Math 211 Math 211 Lecture #28 Phase Plane Portraits November 2, - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #28 Phase Plane Portraits November 2, 2001

2 Procedure to Solve x ′ = A x Procedure to Solve x ′ = A x • Find the eigenvalues of A � the roots of p ( λ ) = det( A − λI ) = 0 • For each eigenvalue λ find the eigenspace � = null( A − λI ) • If λ is an eigenvalue and v is an associated eigenvector, x ( t ) = e λt v is a solution. • Hope that n of these are linearly independent. Return

3 Planar System x ′ = A x Planar System x ′ = A x � a 11 � x 1 ( t ) a 12 � � A = and x ( t ) = a 21 a 22 x 2 ( t ) • The characteristic polynomial is p ( λ ) = λ 2 − Tλ + D. where T = tr A and D = det A Return Procedure

4 • The eigenvalues of A are the roots of p ( λ ) = λ 2 − Tλ + D, √ T 2 − 4 D λ = T ± . 2 • Three cases: � 2 distinct real roots if T 2 − 4 D > 0 � 2 complex conjugate roots if T 2 − 4 D < 0 � Double real root if T 2 − 4 D = 0 Return

5 Procedure in Degenerate Planar Case Procedure in Degenerate Planar Case • Find the (only) eigenvalue λ 1 . • Find an eigenvector v 1 � = 0 . • Find v 2 with ( A − λI ) v 2 = v 1 . � Start with any vector w not a multiple of v 1 � Then ( A − λI ) w = a v 1 with a � = 0 . � Set v 2 = 1 a w . v 2 is not a multiple of v 1 . • x 1 ( t ) = e λt v 1 and x 2 ( t ) = e λt [ v 2 + t v 1 ] form a fundamental set of solutions. Return

6 Example Example � 1 9 � x ′ = A x where A = − 1 − 5 • p ( λ ) = λ 2 + 4 λ + 4 = ( λ + 2) 2 ; λ = − 2 � 3 � − 3 9 � � • A − λI = ; v 1 = − 1 − 3 1 • Eigenspace has dimension 1, with basis v 1 . • One exponential solution: � − 3 � x 1 ( t ) = e λt v 1 = e − 2 t . 1 Return Procedure

7 • Second solution � Start with w = (1 , 0) T . � − 1 � � v 2 = − w = 0 • Fundamental set of solutions: � − 3 � x 1 ( t ) = e λt v 1 = e − 2 t 1 x 2 ( t ) = e λt [ v 2 + t v 1 ] � − 1 − 3 t � = e − 2 t . t Return Example Procedure

8 Examples Examples Solve x ′ = A x , where • � − 2 1 � A = 0 − 2 • � 0 9 � A = − 1 − 6 Procedure

9 Planar System x ′ = A x Planar System x ′ = A x • Equilibrium points for the system � Set of equilibrium points equals null( A ) . � A nonsingular ⇒ only equilibrium point is 0 . • Can we list the types of all possible equilibrium points for planar linear systems? � We will do the six most important cases. ◮ The other cases are Project #3. � Look at solution curves in the phase plane. Return

10 Exponential Solutions Exponential Solutions x ( t ) = Ce λt v • The solution curve is a straight half-line through C v . Sometimes called half-line solutions. • If λ > 0 the solution starts at 0 for t = −∞ , and tends to ∞ as t → ∞ . Unstable solution • If λ < 0 the solution starts at ∞ for t = −∞ , and tends to 0 as t → ∞ . Stable solution Return

11 Distinct Real Eigenvalues Distinct Real Eigenvalues • p ( λ ) = λ 2 − Tλ + D with T 2 − 4 D > 0 . √ √ T 2 − 4 D T 2 − 4 D λ 1 = T − < λ 2 = T + 2 2 • Eigenvectors v 1 and v 2 . General solution x ( t ) = C 1 e λ 1 t v 1 + C 2 e λ 2 t v 2 Return Exponential solution

12 Saddle Point Saddle Point • λ 1 < 0 < λ 2 • General solution x ( t ) = C 1 e λ 1 t v 1 + C 2 e λ 2 t v 2 • Two stable exponential solutions ( C 2 = 0 ) • Two unstable exponential solutions ( C 1 = 0 ). • C 1 � = 0 and C 2 � = 0 . � As t → ∞ , x ( t ) → ∞ , approaching the half-line through C 2 v 2 . � As t → −∞ , x ( t ) → ∞ , approaching the half-line through C 2 v 1 . Return Real eigenvalues

