SOLUTIONS OF A HIGHER ORDER NEUTRAL
DIFFERENCE EQUATION
ZHI-QIANG ZHU, GEN-QIANG WANG, AND SUI SUN CHENG
Received 25 May 2005; Revised 21 September 2005; Accepted 28 September 2005
Nonoscillatory solutions of a nonlinear neutral type higher order difference equations are classified by means of their asymptotic behaviors. By means of the Kranoselskii’s fixed point theorem, existence criteria are then provided for justification of such classification.
Copyright © 2006 Zhi-qiang Zhu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Classification schemes for nonoscillatory solutions of nonlinear difference equations are important since further investigations of some of the qualitative behaviors of nonoscilla-tory solutions can then be reduced to only a number of cases. There are several studies which provide such classification schemes for difference equations, see, for example, [4– 11]. In particular, in [7], a class of nonlinear neutral difference equations of the form
Δmx
n+cnxn−k+ fn,xn−l=0, n=0, 1,..., (1.1)
wherem,kandlare integers such thatm≥2,k >0 andl≥0 is studied and classification schemes are given when{cn}is a nonnegative constant sequence{c0}, and in [10], the
same equation is studied with odd integerm≥1, positive integerk, integerland{cn} =
{−1}.
In this paper, we continue our investigation on the possible types of nonoscillatory solutions when{cn} ⊆(−1, 0] and limn→∞cn=c0(while the case{cn} ⊆(−∞,−1] will be
discussed elsewhere). Besides the assumption that{cn} ⊆(−1, 0], we will assume further
that f is a continuous function defined on{0, 1,...} ×Rsuch that f =f(n,x) is nonde-creasing in the second variablexand satisfiesx f(n,x)>0 forx=0 andn≥0.
We will accomplish two things in this paper: to provide a classification scheme for the nonoscillatory solutions of (1.1) inSection 2and establish inSection 3sufficient and/or necessary criteria for the existence of solutions in each class. There are no overlapping
Hindawi Publishing Corporation Advances in Difference Equations Volume 2006, Article ID 47654, Pages1–19
results between our paper and [4–11], although some proofs are similar. However, the existence proofs are different in that Cheng-Patula existence theorem is applied in [7], monotone method is used in [10] while we use Krasnoselskii fixed point theorem here. We remark further that classification scheme is also provided for neutral differential equa-tions in [2].
Before we go into details, we will need some preparatory terminologies and results. First of all, given initialxifor−max{k,l} ≤i≤0, we may calculate from (1.1)x1,x2,x3,...
in a recursive manner. Such a sequence{xn}is said to be a solution of (1.1). Among the
solutions of (1.1), one is said to be nonoscillatory if it is eventually positive or eventually negative.
Given an integera, it is convenient to set
N(a)= {a,a+ 1,a+ 2,...}. (1.2)
Given an integerα≥0, the generalized factorial functiong(x)=x(α)is defined as
fol-lows
x(α)= ⎧ ⎨ ⎩
x(x−1)(x−2)···(x−α+ 1) α >0
1 α=0. (1.3)
It is well known thatΔn(α)=αn(α−1)forα >0 (see, e.g., [3]).
Let
l∞
N0=
x=xn n≥N0: sup
n≥N0
xn
rn <∞
, (1.4)
where N0>0 is an integer and {rn}n≥N0 is a positive sequence with a uniform posi-tive lower bound. When endowed with the usual linear structure and the normx =
supn≥N0(|xn|/rn), (l∞N0, · ) is a Banach space. A setB⊆l
∞
N0is said to be uniformly Cauchy if for anyε >0 there exists an integerM≥N0such that
xi
ri−
xj
rj
< ε i,j > M (1.5)
for allx= {xn} ∈B.
