DISCRETE SYSTEMS
Saturn’s Satellite Hyperion
6. DISCRETE DYNAMICAL SYSTEMS
6.1
DISCRETE LINEAR SYSTEMS
In some scientific contexts it is natural to regard time as discrete. This is the case in
digital electronics, in parts of economics and finance theory, also in impulsively
driven mechanical systems. Evidently, discrete variables should be used in modelling
animal populations where successive generations do not overlap.
Let us consider discrete time systems. In a general case the system has the form
x ( k  1)  f ( x ( k )), x (0)  x0 .
If the function f is linear with respect to x then one has a linear system. Thus in the
one dimensional case the linear discrete system can be written as
x ( k  1)  a x ( k )  b, x (0)  x0 ,
a and b being given constants.
The graph of the linear function is a straight line. If the line does not go through the
origin of coordinates then the function is called affine.
Let us work out the first few values:
x (0)  x0
x (1)  ax0  b  ax0  b
x (2)  ax (1)  b  a 2 x0  ab  b.
For x ( k ) one has
x ( k )  a k x0  (a k 1  a k 2 ...a  1)b  a k x0 
ak 1
 b,
a 1
provided a  1 and
x ( k )  x0  kb,
when a  1. If a  1 then a k  0 and x ( k ) 
b
regardless of the value of x0 . If
1 a
a  1, then a k   and x( k )  .
In the two or more dimensional case the discrete linear dynamical system can be
written as
x ( k  1)  Ax ( k )  b , x (0)  x0
where x , x0 and b are vectors and A stands for a n  n matrix.
Similarily to the previous case one can compute the iterations x(1), x(2), x(3) etc. to
obtain
x (0)  x0 ,
x (1)  Ax (0)  b  Ax0  b ,
x (2)  Ax (1)  b  A2 x0  Ab  b ,
x (3)  Ax (2)  b  A3 x0  A2b  Ab  b .
In the general case we have
x ( k )  Ak x0  ( Ak 1  Ak 2 ... A  I )b
To simplify this, observe that
( Ak 1  Ak 2 ... A  I )( I  A)  I  Ak
and so, provided I  A is invertible, we have
x ( k )  Ak x0  ( I  Ak )( I  A) 1b .
Now, what happens as k  ? If the absolute values of the eigenvalues of A are
less than 1 (hence I  A is invertible), then A k tends to zero matrix. Hence
x (k )  ~
x  ( I  A) 1b .
Alternatively, if some eigenvalues have absolute value bigger than 1, then A k blows
up, and for most x0 we have x ( k )  .
There are, ofcourse, exceptional values of x0 . For example, if 1 is not an eigenvalue
of A and
x0  ~
x  ( I  A) 1b
then
x (k )  ~
x
for all k.
Finally, if some eigenvalues have absolute value equal to 1 and the other eigenvalues
have absolute value less than 1, we see a range of behaviour. The system might stay
near x or it might blow up.
Consider now the case a  1.
First, if a  1, then x ( k )  x0  kb (in the one dimensional case). So, if b  0, then
x( k )  . Otherwise x ( k ) is stuck at x0 regardless of the value of x0 .
Second, if a  1, then we observe that
x (0)  x0 ,
x (1)   x0  b,
x ( 2 )   (  x0  b )  b  x0 ,
x (3)   x0  b,
x ( 4 )  x0 ,
.
Thus we see that x ( k ) oscillates between two values, x0 and b  x0 . But if
x.
x0  b  x0 , i.e. x0  b / 2  ~
x , then x ( k ) is stuck at the fixed point ~
Exercises 1
1.
Find an exact formula for x ( k ), where
x ( k  1)  ax ( k )  b, x (0)  x0  c
and a , b, c have the following values:
a)
a  1, b  1, c  1;
b)
a  1, b  0, c  2;
c)
a
d)
a  1, b  1, c  4;
e)
1
3
a   , b  1, c  .
2
2
2.
For each of the discrete time systems in the previous problem, determine
3
, b  1, c  0;
2
whether or not x( k )  . Determine if the system has a fixed point and
whether or not the system is approaching that fixed point.
3.
