Section 8.2 : Homogeneous Linear Systems
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1 Section 8.2 : Homogeneous Linear Systems Review: Eigenvalues and Eigenvectors Let A be an n n matrix with constant real components a ij. An eigenvector of A is a nonzero n 1 column vector v such that Av = λv for some scalar λ. A scalar λ is called an eigenvalue of A if there is a nontrivial solution v of Av = λv. Any such v 0 is called an eigenvector of A corresponding to λ. Eigenvalue Problem: Find values of the scalar λ and nonzero n 1 column vectors v such that the matrix equation Av = λv (1) is satisfied. Equation (1) can be written equivalently as the homogeneous matrix equation: (A λi) v = 0 (2) where I is the n n identity matrix. 1
2 (A λi) v = 0 (2) Fact: The homogeneous matrix system (2) has nontrivial solutions if and only if the determinant of the coefficient matrix A λi vanishes, that is, if and only if a 11 λ a 12 a 1n det(a λi) = a 21 a 22 λ a 2n..... = 0. (3). a n1 a n2 a nn λ det(a λi) is a polynomial of degree n in λ, called the characteristic polynomial of A. det(a λi) = 0 is called the characteristic equation of A. The eigenvalues of A are the roots λ of the characteristic equation. Given the eigenvalues of A, the eigenvectors can be determined by finding all nontrivial solutions to Eq. (2). 2
3 Homogeneous Linear System of ODEs with Constant Coefficients We are interested in solving: X = A X (4) where A is an n n matrix with real constant entries, and X = X(t) is an n 1 column vector. For n = 1: Equation (4) becomes the scalar equation: x 1 (t) = a 11 x 1 (t), and the general solution is x 1 (t) = c 1 e a 11t. For n 2: Look for a solution vector X = X(t) of the form X = V e λt (5) where V 0 is a constant n 1 column vector, and λ is a constant. Substituting Eq. (5) into system (4) yields: V λe λt = A V e λt AV e λt V λe λt = 0 (AV λv ) e λt = 0 so that AV = λv since e λt > 0. Therefore, in order to find solutions of system (4), we need to find eigenvalues and eigenvectors of matrix A. 3
4 8.2.1: Distinct Real Eigenvalues Theorem: Let A be a real, constant, n n matrix. If A has k n distinct real eigenvalues λ 1, λ 2,..., λ k with corresponding eigenvectors V 1, V 2,..., V k, then the functions X 1 (t) = V 1 e λ 1t, X 2 (t) = V 2 e λ 2t,..., X k (t) = V k e λ k t are linearly independent on (, ). Here we consider the case of k = n. Theorem 8.2.1: General Solution of Homogeneous Systems Consider the homogeneous system of differential equations X = A X (4) where the coefficient matrix A is a real, constant, n n matrix. Let λ 1, λ 2,..., λ n be n distinct real eigenvalues of A with corresponding eigenvectors V 1, V 2,..., V n. Then X i (t) = V i e λ it, i = 1,..., n, form a fundamental set of solutions of system (4) on (, ), and the general solution on (, ) is: X = c 1 V 1 e λ 1t + c 2 V 2 e λ 2t + + c n V n e λ nt (6) where c i, i = 1,..., n, are arbitrary constants. 4
5 Example: Find the general solution of dx dt dy dt = 9x 3y = 16x 7y 5
6 Example (cont): 6
7 8.2.2: Repeated Eigenvalues Definition: We say that λ 1 is an eigenvalue of (algebraic) multiplicity k, where k is a positive integer, if (λ λ 1 ) k is a factor of the characteristic equation det(a λi) = 0, but (λ λ 1 ) k+1 is not. Fact: An eigenvalue of multiplicity k can have anywhere between 1 and k (inclusive) linearly independent eigenvectors corresponding to it. Let λ 1 be an eigenvalue of A of multiplicity k n. Different cases can occur for the eigenvectors corresponding to λ 1 : 1. There are k linearly independent eigenvectors V 1, V 2,..., V k corresponding to λ There is only one linearly independent eigenvector corresponding to λ There are m linearly independent eigenvectors corresponding to λ 1, where 1 < m < k. Note: We only consider cases 1 and 2. 7
8 Case 1: Eigenvalue of multiplicity k with k linearly independent eigenvector Theorem: Consider the homogeneous system of differential equations X = A X (4) where the coefficient matrix A is a real, constant, n n matrix. If V 1, V 2,..., V k are k linearly independent eigenvectors corresponding to an eigenvalue λ of multiplicity k n, then X 1 (t) = V 1 e λt, X 2 (t) = V 2 e λt,..., X k (t) = V k e λt are linearly independent solutions of system (4) on (, ). In this case, the general solution to system (4) contains the linear combination c 1 V 1 e λt + c 2 V 2 e λt + + c k V k e λt where c i, i = 1,..., k, are arbitrary constants. Note: If k = n, that is, if λ is the only eigenvalue of A, then the functions X i (t), i = 1,..., n, given above form a fundamental set of solutions to system (4) on (, ), and the linear combination above represents the entire general solution on (, ). 8
9 Example: Find the general solution of X = A X, where A =
10 Example (cont): 10