Lemma1.1. A bounded and uniformly Cauchy subsetB⊆l∞
N0is relatively compact. Proof. By assumption, we know that for any suchε >0, there exists an integerM≥N0>0
such that for anyx∈B, we have
xi
ri−
xj
rj <ε
LetΓ>0 be a bound forB. That isx ≤Γfor allx∈B. Choose integersMn,n=N0,N0+
eventually positive or eventually negative, whereiis a nonnegative integer. Suppose further thatzn=xn+cnxn−kandlimn→∞(zn/n(i))=b. Thenlimn→∞(xn/n(i))=b/(1 +c0).
Proof. Without loss of generality, we assume thatxn/n(i)>0 for any positive integern. In caseb is finite, we assert that {xn/n(i)}is bounded. Otherwise, there would exist a
On the other hand, we have
Let lim supn→∞(xn/n(i))=Qand lim infn→∞(xn/n(i))=q. Then 0≤q≤Q <∞.
More-over, there exist{nλ}and{nλ}such that limλ→∞nλ=∞, limλ→∞nλ=∞, limλ→∞(xnλ/nλ(i))= Qand limλ→∞(xnλ/nλ(i))=q. Note that
b=lim
λ→∞
znλ nλ(i) =λlim→∞
xnλ nλ(i)+cnλ
xnλ−k nλ(i)
≥lim
λ→∞
xnλ
nλ(i)+ limλ→∞infcnλ
xnλ−k
nλ−k(i) n
λ−k(i)
nλ(i) ≥Q+c0Q,
b=lim
λ→∞
znλ nλ(i) =λlim→∞
xnλ nλ(i)+cnλ
xnλ−k nλ(i)
≤lim
λ→∞
xnλ
nλ(i)+ limλ→∞supcnλ
xnλ−k
nλ−k(i)
nλ−k(i)
nλ(i) ≤q+c0q,
(1.13)
we have (1 +c0)q≥(1 +c0)Q. It follows that q≥Q. Hence q=Q and it implies that
limn→∞(xn/n(i)) exists. In view ofzn=xn+cnxn−kand limn→∞(zn/n(i))=b, we have
lim
n→∞
xn
n(i) =
b
1 +c0. (1.14)
In casebis infinite, thenb= ∞ orb= −∞. We assert thatb= −∞cannot hold. In fact, for givenc1 with−c0< c1<1, there exists a large integerN0such that−cn≤c1for
n≥N0. Hence, ifb= −∞, thenzn=xn+cnxn−k<0 forn≥Nand
xn<−cnxn−k≤c1xn−k, n≥N, (1.15)
whereN≥N0is some positive integer. It implies that
0< xN+λk< c1xN+(λ−1)k<···< cλ1xN. (1.16)
So that limλ→∞xN+λk=0. Thus
lim
λ→∞zN+λk=0 (1.17)
which implies thatb= −∞is impossible.
Now, for arbitraryM >0, there exists a sufficiently large integerNsuch that zn
n(i)=
xn
n(i)+cn
xn−k
n(i) ≥M, n≥N. (1.18)
It follows that
xn
n(i) ≥M, n≥N. (1.19)
That is limn→∞(xn/n(i))= ∞. The proof is complete.
Lemma1.3. Suppose that the sequence{xn}and{yn}satisfy the following conditions,
(i) yn>0andΔyn>0for all large integersnandlimn→∞yn= ∞, and
(ii) limn→∞(Δxn/Δyn)=b.
Thenlimn→∞(xn/yn)=limn→∞(Δxn/Δyn)=b, wherebcan be finite or infinite.
Lemma1.4. Let u=u(n)be a sequence defined for n∈N(a),u(n)>0 withΔmu(n)of constant sign onN(a)and not identically zero. Then, there exists an integerm∗,0≤m∗≤m
withm+m∗odd forΔmu(n)≤0or,m+m∗even forΔmu(n)≥0and such that m∗≤m−1implies(−1)m∗+i
Δiu(n)>0 ∀n∈N(a), m∗+ 1≤i≤m,
m∗≥1impliesΔiu(n)>0 ∀largen∈N(a), 1≤i≤m∗. (1.20)
Remark 1.5. Ifu(n)<0 in Lemma 1.4, then there existsm∗, 0≤m∗≤mwithm+m∗
odd forΔmu(n)≥0 or,m+m∗even forΔmu(n)≤0 and such that
m∗≤m−1 implies (−1)m∗+i
Δiu(n)<0 ∀n∈N(a), m∗+ 1≤i≤m,
m∗≥1 impliesΔiu(n)<0 ∀largen∈N(a), 1≤i≤m∗. (1.21)
Lemma1.6 (Kranoselskii’s fixed point theorem). SupposeBis a Banach space andΩis a bounded, convex and closed subset ofB. LetU,S:Ω→Bsatisfy the following conditions.
