= Solution
For <flux-driven convective layer cooling>, use $E=C\theta^{4/3}$ with the same unspecified coefficient as in (b). The steady <heat> balance gives
$$
\boxed{\overline T_*=T_s+\theta_*,\qquad
\theta_*=(F/C)^{3/4}=\left(\frac{F}{k}\right)^{3/4}\left(\frac{R_c\eta\kappa}{\rho_0g\alpha}\right)^{1/4}.}
$$
The corresponding boundary-layer thickness is
$$
\delta_*=\frac{k\theta_*}{F}=\left(\frac{kR_c\eta\kappa}{\rho_0g\alpha F}\right)^{1/4}.
$$
Set $\theta=\theta_*+\epsilon$ and expand the <heat> loss to first order. Since $E'(\theta_*)=(4/3)C\theta_*^{1/3}=4F/(3\theta_*)$,
$$
\frac{kd}{\kappa}\dot\epsilon=-\frac{4F}{3\theta_*}\epsilon,
\qquad
\boxed{\epsilon(t)=\epsilon(0)e^{-t/\tau},\quad
\tau=\frac{3kd\theta_*}{4\kappa F}=\frac{3d\delta_*}{4\kappa}.}
$$
This is the \i[near-equilibrium] <relaxation time>. The full equation is nonlinear, so an arbitrary large initial <temperature> departure need not decay as one exponential. Positive <heat> loss <derivative> makes the <steady state> stable. In particular, $\tau$ scales with $d\delta_*$, rather than $d^2$, because <convection> shortens the effective heat-transport distance to the <boundary layer>.
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