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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(1): 163-175

Published online March 31, 2020

### Stability Criterion for Volterra Type Delay Difference Equations Including a Generalized Difference Operator

Murat Gevgesoglu∗ and Yasar Bolat

Department of Mathematics, Kastamonu University, Kastamonu, Turkey
e-mail : mgevgesoglu@kastamonu.edu.tr and ybolat@kastamonu.edu.tr

Received: January 17, 2018; Revised: May 10, 2019; Accepted: November 18, 2019

### Abstract

The stability of a class of Volterra-type difference equations that include a generalized difference operator Δa is investigated using Krasnoselskii’s fixed point theorem and some results are obtained. In addition, some examples are given to illustrate our theoretical results.

Keywords: stability, Volterra diﬀ,erence equations.

### 1. Introduction

Difference equations are the discrete analogues of differential equations and they usually describe certain phenomena over the course of time. Difference equations have many applications in a wide variety of disciplines, such as economics, mathematical biology, social sciences and physics. We refer to [1, 2, 4, 6] for the basic theory and some applications of difference equations. Volterra difference equations are extensively used to model phenomena in engineering, economics, and in the natural and social sciences; their stability has been studied by many authors.

In , Khandaker and Raffoul considered a Volterra discrete system with nonlinear perturbation

$x(n+1)=A(n)x(n)+∑s=0nB(n,s)x(s)+g(n,x(n))$

and obtained necessary and sufficient conditions for stability properties of the zero solution employing the resolvent equation coupled with a variation of parameters formula.

In , Migda et al. investigated the boundedness and asymptotic stability of the zero solution of the discrete Volterra equation

$x(n+1)=a(n)+b(n)x(n)+∑i=n0nK(n,i)x(i)$

using fixed point theory.

In , Islam and Yankson studied the stability and boundedness of the nonlinear difference equation

$x(t+1)=a(t)x(t)+c(t)Δx(t-g(t))+q(x(t),x(t-g(t)))$

using fixed point theorems.

In , Yankson studied the asymptotic stability of the zero solution of the Volterra difference delay equation

$x(n+1)=a(n)x(n)+c(n)Δx(n-g(n))+∑s=n-g(n)n-1k(n,s)h(x(s))$

using Krasnoselskii’s fixed point theorem.

In this paper, motivated by , we investigate the asymptotic stability of the zero solution of neutral and Volterra type difference equations which include a generalized difference operator of the form

$Δa[x(n)-b(n)x(n-σ)]=c(n)x(n)+∑u=n-σn-1k(u,n)h(x(u),x(u-τ))$

using Krasnoselskii’s fixed point theorem. Here b(n) : ℤ → ℝ and c(n) : ℤ → ℝ are discrete bounded functions, k(u, n) : ℤ × ℤ → ℝ+, h : ℝ × ℝ → ℝ, σ and τ are non-negative integers with lim(nσ) = ∞ and lim(nτ) = ∞.

The difference operator Δ and generalized difference operator Δa are defined as

$Δx(n)=x(n+1)-x(n)$

and

$Δa(x)(n)=x(n+1)-ax(n),a>0$

respectively.

We assume that h(0, 0) = 0 and

$∣h(x1,y1)-h(x2,y2)∣≤K max{∣x1-x2∣,∣y1-y2∣}$

for some positive constant K.

### 2. Basic Definitions, Theorems and Lemmas

For any integer n0 ≥ 0 we define Z0 as the set of all integers in the interval [−στ, n0]. Let ω : Z0 → ℝ be a discrete and bounded initial function.

### Definition 2.1

x(n) = x(n, n0, ω) is a solution of (1.1) if x(n) = ω(n) for nZ0 and satisfies (1.1) for nn0.

### Definition 2.2

The zero solution of (1.1) is stable if for any ɛ > 0 and any integer n0 ≥ 0 there exists a δ = δ (ɛ) such that |ω(n)| < δ for nZ0 implies |x(n, n0, ω)| < ɛ for nn0.

### Definition 2.3

The zero solution of (1.1) is asymptotically stable if it is stable and for any integer n0 ≥ 0 there exists a δ = δ (n0) such that |ω(n)| < δ for nZ0 implies $limn→∞x(n)=0$.

