Nonlinear Oscillations and Chaos

CHAPTER 4 Nonlinear Oscillations and Chaos 4-. l l = l + d s d d l l = l + d m θ m (a) (b) (c) The unetended length of each spring is, as shown in (a). In order to attach the mass m, each spring must be stretched a distance d, as indicated in (b). When the mass is moved a distance, as in (c), the force acting on the mass (neglecting gravity) is where ( ) F = k s sinθ and Then, s = + sin θ = () + k k F ( ) = + = + ( d) + + (3) d d = k = k + (4) + 7

8 CHAPTER 4 Epanding the radical in powers of and retaining only the first two terms, we have d F ( ) k The potential is given by so that d d = k + kd k = ( ) U ( d) 3 ( ) ( ) kd 3 = F d ( d) 4 U = + 3 k 4 (5) (6) (7) 4-. Using the general procedure eplained in Section 4.3, the phase diagram is constructed as follows: U() E E E E 3 4 E E 5 6 E E E 4 E 5 E 6 E 3

NONLINEAR OSCILLATIONS AND CHAOS 9 4-3. The potential ( ) ( ) 3 diagram is given in (b): (a) U = λ 3 has the form shown in (a) below. The corresponding phase U() E 5 E 4 E 3 E E E E (b) 4-4. Differentiation of Rayleigh s equation above yields The substitution, ( 3 ) a b +ω = implies that 3b y= y () a a y = 3by a y = 3by a y = 3by (3) When these epressions are substituted in, we find a y a 3ba y y a y a + ω = (4) 3by 3b b y y 3by Multiplying by y 3b a and rearranging, we arrive at van der Pol s equation: a y ( y y ) y + ωy= (5) y

3 CHAPTER 4 4-5. f = tan in the region π shows that there is an intersection (i.e., a solution) for 3π 8. a) A graph of the functions f ( ) = + + and ( ) tan + + O π The procedure is to use this approimate solution as a starting point and to substitute = 3π 8 into f ( ) and then solve for tan f ( ) say 4, of =. If the result is within some specified amount, 3π 8, then this is our solution. If the result is not within this amount of the starting value, then use the result as a new starting point and repeat the calculation. This procedure leads to the following values: f ( ) = + + tan f ( ) Difference.78 3.566.974.93.974 3.986.347.78.347 4.794.334.573.334 4.4.336.8.336 4.47.338.4.338 4.56.339.5 Thus, the solution is =.339. Parts b) and c) are solved in eactly the same way with the results: b) =.95 c) =.97 4-6. For the plane pendulum, the potential energy is u= mg cosθ If the total energy is larger than mg, all values of θ are allowed, and the pendulum revolves continuously in a circular path. The potential energy as a function of θ is shown in (a) below. U mgl mgl Since T = E U(θ), we can write π π O π (a) π θ

NONLINEAR OSCILLATIONS AND CHAOS 3 T = mv = m θ = E mg ( cosθ ) () and, therefore, the phase paths are constructed by plotting θ = E mg m versus θ. The phase diagram is shown in (b) below. E = mgl θ ( cosθ) E = 3mgl E= 3 mgl E= 5 mgl (3) E = mgl π π π π θ (b) 4-7. Let us start with the equation of motion for the simple pendulum: θ θ = ω sin where ω g. Put this in terms of the horizontal component by setting y = sin θ. Solving for θ and taking time derivatives, we obtain yy y θ = + ( y ) y 3 () Since we are keeping terms to third order, we need to get a better handle on the y term. Help comes from the conservation of energy: θ θ cos θ (3) m mg cos = mg where θ is the maimum angle the pendulum makes, and serves as a convenient parameter that describes the total energy. When written in terms of y, the above equation becomes (with the obvious definition for ) y ( y ) y = ω y Substituting (4) into (), and the result into gives y (4)

3 CHAPTER 4 ( ) = (5) y+ ωy 3 y y Using the binomial epansion of the square roots and keeping terms up to third order, we can obtain for the equation of motion 3g 3 + ω + = 3 (6) 4-8. For >, the equation of motion is If the initial conditions are ( ) For the phase path we need = ( ) m = F = A, ( ) =, the solution is F t () = A t () m, so we calculate F ( ) = ± ( A ) (3) m Thus, the phase path is a parabola with a verte on the -ais at = A and symmetrical about both aes as shown below. A t = t = 4 τ Because of the symmetry, the period τ is equal to 4 times the time required to move from = A to = (see diagram). Therefore, from () we have ma τ = 4 (4) F 4-9. The proposed force derives from a potential of the form k < a U( ) = ( + δ) δa > a which is plotted in (a) below.

