International Journal of Mathematics and Mathematical Sciences
Volume 2015 (2015), Article ID 828952, 17 pages
http://dx.doi.org/10.1155/2015/828952

Research Article

Periodic Solutions of Certain Differential Equations with Piecewise Constant Argument

James Guyker

Department of Mathematics, SUNY College at Buffalo, 1300 Elmwood Avenue, Buffalo, NY 14222-1095, USA

Received 30 March 2015; Accepted 27 May 2015

Academic Editor: Harvinder S. Sidhu

Copyright © 2015 James Guyker. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Existence criteria are derived for the eventually periodic solutions of a class of differential equations with piecewise constant argument whose solutions at consecutive integers satisfy nonlinear recurrence relations. The proof characterizes the initial values of periodic solutions in terms of the coefficients of the resulting difference equations. Sufficient conditions for the unboundedness, boundedness, and symmetry of general solutions also follow from the corresponding properties of the difference equations.

1. Introduction

Since the seminal works of Shah and Wiener [1] and Cooke and Wiener [2], differential equations with piecewise constant arguments of the form

where is continuous and is the greatest integer function, have been treated widely in the literature and applied to certain biomedical models (see [3–7] and references therein). Continuity of the solutions of these equations implies recurrence relations for the values of solutions at consecutive integers. Therefore, there is a natural interplay between properties of these differential equations and properties of difference equations.

In this paper, we consider a class of equations of the above form but where is discontinuous: the chaotic and eventually periodic behavior and symmetry of solutions of initial-value problems of the form

on are determined, where , , and are constants; is defined by

for some positive number , and the sequence satisfies a nonlinear difference equation. As in the case when is continuous, by a solution of (2), we mean a function that is defined on with these properties:

(1) is continuous on .(2)The derivative exists at each point in , with the possible exception of the points in , where one-sided derivatives exist.(3) satisfies (2) on for each nonnegative integer .

Specifically, since

it follows that if is a positive integer and , then

Letting increase to , we have the difference equation

where ,

Therefore, the unique solution to (2) is

where

Moreover, satisfies (1)–(3) above since is continuous on with left derivatives

at integers , with right derivatives

at integers , and more generally with derivatives at nonintegral given by

We recall that the solution to (2) is oscillatory if it has arbitrarily large zeros [3]. Accordingly, a sequence is oscillatory if there exists a subsequence such that for all . Moreover, a stationary state of is a term such that [8]. The following holds for solutions of (2).

Proposition 1. Let be given by (8), where satisfies (6).

(1)If and is a stationary state of , then either for all or for all .(2)Let .

(a)If is oscillatory, then is oscillatory with stationary state .(b)If diverges to , then .(c)If diverges to , then .

Proof. (1) Let and let . Suppose first that . Then so since . By induction, and for all integers ; and, by (8), for all .
On the other hand, assume . Then so since . Thus and, by induction, for all integers . Since and , it follows from (8) that for all .
(2) Let . Then and in (8) we have

(a) Suppose by way of contradiction that is oscillatory but is not a stationary state of . It follows that for all positive integers .
Assume first that for some integer . Then and . By induction, and for all .
Let . By (8) and (13), for all ; and thus is not oscillatory in this case, a contradiction.
Therefore, assume that for all positive integers . Let be a positive integer. We show that for all in the interval : note that .
Suppose that . Then and

(i)if , then, by (2),

 for all in ;(ii)if , then, by (2), and, by (8),

 for all in .

If , then is strictly increasing on and , so for all in by continuity.
If , then is strictly decreasing on , but is still positive on by continuity since and are both positive.
Therefore, for all in (and hence for all ). Thus is not oscillatory in this case, contrary to our assumption.
(b) Suppose that . There exists such that for all integers . Let . Then and, by (8) and (13),

Thus, if , then . And if , then since

when and otherwise. Therefore, .
(c) Suppose that . There exists such that for all integers . Let . Then and, by (8) and (13), . Hence, .

Our main results characterize eventually periodic solutions of (2). Since is generally not differentiable at integers but is always differentiable between integers, we restrict our attention to integral periods (see [3, 6]). In this case, by (8), we have the following:

  is eventually periodic with positive integral period if and only if is eventually periodic with period .

