J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
Asymptotic Behavior of 4th order Non-linear Delay Difference Equation DR. B. SELVARAJ and G.GOMATHI JAWAHAR Department of Mathematics, Karunya University, Karunya Nagar, Coimbatore-641114, Tamil Nadu, India E mail: professorselvaraj@rediffmail.com jawahargomathi@yahoo.com
ABSTRACT In this paper some sufficient conditions were obtained for the existence of a bounded oscillatory and non oscillatory solution of the non linear difference equation. ∆4 x n - f (n, x n , x n − k ) = 0, n ∈ N ( n 0 ). Key words:- Oscillatory solution, nonoscillatory solution, nonlinear Delay Difference Equation. INTRODUCTION The importance of the study of oscillation and non oscillation of difference equations has been stressed by many Authors. See, for example1-14 and the references cited therein. Consider the nonlinear difference Equation, ∆4 x n - f (n, x n , x n − k ) = 0 (1.1) Where k is an integer and n ∈ N ( n 0 ). ∆ is the forward difference operator, defined by ∆ x n = x n +1 - x n . By a solution of equation (1.1), we mean a sequence {x n } which is defined for n ≥ n0 − k and which satisfies (1.1) for n ∈ N ( n 0 ). Where N( n 0 ) =
{
n 0 , n 0 + 1 , ……
A solution
{x n }
}.
is said to be
oscillatory if the sequence {x n } are not eventually +ve (or) not eventually -ve.
Otherwise {x n } is called non oscillatory. Let us assume the following conditions. H1: f(n, c1, c2) : N( n 0 ) X R2 → R+ is real valued and continuous as a function of c1, c2 ∈ R. H2: for c ≠ 0, c f(n, c1, c2) > 0, n ∈ N ( n 0 ). Let us assume that equation (1.1) has a unique solution. {x n } Satisfies the condition
xj =
b j , for j = n 0 − k , n 0 − k + 1 ,
………. n 0 , n 0 + 1 , n 0 + 2 . Where n 0 ≥ 0. MAIN RESULTS We use the following Schauder’s fixed point theorem. Let A be a closed convex subset of a banach space and assume there exists a continuous map T sending A to a countably compact subset T(A) of A. Then T has fixed points. By using Schauder’s fixed point
Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
878
theorem we will first establish the existence of a bounded non oscillatory solution for equation (1.1). Banach space is a complete norm linear space. The space l ∞ is the set of all real sequences defined on N ( n 0 ) in which every individual sequence is bounded with respect to the usual supremum norm. There fore we treat l ∞ as a Banach space under the supremum norm. A subset A of a Banach space B is called relatively compact if every sequence in A has a subsequence converging to an element belonging to B. Also, if for every ∈ > 0, we are able to find an integer N ≥ n 0 such at x m − x n < ∈
whenever m, n > N for any sequence {x n } in A. Then we say that a set A of sequences in l ∞ is uniformly Cauchy. Based on the above we use the following lemma, whose proof can be found in2. Lemma 1: A bounded uniformly Cauchy subset S of l ∞ is relatively compact.
if there exists a constant c ≠ 0 such that ∞
∑
f ( r , c, c
<∞.
