We respectively consider two cases \(g_{1}(t)\leq \tau (t)\) and \(g_{1}(t)\geq \tau (t)\) for \(t\geq t_{0}\). Now, we begin with the first case.
Theorem 3.1
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold, and
\(g_{1}(t)\leq \tau (t)\). If there exists
\(\rho (t)\in C^{1}([t _{0},\infty ),(0,\infty ))\)
s.t.
$$\begin{aligned} & \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{\lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s) \gamma (s)\delta _{2}(s,t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds=\infty , \end{aligned}$$
(3.1)
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), where
$$\begin{aligned} &\lambda = \biggl(\frac{\alpha _{1}\alpha _{2}}{\alpha _{3}} \biggr)^{\alpha _{1}\alpha _{2}} \biggl( \frac{1}{\alpha _{1}\alpha _{2}+1} \biggr)^{\alpha _{1} \alpha _{2}+1}, \\ &\gamma (t) =\textstyle\begin{cases} m_{1}(\delta _{3}(t,t_{1}))^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}, \quad m_{1} \textit{ is any positive constant},& \textit{if } \alpha _{1} \alpha _{2}>\alpha _{3}, \\ m_{2}, \quad m_{2} \textit{ is any positive constant},& \textit{if } \alpha _{1}\alpha _{2}\leq \alpha _{3}, \end{cases}\displaystyle \end{aligned}$$
then every solution of (1.2) is either oscillatory or tends to zero.
Proof
Suppose that (1.2) has a nonoscillatory solution \(x(t)\). We may assume that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). So we have that \(z(t)\) is positive and satisfies the two properties for \(t\geq t_{1}\).
We first consider property (I). Define \(\omega (t)\) by
$$ \omega (t)=\rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(t)},\quad t\geq t_{1}. $$
(3.2)
Then \(\omega (t)>0\) and
$$\begin{aligned} \omega '(t) =&\rho '(t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(t)} + \rho (t) \biggl[\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)} \\ &{}-\frac{\alpha _{3} r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}z'(t)}{z^{\alpha _{3}+1}(t)} \biggr] \end{aligned}$$
(3.3)
$$\begin{aligned} = &\frac{\rho '(t)}{\rho (t)}\omega (t) +\rho (t)\frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z ^{\alpha _{3}}(t)} \\ &{}-\alpha _{3} \rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}z'(t)}{z^{\alpha _{3}+1}(t)}. \end{aligned}$$
(3.4)
Based on (2.12), we have
$$ \frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)} \leq -q_{2}(t) \biggl( \frac{z(\tau ^{-1}(g_{1}(t)))}{z(t)} \biggr)^{\alpha _{3}}. $$
(3.5)
Since \(g_{1}(t)\leq \tau (t)\), we get \(\tau ^{-1}(g_{1}(t))\leq t\). Applying (2.15), we obtain
$$ \frac{z(\tau ^{-1}(g_{1}(t)))}{z(t)}\geq \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})}. $$
(3.6)
From (2.13), we have
$$ z'(t)\geq \delta _{2}(t,t_{1}) \bigl(r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}} \bigr)^{\frac{1}{\alpha _{1}\alpha _{2}}}. $$
(3.7)
We combine (3.4)–(3.7) to conclude that
$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\delta _{2}(t,t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)}z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t) \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$
(3.8)
Next, we will compute \(z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\) and consider the following two cases:
Case (i). Assume that \(\alpha _{1}\alpha _{2}>\alpha _{3}\). Since \(z(t)/\delta _{3}(t,t_{1})\) is nonincreasing, due to (2.15), there exist constants \(h_{1}>0\) and \(t_{2}\geq t_{1}\) such that
$$ \frac{z(t)}{\delta _{3}(t,t_{1})}\leq \frac{z(t_{2})}{\delta _{3}(t_{2},t _{1})}=h_{1},\quad t\geq t_{2}, $$
and
$$ z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\geq m_{1}\bigl(\delta _{3}(t,t _{1})\bigr)^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}, $$
(3.9)
where \(m_{1}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{1}\).
