In what follows, all functional inequalities are assumed to hold eventually, that is, they are satisfied for all \(\ell\) large enough. Without loss of generality, in our proofs we only deal with positive solutions of (3).
The following two lemmas provide some inequalities that will be useful in our proofs.
Lemma 1
Let \(0< \omega < 1\) be the ratio of odd positive integers and A, \(B \ge 0\) with \(A \ge B\). Then: \(A^{\omega }  B^{\omega } \le (A  B)^{\omega }\).
Proof
For \(x \ge 1\) let \(f(x) = (x1)^{\omega }  x^{\omega } +1\). Then,
$$\begin{aligned} f'(x) = {\omega }\left[ (x1)^{\omega 1}  x^{\omega 1}\right] = \left[ \frac{x^{1\omega }  (x1)^{1\omega }}{x^{1\omega }(x1)^{1\omega }}\right] \ge 0 \end{aligned}$$
for \(x > 1\). Therefore, \(f(x) \ge f(1) = 0\) for \(x\ge 1\). Letting \(x = A/B\) proves the lemma. \(\square\)
Lemma 2
[15] Suppose that \(\omega >0\) and \(x^{\Delta }\) is of one sign on \([t_{0},\infty )\). Then
$$\begin{aligned} \frac{x^{\Delta }}{(x^{\sigma })^{\omega }}\le \frac{(x^{1\omega })^{\Delta }}{1\omega }\le \frac{x^{\Delta }}{x^{\omega }} \;\, {\mathrm{on }} \; [t_{0},\infty ). \end{aligned}$$
Lemma 3 below can be proved by following the lines of the proof of [20, Lemma 2.1].
Lemma 3
Let u be an eventually positive solution of (3). Then v satisfies one of the following cases:
 (a):

\(v>0\), \(v^{\Delta }>0\;\), and \(\;(a(v^{\Delta })^{\alpha })^{\Delta } \le 0\);
 (b):

\(v<0\), \(v^{\Delta }>0\;\), and \(\;(a(v^{\Delta })^{\alpha })^{\Delta } \le 0\)
for \(\ell \in {\mathbb{T}}\) sufficiently large.
Lemma 4
Let u be an eventually positive solution of (3) such that v satisfies case (b) of Lemma 3. Then
$$\begin{aligned} \lim _{\ell \rightarrow \infty }u(\ell )=0. \end{aligned}$$
Proof
Let u be an eventually positive solution of (3) with \(u(m(\ell )) >0\) and \(u(\tau (\ell )) >0\) and such that Lemma 3(b) holds for \(\ell \ge \ell _{1}\) for some \(\ell _{1} \ge \ell _{0}\). Then \(v(\ell ) <0\) and \(v^{\Delta }(\ell ) >0\) for \(\ell \ge \ell _{1}\), so \(v(\ell )\) is bounded.
We will consider two possibilities. First assume that \(u(\ell )\) is bounded. Then,
$$\begin{aligned} \limsup _{\ell \rightarrow \infty } u(\ell ) = L \quad {\text{with}} \quad 0 \le L < \infty . \end{aligned}$$
To show that \(L=0\), assume that \(L>0\). Then there is a sequence \(\{\ell _{k}\} \rightarrow \infty\) such that \(\{u(\ell _{k})\} \rightarrow L\) as \(\ell \rightarrow \infty\). Let \(\epsilon =  L(1+q_{1})/2q_{1} >0\); then for large k, \(u(m(\ell _{k})) < L + \epsilon\), so
$$\begin{aligned} 0 \ge \lim _{k\rightarrow \infty }v(\ell _{k}) = \lim _{k\rightarrow \infty } [u(\ell _{k}) + q(\ell _{k}) u(m(\ell _{k}))> L + q_{1}(L +\epsilon )> L(1+q_{1})/2 > 0, \end{aligned}$$
which is a contradiction.
