expression $Q$ defined by $$Q =  v^\beta F_\beta [u] -  u^\beta F^*_\beta [v] \equiv v^\beta F_\beta [u] -
 u^\beta  \frac{\delta (v^\gamma F_\gamma [u])}{\delta u^\beta}\,\cdot$$ Applying to the first expression for $Q$ the variational differentiations $\delta/\delta u^\alpha$ we obtain $$\frac{\delta Q}{\delta u^\alpha} = \frac{\delta (v^\beta F_\beta [u])}{\delta u^\alpha}
 - \delta^\beta_\alpha F^*_\beta [v] \equiv F^*_\alpha [v] - \delta^\beta_\alpha F^*_\beta [v] =
0.$$ Applying $\delta/\delta v^\alpha$ to the second expression for $Q$ we obtain $$\frac{\delta Q}{\delta v^\alpha} = \delta^\beta_\alpha F_\beta [u]
 - \frac{\delta}{\delta v^\alpha} \left[u^\beta  \frac{\delta (v^\gamma F_\gamma [u])}{\delta u^\beta}\right]
 \equiv F_\alpha [u]- \frac{\delta}{\delta v^\alpha} \left[u^\beta  \frac{\delta
(v^\gamma F_\gamma [u])}{\delta u^\beta}\right]\,\cdot$$ The reckoning shows that $$\label{lnad:eq1}
 \frac{\delta}{\delta v^\alpha} \left[u^\beta  \frac{\delta (v^\gamma F_\gamma [u])}{\delta
 u^\beta}\right] = F_\alpha [u].$$ Thus $Q$ solves Eq. (\[cladd\]) and hence Eq. (\[cladm\]) is satisfied. This completes the proof.

\[lnad.rem1\] Let us discuss the proof of Eq. (\[lnad:eq1\]) in the case of a econd-order linear operator for one dependent variable: $$F[u] = a^{ij}(x)u_{ij} +b^i(x)u_i +c(x) u.$$ Then we have: $$u \frac{\delta (v F [u])}{\delta u} = u \left[c v - v D_i(b^i) + v D_i D_j(a^{ij}) - b^i v_i
+ 2 v_i D_j(a^{ij}) + a^{ij} v_{ij}\right].$$ Whence, after simple calculations we obtain $$\frac{\delta}{\delta v}\left[ u \frac{\delta (v F [u])}{\delta u}\right] 