CHAPTER 11
Systems of Nonlinear
Differential Equations
Contents
11.1 Autonomous Systems
11.2 Stability of Linear Systems
11.3 Linearization and Local Stability
11.4 Autonomous Systems as Mathematical Models
11.5 Periodic Solutions, Limit Cycles, and Global
Stability
Ch11_2
11.1 Autonomous Systems
Introduction
A system of first-order differential equations is called
autonomous, when it can be written as
dx1
 g1 ( x1 , x2 , , xn )
dt
dx2
 g 2 ( x1 , x2 , , xn )
dt

(1)
dxn
 g n ( x1 , x2 , , xn )
dt
Ch11_3
Example 1
dx1
 x1  3 x2  t 2
dt
dx2
 x1 sin( x2t )
dt
The above is not autonomous, since the presence of t on
the right-hand side.
Ch11_4
Example 2
Consider
d 2 g
 sin   0
2
l
dt
If we let x = , y =  , then
x  y
g
y   sin x
l
is a system of first-order.
Ch11_5
Vector Field Interpretation
A plane autonomous system can be written as
dx
 P ( x, y )
dt
dy
 Q ( x, y )
dt
The vector V(x, y) = (P(x, y), Q(x, y)) defines a vector
field of the plane.
Ch11_6
Example 3
A vector field for the steady-state flow of a fluid around
a cylinder of radius 1 is given by

x2  y 2
 2 xy 

V ( x, y )  V0 1  2
, 2
2 2
2 2
 (x  y ) (x  y ) 
where V0 is the speed of the fluid far from the cylinder.
Ch11_7
Example 3 (2)
If a small cork is released at (−3, 1), the path
X(t) = (x(t), y(t)) satisfies

dx
x2  y 2 

 V0 1  2
2 2
dt
 (x  y ) 
  2 xy 
dy
 V0  2
2 2
dt
 (x  y ) 
subject to X(0) = (−3, 1). See Fig 11.1.
Ch11_8
Fig 11.1
Ch11_9
Types of Solutions
(i) A constant solution x(t) = x0, y(t) = y0 (or X(t) = X0
for all t). The solution is called a critical or stationary
point, and the constant solution is called an
equilibrium solution. Notice that X(t) = 0 means
P ( x, y )  0
Q ( x, y )  0
Ch11_10
(ii) A solution defines an arc – a plane curve that does
not cross itself. See Fig 11.2(a). Referring to
Fig11.2(b), it can not be a solution, since there would
be two solutions starting from point P.
Fig 11.2
Ch11_11
(iii) A periodic solution – is called a cycle. If p is the
period, then X(t + p) = X(t). See Fig 11.3.
Ch11_12
Example 4
Find all critical points of the following:
2
2



 6 (c) x  0.01x (100  x  y )
x
x
y

(a) x   x  y (b)
y  x 2  y
y  x  y
y  0.05 y (60  y  0.2 x)
Solution
(a)
x y 0
x y 0
then y = x. There are infinitely many critical points.
Ch11_13
Example 4 (2)
(b)
x2  y 2  6  0
x y0
2
Since x2 = y, then y2 + y – 6 = (y + 3)(y – 2) = 0. If y = –
3, then x2 = – 3, there are no real solutions. If y = 2,
then x   2 . The critical points are ( 2, 2) and
( 2, 2) .
Ch11_14
Example 4 (3)
(c)
From 0.01x(100 – x – y) = 0, we have x = 0 or x + y =
100. If x = 0, then 0.05y(60 – y – 0.2x) = 0 becomes
y(60 – y) = 0. Thus y = 0 or y = 60, and (0, 0) and (0,
60) are critical points.
If x + y = 100, then 0 = y(60 – y – 0.2(100 – y)) = y(40 –
0.8y). We have y = 0 or y = 50. Thus (100, 0) and (50,
50) are critical points.
Ch11_15
Example 5
Determine whether the following system possesses a
periodic solution. In each case, sketch the graph pf the
solution satisfying X(0) = (2, 0).
(a) x  2 x  8 y
(b) x  x  2 y
y   x  2 y
y   1 / 2 x  y
Solution
(a) In Example 6 of Section 10.2, we have shown
x  c1 (2 cos 2t  2 sin 2t )  c2 (2 cos 2t  2 sin 2t )
y  c1 ( cos 2t )  c2 sin 2t
Ch11_16
Example 5 (2)
Thus every solution is periodic with period . The
solution satisfying X(0) = (2, 0) is
x = 2 cos 2t + 2 sin 2t, y = – sin 2t
See Fig 11.4(a).
Ch11_17
Example 5 (3)
(b) Using the similar method, we have
x  c1 (2et cos t )  c2 (2et sin t )
y  c1 (et sin t )  c 2 (et cos t )
Since the presence of et, there are no periodic solutions.
The solution satisfying X(0) = (2, 0) is
x  2et cos t , y  et sin t
See Fig 11.4(b).
Ch11_18
Fig 11.4(b)
Ch11_19
Changing to Polar Coordinates
Please remember that the transformations are
r2 = x2 + y2 and  = tan–1(y/x),
dr 1  dx
dy  d 1 
dx
dy 
 x  y ,
 2  y  x 
dt r  dt
dt  dt r 
dt
dt 
Ch11_20
Example 6
Find the solution of the following system
2
2

