Chapter 6 - Davidson Physics
Chapter 6 - Davidson Physics Chapter 6 - Davidson Physics
CHAPTER 6. THE CHAOTIC MOTION OF DYNAMICAL SYSTEMS 173m period s m1 1 0.500 000 0002 2 0.809 016 9943 4 0.874 640 4254 8 0.888 660 9705 16 0.891 666 8996 32 0.892 310 8837 64 0.892 448 8238 128 0.892 478 091Table 6.2: Values of the control parameter s m for the superstable trajectories of period 2 m−1 . Ninedecimal places are shown.solve the resultant quadratic equation analytically to find s 2 = (1 + √ 5)/4. Clearly r = s 1 = 1/2solves (6.30) with period 1, because from (6.29), f (1) (x = 1/2) = 4r 1 2 (1 − 1 2) = r = 1/2 only forr = 1/2.1. It is straightforward to adapt the bisection method discussed in Section 6.6. Adapt theclass RecursiveFixedPointApp to find the numerical solutions of (6.29). Good startingvalues for the left-most and right-most values of r are easy to obtain. The left-most value isr = r ∞ ≈ 0.8925. If we already know the sequence s 1 , s 2 , . . . , s m , then we can determine δbyδ m = s m−1 − s m−2. (6.31)s m − s m−1We use this determination for δ m to find the right-most value of r:r (m+1)right= s m − s m−1δ m. (6.32)We choose the desired precision to be 10 −16 . A summary of our results is given in Table 6.2.Verify these results and determine δ.2. Use your values of s m to obtain a more accurate determination of α and δ.Project 6.17. From chaos to orderThe bifurcation diagram of the logistic map (see Figure 6.2) has many interesting features thatwe have not explored. For example, you might have noticed that there are several smooth darkbands in the chaotic region for r > r ∞ . Use BifurcateApp to generate the bifurcation diagram forr ∞ ≤ r ≤ 1. If we start at r = 1.0 and decrease r, we see that there is a band that narrows andeventually splits into two parts at r ≈ 0.9196. If you look closely, you will see that the band splitsinto four parts at r ≈ 0.899. If you look even more closely, you will see many more bands. Whattype of change occurs near the splitting (merging) of these bands)? Use IterateMap to look atthe time series of (6.5) for r = 0.9175. You will notice that although the trajectory looks random,it oscillates back and forth between two bands. This behavior can be seen more clearly if you lookat the time series of x n+1 = f (2) (x n ). A detailed discussion of the splitting of the bands can befound in Peitgen et al.
CHAPTER 6. THE CHAOTIC MOTION OF DYNAMICAL SYSTEMS 174Project 6.18. Calculation of the Lyapunov spectrumIn Section 6.12 we discussed the calculation of the Lyapunov exponent for the logistic map. If adynamical system has a multidimensional phase space, for example, the Hénon map and the Lorenzmodel, there is a set of Lyapunov exponents, called the Lyapunov spectrum, that characterize thedivergence of the trajectory. As an example, consider a set of initial conditions that forms a filledsphere in phase space for the (three-dimensional) Lorenz model. If we iterate the Lorenz equations,then the set of phase space points will deform into another shape. If the system has a fixed point,this shape contracts to a single point. If the system is chaotic, then, typically, the sphere willdiverge in one direction, but become smaller in the other two directions. In this case we can definethree Lyapunov exponents to measure the deformation in three mutually perpendicular directions.These three directions generally will not correspond to the axes of the original variables. Instead,we must use a Gram-Schmidt orthogonalization procedure.The algorithm for finding the Lyapunov spectrum is as follows:(i) Linearize the dynamical equations. If r is the f-component vector containing the dynamicalvariables, then define ∆r as the linearized difference vector. For example, the linearizedLorenz equations ared∆x= −σ∆x + σ∆ydt(6.33a)d∆y= −x∆z − z∆x + r∆x − ∆ydt(6.33b)d∆z= x∆y + y∆x − b∆z.dt(6.33c)(ii) Define f orthonormal initial values for ∆r. For example, ∆r 1 (0) = (1, 0, 0), ∆r 2 (0) = (0, 1, 0),and ∆r 3 (0) = (0, 0, 1). Because these vectors appear in a linearized equation, they do nothave to be small in magnitude.