 Solution

Initially $a=1/2$ and $T=T_{+}/T_0>1$, so $\epsilon T_t=1-T^4<0$. On the fast time $s=t/\epsilon$, albedo changes by only $O(\epsilon )$ andThus temperature rapidly falls towards $1$. At the same time $a_t=a^{\ast }(T)-1/2<0$ while $T>1$, which seeds selection of the warm branch. Afterwards the approximate slow manifold is $T^4=2(1-a)$: decreasing albedo makes temperature rise slowly back towards $T_{+}/T_0$. The preceding figure separates these two time scales; a single time axis would compress the initial fall into a nearly vertical segment.

In the sharp-transition regime, after selection of the warm branch $a^{\ast }(T)\simeq 1/4$. Solving the linear ordinary differential equation with the leading matched value $a(0^{+})=1/2$ givesThese are outer solutions: their initial temperature is $1$, so the fast cooling layer is required to satisfy the actual initial temperature.

There is a useful qualification to the two-limit argument. For fixed finite $\mu >2$, put $x=1-2a$ on the slow manifold. Then $x_t=-x+\frac{1}{2}\tanh (\mu x)$, and $x=0$ is an unstable equilibrium with growth rate $\mu /2-1$. The small positive $O(\epsilon )$ seed from the fast layer can therefore produce a delay of order $\log (1/\epsilon )/(\mu /2-1)$. The simple exponential approximation assumes that the sharp albedo transition is sufficiently fast to make this delay negligible on the displayed slow scale. It is not a uniform approximation for every fixed $\mu >2$ as $\epsilon \to 0$.