C 1 x(t) = e ta C = e C n. 2! A2 + t3

Transcription

1 Matrix Exponential Fundamental Matrix Solution Objective: Solve dt A x with an n n constant coefficient matrix A x (t) Here the unknown is the vector function x(t) x n (t) General Solution Formula in Matrix Exponential Form: C x(t) e ta C e ta C n where C C n are arbitrary constants The solution of the initial value problem is given by dt A x x(t ) x x(t) e (t t )A x Definition (Matrix Exponential): For a square matrix A e ta k t k k! Ak I + ta + t2 2! A2 + t3 3! A3 +

2 Evaluation of Matrix Exponential in the Diagonalizable Case: Suppose that A is diagonalizable; that is there are an invertible matrix P and a diagonal matrix D λ such that A P DP In this case we have e ta P e td P P e λ t e λnt P λ n EXAMPLE Let A (a) Evaluate e ta (b) Find the general solutions of dt A x (c) Solve the initial value probelm dt Solution: A x x() 4 The given matrix A is diagonalized: A P DP with /2 P 2 /2 D 2 Part (a): We have e ta P e td P /2 2 /2 et e 2t 5 3 e t e t e 2t 3e t 3e t e 2t 2e t e t + e t 5e t + 2e 2t + 3e t 3e t + 2e 2t + 2e t e t e t 5e t + e 2t 6e t 3e t + e 2t 4e t e t + 2e t Part (b): The general solutions to the given system are x(t) e ta C C 2 C 3 where C C 2 C 3 are free parameters Part (c): The solution to the initial value problem is e t + e 2t + 8e t x(t) e ta 4 e t 2e 2t 8e t e t e 2t + 6e t 2

3 Evaluation of Matrix Exponential Using Fundamental Matrix: In the case A is not diagonalizable one approach to obtain matrix exponential is to use Jordan forms Here we use another approach We have already learned how to solve the initial value problem dt A x x() x We shall compare the solution formula with x(t) e ta x to figure out what e ta is We know the general solutions of /dt A x are of the following structure: x(t) C x (t) + + C n x n (t) where x (t) x n (t) are n linearly independent particular solutions rewritten as C x(t) x (t) x n (t) C with C C n For the initial value problem C is determined by the initial condition The formula can be x () x n () C x C x () x n () x Thus the solution of the initial value problem is given by x(t) x (t) x n (t) x () x n () x Comparing this with x(t) e ta x we obtain e ta x (t) x n (t) x () x n () In this method of evaluating e ta the matrix M(t) x (t) x n (t) plays an essential role Indeed e ta M(t)M() Definition (Fundamental Matrix Solution): If x (t) x n (t) are n linearly independent solutions of the n dimensional homogeneous linear system /dt A x we call M(t) x (t) x n (t) a fundamental matrix solution of the system (Remark : The matrix function M(t) satisfies the equation M (t) AM(t) Moreover M(t) is an invertible matrix for every t These two properties characterize fundamental matrix solutions) (Remark 2: Given a linear system fundamental matrix solutions are not unique However when we make any choice of a fundamental matrix solution M(t) and compute M(t)M() we always get the same result) 3

4 7 9 9 EXAMPLE 2 Evaluate e ta for A Solution: We first solve /dt A x We obtain 3 x(t) C e t + C 2 e 2t + C 3 e 2t This gives a fundamental matrix solution: 3e t e 2t e 2t M(t) e t e 2t e t e 2t The matrix exponential is 3e t e 2t e 2t 3 e ta M(t)M() e t e 2t e t e 2t 3e t 2e 2t 3e t 3e 2t 3e t + 3e 2t e t + 2e 2t e t + 2e 2t e t e 2t e t e 2t e t e 2t e t + 2e 2t EXAMPLE 3 Evaluate e ta for A Solution: We first solve /dt A x We obtain 3 x(t) C e t + C 2 e 3t This gives a fundamental matrix solution: M(t) + C 3 e 3t 3e t e 3t e 3t 2te 3t e t e 3t 2 e 3t + te 3t e t e 3t te 3t 2t /2 + t t The matrix exponential is 3e t e 3t e 3t 2te 3t 3 e ta M(t)M() e t e 3t 2 e 3t + te 3t /2 e t e 3t te 3t 3e t 2e 3t 8te 3t 6e t 6e 3t 2te 3t 4te 3t e t + e 3t + 4te 3t e t + 3e 3t + te 3t te 3t e t e 3t + 4te 3t 2e t 2e 3t + te 3t e 3t 2te 3t 4

5 EXAMPLE 4 Evaluate e ta for A Solution : (Use Diagonalization) Solving det(a λi) we obtain the eigenvalues of A: λ 7 + 4i λ 2 7 4i Eigenvectors for λ 7 + 4i: are obtained by solving A (7 + 4i)I v : v v i ( ) Eigenvectors for λ 2 7 4i: are complex conjugate of the vectors in ( ) The matrix A is now diagonalized: A P DP with P + i i i D 7 4i We have e ta P e td P + i i 2 2 e (7+4i)t i + i e (7 4i)t i i ( 2 4 i)e(7+4i)t + ( i)e(7 4i)t 5 8 ie(7+4i)t 5 8 ie(7 4i)t 2 ie(7+4i)t + 2 ie(7 4i)t ( i)e(7+4i)t + ( 2 4 i)e(7 4i)t Solution 2: (Use fundamental solutions and complex exp functions) A fundamental matrix solution can be obtained from the eigenvalues and eigenvectors: ( M(t) + 2 i)e(7+4i)t ( 2 i)e(7 4i)t e (7+4i)t e (7 4i)t The matrix exponential is ( e ta M(t)M() + 2 i)e(7+4i)t ( 2 i)e(7 4i)t + i i e (7+4i)t e (7 4i)t 2 2 ( 2 4 i)e(7+4i)t + ( i)e(7 4i)t 5 8 ie(7+4i)t 5 8 ie(7 4i)t 2 ie(7+4i)t + 2 ie(7 4i)t ( i)e(7+4i)t + ( 2 4 i)e(7 4i)t Solution 3: (Use fundamental solutions and avoid complex exp functions) A fundamental matrix solution can be obtained from the eigenvalues and eigenvectors: ( e 7t M(t) cos 4t sin 4t) e ( 7t cos 4t + sin 4t) 2 2 e 7t cos 4t e 7t sin 4t The matrix exponential is ( e e ta M(t)M() 7t cos 4t sin 4t) e ( 7t cos 4t + sin 4t) 2 2 e 7t cos 4t e 7t 2 sin 4t e 7t cos 4t + 2 e7t sin 4t 5 4 e7t sin 4t e 7t sin 4t e 7t cos 4t 2 e7t sin 4t 5

