# Monotonicity of Eventually Positive Solutions for a Second Order Nonlinear Difference Equation

Authors: Huiqin Chen; Zhen Jin; Shugui Kang
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We derive several sufficient conditions for monotonicity of eventually positive solutions on a class of second order perturbed nonlinear difference equation. Furthermore, we obtain a few nonexistence criteria for eventually positive monotone solutions of this equation. Examples are provided to illustrate our main results.


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## Body

## 1. Introduction

The theory of difference equations and their applications have received intensive attention. In the last few years, new research achievements kept emerging (see [1–7]). Among them, in [3], Saker considered the second order nonlinear delay difference equation
(1)Δ(pnΔxn)+qnf(xn-σ)=0,n⩾0.Saker used the Riccati transformation technique to obtain several sufficient conditions which guarantee that every solution of (1) oscillates or converges to zero. In [4], Rath et al. considered the more general second order equations
(2)Δ(rnΔ(yn-pnyn-m))+qnG(yn-k)=0,n⩾0,Δ(rnΔ(yn-pnyn-m))+qnG(yn-k)=fn,n⩾0.They found necessary conditions for the solutions of the above equations to be oscillatory or tend to zero. Following this trend, this paper is concerned with the second order perturbed nonlinear difference equation(3)Δ(anΔxn)+P(n,xn,xn+1)=Q(n,xn,Δxn),n⩾0,
where {an} is a positive sequence, P,Q:N×R2→R are two continuous functions, and Δ is the forward difference operator defined as Δxn=xn+1-xn.In [8], Li and Cheng considered the special case of (3)
(4)Δ(pn-1Δxn-1)+qnf(xn)=0,n⩾0.They got the sufficient conditions for asymptotically monotone solutions of (4). Enlightened by [8, 9], in this paper, we derive several sufficient conditions for monotonicity of eventually positive solutions on (3) and obtain a few nonexistence criteria for eventually positive monotone solutions of (3). Our results improve and generalize results in [8]. We also provide examples to illustrate our main results.For convenience, these essential conditions used in main results are listed as follows:(H1)
there exists a continuous functionf:R→R such that xf(x)>0 for all x≠0;(H2)
f is a derivable function and f′(x)⩾0 for x≠0;(H3)
there exist two sequences{pn} and {qn}, such that P(n,xn,xn+1)/f(xn+1)⩾pn and Q(n,xn,Δxn)/f(xn+1)⩽qn for xn≠0;(H4)
∑n=n0∞‍1/an=+∞, n0 is a positive integral number,wherean, P(n,xn,xn+1) and Q(n,xn,Δxn+1) are all in (3).


