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Consider the Langevin equation ($N$-dimensional) with nonlinear drift term but expressible as a gradient of a function $U(\vec{x})$. Namely, consider the stochastic process described by the set of equations:

$\frac{\partial x_n}{\partial t} = -\frac{\partial}{\partial x_n} U(\vec{x}) + \sqrt{2c} \eta_n$

the problem can be reformulated in terms of the probability distribution $P(\vec{x},t)$, through the following fokker-planck equation:

$\frac{\partial P(\vec{x},t)}{\partial t} = \bigg( \sum_{i=1}^N \frac{\partial}{\partial x_i} \big( \frac{\partial}{\partial x_i}U(\vec{x})\big) + c \sum_{i,j=1}^N \frac{\partial^2}{\partial x_i \partial x_j} \bigg) P(\vec{x},t)$

The equation above admits the following stationary solution:

$P^s(\vec{x}) = \mathcal{N} e^{\frac{-U(\vec{x})}{c}}$

Is there a simple way to convince yourself that, in this case, given any initial distribution I always converge only to above $P^s(\vec{x})$?

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    $\begingroup$ Without suitable assumptions on $U$, there might be no convergence to a stationary solution, e.g., $U \equiv 1$. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Jun 6, 2022 at 20:35

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You can directly verify this by writing the right hand side of your Fokker-Planck equation as $\nabla\cdot\left(P\left(\nabla U + c\nabla \ln P\right)\right)$ where $\nabla$ denotes the Euclidean gradient w.r.t. vector $x$. Now setting the stationarity condition $\frac{\partial P^{s}}{\partial t} = 0$ gives you divergence of $P^{s}\nabla(U + c\ln P^{s}) = 0$, for which to hold for all $x$, we require $\nabla(U + c\ln P^{s}) = 0$, i.e., $U + c\ln P^{s} = \ln k$ for some constant $k$. This simplifies to $P^{s}\propto \exp(-U/c)$ upto a normalization constant.

Also relevant: Intuition/elegant reason for why Langevin diffusion converges to $\exp(-U)$?

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