Existence of a Lagrange multiplier (Euler Lagrange equations + holonomic constraints ) Let $I=[a,b]\subset \mathbb{R}, G:\mathbb{R}^n\to \mathbb{R}^k$ smooth, $0<k<n, M=G^{-1}(0)$. Assume that $DG(x)$ has full rank for all $x\in M$. Fix $p_1,p_2\in M$ and assume $u\in W^{1,\infty}(I,\mathbb{R}^n)$ minimizes the functional given by 
$$J(u)=\int_a^bF(t,u(t),\dot{u}(t))dt $$ 
on the set $$S := \{u\in W^{1,\infty}(I,\mathbb{R}^n): u(a)=p_1,u(b)=p_2, u(t)\in M\}.$$
Show there exists a function $\lambda \in W^{1,\infty}(I,\mathbb{R}^k)$ such that
$$\frac{d}{dt}F_p(t,u(t),\dot{u}(t)) -F_u(t,u(t),\dot{u}(t))= DG(u(t))^T\lambda(t).$$
I got as $\textbf{hint}$: 
Near $p\in M$, there are parameterizations $\psi:V\to U\subset M$ where $V\subset \mathbb{R}^{n-k}$ and $U\subset M$ contains $p$. Assume first $\bigcup_t\{u(t)\}\subset U$. Define $w:I\to\mathbb{R}^{n-k}$ by 
$$w(t) = (\psi^{-1}\circ u)(t) $$
and find a suitable functional $\tilde{J}$ (on a suitable space) which corresponds to $J$ and is minimized by $w$. Use the Euler-Lagrange equations for $\tilde{J}$ and the fact that $DG(\psi(z))D_z\psi(z)=0.$
(for the general case, cover $\bigcup_t\{u(t)\}$ with coordinate patches and localize by subdividing the set into pieces that lie within thise patches). 
I'm simply trying to prove the simpler case, but I have hard time finding such $\tilde{J}$. I appreciate all the help and suggestions. 
 A: It might be useful for you to see how to derive Lagrange multipliers in the finite dimensional setting and then generalize it to the variational setting.
Let's work with a curve in $\mathbb R^3$ for concreteness. Assume the curve is given implicitly by the constraint $G(\vec x) = 0$ where $G: \mathbb R^3 \to \mathbb R^2$ (and the rank of $G$ is 2 along the curve).
Let's assume now that the curve can be parametrized as $\psi: \mathbb R \to \mathbb R^3, \psi: x \mapsto (x, g(x), h(x))$ in these coordinates (such coordinates can always be found because of the maximal rank assumption by the implicit function theorem). Then we have that $G(\psi(x)) = G(x, g(x), h(x)) = 0$ for all $x$ and therefore $D(G \circ \psi)^i = G^i_x + G^i_y g' + G^i_z h' = 0$ for $i = 1,2$ (the argument of $G^i$ is $\psi(x)$ but I suppress it for notational convenience).
Now let's get back to finding extremes of the function $F : \mathbb R^3 \to \mathbb R$ along this curve. This is equivalent to finding extremes of $F(\psi(x))$ on $\mathbb R$. But the condition for the extremes is $D(F\circ \psi) = F_x + F_y g' + G_z h' = 0$. We note that this is very similar for the equation of constraints given by the implicit function theorem and therefore it's enough to solve $\nabla F = \lambda_1 \nabla G^1 + \lambda_2 \nabla G^2$ for two unknown constants $\lambda_1, \lambda_2$. We can suggestively rewrite this equation as $DF = DG  ^T \lambda$ with $\lambda = (\lambda_1, \lambda_2) \in \mathbb R^2$.
Now, returning to the variational setting, everything works out very similarly except that you replace $D$s above with $\delta$s. Applying standard variational arguments you can then reduce integral equations to time-local equation for every $t$ and deduce the existence of a constant $\lambda(t)$. The function $\lambda$ will be in $W^{1,\infty}(I, \mathbb R^k)$ precisely because it must satisfy the Euler-Lagrange equation and all the other functions figuring there ($F\circ u$, $G\circ u$ and their derivatives) belong to $W^{1,\infty}$ spaces.
