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!
 A: 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. 
