Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 11, 1-12; http://www.math.u-szeged.hu/ejqtde/

ON THE OSCILLATION OF SECOND ORDER NONLINEAR NEUTRAL DELAY DIFFERENCE

EQUATIONS

A.K.Tripathy Department of Mathematics Kakatiya Institute of Technology and Science Warangal-506015, INDIA e-Mail: arun tripathy70@rediffmail.com

Abstract

In this paper sufficient conditions are obtained for oscillation of all solutions of a class of nonlinear neutral delay difference equations of the form

Δ^{2}(y(n) + p(n)y(n − m)) + q(n)G(y(n − k)) = 0

under various ranges of p(n). The nonlinear function G, G ∈ C(R, R) is either sublinear or superlinear.

Mathematics Subject classification (2000): 39 A 10, 39 A 12

Keywords : Difference equations; Oscillation; Nonlinear.

1 Introduction

Recently, a good deal of work has been published on the oscillation theory of difference equations. Most of the works in first and higher order neutral delay difference equations are concerned with the study of the behaviour of the solution which oscillates or tends to zero (see [3], [4], [5], [7]). But very few papers are available on oscillatory higher order nonlinear delay difference equations.

In [7], authors Parhi and Tripathy has considered a class of nonlinear neutral delay difference equations of higher order of the form

Δ^{m}[y(n) + p(n)y(n − m)] + q(n)G(y(n − k)) = 0, (E)

where m ≥ 2. They have obtained the results which hold good when G is sub linear only. However, the behaviour of solutions of (E) under the superlinear nature of G is still in progress. In fact, various ranges of p(n) are restricting for all solutions as oscillatory.

In this paper, author has studied the second order nonlinear neutral delay difference equation of the form

Δ^{2}[y(n) + p(n)y(n − m)] + q(n)G(y(n − k)) = 0, n ≥ 0,  (1)

where Δ is the forward difference operator defined by Δ y(n) = y(n + 1) − y(n), p, q are real valued functions defined on N(0) = {0, 1, 2, ...} such that q(n) ≥ 0, G ∈ C(R, R) is nondecreasing and x G(x) > 0 for x ≠ 0 and m > 0, k ≥ 0 are integers. Here, an attempt is made to establish sufficient conditions under which every solution of Eq.(1) oscillates.

The motivation of present work has come under two directions. Firstly, due to the work in [6] and second is due to the work in [7], where G is almost sublinear. It is interesting to observe that unlike differential equation, Eq. (1) is converting immediately into a first order difference inequality and hence study of both are interrelated hypothetically. In this regard the work in [8] provides a good input for the completion of the present work.

By a solution of Eq. (1) we mean a real valued function y(n) defined on N(−r) = {−r, −r + 1, ⋯} which satisfies (1) for n ≥ 0, where r = max{k, m}. If

y(n) = A_{n}, n = −r, −r + 1, ⋯, 0  (2)

are given, then (1) admits a unique solution satisfying the initial condition (2). A solution y(n) of (1) is said to be oscillatory, if for every integer N > 0, there exists an n ≥ N such that y(n)y(n + 1) ≤ 0 : otherwise, it is called nonoscillatory.

The following two results are useful for our discussion in the next sections.

Theorem 1.1 [2]. If q(n) ≥ 0 for n ≥ 0 and

lim inf_{n→∞} ∑_{s=n−k}^{n−1} q(s) > (\frac{k}{k + 1})^{k+1},

then Δ x(n) + q(n) x(n − k) ≤ 0, n ≥ 0 can not have an eventually positive solution.

Theorem 1.2 [8]. Assume that lim inf_{n→∞} ∑_{s=n−k}^{n−1} q(s) > 0. (H) Suppose there are a function g(u) ∈ C(R, R^{+}) and a number ϵ > 0 such that

1. g(u) is nondecreasing in R^{+},

2. g(−u) = g(u) and lim_{n→0} g(u) = 0,

3.  ∫_{0}^{∞}g(e^{−u})du < ∞,
4.  |\frac{G(u)}{u} − 1| ≤ g(u), 0 < |u| < ϵ.

