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Thread: biharmonic equation

  1. #1
    Mar 2010

    biharmonic equation

    verify that the following is a solution to the biharmonic equation

    F(x,y)=(Acosbx+Bsinbc+Cxcosbx+Dxsinbx)(Ecoshby+Fsi nhby)

    i have tried taking the derivates and have gotten up to the laplacian but this is a very long and tedious way,and the result up to second order is really long, like a page long,there must be an easier way. Can someone suggest an alternative method to prove this
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  2. #2
    MHF Contributor
    Jester's Avatar
    Dec 2008
    Conway AR
    The biharmonic eqiation is

    u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0

    Your solution is of the form

    u = A(x)B(y)

    Note that when you substitute into the biharmonic you get

    A^{(4)}(x)B  (y)+ 2 A''(x)B''(y) + A(x) B^{(4)}(y) = 0

    With your B(y) you have B^{(4)}(y) = b^2B''(y) = b^4B(y) so what you really want to show is

    A^{(4)}(x)+ 2b^2 A''(x) + b^4A(x) = 0.

    Next, notice that this is

    \left(A'' + b^2A\right)'' + b^2\left(A'' + b^2A \right) = 0

    so compute C = A'' + b^2A first and then show this satisfies C'' + b^2C = 0. It might make it a little more manageable.
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  3. #3
    Mar 2010
    thanks thats way easier
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