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Boundary value problem involving eigenvalues

  1. Apr 26, 2006 #1
    I need a bit of help with these boundary value problems. I'm trying to find their eigenvalues and eigenfunctions and although I pretty much know how to do it, I want to exactly WHY I'm doing each step. I attached part of my work, and on it I have a little question next to the steps I need explained. The biggest question I have is how do you know what form to choose for the eigenfunctions. Thanks a lot for taking the time to help me out.

    Attached Files:

  2. jcsd
  3. Apr 26, 2006 #2


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    I HATE trying to read those Jpg things!

    In the first problem, you have
    [tex]y"+ \lambda y= 0[/tex]
    with boundary conditions y'(0)= 0, y(0)= 0.
    You first try [itex]\lambda< 0[/itex] and determine that there are no non-trivial functions satisfying that so no negative [itex]\lambda[/itex] is an eigenvalue.

    You then try [itex]\lambda= 0[/itex] so your equation is y"= 0 which has general solution y(x)= C1x+ C2. y'= C1 so the boundary conditions give C1= 0. But C2 can be anything so there exist non-trivial solutions of the form y= C. That is,
    0 is in fact an eigenvalue. Since the set of all such solutions is an "subspace" of the space of all functions, you can find a basis for it. The simplest is y= 1 so that y= C= C(1). Actually, any number (except 0) would have worked.

    If [itex]\lamba> 0[/itex], then you get general solution
    [tex]y= C_1 cos(\sqrt{\lambda}x)+ C_2 sin(\sqrt{\lamba x})[/itex]
    [tex]y'= -C_1/sqrtr{\lambda}sin(\sqrt{\lambda}x)+ C_2\sqrt{\lamba}cos{\lamba x})[/tex]
    The boundary conditions give
    [tex]y'(0)= C_2= 0[/tex]
    and then
    [tex]y'(L)= -C_1\sqrt{\lamba}sin(\sqrt{\lambda}L)= 0[/tex]
    which says that [itex]\sqrt{\lamba}L) must be a multiple of [itex]\pi[/itex] and so we must have
    [tex]\sqrt{\lamba}= \frac{n\pi}{L}[/tex]
    so the eigenvalues are of the form
    [tex]\frac{n^2\pi^2}{L^2}= \(\frac{n\pi}{L}\)^2[/tex]
    for n an integer (that, by the way, includes 0 which you had already found). That means that the "eigenvectors" or solutions are
    [tex]C_1 cos(\frac{n\pix}{L})[/tex]
    Again, any multiple of a single solution is also a solution so we might as well take C1= 1.
    The way you've written it, you have [itex]\lamba_0= 0, y_0(x)= 1[/itex] separate but you could as easily take
    [tex]\lamba_n= \(\frac{n\pi}{L}\)^2 y_n(x)= cos(\frac{n\pix}{L})[/itex]
    with n= 0, 1, 2,... (since cosine is an even function, we don't get anything new taking negative values for n).

    I think your basic question in both problems is why, after having gotten a non-trivial solution, involving an undetermined constant C1, do they then drop the constant. The answer is that the set of all vectors satisfying the equation [itex]Ay= \lamba y[/itex] is a subspace of the set of all vectors and so there exists a basis for it. In particular, in these problems the "eigenspaces" are 1 dimensiona. Since we want a solution yn such that any solution can be written y= Cyn for some number C, and we already have y= C1f(x) we might as well take C1= 1 and use that. In fact, we could have as easily let C1 be any non-zero number.
    Last edited: May 14, 2011
  4. Apr 26, 2006 #3
    That explained it perfectly! Thanks sooo much for taking the time help me out with that...and sorry about the .jpg
  5. May 13, 2011 #4
    I also face a big difficulty in this question! Could anyone help me though, please?

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  6. May 13, 2011 #5


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    Hello DonRico. Welcome to PF. I have quoted a few lines that you will see if you click on the Rules tab above:

    Do not hijack an existing thread with off-topic comments or questions--start a new thread.

    NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums. Once your question or problem has been responded to, do not go back and delete (or edit) your original post.
  7. May 14, 2011 #6
    =) Sorry for breaking rules ! I just read HallsofIvy's solution again wisely ! and I found a way though =) thank you all
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