# Differential operators confussion

I want to solve this PDE: $$u_t-6uu_x+u_{xxx} = 0\,(1)$$ with the Inverse Scattering Method. This method is based on showing that the above equation can be expressed as $$L_t=LB-BL,\,(2)$$ where $L$ and $B$ are the differential operators: $$L=-\frac{\partial^2}{\partial x^2}+u(x)$$ $$B=-4\frac{\partial^3}{\partial x^3}+6u\frac{\partial}{\partial x}+3\frac{\partial u}{\partial x}.$$

I have tried to prove that (1) and (2) are equal composing the differential operators $L$ and $B$. However I don't have experience doing this and I got different results. The way I applied the operators is as if they were just derivatives and by using the chain rule.

This is not a complete solution, just a suggestion for how to do the calculation. Consider the operator $[L,B]$ acting on some function $\phi$, i.e.
All the other terms should cancel and you will have no $\phi_{x}$, $\phi_{xx}$ or $\phi_{xxx}$ in your answer. Notice that the coefficients in front of the $\phi$ term are just the time-independent part of the KdV.
The keyword is Operator Calculus or alternatively Operational Calculus. Here is an introductory (PDF) document . Other references are easily found on the internet, such as Fractional Calculus (Wikipedia), What is operator calculus? (MSE). The key formula you should IMHO use is not the chain rule [ it's rather derived from $(uv)' = u'v + uv'$ ] : $$\frac{d}{dx} f = f \frac{d}{dx} + \frac{df}{dx}$$ Mind that differentiation and multiplying with a function do not commute with Operator Calculus. But, in my experience, that's almost the only thing to take care of.