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

$x²y^{″} + xy^{′} + x²y = 0.$ (28)

Question

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

[latex]x²y^{″} + xy^{′} + x²y = 0.[/latex] (28)

Accepted Answer

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

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

[latex]y^{″} + \frac{1}{x}y^{′} + \frac{x^{2}}{x^{2}} y = 0.[/latex]

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 [latex]p_{0} = 1 and q_{0} = 0[/latex], 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

[latex]y(x) = x^{0} \sum\limits_{n=0}^{\infty}{c_{n} x^{n}}[/latex]

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

[latex]\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.[/latex]

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

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

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

[latex]c^{n} = - \frac{c_{n-2}}{n^{2}}[/latex] for n ≧ 2. (29)

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

[latex]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}.[/latex]

Evidently, the pattern is

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

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

[latex]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} + . . . [/latex] (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.

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

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