ffc team mailing list archive

[Shawn Walker <walker@xxxxxxxxxxxxxxx>]

I've almost got this working.  When I compare the stiffness matrix from my 
code to the one from dolfin the max difference in the entries is machine 
precision.  This is when the mesh is straight (no curves).  If I have a 
slightly curved mesh, the error is 0.0162, but I think this is because the 
quad rule that FFC generated only has 4 points and I only have two 
triangles with diameter approx 0.5 (not small enough resolve the error).

Is there a way to tell FFC to use more quad points than it `should'?  That 
way I can verify that I am computing this properly.

- Shawn

On Tue, 23 Jun 2009, Anders Logg wrote:

> DOLFIN assumes that the mesh is ordered locally so that the local
> vertices on each triangle are ordered increasingly. Take a look at the
> XML files in data/meshes in DOLFIN and you'll see what I mean.
>
> --
> Anders
>
>
> On Tue, Jun 23, 2009 at 12:11:13PM -0400, Shawn Walker wrote:
> > ok.  But I have one more question.
> >
> > Is there any kind of re-ordering of nodes that dolfin does internally?  I
> > thought I remembered seeing this.  I have been trying to compare two
> > matrices, and it looks like the entries are the same, except the ordering
> > is a little different.
> >
> > - Shawn
> >
> > On Tue, 23 Jun 2009, Kristian Oelgaard wrote:
> >
> > > Quoting Shawn Walker <walker@xxxxxxxxxxxxxxx>:
> > >
> > > > Hello.  I need to know how FFC orders P2 Lagrange nodes.  I need to know
> > > > this in order to compare a FEM matrix that I created in another program.
> > > >
> > > > For example, say you have one triangle with the following nodes:
> > > >
> > > > [0 1 2 3 4 5]
> > > >
> > > > Where is the #3 node in relation to the #0, #1, and #2 nodes?  Is #3
> > > > opposite the #0 node?
> > >
> > > Yes.
> > >
> > > I usually look at the tabulate_dofs() function, for the P2 case:
> > >
> > > dofs[0] = c.entity_indices[0][0];
> > > dofs[1] = c.entity_indices[0][1];
> > > dofs[2] = c.entity_indices[0][2];
> > > unsigned int offset = m.num_entities[0];
> > > dofs[3] = offset + c.entity_indices[1][0];
> > > dofs[4] = offset + c.entity_indices[1][1];
> > > dofs[5] = offset + c.entity_indices[1][2];
> > >
> > > So you see that the first 3 dofs are the vertices, the next 3 are the midpoints
> > > of the edges. Then you need to know the UFC ordering of edges to figure out
> > > that e0 is the edge which does not contain vertex 0 (opposite to v0) etc.
> > >
> > > Kristian
> > >
> > > > - Shawn
> > > > _______________________________________________
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