In Theorem 3.1 we use a constant β, the quotient of two odd positive integers with \(\beta >\alpha \), for which
$$ \frac{g(\iota )}{\iota ^{\beta}}\text{ is nondecreasing for }0< \iota . $$
(3.1)
The existence of such a constant can be established by taking \(g(\iota )=|\iota |^{\delta}\operatorname{sgn}(\iota )\) with \(\beta <\delta \).
Theorem 3.1
Let (b)–(f) and (3.1) hold for \(\iota \geq \iota _{0} \). Then every solution of (1.10)–(1.11) is oscillatory if and only if
$$ \int _{0}^{\infty }p^{-1/\alpha}(s) \biggl[ \int _{s}^{\infty }r(\psi ) \,\mathrm{d}\psi +\sum _{\phi _{k}\geq s} \tilde{r}(\phi _{k}) \biggr]^{1/ \alpha} \,\mathrm{d}s =\infty . $$
(3.2)
Proof
Let u be an eventually positive solution of (1.10)–(1.11). Then \(w(\iota )>0\) and there exists \(\iota _{0} \geq 0\) such that \(u(\iota )>0\), \(u(\nu (\iota ))>0\), \(u(\zeta (\iota ))>0\) for all \(\iota \geq \iota _{0} \). Thus, Lemmas 2.1 and 2.2 hold for \(\iota \geq \iota _{1} \). By Lemma 2.1, there exists \(\iota _{2}>\iota _{1} \) such that \(w'(\iota )>0\) for all \(\iota \geq \iota _{2}\). Then there exist \(\iota _{3}>\iota _{2}\) and \(c>0\) such that \(w(\iota )\geq c\) for all \(\iota \geq \iota _{3}\). Next, using Lemma 2.2, we get \(u(\iota )\geq (1-a)w(\iota )\) for all \(\iota \geq \iota _{3}\) and (1.10)–(1.11) become
$$ \begin{gathered} \bigl(p(\iota ) \bigl(w'( \iota ) \bigr)^{\alpha} \bigr)' + R_{1}(\iota ) \leq 0 \quad \text{for } \iota \neq \phi _{k}, \\ \Delta \bigl(p(\phi _{k}) \bigl(w'(\phi _{k}) \bigr)^{\alpha} \bigr) + R_{(1,k)} \leq 0\quad \text{for }k=1,2,\dots . \end{gathered} $$
(3.3)
Integrating (3.3) from ι to ∞, we get
$$\begin{aligned} &\bigl[p(s) \bigl(w'(s) \bigr)^{\alpha}\bigr]_{\iota}^{\infty }+ \int _{\iota}^{ \infty }R_{1}(s) \,\mathrm{d}s+ \sum _{\phi _{k} \geq \iota} R_{(1,k)} \leq 0 . \end{aligned}$$
Since \(p(\iota ) (w'(\iota ) )^{\alpha}\) is positive and nondecreasing, \(\lim_{\iota \to \infty} p(\iota ) (w'(\iota ) )^{\alpha}\) exists, and it is finite and positive. Then
$$ \begin{aligned} p(\iota ) \bigl(w'(\iota ) \bigr)^{\alpha }&\geq \int _{\iota}^{\infty }R_{1}(s) \,\mathrm{d}s+ \sum _{\phi _{k} \geq \iota} R_{(1,k)}, \end{aligned} $$
that is,
$$ \begin{aligned} w'(\iota ) &\geq p^{-1/\alpha}(\iota ) \biggl[ \int _{\iota}^{\infty }R_{1}(s) \,\mathrm{d}s + \sum _{\phi _{k} \geq \iota} R_{(1,k)} \biggr]^{1/\alpha} . \end{aligned} $$
(3.4)
Since
