Chapter 9 diff eq sm

825

9.1 Iterative Equations

First-Order Iterative Equations

For Problems 1-12 we use the fact that the solution of 1n ny ay b+ = + is 0 1 .1n

nn

ay a y ba

= +

1. 112

3n ny y+ = + 0 1y = (b)

00

6

n10

yn

(a),(c) Because 012

, 3, 1,a b y= = = 12

12

11 32

15 62

nn

n

n

y = +

= +

As n increases, the orbit or solution approaches 6 from below. Thus 6 .n ey y =

1 012

3, 1n ny y y+ = + =

2. 112

3n ny y+ = + 0 1y = (b)

n10

yn

1

7

(a),(c) Because 012

, 3, 1,a b y= = = 12

12

11 32

17 62

nn

n

n

y = +

= +

As n increases, the orbit or solution approaches 6 from below. Thus 6 .n ey y =

1 012

3, 1n ny y y+ = + =

CHAPTER 9CHAPTER Discrete Dynamical Systems

826 CHAPTER 9 Discrete Dynamical Systems 3. 1

12

3n ny y+ = + 0 1y = (b)

00

3

n10

y n

(a),(c) Because 01 , 3, 1,2

a b y= = =

( )1232

11 32

1 22

nn

n

n

y = +

= +

As n increases, the orbit approaches 2, with successive ny alternatively above and below this value. Thus

2 .n ey y =

1 012

3, 1n ny y y+ = + =

4. 112

3n ny y+ = + 0 1y = (b)

2

4

n10

yn

(a),(c) Because 01 , 3, 1,2

a b y= = =

1 11 23 32

2

13 22

n

n

n

n

y

= + = +

As n increases, the orbit approaches 2, with successive ny alternatively above and below this value. Thus

2 .n ey y =

1 012

3, 1n ny y y+ = + =

5. 1 2 3n ny y+ = + 0 1y = (b)

00

5000

n10

yn

(a),(c) Because 02, 3, 1,a b y= = = 2 12 3

1

4 2 3

nn

n

n

y = +

=

As n increases, the solution grows without bound.

1 02 3, 1n ny y y+ = + =

SECTION 9.1 Iterative Equations 827

6. 1 2 3n ny y+ = + 0 1y = (b)

2500

0

y

n

n

100

(a),(c) Because 02, 3, 1,a b y= = = 2 12 3

1

2 3

nn

n

n

y = +

=


1 02 3, 1n ny y y+ = + =

7. 1 2 3n ny y+ = + 0 1y = (b)

00

1.25

n10

yn

(a),(c) Because 02, 3, 1,a b y= = = ( 2) 1( 2) 3

31 for all

nn

ny

n

= + =

The orbit starts at 0y and remains there, because this is a fixed point. It is, however, a repelling fixed point, so at any other value for 0y the solution is unbounded. See Problem 8. 1 02 3, 1n ny y y+ = + =

8. 1 2 3n ny y+ = + 0 1y = (b)

1500

n10

yn

2500

(a),(c) Because 02, 3, 1,a b y= = = ( 2) 1( 2) 3

3

2( 2) 1

nn

n

n

y = +

= +

As n increases, the orbit displays larger and larger oscillations, with successive ny alternately above and below zero.

1 02 3, 1n ny y y+ = + =

828 CHAPTER 9 Discrete Dynamical Systems

9. 1 3n ny y+ = + 0 1y = (b)

00

n10

yn35

(a),(c) Because 01, 3, 1,a b y= = = 0

1 3ny y nb

n= += +


1 02 3, 1n ny y y+ = + =

10. 1 3n ny y+ = + 0 1y = (b)

00

n10

yn35

(a),(c) Because 01, 3, 1,a b y= = = 0

1 3ny y nb

n= += +


1 03, 1n ny y y+ = + =

11. 1 3n ny y+ = + 0 1y = (b)

00

2.5

n10

yn

(a),(c) Because 01, 3, 1,a b y= = = ( 1) 1( 1) 3

21 3( 1)2 2

nn

n

n

y = +

= +

As n increases, the solution oscillates between 1 and 2 for all time. there is no steady state or equilibrium, but rather a cycle of period 2. 1 03, 1n ny y y+ = + =


12. 1 3n ny y+ = + 0 1y = (b)

5

n10

yn

2

(a),(c) Because 01, 3, 1,a b y= = = ( 1) 1( 1) 3

25 3( 1)2 2

nn

n

n

y = +

= +

As n increases, the solution oscillates between -1 and 4 for all time. There is no steady state or equilibrium, but rather a cycle of period 2. 1 03, 1n ny y y+ = + =

Fishings End

13. (a) 1 01.04 80,000, 1,000,000n ny y y+ = = has solution

( )(1.04) 11,000,000 (1.04) 80,0001.04 1

1,000,000 (1.04) 2,000,000 (1.04) 1 1,000,000 2 1.04

nn

n

n n n

y =

= =

The negative term will grow until it exceeds the positive term, so extinction will occur.

(b)

00

1,200,000

n20

yn

(b) We note from Figure 9.1.2 that the orbit curves slightly downward, so we expect extinction to occur before

n = 20 . Plotting the next 10 points of the orbit shows

that extinction occurs between n = 17 and n = 18 (see figure).

(c) From the solution in part (a), extinction will occur when

1.04 2n = , or when

ln 2 17.67ln1.04

n = . This is consistent with the graphical estimate

in part (b).

1 03, 1n ny y y+ = + =

830 CHAPTER 9 Discrete Dynamical Systems Lab Problem: Spreadsheet Predictions

In problems 14-21, when there is an equilibrium or steady state solution, its exact value ey is calculated by setting 1 .n ny y+ =

14. 1 0.3 1n ny y+ = 0 0.2y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

0.4

1.6

y

n

n

10

y yn n+ = 1 0 3 1. , y0 02= . (b) The solution decreases monotonically and approaches a steady state near -1.43. Calculation gives 0.7 ye = -1 , or ye = -1.428... .

(c) The coefficient of yn is small so we are not surprised to see there is little effect as n becomes large. The constant -1 contributes to the level of the steady state. The initial value 0.2 > ye causes the orbit to approach the equilibrium from above.

15. yn+1 = 0.3yn - 1 0 1.6y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

0

1.4

1.6

yn

n

10

y yn n+ = 1 0 3 1. , y0 16= . (b) The solution decreases monotonically and approaches a steady state near -1.43. Calculation gives 0.7 ye = -1 , or ye = -1.428... .

(c) The steady state is the same as in Problem 14; the change in initial conditions to

SECTION 9.1 Iterative Equations 831 16. yn+1 = - 0.3yn - 1 0 0.2y =

(a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

0.5

n10

yn

1.5

y yn n+ = 1 03 1. , y0 02= . (b) The solution exhibits a damped oscillation that quickly approaches a steady state near

- 0.7. Calculation gives 1.3 ye = -1 , or ye = - 0.769... .

(c) The small coefficient of yn predicts stability; the negative sign with this coefficient predicts oscillation about the steady state, which depends somewhat on the constant term.

17. yn+1 = 0.3yn + 0.5 0 0.2y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

0.8

n10

yn

00

y yn n+ = +1 0 3 0 5. . , y0 02= . (b) The solution decreases monotonically and approaches a steady state near 0.7. Calculation gives 0.7 ye = 0.5 , or ye = 0.714... .

(c) As in Problem 14 the coefficient of yn is small so we are not surprised to see little effect as n becomes large. The constant 0.5 contributes to the equilibrium level. The initial value y0 = 0.2 < ye causes the solution to approach ye from below.


18. yn+1 = 1.3yn - 1 0 0.2y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

10

n10

y n

50

y yn n+ = 1 13 1. , y0 02= . (b) The solution decreases monotonically and approaches infinity rather than a steady state.

There is an equilibrium for this equation, because calculation gives -0.3ye = - 1 , or ye = 3.33333... ; it is an unstable equilibrium that can only be reached by iterating backward with negative n .

(c) The coefficient of yn is multiplied at each step by 1.3, then -1 is subtracted. Hence once the solution becomes negative it becomes more and more negative and the orbit goes to minus infinity.

19. yn+1 = - 1.3yn - 1 0 0.2y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

4

n10

yn

4

y yn n+ = 1 13 1. , y0 0 2= . (b) The solution oscillates with larger and larger amplitude, not reaching a steady state

(unless iteration goes backward with negative n as in Problem 18). Equilibrium calculation gives ye = 1.3 ye 1, or ye = 0.4348... .

(c) The negative coefficient of yn causes the solution to oscillate on each iteration and for the amplitudes to get larger and larger in absolute value. After a while the constant of -1 that is subtracted on each iteration does not cause much effect.

SECTION 9.1 Iterative Equations 833 20. yn+1 = 1.3yn + 0.5 0 0.2y =

(a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

5

20

n10

yn

y yn n+ = +1 13 0 5. . , y0 0 2= . (b) The solution increases monotonically without bound. Calculation gives

1.3 ye + 0.5 = ye , or ye = 1.667... .

(c) The positive multiplier 1.3 causes yn to increase as soon as it becomes positive, and the positive constant 0.5 makes that happen on the first step.

21. yn+1 = - 1.3yn - 2 0 0.2y = (a) A simple spreadsheet calculation yields the following first 10 iterates and graph.

20

n10

yn

15

y yn n+ = 1 13 2. , y0 02= . (b) Exactly as in Problem 19 the solution oscillates with larger and larger amplitude, not

reaching a steady state (unless iteration goes backward with negative n). Equilibrium calculation 21.3e ey y

= gives ye = - 0.869... . (c) The only difference from Problem 19 is that the constant term is now 2 , so the

equilibrium has moved; the long-term behavior is essentially the same.

834 CHAPTER 9 Discrete Dynamical Systems Closed-Form Sums

22. 1 2 3nS n= + + + + (a) If Sn is the sum of the first n positive integers, this expression requires 0 0S = for the

given expression, which yields the following pattern:

1 1S =

2 1

3 2

4 3

1

1 2 21 2 3 31 2 3 4 4

1 2n n

S SS SS S

S n S n

= + = += + + = += + + + = +

= + + + = +" "

"

and so on. Hence, the iterative equation for Sn is

1 ( 1)n nS S n+ = + + , with 0 0S = . (b) To show that the closed form expression,

( 1)2n

n nS +=

satisfies the preceding iterative equation, we note that it gives 0 0S = , and also that

1( 1)( 2) ( 1) 2( 1) 1

2 2 2n nn n n n nS S n++ + + + = = = +

as predicted.

