Classication schemes for positive solutions of a second-order
nonlinear dierence equation
Guang Zhanga, Sui Sun Chengb;∗, Ying Gaoc a
Department of Mathematics, Datong Advanced College, Datong, Shanxi 037008, Peoples Republic of China
bDepartment of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, R.O.C. c
Department of Mathematics, Yanbei Teacher’s College, Datong, Shanxi 037008, Peoples Republic of China
Received 14 March 1998
Abstract
Classication schemes for eventually positive solutions of a class of second-order nonlinear dierence equations are given in terms of their asymptotic magnitudes, and necessary as well as sucient conditions for the existence of these solutions are also provided. c 1999 Elsevier Science B.V. All rights reserved.
AMS classication:39A10
Keywords:Nonlinear dierence equation; Superlinear function; Sublinear function; Eventually positive solution; !-condition
1. Introduction
In [7], classication schemes for eventually positive solutions of the nonlinear dierence equation (rn(xn−pnxn−)) +f(n; xn−) = 0
are given, and necessary as well as sucient conditions for their existences are also provided. Such schemes are important since further investigations of qualitative behaviors of solutions can then be reduced to only a number of cases.
In this paper, we are concerned with a similar class of nonlinear second-order dierence equations of the form
(rng(xn)) +f(n; xn) = 0; n=K; K+ 1; : : : ; (1)
whereK is a xed integer,{rn}
∞
n=0 is a positive sequence, f(n; x) is a real-valued function dened on {K; K+1; : : :} ×Rwhich is continuous in the second variablex and satisesf(n; x)¿0 forx¿0, and
∗Corresponding author. E-mail: sscheng@math.nthu.edu.tw. Partially supported by the National Science Coucil of ROC.
nally the everywhere continuous real function g is monotone increasing and satises the conditions
g(0) = 0; g−1
(−u) =−g−1
(u) and g−1
(uv)6g−1
(u)g−1
(v) for some ; ¿0 and all u; v¿0. The function g(u) =u satises the above conditions when is a quotient of odd positive integers.
In several cases, we will assume the additional condition that g−1
(uv) =!g−1
(u)g−1
(v) for some
!¿0 and every pairs u and v. Such a function g is said to satisfy the !-condition. As for the functionf, for each xed integern, iff(n; x)=xis nondecreasing inx forx¿0, it is called superlinear. If for each integer n, f(n; x)=x is nonincreasing in x for x¿0, then f is said to be sublinear. Superlinear or sublinear function f will be assumed in some of our later results. But these results will be proved only for superlinear functions because the sublinear cases can be obtained in similar manners. Note that if 0¡a6x6b, then
f(n; a)6f(n; x)6f(n; b)
if f is superlinear, and
a
bf(n; b)6f(n; x)6 b af(n; a)
if f is sublinear. These remarks will be useful later. Since (1) can be written in the recurrence form
xn+2=xn+1+g −1
r
ng(xn)−f(n; xn)
rn+1
;
it is clear that given xK and xK+1; we can successively calculate xK+2; xK+3; : : : in a unique manner.
Such a sequence{xn} will be called a solution of (1). We will be concerned with eventually positive
solutions of (1).
Besides [7], nonlinear dierence equations have also been studied by a number of authors [1–6, 8–10]. In particular, He in [3] (see also [10]) has obtained existence criteria for eventually positive solutions of the equation
(rn(xn)) +f(n; xn) = 0: (2)
When g(u) =u, where is a positive quotient of odd integers, Eq. (1) reduces to the equation
(rn(xn)) +f(n; xn) = 0; (3)
which may be reduced further to the equation
(rn(xn)) +qn+1x
n+1= 0; (4)
where {qn}
∞
n=0 is a positive sequence. The problem of oscillation and nonoscillation of solutions of
(3) and (4) has received a great deal of attention in the last few years, e.g., see [6, 8, 9].
In this paper, we consider classication schemes for all eventually positive solutions of (1) under the assumptions
∞ X
n=K
g−1
1
rn
¡∞ or
∞ X
n=K
g−1
1
rn
For this reason, we will employ the following notations:
Rs; n= n−1
X
i=s
g−1
1
ri
; K6s6n−1
and
Rs=
∞ X
i=s
g−11
ri
; s¿K:
In the following section, we rst give several preparatory lemmas which will be useful for our later results. In Section 3, we will discuss the case RK¡∞, the case RK=∞ will be studied in
Section 4.
