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Found in: Page 303

### 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

# 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 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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