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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 saddlenode bifurcation depends on a single quantity:
.
The bifurcation takes place when
. The saddlenode is a codimension 1 bifurcation.
Example. Let
. Find the saddlenode bifurcation set, that is, the locus
of saddlenode 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 saddlenode 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 saddlenode
crosses . Otherwise the bifurcation takes place when , an unfolding of
. Then
. By the corollary, a bifurcation occurs as crosses zero. For this
, and a saddlenode 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 saddlenode 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 saddlenode 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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 Fall '14
 MarcEvans

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