Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 64, 1-15; http://www.math.u-szeged.hu/ejqtde/

Oscillatory and Asymptotic Behavior of Fourth

order Quasilinear Difference Equations

E.Thandapani and M.Vijaya

Ramanujan Institute for Advanced Study in Mathematics University of Madras Chennai 600 005, India. ethandapani@yahoo.co.in vijayaanbalacan@gmail.com

Abstract

The authors consider the fourth order quasilinear difference equation

Δ^{2} (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) + q_{n}|x_{n+3}|^{β−1}x_{n+3} = 0,

where α and β are positive constants, and {p_{n}} and {q_{n}} are positive real sequences. They obtain sufficient conditions for oscillation of all

solutions when ∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} )^{\frac{1}{α}} < ∞ and ∑_{n=n_{0}}^{∞} (_{\frac{n}{p_{n} ^{\frac{1}{α}}}} ) < ∞. The results are

illustrated with examples.

Keywords: Fourth order difference equation, nonoscillation, oscillation, asymptotic behavior. AMS Subject classification: 39A10.

1 Introduction

In this paper, we are concerned with the fourth order quasilinear difference equation

Δ^{2} (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) + q_{n}|x_{n+3}|^{β−1}x_{n+3} = 0  (1)

where Δ is the forward difference operator defined by Δx_{n} = x_{n+1} − x_{n}, α and β are positive constants, {p_{n}} and {q_{n}} are positive real sequences defined for all n ∈ ℕ(n_{0}) = {n_{0}, n_{0} + 1, ...}, and n_{0} a nonnegative integer.

By a solution of equation (1), we mean a real sequence {x_{n}} that satisfies equation (1) for all n ∈ ℕ(n_{0}). If any four consecutive values of {x_{n}} are given, then a solution {x_{n}} of equation (1) can be defined recursively. A nontrival solution of equation (1) is said to be nonoscillatory if it is either eventually positive or eventually negative, and it is oscillatory otherwise.

The oscillatory and asymptotic behavior of solutions of the nonlinear difference equation of the form (1) has been considered by Thandapani and Selvaraj [11] and Thandapani,Pandian,Dhanasekaran, and Graef [10]. In [11], the equation (1) is discussed under the condition

∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} )^{\frac{1}{α}} = ∞ and ∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}^{\frac{1}{α}}} ) = ∞,  (2)

and in [10] the authors obtain a more detailed analysis of the asymptotic behavior of nonoscillatory solutions of equation (1) under the same condition (2). In [9], the authors considered the equation (1) under the condition (2),

or ∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} )^{\frac{1}{α}} < ∞ and ∑_{n=n_{0}}^{∞} (_{\frac{n}{p \frac{1}{α} ; n}} ) = ∞ or (5), and obtain condition for the

existence of nonoscillatory solutions of equation (1). For the case α = 1, equation (1) becomes

Δ^{2} (p_{n}Δ^{2}x_{n}) + q_{n} |x_{n+3}|^{β−1}x_{n+3} = 0.  (3)

The oscillatory and asymptotic behavior of solutions of equation (3) with

∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} ) = ∞ or ∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} ) < ∞  (4)

was discussed by Yan and Liu [13], and Thandapani and Arockiasamy [12] respectively. Later, the results in [13] were extended to the more general equation

Δ (a_{n}Δ (b_{n}Δ (c_{n}Δx_{n}))) + q_{n} |x_{n}|^{β−1}x_{n} = 0

by Graef and Thandapani [5].

The main objective here is to discuss the oscillatory behavior of solutions of equation (1) under the condition

∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}} )^{\frac{1}{α}} < ∞ and ∑_{n=n_{0}}^{∞} (\frac{n}{p_{n}^{\frac{1}{α}}} ) < ∞.  (5)

In Section 2, we describe the classification of nonoscillatory solutions of equation (1), and in Section 3, we obtain sufficient conditions for the oscillation of all solutions of equation (1). Examples illustrating the results are presented in Section 4.