x  y  x x  y
y  x  y x 2  y 2
satisfying X(0) = (3, 3).
Solution
dr 1
 [ x( y  xr )  y ( x  yr )]  r 2
dt r
d 1
 2 [( y )( y  xr )  x( x  yr )]  1
dt r
Ch11_21
Example 6 (2)
Since (3, 3) is (3 2,  / 4) in polar coordinates, then X(0)
= (3, 3) becomes r (0)  3 2 and (0) =π/4.
Using separation of variables, we have the solution is
1
r
,   t  c2
t  c1
for r  0. Applying the initial conditions, we have
1

r
,  t
t  2/6
4
Ch11_22
Example 6 (3)
The graph of
1
r
  2/6  /4
is shown in Fig 11.5.
Ch11_23
Fig 11.5
Ch11_24
Example 7
Consider the system in polar coordinates:
dr
d
 0.5(3  r ),
1
dt
dt
Find and sketch the solutions satisfying X(0) = (0, 1)
and X(0) = (3, 0) in rectangular coordinates.
Solution
By separation of variables, we have
r  3  c1e
0.5t
,   t  c2
Ch11_25
Example 7 (2)
If X(0) = (0, 1), then r(0) = 1 and (0) = /2. Thus c1
= –2, c2 =/2. The solution curve is the spiral
. Notice that as t →，
r  3  2e0.5(  / 2)
increases without bound and r approaches 3.
If X(0) = (3, 0), then r(0) = 3 and (0) = 0. Thus c1 = c2
= 0 and r = 3,  = t. We have the solution is
x = r cos = 3 cos t and y = r sin  = 3 sin t. It is a
periodic solution. See Fig 11.6.
Ch11_26
Fig 11.6
Ch11_27
11.2 Stability of Linear Systems
Some Fundamental Questions
Suppose X1 is a critical point of a plane autonomous
system and X = X(t) is a solution satisfying X(0) = X0.
Our questions are when X0 is near X1:
(i) Is limt X(t) = X1?
(ii) If the answer of (i) is “no”, does it remain close
to X1 or move away from X1?
See Fig 11.7
Ch11_28
Fig 11.7
Ch11_29
Referring to Fig11.7(a) and (b), we call the critical
point locally stable.
However, if an initial value results in behavior similar
to (c) can be found in nay given neighborhood, we
call the critical point unstable.
Ch11_30
Stability Analysis
Consider
x = ax + by
y = cx + dy
We have the system matrix as
a b 
A

c d 
To ensure that X0 = (0, 0) is the only critical point, we
assume the determinant  = ad – bc  0.
Ch11_31
Then det (A – I) = 0 becomes
2 −  +  = 0
where  = a + d.
Thus
  (   2  4 ) / 2
Ch11_32
Example 1
Find the eigenvalues of the system
x   x  y
y  cx  y
in terms of c, and use a numerical solver to discover the
shape of solutions corresponding to the case c = ¼ , 4, 0
and −9.
Ch11_33
Example 1 (2)
Solution
Since the coefficient matrix is
 1 1 


 c  1
then we have  = −2, and  = 1 – c. Thus
 2  4  4(1  c)

 1  c
2
Ch11_34
Example 1 (3)
 If c = ¼ ,  = −1/2 and −3/2. Fig 11.8(a) shows the
phase portrait of the system.
 When c = 4,  = 1 and 3. See Fig 11.8(b).
Ch11_35
Example 1 (4)
When c = 0,  = −1. See Fig 11.8(c).
When c = −9,  = −1  3i. See Fig 11.8(d).
Ch11_36
Case I: Real Distinct Eigenvalues
According to Sec 10.2, the general solution is
X(t )  c1K1e
1t
 c2 K 2e
2t
 e1t (c1K1  c2 K 2e( 2 1 )t )
(a) Both eigenvalues negative: Stable Node
It is easier to check that under this condition,
X(t)  0 as t  
See Fig 11.9.
Ch11_37
Fig 11.9
Ch11_38
(b) Both eigenvalues positive: Unstable Node
It is easier to check that under this condition,
|X(t)| becomes unbounded as t  
See Fig 11.10
Ch11_39
Fig 11.10
Ch11_40
(c) Eigenvalues have opposite signs
(2 < 0 < 1): Saddle Point
On if c1 = 0,
will approach 0 along the line determined by K2
as t  . This unstable solution is called a
saddle point. See Fig 11.11.
Ch11_41
Fig 11.11
Ch11_42
Example 2
Classify the critical point (0, 0) of each system
X = AX as either a stable node, an unstable node, or a
saddle point.
(a)
(b)
 2 3
  10 6 
A
A


