Page 60 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

50 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and the coefficient of the u term is

(x − 1/2)
q(x) = (4.23)
x 2
2
Both have singularities at x = 0. However multiplying p(x)by x and q(x)by x the singularities
are removed. Thus x = 0 is a regular singular point. Assume a solution in the form of the
∞ n+r

c
Frobenius series: u = n=0 n x , differentiate twice and substitute into (4.21) obtaining
∞ ∞ ∞
n+1 1 n+r n+r+1
(n + r)(n + r − 1)x + (n + r)c n x + 2(n + r)c n x
2
n=0 n=0 n=0
∞ ∞
n+r+1 1 n+r
+ c n x − c n x = 0 (4.24)
2
n=0 n=0
The indices of the third and fourth summations are now shifted as in Example 4.1 and we find

1 1 r 1 1 n+r
∞
r(r − 1) + r − c 0 x + (n + r)(n + r − 1) + (n + r) − c n x
2 2 2 2
n=1
∞
n+r
+ [2(n + r − 1) + 1]c n−1 x = 0 (4.25)
n=1
Each coefficient must be zero for the equation to be true. Thus the coefficient of the c 0 term
must be zero since c 0 itself cannot be zero. This gives a quadratic equation to be solved for r,
and this is called an indicial equation (since we are solving for the index, r).
1 1
r(r − 1) + r − = 0 (4.26)
2 2
with r = 1and r =−1/2. The coefficients of x n+r must also be zero. Thus

[(n + r)(n + r − 1) + 1/2(n + r) − 1/2]c n + [2(n + r − 1) + 1]c n−1 = 0 . (4.27)

The recurrence equation is therefore

2(n + r − 1) + 1
c n =− c n−1 (4.28)
1 1
(n + r)(n + r − 1) + (n + r) −
2 2
For the case of r = 1

2n + 1
3
c n =− c n−1 (4.29)
n n +
2