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; 
face_number<GeometryInfo<dim>::faces_per_cell; ++face_number)

                if (cell->face(face_number)->at_boundary()


                    (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 = 

                            local_rhs(i) += 


                                             shape_value(i,q_point) *





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:
> Jane, 
> > 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? 
> Similarly... 
> > 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. 
> Best 
>   W. 
> -- 
> ------------------------------------------------------------------------ 
> Wolfgang Bangerth          email:                 bang...@colostate.edu 
> <javascript:> 
>                             www: http://www.math.colostate.edu/~bangerth/ 

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