# Coupled 2nd order diff eq's (Bessel functions?)

1. Feb 27, 2014

### julian

I have derived these pair of coupled diff equations for $U_1 (r)$ and $U_2 (r)$:

$r^2 \dfrac{d^2 U_1 (r)}{dr^2} + r \dfrac{d U_1 (r)}{dr} + r^2 U_2(r) = 0$

and

$r^2 \dfrac{d^2 U_2 (r)}{dr^2} + r \dfrac{d U_2 (r)}{dr} - r^2 U_1(r) = 0$

Or written in matrix form

$(r^2 \dfrac{d^2}{dr^2} + r \dfrac{d}{dr}) \begin{pmatrix} U_1(r) \\ U_2(r) \end{pmatrix} + r^2 \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} U_1(r) \\ U_2(r) \end{pmatrix} = 0$

I've tried a couple of things to try to solve them but didn't work. Any tips appreciated! Probably going to involve zeroth-order Bessel / modified Bessel functions? Thanks.

2. Feb 27, 2014

### pasmith

If you set $z = U_1 + iU_2$ then you get
$$r^2 z'' + rz' - ir^2z = 0$$
and if you then set $s = e^{i\pi/4} r$ you obtain
$$s^2 z'' + sz' - s^2z = 0$$
whose solutions are the modified bessel functions $I_0(s)$ and $K_0(s)$.

3. Feb 27, 2014

### julian

Thanks, the general solution of $$r^2 z'' + rz' - ir^2z = 0$$ is

$z (r) = C I_0 (\sqrt{i}r) + D K_0 (\sqrt{i}r)$

Noting $I_0 (x) = J_0 (ix)$ we have $I_0 (\sqrt{i}r) = J_0 (i^{3/2}r)$ and so the general solution is

$z (r) = C J_0 (i^{3/2}r) + D K_0 (\sqrt{i}r)$.

The real and imaginary parts of $J_0 (i^{3/2}r)$ and $K_0 (\sqrt{i}r)$ are the Kelvin functions apparently. Just need to learn a bit about them and apply my boundary conditions....thanks for your help.

Last edited: Feb 27, 2014