Research Article | Open Access

Volume 2013 |Article ID 421545 | https://doi.org/10.1155/2013/421545

E. J. Janowski, M. R. S. Kulenović, "Existence of a Period-Two Solution in Linearizable Difference Equations", Discrete Dynamics in Nature and Society, vol. 2013, Article ID 421545, 9 pages, 2013. https://doi.org/10.1155/2013/421545

# Existence of a Period-Two Solution in Linearizable Difference Equations

Accepted08 Oct 2013
Published25 Nov 2013

#### Abstract

Consider the difference equation where and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it can be rewritten as the nonautonomous linear equation , where and the functions . We give some necessary and sufficient conditions for the equation to have a minimal period-two solution when .

#### 1. Introduction

Consider the difference equation where and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it has the linearization of the form where and the functions . By “(1) has the linearization (2)” we mean that (1) can be rewritten as the nonautonomous linear equation (2); see [1, 2].

The importance of a period-two solution is well known in the case of first order difference equations of the form of (1) with , where the periods of the solutions appear in the well-known Sharkovsky ordering starting with period two. As a consequence of the results on Sharkovsky ordering the nonexistence of the period-two solution implies the nonexistence of periodic solutions of any period; see .

In the case of second order difference equations the following result has been obtained in .

Theorem 1. Let and let be a function which either increases in both variables or decreases in the first variable and increases in the second variable. Then for every solution of (1) with the subsequences and of even and odd terms of the solution do exactly one of the following.(i)Eventually they are both monotonically increasing.(ii)Eventually they are both monotonically decreasing.(iii)One of them is monotonically increasing and the other is monotonically decreasing.

As a consequence of Theorem 1 every bounded solution of (1) with approaches either an equilibrium solution, a period-two solution, or a finite point at the boundary, and every unbounded solution is asymptotic to the point at infinity in a monotonic way. In view of Theorem 1 the results on the nonexistence of period-two solutions are as important as the results on the existence of these solutions. The importance of the existence or nonexistence of period-two solutions is also clear from the fact that one of the two most common local bifurcations for second order monotone autonomous difference equations is period-doubling bifurcation; see [3, 4, 79] for related results. Also the only known global bifurcation for second order monotone autonomous difference equations is period-doubling bifurcation . See  for related results. The nonexistence results for periodic solutions which are the discrete analogue of the Bendixson’s nonexistence result for periodic solutions of differential equations have been obtained in [14, 15].

The results obtained in this paper are applicable to both autonomous and nonautonomous difference equations as the coefficients in (2) are in general functions of and . Some of our examples will reflect this situation. The method of finding period-two solutions in the autonomous case consists of finding the fixed points of the second iterate of the corresponding map. However, in the nonautonomous case this method does not work and the results which will be presented in this paper can be used to find period-two solutions.

Some interesting points of our results can be demonstrated by the following example.

Example 2. The period-two solution , of the difference equation where and are two real sequences, satisfies which implies Conversely, if condition (5) holds, then any possible period-two solution , of (3) must satisfy If for some then and (3) has an infinite number of period-two solutions of the form , . If for every then for every in which case every nonequilibrium solution of (3) is a period-two solution.
Thus, condition (5) is a necessary and sufficient condition for the existence of a period-two solution. This condition is clearly satisfied if, for instance, or if and are period-two sequences which satisfy or if and are period- sequences which satisfy An example of a nonautonomous nonlinear difference equation for which one can find a period-two solution is the following equation The quadratic second order difference equation where are constants, can be linearized as which is of the form of (3) where and condition (5) becomes the first order linear difference equation Since then the period-two solution of (11) has the form , .

#### 2. The Constant Case

In this section we consider the case when the sums of the even indexed functions and the odd indexed functions are both constants.

The following simple result will be a useful technical tool.

