Population extinction in the range r = (0, 1)

We express that function (1) intersects with the 45° diagonal by the following equation

By dividing each side of equation (3) by r, we get a quadratic equation x² - ()x = 0. After factoring out x, we obtain an alternate expression [x - ()]x = 0 which shows

is one solution. This simply says that function (1) and the diagonal start out from the common origin at x = 0. Now, the other solution is x - () = 0, so we can write the second solution as

Now, this is where one expects function (1) will come to cross the diagonal somewhere along the x-axis of figure 2, other than the origin x = 0. As it turns out, when r is less than 1, solution (5) makes no sense because x is negative, so that only the first solution (4) is good in the range r = (0, 1). To see this, two round curves of function (1) for r = 1 and 0.5 are plotted in figure 3, and neither of them can cross the diagonal except for at the origin x = 0. Hence, solution (4) represents the fixed point of population model (1) in the range r = (0, 1)

Figure 3. Intersection of function (1) with the diagonal

For the initial x₀ of table 4, you can verify by Prog#9 that x = 0 is the final population. This is evidenced by that the vertical and horizontal line segments always move toward x = 0, as the iteration proceeds, in the graphical iteration plot of Prog#9. In fact, x = 0 is the stable fixed point of function (1), and we, therefore, conclude population annihilation over the entire range r = (0, 1).

Table 4. Iteration data set for r = (0, 1)

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