208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
1 Limit Cycles
Oscillation is one of the most important phenomena hat occur in
dynamical systems.
A system oscillates when it has a nontrivial periodic solution
x  t  T   x  t  , t  0,
for some T  0. The word “nontrivial” excludes constant solutions.
Periodic orbit or closed orbit means a closed trajectory in the phase
plane.
Center describes an equilibrium point whose nearby trajectories are
periodic orbits. The system has eigenvalues  j  on the imaginary axis.
There are two problems with a system having center equilibrium point.
 First, the oscillator is structurally unstable, that is, an infinitesimal
perturbation of the plant parameter can result in the eigenvalues
falling off of the imaginary axis, and the center could become stable
or unstable focuses.
 Second, the amplitude of oscillation is dependent on the initial
conditions. A different initial condition creates a center with
different radius.
The system that has a center behavior is usually referred to as the
harmonic oscillator. In the case of the harmonic oscillator, there is a
continuum of closed orbits, depending on the initial condition.
Limit cycle describes an isolated periodic orbit. Unlike the center,
the limit cycle is structurally stable and its amplitude is not dependent on
the initial conditions. If all trajectories in the vicinity of the limit cycle
ultimately tend toward the limit cycle, that limit cycle is called stable limit
cycle. If all trajectories in the vicinity of the limit cycle ultimately tend away
from the limit cycle, that limit cycle is called unstable limit cycle.
 Example 1: [1] Show that the negative-resistance oscillator of Lesson 1
has a limit cycle.
1
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
Since h '  0   0, the origin is either an unstable node or unstable focus,
Solution
Recall that the negative-resistance oscillator circuit of Lesson 1 is
given in Figure 1.
depending on the value of
i  h v
1
1
E  CvC2  LiL2 .
2
2
iL
iC
L
C
v
a
0.3V
Knowing that vC  x1 , iL  h  x1  
b v
0.3V
x2 , and   L / C , we can write

The rate of change of the energy is given by


E  C x1 x1   h  x1   x2   h '  x1  x1  x2 
Letting x1  v and x2  v be the two state variables and by changing the
  Cx1h  x1  .
we can obtain the state equations as
We can see that when a  x1  b, the term x1h  x1   0, and the
x1  x2 ,
trajectory gains energy. However, when x1  b and x1  a, the term
x2   x1   h '  x1  x2 ,
where


Figure 1: Negative-resistance oscillator
circuit.
  t / CL ,
1
2
1
E  C x12   h  x1   x2  .
2

time variable from t to
In either case, all trajectories starting
near the origin would diverge away from it.
Next, we consider the total energy stored in the capacitor and
inductor
i

 h '  0.
x1h  x1   0, and the trajectory losses energy. Therefore, we conclude that
there must be a limit cycle for this system.
x1 and x2 represent derivatives with respect to  , and
  L /C.
The system has one equilibrium point at x1  x2  0. The Jacobian
matrix at this point is
A
f
x
x 0
1 
0

.

1


h
'
0




2
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
Solution
Write a Matlab program similar to that of Lesson 2, we obtain the
following phase portraits
 Example 2: [1] Following the previous example, consider a type of the
negative-resistance oscillator circuit with the Van der Pol equation
x1  x2 ,
x2   x1   1  x12  x2 ,
10
x2
for three different values of  : 0.2, 1.0, 5.0, use Matlab to plot three
different phase portraits of each case.
10
x2
8
8
6
6
4
4
2
2
0
0
-2
-2
-4
-4
-6
-6
-8
-8
-10
-10
-10
-10
-8
-6
-4
-2
0
a
2
4
6
8
10
x1
-8
-6
-4
-2
0
b
2
4
6
8
10
x1
10
x2
8
6
4
2
0
-2
-4
-6
-8
-10
-10
-8
-6
-4
-2
0
c
2
4
6
8
10
x1
Figure 2: Phase portrait showing a limit cycle
of a negative-resistance oscillator circuit.
3
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
2 Existence of Periodic Orbits
f  x  V1  x  
Recall that periodic orbit includes center and limit cycle. In this
section, we discuss three methods for detecting the presence or absence
of periodic orbits of a second-order system in the form
x  f  x .
V1  x 
V  x 
f1  x   1
f2  x   0
x1
x2
and
(1)
f  x  V2  x  
The methods apply only to second-order systems and have no
generalizations to higher order systems.
V2  x 
V  x 
f1  x   2
f 2  x   0.
x1
x2
2.1 Poincare-Bendixson Criterion
This criterion gives a condition for the existence of periodic orbits of
(1).
Lemma 1: Poincare-Bendixson criterion
Consider the system (1), and let M be a closed bounded subset of
the plane such that
 M contains no equilibrium points, or contains only one equilibrium
point such that the Jacobian matrix f / x at this point has

