Ch7.6:ComplexEigenvalues光 We consider again a homogeneous system of n first orderlinear equations with constant, real coefficients,X, = aix, +ai2x2 +... +ainxnx, = a21Xj +a22X2 +...+a2nXnx, = anx +an2X2 +... +ammxn,and thus the system can be written as x' = Ax, where(x(t))anr...ainai2x2(t)a21a22a2nA=x(t)=.(,(t))anannan2
Ch 7.6: Complex Eigenvalues We consider again a homogeneous system of n first order linear equations with constant, real coefficients, and thus the system can be written as x' = Ax, where , 1 1 2 2 2 2 1 1 2 2 2 2 1 1 1 1 1 2 2 1 n n n n n n n n n n x a x a x a x x a x a x a x x a x a x a x = + + + = + + + = + + + = = n n n n n n n a a a a a a a a a x t x t x t t 1 2 2 1 2 2 2 1 1 1 2 1 2 1 , ( ) ( ) ( ) x( ) A
Conjugate Eigenvalues and Eigenvectors* We know that x = Eert is a solution of x' = Ax, provided r isan eigenvalue and is an eigenvector of AThe eigenvalues ri,..., rn are the roots of det(A-rl) = O, and上Lthe corresponding eigenvectors satisfy (A-rl) = 0. If A is real, then the coefficients in the polynomial equationdet(A-rl) = O are real, and hence any complex eigenvaluesmust occur in conjugate pairs. Thus if r = a + iu is aneigenvalue, then so is r2 = - iu.X The corresponding eigenvectors E(l), E(2) are conjugates alsoTo see this, recall A and I have real entries, and hence(A-rI)E() =0 = (A-I)E(I) =0 = (A-rI)=(2) =0
Conjugate Eigenvalues and Eigenvectors We know that x = e rt is a solution of x' = Ax, provided r is an eigenvalue and is an eigenvector of A. The eigenvalues r1 ,., rn are the roots of det(A-rI) = 0, and the corresponding eigenvectors satisfy (A-rI) = 0. If A is real, then the coefficients in the polynomial equation det(A-rI) = 0 are real, and hence any complex eigenvalues must occur in conjugate pairs. Thus if r1 = + i is an eigenvalue, then so is r2 = - i. The corresponding eigenvectors (1) , (2) are conjugates also. To see this, recall A and I have real entries, and hence (A − I)ξ = 0 (A − I)ξ = 0 (A − I)ξ = 0 (2) 2 (1) 1 (1) 1 r r r
Conjugate Solutions* It follows from the previous slide that the solutionsx() ==()e', x(2) =≤(2)en2lcorresponding to these eigenvalues and eigenvectors areconjugates conjugates as well, sincex(2) =E(2)e'nt = E()e=x(0)
Conjugate Solutions It follows from the previous slide that the solutions corresponding to these eigenvalues and eigenvectors are conjugates conjugates as well, since (2) (2) (1) (1) 2 2 x = ξ = ξ = x r t r t e e rt r t e e 1 2 (1) (1) (2) (2) x = ξ , x = ξ
Real-Valued Solutions* Thus for complex conjugate eigenvalues r; and r2 , thecorresponding solutions x(1) and x(2) are conjugates also* To obtain real-valued solutions, use real and imaginary partsof either x(1) or x(2). To see this, let E(I) = a + ib. Thenx() = =()e(a+iu)r = (a+ib)e"(cos μt+isin μut)=ert (acos μt -bsin ut)+ie(asin μt+bcos μt)= u(t)+iv(t)whereu(t)=eat(acos μt -bsin μt), v(t) =eat (asin μt +bcos μt)are real valued solutions of x' = Ax, and can be shown to belinearly independent
Real-Valued Solutions Thus for complex conjugate eigenvalues r1 and r2 , the corresponding solutions x (1) and x (2) are conjugates also. To obtain real-valued solutions, use real and imaginary parts of either x (1) or x (2) . To see this, let (1) = a + ib. Then where are real valued solutions of x' = Ax, and can be shown to be linearly independent. ( ) ( ) ( ) ( ) ( ) ( ) ( ) cos sin sin cos cos sin (1) (1) t i t e t t ie t t e i e t i t t t i t t u v a b a b x ξ a b = + = − + + = = + + + (t) e ( cos t sin t), (t) e ( sin t cos t), t t u = a −b v = a +b
General Solution* To summarize, suppose ri = α + iu, r2 = α- iu, and thatr3..., rn are all real and distinct eigenvalues of A. Let thecorresponding eigenvectors be≤0) = a+ib, =(2) =a -ib, (3), s(),., E(n)* Then the general solution of x' = Ax isX = C,u(t)+c,v(t)+c,E(3)e'5* + ...+c,E(n)e'n*whereu(t)=e" (acos μut -bsin ut), v(t)=e^t(asin ut+bcos μt
General Solution To summarize, suppose r1 = + i, r2 = - i, and that r3 ,., rn are all real and distinct eigenvalues of A. Let the corresponding eigenvectors be Then the general solution of x' = Ax is where n r t n r t n c t c t c e c e (3) ( ) 1 2 3 3 x = u( ) + v( ) + ξ ++ ξ (1) (2) (3) (4) ( ) , , , , , n ξ = a +ib ξ = a −ib ξ ξ ξ t e ( t t) t e ( t t) t t u( ) = acos −bsin , v( ) = asin +bcos