Now we are prepared to provide our main oscillatory theorems. By using the Riccati transformation and the integral averaging technique due to Philos [15], we establish new oscillation results for (1.1). Firstly, let us introduce now the class of functions \(\mathcal{P}\) which will be used in this section. Let
$$D_{0} =\bigl\{ (t, s)_{\mathbb{T}}: t>s>t_{0}\bigr\} \quad\text{and}\quad D = \bigl\{ (t, s)_{\mathbb{T}}: t\geq s>t_{0}\bigr\} . $$
A function \(H\in C_{rd}(D, \mathbb{R})\) is said to belong to the class \(\mathcal{P}\) if
$$ \textstyle\begin{cases} H(t, s)>0,& (t, s)_{\mathbb{T}}\in D_{0},\\ H(t,s)=0, &s=t, \end{cases} $$
and \(H(t, s)\) has a continuous and nonpositive partial derivative on \(D_{0}\) with respect to the second variable, and for a positive continuous function h̄,
$$-H^{\Delta_{s}}(t,s)=\bar{h}(t,s)\sqrt{H(t,s)}, \quad(t,s)_{\mathbb{T}}\in D_{0}. $$
When \(H(t,s) = (t-s)^{n}\), \(n\in N\), the Philos-type conditions reduce to the Kamenev-type ones.
Theorem 3.1
Assume that (1.2) (1.3) hold, \(\alpha\geq\beta\). If there exist a function
\(m\in C_{rd}(I, \mathbb{R})\)
such that
\(m(t)>0\)
and a function
\(H(t, s)\in\mathcal{P}\)
satisfying
$$ \begin{gathered} \limsup_{t\rightarrow\infty} \frac{1}{H(t,t_{1})} \int_{t_{1}}^{t} \biggl[k_{2}m(s)q(s)H(t,s) - \frac{P^{2}(t,s)}{4B(s)} \biggr]\Delta s=\infty\quad\textit{for all large } t\geq t_{1}, \end{gathered} $$
(3.1)
where
$$ \begin{aligned} P(t,s)=\bar{h}(t,s)-A(s)\sqrt{H(t,s)}, \end{aligned} $$
and
$$ \textstyle\begin{cases} A(t)=\frac{m^{\Delta}(t)}{m(\sigma(t))}-\frac{k_{1}m(t)p(t)}{m(\sigma (t))r_{1}( a(t))}R_{2}(\sigma(t),t_{1}),\\ B(t)=c^{*}m(t)m^{-2}(\sigma(t))g^{\Delta}(t)(R^{*}(g(\sigma(t)),t_{1}))^{\beta-1} (\frac{R_{2}(g(t),t_{1})}{r_{1}(\xi)} )^{1/\alpha}.\end{cases} $$
Moreover, if every solution of the equation
$$ \begin{gathered} \bigl(r_{2}z^{\Delta}\bigr)^{\Delta}(t)-Q(t)z\bigl(a(t)\bigr) = 0 \end{gathered} $$
(3.2)
is oscillatory, where
$$ \begin{gathered} Q(t)=ck_{2}q(t) \bigl(R_{1}\bigl(a(t),g(t)\bigr)\bigr)^{\beta}- \frac{k_{1}p(t)}{r_{1}( a(t))},\quad t\geq t_{1}, \end{gathered} $$
for all constants
\(c,c^{*}> 0\). Then every solution
\(y(t)\)
of (1.1) or
\(L_{2}y(t)\)
is oscillatory.
Proof
If y is a nonoscillatory solution of (1.1) on \([t_{1},\infty)_{\mathbb{T}}\), \(t_{1}\geq t_{0}\). Assume that \(y(t)>0\) and \(y(g(t))>0\) for \(t\geq t_{1}\). By the proof of Lemma 2.1, we have that two cases of Lemma 2.1 hold. Now, we shall show that in each case we are led to a contradiction.
Case (1). Suppose that (2.1) of Lemma 2.1 holds. Define the following Riccati transformation:
$$ \begin{aligned} w(t)&=m(t)\frac{L_{2}y(t)}{y^{\beta}(g(t))},\quad t \in[t_{1},\infty)_{\mathbb{T}}. \end{aligned} $$
(3.3)
Then \(w(t)>0\), and
$$ \begin{aligned}[b] w^{\Delta}(t)&= \biggl[m(t) \frac{L_{2}y(t)}{y^{\beta}(g(t))} \biggr]^{\Delta}\\ &=m^{\Delta}(t)\frac{L_{2}y(\sigma(t))}{y^{\beta}(g(\sigma(t)))}+m(t)\frac {(L_{2}y)^{\Delta}(t)y^{\beta}(g(t)) -L_{2}y(t)[y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))} \\ &=\frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr)+m(t)\frac{(L_{2}y)^{\Delta}(t)}{y^{\beta}(g(\sigma(t)))} -m(t) \frac{L_{2}y(t)[y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma (t)))} \\ &=\frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr)+\frac{m(t)}{m(\sigma (t))}\frac{(L_{2}y)^{\Delta}(t)}{L_{2}y(\sigma(t))}w \bigl(\sigma(t)\bigr) -m(t)\frac{L_{2}y(t)[y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))}. \end{aligned} $$
