written $$\label{saeq.2}
 F^*\big(x, u, v, u_{(1)}, v_{(1)}, \ldots, u_{(s)}, v_{(s)}\big)  = 0,$$ where $$\label{saeq.3}
 F^*\big(x, u, v,u_{(1)}, v_{(1)},\ldots, u_{(s)}, v_{(s)}\big) =
  \frac{\delta (v F)}{\delta u}\,\cdot$$

Construction of adjoint equations to linear equations {#sa:eveq.3}
-----------------------------------------------------

The following statement has been formulated in [@ibr06a; @ibr07a].

\[lnad.prop1\] In the case of linear differential equations and systems, the adjoint equations determined by Eq. (\[sa:eveq.int.3\]) and by Eq. (\[clad\]) coincide.

**Proof.** The proof is based on the statement (see Proposition \[test:div\] in Section \[test\]) that a function $Q(u, v)$ is a divergence, i.e. $Q = D_i(h^i),$ if and only if $$\label{cladd}
 \frac{\delta Q}{\delta u^\alpha} = 0, \quad  \frac{\delta Q}{\delta v^\alpha} = 0, \quad \alpha = 1, \ldots, m.$$ Let the adjoint operator $F^*$ be constructed according to Eq. (\[clad\]). Let us consider the case of many dependent variables and write Eq. (\[clad\]) as follows: $$\label{cladm}
  v^\beta F_\beta [u] = u^\beta F^*_\beta [v] + D_i(p^i).$$ Applying to (\[cladm\]) the variational differentiations and using Eqs. (\[cladd\]) we obtain $$\frac{\delta (v^\beta F_\beta [u])}{\delta u^\alpha} =
 \delta^\beta_\alpha F^*_\beta [v] \equiv F^*_\alpha [v].$$ Hence, (\[sa:eveq.int.3\]) coincides with $F^*_\alpha [v]$ given by (\[clad\]).

Conversely, let $F^* [v]$ be given by (\[sa:eveq.int.3\]), $$F^*_\beta [v] = \frac{\delta (v^\gamma F_\gamma [u])}{\delta
u^\beta}\,\cdot$$ Consider the 