How is the L2 Qk element basis ranked? Is there a law about its ranking?

1 month 6 days 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

Please Log in or Create an account to join the conversation.

1 month 6 days ago #3666 by mneunteufel
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

Please Log in or Create an account to join the conversation.

1 month 6 days ago #3667 by Younghigh
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.

Please Log in or Create an account to join the conversation.

1 month 6 days ago #3668 by mneunteufel
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
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

Please Log in or Create an account to join the conversation.

© 2019 Netgen/NGSolve