# Asymptotic Properties of Solutions to Second-Order Difference Equations of Volterra Type

Authors: Janusz Migda; Małgorzata Migda; Magdalena Nockowska-Rosiak
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

We consider the discrete Volterra type equation of the formΔ(rnΔxn)=bn+∑k=1nK(n,k)f(xk). We present sufficient conditions for the existence of solutions with prescribed asymptotic behavior. Moreover, we study the asymptotic behavior of solutions. We use o(ns), for given nonpositive real s, as a measure of approximation.


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## Body

## 1. Introduction

In this paper we consider the nonlinear Volterra sum-difference equation of nonconvolution type:EΔrnΔxn=bn+∑k=1nKn,kfxk,(1)rn,bn∈R,rn>0,f:R→R,K:N×N→R.Here N, R denote the set of positive integers and the set of real numbers, respectively. By a solution of E we mean a sequence x:N→R satisfying E for large n.Discrete Volterra equations of different types are widely used in the process of modeling of some real phenomena or by applying a numerical method to a Volterra integral equation. Letm∈N. The general form of a Volterra sum-difference autonomous equation is(2)Δmxn=an+∑k=1nKn,kfxk.Such equations can be regarded as the discrete analogue of Volterra integrodifferential equations of the form(3)xmt=ft+∫0tKt,sfxsds.There are relatively few works devoted to the study of equations of type (2); see, for example, [1–4]. In [5], the asymptotic behaviors of nonoscillatory solutions of the higher-order integrodynamic equation on time scales are presented.In most papers, the following special case of (2) is considered:(4)xn+1=anxn+bn+∑k=1nKn,kxk;see, e.g., [6–9], [10–13], [14], [15], or [16]. For some recent results devoted to nonlinear Volterra equations we refer to [5, 17–22] and references therein.Note, that equationE generalizes the second-order discrete Volterra difference equation of type (2):(5)Δ2xn=bn+∑k=1nKn,kfxk.On the other hand, if K(n,k)=0 for k≠n, then denoting an=K(n,n) equation E takes the form(6)ΔrnΔxn=anfxn+bn.Hence second-order difference equation (6) is a special case of E. The results on asymptotic properties and oscillation of equations of type (6) can be found, i.e., in [23–26].Our main goal is to present sufficient conditions for the existence of a solutionx to equation E such that(7)xn=∑k=1n-1crk+d+ons,where c,d∈R and s∈(-∞,0]. We give also sufficient conditions for a given solution x of equation E to have an asymptotic property (7). Moreover, in Section 5 we show applications of the obtained results to linear Volterra equation of type E. We present also some results for the case when (rn) is a potential sequence.


## 2. Preliminaries

We will denote bySQ the space of all sequences x:N→R. If x,y in SQ, then xy and x denote the sequences defined by xy(n)=xnyn and x(n)=xn, respectively. Moreover,(8)x=sup⁡xn:n∈N.If x∈SQ, s∈R, and limn→∞n-sxn=0, then we write xn=o(ns). Analogously, xn=O(ns) denotes the boundedness of the sequence (n-sxn).The following two lemmas will be useful in the proof of our main results.Lemma 1.
Assumeu∈SQ, n∈N, and(9)∑j=1∞1rj∑i=j∞ui<∞.Then(10)∑j=n∞1rj∑i=j∞ui≤∑j=n∞∑i=1jujri<∞.Proof.
We have(11)∑j=n∞1rj∑i=j∞ui=1rnun+un+1+un+2+⋯+1rn+1un+1+un+2+⋯+⋯=1rnun+1rn+1rn+1un+1+1rn+1rn+1+1rn+2un+2+⋯≤∑j=n∞∑i=1jujri≤∑j=1∞∑i=1jujri=1r1u1+1r1+1r2u2+⋯=1r1u1+u2+⋯+1r2u2+u3+⋯+⋯=∑j=1∞1rj∑i=j∞ui<∞.Lemma 2 ([27, Lemma 4.7]).
Assumey,ρ∈SQ, and limn→∞ρn=0. In the set(12)X=x∈SQ:x-y≤ρwe define a metric by the formula(13)dx,z=x-z.Then any continuous map H:X→X has a fixed point.