V1  x 

f  x
f  x
eigenvalues with positive real parts. (Hence, the equilibrium point is
unstable focus or unstable node.)
 Every trajectory starting in M stays in M for all future time.
Then, M contains a periodic orbit of (1).
V1  x 
A tool for investigating whether every trajectory starting in M
stays in M for all future time is available by investigating Figure 3. If a
closed bounded set M is bounded by two simple closed curves defined by
the equations V1  x  and V2  x  . The gradient vectors V1  x  and
V2  x 
V2  x 
Figure 3: A set M with boundaries V1  x 
and V2  x  .
V2  x  are perpendicular to each curve as shown. Since the vector
f  x   x represents the tangent vector of the trajectory, we have that
trajectories are trapped inside M if the inner product (dot product) of
f  x  and V1  x  is not negative and the inner product (dot product) of
f  x  and V2  x  is not positive, that is,
4
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 4: [1] For a system
 Example 3: [1] Show that there is a periodic orbit of the harmonic
oscillator
x1  x1  x2  x1  x12  x22  ,
x1  x2 ,
x2  2 x1  x2  x2  x12  x22  ,
x2   x1 ,


in the annular region M  c1  V  x   c2 , where V  x   x1  x2 and
find a constant
c2  c1  0.
orbit.
2
2
c for M  V  x   x12  x22  c to contain a periodic
Solution
Since the only equilibrium point is the origin, M does not contain
the equilibrium point.
Furthermore,
f  x  V  x   0
both on V1  x   c1 and V2  x   c2 , therefore from Poincare-Bendixson
criterion, we can conclude that there exists a periodic orbit in M .
We can confirm this result by using Matlab to plot the phase
portrait or simply by inspecting the state equations. By inspecting the state
equations, we see that the vector field f  x   x2 ,  x1  results in a
center.
5
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 5: [1] Consider again the negative-resistance oscillator of
Lesson 1. This time we let x1  v and x2  v   h  v  be the two state
Solution
The system has a unique equilibrium point at the origin. The
Jacobian matrix is
A
f
x
x 0
variables to obtain state equations
1  3x12  x22 1  2 x1 x2 
 1 1





2
2
 2  2 x1 x2 1  x1  3x2  x0  2 1
x1  x2   h  x1  ,
x2   x1.
 is a positive constant and the function h  v  is given by Figure 4.
has eigenvalues 1  j 2. Therefore, the origin is unstable focus.
On the curve V  x   c, we have
i
V
V
f1 
f 2  2 x1  x1  x2  x1  x12  x22  
x1
x2
iC
2 x2  2 x1  x2  x2  x12  x22  
 2 x  x   2 x  x
2
1
2
2
2
1
 2  x12  x22   2  x12  x22
C
iL
L
v
  2x x
  x  x 
2 2
2
2
a v
1 2
2
1
0.3V
2
2
0.3V

 3c  2c 2 ,
Figure 4: A function h  v  .
where we used the fact that 2 x1 x2  x1  x2 . Since
2
i  h v

2
Use the Poincare-Bendixson criterion to show that there exists a periodic
orbit.
3c  2c 2  0,
c  3  2c   0,
3  2c  0,
c  1.5,
we have from the Poincare-Bendixson criterion that there exists a periodic
orbit in M .
6
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
Solution
The first part of the proof will show that there is a closed bounded
set M whose trajectory starting in M stays in M for all future time. It
can be seen from the state equations that a trajectory starting at a point
A   0, p  will move clockwisely pass
ing points

B, C, D,

and intersect the lower half of the x2 -axis at a point E  0,    p  .
We need to show that with
  p   p.
p chosen large enough, then
Consider the function
V  x 
1 2
x1  x22  .