 2 1
 15  19 
Solution
(a) Since the eigenvalues are 4, −1, (0, 0) is a saddle
point. The corresponding eigenvectors are respectively
1
 3
K2   
K1   
  1
 2
Ch11_43
Example 2 (2)
If X(0) lies on the line y = −x, then X(t) approaches 0.
For any other initial conditions, X(t) will become
unbounded in the direction determined by K1. That is, y
= (2/3)x serves an asymptote. See Fig 11.12.
Ch11_44
Fig 11.12＞
Ch11_45
(b) Since the eigenvalues are − 4, −25, (0, 0) is a stable
node. The corresponding eigenvectors are respectively
 1
K1   
 1
 2 
K2   
  5
See Fig 11.13.
Ch11_46
Fig 11.13
Ch11_47
Case II: A Repeated Real Eigenvalue
According to Sec 10.2, we have the following
conditions.
(a) Two linearly independent eigenvectors
The general solution is
X(t )  c1K1e1t  c2K 2e1t  (c1K1  c2K 2 )e1t
If 1 < 0, then X(t) approaches 0 along the line
determined by c1K1 + c2K2 and the critical point is
called a degenerate stable node.
Fig 11.14(a) shows the graph for 1 < 0 and the
arrows are reversed when 1 > 0, and is called a
degenerate unstable node.
Ch11_48
Fig 11.14
Ch11_49
(b) A single linearly independent eigenvector
When we only have a single eigenvector, the general
solution is
X(t )  c1K1e1t  c2 (K1te 1t  Pe1t )
c1
c2
 te (c2K1  K1  P)
t
t
If 1 < 0, then X(t) approaches 0 in one of directions
determined by K1(See Fig 11.14(b)). This critical
point is again called a degenerate stable node.
If 1 > 0, this critical point is again called a
degenerate unstable node.
1t
Ch11_50
Case III: Complex eigenvalues (2 – 4 < 0)
(a) Pure imaginary roots (2 – 4 < 0,  = 0)
We call this critical point a center. See Fig 11.15
Ch11_51
(b) Nonzero real part (2 – 4 < 0,   0)
real part > 0: unstable spiral point (Fig 11.16(a))
real part < 0: stable spiral point (Fig 11.16(b))
Ch11_52
Example 3
Classify the critical point (0, 0) of each system
 3  18 
 1 2
(a) A  
 (b) A  

2  9 
 1 1
Solution
(a) The characteristic equation
2 + 6 + 9 = ( + 3)2= 0
so (0, 0) is a degenerate stable node.
(b) The characteristic equation
2 + 1 = 0
so (0, 0) is a center.
Ch11_53
Example 4
Classify the critical point (0, 0) of each system
3.10 
 1.01
(a) A  

  1.10  1.02 
  axˆ  abxˆ 
(b) A  

  cdyˆ  dyˆ 
for positive constants.
Solution
(a)  = −0.01,  = 2.3789, 2 − 4 < 0: (0, 0) is a stable
spiral point.
Ch11_54
Example 4 (2)
(b)
  (axˆ  dyˆ )  0,
  adxˆyˆ (1  bc)
if bc  1,   0 : a saddle point
if bc  1,   0 : either satble, degenerate stable,
or stable spiral
Ch11_55
THEOREM 11.1
Stability Criteria for Lonear Systems
For a linear plane autonomous system X’ = AX with det A  0,
let X = X(t) denote the solution that satisfies the initial condition
X(0) = X0, where X0  0.
(a) limt→X(t) = 0 if and only if the eigenvalues of A have
negative real parts. This will occur when ∆ > 0 and  < 0.
(b) X(t) is periodic if and only if the eigenvalues of A are pure
imaginary. This will occur when ∆ > 0 and  = 0.
(c) In all other cases, given any neighborhood of the region,
there is at least one X0 in the neighborhood for which X(t)
becomes unbounded as t increases.
Ch11_56
11.3 Linearization and Local Stability
DEFINITION 11.1
Stable Critical Points
Let X1 be a critical point of an autonomous system,
and let X = X(t) denote the solution that satisfies the
initial condition X(0) = X0, where X0  X1. We say
that X1 is a stable critical point when, given any radius
ρ > 0, there is a corresponding radius r > 0 such that
if the initial position X0 satisfies │X0 – X1│< r, then
the corresponding solution X(t) satisfies │X(t) – X1│
< ρ for all t > 0. If, in addition limt→X(t) = X1
whenever │X0 – X1│< r, we call X1 an
asymptotically stable critical point.
Ch11_57
This definition is shown in Fig 11.20(a).
To emphasize that X0 must be selected close to X1,
we also use the terminology locally stable critical
point.
Ch11_58
DEFINITION 11.2
Unstable Critical Points
Let X1 be a critical point of an autonomous system,
and let X = X(t) denote the solution that satisfies the
initial condition X(0) = X0, where X0  X1. We say
that X1 is an unstable critical point if there is a disk
of radius ρ > 0 with the property that, for any
r > 0, there is at least one initial position X0 satisfies
│X0 – X1│< r, yet the corresponding solution X(t)
satisfies │X(t) – X1│ ρ for all t > 0.
Ch11_59
If a critical point X1 is unstable, no matter how small
the neighborhood about X1, an initial position X0 can
be found that the solution will leave some disk at
some time t. See Fig 11.20(b).
Ch11_60
Example 1
Show that (0, 0) is a stable critical point of the system
x'   y  x x 2  y 2
y'   x  y x2  y 2
Solution
In Example 6 of Sec 11.1, we have shown
r = 1/(t + c1),  = t + c2
is the solution.
If X(0) = (r0, 0), then
r = r0/(r0 t + 1),  = t +0
Note that r < r0 for t > 0, and r approaches (0, 0) as t
increase. Hence the critical point (0, 0) is stable and is
in fact asymptotically stable. See Fig 11.21.
Ch11_61
Fig 11.21
Ch11_62
Example 2
Consider the plane system
dr
 0.05r (3  r )
dt
d
 1
dt
Show that (x, y) = (0, 0) is an unstable critical point.
Ch11_63
Example 2 (2)
Solution
Since x = r cos  and y = r sin  , we have
dx
d dr
  r sin 
 cos 
dt
dt dt
dy
d dr
 r cos 
 sin 
dt
dt dt
Since dr/dt = 0.05r(3-r), then r = 0 implies dr/dt = 0.
Thus when r = 0, we have dx/dt = 0, dy/dt = 0. We
conclude that (x, y) = (0, 0) is a critical point.
Ch11_64
Example 2 (3)
Solving the given differential equation with r(0) = r0
and r0  0, we can have
3
r
1  c0e 0.15t
where c0  (3  r0 ) / r0 . Since
3
lim t 
3
0.15t
1  c0e
No matter how close to (0, 0) a solution starts, the
solution will leave (0, 0). Thus (0, 0) is an unstable
critical point. See Fig 11.22.
Ch11_65
Fig 11.22
Ch11_66
Linearization
If we write the systems in Example 1 and 2 as X =
g(X). The process to find a liner term A(X – X1) that
most closely approximates g(X) is called
linearization.
Ch11_67
THEOREM 11.2
Stability Criteria for Linear Systems
Let x1 be a critical point of the autonomous differential
equation x = g(x), where g is differentiable at x1.
(a) If g(x1) < 0, then x1 is an asymptotically stable
critical point.
(b) If g(x1) > 0, then x1 is an unstable critical point.
Ch11_68
Example 3

