Assume we have a particle in $\mathbb{R}^3$, which we will subject to different fields independently. It will have some potential energy $U\in \mathbb{R}$ defined as some constant minus its kinetic energy.
Given a conservative vector field, the potential energy $U$ is given by some function $g:\mathbb{R}^3\to \mathbb{R}$ due to the path-independence of $$\int_a^b\mathbf{F}\cdot d\mathbf{r}.$$
However, given a non-conservative vector field that induces an acceleration $\mathbf{F}\times\mathbf{v}$, it is not immediately obvious what the potential energy function will be given the co-ordinates*. It feels like the (speed and therefore) potential energy will depend strongly on the 'history' of the particle (i.e. how it got to that co-ordinate as well as the co-ordinate itself). It thus seems like the potential energy function $f:S\to \mathbb{R}$ will have its domain $S$ being some sort of manifold.
Is this true, and if it is or I'm not too far off, could someone help in elucidating the matter?
For simplicity, take a non-conservative field $F$ whose divergence is $0$ (e.g. $\mathbf{B}$). For a very simple example, take $\mathbf{F}=(y)\hat{\mathbf{x}}$.
*Obviously it's trivial to give it as a function of velocity, so assume throughout that we're looking for an $f$ .