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@y2 ... @f1
5 (1.2.9) De nition 1.2.5. The norm of a vector y is a scalar kyk such that
1. kyk 0 and kyk = 0 if and only if y = 0,
2. k yk = j jkyk for any scalar , and 3. kx + yk kxk + kyk. We'll identify speci c vector norms when they are needed. For the moment, let's
return to our stability analysis and let A = fy (0). Then, the stability of the nonlinear
autonomous system (1.2.8) is related to that of the linear system y = Ay
0 y(0) = y0 ky0 k : (1.2.10a) The solution of this system is y(t) = eAt y0
where the matrix exponential is de ned by the series
eAt = I + tA + 2! A2 + 3! A3 +
Often, A can be diagonalized as
T 1AT = = 6
4 (1.2.10b) :
5 ; 13 m (1.2.10c) (1.2.11) where 1 2 : : : m are the eigenvalues of A and the columns of T are the corresponding
eigenvectors. In this case, we may easily verify that the solution of (1.2.10) is y(t) = Te tT 1y0 :
; Thus, (1.2.10) is stable if all of the eigenvalues have non-po...
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This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.
- Spring '14