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The Fokker-Planck equation

An alternative way to study stochastic processes is by means of distribution functions . Let $\mathcal{G}(
\vec{r},\vec{v};\vec{r}_{0},\vec{v}_{0};t)$ be the probability density to find a particle at time t at position $
\vec{r}$ with velocity $\vec{v}$, given that at t=0 it was at position $\vec{r}_{0}$ with velocity $\vec{v}_{0}$. We shall demonstrate in Appendix A that this function satisfies

 \begin{displaymath}\frac{\partial }{\partial t}\mathcal{G}(\vec{z};\vec{z}_{0};t...
...\vec{v}}^{2}\frac{\xi kT}{m}\mathcal{G}(\vec{z};\vec{z}_{0};t)
\end{displaymath} (4.14)


 \begin{displaymath}\lim_{t\rightarrow 0}\mathcal{G}(\vec{z};\vec{z}_{0};t)=\delta (\vec{r}-\vec{
r}_{0})\delta (\vec{v}-\vec{v}_{0})
\end{displaymath} (4.15)

where $\mathcal{G}(\vec{z};\vec{z}_{0};t)$ is short hand for $\mathcal{G}(
\vec{r},\vec{v};\vec{r}_{0},\vec{v}_{0};t)$. Eq. (4.14) together with the initial value condition Eq. (4.15) is called the Fokker-Planck equation .

We shall not treat this equation any further here, except for one quick remark. The probability to find the particle at time t with velocity $\vec{v}$, at any position, given it was at t=0 at position $\vec{r}_{0}$with velocity $\vec{v}_{0}$ is given by

 \begin{displaymath}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t)=\int d^{3}r\mathcal{G}(\vec{v},
\vec{r};\vec{v}_{0},\vec{r}_{0};t)
\end{displaymath} (4.16)

where we have assumed the system is translational invariant, i.e. both sides of Eq. (4.16) are independent of $\vec{r}_{0}$. Integrating Eq. (4.14) with respect to $
\vec{r}$ we get
  
$\displaystyle \frac{\partial }{\partial t}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t)$ = $\displaystyle \vec{\nabla}_{\vec{v}}\cdot \xi \vec{v}\tilde{\mathcal{G}}(\vec{v...
...nabla _{\vec{v}}^{2}\frac{\xi kT}{m}\tilde{\mathcal{G}}(\vec{v};
\vec{v}_{0};t)$ (4.17)
$\displaystyle \lim_{t\rightarrow 0}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t)$ = $\displaystyle \delta (
\vec{v}-\vec{v}_{0})$ (4.18)

We may easily obtain the solution to this equation by using the results of the previous section. To this end we notice that $\vec{U}(t)=\vec{v}(t)-\vec{
v}_{0}e^{-\xi t}$ according to Eq. (4.4) is a sum of many random vectors. According to the central limit theorem  therefore it must have a Gaussian distribution. The first and second moments are given by
  
$\displaystyle \langle U_{\alpha }(t)\rangle$ = 0 (4.19)
$\displaystyle \langle U_{\alpha }(t)U_{\beta }(t)\rangle$ = $\displaystyle \delta _{\alpha \beta }
\frac{kT}{m}(1-e^{-2\xi t})$ (4.20)

where we have used the fluctuation-dissipation theorem  in the form

 \begin{displaymath}\langle F_{\alpha }(t)F_{\beta }(t^{\prime })\rangle =\delta _{\alpha \beta
}2
\frac{kT}{m}\xi \delta (t-t^{\prime }).
\end{displaymath} (4.21)

Using Eqs. (4.19) and (4.20) we may write

\begin{displaymath}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t)=\left\{ \frac{m}{2...
...ec{v}-\vec{v
}_{0}e^{-\xi t})^{2}}{2kT(1-e^{-2\xi t})}\right\}
\end{displaymath} (4.22)

Although it is a tedious task, it is not difficult to check that this result indeed is the solution to Eqs. (4.17) and (4.18).


next up previous contents index
Next: The Smoluchowski time scale Up: Stochastic processes Previous: The Langevin equation
W.J. Briels