Equation (2) is similar to recent models related to neural networks ([8–12]). We treat a generalized version of (6) as follows.

For real numbers , , , and , define, for ,

where and are given by (3).

Remark 2. Equation (18) is the difference equation (6) of a differential equation (2) with , , and being arbitrary: if , choose and . And if , then let and . The resulting solution given by (8) satisfies (2) with and .

The next result shows that we may henceforth assume that and in (18).

Proposition 3. Let be defined by (18) for real , , , and , and let satisfy (8), where and are given in (6). In the following, is an arbitrary positive integer.
(1) (a) Suppose that . If , for some , then . On the other hand, if , for all , then

(b) In particular, assume that . If , for some nonnegative integer , then, for all in , one has that . On the other hand, if , for all nonnegative integers , then, for all in ,

(2) (a) Suppose that and . If , then . If , then there exists such that .
(b) In particular, assume that and . If , then for all in . If , then there exists a positive integer such that for all in .
(3) (a) Suppose that . Either is oscillatory or there exists such that .
(b) In particular, assume that . Either is oscillatory or there exists an integer such that for all in .

Proof. (1a) Suppose that . If , for some , then either or . Hence, the desired form of follows by induction in this case.
Assume that for all . Thus, if , then

And if , then . (Since , for all , it follows that in this case.)
(1b) Suppose that . Let for some integer . By (a), and . Let be in . Then and, by (8),

Next, assume that and thus for all integers . Suppose first that and . By (a), . Since and ,

as in the proof of Proposition 1(1). Therefore, .
Finally suppose that . Thus and, by (a), and . As above, for , we have that and

(2a) Suppose that and . If (), then clearly . Assume that . There exists a positive integer such that . Thus, since , , and . As in the proof of (1a), ().
(2b) Suppose that and . If , then by (2a); and hence the result follows as in the proof of (1b).
Assume that . By (2a), there exists a positive integer such that and . Let so that . Then, by (8),

as in the proof of (1b).
(3a) Suppose that . One of the following must be true:

(i)For every integer , there exists an integer such that .(ii)There exists an integer such that for all .

Assume that (i) holds. If , for some , then is oscillatory with stationary state ; thus we further assume that has no zero terms. Therefore, there exists a subsequence of with alternating signs.
We may choose as follows. If , then let . If (), then let .
Next choose , where : by (i), there exists such that . If , then so let . If , then and ; thus, in this case, let .
Since , we have and . By induction, a sequence is constructed such that and for all . Thus is oscillatory.
On the other hand, if (ii) holds, then, as in the proof of (1a), there exists such that .
(3b) Suppose that . By (3a), either is oscillatory (and hence is oscillatory) or (with ) for some integer . The desired result follows in the latter case as in the proof of (2b).

Thus we assume that and ; and therefore Proposition 1 applies to the resulting solution of (2). It will follow from the third section that if is outside the interval , then is either eventually constant or unbounded. Moreover, if and is in , then, by the fourth section, all initial values are derived such that is eventually periodic; and more generally, for any , the eventually periodic solutions of (2) are the bounded solutions with these initializations .

Remark 4. For integers , , , and , periodic solutions of difference equation (18) were used in [13] to determine the real eigenvalues of certain arbitrarily large, sparse matrices.

2. Unbounded Solutions

Let satisfy (18), where and . If is unbounded, then is eventually geometric: we define

The following result shows that we may assume

since otherwise there exists such that for all .

Lemma 5. Assume that is defined by (18), where and .

(1)If , for all , then for all . In particular, if , for some , then for all .(2)If , for all , then for all . In particular, if , for some , then for all .(3)If , for some , then for all . In particular, if either or , for some , then there exists such that for all .

Proof. (1) Assume that for all . Then for all If , then, since , it follows that is not bounded above which contradicts our hypothesis. Thus and .
If (), for some , then clearly for all .
(2) Suppose that for all . Then

and, by induction, for all ,

If , then, since , is not bounded below which is contrary to our hypothesis. Thus and for all .
Assume that for some . Then since , and

where as above.
Similarly, by induction,

and for all .
(3) Assume that for some . Then and

where . By induction, and for all .
Suppose that for some . There exists such that

Thus, by the general case, for all .
Similarly, if , then there is such that for all .