23. 2 31 2 2 2 2nnS = + + + + + (a) This expression requires that 0 1S = , and we observe the general pattern 1 01 2 2S S= + = +

2 22 1

2 3 33 2

21

1 2 2 2

1 2 2 2 2

1 2 2 2 2n nn n

S SS S

S S

= + + = += + + + = +

= + + + = +" "

"

Hence, the iterative equation for Sn is

11 2n

n nS S+

+ = + , with 0 1S = .

(b) To show that the closed form expression 12 1nnS+= satisfies the preceding iterative

equation we note that it gives 0 1S = , and that 2 1 1 1 11 (2 1) (2 1) 2 2 2 2

n n n n nn nS S

+ + + + ++ = = = ,

as predicted.

SECTION 9.1 Iterative Equations 835 24. 21 3 3 3nnS = + + + +

(a) This expression requires 0 1S = , and we observe the general pattern 1 0 3 1 3S S= + = +

2 22 1

2 31

3 1 3 3

1 3 3 3 3 3n nn n

S S

S S

= + = + +

= + + + + = +" "

"

Hence, the iterative equation for Sn is

11 3 ,n

n nS S+

+ = + with 0 1.S =

(b) To show that the closed form expression 11 (3 1)2

nnS

+=

satisfies the preceding iterative equation, we note that it satisfies 0 1S = and that 2 1 1 1 11

1 1 1(3 1) (3 1) (3 3 3 ) 32 2 2

n n n n nn nS S

+ + + + ++ = = = ,

as predicted.

Nonhomogeneous Structure 25. For the general linear iterative equation 1n ny ay b+ = + ,

(a) The function 0n

nu a y= is a solution of the corresponding homogeneous equation 1n ny ay+ = because 11 0 0 0.

n nn nu au a y a a y

++ = =

(b) The function

11

n

nap ba

=

satisfies the nonhomogeneous equation 1n ny ay b+ = + because

1

11 ( 1) ( 1) 11 1 1

n n n

n na a a a aay b ab b b b ya a a

++

+ + = + = = = .

(c) From part (a), nu is a solution of the homogeneous equation, and from part (b), np is a particular solution of the nonhomogeneous equation. Hence the general solution to the nonhomogeneous equation is n n ny u p= + or

011

nn

nay a y ba

= + .

836 CHAPTER 9 Discrete Dynamical Systems Where It Comes From 26. In Problem 22 we found the corresponding iterative IVP for the given sequence to be

1 01, with 0.n nS S n S+ = + = (a) The constant sequence nh c= satisfies the corresponding homogeneous equation 1 0n nS S+ = , because 1 0.n nh h c c+ = = (b) To find a particular solution of the nonhomogeneous equation 1 1n nS S n+ = + we try

2nP An Bn C= + + . Substituting this into the equation yields 2 2( 1) ( 1) 1A n B n C An Bn C n + + + + + + = + . Multiplying out the left side of the equation and collecting like powers of n yields

2 2

2

( 2 1) 1( ) (2 ) ( ) 1

2 ( ) 1.

A n n Bn B C An Bn C nA A n A B B n A B C C n

An A B n

+ + + + + = + + + + + + = +

+ + = +

Equating coefficients of 2n , n, and 1 we get 2 1A = , 1A B+ = . Hence 12

A = , 12

B = ,

and there is no condition on C. However, 1 1S = is given, so 21 1 12 21 1 1p C = + + = .

Hence, C = 0 , and so our particular solution is

21 12 2n

p n n= + . (c) Using the homogeneous solutions from part (a) and a particular solution from part (b)

yields the general solution of the nonhomogeneous equation,

21 12 2n

s c n n= + + where c is an arbitrary constant. Substituting into this expression the initial condition

0 0S = we get 0c = . Hence, the solution to the given IVP is

( ) ( )2 11 1 1 12 2 2 2n n n

n nS h p n n n n

+= + = + = + = .

SECTION 9.1 Iterative Equations 837 Exceptional Case 27. When one eigenvalue of

2 1 0n n nay by cy+ ++ + = is zero, the characteristic equation is

2 0a b + = , or ( ) 0a b + = , and hence the solution is

1 2 .n

nby c ca

= +

For example, the equation 2 12 0n ny y+ + = has the characteristic equation 2 2 0 = and eigenvalues 0 and 2. Hence, 1 2 2

nny c c= + .

Nonreal Eigenvalues 28. In case the roots 1 and 2 of the characteristic equation are complex, 1 and 2 ire = , which yields two solutions

1 ( )n i nre = and 2 ( ) .n i nre =

Rewriting the first solution in Cartesian form using Eulers equation yields

( ) (cos sin ).i n n in nre r e r n i n = = + Recall from Chapter 6.3 that real and complex parts of any complex solution are also solutions;

therefore, cosnr n and sinnr n are solutions. Hence, the general solution for complex eigenvalues ire is

1 2cos sinn n

ny c r n c r n = + .

Second-Order Linear Iterative Equations 29. 2 15 6 0n n ny y y+ + + = The characteristic equation is 2 5 6 0r r + = , which has roots 1 2r = and 2 3r = . Hence, the

general solution is

1 22 3n n

ny c c= + . 30. 2 14 4 0n n ny y y+ + + = The characteristic equation is 2 4 4 0r r + = , which has a double root 1 2 2r r= = . Hence, the

general solution is

1 22 2n n

ny c c n= + .

838 CHAPTER 9 Discrete Dynamical Systems 31. 2 1 2 0n n ny y y+ ++ = The characteristic equation is 2 2 0r r+ = , which has roots 1 2r = and 2 1r = . Hence, the

general solution is

( )1 22 nny c c= + . 32. 2 14 4 0n n ny y y+ + = The characteristic equation is 2 4 4 0r r = , which has roots 1 2 2 2r = + and 2 2 2 2r = .

Hence, the general solution is

1 2 1 2(2 2 2) (2 2 2) (4.828) ( 0.828) .n n n n

ny c c c c= + + +

Lab Problems: More Spreadsheets The reader should experiment with a spreadsheet and then check the results with the following solutions.

33. 2 0n ny y+ = , 0 0y = , 1 1y = (a) The following iterates were found using Excel.

n yn n yn

0 0 6 0

1 1 7 1

2 0 8 0

3 1 9 1

4 0 10 0

5 1

00

1.2

n10

yn

2 0n ny y+ = , 0 0y = , 1 1y =

(b) It is clear from the results of the spreadsheet and the iterative equation itself that the solution will continue to oscillate between 0 and 1.

SECTION 9.1 Iterative Equations 839 34. 2 0n ny y+ + = , 0 0y = , 1 1y = (a) The long-term behavior is indicated by the first 10 iterates of the solution.

n yn n yn

0 0 6 0

1 1 7 1

2 0 8 0

3 1 9 1

4 0 10 0

5 1

1.5

1.5

n10

yn

2 0n ny y+ + = , 0 0y = , 1 1y =

(b) We can solve the IVP to ascertain the long-term behavior of the solution. The characteristic equation is r2 1 0+ = , which has complex roots p iq i = , and so 0p = and 1q = as denoted in the text. Hence,

2 2 2 20 1 11tan .0 2

r p qqp

= + = + == = =

Hence

1 2 1 2( cos sin ) cos sin .2 2n

nn ny r c n c n c c = + = +

Substituting the initial conditions 0 0y = and 1 1y = yields 1 0c = and 2 1c = .Hence, the solution of the IVP is

sin , 0,1,2,2n

ny n= = .


Epidemic Model 35. Setting the solution

( )50 1.1 30 150,000nny = = and solving for n (number of years) yields

( ) 150,0301.1 300050

n = .

Hence, ( ) ( )log 1.1 log 3000n = or log3000 84

log1.1n = years.

Rabbits Again 36. We rewrite the Fibonacci equation as

2 1 0n n ny y y+ + = . The characteristic equation is

2 1 0 = , which has solutions

1 21 5,2 2

= . The general solution can then be written as

1 21 5 1 5 .2 2 2 2

n

ny c c = + +

Using the conditions 0 1y = and 1 1y = yields

( ) ( )1 21 15 1 and 5 1 .2 5 2 5c c= + = Thus, the solution becomes the Binet formula

1 1

1 1 5 1 5 .2 25

n n

ny+ + + =

SECTION 9.1 Iterative Equations 841 Generalized Fibonacci Sequence: More and More Rabbits 37. (a) If each adult rabbit pair has two rabbit pairs every month, then the number of rabbit pairs

after month 2n + will satisfy the discrete initial-value problem 2 1 2n n ny y y+ += + , 0 1y = , and 1 1y = . (b) The characteristic equation for the iterative equation in part (a) is

2 2 0 = , which has roots 1 and 2. Hence, the general solution of the equation is

( ) ( )1 21 2n nny c c= + . Using the initial conditions 0 1y = and 1 1y = yields 1 13c = and 2

23

c = . Hence, the number of rabbit pairs after month n will be

( ) 11 1 23

n nny

+ = + . To give some meaning to this algebraic solution, we compare the regular Fibonacci sequence (see third column) where there is one rabbit pair born each month to the case where each adult has two pairs of offspring each month. Note that in this case there will be 2731 rabbit pairs at the end of the year (after 12 months) compared with 233 in the case of the Fibonacci sequence.

Month 2 pairs 1 pair Month 2 pairs 1 pair 0 1 1 7 85 21 1 1 1 8 171 34 2 3 2 9 341 55 3 5 3 10 683 89 4 11 5 11 1365 144 5 21 8 12 2731 233 6 43 13

3000

n14

yn

00

Two and one rabbit pairs born per month

(c) If each rabbit pair has k rabbit pairs the IVP becomes

2 1 0n n ny y ky+ + = , 0 1y = , and 1 1y = . Its characteristic equation is 2 0k = , which has the two real roots 1 1 41 2 2

k += + and 1 1 42 2 2k += .

The general solution can then be written as

1 21 1 4 1 1 4 .2 2 2 2

n n

nk ky c c

+ += + +

It is left to the reader to evaluate the constants 1c and 2c that satisfy the IC.

