Of which x 0 is the only solution in the phase plane

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of which x = 0 is the only solution. In the phase plane the boundary between the two modes of the phase diagram is the switching curve y = − sgn [ 2 | x | 1 / 2 sgn (x) ] , which is two half parabolas which meet at the origin as shown in Figure 1.42. There are distinct families of phase paths on either side of this curve. 2 | x | 1 / 2 sgn (x) + y > 0. The equation is ¨ x = − 1 so that d y/ d x = − 1 /y and the phase paths are given by the parabolas y 2 = − 2 x + C 1 2 | x | 1 / 2 sgn (x) + y < 0. In this case ¨ x = 1 so that the phase paths are given by y 2 = 2 x + C 2 . When the parabolic paths reach the switching curve their only exit is along the switching curve into the equilibrium point at the origin. 1.28 The relativistic equation for an oscillator is d d t m 0 ˙ x [ 1 ( ˙ x/c) 2 ] + kx = 0, x | < c where m 0 , c and k are positive constants. Show that the phase paths are given by m 0 c 2 [ 1 (y/c) 2 ] + 1 2 kx 2 = constant. If y = 0 when x = a , show that the period, T , of an oscillation is given by T = 4 c ε a 0 [ 1 + ε(a 2 x 2 ) ] d x (a 2 x 2 ) [ 2 + ε(a 2 x 2 ) ] , ε = k 2 m 0 c 2 . The constant ε is small; by expanding the integrand in powers of ε show that T π 2 c ε ( 1 / 2 ) + 3 8 ε 1 / 2 a 2 .
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1 : Second-order differential equations in the phase plane 39 1.28. The equation of the oscillator is d d t m 0 ˙ x [ 1 ( ˙ x/c) 2 ] + kx = 0, which has one equilibrium point at the origin. Also the phase plane is restricted to x | < c . Let y = ˙ x and f (y) = m 0 y [ 1 (y/c) 2 ] . Then the equation of the oscillator is y d f (y) d y + kx = 0, or yf (y) d y d x + kx = 0. This is a separable first-order equation with solution yf (y) d y = − k d x + C , which after integration by parts leads to yf (y) f (y) d y = − 1 2 kx 2 + C , or m 0 y 2 [ 1 (y/c) 2 ] m 0 y d y [ 1 (y/c) 2 ] = − 1 2 kx 2 + C , or m 0 y 2 [ 1 (y/c) 2 ] + m 0 c 2 [ 1 (y/c) 2 ] = − 1 2 kx 2 + C , so that m 0 c 2 [ 1 (y/c) 2 ] = − 1 2 kx 2 + C , ( i ) as required. A sketch of the phase diagram is shown in Figure 1.43. It can be seen that the origin is a centre. The particular path through (a , 0 ) is, from (i), m 0 c 2 [ 1 (y/c) 2 ] = − 1 2 kx 2 + m 0 c 2 + 1 2 ka 2 , or 1 [ 1 (y/c) 2 ] = 1 + ε(a 2 x 2 ) ,
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40 Nonlinear ordinary differential equations: problems and solutions x y 2 1 2 1 Figure 1.43 Problem 1.28: Phase diagram for k = 1, c = 1 and m 0 = 1. where ε = k/( 2 m 0 c 2 ) . Solve this equation for y : y = d x d t = c ε [ a 2 x 2 ] [ 2 + ε(a 2 x 2 ) ] 1 + ε(a 2 x 2 ) . Therefore T = 4 c ε a 0 1 + ε(a 2 x 2 ) d x (a 2 x 2 ) [ 2 + ε(a 2 x 2 ) ] ; ( ii ) the integral is multiplied by 4 since integration between 0 and a covers a quarter of the period, and the time over each quarter is the same by symmetry. Expand the integrand in powers of ε for small ε using a Taylor series. Then 1 + ε(a 2 x 2 ) (a 2 x 2 ) [ 2 + ε(a 2 x 2 ) ] 2 ( 1 / 2 ) 1 (a 2 x 2 ) + 3 4 ε (a 2 x 2 ) .
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