1.4. Lyapunov Exponents 17 suppose exponential stability holds in N(0,γ). Then for x0 Rd the point x1 := γ 2 x0 x0 N(0,γ), and hence ϕ(t, x0) = eAtx0 = 2 x0 γ eAt γ 2 x0 x0 = 2 x0 γ ϕ(t, x1) 2 x0 γ α x1 e−βt = α x0 e−βt, and analogously for asymptotic stability. Clearly, properties (ii), (iii) and (iv) are equivalent and imply (i). Conversely, suppose that one of the Lya- punov exponents is nonnegative. Thus one of the eigenvalues, say μ, has nonnegative real part. If μ is real, i.e., μ 0, the solution corresponding to an eigenvector in Rd does not tend to the origin as time tends to infinity (if μ = 0, all corresponding solutions are fixed points.) If μ is not real, consider the solution (1.3.3) in the two-dimensional eigenspace corresponding to the complex eigenvalue pair μ, ¯. This solution also does not tend to the origin as time tends to infinity. Hence (i) implies (iii). Remark 1.4.9. In particular, the proof above shows that for linear systems the existence of a neighborhood N(0,γ) with limt→∞ ϕ(t, x0) = x∗ whenever x0 N(x∗,γ) implies that one may replace N(0,γ) by Rd. In this sense ‘local stability = global stability’ here. It remains to characterize stability of the origin. If for an eigenvalue μ all complex Jordan blocks are one-dimensional, i.e., a complete set of eigenvectors exists, it is called semisimple. Equivalently, the corresponding real Jordan blocks are one-dimensional if μ is real, and two-dimensional if μ, ¯ C \ R. Theorem 1.4.10. The origin 0 Rd is stable for the linear differential equation ˙ = Ax if and only if all Lyapunov exponents (i.e., all real parts of eigenvalues) are nonpositive and the eigenvalues with real part zero are semisimple. Proof. We only have to discuss eigenvalues with zero real part. Suppose first that λ = 0 spec(A). Then the solution formula (1.3.2) shows that an eigenvector yields a stable solution. For a Jordan block of size m 1, consider y0 = [y1,...,ym] = [0,..., 0, 1] . Then stability does not hold, since y1(t, y0) = eλt m k=1 tk−1 (k 1)! yk = tm−1 (m 1)! for t ∞. Similarly, one argues for a complex-conjugate pair of eigenvalues.
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