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Asymptotic behavior of solutions of neutral differential equations with positive and negative coefficients
J. Math. Anal. Appl. 279 (2003) 326–338 www.elsevier.com/locate/jmaa
Asymptotic behavior of solutions of neutral differential equations with positive and negative coefficients Xiaoping Wang and Liusheng Liao ∗ Department of Mathematics, Chenzhou Teacher’s College, Chenzhou, Hunan 423000, PR China Received 24 June 2002 Submitted by William F. Ames
Abstract Sufficient conditions are established for the asymptotic behavior of solutions of neutral differential equations with positive and negative coefficients x(t) − P (t)x(t − τ ) + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0, t t0 , where τ, σ1 , σ2 ∈ (0, ∞), P , Q1 , Q2 ∈ C([t0 , ∞), R). Some known results are generalized and improved. 2003 Elsevier Science (USA). All rights reserved. Keywords: Asymptotic behavior; Neutral differential equations; Positive and negative coefficients
1. Introduction Consider the neutral delay differential equation with positive and negative coefficients (1) x(t) − P (t)x(t − τ ) + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0, t t0 , where τ, σ1 , σ2 ∈ (0, ∞), P , Q1 , Q2 ∈ C([t0 , ∞), R). When Q2 (t) ≡ 0, Q1 (t) = Q(t) 0, σ1 = σ , Eq. (1) reduces to x(t) − P (t)x(t − τ ) + Q(t)x(t − σ ) = 0, t t0 . * Corresponding author.
E-mail address: [email protected] (L. Liao). 0022-247X/03/$ – see front matter 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-247X(03)00021-0
(2)
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
In [1], it was shown that if ∞ Q(s) ds = ∞
327
(3)
t0
and P (t) ≡ p
t and 2p + lim sup t →∞
Q(s) ds < 1,
(4)
t −σ
then every solution of Eq. (2) tends to zero as t → ∞. By making use of the double iterative method, [2] improves (4) to P (t) p
t and 2p(2 − p) + lim sup t →∞
t −σ
3 Q(s) ds < . 2
(5)
We remark that when p = 0, (5) reduces to the following so-called 3/2-asymptotic behavior condition: t 3 (6) Q(s) ds < , lim sup 2 t →∞ t −σ
for the special form of (2) x (t) + Q(t)x(t − σ ) = 0,
t t0 .
(7)
It was pointed out in [3] that the upper bound 3/2 in (6) is the best possible one for Eq. (7), when Q is a piecewise continuous function. As an application of the above result, it is easy to conclude that if (3) holds and t lim sup t →∞
t −σ
3 Q(s) ds < (1 − p), 2
(8)
then every solution of Eq. (2) with P (t) ≡ p ∈ (0, 1) and τ = 0 tends to zero as t → ∞. Clearly, condition (8) suggests that (5) is far from “the best possible” and there is room for improvement. Therefore it is desirable to establish some better results, and this constitutes the purpose of this paper. In this paper, we will establish a set of conditions which guarantee every solution of Eq. (1) tends to zero as t → ∞. When they are applied to (2), which improve (5) to P (t) p < 1 4
t and 2p + lim sup t →∞
t −σ
3 Q(s) ds < , 2
(9)
or 1 1 p< 4 2
t and
lim sup t →∞
t −σ
Q(s) ds <
2(1 − 2p).
(10)
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Equation (1) has been investigated by many authors, see for example [4–6,8,11,12]. For nonlinear equations, one may refer to [9,10].
2. The main results and proofs Theorem 1. Assume that there is a constant c such that Q(t) = Q1 (t) − Q2 (t + σ2 − σ1 ) 0, t P (t) + sign(σ1 − σ2 )Q2 (s) ds c,
(11) (12)
t +σ2 −σ1
and 1 c< 4
t and 2c + t −σ1
3 Q(s) ds , 2
(13)
or 1 1 c< 4 2
t Q(s) ds
and
2(1 − 2c).
(14)
t −σ1
Then every solution of Eq. (1) is bounded. Proof. If c = 0, it was proved in [7]. In the sequel, we assume that 0 < c < 1/2. Let x(t) be a solution of Eq. (1). Set t z(t) = x(t) − P (t)x(t − τ ) −
Q2 (s)x(s − σ2 ) ds.
(15)
t +σ2 −σ1
Then z (t) = −Q(t)x(t − σ1 ),
t t0 .
(16)
We will prove that x(t) is bounded. Indeed, if x(t) is unbounded, then lim supt →∞ |x(t)| = ∞. Then there exists T > t0 such that |x(T )| > |x(t)|, t0 t < T , and |x(T )| > z(t0 + 2σ1 + 2σ2 + 2τ )/(1 − c). Without of loss of generality, we may assume that x(T ) = |x(T )|. Then T Q2 (s)x(s − σ2 ) ds
z(T ) = x(T ) − P (T )x(T − τ ) − T +σ2 −σ1
> (1 − c)x(T ) > z(t0 + 2σ1 + 2σ2 + 2τ ). t∗
(17) z(t ∗ )
It follows from (17) that there exists ∈ (t0 + 2σ1 + 2σ1 + 2τ, T ] such that = max{z(t): t0 t T } and z(t) < z(t ∗ ) for t0 t < t ∗ . Note that 0 < c < 1/2, it is easy
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
329
to see that x(t ∗ ) z(t ∗ ) − cx(T ) > (1 − 2c)x(T ) 0. Next we prove x(t ∗ − σ1 ) 0. Otherwise, x(t ∗ − σ1 ) > 0. Thus, there is a left neighbor of (t ∗ − σ1 ), which is denoted by (t ∗ − σ1 − h, t ∗ − σ1 ) for some h > 0, such that x(t) > 0 for t ∈ (t ∗ − σ1 − h, t ∗ − σ1 ), and so x(t − σ1 ) > 0 for t ∈ (t ∗ − h, t ∗ ). Therefore by (16), we see that z(t) is not increasing on (t ∗ − h, t ∗ ). This contradicts to the definition of t ∗ and so x(t ∗ − σ1 ) 0. Hence, there exists ξ ∈ [t ∗ − σ1 , t ∗ ) such that x(ξ ) = 0. Integrating (16) from t − σ1 to ξ , we obtain for t0 + 2(τ + σ1 + σ2 ) t t ∗ : t−σ1
−x(t − σ1 ) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ
+ P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds + x(T )
Q(s) ds.
t −σ1
ξ +σ2 −σ1
(18) The proof will be complete, if we conclude that z(T ) z(t ∗ ) (1 − c)x(T ),
(19) z(t ∗ )
which is due to the contradiction to the fact that z(T ) > (1 − c)x(T ). There are three possible cases: t∗ Case 1. 0 < c < 1/4 and ξ Q(s) ds 1. In this case, by integrating (16) from ξ to t ∗ and using (13) and (18), we have z(t ∗ ) = z(ξ ) −
t ∗ Q(t)x(t − σ1 ) dt ξ
ξ Q2 (s)x(s − σ2 ) ds
−P (ξ )x(ξ − τ ) − t ∗
ξ +σ2 −σ1 t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
+
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ Q2 (s)x(s − σ2 ) ds
+ P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ
− P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
t ∗ 1−
Q(s) ds ξ
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ + x(T )
t−σ1
Q(t) P (t − σ1 ) +
sign(σ1 − σ2 )Q2 (s) ds
t +σ2 −2σ1
ξ
ξ +
Q(s) ds dt
t −σ1
ξ
x(T ) P (ξ ) +
sign(σ1 − σ2 )Q2 (s) ds
1−
ξ +σ2 −σ1
t∗
Q(s) ds ξ
ξ
Q(t) c +
+ x(T )
t ∗
Q(s) ds dt
t −σ1
ξ
t ∗
x(T ) c 1 −
Q(s) ds ξ
t ∗ +
t
Q(t) c +
Q(s) ds −
t −σ1
ξ
t ∗
x(T ) c +
Q(s) ds dt ξ
3 − 2c − Q(t) 2
ξ
t
t
Q(s) ds dt ξ
3 − 2c = x(T ) c + 2
t ∗
1 Q(s) ds − 2
ξ
2
t ∗ Q(s) ds ξ
(1 − c)x(T ). t∗ t∗ Case 2. 0 < c < 1/4 and ξ Q(s) ds > 1. Choose η ∈ (ξ, t ∗ ) such that η Q(s) ds = 1. Then integrating (16) first from ξ to η and then from η to t ∗ , and using (13) and (18), we have
z(t ∗ ) = z(ξ ) −
t ∗
η Q(t)x(t − σ1 ) dt −
Q(t)x(t − σ1 ) dt η
ξ
ξ −P (ξ )x(ξ − τ ) − ξ +σ2 −σ1
η Q2 (s)x(s − σ2 ) ds + x(T )
Q(s) ds ξ
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ +
331
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
η
ξ Q2 (s)x(s − σ2 ) ds
+ P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ
− P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds
1−
ξ +σ2 −σ1
+ x(T )
Q(s) ds + x(T )
t ∗
Q(t) c +
η
t ∗ Q(s) ds +
Q(t) dt
Q(s) ds dt
t −σ1
η
= x(T )
ξ
η
ξ
Q(s) ds η
t ∗
η
t ∗
Q(t) c +
Q(s) ds dt
t −σ1
η
ξ
ξ
η = x(T ) c Q(s) ds + Q(t) Q(s) ds dt t∗
t∗
η
x(T )
η
3 −c 2
t ∗
Q(s) ds − η
= x(T )
t −σ1
t∗
Q(t) η
t∗
3 −c 2
t Q(s) ds dt η
2
t∗
1 Q(s) ds − 2
η
Q(s) ds
= (1 − c)x(T ).
η
t∗ √ Case 3. 1/4 c < 1/2 and ξ Q(s) ds 2(1 − 2c). Integrating (16) from ξ to t ∗ and using (14) and (18), we obtain
∗
t ∗
z(t ) = z(ξ ) −
Q(t)x(t − σ1 ) dt ξ
ξ −P (ξ )x(ξ − τ ) − ξ +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ +
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ + P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds
ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ Q2 (s)x(s − σ2 ) ds
− P (ξ )x(ξ − τ ) +
t ∗ + x(T )
ξ +σ2 −σ1 ξ
Q(t) c +
Q(t)
t ∗
Q(s) ds +
t ∗
x(T ) c +
ξ
t
2(1 − 2c) −
ξ
= x(T ) c + 2(1 − 2c)
Q(s) ds dt ξ
Q(s) ds dt ξ
t ∗
Q(s) ds −
t −σ1
t
t
Q(t) c + ξ
Q(s) ds ξ
Q(s) ds dt
t ∗
x(T ) c 1 −
1−
t −σ1
ξ
t ∗
1 Q(s) ds − 2
ξ
2
t ∗ Q(s) ds
(1 − c)x(T ).
ξ
Cases 1, 2, and 3 show that (19) is true. And therefore, the proof is complete. ✷ Theorem 2. Assume that there is a constant c such that (11) and (12) holds, that ∞ Q(s) ds = ∞,
(20)
t0
and 1 c< 4
t and 2c + lim sup t →∞
t −σ1
3 Q(s) ds < , 2
(21)
or 1 1 c< 4 2
t and
lim sup t →∞
Q(s) ds <
2(1 − 2c).
t −σ1
Then every solution of Eq. (1) tends to zero as t → ∞.
(22)
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
333
Proof. Let x(t) be any solution of Eq. (1). We shall prove that lim x(t) = 0,
(23)
t →∞
in two cases where x(t) is nonoscillatory or oscillatory. In the former case, we may assume that x(t) is eventually positive. Let z(t) be defined by (15). It is easy to see that z(t) is eventually nonincreasing. Set β = limt →∞ z(t). Then β is bounded and by (16) ∞ β + Q(s)x(s − σ1 ) ds = z(t0 ), t0
which together with (20) implies lim inft →∞ x(t) = 0. From (16), we have
t lim sup x(t) = β + lim sup P (t)x(t − τ ) + t →∞
t →∞
Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β + lim sup P (t)x(t − τ ) t →∞
t +
sign(σ1 − σ2 )Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β + c lim sup x(t), t →∞
which implies lim sup x(t) t →∞
β . 1−c
On the other hand, lim inf x(t) = β + lim inf P (t)x(t − τ ) + t →∞
t
t →∞
Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β − lim sup P (t)x(t − τ ) t →∞
t
sign(σ1 − σ2 )Q2 (s)x(s − σ2 ) ds
+
t +σ2 −σ1
β − c lim sup x(t) t →∞
1 − 2c β. 1−c
This shows β = 0 and so (23) holds. In the latter case, i.e., x(t) is oscillatory. Set µ = lim supt →∞ |x(t)|. Then Theorem 1 implies 0 µ < ∞. Similarly to (17), we have (24) M = lim supz(t) (1 − c)µ. t →∞
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
The proof will be finished when we prove µ = √ 0. Suppose that µ > 0. Then for any ε ∈ (0, (1 − 2c)µ), there exist A ∈ (1, 3/2), B ∈ (0, 2(1 − 2c) ) and T > t0 such that x(t) < µ + ε, t T − 2(τ + σ1 + σ2 ), and
t Q(s) ds < t −σ
A − 2c, if c < 1/4, B, if 1/4 c < 1/2,
t T.
(25)
Note that z (t) is oscillatory; there is an increasing sequence {Tn } such that Tn > T + τ + 2σ1 , Tn → ∞, |z(Tn )| → M as n → ∞, |z(Tn )| > (1 − c)(µ − ε), and Tn is left local maximum of |z(t)|. We may assume that z(Tn ) > 0. The case when z(Tn ) < 0 is similar and the proof will be omitted. Thus x(Tn ) z(Tn ) − c(µ + ε) > (1 − 2c)µ − ε > 0. It is also easy to see that x(Tn − σ1 ) 0 in view of the definition of Tn . Hence, there exists ξn ∈ [Tn − σ1 , Tn ) such that x(ξn ) = 0. From (16), we have for Tn − σ1 t Tn t−σ1
−x(t − σ1 ) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds.
(26)
t −σ1
Set
A=
max{A − c − 1/2, c + 1/2}, if c < 1/4, B 2 /2 + c, if 1/4 c < 1/2.
Then A < 1 − c. We will prove z(Tn ) Λ(µ + ε).
(27)
To this end, we consider the following three possible cases: T Case 1. 0 < c < 1/4 and ξnn Q(s) ds 1. In this case, by integrating (16) from ξn to Tn and using (25) and (26), we have Tn z(Tn ) = z(ξn ) −
Q(t)x(t − σ1 ) dt ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
335
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Tn + (µ + ε)
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
Q(t) c +
Q(s) ds ξn
Q(s) ds dt
t −σ1
ξn
Tn
(µ + ε) c 1 −
Q(s) ds ξn
Tn
t
ξn
(µ + ε) c +
Q(s) ds −
t −σ1
Tn
t
Q(t) c +
+
1−
ξn
Tn
Q(s) ds dt ξn
t
Q(t) A − 2c − ξn
Q(s) ds dt ξn
Tn 2
1 Q(s) ds = (µ + ε) c + (A − 2c) Q(s) ds − 2 ξn ξn 1 (µ + ε) c + max{A − 2c, 1} − = Λ(µ + ε). 2 T T Case 2. 0 < c < 1/4 and ξnn Q(s) ds > 1. Choose ηn ∈ (ξn , Tn ) such that ηnn Q(s) ds = 1. Then integrating (16) first from ξn to ηn and then from ηn to Tn , and using (25) and (26), we have
Tn
ηn z(Tn ) = z(ξn ) −
Tn Q(t)x(t − σ1 ) dt −
Q(t)x(t − σ1 ) dt ηn
ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
ηn Q2 (s)x(s − σ2 ) ds + (µ + ε)
Q(s) ds ξn
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ηn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
1−
ξn +σ2 −σ1
Q(s) ds + (µ + ε) ηn
ηn
t −σ1
ξn
Q(t) c +
Q(s) ds dt
t −σ1
ηn
ξn
Q(s) ds dt
Tn Q(s) ds +
Q(t) dt
ξn
Q(t) c + ηn
ξn
= (µ + ε)
Tn
Tn
Q(s) ds ηn
ηn + (µ + ε)
Tn
Tn
Tn ηn = (µ + ε) c Q(s) ds + Q(t) Q(s) ds dt ηn
t −σ1
ηn
Tn
Tn Q(s) ds −
(µ + ε) (A − c) ηn
t Q(t)
ηn
Q(s) ds dt ηn
2
1 Q(s) ds = (µ + ε) (A − c) Q(s) ds − 2 ηn ηn
1 Λ(µ + ε). = (µ + ε) A − c − 2 T Case 3. 1/4 c < 1/2 and ξnn Q(s) ds B. Integrating (16) from ξn to Tn and using (25) and (26), we obtain
Tn
Tn
Tn z(Tn ) = z(ξn ) −
Q(t)x(t − σ1 ) dt ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
337
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Tn + (µ + ε)
ξn +σ2 −σ1
1−
Q(s) ds ξn
ξn
Q(t) c +
Q(s) ds dt
t −σ1
ξn
Tn
(µ + ε) c 1 −
Q(s) ds ξn
Tn
t
Q(t) c +
+
Q2 (s)x(s − σ2 ) ds
Tn
Tn
(µ + ε) c +
Q(s) ds −
t −σ1
ξn
ξn
t
Q(t) B −
t
Q(s) ds dt ξn
Q(s) ds dt ξn
Tn
= (µ + ε) c + B
1 Q(s) ds − 2
ξn
2
Tn Q(s) ds ξn
1 c + B 2 (µ + ε) = Λ(µ + ε). 2
Cases 1, 2, and 3 show that (27) is true. Let n → ∞ and ε → 0 in (27). We see that M = lim sup z(Tn ) Λu < (1 − c)µ, n→∞
which contradicts to (24). Therefore µ = 0, and so the proof is complete. ✷ Applying Theorem 2 to Eq. (2), we have immediately Corollary 1. Assume that there is a constant p such that (3) and (9) or (10) hold. Then every solution of Eq. (2) tends to zero as t → ∞.
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Remark 1. Theorem 2 relaxes the usual restrictions on the coefficients (Q1 (t), Q2 (t) 0) and the delays (σ1 σ2 ) in [4–6].
Acknowledgment The authors would like to thank the reviewers and the editor for their valuable suggestions and comments.
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Asymptotic behavior of solutions of neutral differential equations with positive and negative coefficients Xiaoping Wang and Liusheng Liao ∗ Department of Mathematics, Chenzhou Teacher’s College, Chenzhou, Hunan 423000, PR China Received 24 June 2002 Submitted by William F. Ames
Abstract Sufficient conditions are established for the asymptotic behavior of solutions of neutral differential equations with positive and negative coefficients x(t) − P (t)x(t − τ ) + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0, t t0 , where τ, σ1 , σ2 ∈ (0, ∞), P , Q1 , Q2 ∈ C([t0 , ∞), R). Some known results are generalized and improved. 2003 Elsevier Science (USA). All rights reserved. Keywords: Asymptotic behavior; Neutral differential equations; Positive and negative coefficients
1. Introduction Consider the neutral delay differential equation with positive and negative coefficients (1) x(t) − P (t)x(t − τ ) + Q1 (t)x(t − σ1 ) − Q2 (t)x(t − σ2 ) = 0, t t0 , where τ, σ1 , σ2 ∈ (0, ∞), P , Q1 , Q2 ∈ C([t0 , ∞), R). When Q2 (t) ≡ 0, Q1 (t) = Q(t) 0, σ1 = σ , Eq. (1) reduces to x(t) − P (t)x(t − τ ) + Q(t)x(t − σ ) = 0, t t0 . * Corresponding author.
E-mail address: [email protected] (L. Liao). 0022-247X/03/$ – see front matter 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-247X(03)00021-0
(2)
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
In [1], it was shown that if ∞ Q(s) ds = ∞
327
(3)
t0
and P (t) ≡ p
t and 2p + lim sup t →∞
Q(s) ds < 1,
(4)
t −σ
then every solution of Eq. (2) tends to zero as t → ∞. By making use of the double iterative method, [2] improves (4) to P (t) p
t and 2p(2 − p) + lim sup t →∞
t −σ
3 Q(s) ds < . 2
(5)
We remark that when p = 0, (5) reduces to the following so-called 3/2-asymptotic behavior condition: t 3 (6) Q(s) ds < , lim sup 2 t →∞ t −σ
for the special form of (2) x (t) + Q(t)x(t − σ ) = 0,
t t0 .
(7)
It was pointed out in [3] that the upper bound 3/2 in (6) is the best possible one for Eq. (7), when Q is a piecewise continuous function. As an application of the above result, it is easy to conclude that if (3) holds and t lim sup t →∞
t −σ
3 Q(s) ds < (1 − p), 2
(8)
then every solution of Eq. (2) with P (t) ≡ p ∈ (0, 1) and τ = 0 tends to zero as t → ∞. Clearly, condition (8) suggests that (5) is far from “the best possible” and there is room for improvement. Therefore it is desirable to establish some better results, and this constitutes the purpose of this paper. In this paper, we will establish a set of conditions which guarantee every solution of Eq. (1) tends to zero as t → ∞. When they are applied to (2), which improve (5) to P (t) p < 1 4
t and 2p + lim sup t →∞
t −σ
3 Q(s) ds < , 2
(9)
or 1 1 p< 4 2
t and
lim sup t →∞
t −σ
Q(s) ds <
2(1 − 2p).
(10)
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Equation (1) has been investigated by many authors, see for example [4–6,8,11,12]. For nonlinear equations, one may refer to [9,10].
2. The main results and proofs Theorem 1. Assume that there is a constant c such that Q(t) = Q1 (t) − Q2 (t + σ2 − σ1 ) 0, t P (t) + sign(σ1 − σ2 )Q2 (s) ds c,
(11) (12)
t +σ2 −σ1
and 1 c< 4
t and 2c + t −σ1
3 Q(s) ds , 2
(13)
or 1 1 c< 4 2
t Q(s) ds
and
2(1 − 2c).
(14)
t −σ1
Then every solution of Eq. (1) is bounded. Proof. If c = 0, it was proved in [7]. In the sequel, we assume that 0 < c < 1/2. Let x(t) be a solution of Eq. (1). Set t z(t) = x(t) − P (t)x(t − τ ) −
Q2 (s)x(s − σ2 ) ds.
(15)
t +σ2 −σ1
Then z (t) = −Q(t)x(t − σ1 ),
t t0 .
(16)
We will prove that x(t) is bounded. Indeed, if x(t) is unbounded, then lim supt →∞ |x(t)| = ∞. Then there exists T > t0 such that |x(T )| > |x(t)|, t0 t < T , and |x(T )| > z(t0 + 2σ1 + 2σ2 + 2τ )/(1 − c). Without of loss of generality, we may assume that x(T ) = |x(T )|. Then T Q2 (s)x(s − σ2 ) ds
z(T ) = x(T ) − P (T )x(T − τ ) − T +σ2 −σ1
> (1 − c)x(T ) > z(t0 + 2σ1 + 2σ2 + 2τ ). t∗
(17) z(t ∗ )
It follows from (17) that there exists ∈ (t0 + 2σ1 + 2σ1 + 2τ, T ] such that = max{z(t): t0 t T } and z(t) < z(t ∗ ) for t0 t < t ∗ . Note that 0 < c < 1/2, it is easy
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
329
to see that x(t ∗ ) z(t ∗ ) − cx(T ) > (1 − 2c)x(T ) 0. Next we prove x(t ∗ − σ1 ) 0. Otherwise, x(t ∗ − σ1 ) > 0. Thus, there is a left neighbor of (t ∗ − σ1 ), which is denoted by (t ∗ − σ1 − h, t ∗ − σ1 ) for some h > 0, such that x(t) > 0 for t ∈ (t ∗ − σ1 − h, t ∗ − σ1 ), and so x(t − σ1 ) > 0 for t ∈ (t ∗ − h, t ∗ ). Therefore by (16), we see that z(t) is not increasing on (t ∗ − h, t ∗ ). This contradicts to the definition of t ∗ and so x(t ∗ − σ1 ) 0. Hence, there exists ξ ∈ [t ∗ − σ1 , t ∗ ) such that x(ξ ) = 0. Integrating (16) from t − σ1 to ξ , we obtain for t0 + 2(τ + σ1 + σ2 ) t t ∗ : t−σ1
−x(t − σ1 ) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ
+ P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds + x(T )
Q(s) ds.
t −σ1
ξ +σ2 −σ1
(18) The proof will be complete, if we conclude that z(T ) z(t ∗ ) (1 − c)x(T ),
(19) z(t ∗ )
which is due to the contradiction to the fact that z(T ) > (1 − c)x(T ). There are three possible cases: t∗ Case 1. 0 < c < 1/4 and ξ Q(s) ds 1. In this case, by integrating (16) from ξ to t ∗ and using (13) and (18), we have z(t ∗ ) = z(ξ ) −
t ∗ Q(t)x(t − σ1 ) dt ξ
ξ Q2 (s)x(s − σ2 ) ds
−P (ξ )x(ξ − τ ) − t ∗
ξ +σ2 −σ1 t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
+
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ Q2 (s)x(s − σ2 ) ds
+ P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ
− P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
t ∗ 1−
Q(s) ds ξ
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ + x(T )
t−σ1
Q(t) P (t − σ1 ) +
sign(σ1 − σ2 )Q2 (s) ds
t +σ2 −2σ1
ξ
ξ +
Q(s) ds dt
t −σ1
ξ
x(T ) P (ξ ) +
sign(σ1 − σ2 )Q2 (s) ds
1−
ξ +σ2 −σ1
t∗
Q(s) ds ξ
ξ
Q(t) c +
+ x(T )
t ∗
Q(s) ds dt
t −σ1
ξ
t ∗
x(T ) c 1 −
Q(s) ds ξ
t ∗ +
t
Q(t) c +
Q(s) ds −
t −σ1
ξ
t ∗
x(T ) c +
Q(s) ds dt ξ
3 − 2c − Q(t) 2
ξ
t
t
Q(s) ds dt ξ
3 − 2c = x(T ) c + 2
t ∗
1 Q(s) ds − 2
ξ
2
t ∗ Q(s) ds ξ
(1 − c)x(T ). t∗ t∗ Case 2. 0 < c < 1/4 and ξ Q(s) ds > 1. Choose η ∈ (ξ, t ∗ ) such that η Q(s) ds = 1. Then integrating (16) first from ξ to η and then from η to t ∗ , and using (13) and (18), we have
z(t ∗ ) = z(ξ ) −
t ∗
η Q(t)x(t − σ1 ) dt −
Q(t)x(t − σ1 ) dt η
ξ
ξ −P (ξ )x(ξ − τ ) − ξ +σ2 −σ1
η Q2 (s)x(s − σ2 ) ds + x(T )
Q(s) ds ξ
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ +
331
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
η
ξ Q2 (s)x(s − σ2 ) ds
+ P (ξ )x(ξ − τ ) + ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ
− P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds
1−
ξ +σ2 −σ1
+ x(T )
Q(s) ds + x(T )
t ∗
Q(t) c +
η
t ∗ Q(s) ds +
Q(t) dt
Q(s) ds dt
t −σ1
η
= x(T )
ξ
η
ξ
Q(s) ds η
t ∗
η
t ∗
Q(t) c +
Q(s) ds dt
t −σ1
η
ξ
ξ
η = x(T ) c Q(s) ds + Q(t) Q(s) ds dt t∗
t∗
η
x(T )
η
3 −c 2
t ∗
Q(s) ds − η
= x(T )
t −σ1
t∗
Q(t) η
t∗
3 −c 2
t Q(s) ds dt η
2
t∗
1 Q(s) ds − 2
η
Q(s) ds
= (1 − c)x(T ).
η
t∗ √ Case 3. 1/4 c < 1/2 and ξ Q(s) ds 2(1 − 2c). Integrating (16) from ξ to t ∗ and using (14) and (18), we obtain
∗
t ∗
z(t ) = z(ξ ) −
Q(t)x(t − σ1 ) dt ξ
ξ −P (ξ )x(ξ − τ ) − ξ +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
332
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
t ∗ +
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξ
ξ + P (ξ )x(ξ − τ ) +
Q2 (s)x(s − σ2 ) ds
ξ +σ2 −σ1
ξ + x(T )
Q(s) ds dt
t −σ1
ξ Q2 (s)x(s − σ2 ) ds
− P (ξ )x(ξ − τ ) +
t ∗ + x(T )
ξ +σ2 −σ1 ξ
Q(t) c +
Q(t)
t ∗
Q(s) ds +
t ∗
x(T ) c +
ξ
t
2(1 − 2c) −
ξ
= x(T ) c + 2(1 − 2c)
Q(s) ds dt ξ
Q(s) ds dt ξ
t ∗
Q(s) ds −
t −σ1
t
t
Q(t) c + ξ
Q(s) ds ξ
Q(s) ds dt
t ∗
x(T ) c 1 −
1−
t −σ1
ξ
t ∗
1 Q(s) ds − 2
ξ
2
t ∗ Q(s) ds
(1 − c)x(T ).
ξ
Cases 1, 2, and 3 show that (19) is true. And therefore, the proof is complete. ✷ Theorem 2. Assume that there is a constant c such that (11) and (12) holds, that ∞ Q(s) ds = ∞,
(20)
t0
and 1 c< 4
t and 2c + lim sup t →∞
t −σ1
3 Q(s) ds < , 2
(21)
or 1 1 c< 4 2
t and
lim sup t →∞
Q(s) ds <
2(1 − 2c).
t −σ1
Then every solution of Eq. (1) tends to zero as t → ∞.
(22)
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
333
Proof. Let x(t) be any solution of Eq. (1). We shall prove that lim x(t) = 0,
(23)
t →∞
in two cases where x(t) is nonoscillatory or oscillatory. In the former case, we may assume that x(t) is eventually positive. Let z(t) be defined by (15). It is easy to see that z(t) is eventually nonincreasing. Set β = limt →∞ z(t). Then β is bounded and by (16) ∞ β + Q(s)x(s − σ1 ) ds = z(t0 ), t0
which together with (20) implies lim inft →∞ x(t) = 0. From (16), we have
t lim sup x(t) = β + lim sup P (t)x(t − τ ) + t →∞
t →∞
Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β + lim sup P (t)x(t − τ ) t →∞
t +
sign(σ1 − σ2 )Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β + c lim sup x(t), t →∞
which implies lim sup x(t) t →∞
β . 1−c
On the other hand, lim inf x(t) = β + lim inf P (t)x(t − τ ) + t →∞
t
t →∞
Q2 (s)x(s − σ2 ) ds
t +σ2 −σ1
β − lim sup P (t)x(t − τ ) t →∞
t
sign(σ1 − σ2 )Q2 (s)x(s − σ2 ) ds
+
t +σ2 −σ1
β − c lim sup x(t) t →∞
1 − 2c β. 1−c
This shows β = 0 and so (23) holds. In the latter case, i.e., x(t) is oscillatory. Set µ = lim supt →∞ |x(t)|. Then Theorem 1 implies 0 µ < ∞. Similarly to (17), we have (24) M = lim supz(t) (1 − c)µ. t →∞
334
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
The proof will be finished when we prove µ = √ 0. Suppose that µ > 0. Then for any ε ∈ (0, (1 − 2c)µ), there exist A ∈ (1, 3/2), B ∈ (0, 2(1 − 2c) ) and T > t0 such that x(t) < µ + ε, t T − 2(τ + σ1 + σ2 ), and
t Q(s) ds < t −σ
A − 2c, if c < 1/4, B, if 1/4 c < 1/2,
t T.
(25)
Note that z (t) is oscillatory; there is an increasing sequence {Tn } such that Tn > T + τ + 2σ1 , Tn → ∞, |z(Tn )| → M as n → ∞, |z(Tn )| > (1 − c)(µ − ε), and Tn is left local maximum of |z(t)|. We may assume that z(Tn ) > 0. The case when z(Tn ) < 0 is similar and the proof will be omitted. Thus x(Tn ) z(Tn ) − c(µ + ε) > (1 − 2c)µ − ε > 0. It is also easy to see that x(Tn − σ1 ) 0 in view of the definition of Tn . Hence, there exists ξn ∈ [Tn − σ1 , Tn ) such that x(ξn ) = 0. From (16), we have for Tn − σ1 t Tn t−σ1
−x(t − σ1 ) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds.
(26)
t −σ1
Set
A=
max{A − c − 1/2, c + 1/2}, if c < 1/4, B 2 /2 + c, if 1/4 c < 1/2.
Then A < 1 − c. We will prove z(Tn ) Λ(µ + ε).
(27)
To this end, we consider the following three possible cases: T Case 1. 0 < c < 1/4 and ξnn Q(s) ds 1. In this case, by integrating (16) from ξn to Tn and using (25) and (26), we have Tn z(Tn ) = z(ξn ) −
Q(t)x(t − σ1 ) dt ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
335
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Tn + (µ + ε)
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
Q(t) c +
Q(s) ds ξn
Q(s) ds dt
t −σ1
ξn
Tn
(µ + ε) c 1 −
Q(s) ds ξn
Tn
t
ξn
(µ + ε) c +
Q(s) ds −
t −σ1
Tn
t
Q(t) c +
+
1−
ξn
Tn
Q(s) ds dt ξn
t
Q(t) A − 2c − ξn
Q(s) ds dt ξn
Tn 2
1 Q(s) ds = (µ + ε) c + (A − 2c) Q(s) ds − 2 ξn ξn 1 (µ + ε) c + max{A − 2c, 1} − = Λ(µ + ε). 2 T T Case 2. 0 < c < 1/4 and ξnn Q(s) ds > 1. Choose ηn ∈ (ξn , Tn ) such that ηnn Q(s) ds = 1. Then integrating (16) first from ξn to ηn and then from ηn to Tn , and using (25) and (26), we have
Tn
ηn z(Tn ) = z(ξn ) −
Tn Q(t)x(t − σ1 ) dt −
Q(t)x(t − σ1 ) dt ηn
ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
ηn Q2 (s)x(s − σ2 ) ds + (µ + ε)
Q(s) ds ξn
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X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ηn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
1−
ξn +σ2 −σ1
Q(s) ds + (µ + ε) ηn
ηn
t −σ1
ξn
Q(t) c +
Q(s) ds dt
t −σ1
ηn
ξn
Q(s) ds dt
Tn Q(s) ds +
Q(t) dt
ξn
Q(t) c + ηn
ξn
= (µ + ε)
Tn
Tn
Q(s) ds ηn
ηn + (µ + ε)
Tn
Tn
Tn ηn = (µ + ε) c Q(s) ds + Q(t) Q(s) ds dt ηn
t −σ1
ηn
Tn
Tn Q(s) ds −
(µ + ε) (A − c) ηn
t Q(t)
ηn
Q(s) ds dt ηn
2
1 Q(s) ds = (µ + ε) (A − c) Q(s) ds − 2 ηn ηn
1 Λ(µ + ε). = (µ + ε) A − c − 2 T Case 3. 1/4 c < 1/2 and ξnn Q(s) ds B. Integrating (16) from ξn to Tn and using (25) and (26), we obtain
Tn
Tn
Tn z(Tn ) = z(ξn ) −
Q(t)x(t − σ1 ) dt ξn
ξn −P (ξn )x(ξn − τ ) − ξn +σ2 −σ1
Q2 (s)x(s − σ2 ) ds
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Tn +
337
t−σ1
Q(t) −P (t − σ1 )x(t − σ1 − τ ) −
Q2 (s)x(s − σ2 ) ds
t +σ2 −2σ1
ξn
ξn + P (ξn )x(ξn − τ ) +
Q2 (s)x(s − σ2 ) ds
ξn +σ2 −σ1
ξn + (µ + ε)
Q(s) ds dt
t −σ1
ξn
− P (ξn )x(ξn − τ ) +
Tn + (µ + ε)
ξn +σ2 −σ1
1−
Q(s) ds ξn
ξn
Q(t) c +
Q(s) ds dt
t −σ1
ξn
Tn
(µ + ε) c 1 −
Q(s) ds ξn
Tn
t
Q(t) c +
+
Q2 (s)x(s − σ2 ) ds
Tn
Tn
(µ + ε) c +
Q(s) ds −
t −σ1
ξn
ξn
t
Q(t) B −
t
Q(s) ds dt ξn
Q(s) ds dt ξn
Tn
= (µ + ε) c + B
1 Q(s) ds − 2
ξn
2
Tn Q(s) ds ξn
1 c + B 2 (µ + ε) = Λ(µ + ε). 2
Cases 1, 2, and 3 show that (27) is true. Let n → ∞ and ε → 0 in (27). We see that M = lim sup z(Tn ) Λu < (1 − c)µ, n→∞
which contradicts to (24). Therefore µ = 0, and so the proof is complete. ✷ Applying Theorem 2 to Eq. (2), we have immediately Corollary 1. Assume that there is a constant p such that (3) and (9) or (10) hold. Then every solution of Eq. (2) tends to zero as t → ∞.
338
X. Wang, L. Liao / J. Math. Anal. Appl. 279 (2003) 326–338
Remark 1. Theorem 2 relaxes the usual restrictions on the coefficients (Q1 (t), Q2 (t) 0) and the delays (σ1 σ2 ) in [4–6].
Acknowledgment The authors would like to thank the reviewers and the editor for their valuable suggestions and comments.
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