5
Both x  and x 
are critical points of
4
4
x'  cos x  sin x.
predict the behavior of solutions near these two critical
points. Since
g ' ( x)   sin x  cos x
g ' ( 4)   2  0, g ' (5 4)  2  0
Therefore x = /4 is an asymptotically stable critical
point but x = 5/4 is unstable. See Fig 11.23.
Ch11_69
Fig 11.23
Ch11_70
Example 4
Without solving explicitly, analyze the critical points of
the system x = (r/K)x(K – x), where r and K are
positive constants.
Solution
We have two critical points x = 0 and x = K. Since
r
g ' ( x)  ( K  2 x)
K
g ' (0)  r  0, g ' ( K )   r  0
Therefore x = K is an asymptotically stable critical point
but x = 0 is unstable.
Ch11_71
Jacobian Matrix
An equation of the tangent plane to the surface z =
g(x, y) at X1 = (x1, y1) is
g
g
z  g ( x1 , y1 ) 
( x1 , y1 ) ( x  x1 ) 
( x1 , y1 ) ( y  y1 )
x
y
Similarly, when X1 = (x1, y1) is a critical point, then P
(x1, y1) = 0, Q (x1, y1) = 0.
We have
P
P
x '  P ( x, y ) 
( x1 , y1 ) ( x  x1 ) 
( x1 , y1 ) ( y  y1 )
x
y
Q
Q
y '  Q ( x, y ) 
( x1 , y1 ) ( x  x1 ) 
( x1 , y1 ) ( y  y1 )
x
y
Ch11_72
The original system X = g(X) may be approximated
by X = A(X – X1), where
 P

( x1 , y1 )
x

A
 Q
 x ( x1 , y1 )

P

( x1 , y1 ) 
y

Q

(x ,y )
y 1 1 
This matrix is called the Jacobian Matrix at X1 and
is denoted by g(X1).
Ch11_73
THEOREM 11.3
Stability Criteria for Plane
Autonomous Systems
Let X1 be a critical point of the autonomous differential
equation X’ = g(X), where P(x, y) and Q(x, y) have
continuous first partials in a neighborhood of X1.
(a) If the eigenvalues of A = g’(X1) have negative real
part, then X1 is an asymptotically stable critical point.
(b) If A = g’(X1) has an eigenvalue with positive real
part, then X1 is an unstable critical point.
Ch11_74
Example 5
Classify the critical points of each system.
(a) x’ = x2 + y2 – 6
(b) x’ = 0.01x(100 – x – y)
y’ = x2 – y
y’ = 0.05y(60 – y – 0.2x)
Solution
(a) The critical points are ( 2, 2) and ( 2, 2),
 2x 2 y 
g ' ( X)  

 2 x  1
Ch11_75
Example 5 (2)
2 2 4 

A1  g ' (( 2, 2))  
 2 2  1
 2 2 4 

A 2  g ' ((  2, 2))  
  2 2  1
Since the determinant of A1 is negative, A1 has a
positive real eigenvalue. Therefore ( 2, 2) unstable.
A1 has a positive determinant and a negative trace. Both
the eigenvalues have negative real parts. Therefore
( 2, 2) is stable.
Ch11_76
Example 5 (3)
(b) The critical points are (0, 0), (0, 60), (100, 0), (50,
50). The Jacobiam matrix is
 0.01x
 0.01(100  2 x  y )

g ' ( X)  