(3.4)
By (H3) and \(y(g(\sigma(t)))\geq y(g(t))\), we have \(f(t, y(g(t)))\geq k_{2}y^{\beta}(g(\sigma(t)))\). From (1.1) and (2.4), then
$$\begin{aligned}& \frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr)+ \frac{m(t)}{m(\sigma (t))}\frac{(L_{2}y)^{\Delta}(t)}{L_{2}y(\sigma(t))}w\bigl(\sigma(t)\bigr) \\& \quad\leq\frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr)-\frac{m(t)}{m(\sigma (t))} \frac{\frac{k_{1}p(t)}{r_{1}( a(t))}L_{1}y( a(t)) +q(t)f(t,y(g(t)))}{L_{2}y(\sigma(t))}w\bigl(\sigma(t)\bigr) \\& \quad\leq\frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr)-\frac{\frac {k_{1}p(t)}{r_{1}(a(t))}L_{1}y(\sigma(t))m(t)}{m(\sigma(t))L_{2}y(\sigma (t))}w\bigl(\sigma(t) \bigr) -\frac{m(t)q(t)f(t,y(g(t)))}{m(\sigma(t))L_{2}y(\sigma(t))}w\bigl(\sigma(t)\bigr) \\& \quad\leq\frac{m^{\Delta}(t)}{m(\sigma(t))}w\bigl(\sigma(t)\bigr) -\frac{k_{1}m(t)p(t)}{m(\sigma(t))r_{1}(a(t))}R_{2} \bigl(\sigma(t),t_{1}\bigr)w\bigl(\sigma (t)\bigr)-k_{2}m(t)q(t) \\& \quad= \biggl[\frac{m^{\Delta}(t)}{m(\sigma(t))}-\frac{k_{1}m(t)p(t)}{m(\sigma (t))r_{1}( a(t))}R_{2}\bigl( \sigma(t),t_{1}\bigr) \biggr]w\bigl(\sigma(t)\bigr)-k_{2}m(t)q(t) \\& \quad=A(t)w\bigl(\sigma(t)\bigr)-k_{2}m(t)q(t). \end{aligned}$$
Now, according to the method given in [16], and by Lemma 2.3, we have
$$ \bigl(y^{\beta}\bigl(g(t)\bigr)\bigr)^{\Delta}\geq \textstyle\begin{cases} \beta(y(g(\sigma(t))))^{\beta-1}(y(g(t)))^{\Delta}, &0< \beta\leq1,\\ \beta(y(g(t)))^{\beta-1}(y(g(t)))^{\Delta}, &\beta\geq1. \end{cases} $$
(3.5)
Then, if \(\sigma(t) > t\), by Lemma 2.4, we get
$$ \begin{aligned} \bigl(y\bigl(g(t)\bigr)\bigr)^{\Delta}= \frac{y(g(\sigma(t)))-y(g(t))}{\sigma(t)-t} =\frac{y(g(\sigma(t)))-y(g(t))}{g(\sigma(t))-g(t)}g^{\Delta}(t) \geq y^{\Delta}( \xi)g^{\Delta}(t), \end{aligned} $$
where \(\xi\in[g(t),g(\sigma(t)))\). If \(\sigma(t) = t\), we obtain \(g(\sigma(t))=\sigma(g(t))=g(t)\) and
$$\bigl(y^{\beta}\bigl(g(t)\bigr)\bigr)^{\Delta}=y' \bigl(g(t)\bigr)g'(t). $$
Moreover, since \(L_{2}y(t)>0\), which implies that \(r_{1}(t)(y^{\Delta}(t))^{\alpha}\) is increasing, then
$$r_{1}(\xi) \bigl(y^{\Delta}(\xi)\bigr)^{\alpha}\geq r_{1}\bigl(g(t)\bigr) \bigl(y^{\Delta}\bigl(g(t)\bigr) \bigr)^{\alpha}, $$
that is,
$$y^{\Delta}(\xi)\geq \biggl(\frac{r_{1}(g(t))}{r_{1}(\xi)} \biggr)^{1/\alpha }y^{\Delta}\bigl(g(t)\bigr), $$
thus
$$ \begin{aligned} \bigl(y\bigl(g(t)\bigr)\bigr)^{\Delta}\geq \biggl(\frac{r_{1}(g(t))}{r_{1}(\xi)} \biggr)^{1/\alpha}y^{\Delta}\bigl(g(t) \bigr)g^{\Delta}(t). \end{aligned} $$
Then, for \(0<\beta\leq1\),
$$ \begin{aligned} -m(t)\frac{L_{2}y(t)[y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))} &\leq-m\bigl(\sigma(t)\bigr) \frac{m(t)L_{2}y(\sigma(t))[y^{\beta}(g(t))]^{\Delta}}{m(\sigma(t))y^{2\beta}(g(\sigma(t)))} \\ &=-w\bigl(\sigma(t)\bigr)\frac{m(t)[y^{\beta}(g(t))]^{\Delta}}{m(\sigma(t))y^{\beta}(g(\sigma(t)))} \\&\leq-w\bigl(\sigma(t)\bigr) \frac{m(t)\beta y^{\beta-1}(g(\sigma(t)))g^{\Delta}(t) r_{1}^{1/\alpha}(g(t))y^{\Delta}(g(t))}{m(\sigma(t))y^{\beta}(g(\sigma (t)))r_{1}^{1/\alpha}(\xi)} \\ &=-\beta w\bigl(\sigma(t)\bigr)\frac{m(t)g^{\Delta}(t)r_{1}^{1/\alpha}(g(t))y^{\Delta}(g(t))}{ m(\sigma(t))y(g(\sigma(t)))r_{1}^{1/\alpha}(\xi)},\quad 0< \beta\leq1. \end{aligned} $$
