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OSCILLATION OF NONLINEAR DELAY
DIFFERENCE EQUATIONS
JIANCHU JIANG
(Received 16 March 2000)
Abstract.We obtain some oscillation criteria for solutions of the nonlinear delay differ-ence equation of the formxn+1−xn+pnmj=1x
αj n−kj=0.
2000 Mathematics Subject Classification. 39A10.
1. Introduction. Consider the nonlinear delay difference equation
xn+1−xn+pn m
j=1
xαnj−kj=0, n=0,1,2, . . . , (1.1)
wherepn≥0,n=0,1,2, . . . , 0≤k1≤k2≤ ··· ≤km are integers,αj>0 are
ratio-nal numbers with denominator of positive odd integers for eachj=1,2, . . . , mand
m
j=1αj=1.
Equation (1.1) is a discrete analogue of the following first-order nonlinear delay differential equation
x(t)+p(t)
m
j=1
xt−τj
αj=0, (1.2)
wherep(t)∈C([t0,∞), [0,∞)), 0≤τ1≤τ2≤ ··· ≤τm, andαjare the same as in (1.1).
For (1.2), the oscillation of its solutions has been extensively studied in the literature, see, for example [2,4,11,12].
Whenk1=k2= ··· =km=k, (1.1) reduces to the linear delay difference equation
xn+1−xn+pnxn−k=0, n=0,1,2, . . . . (1.3)
Recently, there has been a lot of activity concerning the oscillatory behavior of (1.3). See, for example, [1,3,5,6,7,8,9,10]. In particular, [7] proved that every solution of (1.3) is oscillatory provided
∞
n=0
n+k
i=n
piln
n+k
i=n
pi+1−sign n+k
i=n
pi
− n+k
i=n+1
piln
n+k
i=n+1
pi+1−sign n+k
i=n+1
pi
= ∞.
(1.4)
Our main aim in this note is to generalize condition (1.4) to (1.1).
2. Main results
Theorem2.1. Assume that
∞
n=0
m
j=1
αj n+kj
i=n
piln
m
j=1
αj n+kj
i=n
pi+1−sign
m
j=1
αj n+kj
i=n
pi
− m
j=1
αj n+kj
i=n+1
piln
m
j=1
αj n+kj
i=n+1
pi+1−sign
m
j=1
αj n+kj
i=n+1
pi
= ∞.
(2.1)
Then every solution of (1.1) oscillates.
Proof. Assume, by way of contradiction, that (1.1) has an eventually positive
solutions{xn}. Then there exists an integern1>0 such thatxn−km >0,xn+1≤xn,
n≥n1. We define the functionsp(t)andx(t)as follows:
p(t)=pn, x(t)=xn+(t−n)
xn+1−xn
, n≤t < n+1, n=0,1,2, . . . . (2.2)
Letx(t)denote derivation on the right. Then
x(t) >0, x(t)≤0, t≥n1,
x(t)=xn+1−xn forn≤t < n+1, n=0,1,2, . . . .
(2.3)
Hence, (1.1) can be rewritten as
x(t)+p(t)
m
j=1
xt−kj
αj
=0, t≥0, (2.4)
where here and in the sequel,[·]denotes the greatest integer function.
Setλ(t)= −x(t)/x(t)fort≥n1. Thenλ(t)≥0 fort≥n1, and from (2.4) we have
λ(t)=p(t)exp
m
j=1
αj
t
[t−kj]
λ(s)ds
, t≥n1+km, (2.5)
or
λ(t)
m
j=1
αj
[t+kj+1]
t p(s)ds
≥p(t)
m
j=1
αj
[t+kj+1]
t
λ(s)ds
exp
m
j=1
αj
t
[t−kj]
λ(s)ds
, t≥n1+km.
(2.6)
One can easily show that
φ(r )ex≥φ(r )x+φ(r )ln(er+1−signr ), r≥0, x≥R, (2.7)
nonnegative and right-continuous. Therefore, it follows from
m
j=1
αj
[t+kj+1]
t
p(s)ds=0 (2.8)
thatp(t)=0. Applying inequalities (2.7) to the right side of (2.6), we obtain
λ(t)
m
j=1
αj
[t+kj+1]
t
p(s)ds≥p(t)
m
j=1
αj
t
[t−kj]
λ(s)ds+p(t)lnA(t), n≥n1+km,
(2.9) where
A(t)=e
m
j=1
αj
[t+kj+1]
t p(s)ds+1−sign
m
j=1
αj
[t+kj+1]
t p(s)ds
. (2.10)
Setn2=n1+km. Integrating both sides of (2.9) fromn2toN > n2+2km, we have
m
j=1
αj
N
n2
λ(t)
[t+kj+1]
t p(s)ds dt
≥ m
j=1
αj
N
n2
p(t)
t
[t−kj]
λ(s)ds dt+
N
n2
p(t)lnA(t)dt.
