# Langevin equation with position-dependant damping: existence of an invariant measure?

The usual Langevin equation for a particle in a 1D harmonic potential

$dq(t) = p(t)~dt$

$dp(t) = -q(t)~dt + a ~dW(t) - b~p(t)~dt$

admits as an invariant measure the Gibbs measure ${1\over Z}\exp(-{2b\over a^2}{q^2+p^2\over 2})$. (We assume here $a, b > 0$, and that $W$ is a Brownian motion.)

Now, assume that the damping coefficient $b$ depends on $q$. More precisely, let $b: {\mathbb R} \to [c, d]$ with $0 < c < d < \infty$ be a (smooth) function, and consider the SDE

$dq(t) = p(t)~dt$

$dp(t) = -q(t)~dt + a ~dW(t) - b(q(t))~p(t)~dt$

It seems natural that, since the damping term is bounded below, there exists an invariant probability measure. Is it true? If yes, what is the proof, or what tools should I use to prove it?

For example, if $b(q) = 10 + \sin(q)$, I expect the system to be at least as "well-behaved" as the usual Langevin equation with $b = 9$. But even for this simple example, I fail to find a proof.

Thanks a lot!

• Perhaps look at search results for "nonlinear Langevin equation" – Steve Huntsman Apr 25 '13 at 14:49
• Thanks for the suggestion. I just looked it up, and it does not seem - as far as I can tell - to cover the case described in the question... – Nown Apr 25 '13 at 17:10

Here $q$ is the position and $p$ is the velocity. Let me take $a = \sqrt{2}$ and define $H(q,p) = \frac{q^2+p^2}{2}$.

case 1 - b is a constant.

We agree on the fact that the generator $L$ of $(q(t),p(t))$ is

$L = p \frac{\partial}{\partial q} - (q+bp)\frac{\partial}{\partial p} + \frac{\partial^2}{\partial p^2}$.

Therefore the adjoint operator of $L$, namely $L^\star$, is

$L^\star = -p \frac{\partial}{\partial q} + \frac{\partial}{\partial p} \left ( (q+bp) . \right ) + \frac{\partial^2 .}{\partial p^2}$

and applying $L^\star$ to the function

$m_1(q,p) = \exp(-b H(q,p))$,

we see that $L^\star m_1 = 0$.

case 2 - b is depending on the position q. When $a$ is constant it is not clear if there is an explicit invariant measure.

However, if the function $a$ is also position dependant (satisfying a certain relation with $b$) then your SDE corresponds to Equation (2.39) page 89 of "Free energy computations: A mathematical perspective, 472 pp., Imperial College Press, 2010. of Lelievre, Rousset,Stoltz.

• Thanks for the answer. I agree with what is said, but I don't want to assume that $a$ and $b$ satisfy this relation, since in my case $a$ is constant...But I believe I have found the answer to the question, which I asked two years ago: one can prove it by using a Lyapunov function $V = p^2 + q^2 + k p q$ for some appropriate constant $k$... – Nown Apr 3 '15 at 6:45
• Does your Lyapunov function give uniqueness of the invariant measure? – megaproba Apr 3 '15 at 13:49
• Not directly, but it is easy to show that the system is irreducible, so uniqueness is not really the issue... – Nown Apr 3 '15 at 17:59
• Is your dynamics time reversible? – megaproba Apr 15 '15 at 21:44