And for \(\beta\geq1\),
$$ \begin{aligned} -m(t)\frac{L_{2}y(t)[(y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))} &\leq-m\bigl(\sigma(t)\bigr) \frac{m(t)L_{2}y(\sigma(t))[y^{\beta}(g(t))]^{\Delta}}{m(\sigma(t))y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))} \\ &=-w\bigl(\sigma(t)\bigr)\frac{m(t)[y^{\beta}(g(t))]^{\Delta}}{m(\sigma(t))y^{\beta}(g(t))} \\&\leq-w\bigl(\sigma(t)\bigr) \frac{m(t)\beta y^{\beta-1}(g(t))g^{\Delta}(t)r_{1}^{1/\alpha}(g(t))y^{\Delta}(g(t))}{ m(\sigma(t))y^{\beta}(g(t))r_{1}^{1/\alpha}(\xi)} \\ &=-w\bigl(\sigma(t)\bigr)\frac{m(t)\beta g^{\Delta}(t)r_{1}^{1/\alpha}(g(t))y^{\Delta}(g(t))}{ m(\sigma(t))y(g(t))r_{1}^{1/\alpha}(\xi)} \\&\leq-\beta w\bigl(\sigma(t)\bigr) \frac{m(t) g^{\Delta}(t)r_{1}^{1/\alpha }(g(t))y^{\Delta}(g(t))}{ m(\sigma(t))y(g(\sigma(t)))r_{1}^{1/\alpha}(\xi)},\quad \beta\geq1. \end{aligned} $$
Altogether, for all \(\beta>0\), one has
$$ \begin{aligned} -m(t)\frac{L_{2}y(t)[y^{\beta}(g(t))]^{\Delta}}{y^{\beta}(g(t))y^{\beta}(g(\sigma(t)))} \leq-\beta w \bigl(\sigma(t)\bigr)\frac{m(t)y^{\Delta}(g(t))g^{\Delta}(t)r_{1}^{1/\alpha}(g(t))}{ m(\sigma(t))y(g(\sigma(t)))r_{1}^{1/\alpha}(\xi)}. \end{aligned} $$
Then (3.4) implies that
$$ \begin{aligned} w^{\Delta}(t)\leq A(t)w\bigl( \sigma(t)\bigr)-k_{2}m(t)q(t)-\beta w\bigl(\sigma(t)\bigr) \frac {m(t)y^{\Delta}(g(t))g^{\Delta}(t)r_{1}^{1/\alpha}(g(t))}{ m(\sigma(t))y(g(\sigma(t)))r_{1}^{1/\alpha}(\xi)}. \end{aligned} $$
(3.6)
By (2.4), we have
$$ \begin{aligned} y^{\Delta}\bigl(g(t)\bigr)&= \biggl( \frac{1}{r_{1}(g(t))}L_{1}y\bigl(g(t)\bigr) \biggr)^{1/\alpha} \geq \biggl(\frac{R_{2}(g(t),t_{1})}{r_{1}(g(t))} \biggr)^{1/\alpha} \bigl(L_{2}y\bigl(g(t) \bigr) \bigr)^{1/\alpha} \\ &\geq \biggl(\frac{R_{2}(g(t),t_{1})}{r_{1}(g(t))} \biggr)^{1/\alpha} \bigl(L_{2}y \bigl(\sigma(t)\bigr) \bigr)^{1/\alpha}. \end{aligned} $$
Further,
$$ \begin{aligned} \frac{y^{\Delta}(g(t))}{y(g(\sigma(t)))}&\geq \biggl(\frac {R_{2}(g(t),t_{1})}{m(\sigma(t))r_{1}(g(t))} \biggr)^{1/\alpha} \frac{m^{1/\alpha}(\sigma(t))(L_{2}y)^{1/\alpha}(\sigma(t))}{y^{\beta /\alpha}(g(\sigma(t)))} y^{\beta/\alpha-1}\bigl(g\bigl(\sigma(t) \bigr)\bigr) \\ &\overset{(3.4)}{=} \biggl(\frac{R_{2}(g(t),t_{1})}{m(\sigma (t))r_{1}(g(t))} \biggr)^{1/\alpha}w^{1/\alpha} \bigl(\sigma(t)\bigr) y^{\beta/\alpha-1}\bigl(g\bigl(\sigma(t)\bigr)\bigr). \end{aligned} $$
Then (3.6) implies that
$$ \begin{aligned}[b] w^{\Delta}(t)\leq {}&A(t)w\bigl( \sigma(t)\bigr)-k_{2}m(t)q(t)\\& - \beta w^{1/\alpha +1}\bigl(\sigma(t) \bigr)y^{\beta/\alpha-1}\bigl(g\bigl(\sigma(t)\bigr)\bigr) \frac{m(t)g^{\Delta}(t)R_{2}^{1/\alpha}(g(t),t_{1})}{m^{1/\alpha+1}(\sigma(t))r_{1}^{1/\alpha}(\xi )} . \end{aligned} $$
(3.7)
What is more,