### Lemma 2.1

Where the generalized difference operator Δa is as defined in(1.2), we have

$Δax(n)=an+1Δ(x(n)an).$
Proof

It is obvious.

Now below we state Krasnoselskii’s theorem. For the proof we refer to .

### Theorem 2.1

Let M be a closed convex nonempty subset of a Banach space (B, ||.||). Suppose that A and Q map M into B such that

• x, yM implies Ax + QyM,

• A is continuous and AM is contained in a compact set,

• Q is a contraction mapping.

Then, there exits zM with z = Az + Qz.

### Theorem 2.2.(Ascoli-Arzela Theorem)

Let (X, d) be a compact metric space and C(X) be a vector space consisting of all continuous function f : X → ℝ. A subset F of C(X) is relatively compact if and only if F is equibounded and equicontinuous.

### Lemma 3.1

Assume that (a + c(n)) ≠ 0 for all n ∈ ℤ. Necessary and sufficient condition for x (n) to be the solution of(1.1)are

$x(n)=(x(n0)-b(n0)x(n0-σ))∏u=n0n-1(a+c(u))+b(n)x(n-σ)+∑r=n0n-1[c (r) b (r) x (r-σ)+∑u=r-σr-1k (u,r) h (x (u),x (u-τ))]∏s=r+1n-1(a+c(s)),n≥n0.$
Proof

From (1.1) we can write

$Δax (n)-c (n) x (n)=Δa(b (n) x (n-σ))+∑u=n-σn-1k(u,n) h (x (u),x (u-τ)).$

Using the definition of the operator Δa in the left-hand side of (3.1) and multiplying both sides of (3.1) with $∏s=n0n(a+c(s))-1$ we have

$Δ(x (n)∏s=n0n-1(a+c(s))-1)=[Δa (b (n) x (n-σ))+∑u=n-σn-1k(u,n) h (x (u),x (u-τ))]∏s=n0n(a+c(s))-1.$

By summing both sides of (3.2) from n0 to n − 1, we obtain

$x (n)∏s=n0n-1(a+c(s))-1=x(n0)+∑r=n0n-1[Δa(b (n) x (n-σ))+∑u=n-σn-1k(u,n)h (x (u),x (u-τ))]∏s=n0r(a+c(s))-1$

from this last equality, we write

$x(n)=x(n0)∏s=n0n-1(a+c(s))+{∑r=n0n-1[Δa(b (r) x (r-σ))+∑u=r-σr-1k(u,r)h (x (u),x (u-τ))]∏s=n0r(a+c(s))-1}∏s=n0n-1(a+c(s)).$

Because

$∏s=n0r(a+c(s))-1∏s=n0n-1(a+c(s))=∏s=r+1n-1(a+c(s)),$

we can write

$x(n)=x (n0)∏s=n0n-1(a+c(s))+∑r=n0n-1[Δa(b (r) x (r-σ))+∑u=r-σr-1k(u,r)h (x (u),x (u-τ))]∏s=r+1n-1(a+c(s))$

or

$x(n)=x (n0)∏s=n0n-1(a+c(s))+∑r=n0n-1Δa(b (r) x (r-σ))∏s=r+1n-1(a+c(s))+∑r=n0n-1[∑u=r-σr-1k(u,r)h (x (u),x (u-τ))]∏s=r+1n-1(a+c(s)).$

Now, using Lemma 2.1 in the second term on the right-hand side of (3.3), we have

$∑r=n0n-1Δa(b (r) x (r-σ))∏s=r+1n-1(a+c(s))=∑r=n0n-1ar+1Δ(b (r) x (r-σ)ar)∏s=r+1n-1(a+c(s))=∑r=n0n-1[Δ(b (r) x (r-σ)∏s=rn-1(a+c(s)))-Δ(∏s=rn-1(a+c(s)) ar)b (r) x (r-σ)ar]=∣b (r) x (r-σ)∏s=rn-1(a+c(s))∣r=n0r=n-∑r=n0n-1[Δ(∏s=rn-1(a+c(s))ar)b(r) x (r-σ)ar]=b (n) x (n-σ)-b (n0) x (n0-σ)∏s=n0n-1(a+c(s))-∑r=n0n-1[Δ(∏s=rn-1(a+c(s)) ar)b (r) x (r-σ)ar].$

