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Chapter 5: Q4E (page 303)

Compute and graph the Poincare map with \(t{\bf{ = }}2\pi n,n{\bf{ = }}0,1, \ldots ,20\) for equation\(\left( 4 \right)\), taking\(A{\bf{ = F = }}1,\phi {\bf{ = }}0,\omega {\bf{ = }}1,{\bf{b}} = {\bf{0}}.{\bf{1}}\). Describe the attractor for this system.

Short Answer

Expert verified

The attractor for this Poincare section is a single point that doesn't depend on the initial condition.

Step by step solution

01

Defining the Poincare map

For \(A{\bf{ = F = }}1,\phi {\bf{ = }}0,\omega {\bf{ = }}1,{\bf{b}} = {\bf{0}}.{\bf{1}}\) the Poincare maps are given by:

\(\begin{aligned}{c}{x_n}{\bf{ = }}{e^{{\bf{ - }}\frac{{n\pi }}{{10}}}}\sin \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right){\bf{ + }}10\sin (2n\pi ),\\{v_n}{\bf{ = }}\frac{1}{{20}}{e^{{\bf{ - }}\frac{{n\pi }}{{10}}}}\left( {\sqrt {399} \cos \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right){\bf{ - }}\sin \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right)} \right){\bf{ + }}10\cos (2n\pi )\end{aligned}\)

02

Computing the point \({{\bf{x}}_{\bf{n}}}\)and \({{\bf{v}}_{\bf{n}}}\)

One will compute the points \(\left( {{x_n},{v_n}} \right)\) for\({\bf{n = }}\overline {{\bf{0,20}}} \). The values for \({x_n},{v_n}\) are given in the table below

\(\begin{aligned}{*{20}{c}}{{{\bf{x}}_{\bf{n}}}}&{{{\bf{v}}_{\bf{n}}}}\\{\bf{0}}&{{\bf{10}}{\bf{.9987}}}\\{{\bf{ - 0}}{\bf{.0057401}}}&{{\bf{10}}{\bf{.7298}}}\\{{\bf{ - 0}}{\bf{.00838491}}}&{{\bf{10}}{\bf{.5332}}}\\{{\bf{ - 0}}{\bf{.00918607}}}&{{\bf{10}}{\bf{.3895}}}\\{{\bf{ - 0}}{\bf{.00894539}}}&{{\bf{10}}{\bf{.2846}}}\\{{\bf{ - 0}}{\bf{.00816642}}}&{{\bf{10}}{\bf{.2079}}}\\{{\bf{ - 0}}{\bf{.00715692}}}&{{\bf{10}}{\bf{.1518}}}\\{{\bf{ - 0}}{\bf{.00609785}}}&{{\bf{10}}{\bf{.1109}}}\\{{\bf{ - 0}}{\bf{.00508937}}}&{{\bf{10}}{\bf{.081}}}\\{{\bf{ - 0}}{\bf{.00418122}}}&{{\bf{10}}{\bf{.0592}}}\\{{\bf{ - 0}}{\bf{.00339264}}}&{{\bf{10}}{\bf{.0432}}}\\{{\bf{ - 0}}{\bf{.00272521}}}&{{\bf{10}}{\bf{.0315}}}\\{{\bf{ - 0}}{\bf{.00217094}}}&{{\bf{10}}{\bf{.023}}}\\{{\bf{ - 0}}{\bf{.00171735}}}&{{\bf{10}}{\bf{.0168}}}\\{{\bf{ - 0}}{\bf{.00135047}}}&{{\bf{10}}{\bf{.0123}}}\\{{\bf{ - 0}}{\bf{.00105653}}}&{{\bf{10}}{\bf{.009}}}\\{{\bf{ - 0}}{\bf{.000822876}}}&{{\bf{10}}{\bf{.0065}}}\\{{\bf{ - 0}}{\bf{.000638378}}}&{{\bf{10}}{\bf{.0048}}}\\{{\bf{ - 0}}{\bf{.000493523}}}&{{\bf{10}}{\bf{.0035}}}\\{{\bf{ - 0}}{\bf{.000380351}}}&{{\bf{10}}{\bf{.0025}}}\\{{\bf{ - 0}}{\bf{.000292314}}}&{{\bf{10}}{\bf{.0019}}}\end{aligned}\)

03

Finding the Poincare section’s point

The Poincare section is given in the figure below. Now,

\(\begin{aligned}{c}\mathop {\lim }\limits_{n \to \infty } {x_n}{\bf{ = }}\mathop {\lim }\limits_{n \to \infty } \left( {{e^{{\bf{ - }}\frac{{n\pi }}{{10}}}}\sin \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right){\bf{ + }}10\sin (2n\pi )} \right){\bf{ = }}0\\\mathop {\lim }\limits_{n \to \infty } {v_n}{\bf{ = }}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{20}}{e^{{\bf{ - }}\frac{{n\pi }}{{10}}}}\left( {\sqrt {399} \cos \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right){\bf{ - }}\sin \left( {\frac{{\sqrt {399} }}{{10}}n\pi } \right)} \right)} \right){\bf{ + }}\mathop {\lim }\limits_{n \to \infty } (10\cos (2n\pi )){\bf{ = }}10\end{aligned}\)

So, one sees that only the first few points depending on the initial condition, but the points converge to the point \(\left( {0,10} \right)\) when \(n \to \infty \). So, the attractor for this Poincare section is a single point that doesn't depend on the initial condition.

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Show that the Poincare map for equation (1) is not chaoticby showing that if\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)are two initial values that define the Poincare maps\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\) and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\), respectively, using the recursive formulas in (3), then one can make the distance between\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\)small by making the distance between\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\) and \({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)small. (Hint: Let \({\bf{(A,}}\phi {\bf{)}}\)and \({\bf{(}}{{\bf{A}}^{\bf{*}}}{\bf{,}}{\phi ^ * }{\bf{)}}\) be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by\({{\bf{d}}^{\bf{2}}}{\bf{ = (A - }}{{\bf{A}}^{\bf{*}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 2A}}{{\bf{A}}^{\bf{*}}}{\bf{(1 - cos(}}\phi {\bf{ - }}{\phi ^ * }{\bf{))}}\).)
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