9780199212033

# Given a value of the parameter α the step lengths

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Given a value of the parameter α , the step lengths for a constant δt are determined by the factor + 6 x 2 ) 1 / 2 , and tend to be comparatively shorter when the phase path is closer to the origin. This is illustrated in Figure 1.20 for the branch y = ( 1 2 x 2 ) 1 / 2 . 1.9 On the phase diagram for the equation ¨ x + x = 0, the phase paths are circles. Use (1.13) in the form δt δx/y to indicate, roughly, equal time steps along several phase paths.

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1 : Second-order differential equations in the phase plane 15 x y Figure 1.21 Problem 1.9. 1.9. The phase paths of the simple harmonic oscillator ¨ x + x = 0 are given by d y/ d x = − x/y , which has the general solution x 2 + y 2 = C 2 , C > 0. The paths can be represented parametrically by x = C cos θ , y = C sin θ , where θ is the polar angle. By (1.13), an increment in time is given by δt δx y = C sin θδθ C sin θ = − δθ . This formula can be integrated to give t = − θ + B . Hence equal time steps are equivalent to equal steps in the polar angle θ . All phase paths are circles centred at the origin and the time taken between radii subtending the same angle, say α , at the origin as shown in Figure 1.21. 1.10 Repeat Problem 1.9 for the equation ¨ x + 9 x = 0, in which the phase paths are ellipses. 1.10. The phase paths of ¨ x + 9 x = 0 are given by d y/ d x = − 9 x/y , which has the general solu- tion 9 x 2 + y 2 = C 2 , C > 0. The paths are concentric ellipses. The paths can be represented parametrically by x = 1 3 C cos θ , y = C sin θ , where θ is the polar angle. By (1.13), an increment in time is given by δt δx y = ( 1 / 3 )C sin θδθ C sin θ = − 1 3 δθ . Hence equal time steps are equivalent on all paths to the lengths of segments cut by equal polar angles α as shown in Figure 1.22.
16 Nonlinear ordinary differential equations: problems and solutions x y Figure 1.22 Problem 1.10. 1.11 The pendulum equation, ¨ x + ω 2 sin x = 0, can be approximated for moderate ampli- tudes by the equation ¨ x + ω 2 (x 1 6 x 3 ) = 0. Sketch the phase diagram for the latter equation, and explain the differences between it and Figure 1.2 (in NODE). 1.11. For small | x | , the Taylor expansion of sin x is given by sin x = x 1 6 x 3 + O(x 5 ) . Hence for small | x | , the pendulum equation ¨ x + ω 2 sin x = 0 can be approximated by ¨ x + ω 2 x 1 6 x 3 = 0. If x is unrestricted this equation has three equilibrium points, at x = 0 and x = ± 6 ≈ ± 2.45. The pendulum equation has equilibrium points at x = , (n = 0, 1, 2, . . .) . Obviously, the approximate equation is not periodic in x , and the equilibrium points at x = ± 6 differ con- siderably from those of the pendulum equation. We can put ω = 1 without loss since time can always be rescaled by putting t = ωt . Figure 1.23 shows the phase diagrams for both equations for amplitudes up to 2. The solid curves are phase paths of the approximation and the dashed

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1 : Second-order differential equations in the phase plane 17 x 1 1 y 2 1 1 2 Figure 1.23 Problem 1.11: The solid curves represent the phase paths of the approximate equation ¨ x + x 1 6 x 3 = 0, and the dashed curves show the phase paths of ¨ x + sin x = 0.
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