# Can this two-dimensional process self intersect?

I would like to know more about the two-dimensional processes derived from Brownian motion by the following stochastic differential equation (in the Ito sense)

$$dX_t = f(X_t) dt + \mathcal{R}(f(X_t)) dB_t$$

where

• $X_t$ is the two-dimensional stochastic process
• $f$ is a smooth vector field
• $\mathcal{R}$ is the linear map which rotates a vector through a quarter turn
• $B_t$ is a standard one-dimensional Brownian motion

So $X_t$ follows the field lines of $f$ apart from a lateral "shake" whose intensity is proportional to the strength of the field $f$ at $X_t$.

In particular I am curious about what is known when $f$ is conservative (i.e. $f = \nabla g$ for some real-valued field $g$). In this case, can the trajectory of $X_t$ self intersect?

References to further properties of this process would also be appreciated.

In general, it can self-interesect, as can be seen in the example of the simple rotation $$f(x,y) = (-y,x)\;,\qquad \mathcal{R} f(x,y) = (x,y)\;.$$ In this case, the deterministic solution turns around the origin at unit speed and there is no reason to believe that there won't be any self-intersection for $t > 2\pi$.
In the case where $f = \nabla g$ for some globally defined smooth function $g$ you cannot have self-intersections except in the trivial case when you start on a critical point for $g$, simply because $g$ is strictly increasing along solutions. (Assuming that your stochastic differential is interpreted in the Stratonovich sense.)
• I'm more interested in the Ito sense. Does that still imply $g$ strictly increases? It seems unlikely to me. For example it seems that the process should be able to "cross over" from one side of a ridge to another. But even if $g$ is not strictly increasing, can we can get non-self intersection from other considerations? – Tom Ellis Feb 10 '14 at 18:10