Question

To show that the limit set of the Poincare map given in \(\left( {\bf{3}} \right)\) depends on the initial values, do the following:

(a) Show that when\({\bf{\omega = 2or3}}\), the Poincare map consists of the single point\({\bf{(x,v) = }}\left( {{\bf{Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{,\omega Acos}}\phi } \right)\).

(b) Show that when \({\bf{\omega = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{,}}\) the Poincare map alternates between the two points\(\left( {\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{ \pm Asin}}\phi {\bf{, \pm \omega Acos}}\phi } \right)\).

(c) Use the results of parts \(\left( {\bf{a}} \right)\)and \(\left( {\bf{b}} \right)\)to show that when\({\bf{\omega = 2,3,or}}\frac{1}{2}\), the Poincare map \(\left( {\bf{3}} \right)\)depends on the initial values \(\left( {{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{0}}}} \right){\bf{.}}\)

Short answer

If \({\bf{\omega = 2}}\)or\({\bf{\omega = 3}}\)one has that for every \({\bf{n}} \in {\bf{N}}\) the Poincare map is given by \({{\bf{x}}_{\bf{n}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{,}}\;\;\;{{\bf{v}}_{\bf{n}}}{\bf{ = }}{{\bf{v}}_{\bf{0}}}\); if \({\bf{\omega = }}\frac{1}{2}\)then for even \({\bf{n}}\)one have that \({{\bf{x}}_{\bf{n}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{ = }}{{\bf{v}}_{\bf{0}}}{\bf{,}}\)for odd \({\bf{n}}\)one has that \({{\bf{x}}_{\bf{n}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{ - Asin}}\phi {\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{ = - }}{{\bf{v}}_{\bf{0}}}{\bf{,}}\) so in all three cases one has that the Poincare map depends on the initial values \(\left( {{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{0}}}} \right){\bf{.}}\)

Step-by-step solution

Using the equation \(\left( {\bf{3}} \right)\)

The Poincare map, return map, or time T map for the differential equation \({\bf{\dot x = f(t,x)}}\) is the map\(\phi {\bf{:J}} \to {\bf{R}}\), given by \(\phi \left( {{{\bf{x}}_{\bf{0}}}} \right){\bf{ = }}{{\bf{x}}_{\bf{1}}}\) where \({\bf{x}}\left( {\bf{t}} \right)\) is the solution of the differential equation with\({\bf{x}}\left( {\bf{0}} \right){\bf{ = }}{{\bf{x}}_{\bf{0}}}\), and where\({{\bf{x}}_{\bf{1}}}{\bf{ = x}}\left( {\bf{T}} \right)\).

Here \({\bf{J}} \subset {\bf{R}}\) is the domain of the Poincare map, which consists of those \({{\bf{x}}_{\bf{0}}} \in {\bf{R}}\) for which the solution \({\bf{x}}\left( {\bf{t}} \right)\) of the differential equation exists for\(0 \le {\bf{t}} \le {\bf{T}}\).

(a) The Poincare map given in \(\left( {\bf{3}} \right)\) the textbook for \({\bf{\omega = 2}}\) is

\(\begin{array}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = Asin(2\pi \omega n + }}\phi {\bf{) + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}\\{\bf{ = Asin(4n\pi + }}\phi {\bf{) + }}\frac{{\bf{F}}}{{\bf{3}}}{\bf{, }}\\{{\bf{v}}_{\bf{n}}}{\bf{ = \omega Acos(2\pi \omega n + }}\phi {\bf{)}}\\{\bf{ = 2Acos(4n\pi + }}\phi {\bf{)}}\end{array}\)

Since\({\bf{sin(x + 2k\pi ) = sinxandcos(x + 2k\pi ) = cosx}}\) for \({\bf{k}} \in {\bf{Z}}\)one has that the Poincare map for \({\bf{\omega = 2}}\)is given by

\(\begin{array}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{\bf{3}}}{\bf{ }}\\{{\bf{v}}_{\bf{n}}}{\bf{ = 2Acos}}\phi \end{array}\)

Finding the point \(\left( {{\bf{x,v}}} \right)\)

Similarly, for \({\bf{\omega = 3}}\)one has

\(\begin{array}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = Asin(2\pi \omega n + }}\phi {\bf{) + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{ = Asin(6n\pi + }}\phi {\bf{) + }}\frac{{\bf{F}}}{{\bf{8}}}{\bf{,}}\\{{\bf{v}}_{\bf{n}}}{\bf{ = \omega Acos(2\pi \omega n + }}\phi {\bf{) = 3Acos(6n\pi + }}\phi {\bf{)}}\\{{\bf{x}}_{\bf{n}}}{\bf{ = Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{\bf{8}}}{\bf{,}}\\{{\bf{v}}_{\bf{n}}}{\bf{ = 3Acos}}\phi \end{array}\)

