be rewritten in a strictly self-adjoint equivalent form by using multipliers. The general discussion of this approach will be given in Section \[gsa:6\].

Motivating examples {#samul:1}
-------------------

\[Burg:mult.exa1\] Let us consider the following second-order nonlinear equation $$\label{Burg:mult.eq1}
 u_{xx} +  f(u) u_x - u_t = 0.$$ Its adjoint equation (\[saeq.2\]) is written $$\label{Burg:mult.eq2}
 v_{xx} -  f(u) v_x + v_t = 0.$$ It is manifest that the substitution $v = u$ does not map Eq. (\[Burg:mult.eq2\]) into Eq. (\[Burg:mult.eq1\]). Hence Eq. (\[Burg:mult.eq1\]) is not strictly self-adjoint.

Let us clarify if Eq. (\[Burg:mult.eq1\]) can be written in an equivalent form $$\label{Burg:mult.eq3}
 \mu(u) [u_{xx} +  f(u) u_x - u_t] = 0$$ with a certain multiplier $ \mu(u)\not= 0$ so that Eq. (\[Burg:mult.eq3\]) is strictly self-adjoint. The formal Lagrangian for Eq. (\[Burg:mult.eq3\]) is $${\cal L} = v \mu(u) [u_{xx} +  f(u) u_x - u_t].$$ We have: $$\begin{aligned}
 \frac{\delta {\cal L}}{\delta u} & = D_x^2 [\mu(u) v] - D_x [\mu(u) f(u) v]
 + D_t[\mu(u) v]\notag\\
 & + \mu'(u) v [u_{xx} +  f(u) u_x - u_t] + \mu(u) f'(u) v u_x\,,\notag
 \end{aligned}$$ whence, upon performing the differentiations, $$\frac{\delta {\cal L}}{\delta u} = \mu v_{xx} + 2 \mu' v u_{xx}
 + 2 \mu' u_x  v_x + \mu'' v 