 0.01y
0.05(60  2 y  0.2 x) 

1 0
A1  g ' ((0, 0))  

 0 3
 0 .4 0 
A 2  g ' ((0, 60))  

  0 .6 3 
Ch11_77
Example 5 (4)
  1  1
A3  g ' ((100,0))  

0 2
  0.5  0.5 
A 4  g ' ((50,50))  

  0.5  2.5 
Checking the signs of the determinant and trace of each
matrix, we conclude that (0, 0) is unstable; (0, 60) is
unstable; (100, 0) is unstable; (50, 50) is stable.
Ch11_78
Example 6
Classify each critical point of the system in Example
5(b).
Solution
For the matrix A1 corresponding to (0, 0),  = 3,  = 4,
2 – 4 = 4. Therefore (0, 0) is an unstable node. The
critical points (0, 60) and (100, 0) are saddles since  <
0 in both cases. For A4,  > 0,  < 0, (50, 50) is a stable
node.
Ch11_79
Example 7
Consider the system x + x – x3 = 0. We have
x = y,
y = x3 – x.
Find and classify the critical points.
Solution
x  x  x( x  1)  0, the critical points are
3
2
(0, 0), (1, 0), (-1,0).
Ch11_80
Example 7 (2)
The corresponding matrices are
 0 1
A1  g ' ((0,0))  

 1 0
0 1
A 2  g ' ((1,0))  g ' (( 1,0))  

 2 0
Since det A 2  0, (1, 0) and (-1, 0) are both saddle
points. The eigenvalue s of A1 are  i and the
status of (0, 0) is in doubt.
Ch11_81
Example 8
Use the phase-plane method to classify the sole critical
point (0, 0) of the system
x = y2
y = x2
Solution
The determinant of the Jacobian matrix
 0 2y
g ' ( X)  

 2x 0 
is 0 at (0, 0), and so the nature of (0, 0) is in doubt.
Ch11_82
Example 8 (2)
Using the phase-plane method, we get
dy dy / dt x 2

 2
dx dx / dt y
2
2
3
3
y
dy

x
dx
,
or
y

x
c


If X(0)  (0, y0 ), y  x  y0 or y  x  y0 .
3
3
3
3
3
3
Fig 11.26 shows a collection of solution curves. The
critical point (0, 0) is unstable.
Ch11_83
Fig 11.26
Ch11_84
Example 9
Use the phase-plane method to determine the nature of
the solutions to x + x − x3 = 0 in a neighborhood of (0,
0).
Solution
If we let dx / dt  y then dy / dt  x3  3 x.
dy dy / dt x  3 x


dx dx / dt
y
3
2
4
2
y
x
x
 ydy   ( x  3x)dx, or 2  4  2  c,
2
2
( x  1)
2
thus y 
 c0 .
2
3
Ch11_85
Example 9 (2)
 ( x0  1) 2
If X(0)  ( x0 ,0), 0  x0  1, then c0 
,
2
2
2
2
2
2
2
(
x

1
)
(
x

1
)
(
2

x

x
)(
x

x
)
2
0
0
0
y 


2
2
2
2
2
2
Note that y = 0 when x = −x0 and the right-hand side is
positive when −x0 < x < x0. So each x has two
corresponding values of y. The solution X = X(t) that
satisfies X(0) = (x0, 0) is periodic, and (0, 0) is a center.
See Fig 11.27.
Ch11_86
Fig 11.27
Ch11_87
11.4 Autonomous Systems as Mathematical Models
Nonlinear Pendulum
Consider the nonlinear second-order differential
equation
d 2 g
 sin   0
2
l
dt
When we let x = , y =  , then we can write
x  y
g
y   sin x
l
Ch11_88
The critical points are (k, 0) and the Jacobian
matrix is
0
1


g ' ((  k ,0))  
k 1 g
0
 (1)


l
If k = 2n + 1,  < 0, and so all critical points
((2n +1), 0) are saddle points. Particularly, the
critical point (, 0) is unstable as expected. See Fig
11.28.
Ch11_89
Fig 11.28
Ch11_90
When k = 2n, the eigenvalues are pure imaginary, and
so the nature of these critical points remains in doubt.
Since we assumed that there are no damping forces,
we expect that all the critical points ((2n, 0) are
centers. From
dy dy / dt
g sin y
2g
2


