Discrete Systems

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 2 x(2)  ax(1)  b  a x0  ab  b.

For x(k) one has

a k 1 x(k)  a k x  (a k 1  a k 2 ...a 1)b  a k x  b, 0 0 a 1 provided a  1 and

x(k)  x0  kb,

k b when a  1. If a  1 then a  0 and x(k)  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 , 2 x(2)  Ax(1)  b  A x0  Ab  b , 3 2 x(3)  Ax(2)  b  A x0  A b  Ab  b .

In the general case we have

k k 1 k 2 x(k)  A x0  ( A  A ... 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

k k 1 x(k)  A x0  (I  A )(I  A) b. Now, what happens as k   ? If the absolute values of the eigenvalues of A are less than 1 (hence I  A is invertible), then Ak tends to zero matrix. Hence x(k)  x~  (I  A)1b.

Alternatively, if some eigenvalues have absolute value bigger than 1, then Ak 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

~ 1 x0  x  (I  A) b 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 ~ ~ x0  b  x0 , i.e. x0  b / 2  x , then x(k) is stuck at the fixed point x. 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; 3 c) a  , b  1, c  0; 2 d) a  1, b  1, c  4; 1 3 e) a   , b  1, c  . 2 2

2. For each of the discrete time systems in the previous problem, determine

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 a) x(k  3)  , x(k 1) b) x(k)  x(k 1) 1, c) x(k)  4x(k)  2, d) x(k)  2x(k 1), x(k 1) e) x(k  2)  , x(k) f) x(n  5)  x(n 10)  5, g) x(n  2)  x(n 1)  n2  3. 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

I x(k 1)  GF1 JIx(k)  b H mK 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)  1.1x(k) 100 and mb 100   1000. I 0.1 So x(k)  (1.1)k (1100) 1000.

In particular, x(40)  48785.17 (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

F I Ik F mbI mb x(k)  G1 J Ga0  J . H mKH I K I

When b  0, then F I Ik x(k)  G1 J a0. H mK

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)  1.2P(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)  1.2 P(n 1)  20 and S(n 1)  0.8 P(n) substitution gives 1.2 P(n 1)  20  0.8 P(n). Thus 2 50 P(n 1)   P(n)  , 3 3 or 2 k P(k)  GF JI c 10. H3K

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 s k P(k)  cGF JI  p, Hd K where p is the equilibrium value of this dynamical system

b  a p  . 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)  0.8 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)  1.2 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) x2 (k)  x (k) L1 O L1 2 O. Mx (k 1)PMx (k)  x (k)  2P N2 QN1 2 Q

This may be rewritten as x(k 1)  f (x(k)), where LuOLu2  v O f MP M P. NvQNu  v  2Q

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  , lim x(k)  a. then 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)  x2 (k 1)  2, b) x(k 1)  sin x(k), 1 c) x(k 1)  , x(k) d) x(k 1)  3 x2 (k),

x(k 1)  x2 (k)  y2 (k), e) y(k 1)  x(k)  y(k) 1.

2. 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)  x3 (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.

x(k 1)  3.2x(k)  0.8x2 (k).

After making substitutions x(k 1)  y and x(k)  x we consider a curve

y  3.2x  0.8x2 .

The equilibrium values which are the solutions of the equation a  3.2a  0.8a2 are a  0 and a  2.75. Evidently, f '(x)  3.2 1.6x.

Since f '(2.75)  1.2, if follows that the equilibrium value 2.75 is unstable (since

1.2  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)  0.1), 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)  x2 (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.

x(k 1)  1.4x(k)  0.2 x2 (k)  3

has the two equilibrium values a  3 and a  5. Investigate their stability.

x(k 1)  x3 (k)  x2 (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)  x3  x2 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 2n -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

r  r   lim n n1  4.669... n rn1  rn 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

r  r n n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  n   0 e 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

n n  n  f bx(0)   0 g f bx(0)g we obtain

1 (n) 1 f n x(0)    f n x(0)   ln  ln b 0 g b g. n  0 n  0 Therefore, 1   ln ( f n )' x(0) , n 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:

n1 ( f n )'bx(0)g  f 'bx(i)g. i0 Hence 1 n1 1 n1   ln  f 'bx(i)g  ln f 'bx(i)g. n i0 n i0 If this expression has a limit as n   , we define that limit to be the Lyapunov exponent for the orbit starting at x0:

R1 n1 U   lim ln f 'bx(i)g. nS V Tn i0 W

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

Rrx, 0  x  1/ 2 f (x)  S Tr  rx, 1/ 2  x  1

for 0  r  2 and 0  x  1. 6.8 BIFURCATION DIAGRAMS AND CHAOS

Let us consider a dynamical system x(k 1)  f bx(k)g. We assume that we have a family of functions fa where a is a parameter. In particular, we can think of fa (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.

x(k 1)  (1 b)x(k)  x3 (k) where b is some fixed constant. The fixed points are solutions to the equation

a  (1 b)a  a3.

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  x3 one has that

f '(x)  1 b  3x2 ; 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 x(k 1)  f bx(k)g 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) has a  0.6 as an attracting fixed point. Thus the set S  l0,6q 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)  1.25x(k)  x3 (k) having the attractor S  {0.5,0.5} is not transitive.

The dynamical system x(k 1)  3.58 x(k)[1 x(k)] fulfills the requirements (i)-(iii) and thus exhibits chaos.

Exercises 8

1. Consider the equation

x(k 1)  5x(k)  x2 (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.

x(k 1)  bx(k)  x3 (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.

1 x(k 1)  x(k)  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  0.5 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) Plot the graph of the fixed points on the ba axis with 1.5  b  2.5 and draw arrows in the appropriate directions. Note that this is a transcritical bifurcation.

x(k 1)  x2 (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  ax2 , 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 a   (1 b)2 0 4 and 1 X  b 1 (1 b)2  4a  , Y  bX. 2a

(a, x) There is a turning point ba0 ,(1 b) / 2a0 g in the bifurcation diagram in the 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 a0. Show that this fixed point loses stability in a flip bifurcation (  1) at

2 a1  3/ 4(1 b) .

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  1.06. b) Describe the attractor for a  1.3.