9780199212033

? 1 the phase diagram is shown in figure 137 if the

• Notes
• 62

This preview shows pages 28–32. Sign up to view the full content.

1 will move outwards at a constant rate. λ = 1. The phase diagram is shown in Figure 1.37. If the bead is placed at rest at any point on the wire then it will remain in that position subsequently.

This preview has intentionally blurred sections. Sign up to view the full version.

1 : Second-order differential equations in the phase plane 29 2 1 1 2 x 2 1 1 2 y Figure 1.37 Problem 1.19: Phase diagram for λ = 0. 1.20 A particle is attached to a fixed point O on a smooth horizontal plane by an elastic string. When unstretched, the length of the string is 2 a . The equation of motion of the particle, which is constrained to move on a straight line through O , is ¨ x = − x + a sgn (x) , | x | > a (when the string is stretched), ¨ x = 0, | x | ≤ a (when the string is slack), x being the displacement from O . Find the equilibrium points and the equations of the phase paths, and sketch the phase diagram. 1.20. The equation of motion of the particle is ¨ x = − x + a sgn (x) , ( | x | > a) ¨ x = 0, ( | x | ≤ a) . All points in the interval | x | ≤ a , y = 0 are equilibrium points. The phase paths as follows. (i) x > a . The differential equation is d y d x = x + a y , which has the general solution y 2 + (x a) 2 = C 1 . These phase paths are semicircles centred at (a , 0 ) .
30 Nonlinear ordinary differential equations: problems and solutions x y a a Figure 1.38 Problem 1.20. (ii) a x a . The phase paths are the straight lines y = C 2 . (iii) x < a . The differential equation is d y d x = x a y , which has the general solution y 2 + (x + a) 2 = C 3 . These phase paths are semicircles centred at ( a , 0 ) . A sketch of the phase paths is shown in Figure 1.38. All paths are closed which means that all solutions are periodic. 1.21 The equation of motion of a conservative system is ¨ x + g(x) = 0, where g( 0 ) = 0, and g(x) is strictly increasing for all x , and x 0 g(u) d u → ∞ as x → ±∞ . (i) Show that the motion is always periodic. By considering g(x) = x e x 2 , show that if (i) does not hold, the motions are not all necessarily periodic. 1.21. The equation for the phase paths is d y d x = − g(x) y . The variables separate to give the general solution in the form 1 2 y 2 = − x 0 g(u) d u + C . ( i )

This preview has intentionally blurred sections. Sign up to view the full version.

1 : Second-order differential equations in the phase plane 31 Write x 0 g(u) d u = G(x) . ( ii ) Then (i) defines two families of paths where C > 0; y = 2 { C G(x) } 1 / 2 when G(x) < C ; ( iii ) and the reﬂection in the x axis; y = − 2 { C G(x) } 1 / 2 when G(x) < C . ( iv ) Since g(x) < 0 when x < 0, and g(x) > 0 when x > 0, then G(x) is strictly increasing to +∞ as x → −∞ and x → ∞ . Also G(x) is continuous and G( 0 ) = 0. Therefore, given any value of C > 0, G(x) takes the value C at exactly two values of x , one negative and the other positive. Consider the family of positive solutions (iii). Take any positive value of the constant C . At the two points where G(x) = C , we have y(x) = 0. Between them y(x) > 0, and the graph of the path cuts the x axis at right angles (see Section 1.2). When the corresponding reﬂected curve (iv) (y < 0 ) is joined to this one, we have a smooth closed curve. By varying the parameter C the process generates a family of closed curves nested around the origin (which is therefore a centre), and all motions are periodic.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business â€˜17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. Itâ€™s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania â€˜17, Course Hero Intern

• The ability to access any universityâ€™s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLAâ€™s materials to help me move forward and get everything together on time.

Jill Tulane University â€˜16, Course Hero Intern