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Q10E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 304
Fundamentals Of Differential Equations And Boundary Value Problems

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

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

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