Download existence of nonoscillatory solution of second order neutral

Bulletin of the Institute of Mathematics
Academia Sinica (New Series)
Vol. 2 (2007), No. 3, pp. 785-795
EXISTENCE OF NONOSCILLATORY SOLUTION OF
SECOND ORDER NEUTRAL DIFFERENTIAL
EQUATION
BY
GONG WEIMING, CHENG JINFA AND CHU YUMING
Abstract
Consider the neutral delay differential equation with positive and negative coefficients:
(r(t)(x(t) + px(t − τ )′ )′ + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0,
where p ∈ R and τ ∈ (0, ∞), σ1 , σ2 ∈ [0, ∞) and Q1 (t), Q2 (t),
r(t) ∈ C([t0 , ∞), R+ ). Some sufficient conditions for the existence
of a nonoscillatory solution of the above equation in terms of
R∞
R(s)Qi ds < ∞, i = 1, 2 are obtained.
1. Introduction
Consider the neutral delay differential equation of second order with
positive and negative coefficients:
(r(t)(x(t) + px(t − τ )′ )′ + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0,
(1)
where p ∈ R and
τ ∈ (0, ∞), σ1, σ2 ∈ [0, ∞)
and
Q1 (t), Q2 (t), r(t) ∈ C([t0 , ∞), R+ ).
(2)
Received June 12, 2006 and in revised form February 12, 2007.
AMS Subject Classification: 39A10, 34K11, 34K40.
Key words and phrases: Nonoscillatory solution, delay, Neutral differential equation.
Research supported by the 973 Project(2006CB708304), the NSF of China(10471039)
and Foundation of the Educational Committee of Zhejiang Province(20060306).
785
786
GONG WEIMING, CHENG JINFA AND CHU YUMING
Z
where R(t) =
Rt
[September
∞
R(s)Qi ds < ∞, i = 1, 2,
(3)
r(s)ds.
When r(t) = 1, the equation (1) has been reduced to the following
equation:
d2
[x(t) + px(t − τ )] + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0.
dt2
(4)
When q(t) = 0, the Eq.(4) has been investigated by many authors, see
[1-7] and the reference therein.
For Eq.(4), Recently M. R. S. Kulenović and S.Hadžiomerspahić in [1]
obtained the following result:
Theorem A. Consider the Eq.(1), if condition (2) holds, and
Z ∞
sQi ds < ∞, i = 1, 2,
aQ1 (t) − Q2 (t) ≥ 0,
f or
every
t ≥ T1
and
a > 0,
(5)
(6)
where p 6= ±1, and T1 is large enough, then Eq.(1) has a nonoscillatory
solution.
So far, this is the first global result (with respect to p) in the nonconstant coefficient case, which is a sufficient condition for the existence of a
nonoscillatory solution for all values of p.
But, condition (6) is too restrictive. In [6] the first author of this paper
deleted the strong condition (6), and permitting p=1, obtained the following
global sufficient condition(with respect to p) for the existence of a nonoscillatory solution for equation (1):
Theorem B. Consider equation (4), if condition (2), (5) hold, where
p 6= −1, then equation (1) has a nonoscillatory solution.
For Eq(1), only in special case, for example when p = 0, Q2 (t) = 0,
Hooker and Patula in [7] had investigated the existence of positive solution.
2007]
SECOND ORDER NEUTRAL DIFFERENTIAL EQUATION
787
However results for the existence of nonoscillatory solution of Eq.(1) are
relatives scarce. Motived by the paper [6] and [7], the purpose of this paper
will investigate the existence of nonoscillatory solution of Eq.(1).
Our main result is the following.
Theorem. Consider equation (1), if condition (2), (3) hold, where p 6=
±1, then equation (1) has a nonoscillatory solution.
This result extends the relevant result in [6] for p 6= −1.
2. The Proof of Theorem
The proof of theorem will be divided into four claims, depending on the
four different ranges of the parameter p.
Claim 1. p ∈ (0, 1). Choose t1 > t0 large enough such that
and
t1 ≥ t0 + σ, σ = max{τ, σ1 , σ2 },
Z ∞
R(s)[Q1 (s) + Q2 (s)]ds < 1 − p,
t1
Z ∞
p − (1 − M1 )
R(s)Q1 (s)ds ≤
M1
t1
Z ∞
1 − p − pM2 − M1
R(s)Q2 (s)ds ≤
M2
t1
hold, where M1 and M2 are positive constants which satisfy
1 − M2 < p <
1 − M1
.
1 − M2
Let X be the set of all continuous and bounded functions on [t0 , ∞)
with the sup norm. Set
A = {x ∈ X : M1 ≤ x(t) ≤ M2 , t ≥ t0 }.
Define mapping T : A → X as follows:
788
GONG WEIMING, CHENG JINFA AND CHU YUMING
[September


