appm2360summer2016exam4_sol - APPM 2360 Summer 2016 Exam 4...

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APPM 2360, Summer 2016: Exam 4 July 22, 2016 Instructions : Please show all of your work and make your methods and reasoning clear. Answers out of the blue with no supporting work will receive no credit (unless the directions to a specific problem say otherwise, of course). No calculators or electronic devices are allowed. Please start each numbered problem on a new page in your bluebook and do the problems in order. Please sign the front of your blue book indicating you read and understood these directions in addition to the CU honor code. 1. (24 points) Consider the following system of differential equations: dx dt = x ( y - x - 2) dy dt = x 2 - y You must do parts (a) and (b) correctly, otherwise you stand to lose a lot of points in parts (c)! You don’t have to provide any work in this problem and no partial credit will be given, but to help you start correctly here’s a hint : you should find three equilibrium points in part (b). (a) Determine all horizontal and vertical nullclines of the system. (b) Determine all equilibrium points of this system. (c) Now use the Jacobian to classify the geometry and stability of each equilibrium point you found in part (b), if possible. Your answer should include, for each equilibrium point: the relevant eigen- value(s), what those eigenvalues predict about the stability, and what the eigenvalues predict the phase portrait in the immediate vicinity of each equilibrium “should look like.” If you cannot con- clude anything based on the Jacobian, explain why not in one sentence. Solution : (a) Horizontal nullclines, set dy dt = 0: x 2 - y = 0 or y = x 2 . Vertical nullclines, set dx dt = 0: x ( y - x - 2) = 0 so that x = 0 or y - x - 2 = 0, or y = x + 2 . (b) y = x 2 and x = 0 intersect when y = 0 too so one equilibrium point is at (0 , 0) . y = x 2 and y = x + 2 intersect when x 2 = x +2, or x 2 - x - 2 = 0 or when ( x - 2)( x +1) = 0 so that x = - 1 , 2. The corresponding y -values are 1 and 4 respectively, so we have two more equilibrium points: ( - 1 , 1) and (2 , 4) . (c) With f ( x, y ) = xy - x 2 - 2 x and g ( x, y ) = x 2 - y the Jacobian is J ( x, y ) = f x f y g x g y = y - 2 x - 2 x 2 x - 1 Then J (0 , 0) = - 2 0 0 - 1 which has eigenvalues λ = - 2 , - 1. These are real, distinct and both negative so the equilibrium point at (0 , 0) is an asymptotically stable attracting node.
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