To show that the limit set for the Poincare map \({{\bf{x}}_{\bf{n}}}{\bf{ = x(2\pi n),}}{{\bf{v}}_{\bf{n}}}{\bf{ = x'(2\pi n)}}\) where x(t) is a solution to equation (6), is an ellipse and that this ellipse is the same for any initial values\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\), do the following:

(a)Argue that since the initial values affect only the transient solution to (6), the limit set for the Poincare map is independent of the initial values.

(b)Now show that for nlarge\({{\bf{x}}_{\bf{n}}} \approx {\bf{asin(2}}\sqrt {\bf{2}} {\bf{\pi n + }}\psi {\bf{)}},{{\bf{v}}_{\bf{n}}} \approx {\bf{c + }}\sqrt {\bf{2}} {\bf{acos(2}}\sqrt {\bf{2}} {\bf{\pi n + }}\psi {\bf{)}}\),

Where\({\bf{a = (1 + 2(0}}{\bf{.22}}{{\bf{)}}^{\bf{2}}}{{\bf{)}}^{\frac{{{\bf{ - 1}}}}{{\bf{2}}}}}{\bf{,c = (0}}{\bf{.22}}{{\bf{)}}^{{\bf{ - 1}}}}\), and\(\psi {\bf{ = arctan}}\left\{ {{\bf{ - [(0}}{\bf{.22)}}\sqrt {\bf{2}} {{\bf{]}}^{{\bf{ - 1}}}}} \right\}\).

(c)Use the result of part (b) to conclude that the ellipse\({{\bf{x}}^{\bf{2}}}{\bf{ + }}\frac{{{{{\bf{(v - c)}}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = }}{{\bf{a}}^{\bf{2}}}\)contains the limit set of the Poincare map.

The solution to equation (6) is given by \(x(t) = {x_h}(t) + {x_p}(t)\).where

\(\begin{array}{c}{x_h} = A{e^{ - 0.11t}}\sin \left( {\sqrt {9879} t + \phi } \right)\\{x_p} = \frac{1}{{0.22}}\sin t + \frac{1}{{\sqrt {1 + 2(0.22)} }}\sin \left( {\sqrt 2 t + \phi } \right)\\\tan \varphi = - \frac{1}{{0.22\sqrt 2 }}\end{array}\)

Is a steady state term.

Now, differentiate \(x\left( t \right)\)then:

\(\begin{array}{c}v\left( t \right) = x{'_h} + x{'_p}\\v\left( t \right) = x{'_h} + \frac{1}{{0.22}}\cos t + \frac{{\sqrt 2 }}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\cos \left( {\sqrt 2 t + \phi } \right)\end{array}\)

The steady solution doesn't depend upon the initial conditions. These values affect only constants A and \(\phi \) .

Hence,\(t \to \infty \), the results tend to be zero. So the points \(\left( {x\left( t \right),v\left( t \right)} \right)\) are independent of the initial conditions.

Put \(t = 2\pi n\) then

\(\begin{array}{l}{x_n} = x\left( {2\pi n} \right)\\{x_n} = {x_h}\left( {2\pi n} \right) + \frac{1}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\sin \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{v_n} = v\left( {2\pi n} \right)\\{v_n} = x{'_h}\left( {2\pi n} \right) + \frac{1}{{0.22}} + \frac{{\sqrt 2 }}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\cos \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\end{array}\)

As \(n \to \infty ,{x_h}(2\pi n) \to 0,x{'_h}(2\pi n) \to 0\)

\(\begin{array}{l}{x_n} = \frac{1}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\sin \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{x_n} = a\sin \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{v_n} = \frac{1}{{0.22}} + \frac{{\sqrt 2 }}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\cos \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{v_n} = c + \sqrt 2 a\cos \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\end{array}\)

So, the values are

\(\begin{array}{l}{x_n} = a\sin \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{v_n} = \frac{1}{{0.22}} + \frac{{\sqrt 2 }}{{\sqrt {1 + 2{{\left( {0.22} \right)}^2}} }}\cos \left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\end{array}\)

From the part (b) for large n.

\(\begin{array}{c}{x^2}_n = {a^2}{\sin ^2}\left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{\left( {{v_n} - c} \right)^2} = 2{a^2}{\cos ^2}\left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)\\{x^2}_n + \frac{{{{\left( {{v_n} - c} \right)}^2}}}{2} = {a^2}\left[ {{{\sin }^2}\left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right) + {{\cos }^2}\left( {\sqrt 2 \left( {2\pi n} \right) + \varphi } \right)} \right]\\{x^2}_n + \frac{{{{\left( {{v_n} - c} \right)}^2}}}{2} = {a^2}\end{array}\)

Therefore, the results are\({x^2}_n + \frac{{{{\left( {{v_n} - c} \right)}^2}}}{2} = {a^2}\) having a center\(\left( {0,c} \right)\).