Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Compute and graph the points of the Poincare map with \(t{\bf{ = }}2\pi n,n{\bf{ = }}0,1, \ldots ,20\) for equation\(\left( {\bf{1}} \right)\), taking\(A{\bf{ = F = }}1,\phi {\bf{ = }}0\), and\(\omega {\bf{ = }}\frac{1}{{\sqrt 3 }}\). Describe the limit set for this system.

Step-by-Step Solution

Answer

The limit set for this system is an ellipse.

Step 1: Defining the Poincare map

For \(A{\bf{ = F = }}1,\phi {\bf{ = }}0\) and \(\omega {\bf{ = }}\frac{1}{{\sqrt 3 }}\)one defines the Poincare map as:

\(\begin{aligned}{c}{x_n}{\bf{ = }}\sin \left( {\frac{2}{{\sqrt 3 }}n\pi } \right){\bf{ - }}\frac{3}{2},\\{v_n}{\bf{ = }}\frac{1}{{\sqrt 3 }}\cos \left( {\frac{2}{{\sqrt 3 }}n\pi } \right)\end{aligned}\)

Step 2: Computing \({{\bf{x}}_{\bf{n}}}\) and \({{\bf{v}}_{\bf{n}}}\)

One will compute \({x_n}\) and \({v_n}\) for \({\bf{n = }}\overline {{\bf{0,20}}} \)

 

\(\begin{aligned}{c}{{\bf{x}}_{\bf{0}}}{\bf{ =  - 1}}{\bf{.5,}}\;\;\;{{\bf{x}}_{\bf{1}}}{\bf{ =   - 1}}{\bf{.9671,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ =  - 0}}{\bf{.674,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ =   - 2}}{\bf{.4936,}}\;\;\;{{\bf{x}}_{\bf{4}}}{\bf{ =  - 0}}{\bf{.5688,}}\\{{\bf{x}}_{\bf{5}}}{\bf{ =  - 2}}{\bf{.153,}}\;\;\;{{\bf{x}}_{\bf{6}}}{\bf{ =   - 1}}{\bf{.2764,}}\;\;\;{{\bf{x}}_{\bf{7}}}{\bf{ =  - 1}}{\bf{.2425,}}\;\;\;{{\bf{x}}_{\bf{8}}}{\bf{ =  - 2}}{\bf{.179,}}\;\;\;{{\bf{x}}_{\bf{9}}}{\bf{ =   - 0}}{\bf{.5567,}}\\{{\bf{x}}_{{\bf{10}}}}{\bf{ =  - 2}}{\bf{.4891,}}\;\;\;{{\bf{x}}_{{\bf{11}}}}{\bf{ =  - 0}}{\bf{.6941,}}\;\;\;{{\bf{x}}_{{\bf{12}}}}{\bf{ =  - 1}}{\bf{.936,}}\;\;\;{{\bf{x}}_{{\bf{13}}}}{\bf{ =  - 1}}{\bf{.5349,}}{{\bf{x}}_{{\bf{14}}}}{\bf{ =  - 1}}{\bf{.0023,}}\\{{\bf{x}}_{{\bf{15}}}}{\bf{ =   - 2}}{\bf{.3452,}}\;\;\;{{\bf{x}}_{{\bf{16}}}}{\bf{ =  - 0}}{\bf{.5030,}}\;\;\;{{\bf{x}}_{{\bf{17}}}}{\bf{ =  - 2}}{\bf{.4179,}}\;\;\;{{\bf{x}}_{{\bf{18}}}}{\bf{ =  - 0}}{\bf{.8738,}}\;\;\;{{\bf{x}}_{{\bf{19}}}}{\bf{ =  - 1}}{\bf{.6895,}}\\{{\bf{x}}_{{\bf{20}}}}{\bf{ =   - 1}}{\bf{.7911;}}\\\\{{\bf{v}}_{\bf{0}}}{\bf{ = 0}}{\bf{.5774,}}\;\;\;{{\bf{v}}_{\bf{1}}}{\bf{ =  - 0}}{\bf{.5105,}}\;\;\;{{\bf{v}}_{\bf{2}}}{\bf{ = 0}}{\bf{.3254,}}\;\;\;{{\bf{v}}_{\bf{3}}}{\bf{ =  - 0}}{\bf{.065,}}\;\;\;{{\bf{v}}_{\bf{4}}}{\bf{ =  - 0}}{\bf{.2105,}}\\{{\bf{v}}_{\bf{5}}}{\bf{ = 0}}{\bf{.4373,}}\;\;\;{{\bf{v}}_{\bf{6}}}{\bf{ =  - 0}}{\bf{.56272,}}\;\;\;{{\bf{v}}_{\bf{7}}}{\bf{ = 0}}{\bf{.5579,}}\;\;\;{{\bf{v}}_{\bf{8}}}{\bf{ =  - 0}}{\bf{.4238,}}\;\;\;{{\bf{v}}_{\bf{9}}}{\bf{ = 0}}{\bf{.1916,}}\\{{\bf{v}}_{{\bf{10}}}}{\bf{ = 0}}{\bf{.085,}}\;\;\;{{\bf{v}}_{{\bf{11}}}}{\bf{ =  - 0}}{\bf{.3419,}}\;\;\;{{\bf{v}}_{{\bf{12}}}}{\bf{ = 0}}{\bf{.5196,}}\;\;\;{{\bf{v}}_{{\bf{13}}}}{\bf{ =  - 0}}{\bf{.577,}}\;\;\;{{\bf{v}}_{{\bf{14}}}}{\bf{ = 0}}{\bf{.5008,}}\\{{\bf{v}}_{{\bf{15}}}}{\bf{ =  - 0}}{\bf{.3086,}}\;\;\;{{\bf{v}}_{{\bf{16}}}}{\bf{ = 0}}{\bf{.04492,}}\;\;\;{{\bf{v}}_{{\bf{17}}}}{\bf{ = 0}}{\bf{.2291,}}\;\;\;{{\bf{v}}_{{\bf{18}}}}{\bf{ =  - 0}}{\bf{.4501,}}\;\;\;{{\bf{v}}_{{\bf{19}}}}{\bf{ = 0}}{\bf{.5669,}}\\{{\bf{v}}_{{\bf{20}}}}{\bf{ =  - 0}}{\bf{.5524}}\end{aligned}\)

Step 3:Sketching the points

One sees that any point hasn't been repeated yet. Let's sketch them and try to describe the curve that those points are going to form when\(n \to \infty \).

It looks like those points will eventually create an ellipse. Indeed, one can show that the coordinates of those points satisfy the equation of an ellipse in \({\bf{xv - }}\)plane:

 

\({\left( {{x_n} + \frac{3}{2}} \right)^2} + 3v_n^2 = 1\)          

 

So, the limit set for this system is an ellipse.