Báo cáo hóa học: "A CLASSIFICATION SCHEME FOR NONOSCILLATORY SOLUTIONS OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION" ppt

A CLASSIFICATION SCHEME FOR NONOSCILLATORY SOLUTIONS OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION ZHI-QIANG ZHU, GEN-QIANG WANG, AND SUI SUN CHENG Received 25 May 2005; Revised 21 September 2005; Accepted 28 September 2005 Nonoscillatory solutions of a nonlinear neutral type higher order difference equations are classified by means of their asymptotic behaviors. By means of the Kranoselskii’s fixed point theorem, existence criteria are then provided for justification of such classification. Copyright © 2006 Zhi-qiang Zhu et al. This is an open access article distributed under the Creative Commons Att ribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Classification schemes for nonoscillatory solutions of nonlinear difference equations are important since further investigations of some of the qualitative behaviors of nonoscilla- tory solutions can then be reduced to only a number of cases. There are several studies which provide such classification schemes for difference equations, see, for example, [4– 11]. In particular, in [7], a class of nonlinear neutral difference equations of the form Δ m  x n + c n x n−k  + f  n,x n−l  = 0, n = 0,1, , (1.1) where m, k and l are integers such that m ≥ 2, k>0andl ≥ 0 is studied and classification schemes are given when {c n } is a nonnegative constant sequence {c 0 },andin[10], the same equation is studied with odd integer m ≥ 1, positive integer k,integerl and {c n }= {− 1}. In this paper, we continue our investigation on the p ossible types of nonoscillatory solutions when {c n }⊆(−1,0] and lim n→∞ c n = c 0 (while the case {c n }⊆(−∞,−1] will be discussed elsewhere). Besides the assumption that {c n }⊆(−1,0], we will assume further that f is a continuous function defined on {0,1, }×R such that f = f (n,x)isnonde- creasing in the second variable x and satisfies xf(n,x) > 0forx = 0andn ≥ 0. We will accomplish two things in this paper: to provide a classification scheme for the nonoscillatory solutions of (1.1)inSection 2 and establish in Section 3 sufficient and/or necessary criteria for the existence of solutions in each class. There are no overlapping Hindawi Publishing Corporation Advances in Difference Equations Volume 2006, Ar ticle ID 47654, Pages 1–19 DOI 10.1155/ADE/2006/47654 2 A classification scheme for neutral difference equation results between our paper and [4–11], although some proofs are similar. However, the existence proofs are different in that Cheng-Patula existence theorem is applied in [7], monotone method is used in [10] while we use Krasnoselskii fixed point theorem here. We remark further that classification scheme is also provided for neutral differential equa- tions in [2]. Before we go into details, we will need some preparatory terminologies and results. First of all, given initial x i for −max{k,l}≤i ≤ 0, we may calculate from (1.1) x 1 ,x 2 ,x 3 , in a recursive manner. Such a sequence {x n } is said to be a solution of (1.1). Among the solutions of (1.1), one is said to be nonoscillatory if it is eventually positive or eventually negative. Given an integer a,itisconvenienttoset