$$ \begin{aligned} g\bigl[(1-a)w \bigl(\nu (\iota ) \bigr)\bigr]&= \frac{g[(1-a)w (\nu (\iota ) )]}{(1-a)^{\beta }w^{\beta } (\nu (\iota ) )} (1-a)^{\beta }w^{\beta } \bigl(\nu (\iota ) \bigr) \\ &\geq \frac{g[c(1-a)]}{c^{\beta }(1-a)^{\beta}} (1-a)^{\beta }w^{ \beta } \bigl(\nu (\iota ) \bigr) \\ &= \frac{g[c(1-a)]}{c^{\beta}} w^{\beta } \bigl(\nu (\iota ) \bigr) , \end{aligned} $$
(3.5)
then we use (3.5) in (3.4) to get
$$\begin{aligned} w'(\iota ) &\geq p^{-1/\alpha}(\iota ) \biggl[ \int _{\iota}^{\infty }r(s) \frac{g[c(1-a)]}{c^{\beta}} w^{\beta } \bigl(\nu (s) \bigr) \,\mathrm{d}s \\ &\quad{}+ \sum_{\phi _{k}\geq \iota} \tilde{r} (\phi _{k}) \frac{g[c(1-a)]}{c^{\beta}} w^{\beta } \bigl(\nu (\phi _{k}) \bigr) \biggr]^{1/ \alpha} . \end{aligned}$$
Next, if we set \(K=\frac {g_{0}[c(1-a)]}{c^{\beta}}\), where \(g_{0}[c(1-a)]= \min \{g[c(1-a)]\}\), the above inequality becomes
$$\begin{aligned} &w'(\iota ) \geq K^{1/\alpha} p^{-1/\alpha}(\iota ) \biggl[ \int _{\iota}^{ \infty }r(s)w^{\beta } \bigl(\nu (s) \bigr) \,\mathrm{d}s+ \sum_{\phi _{k} \geq \iota} \tilde{r} (\phi _{k}) w^{\beta } \bigl(\nu (\phi _{k}) \bigr) \biggr]^{1/\alpha} . \end{aligned}$$
Using (b) and the fact that \(w(\iota )\) is nondecreasing, we have
$$\begin{aligned} w'(\iota ) &\geq K^{1/\alpha} p^{-1/\alpha}(\iota ) \biggl[ \int _{\iota}^{ \infty }r(s) \,\mathrm{d}s + \sum _{\phi _{k}\geq \iota} \tilde{r} (\phi _{k}) \biggr]^{1/\alpha} w^{\beta /\alpha}(\iota ), \end{aligned}$$
i.e.,
$$\begin{aligned} \frac{w'(\iota )}{w^{\beta /\alpha}(\iota )} &\geq K^{1/\alpha} p^{-1/ \alpha}(\iota ) \biggl[ \int _{\iota}^{\infty }r(s) \,\mathrm{d}s + \sum _{ \phi _{k}\geq \iota} \tilde{r} (\phi _{k}) \biggr]^{1/\alpha} . \end{aligned}$$
Integrating both sides from \(\iota _{3}\) to ∞, we get
$$\begin{aligned} &K^{1/\alpha} \int _{\iota _{3}}^{\infty }p^{-1/\alpha}(s) \biggl[ \int _{s}^{ \infty }r(\psi ) \,\mathrm{d}\psi + \sum _{\phi _{k}\geq \iota} \tilde{r} ( \phi _{k}) \biggr]^{1/\alpha} \,\mathrm{d}s \leq \int _{\iota _{3}}^{ \infty }\frac{w'(s)}{w^{\beta /\alpha}(s)} \,\mathrm{d}s < \infty \end{aligned}$$
due to \(\beta >\alpha \), which is a contradiction to (3.2) and hence the sufficiency part of the theorem is proved.
Next we prove the necessary part by a contrapositive argument. If (3.2) does not hold, then for every \(\varepsilon >0\) there exists \(\iota \geq \iota _{0} \), for which
$$\begin{aligned} \int _{\iota}^{\infty }p^{-1/\alpha}(s) \biggl[ \int _{s}^{\infty }r( \psi ) \,\mathrm{d}\psi +\sum _{\phi _{k}\geq s} \tilde{r}(\phi _{k}) \biggr]^{1/\alpha} \,\mathrm{d}s< \varepsilon \quad \text{for } \iota \geq Y, \end{aligned}$$
where \(2\varepsilon = [\max \{g(\frac {1}{1-a})\} ]^{-1/\alpha}>0\).
