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

  • $\begingroup$ 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. $\endgroup$
    – kricheli
    Commented Jun 6, 2022 at 17:33
  • $\begingroup$ Your notation of the Fokker-Planck equation is incorrect. On the RHS you need something like $-\nabla \cdot \left(\nabla U P \right) + c \Delta P$. $\endgroup$
    – kricheli
    Commented Jun 6, 2022 at 17:36
  • $\begingroup$ Hi @kricheli, thank you for the correction; by simple I mean a proof that gives you immediatly the intuition behind the details $\endgroup$ Commented Jun 6, 2022 at 18:28
  • $\begingroup$ 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. $\endgroup$
    – Roger V.
    Commented Jun 7, 2022 at 7:55
  • $\begingroup$ Related physics.stackexchange.com/q/357033/226902 $\endgroup$
    – Quillo
    Commented Apr 30 at 12:34

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jun 7, 2022 at 19:36
  • $\begingroup$ The latter: exactly. The non-degeneracy: math.stackexchange.com/questions/4426438/… $\endgroup$
    – kricheli
    Commented Jun 7, 2022 at 20:27

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