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Found in: Page 304

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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 $$\left( {{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{0}}}} \right){\bf{.}}$$

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 $$\left( {{{\bf{x}}_{\bf{0}}}{\bf{,}}{{\bf{v}}_{\bf{0}}}} \right){\bf{.}}$$

See the step by step solution

## Step 1: 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

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

## Step 2: 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,

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

## Step 3: 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$$\left( {{\bf{Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{,\omega Acos}}\phi } \right){\bf{,}}\;\;\;\left( {{\bf{ - Asin}}\phi {\bf{ + }}\frac{{\bf{F}}}{{{{\bf{\omega }}^{\bf{2}}}{\bf{ - 1}}}}{\bf{, - \omega Acos}}\phi } \right){\bf{.}}$$

## Step 5: 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.