2
x2
p
x2   h  x1 
A
B
a
C
x1
D
  p 
E
Figure 5: The orbit ABCDE.
  p  p
implies that V  E   V  A  0 since
V  E   V  A 
7
1 2
  p   p 2 
2
Copyright
  p .
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
Then by the Poincare-Bendixson criterion, we conclude that there is
a closed orbit in M .
The derivative of V  x  is given by
V  x   x1 x1  x2 x2   x1h  x1  .
Thus,
2.2 Bendixson Criterion
V is positive for x1  a and negative for x1  a. Now,
This next result can be used to rule out the existence of periodic
orbits in some cases.
  p   V  E   V  A   V  x  t   dt ,
Lemma 2: Bendixson criterion
If, on a simply connected region D (has no holes!) of the plane,
the expression f1 / x1  f 2 / x2 is not identically zero and does not
AE
where the right-hand-side integral is taken along the arc from A to E. If
p is small, the whole arc will lie inside the strip 0  x1  a, then   p 
change sign, then system (1) has no periodic orbits lying entirely in D.
Proof
Since on any orbit of (1), we have dx2 / dx1  f 2 / f1. Therefore, on
will be positive. As p increases, a piece of the arc will lie outside the strip,
that is, the piece BCD in Figure 5. In this case, we have
  p    V  x  dt  
AB
BCD
any closed orbit
V  x  dt   V  x  dt .
DE
  p
is negative, hence
2
f
x
x 0
1
2
1
1
1
2
2
 0.
By Green’s theorem,
  p   p.
 f1 f 2 
S  x1  x2  dx1 dx2  0,
The proof of a new path in the left-half plane can be done similar to
the above and therefore we can conclude that there exists a closed
bounded set M such that any trajectory starting in M stays in M for all
future time.
The second step is to show that the Jacobian matrix has
eigenvalues with positive real parts. Since the system has a unique
equilibrium point at the origin. The Jacobian given by
A
we have
 f  x , x  dx  f  x , x  dx
It can be shown (in [1]) that, in the equation above, the first and third
integrals are positive and the second integral is negative. Since increasing
p will increase the portion BCD, by choosing p large enough, we can
ensure that
,
where S is the interior of
.
For this to be true, we have
f1 f 2
f
f

 0 or 1  2 must change sign.
x1 x2
x1 x2
1 
0


 1  h '  0 
has eigenvalues with positive real parts since h '  0   0.
8
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 6: [1] Consider the system
2.3 Index Method
The index method is useful also in ruling out the existence of
periodic orbits.
x1  f1  x1 , x2   x2 ,
x2  f 2  x1 , x2   ax1  bx2  x12 x2  x13 .
Lemma 3: (Poincare) Index criterion
Let C be a simple closed curve not passing through any
equilibrium point of (1). Let p be a point on the curve C with a vector
Show that there is no periodic orbit in the whole plane if b  0.
field f  x  . Let p traverse C in the counterclockwise direction, the
vector f  x  rotates continuously and , upon returning to the original
position, must have rotated an angle 2 k for some integer k , where the
angle is measured counterclockwise. The integer k is called the index of
the closed curve C. If C is chosen to encircle a single isolated equilibrium
point x , then k is called the index of x . The following information is
true.
a) The index of a node, a focus, or a center is +1.
b) The index of a saddle is -1.
c) The index of a periodic orbit is +1.
d) The index of a closed curve not encircling any equilibrium points is 0.
e) The index of a closed curve is equal to the sum of the indices of the
equilibrium points within it.
As a corollary to this lemma is as follows.
Corollary 1:
Inside any periodic orbit  , there must be at least one equilibrium
point. Suppose the equilibrium points inside  are hyperbolic, then if N is
the number of nodes and foci and S is the number of saddles, it must be
that N  S  1.
Solution
We have
f1 f 2

