In this section, we establish the oscillation results for Eq. (1).
Theorem 3.1
Suppose that, for
\(t_{0}\in\mathbb{N}_{0}\),
$$ \liminf_{t\rightarrow\infty} \Biggl\{ \sum _{s=0}^{t-\alpha}\frac {(t-s-1)^{(\alpha-1)}}{V(s)} \Biggl[M+\sum _{\xi=t_{0}}^{s-1}g(\xi)V(\xi) \Biggr] \Biggr\} < 0 $$
(9)
and
$$ \limsup_{t\rightarrow\infty} \Biggl\{ \sum _{s=0}^{t-\alpha}\frac {(t-s-1)^{(\alpha-1)}}{V(s)} \Biggl[M+\sum _{\xi=t_{0}}^{s-1}g(\xi)V(\xi) \Biggr] \Biggr\} >0, $$
(10)
where
M
is a constant, and
$$ V(t)=\prod_{s=t_{0}}^{t-1} \bigl(1+p(s)\bigr). $$
(11)
Then every solution
\(x(t)\)
of Eq. (1) is oscillatory.
Proof
Suppose to the contrary that there is a nonoscillatory solution \(x(t)\) of Eq. (1) which has no zero in \(\mathbb{N}_{t_{0}}=\{ t_{0},t_{0}+1,t_{0}+2,\ldots\}\). Then \(x(t)>0\) or \(x(t)<0\) for \(t\in\mathbb {N}_{t_{0}}\).
Case 1. \(x(t)>0\), \(t\in\mathbb{N}_{t_{0}}\). Noting assumption (A), from Eq. (1) we have
$$ \bigl(1+p(t)\bigr)\Delta\bigl(\Delta^{\alpha}x(t)\bigr)+p(t) \Delta^{\alpha }x(t)=-f\bigl(t,x(t)\bigr)+g(t)< g(t). $$
(12)
Therefore, using the fundamental property of Δ and noting the definition of \(V(t)\), we get
$$\begin{aligned} \Delta\bigl(\bigl(\Delta^{\alpha}x(t)\bigr)V(t)\bigr) =& \Delta\bigl(\Delta^{\alpha}x(t)\bigr)V(t+1)+\Delta^{\alpha}x(t)\Delta V(t) \\ =&\Delta\bigl(\Delta^{\alpha}x(t)\bigr) \bigl(1+p(t)\bigr)V(t)+ \Delta^{\alpha}x(t)p(t)V(t) \\ < & g(t)V(t). \end{aligned}$$
(13)
Summing both sides of (13) from \(t_{0}\) to \(t-1\), we obtain
$$\bigl(\Delta^{\alpha}x(t)\bigr)V(t)< \bigl(\Delta^{\alpha}x(t_{0}) \bigr)V(t_{0})+\sum_{s=t_{0}}^{t-1}g(s)V(s)=M+ \sum_{s=t_{0}}^{t-1}g(s)V(s), $$
where \(M=(\Delta^{\alpha}x(t_{0}))V(t_{0})\), that is,
$$ \Delta^{\alpha}x(t)< \frac{M}{V(t)}+\frac{1}{V(t)}\sum _{s=t_{0}}^{t-1}g(s)V(s). $$
(14)
Applying the \(\Delta^{-\alpha}\) operator to inequality (14), we have
$$ \Delta^{-\alpha}\Delta^{\alpha}x(t)< \Delta^{-\alpha} \Biggl[\frac {M}{V(t)}+\frac{1}{V(t)}\sum _{s=t_{0}}^{t-1}g(s)V(s) \Biggr]. $$
(15)
