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International Journal of Mathematics and Mathematical Sciences
Volume 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

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 formwhere is continuous and is the greatest integer function, have been treated widely in the literature and applied to certain biomedical models (see [37] 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 formon are determined, where , , and are constants; is defined byfor 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, sinceit follows that if is a positive integer and , thenLetting increase to , we have the difference equationwhere ,Therefore, the unique solution to (2) iswhere

Moreover, satisfies (1)–(3) above since is continuous on with left derivativesat integers , with right derivativesat 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 sincewhen 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 ([812]). 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 , thenAnd 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 defineThe following result shows that we may assumesince 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 . Thenand, 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 , andwhere 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 thatThus, 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 thatIn this case, writing , we have thatfor ; and if , thenfor all by Lemma 6(2) since .

Example 8. Let and let and assume that . Let be the weighted averageThen andThusand 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 thatso 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 thatand, for any integer , letThen : so . Similarly,        and, for         . Thus andNote 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 thatand let in satisfyDefine for integers and such that . Then and, as in Example 11,Therefore, ,

Note that, for an element in , there exists an integer such thatSimilarly, for in , there exists an integer such thatInequalities (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 thatTherefore, as in Remark 7, there is a smallest integer in such thatIt follows thatand 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 , thenTherefore, 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 showsand 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 ,