For related results on the oscillation of fourth order difference equation, we refer the reader to the monographs of Agarwal, Bohner, Grace, and O’Regan [1] as well as the references ([2], [3], [6], [7], [8]) given below.

2 Classification of Nonoscillatory Solutions

In this section, we state and prove some basic results regarding the classification of nonoscillatory solutions of equation (1). Without any loss of generality, we restrict our attention to the set of positive solutions, because if {x_{n}} satisfies (1), then so does {−x_{n}}.

Lemma 1. If {x_{n}} is an eventually positive solution of equation (1), then one of the following four cases holds for all sufficiently large n:

(I) Δx_{n} > 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(II) Δx_{n} < 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(III) Δx_{n} > 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(IV) Δx_{n} > 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0.

Proof. From equation (1), we have Δ^{2} (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0 for all large n. It follows that {Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n})}, {Δ^{2}x_{n}} and {Δx_{n}} are eventually monotonic and one-signed. Next, we consider the following eight cases:

(a) Δx_{n} > 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(b) Δx_{n} < 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(c) Δx_{n} > 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(d) Δx_{n} < 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) > 0;
(e) Δx_{n} > 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0;
(f) Δx_{n} < 0, Δ^{2}x_{n} > 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0;
(g) Δx_{n} > 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0;
(h) Δx_{n} < 0, Δ^{2}x_{n} < 0, Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0.

If Δx_{n} < 0 and Δ^{2}x_{n} < 0 eventually, then lim_{n→∞} x_{n} = −∞, which contradicts the positivity of the solution {x_{n}}. Hence, the cases (d) and (h) cannot hold. Similarly, since Δ^{2} (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0, if Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) < 0, then lim_{n→∞}(p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) = −∞, that is, Δ^{2}x_{n} < 0 for large n. This observation rule out the cases (e) and (f). This completes the proof of the lemma.

Remark 1. When α = 1, then Lemma 1 reduces to Lemma 3 of [12].

Lemma 2. Let {x_{n}} be a positive solution of equation (1) of type (IV). Then there exists a positive number c such that the following inequality holds for all large n:

x_{n+3} ≥ x_{n} ≥ c n Δx_{n}.  (6)

Proof. Let {x_{n}} be a positive solution of equation (1) of type (IV). Since Δx_{n} is decreasing and positive, we have

x_{n} > x_{n} − x_{N} = ∑_{s=N}^{n−1} Δx_{s} ≥ (n − N)Δx_{n}, n ≥ N.

Then, there is a constant c > 0, and a sufficiently large n, such that (6) holds. This completes the proof.

Lemma 3. Let {x_{n}} be a positive solution of equation (1) of type (II). Then there exists a positive number c such that the following inequalities hold for all large n:

x_{n+3} ≥ p_{n}^{\frac{1}{α}} ψ_{n+3} Δ^{2}x_{n},  (7)
x_{n+3}^{α} ≥ c n ψ_{n+3}^{α} Δ (p_{n} (Δ^{2}x_{n})^{α}) ,  (8)

where ψ_{n} = ∑_{s=n}^{∞} _{\frac{(s−n+1)}{p \frac{1}{α} ; s}}.

Proof. The proof can be modeled as that of Lemma 4(ii) of Thandapani and Arockiasamy [12], and therefore the details are omitted.

Lemma 4. Let α ≤ 1. If {x_{n}} is a positive solution of equation (1) of type (IV), then there exists a positive number c such that the following inequality holds for all large n:

x_{n+3} ≥ c n |Δ (p_{n} |Δ^{2}x_{n}|^{α−1} Δ^{2}x_{n}) |^{\frac{1}{α}} ψ_{n+3}.  (9)

Proof. Since Δ(p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) is decreasing, we find that

Δ(p_{s}|Δ^{2}x_{s}|^{α−1}Δ^{2}x_{s}) ≤ Δ(p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}), s ≥ n.

