Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 19, 1-10; http://www.math.u-szeged.hu/ejqtde/

OSCILLATORY AND ASYMPTOTIC BEHAVIOUR OF A NEUTRAL DIFFERENTIAL EQUATION WITH OSCILLATING

COEFFICIENTS

JULIO G. DIX, NIYATI MISRA, LAXMINARAYAN PADHY, RADHANATH RATH,

Abstract. In this paper, we obtain sufficient conditions so that every solution of

(y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)))′ + ∑_{i=1}^{m} q_{i}(t)y(σ_{i}(t)) = f(t)

oscillates or tends to zero as t → ∞. Here the coefficients p_{i}(t), q_{i}(t) and the forcing term f(t) are allowed to oscillate; such oscillation condition in all coefficients is very rare in the literature. Furthermore, this paper provides an answer to the open problem 2.8.3 in [7, p. 57]. Suitable examples are included to illustrate our results.

1. Introduction

During the previous two decades, oscillation of solutions to neutral delay differential equations has been studied extensively. In this article, we extend some results from equation with fixed-sign coefficients to equations with oscillating coefficients. In particular, we obtain sufficient conditions for every solution of the first-order non-homogeneous nonlinear neutral delay differential equation

(y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)))′ + ∑_{i=1}^{m} q_{i}(t)g_{i}(y(σ_{i}(t))) = f(t),  (1.1)

to oscillate or to tend to zero as t tends to infinity. Here f, g, p_{i}, q_{i}, δ_{i}, σ_{i} are continuous, p_{i}, δ_{i} are differentiable, and f, g, p_{i}, q_{i} can assume positive and negative values.

The main motivation of this work is the open problem [7, Problem 2.8.3, p.57]:

Extend the following result to equations with oscillating coefficients. Theorem 2.3.1 in [7]: Under the assumptions that q(t) ≥ 0 and

lim inf_{t→∞} ∫_{t−τ}^{t} q(s)ds > e^{−1}  (1.2)

every solution of

y′(t) + q(t)y(t − τ) = 0, t ≥ t_{0}  (1.3)

oscillates.

EJQTDE, 2008 No. 19, p. 1

2000 Mathematics Subject Classification. 34c10, 34c15, 34k40.

Key words and phrases. Oscillatory solution; neutral differential equation;

asymptotic behaviour.

In most of the references it is assumed that the coefficients p_{i}, q_{i} are positive [4, 8, 9, 10, 11, 12]. However, in [1, 6], p_{i} oscillates, but the sign of q_{i} remains constant. In [3, Theorem 6(ii)], p oscillates, but the proof is wrong because the conditions needed to apply [7, Lemma 2.2] are not met. In [9, Theorem 2.4], p is periodic, oscillates and is restricted by inequalities similar to (H2).

It seems, that (1.1) is least studied when the functions q_{i} oscillate. This is so because the techniques used in the other cases fail. Ladde [5, Theorem 2.2.2] shows that (1.3) has only oscillatory solutions when q(t) > 0 on a sequence of intervals of length 2τ, whose end points approach +∞, and ∫_{t−τ}^{t} q(s) ds > 1/e on the right half of those intervals.

Our approach here is to separate the positive part and the negative part of the function q_{i}. Our assumptions are stated as follows:

(H1) δ_{i}(t) ≤ t, lim_{t→∞} δ_{i}(t) = ∞, σ_{i}(t) ≤ t, lim_{t→∞} σ_{i}(t) = ∞ for all i.

(H2) There exist constants t_{0} ≥ 0, r_{i} ≤ 0 and R_{i} ≥ 0 such that ∑_{i=1}^{n}(R_{i}−r_{i}) < 1

and r_{i} ≤ p_{i}(t) ≤ R_{i} for t ≥ t_{0}. (H3) The functions g_{i} are bounded. (H4) yg_{i}(y) > 0 for y ≠ 0 and i = 1, … m. (H5) ∫_{0}^{∞} ∑_{i=1}^{m} q_{i}^{+}(s) ds = ∞, where q^{+}(t) = max{q(t), 0}. (H6) ∫_{0}^{∞} ∑_{i=1}^{m} q_{i}^{−}(s) ds < ∞, where q^{−}(t) = max{−q(t), 0}. (H7) ∫_{0}^{∞} |f(s)| ds < ∞.

