Cmjv03i06p0653

Page 1

J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

Oscillation of Solutions of Certain Fifth Order Difference Equations B. SELVARAJ1 and S. KALEESWARI2 1

Dean of Science and Humanities, Nehru Institute of Engineering and Technology, Coimbatore, Tamil Nadu, INDIA. 2 Department of Mathematics, Nehru Institute of Engineering and Technology, Coimbatore, Tamil Nadu, INDIA. (Received on: December 3, 2012) ABSTRACT The objective of this paper is to study the oscillatory behavior of solutions of fifth order difference equation of the form

a  ∆4  n ∆xn  + p n xn +1 = 0, n = 0,1,2,....  pn  Examples are inserted to illustrate the results. Keywords: fifth order, oscillation and difference equation. AMS Subject Classification : 39A11.

1. INTRODUCTION

(i)

In this note, we consider the fifth order difference equation of the form

a  ∆4  n ∆xn  + p n xn +1 = 0, n ≥ n1  pn 

{p n }

and

{p n }

are real sequences and

p n > 0 for infinitely many values of n. n −1 p (ii) rn = ∑ s → ∞ as n → ∞ . s = n1 a s

(1)

where , ∆ is the forward difference operator defined by ∆x n = x n +1 − x n and {a n } and

{a n }

are defined as follows:

We recall that a nontrivial real sequence {x n }, n ≥ 0 is called a solution

of (1), if {x n } satisfies the equation (1) for all

n ≥ 0 . A solution

{x n } is said to be

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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

oscillatory, if for every n0 > 0 , there exists an n ≥ n0 such that x n x n +1 ≤ 0 and it is called non-oscillatory otherwise. Therefore an oscillatory solution is neither eventually positive nor eventually negative. In recent years, there has been an increasing interest in the study of oscillatory and asymptotic behavior of solutions of difference equations of the type (1) and the references cited therein. 2. MAIN RESULTS In this section, some sufficient conditions for the oscillation of all the solutions of (1) have been proved in the following theorems. Theorem: 1 Suppose that conditions (i) and (ii) hold and let 2   a s +1 (∆rs )  =∞ r p − ∑ s s 3  4rs p s +1 (s − n1 )  s = n2  for n2 ≥ n1 . ∞

(2)

Then every solution {x n } of equation (1) is oscillatory. Proof:

Suppose on the contrary that {x n } is a nonoscillatory solution of equation (1). Without loss of generality, we may assume that x n > 0 for n ≥ n1 . From (1), we have

a  ∆4  n ∆x n  = − p n x n +1 ≤ 0, for n ≥ n1 .  pn 

a  ∆ n ∆xn  ,  pn  a  a  ∆2  n ∆x n  , ∆3  n ∆xn  are  pn   pn 

Then

{x n } , {∆xn } ,

eventually non – increasing sequences.

a

3 We shall first show that ∆  n ∆xn  > 0 .  pn 

Suppose to the contrary that

a  ∆3  n ∆xn  ≤ 0, n ≥ n2 for n2 ≥ n1 .  pn   3 a Since ∆  n ∆x n  is non-increasing, there  pn  exists a non-negative

constant

k1 , and

a  n3 ≥ n2 such that ∆3  n ∆x n  ≤ −k1 , for  pn  n ≥ n3 and k1 > 0 . Summing the last inequality from n3 to n − 1 , we obtain n −1  n −1 3  as  ∆ ∆ x (− k1 ) ∑  p s  ≤ s∑ s = n3  s  = n3 That is

 an  a  ∆2  n ∆x n  − ∆2  3 ∆x n3  ≤ −k1 (n − n3 )  pn   pn   3  This implies

 an  a  ∆2  n ∆x n  ≤ ∆2  3 ∆x n3  − k1 (n − n3 )  pn   pn   3  Then

a  ∆2  n ∆xn  → −∞ as n → ∞ .  pn 

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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

Therefore there exists an integer

a

n4 ≥ n3

2 such that ∆  n ∆x n  ≤ − k 2 for n ≥ n 4  pn 

and k 2 > 0 . Summing the last inequality from n 4 to n − 1 , we obtain

 an a  ∆ n ∆x n  − ∆ 4 ∆x n4  pn  pn   4

  ≤ − k 2 (n − n 4 )  

Summing the last inequality from n6 to

n − 1 , from ∆x n ≤ − k 4 n −1

ps . s = n6 a s

Therefore x n → −∞ as n → ∞ , which is a contradiction to the fact that x n is positive.