If

∑_{n=0}^{∞} [∑_{s=n}^{n+k} q(s)ℓn (∑_{s=n}^{n+k} q(s)) − ∑_{s=n+1}^{n+k} q(s)ℓn ( ∑_{s=n+1}^{n+k} q(s))] = ∞,

then every solution of Δ x(n) + q(n) G(x(n − k)) = 0 oscillates. Remark lim inf_{n→∞} ∑_{s=n−k}^{n−1} q(s) > 0 implies that ∑_{n=0}^{∞} q(n) = ∞. Corollary 1.3 If all the conditions of Theorem 1.2 are satisfied, then

Δx(n) + q(n)G(x(n − k)) ≤ 0, n ≥ 0

doesn’t possess any eventually positive solution.

The proof follows from the Theorem 1.2.

2 Sublinear Oscillation

This section deals with the sufficient conditions for the oscillation of all solutions of Eq.(1) when G is sublinear. The following conditions are needed for our use in the sequel.

(H_{1})  ∑_{n=0}^{∞} q(n) = ∞
(H_{2})  ∑_{n=m}^{∞} Q(n) = ∞,
where Q(n) = min{q(n), q(n − m)}

(H_{3}) G(u)G(v) ≥ G(uv) for u > 0, v > 0

(H_{4}) G(−u) = −G(u), u ∈ R

(H_{5}) There exists λ > 0 such that G(u) + G(v) ≥ λ G(u + v), u ∈ R, v ∈ R

(H_{6})  ∫_{0}^{±c} \frac{du}{G(u)} < ∞, c > 0
(H_{7})  lim inf_{|x|→0} \frac{G(x)}{x} ≥ γ > 0
(H_{8})  lim inf_{n→∞} ∑_{s=n+m−k}^{n−1} q(s) > \frac{b}{γ} (\frac{k−m}{k−m+1})^{k−m+1}, b > 0 and k > m.

Theorem 2.1 Let 0 ≤ p(n) ≤ a < ∞. Suppose that (H_{2})-(H_{6}) hold. Then Eq.(1) is oscillatory.

Proof Suppose for contrary that y(n) is a nonoscillatory solution of (1). Then there exists n_{1} > 0 such that y(n) > 0 or < 0 for n ≥ n_{1}. Let the former hold. Setting

z(n) = y(n) + p(n)y(n − m)  (3)

we have from (1)

Δ^{2}z(n) + q(n)G(y(n − k)) = 0,  (4)

that is,

Δ^{2}z(n) = −q(n)G(y(n − k)) ≤ 0,

for n ≥ n_{2} > n_{1} + r. Hence Δz(n) is non-increasing. If Δz(n) < 0, then Eq.(1) becomes

Δz(n + 1) + q(n)G(y(n − k)) = Δz(n) < 0, n ≥ n_{2}.  (5)

Using (5) we get

Δz(n − m + 1) + q(n − m) G (y(n − k − m)) < 0, n ≥ n_{2}^{∗}

and hence for n_{3} > max{n_{2}, n_{2}^{∗}},

0 > Δz(n + 1) + q(n) G(y(n − k)) + G(a) Δz(n − m + 1)
+G(a)q(n − m)G(y(n − k − m)).  (6)

Using (H_{2}), (H_{3}) and (H_{5}), the last inequality implies

0 > Δz(n + 1) + G(a)Δz(n − m + 1) + λQ(n)G(z(n − k)),

where 0 < z(n) ≤ y(n) + a y(n − m). Thus

λQ(n) + \frac{Δz(n + 1)}{G(z(n − k))} + G(a)\frac{Δz(n − m + 1)}{G(z(n − k))} < 0,

that is,

λQ(n) + ∫_{z(n+1)}^{z(n+2)}\frac{du}{G(u)} + G(a) ∫_{z(n−m+1)}^{z(n−m+2)}\frac{dv}{G(v)} < 0,

where z(n + 2) < u < z(n + 1), z(n − m + 2) < v < z(n − m + 1) and n − k < n + 1. Hence for n ≥ n_{3},

