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Math Help - Integration help - Biharmonic Equation

  1. #1
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    Integration help - Biharmonic Equation

    Hi I'm having trouble expanding the following equation for the axially symmetric biharmonic equation in polar co-ordinates.

     \left ( \frac{d^2}{dr^2} + \frac{d}{rdr} \right ) \left ( \frac{d^2\phi}{dr^2} + \frac{d\phi}{rdr} \right ) = 0

    When I expanded it I got:
    \frac{d^4\phi}{dr^2} + \frac{2d\phi}{r^3dr} + \frac{d^3\phi}{rdr^3} + \frac{d^3\phi}{rdr^3} - \frac{d\phi}{r^3dr} + \frac{d^2\phi}{r^2dr^2}

    = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} + \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}

    But it should be:

    = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}

    Any help would be much appreciated regards,

    Callum
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  2. #2
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    Quote Originally Posted by CallumJ View Post
    Hi I'm having trouble expanding the following equation for the axially symmetric biharmonic equation in polar co-ordinates.

     \left ( \frac{d^2}{dr^2} + \frac{d}{rdr} \right ) \left ( \frac{d^2\phi}{dr^2} + \frac{d\phi}{rdr} \right ) = 0

    When I expanded it I got:
    \frac{d^4\phi}{dr^2} + \frac{2d\phi}{r^3dr} + \frac{d^3\phi}{rdr^3} + \frac{d^3\phi}{rdr^3} - \frac{d\phi}{r^3dr} + \frac{d^2\phi}{r^2dr^2}

    = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} + \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}

    But it should be:

    = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}

    Any help would be much appreciated regards,

    Callum
    \frac{d}{dr} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = -\frac{1}{r^2} \frac{d \phi}{dr} + \frac{1}{r} \frac{d^2 \phi}{dr^2}.


    Therefore:


    \frac{d^2}{dr^2} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = \frac{2}{r^3} \frac{d \phi}{dr} - \frac{2}{r^2} \frac{d^2 \phi}{dr^2} + \frac{1}{r} \frac{d^3 \phi}{dr^3}


    and


    \frac{1}{r} \frac{d}{dr} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = -\frac{1}{r^3} \frac{d \phi}{dr} + \frac{1}{r^2} \frac{d^2 \phi}{dr^2}.


    Correctly substitute and simpify and you'll get the correct answer.
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  3. #3
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    Quote Originally Posted by CallumJ View Post

    But it should be:

    = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}

    Any help would be much appreciated regards,

    Callum
    Should it not be:

    \frac{d^4\phi}{dr^4} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}=0

    Then what, multiply throughout by r^4, Euler, then Bingo-bango or no?
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