Consider the Fokker-Planck equation $$\frac{\partial \rho}{\partial t} = \sum_{i,j=1}^2 \mathbf{\Gamma}_{ij}\frac{\partial}{\partial x_i}(x_j \rho) + \mathbf{D}_{ij}\frac{\partial^2 \rho}{\partial x_i \partial x_j}$$ where $$\rho = \begin{pmatrix} x \\ v \end{pmatrix} \quad \mathbf{\Gamma} = \begin{bmatrix} 1 & -1 \\ 0 & \gamma\end{bmatrix} \quad \mathbf{D} = \begin{bmatrix} 0 & 0 \\ 0 & D \gamma^2 \end{bmatrix} \, .$$ These equations represent diffusion where $\gamma$ is the coefficient of friction and $D=k_b T / \gamma m$ is the diffusion constant.

I'm reading Elements of Nonequilibrium Statistical Mechanics by Balakrishnan wherein this equation is studied. The book points out that this Fokker-Planck equation is equivalent to the following Lengevin equation $$\dot{\mathbf{x}} + \mathbf{\Gamma} \mathbf{x} = \frac{1}{m} \begin{pmatrix} 0 \\ \eta(t) \end{pmatrix}$$ where $$\mathbf{x} = \begin{pmatrix} x \\ v \end{pmatrix}$$ and where $\eta(t)$ is white noise satisfying $\langle \eta(t) \eta(t') \rangle = 2 m \gamma k_b T \delta(t-t').$ So far this is all fine.

The author writes

The Langevin equation is solved with the help of the matrix Green function $$\mathbf{G}(t) = \exp(-\mathbf{\Gamma} t) \equiv \sum_{n=0}^\infty \frac{(-\mathbf{\Gamma}t)^n}{n!} \, .$$

This is annoying because the noise $\eta$ is left out. However, the author then writes

We are concerned with sharp initial conditions $\mathbf{x}_0 = (x_0, v_0)$. The average value of $\mathbf{x}$, conditioned upon these initial values, is then given by $$\overline{\mathbf{x}(t)} = \mathbf{G}(t) \mathbf{x}_0 \, . \tag{12.17}$$

So now it's reasonably clear why the noise is left out of $\mathbf{G}$. Note that the noise does two things: 1) it determines the unconditioned probability distribution over $x_0$ and $v_0$, and 2) it leads to randomness in $x(t)$ and $v(t)$. Therefore, for given values of $x_0$ and $v_0$ and taking the average of the conditional time dependent trajectories, the noise plays no role. In other words, we're really just solving $$\dot{\mathbf{x}} + \mathbf{\Gamma} \mathbf{x} = 0 $$ for which (12.17) is obviously the solution. This is annoyingly poorly explained by the book, but at least it's discernibly correct.

Then the author writes

The next step is to define the covariance matrix $$\mathbf{\sigma}(t) = 2 \int_0^t \mathbf{G}(t') \mathbf{D} \mathbf{G}^T (t') \, dt'$$ where $\mathbf{G}^T$ denotes the transpose of the matrix $\mathbf{G}$.

I have no idea where this equation came from. Why is this integral expression the correct one for the covariance?


1 Answer 1


I apologize when I translate this into the stochastic differential equation background I am more familiar with. Relating white noise to a standard Wiener process $W_t$, $$ W_t=\frac{1}{\sqrt{2m\gamma k_bT}}\int_0^t\eta(s)\,ds, $$ we can write the Langevin equation as the system of one ODE and one SDE: \begin{align} dx_t&=(v_t-x_t)\,dt\\ dv_t&=-\gamma\,v_t\,dt-\frac{\sqrt{2m\gamma k_bT}}{m}\,dW_t\,. \end{align} Using $D=k_bT/\gamma m$ the SDE can be written as $$ dv_t=-\gamma\,v_t\,dt-\gamma\sqrt{2D}\,dW_t\,. $$ Using Ito's lemma we can verify that this SDE has the explicit solution $$ v_t=e^{-\gamma t}\,v_0-\gamma\sqrt{2D}\,\,e^{-\gamma t}\int_0^te^{\gamma u}\,dW_u\,. $$ The covariance function of the Gaussian process $v_t$ is \begin{align} {\rm Cov}[v_t,v_s]&=2\gamma^2D\,e^{-\gamma(t+s)}\int_0^{\min(t,s)}e^{2\gamma u}\,du=\gamma D\,e^{-\gamma(t+s)}\Big(e^{2\gamma\min(t,s)}-1\Big)\\ &=\gamma D\Big(e^{-\gamma|t-s|}-e^{-\gamma(t+s)}\Big)\,. \end{align} This is not exactly how the stationary $\sigma(t)$ seems to look like. The reason for this could be that there are two different definitions of Ornstein-Uhlenbeck process.

  • the stationary one does not follow a Langevin SDE

  • the one that does is not stationary.

They are however closely related. See this MSE post.

  • $\begingroup$ Where does $dx_t = (v_t - x_t)dt$ come from? $\endgroup$
    – DanielSank
    Apr 21 at 17:29
  • $\begingroup$ The first component of the vector $\dot{\mathbf{x}}+\mathbf{\Gamma}\mathbf{x}$ is zero according to the Langevin equation. Rewriting this with the components $x$ and $v$ of $\mathbf{x}$ gives $dx_t=(v_t-x_t)dt$. $\endgroup$
    – Kurt G.
    Apr 21 at 21:12
  • $\begingroup$ Oh right, duh. Thank you. $\endgroup$
    – DanielSank
    Apr 21 at 22:16
  • $\begingroup$ I would like to understand how the treatment given in this answer relates to the integral $\sigma(t) = \int_0^t G(t') D G(t')^T \, dt'$ given in the question. Can you give me a hint? Near the end of the answer there is an expression for the covariance in terms of an integral... where does that come from? $\endgroup$
    – DanielSank
    Jun 26 at 16:41
  • $\begingroup$ I believe when you study the linked MSE post in detail you will understand more. In short: there are TWO definitions of OU process, leading to different expressions for $\operatorname{Cov}[v_t,v_s]$, namely eq. (1) and (6) in the linked post. It might be redundant to translate all this into your notation. $\endgroup$
    – Kurt G.
    Jun 27 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.