0 0 such that in the integrand

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Unformatted text preview: 1 ! , so the unstable subspace is the ! ­axis: ! ! = { ! , ! : ! = 0}. The stable eigenvector is !! = 2, −!! 0 the origin: ! , so the stable subspace is a line passing through ! ! ! = { ! , ! : ! = − ! !! 0 ! }. [3] The phase portrait for the linearized system looks like 〈!" ­axes, expansion on ! ! , the line ! ! , contraction of ! ! , other trajectories〉. The skew product form of [1] allows us to integrate these equations exactly. ! ! = !! ! !! . Use an integrating factor to solve for !(!): ! ! = !! ! ! + ! ! ! !! ! !(!! ! !! ) !". ! Make the substitution ! = ! !! in the integral to obtain: !! ! ! !! ! ! = !! ! + ! ! !!! ! !" !" . ! Points (! , !) on the unstable manifold ! ! (0,0) satisfy !! ! , ! → 0 as ! → −∞. Then the form of the first component of !! requires! = 0. If we substitute ! = 0 into the second component and note ! 0 = 0, we see that any ! ∈ ℝ is allowed. Therefore ! ! is the ! ­axis: ! ! 0,0 = { ! , ! : ! = 0}. 6 Ch5Lecs Points ! , ! on the stable manifold ! ! 0,0 satisfy !! ! , ! → 0 as ! → ∞. The first component of !! is compatible with this for any ! ∈ ℝ. From the second component, ! must satisfy lim!→! ! ! (! + Claim ! = − lim!→! ! ! ( ! ! ! !! !" !") = 0. [4] ! !(!") !". To see that this satisfies [4] consider ! ! ! ! !! !" !" − ! ! ! !" !") = − lim!→! ! ! ! !! !(!") !" ! Since ! 0 = 0 and ! is continuous, ∀! > 0∃! > 0 such that !" < ! ⇒ ! !" < ! . In the integrand, !" < ! ! !! . Choose ! > 0 sufficiently large that ! ! !! < ! for ! > !. Then !! ! !! ! ! !" !" < ! . As this can be done for any ! > 0, conclude that [4] is sa...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .

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