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Math Help - Sturm Liouville Eigenfunctions??

  1. #1
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    Sturm Liouville Eigenfunctions??

    Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

    <br />
\displaystyle<br />
\frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0,<br />
\qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0 <br />

    What I did: since x \neq 0 I can multiply by x and rewrite as:

    <br />
\displaystyle<br />
x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0<br />

    I substitute  y = x^k and find k = \pm \sqrt{-\lambda}, so the general solution would be:

    <br />
\displaystyle<br />
y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}<br />

    The condition y(1) = 0 gives A + B = 0. The second condition gives:

    <br />
\displaystyle<br />
y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0<br />

    So i get  2^{\sqrt{\lambda}} = \pm 1 and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??

    Thank you
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  2. #2
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    Quote Originally Posted by DJDorianGray View Post
    Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

    <br />
\displaystyle<br />
\frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0,<br />
\qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0 <br />

    What I did: since x \neq 0 I can multiply by x and rewrite as:

    <br />
\displaystyle<br />
x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0<br />

    I substitute  y = x^k and find k = \pm \sqrt{-\lambda}, so the general solution would be:

    <br />
\displaystyle<br />
y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}<br />

    The condition y(1) = 0 gives A + B = 0. The second condition gives:

    <br />
\displaystyle<br />
y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0<br />

    So i get  2^{\sqrt{\lambda}} = \pm 1 and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??

    Thank you
    You should consider three cases for \lambda: <0, =0, \text{and}\; >0 or \lambda = -\omega^2, \lambda = 0 and \lambda = \omega^2 since the solutions of the ODE are very different for these cases.
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  3. #3
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    Quote Originally Posted by DJDorianGray View Post
    Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

    <br />
\displaystyle<br />
\frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0,<br />
\qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0 <br />

    What I did: since x \neq 0 I can multiply by x and rewrite as:

    <br />
\displaystyle<br />
x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0<br />

    I substitute  y = x^k and find k = \pm \sqrt{-\lambda}, so the general solution would be:

    <br />
\displaystyle<br />
y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}<br />

    The condition y(1) = 0 gives A + B = 0. The second condition gives:

    <br />
\displaystyle<br />
y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0<br />

    So i get  2^{\sqrt{\lambda}} = \pm 1 and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??
    Don't be put off by complex solutions. You can often combine them to form real solutions. In this case, y=x^{\pm k}=e^{\pm k\ln x} solves the equation for every complex value of k. The solution y=Ae^{k\ln x} + Be^{-k\ln x} satisfies the boundary conditions if k=n\pi i/\ln 2, and then y = \sin(n\pi\ln x/\ln2). The eigenvalue for this eigenfunction is \lambda = (n\pi/\ln 2)^2.
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  4. #4
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    Mar 2009
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    My mistake, I did not see that lambda values were actually real thank you guys
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