842 CHAPTER 9 Discrete Dynamical Systems Probabilistic Fibonacci Sequence 38. If each rabbit pair expects to have 0.5 rabbit pairs each month, then the number of rabbit pairs

after the 2n + month will satisfy the equation 2 1 0.5 0n n ny y y+ += + = , 0 1y = , 1 1y = . The characteristic equation is

2 0.5 0 = , which has roots

1 21 32 2

= = . The general solution can then be written as

1 21 3 1 32 2 2 2

n n

ny c c = + +

which using the conditions 0 1y = and 1 1y = yields

1 21 1( 3 1) and ( 3 1).

2 3 2 3c c= + =

Thus, the solution of the IVP is 1

1 1 3 1 3 .2 23

n

ny+ + =

To give some meaning to this awkward formula, we compare the regular Fibonacci sequence (see third column) where there is one rabbit pair born each month to the case where each adult has on average 0.5 pair of offspring each month. Note that in this case there will be 33 rabbit pairs at the end of the year (after 12 months) compared with 233 in the case of the Fibonacci sequence.

Month 0.5 pairs 1 pair Month 0.5 pairs 1 pair

0 1.00 1 7 7.00 21

1 1.00 1 8 9.56 34

2 1.50 2 9 13.06 55

3 2.00 3 10 17.84 89

4 2.75 5 11 24.38 144

5 3.75 8 12 33.30 233

6 5.13 13

250

n12

yn

00

One and one half rabbit pairs born per month

Note that the numbers in the second column are not integers, they represent the expected (average) number of pairs.

SECTION 9.1 Iterative Equations 843 Check This with Your Banker 39. We use the formula for compound interest. Because interest is compounded daily, we have

0.08 0.000219365

r = . Hence, if the initial deposit is 0 $1000y = , then the value in the account will be as follows.

1 day: 1

10.08$1000 1 $1000(1.000219) $1000.22365

y = + = =

10 day: 10

1010

0.08$1000 1 $1000(1.000219) $1002.19365

y = + = =

1 year: 365

365365

0.08$1000 1 $1000(1.000219) $1083.28365

y = + = =

How Much Money Is Enough? 40. (a) As your account earns 8% interest each year and $30,000 is withdrawn each year, the

initial amount is 0 $200,000y = . Then, after n years, you will have 1 1.08 30,000n ny y+ = , 0 $200,000y = . (b) The solution of the initial-value problem in part (a) is

0(1.08) 1(1.08) $30,000 $200,000(1.08) $375,000 (1.08) 1

0.08

$375,000 $175,000(1.08) .

nn n n

n

n

y y = =

=

Clearly, this sequence gets smaller and smaller until you run out of money. To find out when, set 0ny = and solve for n. Doing this yields

( )175,000 1.08 375,000n = which has the solution 9.9n years.

How to Retire a Millionaire 41. (a) We are given 0.08r = , 0 0y = , and d is unspecified. Thus the iterative IVP is 1 1.08n ny y d+ = + , 0 0y = which has the solution

(1.08) 1 12.5 (1.08) 1 .0.08 1

nn

ny d d = =

(b) In order for Sheryl to be worth a million dollars in 50 years, solve

( )5012.5 1.08 1 $1,000,000d = yielding $1742.86d as her annual deposit.

844 CHAPTER 9 Discrete Dynamical Systems Amazing But True

42. (a) If weekly interest is 0.08 0.0015452

(0.154%), and weekly deposits are $25, then the amount of money ny Wei Chen will have in the bank after n weeks satisfies the iterative IVP

1 1.00154 25n ny y+ = + , 0 0y = . (b) The solution of the IVP in part (a) is

( ) ( )25 1.00154 1 $16,237 1.00154 10.00154

n nny = = .

(c) Substituting 208n = (4 years) into the solution found in part (b) yields 208 $6123.94y = .

Amortization Problem 43. (a) We rewrite ( )1 1n np r p d+ = + in the form of a nonhomogeneous linear iterative

equation

( )1 1n np r p d+ + = . We solve it by first solving the corresponding homogeneous equation

( )1 1 0n np r p+ + = whose characteristic equation is ( )1 0r + = which yields 1 r = + . The solution of

the homogeneous equation is then

( )1 nnp c r= + . To find a particular solution of the nonhomogeneous equation ( )1 1n np r p d+ + = , we

try a constant np A= . This gives 1n np p += , so substitution gives ( )1A r A d + = , or dAr

= . The general solution is then given by

( )1 nn dp c r r= + + . Setting 0p S= yields dS c r= + or

dc Sr

= . Hence, we finally have as the IVP solution

(1 ) 1(1 ) (1 ) .n

n nn

d d rP S r S r dr r r

+ = + + = +


(b) The value np represents the outstanding principal, and to have the outstanding principal equal 0 after N periods, set 0Np = , which yields

( ) ( )1 10 1N

N rS r dr

+ = +

Solving this equation for d, yields

( )( )

1

1 1

N

N

Sr rd

r

+= + .

Dividing the top and bottom by ( )1 Nr+ yields

( )1 1 Nrd S

r = +

.

(c) The parameters constant in the equation are $100,000S = , 0.01r = , and 360N = . Substituting these values into the previous expression yields the monthly payment of $1028.61d = .

Fisheries Management 44. (a) Assume the haddock tonnage increases by 2% per year, but diminishes 1000 tons per

year due to fishing. If the initial stock is 100,000 tons of haddock, then the future tonnage (in thousands of tons) of haddock will satisfy the IVP

1 1.02 1n ny y+ = , 0 100y = . (b) We use biological growth with periodic depletion, as in the text Example 3, to find the

solution

(1.02) 1100(1.02) 50 (1.02) 1 .0.02

nn n

ny = = +

(c) If the tonnage of haddock caught every year is changed to ( )5 5,000 tonsd = , then the solution changes to

(1.02) 1100(1.02) 5 250 150(1.02)0.02

nn n

ny = =

.

Hence, the tonnage decreases and reaches 0 when 25.8n = years.


Deer Population

45. (a) Measuring deer in the thousands with 0.10r = , 15d = , and 0 100y = , yields 1 1.10 15n ny y+ = with solution

( ) ( ) ( )15100 1.10 1.10 1 50 1.10 1500.10

n n nny = = + thousands of deer.

(b) In order for the population to be constant, simply harvest the new population. If initially there are 100,000 and they grow by 10% per year, harvest (hunt) 10,000 deer per year. This will keep the population at 100,000 deer.

Save the Whales

46. (a) If the initial population is 0 1000y = whales, the population is increasing at an annual rate of 0.25r = (25%). If 300d = whales are harvested each year, then 1 1.25 300n ny y+ = and after n years the number of whales will be

( ) ( ) ( )3001000 1.25 1.25 1 200 1.25 12000.25

n n nny = = + whales.

(b) The population is clearly decreasing. Setting 0ny = yields 8.03n . If this continues, the whales will be extinct in 8 years.

Drug Therapy

47. (a) We have 1 0.75 100n ny y+ = + and 0 0y = , which has the solution

(0.75) 1100 400 1 (0.75)0.75 1

nn

ny = =

grams of insulin.

(b) As n increases, ( )0.75 n tends to 0, and so the long-term amount of insulin in her body tends to 400 grams.

SECTION 9.1 Iterative Equations 847 Consequence of Periodic Drug Therapy

48. (a) We assume Kashkooli has no drug in his system on day zero. However, 0y is the amount of drug in his body immediately after he takes his dose of 100 mg. Therefore,

0 100 mgy = . Further, because he loses 25% per day and gains 100 mg, the iterative equation that describes the number of mg of drug he has in his body on day n immediately after taking the drug is

1 0.25 100n n ny y y+ = + . Thus, 1 0.75 100n ny y+ = + , 0 100y = . (b) The iterative equation in part (a) can be written

1 0.75 100n ny y+ = and has characteristic equation 0.75 0r = . Hence, the homogeneous solution is

( )0.75 nny c= . Finding a particular solution of the nonhomogeneous equation yields 400ny = , which means the general solution is

( )0.75 400nny c= + . Substituting the initial condition 0 100y = yields 300c = . Thus, the amount of drug he has in his body after n days is

( )100 4 3 0.75 nny = . (c) As n increases ( )0.75 n goes to zero, and hence, ny tends to 400 mg. (d) The homogeneous solution ( )0.75 nc always tends to 0, as n increases no matter what the

dosage. Thus, the particular solution is ny A= for 1 0.75n ny y d+ = , where d is the daily dosage, which is 4A d= . In other words, if the limiting amount of the drug is A, then the daily dosage should be 0.25 mgd A= . If Kashkooli wants the limiting amount of the drug in his body to be 800 mgA = , then his daily dosage should be

( )0.25 800 200 mgd = = .

848 CHAPTER 9 Discrete Dynamical Systems General Growth Problem 49. We are given the equation ( )1 1n n n ny y r y y+ = , which can be rewritten as ( )1 11 0n n ny r y ry+ + + = . This has the characteristic equation ( )2 1 0r r + + = and roots

( )2

1 2

1 41 1 1, , 12 2 2 2

r rr r r r + + + = = = . Hence, the general solution is

1 2n

ny c c r= + . Substituting the initial conditions yields 1 2 0c c y+ = and 1 2 1c c r y+ = , which gives

1 02 1y ycr= and

0 11 1

y r ycr= .

Thus, the population is

( ) ( )1 0 0 11 1 nny y y r y r yr = + .

Chimes in a Day 50. The number of chimes 1ny + on the ( )1n + hour is given by ( )1 1n ny y n+ = + + and 1 1y = . The characteristic equation of this equation is 1 0 = . Hence, the homogeneous solution is a

constant sequence 1 11n

ny c c= = . To find a particular solution, we normally seek a solution of the form ny An B= + . However, the homogeneous solution has the constant solution, so we multiply by n. This yields the trial solution 2ny An Bn= + . Substituting this into the iterative equation yields

( ) ( ) ( )2 21 1 1A n B n An Bn n+ + + = + + + . Comparing coefficients of n2 , n, and 1 yields 1

2A B= = . Hence, the general solution is

( )

1

12n

n ny c

+= + .

Substituting the initial condition gives 1 0c = , and so we get

( )1

2nn n

y+= .

Thus, the total number of chimes over a 24-hour period is

( )122 12 13 156y = = .

