1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Eigenvalues of y''+λy=0
  1. Y' '+ y = 0 (Replies: 10)

  2. Eigenvalue of 0 (Replies: 8)

  3. Help with Y'=0 wanted (Replies: 10)

Loading...