Question 8.6.1: Find a Frobenius series solution for the Bessel–Clifford equ...

With a being a constant, we have p(t ) = 1 − a, so in the series expansion for p, p_{0} = 1−a. Moreover, q(t ) = t, so q_{0}= 0. Thus, for the given DE the indicial equation is

r (r −1)+(1−a)r = 0

Rearranging, we see that r (r −1+1−a) = r (r −a) = 0, and thus the roots of the indicial equation are r = 0 and r = a.

In the case that r = 0, the Method of Frobenius is providing an analytic solution to (8.6.9) of the form

y_{1} =\sum\limits_{k=0}^{∞}{b_{k} t^{k}}

Dividing both sides of (8.6.9) by t and substituting this expression for y using the standard series methods we have already discussed, it follows that

\sum\limits_{k=0}^{∞}{[(k +1)(k +1−a)b_{k+1} +b_{k} ]t^{k}}

from which we obtain the recurrence relation

b_{k+1} =\frac {−1}{(k +1)(k +1−a)}b_{k}           (8.6.10)

It follows from (8.6.10) that the closed form expression for b_{k} is

b_{k}= \frac{(−1)^{k}}{k!(1−a)(2−a) · · · (k −a)}b_{0}, k ≥ 1

so we find that

y_{1}(t ) = b_{0}(1+\sum\limits_{k=1}^{∞}{\frac{(−1)^{k}}{k!(1−a)(2−a) · · · (k −a)}t^{k}})                 (8.6.11)

which is valid for all t provided that a≠1,2, . . .. Note that from this recurrence relation, every b_{n} is a function of b_{0}, and thus there cannot be two linearly independent solutions to the Bessel–Clifford equation that are analytic at 0. Indeed, every solution linearly independent of y_{1}(t ) must be singular at 0. And while the equation has a singular point at the origin, there is an analytic solution there for every a except when a is a positive integer. We now turn to the other root of the indicial equation in search of a second solution to the Bessel–Clifford equation.

Using r = a, we have

ty(t ) =\sum\limits_{k=0}^{∞}{b_{k} t^{k+a+1}}

(1−a)ty′(t ) =\sum\limits_{k=0}^{∞}{(1−a)(k +a)b^{k} t^{k+a}}

t^{2}y″(t ) =\sum\limits_{k=0}^{∞}{(k +a)(k +a −1)b_{k} t^{k+a}}

Adding these equations forms the left side of the differential equation we aspire to solve; doing so and simplifying, we find that

0=t^{2}y″(t ) +(1−a)ty′(t )+ty(t )=\sum\limits_{k=0}^{∞}{k(k +a)b^{k} t^{k+a}}+\sum\limits_{k=0}^{∞}{b_{k} t^{k+a+1}}

Since the first term in the firstsumis zero, if we adjust the index of the summation in the second sum and combine, we have

\sum\limits_{k=1}^{∞}{[k(k +a)b_{k} +b_{k−1}]t^{k+a}}

from which it follows that

k(k +a)b_{k} +b^{k−1} = 0, k ≥ 1

This standard recurrence relation can be solved to write every b_{k}in terms of b_{0}.Indeed, we see

b_{k} = \frac {(−1)^{k}}{k!(1+a)(2+a) · · · (k +a)}b_{0}, k ≥ 1

so that the Frobenius series representation of the solution is

y_{2}(t ) = b_{0}t^{a}(1+\sum\limits_{k=1}^{∞}{ \frac {(−1)^{k}}{k!(1+a)(2+a) · · · (k +a)}t^{k}}                (8.6.12)

We close this example with a few important observations. First, if a =0, then the Frobenius solution y_{2}(t ) is identical to the earlier obtained y_{2}(t ). Moreover, if a is a non-negative integer, then the Method of Frobenius produces a Taylor series expansion that is analytic at t = 0. Thus, the cases for a valid analytic solution excluded by our approach in finding y_{1}(t ) are here reconciled. Finally, if a is not an integer, then y_{2}(t ) is singular at t = 0 and, together with the analytic y_{1}(t ) given by (8.6.11), we have found a linearly independent set of solutions for the Bessel–Clifford equation valid for t > 0.