I am working on Langevin equations:


$m\frac{du}{dt}= -\gamma u + \theta(t)$

where $\theta(t)$ is delta-correlated in time Gauss-distributed noise with zero-mean

$\langle \theta (t) \rangle = 0 $

$\langle \theta (t) \theta (t') \rangle = 2 D \delta(t-t') $

and $\gamma=const$ and $D=const$ are constant coefficients.

I can find the probability of $x$ (position) and probability of $u$ (velocity) given initial position $x_0$ and velocity $u_0$ (see for example [1]):

$F(x,t|u_0,x_0)=\frac{1}{Z_F}\exp\Big[\frac{m \gamma}{2D} \frac{(x-x_0-u_0 (1- e^{-\gamma t})/\gamma)^2}{2\gamma t - 3 + 4 e^{-\gamma t} - e^{-2\gamma t}} \Big]$

$G(u,t|u_0,x_0)=\frac{1}{Z_G}\exp\Big[\frac{m \gamma}{2D} \frac{(u-u_0 e^{-\gamma t})^2}{1 - e^{-2\gamma t}} \Big]$

where $Z_F$ and $Z_G$ are some normalization coeffitients.

But I am missing a joint-probability of $x$ and $u$, meaning a probability for particle to have simultaneously position $x$ and velocity $u$ given $x_0$ and $u_0$ or $P(x,u,t|u_0,x_0)$. Any help to derive $P(x,u,t|u_0,x_0)$ exactly or approximately would be highly appreciated.

[1] - http://dx.doi.org/10.1103/PhysRev.36.823, G. E. Uhlenbeck and L. S. Ornstein, 1930

  • 2
    $\begingroup$ This problem has an explicit pathwise solution. You have $m u'+\gamma u = 2D dW_t$, which you rewrite as $m \frac{d}{dt} \left ( e^{\gamma t/m} u \right ) = 2D e^{\gamma t/m} dW_t$. Now you integrate to get $m e^{\gamma t/m} u(t) - m u(0) = \int_0^t 2D e^{\gamma s/m} dW_s$. Now you can treat the stochastic integral through integration by parts, obtaining $2D e^{\gamma t/m} W_t - \int_0^t 2D (\gamma/m) e^{\gamma s/m} W_s ds$. Finally $x(t)=x(0)+\int_0^t u(s) ds$ of course. You can in principle read off the statistical properties from this explicit solution. $\endgroup$
    – Ian
    May 30, 2016 at 21:04
  • $\begingroup$ Thank you! I will try this way. $\endgroup$
    – baraban55
    May 30, 2016 at 21:07


Your Answer

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