The next result is central to our analysis.

Lemma 6. Let be given by (18), where , , and .

(1)If , then for all . In particular, if , then for all .(2)If , then for all . In this case, if and only if .(3)If , then for all .(4)If , then for all . In this case, if and only if .

Proof. (1) Assume and .
If , then . And if , then . Thus in both cases, and, similarly, by induction, for all .
If , then for all by the above argument.
(2) Suppose that . Then and for all by Lemma 5(2).
If , then . Conversely, if and , then so .
(3) Assume . If , then . And if , then . Thus and, by induction, for all .
(4) Suppose that . Then and for all by Lemma 5(1).
Clearly, if and only if in this case.

Remark 7. Let be defined as in Lemma 6. As in the proof of Lemma 5(3), if , then there is a unique positive integer such that . In this case, for ; and if , then for all by Lemma 6(4) since .
Similarly, if , then there is a unique positive integer such that

In this case, writing , we have that

for ; and if , then

for all by Lemma 6(2) since .

Example 8. Let and let and assume that . Let be the weighted average

Then and

Thus

and is the midpoint of .
If , then for all by Lemma 6(2). And if , then for all by Lemma 6(4).

3. Bounded Solutions

By Lemma 5, solutions of (18) such that , , and are bounded only when for all . In particular, by Lemma 6, if and , then is bounded. It is possible that is bounded but not eventually periodic.

Example 9. Assume that , where and are odd and even integers, respectively; and . If , where and are odd and even integers, respectively, such that , then the solution is bounded but not eventually periodic: by Lemma 6, for all . Note that

so that , where is odd and is even.
Similarly, by induction, for all , , where is odd and is even. It follows that is not eventually periodic since if for some and , then is even.

We now classify the types of solutions that may be bounded. Our results will be stated in terms of the decomposition of into the disjoint union of the intervals ,  ,  ,  , and .

Definition 10. Let , , and . A solution defined by (18) is of

(i)type A if and are in for some such that ,(ii)type B if and are in for some such that ,(iii)type C if is in and either such that is in for some or .

The unbounded solution in Example 8 is of type C.

Let satisfy (18) as in Definition 10. Since , we have that is a stationary state of if and only if or for some , in which case for all : if () or (), then clearly . Conversely, suppose that for some . If , then so since . And if , then so .

The solution is eventually periodic if there are integers and such that (and thus, by (18), for all ). The following eventually periodic solution is either of type A or type C.

Example 11. Let and let and let be an integer such that . (If , then is arbitrary.) Choose such that

and, for any integer , let

Then : so . Similarly,         and, for         . Thus and

Note that if , then, since , it follows that is of type A whenever is in (and ) and is of type C when is in . If , then is of type A since and are in and . The following slight modification is eventually periodic of type C when .

Example 12. Let and let and let be an integer such that

and let in satisfy

Define

for integers and such that . Then and, as in Example 11,

Therefore, ,

Note that, for an element in , there exists an integer such that

Similarly, for in , there exists an integer such that

Inequalities (49) and (50) will be used repeatedly in the next result.

In seeking bounded solutions without stationary states as in Examples 11 and 12, we may further assume that is in by the next result.

Theorem 13. Let be a solution of (18) such that , , and . Then fits one and only one of the following cases.

(1)Let be in .

(a)If and is given by (49), then for all , where is the first integer in such that ; that is, . In this case, if , then ; and when .(b)If , then for all .

(2)Let be in .

(a)If is in and is given by (49), then and one has the following.

(i)If , then is of type A.(ii)If , then for all .

(b) 

(i)If , then is of type A.(ii)If , then for all .

(3)Let be in .

(a)Let be in and let be the golden ratio with the property that .

(i)If , then either (1) or (2) holds for the sequence with initial term .(ii)If and is the least positive integer such that , then there exists an integer in such that is in ; and therefore (1) or (2) applies to .

(b)Let be in .

(i)If , then is in and satisfies either (1) or (2).(ii)If , then is in and satisfies (1), (2), or (3a).

(c)Let be in and let be given by (50). Then and one has the following.

(i)If , then either (and , for all ), or is in and satisfies (1) or (2).(ii)If , then is in and satisfies (1), (2), or (3a).