(i)Ux+Sy∈Ωfor anyx,y∈Ω, (ii)Uis a contraction mapping, and (iii)Sis completely continuous. ThenU+Shas a fixed point inΩ.
2. Classifications of nonoscillatory solutions
In the following discussions, we assume throughout that
lim
n→∞cn=c0∈(−1, 0]. (2.1)
We set
zn=xn+cnxn−k (2.2)
whenever it is defined. Equation (1.1) can now be written as
Δmz
n= −fn,xn−l. (2.3)
We will propose a classification scheme for the nonoscillatory solutions of (1.1). For this purpose, we first note that ifx= {xn}is an eventually negative solution of (1.1), then
y= {yn}defined byyn= −xnwill satisfy
Δmy
n+cyn−k+fn,yn−l=0, (2.4)
where
has the same properties satisfied by f, that is, f is a continuous function defined on
{0, 1,...} ×Rsuch that f= f(n,u) is nondecreasing in the second variableuand satisfies u f(n,u)>0 foru=0 andn≥0. Therefore, we may restrict our attention to the setS+of
all eventually positive solutions of (1.1). Motivated by the classification scheme in [2], we make use of the following notations for classifying our eventually positive solutions:
Ak(α,β)=xn ∈S+: limn→∞n(xkn−1) =α, limn→∞nx(nk) =β
, k≥1,
A0(α)=
xn ∈S+: limn→∞xn=α
. (2.6)
Theorem2.1. (a)Suppose that mis even. Ifx= {xn}is an eventually positive solution
of (1.1), then eitherx∈A0(0)or there are some j∈ {1, 2,...,m/2}anda >0such thatx
belongs toA2j−1(∞,a),A2j−1(∞, 0)orA2j−1(a, 0).(b)Suppose thatmis odd. Ifx= {xn}
is an eventually positive solution of (1.1), then eitherxbelongs toA0(α)for someα≥0, or
there are j∈ {1, 2,..., (m−1)/2}anda >0such thatxbelongs toA2j(∞,a),A2j(∞, 0)or
A2j(a, 0).
Proof. Letmis even andx= {xn}be an eventually positive solution of (1.1). Then, in view of (2.3), there exists some integerN >0 such thatΔmzn<0 forn≥N. Therefore,zn
is eventually of fixed sign. For the sake of simplicity, we may assume that{zn}is of fixed
sign forn≥N.
First of all, supposezn<0 forn≥N. By the same reasoning as in the proof ofLemma
1.2, we may show that
lim
λ→∞zN+λk=0. (2.7)
On the other hand, in view ofLemma 1.4, there exists some evenm∗with 0≤m∗≤m
such that eventuallyΔizn<0 for 0≤i≤m∗ and (−1)m∗+iΔizn<0 form∗+ 1≤i≤m.
There are now two cases to consider. Case 1(m∗=0). Then we have eventually
zn<0, Δzn>0. (2.8)
By (2.8), we can set
lim
n→∞zn=L0≤0. (2.9)
In view of (2.7), we find that limn→∞zn=0. ByLemma 1.2, we have limn→∞xn=0. Hence
xbelongs toA0(0).
Case 2(m∗≥2). Then we have eventually
zn<0, Δzn<0. (2.10)
Now we supposezn>0 forn≥N. Similar to the proof in [7, Theorem 1], we may see
thatxbelongs toA2j−1(∞,a),A2j−1(∞, 0) orA2j−1(a, 0) for some j∈ {1, 2,...,m/2}and
a >0.
