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Math Help - boundary value problem

  1. #1
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    boundary value problem

    Find the eigenvalues , and normalised eigenfunctions of the problem:






    I'm fine of the case but unsure about the other two.



    Let
    DE transforms to
    General solution:





    Hence

    Hence eigenvalues are




    Let
    DE transforms to
    General solution:





    Hence

    Which only has solution w = 0 hence trivial solution and we reject.

    Could some please check this is OK?

    Thanks in advance!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by hunkydory19 View Post
    Find the eigenvalues , and normalised eigenfunctions of the problem:






    I'm fine of the case but unsure about the other two.



    Let
    DE transforms to
    General solution:





    Hence

    Hence eigenvalues are




    Let
    DE transforms to
    General solution:





    Hence

    Which only has solution w = 0 hence trivial solution and we reject.

    Could some please check this is OK?

    Thanks in advance!
    Yeah, that looks okay. One question though. I agree with \lambda_n=\left(n-\frac{1}{2}\right)^2\pi^2, but where are you getting \lambda_n=mu^2? What are u and m?
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    Yeah, that looks okay. One question though. I agree with \lambda_n=\left(n-\frac{1}{2}\right)^2\pi^2, but where are you getting \lambda_n=mu^2? What are u and m?
    It's mu ( \mu)

    \lambda_n = \mu^2 = \left(n-\frac{1}{2}\right)^2\pi^2
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    Thank you so much for the reply.

    I'm also confused on how to get the eigenfunctions. The answer given is  \sqrt 2 sin ((n - \frac{1}{2}) \pi x) but I don't see where this comes from...could you possibly explain?
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  5. #5
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    Quote Originally Posted by hunkydory19 View Post
    Thank you so much for the reply.

    I'm also confused on how to get the eigenfunctions. The answer given is  \sqrt 2 sin ((n - \frac{1}{2}) \pi x) but I don't see where this comes from...could you possibly explain?
    As you said earlier, in order to satisfy the BC's you need B = 0,\; \text{and}\; \mu = \left(n-\frac{1}{2}\right) \pi . Now from your solution

     <br />
y = A \sin \mu x = A\sin \left(n-\frac{1}{2}\right) \pi x<br />

    This satisfies both the equation and BCs (for \lambda = \mu^2). Not sure where the \sqrt{2} came from unless there's another BC floating around.
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    Yes it was the root 2 bit that I was confused about, there are no other boundary conditions so I don't see how they have got this value for A but thank you again for your help!
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by danny arrigo View Post
    It's mu ( \mu)

    \lambda_n = \mu^2 = \left(n-\frac{1}{2}\right)^2\pi^2
    Yikes. I should've realized that. I'm starting to think like Maple. In my defense, the actual symbol \mu was used throughout the rest of the post, so when it got suddenly switched up like that it confused my poor brain.
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  8. #8
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    There is one other possiblity you left out.

    You considered \lambda> 0 and \lambda< 0. What if \lambda= 0?
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  9. #9
    Super Member redsoxfan325's Avatar
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    He said he was fine in the case of \lambda=0.
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    The \sqrt{2} comes from the fact that the question states normalised eigenfunctions hence the square should integrate to 1 over the relevant domain.
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  11. #11
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    Also just noticed you made a mistake with you \lambda < 0 case. You should have found that

    A = 0

    so that

    0 = B \omega \cosh \omega

    which (since the minimum value o cosh function is 1) means that

    B = 0

    so there is no solution for these BCs with \lambda < 0 .
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  12. #12
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    Quote Originally Posted by hunkydory19 View Post
    Find the eigenvalues , and normalised eigenfunctions of the problem:





    Yep - missed that part.
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