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9780199212033

# 138 show that the equation x x εx 3 ε with x a 0

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1.38 Show that the equation ¨ x + x + εx 3 = 0 (ε > 0 ) with x( 0 ) = a , ˙ 0 ) = 0 has phase paths given by ˙ x 2 + x 2 + 1 2 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 + 3 = 0, (ε > 0 )

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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 3 . Given the conditions x( 0 ) = a and ˙ 0 ) = 0, integration of the differential equation gives the phase paths ˙ x 2 + x 2 + 1 2 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) = Z (x + 3 ) d x = 1 2 x 2 + 1 4 4 . Differentiating V twice we obtain V ± = x + 1 2 3 , V ±± = 1 + 3 2 2 . Therefore V ± ( 0 ) = 0 and V ±± ( 0 ) = 1 > 0 which means that V has a minimum at the origin. Locally the origin is a centre. However, V ± (x)< 0 for x< 0 and V ± (x)> 0 for 0, and also V →∞ as x →²∞ . 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 , λ) = 0 which is shown as the curve in Figure 1.51. The function , 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. x 1 ± –1 f ( x , ± )>0 1 –1 Figure 1.51 Problem 1.39: Graph showing equilibrium points on λ = x x 3 .

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1 : Second-order differential equations in the phase plane 55 1.40 Burgers’ equation ∂φ ∂t + φ ∂x = c 2 φ 2 shows diffusion and nonlinear effects in ﬂuid mechanics (see Logan (1994)). Find the equa- tion for permanent waves by putting φ(x , t) = U(x ct) , where c is the constant wave speed. Find the equilibrium points and the phase paths for the resulting equation and interpret the phase diagram. 1.40. Let , = in Burgers’ equation + φ = c 2 φ 2 , so that satis±es the ordinary differential equation cU ± (w) + U(w)U ± (w) = ±± (w) , where w = x ct . All values of w are equilibrium points. Let V = U ± . Then the phase paths in the (U , V) plane are given by c d V d U = U c , which has the general solution cV = 1 2 (U c) 2 + A . ( i )
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138 Show that the equation x x εx 3 ε with x a 0 has...

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