Is Stokes equation a saddle point problem or a minimum problem?

Question

Consider $\Omega \subset \mathbb{R}^2 $ (or $\mathbb{R}^3$). The well known stationary Stokes equations in the incompressible case are

\begin{equation} \begin{cases} - \Delta u + \nabla p = f \text{ in } \Omega\\ \operatorname{div} (u)= 0\\ u|_{\partial \Omega} =0 \end{cases} \end{equation}

with $f \in H^{-1} (\Omega)$, $(u, p ) \in H_0 ^1 (\Omega ) \times L^2 _0 (\Omega )$, where $L^2 _0 (\Omega )$ is made of 0 mean functions. Now, I want to study this equation and looking online I found a humongous amount of information, articles etc. I admit I haven't had time to read even a fraction of them, but in particular I am interested in understanding something. The Stokes equation can be seen as a constrained minimization problem, in the sense that if you look for the minimum of

$$ J(u )= \frac{1}{2} \int_\Omega |\nabla u|^2 - \int_\Omega f u \text{ for } u \in V=\{ v \in H^1_0 \mid \operatorname{div} (v)=0\},$$

that solves the equation and actually the pression is the Lagrange multiplier (this is shown, for example, in Evans's PDE book). However a way the numerical analysts seem to view the equation is as a saddle point problem, viewing $(u, p) \in H_0 ^1 (\Omega ) \times L^2 _0 (\Omega )$ and solving it with a finite element method (see these notes for instance https://finite-element.github.io/L6_stokes.html). I have some background in Calculus of Variations and a little knowledge of numerical analysis, but enough to know that a having a minimum is much better than having a saddle point in almost every case. So, what is the value in studying Stokes equation as a saddle point rather than a minimum? The only reason that comes to my mind is that when you apply the FEM, while in $H^1 _0 \times L^2$ it's easy to construct finite dimensional spaces and base functions, you cannot do it for $V$ due to the divergence free condition. Is that it or am I missing something?

Answer 1

Indeed, the reason is that the construction of divergence-free basis functions is considered very difficult. In nonconforming methods such functions exist and are described in Chapter III.§7 of the book by D. Braess "Finite Elements". More standard methods like Lagrange elements can usually only be enforced to be weakly divergence free, which means that $$ \int_\Omega \operatorname{div}u_h\;q_h = 0 $$ for all $q_h$ from a discrete set. This leads to a well posed problem provided the space hosting $u_h$ and the space hosting $q_h$ are compatible (an "inf-sup stable pair").