Hence the parametric equations for the phase paths

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Hence the parametric equations for the phase paths are x = A e t + B e 2 t , y = ˙ x = − A e t 2 B e 2 t . The node is shown in Figure 1.4. 1.5 1.5 x 1.5 1.5 y Figure 1.4 Problem 1.1(iv): ¨ x + 3 ˙ x + 2 x = 0, stable node.
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4 Nonlinear ordinary differential equations: problems and solutions (v) ¨ x 4 ˙ x + 40 x = 0. In this problem f (x , y) = − 40 x + 4 y , and there is just one equilibrium point, at the origin. From the results in Section 1.4, this equilibrium point is an unstable spiral. The general solution is x = e 2 t [ A cos 6 t + B sin 6 t ] , from which y can be found. Spiral paths are shown in Figure 1.5. 1 1 x 4 2 2 4 y Figure 1.5 Problem 1.1(v): ¨ x 4 ˙ x + 40 x = 0, unstable spiral. (vi) ¨ x + 3 x | + 2 x = 0, f (x , y) = − 2 x 3 | y | . There is a single equilibrium point, at the origin. The phase diagram is a combination of a stable node for y > 0 and an unstable node for y < 0 as shown in Figure 1.6. The equilibrium point is unstable. 1 1 x 1 1 y Figure 1.6 Problem 1.1(vi): ¨ x + 3 x | + 2 x = 0. (vii) ¨ x + k sgn ( ˙ x) + c sgn (x) = 0, c > k . Assume that k > 0 and x 0 > 0. In this problem f (x , y) = − k sgn (y) c sgn (x) , and the system has one equilibrium point, at the origin.
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1 : Second-order differential equations in the phase plane 5 By writing y d y d x = 1 2 d d x (y 2 ) in the equation for the phase paths we obtain y 2 = 2 [− k sgn (y) c sgn (x) ] x + C , where C is a constant. The value of C is assigned separately for each of the four quadrants into which the plane is divided by the coordinate axes, using the requirement that the composite phase paths should be continuous across the axes. For the path starting at (x 0 , 0 ) , its equation in x > 0, y < 0 is y 2 = 2 (k c)x + C 1 . Therefore C 1 = 2 (c k)x 0 . Continuity of the path into x < 0, y < 0 requires y 2 = 2 (k + c)x + 2 (c k)x 0 . On the axis y = 0, x = − (c k)x 0 /(c + k) . The path in the quadrant x < 0, y > 0 is y 2 = 2 (c k)x + C 2 . By continuity, C 2 = (c k) 2 x 0 /(c + k) . Finally the path in the quadrant x > 0, y > 0 is y 2 = − (c + k)x + C 2 . This path cuts the positive x axis at x = x 1 = (c k) 2 x 0 /(c + k) 2 as required. Since c > k , it follows that x 1 < x 0 . After n circuits x n = γ n x 0 where γ = (c k) 2 /(c + k) 2 . Since γ < 1, then x n 0. Hence the phase diagram (not shown) is a stable spiral made by matching parabolas on the axes. (viii) ¨ x + x sgn (x) = 0. The system has a single equilibrium point, at the origin, and f (x , y) = − x sgn (x) . The phase paths are given by y 2 = − x 2 + C 1 , (x > 0 ) , y 2 = x 2 + C 2 , (x < 0 ) . The phase diagram is a centre for x > 0 joined to a saddle for x < 0 as shown in Figure 1.7.
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6 Nonlinear ordinary differential equations: problems and solutions 1 1 x 1 1 y Figure 1.7 Problem 1.1(viii): ¨ x + x sgn (x) = 0. 1.2 Sketch the phase diagram for the equation ¨ x = − x αx 3 , considering all values of α . Check the stability of the equilibrium points by the method of Section 1.7. 1.2. ¨ x = − x αx 3 .
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