Finally, to complete the proof, we need to show that \(u(\ell )\) is not unbounded. If \(u(\ell )\) is unbounded, then there is a sequence \(\{\ell _{j}\} \rightarrow \infty\) such that \(\{u(\ell _{j})\} \rightarrow \infty\) as \(j \rightarrow \infty\) and \(u(\ell _{j}) = \max \{u(\ell ) : \ell _{0} \le \ell \le \ell _{j}\}\). Now \(\{m(\ell _{j})\} \rightarrow \infty\) and \(m(\ell _{j}) \le \ell _{j}\), so
$$\begin{aligned} u(m(\ell _{j})) \le \max \{ u(\ell ) : \ell _{0} \le \ell \le \ell _{j} \} = u(\ell _{j}). \end{aligned}$$
Hence, for large j,
$$\begin{aligned} v(\ell _{j}) = u(\ell _{j}) + q(\ell _{j})u(m(\ell _{j})) \ge u(\ell _{j}) + q_{1}u(m(\ell _{j})) \ge (1+q_{1})u(\ell _{j}) >0, \end{aligned}$$
which contradicts the fact that \(v(\ell ) <0\). This completes the proof of the lemma. \(\square\)
Our first result on the asymptotic behavior of solutions of Eq. (3) is as follows.
Theorem 5
Let \(({\mathcal{H}}_{1})\)–\(({\mathcal{H}}_{3})\) hold and assume that \(\alpha \ge 1\) and there is a constant \(\gamma \in {\mathbb{R}}_{+}\) such that \(\beta< \gamma < \alpha\). Then any solution of (3) either oscillates or satisfies \(\lim _{\ell \rightarrow \infty }u(\ell )=0\) if and only if
 \(({\mathcal{H}}_{4})\):

\(\displaystyle {\int _{\ell _{0}}^{\infty }\Lambda (s){\mathcal{A}}^{\beta }(\tau (s)) \Delta s=\infty }\).
Proof
Necessity: To prove the necessity of the condition, assume that \(({\mathcal{H}}_{4})\) does not hold. Then there exists \(\ell _{1}>\ell _{0}\) such that
$$\begin{aligned} \int _{\ell _{1}}^{\infty }\Lambda (s){\mathcal{A}}^{\beta }(\tau (s)))\Delta s <\infty . \end{aligned}$$
(4)
Let
$$\begin{aligned} \chi = \left\{ u: u\in C_{rd}([\ell _{0},\infty )_{{\mathbb{T}}},{\mathbb{R}}) \ \left \ \sup _{\ell \in [\ell _{0},\infty )_{{\mathbb{T}}}}\frac{u(\ell )}{{\mathcal{A}}(\ell )}<\infty \right. \right\} . \end{aligned}$$
Clearly, \(\chi\) is a Banach space with the norm \(\Vert u\Vert =\sup _{\ell \in [\ell _{0},\infty )_{{\mathbb{T}}}}\frac{u(\ell )}{{\mathcal{A}}(\ell )}\). For any \(\varsigma _{1}>0\), \(\varsigma _{2} >0\), and \(\ell ^{*} \in [\ell _{0},\infty )_{{\mathbb{T}}}\) with \(\varsigma _{1}< (1+q_{1}) \varsigma _{2}\), let \(\Omega _{\varsigma _{1}, \varsigma _{2}} \subset \chi\) be given by
$$\begin{aligned} \Omega _{\varsigma _{1}, \varsigma _{2}} = \{u \in \chi : \varsigma _{1}[{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})] \le u(\ell ) \le \varsigma _{2}[{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})], \ \ell \in [\ell _{0},\infty )_{{\mathbb{T}}}\}. \end{aligned}$$
By (4), we can find \(\ell ^{*}>\ell _{1}\), \(\varsigma _{1}\), \(\varsigma _{2}\), and \(\varsigma _{3}\) such that \((\varsigma _{1})^{\alpha }< \varsigma _{3} < ((1+q_{1})\varsigma _{2})^{\alpha }\) and
$$\begin{aligned} \int _{\ell ^{*}}^{\infty }\Lambda (s){\mathcal{A}}^{\beta }(\tau (s)))\Delta s \le \frac{((1+q_{1})\varsigma _{2})^{\alpha }\varsigma _{3}}{\varsigma _{2}^{\beta }}. \end{aligned}$$
(5)
Define two maps \(\Gamma _{1}\) and \(\Gamma _{2}\) on \(\Omega _{\varsigma _{1}, \varsigma _{2}}\) by
