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
′ =
xy
,
′
=−
()
yx
x
1
To determine whether 0 0
,
a
f
is a spiral or
center, we draw the trajectory starting at 0 05
,.
af
,
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
,
and head off to the upper right, or circle around
the origin 0 0
,
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
spiraling
away from 0 0
,
a
f
; evidence that
both
equilibria are unstable
11.
′ =
′
xx
y
yy
x
3cos
h
-nullcline:
=
3cos
υ
-nullcline:
x
- and
y
-axes
The equilibrium points are located at the points
03
,
a
f
,
±
F
H
I
K
π
2
0
,
,
±
F
H
I
K
3
2
0
,
,
±
F
H
I
K
5
2
0
,
,
…
.
–1
–2
1
2
y
x
3
–3
2
3
–2
–
–3
Nullclines of
′ =
y
,
′
x
The diagram that follows indicates that equilibrium at 0 3
,
is unstable and that the equilibrium
points on the
x
-axis appear to alternate between saddle points and centers or spirals with the
saddle points closest to the origin.
. (The
x
-intercepts at
±
2
,
±
5
2
,
±
9
2
,
…