Page 25 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Here due to R = 0, it can be shown that for some a i.e. 0 x a, J ( ) satisfies the Boundary
n
Condition J ( a) = 0 ...(v)
n
And so the solutions of (iii) form an orthonormal set w.r.t. weight function P = .
So zeros of J ( ) if denoted by i = 1, 2,...
n in
Let
< < ... ....
1a 2a 3a m
So a = a
mn

thus = mn mn
a
d J n
Since both J and are continuous at = 0, therefore for each fixed n = 0, 1, 2... the Bessel
n d

function J ( ) (m = 1, 2,...) with = m , form an orthogonal set on the internal 0 x a w.r.t.
n mn mn a
weight P = i.e.

a
J n ( mn ) ( pn ) 0 for p m
J
n
0
So zeros of J (x) are useful in obtaining orthogonal properties of J (x). The details of the above
n n
discussion will be given in the later units.

Example: Prove that J (x) = 0 has no repeated roots except at x = 0.
n
Solution: If possible let be a repeated root of

J (x) = 0 at x = ...(i)
n
Thus J ( ) = 0 as well as J ( ) = 0 ...(ii)
n n
Now from recurrence formulae I and II,

x J (x) = n J (x) x J (x),
n n n+1
x J (x) + n J (x) = x J (x),
n n n 1
We have

n
J (x) = J (x) J (x) ...(iii)
n+1 x n n

n
J (x) = J (x) + J (x) ...(iv)
n 1 x n n
As J (x) = 0 and J ( ) = 0, we have from III and IV J ( ) = 0 and J ( ) = 0, i.e. for the same value
n n n+1 n 1
of x = , J (x), J (x), J (x) are all zero x, which is absurd as we cannot have two power series
n n+1 n 1
having the same sum function. Then J (x) = 0 cannot have repeated roots except x = 0.
n

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