LIMIT CYCLES
Limit Cycle of the Van der Pol Equation
3. LIMIT CYCLES
3.0
INTRODUCTION
Thus far in our study of dynamical systems the only type of attractors or repellers
encountered have been point attractors or repellers i.e. stable nodes and foci or unstable
nodes and foci. We shall now show that another type of attractor or repeller is possible for
nonlinear systems, the limit cycle.
Polar Coordinates
Often it is advantageous to rewrite a dynamical system in polar coordinates when considering
bg
limit cycles. Recall the relationships between polar coordinates r, and cartesian
bg
coordinates x , y
r 2  x2  y2
tan  
y
x
and the inverse relationships
x  r cos
y  r sin 
Differentiating the former relationships with respect to t we obtain
2rr  2 xx  2 yy
 rr  xx  yy
and
1
xy  yx
sec 2  
x2
 x 2 1  tan 2    xy  yx
c h
|R
Fy I |U
x S
1  GJV
  xy  yx
H
K
x
|
|
T
W
c
x y h
  xy  yx
2
2
2
2
 r 2  xy  yx
Also at an equilibrium point
x  y  0
 rr  0
r 2  0
hence we must have either r    0 or r  r  0 .
Consider the system
c
h
y   x  yc
1 x  y h
x  y  x 1  x 2  y 2
2
2
In order to facilitate the analysis of the system we shall transform it into polar form. Recall
that
rr  xx  yy
r   xy  yx
2
Thus for our system
o c ht  yo x  yc1  r ht
r c
1 r h
rr  x y  x 1  r 2
2
2
2
where r 2  x 2  y 2 . Similarly
2
o c ht  yoy  xc1  r ht
r 2  x  x  y 1  r 2
2
 r 2
and hence the polar form of the system is
b gb g
r  r 1  r 1  r
  1
From these results two obvious solutions are r  0 and r  1 . (Why is r  1 not a
solution?). The first solution corresponds to an unstable focus at the origin. The second
represents the polar form of the equation of a circle of radius one and centre the origin. The
fact that   1 implies that the circle is traversed in a clockwise direction with a constant
angular velocity of one unit. Thus we have a solution of the system which is a closed orbit..
Notice that if r  1 then r  0 and trajectories spiral outwards towards the closed orbit. If
r  1 then r  0 and trajectories spiral inwards towards the closed orbit.
r

Overall the trajectories approach an isolated closed orbit which we call a stable limit cycle.
Other types of limit cycles are possible. For example
b gb g
r  r 1  r r  1
  1
3
represents a system with an unstable limit cycle. In this case if r  1 then r  0 and
trajectories spiral inwards away from the closed orbit. If r  1 then r  0 and trajectories
spiral outwards away from the closed orbit.
r

A third possibility is represented by the system
b g
r  r 1  r
  1
2
In this case we have a semi-stable limit cycle in which if r  1 then r  0 and trajectories
spiral outwards towards the closed orbit. If r  1 then again r  0 and trajectories spiral
outwards away from the closed orbit.
r

4
Exercises 1
1.
Determine the nature of the limit cycles of the following sytems of differential
equations given in polar form
2.
b g
  1
b gb g
  1
b rgbr  1g,
  1
(a)
r  r 1  2r ,
(b)
r  r 1  r 2  r ,
(c)
r  r
(d)
r  r 3  6r 2  11r  6,   2
1
2
c h
Show that the system r  r 1  r 2 ,   1 is equivalent to
c h
y  x  y  yc
x y h
x  x  y  x x 2  y 2
2
3.
2
Show that the dynamical system
c
h
x  ax x 2  x 2  1  x  0
where a  0 has a stable limit and state its equation.
5
3.1
TRAPPING REGIONS
It is not always possible to determine the existence of a limit cycle directly from the polar
form as is illustrated in the following example.

