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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

    $\displaystyle u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$

    Your solution is of the form

    $\displaystyle u = A(x)B(y)$

    Note that when you substitute into the biharmonic you get

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

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

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

    Next, notice that this is

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

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