SECTION 9.1 Iterative Equations 849 Very Interesting 51. Once you make a few sketches, you will convince yourself that the iterative equation 1 1n ny y n+ = + + is correct. Hence, we simply need to solve it. As noted in the hint, the new line divides 1n +

regions into 2 2n + regions, so if there were ny regions before, there are now 1ny n+ + regions. The homogeneous equation has characteristic equation 1 0 = , so the homogeneous solutions are constants 1ny c= . We therefore seek a particular solution of the form 2ny An Bn= + . Substituting this value into the nonhomogeneous equation yields

( )2 1An A B n+ + = + . Setting coefficients equal yields 1

2A B= = . Hence, the general solution is

( )

1

12n

n ny c

+= + . Substituting the initial condition 0 1y = yields 1 1c = ,so

( ) 21 2 21

2 2nn n n ny

+ + += + = distinct regions.

Planters Peanuts Problem 52. (a) Count the number of ways backwards starting at the last letter (the S in the middle of the

bottom row of the pyramid), and count all the paths other than the one that goes straight up. There are 2 nT paths. Now add the path that goes straight, to obtain a total of 2 1nT + paths. Hence, the difference equation

1 2 1n nT T+ = + , 1 1T = . It yields the initial condition 1 1T = , as a pyramid with one letter has one path. (b) The characteristic equation of the homogeneous equation is 2 0r = , so the

homogeneous solution is

12n

nT c= where 1c is an arbitrary constant. To find a particular solution, try ny A= , which yields

1A = . Hence, the general solution is 12 1

nnT c= .

Plugging this into the initial condition yields 1 1c = , so the solution is 2 1nnT = . Because PLANTERS PEANUTS has 15 letters, there is a total of

1515 2 1 32,767T = = ways to spell the word. Suggested Journal Entry 53. Student Project


9.2 Linear Iterative Equations

System Classification 1.

3.5

2.5

0

ny

nx

1

1

0.75 00 0.25

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 0.75 = and 2 0.25 = , or compute Tr 1=A and 0.1875=A Because both eigenvalues are positive and less than 1, the equilibrium point is a sink. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Sink 2.

4

2.5

0

ny

nx

1

1

0.7 00 0.3

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 0.7 = and 2 0.3 = , or compute Tr 1=A and 0.21=A . Because both eigenvalues are positive and less than 1, the equi-librium point is a sink. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Sink 3.

4

4

-4

-4

ny

nx

1

1

0 11 0

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 i = and 2 i = , or compute Tr 0=A and 1=A . From the trace-determinant plane in Figure 9.2.7 the point ( )0, 1 for this matrix A lies in the center region. Hence, the equilibrium is a center, and each iterate remains at the same distance from the origin. In this case the orbit cycles around four points (see figure). Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Center

4.

1200

5

0

ny

nx

1

1

1.9 00 1.1

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.9 = and 2 1.1 = , or compute Tr 3=A and 2.09=A . Because both eigenvalues are positive and greater than 1, the equilibrium point is a source. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Source

SECTION 9.2 Linear Iterative Equations 851 5.

60-40

4

-3

ny

nx

1

1

1.9 00 1.1

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.9 = and 2 1.1 = , or compute Tr 3= A and 2.09=A . Both eigenvalues have an absolute value greater than 1, so the equilibrium point is a source. However, both eigenvalues are negative, so it is a double-flip source. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Double-flip source

6.

40

2.5

-1.5

ny

nx

1

1

1.5 00 0.5

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.5 = and 2 0.5 = , or compute Tr 1=A and 0.75= A . One eigenvalue has an absolute value greater than 1, and the other is less than 1, so the equilibrium is a saddle. Because only one eigenvalue is negative, it is a flip saddle. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Flip saddle

7.

5

3

-3

ny

nx

1

1

1.05 00 1.05

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.05 = and 2 1.05 = , or compute Tr 0=A and 1.1025= A . Both eigenvalues have an absolute value greater than 1, so the equilibrium point is a source. Because only one eigenvalue is negative, it is a flip source. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = . Flip source

8.

40-40

4 ny

nx

1

1

1.5 00 1.1

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.5 = and 2 1.1 = , or compute Tr 0.4= A and 1.65= A . Both eigenvalues have an absolute value greater than 1, so the equilibrium point is a source. Because only one eigenvalue is negative it is a flip source. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = . Flip source

852 CHAPTER 9 Discrete Dynamical Systems 9.

15-15

20

-25

ny

nx

1

1

0 0.91.6 0

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 1.2i = and 2 1.2i = , or compute Tr 0=A and 1.44= A . We see that from the trace-determinant plane in Figure 9.2.7 the point (0, 1.44) of the matrix A lies in the spiral source region. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = . Spiral source

10.

4-3

20

-15

ny

nx

1

1

0.6 00 1.4

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 0.6 = and 2 1.4 = , or compute Tr 2= A and 0.84=A . One eigenvalue has an absolute value greater than 1, and the other is less than 1, so the equilibrium is a saddle. Because both eigenvalues are negative, it is a double-flip saddle. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = .

Double-flip saddle

11.

4-3

2.5

-1.5

ny

nx

1

1

0.8 00 0.6

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 0.8 = and 2 0.6 = , or compute Tr 1.4= A and 0.48=A . Both eigenvalues have an absolute value less than 1, so the equilibrium point is a sink. Because both eigenvalues are negative it is a double-flip sink. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = . Double-flip sink

12.

4-3

2.5

-1.5

ny

nx

1

1

0 1.60.4 0

n n

n n

x xy y

+

+

=

Compute the eigenvalues, 1 0.8i = and 2 0.8i = , or compute Tr 0=A and 0.64=A . We see that from the trace-determinant plane in Figure 9.2.7 the point (0, 0.64) for the matrix A lies in the spiral sink region. Using Excel, we plot the first few points starting at the point ( ) ( )0 0, 3, 2x y = . Spiral sink


00

2.5

2.5

yn

xn

(a) 1 0.5 = , 2 0.3 = The equilibrium (at the origin) is a sink, because

both eigenvalues are less than 1.

(b) ( )( )

0

0

0.5

0.3

nn

nn

x x

y y

==

(c) A few iterates are shown (see figure) starting at ( )2, 2 .

14. 2.5

2

yn

xn2

(a) 1 0.5 = , 2 0.5 = The equilibrium (at the origin) is a flip sink,

because both eigenvalues have absolute values less than 1 and only one eigenvalue is negative.

(b) ( )( )

0

0

0.5

0.5

nn

nn

x x

y y

= =


15. 2.5

2

yn

xn2

1.0

(a) 1 0.5 = , 2 0.3 = The equilibrium (at the origin) is a double-flip

sink, because both eigenvalues are negative and have absolute values less than 1.

(b) ( )( )

0

0

0.5

0.3

nn

nn

x x

y y

= =


16.

20

yn70

00 xn

(a) 1 1.5 = , 2 2 = The equilibrium (at the origin) is a source,

because both eigenvalues are positive and greater than 1.

(b) ( )( )

0

0

1.5

2

nn

nn

x x

y y

==


854 CHAPTER 9 Discrete Dynamical Systems 17.

40

yn300

40xn

(a) 1 1.5 = , 2 2 = The equilibrium (at the origin) is a flip source,

because both eigenvalues have absolute values greater than 1 and only one eigenvalue is negative.

(b) ( )( )

0

0

1.5

2

nn

nn

x x

y y

= =


18.

20

yn200

20xn

600

(a) 1.51 = , 2 3 = The equilibrium (at the origin) is a double-flip

source, because both eigenvalues are negative and have absolute values greater than 1.

(b) ( )( )

1.50

30

nx xnny yn

= =


19.

2.5

yn18

00 xn

(a) 1 0.5 = , 2 2 = The equilibrium (at the origin) is a saddle,

because both eigenvalues are positive and only one eigenvalue is greater than 1.

(b) ( )( )

0

0

0.5

2

nn

nn

x x

y y

==


20. ]

2

yn200

300

xn

(a) 1 0.5 = , 2 2 = The equilibrium (at the origin) is a flip saddle,

because one eigenvalue is positive and less than 1, while the other eigenvalue is negative and has an absolute value greater than 1.

(b) ( )( )

0

0

0.5

2

nn

nn

x x

y y

==



10

2

yn

xn2

20

(a) 1 0.5 = , 2 2 = The equilibrium (at the origin) is a double-flip

saddle, because both eigenvalues are negative and only one eigenvalue has an absolute value greater than 1.

(b) ( )( )

0

0

0.5

2

nn

nn

x x

y y

= =


Iteration by Rotation 22. (a) The eigenvalues are the roots of

( )2 2 2 2 22

cos sincos sin cos 2 cos sin

sin cos

2 cos 1 0

= + = + += + =

which has roots

2

21 2

2cos 4cos 4, cos cos 1 cos sin2

ii e = = = = . (b) The rotation matrix

cos sinsin cos

= R

(which rotates the plane in the counterclockwise direction by an angle ) has eigenvalues ie . Hence, the rotation matrix for a rotation through an angle of n is

cos sinsin cos

n nn n

.

But a rotation through an angle n has the same matrix as the nth power nR .


Spirals or Circles? 23.

1

0 11 0n n+

= x xG G

This is a pure rotation matrix with 2 = . Thus consecutive

points in the solution are rotated in the counterclockwise direction by 90 degrees with no change in the distance from the origin. The orbit is plotted starting at ( )1, 0 (see figure).

1.5

1.5

yn

xn

1.5

1.5

Counterclockwise cyclic trajectory

24. 1

2 2

2 2n n+

= x xG G

We rewrite the coefficient matrix as

2 22 22 2

22 2

2 22 2

=

,

60

60

yn

xn

60

60

Outward spiral, counterclockwise

which is a scalar two times a rotation matrix with 4 = . Hence, the action of the iterative

system is a counterclockwise rotation of 4 at each iteration plus an expansion away from the

origin by a factor of 2. See figure for a few points of the orbit starting at ( )1, 1 . 25.

1.2

1.5

yn

xn1.5

0.2

1

3 10.25

1 3n n+

= x xG G

We rewrite the equation as

1

3 12 2

0.5

1 32 2

n n+

=

x xG G , Spiraling inward,

counterclockwise

which is the scalar (0.5) times a rotation matrix with 6 = . Hence, the action of the iterative

system is a counterclockwise rotation of 6 at each iteration plus a contraction closer to the origin

by a factor of 0.5. See the figure for a few points of the orbit starting at ( )1, 1 .

