Quartic-potential slow-roll solution ($\varphi ={\varphi }_i{e}^{-M_P\sqrt{\lambda /(6\pi )}(t-t_i)}$)

For $V(\varphi )=\lambda {\varphi }^4/4$ with $\lambda >0$, use natural units and the unreduced Planck mass $M_P=G^{-1/2}$. The slow-roll equations give $H\simeq \sqrt{2\pi \lambda /3}{\varphi }^2/M_P$ and $\dot{\varphi }\simeq -M_P\sqrt{\lambda /(6\pi )}\varphi$. Integrating gives the displayed exponential field decay, valid on either sign branch, andThe potential slow-roll parameters require ${\varphi }^2\gg M_P^2$. The expression is valid only during slow-roll inflation; its formal late-time saturation does not describe the subsequent reheating or radiation era.

With unreduced Planck mass $M_P$ and initial potential-dominated density $\lambda {\varphi }_i^4/4\sim M_P^4$, one has ${\varphi }_i^2\sim 2M_P^2/\sqrt{\lambda }$. The slow-roll e-fold count is $N\simeq \pi ({\varphi }_i^2-{\varphi }_e^2)/M_P^2$. Thus small positive $\lambda$ produces a very large expansion factor $e^N$ in this classical quartic model. The order-one endpoint correction is negligible compared with $2\pi /\sqrt{\lambda }$; initial kinetic energy, spatial curvature and the model's validity at trans-Planckian field values are independent assumptions.

For $V=\lambda {\varphi }^4/4$ and unreduced Planck mass $M_P$, the potential slow-roll parameter is ${ϵ}_V=M_P^2/(\pi {\varphi }^2)$. Setting it to one gives the displayed conventional endpoint. The exact end of scalar-dominated accelerating expansion is ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Since slow-roll is failing near that point, a precise endpoint requires evolving the full inflaton equation of motion; potential and kinetic criteria evaluated with a failing approximation need not give the same order-one coefficient.