A prototype of a function satisfying (H3)-(H4) is g(u) = ue^{−u^{2}} which decreases for some values of u; therefore the results in [1, 4, 8, 9, 10, 11, 12] can not be applied here.

From the definitions of the functions q^{+}(t) and q^{−}(t), it follows that q^{+}(t) ≥ 0, q^{−}(t) ≥ 0, and q(t) = q^{+}(t) − q^{−}(t). Then using this decomposition, (1.1) can be rewritten as

(y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)))′ + ∑_{i=1}^{m} q_{i}^{+}(t)g_{i}(y(σ_{i}(t))) − ∑_{i=1}^{m} q_{i}^{−}(t)g_{i}(y(σ_{i}(t))) = f(t) .

(1.4)

By a solution y of (1.1), we mean a real-valued function which is continuous and differentiable on some interval [t_{y}, ∞), such that (1.1) is satisfied. As far as existence and uniqueness of solutions we refer the reader to [7]. In this work we assume the existence of solutions and study only their qualitative behaviour.

A solution of (1.1), is said to be oscillatory if it has arbitrarily large zeros. Otherwise it is said to be non oscillatory. In the sequel, unless otherwise specified, when we write a functional inequality, it will be assumed to hold for all sufficiently large values of t.

2. Main results

Theorem 2.1. Under assumptions (H1)–(H7), every solution of (1.1) oscillates or tends to zero as t → ∞.

Proof. We shall show that for every solutions which does not oscillate, approaches zero as t → ∞.

EJQTDE, 2008 No. 19, p. 2 Case 1: There exits t_{0} such that y(t) > 0 for t ≥ t_{0}. If necessary, increment the t_{0} here to exceed the one in (H2), and by (H1), to have

y(δ_{i}(t)) > 0, y(σ_{i}(t)) > 0, for t ≥ t_{0} .  (2.1)

For simplicity of notation, define

z(t) = y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) .

Using that ∫_{0}^{∞} q_{i}^{−} < ∞ and that g_{i} is bounded, we define

w(t) = z(t) + ∫_{t}^{∞} ∑_{i=1}^{m} q_{i}^{−}(s)g_{i}(y(σ_{i}(s))) ds − ∫_{t_{0}}^{t} f(s) ds .

Then using (1.4),

w′(t) = − ∑_{i=1}^{m} q_{i}^{+}g_{i}(y(σ_{i}(t))) .  (2.2)

Since y > 0, by (H4), w′(t) ≤ 0; so that w(t) is non-increasing. Then

w(t_{0}) ≥ w(t) ≥ y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) − ∫_{t_{0}}^{t} f(s) ds .

By (H7) the function ∫_{0}^{t} f(s) ds is bounded, and by (2.1),

w(t_{0}) + sup_{t≥t_{0}} ∫_{t_{0}}^{t} f(s) ds ≥ y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≥ y(t) − ∑_{i=1}^{n} R_{i}y(δ_{i}(t)) . (2.3)

Using a contradiction argument, we prove that y(t) is bounded above (y is continuous on [0, t_{0}] and is bounded below by zero on [t_{0}, ∞)). Assuming that y is unbounded, we define a sequence {t_{k}}_{k=1}^{∞} such that t_{k} → ∞ and y(t_{k}) = max{y(t) : t_{0} ≤ t ≤ t_{k}}. Then y(t_{k}) → ∞ and by (H1), for each i, y(δ_{i}(t_{k})) → ∞ as k → ∞. Since δ_{i}(t) ≤ t, from (2.3), it follows that for each t_{k},

w(t_{0}) + sup_{t≥t_{0}} ∫_{t_{0}}^{t} f(s) ds ≥ (1 − ∑_{i=1}^{n} R_{i})y(t_{k}) .

