{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Given a value of the parameter α the step lengths

Info iconThis preview shows pages 14–18. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 14

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Background image of page 15
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
Background image of page 16

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Background image of page 17
Image of page 18
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}