## 3. Solutions with Prescribed Asymptotic Behavior

In this section we present sufficient conditions for the existence of a solutionx to equation E such that(14)xn=∑k=1n-1crk+d+ons,where c,d∈R and s∈(-∞,0].Theorem 3.
Assumes∈(-∞,0], t∈[s,∞), c,d∈R, y:N→R, q∈N, α∈(0,∞),(15)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞,yn=d+c∑k=1n-11rk,U=⋃n=q∞yn-α,yn+α,and f is continuous and bounded on U. Then there exists a solution x of E such that(16)xn=yn+ons.Proof.
Forn∈N and x∈SQ let(17)Fxn=bn+∑k=1nKn,kfxk.There exists L>0 such that(18)fu≤Lfor  anyu∈U.Let(19)Y=x∈SQ:x-y≤α.If x∈Y and n≥q, then(20)xn∈yn-α,yn+α⊂U.Choose a positive number Q such that rn-1≤Qnt for any n. Then(21)∑n=1∞1nsrn∑j=n∞∑k=1jKj,k≤Q∑n=1∞nt-s∑j=n∞∑k=1jKj,kSince t-s≥0, we have(22)∑n=1∞nt-s∑j=n∞∑k=1jKj,k≤∑n=1∞∑j=n∞jt-s∑k=1jKj,k.For j∈N let(23)zj=jt-s∑k=1jKj,k.Then we have(24)∑n=1∞∑j=n∞zj=z1+z2+⋯+z2+z3+⋯+⋯=z1+2z2+3z3+⋯=∑n=1∞nzn=∑n=1∞nnt-s∑k=1nKn,k=∑n=1∞n1+t-s∑k=1nKn,k<∞.Hence, using (21) and (22), we get(25)∑n=1∞1nsrn∑j=n∞∑k=1jKj,k<∞.Analogously, replacing ∑k=1j‍Kj,k by bj, we obtain(26)∑n=1∞1nsrn∑j=n∞bj<∞.Using (25) and (26) we get(27)∑j=1∞1jsrj∑i=j∞bi+L∑k=1iKi,k<∞.Since s≤0, we have(28)∑j=1∞1rj∑i=j∞bi+L∑k=1iKi,k<∞.Define a sequence ρ∈SQ by(29)ρn=∑j=n∞1rj∑i=j∞bi+L∑k=1iKi,k.Define w,g∈SQ by(30)wn=bn+L∑k=1nKn,k,gn=∑j=n∞1jsrj∑i=j∞wi.By (27), gn=o(1). We have(31)n-sρn=n-s∑j=n∞1rj∑i=j∞wi=∑j=n∞1nsrj∑i=j∞wi≤∑j=n∞1jsrj∑i=j∞wi=gn.Hence n-sρn=o(1) and we get(32)ρn=nso1=ons.Hence there exists an index p≥q such that(33)ρn≤αfor n≥p. Let(34)X=x∈SQ:x-y≤ρ,xn=ynforn<p,H:Y→SQ,Hxn=ynforn<pyn+∑j=n∞1rj∑i=j∞Fxiforn≥p.We define a metric on X by formula (13). Note that X⊂Y. Let x∈X. By (33) and (20) we have xi∈U for any i≥p. Hence, by (18), fxi≤L for i≥p. Using (17) and (29) we obtain(35)Hxn-yn=∑j=n∞1rj∑i=j∞Fxi≤∑j=n∞1rj∑i=j∞Fxi≤ρnfor n≥p. Therefore HX⊂X. Now we show that the map H is continuous. Using (25) and the assumption s≤0, we have(36)∑n=1∞1rn∑j=n∞∑k=1jKj,k<∞.Hence, by Lemma 1, we get(37)∑n=1∞∑j=1n1rj∑k=1nKn,k<∞.Let ε>0. Choose an index m≥p and a positive constant γ such that(38)L∑n=m∞∑j=1n1rj∑k=1nKn,k<ε,γ∑n=1m∑j=1n1rj∑k=1nKn,k<ε.Let(39)C=⋃n=1myn-α,yn+α.Choose a positive δ such that if t1,t2∈C and t1-t2<δ, then(40)ft1-ft2<γ.Choose x,z∈X such that x-z<δ. Then we have(41)Hx-Hz=supn≥p∑j=n∞1rj∑i=j∞Fxi-Fzi≤supn≥p∑j=n∞1rj∑i=j∞Fxi-Fzi=∑j=p∞1rj∑i=j∞Fxi-Fzi≤∑j=p∞1rj∑i=j∞∑k=1iKi,kfxk-fzk.Using Lemma 1 we obtain(42)Hx-Hz≤∑j=p∞∑i=1j1ri∑k=1jKj,kfxk-fzk.Note that fxj-fzj≤2L for