Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol101.Issue1-2.1999:
Journal of Computational and Applied Mathematics 101 (1999) 39–51
Classi cation schemes for positive solutions of a second-order
nonlinear di erence equation
Guang Zhang a , Sui Sun Cheng b; ∗ , Ying Gao c
a
Department of Mathematics, Datong Advanced College, Datong, Shanxi 037008, Peoples Republic of China
b
Department 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
Classi cation schemes for eventually positive solutions of a class of second-order nonlinear di erence equations are
given in terms of their asymptotic magnitudes, and necessary as well as sucient conditions for the existence of these
c 1999 Elsevier Science B.V. All rights reserved.
solutions are also provided.
AMS classi cation: 39A10
Keywords: Nonlinear di erence equation; Superlinear function; Sublinear function; Eventually positive solution;
!-condition
1. Introduction
In [7], classi cation schemes for eventually positive solutions of the nonlinear di erence equation
(rn (xn − pn xn− )) + 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 di erence equations
of the form
(rn g(xn )) + f(n; xn ) = 0;
n = K; K + 1; : : : ;
(1)
where K is a xed integer, {rn }∞
n=0 is a positive sequence, f(n; x) is a real-valued function de ned on
{K; K +1; : : :} × R which is continuous in the second variable x and satis es f(n; x)¿0 for x¿0, and
∗
Corresponding author. E-mail: [email protected]. Partially supported by the National Science Coucil of ROC.
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 8 9 - 7
40
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
nally the everywhere continuous real function g is monotone increasing and satis es the conditions
g(0) = 0; g−1 (−u) = − g−1 (u) and g−1 (uv)6 g−1 (u)g−1 (v) for some ; ¿0 and all u; v¿0. The
function g(u) = u
satis es 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
function f, for each xed integer n, if f(n; x)=x is nondecreasing in x for x¿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
b
f(n; b)6f(n; x)6 f(n; a)
b
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
rn g(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 di erence 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
= 0;
(rn (xn )
) + qn+1 xn+1
(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 classi cation 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
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
41
For this reason, we will employ the following notations:
Rs; n =
n−1
X
g
−1
i=s
1
;
ri
K6s6n − 1
and
Rs =
∞
X
g−1
i=s
1
;
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 (rn g(xn )) = −f(n; xn )¡0 for all
large n; so that {rn g(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, rn1 g(xn1 )60:
From (1), we have
rn g(xn ) − rn1 g(xn1 ) +
n−1
X
f(i; xi ) = 0:
i=n1
Thus
rn g(xn )6 −
n−1
X
f(i; xi )¡0
i=n1
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 satis es 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
holds and {xn } is an eventually positive solution of (1). Then limn→∞ xn exists.
(5)
42
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Proof. If not, then we have limn→∞ xn = ∞ by Lemma 2.1. On the other hand, we have noted that
{rn g(xn )} is monotone decreasing eventually. Therefore, there exists n1 ¿K such that
rn g(xn )6rn1 g(xn1 );
n¿n1 :
Then
xn 6g
−1
1
1
6 g−1 (rn1 g(xn1 ))g−1
rn1 g(xn1 )
rn
rn
(6)
for n¿n1 , and after summing,
xn − xn1 6 g−1 (rn1 g(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 satis es an additional !-condition, more can be said.
Lemma 2.3. Suppose RK ¡∞. Let {xn } be an eventually positive solution of (1) and g satis es
the !-condition. Then there exist a1 ; a2 ¿0 and N ¿K such that a1 Rn 6xn 6 a2 for n¿N .
Proof. By Lemma 2.2, there exists n0 ¿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 Rn 6xn eventually
because limn→∞ Rn = 0. If xn ¡0 eventually, then since rn g(xn ) is also eventually decreasing,
we may assume that xn ¡0 and rn g(xn ) is monotone decreasing for n¿n1 . By Eq. (6) and the
!-condition, we have
1
;
rn
xn 6!g−1 (rn1 g(xn1 ))g−1
xm − xn 6!g−1 (rn1 g(xn1 ))
m−1
X
n¿n1
and
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 (rn1 g(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
n−1
1X
f(s; xs ) ¡∞:
rn s=K
!
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
43
Proof. In view of Lemma 2.1, we may assume without loss of generality that xn ¿0, and, xn ¿0
or xn ¡0 for n¿K. From (1), we have
rn g(xn ) − rK g(xK ) +
n−1
X
f(s; xs ) = 0:
s=K
Thus
m
X
g
−1
n=K
n−1
m
X
1
1X
f(s; xs ) =
(rK g(xK ) − rn g(xn ))
g−1
rn s=K
rn
n=K
!
for m¿K. Thus, if xn ¿0 for n¿K, we have
m
X
g
−1
n=K
n−1
m
X
1X
1
f(s; xs ) 6 g−1 (rK g(xK ))
g−1
rn s=K
rn
n=K
!
for m¿K, and thus
∞
X
g
−1
n=K
n−1
1X
f(s; xs ) 6 g−1 (rK g(xK ))RK ¡∞:
rn s=K
!
