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Old   March 21, 2003, 06:23
Default Node in the corner
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Suppose I solve the Neumann task in a square (2D) for Poisson's equation by FVM. If one of the grid nodes is in the corner then how to take account of a condition dF/dn=0 (a normal derivative doesn't exist)? Thanks.
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Old   March 21, 2003, 06:51
Default Re: Node in the corner
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At a corner there no unique normal. Hence your question. Mathematically the Neumann problem

-laplace u = f in U du/dn = 0 on boundary of U

has a weak solution which satisfies

int grad(u).grad(v) dx = int fv dx

for all v belonging to H^1(U). There are some extra conditions on f but that is not relevant. What is important is that the Neumann bc is not required in this definition. It is in this sense a non-essential bc. FE methods which use the above definition of weak solution will not have any problem. But if you use a FV method then you have to do something at a corner which cannot justified mathematically. If you are using a cell-centered FV method then again you dont have any problem. Only in cell-vertex scheme you have to solve at the corner. I cannot actually answer your question and will look forward to see what others have to say. In similar situations I have done something REASONABLE and it usually works.
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