What is the corresponding Fokker-Planck equation for,
$\frac{df(t)}{dt}=-kf(t)+\zeta(t)$
where, $\zeta(t)$ is random noise? In particular, how will the Fokker-Planck equation look if $\zeta(t)$ is exponentially correlated (or coloured, see this answer) instead of delta-correlated?
Thanks for the nice problem. First, let us scale the variables $f \rightarrow x$, $\zeta \rightarrow \sqrt{\Gamma} \, \xi$, and consider the Langevin equation $$ \dot x = -k \, x + \sqrt{\Gamma} \, \xi, \qquad (1) $$ where $\xi$ is a colored noise satisfying $\langle \xi(t) \, \xi(t') \rangle = e^{-\gamma \, |t-t'|}$.
We wish to show that the Fokker-Planck equation for $P(x, t)$ is given by $$ \frac{ \partial P } { \partial t } = \frac{ \partial } { \partial x } ( k \, x P) + \frac{ 1 - e^{-(k+\gamma) t} } {k + \gamma} \Gamma \frac{ \partial^2 P } { \partial x^2 }. \qquad (2) $$
The idea van Kampen (page 243) is to treat $\xi$ as a Ornstein-Uhlenbeck process (i.e., the position of a harmonic oscillator under a white-noise Brownian dynamics). Then, the extended equations of motion are $$ \begin{aligned} \dot x &= -k \, x + \sqrt{\Gamma} \, \xi, \\ \dot \xi &= -\gamma \, \xi + \sqrt{2 \gamma} \, d W(t), \end{aligned} \qquad (3) $$ where $W(t)$ is the Wiener process (so $dW(t)$ is a white noise).
The distribution $P(x, \xi, t)$ for Eq. (3) satisfies $$ \frac{ \partial P(x, \xi, t) } { \partial t } = \frac{ \partial } { \partial x } \left[ \left(k \, x - \sqrt{\Gamma} \, \xi \right) \, P(x, \xi, t) \right] + \frac{ \partial } { \partial \xi } \left[ (\gamma \, \xi ) \, P(x, \xi, t) \right] + \gamma \frac{ \partial^2 } { \partial \xi^2} P(x, \xi, t). $$ Since this is a linear Fokker-Planck equation, the solution is a joint Gaussian distribution for $x$ and $\xi$. Further, the deterministic part for $x$ is unaffected by the colored noise $\xi$. So the marginal distribution of $P(x, t)$ must adopt the form of $$ \frac{ \partial P(x, t) } { \partial t } = \frac{ \partial } { \partial x } \left[ k \, x \, P(x, t) \right] + B(t) \frac{ \partial^2 } { \partial x^2} P(x, t), \qquad (4) $$ where $B(t)$ is a function to be determined (which would be a constant if the noise were white).
By multiplying Eq. (4) with $x^2$ and integrating, we get $$ \begin{aligned} \frac{ \partial \langle x^2 \rangle } { \partial t } &= - 2 \, k \, \langle x^2 \rangle + 2 \, B(t). \qquad (5) \end{aligned} $$
On the other hand, the explicit solution of Eq. (1) is (assuming $x(t = 0) = 0$ for simplicity), $$ x = \sqrt{\Gamma} \int_0^t \xi(\tau) \, e^{-k(t-\tau)} \, d\tau. \qquad (6) $$ By multiplying Eq. (1) by $2 \, x$, and averaging, we get $$ \begin{aligned} \frac{ d \langle x^2 \rangle }{ dt } &= - 2 \, k \langle x^2 \rangle + 2 \sqrt{\Gamma} \langle x(t) \, \xi(t) \rangle \\ &= - 2 \, k \langle x^2 \rangle + 2 \, \Gamma \int_0^t \langle \xi(\tau) \xi(t) \rangle \, e^{-k(t-\tau)} \, d\tau \\ &= - 2 \, k \langle x^2 \rangle + 2 \, \Gamma \int_0^t e^{-\gamma (t-\tau)} \, e^{-k(t-\tau)} \, d\tau \\ &= - 2 \, k \langle x^2 \rangle + 2 \, \frac{ \Gamma } { \gamma + k } \left( 1 - e^{-(\gamma + k) \,t} \right). \end{aligned} $$ where we have used (5) on the second line. Comparing this to (5), we get $$ B(t) = \frac{ \Gamma } { \gamma + k } \left( 1 - e^{-(\gamma + k) \,t} \right). $$ So the Fokker-Planck equation for $P(x, t)$ is given by Eq. (2).
This example is instructive, because with a non-white noise, the magnitude $B(t)$ depends on the noise itself, but also the deterministic force (through $k$).
