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Dynamical behaviors of a k-order fuzzy difference equation

Dynamical behaviors of a k-order fuzzy difference equation 1IntroductionAs we all know, the stability theory has always been one of the hotspots in mathematical research, especially the stability theory of time-delay systems [1,2,3]. Difference equations are usually used to describe discrete-time dynamic systems, which can be widely used to establish mathematical models in many areas, such as computer science, ecology, population dynamics, electrical networks, economics, etc. [4,5]. Due to its wide range of applications, the study of the time-delay difference equation and its dynamic behavior has become an important topic in applied mathematics. In practice, more problems encountered are that the conditional parameters fluctuate or change within a certain range. One of the ways to solve this problem is to fuzz the difference equation based on the uncertainty. Fuzzy set theory is an effective tool for modeling uncertainty and processing vague or subjective information [6,7]. Fuzzy difference equations have gradually attracted attention for their practicability [8,11, 12,13,14, 15,16,17, 18,19,20, 21,22,23, 24,25,26, 27,28] and are developing vigorously. Chrysafis et al. [10] studied the fuzzy difference equation of finance, which is an alternative method for studying the time value of money. Deeba and Korvin [11] studied fuzzy difference equation xn+1=xn−ABxn−1+C{x}_{n+1}={x}_{n}-AB{x}_{n-1}+Cabout the level of carbon dioxide (CO2{{\rm{CO}}}_{2}) in the blood.Let us first review the history of the fuzzy difference equations studied in this article. In [14], Zhang and Liu studied the qualitative behavior of the first-order linear fuzzy difference equation, which is xn+1=Axn+B(n=0,1,2,…){x}_{n+1}=A{x}_{n}+B\left(n=0,1,2,\ldots ), where the parameters A,BA,Band initial value are positive fuzzy numbers. In [15], Hatir investigated the existence, oscillatory behavior, and asymptotic behavior of the positive solutions of second-order fuzzy difference equation xn+1=A+Bxn−1(n=0,1,2,…){x}_{n+1}=A+\frac{B}{{x}_{n-1}}\hspace{0.33em}\left(n=0,1,2,\ldots ), where parameters A,BA,Band initial value are positive fuzzy numbers. Subsequently, many researchers used similar methods to study different fuzzy difference equations and came to many conclusions, such as [23,24,25, 26,27].In [16], Papaschinopoulos and Schinas studied the global properties of a high-order nonlinear difference equation system xn+1=A+ynxn−p,yn+1=A+xnyn−q(n=0,1,2,…){x}_{n+1}=A+\frac{{y}_{n}}{{x}_{n-p}},{y}_{n+1}=A+\frac{{x}_{n}}{{y}_{n-q}}\hspace{0.33em}\left(n=0,1,2,\ldots ), where p,qp,qare positive integers, parameter AAand initial values are positive real numbers. Later, he studied m-order fuzzy difference equation xn+1=A+xnxn−m(n=0,1,2,…){x}_{n+1}=A+\frac{{x}_{n}}{{x}_{n-m}}\hspace{0.33em}\left(n=0,1,2,\ldots ), in which parameter AAand initial value are positive fuzzy numbers. In [17], Zhang et al. showed the existence and the asymptotic stability of the positive solutions about first-order fuzzy difference equation xn+1=A+xnB+xn(n=0,1,2,…){x}_{n+1}=\frac{A+{x}_{n}}{B+{x}_{n}}\hspace{0.33em}\left(n=0,1,2,\ldots ), in which parameter A,BA,Band initial value are positive fuzzy numbers.Motivated by the discussions above, we study arbitrary k-order nonlinear fuzzy difference equation in this paper as follows,(1)xn+1=xnA+Bxn−k,n=0,1,2,…,{x}_{n+1}=\frac{{x}_{n}}{A+B{x}_{n-k}},\hspace{1em}n=0,1,2,\ldots ,and also the existence, stability, and other dynamic behaviors of the positive solution of equation (1), where parameters A,BA,Band initial value x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, k∈{0,1,…}k\in \{0,1,\ldots \}are positive fuzzy numbers.For further study of (1), we need some basic definitions and lemmas which are related to fuzzy difference equations, which are obtained from reference resources [14,15,16, 17,18,19, 20,21]. In this paper, R=(−∞,+∞)R=\left(-\infty ,+\infty ), R+=(0,+∞){R}^{+}=\left(0,+\infty ), F+{F}^{+}denote the set of all positive fuzzy numbers.Definition 1.1[18] A function A:R→[0,1]A:R\to {[}0,1]is said to be a fuzzy number if it satisfies the following conditions (i)–(iv): (i)AAis normal, i.e., there exists x∈Rx\in Rsuch that A(x)=1A\left(x)=1;(ii)AAis fuzzy convex, i.e., for all t∈[0,1]t\in {[}0,1]and x1,x2∈R{x}_{1},{x}_{2}\in Rsuch that A(tx1+(1−t)x2)≥min{A(x1),A(x2)};A\left(t{x}_{1}+\left(1-t){x}_{2})\ge \min \left\{A\left({x}_{1}),A\left({x}_{2})\right\};\hspace{1.2em}(iii)AAis upper semicontinuous;(iv)The support of AA, defined as suppA=⋃α∈(0,1][A]α¯={x:A(x)>0}¯{\rm{supp}}A=\overline{{\bigcup }_{\alpha \in (0,1]}{{[}A]}_{\alpha }}=\overline{\{x:A\left(x)\gt 0\}}, is compact.Definition 1.2[18] For α∈(0,1]\alpha \in \left(0,1]we define the α\alpha -cuts of fuzzy number AAwith [A]α={x∈R:A(x)≥α}.{{[}A]}_{\alpha }=\{x\in R:A\left(x)\ge \alpha \}.In particular, for α=0\alpha =0, the support of AAis defined as suppA=[A]0={x∈R∣A(x)>0}¯{\rm{supp}}A={{[}A]}_{0}=\overline{\{x\in R| A\left(x)\gt 0\}}.It is clear that [A]α{{[}A]}_{\alpha }is a closed interval. Fuzzy number AAis positive if min(suppA)>0\min \left({\rm{supp}}A)\gt 0.It is obvious that if AAis a positive real number, then AAis a positive fuzzy number and [A]α=[A,A],α∈[0,1]{{[}A]}_{\alpha }=\hspace{-0.00em}\left[A,A],\alpha \in \left[0,1]. In this case, we say AAis a trivial fuzzy number.Let Bi{B}_{i}(i=0,1,…,k,ki=0,1,\ldots ,k,kis a positive integer) be fuzzy numbers such that [Bi]α=[Bi,l,α,Bi,r,α],i=0,1,…,k,α∈(0,1],{{[}{B}_{i}]}_{\alpha }={[}{B}_{i,l,\alpha },{B}_{i,r,\alpha }],\hspace{1em}i=0,1,\ldots ,k,\hspace{1em}\alpha \in \left(0,1],and for any α∈(0,1]\alpha \in \left(0,1]Cl,α=max{Bi,l,α,i=0,1,…,k},Cr,α=max{Bi,r,α,i=0,1,…,k},{C}_{l,\alpha }={\rm{\max }}\left\{{B}_{i,l,\alpha },i=0,1,\ldots ,k\right\},{C}_{r,\alpha }={\rm{\max }}\left\{{B}_{i,r,\alpha },i=0,1,\ldots ,k\right\},then by ([19], Theorem 2.1), (Cl,α,Cr,α)\left({C}_{l,\alpha },{C}_{r,\alpha })determines a fuzzy number CCsuch that [C]α=[Cl,α,Cr,α],α∈(0,1].{{[}C]}_{\alpha }={[}{C}_{l,\alpha },{C}_{r,\alpha }],\alpha \in \left(0,1].According to [20] and ([21], Lemma 2.3), we can define C=max{Bi,i=0,1,…,k}.C={\rm{\max }}\left\{{B}_{i},i=0,1,\ldots ,k\right\}.Definition 1.3[20] Let A,BA,Bbe fuzzy numbers with [A]α=[Al,α,Ar,α]{{[}A]}_{\alpha }={[}{A}_{l,\alpha },{A}_{r,\alpha }], [B]α=[Bl,α,Br,α]{{[}B]}_{\alpha }={[}{B}_{l,\alpha },{B}_{r,\alpha }], α∈(0,1]\alpha \in (0,1]. The fuzzy number space norm is defined as follows: ∥A∥=supα∈(0,1]max{∣Al,α∣,∣Ar,α∣}.\parallel A\parallel =\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {A}_{l,\alpha }| ,| {A}_{r,\alpha }| \}.The distance between two arbitrary fuzzy numbers AAand BBis defined as follows: D(A,B)=supα∈(0,1]max{∣Al,α−Bl,α∣,∣Ar,α−Br,α∣}.D\left(A,B)=\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {A}_{l,\alpha }-{B}_{l,\alpha }| ,| {A}_{r,\alpha }-{B}_{r,\alpha }| \}.Definition 1.4[22] We say that xn{x}_{n}is a positive solution of (1) if xn{x}_{n}is a sequence of positive fuzzy numbers, which satisfies (1).We say that a sequence of positive fuzzy numbers xn{x}_{n}is persistent (resp. is bounded) if there exists a positive number MM(resp., NN) such that suppxn⊂[M,+∞),(resp.suppxn⊂(0,N]),n=1,2,….{\rm{supp}}{x}_{n}\subset {[}M,+\infty ),\hspace{0.33em}\left({\rm{resp.}}\hspace{0.33em}{\rm{supp}}{x}_{n}\subset (0,N]),\hspace{1em}n=1,2,\ldots .In addition, we say that xn{x}_{n}is bounded and persists if there exist numbers M,N∈(0,+∞)M,N\in \left(0,+\infty )such that suppxn⊂[M,N],n=1,2,….{\rm{supp}}{x}_{n}\subset {[}M,N],\hspace{0.33em}n=1,2,\ldots .Lemma 1.1[18] Let function f:R+×R+×R+→R+f:{R}^{+}\times {R}^{+}\times {R}^{+}\to {R}^{+}be continuous, A,B,CA,B,Cbe fuzzy numbers. Then[f(A,B,C)]α=f([A]α,[B]α,[C]α),α∈(0,1].{{[}f\left(A,B,C)]}_{\alpha }=f\left({{[}A]}_{\alpha },{{[}B]}_{\alpha },{{[}C]}_{\alpha }),\hspace{1em}\alpha \in (0,1].2Main resultsIn this section, we will discuss some dynamical characters of the fuzzy equation (1). We know that the existence of the positive solution is important for equation, which is the basis for discussing all of the dynamic characters. So, let us first prove that for fuzzy difference equation (1) there exists unique positive solution for any positive initial value.Theorem 2.1For any positive fuzzy numbers x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}, fuzzy difference equation (1), there exists a unique positive solution xn{x}_{n}whose initial value is x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}.ProofFor all positive fuzzy numbers x−k,x−k+1,…,x−1,x0,k∈{0,1,2…}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\ldots \}, suppose there exists a fuzzy number sequence that satisfies equation (1) whose initial value is x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}. Consider their α\alpha -cuts, α∈(0,1]\alpha \in \left(0,1], (2)[xn]α=[Ln,α,Rn,α],[A]α=[Al,α,Ar,α],[B]α=[Bl,α,Br,α].n=0,1,2….\left\{\begin{array}{l}{{[}{x}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }],\\ {{[}A]}_{\alpha }={[}{A}_{l,\alpha },{A}_{r,\alpha }],\\ {{[}B]}_{\alpha }={[}{B}_{l,\alpha },{B}_{r,\alpha }].\end{array}\right.\hspace{1em}n=0,1,2\ldots .Following (1), (2), and Lemma 1.1 in the literature [18], we have [xn+1]α=[Ln+1,α,Rn+1,α]=xnA+Bxn−kα=[xn]α[A]α+[B]α×[xn−k]α=[Ln,α,Rn,α][Al,α,Ar,α]+[Bl,α,Br,α]×[Ln−k,α,Rn−k,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ln−k,α,\begin{array}{rcl}\phantom{\rule[-1.5em]{}{0ex}}{{[}{x}_{n+1}]}_{\alpha }& =& {[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]={\left[\frac{{x}_{n}}{A+B{x}_{n-k}}\right]}_{\alpha }=\frac{{{[}{x}_{n}]}_{\alpha }}{{{[}A]}_{\alpha }+{{[}B]}_{\alpha }\times {{[}{x}_{n-k}]}_{\alpha }}\\ & =& \frac{{[}{L}_{n,\alpha },{R}_{n,\alpha }]}{{[}{A}_{l,\alpha },{A}_{r,\alpha }]+{[}{B}_{l,\alpha },{B}_{r,\alpha }]\times {[}{L}_{n-k,\alpha },{R}_{n-k,\alpha }]}\\ & =& \left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}\right],\end{array}where n=0,1,2,…,α∈(0,1]n=0,1,2,\ldots ,\alpha \in \left(0,1], so we obtain the related equation system (3)Ln+1,α=Ln,αAr,α+Br,α×Rn−k,α,Rn+1,α=Rn,αAl,α+Bl,α×Ln−k,α.