2. Preparatory lemmas
In this section, we will give some lemmas which are important in proving our main results. Note rst that if {xn} is an eventually positive solution of (1), then (rng(xn)) =−f(n; xn)¡0 for all
large n; so that {rng(xn)} is eventually decreasing.
Lemma 2.1. Suppose {xn} is an eventually positive solution of(1). Then {xn}is of constant sign
eventually.
Proof. Assume that there exists n0¿K such that xn¿0 for n¿n0. Then f(n; xn)¿0 for n¿n0. If
xn is not eventually positive, then there exists n1¿n0 such that xn160. Therefore, rn1g(xn1)60:
From (1), we have
rng(xn)−rn1g(xn1) +
n−1
X
i=n1
f(i; xi) = 0:
Thus
rng(xn)6− n−1
X
i=n1
f(i; xi)¡0
for n¿n1: This shows that xn¡0 for n¿n1: The proof is complete.
As a consequence, an eventually positive solution {xn} of (1) either satises xn¿0 and xn¿0
for all large n; or, xn¿0 and xn¡0 for all large n.
Lemma 2.2. Suppose that
RK=
∞ X
n=K
g−1
1
rn
¡∞; (5)
Proof. If not, then we have limn→∞xn=∞ by Lemma 2.1. On the other hand, we have noted that
{rng(xn)} is monotone decreasing eventually. Therefore, there exists n1¿K such that
rng(xn)6rn1g(xn1); n¿n1:
Then
xn6g
−1r
n1g(xn1)
1
rn
6g−1(r
n1g(xn1))g −11
rn
(6)
for n¿n1, and after summing,
xn−xn16g −1(r
n1g(xn1))Rn1; n
for n¿n1. But this is contrary to the fact that limn→∞xn=∞ and the assumption that RK¡∞. The
proof is complete.
In case g satises an additional !-condition, more can be said.
Lemma 2.3. Suppose RK¡∞. Let {xn} be an eventually positive solution of (1) and g satises
the !-condition. Then there exist a1; a2¿0 and N¿K such that a1Rn6xn6a2 for n¿N.
Proof. By Lemma 2.2, there existsn0¿K such that xn ≤a2 for some positive number a2. We know
that xn is of constant sign eventually by Lemma 2.1. If xn¿0 eventually, then Rn6xn eventually
because limn→∞Rn= 0. If xn¡0 eventually, then since rng(xn) is also eventually decreasing,
we may assume that xn¡0 and rng(xn) is monotone decreasing for n¿n1. By Eq. (6) and the
!-condition, we have
xn6!g
−1(r
n1g(xn1))g −11
rn
; n¿n1
and
xm−xn6!g
−1(r
n1g(xn1))
m−1 X
s=n
g−1
1
rs
; n¿n1:
Taking the limit as m→ ∞ on both sides of the last inequality, we see that
xn¿−!g
−1(r
n1g(xn1))Rn
for n¿n1. The proof is complete.
Our next result is concerned with necessary conditions for the functions f and g to hold in order that an eventually positive solution of (1) exists.
Lemma 2.4. Suppose that RK¡∞ and {xn} is an eventually positive solution of (1). Then
∞ X
n=K
g−1 1
rn n−1 X
s=K
f(s; xs) !
Proof. In view of Lemma 2.1, we may assume without loss of generality that xn¿0, and, xn¿0
The proof is complete.
We now consider the case where RK=∞.
Lemma 2.5. Suppose that
Thus
xn6!g
−1
(rNg(xN))g
−1 1
rn
; n¿N;
which yields, after summing,
xn−xN6!g
−1
(rNg(xN)) n−1
X
s=N
g−1 1
rs
:
The left-hand side tends to −∞ in view of (7), which is a contradiction. Thus {xn} is eventually
positive, and thus xn¿c1 eventually for some positive constant c1. Furthermore, the same reasoning
just used also leads to
xn6xN+!g
−1(r
Mg(xM)) n−1
X
s=M
g−11
rs
for n¿M where M is an integer such that xn¿0 and xn¿0 for n¿M. SinceRK=∞;thus there
is c2¿0 such that xn6c2RM; n for all large n. The proof is complete.
We need the following result in our subsequent development. Let B be the linear space of all real sequences x={xn}
∞
n=N endowed with the usual operations and
kxk= sup
k¿N
|xk|
hk
¡∞;
where {hk}
∞
k=N is a positive sequence with a uniform positive lower bound. Then B is a Banach
space. A setof sequences inBunder the above norm is said to uniformly Cauchy if for every¿0, there exists an integer M such that wheneveri; j¿M, we have |(xi=hi)−(xj=hj)|¡ forx={xk} ∈.