NONLINEAR OSCILLATIONS AND CHAOS 33 U() E 3 E E 7 E 6 E 5 E 4 E a O (a) a For small deviations from the equilibrium position ( = ), the motion is just that of a harmonic oscillator. For energies E< E 6, the particle cannot reach regions with < a, but it can reach regions of > a if E> E 4. For E < E< E4 the possibility eists that the particle can be trapped near = a. A phase diagram for the system is shown in (b) below. E E3 E 4 E 7 E (b) 4-. The system of equations that we need to solve are y y =.5 y sin +.7 cos ω t The values of ω that give chaotic orbits are.6 and.7. Although we may appear to have chaos for other values, construction of a Poincaré plot that samples at the forcing frequency show that they all settle on a one period per drive cycle orbit. This occurs faster for some values of ω than others. In particular, when ω =.8 the plot looks chaotic until it locks on to the point (.55,.36439). The phase plot for ω =.3 shown in the figure was produced by numerical integration of the system of equations with points per drive cycle. The bo encloses the point on the trajectory of the system at the start of a drive cycle. In addition, we also show Poincaré plot for the case ω =.6 in figure, integrated over 8 drive cycles with points per cycle.

34 CHAPTER 4.5 -.5.5 -.5.5.5 3 3 4 3 3 4 4-. The three-cycle does indeed occur where indicated in the problem, and does turn chaotic near the 8th iteration. This value is approimate, however, and depends on the precision at which the calculations are performed. The behavior returns to a three-cycle near the th iteration, and stays that way until approimately the 7th iteration, although some may see it continue past the 3th..8.6.4. 3 4 5 iteration

NONLINEAR OSCILLATIONS AND CHAOS 35 4-..5 =.4..5..5..4.6.8.5 =.75..5..5..4.6.8 These plots are created in the manner described in the tet. They are created with the logistic equation ( ) = n.9 + n n The first plot has the seed value =.4 as asked for in the tet. Only one additional seed has been done here ( =.75 ) as it is assumed that the reader could easily produce more of these plots after this small amount of practice.

36 CHAPTER 4 4-3..8.6.4. 3 3 34 36 38 4 4 44 46 48 5 iteration =.7 =.7.8.6.4. 3 3 34 36 38 4 4 44 46 48 5 iteration =.7 =.7 The plots are created by iteration on the initial values of (i).7, (ii).7, and (iii).7, using the equation ( ) = n.5 + n n A subset of the iterates from (i) and (ii) are plotted together, and clearly diverge by n = 39. The plot of (i) and (iii) clearly diverge by n = 43. 4-4..8.6.4. 4 6 8 3 3 34 =.9 =.9 fractional difference iteration

NONLINEAR OSCILLATIONS AND CHAOS 37 The given function with the given initial values are plotted in the figure. Here we use the notation =.9 and y =.9, with f ( ) y f y where the function is = and = ( ) n+ n n+ n ( ).5 ( ) f = The fractional difference is defined as y, and clearly eceeds 3% when n = 3. 4-5. A good way to start finding the bifurcations of the function f(α,) = α sin π is to plot its bifurcation diagram..8.6.4...4.6.8 One can epand regions of the diagram to give a rough estimate of the location of a bifurcation. Its accuracy is limited by the fact that the map does not converge very rapidly near the bifurcation point, or more precisely, the Lyapunov eponent approaches zero. One may continue undaunted, however, with the help of a graphical fractal generating software application, to estimate quite a few of the period doublings α n.using Fractint for Windows, and Equation (4.47) to compute the Feigenbaum constant, we can obtain the following: n α δ.7978.8334 4.475 3.85859 4.6 4.8649 4.699 5.8656 4.68 6.8655 4.63 7.865564 4.463 8.865576 One can see that although we should obtain a better value of δ as n increases, numerical precision and human error quickly degrade the quality of the calculation. This is a perfectly acceptable answer to this question. One may compute the α n to higher accuracy by other means, all of which are a great deal more complicated. See, for eample, Eploring Mathematics with Mathematica, which eploits the vanishing Lyapunov eponent. Using their algorithm, one obtains the following: α

