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).

  • $\begingroup$ Maple says this here $$f \left( r \right) =1/4\,{\alpha}^{2}{r}^{2}+1/2\,{n}^{2} \left( \ln \left( r \right) \right) ^{2}+{\it \_C1}\,\ln \left( r \right) +{ \it \_C2} $$ $\endgroup$ – Dr. Sonnhard Graubner Dec 14 '15 at 19:16

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

$$g(r')=AI_n(r')+BK_n(r')=AI_n(\alpha r)+BK_n(\alpha r)=f(r)$$

  • 1
    $\begingroup$ Except for the fact that the $((\alpha r)^2 + n^2)$ term should have a minus sign, thank you, it was very helpful. $\endgroup$ – BowPark Dec 15 '15 at 11:26
  • $\begingroup$ You're welcome. And thank you for carching the typo! +1 for the catch $\endgroup$ – Mark Viola Dec 15 '15 at 14:51

If you scale the $r$ axis by a factor $\alpha$ then the first two terms do not change because the re-scaling of $r$ is compensated by a re-scaling of the derivatives.

  • $\begingroup$ thank you, this is an intuitive way to view this problem $\endgroup$ – BowPark Dec 15 '15 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.