Problems 14 through 17 deal with the predator–prey system

Here each population-the prey population x(t) and the predator population y(t)-is an unsophisticated population (like the alligators of Section 2.1) for which the only alternatives (in the absence of the other population) are doomsday and extinction. Problems 14 through 17 imply that the four critical points (0,0), (0,4), (2,0), and (3, 1) of the system in (5) are as shown in Fig. 9.3.15-a nodal sink at the origin, a saddle point on each coordinate axis, and a spiral source interior to the first quadrant. This is a two-dimensional version of "doomsday versus extinction." If the initial point (x₀, y₀) lies in Region I, then both populations increase without bound (until doomsday), whereas if it lies in Region II, then both populations decrease to zero (and thus both become extinct). In each of these problems use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.15?

Show that the linearization of (5) at (0,4) is u' = -6u, v' = 4u + 4v. Then show that the coefficient matrix of this linear system has the negative eigenvalue λ₁ = -6 and the positive eigenvalue λ₂ = 4. Hence (0,4) is a saddle point for the system in (5).

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dx dt dy dt =x²-2x-xy, = = y² - 4y + xy. (5)