In this section, we discuss the Eq. (1) under the assumptions that \(-1< p(t)\le 0\) and \(0\le p(t)\le 1\), respectively.
Oscillation of Eq. (1) when \(-1<p(t)\le 0\)
Theorem 1
Assume that
\((H_2)-(H_5)\)
hold. If there exists a function
\(\rho \in {\mathrm{C}}^1([t_0, \infty ), (0,\infty ))\)
such that, for any constant
\(K>0\),
$$\begin{aligned} \int _{t_0}^\infty \left[ \rho (t)q(t)-\frac{\left( \rho ^{\prime }(t)\right) ^2}{4\rho ^2(t)\Phi (t)}\right] {\mathrm{d}}t =\infty , \end{aligned}$$
(5)
where
\(\Phi (t)=[\beta \sigma ^{\prime }(t)(\xi (\sigma (t)))^{\beta -1}]/[K^{(1-\beta /\alpha )}\rho (t)(r(\sigma (t)))^{1/\alpha }]\), then all solutions of Eq. (1) are oscillatory or tend to zero as
\(t\rightarrow \infty\).
Proof
Suppose x is a nonoscillatory solution of (1). Without loss of generality, there exists a \(t_1\ge t_0\), such that \(x(t)>0, x(\tau (t))>0,\;\text {and}\;x(\sigma (t))>0,\) for all \(t\ge t_1\). From (1) and the hypothesis \((H_5)\), we get
$$\begin{aligned} \left( r|z^{\prime }|^{\alpha -1}z^{\prime }\right) ^{\prime }\le 0. \end{aligned}$$
(6)
Therefore, \(r|z^{\prime }|^{\alpha -1}z^{\prime }\) is nonincreasing. We claim that \(z^{\prime }>0\). Otherwise, if \(z^{\prime }<0\), using the fact that \(r|z^{\prime }|^{\alpha -1}z^{\prime }\) is nonincreasing, there exists a positive \(k>0\), such that
$$\begin{aligned} -r(t)\left( -z^{\prime }(t)\right) ^{\alpha }\le -k<0. \end{aligned}$$
That is,
$$\begin{aligned} -z^{\prime }(t)\ge \frac{k^{\frac{1}{\alpha }}}{r^{\frac{1}{\alpha }}(t)}. \end{aligned}$$
Integrating the above inequality from \(t_1\) to t, we get
$$\begin{aligned} z(t_1)-z(t) \ge k^{\frac{1}{\alpha }}\int _{t_1}^t r^{-\frac{1}{\alpha }}(s){\mathrm{d}}s. \end{aligned}$$
It follows from \((H_2)\) that
$$\begin{aligned} \lim _{t\rightarrow \infty }z(t)=-\infty . \end{aligned}$$
(7)
We consider the following two cases.
Case 1 If x is unbounded, then there exists a sequence \(\{t_m\}\), such that
$$\begin{aligned} \lim _{m\rightarrow \infty } x(t_m)=\infty , \end{aligned}$$
where \(\{t_m\}\) satisfies \(\lim _{m\rightarrow \infty }t_m=\infty\) and \(x(t_m)=\max _{t_0\le s\le t_m}\{x(s)\}\). By the definition of \(x(t_m)\) and \(\tau (t)\le t\), we have
$$\begin{aligned} x(\tau (t_m))\le x(t_m). \end{aligned}$$
Then we get
$$\begin{aligned} z(t_m)=x(t_m)+p(t_m)x(\tau (t_m))\ge x(t_m)(1+p(t_k))>0, \end{aligned}$$
which contradicts (7).
Case 2 If x is bounded, from the definition of z and \(-1< p(t) \le 0\), z is also bounded, which also contradicts (7).
Hence, it is clear from the above discussion that \(z^{\prime }>0\), and then \(z>0\) or \(z<0\). We consider each of two cases separately.
