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Unformatted text preview: A Catalogue of Phase Portraits of 2D Linear Flows Consider all (real variable) solutions of 2D system bracketleftbigg x ′ ( t ) y ′ ( t ) bracketrightbigg = A bracketleftbigg x ( t ) y ( t ) bracketrightbigg , where A is a constant 2 × 2 real matrix. By sketching the phase portrait of this system, we mean to draw several representative solution curves on the x y plane to display typical asymptotic behavior, especially the solution behavior as t → ∞ and t → −∞ . Equilibria and periodic solutions should be displayed. The stability of equilibria and periodic solutions should be easily read off from the picture. The structure of eigenvalues and (generalized) eigenvectors of matrix A gives the solution formula, completely determines dynamic behavior of the system, and therefore will also guide our classification of phase portraits. u 2 u 1 x K 4 K 2 2 4 y K 4 K 2 2 4 Attractive_improper_node The eigenvalues of A are: λ 1 < λ 2 < 0. Solution Method: • Prepare an eigenvector u 1 for λ 1 : that is, ( A − λ 1 I ) u 1 = 0 , u 1 negationslash = 0. • Prepare an eigenvector u 2 for λ 2 : that is, ( A − λ 2 I ) u 2 = 0 , u 2 negationslash = 0. • The general solutions of the differential system are bracketleftbigg x ( t ) y ( t ) bracketrightbigg = C 1 e λ 1 t u 1 + C 2 e λ 2 t u 2 , where C 1 and C 2 are free parameters. Example. A = 1 19 bracketleftbigg − 58 5 − 4 − 37 bracketrightbigg , λ 1 = − 3 , u 1 = bracketleftbigg 5 1 bracketrightbigg , λ 2 = − 2 , u 2 = bracketleftbigg 1 4 bracketrightbigg . u 1 u 2 x K 4 K 2 2 4 y K 4 K 2 2 4 Repulsive_improper_node The eigenvalues of A are: λ 1 > λ 2 > 0. Solution Method: • Prepare an eigenvector u 1 for λ 1 : that is, ( A − λ 1 I ) u 1 = 0 , u 1 negationslash = 0. • Prepare an eigenvector u 2 for λ 2 : that is, ( A − λ 2 I ) u 2 = 0 , u 2 negationslash = 0. • The general solutions of the differential system are bracketleftbigg x ( t ) y ( t ) bracketrightbigg = C 1 e λ 1 t u 1 + C 2 e λ 2 t u 2 , where C 1 and C 2 are free parameters. Example. A = 1 19 bracketleftbigg 58 − 5 4 37 bracketrightbigg , λ 1 = 3 , u 1 = bracketleftbigg 5 1 bracketrightbigg , λ 2 = 2 , u 2 = bracketleftbigg 1 4 bracketrightbigg . u 1 u 2 x K 4 K 2 2 4 y K 4 K 2 2 4 Saddle The eigenvalues of A are: λ 1 < < λ 2 . Solution Method: • Prepare an eigenvector u 1 for λ 1 : that is, ( A − λ 1 I ) u 1 = 0 , u 1 negationslash = 0....
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This note was uploaded on 01/24/2010 for the course MATH 2403 taught by Professor Wang during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Wang
 Statistics, Probability

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