# Periodic Solution for a Max-Type Fuzzy Difference Equation

Authors: Changyou Wang; Jiahui Li
Journal: Journal of Mathematics
 (2020)

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## Abstract

The paper is concerned with the dynamics behavior of positive solutions for the following max-type fuzzy difference equation system:xn+1=maxA/xn,A/xn−1,xn−2,n=0,1,2,…, where xn is a sequence of positive fuzzy numbers, and the parameter A and the initial conditions x−2,x−1,x0 are also positive fuzzy numbers. Firstly, the fuzzy set theory is used to transform the fuzzy difference equation into the corresponding ordinary difference equations with parameters. Then, the expression for the periodic solution of the max-type ordinary difference equations is obtained by the iteration, the inequality technique, and the mathematical induction. Moreover, we can obtain the expression for the periodic solution of the max-type fuzzy difference equation. In addition, the boundedness and persistence of solutions for the fuzzy difference equation is proved. Finally, the results of this paper are simulated and verified by using MATLAB 2016 software package.


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## Body

## 1. Introduction

With the continuous development of science and technology in the fields of economy, biology, computer science, and so on, the research of nonlinear difference equations has been rapidly pushed forward, (for example, see [1–4] and the relevant reference cited therein). In the recent years, because the max operator has a great importance in automatic control models (see [5, 6]), max-type difference equations which are a special type of difference equations have aroused the concern and attention of many scholars. The present max-type difference equations have evolved into a diverse family of the equation (for example, see [7–10] and the references cited therein).Making a historical flash back for the max-type difference equation we study in this paper, we should mention that, in 2002, Voulov [11] studied positive solutions of the following equation:(1)xn=maxAxn−k,Bxn−m,n=0,1,…,where AandB are any positive real numbers, kandm are any positive integers, and the initial values x−1,x−2,…,x−d, with d=maxk,m, are any positive numbers. This paper proved that every positive solution is eventually periodic with period 2k if either A>B or A=B and m=3k, with period 2m if either A<B or A=B and k=3m, and with period k+m if A=B and neither m=3k nor m=3k. In 1998, Amleh et al. [12] showed that every well-defined positive solution of equation (1) is eventually periodic with period two, three, or four, when k=1,m=2, and A=1. In 2002, Mishev et al. [13] investigated, also, that every solution of equation (1) is eventually periodic, when k=1,m=3.Recently, a number of scholars have made further research about equation (1). In 2010, Gelisken et al. [14] studied the following max-type difference equation:(2)xn=maxAxn−pα,Bxn−kβ,n=0,1,…,where pandk are positive integers, 0<α,β<1, and A,B>0. They proved that every positive solution converges to x¯=maxA1/1+α,B1/1+β. In 2006, Yang et al. [15] proved that every positive solution of equation (2) converges to x¯=1 or, eventually, periodic with period 4, when p=1,k=2,β=1 and A=1. In 2008, Sun [16] showed that every positive solution of equation (2) converges to x¯=maxA1/1+α,B1/1+β, when p=1,k=2. In 2009, Gelisken [17] proved that every positive solution of equation (2) converges to x¯=1 or, eventually, periodic with period 2, when p=1,k=3,α=1, and B=1. In addition, in 1999, Szalkai [18] studied the periodicity of solutions of the following difference equation:(3)xn+1=maxAxn,Axn−1,⋯,Axn−k,n=0,1,…,where k is a positive integer, A is a real constant, and xii=−k0 are real numbers. He proved that every positive solution of (3) is eventually periodic of period k+2. In 2007, Bidwell and Franke [19] extended the form of equation (3) and considered the following difference equation:(4)xn=maxAn−11xn−1,An−12xn−2,⋯,An−1txn−t,n=1,2,…,where t∈1,2,…,Anin=0∞, for all i=1,2,…,t, is a nonnegative periodic sequence with period pi∈z+, and initial conditions are positive numbers. They have showed that if every solution of (4) is bounded, then every solution is eventually periodic. In 2013, Cranston and Kent [20] extended some results obtained by Bidwell in equation (4), and they give both sufficient conditions on the pi′s for the boundedness of all solutions and sufficient conditions for all solutions to be unbounded.It is easy to see that the abovementioned equations are the same type difference equation. Now let us look at another type of equation. In 2012, Qin et al. [21] studied the following max-type difference equation:(5)xn+1=maxAxn,xn−k,n=0,1,…,where k is a positive integer, A is a real constant, and the initial conditions xii=−k0 are real numbers. They show every well-defined solution of this equation is eventually periodic with period k+1. In 2009, Elsayed and Stević [22] considered equation (5), when k=2, and they showed that every well-defined solution of equation (5) is eventually periodic with period 3. In 2010, Iričanin and Elsayed [23] investigated equation (5), when k=3, and proved that every well-defined solution is eventually periodic with period 4. Recently, Xiao and Shi [24] studied equation (5), when k=1, and the result that every well-defined solution is eventually periodic with period 2 is obtained. Because of the necessity for some techniques, in 2014, Sun et al. [25] considered the following max-type equation that extends the form of equation (5):(6)xn=maxAnxn−r,xn−k,n=0,1,…,where Ann=1+∞ is a periodic sequence with period p and k,r∈1,2,… with gcdk,r=1 and k≠r and the initial conditions xii=−d0 are real numbers with d=maxr,k. They showed that if p=1 (or p≥2 and k is odd), then every well-defined solution of this equation is eventually periodic with period k. So far, there are more studies on this kind of difference equations (for example, see [26–30] and the relevant reference cited therein).With the development of the theory of difference equations, it is found that the given information needed to describe many practical problems in a difference equation model is incomplete. One of the effective procedures for considering uncertainty and imprecise real phenomena is the fuzzification of the corresponding difference equation systems. A numbers of studies have been made on the fuzzy difference equations so far, (for example, see [31–41] and the relevant reference cited therein). In recent decades, the max-type fuzzy difference equations have been attracting great increasing attention. In 2004, Stefanidou and Papaschinopoulos [42] extended equation (3) from real number to fuzzy number, where A is a positive fuzzy number and the initial conditions xii=−k0 are positive fuzzy numbers, and they gave a condition so that the solution is eventually periodic, unbounded, and nonpersistent and considered and studied the corresponding fuzzy difference equation (9) in [42] when k=0,m=1, i.e.,(7)xn+1=maxA0xn,A1xn−1,n=0,1,…,where A0andA1 are positive fuzzy numbers and the initial values x−1andx0 are any positive fuzzy numbers, and they proved the positive fuzzy solutions of equation (7) is eventually periodic, unbounded, and nonpersistent.In 2006, Stefanidou and Papaschinopoulos [43], to further extend the difference equation (7), considered the periodic nature of the positive solutions of the following fuzzy difference equation:(8)xn+1=maxA0xn−k,A1xn−m,n=0,1,…,where A0andA1 are positive fuzzy numbers and the initial values x−d,…,x0,d=maxk,m are any positive fuzzy numbers. Some latest related research can be found in [44–49].The fuzzy difference equation we study in this paper is motivated by the abovementioned ordinary difference equation and corresponding fuzzy difference equation. More precisely, we consider the fuzzy difference equation of the following form:(9)xn+1=maxAxn,Axn−1,xn−2,n=0,1,…,where A are positive fuzzy numbers and the initial values x−2,x−1,x0 are any positive fuzzy numbers. This paper aims to study the periodicity of the positive solutions of (9) by using a new iteration method for the more general nonlinear difference equations and inequality skills, as well as the mathematical induction.


