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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:

  1. 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.
  2. The link here kind of answers my question.
Matthew Cassell
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  • Remind me, if you don’t mind, what is the $W_2$ norm again? – Nap D. Lover Jul 16 '21 at 13:30
  • @NapD.Lover Ill edit the question. – orange is the new f Jul 16 '21 at 13:31
  • Sorry all, this kind of answers things : https://www.cedricvillani.org/sites/dev/files/old_images/2012/07/024.CMcV_.pdf – orange is the new f Jul 16 '21 at 14:53
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    Based off your second edit: Might I suggest summarizing your findings, then, into an answer to your own question. This is certainly encouraged and then would give anyone searching this a nice summary before diving deeper. – Nap D. Lover Jul 16 '21 at 15:39
  • I feel like there is a pretty big difference bewteen the paper you shared and the OP. In (1.1) of the paper they have a deterministic equation (in terms of a probability measure sure but still deterministic) but yours has a $W_t$ term in it. Unless (1.1) can be written in a form that's more similar to the OP that i'm not aware of. – Math_Day Dec 28 '23 at 04:51

1 Answers1

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The convergence of the Langevin diffusion is typically studied in total variation. The most recent result in this direction (in fact, a two-sided bound) in terms of $f$-ergodicity (in particular including the total variation norm) is studied in Bresar & Mijatovic in "Subexponential lower bounds for f-ergodic Markov processes". Under some mild regularity assumptions (mostly concerned with the growth rate of $f$ at infinity), the convergence may be polynomial or exponential (this case was studied by Roberts & Tweedie), see Example 3.5, and Remarks 3.7 and 3.10 (in their notation $\pi$ corresponds to your $\rho^\infty$, also note that they analyse more general processes, not just the Langevin diffusion).

If you're interested in Kullback-Leibler, then I'd suggest this reference instead. Finally, a nice summary of some of these results (and implications to the $L^2$-Wasserstein distance and some divergences), you can look at this reference.

In particular, to answer your question more directly: oftentimes the necessary conditions for convergence are mostly concerned with mild regularity of the stationary law and some control on its behaviour at infinity, so that the process will converge for any fixed initial position (and thus for its mixtures, i.e., any initial distribution). For other results, you require the initial distribution to be absolutely continuous with respect to the stationary law and sometimes the corresponding Radon-Nikodym derivative is required to be $L^2$-integrable with respect to the stationary law, see Lemma 1 in this reference.