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Orthogonal eigenfunctions

  1. Dec 2, 2007 #1
    [SOLVED] orthogonal eigenfunctions

    From Sturm-Louisville eigenvalue theory we know that eigenfunctions corresponding to different eigenvalues are orthogonal. For example,

    [tex]\Phi_{xx} + \lambda \Phi = 0[/tex]

    would be of Sturm-Louisville form (note: [itex]\Phi_{xx}[/itex] represents the second derivative of Phi with respect to x)

    [tex]\frac{d}{dx}[p(x) \frac{d \Phi}{dx}] + (q(x) + \sigma(x) \lambda)\Phi = 0[/tex]

    where q = 0, p = 1, and [itex]\sigma[/itex] = 1. The boundary conditions are

    BC1: [tex]\Phi(0) = 0[/tex]

    and

    BC2: [tex]\Phi_x(L) + h \Phi(L,t) = 0 [/tex]

    which means that there can be one eigenfunction corresponding to a negative eigenvalue and the rest are sines relating to the positive eigenvalues, specifically that the eigenfunctions would be

    [tex]\Phi_n(x) = \left\{ {\begin{array}{*{20}c}
    {sinh \sqrt{s_1}x} & {n=1} \\
    {sin\sqrt{\lambda_n}x} & {n > 1} \\
    \end{array}} \right.[/tex]

    Basically all I need to do is show that

    [tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = 0[/tex]

    with integration by parts I have gotten to the point

    [tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = \frac{\sqrt{s_e}cosh(\sqrt{s_e}L)sin(L\sqrt{\lambda}) - \sqrt{\lambda_n}cos(\sqrt{\lambda_n}L)sinh(L\sqrt{s_e})}{s_e+\lambda}[/tex]

    and mathematica gives the same thing. I don't see this being zero any time soon. Any suggestions?
     
  2. jcsd
  3. Dec 2, 2007 #2

    Avodyne

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    I think it is zero. Use the information on lambda and s that you get by imposing the second BC.
     
  4. Dec 2, 2007 #3
    The second BC gives transcendental equations though.

    [tex] tan(\sqrt{\lambda}x) = -\sqrt{\lambda}/h[/tex]

    and

    [tex]tanh(\sqrt{s}L) = -\sqrt{s}/h[/tex]

    which require numerical methods. I'm pretty sure this can be done analytically.
     
  5. Dec 2, 2007 #4

    Avodyne

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    You don't have to solve them, you have to USE them to show that your overlap integral is zero.
     
  6. Dec 3, 2007 #5
    I'm probably going to feel stupid when I finally figure it out, but I still don't really see what to do. The second BC will tell us that the eigenvalues are different, which is why Sturm-Louisville would say that the integral must be zero. Beyond that though, I can't see how it directly influences the integration.
     
  7. Dec 3, 2007 #6

    Avodyne

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    I can write your overlap integral as
    [tex]\frac{\sinh(\sqrt{s_e}L)\sin(\sqrt{\lambda_n}L)}{s_e+\lambda_n}\left[
    \sqrt{s_e}\coth(\sqrt{s_e}L) - \sqrt{\lambda_n}\cot(\sqrt{\lambda_n}L)}\right][/tex]
    Now look at the equations you get from the boundary conditions, and use them to simplify each term in the square brackets ...
     
    Last edited: Dec 3, 2007
  8. Dec 3, 2007 #7
    Shoot dang Avodyne, nice algebra trick and nice foresight. Thanks for the help.
     
  9. Dec 3, 2007 #8

    Avodyne

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    You're welcome!
     
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