Discontinuous growth rate in a first-order dynamical system

Question

Consider the dynamical system defined by

$$\overset{\circ}{x} = x\cdot g(x),$$

where

$$g(x) = \frac{r}{\alpha - x};\quad r,\alpha\in\mathbb{R}^+.$$

I am asked for a biological interpretation of this system, with $x(t)$ representing a population with a growth rate of $g(x)$. Everything would be easier if $g(x)$ were continuous in $[0,+\infty)$, which is clearly false for every value of $\alpha$ (here's a plot of $g(x)$ for $r = 1$, $\alpha = 4$):

Now, I have two main questions:

• It would look like $\alpha$ is not a fixed point, because it is never a root of $\overset{\circ}{x}$. Yet, sketching the corresponding vector field on the real line tells us that, if we start from an initial condition $x_0<\alpha$, we end up approaching $\alpha$ as $t\to\infty$; similarly, $x(t)\to\alpha$ if we start from an initial condition $x_1>\alpha$. Can trajectories really approach a point that's not an equilibrium for the system?
• What on earth happens if our initial condition is exactly equal to $\alpha$? The growth rate $g(x)$ would not even be defined for that value of $x$, so how does this have a biological meaning? In particular,$$\lim_{x\to\alpha^-}g(x) = +\infty;$$doesn't this mean that, if we start from an initial condition that is close enough to $\alpha$ (from the left side), our solution is destined to reach infinity incredibly fast? The growth rate would be pretty big after all, and it would grow even more as we get closer to $\alpha$. But doesn't this contradict the fact that $x(t)\to\alpha$ as $t\to\infty$, for all initial conditions $x_0$?

Answer 1

If the solutions of a model "blow up" or become undefined, that's generally considered to be a bad thing, unless the real situation one is modelling also has a similar sort of blowup. It's the responsibility of the modeller to avoid such bad things.

In this case, it may not be quite as bad as it looks. The solutions of the differential equation do approach $\alpha$, not as $t \to \infty$ but rather as $t$ approaches a finite value from below. The derivative goes to $\pm\infty$ as you approach this point, but the population itself stays finite.

What could happen after that in the reality this is supposed to model? Well, you should first remember that a model is only ever an approximation to the real thing. In particular populations are really discrete rather than continuous, and population changes are not deterministic but have some random fluctuations. If $\alpha$ is not an integer, the population will never actually be equal to $\alpha$. Presumably (and why this could be true is anybody's guess) what this model is saying is that when the population is slightly more than $\alpha$ there is a lot of death or emigration that drives the population down very rapidly, but when the population is slightly less than $\alpha$ there is a lot of birth or immigration that drives the population up very rapidly. So an actual population modelled by this equation might fluctuate very rapidly around $\alpha$.

Answer 2

The answer to the first question is "yes", as you detailed clearly. Of course, your "problem" is caused by the fact that the equation is not defined at $x=\alpha$. Nothing can be done about that, it is your model, so that's it.

There is no possible answer to your second set of questions (other than what you already say yourself), again because the equation is not defined at $x=\alpha$, so whenever you talk about what happens at this point, the discussion will make no sense.