4.4 “End of Inflation” Mechanism 73 T ∝ a − 1 and H ∝ a − $^{2}$. Therefore, if Eq. (4.108) is satisfied, then equivalent constraints for higher T and H are guaranteed to be satisfied as well. In the case of Γ$_{ϕ}$ ≳ H$_{TI}$ , by substituting Eqs. (4.8), (4.10), (4.11) and (4.92) into Eq. (4.108) we obtain the constraint Rearranging this for m$_{0}$ gives the following. For α =1 and any value of n , or for α =2 and n =1 we have For all other α and n combinations, the above inequality is reversed, giving In the case of Γ$_{ϕ}$ ≪ H$_{TI}$ , by substituting Eqs. (4.10), (4.12) and (4.92) into Eq. (4.108) we obtain the constraint Rearranging this for m$_{0}$ gives the following. For the values of n and α given in Table 4.1 we have h ≪ 1 λ 2 α$_{−}$ 1 4 n +4 ( m$_{0}$ M$_{P}$ ) (2 α$_{−}$ 1)( n +2) 2 n +2 ( $_{100}$√$_{H}$$_{∗}$$_{M}$$_{P}$ m$_{0}$ ) 2 α − 2 (4.109) m$_{0}$ ≫  $^{(10}$$^{4}$$^{H}$ ∗ ) (2 α − 2)( n +1) h 2 n $^{+2}$λ 2 α$_{−}$ 1 2 M 2 α + n P   1 2 αn$_{−}$ 3 n$_{−}$ 2 (4.110) m$_{0}$ ≪  $^{(10}$$^{4}$$^{H}$ ∗ ) (2 α − 2)( n +1) h 2 n $^{+2}$λ 2 α$_{−}$ 1 2 M 2 α + n P   1 2 αn$_{−}$ 3 n$_{−}$ 2 (4.111) h ≪ ( m n +2 0 √$_{λM}$$_{P}$ ) 6 α$_{−}$ 3 4 n +4 1 ( M$_{P}$ √ Γ$_{ϕ}$ ) 2 α$_{−}$ 1 2 ( $_{100}$√$_{H}$$_{∗}$$_{M}$$_{P}$ m$_{0}$ ) 2 α − 2 (4.112) m$_{0}$ ≫    (10 $^{$^{4}$H}$∗ M$_{P}$ ) (4 α − 4)( n +1) h 4 n +4 ( $^{√$_{λM}$$_{P}$}$) 6 α − $^{3}$( M 2 P Γ$_{ϕ}$) (2 α − 1)( n $_{+1)}$   1 2 αn$_{−}$ 4 α$_{−}$ 5 n$_{−}$ 2 (4.113)