# 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 $$rho^infty$$.