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Unformatted text preview: Chapter 5 Table 5–G: Poincar´ map for Problem 5. e
n 0 1 2 3 4 5 6 7 8 9 10 xn −1.060065 −0.599847 −1.242301 −1.103418 −0.997156 −1.074094 −1.070300 −1.052491 −1.060495 −1.061795 −1.059271 vn 1.521008 0.037456 0.065170 0.415707 0.251142 0.228322 0.278664 0.264458 0.257447 0.263789 0.263037 n 11 12 13 14 15 16 17 18 19 20 xn −1.059944 −1.060312 −1.059997 −1.060030 −1.060096 −1.060061 −1.060058 −1.060068 −1.060065 −1.060064 vn 0.261743 0.262444 0.262491 0.262297 0.262362 0.262385 0.262360 0.262364 0.262369 0.262366 Therefore,
∗ (vn /ω, x∗ − F/(ω 2 − 1)) → (vn /ω, xn − F/(ω 2 − 1)) n as A∗ → A and φ∗ → φ if A = 0 or as A∗ → 0 (regardless of φ∗ ) if A = 0. Note that the convergence is uniform with respect to n. (One can easily see this from the distance formula in polar coordinates.) This is equivalent to x∗ − F/(ω 2 − 1) → xn − F/(ω 2 − 1), n
∗ vn /ω → vn /ω ⇔ x∗ → xn , n
∗ vn → vn uniformly with respect to n. (ii) On the other hand, A∗ and φ∗ satisfy A∗ sin φ∗ + F/(ω 2 − 1) = x∗ , 0
∗ ωA∗ cos φ∗ = v0 ⇒ A∗ = ∗ (x∗ − F/(ω 2 − 1))2 + (v0 /ω )2 , 0 ∗ cos φ∗ = v0 / (ωA∗ ) . ∗ ∗ Therefore, A∗ is a continuous function of (x∗ , v0 ) and so A∗ → A as (x∗ , v0 ) → (x0 , v0 ). 0 0 ∗ If (x0 , v0 ) is such that A = 0, then φ∗ , as a function of (x∗ , v0 ), is also continuous at 0 ∗ (x0 , v0 ) and, therefore, φ∗ → φ as (x∗ , v0 ) → (x0 , v0 ). 0 328 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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