From master equation to Fokker-Planck equation

Question

For the continuous master equation in real space and time, we have for the distribution $f(x,t)$: $$\frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\mathrm{d}x'$$ In the statistical mechanics book by P.K. Pathria, the right hand side is Taylor expanded to obtain (keeping up to second order) $$\frac{\partial f(x,t)}{\partial t}=-\frac{\partial}{\partial x}[f(x,t)\int_{-\infty}^{\infty}\xi W(x;\xi)\mathrm{d}\xi]+\frac{1}{2}\frac{{\partial}^2}{\partial x^2}[f(x,t)\int_{-\infty}^{\infty}{\xi}^2 W(x;\xi)\mathrm{d}\xi],$$ where $\xi=x'-x$, and $W(x;\xi)=W(x,x')$, which is the transition probability density from $x$ to $x'$. This is pretty much confusing to me and I have the following questions:

1. How is the expansion carried out? It seems that we should expand around $x'$, after which a change of integration variable is done. However, it doesn't seem to lead to the above expression.

2. The difference $\xi$ between $x$ and $x'$ is not necessarily small and in fact as the integration variable it goes all the way to $\infty$. Then in this case, how is the keeping up to second order and neglect all the higher orders justified?

Any help is much appreciated.

Answer 1

For the first question, we start by rewriting $$ \frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\, dx' $$ as $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &=\int_{-\infty}^{\infty}[f(x',t)W(x';x-x')-f(x,t)W(x;x'-x)] \, d (x' - x) \\ &= \int_{-\infty}^{\infty} f(x-\xi,t) W(x-\xi;\xi) \, d \xi -f(x, t) \int_{-\infty}^{\infty} W(x; \xi) \, d \xi, \qquad (1) \end{aligned} $$ where $\xi \equiv x - x'$. For the first term, the change of the minus sign in $d(-\xi) \to d\xi$ is compensated by swapping the lower and upper limits of the integral: $$ \int_{-\infty}^\infty g(\xi) d(-\xi) = -\int_{\infty}^{-\infty} g(\xi) d\xi = \int_{-\infty}^{\infty} g(\xi) d\xi $$

Next we can expand the first term using the Kramers-Moyal expansion $$ f(x-\xi,t) W(x-\xi;\xi) = f(x,t) W(x;\xi) -\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] +\frac{ \xi^2 } {2!} \frac{\partial^2 }{ \partial x^2} \left[ f(x,t) W(x;\xi) \right] + \dots, \qquad (2) $$ in which the first term of right-hand side will cancel the second term of (1). So $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &= \int_{-\infty}^\infty \left( -\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] +\frac{ \xi^2 } {2!} \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right] \right) \, d\xi, \\ &= -\frac{\partial }{ \partial x} \left[f(x,t) \int_{-\infty}^\infty \xi W(x;\xi) \, d\xi\right] + \frac{ 1 } {2!} \frac{\partial^2 }{ \partial x^2}\left[ f(x,t) \int_{-\infty}^\infty \xi^2 W(x;\xi) \, d\xi \right]. \end{aligned} $$

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For the second question, you are absolutely correct to say that we cannot always justify the truncation. Van Kampen has a entire chapter devoted to this (chapter X) among complaints in other places. The formally correct expansion, Kramers-Moyal expansion, yields $$ \begin{aligned} \frac{\partial f(x,t)}{\partial t} &= \sum_{k = 1}^\infty \frac{(-1)^k}{k!} \frac{\partial^k }{ \partial x^k} \left[f(x,t) \int_{-\infty}^\infty \xi^k W(x;\xi) \, d\xi\right]. \end{aligned} $$ The problem is that this expansion is equivalent to the master equation, and it is generally too difficult to solve.