Internat. J. Math. & Math. Sci.
Vol. 22, No. 1 (1999) 147–154
S 0161-17129922147-3
© Electronic Publishing House
OSCILLATION OF A HIGHER ORDER NEUTRAL DIFFERENCE
EQUATION WITH A FORCING TERM
E. THANDAPANI, M. MARIA SUSAI MANUEL, JOHN R. GRAEF,
and PAUL W. SPIKES
(Received 27 May 1997 and in revised form 5 June 1997)
Abstract. The authors obtain oscillation results for the even order forced neutral difference equation
(∗)
∆m yn + pn yn−k + qn f yn− = hn .
Examples illustrating the results are included.
Keywords and phrases. Difference equations, neutral, nonlinear, oscillation, asymptotic
behavior.
1991 Mathematics Subject Classification. 39A10, 39A11.
1. Introduction. In this paper, we consider forced even order nonlinear neutral
difference equations of the form
∆m yn + pn yn−k + qn f yn− = hn ,
(1)
where m ≥ 2 is even, k, ∈ N = {0, 1, 2, . . . }, ∆yn = yn+1 − yn is the usual forward
difference operator, {pn }, {qn }, and {hn } are real sequences, and f : R → R is continuous with uf (u) > 0 for u = 0.
Let σ = max{k, } and let N0 ∈ N be fixed. By a solution of (1), we mean a real
sequence {yn } defined for all n ≥ N0 − σ and satisfying (1) for all n ≥ N0 . Here, we
are concerned only with the nontrivial solutions of (1). Such a solution {yn } of (1)
is said to be oscillatory if, for any N ≥ N0 , there exists n > N such that yn+1 yn ≤ 0.
Otherwise, the solution is said to be nonoscillatory. Throughout the paper, we assume
that the following conditions hold:
(C1 ) qn ≥ 0 for all n ∈ N, and qn is not eventually identically zero;
(C2 ) f is nondecreasing and there exists K > 0 such that
|f (uv)| ≥ K|f (u)| |f (v)|
for all u, v ∈ R,
(2)
and
±c
0
ds
<∞
f (s)
for all c > 0.
(3)
In recent years, the oscillation of delay difference equations, especially unforced
equations, has been studied by a variety of authors. For recent contributions to the
literature, see, for example, the papers [2, 3, 4, 5, 6, 7, 8, 9, 10, 11] and the references
148
E. THANDAPANI ET AL.
contained therein. However, relatively few oscillation results are known for forced
equations (see [5, 6, 7, 8, 9, 10, 11]). In this paper, we give sufficient conditions which
ensure that all solutions of (1) are oscillatory under the influence of certain classes of
forcing terms.
In the sequel, we often make use of the following conditions:
(H1 ) 0 ≤ pn < P1 < 1, where P1 is a constant;
(H2 ) there exists a real sequence
{Fn } such that ∆m Fn = hn ;
n−l (m−1) ∞
= ∞;
(H3 )
n=N0 qn f
2m−1
(H4 ) {Fn } is oscillatory and limn →∞ Fn = 0;
(H5 ) {Fn } is k periodic;
∞
(H6 )
n=N0 qn = ∞;
(H7 ) there exists γ > 0 such that f (u)/u ≥ γ > 0 for u = 0.
We also need the following lemmas whose proof can be found in [1].
Lemma 1 ([1, Thm. 1.7.11]). Let zn > 0 be defined for n ≥ a with ∆m zn of constant
sign for n ≥ a and not identically zero. Then there exists an integer j, 0 ≤ j ≤ m, with
m + j odd for ∆m zn ≤ 0 and m + j even for ∆m zn ≥ 0, such that for n ≥ a
j ≤ m − 1 implies (−1)j+i ∆i zn > 0
i
j ≥ 1 implies ∆ zn > 0
for j ≤ i ≤ m − 1
for 1 ≤ i ≤ j − 1.
(4)
Lemma 2 ([1, Cor. 1.7.12]). Let zn > 0 be defined for n ≥ a with ∆m zn ≤ 0 for n ≥ a
and not eventually identically zero. Then there exists an integer N1 ≥ a such that
zn ≥
(n − N1 )(m−1) m−1
∆
z2m−j−1 n
(m − 1)!
