Let $U:\mathbb R^d \to \mathbb R$ be a continuously differentiable function such that $Z_U:= \int_{\mathbb R^d}e^{-U(x)}dx < \infty$ and consider the Gibbs distribution $\pi(dx) := Z_U^{-1}e^{-U(x)}dx$. Consider the diffusion equation generated by the gradient of $U$, namely $$ dX(t) = -\nabla U(X(t))dt + \sqrt{2}dB_t^d, $$
where $B_t^d$ is Brownian motion on $\mathbb R^d$. It can be shown that if
A1. $U$ is strongly-convex,
A2. $U$ is $L$-smooth, that is $\|\nabla U(x)-\nabla U(y)\| \le L\|x-y\|$ for all $x,y \in \mathbb R^d$,
then $\pi$ is the unique unique stationary distribution of the above diffusion, i.e the unique invariant distribution of the Markov semigroup $(P_t)_{t \in \mathbb R}$ which acts on bounded measurable functions $f: \mathbb R^d \to \mathbb R$ by $P_tf := \mathbb E[f(X(t)) \mid X(0) = x]$.
Question. What can be said if the strong-convexity assumption A1 is removed ?
Convexity of $U$ is irrelevant for convergence to the stationary measure. All that is needed is coercivity: you need to make sure $|U(x)| \rightarrow \infty$ fast enough as $|x| \rightarrow \infty$.
You have a Markov process that is irreducible in a suitable sense, and you can guess the invariant measure by checking that the density $p(x) = C\exp(-U(x))$ satisfies the invariance equation, $$\mathcal{L}^* p = \frac{\partial}{\partial x}\left(\frac{\partial U}{\partial x}p(x)\right) + \frac{\partial^2}{\partial x^2} p(x) = 0,$$ where $\mathcal{L}^*$ is the adjoint of the infinitesimal generator of your process. By standard Markov process theory, the process $X_t$ converges in distribution to your $\pi(dx)$.
For a proof in precisely this setup, see Theorem 2.1 of http://www2.stat.duke.edu/~scs/Courses/Stat376/Papers/ConvergeRates/GeneralState/RobertsTweedie1996.pdf .