# Converting second order DE to Bessel form

I am trying to convert the below second-order equation to Bessel form:

$$(1-s^2)\frac{\partial^2 G}{\partial s^2} + \alpha(1-s)\frac{\partial G}{\partial s} - \beta(1-s) G = 0$$

This looks very close to the second order modified Bessel equation:

$$s^2 \frac{\partial^2 G}{\partial s^2} + s \frac{\partial G}{\partial s} - (s^2-\nu^2)G = 0$$

But I can't quite convert to that form. I also attempted using the Frobenius method but the resulting indicial equation is a horrid mess. Any suggestions on how to transform the problem to something tractable would be welcome.

• As a start, you can factor out $1-s$ from each equation. The only variable coefficient is that of $G''(s)$, which is $1+s$. This in turn suggests substituting $t=1+s$ to make the coefficient as simple as possible. (This still isn't quite enough, though, and I don't have anything more clever to say.) – Semiclassical May 11 '17 at 14:48

$$(1-s^2)\frac{d^2 G}{d s^2} + \alpha(1-s)\frac{d G}{d s} - \beta(1-s) G = 0 \tag 1$$ $$(1+s)\frac{d^2 G}{d s^2} + \alpha\frac{d G}{d s} - \beta G = 0$$ Change of variable : $\quad x=s+1 \quad\to\quad \frac{d^2 G}{d x^2} + \frac{\alpha}{x}\frac{d G}{d x} - \frac{\beta}{x} G = 0$

Comparing to a generalized form of Bessel equation : $$\frac{d^2 y}{d x^2} + \frac{1-2p}{x}\frac{d y}{d x} +\left(\lambda^2 q^2x^{2(q-1)} +\frac{p^2-\nu^2q^2}{x^2} \right) y = 0$$ Which solution is known : $\quad y=c_1 x^p\text{J}_\nu (\lambda x^q)+c_2 x^p\text{J}_{-\nu} (\lambda x^q)$

$\begin{cases} 1-2p=\alpha \\ 2(q-1)=-1\\ p^2-\nu^2q^2=0\\ \lambda^2q^2=-\beta \end{cases} \quad\to\quad \begin{cases} p=\frac{1-\alpha}{2} \\ q=\frac{1}{2}\\ \nu=1-\alpha\\ \lambda=2\sqrt{-\beta} \end{cases}$

If $\beta<0$

$$\quad G=c_1 (s+1)^ {(1-\alpha)/2} \text{J}_ {1-\alpha} (2\sqrt{-\beta} \sqrt{s+1})+c_2 (s+1)^{(1-\alpha)/2}\text{J}_{\alpha-1} (2\sqrt{-\beta} \sqrt{s+1})$$

If $\beta>0$

$$\quad G=c_1 (s+1)^ {(1-\alpha)/2} \text{I}_ {1-\alpha} (2\sqrt{\beta} \sqrt{s+1})+c_2 (s+1)^{(1-\alpha)/2}\text{I}_{\alpha-1} (2\sqrt{\beta} \sqrt{s+1})$$

I and J are the Bessel and Modified Bessel functions respectively.

In other words, from equation $(1)$ the changes $\begin{cases} X=2\sqrt{-\beta} \sqrt{s+1}) \\ y(X)=(s+1)^{(\alpha-1)/2}G(s) \end{cases}$

leads to the Bessel differential equation : $$\frac{d^2y}{dX^2}+\frac{1}{X}\frac{dy}{dX}+\left(1-\frac{(\alpha-1)^2}{X^2} \right)y(X)=0$$

Or the changes $\begin{cases} X=2\sqrt{\beta} \sqrt{s+1}) \\ y(X)=(s+1)^{(\alpha-1)/2}G(s) \end{cases}$ leads to the modified Bessel differential equation : $$\frac{d^2y}{dX^2}+\frac{1}{X}\frac{dy}{dX}+\left(-1-\frac{(\alpha-1)^2}{X^2} \right)y(X)=0$$

Note : In case of $\alpha=1$ the two basic solutions $\text{I}_{1-\alpha}(X)$ and $\text{I}_{\alpha-1}(X)$ are no longer independent. One of them must be replaced by the Bessel function of second kind $\text{Y}_0(X)$. The same for the modified Bessel function of second kind $\text{K}_0(X)$.

• Thanks! Do you have a reference to the solution of the generalized Bessel functions? This would be helpful for later use. – Eweler May 12 '17 at 2:50
• This can be currently found in the mathematical handbooks. For example equation (3) in : mathworld.wolfram.com/BesselDifferentialEquation.html . Of course the symbols used are not always the same from an edition to another. – JJacquelin May 12 '17 at 6:03