set8 - Lecture 21 Linear systems and the Fundamental Matrix...

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Unformatted text preview: Lecture 21 Linear systems and the Fundamental Matrix - continued We now apply the results of the previous lecture to examine the matrix exponential solutions of three classes of linear systems in R 2 : Example 1: x = Ax where A = a b . In the previous lecture, we found e A to be the diagonal matrix with entries e a and e b . To compute e t A , we may simply replace a and b with at and bt so that e t A = e at e bt . (1) This means that the solution to the initial value problem x (0) = a is e t A a = e at e bt a 1 a 2 = a 1 e at a 2 e bt . (2) Of course, the two DEs in x 1 ( t ) and x 2 ( t ) could easily have been solved separately. Since A is diagonal, the two DEs are uncoupled . Nevertheless, it is still worthwhile to see the exponential matrix applied to this problem. Example 2: x = Ax where A = a 1 a . As in the previous lecture, we shall write A as A = a I + N , (3) where N = 0 1 0 0 . (4) Recall that the matrix N is nilpotent since N 2 = N 3 = = . Then e t A = e at I + t N (5) 1 = e at I e N since IN = NI = e at I [ I + t N ] = e at 1 t 0 1 = e at te at e at . The solution to the initial value problem x (0) = a is e t A a = e at te at e at a 1 a 2 = ( a 1 + a 2 t ) e at a 2 e at . (6) Here we see the appearance of solutions with polynomials multiplied by exponentials. This is, of course, reminiscent of the case for second-order linear DEs with constant coefficients for which the eigenvalues of the characteristic polynomial are equal. The matrix A above is the Jordan canonical form for such a situation. Example 3: x = Ax where A = a b b a . As in the previous lecture, we shall write A as A = a I + B , (7) where B = b b , (8) Since I and B commute, e A = e at I + B = e at I e t B = e at e t B . (9) The matrix e B was computed in the previous lecture. We obtain e t B by replacing b with bt . The final result is e t A = e at cos at sin bt cos at sin bt . (10) 2 The solution to the initial value problem x (0) = a is e t A a = e at cos bt e at sin bt e at sin bt e at cos bt a 1 a 2 = e at ( a 1 cos at + a 2 sin bt ) e at ( a 1 sin bt + a 2 cos bt ) . (11) The eigenvalues of A are complex, i.e., = a bi . As we know for linear second order DEs with constant coefficients, the solutions involve exponentials multiplied by trigonometric functions....
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This note was uploaded on 09/20/2010 for the course AMATH 351 taught by Professor Sivabalsivaloganathan during the Spring '08 term at Waterloo.

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set8 - Lecture 21 Linear systems and the Fundamental Matrix...

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