Such 9 ch8lecs since form the higher order term

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Unformatted text preview: ave [3] and . Consider the function . Since , the implicit function theorem guarantees neighborhoods and of the origin such that when there is a unique such that and . Since there must also be an open interval , containing , for which is monotone. On , is either concave up (when ) or concave down (when ). Therefore is the unique extremal value for when , and . determines whether the critical point is a minimum or a maximum. Since when , this remains true by continuity for small enough ; for . Therefore, when , has a minimum at and is positive on . This implies that, if there are no zeros of , if there is a single zero, and if there are two zeros. Similar considerations apply when .□ Picture for -, : -, , For , is a stable equilibrium and is an unstable equilibrium. The bifurcation set is the set of all points in parameter space at which the bifurcation takes place. A bifurcation is codimension on parameters. if the bifurcation set is determined by independent conditions According to theorem 8.3, the saddle-node bifurcation depends on a single quantity: . The bifurcation takes place when . The saddle-node is a codimension 1 bifurcation. Example. Let . Find the saddle-node bifurcation set, that is, the locus of saddle-node bifurcations in space. is defined by . From we have . Then . The bifurcation set corresponds to the single condition .■ 10 Ch8Lecs Corollary 8.4 If f satisfies the hypotheses of theorem 8.3 and there is a single parameter such that the transversality condition holds, then a saddle-node bifurcation occurs as crosses zero. Proof. Recall that defined was defined to satisfy . Thus the chain rule implies . We also . Since , the sign of changes as crosses zero. □ Remark. When there is a vector of parameters , application of the corollary requires that for as crosses . Example. Let bifurcation takes place as Example. Let and system . If , a saddle-node crosses . Otherwise the bifurcation takes place when , an unfolding of . Then . By the corollary, a bifurcation occurs as crosses zero. For this , and a saddle-node bifurcation also takes place when . Example. Let , an unfolding of . Note that does not satisfy the transversality condition. In fact is an equilibrium of . A saddle-node bifurcation clearly does not occur in this unfolding of . ■ , which for all values of Theorem 8.5 Under the hypotheses of theorem 8.3, the saddle-node bifurcation has the miniversal unfolding . Proof. Expand [1] in Taylor series about : . Identifying and gives . Let and [2] . Then . [3] 11 Ch8Lecs Compare [2] and [3] with the form of section 8.3, Eq. [2], which is reproduced below. To obtain that equation, we assumed that the flows of and were related by the diffeomorphism where . T...
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