Lagrangians independent of $x$ In PDE Evans, 2nd edition, the following formula is printed as equation $\text{(9)}$ in §8.6 (on page 514): 

$$\sum_{k=1}^n (L_{p_i}u_{x_k}-L\delta_{ik})_{x_i}=0 \quad (k=1,\ldots,n) \tag{9}$$

(The above formula is found in Example 1 on page 514. This Lagrangian is independent of $x$.)
Then the author goes on to say:

It is a simple exercise to confirm that these formulas follow directly from the Euler-Lagrange equation. The point is that Noether's Theorem provides a systematic procedure for searching for such identities.

The Euler-Lagrange equation is printed in §8.1 (on page 456) as: 

$$-\sum_{i=1}^n (L_{p_i} (Du, u , x))_{x_i}+L_z(Du,u,x)=0 \quad \text{in }U.$$

How can I use the Euler-Lagrange equation to derive $\text{(9)}$? (I do see that the two equations look very similar in form.)
 A: To gain intuition it might be helpful to mention that in physics, if the Lagrangian density $L(z,p,x)$ does not depend explicitly on the "spacetime" point $x$, then the theory is said to possess translational symmetry. In such cases, the Noether's (first) theorem implies that the "canonical stress-energy-momentum tensor" 
$$ T_{ik}~:=~ L_{p_i}u_{x_k}-\delta_{ik}L $$
is "conserved" [i.e. satisfies a continuity equation in the sense of eq. (9)] if the EL eq. 
$$ \sum_i(L_{p_i})_{x_i}~=~L_z $$
is satisfied. In more detail,
$$\sum_i(T_{ik})_{x_i}~=~\sum_i(L_{p_i})_{x_i}u_{x_k}+\sum_iL_{p_i}u_{x_kx_i}
-\sum_i\delta_{ik}\left\{L_z u_{x_i}+\sum_jL_{p_j} (p_j)_{x_i} \right\}~=~0,\tag{9}$$
Here we have used that
$$ p_j~=~u_{x_j}~=~z_{x_j},$$
so that
$$ (p_j)_{x_k}~=~u_{x_jx_k}. $$
A: $\newcommand{\udot}{\dot{u}} \newcommand{\uddot}{\ddot u}$Let us consider the ODE case in which there is only one independent variable $t$, for simplicity. A time-independent Lagrangian is a function 
$$
L=L(\udot, u), $$
and it gives rise to the Euler-Lagrange equation 
$$\tag{EL}
\frac{d}{dt}L_{\dot{u}}-L_u=0,$$
the pedices denoting partial differentiation. If we want to recover the conserved quantity associated to time translation invariance, we can first compute the generator of the symmetry:
$$
m(t)=\left.\frac{\partial }{\partial \epsilon}u(t+\epsilon)\right|_{\epsilon=0}=\udot, $$
then multiply the left hand side of (EL), obtaining 
$$\tag{1}
\udot\uddot L_{\udot\,\udot}+\udot^2 L_{\udot\,u}-\udot L_{u}.$$We know from Noether's principle$^{[1]}$ that this expression must be a derivative, so we proceed to find it.
We use dimensional analysis: we assume that $u$ is dimensionless, that $t$ has the dimension of time $T$ and that $L$ has the dimension of action over time $ST^{-1}$. So $(1)$ has the dimension of $ST^{-2}$. In an expression of the form 
$$
(1)=\frac{d}{dt} F(u, \udot, L), 
$$
the term $F$ in the right hand side must therefore have dimensions $ST^{-1}$, the same as $L$. The most natural functions with those dimensions are $L$, $uL_{u}$ and $\udot L_{\udot}$. So we compute:
$$
\frac{dL}{d t}=\uddot L_\udot+\udot L_u, $$
and
$$
\frac{d}{d t} \udot L_{\udot}=\uddot L_{\udot}+\udot\uddot L_{\udot\,\udot} +\udot^2 L_{\udot\, u}, $$
and there is no need to continue, as we already see that 
$$
(1)=m(t)\left(\frac{d}{dt}L_{\dot{u}}-L_u\right)=\frac{d}{dt}\left(\udot L_{\udot} - L\right).$$
In particular, if $u$ is a solution of (EL), then the quantity $\udot L_{\udot}-L$ is conserved. In this one-dimensional case, this quantity coincides with the Hamiltonian of the system, while in the multidimensional case one gets the energy-momentum tensor, as Qmechanic shows. 

$^{[1]}$ In this case I prefer the use of the word principle rather than theorem.
