VNU Journal of Science, Mathematics - Physics 24 (2008) 133-143 Oscilation and Convergence for a Neutral Difference Equation Dinh Cong Huong* Dept. of Math, Quy Nhon University 170 An Duong Vuong, Quynhon, Binhdinh, Vietnam Received 24 April 2008 Abstract. The oscillation and convergence of the solutions of neutral difference equation ∆(x n + δx n−τ ) + r  i=1 α i (n)F (x n−m i ) = 0, n = 0, 1, · · · are investigated, where m i ∈ N 0 , ∀i = 1, r and F is a function mapping R to R. Keyworks: Neutral difference equation, oscillation, nonoscillation, convergence. 1. Introduction It is well-known that difference equation ∆(x n + δx n−τ ) + α(n)x n−σ = 0, (1) where n ∈ N, the operator ∆ is defined as ∆x n = x n+1 − x n , the function α(n) is defined on N, δ is a constant, τ is a positive integer and σ is a nonnegative integer, was first considered by Brayton and Willoughby from the numerical point of view (see [1]). In recent years, the asymptotic behavior of solutions of this equation has been studied extensively (see [2-7]). In [4, 6, 7], the oscillation of solutions of the difference equation (1) was discussed. Motivated by the work above, in this paper, we aim to study the oscillation and convergence of solutions of neutral difference equation ∆(x n + δx n−τ ) + r  i=1 α i (n)F (x n−m i ) = 0, (2) for n ∈ N, n  a for some a ∈ N, where r, m 1 , m 2 , · · · , m r are fixed positive integers, the functions α i (n) are defined on N and the function F is defined on R. Put A = max{τ, m 1 , · · · , m r }. Then, by a solution of (2) we mean a function which is defined for n  −A and sastisfies the equation (2) for n ∈ N. Clearly, if x n = a n , n = −A, −A + 1, · · · , −1, 0 are given, then (2) has a unique solution, and it can be constructed recursively. A nontrivial solution {x n } na of (2) is called oscillatory if for any n 1  a there exists n 2  n 1 such that x n 2 x n 2 +1  0. The difference equation (2) is called oscillatory if all its solutions are oscillatory. Otherwise, it is called nonoscillatory. ∗ Tel.: 0984769741 E-mail: dconghuong@yahoo.com 133 134 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143 2. Main results 2.1. The Oscillation Consider neutral difference equation ∆(x n + δx n−τ ) + r  i=1 α i (n)x n−m i = 0, (3) for n ∈ N, n  a for some a ∈ N, where r, m 1 , m 2 , · · · , m r are fixed positive integers and the functions α i (n) are defined on N. It is clear that equation (3) is a particular case of (2). We shall establish some sufficient criterias for the oscillation of solutions of the difference equation (3). First of all we have Theorem 1. Assume tha t ( ˜m + 1) ˜m+1 ˜m ˜m r  i=1 lim inf n→∞ α i (n) > 1, (4) where δ = 0, α i (n)  0, n ∈ N, 1  i  r and ˜m = min 1ir m i . Then, (3) is os ci llat ory. Proof. We first prove that the inequality ∆x n + r  i=1 α i (n)x n−m i  0, n ∈ N (5) has no eventually positive solution. Assume, for the sake of contradiction, that (5) has a solution {x n } with x n > 0 for all n  n 1 , n 1 ∈ N. Setting v n = x n x n+1 and dividing this inequality by x n , we obtain 1 v n  1 − r  i=1 α i (n) m i  ℓ=1 v n−ℓ , (6) where n  n 1 + m, m = max 1ir m i . Clearly, {x n } is nonincreasing with n  n 1 + m, and so v n  1 for all n  n 1 + m. From (4) and (6) we see that {v n } is a above bounded sequence. Putting lim inf n→∞ v n = β, we get lim sup n→∞ 1 v n = 1 β  1 − lim inf n→∞ r  i=1 α i (n) m i  ℓ=1 v n−ℓ , or 1 β  1 − r  i=1 lim inf n→∞ α i (n) · β m i . (7) Since β m i  β ˜m , ∀i = 1, r, we have lim inf n→∞ α i (n)β m i  lim inf n→∞ α i (n)β ˜m , ∀i = 1, r and 1 − r  i=1 lim inf n→∞ α i (n)β m i  1 − r  i=1 lim inf n→∞ α i (n)β ˜m . D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 135 From (7) we have lim inf n→∞ r  i=1 α i (n)  β − 1 β ˜m+1 . But β − 1 β ˜m+1  ˜m ˜m ( ˜m + 1) ˜m+1 , so ( ˜m + 1) ˜m+1 ˜m ˜m r  i=1 lim inf n→∞ α i (n)  1, which contradicts condition (4). Hence, (5) has no eventually positive solution. Similarly, we can prove that the inequality ∆x n + r  i=1 α i (n)x n−m i  0, n ∈ N has no eventually negative solution. So, the proof is complete. Corollary. Assume tha t r  r  i=1 lim inf n→∞ α i (n)  1 r > ˆm ˆm ( ˆm + 1) ˆm+1 , (8) where δ = 0, α i (n)  0, n ∈ N, 1  i  r and ˆm = 1 r  r i=1 m i . Then, (3) is oscillator y. Proof. We will prove that the inequality (5) has no eventually positive solution. Assume, for the sake of contradiction, that (5) has a solution {x n } with x n > 0 for all n  n 1 , n 1 ∈ N. Using arithmetic and geometric mean inequality, we obtain r  i=1 lim inf n→∞ α i (n) · β m i  r  r  i=1 lim inf n→∞ α i (n)β m i  1 r , which is the same as r  i=1 lim inf n→∞ α i (n) · β m i  r  r  i=1 lim inf n→∞ α i (n)  1 r β ˆm . This yields 1 − r  i=1 lim inf n→∞ α i (n) · β m i  1 − r  r  i=1 lim inf n→∞ α i (n)  1 r β ˆm . By using the inequality (7) we have r  r  i=1 (li m i nf n→∞ α i (n))  1 r  ˆm ˆm ( ˆm + 1) ˆm+1 , which contradicts condition (8). Hence, (5) has no eventually positive solution. Next, we consider the equation (3) in case δ = 0. We have the following Lemma. Lemma 1. Let α i (n)  0 for all n ∈ N and let {x n } be an eventually positive solution of (3). Put z n = x n + δx n−τ , we have 136 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143 • (a) If −1 < δ < 0, t hen z n > 0 and ∆z n < 0 eventually. • (b) If δ < −1 and  ∞ ℓ=1 [  r i=1 α i (ℓ)] = ∞, then z n < 0 and ∆z n  0 eventually. Proof. (a) Since α i (n) ≡ 0, we have ∆z n = − r  i=1 α i (n)x n−m i < 0 eventually, so z n cannot be eventually identically zero. If z n < 0 eventually, then z n  z N < 0, ∀n  N ∈ N. Since −1 < δ < 0, we get z n = x n + δx n−τ > x n − x n−τ , which implies that x n < z n + x n−τ  z N + x n−τ . Therefore, x N+τn < z N + x N+τn−τ = z N + x N+τ(n−1) < · · · < nz N + x N . Taking n → ∞ in the above inequality, we have x N+τn < 0, which is a contradiction to x n > 0. (b) We