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Sturm-Liouville confusion

  1. Dec 12, 2014 #1
    So the basic idea is that the differential operator acting on the vector space V of functions f:[a,b]->ℂ, with some weight function w(x)
    L≡d/dx[p(x)d/dx]+q(x)
    is Hermitian (self-adjoint) if we have that for any two functions u(x), v(x) in this vector space,
    [u*(x)p(x)dv/dx]ab=0,
    thus L has real eigenvalues and it's eigenfunctions form an orthonormal basis for V.

    Now my notes seem to be suggesting to me that if we write the SL problem, i.e the eigenvalue equation
    Lennen
    for eigenfunctions en and eigenvalues λn, then so long as the eigenfunctions satisfy the above boundary condition, i.e [en*(x)p(x)dem/dx]ab=0, then L is Hermitian. This is confusing me because as I stated above, I'm sure this boundary condition must apply for all u,v in the space of functions V, not just the eigenfunctions.

    There is an example of this form given, which is
    d2endx2=λnen.
    It finds the eigenfunctions as exp[(√λn)x] and exp[-(√λn)x] where the eigenvalues are yet to be determined. Then it says let our interval be [a,b]=[-π,π] and so because p(x)=1, it asks for [e*mden/dx]π=0 (for any combination of the eigenfunctions), which quantizes the eigenvalues as λn=-n2 for integer n. However again, as I said above, we are only asking for the eigenfunctions to satisfy this boundary condition, whilst I was under the impression that any two functions, u,v in the space V (which does include, but is not limited to, pairs of the eigenfunctions) must meet this condition, not just the eigenfunctions.

    Could anyone clear this up for me? Thankyou in advance :D
     
    Last edited: Dec 12, 2014
  2. jcsd
  3. Dec 13, 2014 #2
    Note the example should be d2en/dx2nen. Anybody please?
     
  4. Dec 17, 2014 #3
    ...
     
  5. Dec 17, 2014 #4
    I believe it's the case that the vector space on which L acts is determined by the boundary conditions you place on the problem. Given this, you solve the SL problem and obtain a set of eigenfunctions which form a complete basis for that vector space. So it follows that any function can be written as a sum of eigenfunctions. I hope that makes sense.
     
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