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# hw5_scribe - t ∈ I Since any point on the periodic orbit...

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Solution Scribe: Evan Gawlik CDS 140a Problem Set 5 November 6, 2009 8) The linearization at the origin is given by ˙ x ˙ y = A x y with A = μ - 1 1 μ . The eigenvalues of A are λ ± = μ ± i . When μ < 0, the origin is a stable spiral in the linearized system, in agreement with the full nonlinear phase portrait. When μ > 0, the origin is an unstable spiral in the linearized system, again in agreement with the nonlinear phase portrait. When μ = 0, the linearization has a center at the origin. In contrast, the nonlinear system has a stable spiral at the origin for μ = 0. Note that this does not violate Liapunov’s Theorem, since the eigenvalues of A all have zero real part when μ = 0. 10) Let U R n be an open set containing the periodic orbit γ ([0 , τ ]) and let D X denote the set of all ( x, t ) U × R for which there is an integral curve c : I U through x with c (0) = x and t
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Unformatted text preview: t ∈ I . Since any point on the periodic orbit has an inﬁnite solution lifetime, γ ([0 , τ ]) × R ⊂ D X . Moreover, D X is open in U × R by Proposition 1.3.10(ii). Fix a number T > 0. Then for any t ∈ [0 , τ ], there exists a ﬁnite number b ( t ) > 0 that depends continuously on the parameter t such that B b ( t ) ( γ ( t )) ×{ T } ⊂ D X , where B b ( t ) ( γ ( t )) denotes the open ball of radius b ( t ) centered at γ ( t ). Deﬁne ε = inf t ∈ [0 ,τ ] b ( t ). Since [0 , τ ] is a compact interval, b ( t ) achieves its inﬁmum on [0 , τ ], so ε must be nonzero. This proves that there is a positive number ε such that any point lying within a distance ε from the periodic orbit has a solution lifetime of at least T . 1...
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