Let us define the set
$$ V= \biggl\{ u\in C\bigl([0,\infty)\bigr): \frac{1}{2}\leq u(\iota )\leq \frac{1}{1-a} \text{ for all } \iota \geq Y \biggr\} $$
and \(\Phi : V\rightarrow V\) as
$$ (\Phi u) (\iota )= \textstyle\begin{cases} 0 &\text{if } \iota \leq Y, \\\frac{1+a}{2(1-a)}-q(\iota )u(\zeta (\iota )) \\ \quad {}+ \int _{\iota}^{\iota} p^{-1/\alpha}(s) [\int _{s}^{\infty }r( \psi )g (u(\nu (\psi )) ) \,\mathrm{d}\psi \\ \quad {}+\sum_{\phi _{k}\geq s} \tilde{r}(\phi _{k})g (u(\nu (\phi _{k})) ) ]^{1/\alpha} \,\mathrm{d}s &\text{if } \iota >Y . \end{cases} $$
Now we prove that \((\Phi u)(\iota )\in V\). For \(u(\iota )\in V\),
$$\begin{aligned} (\Phi u) (\iota )&\leq \frac{1+a}{2(1-a)}+ \int _{T}^{\iota} p^{-1/ \alpha}(s) \biggl[ \int _{s}^{\infty }r(\psi )g \biggl(\frac{1}{1-a} \biggr) \,\mathrm{d}\psi \\ &\quad{}+\sum_{\phi _{k}\geq s} \tilde{r}(\phi _{k})g \biggl(\frac{1}{1-a} \biggr) \biggr]^{1/\alpha} \,\mathrm{d}s \\ &\leq \frac{1+a}{2(1-a)}+ \biggl[\max \biggl\{ g \biggl(\frac{1}{1-a} \biggr)\biggr\} \biggr]^{1/ \alpha} .\varepsilon \\ &=\frac{1+a}{2(1-a)} +\frac{1}{2}= \frac{1}{1-a}, \end{aligned}$$
and further, for \(u(\iota )\in V\),
$$\begin{aligned} (\Phi u) (\iota )\geq \frac{1+a}{2(1-a)} - q(\iota ).\frac{1}{1-a}+0 \geq \frac{1+a}{2(1-a)} - \frac{a}{1-a} =\frac{1}{2}. \end{aligned}$$
Hence Φ maps from V to V.
Now we are going to find a fixed point for Φ in V, which will give an eventually positive solution of (1.10)–(1.11).
First we define a sequence of functions in V by
$$\begin{aligned}& u_{0}(\iota )=0 \quad \text{for } \iota \geq _{0}, \\& u_{1}(\iota )=(\Phi u_{0}) (\iota )= \textstyle\begin{cases} 0 &\text{if } \iota < Y, \\ \frac {1}{2} &\text{if } \iota \geq Y, \end{cases}\displaystyle \\& u_{n+1}(\iota ) = (\Phi u_{n}) (\iota )\quad \text{for }n \geq 1, \iota \geq Y. \end{aligned}$$
Here we see \(u_{1}(\iota )\geq u_{0}(\iota )\) for each fixed ι and \(\frac {1}{2}\leq u_{n-1}(\iota )\leq u_{n}(\iota )\leq \frac {1}{1-a}\) for \(\iota \geq Y\) for all \(n\geq 1\). Thus \({u_{n}}\) converges point-wise to a function u. By Lebesgue’s dominated convergence theorem u is a fixed point of Φ in V, which shows that it has a nonoscillatory solution. This completes the proof of the theorem. □
In Theorem 3.2 we take a constant β, the quotient of two odd positive integers with \(\beta < \alpha \), for which
$$ \frac{g(\iota )}{\iota ^{\beta}}\text{ is nonincreasing for }0< \iota . $$
(3.6)
The existence of such a constant can be established by taking \(g(\iota )=|\iota |^{\delta}\operatorname{sgn}(\iota )\) with \(\beta >\delta \). The assumption upon β can be withdrawn by taking \(|u|^{\beta}\operatorname{sgn}(u)\) instead of \(u^{\beta}\).