have ∆z n = − r  i=1 α i (n)x n−m i < 0, for n sufficient large. We shall prove that z n < 0, eventually. Assume, for the sake of a contradiction, that z n = x n + δx n−τ  0, n  N, i.e. x n  −δx n−τ , n  N, which implies that 0 < x N−τ   − 1 δ  x N  · · ·   − 1 δ  j x N+(j−1)τ , j = 1, 2, · · · . On letting j → ∞ in the above inequality, we get x n → ∞ as n → ∞. But ∆z n = − r  i=1 α i (n)x n−m i  −M r  i=1 α i (n), (9) for n sufficient large, where M > 0. Summing (9) from N to n, we obtain z n+1 − z N  −M n  ℓ=N [ r  i=1 α i (ℓ)] , which implies that z n → −∞ as n → ∞. This contradicts the hypothesis that z n  0, n  N. Theorem 2. Suppose that 1 1 + δ ( ˜m + 1) ˜m+1 ˜m ˜m r  i=1 lim inf n→∞ α i (n) > 1, (10) D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 137 where −1 < δ < 0, ˜m = min 1ir m i and α i (n)  0, α i (n) > α i (n − τ), for n sufficient large, 1  i  r. Then, (3) is oscil latory. Proof. Assume the contrary and let {x n } be an eventually positive solution of (3). Let z n = x n +δx n−τ and w n = z n + δz n−τ . Then, by the case (a) of Lemma 1, z n > 0, ∆z n < 0 and w n > 0. We have ∆w n = ∆z n + δ∆z n−τ = − r  i=1 α i (n)x n−m i − δ r  i=1 α i (n − τ )x n−τ−m i ,  − r  i=1 α i (n)x n−m i − δ r  i=1 α i (n)x n−τ−m i , ∆w n  − r  i=1 α i (n)(x n−m i + δx n−τ−m i ), ∆w n  − r  i=1 α i (n)z n−m i  0. Putting lim n→∞ z n = β, we have β  0 and lim n→∞ w n = β + δβ = (1 + δ)β  0. Therefore, w n > 0 for n sufficient large. On the other hand, w n = z n + δz n−τ  (1 + δ)z n , which implies that z n−m i  w n−m i 1 + δ . Hence, we obtain ∆w n  − r  i=1 α i (n)z n−m i  − 1 1 + δ r  i=1 α i (n)w n−m i , or ∆w n + 1 1 + δ r  i=1 α i (n)w n−m i  0. (11) By Theorem 1 and in view of condition (10), the inequality (11) has no eventually positive solution, which is a contradiction. Lemma 2. Assume that −1 < δ < 0 and τ > ˜m + 1, where ˜m = min 1ir m i . Then, the maximum value of f(β) = β−1 β ˜m+2 (1 + δβ τ ) on [1, ∞) is f (β ∗ ), in which β ∗ ∈ (1, (−δ) −1/τ ) is a unique real solution of the equation 1 + δβ τ + (β − 1)[δτβ τ − ( ˜m + 1)(1 + δβ τ )] = 0. Proof. The equation f ′ (β) = 0 is equivalent to 1 + δβ τ + (β − 1)[δτβ τ − ( ˜m + 1)(1 + δβ τ )] = 0. (12) 138 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143 Put ϕ(β) = 1 + δβ τ + (β − 1)[δτβ τ − ( ˜m + 1)(1 + δβ τ )]. It is easy to check that ϕ ′ (β) = δτβ τ −1 + δβ τ [τ − ( ˜m + 1)] − ( ˜m + 1) + (β − 1)δτβ τ −1 [τ − ( ˜m + 1)]. Since τ > ˜m + 1, we get ϕ ′ (β) < 0. On the other hand, we have ϕ(1) = 1 + δ > 0 and lim β→+∞ ϕ(β) = lim β→+∞ {1 + δβ τ + (β − 1)[δβ τ [τ − ( ˜m + 1)] − ( ˜m + 1)]} = −∞. It implies that, ϕ is a decreasing function, starting from a positive value at β = 1, and hence (12) has a unique real solution β ∗ ∈ [1, ∞). Further, it is easy to see that β ∗ ∈ (1, (−δ) −1/τ ) and f(β)  0, ∀β ∈ (1, (−δ) −1/τ ), which implies that f (β ∗ ) is the maximum value of f (β) on [1, ∞). The proof is complete Theorem 