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Thread: 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 $\displaystyle \phi '' - 2 \phi ' + \lambda \phi = 0$ with boundary conditions $\displaystyle \phi(0) = 0$ and $\displaystyle \phi(L) = 0$
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  2. #2
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    Is $\displaystyle \lambda$ a constant or a function? What is $\displaystyle \phi$ 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 $\displaystyle r^2 - 2r + \lambda = 0$

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

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

    For $\displaystyle 4 - 4 \lambda < 0$ though

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

    so the ODE solution is

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

    Plugging in the first boundary condition we will get

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

    Plugging in the second boundary condition we will get

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

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

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

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

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

    Then $\displaystyle \lambda_n = \frac{L^2 + (n \pi)^2}{L^2}$ 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 $\displaystyle \lambda_n = 1 + \bigl(\tfrac{n\pi}L\bigr)^2$ (though you could have made the calculation a bit simpler if you had noticed that $\displaystyle \tfrac{\sqrt{4\lambda-4}}2 = \sqrt{\lambda-1}$).

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