I have the following brownian model:

$$ \dot{x}=v_0cos(\theta(t))+\sqrt{2D_t}\xi_x(t) \\ \dot{y}=v_0sin(\theta(t))+\sqrt{2D_t}\xi_y(t) \\ \dot{\theta}=\sqrt{2D_r}\xi_\theta(t)\\ $$

with $v_0$ constant and $\xi_\theta$ is a white noise, this is: $<\xi_\theta>=0\ and <\xi_i(t)\xi_j(s)>=\delta(t-s)\delta_{ij}$

Now, I want to solve the following integral:

$$ \int_0^t dt \int_0^t ds <cos(\theta(t)+\theta(s))> $$

Unfortunately, I do not know how to solve it. If instead of a sum there is a subtraction inside the cos, I know I can do the following:

$$ \eta=\theta(t)-\theta(s) \\ <cos(\theta(t)-\theta(s))>=<Re(e^\eta)>=Re<e^\eta>=e^{<\eta^2>}\\ <\eta^2>=<(\theta(t)-\theta(s))^2>=2D_rt+2D_rt-2\int dt\int ds<2D_r\xi_\theta(t)\xi_\theta(s)>=2D_rt+2D_rt-2Dr(s\ or\ t) $$ where the last "s or t" means that if you are integrating with t>s then it is an s and if it is s>t then it is a t. If I try to solve the same here, I can see how the solution is not compatible with my simulations results. Any idea on how to solve this? What is wrong if I try to replicate the sum scenario with the subtraction one?

Basically, instead of calculating the MSD ($<x(t)^2+y(t)^2>$), I am calculating $<x(t)^2-y(t)^2>$.

  • $\begingroup$ Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true? $\endgroup$
    – Thomas
    Jan 31, 2019 at 14:48
  • $\begingroup$ Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed. $\endgroup$ Jan 31, 2019 at 14:50
  • $\begingroup$ Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this? $\endgroup$
    – Thomas
    Jan 31, 2019 at 17:08
  • $\begingroup$ One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ? $\endgroup$
    – Thomas
    Feb 1, 2019 at 10:40
  • $\begingroup$ theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian? $\endgroup$ Feb 1, 2019 at 12:50


You must log in to answer this question.