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Forced oscillation of higher-order nonlinear neutral difference equations with positive and negative coefficients

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Abstract

In this paper, we study the forced oscillation of the higher-order nonlinear difference equation of the form

Δ m [ x ( n ) p ( n ) x ( n τ ) ] + q 1 (n) Φ α (n σ 1 )+ q 2 (n) Φ β (n σ 2 )=f(n),

where m1, τ, σ 1 and σ 2 are integers, 0<α<1<β are constants, Φ (u)= | u | 1 u, p(n), q 1 (n), q 2 (n) and f(n) are real sequences with p(n)>0. By taking all possible values of τ, σ 1 and σ 2 into consideration, we establish some new oscillation criteria for the above equation in two cases: (i) q 1 = q 1 (n)0, q 2 = q 2 (n)>0; (ii) q 1 0, q 2 <0.

MSC:39A10.

1 Introduction

Qualitative theory of difference equations has received much attention in recent years due to its extensive applications in computer, probability theory, queuing problems, statistical problems, stochastic time series, combinatorial analysis, number theory, electrical networks, genetics in biology, economics, psychology, sociology, and so on [1, 2].

In this paper, we consider the oscillation of the following m th-order forced nonlinear difference equation of the form

Δ m [ x ( n ) p ( n ) x ( n τ ) ] + q 1 (n) Φ α (n σ 1 )+ q 2 (n) Φ β (n σ 2 )=f(n),
(1)

where m1, τ, σ 1 and σ 2 are integers, Φ (u)= | u | 1 u, p(n), q 1 (n), q 2 (n) and f(n) are real sequences defined on N={0,1,2,} with p(n)>0, 0<α<1<β are constants, and

Δx(n)=x(n+1)x(n), Δ i x(n)=Δ ( Δ i 1 x ( n ) ) ,2im.

As usual, a solution of Eq. (1) is said to be oscillatory, if for every integer N0, there exists nN such that x(n)x(n+1)0; otherwise, it is called nonoscillatory.

For the continuous version of Eq. (1), many authors have studied its oscillation (see monograph [3] and references therein). To the best of our knowledge, little has been known about the forced oscillation of Eq. (1) with positive and negative coefficients ( q 1 0 q 2 >0 or q 1 0 q 2 <0) and mixed nonlinearities (0<α<1β>1). For some particular cases of Eq. (1), there have been many oscillation results in [419], to name a few. Motivated by the work in [2022], we study the forced oscillation of Eq. (1) in this paper.

The main contribution of this paper is that we establish some new oscillation criteria for Eq. (1) with positive and negative coefficients and mixed nonlinearities. Unlike some existing results in the literature, all possible values of delays τ, σ 1 and σ 2 are considered.

2 Main results

Throughout this paper, we denote

ϕ 0 (n,s)=ϕ(n,s)= ( n s ) ( k ) =(ns)(ns+1)(ns+k1),km,
(2)
ϕ i (n,s)= ( 1 ) i Δ s i ϕ(n,s)= C k i ( n s ) ( k i ) ,i=1,2,,m.
(3)

By the straightforward computation, it is not difficult to see that

{ ϕ ( n , s ) = 0 , n s n + k 1 , ϕ i ( n + i + 1 , n + m ) = 0 , i = 0 , 1 , 2 , , m 1 , ϕ m ( n , s ) 0 , 0 s n 1 ,
(4)

and

lim n ϕ i ( n + i + 1 , n 0 ) ϕ ( n + 1 , n 0 ) =o(1),i=1,2,,m,
(5)

where n 0 0 is an integer. We also denote s = l k =0 if k<l.

The following two facts can be easily proved.

Fact 1. Set F(x)=axb x λ , where x0, a0 and b>0. If λ>1, F(x) obtains its maximum F max =(λ1) λ λ 1 λ a λ λ 1 b 1 1 λ .

Fact 2. Set G(x)=cxd x λ , where x0, c>0 and d0. If 0<λ<1, G(x) obtains its minimum G min =(λ1) λ λ 1 λ c λ λ 1 d 1 1 λ .

