Time derivative of the commutator of two operators $P,L$

Question

I am reading a textbook Mathematics for Physics. In page 126, the author defines two operators: $$L=-\partial_x^2+q(x)$$ $$P=\partial_x^3+a(x)\partial_x+\partial_x a(x)$$ Then the author continues to calculate the commutator, which as I understand has the same definition from ring theory: $$[P,L]=PL-LP$$ After some tedious calculations and setting $a(x)=-\frac 34 q(x)$, the following equation is reached. $$[P,L]=\frac 14 q'''-\frac 32 qq'$$ The part where I don't understand is, the author then writes: $$\frac {dL}{dt} = [P,L]$$ or, equivalently $$\dot q = \frac 14 q'''-\frac 32 qq'$$ I think I may lack specific knowledge on this part as why we can set the time derivative $\frac{dL}{dt} = [P,L]$ and what is the author trying to do?

Answer 1

I read the page. The authors do not obtain this result. The give an example (KdV) of an equation written as $$ \frac{dL}{dt}=[P,L],\;\;\;(1) $$ for operators $L$ and $P$. These equations are part of the class of PDEs that have a "Lax pair". Among its properties, as shown in the book, is the fact that the eigenvalues of $L$ are constant in time.

Another way to see (1) is that it comes from the compatibility condition for the system of two linear questions in "$u$": $$ \frac{du}{dt}=Pu\\ \lambda u=Lu. $$ We usually call the system above the "Lax Pair". We compute the compatibility condition of the system above by multiplying the first one by $\lambda$ and taking the derivative wrt time of the second one. One the two LHS's, we get in both cases $\lambda du/dt$. Equating the two right sides, one gets (1).