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

• What does "simple" mean in your book? ;) Usually, a good starting point for almost anything about the Fokker-Planck equation is the book by Risken. In chapter 6.1 he gives a rather technical answer to your question, which I don't think would qualify as simple. Commented Jun 6, 2022 at 17:33
• Your notation of the Fokker-Planck equation is incorrect. On the RHS you need something like $-\nabla \cdot \left(\nabla U P \right) + c \Delta P$. Commented Jun 6, 2022 at 17:36
• Hi @kricheli, thank you for the correction; by simple I mean a proof that gives you immediatly the intuition behind the details Commented Jun 6, 2022 at 18:28
• The eigenvalues of the differential operator in the rhs are all negative (except the zero eigenvalue, corresponding to the stationary solution). I believe there is a proof in Risken's book. Commented Jun 7, 2022 at 7:55
• Commented Apr 30 at 12:34

Okay, so to give some detail to Roger Vadim's comment (a sketch, for full detail cf. Risken... ;) ): $$\partial_t P = L P = \sum_i \partial_i \left(-\partial_i U P + c \partial_i P\right)$$ (I don't follow the convention of letting differential operators acting on everything to their right) has a negative semidefinite operator on the right-hand side. I.e. all eigenvalues of $$L$$ have non-positive real part. And the zero eigenvalue corresponds to the equilibrium/stationary solution.
For a solution decomposed into eigenfunctions of $$L$$, $$P = \sum_i c_i f_i\,,$$ with time-dependent coefficients $$c_i$$ and eigenvalues $$\lambda_i$$, i.e. $$L f_i = \lambda_i f_i\,,$$ we find that $$c_i\left(t\right) = c_{i,0} \text{e}^{\lambda_i t}\,.$$ Now, since all eigenvalues are negative, the exponentials get smaller and smaller at large times, except for the coefficient of the stationary/equilibrium solution, since it is the eigenfunction to the eigenvalue zero.
To see that eigenvalues of $$L$$ have negative (rather, non-positive) real part, we need the fact that $$\Delta$$ is a negative semidefinite operator. You see this easily through its Fourier transform $$-k^2$$. More generally, in case of a non-isotropic diffusion term we would instead need a positive (semi-) definite matrix to form the operator $$\sum_{i,j}\partial_i \left(D_{ij} \partial_j P\right)$$.
Another nice bit of intuition - nothing directly to do with the above - is: If the drift term vanishes, you see easily that the solution approaches the constant equilibrium solution. The Laplace operator on the right sort of smoothens out bumps in the solution: If you have a maximum of $$P$$ somewhere rising above the stationary solution, because it is a maximum there is $$\Delta P < 0$$ at this point and thus $$P$$ at this point is decreasing in time.
• Thank you for the clear answer! :) The only passage not clear to me is how to claim that the "0" eigeinvalue is not degenerate, i.e. there is a unique eigenfunction associated to the equilibrium state. Also, the fact that $c_i(t) = c_i(0) e^{\lambda_i t}$ is due to the fact that {$f_i$} is an orthogonal basis, right? Commented Jun 7, 2022 at 19:36