#### Abstract

A class of difference equations which can be solved in closed form is presented in this paper. Obtained formulas for the solutions are used in the study of the asymptotic behavior of their solutions in a particular case.

#### 1. Introduction

One of the interesting problems in the theory of difference equations is to find the equations which can be solved in closed form. Such kind of equations are among the first ones which attracted attention of scientists. The reason for this is a natural expectation that obtained formulas for solutions can considerably facilitate investigation of their long-term behavior. For some classical results in the area, see, for example, book [1]. In [2], Stević solved, in an elegant way, the difference equation: explaining the formula in [3], which was given without any theoretical explanation. Several authors later considered some generalizations of (1) mostly by using Stević’s ideas and methods in [2] (see, e.g., [49] and the references therein). The idea for solving (1) was to reduce it to a nonhomogeneous linear first-order difference equation, which is solvable. Many equations and systems can be solved in a similar way (see, e.g., [4, 612]). For some related results and applications of linear first-order difference inequalities, see, for example, [1315].

An interesting difference equation is the following: where , and are fixed natural numbers, , and the initial values , , are real numbers.

Special cases of (2) frequently appear in the literature.

Example 1. The next equation is one of them.
Equation (3) is a special case of (2) with or that is, of the equations

Now note that (6) and (7) share a common important feature. Namely, assume that is a well-defined solution of (6) or (7) (i.e., all its terms exist and are finite numbers). Then, if , for every , by using the change of variables (6) and (7) are, respectively, transformed into the following linear difference equations with constant coefficients: which can be solved in closed form. As a consequence, (6) and (7), can be also solved in closed form. In fact, due to the known fact that polynomials of degree five and more need not be explicitly solved, we know that (9) and (10) can be theoretically solved. Nevertheless, we know some methods of how theoretically solutions of (9) and (10) can be found, as well as their form.

Note that for (3) the corresponding equation (9) is The general solution of (11) is , so that

A natural question and a corresponding problem are as follows.

Question 2. Are there any other values of , and except those ones in (4) and (5), such that (2) is solvable in closed form?

Problem 3. In the affirmative case find their solutions in closed form.

The answer to the question is positive, and in recent paper [12], as a byproduct of our results therein, we have essentially shown that, in the case (2) is solvable in closed form, and by using the described method therein, in some subcases of the case in (13) we have presented formulas for their solutions and use them in the study of their long-term behavior.

Here we continue this line of research by showing that there are some cases different from those ones in (4), (5), and (13), for which (2) is solvable in closed form.

#### 2. Case

In this section we consider (2) in the case That is, we consider the next difference equation:

First note that we may assume (the greatest common divisor of natural numbers and ). Otherwise, if , then and for some such that . Since every has the form , for some and , (15) can be written in the form

Using the next change of variables , in (17) we have that , , are independent solutions of the system which is (15) with and instead of and .

Hence, from now on we will assume that condition (16) holds.

##### 2.1. A Case When Some of Initial Values Are Equal to Zero

Now assume that one of the initial values , , is equal to zero. If for some , then is not defined or . If the later holds, then , which along with (15) implies that is not defined. Further, if for some , then is not defined or . If the later holds, then is not defined or . If , then , which along with (15) implies that is not defined.

On the other hand, if for some , then from (15) we see that or . If , then using again (15) we see that or , while if , then or . Repeating the procedure we see that there is an such that .

From all these considerations it follows that all solutions of (15) such that for some are not well defined and that if a solution of (15) has a zero term then it is not defined.

Hence, from now on we will consider only those solutions of (15) such that , for every .

##### 2.2. A Case When None of Initial Values Is Equal to Zero

Assume that is a solution of (15) such that for every . Since in this case for every , then (15) can be written in the following form:

The form suggests the next, natural, change of variables: which transforms (20) into the next difference equation of order :

In fact (22) is essentially an equation of the first order. Indeed, if we use the following decomposition of indices ,  , (22) can be written as where

This means that the subsequences , , are independent solutions of the difference equation of the first order.

Equation (25) is solvable, which along with (21) shows that (15) is solvable, giving another positive answer to Question 2. Indeed, (25) is a special case of the next difference equation:

Equation (26) is classical and can be solved in several ways. Since in the case (26) is a linear first-order difference equation, which is solvable, from now on we will assume that . By using the change of variables into (26), after some calculations, we get

By choosing (here we use the assumption ) we get a second-order difference equation with constant coefficients.