r = n0
Proof: Assume that there exists a constant c > 0. Therefore in view of H1, H2 and (1.2) we ∞
have
∑r
3
f (r , c, c) ≤ c for sufficiently
r = n1
large n1 ∈ N ( n 0 ). Let us consider the equation,
xn =
c + 6
(r − n + 3)(r − n + 2)(r − n + 1) f (r , xr , xr −k ) 6 r = n −1 ∞
∑
(1.3) for x j = b j , j = n1-k-1, n1-k , … n1-1
x n = ∆3 ( x n +1 − x n ) = ∆2 ( x n + 2 − 2 x n +1 + x n ) = ∆ ( x n +3 − 3 x n + 2 + 3 x n +1 − x n ) ∆4 x n = x n + 4 − 4 x n + 3 + 6 x n + 2 − 4 x n +1 + x n . Now ∆4
Theorem 1: Equation (1.1) has a bounded non oscillatory solution {x n } satisfying
lim xn = n→∞
c 6
,
Substituting (1.3) in (1.1) ∆4 x n - f ( n, x n , x n − k ) = -4
c 6
c 6
(r − n − 1)( r − n − 2)( r − n − 3) f (r , x r , x r −k ) 6 r = n+3 ∞
+
∑
( r − n)( r − n − 1)( r − n − 2) f (r , x r , x r −k ) 6 r =n+ 2 ∞
+
(1.2)
∑
Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
+6
-4
+
∞ c ( r − n + 1)( r − n)(r − n − 1) + ∑ f (r , x r , x r −k ) 6 6 r =n +1 c ∞ (r − n + 2)(r − n + 1)(r − n) + ∑ f (r , x r , x r − k ) 6 6 r =n c ∞ (r − n + 3)(r − n + 2)(r − n + 1) + ∑ f (r , x r , x r −k ) 6 6 r =n −1
879
f ( n, x n , x n − k )
3 .2 .1 f ( n + 4, x n + 4 , x n + 4 − k ) 3 .2 4 .3 .2 3 .2 .1 - 4 f (n + 3, x n +3 , x n +3− k ) + f ( n + 4, x n + 4 , x n + 4− k ) 6 3 .2 3 . 2 . 1 4 . 3 . 2 5.4.3. +6 f ( n + 2, x n + 2 , x n + 2 − k ) + f (n + 3, x n +3 , x n +3− k ) + f (n + 4, x n + 4 , x n + 4 − k ) 6 6 6 3 . 2 . 1 4 . 3 . 2 5 . 4 . 3 . 6 . 5 . 4 - 4 f (n +1, xn+1, xn+1−k ) + f (n + 2, xn+2 , xn+2−k ) + f (n +3, xn+3, xn+3−k ) + f (n + 4, xn+4, xn+4−k ) 6 6 6 6 ∆4 x n - f ( n, x n , x n − k ) =
+ 3.2.1 f (n, xn, xn−k )+ 4.3.2 f (n+1, xn+1, xn+1−k ) +5.4.3. f (n+2, xn+2, xn+2−k ) +6.5.4 f (n+3, xn+3, xn+3−k ) +7.6.5 f (n+4, xn+4, xn+4−k ) 6 6 6 6 6 - f ( n, x n , x n − k ). There fore ∆4 x n - f ( n, x n , x n − k ) = 0
Hence {xn } is a solution of (1.1) By using Schauder’s fixed point theorem we will prove that (1.3) has a bounded non oscillatory solution. Define, ∞
c ∈ l ∞ / ≤ xn ≤ c } 1 6
X = { x = {xn }n = n
It is evident that, X is bounded, convex and closed subset of l ∞. Define F: X → X as follows:-
( Fx) n =
c 6
( r − n + 3)( r − n + 2)( r − n + 1) f (r , x r , x r −k ) 6 r = n −1 ∞
+
∑
We shall show that F maps X into X, F is continuous and F is relatively compact. Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
880 i)
c c ≤ 6 6
Since
c ≤ ( Fx) n 6
ii)
( r − n + 3)( r − n + 2)( r − n + 1) f ( r , x r , x r − k ) ≤ c , we have 6 r = n −1 ∞
+
∑
≤c, Hence F maps X into X.
