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

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I have derived these pair of coupled diff equations for [itex]U_1 (r)[/itex] and [itex]U_2 (r)[/itex]:

[itex]r^2 \dfrac{d^2 U_1 (r)}{dr^2} + r \dfrac{d U_1 (r)}{dr} + r^2 U_2(r) = 0[/itex]

and [itex]r^2 \dfrac{d^2 U_2 (r)}{dr^2} + r \dfrac{d U_2 (r)}{dr} - r^2 U_1(r) = 0[/itex]

Or written in matrix form

[itex](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[/itex]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.
 
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julian said:
I have derived these pair of coupled diff equations for [itex]U_1 (r)[/itex] and [itex]U_2 (r)[/itex]:

[itex]r^2 \dfrac{d^2 U_1 (r)}{dr^2} + r \dfrac{d U_1 (r)}{dr} + r^2 U_2(r) = 0[/itex]

and


[itex]r^2 \dfrac{d^2 U_2 (r)}{dr^2} + r \dfrac{d U_2 (r)}{dr} - r^2 U_1(r) = 0[/itex]

Or written in matrix form

[itex](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[/itex]


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

Thanks, the general solution of [tex]r^2 z'' + rz' - ir^2z = 0[/tex] is

[itex]z (r) = C I_0 (\sqrt{i}r) + D K_0 (\sqrt{i}r)[/itex]

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

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

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