# Lecture8

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Unformatted text preview: hen we obtained the relationship between vector fields . Note that the expression for given below [2] is a diffeomorphism, and that Eq. [3] exhibits the correct constant of proportionality . The comparison between [2] and [3] gives . Thus the family of vector fields is induced by a family of form . [4] According to the one-dimensional equivalence theorem, theorem 4.10, there is a neighborhood of the origin for which the dynamics of [4] is topologically equivalent to those of [1] because both systems have two equilibria with the same stability types and arranged in the same order on the line. □ Transcritical Bifurcation Consider , an unfolding of equilibrium point at never disappears. bifurcation. Sketch the bifurcation diagram. -, . This is not a versal unfolding; the is a normal form of the transcritical -, two lines of equilibria with stability exchanged at the origin -, -, with parabolas intersecting the abscissa at stability of equilibrium points and to determine This bifurcation is sometimes referred to as an exchange of stability between the two equilibrium points, and is encountered frequently in applications. To see the relationship with the saddle-node bifurcation, begin by completing squares: . Let and to see that that flow of is homeomorphic to that of . The family of vector fields is induced by using 12 Ch8Lecs . Notice that cannot be positive. As increases through zero, values of take the following path through the saddle-node bifurcation diagram: -, -, curves of equilibria corresponding to SN bifurcation, path of As a bug crawls along this path, it sees the transcritical bifurcation diagram shown above (modulo some distortion due to the coordinate transformations). 8.6 Saddle-Node Bifurcation in Theorem 8.6 (saddle node) Let , and suppose that satisfies . (singularity) Choose coordinates so that is diagonal in the zero eigenvalue and set where corresponds to the zero eigenvalue and are the remaining coordinates. Then , where and . Suppose that . (nondegeneracy) Then there exists an interval containing , functions and , and a neighborhood of such that if there are no equilibria and if there are two. Suppose that has a -dimensional unstable space and an -dimensional stable space. Then, when there are two equilibria, one has a -dimensional unstable and an -dimensional stable manifold and the other has a -dimensional unstable manifold and an -dimensional stable manifold. Proof The equilibria are solutions of , [1] . By assumption, that there is a neighborhood of such that is nonsingular; thus the Implicit Function Theorem ensures where there exists a unique function 13 Ch8Lecs [2] and . Substitute this into to obtain . Consequently, the problem has been reduced to the one-dimensional case; we need only check that satisfies the same criteria as Theorem 8.3, the one-d...
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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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