6 APPENDIX: Common Mistakes Since the years of freshmen calculus we all loved the exponential function e x with scalar variable x There are tons of simple and beautiful formulas for the scalar function e x The matrix exponential is however a quite different beast We need to be a little careful in handling it Here are some common mistakes I have seen people make: e t(a+b) e ta e tb The solutions of dt A(t) x are x(t) eta(t) C The solutions of dt A(t) x are x(t) er t A(s)ds C ( e B(t)) B (t)e B(t) All the above four statements are WRONG! () It is true that e t(a+b) e ta e tb if AB BA But in general e t(a+b) e ta e tb Example: For A B A + B we have e ta e tb + e 2t e 2t e t e t 2 e 2t + e 2t e t + e t 2 e t e t 2 e t e t 2 e t + e t but e t(a+b) e t e t (2) We learned how to solve dt A x where A is a constant matrix Unfortunately there is no general solution method for ( ) In particular dt A(t) x where A(t) is nonconstant x(t) e ta(t) C does not solve ( ) This even fails in the scalar case Example: The solutions of the scalar equation dx dt (sin t)x is given by x(t) e cos t C but not e t sin t C 6

7 (3) Likewise x(t) e R t A(s)ds C is also a wrong solution formula for ( ) This formula is only valid for scalar equations ie when the space dimension is Example: Consider dt 2t x x() The solution of this initial value problem is x(t) e t (t 2 )et + 2 e t The function ( t ) exp A(s)ds does not give the solution Indeed (4) We do have: ( t ) () t exp A(s)ds exp t 2 t e t 2 tet 2 te t e t (e b(t) ) b (t)e b(t) holds for scalar functions b(t) and (e ta ) Ae ta e ta A for constant matrices A But in general for nonconstant matrix function B(t) (e B(t) ) is neither B (t)e B(t) nor e B(t) B (t) t Example: Let B(t) t 2 We have e t B(t) ( e B(t) ) e t t+ 2 et + t 2 e t e t B (t)e B(t) e t 2t 2 tet 2 te t e t e B(t) B e (t) t 2 tet 2 te t e t 2t where A(t) 2t e t 2 tet 2 te t e t 2 tet 2 te t e t and hence e t 3 2 tet + 2 te t e t e t 2 tet te t e t All three matrix functions (e B(t) ) B (t)e B(t) and e B(t) B (t) are different from each other 7

8 EXERCISES 5 3 Evaluate e ta for A (a) Evaluate e ta for A 3 8 (b) Solve 5 A x x() dt 4 3 Evaluate e ta for A Evaluate e ta for A (a) Evaluate e ta for A (b) Solve A x x() dt Evaluate e ta for A 2 7 Evaluate e ta for A See next page for answers 8

9 Answers: 3 e ta 2 e4t 2 e2t 3 2 e4t e2t 2 e4t 2 e2t 2 e4t e2t e 2 (a) e ta 6te 2t 2te 2t 3te 2t e 2t + 6te 2t 5 5e (b) x(t) e ta 42te 2t e 2t 2te 2t 3e t 2e 2t 5e t + 6e 2t e 3t 3e t 4e 2t + e 3t 3 e ta 3e t 3e 2t 5e t + 9e 2t 3e 3t 3e t 6e 2t + 3e 3t 3e t 3e 2t 5e t + 9e 2t 4e 3t 3e t 6e 2t + 4e 3t 3e t 2e t 2e t 2e t e t + e t 4 e ta 6e t + 6e t 4e t + 5e t 2e t 2e t 6e t + 6e t 4e t + 4e t 2e t e t 3e t 2e t + 4te t 2e t 2e t + 3te t e t + e t te t 5 (a) e ta 6e t + 6e t 4te t 4e t + 5e t 3te t 2e t 2e t + te t 6e t + 6e t + 4te t 4e t + 4e t + 3te t 2e t e t te t (b) x(t) e ta 4et 3e t + 6te t 8e t + 9e t 6te t 8e t + 9e t + 6te t cos 2t + sin 2t sin 2t 6 e ta e 3t sin 2t cos 2t sin 2t or equivalently e ta ( i)e (3+2i)t + ( + i)e (3i)t 2ie (3+2i)t 2ie (3i)t 2 ie (3+2i)t + ie (3i)t ( + i)e (3+2i)t + ( i)e (3i)t 7 e ta 2e 3t + 3 cos t + sin t e 3t + 2 cos t sin t e 3t cos t + 3 sin t 4e 3t + 4 cos t 2 sin t 4e 3t + cos t 3 sin t e 3t + 2 cos t + 4 sin t 5 e 3t + 2 cos t 6 sin t 2e 3t 2 cos t 4 sin t e 3t + 6 cos t + 2 sin t 9