First post ever. Let's see how this goes...

My question concerns the commutation of differential operators in the presence of boundary conditions. If it is of any help, this is relevant to me in the context of solving (or writing the propagator of) Fokker-Planck-type equations. I've wondered about this issue several times in various situations, but I will try to distill the essence of it by the simplest nontrivial example to keep things clear.

Consider the equation $$ \frac {\partial \rho} {\partial t} = \nabla^2 \rho - \vec A \cdot \nabla \rho \tag{1} \label{eq}$$ for the evolution of some probability density $\rho (\vec x,t)$ in a finite region $\Omega$ of space, with, say, Neumann conditions on the boundary $\partial \Omega$. Also, the vector $\vec A$ has no $\vec x$-dependence. It follows that the probability density will have the form $$ \rho (\vec x, t) = e^{t \nabla^2 - t \vec A \cdot \nabla} \rho( \vec x,0) \ . \tag{2} \label{sol} $$

Here is what I wonder then: Can I further rewrite this as, for instance, $$ \rho (\vec x, t) = e^{-t \vec A \cdot \nabla} e^{t\nabla^2} \rho (\vec x, 0) \ ? \tag{3} \label{factor} $$

It seems to me that the answer would be yes if the commutator $[\nabla^2, \nabla]=0$. And since it is between an operator and its square, I tend to think the commutator vanishes, without any precise regard for the function space where these operators "live", appropriate to the problem. But I know in my heart of hearts that I'm being naive, even though I don't know what subtlety I'm missing.

Question: So, would the commutator vanish and allow me the factorization of the propagator as in Eq. (\ref{factor})? If not, why? I would also appreciate an accessible reference illuminating these topics (which I'm obviously unaware of) greatly.

Bonus question: A reason for why one may be inclined towards the factorization (\ref{factor}) is that they know what each individual operator does: $e^{t\nabla^2}$ transforms functions into what they would look like after a duration $t$ of diffusion, and $e^{-t \vec A \cdot \nabla}$ simply displaces them by $t \vec A$. But wait. The latter statement seems fishy. The displaced coordinate $\vec x - t\vec A$ may not be even inside the domain $\Omega$... So would it be correct to think of $e^{-t\vec A \cdot \nabla}$ as a translation operator or not? I think I'm plagued by the same lack of understanding here as the main question, and answering one will likely answer the other.

Thanks to anyone who pays attention.

  • $\begingroup$ If $A$ does not depend on ${\bf x}$, you are right, the two terms commute and your Eqn (3) is correct $\endgroup$ – caverac Jan 22 '19 at 16:59
  • $\begingroup$ @caverac, so you don't think I'm being naive and overlooking some subtlety introduced by the boundaries. I would like that to be the case... Do you have some insight on the bonus question? In short, what is the action of $e^{-t \vec A \cdot \nabla}$ on an arbitrary density function $\rho (\vec x)$? $\endgroup$ – Cem Yolcu Jan 24 '19 at 13:11

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