Hence, by putting this last equality in (3.3), we reach

$x(n)=x (n0)∏s=n0n-1(a+c(s))+∑r=n0n-1[∑u=r-σr-1k(u,r)h (x (u),x (u-τ))]∏s=r+1n-1(a+c(s))+b (n) x (n-σ)-b (n0) x (n0-σ)∏s=n0n-1(a+c(s))-∑r=n0n-1[Δ(∏s=rn-1(a+c(s))ar)b(r) x (r-σ)ar].$

Because in the last term on the right-hand side of (3.4)

$Δ(∏s=rn-1(a+c(s)) ar)=∏s=r+1n-1(a+c(s)) ar+1-∏s=rn-1(a+c(s)) ar=-c(r)∏s=r+1n-1(a+c(s)) ar,$

from (3.4) we obtain

$x (n)=[x (n0)-b (n0) x (n0-σ)]∏s=n0n-1(a+c(s))+b (n) x (n-σ)+∑r=n0n-1[c(r)b(r)x(r-σ)+∑u=r-σr-1k(u,r)h (x (u),x (u-τ))]∏s=r+1n-1(a+c(s)), n≥n0.$

This completes the proof.

Now let φ(n) be a real sequence defined on ℤ and define the set S as

$S={φ:ℤ→ℝ∣ ‖φ‖→0, n→∞}$

where

$‖φ‖=max∣φ(n)∣,n∈ℤ.$

Then, we can see that (S, ||.||) is a Banach space. We then define the mapping H : SS on Z0 by

$(Hφ) (n)=ω (n)$

and for nn0 by

$(Hφ) (n)=[ω (n0)-b (n0) ω (n0-σ)]∏s=n0n-1(a+c(s))+b(n) φ (n-σ)+∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s)).$

### Lemma 3.2

Let(1.3)hold. Suppose that

$∏s=n0n-1(a+c(s))→0 as n→∞$

and there exists α ∈ (0, 1) such that for nn0

$∑r=n0n-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣≤α.$

The mapping H defined by(3.5)approaches 0 as n→∞.

Proof

Due to the condition (3.6) the first term of right-hand side of equation (3.5) approaches to zero as n→∞. Because b(n) is bounded and φS is also the second term of right-hand side of equation (3.5) approaches to zero as n → ∞. Now, we show that the last term on the right-hand side of equation (3.5) approaches to zero as n→∞.

Given ɛ1 > 0 and let n1 be a positive integer such that for n > n1 and φS, |φ(nσ)| < ɛ1. Because φ(nσ) → 0, for given ɛ2 > 0 we can find a n2 > n1 such that for n > n2 |φ(nσ)| < ɛ2. Furthermore, because of condition (3.6) we can find a n3 > n2 such that for $n>n3∣∏s=n2n-1(a+c(s))∣<ɛ2αɛ1$.

Hence, for n > n3 from the last term of right-hand side of (3.5) we have

$∣∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ(u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣∣≤∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ(u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣∣≤∑r=n0n2-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ(u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣∣+∑r=n2n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ(u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣≤ɛ1∑r=n0n2-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣+ɛ2∑r=n0n-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣=ɛ1∑r=n0n2-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣+ɛ2α=ɛ1∑r=n0n2-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n2-1(a+c(s))∣ ∣∏s=n2n-1(a+c(s))∣+ɛ2α≤ɛ1α∣∏s=n2n-1(a+c(s))∣+ɛ2α≤ɛ2(1+α).$

This completes the proof.

To use Krasnoselskii’s theorem, we construct two mappings Q and A expressing (3.5) as

$(Hφ) (n)=(Qφ) (n)+(Aφ) (n)$

where Q, A : SS are mappings with

$(Qφ) (n)=[ω (n0)-b (n0) ω (n0-σ)]∏s=n0n-1(a+c(s))+b(n) φ (n-σ)$

and

$(Aφ) (n)=∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s))$

respectively.

