——————————————————————————————————————SCHOOL OF MATHEMATICS AND STATISTICS

MT4508: Dynamical SystemsTutorial Sheet 3 – Solutions: Lyapunov exponents, chaotic orbits and fractals

——————————————– CVT – 2007/2008 ———————————————

1. Consider the map f(r, θ) = (r2, θ − sin θ) on the plane, where r ≥ 0 and 0 ≤ θ < 2π.(i) Calculate the fixed points and determine their stability.(ii) For the sink found in (i) determine the basin of attraction. Also, determine the basin ofinfinity.(iii) For the saddle point found in (i) determine the stable and unstable manifolds.

Solution

Fixed points of the map are solutions of

r2 = r, θ − sin θ = θ.

The first equation yields r = 0, 1 and the second equation yields θ = 0, π. By combiningthese, we obtain the three fixed points (0, 0), (1, 0) and (1, π). It can be seen that (0,0)is a sink attracting the interior of the unit circle r = 1. This circle and the horizontal axisθ = 0, π are invariant. The former is the boundary between the basin of the sink (0,0)and the basin of infinity (which is the exterior of the unit circle). In the upper half planesin θ > 0, so orbits in this region tend to the right because θ − sin θ < θ. In the lower halfplane sin θ < 0, so orbits in this region also tend to the right because θ − sin θ > θ. Thefixed point (1, π) is a source while the fixed point (1, 0) is a saddle. The horizontal half lineθ = 0, excluding the origin, is the unstable manifold of (1, 0). The unit circle, excluding thefixed point (1, π) is the stable manifold of (1, 0) (see figure 1).2. Let f be a map on R. From the definition of Lyapunov exponent show that(i) The Lyapunov exponent of a period-k orbit {p1, p2, · · · , pk} is

h =1

k

k∑

i=1

ln |f ′(pi)|.

(ii) If the Lyapunov exponent of the orbit of x0 is h, then the Lyapunov exponent of theorbit of x0 under the map fk is kh, where k is a positive interger.

Solution

(i) By definition, the Lyapunov number L(p1) (if it exists) is

L(p1) = limn→∞

|n∏

i=1

f ′(pi)|1/n.

If we choose n to be an integer multiple of k, say n = mk, then each derivative f ′(pi), for1 ≤ i ≤ k, in the product appears exactly m times. Hence we can write

L(p1) = limm→∞

|k∏

i=1

[f ′(pi)]m|1/mk = lim

m→∞

|k∏

i=1

[f ′(pi)]|1/k = |k∏

i=1

[f ′(pi)]|1/k.

1

(1, 0)(1, π)ss s- �� -

� R

R �

�

R

R

�

FIGURE 1. A phase space diagram of the map f(r, θ) = (r2, θ − sin θ).

It follows that

h(p1) =1

k

k∑

i=1

ln |f ′(pi)|.

(ii) We denote by {x0, x1, x2, x3, · · ·} the orbit of x0 under f . Then the orbit of x0 under fk

is {x0, xk, x2k, x3k, · · ·}. The Lyapunov number Lk of the latter orbit is

Lk = limn→∞

|n∏

i=0

(fk)′(xik)|1/n.

Each derivative in the product is evaluated by the chain rule to yield

(fk)′(xik) = (ffk−1(xik))′ = f ′(fk−1(xik))(f

k−1(xik))′

= f ′(xik+k−1)(fk−1(xik))

′ = · · · =k−1∏

j=0

f ′(xik+j).

Putting these derivatives together in the above expression of Lk we obtain

Lk = limn→∞

|kn∏

i=0

f ′(xi)|1/n = limn→∞

(

|kn∏

i=0

f ′(xi)|1/kn

)k

= Lk,

where L is the Lyapunov number of the orbit of x0 under f . It follows that

hk = kh,

where hk and h are the Lyapunov exponents of the orbits of x0 by fk and f , respectively.3. The tent map T (x) on [0, 1] is defined by

T (x) =

{

2x if x ≤ 1/2,2(1 − x) if x > 1/2.

By carrying out the steps below, we can show that T (x) has chaotic orbits.(i) For every orbit {x1, x2, · · ·} with xi 6= 1/2 for all i, calculate the Lyapunov number L(x1)and exponent h(x1).

2

(ii) Show that T (x) cannot have any stable periodic orbits (including fixed points). Arguethat any asymptotically periodic orbit must be eventually periodic.(iii) In order to prove that T (x) has chaotic orbits, we need to show that not all of its orbitsare eventually periodic. This can be done as follows. We divide [0, 1] into 2 subintervals[0, 1/2) and (1/2, 1]. For each xi assign the symbol ` if xi ∈ [0, 1/2) and r if xi ∈ (1/2, 1].Each orbit then corresponds to a unique sequence (so-called itinerary) of the two symbols` and r. Graphically or otherwise convince yourselves that each itinerary of k symbolscorresponds to an interval of length 2−k. Explain why each infinite itinerary corresponds toan orbit of exactly one point in [0, 1]. Explain why each eventually periodic orbit must havean eventually repeating itinerary.(iv) Using all the above results show that T (x) has infinitely many chaotic orbits.

