Use Euler' s relation and the corresponding expression for

Consider the nonlinear first-order equation 1 = 2V 1. (a) By separating variables, find a solution xi(t). (b) Your solution should contain one constant of integration k, so you might reasonably expect it to be the general solution. Show, however, that there is another solution, x2(t) = 1, that is not of the form of xi (t) whatever the value of k. (c) Show that although xi (t) and x2(t) are solutions, neither Axi(t), nor Bx2(t), nor x1(t) x2(t) are solutions. (That is, the superposition principle does not apply to this equation.

Here is a different example of the disagreeable things that can happen with nonlinear equations. Consider the nonlinear equation z = 21i. Since this is first-order, one would expect that specification of x (0) would determine a unique solution. Show that for this equation there are two different solutions, both satisfying the initial condition x(0) = 0. [Hint: Find one solution xi (t) by separating variables, but note that x2(t) = 0 is another. Fortunately none of the equations normally encountered in classical mechanics suffer from this disagreeable ambiguity.]

Consider a second-order linear homogeneous equation of the form (12.6). (a) Write out a detailed proof of the superposition principle, that if x JO and x2(t) are solutions of this equation, then so is any linear combination aix a2x2(t), where al and a2 are any two constants. (b) Consider now the nonlinear equation in which the third term of (12.6) is replaced by r (t),Vx (t). Explain clearly why the superposition principle does not hold for this equation.

Consider an inhomogeneous second-order linear equation of the form p(t)x(t) q(t).i(t) r(t)x(t) = f (t). (12.59) Let x (t) denote a solution (a "particular" solution) of this equation and prove that any solution x(t) can be written as x(t) = xp(t) aixi(t) a2x2(t) (12.60) where x1(t) and x2(t) are two independent solutions of the corresponding homogeneous equationthat is, (12.59) with f (t) deleted. [Hint: Write down the equations for x(t) and xp(t) and subtract.] This result shows that to find all solutions of (12.59), we have only to find one particular solution and two independent solutions of the corresponding homogeneous equation. (b) Explain clearly why the result you proved in part (a) is not, in general, true for a nonlinear equation such as p(t).(t) q(t)x(t) r(t)Ax(t) = f (t).

Use Euler' s relation and the corresponding expression for cos 4) (inside the front cover) to prove the identity (12.15).

[Computer] (a) Use appropriate sofware to solve the equation (12.11) numerically, for a DDP with the following parameters: drive strength y = 0.9, drive frequency w = 2n-, natural frequency coo = 1.5o), damping constant p = 0)014, and initial conditions (0) = 4(0) = 0. Solve the equation and plot your solution for six cycles, 0 < t < 6, and verify that you get the result shown in Figure 12.3. (b) and (c) Solve the same equation twice more with the two different initial conditions 4)(0) = +7t/2 both with ..(0) = 0 and plot all three solutions on the same picture. Do your results bear out the claim that, for this drive strength, all solutions (whatever their initial conditions) approach the same periodic attractor?

Computer] Do all the same calculations as in Problem 12.6 but with a drive strength y = 1.06 and for 0 < t < 10. In part (a) verify that your results agree with Figure 12.4. Do your results suggest that there is still a unique attractor to which all solutions (whatever their initial conditions) converge? (At first sight the answer may appear to be "No," but remember that values of 0 that differ by 27r should be considered to be the same.)

Computer] Use a computer to find a numerical solution of the equation of motion (12.11) for a DDP with the following parameters: drive strength y = 1.073, drive frequency w = 27r, natural frequency wo = 1.5w, damping constant fi = wo/4, and initial conditions 0 (0) = 7r/2 and Q)(0) = 0. (a) Solve for 0 < t < 50, and then plot the first ten cycles, 0 < t < 10. (b) To be sure that the initial transients have died out, plot the ten cycles 40 < t < 50. What is the period of the long-term motion (the attractor)?

[Computer] Do the same calculations as in Problem 12.8 with all the same parameters except that 0 (0) = 0. Plot the first 30 cycles 0 < t < 30 and check that you agree with Figure 12.5. Plot the ten cycles 40 < t < 50 and find the period of the long-term motion.

Computer] Explore the behavior of the DDP with the same parameters as in Problem 12.8, but with several different initial conditions. For example, you might keep 4(0) = 0 but try various different values for 0 (0) between n- and 7r. You will find that the initial behavior varies quite a lot according to the initial conditions, but the long-term motion is the same in all cases (as long as you remember that values of 0 that differ by multiples of 27r represent the same position).

