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Let us consider an $\mathbb{R}^d$ diffusion

$$dX_t = dW_t +\mu(X_t)dt.$$

Let further $D\subset \mathbb{R}^d$ be a bounded connected open domain. By $Y^D$ we denote the diffusion $X$ restricted to $D$ (reflected on the boundary or conditioned to stay inside). Then $Y^D$ converges exponentially fast to its stationary distribution, denote the ratio of the convergence by $r^{D,\mu}>0$.

Question: Is it true that there exists $r^{D}>0$ such that for any $\mu$ we have $r^{D,\mu}\geq r^D$.

My impression is that essentially the worse case is $\mu \equiv 0$ and any drift would only help. (I am not interested in pathologies e.g. I am happy to assume some smoothness if necessary.)

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    $\begingroup$ If $\mu$ is not bounded by an a-priori constant then certainly what you want fails. Think of $\mu$ creating two metastable equilibria in $D$ (that is, assume the ODE $dx/dt=\mu(x)$ has two stable points inside $D$), and that the situation is symmetric wrt these two points. Then convergence to equilibrium becomes slower and slower when $\mu$ increases. $\endgroup$ Mar 6, 2015 at 22:15
  • $\begingroup$ You are right. I am happy to assume that $\mu$ is bounded. I guess my question above might be better stated vaguely as follows: "Is there a simple lower-bound for the rate of convergence. By simple I understand something which involves as "little information" about the geometry of the $D$ and "little information" about $\mu$." Probably such an estimate cannot be very sharp but it is a price I am willing to pay :). $\endgroup$ Mar 7, 2015 at 12:27

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To answer the rephrased question: is there a simple lower-bound for the rate of convergence? Yes: assuming certain conditions hold, which I describe below.

For any $t \ge 0$, let $\Pi_t(x, \cdot)$ denote the transition probability of the process $X_t$ defined as: $$ \Pi_t(x,A) = \mathbb{P}(X_t \in A ~\mid~ X_0= x ) $$ for any measurable set $A \subset D$. It is sufficient to establish a Doeblin condition, which is not impractical to obtain since the domain of the diffusion is bounded. This condition requires that there exist $t \ge 0$, a constant $\alpha \in (0,1)$ and a probability measure $\nu$ on $D$ such that $$ \inf_{x \in D} \Pi_t(x, A) \ge \alpha \nu(A) \;. $$ for all measurable sets $A \subset D$. If this Doeblin condition holds, then one can apply a coupling inequality to show that the rate of convergence of $\Pi_t(x,\cdot)$ to equilibrium is determined by the parameter $\alpha$ appearing in this bound.

The usual way to guarantee that this condition holds is to show that $X_t$ is irreducible and its associated semigroup is strongly Feller, see e.g. Section 4 of Hairer 2001. To prove that a diffusion on $\mathbb{R}^d$ is irreducible and its associated semigroup is strongly Feller, it is sufficient to assume that the drift field $\mu$ of the diffusion is locally Lipschitz continuous with at most polynomial growth at infinity, see Chapter 2 of Second Order PDE's in Finite and Infinite Dimension. S. Cerrai. Springer, 2001.

On a bounded domain, and depending on the assumed boundary conditions, other assumptions may be needed to prove irreducibility and the strong Feller property, including a regularity assumption on the boundary, e.g., that the boundary is twice differentiable; for precise conditions see Chapter 4 of Linear and Quasilinear Equations of Parabolic Type. Ladyzensakaja et al. American Mathematical Society, 1967 or other textbooks that discuss existence and properties of classical solutions to linear PDEs of parabolic type on bounded domains.

Note that, in contrast to diffusions on $\mathbb{R}^d$, for bounded diffusions one does not need to assume that the drift is dissipative. In fact, it suffices to assume enough regularity on $\mu$ and $\partial D$ such that the above Doeblin condition holds.

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