Consider eqn iii using v and vi if ε 1 ln ε 1 c vii

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Consider eqn (iii), using (v) and (vi). If ε 1 ln 1 ) < C < ( vii ) there are exactly two values in the range −∞ < y < ε 1 at which the factor C F(y) , and hence x , becomes zero, and between these values x > 0. The corresponding reflected path segment given by (iv) completes a closed path, having parameter C . A representative phase diagram is given in Figure 1.50. The unclosed paths correspond to values of y > ε 1 : their boundary is the singular solution mentioned above. 1.38 Show that the equation ¨ x + x + εx 3 = 0 (ε > 0 ) with x( 0 ) = a , ˙ x( 0 ) = 0 has phase paths given by ˙ x 2 + x 2 + 1 2 εx 4 = ( 1 + 1 2 εa 2 )a 2 . Show that the origin is a centre. Are all phase paths closed, and hence all solutions periodic? 1.38. The differential equation of the phase paths of ¨ x + x + εx 3 = 0, (ε > 0 )
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1 : Second-order differential equations in the phase plane 53 1 1 x 1 1 y Figure 1.50 Problem 1.37: The phase diagram for ¨ x εx ˙ x + x = 0 with ε = 1. is given by y d y d x = − x εx 3 . Given the conditions x( 0 ) = a and ˙ x( 0 ) = 0, integration of the differential equation gives the phase paths ˙ x 2 + x 2 + 1 2 εx 4 = constant = ( 1 + 1 2 εa 2 )a 2 . The equation has a single equilibrium point, at the origin. This is a conservative system (see NODE, Section 1.3) with potential function V (x) = (x + εx 3 ) d x = 1 2 x 2 + 1 4 εx 4 . Differentiating V (x) twice we obtain V (x) = x + 1 2 εx 3 , V (x) = 1 + 3 2 εx 2 . Therefore V ( 0 ) = 0 and V ( 0 ) = 1 > 0 which means that V (x) has a minimum at the origin. Locally the origin is a centre. However, V (x) < 0 for x < 0 and V (x) > 0 for x < 0, and also V (x) → ∞ as x → ± ∞ . These conditions imply that, for every a , x is also zero at x = − a and y is continuous between zero at x = − a and x = a . Hence every path is closed and all solutions periodic.
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54 Nonlinear ordinary differential equations: problems and solutions 1.39 Locate the equilibrium points of the equation ¨ x + λ + x 3 x = 0. in the x , λ plane. Show that the phase paths are given by 1 2 ˙ x 2 + λx + 1 4 λx 4 1 2 x 2 = constant. Investigate the stability of the equilibrium points. 1.39. Consider the parameter-dependent system ¨ x + λ + x 3 x = 0. Using the notation of Section 1.7 (in NODE), let f (x , λ) = x x 3 λ . Equilibrium points occur where f (x , λ) = 0 which is shown as the curve in Figure 1.51. The function f (x , λ) is positive in the shaded region. Points on the curve λ = x x 3 above the shaded areas are stable and all other points are unstable. Treating λ as a function of x , x x 3 has stationary points at x = ± 1 / 3 where λ = ± 2 / 3 as indicated in Figure 1.51. Therefore if 2 / 3 < λ < 2 / 3 the equation has three equilibrium points; if λ = ± 2 / 3 the equation has two; for all other values of λ the equation has one equilibrium point. The phase paths satisfy the differential equation y d y d x = − x 3 + x λ , where y = ˙ x . Integrating, the phase paths are given by 1 2 y 2 + λx + 1 4 x 4 1 2 x 2 = constant.
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