Whenmis odd, the proof is similar to those above and hence is skipped. The proof is
complete.
3. Existence criteria
Eventually positive (and by analog eventually negative) solutions of (1.1) have been clas-sified according toTheorem 2.1. We now justify our classification schemes by finding existence criteria for each type of solutions.
Theorem3.1. Suppose thatmis even. If (1.1) has a solution inA2j−1(∞,a)for some j∈
The converse is also true.
Proof. First of all, we remark that
Summing the above equation again fromN1ton, we obtain
By (3.4), the above equation implies that
∞
Note that (2.1), there are two cases to consider.
Then it is obvious thatΩis a bounded, convex and closed subset oflN∞0, and for anyx∈ Ωandn≥N0+l, we have
fn,xn−l≤f
n,K(n−l)(2j−1). (3.16)
Define operatorsUandSonΩas follows:
(Ux)n=
and, in view of (3.19) and (3.12), we have
(Ux)n+ (Sy)n≤3
4
1−c1
KRn−cnxn−k+
1−c1
K
8 Rn
≤3
4
1−c1
KRn−cnKRn−k+
1−c1
K
8 Rn≤KRn.
(3.22)
That is,Ux+Sy∈Ωfor anyx,y∈Ω. Letx,y∈Ω. In view of (3.11), we have
1 R2
n
(Ux)n−(U y)n=cNxN−k−yN−k
RNRn
=xN−Rk2−yN−k N−k
cNR2 N−k
RNRn ≤c1nsup≥N0
xn−yn
R2 n
(3.23)
forN0≤n < N. And, forn≥N, we have
1 R2
n
(Ux)n−(U y)n≤cnsup n≥N0
xn−yn
R2
n . (3.24)
Therefore, we have
Ux−U y ≤c1x−y (3.25)
for anyx,y∈Ω. Hence,Uis a contraction mapping.
Next, we will prove thatSis a completely continuous mapping. Indeed, it is obvious that (Sx)n≥(K/2)Rnforn≥N0and (Sx)n≤KRnforN0≤n < N. Whenn≥N, by means
of (3.19), we have
(Sx)n≤3
4KRn+
1−c1
K
8 Rn≤KRn. (3.26)
That is, the operatorSmapsΩintoΩ.
Now we consider the continuity ofS. Letx(λ)∈Ωandx(λ)−x →0 whenλ→ ∞, we
assert thatSx(λ)converges toSxby · . Indeed,x(λ)−x →0 implies thatx∈Ωand
|x(λ)
n −xn| →0 whenλ→ ∞for any integern≥N0. Thereby, we have
fn,x(λ) n−l
−fn,xn−l−→0, λ−→ ∞ (3.27)
for any integern≥N0+l. By definition ofS, we have Sx(λ)
n−(Sx)n=0 (3.28)
forN0≤n < Nand
Sx(λ)
forn≥N, where
In view of (3.27) and (3.31), the Lebesque’s dominated theorem [3] then implies limλ→∞(Sxλ)−(Sx) =0. This meansSis continuous.
Finally, we prove thatSΩis relatively compact. We assert thatSΩis uniformly Cauchy. Indeed, for anyε >0, there existsN1> Nsuch that 1/Rn< ε/3Kforn≥N1. For anyx∈Ω
ByLemma 1.1,SΩis relatively compact.
By the Kranoselskii’s fixed point theorem, there then exists x= {xn} ∈Ωsuch that
(Ux)n+ (Sx)n=xn. Therefore, we have
xn=3
4
1−c1
KRn−cnxn−k+F(n), n≥N. (3.35)
It is easy to verify thatxnsatisfy (1.1). Furthermore, we have
Δ2j−1F(n)= ∞ im−2j+1=n
∞
im−2j=im−2j+1
··· ∞
i1=i2
fi1,xi1−l
≤ ∞
im−2j+1=n
∞
im−2j=im−2j+1
··· ∞
i1=i2
fi1,Ki1−l(2j−1)
.