$$ \begin{aligned} r_{1}(t) \bigl(y^{\Delta}\bigr)^{\alpha}(t)&=L_{1}y(t)=L_{1}y(t_{1})+ \int_{t_{1}}^{t}(L_{1}y)^{\Delta}(s) \Delta s \leq L_{1}y(t_{1})+c_{1} \int_{t_{1}}^{t}\frac{\Delta s}{r_{2}(s)} \\ &=L_{1}y(t_{1})+c_{1}R_{2}(t,t_{1})= \biggl[\frac{L_{1}y(t_{1})}{R_{2}(t,t_{1})}+c_{1} \biggr]R_{2}(t,t_{1}) \\ &\leq \biggl[\frac{L_{1}y(t_{1})}{R_{2}(t_{2},t_{1})}+c_{1} \biggr]R_{2}(t,t_{1})= \tilde {c}_{1}R_{2}(t,t_{1}) \end{aligned} $$
holds for all \(t\geq t_{2}\), where \(c_{1}=L_{2}y(t_{1})\) and \(\tilde{c}_{1}=c_{1}+\frac{L_{1}y(t_{1})}{R_{2}(t_{2},t_{1})}\). And
$$ \begin{aligned}[b] y(t)&=y(t_{2})+ \int_{t_{2}}^{t}y^{\Delta}(s)\Delta s\leq y(t_{2})+ \int_{t_{2}}^{t} \biggl(\frac{\tilde{c}_{1}R_{2}(s,t_{1})}{r_{1}(s)} \biggr)^{1/\alpha}\Delta s \\ &\leq y(t_{2})+ \int_{t_{1}}^{t} \biggl(\frac{\tilde {c}_{1}R_{2}(s,t_{1})}{r_{1}(s)} \biggr)^{1/\alpha}\Delta s=y(t_{2})+\tilde {c}_{1}^{1/\alpha}R^{*}(t,t_{1}) \\ &= \biggl[\frac{y(t_{2})}{R^{*}(t,t_{1})}+\tilde{c}_{1}^{1/\alpha} \biggr]R^{*}(t,t_{1}) \leq \biggl[\frac{y(t_{2})}{R^{*}(t_{2},t_{1})}+\tilde{c}_{1}^{1/\alpha} \biggr]R^{*}(t,t_{1})=c_{2}R^{*}(t,t_{1}) \end{aligned} $$
(3.8)
holds for all \(t\geq t_{2}\geq t_{1}\), where \(c_{2}=\frac{y(t_{2})}{R^{*}(t_{2},t_{1})}+\tilde{c}_{1}^{1/\alpha}\). By (3.3) and (2.5), we have
$$ \begin{aligned} w(t)&=m(t)\frac{L_{2}y(t)}{y^{\beta}(g(t))}\leq m(t) \frac{L_{2}y(g(t))}{y^{\beta}(g(t))}\leq m(t) \bigl(R^{*}\bigl(g(t),t_{1}\bigr) \bigr)^{-\alpha }y^{\alpha-\beta}\bigl(g(t)\bigr),\quad t\geq t_{1}. \end{aligned} $$
(3.9)
Using (3.8) in (3.9), we get
$$ \begin{aligned} w(t)\leq c_{2}^{\alpha-\beta} m(t) \bigl(R^{*}\bigl(g(t),t_{1}\bigr)\bigr)^{-\beta},\quad t\geq t_{2}. \end{aligned} $$
(3.10)
Using (3.8) and (3.10) in (3.7), we obtain
$$ \begin{aligned}[b] w^{\Delta}(t)\leq{}& A(t)w\bigl( \sigma(t)\bigr)-k_{2}m(t)q(t) \\ &-w^{2}\bigl(\sigma(t)\bigr) \biggl[\beta c_{2}^{\beta-\alpha} m(t)m^{-2}\bigl(\sigma(t)\bigr) g^{\Delta}(t) \bigl(R^{*}\bigl(g \bigl(\sigma(t)\bigr),t_{1}\bigr)\bigr)^{\beta-1} \biggl( \frac {R_{2}(g(t),t_{1})}{r_{1}(\xi)} \biggr)^{1/\alpha} \biggr] \\ \leq{}& A(t)w\bigl(\sigma(t)\bigr)-k_{2}m(t)q(t)-B(t)w^{2} \bigl(\sigma(t)\bigr),\quad t\geq t_{2}, \end{aligned} $$
(3.11)
where \(c^{*}=\beta c_{2}^{\beta-\alpha}\).
Next,
$$\begin{aligned}& \int_{t_{1}}^{t}k_{2}m(s)q(s)H(t,s)\Delta s \\& \quad\leq \int_{t_{1}}^{t}H(t,s) \bigl\{ -w^{\Delta}(s)+A(s)w \bigl(\sigma (s)\bigr)-B(s)w^{2}\bigl(\sigma(s)\bigr) \bigr\} \Delta s \\& \quad=-H(t,s)w(s)\big|_{s=t_{1}}^{s=t}+ \int_{t_{1}}^{t} \bigl\{ H^{\Delta _{s}}(t,s)w\bigl( \sigma(s)\bigr)+H(t,s) \bigl[A(s)w\bigl(\sigma(s)\bigr)-B(s)w^{2}\bigl( \sigma(s)\bigr) \bigr] \bigr\} \Delta s \\& \quad=H(t,t_{1})w(t_{1})- \int_{t_{1}}^{t} \bigl\{ H(t,s)B(s)w^{2}\bigl( \sigma(s)\bigr)+w\bigl(\sigma (s)\bigr) \bigl[\bar{h}(t,s)\sqrt{H(t,s)}-H(t,s)A(s) \bigr] \bigr\} \Delta s \\& \quad=H(t,t_{1})w(t_{1})- \int_{t_{1}}^{t} \biggl\{ \sqrt{H(t,s)}\sqrt{B(s)}w\bigl( \sigma (s)\bigr)+\frac{P(t,s)}{2\sqrt{B(s)}} \biggr\} ^{2}\Delta s + \int_{t_{1}}^{t}\frac {P^{2}(t,s)}{4B(s)}\Delta s \\& \quad\leq H(t,t_{1})w(t_{1})+ \int_{t_{1}}^{t}\frac{P^{2}(t,s)}{4B(s)}\Delta s. \end{aligned}$$
(3.12)
Therefore,
$$\frac{1}{H(t,t_{1})} \int_{t_{1}}^{t} \biggl[k_{2}m(s)q(s)H(t,s)- \frac {P^{2}(t,s)}{4B(s)} \biggr]\Delta s\leq w(t_{1}), $$
which contradicts with (3.1).
