= Thermal erosion of a two-layer interface
{title2=$h^4=h_0^4-4C^3\ln(1-Pt/D_0)/(\alpha P)$}
Suppose identical <buoyant thermals> deliver buoyancy $B_0$ at rate $f$ to a mixed lower layer of area $A$ before penetrating its interface. Neglect source volume and heat loss. Relative to the upper-layer density, let $D=g h(\rho_L-\rho_2)/\rho_{\rm ref}$, $P=fB_0/A$, and $C=\sqrt{B_0/(2K)}$, where $K=4\pi\alpha^3/3$. The lower-layer buoyancy budget gives $D=D_0-Pt$. A cycle-averaged <fluid entrainment> closure $\dot h/u^*=\mathrm{Ri}^{-1}$, with $u^*=C/h$, $b^*=\alpha h$, yields the displayed law. Neutrality of the thermal at the interface requires $D(t^*)h(t^*)^2=B_0/K$. The closure must incorporate the encounter duty cycle; an isolated-event entrainment law needs an additional averaging prescription.
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