9780199212033

# Consider eqn iii using v and vi if ε 1 ln ε 1 c vii

This preview shows pages 52–55. Sign up to view the full content.

Consider eqn (iii), using (v) and (vi). If ε 1 ln 1 ) < C < ( vii ) there are exactly two values in the range −∞ < y < ε 1 at which the factor C F(y) , and hence x , becomes zero, and between these values x > 0. The corresponding reﬂected path segment given by (iv) completes a closed path, having parameter C . A representative phase diagram is given in Figure 1.50. The unclosed paths correspond to values of y > ε 1 : their boundary is the singular solution mentioned above. 1.38 Show that the equation ¨ x + x + εx 3 = 0 (ε > 0 ) with x( 0 ) = a , ˙ x( 0 ) = 0 has phase paths given by ˙ x 2 + x 2 + 1 2 εx 4 = ( 1 + 1 2 εa 2 )a 2 . Show that the origin is a centre. Are all phase paths closed, and hence all solutions periodic? 1.38. The differential equation of the phase paths of ¨ x + x + εx 3 = 0, (ε > 0 )

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

View Full Document
1 : Second-order differential equations in the phase plane 53 1 1 x 1 1 y Figure 1.50 Problem 1.37: The phase diagram for ¨ x εx ˙ x + x = 0 with ε = 1. is given by y d y d x = − x εx 3 . Given the conditions x( 0 ) = a and ˙ x( 0 ) = 0, integration of the differential equation gives the phase paths ˙ x 2 + x 2 + 1 2 εx 4 = constant = ( 1 + 1 2 εa 2 )a 2 . The equation has a single equilibrium point, at the origin. This is a conservative system (see NODE, Section 1.3) with potential function V (x) = (x + εx 3 ) d x = 1 2 x 2 + 1 4 εx 4 . Differentiating V (x) twice we obtain V (x) = x + 1 2 εx 3 , V (x) = 1 + 3 2 εx 2 . Therefore V ( 0 ) = 0 and V ( 0 ) = 1 > 0 which means that V (x) has a minimum at the origin. Locally the origin is a centre. However, V (x) < 0 for x < 0 and V (x) > 0 for x < 0, and also V (x) → ∞ as x → ± ∞ . These conditions imply that, for every a , x is also zero at x = − a and y is continuous between zero at x = − a and x = a . Hence every path is closed and all solutions periodic.
54 Nonlinear ordinary differential equations: problems and solutions 1.39 Locate the equilibrium points of the equation ¨ x + λ + x 3 x = 0. in the x , λ plane. Show that the phase paths are given by 1 2 ˙ x 2 + λx + 1 4 λx 4 1 2 x 2 = constant. Investigate the stability of the equilibrium points. 1.39. Consider the parameter-dependent system ¨ x + λ + x 3 x = 0. Using the notation of Section 1.7 (in NODE), let f (x , λ) = x x 3 λ . Equilibrium points occur where f (x , λ) = 0 which is shown as the curve in Figure 1.51. The function f (x , λ) is positive in the shaded region. Points on the curve λ = x x 3 above the shaded areas are stable and all other points are unstable. Treating λ as a function of x , x x 3 has stationary points at x = ± 1 / 3 where λ = ± 2 / 3 as indicated in Figure 1.51. Therefore if 2 / 3 < λ < 2 / 3 the equation has three equilibrium points; if λ = ± 2 / 3 the equation has two; for all other values of λ the equation has one equilibrium point. The phase paths satisfy the differential equation y d y d x = − x 3 + x λ , where y = ˙ x . Integrating, the phase paths are given by 1 2 y 2 + λx + 1 4 x 4 1 2 x 2 = constant.

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

View Full Document
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 ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’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 ‘17, Course Hero Intern

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

Jill Tulane University ‘16, Course Hero Intern