On the one hand, applying Lemma 2.3 to the left-hand side of (15), we obtain
$$\begin{aligned} \Delta^{-\alpha}\Delta^{\alpha}x(t) =& \Delta^{-\alpha}\Delta\Delta ^{-(1-\alpha)}x(t) \\ =&\Delta\Delta^{-\alpha}\Delta^{-(1-\alpha)}x(t)- \frac {t^{(\alpha-1)}}{\Gamma(\alpha)}x_{0} \\ =&x(t)- \frac{x_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}. \end{aligned}$$
(16)
On the other hand, using Definition 2.1, it follows from the right-hand side of (15) that
$$\begin{aligned}& \Delta^{-\alpha} \Biggl[\frac{M}{V(t)}+\frac{1}{V(t)} \sum_{s=t_{0}}^{t-1}g(s)V(s) \Biggr] \\& \quad = \frac{1}{\Gamma(\alpha)}\sum_{s=0}^{t-\alpha }(t-s-1)^{(\alpha-1)} \Biggl[\frac{M}{V(s)}+\frac{1}{V(s)}\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr]. \end{aligned}$$
(17)
Combining (15)-(17), we get
$$ x(t)< \frac{x_{0}}{\Gamma(\alpha)}t^{(\alpha-1)}+\frac{1}{\Gamma(\alpha)}\sum _{s=0}^{t-\alpha}(t-s-1)^{(\alpha-1)} \Biggl[ \frac{M}{V(s)}+\frac{1}{V(s)}\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr]. $$
(18)
It follows from (18) that
$$\begin{aligned} \Gamma(\alpha)t^{1-\alpha}x(t) < & x_{0}t^{(\alpha-1)}t^{1-\alpha} \\ &{} + t^{1-\alpha}\sum_{s=0}^{t-\alpha}(t-s-1)^{(\alpha-1)} \Biggl[\frac{M}{V(s)}+\frac{1}{V(s)}\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr]. \end{aligned}$$
(19)
By using the Stirling formula [20]
$$\lim_{t\rightarrow\infty}\frac{\Gamma(t)t^{\varepsilon}}{\Gamma (t+\varepsilon)}=1,\quad \varepsilon>0, $$
we obtain
$$\begin{aligned} \lim_{t\rightarrow\infty}t^{1-\alpha}t^{(\alpha -1)} =& \lim_{t\rightarrow\infty}t^{1-\alpha} \frac{\Gamma(t+1)}{\Gamma(t+1-\alpha+1)} \\ =& \lim_{t\rightarrow\infty}t^{1-\alpha}\frac{t\Gamma (t)}{(t+1-\alpha)\Gamma(t+(1-\alpha))} \\ =& \lim_{t\rightarrow\infty}\frac{t}{t+1-\alpha}\frac {\Gamma(t)t^{1-\alpha}}{\Gamma(t+(1-\alpha))} \\ =&1. \end{aligned}$$
(20)
From (20), taking then limit as \(t\rightarrow\infty\) in (19), we have
$$\liminf_{t\rightarrow\infty} \bigl\{ t^{1-\alpha}x(t) \bigr\} \leq-\infty, $$
which contradicts with \(x(t)>0\).