Theorem 3.2
Let (a), (c)–(f), and (3.6) hold for \(\iota \geq \iota _{0} \). Then every solution of (1.10)–(1.11) is oscillatory if
$$ \begin{aligned} &\frac{1}{(2c)^{\beta}} \Biggl[ \int _{0}^{\infty } r(\psi )g\bigl[c(1-a) P \bigl(\nu ( \psi ) \bigr)\bigr] \,\mathrm{d}\psi \\ &\quad{}+\sum_{k=1}^{\infty }\tilde{r}(\phi _{k})g\bigl[c(1-a)P \bigl(\nu (\phi _{k}) \bigr)\bigr] \Biggr]=\infty \quad \forall c \neq 0 . \end{aligned} $$
(3.7)
Proof
Let \(u(\iota )\) be an eventually positive solution of (1.10)–(1.11). Then, proceeding as in the proof of Theorem 3.1, we have \(\iota _{2}>\iota _{1} >\iota _{0} \) such that inequality (3.4) holds for all \(\iota \geq \iota _{2}\). Using (e), there exists \(\iota _{3}>\iota _{2}\) for which \(P(\iota )-P(\iota _{3})\geq \frac {1}{2} P(\iota )\) for \(\iota \geq \iota _{3}\). Integrating (3.4) from \(\iota _{3}\) to ι, we have
$$ \begin{aligned} w(\iota )-w(\iota _{3})&\geq \int _{\iota _{3}}^{\iota} p^{-1/\alpha}(s) \biggl[ \int _{s}^{\infty }R_{1}(\kappa )\,\mathrm{d} \kappa + \sum_{\phi _{k} \geq s} R_{(1,k)} \biggr]^{1/\alpha} \,\mathrm{d}s \\ &\geq \int _{\iota _{3}}^{\iota} p^{-1/\alpha}(s) \biggl[ \int _{\iota}^{ \infty }R_{1}(\kappa )\,\mathrm{d} \kappa + \sum_{\phi _{k} \geq \iota} R_{(1,k)} \biggr]^{1/\alpha} \,\mathrm{d}s, \end{aligned} $$
(3.8)
that is,
$$\begin{aligned} w(\iota )&\geq \bigl(P(\iota )-P(\iota _{3})\bigr) \biggl[ \int _{\iota}^{\infty }R_{1}( \kappa )\,\mathrm{d} \kappa + \sum_{\phi _{k} \geq \iota} R_{(1,k)} \biggr]^{1/ \alpha} \\ &\geq \frac {1}{2} P(\iota ) \biggl[ \int _{\iota}^{\infty }R_{1}( \kappa )\,\mathrm{d} \kappa + \sum_{\phi _{k} \geq \iota} R_{(1,k)} \biggr]^{1/ \alpha}. \end{aligned}$$
(3.8)
Since \(p(\iota ) (w'(\iota ) )^{\alpha}\) is nonincreasing and positive, then there exist \(c>0\) and \(\iota _{4}>\iota _{3}\) such that \(p(\iota ) (w'(\iota ) )^{\alpha}\leq c^{\alpha}\) for \(\iota \geq \iota _{4}\). Integrating the relation \(w'(\iota ) \leq cp^{-1/\alpha}(\iota )\) from \(\iota _{4}\) to ι, we have
$$\begin{aligned} w(\iota )-w(\iota _{4})&\leq c\bigl(P(\iota )-P(\iota _{4})\bigr), \end{aligned}$$
that is,
$$\begin{aligned} w(\iota )&\leq cP(\iota ) \quad \text{for } \iota \geq \iota _{4} . \end{aligned}$$
(3.9)
Using (3.6) and (3.9), we obtain
