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How can H() return values if itself contains unknowns?

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Old   February 1, 2015, 01:55
Question How can H() return values if itself contains unknowns?
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Chenguang Zhang
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Hi there, I suddenly get quite confused by the H() function . The meaning of H() is explained here:
and in this excellent post:
From where I quote:
Therefore, the linear system becomes (A + H')U = R, which is the same as AU = R - H'U. The right hand side is simply called H, so H = R - H'U. Therefore, we know have AU = H.

In the above explanation we are trying to solve for U, if we define H = R-H'U, then H itself contains the unknown, this doesn't make sense. On the other hand, if the above operation is understood to be Jacobi iteration ( Then this does not make sense either: we don't use Jacobi iteration for anything, and solvers have no error control of it.

In icoFoam, the H() is called after "solve(UEqn == -fvc::grad(p));". There is no return value or assignment for this solve(), then what's the effect of solve() anyway? Does it modify the UEqn itself, then how? If it does modify the matrix, then which U (U at time n-1, U* newly obtained, or a mix of the two) contribute to results of H()?

I would appreciate any explanation from gurus, and people who have the same puzzle are welcomed to post so we can discuss.
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