\left\{\begin{array}{l}{L}_{n+1,\alpha }=\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\\ {R}_{n+1,\alpha }=\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}.\end{array}\right.\hspace{2.5em}Obviously, for any initial value (Li,α,Ri,α)\left({L}_{i,\alpha },{R}_{i,\alpha }), i=−k,−k+1,…,0i=-k,-k+1,\ldots ,0, α∈(0,1]\alpha \in (0,1], system (3), there exists a unique positive solution (Ln,α,Rn,α)\left({L}_{n,\alpha },{R}_{n,\alpha }), α∈(0,1]\alpha \in \left(0,1]. Now we prove that [Ln,α,Rn,α{L}_{n,\alpha },{R}_{n,\alpha }], α∈(0,1]\alpha \in (0,1]determines the solution xn{x}_{n}of (1) whose initial value is x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, where (Ln,α,Rn,α)\left({L}_{n,\alpha },{R}_{n,\alpha })is the positive solution of system (3) with initial value (Li,α,Ri,α)\left({L}_{i,\alpha },{R}_{i,\alpha }), i=−k,−k+1,…,0i=-k,-k+1,\ldots ,0, such that (4)[xn]α=[Ln,α,Rn,α],α∈(0,1],n=0,1,2,….{{[}{x}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }],\alpha \in (0,1],\hspace{1em}n=0,1,2,\ldots .According to the literature [23] and A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}are positive fuzzy numbers, for any α1,α2∈(0,1],α1≤α2{\alpha }_{1},{\alpha }_{2}\in (0,1],{\alpha }_{1}\le {\alpha }_{2}, we have (5)0<Al,α1≤Al,α2≤Ar,α2≤Ar,α1,0<Bl,α1≤Bl,α2≤Br,α2≤Br,α1,0<L−k,α1≤L−k,α2≤R−k,α2≤R−k,α1,0<L−k+1,α1≤L−k+1,α2≤R−k+1,α2≤R−k+1,α1,………………………………,0<L0,α1≤L0,α2≤R0,α2≤R0,α1.\left\{\begin{array}{l}0\lt {A}_{l,{\alpha }_{1}}\le {A}_{l,{\alpha }_{2}}\le {A}_{r,{\alpha }_{2}}\le {A}_{r,{\alpha }_{1}},\\ 0\lt {B}_{l,{\alpha }_{1}}\le {B}_{l,{\alpha }_{2}}\le {B}_{r,{\alpha }_{2}}\le {B}_{r,{\alpha }_{1}},\\ 0\lt {L}_{-k,{\alpha }_{1}}\le {L}_{-k,{\alpha }_{2}}\le {R}_{-k,{\alpha }_{2}}\le {R}_{-k,{\alpha }_{1}},\\ 0\lt {L}_{-k+1,{\alpha }_{1}}\le {L}_{-k+1,{\alpha }_{2}}\le {R}_{-k+1,{\alpha }_{2}}\le {R}_{-k+1,{\alpha }_{1}},\\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\\ 0\lt {L}_{0,{\alpha }_{1}}\le {L}_{0,{\alpha }_{2}}\le {R}_{0,{\alpha }_{2}}\le {R}_{0,{\alpha }_{1}}.\end{array}\right.\hspace{1.7em}By induction and (3), (5), we will show (6)Ln,α1≤Ln,α2≤Rn,α2≤Rn,α1,n=0,1,2,….{L}_{n,{\alpha }_{1}}\le {L}_{n,{\alpha }_{2}}\le {R}_{n,{\alpha }_{2}}\le {R}_{n,{\alpha }_{1}},\hspace{1em}n=0,1,2,\ldots .\hspace{1.2em}By (5), when n=0n=0, it is obvious (6) is true. When n=1n=1, since L1,α1=L0,α1Ar,α1+Br,α1×R−k,α1≤L0,α2Ar,α2+Br,α2×R−k,α2=L1,α2=L0,α2Ar,α2+Br,α2×R−k,α2≤R0,α2Al,α2+Bl,α2×L−k,α2=R1,α2=R0,α2Al,α2+Bl,α2×L−k,α2≤R0,α1Al,α1+Bl,α1×L−k,α1=R1,α1,\hspace{0.3em}\begin{array}{rcl}{L}_{1,{\alpha }_{1}}& =& \frac{{L}_{0,{\alpha }_{1}}}{{A}_{r,{\alpha }_{1}}+{B}_{r,{\alpha }_{1}}\times {R}_{-k,{\alpha }_{1}}}\le \frac{{L}_{0,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{-k,{\alpha }_{2}}}={L}_{1,{\alpha }_{2}}\\ & =& \frac{{L}_{0,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{-k,{\alpha }_{2}}}\le \frac{{R}_{0,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{-k,{\alpha }_{2}}}={R}_{1,{\alpha }_{2}}\\ & =& \frac{{R}_{0,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{-k,{\alpha }_{2}}}\le \frac{{R}_{0,{\alpha }_{1}}}{{A}_{l,{\alpha }_{1}}+{B}_{l,{\alpha }_{1}}\times {L}_{-k,{\alpha }_{1}}}={R}_{1,{\alpha }_{1}},\end{array}so L1,α1≤L1,α2≤R1,α2≤R1,α1{L}_{1,{\alpha }_{1}}\le {L}_{1,{\alpha }_{2}}\le {R}_{1,{\alpha }_{2}}\le {R}_{1,{\alpha }_{1}}. Suppose (6) is true when n≤k,k∈{1,2,…}n\le k,k\in \{1,2,\ldots \}, following (3) and (5), we have Lk+1,α1=Lk,α1Ar,α1+Br,α1×R0,α1≤Lk,α2Ar,α2+Br,α2×R0,α2=Lk+1,α2≤Rk,α2Al,α2+Bl,α2×L0,α2=Rk+1,α2≤Rk,α1Al,α1+Bl,α1×L0,α1=Rk+1,α1,\hspace{3.1em}\begin{array}{rcl}{L}_{k+1,{\alpha }_{1}}& =& \frac{{L}_{k,{\alpha }_{1}}}{{A}_{r,{\alpha }_{1}}+{B}_{r,{\alpha }_{1}}\times {R}_{0,{\alpha }_{1}}}\le \frac{{L}_{k,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{0,{\alpha }_{2}}}={L}_{k+1,{\alpha }_{2}}\\ & \le & \frac{{R}_{k,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{0,{\alpha }_{2}}}={R}_{k+1,{\alpha }_{2}}\le \frac{{R}_{k,{\alpha }_{1}}}{{A}_{l,{\alpha }_{1}}+{B}_{l,{\alpha }_{1}}\times {L}_{0,{\alpha }_{1}}}={R}_{k+1,{\alpha }_{1}},\end{array}so Lk+1,α1≤Lk+1,α2≤Rk+1,α2≤Rk+1,α1,k=1,2,…{L}_{k+1,{\alpha }_{1}}\le {L}_{k+1,{\alpha }_{2}}\le {R}_{k+1,{\alpha }_{2}}\le {R}_{k+1,{\alpha }_{1}},\hspace{0.33em}k=1,2,\ldots . By induction, (6) holds true.Following (3), we have (7)L1,α=L0,αAr,α+Br,α×R−k,α,R1,α=R0,αAl,α+Bl,α×L−k,α.\left\{\begin{array}{rcl}{L}_{1,\alpha }& =& \frac{{L}_{0,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{-k,\alpha }},\\ {R}_{1,\alpha }& =& \frac{{R}_{0,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{-k,\alpha }}.\end{array}\right.\hspace{8.5em}For A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2…}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\ldots \}are positive fuzzy numbers and Lemma 2.2 in the literature [24], we know Al,α,Ar,α,Bl,α,Br,α,L−k,α,R−k,α,…,L−1,α,R−1,α,L0,α,R0,α{A}_{l,\alpha },{A}_{r,\alpha },{B}_{l,\alpha },{B}_{r,\alpha },{L}_{-k,\alpha },{R}_{-k,\alpha },\ldots ,{L}_{-1,\alpha },{R}_{-1,\alpha },{L}_{0,\alpha },{R}_{0,\alpha }are left continuous, by (7) we have L1,α,R1,α{L}_{1,\alpha },{R}_{1,\alpha }are also left continuous. By induction, we have Ln,α,Rn,α(n=1,2,…){L}_{n,\alpha },{R}_{n,\alpha }\hspace{0.25em}\left(n=1,2,\ldots )are left continuous.We assert that the support of xn{x}_{n}, suppxn=⋃α∈(0,1][Ln,α,Rn,α]¯{\rm{supp}}{x}_{n}=\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}is compact. We just need to prove that ⋃α∈(0,1][Ln,α,Rn,α]{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]is bounded. When n=1n=1, since A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}are positive fuzzy numbers, there exist constants MA,NA,MB,NB,Mi,Ni,i=−k,−k+1,…,−1,0{M}_{A},{N}_{A},{M}_{B},{N}_{B},{M}_{i},{N}_{i},i=-k,-k+1,\ldots ,-1,0, such that for all α∈(0,1]\alpha \in (0,1], we have (8)[Al,α,Ar,α]⊂[MA,NA],[Bl,α,Br,α]⊂[MB,NB],[Li,α,Ri,α]⊂[Mi,Ni].{[}{A}_{l,\alpha },{A}_{r,\alpha }]\subset {[}{M}_{A},{N}_{A}],{\rm{}}{[}{B}_{l,\alpha },{B}_{r,\alpha }]\subset {[}{M}_{B},{N}_{B}],{\rm{}}{[}{L}_{i,\alpha },{R}_{i,\alpha }]\subset {[}{M}_{i},{N}_{i}].Following systems (7) and (8) we obtain [L1,α,R1,α]⊂M0NA+NB×N−k,N0MA+MB×M−k{[}{L}_{1,\alpha },{R}_{1,\alpha }]\subset \left[\frac{{M}_{0}}{{N}_{A}+{N}_{B}\times {N}_{-k}},\frac{{N}_{0}}{{M}_{A}+{M}_{B}\times {M}_{-k}}\right](α∈(0,1])\left(\alpha \in (0,1]), so ⋃α∈(0,1][L1,α,R1,α]⊂M0NA+NB×N−k,N0MA+MB×M−k{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]\hspace{-0.08em}\subset \hspace{-0.08em}\left[\hspace{-0.16em},\frac{{M}_{0}}{{N}_{A}+{N}_{B}\times {N}_{-k}},\frac{{N}_{0}}{{M}_{A}+{M}_{B}\times {M}_{-k}}\right]. It implies that ⋃α∈(0,1][L1,α,R1,α]¯\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]}is compact, and ⋃α∈(0,1][L1,α,R1,α]¯⊂(0,+∞).\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]}\hspace{0.25em}\subset \hspace{-0.08em}\left(0,\hspace{-0.08em}+\infty ).By induction, we obtain that ⋃α∈(0,1][Ln,α,Rn,α]¯\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}is compact, and ⋃α∈(0,1][Ln,α,Rn,α]¯⊂(0,+∞)\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}\subset (0,+\infty )for every n=1,2,…n=1,2,\ldots . Following this and (6), and because the left continuous of Ln,α,Rn,α,n=1,2,…{L}_{n,\alpha },{R}_{n,\alpha },\hspace{0.33em}n=1,2,\ldots , [Ln,α,Rn,α]{[}{L}_{n,\alpha },{R}_{n,\alpha }]establishes a sequence of positive fuzzy numbers xn{x}_{n}such that (4) holds.Next, we show that xn{x}_{n}is the solution of (1) with any initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}. Since for all α∈(0,1]\alpha \in (0,1], [xn+1]α=[Ln+1,α,Rn+1,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ln−k,α=xnA+Bxn−kα.{{[}{x}_{n+1}]}_{\alpha }={[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]=\left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}\right]={\left[\frac{{x}_{n}}{A+B{x}_{n-k}}\right]}_{\alpha }.So, xn{x}_{n}is the solution of (1) with any initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}.{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}.That is, the positive solution of fuzzy difference equation (1) exists.Next, we prove that the positive solution of fuzzy difference equation (1) is unique by contradiction. Suppose there exists another solution x¯n{\overline{x}}_{n}of (1) with initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}. Similar to the above proof, we can easily prove that [x¯n+1]α=[Ln+1,α,Rn+1,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ll−k,α=x¯nA+Bx¯n−kα,{{[}{\overline{x}}_{n+1}]}_{\alpha }={[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]=\left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{l-k,\alpha }}\right]={\left[\frac{{\overline{x}}_{n}}{A+B{\overline{x}}_{n-k}}\right]}_{\alpha },which implies [x¯n]α=[Ln,α,Rn,α]{{[}{\overline{x}}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }](n=0,1,2,…n=0,1,2,\ldots ) for any α∈(0,1]\alpha \in (0,1]. Then, by (4) we have [x¯n]α=[xn]α{{[}{\overline{x}}_{n}]}_{\alpha }={{[}{x}_{n}]}_{\alpha }(n=0,1,2,…n=0,1,2,\ldots ) for any α∈(0,1]\alpha \in (0,1]. So x¯n=xn,n=0,1,2,…{\overline{x}}_{n}={x}_{n},n=0,1,2,\ldots , which is contradictory. So the positive solution of fuzzy difference equation (1) is unique.In summary, for fuzzy difference equation (1), there exists a unique positive solution xn{x}_{n}, for every positive initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2,…}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2,\ldots \}.