Biharmonic Operator: Understanding PDEs for Smooth Meshes

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shalayka
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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:
http://www.math.bas.bg/or/NATO_ARW/presentations/Ugail.ppt
 
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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.