MHB What is the compatibility condition for the given operators and constants?

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Condition
Dustinsfl
Messages
2,217
Reaction score
5
Given
\begin{align}
L\phi &= \lambda\phi\\
\phi_t &= M\phi
\end{align}
where \(L\) and \(M\) are operators and \(\lambda\) a constant.
I want to show the compatibily condition is \(L_t + [L,M] = 0\) where \([,]\) is the commutator.
\[
(L\phi)_t = L_t\phi + L\phi_t = \lambda\phi_t = \lambda M\phi
\]
That is, we have \(L_t\phi + L\phi_t - \lambda M\phi = L_t\phi + LM\phi- \lambda M\phi = 0\).
\begin{align}
[L_t + LM - \lambda M]\phi &= 0\\
L_t + LM - \lambda M &= 0
\end{align}
Can I just let \(\lambda = L\) which doesn't make since sense one is an operator and the other a constant? If not, how do I get the commutator \(LM - ML\) part?
 
Physics news on Phys.org
Ah, I see you are doing the Inverse Scattering Transform with Lax pairs. You have $\lambda M \phi$. Since $\lambda$ is a scalar, rearrange thus: $\lambda M \phi=M(\lambda \phi)=M L \phi$. You can go from there.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top