## 2. Main Results

We first state a result which relates a positive sequence and a positive nondecreasing function. Its proof can be found in [8].Lemma 1 (see [8]).
Letf(x) be a positive nondecreasing function defined for x>0. Let {xk} be a real sequence such that xk>0 for i⩽k⩽j+1. Then
(5)∑k=ij‍Δxkf(xk+1)⩽∫xixj+1‍duf(u)⩽∑k=ij‍Δxkf(xk).Theorem 2.
Suppose that conditions(H1)–(H4) hold, pn and qn satisfy the following conditions:(H5)
∑s=n0∞‍(ps-qs)<+∞;(H6)
liminfn→∞∑s=n0n‍(ps-qs)⩾0
for alln0. Then eventually positive solutions of (3) are eventually monotone increasing.Proof.
Suppose that{xn} is a positive solution of (3), say xn>0 for n>N>n0. If conclusion cannot hold, without any loss of generality, assume ΔxN⩽0, in view of (3) and conditions, we have
(6)Δ(anΔxnf(xn))=Δ(anΔxn)f(xn+1)-an(Δxn)2f′(xn+θΔxn)f(xn)f(xn+1)⩽qn-pn(0<θ<1),
by summing (6) from N to n-1, then
(7)anΔxnf(xn)⩽aNΔxNf(xN)-∑s=Nn-1‍(ps-qs).
Making use of condition(H6), we know Δxn<0 for n⩾N. Summing (3) and using (H3), we have
(8)anΔxn⩽aNΔxN-∑s=Nn-1‍f(xs+1)(ps-qs)=aNΔxN-f(xn+1)∑s=Nn-1‍(ps-qs)+∑s=Nn-1‍Δf(xs)(∑t=Ns-1‍(pt-qt))⩽aNΔxN.
By summing (8), we then see that
(9)xn+1⩽xN+aNΔxN∑s=Nn‍1as→-∞(asn→∞),
which contradicts the fact xn>0. The proof is complete.Example 3.
Consider the difference equation(10)Δ(Δxnn2)+xn+1(r(n,xn)+1n2(n+1)-1(n+1)3)=xn+1r(n,xn),n⩾0,
where r(n,xn) is any function of n and xn. By taking f(x)=x, we have
(11)P(n,xn,xn+1)f(xn+1)=r(n,xn)+1n2(n+1)-1(n+1)3=pn,Q(n,xn,Δxn)f(xn+1)=r(n,xn)=qn.
So conditions of Theorem2 hold. By Theorem 2, (10) has a positive monotone increasing solution {xn}={n}.Theorem 4.
If conditions(H1)–(H4) hold, there exist M>0 and j>n0  for n0⩾M such that
(12)limsupn→∞∑k=jn‍1ak∑s=n0k-1‍(ps-qs)>0.
Then eventually positive solutions{xn} of (3) are eventually monotone increasing or liminfn→∞xn=0.Proof.
Suppose{xn} is a positive solution of (3), there exists N>n0 such that xn>0 for n>N. Let ΔxN⩽0, and
(13)limsupn→∞∑k=jn‍1ak∑s=Nk-1‍(ps-qs)>0,j>N.
Ifliminfn→∞xn≠0, then there exist T⩾N and a number α>0 such that xn>α>0 for n⩾T; in view of (7), we get
(14)Δxnf(xn)⩽aNΔxNf(xN)·1an-1an∑s=Nn-1‍(ps-qs).
Summing (14) and making use of Lemma 1, we know
(15)∫xjα‍duf(u)⩽∫xjxn+1‍duf(u)⩽aNΔxNf(xN)∑k=jn‍1ak-∑k=jn‍1ak∑s=Nk-1‍(ps-qs).
By(H4), the right side of (15) tends to -∞ as n→∞, whereas the left side is finite. This contradiction completes our proof.Example 5.
Consider the difference equation(16)Δ(Δxnn)+xn+1n=n+2n+1xn+1,n⩾0.
By takingf(x)=x, we have P(n,xn,xn+1)/f(xn+1)=1/n=pn,Q(n,xn,Δxn)/f(xn+1)=n+2/n+1=qn. So conditions of Theorem 4 hold. By Theorem 4, (16) has a positive monotone increasing solution {xn}={n}.Theorem 6.
If conditions(H1)–(H3) hold, and
(17)liminfn→∞1an∑s=n0n-1‍(ps-qs)>0
holds for all n0. Then eventually positive solutions {xn} of (3) are eventually monotone increasing or eventually monotone decreasing and limn→∞xn=0.Proof.
Suppose{xn} is a positive solution of (3), there exists N>n0 such that xn>0 for n>N. Let ΔxN⩽0 and liminfn→∞1/an∑s=Nn-1‍(ps-qs)>0, then there exists β>0 such that
(18)1an∑s=Nn-1‍(ps-qs)⩾β>0,n⩾N.
From (7), we have
(19)Δxn⩽-f(xn)·1an∑s=Nn-1‍(ps-qs)⩽-βf(xn)<0,n>N.
Iflimn→∞xn≠0, then there exists c>0 such that xn⩾c>0. There is no harm in assumption xn⩾c for n⩾N. Summing (19), we obtain
(20)c⩽xn+1⩽xN-(n+1-N)βf(c)→-∞(n→∞),
which is a contrary. The proof is complete.Example 7.
Consider the difference equation(21)Δ(n2Δxn)+xn+1(r(n,xn)+1n+2)=xn+1r(n,xn),
where r(n,xn) is any function of n and xn. By taking f(x)=x, we have
(22)P(n,xn,xn+1)f(xn+1)=r(n,xn)+1n+2=pn,Q(n,xn,Δxn)f(xn+1)=r(n,xn)=qn.
So conditions of Theorem6 hold. By Theorem 6, (21) has a monotone decreasing positive solution {xn}={1/n}.Theorem 8.
If conditions(H1)–(H3) hold and(H7)
limsupn→∞∑s=kn‍1/as∑t=n0s-1‍(pt-qt)=+∞;(H8)
for allε>0,∫0ε‍du/f(u)<+∞.
Then eventually positive solutions of (3) are eventually monotone increasing.Proof.
Supposexn>0 for n>N>n0, {xn} is a solution of (3), and limsupn→∞∑s=kn‍1/as∑t=Ns-1‍(pt-qt)=∞. If the result does not hold, without any loss of generality, assume ΔxN⩽0. In view of (7), we see that
(23)Δxnf(xn)⩽aNΔxNf(xN)·1an-1an∑t=Nn-1‍(pt-qt)⩽-1an∑t=Nn-1‍(pt-qt).