## 2. Preliminaries and Notations

In order to facilitate the description of this paper, we need the following definitions.Definition 1.
For a setB, we denote by B¯, the closure of B. We say that a function A:R⟶0,1 is a fuzzy number if the following conditions hold:(i)
A is normal, i.e., there exists x∈R such that Ax=1(ii)
A is a fuzzy convex, i.e., Atx1+1−tx2≥minAx1,Ax2,∀t∈0,1,x1,x2∈R(iii)
A is upper semicontinuous on R(iv)
A is compactly supported, i.e., suppA=∪α∈0,1Aα¯=x∈R:Ax>0¯ is compact
Forα∈0,1, the α−cuts of A on R is defined as Aα=x∈R:Ax≥α. It is clear that the Aα is a bounded closed interval in R, and we say that a fuzzy number A is positive if suppA⊂0,∞.
It is obvious that ifA is a positive real number, then A is a positive fuzzy number and Aα=A,A,α∈0,1. At this time, we say that A is a trivial fuzzy number.Definition 2.
([32]). The fuzzy analog of the boundedness and persistence is as follows:(i)
A sequence of positive fuzzy numbersxn persists (resp. is bounded) if there exists a positive real numbers M (resp.N) such that suppxn∈M,∞resp.suppxn∈0,N,n=1,2,…(ii)
A sequence of positive fuzzy numbersxn is bounded and persists if there exist positive real numbers M,N>0 such that suppxn∈M,N,n=1,2,…


