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How is the L2 Qk element basis ranked? Is there a law about its ranking?
3 years 7 months ago #3664
by Younghigh
Hi all,
In DG methods over quadrilateral meshes, we tend to use L2 Qk element, like L2 Q1 basis is
{1, x, x*y, y},
L2 Q2 basis is
{1, x, x*y, x*y**2, x**2, x**2*y, x**2*y**2, y, y**2},
...
We want to know the ranking of L2 Qk element basis in NGSolve, especially for high number k.
Best,
Di Yang
In DG methods over quadrilateral meshes, we tend to use L2 Qk element, like L2 Q1 basis is
{1, x, x*y, y},
L2 Q2 basis is
{1, x, x*y, x*y**2, x**2, x**2*y, x**2*y**2, y, y**2},
...
We want to know the ranking of L2 Qk element basis in NGSolve, especially for high number k.
Best,
Di Yang
- mneunteufel
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3 years 7 months ago #3666
by mneunteufel
Replied by mneunteufel on topic How is the L2 Qk element basis ranked? Is there a law about its ranking?
Hi Di Yang,
Q_k is a tensor product, Q_k = P_k(x) x P_k(y), and as dim(P_k) = k+1there holds dim(Q_k) = (k+1)^2.
Best
Michael
Q_k is a tensor product, Q_k = P_k(x) x P_k(y), and as dim(P_k) = k+1there holds dim(Q_k) = (k+1)^2.
Best
Michael
3 years 7 months ago #3667
by Younghigh
Replied by Younghigh on topic How is the L2 Qk element basis ranked? Is there a law about its ranking?
Thank u. I mean the ranking of degree of freedom in L2 Qk element basis, like Q1,
the order is 1, x, xy, y, but I have no idea about high k.
the order is 1, x, xy, y, but I have no idea about high k.
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3 years 7 months ago #3668
by mneunteufel
Replied by mneunteufel on topic How is the L2 Qk element basis ranked? Is there a law about its ranking?
In fem/l2hofe_impl.hpp in function L2HighOrderFE_Shape<ET_QUAD> ::T_CalcShape about line 430 the code for the computation of the quad basis reads
Code:
LegendrePolynomial (p, xi, polx);
LegendrePolynomial (q, eta, poly);
for (size_t i = 0, ii = 0; i <= p; i++)
for (size_t j = 0; j <= q; j++)
shape[ii++] = polx[i] * poly[j];
The following user(s) said Thank You: Younghigh
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