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:                 bange...@colostate.edu
                           www: http://www.math.colostate.edu/~bangerth/

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