(iii) Iterate the original and linearized equations of motion. One iteration yields a new vectorfrom the original equation of motion and f new vectors ∆r α from the linearized equations.(iv) Find the orthonormal vectors ∆r ′ α from the ∆r α using the Gram-Schmidt procedure. Thatis,∆r ′ 1 = ∆r 1|∆r 1 |∆r ′ 2 = ∆r 2 − (∆r ′ 1 · ∆r 2 )∆r ′ 1|∆r 2 − (∆r ′ 1 · ∆r 2)∆r ′ 1 |∆r ′ 3 = ∆r 3 − (∆r ′ 1 · ∆r 3 )∆r ′ 1 − (∆r ′ 2 · ∆r 3 )∆r∣′ 2∣∆r 3 − (∆r ′ 1 · ∆r 3)∆r ′ 1 − (∆r′ 2 · ∆r 3)∆r ′ ∣ .2(6.34a)(6.34b)(6.34c)It is straightforward to generalize the method to higher dimensional models.(v) Set the ∆r α (t) equal to the orthonormal vectors ∆r ′ α(t).(vi) Accumulate the running sum, S α as S α → S α + log |∆r α (t)|.
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CHAPTER 6. THE CHAOTIC MOTION OF DYNAMICAL SYSTEMS 174Project 6.18. Calculation of the Lyapunov spectrumIn Section 6.12 we discussed the calculation of the Lyapunov exponent for the logistic map. If adynamical system has a multidimensional phase space, for example, the Hénon map and the Lorenzmodel, there is a set of Lyapunov exponents, called the Lyapunov spectrum, that characterize thedivergence of the trajectory. As an example, consider a set of initial conditions that forms a filledsphere in phase space for the (three-dimensional) Lorenz model. If we iterate the Lorenz equations,then the set of phase space points will deform into another shape. If the system has a fixed point,this shape contracts to a single point. If the system is chaotic, then, typically, the sphere willdiverge in one direction, but become smaller in the other two directions. In this case we can definethree Lyapunov exponents to measure the deformation in three mutually perpendicular directions.These three directions generally will not correspond to the axes of the original variables. Instead,we must use a Gram-Schmidt orthogonalization procedure.The algorithm for finding the Lyapunov spectrum is as follows:(i) Linearize the dynamical equations. If r is the f-component vector containing the dynamicalvariables, then define ∆r as the linearized difference vector. For example, the linearizedLorenz equations ared∆x= −σ∆x + σ∆ydt(6.33a)d∆y= −x∆z − z∆x + r∆x − ∆ydt(6.33b)d∆z= x∆y + y∆x − b∆z.dt(6.33c)(ii) Define f orthonormal initial values for ∆r. For example, ∆r 1 (0) = (1, 0, 0), ∆r 2 (0) = (0, 1, 0),and ∆r 3 (0) = (0, 0, 1). Because these vectors appear in a linearized equation, they do nothave to be small in magnitude.(iii) Iterate the original and linearized equations of motion. One iteration yields a new vectorfrom the original equation of motion and f new vectors ∆r α from the linearized equations.(iv) Find the orthonormal vectors ∆r ′ α from the ∆r α using the Gram-Schmidt procedure. Thatis,∆r ′ 1 = ∆r 1|∆r 1 |∆r ′ 2 = ∆r 2 − (∆r ′ 1 · ∆r 2 )∆r ′ 1|∆r 2 − (∆r ′ 1 · ∆r 2)∆r ′ 1 |∆r ′ 3 = ∆r 3 − (∆r ′ 1 · ∆r 3 )∆r ′ 1 − (∆r ′ 2 · ∆r 3 )∆r∣′ 2∣∆r 3 − (∆r ′ 1 · ∆r 3)∆r ′ 1 − (∆r′ 2 · ∆r 3)∆r ′ ∣ .2(6.34a)(6.34b)(6.34c)It is straightforward to generalize the method to higher dimensional models.(v) Set the ∆r α (t) equal to the orthonormal vectors ∆r ′ α(t).(vi) Accumulate the running sum, S α as S α → S α + log |∆r α (t)|.
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