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

user1172131
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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.

kricheli
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