Let $F:\mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}$ be $C^2$ on $[a,b]$ and $u$ be a solution for the Euler-lagrange equations for the functional given by $$J(u) = \int F(u(t),\dot{u}(t)).dt, $$
Show that the function $ M$ on $[a,b]$ given by $$M(t)= \dot{u}(t)F_p(u(t),\dot{u}(t))-F(u(t),\dot{u}(t)) $$ is constant on $[a,b]$.
The Euler lagrange equations tell us that $$\frac{d}{dt}F_p = F_u $$ (where $F_u$ is the vector of partial derivatives w.r.t. the first argument, and $F_p$ w.r.t. the second).
Some light on how tot tackle this problem is very much appreciated.
Just calculate, we have \begin{align*} \dot M(t) &= \ddot u(t) F_p(u(t), \dot u(t)) + \dot u(t)^2 F_{pu}(u(t), \dot u(t)) + \dot u(t)\ddot u(t) F_{pp}(u(t), \dot u(t))\\ &{} \quad - \dot u(t) F_u(u(t), \dot u(t)) - \ddot u(t)F_p(u(t), \dot u(t)) \end{align*} In shorter notation (leaving of the arguments $t$, and $u(t), \dot u(t)$) $$ \dot M = \ddot uF_p + \dot u^2F_{pu} + \dot u \ddot u F_{pp} - \dot u F_u - \ddot u F_p =\dot u^2F_{pu} + \dot u \ddot u F_{pp} - \dot u F_u $$ On the otherhand $\frac{d}{dt} F_p(u(t), \dot u(t)) = F_u(u(t), \dot u(t))$, that is $ F_u = \dot u F_{pu} + \ddot u F_{pp}$, giving \begin{align*} \dot M &=\dot u^2F_{pu} + \dot u \ddot u F_{pp} - \dot u F_u \\ &= \dot u\cdot (\dot u F_{pu} + \ddot u F_{pp}) - \dot u F_u\\ &= \dot u F_u - \dot u F_u\\ &= 0. \end{align*} So $M$ is constant.