Dynamical behaviors of a k-order fuzzy difference equation

Definition 1.1

[18] A function A : R → [ 0 , 1 ] is said to be a fuzzy number if it satisfies the following conditions (i)–(iv):

1. A is normal, i.e., there exists x ∈ R such that A ( x ) = 1 ;

2. A is fuzzy convex, i.e., for all t ∈ [ 0 , 1 ] and x 1 , x 2 ∈ R such that

   A ( t x 1 + ( 1 − t ) x 2 ) ≥ min { A ( x 1 ) , A ( x 2 ) } ;

3. A is upper semicontinuous;

4. The support of A , defined as supp A = ⋃ α ∈ ( 0 , 1 ] [ A ] α ¯ = { x : A ( x ) > 0 } ¯ , is compact.

Definition 1.2

[18] For α ∈ ( 0 , 1 ] we define the α -cuts of fuzzy number A with [ A ] α = { x ∈ R : A ( x ) ≥ α } . In particular, for α = 0 , the support of A is defined as supp A = [ A ] 0 = { x ∈ R ∣ A ( x ) > 0 } ¯ .

It is clear that [ A ] α is a closed interval. Fuzzy number A is positive if min ( supp A ) > 0 .

It is obvious that if A is a positive real number, then A is a positive fuzzy number and [ A ] α = [ A , A ] , α ∈ [ 0 , 1 ] . In this case, we say A is a trivial fuzzy number.

Let B i ( i = 0 , 1 , … , k , k is a positive integer) be fuzzy numbers such that

and for any α ∈ ( 0 , 1 ]

then by ([19], Theorem 2.1), ( C l , α , C r , α ) determines a fuzzy number C such that

According to [20] and ([21], Lemma 2.3), we can define

Definition 1.3

[20] Let A , B be fuzzy numbers with [ A ] α = [ A l , α , A r , α ] , [ B ] α = [ B l , α , B r , α ] , α ∈ ( 0 , 1 ] . The fuzzy number space norm is defined as follows:

The distance between two arbitrary fuzzy numbers A and B is defined as follows:

Definition 1.4

[22] We say that x n is a positive solution of (1) if x n is a sequence of positive fuzzy numbers, which satisfies (1).

We say that a sequence of positive fuzzy numbers x n is persistent (resp. is bounded) if there exists a positive number M (resp., N ) such that

In addition, we say that x n is bounded and persists if there exist numbers M , N ∈ ( 0 , + ∞ ) such that

Lemma 1.1

[18] Let function f : R + × R + × R + → R + be continuous, A , B , C be fuzzy numbers. Then

Theorem 2.1

For any positive fuzzy numbers x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } , fuzzy difference equation (1), there exists a unique positive solution x n whose initial value is x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } .

Proof

For all positive fuzzy numbers x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 … } , suppose there exists a fuzzy number sequence that satisfies equation (1) whose initial value is x − k , x − k + 1 , … , x − 1 , x 0 . Consider their α -cuts, α ∈ ( 0 , 1 ] ,

Following (1), (2), and Lemma 1.1 in the literature [18], we have

where n = 0 , 1 , 2 , … , α ∈ ( 0 , 1 ] , so we obtain the related equation system

Obviously, for any initial value ( L i , α , R i , α ) , i = − k , − k + 1 , … , 0 , α ∈ ( 0 , 1 ] , system (3), there exists a unique positive solution ( L n , α , R n , α ) , α ∈ ( 0 , 1 ] . Now we prove that [ L n , α , R n , α ], α ∈ ( 0 , 1 ] determines the solution x n of (1) whose initial value is x − k , x − k + 1 , … , x − 1 , x 0 , where ( L n , α , R n , α ) is the positive solution of system (3) with initial value ( L i , α , R i , α ) , i = − k , − k + 1 , … , 0 , such that

According to the literature [23] and A , B , x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } are positive fuzzy numbers, for any α 1 , α 2 ∈ ( 0 , 1 ] , α 1 ≤ α 2 , we have

By induction and (3), (5), we will show

By (5), when n = 0 , it is obvious (6) is true. When n = 1 , since

so L 1 , α 1 ≤ L 1 , α 2 ≤ R 1 , α 2 ≤ R 1 , α 1 . Suppose (6) is true when n ≤ k , k ∈ { 1 , 2 , … } , following (3) and (5), we have

so L k + 1 , α 1 ≤ L k + 1 , α 2 ≤ R k + 1 , α 2 ≤ R k + 1 , α 1 , k = 1 , 2 , … . By induction, (6) holds true.

Following (3), we have

For A , B , x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 … } are positive fuzzy numbers and Lemma 2.2 in the literature [24], we know A l , α , A r , α , B l , α , B r , α , L − k , α , R − k , α , … , L − 1 , α , R − 1 , α , L 0 , α , R 0 , α are left continuous, by (7) we have L 1 , α , R 1 , α are also left continuous. By induction, we have L n , α , R n , α ( n = 1 , 2 , … ) are left continuous.