Which of the following systems are dynamical systems; which are linear, affine,
or non-linear
2
,
x ( k  1)
a)
x ( k  3) 
b)
x ( k )  x ( k  1)  1,
c)
x ( k )  4 x ( k )  2,
d)
x ( k )  2 x ( k  1),
e)
x ( k  2) 
f)
x(n  5)  x(n  10)  5,
g)
x(n  2)  x(n  1)  n2  3.
x ( k  1)
,
x( k )
6.2 APPLICATIONS OF DISCRETE DYNAMICAL SYSTEMS
1) Applications to finance
Consider the general situation when we have an account that contains x ( k ) dollars
after k compounding periods. Suppose that account is collecting 100 I per cent
annual interest, compounded m times per year. Assume a constant amount b is added
to the account at the end of each compounding period (or taken from the account if
b  0 ). Let a0 be the initial amount in the account, that is x (0)  a0 .
You should see that the dynamical system
II
F
G
H mJ
Kx(k )  b
x ( k  1)  1 
describes the relationship between the amount in the account at the end of k  1
compounding periods and the amount after k compounding periods. This simple
model applies to a wide range of financial applications.
Let us consider the case when x ( k ) is the amount of saving after k compounding
periods. If, for instance we initially deposit a0  100, plus additional deposit of
b  100 every year thereafter, into a savings account that pays 10 per cent interest,
compounded annually. Then
x ( k  1)  11
. x ( k )  100
and
mb 100

 1000.
I
01
.
So
x( k )  (11
. ) k (1100)  1000.
In particular, x(40)  4878517
. (that is the total amount after 40 years). It is
worthwhile to note that our total deposits have been only 4100 dollars.
It is easy to check that in the general case one has
I I F mb I mb
F
G
H mJ
KG
Ha  I J
K I .
k
x( k )  1 
0
When b  0, then
F1  I IJa .
x( k )  G
H mK
k
0
2) Applications to economics
Let us consider supply and demand as it relates to a product that takes one unit of time
to produce. Let S (n) be the supply of our product, D(n) the demand for our product
and P(n) the price of one unit of our product, all in year n.
To develop a model we need certain assumptions. A reasonable set of assumptions is
the following:
1. The supply of the product in any year depends positively on the price of the
product the previous year;
2. The demand for the product in any year depends negatively on the present price of
the product;
3. Each year, the price of the product is adjusted so that the demand equals the
supply.
Let S (n), D(n) and P(n) represent the supply demand (the amount the consumers
buy) and price per bushel of potatoes, all in year n.
Assume, at first, that the supply equation is presented as
S (n  1)  0.8 P(n).
Thus, if the price is 6 dollars per bushel this year, the producers will grow 4.8 units of
potatoes next year. However, if the price is 12 dollars the production will be 9.6 units
next year.
Assume that the demand equation is
D(n)  12
. P(n)  20.
In this case, if the price is 6 dollars per bushel, the consumers are willing to buy 12.8
units of potatoes, while if the price is 12 dollars, the consumers are only willing to
buy 5.6 units of potatoes.
The third assumption states that the price next year will be adjusted so that the supply
equals the demand, that is
S (n  1)  D(n  1).
Since
D(n  1)  12
. P(n  1)  20
and
S (n  1)  0.8 P(n)
substitution gives
12
. P(n  1)  20  0.8 P(n).
Thus
P(n  1)  
2
50
P ( n)  ,
3
3
or
2I
F
G
H3J
K c  10.
k
P( k )  
From the second assumption, demand in year n 1 depends on the price in year n 1.
Assuming that the relationship is linear, the demand equation for the next year is
D(n  1)   dP(n  1)  b
whereas the supply equation is
S (n  1)  sP(n)  a.
Now one can obtain that
sI
F
G
Hd J
K p,
k
P( k )  c 
where p is the equilibrium value of this dynamical system
p
ba
.
sd
Exercises 2
1.
Set up the dynamical system and compute the amount in your savings account
after
a)
1 year with an initial deposit of 1000 dollars at 8 per cent interest, compounded
quarterly.
b)
5 years with an initial deposit of 200 dollars at 5 per cent interest, compounded
semi-annually.
2.
Suppose you start a savings account which pays 8 per cent interest a year
compounded monthly. You initially deposit 1000 dollars and decide to add an
additional 100 dollars each month thereafter. How much is in your account after
5 years?
3.
Consider the supply and demand equations
S (n  1)  08
. P(n); D(n  1)  0.6 P(n  1)  7.
a)
Find the first order affine dynamical system relating to P(n).
b)
Find the equilibrium value for this equation.
c)
Discuss the behaviour of P( k ); is it stable or unstable?
4.
