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Math Help - Sturm-Liouville

  1. #1
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    Sturm-Liouville

    How does one find the eigenvalues and eigenfunctions of

    -(x^2y')'=lambda *y

    with 1<x<e, and y(1)=y(e)=0

    I'm completely lost. Thank you!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by morganfor View Post
    How does one find the eigenvalues and eigenfunctions of

    -(x^2y')'=lambda *y

    with 1<x<e, and y(1)=y(e)=0

    I'm completely lost. Thank you!
    This equation is of type Cauchy-Euler

    Using the anzant

    y=x^r gives the characteristic equation

    \frac{d}{dx}(x^2\cdot\frac{d}{dx}x^r) = \lambda x^r

    x^r(r^2+r+\lambda)=0

    From the boundary conditions we can see we need periodic solutions so the roots of the characteristic equation must be complex. Solving the above for r gives

    r^2+r+\lambda=0 \iff \left(r +\frac{1}{2}\right)^2=\frac{1-4\lambda}{4} \iff r=\frac{-1 \pm \sqrt{1-4\lambda}}{2}

    This implies that lambda must be positive

    This gives solutions

    y=x^{-\frac{1}{2}}\left( c_1\cos\left( \frac{\sqrt{4\lambda-1}}{2} ln(x) \right)+c_2\sin\left(\frac{\sqrt{4\lambda-1}}{2} \ln(x)\right)\right)

    Using the boundary conditions to solve for lambda gives

    y(1)=c_1=0

    This gives

    y=c_2x^{-\frac{1}{2}}\sin\left(\frac{\sqrt{4\lambda-1}}{2} \ln(x)\right)

    Now with

    y(e)=0=\frac{c_2}{e}\sin\left(\frac{\sqrt{4\lambda-1}}{2}\right)

    This gives

    \pi k =\frac{\sqrt{4\lambda-1}}{2} , k \in \mathbb{Z} \iff \lambda = \frac{4\pi^2 k^2+1}{4}, k=1,2,3...
    Last edited by TheEmptySet; May 8th 2011 at 10:13 PM. Reason: Brain cramp
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  3. #3
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    thanks!!
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