u_{(s)}\big)$$ holds with a certain (in general, variable) coefficient $\lambda.$

\[sa:eveq.KdV\] The Korteweg-de Vries (KdV) equation $$u_t = u_{xxx} + u u_x$$ is strictly self-adjoint [@ibr07a]. Indeed, its adjoint equation (\[saeq.2\]) has the form $$v_t = v_{xxx} + u v_x$$ and coincides with the KdV equation upon the substitution (\[saeq.4\]).

In the case of linear equations the strict self-adjointness is identical with the usual self-adjointness of linear equations.

\[sa:eveq.exa2\] Consider the linear equation $$\label{sa:eveq.exa2:eq1}
 u_{tt} + a(x) u_{xx} + b(x) u_x + c(x) u = 0.$$ According to Eqs. (\[saeq.2\])-(\[saeq.3\]), the adjoint equation to Eq. (\[sa:eveq.exa2:eq1\]) is written $$\frac{\delta}{\delta u} \{v [u_{tt} + a(x) u_{xx} + b(x) u_x + c(x)u]\}
 \equiv
   D^2_t(v) + D_x^2(a v) - D_x(b v) + c v = 0.$$ Upon substituting $v = u$ and performing the differentiations it becomes $$\label{sa:eveq.exa2:eq2}
 u_{tt} + a u_{xx} + (2 a' - b) u_x + (a'' - b' + c) u = 0.$$ According to Definition \[saeq:def\], Eq. (\[sa:eveq.exa2:eq1\]) is strictly self-adjoint if Eq. (\[sa:eveq.exa2:eq2\]) coincides with Eq. (\[sa:eveq.exa2:eq1\]). This is possible if $$\label{sa:eveq.exa2:eq3}
 b(x) = a'(x).$$

Definition \[saeq:def\] is too restrictive. Moreover, it is inconvenient in the case of systems with several dependent variables $u = (u^1, \ldots, 