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X 1 1 f x 0 1 1 figure 151 problem 139 graph showing

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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 ∂φ ∂x = c 2 φ ∂x 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 φ(x , t) = U(x ct) in Burgers’ equation ∂φ ∂t + φ ∂φ ∂x = c 2 φ ∂x 2 , so that U(x ct) satisfies the ordinary differential equation cU (w) + U(w)U (w) = cU (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 ) The phase paths are congruent parabolas all with the axis U = c as shown in Figure 1.52. Phase paths are bounded for V < 0 and unbounded for V > 0: the latter do not have an obvious physical interpretation. 1 2 3 4 U V u 1 u 2 2 1 1 2 Figure 1.52 Problem 1.40: Phase diagram for permanent waves of Burger’s equation.
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56 Nonlinear ordinary differential equations: problems and solutions w U u 1 u 2 Figure 1.53 Problem 1.40: A permanent waveform of Burgers’ equation. Burgers’ equation describes a convection-diffusion process, and a solution U(x ct) is the shape of a wavefront. For U < 0, the wavefront starts at U = u 1 , say, and terminates at U = u 2 . We can obtain an explicit form for the wave if we assume that U u 1 as w → ∞ , and U u 2 as w → −∞ , and U 0 in both cases, with u 2 < u 1 . Hence from (i), A = − 1 2 (u 1 c) 2 = − 1 2 (u 2 c) 2 , so that c = 1 2 (u 1 + u 2 ) and A = − 1 8 (u 1 u 2 ) 2 . Hence (i) becomes c ˙ U = (U u 1 )(U u 2 ) . This is a separable equation with solution w = u 1 + u 2 u 2 u 1 d U U u 1 + d U u 2 U = u 1 + u 2 u 2 u 1 ln u 2 U U u 1 . Solving for U we obtain U(x ct) = u 1 + u 2 u 1 1 + exp [ (u 2 u 1 )(x ct)/(u 1 + u 2 ) ] . The shape of the waveform is indicated in Figure 1.53. 1.41 A uniform rod of mass m and length L is smoothly pivoted at one end and held in a vertical position of equilibrium by two unstretched horizontal springs, each of stiffness k , attached to the other end as shown in Figure 1.37 (in NODE) or Figure 1.54. The rod is free to oscillate in a vertical plane through the springs and the rod. Find the potential energy V (θ) of the system when the rod is inclined at an angle θ to the upward vertical. For small θ confirm that V (θ) (kL 1 4 mg)Lθ 2 , Sketch the phase diagram for small | θ | , and discuss the stability of this inverted pendulum.
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1 : Second-order differential equations in the phase plane 57 ( L sin , LT cos ) L Figure 1.54 Problem 1.41: Inverted pendulum inclined at angle θ .
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