This implies

 an  a  ∆ n ∆x n  ≤ ∆ 4 ∆x n4  − k 2 (n − n4 )  pn   pn   4 

Set

a  ∆ n ∆xn  → −∞ as n → ∞ .  pn  Thus there exists an integer n5 ≥ n 4 such  an  ∆xn  ≤ −k3 for n ≥ n5 and k3 > 0 .  pn 

that ∆

Summing the last inequality from n5 to

n − 1 , we obtain  an  a  ∆ n ∆x n  ≤ ∆ 5 ∆x n5  − k 3 (n − n5 )  pn   pn    5 This implies

Then there exists an integer n6 ≥ n5 such for n ≥ n6

vn =

a  ∆3  n ∆xn  > 0 .  pn 

 rn 3  a n ∆  ∆xn  > 0, n ≥ n1 xn +1  p n 

(3)

a   r  r a  ∆v n = ∆3  n +1 ∆x n +1 ∆ n  + n ∆4  n ∆x n   p n +1   x n +1  x n +1  p n 

From equation (1), we have

a  ∆4  n ∆xn  = − p n and using this in xn +1  p n 

1

last equation, we get a  x ∆r − r ∆x  ∆v n = ∆3  n +1 ∆x n +1  n +1 n n n +1  − rn p n x n +1 x n + 2  p n+1   a  ∆r  r ∆x   a  = ∆3  n+1 ∆xn +1  n −  n n+1 ∆3  n+1 ∆xn +1  − rn p n  p n +1  xn + 2  x n+1 x n+ 2   p n +1 

From equation (3), we get

 an   ∆xn  → −∞ as n → ∞ .   pn a  that  n ∆x n  ≤ −k 4  pn  k4 > 0

pn , we obtain an

xn ≤ x n6 − k 4 ∑

Hence

Then

655

and

 vn +1 1 3  a n +1 = ∆  ∆x n+1  rn +1 xn + 2  p n +1  and substituting this in the last equation, we obtain ∆v n = −rn p n +

 v n+1∆rn  rn ∆x n+1  3  a n+1 ∆  −  ∆x n+1  rn+1  x n+1 xn + 2   p n+1 

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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

By using the fact that {x n } is nonincreasing, we get ∆v n ≤ − rn p n +

Similarly,  a n +1  n −1  a a   ∆2  n +1 ∆x n +1  = ∆2  1 ∆x n1 +1  + ∑ ∆3  s +1 ∆x s +1     p n +1    p n1 +1  s = n1  p s +1

 v n +1 ∆rn  rn ∆x n +1  3  a n +1 ∆  −  2 ∆x n +1  rn +1   x n + 2   p n +1

(4) Consider a  a n +1 a n +1 a n +1 ∆x n +1 = 1 ∆x n1 +1 +  n +1 ∆x n +1 − 1 ∆x n1 +1    p n +1 p n1 +1 p n1 +1  p n +1 

=

a n1 +1 p n1 +1

n −1 a  ∆x n1 +1 + ∑ ∆ s +1 ∆x s +1  s = n1  p s +1 

a  ≥ (n − n1 )∆ n +1 ∆xn +1 , n ≥ n1 + 1  p n +1 

a  ≥ (n − n1 )∆3  n+1 ∆xn +1   p n +1  This implies

 an +1 3 3 a ∆xn +1 ≥ (n − n1 ) ∆  n +1 ∆xn +1  p n +1  p n+1  which implies

∆xn+1 ≥ p n +1 (n − n )3 ∆3  a n +1 ∆x  1 n +1  p a n +1  n +1 

and  an +1  n−1  a a   ∆ n+1 ∆x n+1  = ∆ 1 ∆xn1 +1  + ∑ ∆2  s +1 ∆xs +1    p p p s = n  n+1   s +1  1  n1 +1 

 an +1  ≥ (n − n1 )∆2  1 ∆xn1 +1   p n +1   1 

That is,

− ∆xn+1 ≤ −

  p n +1 (n − n1 )3 ∆3  an+1 ∆xn+1  a n +1  p n +1  (5)

Substituting (5) in (4), we get 3  v ∆r  r (n − n ) p n +1   3  a n +1  ∆   ∆v n ≤ −rn p n + n +1 n −  n 2 1 ∆ x n + 1  p rn +1 x a  n + 2 n +1     n +1

