Question

Investigate the harvesting model in Problem 5 both qualitatively and analytically in the case \({\bf{a = }}\) \({\bf{5,b = 1,h = }}\;\;\;\frac{{{\bf{dP}}}}{{{\bf{dt}}}}{\bf{ = P(a - bP) - h,}}\;\;\;{\bf{P(0) = }}{{\bf{P}}_{\bf{0}}}{\bf{,}}\;\;\;\) Determine whether the population becomes extinct in finite time. If so, find that time.

Short answer

(a)So, the diagram is shown as below:

(b)So, the plot of population is shown below:

(c)So, the required solution is \(t = \frac{2}{5} + \frac{2}{{5 - 2{P_0}}}\).

Step-by-step solution

Definition of derivative

The derivative is a way to show instantaneous rate of change.

Evaluation

(a)

Consider that the model for the population \(P(t)\) of the fishery at time \(t\) is given by \(\frac{{dP}}{{dt}} = P(a - bP) - h.\)

The objective is to determine the long-term behaviour of the population.

Consider the values, \(a = 5,b = 1\), and \(h = 4\).

Substitute the values for constants \(a,b\) and \(h\).

\(\frac{{dP}}{{dt}} = P(5 - P) - \frac{{25}}{4}\)

Find the value of \({\bf{P}}\)

To create the phase portrait, first need to solve for the values of\(P\)such that\(\frac{{dP}}{{dt}} = 0\).

Use the quadratic formula to solve for\(P\).

\(P = \frac{{5 \pm \sqrt {{5^2} - 4(1)\left( {\frac{{25}}{4}} \right)} }}{{2(1)}}\)

\( = \frac{5}{2}\)

Substitute value in differential equation

Substitute values for\(P\)into the differential equation (1) in order to find the stability of the phase portrait.

Let\(P = 1\)

Then

\(\frac{{dP}}{{dt}} = 1(5 - 1) - \frac{{25}}{4}\)

\( = 4 - \frac{{25}}{4}\)

\( = - \frac{9}{4}\)

Let\(P = 4\).

Then

\(\frac{{dP}}{{dt}} = 4(5 - 4) - \frac{{25}}{4}\)

\( = 4 - \frac{{25}}{4}\)

\( = - \frac{9}{4}\)

Make the diagram

The resulting phase diagram and representative solution curves are shown below:

From the figure, note that if \(0 < {P_0} < 5/2\) then \(P(t) \to - \infty \), otherwise \(P(t) \to 5/2\), as \(t \to 00\).

(b)Step 1: Solution of the differential equation

Solve differential equation (1) by distributing the right-hand side and separate the variables.

\(\frac{{dP}}{{dt}} = P(5 - P) - \frac{{25}}{4}\)

\(\frac{{dP}}{{dt}} = - {P^2} + 5P - \frac{{25}}{4}\)

\(\frac{{dP}}{{dt}} = - {\left( {P - \frac{5}{2}} \right)^2}\)

Separate the variables.

\(\frac{{dP}}{{{{\left( {P - \frac{5}{2}} \right)}^2}}} = - dt\)

Integrate on both sides

\(\int {{{\left( {P - \frac{5}{2}} \right)}^{ - 2}}} dP = - \int d t\)

\( - {\left( {P - \frac{5}{2}} \right)^{ - 1}} = - t + e\)

Simplify the solution and solve for\(P\).

\(\frac{1}{{P - \frac{5}{2}}} = t - c\)

\(P - \frac{5}{2} = \frac{1}{{t - c}}\)

\(P = \frac{1}{{t - c}} + \frac{5}{2}\)

Use initial condition

The given initial condition is\(P(0) = P\).

Substitute\(t = 0\)and\(P(0) = {P_0}\)in the solution\(P = \frac{1}{{t - c}} + \frac{5}{2}\).

\({P_0} = \frac{1}{{0 - c}} + \frac{5}{2}\)

\({P_0} = - \frac{1}{c} + \frac{5}{2}\)

\({P_0} - \frac{5}{2} = - \frac{1}{c}\)

\(\frac{{2{P_0} - 5}}{2} = - \frac{1}{c}\)

Solve the equation for\(c\)

\(c = \frac{2}{{5 - 2{p_0}}}\)

Plot the graph

Substitute the value of\(c\)into the solution to get the explicit solution.

\(P = \frac{1}{{t - \frac{2}{{5 - 2{P_0}}}}} + \frac{5}{2}\)

Simplifying the fraction on the right-hand side yields the solution

\(P = \frac{{5 - 2{P_0}}}{{5t - 2{P_0}t - 2}} + \frac{5}{2}\)

Let\({P_0} = 1\)

The following is the plot of\(P(t)\).

\({\rm{ Let }}{{\bf{P}}_{\bf{0}}} = {\bf{4}}{\rm{. The following is the plot of }}{\bf{P}}({\bf{t}}){\rm{. }}\)

(c)Step 1: Verification

The objective is to verify whether the fishery population becomes extinct in finite time or not.

From the phase diagram in part (a), note that the fishery population can reach extinction only in the case\(0 < {P_0} < 5/2\).

In part (b), the obtained intermediate equation is,

\(\frac{1}{{P - \frac{5}{2}}} = t - c\)

The value of\(c\)found in part (b) is\(c = \frac{2}{{5 - 2{P_0}}}\).

Therefore, the solution becomes, \(\frac{1}{{P - \frac{5}{2}}} = t - \frac{2}{{5 - 2{P_0}}}\).

Use exponential and logarithmic rules to solve for time \({\bf{t}}\)

The fishery population is extinct when\(P = 0\).

\(\frac{1}{{0 - \frac{2}{5}}} = t - \frac{2}{{5 - 2{P_0}}}\)

\(\frac{2}{5} = t - \frac{2}{{5 - 2{P_0}}}\)

\(\frac{2}{5} + \frac{2}{{5 - 2{P_0}}} = t\)

The fishery population goes extinct at time \(t\) is \(t = \frac{2}{5} + \frac{2}{{5 - 2{P_0}}}\).