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.
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.