Phase Portraits of 1-D Autonomous Equations In each of the following problems [1]-[5]: (a) find all equilibrium solutions; (b) determine whether each of the equilibrium solutions is stable, asmptoticall stable or unstable; and (c) sketch the phase portrait. [1] [2] [3] [4] [5] dp = P(P 2 1)(P 3). = ( 1)( 3)2. = sin(π). dx (t) = sin2 (πx(t)). = f(), where the function f() is piecewise defined b: 2 0, f() = 0 0 < < 1, 1 1. [6] An equation = f() has the following phase portrait. 1 2 4 (a) Find all equilibrium solutions. (b) Determine whether each of the equilibrium solutions is stable, asmptoticall stable or unstable. (c) Graph the solutions (t) vs t, for the initial values (1.4) = 0, (0) = 0.5, (0) = 1, (0) = 1.1, (0) = 1.5, ( 0.5) = 1.5, (0) = 2, (0) = 2.5, (0) = 3, (0) = 3.5, (0) = 4, (0) = 4.5, ( 1) = 4.5. (Without further quantitative information about the equation and the solution formula, it s clearl impossible to draw accurate graphs of (t) vs t. Here, tr to sketch graphs qualitativel to show the correct namic properties. The point is that a great deal of info about solution namics can be read off from one simple figure of phase portrait.)
[7] Several solution graphs (t) vs t are given below, for an equation = f(). (a) Find all equilibrium solutions in the interval 4 < < 4; (b) Determine whether each of the above equilibrium solutions is stable, asmptoticall stable or unstable; (c) Sketch phase portrait on the interval 4 < < 4. In each of the following problems [8]-[9]: [8] ( ) [9] ( ) (a) find all equilibrium solutions of the equation ( ); (b) for each equilibrium point, write down the linear approximating equation near the equilibrium and determine whether the equilibrium is stable, asmptoticall stable or unstable with respect to the linear approximating equation; (c) tr to use the linear stabilit/instabilit obtained in (b) to determine whether each of the equilibria is stable, asmptoticall stable or unstable with respect to the nonlinear equation ( ); (d) if the linear approximation obtained in (b) was not enough to determine the stabilit of an equilibrium with respect to the nonlinear equation ( ), use other methods to determine whether the equilibrium is stable, asmptoticall stable or unstable with respect to the nonlinear equation ( ) = ( 1)( + 2) = ( 1)( + 2)2
In each of the following problems [10]-[11]: [10] ( ) [11] ( ) (a) verif that = 1 is an equilibrium; (b) give the linear approximating equation for for 1; (c) determine whether = 1 is stable, asmptoticall stable or unstable with respect to the nonlinear equation ( ). = 2 1 + cos(π). = 2 2 + sin(π). (See next page for answers)
Answers: [1] There are four equilibrium solutions: P = 1, 0, 1, 3. The equilibria P = 1 and P = 1 are asmptoticall stable. The equilibria P = 0 and P = 3 are unstable. 1 0 1 3 P [2] There are two equilibrium solutions: = 1, 3. The equilibrium = 1 is asmptoticall stable. The equilibrium = 3 is unstable. 1 3 [3] There are infinitel man equilibrium solutions: an integer is an equilibrium. Among these equilibria, odd integers = ±1, ±3, ±5, are asmptoticall stable, while even integers = 0, ±2, ±4, ±6, are unstable. 3 2 1 0 1 2 3 [4] There are infinitel man equilibrium solutions: an integer is an equilibrium. All equilibria are unstable. x 3 2 1 0 1 2 3 [5] There are infinitel man (actuall a continuum of) equilibrium solutions: each point in the closed interval 0 1 is an equilibrium. The equilibrium = 0 is unstable. All other equilibria 0 < 1 are stable but not asmptoticall stable. 0 1
[6] There are three equilibria: = 1, 2, 4. The equilibrium = 4 is asmptoticall stable. The equilibria = 1 and = 2 are unstable. A rough sketch of the solution graphs is given below. Besides the monotone properties and namic behavior of the solutions, also note that the solution graphs between 2 < < 4 should be all congruent. Indeed, the are horizontal translations of each other. This also holds for each of the following intervals: 1 < < 2, 4 < <, and < < 1. [7] There are three equilibria: = 2, 0, 2. The equilibria = 2 and = 2 are asmptoticall stable. The equilibrium = 0 is unstable. 2 0 2 [8] (a) There are three equilibria: = 2, 0, 1. (b) Near = 2: the linear approximating equation is ( ) = 6( + 2). The equilibrium = 2 is unstable with respect to the lin approx eq ( ). Near = 0: the linear approximating equation is ( ) = 2. The equilibrium = 0 is asmptoticall stable w.r.t. the lin approx eq ( ). Near = 1: the linear approximating equation is ( ) = 3( 1). The equilibrium = 1 is unstable with respect to the lin approx eq ( ). (c) Since each of the linear approximating equations in (b) is non-degenerate, the nonlinear namics near the equilibrium can be qualitativel determined b the linear namics.
The equilibria = 2 and = 1 are unstable with respect to the nonlin eq ( ). The equilibrium = 0 is asmptoticall stable w.r.t. the nonlin eq ( ). (d) No need to consider. [9] (a) There are three equilibria: = 2, 0, 1. (b) (c) Near = 2: the linear approximating equation is ( ) = 0. The equilibrium = 2 is stable but not asmptoticall stable, with respect to the lin approx eq ( ). Near = 0: the linear approximating equation is ( ) = 4. The equilibrium = 0 is unstable with respect to the lin approx eq ( ). Near = 1: the linear approximating equation is ( ) = 9( 1). The equilibrium = 1 is asmptoticall stable w.r.t. the lin approx eq ( ). The linear approximations are sufficient to determine the nonlinear namics near = 0 and near = 1 on the qualitativel level. The equilibrium = 0 is unstable with respect to the nonlin eq ( ). The equilibrium = 1 is asmptoticall stable w.r.t. the nonlin eq ( ). On the other hand, the linear approximating equation near = 2 is degenerate. The linear approximation is insufficient to determine the nonlinear namics near = 2. (d) For = 2, the stabilit/instabilit w.r.t. the nonlinear equation ( ) can be determined b stuing the sign changes of the nonlinear term f() = ( 1)( + 2) 2 for 2. Answer: The equilibrium = 2 is unstable w.r.t. the nonlinear equation ( ). [10] (a) Let f() = 2 1 + cos(π). Verif that f(1) = 0. (b) = 2( 1) (c) = 1 is unstable with respect to the nonlin eq ( ). [11] (a) Let f() = 2 2 + sin(π). Verif that f(1) = 0. (b) = (2 π)( 1) (c) = 1 is asmptoticall stable with respect to the nonlin eq ( ).