$$\begin{aligned} (\Gamma _{1} u)(\ell )= \left\{ \begin{array}{ll} (\Gamma _{1} u)(\ell ^{*}), &{}\quad \ell \in [\ell _{0}, \ell ^{*})_{{\mathbb{T}}},\\ q(\ell )u(m(\ell )), &{}\quad \ell \in [\ell ^{*}, \infty )_{{\mathbb{T}}} \end{array}\right. \end{aligned}$$
and
$$\begin{aligned} (\Gamma _{2} u)(\ell )= \left\{ \begin{array}{ll} (\Gamma _{2} u)(\ell ^{*}), &{}\quad \ell \in [\ell _{0}, \ell ^{*})_{{\mathbb{T}}},\\ \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s, &{}\quad \ell \in [\ell ^{*}, \infty )_{{\mathbb{T}}}. \end{array}\right. \end{aligned}$$
First, we show that for any \(u_{1}\), \(u_{2}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\), we have \(\Gamma _{1}u_{1}+\Gamma _{2}u_{2}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\). To do this, let \(u_{1}, u_{2}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\). Note that \(u(\ell )\le \varsigma _{2} {\mathcal{A}}(\ell )\), so \(u^{\beta }(\tau (\ell ))\le \varsigma _{2}^{\beta } {\mathcal{A}}^{\beta }(\tau (\ell ))\). This, together with (5) implies that for \(\ell \ge \ell ^{*}\),
$$\begin{aligned} (\Gamma _{1} u_{1})(\ell )+(\Gamma _{2} u_{2})(\ell )&=\,q(\ell )u_{1}(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty } \Lambda (\theta )u_{2}^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s\\&\le \,q(\ell )u_{1}(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty }\varsigma _{2}^{\beta } \Lambda (\theta ){\mathcal{A}}^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s\\&\le \,q_{1} \varsigma _{2} [{\mathcal{A}}(m(\ell )){\mathcal{A}}(\ell ^{*})]+ \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left( \varsigma _{3}+((1+q_{1})\varsigma _{2})^{\alpha }\varsigma _{3}\right) \right] ^{1/\alpha }\Delta s\\&\le \,q_{1} \varsigma _{2} [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})]+(1+q_{1})\varsigma _{2}[{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})]\\&\le \varsigma _{2} [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})] \end{aligned}$$
and
$$\begin{aligned} (\Gamma _{1} u_{1})(\ell )+(\Gamma _{2} u_{2})(\ell )&=\,q(\ell )u_{1}(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty } \Lambda (\theta )u_{2}^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s\\&\ge \left[ \int _{\ell ^{*}}^{\ell }\varsigma _{3}\frac{1}{a(s)}\right] ^{1/\alpha }\Delta s\\&=\varsigma _{3}^{1/\alpha } [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})]\\&\ge \varsigma _{1} [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell ^{*})]. \end{aligned}$$
Therefore, \(\Gamma _{1}u_{1}+\Gamma _{2}u_{2} \in \Omega _{\varsigma _{1}, \varsigma _{2}}\).
Next, we show that \(\Gamma _{1}\) is a contraction mapping on \(\Omega _{\varsigma _{1}, \varsigma _{2}}\). Now for \(u_{1}\), \(u_{2}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\) and \(\ell \ge \ell ^{*}\), we have
$$\begin{aligned} (\Gamma _{1}u_{1})(\ell )(\Gamma _{1}u_{2})(\ell )&\le q(\ell ) u_{1}(m(\ell ))u_{2}(m(\ell ))\le \,q_{1}u_{1}(m(\ell ))u_{2}(m(\ell )), \end{aligned}$$
that is,
$$\begin{aligned} \Vert \Gamma _{1}u_{1}\Gamma _{1}u_{2}\Vert \le \,q_{1}\Vert u_{1}u_{2}\Vert . \end{aligned}$$
Since \(0 \le q_{1}<1\), \(\Gamma _{1}\) is a contraction.
