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If we have an inflation model with potential $V(\phi) = V_0 e^{-\sqrt{\frac{2}{\lambda}} \frac{\phi}{M_p}}$, where $V_0$ and $\lambda$ are free parameters, does this lead to eternal inflation for $\lambda > 1$?
The slow roll parameter $\epsilon_V(\phi) = \frac{M_p^2}{2} (\frac{V_{'\phi}}{V})^2 = \frac{1}{\lambda}$ appears to be a constant and so for all $\lambda > 1$, $\epsilon_V(\phi) < 1$. This seems to imply that inflation never breaks down.
Yes, inflation does not end for single field inflation driven by a purely exponential potential. Either one interprets this form of the potential as approximating a different potential when observational scales exit the horizon, or, if taken to be exact, one must introduce some mechanism to end inflation. Non-canonical kinetic terms have been investigated for this purpose. Another thought would be to introduce a second, auxiliary field similar to hybrid inflation.
