Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity?

Question

The spherical Bessel equation is: $$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$

If I seek a Frobenius series solution, I will have: \begin{align*} &\quad y = \sum_{n = 0}^{\infty} a_nx^{n + r} \\ &\implies y' = \sum_{n = 0}^{\infty} (n + r)a_nx^{n + r - 1} \\ &\implies y'' = \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r - 2} \end{align*} Substituting into the ODE of interest: \begin{align*} &\quad x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0 \\ &\equiv \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r} + \sum_{n = 0}^{\infty} 2(n + r)a_nx^{n + r} + \sum_{n = 0}^{\infty} a_nx^{n + r + 2} + \sum_{n = 0}^{\infty} \frac{-5}{16}a_nx^{n + r} = 0 \\ &\equiv \sum_{n = 0}^{\infty} [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_nx^{n + r} + \sum_{n = 2}^{\infty} a_{n - 2}x^{n + r} = 0 \\ &\equiv [r(r-1) + 2r - (5/16)]a_0 + [(r+1)(r) + 2(r+1) - (5/16)]a_1 + \\ &\quad \sum_{n = 2}^{\infty} ([(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2})x^{n + r} = 0 \\ &\implies [r(r-1) + 2r - (5/16)] = 0 \wedge \\ &\quad [(r+1)(r) + 2(r+1) - (5/16)] = 0 \wedge \\ &\quad [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2} = 0 \end{align*}

The first conjunct is the standard indicial equation specifying $r$. The second conjunct is yet another quadratic that $r$ has to satisfy. Is there a mistake? Or should $a_1 = 0$?

Answer 1

You set the value of $r$ found in the first equation into the second equation. It then defines $a_n$ in terms of $a_{n-2}$ for all $n \geq 2$. So, once we know $a_0$, $a_1$ we get $a_2$ from the second equation and so on.

Answer 2

The main indicial is the the first equation. You get two roots from the first equation so you should set $a_1=0$. If there is a common root between the first and second equations, then you can consider that root and set both $a_1$ and $a_0$ nonzero.