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There is no minimum about which this scalar can undergo the damped oscillations normally used for reheating. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.

At the assumed exit, $V_R=m_{\mathrm{pl}}^4{e}^{-30}$. Equating this to the thermal radiation density givesfor $g_{\ast }=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.

Curvature obeys $|\Omega -1|\propto (aH{)}^{-2}$. For constant ${ϵ}_H=A^2/(16\pi )$ and $N=\log (a_R/a_i)$, its suppression during cosmic inflation is $e^{-2(1-{ϵ}_H)N}$. Starting from order-one curvature and requiring present curvature no larger than order one givesAfter instantaneous reheating, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{\ast s,0}/g_{\ast s,R}{)}^{1/3}$, with $H_R\simeq 1.66\sqrt{g_{\ast }}{T}_R^2/m_{\mathrm{pl}}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present curvature tolerance adds its corresponding half-logarithm. In slow-roll approximation this is the conventional requirement of roughly sixty e-folds.

Because the field increases toward ${\varphi }_R$, enough expansion requiresThis is the flatness bound on the initial field in exponential inflation. It is an upper bound on the initial field, not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\mathrm{pl}}^4$, then ${\varphi }_i\ge 0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.