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\({{\bf{x}}_{\bf{0}}}{\bf{ = x(0),}}{{\bf{v}}_{\bf{0}}}{\bf{ = x}}'{\bf{(0)}}\).

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

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

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.