## 3. Main Results

Firstly, we will prove a simple auxiliary result which will be used many times in the rest of the paper.Lemma 1.
Assume thatxnn=−2∞ is a solution of equation (9), and there is k0∈N0∪−2,−1 such that(10)xk0=xk0+3,xk0+1=xk0+4,xk0+2=xk0+5.
Then, this solution is eventually periodic with period three.Proof.
If the following equalities (11) are true, then (10) is true. Now, we need prove that(11)xk0=xk0+3m,xk0+1=xk0+1+3m,xk0+2=xk0+2+3m,for every m∈N, from which the lemma follows.
We use the method of induction. Form=1, (11) becomes (10). Assume that (11) holds for 1≤m≤m0. From this and by using (9) and (10), as well as an iterative method, one can obtain(12)xk0+3m0+1=maxAxk0+3m0+2,Axk0+3m0+1,xk0+3m0=maxAxk0+2,Axk0+1,xk0=xk0+3=xk0,xk0+1+3m0+1=maxAxk0+3m0+3,Axk0+3m0+2,xk0+3m0+1=maxAxk0+3,Axk0+2,xk0+1=xk0+4=xk0+1,xk0+2+3m0+1=maxAxk0+3m0+4,Axk0+3m0+3,xk0+3m0+2=maxAxk0+4,Axk0+3,xk0+2=xk0+5=xk0+2.
The proof is completed.
Next, we will study the existence and uniqueness of the positive solutions of equation (9).Theorem 1.
Consider equation (9), and suppose that A is a positive fuzzy number; then, for every positive fuzzy numbers x−2,x−1,x0, there exists a unique positive solution xn of (9) with initial values x−2,x−1,x0.Proof.
The proof is similar to Proposition 3.1 [42], so we omit the proof of Theorem 1.
Now, we study the periodicity of the positive solutions of (9) when A is a trivial fuzzy number. We need the following lemma.Lemma 2.
Consider the system of difference equations,(13)yn+1=maxAzn,Azn−1,yn−2,zn+1=maxAyn,Ayn−1,zn−2,n≥0,where A is positive real constant and the initial values yi,zi,i=−2,−1,0, are positive real numbers, and then, every positive solution of (13) is eventually periodic with period three.Proof.
Letyn,zn be a positive solution of (12); we can obtain that(14)y1=maxAz0,Az−1,y−2,z1=maxAy0,Ay−1,z−2,and thus, y1≥A/z−1 and z1≥A/y−1. It is easy to see that ynn=−2∞,znn=−2∞ are positive real constants, and then z−1≥A/y1andy−1≥A/z1; therefore,(15)y2=maxAz1,Az0,y−1=maxAz0,y−1,z2=maxAy1,Ay0,z−1=maxAy0,z−1.
From equation (13), if n≥2, then, we have(16)yn≥Azn−2,yn−1≥Azn−2;zn≥Ayn−2,zn−1≥Ayn−2,and thus, we have(17)zn−2≥Ayn,zn−2≥Ayn−1;yn−2≥Azn,yn−2≥Azn−1.
From (17), we have the following equations, when n≥2:(18)yn+1=maxAzn,Azn−1,yn−2=yn−2,zn+1=maxAyn,Ayn−1,zn−2=zn−2.
Hence,(19)y3n=y0,y3n+1=y1,y3n+2=y2,z3n=z0,z3n+1=z1,z3n+2=z2,n≥1.
We can find that the solution of equation (13) has the following form:(20)y=y0,y1,y2,y3,y4,y5,…=y0,y1,y2,y0,y1,y2,…=y0,maxAz0,Az−1,y−2,maxAz0,y−1,y0,maxAz0,Az−1,y−2,maxAz0,y−1,…,z=z0,z1,z2,z3,z4,z5,…=z0,z1,z2,z0,z1,z2,…=z0,maxAy0,Ay−1,z−2,maxAy0,z−1,z0,maxAy0,Ay−1,z−2,maxAy0,z−1,….
The proof is completed.Theorem 2.
Consider equation (9) where A is a trivial fuzzy number and the initial values x−2,x−1,x0 are positive fuzzy numbers. Then, every positive solution of (9) is eventually periodic with period three.Proof.
Letxn be a positive solution of (9) with initial values x−2,x−1,x0, from α−cuts; thus, we have(21)xiα=Li,α,Ri,α,i=−2,−1,0,α∈0,1,xnα=Ln,α,Rn,α,i=1,2,3,…,α∈0,1,and A is a trivial fuzzy number; thus,(22)Aα=Al,α,Ar,α=A,A,α∈0,1.
From Theorem1, Ln,α,Rn,α,i=1,2,3,…,α∈0,1 satisfies system as follows:(23)Ln+1,α=maxARn,α,ARn−1,α,Ln−2,α,Rn+1,α=maxALn,α,ALn−1,α,Rn−2,α,n≥0.
Using Lemma2, we have that(24)L3n,α=L0,α,L3n+1,α=L1,α,L3n+2,α=L2,α,R3n,α=R0,α,R3n+1,α=R1,α,R3n+2,α=R2,α,n≥1,α∈0,1.