We assert that the support of x n , supp x n = ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] ¯ is compact. We just need to prove that ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] is bounded. When n = 1 , since A , B , x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } are positive fuzzy numbers, there exist constants M A , N A , M B , N B , M i , N i , i = − k , − k + 1 , … , − 1 , 0 , such that for all α ∈ ( 0 , 1 ] , we have

Following systems (7) and (8) we obtain [ L 1 , α , R 1 , α ] ⊂ M 0 N A + N B × N − k , N 0 M A + M B × M − k ( α ∈ ( 0 , 1 ] ) , so ⋃ α ∈ ( 0 , 1 ] [ L 1 , α , R 1 , α ] ⊂ M 0 N A + N B × N − k , N 0 M A + M B × M − k . It implies that ⋃ α ∈ ( 0 , 1 ] [ L 1 , α , R 1 , α ] ¯ is compact, and ⋃ α ∈ ( 0 , 1 ] [ L 1 , α , R 1 , α ] ¯ ⊂ ( 0 , + ∞ ) . By induction, we obtain that ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] ¯ is compact, and ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] ¯ ⊂ ( 0 , + ∞ ) for every n = 1 , 2 , … . Following this and (6), and because the left continuous of L n , α , R n , α , n = 1 , 2 , … , [ L n , α , R n , α ] establishes a sequence of positive fuzzy numbers x n such that (4) holds.

Next, we show that x n is the solution of (1) with any initial value x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } . Since for all α ∈ ( 0 , 1 ] ,

So, x n is the solution of (1) with any initial value x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } . 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 of (1) with initial value x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 ⋯ } . Similar to the above proof, we can easily prove that

which implies [ x ¯ n ] α = [ L n , α , R n , α ] ( n = 0 , 1 , 2 , … ) for any α ∈ ( 0 , 1 ] . Then, by (4) we have [ x ¯ n ] α = [ x n ] α ( n = 0 , 1 , 2 , … ) for any α ∈ ( 0 , 1 ] . So x ¯ n = x n , n = 0 , 1 , 2 , … , 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 x n , for every positive initial value x − k , x − k + 1 , … , x − 1 , x 0 , k ∈ { 0 , 1 , 2 , … } .□

Lemma 2.1

Consider the system of difference equations

where the initial value y i , z i ( i = − k , − k + 1 , … , − 1 , 0 ) and parameters a , b , c , d are positive real numbers. If parameters a ≥ 1 and c ≥ 1 , then for all n ≥ k , y n , z n are bounded and

Proof

Let ( y n , z n ) be a positive solution of the difference equation system (9). From (9), the following are concluded by recursion,

It is easy to deduce that y n , z n are bounded and

Theorem 2.2

Consider fuzzy difference equation (1), for all α ∈ ( 0 , 1 ] , if

then every positive solution x n of fuzzy difference equation (1) is bounded.

Proof

Let x n be a positive solution of (1) and satisfy (4). Because A l , α > 1 , so A r , α > 1 . From (3) and Lemma 2.1, for any α ∈ ( 0 , 1 ] we have

So, R n , α ≤ 1 A l , α n R 0 , α and

We have R n , α ≤ 1 A l , α n R 0 , α ≤ R 0 , α , which is because A l , α > 1 . Then since x n is a positive fuzzy number, there exists a constant J > 0 , such that for all α ∈ ( 0 , 1 ] , we have

From (11) and (12), we obtain that [ L n , α , R n , α ] ⊂ ( 0 , J ] for all α ∈ ( 0 , 1 ] , so ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] ⊂ ( 0 , J ] , which implies that ⋃ α ∈ ( 0 , 1 ] [ L n , α , R n , α ] ¯ ⊂ ( 0 , J ] . Thus, the positive solution x n of equation (1) is bounded.□

Lemma 2.2

Consider the difference equation system (9). Then (9) has positive equilibrium point ( y ¯ , z ¯ ) and y ¯ = 1 − c d , z ¯ = 1 − a b when a < 1 , c < 1 .

Theorem 2.3

For all α ∈ ( 0 , 1 ] , there is no positive equilibrium solution of fuzzy difference equation (1).

Proof

Assume there exists a positive fuzzy number x , which satisfies x = x A + B x , [ x ] α = [ L α , R α ] , and L α , R α ≥ 0 , where α ∈ ( 0 , 1 ] . So we derive a related equation system as follows:

Suppose L α ≠ 0 , R α ≠ 0 , following Lemma 2.2 we have L α = 1 − A l , α B l , α , R α = 1 − A r , α B r , α , that is, L α > R α . But [ x ] α = [ L α , R α ] means L α ≤ R α , which is a contradiction. So it can only be L α = R α = 0 .□

Lemma 2.3

Consider the difference equation system (9). Assume that a , c ∈ ( 1 , + ∞ ) , then the solution of system (9) is convergent and lim n → ∞ y n = 0 , lim n → ∞ z n = 0 .

Proof

From Lemma 2.1, we know