(4)Let be in . By Definition 10, if either or and contains a term of the sequence , then is of type C. Therefore, suppose that and is in .

(a)Let be in and let be given by (49). Then is in . If is in , then is of type C. Let be in . Then

(i) for all whenever ;(ii) for all when .

(b)Let be in .

(i)If , then for all .(ii)If , then for all .

(c)Let be in and let be given by (50). Then is in . Thus, if is in , then is of type C. Let be in . Then

(i) for all whenever ;(ii) for all when .

(5)Let be in .

(a)If is in and is given by (50), then and one has the following.

(i)If , then for all .(ii)If , then is of type B.

(b) 

(i)If , then for all .(ii)If , then is of type B.

(6)Let be in .

(a)If and is given by (50), then for all , where is the first integer in such that ; that is, . In this case, if , then and .(b)If , then for all .

(7)Let be in .

(a)Let be in .

(i)If , then either (5) or (6) holds for the sequence with initial term .(ii)If and is the least positive integer such that , then there exists an integer in such that is in and thus (5) or (6) applies to .

(b)Let be in .

(i)If , then is in and satisfies either (5) or (6).(ii)If , then is in and satisfies (5), (6), or (7a).

(c)Let be in and let be given by (49). Then and one has the following.

(i)If , then either (hence for all ) or is in and satisfies (5) or (6).(ii)If , then is in and satisfies (5), (6), or (7a).

Proof. (1a) Assume . By (49), there exists an integer such that

Therefore, as in Remark 7, there is a smallest integer in such that

It follows that

and for all by Lemma 5(1).
Suppose that . Then and so . And if , then

(1b) Lemma 6(4).
(2) Let be in . By (49), if is in , then

Therefore, and (2) follows from Definition 10 and Lemma 6(4).
Note that, for any in , (1) and (2) cover the following cases:

(i) is in whenever .(ii) is in when .

(3) Let be in .
(a) Suppose that is in and . Then and

(i)Suppose that . Then so since . Thus, by (56), is in and (1) or (2) applies to ().(ii)On the other hand, assume that so that . By (56), is in . If is in , then (1) or (2) applies to (). And if is in , then, since by (56), the above argument shows

 and is in .

Note that, in the latter case, if is in , then so .
It follows by induction that, for every , either is in or is in and . Since , there are least positive integers and such that and is in . Since , for , the form of follows from Remark 7.
(b) Let be in .

(i)Let . Then ,

 and is in .(ii)Let . Then ,

 and is in .

(c) Let be in and let be given by (50). Then

so that, for ,

Hence,

where when . Therefore, (3c) follows as above.
Moreover, by Remark 7, .
(4) Assume that is in .
(a) Let be in and . Then and, as in the proof of (2), if is given by (49), then . Thus (4a) follows from Definition 10 and Lemma 6((2), (4)).
(b) Let be in and so that . Hence, (i) and (ii) follow from Lemma 6((2), (4)).
(c) Let be in and . As in the proof of (3c),

Therefore, (4c) follows from Definition 10 and Lemma 6((2), (4)).
(5) Assume that is in .
(a) Let be in . As above, (63) holds for . Moreover,

Thus (5a) follows from Lemma 6(2) and Definition 10.
(b) A direct consequence of Lemma 6(2) and Definition 10.
(6) Let be in .
(a) Assume that and let be the integer satisfying (50):

Thus

and, by Remark 7, there is a smallest integer in such that and

It follows that

if and only if . But, by (50), . Therefore, and for all by Lemma 5(2).
Suppose that . Then

Thus and .
(b) Follows immediately from Lemma 6(2).
Note that, for any in , (5) and (6) cover the following cases:

(i) is in whenever .(ii) is in when .

(7) Let be in .
(a) Let and let be in . Then and

(i)Assume that . Then so and therefore, by (70), is in . Hence, (5) or (6) applies to ().(ii)Suppose that so that . By (70), is in . If is in , then (5) or (6) applies to ().

Assume is in . Then, by (70), and

Hence, is in . Note that if is in , then and .
Continuing by induction, for every , either is in or is in and . Therefore, since , if is the least positive integer such that , then there exists an integer in such that is in .
(b) Let and let be in . Then and . Therefore, is in when ; and is in otherwise.
(c) Let be in and let be given by (49). Then, for ,

Moreover, and

(i)If , then, by (73), either or is in .(ii)If , then and therefore, by (73), is in .