N(a) ={a,a +1,a +2, }. (1.2) Given an integer α ≥ 0, the generalized factorial function g(x) = x (α) is defined as fol- lows x (α) = ⎧ ⎨ ⎩ x(x − 1)(x − 2)···(x − α +1) α>0 1 α = 0. (1.3) It is well known that Δn (α) = αn (α−1) for α>0 (see, e.g., [3]). Let l ∞ N 0 =  x =  x n  n≥N 0 :sup n≥N 0   x n   r n < ∞  , (1.4) where N 0 > 0isanintegerand{r n } n≥N 0 is a positive sequence with a uniform posi- tive lower bound. When endowed with the usual linear structure and the norm x= sup n≥N 0 (|x n |/r n ), (l ∞ N 0 ,·)isaBanachspace.AsetB ⊆ l ∞ N 0 is said to be uniformly Cauchy if for any ε>0 there exists an integer M ≥ N 0 such that     x i r i − x j r j     <ε i, j>M (1.5) for all x ={x n }∈B. Lemma 1.1. A bounded and uniformly Cauchy subset B ⊆ l ∞ N 0 is relatively compact. Proof. By assumption, we know that for any such ε>0, there exists an integer M ≥ N 0 > 0 such that for any x ∈ B,wehave     x i r i − x j r j     < ε 3 , i, j ≥ M. (1.6) Zhi-qiang Zhu et al. 3 Let Γ > 0beaboundforB.Thatis x≤Γ for all x ∈ B. Choose integers M n , n = N 0 ,N 0 + 1, ,M,andnumbersy (1) n <y (2) n < ··· <y (M n ) n such that y (1) n =−r n Γ, y (M n ) n = r n Γ and      y ( j+1) n r n − y ( j) n r n      < ε 3 ,1 ≤ j ≤ M n − 1. (1.7) Now define a sequence {v k } kN 0 as follows. Let v N 0 be one of the values {y (1) N 0 , , y (M N 0 ) N 0 }, v N 0 +1 be one of the values {y (1) N 0 +1 , , y (M N 0 +1 ) N 0 +1 }. In general, for N 0 ≤ k ≤ M,letv k equal one of the values {y (1) k , , y (M k ) k }.Fork>M,letv k = (r k /r M )v M . It is clear that the sequence {v k } kN 0 belongs to l ∞ N 0 .LetL be the set of all possible sequences {v k } k≥N 0 defined as above. Note that L has M N 0 M N 0 +1 ···M M elements. We assert that L is a finite ε-net for B.Itissufficient to show that for any x ={x k } kN 0 ∈ B, L contains a sequence v ={v k } kN 0 such that x − v= sup n≥N 0   x n − v n   r n <ε. (1.8) Indeed, by definition of L, we can choose a sequence {v k } kN 0 in L such that     x k r k − v k r k     < ε 3 , N 0 ≤ k ≤ M. (1.9) Furthermore, by (1.6), (1.9), and the definition of v ={v k } kN 0 ,fork>M,wehave     x k r k − v k r k     =     x k r k − v M r M     ≤     x k r k − x M r M     +     x M r M − v M r M     ≤ ε 3 + ε 3 = 2ε 3 . (1.10) From (1.9)and(1.10), we see that (1.8) holds. The proof is complete.  Lemma 1.2. Suppose that lim n→∞ c n = c 0 with c 0 ∈ (−1,0] and the sequence {x n /n (i) } is eventually positive or eventually negative, where i is a nonnegative integer. Suppose further that z n = x n + c n x n−k and lim n→∞ (z n /n (i) ) = b. Then lim n→∞ (x n /n (i) ) = b/(1 + c 0 ). Proof. Without loss of generality, we assume that x n /n (i) > 0 for any positive integer n. In case b is finite, we assert that {x n /n (i) } is bounded. Otherwise, there would exist a sequence {n λ } of integers with n λ →∞for λ →∞such that lim λ→∞ x n λ n (i) λ =∞, x n ≤ x n λ ,0<n≤ n λ . (1.11) On the other hand, we have z n λ n (i) λ = x n λ n (i) λ + c n λ x n λ −k n (i) λ ≥  1+c n λ  x n λ n (i) λ −→ ∞ (1.12) as λ →∞. This is contrary to the fact that b is finite. 