A population of fish is modeled by the logistic equation modified to account for harvesting: 4

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1 Partial solution key A population of fish is modeled by the logistic equation modified to account for harvesting: dddd dddd =.pp pp aa or equivalently dddd dddd = 6 pp + pp aa. What are the equilibrium solutions? Note: your answer(s) will depend on a. SSSSSSSSSS 6 pp + pp aa = tttt ffffffff pp = ± aa. Determine the range of values of a for which this model has different equilibrium values. When does it have equilibrium? equilibria? Carefully interpret each of these scenarios in the context of this model.. Suppose a has the value for which there is equilibrium. What is the value of this equilibrium? Is it an attractor, a repellor, or a node? What does this mean to our population? (answer to and ) If a</ the above formula yields equilibria. If a=/ there is one, and if a>/ there are none. If a>/, then dp/dt is never, and in fact it is always negative. So no matter how large the population is initially, it will always decrease, and never level off (until the population goes extinct). If a=/ there is an equilibrium at p= which is a node. Any population size greater than would decrease toward, any population smaller than would die off. This is not a good situation for the fish! No matter how large the population starts it will quickly approach a level of, and from there any slight perturbation that sends the size below dooms it to extinction. If a</ the population has stable values, a smaller one (source) and a larger one (sink). The population will stabilize at the sink regardless of initial size so long as it is larger than the source. If it is smaller than the source it is again doomed to die off. For # and #, let a =.6. What will happen to the population in the long run? If your answer depends on the initial population, be sure to include all possible scenarios. There is a sink at p=. and a source at p=.8. If p is initially one of these values it remains constant. The other scenarios are as described in #.. a. Use MATLAB to find the analytical solution to this differential equation. Type dddddddddddd( DDDD = /6 pp^ + / pp. 6 ) b. Use MATLAB to sketch a graph which shows solution curves for this system with several different initial values. Be sure to show all of the scenarios found in #.

2 This graph is generated with our Euler code, using multiple ICs, including.8,., and several others above and below it. 6. Now choose new values for a; one in which a equals the value that you found in #, and one when a is slightly larger than that. Repeat both parts of question # for each of these values of a. Solve with MATLAB as above. The graph should look like this for a=.:

3 And this for a=. (your choice of a value may be slightly different) Notice that all of the curves constantly decrease, and eventually dip below and then on to Here is the same graph on a longer time scale, more clearly showing the descent toward : -

4 Now assume that the harvesting is not done at a constant rate, but rather at different rates at different times of the year. This can be modeled by dddd =.pp pp cc( + sin(tt)). This dddd problem can no longer be solved analytically, but can still be analyzed graphically in much the same way as you did above. 7. Let c=.6. Use MATLAB to graph approximate solutions for several different values of p(), and interpret what you see. Turn in the graphs together with your analysis. Note: <p<, and <t< is a reasonable viewing window to start with, though you may want to change it as you proceed. - A second graph, again with c=.6, picking IC s only between.8 (blue above, which dies off) and (yellow above, which survives) looks like: -

5 And again, this time looking between.9 and, we see the critical threshold is around.96 - There is no equilibrium (all solutions change over time), but if the population is below.96 initially it will die off, while if above this amount, regardless of initial quantity, will settle into a repeating oscillation at around -. (more detail can be determined by zooming in to determine period and amplitude) 8. Repeat question 7 with c=. This time all initial conditions oscillate but trend downward, and eventually the population will die off, regardless of IC. -