(c) The fixed points for the system are found by solving ˙
x
= 0, ˙
y
= 0.
Their values are (1
,
1),
(

1
,

1), (1
,

1) and (

1
,
1).
The linearized system about these fixed points is obtained by
evaluating the Jacobian at those points, and the eigenvalues and eigenvectors for the each of the
fixed points in which we are interested are found.
The fixed points of interest are (

1
,
1) and (

1
,

1). The Jacobian of the system is
bracketleftbigg
2
xy
1

x
2
0

2
y
bracketrightbigg
.
The Jacobian at the fixed point (

1
,
1) is (

1
,
1) :
bracketleftbigg
2
0
2

2
bracketrightbigg
.
Eigenvalues are
λ
1
=

2 and
λ
2
= 2, and the eigenvectors are e
1
= [0
,
1] and e
2
= [1
,
0].
The Jacobian at the fixed point (

1
,

1) is (

1
,

1) :
bracketleftbigg

2
0
2
2
bracketrightbigg
.
Eigenvalues are
λ
1
=

2
and
λ
2
= 2, and the eigenvectors are e
1
= [1
,
0] and e
2
= [0
,
1].
(d) At the fixed point (

1
,

1) the signs of ˙
x
are such that the solutions are kept in towards the
nullcline
x
=

1 and since the signs of ˙
y >
0 the solutions move upwards, for all

1
< y <
0.
The nullcline of ˙
x
,
y
= 0 make the solutions move vertically so solutions continue vertically
across the axis. Therefore the manifold intersects the
x
axis.
To show that the system is reversible, consider ˙
x
=
y
(1

x
2
) =
f
(
x, y
)
,
˙
y
= 1

y
2
=
g
(
x, y
).
Now
f
(
x,

y
) =
y

x
2
y
=

y
+
x
2
y
=

y
(1

x
2
) =

f
(
x, y
)
g
(
x,

y
) = 1

(

y
)
2
= 1

y
2
=
g
(
x, y
)
which illustrates that
f
is odd in
y
and
g
is even in
y
. Thus there is existence of a heteroclinic
trajectory connecting (

1
,

1) to (

1
,
1).
(e) Similarly, at the fixed point (1
,

1), the signs of ˙
x
and ˙
y
are such that the solutions move
vertically from the fixed point toward the fixed point (1
,
1). As already shown above,
f
is odd
in
y
and
g
is even in
y
so there is the existence of a heteroclinic trajectory connecting (1
,

1)
to (1
,
1). The phase portrait of the system can be seen in Figure 20.
–2
–1
0
1
2
y
–2
–1
1
2
x
Figure 20: Phase portrait of the system in Exercise 4.
19