# Diff of /trunk/doc/user/stokessolver.tex

revision 2719 by gross, Wed Oct 14 06:38:03 2009 UTC revision 2748 by gross, Tue Nov 17 07:32:59 2009 UTC
# Line 43  G \\ Line 43  G \\
43  \end{array} \right]  \end{array} \right]
45
46  where $A$ is coercive (assuming $A$ is not depending on $v$ or $p$), self-adjoint linear operator in a suitable Hilbert space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is it adjoint operator (=gradient operator). For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.  where $A$ is coercive (assuming $A$ is not depending on $v$ or $p$), self-adjoint linear operator in a suitable Hilbert space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is it adjoint operator (=gradient operator).
47    For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
48  We use iterative techniques to solve this problem: Given an approximation $v$ and $p$ for  We use iterative techniques to solve this problem: Given an approximation $v$ and $p$ for
49  velocity and pressure we perform the following steps in the Uzawa scheme \index{Uzawa scheme} style:  velocity and pressure we perform the following steps in the Uzawa scheme \index{Uzawa scheme} style:
50  \begin{enumerate}  \begin{enumerate}
# Line 88  A w = B^{*} s Line 89  A w = B^{*} s
89
90  with sufficient accuracy to return $q=Bw$. Notice that the residual $r$ is given as  with sufficient accuracy to return $q=Bw$. Notice that the residual $r$ is given as
91   \label{STOKES RES }   \label{STOKES RES }
92   r= B (v\hackscore{1} - B A^{-1} B^{*} dp) =  B (v\hackscore{1} - A^{-1} B^{*} dp) = B (v\hackscore{1}-dv\hackscore{2}) = B v\hackscore{2}   r= B (v\hackscore{1} -  A^{-1} B^{*} dp) =  B (v\hackscore{1} - A^{-1} B^{*} dp) = B (v\hackscore{1}-dv\hackscore{2}) = B v\hackscore{2}
93
94  so in fact the residual $r$ is represented by the updated velocity $v\hackscore{2}$. This saves the recovery of  so in fact the residual $r$ is represented by the updated velocity $v\hackscore{2}$. This saves the recovery of
95  $dv\hackscore{2}$ in~\ref{SADDLEPOINT ITER STEP 2} after $dp$ has been calculated as iterative method such as PCG calculate the solution approximations along with the their residual. In PCG the iteration is terminated if  $dv\hackscore{2}$ in~\ref{SADDLEPOINT ITER STEP 2} after $dp$ has been calculated as iterative method such as PCG calculate the solution approximations along with the their residual. In PCG the iteration is terminated if

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