If xn ¡0 for n¿K, we have
∞
X
g−1
n=K
n−1
∞
∞
∞
X
1X
1X
1X
1
−1
−1
f(s; xs ) 6
g (−g(xn )) = −
g (g(xn )) = −
xn 6 xK ¡∞:
rn s=K
n=K
n=K
n=K
!
The proof is complete.
We now consider the case where RK = ∞.
Lemma 2.5. Suppose that
RK =
∞
X
n=K
g−1
1
=∞
rn
(7)
and g satis es the !-condition. Let {xn } be an eventually positive solution of (1). Then {xn } is
eventually positive and there exist c1 ¿0; c2 ¿0 and M ¿K such that c1 6 xn 6c2 RN; n for n¿M .
Proof. In view of Lemma 2.1, {xn } is of constant sign eventually. If xn ¿0 and xn ¡0 for n¿N ,
then we have
rn g(xn )6rN g(xN )¡0:
44
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Thus
xn 6!g−1 (rN g(xN ))g−1
1
;
rn
n¿N;
which yields, after summing,
xn − xN 6!g−1 (rN g(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
xn 6xN + !g−1 (rM g(xM ))
n−1
X
s=M
g−1
1
rs
for n¿M where M is an integer such that xn ¿0 and xn ¿0 for n ¿ M . Since RK = ∞; thus there
is c2 ¿0 such that xn 6c2 RM; 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 set
of sequences in B under the above norm is said to uniformly Cauchy if for every ¿0,
there exists an integer M such that whenever i; j¿M , we have |(xi =hi ) − (xj =hj )|¡ for x = {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 {(rn g(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 {rn g(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 and S+ respectively. But for any x = {xn } in S0 , since {rn g(xn )} either tends to a nite limit
or to −∞, we can further partition S+ into S+; ∗ and S+; −∞ .
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
45
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};
and
n
o
S+; ∗ = {xn } tS lim xn ∈ (0; ∞); lim rn g(xn ) ∈ R ;
n→∞
n
n→∞
o
S+; −∞ = {xn } tS lim xn ∈ (0; ∞); lim rn g(xn ) = −∞ :
n→∞
n→∞
To justify the above classi cation scheme, we will derive several existence theorems.
Lemma 3.2. Suppose RK ¡∞. Suppose further that f is superlinear. Then a necessary and sucient condition for (1) to have an eventually positive solution {xn } which belong to S+ is that
∞
X
g
−1
n=K
n−1
1X
f(s; C) ¡∞
rn s=K
!
(8)
for some C¿0.
Proof. Let {xn } be any eventually positive solution of (1) such that limn→∞ xn = c¿0. Thus, there
exist C1 ¿0; C2 ¿0 and N ¿K such that C1 6xn 6 C2 for n¿N . On the other hand, using Lemma 2.4
we have
∞
X
g−1
n=K
n−1
1X
f(s; xs ) ¡∞:
rn s=K
!
Since f is superlinear, thus we have
∞
X
g
−1
n=K
n−1
1X
f(s; C1 ) ¡∞:
rn s=K
!
Conversely, let a = C=2. In view of (8), we may choose an integer N so large that
∞
X
n=N
g−1
n−1
1X
a
f(s; C) ¡ :
rn s=K
!
(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 de ne a subset
of X as follows:
= {{xn }∞
n=N ∈ X : a6xn 62a; n¿N }:
Then
is a bounded, convex and closed subset of X . Let us further de ne an operator T :
→ X
as follows:
(T x)n = a −
∞
X
s=n
g−1
s−1
−1 X
f(i; xi ) :
rs i=K
!
(10)
46
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
The mapping T has the following properties. First of all, T maps
into
. Indeed, if x = {xn }∞
n=N
belong to
, then
a6(Tx)n = a +
∞
X
g
−1
s=n
s−1
s−1
∞
X
1X
1X
f(i; xi ) 6a +
f(i; C) 62a:
g−1
rs i=K
rs i=K
s=n
!
!
Next, we show that T is continuous. To see this, let ¿0. Choose M ¿N so large that
∞
X
s=n
g
−1
s−1
1X
f(i; C) 6 :
rs i=K
2
!
(11)
Let {x(v) } be a sequence in
such that x(v) → x. Since
is closed, x ∈
. Furthermore, for all large
v,
∞
s−1
X
1X
(v)
f(i; xi ) +
g−1
rs i=K
s=n
!
