### Saddle point problem in Navier-Stokes equation

It is well known that the the optimization problem

$\min J(x)= \frac{1}{2}(Ax, x)-(f, x)$

is equal to solving a linear operator equation

$Ax=f$

provided that $A$ is symmetric and positive definite. Here $x\in X$ which is a Hilbert space, and $f\in X'$ while $A$ is a bilinear operator.

This is a very general problem considered widely in many areas. However, in the real world, some further condition are to be applied on the space where $x$ lies on. If this is described as constraint equation

$Bx=g$

then we are dealing with constraint optimization problem ,to which Lagrange multiplier is usually the first one can come up. We then try to solve the stationary point of Lagrangian

$L(x,\lambda)=\frac{1}{2}(Ax, x)-(f, x)+\lambda (Bx-g)$

Now supposing we set $X=R^n$, using stationary point condition on above equation, we arrive at

$\left[\begin{array}{cc} A & B^T\\ B & O \end{array}\right]\left[\begin{array}{c} x\\ y\end{array}\right]=\left[\begin{array}{c} f\\ g\end{array}\right]$

where $\lambda$ is replaced by $y$ indicating the parallel unknown as $x$ in the problem. The stationary point nature of this solution gives the name of saddle point problem, whose general case is

$L(x,v)\le L(x,y) \le L(u,y), \forall u,v \Leftrightarrow \min_u\max_v L(u,v)=L(x,y)=\max_v\min_u L(u,v)$

So, putting a constraint to a minimization problem, one try to solve a saddle problem instead of a simple linear system with SPD matrix.

For a linear system, one can always decompose it in to the form of

$\left[\begin{array}{cc} A & B_1^T\\ B_2 & -C \end{array}\right]\left[\begin{array}{c} x\\ y\end{array}\right]=\left[\begin{array}{c} f\\ g\end{array}\right]$

and this is the most general form of saddle problem.

For incompressible Navier-Stokes equation, we have (without mentioning the boundary conditions)

$\frac{\partial\mathbf{u}}{\partial t}+\mathbf{u}\cdot\nabla\mathbf{u}-\nu\Delta \mathbf{u}+\nabla p/\rho=\mathbf{f}$

$\nabla\cdot\mathbf{u}=0$

Following $\theta$-scheme by Glowinski, one has the time discretized plan as:

$(\mathbf{u^{n+\theta}}-\mathbf{u}^n)/(\theta\Delta t)-\alpha\nu\Delta\mathbf{u}^{n+\theta}+\nabla p^{n+\theta}=\mathbf{f}^{n+\theta}+\beta\nu\Delta \mathbf{u}^n-(\mathbf{u}^n\cdot\Delta)\mathbf{u}^n$

$\nabla\cdot\mathbf{u}^{n+\theta}=0$

then

$(\mathbf{u^{n+1-\theta}}-\mathbf{u}^{n+\theta})/((1-2\theta)\Delta t)-\beta\nu\Delta\mathbf{u}^{n+1-\theta}+(\mathbf{u}^{n+1-\theta}\cdot\Delta)\mathbf{u}^{n+1-\theta}=\mathbf{f}^{n+\theta}+\alpha\nu\Delta \mathbf{u}^{n+\theta}-\nabla p^{n+\theta}$

then

$(\mathbf{u}^{n+\theta}-\mathbf{u}^{n+1-\theta})/(\theta\Delta t)-\alpha\nu\Delta\mathbf{u}^{n+1}+\nabla p^{n+1}=\mathbf{f}^{n+1}+\beta\nu\Delta \mathbf{u}^{n+1-\theta}-(\mathbf{u}^{n+1-\theta}\cdot\Delta)\mathbf{u}^{n+1-\theta}$

$\nabla\cdot\mathbf{u}^{n+1}=0$

The first and last step in above scheme are to solve Stokes problem like

$-\Delta\mathbf{u}+\nabla p=\mathbf{f}$

$\nabla\cdot \mathbf{u}=0$

Multiply $\mathbf{u}$ on both sides of the first equation, by using Green’s formula, we can see it is just the saddle point problem, with $A$ as $-\Delta$ and $B$ as $\nabla$:

$\left[\begin{array}{cc} -\Delta & \nabla\\ \nabla & O \end{array}\right]\left[\begin{array}{c} \mathbf{u}\\ p\end{array}\right]=\left[\begin{array}{c} \mathbf{f}\\ 0\end{array}\right]$

Here pressure acts as Lagrange multiplier, and the Lagrangian is

$L(\mathbf{u}, p)=\frac{1}{2}|\Delta \mathbf{u}|_{\Omega}^2-(\mathbf{f}, \mathbf{u})+p(\nabla\cdot\mathbf{u}) \Longrightarrow$

$L(\mathbf{u})=\frac{1}{2}|\Delta \mathbf{u}|_{\Omega}^2-(\mathbf{f}, \mathbf{u})$

where $|\cdot|_{\Omega}$ is the seminorm on $\Omega$ and $\mathbf{u}$ is in $H_0^1(\Omega)$.