Worked Example 1
Show that the system
c h
y  x  y  yc
x y h
x  x  y  x x 2  2 y 2
2
2
has a stable limit cycle which lies in the annular region
2
5
 r 1
As before transform the equations into polar form to obtain
c
r  r  r 3 1  41 sin 2 2
h
  1  21 r 2 sin 2  sin 2
Note that the polar form does not immediately give a limit cycle. However consider r  0 .
This implies that
c
h
1  r 2 1  41 sin 2 2  0
r2 
1
1  sin 2 2
c
1
4
But this must be true for all  . This means that
r2 
6
4
5
h
since the maximum value of the expression
1  41 sin 2 2
is
5
4
when sin 2 2  1 .
Now consider r  0 . This implies that
c
h
1  r 2 1  41 sin 2 2  0
r2 
1
1  sin 2 2
c
1
4
h
But this must be true for all  . This means that
r 2  1
since the minimum value of the expression
1  41 sin 2 2
is 1 when sin 2 2  0 . These results imply that r  0 when r 
Now consider a trajectory which starts inside the annular region
r  0
2
5
and r  0 when r  1 .
2
5
r 1
r  0
r
7
2
5
 r  1 shown below.
Suppose that the trajectory touches the outer boundary. Then immediately r  0 and the
trajectory is forced to remain in the annulus. Similarly if the trajectory touches the inner
boundary then r  0 and once more the trajectory is forced to remain in the annulus. Thus
any trajectory which starts inside the annulus is trapped within it. Such a region is called a
trapping region.
The equilibrium points of the system are given by
rr  0
r 2  0
Hence either r and r  0 or both r  0 and   0. But   0 implies that
1  21 r 2 sin 2  sin 2  0
This cannot hold inside the region since
2
5
 r  1 . Hence there are no equilibrium points
within the annulus. Intuitively since a trajectory cannot tend to an equilibrium point and
cannot cross itself the trajectory must either be a closed orbit or eventually tend to a closed
orbit. Hence there must be at least one limit cycle within the annulus.
Exercises 2
1.
2.
Find the polar form of the following systems of differenial equations.
c
h
c
(a)
x  y  x 1  x 2  y 2 , y   x  y 1  x 2  y 2
(b)
x  x  y  x x 2  2 y 2 ,
(c)
x  y  x 3  x x 2  y 2 ,
h
c
h
y   x  y  y x 2  2 y 2
c
h
y   x  y 3  y x 2  y 2
2
c
h
c
h
2
Find an annular trapping region for each of the systems in Q1.
8
3.2
POINCARÉ-BENDIXSON THEORY
The results of Worked Example 1 can be formalised in the following theorem.
Theorem - Poincaré-Bedixson
Let D be a closed bounded region of the x-y plane and
x  f ( x , y )
y  g ( x , y )
be a dynamical system in which f and g are continuously differentiable. If a trajectory of the
dynamical system is such that it remains in D for all t  0 then the trajectory must be
(i)
a closed orbit,
(ii)
approach a closed orbit or
(iii)
approach an equilibrium point as t   .
The implication of this theorem is that if we can find a trapping region for a dynamical
system which does not contain an equilibrium point then there must be at least one limit cycle
within the region. The technique of transforming to polar coordinates although often
employed does not always work. The following examples demonstrates other methods of
obtaining trapping regions.

Worked Example 2
In the biochemical process of glycolis, living cells obtain energy by breaking down sugar
molecules. Often this process turns out to be oscillatory in nature. A simple model of the
process is described by the equations
x   x  ay  x 2 y
y  b  ay  x 2 y
where x and y are concentrations of reactants and a and b are positive parameters.
9
The diagram below shows the graphs of the expressions
y
x
a  x2
y
b
a  x2
and
on which x  0 and y  0 respectively. We shall show that the region within the dotted lines
is a trapping region for the system.
y
b
a
bb, b ag
b
k
x
On the x-axis
x   x
y  b
and hence x  0, y  0 and thus the trajectories must flow in the direction indicated by the
arrow.
10
On the y-axis
x  ay
y  b  ay
and hence since y 
When y 
b
, x  0, y  0 and the flow of the trajecories is as shown.
a
b
a
x  b  x  x 2
y   x 2
b
a
b
a
and hence x  0, y  0 for x  b giving the direction of flow shown.
On the small vertical line x  k hence
k I
y
G
c hF
H ak J
K
b I
F
y  b  ay  k y  c
ak h
y
G
H ak J
K
x   k  ay  k 2 y  a  k 2
2
2
2
2
Thus x  0 since the line is below the curve y 
y
x
. Also y  0 above the curve
a  x2
b
b
and y  0 below the curve y 
. In both cases the flow is into the region.
2
ax
a  x2
Now choose k so that the diagonal line has slope -1. Consider
x  y   x  ay  x 2 y  b  ay  x 2 y
b x
11
Now on the line x  b and hence
x  y  0
y dy
 
 1
x dx
and hence the trajectories are steeper than the line and hence must point into the region.
These results show that the region is a trapping region. However we cannot conlude that there
b I
F
G
H a b J
Klies within the region.
is a limit cycle within region since the equilibrium point b,
2
The Jacobian matrix of the linearzation is
L1  2 xy
JM
N2 xy
L
O
b a
b a P
M
ax O
b a
M
P
P
2b
a  x Q
F
IJ M
P
G

b

a
c
h
H K M
P
b a
N
Q
2
2
2
2
2
2
b,
b
a b 2
2
2
bg
Now det J  a  b 2 and
bg
tr J 
c h
b2  a  b2  a
2
b2  a
bg
If we choose the parameters a and b so that tr J  0 then the equilibrium point will be a
repeller. In this case suuround the equilibrium point by a small circular region as shown
below.
12
y
b
a
bb, b ag
b
k
x
Since the equilubrium point is a repeller, trajectories crossing the boundary must point into
the region defined by the dotted lines. In this case the new region is a trapping region which
does not include an equilibrium point. Thus the Poincare´-Bendixson theorem guarantees that
the region must contain at least one limit cycle. We can verify this result using Maple to
obtain the phase portrait. Suitable parameter values are a  0.08, b  0.6.
Limit Cycle
13

