Hi Wolfgang, thanks so much for getting back to me
1. I was wondering re dimensions because I could find a component mask
function or something similar to fe_values[velocities], eg, when using
fe_face_values which you need to apply the neumann conditions.
2. This is my fault. I meant working in 2d with a solution that only acts
vertically, hence choosing a solution that is only dependent on y and that
is 0 in the xdirection.
3. indeed i was showing the full system right hand side component
4. for the top and bottom boundaries, you are right, i do mean the stress
as mentioned in previous messages. Either way, I am now implementing things
in the neumann BC way as in step-22. with the solution as before, i add the
following to my top boundary:
for (unsigned int face_number=0;
(cell->face(face_number)->boundary_id() == 1))
fe_face_values.reinit (cell, face_number);
for (unsigned int q_point=0; q_point<n_face_q_points;
for (unsigned int i=0; i<dofs_per_cell; ++i)
const unsigned int component_i =
where the top stress is g_N as in step-22 (where i calculate the normal as
plus or minus (0,1) depending on whether it's the top or bottom boundary).
I am running this off exactly using the equatoins in step-22 with the right
hand sides modified and using a direct solver instead.
The thing i have worked out is that as my program runs currently, but with
DIRICHLET conditions on top and bottom, it works. but when i try to use the
nuemann conditions as i have shown in the code snippet above does not (with
everything else kept the same). So i am wondering whether the snippet i
have shown here contains an error I have not figured out.
Thanks for your help
On Monday, February 19, 2018 at 5:44:39 PM UTC-5, Wolfgang Bangerth wrote:
> > Firstly, I decided not to use the normal vector for now. Since the
> > normal vector is 2D, i wasn't sure how to implement the rest so that it
> > is a 'double' since my g (neumann condition vector) has 3 components
> > when the normal vector will only have 2?
> I'm not sure I understand this -- in 2 space dimensions, the normal
> vector has 2 components; in 3 space dimensions, it has 3 components.
> > I went back to the beginning - step-22 tutorial and started from
> > absolute basics. Testing with 1D, i have started with exact solution
> > u=(0,-y^2)^T and p = y^3, ie purely vertical.
> I also don't understand this, I'm afraid: in d space dimensions, the
> velocity vector has d components. So if you're in 1d, how come your
> velocity vector has 2 components?
> > Doing a manufactured
> > solution type, i get the right hand side vector f = (0, 4+3z^2, 2z)^T.
> the right hand side has d components. I suspect you are here showing the
> right hand side of the whole system, which should then have d+1
> components (which in 1d should be 2, not 3).
> Or do you want to suggest in all of this that you really have a
> two-dimensional domain but that the flow is one-dimensional?
> > with my solution, on a rectangular domain, i have no tangential stresses
> > anywhere, no flux on sides (unit normal being (/pm 1,0)), and plus and
> > minus (0, z^3 + 4z) at the top.
> Your domain is [0,1]^2, so I understand how there is no flux through the
> bottom because there u=[0,0]. But how do you arrive at the flux you show
> here? The velocity field is quadratic in z (y?), and so should the flux,
> but you show something that is cubic with the distance.
> If instead of 'flux' you really mean 'stress', then it should actually
> only be linear in z since it involves the *derivative* of the velocity.
> Wolfgang Bangerth email: bang...@colostate.edu
> www: http://www.math.colostate.edu/~bangerth/
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