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# De nition 124 the jacobian matrix of a vector valued

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Unformatted text preview: 6 6 4 @f1 @y1 @f2 @y1 @f1 @y2 @f2 @y2 @fm @y1 @fm @y2 ... @f1 @ym @f2 @ym @fm @ym 3 7 7: 7 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 t2 t3 eAt = I + tA + 2! A2 + 3! A3 + Often, A can be diagonalized as 2 1 6 2 T 1AT = = 6 6 ... 4 (1.2.10b) : 3 7 7 7 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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