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. Let m∈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 equation E 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 solution x 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.

3. Solutions with Prescribed Asymptotic Behavior In this section we present sufficient conditions for the existence of a solution x to equation E such that(14)xn=∑k=1n-1crk+d+ons,where c,d∈R and s∈(-∞,0].

Theorem 3. Assume s∈(-∞,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. For n∈N and x∈SQ let(17)Fxn=bn+∑k=1nKn,kfxk.There exists L>0 such that(18)fu≤L for any u∈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=yn for n<p,H:Y→SQ,Hxn=ynfor n<pyn+∑j=n∞1rj∑i=j∞Fxifor n≥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≤γ for j∈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 function f is continuous, then from Theorem 3 we get the following two results.

Corollary 4. Assume s∈(-∞,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. Taking c=0, q=1, and α=1 in Theorem 3, we obtain the result.

Corollary 5. Assume t∈(-∞,-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. Assume c,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 Corollaries 4 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. Assume rn=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 solution x of equation E to have an asymptotic property(67)xn=∑k=1n-1crk+d+ons,where c,d∈R and s∈(-∞,0].

Theorem 7. Assume s∈(-∞,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. Assume s∈(-∞,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. Since x is bounded and f is locally bounded, the sequence (f(xn)) is bounded. Hence the assertion is a consequence of Theorem 7.

Corollary 9. Assume t∈[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. Assume x 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. Assume s∈(-∞,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 Theorem 7.

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 type E. From Corollary 4 we get the following result.

Corollary 11. Assume s∈(-∞,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 Corollary 5 we get the following.

Corollary 12. Assume t∈(-∞,-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. Assume s=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. Assume s∈(-∞,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 Corollaries 4, 5, 8, and 9, respectively, we obtain following results.

Corollary 15. Assume s∈(-∞,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. Assume t∈(-∞,-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. Assume s∈(-∞,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. Assume t∈[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. If u:N→R and lim supn→∞unn<1, then, by the root test,(108)∑n=1∞nλun<∞for any λ∈R.

Corollary 20. Assume t∈R, f is continuous, and(109)rn-1=Ont,lim supn→∞∑i=1nKn,in<1,lim supn→∞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. Let d∈R. Choose s∈(-∞,mint,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 Corollary 5, we get the following.

Corollary 21. Assume t∈(-∞,-1), f is continuous, and(113)rn-1=Ont,lim supn→∞∑i=1nKn,in<1,lim supn→∞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. If ω∈(1,∞), then(115)∑k=1n-11kω=∑n=1∞1nω+On1-ω.

Proof. Define u∈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. Assume t∈(-∞,-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. Assume μ∈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. Assume t∈(-∞,-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 Corollary 8 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, where μ=d+c∑k=1∞kt.