On the qualitative behavior of the solutions to second-order neutral delay differential equations

Abstract

Differential equations of second order appear in numerous applications such as fluid dynamics, electromagnetism, quantum mechanics, neural networks and the field of time symmetric electrodynamics. The aim of this work is to establish necessary and sufficient conditions for the oscillation of the solutions to a second-order neutral differential equation. First, we have taken a single delay and later the results are generalized for multiple delays. Some examples are given and open problems are presented.

1 Introduction

Consider the class of nonlinear neutral delay differential equations of the form

$$ \bigl(a\bigl(w'\bigr)^{\mu } \bigr)'(y)+c(y)g \bigl(u\bigl(\varsigma (y)\bigr) \bigr)=0, $$

(1)

where $w(y)=u(y)+b(y)u(\vartheta (y))$ and μ is the ratio of two odd positive integers. We assume the following conditions hold.

• (A1) $a, c, \vartheta, \varsigma \in C (\mathbb{R_{+}},\mathbb{R_{+}})$ such that $\vartheta (y)\leq y$, $\varsigma (y)\leq y$ for $y \geq y_{0}$, $\vartheta (y) \to \infty$, $\varsigma (y) \to \infty$ as $y \to \infty$.

• (A2) $g \in C(\mathbb{R,\mathbb{R}})$ is non-decreasing and odd with $ug(u)>0$ for $u\neq 0$.

• (A3) $a(y)>0$ and $\int _{0}^{\infty } (a(\eta ) )^{-1/\mu }\,d\eta =\infty$. By letting $A(y)=\int _{0}^{y} (a(\eta ) )^{-1/\mu }\,d\eta$, we have $\lim_{y \to \infty } A(y)=\infty$.

• (A4) $b \in C(\mathbb{R_{+}},\mathbb{R_{-}})$ with $-1+(2/3)^{1/\mu } \leq -b_{0} \leq b(y) \leq 0$ for $y \in \mathbb{R_{+}}$.

• (A5) $b \in C(\mathbb{R_{+}},\mathbb{R_{-}})$ with $-1 <-b_{0} \leq b(y) \leq 0$ for $y \in \mathbb{R_{+}}$.

In 1978, Brands [1] showed that the solutions to

$$ u''(y)+c(y)u\bigl(y-\varsigma (y)\bigr) =0 $$

are oscillatory, if and only if, the solutions to $u''(y)+c(y)u(y) =0$ are oscillatory. Baculikova et al. [2] considered (1) and studied the oscillatory behavior of (1) for $g(u)=u$, $0\leq {}b(y)\leq {}b_{0}<\infty$ and (A3). They obtained sufficient conditions for the oscillation of the solutions of the linear counterpart of (1), using comparison techniques. Chatzarakis et al. [3] considered the equation

$$\begin{aligned} \bigl(a\bigl(u^{\prime }\bigr)^{\mu _{2}} \bigr)^{\prime }(y)+c(y)u^{{\mu _{2}}}\bigl( \varsigma (y)\bigr)=0. \end{aligned}$$

(2)

Also, Chatzarakis et al. [4] studied (2) to obtain new oscillation criteria. Džurina [5] studied the linear counterpart of (1) when $0\leq b(y)\leq b_{0}<\infty$ and (A3) and established sufficient conditions for the oscillation of the solutions of the linear counterpart of (1) by comparison techniques. Karpuz et al. [6] studied (1) for various ranges of the neutral coefficient b. Pinelas and Santra [7] studied necessary and sufficient conditions for the solutions of

$$ \bigl(u(y)+b(y)u(y-\vartheta ) \bigr)^{\prime }+\sum _{i=1}^{m}c_{j}(y)g \bigl(u(y-\varsigma _{j}) \bigr)=0. $$

Wong [8] obtained necessary and sufficient conditions for the oscillation of

$$ \bigl(u(y)+bu(y-\vartheta ) \bigr)''+ c(y)g(y- \varsigma )=0, $$

where the constant b satisfies $-1< b<0$. Grace et al. [9] studied (1) and established sufficient conditions for $0 \leq b(y) <1$. For further work on this type of equations, we refer the reader to [10–36] and the references cited therein. We may note that most of the authors considered only sufficient conditions, and only a few considered necessary and sufficient conditions. Hence, the objective of this work is to establish both necessary and sufficient conditions for oscillation of (1) without using comparison techniques.

In Sect. 2 some preliminary results are presented, Sect. 3 deals with main results, Sect. 4 represents the conclusion and the final section includes open problems.

2 Preliminary results

In this section, two lemmas are presented which we need for our work in the sequel.

Lemma 2.1

Under the assumptions (A1)–(A3) and (A4) or (A5) and the solution u of (1) is an eventually positive solution, we have

• (i) $w(y)<0$, $w^{\prime }(y)>0$ and $(a(w^{\prime })^{\mu })^{\prime }(y)<0$;

• (ii) $w(y)>0$, $w^{\prime }(y)>0$ and $(a(w^{\prime })^{\mu })^{\prime }(y)<0$,

for sufficiently large y.

