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Therefore, in this paper we study the oscillation of solution of the neutral type ...... Ramanujan Institute for Advanced Study in Mathematics, University of Madras,.
ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

OSCILLATION THEOREMS FOR SECOND ORDER NONLINEAR NEUTRAL TYPE DIFFERENCE EQUATIONS WITH POSITIVE AND NEGATIVE COEFFICIENTS S.SELVARANGAM1 , M.MADHAN2 AND E.THANDAPANI3 Abstract. In this paper, the authors using two inequalities and Ricatti type transformation obtained some new oscillation results for the second order nonlinear neutral type difference equations of the form ∆(an ∆(xn + cn xn−k )) + pn f (xn+1−l ) − qn g(xn+1−m ) = 0, and ∆(an ∆(xn − cn xn−k )) + pn f (xn+1−l ) − qn g(xn+1−m ) = 0. The obtained results improve, extend and generalize some of the known results. Further examples are provided to illustrate the importance of the main results. Mathematics Subject Classification (2010): 39A10, 39A21. Key words: Oscillation, neutral, difference equation, positive and negative coefficients.

Article history: Received 20 August 2016 Received in revised form 17 January 2017 Accepted 24 January 2017 1. Introduction Neutral type difference and differential equations arise in many areas of applied mathematics, such as population dynamics [5], bifurcation analysis [2], circuit theory [3], dynamic behavior of delayed network systems [20], and so on. Hence these equations have attracted a great interest during last few decades. Therefore, in this paper we study the oscillation of solution of the neutral type difference equations of the form (1.1)

∆(an ∆(xn + cn xn−k )) + pn f (xn+1−l ) − qn g(xn+1−m ) = 0

and (1.2)

∆(an ∆(xn − cn xn−k )) + pn f (xn+1−l ) − qn g(xn+1−m ) = 0

where n ∈ N(n0 ) = {n0 , n0 + 1, ...}, n0 is a nonnegative integer k, l, m are nonnegative integers, {an }, {cn }, {pn }, {qn } are real sequences, f and g : R → R are continuous and nondecreasing functions with uf (u) > 0, and ug(u) > 0 for u 6= 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 ). It is well known that equation (1.1)((1.2)) has a unique solution {xn } if an initial sequence {x0 (n)} is given to hold for 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 nonoscillatory otherwise. In [4, 6, 7, 10, 11, 8, 9, 18, 17, 19], the authors obtained some sufficient conditions for the existence of nonoscillatory solutions and oscillation of all solutions of equations (1.1) and (1.2) when f (u) = g(u) = u, and an ≡ 1. In [15], the authors established some sufficient conditions for the oscillation of equations

1

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(1.1) and (1.2) with f ≡ g, and f (u) u ≥ M1 > 0 for u 6= 0, and in [13], the authors discussed oscillatory behavior of solutions of equations (1.1) and (1.2) with an ≡ 1. Motivated by these results, in this paper we established sufficient conditions for the oscillation of all solutions of equations (1.1) and (1.2) without these types of restrictions. Our results extend and generalize some of the results in [1, 4, 7, 8, 9, 10, 11, 13, 15, 17], and the references cited therein. In Section 2, we present our main results for equations (1.1) and (1.2), and in Section 3, we present some examples to illustrate our theorems. 2. Oscillation Results In this section, we obtain some oscillation criteria for equations (1.1) and (1.2), subject to the following conditions: P∞ (H1 ) {an } is a positive real sequence such that n=n0 a1n = ∞; (H2 ) {cn }, {pn } and {qn } are nonnegative real sequences; (H3 ) there exists β, ratio of odd positive integers, and a positive constant M1 such that fu(u) ≥ M1 for β u 6= 0; g(u) (H4 ) there are positive constants M and M2 such that 0 ≤ g(u) u ≤ M2 , and 0 < f (u) ≤ M for u 6= 0; (H5 ) there is a constant M3 such that pn − M qn−m+l ≥ M3 > 0 for all n ∈ N(n0 ). Lemma 2.1. If b1 and b2 are nonnegative, then (b1 + b2 )β ≤ 2β−1 (bβ1 + bβ2 ) for β ≥ 1, and (b1 + b2 )β ≤ (bβ1 + bβ2 ) for 0 < β < 1. Proof. The proof can be found in [16].



