )dx ...(i)
x
dx n 1
, ,
Integrating (i) w.r.t. x between the limits 0 and x, we get
x
x
n
n
x
x J n ( ) = x J ( )dx
x
0 n 1
0
x
x
J n ( ) n
n
x
x J ( ) lim x j ( )dx ...(ii)
x
n n n 1
x 0 x
0
J ( ) 1 x n x 2 1
x
But lim n lim . 1 ...
x 0 x n x 0 x n 2 n (n 1) 2.2.(n 1) 2 n (n 1)
Hence (ii) may be written as
x
n
n
x J ( )dx 1 x J ( )
x
x
n 1 n n
2 (n 1)
0
, ,
(b) Integrating (i) w.r.t. from 0 to we get
n
n
x
x J ( ) x J ( )dx
x
n
0 n 1
0
x
J ( ) J ( )
x
n
x
lim n lim n x J n 1 ( )dx ...(iii)
x 0 x n x 0 x n
0
x
J ( ) 1
As in part (a), lim n 1 n n ...(iv)
x 0 x 2 (n 1)
We know that for large values of x the approximate value of J n ( )is given by
x
1
2 2 1 1
J n ( ) ~ cos x n ,n
x
x 2 2 2
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