convergence divergence – Langevin equation and stationary solutions.

Let $fgeq 0$ 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~~~~~~~~~~~~~~~~~~(1)
end{equation}

has associated free energy
begin{equation}
F(mu)=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 $rho^infty$.