# Modified Bessel differential equation

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

$$f(r) = AI_n(r) + BK_n(r)$$

$$r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} - (\alpha^2 r^2 + n^2)f(r) = 0$$

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).

• Maple says this here $$f \left( r \right) =1/4\,{\alpha}^{2}{r}^{2}+1/2\,{n}^{2} \left( \ln \left( r \right) \right) ^{2}+{\it \_C1}\,\ln \left( r \right) +{ \it \_C2}$$ – 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

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

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

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