can be satisfied by requiring 2 and 4

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: ! ! ! !". This must be true for all !. Replace ! by ! and solve for !! ! (!) to obtain !! ! ! = − ! (! ! ! )! ! !! ! ! ! !". [9] When these last two expressions are substituted back into [8] an identity is obtained, showing that it is a solution for [8]. This expression for !! ! ! is also bounded since, using [3]: ! ! ! !! ! ! !! ! ! ! !" ≤ ! ! ! !! ! !! ! ≤ !" ! !" !! ! ! !" ! !!" ! !" ! 10 Ch5Lecs ≤ !" /!. Adding [7] and [9] gives [5], the promised bounded solution. Note that [9] is required in order that !! ! (!) be bounded, so our solution is unique. □ To construct the integral operator !: !0 ℝ+ , ℝ! → !0 ℝ+ , ℝ! whose fixed point will give us the local stable manifold, replace !(!) by !(! ! ) in [5]: ! ! ! = ! !" ! + ! ! !! ! ! !! ! ! ! ! !" − ! ! !! ! ! ! !! ! ! ! !", [10] where ! = !! ! (0). Exercise 5 shows that a fixed point of [10] is also a solution of [1]. Theorem 5.3 (Local Stable Manifold) Let ! be hyperbolic, ! ∈ ! ! ! , ! ≥ 1, for some neighborhood ! of 0, and ! ! = !(! ) as ! → 0. Denote the linear stable and unstable subspaces of ! by ! ! and ! ! . Then there is a ! ⊂ ! such that the local stable manifold of [1] ! !!"# 0 = {! ∈ ! ! 0 :!! ! ⊂ !, ! ≥ 0} , is a Lipschitz graph over ! ! that is tangent to ! ! at 0. Moreover, ! ! (0) is a ! ! manifold. Note that the phrase “Lipschitz graph over ! ! ” only means that the unstable component of ! !!"# is a Lipschitz function of the stable component. ! Remark The existence and pro...
View Full Document

This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .

Ask a homework question - tutors are online