∞
X
Tx(v) − (Tx)n 6
g−1
n
s=n
6 2
∞
X
g−1
s=n
s−1
1X
f(i; C) ¡:
rs i=K
!
s−1
1X
f(i; xi )
rs i=K
!
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
|(Tx)m − (Tx)n |6
∞
X
s=n
g−1
s−1
∞
X
1X
f(i; xi ) +
g−1
rs i=K
s=m
!
s−1
1X
f(i; xi ) ¡
rs i=K
!
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
for some C¿0 and
∞
X
(12)
f(n; D)¡∞
n=K
for some D¿0:
Indeed, if {xn } is an eventually positive solution in S+; ∗ ; then as in the proof of Theorem 3.3,
0¡C1 6xn 6 C2 for n¿N: In view of (1), we see that
∞
X
n=N
f(n; C1 )6
∞
X
n=N
f(n; xn ) = rN g(xN ) − lim rm g(xm )¡∞:
m→∞
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
47
Conversely, the solution x∗ = {xn∗ } obtained in the proof of Lemma 3.2 satis es
∞
X
rn g(xn∗ ) =
f(s; xs∗ );
s¿N:
s=n
Hence limn → ∞ rn g(xn∗ ) = 0 as required.
We remark that if we let
a
6a;
n¿N
∈
X
:
= {xn }∞
6x
n
n=N
2
and
(Tx)n = a −
∞
X
g−1
s=n
!
∞
1X
f(i; xi ) :
+
rs rs i=s
(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 satis es limn→∞ un = a¿0 and
rn g(un ) = +
∞
X
f(s; xs );
n¿N;
s=n
so that limn→∞ rn g(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
f(n; D) = ∞
(14)
n=K
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 satis es !-condition
and g−1 (u) ¿0 for u¿0. If
∞
X
f(n; LRn )¡∞;
(15)
n=K
where
L = max(1; ! g−1 (2) + ! M );
M = max (g−1 (u))′ ;
16u62
then (1) has an eventually positive solution in S0 : Conversely; if (1) has an eventually positive
solution {xn } such that xn → 0 and limn→∞ rn g(xn ) = d 6= 0; then
∞
X
n=K
f(n; CRn )¡∞ for some C¿0:
48
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Proof. Suppose (15) holds. Then there exists N such that
∞
X
f(i; LRi )¡1 for n¿N:
i=n
Let X be the Banach space of all real sequences x = {xn }∞
n=N endowed with the usual operations and
the supremum norm. De ne a subset
of X as follows:
−1
= {{xn }∞
n=N ∈ X : ! g (1)Rn 6xn 6LRn ; n¿N }:
Then
is a bounded, convex and closed subset of X . Let us further de ne an operator T :
→ X
as follows:
(Tx)n = ! Rn g
−1
1+
∞
X
!
f(s; xs ) + !
s=n
∞
X
Rs+1 g
−1
1+
s=n
∞
X
!
f(i; xi ) ;
i=s
n¿N;
(16)
where
g
−1
1+
∞
X
f(i; xi )
i=s
!
=g
−1
1+
∞
X
!
f(i; xi ) − g
i=s+1
−1
1+
∞
X
!
f(i; xi ) :
i=s
The mapping T has the following properties. First of all, T maps
into
. Indeed, if x = {xn }∞
n=N
belong to
, then
(Tx)n ¿! g−1 (1)Rn ;
and by means of the mean value theorem,
(Tx)n 6 ! g−1 (2)Rn + ! MRn
∞
X
f(s; xs )
s=n
6 (! g−1 (2) + ! M )Rn ;
n¿N:
Next, we show that T is continuous. To see this, let ¿0. Choose N ∗ ¿N so large that
∞
X
s=n
f(s; LRs )¡
;
4! MRK
n¿N ∗ :
Let {x(v) } be a sequence in
such that x(v) → x. Since
is closed, x ∈
. Furthermore, for all
large v,
Tx(v) − (Tx)n 6! Rn g−1
n
+ !
∞
X
s=n
62! MRn
1+
∞
X
∞
X
s=n
(v)
f s; xs
s=n
Rs+1 g−1
s=n
6 4! MRn
∞
X
1+
!
∞
X
i=s
− g−1 1 +
f
s=n
i; xi(v)
f s; x(v) − f(s; xs )
s
f(s; LRs )¡;
∞
X
n ¿ N ∗:
!
!
f (s; xs )
− g
−1
1+
∞
X
i=s
!
f (i; xi )
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
49
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 satis es that limn→∞ xn∗ = 0 and limn→∞ rn g(xn∗ ) = 1.
