Math 211 Math 211 Lecture #31 Exponential of a Matrix Stability - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #31 Exponential of a Matrix Stability of Solutions November 8, 2002

2 Exponential of a Matrix Exponential of a Matrix The exponential of the n × n matrix A is Definition: the n × n matrix ∞ e A = I + A + 1 2! A 2 + 1 1 3! A 3 + · · · = � n ! A n . 0 The solution to the initial value problem Theorem: x ′ = A x with x (0) = v is x ( t ) = e tA v . • Can we compute e tA v for enough vectors to find a fundamental set of solutions? Return

3 Key to Computing e tA or e tA v Key to Computing e tA or e tA v Suppose that A an n × n matrix, and λ a number (an eigenvalue). Then e tA = e λt · [ I + t ( A − λI ) + t 2 2!( A − λI ) 2 + · · · ] e tA v = e λt · [ v + t ( A − λI ) v + t 2 2!( A − λI ) 2 v + · · · ] • If λ is an eigenvalue and v is an associated eigenvector, then e tA v = e λt v . • If ( A − λI ) 2 v = 0 , then e tA v = e λt [ v + t ( A − λI ) v ] . Return

4 Example 2, Reprise Example 2, Reprise 1 2 − 1   A = − 4 − 7 4   − 4 − 4 1 • p ( λ ) = ( λ + 3)( λ + 1) 2 • λ 1 = − 3 , with algebraic multiplicity 1. � null( A − λ 1 I ) has basis v 1 = ( − 1 / 2 , 3 / 2 , 1) T , so the geometric multiplicity is 1. � There is one exponential solution x 1 ( t ) = e λ 1 t v 1 = e − 3 t ( − 1 / 2 , 3 / 2 , 1) T . Return

5 • λ 2 = − 1 , with algebraic multiplicity 2. � null( A − λ 2 I ) has basis v 2 = ( − 1 / 2 , 1 , 1) T , so the geometric multiplicity is 1. � So there is only one exponential solution x 2 ( t ) = e λ 2 t v 2 = e − t ( − 1 / 2 , 1 , 1) T . � However, null(( A − λ 2 I ) 2 ) has dimension 2, with basis (0 , 1 , 1) T ) and (1 , 0 , 0) T . With v 3 = (1 , 0 , 0) T we get the third solution x 3 ( t ) = e tA v 3 = e − t [ v 3 + t ( A + I ) v 3 ] = e − t (1 + 2 t, − 4 t, − 4 t ) T . • x 1 , x 2 , and x 3 are a fundamental set of solutions. Return Example 2

6 Generalized Eigenvectors Generalized Eigenvectors If λ is an eigenvalue of A and Definition: ( A − λI ) p v = 0 for some integer p ≥ 1 , then v is called a generalized eigenvector associated with λ. • Then v + + t ( A − λI ) v + t 2 � e tA v = e λt 2!( A − λI ) 2 v t p − 1 � ( p − 1)!( A − λI ) p − 1 v + · · · + • We can compute e tA v for any generalized eigenvector. Return

7 Solution Strategy Solution Strategy If λ is an eigenvalue of A with algebraic Theorem: multiplicity q , then there is an integer p ≤ q such that null(( A − λI ) p ) has dimension q . • Thus, we can find q linearly independent solutions associated with the eigenvalue λ . • Since the sum of the algebraic multiplicities is n , we can find a fundamental set of solutions. Return Key

8 Procedure for Solving x ′ = A x Procedure for Solving x ′ = A x • Find the eigenvalues of A . • For each eigenvalue λ : � Find the algebraic multiplicity q . � Find the smallest integer p such that null(( A − λI ) p ) has dimension q. � Find a basis v 1 , v 2 , . . . , v q of null(( A − λI ) p ) . � For j = 1 , 2 , . . . , q , set x j ( t ) = e tA v j . � If λ is complex , find real solutions. e tA v Return Key

9 Examples Examples • Use M ATLAB .

10 Procedure for a Complex Eigenvalue Procedure for a Complex Eigenvalue If λ is a complex eigenvalue of algebraic multiplicity q . Then λ also has algebraic multiplicity q . • Find the smallest integer p such that null(( A − λI ) p ) has dimension q. • Find a basis w 1 , w 2 , . . . , w q of null(( A − λI ) p ) . • For j = 1 , 2 , . . . , q , set z j ( t ) = e tA w j . z 1 , . . . , z q . Together with z 1 , . . . , z q , these are 2 q linearly independent complex valued solutions. • For j = 1 , 2 , . . . , q, set x j ( t ) = Re( z j ( t )) and y j ( t ) = Im( z j ( t )) . These are 2 q linearly independent real valued solutions. Return Procedure

11 Stability Stability Autonomous system x ′ = f ( x ) with an equilibrium point at x 0 . • Basic question: What happens to all solutions as t → ∞ ? • x 0 is stable if for every ǫ > 0 there is a δ > 0 such that a solution x ( t ) with | x (0) − x 0 | < δ ⇒ | x ( t ) − x 0 | < ǫ for all t ≥ 0 . � Every solution that starts close to x 0 stays close to x 0 . Return

12 • x 0 is asymptotically stable if it is stable and there is an η > 0 such that if x ( t ) is a solution with | x (0) − x 0 | < η , then x ( t ) → x 0 as t → ∞ . � x 0 is called a sink. � Every solution that starts close to x 0 approaches x 0 . • x 0 is unstable if there is an ǫ > 0 such that for any δ > 0 there is a solution x ( t ) with | x (0) − x 0 | < δ with the property that there are values of t > 0 such that | x ( t ) − x 0 | > ǫ . � There are solutions starting arbitrarily close to x 0 that move away from x 0 . Return

13 Examples D = 2 Examples D = 2 • Sinks are asymptotically stable. � The eigenvalues have negative real part. • Sources are unstable. � The eigenvalues have positive real part. • Saddles are unstable. � One eigenvalue has positive real part. • Centers are stable but not asymptotically stable. � The eigenvalues have real part = 0 . Return

14 Let A be an n × n real matrix. Theorem: • Suppose the real part of every eigenvalue of A is negative. Then 0 is an asymptotically stable equilibrium point for the system x ′ = A x . • Suppose A has at least one eigenvalue with positive real part. Then 0 is an unstable equilibrium point for the system x ′ = A x . Return D = 2 Procedure

15 Examples Examples • D = 2 � T 2 − 4 D = 0 . ◮ T < 0 ⇒ sink. T > 0 ⇒ source. • y ′ = A y , − 2 − 18 − 7 − 14   1 6 2 5   A =  .   2 2 − 3 0  − 2 − 8 − 1 − 6 � A has eigenvalues − 1 , − 2 , & − 1 ± i . � 0 is asymptotically stable. Theorem

16 Multiplicities Multiplicities A an n × n matrix with distinct eigenvalues λ 1 , . . . , λ k . • The characteristic polynomial has the form p ( λ ) = ( λ − λ 1 ) q 1 ( λ − λ 2 ) q 2 · . . . · ( λ − λ k ) q k . • The algebraic multiplicity of λ j is q j . • The geometric multiplicity of λ j is d j , the dimension of the eigenspace of λ j . • q 1 + q 2 + . . . + q k = n. • 1 ≤ d j ≤ q j . Return