# Lagrange multipliers in minimization problem with bilinear forms

Let $V$ a Hilbert space, $a:V\times V\rightarrow \mathbb{R}$ a bounded, symmetric and positive bilinear form and $f:V\rightarrow\mathbb{R}$ bounded.

Is well known that problem

$$\left\lbrace\begin{array}{l}\textrm{Find }u\textrm{ such that}\\ a(u,v)=f(v)\quad\forall v\in V\end{array}\right.$$

is equivalent to minimize the following (energy) functional:

$$J(v)=\dfrac{1}{2}a(v,v)-f(v).$$

If now I have some extra condition I get the problem

$$\left\lbrace\begin{array}{l}\textrm{min }J(v)\textrm{ with }v\in V\\ \textrm{ subjetc to }u=g\quad\forall v\in \tilde V\subset V\end{array}\right.$$

I'm thinking in how to get the corresponding Lagrange multipliers in Stokes-like problems, for example, where the problem is

$$\left\lbrace\begin{array}{l}\textrm{min }J(u,p)\textrm{ with }(u,q)\in V\times Q\\ \textrm{ subjetc to }\int_\Omega p=0\quad\end{array}\right.$$

but when I try to use Lagrange multipliers (learned in my first course of calculus in several variables) I need to calculate things like:

$$\nabla f=\lambda\nabla g$$

but what is "$\nabla$" operator in my context (with bilinear forms)? In other words, How can I use Lagrange multipliers in this context?

• This is probably too difficult to answer on a forum like this. You should read a book on the calculus of variations that explains exactly this sort of thing. – Wolfgang Bangerth Apr 19 '18 at 19:53
• Thanks @WolfgangBangerth It's not just a method? (the "Lagrange multipliers method" applied to bilinear forms). (I'm not looking for proofs). Can you suggest a book about it? I have not found anything surfing the internet – yemino Apr 19 '18 at 20:03
• It's just a method, but it's a non-trivial one if you operate in function spaces. In particular, you need to learn what $\nabla$ means in this context, namely the variation of a functional with respect to a test function. – Wolfgang Bangerth Apr 20 '18 at 3:06
• There is something strange with your Stokes-like formulation. It is well known that Stokes problem is a saddle point (minmax) problem and not a direct minimization problem. Stokes is a constrained minimization problem with the constraint $\mathrm{div}\,u = 0$. – knl Apr 20 '18 at 6:21

Let's assume that your functions $u$ and $p$ can be expanded using a finite discrete basis in the subspaces $V$ and $Q$, e.g. $\phi_i\in V$ and $\psi_i \in Q$ $$u = \sum_iu_i\phi_i\qquad p = \sum_ip_i\psi_i \tag{*}$$. The Stokes problem fits into the frame of the following stationary problem: $$\left\{\begin{array}{l}% find \quad stationary \quad point \quad of\quad J(v)=\frac{1}{2}a(v,v)-f(v) \\ s.t.\qquad \textrm{div}(v) = 0 \end{array}\right.$$

with appropiate B.C.for $v$ and where

$$a(u,v) = \int_\Omega \nabla u\cdot \nabla v dV$$ $$f(v) = \int_\Omega{fv\,dV}$$

Bearing in mind that $a(\cdot,\cdot)$ is a symmetric bilinear form and $f(\cdot)$ is a linear form inserting the subspace projections $(*)$ you'll find that the previous problem my be rewritten a: $$\left\{\begin{array}{l}% J(v)=\frac{1}{2}\sum_{i,k}a(\phi_i,\phi_k)u_iu_k-\sum_{i}f(\phi_i)u_i=\frac{1}{2}u^TAu-f^Tu \\ s.t.\qquad \sum_i \textrm{div}(u_i\phi_i) =0 \end{array}\right.$$

Since the restriction is supposed to be fulfilled in every point, we define a lagrange multiplier $p$ such as $(*)$ to impose this restriction as: $$\sum_i\int_\Omega p\textrm{div}(u_i\phi_i)\,dV =\sum_{i,k}\int_\Omega p_k\psi_k\textrm{div}(u_i\phi_i)\,dV =- \sum_{i,k}\int_\Omega \textrm{grad}(p_k\psi_k)\cdot u_i\phi_i\,dV =-p^TBu$$ Where $B$ coincides with $$B = \int_\Omega \textrm{grad}(\psi_k)\phi_i\,dV$$ Therefore your augmented lagrangian will be a function of $v_i$ and the Lagrange multipliers $p_i$: $$L(v,p)=\frac{1}{2}u^TAu-f^Tu-p^TBu$$

Now, you can apply the usual rule for Lagrange multipliers, that permitts yuo to obtain the equations that solve the Stokes problem: $$\frac{\partial L}{\partial u}=Au-f-B^Tp=0$$ You can see this expression as (using a slightly different notation for $\nabla F = \lambda \nabla G$): $$\nabla F = Au-f, \qquad \lambda = p^T, \qquad \nabla G = B$$

And you restriction equations: $$\frac{\partial L}{\partial p} = Bu = 0$$

All this resultsin the following saddle point optimisation problem: $$\left[\begin{matrix} % A & B^T\\ B & 0\end{matrix}\right]\left[\begin{matrix} u\\p\end{matrix}\right]=\left[\begin{matrix}{f\\0}\end{matrix}\right]$$

Note: This does not pretent to be rigurous, only is a shallow explanation of which is inside the problem you asked for.