Test how well the values of the thresholds yn given in Table 12.1 fit the Feigenbaum relation (12.17) as follows: (a) Assuming that the Feigenbaum relation is exactly true use it to prove that (Yn+1 Yn) = (1/6)n-1(Y2 Y1) and, hence, that a plot of lnly s n+1 yn) against n should be a straight line with slope In 8. (b) Make this plot for the three differences of Table 12.1. How well do your points seem to bear out our prediction? Fit a line to your plot (either graphically or using a least squares fit) and find the slope and hence the Feigenbaum number 6. [You would not expect to get very good agreement with the known value (12.18) for two reasons: You have only three points to plot and the Feigenbaum relation (12.17) is only approximate, except in the limit of large n. Under the circumstances, you will find the agreement is remarkable.]

Here is another way to look at the Feigenbaum relation (12.17): (a) Assuming that (12.17) is exactly true, prove that the thresholds )6, approach a finite limit y, and then prove that yp, = y, where K is a constant. This means that a plot of yn against 6' should be a straight line. (b) Using the known value (12.18) of 8 and the four values of yn in Table 12.1 make this plot. Does it seem to fit our prediction? The vertical intercept of your graph should be yc. What is your value and how well does it agree with (12.20)?

You can see in Figure 12.13 that for y = 1.105, the separation of two identical pendulums with slightly different initial conditions increases exponentially. Specifically AO starts out at 10-4 and by t = 14.5 it has reached about 1. Use this to estimate the Liapunov exponent X as defined in (12.26), AO (t) K eAt . Your answer should confirm that A > 0 for chaotic motion.

[Computer] Numerically solve the equation of motion (12.11) for a DDP with drive strength y = 1.084, and the following other parameters: drive frequency w = 27r, natural frequency wo = 1.5w, damping constant 13 = wo/4, and initial conditions 0 (0) = 0(0) = 0. Solve for the first seven drive cycles (0 < t < 7) and call your solution 01(t) . Solve again for all the same parameters except that 0 (0) = 0.00001 and call this solution 02(t) . Let A0 (t) 02(t) 01(0 and make a plot of log I AO (t)i against t. With this drive strength the motion is chaotic. Does your plot confirm this? In what sense?

[Computer] Numerically solve the equation of motion (12.11) for a DDP with the following parameters: drive strength y = 0.3, drive frequency w = 27, natural frequency wo = 1.5w, damping constant 18 = woI4, and initial conditions 0 (0) = (i)(0) = 0. Solve for the first five drive cycles (0 < t < 5) and call your solution 01(t). Solve again for all the same parameters except that 0 (0) = 1 (that is, the initial angle is one radian) and call this solution 02(t) . Let AO (t) = 02(t) - 01(t) and make a plot of log I AO (t)I against t. Does your plot confirm that A0 (t) goes to zero exponentially? Note: The exponential decay continues indefinitely, but A0 (t) eventually gets so small that it is smaller than the rounding errors, and the exponential decay cannot be seen. If you want to go further, you will probably need to crank up your precision.

Consider the chaotic motion of a DDP for which the Liapunov exponent is A = 1, with time measured in units of the drive period as usual. (This is very roughly the value found in Problem 12.13.) (a) Suppose that you need to predict 0(t) with an accuracy of 1/100 rad and that you know the initial value 0 (0) within 10-6 rad. What is the maximum time tmax for which you can predict 0 (t) within the required accuracy? This tmax is sometimes called the time horizon for prediction within a specified accuracy. (b) Suppose that, with a vast expenditure of money and labor, you manage to improve the accuracy of your initial value to 10-9 radians (a thousand-fold improvement). What is the time horizon now (for the same required accuracy of prediction)? By what factor has tmax improved? Your results illustrate the difficulty of making accurate long-term predictions for chaotic motion.

Computer] In Figure 12.15, you can see that for y = 1.503 the DDP "tries" to execute a steady rolling motion changing by 27 once each cycle, but that there is superposed an erratic wobbling and that the direction of the rolling reverses itself from time to time. For other values of y, the pendulum actually does approach a steady, periodic rolling. (a) Solve the equation of motion (12.11) for a drive strength y = 1.3 and all other parameters as in the first part of Problem 12.14, for 0 < t < 8. Call your solution 01(t) and plot it as a function of t. Describe the motion. (b) It is hard to be sure that the motion is periodic based on this graph, because of the steady rolling through -27 each cycle. As a better check, plot 01(t) 2nt against time. Describe what this shows. This kind of periodic rolling motion is sometimes described as phase-locked.

[Computer] Since the rolling motion of Problem 12.17 is periodic (and hence not chaotic) we would expect the difference AO (t) between neighboring solutions (solutions of the same equation, but with slightly different initial conditions) to decrease exponentially. To illustrate this, do part (a) of Problem 12.17 and then find the solution of the same problem except that 0 (0) = 1. Call this second solution 02(0 and let AO (t) = 02(t) - 01(t) . Make a plot of log I AO (t)I against t and comment.