(2.11)
Interchanging the order of integration, we get
N
n2
p(t)
t
[t−kj]
λ(s)ds dt≥
N−kj
n2
λ(s)
[s+kj+1]
s
p(t)dt ds
=
N−kj
n2
λ(t)
[t+kj+1]
t p(s)ds dt.
(2.12)
Substituting this into (2.11), we have
m
j=1
αj
N
N−kj
λ(t)
[t+kj+1]
t
p(s)ds dt≥
N
n2
p(t)lnA(t)dt. (2.13)
From (1.1) we have
xn+1−xn+pnx1n−αmx αm
n−km≤0, n≥n1. (2.14)
Setyn=xαmn forn≥n1. Then
yn+1−yn+αmpnyn−km≤0, n≥n1+km. (2.15)
It follows thatαmni=n−kmpi≤1, and so
αm
[t+km+1]
t
p(s)ds≤αm [t]+km
i=[t]
Recall thatk1≤k2≤ ··· ≤km, therefore
αj
[t+kj+1]
t p(s)ds≤
αj
αm
, t≥n2, j=1,2, . . . , m. (2.17)
Substituting this into (2.13), we obtain
m
j=1
αj
αm
N
N−kjλ(t)dt≥
N
n2
p(t)lnA(t)dt, (2.18)
or
ln
m
j=1
xN−kj
x(N)
αj
≥αm
N
n2
p(t)lnA(t)dt. (2.19)
It follows that
lim
N→∞ m
i=1
xN−kj
x(N)
αj ≥exp
αm
∞
n2
p(t)lnA(t)dt
. (2.20)
LetE= {n≥n2|pn>0}. Then
∞
n2
p(t)lnA(t)dt
= ∞
n=n2
n+1
n
p(t)ln
e
m
j=1
αj
[t+kj+1]
t
p(s)ds+1−sign
m
j=1
αj
[t+kj+1]
t
p(s)ds
dt
= ∞
n=n2
pn
n+1
n
ln
e
m
j=1
αj
n+kj+1
n
p(s)ds−
t
n
p(s)ds
+1
−sign
m
j=1
αj
n+kj+1
n
p(s)ds−
t
n
p(s)ds
dt
= ∞
n=n2
pn
n+1
n
ln
e
m
j=1
αj
n+kj
i=n
pi−pn(t−n)
+1
−sign
m
j=1
αj
n+kj
i=n
pi−pn(t−n)
dt
= E
pn
n+1
n ln
e
m
j=1
αj
n+kj
i=n
pi−pn(t−n)
= E
m
j=1
αj n+kj
i=n
piln
m
j=1
αj n+kj
i=n
pi
− m
j=1
αj n+kj
i=n+1
piln
m
j=1
αj n+kj
i=n+1
pi+1−sign
m
j=1
αj n+kj
i=n+1
pi
= ∞
n=n2
m
j=1
αj n+kj
i=n
piln
m
j=1
αj n+kj
i=n
pi+1−sign
m
j=1
αj n+kj
i=n
pi
− m
j=1
αj n+kj
i=n+1
piln
m
j=1
αj n+kj
i=n+1
pi+1−sign
m
j=1
αj n+kj
i=n+1
pi
.
(2.21)
From (2.1) and (2.20), we have
lim
N→∞ m
j=1
xN−kj
x(N)
αj
= ∞. (2.22)
On the other hand, it follows from (2.1) that
lim sup
n→∞ pn>0. (2.23)
By (2.15), we have
lim sup
n→∞ pn≤lim supn→∞
yn
αmyn−km =
1 αm
1
lim infn→∞yn−km/yn
, (2.24)
which, together with (2.23) yields
lim inf
n→∞
yn−km
yn
<∞, (2.25)
that is,
lim inf
N→∞
xN−km
x(N)
αm
<∞, (2.26)
and so
lim inf
N→∞ m
j=1
xN−kj
x(N)
αj
≤lim inf
N→∞
xN−km
x(N) <∞, (2.27)
FromTheorem 2.1we have immediately.
Corollary2.2. Assume that there exists an integerN≥0such that
m
j=1
αj n+kj
i=n
pi>0 forn≥N, (2.28)
and that
∞
n=N
m
j=1
αj n+kj
j=n
piln
m
j=1
αj n+kj
i=n
pi
−m
j=1
αj n+kj
i=n+1
piln
m
j=1
αj n+ki
i=n+1
pi
= ∞. (2.29)
Then every solution of (1.1) oscillates.
Clearly, whenk1=k2= ··· =km, condition (2.1) reduces to (1.4).
Acknowledgement. This work was supported by the Science Foundation of
Hunan Educational Committee of China.
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Jianchu Jiang: Department of Mathematics, Loudi Teachers’ College, Loudi, Hunan 417000, China