Case (ii). Assume that \(\alpha _{1}\alpha _{2}\leq \alpha _{3}\). Since \(z'(t)>0\), there exists \(h_{2}>0\) such that
$$ z(t)\geq z(t_{1})=h_{2},\quad t\geq t_{1}, $$
and
$$ z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\geq m_{2},\quad t\geq t _{1}, $$
(3.10)
where \(m_{2}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{2}\). We combine (3.8) with (3.9) and (3.10) to have
$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$
(3.11)
By using the inequality (see [18])
$$ Bu-Au^{\frac{1}{\alpha }+1}\leq \frac{\alpha ^{\alpha }}{(\alpha +1)^{ \alpha +1}} \frac{B^{\alpha +1}}{A^{\alpha }},\quad A>0, $$
(3.12)
where
$$ B=\frac{\rho '_{+}(t)}{\rho (t)},\qquad A=\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)},\quad \alpha = \alpha _{1}\alpha _{2}, u=\omega (t), $$
we obtain
$$\begin{aligned} &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t)-\frac{\alpha _{3}\gamma (t) \delta _{2}(t,t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t) \\ &\quad \leq \biggl(\frac{\alpha _{1}\alpha _{2}}{\alpha _{3}\rho (t)\gamma (t) \delta _{2}(t,t_{1})} \biggr)^{\alpha _{1}\alpha _{2}} \biggl( \frac{ \rho '_{+}(t)}{\alpha _{1}\alpha _{2}+1} \biggr)^{\alpha _{1}\alpha _{2}+1} \\ &\quad = \frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t) \gamma (t)\delta _{2}(t,t_{1}))^{\alpha _{1}\alpha _{2}}}. \end{aligned}$$
(3.13)
We combine (3.11) and (3.13) to conclude that
$$ \omega '(t)\leq -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} +\frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t)\gamma (t)\delta _{2}(t,t_{1}))^{\alpha _{1}\alpha _{2}}}. $$
We integrate the latter inequality to make
$$\begin{aligned} & \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma (s)\delta _{2}(s,t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds \\ &\quad \leq \omega (t_{2})-\omega (t)< \omega (t_{2}), \end{aligned}$$
which contradicts (3.1).
Secondly, we investigate property (II) and deduce \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.2
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that
\(g_{1}(t)\leq \tau (t)\)
and
\(\alpha _{1} \alpha _{2}=\alpha _{3}\). If there exists
\(\rho (t)\)
s.t.
$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\rho '_{+}(s)}{(\delta _{3}(s,t_{1}))^{\alpha _{3}}} \biggr]\,ds=\infty , $$
(3.14)
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem
3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). From (2.13) and (2.14), we have
$$\begin{aligned} r_{1}(t)z^{\alpha _{1}}(t) &\geq \biggl(\frac{\delta _{3}(t,t_{1})}{\delta _{2}(t,t_{1})} \biggr)^{\alpha _{1}}r_{1}(t) \bigl(z'(t) \bigr)^{\alpha _{1}} \\ &\geq \biggl(\frac{\delta _{3}(t,t_{1})}{\delta _{2}(t,t_{1})} \biggr)^{ \alpha _{1}} \delta _{1}(t,t_{1})r^{\frac{1}{\alpha _{2}}}_{2}(t) \bigl(r_{1}(t) \bigl(z'(t)\bigr)^{ \alpha _{1}} \bigr)', \end{aligned}$$
and
$$ z^{\alpha _{3}}(t)\geq \bigl(\delta _{3}(t,t_{1}) \bigr)^{\alpha _{3}}r_{2}(t) \bigl(\bigl(r _{1}(t) \bigl(z'(t)\bigr)^{\alpha _{1}}\bigr)' \bigr)^{\alpha _{2}}. $$
(3.15)
Define \(\omega (t)\) by (3.2). As in the proof of Theorem 3.1, we get (3.3). Since \(E_{2}(t)>0\), (3.3) indicates
$$ \omega '(t)\leq \rho '_{+}(t) \frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}}{z^{\alpha _{3}}(t)} +\rho (t)\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(t)}. $$
Combining the latter inequality with (3.5), (3.6), and (3.15), we see that
$$ \omega '(t)\leq \frac{\rho '_{+}(t)}{(\delta _{3}(t,t_{1}))^{\alpha _{3}}} -\rho (t)q_{2}(t) \biggl(\frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t _{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}}. $$
(3.16)
An integration of (3.16) from \(t_{2}\) (\(t_{2}\geq t_{1}\)) to t leads to
$$ \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} -\frac{\rho '_{+}(s)}{( \delta _{3}(s,t_{1}))^{\alpha _{3}}} \biggr]\,ds\leq \omega (t_{2})-\omega (t)< \omega (t_{2}), $$
for all sufficiently large t, which contradicts (3.14).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.3
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that
\(g_{1}(t)\leq \tau (t)\)
and
\(\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1\). If there exists
\(\rho (t)\)
s.t.