### Lemma 3.3

Assume that(1.3), (3.6) and(3.7)hold and suppose that there exists a positive constant ξ such that

$a+c(n)≤1 and maxn∈ℤ∣a+c(n)∣=ξ$

Then, the mapping A defined by(3.9)is continuous and compact.

Proof

First, we show that the mapping A defined by (3.9) is continuous. Let φ, φ̄S. For a given ɛ > 0 choose $δ=ɛα$ such that ||φφ̄|| < δ holds. Then, we have

$‖(Aφ)-(Aφ¯)‖=maxn∈ℤ∣{∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ(u),φ (u-τ))]∏s=r+1n-1(a+c(s))}-{∑r=n0n-1[c(r)b(r)φ¯(r-σ)+∑u=r-σr-1k(u,r)h (φ¯ (u),φ¯ (u-τ))]∏s=r+1n-1(a+c(s))}∣≤∑r=n0n-1[∣c(r)b(r)∣ ∣φ(r-σ)-φ¯(r-σ)∣] ∣∏s=r+1n-1(a+c(s))∣+∑r=n0n-1∣∑u=r-σr-1k(u,r)h (φ (u),φ (u-τ))-∑u=r-σr-1k(u,r)h (φ¯ (u),φ¯ (u-τ))∣ ∣∏s=r+1n-1(a+c(s))∣≤∑r=n0n-1∣c(r)b(r)∣ ∣∏s=r+1n-1(a+c(s))∣ ‖φ-φ¯‖+∑r=n0n-1∣∑u=r-σr-1k(u,r) [h (φ (u),φ (u-τ))-h (φ¯ (u),φ¯ (u-τ))]∣ ∣∏s=r+1n-1(a+c(s))∣≤∑r=n0n-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣ ‖φ-φ¯‖≤α‖φ-φ¯‖≤ɛ$

which shows that the mapping A is continuous. Now we show that A is compact. For this we use Arzela-Ascoli theorem. Let {φn} ⊂ S be a sequence of uniformly bounded functions where ||φn|| ≤ m for m > 0 and n is a positive integer. Then using (1.3) we have

$‖Aφn‖=maxn∈ℤ∣∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣≤∑r=n0n-1∣[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r) h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣≤∑r=n0n-1∣[∣c(r)b(r)∣+L∑u=r-σr-1k(u,r)]∣ ∣∏s=r+1n-1(a+c(s))∣ ‖φ‖≤α‖φ‖≤αm$

which shows that (n) is uniformly bounded. Furthermore,

$‖Δ (Aφ)‖=maxn∈ℤ∣(Aφ) (n+1)-(Aφ) (n)∣≤∣a+c(n)∣ ∣c(n)b(n)φ(n-σ)+∑u=n-σn-1k(u,r)h (φ(u),φ (u-τ))∣ ∣∏s=r+1n-1(a+c(s))∣≤ξ(∣c(n)b(n)∣+K∑u=n-σn-1k(u,r)) ‖φ‖≤ξαm≤γ$

for some positive constant γ. This shows that (n) is equi-continuous. Hence, by Arzela-Ascoli’s theorem, the mapping A is compact.

### Lemma 3.4

Consider the mapping Q defined by(3.8)and assume that

$∣b(n)∣≤μ<1$

holds for some positive constant μ. Then, Q is a contraction.

Proof

Take any two functions φ, φ̄S. We then have

$‖(Qφ)-(Qφ¯)‖=maxn∈ℤ∣[ω (n0)-b (n0) ω (n0-σ)]∏s=n0n-1(a+c(s))+b (n) φ (n-σ)-[ω (n0)-b (n0) ω (n0-σ)]∏s=n0n-1(a+c(s))-b (n) φ (n-σ)∣≤∣b(n)∣ ‖φ-φ¯‖≤μ‖φ-φ¯‖$

which shows that Q is a contraction mapping.

### Theorem 3.1

Suppose that(1.3), (3.6), (3.7), (3.10) and(3.11)hold. Also suppose that there exists positive constants c and β ∈ (0, 1) such that

$∣∏s=n0n-1(a+c(s))∣≤c$

and

$∣b(n)∣+∑r=n0n-1[∣c(r)b(r)∣+K∑u=r-σr-1k(u,r)] ∣∏s=r+1n-1(a+c(s))∣≤β, n≥n0$

hold. Then, the zero solution of(1.1)is asymptotically stable.

