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.


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


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


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – danny
    May 1, 2017 at 13:31
  • $\begingroup$ 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 ? $\endgroup$
    – danny
    May 1, 2017 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.