Abstract

In this work, we derive the solution formulas and study their behaviors for the difference equations and with real initials and positive parameters. We show that there exist periodic solutions for the second equation under certain conditions when . Finally, we give some illustrative examples.

1. Introduction

In [15], the first author ([1] together with Kamal) solved and studied the solutions for the difference equationswhere , with real initials and positive parameters.

In [6], the authors explored the dynamics of the difference equationwhere and are positive real numbers. They provided the solution of the mentioned equation when , , , and .

By virtue of the wide applications in the last few decades, difference equations turned into one of the major areas of research. There are many books dealing with difference equations through the qualitative behavior of nonlinear equations (see [711]).

Closed-form solutions for nonlinear difference are not available, except for some few equations (see, for example, [6, 7, 10, 1226]). In [27], the authors studied the two recursive equationswith real initials and positive parameters. They provided the solutions for the two mentioned equations when .

Motivated by [27], we shall solve, find the forbidden set, and study ( positive real values of ) global behavior of the admissible solutions for equations (3) and (4) where are positive real numbers and , , , are nonzero real numbers.

If we setthen equations (3) and (4) are reduced to

2. The Difference Equation

During this section, we suppose thatwhere

2.1. Solution of Equation (6)

According with the transformation,

Equation (6) becomes

By solving equation (11) and after some calculations, the solution of equation (6) can be obtained.

Theorem 1. Suppose is an admissible solution of equation (6). The solution of equation (6) iswhere and

Proof. We can write the solution formula (12) asThe proof is by using the mathematical induction on .
When ,Suppose that the equalities (14) are true for .
Then,But, we can see thatThen,Therefore,Similarly, we can obtain and .
This completes the proof.

For equation (6), it is clear that if and , then is not defined. Let and , then is not defined. Also, if and , then is not defined. Now, if and , then is undefined.

This implies that the point with belongs to the forbidden point of equation (6).

Theorem 2. Equation (6) has the forbidden set

2.2. Global Dynamics of Equation (6)

Here, we illustrate the global behavior result and provide some examples.

Theorem 3. Assume that is an admissible solution of equation (6). Then, the following hold:(1)If , then the solution converges to zero.(2)If , then the solution converges to period-2 solution.(3)If , then the solution is unbounded.

Proof. We can write .(1)If , then . It follows that and as .Therefore, converges to zero.(2)Assume that , then .This implies thatAlso,Similarly,Using formula (14), we can writewhere .Similarly, we can get for in terms of .The result is obtained by notingand as , we have that(3)If , then . That is, as . Also, , , and are converging to zero.This implies that are unbounded, .

Example 1. Consider the solution of equation (6) such that , (), with initial values , , , and . Figure 1 shows that solution is unbounded.

Example 2. Consider of equation (6) such that , (), with , , , and . Figure 2 shows that converges to zero.

Example 3. Consider of equation (6) such that , (), with , , , and . Figure 3 shows that converges to the period-2 solutionwhere

Example 4. Consider of equation (6) such that , (), with , , , and . Figure 4 shows that the solution converges to the period-2 solutionwhere

3. The Difference Equation

We discuss the behaviors of the solutions of equation (7). The transformation (10) reduces equation (7) into the recursive equation

During this section, we suppose that

Clear that and are the roots of the equation

3.1. Case

In this subsection, we assume that . Clear that

During this subsection, we suppose that

Theorem 4. Assume that is an admissible solution of equation (7), thenwhere and

Proof. Its proof is same as the proof of Theorem 1 and is omitted.

Theorem 5. Let be an admissible solution of recursive equation (7). Then, the following hold:(1)If , then one has the following:(a)If , then the solution is unbounded.(b)If , then the solution converges to zero.(2)If , then one has the following:(a)If , then the solution converges to a finite limit.(b)If , then the solution converges to zero.(3)If , then the solution converges to zero.

Proof. We can write .(1)If , then either or .(a)If , then . That is, as . This implies that , , and are converging to zero. Therefore, is unbounded.(b)If , then . That is, as . This implies that , , and are unbounded. Therefore, converges to zero.(2)Suppose that , then either or .(a)If , then . It follows that as . This implies thatThen,whereSimilarly, we can show that and as .Therefore, converges to as .(b)If , then . That is, as .This implies that , , and are unbounded.Therefore, converges to zero.(3)If , then . That is, as .This implies that , , and are unbounded.
Therefore, converges to zero.

Example 5. Consider of recursive equation (7) where , ( and ), with , , , and . Figure 5 shows the unbounded solution .

Example 6. Consider of recursive equation (7) such that , (), with , , , and . Figure 6 shows that converges to zero.

Example 7. Consider of recursive equation (7) such that , ( and ), with , , , and . Figure 7 shows that converges to where

Example 8. Consider of recursive equation (7) such that , ( and ), with , , , and . Figure 8 shows that the solution converges to zero.

3.2. Case

In this subsection, we assume that . That is, .

Theorem 6. Assume that is an admissible solution of recursive equation (7). Then,where and

Proof. It is enough to see thatand its proof is same as of Theorem 1 and is omitted.

Theorem 7. Let be an admissible solution of recursive equation (7). Then, the following hold:(1)If , then solution converges to zero.(2)If , then is unbounded.

Proof. Clear that as , .(1)If , then as . This implies that the solution converges to zero.Now, suppose that . Then,Similarly, for and .Therefore, converges to zero.(2)If , then as . Then,By applying L’Hospital’s rule, we getSimilarly, for and .
Hence, is unbounded.

Example 9. Consider of recursive equation (7) such that , ( and ), with , , , and . Figure 9 shows that the unbounded solution .

Example 10. Consider of recursive equation (7) such that , ( and ), with , , , and . Figure 10 shows that converges to zero.

3.3. Case

Hereafter, we study the final case when .

During this subsection, we suppose that

That is,

Theorem 8. Let be an admissible solution of recursive equation (3). Then,where , , .

Theorem 9. Assume that is an admissible solution of equation (7). Then, we have the following:(1)If , then the solution converges to zero.(2)If , then the solution is unbounded.

Proof. Its proof is direct consequence and is omitted.

Theorem 10. Let be an admissible solution of recursive equation (7) and let . If (with ), then is periodic having prime period-.

Proof. Using formula (50), we can writeBut, for each , we haveThen, for , we haveSimilarly, we can show that for , we haveThis completes the proof.

Example 11. Consider of recursive equation (7) with , , , and . If ( and ), then is periodic with prime period (see Figure 11).

Example 12. Consider of recursive equation (7) with , , , and . If ( and ), then is periodic having prime period- (see Figure 12).

3.4. Forbidden Sets

In this subsection, we give the forbidden set of recursive equation (7) when , , and .

Clear that, if and , then is undefined. If and , then is undefined. If and , then is undefined. If and , then is undefined

The following result gives the forbidden sets of equation (7) for all values of and .

Theorem 11. We have the following statements:(1)If , then equation (3) has the forbidden set(2)If , then equation (3) has the forbidden set(3)If , then equation (3) has the forbidden set