# constrained minimization problem for mixed finite element method

I'm trying to prove that a mixed finite element formulation's solution is equivalent to a constrained minimization problem's solution.

Let $$V$$ and $$Q$$ be Hilbert spaces, and let $$a : V \times V \to \mathbb{R}$$, $$b : V \times Q \to \mathbb{R}$$ be bilinear forms. Let $$f \in V^*$$, $$g \in Q^*$$ where $$V^*$$, $$Q^*$$ are the dual spaces.

Suppose that $$a$$ and $$b$$ are both continuous bilinear forms, and moreover that $$a$$ is coercive on the kernel of $$b$$:

$$a(v, v) \ge \alpha\|v\|_V^2$$

for all $$v$$ such that $$b(v, q) = 0$$ for all $$q \in Q$$. $$b$$ satisfies the '''inf–sup''' or '''Ladyzhenskaya–Babuška–Brezzi''' condition:

$$\sup_{v \in V, v\neq 0}\frac{b(v, q)}{\|v\|_V} \ge \beta\|q\|_Q$$

Use the notations and the assumptions of Theorem 101 (Brezzi). We assume that the bilinear form $$a$$ is symmetric. Consider the constrained minimization problem \begin{align} \min_{\substack{v\in V; \\ b(v,q)=g(q)\forall q\in Q}} \frac{1}{2}a(v,v)-f(v) \end{align} Show that there exists a unique solution and that it is given by the variational problem: Find $$u\in V$$ and $$p\in Q$$ such that \begin{align*} a(u,v)+b(v,p)&=f(v) \ \forall v\in V \\ b(u,q)&=g(q) \ \forall q \in Q \end{align*}

The assumptions of Theorem 101 (Brezzi) are given in the following link if you search for the Theorem 101: http://www.asc.tuwien.ac.at/~schoeberl/wiki/lva/numpde18/numpde.pdf

So far i tried to prove that from the minimization problem follows the mixed formulation stated above:

Since Th. 101 Brezzi's assumptions are fulfilled we know there exists $$u\in V,p\in Q$$ unique st.: \begin{align*} a(u,v)+b(v,p)&=f(v) \ \forall v\in V \\ b(u,q)&=g(q) \ \forall q \in Q \end{align*} Now consider the minimization problem find $$min_{v\in V} \frac{1}{2}a(v,v)-f(v)$$ under the constraint $$b(v,q)=g(q)\ \forall q\in Q$$. To apply the method of Lagrange multipliers consider the variation $$u+wt$$ where u is the minimal solution and $$w\in V, t\in \mathbb{R}$$ are arbitrary. Next define the Lagrange multiplier by $$q+\lambda \bar{w}$$ for a $$q,w\in Q$$ arbitrary and $$\lambda \in R$$. Then the Lagrange functional is given by: \begin{align*} L(t,\lambda) = \frac{1}{2}a(u+wt,u+wt)-f(u+wt)+(b(u+wt,q+\lambda\bar{w})-g(q+\lambda\bar{w})) \end{align*} Then we get for the derivatives \begin{align*} \partial_t L &= ta(w,w)+a(u,w)-f(w)+b(w,q+\lambda\bar{w})\\ \partial_\lambda L &= b(u+wt,\bar{w})-g(\bar{w}) \end{align*} For the extremum we equate the derivatives to 0 and to find the minimum we note that $$t\rightarrow 0$$ which yields \begin{align*} a(u,w)-f(w)+b(w,q+\lambda\bar{w})=0\\ b(u,\bar{w})-g(\bar{w})=0 \end{align*} Since we know that the mixed formulation above has a unique solution and $$q\in Q$$ is arbitrary we conclude $$q+\lambda\bar{w}=p$$ and therefore the minimization problem above yields a minimum that is also a solution to the mixed formulation. (Not sure how to prove the backwards direction to get equivalence...)

Now I have following questions:

1) Is the way that I try to prove that the minimization problem implies the mixed formulation correct? If not can you help me to show it correctly (by using Lagrange multipliers).

2) My instructor told me that this is a saddle point problem. How can I see this?

3) Is it possible to also prove that the solution of the mixed formulation implies the solution of the minimization Problem? So that we actually get an equivalence of the solution in the mixed formulation and the minimum in the minimization Problem.

• If $a=b=g=f=0$ then you are claiming the minimum of $-f(v)$ over $v \in V$ is unique, but, all elements of $V$ are minimizers of the always-zero function. – Michael May 22 at 23:38
• You are right! I think the additional assumptions from Theorem 101 that I mentioned above are needed to exclude the trivial case. I added them now. – Pepe May 23 at 0:19