The modified Bessel differential equation is always presented as
$$r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} – (r^2 + n^2)f(r) = 0$$
with solutions
$$f(r) = AI_n(r) + BK_n(r)$$
But if I had
$$r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} – (\alpha^2 r^2 + n^2)f(r) = 0$$
What form should have the solutions and how to prove it? It seems not to be a simple substitution $r' = \alpha r$, because $\alpha$ appears just one time in the second equation and not in all the $r$ terms (as I would expect instead).
Best Answer
Your intuition is correct. Let $r'= \alpha r$ and $f(r)=g(r')$. Then, using the chain rule yields
$$\begin{align} \frac{\partial f(r)}{\partial r}&=\frac{dr'}{dr}\frac{\partial g(r')}{\partial r'}\\\\ &=\alpha \frac{\partial g(r')}{\partial r'} \tag 1 \end{align}$$
Then, using $(1)$ we obtain
$$\begin{align} 0=&r^2\frac{\partial^2 f(r)}{\partial r^2}+r\frac{\partial f(r)}{\partial r}-\left(\left(\alpha r\right)^2+n^2\right)\\\\ &=\left(r'/\alpha\right)^2(\alpha^2)\frac{\partial^2 g(r')}{\partial r'^2}+(r'/\alpha)(\alpha)\frac{\partial g(r')}{\partial r'}-\left(\left(\alpha r'/\alpha\right)^2+n^2\right)g(r')\\\\ &=r'^2\frac{\partial^2 g(r')}{\partial r'^2}+r'\frac{\partial g(r')}{\partial r'}-\left(r'^2+n^2\right)g(r')\tag 2 \end{align}$$
The solution to $(2)$ is
$$g(r')=AI_n(r')+BK_n(r')=AI_n(\alpha r)+BK_n(\alpha r)=f(r)$$