3. Assume that −1 < δ < 0; τ > ˜m + 1; α i (n)  0, α i (n) > α i (n − τ ), for n suffi ci ent large, 1  i  r, ˜m = min 1ir m i and r  i=1 lim inf n→∞ α i (n) > β ∗ − 1 β ∗ ˜m+2 (1 + δβ ∗ τ −1 ), (13) where β ∗ ∈ [1, ∞) is defined as in Lemma 2. Then, (3) is oscillatory. Proof. Suppose to the contrary, and let {x n } be an eventually positive solution of (3). By the case (a) of Lemma 1, we get z n > 0, ∆z n < 0 eventually. On the other hand, ∆w n = ∆( z n + δz n−τ )  − r  i=1 α i (n)z n−m i  0. (14) Putting γ n = z n−1 z n , we have γ n  1 for n sufficient large. Dividing (14) by z n , we get 1 γ n+1  1 + δ  z n−τ z n − z n−τ+1 z n  − r  i=1 α i (n) z n−m i z n , or 1 γ n+1  1 + δ  γ n−τ+1 · · ·γ n − γ n−τ+2 · · · γ n  − r  i=1 α i (n) m i  ℓ=0 γ n−ℓ . (15) Setting lim inf n→∞ γ n = β, we get β  1. It is clear that β is finite. From (15) we have lim sup n→∞ 1 γ n+1 = 1 β  1 + δβ τ −1 (β − 1) − r  i=1 lim inf n→∞ α i (n) · β m i , r  i=1 lim inf n→∞ α i (n) · β m i +1  1 + δβ τ −1 (β − 1) − 1 β = (β − 1)[ 1 β + δβ τ −1 ], r  i=1 lim inf n→∞ α i (n)  β − 1 β ˜m+2 (1 + δβ τ ) = f(β). D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 139 By Lemma 2, we have r  i=1 lim inf n→∞ α i (n)  f(β ∗ ) = β ∗ − 1 β ∗ ˜m+2 (1 + δβ ∗ τ ), which contradicts condition (13). Hence, (3) has no eventually positive solution. Theorem 4. Suppose that − 1 δ + 1 (τ − m ∗ ) τ −m ∗ (τ − m ∗ − 1) τ −m ∗ −1 r  i=1 lim inf n→∞ α i (n) > 1, (16) where α i (n)  α i (n − τ) for n sufficient large; δ < −1, m ∗ = ma x 1ir m i , τ > m ∗ + 1 and  ∞ ℓ=1 [  r i=1 α i (ℓ)] = ∞. Then, (3) is oscillatory. Proof. Assume the contrary. Without loss of generality, let {x n } be an eventually positive solution of (3). By the case (b) of Lemma 1, we have z n < 0 and ∆z n  0. Putting w n = z n + δz n−τ , we have w n = z n + δz n−τ  (1 + δ)z n−τ , which is the same as z n−τ  1 δ + 1 w n . Therefore, it follows that ∆w n = ∆z n + δ∆z n−τ = − r  i=1 α i (n)x n−m i − δ r  i=1 α i (n − τ )x n−τ−m i ,  − r  i=1 α i (n)x n−m i − δ r  i=1 α i (n)x n−τ−m i , ∆w n  − r  i=1 α i (n)(x n−m i + δx n−τ−m i ), ∆w n  − r  i=1 α i (n)z n−m i  0, so we get 0  ∆w n + r  i=1 α i (n)z n−m i  ∆w n + 1 δ + 1 r  i=1 α i (n)w n−m i +τ . Setting γ n = w n+1 w n , we obtain γ n  1 − 1 δ + 1 r  i=1 α i (n) τ −m i  ℓ=1 γ n−m i +τ−ℓ . (17) 140 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143 Putting β = lim inf n→∞ γ n , we have β  1. Taking lower limit on both sides of (17), we obtain β  1 − 1 δ + 1 r  i=1 lim inf n→∞ α i (n) · β τ −m i , or β − 1  − 1 δ + 1 r  i=1 lim inf n→∞ α i (n) · β τ −m i . (18) Since β τ −m i  β τ −m ∗ , ∀i = 1, r, − 1 δ + 1 lim inf n→∞ α i (n)β τ −m i  − 1 δ + 1 lim inf n→∞ α i (n)β τ −m ∗ , ∀i = 1, r. From (18) we get − 1 δ + 1 r  i=1 lim inf n→∞ α i (n)  β − 1 β τ −m ∗ . But β − 1 β τ −m ∗  (τ − m ∗ − 1) τ −m ∗ −1 (τ − m ∗ ) τ −m ∗ , so − 1 δ + 1 (τ − m ∗ ) τ −m ∗ (τ − m ∗ − 1) τ −m ∗ −1 r  i=1 lim inf n→∞ α i (n)  