# Modal representations of nodal tensor product Galerkin elements

Nodal discontinuous Galerkin methods on simplices, like those described in Hesthaven and Warburton, have the nice property that the number of nodes is equal to the minimum number needed to represent a unisolvent order $$N$$ polynomial in $$d$$ dimensions,

$$N_p = \frac{\prod_{i=1}^{d}(N+i)}{d!}$$

And we can construct $$N_p$$ unisolvent, hierarchical, orthonormal basis functions, and define an $$N_p \times N_p$$ Vandermonde matrix to uniquely map between nodal values and modal coefficients,

$$V \hat{u} = u$$

This is useful in many contexts:

1. Interpolation from solution nodes to other arbitrary nodes happens via a matrix multiplication like $$u^\star = V^\star V^{-1} u$$ where $$()^\star$$ is defined on the other nodes but with the same basis functions
2. P-multigrid restriction and prolongation matrices are just a matter of removing the highest modes or adding higher zero modes.
3. Spectral filtering operations are accomplished by mapping to modal space, truncating higher order coefficients, and mapping back to nodal space.

On tensor-product elements, we now have more nodes/coefficient than are required to be unisolvent, and we no longer have hierarchical basis functions (correct me if I'm wrong). How does one perform the above types of operations for tensor product elements? Do we form a Vandermonde matrix with the multidimensional monomials, and just not worry about the ill-conditioning and lack of orthogonality?

• Can you elaborate what you mean by "On tensor-product elements, we now have more nodes/coefficient than are required to be unisolvent"? For example, on quads, you have four vertices and four shape functions whose coefficients you uniquely get by interpolation at the vertices. The same applies to higher-degree polynomials. Sep 22 at 3:59
• The tensor elements simply have a larger basis, but don't completely fill the space to be considered a $N_{P+1}$ basis. For example a 9 node element has the basis $1$, $x$, $x^2$, $y$, $y^2$, $xy$, $xy^2$, $x^2 y$, and $x^2 y^2$. It's at least complete up to $N_P = 2$, however there are some modes in $N_P = 3$ which are missing, for example $x^3$. This doesn't mean you can't uniquely solve for all the modes though or apply a Vandermonde matrix to convert between nodal and modal values. Sep 22 at 5:10
• @WolfgangBangerth, what helloworld922 said. helloworld922, that's fair, I can always build a Vandermonde matrix from the monomials. One concern there is it quickly becomes ill-conditioned at higher orders. The other is point 3 in my question: it's not an orthogonal basis, so the $x^3$ mode has some $x^2$ content in it, so it's not clear to me how to apply the common spectral filters to this. e.g. clipping that $x^3$ coefficient will have an effect on the 2nd order modes. Sep 22 at 11:20
• @Aurelius It's all a question about how you define "modes" :-) Just like when you build the tensor product basis for quads as, well, a tensor product, you define what modes you want to consider not by the total polynomial degree, but as a tensor product of things with a specific degree in each coordinate direction. Sep 22 at 14:34
• You can turn any monomial series (even one which doesn't span all of $N_{P+1}$) into an orthogonal basis. For the tensor elements, these are just a tensor product of the 1D orthogonal basis. You can try to prove to yourself that this basis is indeed orthogonal and has the same span as the monomial one. Sep 22 at 15:45