1 − p − px(t − τ )

R


 +R(t) ∞ [Q1 (s)x (s − σ1 ) − Q2 (s)x(s − σ2 )]ds
Rt t
(T x)(t) =
+ t1 R(s)[Q1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds, t ≥ t1 ;





(T x)(t1 ),
t0 ≤ t ≤ t1 .
We have
Z
(T x)(t) ≤ 1 − p + R(t)
Z
≤ 1 − p + M2
∞
M2 Q1 (s)ds +
t
∞
Z
t
R(s)M2 Q1 (s)ds
t1
R(s)Q1 (s)ds ≤ M2
t
and
Z
∞
Q2 (s)x (s − σ2 ) ds
(T x)(t) ≥ 1 − p − pM2 − R(t)
t
Z t
−
R(s)[Q2 (s)x(s − σ2 )]ds
t1
Z ∞
≥ 1 − p − pM2 − M2
R(s)Q2 (s)ds ≥ M1 ,
t
so TA ⊆ A .
Now for x1 , x2 ∈ A and t ≥ t1 ,we have
|(T x1 )(t) − (T x2 )(t)| ≤ p|x1 (t − τ ) − x2 (t − τ )|
Z ∞
Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )|ds
+R(t)
t
Z ∞
Q2 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
+R(t)
t
Z t
+
R(s)Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )|ds
+
Z
t1
t
R(s)Q2 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
Z ∞
R(s)[Q1 (s) + Q2 (s)]ds
≤ p||x1 − x2 || + ||x1 − x2 ||{
t
Z t
R(s)[Q1 (s) + Q2 (s)]ds}
+
t1
Z ∞
R(s)[Q1 (s) + Q2 (s)]ds}
= ||x1 − x2 ||{p +
t1
t1
2007]
SECOND ORDER NEUTRAL DIFFERENTIAL EQUATION
= q1 ||x1 − x2 ||,
789
q1 < 1.
Thus we know that T is a contraction mapping . Consequently T has
the unique fixed point x, which is obviously a positive solution of Eq.(1).
This completes the proof of Claim 1.
Claim 2. p ∈ (1, ∞). Choose t1 ≥ t0 large enough such that
Z ∞
R(s)[Q1 (s) + Q2 (s)]ds < p − 1,
t1
Z ∞
1 − p(1 − N1 )
R(s)Q1 (s)ds ≤
N1
t1
and
Z
∞
R(s)Q2 (s)ds ≤
t1
(1 − N1 )p − (1 + N2 )
,
N2
(7)
where N1 , N2 are positive constants which satisfy
(1 − N1 )p ≥ 1 + N2
and
p(1 − N2 ) < 1.
Let X be the same set as in Claim 1.Set
A = {x ∈ X : N1 ≤ x(t) ≤ N2 , t ≥ t0 }.
Define mapping T : A → X as follows:

1 − 1p − p1 x(t + τ )



R

 + R(t+τ ) ∞ [Q1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds
p
t+τ
R
(T x)(t) =
1 t+τ
+
R(s)[Q

