368 Chapter 16 Homoclinic Phenomena
points whose forward orbits never enter D
1
and so lie completely in S. Let
+
={X ∈ S |F
n
(X) ∈ S for n = 0, 1, 2, ...}.
We claim that
+
has properties similar to the corresponding set for the
one-dimensional logistic map described in Chapter 15.
If X ∈
+
, then F (X) ∈ S, so we must have that either X ∈ H
0
or X ∈ H
1
,
for all other points in S are mapped into D
1
or D
2
. Since F
2
(X) ∈ S as well,
we must also have F(X ) ∈ H
0
∪H
1
, so that X ∈ F
−1
(H
0
∪H
1
). Here F
−1
(W )
denotes the preimage of a set W lying in D. In general, since F
n
(X) ∈ S,we
have X ∈ F
−n
(H
0
∪ H
1
). Thus we may write
+
=
∞
n=0
F
−n
(H
0
∪ H
1
).
Now if H is any horizontal strip connecting the left and right boundaries of
S with height h, then F
−1
(H) consists of a pair of narrower horizontal strips
of height δh, one in each of H
0
and H
1
. The images under F of these narrower
strips are given by H ∩ V
0
and H ∩ V
1
. In particular, if H = H
i
, F
−1
(H
i
)isa
pair of horizontal strips, each of height δ
2
, with one in H
0
and the other in H
1
.
Similarly, F
−1
(F
−1
(H
i
)) = F
−2
(H
i
) consists of four horizontal strips, each of
height δ
3
, and F
−n
(H
i
) consists of 2
n
horizontal strips of width δ
n+1
. Hence
the same procedure we used in Section 15.5 shows that
+
is a Cantor set of
line segments, each extending horizontally across S.
The main difference between the horseshoe and the logistic map is that, in
the horseshoe case, there is a single backward orbit rather than infinitely many
such orbits. The backward orbit of X ∈ S is {F
−n
(X) |n = 1, 2, ...}, provided
F
−n
(X) is defined and in D.IfF
−n
(X) is not defined, then the backward
orbit of X terminates. Let
−
denote the set of points whose backward orbit
is defined for all n and lies entirely in S.IfX ∈
−
, then we have F
−n
(X) ∈ S
for all n ≥ 1, which implies that X ∈ F
n
(S) for all n ≥ 1. As above, this forces
X ∈ F
n
(H
0
∪ H
1
) for all n ≥ 1. Therefore we may also write
−
=
∞
n=1
F
n
(H
0
∪ H
1
).
On the other hand, if X ∈ S and F
−1
(X) ∈ S, then we must have X ∈
F(S) ∩ S, so that X ∈ V
0
or X ∈ V
1
. Similarly, if F
−2
(X) ∈ S as well, then
X ∈ F
2
(S) ∩ S, which consists of four narrower vertical strips, two in V
0
and
two in V
1
. In Figure 16.8 we show the image of D under F
2
. Arguing entirely
analogously as above, it is easy to check that
−
consists of a Cantor set of
vertical lines.