By (H2), (1−∑_{i=1}^{n} R_{i}) > 0; so that the right-hand side approaches +∞, as k → ∞. This is a contradiction that proves y being bounded.

Since y and p_{i} are bounded functions, so is z. Then by (H6), (H7), it follows that w is bounded. Since w is bounded and non-increasing, it must converge as t → ∞. Also by the definition of w(t), the function z(t) converges. Let

l := lim_{t→∞} z(t) = lim_{t→∞} (y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t))).  (2.4)

Now, using a contradiction argument, we show that lim inf_{t→∞} y(t) = 0. Suppose lim inf_{t→∞} y(t) > 0. Then by (H1), lim inf_{t→∞} y(σ_{i}(t)) > 0. From the definition of lim inf, there exist constants y_{1} and t_{1} such that y(σ_{i}(t)) ≥ y_{1} > 0 for all t ≥ t_{1}, and all 1 ≤ i ≤ m. Let y_{2} be an upper bound for y. Because g_{i}’s are

EJQTDE, 2008 No. 19, p. 3 continuous, by (H4), there exists a positive lower bound m_{1} for all g_{i}’s on [y_{1}, y_{2}]; i.e., 0 < m_{1} ≤ g_{i}(y(σ_{i}(s))) for all s ≥ t_{1}, i = 1, …, m. Then integrating (2.2),

w(t_{1}) − w(t) = ∫_{t_{1}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s)g_{i}(y(σ_{i}(s))) ds ≥ m_{1} ∫_{t_{1}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s) ds .

Since the left-hand side is a bounded function while, by (H5), the right-hand side approaches +∞, we have a contradiction. Therefore, lim inf_{t→∞} y(t) = 0.

Now we prove that lim sup_{t→∞} y(t) = 0. Since y ≥ 0, from assumption (H2), it follows that y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≥ y(t) − ∑_{i=1}^{n} R_{i}y(δ_{i}(t)). Recall that for bounded functions, lim sup{f(t)+g(t)} ≥ lim sup{f(t)}+lim inf{g(t)}. Taking the lim sup in (2.4), we have

l ≥ lim sup_{t→∞}{y(t) + ∑_{i=1}^{n} −R_{i}y(δ_{i}(t))}

≥ lim sup_{t→∞}{y(t)} + ∑_{i=1}^{n} lim inf_{t→∞}{−R_{i}y(t)}

= lim sup_{t→∞}{y(t)} − ∑_{i=1}^{n} R_{i} lim sup_{t→∞}{y(t)}

= (1 − ∑_{i=1}^{n} R_{i}) lim sup_{t→∞}{y(t)} .

(2.5)

In the equality above, we use that −R_{i} ≤ 0.

Since y ≥ 0, from assumption (H2), it follows that y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≤ y(t) − ∑_{i=1}^{n} r_{i}y(δ_{i}(t)). Recall that for bounded functions, lim inf{f(t) + g(t)} ≤ lim inf{f(t)} + lim sup{g(t)}. Taking lim inf in (2.4), we have

l ≤ lim inf_{t→∞}{y(t) + ∑_{i=1}^{n} −r_{i}y(δ_{i}(t))}

≤ lim inf_{t→∞}{y(t)} + ∑_{i=1}^{n} lim sup_{t→∞}{−r_{i}y(t)}

= 0 − ∑_{i=1}^{n} r_{i} lim sup_{t→∞}{y(t)} .

In the the equality above, we use that −r_{i} ≥ 0. From (2.5) and the above inequality,

(1 − ∑_{i=1}^{n}(R_{i} − r_{i})) lim sup_{t→∞}{y(t)} ≤ 0 .

