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Logistic Growth with a Threshold. As we mentioned in the last subsection, the threshold model (14) may need to be modified so that unbounded growth does not occur when y is above the threshold T. The simplest way to do this is to introduce
another factor that will have the effect of making dy/ dt negative when y is large. Thus
| = -r(l - T)(1 - K) y (17)
where r > 0 and 0 < T < K.
The graph of f (y) versus y is shown in Figure 2.5.8. In this problem there are three critical points: y = 0, y = T, and y = K, corresponding to the equilibrium solutions 0l(t) = 0, 02(t) = T, and 03(t) = K, respectively. From Figure 2.5.8 it is clear that dy/dt > 0 for T < y < K, and consequently y is increasing there. The reverse is true for y < T and for y > K. Consequently, the equilibrium solutions 01 (t) and 03 (t) are asymptotically stable, and the solution 02(t) is unstable. Graphs of y versus t have the qualitative appearance shown in Figure 2.5.9. If y starts below the threshold T, then y declines to ultimate extinction. On the other hand, if y starts above T, then y eventually approaches the carrying capacity K. The inflection points on the graphs of y versus t in Figure 2.5.9 correspond to the maximum and minimum points, y1 and y2, respectively, on the graph of f (y) versus y in Figure 2.5.8. These values can be obtained by differentiating the right side of Eq. (17) with respect to y, setting the result equal to zero, and solving for y. We obtain
y12 = (K + T ±vK2 - KT + T2)/3, (18)
where the plus sign yields y1 and the minus sign y2.
A model of this general sort apparently describes the population of the passenger pigeon,7 which was present in the United States in vast numbers until late in the nineteenth century. It was heavily hunted for food and for sport, and consequently its
i y T y K y V i y \
FIGURE 2.5.8 f (y) versus y for dy/dt = -r(1 - y/ T)(1 - y/K)y.
7See, for example, Oliver L. Austin, Jr., Birds of the World (New York: Golden Press, 1983), pp. 143-145.
Chapter 2. First Order Differential Equations
FIGURE 2.5.9 y versus t for dy/dt = r(1 y/ T)(1 y/K)y.
numbers were drastically reduced by the 1880s. Unfortunately, the passenger pigeon could apparently breed successfully only when present in a large concentration, corresponding to a relatively high threshold T. Although a reasonably large number of individual birds remained alive in the late 1880s, there were not enough in any one place to permit successful breeding, and the population rapidly declined to extinction. The last survivor died in 1914. The precipitous decline in the passenger pigeon population from huge numbers to extinction in scarcely more than three decades was one of the early factors contributing to a concern for conservation in this country.
PROBLEMS Problems 1 through 6 involve equations of the form dy/dt = f (y). In each problem sketch the
I graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable.
1. dy/dt = ay + by2, a > 0, b > 0, y0 > 0
2. dy/dt = ay + by2, a > 0, b > 0, to < y0 < to
3. dy/dt = y(y 1)(y 2), y0 > 0
4. dy/dt = ey 1, to < y0 < to
5. dy/dt = e~y 1, to < y0 < to
6. dy/dt = 2(arctany)/(1 + y2), to < y0 < to
7. Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it (see Figure 2.5.10). In this case the equilibrium solution is said to be semistable.
(a) Consider the equation
dy/dt = k(1 y)2, (i)
where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution 0(t) = 1.
(b) Sketch f (y) versus y. Show that y is increasing as a function of t for y < 1 and also for y > 1. Thus solutions below the equilibrium solution approach it, while those above it grow farther away. Therefore 0(t) = 1 is semistable.
2.5 Autonomous Equations and Population Dynamics
FIGURE 2.5.10 In both cases the equilibrium solution$(t) = k is semistable. (a) dy/dt < 0;
(b) dy/dt > 0.
(c) Solve Eq. (i) subject to the initial condition /(0) = y0 and confirm the conclusions reached in part (b).
Problems 8 through 13 involve equations of the form dy/dt = f (y). In each problem sketch the graph of f (y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 7).
8. dy/dt =k(y 1)2, k > 0, to < y0 < to
9. dy/dt = y2(y2 1), to < y0 < to
10. dy/dt = y(1 y2), to < y0 < to
11. dy/dt = ay b^y, a > 0, b > 0, y0 > 0
12. dy/dt = y2(4 y2), to < y0 < to
13. dy/dt = y2(1 y)2, to < y0 < to
14. Consider the equation dy/dt = f (y) and suppose that y1 is a critical point, that is, f (y1) = 0. Show that the constant equilibrium solution 0 (t) = y1 is asymptotically stable if f'(y1) < 0 and unstable if f'(y1) > 0.