LECTURE 31: LINEARIZATION FOR SYSTEM AND THE CLASSIFICATION OF CRITICAL
POINTS
MINGFENG ZHAO
November 27, 2015
Let λ = m + ik be a complex eigenvalue of the real 2 × 2 matrix A, then we can take an eigenvector of the form


1
, so we know that
~v = 
a + ib


cos(kt)
.
Re(eλt~v ) = emt 
A cos(kt) + B sin(kt)
To draw the correct direction of the above spiral or ellipse, you just need to know the positions of five points at t = 0,
π π 3π
2π
, ,
and
.
2k k 2k
k
Linearization of a system

 x0 = f (x, y)
Definition 1. Let (x0 , y0 ) be a critical point to the autonomous system
, that is, f (x0 , y0 ) = g(x0 , y0 ) =
 y 0 = g(x, y)

 x0 = f (x, y)
0. Let u = x − x0 and v = y − y0 , the linearization at (x0 , y0 ) of the autonomous system
is:
 y 0 = g(x, y)

in which the matrix 

u0

v0

=
fx (x0 , y0 ) fy (x0 , y0 )
gx (x0 , y0 )

fx (x0 , y0 ) fy (x0 , y0 )

u

gx (x0 , y0 )
gy (x0 , y0 )

,
v


 is called the Jacobian matrix of the vector function 
gy (x0 , y0 )
f (x, y)

 at the
g(x, y)
point (x0 , y0 ).

 x0 = f (x, y)
Remark 1. The Linearization of an autonomous system
 y 0 = g(x, y)
at a critical point (x0 , y0 ) is the just the
linearizations of each component f (x, y) and g(x, y) at this critical point (x0 , y0 ).

 x0 = y
.
Example 1. Find all critical points and linearizations at these critical points for the system
 y 0 = −x + x2
1
2
MINGFENG ZHAO
It’s easy to see that there are two critical points: (0, 0) and (1, 0). Let f (x, y) = y and g(x, y) = −x + x2 , then the


f (x, y)
 is:
Jacobian matrix of 
g(x, y)


fx (x, y) fy (x, y)
=

gx (x, y)
gy (x, y)

 x0 = y
• At (0, 0) the linearization of
 y 0 = −x + x2

u0

v0


=
0
1
−1
0
u0

v
0


=
0
1

0
−1 + 2x 0

.
,
with u = x and v = y.
v
is:

u
with u = x − 1 and y = y.
,

1
1

u


0
is:

 x0 = y
• At (1, 0) the linearization of
 y 0 = −x + x2


v
Example 2. Find all critical points and linearizations at these critical points for the system x0 = (1 − y)(2x − y),
y 0 = (2 + x)(x − 2y).
Let’s solve
(1 − y)(2x − y) = 0,
and
(2 + x)(x − 2y) = 0.
Then there are four critical points:
(−2, 1),
(2, 1),
(−2, −4)
and
(0, 0).
Let f (x, y) = (1 − y)(2x − y) = 2x − y − 2xy + y 2 and g(x, y) = (2 + x)(x − 2y) = 2x − 4y + x2 − 2xy, then the


f (x, y)
 is:
Jacobian matrix of 
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)
Therefore, we know that
gy (x, y)


=
2 − 2y
−1 − 2x + 2y
2 + 2x − 2y
−4 − 2x

.
LECTURE 31: LINEARIZATION FOR SYSTEM AND THE CLASSIFICATION OF CRITICAL POINTS
3

 x0 = (1 − y)(2x − y)
I. At (−2, 1) the linearization of
is:
 y 0 = (2 + x)(x − 2y)


u0

v

=
0
0
5

u

−4

with u = x + 2 and v = y − 1.

0
v

 x0 = (1 − y)(2x − y)
II. At (2, 1) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v0


=
0
−3
4
−4

u


with u = x − 2 and v = y − 1.

v

 x0 = (1 − y)(2x − y)
III. At (−2, −4) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v0


=
10 −5

u

6
0


with u = x + 2 and v = y + 4.
v

 x0 = (1 − y)(2x − y)
IV. At (0, 0) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v
0


=
2
2
−1
−4

u



with u = x and v = y.
v
Isolated critical points and almost linear systems

 x0 = f (x, y)
Definition 2. Let (x0 , y0 ) be a critical point of the system
, that is, f (x0 , y0 ) = g(x0 , y0 ) = 0, then
 y 0 = g(x, y)
I. We say that the critical point (x0 , y0 ) is isolated if it is the only critical point in a small “neighborhood” of
(x0 , y0 ).