Worked Example 3
By considering the rate of change with respect to time of the function
c
V  1 4x2  y2
h
2
on the trajectories of the dynamical system
c
y  yc
1 4x
h b g
y h
 2 xb
1  xg
x  x 1  4 x 2  y 2  21 y 1  x
2
2
show that the dynamical system has a stable elliptical limit cycle.
It is easily seen that the only equilibrium point of the system occurs at the origin.
Now
V
V
V 
x 
y
x
y
c
hc
h b gt
o
4 yc
1  4 x  y hc
oy 1  4 x  y h 2 xb1  xgt
 4c
4x  y h
c1  4 x  y h
 16 x 1  4 x 2  y 2 x 1  4 x 2  y 2  21 y 1  x
2
2
2
2
2
2
2
2 2
Thus we see that V decreases on the trajectories except at the origin where it clearly has a
relative maximum, and on the ellipse 4 x 2  y 2  1 where it is identically zero.
14
Now consider the plot of V shown below.
Outside the ellipse the function is decreasing towards the ellipse. Inside the ellipse the
function is decreasing towards the ellipse again. Thus the trajectories of the system must
move towards the ellipse from points outside and inside for increasing t. Now consider a
contour plot of the function.
We can construct a trapping region by taking any contour inside the ellipse as the inner
boundary of an annular region together with any contour outside the ellipse as the outer
15
boundary. Hence by the Poincaré-Bendixson theorem since there are no equilibrium points
within the annular region there must be a stable limit cycle within the region.
Note also that as the contours can be taken as close to the ellipse as we like the limit cycle
must be the ellipse.
16
Exercises 3
1.
Use the Poincaré-Bendixson theorem to show that the systems of Exercise 1 of the
previous examples have at least one lmit cycle.
2.
Show that the system
x  x  y  x 3
y  x  y  y 3
has a limit cycle.
3.
By considering the level curves of the positive definite function V  41 x 4  21 y 2
show that the nonlinear oscillator
c
h
x  x 2  x 2  1 x  x 3  0 .
has a stable limit cycle.
4.
The following dynamical system is a simple model of a chemical oscillator
b g
x  1  b  1 x  ax 2 y
y  bx  ax 2 y
where a and b are positive parameters and x , y  0 are dimensionless concentrations.
b g
(a)
Show that the system has an equilibriun point at 1,b a which is a repeller
(b)
when b  1  a ;
by considering the curves given by x  0 and y  0 show that there exists a
(c)
trapping region for the system;
use the Poincaré-Bendixson theorem to show that the system has at least one
limit cycle.
17
3.3
LIÉNARD SYSTEMS
A great number of mathematical models of physical systems give rise to differential
equations of rhe type
bg bg
x  f x x  g x  0
This is known as Liénard's equation. The system is a generalization of the mass-springdamper system
x  bx  kx  0
bg
bg
previously encountered. In Liénard's equation f x x is the damping term and g x is the
spring term. The equation can be written as the planar system
x  y
bg bg
y   g x  f x y
Under certain conditions on f and g it can be shown that Liénard's equation has a limit cycle.
This result is known as Liénard's Theorem.
Theorem - Liénard's Theorem
Suppose that f and g satisfy the conditions
(i)
f and g are continuously differentiable;
(ii)
g is an odd function i.e. g  x   g x ;
(iii)
g x  0 for x  0 ;
(iv)
f is an even function i.e. f  x  f x ;
(v)
b g bg
bg
b g bg
x  0 for 0 < x  a ,
let F bg
x  zf bg
u du and F bg
F bg
x  0 and increasing for x > a , (  F bg
0  Fb
a g
 0 );
x
0
then Liénard's equation has a unique stable limit cycle surrounding the origin.
18
The properties of the odd function F are indicated in the diagram below.
bg
F x
a
a
Use the graph to check the properties of F listed above.