Since from (3),

Consequently,

 vn +1 1 3  a n +1 = ∆  ∆xn +1  rn+1 xn + 2  p n +1 

2

a

3 we have, ∆  n+1 ∆x n+1  =  p n +1 

vn +1 xn+ 2 rn +1

3 v n +1∆rn  rn (n − n1 ) p n +1  v n2+1   2 ∆v n ≤ −rn p n + − r rn +1 a n +1   n +1

Journal of Computer and Mathematical Sciences Vol. 3, Issue 6, 31 December, 2012 Pages (557-663)

(6)


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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

Consider

∆v n ≤ − rn p n + +

3 v n +1 ∆rn  rn (n − n1 ) p n +1  v n2+1  −   r2 rn +1 a n + 1   n +1

(∆rn )2 a n +1 3 4rn p n +1 (n − n1 )

(∆rn )2 a n +1 3 4rn p n +1 (n − n1 )

(adding and subtracting last two terms) That is

∆v n ≤ −rn p n +

(∆rn )2 an+1 3 4rn p n+1 (n − n1 )

 (n − n1 ) n − n1 − rn +1 

∆rn rn p n +1 v n +1 − a n+1 2(n − n1 ) n − n1

a n +1   rn p n +1 

2

This implies that

(∆rn )2 an+1 ∆vn < −rn p n + 3 4rn p n+1 (n − n1 ) That is

 (∆rn )2 a n+1   ∆v n < − rn p n − 3 4rn p n +1 (n − n1 )  

Summing the last inequality from n2 to n − 1 , we have n −1  (∆rs )2 a s +1  v n ≤ v n 2 − ∑  rs p s − 3 4rs p s +1 (s − n1 )  s = n2 

In view of (2), we see that

vn → −∞ as n → ∞ which is a contradiction to the fact that v n is positive and hence the proof. Next, we introduce a double sequence oscillation criteria of equation (1). Consider a double sequence such that

{H (m, n )}

by means of which we will study the

{H (m, n ) / m ≥ n ≥ 0}

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658 (i) (ii)

B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

H (m, m ) = 0 for m ≥ 0 H (m, n ) > 0 for m > n ≥ 0

∆ 2 H (m, n) = −h(m, n) H (m, n) for m > n ≥ 0 where ∆ 2 H (m, n ) = H (m, n + 1) − H (m, n ) ≤ 0 for m ≥ n ≥ 0 (iii)

Theorem: 2 Suppose that conditions (i) and (ii) hold. If there exists a double sequence H (m, n ) such that

lim sup n→∞

1 × H (n, n2 )

2     as +1rs2+1 ∆rs ∑ H (n, s )rs ps − 4r p (s − n )3  h(n, s ) − r H (n, s )   = ∞, s =n2  s +1    s s +1 1  for n2 ≥ n1 . Then every solution {x n } of equation (1) is oscillatory. n−1

(7)

Proof: Proceeding as in the proof of theorem 1 with the assumption that equation (1) has a non-oscillatory solution, say x n > 0 for n ≥ n1 and by taking v n as in equation (3), we get

vn > 0 and equation (6) holds and so we obtain from the equation (6), for n ≥ n2 , rn pn ≤ −∆vn +

3 vn +1∆rn  rn (n − n1 ) p n+1  vn2+1  2 −  r rn+1 a n +1   n+1

This implies n −1

n −1

n −1

s = n2

s = n2

s = n2

∑ H (n, s )rs p s ≤ − ∑ H (n, s )∆v s + ∑ H (n, s )

v s +1 ∆rs rs +1

 rs (s − n1 )3 p s +1  v s2+1  − ∑ H (n, s )  r2 a s = n2 s +1   s +1 n −1

Using summing by parts, we get n −1

∑ H (n, s )r p

s = n2

s

s

n −1   n ≤ −  (H (n, s )v s )n2 − ∑ ∆ 2 H (n, s )∆v s +1  s = n2   n −1 n −1  r (s − n1 )3 p s +1  v s2+1 v ∆r  + ∑ H (n, s ) s +1 s − ∑ H (n, s ) s   r2 rs +1 a s = n2 s = n2 s + 1   s +1

