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9780199212033

# 125 the kinetic and potential energies are given by t

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1.25. The kinetic and potential energies are given by T = T 0 1 x 2 T 2 , V = V . Applying Lagrange’s equation d d t ± T ˙ x T V , the equation of motion is d d t ( 2 T 2 ˙ x + T 1 ) (T ± 2 ˙ x 2 + T ± 1 ˙ x + T ± 0 ) V ± , or 2 T 2 ¨ x + T ± 2 ˙ x 2 T ± 0 V ± . ( i )

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36 Nonlinear ordinary differential equations: problems and solutions Equilibrium points, where ¨ x x = 0, occur where T ± 0 V ± = 0, that is, at the stationary points of the energy function T 0 (x) V . Let y x . Equation (i) can be expressed in the form d d x (T 2 (x)y 2 ) T ± 0 + V ± = 0, which can be integrated to give the phase paths, namely T 2 (x)y 2 T 0 + V = C . 1.26 Sketch the phase diagram for the equation ¨ x =− f(x x) , where f(u) = f 0 u c , f 0 u/c | u |≤ c , f 0 u ≤− c where f 0 , c are constants, f 0 > 0, and c> 0. How does the system behave as c 0? 1.26. The system is governed by the equation ¨ x , where = f 0 u>c f 0 u/c | u c f 0 u< c Let y x . The phase paths are as follows. x + y>c , ¨ x f 0 . The equation for the phase paths is d y d x f 0 y 1 2 y 2 f 0 x + C 1 . The phase paths are parabolas with their axes along the x axis. •| x + y c , ¨ x f 0 (x x)/c . It is easier to solve the linear equation c ¨ x + f 0 ˙ x + f 0 x = 0 parametrically in terms of t . The characteristic equation is cm 2 + f 0 m + f 0 = 0. which has the roots m 1 , m 2 = 1 2 c [− f 0 ² (f 2 0 4 cf 0 ) ] . Therefore x = A e m 1 t + B e m 2 t
1 : Second-order differential equations in the phase plane 37 –6 –5 –4 –3 –2 –1 1 2 3 x –1 1 y x+y =1 = –1 Figure 1.41 Problem 1.26: The spirals are shown for the parameter values f 0 = 0.25 and c = 1. Note that scales on the axes are not the same in the drawing. The roots are both real and negative if f 0 > 4 c , which means that the phase diagram between the lines x + y = c and x + y =− c is a stable node. If f 0 < 4 c , then the phase diagram is a stable spiral. x + y< c , ¨ x = f 0 . The phase paths are given by 1 2 y 2 = f 0 x + C 2 , which again are parabolas but pointing in the opposite direction. Figure 1.41 shows a phase diagram for the spiral case. The spiral between the lines x = y = 1 and x + y 1 is linked with the parabolas on either side of the two lines. The total picture is a stable spiral. A similar matching occurs with the stable node. As c 0, the lines x + y = c and x + c 1 merge and the spiral disappears leaving a centre created by the joining of the parabolas. 1.27 Sketch the phase diagram for the equation ¨ x = u , where u sgn ( 2 | x | 1 / 2 sgn (x) x) . ( u is an elementary control variable which can switch between + 1 and 1. The curve 2 | x | 1 / 2 sgn + y = 0 is called the switching curve.) 1.27. The control equation is ¨ x sgn [ 2 | x | 1 / 2 sgn x ] . The equilibrium point satis±es sgn [ 2 | x | 1 / 2 sgn ]= 0, or | x | 1 / 2 sgn = 0,

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38 Nonlinear ordinary differential equations: problems and solutions –6 –4 –2 2 4 6 x –4 –2 2 4 y Figure 1.42 Problem 1.27.
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125 The kinetic and potential energies are given by T T 1 x...

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