λ ∑_{s=n_{3}}^{n} Q(s) + ∑_{s=n_{3}}^{n} ∫_{z(s+1)}^{z(s+2)}\frac{du}{G(u)} + G(a) ∑_{s=n_{3}}^{n} ∫_{z(s−m+1)}^{z(s−m+2)}\frac{dv}{G(v)} < 0,

that is,

λ ∑_{s=n_{3}}^{n} Q(s) + ∫_{z(n_{3}+1)}^{z(n+2)} \frac{du}{G(u)} + G(a) ∫_{z(n_{3}−m+1)}^{z(n_{3}−m+2)}\frac{dv}{G(v)} < 0.

Since lim_{n→∞} z(n) exists, then the above inequality implies that

∑_{s=n_{3}}^{∞} Q(s) < ∞,

a contradiction to (H_{2}). If Δz(n) > 0 for n ≥ n_{2}, then z(n) is nondecreasing and hence there exists a constant α > 0 such that z(n) > α, n ≥ n^{∗}. Application of Eq.(4) gives

Δ^{2}z(n) + G(a)Δ^{2}z(n − m) + λQ(n)G(z(n − k)) ≤ 0.

Consequently, for n ≥ n_{3} > max{n_{2}, n^{∗}},

λG(a)Q(n) < −Δ^{2}z(n) − G(a)Δ^{2}z(n − m).

Thus,

∑_{n=n_{3}}^{∞} Q(n) < ∞,

a contradiction to (H_{2}).

Suppose the later holds. Then setting x(n) = −y(n) > 0, for n ≥ n_{1} and using (H_{4}), Eq.(1) can be written as

Δ^{2}(x(n) + p(n)x(n − m)) + q(n)G(x(n − k)) = 0.  (7)

Following the above procedure to Eq.(7), similar contradictions can be obtained. Hence the proof of the theorem is complete.

Remark The prototype of G satisfying (H_{3}), (H_{4}) and (H_{5}) is

G(u) = (α + β |u|^{λ}) |u|^{μ} sqn u,

where α ≥ 1, β ≥ 1, λ ≥ 0 and μ ≥ 0.

Example Consider

Δ^{2}[y(n) + (3 + (−1)^{n})y(n − 1)] + 8 y^{\frac{1}{3}} (n − 2) = 0, n ≥ 0.

Clearly, the above equation satisfies all the conditions of Theorem 2.1 and hence it is oscillatory. In particular, y(n) = (−1)^{3n} is such an oscillatory solution.

Theorem 2.2 Let −1 < b ≤ p(n) ≤ 0. If (H_{1}), (H_{4}), (H_{6}) and m < k hold, then every solution of (1) oscillates.

Proof Proceeding as in Theorem 2.1, we get the inequality (5), where Δz(n) < 0 and z(n) > 0 for n ≥ n_{2}. Since z(n) ≤ y(n), then (5) becomes

Δz(n + 1) + q(n)G(z(n − k)) < 0, n ≥ n_{2}.

Following the similar steps of Theorem 2.1, we have a contradiction to (H_{1}). If z(n) < 0, for n ≥ n_{2}, then y(n) < y(n − m), that is, y(n) is bounded. Consequently, z(n) is bounded, a contradiction to the fact that Δz(n) < 0 and z(n) < 0 for n ≥ n_{2}. Hence Δz(n) > 0 for n ≥ n_{2}. If z(n) > 0, then using the similar argument as in Theorem 2.1, the contradiction is obtained to (H_{1}). Suppose that z(n) < 0, for n ≥ n_{2}. Then y(n − k) > (1/b)z(n + m − k). Thus Eq.(1) becomes

