With the solution of the Euler-Lagrange equation prove the following equation Let $\Omega \subset \mathbb{R}^n$ and $F=F(z,p):\Omega \times \mathbb{R} \times \mathbb{R}^n$ be smooth and independent of $x\in\Omega$. Let $u$ be a solution of the Euler-Lagrange equation of $\mathcal{F}(u)= \int_\Omega F(u,Du)dx$. I'd like to show that $\sum_i (\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j})_{x_i}=0$.
Since $u$ is a solution of the Euler-Lagrange equation and we've defined the Euler-Lagrange operator $L_F(u)=-\sum_i \partial_i (\partial_{p_i}F + \partial_zF)$ it follows that $L_F(u)=0$. I've been told that one should simply transform and compute the left side of the equation to prove that it equals $0$. I tried to transform the equation in order to use the fact that $u$ is a solution of the ELE of $\mathcal{F}(u)$. We have
$\sum_i (\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j})_{x_i} \\
= \sum_i \partial_{x_i}(\partial_{p_i}F \partial_{x_j}u-F \delta_{i,j}) \\
= \sum_i \partial_{x_i}(\partial_{p_i}F \partial_{x_j}u) - \sum_i \partial_{x_i}F \delta_{i,j} \\
= \sum_i \partial_{x_i}(\partial_{p_i}F) \partial_{x_j}u + \sum_i \partial_{p_i}F \partial_{x_i}( \partial_{x_j}u) -\partial_{x_j}F \; \mathrm{(product \, rule)} \\
= \sum_i div(\partial_{p_i}F \partial_{x_j}u) - \sum_i \partial_{p_i}F \partial_{x_i} (\partial_{x_j}u) + \sum_i \partial_{p_i}F \partial_{x_i}( \partial_{x_j}u) -\partial_{x_j}F \; \mathrm{(chain \, rule \, for \, divergence)}\\
= \sum_i div(\partial_{p_i}F \partial_{x_j}u) -\partial_{x_j}F.$
But I don't see how I can transform this expression any further or how I could use that $u$ is a solution of the Euler-Lagrange equation. I would appreciate if someone could give me tips so that I will be able to prove this assumption.
 A: Hint: One has to distinguish between explicit $x$-derivatives $\partial_{x^i}$ and total $x$-derivatives $$d_{x^i}~=~\partial_{x^i} +~ d_{x^j}u ~\partial_{z}+d_{x^j}\partial_{x^i}u~\partial_{p_i}\tag{0}.$$ No explicit $x$-dependence 
$$ \partial_{x^i}F~=~0 \tag{1}$$
of the Lagrangian $F$ leads (via Noether's theorem) to the on-shell continuum equations for the energy-momentum tensor:
$$\begin{align}
 (\partial_{p_i}F ~\partial_{x^j}u - F \delta_i^j)_{x^i} 
&~~~=~  d_{x^i}(\partial_{p_i}F ~\partial_{x^j}u - F \delta_i^j) \cr
&~~~=~  d_{x^i}\partial_{p_i}F~ \partial_{x^j}u 
+ \partial_{p_i}F~d_{x^i}\partial_{x^j}u - d_{x^j}F   \cr
&~~~\stackrel{(0)}{=}~  d_{x^i}\partial_{p_i}F~ \partial_{x^j}u 
+ \partial_{p_i}F~\partial_{x^i}\partial_{x^j}u \cr
&~~~-~ (\partial_{x^i}F +\partial_{z}F~ d_{x^j}u +\partial_{p_i}F~d_{x^j}\partial_{x^i}u)~\cr
&\stackrel{(1)+(3)}{=}~0. 
\end{align}\tag{2}$$
The last equality follows from eq. (1) and the Euler-Lagrange (EL) equation
$$ d_{x^i}\partial_{p_i}F~=~\partial_{z}F.\tag{3}$$
The word on-shell here means that the EL eq. (3) is satisfied. It is implicitly understood in eq. (2) that the Lagrangian $F$ is applied $F(u,\partial u)$ to a solution $u$ of the  EL eq. (3). In contrast, the  Lagrangian $F$ in eq. (1) is not applied to a solution $u$.