Suppose first that \(z>0\). Considering the definition of z and \(-1<p(t)\le 0\), we get
$$\begin{aligned} z(t)=x(t)+p(t)x(\tau (t))\le x(t). \end{aligned}$$
(8)
From \(\sigma (t)\le t\) and the fact that \(r(z^{\prime })^{\alpha }\) is nonincreasing, we obtain
$$\begin{aligned} \left( r(\sigma (t))\right) ^{\frac{1}{\alpha }}z^{\prime }(\sigma (t))\ge \left( r(t)\right) ^{\frac{1}{\alpha }}z^{\prime }(t) \end{aligned}$$
(9)
and there exist a positive constant K and a \(t_2\ge t_1\), such that
$$\begin{aligned} r(t)\left( z^{\prime }(t)\right) ^{\alpha }\le K,\quad t\ge t_2. \end{aligned}$$
(10)
From the fact that \(r(z^{\prime })^{\alpha }\) is nonincreasing, we get
$$z(t) = {} z(t_1)+\int _{t_1}^t \frac{\left( r(s)\left( z^{\prime }(s)\right) ^{\alpha }\right) ^{\frac{1}{\alpha }}}{r^{\frac{1}{\alpha }}(s)} {\mathrm{d}} s \ge r^{\frac{1}{\alpha }}(t)z^{\prime }(t)\int _{t_1}^t r^{-\frac{1}{\alpha }}(s){\mathrm{d}}s = \xi (t)r^{\frac{1}{\alpha }}(t)z^{\prime }(t),$$
(11)
where \(\xi (t)=\int _{t_1}^t r^{-1/{\alpha }}(s) {\mathrm{d}}s\). Define a function \(\omega\) by
$$\begin{aligned} \omega (t)=\rho (t)\frac{r(t)\left( z^{\prime }(t)\right) ^{\alpha }}{z^{\beta }(\sigma (t))}, \end{aligned}$$
then \(\omega (t)>0\). Differentiating \(\omega\), we get
$$\omega ^{\prime }(t) = \frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)+\rho (t)\frac{\left( r(t)z^{\prime }(t)\right) ^{\prime }}{z^{\beta }(\sigma (t))} - \beta \sigma ^{\prime }(t)\rho (t)\frac{r(t)\left( z^{\prime }(t)\right) ^{\alpha }z^{\beta -1}(\sigma (t))z^{\prime }(\sigma (t))}{z^{2\beta }(\sigma (t))}.$$
From (1) and (8), we conclude that
$$\begin{aligned} \omega ^{\prime }(t) \le -\rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)-\beta \sigma ^{\prime }(t)\rho (t)\frac{r(t)\left( z^{\prime }(t)\right) ^{\alpha }}{z^{2\beta }(\sigma (t))} z^{\prime }(\sigma (t))z^{\beta -1}(\sigma (t)). \end{aligned}$$
Taking into account (11), the last inequality implies
$$\begin{aligned}&\omega ^{\prime }(t)\le - \rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)-\beta \sigma ^{\prime }(t)(\xi (\sigma (t)))^{\beta -1}\nonumber \\&\quad \qquad \rho (t) \frac{r(t)\left( z^{\prime }(t)\right) ^{\alpha }}{z^{2\beta }(\sigma (t))} \left( r(\sigma (t))\right) ^{\frac{\beta -1}{\alpha }}\left( z^{\prime }(\sigma (t))\right) ^{\beta }. \end{aligned}$$
(12)
It follows from (9), (10), and (12) that
$$\begin{aligned} \omega ^{\prime }(t)&\le - \rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)- \frac{\beta \sigma ^{\prime }(t)\left( \xi (\sigma (t))\right) ^{\beta -1}}{\left( r(\sigma (t))\right) ^{1/\alpha }} \rho (t)\frac{r(t)\left( z^{\prime }(t)\right) ^{\alpha }}{z^{2\beta }(\sigma (t))} \left( r(t)\right) ^{\frac{\beta }{\alpha }}\left( z^{\prime }(t)\right) ^{\beta }\\&=-\rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)- \frac{\beta \sigma ^{\prime }(t)\left( \xi (\sigma (t))\right) ^{\beta -1}\left( r(t)\right) ^{\frac{\beta }{\alpha }}}{\left( r(\sigma (t))\right) ^{1/\alpha }\rho (t)r(t)} \omega ^2(t)\frac{1}{\left( z^{\prime }(t)\right) ^{\alpha -\beta }}\\&\le -\rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)- \frac{\beta \sigma ^{\prime }(t)\left( \xi (\sigma (t))\right) ^{\beta -1}}{K^{1-\frac{\beta }{\alpha }}\left( r(\sigma (t))\right) ^{1/\alpha }\rho (t)} \omega ^2(t)\\&= -\rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)-\Phi (t)\omega ^2(t)\\&\le -\rho (t)q(t)+\frac{(\rho ^{\prime }(t))^2}{4\rho ^2(t)\Phi (t)},\\ \end{aligned}$$
(13)
where \(\Phi (t)=[\beta \sigma ^{\prime }(t)(\xi (\sigma (t)))^{\beta -1}]/[K^{(1-\beta /\alpha )}\rho (t)(r(\sigma (t)))^{1/\alpha }].\) Integrating (13) from \(t_2\) to t, we get
$$\begin{aligned} 0<\omega (t)\le \omega (t_2)-\int ^t_{t_2}\left[ \rho (s)q(s)-\frac{\left( \rho ^{\prime }(s)\right) ^2}{4\rho ^2(s)\Phi (s)}\right] {\mathrm{d}}s, \end{aligned}$$
which contradicts (5).
If \(z<0\), we claim that \(\lim _ {t\rightarrow \infty }{x(t)}=0\). Using \(z<0\) and \(z^{\prime }>0\), we deduce that
$$\begin{aligned} \lim _{t\rightarrow \infty }z(t)=l\le 0, \end{aligned}$$
where l is a constant. That is, for all sufficiently large t, z is bounded. We can easily prove that x is also bounded. From the fact that x is bounded, we get
$$\begin{aligned} \limsup _{t\rightarrow \infty } x(t)=a, \end{aligned}$$
where a is a constant. We claim that \(a=0\). Otherwise, if \(a>0\), there exists a sequence \(\{t_n\}\), such that \(\lim _{n\rightarrow \infty }t_n=\infty\) and \(\lim _{n\rightarrow \infty } x(t_n)=a.\) Letting
$$\begin{aligned} \varepsilon =\frac{a(1-p_0)}{2p_0}, \end{aligned}$$
for large enough n, we obtain
$$\begin{aligned} x(\tau (t_n))<a+\varepsilon . \end{aligned}$$
Then, from the definition of z(t) and \(p(t)\ge -p_0\), we have
$$\begin{aligned} \lim _{n\rightarrow \infty }z(t_n)\ge \lim _{n\rightarrow \infty }x(t_n)-p_0(a+\varepsilon )=\frac{a(1-p_0)}{2}>0, \end{aligned}$$
which contradicts \(z<0\). Thus \(\limsup _{t\rightarrow \infty } x(t)=0.\) By \(x>0\), we get
$$\begin{aligned} \lim _ {t\rightarrow \infty }{x(t)}=0. \end{aligned}$$
The proof is complete. \(\square\)
From Theorem 1, letting \(\rho =1\), we get the following corollary.
Corollary 1
Assume that
\((H_2)-(H_5)\)
hold. If
$$\begin{aligned} \int _{t_0}^{\infty }q(t) {\mathrm{d}}t=\infty , \end{aligned}$$
then all solutions of Eq. (1) are oscillatory or tend to zero as
\(t\rightarrow \infty\).
Theorem 2
Assume that
\((H_2)-(H_5)\)
hold. If there exist two functions
\(H\in P\)
and
\(\rho \in {\mathrm{C}}^1([t_0, \infty ), (0,+\infty ))\), such that
$$\begin{aligned} \limsup _{t\rightarrow \infty } \frac{1}{H(t,t_0)}\int _{t_0}^t\left[ H(t,s)\rho (s)q(s)-\frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s) -h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right] {\mathrm{d}}s=\infty , \end{aligned}$$
(14)
where
\(\Phi\)
is as in Theorem 1, then the conclusion of Theorem 1 remains intact.