□We have proved that the fuzzy difference equation (1) has a unique positive solution, and then we will study the dynamic behavior of the solution. First, we discuss the boundedness of positive solutions of equation (1), which needs to rely on related equation system from (3).Lemma 2.1Consider the system of difference equations(9)yn+1=yna+bzn−k,zn+1=znc+dyn−k,n=0,1,2,…,\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}},\end{array}\right.\hspace{1em}n=0,1,2,\ldots ,where the initial value yi,zi(i=−k,−k+1,…,−1,0){y}_{i},{z}_{i}\hspace{1em}\left(i=-k,-k+1,\ldots ,-1,0)and parameters a,b,c,da,b,c,dare positive real numbers. If parameters a≥1a\ge 1and c≥1c\ge 1, then for all n≥kn\ge k, yn,zn{y}_{n},{z}_{n}are bounded and1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n=0,1,2,….\hspace{4.25em}\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots .ProofLet (yn,zn)({y}_{n},{z}_{n})be a positive solution of the difference equation system (9). From (9), the following are concluded by recursion, yn+1=yna+bzn−k≤1ayn≤1a2yn−1≤⋯≤1an+1y0,zn+1=znc+dyn−k≤1czn≤1c2zn−1≤⋯≤1cn+1z0,n=0,1,2,…,\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}}\le \frac{1}{a}{y}_{n}\le \frac{1}{{a}^{2}}{y}_{n-1}\le \cdots \le \frac{1}{{a}^{n+1}}{y}_{0},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}}\le \frac{1}{c}{z}_{n}\le \frac{1}{{c}^{2}}{z}_{n-1}\le \cdots \le \frac{1}{{c}^{n+1}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots ,yn+1=yna+bzn−k≥yna+bz0≥1(a+bz0)2yn−1≥⋯≥1(a+bz0)n+1−kyk,zn+1=znc+dyn−k≥znc+dy0≥1(c+dy0)2zn−1≥⋯≥1(c+dy0)n+1−kzk,n≥k.\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}}\ge \frac{{y}_{n}}{a+b{z}_{0}}\ge \frac{1}{{\left(a+b{z}_{0})}^{2}}{y}_{n-1}\ge \cdots \ge \frac{1}{{\left(a+b{z}_{0})}^{n+1-k}}{y}_{k},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}}\ge \frac{{z}_{n}}{c+d{y}_{0}}\ge \frac{1}{{\left(c+d{y}_{0})}^{2}}{z}_{n-1}\ge \cdots \ge \frac{1}{{\left(c+d{y}_{0})}^{n+1-k}}{z}_{k},\end{array}\right.\hspace{1em}n\ge k.It is easy to deduce that yn,zn{y}_{n},{z}_{n}are bounded and 1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n≥k.□\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n\ge k.\square Theorem 2.2Consider fuzzy difference equation (1), for all α∈(0,1]\alpha \in (0,1], if(10)Al,α>1,{A}_{l,\alpha }\gt 1,then every positive solution xn{x}_{n}of fuzzy difference equation (1) is bounded.ProofLet xn{x}_{n}be a positive solution of (1) and satisfy (4). Because Al,α>1{A}_{l,\alpha }\gt 1, so Ar,α>1{A}_{r,\alpha }\gt 1. From (3) and Lemma 2.1, for any α∈(0,1]\alpha \in (0,1]we have 1(Ar,α+Br,α×R0,α)n−kLk,α≤Ln,α≤1Ar,αnL0,α,1(Al,α+Bl,α×L0,α)n−kRk,α≤Rn,α≤1Al,αnR0,α.n=0,1,2,….\hspace{6.65em}\left\{\begin{array}{r}\frac{1}{{\left({A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{0,\alpha })}^{n-k}}{L}_{k,\alpha }\le {L}_{n,\alpha }\le \frac{1}{{A}_{r,\alpha }^{n}}{L}_{0,\alpha },\\ \frac{1}{{\left({A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{0,\alpha })}^{n-k}}{R}_{k,\alpha }\le {R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }.\end{array}\right.\hspace{1em}n=0,1,2,\ldots .So, Rn,α≤1Al,αnR0,α{R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }and (11)Ln,α>0.{L}_{n,\alpha }\gt 0.We have Rn,α≤1Al,αnR0,α≤R0,α{R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }\le {R}_{0,\alpha }, which is because Al,α>1{A}_{l,\alpha }\gt 1. Then since xn{x}_{n}is a positive fuzzy number, there exists a constant J>0J\gt 0, such that for all α∈(0,1]\alpha \in \left(0,1], we have (12)R0,α≤J.{R}_{0,\alpha }\le J.From (11) and (12), we obtain that [Ln,α,Rn,α]⊂(0,J]{[}{L}_{n,\alpha },{R}_{n,\alpha }]\subset \left(0,J]for all α∈(0,1]\alpha \in (0,1], so ⋃α∈(0,1][Ln,α,Rn,α]⊂(0,J]{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]\subset \left(0,J], which implies that ⋃α∈(0,1][Ln,α,Rn,α]¯⊂(0,J]\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}\subset \left(0,J]. Thus, the positive solution xn{x}_{n}of equation (1) is bounded.□Next, we consider the existence of equilibrium solution of (1). It is obvious that x¯=0\bar{x}=0is an equilibrium solution of (1).Let y¯,z¯\overline{y},\overline{z}be real numbers and satisfy equation systems y¯=y¯a+bz¯,z¯=z¯c+dy¯\overline{y}=\frac{\overline{y}}{a+b\overline{z}},\overline{z}=\frac{\overline{z}}{c+d\overline{y}}. Solving it we have y¯=1−cd,z¯=1−ab\overline{y}=\frac{1-c}{d},\overline{z}=\frac{1-a}{b}. It is easy to get the following lemma.Lemma 2.2Consider the difference equation system (9). Then (9) has positive equilibrium point (y¯,z¯)(\overline{y},\overline{z})and y¯=1−cd,z¯=1−ab\overline{y}=\frac{1-c}{d},\overline{z}=\frac{1-a}{b}when a<1,c<1a\lt 1,c\lt 1.Theorem 2.3For all α∈(0,1]\alpha \in (0,1], there is no positive equilibrium solution of fuzzy difference equation (1).ProofAssume there exists a positive fuzzy number xx, which satisfies x=xA+Bxx=\frac{x}{A+Bx}, [x]α=[Lα,Rα]{{[}x]}_{\alpha }={[}{L}_{\alpha },{R}_{\alpha }], and Lα,Rα≥0{L}_{\alpha },{R}_{\alpha }\ge 0, where α∈(0,1]\alpha \in (0,1]. So we derive a related equation system as follows: Lα=LαAr,α+Br,α×Rα,Rα=RαAl,α+Bl,α×Lα.{L}_{\alpha }=\frac{{L}_{\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{\alpha }},\hspace{1em}{R}_{\alpha }=\frac{{R}_{\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{\alpha }}.Suppose Lα≠0,Rα≠0{L}_{\alpha }\ne 0,{R}_{\alpha }\ne 0, following Lemma 2.2 we have Lα=1−Al,αBl,α,Rα=1−Ar,αBr,α{L}_{\alpha }=\frac{1-{A}_{l,\alpha }}{{B}_{l,\alpha }},{R}_{\alpha }=\frac{1-{A}_{r,\alpha }}{{B}_{r,\alpha }}, that is, Lα>Rα{L}_{\alpha }\gt {R}_{\alpha }. But [x]α=[Lα,Rα]{\left[x]}_{\alpha }=\left[{L}_{\alpha },{R}_{\alpha }]means Lα≤Rα{L}_{\alpha }\le {R}_{\alpha }, which is a contradiction. So it can only be Lα=Rα=0{L}_{\alpha }={R}_{\alpha }=0.□Therefore, there is no positive equilibrium solution of fuzzy difference equation (1), but have equilibrium solution x¯=0\bar{x}=0.Next, we discuss the positive solution’s convergence and asymptotic stability of fuzzy difference equation (1).Lemma 2.3Consider the difference equation system (9). Assume that a,c∈(1,+∞)a,c\in \left(1,+\infty ), then the solution of system (9) is convergent and limn→∞yn=0,limn→∞zn=0{\mathrm{lim}}_{n\to \infty }{y}_{n}=0,{\mathrm{lim}}_{n\to \infty }{z}_{n}=0.ProofFrom Lemma 2.1, we know 1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n=0,1,2,….\hspace{1.4em}\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots .If a,c∈(1,+∞)a,c\in \left(1,+\infty ), then limn→∞1(a+bz0)n−kyk=0,limn→∞1any0=0,limn→∞1(c+dy0)n−kzk=0,limn→∞1cnz0=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{a}^{n}}{y}_{0}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{c}^{n}}{z}_{0}=0.It is obvious that the solution of system (9) (yn,zn)({y}_{n},{z}_{n})is convergent and limn→∞yn=0,limn→∞zn=0{\mathrm{lim}}_{n\to \infty }{y}_{n}=0,{\mathrm{lim}}_{n\to \infty }{z}_{n}=0.□Theorem 2.4Consider fuzzy difference equation (1). Assume Al,α>1{A}_{l,\alpha }\gt 1, then every positive solution of (1) converges to equilibrium x¯=0\bar{x}=0when n→∞n\to \infty .ProofSince Ln+1,α=LnAr,α+Br,α×Rn−k,Rn+1,α=RnAl,α+Bl,α×Ln−k,{L}_{n+1,\alpha }=\frac{{L}_{n}}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k}},\hspace{1em}{R}_{n+1,\alpha }=\frac{{R}_{n}}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k}},following Lemma 2.3 we have limn→∞Ln,α=0,limn→∞Rn,α=0{\mathrm{lim}}_{n\to \infty }{L}_{n,\alpha }=0,{\mathrm{lim}}_{n\to \infty }{R}_{n,\alpha }=0, and according to the distance between two arbitrary fuzzy numbers (Definition 1.3 ([19])), we obtain limn→∞D(xn,x)=limn→∞D(xn,0)=limn→∞supα∈(0,1]max{∣Ln,α−0∣,∣Rn,α−0∣}=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }D({x}_{n},x)=\mathop{\mathrm{lim}}\limits_{n\to \infty }D({x}_{n},0)=\mathop{\mathrm{lim}}\limits_{n\to \infty }\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {L}_{n,\alpha }-0| ,| {R}_{n,\alpha }-0| \}=0.So limn→∞xn=x¯=0{\mathrm{lim}}_{n\to \infty }{x}_{n}=\bar{x}=0. That is every positive solution of equation (1) converges to equilibrium x¯=0\bar{x}=0when n→∞n\to \infty .□Lemma 2.4For the equilibrium point (0,0)\left(0,0)of equation system (9), if a,c∈(1,+∞)a,c\in \left(1,+\infty ), then the equilibrium point (0,0)\left(0,0)is locally asymptotically stable.ProofThe linearized equation of equation system (9) about the equilibrium point (0,0)\left(0,0)is (13)φn+1=Aφn,{\varphi }_{n+1}=A{\varphi }_{n},where φn=(yn,yn−1,…,yn−k,zn,zn−1,…,zn−k)T{\varphi }_{n}={({y}_{n},{y}_{n-1},\ldots ,{y}_{n-k},{z}_{n},{z}_{n-1},\ldots ,{z}_{n-k})}^{T}, and A=1/a0⋯0010⋯00⋯⋯⋯⋯⋯00⋯1000⋯0000⋯00⋯⋯⋯⋯⋯00⋯0000⋯0000⋯00⋯⋯⋯⋯⋯00⋯001/c0⋯0010⋯00⋯⋯⋯⋯⋯00⋯10.A=\left(\begin{array}{cc}\begin{array}{ccccc}1\hspace{-0.08em}\text{/}\hspace{-0.08em}a& 0& \cdots \hspace{0.33em}& 0& 0\\ 1& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 1& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 0& 0\end{array}& \begin{array}{ccccc}0& 0& \cdots \hspace{0.33em}& 0& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ 1\hspace{-0.08em}\text{/}\hspace{-0.08em}c& 0& \cdots \hspace{0.33em}& 0& 0\\ 1& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 1& 0\end{array}\end{array}\right).The characteristic equation with equation (13) is P(λ)=(−λ)k−21a−λ1c−λP\left(\lambda )={\left(-\lambda )}^{k-2}\left(\frac{1}{a}-\lambda \right)\left(\frac{1}{c}-\lambda \right), and this shows that all eigenvalues are ∣λ∣<1| \lambda | \lt 1. Thus, the equilibrium (0,0) is locally asymptotically stable.