If ? 1 there are two equilibrium point at θ π or π

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If λ 1, there are two equilibrium point at θ = π (or π ). The phase diagram is shown in Figure 1.33 with λ = 2. The origin now becomes a stable centre but θ = π remains a saddle.
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26 Nonlinear ordinary differential equations: problems and solutions 1 1 . Figure 1.32 Problem 1.17: Phase diagram for λ = 0.4 < 1. 1 1 . Figure 1.33 Problem 1.17: Phase diagram for λ = 2 > 1. 1.18 Investigate the stability of the equilibrium points of the parameter-dependent system ¨ x = (x λ)(x 2 λ) . 1.18. The equation is ¨ x = (x λ)(x 2 λ) = f (x , λ) in the notation of NODE, Section 1.7. The system is in equilibrium on the line x = λ and the parabola x 2 = λ . These boundaries are shown in Figure 1.34 together with the shaded regions in which f (x , λ) > 0. λ 0. There is one equilibrium point, an unstable saddle at x = λ . 0 < λ < 1. There are three equilibrium points: at x = − λ (saddle), x = λ (centre) and x = λ (saddle). λ = 1. This is a critical case in which f (x , λ) is positive on both sides of x = 1. The equilibrium point is an unstable hybrid centre/saddle. λ > 1. There are three equilibrium points: at x = − λ (saddle), x = λ (centre) and x = λ (saddle).
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1 : Second-order differential equations in the phase plane 27 2 1 1 2 x x = x 2 = 2 1 1 2 Figure 1.34 Problem 1.18. 1.19 If a bead slides on a smooth parabolic wire rotating with constant angular velocity ω about a vertical axis, then the distance x of the particle from the axis of rotation satisfies ( 1 + x 2 ) ¨ x + (g ω 2 + ˙ x 2 )x = 0. Analyse the motion of the bead in the phase plane. 1.19. The differential equation of the bead is ( 1 + x 2 ) ¨ x + (g ω 2 + ˙ x 2 )x = 0. The equation represents the motion of a bead sliding on a rotating parabolic wire with its lowest point at the origin. The variable x represents distance from the axis of rotation. Put ˙ x = y and g ω 2 = λ ; then equilibrium points occur where y = 0 and + y 2 )x = 0. If λ = 0, all points on the x axis of the phase diagram are equilibrium points, and if λ = 0 there is a single equilibrium point, at the origin. The differential equation of the phase paths is d y d x = − + y 2 )x ( 1 + x 2 )y , which is a separable first-order equation. Hence, separating the variables and integrating y d y λ + y 2 = − x d x 1 + x 2 + C , or 1 2 ln | λ + y 2 | = − 1 2 ln ( 1 + x 2 ) + C , or + y 2 )( 1 + x 2 ) = A .
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28 Nonlinear ordinary differential equations: problems and solutions 2 1 1 2 x y 2 1 1 2 Figure 1.35 Problem 1.19: Phase diagram for λ = 1. 2 1 1 2 x 2 1 1 2 y Figure 1.36 Problem 1.19: Phase diagram for λ = − 1. λ > 0. The phase diagram for λ = 1 is shown in Figure 1.35 which implies that the origin is a centre. In this mode, for low angular rates, the bead oscillates about the lowest point of the parabola. λ < 0. The phase diagram for λ = − 1 is plotted in Figure 1.36 which shows that the origin is a saddle. For higher angular rates the origin becomes unstable and the bead will theoret- ically go off to infinity. Note that y = ± 1 are phase paths which means, for example, that the bead starting from x = 0 with velocity y = 1 will move outwards at a constant rate.
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