(5)
for n ≥ N1 , where j is defined in Lemma 1.
Remark 1. Observe that under the hypotheses of Lemma 1, if zn is increasing,
then
(m−1)
n
1
(6)
∆m−1 zn
zn ≥
m−1
(m − 1)! 2
for n ≥ 2m−1 N1 .
2. Main results. Our first theorem is a new result for unforced equations, but the
technique of proof will be used in subsequent theorems for forced equations.
Theorem 1. Let hn ≡ 0 for all n ∈ N, and let (H1 ) and (H3 ) hold. Then all solutions
of (1) are oscillatory.
Proof. Let {yn } be a solution of (1) with yn > 0, yn−k > 0, and yn− > 0 for
n ≥ N1 ≥ N0 . Setting
zn = yn + pn yn−k ,
(7)
∆m zn = −qn f yn− ≤ 0
(8)
we obtain zn ≥ yn > 0 and
OSCILLATION OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION . . .
149
for n ≥ N1 . By Lemma 1, there exists an odd integer j with 0 ≤ j ≤ m such that
∆i zn > 0
for i = 1, . . . , j − 1
and
(−1)j+i ∆i zn > 0
for i = j, j + 1, . . . , m − 1
(9)
for n ≥ N2 for some N2 ≥ N1 .
Since m is even, ∆zn > 0 and ∆m−1 zn > 0 for n ≥ N2 . From (7), we have
zn − pn yn−k = yn ,
(10)
so zn ≥ yn and {zn } increasing imply that
0 < (1 − P1 )zn ≤ (1 − pn )zn ≤ yn .
(11)
Again, since zn is increasing, Remark 1 and (11) imply that there exists N3 ≥ N2 such
that
n (m−1) m−1
(1 − P1 )
∆
zn
(m − 1)! 2m−1
yn ≥ (1 − P1 )zn ≥
(12)
for n ≥ 2m−1 N3 . Applying (C2 ) to (12) yields
n − (m−1)
(1 − P1 )
f
f ∆m−1 zn−
f (yn− ) ≥ K f
(m − 1)!
2m−1
n − (m−1)
f ∆m−1 zn
≥ K1 f
2m−1
2
for n ≥ N4 ≥ 2m−1 N3 , where K1 = K 2 f
∆m zn + K1 qn f
(1−P1 )
(m−1)!
n−
2m−1
(13)
> 0. Combining (8) and (13), we obtain
(m−1) f ∆m−1 zn ≤ 0
(14)
for n ≥ N4 and summing, we get
K1
n−1
qs f
s=N4
s −
2m−1
(m−1) ≤−
n−1
s=N4
∆m zs
≤
f ∆m−1 zs
∆m−1 zN
∆m−1 z
n
4
du
.
f (u)
(15)
Letting n → ∞ and using (C2 ), we get
∞
n=N4
qn f
n−
2m−1
(m−1) < ∞,
(16)
which contradicts (H3 ).
Theorem 2. If (H1 ) and (H2 )–(H4 ) holds, then all the solutions of (1) are oscillatory.
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E. THANDAPANI ET AL.
Proof. Let {yn } be a nonoscillatory solution of (1) with yn > 0, yn−k > 0, and
yn− > 0 for all n ≥ N1 ≥ N0 . For n ≥ N1 , let
xn = yn + pn yn−k − Fn .
(17)
∆m xn = −qn f (yn− ) ≤ 0.
(18)
Then from (1) and (H2 ),
Hence, xn > 0 or xn < 0 for n ≥ N2 for some N2 ≥ N1 . But xn < 0 implies that 0 <
yn < Fn for n ≥ N2 which is impossible since {Fn } oscillates. Thus, xn > 0 for n ≥ N2 .
From Lemma 1, it follows that there is an odd integer j with 0 ≤ j ≤ m such that
∆i xn > 0,
for i = 1, . . . , j − 1
and
(−1)j+i ∆i xn > 0,
for i = j, j + 1, . . . , m − 1
(19)
for n ≥ N3 ≥ N2 .