 x0 = f (x, y)
II. We say that the system
is almost linear at critical point (x0 , y0 ) if the critical point ~a is isolated,
 y 0 = g(x, y)


f (x, y)
 at (x0 , y0 ) is invertible.
and the Jacobian matrix of 
g(x, y)
Remark 2. If the Jacobian matrix at ~a is invertible, then ~a is isolated. Therefore, it suffices to verify that the Jacobian
at ~a is invertible.
4
MINGFENG ZHAO
NonExample 1. It’s easy to see (0, 0) is a critical point of the system x0 = x2 , y 0 = y 2 , but the system is


0 0
 which is not invertible.
not almost linear at (0, 0), because the Jacobian matrix at (0, 0) is 
0 0
Stability and classification of isolated critical points

 x0 = f (x, y)
In the following, we assume that (x0 , y0 ) is a critical point of the system
, that is, f (x0 , y0 ) =
 y 0 = g(x, y)


f (x, y)
 at (x0 , y0 ) is invertible. Also we assume that the Jacobian matrix
g(x0 , y0 ) = 0, and the Jacobian matrix of 
g(x, y)


f (x, y)
 at (x0 , y0 ) has two different nonzero eigenvalues λ1 and λ2 . Notice that the linearization of the system
of 
g(x, y)

 x0 = f (x, y)
at (x0 , y0 ) is:
 y 0 = g(x, y)

u0

v0


=
fx (x0 , y0 ) fy (x0 , y0 )

u

gx (x0 , y0 )
gy (x0 , y0 )


with u = x − x0 and v = y − y0 .
v

 x0 = f (x, y)
The following table shows the behavior of the system
near this isolated critical point (x0 , y0 ):
 y 0 = g(x, y)
Eigenvalues of the Jacobian matrix at (x0 , y0 )
Type of (x0 , y0 )
Stability of (x0 , y0 )
I.
λ1 and λ2 are real and both positive
source
unstable
II.
λ1 and λ2 are real and both negative
sink
asymptotically stable
III. λ1 and λ2 are real and opposite signs
saddle point
unstable
IV.
λ1 and λ2 are complex with positive real part
spiral source
unstable
V.
λ1 and λ2 are complex with negative real part
spiral sink
asymptotically stable
VI.
λ1 and λ2 are complex with zero real part
center
Example 3. Find all critical points and classify the stabilities of these critical points for the system x0 = −y − x2 ,
y 0 = −x + y 2 .
Let’s solve −y − x2 = 0 and −x + y 2 = 0, then
(x, y) = (0, 0),
and
(x, y) = (1, −1).
LECTURE 31: LINEARIZATION FOR SYSTEM AND THE CLASSIFICATION OF CRITICAL POINTS
5
That is, all critical points of the system x0 = −y − x2 , y 0 = −x + y 2 are:
(0, 0) and (1, −1) .

Let f (x, y) = −y − x2 and g(x, y) = −x + y 2 , then Jacobian matrix of 
f (x, y)

 is
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)


=
gy (x, y)
−2x
−1
−1
2y

.
Therefore, we know that

I. At (0, 0) the linearization is 
u0
v0


=
0
−1
−1
0

u



 with u = x and v = u. Notice that 
v

is invertible, then the system is almost linear at (0, 0). Since the eigenvalues of 
0
−1
−1
0
0
−1
−1
0



 are λ1 = −1 and
λ2 = 1, then (0, 0) is an unstable critical point of the system, solutions behaves like a saddle point near (0, 0).





u0
2 −1
u
 = 

 with u = x − 1 and v = y + 1. Notice that
II. at (1, −1) the linearization is 
0
v
−1 −2
v




2 −1
2 −1

 is invertible, then the system is almost linear at (1, −1) . Since the eigenvalues of 

−1 −2
−1 −2
are λ1 = −1 and λ2 = −3, then (1, −1) is an asymptotically stable critical point of the system, solutions
behaves like a sink near (0, 0).
Problems you can do:
Lebl’s Book [2]: All Exercises on Page 305, Page 306, Page 312, Page 313, Page 321 and Page 322.
Braun’s Book [1]: All exercises on Page 393 and Page 398. Read all materials in Section 4.3 and Section 4.4.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
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