What is the relation between Langevin and Fokker planck equation? What is 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 will look like if $\zeta(t)$ is exponentially correlated instead of delta correlated?
Nitin
 A: 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
$$
\frac{ \partial P } { \partial t } = \frac{ \partial } { \partial x } (  k \, x P)
+
\frac{ 1 - e^{-(k+\gamma) t} } {k + \gamma} \Gamma
\frac{ \partial^2 P } { \partial x^2 }.
\qquad (2)
$$
Solution is a Gaussian
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).
Determining $B(t)$
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).
Discussion
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 $k$).
