CHAPTER 8: NONLINEAR SYSTEMS
MINGFENG ZHAO
November 26, 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

Definition 3. The Jacobian matrix at point (x0 , y0 ) of vector function 
f (x, y)
g(x, y)

fx (x0 , y0 ) fy (x0 , y0 )

gx (x0 , y0 )
gy (x0 , y0 )

.

 is:
CHAPTER 8: NONLINEAR SYSTEMS
3

 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)
and v = y − y0 , the linearization at (x0 , y0 ) of the autonomous system
is:
 y 0 = g(x, y)

u0

v0



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)

 is:
g(x, y)


0
1
−1 + 2x
0

.
Therefore, at (0, 0) the linearization is:


u0

v

=
0
0
−1
1

u

0

.
v
Therefore, at (1, 0) the linearization is:

u0

v
0


=
0
1

u

1
0

.
v
Definition 5. Let f~(~x) be a vector function and ~a be a critical point of f~, that is, f~(~a) = 0, then
I. We say that the critical point ~a is isolated if it is the only critical point in a small “neighborhood” of ~a.
II. We say that the system ~x0 = f~(~x) is almost linear at critical point ~a if the critical point ~a is isolated, and the
Jacobian at ~a is invertible.
4
MINGFENG ZHAO
Remark 3. If the Jacobian matrix at ~a is invertible, then ~a is isolated. Therefore, it suffices to verify that the Jacobian
at ~a is invertible.
Definition 6. Let ~a be a critical point of f~, that is, ~x(t) ≡ a is an equilibrium solution to the system ~x0 = f~(~x. Then
I. We say that the critical point ~a is stable if for any given > 0, there exists some δ > 0 such that for any
solutions ~x(t) to the initial value problem ~x0 = f~(~x), ~x(0) = ~x0 with |~x0 − ~a| < δ, we have
|~x(t) − ~x0 | < ,
for all t ≥ 0.
II. We say that the critical point ~a is unstable if ~a is not stable.
III. We say that the critical point ~a is asymptotically stable if ~a is stable, and there exists some δ > 0 such that for
any solutions ~x(t) to the initial value problem ~x0 = f~(~x), ~x(0) = ~x0 with |~x0 − ~a| < δ, we have
lim ~x(t) = ~a.
t→∞
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
III. real and opposite signs
saddle
unstable
V.
complex with positive real part
spiral source
unstable
VI.
complex with negative real part
spiral sink
asymptotically stable
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).
That is, all critical points of the system x0 = sin(πy) + (x − 1)2 , y 0 = y 2 − y 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)
g(x, y)
know that


 is 
−2x
−1
−1
2y

. Therefore, we
CHAPTER 8: NONLINEAR SYSTEMS

I. At (0, 0) the linearization is 
u0
v
0


=
0
−1

u

−1
0
5


. Notice that 
v

system is almost linear at (0, 0). Since the eigenvalues of 
0
−1
−1
0
0
−1
−1
0

 is invertible, then the

 are λ1 = −1 and λ2 = 1, then (0, 0) is
an unstable equilibrium point of the system, solutions behaves like a saddle near (0, 0).

 




u0
2 −1
u
2 −1
=

. Notice that 
 is invertible, then the
II. at (1, −1) the linearization is 
0
v
−1 −2
v
−1 −2


2 −1
 are λ1 = −1 and λ2 = −3, then
system is almost linear at (1, −1) . Since the eigenvalues of 
−1 −2
(1, −1) is an asymptotically stable equilibrium point of the system, solutions behaves like a sink near (0, 0).
Figure 2. Phase portrait with some trajectories of x0 = −y − x2 , y 0 = −x + y 2
Conservative equations
For a conservative equation x00 + f (x) = 0, let y = x0 , then we have the system:
x0 = y,
and y 0 = −f (x).
Then there are never any asymptotically stable points for the system x0 = y, y 0 = −f (x), and the trajectories
satisfy
1 2
y +
2
Z
which is called the Hamiltonian or energy of the system.
f (x) dx = C,
6
MINGFENG ZHAO
Pendulum
Suppose we have a mass m (in kilograms) on a pendulum of length L (in meters).
Figure 3. Pendulum Problem
Let θ(t) be the angle between the vertical line and the pendulum at time t (in seconds), then
θ00 +
g
sin(θ) = 0 .
L
Let ω = θ0 , then we can study the following two dimensional system:

θ

0

 =
ω

ω

− Lg sin(θ)
The trajectories satisfy:
1 2 g
ω − cos(θ) = C.
2
L
Critical points are:
(kπ, 0),
k ∈ Z.
Then
I. At (2kπ, 0), the linearization is:

u

v
0

 =
0
− Lg
1

u

0
v

.
CHAPTER 8: NONLINEAR SYSTEMS

7


r
r
0
1
g
g
 has eigenvalues λ1 = i
 has eigenvalues
Since 
and λ2 = −i
. Since 
L
L
g
g
− L cos(θ) 0
−L 0
r
r
g
g
λ1 = i
cos(θ) and λ2 = −i
cos(θ) when θ is close to (2kπ, 0). So we know that the system has a
L
L
stable center at critical point (2kπ, 0).
0
1

II. At ((2k + 1)π, 0), the linearization is:

u

0

 =
v

Since 
0
g
L
1

r
 has eigenvalues λ1 =
0
critical point ((2k + 1)π, 0).
0
1
g
L
0

u


.
v
g
and λ2 = −
L
r
g
. Then the system has an unstable saddle at
L
The period is:
s
T =4
L
2g
Z
θ0
0
1
dθ .
·p
cos(θ) − cos(θ0 )
Also we have
lim T = ∞ .
θ0 →π
The general solution to the linearized pendulum equation θ00 +
g
√
√
θ = 0 is: θ(t) = C1 cos ( gLt) + C2 sin ( gLt). So
L
the period Tlinear for the the linearized pendulum equation is:
Tlinear
2π
= p g = 2π
L
s
L
.
g
Predator-prey or Lotka-Volterra systems
Let
x =
number of the prey(e.g., hares)
y
number of the predator(e.g., foxes)
=
Then we have
x0
= ax − (by)x = (a − by)x
y0
=
(cx)y − dy = (cx − d)y
8
MINGFENG ZHAO
for some positive constants a, b, c, d > 0.
Critical points are:
(x, y) = (0, 0),
and
(x, y) =
d a
,
c b
.
Then
I. At (0, 0), the linearization is:

u

0
 =
v

Since det 
a
0
0
−d

a
0
0
−d

u


.
v


 = −ad 6= 0, then the system is almost linear at (0, 0). Since 
a
0
0
−d

 has eigen-
values λ1 = a and λ2 = −d, then the solutions behaves like a saddle near (0, 0), which is unstable .
d a
II. At
,
, the linearization is:
c b



0 
u
u
0 − bd
c 
.

 =
ac
0
v
v
b



0 − bd
a
d
c 
 = −ad 6= 0, then the system is almost linear at
Since det 
,
has
. Since 
c
b
ac
ac
0
0
b
b
√
√
d a
eigenvalues λ1 = i ad and λ2 = −i ad, then the solutions behaves like a center near
,
, which is a
c b
stable center
0
− bd
c

The trajectories satisfy:
C=
y a xd
= y a xd e−cx−by .
ecx+by
This constant C is called the constant of motion. That is, the trajectories for the predator-prey system are the level
y a xd
curves of cx+by .
e
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