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I am studying some Langevin type dynamics and have an equation of the following form:

$$m\dot v= -\gamma v \ +\xi(t) + F_0\cos(\omega t) \\ \langle\xi_0\xi_t\rangle =2D\ \delta(t)$$

And I am asked to find the evolution for the cumulants. I am familiar with the cumulants as being generated by $G(\lambda)=\ln \int dx \ f(x)\ e^{\lambda x}$ and then $\kappa_n= d^nG/d\lambda^n$ at $\lambda=0$

So where I get stuck at is how to get $f(x)$, which would actually be $f(v)$ for this case. I know I need some kind of Fokker-Planck equation from this, but I am still a bit uncertain as to where that comes from (and if I need it in this scenario).

Once I know $f$, then I just need to differentiate $G$ then the cumulants are still the nth derivatives.

Jean Marie
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yankeefan11
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1 Answers1

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The is an Ornstein-Uhlebeck process with time dependent drift and you can solve this linear ODE with inhomogenous forcing term formally as (i'm going to do it with unit mass but it is straight forward to adapt to the general case) $$ \begin{align} v(t) = e^{-\gamma t} v_0 + \int_0^t e^{-\gamma(t-\tau)} \xi(\tau) \operatorname{d}\tau + F_0 \int_0^te^{-\gamma(t- \tau)} \cos (\omega \tau)\operatorname{d}\tau, \end{align} $$ this is just a Gaussian random variable - all of the stochastic behaviour coming from $\int_0^t e^{-\gamma(t-\tau)}\xi(\tau)\operatorname{d\tau}$ which being a linear transformation of a Gaussian random variable is itself a Gaussian random variable - there are existing questions justifying this fact.

Now the moment generating function of a Gaussian random variable is known to be $$ M(\vartheta)=e^{\mu_t \vartheta + \frac{\vartheta^2\sigma^2_t }{2}}, $$ where $$ \mu_t := \mathbb{E}\left[v(t)\right], \qquad \sigma_t^2 :=\mathbb{E}\left[ (v(t)-\mu_t)^2\right], $$ where the subscript indicates the time dependence of these values. Taking the logarithm of this we get the cumulant generating function and we also see that it terminates after the second order term.

It is straight forward to calculate $\mu_t$ and $\sigma_t^2$ from the explicit solution above, and this is what makes the OU process so tractable in comparison with the those cases where the Gaussian force acts multiplicatively with the state or more general nonlinear diffusions. Doing this we get $$ \begin{align} \mu_t &= \mathbb{E}\left[e^{-\gamma t}v_0 + \int_0^t e^{-\gamma(t-\tau)} \xi(\tau) \operatorname{d}\tau + F_0 \int_0^te^{-\gamma(t- \tau)} \cos (\omega \tau)\operatorname{d}\tau\right] \\ &= e^{-\gamma t}v_0 + F_0 \int_0^te^{-\gamma(t- \tau)} \cos (\omega \tau)\operatorname{d}\tau + \mathbb{E}\left[ \int_0^t e^{-\gamma(t-\tau)} \xi(\tau) \operatorname{d}\tau \right] \\ &= e^{-\gamma t}v_0 + F_0 \int_0^te^{-\gamma(t- \tau)} \cos (\omega \tau)\operatorname{d}\tau \end{align} $$ and also to calculate $\sigma_t^2$ we have the more general result that the covariance is given by $$ \begin{align} \operatorname{Cov}\left\{\int_0^s e^{-\gamma(t-s)}\operatorname{d}\sigma, \int_0^te^{-\gamma(t-\tau)}\operatorname{d}\tau\right\} &= \int_0^s \int_0^te^{-\gamma(t+s - (\tau + \sigma)}\langle \xi(\tau)\xi(\sigma)\rangle \operatorname{d}\tau \operatorname{d}\sigma \\ &= \frac{D}{\gamma}\left(e^{-\gamma|t-s|}-e^{-\gamma(t+s)} \right), \end{align} $$ and in particular $$ \sigma_t^2 = \frac{D}{\gamma}\left(1-e^{-2\gamma t} \right). $$

Nadiels
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  • So even though we solve for v, this will allow for solving for the moments. Not sure how to use this to find the moment generating function since I need the pdf of v. Actually, thinking about this since I have exp(lambda v) for the generating function I can expand that so that G=ln(1++...). Is that the idea? – yankeefan11 Mar 24 '18 at 14:42
  • You get the moments from the claim that $v(t)$ us Gaussian with known mean and variance. I will expand on this later though if it isn't clear enough – Nadiels Mar 24 '18 at 15:37
  • I can kind of see why it is gaussian, but it would have a time dependent mean? Are the moments of a gaussian equal to the cumulants? – yankeefan11 Mar 24 '18 at 15:51
  • Thanks for the recent edit! That helps a lot! My remaining question is why we can say it is a gaussian random variable with the time dependent drift. Is that just an assumption? Does the additional drive not really matter since it is non stochastic? – yankeefan11 Mar 24 '18 at 17:58
  • It is a Gaussian random variable because the "random part" is all down to the integral $\int_0^t e^{-\gamma(t-\tau)}\xi(\tau)\operatorname{d}\tau$ - all of the rest, the $\cos(\omega t)$ and the initial condition term are deterministic - when looking at the variable $v(t)$ they are exactly analogous to adding a deterministic constant to a Gaussian, which is again a Gaussian with shifted mean. – Nadiels Mar 24 '18 at 18:28
  • Finally take a look at this question for the claim that the integral of a Gaussian is again a Gaussian – Nadiels Mar 24 '18 at 18:32
  • So I am trying to see how $sigma_t$ end up being that. IF I take $<p^2>$, I get stuff that looks like $2p_0e^{-2\gamma t}\int_0^t F_0 \cos(\omega t1)e^{\gamma t1}\ dt_1$ (one of the cross terms) and another that is that integral squared. So I fail to see where D/gamma comes from. – yankeefan11 Mar 26 '18 at 18:15
  • You won't get any cross terms if you calculate the variance, i.e. $\langle (v(t) - \langle v(t)\rangle)^2 \rangle$ because the deterministic terms, those involving the initial condition and $\cos$ etc. will vanish leaving only a double integral involving the $\delta$ function and the exponential terms. This part would be well covered in a standard reference for the OU process - if you still aren't getting it then I can edit my answer or you can ask a new question covering exactly the parts you are unsure of else this comment section will get very cluttered. – Nadiels Mar 26 '18 at 22:11