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Orthogonal (Modal) Bases

The results we now present for the triangle and the tetrahedron are special cases of the previous section, but for clarity we will outline some of the steps used to find these results.

We first consider the tetrahedron as this has the most complex mapping between the local cartesian coordinates and the tensor coordinates. Using the scaled coordinates $(a,b,c) \in
[0,1]\times[0,1]\times[0,1]$ for the tetrahedron (for simplicity) we can express the local orthogonal coordinates $(r,s,t) \in \{0 \leq
r,s,t , r+s+t \leq 1\}$ as:

We consider the operator:

Using the following identities it is straightforward to express the operator with respect to the (a,b,c) coordinates:

After some manipulation the operator can be expressed in terms of the (a,b,c) coordinates and is:

This demonstrates that the operator maintains tensor form in the (a,b,c) coordinate system. It is now trivial to show that this is a self-adjoint operator by applying one-dimensional integration by parts to each of the three tensor parts. Also, by using the definition of the Jacobi polynomials we can show that the orthogonal basis is a set of eigenfunctions of LTet and find their eigenvalues.

We will now show that the polynomial functions $\phi_{ijk}$ defined by:

are eigenfunctions of the LTet operator. We consider the first part of the operator. The definition of the Jacobi polynomial directly implies the following relationship:

\begin{displaymath}
\partial_a(a(1-a)\partial_a\phi_{ijk}) = -i(i+1)\phi_{ijk}\end{displaymath}

We now consider the first two terms of the operator. Using the previous result we can remove the dependency on a and then use the definition of the Jacobi polynomials with non-zero $(\alpha,\beta)$ to show that the polynomials are indeed eigenfunctions of the first two terms of the operator:

Applying the same technique again we come to the relationship:

Thus, the tensor product of Jacobi polynomials $\phi_{ijk}$ are eigenfunctions of the total operator LTet with eigenvalues:

\begin{displaymath}
\lambda_{ijk} = -(i+j+k)(i+j+k+3) \end{displaymath}

This approach clearly applies to the triangular, quadrilateral, prismatic and hexahedron elements as well.



 
next up previous contents
Next: The Orthogonal Triangle Basis Up: Basis Functions Previous: Singular Sturm-Liouville Problems on
T. Warburton
10/24/1998