We now present the main results of this paper as follows.

Theorem 1 Assume that q 1 (n)0, q 2 (n)>0, σ 1 mand σ 2 τm. If

lim n sup 1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) Q 2 ( n , s ) ] + s = n 0 n σ 1 1 Q 1 ( n , s ) } =+,
(6)
lim n inf 1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) + Q 2 ( n , s ) ] s = n 0 n σ 1 1 Q 1 ( n , s ) } =,
(7)

where

Q 1 (n,s)=(α1) α α 1 α [ ϕ m ( n + m , s ) ] α α 1 [ ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | ] 1 1 α ,
(8)
Q 2 (n,s)=(β1) β β 1 β [ ϕ m ( n + m , s ) p ( s ) ] β β 1 [ ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) ] 1 1 β ,
(9)

all solutions of Eq. (1) are oscillatory.

Proof Assume to the contrary that there exists a nontrivial solution x(n) of Eq. (1) such that x(n) is nonoscillatory. That is, x(n) does not change sign eventually. Without loss of generality, let x(n σ 1 )0, x(n σ 2 )0, x(nτ)0 for n n 0 , where n 0 0 is sufficiently large. By the straightforward computation, we have

F 1 (n,s) F 2 (n,s)= s = n 0 n + m 1 ϕ(n,s)f(s),
(10)

where

F 1 ( n , s ) = s = n 0 n + m 1 ϕ ( n , s ) Δ m x ( s ) s = n 0 n + m 1 ϕ ( n , s ) | q 1 ( s ) | x α ( s σ 1 ) , F 2 ( n , s ) = s = n 0 n + m 1 ϕ ( n , s ) Δ m [ p ( s ) x ( n τ ) ] s = n 0 n + m 1 ϕ ( n , s ) q 2 ( s ) x β ( s σ 2 ) .

Noting that

ϕ ( n , s ) Δ x ( s ) = Δ [ ϕ ( n , s 1 ) x ( s ) ] + ϕ 1 ( n , s 1 ) x ( s ) = Δ [ ϕ ( n + 1 , s ) x ( s ) ] + ϕ 1 ( n + 1 , s ) x ( s ) ,

we can get from (2), (3) and (4) that

s = n 0 n + m 1 ϕ ( n , s ) Δ m x ( s ) = s = n 0 n + m 1 Δ [ ϕ ( n + 1 , s ) Δ m 1 x ( s ) ] + s = n 0 n + m 1 ϕ 1 ( n + 1 , s ) Δ m 1 x ( s ) = ϕ ( n + 1 , n 0 ) Δ m 1 x ( n 0 ) + s = n 0 n + m 1 ϕ 1 ( n + 1 , s ) Δ m 1 x ( s ) = ϕ ( n + 1 , n 0 ) Δ m 1 x ( n 0 ) + s = n 0 n + m 1 Δ [ ϕ 1 ( n + 2 , s ) Δ m 2 x ( s ) ] + s = n 0 n + m 1 ϕ 2 ( n + 2 , s ) Δ m 2 x ( s ) = ϕ ( n + 1 , n 0 ) Δ m 1 x ( n 0 ) ϕ 1 ( n + 2 , n 0 ) Δ m 2 x ( n 0 ) + s = n 0 n + m 1 ϕ 2 ( n + 2 , s ) Δ m 2 x ( s ) = i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) + s = n 0 n + m 1 ϕ m ( n + m , s ) x ( s ) .
(11)

Since ϕ(n,s)=0 for nsn+m1 due to (4), we get from (11) that

F 1 ( n , s ) = s = n 0 n + m 1 ϕ m ( n + m , s ) x ( s ) s = n 0 n 1 ϕ ( n , s ) | q 1 ( s ) | x α ( s σ 1 ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) = s = n 0 n + m 1 ϕ m ( n + m , s ) x ( s ) s = n 0 σ 1 n σ 1 1 ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) .
(12)

Noting that σ 1 m, we have that n+m1n σ 1 1. Therefore, we get from (12) that

F 1 ( n , s ) s = n 0 n σ 1 1 [ ϕ m ( n + m , s ) x ( s ) ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) ] s = n 0 σ 1 n 0 1 ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) .
(13)

By Fact 2 and (13), it is not difficult to see that

F 1 ( n , s ) s = n 0 n σ 1 1 Q 1 ( n , s ) s = n 0 σ 1 n 0 1 ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) ,
(14)

where Q 1 (n,s) is defined by (8).

