Equivalent to those for the first cycle and similarly

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equivalent to those for the first cycle, and similarly for all the subsequent cycles. The motion is therefore periodic, with period T = 4 π/ 3. The turning points of x(t) occur where ˙ x(t) = 0; that is, where tan ( 3 / 2 ) = 3 (from (ii)). This has two solutions in the range 0 and 2 π . These are t = 2 π 3 3 and t = 4 π 3 3 (by noting that tan 1 3 = 1 3 π ). From (i) the corresponding values of x are x = v e π/( 3 3 ) and x = − v e 2 π/( 3 3 ) . The overall maximum of x(t) is therefore v e π/( 3 3 ) . 1.24 Show how phase paths of Problem 1.23 having arbitrary initial conditions spiral on to a limit cycle. Sketch the phase diagram.
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34 Nonlinear ordinary differential equations: problems and solutions x y v n 1 v n Figure 1.40 Problem 1.24: The limit cycle with the jump along the y axis. 1.24. (Refer to Problem 1.23.) The system is the same as that of Problem 1.23, but with the initial conditions x( 0 ) = 0, ˙ x( 0 ) = v 0 > 0, where v 0 is arbitrary. Suppose the impulsive energy increment at the end of every cycle is E , an arbitrary positive constant. v n will represent the value of ˙ x at the end of the n th cycle, following the energy increment delivered at the end of that cycle, and it serves as the initial condition for the next cycle (see Figure 1.40). For the first cycle, starting at x = 0, ˙ x( 0 ) = v 0 , we have (as in Problem 1.23) 1 2 v 2 1 1 2 v 2 0 e 4 π/ 3 = E , or v 2 1 = 2 E + v 2 0 e 4 π/ 3 . ( i ) For the second cycle (starting at v 1 ) v 2 2 = 2 E + v 2 1 e 4 π/ 3 , ( ii ) and so on. For the n th cycle v 2 n = 2 E + v 2 n 1 e 4 π/ 3 . ( iii ) By successive substitution we obtain v 2 n = 2 E( 1 + e ρ + · · · + e (n 1 + v 2 0 e ) , ( iv ) in which we have written for brevity ρ = 4 π/ 3. By using the usual formula for the sum of a geometric series (iv) reduces to v 2 n v 2 0 = ( 1 e ) 2 E 1 e ρ v 2 0 for all n 1. ( v )
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1 : Second-order differential equations in the phase plane 35 (A) In the special case when E = 1 2 v 2 0 ( 1 e ρ ) , the right-hand side of (v) is zero, so v 2 0 = v 2 1 = · · · = v 2 n , which corresponds to the periodic solution in Problem 1.23. (B) If v 2 0 < 2 E/( 1 e ρ ) , the sequence v 0 , v 1 , . . . , v n is strictly increasing, and lim n →∞ v 2 n = 2 E 1 e ρ . The limit cycle in (A) is approached from inside. (C) If v 2 0 = 2 E/( 1 e ρ ) , the sequence is strictly decreasing and lim n →∞ v 2 n = 2 E 1 e P ρ ; so the limit cycle in (A) is approached from the outside. The more general initial conditions x( 0 ) = X , ˙ x( 0 ) = V , where X and V are both arbitrary, correspond to one of the categories (A), (B) or (C); so the same limit (A) is approached. 1.25 The kinetic energy, T , and the potential energy, V , of a system with one degree of freedom are given by T = T 0 (x) + ˙ xT 1 (x) + ˙ x 2 T 2 (x) , V = V (x) . Use Lagrange’s equation d d t T ˙ x T ∂x = − V ∂x to obtain the equation of motion of the system. Show that the equilibrium points are stationary points of T 0 (x) V (x) , and that the phase paths are given by the energy equation T 2 (x) ˙ x 2 T 0 (x) + V (x) = constant.
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