 b  x12 .
x1 x2
If b  0, there can be no periodic orbit in the whole plane.
9
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
3 Bifurcation
 Example 7: [1] Find the possible locations of the periodic orbits of the
system
In this section, we are interested in perturbations that will change
the equilibrium points or periodic orbits of the system or change their
stability properties.
Bifurcation is a change in the equilibrium points or periodic orbits,
or in their stability properties, as a parameter is varied. The parameter is
called a bifurcation parameter, and the values at which changes occur are
called bifurcation points. The bifurcation diagram is a sketch between the
values of the bifurcation parameter versus the amplitude of the equilibrium
points.
x1   x1  x1 x2 ,
x2  x1  x2  2 x1 x2 .
 Example 8: [1] Sketch a bifurcation diagram of the system
x1    x12 ,
x2   x2 ,
which depends on a parameter
Solution
The system has two equilibrium points at
 0, 0
.
and 1,1 . The
Jacobian matrices at these points are
 1 0  f 
0 1
 f 

,



 1 1 .
 x 
 x 
1
1



 0, 0
1, 1 
Hence,  0, 0  is a saddle, while 1,1 is a stable focus. Therefore the only
possible location of a periodic orbit is around the point  0, 0  .
10
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2007 by Withit Chatlatanagulchai
Solution
  0, the system has two equilibrium points at
For


u , 0 . Linearization at
 2 

 0




u , 0 and
 results in the Jacobian matrix
that   , 0  is a stable node,

, 0
0
 , which shows
1
linearization at


2 
u , 0 yields the Jacobian matrix 
 0
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
the solid lines represent the stable nodes, stable focuses, and
stable limit cycles
the dashed lines represent unstable nodes, unstable focuses,
unstable limit cycles, and saddles.
while
0
 which
1



shows that  u , 0 is a saddle.
 decreases, the saddle and node approach each other, collide
at   0, and disappear for   0. The phase portraits of the three cases
(a)
As

(b)
is given in Figure 6.
x2
x2
x2

(c)
x1
(a)
x1
(b)

(d)
x1

(c)
(e)
Figure 6: Phase portrait of the saddle-node
bifurcation. (a)   0. (b)   0. (c)   0.

(f)
Figure 7: Bifurcation diagrams of various
examples. (a) Saddle-node bifurcation. (b)
Transcritical bifurcation. (c) Supercritical
pitchfork bifurcation. (d) Subcritical pitchfork
bifurcation. (e) Supercritical Hopf bifurcation.
(f) Subcritical Hopf bifurcation.
The bifurcation diagram is given in Figure 7. The bifurcation
diagram has following attributes:
 the bifurcation diagram is a sketch between the values of the
bifurcation parameter versus the amplitude (norm) of the
equilibrium points
11
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 10: [1] Show that the bifurcation diagram of the system
 Example 9: [1] Show that the bifurcation diagram of the system
x1   x1  x12 ,
x1   x1  x13 ,
x2   x2 ,
x2   x2 ,
is given as in Figure 7(b).
is given as in Figure 7(c).
Solution
The system has two equilibrium points at
Jacobian at  0, 0  is

0

 0, 0 is a stable
  0. The Jacobian at   , 0  is
which shows that

 0

which shows that
  0.
  , 0
 0, 0
and
  , 0.
The
0
1
node for
  0 and a saddle for
0
1
is a saddle for
  0 and a stable node for
12
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 12: [1] Show that the bifurcation diagram of the system
 Example 11: [1] Show that the bifurcation diagram of the system
x1  x1    x12  x22   x2 ,
x1   x1  x13 ,
x2   x2 ,
x2  x2    x12  x22   x1 ,
is given as in Figure 7(d).
is given as in Figure 7(e).
13
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 3
 Example 13: [1] Show that the bifurcation diagram of the system
2
x1  x1     x12  x22    x12  x22    x2 ,


2
x2  x2     x12  x22    x12  x22    x1 ,


is given as in Figure 7(f).
Lesson 3 Homework Problems
2.17 (no. 2 only), 2.20 (no. 2 only), 2.26 (no. 1 only), 2.27 (no. 1,
3), 2.30
Homework problems are from the required textbook (Nonlinear
Systems, by Hassan K. Khalil, Prentice Hall, 2002)
References
[1]
14
Nonlinear Systems, by Hassan K. Khalil, Prentice Hall, 2002.
Copyright
2007 by Withit Chatlatanagulchai