It is interesting that, by using the change (30) is transformed into the equation

Let and be the characteristic roots of the characteristic polynomial: associated with (32). Then by a well-known result on linear difference equations with constant coefficients and above transformations we get if , and if .

In our case we have that , , , and , so that (30) becomes while the constant in (29) is equal to . The characteristic roots of the characteristic polynomial associated with (36) are Hence, the general solution of (36) is if , and if .

If we choose initial values and such that then from (40) and (38) we get if , while from (40) and (39) we get if .

Using (41) and (42) in (27) we get if , and if .

Using the fact that the sequences , , , are solutions of (25), from formulas (43) and (44) it follows thatfor , if , and for , if .

From (21) we have that which implies that for and .

Although formulas (45)–(48) give solutions to (15) they are not easily applicable due to the assumption . Namely, subsequences , , appearing in (45) and (46), and the subsequences , , appearing in (48), are different decompositions of the sequence , so they are not easily connected to each other.

However, if in (15), then they give complete solution of the equation. Namely, in this case (45) and (46) become if , and if .

Using (49) and (50) in (48), we getfor each , and every , if , andfor each , and every , if .

#### 3. Asymptotic Behavior of Solutions of (15) When

In this section we will use formulas (51) and (52) in studying of the asymptotic behavior of solutions of (15) for the case when .

The results are incorporated into two theorems. Before we formulate and prove the results we will present two auxiliary results.

Lemma 4. Assume that , , and . Then difference equation (15) has a (not necessarily prime) -periodic solution.

Proof. Since it follows that (this is equivalent to ). Further, in this case (25) has one or two stationary solutions. Really, if , , then it follows that . That is, , from which it follows that if , that is, if , and if , that is, if . Hence, in both cases is a stationary solution of (25).
By using (21), we see that, for the stationary solution of (25) given in (55), we have a solution of (15) such that from which the lemma follows.

Lemma 5. Assume that , , and . Then difference equation (15) has a (not necessarily prime) -periodic solution.

Proof. First note that from the condition we have also that (it is also equivalent to ). On the other hand, in this case (25) also has one or two stationary solutions. Namely, if , , then it follows that . That is, , from which we obtain if , that is, if , and if , that is, if . Hence, in both cases is a stationary solution of (25).
By using (21), we see that the stationary solution of (25) given in (59) produces a solution of (15) such that , , which implies that from which the lemma follows.

Remark 6. By some calculation it is easy to see that if and only if and , while if and only if and .
Also, if and only if and , while if and only if and .

Theorem 7. Consider (15) with . Assume that , , and . Let , where . Then the following statements are true.(a)If   and , for , then , as .(b)If  , , and , then , as .(c)If  , , and , then , as .(d)If   and , for , then , as .(e)If  , , and , then , as .(f)If  , , and , , as .(g)If   and , for , then converges to a (not necessarily prime) -periodic solution of (15), as .(h)If  , , and , then is a (not necessarily prime) -periodic solution of (15).(i)If  , , and , then is a (not necessarily prime) -periodic solution of (15).(j)If   and , for , then converges to a (not necessarily prime) -periodic solution of (15), as .(k)If  , , and , then is a (not necessarily prime) -periodic solution of (15).(l)If  , , and , then is a (not necessarily prime) -periodic solution of (15).

Proof. First note that since , , we have that for every . Further note that the condition implies that , . If it were , then it would be , the case which is excluded. Hence we have that , from which it follows that
Letfor and .
Then if , for , we have that for each From (51) and (63) the statements in (a) and (d) easily follow.
If and , then we have that for each from which along with (51) the statements in (b) and (e) easily follow.
If and , then we have that for each from which along with (51) the statements in (c) and (f) follow.
(g), (j) in this case we have that for sufficiently large for each , from which along with (51), Lemmas 4 and 5, and Remark 6 these two statements follow.
(h), (k) in this case we have that (64) holds, from which along with (51) these two statements follow.
(i), (l) in this case we have that (65) holds, from which along with (51) these two statements follow.

Theorem 8. Consider (15) with . Assume that , , and , . Then the following statements are true.(a)If  , then , as .(b)If  , then , as .(c)If  , then , as .

Proof. In this case we have that . Letfor and .
Then for each we have that if , while for and , if .
From (52), (68), and (69) the statements in (a) and (b) easily follow.
(c) in this case we have that for sufficiently large for each , from which along with (52) and the fact that as , the statement easily follows.

#### Acknowledgments

This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant no. (11-130/1433 HiCi). The authors, therefore, acknowledge technical and financial support of KAU.