(3.36)
In view of (3.1) and (3.10), we have
lim
n→∞Δ
2j−1F(n)=0, (3.37)
so that
lim
n→∞
F(n)
n(2j−1)=0. (3.38)
Now we turn to (3.35) and obtain
lim
n→∞
zn
n(2j−1)=
3 4
1−c1
K. (3.39)
ByLemma 1.2, we have
lim
n→∞
xn
n(2j−1) =
31−c1
K 41 +c0
, (3.40)
which infers that limn→∞(xn/n(2j−2))= ∞. In summary, (1.1) has a solution inA2j−1(∞,a)
when−1< c0<0.
In casec0=0, takec1so that 0< c1≤1/3. Then, there exists an integerN > k+lsuch
that whenn≥N, (3.11) to (3.14) hold. Take operatorsUandSto be the same operators as above. Then we may prove in similar manners that (1.1) has a solution inA2j−1(∞,a).
The proof is complete.
A similar theorem holds whenmis odd, the proof is similar to that ofTheorem 3.1 and hence is skipped.
Theorem3.2. Supposemis odd. If (1.1) has a solution inA2j(∞,a)for some j∈ {1, 2,...,
(m−1)/2}anda >0, then there exists someK >0such that
∞
i=0
(i+m−2j−1)(m−2j−1)
(m−2j−1)! f
i,K(i−l)(2j)<∞. (3.41)
Theorem3.3. Suppose thatmis even. If (1.1) has an eventually positive solution inA2j−1(a,
0)for somej∈ {1,...,m/2}anda >0, then there is someK >0such that
∞
i=0
(i+m−2j+ 1)(m−2j+1)
(m−2j+ 1)! f
i,K(i−l)(2j−2)<∞. (3.42)
The converse is also true.
The proof is similar to that ofTheorem 3.1by takingRn=n(2j−2).
Theorem3.4. Supposemis odd. If (1.1) has a solution inA2j(a, 0)for some j∈ {1, 2,...,
(m−1)/2}anda >0, then there is someK >0such that
∞
i=0
(i+m−2j)(m−2j)
(m−2j)! f
i,K(i−l)(2j−1)<∞. (3.43)
The converse also holds.
Theorem3.5. Suppose thatmis even. If (1.1) has a solution inA2j−1(∞, 0)for some j∈
{1, 2,...,m/2}, then
∞
i=0
(i+m−2j)(m−2j)
(m−2j)! f
i, (i−l)(2j−2)<∞, (3.44)
∞
i=0
(i+m−2j+ 1)(m−2j+1)
(m−2j+ 1)! f
i, (i−l)(2j−1)= ∞. (3.45)
Conversely, if there is somej∈ {1, 2,...,m/2}such that
∞
i=0
(i+m−2j)(m−2j)
(m−2j)! f
i, (i−l)(2j−1)<∞, (3.46)
∞
i=0
(i+m−2j+ 1)(m−2j+1)
(m−2j+ 1)! f
i, (i−l)(2j−2)= ∞, (3.47)
then (1.1) has a solution inA2j−1(∞, 0).
Proof. Letx= {xn}be an eventually positive solution of (1.1) inA2j−1(∞, 0). Note that
limn→∞(xn/n(2j−1))=0, limn→∞(xn/n(2j−2))= ∞ and (2.3) holds. Therefore there exists an integerN0>0 such that
Δmzn<0, n≥N
0, (3.48)
xn≤n(2j−1), n≥N0, (3.49)
Define two operators onΩas follows:
(Ux)n= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
1
2Ln N0≤n < N
−3
2c1Ln−cnxn−k n≥N,
(Sx)n= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
1
2Ln N0≤n < N 3
2Ln+F(n) n≥N,
(3.59)
where
F(n)=
n−1
im=N
···
im−2j+4−1
im−2j+3=N
im−2j+3−1
im−2j+2=N
∞
im−2j+1=im−2j+2
∞
im−2j=im−2j+1
··· ∞
i1=i2
fi1,xi1−l
. (3.60)
Analogous to the discussions inTheorem 3.1, there existsx= {xn} ∈Ωsuch that
xn=3
2
1−c1Ln−cnxn−k+F(n), n≥N. (3.61)
From (3.61) and (3.46), we see that
lim
n→∞
zn
n(2j−1)=0. (3.62)
ByLemma 1.2, we have
lim
n→∞
xn
n(2j−1)=0. (3.63)
Note that (3.47) implies
lim
n→∞Δ
2j−2F(n)= ∞. (3.64)
Hence, in view of (3.61) andLemma 1.2, we see that
lim
n→∞
xn
n(2j−2) = ∞. (3.65)
This means (1.1) has an eventually positive solution inA2j−1(∞, 0) when−1< c0<0.
