Appendix: A Computer-Generated Portrait Gallery

Transcription

1 Appendi: A Computer-Generated Portrait Galler There are a number of public-domain computer programs which produce phase portraits for 2 2 autonomous sstems. One has the option of displaing the trajectories with or without the little" arrows of the vector direction field. We first give an eample where the direction field is included, and then a portrait galler with onl the tractories themselves. In the pictures which illustrate these cases, onl a few trajectories are shown, but these are sufficent to shown the qualitative behavior of the sstem. A much richer understanding of this galler can be achieved using the Mathlets Linear Phase Portraits: Cursor Entr and Linear Phase Portraits: Matri Entr. Eample. (direction field included). ( ) ( ) =. General solution: = c 1 cos t + c 2 sin t, = c 1 sin t + c 2 cos t As we saw before, the trajectories are circles. We know the direction is clockwise b looking at a single tangent vector. The portrait galler. 1. A has real eigenvalues with two independent eigenvectors. Let λ 1, λ 2 be the eigenvalues and v 1 and v 2 the corresponding eigenvectors. general solution to ( ) is = c 1 e λ1t v 1 + c 2 e λ2t v Vector field: ' =, ' = i) λ 1 > λ 2 > 0. Unstable nodal source. As t the term e λ 1t v 1 dominates and. As t the term e λ 2t v 2 dominates and 0.

2 Appendi: A Computer-Generated Portrait Galler ii) λ 1 < λ 2 < 0. Stable nodal sink (asmptoticall stable). (Simpl reverse the arrows on case (i).) As t 0 the term e λ 2t v 2 dominates and. As t the term e λ 1t v 1 dominates and. iii) λ 1 > 0 > λ 2. Saddle, unstable. As t the term e λ 1t v 1 dominates and. As t the term e λ 2t v 2 dominates and. iv) λ 1 = 0 > λ 2. Line of critical points. The critical points are not isolated the lie on the line through 0 with direction v 1. = c 1 v 1 + c 2 e λ 2t v 2 As t c 1 v 1 along a line parallel to v 2. v) λ 1 = 0 < λ 2. Line of critical points. (Simpl reverse the arrows in case (iv).) The critical points are not isolated the lie on the line through 0 with direction v 1. = c 1 v 1 + c 2 e λ 2t v 2 As t along a line parallel to v 2. 2

3 Appendi: A Computer-Generated Portrait Galler vi) λ 1 = λ 2 > 0. Star nodal source (unstable). Let λ = λ 1 = λ 2. Two independent eigenvectors A is a scalar matri = e λt c. That is all trajectories are straight ras. As t along a line from 0. As t 0 vii) λ 1 = λ 2 < 0. Star nodal sink (asmptoticall stable). (Simpl reverse the arrows in case (vi).) viii) λ 1 = λ 2 = 0. Non-isolated critical points. Ever point is a critical point, ever trajector is a point. 2. Real defective case (repeated eigenvalue, onl one eigenvector). Let λ be the eigenvalue and v 1 the corresponding eigenvector. Let v 2 be a generalized eigenvector associated with v 1. general solution to ( ) is = e λt (c 1 v 1 + c 2 (tv 1 + v 2 )). i) λ > 0. Defective source node (unstable). As t the term tv 1 dominates and. As t the term tv 1 dominates and 0. ii) λ < 0. Defective sink node (asmptoticall stable). (Simpl reverse the arrows in case (i).) 3

4 Appendi: A Computer-Generated Portrait Galler iii) λ = 0. Line of critical points. = c 1 v 1 + c 2 (tv 1 + v 2 ). Trajectories are parallel to v Complet roots (pair of comple conjugate eigenvalues and vectors). Let an eigenvalue be λ = α + β i, with eigenvector v + i w. = e αt (c 1 (cos βt v sin βt w) + c 2 (sin βt v + cos βt w)). The sines and cosines in the solution will cause the trajector to spiral around the critical point. i) Re(λ) > 0 (i.e. α > 0). Spiral source (unstable). Trajectories can spiral clockwise or counterclockwise. As t,. As t, 0. ii) Re(λ) < 0 (i.e. α < 0). Spiral sink (asmptoticall stable). (Reverse arrows from case (i).) Trajectories can spiral clockwise or counterclockwise. As t, 0. As t,. iii) Re(λ) = 0 (i.e. α = 0). Stable center. Trajectories can turn clockwise or counterclockwise. As t ±, follows an ellipse. For the comple case ou can find the direction of rotation b checking the tangent vector at one point. 4

5 Appendi: A Computer-Generated Portrait Galler ( ) 2 3 Eample. A = has eigenvalues 2 ± 3i. the critical point is 3 2 a spiral source. ( ) ( ) 1 2 The tangent vector at the point 0 = is A 0 =. 0 3 spiral is clockwise. 5

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