Case (2). Suppose that (2.2) of Lemma 2.1 holds. Now, for \(v\geq u\geq t_{2}\), we have
$$ \begin{aligned} y(u)&>y(u)-y(v)\\&=- \int_{u}^{v} r_{1}^{-1/\alpha}(\tau) \bigl(r_{1}(\tau) \bigl(y^{\Delta}(\tau )\bigr)^{\alpha}\bigr)^{1/\alpha}\Delta\tau\geq \biggl( \int_{u}^{v}r_{1}^{-1/\alpha}(\tau ) \Delta\tau \biggr) \bigl(-L_{1}y(v)\bigr)^{1/\alpha} \\ &=R_{1}(v,u) \bigl(-L_{1}y(v)\bigr)^{1/\alpha}. \end{aligned} $$
Letting \(u=g(t)\) and \(v=a(t)\),
$$y\bigl(g(t)\bigr)>R_{1}\bigl(a(t),g(t)\bigr) \bigl(-L_{1}y \bigl(a(t)\bigr)\bigr)^{1/\alpha}=R_{1}\bigl(a(t),g(t)\bigr)x \bigl(a(t)\bigr) $$
for \(a(t)\geq g(t)\geq t_{2}\), where \(x(t)=(-L_{1}y(t))^{1/\alpha}>0\) for \(t\geq t_{2}\). By (H3) and \(y(g(t))\geq y(g(\sigma(t)))\), we have \(f(t, y(g(t)))\geq k_{2}y^{\beta}(g(t))\). Then from (1.1) and combined with the fact that \(x(t)\) is decreasing, we get
$$\bigl(r_{2}z^{\Delta}\bigr)^{\Delta}(t)+ \frac{k_{1}p(t)}{r_{1}( a(t))}z\bigl(a(t)\bigr) \geq k_{2}q(t) \bigl(R_{1} \bigl(a(t),g(t)\bigr)\bigr)^{\beta}z\bigl(a(t)\bigr)z^{\beta/\alpha-1} \bigl(a(t)\bigr), $$
where \(z(t)= x^{\alpha}(t)>0\). Since \(z(t)\) is decreasing and \(\alpha\geq \beta\), there exists a constant \(c_{4}>0\) such that \(z^{\beta/\alpha-1}(t)\geq c_{4}\) for \(t\geq t_{2}\). Then
$$ \begin{aligned} \bigl(r_{2}z^{\Delta}\bigr)^{\Delta}(t)&\geq k_{2}q(t) \bigl(R_{1} \bigl(a(t),g(t)\bigr)\bigr)^{\beta}z\bigl(a(t)\bigr)z^{\beta/\alpha-1} \bigl(a(t)\bigr)-\frac{k_{1}p(t)}{r_{1}( a(t))}z\bigl(a(t)\bigr) \\ &\geq \biggl[c_{4}k_{2}q(t) \bigl(R_{1} \bigl(a(t),g(t)\bigr)\bigr)^{\beta}-\frac{k_{1}p(t)}{r_{1}( a(t))} \biggr]z\bigl(a(t) \bigr). \end{aligned} $$
This gives
$$ \begin{aligned} \bigl(r_{2}z^{\Delta}\bigr)^{\Delta}(t)\geq Q(t)z\bigl(a(t)\bigr), \end{aligned} $$
(3.13)
then
$$ \begin{aligned} \bigl(r_{2}z^{\Delta}\bigr)^{\Delta}(t)-Q(t)z\bigl(a(t)\bigr)\geq0. \end{aligned} $$
(3.14)
And \(z(t)\) is an eventually positive solution of inequation (3.14). Integrating \(y(t)=-z^{\Delta}(t)>0\) from \(t_{1}\) to \(t\geq t_{1}\), we obtain
$$z(t)=z(t_{1})- \int_{t_{1}}^{t}y(s)\Delta s, $$
then we have
$$z\bigl( a(t) \bigr)=z(t_{1})- \int_{t_{1}}^{ a(t) }y(s)\Delta s, $$
and (3.14) can be written as
$$ \begin{aligned} (r_{2}y)^{\Delta}(t)+Q(t) \biggl(z(t_{1})- \int_{t_{1}}^{ a(t) }y(s)\Delta s \biggr)\leq0,\quad t\geq t_{1}. \end{aligned} $$
(3.15)
Integrating (3.15) from t to \(u\geq t\geq t_{1}\) and \(u\rightarrow\infty\), we obtain
$$ \begin{aligned} y(t)\geq\frac{1}{r_{2}(t)} \int_{t}^{+\infty} Q(s) \biggl(z(t_{1})- \int_{t_{1}}^{ a(s) }y(\tau)\Delta\tau \biggr)\Delta s. \end{aligned} $$
(3.16)
Now define the sequence \(\{x_{j}(t)\}_{j\in N_{0}}\): \(x_{0}(t)=y(t)\):
$$ \begin{aligned} x_{j+1}(t)=\frac{1}{r_{2}(t)} \int_{t}^{+\infty} Q(s) \biggl(z(t_{1}) - \int_{t_{1}}^{a(s)}x_{j}(\tau)\Delta\tau \biggr)\Delta s,\quad j\in N_{0}, t\geq t_{1}. \end{aligned} $$
(3.17)
Then by (3.16) we get
$$0< x_{j}(t)\leq y(t),\quad \text{and}\quad x_{j+1}(t)\leq x_{j}(t),\quad j\in N_{0}, t\geq t_{1}. $$
So we obtain that the sequence \(\{x_{j}(t)\}_{j\in N_{0}}\) is positive and nonincreasing on j. Then we define
$$x(t)=\lim_{j\rightarrow\infty}x_{j}(t)\geq0. $$
By the Lebesgue control convergence theorem [17], from (3.17), we have
