Consider the over damped Langevin dynamics: $d X_{t} = d B_{t} - \nabla U(X_{t}) dt $ on $\mathbb{R}^{d}$ where $B_t$ is a standard Brownian motion. On pages 29 and 30 of the following book
Royer, Gilles, An introduction to logarithmic Sobolev inequalities, Cours Spécialisés (Paris). 5. Paris: Société Mathématique de France. 114 p. (1999). ZBL0927.60006.
Royer, Gilles, An initiation to logarithmic Sobolev inequalities. Transl. from the French by Donald Babbitt, SMF/AMS Texts and Monographs 14; Cours Spécialisés (Paris) 5. Providence, RI: American Mathematical Society (AMS); Paris: Société Mathématique de France (ISBN 978-0-8218-4401-4/pbk). viii, 119 p. (2007). ZBL1138.60007.
the author says that if
- $U$ is $C^{2}$,
- the corresponding SDE doesn't explode in finite time almost surely, and
- $\exp(-U)$ is integrable on $\mathbb{R}^{d}$.
then the corresponding Boltzmann-Gibbs measure defined by $d\mu(x)\propto \exp(-U(x)) dx$ is reversible for the process (and consequently the stationary distribution for the process).
Now, consider $U$ which doesn't satisfy the third assumption (integrability). Then, according to the theory above, we can't deduce stationarity of the Boltzmann-Gibbs measure written above.
In this context, my question is:
What can we say now about the stationary distribution of the process? Can this SDE have a stationary distribution of some other form (i.e. other than the Boltzmann-Gibbs)? OR does the SDE have no stationary distribution at all (implying that if the SDE has a stationary distribution then it has to be of the Boltzmann-Gibbs form)?
I would really appreciate some help here because I am confused.
