regular function $g$ on $\Gamma$, find the solution $(\textbf{u},p)$ to
@f{eqnarray*}
-{
-\Delta \mathbf{u} + \nabla p &=& 0,\\
-\nabla \cdot \textbf{u} &=& 0,\\
\textbf{u} &=& \textbf{g} \text{ in } \Gamma,\\
\end{pmatrix}
=
\begin{pmatrix}
- {0
+ 0
\\
0
\end{pmatrix},
boundary condition on $\Gamma$ is done through Nitsche method. This is achieved by using the following modified formulation :
-@f{eqnarray*}{
+@f{eqnarray*}
(\nabla \textbf{v}, \nabla \textbf{u})_{\Omega} + (\nabla \cdot \textbf{v}, p)_{\Omega}
+ (q, \nabla \cdot \textbf{u})_{\Omega} - (\nabla\textbf{v}\cdot \textbf{n},\textbf{u})_{\Gamma}
+ \beta (\textbf{v}},\textbf{u})_{\Gamma} &=& -(\nabla\textbf{v}\cdot \textbf{n},\textbf{g})_{\Gamma}
and right-hand side are equal since $\textbf{u}=\textbf{g}\text{ in } \Gamma$.
It follows that :
-@f{eqnarray*}{
+@f{eqnarray*}
(\nabla\textbf{v}\cdot \textbf{n},\textbf{u})_{\Gamma}
+ \beta (\textbf{v},\textbf{u})_{\Gamma} &=& -(\nabla\textbf{v}\cdot \textbf{n},\textbf{g})_{\Gamma}
+ \beta (\textbf{v},\textbf{g})_{\Gamma}
We note that an alternative formulation can be used :
-@f{eqnarray*}{
+@f{eqnarray*}
(\nabla \textbf{v}, \nabla \textbf{u})_{\Omega} + (\nabla \cdot \textbf{v}, p)_{\Omega}
+ (q, \nabla \cdot \textbf{u})_{\Omega} + (\nabla \textbf{v}\cdot \textbf{n},\textbf{u})_{\Gamma}
+ \beta (\textbf{v},\textbf{u})_{\Gamma} &=& (\nabla \textbf{v}\cdot \textbf{n},\textbf{g})_{\Gamma}
In the case of $\mathcal{L}^2$ penalization, an additional Darcy term is added
within $\Gamma$ resulting in :
-@f{eqnarray*}{
+@f{eqnarray*}
(\nabla \textbf{v}, \nabla \textbf{u})_{\Omega} + (\nabla \cdot \textbf{v}, p)_{\Omega}
+ (q, \nabla \cdot \textbf{u})_{\Omega}
+ \beta_1 (\textbf{v}},\textbf{u})_{\Gamma} &=&