Question 11.3.5: Find a Frobenius solution of Bessel’s equation of order zero...

Find a Frobenius solution of Bessel’s equation of order zero,

x²y^{″} + xy^{′} + x²y = 0.                     (28)

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In the form of (17), Eq. (28) becomes

y^{″} + \frac{p_{0} + p_{1} x + p_{2}x² + . . .}{x}y^{′} + \frac{q_{0} +  q_{1} x + q_{2}x² + . . .}{x²} y = 0.           (17)

y^{″} + \frac{1}{x}y^{′} + \frac{x^{2}}{x^{2}} y = 0.

Hence x = 0 is a regular singular point with p(x) ≡ 1 and q(x) = x², so our series will
converge for all x > 0. Because p_{0} = 1  and  q_{0} = 0, the indicial equation is

r(r – 1) + r = r² = 0.

Thus we obtain only the single exponent r = 0, and so there is only one Frobenius series solution

y(x) = x^{0} \sum\limits_{n=0}^{\infty}{c_{n} x^{n}}

of Eq. (28); it is in fact a power series.
Thus we substitute y = \sum{c_{n} x^{n}} in (28); the result is

\sum\limits_{n=0}^{\infty}{n(n – 1)c_{n} x^{n}} + \sum\limits_{n=0}^{\infty}{n c_{n} x^{n}} + \sum\limits_{n=0}^{\infty}{c_{n} x^{n+2}} = 0.

We combine the first two sums and shift the index of summation in the third by -2 to obtain

\sum\limits_{n=0}^{\infty}{n^{2} c_{n} x^{n}} + \sum\limits_{n=2}^{\infty}{c_{n-2} x^{n} }= 0.

The term corresponding to x^{0} gives 0 = 0: no information. The term corresponding to x^{1} gives c_{1} = 0, and the term for x^{n} yields the recurrence relation

c^{n} = – \frac{c_{n-2}}{n^{2}}              for n ≧ 2.          (29)

Because c_{1} = 0, we see that c_{n} = 0 whenever n is odd. Substituting n = 2, 4, and 6 in  Eq. (29), we get

c_{2} = – \frac{c_{0}}{2²},        c_{4} = – \frac{c_{2}}{4²} = \frac{c_{0}}{2² · 4²},       and   c_{6} = – \frac{c_{4}}{6²} = – \frac{c_{0}}{2² · 4² · 6²}.

Evidently, the pattern is

c_{2n} = \frac{(-1)^{n} c_{0}}{2² · 4². . . (2n)^{2}} = \frac{(-1)^{n}c_{0}}{2^{2n}(n!)²}.

The choice c_{0} = 1 gives us one of the most important special functions in mathematics, the Bessel function of order zero of the first kind, denoted by J_{0}(x). Thus

J_{0}(x) = \sum\limits_{n=0}^{\infty}{\frac{(-1)^{n} x^{2n}}{2^{2n} (n!)^{2}}} = 1 – \frac{x^{2}}{4} + \frac{x^{4}}{64} – \frac{x^{6}}{2304} + . . .                                     (30)

In this example we have not been able to find a second linearly independent solution of Bessel’s equation of order zero. We will derive that solution in Section 11.4; it will not be a Frobenius series.

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