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659

B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

That is, by using the fact that H (n, n ) = 0 , we obtain n −1

∑ H (n, s )rs p s ≤ H (n, n 2 )v n2 +

s = n2

n −1

∑ ∆ H (n, s )∆v 2

s = n2

s +1

 rs (s − n1 )3 p s +1  v s2+1 v s +1 ∆rs n −1  + ∑ H (n, s ) − ∑ H (n, s )   r2 r a s = n2 s = n2 s +1 s +1   s +1 n −1

= H (n, n 2 )v n2 + −

 ∆r ∑  ∆ H (n, s ) + H (n, s ) r

 v s +1 s +1 

n −1

s

s = n2

2

 rs (s − n1 )3 p s +1  v s2+1   ( ) H n s , ∑   r2 a s +1 s = n2   s +1 n −1

In view of ∆ 2 H (n, s ) = −h(n, s ) H (n, s ) , the last equation becomes n −1

s = n2

n −1

∑ H (n, s )rs p s ≤ H (n, n2 )v n2 +

s = n2

∑  − h(n, s )

H (n, s ) + H (n, s )

∆rs  v s +1 rs +1 

 rs (s − n1 )3 p s +1  v s2+1  − ∑ H (n, s )  r2 a s = n2 s +1   s +1 n −1  ∆r  = H (n, n 2 )v n2 − ∑  h(n, s ) H (n, s ) − H (n, s ) s v s +1 rs +1  s = n2  n −1  r (s − n1 )3 p s +1  v s2+1  − ∑ H (n, s ) s  r2 a s = n2 s + 1   s +1 n −1

Consider n −1

∑ H (n , s )r

s = n2

s

p s ≤ H (n , n 2 )v n 2 − −

s = n2

∑ H (n , s )

1 + 4 1 4

n −1

a s +1 rs2+1

∑ r (s − n )

s = n2

3

s

n −1

1

a

s

1

∆ rs rs + 1

 v s +1 

 v s2+1  2 r  s +1

  h (n , s ) − p s +1 

∆r H (n , s ) s  

2

rs + 1 

 ∆r  h (n , s ) H (n , s ) − H (n , s ) s rs + 1 H (n , s ) p s +1 

2 s +1 s +1 3

∑ r (s − n )

s = n2

H (n , s ) − H (n , s )

 rs (s − n 1 )3 p s +1 a s +1 

n −1

s = n2

n −1

∑  h (n , s )

r

That is, Journal of Computer and Mathematical Sciences Vol. 3, Issue 6, 31 December, 2012 Pages (557-663)

  

2


660

B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012) n−1

∑ H (n, s)r p s

s = n2

s

≤ H (n, n2 )vn2 +

as+1rs2+1  ∆r  1 n−1  h(n, s ) − H (n, s ) s  ∑ 3 4 s=n2 rs (s − n1 ) ps+1  rs+1 

2

 (s − n1 ) (s − n1 ) H (n, s ) ps+1rs  vs+1   n−1 rs+1 as+1   − ∑   s =n2 ∆r  rs+1 as+1 +  h(n, s ) H (n, s ) − H (n, s ) s   2(s − n1 ) (s − n1 ) H (n, s ) ps+1rs  rs+1  

2

which implies n −1

∑ H (n, s )r p s

s = n2

s

< H (n, n 2 )v n2  ∆r  a s +1 rs2+1 1 n −1  h(n, s ) − H (n, s ) s  + ∑ 3 4 s = n2 rs (s − n1 ) p s +1  rs +1 

2

Therefore, 2 n −1   a s +1rs2+1 ∆rs   1 ∑  H (n, s )rs ps − r (s − n )3 p  h(n, s ) − H (n, s ) r   < vn2 H (n, n2 ) s = n2  s +1   s s +1  1 

which contradicts the equation (7). Hence the theorem is proved. 3. Examples: 1. Consider the difference equation

∆4 ((n + 3)∆ x n ) + (n + 7 ) x n +1 = 0 . 2

Here a n = (n + 3 )(n + 7 )

2

(8)

and p n = (n + 7 )

2

n −1 p s n−1 (s + 7 )2 1 =∑ = → ∞ as ∑ 2 s = n1 a s s = n1 (s + 3)(s + 7 ) s = n1 s + 3 n −1

Then,

rn = ∑

And

2   a s +1 (∆rs )  =∞ r p − ∑ s s 3  4rs p s +1 (s − n1 )  s = n2 

n→∞

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B. Selvaraj, et al., J. Comp. & Math. Sci. Vol.3 (6), 653-663 (2012)

Hence all conditions of theorem 1 are satisfied. equation (8) are oscillatory.

