(\[nsa.linearized\]) is identical with Eq. (\[adad\]).

Let us illustrate Eq. (\[nsa.linearized\]) by the equation $$\label{nsa.linearize2}
 F \equiv  u_{xy} - \sin u = 0.$$ Eq. (\[sa:eveq.int.3\]) yields $$\label{nsa.linearize3}
 F^* \equiv   \frac{\delta}{\delta u}[v(u_{xy} - \sin u)]
 = v_{xy} - v \cos u$$ and $$\label{nsa.linearize4}
 \left(F^*\right)^* \equiv  \frac{\delta}{\delta v}[w(v_{xy} - v \cos u)] = w_{xy} - w \cos u.$$ Let us find $\hat F$ by using Eq. (\[nsa.linearize1\]). Since $\sin w \approx w, \ \cos w \approx 1$ when $w \ll 1,$ we have $$\begin{aligned}
 F [u + w] \equiv & \ (u + w)_{xy} - \sin (u + w)\notag\\[.5ex]
 & = u_{xy} + w_{xy} - \sin u \cos w - \sin w \cos u\notag\\[.5ex]
 & \approx u_{xy} - \sin u + w_{xy} -  w \cos u,\notag\\[.5ex]
 & = F[u] + w_{xy} -  w \cos u.\notag
 \end{aligned}$$ Hence, by (\[nsa.linearize1\]) and (\[nsa.linearize4\]), we have $$\label{nsa.linearize5}
 \hat F[w] = w_{xy} -  w \cos u = \left(F^*\right)^*$$ in accordance with Eq. (\[nsa.linearized\]).

The case of one dependent variable {#sa:eveq.2}
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Let us consider the differential equation $$\label{saeq.1}
 F\big(x, u, u_{(1)}, \ldots, u_{(s)}\big) = 0$$ with one dependent variable $u$ and any number of independent variables. In this case Definition \[sa:adj\] of the adjoint equation is 