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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.


Answered 2 years ago
Answered 2 years ago
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The general solution for the fish population is

p(t)=0.81+0.8Ce0.8tp(t) = \frac{0.8}{1+0.8Ce^{-0.8t}}

while when there is no fishing it is

pnp(t)=0.81+Cetp_{np}(t) = \frac{0.8}{1+C'e^{-t'}}

where tt' is the time measured from the beginning of banned fishing period. Obviously t=t3t'=t-3 (in years) so we can write

pnp(t)=0.81+Ce(t3).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

p1(t)=0.810.6e0.8t.p_1(t) = \frac{0.8}{1-0.6e^{-0.8t}}.

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

p1(3)=pnp(3)p_1(3) = p_{np} (3)

and we will find CC' from this condition. We have

11+C=0.810.6e2.40.20.6e2.4=0.8CC=0.182\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


pnp=11+0.182e(t3).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

p2(t)=0.81+0.8Ce0.8tp_2(t) = \frac{0.8}{1+0.8C''e^{-0.8t''}}

where t=t6t''=t-6 is the time in years measured from the moment ban has expired. From the continuity argument pnp(6)=p2(6)p_{np}(6) = p_2(6) which yields the conditions for CC''

11+0.182e3=0.81+0.8C0.8+0.8×0.182e3=1+0.8CC=0.24\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

p2(t)=0.810.80.24e0.8(t6)=0.810.192e0.8(t6).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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