From now on we will assume that the system (1.1) has a unique solution for a given initial condition on
and without loss of generality
, thus
.
We will make the following assumptions on (1.1).
-
(H1)
is continuous in the second variable for any fixed
.
-
(H2)
System (1.1) has a bounded solution
, passing through
,
, that is,
.
For this bounded solution
, there is an
such that
for all
. So, we will have to assume that
for all
, and
. Next, we will point out the definitions of stability for functional difference equations adapting it from the continuous case according to Hino et al. in [3].
Definition 4.1.
A bounded solution
of (1.1) is said to be:
-
(i)
stable, if for any
and any integer
, there is
such that
implies that
for all
, where
is any solution of (1.1);
-
(ii)
uniformly stable, abbreviated as "
'', if for any
and any integer
, there is
(
does not depend on
) such that
implies that
for all
, where
is any solution of (1.1);
-
(iii)
uniformly asymptotically stable, abbreviated as "
'', if it is uniformly stable and there is
such that for any
, there is a positive integer
such that if
and
, then
for all
, where
is any solution of (1.1);
-
(iv)
globally uniformly asymptotically stable, abbreviated as "
'', if it is uniformly stable and
as
, whenever
is any solution of (1.1).
Remark 4.2.
It is easy to see that an equivalent definition for
, being
, is the following:
-
(iii)
is
, if it is uniformly stable, and there exists
such that if
and
, then
as
, where
is any solution of (1.1).
4.1. The Periodic Case
Here, we will assume what follows.
-
(H3)
The function
in (1.1) is periodic in
, that is, there exists a positive integer
such that
for all
.
Moreover, we will assume what follows.
Lemma 4.3.
Suppose that condition (
) holds. If
is a bounded solution of (1.1) such that
, then
is also bounded in
.
Proof.
Let us say that
for all
. Then by Axiom (A)(iii) and hypothesis (
) we have
Lemma 4.4.
Suppose that condition (
) holds. Let
be a sequence in
such that
for all
. Assume that
as
for every
and
, then
in
as
for each
. In particular, if
as
uniformly in
, then
in
as
uniformly in
.
Proof.
By Axiom (A)(iii) and hypotheses we have that
In the particular case
we obtain
and so
as
. On the other hand, since
is fixed, it follows that
for each
. Therefore, we have concluded the proof.
Theorem 4.5.
Suppose that condition (
) and (H1)–(H3) hold. If the bounded solution
of (1.1) is
, then
is an a.a.p. sequence in
, equivalently, (1.1) has an a.a.p. solution.
Proof.
By Lemma 4.3 there exists
such that
for all
, and a bounded (or compact) set
such that
for all
. Let
be any integer sequence such that
and
as
. For each
, there exists a nonnegative integer
such that
. Set
. Then
for all
. Since
is a bounded set, we can assume that, taking a subsequence if necessary,
for all
, where
. Now, set
. Thus,
which implies that
is a solution of the system,
through
. It is clear that if
is
, then
is also
with the same pair
as the one for
.
Since
is bounded for all
and
, we can use the diagonal method to get a subsequence
of
such that
converges for each
as
. Thus, we can assume that the sequence
converges for each
as
. Since
, by Lemma 4.4 it follows that
is also convergent for each
. In particular, for any
there exists a positive integer
such that if
(
is the constant given in Axiom A(ii), then
where
is the number given by the uniform stability of
. Since
, it follows from Definition 4.1 and (4.7) that
and by Axiom A(ii) it follows that
This implies that for any positive integer sequence
,
as
, there is a subsequence
of
for which
converges uniformly on
as
. Thus, the conclusion of the theorem follows from Lemma 3.9(d).
Before proving our following result we remark that if
is a.a.p. then there are unique sequences
such that
, with
a.p. and
as
as
. By Lemma 3.9(a) it follows that
is bounded and thus
. Hence, by Axiom (C) we must have that
for all
. In particular,
for all
.
Theorem 4.6.
Suppose that
and (H1)–(H3) hold and the bounded solution
of (1.1) is
, then system (1.1) has an a.p. solution, which is also
.
Proof.
It follows from Theorem 4.5 that
is an a.a.p. Set
(
), where
is a.p. sequence and
as
. For the positive integer sequence
, by Lemma 3.9(b)–(d) and arguments of the previous theorem, we can choice a subsequence
of
such that
converges uniformly in
and
uniformly on
as
and
is also a.p. Then,
uniformly in
, and thus by Lemma 4.4
uniformly in
on
as
and
. Since
as
, we have
for
, that is, the system (1.1) has an almost periodic solution, and so we have proved the first statement of the theorem.
In order to prove the second affirmation, notice that
since
. For any
, let
be a solution of (1.1) such that
and
. Again, by Lemma 4.4
as
for each
, so there is a positive integer
such that if
, then
Thus, for
, we have
Then,
Therefore, there is
such that if
, then
and hence,
for all
, where
is a pair for the uniform stability of
. This shows that if
, then
for all
, which implies that
for all
if
because
is arbitrary. This proves that
is
.
In the case when we have an asymptotically stable solution of (1.1) we obtain the following result.
Theorem 4.7.
Suppose that
and (H1)–(H3) hold and the bounded solution
of (1.1) is
, then the system (1.1) has a periodic solution of period
for some positive integer
, which is also
.
