= Solution
Using the constants \b[as printed], with <volume flux> measured in <Sverdrups>, gives
$$
\boxed{A=(7\times10^6)(2\times10^{-7})(15)=21\ \mathrm{Sv},\qquad
B=(7\times10^6)(8\times10^{-7})(35)=196\ \mathrm{Sv}.}
$$
In 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 are
$$
\boxed{q_\pm(E)=\frac{A\pm\sqrt{A^2-4BE}}2
\quad\left(0<E\leq\frac{A^2}{4B}\right),\qquad
q_{\rm rev}(E)=\frac{A-\sqrt{A^2+4BE}}2<0.}
$$
The upper positive branch $q_+$ is stable, the lower positive branch $q_-$ is unstable, and the reversed branch $q_{\rm rev}$ 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.
\Image[/past-exam-of-the-geophysical-fluid-dynamics-course-of-the-university-of-oxford/2025-overturning-bifurcation.png]
{title=Steady overturning branches, their stability, and the effect of reducing the density-to-transport coefficient}
At $E=0.3\ \mathrm{Sv}$, the three steady <volume fluxes> are
$$
\boxed{q_+=17.67\ \mathrm{Sv},\qquad q_-=3.327\ \mathrm{Sv},\qquad
q_{\rm rev}=-2.502\ \mathrm{Sv}.}
$$
The present thermally dominated, northward-upper-flow <Atlantic meridional overturning circulation> corresponds to $q_+\simeq17.7\ \mathrm{Sv}$. 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 $\Delta S=ES_0/q_+\simeq0.594\ \mathrm{g\,kg^{-1}}$, consistent with the reference-salinity approximation. The weaker branches give larger contrasts, approximately $3.16$ and $4.20\ \mathrm{g\,kg^{-1}}$, so the approximation is less accurate there.
The <saddle-node bifurcation> occurs at
$$
\boxed{E_c=\frac{A^2}{4B}=0.5625\ \mathrm{Sv},\qquad q_c=\frac A2=10.5\ \mathrm{Sv}.}
$$
For $E>E_c$, 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 has
$$
q_{\rm rev}(E_c)=\frac A2(1-\sqrt2)\simeq-4.35\ \mathrm{Sv}.
$$
For 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 $E\to0^+$. At $E=0$ this is an equal-density, no-exchange equilibrium with $\Delta S=\alpha\Delta T/\beta=3.75\ \mathrm{g\,kg^{-1}}$, 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).
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