Therefore, we have thatxn is eventually periodic of period three. The proof of Theorem 2 is completed.
In the following, we study the periodicity of the positive solutions of (9) when A is a nontrivial fuzzy number. We need the following lemma.Lemma 3.
Consider the system of difference equations,(25)yn+1=maxBzn,Bzn−1,yn−2,zn+1=maxCyn,Cyn−1,zn−2,n≥0,where BandC are positive real constants and B<C, the initial values yi,zi,i=−2,−1,0 are positive real numbers, and then, every positive solution of (25) is eventually periodic with period three.Proof.
Letyn,zn be a positive solution of (25), and we can obtain that(26)y1=maxBz0,Bz−1,y−2,z1=maxCy0,Cy−1,z−2,and thus, z1≥C/y−1≥B/y−1. It is easy to see that ynn=−2∞,znn=−2∞ are positive real constants, and then, y−1≥B/z1; therefore,(27)y2=maxBz1,Bz0,y−1=maxBz0,y−1,z2=maxCy1,Cy0,z−1.
From equation (25), if n≥2, then we have(28)zn≥Cyn−2≥Byn−2,zn−1≥Cyn−2≥Byn−2,and thus, we have(29)yn−2≥Bzn,yn−2≥Bzn−1.
From (29), if n≥2, we have(30)yn+1=maxBzn,Bzn−1,yn−2=yn−2.
Hence,(31)y3n=y0,y3n+1=y1,y3n+2=y2,n≥1.
Now, we consider the period nature ofznn=3∞. There are three big cases to be considered.
(a1) Assume that B/z0≥B/z−1 and B/z0≥y−2, and then, y1=B/z0. (a11) Assume that C/y0≥C/y−1 and C/y0≥z−2, and then, z1=C/y0. (a111) Assume that B/z0≥y−1, and then, y2=B/z0. Thus, we have(32)z3=maxCy2,Cy1,z0=maxCy1,Cy1,By1=Cy1,z4=maxCy3,Cy2,z1=maxCy0,Cy1,Cy0=maxCy1,Cy0,z5=maxCy4,Cy3,z2=maxCy1,Cy0,z2=z2,z6=maxCy5,Cy4,z3=maxCy2,Cy1,z3=z3,z7=maxCy6,Cy5,z4=maxCy0,Cy2,z4=z4,z8=maxCy7,Cy6,z5=maxCy1,Cy0,z2=z5.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(33)z3n=z3=Cy1=CBz0,z3n+1=z4=maxCBz0,Cy0,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(a112) Assume that y−1≥B/z0, and then y2=y−1; thus, we have(34)z3=maxCy2,Cy1,z0=maxCy−1,Cy1,By1=Cy1,z4=maxCy3,Cy2,z1=maxCy0,Cy−1,z1=z1,z5=maxCy4,Cy3,z2=maxCy1,Cy0,z2=z2,z6=maxCy5,Cy4,z3=maxCy2,Cy1,z3=z3,z7=maxCy6,Cy5,z4=maxCy0,Cy−1,z4=z4,z8=maxCy7,Cy6,z5=maxCy1,Cy0,z2=z5.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(35)z3n=z3=CBz0,z3n+1=z4=Cy0,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(a12) Assume that C/y−1≥C/y0 and C/y−1≥z−2, and then, z1=C/y−1. Thus, we have (27) hold. (a121) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z3,z5,z6,z7,andz8 are the same with (32), and(36)z4=maxCy3,Cy2,z1=maxCy0,Cy1,Cy−1=Cy−1.
Hence,z6=z3,z7=z4,andz8=z5, From this and by Lemma 1, from the induction and iterative method, we have(37)z3n=z3=CBz0,z3n+1=z4=Cy−1,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(a122) Assume that y−1≥B/z0, then y2=y−1; thus, we find that z3,z5,z6,z7,andz8 are the same with (34), and(38)z4=maxCy3,Cy2,z1=maxCy0,Cy−1,Cy−1=Cy−1.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(39)z3n=z3=CBz0,z3n+1=z4=Cy−1,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(a13) Assume that z−2≥C/y0 and z−2≥C/y−1, and then, z1=z−2. Thus, we have (27) hold. (a131) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z3,z5,z6,z7,andz8 are the same with (32), and(40)z4=maxCy3,Cy2,z1=maxCy0,Cy1,z−2=z−2.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(41)z3n=z3=CBz0,z3n+1=z4=z−2,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(a132) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z3,z5,z6,z7,andz8 are the same with (34), and(42)z4=maxCy3,Cy2,z1=maxCy0,Cy−1,z−2=z−2.
Hence,z6=z3,z7=z4,z8=z5. From this and by Lemma 1, from the induction and iterative method, we have(43)z3n=z3=CBz0,z3n+1=z4=z−2,z3n+2=z5=maxCBz0,Cy0,z−1,n≥2.
