Question

Constant-Harvest Model A model that describes the population of a fishery in which harvesting takes place at a constant rate is given by

$\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\mathit{P}\mathbf{-}\mathit{h}$

where k and h are positive constants.

(a) Solve the DE subject to $\mathit{P}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{P}}_{\mathbf{0}}$.

(b) Describe the behavior of the population P(t) forincreasing_time in the three cases ${\mathit{P}}_{\mathbf{0}}\mathbf{>}\frac{\mathbf{h}}{\mathbf{k}}\mathbf{,}{\mathit{P}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{h}}{\mathbf{k}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{0}\mathbf{<}{\mathit{P}}_{\mathbf{0}}\mathbf{<}\frac{\mathbf{h}}{\mathbf{k}}$

(c) Use the results from part (b) to determine whether the fish population will ever go extinct in finite time, that is, whether there exists a time $\mathit{T}\mathbf{>}\mathbf{0}$

such that $\mathit{P}\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{=}\mathbf{0}$. If the population goes extinct, then find T.

Short answer

Therefore, the result is

$a)P=\left(P_0-\frac{h}{k}\right)e^{kt}+\frac{h}{k}$

b) The population of the fishery continues to grow for $P_0>\frac{h}{k}$. The population becomes constant when $P_0=\frac{h}{k}$. The population of$0<P_0<\frac{h}{k}$ continues to decline.

c)$t_e=\frac{1}{k}\ln \left[\frac{h}{k{P}^{\text{'}}}\right]$

Step-by-step solution

Given Information

The given value is:

$\frac{dp}{dt}=kP-h$

Determining the fishery's population at time t

We have a population model defined by the differential equation

$\frac{dp}{dt}=kP-h$

with the predicament

$P\left(0\right)=P_0$

We must address the problem as follows:

Because this is a first-order and separable differential equation, we can solve it as follows:

$\begin{array}{c}\frac{\mathrm{dP}}{\mathrm{kP}-\mathrm{h}}=\mathrm{dt}\int \frac{1}{\mathrm{kP}-\mathrm{h}}\mathrm{dP}=\int \mathrm{dt}\\ \mathrm{dfrac}1\mathrm{k}\int \frac{\mathrm{k}}{\mathrm{kP}-\mathrm{h}}\mathrm{dP}=\int \mathrm{dt}\\ \ln (\mathrm{kP}-\mathrm{h})=\mathrm{kt}+{\mathrm{c}}_1\\ {\mathrm{e}}^{\ln (\mathrm{kP}-\mathrm{h})}={\mathrm{e}}^{\mathrm{kt}+{\mathrm{c}}_1}\\ \mathrm{kP}-\mathrm{h}={\mathrm{e}}^{{\mathrm{c}}_1}{\mathrm{e}}^{\mathrm{kt}}\\ \mathrm{kp}={\mathrm{ce}}^{\mathrm{kt}}+\mathrm{h}\end{array}$

Then there's

$P=n{e}^{kt}+\frac{h}{k}$

Then there's

Substituting the value of the constant n into equation(1), we get

$P=\left(P_0-\frac{h}{k}\right)e^{kt}+\frac{h}{k}$

is the fishery's population at time t.

Analyzing the population at the given cases

For the population of the fishery at time t described in equation (2), we can predict its behaviour in the following three scenarios:

1. If we observe that the R.H.S is increasing for $P_0>\frac{h}{k}$, then the fishery's population will continue to grow.
2. We discover that when $P_0=\frac{h}{k}$, the R.H.S equals h, and the population is constant, $P=P_0$.

3. If we observe that the R.H.S is falling for $0<P_0<\frac{h}{k}$, then the fishing population will continue to decline.

to see if the fish population would ever go extinct in a finite amount of time

For the third situation (oPh), the fish population will go extinct, and we can calculate the time when the population will become extinct as follows:

allow us to have

$P_0-\frac{h}{k}=P^{\text{'}}$

Equation (2) is then obtained by putting P-0 into equation (a).

$\begin{array}{c}0=P^{\text{'}}{e}^{k{t}_e}+\frac{h}{k}\\ P^{\text{'}}{e}^{k{t}_e}=\frac{h}{k}{e}^{k{t}_e}=\frac{1}{P^{\text{'}}}\frac{h}{k}\\ \ln e^{k{t}_e}=\ln \left(\frac{h}{k{P}^{\text{'}}}\right)\end{array}$

Then there's

$t_e=\frac{1}{k}\ln \left(\frac{h}{k{P}^{\text{'}}}\right)$

For the third situation, this is the time when the population goes extinct.

Conclusion

$a)P=\left(P_0-\frac{h}{k}\right)e^{kt}+\frac{h}{k}$

b) The population of the fishery continues to grow for $P_0>\frac{h}{k}$. The population becomes constant when $P_0=\frac{h}{k}$. The population of $0<P_0<\frac{h}{k}$continues to decline.

c)$t_e=\frac{1}{k}\ln \left[\frac{h}{k{P}^{\text{'}}}\right]$