Since ε is small the solutions are both real and

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Since ε is small, the solutions are both real and negative. The general solution is x = A e m 1 t + B e m 2 t + v 0 ε . which is the solution for a stable node centred at (v 0 , 0 ) . y < v 0 ε . With F = − 1, the phase paths are given by (x 1 ) 2 + y 2 = C 2 , which are arcs of circles centred at ( 1, 0 ) . Figure 1.46 shows a computed phase diagram for the oscillator with the parameters ε = 0.2, v 0 = 1. The equilibrium point at ( 1, 0 ) is a centre. The phase paths between y = v 0 + ε and y = v 0 ε are parts of those of a stable node centred at x = v 0 = 5, y = 0. All paths which start outside the circle (x 1 ) 2 + y 2 = (v 0 ε) 2 = 0.8 2 , eventually approach this periodic solution. 1.31 The system ¨ x + x = F( ˙ x) , where F( ˙ x) = k ˙ x + 1 ˙ x < v 0 0 ˙ x = v 0 k ˙ x 1 ˙ x > v 0 , and k > 0, is a possible model for Coulomb dry friction with damping. If k < 2, show that the equilibrium point is an unstable spiral. Compute the phase paths for, say, k = 0.5, v 0 = 1. Using the phase diagram discuss the motion of the system, and describe the limit cycle.
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44 Nonlinear ordinary differential equations: problems and solutions 1.31. The equation for the Coulomb friction is ¨ x + x = F( ˙ x) , where F( ˙ x) = k ˙ x + 1 ˙ x < v 0 0 ˙ x = v 0 k ˙ x 1 ˙ x > v 0 For y < v 0 , the equation of motion is ¨ x k ˙ x + x = 1. The system has one equilibrium point at x = 1 which is unstable, a spiral if k < 2 and a node if k > 2. For y > v 0 , the equation of motion is ¨ x + k ˙ x + x = − 1, which as part of a phase diagram of a stable spiral or node centred at x = − 1. These families of paths meet at the line y = v 0 . Assume that k < 2. For y < v 0 , the phase paths have zero slope on the line x + ky + 1 = 0, which meets the line y = v 0 at x = kv 0 + 1. Similarly, the phase paths for y > v 0 have zero slope along the line x + ky + 1 = 0 which meets the line y = v 0 at x = − kv 0 1. On the phase diagram insert a phase path on y = v 0 between x = − kv 0 1 and x = kv 0 + 1 along which phase paths meet pointing in the direction of positive x . In this singular situation the only exit is along the line until x = kv 0 + 1 is reached where the path continues for y < v 0 . See Figure 1.47. This particular path continues as the limit cycle. Paths spiral into the limit cycle from external and internal points. The section of phase path on y = v 0 corresponds to dry friction in which two surfaces stick for a time. This occurs in every period of the limit cycle. 3 2 1 1 2 3 x y y = v 0 2 1 1 2 Figure 1.47 Problem 1.31: The phase diagram with k = 0.5 and v = 1. The thickest curve is the limit cycle.
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1 : Second-order differential equations in the phase plane 45 1.32 A pendulum with magnetic bob oscillates in a vertical plane over a magnet, which repels the bob according to the inverse square law, so that the equation of motion is (Figure 1.36 in NODE) ma 2 ¨ θ = − mgasinθ + Fh sin φ , where h > a and F = c/(a 2 + h 2 2 ah cos θ) and c is a constant. Find the equilibrium posi- tions of the bob, and classify them as centres and saddle points according to the parameters of the problem. Describe the motion of the pendulum.
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