Lemma 3. Suppose that (1) has the linearization (2). Let Assume that (1) has a minimal period-two solution where ,   for . (a)For , (b)For ,

Proof. By plugging and for in (2) and assuming that is odd we obtain immediately that when is even (15) holds, while in the case when is odd (16) holds. Similarly assuming that is even we obtain immediately that in the case when is even (17) holds, while when is odd (18) holds.

Theorem 4. Suppose that (1) has the linearization (2) with and that are given by (14). Then (1) has a minimal period-two solution where for if and only if the following hold: (a)if is even, then ;(b)if is odd, then .

Proof. The necessary part of the proof follows from part (a) of Lemma 3.
For the proof of the sufficient part choose the initial conditions = and . Setting we get that, by part (a), By using induction we get that for .

Theorem 5. Suppose that (1) has the linearization (2) and that are given by (14). Assume that (1) has a minimal period-two solution where for . (1)If  , then (a) if and only if ; (b)if , then and . (2)If  , then (a) if and only if ; (b)if , then and .

Proof. The proof is as follows. (1) Let . In view of Lemma 3 part (a) the identities (15) and (16) are satisfied. (a) Assume that . Then by (15) . Hence either or . If , then by (16) , and since , we have . Assume that . Then by (15) . Hence either or . If , then by (16) , and since , we obtain that . (b) Assume that . Then by (15) and (16) . Thus and so . By (15) , and since , then . (2) Let . In view of Lemma 3 part (b) the identities (17) and (18) are satisfied. (a) Assume that . Then by (17) . Hence either or . If , then by (18) , and since , we have . Conversely, assume that . Then by (17) . Hence either or . If , then by (18) , and since , we obtain that . (b) Assume that . Then by (17) and (18) . Thus and so . By (17) , and since , then .

Theorem 6. Suppose that (1) has the linearization (2) with and that are given by (14). Then (1) has a minimal period-two solution if and only if

Proof. If (1) has a minimal period-two solution then by Theorem 5 part the necessary condition follows.
Conversely, assume that . Then . Choose the initial conditions Then for we get and for we get , which shows that is a minimal period-two solution.
Now suppose that . Choose the initial conditions Then for we get and for we get , which shows that is a minimal period-two solution.

An immediate consequence of Theorem 6 is this result.

Corollary 7. (a) If and , then the minimal period-two solution of (1) is .
(b) If and , then the minimal period-two solution of (1) is either or .

#### 3. The Nonconstant Case

In this section we consider the case when the sums of the even indexed functions and the odd indexed functions are both nonconstants.

Theorem 8. Suppose that (1) has the linearization (2) with . Let Then (1) has a minimal period-two solution where for if and only if .

Proof. The proof follows from the same reasoning as in Example 2.

Note that it is possible for (3) to have a minimal period-two solution other than , when for all (see Example 12). In order to handle the cases not covered by Theorem 8 we establish the following results.

The following simple result will be a useful technical tool.

Lemma 9. Suppose that (1) has the linearization (2) and that are given by (23). Assume that (1) has a minimal period-two solution where for . Then for (a)if  , then (b)if  , then

Proof. Assume that is odd. By plugging for in (2) and setting we obtain immediately that Now simple induction completes the proof of (24) and (25) and so of part (a). The proof of part (b) is similar.

Theorem 10. Suppose that (1) has the linearization (2) with and that are given by (23). Then (1) has a minimal period-two solution where for if and only if (24) and (25) hold.

Proof. The necessary part of the proof follows from part (a) of Lemma 9.
For the proof of sufficient part choose the initial conditions , and .(1)Assume . Setting we get that By using induction we get that for . (2)Assume . We obtain from (24) and (25) that and for    which implies that and (3)Assume . We obtain from (24) and (25) that and for   which implies that and

When (1) has been embedded into a higher order equation, the following results can be used to establish either the nonexistence or the necessary conditions for existence of a minimal period-two solution.

Theorem 11. Let . Suppose that (1) has the linearization (2) and that are as in (23). Assume that (1) has a minimal period-two solution where ,   for .(1) Let either or   for ; then (a)(b)(2) Let for .(a) If   then .(b) If   then the following are true: (i) if and only if for (ii) if and only if for (iii) either or . (3) Let for . (a) If   then (b) If   then the following are true: (i) if and only if for (ii) if and only if for (iii) either or .

