Problem III (The double pendulum.) We consider a system made of two successive pendulums of lengths...

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Problem III (The double pendulum.) We consider a system made of two successive pendulums of lengths > 0 and 12 > 0 each attached to a mass m. We denote by 0 (t) and 02(t) their respective angles with respect to the vertical axis at time t, as shown in the figure below. 0(t) M M The equations of motion, derived from Newton's principles, are given for t = R by: 01(t) = 2(t) 3g sin(0 ()) g sin(0 () 202(t)) 2 sin(0 (t) 2(t)) [12(t) + 1(t) cos(01(t) 02(t))] 11(3-cos(201(t) 202(t))) 2 sin(0 (t) 2(t)) [21 () + 2g cos( ()) + () cos(0(t) 02(t))] 12(3 cos(201(t) 202(t))) where g is the gravity constant of the earth. Because the r.h.s. is a continuous function of 01,02, the Cauchy-Lipschitz theorem can be applied to show that maximal solutions exist for any choice of initial conditions (0 (0), 0(0), 02(0), 02(0)). Using the fact that the total energy of this system E, given by the following expression, E = m[2l(t) + l2(t) + 2 ()() cos(01(t) 2(t))] - mg(21 cos(0 (t)) + 12 cos(#2(t))). is conserved along any solution, one can also show that solutions are global. The proof of these statements is not required, but feel free to try. == We fix the parameters to = 1m, l2 = 0.5m, m = 1m, l = 0.5m, m = 1kg, g = 9.8m/s, and take as initial conditions 0(0) = 0, 0(0) = 0, 02(0) = 0, 02(0) = 15. = (1) Use an RK4(5) solver to compute the solution up to final time T 10, limiting the step size to 0.01 to ensure enough accuracy. Show in a figure the full trajectory of the end mass (attached to the second pendulum) and provide the maximum and minimum values of the energy E above using four digits after the point. (2) Consider now a slightly perturbed initial condition where, instead of 02 (0) = 0, we have 02(0) = 15.05, all other initial conditions being identical. Show again the resulting trajectory of the end point and compare to the result of the previous question. Remark: this is an example of a chaotic system in the sense that the high dependency in the initial conditions makes it particularly difficult to predict its long time behaviour. Problem III (The double pendulum.) We consider a system made of two successive pendulums of lengths > 0 and 12 > 0 each attached to a mass m. We denote by 0 (t) and 02(t) their respective angles with respect to the vertical axis at time t, as shown in the figure below. 0(t) M M The equations of motion, derived from Newton's principles, are given for t = R by: 01(t) = 2(t) 3g sin(0 ()) g sin(0 () 202(t)) 2 sin(0 (t) 2(t)) [12(t) + 1(t) cos(01(t) 02(t))] 11(3-cos(201(t) 202(t))) 2 sin(0 (t) 2(t)) [21 () + 2g cos( ()) + () cos(0(t) 02(t))] 12(3 cos(201(t) 202(t))) where g is the gravity constant of the earth. Because the r.h.s. is a continuous function of 01,02, the Cauchy-Lipschitz theorem can be applied to show that maximal solutions exist for any choice of initial conditions (0 (0), 0(0), 02(0), 02(0)). Using the fact that the total energy of this system E, given by the following expression, E = m[2l(t) + l2(t) + 2 ()() cos(01(t) 2(t))] - mg(21 cos(0 (t)) + 12 cos(#2(t))). is conserved along any solution, one can also show that solutions are global. The proof of these statements is not required, but feel free to try. == We fix the parameters to = 1m, l2 = 0.5m, m = 1m, l = 0.5m, m = 1kg, g = 9.8m/s, and take as initial conditions 0(0) = 0, 0(0) = 0, 02(0) = 0, 02(0) = 15. = (1) Use an RK4(5) solver to compute the solution up to final time T 10, limiting the step size to 0.01 to ensure enough accuracy. Show in a figure the full trajectory of the end mass (attached to the second pendulum) and provide the maximum and minimum values of the energy E above using four digits after the point. (2) Consider now a slightly perturbed initial condition where, instead of 02 (0) = 0, we have 02(0) = 15.05, all other initial conditions being identical. Show again the resulting trajectory of the end point and compare to the result of the previous question. Remark: this is an example of a chaotic system in the sense that the high dependency in the initial conditions makes it particularly difficult to predict its long time behaviour.