= Solution
The exact mixture <equation of state> is $\rho=(1-\phi)\rho_f+\phi\rho_p$. Keeping first-order terms in the small <particle volume fraction> and <thermal expansion> gives
$$
\boxed{\rho=\rho_0(1+\gamma\phi-\theta)}.
$$
The omitted term is $\rho_0\phi\theta$. The downward solid-volume <particle deposition flux> is $J_v=W_s\phi$; the corresponding particle <mass flux> is $J_m=\rho_pW_s\phi$. Uniform vertical mixing and constant layer depth imply
$$
H\dot\phi=-W_s\phi,\qquad \dot\theta=\beta\phi.
$$
For $W_s>0$, set $r=W_s/H$. Then
$$
\boxed{\phi(t)=\phi_0e^{-rt},\qquad
\theta(t)=\frac{\beta\phi_0}{r}(1-e^{-rt})},
$$
and hence
$$
\frac{\rho(t)-\rho_0}{\rho_0}
=\phi_0\left[\left(\gamma+\frac\beta r\right)e^{-rt}-\frac\beta r\right].
$$
The <heated particle-laden layer> loses its <stable density stratification> when this contrast vanishes. For $\beta>0$, solving for the neutral-buoyancy time gives
$$
\boxed{T_s=\frac H{W_s}\log\left(1+\frac{\gamma W_s}{\beta H}\right)}.
$$
\b[The layer is neutral at $T_s$ and statically unstable for $t>T_s$.] The subsequent uniformly mixed lower-layer solution cannot represent the resulting overturning.
If $W_s=0$, then $\phi=\phi_0$, $\theta=\beta\phi_0t$, and the continuous limit is $T_s=\gamma/\beta$. If $\beta=0$, the layer stays denser than its surroundings at every finite time and becomes neutral only asymptotically when $W_s>0$; thus $T_s=\infty$. These limiting cases must replace the printed formula when its denominator vanishes.
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