1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds, t ≥ t1 ;
p
t

1



(T x)(t1 ),
t0 ≤ t ≤ t1 .
Clearly, T x is continuous. For every x ∈ A and t ≥ t1 , we get
Z
Z
1 R(t + τ ) ∞
1 ∞
(T x)(t) ≤ 1 − +
N2 Q1 (s)ds +
N2 R(s)Q1 (s)ds
p
p
p t1
t+τ
Z
1 N2 ∞
R(s)Q1 (s)ds ≤ N2
≤ 1− +
p
p t1
790
GONG WEIMING, CHENG JINFA AND CHU YUMING
[September
and
Z
1 N2 R(t + τ ) ∞
(T x)(t) ≥ 1 − −
+
(−N2 Q2 (s))ds
p
p
p
t+τ
Z
1 t+τ
+
(−N2 R(s)Q2 (s))ds
p t1
Z
1 N2 N2 ∞
−
R(s)Q2 (s)ds ≥ N1 .
≥ 1− −
p
p
p t1
Thus we know that T A ⊂ A. Since A is a bounded, closed, and convex
subset of X, hence we can prove that T is a contraction mapping on A by
the contraction principle.
In fact, for x1 , x2 ∈ A, we have
|(T x1 )(t) − (T x2 )(t)|
1
≤ |x1 (t + τ ) − x2 (t + τ )|
p
Z ∞
R(t + τ )
+
Q1 (s)|x1 (s − σ1 ) − x2 (s − σ2 )|ds
p
t+τ
Z ∞
Q2 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
+
t+τ
Z t+τ
1
+
R(s)Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )|ds
p t1
Z t+τ
R(s)Q1 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
+
t1
Z ∞
1
1
R(s)[Q1 (s) + Q2 (s)]ds
≤ ||x1 − x2 || + ||x1 − x2 ||
p
p
t+τ
Z t+τ
+
R(s)[Q1 (s) + Q2 (s)]ds
t1
Z ∞
1
= ||x1 − x2 ||{1 +
R(s)[Q1 (s) + Q2 (s)]ds}
p
t1
= q2 ||x1 − x2 ||,
q2 < 1.
This implies that
||T x1 − T x2 || ≤ q2 ||x1 − x2 ||.
2007]
SECOND ORDER NEUTRAL DIFFERENTIAL EQUATION
791
Thus we know that T is a contraction mapping. Consequently T has
the unique fixed point x, which is obviously a positive solution of Eq.(1).
This completes the proof of Claim 2.
Claim 3. p ∈ (−1, 0). Choose t1 > t0 large enough such that the
inequalities
Z
∞
Z
∞
R(s)[Q1 (s) + Q2 (s)]ds < p + 1,
Z ∞
M3 (1 + p) − (1 + p)
0≤
R(s)Q1 (s)ds ≤
M3
t1
t1
and
R(s)Q2 (s)ds ≤
t1
(1 + p) − M3 (1 + p)
.
M4
(8)
hold, where the positive constants M3 and M4 satisfy
0 < M3 < 1 < M4 .
Let X be the same set as in Claim 1. Set
A = {x ∈ X : M3 ≤ x(t) ≤ M4 , t ≥ t0 }.
(9)
Define mapping T : A → X as follows:


1 + p − px(t − τ )


R

 +R(t) ∞ [Q1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds
Rt t
(T x)(t) =
+

t1 R(s)[Q1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds, t ≥ t1 ;




(T x)(t1 ),
t0 ≤ t ≤ t1 .
Clearly, T x is continuous. For every x ∈ A and t ≥ t1 , by (8) we get
Z
(T x)(t) ≤ 1 + p − pM4 + R(t)
Z
≤ 1 + p − pM4 + M4
∞
t
∞
t1
M4 Q1 (s)ds +
Z
t
R(s)M4 Q1 (s)ds
t1
R(s)Q1 (s)ds ≤ M4
792
GONG WEIMING, CHENG JINFA AND CHU YUMING
[September
and
Z
(T x)(t) ≥ 1 + p − pM3 − R(t)
Z
= 1 + p − pM3 − M4
∞
M4 Q2 (s)ds −
t
∞
Z
t
R(s)M4 Q2 (s)ds
t1
R(s)Q2 (s)ds ≥ M3 .
t
Thus we know that T A ⊂ A. Since A is a bounded, closed, and convex
subset of X, hence we can prove that T is a contraction mapping on A by
the contraction principle.
In fact, for x1 , x2 ∈ A, we have
|(T x1 )(t) − (T x2 )(t)|
≤ −p|x1 (t − τ ) − x2 (t − τ )|
Z ∞
Q1 (s) |x1 (s − σ2 ) − x2 (s − σ2 ) |ds
+R(t)
Zt ∞
Q2 (s) |x1 (s − σ2 ) − x2 (s − σ2 ) |ds
+R(t)
t
Z t
+
R(s)Q1 (s) |x1 (s − σ1 ) − x2 (s − σ1 ) |ds
+
Z
t1
t
R(s)Q2 (s) |x1 (s − σ2 ) − x2 (s − σ2 ) |ds
Z ∞
R(s) [Q1 (s) + Q2 (s)] ds
≤ −p||x1 − x2 || + ||x1 − x2 ||
t
Z ∞
= ||x1 − x2 ||{−p +
R(s) [Q1 (s) + Q2 (s)] ds}
t1
t1
= q3 ||x1 − x2 ||,
q3 < 1.
This implies that
||T x1 − T x2 || ≤ q3 ||x1 − x2 ||.
Thus we know that T is a contraction mapping. Consequently T has
the unique fixed point x, which is obviously a positive solution of Eq.(1).
This completes the proof of Claim 3.
Claim 4. p ∈ (−∞, −1). Choose t1 > t0 large enough such that the
2007]
SECOND ORDER NEUTRAL DIFFERENTIAL EQUATION
793
inequalities
Z
∞
R(s) [Q1 (s) + Q2 (s)] ds < −(p + 1),
t
Z 1∞
R(s)Q2 (s)ds <
t1
and
Z
∞
R(s)Q1 (s) <
t1
−(p + 1)(N3 − 1)
N3
−(1 + p)(1 − N3 )
N4
(10)
(11)
(12)
hold, where the positive constants N3 and N4 satisfy
0 < N3 < 1 < N4 .
Let X be the same set as in Claim 1. Set
A = {x ∈ X : N3 ≤ x(t) ≤ N4 , t ≥ t0 }.
(13)
Define mapping T : A → X as follows:

1 + 1p − p1 x(t + τ )



R

 + R(t+τ ) ∞ [Q1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds
t+τ
Rp
(T x)(t) =
1 t+τ
+
R(s)[Q

1 (s)x(s − σ1 ) − Q2 (s)x(s − σ2 )]ds, t ≥ t1 ;
p
t1




(T x)(t1 ),
t0 ≤ t ≤ t1 .
Clearly, T x is continuous. For every x ∈ A and t ≥ t1 , using (10) and
(12) we get
Z
1 N4 N4 ∞
−
R(s)Q2 (s)ds ≤ N4 .
(T x)(t) ≤ 1 + −
p
p
p t
Furthermore, in view of (11) and (12) we have
Z
1 N3 N4 ∞
(T x)(t) ≥ 1 + −
+
R(s)Q1 (s)ds ≥ N3 .
p
p
p t
Thus we know that T A ⊂ A. Since A is a bounded, closed, and convex
subset of X, hence we can prove that T is a contraction mapping on A by
the contraction principle.
794
GONG WEIMING, CHENG JINFA AND CHU YUMING
[September
In fact, for x1 , x2 ∈ A, we have
|(T x1 )(t) − (T x2 )(t)|
1
≤ − |x1 (t + τ ) − x2 (t + τ )|
p
Z ∞
R(t + τ )
Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )|ds
−
p
t+τ
Z ∞
Q2 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
+
t+τ
Z t+τ
1
−
R(s)Q1 (s)|x1 (s − σ1 ) − x2 (s − σ1 )|ds
p t1
Z t+τ
+
R(s)Q2 (s)|x1 (s − σ2 ) − x2 (s − σ2 )|ds
t1
Z ∞
1
1
R(s)[Q1 (s) + Q2 (s)]ds
≤ − ||x1 − x2 || − ||x1 − x2 ||
p
p
t+τ
Z t+τ
+
R(s)[Q1 (s) + Q2 (s)]ds
t1
Z ∞
1
= − ||x1 − x2 ||{1 +
R(s)[Q1 (s) + Q2 (s)]ds}
p
t1
= q4 ||x1 − x2 ||,
q4 < 1.
This implies that
||T x1 − T x2 || ≤ q4 ||x1 − x2 ||.
Thus we know that T is a contraction mapping. Consequently T has the
unique fixed point x, which is obviously a positive solution of Eq.(1). This
completes the proof of Claim 4.
References
1. M. R. S. Kulenović and S.Hadz̆iomerspahić, Existence of nonoscillatory solution
of second order linear neutral delay equation, J. Math. Anal. Appl., 228(1998), no.2,
436-448.
2. I. Györi and G. Ladas, Oscillation Theory of Delay Differential Equations, Oxford
Univ. Press, New York, 1991.
3. D. D. Baǐnov and D. P. Mishev, Oscillation Theory for Neutral Differential Equations with Delay, Adam Hilger, Bristol, 1991.
2007]
SECOND ORDER NEUTRAL DIFFERENTIAL EQUATION
795
4. K. Gopalsamy, Stability and Oscillations in Delay Differential Equations of Population Dynamics, Kluwer Academic, Dordrecht, 1992.
5. B. G. Zhang and J. S. Yu, Existence of positive solutions for neutral differential
equations, Sci. China Ser. A, 35(1992), no.11, 1306-1313.
6. J. F. Cheng, Existence of nonoscillatory solution of second order linear neutral
difference equation, Appl. Math. Lett., 20(2007), no.8, 892-899.
7. J. W. Hooker and W. T. Patula, A second-order nonlinear difference equation:
oscillation and asymptotic behavior, J. Math. Anal. Appl., 91(1983), no.1, 9-29.
Department of mathematics, Hunan City University, Yiyang 413000, P.R. China.
E-mail: [email protected]
Department of mathematics, Xiamen University, Xiamen 361005, P.R. China.
E-mail: [email protected]
Department of mathematics, Huzhou Teachers College, Huzhou 313000, P.R. China.
E-mail: [email protected]