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final_review (dragged) 5 - 6 MA 36600 FINAL REVIEW §7.7:...

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Unformatted text preview: 6 MA 36600 FINAL REVIEW §7.7: Fundamental Matrices. • Consider the initial value problem d x = P(t) x, x(t0 ) = x0 . dt ￿ ￿ Say that x(1) , x(2) , . . . , x(n) is a fundamental set of solutions. Then the general solution is x(t) = c1 x(1) (t) + c2 x(2) (t) + · · · + cn x(n) (t) = Ψ(t) c in terms of the matrices x11 (t x21 (t) Ψ(t) = . . . x12 (t) x22 (t) . . . x1n (t) x2n (t) . . . ··· ··· .. . c1 c2 c = . . . . and xn1 (t) xn2 (t) · · · xnn (t) cn ￿ ￿ We call Ψ(t) = xij (t) a fundamental matrix. It satisfies the matrix equation d Ψ = P(t) Ψ. dt Since the Wronskian is W = det Ψ, we see that Ψ(t) is a nonsingular matrix. Hence we can choose the constant c in order to solve the initial value problem: x0 = Ψ(t0 ) c =⇒ c = Ψ(t0 )−1 x0 . • Say P(t) = A is an n × n constant matrix. The solution to the initial value problem is x(t) = Φ(t) x0 in terms of the matrix exponential Φ(t) = exp (A t) = ∞ ￿ k=0 In general, the initial value problem d Ψ = A Ψ, dt Ψ(0) = Ψ0 Ak tk t2 = I + A t + A2 + · · · . k! 2 has the unique solution Ψ(t) = exp (A t) Ψ0 . The order does matter! The matrix Ψ0 exp (A t) is not a solution to the initial value problem. • We say that an n × n matrix is a diagonal matrix if it is in the form rt r1 0 · · · 0 e1 0 ··· 0 0 r2 · · · 0 0 er2 t · · · 0 D=. =⇒ exp (D t) = . . .. . . . . .. . . . . . . .. . . . . . . 0 0 · · · rn 0 0 · · · ern t An n × n matrix A is said to be diagonalizable there exists a nonsingular n × n matrix T such that T−1 A T = D is an n × n diagonal matrix. The exponential of such a matrix is exp (A t) = T exp (D t) T−1 . • Say A is a 2 × 2 matrix with distinct eigenvalues r1 and r2 having corresponding eigenvectors ￿￿ ￿￿ ξ ξ ξ (1) = 11 and ξ (2) = 12 . ξ21 ξ22 Denote the 2 × 2 matrices ￿ ￿ r1 0 D= 0 r2 and ￿ ξ T = 11 ξ21 ξ12 ξ22 ￿ =⇒ T−1 A T = D. ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University.

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