2.
Purpose:
To study fixed points and linearizations.
From [Str94], 6.3
.
Excercise:
For each of the following systems, find the fixed points, classify them, sketch the neigh
boring trajectories, and try to fill in the rest of the phase portrait.
(a)
˙
x
=
x

y,
˙
y
=
x
2

4
(b)
˙
x
= sin
y,
˙
y
=
x

x
3
(c)
˙
x
= 1 +
y

e
−
x
,
˙
y
=
x
3

y
(d)
˙
x
=
y
+
x

x
3
,
˙
y
=

y
(e)
˙
x
= sin
y,
˙
y
= cos
x
(f)
˙
x
=
xy

1
,
˙
y
=
x

y
3
Solution:
For the following systems, the fixed points are found by solving ˙
x
= 0, ˙
y
= 0. The linearized system
about the fixed point(s) is obtained by evaluating the Jacobian at those point(s). The eigenvalues
of the resulting matrix are used to classify the fixed point(s). The phase portrait for each system in
shown in the solution to Exercise 3.
(a)
˙
x
=
x

y,
˙
y
=
x
2

4
The fixed points are (2
,
2) and (

2
,

2). The Jacobian of the system is
bracketleftbigg
1

1
2
x
0
bracketrightbigg
. Classifying
the fixed points using the eigenvalues,
(2
,
2) :
bracketleftbigg
1

1
4
0
bracketrightbigg
.
Eigenvalues are
1
2
±
1
2
I
√
15, and thus (2
,
2) is an
unstable spiral
.
(

2
,

2) :
bracketleftbigg
1

1

4
0
bracketrightbigg
.
Eigenvalues are 2
.
5616 and

1
.
5616, and thus (

2
,

2) is a
saddle
point
.
(b)
˙
x
= sin
y,
˙
y
=
x

x
3
The fixed points are (0
, nπ
), (0
,
(2
n
+ 1)
π
), (1
, nπ
), (1
,
(2
n
+ 1)
π
), (

1
, nπ
), and (

1
,
(2
n
+
1)
π
). The Jacobian of the system is
bracketleftbigg
0
cos
y
1

3
x
2
0
bracketrightbigg
. Classifying the fixed points using the
eigenvalues,
(0
, nπ
) :
bracketleftbigg
0
1
1
0
bracketrightbigg
.
Eigenvalues are 1 and

1, and thus (0
, nπ
) is a
saddle point
.
(1
, nπ
)
,
(

1
, nπ
) :
bracketleftbigg
0
1

2
0
bracketrightbigg
.
Eigenvalues are
I
√
2 and

I
√
2, and thus a fixed point of this
form is a
center
.
(0
,
(2
n
+ 1)
π
) :
bracketleftbigg
0

1
1
0
bracketrightbigg
.
Eigenvalues are
I
and

I
, and thus (0
,
(2
n
+ 1)
π
) is a
center
.
(1
,
(2
n
+ 1)
π
)
,
(

1
,
(2
n
+ 1)
π
) :
bracketleftbigg
0

1

2
0
bracketrightbigg
.
Eigenvalues are
√
2 and

√
2, and thus a fixed
point of this form is a
center
.
(c)
˙
x
= 1 +
y

e
−
x
,
˙
y
=
x
3

y
The only fixed point is (0
,
0). The Jacobian of the system is
bracketleftbigg
e
−
x
1
3
x
2

1
bracketrightbigg
. Classifying the fixed
points using the eigenvalues,
(0
,
0) :
bracketleftbigg
1
1
0

1
bracketrightbigg
.
Eigenvalues are 1 and

1, and thus (0
,
0) is a
saddle point
.
(d)
˙
x
=
y
+
x

x
3
,
˙
y
=

y
The fixed points are (0
,
0), (1
,
0), and (

1
,
0). The Jacobian of the system is
bracketleftbigg
1

3
x
2
1
0

1
bracketrightbigg
.
Classifying the fixed points using the eigenvalues,
(0
,
0) :
bracketleftbigg
1
1
0

1
bracketrightbigg
.
Eigenvalues are 1 and

1, and thus (0
,
0) is a
saddle point
.
13