□From Lemmas 2.3 and 2.4, we have that the equilibrium point (0,0)\left(0,0)of equation (9) is global asymptotically stable. Combined with Theorem 4, we can easily obtain the following theorem:Theorem 2.5Consider fuzzy difference equation (1). If Al,α>1{A}_{l,\alpha }\gt 1, then the equilibrium point x¯=0\bar{x}=0is global asymptotically stable.3Numerical exampleIn this section, we give numerical examples on the main conclusions of Section 2.Model. Take k=2k=2in fuzzy difference equation (1). Considering the following fuzzy difference equation: (14)xn+1=xnA+Bxn−2,n=3,4,5…,{x}_{n+1}=\frac{{x}_{n}}{A+B{x}_{n-2}},\hspace{1em}n=3,4,5\ldots ,in which parameters A,BA,Band initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are positive fuzzy numbers. According to Theorems 2.1–2.3, we know the following conclusions.For any positive solution xn{x}_{n}of (14), [xn]α=[Ln,α,Rn,α]{\left[{x}_{n}]}_{\alpha }=\left[{L}_{n,\alpha },{R}_{n,\alpha }], a difference equation system with a parameter α\alpha is obtained (15)Ln+1,α=Ln,αAr,α+Br,α×Rn−2,α,Rn+1,α=Rn,αAl,α+Bl,α×Ln−2,α,α∈(0,1],{L}_{n+1,\alpha }=\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-2,\alpha }},\hspace{1em}\hspace{1em}{R}_{n+1,\alpha }=\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-2,\alpha }},\hspace{1em}\alpha \in \left(0,1],and we know the unique equilibrium solution is x¯=0\bar{x}=0.Example 1Take A,BA,Band initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}as follows: A(t)=3t−7,73≤t≤83,−3t+9,83≤t≤3,B(t)=2t−5,52≤t≤3,−2t+7,3≤t≤72,x1(t)=t−4,4≤t≤5,−12t+72,5≤t≤7,x2(t)=12t−4,8≤t≤10,−12t+6,10≤t≤12,x3(t)=13t−2,6≤t≤9,−13t+4,9≤t≤12.\begin{array}{ll}A\left(t)=\left\{\begin{array}{ll}3t-7,& \frac{7}{3}\le t\le \frac{8}{3},\\ -3t+9,& \frac{8}{3}\le t\le 3,\end{array}\right.& B\left(t)=\left\{\begin{array}{ll}2t-5,& \frac{5}{2}\le t\le 3,\\ -2t+7,& 3\le t\le \frac{7}{2},\end{array}\right.\\ {x}_{1}\left(t)=\left\{\begin{array}{ll}t-4,& 4\le t\le 5,\\ -\frac{1}{2}t+\frac{7}{2},& 5\le t\le 7,\end{array}\right.& {x}_{2}\left(t)=\left\{\begin{array}{ll}\frac{1}{2}t-4,& 8\le t\le 10,\\ -\frac{1}{2}t+6,& 10\le t\le 12,\end{array}\right.\\ {x}_{3}\left(t)=\left\{\begin{array}{ll}\frac{1}{3}t-2,& 6\le t\le 9,\\ -\frac{1}{3}t+4,& 9\le t\le 12.\end{array}\right.& \end{array}\hspace{2.5em}From function A(x),B(x),x1(x),x2(x),x3(x)A\left(x),B\left(x),{x}_{1}\left(x),{x}_{2}\left(x),{x}_{3}\left(x), for all α∈(0,1]\alpha \in \left(0,1], we have [A]α=73+α3,3−α3,[B]α=52+α2,72−α2,[x1]α=[4+α,7−2α],[x2]α=[8+2α,12−2α],[x3]α=[6+3α,12−3α].\hspace{4.8em}\begin{array}{l}\phantom{\rule[-1.05em]{}{0ex}}{\left[A]}_{\alpha }=\left[\frac{7}{3}+\frac{\alpha }{3},3-\frac{\alpha }{3}\right],\hspace{1em}{\left[B]}_{\alpha }=\left[\frac{5}{2}+\frac{\alpha }{2},\frac{7}{2}-\frac{\alpha }{2}\right],\\ {\left[{x}_{1}]}_{\alpha }=\left[4+\alpha ,7-2\alpha ],\hspace{2em}{\left[{x}_{2}]}_{\alpha }=\left[8+2\alpha ,12-2\alpha ],\hspace{1em}{\left[{x}_{3}]}_{\alpha }=\left[6+3\alpha ,12-3\alpha ].\end{array}So the support of them are ⋃α∈(0,1][A]α¯=73,3,⋃α∈(0,1][B]α¯=52,72,⋃α∈(0,1][x0]α¯=[4,7],⋃α∈(0,1][x−1]α¯=[8,12],⋃α∈(0,1][x−2]α¯=[6,12].\begin{array}{l}\phantom{\rule[-1.25em]{}{0ex}}\overline{\bigcup _{\alpha \in (0,1]}{\left[A]}_{\alpha }}=\left[\frac{7}{3},3\right],\hspace{1em}\overline{\bigcup _{\alpha \in (0,1]}{\left[B]}_{\alpha }}=\left[\frac{5}{2},\frac{7}{2}\right],\\ \overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{0}]}_{\alpha }}=\left[4,7],\hspace{1.6em}\overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{-1}]}_{\alpha }}=\left[8,12],\hspace{1em}\overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{-2}]}_{\alpha }}=\left[6,12].\end{array}\hspace{0.7em}It is clear that Al,α>1{A}_{l,\alpha }\gt 1and (14) satisfies Theorems 2.1–2.5. So, every positive solution of (14) is bounded and persists, and converges to equilibrium x¯=0\bar{x}=0when n→+∞n\to +\infty . Moreover, the unique equilibrium x¯=0\bar{x}=0is global asymptotically stable (Figures 1–3).Figure 1In Example 1, the solution of system (15) when a=0a=0.Figure 2In Example 1, the solution of system (15) when a=0.5a=0.5.Figure 3In Example 1, the solution of system (15) when a=1a=1.Remark 1There exists positive fuzzy number A with Al,α<1{A}_{l,\alpha }\lt 1, such that the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable (see the following Examples 2–3).Example 2Let us still consider equation (14), whose parameter BBand initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are the same as those in Example 1, but parameter AAchanges to (16)A(t)=8t−3,38≤t≤12,−8t+5,12≤t≤58.A\left(t)=\left\{\begin{array}{ll}8t-3,& \frac{3}{8}\le t\le \frac{1}{2},\\ -8t+5,& \frac{1}{2}\le t\le \frac{5}{8}.\end{array}\right.So, we have [A]α=38+α8,58−α8{\left[A]}_{\alpha }=\left[\frac{3}{8}+\frac{\alpha }{8},\frac{5}{8}-\frac{\alpha }{8}\right]and ⋃α∈(0,1][A]α¯=38,58\overline{{\bigcup }_{\alpha \in (0,1]}{\left[A]}_{\alpha }}=\left[\frac{3}{8},\frac{5}{8}\right]. It is clear that Al,α<1,AR,α<1{A}_{l,\alpha }\lt 1,{A}_{R,\alpha }\lt 1, so Example 2 does not satisfy the conditions of Theorems 2.2 and 2.4. Let us see the behaviors of the solution through the following graph (Figure 4).Figure 4In Example 2, the solution of system (15) when a=0.5a=0.5.In Figure 4, it is obvious that Ln{L}_{n}tends to 0 and Rn{R}_{n}tends to positive infinity when nnis sufficiently large. So, the solutions of Example 2 do not converge and the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable.Example 3Let us still consider equation (14), where the initial values x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are the same as those in Example 1, but parameters A,BA,Bchange to (17)A(t)=5t−3,35≤t≤45,1,45≤t≤2,B(t)=5t−1,15≤t≤25,95−2t,25≤t≤910.A\left(t)=\left\{\begin{array}{ll}5t-3,& \frac{3}{5}\le t\le \frac{4}{5},\\ 1,& \frac{4}{5}\le t\le 2,\end{array}\right.\hspace{1em}B\left(t)=\left\{\begin{array}{ll}5t-1,& \frac{1}{5}\le t\le \frac{2}{5},\\ \frac{9}{5}-2t,& \frac{2}{5}\le t\le \frac{9}{10}.\end{array}\right.So we have [A]α=0.6+α5,2,⋃α∈(0,1][A]α¯=[0.6,2],[B]α=0.2+α5,0.9−α2{\left[A]}_{\alpha }=\left[\hspace{-0.08em},0.6\hspace{-0.08em}+\hspace{-0.08em}\frac{\alpha }{5},2,\hspace{-0.08em}\right],\overline{{\bigcup }_{\alpha \in \left(0,1]}{\left[A]}_{\alpha }}\hspace{-0.08em}=\hspace{-0.08em}\left[0.6,2],{\left[B]}_{\alpha }\hspace{-0.08em}=\hspace{-0.08em}\left[\hspace{-0.08em},0.2\hspace{-0.08em}+\hspace{-0.08em}\frac{\alpha }{5},0.9\hspace{-0.08em}-\hspace{-0.08em}\frac{\alpha }{2},\hspace{-0.08em}\right], and ⋃α∈(0,1][B]α¯=[0.2,0.9]\overline{{\bigcup }_{\alpha \in \left(0,1]}{\left[B]}_{\alpha }}\hspace{-0.08em}=\hspace{-0.08em}\left[0.2,0.9]. It is clear that Al,α<1,AR,α>1{A}_{l,\alpha }\lt 1,{A}_{R,\alpha }\gt 1, Example 3 does not satisfy the conditions of Theorems 2.2 and 2.4. Let us see the behaviors of the solution through the following graph (Figure 5).Figure 5In Example 3, the solution of system (15) when a=0.5a=0.5.In Figure 5, it is obvious that Ln{L}_{n}tends to 0 and Rn{R}_{n}tends to positive infinity when nnis sufficiently large. So, the solutions of Example 3 do not converge, the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable.In [17], Zhang studied a first-order fuzzy difference equation xn+1=A+xnB+xn{x}_{n+1}=\frac{A+{x}_{n}}{B+{x}_{n}}. Later, he studied second-order exponential-type fuzzy difference equation xn+1=A+Be−xnC+xn−1{x}_{n+1}=\frac{A+B{e}^{-{x}_{n}}}{C+{x}_{n-1}}in [27]. Similarly, we have also studied kk-order exponential-type fuzzy difference equation which is related to equation (1):Remark 2We amend equation (1) to be exponential-type fuzzy difference equation xn+1=xne−xn−mA+Bxn−k,n=0,1,2,…,{x}_{n+1}=\frac{{x}_{n}{e}^{-{x}_{n-m}}}{A+B{x}_{n-k}},\hspace{1em}n=0,1,2,\ldots ,where m∈{0,1,…k}m\in \left\{0,1,\ldots k\right\}, parameters A,BA,Band initial value x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, k∈{0,1,…}k\in \{0,1,\ldots \}are positive fuzzy numbers. We can obtain the same conclusion as equation (1) on the existence, boundedness, convergence, and global asymptotic stability of the solution.The proof method and process are similar to the conclusion proof of equation (1).Remark 3The existence and uniqueness of the positive solution is important for equation, which is the basis for discussing all of the dynamic characters of solutions. The commonly used method of proof is the definition method, and the proof of this article also uses the same method.4ConclusionIn this paper, we investigate the dynamical behaviors of positive solution of the fuzzy difference equation (1).First, we have proved that for the fuzzy difference equation (1) there exists a unique positive solution. Then, using the relationship between the fuzzy difference equation and the relative real number difference equation system, we have proved that if Al,α>1{A}_{l,\alpha }\gt 1then every positive solution of (1) is bounded and converges to equilibrium x=0x=0when n→∞n\to \infty . Moreover, we obtain that the equilibrium x=0x=0is global asymptotically stable. At last, we give numerical examples about equation (1) where k=2k=2draw trajectories of the solutions to reflect the global behaviors of fuzzy difference equation intuitively. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Open Mathematics de Gruyter

Dynamical behaviors of a k-order fuzzy difference equation

Han, Caihong; Li, Lue; Su, Guangwang; Sun, Taixiang
Open Mathematics , Volume 20 (1): 13 – Jan 1, 2022
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de Gruyter
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© 2022 Caihong Han et al., published by De Gruyter
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2391-5455
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2391-5455
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10.1515/math-2022-0020
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Abstract

