I'm trying to understand the following:
Let $\Omega \subset \mathbb{R}^2,\ V$ = space of (vectorvalued) continuous piecewise linear functions with zero boundary, $W = $ space of continuous piecewise linear (scalar) function.
We're looking for solutions $(u,p) \in V \times W$ of the following discrete formulation of the Stokes equations:
$\begin{eqnarray} a(u,v) - (p,\text{div}v) &=& (f,v) \ \ \ \forall v \in V \\ (q, \text{div} u) &=& 0\ \ \ \forall q \in W \end{eqnarray}$
with $a(u,v):= \int_\Omega \sum\limits_i Du^iDv^idx$ and the $L^2$-scalar product $(\cdot,\cdot).$ Now the notes say, these equations are a necessary condition for the minimization problem
$E(v) := \frac{1}{2}a(v,v) - (f,v) \rightarrow \text{Min}$ in $V_\text{div}$
with $\ \ \ V_\text{div}:= \{ v \in V : (q, \text{div}v) = 0 \ \ \ \forall q \in W \}$, since the Lagrangian of the problem is given by
$L(v,q) = \frac{1}{2}a(v,v) - (f,v) - (q,\text{div} v)$
and the above equations are a necessary condition for a stationary point in this function. Now my questions are:
1.) Why is this the Lagrangian for the Minimization problem? I only know the Lagrange multipliers for functions of real numbers and I'm not sure how to apply this to such a functional. The first two terms are obvious, but I can't see how exactly we obtain the term $-(q, \text{div}u)$ from the above constraint. Is $q$ the Lagrange mulitplier?
2.) Why are the equations for $u$ and $p$ the conditions for a stationary point? Here I have no idea, in what way do we take the derivatives of $L$?
3.) The lecture notes say further, that this is the reason, why this discretization does not work, because $V_\text{div}$ won't contain a good approximation for the classical solution $u$ anymore. This seems logic, but i don't see why we need the minimization problem to see this, since we already have the same restriction $(q,\text{div}v) = 0 \ \ \ \forall q \in W$ in the discrete formulation above?
I would be very thankful for any hint to these questions!