Geometry
of
the Melnikov Vector
107
defined by
(6.1)
b(7)
=
dim
Wa(z-)
-
dim
Wa(z+).
Notice from (5.10) and diagram
1
in
$5
that the dimension
m
of the Melnikov vector
is given by
m
=
n
-
[{n
-
dimWa(z-)}
+
dimWa(z+)
-
k].
Thus
m,
k
and
b(7)
satisfy the following relation.
(6.2)
m
=
k
+
b(7).
The splitting index
6(7)
expresses the 'degeneracy' of
7
in the following sense.
Here we distinguish the homoclinic and heteroclinic cases.
(i) Suppose that
7
is
a
homoclinic orbit. Then the splitting index
b(y)
is always
zero. Thus by theorem 5.5(ii), all of
k
column vectors in
[&
n
M&M]
must be
nonzero to guarantee the transversal intersection. This situation is only the
case
in
the homoclinic case.
(ii) Suppose that
7
is
a
heteroclinic orbit in
R".
In this case we have three
different situations.
(ii-1)
b(7)
>
0.
Then
m
=
k
+
b(7)
>
k.
Thus theorem 5.5(ii) implies that
there is no transversal intersection because the matrix
[&!&$A41
is
of
size
m
x
k.
A
reason for this is that dimWa(z+)
<
dimWa(z-) is equivalent to saying that
dimWU(z-)
+
dimWa(z+)
<
n.
(ii-2)
b(7)
=
0.
In this case we have the same situation
as
in the homoclinic case.
(iii-3)
6(7)
<
0.
Then
m
=
k+b(~)
<
k.
Thus transversal intersection is possible.
In this way, we can classify the possibility and impossibility of transversal intersection
by using the splitting index
b(7).