0
$\begingroup$

I have the following ODE:

$$ r^2 \frac{f''}{f} + r \frac{f'}{f} - (kr)^2 = n^2 $$

And I would like to transform it into the modified bessel function for $z=kr$, so

$$ z^2 u'' + z u' - (z + \nu)u = 0 $$

I can't seem to find an appropriate transformation though. Any ideas?

$\endgroup$
1
  • $\begingroup$ Your equation is not a Bessel's equation $z^2 u'' + z u' - (z + \nu)u = 0$ $\endgroup$
    – user577215664
    Commented Jan 9, 2020 at 13:55

1 Answer 1

Reset to default
0
$\begingroup$

$$r^2 \frac{f''}{f} + r \frac{f'}{f} - (kr)^2 = n^2$$ $$r^2 {f''} + r {f'} -( (kr)^2 + n^2)f=0$$ Substitute the variable $z=kr$: $$z=kr \implies f'=\frac{df}{dr}=\frac{df}{dz}\frac{dz}{dr}=k\frac{df}{dz}$$ $$(rk)^2 {f''} + rk {f'} -( (kr)^2 + n^2)f=0$$ $$z^2 {f''} + z {f'} -( z^2 + n^2)f=0$$ And this is Bessel's modified equation. Not that $z^2 u'' + z u' - (z + \nu)u = 0$ is not a Bessel's equation. Maybe you made a typo ?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .