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An oscillation theorem
Applied Mathematics Letters 20 (2007) 238–240 www.elsevier.com/locate/aml
An oscillation theorem Chung-Fen Lee a , Cheh-Chih Yeh b,∗ a National United University, No.1 Lein Kung, Kung-Chin Li, Miaoli City, Taiwan, ROC b Department of Information Management, Lunghwa University of Science and Technology, Kueishan Taoyuan 333, Taiwan, ROC
Received 27 March 2006; accepted 13 April 2006
Abstract In this paper, we discuss the oscillation criterion of y (t) for the solution y(t) of the more general functional differential equation. r (t)ψ(y(t))ϕ(y (t)) + p(t) f (y(g(t))) = 0 (E) where ψ, f and g satisfy some suitable conditions. c 2006 Elsevier Ltd. All rights reserved. Keywords: Oscillatory solution; Functional differential equation
The purpose of this work is to extend a result of Travis [1] concerning the oscillation criterion of y (t) for the solution y(t) of the second-order functional differential equation y (t) + p(t) f (y(g(t))) = 0 to the more general functional differential equation r (t)ψ(y(t))ϕ(y (t)) + p(t) f (y(g(t))) = 0
(E)
where (C1 ) r, p, g ∈ C([a, ∞), R) with r (t) > 0 on [a, ∞) for some a ≥ 0 and limt →∞ g(t) = ∞; (C2 ) ϕ, ψ ∈ C(R, R), xϕ(x) > 0 for x = 0, ϕ is strictly increasing on R and there exist two constants k1 , k2 ∈ (0, ∞) such that k1 ≤ ψ(x) ≤ k2 for all x = 0; (C3 ) f ∈ C 1 (R, R), x f (x) > 0 and f (x) ≥ 0 for x = 0. For related results, we refer the reader to [2,3]. We now state and prove our main result as follows: Theorem 1. Let g ∈ C 1 ([a, ∞), R) and g (t) ≥ 0 on [a, ∞). Assume that ∞ p(t)dt = ∞ ∗ Corresponding author.
E-mail addresses: [email protected] (C.-F. Lee), [email protected] (C.-C. Yeh). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.04.005
(H 1 )
C.-F. Lee, C.-C. Yeh / Applied Mathematics Letters 20 (2007) 238–240
and
∞
ϕ −1
c ds = −∞ r (s)
239
for every c < 0.
(H 2 )
If y(t) is a solution of (E), then y (t) is oscillatory. Proof. If y(t) is an oscillatory solution of (E) on [a, ∞), then y (t) is oscillatory on [a, ∞). Without loss of generality, we may assume that y(t) is an eventually positive solution of (E) on [a, ∞); then it follows from condition (C1 ) that y(g(t)) is also eventually positive. Hence f (y(g(t))) > 0. We can divide our proof into the following three cases: (A) y (t) > 0, for t large enough, (B) y (t) < 0, for t large enough, (C) y (t) is oscillatory. Case (A). In this case, let w(t) = then
r (t)ϕ(y (t))ψ(y(t)) ; f (y(g(t)))
r (t)ϕ(y (t))ψ(y(t)) r (t)ϕ(y (t))ψ(y(t)) f (y(g(t)))y (g(t))g (t) w (t) = − f (y(g(t))) f 2 (y(g(t))) ≤ − p(t)
on [T, ∞), where T ≥ a is large enough. Integrating the above inequality from T to t (>T ), we get t p(s)ds. w(t) − w(T ) ≤ − T
This and (H 1 ) imply w(t) < 0, and hence y (t) < 0 for t large enough. This is a contradiction. Case (B). y (t) < 0 for t large enough. It follows from (H 1 ) that there exists a number T ≥ a such that t p(s)ds ≥ 0 for t ≥ T. T
Hence,
t T
s=t s t p(s) f (y(g(s)))ds = f (y(g(s))) p(u)du − f (y(g(s)))y (g(s))g (s) p(u)du ds T T T s=T s t t = f (y(g(t))) p(u)du − f (y(g(s)))y (g(s))g (s) p(u)du ds
s
T
T
≥ 0. Integrating (E) from T to t (>T ) and using the above inequality, we have t t p(s) f (y(g(s)))ds ≤ 0, r (s)ψ(y(s))ϕ(y (s)) ds = − T
T
which implies r (t)ψ(y(t))ϕ(y (t)) ≤ r (T )ψ(y(T ))ϕ(y (T )) < 0. Thus, it follows from condition (C2 ) that ϕ(y (t)) ≤
k <0 r (t)
for some constants k < 0, and hence, k −1 y (t) ≤ ϕ . r (t)
T
240
C.-F. Lee, C.-C. Yeh / Applied Mathematics Letters 20 (2007) 238–240
Integrating the above inequality from T to t (≥ T ), we get t k −1 y(t) − y(T ) ≤ ϕ ds. r (s) T This and (H 2 ) imply limt →∞ y(t) = −∞. This contradicts y(t) being eventually positive on [a, ∞). Thus, we have proved that y (t) is oscillatory. Letting ψ(s) = 1 and ϕ(y) = f (y) = |y| p−2 y, (E) is reduced to r (t)|y (t)| p−2 y (t) + p(t)|y(g(t))| p−2 y(g(t)) = 0, where p > 1.
(E1 )
1
Clearly, the inverse of ϕ is ϕ −1 (s) = |s| p−1 sgn(s). Thus we have the following: Theorem 2. Let g ∈ C 1 ([a, ∞), R) and g (t) ≥ 0 on [a, ∞). Assume that (H 1 ) and 1 ∞ c p−1 ds = ∞ for every c < 0 r (s)
(H 3 )
hold. If y(t) is a solution of (E1 ), then y (t) is oscillatory. Example 3. Consider the second-order functional differential equations 1 1 3 (y (t)) + y 3 (2t) = 0, t ≥ 1. t t
(E2 )
Clearly, f (x) = x 3 , ψ(x) = 1, ϕ(x) = x 3 , r (t) = 1t , p(t) = 1t , f (x) = 3x 2 ≥ 0 and ϕ −1 (x) = x 1/3 for x ∈ R.
∞
∞ k 1/3 Since p(t)dt = ∞ and ( r(s) ) ds = −∞ for k < 0, it follows from Theorem 1 that y (t) is oscillatory if y(t) is a solution of (E2 ). Clearly, Theorem 1 of Travis [1] cannot be applied to Eq. (E2 ). References [1] C.C. Travis, Oscillation theorems for second order differential equation with functional arguments, Proc. Amer. Math. Soc. 31 (1972) 199–202. [2] H.B. Hsu, W.C. Lian, C.C. Yeh, Oscillation criteria for second order quasilinear functional differential equations, Nonlinear World 3 (1996) 835–848. [3] R.P. Agarwal, S.R. Grace, D. O’Regan, On the oscillation of second order functional differential equations, Adv. Math. Sci. Appl. 12 (2002) 257–272.
An oscillation theorem Chung-Fen Lee a , Cheh-Chih Yeh b,∗ a National United University, No.1 Lein Kung, Kung-Chin Li, Miaoli City, Taiwan, ROC b Department of Information Management, Lunghwa University of Science and Technology, Kueishan Taoyuan 333, Taiwan, ROC
Received 27 March 2006; accepted 13 April 2006
Abstract In this paper, we discuss the oscillation criterion of y (t) for the solution y(t) of the more general functional differential equation. r (t)ψ(y(t))ϕ(y (t)) + p(t) f (y(g(t))) = 0 (E) where ψ, f and g satisfy some suitable conditions. c 2006 Elsevier Ltd. All rights reserved. Keywords: Oscillatory solution; Functional differential equation
The purpose of this work is to extend a result of Travis [1] concerning the oscillation criterion of y (t) for the solution y(t) of the second-order functional differential equation y (t) + p(t) f (y(g(t))) = 0 to the more general functional differential equation r (t)ψ(y(t))ϕ(y (t)) + p(t) f (y(g(t))) = 0
(E)
where (C1 ) r, p, g ∈ C([a, ∞), R) with r (t) > 0 on [a, ∞) for some a ≥ 0 and limt →∞ g(t) = ∞; (C2 ) ϕ, ψ ∈ C(R, R), xϕ(x) > 0 for x = 0, ϕ is strictly increasing on R and there exist two constants k1 , k2 ∈ (0, ∞) such that k1 ≤ ψ(x) ≤ k2 for all x = 0; (C3 ) f ∈ C 1 (R, R), x f (x) > 0 and f (x) ≥ 0 for x = 0. For related results, we refer the reader to [2,3]. We now state and prove our main result as follows: Theorem 1. Let g ∈ C 1 ([a, ∞), R) and g (t) ≥ 0 on [a, ∞). Assume that ∞ p(t)dt = ∞ ∗ Corresponding author.
E-mail addresses: [email protected] (C.-F. Lee), [email protected] (C.-C. Yeh). c 2006 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2006.04.005
(H 1 )
C.-F. Lee, C.-C. Yeh / Applied Mathematics Letters 20 (2007) 238–240
and
∞
ϕ −1
c ds = −∞ r (s)
239
for every c < 0.
(H 2 )
If y(t) is a solution of (E), then y (t) is oscillatory. Proof. If y(t) is an oscillatory solution of (E) on [a, ∞), then y (t) is oscillatory on [a, ∞). Without loss of generality, we may assume that y(t) is an eventually positive solution of (E) on [a, ∞); then it follows from condition (C1 ) that y(g(t)) is also eventually positive. Hence f (y(g(t))) > 0. We can divide our proof into the following three cases: (A) y (t) > 0, for t large enough, (B) y (t) < 0, for t large enough, (C) y (t) is oscillatory. Case (A). In this case, let w(t) = then
r (t)ϕ(y (t))ψ(y(t)) ; f (y(g(t)))
r (t)ϕ(y (t))ψ(y(t)) r (t)ϕ(y (t))ψ(y(t)) f (y(g(t)))y (g(t))g (t) w (t) = − f (y(g(t))) f 2 (y(g(t))) ≤ − p(t)
on [T, ∞), where T ≥ a is large enough. Integrating the above inequality from T to t (>T ), we get t p(s)ds. w(t) − w(T ) ≤ − T
This and (H 1 ) imply w(t) < 0, and hence y (t) < 0 for t large enough. This is a contradiction. Case (B). y (t) < 0 for t large enough. It follows from (H 1 ) that there exists a number T ≥ a such that t p(s)ds ≥ 0 for t ≥ T. T
Hence,
t T
s=t s t p(s) f (y(g(s)))ds = f (y(g(s))) p(u)du − f (y(g(s)))y (g(s))g (s) p(u)du ds T T T s=T s t t = f (y(g(t))) p(u)du − f (y(g(s)))y (g(s))g (s) p(u)du ds
s
T
T
≥ 0. Integrating (E) from T to t (>T ) and using the above inequality, we have t t p(s) f (y(g(s)))ds ≤ 0, r (s)ψ(y(s))ϕ(y (s)) ds = − T
T
which implies r (t)ψ(y(t))ϕ(y (t)) ≤ r (T )ψ(y(T ))ϕ(y (T )) < 0. Thus, it follows from condition (C2 ) that ϕ(y (t)) ≤
k <0 r (t)
for some constants k < 0, and hence, k −1 y (t) ≤ ϕ . r (t)
T
240
C.-F. Lee, C.-C. Yeh / Applied Mathematics Letters 20 (2007) 238–240
Integrating the above inequality from T to t (≥ T ), we get t k −1 y(t) − y(T ) ≤ ϕ ds. r (s) T This and (H 2 ) imply limt →∞ y(t) = −∞. This contradicts y(t) being eventually positive on [a, ∞). Thus, we have proved that y (t) is oscillatory. Letting ψ(s) = 1 and ϕ(y) = f (y) = |y| p−2 y, (E) is reduced to r (t)|y (t)| p−2 y (t) + p(t)|y(g(t))| p−2 y(g(t)) = 0, where p > 1.
(E1 )
1
Clearly, the inverse of ϕ is ϕ −1 (s) = |s| p−1 sgn(s). Thus we have the following: Theorem 2. Let g ∈ C 1 ([a, ∞), R) and g (t) ≥ 0 on [a, ∞). Assume that (H 1 ) and 1 ∞ c p−1 ds = ∞ for every c < 0 r (s)
(H 3 )
hold. If y(t) is a solution of (E1 ), then y (t) is oscillatory. Example 3. Consider the second-order functional differential equations 1 1 3 (y (t)) + y 3 (2t) = 0, t ≥ 1. t t
(E2 )
Clearly, f (x) = x 3 , ψ(x) = 1, ϕ(x) = x 3 , r (t) = 1t , p(t) = 1t , f (x) = 3x 2 ≥ 0 and ϕ −1 (x) = x 1/3 for x ∈ R.
∞
∞ k 1/3 Since p(t)dt = ∞ and ( r(s) ) ds = −∞ for k < 0, it follows from Theorem 1 that y (t) is oscillatory if y(t) is a solution of (E2 ). Clearly, Theorem 1 of Travis [1] cannot be applied to Eq. (E2 ). References [1] C.C. Travis, Oscillation theorems for second order differential equation with functional arguments, Proc. Amer. Math. Soc. 31 (1972) 199–202. [2] H.B. Hsu, W.C. Lian, C.C. Yeh, Oscillation criteria for second order quasilinear functional differential equations, Nonlinear World 3 (1996) 835–848. [3] R.P. Agarwal, S.R. Grace, D. O’Regan, On the oscillation of second order functional differential equations, Adv. Math. Sci. Appl. 12 (2002) 257–272.