= F.$$ In other words, the adjointness of linear operators is a *symmetric relation*.

The linear operator $F$ is said to be self-adjoint if $F^* = F.$ In this case we say that the equation $F[u] = 0$ is self-adjoint. Thus, the self-adjointness of a linear equation $F[u] = 0$ can be expressed by the equation $$\label{sad-lin}
 F^*[v]\Big|_{v = u} =  F[u].$$

Adjoint equations to nonlinear differential equations {#sa:eveq.1}
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Let us consider a system of $m$ differential equations (linear or nonlinear) $$\label{sa:eveq.int.1}
 F_\alpha \big(x, u, u_{(1)}, \ldots, u_{(s)}\big) = 0, \quad
 \alpha = 1, \ldots, m,$$ with $m$ dependent variables $u = (u^1, \ldots, u^m).$ Eqs. (\[sa:eveq.int.1\]) involve the partial derivatives $u_{(1)}, \ldots, u_{(s)}$ up to order $s.$

\[sa:adj\] The [*adjoint equations*]{} to Eqs. (\[sa:eveq.int.1\]) are given by $$\label{sa:eveq.int.2}
 F^*_\alpha \big(x, u, v, u_{(1)}, v_{(1)}, \ldots, u_{(s)}, v_{(s)}\big)  = 0, \quad
 \alpha = 1, \ldots, m,$$ with $$\label{sa:eveq.int.3}
 F^*_\alpha \big(x, u, v,u_{(1)}, v_{(1)}, \ldots, u_{(s)}, v_{(s)}\big) =
  \frac{\delta {\cal L}}{\delta u^\alpha}\,,$$ where ${\cal L}$ is the *formal Lagrangian* for Eqs. (\[sa:eveq.int.1\]) defined by [^1] $$\label{sa:Lag}
 {\cal L} = v^\beta F_\beta \equiv \sum_{\beta= 1}^m v^\beta
 F_\beta.$$ Here $v = (v^1, \ldots, v^m)$ are new dependent variables, $v_{(1)}, \ldots, v_{(s)}$ are 