Whenc0=0, we takec1so that 0< c1≤1/3 and the rest of proof is the same as the
above and is thus skipped. The proof is complete.
Theorem3.6. Supposemis odd. If (1.1) has a solution inA2j(∞, 0)for some j∈ {1, 2,...,
Proof. First note that there exists integerN > Msuch that
e−αn< 1
Define two operators onΩas follows:
where
F(n)= ∞
im=n
∞
im−1=im
···∞
i2=i3
∞
i1=i2
fi1,xi1−l
. (3.73)
By arguments similar to those in the proof ofTheorem 3.1, we may prove that there exists x= {xn} ∈Ωsuch that
xn= −cnxn−k+F(n),n≥N. (3.74)
In view of the definition ofΩ, we see thatxis a solution of (1.1) inA0(0). The proof is
complete.
A variant ofTheorem 3.7is the following and its proof is omitted.
Theorem3.8. Supposemis odd andc0<0. If there exist constantsα >0,c1,c2 with0<
c1<−c0< c2<1and integerM > l+ksuch that
c1eαk>1 (3.75)
as well as
∞
i=n
(i−n+m−1)(m−1)
(m−1)! f
i,i1
−l
≤n1−nc2
−k, n≥M, (3.76)
then (1.1) has a solution inA0(0).
Theorem3.9. Suppose thatmis odd. If (1.1) has a solution inA0(a)for somea >0, then
there exists someK >0such that
∞
i=0
(i+m−1)(m−1)
(m−1)! f(i,K)<∞. (3.77)
The converse also holds.
The proof is similar to that ofTheorem 3.1and is skipped.
Example 3.10. Consider the equation
Δ2x n−3
4xn−1
+ 1
16xn−1=0, (3.78)
here f(n,x)=(1/16)x. It is clear that
∞
i=0
fi,K(i−1)= ∞,
∞
i=0
(i+ 1)f(i,K)= ∞ (3.79)
for anyK >0 and
∞
i=0
Hence by Theorems3.1and3.3, an eventually positive solution of (3.78) cannot be in A1(∞,a) nor inA1(a, 0). In addition, byTheorem 3.5, an eventually positive solution of
(3.78) cannot be inA1(∞, 0). However, byTheorem 2.1, (3.78) has a solution inA0(0)
if it has some eventually positive solution. Indeed, {xn} = {1/2n} satisfies (3.78) and
limn→∞xn=0.
Consider another equation
Δ3
xn−1
5xn−2
+ 1
80xn−1=0, (3.81)
here f(n,x)=(1/80)x. Then, it is easy to see that
∞
i=0
fi,K(i−1)(2)= ∞,
∞
i=0
(i+ 1)fi,K(i−1)= ∞, (3.82)
as well as
∞
i=0
(i+ 2)(2)
2 f(i,K)= ∞ (3.83)
for anyK >0, and
∞
i=0
f(i,i−1)= ∞. (3.84)
By the similar reasons to the above, (3.81) has a solution inA0(0) if it has some eventually
positive solution. Indeed,{xn} = {1/2n}is such a solution of (3.81).
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Zhi-qiang Zhu: Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China
Gen-qiang Wang: Department of Computer Science, Guangdong Polytechnic Normal University, Guangzhou, Guangdong 510665, China