Proof
Suppose that x is a nonoscillatory solution of (1). Proceeding as in the proof of Theorem 1, we get \(z>0\) or \(z<0\).
Firstly, we consider \(z>0\). As the proof above (13) holds. That is
$$\begin{aligned} \omega ^{\prime }(t) \le -\rho (t)q(t)+\frac{\rho ^{\prime }(t)}{\rho (t)}\omega (t)-\Phi (t)\omega ^2(t). \end{aligned}$$
Multiplying this inequality by H(t, s) and integrating it from \(t_2\) to t, we have
$$\begin{aligned} \int _{t_2}^t H(t,s)\omega ^{\prime }(s){\mathrm{d}}s \le \int _{t_2}^t H(t,s)\left[ -\rho (s)q(s)+ \frac{\rho ^{\prime }(s)}{\rho (s)}\omega (s)-\Phi (s)\omega ^2(s)\right] {\mathrm{d}}s. \end{aligned}$$
(15)
By the property of \(\partial H(t, s)/\partial s=-h_2(t, s)\sqrt{H(t, s)}<0\), we conclude that
$$\begin{aligned}&\int _{t_2}^t\left[ H(t,s)\rho (s)q(s)-\frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s) -h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right] {\mathrm{d}}s\nonumber \\&\quad \le H(t,t_2)\omega (t_2)\le H(t,t_0)\omega (t_2) . \end{aligned}$$
Adding \(\int _{t_0}^{t_2}\left[H(t,s)\rho (s)q(s)- \frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s)-h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right]{\mathrm{d}}s\) to the latter inequality and multiplying this inequality by \(1/H(t,t_0)\), we get
$$\begin{aligned} &\frac{1}{H(t,t_0)}\left[ \int _{t_0}^{t_2}+\int _{t_2}^t\right] \left[ H(t,s)\rho (s)q(s)-\frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s)-h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right] {\mathrm{d}}s\\&\quad \quad \le \frac{1}{H(t,t_0)}\int _{t_0}^{t_2}\left[ H(t,s)\rho (s)q(s)- \frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s)-h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right] {\mathrm{d}}s + \omega (t_2).\\&\quad \quad \le \omega (t_2)+\int _{t_0}^{t_2}\left[ \frac{H(t,s)}{H(t,t_0)}\rho (s)q(s)- \frac{\left( \frac{\rho ^{\prime }(s)}{\rho (s)}\sqrt{H(t,s)}-h_2(t,s)\right) ^2}{4H(t,t_0)\Phi (s)}\right] {\mathrm{d}}s\\&\quad \quad \le \omega (t_2)+\int _{t_0}^{t_2}\rho (s)q(s) {\mathrm{d}}s.\\ \end{aligned}$$
(16)
Letting \(t\rightarrow \infty\) in (16), we can get a contradict to (14).
If \(z<0\), repeating the proof of Theorem 1, we have \(\lim _ {t\rightarrow \infty }{x(t)}=0\). This completes the proof. \(\square\)
From Theorem 2, letting \(H(t,s)=(t-s)^{\lambda } (\lambda >0)\) and \(\rho (t)=1\), we may get the following corollary.
Corollary 2
Assume that
\((H_2)-(H_5)\)
hold. If
$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{1}{(t-t_0)^{\lambda }}\int _{t_0}^t (t-s)^{\lambda -2}\left[ (t-s)^2q(s)-\frac{\lambda ^2}{4\Phi (s)}\right] {\mathrm{d}}s=\infty , \end{aligned}$$
(17)
then the conclusion of Theorem 1 remains intact.
