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Thread: Eigenvalues and eigenfunctions

  1. #1
    Senior Member chella182's Avatar
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    Eigenvalues and eigenfunctions

    Find the eigenvalues and eigenfunctions for

    $\displaystyle y''+\lambda y=0$, $\displaystyle 0\leq x\leq1$

    with the homogenous boundary conditions $\displaystyle y(0)=y'(1)=0$.



    So I subbed in $\displaystyle y=e^{\alpha x}$ et cetera, and somewhere down the line ended up needing to solve

    $\displaystyle \sin{\omega}+\omega\cos{\omega}=0$

    and now I'm stuck and not even sure if this can be done. Obviously $\displaystyle \omega=0$ would work, but wouldn't that just give a trivial solution?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by chella182 View Post
    Find the eigenvalues and eigenfunctions for

    $\displaystyle y''+\lambda y=0$, $\displaystyle 0\leq x\leq1$

    with the homogenous boundary conditions $\displaystyle y(0)=y'(1)=0$.



    So I subbed in $\displaystyle y=e^{\alpha x}$ et cetera, and somewhere down the line ended up needing to solve

    $\displaystyle \sin{\omega}+\omega\cos{\omega}=0$

    and now I'm stuck and not even sure if this can be done. Obviously $\displaystyle \omega=0$ would work, but wouldn't that just give a trivial solution?
    I assume that you have put $\displaystyle \omega = \sqrt{-\lambda}$, so that the general solution of the equation is $\displaystyle y = A\cos\omega x + B\sin\omega x$. The boundary condition y(0)=0 tells you that A=0, so that $\displaystyle y = B\sin\omega x$. Then you differentiate to get the other boundary condition in the form $\displaystyle \omega\cos\omega = 0$. You can disregard the solution $\displaystyle \omega=0$, because that leads to the trivial solution y=0. So the eigenvalues are given by $\displaystyle \cos\omega=0$, which has an infinite family of solutions.
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  3. #3
    Senior Member chella182's Avatar
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    Ignore this, I've figured out why you don't have $\displaystyle e^{x}$ in your solution. Damn shoddy notes :@
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