= Solution
Let $\overline T=T_0+\vartheta(t)$, with $0\leq\vartheta\leq\Delta T$. 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 gives
$$
\boxed{\rho c_ph\frac{d\vartheta}{dt}=K\left[(\Delta T-\vartheta)^{4/3}-\vartheta^{4/3}\right],\qquad \vartheta(0)=0}.
$$
Here $\rho c_p$ 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.
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