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Math Help - Solving a linear difference equation subject to boundary conditions

  1. #1
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    Question Solving a linear difference equation subject to boundary conditions

    Hi everyone,

    I read this on a book, which says we have a linear difference equation:

    pk= 1/2 (pk+1 + pK-1) if 0<k<N,

    s.t. boundary conditions: p0= 1, pN= 0.

    The solution is also provided however I don't get it, it says:

    Put bk = pk - pk-1, to obtain bk = bk-1 and hance bk = b1 for all k

    [How do we get bk = bk-1 ?]

    Thus, pk = b1 + pk-1 = 2b1 + pk-2 =...= kb1 + p0

    Boundary conditions imply p0 = 1, b1 = -1/N

    so we have pk = 1 - (k/N)

    At first glance everything looks fine...but when I read it again I don't quite understand how does bk = bk-1? Could anyone explain why it is the case?

    Thank you very much for your help
    Last edited by sweetadam; March 16th 2012 at 08:22 PM.
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  2. #2
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    Re: Solving a linear difference equation subject to boundary conditions

    alrite seems no one wants to answer this Q probably because it is too easy...

    after read it a few more times I think I see it

    so I just answer myself:

    to see this: split LHS pk in to 1/2(pk + pk) and the first equation becomes 1/2(pk + pk)= 1/2 (pk+1 + pK-1)

    rewrite above in to: 1/2(pk+1-pk) = 1/2(pk-pK-1) which is: 1/2bk+1 = 1/2bk gives us bk = bk-1
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