Theorem 2.2. Let assumptions (H1 ) − (H5 ) hold. Further assume that there are constants α1 and α2 such that 0 ≤ α1 ≤ cn ≤ α2 for all n ∈ N(n0 ). If l ≥ m + 1 ≥ k, and ! ∞ n−1 X X 1 (2.1) qs < ∞, an n=n s=n−l+m

0

then every solution of equation (1.1) is oscillatory. Proof. Assume that {xn } is a nonoscillatory solution of equation (1.1). Without loss of generality, we may assume that xn > 0 and xn−θ > 0 for all n ≥ n1 ∈ N(n0 ). The proof for the case xn < 0 is similar and is omitted. Choose an integer N > n1 so that ! n−1 ∞ X X α1 1 qs < . (2.2) an M2 n=N

s=n−l+m

Set zn = xn + cn xn−k −

(2.3)

n−1 X s=N

1 as

s−1 X

qt g(xt+1−m )

t=s−l+m

for all n ≥ N. Then from equation (1.1) and conditions (H3 ) and (H4 ), we have ∆(an ∆zn )

(2.4)

=

∆(an ∆(xn + cn xn−k )) − pn f (xn+1−m ) + qn−l+m g(xn+1−l )

=

−pn f (xn+1−l ) + qn−l+m g(xn+1−l )



−M1 [pn − M qn−l+m ]xβn+1−l



−M3 M1 xβn+1−l ≤ 0

for all n ≥ N. Hence {an ∆zn } is eventually nondecreasing. So either ∆zn < 0 or ∆zn ≥ 0 for all n ≥ N1 for some integer N1 ≥ N.

2

If ∆zn < 0 for all n ≥ N1 , then (2.4) and (H1 ) imply limn→∞ zn = −∞. We claim that {xn } is bounded from above. If this is not the case, then there is an integer N2 ≥ N1 + k such that (2.5)

zN2 < 0, and

max

N1 ≤n≤N2

xn = xN2 .

Then, we have

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

0 > zN2

= xN2 + cN2 xN2 −k −

NX 2 −1 s=N

≥ α1 xN2 −k − M2 xN2 −k

1 as

NX 2 −1 s=N

≥ [α1 − M2

∞ X 1 an

n=N

n−1 X

s−1 X

qt g(xt+1−m )

t=s−l+m

1 as

s−1 X

qt

t=s−l+m

qs ]xN2 −k ≥ 0.

t=n−l+m

This contradiction shows that {xn } must be bounded so there exists a constant L > 0 such that xn ≤ L for all n ≥ N1 . It follows from (2.3) that zn ≥ −LM2

n−1 X s=N1

s−1 X

1 as

qt ≥ −Lα1 > −∞,

t=s−l+m

which contradicts the fact that limn→∞ zn = −∞. Therefore, we have ∆zn ≥ 0 for all n ≥ N1 . Now, summing (2.3) from N1 to n − 1, we obtain n−1 X

∞ > aN1 ∆zN1 ≥ −an+1 ∆zn+1 + aN ∆zN ≥ M3 M1

xβs+1−l ,

s=N1

and therefore {xβn } is summable for n ∈ N(N1 ). Then, by Lemma 2.1, we have ynβ = (xn + cn xn−k )β ≤ 2β−1 (xβn + α2β xβn−k ) f or β ≥ 1, and ynβ = (xn + cn xn−k )β ≤ (xβn + α2β xβn−k ) f or 0 < β < 1, so {ynβ } is also summable. On the other hand, from equation (2.3), we obtain ∆yn = ∆zn +

1 an

n−1 X

qs g(xs+1−m ) ≥ 0

s=n−l+m

β so that ∆yn is nondecreasing for all n ≥ N1 . But then ynβ ≥ yN for all n ≥ N1 implies that {ynβ } is not 1 summable, a contradiction. This completes the proof of the theorem. 