$$\begin{aligned} g\bigl[(1-a)w \bigl(\nu (\iota ) \bigr)\bigr]&= \frac{g[(1-a)w (\nu (\iota ) )]}{(1-a)^{\beta }w^{\beta } (\nu (\iota ) )} (1-a)^{\beta }w^{\beta } \bigl(\nu (\iota ) \bigr) \\ &\geq \frac{g[c(1-a)P (\nu (\iota ) )]}{c^{\beta }(1-a)^{\beta }P^{\beta } (\nu (\iota ) )} (1-a)^{\beta }w^{\beta } \bigl(\nu (\iota ) \bigr) \\ &= \frac{g[c(1-a)P (\nu (\iota ) )]}{c^{\beta }P^{\beta } (\nu (\iota ) )} w^{\beta } \bigl(\nu (\iota ) \bigr) \quad \forall \iota \geq \iota _{4} . \end{aligned}$$
(3.10)
Using (3.10) in (3.8), we obtain
$$\begin{aligned} w(\iota )&\geq \frac{1}{2} P(\iota ) \biggl[ \int _{\iota}^{\infty }r( \kappa ) \frac{g[c(1-a)P (\nu (\kappa ) )]}{c^{\beta }P^{\beta } (\nu (\kappa ) )} w^{\beta } \bigl(\nu (\kappa ) \bigr) \,\mathrm{d}\kappa \\ &\quad{}+\sum_{\phi _{k}\geq \iota} \tilde{r}(\phi _{k}) \frac{g[c(1-a)P (\nu (\phi _{k}) )]}{c^{\beta }P^{\beta } (\nu (\phi _{k}) )} w^{\beta } \bigl(\nu (\phi _{k}) \bigr) \biggr]^{1/\alpha} . \end{aligned}$$
Hence,
$$\begin{aligned} w(\iota )\geq \frac{1}{2}P(\iota )U^{1/\alpha}(\iota ) \quad \text{for } \iota \geq \iota _{4}, \end{aligned}$$
where
$$\begin{aligned} U(\iota )&=\frac{1}{c^{\beta}} \biggl[ \int _{\iota}^{\infty }r(\kappa )g\bigl[c(1-a)P \bigl(\nu ( \kappa ) \bigr)\bigr] \frac{w^{\beta } (\nu (\kappa ) )}{P^{\beta } (\nu (\kappa ) )} \,\mathrm{d}\kappa \\ &\quad{}+\sum_{\phi _{k}\geq \iota} \tilde{r}(\phi _{k})g \bigl[c(1-a)P \bigl(\nu ( \phi _{k}) \bigr)\bigr] \frac{w^{\beta } (\nu (\phi _{k}) )}{P^{\beta } (\nu (\phi _{k}) )} \biggr] . \end{aligned}$$
Now,
$$\begin{aligned} U'(\iota )&=-\frac {1}{c^{\beta}} r(\iota )g \bigl[c(1-a)P \bigl(\nu (\iota ) \bigr)\bigr] \frac{w^{\beta } (\nu (\iota ) )}{P^{\beta } (\nu (\iota ) )} \\ &\leq -\frac {1}{(2c)^{\beta}} r(\iota )g\bigl[c(1-a)P \bigl(\nu (\iota ) \bigr) \bigr]U^{\beta /\alpha}\bigl(\nu (\iota ) \bigr)\leq 0 \end{aligned}$$
(3.11)
and
$$\begin{aligned} \quad \Delta U(\phi _{k})= -\frac {1}{(2c)^{\beta}} r(\phi _{k})g\bigl[c(1-a)P \bigl(\nu (\phi _{k}) \bigr) \bigr]U^{\beta /\alpha}\bigl(\nu (\phi _{k}) \bigr)\leq 0, \end{aligned}$$
(3.12)
which shows that \(U(\iota )\) is nonincreasing on \([\iota _{4},\infty )\) and \(\lim_{\iota \to \infty}U(\iota )\) exists. Using (3.11) and (a), we find
