Problem 2. [Lotke-Volterra equation.] In class we considered the competition between two species x ('prey') and y ('pred

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Problem 2 Lotke Volterra Equation In Class We Considered The Competition Between Two Species X Prey And Y Pred 1 (172.36 KiB) Viewed 42 times

Problem 2. [Lotke-Volterra equation.] In class we considered the competition between two species x ('prey') and y ('predator') mathematically described by the Lotke-Volterra model as (1) dr = ar - Bry dt dy = 0xy — уу, dt where a, b, 7 and 8 are constants (see the linked articles on Canvas for their meaning). If, instead of the time-dependence of each species, we are interested in the amount predators as a function of the amount of prey, by dividing the two equations we get dy dc y dr – 7 x By - α We can then transform this equation to (8-9) dy=-(6-3) dr , B- = which can be directly integrated. We get the implicit equation By - alny= -83 + In x +C where C is the integration constant. Suppose that at the initial time there is to prey animals and yo predator animals. The initial conditions are then imply that C = Byo - aln yo + 8x0 - Inco , and substituting the constant into the above implic equations we get y y B - a In = -8 + In 1-() (2) Yo To Το The boxed equation is the starting point of this problem. To be specific, for the rest of the problem we will choose the constants to have values a = 2, B = 1/2, 7 = 1, 8=1/4.
30 20 10 The easiest way to make plots for this problem it is to use the ready-made commands for plotting implicit functions in MatLab or Mathematica - the image above is pro- duced using Mathematica's ContourPlot. MatLab's function is called fimplicit (a) Plot the boxed equation (2) with initial points zo = 3, yo = 4. The resulting image should be like the one shown above. The closed loop signifies that the function is periodic. Note that the steep decline in y that occurs once the x- component becomes small represents the collapse of the predator population once the prey becomes scarce. The plot of this type in mathematics is called phase space plot. (b) Keep the initial number of predators the same as before yo = 4, but change the initial number of prey animals as Xo = 1,1.5, 2, 2.5,3. Try to make all these plots on the same graph, so you can compare them. You should see that, while the maximal amount of prey changes significantly, the maximum number of predators stays the same. Also, note the more rapid collapse of the predator population with the smaller initial amount of prey. (c) Consider equations ). Suppose that you have an equilibrium between the preda- tor and the prey, i.e. that there is no time dependence on the right hand side of the equation. Find x and y that correspond to that condition. One solution will be trivial x = 0 and y=0, but what is the other one? Find the other equilibrium solution in symbolic form expressed of constants a, b, 7 and 8, and find their numerical values with the constants given before. (d) (optional) If you set the initial values X, and yo to be equal to the equilibrium values you found in (c) you are still not going to get only one point in the phase space plot. Why?