11 Case 2: Eigenvalue of multiplicity k with 1 linearly independent eigenvector Let λ be an eigenvalue of multiplicity k with 1 linearly independent eigenvector. (A) Suppose k = 2: We are looking for 2 linearly independent solutions to system (4) from this eigenvalue λ. One solution: X 1 = V 1 e λt, where V 1 is an eigenvector corresponding to λ. To find a second solution: Try solution of the form X 2 = W 1 t e λt + W 2 e λt (7) Substituting this X 2 into system (4), X = A X, simplifying, and rearranging terms yields: (A W 1 λw 1 ) t e λt + (A W 2 λw 2 W 1 ) e λt = 0 This equation holds for all t if and only if each term in parenthesis is 0. This gives: (A λi) W 1 = 0 (8) (A λi) W 2 = W 1 (9) Therefore, solving Eq. (8) for W 1 and then solving Eq. (9) for W 2 yields the second solution X 2 in Eq. (7). 11
12 Note: From Eq. (8), W 1 is an eigenvector of A corresponding to eigenvalue λ. Therefore, set W 1 = V 1. For consistency, also set W 2 = V 2. We can now write the second solution X 2 as: where X 2 = V 1 t e λt + V 2 e λt (10) (A λi) V 1 = 0 (11) (A λi) V 2 = V 1 (12) That is, V 1 is an eigenvector of A corresponding to eigenvalue λ, and V 2, which solves Eq. (12), is called a generalized eigenvector. (B) Suppose k = 3: We are looking for 3 linearly independent solutions to system (4) from this eigenvalue λ. We know two solutions: X 1 X 2 = V 1 e λt = V 1 t e λt + V 2 e λt where V 1 and V 2 solve Eqs. (11) and (12). 12
13 To find a third solution: Try solution of the form X 3 = W 1 t 2 2 eλt + W 2 t e λt + W 3 e λt (13) Substituting this X 3 into system (4), X = A X, simplifying, and rearranging terms, we find that the vectors W 1, W 2 and W 3 must satisfy: (A λi) W 1 = 0 (14) (A λi) W 2 = W 1 (15) (A λi) W 3 = W 2 (16) Note: Equations (14) and (15) are the same as Eqs. (8) and (9), respectively. Therefore, set W 1 = V 1 and W 2 = V 2. For consistency, also set W 3 = V 3. We can now write the third solution X 3 as: where X 3 = V 1 t 2 2 eλt + V 2 t e λt + V 3 e λt (17) (A λi) V 1 = 0 (18) (A λi) V 2 = V 1 (19) (A λi) V 3 = V 2 (20) That is, V 1 is an eigenvector of A corresponding to eigenvalue λ, and V 2 and V 3, which solve Eqs. (19) and (20), are generalized eigenvectors. 13
14 (C) In general, for multiplicity k: We are looking for k linearly independent solutions to system (4) from this eigenvalue λ. They are: X 1 X 2 X 3 = V 1 e λt = V 1 t e λt + V 2 e λt = V 1 t 2 2 eλt + V 2 t e λt + V 3 e λt. X k = V 1 t k 1 (k 1)! eλt + V 2 t k 2 (k 2)! eλt + + V k e λt where V 1, V 2,..., V k is a chain of generalized eigenvectors satisfying: (A λi) V 1 = 0 (A λi) V 2 = V 1 (A λi) V 3 = V 2. (A λi) V k = V k 1 14
15 Example: Find the general solution of X = A X, where A = [ ] 15
16 Example (cont): 16
17 Example: Suppose a 3 3 matrix A has eigenvalue λ = 2 of multiplicity 3 with only one linearly independent eigenvector V 1. Let V 1 = 2 7, V 2 = 1 9, V 3 = 2 7 be a chain of generalized eigenvectors. Find the general solution to X = A X. 17
18 8.2.3: Complex Eigenvalues Notes: 1. Given a complex number z = α + i β: The conjugate of z is z = α i β. The real part of z is Re(z) = α. The imaginary part of z is Im(z) = β. 2. Given a vector V with complex entries v k = α k + i β k : The conjugate of V is denoted by V and has kth entry v k = α k i β k. The real part of V is denoted by Re(V ) and has kth entry Re(v k ) = α k. The real part of V is denoted by Im(V ) and has kth entry Im(v k ) = β k. 3. Given a square matrix A whose characteristic equation has real coefficients: Complex eigenvalues always appear in conjugate pairs. That is, if λ = α + i β is an eigenvalue of A, then so is λ = α i β. If V is an eigenvector corresponding to a complex eigenevalue λ, then V is an eigenvector corresponding to λ. 18
19 Theorem 8.2.2: Solutions Corresponding to a Complex Eigenvalue Given the homogeneous system X = A X (Eq. (4)), where A is an n n matrix with real entries. Suppose A has the complex eigenvalue λ 1 = α + i β (α and β are real) with corresponding eigenvector V 1. Then: V 1 e λ 1t and V 1 e λ 1t (21) are solutions to system (4). We can use Euler s formula to derive real solutions to system (4) from the complex solutions in Eq. (21). Theorem 8.2.3: Real Solutions Corresponding to a Complex Eigenvalue Given the homogeneous system X = A X (Eq. (4)), where A is an n n matrix with real entries. Suppose A has the complex eigenvalue λ 1 = α + i β (α and β are real) with corresponding eigenvector V 1. Let B 1 = Re(V 1 ) and B 2 = Im(V 1 ). Then: X 1 = [ B 1 cos(βt) B 2 sin(βt) ] e αt (22) X 2 = [ B 2 cos(βt) + B 1 sin(βt) ] e αt (23) are linearly independent solution to system (4) on (, ). 19
20 Example: Find the general solution of dx dt dy dt = 4x + 5y = 2x + 6y 20
21 Example (cont): 21
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