On the other hand, similar to the above analysis, we have that

F 2 ( n , s ) = s = n 0 n + m 1 ϕ m ( n + m , s ) p ( s ) x ( s τ ) s = n 0 n + m 1 ϕ ( n , s ) q 2 ( s ) x β ( s σ 2 ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] = s = n 0 n + m 1 ϕ m ( n + m , s ) p ( s ) x ( s τ ) s = n 0 n 1 ϕ ( n , s ) q 2 ( s ) x β ( s σ 2 ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ x ( n 0 τ ) p ( n 0 ) ] = s = n 0 n + m 1 ϕ m ( n + m , s ) p ( s ) x ( s τ ) s = n 0 σ 2 + τ n σ 2 + τ 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] .
(15)

Since σ 2 τm, we have that n σ 2 +τ1n+m1. By (15), we get

F 2 ( n , s ) s = n 0 n + m 1 [ ϕ m ( n + m , s ) p ( s ) x ( s τ ) ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) ] + s = n 0 n 0 σ 2 + τ 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] .
(16)

By Fact 1 and (16), we have that

F 2 ( n , s ) s = n 0 n + m 1 Q 2 ( n , s ) + s = n 0 n 0 σ 2 + τ 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] ,
(17)

where Q 2 (n,s) is defined by (9).

Multiplying 1 ϕ ( n + 1 , n 0 ) on both sides of (10), by (14), (17) and (5), we have that there exists a constant M 1 such that

1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) + Q 2 ( n , s ) ] s = n 0 n σ 1 1 Q 1 ( n , s ) } M 1 ,

which contradicts (7). For the case when x(n) is eventually negative, we can similarly get a contradiction to (6). This completes the proof of Theorem 1. □

Theorem 2 Assume that q 1 (n)0, q 2 (n)<0, σ 1 τmand σ 2 m. If

lim n sup 1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) P 1 ( n , s ) ] + s = n 0 n σ 2 + τ 1 P 2 ( n , s ) } =+,
(18)
lim n inf 1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) + P 1 ( n , s ) ] s = n 0 n σ 2 + τ 1 P 2 ( n , s ) } =,
(19)

where

P 1 (n,s)=(β1) β β 1 β [ ϕ m ( n + m , s ) ] β β 1 [ ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | ] 1 1 β ,
(20)
P 2 (n,s)=(α1) α α 1 α [ ϕ m ( n + m , s ) p ( s ) ] α α 1 [ ϕ ( n , s + σ 1 τ ) q 1 ( s + σ 1 τ ) ] 1 1 α ,
(21)

all solutions of Eq. (1) are oscillatory.

Proof Suppose to the contrary that there exists a nontrivial solution x(n) of Eq. (1) such that x(n) is nonoscillatory. We may let x(n σ 1 )0, x(n σ 2 )0, x(nτ)0 for n n 0 , where n 0 0 is sufficiently large. By the straightforward computation, we get from Eq. (1) that

G 1 (n,s) G 2 (n,s)= s = n 0 n + m 1 ϕ(n,s)f(s),
(22)

where

G 1 ( n , s ) = s = n 0 n + m 1 ϕ ( n , s ) Δ m x ( s ) s = n 0 n + m 1 ϕ ( n , s ) | q 2 ( s ) | x β ( s σ 2 ) , G 2 ( n , s ) = s = n 0 n + m 1 ϕ ( n , s ) Δ m [ p ( s ) x ( n τ ) ] s = n 0 n + m 1 ϕ ( n , s ) q 1 ( s ) x α ( s σ 1 ) .