Remark 2.3. If f (u) = g(u), then Theorem 2.2 reduced to Theorem 2.1 of [15]. Next, we establish an oscillation result for equation (1.1) when l = m, and for this case the condition 0 < g(u) u ≤ M2 is not required. Theorem 2.4. Let assumptions (H1 ) − (H4 ) hold. Further assume that β ≥ 1, and l = m, (2.6)

1 − cn+1−l > 0 f or all n ∈ N(n0 ),

and (2.7)

Qn = pn − M qn > 0 f or all n ∈ N(no ).

3

If there exists a positive and nondecreasing sequence {ρn } such that  n−1 X (∆ρs )2 as−l =∞ (2.8) lim sup M1 ρs Qs (1 − cs+1−l )β − n→∞ 4Lβρs s=N

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for any L > 0, then every solution of equation (1.1) is oscillatory. Proof. Assume that {xn } is a nonoscillatory solution of equation (1.1). Without loss of generality, we may assume that xn > 0 for all n ≥ n0 + θ. The proof for the case xn < 0 is similar and is omitted. Define zn = xn + cn xn−k , n ≥ N ∈ N(n0 ), then zn > 0, and from equation (1.1), and conditions (H3 ) and (H4 ), we have ∆(an ∆zn ) + M1 Qn xβn+1−l ≤ 0, n ≥ N.

(2.9)

From (2.7) and (2.9), we obtain ∆(an ∆zn ) ≤ 0 for all n ≥ N1 . Therefore ∆zn ≤ 0 or ∆zn > 0 for all n ≥ N. If ∆zn ≤ 0 for all n ≥ N1 ≥ N then by (H1 ), we obtain zn → −∞ as n → ∞, which is a contradiction. Hence ∆zn > 0 for all n ≥ N. From the definition zn , we obtain xn ≥ (1 − cn )zn , and using this in (2.9) we have β ∆(an ∆zn ) + M1 Qn (1 − cn+1−l )β zn+1−l ≤ 0, n ≥ N.

(2.10) Define

wn =

ρn an ∆zn β zn−l

, n ≥ N,

then wn > 0, and from (2.10), we obtain (2.11)

∆wn ≤ −M1 ρn Qn (1 − cn+1−l )β +

∆ρn ρn wn+1 − Lβ 2 w2 , n ≥ N ρn+1 ρn+1 an−l n+1

β−1 where we have used {an ∆zn } is positive and nonincreasing and L = zN −l . Summing the inequality (2.11) from N to n − 1 and using completing the square, we have  n−1 X (∆ρs )2 as−l β M1 ρs Qs (1 − cs+1−l ) − ≤ wN − wn ≤ wN . 4Lβρs s=N

Taking lim sup in the last inequality, we obtain a contradiction with (2.8), and the proof of the theorem is complete.  Remark 2.5. Let qn ≡ 0 in equation (1.1). Then Theorem 2.2 reduced to the known oscillation criterion for the equation ∆(an ∆(xn + cn xn−k )) + pn f (xn+1−l ) = 0 given in [1], and the references cited therein. In the following we establish oscillation results for equation (1.2). Theorem 2.6. Let assumptions (H1 ) − (H5 ) hold. Further assume that there is a constant α3 such that 0 ≤ cn ≤ α3 < 1, for all n ∈ N(n0 ). If l ≥ m + 1, and (2.12)

α3 + M2

∞ X 1 an

n=N

n−1 X

qs ≤ 1

s=n−l+m

then any solution {xn } of equation (1.2) is either oscillatory or satisfies limn→∞ xn = 0.