Proof

Assume there exists a $y_{1} \geq {}y_{0}$ such that $u(y)>0$, $u(\vartheta (y))$, and $u(\varsigma (y))>0$ for $y\geq {}y_{1}$. From (1) and (A2), we have

$$ \bigl(a\bigl(w^{\prime }\bigr)^{\mu } \bigr)^{\prime }(y)=-c(y)g \bigl(u\bigl(\varsigma (y)\bigr) \bigr)< 0 \quad\text{for } y\geq {}y_{1}, $$

(3)

which implies that $(a(w^{\prime })^{\mu } )(y)$ is non-increasing on $[y_{1},\infty )$. We have $a(y)>0$, and thus either $w^{\prime }(y)<0$ or $w^{\prime }(y)>0$ for $y\geq {}y_{2}$, where $y_{2}\geq {}y_{1}$.

If $w^{\prime }(y)>0$ for $y\geq {}y_{2}$, then we have (i) and (ii). We prove now that $w^{\prime }(y)<0$ cannot occur.

If $w^{\prime }(y)<0$ for $y\geq {}y_{2}$, then there exists $\kappa _{1}>0$ such that $(a(w^{\prime })^{\mu } )(y)\leq -\kappa _{1}$ for $y\geq {}y_{2}$, which yields upon integration over $[y_{2},y)\subset [y_{2},\infty )$ after dividing through by a

$$ w(y)\leq {}w(y_{2})-\kappa _{1}^{1/\mu } \int _{y_{2}}^{y} \bigl(a(\eta ) \bigr)^{-1/\mu }\,d \eta \quad\text{for } y\geq {}y_{2}. $$

(4)

By virtue of condition (A3), $\lim_{t\to \infty }w(y) =-\infty$. We consider the following possibilities:

Let the solution u be unbounded. There exists a sequence $\{y_{k}\}$ such that $\lim_{k \to \infty } y_{k} = \infty$ and $\lim_{k\to \infty } u(y_{k}) =\infty$, where $u(y_{k}) = \max \{u(\eta ): y_{0} \leq \eta \leq y_{k}\}$. Since $\lim_{y \to \infty } \vartheta (y) = \infty$, $\vartheta (y_{k}) > y_{0}$ for all sufficiently large k. By $\vartheta (y) \leq y$,

$$ u \bigl(\vartheta (y_{k}) \bigr) = \max \bigl\{ u(\eta ): y_{0} \leq \eta \leq \vartheta (y_{k})\bigr\} \leq \max \bigl\{ u(\eta ): y_{0} \leq \eta \leq y_{k} \bigr\} = u(y_{k}). $$

Therefore, for all large k,

$$ w(y_{k}) = u(y_{k}) + b(y_{k})u \bigl(\vartheta (y_{k}) \bigr) \geq \bigl(1+ b(y_{k})\bigr)u(y_{k}) > 0, $$

which contradicts $\lim_{y \to \infty } w(y) = -\infty$.

Let the solution u be bounded, then w is bounded, from which one concludes $\lim_{y \to \infty } w(y) = -\infty$, a contradiction. Hence, w satisfies one of the cases (i) or (ii). This completes the proof. □

Lemma 2.2

Under the assumptions (A1)–(A3), (A4) or (A5), (i) and u is an eventually positive solution of (1), we have $\lim_{ y \to \infty }u(y)=0$.

Proof

Assume that there exists a $y_{1} \geq {}y_{0}$ such that $u(y)>0$, $u(\vartheta (y))$, and $u(\varsigma (y))>0$ for $y\geq {}y_{1}$. Then Lemma 2.1 holds and w satisfies one of the cases (i) or (ii) for $y_{2} \geq y_{1}$, where $y \geq y_{2}$. Let w satisfy (i) for $y \geq y_{2}$. Therefore,

$$\begin{aligned} 0&\geq \lim_{y \to \infty }w(y)= \limsup_{y\to \infty } w(y) \geq \limsup_{y \to \infty } \bigl(u(y)-b_{0} u\bigl(\vartheta (y)\bigr) \bigr) \\ &\geq \limsup_{y \to \infty } u(y)+\liminf_{t\to \infty } \bigl(-b_{0} u\bigl(\vartheta (y)\bigr) \bigr) = (1-b_{0}) \limsup_{y \to \infty } u(y), \end{aligned}$$

which implies that $\limsup_{y \to \infty } u(y)=0$ and hence $\lim_{y \to \infty }u(y)=0$. □

Remark 1

In view of (ii) of Lemma 2.1, it is obvious that $\lim_{y\to \infty }w(y)>0$, i.e., there exists $\kappa _{1}>0$ such that $w(y)\geq \kappa _{1}$ for all large y.

3 Main results

In this section, we establish the necessary and sufficient conditions for the oscillation of the solution of (1) by considering the two cases when $g(v)/v^{\mu _{1}}$ is non-increasing and $g(v)/v^{\mu _{1}}$ is non-decreasing.

3.1 The case when $g(v)/v^{\mu _{1}}$ is non-increasing

Suppose that there exists ${\mu _{1}}$ such that $0<{\mu _{1}}<\mu$ and

$$ \frac{g(v)}{v^{\mu _{1}}} \geq \frac{g(u)}{u^{\mu _{1}}} \quad\text{for } 0< v \leq u. $$

(5)

For example the function $g(u)=|u|^{\mu _{2}} \operatorname{sgn}(u)$ with $0<{\mu _{2}}<{\mu _{1}}<\mu$ satisfying (5).

Theorem 3.1

Assume that (A1)–(A4) and (5) hold. Then each unbounded solution of (1) is oscillatory if and only if

$$\begin{aligned} \int _{Y}^{\infty }c(\eta )g \bigl(\kappa ^{1/ \mu } A\bigl(\varsigma (\eta )\bigr) \bigr)\,d\eta =+\infty \quad\forall Y>0\textit{ and }\kappa >0. \end{aligned}$$

(6)

Proof

On the contrary, we assume that there exists a nonoscillatory unbounded solution $u(y)$ of (1). Suppose that the solution $u(y)$ is eventually positive. Then there exists $y_{1} \geq y_{0}$ such that $u(y) > 0$, $u(y)>0$, $u(\vartheta (y))>0$ and $u(\varsigma (y))>0$ for $y\geq {}y_{1}$. Proceeding as in the proof of Lemma 2.1, we see that $(a(w')^{\mu } )(y)$ is non-increasing, and w satisfies one of the cases (i) or (ii) on $[y_{2},\infty )$, where $y_{2}\geq {}y_{1}$. Then we have the following two possible cases.

Case 1. Let w satisfy (i) for $y\geq {}y_{2}$. As u is the unbounded solution, there exists $y\geq {}y_{2}$ such that $u(y)=\max \{u(s): y_{2}\leq s\leq {}T\}$. Since $w(y)=u(y)+b(y)u(\vartheta (y))$, we have $u(y)\leq {}w(y)+\{1-(2/3)^{1/\mu }\}u(\vartheta (y))< u(y)$, which leads a contradiction.

Case 2. Let w satisfy (ii) for $y\geq y_{2}$. Note that $\lim_{y \to \infty } (a(w')^{\mu } )(y)$ exists. Using $w(y) \leq u(y)$ in (1) and integrating the new inequality from y to +∞, we obtain

$$ \int _{y}^{\infty }c(\eta )g \bigl(w\bigl(\varsigma (\eta ) \bigr) \bigr)\,d\eta \leq \bigl(a\bigl(w'\bigr)^{\mu } \bigr) (y). $$

That is,

$$ w'(y)\geq \biggl[\frac{1}{a(y)} \int _{y}^{\infty }c(\eta )g \bigl(w\bigl( \varsigma (\eta )\bigr) \bigr)\,d\eta \biggr]^{1/\mu } $$

(7)

for $y\geq y_{3}$. Let $y_{4}> y_{3}$ be a point such that

$$ A(y)-A(y_{3})\geq \frac{1}{2}A(y),\quad y \geq y_{4}. $$

Then integrating (7) from $y_{3}$ to y, we get

$$\begin{aligned} w(y)-w(y_{3})& \geq \int _{y_{3}}^{y} \biggl[\frac{1}{a(\eta )} \int _{ \eta }^{\infty }c(\zeta )g \bigl(w\bigl(\varsigma ( \zeta )\bigr) \bigr)\,d\zeta \biggr]^{1/ \mu } \,d\eta \\ & \geq \int _{y_{3}}^{y} \biggl[\frac{1}{a(\eta )} \int _{y}^{\infty }c( \zeta )g \bigl(w\bigl(\varsigma ( \zeta )\bigr) \bigr)\,d\zeta \biggr]^{1/\mu }\,d\eta, \end{aligned}$$

i.e.,

$$\begin{aligned} w(y) & \geq \bigl(A(y)-A(y_{3}) \bigr) \biggl[ \int _{y}^{\infty } c(\zeta )g \bigl(w\bigl(\varsigma ( \zeta )\bigr) \bigr)\,d\zeta \biggr]^{1/\mu } \\ &\geq \frac{1}{2}A(y) \biggl[ \int _{y}^{\infty }c(\zeta )g \bigl(w\bigl( \varsigma ( \zeta )\bigr) \bigr)\,d\zeta \biggr]^{1/\mu }. \end{aligned}$$

(8)

Since $(a(w')^{\mu } )(y)$ is non-increasing on $[y_{4},\infty )$, there exist $\kappa >0$ and $y_{5}> y_{4}$ such that $(a(w')^{\mu } )(y) \leq \kappa$ for $y\geq y_{5}$. Integrating the inequality $w'(y) \leq (\kappa / a(y))^{1/\mu }$, we have

$$ w(y)\leq w(y_{5})+\kappa ^{1/\mu } \bigl(A(y)-A(y_{5}) \bigr). $$

Since $\lim_{t\to \infty }A(y)=\infty$, the last inequality becomes

$$\begin{aligned} w(y) \leq \kappa ^{1/\mu } A(y) \quad\text{for } y\geq y_{5}. \end{aligned}$$

On the other hand, (5) implies that

$$\begin{aligned} g \bigl(w\bigl(\varsigma (\zeta )\bigr) \bigr)= \frac{g (w(\varsigma (\zeta )) )}{w^{{\mu _{1}}} (\varsigma (\zeta ) )} w^{{\mu _{1}}} \bigl(\varsigma (\zeta ) \bigr)\geq \frac{ g (\kappa ^{1/\mu } A(\varsigma (\zeta )) )}{ (\kappa ^{1/\mu }A(\varsigma (\zeta )) )^{{\mu _{1}}}} {w^{{\mu _{1}}} \bigl(\varsigma (\zeta ) \bigr)}. \end{aligned}$$