Determine the dynamical system for P(n 1) in terms of P(n), find the
equilibrium value, and classify it as stable, unstable, or unsure, if
S (n  1)  12
. P(n)  1
and
D(n  1)  0.8 P(n  1)  7.
6.3
EQUILIBRIUM POINTS
Let the vector x be the state of the dynamical system. The function f tells us how
the system moves, e.g.
x ( k  1)  f ( x ( k )).
In special circumstances, however, the system does not move. The system can be
stuck (we shall say fixed) in a special state. We call these states fixed (equilibrium)
points of the dynamical system.
For example, consider the non-linear discrete time system
x( k  1)  [ x( k )]2  6.
Suppose the system is in the state x ( k )  3. Evidently,
x( k  1)  [ x( k )]2  6  3.
Thus the system is again at the state x  3.
The equilibrium point of a dynamical system is a state with the property that if the
system is ever in this state it will remain in that state for all time.

Worked Example 1
Consider the system
x ( k  1) OL
x (k )  x (k ) O
L
M
P.
M
P
x ( k  1)QN
x ( k )  x ( k )  2Q
N
1
2
1
2
2
1
2
This may be rewritten as x ( k  1)  f ( x ( k )), where
uOL
u v O
L
M
P.
M
P
v QN
u  v  2Q
N
2
f
In order to find equilibrium points we solve the equations
u  u2  v
v  uv2
from which one can find u  2, v  2.
The equilibrium points can be stable, unstable or semistable. Suppose a first order
dynamical system has an equilibrium value a. This equilibrium value is said to be
stable or attracting if there is a number  , unique to each system, such that, when
x(0)  a   ,
then lim x ( k )  a.
k 
An equilibrium value is unstable or repelling if there is a number  such that, when
0  x(0)  a  
then
x( k )  a  
for some, but not necessarily all, values of k. Note that x( k )  a is a measure of the
distance between the two values a and x ( k ). The definition states that an equilibrium
value is stable if whenever x(0) is sufficiently close to a then x ( k ) tends to a.
The equilibrium point is called marginally stable or neutral provided the following:
for all starting values near a the system stays near a but does not converge to a .
Consider a dynamical system (called a general affine system)
x ( k  1)  r x ( k )  b, r  1.
Theorem: The equilibrium value a  b / (1  r ) for this dynamical system is stable, if
r  1 for any value of x(0). Also, if r  1 then the equilibrium value a is unstable
and x ( k ) goes to infinity for any x(0)  a.
The equilibrium value is called semistable if for some values of x (0), x ( k ) tends to a,
while for others x ( k ) goes away from a.
Exercises 3
1.
Find all fixed points of the following discrete time systems
a)
x( k  1)  x 2 ( k  1)  2,
b)
x ( k  1)  sin x ( k ),
c)
x ( k  1) 
d)
x ( k  1)  3 x 2 ( k ) ,
e)
2.
1
,
x( k )
x ( k  1)  x 2 ( k )  y 2 ( k ),
y ( k  1)  x ( k )  y ( k )  1.
Do numerical experiments near each of the fixed points you found in the
previous problem to determine their stability.
3.
Explain why it is impossible for a linear system to have exactly two fixed points.
6.4
GEOMETRIC APPROACH TO STABILITY – COBWEB DIAGRAMS
Consider the dynamical system
x ( k  1)  f ( x ( k )), x (0)  x0 .
Let the function f be depected in the figure.
Draw a vertical line from the point lying on the horizontal axis until it intersects the
graph of f ; that height is the output x(1). At this stage we could return to the
horizontal axis and repeat the procedure to get x(2) from x(1), but it is more
convenient simply to trace a horizontal line till it intersects the diagonal line
x ( k  1)  x ( k ), and then move vertically to the curve again. Repeat the process n
times to generate the first n points in the orbit.
Cobwebs are useful because they allow us to see global behaviour at a glance, thereby
supplementing the local information available from the linearisation.
Cobwebs become even more valuable when linear analysis fails.
Exercises 4
1.
Consider the map
x ( k  1)  sin x ( k ).
Use a cobweb to show that x *  0 is stable, in fact, globally stable.
2.
Consider the dynamical system
x( k  1)  3x( k )  x 3 ( k ).
a)
Find all the fixed points and classify their stability.
b)
Draw a cobweb starting at x0  1.9.
c)
Draw a cobweb starting at x0  2.1.
d)
Try to explain the dramatic difference between the orbits found in parts (b) and
(c). Can you prove that the orbit found in (b) will remain bounded for all n ?