4

Proof. Suppose that {xn } is a nonoscillatory solution of equation (1.2), say xn > 0 for n ≥ N ≥ n0 + θ. Define n−1 s−1 X 1 X (2.13) wn = xn − cn xn−k − qt g(xt+1−m ). as s=N

t=s−l+m

Then as in the proof of Theorem 2.2, we have ∆(an ∆wn ) ≤ −M3 M1 xβn+1−l ≤ 0

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

(2.14)

for all n ≥ N, and conclude that {an ∆wn } is eventually nonincreasing. Therefore ∆wn < 0 or ∆wn ≥ 0 for all n ≥ N1 ≥ N. Assume that ∆wn < 0 for all n ≥ N1 , then by (H1 ) we have limn→∞ wn = −∞. We claim that {xn } is bounded from above. If not, there exists an integer N2 ≥ N1 + k such that (2.15)

wN2 < 0, and

max

N1 ≤n≤N2 −k

xn = xN2 −k ,

and we have 0 > wN2

= xN2 − cN2 xN2 −k −

NX 2 −1 s=N

" ≥

1 − α3 − M2

NX 2 −1 s=N

1 as

1 as

s−1 X

qt g(xs+1−m )

t=s−l+m

s−1 X

# qt xN2 −k ≥ 0.

t=s−l+m

This contradiction shows that {xn } must be bounded from above, so there exists a constant L > 0 such that xn ≤ L for all n ≥ N1 . It follows from (2.12) and (2.13) that " # NX s−1 2 −1 X 1 wn ≥ −L α3 + M2 qt ≥ −L > −∞, as s=N

t=s−l+m

which contradicts the fact that limn→∞ wn = −∞. Hence ∆w > 0 for n ≥ N1 . In this case, we see that L is a nonnegative constant , where L = limn→∞ an ∆wn . Summing (2.14) from N1 to ∞, we obtain ∞ X ∞ > aN1 ∆wN1 − L ≥ M3 M1 xβn+1−l n=N1

which implies that

{xβn }

is summable, and thus limn→∞ xn = 0. This completes the proof.



Finally we obtain oscillation results for equation (1.2) when l = m, and for this case the condition 0 < g(u) u ≤ M2 , is not needed. Theorem 2.7. Let assumptions (H1 ) − (H4 ) hold. Further assume that l ≥ k + 1, (2.16)

0 ≤ cn ≤ α3 < 1 f or n ∈ N(N0 ),

(2.17)

0 < β ≤ 1, ∞ X

(2.18)

Qn = ∞,

n=N

and (2.19)

lim sup

n→∞

n−1 X s=n−l+k

Then every solution of equation (1.2) is oscillatory.

5

n−1 1 X Qt = ∞. as t=s

Proof. Let {xn } be a nonoscillatory solution of equation (1.2). Without loss of generality, we may assume that xn > 0 and xn−θ > 0 for all n ≥ n1 ∈ N(n0 ). The proof for the case xn < 0 is similar and is omitted. Define zn = xn − cn xn−k , n ≥ n1 . From equation (1.2), and conditions (H3 ) and (H4 ) we have

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

(2.20)

∆(an ∆zn ) ≤ −M1 Qn xβn+1−l ≤ 0, n ≥ N ≥ n1 .

Hence {zn } and {an ∆zn } are eventually of one sign for all n ≥ N. Then by Lemma 2.1 of [12], and (H1 ) the sequence {zn } satisfies one of the following two cases for all n ≥ N : (i) zn > 0, an ∆zn > 0, ∆(an ∆zn ) ≤ 0; (ii) zn < 0, an ∆zn > 0, ∆(an ∆zn ) ≤ 0. Case (i): From the definition of zn , we have xn ≥ zn , and using this in (2.20), we obtain (2.21)

β ∆(an ∆zn ) + M1 Qn zn+1−l ≤ 0, n ≥ N.

