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Homework Help: Eigenvalues of y''+λy=0

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data

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


    2. Relevant equations



    3. The attempt at a solution

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

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

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

    Thoughts?
     
    Last edited: Mar 31, 2012
  2. jcsd
  3. Mar 31, 2012 #2

    LCKurtz

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    That looks well done. You don't need the ##\sqrt{1+\lambda}## in the answer though. You can just write your final eigenfunction$$
    y_n = \sin(\frac{n\pi - \delta}{\pi}x)$$When you write something like this up, it smooths things out a bit if you do this case as ##\lambda = \mu^2 >0## and carry the ##\mu## through to the end. It reads nicer without all the square root signs.
     
  4. Mar 31, 2012 #3

    Thanks for your input! I'm still a bit uncomfortable with [itex]u_n = \sin\left(\dfrac{n\pi - \delta}{\pi}x\right)[/itex] as [itex]\delta[/itex] depends on [itex]\lambda[/itex], though. I tried a few additional approaches, but I haven't been able to get anything "better" than that.
     
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