Noticing that ϕ(n,s)=0 for nsn+m1, we get from (11) that

G 1 ( n , s ) = s = n 0 n + m 1 ϕ m ( n + m , s ) x ( s ) s = n 0 n 1 ϕ ( n , s ) | q 2 ( s ) | x β ( s σ 2 ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) = s = n 0 n + m 1 ϕ m ( n + m , s ) x ( s ) s = n 0 σ 2 n σ 2 1 ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | x β ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) .
(23)

Since σ 2 m, we have that n+m1n σ 2 1. Thus, we can get from (23) that

G 1 ( n , s ) s = n 0 n + m 1 [ ϕ m ( n + m , s ) x ( s ) ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | x β ( s ) ] + s = n 0 n 0 σ 2 1 ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | x β ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) .
(24)

By Fact 1 and (24), it is easy to see that

G 1 ( n , s ) s = n 0 n + m 1 P 1 ( n , s ) + s = n 0 n 0 σ 2 1 ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | x β ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) ,
(25)

where P 1 (n,s) is defined by (20).

On the other hand, similar to the computation of (11), we can get

G 2 ( n , s ) = s = n 0 n + m 1 ϕ m ( n + m , s ) p ( s ) x ( s τ ) s = n 0 n 1 ϕ ( n , s ) q 1 ( s ) x α ( s σ 1 ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] = s = n 0 n + m 1 ϕ m ( n + m , s ) p ( s ) x ( s τ ) s = n 0 σ 1 + τ n σ 1 + τ 1 ϕ ( n , s + σ 1 τ ) q 1 ( s + σ 1 τ ) x α ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] .
(26)

Noting that σ 1 τm implies n+m1n σ 1 +τ1, we get from (26) that

G 2 ( n , s ) s = n 0 n σ 1 + τ 1 [ ϕ m ( n + m , s ) p ( s ) x ( s τ ) ϕ ( n , s + σ 1 τ ) q 1 ( s + σ 1 τ ) x α ( s τ ) ] s = n 0 σ 1 + τ n 0 1 ϕ ( n , s + σ 1 τ ) q 1 ( s + σ 1 τ ) x α ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] .
(27)

By Fact 2 and (27), we have that

G 2 ( n , s ) s = n 0 n σ 1 + τ 1 P 2 ( n , s ) s = n 0 σ 1 + τ n 0 1 ϕ ( n , s + σ 1 τ ) q 1 ( s + σ 1 τ ) x α ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] ,
(28)

where P 2 (n,s) is defined by (21).

Multiplying 1 ϕ ( n + 1 , n 0 ) on both sides of (22), from (25), (28) and (5), we have that there exists a constant M 2 such that

1 ϕ ( n + 1 , n 0 ) { s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) P 1 ( n , s ) ] + s = n 0 n σ 2 + τ 1 P 2 ( n , s ) } M 2 .

This is a contradiction to (18). For the case when x(n) is eventually negative, we can similarly get a contradiction to (19). This completes the proof of Theorem 2. □

By Theorems 1 and 2, the following two corollaries are immediate.

Corollary 1 Assume that q 1 (n)0, q 2 (n)>0and σ 2 τm. If

lim n sup 1 ϕ ( n + 1 , n 0 ) s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) Q 2 ( n , s ) ] = + , lim n inf 1 ϕ ( n + 1 , n 0 ) s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) + Q 2 ( n , s ) ] = ,

where Q 2 (n,s)is defined by (9), all solutions of Eq. (1) are oscillatory for any constant σ 1 .

Proof In fact, we have that F 1 (n,s)0 for any constant σ 1 since q 1 (n,s)0. So, we can drop F 1 (n,s) in the estimation of (10). The other proof runs as that of Theorem 1, and hence it is omitted. □

Corollary 2 Assume that q 1 (n)0, q 2 (n)<0and σ 2 m. If

lim n sup 1 ϕ ( n + 1 , n 0 ) s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) P 1 ( n , s ) ] = + , lim n inf 1 ϕ ( n + 1 , n 0 ) s = n 0 n + m 1 [ ϕ ( n , s ) f ( s ) + P 1 ( n , s ) ] = ,

where P 1 (n,s)is defined by (20), all nontrivial solutions of Eq. (1) are oscillatory.