Proof. The proof is as follows.(1) Assume that for . Then . Otherwise, suppose that . Then and in view of (24) . This implies for , which is a contradiction. Now, suppose that . Then and in view of (25) . This implies for , which is a contradiction. Thus , which in view of Lemma 9, implies for . (a) By (24) . (b) By (25) . Next, assume that for . Then . Otherwise, suppose that . Then and in view of (25) . This implies for , which is a contradiction. Now, suppose that . Then and in view of (24) . This implies for , which is a contradiction. Thus , which in view of Lemma 9, implies for . The rest of the proof is similar to the first part of the proof and will be omitted. (2) In view of (24) and (25) the condition for implies for .(a) Assume that .If for some , then for this and so , which is a contradiction. Thus for , and so .(b) Assume that for .(i) If for , then for . Now suppose that for ; then from we get that for . Since , then . Thus for . (ii) The proof is similar to part (i) and will be omitted. (iii) By (24) for . Now suppose that for , then . If for ; then for . In view of (25) for and so . (3) In view of (24) and (25) condition for implies for . (a) Assume that .If for some , then for this and so , which is a contradiction. Thus for and consequently . (b) Assume that for . (i) If for , then for . Now suppose that for ; then from we get that / for . Since , then . Thus for . (ii) The proof is similar to part (i) and will be omitted. (iii) By (24) for . Now suppose that for , then . If for , then for . In view of (25) for and so .(i) If for ; then for . Now suppose that for ; then from we get that for . Since , then . Thus for .(ii) The proof is similar to the proof of part (i).(iii) By (26) for . If for , then . Now if for , then for . In view of (27) for and so .

Example 12. The difference equation where is a period-two sequence such that , , has an infinite number of period-two solutions of the form , which can be seen by immediate checking.
This equation is an illustration of Theorem 11 part 1. In this case and .

Theorem 13. Let . Suppose that (1) has the linearization (2) and are as in (23). Assume that (1) has a minimal period-two solution where ,   for . (1) Let either or for ; then (a)(b).(2) Let for . (a) If   then . (b) If then the following are true: (i) if and only if for (ii) if and only if for (iii) either or . (3) Let for . (a) If then . (b) If then the following are true: (i) if and only if for (ii) if and only if for (iii)either or .

Proof. The proof is as follows. (1) Assume that for . Then . Otherwise, if then . By (26) and so for , which is a contradiction. In the other case, if then . By (27) and so for , which is a contradiction. Hence for .(a) By (26) for . (b) By (27) for . Now assume that for . Then . Otherwise, if then . By (26) and so for , which is a contradiction. In the other case, if then . By (27) and so for , which is a contradiction. Hence for and the proof follows similarly to the previous part.(2) In view of (26) and (27) condition for implies for . (a) Assume that .If for some , then for this and so , which is a contradiction. Hence for . Thus . (b) Assume that for . (i) If for ; then for . Now suppose that for ; then from we get that for . Since , then . Thus for .(ii) The proof is similar to the proof of part (i).(iii) By (26) for . If for , then . Now if for , then for . In view of (27) for and so . (3) In view of (26) and (27) condition for implies for . (a) Assume that .If for some , then for this and so , which is a contradiction. Hence for . Thus .(b) Assume that for . (i) If for , then for . Now assume that for ; then from we get that / for . Since , then . Thus for .(ii) The proof is similar to the proof of part (i).(iii) By (26) for . If for , then . Now if for , then for . In view of (27) for and so .

Corollary 14. Suppose that (1) has the linearization (2) and that are given by (23). Assume that (1) has a minimal period-two solution , where for . (1)Let . (a) if and only if .(b)If   and either or for   then (i) and ;(ii). (2)Let . (a) if and only if .(b)If   and either or for   then (i) and (ii).