Consider an undamped, undriven simple harmonic oscillator a mass m on the end of a spring whose force constant is k. (a) Write down the general solution x(t) for the position as a function of time t. Use this to sketch the state-space orbit, showing the motion of the point [x(t), .i(t)] in the two-dimensional state space with coordinates (x, Explain the direction in which the orbit is traced as time advances. (b) Write down the total energy of the system and use conservation of energy to prove that the state- space orbit is an ellipse.

Consider a weakly-damped, but undriven oscillator as described by Equation (5.28). The general motion is given by (5.38). (a) Use this to sketch the state-space orbit, showing the movement of (x, 5c) for 0 < t < 10 with the parameters b = 0, col = 27r, and ,8 = 0.5. (b) This system has a unique stationary attractor. What is it? Explain it, both with reference to your sketch and in terms of energy conservation.

Here is an iterated map that is easily studied with the help of your calculator: Let xt+1 = f (xt) where f (x) = cos(x). If you choose any value for xo, you can find x1, x2, x3, by simply pressing the cosine button on your calculator over and over again. (Be sure the calculator is in radians mode.) (a) Try this for several different choices of xo, finding the first 30 or so values of xt. Describe what happens. (b) You should have found that there seems to be a single fixed attractor. What is it? Explain it, by examining (graphically, for instance) the equation for a fixed point f (x*) = x* and applying our test for stability [namely, that a fixed point x* is stable if I f (x*)1 < 1].

Consider the iterated map xt+i = f (xt) where f (x) = x2. (a) Show that it has exactly two fixed points of which just one is stable. What are they? (b) Show that xt approaches the stable fixed point if and only if 1 < xo < 1. The interval 1 < xo < 1 is called a basin of attraction since all sequences xo, x1, that start in the "basin" are attracted to the same attractor. (c) Show that xt co if and only if Ixol > 1. (Thus we could say the map has a second stable fixed point at x = oo and the basin of attraction for this fixed point is the set Ixo > 1.) For chaotic systems, the basins of attraction can be much more complicated than these examples and are often fractals.

[Computer] Consider the sine map xti = f (xt) where f (x) = r sin(rrx). The interesting behavior of this map is for 0 < x < 1 and 0 < r < 1, so restrict your attention to these ranges. (a) Using a plot analogous to Figure 12.35, discuss the fixed points of this map. Show that the map has either one or two fixed points, depending on the value of r. Show that when r is small there is just one fixed point, which is stable. (b) At what value of r (call it ro) does the second fixed point appear? Show that ro is also the value of r at which the first fixed point becomes unstable. (c) As r increases, the second fixed point eventually becomes unstable. Find numerically the value ri at which this occurs.

[Computer] Consider the sine map of Problem 12.23. Using a progammable calculator (or a computer) you can easily find the first ten or twenty values of xt for any chosen inital value xo. Taking xo = 0.3, calculate the first 10 values of xt for each of the following values of the parameter r: (a) r = 0.1, (b) r = 0.5, (c) r = 0.78. In each case plot your results (xt against t) and describe the long-term attractor. If you did Problem 12.23, are your results here consistent with what you proved there?

[Computer] The sine map of Problem 12.23 exhibits period-doubling cascades just like the logistic map. To illustrate this, take xo = 0.8 and find the first twenty values xt for each of the following values of the parameter r: (a) r = 0.60, (b) r = 0.79, (c) r = 0.85, and (d) r = 0.865. Plot your results (as four separate plots) and comment.

The appearance of a two-cycle of the logistic map at the exact moment when the one-cycle becomes unstable follows directly from the behavior of the graphs of f (x) and g(x) = f (f (x)) as shown in Figures 12.38 and 12.39. The crucial point is that the function f (x) is a simple arch that gets steadily higher as we increase the control parameter r from 0; at the same time, f (f (x)) starts out as a simple arch which is lower than f (x), but developes two maxima [higher than f (x)] with a minimum [lower than f (x)] in between. Explain clearly in words why this behavior off (f (x)) follows inevitably from that of f (x). [Hint: Since f (x) is symmetric about x = 0.5, you need only consider its behavior as x runs from 0 to 0.5. The advantage of this argument is that it applies to any map of the form f (x) = r0 (x) where (/)(x) has a single symmetric arch (such as the sine map of Problem 12.23).]