j≥p and(43)fxj-fzj≤γforj∈p,p+1,…,m.Hence we obtain(44)Hx-Hz≤γ∑n=1m∑j=1n1rj∑k=1nKn,k+2L∑n=m∞∑j=1n1rj∑k=1nKn,k<3ε.Therefore H:X→X is continuous. By Lemma 2 there exists a point x∈X such that x=Hx. Then, for n≥p, we have(45)xn=yn+∑j=n∞1rj∑i=j∞Fxi.Note that(46)ΔrnΔyn=ΔrnΔd+c∑k=1n-11rk=cΔrnΔ∑k=1n-11rk=cΔ1=0for any n. Hence, for n≥p, we get(47)ΔrnΔxn=ΔrnΔ∑j=n∞1rj∑i=j∞Fxi=-Δrn1rn∑i=n∞Fxi=Fxn=bn+∑k=1nKn,kfxk.Therefore x is a solution of E. Since x∈X we have xn=yn+o(ns).If the functionf is continuous, then from Theorem 3 we get the following two results.Corollary 4.
Assumes∈(-∞,0], t∈[s,∞), f is continuous, and(48)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any d∈R there exists a solution x of E such that xn=d+o(ns).Proof.
Takingc=0, q=1, and α=1 in Theorem 3, we obtain the result.Corollary 5.
Assumet∈(-∞,-1), s∈(-∞,t], f is continuous, and(49)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any c,d∈R there exists a solution x of E such that(50)xn=d+c∑k=1n-11rk+ons.Proof.
Assumec,d∈R and a sequence y∈SQ is defined by(51)yn=d+c∑k=1n-11rk.Since rn-1=O(nt) and t<-1, we see that y is bounded. Now, taking q=1 and α=1 in Theorem 3, we obtain the result.Note that Corollaries4 and 5 concern convergent solutions. However, Theorem 3 includes also divergent solutions. For example, if f(x)=x-1 for x≠0, s∈(-∞,0], t=0, rn-1=O1, and(52)∑k=1∞rk-1=∞,∑n=1∞n1-s∑i=1nKn,i<∞,∑n=1∞n1-sbn<∞.then, by Theorem 3, for any nonzero c∈R and any d∈R there exists a solution x of E such that(53)xn=d+c∑k=1n-11rk+ons.Now we present an example that proves the assumption(54)∑n=1∞n1+t-s∑i=1nKn,i<∞,in Theorem 3, is essential.Example 6.
Assumern=n, bn=0,(55)Kn,k=1n2,fx=1x+1+1,s=0, and t=0. Then equation E takes the form(56)ΔnΔxn=1n2∑k=1n1xk+1+1.Let c,d∈R and(57)yn=d+c∑k=1n-11rk=d+c∑k=1n-11k.Notice that f is continuous and bounded on R. Moreover,(58)∑n=1∞n1+t-s∑k=1nKn,k=∑n=1∞n1n=∞,and(59)Δyn=cn,ΔnΔyn=0.Assume x is a solution of (56) such that(60)xn=yn+zn,zn=ons=o1.Since Δ(nΔyn)=0, we have(61)ΔnΔxn=ΔnΔyn+ΔnΔzn=ΔnΔzn.Hence(62)ΔnΔzn=ΔnΔxn=∑k=1n1n21xk+1+1>0for large n. Therefore, the sequence nΔzn is eventually increasing and there exists the limit(63)λ=limn→∞nΔzn>-∞.If λ<∞, then the sequence nΔzn is convergent in R. Hence the series(64)∑n=1∞ΔnΔxn=∑n=1∞ΔnΔznis convergent. On the other hand(65)ΔnΔxn=∑k=1n1n21xk+1+1>1nfor large n. Hence λ=∞. Therefore nΔzn>1 for large n and we get(66)∑n=1∞Δzn≥∑n=1∞1n=∞.But since zn→0, the series ∑n=1∞‍Δzn is convergent.