Proof

Given ɛ > 0. Choose δ such that

$∣1-b(n0)∣δc<ɛ(1-β)$

Let ω be a given initial function such that |ω(n)| < δ. Let us define the set M as

$M={φ∈S:‖φ‖<ɛ}$

and take any φ, ϕM. Then, we have

$‖(Qϕ)+(Aφ)‖=maxn∈ℤ∣[ω (n0)-b (n0) ω (n0-σ)]∏u=n0n-1(a+c(u))+b (n) ϕ (n-σ)+∑r=n0n-1[c(r)b(r)φ(r-σ)+∑u=r-σr-1k(u,r)h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣≤∣[ω (n0)-b (n0) ω (n0-σ)]∏u=n0n-1(a+c (u))∣+∣b (n) ϕ (n-σ)∣+∑r=n0n-1∣[c(r) b (r) φ (r-σ)+∑u=r-σr-1k(u,r) h (φ (u),φ (u-τ))]∏s=r+1n-1(a+c(s))∣≤∣1-b(n0)∣ δc+∣b(n)∣ɛ+ɛ∑r=n0n-1[∣c (r) b (r)∣+K∑u=r-σr-1k (u,r)] ∣∏s=r+1n-1(a+c(s))∣≤∣1-b(n0)∣ δc+{∣b(n)∣+∑r=n0n-1[∣c (r) b (r)∣+K∑u=r-σr-1k (u,r)] ∣∏s=r+1n-1(a+c(s))∣}ɛ≤∣1-b(n0)∣ δc+βɛ<ɛ$

which shows that () + () ∈ M.

By the last result, Lemma 4 and Lemma 5 all conditions of Theorem 1 are satisfied onM. Consequently, there exits a fixed point xM such that x = Qx+Ax holds. Lemma 2 implies that this fixed point x(n) is a solution of (1.1). Furthermore the solution x(n) is stable because ||x|| < ɛ for a given ɛ > 0. By Lemma 3 the solution x(n) is asymptotically stable.

### Example 3.1

Consider the difference equation

$Δ2[x (n)-132 (n+1)!x(n-2)]=-2nn+1x(n)+∑u=n-2n-12n16 (n+1)! (u2+4)h (x (u),x (u-3)),n≥1$

Here,

$a=σ=2, τ=3, n0=1,c(n)=-2nn+1, b(n)=132 (n+1)!,K(u,n)=2n16 (n+1)! (u2+4).$

We see that

$∏s=1n-1(2-2nn+1)=2n-1n!→0 as n→∞,$

so (3.6) holds. Because

$∑r=1n-1[2rr+1132 (n+1)!+∑u=r-2r-12r16 (r+1)! (u2+4)] ∣∏s=r+1n-1(2s+1)∣≤316<1,$

(3.7) holds. Because

$a+c(n)=2-2nn+1≤1 and maxn∈ℤ∣a+c(n)∣=maxn∈ℤ∣2-2nn+1∣=1$

(3.10) holds. Because

$∣132 (n+1)!∣≤132<1,$

(3.11) holds. Because

$∣∏s=1n-1(2-2ss+1))∣=∣∏s=1n-1(2s+1))∣≤1,$

(3.12) holds. Also, because

$∣132 (n+1)!∣+∑r=1n-1[2rr+1132 (n+1)!+∑u=r-2r-12r16 (r+1)! (u2+4)] ∣∏s=r+1n-1(2s+1)∣≤1364<1,$

(3.13) holds. So, by Theorem 3 the zero solution of (3.14) is asymptotically stable.

The solution is of the form

$x(n)=(x(1)-164x (-1))∏u=1n-1(2-2uu+1)+132 (n+1)!x (n-2)+∑r=1n-1[-2rr+1132 (r+1)!x (r-2)+∑u=r-2r-12r16 (r+1)! (u2+4)h (x (u),x (u-3))]∏s=r+1n-1(2-2ss+1),n≥1.$

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