4. Periodic Solutions

The following sets are basic components of any eventually periodic solution of the difference equation (18).

Definition 14. Let and, for , let be the set of all polynomials in with degree at most and with all coefficients either or .

A consequence of the next result is that if , then there are possibilities for .

Lemma 15. Let be defined by (18) with , , and . If , for some integers and , then there exist polynomials in and in such that

Proof. Note that and, for ,

We solve the equation for general initial value .
The possible expressions for may be written as follows:

Similarly,

where .
Continuing in this manner, we have

where is an arbitrary element of .
In particular,

Therefore, is equivalent to

and thus the desired form (74) for follows.

Note that the initial value of the unbounded solution of Example 8 is not of form (74).

Our main results are converses of Lemma 15. If is defined by (74), then

Furthermore, we have the following refinement.

Lemma 16. Let and let and suppose that satisfies (74) for integers and and polynomials in and in .
(1) If

then .
(2) If

then .
Moreover, the converses of (1) and (2) hold whenever is not in .

Proof. Suppose that and . Then and and thus, by (74),

Similarly, if and , then and, by (74) and (81),

Suppose that . Then . If , then, by (74),

And if , then

Assume next that and . Then . If , then and

since . And if , then and

Finally assume that and . Then . If , then . And if , then .
Conversely, suppose that and   but is not in . Either or . If , then by (2). Thus, in this case, if , then is in ; and if , then and is again in . Hence, .
The other converses follow similarly.

The converses in Lemma 16 may fail when is in : in Example 12, we have , is in , , and satisfies (74) with and but .

If is in , then, by Theorem 13, either has a stationary state or is unbounded. Using Lemma 16, we may extend this result to the other cases of .

Theorem 17. Let be a solution of (18) such that , , and satisfies (74) for some integers and and some polynomials in and in .

(1)If is in , then either or there exists a positive integer such that for all .(2)Let be in . If or , then . On the other hand, if and , then either or there exists a positive integer such that for all , where .(3)If is in , then either or there exists a positive integer such that for all .

Proof. (2) Let be in and suppose that either or . We first verify that, for ,

Suppose that . Then

since and is in . Thus and similarly for . Hence, satisfies (90).
Next assume and . Since and is in , it follows that

Additionally, since or , at least one of the inequalities is strict and . Thus satisfies (90).
Hence, assume that and . There exists a nonempty subset of such that

where . Let and . We will show that

and therefore (90) will hold as in the initial case .
We begin by showing that, for ,

Let . If , then and therefore is not in . On the other hand, if , then and, by Lemma 16, is again not in .
Since , , and is not in , it follows that and, by Lemma 16, .
Conversely, let . Then and

Since or , at least one of the latter inequalities is strict and thus . Hence, (95) holds.
Next consider , where and is given by (93). By (95),

where such that

which is a subset of . In addition, since , it follows that

in which case (94) holds for .
Suppose that . Then , is in , and is of form (93) with replaced by . By (95),

Moreover, as above, (94) holds for if .
By induction, for , we may assume

where such that

which is a subset of , and, by (95),

Moreover, after steps, we conclude that and satisfies (94). Thus (90) follows as in the case .
Finally, we verify that , where is given by (90). If , then is or ; so . Thus assume . In this case, we prove that

Let . Suppose first that is in . Since , we have ; and, by Lemma 16, and . Hence, by (90), , which is impossible since and is a positive integer but and are relatively prime.
Thus assume is not in . Then is in since is in and . Hence, by Lemma 16, since .
Conversely, let (and ). Then by Lemma 16 and therefore . Thus (104) follows.
As in (93), by (90),

for some nonempty subset of . By (104),

where is the mapping on the set defined by

Similarly, for ,

and since , for all , it follows that .
We now turn to the second part of (2). Assume that , , and and thus . We prove the following:

Suppose that , where satisfies the above hypotheses. Consider two cases for .
(i) Assume . Then : since

