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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.