1 Introduction

In this article we consider the linear delay differential equation

$$ x'(t)+p(t)x\bigl(\tau (t)\bigr)=0,\quad t\ge t_{0}, $$
(1)

where the functions \(p, \tau \in C([t_{0},\infty),(0,\infty)), \tau (t)\) is nondecreasing, \(\tau (t)< t\) for \(t\ge t_{0}\) and \(\lim_{t\to \infty }\tau (t)=\infty \).

Our aim is to establish new sufficient conditions for the oscillation of all solutions of Eq. (1). This problem has been recently investigated by many authors. See, for example, [1–14] and the references cited therein.

A continuously differentiable function defined on \([\tau (T _{0}),\infty)\) for some \(T_{0}\ge t_{0}\) and satisfying Eq. (1) for \(t\ge T_{0}\) is called a solution of Eq. (1). Such a solution is called oscillatory if it has arbitrarily large zeros. Otherwise, it is called nonoscillatory.

We will denote by k the lower limit of the average

$$\int_{\tau (t)}^{t} p(s)\,ds $$

as \(t\to \infty \), i.e.,

$$k=\liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\,ds, $$

and constant L is defined by

$$L=\limsup_{t\to \infty } \int_{\tau (t)}^{t} p(s)\,ds. $$

We refer the readers to the papers [6, 13] for the historical and chronological review of the results. We mention some results in the literature (cf. [6, 13]) for the purpose of this article. In 1972 Ladas, Lakshmikantham, and Papadakis [11] proved that if the following holds

$$ L=\limsup_{t\to \infty } \int_{\tau (t)}^{t} p(s)\,ds>1, $$
(2)

then every solution of Eq. (1) oscillates.

In 1979 Ladas [10] and in 1982 Koplatadze and Chanturija [7] showed that the same conclusion holds if

$$ k=\liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\,ds>\frac{1}{e}. $$
(3)

We point out that if the inequality

$$\int_{\tau (t)}^{t} p(s)\,ds\le \frac{1}{e} $$

holds eventually, then, according to a result in [7], Eq. (1) has a nonoscillatory solution.

It is obvious that there is a gap between conditions (2) and (3) when the limit

$$\lim_{t\to \infty } \int_{\tau (t)}^{t} p(s)\,ds $$

does not exist. How to fill this gap is an interesting problem which has been recently investigated by several authors, e.g., [1–6, 8, 9, 12–15].

In the case \(0< k\le 1/e\), all conditions in the papers [4–6, 8, 9, 12–14] are dependent on \(0< L<1\). The aim of this article is to establish such conditions for oscillation of solutions of Eq. (1) which are independent of L.

We assume for the analysis of asymptotic behavior of the function

$$w(t)=\frac{x(\tau (t))}{x(t)} $$

that Eq. (1) has a solution \(x(t)\) which is positive for all large t.

In the second section we will use the next lemma by Jaroš and Stavroulakis [5].

Lemma 1.1

([5])

Suppose that \(k>0\) and Eq. (1) has an eventually positive solution \(x(t)\). Then \(k\le 1/e\) and

$$\lambda_{1}\le \liminf_{t\to \infty }w(t)\le \lambda_{2}, $$

where \(\lambda_{1}\) is the smaller and \(\lambda_{2}\) is the greater root of the equation \(\lambda =e^{k\lambda }\).

2 Oscillatory properties

In this section we will study the oscillatory properties of Eq. (1).

Lemma 2.1

Let \(x(t)\) be an eventually positive solution of Eq. (1) and \(0< k\le \frac{1}{e}\). Suppose that

$$ \liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda_{1} \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds>1-\frac{1}{\beta }, $$
(4)

where \(\beta \in [\lambda_{1}, \lambda_{2}]\), and \(\lambda_{1}\) is the smaller and \(\lambda_{2}\) is the greater root of the equation \(\lambda =e^{k\lambda }\). Then

$$\liminf_{t\to \infty }\frac{x(\tau (t))}{x(t)}>\beta. $$

Proof

Let \(t>t_{0}\) be large enough so that \(\tau (t)>t_{0}\). Integrating (1) from \(\tau (t)\) to t, we obtain

$$ x\bigl(\tau (t)\bigr)=x(t)+ \int_{\tau (t)}^{t} p(s)x\bigl(\tau (s)\bigr)\,ds. $$
(5)