Summing the last inequality from n to s-1, and using the fact Δ^{2}x_{n} < 0 , we obtain

Δ^{2}x_{s} ≤ −|Δ(p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n})|^{\frac{1}{α}} (\frac{s − n + 1}{p_{s}})^{\frac{1}{α}} , s ≥ n.

Since the lim_{n→∞} Δx_{n} = η ≥ 0 is finite, and noting that \frac{1}{α} ≥ 1, summing the last inequality from n to ∞, we have

Δx_{n} ≥ |Δ(p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n})|^{\frac{1}{α}} ∑_{s=n}^{∞} (\frac{s − n + 1}{p_{s}^{\frac{1}{α}}})
= |Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) |^{\frac{1}{α}} ψ_{n}
≥ |Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) |^{\frac{1}{α}} ψ_{n+3}.

Combining the last inequality and the inequality (6), we obtain (9). This completes the proof.

Lemma 5. Let β < 1 ≤ α. Then the condition

∑_{n=n_{0}}^{∞} (\frac{1}{p_{n}} ∑_{s=n_{0}}^{n−1} (n − s) s^{β}q_{s})^{\frac{1}{α}} = ∞,  (10)

implies

∑_{n=n_{0}}^{∞} n^{\frac{β}{α}} ψ_{n+1}^{β} q_{n} = ∞.  (11)

Proof. If (10) holds, then for any N > n_{0}, we have

∑_{n=N}^{∞} p_{n}^{\frac{−1}{α}} (∑_{s=N}^{n−1} (n − s) s^{β}q_{s})^{\frac{1}{α}} = ∞,  (12)

and choose N > 1, such that ψ_{n} ≤ 1 for n ≥ N. In view of (5), the equation (12) implies that

lim_{n→∞} ∑_{s=N}^{n−1} (n − s) s^{β}q_{s} = ∞.

Since {n} is increasing and lim_{n→∞} n = ∞, then by Stolz’s theorem [4], we have

lim_{n→∞} \frac{∑_{s=N}^{n−1} (n − s) s^{β}q_{s}}{n} = lim_{n→∞} \frac{Δ ( ∑_{s=N}^{n−1} (n − s) s^{β}q_{s})}{(Δn)}
= lim_{n→∞} ∑_{s=N}^{n} s^{β}q_{s} ∈ (0, ∞].

Therefore, there exists a constant d > 0, and an integer N_{1} > N, such that

∑_{s=N}^{n} (n − s) s^{β}q_{s} ≥ d n, for every n ≥ N_{1}.  (13)

For all n ≥ N_{1}, using summation by parts, we obtain

∑_{s=N_{1}}^{n−1} p_{s}^{\frac{−1}{α}} (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} = ∑_{s=N_{1}}^{n−1} Δ^{2}ψ_{s} (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}}

= [Δψ_{s} (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} ]_{N_{1}}^{n} − ∑_{s=N_{1}}^{n−1} Δψ_{s+1} Δ (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} . (14)

By Mean value theorem, we have

Δ (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} ≤ \frac{1}{α} (∑_{t=N}^{s} (s − t) t^{β}q_{t})^{\frac{1}{α} −1} ∑_{t=N}^{s} t^{β}q_{t}.  (15)

From the inequalities (14) and (15), using (13), and the fact that \frac{1}{α} ≤ 1 as well as that Δψ_{n} is a negative function , we obtain

∑_{s=N_{1}}^{n−1} p_{s}^{\frac{−1}{α}} (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} ≤ d_{1} − d_{2} ∑_{s=N_{1}}^{n−1} Δψ_{s+1} s^{\frac{1}{α} −1} ∑_{t=N}^{s} t^{β}q_{t}, n ≥ N_{1};

(16)

where

d_{1} = −Δψ_{N_{1}} ( ∑_{t=N}^{N_{1}−1} (N_{1} − t) t^{β}q_{t})^{\frac{1}{α}} > 0, d_{2} = \frac{1}{α}d^{\frac{1}{α} −1} > 0.

