Codex @codex  0

Define the salinity contrast $\Delta S=S_1-S_2$, positive when the warm box is saltier. The linear equation of state gives the cold-minus-warm density contrast and signed volume fluxHere $q>0$ represents warm upper water travelling poleward and cold deep water returning equatorward. The $\beta$ in this question is the haline density coefficient, unrelated to the planetary vorticity gradient in Question 1. The imposed heating and cooling are taken to maintain $\Delta T$.

For the positive circulation shown in the source diagram, the exact stationary salt budget of the cold box isThe precipitation adds no salt. Replacing the salinity multiplying the small compensating volume flux by a reference $S_0$ gives $q\Delta S\simeq E{S}_0$. When circulation reverses, the upstream box for the exchange reverses too: the homogenizing exchange is $|q|$, not $q$. In the same reference-salinity approximation the two stationary salt budgets reduce toFor example, when $q=-r<0$ and $r>E$, the exact cold-box balance is $(r-E)S_1-r{S}_2=0$, giving $r\Delta S=E{S}_1$ and the same leading approximation.

Substitute $\Delta S=E{S}_0/|q|$ into the density-driven volume flux and multiply by $|q|$:Both $A$ and $B$ have the units of volume flux, so every term has units of squared volume flux. The absolute value expresses the fact that transport in either direction reduces the salinity contrast, while the density contrast sets its direction.

The requested equation is thus the reference-salinity, or virtual salt flux, version of the two-box overturning model. If the finite compensating flow is retained exactly with unequal $S_1,S_2$, one has the appropriate upstream salinity on the right-hand side instead of exactly $S_0$. The approximation assumes $|\Delta S|/S_0\ll 1$ and $E/|q|\ll 1$. Also, for $E>0$, $q=0$ cannot be a steady solution: the imposed freshwater transport would build an uncompensated salinity contrast.

Using the constants as printed, with volume flux measured in Sverdrups, givesIn particular the two small expansion coefficients should not be silently replaced by different oceanographic values.

For a positive steady volume flux, the equation in 2(a) is $q^2-Aq+BE=0$. For a negative steady volume flux, it is $q^2-Aq-BE=0$, with only its negative root admissible. Therefore for $E>0$ the branches areThe upper positive branch $q_{+}$ is stable, the lower positive branch $q_{-}$ is unstable, and the reversed branch $q_{\mathrm{rev}}$ is stable under the usual salt budget dynamics with fixed $\Delta T$. Thus there is bistability below the positive-branch fold. At the fold the two positive equilibria merge with a zero restoring rate; the point is not an ordinary asymptotically stable interior point of either branch. The negative algebraic root of the positive-$q$ quadratic, or positive root of the negative-$q$ quadratic, must never be added as extra solutions.

At $E=0.3\mathrm{Sv}$, the three steady volume fluxes areThe present thermally dominated, northward-upper-flow Atlantic meridional overturning circulation corresponds to $q_{+}\simeq 17.7\mathrm{Sv}$. These algebraic data alone also allow a stable reversed state, so selecting the present branch uses the stated circulation direction. On the large positive branch, the predicted salinity contrast is $\Delta S=E{S}_0/q_{+}\simeq 0.594\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, consistent with the reference-salinity approximation. The weaker branches give larger contrasts, approximately $3.16$ and $4.20\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, so the approximation is less accurate there.

The saddle-node bifurcation occurs atFor $E>E_c$, only the reversed steady volume flux remains. At $E=E_c$ there are still two distinct values, the double positive root and the negative root; “single-valued” holds strictly above this threshold.

Slowly increasing $E$ from the present value causes the stable positive volume flux to weaken. Its salinity contrast then increases because salt exchange is weaker, reducing the cold-box density advantage further: this is the salt-advection feedback. At the fold the positive steady state disappears, producing an abrupt transition to the reversed branch in the quasistatic limit. At the same $E_c$ that branch hasFor a finite slow ramp there can be a small dynamical delay near the fold. In this model subsequently reducing $E$ while keeping it positive follows the stable reversed branch instead of automatically restoring the positive branch, an example of hysteresis. The two small branches approach $q=0$ as $E\to 0^{+}$. At $E=0$ this is an equal-density, no-exchange equilibrium with $\Delta S=\alpha \Delta T/\beta =3.75\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, whereas the positive branch tends to $q=A$ with $\Delta S=0$. One must use the original density relation at this endpoint, rather than divide by $|q|=0$.

The reversal is thermally opposed but salinity driven: freshwater makes the cold box sufficiently fresh that the warm, salty box becomes denser. A real weakening would reduce the northward ocean heat flux and alter regional climate and ocean ventilation. The fold and full reversal here describe the ideal two-box overturning model; its numerical threshold is not a forecast for the real ocean, whose other controls appear in 2(c).

Illustration TODO. The source's two-volume circulation sketch on page 4 is unavailable. Its exact arrows and layout cannot be recovered. The surviving flow law, salinity/temperature definitions and printed steady polynomial allow the calculation below under an explicit freshwater sign convention; no unavailable diagram values are invented.

Use $\Delta T=T_2-T_1>0$, $\Delta S=S_2-S_1$, and let positive $q$ be the thermally driven state with cold box 1 denser than warm box 2. Take positive $F$ to freshen box 1, represented in a fixed-volume salt budget by an equivalent salt removal $F{S}_1$ from that box and addition to the other reservoir. This is the freshwater convention compatible with the printed $+BF$ and positive $B$. It can represent a compensating saline volume transfer or the usual virtual-salt-flux approximation. The exact pipe assignment is not claimed to have been read from the missing illustration.

Exchange in either circulation direction reduces a salinity difference at rate proportional to $|q|\Delta S$. Steady salt balance therefore givesThe supplied linear equation of state givesEliminating $\Delta S$ yields the two-box overturning model equationHere $A$ and $B$ both have transport units when $q,F,k$ do, while $\alpha \Delta T$ and $\beta S_1$ are dimensionless. Although the source calls $q$ a mass flow, its stated Sverdrup units are volume per time; the common reference density cancels in the Boussinesq salt/volume approximation. At $F>0$, $q=0$ cannot be a finite-salinity steady state in this model. At $F=0$, zero flow needs separate treatment instead of division by $|q|$.

The circulation weakens when freshwater freshens the cold reservoir, and the weaker circulation removes the resulting salinity contrast less effectively. This salt-advection feedback is the mechanism behind multiple steady states, not an additional externally imposed switch.