Consequently, (8) becomes

$$ w(y)\geq \frac{A(y)}{2} \biggl[ \int _{y}^{\infty } \frac{c(\zeta )g (\kappa ^{1/\mu } A(\varsigma (\zeta )) )w^{\mu _{1}}(\varsigma (\zeta ))}{ (\kappa ^{1/\mu }A(\varsigma (\zeta )) )^{{\mu _{1}}}}\,d \zeta \biggr]^{1/\mu }. $$

If we define

$$ \Upsilon (y)= \int _{y}^{\infty } \frac{c(\zeta )g (\kappa ^{1/\mu } A(\varsigma (\zeta )) )w^{\mu _{1}}(\varsigma (\zeta ))}{ (\kappa ^{1/\mu }A(\varsigma (\zeta )) )^{{\mu _{1}}}}\,d \zeta, $$

then $w^{\mu _{1}} / (\kappa ^{1/\mu }A )^{\mu _{1}} \geq \Upsilon ^{{ \mu _{1}}/\mu }/ (2\kappa ^{1/\mu } )^{\mu _{1}}$. Taking the derivative of ϒ we get

$$\begin{aligned} \Upsilon '(y) \leq - \frac{g (\kappa ^{1/\mu } A(\varsigma (y)) )c(y)w^{{\mu _{1}}}(\varsigma (y))}{ (\kappa ^{1/\mu }A(\varsigma (y)) )^{{\mu _{1}}}} \leq - \frac{c(y)g (\kappa ^{1/\mu } A(\varsigma (y)) )}{ (2\kappa ^{1/\mu } )^{\mu _{1}}} \Upsilon ^{{\mu _{1}}/\mu }\bigl(\varsigma (y)\bigr) \leq 0. \end{aligned}$$

Therefore, $\Upsilon (y)$ is non-increasing on $[y_{5}, \infty )$ so $\Upsilon ^{{\mu _{1}}/\mu }(\varsigma (y))/\Upsilon ^{{\mu _{1}}/\mu }(y) \geq 1$, and

$$\begin{aligned} \bigl(\Upsilon ^{1-{\mu _{1}}/\mu }(y) \bigr)' & \leq - (1-{\mu _{1}}/ \mu ) \Upsilon ^{-{\mu _{1}}/\mu }(y) \frac{c(y)g (\kappa ^{1/\mu } A(\varsigma (y)) )}{ (2\kappa ^{1/\mu } )^{\mu _{1}}} \Upsilon ^{{\mu _{1}}/\mu } \bigl(\varsigma (y) \bigr) \\ & \leq -(1-{\mu _{1}}/\mu ) \frac{c(y)g (\kappa ^{1/\mu } A(\varsigma (y)) )}{ (2\kappa ^{1/\mu } )^{\mu _{1}}}. \end{aligned}$$

We have ${\mu _{1}}/\mu <1$ and $\Upsilon (y)$ is positive and non-increasing. Integrating the last inequality, from $y_{5}$ to y, we have

$$\begin{aligned} \frac{(1-{\mu _{1}}/\mu )}{(2 \kappa ^{1/\mu })^{{\mu _{1}}}} \int _{t^{5}}^{y}c( \eta )g \bigl(\kappa ^{1/\mu } A\bigl(\varsigma (\eta )\bigr) \bigr)\,d\eta \leq - \bigl[\Upsilon ^{1-{\mu _{1}}/\mu }(\eta ) \bigr]_{y_{5}}^{y} < \Upsilon ^{1-{ \mu _{1}}/\mu }(y_{5})< \infty, \end{aligned}$$

which contradicts (6).

If $u(y)<0$ for $y\geq {}y_{1}$, then we set $y(y):=-u(y)$ for $y\geq {}y_{1}$ in (1). Using (A2), we find

$$ \bigl(a(y) \bigl(\overline{w}'(y)\bigr)^{\mu } \bigr)+c(y) \overline{g} \bigl(y\bigl( \varsigma (y)\bigr) \bigr)=0 \quad\text{for } y\geq {}y_{1}, $$

where $\overline{w}(y)=y(y)+b(y)y(\vartheta (y))$ and $\overline{g}(u):=-g(-u)$ for $u\in \mathbb{R}$. Clearly, g̅ satisfies (A2). Then, proceeding as above, we can find the same contradiction.

To prove the condition (6) is necessary, assume that (6) does not hold; so for some $\kappa > 0$ and $y \geq y_{0}$ we have

$$ \int _{Y}^{\infty }c(\eta )g \bigl(\kappa ^{1/\mu } A \bigl(\varsigma (\eta )\bigr) \bigr)\,d\eta \leq \frac{\kappa }{3}. $$

We set

$$\begin{aligned} S={} &\biggl\{ u: u \in C\bigl([y_{0},\infty),\mathbb{R}\bigr), u(y)=0 \text{ for } y \in [y_{0}, Y] \text{ and} \\ & \biggl(\frac{\kappa }{3} \biggr)^{1/\mu }\bigl[A(y)-A(Y)\bigr]\leq u(y) \leq \kappa ^{1/ \mu } \bigl[A(y)-A(Y)\bigr] \text{ for } y \geq y_{0}\biggr\} . \end{aligned}$$

