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 $<x(t)^2-y(t)^2>$.