Conversely, let {xn } be an eventually positive solution of (1) such that xn → 0 and rn g(xn ) →
d¡0 (the proof of the case d¿0 being similar). Then there exist C1 ; C2 ¡0 and N ¿K such that
C1 ¡rn g(xn )¡C2 for n¿N . Hence,
!g−1 (C1 )g−1
1
1
¡xn ¡!g−1 (C2 )g−1
rn
rn
and, after summing,
!g−1 (C1 )Rn ¡ − xn ¡!g−1 (C2 )Rn
for n¿N: Let a2 = − !g−1 (C1 ) and a1 = − !g−1 (C2 ), then we see that
0¡a1 Rn 6xn 6a2 Rn ; n¿N:
On the other hand, in view of (1) we have
∞
X
f(n; xn ) = rN g(xN ) − d¡∞:
n=N
Thus, we see that
∞
X
f(n; a1 Rn )6
n=N
∞
X
f(n; xn )¡∞:
n=N
The proof is complete.
4. The case RK = ∞
In this section, we assume that RK = ∞ and g satis es the !-condition. Let S denotes the set of
all eventually positive solutions of (1). Recall that if {xn } belongs to S; then {rn g(xn )} is eventually decreasing. Furthermore, in view of Lemma 2.5, we see that {xn }; and hence {rn g(xn )};
are eventually positive. Hence {xn } either tends to a positive constant or to positive in nity, and
{rn g(xn )} tends to a nonnegative constant. Note that if {xn } tends to a positive constant, then
{rn g(xn )} must tend to zero. Otherwise rn g(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
which is a contradiction.
g−1
1
rn
→ ∞;
50
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Theorem 4.1. Suppose that RK = ∞ and that g satis es the !-condition. Then any eventually
positive solution {xn } of (1) must belong to one of the following two classes:
and
n
o
S +; 0 = {xn } ∈ S lim xn ∈ (0; ∞); lim rn g(xn ) = 0 ;
n
n→∞
n→∞
o
S ∞ = {xn } ∈ S lim xn = ∞ :
n→∞
In order to justify our classi cation scheme, we derive the following two results.
Theorem 4.2. Suppose that RK = ∞ and suppose g satis es 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
∞
X
g
−1
n=K
!
∞
1 X
f(s; C) ¡∞ for some C¿0:
rn s=n
(17)
Proof. Let {xn } be an eventually positive solution of (1) which belong to S +; 0 , i.e., limn→∞ xn = c¿0
and limn→∞ rn g(xn ) = 0. Then there exist two positive constants C1 ; C2 and N ¿K such that
C1 6 xn 6C2 for n¿N: On the other hand, in view of (1) we have
rn g(xn ) =
∞
X
f(s; xs )
s=n
for n¿N: After summing, we see that
∞
X
g
−1
s=N
!
!
∞
∞
X
1X
f(i; C1 ) 6
g−1
rs i=s
s=N
∞
1X
f(i; xi ) 6c − xN :
rs i=s
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
∞
X
n=N
g−1
!
∞
1 X
a
f(s; C) ¡
rn s=n
(18)
Let X be the Banach space of all bounded sequences x = {xn }∞
n=N endowed with the usual operations
and the norm kxk = supn¿N |xn |. De ne a bounded, convex, and closed subset
of X and an
operator T :
→
as
= {{xn }∞
n=N ∈ X : a6xn 62a; n¿N } ;
and
(Tx)n = a −
∞
X
s=n
g−1
!
∞
−1 X
f(i; xi ) ;
rs i=s
n¿N
51
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
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
satis es (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 R
′K = ∞ and f is superlinear. Suppose further that g satis es the
!-condition and g−1 (u) ¿0 for u¿0. If
∞
X
f(n; CRK;n )¡∞
for some C¿0;
(19)
n=K
then (1) has a solution in S ∞ : Conversely; if (1) has a solution {xn } in S ∞ such that {rn g(xn )}
tends to a positive limit; then (19) holds.
References
[1] S.S. Cheng, H.J. Li, W.T. Patula, Bounded and zero convergent solutions of second order di erence equations,
J. Math. Anal. Appl. 141 (1989) 463– 484.
[2] S.S. Cheng, W.T. Patula, An existence theorem for a nonlinear di erence equation, Nonlinear Analy. 20 (1993)
193–203.
[3] X.Z. He, Oscillatory and asymptotic behavior of second order nonlinear di erence equations, J. Math Anal. Appl.
175 (1993) 482– 498.
[4] J.W. Hooker, W.T. Patula, Riccati type transformations for second order linear di erence equations, J. Math. Anal.
Appl. 82 (1981) 451– 462.
[5] J.W. Hooker, W.T. Patula, A second order nonlinear di erence equation: oscillation and asymptotic behavior,
J. Math. Anal. Appl. 91 (1983) 9 –29.
[6] H.J. Li, S.S. Cheng, Asymptotically monotone solutions of a nonlinear di erence equation, Tamkang J. Math. 24
(1993) 269 –282.