We define the operator $\Omega: S \to C([y_{0},+\infty ),\mathbb{R})$ by

$$ (\Omega u) (y)= \textstyle\begin{cases} 0,& y \in [y_{0}, Y], \\ -b(y)u (\vartheta (y) )+\int _{Y}^{y} [\frac{1}{a(\eta )} [\frac{\kappa }{3}+\int _{\eta }^{\infty }c(\zeta )g (u( \varsigma (\zeta )) )\,d\zeta ] ]^{1/\mu } \,d\eta, & y\geq Y. \end{cases} $$

For every $u \in S$ and $y \geq Y$, we have

$$\begin{aligned} (\Omega u) (y) &\geq \int _{Y}^{y} \biggl[\frac{1}{a(\eta )} \biggl[ \frac{\kappa }{3}+ \int _{\eta }^{\infty }c(\zeta )g \bigl(u\bigl(\varsigma ( \zeta )\bigr) \bigr)\,d\zeta \biggr] \biggr]^{1/\mu }\,d\eta \\ &\geq \int _{Y}^{y} \biggl[\frac{1}{a(\eta )} \frac{\kappa }{3} \biggr]^{1/ \mu }\,d\eta = \biggl(\frac{\kappa }{3} \biggr)^{1/\mu }\bigl[A(y)-A(Y)\bigr]. \end{aligned}$$

For every $u \in S$ and $y \geq Y$, we have $u(y)\leq \kappa ^{1/\mu } A(y)$ and $g(u(y))\leq g(\kappa ^{1/\mu } A(y))$. Then

$$\begin{aligned} (\Omega u) (y)&\leq - b(y)u \bigl(\vartheta (y) \bigr)+ \int _{T}^{y} \biggl[\frac{1}{a(\eta )} \biggl( \frac{\kappa }{3}+\frac{\kappa }{3} \biggr) \biggr]^{1/\mu }\,d\eta \\ &\leq b_{0} \kappa ^{1/\mu } \bigl[A\bigl(\vartheta (y) \bigr)-A(Y) \bigr]+(2 \kappa /3)^{1/\mu } \bigl[A(y)-A(Y) \bigr] \\ &\leq b_{0} \kappa ^{1/\mu } \bigl[A(y)-A(Y) \bigr]+ (2\kappa /3)^{1/ \mu } \bigl[A(y)-A(Y) \bigr] \\ &= \bigl(b_{0}+(2/3)^{1/\mu } \bigr)\kappa ^{1/\mu } \bigl[A(y)-A(Y) \bigr] \leq \kappa ^{1/\mu } \bigl[A(y)-A(Y) \bigr], \end{aligned}$$

which implies that $(\Omega u)(y) \in S$. Let us define now a sequence of continuous function $v_{n}: [y_{0}, +\infty )\to \mathbb{R}$ by the recursive formula

$$\begin{aligned} &u_{0}(y)= \textstyle\begin{cases} 0,& y \in [y_{0}, Y], \\ \frac{\kappa }{3}[A(y)-A(Y)],& y\geq Y, \end{cases}\displaystyle \\ &u_{n}(y)= (\Omega u_{n-1} ) (y),\quad n\geq 1. \end{aligned}$$

Inductively, it is easy to verify that, for $n>1$,

$$ \biggl(\frac{\kappa }{3} \biggr)^{1/\mu } \bigl[A(y)-A(Y) \bigr]\leq u_{n-1}(y) \leq u_{n}(y) \leq \kappa ^{1/\mu } \bigl[A(y)-A(Y) \bigr]. $$

Therefore the point-wise limit of the sequence exists. Let $\lim_{y \to \infty }u_{n}(y)=v(y)$ for $y \geq y_{0}$. By Lebesgue’s dominated convergence theorem, $u \in S$ and $(\Omega u)(y) =u(y)$, where $u(y)$ is a solution of (1) on $[Y,\infty )$ such that $u(y)>0$. Hence, (6) is necessary. This completes the proof. □

Example 3.2

Consider the delay differential equation

$$ \bigl(e^{-y} \bigl(\bigl(u(y)-e^{-y}u(y-1) \bigr)' \bigr)^{3/5}\bigr)'+y\bigl(u(y-2) \bigr)^{1/3}=0,\quad y\geq 0. $$

(9)

Here $\mu = 3/5$, $a(y)=e^{-y}$, $-1 < b(y)=-e^{-y} \leq 0$, $\vartheta (y)=y-1$, $\varsigma (y)=y-2$, $A(y)=\int _{0}^{y} e^{5s/3} \,ds= \frac{3}{5} (e^{5y/3}-1 )$, $g(v)=v^{1/3}$. For ${\mu _{1}}=1/2$, we have a decreasing function $g(v)/v^{\mu _{1}}=v^{-1/6}$. Now

$$\begin{aligned} \int _{0}^{\infty } c(\eta )g \bigl(\kappa ^{1/\mu } A\bigl(\varsigma (\eta )\bigr) \bigr)\,d\eta = \int _{0}^{\infty }\eta \biggl(\kappa ^{5/3} \frac{3}{5} \bigl(e^{5(\eta -2/3}-1 \bigr) \biggr)^{1/3} \,d\eta =\infty \quad\forall \kappa >0. \end{aligned}$$

So, all the conditions of Theorem 3.1 hold, and therefore every unbounded solution of (9) is oscillatory.

