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Solving Sturn-Liousville problem

  1. May 3, 2004 #1
    My problem is:

    [tex] \frac{d^2\phi}{dx^2} + 6\frac{d\phi}{dx} + \lambda\phi = 0; \\ \frac{d\phi}{dx}(0) = 0, \frac{d\phi}{dx}(L) = 0. [/tex]

    I am told to begin by finding every eigenvalue and corresponding eigenfunction. I missed the last class where we went over this and the book is not giving me much advice on how to begin. Can someone just provide some help on how I begin solving this? I would show what I have but I have nothing yet ;).

    Thanks,
    -Jason
     
    Last edited: May 3, 2004
  2. jcsd
  3. May 3, 2004 #2
    Try a solution of the form

    [tex]\phi = e^{rx}[/tex]

    then consider cases for lambda, i.e. positive, zero, negative, and match boundary conditions.

    cookiemonster
     
  4. May 3, 2004 #3
    Sorry for seeming so dense, but I understand how to apply that to normal O.D.E's, but the eigenvalue is throwing me off. What I am doing is:

    [tex]
    Let \ \phi = e^{rx}
    [/tex]

    [tex]
    \frac{d\phi}{dx} = re^{rx}
    [/tex]

    [tex]
    \frac{d^2\phi}{dx^2} = r^2e^{rx}
    [/tex]

    After substituting this back in for [tex]\phi(x)[/tex] I get:

    [tex]
    r^2 + 6r + \lambda = 0
    [/tex]

    This is easily solved for [tex] \lambda = 0 [/tex], but how do I go about solving it for non-zero eigenvalues? I am pretty embarrased for having to ask questions on such a simple problem, but thank you for helping!
     
    Last edited: May 3, 2004
  5. May 3, 2004 #4
    How about the quadratic formula? =]

    cookiemonster
     
  6. May 3, 2004 #5
    Yeah I did that originally, and I think i was looking for some sort of fourier series like lambda to come about, and getting confused by that. I got it now though, thanks man.

    -Jason
     
    Last edited: May 3, 2004
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