Mathematical Methods for Physics and Engineering : A Comprehensive Guide

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT

one solution in the form of a Frobenius series. We therefore substitutey=zσ

∑∞

n=0anz

n

into (16.21) and, using (16.13) and (16.14), we obtain

∑∞

n=0

(n+σ)(n+σ−1)anzn+σ−^2 +

3

z− 1

∑∞

n=0

(n+σ)anzn+σ−^1

+

1

z(z−1)

∑∞

n=0

anzn+σ=0,

which, on dividing through byzσ−^2 ,gives

∑∞

n=0

[

(n+σ)(n+σ−1) +

3 z

z− 1

(n+σ)+

z

z− 1

]

anzn=0.

Although we could use this expression to find the indicial equation and recurrence relations,
the working is simpler if we now multiply through byz−1togive

∑∞

n=0

[(z−1)(n+σ)(n+σ−1) + 3z(n+σ)+z]anzn=0. (16.22)

If we setz= 0 then all terms in the sum with the exponent ofzgreater than zero vanish,
and we obtain the indicial equation

σ(σ−1) = 0,

which has the rootsσ=1andσ= 0. Since the roots differ by an integer (unity), it may not
be possible to find two linearly independent solutions of (16.21) in the form of Frobenius
series. We are guaranteed, however, to find one such solution corresponding to the larger
root,σ=1.
Demanding that the coefficients ofznvanish separately in (16.22), we obtain the
recurrence relation

(n−1+σ)(n−2+σ)an− 1 −(n+σ)(n+σ−1)an+3(n−1+σ)an− 1 +an− 1 =0,

which can be simplified to give

(n+σ−1)an=(n+σ)an− 1. (16.23)

On substitutingσ= 1 into this expression, we obtain

an=

(

n+1

n

)

an− 1 ,

andonsettinga 0 = 1 we findan=n+ 1; so one solution to (16.21) is given by

y 1 (z)=z

∑∞

n=0

(n+1)zn=z(1+2z+3z^2 +···)

=

z

(1−z)^2

. (16.24)

If we attempt to find a second solution (corresponding to the smaller root of the indicial
equation) by settingσ= 0 in (16.23), we find

an=

( n

n− 1

)

an− 1.

But we requirea 0 =0,soa 1 is formally infinite and the method fails. We discuss how to
find a second linearly independent solution in the next section.

One particular case is worth mentioning. If the point about which the solution