To show that \(\Gamma _{2}\) is completely continuous, we will first show that \(\Gamma _{2}\) is continuous. So fix \(\ell \ge \ell ^*\) and let \(u_{k}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\) be such that \(u_{k}(\ell )\rightarrow u(\ell )\) as \(k\rightarrow \infty\). By taking a subsequence if necessary and again calling it \(u_{k}(\ell )\), we can assume that \(u_{k}(\ell )  u(\ell )\) is of fixed sign, say \(u_{k}(\ell ) \ge u(\ell )\) for \(k = 1, 2, \dots\). Since \(\Omega _{\varsigma _{1}, \varsigma _{2}}\) is closed, \(u(\ell )\in \Omega _{\varsigma _{1}, \varsigma _{2}}\). By Lemma 1 with \(\omega = 1/ \alpha \le 1\), we obtain
$$\begin{aligned} (\Gamma _{2}u_{k})(\ell )(\Gamma _{2}u)(\ell )&= \left \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty }\Lambda (\theta )u_{k}^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s\right. \\&\left. \quad \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s\right \\&\le \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)} \int _{s}^{\infty } \Lambda (\theta ) \left u_{k}^{\beta }(\tau (\theta ))u^{\beta }(\tau (\theta ))\right \Delta \theta \right] ^{1/\alpha }\Delta s. \end{aligned}$$
Since \(u_{k}^{\beta }(\tau (\theta ))u^{\beta }(\tau (\theta ))\rightarrow 0\) as \(k\rightarrow \infty\), an application of Lebesgue’s dominated convergence theorem shows that \(\lim _{k\rightarrow \infty }(\Gamma _{2}u_{k})(\ell )(\Gamma _{2}u)(\ell )\rightarrow 0\), so \(\Gamma _{2}u\) is continuous.
To show that \(\Gamma _{2}\) is relatively compact, it suffices to show that the family of functions \(\{\Gamma _{2}u:u\in \Omega _{\varsigma _{1}, \varsigma _{2}}\}\) is uniformly bounded and equicontinuous on \([\ell ^{*},\infty )_{{\mathbb{T}}}\). Clearly, \(\Gamma _{2}u\) is uniformly bounded. To see that \(\Gamma _{2}\) is equicontinuous, let \(\epsilon >0\) be given and choose \(\delta > 0\) such that \(\ell _{3}> \ell _{2}\ge \ell ^{*}\) and \(\ell _{2}  \ell _{1} < \delta\) implies \({\mathcal{A}}(\ell _{3}){\mathcal{A}}(\ell _{2}) < \epsilon \left\{ \frac{1}{[(1+q_{1})\varsigma _{2}]^{\alpha }\varsigma _{3}}\right\} ^{1/\alpha }\). Then,
$$\begin{aligned}&(\Gamma _{2}u)(\ell _{3})(\Gamma _{2}u)(\ell _{2})\\&\quad = \left \int _{\ell ^{*}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s  \int _{\ell ^{*}}^{\ell _{2}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s \right \\&\quad = \left \int _{\ell _{2}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s \right \\&\quad \le \left \int _{\ell _{2}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )\varsigma _{2}^{\beta } {\mathcal{A}}^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s \right \\&\quad \le [{\mathcal{A}}(\ell _{3}){\mathcal{A}}(\ell _{2})]\left[ \int _{s}^{\infty }\Lambda (\theta )\varsigma _{2}^{\beta } {\mathcal{A}}^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s < \epsilon . \end{aligned}$$
Thus, \(\{\Gamma _{2}u : u\in \Omega _{\varsigma _{1}, \varsigma _{2}}\}\) is uniformly bounded and equicontinuous on \([\ell ^{*},\infty )_{{\mathbb{T}}}\), and so \(\Gamma _{2}u\) is relatively compact. By Krasnosel’skii’s fixed point theorem [29, Lemma 5], \(\Gamma _{1} + \Gamma _{2}\) has a unique fixed point \(u\in \Omega _{\varsigma _{1}, \varsigma _{2}}\), i.e., \(\Gamma _{1}u+\Gamma _{2}u=u\). That is,
$$\begin{aligned} u(\ell )= q(\ell )u(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\left[ \varsigma _{3}+\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] \right] ^{1/\alpha }\Delta s, \ \ \ell \in [\ell ^{*},\infty )_{{\mathbb{T}}}. \end{aligned}$$
is a nonoscillatory solution of (3).