Proof. Part (1a) follows from the proof of Theorem 11 part . Part (1b) follows from the proof of Theorem 11 part and , which implies . Hence and so . Thus which implies and for .
Part (2a) follows from the proof of Theorem 13 part and the proof of part (2b) follows in a similar way as the proof of part (1b).

The next result gives a necessary and sufficient condition for the existence of a minimal period-two solution in the special case when in (2).

Theorem 15. Suppose that (1) has the linearization (2) with and that are as in (23). Then (1) has a minimal period-two solution if and only if one of the following holds. (1)All the conditions of Theorem 11 part are satisfied. Furthermore the period-two solution is (2) for and one of the following conditions is satisfied for : (a),(b) and ,(c) and . (3) for and one of the following conditions is satisfied for : (a),(b) and ,(c) and .

Proof. The necessary part follows from Theorem 11. We will now prove the sufficient part.(1)Assume that conditions (a) and (b) of Theorem 11 part are satisfied. Then by (a) for and by (b) / for . Choose the initial conditions ,  , and . Then by using these equalities we get Simple induction completes the proof.(2)Next, suppose that for .(a)Assume that for . By choosing the initial conditions , and , an immediate calculation shows that is a minimal period-two solution.(b)Assume that and for . By choosing the initial conditions , and , we obtain and straightforward induction shows that is the minimal period-two solution.(c)Assume that and for . Choose the initial conditions , and . By straightforward induction we obtain that is the minimal period-two solution. (3)Next, suppose that for .(a)Assume that for . By choosing the initial conditions = , and , an immediate calculation shows that is the minimal period-two solution.(b)Assume that and for . By choosing the initial conditions , and , we obtain and straightforward induction shows that is the minimal period-two solution. (c)Assume that and for . By choosing the initial conditions , and , and using a straightforward induction we obtain that is the minimal period-two solution.

Example 16. The difference equation where , has a minimal period-two solutions of the form , if and only if and .
This equation is an illustration of Theorem 15. The linearization of (34) gives where . Since for all , then from Theorem 15 part (3a) a period-two solution exists if and only if Observe that if , then for , which is a contradiction.
Equation (36) gives the first order equation which has a period-two solution if and only if and as Theorem 15 part is satisfied.
Therefore, (34) has a period-two solution of the form if and only if .
This example can be extended to a more general equation of the form where . Similar reasoning gives the necessary and sufficient conditions for the existence of a period-two solution to be In this case there is an infinite number of period-two solutions of the form , .

Corollary 17. Suppose that (1) has the linearization (2) with and that are as in (23). Assume that and either or for . Then (1) has a minimal period-two solution if and only if for .

Proof. The necessary part follows from Corollary 14 part (1b).
For the proof of sufficient part choose the initial conditions , and . Setting we obtain Simple induction completes the proof.

So far we have considered the cases when and is either equal or not equal to one for all . But what happens when and for some ’s and for other ’s or when for some ’s and for other ’s? We will now investigate these cases along with the cases when and for some ’s and for other ’s or when for some ’s and for other ’s.

Theorem 18. Let . Suppose that (1) has the linearization (2) and that are as in (23). Assume that (1) has a minimal period-two solution , where for . (1)Let . Then(a)for some if and only if ;(b)if some are even and , then ;(c)if some are odd and , then .(2)If  , then for all and .(3)If  , then for all and .

Proof. The proof is as follows.(1)The result follows from Lemma 9 part (a).(2)Since , then and the result follows from Lemma 9 part (a).(3)Since , then and the result follows from Lemma 9 part (a).

Remark 19. Note that in part of Theorem 18   and for all . Similarly in part of Theorem 18   and for all .

Theorem 20. Let . Suppose that (1) has the linearization (2) and that are as in (23). Assume that (1) has a minimal period-two solution where for . (1)Let . Then(a)for some if and only if ;(b)if some are even and , then ;(c)if some are odd and , then . (2)If  , then for all and .(3)If