Δ^{2}z(n) + q(n)G (\frac{1}{b}z(n)) < 0

due to increasing z(n). Consequently, the last inequality can be made as

−Δz(n) + q(n)G (\frac{1}{b} z(n + 1)) < 0

that is,

Δx(n) − \frac{1}{b}q(n)G(x(n + 1) < 0,

where x(n) = \frac{1}{b}z(n) > 0. Hence

−\frac{1}{b} q(n) < −\frac{Δx(n)}{G(x(n + 1))} = − ∫_{x(n)}^{x(n+1)}\frac{du}{G(x(n + 1))}

For x(n + 1) < u < x(n), it is immediate to get

\frac{1}{b} ∑_{n=n_{2}}^{N} q(n) < − ∑_{n=n_{2}}^{N} ∫_{x(n)}^{x(n+1)}\frac{du}{G(u)} = − ∫_{x(n_{2})}^{x(N+1)}\frac{du}{G(u)}

that is,

−\frac{1}{b} ∑_{n=n_{2}}^{∞} q(n) < − lim_{N→∞} ∫_{x(n_{2})}^{x(N+1)}\frac{du}{G(u)} < ∞,

a contradiction.

The case y(n) < 0 for n ≥ n_{1} is similar. This completes the proof of the theorem.

Theorem 2.3 Let −∞ < −b ≤ p(n) < −1, b > 0. If (H_{1}), (H_{4}), (H_{7}) and (H_{8}) hold, then every bounded solution of (1) oscillates.

Proof Let y(n) be a bounded nonoscillatory solution of (1). Then there exists n_{1} > 0 such that y(n) > 0 for n ≥ n_{1}. The case y(n) < 0 for n ≥ n_{1} can similarly be dealt with. Setting z(n) as in (3), we get (4). Hence Δ^{2}z(n) ≤ 0, for n ≤ n_{2} > n_{1} + r implies that Δz(n) in nonincreasing. Let Δz(n) < 0. Then z(n) > 0 or < 0, for n ≥ n_{2}. Suppose the former holds. Then lim_{n→∞} z(n) exists. On the other hand, Δ^{2}z(n) < 0 and Δz(n) < 0 implies that there exists L > 0 such that Δz(n) < −L, for n ≥ n^{∗} that is, z(n) < −L_{1}(n), where L_{1} > 0 is a constant. Hence lim_{n→∞} z(n) < −∞, a contradiction. Consequently, the later holds and hence lim_{n→∞} z(n) = −∞, a contradiction to the fact that z(n + m − k) ≥ −by(n − k) which is bounded. Assume that Δz(n) > 0, for n ≥ n_{2}. If z(n) > 0, then there exists a constant α > 0 such that z(n) > α, for n ≥ n^{∗}. Thus z(n) = y(n) + p(n)y(n − m) < y(n) implies that

Δ^{2}z(n) + G(α)q(n) < 0,

for n ≥ n_{3} > max{n_{2}, n^{∗}}. Hence

∑_{n=n_{3}}^{∞} Q(n) < ∞,

a contradiction to (H_{1}). Ultimately, z(n) < 0, for n ≥ n_{2}. In this case lim_{n→∞} z(n) exists. Let it be β, 0 ≤ β < ∞. Suppose that β = 0. Due to Δz(n) > 0, Eq.(1) becomes

−Δz(n) + q(n)G (−\frac{1}{b}z(n + m − k)) < 0

where y(n − k) ≥ −\frac{1}{b}z(n + m − k). Setting −\frac{1}{b}z(n) = x(n), the last inequality can be written as

Δx(n) + (\frac{1}{b}) q(n)G(x(n + m − k)) < 0

for n ≥ n_{3} > n_{2} and hence using (H_{7}), we get

Δx(n) + (\frac{γ}{b} ) q(n)x(n + m − k) < 0

which has no positive solution due to (H_{8}), a contradiction to the fact that x(n) > 0 is a solution. If 0 < β < ∞, then the contradiction is obivious due to (H_{1}). Hence the theorem is proved.

3 Superlinear Oscillation

This section deals with the oscillation of all solutions of Equation (1) such that G is superlinear.

