# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

**Download**(direct link)

**:**

**59**> 60 61 62 63 64 65 .. 486 >> Next

Nonlinear Equations. Nonlinear difference equations are much more complicated and have much more varied solutions than linear equations. We will restrict our attention to a single equation, the logistic difference equation

Yn+i = pyn (1 - y) , (19)

which is analogous to the logistic differential equation

- = ry( i - 4)

dt Ki

(20)

2.9 First Order Difference Equations

119

that was discussed in Section 2.5. Note that if the derivative dy/ dt in Eq. (20) is replaced by the difference (yn+1 — yn)/h, then Eq. (20) reduces to Eq. (19) with p = 1 + hr and k = (1 + hr)K/ hr. To simplify Eq. (19) a little more, we can scale the variable yn by introducing the new variable un = yn/ k. Then Eq. (19) becomes

Un+1 = p un(1 — un). (21)

where p is a positive parameter.

We begin our investigation of Eq. (21) by seeking the equilibrium, or constant solutions. These can be found by setting u 1 equal to un in Eq. (21), which corresponds to setting dy/dt equal to zero in Eq. (20). The resulting equation is

un = Pun — P u2n’ (22)

so it follows that the equilibrium solutions of Eq. (21) are

un = 0 un = P? (23)

p

The next question is whether the equilibrium solutions are asymptotically stable or unstable; that is, for an initial condition near one of the equilibrium solutions, does the resulting solution sequence approach or depart from the equilibrium solution? One way to examine this question is by approximating Eq. (21) by a linear equation in the neighborhood of an equilibrium solution. For example, near the equilibrium solution un = 0, the quantity u2n is small compared to un itself, so we assume that we can neglect the quadratic term in Eq. (21) in comparison with the linear terms. This leaves us with the linear difference equation

un+1 = P un, (24)

which is presumably a good approximation to Eq. (21) for un sufficiently near zero. However, Eq. (24) is the same as Eq. (7), and we have already concluded, in Eq. (9), that un ^ 0 as n if and only if Ipl < 1, or since p must be positive, for 0 < p < 1. Thus the equilibrium solution un = 0 is asymptotically stable for the linear approximation (24) for this set of p values, so we conclude that it is also asymptotically stable for the full nonlinear equation (21). This conclusion is correct, although our argument is not complete. What is lacking is a theorem stating that the solutions of the nonlinear equation (21) resemble those of the linear equation (24) near the equilibrium solution un = 0. We will not take time to discuss this issue here; the same question is treated for differential equations in Section 9.3.

Now consider the other equilibrium solution un = (p — 1)/p. To study solutions in the neighborhood of this point, we write

un = p—1 + V (25)

p

where we assume that vn is small. By substituting from Eq. (25) in Eq. (21) and simplifying the resulting equation, we eventually obtain

Vn+1 = (2 — p)vn — pvl- (26)

Since vn is small, we again neglect the quadratic term in comparison with the linear terms and thereby obtain the linear equation

vn+1 = (2 - P)vn.

(27)

120

Chapter 2. First Order Differential Equations

Referring to Eq. (9) once more, we find that vn ^ 0as n ^ <x for 12 — pi < 1, that is, for 1 < p < 3. Therefore we conclude that, for this range of values of p, the equilibrium solution un = (p — 1)/p is asymptotically stable.

Figure 2.9.1 contains the graphs of solutions of Eq. (21) for p = 0.8, p = 1.5, and p = 2.8, respectively. Observe that the solution converges to zero for p = 0.8 and to the nonzero equilibrium solution for p = 1.5 and p = 2.8. The convergence is monotone for p = 0.8 and p = 1.5 and is oscillatory for p = 2.8. While the graphs shown are for particular initial conditions, the graphs for other initial conditions are similar.

Another way of displaying the solution of a difference equation is shown in Figure

2.9.2. In each part of this figure the graphs of the parabola y = px(1 — x) and of the straight line y = x are shown. The equilibrium solutions correspond to the points of intersection of these two curves. The piecewise linear graph consisting of successive vertical and horizontal line segments, sometimes called a stairstep diagram, represents the solution sequence. The sequence starts at the point u0 on the x-axis. The vertical line segment drawn upward to the parabola at u0 corresponds to the calculation of pu0(1 — u0) = u1. This value is then transferred from the y-axistothe x-axis; this step is represented by the horizontal line segment from the parabola to the line y = x. Then the process is repeated over and over again. Clearly, the sequence converges to the origin in Figure 2.9.2a and to the nonzero equilibrium solution in the other two cases.

To summarize our results so far: The difference equation (21) has two equilibrium solutions, un = 0and un = (p — 1)/p; the former is stable for 0 < p < 1, and the latter is stable for 1 < p < 3. This can be depicted as shown in Figure 2.9.3. The parameter p is plotted on the horizontal axis and u on the vertical axis. The equilibrium solutions u = 0 and u = (p — 1)/p are shown. The intervals in which each one is stable are indicated by the heavy portions of the curves. Note that the two curves intersect at p = 1, where there is an exchange of stability from one equilibrium solution to the other.

**59**> 60 61 62 63 64 65 .. 486 >> Next