Proof For this case, we have that G 2 (n,s)0 for any constant σ 1 since q 1 (n,s)0. Therefore, we can drop G 2 (n,s) in the estimation of (22). The other proof runs as that of Theorem 2. □

For other cases of σ 1 and σ 2 that are not covered by Theorem 1 and Theorem 2, the above method usually does not give sufficient conditions for the oscillation of all solutions of Eq. (1). However, when assuming that the solutions of Eq. (1) satisfy appropriate conditions, sufficient conditions for such solutions can also be derived. In the following, we are focused on the oscillation of all solutions of Eq. (1) satisfying x(n)=O( n r ) for some r>0. Here, x(n)=O( n r ) means that there exists a constant c>0 such that |x(n)|c n r for n n 0 .

Theorem 3 Assume that q 1 (n)0, q 2 (n)>0, and (6) and (7) hold. All solutions satisfyingx(n)=O( n r )are oscillatory if one of the following conditions holds:

  1. (i)

    σ 1 <m, σ 2 τm, and

    lim n sup 1 ϕ ( n + 1 , n 0 ) s = n + m n σ 1 [ ϕ m ( n + m , s ) s r ] <,
    (29)
  2. (ii)

    σ 1 m, σ 2 τ>m, and

    lim n sup 1 ϕ ( n + 1 , n 0 ) s = n σ 2 + τ n + m 1 [ ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) s r β ] <,
    (30)
  3. (iii)

    σ 1 <m, σ 2 τ>m, (29) and (30) hold.

Proof Assume that there exists a nontrivial solution x(n) of Eq. (1) such that x(n) is nonoscillatory. Without loss of generality, let x(n σ 1 )0, x(n σ 2 )0, x(nτ)0 for n n 0 , where n 0 0 is sufficiently large.

  1. (i)

    For the case σ 1 <m, we have that n+m1<n σ 1 1. Therefore, we get from (12) that

    F 1 ( n , s ) = s = n 0 n σ 1 1 [ ϕ m ( n + m , s ) x ( s ) ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) ] s = n + m n σ 1 1 ϕ m ( n + m , s ) x ( s ) + s = n 0 n 0 σ 1 1 ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | x α ( s ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) .
    (31)

By Fact 2 and (31), and noting that x(n)c n r for n n 0 and some constant c>0, we get

F 1 ( n , s ) s = n 0 n σ 1 1 Q 1 ( n , s ) c s = n + m n σ 1 1 ϕ ( n , s + σ 1 ) | q 1 ( s + σ 1 ) | s r i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 x ( n 0 ) ,
(32)

where Q 1 (n,s) is defined by (8). Multiplying 1 ϕ ( n + 1 , n 0 ) on both sides of (10), from (32), (17), (5) and (29), we get a contradiction to (7).

  1. (ii)

    For the case σ 2 τ>m, we have that n σ 2 +τ1<n+m1. By (15), we get

    F 2 ( n , s ) = s = n 0 n + m 1 [ ϕ m ( n + m , s ) p ( s ) x ( s τ ) ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) ] + s = n σ 2 + τ n + m 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) s = n 0 σ 2 + τ n 0 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) x β ( s τ ) i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] .
    (33)

By Fact 1 and (33), and noting that x(n)c n r for n n 0 , we have that

F 2 ( n , s ) s = n 0 n + m 1 Q 2 ( n , s ) + c β s = n σ 2 + τ n + m 1 ϕ ( n , s + σ 2 τ ) q 2 ( s + σ 2 τ ) ( s τ ) β r i = 0 m 1 ϕ i ( n + i + 1 , n 0 ) Δ m i 1 [ p ( n 0 ) x ( n 0 τ ) ] ,
(34)

where Q 2 (n,s) is defined by (9). Multiplying 1 ϕ ( n + 1 , n 0 ) on both sides of (10), from (14), (34) and (5), we can get a contradiction to (7).