[7] W.T. Li, S.S. Cheng, Classi cations and existence of positive solutions of second order nonlinear neutral di erence
equations, Funkcialaj Ekvacioj, 40 (1997) 371–393.
[8] P.J.Y. Wong, R.P. Agarwal, Oscillation theorems and existence of positive monotone solutions for second order
nonlinear di erence equations, Math. Comput. Modelling 21 (1995) 63– 84.
[9] P.J.Y. Wong, R.P. Agarwal, Oscillation theorems for certain second order nonlinear di erence equations, J. Math.
Anal. Appl. 204 (1996) 813– 829.
[10] X.B. Wu, Some notes “Oscillatory and asymptotic behavior of second order nonlinear di erence equations”, J. Math
Anal. Appl. 189 (1995) 310 –312.
Classi cation schemes for positive solutions of a second-order
nonlinear di erence equation
Guang Zhang a , Sui Sun Cheng b; ∗ , Ying Gao c
a
Department of Mathematics, Datong Advanced College, Datong, Shanxi 037008, Peoples Republic of China
b
Department 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
Classi cation schemes for eventually positive solutions of a class of second-order nonlinear di erence equations are
given in terms of their asymptotic magnitudes, and necessary as well as sucient conditions for the existence of these
c 1999 Elsevier Science B.V. All rights reserved.
solutions are also provided.
AMS classi cation: 39A10
Keywords: Nonlinear di erence equation; Superlinear function; Sublinear function; Eventually positive solution;
!-condition
1. Introduction
In [7], classi cation schemes for eventually positive solutions of the nonlinear di erence equation
(rn (xn − pn xn− )) + 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 di erence equations
of the form
(rn g(xn )) + f(n; xn ) = 0;
n = K; K + 1; : : : ;
(1)
where K is a xed integer, {rn }∞
n=0 is a positive sequence, f(n; x) is a real-valued function de ned on
{K; K +1; : : :} × R which is continuous in the second variable x and satis es f(n; x)¿0 for x¿0, and
∗
Corresponding author. E-mail: [email protected]. Partially supported by the National Science Coucil of ROC.
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 8 9 - 7
40
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
nally the everywhere continuous real function g is monotone increasing and satis es the conditions
g(0) = 0; g−1 (−u) = − g−1 (u) and g−1 (uv)6 g−1 (u)g−1 (v) for some ; ¿0 and all u; v¿0. The
function g(u) = u
satis es 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
function f, for each xed integer n, if f(n; x)=x is nondecreasing in x for x¿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
b
f(n; b)6f(n; x)6 f(n; a)
b
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
rn g(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 di erence 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
= 0;
(rn (xn )
) + qn+1 xn+1
(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 classi cation 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
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
41
For this reason, we will employ the following notations:
Rs; n =
n−1
X
g
−1
i=s
1
;
ri
K6s6n − 1
and
Rs =
∞
X
g−1
i=s
1
;
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 (rn g(xn )) = −f(n; xn )¡0 for all
large n; so that {rn g(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, rn1 g(xn1 )60:
From (1), we have
rn g(xn ) − rn1 g(xn1 ) +
n−1
X
f(i; xi ) = 0:
i=n1
Thus
rn g(xn )6 −
n−1
X
f(i; xi )¡0
i=n1
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 satis es 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
holds and {xn } is an eventually positive solution of (1). Then limn→∞ xn exists.
(5)
42
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Proof. If not, then we have limn→∞ xn = ∞ by Lemma 2.1. On the other hand, we have noted that
{rn g(xn )} is monotone decreasing eventually. Therefore, there exists n1 ¿K such that
rn g(xn )6rn1 g(xn1 );
n¿n1 :
Then
xn 6g
−1
1
1
6 g−1 (rn1 g(xn1 ))g−1
rn1 g(xn1 )
rn
rn
(6)
for n¿n1 , and after summing,
xn − xn1 6 g−1 (rn1 g(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 satis es an additional !-condition, more can be said.
Lemma 2.3. Suppose RK ¡∞. Let {xn } be an eventually positive solution of (1) and g satis es
the !-condition. Then there exist a1 ; a2 ¿0 and N ¿K such that a1 Rn 6xn 6 a2 for n¿N .
Proof. By Lemma 2.2, there exists n0 ¿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 Rn 6xn eventually
because limn→∞ Rn = 0. If xn ¡0 eventually, then since rn g(xn ) is also eventually decreasing,
we may assume that xn ¡0 and rn g(xn ) is monotone decreasing for n¿n1 . By Eq. (6) and the
!-condition, we have
1
;
rn
xn 6!g−1 (rn1 g(xn1 ))g−1
xm − xn 6!g−1 (rn1 g(xn1 ))
m−1
X
n¿n1
and
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 (rn1 g(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
n−1
1X
f(s; xs ) ¡∞:
rn s=K
!
