Let $f\geq 0$ be Lipschtiz. The overdamped Langevin equation
\begin{equation}\label{eq overdamped Langevin SDE} dX=-\nabla f(X)dt+\sqrt{2} dW_t \end{equation} with Kolmogorov forward equation \begin{equation} ~~~~~~~~~~~~ \partial_t \rho = \text{div} (\rho \nabla f)+\Delta \rho \tag 1 \end{equation} has associated free energy \begin{equation} F(\rho)=\int f \rho + \int \rho \log \rho. \end{equation}
It is easy to see that \begin{equation} \rho^\infty(x) := \frac{e^{-f(x)}}{\int e^{-f(x)}dx}, \end{equation}
is a stationary distribution of the overdamped dynamics, since
\begin{align*} \int e^{-f(x)}dx \big( \text{div}(\rho^\infty \nabla f )+ \Delta \rho^\infty \big) &= \text{div}(e^{-f} \nabla f)+ \text{div}\nabla e^{-f} \\ &= \text{div} (e^{-f} \nabla f)+ \text{div}(- \nabla f e^{-f}) = 0 \end{align*}
My question is
Consider $(1)$, coupled with the initial condition that $\rho(0,x)=\rho_0(x)$. Under what conditions on the initial distribution $\rho_0$ and on the function $f$ do we need for the solution $\rho$ to actually converge (in some norm like $L^1, L^2$ or $W_2$ ) to the stationary distribution $\rho^\infty$? Moreover it is obvious that $\rho^\infty$ is a minimiser of the free energy $F$?
EDIT:
- Here we are considering dynamics in $\mathbb{R}^d$, where $W_t$ is a d-dimensional Brownian motion and $W_2$ is the 2-Wasserstein norm.
- The link here kind of answers my question.