$$\begin{aligned} &\limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{(\rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (s)\delta _{2}(s,t _{1})(\delta _{3}(s,t_{1}))^{\alpha _{3}-1}} \biggr]\,ds=\infty , \end{aligned}$$
(3.17)
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), where
\(\gamma (t)\)
is given in Theorem
3.1, then we get the same conclusion as in Theorem
3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Proceeding as in the proof of Theorem 3.1, we get (3.11), and
$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})\omega ^{2}(t)}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t). \end{aligned}$$
(3.18)
From (3.2) and (3.15), one has
$$\begin{aligned} \omega ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) &= \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) \biggl( \frac{r_{2}(t)((r_{1}(t)(z'(t))^{ \alpha _{1}})')^{\alpha _{2}}}{z^{\alpha _{3}}(t)} \biggr)^{\frac{1}{\alpha _{1}\alpha _{2}}-1} \\ &\geq \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}-1}(t) \bigl(\delta _{3}(t,t_{1}) \bigr)^{ \alpha _{3}-1}. \end{aligned}$$
(3.19)
We substitute (3.19) into (3.18) to see that
$$\begin{aligned} \omega '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\omega (t) -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g_{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} \\ &{}-\frac{\alpha _{3}\gamma (t)\delta _{2}(t,t_{1})(\delta _{3}(t,t_{1}))^{ \alpha _{3}-1}}{\rho (t)} \omega ^{2}(t), \end{aligned}$$
from which one gets
$$ \omega '(t)\leq -\rho (t)q_{2}(t) \biggl( \frac{\delta _{3}(\tau ^{-1}(g _{1}(t)),t_{1})}{\delta _{3}(t,t_{1})} \biggr)^{\alpha _{3}} +\frac{( \rho '_{+}(t))^{2}}{4\alpha _{3}\rho (t)\gamma (t)\delta _{2}(t,t_{1})( \delta _{3}(t,t_{1}))^{\alpha _{3}-1}}, $$
by completing the square with respect to \(\omega (t)\). We integrate the latter inequality from \(t_{2}\) (\(t_{2}\geq t_{1}\)) to t to obtain
$$\begin{aligned} & \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) \biggl(\frac{\delta _{3}(\tau ^{-1}(g _{1}(s)),t_{1})}{\delta _{3}(s,t_{1})} \biggr)^{\alpha _{3}} \\ &\quad {}-\frac{(\rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (s)\delta _{2}(s,t _{1})(\delta _{3}(s,t_{1}))^{\alpha _{3}-1}} \biggr]\,ds\leq \omega (t_{2}), \end{aligned}$$
for all sufficiently large t, which contradicts (3.17).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Next, we consider \(g_{1}(t)\geq \tau (t)\) for \(t\geq t_{0}\).
Theorem 3.4
Assume that conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, and
\(g_{1}(t)\geq \tau (t)\). If there exists
\(\rho (t)\)
s.t.
$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{ \lambda (\rho '_{+}(s))^{\alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma ( \tau (s))\delta _{2}(\tau (s),t_{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds= \infty , $$
(3.20)
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem
3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Define \(\nu (t)\) by
$$ \nu (t)=\rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))},\quad t\geq t_{1}. $$
(3.21)
Then \(\nu (t)>0\) and
$$\begin{aligned} \nu '(t) = &\rho '(t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))} +\rho (t) \biggl[\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}( \tau (t))} \\ &{}-\frac{\alpha _{3} r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}}(z(\tau (t)))'}{z^{\alpha _{3}+1}(\tau (t))} \biggr] \end{aligned}$$
(3.22)
$$\begin{aligned} = &\frac{\rho '(t)}{\rho (t)}\nu (t) +\rho (t)\frac{ [r _{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(\tau (t))} \\ &{}-\alpha _{3} \rho (t)\frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}(z(\tau (t)))'}{z^{\alpha _{3}+1}(\tau (t))}. \end{aligned}$$
(3.23)
