Selçuk J. Appl. Math. Selçuk Journal of Vol. 9. No.1. pp. 23-29 , 2008 Applied Mathematics
Oscillation of Higher-Order of System of Difference Equations With Continuous Arguments
Özkan Öcalan and Umut Mutlu Özkan
Department of Mathematics, Faculty of Science and Arts, Kocatepe University, Afyonkarahisar-TURKEY
e-mail:ozkan@ aku.edu.tr,umut_ ozkan@ aku.edu.tr
Received : October 15, 2007
Summary. This paper is concerned with the oscillatory behavior of all so-lutions of the higher-order of system of difference equations with continuous arguments ∇() + P =1 ( − ) = 0
where ∈ R× and ∈ R for = 1 2 and ∇ is a backward
difference operator, defined by ∇() = () − ( − ).
Key words:Oscillation, difference equations, characteristic equation, logarith-mic norm.
2000 Mathematical Subject Classification: 39A10 1. Introduction
There has been a lot of activity concerning the oscillation of difference equations with continuous arguments. See, for example, [2, 3, 5, 6, 7, 9, 10].
In [4], Ladas investigated the oscillatory behavior of the following difference equation
∆+ −= 0 = 1 2
where ∈ R and ∈ Z. In [1], Chuanxi et al. obtained necessary and sufficient conditions for the oscillation of all solutions of linear autonomous system of difference equations
where ∈ R× and ∈ Z and sufficient conditions for the oscillation of all solutions of the difference equation
∆+
P
=1
− = 0 = 1 2
where ∈ R× and ∈ Z for = 1 2
In [8], Öcalan and Akin provided necessary and sufficient conditions for the oscillation of all solutions of system of difference equations
∆+ −= 0 = 1 2
where ∈ R× and ∈ Z. Furthermore, they obtained sufficient conditions
for the oscillation of all solutions of the difference equation ∆+
P
=1
−= 0 = 1 2
where ∈ R× and ∈ Z for = 1 2
Recently, in [6] Meng et al. have obtained sufficient conditions for the oscillation of all solutions of the system of difference equations with continuous arguments
() − ( − ) +
X
=1
( − ) = 0
where ∈ R× and ∈ R+ for = 1 2 . Also, they have established
necessary and sufficient conditions for the oscillation of all solutions of system () − ( − ) + ( − ) = 0
in terms of the eigenvalues of coefficient matrices .
For an × matrix the logarithmic norm of is denoted ( ) and is defined to be
( ) = max
kk=1( )
where ( ) is an inner product in Rand k k= ( )1 2.
In this paper we establish sufficient conditions for the oscillation of all solutions of the higher-order of system of difference equations with continuous arguments
(1.1) ∇() +P
=1
( − ) = 0
where ∈ R× and ∈ R for = 1 2 and is a positive integer.
Moreover, we obtain necessary and sufficient conditions for the oscillation of all solutions of the system of difference equations with continuous arguments
where ∈ R× and ∈ R and is a positive integer. Our result improves the known results in the literature. 2. Explicit conditions for oscillation
Lemma 1.Assume that ∈ R× ( = 1 2 ) The following statements
are equvalent.
() Every solution of eq. (11) oscillates (componentwise). () The characteristic equation of (11) has no real roots.
Proof. The proof can easily be made by using Laplace transform as in [6, Theorem 1] and is omitted it here.
Theorem 1. Assume that ∈ R×, is an odd positive integer and 0.
Then every solution of eq. (12) oscillates (componentwise) if and only if has no eigenvalues in the interval
µ
−∞³( )( − )−
´1 ¸
. Proof. The characteristic equation of eq. (12) is
(2.1) deth¡1 − −¢ + −i= 0, which can be also written as
deth¡−− 1¢ − i= 0. Set
() = ¡−− 1¢. Note that is a contiuous function on (−∞ ∞) and
max ∈(−∞∞)() = µ ( ) ( − ) − ¶1 , lim →∞() = −∞.
Thus the image of (−∞ ∞) under is µ
−∞³( )( − )−
´1 ¸
. There-fore eq. (21) has no real roots if and only if has no eigenvalues in
µ
−∞³( )( − )−
´1 ¸
. So, the proof is complete.
Theorem 2. Assume that ∈ R×, is an even positive integer and
0 . Then every solution of eq. (12) oscillates (componentwise) if and only if has no eigenvalues in the interval∙³( )( − )−
´1
∞ ¶
. Proof. The characteristic equation of eq. (12) is (21). So,
min ∈(−∞∞)() = µ( ) ( − ) − ¶1 , lim →∞() = ∞,
the image of (−∞ ∞) under is∙³( )( − )− ´1 ∞ ¶ . Therefore eq. (21) has no real roots if and only if has no eigenvalues in∙³( )( − )
−´ 1 ∞ ¶ . So, the proof is complete.
