Compute and graph the points of the Poincare map with \({\bf{t = 2\pi n,n = 0,1, \ldots ,20}}\)for equation (1), taking \({\bf{A = F = 1,f = 0,\omega = }}\frac{3}{2}\). Repeat, taking \({\bf{\omega = }}\frac{3}{5}\). Do you think the equation has a \({\bf{2\pi }}\)-periodic solution for either choice of \({\bf{\omega }}\)? A subharmonic solution?

The Poincare map for \({\bf{A = F = 1,f = 0}}\) and \({\bf{\omega = 3/2}}\)is given by:

\(\begin{aligned}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = sin}}\left( {{\bf{2\pi \times }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{n}}} \right){\bf{ + }}\frac{{\bf{1}}}{{{{{\bf{(3/2)}}}^{\bf{2}}}{\bf{ - 1}}}}\\{\bf{ = sin(3n\pi ) + }}\frac{{\bf{4}}}{{\bf{5}}}{\bf{, }}\\{{\bf{v}}_{\bf{n}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{cos}}\left( {{\bf{2\pi \times }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{n}}} \right)\\{\bf{ = }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{cos(3n\pi )}}\end{aligned}\)

One has to compute and graph points \(\left( {{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{v}}_{\bf{n}}}} \right)\)for \({\bf{n = }}\overline {{\bf{0,20}}} {\bf{.}}\) Since \({\bf{sin(3n\pi ) = 0}}\)for everyone has that\({{\bf{x}}_{\bf{n}}}{\bf{ = }}\frac{{\bf{4}}}{{\bf{5}}}{\bf{,}}\)for\({\bf{n = }}\overline {{\bf{0,20}}} \).

When \({\bf{n}}\) is even, \({\bf{cos(3n\pi ) = 1}}\) and when \({\bf{n}}\) is odd,\({\bf{cos(3n\pi ) = - 1}}\), so one has that

\({{\bf{v}}_{\bf{n}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{,}}\)for \({\bf{n = 0,2, \ldots ,20}}\)

\({{\bf{v}}_{\bf{n}}}{\bf{ = - }}\frac{{\bf{3}}}{{\bf{2}}}{\bf{,}}\)for \({\bf{n = 1,3, \ldots ,19}}\)

One can now conclude from the Poincare section that a solution is \({\bf{4\pi }}\)-periodic and it alternates between points \(\left( {\frac{{\bf{4}}}{{\bf{5}}}{\bf{,}}\frac{{\bf{3}}}{{\bf{2}}}} \right)\)and \(\left( {\frac{{\bf{4}}}{{\bf{5}}}{\bf{, - }}\frac{{\bf{3}}}{{\bf{2}}}} \right)\) for every step of\({\bf{2\pi }}\), so it is subharmonic.

When \({\bf{A = F = 1,f = 0}}\)and\({\bf{\omega = 3/5}}\), Poincare map is given by

\(\begin{aligned}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = sin}}\left( {{\bf{2\pi \times }}\frac{{\bf{3}}}{{\bf{5}}}{\bf{n}}} \right){\bf{ + }}\frac{{\bf{1}}}{{{{{\bf{(3/5)}}}^{\bf{2}}}{\bf{ - 1}}}}{\bf{ = sin}}\left( {\frac{{\bf{6}}}{{\bf{5}}}{\bf{n\pi }}} \right){\bf{ - }}\frac{{{\bf{16}}}}{{{\bf{25}}}}{\bf{, }}\\{{\bf{v}}_{\bf{n}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{5}}}{\bf{cos}}\left( {{\bf{2\pi \times }}\frac{{\bf{3}}}{{\bf{5}}}{\bf{n}}} \right){\bf{ = }}\frac{{\bf{3}}}{{\bf{5}}}{\bf{cos}}\left( {\frac{{\bf{6}}}{{\bf{5}}}{\bf{n\pi }}} \right){\bf{.}}\end{aligned}\)

So, for\({\bf{\omega = }}\frac{3}{5}\)one has;