The following discrete Schauder type xed point theorem is obtained by Cheng and Patula [1].
Lemma 2.6. Let be a closed; bounded and convex subset of B. Suppose T is a continuous mapping such that T() is contained in , and suppose the T() is uniformly Cauchy. Then T
has a xed point in .
3. The case RK¡∞
We have shown in the previous section that when {xn} is an eventually positive solution of (1),
then {(rng(xn))} is eventually decreasing and{xn} is eventually of constant sign. We have also
shown that under the assumption that RK¡∞; {xn} must converge to some (nonnegative) constant.
As a consequence, under the condition RK¡∞, we may now classify an eventually positive solution
{xn} of (1) according to the limits of the sequences {xn} and {rng(xn)}. For this purpose, we rst
denote the set of eventually positive solutions of (1) by S. We then single out eventually positive solutions of (1) which converge to zero or to positive constants, and denote the corresponding subsets by S0 andS+ respectively. But for any x={xn} in S0, since {rng(xn)} either tends to a nite limit
Theorem 3.1. Suppose RK¡∞. Then any eventually positive solutions of (1) must belong to one
of the following classes:
S0={{xn} ∈S|xn→0};
S+;∗= n
{xn}tS
nlim→∞xn∈(0;∞);nlim→∞rng(xn)∈R o
;
and
S+;−∞= n
{xn}tS
nlim→∞xn∈(0;∞);nlim→∞rng(xn) =−∞ o
:
To justify the above classication scheme, we will derive several existence theorems.
Lemma 3.2. Suppose RK¡∞. Suppose further that f is superlinear. Then a necessary and
su-cient condition for (1) to have an eventually positive solution {xn} which belong to S+ is that
∞ X
n=K
g−1 1
rn n−1
X
s=K
f(s; C)
!
¡∞ (8)
for some C¿0.
Proof. Let {xn} be any eventually positive solution of (1) such that limn→∞xn=c¿0. Thus, there
existC1¿0; C2¿0 andN¿K such thatC16xn6C2 forn¿N. On the other hand, using Lemma 2.4
we have
∞ X
n=K
g−1 1
rn n−1
X
s=K
f(s; xs) !
¡∞:
Since f is superlinear, thus we have
∞ X
n=K
g−1 1
rn n−1 X
s=K
f(s; C1)
!
¡∞:
Conversely, let a=C=2. In view of (8), we may choose an integer N so large that
∞ X
n=N
g−1 1
rn n−1
X
s=K
f(s; C)
!
¡a
: (9)
Let X be the set of all real sequences x={xn}
∞
n=N endowed with the usual operations and the
supremum norm. Then X is a Banach space. We dene a subset of X as follows:
={{xn}
∞
n=N∈X :a6xn62a; n¿N}:
Then is a bounded, convex and closed subset of X. Let us further dene an operator T:→X
as follows:
(T x)n=a−
∞ X
s=n
g−1 −1
rs s−1 X
i=K
f(i; xi) !
The mapping T has the following properties. First of all, T maps into . Indeed, if x={xn}
This shows that
Tx(v)−Tx
tends to zero, i.e., T is continuous.
When m; n¿M, by our assumptions on g and (11) we have
which holds for any x∈: Therefore, T is uniformly Cauchy.
In view of Lemma 6, we see that there is an x∗ ∈ such that Tx∗=x∗. It is easy to check that
x∗ is an eventually positive solution of (1). The proof is complete.
We remark that instead of assuming a superlinear function f; we may assume that f(n; x) is nondecreasing in the second variable for all n; and the proof still goes through.
Theorem 3.3. Suppose RK¡∞: Suppose further that f is superlinear. A necessary and sucient
condition for(1) to have an eventually positive solution{xn}which belong to S+;∗ is that(8)holds
Conversely, the solution x∗={xn∗} obtained in the proof of Lemma 3.2 satises
rng(x∗n) =
∞ X
s=n
f(s; x∗s); s¿N:
Hence limn→ ∞rng(x∗n) = 0 as required.
We remark that if we let
=
{xn}
∞
n=N∈X:
a
26xn6a; n¿N
and
(Tx)n=a−
∞ X
s=n
g−1
rs
+ 1
rs
∞ X
i=s
f(i; xi) !
: (13)
Then under the same conditions (8) and (12), arguments similar to those in the proof of Theorem 3.3 show that T has a xed point u={un} which satises limn→∞un=a¿0 and
rng(un) =+
∞ X
s=n
f(s; xs); n¿N;
so that limn→∞rng(un) =¿0: The following is now clear.
