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Boundary value ODE

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the eigenvalues and eigenfunctions of the following boundary-value problem.
    ## y''+\lambda y = 0 ; y(0) = 0, y'(L) = 0 ##

    2. Relevant equations


    3. The attempt at a solution
    So, we have to test when lambda is equal to, less than and greater than 0.
    Let ## \lambda = 0## thus, the ODE becomes ## y'' = 0 ## which implies solutions of the form ## y(t) = C_1t+C_2 ## which would make the derivative ## y'(t) = C_1 ##. When we apply our boundary conditions we see that ## y(0) = C_2 = 0 ## and ##y'(L) = C_1 = 0 ## which are trivial solutions.

    Let ## \lambda < 0 ## thus, the ODE becomes ## y''- \lambda y = 0 ## which implies solutions of the form ## y(t) = C_1e^{\sqrt{\lambda}t}+C_2e^{-\sqrt{\lambda}t} ## which would make the derivative ## y'(t) = \sqrt{\lambda}C_1e^{\sqrt{\lambda}t}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}t} ##. When we apply our boundary conditions we see that ## y(0) = C_1+C_2 = 0 ## therefore ## C_2=-C_1 ## and ## y'(L) = \sqrt{\lambda}C_1e^{\sqrt{\lambda}L}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}L} = \sqrt{\lambda}C_1e^{\sqrt{\lambda}L}+\sqrt{\lambda}C_1e^{-\sqrt{\lambda}L} = C_1[e^{\sqrt{\lambda}L}+e^{-\sqrt{\lambda}L} = 0 ## Since the exponentials can't ever be zero, it implies that ## C_1 = 0 ## this, these solutions are trivial.

    Let ## \lambda > 0 ## thus, the ODE becomes ## y''+\lambda y = 0 ## which implies solutions of the form ## y(t) = C_1\cos{\sqrt{\lambda}t}+ C_2\sin{\sqrt{\lambda}t}. ##Which would make the derivative ## y'(t) = -\sqrt{\lambda}C_1\sin{\sqrt{\lambda}t} + \sqrt{\lambda}C_2\cos{\sqrt{\lambda}t} ##. When I apply my boundary conditions, we see that ## y(0) = C_1 = 0 ## and ## y'(L) = C_2\cos\sqrt{\lambda}L = 0 ## Now we solve for ## \lambda ## to see that ## \sqrt{\lambda} L = \frac{n\pi}{2} ## which makes ## \lambda = \frac{n^2\pi^2}{4L^2} ##

    Thus, our solution for this boundary problem would be ## y(t) = C\sin\frac{n\pi}{2L}t ## for ## \lambda = \frac{n^2\pi^2}{4L^2}## but the solution in the book is ## y(t) = C\sin\frac{(2n+1)\pi}{2L}t ## for ## \lambda = \frac{(2n+1)^2\pi^2}{4L^2}## which I don't understand why we are limiting our n to be odd?

    Any help would be appreciated.
     
  2. jcsd
  3. Aug 6, 2017 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    ##\cos\sqrt{\lambda}L = 0 ## is what you want. But the cosine is not zero for all integer multiples of ##\frac \pi 2##, just for odd multiples. Try an even multiple for example.

     
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