Can you prove that x( k )   in the case of the problem (c)?
6.5
ANALYTICAL APPROACH TO STABILITY - LINEARISATION
Consider a discrete time dynamical system
x ( k  1)  f ( x ( k ))
with an equilibrium value a.
Theorem: The equilibrium value a is stable or attracting if
f '(a )  1
and is unstable or repelling if
f '(a)  1.
If f '(a )  1, our work is in-conclusive.
The derivative of a curve at a point is, in some sense, the best linear approximation of
the curve at that point.

Worked Example 2
Consider the dynamical system
x( k  1)  3.2 x( k )  0.8x 2 ( k ).
After making substitutions x ( k  1)  y and x ( k )  x we consider a curve
y  3.2 x  0.8 x 2 .
The equilibrium values which are the solutions of the equation a  3.2a  0.8a 2 are
a  0 and a  2.75. Evidently,
f ' ( x )  3.2  16
. x.
. , if follows that the equilibrium value 2.75 is unstable (since
Since f ' (2.75)  12
12
.  1 ).
Likewise a  0 is unstable since f ' (0)  3.2  1. If you were to construct a cobweb
starting with a small positive initial value x(0) (say x(0)  01
. ), you would find that
the solution x ( k ) would be repelled from a  0.
Exercises 5
1.
The dynamical system
x( k  1)  3x( k )  x 2 ( k )  3
has two equilibrium values a  1 and a  3.
Show that
a)
The first value a  1 is unstable,
b)
The second value a  3 is unstable.
2.
The dynamical system
x( k  1)  14
. x( k )  0.2 x 2 ( k )  3
has the two equilibrium values a  3 and a  5. Investigate their stability.
3.
The dynamical system
x( k  1)  x 3 ( k )  x 2 ( k )  1
has two equilibrium values a  1 and a  1.
a)
Show that a  1 is unstable.
b)
Show that no conclusion can be made for the fixed point a  1 resorting to the
theorem given above.
c)
Draw a cobweb using the curve
f ( x)  x 3  x 2  1
and show that a  1 is semistable.
6.6
PERIODIC POINTS AND FEIGENBAUM’S CONSTANT
Around 1975, Feigenbaum began to study period-doubling in the logistic map. First
he developed a complicated generating function theory to predict rn, the value of r
where a 2 n -cycle first appears. To check his theory numerically he programmed his
hand-held computer to calculate the first several rn . He noticed a simple rule: the rn
converged geometrically, with the distance between successive transitions shrinking a
constant factor of about 4.669.
Perhaps a month later he decided to compute the rn in the trancendental case
numerically. Again, it became apparent that rn converged geometrically, the
convergence rate was the same 4.669.
In fact, the same convergence rate appears no matter what unimodal map is iterated.
In this sense, the number
rn  rn 1
 4.669...
n  r
n 1  rn
  lim
is universal. It is a new mathematical constant, as basic to periodic-doubling as  is
to circles.
Exercises 6
1.
Consider the map
x (n  1)  r sin  x (n)
where 0  r  1 and x [0,1]. At each of 200 equally spaced r values, plot x700
through x1000 vertically above r starting from some random initial condition x0 .
2.
In the previous problem go to finer resolution near the period-doubling
bifurcations and estimate rn for n  1,2,3,4,5,6. Try to achieve five significant
figures of accuracy. Use these numbers to estimate the Feigenbaum ratio
rn  rn 1
.
rn 1  rn
6.7
LYAPUNOV EXPONENTS
To be called chaotic a system should also show sensitive dependence on initial
conditions in the sense that neighboring orbits separate exponentially fast, on average.
This sensitive dependence could be quantified by defining the Lyapunov exponent for
a chaotic differential equation.
Let us consider two neighboring trajectories. Assume that nearby to the initial point
x0 lies a point x(0)   0 where the initial separation is extremely small. Let  n be the
separation after n iterations. If
 n   0 e n
then  is called the Lyapunov exponent. A positive Lyapunov exponent is a signature
of chaos.
A more precise and computationally useful formula for  can be derived. By taking
logarithms and noting that
b
g b g
 n  f n x (0)   0  f n x (0)
we obtain
1
n
  ln
b
g b g
 (n) 1 f n x (0)   0  f n x (0)
 ln
.
0
n
0
Therefore,
1
n
  ln ( f n )' x (0) ,
where we have taken the limit  0  0 in the last step.