$$\begin{aligned} \bigl[U^{1-\beta /\alpha}(\iota ) \bigr]'&= (1-\beta / \alpha )U^{-\beta / \alpha}(\iota )U'(\iota ) \\ &\leq -\frac{1-\beta /\alpha}{(2c)^{\beta}} r(\iota )g\bigl[c(1-a)P \bigl( \nu (\iota ) \bigr) \bigr]U^{\beta /\alpha}\bigl(\nu (\iota ) \bigr)U^{-\beta / \alpha}(\iota ) \\ &\leq -\frac{1-\beta /\alpha}{(2c)^{\beta}} r(\iota )g\bigl[c(1-a)P \bigl( \nu (\iota ) \bigr) \bigr]. \end{aligned}$$
(3.13)
To estimate the discontinuity of \(U^{1-\beta /\alpha}\), we use a Taylor polynomial of order 1 from the function \(h(u)=u^{1-\beta /\alpha}\), with \(0<\beta <\alpha \), about \(u=a\):
$$\begin{aligned} b^{1-\beta /\alpha}-a^{1-\beta /\alpha}\leq (1-\beta /\alpha )a^{- \beta /\alpha}(b-a). \end{aligned}$$
Then
$$\begin{aligned} \Delta U^{1-\beta /\alpha}(\phi _{k})&\leq (1-\beta /\alpha )U^{- \beta /\alpha}(\phi _{k})\Delta U(\phi _{k}) \\ &\leq -\frac{1-\beta /\alpha}{(2c)^{\beta}} r(\phi _{k})g\bigl[c(1-a)P \bigl(\nu (\phi _{k}) \bigr)\bigr]. \end{aligned}$$
Now, integrating (3.13) from \(\iota _{4}\) to ι, we have
$$\begin{aligned} &\bigl[U^{1-\beta /\alpha}(s) \bigr]_{\iota _{4}}^{\iota} -\sum _{\phi _{k} \geq \iota}\Delta \bigl[U^{1-\beta /\alpha}(\phi _{k}) \bigr] \\ &\quad \leq -\frac{1-\beta /\alpha}{(2c)^{\beta}} \int _{\iota _{4}}^{\iota} r(s)g\bigl[c(1-a)P \bigl(\nu (s) \bigr)\bigr]\,\mathrm{d}s, \end{aligned}$$
that is,
$$\begin{aligned} &\frac{1-\beta /\alpha}{(2c)^{\beta}} \Biggl[ \int _{0}^{\infty } r(s)g\bigl[c(1-a) P \bigl(\nu (s) \bigr)\bigr] \,\mathrm{d}s+\sum_{k=1}^{\infty } \tilde{r}(\phi _{k})g\bigl[c(1-a)P \bigl(\nu (\phi _{k}) \bigr)\bigr] \Biggr] \\ &\quad \leq - \bigl[U^{1-\beta /\alpha}(s) \bigr]_{\iota _{4}}^{\iota}< U^{1- \beta /\alpha}( \iota _{4})< \infty , \end{aligned}$$
which contradicts (3.7). This completes the proof. □
Example 3.1
Consider the neutral differential equations
$$\begin{aligned}& \bigl( \bigl( \bigl(u(\iota )+e^{-\iota}u\bigl(\zeta (\iota )\bigr) \bigr)' \bigr)^{1/3} \bigr)'+\iota \bigl(u(\iota -2)\bigr)^{7/3} =0 , \end{aligned}$$
(3.14)
$$\begin{aligned}& \bigl( \bigl( \bigl(u\bigl(3^{k}\bigr)-e^{-3^{k}}x \bigl(\zeta \bigl(3^{k}\bigr)\bigr) \bigr)' \bigr)^{1/3} \bigr)' +(\iota +2) \bigl(u \bigl(3^{k}-2\bigr)\bigr)^{7/3} =0 . \end{aligned}$$
(3.15)
