Case i α 0 the equation has a single equilibrium

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Case (i). α > 0. The equation has a single equilibrium point, at x = 1. The phase paths are given by d y d x = − x( 1 + αx 2 ) y , which is a separable first-order equation. The general solution is given by y d y = − x( 1 + αx 2 ) d x + C , ( i ) so that 1 2 y 2 = − 1 2 x 2 1 4 x 4 + C . The phase diagram is shown in Figure 1.8 with α = 1, and the origin can be seen to be a centre. Case (ii). α < 0. There are now three equilibrium points: at x = 0 and at x = ± 1 / α . The phase paths are still given by (i), but computed in this case with α = − 1 (see Figure 1.9). There is a centre at ( 0, 0 ) and saddles at ( ± 1, 0 ) . This equation is a parameter-dependent system with parameter α as discussed in Section 1.7. As in eqn (1.62), let f (x , α) = − x αx 3 . Figure 1.10 shows that in the region above x = 0, f (x , α) is positive for all α , which according to Section 1.7 (in NODE) implies that the origin is stable. The other equilibrium points are unstable.
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1 : Second-order differential equations in the phase plane 7 2 1 1 2 x 3 2 1 1 2 3 y Figure 1.8 Problem 1.2: Phase diagram for α = 1. 2 1 1 2 x 1 1 y Figure 1.9 Problem 1.2: Phase diagram for α = − 1. 3 2 1 1 2 3 1 2 x a 2 1 Figure 1.10 Problem 1.2: The diagram shows the boundary x( 1 αx 2 ) = 0; the shaded regions indicate f (x , α) > 0.
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8 Nonlinear ordinary differential equations: problems and solutions 1.3 A certain dynamical system is governed by the equation ¨ x + ˙ x 2 + x = 0. Show that the origin is a centre in the phase plane, and that the open and closed paths are separated by the path 2 y 2 = 1 2 x . 1.3. ¨ x + ˙ x 2 + x = 0. The phase paths in the (x , y) plane are given by the differential equation d y d x = y 2 x y . By putting y d y d x = 1 2 d d x (y 2 ) , the equation can be expressed in the form d (y 2 ) d x + 2 y 2 = − 2 x , which is a linear equation for y 2 . Hence y 2 = C e 2 x x + 1 2 , which is the equation for the phase paths. The equation has a single equilibrium point, at the origin. Near the origin for y small, ¨ x + x 0 which is the equation for simple harmonic motion (see Example 1.2 in NODE). This approximation indicates that the origin is a centre. If the constant C < 0, then C e 2 x → −∞ as x → ∞ , which implies that x + 1 2 + C e 2 x must be zero for a negative value of x . There is also a positive solution for x The paths are closed for C < 0 since any path is reflected in the x axis. If C 0, then the equation x + 1 2 + C e 2 x = 0 has exactly one solution and this is positive. To see this sketch the line z = x 1 2 and the exponential curve z = C e 2 x for positive and negative values for C and see where they intersect. The curve bounding the closed paths is the parabola y 2 = − x + 1 2 . The phase diagram is shown in Figure 1.11. 1.4 Sketch the phase diagrams for the equation ¨ x + e x = a , for a < 0, a = 0, and a > 0.
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