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Homework Help: Boundary Value Problem; Eigenvalues and Eigenfunctions

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the eigenvalues and eigenfunction for the BVP:

    y(0)=0, y'(0)=0, y'(L)=0

    2. Relevant equations

    m^3+[itex]\lambda[/itex]m=0, auxiliary equation

    3. The attempt at a solution

    3 cases [itex]\lambda[/itex]=0, [itex]\lambda[/itex]<0, [itex]\lambda[/itex]>0
    this first 2 give y=0 always, as the only solution.

    [itex]\lambda[/itex]>0 solution attempt

    m=0, and +/- [itex]\lambda[/itex]i

    general solution:
    Where A, B, and C are constants


    y(0)=0 gives
    0=A+B, or A=-B

    y'(0)=0 gives
    0=[itex]\lambda[/itex]C, so C=0

    y'(L)=0 gives

    The only solution I find from these data is y=0, which seems kind of off since no eigenfunction/values are found. From what I've read/studied so far when [itex]\lambda[/itex]>0 there is always an eigen function/value.

    The alternative I've considered is to consider B≠0 and having the eigenvalue be
    [itex]\lambda[/itex]L=n[itex]\pi[/itex] giving [itex]\lambda[/itex]=n[itex]\pi[/itex]/L

    which then gives the eigen function
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 22, 2012 #2


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    Science Advisor
    Homework Helper
    Gold Member

    That's wrong, your next paragraph is the correct procedure
    And since A = -B, you have ##B(-1+\cos(\frac {n\pi x} L))##. You could leave off the B and write ##y_n =-1+\cos(\frac {n\pi x} L)##.
  4. Apr 22, 2012 #3
    OK, thank you!
    I had thought that there always had to be a function for [itex]\lambda[/itex]>0, but I wasn't sure and I couldn't find any literature specifically mentioning it.
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