Proof.
Set
,
. By the proof of Theorem 4.5, there is a subsequence
which converges to a solution
of (4.6) for each
and hence by Lemma 4.4,
as
. Thus, there is a positive integer
such that
(
), where
is obtained from the uniformly asymptotic stability of
. Let
, and notice that
is a solution of (1.1). Since
for
, that is,
, we have
and hence,
because
is
(see also Remark 4.2). On the other hand,
is a.a.p. by Theorem 4.5, then
where
is a.p. and
as
. It follows from (4.17) and (4.18) that
which implies that
for all
because
is a.p.
For the integer sequence
,
, we have
. Then
uniformly for all
as
, and again by Lemma 4.4,
uniformly in
as
. Since
, we have
for
, which implies that (1.1) has a periodic solution
of period
.
Now, we will proceed to prove that
by the use of definition
in Remark 4.2. Notice that since
then
is a
solution of (1.1) with the same
as the one for
. Let
be any solution of (1.1) such that
. Set
. Again, for sufficient large
, we have the similar relations (4.12) and (4.14) with
and
. Thus,
as
if
, because
,
, and
satisfy (1.1). This completes the proof.
Finally, if the particular solution is
, we will prove that system (1.1) has a periodic solution.
Theorem 4.8.
Suppose that
and (H1)–(H3) hold and that the bounded solution
of (1.1) is
, then the system (1.1) has a periodic solution of period
.
Proof.
By Theorem 4.5,
is a.a.p. Then 
), where
(
) is an a.p. sequence and
as
. Notice that
is also a solution of (1.1) satisfying
. Since
is
, we have that
as
, which implies that
for all
. Using same technique as in the proof of Theorem 4.7, we can show that
is a
-periodic solution of (1.1).
4.2. The Almost Periodic Case
Here, we will assume that
-
(H4)
the function
in (1.1) is almost periodic in
uniformly in the second variable.
By
we denote the uniform closure of
, that is,
. Note that
by Lemma 3.12 and
by Lemma 3.13.
Lemma 4.9.
Suppose that Axiom (C) is true, and that
is an a.p. sequence with
, then
is a.p.
Proof.
We know that, given
, there exists an integer
such that each discrete interval of length
contains a
such that
By Axiom (C) we have
Lemma 4.10.
Suppose that
is a fading memory space and
is a.a.p. with
, then
is a.a.p.
Proof.
Since
is a.a.p. there are unique sequences
and
such that
is a.p. and
as
. Then by Lemma 4.9 it follows that
is a.p., and by Lemma 2.3 it follows that
as
. Therefore,
is a.a.p.
Theorem 4.11.
Suppose that conditions
, (H1)-(H2), and (H4) hold and that
is a fading memory space. If the bounded solution
of (1.1) is an a.a.p. sequence, then the system (1.1) has an a.p. solution.
Proof.
Since the solution
is a.a.p., it follows from Lemma 3.10 that
has a unique decomposition
, where
is a.p. and
as
. Notice that
is bounded. By Lemma 4.3 there is a compact set
in
such that
for all
. By Lemma 3.13, there is an integer sequence
,
, such that
as
and
uniformly on
as
. Taking a subsequence if necessary, we can also assume that
uniformly on
, and by Lemma 3.9(b) we have that
is also an a.p. sequence. For any
, there is a positive integer
such that if
, then
. In this case, we see that
uniformly for all
as
, and hence by Lemma 4.4
in
in
as
. Since
and from the previous considerations the first term of the right-hand side of (4.23) tends to zero as
and since
as
, we have that
for all
, which implies that (1.1) has an a.p. solution
passing through
, where
for
.
We are now in a position to prove the following result.
Theorem 4.12.
Suppose that the assumptions
, (H1), (H2), and (H4) hold, and that
is a fading memory space. If the bounded solution
of (1.1) is
, then
is a.a.p. Consequently, (1.1) has an a.p. solution which is
.
Proof.
Let the bounded solution
of (1.1) be
with the triple
. Let
be any positive integer such that
as
. Set
. As previously
is a solution of
and
is
with the same triple
. By Lemma A.2, for the set
and any
there exists
such that
and
for some
implies that
for all
, where
is a bounded solution of
passing through
and
for
. Since
is uniformly bounded for all
and
, taking a subsequence if necessary, we can assume that
is convergent for each
and
uniformly on
, for some a.p. function
. In this case, by Lemma 4.4 there is a positive integer
such that if
, then
On the other hand,
for
is a solution of (4.25) with
, that is,
where
is defined by the relation
To apply Lemma A.2 to (4.24) and its associated equation (4.27), we will point out some properties of the sequence
. Since
uniformly on
, for the above
, there is a positive integer
such that if
, then
which implies that
for all
. Applying Lemma A.2 to (4.24) and its associated equation (4.27) with the above arguments and condition (4.26), we conclude that for any positive integer sequence
,
as
, and
, there is a positive integer
such that
and hence by Axiom A(ii)
for all
if
. This implies that the bounded solution
of (1.1) is a.a.p. by Lemma 3.9(d). Furthermore, (1.1) has an a.p. solution, which is
by Theorem 4.11. This ends the proof.