u^m)$ because in this case Eq. (\[saeq.4\]) is not uniquely determined as it is clear from the following example.

\[sa:eveq.exa3\] Let us consider the system of two equations $$\begin{aligned}
 & u^1_y + u^2 u^2_x - u^2_t = 0, \notag \\[1ex]
 & u^2_y - u^1_x = 0 \label{sa:eveq.exa3.eq1}
 \end{aligned}$$ with two dependent variables, $u = (u^1, u^2),$ and three independent variables $t, x, y.$ Using the formal Lagrangian (\[sa:Lag\]) $${\cal L} = v^1 (u^1_y + u^2 u^2_x - u^2_t) + v^2 (u^2_y - u^1_x)$$ and Eqs. (\[sa:eveq.int.3\]) we write the adjoint equations (\[sa:eveq.int.2\]), changing their sign, in the form $$\begin{aligned}
 & v^2_y + u^2 v^1_x - v^1_t = 0, \notag \\[1ex]
 & v^1_y - v^2_x = 0.\label{sa:eveq.exa3.eq2}
 \end{aligned}$$ If we use here the substitution (\[saeq.4\]), $v = u$ with $v = (v^1, v^2),$ i.e. let $$v^1 = u^1, \quad v^2 = u^2,$$ then the adjoint system (\[sa:eveq.exa3.eq2\]) becomes $$\begin{aligned}
 & u^2_y + u^2 u^1_x - u^1_t = 0, \notag \\[1ex]
 & u^1_y - u^2_x = 0,\notag
 \end{aligned}$$ which is not connected with the system (\[sa:eveq.exa3.eq1\]) by the equivalence relation (\[saeq.5\]). But if we set $$v^1 = u^2, \quad v^2 = u^1,$$ the adjoint system (\[sa:eveq.exa3.eq2\]) coincides with the original system (\[sa:eveq.exa3.eq1\]).