Now, we obtain sufficient conditions for the oscillation of eq. (11). The con-ditions will be given in terms of the ∈ R and ∈ R× matrices for
= 1 2 . We need the following lemma.
Lemma 2. Assume that is an odd positive integer, ∈ R×and 0
for = 1 2 . Then every solution of eq. (11) oscillates (componentwise) provided that () P =1 −(− ) 0 for 0 or () sup 0 ∙ 1 (1−−) P =1 −(− ) ¸ 1 and P =1(− ) 0 () inf 0 ∙ 1 (1−−) P =1 −(− ) ¸ 1
Proof. Assume that eq. (11) has a non-oscillatory solution. Then by Lemma 1 the characteristic equation of (11)
det ∙¡ 1 − −¢ + P =1 − ¸ = 0
has a real root . Therefore, there exists a non-zero vector ∈ R, with kk = 1,
such that ∙ ¡ 1 − −¢ + P =1 − ¸ = 0. Hence, ¡ 1 − −¢= P =1 −(− ) and so (2.2) ¡1 − −¢≤ P =1 −(− ).
It follows from () that (22) cannot hold. Therefore () holds. Now if 0, then from (22) we get
1 ≤ 1 (1 − −) P =1 − (−) and so 1 ≤ sup 0 ∙ 1 (1 − −) P =1 −(− ) ¸ ,
which contradicts the first part of (). Also if = 0, then from (22) 0 ≤ P =1(− )
which contradicts the second part of (). Therefore 0 and (22) yields
1 ≥ 1 (1 − −) P =1 − (−). Hence 1 ≥ inf 0 1 (1 − −) P =1 −(− ),
which contradicts () and the proof is complete.
Theorem 3. Assume that is an odd positive integer, ∈ R× and
0 for = 1 2 . Suppose that for = 1 2 , (−) ≤ 0.
Then every solution of eq.(11) oscillates (componentwise) provide that one of the following two conditions is satisfied:
() P =1(−(− )) ³ ( )( −)− ´1 1, () ∙ Q =1(−(− )) ¸1 ³ ( )(−)− ´1 1, where = 1 P =1 .
Proof. We employ Lemma 2. As (−) ≤ 0, Lemma 2 () is satisfied so it
suffices to show that each of () and () implies Lemma 2 (). Now, assume that () holds. Then, we get
sup 0 − (1 − −) = µ ( )( − )− ¶1 . Hence we have for 0,
1 (1−−) P =1 −(− ) = (−1−1) P =1 −(−(− )) ≥ P =1(−(− )) ³ ( )(− )− ´1 , so Lemma 2 () holds.
mean inequality we see that for 0, 1 (1−−) P =1 −(− ) = (−1−1) P =1 −(−(− )) ≥ ∙ Q =1(−(− )) − (1−−) ¸1 = ∙ Q =1(−(− )) ¸1 − (1−−) ≥ ∙ Q =1(−(− )) ¸1 ³ ( )(− )− ´1 , so Lemma 2 () holds. The proof is complete.
Theorem 4. Assume that is an even positive integer, ∈ R× and
0 for = 1 2 . Then every solution of eq. (11) oscillates
(componentwise) provided that (2.3) P =1 −(− ) 0 for 0 or (2.4) sup ∈(−∞∞) ∙ 1 (1 − −) P =1 −(− ) ¸ 1 for 6= 0 and P =1(− ) 0
Proof. Assume that eq. (11) has a non-oscillatory solution. Then by Lemma 1 the characteristic equation of (11)
det ∙¡ 1 − −¢ + P =1 − ¸ = 0
has a real root . Therefore, there exists a non-zero vector ∈ R, with kk = 1,
such that ∙ ¡ 1 − −¢ +P =1 − ¸ = 0. Hence, ¡ 1 − −¢= P =1 −(− ) and so (2.5) ¡1 − −¢≤ P =1 − (−).
It follows from (23) that (25) cannot hold. Therefore (24) holds. If ∈ (−∞ ∞) and 6= 0, then from (25) we get
1 ≤ 1 (1 − −) P =1 −(− )
and so 1 ≤ sup ∈(−∞∞) ∙ 1 (1 − −) P =1 − (−) ¸ , which contradicts the first part of (24).
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