1, which contradicts condition (16). Hence, (3) has no eventually positive solution. Theorem 5. Suppose that − 1 δ (τ − m ∗ ) τ −m ∗ (τ − m ∗ − 1) τ −m ∗ −1 r  i=1 lim inf n→∞ α i (n) > 1, (19) where δ < −1, m ∗ = max 1ir m i , τ > m ∗ + 1 and  ∞ ℓ=1 [  r i=1 α i (ℓ)] = ∞. Then, (3) is oscilla tory. Proof. Suppose to the contrary, and let {x n } be an eventually positive solution of (3). Put z n = x n + δx n−τ . By the case (b) of Lemma 1, we obtain z n < 0 and ∆z n  0. On the other hand, we have z n > δx n−τ or x n−τ > 1 δ z n , which implies that x n−m i > 1 δ z n+τ−m i . Hence, ∆z n  − 1 δ r  i=1 α i (n)z n+τ−m i . (20) Setting v n = z n+1 z n and dividing (20) by z n , we obtain v n  1 − 1 δ r  i=1 α i (n) z n+τ−m i z n , or v n  1 − 1 δ r  i=1 α i (n) τ −m i −1  ℓ=0 z n+τ−m i −ℓ z n+τ−m i −ℓ−1 . (21) D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 141 Taking lower limit on both sides of (21) and putting β = lim inf n→∞ v n , we have β  1 and β − 1  − 1 δ r  i=1 lim inf n→∞ α i (n) · β τ −m i . We can prove − 1 δ (τ − m ∗ ) τ −m ∗ (τ − m ∗ − 1) τ −m ∗ −1 r  i=1 lim inf n→∞ α i (n)  1 similarly as the proof of Theorem 4, which contradicts condition (19). Hence, (3) has no eventually positive solution. 2.2. The Convergence We give conditions implying that every nonoscillatory solution is convergent. To begin with, we have Lemma 3. Let {x n } be a nonosci llat ory solution of (2). Put z n = x n + δx n−τ . • (a) If {x n } is eventually positive (negative), then {z n } is eventually nonincreasing (nonde- creasing). • (b) If {x n } is eventually positive (negative) and there exists a con stant γ such that −1 < γ  δ, (22) then eventually z n > 0 (z n < 0). Proof. Let {x n } be an eventually positive solution of (2). The case {x n } is an eventually negative solution of (2) can be considered similarly. (a) We have ∆z n = − r  i=1 α i (n)F (x n−m i )  0 for all large n. Thus, {z n } is eventually nonincreasing. (b) Suppose the conclusion does not hold, then since by (a) {z n } is nonincreasing, it follows that eventually either z n ≡ 0 or z n < 0. Now z n ≡ 0 implies that ∆z n = − r  i=1 α i (n)F (x n−m i ) ≡ 0, but this contradicts the fact that α i (n) ≡ 0 for infinitely many n. If z n < 0, then x n < −δx n−τ so δ < 0. From (22) it follows that −1 < γ < 0 and x n < −γx n−τ . Thus, by induction, we obtain x n+jτ  (−γ) j x n for all positive integers j. Hence, x n → 0 as n → ∞. It implies that {z n } decreases to zero as n → ∞. This contradicts the fact that z n < 0. Theorem 6. Assume that ∞  ℓ=1 r  i=1 α i (ℓ) = ∞, (23) and there exists a constant η such that −1 < η  δ  0. (24) Suppos e further that, if |x|  c th en |F(x)|  c 1 where c and c 1 are positi ve constants. Then, every nonoscillatory solution of (2) tends to 0 as n → ∞. 