unsure how to expand a LAX pair for KDV equation

Question

This is my first post so im not sure how to make it all mathsy so im going to write it on here, I know that to find the lax equation you find [LM-ML]=0 but im struggling to follow the expansion,

for eg $$L=\left(\frac{d}{dx}\right)^2+u$$ $$M=a\left(\frac{d}{dx}\right)^3+bu\frac{d}{dx}+c\frac{du}{dx}$$

so LM would be $=a(d/dx)^5+$(on the book it says $b(d/dx)(u(d/dx)^2+(du/dx)(d/dx)$) i am unsure how this part comes about, ive tried to expand it and do it all but it doesn't simplify to that

im sorry if it doesnt look clear, again this is my first post.

Answer 1

The composition of $L$ and $M$ involves the application of differential operators in a certain order. It may be helpful to think about each composition $LM$ or $ML$ as an operator acting on a differentiable function $f(x)$, and just carry out the operations systematically by using the usual rules of calculus.

For example, $LM(f)$ would be $$ \left(\frac{d^2}{dx^2}+u\right)\left(a\frac{d^3f}{dx^3}+bu\frac{df}{dx}+c\frac{du}{dx}f\right). $$ Assuming $a$, $b$, $c$ are constants, the first part of the result (from applying $d^2/dx^2$) is $$ a\frac{d^5f}{dx^5}+b\frac{d^2}{dx^2}\left(u\frac{df}{dx}\right)+c \frac{d^2}{dx^2}\left(\frac{du}{dx}f\right). $$ Note that this expands out to give $$ a\frac{d^5f}{dx^5}+b \left(\frac{d^2u}{dx^2}\frac{df}{dx}+2\frac{du}{dx}\frac{d^2f}{dx^2}+u\frac{d^3f}{dx^3}\right)+c\left(\frac{d^3u}{dx^3}f+2\frac{d^2u}{dx^2}\frac{df}{dx}+\frac{du}{dx}\frac{d^2f}{dx^2}\right). $$ And, the second part is obtained by multiplication alone $$ au\frac{d^3f}{dx^3}+bu^2\frac{df}{dx}+cuf\frac{du}{dx}. $$

Now if you remove the placeholder $f$, these would give the answer: $$ LM = a\left(\frac{d^5}{dx^5} + u\frac{d^3}{dx^3} \right) +b\left(\frac{d^2u}{dx^2}\frac{d}{dx}+2\frac{du}{dx}\frac{d^2}{dx^2}+u\frac{d^3}{dx^3} +u^2\frac{d}{dx}\right) +c\left(\frac{d^3u}{dx^3}+2\frac{d^2u}{dx^2}\frac{d}{dx}+\frac{du}{dx}\frac{d^2}{dx^2}+ u\frac{du}{dx}\right). $$ And, you can combine some of these terms again to get a more concise expression like the one you are quoting in your post.

But, note that you may not be quoting all the terms in $LM$ in the reference you are reading. For example, there are clearly two parts multiplied by $a$ in $LM$, given by $$ a\left(\frac{d^5}{dx^5} + u\frac{d^3}{dx^3} \right), $$ but this is not in your post. Actually, the second term cancels out with a corresponding term in $ML$.

Answer 2

The commutator, like any product, is distributive. That is, you can compute the commutators of the terms and then add all up in the end. This might be simpler than herding the long total expression, also cancellations occur earlier. Assuming again that $a,b,c$ are constants, one gets using $D=\frac{d}{dx}$ \begin{align} [D^2,aD^3]&=0\\ [u,aD^3]&=-a(D^3u) -3a(D^2u)D-3a(Du)D^2\\ [D^2,buD]&=b(D^2u)D+2b(Du)D^2\\ [u,buD]&=-bu(Du)\\ [D^2,c(Du)]&=c(D^3u)+2c(D^2u)D\\ [u,c(Du)]&=0 \end{align} To get $[L,M]=0$ would require, by comparing the coefficients of the free derivative powers, \begin{align} D^2:&&0&=(-3a+2b)Du\\ D: && 0&=(-3a+b+2c)D^2u\\ 1:&& 0&=-aD^3u -bu(Du) \end{align} so that $b=\frac32a$, $c=\frac34a$, $u'''+\frac32uu'=0$.