Now, from (16), using the fact that

0 ≤ \frac{(1 − α)(β − 1)}{α} = \frac{β}{α} − (\frac{1}{α} − 1 + β)

we have

∑_{s=N_{1}}^{n−1} p_{s}^{\frac{−1}{α}} (∑_{t=N}^{s−1} (s − t) t^{β}q_{t})^{\frac{1}{α}} ≤ d_{1} − d_{2} ∑_{s=N_{1}}^{n−1} Δψ_{s+1} (∑_{t=N}^{s} t^{\frac{β}{α}} q_{t})

= d_{1} + d_{2} [(−ψ_{s+1} ∑_{t=N}^{s−1} t^{\frac{β}{α}} q_{t})_{N_{1}}^{n} + ∑_{s=N_{1}}^{n−1} ψ_{s+1} s^{\frac{β}{α}} q_{s}]

≤ d_{3} + d_{2} ∑_{s=N_{1}}^{n−1} ψ_{s+1} s^{\frac{β}{α}} q_{s}

≤ d_{3} + d_{2} ∑_{s=N_{1}}^{n−1} s^{\frac{β}{α}} ψ_{s+1}^{β} q_{s}, n ≥ N_{1},

where d_{3} = d_{1} + d_{2} ψ_{N_{1}+1} ∑_{t=N_{1}}^{N_{1}−1} t^{\frac{β}{α}} q_{t}. Letting n → ∞, we conclude that (10)

implies (11). The proof is now complete.

In the following, we state and prove some useful propositions that play an important role in proving the main results given in Section 3.

Proposition 1. Let β > α. If there exists a positive solution {x_{n}} of the equation (1) of type (II), then

∑_{n=n_{0}}^{∞} n q_{n} ψ_{n+3}^{β} < ∞.  (17)

Proof. Let {x_{n}} be a positive solution of equation (1) of type (II). Then we have x_{n} ∼ cψ_{n} as n → ∞ (0 < c < ∞). Otherwise if {x_{n}} is of type (I) , (III), or (IV) then lim_{n→∞} x_{n} = c_{0} ∈ (0, ∞]. Moreover lim_{n→∞} ψ_{n} = 0. Hence we have lim_{n→∞} \frac{x_{n}}{ψ_{n}} = ∞, a contradiction . Therefore there is an integer N ≥ n_{0} such that

d_{4}ψ_{n} ≤ x_{n} ≤ 2 d_{4}ψ_{n},  (18)

for all n ≥ N. Now multiply the equation (1) by n, and summing from N to n-1, we obtain

n Δ (p_{n} (Δ^{2}x_{n})^{α}) − p_{n+1} (Δ^{2}x_{n+1})^{α} + ∑_{s=N}^{n−1} s q_{s} x_{s+3}^{β} = c,  (19)

where c is a constant. From (7) and (18), we have

p_{n} (Δ^{2}x_{n})^{α} ≤ 2^{α} dα 4  (20)

for n ≥ N. From (19) and (20) and the fact that Δ (p_{n} (Δ^{2}x_{n})^{α}) > 0, we have

d_{4}^{β} ∑_{n=n_{0}}^{∞} n q_{n} ψ_{n+3}^{β} < ∞,

which gives (17). This completes the proof.

Proposition 2. Let β < α. If there exists a positive solution {x_{n}} of the equation (1) of type (II), then

∑_{n=n_{0}}^{∞} n^{\frac{β}{α}} q_{n} ψ_{n+3}^{β} < ∞.  (21)

Proof. For a positive solution {x_{n}} of equation (1), let us choose N ≥ n_{0} such that type(II) and (8) hold for all n ≥ N. Denote by

A_{n} = Δ (p_{n} (Δ^{2}x_{n})^{α}) .