The two-cycles of the map f (x) correspond to fixed points of the second-iterate g(x) = f (f (x)). Thus the two values shown as xa and xb in Figure 12.37 are roots of the equation f (f (x)) = x. In the case of the logistic map this is a quartic equation, which is not too hard to solve: (a) Verify that for the logistic function f (x) x f(f(x))=rxlx r 1) [r2X2 r (r 1)x r l]. (12.61) Thus the fixed-point equation x f (f (x)) = 0 has four roots. The first two are x = 0 and x = (r 1)/r. Explain these and show that the other two roots are r + 1 ,Ar 1)(r 3) Xa, xb 2r (12.62) Explain how you know that these are the two points of a two-cycle. (b) Show that for r < 3 these roots are complex and hence that there is no real two- cycle. (c) For r > 3 these two roots are real and there is a real two-cycle. Find the values of xa and xb for the case r = 3.2 and verify the values shown in Figure 12.37.

Equation (12.62) in Problem 12.27 gives the two fixed points of the two-cycle of the logistic map. One can observe this two-cycle only if it is stable, which will happen if I g' (x)I < 1, where g(x) is the double map g(x) = f (f (x)) and x is either of the values xa or xb. (a) Combine (12.62) with Equation (12.57) to find g' (xa). [Notice that because (12.57) is symmetric in xa and xb, you will get the same result whether you use xa or xb; that is, the two points necessarily become stable or unstable at the same time.] (b) Show that the two-cycle is stable for 3 < r < 1 + Ni6-. This establishes that the threshold at which period 2 is replaced by period 4 is r2 = 1 + = 3.449.

[Computer] The thresholds rn for period doubling of the logistic map are given by Equation (12.58). These should satisfy the Feigenbaum relation (12.17), at least in the limit that n cc (with y replaced by r, of course). Test this claim as follows: (a) If you have not done Problem 12.11 prove that the Feigenbaum relation (if exactly true) implies that (r,1 rn) = K / 6n . (b) Make a plot of ln(ri rn) against n. Find the best-fit straight line to the data and from its slope predict the Feigenbaum constant. How does your answer compare with the accepted value 6 = 4.67?

[Computer] The chaotic evolution of the logistic map shows the same sensitivity to initial conditions that we met in the DDP. To illustrate this do the following: (a) Using a growth rate r = 2.6, calculate xt for 1 < t < 40 starting from xo = 0.4. Repeat but with the inital condition 4 = 0.5 (the prime is just to distinguish this second solution from the first it does not denote differentiation) and then plot log Ix; xt I against t. Describe the behavior of the difference x; xt. (b) Repeat part (a) but with r = 3.3. In this case, the long term evolution has period 2. Again describe the behavior of the difference x; xt. (c) Repeat parts (a) and (b), but with r = 3.6. In this case, the evolution is chaotic, and we expect the difference to grow exponentially; therefore, it is more interesting to take the two inital values much closer together. To be definite take xo = 0.4 and xo = 0.400001. How does the difference behave?

[Computer] When the evolution of the logistic map is non-chaotic, two solutions with the same r that start out sufficiently close, will converge exponentially. (This was illustrated in Problem 12.30.) This does not mean that any two solutions with the same r will converge. (a) Repeat Problem 12.30(a), with all the same parameters except r = 3.5 (a value for which we know the long-term motion has period 4). Does x; xt approach zero? (b) Now do the same exercise but with the initial conditions x0 = 0.45 and xo = 0.5. Comment. Can you explain why, if the period is greater than 1, it is impossible that the difference x; xt go to zero for all choices of initial conditions?

[Computer] Make a bifurcation diagram for the logistic map, in the style of Figure 12.41 but for the range 0 < r < 3.55. Take x0 = 0.1. Comment on its main features. [Hint: Start by using a very small number of points, perhaps just r going from 0 to 3.5 in steps of 0.5 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 0 to 3.55 in steps of 0.025, and t from 51 to 60, perhaps), and you will certainly need to automate the calculation of the large number of points.]

[Computer] Reproduce the logistic bifurcation diagram of Figure 12.41 for the range 2.8 < r < 4. Take x0 = 0.1. [Hint: To make Figure 12.41 I used about 50,000 points, but you cer-tainly don't need to use that many. In any case, start by using a very small number of points, perhaps just r going from 2.8 to 3.4 in steps of 0.2 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 2.8 to 4 in steps of 0.025, and t from 500 to 600, perhaps), and you will certainly need to automate the calculation of the large number of points.]

[Computer] Make a bifurcation diagram for the sine map of Problems 12.23 and 12.25. This should resemble Figure 12.41 but for the range 0.6 < r < 1. Take x0 = 0.1. Comment on its main features. [Hint: Start by using a very small number of points, perhaps just r going from 0.6 to 0.8 in steps of 0.05 and t going from 51 to 54. This will let you calculate for each of the values of r individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points (r going from 0.6 to 1 in steps of 0.005, and t from 400 to 500, perhaps), and you will certainly need to automate the calculation of the large number of points.]