u}{\partial x^n}
  = 0,  \quad f(u) \not= 0,$$ with one spatial variable $x$ is not strictly self-adjoint. The following statement shows that it becomes strictly self-adjoint after using an appropriate multiplier.

\[Burg:mult.prop2\] Eq.(\[Burg:mult.eq12\]) becomes strictly self-adjoint upon rewriting it in the following equivalent form: $$\label{Burg:mult.eq13}
 \frac{1}{u f(u)}\,\left[ \frac{\partial u}{\partial t}
- f(u) \frac{\partial^n u}{\partial x^n}\right] = 0.$$ **Proof.** Multiplying Eq. (\[Burg:mult.eq12\]) by $\mu(u)$ and taking the formal Lagrangian $${\cal L} = v \mu(u) [u_t - f(u) u_n],$$ where $u_n = D_x^n(u),$ we have: $$\frac{\delta {\cal L}}{\delta u} = - D_t[\mu(u) v] - D_x^n [\mu(u)f(u) v] +
 v \mu'(u) u_t - v [\mu(u)f(u)]' u_n.$$ Noting that $- D_t[\mu(u) v] + v \mu'(u) u_t = - \mu(u) v_t$ and letting $v = u$ we obtain $$\frac{\delta {\cal L}}{\delta u}\bigg|_{v = u} = - \mu(u) u_t
 - D_x^n [\mu(u)f(u) u] - [\mu(u)f(u)]' u u_n .$$ If we take $\mu(u) = [u f(u)]^{-1},$ then $\mu(u)f(u) u = 1, \ \mu(u)f(u) = u^{-1},$ and hence $$\frac{\delta {\cal L}}{\delta u}\bigg|_{v = u}
= - \frac{1}{u f(u)}\, [u_t - f(u) u_n].$$ Thus, Eq. (\[Burg:mult.eq13\]) satisfies the strict self-adjointness condition (\[saeq.5\]) with $\lambda = - 1.$

Linear heat equation {#samul:2}
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Taking in (\[Burg:mult.eq5\]) $f (u) = 0,$ 