## Related questions with answers

Assume that you fish for 3 years, then fishing is banned for the next 3 years. Thereafter you start again. And so on. This is called intermittent harvesting. Describe qualitatively how the population will develop if intermitting is continued periodically. Find and graph the solution for the first 9 years, assuming that A=B=1, H=0.2, and y(0)=2.

Solution

VerifiedThe general solution for the fish population is

$p(t) = \frac{0.8}{1+0.8Ce^{-0.8t}}$

while when there is no fishing it is

$p_{np}(t) = \frac{0.8}{1+C'e^{-t'}}$

where $t'$ is the time measured from the beginning of banned fishing period. Obviously $t'=t-3$ (in years) so we can write

$p_{np}(t) = \frac{0.8}{1+Ce^{-(t-3)}}.$

If we respect same initial conditions for the first period when fishing is not banned we get

$p_1(t) = \frac{0.8}{1-0.6e^{-0.8t}}.$

Since function $p(t)$ which describes the population of fishes has to be continuous

$p_1(3) = p_{np} (3)$

and we will find $C'$ from this condition. We have

$\frac{1}{1+C'} = \frac{0.8}{1-0.6e^{-2.4}}\Rightarrow 0.2-0.6e^{-2.4}=0.8C'\Rightarrow C'=0.182$

so

$p_{np} = \frac{1}{1+0.182e^{-(t-3)}}.$

The function describing fish population after the ban is over has to also be of form

$p_2(t) = \frac{0.8}{1+0.8C''e^{-0.8t''}}$

where $t''=t-6$ is the time in years measured from the moment ban has expired. From the continuity argument $p_{np}(6) = p_2(6)$ which yields the conditions for $C''$

$\frac{1}{1+0.182e^{-3}} = \frac{0.8}{1+0.8C''}\Rightarrow 0.8+0.8\times0.182e^{-3} = 1+0.8C''\Rightarrow C''=-0.24$

So now we have

$p_2(t) = \frac{0.8}{1-0.8*0.24e^{-0.8(t-6)}} = \frac{0.8}{1-0.192e^{-0.8(t-6)}}.$

Solutions are graphed on the following figure:

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