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How to use BDM element in a quadrilateral mesh?
3 years 7 months ago #3688
by Younghigh
How to use BDM element in a quadrilateral mesh? was created by Younghigh
Hi,
When running the following code:
I find there are 24 vtk documents shown, which should be RT2 basis in a rectangle. But I have used
, my question is: how to use BDM element in a quadrilateral mesh?
Best,
Di Yang
When running the following code:
Code:
from ngsolve import *
import ngsolve.meshes as ngm
mesh = ngm.MakeStructured2DMesh(nx=1,ny=1)
V = HDiv(mesh, order=2, RT=False)
gfu = GridFunction(V)
vtk = VTKOutput(ma=mesh, coefs=[gfu], names=["bdm-basis"], filename="bdm2basis", subdivision=3)
for i in range(gfu.vec.size):
gfu.vec[:] = 0
gfu.vec[i] = 1
vtk.Do()
Code:
RT=False
Best,
Di Yang
- mneunteufel
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3 years 7 months ago #3689
by mneunteufel
Replied by mneunteufel on topic How to use BDM element in a quadrilateral mesh?
Hi Di Yang,
to check the number of degrees of freedom on a single quad for different polynomial orders you can use the following code
which gives the output
RT: 0 : 4, 1 : 12, 2 : 24, 3 : 40, 4 : 60, 5 : 84
BDM: 0 : 12, 1 : 24, 2 : 40, 3 : 60, 4 : 84, 5 : 112
As you can see they coincide (RT is "shifted"). The reason for this is that RT elements are available for triangles and tetrahedra, but not for quadrilaterals, hexahedra, or prisms. So there are only BDM elements for quads.
Best
Michael
to check the number of degrees of freedom on a single quad for different polynomial orders you can use the following code
Code:
from ngsolve import *
from ngsolve.meshes import MakeStructured2DMesh
mesh = MakeStructured2DMesh(quads=True,nx=1,ny=1)
Draw(mesh)
print("RT")
for p in range(6):
fes = HDiv(mesh, order=p, RT=True)
print(p,":",fes.ndof)
print("BDM")
for p in range(6):
fes = HDiv(mesh, order=p+1)
print(p,":",fes.ndof)
RT: 0 : 4, 1 : 12, 2 : 24, 3 : 40, 4 : 60, 5 : 84
BDM: 0 : 12, 1 : 24, 2 : 40, 3 : 60, 4 : 84, 5 : 112
As you can see they coincide (RT is "shifted"). The reason for this is that RT elements are available for triangles and tetrahedra, but not for quadrilaterals, hexahedra, or prisms. So there are only BDM elements for quads.
Best
Michael
3 years 7 months ago - 3 years 7 months ago #3690
by Younghigh
Replied by Younghigh on topic How to use BDM element in a quadrilateral mesh?
Thank u first for your specific replies. However, the dofs of BDMk on any rectangle K are devised as follows.
[tex]
(\mathbf{w}\cdot\mathbf{n}, p_k)_{e_i},\quad \forall\,p_k\in P_k(e_i),\ i=1,2,3,4;
[/tex]
[tex]
(\mathbf{w}, \mathbf{p}_{k-2})_K,\quad \forall\,\mathbf{p}_{k-2}\in (P_{k-2}(K))^2.
[/tex]
So, it should be
[tex]
\mathrm{dim}(BDMk)=k^2+3k+4,
[/tex]
i.e., BDMk
k=1: 8;
k=2: 14;
k=3: 22;
k=4: 32;
...
But if running
the number of dofs of V is 24. I think it is not a BDM2.
Best,
Di Yang
[tex]
(\mathbf{w}\cdot\mathbf{n}, p_k)_{e_i},\quad \forall\,p_k\in P_k(e_i),\ i=1,2,3,4;
[/tex]
[tex]
(\mathbf{w}, \mathbf{p}_{k-2})_K,\quad \forall\,\mathbf{p}_{k-2}\in (P_{k-2}(K))^2.
[/tex]
So, it should be
[tex]
\mathrm{dim}(BDMk)=k^2+3k+4,
[/tex]
i.e., BDMk
k=1: 8;
k=2: 14;
k=3: 22;
k=4: 32;
...
But if running
Code:
V = HDiv(mesh, order=2, RT=False)
Best,
Di Yang
Last edit: 3 years 7 months ago by Younghigh.
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3 years 7 months ago #3691
by mneunteufel
Replied by mneunteufel on topic How to use BDM element in a quadrilateral mesh?
Could be that it is exactly the other way round, sorry.
The Quad-HDiv basis in NGSolve is implemented as described in this thesis .
The basis span Q_{k+1,k} x Q_{k,k+1} if I remember correctly and thus has dimension 2*(k+2)*(k+1)=2k**2+6k+4.
Best
Michael
The Quad-HDiv basis in NGSolve is implemented as described in this thesis .
The basis span Q_{k+1,k} x Q_{k,k+1} if I remember correctly and thus has dimension 2*(k+2)*(k+1)=2k**2+6k+4.
Best
Michael
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