Study of asymptotic behavior of solutions of neutral mixed type difference equations

In this paper, we consider a neutral mixed type difference equation, and obtain explicitly sufficient conditions for asymptotic behavior of solutions. A necessary condition is provided as well. An example is given to illustrate our main results.

In this paper, we consider the following mixed type neutral difference equation with an assumed initial condition where ψ : [m (t 0 ) , t 0 ] ∩ Z → R is a bounded sequence and for t 0 ≥ 0 Here ∆ denotes the forward difference operator ∆x(t) = x(t + 1) − x(t) for any sequence {x (t) , t ∈ Z + }. For more details on the calculus of difference equations, we refer the reader to [11] and [24]. Throughout this paper, we assume that a, b i and c j are bounded sequences, and τ, σ i and τ j are non-negative sequences such that Equation (1) can be viewed as a discrete analogue of the mixed type neutral differential equation; In [25], Bicer investigated (3) and obtained the asymptotic behavior of solutions. Our purpose here is to show the asymptotic behavior of solutions for (1). An asymptotic stability theorem with a necessary and sufficient condition is proved by using the contraction mapping theorem. For details on contraction mapping principle we refer the reader to [33] . An example is given to illustrate our main results.

Main results
Theorem 1. Let a, b i and c j non positive sequences. Assume that the following inequality has a nonnegative solution −a (t) λ (τ (t)) with λ (t) < 1. Then, (1) has a positive solution.

Theorem 2. Let a, b i and c j be non positive sequences and let
and x is a eventually positive solution of (1), then x(t) → ∞ as t → ∞.
From this, we can write which implies So, we get Theorem 3. Let a(t) > 0, b i and c j be nonnegative sequences and let ∆a(t) < 0, and x is a eventually positive solution of (1), then x(t) → 0 as t → ∞.
From this, we can write which implies So, we get Now, we investigate the asymptotic behavior of solutions of (1), free of the sign of the coefficients. During the process of inverting (1), an summation by parts will have to performed on the term involving ∆x(τ(t)).

Lemma 1.
A sequence x is a solution of (1)- (2) if and only if for t ≥ t 0 , where Proof. Since We can rewrite (1) as Multiplying both sides of (6) with t ∏ s=t 0 (1 − B(s)) −1 , by summing from t 0 to t − 1, we obtain By dividing both sides of the above expression by By performing an summation by parts, we get where h is given by (5). We obtain (4) by replacing (8) into (7). Since each step is reversible, the converse follows easily. This completes the proof.
Theorem 4. Assume that 0 < B (t) < 1 and the following conditions hold and Then for each initial condition (2), every solution of (1) converges to zero.
It is clear that for x ∈ M, φx is bounded. Now, we will show that φ is a contraction. Let x and y be two bounded sequences on [m (t 0 ) , ∞) ∩ Z and satisfying same initial condition (2). Then for t ≥ t 0 , we get Thus, the operator φ has a unique fixed point in M, which solves ( 1). Now, we will show that, (φx) (t) → 0 as t → ∞. Actually, for x ∈ M, we have Note that by (9), Moreover, since x(t) → 0 as t → ∞, for each ε > 0, there exists T 1 > t 0 such that u ≥ T 1 implies that |x(τ(u))| < ε 2 . Thus, for t ≥ T 1 , the third term I 3 in (12) satisfies Thus I 3 → 0 as t → ∞. By a similar technique, we can prove that the rest of terms in (12) tend zero as t → ∞. Therefore (φx)(t) → 0 as t → ∞. This completes the proof.
Proof. Suppose that (9) does not holds. That is, So, from (13), we can write δ = 0. Then, there exists a sequence {t n } approaching ∞, such that For x(t 0 ) = 0, let x be a solution. Then, From Lemma 1, x(t n ) satisfies (4). On the other hand, we know that Since all solutions tend zero, from (4), (14) and (15), we get which contradicts all solutions of (1) converge to zero. The proof is completed.
We end the paper with the following example.

Concluding remarks
In this article, a neutral mixed type difference equation is considered. The asymptotic behavior of solutions is obtained with a necessary and sufficient condition by using fixed point theorems. The results are supported with a suitable illustrative example.