LECTURE 30: VECTOR FIELDS AND LINEARIZATION FOR SYSTEM
MINGFENG ZHAO
November 25, 2015
Two dimensional linear
systems
and their vector fields


In the following, let A = 
a
b
c
d
 be a 2 × 2 matrix, λ1 and λ2 be two nonzero different eigenvalues of A. Notice
that det(A − λI2 ) = λ2 − (a + d)λ + (ad − bc). Since λ1 6= 0 and λ2 6= 0, then det A 6= 0, which implies that the origin
(0, 0) is the only critical point to the system ~x0 = A~x.
For the behaviors of solutions to ~x0 = A~x near the origin (0, 0), we have the following six cases:
Eigenvalues
Type of Critical Point (0,0)
Stability
I.
λ1 , λ2 are real and both positive
source
unstable
II.
λ1 , λ2 are real and both negative
sink
asymptotically stable
saddle point
unstable
III. λ1 , λ2 are real and opposite signs
IV.
λ1 , λ2 are complex with positive real part
spiral source
unstable
V.
λ1 , λ2 are complex with negative real part
spiral sink
asymptotically stable
VI.
λ1 , λ2 are complex with zero real part
center point
stable
For the Case IV, V and VI, we have:
IV. Both eigenvalues λ1 and λ2 are complex, and have positive real parts, then the solutions grow in magnitude while spinning around the origin, we say the critical point (0, 0) is a spiral source and unstable.

Example 1. Let A = 

of A, and ~v1 = 
1
2i

1
1
−4
1

, then it’s easy to know that λ1 = 1 + 2i and λ2 = 1 − 2i are two eigenvalues

 and ~v2 = 
1
−2i

 are eigenvectors corresponding to λ1 = 1 + 2i and λ2 = 1 − 2i,
1
2
MINGFENG ZHAO
respectively. Notice that




cos(2t)
1
,
 e(1+2i)t = et 
Re 
−2 sin(2t)
2i
So the general solution to ~x0 = A~x is:

~x(t) = C1 et 

and
cos(2t)
−2 sin(2t)
Im 
1


 e(1+2i)t = et 
2i


 + C2 et 
sin(2t)

.
2 cos(2t)
sin(2t)

.
2 cos(2t)
Figure 1. Example spiral source vector field
V. Both eigenvalues λ1 and λ2 are complex, and have negative real parts, then the solutions shrink in magnitude while spinning around the origin, we say the critical point (0, 0) is a spiral sink and asymptotically
stable.

Example 2. Let A = 
−1
−1

, then it’s easy to know that λ1 = −1 − 2i and λ2 = −1 + 2i are two
4 −1




1
1
 and ~v2 = 
 are eigenvectors corresponding to λ1 = −1 − 2i and
eigenvalues of A, and ~v1 = 
2i
−2i
λ2 = −1 + 2i, respectively. Notice that








1
cos(2t)
1
−
sin(2t)
 e(−1−2i)t = e−t 
 , and Im 
 e(−1−2i)t = e−t 
.
Re 
2i
2 sin(2t)
2i
2 cos(2t)
So the general solution to ~x0 = A~x is:

~x(t) = C1 e−t 
cos(2t)
2 sin(2t)


 + C2 e−t 
− sin(2t)
2 cos(2t)

.
LECTURE 30: VECTOR FIELDS AND LINEARIZATION FOR SYSTEM
3
Figure 2. Example spiral sink vector field
VI. Both eigenvalues λ1 and λ2 are purely imaginary, that is, the eigenvalues are ±ik, then we get ellipses of
solutions, we say the critical point (0, 0) is a center point and stable
Let ~v be an eigenvector corresponding to eik , then the general solution is:
~x(t) = C1 Re(eik~v ) + C2 Im(eik~v ).

Example 3. Let A = 

A, and ~v1 = 
1

0
1
−4
0

 and ~v2 = 
2i

, then it’s easy to know that λ1 = 2i and λ2 = −2i are two eigenvalues of

1
−2i
 are eigenvectors corresponding to λ1 = 2i and λ2 = −2i, respectively.
Notice that

Re 
1
2i


 ei2t = 
cos(2t)
−2 sin(2t)

,

Im 
1


 ei2t = 
2i
Figure 3. Example center vector field
sin(2t)
2 cos(2t)

.
4
MINGFENG ZHAO
Trajectories for the two dimensional
nonlinear systems

Recall that trajectories for the system
 x0 = f1 (x1 , x2 )
1
will satisfy
 x0 = f (x , x )
2
1
2
2
f2 (x1 , x2 )
dx2
=
.
dx1
f1 (x1 , x2 )
(1)
Remark 1. In Chapter 1, we know that for the general first order differential equation
dx2
= g(x1 , x2 ), by using
dx1
dx2
the integrating factor method, we can get solutions to
= g(x1 , x2 ) to be of the form F (x1 , x2 ) = C. That is, the
dx1

 x0 = f1 (x1 , x2 )
1
trajectories to the system
are the level sets of some function F (x1 , x2 ).
 x0 = f (x , x )
2
2
1
2
Example 4. Consider the second order equation x00 = −x + x2 .

 x0 = y
Rewrite this equation as a first order system, we get
. It’s easy to see that (0, 0) and (1, 0) are only
 y 0 = −x + x2
equilibrium solutions, and the trajectory equation is:
dy
−x + x2
=
.
dx
y
The above equation is separable, then
1 2
1
1
y = − x2 + x3 + C.
2
2
3
Figure 4. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2
Linearization of a system
Recall that
LECTURE 30: VECTOR FIELDS AND LINEARIZATION FOR SYSTEM
5
I. The linear approximation of a function f (x) at a is:
L(x) = f (a) + f 0 (a)(x − a).
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).

 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)

u0

v

in which the matrix 
0

=
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 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 component f (x, y) and g(x, y) at this critical point (x0 , y0 ).
Problems you can do:
Lebl’s Book [2]: All Exercises on Page 305 and Page 306.
Braun’s Book [1]: All exercises on Page 383, Page 384 and Page 385. Read all materials in Section 4.2.
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