Page 69 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 59
summation to vanish
−1 −1
c j = c j−2 = c j−2 , r = n j = 2, 3, 4, ··· (4.73)
(r + j − n)(r + j + n) j(2n + j)
This is the recurrence relation. Since c 1 = 0, all c j = 0 when j is an odd number. It is therefore
convenient to write j = 2k and note that

−1
c 2k = c 2k−2 (4.74)
2
2 k(r + k)
so that
(−1) k
c 2k = c 0 (4.75)
k!(n + 1)(n + 2) ... (n + k)2 2k
The Frobenius series is

∞ (−1) k 2k
x

u = c 0 x n 1 + (4.76)
k!(n + 1)(n + 2) ....(n + k) 2
k=1
n
Now c 0 is an arbitrary constant so we can choose it to be c 0 = 1/n!2 in which case the above
equation reduces to
∞ k
x
(−1) n+2k

J n = u = (4.77)
k!(n + k)! 2
k=0
The usual notation is J n and the function is called a Bessel function of the first kind of order n.
Note that we can immediately conclude from (4.77) that
n
J n (−x) = (−1) J n (x) (4.78)
Note that the roots of the indicial equation differ by an integer. When r =−n (4.72) does not
yield a useful second solution since the denominator is zero for j = 0or2n. In any case it is easy
n
to show that J n (x) = (−1) J −n , so when r is an integer the two solutions are not independent.
A second solution is determined by the methods detailed above and involves natural
logarithms. The details are very messy and will not be given here. The result is
∞
2 (−1) k+1 [φ(k) + φ(k + 1)] 2k+n
x
Y n (x) = J n (x) ln + γ + x
π 2 2 2k+n+1 k!(k + n)!
k=1
n−1
2 (n − k − 1)! 2k−n
− x (4.79)
π 2 2k−n+1 k!
k=0
In this equation (k) = 1 + 1/2 + 1/3 + ··· + 1/k and γ is Euler’s constant 0.5772156649
......