Theorem 3
Assume that
\((H_2)-(H_5)\)
hold. If there exist three functions
\(H\in P\), \(\rho \in ([t_0, \infty ), (0,\infty ))\), and
\(A\in {\mathrm {C}}([t_0,\infty ), {\mathbb {R}})\), such that
$$\begin{aligned} \int _{t_0}^{\infty }A_{+}^2(t)\Phi (t){\mathrm{d}} t=\infty \end{aligned}$$
(18)
and
$$\begin{aligned}&\limsup _{t\rightarrow \infty }\frac{1}{H(t,T)}\nonumber \\&\quad \int _T^t\left[ h(t,s)\rho (s)q(s)- \frac{\theta \left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s)- h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4H(t,s)\Phi (s)}\right] {\mathrm{d}}s\ge A(T), \end{aligned}$$
(19)
for all
\(T\ge t_0\)
and for some constant
\(\theta >1\), where
\(A_{+}(t)=\max \{A(t),0\}\), \(\Phi\)
is as in Theorem 1, and
H
satisfies
$$\begin{aligned} 0<\inf _{s\ge t_0}\left\{ \limsup _{t\rightarrow \infty }\frac{H(t,s)}{H(t,t_0}\right\} \le \infty , \end{aligned}$$
(20)
then every solution of Eq. (1) is oscillatory or tends to zero as
\(t\rightarrow \infty\).
Proof
Suppose that x is a nonoscillatory solution of Eq. (1). Then, as in the proof of Theorem 1, \(z>0\) or \(z<0\).
We consider \(z>0\) firstly. By virtue of Theorem 2, (15) holds. That is, for all \(T\ge t_2\ge t_1,\)
$$\begin{aligned} &\frac{1}{H(t,T)}\int _{T}^t H(t,s)\rho (s)q(s){\mathrm{d}}s\\&\quad \le \omega (T)+\frac{1}{H(t,T)} \int _{T}^t \frac{\theta \left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s) -h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4\Phi (s)H(t,s)}{\mathrm{d}}s\\&\quad -\frac{1}{H(t,T)}\int _T^t\frac{\theta -1}{\theta }H(t,s)\Phi (s)\omega ^2(s){\mathrm{d}}s.\\ \end{aligned}$$
Thus,
$$\begin{aligned} &\frac{1}{H(t,T)}\int _T^t \left[ H(t,s)\rho (s)q(s)-\frac{\theta \left( \frac{\rho ^{\prime }(s)}{\rho (s)}H(t,s)- h_2(t,s)\sqrt{H(t,s)}\right) ^2}{4\Phi (s)H(t,s)}\right] {\mathrm{d}}s\\&\quad \quad \le \omega (T)-\frac{1}{H(t,T)}\int _T^t \frac{\theta -1}{\theta }H(t,s)\Phi (s)\omega ^2(s){\mathrm{d}}s.\\ \end{aligned}$$
(21)
Taking into account (19) and (21), we deduce that
$$\begin{aligned} A(T)+\liminf _{t\rightarrow \infty }\frac{1}{H(t,T)} \int _T^t \frac{\theta -1}{\theta }H(t,s)\Phi (s)\omega ^2(s){\mathrm{d}}s\le \omega (T), \end{aligned}$$
then
$$A(T)\le \omega (T),\quad \text {for all}\quad T\ge t_2,$$
(22)
and
$$\begin{aligned} \liminf _{t\rightarrow \infty }\frac{1}{H(t,T)} \int _T^t H(t,s)\Phi (s)\omega ^2(s){\mathrm{d}}s\le \frac{\theta }{\theta -1}\left( \omega (T)-A(T)\right) <\infty . \end{aligned}$$
(23)
Now we will prove that
$$\begin{aligned} \lim _{t\rightarrow \infty }\int _{T}^t\Phi (s)\omega ^2(s){\mathrm{d}}s<\infty . \end{aligned}$$
On the contrary, if
$$\begin{aligned} \lim _{t\rightarrow \infty }\int _{T}^t\Phi (s)\omega ^2(s){\mathrm{d}}s=\infty . \end{aligned}$$
(24)
By the condition (20), there exists a positive constant c, such that
