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A well-mixed layer of depth $h$ losing heat through an unstable cold upper boundary has bulk excess temperature $\theta =T-T_0$. With four-thirds convective heat-transfer law flux $q=Ck[g\beta /(\nu \kappa ){]}^{1/3}{\theta }^{4/3}$, its heat capacity per area is $kh/\kappa$. Thus $\dot{\theta }=-K{\theta }^{4/3}$ with $K=C\kappa [g\beta /(\nu \kappa ){]}^{1/3}/h$, giving the displayed cooling law. It excludes the initial conductive transient and the late low-Rayleigh conduction regime.

The Rayleigh number compares destabilizing thermal buoyancy with viscous and thermal diffusion:The conductive heat flux is $k\Delta T/h$, and the Nusselt number is the actual upward heat flux divided by this value. The four-thirds convective heat-transfer law is the empirical large-Rayleigh closure $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$. It givesThe layer depth cancels: thin thermal boundary layers control the resistance while convection mixes the interior. The dimensionless coefficient $C$ depends on the boundary and material-property regime; the two boundary-layer temperature drops can be absorbed into its convention. This is not a universal law for all Rayleigh-Bénard convection. Below convective onset the stable conductive solution instead has $\mathrm{Nu}=1$, and an impulsive initial conductive transient is not described by the steady large-Rayleigh closure.

Cooling from above makes dense cold fluid overlie lighter warm fluid, so the upper boundary layer is unstable and can mix the interior. Assume high Rayleigh number, rapid interior mixing, constant properties, negligible wall heat capacity and a quasi-steady cooling flux obeying the four-thirds convective heat-transfer law. Write $\theta (t)=T_{\mathrm{bulk}}(t)-T_0$ and use a coefficient $C$ appropriate to this single cooling boundary. The volumetric heat capacity is $\rho c_p=k/\kappa$, so the area-wise heat balance isWith $K=C\kappa [g\beta /(\nu \kappa ){]}^{1/3}/h$, integration gives convective cooling under the four-thirds heat-transfer law:The power-law decay is a bulk approximation between the initial boundary transient and the later stage when the decreasing temperature contrast no longer maintains strong convection. It is not an exact spatially uniform solution satisfying the cold boundary itself.

Cooling from below is different: dense cold fluid now lies below warm fluid and stabilizes the layer. With no other imposed motion, cooling is by thermal conduction, and temperature becomes depth dependent. Let $z=0$ be the cooled bottom and $z=h$ the insulating top. Solve ${\theta }_t=\kappa {\theta }_{zz}$ with $\theta (0,t)=0$, ${\theta }_z(h,t)=0$ and $\theta (z,0)=\Delta T$. The mixed-boundary eigenfunctions have ${\lambda }_n=(n+1/2)\pi /h$, givingIts volume-averaged temperature isThe late conductive decay is exponential on the timescale $4h^2/({\pi }^2\kappa )$, rather than the convective $t^{-3}$ bulk law. This assumes the usual positive thermal expansion coefficient and no destabilizing compositional stratification.

Write $\beta$ for the thermal expansion coefficient, $\nu$ for kinematic viscosity, $\kappa$ for thermal diffusivity, and $k_T=\rho c_p\kappa$ for thermal conductivity. For a temperature difference $\Delta >0$ across depth $h$,The four-thirds convective heat-transfer law posits $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ at high Rayleigh number, and thereforeThe layer depth cancels. A useful physical justification is a mixed interior with thin conductive boundary layers of thickness $\delta$, each maintained near a fixed critical local Rayleigh number: $g\beta {\Delta }_b{\delta }^3/(\nu \kappa )\sim {\mathrm{Ra}}_{\mathrm{crit}}$ and $q\sim k_T{\Delta }_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one boundary layer changes the coefficient, not the exponent.

Assume Boussinesq approximation, constant material properties, a laterally extensive statistically turbulent Rayleigh-Bénard convection layer, negligible side-wall heat transfer, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified Prandtl number regime when choosing $C$. The closure is empirical and regime dependent; high Rayleigh number alone does not make it exact or extend it to the impulsive conductive initial transient.

Let $\overline{T}=T_0+\vartheta (t)$, with $0\le \vartheta \le \Delta T$. Apply the same one-boundary four-thirds convective heat-transfer law coefficient $K$ to both boundary layers. The lower inward heat flux is $K(\Delta T-\vartheta {)}^{4/3}$ and the upper outward heat flux is $K{\vartheta }^{4/3}$. The layer's conservation of energy per unit area givesHere $\rho c_p$ is the volumetric heat capacity. This uses a uniform bulk temperature, negligible boundary layer heat storage, constant depth and material properties, prescribed boundary temperatures, no lateral losses or internal heat generation, and identical upper and lower closure coefficients. It deliberately extrapolates the turbulent closure to $t=0$, as the model requires; an actual impulsive start first has a conductive transient. This is the requested evolution equation, without solving it.