= Solution
With the same prescribed $\Delta T$, $S_0$, $\alpha$ and $\beta$, replacing $k$ by $\lambda k$ changes both transport coefficients:
$$
A_\lambda=\lambda A,\qquad B_\lambda=\lambda B.
$$
Hence the positive steady <volume fluxes> become
$$
q_{\pm,\lambda}(E)=\frac{\lambda A\pm\sqrt{\lambda^2A^2-4\lambda BE}}2,
$$
and the negative steady <volume flux> is
$$
q_{{\rm rev},\lambda}(E)=\frac{\lambda A-\sqrt{\lambda^2A^2+4\lambda BE}}2.
$$
The <bifurcation diagram> undergoes a particularly simple rescaling:
$$
\boxed{q_\lambda(E)=\lambda q_1(E/\lambda),\qquad
E_{c,\lambda}=\lambda E_c,\qquad q_{c,\lambda}=\lambda q_c.}
$$
Thus it contracts horizontally and vertically by the same factor. Its branch stability and <saddle-node bifurcation> structure persist, but the range of freshwater forcing that admits the stable positive circulation shrinks. At a fixed subcritical $E$, the upper stable positive branch weakens, the lower unstable positive branch moves upward, and the reversed branch decreases in magnitude. These changes reduce the separation of the positive <equilibria> and the margin against an overturning transition. The right panel of the figure in 2(b) shows this for $\lambda=0.7$ and $0.45$.
For the present $E=0.3\ \mathrm{Sv}$, a positive steady state exists only if
$$
0.3\leq0.5625\lambda,
\qquad\text{or}\qquad
\boxed{\lambda\geq\frac8{15}\simeq0.5333.}
$$
At equality the positive branches merge at $q=\lambda A/2=5.6\ \mathrm{Sv}$; below it only the reversed steady state remains. If a discovery merely revises our estimate of $k$, it revises the inferred branch strengths and threshold, rather than itself causing an instantaneous physical transition. If the real effective $k$ were to decrease slowly through the critical value while $E$ stayed fixed, this model predicts loss of the positive steady circulation and transition to the reversed branch. Uncertainty in $k$ therefore translates directly into uncertainty in the model's freshwater threshold and its inferred proximity to a transition.
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