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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 ² = A ± x y ² with A = ± μ - 1 1 μ ² . 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
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Unformatted text preview: t I . Since any point on the periodic orbit has an innite solution lifetime, ([0 , ]) R D X . Moreover, D X is open in U R by Proposition 1.3.10(ii). Fix a number T &gt; 0. Then for any t [0 , ], there exists a nite number b ( t ) &gt; 0 that depends continuously on the parameter t such that B b ( t ) ( ( t )) { T } D X , where B b ( t ) ( ( t )) denotes the open ball of radius b ( t ) centered at ( t ). Dene = inf t [0 , ] b ( t ). Since [0 , ] is a compact interval, b ( t ) achieves its inmum on [0 , ], so must be nonzero. This proves that there is a positive number such that any point lying within a distance from the periodic orbit has a solution lifetime of at least T . 1...
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This note was uploaded on 02/28/2011 for the course MATH 140A taught by Professor Marsden during the Fall '09 term at Caltech.

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