SECTION 9.2 Linear Iterative Equations 857 Moose Extinction For Problems 26-28 the iterative equations (of the introductory example) are

1 10.72 0.24 and 0.16 1.28n n n n n nw w m m w m+ += + = + 26. (a) 0 0700, 225w m= =

We note that the moose population becomes extinct. See figure for Excel graph.

00

800

n10

w nm n

Moose extinctions

(b) 0 0700, 300w m= = It is interesting to see that if the initial moose

population were increased, the results are much different. The moose can now hold on.

00

1500

n15

w nm n

Wolf and moose both survive

Wolf Extinction 27. 0 0300, 0w m= =

If there are no moose present and the wolf population is initially 0w , 0w will obey the IVP 1 0.72n nw w+ = and

0 300w = , whose solution is ( )300 0.72 nnw = . Without a replacement food supply, the wolf population dies out within 15 years (see figure). n15

w n350

0

Wolf extinction

Moose and Wolves Together 28. 0 0100, 300w m= =

We used Excel obtain the curve shown in the figure. If 0n = corresponds to 1990, then the 6th value corresponds to 1995 and the 12th value to 2001. The x-axis is the wolf population and the y-axis is the moose population. In the year 2001 (last point at the upper right) the wolf population is roughly 1000 and the moose population is roughly 2000.

1200

m n2500

0 w n

Moose versus wolf population

858 CHAPTER 9 Discrete Dynamical Systems System Analysis

29. 0.92 0.360.06 0.98

= A

-12

-12

12

12

x

y

v1

v2

(a) The eigenvalues are 1 0.8 = and 2 1.1 = with corresponding eigenvectors

[ ] [ ]1 23, 1 and 2, 1= =G Gv v . (b) Tr 1.9= A , 0.88=A The equilibrium (at the origin) is a double-flip saddle,

because both eigenvalues are negative and only one eigenvalue has an absolute value greater then 1.

(c) The line with negative slope is the eigenvector corresponding to 1 0.8 = , and the line with positive slope is the eigenvector corresponding to 2 1.1 = .

The figure shows that the solution starting at ( )5, 0 gets closer and closer to 2

Gv and that the points flip back and

forth about 2Gv while moving further and further away

from the origin.

Typical double-flip saddle solution

(d) ( ) ( )1 23 20.8 1.11 1n n

n c c = +

x

The first term in the solution goes to zero as a result of the factor ( )0.8 n while the second gets larger due to the factor (1.1). Hence, the orbit gets closer and closer to 2

Gv

and the negative value of 1 causes the solution to oscillate around 2Gv with smaller and smaller amplitudes; at the same time the negative value of 2 causes the orbit to flip back and forth about the origin with larger and larger amplitude.

30. 0.04 2.28

0.38 0.34 =

A

(a) The eigenvalues are computed as 1 0.8 = and 2 1.1 =with corresponding eigenvectors

[ ] [ ]1 23, 1 and 2, 1= =G Gv v . (b) Tr 0.3=A , 0.88= A (c) The line with negative slope is the eigenvector

corresponding to 1 0.8 = and the line with positive slope is the eigenvector corresponding to 2 1.1 = .

-12

-12

12

12

x

y

v1

v2

Single-flip saddle solution

SECTION 9.2 Linear Iterative Equations 859 The second figure shows the solution starting from

0 5x = , 0 0y = , which gets closer and closer to the eigenvector 2

Gv . Although iterates flip back and forth

around 2Gv (due to the negative eigenvalue) for small n,

they will just move out along the eigenvector going away from the origin.

(d) ( ) ( )1 23 20.8 1.11 1n n

n c c = +

Gx

The first term in the solution goes to zero as a result of the factor ( )0.8 n . Hence, the orbit gets closer and closer to 2

Gv and oscillates around this eigenvector with smaller and

smaller amplitudes as a result of the negative value of 1 . Meanwhile the factor (1.1)n causes the orbit to move steadily away from the origin.

Owls and Rats 31. 1

1

0.5 0.40.1 1.1

n n n

n n n

O O RR O R

+

+

= += +

(a) The eigenvalues and eigenvectors of the matrix

0.5 0.40.1 1.1

= A

are [ ][ ]1 12 2

0.58 5.236, 11.02 0.764, 1 .

= == =

GGvv

Hence the solution

( ) ( )1 25.236 0.7640.58 1.021 1n nn

n

Oc c

R = +

.

(b) We start with 2000 rats and 250 owls; the solution is shown in the figure. Because 1.6; 0.59= =TrA A , we can see from Figure 9.2.7 that the origin is a saddle point, which is what the trajectory indicates.

(c) We see that not only do both species survive they both thrive! The long-term ratio will be along 2

Gv , so

owls 0.764rats

= which fits the slope of the eigenvector from the right

end of the solution through the origin.

0 5

5

x

y

rats

in th

ousa

nds

owls in hundreds

v1v2

Rats versus owls

860 CHAPTER 9 Discrete Dynamical Systems Diabetes Mode 32. (a) The coefficient matrix

0.978 0.0060.004 0.992

= A

has eigenvalues 1 2, 0.98, 0.99 = , which tells us that all solutions approach the origin. Corresponding eigenvectors are

[ ]1 3, 1= Gv and [ ]2 1, 2= Gv . (b) The general solution is

( ) ( )1 23 10.98 0.991 2n n

n c c = +

Gx .

Substituting 0 100x = and 0 0y = yields

1 21 2

100 0.95 0.630 0.32 1.26

c cc c

= +=

which has solution 1 127c = and 2 32c = . So the solution of the IVP is

( ) ( )3 1127 0.98 32 0.991 2

n nn

= Gx .

The solution will approach zero ( )0, 0 , but very slowly. (c) We plot the solution using Excel and see from the

spreadsheet that the glucose level becomes negative when 177n = minutes (nearly 3 hours).

120

12

0 xnglucose level

yn

Solution to Diabetes Problem

SECTION 9.2 Linear Iterative Equations 861 Conversion Job 33. Using the equation numbers in the text, the two equations are

1n n nx ax by+ = + (11) 1n n ny cx dy+ = + . (12) Rewrite Equation (12) as

2 1 1n n ny cx dy+ + += + (13) and then work with equations (11) and (12) to get all the terms expressed in ys. E.g., multiply Equation (11) by c, to get

1n n ncx cax cby+ = + . (14) Now solve, respectively, Equation (12) for ncx and Equation (13) for 1ncx + , we get

11 2 1 .n n n

n n n

cx y dycx y dy

+

+ + +

= =

Substituting these values in Equation (14) yields

2 1 1n n n n ny dy ay ady cby+ + + = + or

( ) ( )2 1 0n n ny a d y ad bc y+ + + + = . Decomposition Job 34. Given the equation

2 1 0n n npx qx rx+ ++ + = , we let 1n ny x += . Then,

1 2n n n nr qy x x yp p+ +

= = .

We can, therefore, write the two first-order equations as

1

1 .

n n

n n n

x yr qy x yp p

+

+

==

862 CHAPTER 9 Discrete Dynamical Systems The Lilac Bush 35. (a) We call number of new stems

number of old stems.n

n

xy

==

Because each old stem grows two new stems every year, we have 1 2n nx y+ = . Also because every new stem becomes an old stem the next year, the number of old stems every year will be the sum of the old and new stems on the previous year, or

1n n ny x y+ = + . Hence, the iterative system is 1 0

1 0

2 10

n n

n n n

x y xy x y y

+

+

= == + =

or in matrix form

1 01 0

0 2 1;

1 1 0n n

n n

x x xy y y

+

+

= = .

(b) The eigenvalues and eigenvectors of the coefficient matrix are

1 1

2 2

12, ;

1

21, .

1

= =

= =

G

G

v

v

(c) Hence, the general solution is

( )1 21 22 11 1nn n

n

xc c

y = +

.

Substituting in the initial values 0 1x = and 0 0y = yields

0 1 20 1 2

2 10

x c cy c c

= + == =

whose solution is 113

c = and 2 13c = . Hence, the number of stems on the plant on year n is

( )1 21 12 11 13 3

nn n

n

xy = = +

Gxn .

(d) On the 6th year the number of new and old plants is

( )666 1 21 2 11 1366 221 .63 213

= + = =

Gx

In other words, the next lilac bush has 22 new stems and 21 old stems (see figure). Lilac bush

SECTION 9.2 Linear Iterative Equations 863 Saddle Regions 36. (a) The characteristic equation of the coefficient matrix of

11

n n n

n n n

x ax byy cx dy

+

+

= += +

is

2 Tr 0a b

c d

= + = A A .

This is a quadratic equation in , so we have 1 2Tr = +A and 1 2 =A where 1 and 2 are the eigenvalues of the coefficient matrix. Hence, because the region S is defined by

Tr 1 Tr 1 < < A A A , this inequality provides the relationship between the eigenvalues as

( )1 2 1 2 1 21 1 + < < + or equivalently

1 2 1 2 1 0 + < and 1 2 1 2 1 0 + + + > .

(b) We can factor each of the left-hand sides of the inequalities in part (a) getting

( )( )1 2 1 2 1 21 1 1 + = and ( )( )1 2 1 2 1 21 1 1 + + + = + + . Thus, the inequalities become

( )( )1 21 1 0 < and ( )( )1 21 1 0 + + > .

But it is easy to see that both of the previous inequalities hold if either of the following sets of inequalities hold:

1 21 1 and 1 < < > or 2 11 1 and 1 < < > . The details of this simple verification are left for the student.

(c) As we obtained in part (b) one eigenvalue is positive and greater than 1, and the other eigenvalue has an absolute value lower then 1. One can see that the origin is a saddle point.

864 CHAPTER 9 Discrete Dynamical Systems Source and Sink Bifurcation

37. On the trace-determinant plane, inside the parabola ( )21 Tr4

=A A the eigenvalues of A are complex conjugates, 1 2, i = , so we consider the matrix

0

0i

i

+ =

A .

Note that 2 2 = +A , but it is also known that ( )2 2 = + . Thus, 2=A . If 1=A , then 1 = , which will separate the sources ( )1 > from the sinks ( )1 < . Hence, 1=A is indeed the line of bifurcation within the region of complex eigenvalues.

Lab Exercise 38. Student Project

Suggested Journal Entry 39. Student Project

SECTION 9.3 Linear Iterative Equations Chaos Again 865 9.3 Linear Iterative Equations Chaos Again

Attractors and Repellers 1. 2

1n nx x+ = Set 2x x= , which yields 2 0x x = , so we have two fixed points

0, 1ex = . The cobweb diagram shows that the equilibrium at 0 is attracting and that at 1 is repelling.

