4. This is based on the Chirikov map problem 4-23 in T-M. Consider the two dimensional Chirikov map 7 Pn+1 = Pn - K sin

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answerhappygod
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4. This is based on the Chirikov map problem 4-23 in T-M. Consider the two dimensional Chirikov map 7 Pn+1 = Pn - K sin

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4 This Is Based On The Chirikov Map Problem 4 23 In T M Consider The Two Dimensional Chirikov Map 7 Pn 1 Pn K Sin 1
4 This Is Based On The Chirikov Map Problem 4 23 In T M Consider The Two Dimensional Chirikov Map 7 Pn 1 Pn K Sin 1 (27.85 KiB) Viewed 44 times
4. This is based on the Chirikov map problem 4-23 in T-M. Consider the two dimensional Chirikov map 7 Pn+1 = Pn - K sin n. 4n+1 = 4n + Pn+1, where - <p <n and -1 < 51, with periodic boundary conditions assumed, i.e. PnPn + 21, and likewise with an. Assuming three sets of K = 0.5, 1., 2. and taking two sets of initial conditions Po = 0,40 = n/4 and go = 0.Po = 1/4, iterate up to n= 100 and plot the results. Inspect these results to make a (very) brief comment on the influence of K on the results, and also the effect of asymmetry between p and q in the maps.
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