1IntroductionAs we all know, the stability theory has always been one of the hotspots in mathematical research, especially the stability theory of time-delay systems [1,2,3]. Difference equations are usually used to describe discrete-time dynamic systems, which can be widely used to establish mathematical models in many areas, such as computer science, ecology, population dynamics, electrical networks, economics, etc. [4,5]. Due to its wide range of applications, the study of the time-delay difference equation and its dynamic behavior has become an important topic in applied mathematics. In practice, more problems encountered are that the conditional parameters fluctuate or change within a certain range. One of the ways to solve this problem is to fuzz the difference equation based on the uncertainty. Fuzzy set theory is an effective tool for modeling uncertainty and processing vague or subjective information [6,7]. Fuzzy difference equations have gradually attracted attention for their practicability [8,11, 12,13,14, 15,16,17, 18,19,20, 21,22,23, 24,25,26, 27,28] and are developing vigorously. Chrysafis et al. [10] studied the fuzzy difference equation of finance, which is an alternative method for studying the time value of money. Deeba and Korvin [11] studied fuzzy difference equation xn+1=xn−ABxn−1+C{x}_{n+1}={x}_{n}-AB{x}_{n-1}+Cabout the level of carbon dioxide (CO2{{\rm{CO}}}_{2}) in the blood.Let us first review the history of the fuzzy difference equations studied in this article. In [14], Zhang and Liu studied the qualitative behavior of the first-order linear fuzzy difference equation, which is xn+1=Axn+B(n=0,1,2,…){x}_{n+1}=A{x}_{n}+B\left(n=0,1,2,\ldots ), where the parameters A,BA,Band initial value are positive fuzzy numbers. In [15], Hatir investigated the existence, oscillatory behavior, and asymptotic behavior of the positive solutions of second-order fuzzy difference equation xn+1=A+Bxn−1(n=0,1,2,…){x}_{n+1}=A+\frac{B}{{x}_{n-1}}\hspace{0.33em}\left(n=0,1,2,\ldots ), where parameters A,BA,Band initial value are positive fuzzy numbers. Subsequently, many researchers used similar methods to study different fuzzy difference equations and came to many conclusions, such as [23,24,25, 26,27].In [16], Papaschinopoulos and Schinas studied the global properties of a high-order nonlinear difference equation system xn+1=A+ynxn−p,yn+1=A+xnyn−q(n=0,1,2,…){x}_{n+1}=A+\frac{{y}_{n}}{{x}_{n-p}},{y}_{n+1}=A+\frac{{x}_{n}}{{y}_{n-q}}\hspace{0.33em}\left(n=0,1,2,\ldots ), where p,qp,qare positive integers, parameter AAand initial values are positive real numbers. Later, he studied m-order fuzzy difference equation xn+1=A+xnxn−m(n=0,1,2,…){x}_{n+1}=A+\frac{{x}_{n}}{{x}_{n-m}}\hspace{0.33em}\left(n=0,1,2,\ldots ), in which parameter AAand initial value are positive fuzzy numbers. In [17], Zhang et al. showed the existence and the asymptotic stability of the positive solutions about first-order fuzzy difference equation xn+1=A+xnB+xn(n=0,1,2,…){x}_{n+1}=\frac{A+{x}_{n}}{B+{x}_{n}}\hspace{0.33em}\left(n=0,1,2,\ldots ), in which parameter A,BA,Band initial value are positive fuzzy numbers.Motivated by the discussions above, we study arbitrary k-order nonlinear fuzzy difference equation in this paper as follows,(1)xn+1=xnA+Bxn−k,n=0,1,2,…,{x}_{n+1}=\frac{{x}_{n}}{A+B{x}_{n-k}},\hspace{1em}n=0,1,2,\ldots ,and also the existence, stability, and other dynamic behaviors of the positive solution of equation (1), where parameters A,BA,Band initial value x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, k∈{0,1,…}k\in \{0,1,\ldots \}are positive fuzzy numbers.For further study of (1), we need some basic definitions and lemmas which are related to fuzzy difference equations, which are obtained from reference resources [14,15,16, 17,18,19, 20,21]. In this paper, R=(−∞,+∞)R=\left(-\infty ,+\infty ), R+=(0,+∞){R}^{+}=\left(0,+\infty ), F+{F}^{+}denote the set of all positive fuzzy numbers.Definition 1.1[18] A function A:R→[0,1]A:R\to {[}0,1]is said to be a fuzzy number if it satisfies the following conditions (i)–(iv): (i)AAis normal, i.e., there exists x∈Rx\in Rsuch that A(x)=1A\left(x)=1;(ii)AAis fuzzy convex, i.e., for all t∈[0,1]t\in {[}0,1]and x1,x2∈R{x}_{1},{x}_{2}\in Rsuch that A(tx1+(1−t)x2)≥min{A(x1),A(x2)};A\left(t{x}_{1}+\left(1-t){x}_{2})\ge \min \left\{A\left({x}_{1}),A\left({x}_{2})\right\};\hspace{1.2em}(iii)AAis upper semicontinuous;(iv)The support of AA, defined as suppA=⋃α∈(0,1][A]α¯={x:A(x)>0}¯{\rm{supp}}A=\overline{{\bigcup }_{\alpha \in (0,1]}{{[}A]}_{\alpha }}=\overline{\{x:A\left(x)\gt 0\}}, is compact.Definition 1.2[18] For α∈(0,1]\alpha \in \left(0,1]we define the α\alpha -cuts of fuzzy number AAwith [A]α={x∈R:A(x)≥α}.{{[}A]}_{\alpha }=\{x\in R:A\left(x)\ge \alpha \}.In particular, for α=0\alpha =0, the support of AAis defined as suppA=[A]0={x∈R∣A(x)>0}¯{\rm{supp}}A={{[}A]}_{0}=\overline{\{x\in R| A\left(x)\gt 0\}}.It is clear that [A]α{{[}A]}_{\alpha }is a closed interval. Fuzzy number AAis positive if min(suppA)>0\min \left({\rm{supp}}A)\gt 0.It is obvious that if AAis a positive real number, then AAis a positive fuzzy number and [A]α=[A,A],α∈[0,1]{{[}A]}_{\alpha }=\hspace{-0.00em}\left[A,A],\alpha \in \left[0,1]. In this case, we say AAis a trivial fuzzy number.Let Bi{B}_{i}(i=0,1,…,k,ki=0,1,\ldots ,k,kis a positive integer) be fuzzy numbers such that [Bi]α=[Bi,l,α,Bi,r,α],i=0,1,…,k,α∈(0,1],{{[}{B}_{i}]}_{\alpha }={[}{B}_{i,l,\alpha },{B}_{i,r,\alpha }],\hspace{1em}i=0,1,\ldots ,k,\hspace{1em}\alpha \in \left(0,1],and for any α∈(0,1]\alpha \in \left(0,1]Cl,α=max{Bi,l,α,i=0,1,…,k},Cr,α=max{Bi,r,α,i=0,1,…,k},{C}_{l,\alpha }={\rm{\max }}\left\{{B}_{i,l,\alpha },i=0,1,\ldots ,k\right\},{C}_{r,\alpha }={\rm{\max }}\left\{{B}_{i,r,\alpha },i=0,1,\ldots ,k\right\},then by ([19], Theorem 2.1), (Cl,α,Cr,α)\left({C}_{l,\alpha },{C}_{r,\alpha })determines a fuzzy number CCsuch that [C]α=[Cl,α,Cr,α],α∈(0,1].{{[}C]}_{\alpha }={[}{C}_{l,\alpha },{C}_{r,\alpha }],\alpha \in \left(0,1].According to [20] and ([21], Lemma 2.3), we can define C=max{Bi,i=0,1,…,k}.C={\rm{\max }}\left\{{B}_{i},i=0,1,\ldots ,k\right\}.Definition 1.3[20] Let A,BA,Bbe fuzzy numbers with [A]α=[Al,α,Ar,α]{{[}A]}_{\alpha }={[}{A}_{l,\alpha },{A}_{r,\alpha }], [B]α=[Bl,α,Br,α]{{[}B]}_{\alpha }={[}{B}_{l,\alpha },{B}_{r,\alpha }], α∈(0,1]\alpha \in (0,1]. The fuzzy number space norm is defined as follows: ∥A∥=supα∈(0,1]max{∣Al,α∣,∣Ar,α∣}.\parallel A\parallel =\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {A}_{l,\alpha }| ,| {A}_{r,\alpha }| \}.The distance between two arbitrary fuzzy numbers AAand BBis defined as follows: D(A,B)=supα∈(0,1]max{∣Al,α−Bl,α∣,∣Ar,α−Br,α∣}.D\left(A,B)=\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {A}_{l,\alpha }-{B}_{l,\alpha }| ,| {A}_{r,\alpha }-{B}_{r,\alpha }| \}.Definition 1.4[22] We say that xn{x}_{n}is a positive solution of (1) if xn{x}_{n}is a sequence of positive fuzzy numbers, which satisfies (1).We say that a sequence of positive fuzzy numbers xn{x}_{n}is persistent (resp. is bounded) if there exists a positive number MM(resp., NN) such that suppxn⊂[M,+∞),(resp.suppxn⊂(0,N]),n=1,2,….{\rm{supp}}{x}_{n}\subset {[}M,+\infty ),\hspace{0.33em}\left({\rm{resp.}}\hspace{0.33em}{\rm{supp}}{x}_{n}\subset (0,N]),\hspace{1em}n=1,2,\ldots .In addition, we say that xn{x}_{n}is bounded and persists if there exist numbers M,N∈(0,+∞)M,N\in \left(0,+\infty )such that suppxn⊂[M,N],n=1,2,….{\rm{supp}}{x}_{n}\subset {[}M,N],\hspace{0.33em}n=1,2,\ldots .Lemma 1.1[18] Let function f:R+×R+×R+→R+f:{R}^{+}\times {R}^{+}\times {R}^{+}\to {R}^{+}be continuous, A,B,CA,B,Cbe fuzzy numbers. Then[f(A,B,C)]α=f([A]α,[B]α,[C]α),α∈(0,1].{{[}f\left(A,B,C)]}_{\alpha }=f\left({{[}A]}_{\alpha },{{[}B]}_{\alpha },{{[}C]}_{\alpha }),\hspace{1em}\alpha \in (0,1].2Main resultsIn this section, we will discuss some dynamical characters of the fuzzy equation (1). We know that the existence of the positive solution is important for equation, which is the basis for discussing all of the dynamic characters. So, let us first prove that for fuzzy difference equation (1) there exists unique positive solution for any positive initial value.Theorem 2.1For any positive fuzzy numbers x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}, fuzzy difference equation (1), there exists a unique positive solution xn{x}_{n}whose initial value is x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}.ProofFor all positive fuzzy numbers x−k,x−k+1,…,x−1,x0,k∈{0,1,2…}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\ldots \}, suppose there exists a fuzzy number sequence that satisfies equation (1) whose initial value is x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}. Consider their α\alpha -cuts, α∈(0,1]\alpha \in \left(0,1], (2)[xn]α=[Ln,α,Rn,α],[A]α=[Al,α,Ar,α],[B]α=[Bl,α,Br,α].n=0,1,2….\left\{\begin{array}{l}{{[}{x}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }],\\ {{[}A]}_{\alpha }={[}{A}_{l,\alpha },{A}_{r,\alpha }],\\ {{[}B]}_{\alpha }={[}{B}_{l,\alpha },{B}_{r,\alpha }].\end{array}\right.\hspace{1em}n=0,1,2\ldots .Following (1), (2), and Lemma 1.1 in the literature [18], we have [xn+1]α=[Ln+1,α,Rn+1,α]=xnA+Bxn−kα=[xn]α[A]α+[B]α×[xn−k]α=[Ln,α,Rn,α][Al,α,Ar,α]+[Bl,α,Br,α]×[Ln−k,α,Rn−k,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ln−k,α,\begin{array}{rcl}\phantom{\rule[-1.5em]{}{0ex}}{{[}{x}_{n+1}]}_{\alpha }& =& {[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]={\left[\frac{{x}_{n}}{A+B{x}_{n-k}}\right]}_{\alpha }=\frac{{{[}{x}_{n}]}_{\alpha }}{{{[}A]}_{\alpha }+{{[}B]}_{\alpha }\times {{[}{x}_{n-k}]}_{\alpha }}\\ & =& \frac{{[}{L}_{n,\alpha },{R}_{n,\alpha }]}{{[}{A}_{l,\alpha },{A}_{r,\alpha }]+{[}{B}_{l,\alpha },{B}_{r,\alpha }]\times {[}{L}_{n-k,\alpha },{R}_{n-k,\alpha }]}\\ & =& \left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}\right],\end{array}where n=0,1,2,…,α∈(0,1]n=0,1,2,\ldots ,\alpha \in \left(0,1], so we obtain the related equation system (3)Ln+1,α=Ln,αAr,α+Br,α×Rn−k,α,Rn+1,α=Rn,αAl,α+Bl,α×Ln−k,α.