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

  1. Feb 27, 2014 #1


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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]


    [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.
  2. jcsd
  3. Feb 27, 2014 #2


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    If you set [itex]z = U_1 + iU_2[/itex] then you get
    r^2 z'' + rz' - ir^2z = 0
    and if you then set [itex]s = e^{i\pi/4} r[/itex] you obtain
    s^2 z'' + sz' - s^2z = 0
    whose solutions are the modified bessel functions [itex]I_0(s)[/itex] and [itex]K_0(s)[/itex].
  4. Feb 27, 2014 #3


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    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.
    Last edited: Feb 27, 2014
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