I am puzzled by the following, but I have a feeling that this is standard stuff.

Suppose that the position $x$ and the velocity $v$ of a particle is described by $$ \dot{x} = v \\ \dot{v} = -f(x) v + \xi_1(t), $$ where $\xi_1$ is white noise and $ f $ is a given function of the particle's position.

Then under what assumptions (if any) can I say that the position $ x$ is described by

$$ \dot{x} = -g(x) + \xi_2(t),$$

where $\xi_2$ is again white noise (but potentially with different magnitude)? Also, what is the relationship between $f$ and $g$?

Please let me know if I need (how) to make my question more accurate.

  • $\begingroup$ Since $f(x)$ represents some sort of damping, it must be positive. I should think that the jitter amplitude ${{\xi }_{2}}(t)$ will be position-dependent, $\tilde{\ }1/f(x)$. The low-frequency components of truly white noise ${{\xi }_{1}}(t)$ will make $x(t)$ take walks of infinite variance characterized by so-called “1/f” noise (where f means frequency). You may need to apply a lower frequency cutoff if this seems unrealistic. $\endgroup$ – Bert Barrois Apr 2 '18 at 11:32
  • $\begingroup$ I think having f(x)v complicates things a bit. What you'd typically have is a v dependent term (for friction) and an x dependent term (for potential), but not a term that mixes them. If you need the mixing term, I guess you could try e.g. writing out the Fokker-Planck equation and integrate out the v terms or I guess you could try something like Mori-Zwanzig, but not sure either'll work. In general I don't think x'll end up having white noise: e.g. a damped harmonic oscillator I think should have exponential correation in time in the noise if you wrote it in terms of x. $\endgroup$ – alarge Apr 2 '18 at 12:58
  • $\begingroup$ @adam Upon further reflection ... If f is positive and roughly constant, then $\left\langle {{v}^{2}} \right\rangle \to C/f$, once the initial velocity has decayed. As for $\dot{x}(t)=v(t)=\zeta (t)$, note that it is no longer white noise but rather low-pass filtered Gaussian noise. If f varies, then $x(t)$ will wander off into regions with $f(x(t))\ne f({{x}_{0}})$, with the baleful consequence that neither $v(t)$ nor $x(t)$ will be a stationary random process. $\endgroup$ – Bert Barrois Apr 4 '18 at 14:17

Your Answer

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

Browse other questions tagged or ask your own question.