Then, from the equation (1), and using Mean value theorem, we have

Δ (A_{n}^{(1− \frac{β}{α})}) = (1 − \frac{β}{α}) t^{\frac{−β}{α}} Δ (A_{n}) , A_{n+1} < t < A_{n},
≤ −\frac{(α − β)}{α} q_{n} x_{n+3}^{β} An^{\frac{−β}{α}}
≤ −\frac{(α − β)}{α} c^{\frac{β}{α}} n^{\frac{β}{α}} ψ_{n+3}^{β} q_{n},

where we have used the inequality (8). Summing the last inequality from N to n-1, we obtain

\frac{(α − β)}{α} c^{\frac{β}{α}} ∑_{s=N}^{n−1} s^{\frac{β}{α}} q_{s} ψ_{s+3}^{β} < A_{N}^{1− \frac{β}{α}} < ∞,

which gives (21). This completes the proof.

Proposition 3. Let α < 1 ≤ β. If there exists a positive solution {x_{n}} of the equation (1) of type (IV), then

∑_{n=n_{0}}^{∞} n q_{n} ψ_{n+3}^{β} < ∞.  (22)

Proof. For a positive solution {x_{n}} of equation (1), let us choose an integer N ≥ n_{0} such that type (IV) and (9) hold for all n ≥ N. Denote by A_{n} = Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) . Then, from the equation (1), and using Mean value theorem, we have

−Δ (|A_{n}|^{1− \frac{β}{α}} ) ≥ −\frac{(α − β)}{α} |A_{n}|^{\frac{−β}{α}} (−ΔA_{n})
= −\frac{(α − β)}{α} |A_{n}|^{\frac{−β}{α}} q_{n} x_{n+3}^{β}.

From the last inequality and (9), we see that c_{2} > 0 exists such that

−Δ (|A_{n}|^{1− \frac{β}{α}} ) ≥ \frac{(β − α)}{α}|A_{n}|^{\frac{−β}{α}} c_{2}^{β} n^{β} q_{n} |A_{n}|^{\frac{β}{α}} ψ_{n+3}^{β}, n ≥ N.

Using the fact that β ≥ 1, that is, n^{β} ≥ n for all n ≥ max{1, N} = N_{1}, we have

−Δ (|A_{n}|^{1− \frac{β}{α}} ) ≥ \frac{(β − α)}{α} c_{2}^{β} n q_{n} ψ_{n+3}^{β}, n ≥ N_{1}.

Summing the last inequality from N_{1} to n − 1, we obtain

−|A_{n}|^{1− \frac{β}{α}} + |A_{N_{1}}|^{1− \frac{β}{α}} ≥ \frac{(β − α)}{α} c_{2}^{β} ∑_{s=N_{1}}^{n−1} s q_{s} ψ_{s+3}^{β},

which gives (22). This completes the proof.

Proposition 4. Let β < 1 ≤ α. If there exists a positive solution {x_{n}} of equation (1) of type (IV), then

∑_{n=n_{0}}^{∞} p_{n}^{\frac{−1}{α}} (∑_{s=n_{0}}^{n−1} (n − s) s^{β}q_{s})^{\frac{1}{α}} < ∞.  (23)

Proof. Assume that for a positive solution {x_{n}} of equation (1), type (IV) holds for all n ≥ N ∈ ℕ(n_{0}). By Lemma 2, there exists a constant c_{1} > 0, and an integer N_{1} ≥ N, such that (6) holds for all n ≥ N_{1}. Summing the equation (1) from N_{1} to n − 1, we see that

−Δ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) ≥ ∑_{s=N_{1}}^{n−1} q_{s} x_{s+3}^{β}, n ≥ N_{1}.

Summing the last inequality from N_{1} to n-1, we have

−Δ^{2}x_{n} ≥ \frac{1}{pn^{\frac{1}{α}}} ( ∑_{s=N_{1}}^{n−1}(n − s) q_{s} x_{s+3}^{β})^{\frac{1}{α}} , n ≥ N_{1}.

From the last inequality, we see that

−Δ ((Δx_{n})^{1−β}) ≥ (1 − β)(Δx_{n})^{−β}(−Δ^{2}x_{n})

≥ (1 − β)(Δx_{n})^{−β} \frac{1}{pn^{\frac{1}{α}}} ( ∑_{s=N_{1}}^{n−1}(n − s) q_{s} x_{s+3}^{β})^{\frac{1}{α}} , n ≥ N_{1}.

