11. The Stokes equations

Considering a stationary creeping flow, i.e. neglecting the time derivative and the non-linear convective term, we get the Stokes equations

\[\begin{alignat*}{2} - \nu \Delta u + \nabla p &=f \quad && \text{in } \Omega, \\ \operatorname{div} u &=0 \quad && \text{in } \Omega. \end{alignat*}\]

Let \(V = [H^1(\Omega)]^d\) and \(Q = L^2(\Omega)\), then the weak formulation is given by: Find \((u, p) \in V \times Q\), \(u = u_{in}\) on \(\Gamma_{in}\) s.t.

\[\begin{alignat*}{2} \nu \int_\Omega \nabla u: \nabla v - \int_\Omega \operatorname{div} v \,p &=\int_\Omega f v, \qquad && \forall v \in V, v = 0 \text{ on } \Gamma_{in} \cup \Gamma_{wall}, \\ -\int_\Omega \operatorname{div} u \, q &=0, \qquad && \forall q \in Q, \end{alignat*}\]

or

\[ K((u,p), (v,q)) = a(u,v) - b(v,p) -b(u,q) = (f,v), \qquad \forall v \in V, v = 0 \text{ on } \Gamma_{in} \cup \Gamma_{wall}, \quad \forall q \in Q, \]

with

\[ a(u,v) = \nu \int_\Omega \nabla u: \nabla v \quad \text{and} \quad b(v,q) = \int_\Omega \operatorname{div} v \, p \quad \text{and} \quad (f, v) = \int_\Omega f\, v. \]

Note

In the case \(\Gamma_{out} = \emptyset\) we have to add an additional constraint for \(Q\) to make the pressure unique. A common approach is to choose \(Q = \{q \in L^2(\Omega): \int_\Omega q = 0 \}\).