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Sturm-Liouville Theory Question

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Given the ODE:
    [tex] (1-x^2) \frac{d^2y}{d^2x} - x \frac{dy}{dx} + n^2 y = 0[/tex]

    and that the weight function of the differential operator [tex] -(1-x^2) \frac{d^2y}{d^2x} + x \frac{dy}{dx} [/tex] is [tex] w = \frac{1}{\sqrt{1-x^2}} [/tex] to turn it into an SL operator, prove that:

    [tex] \int_{-1}^{1} \frac{dy_m}{dx} \frac{dy_n}{dx} \sqrt{1-x^2} dx = 0 [/tex]

    where [tex] y_m [/tex] and [tex] y_n [/tex] are orthogonal eigenfunctions of the ODE.

    2. Relevant equations

    All above.

    3. The attempt at a solution

    I attempted integration by parts of the LHS of the proposition, but that didn't seem to go anywhere.
    Last edited: Mar 24, 2014
  2. jcsd
  3. Mar 24, 2014 #2


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    Integration by parts worked for me.
  4. Mar 24, 2014 #3
    I tried integrating one of the derivatives, and differentiating everything else, but this ended with a 2nd derivative multiplied with a quotient.

    I also tried integrating 1 and differentiating the integrand. To get rid of the second derivatives, I use the ODE to substitute, but that still gave me a quotient term I couldn't get rid of.

    Can you point me in a specific direction?
  5. Mar 24, 2014 #4


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    Not really other than to say do the integration by parts correctly. Apparently you're doing something wrong or not seeing something, but without seeing your actual work, no one can say what it is. Vague descriptions aren't very helpful to us.
  6. Mar 24, 2014 #5
    How did you start off, then? Which part of the integrand did you choose to integrate and differentiate?
  7. Mar 24, 2014 #6


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    Clearly one doesn't want to integrate [itex]y_n'\sqrt{1 - x^2}[/itex], so one must differentiate [itex]y_n'\sqrt{1 - x^2}[/itex] and integrate [itex]y_m'[/itex]. You want to end up with a multiple of
    \int_{-1}^1 y_n(x) y_m(x) w(x)\,dx = \int_{-1}^1 \frac{y_n(x) y_m(x)}{\sqrt{1 - x^2}}\,dx
    which is zero if [itex]y_n[/itex] and [itex]y_m[/itex] are orthogonal.
  8. Mar 24, 2014 #7
    Ah, thanks.

    Finally figured it out!
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