Worked Example 4
Show that the Van der Pol equation
c h
x   x 2  1 x  x  0
where   0 , has a stable limit cycle.
This is an example of Liénard's equation with
bg c h
gbg
x x
f x   x2  1
Checking the conditions of Liénard's Theorem we find that
(i)
both f and g are continuosly differentiable;
(ii)
g x  x is odd and obviously g x  0 for x  0 ;
bg
bg
19
(iii)
(iv)
b g eb g j c h bg
F bg
x  z
cu  1hdu  cx  xh
F bg
x   cx  x h
 0 for 0  x  3 ,
F bg
x   cx  x h
 0 and increasing for x 
f  x    x  1   x 2  1  f x and so f is even;
2
x
2
1
3
0
1
3
3
1
3
3
3
bg d i
3 , ( F 0  F  3  0 ).
The conditions are satisfied hence there exists a unique stable limit cycle surrounding the
origin.
Van der Pol's equation arose from a 1926 mathematical model of an electrical circuit
containing a thermionic valve. Valves were used in early radios and TVs. Careful observation
c h
of the damping term  x 2  1 x shows that for small x the oscillations are negatively damped
and hence increase in amplitude. For large x the oscillations are positively damped and hence
decrease in amplitude. Intuitively somewhere between these two extremes the damping must
be zero resulting in an oscillation with constant amplitude. This is an isolated periodic
solution and hence a limit cycle. The most interesting behaviour occurs for relatively large
values of the parameter  . The diagram below shows the phase portrait for the case   5 .
Notice that the limit cycle is very different from one resulting from sinusoidal oscillations.
This can be seen clearly in the time series plots of x and y shown below.
20
21
Exercises 4
1.
Show that the equation
c h
x   x 2  1 x  tanh x  0
for   0 has exactly one periodic solution and classify its stability.
2.
Consider the equation
c h
x   x 4  1 x  x  0
(a)
(b)
(c)
Prove that the system has a unique limit cycle if   0 ;
plot the phase portrait for the case   1 ;
if   0 does the system still have a limit cycle?
22
3.4
DULAC’S CRITERION
It is often useful to be able to rule out the existence of a limit cycle in a region of the plane.
The following result due to Dulac sometimes enables us to do this.
Dulac's Criterion
Let D be a simply connected region of the phase plane. If there exists a continuously
differentiable function  x , y such that
bg


 x, y f x, y 
 x, y g x, y
x
y
mb gb gr mb gb gr
is of constant sign in D then the dynamical system
bg
y  gb
x, yg
x  f x , y
has no closed orbits wholly contained in D.
This method suffers from the same problem as finding Lyapunov functions in that it is often
difficult to find a suitable function  x , y . Some possible choices for  x , y are
bg
bg
 x , y  1,
bg
1
, ex , ey

x y

23

Worked Example 5
Show that the system
x  y
y   x  y  x 2  y 2
has no closed orbits anywhere.
bg
Let us try  x, y  1 .


y 
 x  y  x 2  y 2  1  2 y
x
y
bg c
h
Although this shows that there is no closed orbit contained in either half-plane
y  21 , or y  21 it does not rule out the existence of a closed orbit in the whole plane since
there may be such an orbit which crosses the line y  21 .
bg
Now let us try  x , y  ex .
 x
 x
e y 
e  x  y  x2  y2
x
y
c h oc
ht
 yex  ex  2 yex
b g r
m
 ex   2 y  1
Choosing   2 reduces the expression to e 2 x which is negative everywhere. Hence
there are no closed orbits.
24
Exercises 5
1.
Show that the system
x   x  y  x 2
y  x
has no closed orbits.
2.
Show that the competing species model
b g
y  cyb
1  yg
 dxy
x  ax 1  x  bxy
where a , b, c, d are positive parameters and x , y  0 has no periodic solutions.
25
3.5
THE HOPF BIFURCATION
The behaviour of a one-dimensional dynamical system may depend on the value of a certain
parameter. As the parameter value passes through a critical value the system dynamics can
change substantially. This is termed a bifurcation. Consider the following example.

Worked Example 6
Investigate the effect of changing the parameter  on the dynamical system
b gc h
y  x  y  bx  y g
cx  y h
x  x  y  x  23 y x 2  y 2
2
3
2
2
It can be shown that the polar form of the system is
r  r  r 3
  1  3 r 2
2
from which it is clear that there is a single equilibrium point at the origin. The first polar
equation can be written as
r  r
d  rid  ri,
Clearly there is a limit cycle of radius
0
 in this case. Also the origin is unstable since for
small r
r  r
which is positive.
If   0 then r  0 and the origin is globally asymptotically stable. This behaviour is
illustrated below for the cases   4 and   4 respectively.
26
Stable Focus
Unstable Focus + Stable Limit Cycle
We see that as the parameter  passes through zero from negative to positive a stable focus
gives way to an unstable focus surrounded by a stable limit cycle whose radius increases with
 . This is an example of a Hopf bifurcation.
In order to gain further insight into the behaviour of this system and to set the scene for the
introduction of the principal result of this section let us carry out a linear stability analysis of
the cartesian form of the equations. The linearization about the origin is obviously
x  x  y
y  x  y
The Jacobian matrix is therefore
27
 1O
L
M
N1  P
Q
The eigenvalues are given by
2  2  1   2  0
Solving gives
   i
Thus if   0 we have a stable focus, if   0 an unstable focus and if   0 we cannot
decide (although we know from the polar form that we have a stable focus). As the
bifurcation from a stable focus to an unstable focus surrounded by a stable limit cycle occurs
the real part of the eigenvalues changes from negative to positive being zero at the bifurcation
point. This is typical of this type of Hopf bifurcation which is said to be supercritical. The
following theorem generalises these results.
Theorem - The Hopf Bifurcation Theorem
Suppose that the system
b g
y  gb
x, y,  g
x  f x, y , 
has an equilibrium point at the origin for all  . In addition suppose that the eigenvalues
 1  ,  2  of the linearization are purely imaginary when    c . If the real part of the
bg bg
eigenvalues satisfy
c bgh
d
Re  1,2 
d
c
0
and the origin is asymptotically stable when    c then
(i)
(ii)
(iii)
   c is a bifurcation point of the system;
for some  1 such that  1     c the origin is a stable focus;
for some  2 such that  c     2 the origin is an unstable focus surrounded by a
stable limit cycle whose size increases with  .
28
In the previous example  1   and  2   . Also
d

