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

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

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

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

  

$$\frac{\partial }{\partial t}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t) \vec{\nabla}_{\vec{v}}\cdot \xi \vec{v}\tilde{\mathcal{G}}(\vec{v... ...nabla _{\vec{v}}^{2}\frac{\xi kT}{m}\tilde{\mathcal{G}}(\vec{v}; \vec{v}_{0};t)$$ = (4.17)

$$\lim_{t\rightarrow 0}\tilde{\mathcal{G}}(\vec{v};\vec{v}_{0};t) \delta ( \vec{v}-\vec{v}_{0})$$ = (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

  

$$\langle U_{\alpha }(t)\rangle$$ = 0 (4.19)

$$\langle U_{\alpha }(t)U_{\beta }(t)\rangle \delta _{\alpha \beta } \frac{kT}{m}(1-e^{-2\xi t})$$ = (4.20)

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

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

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

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