# Differential Equations Population Problem

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.

• Let it be noted that your equation decays when $\beta < 0$, and when $\delta = 0$ your function will decay to zero (i.e. extinction) – Brevan Ellefsen Feb 1 '17 at 4:13
• @BrevanEllefsen Can a birth rate be negative? – Grizzly0111 Feb 1 '17 at 14:49
• 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. – 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)
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 $\beta < 0$)