Solution

(i) We have |T ′(x)| = 2, for x 6= 1/2. The Lyapunov number L for a general orbit{x1, x2, x3, · · ·} with xi 6= 1/2 for all i is

L = limn→∞

|T ′(x1)T′(x1) · · · T ′(xn)|1/n = lim

n→∞

(2n)1/n = 2.

The Lyapunov exponent is ln 2.(ii) Since the Lyapunov exponent is ln 2 > 0 for every orbit that does not contain 1/2,

all periodic orbits (including fixed points) are unstable, repelling nearby orbits. This meansthat there exist no asymptotically periodic orbits. However, there exist eventually periodicorbits. In fact, there are infinitely many of them. To see this, let us consider the orbits ofx0 = 1/(2n + 1). We have

T (x0) =2

2n + 1= x1,

T 2(x0) =22

2n + 1= x2,

T n(x0) =2n

2n + 1= xn,

T n+1(x0) = 2(1 − 2n

2n + 1) =

2

2n + 1= x1.

This means that {x1, x2, · · · , xn} is a period-n orbit. So we indeed have infinitely manyperiodic orbits and infinitely many asymptotically periodic orbits landing precisely on them.

(iii) An orbit is chaotic if it has positive Lyapunov exponent and is not asymptoticallyperiodic (and of course not eventually periodic). The first requirement is satisfied, so theproof of existence of infinitely many chaotic orbits reduces to showing that there are infinitelymany orbits that are not eventually periodic. To show this we resort to symbolic dynamicsof the tent map.

Let ` = [0, 1/2] and r = [1/2, 1]. We label a subinterval by ` if it is contained in ` andby r if it is contained in r. We now divide [0, 1] into 2n closed subintervals, denoted by Ii,of equal width 2−n. We consider successive iterations of T on these subintervals. We labelT k(Ii), for k = 0, 1, 2, · · · , n and i = 1, 2, · · · , 2n, by ` if T k(Ii) ⊆ ` and by r if T k(Ii) ⊆ r.(Convince yourselves that no T k(Ii) straddles the point 1/2, i.e. T k(Ii) is either entirely in` or entirely in r.) The association of T k(Ii) with ` or r enables us to symbolically representthe orbit of Ii (more precisely the orbit of each point in Ii), after n iterations of T , by a

3

sequence of n symbols X1X2X3 · · · (so-called itinerary), where Xi is either ` or r. This is abookkeeping device that allows some information of an orbit to be coded in terms of discretesymbols. The figure below shows the first 4 cases n = 0, 1, 2, 3.

