Description

$\newcommand{\R}{\mathbb{R}} \newcommand{\dx}{\mathrm{d}}$
The Fokker Planck equation has the form

$$\partial_t \varrho(x,t) = \nabla \cdot \bigl(\beta^{-1} \nabla \varrho(x,t) + \varrho(x,t) \nabla H(x)\bigr) ,\qquad \varrho(x,0)=\varrho_0(x) ,$$ where $H:\R^n\to \R$ is a smooth function, $\beta>0$ some parameter and $\varrho_0$ a probability density on $\R^n$. The partial differential equation is in divergence form and conserves mass. Hence, also $\varrho(\cdot,t)$ is a probability density on $\R^n$. In the case, where $H$ has some growth at $\infty$, the equilibrium solutions $\varrho_\infty:\R^n \to \R$ are characterized by
$$\beta \nabla \varrho_\infty + \varrho_\infty \nabla H =0$$ leading to the solution
$$\varrho_\infty(x) = Z^{-1} e^{-\beta H(x)} \quad \text{with} \quad Z = \int e^{-\beta H(x)} \; \dx x$$
The density $\varrho_\infty$ is called Gibbs distribution, which accounts to the fact, that it minimizes the (Gibbs) free energy
$$F(\varrho) = E(\varrho) + \beta^{-1} S(\varrho)$$ where $E$ is the energy and $S$ the (Gibbs-Boltzmann) entropy
$$E(\varrho) := \int H \varrho\;\dx x \quad\text{and}\quad S(\varrho) := \int \varrho \log \varrho \; \dx{x} .$$ By this formulation  $\beta$ plays the role of the inverse temperature. Note, that the Euler-Lagrange equation is given by
$$H(x) + \beta^{-1} \bigl(\varrho(x) + 1\bigr) = 0 ,$$ which is solved by $\varrho_\infty$.

Pathwise formulation via stochastic differential equations

The stochastic process $X(t)$ defined by

$$\dx X(t) = – \nabla H(X(t))\;\dx t + \sqrt{2\beta^{-1}} \;\dx W(t) , \quad X(t) = X_0 ,$$ where $W(t)$ is an $n$-dimensional Brownian motion, is the pathwise formulation of the Fokker Planck equation. If distribution of the initial random vector $X_0$ is given by $\mathrm{law}\; X_0 = \varrho_0$, then at later times the solution $X(t)$ has distribution $\mathrm{law}\; X(t) = \varrho(\cdot,t)$, where $\varrho$ is solution of the Fokker Planck equation.

The dynamic of the pathwise formulation undergoing diffusion in the potential $H$. The particle is solely described by its position $X(t)\in \R^n$, i.e. inertia is neglected. Then, the particle chooses the steepest descent with respect ot $H$, which is perturbed by random fluctuation, which are assumed to be normal distributed. The coupling parameter $\beta$ describing the strength of the fluctuations plays again the role of inverse temperature. In particular, the limit $\beta\to \infty$ describes the behaviour of the system without fluctuations.

Ornstein-Uhlenbeck process

A prominent example, which can be explicitly solved, is the Ornstein-Uhlenbeck process, corresponding to the choice

$$H(x) = \frac{\gamma}{2} | x |^2 .$$ The stationary distribution is given by the Gaussian
$$\varrho_\infty(x) = \left( \frac{2\pi}{\gamma\beta} \right)^{-\frac{n}{2}} \exp\left(-\frac{\gamma \beta |x|^2}{2}\right) .$$ The underlying picture becomes particular simple in one dimension. There, the leading order behaviour of the process can be characterized by its mean $m(t)$ and variance $\sigma(t)$, which are given by
$$m(t) := \int x \varrho(x,t) \;\dx x \qquad\text{and}\qquad \sigma(t) := \int (x-m(t))^2 \varrho(x,t) \;\dx{x} = \int x^2 \varrho(x,t)\; \dx{x} – m(t)^2 .$$ Both, $m(t)$ and $\sigma(t)$, satisfy an ordinary differential equation not depending on $\varrho$
$$\dot m(t) = -\gamma m(t) \qquad \text{and}\qquad \dot sigma(t) = 2\beta^{-1} – 2\gamma \sigma(t) .$$ Hence, the mean converges exponentially fast to $0$ with rate $\gamma$ and the variance converges exponentially fast from the initial variance $\sigma(0)$ to $(\beta \gamma)^{-1}$ with rate $2 \gamma$
$$m(t) = m(0) \; e^{-2\gamma t} \qquad\text{and}\qquad \sigma(t) = \sigma(0) e^{-2\gamma t} + \frac{1}{\beta \gamma} \bigl(1-e^{-2\gamma t}\bigr) .$$

Numerical example

For general potentials $H$ explicit solutions of the Fokker Planck equation are not available. However, the pathwise formulation allows to do Monte-Carlo simulation of single realisations of the process $X(t)$. The case, where $H$ is not convex is especially interessting, because at low temperature, i.e. $\beta \gg 1$, the process shows metastable behaviour.

Therefore, we consider the potential function $H:\R^2 \to \R$ looking like

The realization of the process $X(t)$ for increasing values of $\beta$:

For increasing values of $\beta$ the particle goes quickly to the local minima and is trapped there longer and longer times. The transition to the global energy minimizing state happens along a path crossing the saddle. This behaviour is called metastable. For the limiting dynamic $\beta=\infty$ without any fluctuation a transition is not possible and the particle is stuck in the local minima.

References

1. R. Jordan, D. Kinderlehrer, and F. Otto, “The Variational Formulation of the Fokker-Planck Equation,” SIAM Journal on Mathematical Analysis, vol. 29, no. 1, 1998.