Question

(a) If a constant number $\mathbf{h}$of fish are harvested from a fishery per unit time, then a model for the population $\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$of the fishery at time$\mathit{t}$ is given by

$\frac{\mathrm{dP}}{\mathrm{dt}}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{bP}\mathbf{)}\mathbf{-}\mathbf{h}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{P}}_0$

Where$\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{h}$ , and P0 are positive constants. Suppose $\mathbf{a}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{1}$, and $\mathbf{h}\mathbf{=}\mathbf{4}$. Since the DE is autonomous, use the phase portrait concept of Section to sketch representative solution curves corresponding to the cases${\mathbf{P}}_0\mathbf{>}\mathbf{4}\mathbf{,}\mathbf{1}\mathbf{<}{\mathbf{P}}_0\mathbf{<}\mathbf{4}$, and$\mathbf{0}\mathbf{<}{\mathbf{P}}_0\mathbf{<}\mathbf{1}$ . Determine the long-term behavior of the population in each case.

(b) Solve the IVP in part (a). Verify the results of your phase portrait in part (a) by using a graphing utility to plot the graph of $\mathbf{P}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$with an initial condition taken from each of the three intervals given.

(c) Use the information in parts (a) and (b) to determine whether the fishery population becomes extinct in finite time. If so, find that time.

Short answer

a.Long term behavior of the population is showing in the following graph.

b.We have verified the solution.

c.So, the required solution is $t=-\frac{1}{3}\ln \left(\frac{4(P_0-1)}{P_0-4}\right)$.

Step-by-step solution

Definition of derivative

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

(a)Step 2: Find the behavior of population

If a constant number$h$of fish are harvested from a fishery per unit time, then a model for the population$P\left(t\right)$of the fishery at time$t$is given by,

role="math" localid="1667891499632" $\frac{dP}{dt}=P(a-bP)-h,P\left(0\right)=P_0\dots \dots \text{(i)}$

Where$a,b,h$, and$P_0$are positive constants.

Suppose$a=5,b=1$, and$h=4$.

The objective is to use the concept of phase portrait to sketch the solution curves corresponding to the cases$P_0>4,1<P_0<4$, and$0<P_0<1$.

Also determine the long-term behavior of the population in each case.

Compare the equation

For$a=5,b=1$, and$h=4$, differential equation (i) becomes,

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

Compare it withrole="math" localid="1667891857943" $\frac{dP}{dt}=f\left(t\right)$to get$f\left(t\right)=P(5-P)-4$

Critical points of the differential equation are solutions of the equation$\frac{dP}{dt}=0$

$\Rightarrow P(5-P)-4=0$

$5P-P^2-4=0$

$\begin{array}{r}{P}^2-5P+4=0\\ P^2-4P-P+4=0\\ P(P-4)-1(P-4)=0\end{array}$

$(P-1)(P-4)=0P$

$=1,4$

So, the critical points are 1 and 4 .

Therefore, equilibrium solutions are $P\left(t\right)=1$ and $P\left(t\right)=4$.

Substitution

By putting the critical points on a vertical line, divide the line into three intervals:

$-\mathrm{\infty }<\mathrm{P}<1,1<\mathrm{P}<4,4<\mathrm{P}<\mathrm{\infty }$

Compute the algebraic sign of$f\left(t\right)=P(5-P)-4$on these intervals.

If the algebraic sign is negative,$P\left(t\right)$ is decreasing and the arrow downward on the phase portrait of $dP/dt=f\left(P\right)$. If the algebraic sign is positive, $P\left(t\right)$is increasing and the arrow upward on the phase portrait of $dP/dt=f\left(P\right)$.

Plot the graph

For clarity, the original phase line from Figure-1 is reproduced to the left of the plane in which the sub-regions $R_1,R_2$, and $R_3$ are shaded.

From the above $tP$-plane, observe that, the solution curve approaches to $P\left(t\right)=1$ as $t\to -\infty$, approaches to $P\left(t\right)=4$ when$1<t<4$ and approaches to $P\left(t\right)=4$ when $t>4$.

