Differential commutator expression stuck

[binbagsss]

Homework Statement

I am trying to show that $a(x)[u(x),D^{3}]=-au_{xxx}-3au_{xx}D-3au_{x}D^{2}$, where $D=d/dx$, $D^{2}=d^{2}/dx^{2}$ etc.

Homework Equations

[/B]
I have the known results :

$[D,u]=u_{x}$
$[D^{2},u]=u_{xx}+2u_{x}D$

The property: $[A,BC]=[A,B]C+B[A,C]$*

The Attempt at a Solution

Let me drop the $a(x)$ and consider $[u(x),D^{3}]$

$=[u(x),D^{2}(D)] = [u(x),D^{2}]D+D^{2}[u(x),D]$ using *

$=-[D^{2},u(x)]D-D^{2}[D,u(x)]=-u_{xx}D-2u_{x}D-D^{2}u_{x}=-u_{xx}D-2u_{x}D-u_{xxx}$, using the 2 results quoted above.

Multiplying by $a(x)$ doesn't give me the correct answer.

Thanks in advance.

 

[vela]

I suggest you use parentheses and take things one step at a time. You seem to have the basic idea, but you're messing up the algebra. I'd also write $D^3$ as $D(D^2)$. That way the $D^2$ ends up to the right of the commutator, so you don't have to differentiate a product twice, i.e., $D^2[u,D]$ isn't just $D^2 u_x = u_{xxx}$ because $D^2[u,D]f = D^2(u_x f)$.