d
 0
 1 0
and hence the theorem predicts the existence of the stable limit cycle.

Worked Example 7
Investigate the behaviour of the system
x  x  y  xy 2
y  x  y  y 3
as the parameter  is varied.
This system has a single equilibrium point at the origin as you should verify. The
linearization is clearly
x  x  y
y  x  y
which is identical to that found in Worked Example 6. This means that all the conditions of
the Hopf bifurcation theorem are satisfied apart from the asymptotic stability of the origin at
  0 . This suggests that there may be a Hopf bifurcation. However the polar form of the
equation for r is
r  r  ry 2
which shows that the origin is unstable. Hence the theorem does not apply. Using Maple we
obtain the following plots for   0.2 and   0.2 respectively.
29
Stable Focus + Unstable Limit Cycle
Unstable Focus
The first plot shows the existence of an unstabe limit cycle surrounding the stable focus at the
origin The second plot shows the unstable focus which occurs when  is positive. We have a
different type of Hopf bifurcation which is referred to as subcritical.
30

Worked Example 8
Show that the dynamical system
c hc
y  x  ay  yc
x y h
c2  x
h
y h
x  ax  y  x x 2  y 2 2  x 2  y 2
2
2
2
2
undergoes a subcritical Hopf bifurcation at a  0 for a certain range of values of the
parameter a.
Clearly an E.P. at the origin.
Transforming to polar coordinates
rr  xx  yy
c hc
 ac
x  y hc
 x  y hc
2 x
 ar  r c
2r h
h
y h
c
hc
h
h
c
hc
h
 ax 2  xy  x 2 x 2  y 2 2  x 2  y 2  xy  ay 2  y 2 x 2  y 2 2  x 2  y 2
2
2
2
4
2
2 2
2
2
2
 ar 2  2r 4  r 6
r 2  xy  yx
c
hc
 x 2  axy  xy x 2  y 2 2  x 2  y 2  axy  y 2  xy x 2  y 2 2  x 2  y 2
x y
2
2
 r2
Thus
r  ar  2r 3  r 5
  1
Thus for ‘small’ values of r
r  ar
31
and hence the equilibrium point is a stable focus if a  0 and an unstable focus if a  0 .
For a  0
r  2r 3  r 5  2r 3  0
for ‘small’ r. Thus the origin is unstable.
Now we can factorise the expression for r to obtain
d
id
r  r r 2  1  1  a r 2  1  1  a
i
If 1  a  0 then 1  1  a , 1  1  a are both real and positive so we can further factorise
to obtain
e
je
je
je
r  r r  1  1  a r  1  1  a r  1  1  a r  1  1  a
and thus we obtain two limit cycles of radii r1 
j
1  1  a and r2  1  1  a .
Examining the signs of the terms in the factorisation
0  r  r1  r  0
r1  r  r2  r  0
r  r2  r  0
Thus r1  1  1  a is unstable and r2  1  1  a is stable.
If a  0 then 1  1  a is real and positive but 1  1  a is real and negative. In this case we
can only factorise into the form
e
jd
je
i
r   r r  1  1  a r  1  1  a r 2  1  a  1
32
and thus there is a single limit cycle of radius r2  1  1  a . Examining the signs of the
terms in the factorisation
0  r  r2  r  0
r  r2  r  0
and thus the limit cycle is stable.
Summarising these results we see that if 1  a  0 the origin is a stable focus surrounded by
an unstable limit cycle of radius r1  1  1  a and a stable limit cycle of radius
r2  1  1  a . If a  0 the origin is an unstable focus surrounded by a stable limit cycle
of radius r2  1  1  a .
Thus there is a subcritical Hopf bifurcation. The existence of the second stable limit cycle
does not affect this conclusion.
33
Exercises 6
1.
By calculating the linearisation at the origin show that the system
x   y  x  xy 2
y  x  y  x 2
has pure imaginary eigenvalues when   0 . By plotting phase portraits show that the
system undergoes a Hopf bifurcation at   0 and state whether it is subcritical or
supercritical.
2.
Show that the dynamical system
x  y  x,
y   x  y  x 2 y
undergoes a supercritical Hopf bifurcation at   0 .
3.
The following system undergoe a Hopf bifurcation at the origin when   0 . By
plotting the phase portrait determine whether the Hopf bifurcation is subcritical or
supercritical.
x   y  x  xy 2
y  x  y  x 2
4.
Consider again the chemical reaction equations
b g
x  1  b  1 x  ax 2 y
y  bx  ax 2 y
considered in Exercises 3 Q4. Show that a Hopf bifurcation occurs for some
parameter value b  bc to be determined. Does the limit cycle exist for b  bc or
b  bc ? Explain using the Poincaré-Bendixson theorem.
34
3.6
CASE STUDY
ECONOMIC INVESTMENT
A simplified economic model of a national economy with investment is represented by the
dynamical system
bg b g
x  rv x  2 x  y
y  x
bg
where y is the national product and v x  arctan x is the investment function.
Stability Analysis
It is clear that the system has a single equilibrium point at the origin for all r. The Jacobian
matrix is given by
J
bg
r 2
L
M
N1
2
0
O
P
Q
bg
and tr J  r  2 , det J  2 and hence the equilibrium point is an attractor if r  rc and a
repeller if r  rc where rc  2 .
The eigenvalues of the Jacobian matrix are given by
 1,2 
r 2
br  2g 8
2
2
which are given by  2i at r  rc . Also
n c hs
d
Re  1,2
dr
r  rc

