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= − y x 2 2 1 1 at(1 1 the linearized system has

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Unformatted text preview: = − y x 2 2 1 1 . At (1, 1), the linearized system has coefficient matrix: A = − 2 2 1 1 . The eigenvalues are 2 7 3 i r ± = . The critical point is an unstable spiral point. At (−1, −1), the linearized system has coefficient matrix: A = − − − 2 2 1 1 . The eigenvalues are 2 17 1 ± − = r . The critical point is an unstable saddle point. The phase portrait is shown on the next page. © 2008 Zachary S Tseng D-2 - 25 (1, 1) is an unstable spiral point. (−1, −1) is an unstable saddle point. © 2008 Zachary S Tseng D-2 - 26 Example : x ′ = x − xy y ′ = y + 2 xy The critical points are at (0, 0) and (−1/2, 1). The Jacobian matrix is J = + − − x y x y 2 1 2 1 At (0, 0), the linearized system has coefficient matrix: A = 1 1 . There is a repeated eigenvalue r = 1. A linear system would normally have an unstable proper node (star point) here. But as a nonlinear system it actually has an unstable node. (Didn’t I say that this approximation using linearization is not always 100% accurate?) At (−1/2, 1), the linearized system has coefficient matrix: A = 2 2 / 1 . The eigenvalues are r = 1 and −1. Thus, the critical point is an unstable saddle point. The phase portrait is shown on the next page. © 2008 Zachary S Tseng D-2 - 27 (0, 0) is an unstable node. (−1/2, 1) is an unstable saddle point. © 2008 Zachary S Tseng D-2 - 28 Exercises D-2.2 : Find all the critical point(s) of each nonlinear system given. Then determine the type and stability of each critical point. 1. x ′ = xy + 3 y y ′ = xy − 3 x 2. x ′ = x 2 − 3 xy + 2 x y ′ = x + y − 1 3. x ′ = x 2 + y 2 − 13 y ′ = xy − 2 x − 2 y + 4 4. x ′ = 2 − x 2 − y 2 y ′ = x 2 − y 2 5. x ′ = x 2 y + 3 xy – 10 y y ′ = xy − 4 x Answer D-2.2 : 1. Critical points are (0, 0) and (−3, 3). (0, 0) is a stable center, and (−3, 3) is an unstable saddle point. 2. Critical points are (0, 1) and (1/4, 3/4). (0, 1) is an unstable saddle point, and (1/4, 3/4) is an unstable spiral point. 3. Critical points are (2, 3), (2, −3), (3, 2), and (−3, 2). (2, 3) is an unstable saddle point, (2, −3) is an unstable saddle point, (3, 2) is an unstable node, and (−3, 2) is an asymptotically stable node. 4. Critical points are (1, 1), (1, −1), (−1, 1), and (−1, −1). (1, 1) is an asymptotically stable spiral point, (1, −1) and (−1, 1) both are unstable saddle points, and (−1, −1) is an unstable spiral point. 5. Critical points are (0, 0), (2, 4), and (−5, 4). (0, 0) is an unstable saddle point, (2, 4) is an unstable node, and (−5, 4) is an asymptotically stable node....
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= − y x 2 2 1 1 At(1 1 the linearized system has...

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