Question

Question: With the identifications\(a = r,b = r/K\), and\(a/b = K\), Figures 2.1.7 and 3.2.2 show that the logistic population model, (3) of Section\(3.2\), predicts that for an initial population\({P_0},0 < {P_0} < K\), regardless of how small \({P_0}\)is, the population increases over time but does not surpass the carrying capacity\(K\).Also, for \({P_0} > K\)the same model predicts that a population cannot sustain itself over time, so it decreases but yet never falls below the carrying capacity \(K\)of the ecosystem. The American ecologist Warder Clyde Alle (1885-1955) showed that by depleting certain fisheries beyond a certain level, the fish population never recovers. How would you modify the differential equation (3) to describe a population \(P\)that has these same two characteristics of (3) but additionally has a threshold level\(A,0 < A < K\), below which the population cannot sustain itself and approaches extinction over time. (Hint: Construct a phase portrait of what you want and then form a differential equation.)

Short answer

\(\frac{{dP}}{{dt}} = (P - A)\left( {r - \frac{r}{K}P} \right)\)

Step-by-step solution

Procedure

We want to modify the well known logistic equation\(\frac{{dP}}{{dt}} = P\left( {r - \frac{r}{k}P} \right)\)

to include a threshold\(A,0 < A < K\)such that if the population drops below\(A\), then the population will vanish. As with the equation above all considerations can be made without actually solving the equation. We basically want the following, if\(P < A\), then\(\frac{{dP}}{{dt}} < 0\), which will guarantee that\(P(t)\) will keep decreasing, and eventually vanish:

i : We want \(\frac{{dP}}{{dt}} < 0\)for\(P < A\) (this tells us that he population cannot recover and will vanish)

ii: We want \(\frac{{dP}}{{dt}} > 0\) for\(A < P < K\) (this tells us that the population will increase and approach the equilibrium\(K\) if\(P > A\))

iii: We want \(\frac{{dP}}{{dt}} < 0\)if\(P > K\) (this tells us that the population will decrease and approach the equilibrium \(K\)if\(P > A\))

Observation

The factor \(r - \frac{r}{K}P\)is sufficient so the second and third conditions are satisfied. So instead of\(P\), we will take\(P - A\)to be the first factor in the equation. Since\(r - \frac{r}{K}P\)is positive for\(P < K\)the product \((P - A)\left( {r - \frac{r}{K}P} \right)\)will be negative for \(P < A < K\).

Next, for\(A < P < K\) both factors will be positive which will result in\(\frac{{dP}}{{dt}} > 0\).

Finally, if\(P > K\)then the first factor is positive while the second is negative so we get\(\frac{{dP}}{{dt}} < 0\), which is what we wanted. So, the modified equation is\(\frac{{dP}}{{dt}} = (P - A)\left( {r - \frac{r}{K}P} \right)\).