4 A classification scheme for neutral difference equation Let limsup n→∞ (x n /n (i) ) = Q and liminf n→∞ (x n /n (i) ) = q.Then0≤ q ≤ Q<∞.More- over, there exist {n λ } and {n λ } such that lim λ→∞ n λ =∞,lim λ→∞ n λ =∞,lim λ→∞ (x n λ /n λ (i) )= Q and lim λ→∞ (x n λ /n λ (i) ) = q.Notethat b = lim λ→∞ z n λ n λ (i) = lim λ→∞  x n λ n λ (i) + c n λ x n λ −k n λ (i)  ≥ lim λ→∞ x n λ n λ (i) +lim λ→∞ inf c n λ x n λ −k  n λ − k  (i)  n λ − k  (i) n λ (i) ≥ Q + c 0 Q, b = lim λ→∞ z n λ n λ (i) = lim λ→∞  x n λ n λ (i) + c n λ x n λ −k n λ (i)  ≤ lim λ→∞ x n λ n λ (i) +lim λ→∞ supc n λ x n λ −k  n λ − k  (i)  n λ − k  (i) n λ (i) ≤ q + c 0 q, (1.13) we have (1 + c 0 )q ≥ (1 + c 0 )Q. It follows that q ≥ Q.Henceq = Q and it implies that lim n→∞ (x n /n (i) ) exists. In view of z n = x n + c n x n−k and lim n→∞ (z n /n (i) ) = b,wehave lim n→∞ x n n (i) = b 1+c 0 . (1.14) In case b is infinite, then b =∞or b =−∞. We assert that b =−∞cannot hold. In fact, for given c 1 with −c 0 <c 1 < 1, there exists a large integer N 0 such that −c n ≤ c 1 for n ≥ N 0 .Hence,ifb =−∞,thenz n = x n + c n x n−k < 0forn ≥ N and x n < −c n x n−k ≤ c 1 x n−k , n ≥ N, (1.15) where N ≥ N 0 is some positive integer. It implies that 0 <x N+λk <c 1 x N+(λ−1)k < ··· <c λ 1 x N . (1.16) So that lim λ→∞ x N+λk = 0. Thus lim λ→∞ z N+λk = 0 (1.17) which implies that b =−∞is impossible. Now, for arbitrary M>0, there exists a sufficiently large integer N such that z n n (i) = x n n (i) + c n x n−k n (i) ≥ M, n ≥ N. (1.18) It follows that x n n (i) ≥ M, n ≥ N. (1.19) That is lim n→∞ (x n /n (i) ) =∞.Theproofiscomplete.  The following two propositions are respectively in [1, Theorems 1.7.9 and 1.7.11]. Zhi-qiang Zhu et al. 5 Lemma 1.3. Suppose that the sequence {x n } and {y n } satisfy the following conditions, (i) y n > 0 and Δ y n > 0 for all large integers n and lim n→∞ y n =∞,and (ii) lim n→∞ (Δx n /Δy n ) = b. Then lim n→∞ (x n /y n ) = lim n→∞ (Δx n /Δy n ) = b,whereb can be finite or infinite. Lemma 1.4. Let u = u(n) beasequencedefinedforn ∈ N(a), u(n) > 0 with Δ m u(n) of constant sign on N(a) and not identically zero. Then, there exists an integer m ∗ , 0 ≤ m ∗ ≤ m with m+ m ∗ odd for Δ m u(n) ≤ 0 or, m + m ∗ even for Δ m u(n) ≥ 0 and such that m ∗ ≤ m − 1 implies (−1) m ∗ +i Δ i u(n) > 0 ∀n ∈ N(a), m ∗ +1≤ i ≤ m, m ∗ ≥ 1 implies Δ i u(n) > 0 ∀ large n ∈ N(a), 1 ≤ i ≤ m ∗ . (1.20) Remark 1.5. If u(n) < 0inLemma 1.4, then there exists m ∗ ,0≤ m ∗ ≤ m with m + m ∗ odd for Δ m u(n) ≥ 0or,m + m ∗ even for Δ m u(n) ≤ 0 and such that m ∗ ≤ m − 1 implies (−1) m ∗ +i Δ i u(n) < 0 ∀n ∈ N(a), m ∗ +1≤ i ≤ m, m ∗ ≥ 1 implies Δ i u(n) < 0 ∀ large n ∈ N(a), 1 ≤ i ≤ m ∗ . (1.21) Lemma 1.6 (Kranoselskii’s fixed point theorem). Suppose B is a Banach space and Ω is a bounded, convex and closed subset of B.LetU,S : Ω → B satisfy the following conditions. (i) Ux+ Sy ∈ Ω for any x, y ∈ Ω, (ii) U is a contraction mapping, and (iii) S is completely continuous. Then U + S has a fixed point in Ω. 2. Classifications of nonoscillatory solutions In the following discussions, we assume throughout that lim n→∞ c n = c 0 ∈ (−1, 0]. (2.1) We set z n = x n + c n x n−k (2.2) whenever it is defined. Equation (1.1) can now be written as Δ m z n =−f  n,x n−l  . (2.3) We will propose a classification scheme for the nonoscillatory solutions of (1.1). For this purpose, we first note that if x ={x n } is an