  1. (iii)

    Multiplying 1 ϕ ( n + 1 , n 0 ) on both sides of (10), from (32), (34), (29) and (30), we derive a contradiction. The proof of Theorem 3 is complete. □

Theorem 4 Assume that q 1 (n)0, q 2 (n)<0, (18) and (19) hold. All solutions satisfyingx(n)=O( n r )are oscillatory if one of the following conditions holds:

  1. (i)

    σ 1 τm, σ 2 >m, and

    lim n sup 1 ϕ ( n + 1 , n 0 ) s = n σ 2 n + m 1 [ ϕ ( n , s + σ 2 ) | q 2 ( s + σ 2 ) | s r β ] <,
    (35)
  2. (ii)

    σ 1 τm, σ 2 m, and

    lim n sup 1 ϕ ( n + 1 , n 0 ) s = n + m n σ 1 + τ 1 [ ϕ m ( n + m , s ) p ( s ) ( s τ ) r ] <,
    (36)
  3. (iii)

    σ 1 τ<m, σ 2 >m, (35) and (36) hold.

Proof The proof is similar to that of Theorem 2 and Theorem 3, and hence it is omitted. □

3 Examples

We here work out two simple examples to illustrate the importance of Theorem 1 and Theorem 2.

Example 1 Consider the following third-order neutral difference equation:

Δ 3 [ x ( n ) x ( n 1 ) ] Φ 1 / 2 (n σ 1 )+ Φ 2 (n σ 2 )= n k sinn,n0,
(37)

where k>0 is a constant. It is evident that m=3, σ 1 =τ=1, σ 2 =2, α=1/2, β=2, p(n)1, q 1 (n)1, q 2 (n)1 and f(n)= n k sinn. If we choose ϕ(n,s)= ( n s ) ( 3 ) , we have ϕ 3 (n,s)=6. By the straightforward computation, we have that

Q 1 (n,s)= [ ( n s 1 ) ( 2 ) ] 2 /24, Q 2 (n,s)= [ ( n s + 3 ) ( 3 ) ] 1 .

By Theorem 1, we have that every solution of Eq. (37) is oscillatory if

lim n sup 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ s = 0 n + 2 ( n s ) ( 3 ) s k sin s + s = 0 n 2 Q 1 ( n , s ) ] = + , lim n inf 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ s = 0 n + 2 ( n s ) ( 3 ) s k sin s s = n 0 n 2 Q 1 ( n , s ) ] = .

It is not difficult to see that the above two inequalities hold for appropriate k>0.

Example 2 Consider the following third-order neutral difference equation:

Δ 3 [ x ( n ) x ( n 1 ) ] + Φ 1 / 2 (n σ 1 ) Φ 2 (n σ 2 )= n k cosn,n0,
(38)

where k>0 is a constant. It is obvious that m=3, σ 1 =2, σ 2 =τ=1, α=1/2, β=2, p(n)1, q 1 (n)1, q 2 (n)1 and f(n)= n k cosn. We also choose ϕ(n,s)= ( n s ) ( 3 ) . By the straightforward computation, we have that

P 1 (n,s)= [ ( n s + 3 ) ( 3 ) ] 1 , P 2 (n,s)= [ ( n s 1 ) ( 3 ) ] 2 /24.

By Theorem 2, we have that every solution of Eq. (38) is oscillatory if

lim n sup 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ s = 0 n + 2 ( n s ) ( 3 ) s k cos s + s = 0 n 1 P 2 ( n , s ) ] = + , lim n inf 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) [ s = 0 n + 2 ( n s ) ( 3 ) s k cos s s = 0 n 1 P 2 ( n , s ) ] = .

It is not difficult to see that the above two inequalities hold for appropriate k>0.

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Acknowledgements

The authors thank the reviewers for their helpful and valuable comments on this paper. This work was supported by the Natural Science Foundation of Shandong Province (grant nos. ZR2010AL002 and JQ201119) and the National Natural Science Foundation of China (grant no. 61174217).

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Correspondence to Yuangong Sun.

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Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

YS framed the problem. YG solved the problem. BZ and HL made necessary changes in the proof of the theorems. All authors read and approved the manuscript.

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Keywords

  • forced oscillation
  • neutral difference equation
  • positive and negative coefficients
  • higher-order