I consider a one-dimensional SDE $$ \mathrm d X_t = F(X_t) \mathrm d t + \sigma(X_t)\mathrm d B_t $$ where $\mathbf P^{X_0} = \mu$ is the starting distribution, $F, \sigma: \mathbf R \to \mathbf R$. Let us assume that $X_t$ satisfies this SDE and is a timeinvariant Markov process. The main part of this question is:
Can one determine under which conditions on $F$ and $\sigma$ there exists a $\mu$ such that $\mathbf P^{X_t} = \mu$ for all $t \geq 0$?
In addition I am also interested in determining $\mu$ (or rather: its density $\pi$ w.r.t. Lebesgue measure on $\mathbf R$) from $F$ and $\sigma$. If I understand correctly this density would have to satisfy the following second-order differential equation given by the adjoint of the generator of $X_t$, i.e.
$$ 0 = (\frac 1 2\sigma^2(x) \pi(x))'' - (F(x)\pi(x))' $$ so I speculate that, given $\sigma$ and $F$, we can just solve this differential equation and ensure that $\pi$ is positive and integrates to $1$.
If we consdier for example the Ornstein-Uhlenbeck process $$ \mathrm d X_t = -\alpha X_t \mathrm dt + \sigma dB_t $$ then we can easily guess and verify that $\mu = \mathcal N(0, \frac{\sigma^2}{2 \alpha})$ is such an invariant distribution and that its density fulfils above differential equation. However I have not been able to come up with another stationary Markov process based on playing around with the functions $F$ and $\sigma$ so examples of these would also be of interest to me.