Since y ≥ 0, by (H2), it follows that lim sup_{t→∞}{y(t)} = 0. The proof of case 1 is complete. Case 2: There exists t_{0} such that y(t) < 0 for all t ≥ t_{0}. If necessary increment the t_{0} here to exceed the one in (H2), and by (H1) to have

y(δ_{i}(t)) < 0, y(σ_{i}(t)) < 0, for t ≥ t_{0} .  (2.6)

EJQTDE, 2008 No. 19, p. 4 We define z(t) and w(t) as in case 1. Then w′(t) ≥ 0 and w(t) is non-decreasing; so that

w(t_{0}) ≤ w(t) ≤ y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) − ∫_{t_{0}}^{t} f(s) ds .

By (H2),

w(t_{0}) + inf_{t≥t_{0}} ∫_{t_{0}}^{t} f(s) ds ≤ y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≤ y(t) − ∑_{i=1}^{n} R_{i}y(δ_{i}(t)) . (2.7)

We claim that y(t) is bounded (y is continuous on [0, t_{0}] and is bounded above by zero on [t_{0}, ∞)). On the contrary suppose that y is unbounded, we define a sequence {t_{k}}_{k=1}^{∞} such that t_{k} → ∞ and y(t_{k}) = min{y(t) : t_{0} ≤ t ≤ t_{k}}. Then y(t_{k}) → −∞ and by (H1), for each i, y(δ_{i}(t_{k})) → −∞ as k → ∞. From (2.7), it follows that for each t_{k},

w(t_{0}) + sup_{t≥t_{0}} F (t) ≤ (1 − ∑_{i=1}^{n} R_{i})y(t_{k}) .

By (H2), (1 − ∑_{i=1}^{n} R_{i}) > 0, so that the right-hand side approaches −∞. This contradiction implies y being bounded.

Since y and p_{i} are bounded functions, so is z. Then by (H6), (H7), it follows that w is bounded. Since w is bounded and non-decreasing, it must converge as t → ∞. Also by the definition of w(t), it follows that z(t) converges. Let

l := lim_{t→∞} z(t) = lim_{t→∞} (y(t) − ∑_{i=1}^{n} −p_{i}(t)y(δ_{i}(t))) .  (2.8)

Now, using a contradiction argument, we prove lim sup_{t→∞} y(t) = 0. Suppose lim sup_{t→∞} y(t) < 0. Then by (H1), lim sup_{t→∞} y(σ_{i}(t)) < 0. From the definition of lim sup, there exist constants y_{2} and t_{1} such that y(σ_{i}(t)) ≤ y_{2} < 0 for all t ≥ t_{1}, and all 1 ≤ i ≤ n. Let y_{1} be a lower bound for y. Then there exists a negative upper bound m_{2} for all g_{i}’s on [y_{2}, y_{1}]; i.e., g_{i}(y(σ_{i}(s))) ≤ m_{2} < 0 for all s ≥ t_{1}, i = 1, … m. Then integrating on (2.2),

w(t) − w(t_{1}) = − ∫_{t_{1}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s)g_{i}(y(σ_{i}(s))) ds ≥ −m_{2} ∫_{t_{1}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s) ds .

The left-hand side is a bounded function and, by (H5), the right-hand side approaches +∞ as t → ∞. This contradiction implies lim sup_{t→∞} y(t) = 0.

Now we prove that lim inf_{t→∞} y(t) = 0. Since y ≤ 0, from assumption (H2), it follows that y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≥ y(t) − ∑_{i=1}^{n} r_{i}y(δ_{i}(t)). Recall that for bounded functions, lim sup{f(t)+g(t)} ≥ lim sup{f(t)}+lim inf{g(t)}. Taking the

EJQTDE, 2008 No. 19, p. 5 lim sup in (2.8), we have

l ≥ lim sup_{t→∞}{y(t) + ∑_{i=1}^{n} −r_{i}y(δ_{i}(t))}

≥ lim sup_{t→∞}{y(t)} + ∑_{i=1}^{n} lim inf_{t→∞}{−r_{i}y(t)}

= 0 − ∑_{i=1}^{n} r_{i} lim inf_{t→∞}{y(t)} .

(2.9)

In the equality above, we use that −r_{i} ≥ 0.