(b)Step 6: Using partial fraction

The objective is to find the solution of the initial value problem,

$\frac{dP}{dt}=P(5-P)-4,P\left(0\right)=P_0$

Rewrite the differential equation as,

$\frac{dP}{dt}=5P-P^2-4$

$=-(P-1)(P-4)$

Separating the variables,

$\frac{dP}{(P-1)(P-4)}=-dt$

Use partial fraction;

$\frac{1}{(\mathrm{P}-1)(\mathrm{P}-4)}=\frac{\mathrm{A}}{\mathrm{P}-1}+\frac{\mathrm{B}}{\mathrm{P}-4}\frac{1}{(\mathrm{P}-1)(\mathrm{P}-4)}$

$=\frac{A(P-4)+B(P-1)}{(P-1)(P-4)}1$

$=A(P-4)+B(P-1)1$

$=P(A+B)-(4A+B)$

Compare the coefficients

Comparing the coefficients, get$A+B=0,-4A+B=-1$

By solving these two equations get,$A=-\frac{1}{3},B=\frac{1}{3}$

So,$\frac{1}{(P-1)(P-4)}=-\frac{1}{3}\frac{1}{P-1}+\frac{1}{3}\frac{1}{P-4}$

Equation (ii) becomes,

$\left(-\frac{1}{3}\frac{1}{\mathrm{P}-1}+\frac{1}{3}\frac{1}{\mathrm{P}-4}\right)\mathrm{dP}=-\mathrm{dt}$

Integrate on both sides,

$\frac{1}{3}\int \frac{1}{P-1}dP+\frac{1}{3}\int \frac{1}{P-4}dP=-\int dt$

$-\frac{1}{3}\ln |P-1|+\frac{1}{3}\ln |P-4|=-t+c$

$\frac{1}{3}\ln \left|\frac{P-4}{P-1}\right|=-t+c$

$\ln \left|\frac{P-4}{P-1}\right|=-3t+3c$

Evaluation

To evaluate the constant$C$, use the initial condition$P\left(0\right)=P_0$

$\frac{P_0-4}{P_0-1}=C{e}^{-30}$

$\frac{P_0-4}{P_0-1}=C$

Put the value of $C$ in (iii).

$\left(\frac{P-4}{P-1}\right)=\left(\frac{P_0-4}{P_0-1}\right)e^{-3t}$

$P[(P_0-1)-(P_0-4)e^{-3t}]=4(P_0-1)-(P_0-4)e^{-3t}$

$\mathrm{P}\left(\mathrm{t}\right)=\frac{4({\mathrm{P}}_0-1)-({\mathrm{P}}_0-4){\mathrm{e}}^{-3\mathrm{t}}}{({\mathrm{P}}_0-1)-({\mathrm{P}}_0-4){\mathrm{e}}^{-3\mathrm{t}}}$

Therefore, solution of the initial value problem is $\mathrm{P}\left(\mathrm{t}\right)=\frac{4({\mathrm{P}}_0-1)-({\mathrm{P}}_0-4){\mathrm{e}}^{-3\mathrm{t}}}{({\mathrm{P}}_0-1)-({\mathrm{P}}_0-4){\mathrm{e}}^{-3\mathrm{t}}}$.

(C)Step 9:  Find fishery population

The objective is to find, whether the fishery population becomes extinct in finite using the information in subparts (a) and (b). if so, find that time.

From the phase diagram in part (a), we see the fishery population can reach extinction only in the case$0<P_0<1$.

From subpart (b), we have that,

$\left(\frac{P-4}{P-1}\right)=\left(\frac{P_0-4}{P_0-1}\right)e^{-3t}\dots \dots \text{(iv)}$

The fishery population is extinct when$P=0$

With $P=0$equation (iv) becomes,

$\left(\frac{0-4}{0-1}\right)=\left(\frac{P_0-4}{P_0-1}\right)e^{-3t}$

$4=\left(\frac{P_0-4}{P_0-1}\right)e^{-3t}$

$\frac{P_0-4}{P_0-1}=4e^{3t}$

$\frac{1}{4}\left(\frac{P_0-4}{P_0-1}\right)=e^{3t}$

$3t=\ln \left(\frac{P_0-4}{4(P_0-1)}\right)$

$t=\frac{1}{3}\ln \left(\frac{P_0-4}{4(P_0-1)}\right)$

$=-\frac{1}{3}\ln \left(\frac{4(P_0-1)}{P_0-4}\right)$

Therefore, for $0<P_0<1$, the time of extinction is $t=-\frac{1}{3}\ln \left(\frac{4(P_0-1)}{P_0-4}\right)$.