eventually negative solution of (1.1), then y ={y n } defined by y n =−x n will satisfy Δ m  y n + cy n−k  +  f  n, y n−l  = 0, (2.4) where  f (n,u) =−f (n,−u), n ∈ N(0), u ∈ R (2.5) 6 A classification scheme for neutral difference equation has the same properties satisfied by f , that is,  f is a continuous function defined on {0,1, }×R such that  f =  f (n,u) is nondecreasing in the second variable u and satisfies u  f (n, u) > 0foru = 0andn ≥ 0. Therefore, we may restrict our attention to the set S + of all eventually positive solutions of (1.1). Motivated by the classification scheme in [2], we make use of the following notations for classifying our eventually positive solutions: A k (α,β) =   x n  ∈ S + :lim n→∞ x n n (k−1) = α,lim n→∞ x n n (k) = β  , k ≥ 1, A 0 (α) =   x n  ∈ S + :lim n→∞ x n = α  . (2.6) Theorem 2.1. (a) Suppose that m is even. If x ={x n } is an eventually positive solution of (1.1), then either x ∈ A 0 (0) or there are some j ∈{1,2, ,m/2} and a>0 such that x belongs to A 2 j−1 (∞,a), A 2 j−1 (∞,0) or A 2 j−1 (a,0). (b) Suppose that m is odd. If x ={x n } is an eventually positive solution of (1.1), then either x belongs to A 0 (α) for some α ≥ 0,or there are j ∈{1, 2, ,(m − 1)/2} and a>0 such that x belongs to A 2 j (∞,a), A 2 j (∞,0) or A 2 j (a,0). Proof. Let m is even and x ={x n } be an eventually positive solution of (1.1). Then, in view of (2.3), there exists some integer N>0suchthatΔ m z n < 0forn ≥ N. Therefore, z n is eventually of fixed sign. For the sake of simplicity, we may assume that {z n } is of fixed sign for n ≥ N. First of all, suppose z n < 0forn ≥ N. By the same reasoning as in the proof of Lemma 1.2, we may show that lim λ→∞ z N+λk = 0. (2.7) On the other hand, in view of Lemma 1.4, there exists some even m ∗ with 0 ≤ m ∗ ≤ m such that eventually Δ i z n < 0for0≤ i ≤ m ∗ and (−1) m ∗ +i Δ i z n < 0form ∗ +1≤ i ≤ m. There are now two cases to consider. Case 1 (m ∗ = 0). Then we have eventually z n < 0, Δz n > 0. (2.8) By (2.8), we can set lim n→∞ z n = L 0 ≤ 0. (2.9) In view of (2.7), we find that lim n→∞ z n = 0. By Lemma 1.2,wehavelim n→∞ x n = 0. Hence x belongs to A 0 (0). Case 2 (m ∗ ≥ 2). Then we have eventually z n < 0, Δz n < 0. (2.10) It implies lim n→+∞ z n < 0 which is contrary to (2.7). Hence m ∗ ≥ 2 does not hold. Zhi-qiang Zhu et al. 7 Now we suppose z n > 0forn ≥ N. Similar to the proof in [7, Theorem 1], we may see that x belongs to A 2 j−1 (∞,a), A 2 j−1 (∞,0) or A 2 j−1 (a,0) for some j ∈{1,2, ,m/2} and a>0. When m is odd, the proof is similar to those above and hence is skipped. The proof is complete.  3. Existence criteria Eventually positive (and by analog eventually negative) solutions of (1.1)havebeenclas- sified according to Theorem 2.1. We now justify our classification schemes by finding existence criteria for each type of solutions. Theorem 3.1. Suppose that m is even. If (1.1)hasasolutioninA 2 j−1 (∞,a) for some j ∈ { 1,2, ,m/2} and a>0,thenthereexistssomeK>0 such that ∞  i=0 (i + m − 2 j) (m−2 j) (m − 2 j)! f  i,K(i − l) (2 j−1)  < ∞. (3.1) Theconverseisalsotrue. Proof. Firstofall,weremarkthat ∞  i λ =n ∞  i λ−1 =i λ ··· ∞  i 2 =i 3 ∞  i 1 =i 2 f  i 1 ,x i 1 −l  = ∞  i=n (i − n + λ − 1) (λ−1) (λ − 1)! f  i,x i−l  . (3.2) Let x ={x n } be an eventually positive solution of (1.1)inA 2 j−1 (∞,a). Then we may suppose that there exists an integer N 0 > 0suchthatx n > 0andx n−l > 0forn>N 0 .In view of (2.3), we have Δ m z n < 0forn>N 0 .Thereby{Δ i z n } is eventually monotonic for i = 0,1,2, ,m − 1. Since lim n→∞ (x n /n (2 j−1) ) = a>0,thereexistssomeintegerN 1 >N 0 such that 1 2 an (2 j−1) ≤ x n ≤ 3 2 an (2 j−1) , n ≥ N 1 . (3.3) Note that lim n→∞ (z n /n (2 j−1) ) = (1 + c 0 )a implies lim n→∞ Δ 2 j−1 z n =  1+c 0  a  2 j − 1  !. (3.4) By (3.4) and the monotonicity of Δ i z n ,wehave lim n→∞ Δ i z n = 0, i = 2 j,2j +1, ,m − 1. (3.5) Summing (2.3) m − 2j times from n to N 1 and invoking (3.5) in each time, we obtain Δ 2 j z n =− ∞  i m−2j =n ··· ∞  i 2 =i 3 ∞  i 1 =i 2 f  i 1 ,x i 1 −l  =− ∞  i=n (i − n + m − 2 j − 1) (m−2 j−1) (m − 2 j − 1)! f  i,x i−l  , n ≥ N 1 . (3.6) 8 A classification scheme for neutral difference equation Summing the above equation again from N 1 to n,weobtain Δ 2 j−1 z n+1 = Δ 2 j−1 z N 1 − n  i 2 =N 1 ∞  i 1 =i 2  i 1 − i 2 + m − 2j − 1  (m−2 j−1) (m − 2 j − 1)! f  i 1 ,x i 1 −l  . (3.7) By (3.4), the above equation implies that ∞  i 2 =n ∞  i 1 =i 2  i 1 − i 2 + m − 2j − 1  (m−2 j−1) (m − 2 j − 1)! f  i 1 ,x i 1 −l  < ∞, n ≥ N 1 . (3.8) That is, ∞  i=n (i − n + m − 2 j) (m−2 j) (m − 2 j)! f  i,x i−l  < ∞, n ≥ N 1 . (3.9) Let K = a/2. In view of (3.3), (3.9) and the monotonicity of f (n,x)inx,weseethat(3.1) holds. Conversely, suppose (3.1)holdsforsomeK>0. Set R n = n (2 j−1) .Inviewof(3.2), we have ∞  i m−2j+1 =n ∞  i m−2j =i m−2j+1 ··· ∞  i 2 =i 3 ∞  i 1 =i 2 f  i 1 ,K  i 1 − l  (2 j−1)  = ∞  i=n (i − n + m − 2 j) (m−2 j) (m − 2 j)! f  i,K(i − l) (2 j−1)  . (3.10) Note that (2.1), there are two cases to consider. In case −1 <c 0 < 0, take c 1 so that −c 0 <c 1 < (1 − 4c 0 )/5 < 1. Then (1 − 5c 1 )/(4c 0 ) < 1. Note that lim n→∞ (|c n |R n /R n−k−l ) =|c 0 |,lim n→∞ (R n−k /R n ) = 1and(3.1)holds.Thus thereexistsanintegerN>k+ l such that when n ≥ N,wehave   c n   R n R n−k−l ≤ c 1 , (3.11) −c n ≤ c 1 , (3.12) R n−k R n ≥ 1 − 5 c 1 4c n , (3.13) ∞  i m−2j+1 =N ∞  i m−2j =i m−2j+1 ··· ∞  i 1 =i 2 f  i 1 ,K  i 1 − l  (2 j−1)  <  1 − c 1  K 8 . (3.14) Take N 0 = N − k − l, r n = R 2 n and define the Banach space l ∞ N 0 as in (1.4). Let Ω =  x ∈ l ∞ N 0 : 1 2 KR n ≤ x n ≤ KR n  . (3.15) Zhi-qiang Zhu et al. 9 Then it is obvious that Ω is a bounded, convex and closed subset of l ∞ N 0 ,andforanyx ∈ Ω and n ≥ N 0 + l,wehave f  n,x n−l  ≤ f  n,K(n − l) (2 j−1)  . (3.16) Define operators U and S on Ω as follows: (Ux) n = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ − 3 4 c 1 KR n − c N x N−k R N R n N 0 ≤ n<N − 3 4 c 1 KR n − c n x n−k n ≥ N, (Sx) n = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 3 4 KR n N 0 ≤ n<N 3 4 KR n + F(n) n ≥ N, (3.17) where F(n) = n−1  i m =N ··· i m−2j+4 −1  i m−2j+3 =N i m−2j+3 −1  i m−2j+2 =N ∞  i m−2j+1 =i m−2j+2 ∞  i m−2j =i m−2j+1 ··· ∞  i 1 =i 2 f  i 1 ,x i 1 −l  . (3.18) In view of (3.16)and(3.14),we have F(n) ≤ n−1  i m =N ··· i m−2j+4 −1  i m−2j+3 =N i m−2j+3 −1  i m−2j+2 =N  1 − c 1  K 8 =  1 − c 1  K(n − N) (2 j−1) 8(2 j − 1)! ≤  1 − c 1  K 8 R n (3.19) for n ≥ N. Next, we will show that the operators U and S satisfy the conditions of