I'm a physics graduate student, it's when my professor for mathematical physics mentioned KdV equations that I encountered the following problem.
This is what our professor wrote: $\mathbb{L}:=-d^2+q(t,x)$, $\mathbb{P}:=d^3-\frac{3}{4}q(t,x)$. And $$\frac{d \mathbb{L}(t)}{d t}=[\mathbb{P},\mathbb{L}],$$ with the evolution operator $U(t)$ defined such that $\mathbb{L}(t)=U(t)\mathbb{L}_0U(t)^{-1}$. If one tries to differentiate the above equation, they would normally arrive at terms like $\partial_t U(t)$.
And here is where I'm confused. How do I understand the inverse of an operator containing the differential operation, and what's worse, how to make sense of the evolution and time derivative of an operator containing derivative?
P.S.: I think those are typical things a physics student does not understand (because I have tried to discuss with my mates, and nobody can really give a good answer). So anybody can try to modify the title or even description to make it general enough.
It is evident that you are working in the context of the Inverse Scattering Transform, as evidenced by quoting Lax's equation $\frac{d \mathbb{L}(t)}{d t}=[\mathbb{P},\mathbb{L}].$
The non-intuitive fact about $\mathbb{L}_t$ is that we only take the time derivative where $t$ explicitly appears in $\mathbb{L}$. The reason is that we would like to have $$(\mathbb{L}y)_t=\mathbb{L}_ty+\mathbb{L}y_t. $$ On the LHS, we have $$\frac{\partial}{\partial t}((-D^2+q)y)=-\frac{\partial}{\partial t}\,D^2y+q_ty+qy_t=-D^2y_t+q_ty+qy_t=\mathbb{L}y_t+q_ty.$$ If we compare this calculation with the one above, we see that $\mathbb{L}_t=q_t,$ by definition. This is tantamount to setting the $\frac{\partial}{\partial t}\,D^2$ term in $\frac{\partial}{\partial t}\,\mathbb{L}$ equal to zero.