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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.

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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.

\begin{align} [L,B]\phi & = (-\partial_{xx} + u)(-4\partial_{xxx} + 6u\partial_{x} + 3u_{x})\phi - (-4\partial_{xxx} + 6u\partial_{x} + 3u_{x})(-\partial_{xx} + u)\phi \\ & = \text{a long and messy computation} \\ & = (u_{xxx} - 6uu_{x})\phi\end{align}

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.

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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.

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  • $\begingroup$ What has to do fractional calculus with this? $\endgroup$ – Ambesh Nov 25 '13 at 11:43

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