0
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
    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
    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 $\endgroup$ 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.
    Jun 7, 2022 at 7:55

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Jun 7, 2022 at 19:36
  • $\begingroup$ The latter: exactly. The non-degeneracy: math.stackexchange.com/questions/4426438/… $\endgroup$
    – kricheli
    Jun 7, 2022 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.