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Solve the eigenvalue problem

  1. Jun 30, 2008 #1
    Solve the eigenvalue problem

    [tex]\frac{d^2 \phi}{dx^2} = -\lambda \phi[/tex]

    subject to

    [tex]\phi(0) = \phi(2\pi)[/tex]


    [tex] \frac{d \phi}{dx} (0) = \frac{d \phi}{dx} (2 \pi).[/tex]

    I had the solution already, but am looking for a much simpler way, if any.


    Sorry that I accidentally posted this twice; I meant to edit the post, but not sure why the edited post becomes a new one...
  2. jcsd
  3. Jul 1, 2008 #2


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    How do we know that our suggested solution is simpler than yours if you dont demonstrate your attempt?
  4. Jul 4, 2008 #3
    Here's what I had:

    after solving the ODE, we have the general solution

    [tex]\phi = C_1 \sin(\sqrt{\lambda}x) + C_2 \cos(\sqrt{\lambda}x)[/tex]

    applying the conditions we have the system

    [tex]C_2 = C_1 \sin(\sqrt{\lambda}2\pi) + C_2 \cos(\sqrt{\lambda}2\pi)[/tex]

    [tex] C_1 \sqrt{\lambda} = C_1 \sqrt{\lambda} \cos(\sqrt{\lambda}2\pi) - C_2 \sqrt{\lambda} \sin(\sqrt{\lambda}2\pi)[/tex]

    then using a little knowledge of linear algebra and determinant, we get, in order not to have trivial solutions,

    [tex]C_1 C_2 u^2 + C_1 C_2 (v-1)^2 = 0[/tex]

    where [tex]u = \sin(\sqrt{\lambda}2\pi)[/tex] and
    [tex]v = \cos(\sqrt{\lambda}2\pi)[/tex]

    or simply

    [tex] u^2 + (v-1)^2 = 0[/tex]

    So [tex]u^2 = (v-1)^2 = 0[/tex] or
    [tex] u = 0; v = 1[/tex]

    [tex] \sqrt{\lambda_n}2\pi = 2n\pi[/tex]
    [tex]\lambda_n = n^2[/tex]

    So the eigenfunction is

    [tex]\phi_n = C_1 \sin(nx) + C_2 \cos(nx)[/tex]

    I don't think this is complete because we haven't determined [tex]C_1 [/tex] and [tex] C_2[/tex] yet.
  5. Jul 5, 2008 #4
    Use the BC's to get your constants.
  6. Jul 5, 2008 #5
    I thought that I've already used them in first determining the eigenvalue.
  7. Sep 28, 2008 #6
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