Case 2. \(x(t)<0\), \(t\in\mathbb{N}_{t_{0}}\). By assumption (A), from Eq. (1) we have
$$ \bigl(1+p(t)\bigr)\Delta\bigl(\Delta^{\alpha}x(t)\bigr)+p(t) \Delta^{\alpha }x(t)=-f\bigl(t,x(t)\bigr)+g(t)>g(t). $$
(21)
Therefore,
$$ \Delta\bigl(\bigl(\Delta^{\alpha}x(t)\bigr)V(t) \bigr)>g(t)V(t). $$
(22)
Summing both sides of (22) from \(t_{0}\) to \(t-1\), we obtain
$$\bigl(\Delta^{\alpha}x(t)\bigr)V(t)>\bigl(\Delta^{\alpha}x(t_{0}) \bigr)V(t_{0})+\sum_{s=t_{0}}^{t-1}g(s)V(s)=M+ \sum_{s=t_{0}}^{t-1}g(s)V(s), $$
where \(M=(\Delta^{\alpha}x(t_{0}))V(t_{0})\), that is,
$$ \Delta^{\alpha}x(t)>\frac{M}{V(t)}+\frac{1}{V(t)}\sum _{s=t_{0}}^{t-1}g(s)V(s). $$
(23)
Using the procedure of the proof of Case 1, we conclude that
$$\begin{aligned} \Gamma(\alpha)t^{1-\alpha}x(t) >& x_{0}t^{(\alpha-1)}t^{1-\alpha} \\ &{} + t^{1-\alpha}\sum_{s=0}^{t-\alpha}(t-s-1)^{(\alpha-1)} \Biggl[\frac{M}{V(s)}+\frac{1}{V(s)}\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr]. \end{aligned}$$
(24)
By (20), taking the limit as \(t\rightarrow\infty\) in (24), we have
$$\limsup_{t\rightarrow\infty} \bigl\{ t^{1-\alpha}x(t) \bigr\} \geq\infty, $$
which contradicts with \(x(t)<0\). The proof of Theorem 3.1 is complete. □
Theorem 3.2
Suppose that, for
\(t_{0}\in\mathbb{N}_{0}\),
$$ \liminf_{t\rightarrow\infty}\sum_{s=t_{0}}^{t-1} \frac{1}{V(s)} \Biggl\{ M+\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr\} =-\infty $$
(25)
and
$$ \limsup_{t\rightarrow\infty}\sum_{s=t_{0}}^{t-1} \frac{1}{V(s)} \Biggl\{ M+\sum_{\xi=t_{0}}^{s-1}g( \xi)V(\xi) \Biggr\} =\infty, $$
(26)
where
M
is a constant, and
\(V(t)\)
is defined by (11). Then every solution
\(x(t)\)
of Eq. (1) is oscillatory.
Proof
Suppose to the contrary that there is a nonoscillatory solution \(x(t)\) of Eq. (1) that has no zero in \(\mathbb{N}_{t_{0}}\). Then \(x(t)>0\) or \(x(t)<0\) for \(t\in\mathbb{N}_{t_{0}}\).
Case 1. \(x(t)>0\), \(t\in\mathbb{N}_{t_{0}}\). As in the proof of Case 1 in Theorem 3.1, we obtain (14). By Lemma 2.4 it follows from (14) that
$$ \Delta E(t)< \frac{\Gamma(1-\alpha)}{V(t)} \Biggl\{ M+\sum _{s=t_{0}}^{t-1}g(s)V(s) \Biggr\} . $$
(27)
Summing both sides of (27) from \(t_{0}\) to \(t-1\), we have
$$ E(t)< E(t_{0})+\Gamma(1-\alpha)\sum _{s=t_{0}}^{t-1}\frac{1}{V(s)} \Biggl\{ M+\sum _{\xi=t_{0}}^{s-1}g(\xi)V(\xi) \Biggr\} . $$
(28)
Letting \(t\rightarrow\infty\) in (28), we obtain a contradiction with \(E(t)>0\).
Case 2. \(x(t)<0\), \(t\in\mathbb{N}_{t_{0}}\). As in the proof of Case 2 in Theorem 3.1, we obtain (23). By Lemma 2.4 it follows from (23) that
$$ \Delta E(t)>\frac{\Gamma(1-\alpha)}{V(t)} \Biggl\{ M+\sum _{s=t_{0}}^{t-1}g(s)V(s) \Biggr\} . $$
(29)
Summing both sides of (29) from \(t_{0}\) to \(t-1\), we have
$$ E(t)>E(t_{0})+\Gamma(1-\alpha)\sum _{s=t_{0}}^{t-1}\frac{1}{V(s)} \Biggl\{ M+\sum _{\xi=t_{0}}^{s-1}g(\xi)V(\xi) \Biggr\} . $$
(30)
Letting \(t\rightarrow\infty\) in (30), we obtain a contradiction with \(E(t)<0\). This completes the proof of Theorem 3.2. □