# Re: [deal.II] Re: step-22 partial boundary conditions

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Jane,```
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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?
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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.
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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.
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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?
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Similarly...

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Doing a manufactured solution type, i get the right hand side vector f = (0, 4+3z^2, 2z)^T.
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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).
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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?
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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.
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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.
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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.
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Best
W.

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

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