Theorem 3.3

Let assumptions (A1)–(A4) hold. Then each unbounded solution of (1) oscillates if and only if (6) holds for every $\kappa >0$.

Proof

To prove sufficiency by contradiction, assume that the solution u of (1) is eventually positive and unbounded. So, there exists $y_{1}\geq {}y_{0}$ such that $u(y)>0$, $u (\vartheta (y) )>0$ and $u (\varsigma (y) )>0$ for $y\geq {}y_{1}$. Proceeding as in the proof of Lemma 2.1, $(a(w')^{\mu } )(y)$ is non-increasing, w satisfies one of the cases (i) or (ii) on $[y_{2},\infty )$, where $y_{2}\geq {}y_{1}$. We have the following two possible cases.

Case 1. Let w satisfy (i) for $y \geq y_{2}$. This case is similar to the proof of Theorem 3.1.

Case 2. Let w satisfy (ii) for $y \geq y_{2}$. Since $w(y)$ is unbounded and monotonically increasing, it follows that

$$ \lim_{y \to \infty }\frac{w^{\mu }(y)}{A^{\mu }(y)}=\lim_{y \to \infty } \frac{(w'(y))^{\mu }}{(A'(y))^{\mu }}=\lim_{y\to \infty } \bigl(a\bigl(w' \bigr)^{\mu } \bigr) (y)=c < \infty. $$

If $c =0$, then $\lim_{t\to \infty }A(y)=+\infty$ implies that $\lim_{t\to \infty }w(y)< +\infty$, which is invalid ($\because w(y)$ is unbounded). Hence $c\neq 0$. Therefore, there exist a constant $\kappa > 0$ and a $y_{2} > y_{1}$ such that $w(y)\geq \kappa ^{1/\mu } A(y)$ for $y\geq y_{2}$. Consequently, $u(y) \geq w(y) \geq \kappa ^{1/\mu } A(y)$ for $y \geq y_{2}$. Using $u(y)\geq \kappa ^{1/\mu } A(y)$ in (1) and then integrating the final inequality from $y_{2}$ to +∞, we obtain a contradiction to (6) for every $\kappa >0$.

By using the same transformation as in the proof of Theorem 3.1 we can get a contradiction for an eventually negative unbounded solution, so we omit it here.

One can prove the necessary part by following the proof of Theorem 3.1. So we omit it here. The proof of the theorem is complete. □

Theorem 3.4

Assume that (A1)–(A4) and (5) hold. Then each solution of (1) is oscillatory or $\lim_{y \to \infty }u(y)=0$ if and only if (6) holds for every $\kappa >0$.

Proof

On the contrary, we assume that the solution u of (1) is eventually positive. Then there exists $y_{1}\geq {}y_{0}$ such that $u(y)>0$, $u(\vartheta (y))>0$ and $u(\varsigma (y))>0$ for $y\geq {}y_{1}$. Proceeding as in the proof of Lemma 2.1, we see $(a(w')^{\mu } )(y)$ is non-increasing, and w satisfies one of the cases (i) or (ii) on $[y_{2},\infty )$, where $y_{2}\geq {}y_{1}$. Thus, we have the following two possible cases.

Case 1. Let w satisfy (i) for $y\geq y_{2}$. Then, by Lemma 2.2, we have $\lim_{y \to \infty }u(y)=0$.

Case 2. Let w satisfy (ii) for $y\geq y_{2}$. The case follows from the proof of Theorem 3.1.

The necessary part is similar to Theorem 3.1. The proof of the theorem is complete. □

3.2 The case when $g(u)/u^{\mu _{1}}$ is non-decreasing

Suppose that there exists ${\mu _{1}}>\mu$ such that

$$ \frac{g(v)}{v^{\mu _{1}}} \leq \frac{g(u)}{u^{\mu _{1}}} \quad\text{for }0< v \leq u. $$

(10)

For example we might consider the function $g(u)=|u|^{\mu _{2}} \operatorname{sgn}(u)$ with $\mu <{\mu _{1}}<{\mu _{2}}$ satisfying (10).

Theorem 3.5

Assume that (A1)–(A3), (A5), (10), $\varsigma ^{\prime }(y) \geq 1$ hold. Then each solution of (1) oscillates or $\lim_{y \to \infty }u(y)=0$ if and only if

$$\begin{aligned} \int _{Y}^{\infty } \biggl[\frac{1}{a(\zeta )} \biggl[ \int _{\zeta }^{\infty }c( \eta )\,d\eta \biggr] \biggr]^{1/\mu }\,d\zeta =+\infty \quad\forall y>0. \end{aligned}$$

(11)

Proof

Proceeding in the proof of Theorem 3.4, we can conclude that $\lim_{y \to \infty }u(y)=0$ when z satisfies (i). Let us consider Case 2, for $y\geq y_{2}$. By Remark 1, there exist a constant $\kappa > 0$ and $y_{2} >y_{1}$ such that $z (\varsigma (y) )\geq \kappa$ for $y\geq y_{2}$. Consequently,

$$\begin{aligned} g \bigl(w\bigl(\varsigma (y)\bigr) \bigr)= \frac{g (w(\varsigma (y)) )}{w^{{\mu _{1}}} (\varsigma (y) )}w^{{ \mu _{1}}} \bigl(\varsigma (y) \bigr) \geq \frac{g(\kappa )}{\kappa ^{{\mu _{1}}}}w^{{\mu _{1}}} \bigl( \varsigma (y) \bigr) \end{aligned}$$