Since {Δx_{n}} is decreasing, that is, (Δx_{n})^{−β} ≥ (Δx_{s})^{−β} for n ≥ s, we have

−Δ ((Δx_{n})^{1−β}) ≥ (1 − β)\frac{1}{pn^{\frac{1}{α}}} ( ∑_{s=N_{1}}^{n−1}(n − s) q_{s} (Δx_{s})^{−βα} x_{s+3}^{β})^{\frac{1}{α}}

= (1 − β)\frac{1}{pn^{\frac{1}{α}}} ( ∑_{s=N_{1}}^{n−1}(n − s) q_{s} (sΔx_{s})^{−βα} s^{βα} x_{s+3}^{β})^{\frac{1}{α}} , n ≥ N_{1}.

Now using (6), we obtain

−Δ ((Δx_{n})^{1−β}) ≥ (1 − β) \frac{c_{1}^{β}}{pn^{\frac{1}{α}}} ( ∑_{s=N_{1}}^{n−1}(n − s) q_{s} x_{s+3}^{−βα} s^{βα} x_{s+3}^{β})^{\frac{1}{α}} , n ≥ N_{1}.

Since Δx_{n+3} ≤ Δx_{N_{1}}, n ≥ N_{1}, there exists some constant c > 0, and an integer N_{2} ≥ N_{1}, such that x_{n+3} ≤ c n for n ≥ N_{2}. Therefore, the fact that β(1 − α) ≤ 0 implies x_{n+3}^{β(1−α)} ≥ c^{β(1−α)} n^{β(1−α)} for n ≥ N_{2}. Hence we have

−Δ ((Δx_{n})^{1−β}) ≥ (1−β)c^{\frac{β(1−α)}{α}} \frac{c_{1}^{β}}{pn^{\frac{1}{α}}} ( ∑_{s=N_{2}}^{n−1}(n − s) q_{s} s^{β(1−α)} s^{βα})^{\frac{1}{α}} , n ≥ N_{2}.

Summing the last inequality from N_{2} to ∞, we obtain

(Δx_{N_{2}})^{1−β} ≥ K ∑_{n=N_{2}}^{∞} \frac{1}{pn^{\frac{1}{α}}} ( ∑_{s=N_{2}}^{n−1}(n − s) q_{s} s^{β})^{\frac{1}{α}} ,

where K = (1 − β)c^{\frac{β(1−α)}{α}} c_{1}^{β} > 0, which gives (23). This completes the proof.

3 Oscillation Theorems

In this section, we state and prove criteria for the oscillation of all solutions of equation (1).

Theorem 1. Let β ≥ 1 > α. If

∑_{n=n_{0}}^{∞} nq_{n}ψ_{n+3}^{β} = ∞,  (24)

then every solution of equation (1) is oscillatory.

Proof. Assume, to contrary, that {x_{n}} is a positive solution of equation(1). Then {x_{n}} falls into one of the four types (I)-(IV) mentioned in Lemma 1. Therefore it is enough to show that in each case, we are led to a contradiction to (24).

Case (I): Let {x_{n}} be a positive solution of equation (1) of type (I) for all n ≥ N. Since Δx_{n} ≥ Δx_{N}, n ≥ N, there exists some constant c > 0, and N_{1} ≥ N, such that x_{n} ≥ cn for n ≥ N_{1}. Summing the equation (1) from N_{1} to ∞ and using the last inequality we have

c^{β} ∑_{n=N_{1}}^{∞} n^{β}q_{n} ≤ ∑_{n=N_{1}}^{∞} x_{n}^{β}q_{n} ≤ Δ (p_{N_{1}}|Δ^{2}x_{N_{1}}|^{α−1}Δ^{2}x_{N_{1}}) .

