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Math Help - Eigenvalues of y''+λy=0

  1. #1
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    Eigenvalues of y''+λy=0

    The problem is stated as follows:

    "Show that y''+\lambda y=0 with the initial conditions y(0)=y(\pi)+y'(\pi)=0 has an infinite sequence of eigenfunctions with distinct eigenvalues. Identify the eigenvalues explicitly."

    ---


    \lambda \le 0 seems to yield the trivial solution, so \lambda > 0. The general solution is then y=A\sin{(\sqrt{\lambda} x)} + B\cos{(\sqrt{\lambda} x)}. The first initial condition gives y=A\sin{(\sqrt{\lambda} x)}, and then the second gives A(\sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}) = 0.


    I've tried to solve \sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}= 0 for \lambda by writing the LHS as a single sine function, and separately by dividing with \cos{(\sqrt{\lambda} \pi)}, but neither approach seems to give a good way of giving an explicit formula for \lambda.


    (The best I've been able to do with the single sine function is to write \sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)} as \sqrt{1+\lambda} \sin{(\sqrt{\lambda} x + \delta)}. Consequently, \lambda = \left(\dfrac{\pi k - \delta}{\pi}\right)^2, k \in \mathbb{N}. But then \delta = \arccos \frac{1}{\sqrt{1+\lambda}} = \arcsin \frac{\sqrt{\lambda}}{\sqrt{1+\lambda}}, so I wouldn't call this explicit.)


    Thoughts?
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  2. #2
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    Re: Eigenvalues of y''+λy=0

    Try this:
    \sin (\sqrt{\lambda}\pi)+\sqrt{\lambda}\cos (\sqrt{\lambda}\pi)=0
    \sqrt{\lambda}=-\frac{\sin (\sqrt{\lambda}\pi)}{\cos (\sqrt{\lambda}\pi)}

    Your eigenvalues are given by the intersection of this curves.
    Last edited by Ulysses; April 1st 2012 at 05:06 PM.
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