This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 D2  25 (1, 1) is an unstable spiral point. (−1, −1) is an unstable saddle point. © 2008 Zachary S Tseng D2  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 D2  27 (0, 0) is an unstable node. (−1/2, 1) is an unstable saddle point. © 2008 Zachary S Tseng D2  28 Exercises D2.2 : 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 D2.2 : 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....
View
Full Document
 Spring '13
 MRR
 Math, Differential Equations, Equations, Critical Point, Linear Systems, Stationary point, Stability theory, Zachary S Tseng

Click to edit the document details