Saddle point problem in Navier-Stokes equation

by Yi Zhang

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).

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