T 3 n = 4++++ ++++ ++ ++ ++ +````

```r

``rr

``r`

`rr`

`rrr

`r`r

`r``

rr``

rr`r

rrrr

rrr`

r`r`

r`rr

r``r

r```

T 2 n = 3++ ++ + + +``` ``r `rr `r` rr` rrr r`r r``

T 1 n = 2++ +`` `r rr r`

T 0 n = 1+` r

It is noted that at each and every step n, no two itineraries are the same. The 2n

subintervals have 2n different itineraries. Now consider a sequence of n letters to be filledwith `, r. There are 2n possible ways that the sequence can be filled. This is because there are2 choices for each letter in the sequence, so that there are 2n choices in total, correspondingto 2n different sequences. This means that the itineraries of the 2n subintervals exhaust allof these possibilities.

(iv) In the limit n → ∞, each Ii reduces to a single point and the corresponding infiniteitinerary represents the orbit of the limiting point. (The limit set is no empty and containsonly one point. This is the result of a nested intersection of the closed subintervals of decreas-ing width.) Now an eventually periodic orbit must have a repeating itinerary. But amongall possible arrangements of ` and r to form an infinite sequence, there exist infinitely manyways to make a non-repeating sequence. We therefore conclude that there exist infinitelymany chaotic orbits.4. Consider the tent map T (x) on R defined by

T (x) =

{

3x if x ≤ 1/2,3(1 − x) if x > 1/2.

(i) Check that T (1/3, 2/3) [the image of the interval (1/3, 2/3)], T 2(1/9, 2/9) and T 2(7/9, 8/9),T 3(1/27, 2/27) and T 2(25/27, 26/27), etc., have no points in [0, 1]. Hence deduce that if x isin the middle-third Cantor set then T n(x) remains in the Cantor set for all n.(ii) By considering the orbits of 1/(3n + 1), n = 1, 2, 3, · · ·, show that there exist periodicorbits of all periods.(iii) What about the orbits of 1/(3n + 2), n = 1, 2, 3, · · ·.(iv) Can you find a period-three orbit different from what you obtain in (ii)?

Solution

(i) It is straightforward to verify that

T (1/3, 2/3) = T 2(1/9, 2/9) = T 2(7/9, 8/9) = T 3(1/27, 2/27) = T 3(25/27, 26/27) = · · ·

has no points in common with the unit interval [0, 1]. This means that points not in themiddle-third Cantor set eventually map out of [0, 1]. The above equation also means thatpoints in the middle-third Cantor set forever remain in the middle-third Cantor set.(ii) Let us denote by x0 = 1/(3n + 1) the point under consideration. For n > 1, we have

T (x0) =3

3n + 1= x1,

4

T 2(x0) =32

3n + 1= x2,

T n(x0) =3n

3n + 1= xn,

T n+1(x0) = 3

(

1 − 3n

3n + 1

)

=3

3n + 1= x1.

This means that {x1, x2, · · · , xn} is a period-n orbit.(iii) For x0 = 1/(3n + 2) and n > 1, we have

T (x0) =3

3n + 2= x1,

T 2(x0) =32

3n + 2= x2,

T n(x0) =3n

3n + 2= xn,

T n+1(x0) = 3

(

1 − 3n

3n + 2

)

= 23

3n + 2,

T 2n−1(x0) = 23n−1

3n + 2=

1

2

(3 + 1)3n−1

3n + 2=

1

2

3n + 3n−1

3n + 2>

1

2,

T 2n(x0) = 3

(

1 − 23n−1

3n + 2

)

= 33n−1 + 2

3n + 2=

3n + 6

3n + 2> 1.

Hence, the orbit of x0 = 1/(3n + 2) escapes the unit interval after 2n iterates of the map.We can see that T 2n+1(x0) < 0, and the orbit eventually diverges to −∞.(iv) From (ii) we have a period-three orbit: {3/28, 9/28, 27/28}. We denote by {p, T (p), T 2(p)}a possible period-three orbit and try to solve for p with p < 1/2, T (p) > 1/2 and T 2(p) > 1/2.This is just an educated guess, avoiding possible rediscovery of the other orbit. With thischoice we have

T (p) = 3p,

T 2(p) = 3(1 − 3p),

T 3(p) = 3(1 − 3(1 − 3p)) = 3(9p − 2) = p.

Solving the last equation for p yields p = 3/13. Hence the period-three orbit is {3/13, 9/13, 12/13}.5. The cat map is defined by

f(x, y) =

(

2 11 1

) (

xy

)

(mod 1) .

Calculate the Lyapunov number and exponent for a general orbit. What can you concludeabout the possibility of chaotic orbits?

Solution

Since the map is linear, the Jacobian matrix is the linear matrix

A =

(

2 11 1

)

.

5

It follows that

Dfn = An =

(

2 11 1

)n

.

Since A is symmetric, i.e, A = AT , we have An = (An)T . It follows that

Dfn(Dfn)T = A2n =

(

2 11 1

)2n

.

Now if λ is an eigenvalue of A corresponding to an eigenvector e, then

A2ne = A2n−1Ae = λA2n−1e = · · · = λ2ne.

That means that e is an eigenvector of A2n with an eigenvalue λ2n. So the problem ofdetermining the eigenvalues of A2n reduces to that of A. By straightforward calculationswe obtain the eigenvalues of A: (3 ±

√5)/2. Hence the eigenvalues of A2n are λ± = (3 ±√

5)2n/22n. The Lyapunov numbers and exponents are

L1 = limn→∞

(λ1/2

+ )1/n =3 +

√5

2⇒ h1 = ln

3 +√

5

2> 0,

L2 = limn→∞

(λ1/2

− )1/n =3 −

√5

2⇒ h1 = ln

3 −√

5

2< 0.

There is one positive Lyapunov exponent (and no zero exponent), so orbits that are notasymptotically periodic are chaotic.6. The skinny baker map on the unit square [0, 1] × [0, 1] is defined by