Kranoselskii’s fixed point theorem. First, we claim that Ux+ Sy ∈ Ω for any x, y ∈ Ω. Indeed, for N 0 ≤ n<N,inviewof (3.13)and(3.12), we have (Ux) n +(Sy) n =  3 4  1 − c 1  K − c N x N−k R N  R n ≥  3 4  1 − c 1  K − c N K R N−k 2R N  R n ≥ 1 2 KR n , (Ux) n +(Sy) n ≤  3 4  1 − c 1  K − c N K R N−k R N  R n ≤  3 4  1 − c 1  + c 1  KR n ≤ KR n . (3.20) When n ≥ N,invoking(3.13)again,wehave (Ux) n +(Sy) n ≥ 3 4  1 − c 1  KR n − c n x n−k ≥ 3 4  1 − c 1  KR n − c n 1 2 K R n−k R n R n ≥ 1 2 KR n (3.21) 10 A classification scheme for neutral difference equation and, in view of (3.19)and(3.12), we have (Ux) n +(Sy) n ≤ 3 4  1 − c 1  KR n − c n x n−k +  1 − c 1  K 8 R n ≤ 3 4  1 − c 1  KR n − c n KR n−k +  1 − c 1  K 8 R n ≤ KR n . (3.22) That is, Ux+ Sy ∈ Ω for any x, y ∈ Ω. Let x, y ∈ Ω.Inviewof(3.11), we have 1 R 2 n   (Ux) n − (Uy) n   =   c N     x N−k − y N−k   R N R n =   x N−k − y N−k   R 2 N −k   c N   R 2 N −k R N R n ≤ c 1 sup n≥N 0   x n − y n   R 2 n (3.23) for N 0 ≤ n<N.And,forn ≥ N,wehave 1 R 2 n   (Ux) n − (Uy) n   ≤   c n   sup n≥N 0   x n − y n   R 2 n . (3.24) Therefore, we have Ux− Uy≤c 1 x − y (3.25) for any x, y ∈ Ω.Hence,U is a contraction mapping. Next, we will prove that S is a completely continuous mapping. Indeed, it is obvious that (Sx) n ≥ (K/2)R n for n ≥ N 0 and (Sx) n ≤ KR n for N 0 ≤ n<N.Whenn ≥ N, by means of (3.19), we have (Sx) n ≤ 3 4 KR n +  1 − c 1  K 8 R n ≤ KR n . (3.26) That is, the operator S maps Ω into Ω. Now we consider the continuity of S.Letx (λ) ∈ Ω and x (λ) − x→0whenλ →∞,we assert that Sx (λ) converges to Sx by ·. Indeed, x (λ) − x→0 implies that x ∈ Ω and |x (λ) n − x n |→0whenλ →∞for any integer n ≥ N 0 .Thereby,wehave    f  n,x (λ) n −l  − f  n,x n−l     −→ 0, λ −→ ∞ (3.27) for any integer n ≥ N 0 + l. By definition of S,wehave    Sx (λ)  n − (Sx) n   = 0 (3.28) for N 0 ≤ n<N and    Sx (λ)  n − (Sx) n   ≤ H λ (n) (3.29) [...]... Cheng, and G Zhang, A classification scheme for nonoscillatory solutions of a higher order neutral nonlinear difference equation, Journal of the Australian Mathematical Society Series A 67 (1999), no 1, 122–142 [8] X Tang and J Yan, Oscillation and nonoscillation of an odd -order nonlinear neutral difference equation, Functional Differential Equations 7 (2000), no 1-2, 157–166 (2001) [9] G Zhang and S S... Oscillation criteria for a neutral difference equation with delay, Applied Mathematics Letters 8 (1995), no 3, 13–17 Zhi-qiang Zhu et al 19 [10] B Zhang and Y J Sun, Classification of nonoscillatory solutions of a higher order neutral difference equation, Journal of Difference Equations and Applications 8 (2002), no 11, 937–955 [11] B G Zhang and B Yang, Oscillation in higher- order nonlinear difference equations,... Applied Mathematics, vol 155, Marcel Dekker, New York, 1992 [2] Y S Chen, Asymptotic behavior of nonoscillatory solutions of higher order neutral delay differential equations, Annals of Differential Equations 9 (1993), no 3, 270–286 [3] S S Cheng, Partial Difference Equations, Taylor & Francis, London, 2003 [4] S S Cheng, G Zhang, and W.