## 4. Asymptotic Behavior of Solutions

In this section we present sufficient conditions for a given solutionx of equation E to have an asymptotic property(67)xn=∑k=1n-1crk+d+ons,where c,d∈R and s∈(-∞,0].Theorem 7.
Assumes∈(-∞,0], t∈[s,∞),(68)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞,and x is a solution of E such that the sequence (f(xn)) is bounded. Then there exist c,d∈R such that(69)xn=∑k=1n-1crk+d+ons.Proof.
We have(70)ΔrnΔxn=bn+∑k=1nKn,kfxkfor large n. Using boundedness of the sequence (f(xn)) and (68) we get(71)∑n=1∞n1+t-sΔrnΔxn<∞.Define w,u∈SQ by(72)wn=ΔrnΔxn,un=nt-swn.Choose a positive L such that rn-1≤Lnt for any n. Since t-s≥0, we have(73)∑n=1∞1nsrn∑j=n∞wj≤L∑n=1∞nt-s∑j=n∞wj=L∑n=1∞∑j=n∞nt-swj≤L∑n=1∞∑j=n∞jt-swj.Moreover,(74)∑n=1∞n1+t-swn=∑n=1∞nun=u1+u2+u2+u3+u3+u3+⋯=∑j=1∞uj+∑j=2∞uj+∑j=3∞uj+⋯=∑n=1∞∑j=n∞uj=∑n=1∞∑j=n∞jt-swj.Hence, by (73)(75)∑n=1∞1nsrn∑j=n∞wj<∞.Since s≤0, we have(76)∑n=1∞1rn∑j=n∞wj<∑n=1∞1nsrn∑j=n∞wj<∞.Let(77)zn=∑j=n∞1rj∑i=j∞wi.Then(78)n-szn=n-s∑j=n∞1rj∑i=j∞wi≤n-s∑j=n∞1rj∑i=j∞wi=∑j=n∞1nsrj∑i=j∞wi≤∑j=n∞1jsrj∑i=j∞wi=o1.Thus zn=o(ns). Let(79)yn=xn-zn.Then(80)Δyn=Δxn-Δzn=Δxn+1rn∑i=n∞wi.Hence(81)rnΔyn=rnΔxn+∑i=n∞wiand we get(82)ΔrnΔyn=ΔrnΔxn-wn=ΔrnΔxn-ΔrnΔxn=0for any n∈N. Therefore, there exists a real constant c such that rnΔyn=c. Thus(83)yn-y1=Δy1+⋯+Δyn-1=cr1+⋯+crn-1.Hence(84)xn=yn+zn=∑k=1n-1crk+d+onswhere d=y1.Corollary 8.
Assumes∈(-∞,0], t∈[s,∞), rn-1=O(nt), f is locally bounded,(85)∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞,and x is a bounded solution of E. Then there exist c,d∈R such that(86)xn=∑k=1n-1crk+d+ons.Proof.
Sincex is bounded and f is locally bounded, the sequence (f(xn)) is bounded. Hence the assertion is a consequence of Theorem 7.Corollary 9.
Assumet∈[0,∞), rn-1=O(nt), f is locally bounded, and(87)∑n=1∞n1+t∑i=1nKn,i<∞,∑n=1∞n1+tbn<∞.Then any bounded solution of E is convergent.Proof.
Assumex is a bounded solution of E. Let s=0. By Corollary 8, there exist c,d∈R such that(88)xn=c∑k=1n-11rk+d+o1.Define a sequence u∈SQ, by un=r1-1+r2-1+⋯+rn-1-1. Then u is increasing and bounded. Hence u is convergent. Therefore xn=cun+d+o(1) is convergent.Corollary 10.
Assumes∈(-∞,0], t∈[s,∞), rn-1=O(nt), f is bounded,(89)∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞,and x is an arbitrary solution of E. Then there exist c,d∈R such that(90)xn=∑k=1n-1crk+d+ons.Proof.
The assertion is an immediate consequence of Theorem7.


