= \left[c u + b^i u_i + a^{ij} u_{ij}\right]
 + \left\{D_i D_j(a^{ij} u) - D_i(a^{ij} u_j) - D_i [u
D_j(a^{ij})]\right\}$$ and, noting that the expression in the braces vanishes, arrive at Eq. (\[lnad:eq1\]).

Let us illustrate Proposition \[lnad.prop1\] by the following simple example.

\[sa:eveq.exa1\] Consider the heat equation $$\label{sa:eveq.exa1:eq1}
 F[u] \equiv u_t - u_{xx}= 0$$ and construct the adjoint operator to the linear operator $$\label{sa:eveq.exa1:eq2}
 F = D_t - D_x^2$$ by using Eq. (\[clad\]). Noting that $$\begin{aligned}
 & v u_t = D_t(u v) - u v_t, \notag \\[1ex]
 & v u_{xx} = D_x(v u_x) - v_x u_x
 = D_x(v u_x - u v_x) + u v_{xx}\notag
 \end{aligned}$$ we have: $$v F[u] \equiv v(u_t - u_{xx}) = u(- v_t - v_{xx}) + D_t(u v) + D_x(u v_x - v
 u_x).$$ Hence, $$v F[u] - u (- v_t - v_{xx}) = D_t(u v) + D_x(u v_x - v u_x).$$ Therefore, denoting $t = x^1, \ x = x^2,$ we obtain Eq. (\[clad\]) with $F^*[v] = - v_t - v_{xx}$ and $p^1 = u v, \ p^2 = u v_x - v u_x.$ Thus, the adjoint operator to the linear operator (\[sa:eveq.exa1:eq2\]) is $$\label{sa:eveq.exa1:eq3}
 F^* = - D_t - D_x^2$$ and the adjoint 