I'm doing some calculations from the article of Stommel $1961$: Thermohaline convection with two stable regimes of flow. At a certain point he write down a system of two nonlinear ODEs which has the following form: $$ \frac{dy}{dt} = 1-y - \frac{y}{\lambda}\left| -y + Rx \right| $$ $$ \frac{dx}{dt} = \delta\left(1-x\right) - \frac{x}{\lambda}\left| -y + Rx \right| $$ And then he says: "The points of equilibrium correspond to those values of $x$ and $y$ for which $\frac{dy}{dt}$ and $\frac{dx}{dt}$ vanish, thus leading to a cubic for $y$ in terms of $x$. I really don't get this last point.
Can someone show me the steps to gain this mentioned cubic for $y$ in terms of $x$? It's a crucial point without which the entire article have no sense at all.
Analytically:
For equilibrium, we set $\frac{dx}{dt}$ and $\frac{dy}{dt}$ to be zero. That gives us $$1 - y - \frac{y}{\lambda}|-y + Rx| = 0$$ $$\delta(1-x) - \frac{x}{\lambda}|-y+Rx| =0$$
Since the equation has $|-y + Rx|$ in both of the equations, we can try to write each equation in terms of that and equate.
So we get $ y = \frac{x}{x + \delta(1-x)}$
In order to find the $x$ value, we substitute back the above value of $y$ in suppose say to the equation $\delta(1-x) - \frac{x}{\lambda}|-y+Rx| =0$.
This gives us $$\delta(1-x) -\frac{x}{\lambda}\left|\frac{x + Rx^2 + R\delta x(1-x)}{x + \delta(1-x)}\right| = 0$$ From which we can see that we will get a cubic equation in $x$.
So there are chances that the system of ODE has three equilibrium points.
You can see that graphically also (for example in DESMOS).