Let $f$ be a $\mathcal{C}^1$ vector field on a compact subset $M \subset \mathbb{R}^n$. We define a dynamical system by $$\dot{x}(t)=f(x(t))$$
In ergodic theory, the occupation measure is $$\mu_{x, T}(B)=\frac{1}{T} \int_{0}^{T} \mathbb{1}_{B}(x(s)) d s$$ where $B$ is a Borel-measurable set and $\mathbb{1}_{B}$ is its indicator function and $x$ is a trajectory.
We say that a measure $\mu$ is physical if for Lebesgue almost every $x \in M$, as $T \rightarrow \infty$, the occupation measure $\mu_{x, T}$ converges weakly to $\mu$.
We can think from a PDE-viewpoint of the Frobenius-Perron operator $S(t)g=p(\cdot |t)$ as "the solution of" $$\left\{\begin{array}{ll}p_{t}+\operatorname{div}(f p)=0 & x \in \mathbb{R}^{d}, t>0 \\ p(x \mid 0)=g & x \in \mathbb{R}^{d}\end{array}\right.$$ I.e. $S$ evolves the initial density $g$ by some time $t$. An invariant measure (its density) solves $\operatorname{div}(fp)=0$. A physical measure is invariant. But generally, a DS can have many invariant measures.
We assume that the physical measure can be written as density $\mu = p(x)dx$ with the Lebesgue-measure $dx$.
My question is the following: How can we find the physical measure by solving $\operatorname{div}(fp)=0$? (I.e. which boundary values do we need to provide on $p$ in order to get the physical measure, or what are generally the conditions such that the solution of $\operatorname{div}(fp)=0$ is a physical measure (its density))
