J. Comp. & Math. Sci. Vol.2 (3), 531-536 (2011)
Oscillatiory Behavior of Solutions of the Third-Order Neutral Difference Equations with Positive and Negative Coefficients B. SELVARAJ1 and J. DAPHY LOUIS LOVENIA2 1
Dean of Science and Humanities, Nehru Institute of Engineering and Technology, Coimbatore, Tamil Nadu, India. 2 Department of Mathematics, Karunya University, Karunya Nagar, Coimbatore, Tamil Nadu, India. ABSTRACT In this paper, oscillation criteria for solutions of third order neutral difference equation with positive and negative coefficients are established. Examples are provided to illustrate the results. Keywords: Oscillation, Neutral difference equations, positive and negative coefficients. AMS Classification: 39A10
1. INTRODUCTION Neutral difference equations exist in applied Mathematics, for example stability theory, circuit theory, network systems and so on. Consider the third –order neutral type delay difference equation ∆ (a n ∆2 ( xn + cn x n − k ) + pn f ( xn −l ) − qn f ( xn− m ) = 0
(1.1)
∆ (a n ∆2 ( xn − cn xn − k ) + p n f ( x n −l ) − qn f ( xn− m ) = 0
(1.2)
where n ∈ N (no ) = {n0 , n0 + 1, n0 + 2,...}, n0 is a nonnegative integer, k, l, m are positive
integers, {a n }, {cn }, { pn }, {qn } are real sequences, f : R → R is continuous and nondecreasing with y f(y) > 0 for y ≠ 0. Let θ = max{k , l , m}. By a solution of equation (1.1), (1.2), we mean a real sequence {xn } which is defined for all n ≥ n0 − θ and satisfies equation (1.1), (1.2) for all n ∈ N (n0 ). Equation (1.1), (1.2), has a unique solution {xn } if an initial sequence {x0 (n)} is given to hold xn = x0 (n), n = n0 − θ , n0 − θ + 1,...n0 . A nontrivial solution {xn } of equation (1.1), (1.2) is said to be oscillatory if it is neither eventually positive nor eventually negative and it is non-oscillatory otherwise.
Journal of Computer and Mathematical Sciences Vol. 2, Issue 3, 30 June, 2011 Pages (399-580)
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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (3), 531-536 (2011)
In the past few years there has been an increasing interest in the study of oscillatory behavior of solutions of difference equations see3-14 and references cited therein. Sufficient conditions for oscillation of solutions of first and second order neutral delay difference equations with positive and negative coefficients have been investigated by many authors. In this paper, the sufficient conditions for every solution of equations (1.1) and (1.2) to be oscillatory are obtained. 2. OSCILLATION RESULTS FOR EQUATIONS (1.1) and (1.2) In this section, we submit the sufficient conditions for oscillation of all solutions of equations (1.1) and (1.2). The following assumptions are used: (A1) {a n } is a positive sequence such that ∞
1 =∞ ∑ n = n0 an
(A2) {cn }, { pn }and{qn } are nonnegative real sequences: (A3) l ≥ m (A4) pn − qn +l − m ≥ b > 0, where b is a constant. (A5) there exist positive constants G and H f ( y) such that G ≤ ≤H y
for y
0.
Then every solution of (1.1) is oscillatory. Proof : Suppose that {xn } is a nonoscillatory solution of (1.1). We assume that xn > 0 and x n −θ > 0 n ≥ n1 ∈ N (n0 ). , Let n −1
1 a t = n1 t
z n = xn + cn xn − k − ∑
t −1
s −1
∑ qv s =t − m v = s −l + m ∑
f ( xv − m )
for n ≥ n1 + θ , then ∆ ( a n ∆2 z n ) = ∆ ( a n ∆2 ( x n + c n x n − k )) − q n f ( x n − m ) + q n −l + m f ( x n −l )
∆(a n ∆2 z n ) = − pn f ( xn −l ) + qn −l + m f ( xn−l ) = −( p n − q n −l + m ) f ( x n − l ) ≤ −bGxn −l (2.3)
for n ≥ n1 + θ . Now we have {an ∆2 z n } nonincreasing and ∆2 z n ≥ 0 or ∆2 z n < 0, n ≥ N for some N ≥ n1 + θ . Let us discuss the following two possible cases: Case 1: ∆2 z n ≥ 0 or all n ≥ N. Summing (2.3) from N to n, we obtain ∆(a N ∆2 z N + a N +1∆2 z N +1 + ... + a n ∆2 z n ) ≤ ( − b G x N − l − b G x N + 1 − l − ... − b G x n − l ) = − b G ( x N −l + xN
+1− l
+ ... + x n − l )
n
= − bG ∑ x s − l s= N
n
Therefore
− a n +1 ∆ 2 z n +1 + a N ∆ 2 z N ≥ b G ∑ x s − l s= N
We obtain n
bG ∑ x s − l ≤ a N ∆ 2 z N − a n + 1 ∆ 2 z n + 1 s= N
Theorem 2.1 Consider the difference equation (1.1) and assume (A1)-(A5). If m + 1 ≥ k ,0 ≤ cn ≤ c for n ∈ N (n0 ) ∞
∑
n = n0
1 an
n −1
∑
t −i
∑ qs ≤
t = n − m s =t − l + m
1 + cn H
(2.1) (2.2)
≤ a N ∆2 z N < ∞
and {xn } is summable for From the condition (2.1), m + 1 ≥ k , 0 ≤ c n ≤ c,
we have
for
n ∈ N (N ).
