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revision 3331 by caltinay, Mon Nov 1 05:21:18 2010 UTC revision 3501 by gross, Wed Apr 13 04:07:53 2011 UTC
# Line 41  $\alpha_{1}$ which are independent from Line 41  $\alpha_{1}$ which are independent from
41  \end{equation}  \end{equation}
42  for all $x_{i}$.  for all $x_{i}$.
43    
44    We set $u=\hat{u}+u^{N}$ and
45    $p=\hat{p}+p^D$ where $\hat{u}$ and $\hat{p}$ fullfill the homogeneous boundary conditions~\label{DARCY BOUNDARY} and
46    the PDE
47    \begin{equation}\label{DARCY PROBLEM 2}
48    \begin{array}{rcl}
49    \hat{u}_{i} + \kappa_{ij} \hat{p}_{,j} & = & \hat{g}_{i} = g_{i} -u^{N} - \kappa_{ij} p^D_{,j} \\
50    \hat{u}_{k,k} & = & f - u^N_{k,k} = \hat{f}
51    \end{array}
52    \end{equation}
53    In the following the hat is dropped.
54    
55    
56    \subsection{Solution Method \label{DARCY SOLVE}}
57    We set
58    \begin{align}
59    Q=\{ q: \Omega \to R | q = 0 \mbox{ on } \Gamma_{D} \} \label{DARCY SPACE P} \\
60    V=\{ v: \Omega \to R^3 |v_{i}  \; n_{i}  =0 \mbox{ on }  \Gamma_{N}  \} \label{DARCY SPACE P}
61    \end{align}
62    We define the gardient operator $G: Q \to V$ as
63    \begin{equation}
64    (Gp)_{i} =  p_{,j}
65    \end{equation}
66    for all $p \in Q$ and the
67    divergence operator $D: V \to Q$
68    and for velocity $v$ we set
69    \begin{equation}
70    Dv = v_{k,k}
71    \end{equation}
72    As $\Gamma_{D}, \Gamma_{N}$ are a subdivision of the boundary of $\Omega$ we have
73    \begin{equation}
74    D^*= - G
75    \end{equation}
76    With $K=(\kappa_{ij})$ we can write the Darcy problem~\ref{DARCY PROBLEM} as
77    \begin{equation}
78    \begin{array}{rcl}
79    K^{-1} u - D^* p & = & K^{-1} g \\
80    Du & = & f
81    \end{array}
82    \end{equation}
83    This problem needs to be stabalized in the following way:
84    \begin{equation}
85    \begin{array}{rcl}
86    K^{-1} u - D^* p - \frac{1}{2} \left( K^{-1} u +G  p \right) & = & K^{-1} g - \frac{1}{2} K^{-1} g \\
87    Du + \frac{1}{2} G^* K \left( K^{-1} u +G  p \right) & = & f + \frac{1}{2} G^* K K^{-1} g
88    \end{array}
89    \end{equation}
90    which can be simplified to
91    \begin{equation}
92    \begin{array}{rcl}
93    \frac{1}{2}  K^{-1} u - D^* p - \frac{1}{2} G  p & = & \frac{1}{2} K^{-1} g \\
94    Du + \frac{1}{2} G^* u + \frac{1}{2} G^* K G  p  & = & f + \frac{1}{2} G^* g
95    \end{array}
96    \end{equation}
97    
98    
99    
100    %===========================================================================
101  \subsection{Solution Method \label{DARCY SOLVE}}  \subsection{Solution Method \label{DARCY SOLVE}}
102  It is useful to write \eqn{DARCY PROBLEM} in operator form.  It is useful to write \eqn{DARCY PROBLEM} in operator form.
103  For any pressure $p$ we set  For any pressure $p$ we set

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