Since y ≤ 0, from assumption (H2), it follows that y(t) − ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) ≤ y(t) − ∑_{i=1}^{n} R_{i}y(δ_{i}(t)). Recall that for bounded functions, lim inf{f(t) + g(t)} ≤ lim inf{f(t)} + lim sup{g(t)}. Taking lim inf in (2.4), we have

l ≤ lim inf_{t→∞}{y(t) + ∑_{i=1}^{n} −R_{i}y(δ_{i}(t))}

≤ lim inf_{t→∞}{y(t)} + ∑_{i=1}^{n} lim sup_{t→∞}{−R_{i}y(t)}

= lim inf_{t→∞}{y(t)} + ∑_{i=1}^{n} −R_{i} lim inf_{t→∞}{y(t)}

= (1 − ∑_{i=1}^{n} R_{i}) lim inf_{t→∞}{y(t)} .

For the equality above, we use that −R_{i} ≤ 0. From the (2.9) and the above inequality,

0 ≤ (1 − ∑_{i=1}^{n}(R_{i} − r_{i})) lim inf_{t→∞}{y(t).

Since y ≤ 0, by (H2), it follows that lim inf_{t→∞}{y(t)} = 0. The proof of case 2 is complete.

In summary, every solution does not oscillate approaches zero. □

Note that in the above theorem, (H3) requires g_{i} being bounded. However, the open problem in [7] does not satisfy this condition. To address this shortcoming, we introduce the following hypotheses, and state another theorem.

(H8) There exists a positive constant τ such that δ_{i}(t) ≤ t − τ and σ_{i}(t) ≤ t − τ

for all t ≥ 0 and all i’s. (H9) There exist non-negative constants a, b such that

|g_{i}(u)| ≤ a|u| + b for all u, 1 ≤ i ≤ m .

Theorem 2.2. Assume (H1)-(H2), (H4)-(H9) hold. Then every solution of (1.1) oscillates or tends to zero as t → ∞.

Proof. As in Theorem 2.1, we prove that every solution which does not oscillate, approaches zero as t → ∞.

EJQTDE, 2008 No. 19, p. 6 Case 1 There exists a t_{0} such that y(t) > 0 for t ≥ t_{0}. If necessary increment t_{0} so that (2.1) is satisfied, and, by (H6),

α := (a + b) ∫_{t_{0}}^{∞} ∑_{i=1}^{m} q_{i}^{−}(s) ds < 1 − ∑_{i=1}^{n} R_{i} .  (2.10)

Using that δ_{i}(t) and σ_{i}(t) are continuous and both tend to ∞ as t → ∞, we define the values δ_{0} = inf{δ_{i}(t) : t ≥ t_{0}, 1 ≤ i ≤ n} and σ_{0} = inf{σ_{i}(t) : t ≥ t_{0}, 1 ≤ i ≤ m}. Select a constant M large enough such that

\begin{matrix}1 ≤ M , \\ |y(t)| ≤ M for min\{δ_{0}, σ_{0}\} ≤ t ≤ t_{0} , \\ α + ∑_{i=1}^{n} R_{i} ≤ \frac{M}{M + |y(t_{0})| + | ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))| + ∫_{t_{0}}^{∞} |f|} . \end{matrix}  (2.11)

Such M exists that

0 ≤ y(t) ≤ M + |y(t_{0})| + | ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))| + ∫_{t_{0}}^{∞} |f| := M_{2} for t ≤ t_{0} (2.12)

by (H2), (2.10) and 0 ≤ α + ∑_{i=1}^{n} R_{i} < 1. Then Now for t_{0} ≤ t ≤ t_{0} + τ, we integrate (1.4), to obtain

\begin{aligned}y(t) & = ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t)) + y(t_{0}) − ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0})) \\ & − ∫_{t_{0}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s)g_{i}(y(σ_{i}(s))) ds + ∫_{t_{0}}^{t} ∑_{i=1}^{m} q_{i}^{−}(s)g_{i}(y(σ_{i}(s))) ds + ∫_{t_{0}}^{t} f(s) ds . \end{aligned}

(2.13)

Because of δ_{i}(t) ≤ t − τ and σ_{i}(t) ≤ t − τ, we can use (2.12) to estimate each term in the above expression. Using p_{i}(t) ≤ R_{i}, we obtain

| ∑_{i=1}^{n} p_{i}(t)y(δ_{i}(t))| ≤ ∑_{i=1}^{n} R_{i}M_{2} .

