lecture_42

# lecture_42 - MA 265 LECTURE NOTES FRIDAY APRIL 25 Matrix...

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Unformatted text preview: MA 265 LECTURE NOTES: FRIDAY, APRIL 25 Matrix Exponentials Dynamical Systems. In this lecture, we focus on solutions to a first-order homogeneous system in just two variables: dx dx = ax + by x (0) = x dy dt = cx + dy y (0) = y If a , b , c , and d are constants, we call such a system a dynamical system . Recall that we can express this initial value problem in the form d x dt = A x ( t ) where x (0) = x y in terms of the matrices A = a b c d and x ( t ) = x ( t ) y ( t ) . We perform three steps to find the solution x = x ( t ) of this dynamical system. The solution will be a little different depending on the eigenvalues of the system. Step #1: Eigenvalues and Eigenvectors. First, we compute the eigenvalues { λ 1 , λ 2 } and their corre- sponding eigenvectors { p 1 , p 2 } for the 2 × 2 matrix A . The characteristic polynomial of A is p ( λ ) = det[ λI 2- A ] = λ- a- b- c λ- d = ( λ- a )( λ- d )- bc = λ 2- Tr( A ) λ + det( A ) in terms of the trace Tr( A ) = a + d and the determinant det( A ) = ad- bc . Using the Quadratic Formula, will be useful for us to write the roots λ 1 = α + β and λ 2 = α- β in terms of α = Tr( A ) 2 = a + d 2 and β = p Tr( A ) 2- 4 det( A ) 2 = p ( a- d ) 2 + 4 bc 2 . Step #2: Compute the General Solution. Next, we find the general solution to the homogeneous system. We make an observation which will be useful later. Consider a “diagonal” system: d u dt = D u ( t ) where D = λ 1 λ 2 . We know that the general solution of this diagonal system is u ( t ) = b 1 e λ 1 t b 2 e λ 2 t = b 1 1 e λ 1 t + b 2 1 e λ 2 t = e λ 1 t e λ 2 t b 1 b 2 ....
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## This document was uploaded on 12/11/2009.

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lecture_42 - MA 265 LECTURE NOTES FRIDAY APRIL 25 Matrix...

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