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Consider the continuous-time Markov process on ${\mathbb R}^n$ described by the SDE

$\dot{x}(t) = F(x(t)) + \xi(t)$

where $F:{\mathbb R}^n \to {\mathbb R}^n$ is a smooth mapping, and $\xi(t)$ is a white noise acting on some of the variables.

My question is really about Markov processes, but it turns out that if somebody has an answer for the case where $\xi(t) \equiv 0$ (simple deterministic dynamical system) it might be of help, too.

Let us assume that $F$ is such that the SDE above admits a unique invariant probability measure. One might if necessary assume that this invariant measure has an everywhere-positive smooth density which decays exponentially fast at infinity.

Now, consider the SDE

$\dot{y}(t) = F(y(t)) - \alpha\ y_n(t)\ \hat{e_n} + \xi(t)$

with $\alpha > 0$, and $\hat{e_n}$ the $n$th vector of the canonical basis of ${\mathbb R}^n$. This SDE is the same as before, except for the addition of a damping term on the variable $y_n$.

The question is: can we say in general that there exists an invariant probability measure for the "damped" SDE?

In a sense, it seems intuitive that the damping term only increases the stability of the system. Thus, at least for some class of SDEs, the answer should be yes. (Of course, one might lose uniqueness however.)

One can also state the problem in terms of generators. The generator of the semigroup for the damped Markov process is formally

$L = L_0 - \alpha\ y_n \partial_{y_n}$

where $L_0$ is the generator for the original, undamped semigroup. Its formal adjoint (which acts on densities of measures : $\partial_t \rho = L^* \rho$) is formally

$L^* = L_0^* + \alpha\ ( 1 + y_n \partial_{y_n})$.

In these terms, what I would like to show is that if the kernel of $L^*_0$ is non-empty, then so is that of $L^*$.

Any hint or reference would be most appreciated... Thanks a lot.

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I wouldn't be surprised if things could break. Imagine a random walk on a graph induced by nearest integer neighbors in a dumbbell-shaped region of the plane. Suppose that the walk is uniform except on the bar, where it's heavily biased in one direction. Convergence to the invariant measure will be very slow because one of the ends of the dumbbell will take a long time to reach. With damping, we might be able to break ergodicity. – Steve Huntsman Mar 28 '13 at 14:33
In the deterministic case, it is clear that in many cases, adding damping simply reduces the invariant measure to a point measure. E.g. a simple damped pendulum. – Piyush Grover Mar 28 '13 at 14:56
Thank you both for your comments. The dumbbell example indeed shows that convergence to the invariant measure – if there is one – is an issue. But for now, I am merely interested in existence of an invariant measure. Secondly, the damping term may indeed make the system collapse to a lower-dimensional attractor (for example a point), but I don't mind. – Nown Mar 28 '13 at 15:13

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