Hence, (109) holds since , where so let .
(ii) Suppose that . If , then so assume that since . In this case, (109) also holds since and hence , where and .
Thus, starting with , , , and, applying (109) to , we conclude that for some in . Therefore,

and for all by Lemma 5(2).
(1) Let be in . The condition “ for all” is equivalent to “.” Thus we further assume that for and show that following the outline of the proof of (2).
We first verify that satisfies (90) for . Suppose that . Then, since and ,

If , then by Lemma 6(4). Therefore, and

Similarly, for and hence satisfies (90).
Suppose next that and . Since is in ,

and satisfies (90).
Thus assume that and . Then is given by (93), where for some nonempty subset of . We will show that (94) holds, where , and therefore (90) will hold as in the case .
Let . We begin by showing (95) for .
Let . If is in , then by Lemma 6(4). Since by assumption and , it follows that is in ; and therefore by Lemma 16 since .
Conversely, let . By Lemma 16,

and . Thus (95) holds.
Equation (94) and consequently (90) follow from (95) as in the proof of (2).
Finally, we verify that , where is given by (90). We may assume since otherwise is or , and therefore . We start by proving (104): since is in and by assumption, it follows by Lemma 6(4) that is not in .
Let . Then is in so by Lemma 16 since is not in and .
Conversely, assume that . By Lemma 16, , but is not in . Therefore and (104) holds.
By (90) and (104), the desired result follows as in the proof of (2).
(3) Let be in . The condition “ for all ” is equivalent to “.” Thus we assume that for and show that following the outline of the proof of (2).
We first verify that satisfies (90) for . Suppose that . Then, since , , and is in ,

Thus and similarly for . Hence, satisfies (90).
Next, we assume and . Then

If , then by Lemma 6(2), contrary to our hypothesis. Thus and satisfies (90).
Therefore assume that and . Then is given by (93), where for some nonempty subset of . We will show that (94) holds, where , and hence (90) will follow as in the case .
Let . We begin by showing (95) for : if is in , then by Lemma 6(2). Thus is not in by our hypothesis.
Let . Then is in ,  , and hence by Lemma 16 since is not in .
Conversely, let (and ). By Lemma 16, . Since is not in , we have that ; and (95) holds.
Equation (94) and consequently (90) follow from (95) as in the proof of (2).
Finally, we verify that , where is given by (90). We may assume since otherwise is or , and therefore . We start by proving (104).
Let . Then is not in by Lemma 6(2) since . Thus is in and therefore by Lemma 16 since is not in and .
Conversely, if (and ), then by Lemma 16, and (104) follows.
Finally, by (90) and (104) as in the proof of (2).

The following two examples illustrate symmetry between type C solutions about the point that will be generalized in the next section. In particular, type B solutions will be shown to be reflections of type A solutions.

Example 18. Let and let and let be in . For positive integer , consider the weighted average

in (that converges to as tends to infinity). Then

so by Theorem 17(2).

Example 19. For integer , redefine in Example 18 to be the weighted average

in (that converges to as ). Then as above since

in form (74). Note that, in this case, is the initial value in Example 18.

Theorem 17 may be applied to cryptography.

Example 20. Arbitrary vectors of zeros and ones may be encoded by selecting any real numbers , , and in and defining

in form (74). By Theorem 17(2), the “cipher-set” may then be decoded for as follows: if , then is a zero vector of indeterminate length. If , then is a vector of all ones of indeterminate length.
Assume . Compute enough terms of the periodic sequence to determine its period . Then is the length of and the proof of Theorem 17(2) shows that

The following eventually periodic solution with of form (74) is of type A.

Example 21. Let and let and, for integers and , let be the weighted average

in . Then is of form (74) with and . Choose in such that . Then :

so the claim is true for .
Assume . We prove by induction that

for . By the form of above,

since ; so (126) holds for .
Suppose that (126) is true for some . Then

as in the case since . Thus (126) holds for .
In particular, .
Finally :    since , and, by inducton, for . Hence, by the form of and the choice of . Therefore, .

Unfortunately, not all type A solutions with of form (74) are bounded.

Example 22. Let and let and

Then is of form (74) with , , , and . Moreover,

since . Therefore, for all .

Suppose that is defined by (74). By Theorems 13 and 17, if and is bounded, then is eventually periodic. In fact, if and is in , then by treating the cases and separately. However, if and is in , then is bounded (Lemma 6) but may not be eventually periodic.

