SECTION 7.1
Nonlinear Systems
519
To better understand how solutions
behave, we draw sample trajectories in the phase
plane. Note from the figure, that we still cannot
determine whether 0 0
,
a
f
is a center or a spiral
point.
–2
2
y
2
–2
x
Phase portrait of
′ =
x
y
,
′ =
−
(
)
y
x x
1
To determine whether 0 0
,
a
f
is a spiral or
center, we draw the trajectory starting at 0 05
, .
a
f
,
which shows that 0 0
,
a
f
is a stable center. Note
that the solution moves around the origin.
Examining the previous figures, we con-
clude the solutions come in from the 4th quad-
rant, either cross the
x
-axis to the right of 1 0
,
a f
and head off to the upper right, or circle around
the origin 0 0
,
a
f
and then head off to the upper
right. Points near the origin 0 0
,
a
f
are circling
around the origin.
–0.6
0.6
y
1.5
–1.5
x
Trajectory starting at 0 05
, .
a
f
spiraling
away from 0 0
,
a
f
; evidence that
both
equilibria are unstable
11.
′ =
′ =
−
x
xy
y
y
x
3cos
h
-nullcline:
y
x
=
3cos
υ
-nullcline:
x
- and
y
-axes
The equilibrium points are located at the points
0 3
,
a
f
,
±
F
H
I
K
π
2
0
,
,
±
F
H
I
K
3
2
0
π
,
,
±
F
H
I
K
5
2
0
π
,
,
…
.