-T Li, On a higher order neutral difference equation, Mathematical Analysis... Analysis and Applications (Th M Rassias, ed.), Hadronic Press, Florida, 2000, pp 37–64 [5] X Z He, Oscillatory and asymptotic behaviour of second order nonlinear difference equations, Journal of Mathematical Analysis and Applications 175 (1993), no 2, 482–498 [6] B S Lalli, Oscillation theorems for neutral difference equations Advances in difference equations, Computers & Mathematics with Applications 28... exists an integer N > k + l such that when n ≥ N, (3.11) to (3.14) hold Take operators U and S to be the same operators as above Then we may prove in similar manners that (1.1) has a solution in A2 j −1 (∞ ,a) The proof is complete A similar theorem holds when m is odd, the proof is similar to that of Theorem 3.1 and hence is skipped Theorem 3.2 Suppose m is odd If (1.1) has a solution in A2 j (∞ ,a) for. .. Chinese Annals of Mathematics Series A 20 (1999), no 1, 71–80 (Chinese) Zhi-qiang Zhu: Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China Gen-qiang Wang: Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China Sui Sun Cheng: Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043,... well as ∞ (i + 2)(2) f (i,K) = ∞ 2 i=0 (3.83) for any K > 0, and ∞ f (i,i − 1) = ∞ (3.84) i =0 By the similar reasons to the above, (3.81) has a solution in A0 (0) if it has some eventually positive solution Indeed, {xn } = {1/2n } is such a solution of (3.81) References [1] R P Agarwal, Difference Equations and Inequalities Theory, Methods, and Applications, Monographs and Textbooks in Pure and Applied... equation 1 3 Δ2 xn − xn−1 + xn−1 = 0, 4 16 (3.78) here f (n,x) = (1/16)x It is clear that ∞ f i,K(i − 1) = ∞, i =0 ∞ (i + 1) f (i,K) = ∞ (3.79) i=0 for any K > 0 and ∞ i =0 (i + 1) f i,i − 1 = ∞ (3.80) 18 A classification scheme for neutral difference equation Hence by Theorems 3.1 and 3.3, an eventually positive solution of (3.78) cannot be in A1 (∞ ,a) nor in A1 (a, 0) In addition, by Theorem 3.5, an... solution in A2 j −1 (∞,0) when −1 < c0 < 0 When c0 = 0, we take c1 so that 0 < c1 ≤ 1/3 and the rest of proof is the same as the above and is thus skipped The proof is complete A result similar to Theorem 3.5 is the following 16 A classification scheme for neutral difference equation Theorem 3.6 Suppose m is odd If (1.1) has a solution in A2 j (∞,0) for some j ∈ {1,2, , (m − 1)/2}, then ∞ (i + m − 2 j −... that there exists x = {xn } ∈ Ω such that xn = −cn xn−k + F(n), n ≥ N (3.74) In view of the definition of Ω, we see that x is a solution of (1.1) in A0 (0) The proof is complete A variant of Theorem 3.7 is the following and its proof is omitted Theorem 3.8 Suppose m is odd and c0 < 0 If there exist constants α > 0, c1 , c2 with 0 < c1 < −c0 < c2 < 1 and integer M > l + k such that c1 eαk > 1 (3.75) as . Cheng, and G. Zhang, A classification scheme for nonoscillatory solutions of a higher order neutral nonlinear difference equation, Journal of the Australian Mathematical Society. Series A 67 (1999),. A CLASSIFICATION SCHEME FOR NONOSCILLATORY SOLUTIONS OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION ZHI-QIANG ZHU, GEN-QIANG WANG, AND SUI SUN CHENG Received 25 May 2005; Revised. behavior of nonoscillatory solutions of higher order neutral delay differential equations, Annals of Differential Equations 9 (1993), no. 3, 270–286. [3] S. S. Cheng, Partial Difference Equations, Taylor