9780199212033

# Therefore the paths in the interval x 1 are the

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Therefore the paths in the interval | x | ≤ 1 are the straight lines y = C . Since V (x) is strictly increasing for x < 1, the paths must resemble the left-hand half of a centre at x = − 1. In the same way the paths for x > 1 must be the right-hand half of a centre. A schematic phase diagram is shown in Figure 1.17. 2 1 1 2 x 1 y 2 1 1 2 x 0.2 0.1 0.1 0.2 v ( x ) 1 Figure 1.17 Problem 1.6: This diagram shows some phase paths for the equation with V (x) = x + 1, (x < 1 ) , V (x) = − x + 1, (x > 1 ) . 1.7 Figure 1.33 (in NODE) shows a pendulum striking an inclined wall. Sketch the phase diagram of this ‘impact oscillator’, for α positive and α negative, when (i) there is no loss of energy at impact, (ii) the magnitude of the velocity is halved on impact. 1.7. Assume the approximate pendulum equation (1.1), namely ¨ θ + ω 2 θ = 0, and assume that the amplitude of the oscillations does not exceed θ = 1 2 π (thus avoiding any complications arising from impacts above the point of suspension).

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1 : Second-order differential equations in the phase plane 13 u u u a a . u . Figure 1.18 Problem 1.7: Perfect rebound for α > 0 and α < 0. u u u . u . Figure 1.19 Problem 1.7: The rebound speed is half that of the impact speed. (i) Perfect rebound with no loss of energy. In all cases the phase diagram consists of segments of a centre cut off at θ = α , and the return path after impact will depend on the rebound velocity after impact. The dashed lines in Figure 1.18 indicate the rebound velocity which has the same magnitude as the impact velocity. (ii) In these phase diagrams (see Figure 1.19) the rebound speed is half that of the impact speed. 1.8 Show that the time elapsed, T , along a phase path C of the system ˙ x = y , ˙ y = f (x , y) is given, in a form alternative to (1.13), by T = C (y 2 + f 2 ) ( 1 / 2 ) d s , where d s is an element of distance along C . By writing δs (y 2 + f 2 ) 1 2 δt , indicate, very roughly, equal time intervals along the phase paths of the system ˙ x = y , ˙ y = 2 x .
14 Nonlinear ordinary differential equations: problems and solutions 1 1 x 1 2 y Figure 1.20 Problem 1.8: The phase path y = ( 1 + 2 x 2 ) 1 / 2 is shown with equal time steps δt = 0.1. 1.8. Let C be a segment of a phase path from A to B , traced out by a representative point P between times T A and t B , and s(t) be the arc length along C measured from A to P . Along the path δs ≈ [ (δx) 2 + (δy) 2 ] 1 / 2 , so δs δt δx δt 2 + δy δt 2 1 / 2 . In the limit δt 0, the velocity of P is d s/ d t given by d s d t = [˙ x 2 + ˙ y 2 ] 1 / 2 . ( i ) The transient time T is given by T = t B t A = t B t A d t = C d s d s/ d t = C d s x 2 + ˙ y 2 ] 1 / 2 = C (y 2 + f 2 ) ( 1 / 2 ) d s , ( ii ) since ˙ x = y and ˙ y = f . For the case ˙ x = y , ˙ y = 2 x the phase paths consist of the family of hyperbolas y 2 2 x 2 = α , where α is an arbitrary constant. From (i), a small time interval δt corresponds to a step length δs along a phase path given approximately by δs ≈ [˙ x 2 + ˙ y 2 ] 1 / 2 δt = (y 2 + 4 x 2 ) 1 / 2 δt = + 6 x 2 ) 1 / 2 δt .

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