B(x, y) =

(

1/3 00 2

) (

xy

)

if 0 ≤ y ≤ 1/2,(

1/3 00 2

) (

xy

)

+

(

2/3−1

)

if 1/2 < y ≤ 1.

(i) Sketch the image of the unit square. What images do you expect after two and moreiterates?(ii) Calculate the Lyapunov number and exponent of an orbit which never maps onto theline y = 1/2(iii) What can you conclude about the possibility of chaotic orbits?

Solution

The map B is basically a combination of two simple linear operations: a shrinking-stretching map and a translation map. First the stretching of the vertical dimension by afactor of 2 and the shrinking of the horizontal dimension by a factor of 1/3 are applied. Afterthat points that are mapped outside of the unit square are translated by the vector (2/3,−1)back to the square. This observation helps to construct the images of the unit square bysuccessive iterates of B more easily. The calculation below is standard for constructing theimage of an extended object. We basically identify the images of special points on the objectand extrapolate the results in the regions of continuity of the map.

6

(i) The line y = 1/2 divides the unit square into two subsets. The corner points of the lowerhalf are mapped as follows.

B

(

00

)

=

(

1/3 00 2

)(

00

)

=

(

00

)

,

B

(

01/2

)

=

(

1/3 00 2

)(

01/2

)

=

(

01

)

,

B

(

11/2

)

=

(

1/3 00 2

)(

11/2

)

=

(

1/31

)

,

B

(

10

)

=

(

1/3 00 2

)(

10

)

=

(

1/30

)

.

The two upper corners of the upper half and points near the line y = 1/2 are mapped asfollows.

B

(

01

)

=

(

1/3 00 2

)(

01

)

+

(

2/3−1

)

=

(

2/31

)

,

B

(

11

)

=

(

1/3 00 2

)(

11

)

+

(

2/3−1

)

=

(

11

)

,

B

(

x1/2 + ε

)

=

(

1/3 00 2

)(

x1/2 + ε

)

+

(

2/3−1

)

=

(

x/31 + 2ε

)

+

(

2/3−1

)

→(

(x + 2)/30

)

as ε → 0.

These results allow us to construct the following figure, depicting the image of the unitsquare by B.

**

*

++

+

-

1/3 2/3

*

*

*

+

+

+

It is seen that the middle third of the unit square is removed! The next application of Bremoves the middle third of each of the two strips. The image after B2 is depicted by thefigure below.

1/3 2/3

**

*

*

*

*

+

++

+

+

+

-

1/9

2/9

7/9

8/9*

*

*

+

+

+

*

*

*

+

+

+

7

The image of the unit square after Bn consists of 2n vertical strips each of width 1/3n.The set of points that remain in the limit n → ∞ is the middle-third Cantor set of verticalline segments of unit length.(ii) The Jacobians of B and of Bn are

DB =

(

1/3 00 2

)

, DBn =

(

1/3n 00 2n

)

.

The matrix

DBn(DBn)T =

(

1/32n 00 22n

)

has two eigenvalues 22n and 1/32n. So the Lyapunov numbers and exponents are, respectively,given by L1 = 2, L2 = 1/3 and h1 = ln 2 > 0, h2 = − ln 3 < 0.(iii) As h1 > 0 > h2 (a positive exponent and no zero exponents), all orbits that are notasymptotically periodic are chaotic.7. Calculate the box-counting dimension of the following sets.(i) The middle-fifth Cantor set.(ii) The Sierpinski gasket, constructed by removing from an equilateral triangle of size ` theequilateral triangle of size `/2 in the middle, leaving 3 equilateral triangles at the vertices,each having size `/2. This process is then repeated infinitely many times with the remainingtriangles.(iii) The Sierpinski carpet, constructed by removing from a square of size ` the central cross,leaving 4 squares at the corners, each having size `/3. This process is then repeated infinitelymany times, with the remaining squares.

Solution

(i) Let Cn denote the set obtained at step n of the construction of the Cantor set. C1 consistsof 2 intervals each of length 2/5. C2 consists of 4 intervals each of length (2/5)2. Cn consistsof 2n intervals each of length (2/5)n. We can therefore cover Cn by N(ε) = 2n intervals eachof length ε = (2/5)n. Hence the box-counting dimension dB of the Cantor set C∞ is givenby

dB = limε→0

ln N(ε)

ln(1/ε)= lim

n→∞

ln 2n

ln(5/2)n=

ln 2

ln(5/2).

(ii) At step 1 of the construction of the gasket, we have 3 triangles each of size `/2. At stepn, the resulting set consists of 3n triangles each of size `/2n. We can cover this set by 3n

squares each of size `/2n. Hence the box-counting dimension dB of the gasket is given by

dB = limn→∞

ln 3n

ln(2n/`)= lim

n→∞

n ln 3

n ln 2 − ln `=

ln 3

ln 2.

(iii) At step 1 of the construction of the carpet, we have 4 squares each of size `/3. At stepn, the resulting set consists of 4n squares each of size `/3n. We can cover this set by 4n

squares each of size `/3n. Hence the box-counting dimension dB of the carpet is given by

dB = limn→∞

ln 4n

ln(3n/`)= lim

n→∞

n ln 4

n ln 3 − ln `=

ln 4

ln 3.

Tutorial Sheets and Solutions for MT4508 can be found athttp://www-vortex.mcs.st-and.ac.uk/~chuong/MT4508/index.html

8