(b1) Assume that B/z−1≥B/z0 and B/z−1≥y−2, and then, y1=B/z−1. (b11) Assume that C/y0≥C/y−1 and C/y0≥z−2, and then, z1=C/y0. Thus, we have(44)y2=maxBz1,Bz0,y−1=maxBz0,y−1,z2=maxCy1,Cy0,z−1=maxCy1,Cy0=maxCBz−1,Cy0.
(b111) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z3,z5,z6,z7,andz8 are the same with (32), and(45)z3=maxCy2,Cy1,z0=maxCy2,By2=Cy2=CBz0,z4=maxCy3,Cy2,z1=maxCy0,Cy2,Cy0=maxCBz0,Cy0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(46)z3n=z3=CBz0,z3n+1=z4=maxCBz0,Cy0,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(b112) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z4,z5,z6,z7,andz8 are the same with (34), and(47)z3=maxCy2,Cy1,z0=maxCy−1,CBz−1,z0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(48)z3n=z3=maxCy−1,CBz−1,z0,z3n+1=z4=Cy0,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(b12) Assume that C/y−1≥C/y0 and C/y−1≥z−2, and then, z1=C/y−1. Thus, we have (44) hold. (b121) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z3,z5,z6,z7,andz8 are the same with (32), and(49)z3=maxCy2,Cy1,z0=maxCy2,Cy1=Cy2=CBz0,z4=maxCy3,Cy2,z1=maxCy0,Cy2,Cy−1=maxCBz0,Cy−1.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(50)z3n=z3=CBz0,z3n+1=z4=maxCBz0,Cy−1,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(b122) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z4,z5,z6,z7,andz8 are the same with (34), and(51)z3=maxCy2,Cy1,z0=maxCy−1,CBz−1,z0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(52)z3n=z3=maxCy−1,CBz−1,z0,z3n+1=z4=Cy−1,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(b13) Assume that z−2≥C/y0 and z−2≥C/y−1, and then, z1=z−2. Thus, we have (44) hold. (b131) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we can find that z5,z6,z7,andz8 are the same with (32), and(53)z3=maxCy2,Cy1,z0=Cy2=CBz0,z4=maxCy3,Cy2,z1=maxCy0,CBz0,z1=maxCBz0,z−2.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(54)z3n=z3=CBz0,z3n+1=z4=maxCBz0,z−2,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(b132) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z4,z5,z6,z7,andz8 are the same with (34), and(55)z3=maxCy2,Cy1,z0=maxCy−1,CBz−1,z0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(56)z3n=z3=maxCy−1,CBz−1,z0,z3n+1=z4=z−2,z3n+2=z5=maxCBz−1,Cy0,n≥2.
(c1) Assume that y−2≥B/z−1 and y−2≥B/z0, and then, y1=y−2. (c11) Assume that C/y0≥C/y−1 and C/y0≥z−2, and then, z1=C/y0. Thus, we have(57)y2=maxBz1,Bz0,y−1=maxBz0,y−1,z2=maxCy1,Cy0,z−1=maxCy−2,Cy0,z−1.
(c111) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z5,z6,z7,andz8 are the same with (32), and(58)z3=maxCy2,Cy1,z0=maxCy2,Cy1=maxCBz0,Cy−2,z4=maxCy3,Cy2,z1=maxCy0,Cy2,Cy0=maxCBz0,Cy0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(59)z3n=z3=maxCBz0,Cy−2,z3n+1=z4=maxCBz0,Cy0,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
(c112) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z4,z5,z6,z7,andz8 are the same with (34), and(60)z3=maxCy2,Cy1,z0=maxCy−1,Cy−2,z0.
Hence,z6=z3,z7=z4,andz8=z5, From this and by Lemma 1, from the induction and iterative method, we have(61)z3n=z3=maxCy−1,Cy−2,z0,z3n+1=z4=Cy0,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
(c12) Assume that C/y−1≥C/y0 and C/y−1≥z−2, and then, z1=C/y−1. Thus, we have (57) hold. (c121) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z5,z6,z7,andz8 are the same with (32), and(62)z3=maxCy2,Cy1,z0=maxCBz0,Cy−2,z4=maxCy3,Cy2,z1=maxCy2,Cy−1=maxCBz0,Cy−1.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(63)z3n=z3=maxCBz0,Cy−2,z3n+1=z4=maxCBz0,Cy−1,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
