No adaptivity.
I suspect it would not be about the weak form directly. So, every time I
give an set of initial cells, I need to make sure total initial cell number
N_cell_tot % n_process == 0. If this condition is fulfilled, solution will
always be correct. Say, I want to use 5 processors, then I have to give
N_cell = 5 or 10 or 5 * some integer.
You see what's going wrong?
Here is my mesh generation
template <int dim>
void MeshGenerator<dim>::generate_initial_grid
(parallel::distributed::Triangulation<dim> &tria)
{
Point<dim> origin;
Point<dim> diagonal;
switch (dim)
{
case 1:
{
diagonal[0] = axis_max_values[0];
break;
}
case 2:
{
diagonal[0] = axis_max_values[0];
diagonal[1] = axis_max_values[1];
break;
}
case 3:
{
diagonal[0] = axis_max_values[0];
diagonal[1] = axis_max_values[1];
diagonal[2] = axis_max_values[2];
break;
}
default:
break;
}
GridGenerator::subdivided_hyper_rectangle (tria,
ncell_per_dir,
origin,
diagonal);
}
在 2017年7月6日星期四 UTC-7上午8:26:06，Wolfgang Bangerth写道：
>
> On 07/06/2017 09:02 AM, Weixiong Zheng wrote:
> > Both are generated with subdivided hyper rectangle with same reps and
> diagonals.
>
> And no adaptive refinement?
>
> In that case, the solution *should* be the same, assuming your linear
> solver
> tolerance is small enough. If it isn't, there is probably a bug somewhere.
> But
> it should be possible to debug this on a relatively coarse mesh, since you
> have such a nice invariant (the symmetry) that you can check on a very
> coarse
> mesh as well as in your matrix and rhs vectors. For example, you could
> look
> graphically at the rhs vector to see whether the quantity it represents is
> symmetric or not.
>
> Best
> W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bang...@colostate.edu
> <javascript:>
> www: http://www.math.colostate.edu/~bangerth/
>
>
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