1550078481-Ordinary_Differential_Equations__Roberts_

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Applications of Systems of Equations 431

Calculating first partial derivatives, we find fx = -3x^2 , fy = -1, 9x = 1,

and gy = -3y^2. Evaluating these partial derivatives at (0, 0), we find the

associated linear system is

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


y' = 9x(O, O)x + gy(O, O)y = lx + Oy = x.
The eigenvalues of this linear system are A = ±i, so (0, 0) is a neutrally stable
center of this linear system. Since this is one of the exceptional cases, we
have no information regarding the stability characteristics at the origin of the
nonlinear system (13).


To further analyze system (13), we let r^2 (t) = x^2 (t) + y^2 (t). Thus, r(t)
is the distance of a particle from the origin at time t. Differentiating with
respect to t, we find 2rr' = 2xx' + 2yy'. Dividing this equation by 2 and
substituting for x' and y' from (13), we find


rr' = x(-y - x^3 ) + y(x - y^3 ) = -x^4 - y^4 = -(x^4 + y^4 ).


Since for (x, y) -/= (0, 0), r = Jx^2 + y^2 > 0 and x^4 + y^4 > 0, the derivative


r' = -(x^4 + y^4 )/r < 0 for all (x, y) -/= (0, 0). That is, for a particle which


is not at the origin, r' < 0 for all t. So the particle always moves toward

the origin as t increases. Hence, the origin is an asymptotically stable critical
point of the nonlinear system (13) and consequently the origin is a node or
spiral point. From the direction field shown in Figure 10.9 for


dy y'
dx x'

x -y3


-y-x3


it is fairly obvious that the origin is a spiral point and that the trajectories
spiral counterclockwise and inward toward the origin.


0 .5

y 0

-0.5

-1
-1 -0.5 0 0.5
x
Figure 10.9 Direction Field for System (13)
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