So, in both cases, the Poincare map consists of the point,

\({\bf{(x,v) = }}\left( {{\bf{Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{,\omega Acos}}\phi } \right)\)

Substitute the value for \({\bf{n = 2\pi }}\)

(b) For \({\bf{\omega = }}\frac{1}{2}\)one has

\(\begin{array}{c}{{\bf{x}}_{\bf{n}}}{\bf{ = Asin(2\pi \omega n + }}\phi {\bf{) + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}\\{\bf{ = Asin(n\pi + }}\phi {\bf{) - }}\frac{{{\bf{4F}}}}{{\bf{3}}}{\bf{ }}\\{{\bf{v}}_{\bf{n}}}{\bf{ = \omega Acos(2\pi \omega n + }}\phi {\bf{)}}\\{\bf{ = }}\frac{{\bf{A}}}{{\bf{2}}}{\bf{cos(n\pi + }}\phi {\bf{)}}\end{array}\)

Since\({\bf{n}} \in {{\bf{N}}_{\bf{0}}}\), one has that when \({\bf{n = 2 k}}\)for\({\bf{k}} \in {{\bf{N}}_{\bf{0}}}\), then

\(\begin{array}{c}{{\bf{x}}_{{\bf{2k}}}}{\bf{ = Asin(2k\pi + }}\phi {\bf{) - }}\frac{{{\bf{4F}}}}{{\bf{3}}}{\bf{ = Asin}}\phi {\bf{ - }}\frac{{{\bf{4F}}}}{{\bf{3}}}{\bf{ }}\\{{\bf{v}}_{{\bf{2k}}}}{\bf{ = }}\frac{{\bf{A}}}{{\bf{2}}}{\bf{cos(2k\pi + }}\phi {\bf{) = }}\frac{{\bf{A}}}{{\bf{2}}}{\bf{cos}}\phi \end{array}\)

And when \({\bf{n = 2 k + 1}}\) for \({\bf{k}} \in {{\bf{N}}_{\bf{0}}}\)

Step 4:Finding the two points

Then,

\(\begin{array}{c}{{\bf{x}}_{{\bf{2k + 1}}}}{\bf{ = Asin((2k + 1)\pi + }}\phi {\bf{) - }}\frac{{{\bf{4F}}}}{{\bf{3}}}\\{\bf{ = - Asin}}\phi {\bf{ - }}\frac{{{\bf{4F}}}}{{\bf{3}}}{\bf{ }}\\{{\bf{v}}_{{\bf{2k + 1}}}}{\bf{ = }}\frac{{\bf{A}}}{{\bf{2}}}{\bf{cos((2k + 1)\pi + }}\phi {\bf{)}}\\{\bf{ = - }}\frac{{\bf{A}}}{{\bf{2}}}{\bf{cos}}\phi \end{array}\)

So, now one has two points, one when \({\bf{n}}\) is even and one when it is odd. Those dots are\(\left( {{\bf{Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{,\omega Acos}}\phi } \right){\bf{,}}\;\;\;\left( {{\bf{ - Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{, - \omega Acos}}\phi } \right){\bf{.}}\)

Checking the Poincare maps

(c) If \({\bf{\omega = 2}}\)or \({\bf{\omega = 3}}\) the Poincare map is the single point, so if \(\left( {{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{0}}}} \right)\) are the initial values, then for everyone has that\({{\bf{x}}_{\bf{n}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{,}}\;\;\;{{\bf{v}}_{\bf{n}}}{\bf{ = }}{{\bf{v}}_{\bf{0}}}\)when \({\bf{\omega = }}\frac{1}{2}\) then for \({\bf{n = 2k,}}\;\;{\bf{k}} \in {\bf{N}}\)one has that\({{\bf{x}}_{{\bf{2k}}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{,}}\;\;{{\bf{v}}_{{\bf{2k}}}}{\bf{ = }}{{\bf{v}}_{\bf{0}}}\)and when \({\bf{n = 2k + 1,}}\;\;\;{\bf{k}} \in {{\bf{N}}_{\bf{0}}}\)one has that\({{\bf{x}}_{{\bf{2k + 1}}}}{\bf{ = }}{{\bf{x}}_{\bf{0}}}{\bf{ - 2Asin}}\phi {\bf{,}}\;\;\;{{\bf{v}}_{{\bf{2k + 1}}}}{\bf{ = - }}{{\bf{v}}_{\bf{0}}}\)

So, one sees that if \({\bf{\omega = 2,3,}}\frac{1}{2}\) every point of the Poincare map depends on the initial values.