Question 8.6.2: Find a Frobenius series solution of Bessel’s equation, t^2y...

n section 8.5.3, we derived a solution to (8.6.13) in the case where λ is an integer. Thus, in what follows we assume that λ>0 is not an integer. Since p(t ) = 1 and q(t )=−λ^{2} +t^{2}, we have p_{0} = 1 and q_{0} =−λ^{2}, which tells us that the indicial equation is

r (r −1)+r −λ^{2} = r^{2} −λ^{2} = 0

Thus, r =±λ. Choosing r = λ and using (8.6.4), (8.6.5), and (8.6.6), we find that the three relevant series for the differential equation (8.6.13) are

(t^{2} −λ^{2})y(t ) =\sum\limits_{k=0}^{∞}{b_{k} t^{k+λ+2}}-\sum\limits_{k=0}^{∞}{b_{k} t^{k+λ+2}}

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

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

From the form of Bessel’s equation, the sum of these three expressions vanishes; adding and simplifying, we observe that

\sum\limits_{k=0}^{∞}{k(k +2λ)b_{k} t^{k+λ}}-\sum\limits_{k=0}^{∞}{b_{k} t^{k+λ+2}}=0

To combine the sums, we step up the index in the second summation by 2 and find

(1+2λ)b_{1}t^{1+λ}- \sum\limits_{k=2}^{∞}{[k(k +2λ)b_{k} +b_{k−2}]t^{k+λ}}=0

So, (1+2λ)b_{1} = 0, and

k(k +2λ)b_{k} +b_{k−2} = 0, k ≥ 2            (8.6.14)

One solution to this recurrence relation is obtained by setting b_{0} =1 and b_{1}=0. Then, since we are assuming that λ is not an integer and b_{1}=0, (8.6.14) implies that all odd-subscripted coefficients are zero and that

b_{k} =\frac {−1}{k(2λ+k)}b_{k−2}, k = 2,4, . . .

Therefore, it follows that in closed form we have

b_{2k} = \frac {(−1)^{k}2^{−2k}}{k!(1+λ)(2+λ) · · · (k +λ)}

and thus a Frobenius solution to the Bessel equation is

y(t ) = t^{λ }+\sum\limits_{k=2}^{∞}{\frac {(−1)^{k}2^{−2k}}{k!(1+λ)(2+λ) · · · (k +λ)}t^{2k+λ}}

Note that since λ > 0, the ratio test can be applied to show that this series converges for all values of t .