The term inside the logarithm can be expanded by the chain rule derived earlier:
b g
n 1
bg
( f n )' x (0)   f ' x (i ) .
i 0
Hence
1 n 1
1 n 1
  ln  f ' x (i )   ln f ' x (i ) .
n i 0
n i 0
b g
b g
If this expression has a limit as n  , we define that limit to be the Lyapunov
exponent for the orbit starting at x0 :
R1
U
  limS ln f ' b
x (i )g
.
V
Tn
W
n 1
n
i 0
Note that  depends on the initial state x0 . However, it is the same for all x0 in the
basin of attraction of a given attractor.
For stable fixed points and cycle,  is negative; for chaotic attractors  is positive.
Exercises 7
1.
Calculate the Lyapunov exponent for linear map
x(n  1)  rx(n).
2.
Calculate the Lyapunov exponent for the decimal shift map
x(n  1)  10 x(n).
3.
Using a computer compute and plot the Lyapunov exponent as a function of r
for the sine map
x (n  1)  r sin  x (n)
for 0  x(n)  1 and 0  r  1.
4.
Plot the orbit diagram for the tent map, defined by
f ( x) 
for 0  r  2 and 0  x  1.
rx , 0  x  1 / 2
R
S
Tr  rx, 1 / 2  x  1
6.8
BIFURCATION DIAGRAMS AND CHAOS
b g
Let us consider a dynamical system x( k  1)  f x( k ) . We assume that we have a
family of functions f a where a is a parameter. In particular, we can think of f a ( x )
as a function of two numbers: a and x.
A bifurcation is a sudden change in the number or nature of the fixed and periodic
points of the system. Fixed points may appear or disappear, change their stability or
even break apart into periodic points.
Bifurcation diagrams
For a dynamical system involving a parameter b, find all fixed points a as functions
of b. Plot these functions on the ba axis. Find ranges of b for which each of these
fixed points is attracting and draw vertical arrows towards them. In those same
ranges, draw arrows away from repelling fixed points, and appropriate arrows for
semistable fixed points. Also draw arrows either up or down for values of b for which
these are no fixed points.

Worked Example 3
Consider the dynamical system
x( k  1)  (1  b) x( k )  x 3 ( k )
where b is some fixed constant. The fixed points are solutions to the equation
a  (1  b)a  a 3 .
Notice that, for all b values, a  0 is a fixed point, while a   b are fixed points
only when b  0. Thus, if b  0 we have one fixed point, while if b  0 there are
three fixed points.
Since
f ( x)  (1  b) x  x 3
one has that
f '( x)  1  b  3x 2 ; f ''( x)  6x; f '''( x)  6.
Thus a  0 is attracting when 2  b  0. Likewise a  0 is repelling when b  2
and b  0.
To find the stability of a   b we compute
f ' (  b )  1  b  3b  1  2b.
It follows that both fixed points a   b are attracting when 0  b  1.
The bifurcation diagram is presented in the figure.
The critical value is called in this case a pitchfork bifurcation. The other important
types of bifurcations are transcritical bifurcations (the fixed points lying on two
intersecting curves) and saddle node bifurcations. In the latter case at the bifurcation
value b0 the fixed points form a U-shaped curve.
Chaos
Sometimes it happens that all the fixed points and cycles are repelling.
Suppose that a dynamical system
(i)
is transitive on its attractor S,
(ii)
has sensitive dependence of initial values,
(iii) has repelling cycles that are close to the attractor S. Then this dynamical system
exhibits chaos.
Intuitively, a dynamical system exhibits chaos if in one sense there is unpredictability
(sensitive dependence on initial values says we can not make precise predictions), but
in another sense there is predictability (transitivity says we will be at a point, we just
don't know when).
A set of points S is called an attracting set or simply attractor for a dynamical system
b g
x( k  1)  f x( k ) if there is a number  such that if a0  s   for some point
s  S then
lim x ( k )  S  0.
k 
For instance, when r  2.5 the dynamical system
x( k  1)  2.5x( k ) 1  x( k )
l q
has a  0.6 as an attracting fixed point. Thus the set S  0,6 is an attractor.
The dynamical system is said to be transitive if, when a0 is close to some point in an
attractor, then for every point s in the attractor there is a subsequence x ( k j ) of x ( k )
values that converge to s, that is
lim x ( k j )  s.
j 
For instance, the dynamical system
x( k  1)  125
. x( k )  x 3 ( k )
having the attractor S  {0.5,0.5} is not transitive.
