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Unformatted text preview: CEIC3000 Process Modelling and Analysis Week 5: Nonlinear Analysis Session 1, 2012 Lecturer: A/Prof. J. Bao 1 Phaseplane Analysis (continued) If the eigenvalues of matrix A are complex, the phase plane plots will have the behaviors called “focus” or “center”. Example 1 (Unstable Focus (Spiral Source)) Consider the following system of equations: ˙ x 1 = x 1 + x 2 ˙ x 2 = − 2 x 1 + x 2 (1) We have: A = [ 1 2 − 2 1 ] (2) with eigenvalues λ 1 = 1 + 2 i and λ 2 = 1 − 2 i. This system is unstable because the real portion of the complex eigenvalues is positive. Example 2 (Center) Consider the following system of equations: ˙ x 1 = − x 1 − x 2 ˙ x 2 = 4 x 1 + x 2 (3) We have: A = [ − 1 − 1 4 1 ] (4) with eigenvalues λ 1 = 0 + 1 . 732 i and λ 2 = 0 − 1 . 732 i. Since the real part of the eigenvalues is , this system has a periodic solution, resulting in a phaseplane plot where the equilibrium point is a center, as shown in Figure . The phaseplane plots of second order systems will have the following behaviors: Sinks (stable nodes): λ 1 < 0 and λ 2 < 0, where λ 1 , λ 1 ∈ R Saddles (unstable): λ 1 < 0 and λ 2 > 0, where λ 1 , λ 1 ∈ R Source (unstable nodes): λ 1 > 0 and λ 2 > 0, where λ 1 , λ 1 ∈ R Spirals: λ 1 and λ 2 are complex conjugates. If Re( λ 1 ) < 0 then stable otherwise unstable Center: λ 1 and λ 2 are complex conjugates with Re( λ 1 ) = 0 1 Figure 1: Phaseplane plot for Example 1 Figure 2: Phaseplane plot for Example 2 2 1.1 Generalisation of Phaseplane Behaviour Let us generalise the phaseplane analysis results for secondorder linear systems: ˙ x = Ax (5) where A = [ a 11 a 12 a 21 a 22 ] (6) The eigenvalues can be found by solving det ( λI − A ) = 0 : det ( λI − A ) = ( λ − a 11 ) ( λ − a 22 ) − a 12 α 21 = λ 2 − ( a 11 + a 22 ) λ + a 11 a 22 − a 12 α 21 = λ 2 − tr( A ) λ + det ( A ) (7) Therefore the quadratic formula can be used to find the eigenvalues: λ 1 = tr ( A ) − √ (tr ( A )) 2 − 4 det ( A ) 2 (8) λ 2 = tr ( A ) + √ (tr ( A )) 2 − 4 det ( A ) 2 (9) We have: 1. At least one eigenvalue will be negative if tr ( A ) < 0. 2. If tr ( A ) 2 > 4 det ( A ) , the eigenvalues will be real. (a) If det ( A ) > 0: i. If tr ( A ) > 0, λ 1 > , λ 2 > 0 : Node source ii. If tr ( A ) < 0, λ 1 < , λ 2 < 0 : Node sink (b) If det ( A ) < , then one of the eigenvalues is positive and one is negative: saddle 3. If tr ( A ) 2 < 4 det ( A ) , the eigenvalues will be complex In this case, Re ( λ i ) = tr ( A ) . (a) If tr ( A ) > 0: Spiral source (b) If tr ( A ) < 0 : Spiral sink Example 3 A = [ − 1 − 4 ] tr ( A ) = − 5 det ( A ) = 4 Therefore, it is a stable node....
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This note was uploaded on 03/26/2012 for the course CHEM ENG CEIC at University of New South Wales.
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