13 Nodal Sink Nodal Sink • λ 1 < λ 2 < 0 • General solution x ( t ) = C 1 e λ 1 t v 1 + C 2 e λ 2 t v 2 • Four stable exponential solutions. • All solutions → 0 as t → ∞ . (Stable) � Tangent to C 2 v 2 if C 2 � = 0 . • All solutions → ∞ as t → −∞ . � � to the half line through C 1 v 1 if C 1 � = 0 . Return Real eigenvalues

14 Nodal Source Nodal Source • 0 < λ 1 < λ 2 • General solution x ( t ) = C 1 e λ 1 t v 1 + C 2 e λ 2 t v 2 • Four unstable exponential solutions. • All solutions → 0 as t → −∞ . � Tangent to C 1 v 1 if C 1 � = 0 . • All solutions → ∞ as t → ∞ . (Unstable) � � to the half line through C 2 v 2 if C 2 � = 0 . Return Real eigenvalues Nodal Sink

15 Complex Eigenvalues Complex Eigenvalues • p ( λ ) = λ 2 − Tλ + D with T 2 − 4 D < 0 λ = α + iβ and λ = α − iβ. • Eigenvector w = v 1 + i v 2 associated to λ . • Complex solutions z ( t ) = e λt w = e t ( α + iβ ) [ v 1 + i v 2 ] z ( t ) = e λt w = e t ( α − iβ ) [ v 1 − i v 2 ] Return

16 • Real solutions x 1 ( t ) = Re( z ( t )) = e αt [cos βt · v 1 − sin βt · v 2 ] x 2 ( t ) = Im( z ( t )) = e αt [sin βt · v 1 + cos βt · v 2 ] • General solution x ( t ) = C 1 e αt [cos βt · v 1 − sin βt · v 2 ] + C 2 e αt [sin βt · v 1 + cos βt · v 2 ] Return

17 Center Center • α = Re( λ ) = 0 • General real solution x ( t ) = C 1 [cos βt · v 1 − sin βt · v 2 ] + C 2 [sin βt · v 1 + cos βt · v 2 ] • Every solution is periodic with period T = 2 π/β. • All solution curves are ellipses. Return

18 Spiral Sink Spiral Sink • α = Re( λ ) < 0 • General real solution x ( t ) = C 1 e αt [cos βt · v 1 − sin βt · v 2 ] + C 2 e αt [sin βt · v 1 + cos βt · v 2 ] • All solutions spiral into 0 as t → ∞ . Return

19 Spiral Source Spiral Source • α = Re( λ ) > 0 • General real solution x ( t ) = C 1 e αt [cos βt · v 1 − sin βt · v 2 ] + C 2 e αt [sin βt · v 1 + cos βt · v 2 ] • All solutions spiral into 0 as t → −∞ . Return

20 Planar Systems Planar Systems � a 11 a 12 � A = a 21 a 22 • Char. polynomial p ( λ ) = λ 2 − Tλ + D . • Eigenvalues √ T 2 − 4 D λ 1 , λ 2 = T ± . 2 Return

21 • λ 1 & λ 2 are the roots of p ( λ ) , so p ( λ ) = λ 2 − Tλ + D = ( λ − λ 1 )( λ − λ 2 ) = λ 2 − ( λ 1 + λ 2 ) λ + λ 1 λ 2 • T = λ 1 + λ 2 and D = λ 1 λ 2 . • Duality between ( λ 1 , λ 2 ) and ( T, D ) . • Represent systems by location of ( T, D ) in the TD -plane. Return Characteristic polynomial

22 Trace-Determinant Plane Trace-Determinant Plane • T 2 − 4 D > 0 � ⇒ distinct real eigenvalues λ 1 & λ 2 � D = λ 1 λ 2 < 0 ⇒ Saddle point. � D = λ 1 λ 2 > 0 ⇒ Eigenvalues have the same sign. ◮ T = λ 1 + λ 2 > 0 ⇒ Nodal source. ◮ T = λ 1 + λ 2 < 0 ⇒ Nodal sink. Return Duality

23 • T 2 − 4 D < 0 ⇒ complex eigenvalues λ = α + iβ and λ = α − iβ. � T = λ + λ = 2 α > 0 ⇒ Spiral source. � T = λ + λ = 2 α < 0 ⇒ Spiral sink. � T = λ + λ = 2 α = 0 ⇒ Center. Return Duality TD plane

24 Types of Equilibrium Points Types of Equilibrium Points • Generic types � Saddle, nodal source, nodal sink, spiral source, and spiral sink. � All occupy large open subsets of the trace-determinant plane. • Nongeneric types � Center and many others. Occupy pieces of the boundaries between the generic types. Return