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
43
Proof. In view of Lemma 2.1, we may assume without loss of generality that xn ¿0, and, xn ¿0
or xn ¡0 for n¿K. From (1), we have
rn g(xn ) − rK g(xK ) +
n−1
X
f(s; xs ) = 0:
s=K
Thus
m
X
g
−1
n=K
n−1
m
X
1
1X
f(s; xs ) =
(rK g(xK ) − rn g(xn ))
g−1
rn s=K
rn
n=K
!
for m¿K. Thus, if xn ¿0 for n¿K, we have
m
X
g
−1
n=K
n−1
m
X
1X
1
f(s; xs ) 6 g−1 (rK g(xK ))
g−1
rn s=K
rn
n=K
!
for m¿K, and thus
∞
X
g
−1
n=K
n−1
1X
f(s; xs ) 6 g−1 (rK g(xK ))RK ¡∞:
rn s=K
!
If xn ¡0 for n¿K, we have
∞
X
g−1
n=K
n−1
∞
∞
∞
X
1X
1X
1X
1
−1
−1
f(s; xs ) 6
g (−g(xn )) = −
g (g(xn )) = −
xn 6 xK ¡∞:
rn s=K
n=K
n=K
n=K
!
The proof is complete.
We now consider the case where RK = ∞.
Lemma 2.5. Suppose that
RK =
∞
X
n=K
g−1
1
=∞
rn
(7)
and g satis es the !-condition. Let {xn } be an eventually positive solution of (1). Then {xn } is
eventually positive and there exist c1 ¿0; c2 ¿0 and M ¿K such that c1 6 xn 6c2 RN; n for n¿M .
Proof. In view of Lemma 2.1, {xn } is of constant sign eventually. If xn ¿0 and xn ¡0 for n¿N ,
then we have
rn g(xn )6rN g(xN )¡0:
44
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Thus
xn 6!g−1 (rN g(xN ))g−1
1
;
rn
n¿N;
which yields, after summing,
xn − xN 6!g−1 (rN g(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
xn 6xN + !g−1 (rM g(xM ))
n−1
X
s=M
g−1
1
rs
for n¿M where M is an integer such that xn ¿0 and xn ¿0 for n ¿ M . Since RK = ∞; thus there
is c2 ¿0 such that xn 6c2 RM; 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 set
of sequences in B under the above norm is said to uniformly Cauchy if for every ¿0,
there exists an integer M such that whenever i; j¿M , we have |(xi =hi ) − (xj =hj )|¡ for x = {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 {(rn g(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 {rn g(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 and S+ respectively. But for any x = {xn } in S0 , since {rn g(xn )} either tends to a nite limit
or to −∞, we can further partition S+ into S+; ∗ and S+; −∞ .
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
45
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};
and
n
o
S+; ∗ = {xn } tS lim xn ∈ (0; ∞); lim rn g(xn ) ∈ R ;
n→∞
n
n→∞
o
S+; −∞ = {xn } tS lim xn ∈ (0; ∞); lim rn g(xn ) = −∞ :
n→∞
n→∞
To justify the above classi cation scheme, we will derive several existence theorems.
Lemma 3.2. Suppose RK ¡∞. Suppose further that f is superlinear. Then a necessary and sucient condition for (1) to have an eventually positive solution {xn } which belong to S+ is that
∞
X
g
−1
n=K
n−1
1X
f(s; C) ¡∞
rn s=K
!
(8)
for some C¿0.
Proof. Let {xn } be any eventually positive solution of (1) such that limn→∞ xn = c¿0. Thus, there
exist C1 ¿0; C2 ¿0 and N ¿K such that C1 6xn 6 C2 for n¿N . On the other hand, using Lemma 2.4
we have
∞
X
g−1
n=K
n−1
1X
f(s; xs ) ¡∞:
rn s=K
!
Since f is superlinear, thus we have
∞
X
g
−1
n=K
n−1
1X
f(s; C1 ) ¡∞:
rn s=K
!
Conversely, let a = C=2. In view of (8), we may choose an integer N so large that
∞
X
n=N
g−1
n−1
1X
a
f(s; C) ¡ :
rn s=K
!
(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 de ne a subset
of X as follows:
= {{xn }∞
n=N ∈ X : a6xn 62a; n¿N }:
Then
is a bounded, convex and closed subset of X . Let us further de ne an operator T :
→ X
as follows:
(T x)n = a −
∞
X
s=n
g−1
s−1
−1 X
f(i; xi ) :
rs i=K
!
