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Orthogonality and Weighting Function of Sturm-Liouville Equation

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A set of eigenfunctions yn(x) satisfies the following Sturm-Liouville equation:
    [itex] \frac{d(f(x)*y'_{m})}{dx}+\lambda*\omega*y_{m}=0[/itex]
    with following boundary conditions:
    [itex] \alpha_{1}y+\beta_{1}y'=0[/itex]
    at x=a
    [itex] \alpha_{2}y+\beta_{2}y'=0[/itex]
    at x=b
    Show that the derivatives un(x)=y'n(x) are orthogonal functions.
    Determine the weighting function for these functions.
    What boundary conditions are required for orthogonality?

    2. Relevant equations

    Orthogonal functions:
    [itex]\int(dx*\omega*y_{n}(x)*y_{m}(x)=0[/itex]
    Integrate from a to b.
    3. The attempt at a solution
    I'm not sure how to start this problem, can someone point me in the right direction?
     
  2. jcsd
  3. Oct 30, 2012 #2

    Mark44

    Staff: Mentor

    Problems that involve differential equations should be posted in the Calculus & Beyond section, not in the Precalculus Math section. I am moving this thread to that section.
     
  4. Oct 30, 2012 #3
    Sorry about that...
    Anyway, well I found a way to prove orthogonality and ended up with:
    [itex] (\lambda_{m}-\lambda_{n})\int(w*y_{n}*y_{m}) =0[/itex]
    (integral from a to b)
    Now how do I find the weighting function?
     
  5. Oct 30, 2012 #4

    LCKurtz

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    Of course you must mean the definite integral$$
    (\lambda_{m}-\lambda_{n})\int_a^b(w*y_{n}*y_{m}) =0$$That is a standard result in S-L theory, and the weight function is the ##w## in the integrand. However, your original post asked you to show the derivatives of the ##y_n## were orthogonal. I wondered when I saw your OP whether that was a typo or whether it was true. In any case, it isn't what you found the proof for.
     
    Last edited: Oct 30, 2012
  6. Oct 30, 2012 #5
    I don't quiet understand what it means then to show if the derivatives are orthogonal...
     
  7. Oct 30, 2012 #6

    LCKurtz

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    Above, if ##\lambda_m\ne \lambda_n## then ##\int_a^b(w\cdot y_{n}\cdot y_{m})\, dx =0##, which is what it means for ##y_m## and ##y_n## to be orthogonal with respect to the weight function ##w##. For the derivatives to be orthogonal with respect to some weight function ##f(x)##would mean ##\int_a^bf\cdot y'_{n}\cdot y'_{m}\, dx =0## if ##\lambda_m\ne \lambda_n##.
     
  8. Oct 30, 2012 #7
    Ok, so looking at the equation:
    [itex](\lambda_{m}-\lambda_{n})\int(f(x))y'_{n}y'_{m}=0[/itex]
    the only possibility is f(x)=0 because no matter what I do, I can't get terms to separate and moved to the right hand side.
     
  9. Oct 30, 2012 #8

    LCKurtz

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    Assuming your mathematics is at an advanced enough level to be studying S-L problems and orthogonality, that comment is just silly. Above you stated that you understand how to get the orthogonality of the eigenfunctions ##y_n(x)## so presumably you have some idea of what is involved to prove orthogonality.

    I have shown you what you need to prove. I don't know offhand how to solve your problem and I'm not inclined to spend the evening figuring it out. It might even be easy; I don't know. But I'm sure you need to use the given DE and boundary conditions somehow and maybe even use the orthogonality of the ##y_n(x)## themselves. You don't start with the conclusion. Good luck with it.

    [Edit, added]:You might try mimicking the standard proof but changing it where required because your boundary conditions are different.
     
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