$$r_{2}(t)x(t)= \int_{t}^{+\infty} Q(s) \biggl(z(t_{1})- \int_{t_{1}}^{a(s)}x(\tau )\Delta\tau \biggr)\Delta s, $$
then
$$ \begin{aligned} \bigl(r_{2}(t)x(t) \bigr)^{\Delta}=-Q(t) \biggl(z(t_{1})- \int_{t_{1}}^{ a(t) }x(s)\Delta s \biggr). \end{aligned} $$
(3.18)
Let
$$v(t)=z(t_{1})- \int_{t_{1}}^{t}x(s)\Delta s>0 $$
and
$$ \begin{aligned} v^{\Delta}(t)=-x(t). \end{aligned} $$
(3.19)
From (3.18) we get
$$ \begin{aligned} \bigl(r_{2}(t)v^{\Delta}(t) \bigr)^{\Delta}=-\bigl(r_{2}(t)x(t)\bigr)^{\Delta}=Q(t)v \bigl(a(t)\bigr), \end{aligned} $$
(3.20)
where
$$v\bigl(a(t)\bigr)=z(t_{1})- \int_{ t_{1} }^{a(t)}x(s)\Delta s. $$
By (3.19), (3.20), we get
$$\bigl(r_{2}v^{\Delta}\bigr)^{\Delta}(t)-Q(t)v\bigl(a(t) \bigr)=0. $$
So v is a positive solution of (3.2), which contradicts with (3.2) is oscillatory. This completes the proof. □
Theorem 3.2
Assume that the hypotheses of Theorem 3.1
hold, except (3.1). Moreover, suppose that, for all
\(t\in I\),
$$ \begin{aligned} \limsup_{t\rightarrow\infty} \int_{t_{1}}^{t} \biggl[k_{2}m(s)q(s)- \frac {A^{2}(s)}{4B(s)} \biggr]\Delta s=\infty. \end{aligned} $$
(3.21)
Then every solution
\(y(t)\)
of (1.1) or
\(L_{2}y(t)\)
is oscillatory.
Proof
If y is a nonoscillatory solution of (1.1) on \([t_{1},\infty )_{\mathbb{T}}\). Assume that \(y(t)>0\) and \(y(g(t))> 0\) for \(t\geq t_{1}\). By the proof of Lemma 2.1, we have that two cases of Lemma 2.1 hold.
Case (1). Suppose that (2.1) of Lemma 2.1 holds, then proceeding as in the proof of Theorem 3.1, we obtain (3.11), then
$$ \begin{aligned}[b] w^{\Delta}(t) \leq{}&A(t)w\bigl( \sigma(t)\bigr)-k_{2}m(t)q(t)-B(t)w^{2}\bigl(\sigma(t)\bigr) \\ ={}&{-}k_{2}m(t)q(t)- \biggl(\sqrt{B(t)}w\bigl(\sigma(t)\bigr)- \frac{A(t)}{2\sqrt {B(t)}} \biggr)^{2}+\frac{A^{2}(t)}{4B(t)} \\ \leq{}& {-}k_{2}m(t)q(t)+\frac{A^{2}(t)}{4B(t)},\quad t\geq t_{2}. \end{aligned} $$
(3.22)
Integrating (3.22) from \(t_{2}\) to t, we get
$$\int_{t_{2}}^{t} \biggl[k_{2}m(s)q(s)- \frac{A^{2}(s)}{4B(s)} \biggr]\Delta s \leq w(t_{2})-w(t)\leq w(t_{2}), $$
which contradicts with (3.21).
Case (2). The proof of the case if (2.2) of Lemma 2.1 holds is similar to the proof of Theorem 3.1 and hence it is omitted. □
Theorem 3.3
Assume that the hypotheses of Theorem 3.1
hold, except (3.1). Moreover, suppose that, for every
\(t_{1}>t_{0}\),
$$ \begin{gathered} \mathrm{(I)}\quad 0< \inf_{s\geq t_{1}} \biggl[\liminf _{t\rightarrow\infty}\frac {H(t,s)}{H(t,t_{1})} \biggr]< \infty, \\ \mathrm{(II)}\quad \limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})} \int_{t_{1}}^{t}\frac {P^{2}(t,s)}{B(s)}\Delta s< \infty, \end{gathered} $$
and there exists
\(\psi\in C_{rd}(I)\)
such that
$$ \begin{gathered} \mathrm{(III)}\quad \int_{t_{1}}^{\infty}\psi^{2}_{+}\bigl(\sigma(s) \bigr)B(s)\Delta s=\infty,\qquad \psi _{+}(s)=\max\bigl\{ \psi(s),0\bigr\} , \\ \mathrm{(IV)}\quad \limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})} \int_{t_{1}}^{t} \biggl[k_{2}m(s)q(s)H(t,s)- \frac{P^{2}(t,s)}{4B(s)} \biggr]\Delta s\geq\psi (t_{1}). \end{gathered} $$
Then every solution
\(y(t)\)
of (1.1) or
\(L_{2}y(t)\)
is oscillatory.
Proof
If y is a nonoscillatory solution of (1.1) on \([t_{1},\infty )_{\mathbb{T}}\). Assume that \(y(t)>0\) and \(y(g(t))> 0\) for \(t\geq t_{1}\). By the proof of Lemma 2.1, we have that two cases of Lemma 2.1 hold.