\left\{\begin{array}{l}{L}_{n+1,\alpha }=\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\\ {R}_{n+1,\alpha }=\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}.\end{array}\right.\hspace{2.5em}Obviously, for any initial value (Li,α,Ri,α)\left({L}_{i,\alpha },{R}_{i,\alpha }), i=−k,−k+1,…,0i=-k,-k+1,\ldots ,0, α∈(0,1]\alpha \in (0,1], system (3), there exists a unique positive solution (Ln,α,Rn,α)\left({L}_{n,\alpha },{R}_{n,\alpha }), α∈(0,1]\alpha \in \left(0,1]. Now we prove that [Ln,α,Rn,α{L}_{n,\alpha },{R}_{n,\alpha }], α∈(0,1]\alpha \in (0,1]determines the solution xn{x}_{n}of (1) whose initial value is x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, where (Ln,α,Rn,α)\left({L}_{n,\alpha },{R}_{n,\alpha })is the positive solution of system (3) with initial value (Li,α,Ri,α)\left({L}_{i,\alpha },{R}_{i,\alpha }), i=−k,−k+1,…,0i=-k,-k+1,\ldots ,0, such that (4)[xn]α=[Ln,α,Rn,α],α∈(0,1],n=0,1,2,….{{[}{x}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }],\alpha \in (0,1],\hspace{1em}n=0,1,2,\ldots .According to the literature [23] and A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}are positive fuzzy numbers, for any α1,α2∈(0,1],α1≤α2{\alpha }_{1},{\alpha }_{2}\in (0,1],{\alpha }_{1}\le {\alpha }_{2}, we have (5)0<Al,α1≤Al,α2≤Ar,α2≤Ar,α1,0<Bl,α1≤Bl,α2≤Br,α2≤Br,α1,0<L−k,α1≤L−k,α2≤R−k,α2≤R−k,α1,0<L−k+1,α1≤L−k+1,α2≤R−k+1,α2≤R−k+1,α1,………………………………,0<L0,α1≤L0,α2≤R0,α2≤R0,α1.\left\{\begin{array}{l}0\lt {A}_{l,{\alpha }_{1}}\le {A}_{l,{\alpha }_{2}}\le {A}_{r,{\alpha }_{2}}\le {A}_{r,{\alpha }_{1}},\\ 0\lt {B}_{l,{\alpha }_{1}}\le {B}_{l,{\alpha }_{2}}\le {B}_{r,{\alpha }_{2}}\le {B}_{r,{\alpha }_{1}},\\ 0\lt {L}_{-k,{\alpha }_{1}}\le {L}_{-k,{\alpha }_{2}}\le {R}_{-k,{\alpha }_{2}}\le {R}_{-k,{\alpha }_{1}},\\ 0\lt {L}_{-k+1,{\alpha }_{1}}\le {L}_{-k+1,{\alpha }_{2}}\le {R}_{-k+1,{\alpha }_{2}}\le {R}_{-k+1,{\alpha }_{1}},\\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ,\\ 0\lt {L}_{0,{\alpha }_{1}}\le {L}_{0,{\alpha }_{2}}\le {R}_{0,{\alpha }_{2}}\le {R}_{0,{\alpha }_{1}}.\end{array}\right.\hspace{1.7em}By induction and (3), (5), we will show (6)Ln,α1≤Ln,α2≤Rn,α2≤Rn,α1,n=0,1,2,….{L}_{n,{\alpha }_{1}}\le {L}_{n,{\alpha }_{2}}\le {R}_{n,{\alpha }_{2}}\le {R}_{n,{\alpha }_{1}},\hspace{1em}n=0,1,2,\ldots .\hspace{1.2em}By (5), when n=0n=0, it is obvious (6) is true. When n=1n=1, since L1,α1=L0,α1Ar,α1+Br,α1×R−k,α1≤L0,α2Ar,α2+Br,α2×R−k,α2=L1,α2=L0,α2Ar,α2+Br,α2×R−k,α2≤R0,α2Al,α2+Bl,α2×L−k,α2=R1,α2=R0,α2Al,α2+Bl,α2×L−k,α2≤R0,α1Al,α1+Bl,α1×L−k,α1=R1,α1,\hspace{0.3em}\begin{array}{rcl}{L}_{1,{\alpha }_{1}}& =& \frac{{L}_{0,{\alpha }_{1}}}{{A}_{r,{\alpha }_{1}}+{B}_{r,{\alpha }_{1}}\times {R}_{-k,{\alpha }_{1}}}\le \frac{{L}_{0,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{-k,{\alpha }_{2}}}={L}_{1,{\alpha }_{2}}\\ & =& \frac{{L}_{0,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{-k,{\alpha }_{2}}}\le \frac{{R}_{0,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{-k,{\alpha }_{2}}}={R}_{1,{\alpha }_{2}}\\ & =& \frac{{R}_{0,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{-k,{\alpha }_{2}}}\le \frac{{R}_{0,{\alpha }_{1}}}{{A}_{l,{\alpha }_{1}}+{B}_{l,{\alpha }_{1}}\times {L}_{-k,{\alpha }_{1}}}={R}_{1,{\alpha }_{1}},\end{array}so L1,α1≤L1,α2≤R1,α2≤R1,α1{L}_{1,{\alpha }_{1}}\le {L}_{1,{\alpha }_{2}}\le {R}_{1,{\alpha }_{2}}\le {R}_{1,{\alpha }_{1}}. Suppose (6) is true when n≤k,k∈{1,2,…}n\le k,k\in \{1,2,\ldots \}, following (3) and (5), we have Lk+1,α1=Lk,α1Ar,α1+Br,α1×R0,α1≤Lk,α2Ar,α2+Br,α2×R0,α2=Lk+1,α2≤Rk,α2Al,α2+Bl,α2×L0,α2=Rk+1,α2≤Rk,α1Al,α1+Bl,α1×L0,α1=Rk+1,α1,\hspace{3.1em}\begin{array}{rcl}{L}_{k+1,{\alpha }_{1}}& =& \frac{{L}_{k,{\alpha }_{1}}}{{A}_{r,{\alpha }_{1}}+{B}_{r,{\alpha }_{1}}\times {R}_{0,{\alpha }_{1}}}\le \frac{{L}_{k,{\alpha }_{2}}}{{A}_{r,{\alpha }_{2}}+{B}_{r,{\alpha }_{2}}\times {R}_{0,{\alpha }_{2}}}={L}_{k+1,{\alpha }_{2}}\\ & \le & \frac{{R}_{k,{\alpha }_{2}}}{{A}_{l,{\alpha }_{2}}+{B}_{l,{\alpha }_{2}}\times {L}_{0,{\alpha }_{2}}}={R}_{k+1,{\alpha }_{2}}\le \frac{{R}_{k,{\alpha }_{1}}}{{A}_{l,{\alpha }_{1}}+{B}_{l,{\alpha }_{1}}\times {L}_{0,{\alpha }_{1}}}={R}_{k+1,{\alpha }_{1}},\end{array}so Lk+1,α1≤Lk+1,α2≤Rk+1,α2≤Rk+1,α1,k=1,2,…{L}_{k+1,{\alpha }_{1}}\le {L}_{k+1,{\alpha }_{2}}\le {R}_{k+1,{\alpha }_{2}}\le {R}_{k+1,{\alpha }_{1}},\hspace{0.33em}k=1,2,\ldots . By induction, (6) holds true.Following (3), we have (7)L1,α=L0,αAr,α+Br,α×R−k,α,R1,α=R0,αAl,α+Bl,α×L−k,α.\left\{\begin{array}{rcl}{L}_{1,\alpha }& =& \frac{{L}_{0,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{-k,\alpha }},\\ {R}_{1,\alpha }& =& \frac{{R}_{0,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{-k,\alpha }}.\end{array}\right.\hspace{8.5em}For A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2…}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\ldots \}are positive fuzzy numbers and Lemma 2.2 in the literature [24], we know Al,α,Ar,α,Bl,α,Br,α,L−k,α,R−k,α,…,L−1,α,R−1,α,L0,α,R0,α{A}_{l,\alpha },{A}_{r,\alpha },{B}_{l,\alpha },{B}_{r,\alpha },{L}_{-k,\alpha },{R}_{-k,\alpha },\ldots ,{L}_{-1,\alpha },{R}_{-1,\alpha },{L}_{0,\alpha },{R}_{0,\alpha }are left continuous, by (7) we have L1,α,R1,α{L}_{1,\alpha },{R}_{1,\alpha }are also left continuous. By induction, we have Ln,α,Rn,α(n=1,2,…){L}_{n,\alpha },{R}_{n,\alpha }\hspace{0.25em}\left(n=1,2,\ldots )are left continuous.We assert that the support of xn{x}_{n}, suppxn=⋃α∈(0,1][Ln,α,Rn,α]¯{\rm{supp}}{x}_{n}=\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}is compact. We just need to prove that ⋃α∈(0,1][Ln,α,Rn,α]{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]is bounded. When n=1n=1, since A,B,x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}A,B,{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}are positive fuzzy numbers, there exist constants MA,NA,MB,NB,Mi,Ni,i=−k,−k+1,…,−1,0{M}_{A},{N}_{A},{M}_{B},{N}_{B},{M}_{i},{N}_{i},i=-k,-k+1,\ldots ,-1,0, such that for all α∈(0,1]\alpha \in (0,1], we have (8)[Al,α,Ar,α]⊂[MA,NA],[Bl,α,Br,α]⊂[MB,NB],[Li,α,Ri,α]⊂[Mi,Ni].{[}{A}_{l,\alpha },{A}_{r,\alpha }]\subset {[}{M}_{A},{N}_{A}],{\rm{}}{[}{B}_{l,\alpha },{B}_{r,\alpha }]\subset {[}{M}_{B},{N}_{B}],{\rm{}}{[}{L}_{i,\alpha },{R}_{i,\alpha }]\subset {[}{M}_{i},{N}_{i}].Following systems (7) and (8) we obtain [L1,α,R1,α]⊂M0NA+NB×N−k,N0MA+MB×M−k{[}{L}_{1,\alpha },{R}_{1,\alpha }]\subset \left[\frac{{M}_{0}}{{N}_{A}+{N}_{B}\times {N}_{-k}},\frac{{N}_{0}}{{M}_{A}+{M}_{B}\times {M}_{-k}}\right](α∈(0,1])\left(\alpha \in (0,1]), so ⋃α∈(0,1][L1,α,R1,α]⊂M0NA+NB×N−k,N0MA+MB×M−k{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]\hspace{-0.08em}\subset \hspace{-0.08em}\left[\hspace{-0.16em},\frac{{M}_{0}}{{N}_{A}+{N}_{B}\times {N}_{-k}},\frac{{N}_{0}}{{M}_{A}+{M}_{B}\times {M}_{-k}}\right]. It implies that ⋃α∈(0,1][L1,α,R1,α]¯\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]}is compact, and ⋃α∈(0,1][L1,α,R1,α]¯⊂(0,+∞).\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{1,\alpha },{R}_{1,\alpha }]}\hspace{0.25em}\subset \hspace{-0.08em}\left(0,\hspace{-0.08em}+\infty ).By induction, we obtain that ⋃α∈(0,1][Ln,α,Rn,α]¯\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}is compact, and ⋃α∈(0,1][Ln,α,Rn,α]¯⊂(0,+∞)\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}\subset (0,+\infty )for every n=1,2,…n=1,2,\ldots . Following this and (6), and because the left continuous of Ln,α,Rn,α,n=1,2,…{L}_{n,\alpha },{R}_{n,\alpha },\hspace{0.33em}n=1,2,\ldots , [Ln,α,Rn,α]{[}{L}_{n,\alpha },{R}_{n,\alpha }]establishes a sequence of positive fuzzy numbers xn{x}_{n}such that (4) holds.Next, we show that xn{x}_{n}is the solution of (1) with any initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}. Since for all α∈(0,1]\alpha \in (0,1], [xn+1]α=[Ln+1,α,Rn+1,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ln−k,α=xnA+Bxn−kα.{{[}{x}_{n+1}]}_{\alpha }={[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]=\left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k,\alpha }}\right]={\left[\frac{{x}_{n}}{A+B{x}_{n-k}}\right]}_{\alpha }.So, xn{x}_{n}is the solution of (1) with any initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}.{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}.That is, the positive solution of fuzzy difference equation (1) exists.Next, we prove that the positive solution of fuzzy difference equation (1) is unique by contradiction. Suppose there exists another solution x¯n{\overline{x}}_{n}of (1) with initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2⋯}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2\cdots \}. Similar to the above proof, we can easily prove that [x¯n+1]α=[Ln+1,α,Rn+1,α]=Ln,αAr,α+Br,α×Rn−k,α,Rn,αAl,α+Bl,α×Ll−k,α=x¯nA+Bx¯n−kα,{{[}{\overline{x}}_{n+1}]}_{\alpha }={[}{L}_{n+1,\alpha },{R}_{n+1,\alpha }]=\left[\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k,\alpha }},\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{l-k,\alpha }}\right]={\left[\frac{{\overline{x}}_{n}}{A+B{\overline{x}}_{n-k}}\right]}_{\alpha },which implies [x¯n]α=[Ln,α,Rn,α]{{[}{\overline{x}}_{n}]}_{\alpha }={[}{L}_{n,\alpha },{R}_{n,\alpha }](n=0,1,2,…n=0,1,2,\ldots ) for any α∈(0,1]\alpha \in (0,1]. Then, by (4) we have [x¯n]α=[xn]α{{[}{\overline{x}}_{n}]}_{\alpha }={{[}{x}_{n}]}_{\alpha }(n=0,1,2,…n=0,1,2,\ldots ) for any α∈(0,1]\alpha \in (0,1]. So x¯n=xn,n=0,1,2,…{\overline{x}}_{n}={x}_{n},n=0,1,2,\ldots , which is contradictory. So the positive solution of fuzzy difference equation (1) is unique.In summary, for fuzzy difference equation (1), there exists a unique positive solution xn{x}_{n}, for every positive initial value x−k,x−k+1,…,x−1,x0,k∈{0,1,2,…}{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0},k\in \{0,1,2,\ldots \}.