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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 303
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Show that for\({\bf{b > 0}}\), the Poincare map for the equation \(\left( 4 \right)\) is not chaotic by showing that as \({\bf{t}}\) gets large

\(\begin{aligned}{c}{x_n}{\bf{ = }}x(2\pi n) \approx \frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\sin (2\pi n{\bf{ + }}\theta )\\{v_n}{\bf{ = x}}'(2\pi n) \approx \frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\cos (2\pi n{\bf{ + }}\theta )\end{aligned}\)

Independent of the initial values\({{\bf{x}}_{\bf{0}}}{\bf{ = x(0),}}{{\bf{v}}_{\bf{0}}}{\bf{ = x}}'{\bf{(0)}}\).

Use that \(Sine\) and \(Cosine\) functions are \(2\pi {\bf{ - }}\)periodic\({x_n}{\bf{ = }}\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\sin \theta \)and\({v_n}{\bf{ = }}\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\cos \theta \)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Finding sine periodic

Since\(Sine\)the function is \(2\pi {\bf{ - }}\)periodic.

\({x_n}{\bf{ = }}x(2\pi n) \approx \frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\sin (2\pi n{\bf{ + }}\theta ){\bf{ = }}\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\sin \theta \)

Step 2: Finding cosine’s periotic

Since \(Cosine\) is also \(2\pi {\bf{ - }}\)periodic.

\({v_n}{\bf{ = }}x'(2\pi n) \approx \frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\cos (2\pi n{\bf{ + }}\theta ){\bf{ = }}\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\cos \theta \)

Therefore,\(\left( {{x_n},{v_n}} \right)\) is one point \(\left( {\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\sin \theta ,\frac{F}{{\sqrt {{{\left( {{\omega ^2}{\bf{ - }}1} \right)}^2}{\bf{ + }}{b^2}} }}\cos \theta } \right)\) for all \(n = 1,2,3, \ldots \)i.e., the Poincare map is not chaotic.

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