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

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