# Fokker-Planck: uniqueness and convergence to stationary distribution

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 $$P^s(\vec{x})$$?

• Without suitable assumptions on $U$, there might be no convergence to a stationary solution, e.g., $U \equiv 1$. Jun 6, 2022 at 20:35

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.