1

1xn

1

1

xn+1

One repelling and one attracting fixed point

2. 21 1n nx x+ =

Set 2 1x x= , which yields 2 1 0x x = , so we have two fixed points

1 52 2e

x = . The cobweb diagram shows that both fixed points are repelling. Note that there is an attracting cycle that passes between them.

2

2xn

2

2

xn +1

Two repelling fixed points and one attracting cycle

3. 31n nx x+ =

Set 3x x= , which yields 3 0x x = , so we have three fixed points

1, 0, 1ex = . The cobweb diagram shows that 0 is attracting and that 1 and 1 are repelling.

2

2xn

2

2

xn+1

One attracting and two repelling fixed points

4. 1 cosn nx x+ = Set cos ,x x= which yields cos 0.x x = The roots of this equation can be found numerically using a computer. Graph y x= and cosy x= to discover that they intersect only once. This value is approximately

0.74ex (radians). The cobweb diagram shows this to be an attracting fixed point.

2

2xn

2

2

xn+1

Attracting fixed point

866 CHAPTER 9 Discrete Dynamical Systems Hindsight 5. (a) Cobweb diagrams for 1n ny ay+ = are shown in the following figures for different values

of a and for different initial conditions. In each case 0y = is an equilibrium point.

(i) 1a > . The line 1n ny ay+ = has slope > 1, which means the orbit goes monotonically toward (diverges) for every starting point 0ny . The cobweb diagram indicates this fact for 2a = .

4

4 4 yny0

y4

n +1

(ii) 1a = . The line 1n ny y+ = coincides with the diagonal, which means that every solution is a constant, as indicated by the cobweb diagram. In this case all points are stable equilibrium points.

4

4

y4

n +1

4 yny0

(iii) 0 1a< < . The line 1 0.5n ny y+ = has slope < 1, which means that solutions converge to zero from all starting points, as shown for a = 0.5.

4

4 yn4

y4

n +1

y0

(iv) 0a = . The line 1 0ny + = is a horizontal line with slope 0. All initial points converge to zero on the first iteration as indicated by the cobweb diagram.

4

4

y4

n +1

4 yny0

SECTION 9.3 Linear Iterative Equations Chaos Again 867 (v) 1 0a < < . The line 1n ny a y+ = has negative slope less

steep than 1. Orbits from all initial points converge to zero, as indicated by the cobweb diagram for 0.5a = .

4

4

y4

n+1

4 yny0

(vi) 1a = . The line 1n ny y+ = has slope 1. The orbit from every initial point cycles between

0 1 0 1, , , , y y y y " , as indicated by the cobweb diagram for 0 3y = .

4

4

y4

n+1

4 yny

0

(vii) 1a < . The line 1n ny a y+ = has slope steeper than 1, which means the solution diverges for every starting point 0y , as indicated by the cobweb diagram for

2a = .

4

4

y4

n+1

4 yny0

(b) When 1a , 0y = is a stable equilibrium point. When 1a then 0y = is an unstable equilibrium point.

(c) When a is positive, iteration is monotonic and cobweb diagrams look like stair steps. When a is negative, iterations oscillate in value, and cobwebs wind around the fixed point.

(d) The value of b plays a role in locating the fixed point. That is, 1e

bxa

= .

868 CHAPTER 9 Discrete Dynamical Systems Stability of Fixed Points 6.

(a) 3 2( )2

xf xx+= +

The slope of ( )f x at the left-hand equilibrium is unstable, and the right-hand equilibrium is stable. Note: the text Figure 9.3.16(a) shows only the part of

( )f x to the right of 2x = , which is an asymptote for ( )f x .

If we try a cobweb from an 0x just to the left of 2, we discover we must graph ( )f x to left of 2 as well.

(b) ( ) 3 (1 )f x x x= Eyeballing the slopes of ( )f x at the equilibria shows

that the origin is repelling and unstable, but at the right-hand fixed point 2 / 3ex = the slope appears close to 1, so we take the derivative to check it out. ( ) 3 6f x x = , so (2 / 3) 1f = .

As shown in Problem 5(a)(vi), a straight line with slope 1 produces a cyclic orbit of period 2; however the curvature of ( ) 3 (1 )f x x x= spoils the possibility of a cycle. In fact, if the orbit is extended you will see it actually converges, very very slowly, to 2 / 3ex = .

Analyzing the Data There are many ways to show a solution sequence for an iterative equation. In problems 7-12 we chose a time series with a typical seed, usually 0 0.5x = . A list of values, or sequences from other seeds (not at an equilibrium) should give the same information. Excel is an excellent source of all these options.

7. ( )1 0.5 1n n nx x x+ = (a) Starting at 0 0.5x = , the function iterates toward an

equilibrium of zero, as shown in the figure. (b) The equilibrium point(s) ex of this iteration are the

root(s) of ( ) ( )0.5 1x f x x x= = ; i.e., when 20.5 0.5 0x x+ = , which yields 0ex = and 1ex = .

n

nx

200

.6

0

( )1 0.5 1n n nx x x+ = (c) Because ( ) ( )0.5 1 2f x x = yields ( ) ( )( )0 0.5 1 2 0 0.5 1f = = < , the origin is asymptotically stable. The other fixed point 1ex = is unstable because ( ) ( )1 0.5 1 2 1.5 1f = + = >

SECTION 9.3 Linear Iterative Equations Chaos Again 869 8. ( )1 2.8 1n n nx x x+ =

(a) Starting at 0 0.5x = , the values oscillate about 0.65, getting closer and closer. (b) The equilibrium point(s) ex of this iteration are the

root(s) of

( ) ( )2.8 1x f x x x= = or

22.8 1.8 0x x = , which yields 0ex = and 0.643ex = .

00

0.8

n10

xn

( )1 2.8 1n n nx x x+ = (c) Because ( ) ( )2.8 1 2f x x = yields ( ) ( )( )0 2.8 1 2 0 2.8 1f = = > ( ) ( )( )0.643 2.8 1 2 0.643 0.8 1,f = = < the origin is unstable and 0.643 is asymptotically stable as the graph shows.

9. ( )1 3.2 1n n nx x x+ = (a) Starting at 0 0.5x = , the function iterates in a cycle, shown in gray.

Starting at 0 0.2x = , the function iterates toward the same cycle, as shown in black.

00

0.8

n10

xn

(b) The equilibrium point(s) ex of this iteration are the root(s) of

( ) ( )3.2 1x f x x x= = or

23.2 2.2 0x x = , which yields 0ex = and 2.2 0.6875.3.2ex = . ( )1 3.2 1n n nx x x+ = (c) ( )0 3.2 1f = > , and 2.2 2.23.2 6.4 3.2 4.4 1.2 1

3.2 3.2f = = = > .

Hence both fixed points are unstable or repelling, with an attracting cycle passing between them. See Problem 30.

870 CHAPTER 9 Discrete Dynamical Systems 10. ( )1 4 1n n nx x x+ = (a) If 0 0.5x = , the iterates are 1 1, 0 for all 2nx x n= =

does that always happen? No! E.g. starting at 0 0.3x = , the motion appears chaotic.


( ) ( )4 1x f x x x= = or

24 3 0x x = , which yields 0ex = and 0.75ex = .

00

1.2

n25

xn

( )1 4 1n n nx x x+ =

(c) Because ( ) ( )4 1 2f x x = , ( ) ( )( )0 4 1 2 0 4 1f = = > , and ( ) ( )( )0.75 4 1 2 0.75 2 1.f = = > Hence, both equilibrium points are unstable. In this case, there is no attracting cyclic

point between the two repelling points, as there is in Problems 6(b) and 9. This lack of anything to attract is what causes the chaotic motion shown in part (a).

Oddly enough, however, there are certain seeds (e.g., 0 0x = , 0.5, or 1) that iterate to zero and stay there! But an orbit with the slightest change from these initial values is chaotic.

11. 1 2sinn nx x+ = (a) Starting at 0 0.5x = , the iterates quickly reach a cycle.

(b) The equilibrium point(s) ex of this iteration are the root(s) of ( ) 2sinx f x x= = , or 0ex = .

(c) Because ( ) ( )2cosf x x = yields ( ) ( )0 2cos 0 2 1f = = > , the fixed point zero is unstable, repelling toward a cycle

period 2 as the graph shows. 3

3

n25

xn

1 2sinn nx x+ =

A question remains: How does one analytically find the two-cycle shown in the graph? The answer will be found in Problems 31 and 32.

SECTION 9.3 Linear Iterative Equations Chaos Again 871 12. 1 cosn nx x+ =

00

1.0

n25

xn

(a) Starting at 0 0.5x = , the function iterates in an oscillatory fashion to an equilibrium ex 0.75.

(b) The equilibrium point(s) ex of this iteration are the root(s) of ( ) cosx f x x= = , or

0.739ex = . (c) Because ( ) ( )sinf x x = yields ( ) ( )0.739 sin 0.739 0.674 1f = = < , 1 cosn nx x+ =

0.739 is an asymptotically stable equilibrium, as the graph shows.

13. 21 0.1n nx x+ = + (a) Starting at 0 0.5x = , the function iterates asymptotically to an equilibrium just above 0.1.

00

0.6

n10

xn

(b) The equilibrium point(s) ex of this iteration are the root(s) of ( ) 2 0.1x f x x= = + , or

2 0.1 0x x + = , which yields 0.113ex = and 0.887ex = . (c) Because ( ) 2f x x = yields

( ) ( )( ) ( )0.113 2 0.113 0.226 1,

0.887 2 0.887 1.774 1,

f

f

= = < = = >

21 0.1n nx x+ = +

0.113 is asymptotically stable and 0.887 is unstable, as the graph shows.

14. 21 0.1n nx x+ =

0.1

0.5

n25

xn

(a) Starting at 0 0.5x = , the function iterates asymptotically to an equilibrium of about 0.1.

(b) The equilibrium point(s) ex of this iteration are the root(s) of ( ) 2 0.1x f x x= = , or

2 0.1 0x x = , which yields 0.092ex = and 1.092ex = . (c) Because ( ) 2f x x = yields 21 0.1n nx x+ = ( ) ( )0.092 2 0.092 0.184 1f = = < , ( ) ( )1.092 2 1.092 2.184 1,f = = > 0.092 is asymptotically stable and 1.092 is unstable, as the graph shows.

