Applications of Systems of Equations 413Case la. In this case, we are assuming r = s = 0 and the slope of £ 1 is
greater than the slope of L2 ( C /A > B / D). Regions I , II, and III refer to
the regions shown in Figure 10.2 a. In region I, dx/dt < 0 and dy/dt > 0.
So if a particle is in region I, it will tend to move to the left and up untilit reaches the line L2. On L2, dx/dt < 0 and dy/dt = 0, so a particle from
region I crosses L2 into region II. If a particle is in region III, dx / dt > 0 and
dy / dt < 0. So a particle in region III tends to move to the right and down
until it reaches the line L1. On £ 1 , dx/dt = 0 and dy/dt < 0, so a particle
from region III crosses £ 1 into region II. In region II, dx/dt < 0 and dy/dt < 0.
So a particle in region II moves to the left and down. If at some instant in
time the particle were to move from region II to the line £ 1 , then dx / dt = 0
and dy/dt < 0 and the particle would move downward into region II again. A
similar argument holds for any particle in region II which approaches the line
L2. Such a particle would move to the left and return to region II. Thus, any
particle in region II or any particle crossing into region II from either region I
or region III remains in region II and approaches the origin as t -> oo. Thus,
for any initial conditions x(to) = xo 2': 0, y(to) = Yo 2': 0- that is, for any
initial point in the first quadrant- x(t) -> 0 and y(t) -> 0 as t -> oo. Hence,
case la. results in mutual disarmament regardless of the initial conditions (the
initial expenditures for arms).Case lb. This case is left as an exercise.
Case 2. Supposer< 0, s < 0 and £ 1 and L2 are not parallel. Since r < 0
and s < 0 both nations have a permanent underlying feeling of "goodwill"
toward the other nation. Since £ 1 and £ 2 are not parallel, they intersect at
an equilibrium point (x,y) and either (a.) the slope of £ 1 , namely C/A, is
greater than the slope of £ 2 , namely B / D, or (b .) the slope of £ 1 is less than
the slope of L2 (C/A < B/D).
Case 2a. This case is left as an exercise.Case 2b. We now consider case where C /A < B / D. Multiplying this in-
equality by AD which is positive, since A and D are assumed to be positive,
we find CD< AB or E =CD -AB< 0. Since A> 0 , B > 0, C > 0, D > 0,
r < 0, s < 0 and E< 0, we see from equations (4) that x* = (rD+sA)/E > 0
and y =(Cs+ Br)/E > 0. Thus, the equilibrium point (x,y*) li es in the
first quadrant. A sketch of the situation for this particular case and a table
indicating the sign of dx/dt and dy/dt in each region is shown in Figure 10.3.
The arrows in each region indicate the horizontal and vertical direction a
particle in the region will take. Notice that all horizontal arrows point toward
the line £ 1 while all vertical arrows point toward the line L2.
Let the initial values for arms expenditures be (xo, Yo) and assume (xo, Yo)
lies in region I of Figure 10.3. In region I, dx/dt > 0 and dy/dt > 0. So a
particle in region I will tend to move to the right and up. That is, the particle
will tend to move farther out in region I. If the particle were to reach the
line £ 1 , dx/dt would be zero but dy/dt would be positive. Hence, at L1 the