n ∈ N (n0 ).
y n = xn + cn xn − k
Journal of Computer and Mathematical Sciences Vol. 2, Issue 3, 30 June, 2011 Pages (399-580)
(2.4)
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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (3), 531-536 (2011)
N1 ≥ N + 1 such that z N1 < 0 and
is also summable. Also for n ≥ N
integer
∆ y n = ∆ ( xn + cn xn − k )
max N ≤ n ≤ N1 xn = xN1 .
2
2
1 n −1 = ∆ zn + ∑ qv f ( xv − m ) ≥ 0 an v = n −l + m which implies that is ∆ 2 y n > k , k > 0 . Summing the last inequality from m to (n-1), we get ∆ y n ≥ ∆ y m + k ( n − m ). There are
(2.5)
2
two cases :
∆yn < 0. If yn ≥ yN , n ≥ N , which
∆yn > 0
and
∆yn > 0 , we get yields that { yn } is not summable, a contradiction. If ∆yn < 0 which implies , ∆yn < −l , l > 0, summing the last inequality from m1 to (n-1), we get y n → −∞ as n → ∞ , which is a contradiction. 2 Case 2: ∆ zn < 0 for all n ≥ N . Summing
a n ∆2 z n ≤ a N ∆2 z N < 0 .from N to (n-1), we
We have 0 > z N1 = x N1 + c N1 x N1 − k N1 −1
− ∑
t=N
1 at
t −1
s −1
∑ q v f ( xv − m )
∑
.
s = t − m v = s −l + m N1 −1
≥ {1 + c N1 − H ∑
t=N ∞
≥ {1 + c N1 − H ∑
n = n0
1 at
t −1
s −1
∑ qv }x N1 − k s =t − m v = s −l + m ∑
1 n −1 t −1 ∑ ∑ q s }x N1 − k ≥ 0 an t = n − m s =t −l + m
which is a contradiction, therefore {xn } is bounded from above. Hence for every L>0 there exists an in integer N 2 ≥ N1 such that
xn ≤ L
for ∞
n ≥ N2 .
all n −1
Therefore
t −1
get 1 z ≥ − HL ∑ ∑ ∑ qs ≥ − L > −∞, n ≥ N 2 . ∆ ( a N ∆ 2 z N + a N +1∆ 2 z N +1 + ... + an −1∆ 2 z n −1 ) ≤ ( −bGx N −nl − bGxnN=+n10− la−n ... t = n− s =t −nl +−1m− l ) − mbGx This contradicts the fact that lim n →∞ z n = −∞ z N +1 + ... + an −1∆ z n −1 ) ≤ ( −bGx N − l − bGx N +1− l − ... − bGxn −1− l ) . Hence the proof is complete. = −bG ( xN −l + xN +1−l + ... + xn −1−l ) n −1
= −bG ∑ xs −l ,
Theorem 2.2
s=N
With respect to the
which implies
difference
equation
(1.2), assume (A1) – (A5). If n −1
a n ∆ 2 z n − a N ∆ 2 z N ≤ − bG ∑ x s − l .
0 ≤ cn ≤ c < 1
Therefore an ∆ 2 zn ≤ a N ∆ 2 z N < 0.
and c + H
s=N
∞
∑
n = n0
(2.6) 1 an
n −1
∑
t −1
∑q
t=n−m s=t−l+ m
s
≤1
From (A1) we see that lim n →∞ z n = −∞
. We claim that {xn } is bounded from above. If this is not the case, then there exists an
(2.7)
Then every solution of (1.2) oscillates or satisfies lim n→∞ xn = 0.