Here \(\alpha = 1/3\), \(p(\iota )=1\), \(0< q(\iota )=e^{-\iota}<1\) \(\nu (\iota )= \iota -2\), \(\phi _{k}=3^{k}\) for , \(g(\iota )=\iota ^{7/3}\). For \(\beta =5/3\), we have \(\delta =7/3>\beta =5/3>\alpha =1/3\) and \(g(\iota )/\iota ^{\beta}=\iota ^{2/3}\), which are increasing functions. Now we check (3.2). We have
$$\begin{aligned} & \int _{\iota _{0} }^{\infty} \biggl[ \frac{1}{p(s)} \biggl[ \int _{s}^{ \infty }r(\psi ) \,d\psi +\sum _{\phi _{k}\geq s} \tilde{r}(\phi _{k}) \biggr] \biggr]^{1/\alpha} \,\mathrm{d}s \\ &\quad \geq \int _{\iota _{0} }^{\infty} \biggl[ \frac{1}{p(s)} \biggl[ \int _{s}^{ \infty }r(\psi ) \,d\psi \biggr] \biggr]^{1/\alpha} \,\mathrm{d}s \\ &\quad= \int _{2}^{\infty} \biggl[ \int _{s}^{\infty }\psi \,d\psi \biggr]^{3} \,\mathrm{d}s=\infty . \end{aligned}$$
So, all the conditions of Theorem 3.1 hold. Thus, each solution of (3.14)–(3.15) is oscillatory.
Example 3.2
Consider the neutral differential equations
$$\begin{aligned}& \bigl(e^{-\iota} \bigl( \bigl(u(\iota )+e^{-\iota}u \bigl(\zeta (\iota )\bigr) \bigr)' \bigr)^{11/3} \bigr)' +\frac{1}{\iota +1}\bigl(u(\iota -2)\bigr)^{1/3}=0 , \end{aligned}$$
(3.16)
$$\begin{aligned}& \bigl(e^{-k} \bigl( \bigl(u(k)+e^{-k}u\bigl(\zeta (k)\bigr) \bigr)' \bigr)^{11/3} \bigr)' + \frac{1}{\iota +4}\bigl(u(k-2)\bigr)^{1/3}=0 . \end{aligned}$$
(3.17)
Here \(\alpha = 11/3\), \(p(\iota )=e^{-\iota}\), \(0< q(\iota )=e^{-\iota}<1\), \(\nu (\iota )= \iota -2\), \(\phi _{k}=k\) for , \(P(\iota )= \int _{0}^{\iota }e^{3s/11} \,ds=\frac{11}{3}(e^{3\iota /11}-1)\), \(g(\iota )=\iota ^{1/3}\). For \(\beta =7/3\), we have \(\delta =1/3<\beta =7/3<\alpha =11/3\) and \(g(\iota )/\iota ^{\beta}=\iota ^{-2}\), which are decreasing functions. Now we check (3.7). We have
$$\begin{aligned} &\frac{1}{(2c)^{\beta}} \biggl[ \int _{0}^{\infty }r(\psi )gc(1-a) P \bigl(\nu (\psi ) \bigr)\biggr] \,\mathrm{d}\psi +\sum_{k=1}^{\infty } \tilde{r}( \phi _{k})g\bigl[c(1-a)P \bigl(\nu (\phi _{k}) \bigr)\bigr] ] \\ &\quad\geq \frac{1}{(2c)^{7/3}} \int _{0}^{\infty }r(\psi )g\bigl[c(1-a) P \bigl( \nu ( \psi ) \bigr)\bigr] \,\mathrm{d}\psi \\ &\quad= \frac{1}{(2c)^{7/3}} \int _{0}^{\infty}\frac{1}{\psi +1} \biggl[c(1-a) \frac{11}{3} \bigl(e^{3(\psi -2)/11}-1 \bigr) \biggr]^{1/3} \,\mathrm{d}\psi = \infty \quad \forall c>0. \end{aligned}$$
So, all the conditions of Theorem 3.2 hold, and therefore each solution of (3.16)–(3.17) is oscillatory.