Practice Problems for Final Exam

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1 Math 1280 Spring 2016 Practice Problems for Final Exam Part 2 (Sections 6.6, 6.7, 6.8, and chapter 7) S o l u t i o n s 1. Show that the given system has a nonlinear center at the origin. ẋ = 9y 5y 5, ẏ = 4x + x 3 xy 4 Solution: Denote f(y) = 9y 5y 5, g(y) = 4x + x 3 xy 4. f( y) = 9y + 5y 5 = f(y), f(y) is odd; g( y) = 4x + x 3 x( y) 4 = 4x + x 3 xy 4 = g(y), g(y) is even. Therefore, the system is reversible. It is easy to check that the origin is a FP because both right hand sides vanish when x = 0 and y = 0. [ y Jacobian is A = x 2 y 4 4xy 3 [ 0 9 A(0, 0) = T = 0, = 36 > 0 Hence, there is a center at the origin. 4 0 The system has a center at the origin and it is reversible. Hence it has a nonlinear center at the origin. 2. Show that the given system is reversible but not conservative. ẋ = 4y cos y + 3x sin y, ẏ = 5 cos y + 4 sin x Hint: Show that the point ( 0, π 2 ) is a FP and determine its type. Solution: Denote f(y) = 4y cos y + 3x sin y, g(y) = 5 cos y + 4 sin x. f( y) = 4( y) cos( y) + 3x sin( y) = 4y cos y 3x sin y = f(y), f(y) is odd; g( y) = 5 cos( y) + 4 sin x = 5 cos y + 4 sin x = g(y), g(y) is even. Therefore, the system is reversible. It is easy to check that the point ( 0, π 2 ) is a FP because both right hand sides vanish when x = 0 and y = π 2.

2 [ 3 sin y 4 cos y 4y sin y + 3x cos y Jacobian is A = 4 cos x 5 sin y [ 3 2π A(0, π) = T = 1 < 0, = 15+8π > 0 The FP ( ) 0, π 2 is an attractor. A conservative system can never have an attracting FP. Therefore, the system cannot be conservative. 3. Consider the system ẋ = x y x(x 2 + y 2 ), ẏ = x + y y(x 2 + y 2 ) (a) Write the system in polar coordinates Solution: ṙ = 1 r (xẋ + yẏ) = 1 r (x2 xy x 2 r 2 + xy + y 2 y 2 r 2 ) = 1 r (r2 r 4 ) ṙ = r(1 r 2 ) θ = 1 r (xẏ yẋ) = 1 2 r 2 (x2 + xy xyr 2 xy + y 2 + xyr 2 ) = 1 r 2 (r2 ) = 1 The system in polar coordinates is ṙ = r(1 r 2 ), θ = 1 (b) Show that a limit cycle exists and find it. Solution: If r(1 r 2 ) = 0, then r = 0 or r = 1. The circle r = 1 is the limit cycle. 4. Consider the system ẋ = x + y 2, ẏ = y(2 + 2x y 2 ) (a) Use index theory to show that the system has no closed orbits Solution: x-nullcline: x = y 2, y-nullcline: y = 0, x = 1 2 y2 1. The curves x = y 2 and x = 1 2 y2 1 don t intersect. Indeed, assume they intersect. Then we have y 2 = 1 2 y2 1 or 1 2 y2 = 1 which is impossible. Therefore, the only FP is the intersection of the curve x = y 2 and the line y = 0, Page 2

3 which is the point (0, 0). [ 1 2y Jacobian is A = 2y 2 + 2x 3y 2 [ 1 0 A(0, 0) = 0 2 = 2 < 0 and the FP (0, 0) is saddle. If there is a closed orbit C then it either encloses the saddle FP at the origin or does not. An index of a saddle point is 1 and an index of a closed orbit according to the index theory has to be +1. If C encloses the origin its index is 1. If C doesn t enclose it its index is 0. In either case the index of C is not +1. Therefore, the system cannot have a closed orbit. (b) Use a method different from index theory and Dulac s criterion to show that the system has no closed orbits Solution: Denote f(x, y) = x + y 2, g(x, y) = y(2 + 2x y 2 ). Then f y = 2y = g x and the system is gradient. Gradient systems have no closed orbits. Therefore, the system has no closed orbits. (c) Use Dulac s criterion to show that the system has no closed orbits. You may use a Dulac s function g(x, y) = 1 y. Solution: (g(x, y)(ẋ, ẏ)) = x (gẋ) + y (gẏ) = ( xy ) x + y + y (2 + 2x y2 ) = 1 y 2y. Now let s look at the x-axis. Suppose (x 0, 0) is a point on the axis. Let x(t) be a solution to the equation ẋ = x + y 2 with the initial condition x(0) = x 0, and let y(t) = 0. Then (x(t), y(t)) = (x(t), 0) is a solution to the given system. This means that the x-axis is a trajectory and cannot be crossed by another trajectory by existence and uniqueness theorem. The x-axis divides the xy-plane into two parts: the upper half-plane (y > 0) and the lower half-plane (y < 0). In each halves (g(x, y)(ẋ, ẏ)) preserves its sign ( (g(x, y)(ẋ, ẏ)) < 0 when y > 0 and (g(x, y)(ẋ, ẏ)) > 0 when y < 0 ). Each half is a simply connected region. Hence in each half conditions for Dulac s criterion hold and there are no closed orbits. Closed orbits being trajectories cannot cross the x-axis. Therefore, the system has no closed orbits in the xy-plane. Page 3