. x ( k )[1  x ( k )] fulfills the requirements (i)-(iii)
The dynamical system x ( k  1)  358
and thus exhibits chaos.
Exercises 8
1.
Consider the equation
x( k  1)  5x( k )  x 2 ( k )  b
a)
Find fixed points
b)
Do the x ( k ) values go to positive or negative infinity?
c)
For what values of b are the two fixed points attracting? Semistable?
d)
Plot the graph of the parabola given by the two fixed points on the ba axis.
e)
Draw arrows in the appropriate directions to complete the bifurcation diagram.
This diagram should be drawn for 3  b  5. Note that you have a saddle node
bifurcation.
2.
Consider the dynamical system
x( k  1)  bx( k )  x 3 ( k )
a)
Find the three equilibrium values.
b)
Plot the graph of the equilibrium values, letting the b axis go from -3 to 1.
c)
For what values of b is the constant fixed point attracting, semistable, repelling?
d)
Show that the fixed points on the parabola are repelling for all values of b for
which they exist.
e)
Draw arrows in the appropriate directions on the graph. You should have a
pitchfork bifurcation.
3.
Consider the dynamical system
x ( k  1)  x ( k ) 
1
1  0.5x ( k ) x ( k )  bx ( k )
2
which corresponds to a fixed proportion harvesting policy for a population in which
the unrestricted growth rate is r  05
. and carrying capacity is L  2.
a)
Find the fixed points for this equation and determine for what values of b they
are attracting.
b)
.  b  2.5 and draw
Plot the graph of the fixed points on the ba axis with 15
arrows in the appropriate directions. Note that this is a transcritical bifurcation.
4.
Consider the dynamical system
x( k  1)  x 2 ( k )  (b  1) x( k )  2b2 .
a)
Find the fixed points for this equation and determine for what values of b they
are attracting.
b)
Plot the graph of the fixed points on the ba axis with 2 / 3  b  2 / 3. Draw
arrows in appropriate directions.
6.9 THE HENON MAPPING
Henon considered the pair of difference equations with
f ( x, y)  1  y  ax 2 , g( x, y)  bx
for given parameters a and b.
This is not a horseshop map, but it is somewhat similar geometrically, and it allows us
to follow the details closely. It is easy to show that there are two fixed points
X  ( X , Y ), if a  a0 where
1
a0   (1  b) 2
4
and
X  b  1  (1  b) 2  4a 
b
1
, Y  bX .
2a
g
There is a turning point a0 , (1  b) / 2a0 in the bifurcation diagram in the (a , x )
plane for fixed b  1.
Henon and Pomeau interpreted the map geometrically as the product of successively a
folding, a contraction (when 1  b  1 ) and a reflection in the line y  x , and
showed that the map is a canonical form to which any quadratic map with a constant
Jacobian may be reduced by a linear transformation.
The composite transformation yields the Henon mapping.
Properties of the Henon Map
1. The Henon map is invertible. This property is the counterpart of the fact that in the
Lorenz system, there is a unique trajectory through each point in phase space. In
this respect the Henon map is superior to the logistic map, its one-dimensional
analog. The logistic map stretches and folds the unit interval, but it is not
invertible since all points (except the maximum) come from two pre-images.
2. The Henon map is dissipative. It contracts areas, and does so at the same rate
everywhere in phase space. This property is the analog of constant negative
divergence in the Lorenz system.
3. For certain parameter values, the Henon map has a trapping region.
In other words, there is a region R that gets mapped inside itself.
4. Some trajectories of the Henon map escape to infinity.
In contrast, all trajectories of the Lorenz system are bounded, they all eventually
enter and stay inside a certain large ellipsoid.
Exercises 8
1.
Find all the fixed points of the Henon map and show that they exist only if
a  a0 , where a0 is to be determined.
2.
Calculate the Jacobian matrix of the Henon map and find its eigenvalues.
3.
A fixed point of a map is linearly stable if and only if all eigenvalues of the
Jacobian matrix satisfy the condition   1. Determine the stability of the fixed
points of the Henon map, as a function of a and b. Show that one fixed point is
always unstable, while the other is stable for a slightly larger than a 0 . Show
that this fixed point loses stability in a flip bifurcation (   1) at
a1  3 / 4 (1  b) 2 .
4.
Explore numerically what happens in the Henon map for other values of a, still
keeping b  0.3.
a)
Show that period-doubling can occur, leading to the onset of chaos at a  106
. .
b)
Describe the attractor for a  13
..