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Eigenvalue problem.

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a problem

    u'' + lambda u = 0

    with BCs: u'(0) = b*u'(pi), u(0) = u(pi).

    where b is a constant.

    I have to find b which makes the BCs and problem self-adjoint.



    2. Relevant equations

    see below


    3. The attempt at a solution

    I see in my notes that it says that when there are periodic BCs, then the problem is self-adjoint. I think b = 1, and not any other number. Won't the value of the derivative of the solution change if b does not equal one? And therefore the BCs will not be periodic any more? Thoughts? Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 12, 2012 #2
    For a Sturm-Liouville problem(Self-adjoint) on the form:
    [itex] \dfrac{d}{dx}\left(p(x)\dfrac{dU}{dx}\right) +q(x)U +\lambda w(x)U = 0[/itex]
    the following integrated value should vanish, then the problem is self-adjoint:
    [itex] \left[p\left(u^*\dfrac{dv}{dx}-\dfrac{du^*}{dx}v\right)\right]_{0}^{\pi}[/itex]
    Where u and v are solutions to the problem. so in your case:
    [itex] p(x)=1, q(x)=0, w(x)=1\\[/itex]
    With periodic boundary condiitions:
    [itex] u\dfrac{du}{dx} |_0 = u\dfrac{du}{dx}|_\pi = 0[/itex]
    Only holds true for b=1 in my opinion.
    But maybe diriclet or legendre or neumann or mixed BCS can lead to another value of b?
     
  4. Apr 12, 2012 #3
    actually i think that mixed BCS are the only other option in your case.
     
  5. Apr 12, 2012 #4
    Ah yes, now I comprehend the method. Thank you.
     
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