(c122) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z5,z6,z7,andz8 are the same with (34), and(64)z3=maxCy2,Cy1,z0=maxCy−1,Cy1,z0=maxCy−1,Cy−2,z0,z4=maxCy3,Cy2,z1=maxCy0,Cy−1,Cy−1=Cy−1.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(65)z3n=z3=maxCy−1,Cy−2,z0,z3n+1=z4=Cy−1,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
(c13) Assume that z−2≥C/y0 and z−2≥C/y−1, and then, z1=z−2. Thus, we have (57) hold. (c131) Assume that B/z0≥y−1, and then, y2=B/z0; thus, we find that z5,z6,z7,andz8 are the same with (32), and(66)z3=maxCy2,Cy1,z0=maxCBz0,Cy−2,z4=maxCy3,Cy2,z1=maxCBz0,z−2.
Hence,z6=z3,z7=z4,andz8=z5, From this and by Lemma 1, from the induction and iterative method, we have(67)z3n=z3=maxCBz0,Cy−2,z3n+1=z4=maxCBz0,z−2,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
(c132) Assume that y−1≥B/z0, and then, y2=y−1; thus, we find that z5,z6,z7,andz8 are the same with (34), and(68)z3=maxCy2,Cy1,z0=maxCy−1,Cy−2,z0.
Hence,z6=z3,z7=z4,andz8=z5. From this and by Lemma 1, from the induction and iterative method, we have(69)z3n=z3=maxCy−1,Cy−2,z0,z3n+1=z4=z−2,z3n+2=z5=maxCy−2,Cy0,z−1,n≥2.
The proof is complete.Theorem 3.
Consider equation (9) where A is a nontrivial fuzzy number and the initial values x−2,x−1,x0 are positive fuzzy numbers. Then, every positive solution of (9) is eventually periodic with period three.Proof.
Letxn be a positive solution of (9) with initial values x−2,x−1,x0, such that we have (21) hold, and A is a nontrivial fuzzy number; thus,(70)Aα=Al,α,Ar,α=B,C,α∈0,1.
From Theorem1, Ln,α,Rn,α,i=1,2,3,…,α∈0,1 satisfies the system as follows:(71)Ln+1,α=maxBRn,α,BRn−1,α,Ln−2,α,Rn+1,α=maxCLn,α,CLn−1,α,Rn−2,α,n≥0.
Using Lemma3, we have that(72)L3n,α=L3,α,L3n+1,α=L4,α,L3n+2,α=L5,α,R3n,α=R3,α,R3n+1,α=R4,α,R3n+2,α=R5,α,n≥2,α∈0,1.
Therefore, we have thatxn is eventually periodic of period three. The proof is completed.Theorem 4.
Consider equation (9) where A is a positive fuzzy number and the initial values x−2,x−1,x0 are positive fuzzy numbers. Then, every positive solution of (9) is bounded and persists.Proof.
From Theorem3, Ln,α,Rn,α,i=1,2,3,…,α∈0,1 satisfies the system as follows:(73)Ln+1,α=maxAl,αRn,α,Al,αRn−1,α,Ln−2,α,Rn+1,α=maxAr,αLn,α,Ar,αLn−1,α,Rn−2,α,n≥0,and(74)L3n,α=L3,α,L3n+1,α=L4,α,L3n+2,α=L5,α,R3n,α=R3,α,R3n+1,α=R4,α,R3n+2,α=R5,α,n≥2,α∈0,1.
Form Lemma3, we find Ln,α are equal to one of (i)–(iii):(i)
L3n,α=L0,α(ii)
L3n+1,α=L1,α=maxAl,α/R0,α,Al,α/R−1,α,L−2,α(iii)
L3n+2,α=L2,α=maxminAl,α/Ar,αL0,α,Al,α/Ar,αL−1,α,Al,α/R−2,α,Al,α/R0,α,L−1,α
Thus, there exist numbersM,N∈0,∞ such that(75)suppLn,α∈M,N,n=1,2,3,….
Similarly,Rn,α are equal to one of R3,α,R4,α, and R5,α. From (33), (35)–(43), (46)–(56), and (59)–(69), we have(1)
R3,α is equal to one of ai−aiv:(76)aiAr,αAl,αR0,α,aiimaxAr,αL−1,α,Ar,αAl,αR−1,α,R0,α,aiiimaxAr,αAl,αR0,α,Ar,αL−2,α,aivmaxAr,αL−1,α,Ar,αL−2,α,R0,α.(2)
R4,α is equal to one of bi−bvi:(77)biAr,αL0,α,biimaxAr,αAl,αR0,α,Ar,αL0,α,biiiAr,αL−1,α,bivR−2,α,bvmaxAr,αAl,αR0,α,Ar,αL−1,α,bvimaxAr,αAl,αR0,α,R−2,α.(3)
R5,α is equal to one of ci−ciii:(78)cimaxAr,αAl,αR0,α,Ar,αL0,α,R−1,α,ciimaxAr,αAl,αR−1,α,Ar,αL0,α,ciiimaxAr,αL−2,α,Ar,αL0,α,R−1,α.
Thus, there exist numbersP,Q∈0,∞,P≥N such that(79)suppRn,α∈P,Q,n=1,2,3,…
From (75) and (79), for any xn, there exists μ,ν∈0,∞ such that(80)suppxn∈μ,ν.
The proof is completed.


