9780199212033

# 1 and a path eventually meets the x axis between x 1

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1), and a path eventually meets the x axis between x = − 1 and x = 1 either from above or below depending on the initial value x 0 . We have to insert a path from ( 1, 0 ) to the origin and a path from ( 1, 0 ) to the origin to complete the phase diagram. Let the path which starts at (x 0 , 0 ) next cut the x axis at (x 1 , 0 ) . From (ii) the path is (x F 0 ) 2 + y 2 = (x 0 F 0 ) 2 , (y < 0 ) .

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24 Nonlinear ordinary differential equations: problems and solutions 3 2 1 1 2 3 x 3 2 1 1 2 3 y Figure 1.29 Problem 1.16. from which it follows that x 1 = 2 F 0 x 0 . Assume that 2 F 0 x 0 < F 0 , that is, x 0 > 3 F 0 so that the path continues. The continuation of the path lies on the semicircle (x + F 0 ) 2 + y 2 = (x 1 + F 0 ) 2 = ( 3 F 0 x 0 ) 2 , (y > 0 ) . Assume that it meets the x axis again at x = x 2 . Hence x 2 = x 0 4 F 0 . The spiral will continue if x 2 > F 0 or x 0 > 5 F 0 and terminate if x 0 < 5 F 0 . Hence just one cycle of the spiral occurs if 3 F 0 < x 0 < 5 F 0 . If the spiral continues then x = x 2 becomes the new initial point and a further spiral occurs if 3 F 0 < x 2 < 5 F 0 or 7 F 0 < x 0 < 9 F 0 . Continuing this process, a phase path will spiral just n times if ( 4 n 1 )F 0 < x 0 < ( 4 n + 1 )x 0 . 1.17 A pendulum of length a has a bob of mass m which is subject to a horizontal force 2 a sin θ , where θ is the inclination to the downward vertical. Show that the equation of motion is ¨ θ = ω 2 ( cos θ λ) sin θ , where λ = g/(ω 2 a) . Investigate the stability of the equilib- rium states by the method of NODE, Section 1.7 for parameter-dependent systems. Sketch the phase diagrams for various λ . 1.17. The forces acting on the bob are shown in Figure 1.30. Taking moments about O 2 a sin θ · a cos θ mga sin θ = ma 2 ¨ θ ,
1 : Second-order differential equations in the phase plane 25 O a mg m v 2 a sin Figure 1.30 Problem 1.17. 1 2 2 2 Figure 1.31 Problem 1.17: The graph of f (θ , λ) = 0 where the regions in which f (θ , λ) > 0 are shaded. or ¨ θ = ω 2 ( cos θ λ) sin θ = f (θ , λ) , in the notation of NODE, Section 1.7, where λ/(ω 2 a) . Let ω = 1: time can be scaled to eliminate ω . The curves in the , λ) given by f (θ , λ) = 0 are shown in Figure 1.31 with the regions where f (θ , λ) > 0 are shaded. If λ < 1, then the pendulum has four equilibrium points at θ = ± cos 1 λ , θ = 0 and θ = π . The diagram is periodic with period 2 π in θ so that the equilibrium point at θ = − π is the same as that at θ = π . Any curves above the shaded regions indicate stable equilibrium points (centres) and any curves below shaded regions indicate unstable equilibrium points (saddles). Hence, for λ < 1, θ = ± cos 1 λ are stable points, whilst θ = 0 and θ = π are both unstable. The equations of the phase paths can be found by integrating ˙ θ d ˙ θ d θ = sin θ cos θ λ sin θ . The general solution is ˙ θ 2 = sin 2 θ + 2 cos θ + C . The phase diagram is shown in Figure 1.32 for λ = 0.4. As expected from the stability diagram, there are centres at θ = ± cos 1 λ and saddles at x = 0 and x = π .

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