Theorem 3.1 Let 0 ≤ p(n) ≤ a < ∞. Assume that (H), (H_{3})-(H_{5}) and the following conditions

(H_{9})  lim inf_{n→∞} ∑_{s=n−k+1}^{n} Q(s) > 0,
where Q(n + 1) = min{q(n), q(n − m)} .

(H_{10})  ∑_{n−0}^{∞} [ ∑_{s=n+1}^{n+k+1} Q(s)ℓn ( ∑_{s=n+1}^{n+k+1} \frac{λQ(s)}{(1+G(a))}) − ∑_{s=n+2}^{n+k+1} q(s)ℓn ( ∑_{s=n+2}^{n+k+1} \frac{λQ(s)}{(1+G(a))})] = ∞

hold. Then Eq.(1) is oscillatory.

Proof Proceeding as in Theorem 2.1, we get the inequality (6). Using the fact that both z(n) and Δz(n) are non-increasing, inequality (6) can be written as

0 > (1 + G(a))Δz(n + 1) + λQ(n + 1)G(z(n + 1 − k))

that is,

Δz(n + 1) + \frac{λ}{(1 + G(a))}Q(n + 1)G(z(n + 1 − k)) < 0

for n ≥ n_{3} > n_{2}. In view of (H_{10}) and Theorem 1.2, the last inequality has no positive solution, a contradiction. If Δz(n) > 0 for n ≥ n_{2}, then proceeding as in Theorem 2.1, we get

∑_{n=n_{3}}^{∞} Q(n + 1) < ∞.

On the otherhand (H_{9}) implies that

∑_{n=0}^{∞} Q(n + 1) = ∞,

a contradiction. Hence the theorem is proved.

Theorem 3.2 Let −1 < b ≤ p(n) ≤ 0. Assume that (H_{1}), (H_{4}) and the following conditions

(H_{11})  lim inf_{n→∞} ∑_{s=n+m−k}^{n−1} q(s) > 0, k > m

(H_{12})  ∑_{n=0}^{∞} [ ∑_{s=n}^{n+k−m} q(s)ℓn ( ∑_{s=n}^{n+k−m} (\frac{−1}{b}) q(s)) − ∑_{s=n+1}^{n+k−m} q(s)ℓn ( ∑_{s=n+1}^{n+k−m} (\frac{−1}{b}) q(s))] = ∞

hold. Then every solution of (1) oscillates.

Proof Clearly, (H_{11}) implies that ∑_{n=0}^{∞} q(n) = ∞ . Let y(n) be a non-oscillatory

solution of (1) such that y(n) > 0 for n ≥ n_{0}. Proceeding as in Theorem 2.1, we get the inequality (4). Consequently, Δ^{2}z(n) ≤ 0 for n ≥ n_{1} > n_{0} + r shows that there are four possible cases:

1. z(n) > 0, Δz(n) < 0,
2. z(n) < 0, Δz(n) < 0,
3. z(n) > 0, Δz(n) > 0,
4. z(n) < 0, Δz(n) > 0,

Using the same type of reasoning as in the proof of the Theorem 2.2 we may obtain respective contradictions for the cases (1), (2) and (3). Consider the case (4). Since z(n) < 0. Then substituting y(n − k) > (\frac{1}{b})z(n + m − k) for n ≥ n_{2} . Eq.(1) can be written as

Δx(n) − \frac{1}{b}q(n)G(x(n + m − k)) < 0,

where x(n) = (\frac{1}{b}z(n)) > 0. In view of Theorem 1.2 and (H_{12}), the last inequality has no positive solution, a contradiction. This completes the proof of the theorem.

Theorem 3.3 Let −∞ < b ≤ p(n) < −1. Assume that all the conditions of Theorem 3.2 hold. Then every bounded solution of (1) oscillates.

