A deer population grows logistically and is harvested at a rate proportional to its population size. The dynamics of population growth is modeled by

$P' = rP (1-\frac{P}{K}) - hP$

where $r$ (the rate of growth), $K$ (capacity of the environment) and $h$ (harvesting rate) are constants

For $h > 0$, use a bifurcation diagram to explain the effects on the equilibrium deer population when h is slowly increased from a small value to a large value.


How do I use changing variables to simplify the model (only left with one variable and the constant $h$) so that I am able to find the equilibria?

For example in the following model: $P' = rP (1-\frac{P}{K}) - H$

I am able to change varibles via $x = \frac{P}{K}$ and $ \mu = rt$, and got $\frac{dx}{d\mu } = x(1-x) - h$ where h is a constant $h = \frac{H}{rK}$ in order to find its equilibria.

Or, there might be other approaches for this question.


1 Answer 1


I prefer to think about this through the relatively systematic procedure of nondimensionalization.

Introduce $P=P_C R,t=t_C s$, for unknown quantities $P_C$ with units of population and $t_C$ with units of time. Insert this into the equation to get

$$\frac{P_C}{t_C} \frac{dR}{ds} = r P_C R \left ( 1 - \frac{P_C R}{K} \right ) - h P_C R.$$

Cancel coefficients:

$$\frac{dR}{ds}=r t_C R \left ( 1 - \frac{P_C R}{K} \right ) - h t_C R$$

Now you want to choose $P_C$ and $t_C$ to simplify this equation. You can do that by making some of the coefficients be $1$; there are two reasonable ways to do this for this equation. For your bifurcation study, one of them is more convenient than the other, and that is what you almost correctly did, namely to have $t_C=\frac{1}{r}$ and $P_C=K$. Then you have


where $a=\frac{h}{r}$ is a bifurcation parameter.

  • $\begingroup$ Thank you very much! I have one more stupid question... The problem wants me explain the effects on the equilibrium deer population when $h$ is slowly increased from a small value to a large value, not $a$. How can I do that when I don't know the value of $r$? $\endgroup$
    – Melisaa
    Commented Feb 8, 2016 at 2:26
  • $\begingroup$ @Melisaa Well, $a$ is proportional to $h$, so if you can draw the bifurcation diagram in $a$ then you can rescale the axis by $r$ in order to get the bifurcation diagram in $h$. The $h$ bifurcation diagram will necessarily involve $r$, though, so if you don't know what $r$ is, it will have to be present just as a variable. $\endgroup$
    – Ian
    Commented Feb 8, 2016 at 2:28
  • $\begingroup$ Thank you. For the equilibria, I got $x*=0$ and $x*=1-\frac{h}{r}$. Therefore, the bifurcation diagram (on $x*h$ plane) will be one horizontal line on $x*=0$ and a decreasing linear graph which intercts (0,1) and (r,0). Two branches will meet at (r,0). $\endgroup$
    – Melisaa
    Commented Feb 8, 2016 at 2:41
  • $\begingroup$ @Melisaa I had a minor algebraic error before, which I've edited now. It doesn't change your diagram, though, which you have indeed done correctly. $\endgroup$
    – Ian
    Commented Feb 8, 2016 at 14:33

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