From: Roy Stogner <roystgnr@ic...>  20060723 03:02:49

On Sat, 22 Jul 2006, David Xu wrote: > On 7/22/06, Roy Stogner <roystgnr@...> wrote: >> >> On Sat, 22 Jul 2006, David Xu wrote: >> >> > Are you saying HEX27 and HEX8 will give same level of accuracy in the >> > solutions if I don't need quadratic mapping functions? >> >> Yes. > > Just ccurious. Does the same rule apply to other types of element? What > about Tet, Tri, Quad and Prism. So, is the level of solution accuracy > independent with the number of node within the same type of element? The level of solution accuracy is independent of the number of geometric nodes... but libMesh reuses geometric nodes to store degrees of the freedom that have the same topological connectivity, so you usually still need secondorder nodes even if you aren't fitting a secondorder geometry. As far as I know, the HERMITE elements and the two discontinuous elements are the only way to get better than linear approximations on linear geometric elements. If you try to use finite elements on geometric elements that don't support them, however, you won't just get reduced accuracy, your code will exit with an error. > What about the difference between different element types in terms > of the effect on the quality of the solutions? You can get better solutions (better conditioned matrices, at least) from quadratic elements if you use a mesh smoother that takes advantage of them. Mostly, though, you only need higher order geometric elements to better fit curved domain boundaries. >> Almost certainly it is. None of our finite element classes are as >> optimized as they should be, but I think the Hermite elements may be >> worse than average. >> >> Keep in mind, too, that even if they were equally optimized, the >> Hermite assembly would be more expensive. If you're using the default >> quadrature order (which is designed for nonlinear problems and may be >> gross overkill for you) then I think quadratic hexes will be >> calculating at 27 points and cubic hexes will be calculating at 64. > > That explains why the ouput filesize from hermite is much larger than > lagrange. No, it doesn't. I'm talking about quadrature points here, and the size of your final matrix is (with few exceptions) independent of the quadrature rule you use to calculate it. I can see why that's confusing, though: by coincidence the number of quadrature points is the same as the number of local DoFs for both elements here. Of course, it's not just the quadrature rule that's important. Having 64 local DoFs instead of 27 also increases calculation time. Finally, on uniform meshes Hermite cube DoFs usually couple to 216 DoFs rather than 27, 45, or 125, which is probably what's increasing your output file size. > Does that mean, even the matrix dimension is the same, but hermit > produces more entries in the matrix, thus it's less sparse? Yes. Increasing polynomial order requires more bandwidth, so does increaing continuity, and going from quadratic Lagrange to cubic Hermite does both at once.  Roy 