In: Advanced Math
In this project we explore how two populations develop when one preys on the other. Clearly if there are no predators, the prey population will keep growing, whereas if there are no prey, the predators will go extinct. Suppose x and y denote the populations of the prey and predators respectively.
If y = 0, we will assume that
dx/dt = ax, a > 0.
If y does not equal 0, it is natural to assume that the number of encounters between predators and prey is jointly proportional to x and y. If we further assume a proportion of these encounters leads to the prey being eaten, we have
dx/dt = dx − bxy, a, b > 0.
Similarly, we have
dy/dt = −cy + dxy, c, d > 0.
This system of equations is called Volterra’s predator-prey equations.
Part a) Solve this system of equations to find solutions in the form g(y) = f(x). You will see that we cannot explicitly find y in terms of x, so our solutions are implicit. We can still, however, study these solutions.
Part b) Suppose g(y) = C1, where C1 is a constant. Determine how many solutions there are to this equation by using calculus techniques. Note this may well depend on the value of C1. Do the same thing for f(x) = C2, where C2 is a constant.
Part c) Hence determine the shape of the trajectories in the x, y-plane (do a sketch!), and their directions.
Part d) Clearly the system has a rest point at x = c/d and y = a/b. By making the substitutions x = c/d + X and y = a/b + Y , assuming X and Y are small enough that we can neglect any second order terms in X and Y , show that near the rest point, trajectories are approximately ellipses.
Part e) Finally, sketch graphs of x(t) and y(t) against t on the same axes. To help, show that d^2 * y / dt^2 > 0 whenever dx/dt > 0 are think about what this means in terms of the shapes of the graphs.