Therefore, all solutions of

2. Consider the difference equation

∆4 (n∆xn ) +

1 xn+1 = 0 . n3

(9) and H (n, s ) = n − s.

Here a n =

1 1 , pn = 3 2 n n

Then

1 p s n −1 s 3 n −1 1 rn = ∑ =∑ = → ∞ as 1 2 s∑ s = n1 a s s = n1 = n1 s s n −1

n→∞

And n −1   a s +1 rs2+1 ∆r 1  h(n, s ) − s lim sup × ∑  H (n, s )rs p s − 3 n→∞ H (n, n2 ) s = n2  rs +1 4rs p s +1 (s − n1 )  

 H (n, s )  

2

  = ∞. 

Hence all conditions of theorem 2 are satisfied. Therefore, all solutions of equation (9) are oscillatory. REFERENCES 1. R.P. Agarwal-‘ Difference Equation and Inequalities’. Theory, Methods and Applications, 2nd edition. 2. R.P. Agarwal, Martin Bohner, Said R. Grace, Donal O’Regan. ‘Discrete Oscillation Theory’- CMIA Book Series, Volume 1, ISBN:977-5945-19-4. 3. E. Thandapani and B. Selvaraj. ‘Existence and Asymptotic Behavior of Non Oscillatory Solutions of Certain Nonlinear Difference Equations’ – Far East Journal of Mathematical Sciences (FJMS), 14 (1), 9-25 (2004). 4. E. Thandapani and B. Selvaraj. ‘Oscillatory Behavior of Solutions of Three Dimensional Delay Difference Systems’-Radovi MateMaticki, Vol. 13, 39–52 (2004).

5. E. Thandapani and B. Selvaraj. ‘Oscillatory and Non Oscillatory Behavior of Fourth Order Quasilinear Difference Equations’-Fast East Journal of Mathematical Sciences (FJMS), 17 (3), 287–307(2004). 6. E. Thandapani and B. Selvaraj. ‘Behavior of Oscillatory and Non Oscillatory Solutions of Certain Fourth Order Quasilinear Difference Equations’. The Mathematics Education, Vol. XXXIX (4), 214–232 (2005). 7. E. Thandapani and B. Selvaraj. ‘Oscillation of Fourth Order Quasilinear Difference Equations’-Fasciculi Mathematici, Nr. 37, 109–119 (2007). 8. B. Selvaraj and J. Daphy Louis Lovenia‘Oscillation Behavior of Fourth Order Neutral Difference Equations with Variable Coefficients’-Far East Journal

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16. B. Selvaraj and I. Mohammed Ali Jaffer. ‘Oscillation Theorems of Solutions for Certain Third Order Functional Difference Equations with Delay. Bulletin of Pure and Applied Sciences, Volume 29E, Issue 2, P. 207–216 (2010). 17. B. Selvaraj and I.Mohammed Ali Jaffer. ‘Solutions on the Oscillation to of Third Order Difference Equations’ J. Comp. & Maths. Sci. Vol. 1(7), 873– 876 (2010). 18. B.Selvaraj and G.Gomathi Jawahar. ‘Certain Oscillation Criteria for Second Order Delay Difference Equations‘ Advances in Theoretical and Applied Mathematics, Volume 6, Number 2, pp. 147–151 (2011). 19. B.Selvaraj and I.Mohammed Ali Jaffer. ‘Oscillation Behavior of Certain Fourth order Linear and Nonlinear Equations’ Advances in Theoretical and Applied Mathematics, Volume 6, Number 2, pp. 203–211 (2011). 20. B. Selvaraj and I. Mohammed Ali Jaffer. ‘Oscillation Theorems of Solutions for Certain Third Order Functional Difference Equations with Delay’ Bulletin of Pure and Applied Sciences, Volume 29E, Issue 2, P. 207–216 (2010). 21. B. Selvaraj and I. Mohammed Ali Jaffer. ‘Oscillation Behavior of Certain fourth order Linear and Nonlinear Difference Equations’ Advances in Theoretical and Applied Mathematics, Volume 6, Number 2, pp. 191–201 (2011). 22. B. Selvaraj and G. Gomathi Jawahar. ‘New Oscillation Criteria for First Order Neutral Delay Difference Equations’ Bulletin of Pure and Applied Sciences,

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