{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 0 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 0 0 e bt . (1) This means that the solution to the initial value problem x (0) = a is e t A a = e at 0 0 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 0 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 = · · · = 0 . Then e t A = e at I + t N (5) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
= e at I e N since IN = NI = e at I [ I + t N ] = e at 1 t 0 1 = e at te at 0 e at . The solution to the initial value problem x (0) = a is e t A a = e at te at 0 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 = 0 b b 0 , (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
Background image of page 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., λ = a ± bi . As we know for linear second order DEs with constant coefficients, the solutions involve exponentials multiplied by trigonometric functions. The above three examples actually take care of all situations that would be encountered in R 2 . Why is this? Because they are the three fundamental Jordan canonical (or normal) forms of 2 × 2 matrices into which all 2 × 2 matrices can be converted by means of a similarity transformation.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}