# Biharmonic operator

1. Mar 6, 2008

### shalayka

While looking into higher-order PDEs, I came across the biharmonic.

Where the biharmonic equation is:

$$\left(\frac{\partial^2}{\partial {x}^2} + \frac{\partial^2}{\partial {y}^2} + \frac{\partial^2}{\partial {z}^2}\right)\left(\frac{\partial^2}{\partial {x}^2} + \frac{\partial^2}{\partial {y}^2} + \frac{\partial^2}{\partial {z}^2}\right).$$

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

$$\frac{\partial^4}{\partial {x}^4} + \frac{\partial^4}{\partial {y}^4} + \frac{\partial^4}{\partial {z}^4} + \frac{\partial^4}{\partial {x}^2 \partial{y}^2} + \frac{\partial^4}{\partial {x}^2 \partial{y}^2} + \frac{\partial^4}{\partial {y}^2 \partial{z}^2} + \frac{\partial^4}{\partial {y}^2 \partial{z}^2} + \frac{\partial^4}{\partial {x}^2 \partial{z}^2} + \frac{\partial^4}{\partial {x}^2 \partial{z}^2}.$$

Why would one use this instead of:

$$\frac{\partial^4}{\partial {x}^4} + \frac{\partial^4}{\partial {y}^4} + \frac{\partial^4}{\partial {z}^4}?$$

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

Last edited: Mar 6, 2008
2. Mar 6, 2008

### HallsofIvy

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