Let \(0<\lambda <\lambda_{1}\). Then the function

$$ \varphi (t)=x(t)\exp \biggl(\lambda \int_{t_{0}}^{t} p(s)\,ds \biggr), \quad t\ge t_{1}, $$
(6)

is decreasing for appropriate \(t_{1}\ge t_{0}\) (cf. [6, 13]). Indeed by Lemma 1.1

$$\frac{x(\tau (t))}{x(t)}>\lambda $$

for \(t\ge t_{2}\), where \(t_{2}\ge t_{1}\) is sufficiently large, and consequently,

$$0=x'(t)+p(t)x\bigl(\tau (t)\bigr)>x'(t)+\lambda p(t)x(t), $$

which implies \(\varphi '(t)<0\) for \(t\ge t_{2}\). Substituting (6) into (5), we derive for \(t\ge t_{2}\) that

$$\begin{aligned}& \begin{aligned} &x\bigl(\tau (t)\bigr) = x(t)+ \int_{\tau (t)}^{t} p(s)\varphi \bigl(\tau (s)\bigr)\exp \biggl(-\lambda \int_{t_{0}}^{\tau (s)} p(u)\,du \biggr)\,ds, \\ &x\bigl(\tau (t)\bigr) \ge x(t)+\varphi \bigl(\tau (t)\bigr) \int_{\tau (t)}^{t} p(s) \exp \biggl(-\lambda \int_{t_{0}}^{\tau (s)} p(u)\,du \biggr)\,ds, \\ &x\bigl(\tau (t)\bigr) \ge x(t)+x\bigl(\tau (t)\bigr)\exp \biggl(\lambda \int_{t_{0}}^{ \tau (t)}p(u)\,du \biggr) \\ &\hphantom{x\bigl(\tau (t)\bigr) \ge}{}\times \int_{\tau (t)}^{t} p(s)\exp \biggl(-\lambda \int_{t_{0}}^{ \tau (s)} p(u)\,du \biggr)\,ds, \\ &x\bigl(\tau (t)\bigr) \ge x(t)+x\bigl(\tau (t)\bigr) \int_{\tau (t)}^{t} p(s)\exp \biggl( \lambda \int_{\tau (s)}^{\tau (t)} p(u)\,du \biggr)\,ds, \\ &0 \ge x(t)+x\bigl(\tau (t)\bigr) \biggl(-1+ \int_{\tau (t)}^{t} p(s)\exp \biggl( \lambda \int_{\tau (s)}^{\tau (t)} p(u)\,du \biggr)\,ds \biggr). \end{aligned} \end{aligned}$$
(7)

From (4) it follows that there exists a constant c such that \(c>1-\frac{1-\varepsilon }{\beta }\), where \(0<\varepsilon <\beta [c-(1-\frac{1}{ \beta })]\leq 1\), and

$$\begin{aligned} \liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda_{1} \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds\geq c>1- \frac{1-\varepsilon }{\beta } . \end{aligned}$$
(8)

Then, for λ sufficiently close to \(\lambda_{1}\), we get

$$\int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda \int_{\tau (s)}^{\tau (t)} p(u)\,du \biggr)\,ds\ge 1- \frac{1-\varepsilon }{\beta },\quad t\ge t_{3}, $$

where \(t_{3}\ge t_{2}\) is sufficiently large. If it is not true, then for all \(0<\lambda <\lambda_{1}\) we have

$$\liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds\leq {1- \frac{1-\varepsilon }{\beta }}< c. $$

By letting \(\lambda \rightarrow \lambda_{1}\), the last inequality leads to

$$\liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda_{1} \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds< c. $$

This inequality contradicts (8). Thus we obtain from (7)

$$\begin{aligned}& 0 \ge x(t)-\frac{1-\varepsilon }{\beta }x\bigl(\tau (t)\bigr), \\& 0\ge 1-\frac{1-\varepsilon }{\beta } \frac{x(\tau (t))}{x(t)}, \\& \frac{x(\tau (t))}{x(t)} \ge \frac{\beta }{1-\varepsilon }> \frac{(1- \varepsilon)\beta }{1-\varepsilon }=\beta,\quad t\ge t_{3}. \end{aligned}$$

Therefore, we have

$$\liminf_{t\to \infty }\frac{x(\tau (t))}{x(t)}>\beta. $$

The proof is complete. □

Theorem 2.1

Let \(0< k\le \frac{1}{e}\). Suppose that

$$\liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda_{1} \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds>1-\frac{1}{\lambda_{2}} . $$

Then all solutions of Eq. (1) oscillate.