$$\begin{aligned} \inf _{s\ge t_0}\left\{ \liminf _{t\rightarrow \infty }\frac{H(t,s)}{H(t,t_0)}\right\}>c>0. \end{aligned}$$
(25)
On the other hand, using (24), for arbitrary positive M, there exists a \(t_3\ge T\), such that
$$\begin{aligned} \int _{T}^t \Phi (s)\omega ^2(s) {\mathrm{d}}s\ge \frac{M}{c}, \quad t\ge t_3. \end{aligned}$$
From the property (ii) of H and (25), we have
$$\begin{aligned} &\frac{1}{H(t,T)} \int _{T}^t H(t,s)\Phi (s)\omega ^2(s) {\mathrm{d}}s\\&\quad =\frac{1}{H(t,T)}\int _{T}^t\left[ -\frac{\partial H(t,s)}{\partial s}\right] \left[ \int _{T}^s \Phi (u)\omega ^2(u){\mathrm{d}}u\right] {\mathrm{d}}s\\&\quad \ge \frac{1}{H(t,T)}\int _{t_3}^t\left[ -\frac{\partial H(t,s)}{\partial s}\right] \left[ \int _{T}^s \Phi (u)\omega ^2(u){\mathrm{d}}u\right] {\mathrm{d}}s\\&\quad \ge \frac{M}{c}\frac{H(t,t_3)}{H(t,T)}\ge \frac{M}{c}\frac{H(t,t_3)}{H(t,t_0)}\ge M.\\ \end{aligned}$$
(26)
Taking account into the fact that M is arbitrary positive constant, (26) implies that
$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{H(t,T)} \int _{T}^t H(t,s)\Phi (s)\omega ^2(s) {\mathrm{d}}s=\infty , \end{aligned}$$
which contradicts (23). Thus,
$$\begin{aligned} \lim _{t\rightarrow \infty }\int _{T}^t\Phi (s)\omega ^2(s){\mathrm{d}}s<\infty . \end{aligned}$$
Using (22) and \(\Phi (t)>0\), we have
$$\begin{aligned} \int _{T}^{\infty }A_{+}^2(t)\Phi (t){\mathrm{d}}t \le \int _{T}^{\infty }\omega ^2(t)\Phi (t){\mathrm{d}}t <\infty , \end{aligned}$$
which contradicts (18).
If \(z<0\), repeating the proof in Theorem 1, we get \(\lim _{t\rightarrow \infty }x(t)=0\). The proof is complete. \(\square\)
Remark 1
Theorem 1–3 and Corollaries 1 and 2 are the oscillation and asymptotic results of (1) under the assumption that \(-1<p(t)\le 0\). However, the results in Liu et al. (2012), Shi et al. (2016) are established in the case where \(0\le p(t)\le 1\). In the hypothesis \((H_5)\), there is another parameter \(\beta\) and the condition on function f in Li et al. (2015), Erbe et al. (2009) does not satisfy \((H_5)\). Therefore, the results of Li et al. (2015), Erbe et al. (2009) can not apply to Eq. (1).
Oscillation of Eq. (1) when \(0\le p(t)\le 1\)
Theorem 4
Assume that
\((H_1)\)
and
\((H_3)-(H_5)\)
hold. If there exists a function
\(\rho (t)\in {\mathrm{C}}^1([t_0,\infty ),(0,\infty ))\) such that for any positive number M,
$$\begin{aligned} \int _{t_0}^\infty \left[ \rho (t){\overline{p}}(t)- \frac{M^{1-\frac{\beta }{\alpha }}\left( \rho ^{\prime }(t)\right) ^2(r(\sigma (t)))^{\frac{1}{\alpha }}}{4\beta \sigma ^{\prime }(t)\rho (t)(\xi (t))^{\beta -1}}\right] {\mathrm{d}}t =\infty , \end{aligned}$$
(27)
where
\({\overline{p}}(t)=q(t)(1-p(\sigma (t)))^\beta\)
and
\(\xi (t)=\int _{t_1}^t r^{-1/{\alpha }}(s) {\mathrm{d}}s\), then the Eq. (1) is oscillatory.