□We have proved that the fuzzy difference equation (1) has a unique positive solution, and then we will study the dynamic behavior of the solution. First, we discuss the boundedness of positive solutions of equation (1), which needs to rely on related equation system from (3).Lemma 2.1Consider the system of difference equations(9)yn+1=yna+bzn−k,zn+1=znc+dyn−k,n=0,1,2,…,\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}},\end{array}\right.\hspace{1em}n=0,1,2,\ldots ,where the initial value yi,zi(i=−k,−k+1,…,−1,0){y}_{i},{z}_{i}\hspace{1em}\left(i=-k,-k+1,\ldots ,-1,0)and parameters a,b,c,da,b,c,dare positive real numbers. If parameters a≥1a\ge 1and c≥1c\ge 1, then for all n≥kn\ge k, yn,zn{y}_{n},{z}_{n}are bounded and1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n=0,1,2,….\hspace{4.25em}\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots .ProofLet (yn,zn)({y}_{n},{z}_{n})be a positive solution of the difference equation system (9). From (9), the following are concluded by recursion, yn+1=yna+bzn−k≤1ayn≤1a2yn−1≤⋯≤1an+1y0,zn+1=znc+dyn−k≤1czn≤1c2zn−1≤⋯≤1cn+1z0,n=0,1,2,…,\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}}\le \frac{1}{a}{y}_{n}\le \frac{1}{{a}^{2}}{y}_{n-1}\le \cdots \le \frac{1}{{a}^{n+1}}{y}_{0},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}}\le \frac{1}{c}{z}_{n}\le \frac{1}{{c}^{2}}{z}_{n-1}\le \cdots \le \frac{1}{{c}^{n+1}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots ,yn+1=yna+bzn−k≥yna+bz0≥1(a+bz0)2yn−1≥⋯≥1(a+bz0)n+1−kyk,zn+1=znc+dyn−k≥znc+dy0≥1(c+dy0)2zn−1≥⋯≥1(c+dy0)n+1−kzk,n≥k.\left\{\begin{array}{l}{y}_{n+1}=\frac{{y}_{n}}{a+b{z}_{n-k}}\ge \frac{{y}_{n}}{a+b{z}_{0}}\ge \frac{1}{{\left(a+b{z}_{0})}^{2}}{y}_{n-1}\ge \cdots \ge \frac{1}{{\left(a+b{z}_{0})}^{n+1-k}}{y}_{k},\\ {z}_{n+1}=\frac{{z}_{n}}{c+d{y}_{n-k}}\ge \frac{{z}_{n}}{c+d{y}_{0}}\ge \frac{1}{{\left(c+d{y}_{0})}^{2}}{z}_{n-1}\ge \cdots \ge \frac{1}{{\left(c+d{y}_{0})}^{n+1-k}}{z}_{k},\end{array}\right.\hspace{1em}n\ge k.It is easy to deduce that yn,zn{y}_{n},{z}_{n}are bounded and 1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n≥k.□\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n\ge k.\square Theorem 2.2Consider fuzzy difference equation (1), for all α∈(0,1]\alpha \in (0,1], if(10)Al,α>1,{A}_{l,\alpha }\gt 1,then every positive solution xn{x}_{n}of fuzzy difference equation (1) is bounded.ProofLet xn{x}_{n}be a positive solution of (1) and satisfy (4). Because Al,α>1{A}_{l,\alpha }\gt 1, so Ar,α>1{A}_{r,\alpha }\gt 1. From (3) and Lemma 2.1, for any α∈(0,1]\alpha \in (0,1]we have 1(Ar,α+Br,α×R0,α)n−kLk,α≤Ln,α≤1Ar,αnL0,α,1(Al,α+Bl,α×L0,α)n−kRk,α≤Rn,α≤1Al,αnR0,α.n=0,1,2,….\hspace{6.65em}\left\{\begin{array}{r}\frac{1}{{\left({A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{0,\alpha })}^{n-k}}{L}_{k,\alpha }\le {L}_{n,\alpha }\le \frac{1}{{A}_{r,\alpha }^{n}}{L}_{0,\alpha },\\ \frac{1}{{\left({A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{0,\alpha })}^{n-k}}{R}_{k,\alpha }\le {R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }.\end{array}\right.\hspace{1em}n=0,1,2,\ldots .So, Rn,α≤1Al,αnR0,α{R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }and (11)Ln,α>0.{L}_{n,\alpha }\gt 0.We have Rn,α≤1Al,αnR0,α≤R0,α{R}_{n,\alpha }\le \frac{1}{{A}_{l,\alpha }^{n}}{R}_{0,\alpha }\le {R}_{0,\alpha }, which is because Al,α>1{A}_{l,\alpha }\gt 1. Then since xn{x}_{n}is a positive fuzzy number, there exists a constant J>0J\gt 0, such that for all α∈(0,1]\alpha \in \left(0,1], we have (12)R0,α≤J.{R}_{0,\alpha }\le J.From (11) and (12), we obtain that [Ln,α,Rn,α]⊂(0,J]{[}{L}_{n,\alpha },{R}_{n,\alpha }]\subset \left(0,J]for all α∈(0,1]\alpha \in (0,1], so ⋃α∈(0,1][Ln,α,Rn,α]⊂(0,J]{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]\subset \left(0,J], which implies that ⋃α∈(0,1][Ln,α,Rn,α]¯⊂(0,J]\overline{{\bigcup }_{\alpha \in (0,1]}{[}{L}_{n,\alpha },{R}_{n,\alpha }]}\subset \left(0,J]. Thus, the positive solution xn{x}_{n}of equation (1) is bounded.□Next, we consider the existence of equilibrium solution of (1). It is obvious that x¯=0\bar{x}=0is an equilibrium solution of (1).Let y¯,z¯\overline{y},\overline{z}be real numbers and satisfy equation systems y¯=y¯a+bz¯,z¯=z¯c+dy¯\overline{y}=\frac{\overline{y}}{a+b\overline{z}},\overline{z}=\frac{\overline{z}}{c+d\overline{y}}. Solving it we have y¯=1−cd,z¯=1−ab\overline{y}=\frac{1-c}{d},\overline{z}=\frac{1-a}{b}. It is easy to get the following lemma.Lemma 2.2Consider the difference equation system (9). Then (9) has positive equilibrium point (y¯,z¯)(\overline{y},\overline{z})and y¯=1−cd,z¯=1−ab\overline{y}=\frac{1-c}{d},\overline{z}=\frac{1-a}{b}when a<1,c<1a\lt 1,c\lt 1.Theorem 2.3For all α∈(0,1]\alpha \in (0,1], there is no positive equilibrium solution of fuzzy difference equation (1).ProofAssume there exists a positive fuzzy number xx, which satisfies x=xA+Bxx=\frac{x}{A+Bx}, [x]α=[Lα,Rα]{{[}x]}_{\alpha }={[}{L}_{\alpha },{R}_{\alpha }], and Lα,Rα≥0{L}_{\alpha },{R}_{\alpha }\ge 0, where α∈(0,1]\alpha \in (0,1]. So we derive a related equation system as follows: Lα=LαAr,α+Br,α×Rα,Rα=RαAl,α+Bl,α×Lα.{L}_{\alpha }=\frac{{L}_{\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{\alpha }},\hspace{1em}{R}_{\alpha }=\frac{{R}_{\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{\alpha }}.Suppose Lα≠0,Rα≠0{L}_{\alpha }\ne 0,{R}_{\alpha }\ne 0, following Lemma 2.2 we have Lα=1−Al,αBl,α,Rα=1−Ar,αBr,α{L}_{\alpha }=\frac{1-{A}_{l,\alpha }}{{B}_{l,\alpha }},{R}_{\alpha }=\frac{1-{A}_{r,\alpha }}{{B}_{r,\alpha }}, that is, Lα>Rα{L}_{\alpha }\gt {R}_{\alpha }. But [x]α=[Lα,Rα]{\left[x]}_{\alpha }=\left[{L}_{\alpha },{R}_{\alpha }]means Lα≤Rα{L}_{\alpha }\le {R}_{\alpha }, which is a contradiction. So it can only be Lα=Rα=0{L}_{\alpha }={R}_{\alpha }=0.□Therefore, there is no positive equilibrium solution of fuzzy difference equation (1), but have equilibrium solution x¯=0\bar{x}=0.Next, we discuss the positive solution’s convergence and asymptotic stability of fuzzy difference equation (1).Lemma 2.3Consider the difference equation system (9). Assume that a,c∈(1,+∞)a,c\in \left(1,+\infty ), then the solution of system (9) is convergent and limn→∞yn=0,limn→∞zn=0{\mathrm{lim}}_{n\to \infty }{y}_{n}=0,{\mathrm{lim}}_{n\to \infty }{z}_{n}=0.ProofFrom Lemma 2.1, we know 1(a+bz0)n−kyk≤yn≤1any0,1(c+dy0)n−kzk≤zn≤1cnz0,n=0,1,2,….\hspace{1.4em}\left\{\begin{array}{l}\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}\le {y}_{n}\le \frac{1}{{a}^{n}}{y}_{0},\\ \frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}\le {z}_{n}\le \frac{1}{{c}^{n}}{z}_{0},\end{array}\right.\hspace{1em}n=0,1,2,\ldots .If a,c∈(1,+∞)a,c\in \left(1,+\infty ), then limn→∞1(a+bz0)n−kyk=0,limn→∞1any0=0,limn→∞1(c+dy0)n−kzk=0,limn→∞1cnz0=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{\left(a+b{z}_{0})}^{n-k}}{y}_{k}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{a}^{n}}{y}_{0}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{\left(c+d{y}_{0})}^{n-k}}{z}_{k}=0,\hspace{1em}\mathop{\mathrm{lim}}\limits_{n\to \infty }\frac{1}{{c}^{n}}{z}_{0}=0.It is obvious that the solution of system (9) (yn,zn)({y}_{n},{z}_{n})is convergent and limn→∞yn=0,limn→∞zn=0{\mathrm{lim}}_{n\to \infty }{y}_{n}=0,{\mathrm{lim}}_{n\to \infty }{z}_{n}=0.□Theorem 2.4Consider fuzzy difference equation (1). Assume Al,α>1{A}_{l,\alpha }\gt 1, then every positive solution of (1) converges to equilibrium x¯=0\bar{x}=0when n→∞n\to \infty .ProofSince Ln+1,α=LnAr,α+Br,α×Rn−k,Rn+1,α=RnAl,α+Bl,α×Ln−k,{L}_{n+1,\alpha }=\frac{{L}_{n}}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-k}},\hspace{1em}{R}_{n+1,\alpha }=\frac{{R}_{n}}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-k}},following Lemma 2.3 we have limn→∞Ln,α=0,limn→∞Rn,α=0{\mathrm{lim}}_{n\to \infty }{L}_{n,\alpha }=0,{\mathrm{lim}}_{n\to \infty }{R}_{n,\alpha }=0, and according to the distance between two arbitrary fuzzy numbers (Definition 1.3 ([19])), we obtain limn→∞D(xn,x)=limn→∞D(xn,0)=limn→∞supα∈(0,1]max{∣Ln,α−0∣,∣Rn,α−0∣}=0.\mathop{\mathrm{lim}}\limits_{n\to \infty }D({x}_{n},x)=\mathop{\mathrm{lim}}\limits_{n\to \infty }D({x}_{n},0)=\mathop{\mathrm{lim}}\limits_{n\to \infty }\mathop{\sup }\limits_{\alpha \in (0,1]}\max \{| {L}_{n,\alpha }-0| ,| {R}_{n,\alpha }-0| \}=0.So limn→∞xn=x¯=0{\mathrm{lim}}_{n\to \infty }{x}_{n}=\bar{x}=0. That is every positive solution of equation (1) converges to equilibrium x¯=0\bar{x}=0when n→∞n\to \infty .□Lemma 2.4For the equilibrium point (0,0)\left(0,0)of equation system (9), if a,c∈(1,+∞)a,c\in \left(1,+\infty ), then the equilibrium point (0,0)\left(0,0)is locally asymptotically stable.ProofThe linearized equation of equation system (9) about the equilibrium point (0,0)\left(0,0)is (13)φn+1=Aφn,{\varphi }_{n+1}=A{\varphi }_{n},where φn=(yn,yn−1,…,yn−k,zn,zn−1,…,zn−k)T{\varphi }_{n}={({y}_{n},{y}_{n-1},\ldots ,{y}_{n-k},{z}_{n},{z}_{n-1},\ldots ,{z}_{n-k})}^{T}, and A=1/a0⋯0010⋯00⋯⋯⋯⋯⋯00⋯1000⋯0000⋯00⋯⋯⋯⋯⋯00⋯0000⋯0000⋯00⋯⋯⋯⋯⋯00⋯001/c0⋯0010⋯00⋯⋯⋯⋯⋯00⋯10.A=\left(\begin{array}{cc}\begin{array}{ccccc}1\hspace{-0.08em}\text{/}\hspace{-0.08em}a& 0& \cdots \hspace{0.33em}& 0& 0\\ 1& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 1& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 0& 0\end{array}& \begin{array}{ccccc}0& 0& \cdots \hspace{0.33em}& 0& 0\\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 0& 0\\ 1\hspace{-0.08em}\text{/}\hspace{-0.08em}c& 0& \cdots \hspace{0.33em}& 0& 0\\ 1& 0& \cdots \hspace{0.33em}& 0& 0\\ \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \hspace{0.33em}& \cdots \\ 0& 0& \cdots \hspace{0.33em}& 1& 0\end{array}\end{array}\right).The characteristic equation with equation (13) is P(λ)=(−λ)k−21a−λ1c−λP\left(\lambda )={\left(-\lambda )}^{k-2}\left(\frac{1}{a}-\lambda \right)\left(\frac{1}{c}-\lambda \right), and this shows that all eigenvalues are ∣λ∣<1| \lambda | \lt 1. Thus, the equilibrium (0,0) is locally asymptotically stable.