Math 211 Math 211 Lecture #26 Solutions of a Planar System March - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #26 Solutions of a Planar System March 25, 2001

2 Polar Representation Polar Representation • z = x + iy = r [cos θ + i sin θ ] . ⋄ θ is the argument of z : tan θ = y/x. ⋄ r = | z | . • Euler’s formula: e iθ = cos θ + i sin θ. ⋄ z = | z | e iθ . ⋄ z = | z | e − iθ . Return

3 Multiplication Multiplication • Two complex numbers z = | z | e iθ w = | w | e iφ and • The product is zw = | z | e iθ · | w | e iφ = | z || w | e i ( θ + φ ) . • | zw | = | z || w | . • The argument of zw is the sum of the arguments of z and w . Return

4 Complex Exponential Complex Exponential For z = x + iy we define Definition: e z = e x + iy = e x · e iy = e x [cos y + i sin y ] . Properties: • e z + w = e z · e w ; e z − w = e z · e − w = e z /e w • e z = e z • | e z | = e x = e Re z • If λ is a complex number, then d dte λt = λe λt Return

5 Complex Matrices Complex Matrices Matrices (or vectors) with complex entries inherit many of the properties of complex numbers. • M = A + iB where A = Re M and B = Im M are real matrices. • M = A − iB ; M = M ⇔ M is real. • Re M = 1 Im M = 1 2 ( M + M ) ; 2 i ( M − M ) • M + N = M + N • M z = M z Return

6 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. • Show that n of these are linearly independent. Return

7 Cases Cases • Distinct real eigenvalues. ⋄ In this case the method works as described. • Complex eigenvalues. ⋄ The method yields complex solutions. • Repeated eigenvalues. ⋄ The method does not always give enough solutions. Return

8 Complex Eigenpairs Complex Eigenpairs A a real matrix • λ a complex eigenvalue with associated eigenvector w , so A w = λ w . A w = A w = A w λ w = λ w • A w = λ w ⇒ A w = λ w ⇒ A w = λ w • ⇒ λ is an eigenvalue of A with associated eigenvector w Return Complex Matrices

9 • Thus complex eigenvalues come in conjugate pairs λ and λ. • The associated eigenvectors also come in conjugate pairs w and w . • λ � = λ ⇒ w and w are linearly independent. Return Complex Matrices

10 • Complex exponential solutions z ( t ) = e λt w z ( t ) = e λt w . and • z and z are linearly independent complex valued solutions to x ′ = A x . Return Complex Matrices Complex eigenpairs

11 z ( t ) = x ( t ) + i y ( t ) & z ( t ) = x ( t ) − i y ( t ) x ( t ) = Re( z ( t )) = z ( t ) + z ( t ) 2 y ( t ) = Im( z ( t )) = z ( t ) − z ( t ) 2 i • x ( t ) and y ( t ) are real valued solutions. • x ( t ) and y ( t ) are linearly independent. Return Complex Matrices Complex eigenpairs Complex solutions

12 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 ) • Characteristic polynomial: p ( λ ) = λ 2 − Tλ + D. ⋄ T = tr A = a 11 + a 22 ; D = det A ⋄ The eigenvalues of A are the roots of p . Return

13 Eigenvalues of A Eigenvalues of A • Roots of p ( λ ) = λ 2 − Tλ + D = 0 . √ 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

14 Example Example � − 5 20 � x ′ = A x where A = − 2 7 • p ( λ ) = λ 2 − 2 λ + 5 . • Eigenvalues: λ = 1 + 2 i and λ = 1 − 2 i Return

15 λ = 1 + 2 i � − 6 − 2 i 20 � • A − λI = . − 2 6 − 2 i � 3 − i � • Eigenvector: w = 1 Return Example

16 • Complex Solutions � � 3 − i z ( t ) = e λt w = e (1+2 i ) t 1 � � 3 + i z ( t ) = e λt w = e (1 − 2 i ) t 1 • Real Solutions � � 3 cos 2 t + sin 2 t x ( t ) = Re( z ( t )) = e t cos 2 t � � 3 sin 2 t − cos 2 t y ( t ) = Im( z ( t )) = e t sin 2 t Return Example λ

17 Initial Value Problem Initial Value Problem Solve � � − 5 20 x ′ = A x A = where − 2 7 with the intial condition � � 5 x (0) = . 3 Return Fundamental set of solutions

18 Initial Value Problem Initial Value Problem Solution is � � 3 cos 2 t + sin 2 t u ( t ) = 3 e t cos 2 t � � 3 sin 2 t − cos 2 t + 4 e t sin 2 t � � 5 cos 2 t + 15 sin 2 t = e t 3 cos 2 t + 4 sin 2 t Return Real solutions

19 Summary — Complex Eigenvalues Summary — Complex Eigenvalues Suppose A is a real 2 × 2 matrix with • complex conjugate eigenvalues λ and λ , and • associated nonzero eigenvectors w and w . Then • z ( t ) = e λt w and z ( t ) = e λt w form a complex valued fundamental set of solutions, and • x ( t ) = Re( z ( t )) and y ( t ) = Im( z ( t )) form a real valued fundamental set of solutions. Return