## 4. An Illustrative Example

To illustrate our results, we give an example in which the conditions of our theorem hold.Example 1.
Consider the following fuzzy difference equation:(81)xn+1=maxAxn,Axn−1,xn−2,n=0,1,2,…,where A is a positive fuzzy number. Moreover, these are defined in here by(82)Ax=12x−32,3≤x≤5,−13x+83,5≤x≤8.
In addition, we take the initial conditionsx−2,x−1,x0 such that(83)x0x=x−4,4≤x≤5,−13x+83,5≤x≤8.x−1x=13x−23,2≤x≤3.4,−12x+72,3.4≤x≤7.x−2x=x−32,1.5≤x≤2.5,−2x+6,2.5≤x≤3.
From (82), we get Aα=3+2α,8−3α,α∈0,1. From (83), we get(84)x0α=4+α,8−3α,x−1α=2+3α,7−2α,x−2α=32+α,3−12α.
From (81), it results in a coupled system of the difference equation with parameter α,(85)Ln+1,α=max3+2αRn,α,3+2αRn−1,α,Ln−2,α,Rn+1,α=max8−3αLn,α,8−3αLn−1,α,Rn−2,α,α∈0,1,n=0,1,….
It is easy to show that systems (81)–(83) satisfy the conditions of Theorem 3. It follows from Theorem 3 that every positive solution of systems (81)–(83) is eventually periodic with period three, bounded, and persists (see Figures 1–3).Figure 1
The solution of system (85) when α=0.Figure 2
The solution of system (85) when α=0.5.Figure 3
The solution of system (85) when α=1.



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