Example 23. Assume that , where and are odd and even integers, respectively; and . Then is of form (74) with , , and ; and is in since . Choose in such that . Then and , but is not eventually periodic: with and , we have as in Example 9, for ,

where is odd. If is eventually periodic, then for some and ; and therefore is both even and odd.

5. Symmetric Solutions

Theorems 13 and 17 indicate symmetry about the midpoint of between pairs of solutions of (18). Moreover, if satisfies (74), then so does : if

then

Therefore, in view of Theorem 17, a natural question is, if and are given by (18), where , does it follow that for all ? We show affirmative answers in general for at least two choices of , the first of which requires that is not a term of .

Theorem 24. Let satisfy difference equation (18) with respect to , , , and ; and let be defined by (18) in terms of , , , , and the corresponding denoted by . Then

Moreover, if and , then for all .

Proof. Suppose first that . Then and so that .
Next, assume that . Then . If , then and therefore . And if , then and .
Finally, if , then by Lemma 6(2).

For the solutions in the following examples (with the conditions imposed), for all by Theorem 24 since for all : Examples 8 (if , 12, 21 (if ), and 22.

Let and be given as in Theorem 24 and suppose further that for all . By Remark 2, , where satisfies

for constants , , and such that . Therefore, , where is the symmetric solution about the line of the differential equation

with ; and since , and hence if and only if .

For type C solutions, where , there may be another symmetric solution obtained by reflecting about .

Theorem 25. Let satisfy (18) with respect to , , and such that either

(1) and is bounded below or(2) and is bounded above.

If satisfies (18) in terms of , , , and , then for all .

Proof. Suppose that (1) holds. If , then and . Thus, in this case.
Next, assume that . Then . We claim that . By way of contradiction, assume that . Then . Hence, and is not bounded below by Lemma 5(1), contrary to (1). Therefore and .
It follows in either case that ; and, by induction, for all .
Suppose now that (2) holds. If , then and . Thus in this case.
Finally assume that . We claim that . Suppose otherwise that . Then and ; so is not bounded above by Lemma 5(2), contradicting (2). Therefore and .
Hence, in both cases; and, by induction, for all .

Note that Theorem 25 may fail when : let and let and let . Then and . Thus .

The solution in Example 11 satisfies for all whenever by Theorem 24 since is thus not a term of . Furthermore, in this example, if and is in , then for all by Theorem 25 since is eventually periodic and therefore bounded.

Let and be given as in Theorem 25. By Remark 2, , where satisfies (2) for constants , , and such that . Therefore, , where is the symmetric solution about of the same differential equation

but with , since, by the proof of Theorem 25, (i.e., if and only if ).

If is given by (74), then is given by (133) and Theorem 17 may be applied to Theorems 24 and 25.

Corollary 26. Let satisfy (18), where , , is in , and is given by (74) for some integers and and some polynomials in and in .

(1)If or or , then . Moreover, if is bounded with no stationary states, then for all . In particular, if , then for all .(2)If or , then . Moreover, if , , and is bounded, then for all .

Proof. Suppose that , , is in , and is given by (74). The first lines of (1) and (2) follow directly from (133) and Theorem 17(2).
(1) Assume is bounded with no stationary states. Then and by (81). By Theorem 13, is not in so for all . Thus by Theorem 24.
In particular, suppose that . Then is eventually periodic, and hence bounded, by Theorem 17(2); and, by the proof of this theorem, the only possible stationary states are when

(i) and () or(ii) is or ,

that is, when is or , which are ruled out. Thus, by the previous case, for all .
(2) Next, assume that , , and is bounded. By (81), . If , then for all ; and hence and thus for all . Similarly, if , then for all .
Therefore, we may assume and the desired result is now immediate from Theorem 25.

Example 18 (revisited). In this example, , , is in , and is of form (74) with , , , and . Moreover, . By Corollary 26, for all and . Furthermore, if and , then for all .

Example 27. Let and let and let be in . For positive integer , consider the weighted average

in (that converges to as tends to infinity). Then is of form (74) with , , , and so by Theorem 17(2). Therefore, as above,

(i) for all ;(ii);(iii)if and , then for all .

Moreover,

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author would like to thank the referee(s) for several helpful suggestions.

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