Draw a stability diagram / phase diagram for the logistic equation

Question

Draw a stability diagram / phase diagram for the logistic equation with a constant harvesting term: $ \ \ \frac{dP}{dt} = rP(1 - \frac{P}{k}) - h \ $

Answer:

Here the harvesting term $ \ h \ $ is constant.

So let $ \ h=400 \ $

But how to draw phase diagram with this situation ?

I tried in this way:

$ \frac{dP}{dt}=0 \\ \Rightarrow rP(1-\frac{P}{k})-400=0 \\ \Rightarrow rP^2-rkP+400r=0 \\ \Rightarrow P= \large \frac{rk \pm \sqrt{r^2k^2+1600r}}{2r} $

These are the two fixed points.

Thus,

$ \frac{dP}{dt}= \ (P- \large \frac{rk + \sqrt{r^2k^2+1600r}}{2r})(P-\large \frac{rk - \sqrt{r^2k^2+1600r}}{2r}) $

If we do not fix values $ \ r , \ k \ $ , how can we draw the phase diagram?

Help me out

Answer 1

You made some calculation errors, so here is the correct factorization:$$\frac{dP}{dt} = rP(1 - \frac{P}{k}) - h =0$$ Multiplying by $-k$ we obtain $$rP^2- rkP +hk =0$$ and thus $$P=\frac{rk \pm \sqrt{r^2k^2-4hkr}}{2r}.$$

Now first consider $h<\frac{rk}{4}$. This would give us $$\frac{dP}{dt}=-\frac{r}{k}\left(P-\frac{rk + \sqrt{r^2k^2-4hkr}}{2r}\right)\left(P-\frac{rk - \sqrt{r^2k^2-4hkr}}{2r}\right).$$

Since the only two zeroes of $\frac{dP}{dt}$ are $\frac{rk \pm \sqrt{r^2k^2-4hkr}}{2r}$, we have that the sign of $\frac{dP}{dt}$ can only change going through these zeroes, so letting $P\to\infty$ and noting that $-\frac{r}{k}P^2$ is the dominant term, we see that the right side must be $\text{sgn}(-\frac{r}{k})$, so between $\frac{rk + \sqrt{r^2k^2-4hkr}}{2r}$ and $\frac{rk - \sqrt{r^2k^2-4hkr}}{2r}$ it must be $-\text{sgn}(-\frac{r}{k})$ and on the left we have $\text{sgn}(-\frac{r}{k})$.

Now consider $h=\frac{rk}{4}$, then we get $$\frac{dP}{dt}=-\frac{r}{k}\left(P-\frac{k }{2}\right)^2$$ which has one equilibrium $\frac{k}{2}$ with sign $\text{sgn}(-\frac{r}{k})$ everywhere outside the equilibrium.

Now if $h>\frac{rk}{4}$, then there is no equilibrium.

We would also like the population density $P$ to be nonnegative (thanks to user539887 for noting this), so we might want to assume that $h,k,r$ are nonnegative (as the model is set up like that) and then we see that in each case all zeroes of $\frac{dP}{dt}$ happen at nonnegative values of $P$.

Answer 2

The $h$ can be absorbed into a quadratic in $P$ for a form with roots $(a,b)$:

$$\frac{dP}{dt}=- c\, (P-a)(P-b)$$

resulting in the set of parabolic phase portraits of varying $(a,b)$ intersection on $P$ axis when the constants $(r,k)$ are varied.

EDIT1:

When this is merged that way the logistic equation is no different from phase portraits serving the purpose.. and so needs no further going in their definition.