, then y 
cos x  c
dx dx / dt
l y
l
2g
2
If X(0)  ( x0 , 0), then y 
(cos x  cos x0 )
l
Ch11_91
Note that y = 0 when x = −x0, and that
(2g/l)(cos x – cos x0) > 0 for |x| < |x0| < . Thus each
such x has two corresponding values of y, and so the
solution X = X(t) that satisfies X(0) = (x0, 0) is
periodic. We may conclude that (0, 0) is a center. See
Fig 11.29.
Ch11_92
Fig 11.29
Ch11_93
Example 1
A pendulum is an equilibrium position with  = 0 is
given an initial velocity of 0 rad/s. Determine under
what conditions the resulting motion is periodic.
Solution
The initial condition is X(0) = (0, 0).
2g
From y 
(cos x  c), it follows that
l
2g
l
2
2
y 
(cos x  1  0 )
l
2g
2
Ch11_94
Example 1 (2)
To establish that the solution X(t) is periodic it is
sufficient to show that there are two x-intercepts x = x0
between − and  and that the right-hand side is
positive for |x| < |x0|. Each such x then has two
corresponding values of y.
If y = 0, cos x = 1 – (l/2g)02, and this equation has two
solutions x = x0 between − and , provided
1 – (l/2g)02 > −1. Note that (l/2g)(cos x – cos x0) is
positive for |x| < |x0|. The restriction on the initial
angular velocity may be written as
g
0  2
l
Ch11_95
Nonlinear Oscillations: The Sliding Bead
See Fig 11.30. The tangential force F has the
magnitude mg sin , thus Fx = − mg sin  cos .
Since tan  = f (x), then
f ' ( x)
Fx   mg sin  cos    mg
1  [ f ' ( x)]2
dx
Assume there is a damping force D, and Dx    ,
dt
f ' ( x)
From Newton' s law : mx"   mg
 x'
2
1  [ f ' ( x)]
Ch11_96
and the corresponding plane autonomous system is
x'  y
f ' ( x)

y'   g
 y
2
1  [ f ' ( x)] m
If X1 = (x1, y1) is a critical point, then y1 = 0 and f (x1)
= 0. The bead must be at rest at a point on the wire
where the tangent line is horizontal.
Ch11_97
The Jacobian matrix at X1 is
0
1 

 , and so
g' ( X1 )  
  gf " ( x1 )  

m

   ,   gf " ( x1 ),   4 
m
2

2
m
2
 4 gf " ( x1 ).
Ch11_98
We can make the following conclusions.
(i) f ”(x1) < 0 :
A relative maximum occurs at x = x1 and since  < 0,
an unstable saddle point occurs at X1 = (x1, 0).
(ii) f ”(x1) > 0 and  > 0:
A relative minimum occurs at x = x1 and since  < 0
and  > 0, X1 = (x1, 0) is a stable critical point. If 2 >
4gm2f (x1), the system is overdamped and the critical
point is a stable node.
Ch11_99
If  2 < 4gm2f (x1), the system is underdamped and
the critical point is a stable spiral point. The exact
nature of the stable critical point is still in doubt if
 2 = 4gm2f (x1).
(iii) f ”(x1) > 0 and the system is undamped ( = 0):
In this case the eigenvalues are pure imaginary, but
the phase plane method can be used to show that the
critical point is a center. Thus solutions with X(0) =
(x(0), x(0)) near X1 = (x1, 0) are periodic.
Ch11_100
Example 2
A 10-gram bead slides along the graph z = sin x. The
relative minima at x1 = −/2 and x2 = 3/2 are stable
critical points. See Fig 11.31.
Ch11_101
Fig 11.32
Fig 11.32 shows the motions when the critical points
are stable spiral points.
Ch11_102
Fig 11.33
Fig 11.33 shows a collection of solution curves for
the undamped case.
Ch11_103
Lotka-Volterra Predator-Prey Model
Recall the predator-prey model:
x'  ax  bxy  x( a  by )
y '  cxy  dy  y (cx  d )
The critical points are (0,0) and (d/c,a/b), then
 a 0
A1  g ' ((0,0))  
 and
 0 d
bd / c 
 0
A 2  g ' (( d / c, a / b))  
.
0 
  ac / b
Ch11_104
Fig 11.34
The critical point (0, 0) is a saddle point. See Fig
11.34.
Ch11_105
Since A2 has the pure imaginary eigenvalues, the
critical point may be a center. Since
dy y (cx  d )

, then
dx x( a  by )
 a  by
 cx  d
 y dy   x dy,
 a ln y  by  cx  d ln x  c1 , or
d cx
(x e
a by
)( y e
)  c0
Ch11_106
Fig 11.35
Typical graphs are shown in Fig 11.35.
Ch11_107
1. If y = a/b, the equation F(x)G(y) = c0 has exactly
two solutions xm and xM that satisfy xm < d/c < xM.
2. If xm < x1 < xM and x = x1, then F(x)G(y) = c0 has
exactly two solutions y1 and y2 that satisfy y1 < a/b <
y2.
3. If x is outside the interval [xm, xM], then F(x)G(y) =
c0 has no solutions.
 The graph of a typical periodic solution is shown in
Fig 11.36.
Ch11_108
Fig 11.36
Ch11_109
Example 3
If we let a = 0.1, b = 0.002, c = 0.0025, d = 0.2, the
critical point in the first quadrant is (d/c, a/b) = (80,
50), and we know it is a center. See Fig 11.37.
Ch11_110
Fig 11.37
Ch11_111
Lotka-Volterra Competition Model
Consider the model:
r1
x' 
x( K1  x  12 y )
K1
r2
y' 
y ( K 2  y   21 x)
(1)
K2
This system has critical points at (0, 0), (K1, 0) and (0,
K2).
Ch11_112
Example 4
Consider the model
x'  0.004 x(50  x  0.75 y )
y '  0.001y (100  y  3.0 x)
Find and classify all critical points.
Solution
Critical points are (0, 0), (50, 0), (0, 100), (20, 40).
Since 1221 = 2.25 > 1, and so the critical point (20, 40)
is a saddle point. The Jacobian matrix is
Ch11_113
Example 4 (2)
 0.003x
 0.2  0.008x  0.003 y

g ' ( X)  