## 5. Additional Results

In this section we present some additional results. First, we give some applications of our results to linear discrete Volterra equations of typeE. From Corollary 4 we get the following result.Corollary 11.
Assumes∈(-∞,0], t∈[s,∞),(91)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any d∈R there exists a solution x of equation(92)ΔrnΔxn=bn+∑k=1nKn,kxksuch that xn=d+o(ns).From Corollary5 we get the following.Corollary 12.
Assumet∈(-∞,-1), s∈(-∞,t], and(93)rn-1=Ont,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any c,d∈R there exists a solution x of (92) such that(94)xn=d+c∑k=1n-11rk+ons.Example 13.
Assumes=0, t=1, and(95)rn=1n-1,bn=-3n+2n+1nn-1-1n4,Kn,k=2kn6.Then (92) takes the form(96)Δ1n-1Δxn=-3n+2n+1nn-1-1n4+∑k=1n2kn6xk.It is easy to check that all assumptions of Corollary 11 hold. Indeed, we have(97)∑n=1∞n2∑k=1nKn,k=∑n=1∞n2∑k=1n2kn6=∑n=1∞n+1n3<∞,and(98)∑n=1∞n2bn<∞.So, for every d∈R, there exists a solution x of (96) such that limn→∞xn=d. One such solution is xn=1-1/n.In our investigations the condition(99)∑n=1∞n1+t-s∑i=1nKn,i<∞plays an important role. In practice, this condition can be difficult to verify. In the following remark we present the condition, which is a little stronger but easier to check.Remark 14.
Assumes∈(-∞,0], t∈R, λ∈(-∞,s-t-2), and un=O(nλ). Let ε=s-t-2-λ, L>0, un≤Lnλ for any n. Then λ=s-t-2-ε and(100)∑n=1∞n1+t-sun≤L∑n=1∞n1+t-snλ=∑n=1∞1n1+ε<∞.Applying this remark to Corollaries4, 5, 8, and 9, respectively, we obtain following results.Corollary 15.
Assumes∈(-∞,0], t∈[s,∞), λ∈(-∞,s-t-2), f is continuous, and(101)rn-1=Ont,∑i=1nKn,i=Onλ,bn=Onλ.Then for any d∈R there exists a solution x of E such that xn=d+o(ns).Corollary 16.
Assumet∈(-∞,-1), s∈(-∞,t], λ∈(-∞,s-t-2), f is continuous, and(102)rn-1=Ont,∑i=1nKn,i=Onλ,bn=Onλ.Then for any c,d∈R there exists a solution x of E such that(103)xn=d+c∑k=1n-11rk+ons.Corollary 17.
Assumes∈(-∞,0], t∈[s,∞), λ∈(-∞,s-t-2), f is locally bounded,(104)rn-1=Ont,∑i=1nKn,i=Onλ,bn=Onλ,and x is a bounded solution of E. Then there exist c,d∈R such that(105)xn=∑k=1n-1crk+d+ons.Corollary 18.