Theorem 3.4. Suppose RK¡∞ and f is superlinear. A necessary and sucient condition for (1)
to have an eventually positive solution {xn} which belong to S+;∞ is that (8) holds for some C¿0 and
∞ X
n=K
f(n; D) =∞ (14)
for some D¿0:
Our nal result is concerned with the existence of eventually positive solutions in S0:
Theorem 3.5. Suppose RK¡∞ and f is superlinear. Suppose further that g satises !-condition
and g−1(u)′
¿0 for u¿0. If ∞
X
n=K
f(n; LRn)¡∞; (15)
where
L= max(1; !g−1
(2) +!M); M= max
16u62(g −1
(u))′
;
then (1) has an eventually positive solution in S0: Conversely; if (1) has an eventually positive solution {xn} such that xn→0 and limn→∞rng(xn) =d6= 0; then
∞ X
n=K
Proof. Suppose (15) holds. Then there exists N such that
n=N endowed with the usual operations and
the supremum norm. Dene a subset of X as follows:
={{xn}
and by means of the mean value theorem,
Therefore T is continuous. The fact, that T is uniformly Cauchy, is similarly proved. In view of Lemma 2.6, we see that there is x∗ ∈ such that Tx∗=x∗. It is easy to check that x∗ is an eventually positive solution of (1) which satises that limn→∞x∗n = 0 and limn→∞rng(x∗n) = 1.
Conversely, let {xn} be an eventually positive solution of (1) such that xn→0 and rng(xn)→
d¡0 (the proof of the case d¿0 being similar). Then there exist C1; C2¡0 and N¿K such that
C1¡rng(xn)¡C2 for n¿N. Hence,
!g−1
(C1)g −1
1
rn
¡xn¡!g
−1
(C2)g −1
1
rn
and, after summing,
!g−1(C
1)Rn¡−xn¡!g
−1(C 2)Rn
for n¿N: Let a2= −!g −1(C
1) and a1= −!g −1(C
2), then we see that
0¡a1Rn6xn6a2Rn; n¿N:
On the other hand, in view of (1) we have
∞ X
n=N
f(n; xn) =rNg(xN)−d¡∞:
Thus, we see that
∞ X
n=N
f(n; a1Rn)6
∞ X
n=N
f(n; xn)¡∞:
The proof is complete.
4. The case RK=∞
In this section, we assume that RK=∞ and g satises the !-condition. Let S denotes the set of
all eventually positive solutions of (1). Recall that if {xn} belongs to S; then {rng(xn)} is
even-tually decreasing. Furthermore, in view of Lemma 2.5, we see that {xn}; and hence {rng(xn)};
are eventually positive. Hence {xn} either tends to a positive constant or to positive innity, and
{rng(xn)} tends to a nonnegative constant. Note that if {xn} tends to a positive constant, then
{rng(xn)} must tend to zero. Otherwise rng(xn)}¿d¿0 for n larger than or equal to N; so that
xn¿!g
−1(d)g−1
1
rn
and
xn+1¿xN +!g
−1(d)
n X
n=N
g−11
rn
→ ∞;
Theorem 4.1. Suppose that RK=∞ and that g satises the !-condition. Then any eventually
positive solution {xn} of (1) must belong to one of the following two classes:
S+;0=n{x
In order to justify our classication scheme, we derive the following two results.
Theorem 4.2. Suppose that RK=∞ and suppose g satises the !-condition and f is superlinear.
A necessary and sucient condition for (1) to have an eventually positive solution {xn} which
belong to S+;0 is that
The proof of the converse is similar to that of Lemma 3.2 and hence is sketched. Let a=C=2. In view of (17), we may choose an integer N so large that
∞
n=N endowed with the usual operations
respectively. As seen in the proofs of Theorem 3.3, we may prove that T maps into ; that T
is continuous, and that T is uniformly Cauchy. The xed point {xn∗} of T will converge to a and satises (1). The proof is complete.
The proof of the following result is again similar to that of Theorem 3.3 and hence is omitted.
Theorem 4.3. Suppose RK=∞ and f is superlinear. Suppose further that g satises the
!-condition and g−1
(u)′
¿0 for u¿0. If ∞
X
n=K
f(n; CRK;n)¡∞ for some C¿0; (19)
then (1) has a solution in S∞
: Conversely; if (1) has a solution {xn} in S
∞
such that {rng(xn)}
tends to a positive limit; then (19) holds.
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