(12)

for $y\geq y_{2}$. Using $w(y) \leq u(x)$ and (12) in (1), and then integrating the final inequality we have

$$ \lim_{A \to \infty } \bigl[ \bigl(a\bigl(w' \bigr)' \bigr) (\eta ) \bigr]_{y}^{A}+ \frac{g(\kappa )}{\kappa ^{{\mu _{1}}}} \int _{y}^{\infty }c(\zeta )w^{{ \mu _{1}}} \bigl( \varsigma (\zeta ) \bigr)\,d\zeta \leq 0. $$

Since $(a(w')' )(y)$ is non-increasing and positive, we have

$$\begin{aligned} \frac{g(\kappa )}{\kappa ^{{\mu _{1}}}} \int _{y}^{\infty }c(\eta )w^{{ \mu _{1}}} \bigl(\varsigma (\eta ) \bigr)\,d\eta \leq \bigl(a\bigl(w'\bigr)^{\mu } \bigr) (y) \leq \bigl(a\bigl(w'\bigr)^{\mu } \bigr) \bigl( \varsigma (y)\bigr) \leq a(y) \bigl(\bigl(w'\bigr)^{\mu } \bigr) \bigl(\varsigma (y)\bigr) \end{aligned}$$

for all $y \geq y_{2}$. Therefore,

$$\begin{aligned} \biggl(\frac{g(\kappa )}{\kappa ^{{\mu _{1}}}} \biggr)^{1/\mu } \biggl[ \frac{1}{a(y)} \biggl[ \int _{y}^{\infty }c(\zeta )w^{\mu _{1}} \bigl( \varsigma (\zeta ) \bigr)\,d\zeta \biggr] \biggr]^{1/\mu } \leq w' \bigl( \varsigma (y) \bigr) \end{aligned}$$

implies that

$$\begin{aligned} \biggl(\frac{g(\kappa )}{\kappa ^{{\mu _{1}}}} \biggr)^{1/\mu } \biggl[ \frac{1}{a(y)} \biggl[ \int _{y}^{\infty }c(\zeta )\,d\zeta \biggr] \biggr]^{1/ \mu } \leq \frac{w' (\varsigma (y) )}{w^{{\mu _{1}}/\mu } (\varsigma (y) )} \leq \frac{w' (\varsigma (y) )\varsigma '(y)}{w^{{\mu _{1}}/\mu } (\varsigma (y) )}. \end{aligned}$$

Integrating the final inequality from $y_{2}$ to +∞, we have

$$\begin{aligned} \biggl(\frac{g(\kappa )}{\kappa ^{{\mu _{1}}}} \biggr)^{1/\mu } \int _{y_{2}}^{ \infty } \biggl[\frac{1}{a(\zeta )} \biggl[ \int _{\zeta }^{\infty }c(\eta )\,d \eta \biggr] \biggr]^{1/\mu }\,d\zeta &< \int _{y_{2}}^{\infty } \frac{w' (\varsigma (\eta ) )\varsigma '(\eta )}{w^{{\mu _{1}}/\mu } (\varsigma (\eta ) )}\,d \eta \\ &\leq \frac{w^{1-{\mu _{1}}/\mu }(\varsigma (y_{2}))}{{\mu _{1}}/\mu -1}< \infty, \end{aligned}$$

which contradicts (11).

Next, we show that (11) is necessary. Assume that (11) does not hold and let there exist $y \geq y_{0}$ such that

$$ \int _{Y}^{y} \biggl[\frac{1}{a(\zeta )} \biggl[ \int _{\zeta }^{\infty }c( \eta )\,d\eta \biggr] \biggr]^{1/\mu }\,d\zeta \leq \frac{(1-b_{0}) (g(1) )^{-1/\mu }}{5}, $$

where $\kappa > 0$ is a constant. We set

$$ S= \biggl\{ u \in C\bigl([y_{0},\infty),\mathbb{R}\bigr): u(y)= \frac{1-b_{0}}{5}, y\in [y_{0},Y] \frac{1-b_{0}}{5}\leq u(y)\leq 1 \text{ for } y \geq Y \biggr\} . $$

We define the operator $\Omega: S \to C([y_{0},\infty ),\mathbb{R})$ by

$$ (\Omega u) (y)= \textstyle\begin{cases} \frac{1-b_{0}}{5}, & y \in [y_{0}, Y], \\ -b(y)u (\vartheta (y) )+\frac{1-b_{0}}{5}+\int _{T}^{y} [ \frac{1}{a(\eta )} [\int _{\eta }^{\infty }c(\zeta )g (u( \varsigma (\zeta )) )\,d\zeta ] ]^{1/\mu } \,d\eta, & y \geq T. \end{cases} $$

For every $u \in S$ and $y \geq Y$, $(\Omega u)(y)\geq \frac{1-b_{0}}{5}$ and

$$\begin{aligned} (\Omega u) (y)&\leq b_{0}+\frac{1-b_{0}}{5}+ \bigl(g(1) \bigr)^{1/\mu } \int _{Y}^{y} \biggl[\frac{1}{a(\eta )} \biggl[ \int _{\eta }^{\infty }c( \zeta )\,d\zeta \biggr] \biggr]^{1/\mu } \,d\eta \\ &\leq b_{0}+\frac{1-b_{0}}{5}+\frac{1-b_{0}}{5}= \frac{3b_{0}+2}{5} < 1, \end{aligned}$$

which implies that $\Omega u \in S$. The remaining proof follows from Theorem 3.1. This completes the proof. □

