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178 5. Approximation and Location of Zeros

Sketch the graph of the equation y = f(x) when a = l/2, 1, 2, 3, 4, 8. On
the same axes, sketch the graph of y = x.
If a is sufficiently small, then f(x) < x for all x > 0, and the population
modelled by this function is headed for extinction. Show that, if a > 1,
then the function f(x) has a unique fixed point u for which ‘1~ = f(u) and
O<u<l.
Is this fixed point stable or unstable? It will be stable if, when we start
the population size at a value close to u, it will, as the years pass, approach
ever more closely to 21. Otherwise it will be unstable. There is a convenient
way of picturing the situation.
Consider the example a = 3/2. In this case, the fixed point of f(x) is
l/3, and we can sketch the graph below.

Suppose that we begin with a population density xc, which we will mark
on the x-axis. To find the point on the axis corresponding to the density
xi = f(x0) one year later,

(i) draw the vertical line x = xc; it will meet the graph of the equation

y = f(x) at the point (xc, xi);

(ii) draw the horizontal line y = zi through (xe,xi); it will meet the

graph of the equation y = x at the point (xi, xi);

(iii) draw a perpendicular from (xi, xi) to meet the x-axis at (xi, 0).

By repeating this procedure, we can mark on the axes the population den-
sities 22, x3,... for successive years. Perform this for initial population
densities l/4 and l/2, and verify in both cases that the population moves
monotonically (either steadily increasing or steadily decreasing) towards
l/3.
When there is stability, it is not always the case that the population
varies monotonically. Sometimes it oscillates around the stable position
while approaching it. Sketch on the same axes the graphs of y =
(2.5)x(1-z) and y = x. With an initial population density of 0.5, compute
the density for the next five years. Using your graphs, make a diagram to
illustrate how the population oscillates around the fixed point 0.6.