Assumet∈[0,∞), λ∈(-∞,-t-2), f is locally bounded, and(106)rn-1=Ont,∑i=1nKn,i=Onλ,bn=Onλ.Then any bounded solution of E is convergent.Now we present some results for the case when the series(107)∑n=1∞n1+t-s∑i=1nKn,i,∑n=1∞n1+t-sbnare strongly convergent.Remark 19.
Ifu:N→R and limsupn→∞unn<1, then, by the root test,(108)∑n=1∞nλun<∞for any λ∈R.Corollary 20.
Assumet∈R, f is continuous, and(109)rn-1=Ont,limsupn→∞∑i=1nKn,in<1,limsupn→∞bnn<1.Then for any d∈R and any λ∈(-∞,0] there exists a solution x of E such that(110)xn=d+onλ.Proof.
Letd∈R. Choose s∈(-∞,min⁡t,λ). By Remark 19, we have(111)∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.By Corollary 4, there exists a solution x of E such that(112)xn=d+ons=d+onλ.Analogously, using Corollary5, we get the following.Corollary 21.
Assumet∈(-∞,-1), f is continuous, and(113)rn-1=Ont,limsupn→∞∑i=1nKn,in<1,limsupn→∞bnn<1.Then for any c,d∈R and any λ∈(-∞,0] there exists a solution x of E such that(114)xn=d+∑k=1n-1crk+onλ.To the end we consider the case when(rn) is a potential sequence.Lemma 22.
Ifω∈(1,∞), then(115)∑k=1n-11kω=∑n=1∞1nω+On1-ω.Proof.
Defineu∈SQ and λ∈R by(116)un=∑k=1n-11kω,λ=∑k=1∞1kω.By [28, Theorem 2.2], we have(117)Δn1-ω=1-ωn-ω+on-ω.Since Δun=n-ω, we get(118)Δun-λΔn1-ω=n-ω1-ωn-ω+on-ω=11-ω+o1→11-ω.Note that n1-ω→0 and (un-λ)→0. Hence, by discrete L’Hospital’s Rule,(119)un-λn1-ω=11-ω+o1.Therefore(120)un=λ+11-ωn1-ω+on1-ω=∑k=1∞1kω+On1-ω.Corollary 23.
Assumet∈(-∞,-1), s∈(-∞,t], f is continuous, and(121)rn=n-t,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any μ∈R there exists a solution x of E such that(122)xn=μ+Ont+1.Proof.
Assumeμ∈R. Let(123)d=0,λ=∑n=1∞nt,c=μλ.By Corollary 5, there exists a solution x of E such that(124)xn=d+c∑k=1n-1kt+ons.By Lemma 22(125)∑k=1n-1kt=λ+Ont+1.Hence(126)xn=μ+Ont+1+ons=μ+Ont+1.Corollary 24.
Assumet∈(-∞,-1), s∈(-∞,t], f is locally bounded, and(127)rn=n-t,∑n=1∞n1+t-s∑i=1nKn,i<∞,∑n=1∞n1+t-sbn<∞.Then for any bounded solution x of E there exists a real number μ such that(128)xn=μ+Ont+1.Proof.
By Corollary8 there exist c,d∈R such that(129)xn=d+c∑k=1n-1kt+ons.By Lemma 22 we obtain(130)xn=d+c∑k=1∞kt+Ont+1+ons=μ+Ont+1,whereμ=d+c∑k=1∞kt.



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