Since yg_{i}(y) > 0, the fourth term on the right-hand side of (2.13) can be estimated by zero. Using that y(σ_{i}(s)) ≤ M_{2} and M_{2} ≥ 1, by (H9), we have |g(y(σ_{i}(s)))| ≤ a|y(σ_{i}(s))| + b ≤ aM_{2} + b ≤ (a + b)M_{2}. Then by (2.10),

| ∫_{t_{0}}^{t} ∑_{i=1}^{m} q_{i}^{−}(s)g_{i}(y(σ_{i}(s))) ds| ≤ αM_{2} .

From the two inequalities above, (2.13), and (2.11) we obtain

0 ≤ y(t) ≤ (α + ∑_{i=1}^{n} R_{i})M_{2} + |y(t_{0})| + | ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))| + ∫_{t_{0}}^{∞} |f| ≤ M_{2}

for t_{0} ≤ t ≤ t_{0} + τ. Recursively, we show that 0 ≤ y(t) ≤ M_{2} on the intervals [t_{0} + τ, t_{0} + 2τ], [t_{0} + 2τ, t_{0} + 3τ], ….

Next we define z(t) and w(t) as in Theorem 2.1. Then prove that lim_{t→∞} y(t) = 0 by the same methods as in the proof of Theorem 2.1.

EJQTDE, 2008 No. 19, p. 7 Case2 There exists t_{0} such that y(t) < 0 for t ≥ t_{0}. The proof is similar to the proof of case 1; so we just sketch it. If necessary increment t_{0} so that (2.6) and (2.10) are satisfied. Let M be defined as in (2.11). Then

0 ≤ −y(t) ≤ M +|y(t_{0})|+| ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))|+∫_{t_{0}}^{∞} |f| := M_{2} for t ≤ t_{0} . (2.14)

Now for t_{0} ≤ t ≤ t_{0} + τ, from (1.4), we have

−y(t) = ∑_{i=1}^{n} p_{i}(t)(−y(δ_{i}(t))) − y(t_{0}) + ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))

+ ∫_{t_{0}}^{t} ∑_{i=1}^{m} q_{i}^{+}(s)g_{i}(y(σ_{i}(s))) ds − ∫_{t_{0}}^{t} ∑_{i=1}^{m} q_{i}^{−}(s)g_{i}(y(σ_{i}(s))) ds − ∫_{t_{0}}^{t} f(s) ds .

Because yg_{i}(y) > 0, the fourth term on the right-hand side can be estimated by zero. Because δ_{i}(t) ≤ t − τ and σ_{i}(t) ≤ t − τ, we can use (2.14) to obtain

0 ≤ −y(t) ≤ (α + ∑_{i=1}^{n} R_{i})M_{2} + |y(t_{0})| + | ∑_{i=1}^{n} p_{i}(t_{0})y(δ_{i}(t_{0}))| + ∫_{t_{0}}^{∞} |f| ≤ M_{2},

for t_{0} ≤ t ≤ t_{0} + τ. Recursively, we show that 0 ≤ −y(t) ≤ M_{2} on the intervals [t_{0} + τ, t_{0} + 2τ], [t_{0} + 2τ, t_{0} + 3τ], …. For the rest of the proof, we proceed as in Theorem 2.1. □

The results in Theorems 2.1 and 2.2 hold for bounded solutions, as follows.

Theorem 2.3. Under assumptions (H1)-(H2), (H4)–(H7), every bounded solution of (1.1) oscillates or tends to zero as t → ∞.