1
0
2
Thus all the conditions of the Hopf Bifurcation theorem are satisfied except for the
asymptotic stability of the origin at r  rc .
35
Asymptotic Stability of Origin
bg
Introduce the Lyapunov function V x , y  px 2  qy 2 then on the trajectories
V  2 pxx  2qyy
mbg b gr
 2 prxvbg
x  4 px  4 pxy  2qxy
 2 px rv x  2 x  y  2qxy
2
Now choose p  1, q  2 to eliminate the cross term and
bg
V  2rxv x  4 x 2
Thus at r  rc
V  4 x arctan x  4 x 2
b
 4 x arctan x  x
g
But
l
q
d
1
arctan x  x 
1 0
dx
1 x2
hence we have the plot shown below.
yx
y
y  arctan x
x
36
Thus, if x  0 , arctan x  x  0 and if x  0 , arctan x  x  0 and thus V  0 with
inequalityonly on the line x  0 which is not a trajectory of the system. Hence the origin is
asymptotically stable and there must be a supercritical Hopf bifurcation at r  rc .
Limit Cycle
The limit cycle is shown below for parameter value r  3 .
Unstable Focus + Stable Limit Cycle
Limit Cycle
37
APPENDIX 1
CASE STUDIES
38
Economic Investment
A simplified economic model of a national economy with investment is represented by the
dynamical system
bg b g
x  rv x  2 x  y
y  x
bg
where y is the national product and v x  arctan x is the investment function.
(a)
Show that the system has a single equilibrium point at the origin.
(b)
By using the trace-determinant diagram or otherwise show that the equilibrium point
cannot be a saddle point. Show further that the equilibrium point is a repeller if
r  rc  2
(c)
Show that all the conditions of the Hopf bifurcation theorem are satisfied except the
asymptotic stability of the origin at the critical value of r.
(d)
bg
Show using a Lyapunov function of the form V x , y  mx 2  ny 2 , where m and n are
to be determined, that the origin is asymptotically stable at the critical value of r.
(e)
Demonstrate computationally the existence of a supercritical Hopf bifurcation at
r  rc .
39
Hopf Bifurcation in a Chemical Reaction
A chemical reaction is governed by the system of differential equations
4 xy
1 x2
y
y  bx 1 
1 x2
x  a  x 
F
G
H
IJ
K
where x and y are concentrations of the reactants and a and b are positive parameters.
F
a
a I
,1  J
G
H5 25K
2
(a)
Show that the system has a single equilibrium point at
(b)
By using the trace-determinant diagram or otherwise show that the equilibrium point
cannot be a saddle point. Show further that the equilibrium point is a repeller if
b  bc 
(c)
3a 25