38 CHAPTER 4 n α δ.7996.83366 4.4789 3.85869 4.687 4.86484 4.6698 5.86559 4.65633 6.8655 5.345 7.86556 Note that these are shown here only as reference, and the student may not necessarily be epected to perform to this degree of sophistication. The above values are only good to about 6, but this time only limited by machine precision. Another alternative in computing the Feigenbaum constant, which is not requested in the problem, is to use the so-called supercycles, or super-stable points, which are defined by The values R n R n n ( ) f R n, = obey the same scaling as the bifurcation points, and are much easier to compute since these points converge faster than for other α (the Lyapunov eponent goes to ). See, for eample, Deterministic Chaos: An Introduction by Heinz Georg Schuster or Chaos and Fractals: New Frontiers of Science by Peitgen, Jürgens and Saupe. As a result, the estimates for δ obtained in this way are more accurate than those obtained by calculating the bifurcation points. 4-6. The function y = f() intersects the line y = at =, i.e. is defined as the point where = f. Now epand f() in a Taylor series, so that near we have where ( ) ( ) ( ) + β( ) = + β( ) f f df β () d Now define εn n. If we have very close to, then ε should be very small, and we f becomes may use the Taylor epansion. The equation of iteration ( ) ε n+ = n n+ βεn (3) n If the approimation remains valid from the initial value, we have ε n+ βε. a) The values n = ε n form the geometric sequence ε, βε, β ε,. b) Clearly, when β < we have stability since lim ε = n n Similarly we have a divergent sequence when β >, although it will not really be eponentially divergent since the approimation becomes invalid after some number of iterations, and normally the range of allowable is restricted to some subset of the real numbers. n

NONLINEAR OSCILLATIONS AND CHAOS 39 4-7..6.4. 4 6 8 4 6 8 α =.4 α =.7 iteration The first plot (with α =.4) converges rather rapidly to zero, but the second (with α =.7) does appear to be chaotic. 4-8..8.6.4...4.6.8 The tent map always converges to zero for α <.5. Near α =.5 it takes longer to converge, and that is the artifact seen in the figure. There eists a hole in the region.5 < α <.7 (.7 is approimate), where the iterations are chaotic but oscillate between an upper and lower range of values. For α >.7, there is only a single range of chaos, which becomes larger until it fills the range (,) at α =. α 4-9. From the definition in Equation (4.5) the Lyapunov eponent is given by The tent map is defined as n df λ = lim ln n n d i= i This gives df d = α, so we have α for < < f ( ) = α ( ) for < < ()

4 CHAPTER 4 n λ = lim ln( α) = ln( α) (3) n n As indicated in the discussion below Equation (4.5), chaos occurs when λ is positive: α > for the tent map. 4-..4. y..4.5.5.5.5 4-..4. y..4.5.5.5.5 The shape of this plot (the attractor) is nearly identical to that obtained in the previous problem. In Problem 4-, however, we can clearly see the first few iterations (,), (,), (.4,.3),

NONLINEAR OSCILLATIONS AND CHAOS 4 whereas the net iteration (.76,.) is almost on the attractor. In this problem the initial value is taken to be on the attractor already, so we do not see any transient points. 4-. The following. system of differential equations were integrated numerically y 3 y =. y + Bcos t using different values of B in the range [9.8,3.4], and with a variety of initial conditions. The integration range is over a large number of drive cycles, throwing away the first several before starting to store the data in order to reduce the effects of the transient response. For the case B = 9.8, we have a one period per three drive cycle orbit. The phase space plot (line) and Poincaré section (boes) for this case are overlaid and shown in figure (a). All integrations are done here with points per drive cycle. One can eperiment with B and determine that the system becomes chaotic somewhere between 9.8 and 9.9. The section for B =., created by integrating over 8 drive cycles, is shown in figure (b). If one further eperiments with different values of B, and one is also lucky enough to have the right initial conditions, (,) is one that works, then a transition will be found for B in the range (.6,.7). As an eample of the different results one can get depending on the initial conditions, we show two plots in figure (c). One is a phase plot, overlaid with its section, for B =. and the initial condition (,). Eamination of the time evolution reveals that it has one period per cycle. The second plot is a Poincaré section for the same B but with the initial condition (,), clearly showing chaotic motion. Note that the section looks quite similar to the one for B =.. Another transition is in the range (3.3,3.4), where the orbits become regular again, with one period per drive cycle, regardless of initial conditions. The phase plot for B = 3.4 looks similar to the one with B =. and initial condition (,). To summarize, we may enumerate the above transition points by B, B, and B3. Circumventing the actual task of computing where these transition points are, we do know that 9.8 < B < 9.9,.6 < B <.7, and 3.3 < B3 < 3.4. We can then describe the behavior of the system by region. B< B : one period per three drive cycles B < B< B : chaotic B < B< B : mied chaotic/one period per drive cycle (depending on initial conditions) 3 B < B: one period per drive cycle 3 We should remind ourselves, though, that the above list only applies for B in the range we have eamined here. We do not know the behavior when B < 9.8 and B > 3.4, without going beyond the scope of this problem.