Case (1). Suppose that (2.1) of Lemma 2.1 holds, then proceeding as in the proof of Theorem 3.1, we obtain (3.12), then
$$ \begin{gathered} \int_{t_{1}}^{t}k_{2}m(s)q(s)H(t,s)\Delta s \\ \quad\leq H(t,t_{1})w(t_{1})+ \int_{t_{1}}^{t}\frac{P^{2}(t,s)}{4B(s)}\Delta s- \int _{t_{1}}^{t} \biggl[\sqrt{H(t,s)}\sqrt{B(s)}w\bigl( \sigma(s)\bigr)+\frac{P(t,s)}{2\sqrt {B(s)}} \biggr]^{2}\Delta s. \end{gathered} $$
By (IV), we get
$$ \begin{aligned} \psi(t_{1})&\leq\limsup _{t\rightarrow\infty}\frac{1}{H(t,t_{1})} \int _{t_{1}}^{t} \biggl[k_{2}m(s)q(s)H(t,s)- \frac{P^{2}(t,s)}{4B(s)} \biggr]\Delta s \\ &\leq w(t_{1})\\&\quad{}-\liminf_{t\rightarrow\infty}\frac{1}{H(t,t_{1})} \int _{t_{1}}^{t} \biggl[\sqrt{H(t,s)}\sqrt{B(s)}w\bigl( \sigma(s)\bigr)+\frac{P(t,s)}{2\sqrt {B(s)}} \biggr]^{2}\Delta s\quad \text{for all } t_{1}\geq t_{0}, \end{aligned} $$
which implies that
$$ \begin{aligned} \psi(t)\leq w(t),\quad t\geq t_{0}, \end{aligned} $$
(3.23)
and
$$ \begin{aligned} \liminf_{t\rightarrow\infty} \frac{1}{H(t,t_{1})} \int_{t_{1}}^{t} \biggl[\sqrt {H(t,s)}\sqrt{B(s)}w\bigl( \sigma(s)\bigr)+\frac{P(t,s)}{2\sqrt{B(s)}} \biggr]^{2}\Delta s< \infty. \end{aligned} $$
(3.24)
Now, define
$$ \textstyle\begin{cases} c_{1}(t)=\frac{1}{H(t,t_{1})}\int_{t_{1}}^{t}H(t,s)B(s)w^{2}(\sigma(s))\Delta s, &t>t_{1},\\ c_{2}(t)=\frac{1}{H(t,t_{1})}\int_{t_{1}}^{t}\sqrt{H(t,s)}P(t,s)w(\sigma (s))\Delta s, &t>t_{1}. \end{cases} $$
It follows from (3.24) that
$$ \begin{aligned} \liminf_{t\rightarrow\infty} \bigl[c_{1}(t)+c_{2}(t) \bigr]< \infty. \end{aligned} $$
(3.25)
Suppose that
$$ \begin{aligned} \int^{\infty}_{t_{1}}w^{2}\bigl(\sigma(s) \bigr)B(s)\Delta s=\infty, \end{aligned} $$
(3.26)
i.e.,
$$ \begin{aligned} \lim_{t\rightarrow\infty}c_{1}(t)= \infty. \end{aligned} $$
(3.27)
In fact, let l be an arbitrary positive number. By condition (I) we can take a constant δ with
$$ \begin{aligned} \inf_{s\geq t_{1}} \biggl[\liminf _{t\rightarrow\infty}\frac {H(t,s)}{H(t,t_{1})} \biggr]>\delta>0. \end{aligned} $$
Since (3.26), there exists \(T_{1}>t_{1}\) such that
$$\int^{t}_{t_{1}}w^{2}\bigl(\sigma(\xi)\bigr)B( \xi)\Delta\xi\geq\frac{l}{\delta} \quad\text{for all } t\geq T_{1}. $$
Then, for every \(t>t_{1}\), we have
$$ \begin{aligned} c_{1}(t)&=\frac{1}{H(t,t_{1})} \int_{t_{1}}^{t}H(t,s) \biggl[ \int_{t_{1}}^{s}w^{2}\bigl(\sigma (\xi) \bigr)B(\xi)\Delta\xi \biggr]^{\Delta_{s}}\Delta s \\ &=\frac{1}{H(t,t_{1})} \int_{t_{1}}^{t} \biggl[ \int_{t_{1}}^{\sigma(s)}w^{2}\bigl(\sigma (\xi) \bigr)B\bigl(\sigma(\xi)\bigr)\Delta\xi \biggr] \bigl[-H^{\Delta_{s}}(t,s) \bigr]\Delta s, \end{aligned} $$
and consequently we have, for \(t\geq T_{1}>t_{1}\),
$$ \begin{aligned} c_{1}(t)&\geq\frac{1}{H(t,t_{1})} \int_{T_{1}}^{t} \biggl[ \int_{t_{1}}^{\sigma (s)}w^{2}\bigl(\sigma(\xi)\bigr)B \bigl(\sigma(\xi)\bigr)\Delta\xi \biggr] \bigl[-H^{\Delta _{s}}(t,s) \bigr]\Delta s \\ &\geq\frac{l/\delta}{H(t,t_{1})} \int_{T_{1}}^{t} \bigl[-H^{\Delta _{s}}(t,s) \bigr] \Delta s =\frac{l}{\delta}\frac{H(t,T_{1})}{H(t,t_{1})}. \end{aligned} $$
But
$$\liminf_{t\rightarrow\infty}\frac{H(t,T_{1})}{H(t,t_{1})}>\delta, $$
we can choose \(T'_{1}\geq T_{1}>t_{1}\) so that
$$ \begin{aligned} \frac{H(t,T_{1})}{H(t,t_{1})}>\delta \end{aligned} $$
(3.28)
for every \(t\geq T'_{1}\). Thus
$$c_{1}(t)\geq l\quad \text{for all } t\geq T'_{1}, $$
which proves (3.27), since \(l>0\) is arbitrary.