Journal of Computer and Mathematical Sciences Vol. 2, Issue 3, 30 June, 2011 Pages (399-580)
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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (3), 531-536 (2011)
Proof: Let {x n } be a non -oscillatory solution of (1.2), we assume that xn > 0 and x n −θ > 0
,
n ≤ n1 ∈ N ( n0 ), we set
bounded from above. If it is not the case, then there exists an integer N1 ≥ N such that
z N1 < 0 and max N ≤ n ≤ N1 x n = x N1. Then we .
have
n−1
1 t −1 s−1 ∑ ∑ qv f ( xv−m ) (2.8) . t =n1 at s =t −m v=s −l +m
zn = xn − cn xn−k − ∑
0 > z N11 = x N1 − c N1 x N1 − k
Then as in the proof of Theorem 2.1, we
− ∑
N1 −1
have for
∆ ( an ∆2 z n ) = −( pn − qn −l + m )
∞
(2.9)
f ( xn −l ) ≤ −bGx n −l
t=N
1 t −1 s −1 ∑ ∑ q v f ( xv − m ) a t s =t − m v = s − l + m
≥ {1 − c − H ∑
n ≥ n1 + θ , Now we have {an ∆ z n }
n = n0
1 n −1 t −1 ∑ ∑ q s }x N1 ≥ 0 an t = n − m s =t −l + m
2
or
which is a contradiction, so {xn } is bounded
∆2 z n < 0, n ≥ N for some N ≥ n1 + θ . Let
from above .We see from (2.6) to (2.8) {z n } is bounded which contradicts the fact that
non-increasing
∆2 z n ≥ 0
and
us discuss the following two possible cases; Case 1: ∆2 z n ≥ 0 for all n ≥ n1 . Here L is a nonnegative constant, where L = lim n→∞ an ∆z n . Using (A4) and summing (2.9) from n1 to ∞ we get,
lim n→∞ z n = −∞
. Hence the proof is complete 3. EXAMPLES 1. Consider the difference equation ∆ (n∆2 ( xn + 2 xn −1 )) + (12n + 6 +
∞
∞ > an1 ∆ z n1 − L = ∑ ( p n − q n−l + m ) f ( xn −l ) 2
−
n = n1
2 n+ 2
xn −1(1 + x 2+ x
3
2
2
n −1 )
n −1
2
) 3n + 2
xn −3 (1 + x 2 n −3 )
= 0, n ≥ 1
(3.1)
∞
Here
≥ G ∑ ( p n − q n − l + m ) x n −l
an = n, cn = 2, l = 3, m = 1, pn = 12n + 6 +
n = n1
∞
≥ Gb ∑ xn −l ,
k = 1, qn =
n = n1
which implies that {x n } is summable
∆ zn < 0
for all
lim n→∞ xn = −∞. we claim that
n≥N
With G = .
{xn } is
∞
1 2
and
H =1,
1 ∞1 =∑ =∞ n =1 an 1 n ∑
3
,
2
the conditions (A1)-
(A5) hold. Case 2:
2 n+2
y (1 + y 2 ) , f ( y ) = . 3n + 2 2 + y2
and
lim n→∞ x n = 0. 2
2 + x 2 n −3
and
Journal of Computer and Mathematical Sciences Vol. 2, Issue 3, 30 June, 2011 Pages (399-580)
B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.2 (3), 531-536 (2011) ∞
1 n −1 t −1 2 ∑ ∑ s + 2 < 3. n =1 n t = n −1s =t − 2 3
∑
Hence by Theorem 2.1, all the solutions of equation (3.1) are oscillatory. {xn} = {(−1)n } is one solution of equation (3.1). 2.
Consider the difference equation 1 1 x + x 3n − 4 xn − 2 )) + (6n + 3 + n + 6 )( n − 4 2 ) 2 3 2 + x n−4
∆ (n∆2 ( xn − −
1 n+6
(
xn − 2 + x3n − 2
3
2 + x2n −2
) = 0, n ≥ 1.
(3.2) 1 , k = 2, l = 4, m = 2, pn 2
an = n, cn = = 6n + 3 +
1 y (1 + y 2 ) , q = , f ( y ) = n 3n + 6 3n + 6 2 + y2
With G =
1
1 2
and
H =1,
.
the conditions (A1)-
(A5) hold. ∞
1 ∞1 =∑ =∞ n=1 an 1n ∑
And ∞
1 n−1 t −1 2 ∑ ∑ s +6 < 1 n=1 n t =n=2 s =t −2 3
c+ ∑
By Theorem (2.2), all solutions of equation (3.2) are oscillatory. {xn} = {(−1)n } is one solution of equation (3.2). REFERENCES 1. R. P. Agarwal; Difference Equations and Inequalities, Marcel Dekker, New York, (2000).
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Journal of Computer and Mathematical Sciences Vol. 2, Issue 3, 30 June, 2011 Pages (399-580)