872 CHAPTER 9 Discrete Dynamical Systems 15. 21 2n nx x+ =

(a) Starting at 0 0.5x = , the function seems to show chaotic iterative behavior, and extending the orbit does not change that view.

3

3

n25

xn


( ) 2 2x f x x= = , or

2 2 0x x = , which yields 1ex = and 2ex = . (c) Because ( ) 2f x x = yields 21 2n nx x+ = ( ) ( )1 2 1 2 1f = = > , ( ) ( )2 2 2 4 1f = = > , both fixed points are unstable, with no evidence of anything attracting between. This

gives rise to the chaos we see in the time series.

16. 21 1n nx x+ = (a) Starting at 0 0x = , the function cycles between 0 and -1.

(b) The equilibrium point(s) ex of this iteration are the root(s) of ( ) 2 1x f x x= = , or 2 1 0x x = , which yields 1.618ex = and 0.618ex = .

(c) ( ) ( )1.618 2 1.618 1f = > , ( ) ( )0.618 2 0.618 1f = > . Both fixed points are repelling, but an attracting cycle

can be seen that passes between them.

n

nx

200

0.6

-1.2

21 1n nx x+ =

Pete Repeats 17. Pete is using degrees instead of radian measure. To find the equilibrium of the sequence

1 cos 180n

nxx + = , we must solve the equation cos180

xx = . Using Maple yields 0.999848x = , which was the limiting value Pete got.

SECTION 9.3 Linear Iterative Equations Chaos Again 873 Repeat Petes Repeat 18.

00

0.6

n50

xn

1 sinn nx x+ = Starting at 0 0.5x = , the solution appears to be converging ever so slowly to 0. As we saw in Problem 5, for a linear iterative equation, an equilibrium or fixed point ex of ( )1n nx f x+ = is asymptotically stable if ( ) 1ef x < and unstable if

( ) 1ef x > . The only equilibrium point ex of this iteration is the root of ( ) sinx f x x= = or 0ex = . Because ( ) cosf x x = yields

( ) ( )cos 0 1ef x = = , so the derivative test is inconclusive.

time series 21 0.1n nx x+ =

Petes Parameter 19. 1 sinn nx r x+ = An orbit diagram for this function is given in the text as Figure 9.3.13. You will see that the

range of parameter values shown there is 2 r ; we limited our experiments to that region*.

We show just a few of the figures from those experiments, and discuss how they relate to the orbit diagram.

(a) r = 2.65

4-cycle

(b) r = 2.9

chaotic orbit

(c) r = 2.95

3-cycle

The cycles shown match vertical windows in the orbit diagram cited; the chaotic orbit lands in a pretty black vertical band.

* (Apology: the first printing of the text suggested experimentation for 0 r 1 , but as you can verify, those values give nothing but fixed points.)

874 CHAPTER 9 Discrete Dynamical Systems Petes Got It Down Pat 20. 1n nx x+ = A few sample orbits are given in the table, starting at the points 0 0.5x = , 0 1.5x = , and 0 2x = .

=0x 0.5 =0x 1.5 =0x 2 0.5 1.5 2

0.707107 1.224745 1.414214

0.840896 1.106682 1.189207

0.917004 1.05199 1.090508

0.957603 1.025665 1.044274

0.978572 1.012751 1.021897

0.989228 1.006356 1.010889

0.994599 1.003173 1.00543

0.997296 1.001585 1.002711

0.998647 1.000792 1.001355

The cobweb graph shows that the orbit that starts with these initial values approaches one.

The fixed point of this iteration is the single root of x x= , which is 1ex = . Use the derivative test described in Problem 5 to test its

stability. Here, ( ) 12

f xx

= , so ( ) 112

f = < 1.

Therefore, 1ex = is asymptotically stable, as the graph shows.

Iterates of 1n nx x+ =

The Bernoulli Mapping 21.

1

2 0 0.52 1 0.5 1

n nn

n n

x xx

x x+

SECTION 9.3 Linear Iterative Equations Chaos Again 875 Stretch and Fold 22. 1 0.98sinn nx x+ = (a) Start at 0 0.5x = , the next point is ( )1 0.98sin 0.5 0.98x = = , which is the maximum

value for the iterative function. For 0.5nx > , the sine function is decreasing and hits 0 at 1x = . This means that the next points are folded over 0.5 and become values less than

0.5. In other words, points near 0.5 get stretched out to near 1 and then folded back over 0.5. In this case, once the points are less than 0.5, stretching only compresses the points, so they become smaller and smaller and approach 0.

(b) A cobweb graph of siny x= is shown, and we observe the same general shape as ( )1y x x= . Thus, it is expected that the iteration

1 sinn nx r x+ = will exhibit similar properties to

( )1 1n n nx rx x+ = . For r = 0.98, the slope at both fixed points has absolute

value >1, so both fixed points are repelling. Because we find no attracting cyclic behavior between them, the orbit is chaotic.

Iterating 0.98sin x

23. ( )11.8 0 0.51.8 1 0.5 1

n nn

n n

x xx

x x+ get mapped into points ( )1.8 1 x , which being a decreasing function, means the larger points get mapped to points near 0. This can be interpreted as a folding. Points at 0.5x < get mapped to 1.8x, which being an increasing function with slope 1.8, means the points get stretched to the right.

(b) Iterates (100) for tent map


Orbit Diagram of the Tent Mapping 24. It is an easy matter to write a computer program to find the

( )12 0 0.52 1 0.5 1

n nn

n n

rx xx

r x x+ 0.5 THEN GO TO 200

50 REM RSTEPS = # OF R VALUES 180 X = 2*R*X

60 SCREEN 2 190 GOTO 210

70 WINDOW (0, 0.4) - (1, 1) 200 X = 2*R*(1 - X)

80 LET N = 3150 210 IF J < 3000 THEN GOTO 230

90 LET MINR = 0.4 220 PLOT (R, X)

100 LET MAXR = 1.0 230 NEXT J

110 LET RSTEPS = 101 240 NEXT I

120 LET D = (MAXR - MINR)/(RSTEPS - 1) 250 END

130 FOR I = 1 TO RSTEPS

It is easy to change this program to draw the orbit diagram for any other iteration.

To change the range of the parameter r, simply change statements 90 and 100. (Note: For some iterations the parameter is not called r. It is suggested that you simply call it R and not change the program to a new name.)

To change the iteration function simply replace the statements 170, 180, 190, and 200 by the new iteration function. (Note: Most iteration functions will only take one line; the tent mapping is a conditional function that requires more than one line.)

Line 140 gives the initial condition, which can be changed. (For the tent mapping, the diagram comes more quickly if 0x = 0.)

In line 210, J < # transient points to skip in the plotting. This number is adjustable. (For some functions, including the tent mapping, the number of transients to skip must be large to avoid extraneous patterns.)

SECTION 9.3 Linear Iterative Equations Chaos Again 877 25.

Chaotic Numerical Iterations

Using Newtons formula to approximate a real root of

( ) 3 1 0f x x rx= + + = yields the formula

3

1 2

13

n nn n

n

x rxx xx r++ += + . Orbit diagram for

Newtons method

We adapt the BASIC program in Problem 24 by replacing statements 170200 by the single line

^ 3 13 ^ 2

X R XX XX R+ += +

and changing the range of the parameter r and the window bounds to generate the orbit diagram shown in the figure. The resulting computer program is given in the Table below.

Note that the orbit diagram exhibits either cyclic or chaotic behavior, depending on r. This shares features of orbit diagrams for other functions studied. Furthermore, the orbit diagram looks like distorted copies of the logistic bifurcation diagram, stretched horizontally and shrunk, but different factors, in the vertical direction.

BASIC Program to Compute Orbit Diagrams 10 REM ORBIT DIAGRAM 140 LET X = 0.5

20 REM N = ITERATIONS FOR EACH R 150 LET R = MINR + (I 1)*D

30 REM MINR = MINIMUM R 160 FOR J = 1 TO N

40 REM MAXR = MAXIMUM R 170 LET X = X (X^3 + R

* X + 1) / (3 * X^2 + R)

50 REM RSTEPS = # OF R VALUES 180 IF J < 1000 THEN 200

60 SCREEN 2 190 PLOT (R, X)

70 WINDOW (1.3, 0.2) (1.25, 1) 200 NEXT J

80 LET N = 3200 210 NEXT I

90 LET MINR = 1.30 220 END

100 LET MAXR = 1.25 230

110 LET RSTEPS = 101 240

120 LET D = (MAXR - MINR)/(RSTEPS - 1) 250


Extremum Problem 26. The values of the function ( ) ( )1f x rx x= are greater or equal to zero in the interval [ ]0, 1 . It

describes an upside-down parabola that crosses the x-axis at 0x = , 1 (thus attains a maximum value of ( )0.5

4rf = at 0.5x = , ( )0 f x r< < . If ( )f x is to remain in the interval [ ]0, 1 , then

max 4r = . Hence, ( )0 4f x r < .

878 CHAPTER 9 Discrete Dynamical Systems Sequential Analysis 27. (a) Because 1nx + is a monotone decreasing sequence bounded below by 0 we know that it

converges to its greatest lower bound L. Also the sequence 1nx + is always decreasing because if 0 1r , then ( )1 1nr x < . Multiplying by nx yields ( )1n n nrx x x < or, in other words, 1n nx x+ < .

(b) We have

( )1 1 0n n nx rx x+ = , 1, 2, n = for 00 1x and 0 1r , hence the greatest lower bound L of the sequence is greater than or equal to zero also.

(c) The greatest lower bound L of the monotone decreasing sequence { }1nx + is 0 inasmuch as it will eventually lie below any positive number. Hence 0L = .

Matter of Size 28. (a) We have a fixed point x of

( )1 1n n nx rx x+ = when ( )1x rx x = . Solving for x yields 1 11 rx

r r = = .

(b) Computing the derivative of

( ) ( )1f x rx x= yields ( ) ( )1 2f x r x = . Evaluating the derivative at 1rx

r = we find

( ) ( ) 1 2 21 2 1 2 2r r rf x r x r r rr r + = = = = . Hence for 1 3r< we have ( ) 1f x . If the slope at a fixed point has absolute value less than one, then it is stable.