Sufficiency: Now assume that v is a nonoscillatory solution of (3). Then Lemma 3 holds for \(\ell \in [\ell _{1}, \infty )_{{\mathbb{T}}}\) for some \(\ell _{1} \ge \ell _{0}\), and there are two possible cases.
Case a Since \(a(v^{\Delta })^{\alpha }\) is nonincreasing and positive for \(\ell \in [\ell _{1}, \infty )_{{\mathbb{T}}}\), we can find \({\mathcal{C}}>0\) and \(\ell _{2} > \ell _{0}\) such that
$$\begin{aligned} a(\ell )(v^{\Delta }(\ell ))^{\alpha } \le {\mathcal{C}} \ \ {\text{for}} \ \ \ell \in [\ell _{2}, \infty )_{{\mathbb{T}}}. \end{aligned}$$
Integrating from \(\ell _{2}\) to \(\ell\) gives
$$\begin{aligned} v(\ell )\le v(\ell _{2})+{\mathcal{C}}^{1/\alpha }\int _{\ell _{2}}^{\ell }\frac{\Delta s}{a^{1/\alpha }(s)}= v(\ell _{2})+{\mathcal{C}}^{1/\alpha }\left( {\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{2})\right) . \end{aligned}$$
Since \(\lim _{\ell \rightarrow \infty }{\mathcal{A}}(\ell )=\infty\),
$$\begin{aligned} v(\ell )\le {\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\ell ) \end{aligned}$$
(6)
for \(\ell\) sufficiently large, say \(\ell \ge \ell _{3}\). Now \(\beta <\gamma\) and (6) imply
$$\begin{aligned} v^{\beta }(\tau (\ell ))= v^{\beta \gamma }(\tau (\ell ))v^{\gamma }(\tau (\ell )) \ge [{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (\ell ))]^{\beta \gamma } v^{\gamma }(\tau (\ell )). \end{aligned}$$
Therefore, (3) becomes
$$\begin{aligned} {[}a(t)(v^{\Delta }(\ell ))^{\alpha }]^{\Delta } +\Lambda (\ell )[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (\ell ))]^{\beta \gamma } v^{\gamma }(\tau (\ell ))\le 0. \end{aligned}$$
Integrating the last inequality from \(\ell \ge \ell _{3}\) to \(\infty\) gives
$$\begin{aligned} \lim _{t\rightarrow \infty }a(t)(v^{\Delta }(t))^{\alpha }a(\ell )(v^{\Delta }(\ell ))^{\alpha }+ \int _{\ell }^{\infty } \Lambda (s)[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (s))]^{\beta \gamma } v^{\gamma }(\tau (s))\Delta s\le 0, \end{aligned}$$
which implies
$$\begin{aligned} a(\ell )(v^{\Delta }(\ell ))^{\alpha } \ge \int _{\ell }^{\infty } \Lambda (s)[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (s))]^{\beta \gamma } v^{\gamma }(\tau (s))\Delta s. \end{aligned}$$
As a result,
$$\begin{aligned} v^{\Delta }(\ell ) \ge \left[ \frac{1}{a(\ell )}\int _{\ell }^{\infty } \Lambda (s)[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (s))]^{\beta \gamma } v^{\gamma }(\tau (s))\Delta s\right] ^{1/\alpha }. \end{aligned}$$
(7)
Integrating this from \(\ell _{3}\) to \(\ell\), we have