Hence, we conclude that

∑_{n=N_{1}}^{∞} n^{β}q_{n} < ∞.  (25)

On the other hand, from (5),there exists an integerN_{2} ≥ N_{1}, such that ψ_{n+3} ≤ 1 for n ≥ N_{2}. The fact that β ≥ 1, implies n^{β} ≥ n for n ≥ N_{3} = max{N_{2}, 1}, and consequently, we have

∑_{n=N_{3}}^{∞} n q_{n} ψ_{n+3}^{β} ≤ ∑_{n=N_{3}}^{∞} n q_{n} ≤ ∑_{n=N_{3}}^{∞} n^{β}q_{n} < ∞.  (26)

by (25), a contradiction to (24).

Case(II): If there exists a positive solution {x_{n}} of type(II), then from Proposition 1, we see that condition (24) fails to hold.

Case(III): Let {x_{n}} be a positive solution of equation (1) of type (III) for all n ≥ N. Multiplying equation (1) by n, and summing the resulting equation from N to n-1, using summation by parts, we have

∑_{s=N}^{n−1} s q_{s} x_{s+3}^{β} = c_{3} − nΔ (p_{n}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) + p_{n+1}|Δ^{2}x_{n+1}|^{α−1}Δ^{2}x_{n+1} < c_{3},

where c_{3} = NΔ (p_{N}|Δ^{2}x_{N}|^{α−1}Δ^{2}x_{N}) − p_{N+1}|Δ^{2}x_{N+1}|^{α−1}Δ^{2}x_{N+1} > 0 is constant. Hence, we conclude that

∑_{n=n_{0}}^{∞} n q_{n} x_{n+3}^{β} < ∞.

Since {x_{n}} is increasing, this implies that

∑_{n=n_{0}}^{∞} n q_{n} < ∞.  (27)

Combining (26) and (27) we are led to contradiction with the assumption (24).

Case (IV): If there exists a positive solution {x_{n}} of type (IV), then from Proposition 3, we conclude that (24) is not satisfied. This completes the proof of the theorem.

Theorem 2. Let β < 1 ≤ α. If

∑_{n=n_{0}}^{∞} (\frac{1}{p_{n}} ∑_{s=n_{0}}^{n−1}(n − s)s^{β}q_{s})^{\frac{1}{α}} = ∞,  (28)

then every solution of equation (1) is oscillatory .

Proof. Assume, to contrary, that {x_{n}} is a positive solution of equation (1). Then {x_{n}} falls into one of the four types (I)-(IV) mentioned in Lemma 1. Therefore it is enough to show that in each case, we are led to a contradiction to (28).

Case (I): Let {x_{n}} be a positive solution of equation (1) of type (I) for all n ≥ N. As in the proof of Theorem 1, we can obtain (25). Then

∑_{s=n_{0}}^{n−1}(n − s)s^{β}q_{s} ≤ (n − n_{0}) ∑_{s=n_{0}}^{n−1} s^{β}q_{s} ≤ Q n for n ≥ N,

where Q = ∑_{n=n_{0}}^{∞} n^{β}q_{n}. Using the last inequality, we see that

∑_{n=N}^{∞} \frac{1}{pn^{\frac{1}{α}}} [∑_{s=n_{0}}^{n−1}(n − s)s^{β}q_{s}]^{\frac{1}{α}} ≤ Q^{\frac{1}{α}} ∑_{n=N}^{∞} (\frac{n}{p_{n}} )^{\frac{1}{α}} .  (29)

In view of (5), (29) implies that (28) fails to hold.

Case(II): If there exists a positive solution {x_{n}} of type (II), then from Proposition 2, we conclude that (28) is not satisfied.

Case (III): Let {x_{n}} be a positive solution of equation (1) of type (III) for all n ≥ N. As in the proof of Theorem 1, we can obtain (27). Using (27) and the fact that β < 1, implies n^{β} < n for all n ≥ max{1, n_{0}} = N_{1}, we get (25). Now, following the same arguments as in case (I), using (25), we are led to a contradiction to (28).

Case (IV): If there exists a positive solution {x_{n}} of type (IV), then from Proposition 4, we find that condition (28) fails to hold. This completes the proof of the theorem.