 0.003 y
0.1  0.002 y  0.003x 

 0.2 0 
A1  g ' ((0,0))  

 0 0. 1 
  0.2  0.15 
A 2  g ' ((50,0))  

0.05 
 0
  0.08  0.12 
A3  g' (( 20,40))  

  0.06  0.04 
0 
  0.1
A 4  g' ((0,100))  

  0.3  0.1
Ch11_114
Example 4 (3)
Therefore (0, 0) is unstable, whereas both (50, 0) and (0,
100) are stable nodes and (20, 40) is a saddle point.
Ch11_115
11.5 Periodic Solutions, Limit Cycles and Global
Stability
THEOREM 11.4
Cycles and Critical Points
If a plane autonomous system has a periodic solution
X = X(t) in a simply connected region R, then the system
has at least one critical point inside the corresponding
simple closed curve C. If there is a single critical point
inside C, then that critical point cannot be a saddle point.
COROLLARY
If a simply connected region R either contains no critical
points of a plane autonomous system or contains a single
saddle point, then there are no periodic solutions in R.
Ch11_116
Example 1
Show that the plane autonomous system
x’ = xy
y’ = −1 – x2 – y2
has no periodic solutions.
Solution
If (x, y) is a critical point, then either x = 0 or y = 0. If x
= 0, then −1 – y2 = 0, y2 = –1. Likewise, y = 0 implies
x2 = –1. Thus this system has no critical points and has
no periodic solutions.
Ch11_117
Example 2
Show that
x'  0.004 x(50  x  0.75 y )
y '  0.001y (100  y  3.0 x)
has no periodic solutions in the first quadrant.
Solution
From Example 4 of Sec 11.4, we knew only (20, 40) lies
in the first quadrant and (20, 40) is a saddle point. By
the corollary, there are no periodic solutions in the first
quadrant.
Ch11_118
THEOREM 11.5
Bendixson Negative Criterion
If div V =  P/y + Q/ y does not change sign in a
connected region R, then the plane autonomous system
has no periodic solution in R.
Ch11_119
Example 3
Investigating possible periodic solutions of each system.
(a) x'  x  2 y  4 x3  y 2 ,
y '   x  2 y  yx 2  y 3
(b) x'  y  x(2  x 2  y 2 ), y '   x  y (2  x 2  y 2 )
Solution
(a) div V  P / x  Q / y
 1  12 x 2  2  x 2  3 y 2  3, and so
there are no periodic solutions.
Ch11_120
Example 3 (2)
(b) div V  (2  3x 2  y 2 )  (2  x 2  3 y 2 )
 4  4( x 2  y 2 )
If R is the interior of the given circle, div V > 0 and so
there are no periodic solutions inside the disk. Note that
div V < 0 on the exterior of the circle. If R is any simply
connected subset of the exterior, then there are no
periodic solutions in R. If there is a periodic solution in
the exterior, it must enclose the circle x2 + y2 = 1.
Ch11_121
Example 4
The sliding bead in Sec 11.4 satisfies
f ' ( x)
mx"  mg
 x'
2
1  [ f ' ( x)]
Show that there are no periodic solutions.
Solution
f ' ( x)

x'  y, y '   g
 y
2
1  [ f ' ( x)] m
P Q

div V 

 0
x y
m
Ch11_122
THEOREM 11.6
Dulac Negative Criterion
If (x, y) has continuous first derivatives in a simply
 ( P )  ( Q )
connected region R and
does not change

x
y
sign in R, then the plane autonomous system has no
periodic solution in R.
Ch11_123
Example 5
Show that
x"  x 2 ( x') 2  x  x'
has no periodic solutions.
Solution
x'  y, y '  x 2  y 2  x  y.
Letting δ ( x, y )  e axby ,
 (P )  (Q)

x
y
 e axby (ay  2 y  1)  e axbyb( x 2  y 2  x  y )
Ch11_124
Example 5 (2)
If we set a = −2, b = 0, then
 (P )  (Q )

 e ax by
x
y
which is always negative. The system has no periodic
solutions.
Ch11_125
Example 6
Use (x, y) = 1/(xy) to show
r1
x' 
x( K1  x  12 y )
K1
r2
y' 
y ( K 2  y   21 x)
K2
have no periodic solutions in the first quadrant.
Ch11_126
Example 6 (2)
Solution
r1 K1 x
r2 K 2 y
P  (   12 ), Q  (    21 )
K1 y y
K2 x x
 (P )  (Q ) r1 1
r2 1

