Theoretical Basis for Eigenvalue transformation on Bessel's Equation

Question

The method I've been taught for finding all of the eigenvalue solutions to Bessel's operator $$b(f)\equiv f''(x)+\frac{1}{x}f'(x)$$ goes as follows. Let $g(a)=f(\sqrt{\lambda}x)$. Then $$b(g)=\lambda g(a)$$ $$\lambda \frac{d^2g}{da^2}+\frac{\sqrt{\lambda}}{x}\frac{dg}{da} -\lambda g(a)=0$$ $$\frac{d^2g}{da^2}+\frac{1}{a}\frac{dg}{da}-g(a)=0$$ We can then solve this equation for $g(a)$ and apply the inverse change of variables to yield all of our eigenfunctions. My question is: what kind of transformation is this change of variables applying to our Hilbert Space? I would think that, given the nice algebraic properties of the operator (call it $M_a(\cdot)$) which when applied to some function $h(x)$ returns $h(a)$, it would be fairly straightforward to describe, but I'm not seeing it. It almost feels like we're applying a Mobius transform on our Hilbert Space where every for every function $f(x) \in H$, there are two subspaces in $H$, one which corresponds to the scaling of $f(x)$ and another which corresponds to evaluating $f(x)$. By extension, what kind of linear operators can have their eigenfunctions found in this way? I would think that there is something analogous to the characteristic polynomial method in finite dimensional linear algebra going on here, but I can't put my finger on it. Any and all insights are welcome, although geometric arguments are at somewhat of a premium here.

Answer 1

The Bessel operator $Lf = -(xf')'$ applied to functions on $(0,\infty)$ has a nice scale property. Define $S_{\alpha}$ acting on such functions by $(S_{\alpha}f)(x)=f(\alpha x)$ for $0 < \alpha < \infty$. Then \begin{align} LS_{\alpha}f & = -x\frac{d^{2}}{dx^{2}}f(\alpha x) -\frac{d}{dx}f(\alpha x) \\ & = -x \alpha^{2}f''(\alpha x)-\alpha f'(\alpha x) \\ & = \alpha S_{\alpha}\{-xf''(x)-f'(x)\} \\ & = \alpha S_{\alpha} Lf. \end{align} Therefore, if you have a solution $Lf = \lambda f$ on $(0,\infty)$, then $$ LS_{\alpha} f = \alpha\lambda S_{\alpha}f $$ So the one solution of $Lf=\lambda f$ can be used to get a solution of $Lf=\alpha\lambda f$ for every $\alpha > 0$. This has to do with the radial scaling properties of the Laplacian, which are inherited by the Bessel operator in the radial variable. If you have the two linearly independent solutions of $Lf=f$ on $(0,\infty)$, then you have all the solutions of $Lf=\alpha f$ for all any $\alpha > 0$.

Knowing two linearly independent solutions of $Lf=\lambda f$ on $(0,\infty)$, you can get them all for all parameters $\lambda$. You can then restrict to an interval if you want. If you want to find an eigenvalue function that vanishes at $x=b$, then you can apply $S_{\alpha}$ to the one eigenfunction $Lf=\lambda f$ and adjust $\alpha$ until $f(\alpha b)$ hits a zero of $f$. And this is general method for dealing with Bessel's equation. The action of the radial scaling group must act on the eigenspaces; this is a general principle.