□From Lemmas 2.3 and 2.4, we have that the equilibrium point (0,0)\left(0,0)of equation (9) is global asymptotically stable. Combined with Theorem 4, we can easily obtain the following theorem:Theorem 2.5Consider fuzzy difference equation (1). If Al,α>1{A}_{l,\alpha }\gt 1, then the equilibrium point x¯=0\bar{x}=0is global asymptotically stable.3Numerical exampleIn this section, we give numerical examples on the main conclusions of Section 2.Model. Take k=2k=2in fuzzy difference equation (1). Considering the following fuzzy difference equation: (14)xn+1=xnA+Bxn−2,n=3,4,5…,{x}_{n+1}=\frac{{x}_{n}}{A+B{x}_{n-2}},\hspace{1em}n=3,4,5\ldots ,in which parameters A,BA,Band initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are positive fuzzy numbers. According to Theorems 2.1–2.3, we know the following conclusions.For any positive solution xn{x}_{n}of (14), [xn]α=[Ln,α,Rn,α]{\left[{x}_{n}]}_{\alpha }=\left[{L}_{n,\alpha },{R}_{n,\alpha }], a difference equation system with a parameter α\alpha is obtained (15)Ln+1,α=Ln,αAr,α+Br,α×Rn−2,α,Rn+1,α=Rn,αAl,α+Bl,α×Ln−2,α,α∈(0,1],{L}_{n+1,\alpha }=\frac{{L}_{n,\alpha }}{{A}_{r,\alpha }+{B}_{r,\alpha }\times {R}_{n-2,\alpha }},\hspace{1em}\hspace{1em}{R}_{n+1,\alpha }=\frac{{R}_{n,\alpha }}{{A}_{l,\alpha }+{B}_{l,\alpha }\times {L}_{n-2,\alpha }},\hspace{1em}\alpha \in \left(0,1],and we know the unique equilibrium solution is x¯=0\bar{x}=0.Example 1Take A,BA,Band initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}as follows: A(t)=3t−7,73≤t≤83,−3t+9,83≤t≤3,B(t)=2t−5,52≤t≤3,−2t+7,3≤t≤72,x1(t)=t−4,4≤t≤5,−12t+72,5≤t≤7,x2(t)=12t−4,8≤t≤10,−12t+6,10≤t≤12,x3(t)=13t−2,6≤t≤9,−13t+4,9≤t≤12.\begin{array}{ll}A\left(t)=\left\{\begin{array}{ll}3t-7,& \frac{7}{3}\le t\le \frac{8}{3},\\ -3t+9,& \frac{8}{3}\le t\le 3,\end{array}\right.& B\left(t)=\left\{\begin{array}{ll}2t-5,& \frac{5}{2}\le t\le 3,\\ -2t+7,& 3\le t\le \frac{7}{2},\end{array}\right.\\ {x}_{1}\left(t)=\left\{\begin{array}{ll}t-4,& 4\le t\le 5,\\ -\frac{1}{2}t+\frac{7}{2},& 5\le t\le 7,\end{array}\right.& {x}_{2}\left(t)=\left\{\begin{array}{ll}\frac{1}{2}t-4,& 8\le t\le 10,\\ -\frac{1}{2}t+6,& 10\le t\le 12,\end{array}\right.\\ {x}_{3}\left(t)=\left\{\begin{array}{ll}\frac{1}{3}t-2,& 6\le t\le 9,\\ -\frac{1}{3}t+4,& 9\le t\le 12.\end{array}\right.& \end{array}\hspace{2.5em}From function A(x),B(x),x1(x),x2(x),x3(x)A\left(x),B\left(x),{x}_{1}\left(x),{x}_{2}\left(x),{x}_{3}\left(x), for all α∈(0,1]\alpha \in \left(0,1], we have [A]α=73+α3,3−α3,[B]α=52+α2,72−α2,[x1]α=[4+α,7−2α],[x2]α=[8+2α,12−2α],[x3]α=[6+3α,12−3α].\hspace{4.8em}\begin{array}{l}\phantom{\rule[-1.05em]{}{0ex}}{\left[A]}_{\alpha }=\left[\frac{7}{3}+\frac{\alpha }{3},3-\frac{\alpha }{3}\right],\hspace{1em}{\left[B]}_{\alpha }=\left[\frac{5}{2}+\frac{\alpha }{2},\frac{7}{2}-\frac{\alpha }{2}\right],\\ {\left[{x}_{1}]}_{\alpha }=\left[4+\alpha ,7-2\alpha ],\hspace{2em}{\left[{x}_{2}]}_{\alpha }=\left[8+2\alpha ,12-2\alpha ],\hspace{1em}{\left[{x}_{3}]}_{\alpha }=\left[6+3\alpha ,12-3\alpha ].\end{array}So the support of them are ⋃α∈(0,1][A]α¯=73,3,⋃α∈(0,1][B]α¯=52,72,⋃α∈(0,1][x0]α¯=[4,7],⋃α∈(0,1][x−1]α¯=[8,12],⋃α∈(0,1][x−2]α¯=[6,12].\begin{array}{l}\phantom{\rule[-1.25em]{}{0ex}}\overline{\bigcup _{\alpha \in (0,1]}{\left[A]}_{\alpha }}=\left[\frac{7}{3},3\right],\hspace{1em}\overline{\bigcup _{\alpha \in (0,1]}{\left[B]}_{\alpha }}=\left[\frac{5}{2},\frac{7}{2}\right],\\ \overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{0}]}_{\alpha }}=\left[4,7],\hspace{1.6em}\overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{-1}]}_{\alpha }}=\left[8,12],\hspace{1em}\overline{\bigcup _{\alpha \in (0,1]}{\left[{x}_{-2}]}_{\alpha }}=\left[6,12].\end{array}\hspace{0.7em}It is clear that Al,α>1{A}_{l,\alpha }\gt 1and (14) satisfies Theorems 2.1–2.5. So, every positive solution of (14) is bounded and persists, and converges to equilibrium x¯=0\bar{x}=0when n→+∞n\to +\infty . Moreover, the unique equilibrium x¯=0\bar{x}=0is global asymptotically stable (Figures 1–3).Figure 1In Example 1, the solution of system (15) when a=0a=0.Figure 2In Example 1, the solution of system (15) when a=0.5a=0.5.Figure 3In Example 1, the solution of system (15) when a=1a=1.Remark 1There exists positive fuzzy number A with Al,α<1{A}_{l,\alpha }\lt 1, such that the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable (see the following Examples 2–3).Example 2Let us still consider equation (14), whose parameter BBand initial value x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are the same as those in Example 1, but parameter AAchanges to (16)A(t)=8t−3,38≤t≤12,−8t+5,12≤t≤58.A\left(t)=\left\{\begin{array}{ll}8t-3,& \frac{3}{8}\le t\le \frac{1}{2},\\ -8t+5,& \frac{1}{2}\le t\le \frac{5}{8}.\end{array}\right.So, we have [A]α=38+α8,58−α8{\left[A]}_{\alpha }=\left[\frac{3}{8}+\frac{\alpha }{8},\frac{5}{8}-\frac{\alpha }{8}\right]and ⋃α∈(0,1][A]α¯=38,58\overline{{\bigcup }_{\alpha \in (0,1]}{\left[A]}_{\alpha }}=\left[\frac{3}{8},\frac{5}{8}\right]. It is clear that Al,α<1,AR,α<1{A}_{l,\alpha }\lt 1,{A}_{R,\alpha }\lt 1, so Example 2 does not satisfy the conditions of Theorems 2.2 and 2.4. Let us see the behaviors of the solution through the following graph (Figure 4).Figure 4In Example 2, the solution of system (15) when a=0.5a=0.5.In Figure 4, it is obvious that Ln{L}_{n}tends to 0 and Rn{R}_{n}tends to positive infinity when nnis sufficiently large. So, the solutions of Example 2 do not converge and the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable.Example 3Let us still consider equation (14), where the initial values x1,x2,x3{x}_{1},{x}_{2},{x}_{3}are the same as those in Example 1, but parameters A,BA,Bchange to (17)A(t)=5t−3,35≤t≤45,1,45≤t≤2,B(t)=5t−1,15≤t≤25,95−2t,25≤t≤910.A\left(t)=\left\{\begin{array}{ll}5t-3,& \frac{3}{5}\le t\le \frac{4}{5},\\ 1,& \frac{4}{5}\le t\le 2,\end{array}\right.\hspace{1em}B\left(t)=\left\{\begin{array}{ll}5t-1,& \frac{1}{5}\le t\le \frac{2}{5},\\ \frac{9}{5}-2t,& \frac{2}{5}\le t\le \frac{9}{10}.\end{array}\right.So we have [A]α=0.6+α5,2,⋃α∈(0,1][A]α¯=[0.6,2],[B]α=0.2+α5,0.9−α2{\left[A]}_{\alpha }=\left[\hspace{-0.08em},0.6\hspace{-0.08em}+\hspace{-0.08em}\frac{\alpha }{5},2,\hspace{-0.08em}\right],\overline{{\bigcup }_{\alpha \in \left(0,1]}{\left[A]}_{\alpha }}\hspace{-0.08em}=\hspace{-0.08em}\left[0.6,2],{\left[B]}_{\alpha }\hspace{-0.08em}=\hspace{-0.08em}\left[\hspace{-0.08em},0.2\hspace{-0.08em}+\hspace{-0.08em}\frac{\alpha }{5},0.9\hspace{-0.08em}-\hspace{-0.08em}\frac{\alpha }{2},\hspace{-0.08em}\right], and ⋃α∈(0,1][B]α¯=[0.2,0.9]\overline{{\bigcup }_{\alpha \in \left(0,1]}{\left[B]}_{\alpha }}\hspace{-0.08em}=\hspace{-0.08em}\left[0.2,0.9]. It is clear that Al,α<1,AR,α>1{A}_{l,\alpha }\lt 1,{A}_{R,\alpha }\gt 1, Example 3 does not satisfy the conditions of Theorems 2.2 and 2.4. Let us see the behaviors of the solution through the following graph (Figure 5).Figure 5In Example 3, the solution of system (15) when a=0.5a=0.5.In Figure 5, it is obvious that Ln{L}_{n}tends to 0 and Rn{R}_{n}tends to positive infinity when nnis sufficiently large. So, the solutions of Example 3 do not converge, the equilibrium point x¯=0\bar{x}=0of (1) is not global asymptotically stable.In [17], Zhang studied a first-order fuzzy difference equation xn+1=A+xnB+xn{x}_{n+1}=\frac{A+{x}_{n}}{B+{x}_{n}}. Later, he studied second-order exponential-type fuzzy difference equation xn+1=A+Be−xnC+xn−1{x}_{n+1}=\frac{A+B{e}^{-{x}_{n}}}{C+{x}_{n-1}}in [27]. Similarly, we have also studied kk-order exponential-type fuzzy difference equation which is related to equation (1):Remark 2We amend equation (1) to be exponential-type fuzzy difference equation xn+1=xne−xn−mA+Bxn−k,n=0,1,2,…,{x}_{n+1}=\frac{{x}_{n}{e}^{-{x}_{n-m}}}{A+B{x}_{n-k}},\hspace{1em}n=0,1,2,\ldots ,where m∈{0,1,…k}m\in \left\{0,1,\ldots k\right\}, parameters A,BA,Band initial value x−k,x−k+1,…,x−1,x0{x}_{-k},{x}_{-k+1},\ldots ,{x}_{-1},{x}_{0}, k∈{0,1,…}k\in \{0,1,\ldots \}are positive fuzzy numbers. We can obtain the same conclusion as equation (1) on the existence, boundedness, convergence, and global asymptotic stability of the solution.The proof method and process are similar to the conclusion proof of equation (1).Remark 3The existence and uniqueness of the positive solution is important for equation, which is the basis for discussing all of the dynamic characters of solutions. The commonly used method of proof is the definition method, and the proof of this article also uses the same method.4ConclusionIn this paper, we investigate the dynamical behaviors of positive solution of the fuzzy difference equation (1).First, we have proved that for the fuzzy difference equation (1) there exists a unique positive solution. Then, using the relationship between the fuzzy difference equation and the relative real number difference equation system, we have proved that if Al,α>1{A}_{l,\alpha }\gt 1then every positive solution of (1) is bounded and converges to equilibrium x=0x=0when n→∞n\to \infty . Moreover, we obtain that the equilibrium x=0x=0is global asymptotically stable. At last, we give numerical examples about equation (1) where k=2k=2draw trajectories of the solutions to reflect the global behaviors of fuzzy difference equation intuitively.

Journal

Open Mathematics – de Gruyter

Published: Jan 1, 2022

Keywords: fuzzy difference equation; solution; boundedness; convergence; 39A10; 65K10

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