LECTURE 32: STABILITY AND CLASSIFICATION OF ISOLATED CRITICAL POINTS
MINGFENG ZHAO
November 21, 2014
Stability and classification of isolated critical points
The following table shows the behavior of an almost linear system near an isolated critical point:
Eigenvalues of the Jacobian matrix
Behavior
Stability
I.
real and both positive
source(unstable node)
unstable
II.
real and both negative
sink (stable node)
asymptotically stable
saddle
unstable
III. real and opposite signs
V.
complex with positive real part
spiral source
unstable
VI.
complex with negative real part
spiral sink
asymptotically stable
Example 1. Find all critical points and classify the stabilities of these critical points for x0 = 2 sin(x) cos(y), y 0 =
− cos(x) sin(y).
Let’s solve
2 sin(x) cos(y) = 0,
and
− cos(x) sin(y) = 0.
Since 2 sin(x) cos(y) = 0, then either sin(x) = 0 or cos(y) = 0. We have the following two cases:
I. If sin(x) = 0, then x = mπ for m ∈ Z. Since sin2 (x) + cos2 (x) = 1, then cos(x) 6= 0. Since cos(x) sin(y) = 0,
then sin(y) = 0, which implies that y = nπ for n ∈ Z.
π
II. If cos(y) = 0, then y = + mπ for m ∈ Z. Since sin2 (y) + cos2 (y) = 1, then sin(y) 6= 0. Since cos(x) sin(y) = 0,
2
π
then cos(x) = 0, which implies that x = + nπ for n ∈ Z.
2
In summary, all critical points are:
(mπ, nπ),
and
π
2
+ mπ,
1
π
+ nπ ,
2
∀m, n ∈ Z.
2
MINGFENG ZHAO

Let f (x, y) = 2 sin(x) cos(y) and g(x, y) = cos(x) sin(y), then the Jacobian matrix of 
f (x, y)

 is:
g(x, y)

2 cos(x) cos(y) −2 sin(x) sin(y)

− cos(x) cos(y)
sin(x) sin(y)

.
Then
I. At (mπ, nπ), linearization is:

u

0

 =

2 cos(mπ) cos(nπ)
0
0
− cos(mπ) cos(nπ)
v
u


.
v
Notice that

det 

Then 

2 cos(mπ) cos(nπ)
0
0
− cos(mπ) cos(nπ)
 = −2 cos2 (mπ) cos2 (nπ) = −2 6= 0.

2 cos(mπ) cos(nπ)
0
0
cos(mπ) cos(nπ)
 is invertible, which implies that the system is almost linear

at (mπ, nπ). It’s easy to see that the eigenvalues of 
2 cos(mπ) cos(nπ)
0
0
cos(mπ) cos(nπ)

 are:
and λ2 = − cos(mπ) cos(nπ).
λ1 = 2 cos(mπ) cos(nπ),
Therefore, we know that (mπ, nπ) is an unstable equilibrium point of the system, solutions behaves like a
saddle near (mπ, nπ).
π
π
II. At
+ mπ, + nπ , linearization is:
2
2

0 
u
0

 =
v
sin π2 + mπ sin
Notice that

det 
sin
π
2
0
+ mπ sin
−2 sin
π
2
+ nπ
−2 sin
π
2
+ nπ
π
2
+ mπ sin
π
2
+ mπ sin
π
2
+ nπ
0
π
2
+ nπ

 =
2 sin2
0
=
2
6=
0.


u

.
v
π
π
+ mπ sin2
+ nπ
2
2
LECTURE 32: STABILITY AND CLASSIFICATION OF ISOLATED CRITICAL POINTS

Then 
sin
π
2
0
+ mπ sin
−2 sin
π
2
+ nπ
π
2
+ mπ sin
π
2
+ nπ
3

 is invertible, which implies that
0
π
the system is almost linear at
+ mπ, + nπ .
It’s easy to see that the eigenvalues of
2
2


0
−2 sin π2 + mπ sin π2 + nπ

 are:
π
π
0
sin 2 + mπ sin 2 + nπ
π
λ1 =
Therefore, near critical points
π
2
√
+ mπ,
√
and λ2 = − 2i.
2i,
π
+ nπ , solutions behaves like a center , we will have trouble to
2
determine the stability.
The trouble with centers
Recall, a linear system with a center meant that trajectories traveled in closed elliptical orbits in some direction around
the critical point. It would not be an asymptotically stable critical point, as the trajectories would never approach the
critical point, but at least if you start sufficiently close to the critical point, you will stay close to the critical point.
The trouble with a center in a nonlinear system is that whether the trajectory goes towards or away from the critical
point is governed by the sign of the real part of the eigenvalues of the Jacobian. Since this real part is zero at the critical
point itself, it can have either sign nearby, meaning the trajectory could be pulled towards or away from the critical
point.
Example 2. Consider the system x0 = y, y 0 = −x + y 3 . It’s easy to see that (0, 0) is the only critical point. Let


f
(x,
y)
 is
f (x, y) = y and g(x, y) = −x + y 3 , then the Jacobian matrix of 
g(x, y)



So at (0, 0), the Jacobian matrix is 
0
1
−1
0
0
−1

1
3y
2
.

 which has eigenvalues ±i, that is, the linearization at (0, 0) has a
center.
On the other hand, the eigenvlaues of the Jacobian matrix at any point (x, y) are
p
3
4 − 9y 4
.
λ1,2 = y 2 ± i
2
2
4
MINGFENG ZHAO
For any point where y 6= 0, the eigenvalues have a positive real part, which will pull the trajectory away from the
origin.
Figure 1. An unstable critical point (spiral source) at the origin for x0 = y, y 0 = −x + y 3
Conservative equations
Definition 1. An equation of the form x00 + f (x) = 0 is called a conservative equation.
Let y = x0 , then we have the system:
x0 = y,
and y 0 = −f (x).
Then
f (x)
dy
=−
.
dx
y
Solve the above equation, we get
1 2
y +
2
Z
f (x) dx = C,
which is called the Hamiltonian or energy of the system.


 
p
y
0
1
 which has eigenvalues λ1,2 = ± −f 0 (x).
 is 
It’s easy to see that the Jacobian matrix of 
−f (x)
−f 0 (x) 0
Therefore, there are never any asymptotically stable points for the system x0 = y, y 0 = −f (x).
LECTURE 32: STABILITY AND CLASSIFICATION OF ISOLATED CRITICAL POINTS
5
Example 3. Find the trajectories for the equation x00 + x − x2 = 0.
Let y = x0 , then
x0 = y,
y 0 = x00 = x2 − x.
So we get
x2 − x
dy
=
.
dx
y
Solve the above differential equation, we know that the trajectories satisfy
1 2
1
1
y = x3 − x2 + C .
2
3
2
Figure 2. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca