Uzawa method

We want to solve the same following discrete problem by the Uzawa method, see for example [RT93]:

\[\begin{eqnarray} \frac{\rho}{\Delta t}\int_{\Omega}{\mathbf{u}_h^{n+1}\cdot\mathbf{v}_h} + \mu\int_{\Omega}{\nabla\mathbf{u}_h^{n+1}\cdot\nabla\mathbf{v}_h} - \int_{\Omega}{p_h^{n+1}\mathrm{div}(\mathbf{v}_h)} &&\nonumber\\ = \int_{\Omega}{\textbf{f}^{n+1}\cdot\mathbf{v}_h} &+& \frac{\rho}{\Delta t}\int_{\Omega}{(\mathbf{u}_h^n\circ\textbf{X}_h^n)\cdot\mathbf{v}_h}\\ \int_{\Omega}{\mathrm{div}(\mathbf{u}_h^{n+1})q_h} &=& 0 \end{eqnarray}\]

Let \(a:V\times V\rightarrow\mathbb{R}\) and \(b:V\times Q\rightarrow\mathbb{R}\) be bilinear forms defined by :

\[\begin{eqnarray} a(\mathbf{u}, \mathbf{v}) &=& \frac{\rho}{\Delta t}\int_{\Omega}{\mathbf{u}^{n+1}\mathbf{v}} +\mu\int_{\Omega}{\nabla\mathbf{u}^{n+1}\cdot\mathbf{v}}\\ b(\mathbf{v}, q) &=& -\int_{\Omega}{p^{n+1}\mathrm{div}(\mathbf{v})} \end{eqnarray}\]
\[\begin{eqnarray} a(\mathbf{u}, \mathbf{v}) &=& \frac{\rho}{\Delta t}\int_{\Omega}{\mathbf{u}^{n+1}\mathbf{v}} +\mu\int_{\Omega}{\nabla\mathbf{u}^{n+1}\cdot\mathbf{v}}\\ b(\mathbf{v}, q) &=& -\int_{\Omega}{p^{n+1}\mathrm{div}(\mathbf{v})} \end{eqnarray}\]

And let \(f:V\rightarrow\mathbb{R}\) be a linear functionnal defined by:

\[f(\mathbf{u}, \mathbf{v}) = \int_{\Omega}{\textbf{f}^{n+1}\mathbf{v}} + \frac{\rho}{\Delta t}\int_{\Omega}{(\mathbf{u}^n\circ\textbf{X}^n)\mathbf{v}}\]

The discrete problem becomes:

\[\begin{eqnarray} a(\mathbf{u}_h, \mathbf{v}_h) + b(\mathbf{v}_h, p_h) &=& f(\mathbf{v}_h)\\ b(\mathbf{u}_h, q_h) &=& 0 \end{eqnarray}\]

This problem is equivalent to the algrebraic system:

\[\begin{equation} \left( \begin{array}{cc} A & B^T\\ B & 0 \end{array} \right) \left( \begin{array}{c} \mathbf{u}_h\\ p_h \end{array} \right) = \left( \begin{array}{c} F\\ 0 \end{array} \right) \label{eqn:algebraic} \end{equation}\]

Where A, B and F are respectively the representative matrices of a, b and f.

Let introduce the functionnal:

\[J(\mathbf{u}) = \frac{1}{2}(A,\mathbf{u})-(F, \mathbf{u})\]

Then, the problem [eqn:algebraic] is equivalent to the contrained minimization problem:

\[\left\{ \begin{array}{rcl} J(\mathbf{u}) & = & \inf_{\mathbf{v}\in X}{J(\mathbf{v})}\\ X & = & \{\mathbf{v}\in\mathbb{R}, B\mathbf{u}=0\} \label{eqn:minimmization} \end{array} \right.\]

We associate the Lagrangian function:

\[\mathcal{L}(\mathbf{u},q) = J(\mathbf{u})-(q,B\mathbf{u})\]

A saddle-point \((\mathbf{u},p)\in V\times\mathbb{R}^+\) of L is characterized by the min-max problem:

\[\forall q\in\mathbb{R}^+, \mathcal{L}(\mathbf{u},q)\leq\mathcal{L}(\mathbf{u},p)\leq\mathcal{L}(\mathbf{v},p), \forall \mathbf{v}\in V\]

We assume that functions \(J\) and \(b\) are differentiable and convex. Since Karush-Kuhn-Tucker conditions are verified, a saddle-point \((\mathbf{u},p)\) of \(\mathcal{L}\) exists and the problem [eq:const-min] admits a solution.

To find the saddle-point \((\mathbf{u},p)\), we will solve two easiest problems:

  • The primal problem consists of finding the minimum of the function \(\mathbf{v}\rightarrow\mathcal{L}(\mathbf{v},p)\)

\[\mbox{Find }\mathbf{u}_h^n: A\mathbf{u}_h^n + B^Tp_h^n =F\]
  • The dual problem consists of finding the maximum of the function \(q\rightarrow\mathcal{L}(\mathbf{u},q)\) s.t. \(q\geq 0\). We introduce the dual function:

\[G(q) = \inf_{v\in V}{\mathcal{L}(\mathbf{v},q)}\]

Then we search \(p\in\mathbb{R}^+\) which maximize \(G\). Given that \(G(q)=J(\mathbf{u}_q)-(q,B\mathbf{u}_q)\), by derivation we have \(\nabla G(q) = -B\mathbf{u}_q\). To solve this constrained maximization problem we can use the projected gradient method:

\[\mbox{Find }p_h^n:p_h^{n+1}=P_{\mathbb{R}^+}(p_h^n-\rho\nabla G(p_h^n))=\max(0,p_h^n+\rho B\mathbf{u}_h^n)\]