From 1.2, given ∈> 0 we can find n2 ∈ N ( n 0 ) satisfying,
1 6
∞
∑ (r + 1)(r + 2)(r + 3) f (r , c, c) <∈ .
r = n2
{ }
Choose a sequence x i
∞
i =1
belonging to X such that lim x i − x = 0 since X is closed n →∞
x ∈ X. Also from (H1), (H2), we get for n-1 ≥ n2,
( Fxi ) n − ( Fx) n 1 6 1 ≤ 6 ≤
≤
( r − n + 3)( r − n + 2 )( r − n + 1) f ( r , x r , x ri − k ) − f ( r , x r , x r − k ) 6 r = n −1 ∞
∑
∞
∑ (r + 1)(r + 2)(r + 3) f (r , x r =n
r
, x ri − k ) − f ( r , x r , x r − k )
∞
∑ (r + 1)(r + 2)(r + 3) f (r , c, c) <∈
r = n2
⇒ lim ( Fx i ) n − ( Fx ) n = 0 i →∞
iii) In order to prove FX is relatively compact, it is enough to prove FX is a uniformly Cauchy subset of l ∞ . from (1.2) for any ∈> 0 , ∃ n3 ∈ N ( n 0 ) ∋
1 6
∞
∑ (r + 1)(r + 2)(r + 3) f (r , c, c) <∈
k = n3
Let x ∈ X, n, m ∈ N ( n 3 ), n > m
( r − n + 3)( r − n + 2)( r − n + 1) f (r , x r , x r − k ) 6 r = n −1 ∞ ( r − m + 3)( r − m + 2)( r − m + 1) - ∑ f (r , x r , x r − k ) 6 r = m −1 n−2 ( r − m + 3)( r − m + 2)( r − m + 1) = ∑ f (r , x r , x r −k ) 6 r = m −1
( Fx) n − ( Fx) m ≤
∞
∑
Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
(1.4)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
881
1 ∞ ∑ (r + 3)(r + 2)(r + 1) f (r , c, c) <∈ from (1.4). 6 r =m ⇒ FX is uniformly Cauchy subset of l ∞.
≤
Since FX is a bounded uniformly Cauchy subset of l ∞ , by lemma (1), FX is relatively compact. Hence by Schauder’s fixed point theorem, there exists x ∈ X such that x = Fx . Thus {x n } is a solution for (1.3). Therefore, it is a bounded non oscillatory solution of the
xn = n→∞
equation (1.1) with the condition lim
c 6
This completed the proof of the theorem. Example: Consider the difference equation, ∆4 x n -
24(n − 1) 4 ( x n −1 ) 4 = 0, n 5 (n + 1)(n + 2)(n + 3)(n + 4)
n ∈ N ( n 0 ).
(1.5)
All the conditions of theorem (1) are satisfied with respect to the equation (1.5). Hence every solution of (1.5) is non oscillatory. One such solution is
{x n } = 1 + 1 .
n
Theorem 2: Every bounded solution {x n } of (1.1) is either oscillatory or {x n } , {∆x n } , ∆2 x n ,
{
∞
{∆ x } → 0 as n → ∞, if s = n
s3 ∑ o
3
n
f ( s , c, c ) = ∞
}
(1.6)
for all c ≠ 0. Proof: Suppose there exists bounded non oscillatory solution {x n } . Let x n , x n −k > 0 . n −1
Summing(1.1) from m to n-1, we have n −1 s=m
n −1
n −1
∆ 4 xs − S = m ∑ f ( s , xs , xs − k ) = 0 S =m ∑
∑ ∆3 ( xs +1 − xs ) − S =m ∑ f (s, xs , xs −k ) = 0 Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
882
n −1
∆3 x m +1 − ∆3 x m + ∆3 x m + 2 − ∆3 x m +1 + ∆3 x m + 3 − ∆3 x m + 2 + ... + ∆3 x n − ∆3 x n −1 − ∑ f ( s, x s , x s − k ) = 0 s=m
n −1
∆ x n - ∆ x m − S =m 3
3
∑ f ( s, x , x s
s −k
)=0
(1.7)
From (1.1) & H2, ∆4 x n > 0, for n ∈ N ( n1 ). Therefore ∆3 x n is strictly increasing. Now, there are 2 possible cases. ∆3 x n > 0, ∆3 x n < 0, n ∈ N ( n1 ). Suppose i) ∆3 x n > 0, n ∈ N ( n1 ). There are 2 possible cases. n ∈ N ( n 2 ) n2 ≥ n1 . ∆2 x n > 0, ∆2 x n < 0, Suppose ∆2 x n > 0,
⇒ ∆ x n < 0, ∆ x n > 0, n ∈ N ( n3 ) n3 ≥ n1.
