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Math Help - Eigenfunctions and Eigenvalues

  1. #1
    Junior Member BrooketheChook's Avatar
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    Eigenfunctions and Eigenvalues

    Find the eigenfunctions and eigenvalues of the equation \phi '' - 2 \phi ' + \lambda \phi = 0 with boundary conditions \phi(0) = 0 and \phi(L) = 0
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    Is \lambda a constant or a function? What is \phi a function of?
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  3. #3
    Junior Member BrooketheChook's Avatar
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    This is a Sturm-Liouville Problem.

    The aux equation is r^2 - 2r + \lambda = 0

    r = \frac{2 \pm \sqrt{4 - 4 \lambda}}{2}

    I have found that for 4 - 4 \lambda = 0 and 4 - 4 \lambda > 0 the solutions are trivial.

    For 4 - 4 \lambda < 0 though

    we get complex roots r = 1 \pm \ i \frac{\sqrt{4-4 \lambda}}{2}

    so the ODE solution is

    \phi = c_1 e^x \text{sin}(\frac{\sqrt{4 \lambda - 4}}{2}x) + c_2 e^x \text{cos} (\frac{\sqrt{4 \lambda - 4}}{2}x)

    Plugging in the first boundary condition we will get

    \phi(0) = c_2 = 0 so c_2 = 0

    Plugging in the second boundary condition we will get

    \phi(L) = c_1 e^L \text{sin}(\frac{\sqrt{4 \lambda - 4}}{2}L) = 0

    Now
    For a non-trivial solution we assume the c_2 \neq 0 which implies

    \text{sin}(\frac{\sqrt{4 \lambda - 4}}{2}L) = 0

    Since zeros of Sin only occur at integral multiples of \pi, to satisfy the boundary condition we must have

    \frac{\sqrt {4 \lambda-4}}{2}L= n \pi for n = 1,2....

    Then \lambda_n = \frac{L^2 + (n \pi)^2}{L^2} is the eigenvalue, but Im having trouble with the eigenfunction......

    Can anyone please help?
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  4. #4
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    You have done the hard part, which is to find the eigenvalues, namely \lambda_n = 1 + \bigl(\tfrac{n\pi}L\bigr)^2 (though you could have made the calculation a bit simpler if you had noticed that \tfrac{\sqrt{4\lambda-4}}2 = \sqrt{\lambda-1}).

    For the eigenfunctions, just substitute the value \lambda = \lambda_n into the equation \phi(x) = c_1 e^x \sin(\sqrt{ \lambda - 1}\,x) to get \boxed{\phi_n(x) = c_1 e^x \sin\bigl(\tfrac{n\pi x}L\bigr)}.
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