Summing (23) and using Lemma 1, we know
(24)∫xkxn+1‍duf(u)⩽∑s=kn‍Δxsf(xs)⩽-∑s=kn‍1as∑t=Ns-1‍(pt-qt).
This is a contradiction. The proof is complete.Remark 9.
In Theorems2 and 4, condition (H4) is essential; that is, the series with positive terms ∑n=n0∞‍1/an is divergent, but it is not required in Theorems 6 and 8.Remark 10.
The eventually positive solutions in Theorems4 and 8 are increasing it is not necessarily so in Theorem 6.
Next, we will derive several nonexistence criteria for eventually positive monotone solutions of (3).Theorem 11.
If conditions(H1)–(H3) hold and
(25)limsupn→∞∑k=1n‍(pk-qk)=+∞.
Then, (3) cannot have any eventually positive monotone increasing solutions.Proof of Theorem11 is obvious. If xn>0 is an eventually positive increasing solution, by means of conditions, (7) is a contrary.Theorem 12.
If conditions(H1)–(H3) hold, and there is a nonnegative and nondegenerate sequence {φn} such that
(26)limsupn→∞∑k=n0n‍φk+1/ak∑s=n0k-1‍(ps-qs)∑k=n0n‍φk+1/ak=∞
holds for all n0. Then, (3) cannot have any eventually positive nondecreasing solutions.Proof.
Suppose that{xn} is a positive solution of (3), there exists N>n0 such that xn>0 and Δxn⩾0 for n>N. Multiplying (7) by φn+1/an, we have
(27)φn+1Δxnf(xn)+φn+1an∑s=Nn-1‍(ps-qs)⩽φn+1an·aNΔxNf(xN).
So we obtain(28)∑k=Nn‍φk+1ak∑s=Nk-1‍(ps-qs)⩽aNΔxNf(xN)∑k=Nn‍φk+1ak.
This is contrary to our condition. The proof is complete.Theorem 13.
If(H1) and (H3) hold, {an} is a nondecreasing sequence, f(x) is a nondecreasing function, and there is a nonnegative sequence {φn}, where {Δφn} is bounded, and(H9)
limn→∞∑s=n0n-1‍φs+1(ps-qs)/as+1=+∞ for all n0;(H10)
0<∫ε+∞‍du/f(u)<+∞, ε>0.
Then, (3) cannot have any eventually positive monotone increasing solutions.Proof.
Suppose that{xn} is a solution of (3), and there exists N>n0 such that xn>0 and Δxn>0 for n>N. Multiplying (3) by φn+1/an+1f(xn+1) and summing from N to n-1 again, we have
(29)∑s=Nn-1‍φs+1as+1f(xs+1)Δ(asΔxs)⩽∑s=Nn-1‍φs+1as+1(qs-ps).
Namely,(30)φnΔxnf(xn)-φNΔxNf(xN)-∑s=Nn-1‍asΔxsΔ(φsasf(xs))⩽∑s=Nn-1‍φs+1as+1(qs-ps).
As{an} is a nondecreasing sequence, we get
(31)Δ(φsasf(xs))=φs+1as+1f(xs+1)-φsasf(xs)⩽Δφsas+1f(xs+1).
Thus(32)asΔxs·Δ(φsasf(xs))⩽ΔxsΔφsf(xs+1).
From (30), we obtain
(33)φnΔxnf(xn)-φNΔxNf(xN)+∑s=Nn-1‍φs+1as+1(ps-qs)⩽∑s=Nn-1‍ΔxsΔφsf(xs+1),
using Lemma 1 and conditions, we have
(34)∑s=Nn-1‍ΔxsΔφsf(xs+1)⩽M∑s=Nn-1‍Δxsf(xs+1)⩽M∫xN+∞‍duf(u),M>0.
By lettingn→∞, we see that the left-hand side of (33)  is bounded, this is contrary to our condition (H9). The proof is complete.By means of proof of Theorem13, we getCorollary 14.
If(H1), (H3), and (H9) hold, {an} is a nondecreasing sequence, f(x) is a nondecreasing function, and there is a nonnegative sequence {φn}, {Δφn} is bounded, and 0<∫0ε‍du/f(u)<+∞ for ε>0. Then, (3) cannot have any eventually positive nondecreasing bounded solutions.Corollary 15.
Suppose(H1), (H3), and (H9) hold, {an} is a nondecreasing sequence, f(x) is a nondecreasing function, and there is a nonnegative nonincreasing sequence {φn}. Then, (3) cannot have any eventually positive monotone increasing solutions.Theorem 16.
Suppose(H1), (H3), and (H5) hold, f(x) is a nondecreasing function, and(H11)
limsupn→∞∑k=n0n‍1/ak∑s=k∞‍(ps-qs)=+∞ for all n0;(H12)
0<∫ε+∞‍du/f(u)<+∞,ε>0.
Then, (3) cannot have any eventually positive nondecreasing solutions.Proof.
Assume to the contrary that there existsN>n0 such that xn>0 and Δxn>0 for n>N.{xn} is a solution of (3). By means of (3) and (H3), we get
(35)Δ(anΔxn)f(xn+1)⩽(qn-pn),
by summing (35) from N to n-1, thus
(36)anΔxnf(xn+1)-aNΔxNf(xN+1)-∑s=Nn-1‍as+1Δxs+1Δ(1f(xs+1))⩽∑s=Nn-1‍(qs-ps).
Asf(x) is a nondecreasing function, we know Δxs+1Δ(1/f(xs+1))⩽0, so
(37)∑s=N∞‍(ps-qs)⩽aNΔxNf(xN+1).
In view of Lemma1, we see that
(38)∑N=Tn‍1aN∑s=N∞‍(ps-qs)⩽∑N=Tn‍ΔxNf(xN+1)⩽∫xTxn+1‍duf(u).
This contradiction establishes our assertion.By means of proof of Theorem16, we obtain the following.Corollary 17.
Suppose(H1), (H3), (H5), and (H11) hold, f(x) is a nondecreasing function. Then (3) cannot have any eventually positive nondecreasing bounded solutions.



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