A proof of the Ibragimov-Kara-Mahomed commutation relation Let $u_a(x),\,a=1,2,\ldots n$ be a $n$-component field in Minkowski spacetime
$x^\mu,\,\mu=0,1,2,3$ and  let $u_{a,\,\mu}=\frac{du_a}{dx^\mu}$. Let us introduce two operators (we use Einstein summation convention that repeated indices are implicitly summed over)
$$\hat G=\tau^\mu\,\frac{d}{dx^\mu}+\sigma_a\,\frac{\partial}{\partial u_a}+
\sigma_{a,\mu}\,\frac{\partial}{\partial u_{a,\,\mu}}$$
and
$$\hat N^\mu=\tau^\mu+\sigma_a\,\frac{\delta}{\delta u_{a,\,\mu}}+
\sigma_{a,\nu}\,\frac{\delta}{\delta u_{a,\,\mu\nu}},$$
where $$\sigma_a=\xi_a-\tau^\mu\,u_{a,\,\mu},$$ $\xi_a(x,u)$ and $\tau^\mu(x,u)$
do not depend on field derivatives, and
$$\frac{\delta}{\delta u_{a,\,\mu}}=\frac{\partial}{\partial u_{a,\,\mu}}-
\frac{d}{dx^\nu}\frac{\partial}{\partial u_{a,\,\mu\nu}},$$
$$\frac{\delta}{\delta u_{a,\,\mu\nu}}=\frac{\partial}{\partial u_{a,\,\mu\nu}}-
\frac{d}{dx^\alpha}\frac{\partial}{\partial u_{a,\,\mu\nu\alpha}}.$$
It is claimed in  http://link.springer.com/article/10.1023%2FA%3A1008240112483 (Lie–Bäcklund and Noether Symmetries with Applications, by   N.H. Ibragimov,
A.H. Kara and F.M. Mahomed) that
$$[\hat G+\tau^\nu_{,\,\nu},\hat N^\mu]=\tau^\mu_{,\,\nu}\,\hat N^\nu. \tag{1}$$ The authors write that "the relation is proved by straightforward, albeit tedious, computation". 
Is there a simpler way to prove the commutation relation (1)? I'm interested to prove it only in the first jet space $(x^\mu,u_a,u_{a,\,\nu})$, so only the first few terms from the cited article are presented in the definitions of $\hat G$ and $\hat N^\mu$. 
 A: Now I have found a proof. In the first jet space, it is sufficient to take $$\hat N^\mu=\tau^\mu+\sigma_a\,\frac{\partial}{\partial u_{a,\,\mu}},$$ and the proof proceeds as follows. We have
\begin{equation}
[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\hat G(\tau^\mu)+\hat G(\sigma_a)\,
\frac{\partial}{\partial u_{a,\,\mu}}+\sigma_a\left [\hat G,\,\frac{\partial}
{\partial u_{a,\,\mu}}\right]+\sigma_a\left [\tau^\nu_{,\,\nu},\,
\frac{\partial}{\partial u_{a,\,\mu}}\right].
\tag{1}
\end{equation}
But
\begin{equation}
\left [\tau^\nu_{,\,\nu},\,\frac{\partial}{\partial u_{a,\,\mu}}\right]=
-\frac{\partial \tau^\nu_{,\,\nu}}{\partial u_{a,\,\mu}}=-\frac{\partial 
\tau^\mu}{\partial u_a},
\tag{2}
\end{equation}
because
\begin{equation}
\tau^\nu_{,\,\nu}=\frac{\partial \tau^\nu}{\partial x^\nu}+u_{b,\,\nu}\,
\frac{\partial \tau^\nu}{\partial u_b}.
\tag{3}
\end{equation}
On the other hand, as $\tau^\mu$ doesn't depend on field derivatives,
\begin{equation}
\hat G(\tau^\mu)=\tau^\nu\,\tau^\mu_{,\,\nu}+\sigma_a\,\frac{\partial 
\tau^\mu}{\partial u_a}.
\tag{4}
\end{equation}
Further we have
\begin{equation}
\hat G(\sigma_a)=\tau^\nu\,\sigma_{a,\,\nu}+\sigma_b\,\frac{\partial 
\sigma_a}{\partial u_b}+\sigma_{b,\,\nu}\,\frac{\partial \sigma_a}
{\partial u_{b,\,\nu}}.
\tag{5}
\end{equation}
But 
$$\sigma_a=\xi(x,u)-\tau^\mu(x,u)\,u_{a,\,\mu},$$
and, therefore,
\begin{equation}
\frac{\partial \sigma_a}{\partial u_{b,\,\nu}}=-\delta_a^b\,\tau^\nu.
\tag{6}
\end{equation}
Substituting this into (5), we get
\begin{equation}
\hat G(\sigma_a)=\sigma_b\,\frac{\partial \sigma_a}{\partial u_b}.
\tag{7}
\end{equation}
It remains to calculate the commutator
\begin{equation}
\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]=
\left [\tau^\nu \,\frac{\partial }{\partial x^\nu}+\xi_b\,\frac{\partial }
{\partial u_b}+\eta_{b\nu}\,\frac{\partial }{\partial u_{b,\,\nu}},\,
\frac{\partial} {\partial u_{a,\,\mu}}\right]=-\frac{\partial \eta_{b\nu}}
{\partial u_{a,\,\mu}}\,\frac{\partial }{\partial u_{b,\,\nu}},
\tag{8}
\end{equation}
where we have used the fact that $\tau^\nu$ and $\xi_a$ do not depend on
field derivatives. Using
$$\eta_{b\nu}=\xi_{b,\,\nu}-\tau^\alpha_{,\,\nu}\,u_{b,\,\alpha},$$
along with
\begin{equation}
\frac{\partial \xi_{b,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,
\frac{\partial \xi_b}{\partial u_a},\;\;\;\frac{\partial 
\tau^\alpha_{,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,
\frac{\partial \tau^\alpha}{\partial u_a},
\tag{9}
\end{equation}
we get
\begin{equation}
\frac{\partial \eta_{b\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,\left (
\frac{\partial \xi_b}{\partial u_a}-u_{b,\,\alpha}\,\frac{\partial 
\tau^\alpha}{\partial u_a}\right )-\delta_a^b\,\tau^\mu_{,\,\nu}=
\delta^\mu_\nu\,\frac{\partial \sigma_b}{\partial u_a}-\delta_a^b\,
\tau^\mu_{,\,\nu}.
\tag{10}
\end{equation}
Therefore
\begin{equation}
\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]=
\tau^\mu_{,\,\nu}\,\frac{\partial }{\partial u_{a,\,\nu}}-
\frac{\partial \sigma_b}{\partial u_a}\,\frac{\partial }
{\partial u_{b,\,\mu}}.
\tag{11}
\end{equation}
Now (2), (4), (7) and (11), in combination with (1), imply the desired result:
\begin{equation}
[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\tau^\mu_{,\,\nu}\,\left (\tau^\nu+
\sigma_a\,\frac{\partial }{\partial u_{a,\,\nu}}\right )=\tau^\mu_{,\,\nu}\,
\hat N^\nu.
\end{equation}
