# Orthonormal basis for hexahedron

Orthogonal polynomials are often preferred as basis functions. Recently I learned selecting orthonormal basis further simplifies the mass matrix from diagonal to simply the identity matrix when used as modal basis. But I am confused how the normalizing factors are obtained.

Say the orthonormal Legendre polynomials in each direction of a reference hexahedron are $P_1$, $P_2$, $P_3$. Then using tensor-product approach, is the orthonormal basis for hex simply $P_1\times P_2\times P_3$? What about for quads, simply $P_1\times P_2$, or do we need to multiply by some constant to keep the orthonormality of the basis? I learned from the text book (Hesthaven & Warburton) that orthonormal basis for triangles and tetrahedron have different normalizing factors, i.e. simply cancelling one of the directions of an orthonormal basis in tetrahedron does not result in a basis for the triangle.

Triangle:

$$\sqrt{2} P_i^{(0,0)}(a) P_j^{(2i+1,0)}(b) (1-b)^i$$

Tetrahedron:

$$2\sqrt{2} P_i^{(0,0)}(a) P_j^{(2i+1,0)}(b) P_k^{(2i+2j+2,0)}(c) (1-b)^i (1-c)^{(i+j)}$$

On quad/hex you can use tensor product polynomials. For example in 2d, you map the cell to a reference cell $[-1,1] \times [-1,1]$ and if $N$ is the degree, you would use $$P_i(\xi) P_j(\eta), \qquad 0 \le i,j \le N$$ as the basis functions, where $\xi,\eta$ are coordinates on the reference cell and $P_i$ is the Legendre polynomial of degree $i$. These are orthogonal. To make it orthonormal, you just calculate $$m_i = \int_{-1}^{+1} P_i^2(\xi) d\xi$$ and redefine $$P_i \rightarrow \frac{1}{\sqrt{m_i}} P_i$$ You do the same thing in 3-d.
• Thank you! Question: For the formulas I gave in tri/tets, If I set c=0 in the tet's orthonormal basis, I recover all terms of the orthonormal tri basis BUT constants are different $\sqrt(2)$ vs $2\sqrt(2)$. Where is that coming from? May 1 '17 at 13:31
• I am thinking that the orthonormal basis are: Line = $P1$, Quad = $\sqrt(2) x P1xP2$, and Hex = $2 x P1 x P2 x P3$, because the orthonormal jacobi polynomials of degree 0 are $\sqrt(2)/2$ instead of $1$. Is that correct ? May 1 '17 at 13:53