Clearly, ∆xn > 0 for n ≥ N3 . For 0 < " < (1−P1 )xN3 , (H4 ) implies that there exists an
integer N4 > N3 such that |Fn | < "/2 for n ≥ N4 . From (17), we have yn ≤ xn + Fn . So
xn − pn xn−k ≤ yn − Fn + pn Fn−k < yn +
" "
+ pn .
2 2
(20)
Hence,
0 < (1 − P1 )xN3 − " < (1 − P1 )xn − " < yn
(21)
for n ≥ N4 . Setting rn = (1 − P1 )xn − " for n ≥ N4 , we get 0 < rn < yn , ∆rn > 0, and
∆m rn = −(1 − P1 )qn f (yn− ) ≤ 0. Now, proceeding as in the proof of Theorem 1, we
again obtain a contradiction.
We can remove the “oscillatory” part in condition (H4 ) and obtain the weaker conclusion that the solutions either oscillate or converge to zero.
Corollary 3. If (H1 ), (H2 ), and (H3 ) hold and limn →∞ Fn = 0, then all the solutions
of (1) are either oscillatory or converge to zero.
Proof. Proceeding as in the proof of Theorem 2, we again obtain that xn > 0 or
xn < 0 for n ≥ N2 . If xn < 0, then 0 < yn < Fn . So, {yn } → 0 as n → ∞. The remainder
of the proof is the same as proof of Theorem 2.
Our next result replaces condition (H4 ) with a periodicity condition on forcing term.
Theorem 4. If (H1 )–(H3 ), and (H5 ) hold, then every solution of (1) is oscillatory.
Proof. Let {yn } be a nonoscillatory solution of (1) with yn > 0, yn−k > 0, and
yn− > 0 for all n ≥ N1 ≥ N0 . Defining xn as in (17), we have that (18) holds and so
either xn > 0 or xn < 0 for n ≥ N2 for some N2 ≥ N1 .
OSCILLATION OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION . . .
151
We claim that {yn } is bounded. If not, then {yn } is unbounded and since 0 <
yn < xn + Fn and {Fn } is bounded, {xn } must be unbounded and eventually positive. Clearly, ∆xn > 0 for large n since ∆xn < 0 implies that {xn } is bounded. From
(17), we have
xn − pn xn−k = yn − Fn − pn pn−k yn−2k + pn Fn−k ,
(22)
for n ≥ N3 for some N3 ≥ N2 . That is,
(1 − pn )xn ≤ yn − (1 − pn )Fn ,
(23)
0 < (1 − P1 )(xn + Fn ) ≤ yn .
(24)
or
Since {Fn } is periodic, there exist real numbers c1 and c2 and two increasing se
quences {ni } and {n
i } ⊂ N such that limi →∞ ni = limi →∞ ni = ∞, Fni = c1 , Fni = c2 ,
and c1 ≤ Fn ≤ c2 for all n ≥ N0 . Hence, for n ≥ ni , i ≥ 1, we have
xn + c1 ≥ xn + c1 = xn + Fn ≥ yn > 0.
(25)
0 < (1 − P1 )(xn + c1 ) ≤ (1 − P1 )(xn + Fn ) ≤ yn
(26)
i
i
i
i
Thus,
for n ≥ ni . Setting rn = (1 − P1 )(xn + c1 ) for n ≥ ni , and i ≥ 1, we obtain 0 < rn ≤ yn ,
∆rn > 0, and
∆m rn = −(1 − P1 )qn f (yn− ) ≤ 0.
(27)
Now, applying Lemma 1 and proceeding as in the proof of Theorem 1, we arrive at a
contradiction. Thus, our claim holds, that is, {yn } is bounded.
The boundedness of {yn } implies that {xn } is bounded. Since m is even, j is odd.
So (19) implies that ∆xn > 0 for n ≥ N2 . Again, proceeding as the proof of Theorem 1,
we arrive at a contradiction. Hence, {yn } is oscillatory.
Remark 2. With appropriate modifications in condition (C1 ), (C2 ), and (H3 ), Theorems 1, 2, and 4 and Corollary 3 also hold for the more general equation
m
∆m yn + pn yn−k +
qj,n fj yn−j = hn .
(28)
j=1
Our final result, in this paper, is for the case pn ≡ 1.