SECTION 9.3 Linear Iterative Equations Chaos Again 879 Not Quite a Two-Cycle 29. 3r = (a) ( )1 3 1n n nx x x+ = The time series for the first 20 iterates looks very much

like it settles into a two-cycle

n

nx

200

0.8

0

(b) ( )f x has fixed points 20, 3e

x = . The first at 0 is repelling, but at 23

the derivative test

2 13

f = is inconclusive because 2 13

f = . The second iterate function ( )( )f f x has only two fixed points, 0 and 2

3 (the root 2

3 has multiplicity 3), so there is not in fact

a 2-cycle. (c) The lack of a 2-cycle means the time series in part (a) is in

fact converging, extremely slowly, to the fixed point 23

,

between the highs and lows of the orbit shown in part (a).

(d) The extremely slow convergence to the fixed point 23

causes the cobweb to appear solid black in the vicinity.

Finding Two-Cycle Values 30. If ( ) ( )1f x rx x= , then ( )( ) ( ) ( ) ( ) ( )1 1 1 1f f x rf x f x r rx x rx x = = .

If we set this value to x we get

( ) ( )3.2 3.2 1 1 3.2 1x x x x x = . This fourth-degree equation was solved with Maple, yielding the four (approximate) roots 0, 0.513, 0.687, and 0.799. The four roots can been seen in the figure as the intersection of the curves

( )( )y f f x= and y x= . Graph of ( )( )y f f x=

Of the four fixed points for the second iterative function ( )( )y f f x= , 0 and 0.687 are unstable and 0.513 and 0.799 are stable. The first two are the fixed points (repelling) of ( )f x . The latter two are cyclic points (attracting) of period 2

for ( )f x .

880 CHAPTER 9 Discrete Dynamical Systems Four-Cycle Values 31. (a) Using Maple with parameter 3.2r = , find the 16th order

polynomial for the fourth iterate function:

( )( )( )( ) ( )( )( )( ) ( )( )( )( ) ( )( )(

( ) ( )( )( ))

104.8576 1

1 3.2 1

1 10.24 1 1 3.2 1

1 32.768 1 1 3.2 1

1 10.24 1 1 3.2 1

f f f f x x x

x x

x x x x

x x x x

x x x x

=

The graph is on the interval [ ]0, 1 .

Graph of ( )( )( )( )f f f f x for 3.2r =

Solving the equation ( )( )( )( )f f f f x x= yields four real roots (to three places) 0, 0.513, 0.687, and 0.799. We see from the graph that the absolute value of the slope of

( )( )( )( )f f f f x is greater than 1 at 0 and 0.687 and less than 1 at 0.513 and 0.799. Thus, 0.513 and 0.799 cycle with period four. But because there are only two such points, the cycle is actually period two, as seen in Problem 30. That is, the iterates of

( )1 3.2 1n n nx x x+ = will tend to oscillate between these two values.

With 3.5r = we get

( )( )( )( ) ( )( )( )( ) ( )( )( )( ) ( )( )(

( ) ( )( )( ))

150.0625 1

1 3.5 1

1 12.25 1 1 3.5 1

1 42.875 1 1 3.5 1

1 12.25 1 1 3.5 1

f f f f x x x

x x

x x x x

x x x x

x x x x

=

This graph is on the interval [ ]0, 1 . Note: As should be expected, between every pair of attracting fixed points there is a repelling fixed point, and vice versa.

Graph of ( )( )( )( )f f f f x for 3.5r =

Continued on next page.

SECTION 9.3 Linear Iterative Equations Chaos Again 881

Continued from previous page.

(b) Solving ( )( )( )( )f f f f x x= for 3.5r = yields eight real (approximate) roots 0, 0.383, 0.428, 0.501, 0.714, 0.827, 0.857, and 0.874. From close inspection of the graph, we see that the absolute value of ( )( )( )( )f f f f x is greater than 1 at 0, 0.428, 0.714, and 0.874, whereas the absolute value is less than 1 at 0.383, 0.501, 0.827, and 0.874. Hence, the points 0.383, 0.827, 0.501, and 0.874 form a four-cycle.

The following table shows the values (from Excel with 6 decimal places) for iterates 92 through 99 in the sequence ( )1 3.5 1n n nx x x+ = , starting at 0 0.3x = to illustrate two periods of this four-cycle.

n nx n nx

92 0.38282 96 0.38282

93 0.826941 97 0.826941

94 0.500884 98 0.500884

95 0.874997 99 0.874997

r = 3.2 2-cycle

r = 3.5 4-cycle

882 CHAPTER 9 Discrete Dynamical Systems Role of the nth Iterates

r values n-cycles

3.628 6-cycle

3.702 7-cycle

3.74 5-cycle

3.88615 7-cycle

3.8995 8-cycle

3.9057 5-cycle

3.91205 8-cycle

32. (a) For the logistic function ( )1rx x the given r-values produce the n-cycles listed in the table in order of increasing r. We observe that the n-values do not occur in any such order.

The n-cycles are all illustrated by cobweb diagrams in part (b).

NOTE: All r-values were chosen so that x = 0.5 is on the cycle. This makes it easy to find the cycle without transients.

3.9605 4-cycle

(b) Each row of figures shows the iteration n-cycle for ( )1rx x , then adds the nth iterate graph, which meets the diagonal with a shallow slope at the n-cycle points.

r=3.628 6-cycle

r=3.702 7-cycle

r=3.74 5-cycle


SECTION 9.3 Linear Iterative Equations Chaos Again 883 Continued from previous page.

(b)

r=3.9057

5-cycle

r=3.9605

4-cycle

For the other given r-values we show here only the n-cycle cobwebs because the nth iterate graphs have too much detail to decipher on this small scale.

r = 3.88615 7-cycle

r = 3.8995 8-cycle

r = 3.91205 8-cycle

884 CHAPTER 9 Discrete Dynamical Systems Windows in the Orbit Diagram 33. (a) The first 15 computed values of the iteration ( )1 1n n nx r x x+ = with r = 3.84 are shown

in the table. Note the three-cycle of approximately 0.959, 0.148, and 0.485, which fits with the cobweb diagram shown.

n nx n nx

0 0.5 8 0.14801 1 0.96 9 0.484237 2 0.147456 10 0.959046 3 0.482737 11 0.150823 4 0.958856 12 0.491811 5 0.151494 13 0.959742 6 0.493607 14 0.148366 7 0.959843 15 0.485196

Cobweb diagram showing three-cycle ( )3.84r =

(b) (i) When 3.628r = the solution approaches a six-cycle with values 0.306, 0.771, 0.641, 0.834, 0.501, as shown in the cobweb diagram.

Cobweb diagram showing six-cycle ( )3.628r =

When 3.85r = the solution approaches a six-cycle with values 0.155, 0.506, 0.962, 0.139, 0.462, and 0.956. as shown in the cobweb diagram.

Cobweb diagram showing six-cycle ( )3.85r =

(ii) The cobweb diagram for 3.85r = illustrates the period doubling phenoenon tearing apart from the 3-cycle for 3.84r = . Compare with figure in part (a), the cobweb diagram for 3.628r = shows an entirely different six-cycle behavior.

Suppose we start an orbit at the lower left most cycle point, and label the points along the diagonal in the order they are visited. For 3.628r = , the visitation labels read from the left along the left line.


SECTION 9.3 Linear Iterative Equations Chaos Again 885 Continued from previous page.

(c) When 3.74r = the solution approaches a five-cycle with values 0.935, 0.227, 0.657, 0.842, and 0.496, as shown in the cobweb diagram.

Another 5-cycle occurs for 3.9057r = . See the solution for Problem 32 to compare the cobwebs for these two cycles.

Cobweb diagram showing five-cycle ( )3.74r =

The Square-and-Add System 34. For each equation in (a), (b), and (c) the first 15 values have been computed, using EXCEL. Note

from the table and figure that the numbers of the first sequence (series 1) settle into a cycle of period four, the second series is periodic with period 3, and the third sequence settles into the constant value of 2.

+ = 2n 1 nx x a (a) (b) (c) =a 1.3 =a 1.755 =a 2 =a 1.3 =a 1.755 =a 2 n (series 1) (series 2) (series 3) n (series 1) (series 2) (series 3)

0 0.00000 0.00000 0 8 0.01913 1.32502 2

1 1.30000 1.75500 2 9 1.29963 0.00069 2

2 0.39000 1.32503 2 10 0.38905 1.75500 2

3 1.14790 0.00069 2 11 1.14864 1.32502 2

4 0.01767 1.75500 2 12 0.01938 0.00069 2

5 1.29969 1.32502 2 13 1.29962 1.75500 2

6 0.38919 0.00069 2 14 0.38902 1.32502 2

7 1.14853 1.75500 2 15 1.14866 0.00069 2

3

3

xn (series 1)(series 2)(series 3)

a = 2 constant solutiona = 1 755 3 - cycle.

a = 1 3 4 -cycle.n

Quadruply periodic, triply periodic, and constant solution for different c values

886 CHAPTER 9 Discrete Dynamical Systems 35. To draw the orbit diagram for 2x c+ , we adapt the BASIC

program in Problem 25, continuing to use R for the parameter c.

We replace statement 170 by the single line

^ 2X X R= + ; we also change the window bounds and the range of the parameter r to generate the orbit diagram shown in the figure. The resulting computer program is given in the Table below.

Note the bifurcation points at c = -1.0 , -1.5, -1.75, and so on. Orbit diagram for 21n nx x c+ = +

BASIC Program to Compute Orbit Diagrams

10 REM ORBIT DIAGRAM 140 LET X = 0.5

20 REM N = ITERATIONS FOR EACH R 150 LET R = MINR + (I 1)*D

30 REM MINR = MINIMUM R 160 FOR J = 1 TO N

40 REM MAXR = MAXIMUM R 170 LET X = X^2 + R

* X + 1) / (3 * X^2 + R)

50 REM RSTEPS = # OF R VALUES 180 IF J < 1000 THEN 200

60 SCREEN 2 190 PLOT (R, X)

70 WINDOW ( 2, 2) ( 0.5, 2) 200 NEXT J

80 LET N = 1300 210 NEXT I

90 LET MINR = 2 220 END

100 LET MAXR = 0.5 230

110 LET RSTEPS = 101 240

120 LET D = (MAXR - MINR)/(RSTEPS - 1) 250


Class Project: The Real Bifurcation Diagram 36. This problem is left as a project for a group of students. The larger t

Chapter 9 diff eq sm

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