Stationary Stokes equation for homogeneous, incompressible and viscose fluid Let $\Omega \subset \mathbb{R}^2$ be a convex and bounded domain  with smooth boundary. For $u : \Omega \rightarrow \mathbb{R}^2$, $x \mapsto u(x) = (u_1,u_2)(x)$ define
\begin{equation*}
\mbox{curl}(u) = \partial_{x_1} u_2 - \partial_{x_2}u_1, \Delta u = (\Delta u_1, \Delta u_2).
\end{equation*}
The stationary Stokes-equation for a homogeneous, incompressible and viscose fluid is
\begin{equation}\label{navier_stoke}
\begin{cases}
-\Delta u + \nabla p = 0 &,\mbox{in } \Omega, \\
\mbox{div } u = 0 &, \mbox{in } \Omega, \\
u = 0 &, \mbox{on } \Omega. 
\end{cases}
\end{equation}
I want to show that for smooth functions $u$ and $p$ with $u = 0$ on $\partial \Omega$ the stoke-equations above are equivalent to the Euler-Lagrange-equation of the minimization problem
\begin{equation*}
\frac{1}{2} \int_{\Omega}{|\mbox{curl}(u)|^2 dx} \rightarrow \mbox{min!}
\end{equation*}
under the "holonomic" condition $\mbox{div}(u) = 0$.
Formally treating the condition as a holonomic condition gives me for the first variation
with $u,v \in C^2(\Omega)$
\begin{equation}
\delta \mathcal{F}(u,v) = \int_{\Omega}{\operatorname{curl}(v) \mbox{curl}(u) dx}.
\end{equation}
The Euler-Lagrange equation with holonomic condition would be for $\lambda \in C^0(\Omega)$ and $G(x,z,p) := \mbox{div}z$
\begin{equation}
\nabla_z(F + \lambda G) + \mbox{div}_x(\nabla_p F) = 0
\end{equation}
but what is $\nabla_zF = \nabla_z(\mbox{curl}(z))$ in this expression and $\nabla_zG = \nabla_z \mbox{div } z$.
I can't see how the integrand is linked to an Euler-Lagrange equation with holonomic condition equivalent to the Navier-Stokes equation in \ref{navier_stoke}. I tried to use some identities for the $\mbox{curl}$ operator without any success.
 A: For anyone who is interested:
With the energy
\begin{equation*}
\mathcal{F}(u) := \frac{1}{2}\int_{\Omega}{|Du|^2 d x}
\end{equation*}
we get the first variation
\begin{equation*}
\int_{\Omega}{D u :Dv  dx} = \int_{\Omega}{-\Delta u v dx}
\end{equation*}
and with $p \in C^1(\Omega) \cap L^2_{loc}(\Omega)$ as a Lagrange-Multiplicator which comes from the equation
\begin{equation*}
0 = \int_{\Omega}{p \mbox{div } v dx} = - \int_{\Omega}{\nabla p \cdot v dx}
\end{equation*}
which motivates the existence of a $p$ such that $\Delta u = \nabla p$ we get
\begin{equation}\label{euler_lag}
\int_{\Omega}{D u : Dv + p \mbox{div}v dx} = \int_{\Omega}{(-\Delta u  + \nabla p)v d x}
\end{equation}
with
\begin{equation*}
\int_{\Omega}{\mbox{div }u dx} = 0
\end{equation*}
for all $v \in H^1(\Omega)^2$. For more regularity of $u \in C^2(\Omega)^2 \cap C_0(\Omega)^2$ we get the stationary stoke-equation
\begin{equation*}
-\Delta u + \nabla p = 0.
\end{equation*}
With $u \in H^2(\Omega)$ and $v \in H^2(\Omega)^2$
\begin{equation*}
\mathbf{curl}(u) = \begin{pmatrix}
-\partial_{x_2}u \\
\partial_{x_1} u
\end{pmatrix}, \mbox{curl}(v) = \partial_{x_2}v_1 - \partial_{x_1}v_2
\end{equation*}
we get
\begin{equation*}
\mbox{curl}(\mathbf{curl} u) = - \Delta u, \mathbf{curl}(\mbox{curl}(v)) = - \Delta v + \nabla(\mbox{div}(v))
\end{equation*}
and with the divergence theorem and stokes theorem
\begin{equation*}
\int_{\Omega}{v \cdot \mathbf{curl}(u) d x} = \int_{\Omega}{\mbox{curl}(v) u dx} + \int_{\partial \Omega}{u v \cdot t dx}
\end{equation*}
where $t$ is a unit tangent field on $\partial \Omega$. We get
\begin{equation*}
\int_{\Omega}{\mbox{curl}(u) \mbox{curl}(v)dx} = \int_{\Omega}{D u : Dv dx}
\end{equation*}
for all $u \in H^1_0(\Omega)^2$ with $\mbox{div}(u) = 0$ and $v \in H^1(\Omega)$. For classical solutions $u \in C^2(\Omega)^2 \cap C_0(\Omega)^2$ and $p \in C^1(\Omega) \cap L^2_{loc}(\Omega)$ the euler lagrange equation of the energy integral is the stoke-equation stated in the problem.
