# Ito diffusion: Connection between backward Kolmogorov equation and stationary distribution

Suppose we have an Ito diffusion $$dX_t = b(X_t)dt + \sigma(X_t) dB_t, \tag{1}$$ where $$dB_t$$ is Brownian motion. Also assume we know that this diffusion process converges to a stationary distribution $$\pi$$. I am interested in the quantity $$u_{\infty}(x) = \lim_{t \rightarrow \infty} \mathbb E[\psi(X_t)] = \mathbb E_\pi[\psi(X)],$$ for some function $$\psi : \mathbb R \rightarrow \mathbb R$$; assuming the expectation exists.

I am wondering if there is a way to connect this quantity with the backward Kolmogorov equation, i.e. define $$u(x,t) = \mathbb E^x[\psi(X_t)]$$; then the backward Kolmogorov equation says that $$u$$ solves \begin{align} \frac{\partial u}{\partial t} &= \mathcal A u, \tag{2}\\ u(0,x) &= \psi(x) \end{align} where $$\mathcal A$$ is the infitesimal generator for (1).

To me it seems like $$u_\infty(x)$$ should be the stationary solution of (2), so we might be able to set $$\frac{\partial u}{\partial t} = 0$$ to reduce things to an ODE. But I can't wrap my mind around the boundary conditions that should be in place for this to work, or if I've missed an obvious reason for why this wouldn't work.

The stationary distribution is a time-invariant solution of the Fokker--Planck (forward) equation, which is probably an easier way of looking at it, as it comes out as an exponential-integral (of $$b(x)/\sigma^2(x)$$).