hw2_scribe

hw2_scribe - CDS 140A – Solutions to Problems 1,9,10 from...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CDS 140A – Solutions to Problems 1,9,10 from Homework 2 Ragavendran Gopalakrishnan October 19, 2009 1. Problem 1. First, verify that the given orbit satisfies the differential equation of the system. φ ( t ) = ± 2 tan- 1 (sinh t ) ˙ φ ( t ) = ± 2 1 1 + sinh 2 t cosh t = ± 2 (cosh t )- 1 ¨ φ ( t ) = ∓ 2 1 cosh 2 t sinh t Now, given that tan( φ ( t ) 2 ) = ± sinh t , we get, sinh t = ± tan φ ( t ) 2 cosh 2 t = 1 + sinh 2 t = 1 + tan 2 φ ( t ) 2 = sec 2 φ ( t ) 2 Therefore, we have, ¨ φ ( t ) = ∓ ± 2 tan φ ( t ) 2 sec 2 φ ( t ) 2 =- sin φ ( t ) This finally gives us ¨ φ ( t ) + sin φ ( t ) = 0. Therefore, the given orbits are solutions to the system. To verify that they are homoclinic, note that the equilibrium points of this system (in the range- π ≤ ≤ + π ) are at- π , 0 and + π . And, clearly, lim t → + ∞ ± 2 tan- 1 (sinh t ) = ± π , and lim t →-∞ ± 2 tan- 1 (sinh t ) = ∓ π , which are equilibrium points. By definition, therefore, these orbits are homoclinic., which are equilibrium points....
View Full Document

This note was uploaded on 02/28/2011 for the course MATH 140A taught by Professor Marsden during the Fall '09 term at Caltech.

Page1 / 2

hw2_scribe - CDS 140A – Solutions to Problems 1,9,10 from...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online