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**Unformatted text preview: **Exercises 5.4
(c) To ﬁnd critical points for the system in (a), we solve v = 0, −x + 1 =0 λ−x ⇒ v = 0, x2 − λx + 1 = 0 ⇒ v = 0, √ λ ± λ2 − 4 . x= 2 For 0 < λ < 2, λ2 − 4 < 0 and so both roots are complex numbers. However, for λ > 2 there are two distinct real solutions, √ λ − λ2 − 4 x1 = 2 and the critical points are √ λ − λ2 − 4 ,0 2 √ λ2 − 4 , 2 and x2 = λ+ and λ+ √ λ2 − 4 ,0 . 2 (d) The phase plane diagrams for λ = 1 and λ = 3 are shown in Figures B.40 and B.41 in the answers section of the text. (e) From Figures B.40 we conclude that, for λ = 1, all solution curves approach the vertical line x = 1(= λ). This means that the bar is attracted to the magnet. The case λ = 3 is more complicated. The behavior of the bar depends on the initial displacement x(0) and the initial velocity v (0) = x (0). From Figure B.41 we see that (with v (0) = 0) if x(0) is small enough, then the bar will oscillate about the position x = x1 ; if x(0) is close enough to λ, then the bar will be attracted to the magnet. It is also possible that, with an appropriate combination of x(0) and v (0), the bar will come to rest at the saddle point (x2 , 0). 37. (a) Denoting y = v , we have y = v , and (with m = µ = k = 1) (16) can be written as a system y = v, v = −y + y, 0, = if |y | < 1, v = 0, if |y | < 1, v = 0, sign(y ), if |y | ≥ 1, v = 0, −sign(v ), if v = 0 −y + sign(y ), if |y | ≥ 1, v = 0, −y − sign(v ), if v = 0. 305 ...

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