4 Example

In this section, we present some examples to illustrate the results given in the previous section.

Example 1. Consider the equation

Δ^{2} (n^{μ}|Δ^{2}x_{n}|^{α−1}Δ^{2}x_{n}) + n^{−λ}|x_{n+3}|^{β−1}x_{n+3} = 0, n ∈ ℕ(n_{0}).  (30)

Let β ≥ 1 > α and μ > 1 + α. The assumption μ > 1 + α ensures that (5) is satisfied for the function p_{n} = n^{μ}. Then by using Theorem 1, we obtain that the equation (1) is oscillatory if (24) is satisfied .

Since ψ_{n} = ∑_{s=n}^{∞} _{\frac{(s−n+1)}{p \frac{1}{α} ; s}} ∼ n^{2− \frac{μ}{α}} , n → ∞,it follows that

∑_{n=n_{0}}^{∞} n q_{n} ψ_{n+3}^{β} ∼ ∑_{n=n_{0}}^{∞} n^{1−λ+β(2− \frac{μ}{α})}, n → ∞.

Thus, the condition (24) holds if 2 − λ + β(2 − \frac{μ}{α}) > 0. Therefore the equation (30) is oscillatory, if λ < 2 + β(2 − \frac{μ}{α}). Using the assumptions β ≥ 1 > α and μ > 1 + α, we have λ < 2 + (1 − \frac{1}{α})β < 2.

Let β < 1 ≤ α. The assumption μ > 2α ensures that (5) is satisfied.Then by using Theorem 2, we obtain that the equation (30) is oscillatory, if condition (28) is satisfied. For q_{n} = n^{−λ}, it is easy to see that

∑_{s=n_{0}}^{n−1} ∑_{t=n_{0}}^{s−1} t^{β}q_{t} ∼ n^{β−λ+2}, n → ∞,

and so ,

∑_{n=n_{0}}^{∞} (\frac{1}{p_{n}} ∑_{s=n_{0}}^{n−1}(n − s)s^{β}q_{s})^{\frac{1}{α}} ∼ ∑_{n=n_{0}}^{∞} n^{\frac{β−λ+2−μ}{α}} , n → ∞.

Then the condition (28) holds if λ < α + β − μ + 2. Thus, the equation (30) is oscillatory if λ < α + β − μ + 2. Using the assumptions β < 1 ≤ α and μ > 2α, we obtain that λ < β − α + 2 < 2.

Example 2. Consider the difference equation

Δ^{2} (2^{n}(Δ^{2}x_{n})^{\frac{1}{3}} ) + 2^{3n+9}9^{\frac{2}{3}} x_{n+3} = 0, n ≥ 1.  (31)

Here p_{n} = 2^{n}, q_{n} = 2^{3n+9}9^{\frac{2}{3}} , α = \frac{1}{3}, β = 1. It is easy to see that all conditions of Theorem 1 are satisfied and hence every solution of equation (31) is oscillatory. In fact \{x_{n}\} = \{\frac{(−1)^{n}}{2^{3n}} \} is one such solution of equation (31).

Example 3. Consider the difference equation

Δ^{2} (n^{7}(Δ^{2}x_{n})^{3}) + 64 ((n + 2)^{7} + 2(n + 1)^{7} + n^{7}) xn+3^{\frac{1}{3}} = 0, n ≥ 1. (32)

Here p_{n} = n^{7}, q_{n} = 64 ((n + 2)^{7} + 2(n + 1)^{7} + n^{7}) , α = 3, β = \frac{1}{3}. It is easy to see that all conditions of Theorem 2 are satisfied and hence all solutions of equation (32) are oscillatory. In fact {x_{n}} = {(−1)^{n}} is one such solution of equation (32). We conclude this paper with the following remark.

Remark 2. The results obtained in this paper gives a partial answer to the problem given in [10]. Further the results reduces to that of in [12] when α = 1.

Acknowledgement: The authors thank the referee for his/her suggestions which improve the content of the paper.

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(Received August 8, 2009)