(10)
46
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
The mapping T has the following properties. First of all, T maps
into
. Indeed, if x = {xn }∞
n=N
belong to
, then
a6(Tx)n = a +
∞
X
g
−1
s=n
s−1
s−1
∞
X
1X
1X
f(i; xi ) 6a +
f(i; C) 62a:
g−1
rs i=K
rs i=K
s=n
!
!
Next, we show that T is continuous. To see this, let ¿0. Choose M ¿N so large that
∞
X
s=n
g
−1
s−1
1X
f(i; C) 6 :
rs i=K
2
!
(11)
Let {x(v) } be a sequence in
such that x(v) → x. Since
is closed, x ∈
. Furthermore, for all large
v,
∞
s−1
X
1X
(v)
f(i; xi ) +
g−1
rs i=K
s=n
!
∞
X
Tx(v) − (Tx)n 6
g−1
n
s=n
6 2
∞
X
g−1
s=n
s−1
1X
f(i; C) ¡:
rs i=K
!
s−1
1X
f(i; xi )
rs i=K
!
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
|(Tx)m − (Tx)n |6
∞
X
s=n
g−1
s−1
∞
X
1X
f(i; xi ) +
g−1
rs i=K
s=m
!
s−1
1X
f(i; xi ) ¡
rs i=K
!
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
for some C¿0 and
∞
X
(12)
f(n; D)¡∞
n=K
for some D¿0:
Indeed, if {xn } is an eventually positive solution in S+; ∗ ; then as in the proof of Theorem 3.3,
0¡C1 6xn 6 C2 for n¿N: In view of (1), we see that
∞
X
n=N
f(n; C1 )6
∞
X
n=N
f(n; xn ) = rN g(xN ) − lim rm g(xm )¡∞:
m→∞
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
47
Conversely, the solution x∗ = {xn∗ } obtained in the proof of Lemma 3.2 satis es
∞
X
rn g(xn∗ ) =
f(s; xs∗ );
s¿N:
s=n
Hence limn → ∞ rn g(xn∗ ) = 0 as required.
We remark that if we let
a
6a;
n¿N
∈
X
:
= {xn }∞
6x
n
n=N
2
and
(Tx)n = a −
∞
X
g−1
s=n
!
∞
1X
f(i; xi ) :
+
rs rs i=s
(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 satis es limn→∞ un = a¿0 and
rn g(un ) = +
∞
X
f(s; xs );
n¿N;
s=n
so that limn→∞ rn g(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
f(n; D) = ∞
(14)
n=K
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 satis es !-condition
and g−1 (u) ¿0 for u¿0. If
∞
X
f(n; LRn )¡∞;
(15)
n=K
where
L = max(1; ! g−1 (2) + ! M );
M = max (g−1 (u))′ ;
16u62
then (1) has an eventually positive solution in S0 : Conversely; if (1) has an eventually positive
solution {xn } such that xn → 0 and limn→∞ rn g(xn ) = d 6= 0; then
∞
X
n=K
f(n; CRn )¡∞ for some C¿0:
48
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Proof. Suppose (15) holds. Then there exists N such that
∞
X
f(i; LRi )¡1 for n¿N:
i=n
Let X be the Banach space of all real sequences x = {xn }∞
n=N endowed with the usual operations and
the supremum norm. De ne a subset
of X as follows:
−1
= {{xn }∞
n=N ∈ X : ! g (1)Rn 6xn 6LRn ; n¿N }:
Then
is a bounded, convex and closed subset of X . Let us further de ne an operator T :
→ X
as follows:
(Tx)n = ! Rn g
−1
1+
∞
X
!
f(s; xs ) + !
s=n
∞
X
Rs+1 g
−1
1+
s=n
∞
X
!
f(i; xi ) ;
i=s
n¿N;
(16)
where
g
−1
1+
∞
X
f(i; xi )
i=s
!
=g
−1
1+
∞
X
!
f(i; xi ) − g
i=s+1
−1
1+
∞
X
!
f(i; xi ) :
i=s
The mapping T has the following properties. First of all, T maps
into
. Indeed, if x = {xn }∞
n=N
belong to
, then
(Tx)n ¿! g−1 (1)Rn ;
and by means of the mean value theorem,
(Tx)n 6 ! g−1 (2)Rn + ! MRn
∞
X
f(s; xs )
s=n
6 (! g−1 (2) + ! M )Rn ;
n¿N:
Next, we show that T is continuous. To see this, let ¿0. Choose N ∗ ¿N so large that
∞
X
s=n
f(s; LRs )¡
;
4! MRK
n¿N ∗ :
Let {x(v) } be a sequence in
such that x(v) → x. Since
is closed, x ∈
. Furthermore, for all
large v,
Tx(v) − (Tx)n 6! Rn g−1
n
+ !