If ∆ x n > 0, for n ∈ N ( n3 ) n3 ≥ n1. then ∆ x n ≥ ∆ x n3 > 0 ⇒ {x n } is unbounded. Which is a contradiction to the hypothesis. If ∆ x n < 0, for n ∈ N ( n 2 ), Then ∆ x n ≤ ∆ x n3 < 0 ,
That is {x n } is bounded and decreasing. Therefore lim x n = c1 ≥ 0, n →∞
2
3
⇒ lim ∆ x n = 0, ⇒ lim ∆ x n = 0, ⇒ lim ∆ x n = 0, n →∞
n →∞
n →∞
To prove c1 = 0: If c1> 0, then x n > c1, x n − k > c1, for n ≥ n3. similarly we can proceed for ∆2 x n < 0. from (1.7) ∆3 x n ≥
∞ S =n
∑ f (s, xs , xs −k )
ie ∆3 x n ≥ f (n, x n , x n − k ) + f ( n + 1, x n +1 , x n +1− k ) + f (n + 2, xn+ 2 , xn+ 2−k ) + f (n + 3, xn+3 , xn +3− k ) + ........... =
(1.8)
∞ ∞ − s = n ∑ ( s − n) f ( s, xs , x s − k )− s =n −1 ∑ ( s − n + 1) f ( s, x s , xs −k )
Therefore
∞
∆3 x n ≥ − s = n
∑ ( s − n) f ( s , x , x s
s−k
∞ )− s =n −1 ∑ ( s − n + 1) f ( s, x s , xs −k )
Summing the L.H.S. of the inequality (1.9) from n to ∞, ∞ n
∑∆ x
∞
3
n
= n ∑ ∆2 ( x n +1 − x n )
Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
(1.9)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
= ∆2 x n +1 − ∆2 x n + ∆2 x n + 2 − ∆2 x n +1 + ∆2 x n + 3 − ∆2 x n + 2 + ........
883 (2.0)
Summing the R.H.S of the inequality (1.9) from n to ∞, ∞ s=n
=
∑ − ∑ ( s − n) f ( s, x , x ∞
s=n
f (n, x n , x n − k ) +
∞ )− s =n−1 ∑ ( s − n + 1) f ( s, xs , xs −k ) f (n + 1, x n +1 , x n +1− k ) + f ( n + 2, xn + 2 , xn + 2− k ) + f ( n + 3, xn + 3 , xn + 3− k ) + ........... s
s −k
+ f (n + 1, x n +1 , x n +1− k ) + f (n + 2, x n + 2 , x n + 2 − k ) + f (n + 3, x n + 3 , x n + 3− k ) + .........
