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

[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} = 0I'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.

 

[pasmith]

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.

If you set z = U_1 + iU_2 then you get
<br /> r^2 z&#039;&#039; + rz&#039; - ir^2z = 0<br />
and if you then set s = e^{i\pi/4} r you obtain
<br /> s^2 z&#039;&#039; + sz&#039; - s^2z = 0<br />
whose solutions are the modified bessel functions I_0(s) and K_0(s).

 

[julian]

If you set z = U_1 + iU_2 then you get
<br /> r^2 z&#039;&#039; + rz&#039; - ir^2z = 0<br />
and if you then set s = e^{i\pi/4} r you obtain
<br /> s^2 z&#039;&#039; + sz&#039; - s^2z = 0<br />
whose solutions are the modified bessel functions I_0(s) and K_0(s).

Thanks, the general solution of r^2 z&#039;&#039; + rz&#039; - 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.