 Solution

Let $T_0=T_i-T_{\infty }>0$, $S_0=S_i-S_{\infty }>0$ and $R_0=\beta S_0/(\alpha T_0)$. At steady volume, the weir outflow equals the inflow $Q$. The well-mixed heat and salt balances giveWith $\theta =(T_i-T)/T_0$ and $\eta =(S_i-S)/S_0$, the flux ratio implies ${\eta }_s=q{\theta }_s$, where $q=R_F/R_0$. The actual interface ratios are $\Delta T=T_0(1-\theta )$, $\Delta S=S_0(1-\eta )$ and $R_{\rho }=R_0(1-\eta )/(1-\theta )$.

The original PDF uses $R_{\rho }^{-2}$ in the heat-flux law; the converted TeX's exponent $-3$ is a transcription error. Using the PDF yieldsSubstitute this into the heat balance to obtain the double-diffusive overflow reservoir relationsFor the usual heat-dominated positive upward buoyancy flux, $0\le R_F<1$ and $R_0>1$ give $0\le q<1$. Then the right-hand side after substituting ${\eta }_s=q{\theta }_s$ decreases from $C$ to zero as ${\theta }_s$ runs from zero to one; there is a unique physical steady root. The net upward buoyancy flux is $B_0=g\alpha F_T(1-R_F)$.