Define wn =

an ∆zn β zn−l

, n ≥ N,

then wn > 0 for n ≥ N, and from (2.21), we obtain ∆wn

= −M1 Qn − ≤ ≤

an ∆zn β ∆zn−l β β zn+1−l zn−l

−M1 Qn −

−βan ∆zn

β β zn−l zn+1−l −M1 Qn , n ≥ N.

∆zn−l

Summing the last inequality from N to n − 1, we have M1

n−1 X

Qs ≤ wN − wn ≤ wN .

s=N

Letting n → ∞, we obtain a contradiction with (2.18). Case(ii): From the definition of zn and (2.16), we have   −zn (2.22) xn−k ≥ . α3 Using (2.22) in (2.20), we obtain ∆(an ∆zn ) −

M1 Qn α3β

β zn+1−l+k ≤ 0, n ≥ N.

Summing the last inequality from s to n − 1 for n > s + 1, we have (2.23)

n−1 M1 X

an ∆zn − as ∆zs −

α3β

β Qt zt+1−l+k ≤ 0.

t=s

Summing again the last inequality from n − l + k to n − 1 for s, we have zn−l+k − zn ≤

M1 α3β

n−1 X

β zn−l+k

s=n−l+k

n−1 1 X Qt as t=s

or zn−l+k β zn−l+k



M1

n−1 X

α3β

s=n−l+k

6

n−1 1 X Qt . as t=s

If β = 1, then from the last inequality, we obtain α3 ≥ lim sup n→∞ M1

n−1 X s=n−l+k

n−1 1 X Qt as t=s

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

which is a contradiction with (2.19). Next, assume 0 < β < 1. Since zn is negative and increasing, we have limn→∞ zn = δ ≤ 0. If δ = 0, then from (2.23), we obtain n−1 X

lim sup

n→∞

s=n−l+k

n−1 1 X Qt ≤ 0 as t=s

since 1 − β > 0, which is a contradiction with (2.19). Now assume that δ < 0. From (2.23) we have ∆zs +

n−1 M1 znβ X

α3β as

Qt ≥ 0.

t=s

Summing the last inequality from N to n − 1, and rearranging we obtain α3β zN M1 znβ In view of δ < 0,

αβ 3 zN β M1 zn



n−1 X s=N

n−1 1 X Qt . as t=s

has an upper bound, so lim

n→∞

n−1 X s=N

n−1 1 X Qt < ∞ as t=s

which again contradicts (2.19). This completes the proof of the theorem.



Remark 2.8. Let qn ≡ 0 in equation (1.2), then Theorem 2.5 reduced to the known oscillation criteria for the equation ∆(an ∆(xn − cn xn−h )) + pn f (xn+1−l ) = 0 given in [1, 14], and the references cited therein. 3. Examples In this section, we present some examples to illustrate the main results. Example 3.1. Consider the second order nonlinear neutral difference equation of the form    1 4 x3n (3.1) ∆ (n∆ (xn + 2xn−1 )) + 2n + 1 + n x3n−2 1 + x2n−2 − n = 0, n ≥ 2. 3 3 (1 + x2n ) Here an = n, cn = 2, pn = 2n + 1 + 31n , qn = 34n , k = 1, l = 3, m = 1, f (u) = u3 (1 + u2 ), and u3 g(u) = 1+u 2 . By taking β = 3, and M1 = M2 = M = 1, we see that conditions (H1 ) − (H4 ) hold. Further, pn − qn−m+l = 2n + 1 +

1 4 − n+2 > 1, 3n 3

and   ∞ n−1 ∞ X X 1 1 X 4 4 1 = + < ∞. n s=n−2 3s n 3n−2 3n−1 n=2 n=2 Therefore all the conditions of Theorem 2.2 are satisfied, and hence every solution of equation (3.1) is oscillatory. In fact, {xn } = {(−1)n } is one such oscillatory solution of equation (3.1).