Theorem 5. If pn ≡ 1 and the conditions (H2 ) and (H5 )–(H7 ) hold, then all the solutions of (1) are oscillatory.
Proof. Let {yn } be a nonoscillatory solution of (1) with yn > 0, yn−k > 0, and
yn− > 0 for all n ≥ N1 ≥ N0 . Since {Fn } is periodic, there is a real number ω such
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E. THANDAPANI ET AL.
that the sequence {Fn − ω} is oscillatory. For n ≥ N1 , let wn = yn + yn−k − (Fn − ω).
Then
∆m wn = −qn f (yn− ) ≤ 0,
(29)
and so {wn } is monotonic. If wn < 0 eventually, then 0 < yn < Fn −ω for large n which
is impossible since {Fn − ω} oscillates. Thus, wn > 0 for n ≥ N2 for some N2 ≥ N1 .
By Lemma 1, we have ∆m−1 wn > 0 for n ≥ N2 . Summing (29) from N2 to n − 1 and
applying (H7 ), we obtain
∆m−1 wN2 =
n−1
qs f (ys− ) + ∆m−1 wn >
s=N2
n−1
qs f (ys− ) > γ
s=N2
n−1
qs ys− ,
(30)
s=N2
which yields
∞
qs ys− < ∞.
(31)
s=N2
From Lemma 1, we see that j is odd, and, hence, ∆wn > 0 for n ≥ N2 . This means that
for n ≥ N2 ,
wn − wn−k = yn − yn−2k − (Fn − Fn−k ),
(32)
which, in view of (H5 ), yields
wn − wn−k = yn − yn−2k > 0,
or yn > yn−2k for n ≥ N2 . Therefore, lim inf n →∞ yn > 0 and so
contradicts (H6 ).
(33)
∞
s=N2
qs < ∞, which
It should be pointed out that whether results analogous to Theorems 1, 2, 4, and
5 and Corollary 3 hold when m is odd remains an open question. We conclude this
paper with some examples of the above theorems.
Example 1. Consider the difference equation
α
= 0,
∆m yn + 12 yn−k + 3(2)m−1 yn−
(E1 )
where α ∈ (0, 1) is a ratio of odd positive integers, k is any positive even integer, and
is any nonnegative integer such that α is an odd integer. It is easy to see that all
the conditions of Theorem 1 are satisfied. In fact, {yn } = {(−1)n } is an oscillatory
solution of (E1 ).
Example 2. In the equation
3m
(−1)n 3m
1
α
=
,
∆m yn + 2 yn−k + 3(2)m−1 − n+m yn−
2
2n+m
(E2 )
let α ∈ (0, 1) be the ratio of odd positive integer, k an even positive integer, and any
nonnegative integer such that α is an odd integer. If we let {Fn } = {(−1)n /2n }, then
all the conditions of Theorem 2 are satisfied and, in fact, {yn } = {(−1)n } is oscillatory
solution of (E2 ).
OSCILLATION OF A HIGHER ORDER NEUTRAL DIFFERENCE EQUATION . . .
Example 3. Consider the difference equation
1
α
= 3(2)m−1 (−1)n ,
∆m yn + 4 yn−k + 2m−2 yn−
153
(E3 )
where α ∈ (0, 1) is a ratio of odd positive integer, k is an even positive integer, and
is any nonnegative integer such that α is an even integer. Here, we take {Fn } =
{3/2(−1)n }. Then all the conditions of Theorem 4 are satisfied and {yn } = {(−1)n }
is an oscillatory solution of (E3 ).
Example 4. The difference equation
∆m (yn + yn−k ) + 2m+1 yn− = 2m+2 (−1)n = 0,
(E4 )
where k and are positive even integers and {Fn } = {4(−1)n }, satisfies all the conditions of Theorem 5. Here, {yn } = {(−1)n } is an oscillatory solution of (E4 ).
Acknowledgement. The research by J. R. Graef was supported by the Mississippi
State University of Biological and Physical Sciences Research Institute.
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Thandapani: Department of Mathematics, Periyar University, Salem 636 011, Tamil
Nadu, India
Manuel: Department of Mathematics, Sarred Heart College, Tiruppattur 635 601,
India
Graef: Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762, USA
Spikes: Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS 39762, USA