∞
X
s=n
62! MRn
1+
∞
X
∞
X
s=n
(v)
f s; xs
s=n
Rs+1 g−1
s=n
6 4! MRn
∞
X
1+
!
∞
X
i=s
− g−1 1 +
f
s=n
i; xi(v)
f s; x(v) − f(s; xs )
s
f(s; LRs )¡;
∞
X
n ¿ N ∗:
!
!
f (s; xs )
− g
−1
1+
∞
X
i=s
!
f (i; xi )
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
49
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 satis es that limn→∞ xn∗ = 0 and limn→∞ rn g(xn∗ ) = 1.
Conversely, let {xn } be an eventually positive solution of (1) such that xn → 0 and rn g(xn ) →
d¡0 (the proof of the case d¿0 being similar). Then there exist C1 ; C2 ¡0 and N ¿K such that
C1 ¡rn g(xn )¡C2 for n¿N . Hence,
!g−1 (C1 )g−1
1
1
¡xn ¡!g−1 (C2 )g−1
rn
rn
and, after summing,
!g−1 (C1 )Rn ¡ − xn ¡!g−1 (C2 )Rn
for n¿N: Let a2 = − !g−1 (C1 ) and a1 = − !g−1 (C2 ), then we see that
0¡a1 Rn 6xn 6a2 Rn ; n¿N:
On the other hand, in view of (1) we have
∞
X
f(n; xn ) = rN g(xN ) − d¡∞:
n=N
Thus, we see that
∞
X
f(n; a1 Rn )6
n=N
∞
X
f(n; xn )¡∞:
n=N
The proof is complete.
4. The case RK = ∞
In this section, we assume that RK = ∞ and g satis es the !-condition. Let S denotes the set of
all eventually positive solutions of (1). Recall that if {xn } belongs to S; then {rn g(xn )} is eventually decreasing. Furthermore, in view of Lemma 2.5, we see that {xn }; and hence {rn g(xn )};
are eventually positive. Hence {xn } either tends to a positive constant or to positive in nity, and
{rn g(xn )} tends to a nonnegative constant. Note that if {xn } tends to a positive constant, then
{rn g(xn )} must tend to zero. Otherwise rn g(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
which is a contradiction.
g−1
1
rn
→ ∞;
50
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
Theorem 4.1. Suppose that RK = ∞ and that g satis es the !-condition. Then any eventually
positive solution {xn } of (1) must belong to one of the following two classes:
and
n
o
S +; 0 = {xn } ∈ S lim xn ∈ (0; ∞); lim rn g(xn ) = 0 ;
n
n→∞
n→∞
o
S ∞ = {xn } ∈ S lim xn = ∞ :
n→∞
In order to justify our classi cation scheme, we derive the following two results.
Theorem 4.2. Suppose that RK = ∞ and suppose g satis es 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
∞
X
g
−1
n=K
!
∞
1 X
f(s; C) ¡∞ for some C¿0:
rn s=n
(17)
Proof. Let {xn } be an eventually positive solution of (1) which belong to S +; 0 , i.e., limn→∞ xn = c¿0
and limn→∞ rn g(xn ) = 0. Then there exist two positive constants C1 ; C2 and N ¿K such that
C1 6 xn 6C2 for n¿N: On the other hand, in view of (1) we have
rn g(xn ) =
∞
X
f(s; xs )
s=n
for n¿N: After summing, we see that
∞
X
g
−1
s=N
!
!
∞
∞
X
1X
f(i; C1 ) 6
g−1
rs i=s
s=N
∞
1X
f(i; xi ) 6c − xN :
rs i=s
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
∞
X
n=N
g−1
!
∞
1 X
a
f(s; C) ¡
rn s=n
(18)
Let X be the Banach space of all bounded sequences x = {xn }∞
n=N endowed with the usual operations
and the norm kxk = supn¿N |xn |. De ne a bounded, convex, and closed subset
of X and an
operator T :
→
as
= {{xn }∞
n=N ∈ X : a6xn 62a; n¿N } ;
and
(Tx)n = a −
∞
X
s=n
g−1
!
∞
−1 X
f(i; xi ) ;
rs i=s
n¿N
51
G. Zhang et al. / Journal of Computational and Applied Mathematics 101 (1999) 39–51
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
satis es (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 R
′K = ∞ and f is superlinear. Suppose further that g satis es the
!-condition and g−1 (u) ¿0 for u¿0. If
∞
X
f(n; CRK;n )¡∞
for some C¿0;
(19)
n=K
then (1) has a solution in S ∞ : Conversely; if (1) has a solution {xn } in S ∞ such that {rn g(xn )}
tends to a positive limit; then (19) holds.
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