Proof
See “Appendix”. \(\square\)
Letting \(\rho (t)=1\), we can get the following result.
Corollary 3
Assume that
\((H_1)\)
and
\((H_3)-(H_5)\)
hold. If
$$\begin{aligned} \int _{t_0}^\infty q(t)(1-p(\sigma (t)))^\beta {\mathrm{d}}t =\infty , \end{aligned}$$
then the Eq. (1) is oscillatory.
Example 1
Consider the second-order nonlinear neutral delay differential equation
$$\begin{aligned} \left( \left| \left( x(t)+\left( 1-t^{-\frac{1}{2}}\right) x(\tau (t))\right) ^{\prime }\right| ^{\alpha -1}\left[ x(t)+\left( 1-t^{-\frac{1}{2}}\right) x(\tau (t))\right] ^{\prime }\right) ^{\prime }+\frac{\gamma }{t^2}|x(t)|x(t)=0, \end{aligned}$$
(28)
where \(\beta =2, \alpha \ge 2, r(t)=1, p(t)=1-t^{-\frac{1}{2}}, \sigma (t)=t,\; \text {and} \;q(t)=\frac{\gamma }{t^2}\), where \(\gamma\) is a positive constant.
Letting \(\rho (t)=t^2\), we have
$$\begin{aligned} \xi (t)=\int _{t_1}^t r^{-1/{\alpha }}(s) {\mathrm{d}}s=\int _{t_1}^t 1 {\mathrm{d}}s=t-t_1 \end{aligned}$$
and
$$\begin{aligned} &\int _T^t\left[ \rho (t){\overline{p}}(t)- \frac{\left( \rho ^{\prime }(t)\right) ^2(r(\sigma (t)))^{\frac{1}{\alpha }})}{4\beta \sigma ^{\prime }(t)M^{1-\frac{\beta }{\alpha }} \rho (t)(\xi (t))^{\beta -1}}\right] {\mathrm{d}}s= \int _T^t \left[ \frac{\gamma }{s}-\frac{1}{2K^{1-\frac{2}{\alpha }}(s-t_1)}\right] {\mathrm{d}} s\\&\quad =\gamma \ln t-\frac{1}{2M^{1-\frac{2}{\alpha }}}\ln (t-t_1)-\gamma \ln T+\frac{1}{2M^{1-\frac{2}{\alpha }}}\ln (T-t_1). \end{aligned}$$
Therefore, if \(\gamma > \frac{1}{2M^{1-(2/{\alpha })}}\), then
$$\begin{aligned} \int _T^{\infty }\left[ \rho (t){\overline{p}}(t)- \frac{\left( \rho ^{\prime }(t)\right) ^2(r(\sigma (t)))^{\frac{1}{\alpha }})}{4\beta \sigma ^{\prime }(t)M^{1-\frac{\beta }{\alpha }} \rho (t)(\xi (t))^{\beta -1}}\right] {\mathrm{d}}s=\infty . \end{aligned}$$
It follows from Theorem 4 that all solutions of (28) are oscillatory if \(\gamma > \frac{1}{2M^{1-(2/{\alpha })}}\). However, the Eq. (28) is oscillatory when \(\gamma > \frac{1}{M^{1-(2/{\alpha })}}\) from Theorem 1 in Liu et al. (2012). That is, if \(\frac{1}{2M^{1-(2/{\alpha })}}<\gamma \le \frac{1}{M^{1-(2/{\alpha })}}\), then Theorem 1 in Liu et al. (2012) can not apply to (28).
Remark 2
From Theorem 4, the Eq. (28) is oscillatory if \(\gamma > \frac{1}{2M^{1-(2/{\alpha })}}\). However, from Theorem 1 in Liu et al. (2012), when \(\gamma > \frac{1}{M^{1-(2/{\alpha })}}\), the Eq. (28) is oscillation. Thus, Theorem 4 improves Theorem 1 in Liu et al. (2012) in the case where \(0\le p(t)\le 1\).