Regarding the open problem in [7], we have the following result.

Corollary 2.4. Assume that

∫_{0}^{∞} q^{+}(s) ds = ∞, and ∫_{0}^{∞} q^{−}(s) ds < ∞ .

Then every solution of (1.3) oscillates or tends to zero as t → ∞.

Proof. The delay equation (1.3) is a particular case of (1.1) where n = m = 1, p_{1}(t) = 0, q_{1}(t) = q(t), g_{1}(u) = u, σ_{1}(t) = t − τ, and f(t) = 0. Condition (H3) is not satisfied, but (H9) is satisfied with a = 1, b = 0. Since (H1)-(H2), (H4)-(H9) are satisfied, we apply Theorem 2.2 and obtain the desired result. □

Remark 2.5. Condition (1.2) implies (H5). In fact, from the definition of lim inf, there exists t_{0} such that for t ≥ t_{0},

∫_{t−τ}^{t} q^{+}(s) ds ≥ ∫_{t−τ}^{t} q(s) ds ≥ \frac{1}{2e} .

Partitioning the interval of integration [t_{0}, ∞) in intervals of length τ, we have

∫_{t_{0}}^{∞} q^{+}(s) ds = ∫_{t_{0}}^{t_{0}+τ} q^{+}(s) ds + ∫_{t_{0}+τ}^{t_{0}+2τ} q^{+}(s) ds + ⋯ ≥ ∑_{i=1}^{∞} \frac{1}{2e} = ∞

which implies (H5). Also note that (1.2) does not imply (H6).

EJQTDE, 2008 No. 19, p. 8 Example. To present an equation where Theorem 2.2 applies, we define

q(t) = ( sin(t))^{+} − \frac{1}{t^{2} + 1}( sin(t))^{−}.

Then q^{+}(t) = ( sin(t))^{+} and ∫_{0}^{∞} q^{+} = ∞; so (H5) is satisfied. Also 0 ≤ q^{−}(t) = \frac{1}{t^{2}+1}( sin(t))^{−} ≤ \frac{1}{t^{2}+1} and ∫_{0}^{∞} q^{−} ≤ π/2; so (H6) is satisfied. Consider the delay equation

y′(t) + q(t)y(t − 1) = e^{−t}(q(t)e − 1) .

Since q is bounded, the right-hand side is integrable, in absolute value, and (H7) is satisfied. In fact, (H1)-(H2), (H4)-(H9) are satisfied and the solution to the above equation is y(t) = e^{−t} which approaches zero as t → ∞.

Example. To emphasize the need for (H3), or for (H6), we present the delay equation

y′(t) − cos(t)y(t − 2π) = 0 ,

where q(t) = − cos(t) which does not satisfy (H6), and g(y) = y which does not satisfy (H3). Note that (H1), (H2), (H4), (H5), (H7)–(H9) hold, but we can not apply Theorem 2.1 or Theorem 2.2. Also note that the solution is y(t) = exp sin(t) which does not oscillate and does not tend to zero as t → ∞.

Acknowledgments. The authors are thankful to the anonymous referee for his/her helpful comments to improve the presentation of this paper.

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Indian Math. Soc. 71 (2004), 175-188.

EJQTDE, 2008 No. 19, p. 9

(Received February 11, 2008)

Julio G. Dix Department of Mathematics, Texas State University, San Marcos Texas 78666, USA

E-mail address: julio@txstate.edu

Niyati Misra Department of Mathematics, Berhampur University, Berhampur 760007, Orissa, India

E-mail address: niyatimath@yahoo.co.in

Laxmi Narayan Padhi Department Of Computer science and Engineering, K.I.S.T, Bhubaneswar Orissa, India

E-mail address: ln padhy 2006@yahoo.co.in

Radhanath Rath Department of Mathematics, Khallikote Autonomous College, Berhampur, 760001 Orissa, India

E-mail address: radhanathmath@yahoo.co.in

EJQTDE, 2008 No. 19, p. 10