$1+w_{\phi}$ becomes ${\cal O}(\epsilon)$ as usual. Then, the energy density of the field decays slowly unless the potential becomes steep again at a later stage.

In contrast, if the potential is steep at the nucleation point, i.e. $\epsilon_{*}\geq 1$, the situation is different. As one can see from Eq. , $1+w_{\phi}$ becomes ${\cal O}(1)$ at $t\approx \epsilon_{*}^{-1/2}H_*^{-1}$ before the potential starts to dominate. The energy density of the field starts to decay rapidly proportional to $a_{\rm R}^{-3(1+w_{\phi})}$. During this rapid-roll phase, $1+w_{\phi}$ is estimated as $$\begin{aligned}
1+w_{\phi}={\cal O}\left(\epsilon\frac{\kappa V}{H^2}\right)\,.\end{aligned}$$ where we have used $\dot{\phi}={\cal O}(V'/H)$. One can see that the rapid-roll phase lasts as far as $\epsilon\gg H^2/\kappa V(\gtrsim 1)$. When the field falls down to a plateau of the potential, the velocity of the field eventually decreases. Then, the slow-roll inflation starts and the curvature term becomes completely irrelevant for the background expansion of the universe. We denote the Hubble parameter at the onset of this slow-roll phase by $H_{\rm R}$.

An interesting feature of the dynamics for a steep slope is “tracking”, which means that the ratios among the kinetic term, the potential term and the curvature term are approximately constant. Suppose $\epsilon (\phi )$ is a constant 