5
a
By considering the nullclines
ba  xgc1  x h
y
2
4x
y  1 x
2
show that the rectangle bounded by the lines x  0 , y  0 , x  a , y  1  a 2 is a
trapping region for the trajectories of the system.
(d)
Deduce using the results of (b) and (c) together with the Poincaré-Bendixson theorem
that the system has a stable limit cycle within the rectangle defined in (c).
(e)
Show by computation that the system undergoes a Hopf bifurcation at b  bc and
state whether it is sub- or super-critical.
40
APPENDIX 2
ANSWERS AND HINTS
FOR THE EXERCISES
41
Introduction
Exercises 1
1. (a) r  21 , stable.
(b) r  1 , stable; r  2 , unstable.
(c) r  21 , unstable; r  1 , stable.
(d) r  1 , unstable; r  2 , stable; r  3 , unstable
2. Use x  r cos , y  r sin  , x  r cos  r sin  , y  r sin   r cos .
3. Write as a first order system, transform to polars, and show that r  1 is a solution.
Trapping Regions
Exercises 2
c h
1  sin  h
,   1.
(b) r  r  r c
1  sin 2 h
r  r ,   1 
(c) r  c
1. (a) r  r 1  r 2 ,   1.
3
1
2
2
2
3
5
1
4
r 2 sin 4 .
2. (a) Any annular region for which r1  r  r2 with r1  1, r2  1 .
(b)
1
2
 r  1.
(c)
1
2
 r  1.
Poincaré-Bendixson Theory
Exercises 3
1. Show that in each case the only equilibrium point is at the origin.
c
h
2. r  r  r 3 1  21 sin 2 2 ,   1  41 r 2 sin 4 , 1  r  2 .
c
h
3. Show that V   y 2 x 2  y 2  1 . Choose level curves V  C1 , V  C2 such that for C1 ,
V  0 and for C2 , V  0 .
42
bg
bg
4. tr J  b  1  a, det J  a . Use the same idea as in Worked Example 2.
Liénard Systems
Exercises 4
1. Use Liénard’s theorem.
2. (a) Use Liénard’s theorem;
(d) Globally asymptotically stable for   0 and hence cannot have a limit cycle.
Dulac’s Criterion
Exercises 5
bg
1
Use weighting function  b
.
x, yg

