 Solution

A comoving wavenumber crosses the inflationary Hubble scale at $k=aH$. Consequently $d\ln k=d\ln a+d\ln H\simeq Hdt$ when $H$ changes slowly. Combining with the slow-roll approximation givesBoth $k$ and $\varphi$ increase along the rolling branch. If $N$ denotes the remaining number of e-folds rather than elapsed expansion, then $d\ln k\simeq -dN$. More precisely $d\ln k=(1-{ϵ}_H)d\ln a$; the correction is beyond the constant-Hubble approximation used here.