Proof

Assume that Eq. (1) eventually has a positive solution \(x(t)\). It follows from Lemma 2.1 that

$$\liminf_{t\to \infty }\frac{x(\tau (t))}{x(t)}>\lambda_{2}. $$

This contradicts the result of Lemma 1.1 and completes the proof of the theorem. □

Theorem 2.2

Let \(0< k\le \frac{1}{e}\). Suppose that there exists \(\gamma \in ( \lambda_{1},\lambda_{2})\) such that

$$\begin{aligned}& \liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\lambda_{1} \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds>1-\frac{1}{\gamma } , \end{aligned}$$
(9)
$$\begin{aligned}& \liminf_{t\to \infty } \int_{\tau (t)}^{t} p(s)\exp \biggl(\gamma \int_{\tau (s)}^{\tau (t)}p(u)\,du \biggr)\,ds>1-\frac{1}{\lambda_{2}} . \end{aligned}$$
(10)

Then all solutions of Eq. (1) oscillate.

Proof

Assume that Eq. (1) has an eventually positive solution \(x(t)\). By Lemma 2.1, condition (9) implies that

$$\liminf_{t\to \infty }\frac{x(\tau (t))}{x(t)}>\gamma. $$

With regard to condition (10) and repeating the procedure as in the proof of Lemma 2.1, we get

$$\liminf_{t\to \infty }\frac{x(\tau (t))}{x(t)}>\lambda_{2}. $$

This contradicts the result of Lemma 1.1. The proof is complete. □

In the next example we observe the case \(k=1/e\). Then \(\lambda_{1}= \lambda_{2}=e\).

Example

([6, 13])

Consider the linear delay differential equation

$$ x'(t)+px \biggl(t-a\sin^{2}\sqrt{t}-\frac{1}{pe} \biggr)=0,\quad t\ge t _{0}, $$
(11)

where \(p>0\), \(a>0\).

Then

$$\begin{aligned} k =& \liminf_{t\to \infty } \int_{\tau (t)}^{t} p \,ds= \liminf_{t\to \infty }p \biggl(a\sin^{2}\sqrt{t}+\frac{1}{pe} \biggr) \\ =&\liminf_{t\to \infty } \biggl(pa\sin^{2}\sqrt{t}+ \frac{1}{e} \biggr)= \frac{1}{e} . \end{aligned}$$

We get

$$\begin{aligned} \int_{\tau (t)}^{t} p \exp \biggl(e \int_{\tau (s)}^{\tau (t)}p \,du \biggr)\,ds =&p \int_{\tau (t)}^{t}\exp \bigl(pe\bigl(\tau (t)-\tau (s) \bigr) \bigr)\,ds \\ =&p \exp \bigl(pe\tau (t) \bigr) \int_{\tau (t)}^{t}\exp \bigl(-pe\tau (s) \bigr)\,ds \\ =&p \exp \bigl(pe\tau (t) \bigr) \int_{\tau (t)}^{t}\exp \biggl(-pe \biggl(s-a \sin^{2}\sqrt{s}-\frac{1}{pe} \biggr) \biggr)\,ds \\ =&ep \exp \bigl(pe\tau (t)\bigr) \int_{\tau (t)}^{t}\exp (-pes)\exp \bigl(ape\sin ^{2}\sqrt{s}\bigr)\,ds. \end{aligned}$$

Then we have

$$\begin{aligned}& \liminf_{t\to \infty }ep \exp \bigl(pe\tau (t)\bigr) \int_{\tau (t)}^{t} \exp (-pes)\exp \bigl(ape \sin^{2}\sqrt{s}\bigr)\,ds \\& \quad >\liminf_{t\to \infty }ep \exp \bigl(pe\tau (t)\bigr) \int_{\tau (t)}^{t} \exp (-pes)\,ds \\& \quad =\liminf_{t\to \infty }\exp \bigl(pe\tau (t)\bigr) \bigl(\exp \bigl(-pe\tau (t)\bigr)-\exp (-pet)\bigr) \\& \quad =\liminf_{t\to \infty }\bigl(1-\exp \bigl(pe\bigl(\tau (t)-t\bigr) \bigr)\bigr)=\liminf_{t\to \infty }\bigl(1-\exp \bigl(-ape \sin^{2}\sqrt{t}-1\bigr)\bigr) \\& \quad =\liminf_{t\to \infty }\biggl(1-\frac{1}{e}\exp \bigl(-ape \sin^{2}\sqrt{t}\bigr)\biggr)= 1-\frac{1}{e} . \end{aligned}$$

We conclude that

$$\liminf_{t\to \infty } \int_{\tau (t)}^{t} p \exp \biggl(e \int_{\tau (s)} ^{\tau (t)}p \,du \biggr)\,ds>1-\frac{1}{e} , $$

and by Theorem 2.1 all solutions of Eq. (11) oscillate.

We point out that the results in [6, 13] are dependent on the constant \(0< L<1\), while our results do not depend on the constant L.