u_x^2 - \mu f v_x + \mu v_t\,.$$ The strict self-adjointness requires that $$\frac{\delta {\cal L}}{\delta u}\bigg|_{v = u} = \lambda [u_{xx} +  f(u) u_x -
u_t].$$ This provides the following equation for the unknown multiplier $\mu (u):$ $$\label{Burg:mult.eq4}
 (\mu + 2 u \mu') u_{xx}
 + (2 \mu' + u \mu'') u_x^2 - \mu f u_x + \mu u_t
 = \lambda [u_{xx} +  f(u) u_x - u_t].$$ Since the right side of Eq. (\[Burg:mult.eq4\]) does not contain $u_x^2$ we should have $2 \mu' + u \mu'' = 0,$ whence $\mu = C_1 u^{-1} + C_2.$ Furthermore, comparing the coefficients of $u_t$ in both sides of Eq. (\[Burg:mult.eq4\]) we obtain $\lambda = - \mu.$ Now Eq. (\[Burg:mult.eq4\]) takes the form $$(C_2 - C_1 u^{-1}) u_{xx} - (C_1 u^{-1} + C_2) f u_x
 = - (C_1 u^{-1} + C_2) [u_{xx} +  f(u) u_x]$$ and yields $C_2 = 0.$ Thus, $\mu = C_1 u^{-1}.$ We can let $C_1= - 1$ and formulate the result.

\[Burg:mult.prop\] Eq. (\[Burg:mult.eq1\]) becomes strictly self-adjoint if we rewrite it in the form $$\label{Burg:mult.eq5}
 \frac{1}{u}\,[u_t - u_{xx} -  f(u) u_x] = 0.$$

\[Burg:mult.exa4\] One can verify that the $n$th-order nonlinear evolution equation $$\label{Burg:mult.eq12}
 \frac{\partial u}{\partial t} - f(u) \frac{\partial^n 