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Unformatted text preview: ≤ !" ! − ! 2/! Thus ! is a contraction when ! ≤ !/4! , which we have already assumed. Then the Contraction Mapping Theorem guarantees that ! has a unique fixed point in !! . This unique fixed point is a function of ! ∈ ! ! and is denoted ! (!; !). The set ! (0; !) is a graph over ! ! . □
We now know that the integral operator [10] has a unique solution in !! . To further explore the properties of this solution we need a generalization of Grönwall’s inequality. Lemma 5.4 (Generalized Grönwall) Suppose !, ! and ! are nonnegative, ! < !/2, and there is a nonnegative, bounded, continuous function !: ℝ! → ℝ! satisfying ! ! ≤ ! !!" ! + !
Then ! ! ≤ ! !
! ! !! ! !!
!
! ! !! !
! ! ! ! !" + ! ! ! ! !!
!
! ! ! !"; [12] , where ! = 1 − 2!/!. Proof. Although ingenious this seems specialized and won’t be given here. Remark. The first two terms have the form given in the generalization of Grönwall’s inequality discussed by problem 9 in section 3.6. That is ! ! ≤! ! + !
!
! ! !(!) !", where ! ! = ! !!" ! and ! ! = !! !!
! ! ≤! ! ! !
!
! ! !" ! !! ≤ ! !!" !! !/! . The last term of [12] has modified the result. . Then according to the result of that problem: 12 Ch5Lecs Proof of theorem 5.3 (Part 2). In part 1 of the proof we showed that the integral operator [10] has a unique fixed point ! (!; !) for ! sufficiently close to the origin. To show that ! (!; !) is a point on the stable manifold we must show that ! !; ! → 0 as ! → ∞. From [11], ! !; ! satisfies ! (!; !) ≤ ! ! !!" ! + !" ! ! ! !! !
!
! ! ! !! !
!
!
! ! !; ! !" + !" !; ! !". This has the form of [12] with ! ! = ! (!; !) ,...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .
 Fall '14
 MarcEvans
 Sets

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