Proof Let y(n) be a bounded solution of (1) such that y(n) > 0 for n ≥ n_{0}. Then proceeding as in the proof of the Theorem 3.2, we have four cases, cases (1), (2) and (3) follows from the Theorem 2.3 and case (4) follows from the Theorem 3.2. Hence the details are omitted. This completes the proof of the theorem.

Theorem 3.4 Let −∞ < b ≤ p(n) < −1. Assume that m ≥ k + 1 and the following conditions

(H_{13})  ∫_{0}^{∞} \frac{dx}{G(x)} < ∞
(H_{14})  ∑_{j=0}^{∞} q(n_{j}) = ∞ for every sequence {n_{j}} of {n}

hold. Then every unbounded solution of (1) oscillates.

Proof Let y(n) be an unbounded solution of (1) such that y(n) > 0 for n ≥ n_{0}. The case y(n) < 0 for n ≥ n_{0} is similar. Using the same type of reasoning as in the proof of Theorem 3.2, we consider the four cases (1), (2), (3) and (4). For the cases (1) and (3), the discussion is same which can be followed from the Theorem 3.2 directly. Consider the case (2). Here Eq.(1) reducess to

Δz(n + 1) + q(n)G (\frac{1}{b}z(n + 1)) = Δz(n) < 0

that is,

q(n) < −\frac{Δz(n + 1)}{G (\frac{1}{b}z(n + 1))} = − ∫_{z(n+1)}^{z(n+2)}\frac{du}{G (\frac{1}{b}z(n + 1))}

< − ∫_{z(n+1)}^{z(n+2)}\frac{du}{G (\frac{1}{b}u)} = −b ∫_{z(n+1)}^{z(n+2)}\frac{d (\frac{1}{b}u)}{G (\frac{1}{b}u)}

where z(n + 2) < u < z(n + 1). Hence

∑_{n=n_{2}}^{N} q(n) < −b ∑_{n=n_{2}}^{N} ∫_{z(n+1)}^{z(n+2)}\frac{d (\frac{1}{b}u)}{G (\frac{1}{b}u)},

= −b ∫_{z(n_{2}+1)}^{z(N+2)} \frac{d (\frac{1}{b}u)}{G (\frac{1}{b}u)}.

Consequently,

∑_{n=n_{2}}^{∞} q(n) < −b lim_{N→∞} ∫_{z(n_{2}+1)}^{z(N+2)} \frac{d (\frac{1}{b}u)}{G (\frac{1}{b}u)} < ∞,

a contradiction to (H_{14}). This is because (H_{14}) implies that (H_{1}) hold. Next, we consider case (4). Since y(n) is unbounded, there exists a sequence {n_{j}} of {n} such that y(n_{j}) → ∞ as j → ∞. Hence for every M > 0, it is possible to find n_{2} > n_{1} such that n_{j} ≥ n_{2} implies y(n_{j}) > M. Let n_{3} ≥ n_{2} + K. Then Eq.(1) yields

∑_{n_{j}=n_{3}}^{s−1} q(n_{j})G(y(n_{j} − k)) = − ∑_{n_{j}=n_{3}}^{s−1} Δ^{2}z(n_{j}) = −Δz(s) + Δz(n_{3}).

Thus

G(M) ∑_{n_{j}=n_{3}}^{∞} q(n_{j}) < ∑_{n_{j}=n_{3}}^{∞} q(n_{j})G(y(n_{j} − k)) < ∞,

a contradiction to (H_{14}). Hence the proof of the theorem is complete.

Remark The prototype of G satisfying the Theorems 3.1 - 3.3 may be of the form

G(u) = \{ \begin{matrix}u [1 + (a + ℓn^{2}|u|)^{−1}] , u ≠ 0 \\ 0, u = 0 \end{matrix}

and

g(u) = \{ \begin{matrix}a, |u| > 1 \\ (a + ℓn^{2}|u|)^{−1}, 0 < |u| ≤ 1 \\ 0, u = 0. \end{matrix}

The author is thankful to the referee, for helpful suggestions and necessary corrections in completion of this paper.

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(Received November 10, 2007)