I report below (part of) page 73 of the book The Fokker-Planck Equation, by H. Risken

We now derive an expression for the transition probability density for small $\tau$ \begin{equation}\tag{1} p(x,t+\tau|x', t)=\bigg(1+L_{\mathrm{FP}}(x,t)+O(\tau^2)\bigg)\delta(x-x') \end{equation} with \begin{equation}\tag{2} L_{\mathrm{FP}}(x,t):=-\frac{\partial}{\partial t}D_1(x,t)+\frac{\partial^2}{\partial x^2}D_2(x, t) \end{equation} We get up to corrections of the order $\tau^2$: \begin{equation} \begin{aligned} p(x,t+\tau|x', t)&=\bigg(1-\frac{\partial}{\partial t}D_1(x',t)\tau+\frac{\partial^2}{\partial x^2}D_2(x', t)\tau\bigg)\delta(x-x')\\ &=\exp\bigg(-\frac{\partial}{\partial t}D_1(x',t)\tau+\frac{\partial^2}{\partial x^2}D_2(x', t)\tau\bigg)\delta(x-x') \end{aligned} \tag{3} \end{equation}

I don't get the last equality in equation $\mathrm{(3)}$. At first glance, it would appear to be just a replacement. After all $\exp(L_{\mathrm{FP}}\,\tau)\simeq 1+L_{\mathrm{FP}}\,\tau$ for $\tau\to 0$. But is it?

After a few other steps, he writes the solution as

\begin{equation}\tag{4} p(x,t+\tau|x',t)=\frac{1}{\sqrt{4\pi\,D_2(x',t)}}\exp\left(\frac{[(x-x')-D_1(x',t)\tau]^2}{4D_2(x', t)\tau}\right) \end{equation}

and then he says

for drift and diffusion coefficients independent of $x$ and $t$, $\mathrm{(4)}$ is not only valid for small $\tau$, but for arbitrary $\tau>0$ (the last line in equation $\mathrm{(3)}$ is then the formal solution).

So... did he just replace the approximate solution with the exact one? One is not supposed to know it ahead.

  • $\begingroup$ Hi was my answer helpful at all? $\endgroup$ Feb 10, 2023 at 22:37

1 Answer 1


It is more than just a replacement of $1+Lt\rightarrow \exp (Lt)$. It is the following: within a region of $x$ and $t$ that $D_{1,2}(x,t)$ can be reliably approximated as a constant, the transition probability will be given by $(4)$. In other words, if we isolate our attention to such a confined region, we can reliably approximate the transition probability as $(4)$.

An analogous situation would be solving the differential equation $y'(x) = \sin(x) y(x)$ near the point $x=\pi/2$. The exact solution to this differential equation is $y(x)=C e^{-\cos(x)}$, and if we expand it about $x=\pi/2$ we get

$$y(x)\approx C e^{-\left(\frac{\pi}{2}-x\right)}$$

However in the vicinity of $x=\pi/2$ we can reliably approximate $\sin (x)\approx 1$, and therefore $y(x)\approx C' e^{x}$ near this point. Notice that from this approximate solution near $x=\pi/2$, you cannot deduce anything about $y(x)$ outside this region of validity.


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