LECTURE 30: LINEARIZATION, CRITICAL POINTS, AND EQUILIBRIA
MINGFENG ZHAO
November 17, 2014
Autonomous systems

 x0 = f1 (x1 , x2 )
1
The vector field/phase portrait/phase diagram of
is the projection on the x1 x2 -plane of its three
 x0 = f (x , x )
2 1
2
2

 x0 = f1 (x1 , x2 )
1
dimensional vector field in the tx1 x2 -space. To draw the vector field of
:
 x0 = f (x , x )
2 1
2
2
I. Plot the x1 x2 -plane.
II. Select points as many as possible in the plane, say P1 , P2 , · · · , Pn .
III. At each point Pi , draw a short arrow with direction (f1 (Pi ), f2 (Pi )).
Definition 1. Consider the autonomous system ~x0 = f~(~x), let ~a be a constant vector, we say that ~a is a critical point
of f~ if f~(~a) = 0.
In this case, we say that ~x(t) ≡ ~a is an equilibria/ equilibrium solution to the system ~x0 = f~(~x).

 x0 = f1 (x1 , x2 )
1
Definition 2. If we find all solutions ~x(t) to the system
, we draw the trajectories by plotting all
 x0 = f (x , x )
2 1
2
2
points (x1 (t), x2 (t)) for a certain range of t. That is, we should find all solutions to the following equation:
dx2
f2 (x1 , x2 )
.
=
dx1
f1 (x1 , x2 )
dx2
= g(x1 , x2 ), we only can
dx1
dx2
use the integrating factor methods to solve an exact equation, that is, we can get solutions to
= f (x1 , x2 ) to be of
dx1
the form:
Remark 1. In Chapter 1, we know that for the general first order differential equation
F (x1 , x2 ) = C.

 x0 = f1 (x1 , x2 )
1
That is, the trajectories to the system
are the level sets of some function F (x1 , x2 ).
 x0 = f (x , x )
2 1
2
2
1
2
MINGFENG ZHAO
Example 1. Consider the second order equation x00 = −x + x2 .
Write this equation as a first order system:
x0 = y,
y 0 = −x + x2 .
Then (0, 0) and (1, 0) are only equilibrium solutions, and
−x + x2
dy
=
.
dx
y
The above equation is separable, then
y dy = (−x + x2 ) dx.
That is, we have
1
1
1 2
y = − x2 + x3 + C.
2
2
3
Figure 1. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2
Linearization
Recall that
I. The linear approximation of a function f (x) at a is:
L(x) = f (a) + f 0 (a)(x − a).
LECTURE 30: LINEARIZATION, CRITICAL POINTS, AND EQUILIBRIA
3
II. The linear approximation of a function f (x, y) at (a, b) is:
L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).

Definition 3. The Jacobian matrix at point (x0 , y0 ) of vector function 
f (x, y)

 is:
g(x, y)

fx (x0 , y0 ) fy (x0 , y0 )

gx (x0 , y0 )

.
gy (x0 , y0 )

 x0 = f (x, y)
Definition 4. Suppose (x0 , y0 ) is an equilibrium solution to the autonomous system
. Let u = x − x0
 y 0 = g(x, y)

 x0 = f (x, y)
is:
and v = y − y0 , the linearization at (x0 , y0 ) of the autonomous system
 y 0 = g(x, y)

u0

v
0


=

fx (x0 , y0 ) fy (x0 , y0 )
u

gx (x0 , y0 )
gy (x0 , y0 )

.
v

 x0 = f (x, y)
Remark 2. The linearization of an autonomous system
 y 0 = g(x, y)
at a critical point (x0 , y0 ) is the just the
linearizations of each compoenents f (x, y) and g(x, y) at this critical point (x0 , y0 ).
Example 2. Find all critical points and linearizations at these critical points for the system x0 = y, y 0 = −x + x2 .
By Example 1, there are two critical points:
(0, 0)
and
(1, 0).

Let f (x, y) = y and g(x, y) = −x + x2 , then the Jacobian matrix of 
f (x, y)
g(x, y)


0
1
−1 + 2x
0

.
Therefore, at (0, 0) the linearization is:

u0

v
0


=
0
−1
1

u

0
v

.

 is:
4
MINGFENG ZHAO

Notice that eigenvalues of 
0
1
−1
0

 are: λ1 = i and λ2 = i.
Figure 2. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2 near (0, 0)
Therefore, at (1, 0) the linearization is:

u0

v

Notice that eigenvalues of 
0
1
1
0
0


=
0
1

u

1
0

.
v

 are: λ1 = 1 and λ2 = −1.
Figure 3. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2 near (1, 0)
LECTURE 30: LINEARIZATION, CRITICAL POINTS, AND EQUILIBRIA
5
Example 3. Find all critical points and linearizations at these critical points for the system x0 = sin(πy) + (x − 1)2 ,
y 0 = y 2 − y.
Let’s solve
sin(πy) + (x − 1)2 = 0,
and y 2 − y = 0.
Then
(x, y) = (1, 0),
and
(x, y) = (1, 1).
That is, all critical points of the system x0 = sin(πy) + (x − 1)2 , y 0 = y 2 − y are:
(1, 0) and (1, 1) .

Let f (x, y) = sin(πy) + (x − 1)2 and g(x, y) = y 2 − y, then Jacobian matrix of 
f (x, y)

 is:
g(x, y)

2(x − 1) π cos(πy)

2y − 1
0

.
Therefore, at (1, 0) the linearization is:

u0

v
0


=
0
π
0
−1
0
−π
0
1

u


.
v
Therefore, at (1, 1) the linearization is:

u0

v0


=

u


.
v
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