 Solution

During the addition interval, subtracting the formulas in (ii) and (iii) cancels the common unforced temperature:This is zero precisely when $e^{-a(t-1)}=1/(1+a)$, soThe inequalities $0<\log (1+a)<a$ give $1<t^{\ast }<2$, so using the active-interval formula was justified. The difference has derivative $(1+a)e^{-a(t-1)}>0$, making this crossing unique within that interval. Immediately after time one it is minus one; after time two it is strictly positive by (iv). Hence there is no other equality at any finite time after time one. Both temperatures approach the same ambient limit, but that limiting equality is not an additional crossing.