Since \(\tau ^{-1}(g_{1}(t))\geq t\geq \tau (t)\) and \(z'(t)>0\), we have
$$ \frac{z(\tau ^{-1}(g_{1}(t)))}{z(\tau (t))}\geq 1,\quad t\geq t_{1}, $$
which indicates that
$$ \frac{ [r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}(\tau (t))} \leq -q_{2}(t), $$
(3.24)
due to (2.12). Based on (3.7), \(E'_{2}(t)\leq 0\) and \(\tau (t)\leq t\), so one has \(\tau (t)\geq t_{1}\) and
$$ \bigl(z\bigl(\tau (t)\bigr)\bigr)'\geq \delta _{2}\bigl(\tau (t),t_{1}\bigr) \bigl(r_{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t)\bigr)^{ \alpha _{1}} \bigr)'\bigr)^{\alpha _{2}} \bigr)^{\frac{1}{\alpha _{1}\alpha _{2}}}, $$
(3.25)
for \(t\geq t_{2}>t_{1}\). Combining (3.9), (3.10), (3.23)–(3.25), we conclude that
$$\begin{aligned} \nu '(t) \leq &\frac{\rho '_{+}(t)}{\rho (t)}\nu (t)-\rho (t)q_{2}(t) \\ &{}-\frac{\alpha _{3}\gamma (\tau (t))\delta _{2}(\tau (t),t_{1})}{ \rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)} \nu ^{\frac{1}{\alpha _{1}\alpha _{2}}+1}(t). \end{aligned}$$
(3.26)
Using (3.12) and (3.26) with
$$ B=\frac{\rho '_{+}(t)}{\rho (t)},\qquad A=\frac{\alpha _{3}\gamma (\tau (t)) \delta _{2}(\tau (t),t_{1})}{\rho ^{\frac{1}{\alpha _{1}\alpha _{2}}}(t)}, $$
one gets
$$ \nu '(t)\leq -\rho (t)q_{2}(t) + \frac{\lambda (\rho '_{+}(t))^{\alpha _{1}\alpha _{2}+1}}{(\rho (t)\gamma (\tau (t))\delta _{2}(\tau (t),t _{1}))^{\alpha _{1}\alpha _{2}}}. $$
Integrating the latter inequality from \(t_{2}\) to t, we have
$$ \int ^{t}_{t_{2}} \biggl[\rho (s)q_{2}(s) -\frac{\lambda (\rho '_{+}(s))^{ \alpha _{1}\alpha _{2}+1}}{(\rho (s)\gamma (\tau (s))\delta _{2}(\tau (s),t _{1}))^{\alpha _{1}\alpha _{2}}} \biggr]\,ds\leq \omega (t_{2}), $$
for all sufficiently large t, which contradicts (3.20).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.5
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that
\(g_{1}(t)\geq \tau (t)\)
and
\(\alpha _{1} \alpha _{2}=\alpha _{3}\). If there exists
\(\rho (t)\)
s.t.
$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{ \rho '_{+}(s)}{(\delta _{3}(\tau (s),t_{1}))^{\alpha _{3}}} \biggr]\,ds= \infty , $$
(3.27)
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem
3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Proceeding as in the proof of Theorem 3.4, we get (3.22), which implies
$$ \nu '(t)\leq \rho '_{+}(t) \frac{r_{2}(t)((r_{1}(t)(z'(t))^{\alpha _{1}})')^{ \alpha _{2}}}{z^{\alpha _{3}}(\tau (t))} +\rho (t)\frac{ [r_{2}(t)((r _{1}(t)(z'(t))^{\alpha _{1}})')^{\alpha _{2}} ]'}{z^{\alpha _{3}}( \tau (t))}. $$
(3.28)
Applying (3.15), the monotonicity of \(E_{2}(t)\) and the fact that \(\tau (t)\leq t\), one has \(\tau (t)\geq t_{1}\) and
$$ z^{\alpha _{3}}\bigl(\tau (t)\bigr)\geq \bigl(\delta _{3}\bigl(\tau (t),t_{1}\bigr)\bigr)^{\alpha _{3}}r _{2}(t) \bigl(\bigl(r_{1}(t) \bigl(z'(t) \bigr)^{\alpha _{1}}\bigr)'\bigr)^{\alpha _{2}}, $$
(3.29)
for \(t\geq t_{2}>t_{1}\). Combining (3.24), (3.28), and (3.29), one gets
$$ \nu '(t)\leq \frac{\rho '_{+}(t)}{(\delta _{3}(\tau (t),t_{1}))^{\alpha _{3}}} -\rho (t)q_{2}(t). $$
(3.30)
Upon integrating (3.30) from \(t_{2}\) to t, we obtain a contradiction to (3.27).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.6
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that
\(g_{1}(t)\geq \tau (t)\)
and
\(\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1\). If there exists
\(\rho (t)\)
s.t.
$$ \limsup_{t\rightarrow \infty } \int ^{t}_{t_{*}} \biggl[\rho (s)q_{2}(s) -\frac{( \rho '_{+}(s))^{2}}{4\alpha _{3}\rho (s)\gamma (\tau (s))\delta _{2}( \tau (s),t_{1})(\delta _{3}(\tau (s),t_{1}))^{\alpha _{3}-1}} \biggr]\,ds= \infty , $$
for
\(t_{1}\)
and
\(t_{*}\)
with
\(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem
3.1.
We omit the proof of Theorem 3.6 here, since it is similar to that of Theorem 3.3.