Page 26 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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Unit 1: Bessel s Functions

1.7 Illustrative Examples Notes

Example 1: Show that

2
2
2
(i) x sin x = 2 (2 J 4 J + 6 J ...)
2 4 6
2
2
2
(ii) x cos x = 2 (1 J 3 J + 5 J ...)
1 3 5
Solution: (i) We know that
cos (x sin ) = J + 2 J cos 2 + 2 J cos 4 + ... ...(i)
0 2 4
,
Differentiating w.r.t. ‘ we get
sin (x sin ). x cos = 0 2.2 J sin 2 2.4 J sin 4 ... ...(ii)
2 4
, ,
Differentiating (ii) w.r.t. , we have
2
cos (x sin ). (x cos ) + sin (x sin ) (x sin )
2
2
2
= 2.2 .J cos 2 2.4 J cos 4 2.6 J cos 6 ... ...(iii)
2 4 0
Replacing by /2 in (iii), we get
2
2
2
x sin x = 2 (2 J 4 J + 6 J ...)
2 4 6
(ii) Start with
sin (x sin ) = 2J sin + 2J sin 3 + ...
1 3
, ,
Differentiate this twice w.r.t. as in part (i) and then replace by /2. Thus we can get the
required answer.
Example 2: Show that when n is integral
(a) J = cos(n x sin )d
n
0
(b) J = cos( cos )d
x
0
0
x
= cos( sin )d
0
and hence deduce that

x 2 x 4 x 6
J (x) = 1 + ....
2
2
2
0 2 2 2 .4 2 2 .4 .6 2
r
( 1) x 2r
= 2
r
r 0 2 . !r
Solution: We know that
cos (x sin ) = J + 2J cos 2 +... + 2J cos 2m +... ...(i)
0 2 2m
and sin (x sin ) = 2 sin .J + 2 sin 3 J + ...
1 3
+2J sin (2m + 1) +... ...(ii)
2m + 1

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