+ f (n + 3, xn+3 , xn +3−k ) + ......... = f (n, x n , x n − k ) + 2 f (n + 1, x n +1 , x n +1− k ) + 3 f (n + 2, x n + 2 , x n + 2 − k )
+ 4 f (n + 3, x n + 3 , x n +3− k ) + ........... =
∞ s = n −1
∑ ( s − n + 1) f ( s, x , x s
s −k
(2.1)
)
(2.2)
In view of the equations (2.0) and (2.2), the inequality (1.9) becomes, ∞
- ∆2 x n ≥
S = n −1
∑ (s − n + 1) f ( s, x , x s
s −k
)
= f (n, x n , x n − k ) + 2 f (n + 1, x n +1 , x n +1− k ) + 3 f (n + 2, x n + 2 , x n + 2− k ) + .......... ∞ ∞ ( s − n + 1)( s − n) ( s − n + 2)( s − n + 1) = − ∑ f ( s, x s , x s − k ) − f ( s , x , x ) ∑ s s −k 2 2 s = n −1 s = n
Therefore
(− ∆2 x n ) ≥ ∞ ∞ ( s − n + 1)( s − n) ( s − n + 2)( s − n + 1) f ( s, x s , x s − k ) − f ( s , x , x ) − ∑ ∑ s s −k 2 2 s = n −1 s = n
(2.3) Summing the L.H.S. of inequality (2.3) from n to ∞, ∞
∑ (−∆ x = ∑ ∆( x
∞
2
n
n
∞
n
=
∞ n
∑ (∆x
) = n ∑ ∆ ( − ∆x n )
n
−x n +1 )
n
− ∆x n +1 )
Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
884
= ∆ x n − ∆ x n + 1 + ∆ x n + 1 − ∆ x n + 2 + ∆ x n + 2 − ∆ x n + 3 + ∆ x n + 3 − ∆ x n + 4 + .......... (2.4) Summing the R.H.S. of inequality (2.3) from n to ∞,
∞ ( s − n + 1)( s − n) f ( s, xs , xs − k ) + ∑ − s =n ∑ 2 =
∞ n
∑ [ f (n, x , x n
n −k
( s − n + 2)( s − n + 1) f s x x ( , , ) ∑ s s−k 2 s = n +1 ∞
) + 2 f (n + 1, xn+1 , xn+1−k ) + 3 f (n + 2, xn+ 2 , xn+ 2−k ) + ................]
= f (n, x n , x n − k ) + 3 f (n + 1, x n +1 , x n +1− k ) + 6 f (n + 2, x n + 2 , x n + 2 − k ) + ................ ∞
= s = n −1
∑
( s − n + 2)( s − n + 1) f ( s, x s , x s − k ) 2
(2.5)
In view of the equations (2.4) & (2.5), the inequality (2.3) becomes, ( ∆x n ) ≥
∞ s = n −1
∑
( s − n + 2)( s − n + 1) f ( s, x s , xs − k ) 2
(2.6)
Summing the L.H.S of the inequality (2.6) from n to ∞, ∞ n
∑ ∆x
∞
= n ∑ x n +1 − x n = x n +1 − x n + x n + 2 − x n +1 + x n + 3 − x n + 2 + ...........
n
(2.7) Summing the R.H.S of the inequality (2.6) from n to ∞, ∞
∞ ( s − n + 2)( s − n + 1) f ( s, x s , x s − k ) ∑ s =n −1 ∑ 2 n
=
∞ n
∑ [ f (n, x
n
, x n − k ) + 3 f (n + 1, x n +1 , x n +1− k ) + ...............]
= f (n, x n , x n − k ) + 4 f (n + 1, x n +1 , x n +1− k ) + 10 f (n + 2, x n + 2 , x n + 2 − k ) + ................ ∞
= s = n −1
∑
( s − n + 3)( s − n + 2)( s − n + 1) f ( s, x s , x s − k ) 6
In view of the equations (2.7) and (2.8), the inequality (2.6) becomes, Journal of Computer and Mathematical Sciences Vol. 1, Issue 7, 31 December, 2010 Pages (769-924)
(2.8)
B. Selvaraj et al., J. Comp. & Math. Sci. Vol. 1(7), 877-886 (2010)
−x n >
∞ s = n −1
That is
∞ s = n −1
Therefore Hence
∑
( s − n + 3)( s − n + 2)( s − n + 1) f ( s, x s , x s − k ) 6
∑ (s − n + 3)(s − n + 2)(s − n + 1) f (s, x , x s
∞ s = n −1 ∞
s = n −1
885
∑s
∑s
3
3
s −k
)< ∞
f ( s, x s , x s − k ) < ∞
f ( s, c1 , c1 ) < ∞
Which is a contradiction to (1.6). Therefore c1 = 0. The proof for ∆3 xn < 0 , is similar and hence omitted. Example: Consider the difference equation
∆4 x n − 16( x n − 2 ) 3 = 0 All the conditions of theorem (2) are satisfied with respect to the equation (2.9). Hence every solution of (2.9) is oscillatory. One such solution is {x n } = (−1) n .
5.
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