\(\begin{aligned}{c}{{\bf{x}}_{\bf{0}}}{\bf{ = - 1}}{\bf{.5625,}}\;\;\;{{\bf{x}}_{\bf{1}}}{\bf{ = - 2}}{\bf{.1503,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = - 0}}{\bf{.6114,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = - 2}}{\bf{.5136,}}\;\;\;{{\bf{x}}_{\bf{4}}}{\bf{ = - 0}}{\bf{.9747,}}\\{{\bf{x}}_{\bf{5}}}{\bf{ = - 1}}{\bf{.5625,}}\;\;\;{{\bf{x}}_{\bf{6}}}{\bf{ = - 2}}{\bf{.1503,}}\;\;\;{{\bf{x}}_{\bf{7}}}{\bf{ = - 0}}{\bf{.6114,}}\;\;\;{{\bf{x}}_{\bf{8}}}{\bf{ = - 2}}{\bf{.5136,}}\;\;\;{{\bf{x}}_{\bf{9}}}{\bf{ = - 0}}{\bf{.9747,}}\\{{\bf{x}}_{{\bf{10}}}}{\bf{ = - 1}}{\bf{.5625,}}\;\;\;{{\bf{x}}_{{\bf{11}}}}{\bf{ = - 2}}{\bf{.1503,}}\;\;\;{{\bf{x}}_{{\bf{12}}}}{\bf{ = - 0}}{\bf{.6114,}}\;\;\;{{\bf{x}}_{{\bf{13}}}}{\bf{ = - 2}}{\bf{.5136,}}\;\;\;{{\bf{x}}_{{\bf{14}}}}{\bf{ = - 0}}{\bf{.9747,}}\\{{\bf{x}}_{{\bf{15}}}}{\bf{ = - 1}}{\bf{.5625,}}\;\;\;{{\bf{x}}_{{\bf{16}}}}{\bf{ = - 2}}{\bf{.1503,}}\;\;\;{{\bf{x}}_{{\bf{17}}}}{\bf{ = - 0}}{\bf{.6114,}}\;\;\;{{\bf{x}}_{{\bf{18}}}}{\bf{ = - 2}}{\bf{.5136,}}\;\;\;{{\bf{x}}_{{\bf{19}}}}{\bf{ = - 0}}{\bf{.9747,}}\\{{\bf{x}}_{{\bf{20}}}}{\bf{ = - 1}}{\bf{.5625;}}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{{\bf{v}}_{\bf{0}}}{\bf{ = 0}}{\bf{.6,}}}&{{{\bf{v}}_{\bf{1}}}{\bf{ = - 0}}{\bf{.4854,}}}&{{{\bf{v}}_{\bf{2}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{\bf{3}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{\bf{4}}}{\bf{ = - 0}}{\bf{.4854}}}\\{{{\bf{v}}_{\bf{5}}}{\bf{ = 0}}{\bf{.6,}}}&{{{\bf{v}}_{\bf{6}}}{\bf{ = - 0}}{\bf{.4854,}}}&{{{\bf{v}}_{\bf{7}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{\bf{8}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{\bf{9}}}{\bf{ = - 0}}{\bf{.4854}}}\\{{{\bf{v}}_{{\bf{10}}}}{\bf{ = 0}}{\bf{.6,}}}&{{{\bf{v}}_{{\bf{11}}}}{\bf{ = - 0}}{\bf{.4854,}}}&{{{\bf{v}}_{{\bf{12}}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{{\bf{13}}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{{\bf{14}}}}{\bf{ = - 0}}{\bf{.4854}}}\\{{{\bf{v}}_{{\bf{15}}}}{\bf{ = 0}}{\bf{.6,}}}&{{{\bf{v}}_{{\bf{16}}}}{\bf{ = - 0}}{\bf{.4854,}}}&{{{\bf{v}}_{{\bf{17}}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{{\bf{18}}}}{\bf{ = 0}}{\bf{.1854,}}}&{{{\bf{v}}_{{\bf{19}}}}{\bf{ = - 0}}{\bf{.4854}}}\\{{{\bf{v}}_{{\bf{20}}}}{\bf{ = 0}}{\bf{.6}}}&{}&{}&{}&{}\end{aligned}\)

Now, one can conclude that for \({\bf{\omega = 3/5}}\)the Poincare map cycles through five points,

\(\begin{aligned}{*{20}{l}}{\left( {{\bf{ - 1}}{\bf{.5625,0}}{\bf{.6}}} \right){\bf{,}}\left( {{\bf{ - 2}}{\bf{.1503, - 0}}{\bf{.4854}}} \right){\bf{,}}\left( {{\bf{ - 0}}{\bf{.6114,0}}{\bf{.1854}}} \right){\bf{ }}}\\{\left( {{\bf{ - 2}}{\bf{.5136,0}}{\bf{.1854}}} \right){\bf{,}}\left( {{\bf{ - 0}}{\bf{.9747, - 0}}{\bf{.4854}}} \right)}\end{aligned}\)

And a solution is subharmonic of period \({\bf{10\pi }}\)

The graph of the Poincare map is on the figure below for both cases.