Next, we consider a sequence \((\varphi_{v})_{v=1,2,3,\ldots}\) in the interval \((t_{1},\infty)_{\mathbb{T}}\) with \(\lim_{v\rightarrow\infty}\varphi_{v}=\infty\) and
$$\lim_{v\rightarrow\infty} \bigl[c_{1}(\varphi_{v})+c_{2}( \varphi_{v}) \bigr]=\liminf_{t\rightarrow\infty} \bigl[c_{1}(t)+c_{2}(t) \bigr]. $$
Since (3.25), there exists a constant M so that
$$ \begin{aligned}c_{1}(\varphi_{v})+c_{2}( \varphi_{v})\leq M\quad(v=1,2,3,\ldots). \end{aligned} $$
(3.29)
Furthermore, (3.27) guarantees that
$$ \begin{aligned} \lim_{v\rightarrow\infty}c_{1}( \varphi_{v})=\infty. \end{aligned} $$
(3.30)
Hence (3.29) implies
$$ \begin{aligned} \lim_{v\rightarrow\infty}c_{2}( \varphi_{v})=-\infty. \end{aligned} $$
(3.31)
From (3.29) and (3.30), for sufficiently large v, we derive
$$1+\frac{c_{2}(\varphi_{v})}{c_{1}(\varphi_{v})}\leq\frac{M}{c_{1}(\varphi _{v})}< \frac{1}{2}. $$
Thus
$$ \begin{aligned} \frac{c_{2}(\varphi_{v})}{c_{1}(\varphi_{v})}< -\frac{1}{2} \quad\text{for all large } v, \end{aligned} $$
from (3.31),
$$ \begin{aligned} \lim_{v\rightarrow\infty} \frac{c_{2}^{2}(\varphi_{v})}{c_{1}(\varphi _{v})}=\infty. \end{aligned} $$
(3.32)
On the other hand, by Hölder’s inequality [18], for any positive integer v, we have
$$ \begin{aligned} c_{2}^{2}(\varphi_{v})&= \frac{1}{H^{2}(\varphi_{v},t_{1})} \biggl\{ \int_{t_{1}}^{\varphi _{v}}P(\varphi_{v},s)\sqrt{H( \varphi_{v},s)}w\bigl(\sigma(s)\bigr)\Delta s \biggr\} ^{2} \\ &\leq \biggl[\frac{1}{H(\varphi_{v},t_{1})} \int_{t_{1}}^{\varphi_{v}}P^{2}(\varphi _{v},s)\frac{1}{B(s)}\Delta s \biggr] \biggl[\frac{1}{H(\varphi_{v},t_{1})} \int _{t_{1}}^{\varphi_{v}}H(\varphi_{v},s)w^{2} \bigl(\sigma(s)\bigr)B(s)\Delta s \biggr] \\ &\leq \biggl[\frac{1}{H(\varphi_{v},t_{1})} \int_{t_{1}}^{\varphi_{v}}P^{2}(\varphi _{v},s)\frac{1}{B(s)}\Delta s \biggr]c_{1}( \varphi_{v}), \end{aligned} $$
then
$$\frac{c_{2}^{2}(\varphi_{v})}{c_{1}(\varphi_{v})}\leq\frac{1}{H(\varphi _{v},t_{1})} \int_{t_{1}}^{\varphi_{v}}P^{2}(\varphi_{v},s) \frac{1}{B(s)}\Delta s\quad \text{for all large } v. $$
By (3.28), we obtain
$$\frac{H(\varphi_{v},T_{1})}{H(\varphi_{v},t_{1})}>\delta\quad\text{for sufficiently large } v, $$
therefore
$$\frac{c_{2}^{2}(\varphi_{v})}{c_{1}(\varphi_{v})}\leq\frac{1}{\delta}\frac {1}{H(\varphi_{v},t_{1})} \int_{t_{1}}^{\varphi_{v}}P^{2}(\varphi_{v},s) \frac {1}{B(s)}\Delta s \quad\text{for all large } v. $$
Because of (3.32), we get
$$ \begin{aligned} \lim_{v\rightarrow\infty} \frac{1}{H(\varphi_{v},t_{1})} \int_{t_{1}}^{\varphi _{v}}P^{2}(\varphi_{v},s) \frac{1}{B(s)}\Delta s=\infty. \end{aligned} $$
(3.33)
Thus
$$\limsup_{t\rightarrow\infty}\frac{1}{H(t,t_{1})} \int _{t_{1}}^{t}P^{2}(t,s)\frac{1}{B(s)} \Delta s=\infty, $$
which contradicts with condition (II). We have thus proved that (3.26) fails. So, it holds that
$$\int_{T_{0}}^{\infty}w^{2}\bigl(\sigma(s) \bigr)B(s)\Delta s< \infty. $$
By (3.23) and \(\psi_{+}(s)=\max\{\psi(s),0\}\), we get
$$\int_{T_{0}}^{\infty}\psi_{+}^{2}\bigl(\sigma(s) \bigr)B(s)\Delta s\leq \int_{T_{0}}^{\infty}w^{2}\bigl(\sigma(s) \bigr)B(s)\Delta s< \infty, $$
which yields a contradiction to condition (III). This completes the proof.
Case (2). The proof of the case if (2.2) of Lemma 2.1 holds is similar to the proof of Theorem 3.1 and hence it is omitted. □