Polynomial expansion sets. More...
Functions | |
| xt::xtensor< double, 3 > | tabulate (cell::type celltype, int d, int n, const xt::xtensor< double, 2 > &x) |
| void | tabulate (xt::xtensor< double, 3 > &P, cell::type celltype, int d, int n, const xt::xtensor< double, 2 > &x) |
| int | dim (cell::type cell, int d) |
| int | nderivs (cell::type cell, int d) |
Polynomial expansion sets.
In Basix, the bases of the polynomial sets spanned by finite elements are represented by a set of coefficients of orthonormal polynomials on the cell. By orthogonality, we mean that
\[\int_R p_ip_j=\begin{cases}1&i=j\\0&i\not=j\end{cases}\]
Legendre polynomials form a set of orthogonal polynomials on an interval. Legrendre polynomials are a special case of the Jacobi polynomials, \(P_n^{\alpha, \beta}\), with the weights \((\alpha, \beta) = (0, 0)\). Using the Jacobi recurrence relation, we have
\[ n P^{0,0}_n(x) = (2n - 1) x P^{0,0}_{n - 1}(x) - (n-1) P^{0,0}_{n-2}(x),\]
starting with \(P^{0,0}_0 = 1\). Legendre polynomials are orthogonal when integrated over the interval [-1, 1].
In Basix, the interval [0, 1] is used as the reference, so we apply the transformation \(x\mapsto 2x-1\) to the standard Legendre polynomials. The orthogonal polynomials we use are therefore given by the recurrence relation
\[n p_n(x)=(2n-1)(2x-1)p_{n-1}(x)-(n-1)p_{n-2}(x).\]
For a given set of points, the recurrence relation can be used to compute the polynomial set up to arbitrary order.
See Sherwin & Karniadakis (1995).
An orthogonal set over a triangle can be obtained with a change of variable to \(\zeta = 2\frac{1+x}{1 - y} - 1\), and a modified polynomial. The polynomial basis becomes:
\[ Q_{p,q} = P^{0,0}_p(\zeta) \left(\frac{1-y}{2}\right)^p\ P_q^{2p+1, 0}(y), \]
with a Legendre Polynomial of \(\zeta\) and a weighted Jacobi Polynomial of \(y\). In order to calculate these without actually multiplying together the two polynomials directly, we can first compute \(Q_{p, 0}\) using a recurrence relation (since \(P_0^{2p + 1, 0} = 1\)):
\[ p Q_{p,0} = (2p-1)(2x + y+1)Q_{p-1,0} - (p-1)(1-y)^2 Q_{p-2, 0}.\]
Subsequently, we can calculate \(Q_{p,q}\) by building up from \(Q_{p,0}\) with another recurrence relation:
\[ Q_{p,q} = (a_0 y + a_1) Q_{p, q-1} + a_2 Q_{p, q-2}, \]
where \(a_0, a_1, a_2\) are the coefficients in the Jacobi recurrence relation with \((\alpha,\beta) = (2p+1, 0)\), see Wikipedia. Note that \(a_2 = 0\) when \(q < 2\).
See Sherwin & Karniadakis (1995).
Let \(\zeta = 2\frac{1+x}{y+z} + 1\) and \(\xi = 2\frac{1+y}{1-z} - 1\). Orthogonal polynomials on a tetrahedron are then given by
\[ Q_{p, q, r} = P^{0,0}_p(\zeta)\left(\frac{y+z}{2}\right)^p\ P_q^{2p+1,0}(\xi)\left(\frac{1-z}{2}\right)^q\ P_r^{2(p+q+1), 0}(z).\]
This can similarly be built up by first computing \(Q_{p,0,0}\), then \(Q_{p, q, 0}\) and finally \(Q_{p, q, r}\) with recurrence relations on each axis.
The polynomial sets for quadrilateral and hexahedral elements can be formed by applying the recurrence relation for intervals in each direction.
The polynomial set for a prism can be computed by using using the recurrence relation for the triangle to get the polynomials that are constant with repect to \(z\), then applying the recurrence relation for the interval to compute further polynomials with \(z\)s.
Orthogonal polynomials on the pyramid element are best calculated in the same way as the tetrahedron, using recurrence relations on each axis. Let \(\zeta_x = 2\frac{1+x}{1-z} - 1\), \(\zeta_y = 2\frac{1+y}{1-z} - 1\). The recurrence relation is then
\[Q_{p, q, r} = P^{0,0}_p(\zeta_x) P^{0,0}_q(\zeta_y) \left(\frac{1-z}{2}\right)^{(p+q)} P_r^{2(p+q+1), 0}(z).\]
For each cell type, we obtain an orthonormal set of polynomials by scaling each of the orthogonal polynomials.