142 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143 Proof. Let {x n } be an eventually positive solution of (2), say x n > 0, x n−τ > 0 and x n−m i > 0 for n  n 0 ∈ N. Put z n = x n +δx n−τ . We first prove that z n → 0 as n → ∞. Note that (24) implies (22) with γ replace by η. By Lemma 3 we have {z n } is eventually positive and nonincreasing. Therefore, there exists lim n→∞ z n . Put lim n→∞ z n = β. Now, suppose that β > 0. By (24), we have z n  x n . Thus, there exists an integer n 1  n 0 ∈ N such that β  z n−m i  x n−m i , ∀n  n 1 , i = 1, · · · , r. Hence, ∆z n = − r  i=1 α i (n)F (x n−m i )  −M r  i=1 α i (n), ∀n  n 1 for some positive constant M. Summing the last inequality, we obtain z n  z n 1 − M n−1  ℓ=n 1 r  i=1 α i (ℓ), which as n → ∞, in view of (23), implies that z n → −∞. This is a contradiction. Since lim n→∞ z n = 0, there exists a positive constant A such that 0 < z n  A and so, by (24) we have x n  −ηx n−τ + A. (25) Assume that {x n } is not bounded. Then, there exists a subsequence {n k } of N, so that lim k→∞ x n k = ∞ and x n k = max n 0 jn k x j , k = 1, 2, · · · . From (25), for k sufficiently large, we get x n k  −ηx n k + A and so (1 + η)x n k  A, which as k → ∞ leads to a contradiction. Now suppose that lim sup n→∞ x n = α > 0. Then, there exists a subsequence {n k } of N, with n 1 large enough so that x n > 0 for n > n 1 − τ and x n k → α as k → ∞. Then, from (24), we have z n k  x n k + ηx n k −τ and so x n k −τ  − 1 η (x n k − z n k ). As k → ∞, we obtain α  lim k→∞ x n k −τ  − α η . Since −η ∈ (0, 1), it follows that α = 0, i.e. x n → 0 as n → ∞. The arguments when {x n } is an eventually negative solution of (2) is similar. Theorem 7. Suppose there exists positi ve const ants M , α i , i = 1, 2, · · · , r such that α i (n)  α i , i = 1, 2, · · · , r, ∀n ∈ N, (26) |F (x)|  M|x|, ∀x ∈ R, (27) δ  0. (28) [...]... 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(26) and (27), we get ∞ r ∞ r αi (ℓ)xℓ−mi αM ℓ=n i=1 αi (ℓ)F (xℓ−mi ) zn − β < ∞, ℓ=n i=1 which implies that xn → 0 as n → ∞ The proof is similar when {xn } is eventually negative References [1] R K Brayton and R A Willoughby, On the numerical intagration of a symetric system of difference differential equations of neutral type, J Math Anal Appl, Vol 18 (1967) [2] L H Huang and J S Ju, Asymptotic behavior... VNU Journal of Science, Mathematics - Physics 24 (2008) 133-143 143 Then, every nonoscillatory solution of (2) tends to 0 as n → ∞ Proof Let {xn } be an eventually positive solution of (2), say xn > 0, xn−τ > 0 and xn−mi > 0 for n n0 ∈ N By Lemma 3, {zn } is eventually positive and nonincreasing, so there exists lim zn n→∞ Put lim zn = β Summing the equation (2) from n to ∞ for n n0 , we obtain n→∞ . (1992). [6] B. S. Lalli and B. G. Zhang, Oscillation and comparison theorems for certain neutral delay difference equation, J. Aus.tral. Math. Soc. Vol. 34 (1992). [7]. intagration of a symetric system of difference differential equations of neutral type, J. Math. Anal. Appl, Vol. 18 (1967). [2] L. H. Huang and J. S. Ju, Asymptotic