xy
1. Use weighting function  x , y  ey .
2.
The Hopf Bifurcation
Exercises 6
1. Subcritical.
2. Use Hopf bifurcation theorem and V  x 2  y 2 to confirm asymptotic stability.
3. Choose   1 ; subcritical.
4. bc  1  a ; b  bc .
43
APPENDIX 3
MAPLE WORK SHEETS
44
Worked Example 2
> restart:
> with(linalg):
> with(DEtools):
Plot of the nullclines
> yf:=x/(a+x^2);
> yg:=b/(a+x^2);
Need to choose values for a and b
> a:=0.08:b:=0.6:
> plot({yf,yg},x=0..3);
Stability Analysis
Define the dynamical system
> a:='a':b:='b':
> f:=-x+a*y+x^2*y;
> g:=b-a*y-x^2*y;
Obtain the equilibrium point
> ep:=solve({f,g},{x,y});
Find the Jacobian matrix
> u:=[f,g];
> J:=jacobian(u,[x,y]);
> J1:=map(simplify,subs(ep,evalm(J)));
Find the trace and determinant
45
> trJ1:=trace(J1);
> detJ1:=det(J1);
Now plot the phase portrait
> f1:=-x(t)+a*y(t)+x(t)^2*y(t);
> g1:=b-a*y(t)-x(t)^2*y(t);
> a:=0.08:b:=0.6:
> de1:=diff(x(t),t)=f1;
> de2:=diff(y(t),t)=g1;
> ics:=[[0,1,1],[0,0.3,0.3]];
> DEplot([de1,de2],[x(t),y(t)],t=-10..30,ics,x=0..2,y=0..3,stepsize=0.1,linecolour=blue);
Plot the limit cycle
> DEplot([de1,de2],[x(t),y(t)],t=20..50,ics,x=0..2,y=0..3,stepsize=0.1,linecolour=blue);
Worked Example 3
> restart:
> with(plots):
Define V
> V:=(1-4*x^2-y^2)^2;
Define the system
> f:=x*(1-4*x^2-y^2)-y*(1+x)/2;
> g:=y*(1-4*x^2-y^2)+2*x*(1+x);
Find dV/dt
> dV:=factor(diff(V,x)*f+diff(V,y)*g);
Show contour plots
> contourplot3d(V,x=-0.6..0.6,y=-1.2..1.2,contours=50);
> p1:=contourplot(V,x=-0.8..0.8,y=-1.5..1.5,contours=20):
> p2:=implicitplot(1-4*x^2-y^2,x=-0.8..0.8,y=-1.5..1.5,colour=blue):
> display({p1,p2});
46
Worked Example 4
> restart:
> with(DEtools):
Define the system
> f:=y(t);
> g:=-x(t)+epsilon*(1-x(t)^2)*y(t);
Set the value of epsilon;
> epsilon:=5:
Obtain the phase portrait
> de1:=diff(x(t),t)=f;
> de2:=diff(y(t),t)=g;
> ics:=[[0,0,0.5]]:
> DEplot([de1,de2],[x(t),y(t)],t=0..30,ics,x=-2..2,y=8..8,stepsize=0.01,linecolour=blue,maxfun=-1);
and the time series plots for x and y
> DEplot([de1,de2],[x(t),y(t)],t=0..50,ics,linecolour=blue,scene=[t,x],
stepsize=0.01,method=classical[rk4]);
> DEplot([de1,de2],[x(t),y(t)],t=0..50,ics,linecolour=blue,scene=[t,y],
stepsize=0.01,method=classical[rk4]);
Worked Example 6
> restart:
> with(DEtools):
Define the system
> f:=mu*x(t)-y(t)-(x(t)+2/3*y(t))*(x(t)^2+y(t)^2);
> g:=x(t)+mu*y(t)+(3/2*x(t)-y(t))*(x(t)^2+y(t)^2);
Set the value of mu;
> mu:=-4:
Obtain the phase portrait
> de1:=diff(x(t),t)=f;
> de2:=diff(y(t),t)=g;
> ics:=[[0,1,1],[0,-1,1],[0,1,-1],[0,-1,-1]]:
> DEplot([de1,de2],[x(t),y(t)],t=-5..5,ics,x=-3..3,y=-3..3,stepsize=0.01,
47
method=classical[rk4],linecolour=blue);
Choose another value for mu;
> mu:=4:
Obtain the phase portrait
> de1:=diff(x(t),t)=f;
> de2:=diff(y(t),t)=g;
> ics:=[[0,1,1],[0,-1,1],[0,1,-1],[0,-1,-1]]:
> DEplot([de1,de2],[x(t),y(t)],t=-5..5,ics,x=-3..3,y=-3..3,stepsize=0.01,
method=classical[rk4],linecolour=blue);
Worked Example 7
> restart:
> with(DEtools):
Define the system
> f:=mu*x(t)-y(t)+x(t)*y(t)^2;
> g:=x(t)+mu*y(t)+y(t)^3;
Choose a positive value for mu;
> mu:=0.2:
Plot the phase portrait
> de1:=diff(x(t),t)=f;
> de2:=diff(y(t),t)=g;
> ics:=[[0,1,1],[0,-1,1],[0,1,-1],[0,-1,-1]]:
> DEplot([de1,de2],[x(t),y(t)],t=-5..5,ics,x=-2..2,y=-1.5..1.5,stepsize=0.1,
method=classical[rk4],linecolour=blue);
Choose a negative value for mu;
> mu:=-0.2:
Obtain the phase portrait
> de1:=diff(x(t),t)=f;
> de2:=diff(y(t),t)=g;
> ics:=[[0,1,1],[0,-1,-1],[0,0.2,0.2],[0,-0.2,-0.2]]:
> DEplot([de1,de2],[x(t),y(t)],t=-10..10,ics,x=-2..2,y=-1.5..1.5,
stepsize=0.1,method=classical[rk4],linecolour=blue);
48
Case Study - Economic Investment
> restart:
> with(linalg):
> with(DEtools):
Define the investment function
> v:=x->arctan(x):
Define the system
> f:=r*v(x)-2*(x+y);
> g:=x;
Find the equilibrium points
> ep:=solve({f,g},{x,y});
Find the Jacobian matrix
> u:=[f,g];
> J:=jacobian(u,[x,y]);
> J0:=subs(ep,evalm(J));
Find the trace and determinant and discriminant
> trJ0:=trace(J0);
> detJ0:=det(J0);
> discJ0:=simplify(trJ0^2-4*detJ0);
> solve(discJ0>0,r);
From these results we have an attractor if r < 2; and a repeller if 2 < r; which is a focus if r <
2+2*sqrt(2); and a node otherwise.
Find the eigenvalues of the Jacobian matrix
> evs:=eigenvals(J0);
> ev1:=evs[1];
> ev2:=evs[2];
> subs(r=2,ev1);
> subs(r=2,ev2);
Thus eigenvalues are purely imaginary at r = 2;.
Also derivative of real part =1/2>0.
To investigate A.S. of the origin at r = 2; use the Lyapunov function
> V:=m*x^2+n*y^2;
49
Find dV/dt
> dV:=collect(diff(V,x)*f+diff(V,y)*g,[x,y],distributed);
Choose m = 1; and n = 2; to eliminate cross term and let r = 2;.
> dV1:=subs(m=1,n=2,r=2,dV);
This expression is negative except on the line x = 0; which is not a trajectory.
> plot(dV1,x=-2..2);
Thus all the conditions of the Hopf bifurcation theorem are satisfied.
Now obtain the phase portrait to show the bifurcation
> f:=r*v(x(t))-2*(x(t)+y(t)):
> g:=x(t):
> r:=3:
> de1:=diff(x(t),t)=f:de2:=diff(y(t),t)=g:
> ics:=[[0,2,0],[0,0.5,0]]:
> DEplot([de1,de2],[x(t),y(t)],t=-10..10,ics,x=-2.5..2.5,y=-2..2,
stepsize=0.1,linecolour=blue);
> DEplot([de1,de2],[x(t),y(t)],t=20..30,ics,x=-2.5..2.5,y=-2..2,
stepsize=0.1,linecolour=blue);
50