Define
for
. Then
for
if
. Equation (1.1) can be rewritten as
Set
then (2.2) reduces to
And it is easy to see that
To prove Theorems 1.4 and 1.6, we need the following lemma.
Lemma 2.1.
Let
,
be a solution of (1.1),
is defined by (2.3),
, and
. If
holds for some
and either
or
then
.
Proof.
We assume that
and
. The case where
and
is similar and is omitted.
It is easy to see that Lemma 2.1 holds for
.
If
, then
and
for all
by (2.6). Thus we only need to prove that
. Since
,
, by (1.7) we have
that is,
So
which admits that
. Hence
which implies that
, so
by the definition of
.
Now, assume that
,
and
. By (1.7), we also have
So,
Hence, there is
such that
and
for all
. So,
and
for all
. For
, by (1.7) and (2.4), one has
provided that
contains no impulsive points. And since
if
meets one of impulsive points, we always have
for
, where
.
By the choice of
, there is a real number
such that
Next, we will show that, for any
,
In fact, for any
,
which shows that (2.17) holds.
Substituting (2.17) into (2.4), it is easy to get
Let
There are two cases to consider.
Case 1 (
).
We have by (2.6), (2.16), and (2.19)
Since
we have
Case 2 (
).
In this case, there exists
such that
Therefore, we may choose a real number
such that
Notice that
So, by (2.6), (2.15), (2.19), and (2.25), we have
The proof is completed by combining Cases 1 and 2.
Proof of Theorem 1.4.
By Lemma 2.1, (1.9) implies that (2.6) holds with
. For any
, assume that
contains
impulsive points:
. Set
We will prove that
implies that
To this end, we first prove that
If
, then
. If
, then, for
, we claim that
In fact,
In general, we can obtain (2.31) by induction. And so
for either case with
. For
, we have
And so
In general, we have
Now for
,
And so
which has proved that (2.30) holds. Now, we will prove that
Thus we only need to prove that
For any
and
we claim that
In fact, we can assume that
. The case where
is similar and is omitted. If
, by the definition of
, (2.40) holds. If
, then by Lemma 2.1 we have that (2.40) holds.
Since
by (2.40) we have
By repeatedly using (2.40) we get that (2.39) holds.
Combining (2.30) and (2.39), we find that (2.29) holds and the proof of Theorem 1.4 is complete.
Proof of Theorem 1.6.
In view of Theorem 1.4, we see that the zero solution of (1.1) is uniformly stable. Therefore, for any
, there exists
such that
implies that
Next, we will prove that
There are two cases to consider.
Case 1.
is eventually positive, that is, there exists
such that
for all
. Hence, by (1.7) and (1.8), we have
for
. That is,
is eventually nonincreasing and hence
. Thus, by (1.11) we see that (2.44) holds. The case when
is eventually negative is similar and will be omitted.
Case 2.
is oscillatory in the sense that
is neither eventually positive nor eventually negative, so
is also oscillatory. To prove (2.44), we only need to prove that
By the proof of Theorem 1.4, we have
Let
then
. It suffices to show that
In fact, if (2.48) does not hold, we assume that
and
. The case where
and
is similar and is omitted.
Let
, where
is given in Lemma 2.1. Then there exists an integer
such that
Since
and
is oscillatory, there must exist an integer
such that
By Lemma 2.1 and (2.49) we have
Equations (2.49) and (2.51) imply that
, which contradicts the fact that
; thus (2.48) holds and the proof is complete.