$$\begin{aligned} v(\ell ) \ge \int _{\ell _{3}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty } \Lambda (\theta )[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (\theta ))]^{\beta \gamma } v^{\gamma }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s. \end{aligned}$$
Consequently,
$$\begin{aligned} v(\ell )\ge [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{1})]\left[ \int _{s}^{\infty } \Lambda (\theta )[{\mathcal{C}}^{1/\alpha } {\mathcal{A}}(\tau (\theta ))]^{\beta \gamma } v^{\gamma }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }. \end{aligned}$$
(8)
Clearly, \(\displaystyle {\int _{\ell _{3}}^{\ell }\frac{1}{a(s)}\Delta s = {\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{3}) =\pi (\ell ) {\mathcal{A}}(\ell )}\), where \(\pi (\ell )=\frac{{\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{3})}{{\mathcal{A}}(\ell )}\). In view of \(({\mathcal{H}}_{3})\), we have \(\lim _{\ell \rightarrow \infty }\pi (\ell )=1\), so there exists \(\ell _{4} \ge \ell _{3}\) and \(\pi ^{*}\in (0,1)\) such that \(\pi (\ell )\ge \pi ^{*}\), that is,
$$\begin{aligned} {\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{3})\ge \pi ^{*} {\mathcal{A}}(\ell ) \ {\text{for}} \ \ell \in [\ell _{4},\infty )_{{\mathbb{T}}}. \end{aligned}$$
(9)
Setting
$$\begin{aligned} \Psi (\ell )=\int _{\ell }^{\infty }\Lambda (s)({\mathcal{C}}^{1/\alpha }{\mathcal{A}}(\tau (s))^{\beta \gamma }v^{\gamma }(\tau (s))\Delta s, \end{aligned}$$
(10)
in (8), we have
$$\begin{aligned} v(\ell )\ge [{\mathcal{A}}(\ell ){\mathcal{A}}(\ell _{1})]\Psi ^{1/\alpha }(\ell ). \end{aligned}$$
and in view of (9),
$$\begin{aligned} v(\ell )\ge \pi ^{*} {\mathcal{A}}(\ell )\Psi ^{1/\alpha }(\ell ) \end{aligned}$$
for \(\ell \in [\ell _{4},\infty )_{{\mathbb{T}}}\). From the preceding inequality, it is easy to verify that
$$\begin{aligned} \frac{v^{\gamma }(\ell )}{{\mathcal{C}}^{\gamma /\alpha }{\mathcal{A}}^{\gamma }(\ell )} \ge \left( \frac{\pi ^{*}}{{\mathcal{C}}^{1/\alpha }}\right) ^{\gamma }\Psi ^{\gamma /\alpha }(\ell ) \end{aligned}$$
which implies that
$$\begin{aligned} \frac{v^{\gamma }(\tau (\ell ))}{{\mathcal{C}}^{\gamma /\alpha }{\mathcal{A}}^{\gamma }(\tau (\ell ))} \ge \left( \frac{\pi ^{*}}{{\mathcal{C}}^{1/\alpha }}\right) ^{\gamma }\Psi ^{\gamma /\alpha }(\tau (\ell )) \end{aligned}$$
for \(\ell \in [\ell _{5},\infty )_{{\mathbb{T}}}\subset [\ell _{4},\infty )_{{\mathbb{T}}}\). From (10), we have
