Let $$dX_t=(a+bX_t)dt+dW_t,$$ where $a,b\in\mathbb R$ and $X_0=0$ a.s. I read on wikipedia that contrary to the Brownian motion, $(X_t)$ has a stationnary solution. So, I guess that $\partial _tp(x,t)=0$ where $p(x,t)$ is the density of $X$ at time $t$. I know that such a distribution has to satisfy fokker planck equation, i.e. $$\partial _tp(x,t)=-\partial _x((a+bx)p(x,t))+\partial _{xx}p(x,t),\tag{E}$$ where $p(0,t)=\delta _0$.
Q1) How can I prove that (E) has a stationary solution ?
Q2) I know that the solution of such an equation is unique. So I don't really understand how it can have either a stationary solution and a non stationary solution. Maybe Ornstein Uhlenbeck process has always stationary distribution ?
By stationary you define, I guess you mean to show $p(x, t) = p(x)$?
Ornstein–Uhlenbeck process is a Gaussian process, which has a Gaussian probability density. Thus you can show its mean and covariance function do not depend on $t$.
You can verify that the mean and covariance are Wiki
$$ \mathbb{E}[X_t] = X_0 \, e^{bt},\\ \mathrm{cov}[X_t, X_t] = -\frac{1}{2b}, $$
provided that your $b<0$, $a=0$, and you start from initial stationary condition $X_0\sim N(0, -\frac{1}{2b})$.