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Biharmonic operator

  1. Mar 6, 2008 #1
    While looking into higher-order PDEs, I came across the biharmonic.

    Where the biharmonic equation is:

    [tex]\left(\frac{\partial^2}{\partial {x}^2} + \frac{\partial^2}{\partial {y}^2} + \frac{\partial^2}{\partial {z}^2}\right)\left(\frac{\partial^2}{\partial {x}^2} + \frac{\partial^2}{\partial {y}^2} + \frac{\partial^2}{\partial {z}^2}\right).[/tex]

    Using basic algebra for the multiplication, this works out to include a bunch of terms involving mixed axes:

    [tex]\frac{\partial^4}{\partial {x}^4} + \frac{\partial^4}{\partial {y}^4} + \frac{\partial^4}{\partial {z}^4} + \frac{\partial^4}{\partial {x}^2 \partial{y}^2} + \frac{\partial^4}{\partial {x}^2 \partial{y}^2} + \frac{\partial^4}{\partial {y}^2 \partial{z}^2} + \frac{\partial^4}{\partial {y}^2 \partial{z}^2} + \frac{\partial^4}{\partial {x}^2 \partial{z}^2} + \frac{\partial^4}{\partial {x}^2 \partial{z}^2}.[/tex]

    Why would one use this instead of:

    [tex]\frac{\partial^4}{\partial {x}^4} + \frac{\partial^4}{\partial {y}^4} + \frac{\partial^4}{\partial {z}^4}?[/tex]

    Thanks for any help on clarification.

    I've found this presentation which shows how the smoothness of meshes is obtained using the biharmonic equation:
    Last edited: Mar 6, 2008
  2. jcsd
  3. Mar 6, 2008 #2


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    (I was sorely tempted to say "For the same reason we would use (x+ y)2 instead of x2+ y2, but I will behave!)

    Well, one would use one instead of the other because the are different!

    In particular, the "harmonic" operator, [itex]\nabla^2[/itex] is [itex]\partial^2/\partial x^2+ \partial^2/\partial y^2+ \partial^2/\partial z^2[/itex] has the nice property that it is "invariant under rigid motions" and, therefore, so is [itex]\nabla^2(\nabla^2 )[/itex] is also "invariant under rigid motions". The second formula you give is not.
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