4 CHAPTER 4 (a) 5 y 5 3 3 (b) 5 y (c) 5 5.4.6.8 3 3. 3.4 3.6 y (d) 5 3 3 5 y 5.4.6.8 3 3. 3.4 3.6 3.8

NONLINEAR OSCILLATIONS AND CHAOS 43 4-3. The Chirikov map is defined by pn+ = pn K sin qn qn + qn pn + = () The results one should get from doing this problem should be some subset of the results shown in figures (a), (b), and (c) (for K =.8, 3., and 6.4, respectively). These were actually generated using some not-so-random initial points so that a reasonably complete picture could be made. What look to be phase paths in the figures are actually just different points that come from iterating on a single initial condition. For eample, in figure (a), an ellipse about the origin (just pick one) comes from iterating on any one of the points on it. Above the ellipses is chaotic orbit, then a five ellipse orbit (all five come from a single initial condition), etc. The case for K = 3. is similar ecept that there is an orbit outside of which the system is always undergoing chaotic motion. Finally, for K = 6.4 the entire space is filled with chaotic orbits, with the eception of two small lobes. Inside of these lobes are regular orbits (the ones in the left are separate from the ones in the right). (a).5 p π.5.5.5 q π (b).5 p π.5.5.5 q π

44 CHAPTER 4 (c).5 p π.5 4-4. a) The Van de Pol equation is Now look for solution in the form we have and.5.5 d dt q π ω µ ( a ) + = d dt t () = bcos ω t+ ut () d du = bω sin ωt+ dt dt d dt du dt = bω cosωt+ Putting these into the Van de Pol equation, we obtain dut () () + ω ut ( ) = µ b cos ωt+ u( t) + but ( ) cos ωt a bω sin ωt+ dt dut { }{ dt } From this one can see that u(t) is of order µ (i.e. u~ O ( µ ) ), which is assumed to be small here. Keeping only terms up to order µ, the above equation reads () + ω ut ( ) = µ bω sin ωtcos ωt+ abω sin ω t dt dut 3 { } b b = µω b a sin ωt sin 3ωt 4 4 (where we have used the identity 4 sin ω tcos ω t= sin ω t+ sin 3ω ) This equation has frequencies ( ω and 3ω ), and is complicated. However, if term sinω t disappears and the above equation becomes t b= a then the

NONLINEAR OSCILLATIONS AND CHAOS 45 dut () dt b 4 3 + ω ut () = µω sin3ω We let b= a, and the solution for this equation is 3 3 µ b µ a ut () = sin3ω t= sin3ω t 3ω 4 ω So, finally putting this form of u(t) into, we obtain one of the eact solutions of Van de Pol equation: 3 µ a ut () = acosω t sin3ω t 4ω b) See phase diagram below. Since µ =.5 is very small, then actually the second term in the epression of u(t) is negligible, and the phase diagram is very close to a circle of radius b = a =.. t 4-5. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because µ =.7 is a very small value, the limit cycle is very close to a circle of radius b = a =. a) In this case, see figure a), the phase diagram starts at the point ( =, = ) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-4 for eact epression of stable solution).

46 CHAPTER 4. b) In this case, see figure b), the phase diagram starts at the point ( = 3, = ) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-4 for eact epression of stable solution).. 3 4-6. We have used Mathematica to numerically solve and plot the phase diagram for the van de Pol equation. Because µ =.5 is not a small value, the limit cycle is NOT close to a circle (see problem 4-4 above). a) In this case, see figure a), the phase diagram starts at the point ( =, = ) inside the limit cycle, so the phase diagram spirals outward to ultimately approach the stable solution presented by the limit cycle (see problem 4-4 for eact epression of stable solution).

NONLINEAR OSCILLATIONS AND CHAOS 47. b) In this case (see figure below), the phase diagram starts at the point ( = 3, = ) outside the limit cycle, so the phase diagram spirals inward to ultimately approaches the stable solution presented by the limit cycle (see problem 4-4 for eact epression of stable solution).. 3

48 CHAPTER 4