I had to find a function for the population of a species that reproduces at a rate proportional to the current population and that dies at a rate proportional to the square root of the current population. Therefore, I assumed that this meant I had to solve

$$ \frac{dP}{dt}=\beta P-\delta \sqrt{P} $$

where $P$ is the population (a function of $t$, time) and $\beta$ and $\delta$ the birth and death rates repectively. Solving this differential equation gave me

$$ P(t)=\left(\frac{e^\left(\frac{\beta t}{2}+C \right)+\delta}{\beta}\right)^2 $$

which is correct, according to Wolfram. The next part of the questionis to show that for some value of $P_0$ (the initial population at $t=0$), the population will be extinct in the long run. However, it is clear that my formula for $P(t)$ is always increasing and that it is never equal to zero. Did I miss anything?Like I said, I checked with Wolfram and it is a solution to my differential equation.

  • $\begingroup$ Let it be noted that your equation decays when $\beta < 0$, and when $\delta = 0$ your function will decay to zero (i.e. extinction) $\endgroup$ – Brevan Ellefsen Feb 1 '17 at 4:13
  • $\begingroup$ @BrevanEllefsen Can a birth rate be negative? $\endgroup$ – Grizzly0111 Feb 1 '17 at 14:49
  • 1
    $\begingroup$ depends on your definition of birth rate I suppose. I would generally say no, but I was pointing out that you should explicitly assume that $\beta$ is positive to have your analysis be correct. $\endgroup$ – Brevan Ellefsen Feb 1 '17 at 15:56

With $t=0$ we have $e^C=\beta\sqrt{P_0}-\delta$ and the population after $t$ (usually in years) is $$ P(t)=\left(\frac{(\beta\sqrt{P_0}-\delta)e^\left(\frac{\beta t}{2} \right)+\delta}{\beta}\right)^2 $$ this makes the equation easier.

In these types problems we can't say when population will be extinct, scince as you mention, $P(t)$ is never equal to zero. Instead of it we could say for which times the population will be smaller than an amount for example $P(t)<\dfrac{1}{1000}P(t_0)$.


However, it is clear that my formula for $P(t)$ is always increasing and that it is never equal to zero.

The problem with your analysis is that your function is NOT always increasing. Whenever $\beta < 0$ your function decays, and when $\delta = 0 $ your function decays to zero (extinction)

All we can is that your function must be non-negative because of the squaring operation

Note: the decay is not monotonic unless $\delta \geq 0,$ although $\lim x \to -\infty$ will still be unbounded towards $+\infty$ and $\lim x \to +\infty$ will still be bounded (assuming still that $\beta < 0$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.