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Say I have a function $f : \mathbf{R}^3 \to \mathbf{R}$ that I wish to integrate over a tetrahedron $T \subset \mathbf{R}^3$. If $f$ was arbitrary, Gauss quadrature would be a good solution, but I happen to know that $f$ is harmonic. How much can Gauss quadrature be accelerated using this information?

For example, if $T$ was instead a sphere, evaluating $f$ once at the center of the sphere gives the exact answer by the mean value property.

A search turned up the following paper, which is interesting but generalizes the sphere case in a different direction (to polyharmonic instead of away from spheres):

Bojanov and Dimitrov, Gaussian extended cubature formulae for polyharmonic functions

Geoffrey Irving
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I found something that might be interesting. http://www.math.kth.se/~gbjorn/exact.pdf

I hope this helps, Tom

Tom
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  • That's an interesting paper, but it looks like it and its references only treat integrals of differential operators of harmonic functions. Do you know if they can be used for straight integrals? – Geoffrey Irving Mar 04 '13 at 17:38
  • I'm wondering if introducing a quadrature formula with the so-called "Poisson kernel" (http://en.wikipedia.org/wiki/Poisson_kernel) could help... Otherwise I know that some xfem techiques use harmonic functions to enrich the FE space, and therefore should use specific quadrature methods for integrating the variational forms (?). – Tom Mar 04 '13 at 20:46