4 5. For ẋ = y + 3xy 2, ẏ = x 4x 2 y use Lyapunov function to show that the system has no closed orbits. Specifically, show that V (x, y) = x 2 + y 2 is Lyapunov function for the system and hence no closed orbits are allowed. Solution: Consider a Lyapunov function V (x, y) = x 2 + y V (0, 0) = V (x, y) > 0 for all (x, y) V = Vx ẋ + V y ẏ = 2x( y + 3xy 2 ) + 2y(x 4x 2 y) = 2xy + 6x 2 y 2 + 2yx 8x 2 y 2 = 2x 2 y 2 < 0 for all (x, y), a < b. V (x, y) = x 2 + y 2 and the system has no closed orbits. satisfies all requirements for a Lyapunov function 6. Determine if the given system is gradient. If it is, find its potential function V. ẋ = 2xy 3 + 3y 2 cos x, ẏ = 3(x 2 1)y 2 + 6y sin x Solution: Denote f(x, y) = 2xy 3 + 3y 2 cos x, g(x, y) = 3(x 2 1)y 2 + 6y sin x. f y = 6xy 2 + 6y cos x, g x = 6xy 2 + 6y cos x. f y = g x. Therefore, the system is gradient. V x (x, y) = f(x, y) = 2xy 3 + 3y 2 cos x, V (x, y) = x 2 y 3 + 3y 2 sin x + φ(y). V y (x, y) = g(x, y). 3x 2 y 2 + 6y sin x + φ (y) = 3(x 2 1)y 2 + 6y sin x. φ (y) = 3y 2, φ(y) = ( 3y 2 ) dy = y 3 + C. V (x, y) = x 2 y 3 + 3y 2 sin x y 3 + C. We can put C = 0. V (x, y) = (x 2 1)y 3 + 3y 2 sin x. 7. Consider the system ẋ = 3x 5y x(x 2 + 4y 2 ), ẏ = 5x + 3y y(2x 2 + y 2 ) Page 4

5 (a) Classify the FP at the origin. Solution: Jacobian: A = A(0, 0) = [ (0, 0) is an unstable spiral. [ 3 3x 2 4y 2 5 8xy 5 4xy 3 2x 2 3y 2, T = 3, = 34, > T 2. (b) Switch to polar coordinates and write an equation for ṙ Solution: ṙ = 1 r (xẋ + yẏ) = 1 r (3x2 5xy x 4 4x 2 y 2 + 5xy + 3y 2 2x 2 y 2 y 4 ) = 1 r (3r2 (x 4 + 2x 2 y 2 + y 4 ) 4x 2 y 2 ) = 1 ( 3r 2 (x 2 + y 2 ) 2 4r 4 cos 2 θ sin 2 θ ) = 1 ( 3r 2 r 4 r 4 sin 2 2θ ) r r ṙ = r ( 3 r 2 r 2 sin 2 2θ ) (c) Determine the circle of maximum radius r 1, centered at the origin such that all trajectories have a radially outward component on it. Solution: On this circle ṙ > 0. To find its radius we require 3 r 2 r 2 sin 2 2θ = 3 ( 1 + sin 2 2θ ) r 2 > 0. 0 sin 2 2θ 1, sin 2 2θ 2. Then a sufficient condition is 3 2r 2 > 0 3 Which gives r <. We take r 2 1 = (d) Determine the circle of minimum radius r 2, centered at the origin such that all trajectories have a radially inward component on it. Solution: On this circle ṙ < 0. To find its radius we require 3 ( 1 + sin 2 2θ ) r 2 < 0. As before, sin 2 2θ 2. Then a sufficient condition is 3 r 2 < 0 Page 5

6 Which gives r > 3. We take r 2 = (e) There is no FPs in the region r 1 r r 2. You don t have to show that. Prove that the system has a limit cycle somewhere in the trapping region r 1 r r 2. Solution: There is a source at the origin. All trajectories go inside the trapping region through both circles of radii r 1 and r 2 centered at the origin. There are no FPs inside the region. By Poincare-Bendixson theorem there must be a limit cycle inside the region r 1 r r 2. Page 6

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