Question

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.

Short answer

1. As \(t \to \infty \)the results tend to zero. So the points \(\left( {x\left( t \right),v\left( t \right)} \right)\) are independent of the initial conditions
2. The values are

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

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

Step-by-step solution

(a)Step 1: Show that the map is independent of initial conditions

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.

(b) Step 2: Find the value of \({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{v}}_{\bf{n}}}{\bf{)}}\)

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

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

(c) Step 3: Verify the given result.

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\(\left( {0,c} \right)\).