Long time behavior of noise process with flat spectral density and upper cutoff frequency

Question

From this question we find that given a noisy process $x(t)$ with known spectral density $S_x(\omega)$, the mean square displacement is $$ \langle x(t)^2 \rangle = t^2 \int_0^\infty S_\dot{x}(\omega) \left( \frac{\sin(\omega t / 2)}{\omega t / 2} \right)^2 \frac{d \omega}{2\pi} \, . \tag{1} $$ where $S_\dot{x}$ is the spectral density of the velocity of $x$. Suppose $$ S_x(\omega)= \begin{cases} S \quad \text{for }\omega \leq \Omega \\ 0 \quad \text{otherwise} \end{cases} $$ In that case, and assuming that $S_\dot{x} = \omega^2 S_x$, we find $$ \langle x(t)^2 \rangle = 2 S \Omega \left( 1 - \frac{\sin(\Omega t)}{\Omega t} \right) \, . \tag{2} $$ I don't like that in the limit $t \rightarrow \infty$ it seems that $\langle x(t)^2 \rangle$ goes to a constant. Intuitively, I would think that if we allow $x$ to wander around for a long time, it should drift increasingly far away from the starting point.

One of the following must be true:

1. Eq. (1) is incorrect.
2. I made a mistake doing the integral and expression (2) is incorrect.
3. My assumption that $S_\dot{x}(\omega) = \omega^2 S_x(\omega)$ is incorrect.
4. Everything was done correctly and it's true that the mean displacement goes to a constant in the long time limit.

Which is it?

Answer 1

It looks fine to me. 2. and 3. are surely correct. 1 seems good too from your link and works well for a Wiener process where you indeed find that from your formula that MSD$\sim t$.

Concerning your example, it's pretty far from a regular Brownian motion. If we have the following Brownian motion, with $\eta$ a Gaussian random noise:

$$\dot x(t) = \eta(t)\tag 1$$

We obtain: $S_{x}(w) = S_\eta(w)/w^2$. In your case, this would imply that the noise has a spectrum:

$$S_\eta(w) = S\Theta(\Omega-|w|)w^2 \tag 2$$

And thus an auto-correlation (for $t>0$):

$$S_{\eta}(t) = S\dfrac{4t\Omega \cos(t\Omega)+2((t\Omega)^2-2)\sin(t\Omega)}{t^3} \tag 3$$

I'm sure the double integral required to obtain the MSD from (3) can be done, but I'm lazy so I just asked python to do it numerically (see edit). And here is what we obtain ($S = 1$ and $\Omega = 1$):

So your formula is indeed right! As a rule of thumb, any system driven by a noise with power law autocorrelation will display highly non diffusive behavior (usually you obtain sub or super diffusion, not "no diffusion at all"). Here the finite MSD, I think is due to the fact that your noise can easily be anticorrelated which implies that if you start to explore some region, it is highly likely that in the future you will just go back to where you were before.

Edit

Just for completeness, I'll do the integral (which in any case is already done in your linked post):

$$MSD(t) = \int_0^t\int_0^tS_\eta(|t'-t''|)dt'dt''.$$

By symmetry:

$$MSD(t) = 2\int_0^tS_\eta(u)(t-u)du=2S\Omega(1-\text{sinc}(\Omega t))$$

as expected.