Example 3.6

Consider the differential equation

$$ \bigl( \bigl( \bigl(u(y)-e^{-y}u\bigl(\vartheta (y) \bigr) \bigr)' \bigr)^{1/5} \bigr)'+(y+1) \bigl(u(y-2)\bigr)^{\frac{7}{3}}=0,\quad y\geq 0. $$

(13)

Here $\mu = 1/5$, $a(y)=1$, $\varsigma (y)=y-2$, $g(v)=v^{\frac{7}{3}}$. For ${\mu _{1}}=4/3$, we have $g(v)/v^{\mu _{1}}=v$, which is an increasing function. To check (11) we have

$$ \int _{2}^{\infty } \biggl[ \int _{\zeta }^{\infty }(\eta +1)\,d\eta \biggr]^{5} \,d\zeta =\infty. $$

So, all conditions of Theorem 3.5 hold, and therefore each solution of (13) oscillates or converges to zero.

4 Conclusion

It is worth noting that we have established the necessary and sufficient conditions when $-1 < b(y) \leq 0$. These conditions do not hold in all ranges of $b(y)$.

Remark 2

Theorems 3.1–3.5 also hold for the following equation:

where $b, a, c_{j}, g_{j}, \varsigma _{j}$ $(j =1,2,\dots,m)$ satisfy assumptions (A1)–(A5). In order to extend Theorems 3.1–3.5, we can find an index i so that $c_{j}, g_{j}, \varsigma _{j}$ satisfies (6) and (11).

Example 4.1

Consider the neutral differential equation

$$ \bigl(e^{-y} \bigl( \bigl(u(y)-e^{-y}u\bigl( \vartheta (y)\bigr) \bigr)' \bigr)^{3/5} \bigr)'+\frac{1}{y+1}\bigl(u(y-2)\bigr)^{1/3} + \frac{1}{y+2}\bigl(u(y-1)\bigr)^{1/5}=0, \quad y\geq 0. $$

(14)

Here $\mu = 3/5$, $a(y)=e^{-y}$, $b(y)=-e^{-y}$, $\varsigma _{1}(y)=u-2$, $\varsigma _{2}(y)=u-1$, $A(y)=\int _{0}^{y} e^{5s/3} \,ds= \frac{3}{5} (e^{5y/3}-1 )$, $g_{1}(v)=v^{1/3}$ and $g_{2}(v)=v^{1/5}$. For ${\mu _{1}}=1/2$, we have decreasing functions $g_{1}(v)/v^{\mu _{1}}=v^{-1/6}$ and $g_{2}(v)/v^{\mu _{1}}=v^{-3/10}$. Now,

$$\begin{aligned} & \int _{0}^{\infty } \sum_{i=1}^{m}c_{j}( \eta )g_{j} \bigl(\kappa ^{1/ \mu } A\bigl(\varsigma _{j}(\eta )\bigr) \bigr)\,d\eta \\ &\quad \geq \int _{0}^{\infty } g_{1}( \eta )f_{1} \bigl(\kappa ^{1/\mu } A\bigl(\varsigma _{1}( \eta )\bigr) \bigr)\,d\eta \\ &\quad = \int _{0}^{\infty }\frac{1}{\eta +1} \biggl(\kappa ^{5/3}\frac{3}{5} \bigl(e^{5(\eta -2)/3}-1 \bigr) \biggr)^{1/3} \,d\eta =\infty \quad\forall \kappa >0. \end{aligned}$$

So, all the conditions of Theorem 3.1 hold, and therefore every unbounded solution of (14) is oscillatory.

Example 4.2

Consider the differential equation

$$ \bigl( \bigl( \bigl(u(y)-e^{-y}u\bigl(\vartheta (y) \bigr) \bigr)' \bigr)^{5/7} \bigr)'+t \bigl(u(y-2)\bigr)^{5/3} + (y+1) \bigl(u(y-1)\bigr)^{3}=0,\quad y \geq 0. $$

(15)

Here $\mu = 5/7$, $a(y)=1$, $\varsigma _{1}(y)=y-2$, $\varsigma _{2}(y)=y-1$, $g_{1}(v)=v^{5/3}$ and $g_{2}(v)=v^{3}$. For ${\mu _{1}}=4/3$, we have decreasing functions $g_{1}(v)/v^{\mu _{1}}=v^{1/3}$ and $g_{2}(v)/v^{\mu _{1}}=v^{5/3}$. Clearly, all the conditions of Theorem 3.5 hold. Thus, each solution of (15) oscillates or $\lim_{y \to \infty }u(y)=0$.

Remark 3

Examples 4.1 and 4.2 prove the feasibility and effectiveness of Remark 2.

5 Open problem

This work leads to some open problems:

• 1. Can we find necessary and sufficient conditions for the oscillation of solutions to second-order differential equation (1) for the other ranges of the neutral coefficient b?

• 2. Is it possible to generalize this work to fractional order?