Recurrence relations can also be used to find the derivatives of the polynomials at given points. For example, on the interval, the first derivatives are given by
\[ n P'_n(x) = (2n - 1) \left(P_{n-1}(x) + x P'_{n - 1}(x)\right) + (n-1)P'_{n-2}(x). \]
More generally, the \(k\)-th derivative is given by
\[ n P^k_n(x) = (2n - 1) \left(k P^{k-1}_{n-1}(x) + x P^k_{n - 1}(x)\right)+ (n-1) P^k_{n-2}(x) \]
This is now a recurrence relation in both \(n\) and \(k\). Similar recurrence relations can be obtained for the derivatives of all the polynomial sets on the other shapes. Care must be taken with quadratic terms, and cross-terms in two and three dimensions.
| int basix::polyset::dim | ( | cell::type | cell, |
| int | d | ||
| ) |
Dimension of a polynomial space
| [in] | cell | The cell type |
| [in] | d | The polynomial degree |
d | int basix::polyset::nderivs | ( | cell::type | cell, |
| int | d | ||
| ) |
Number of derivatives that the orthonormal basis will have on the given cell
| [in] | cell | The cell type |
| [in] | d | The highest derivative order |
d | xt::xtensor< double, 3 > basix::polyset::tabulate | ( | cell::type | celltype, |
| int | d, | ||
| int | n, | ||
| const xt::xtensor< double, 2 > & | x | ||
| ) |
Tabulate the orthonormal polynomial basis, and derivatives, at points on the reference cell.
All derivatives up to the given order are computed. If derivatives are not required, use n = 0. For example, order n = 2 for a 2D cell, will compute the basis \(\psi, d\psi/dx, d\psi/dy, d^2 \psi/dx^2, d^2\psi/dxdy, d^2\psi/dy^2\) in that order (0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0 ,2).
For an interval cell there are nderiv + 1 derivatives, for a 2D cell, there are (nderiv + 1)(nderiv + 2)/2 derivatives, and in 3D, there are (nderiv + 1)(nderiv + 2)(nderiv + 3)/6. The ordering is 'triangular' with the lower derivatives appearing first.
| [in] | celltype | Cell type |
| [in] | d | Polynomial degree |
| [in] | n | Maximum derivative order. Use n = 0 for the basis only. |
| [in] | x | Points at which to evaluate the basis. The shape is (number of points, geometric dimension). |
(number of derivatives computed, number of points, basis index).(p, q) denotes p order dervative with repsect to x and q order derivative with respect to y, [0] -> (0, 0), [1] -> (1, 0), [2] -> (0, 1), [3] -> (2, 0), [4] -> (1, 1), [5] -> (0, 2), [6] -> (3, 0),... The function basix::indexing::idx maps tuples (p, q, r) to the array index.i correspondign to the point in row i of x.| void basix::polyset::tabulate | ( | xt::xtensor< double, 3 > & | P, |
| cell::type | celltype, | ||
| int | d, | ||
| int | n, | ||
| const xt::xtensor< double, 2 > & | x | ||
| ) |
Tabulate the orthonormal polynomial basis, and derivatives, at points on the reference cell.
All derivatives up to the given order are computed. If derivatives are not required, use n = 0. For example, order n = 2 for a 2D cell, will compute the basis \(\psi, d\psi/dx, d\psi/dy, d^2 \psi/dx^2, d^2\psi/dxdy, d^2\psi/dy^2\) in that order (0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0 ,2).
For an interval cell there are nderiv + 1 derivatives, for a 2D cell, there are (nderiv + 1)(nderiv + 2)/2 derivatives, and in 3D, there are (nderiv + 1)(nderiv + 2)(nderiv + 3)/6. The ordering is 'triangular' with the lower derivatives appearing first.
| [in,out] | P | Polynomial sets, for each derivative, tabulated at points. The shape is (number of derivatives computed, number of points, basis index). |
(p, q) denotes p order dervative with repsect to x and q order derivative with respect to y, [0] -> (0, 0), [1] -> (1, 0), [2] -> (0, 1), [3] -> (2, 0), [4] -> (1, 1), [5] -> (0, 2), [6] -> (3, 0),... The function basix::indexing::idx maps tuples (p, q, r) to the array index.i correspondign to the point in row i of x.| [in] | celltype | Cell type |
| [in] | d | Polynomial degree |
| [in] | n | Maximum derivative order. Use n = 0 for the basis only. |
| [in] | x | Points at which to evaluate the basis. The shape is (number of points, geometric dimension). |