7

Example 3.2. Consider the second order nonlinear neutral difference equation of the form       5/3  xn−1 1 1 1 1 5/3 4 (3.2) ∆ n ∆ xn + xn−1 + n + + n+2 xn−1 1 + xn−1 − n = 0, n ≥ 1. 2 2 2 2 (1 + x2n−1 )

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2017, VOLUME 7, ISSUE 1, p.1-10

5/3

1 u Here an = n, cn = 12 , pn = n + 12 + 2n+2 , qn = 21n , f (u) = u5/3 (1 + u4 ), g(u) = 1+u 4 , k = 1, and l = 2. 5 With β = 3 , M = 1, and M1 = 1 we see that the conditions (H1 ) − (H4 ) are satisfied. Further we see that 1 1 − cn+1−l = > 0, 2 and 9 1 3 Qn = pn − M qn = n + − n+2 ≥ > 0. 2 2 8 By taking ρn ≡ 1, we see that the condition (2.8) becomes  n−1 ∞  5/3  i X Xh 1 3 1 β lim sup n + − n+2 = ∞. Qs (1 − cs+1−l ) = n→∞ 2 2 2 s=1 1

Hence all conditions of Theorem 2.4 are satisfied, and therefore every solution of equation (3.2) is oscillatory. In fact, {xn } = {(−1)n } is one such oscillatory solution of equation (3.2). Example 3.3. Consider the second order nonlinear neutral difference equation of the form      3  xn−1 2 + x2n−1 1 1 1 x3n 1 (3.3) ∆ n∆ xn − xn−2 4n + 2 − n+3 − n+3 = 0, n ≥ 1. + 2 2 3 4 (1 + xn−1 ) 4 (1 + x2n ) 3

2

3

(2+u ) 1 1 u Here an = n, cn = 12 , pn = 13 (4n + 2 − 4n+3 ), qn = 4n+3 , f (u) = u(1+u 2 ) , g(u) = (1+u2 ) , k = 2, l = 2, m = 1. With β = 3, M1 = 1, M2 = 1, and M = 1, we see that the conditions (H1 ) − (H5 ) hold. Further we see that ∞ ∞ X X 1 1 = =∞ a n n 1 1

and α3 + M2

∞ X 1 a n=1 n

n−1 X



qs

=

s=n−l+m

1 X1 1 + ( ) 2 n 4n+2 1

1 1 + < 1. 2 48 Hence by Theorem 2.4, every solution of equation (3.3) is either oscillatory or tends to zero as n → ∞. In fact {xn } = {(−1)n } is one such oscillatory solution of equation (3.3).
43 ∞  X n=1

2n+1 3

+

1 n 43

], qn =

2 , 4n/3

k = 1, l = 2, and Qn =

0. Further, we see that

 2n+1 n 15 n 3 3 (2 + 2 )(3n + 2)2 +2 = ∞, 8

and lim sup

n→∞

n−1 X s=n−l+k

n−1 2n−1 n−1 1 X 1 15 )[ (2 + 2n−1 )(3n − 1)2 3 + 2 3 ] = ∞. Qt = lim sup( n→∞ as t=s n−1 8

8

Hence all conditions of Theorem 2.7 are satisfied, and therefore every solution of equation (3.4) is oscillatory. In fact {xn } = {(−1)n 2n } is one such oscillatory solution of equation (3.4).

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4. Conclusion In this study, we have obtained new sufficient conditions for the oscillation of all solutions of equations (1.1) and (1.2) via Ricatti transformation and inequalities. Further the oscillation criteria obtained here are applicable to deduce oscillation results for the equations (3.1) to (3.4). The same cannot be deducible for equations (3.1) to (3.4) from any previously known oscillation criteria given in [4, 7, 8, 9, 10, 11, 13, 15, 17], since an 6≡ 1, and f 6= g. Therefore the results obtained here improve, extend and generalize the existing results. References

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Department of Mathematics, Presidency College, Chennai - 600 005, India. E-mail address: [email protected] E-mail address: [email protected] 3

Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai - 600 005, India. E-mail address: [email protected]

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