1550078481-Ordinary_Differential_Equations__Roberts_

428 Ordinary Differential Equations

EXAMPLE 7 Determining Critical Points and Type of Stability

for a Nonlinear System

Find all critical points of the nonlinear system

x' = x - xy = f(x, y)

(11)

y' = y - xy = g(x, y).

For each critical point, determine the type of stability (asymptotically stable,
neutrally stable, or unstable) and the type of critical point (node, saddle point,
spiral point, or center).

SOLUTION

Simultaneously solving

(12a)

(12b)

x - xy = x(l - y) = 0

y - xy = y(l - x) = 0 ,

we find from (12a) that x = 0 or y = l. Substituting x = 0 into (12b), we

get y = 0. So (0, 0) is a critical point of system (11). And substituting y = 1

into (12b) and solving for x, we ge t x = l. So (1, 1) is also a critical point

of system (11). Since f(x, y) = x - xy, f x = 1 - y and fy = -x. And since

g(x, y) = y - xy, gx = -y and gy = 1 - x.

At (0, 0) t he linear system associated with the nonlinear system (11) is

x ' = f x(O, O)(x - 0) + fy(O, O)(y - 0) = l x + Oy = x

y' = gx(O, O)(x - 0) + gy(O, O)(y - 0) = Ox+ ly = y.

The eigenvalues of this system, which are the eigenvalues of the matrix

A= G ~)

are >- 1 = >-2 = 1 > 0. So, from Table 10.l the origin is an unstable node of
t his linear system and an unstable critical point of the nonlinear system (11).

At (1, 1) t he linear system associated with the nonlinear system (11) is

x' = fx(l, l)(x - 1) + fy(l, l)(y-1) = O(x - 1) + (-l)(y - 1) = (-l)(y -1)

y' = gx(l, l)(x - 1) + gy(l, l)(y-1) = (-l)(x - 1) + O(y-1) = (-l)(x - 1).