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LECTURE 24: EIGENVALUE METHOD MINGFENG ZHAO October 31, 2014 Theorem 1. Let A be a 2 × 2 matrix, ~y (t) and ~z(t) be two linearly independent solutions to ~x0 = A~x, then the general solution to ~x0 = A~x is ~x(t) = C1 ~y (t) + C2 ~z(t). Theorem 2 (Eigenvalue Method). Let λ be an eigenvalue of A and ~v be an eigenvector corresponding to λ, then ~x(t) = eλt~v is a solution to ~x0 = A~x. For find eigenvalues of A, we need to solve the characteristic polynomial of A, that is: a b = (λ − a)(λ − d) − bc = λ2 − (a + d)λ + (ad − bc) = 0. det (λI2 − A) = det c d Decoupled system Definition 1. Let A = a b c d , we say ~x0 = A~x + f~(t) is a decoupled system if bc = 0. For a decoupled system, we can solve x1 and x2 separately. For example, if b = 0, then we have (1) x01 = ax1 + f1 (t) (2) x02 = cx1 + dx2 + f2 (t). For the first equation (1), it’s a first order linear equation of x1 , by using the method of variation of parameters, we can solve x1 from (1). Then plug x1 (t) into (2), we get a first order linear equation of x2 , by using the method of variation of parameters, we can solve x2 . x0 = x1 , 1 . Example 1. Solve the system x0 = x − x . 1 2 2 1 2 MINGFENG ZHAO C1 t Since x01 = x1 , then x1 (t) = C1 et . Since x02 = x1 − x2 = −x2 + C1 et , solve x2 , we get x2 = e + C2 e−t . Then the 2 x0 = x1 , 1 solution to the system is: x0 = x − x . 1 2 2 ~x(t) = C 1 et C1 t 2 e + C2 e −t . Eigenvalue method with distinct real eigenvalues Theorem 3. Let A be a 2 × 2 matrix with two distinct real eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors corresponding to λ1 and λ2 , respectively, then the general solution to ~x0 = A~x is: ~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 . Example 2. Find the general solution to the system: Let ~x(t) = x1 (t) , A = x2 (t) 2 1 0 1 x01 = 2x1 + x2 x02 = x2 . , then we need to solve ~x0 = A~x. 2 1 and corresponding eigenvectors. First, we need to find the eigenvalues of A = 0 1 2 1 λ−2 −1 = det = (λ − 2)(λ − 1) = 0, then Since det λI2 − 0 1 0 λ−1 λ1 = 1, and λ2 = 2. I. When λ1 = 1, let’s solve 2 1 0 1 1 1 x1 = x2 x1 . x2 That is, x1 0 0 x2 = 0 0 . LECTURE 24: EIGENVALUE METHOD 3 Then we know that x1 = x1 x2 So 1 . −1 1 is an eigenvector corresponding to λ1 = 1. −1 II. When λ1 = 2, let’s solve 2 1 x1 x1 = 2 . 0 1 x2 x2 That is, 0 1 0 −1 x1 0 = x2 . 0 Then we know that x1 = x1 x2 So 1 1 . 0 is an eigenvector corresponding to λ1 = 2. 0 In summary, the eigenvalues and corresponding eigenvectors of λ1 = 1 with 1 −1 2 1 0 1 are: and λ2 = 2 with 1 . 0 Therefore, the general solution to ~x0 = A~x is: x1 (t) x2 (t) = C 1 et 1 −1 + C2 e2t 1 = 0 C1 et + C2 e2t t −C1 e . Eigenvalue method with distinct complex eigenvalues Theorem 4. Let A be a 2 × 2 matrix with two distinct complex eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors corresponding to λ1 and λ2 , respectively, then λ2 = λ1 , ~v2 = ~v1 , and the general solution to ~x0 = A~x is: ~x = C1 Re eλ1 t~v1 + C2 Im eλ1 t~v1 . 4 MINGFENG ZHAO Recall the Euler’s identity: ea+ib = ea cos(b) + iea sin(b). 1 1 −1 1 λ−1 −1 1 λ−1 Example 3. Find the general solution to the system ~x0 = First, let’s find the eigenvalues of A = 1 1 −1 1 ~x. , then det (λI2 − A) = det = (λ − 1)2 + 1 = 0. Then and λ2 = 1 − i. λ1 = 1 + i, For λ = 1 + i, let’s solve A~v = λ~v , that is, 1 −1 1 x1 1 x1 = (1 + i) x2 . x2 So we get −1 i x1 1 i 0 = x2 . 0 So we have x1 = x1 x2 That is, 1 1 . i is an eigenvalue corresponding to 1 + i, then e(1+i)t i identity, we have 1 is a solution to ~x0 = A~x. By Euler’s i e(1+i)t 1 i = e(1+i)t ie (1+i)t = et cos(t) + iet sin(t) t t = i [e cos(t) + ie sin(t)] et cos(t) t −e sin(t) Therefore, the general solution to ~x0 = A~x is: ~x(t) = C1 et cos(t) t −e sin(t) + C2 et sin(t) t e cos(t) . + i et sin(t) t e cos(t) . LECTURE 24: EIGENVALUE METHOD 5 Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C. Canada V6T 1Z2 E-mail address: [email protected]