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.