 ( ) 
( )
x
y
K1 y K 2 x
For (x, y) in the first quadrant, the last expression is
always negative.
Ch11_127
DEFINITION 11.3
Invariant Region
A region R is called an invariant region for a plane
autonomous system if, whenever X0 is in R, the
X = X(t) satisfying X(0) = X0 remains in R.
Fig 11.40 shows two standard types of invariant
regions.
Ch11_128
Fig 11.40
Ch11_129
THEOREM 11.7
Normal Vector and Invariant Regions
If n(x, y) denote a normal vector on the boundary that
point inside the region, then R will be an invariant
region for the plane autonomous system provided
V(x, y)‧n(x, y)  0 for all points (x, y) on the
boundary.
Ch11_130
Example 7
Find a circular region with center at (0, 0) that serves an
invariant region for the system
x'   y  x3
y'  x  y3
Solution
For the circle x2 + y2 = r2, n = (−2x, −2y) is a normal
vector that points toward the interior of the circle.
Since
V  n  ( y  x3 , x  y3 )  (2 x,2 y)  2( x 4  y 4 )
we may conclude that Vn  0 on the circle x2 + y2 = r2.
By Theorem 11.7, the circular region x2 + y2  r2 serves
as an invariant region for the system for any r > 0. Ch11_131
Example 8
Find an annular region bounded by circles that serves as
an invariant region for the system
2
2
5
x'  x  y  5 x( x  y )  x
y'  x  y  5 y( x2  y 2 )  y5
Solution
As in Example 7, the normal vector n1 = (−2x, −2y)
points inside the circle x2 + y2 = r2, while the normal
vector n2 = − n1 points outside the circle.
V  n1  2(r 2  5r 4  x6  y 6 )
Ch11_132
Example 8 (2)
If r = 1, V‧n1 = 8 – 2(x6 + y6)  0.
If r = 1/4, V‧n1  – 2(r2 – 5r4) < 0 and so V‧n2 > 0.
The annular 1/16  x2 + y2  1 is an invariant region.
Ch11_133
Example 9
The Van der Pol equation is a nonlinear second-order
differential equation that arise in electronics,
x'  y
y '    ( x 2  1) y  x
Fig 11.41 shows the vector field for  = 1, together with
the curves y = 0 and (x2 – 1)y = −x along which the
vectors are vertical and horizontal, respectively.
Ch11_134
Fig 11.41
It is not possible to find a simple invariant region
whose boundary consists of lines or circles.
Ch11_135
THEOREM 11.8
Poincare-Bendixson I
Let R be an invariant region for a plane autonomous system and
suppose that R has no critical points on its boundary.
(a) If R is a Type I region that has a single unstable node or an
unstable spiral point in its interior, then there is at least one
periodic solution in R.
(b) If R is a Type II region that contains no critical points of the
system, then there is at least one periodic solution in R.
In either of the two cases, if X = X(t) is a nonperiodic solution
in R, then X(t) spirals toward a cycle that is a solution to the
solution to the system. This periodic solution is called a limit
cycle.
Ch11_136
Example 10
Use Theorem 11.8 to show that
x'   y  x(1  x 2  y 2 )  y ( x 2  y 2 )
y '  x  y (1  x 2  y 2 )  x( x 2  y 2 )
has at least one periodic solution.
Solution
We first construct an invariant region bounded by
circles. If n1 = (−2x, −2y) then
V  n1  2r 2 (1  r 2 )
Ch11_137
Example 10 (2)
If we let r = 2 and then r = ½, we conclude that the
annular region R: ¼  x2 + y2  4 is invariant. If (x1, y1)
is a critical point, then V‧n1 = (0, 0)‧n1 = 0. Therefore r
= 0 or r = 1. If r = 0, then (x1, y1) = (0, 0) is a critical
point.
If r = 1, the system reduces to −2y = 0, 2x = 0 and we
have reached a contradiction. Therefore (0, 0) is the
only critical point and is not in R. Thus the system has
at least one periodic solution in R.
Ch11_138
Example 11
Show that the Van der Pol equations has a periodic
solution when  > 0.
Solution
We found that the only critical point is (0, 0) and the
Jacobian matrix is
 0 1
g ' ((0, 0))  
, then
 1  
   ,   1,  2  4   2  4
Ch11_139
Example 11 (2)
Since  > 0, the critical point is either an unstable spiral
point or an unstable node. Bu part (i) of Theorem 11.8
the system has at least one periodic solution in R. See
Fig 11.42.
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Fig 11.42
Ch11_141
THEOREM 11.8
Poincare-Bendixson II
Let R be a Type I invariant region for a plane
autonomous system that has no periodic solution in R.
(a) If R has a finite number of nodes or spiral points,
then given any initial position X0 in R,
limt→X(t) = X1 for some critical point X1.
(b) If R has a single stable node or stable spiral point
X1 in its interior and no critical points on its
boundary, the limt→X(t) = X1 for all initial position
X0 in R.
Ch11_142
Example 12
Investigate global stability for the system in Example 7.
x'   y  x3
y'  x  y3
Solution
It is not hard to show that the only critical point is (0, 0)
and the Jacobian matrix is
 0  1
g' ((0, 0))  
, and   0,   1
1 0 
Ch11_143
Example 12 (2)
(0, 0) may be either a stable or an unstable spiral.
Theorem 11.9 guarantees that
lim X(t )  X1 for some critical points X1.
t 
Since (0, 0) is the only critical point, we must have
lim X(t )  (0, 0) for any initial position X(0) in the plane.
t 
The critical point is therefore a globally stable spiral
point. See Fig 11.43.
Ch11_144
Fig 11.43
Ch11_145