$$\begin{aligned} \Psi ^{\Delta }(\ell )&=\left( \int _{\ell }^{\infty }\Lambda (s)({\mathcal{C}}^{1/\alpha }{\mathcal{A}}(\tau (s))^{\beta \gamma }v^{\gamma }(\tau (s))\Delta s\right) ^{\Delta }\\&=\Lambda (\ell )({\mathcal{C}}^{1/\alpha }{\mathcal{A}}(\tau (\ell ))^{\beta \gamma }v^{\gamma }(\tau (\ell ))\\&=\Lambda (\ell )({\mathcal{C}}^{1/\alpha }{\mathcal{A}}(\tau (\ell ))^{\beta }\left( \frac{v(\tau (\ell ))}{{\mathcal{C}}^{1/\alpha }{\mathcal{A}}(\tau (\ell ))}\right) ^{\gamma }\\&\le  (\pi ^{*})^{\gamma }{\mathcal{C}}^{(\beta \gamma )/\alpha }\Lambda (\ell ){\mathcal{A}}^{\beta }(\tau (\ell ))\Psi ^{\gamma /\alpha }(\tau (\ell )). \end{aligned}$$
From Lemma 2 with \(\omega = \gamma / \alpha\) and \(x = \Psi (\ell )\) and the fact that \(\gamma < \alpha\), it follows that
$$\begin{aligned} [\Psi ^{1\gamma /\alpha }(\ell )]^{\Delta }&\ge (1\gamma /\alpha )\Psi ^{\gamma /\alpha }(\ell )\Psi ^{\Delta }(\ell ) \nonumber \\&\ge (\pi ^{*})^{\gamma }{\mathcal{C}}^{(\beta \gamma )/\alpha }(1\gamma /\alpha )\Psi ^{\gamma /\alpha }(\ell )\Lambda (\ell ){\mathcal{A}}^{\beta }(\tau (\ell ))\Psi ^{\gamma /\alpha }(\tau (\ell ))\nonumber \\&= (\pi ^{*})^{\gamma }{\mathcal{C}}^{(\beta \gamma )/\alpha }(1\gamma /\alpha )\Lambda (\ell ){\mathcal{A}}^{\beta }(\tau (\ell )) \end{aligned}$$
(11)
for \(\ell \in [\ell _{5},\infty )_{{\mathbb{T}}}\). Integrating (11) from \(\ell _{5}\) to \(\ell\),
$$\begin{aligned} \Psi ^{1\gamma /\alpha }(\ell )+\Psi ^{1\gamma /\alpha }(\ell _{5})\ge (\pi ^{*})^{\gamma }{\mathcal{C}}^{(\beta \gamma )/\alpha }(1\gamma /\alpha )\int _{\ell _{5}}^{\ell }\Lambda (s){\mathcal{A}}^{\beta }(\tau (s))\Delta s \end{aligned}$$
so
$$\begin{aligned} \int _{\ell _{5}}^{\ell }\Lambda (s){\mathcal{A}}^{\beta }(\tau (s))\Delta s\le \frac{{\mathcal{C}}^{(\gamma \beta )/\alpha }}{(\pi ^{*})^{\gamma }(1\gamma /\alpha )}\Psi ^{1\gamma /\alpha }(\ell _{5}) \end{aligned}$$
contradicting \(({\mathcal{H}}_{4})\).
Case b Now suppose \(v<0\) for \(\ell \in [\ell _{0},\infty )_{{\mathbb{T}}}\). Then \(u(\ell ) \rightarrow 0\) as \(\ell \rightarrow \infty\) by Lemma 4. This completes the proof of the theorem. \(\square\)
The following corollary is immediate.
Corollary 6
Under the assumption of Theorem 5, every unbounded solution of (3) oscillates if and only if \(({\mathcal{H}}_{4})\) holds.
Theorem 7
Let \(({\mathcal{H}}_{1})\)–\(({\mathcal{H}}_{3})\) hold, \(\sigma (\tau (\ell ))=\tau (\sigma (\ell ))\), \(a^{\Delta }(\ell )\ge 0\), and there is a constant \(\gamma \in {\mathbb{R}}_{+}\) such that \(\alpha<\gamma <\beta\). Then any solution \(u(\ell )\) of (3) is either oscillatory or satisfies \(\lim _{\ell \rightarrow \infty }u(\ell )=0\) if and only if
 \(({\mathcal{H}}_{5})\):

\(\displaystyle {\lim _{\ell \rightarrow \infty }\int _{\ell _{0}}^{\ell } \int _{s}^{\infty }\left( \frac{\Lambda (\theta )}{a(s)} \right) ^{1/\alpha }\Delta \theta \Delta s =\infty }\) .
Proof
Necessity: Assume that \(({\mathcal{H}}_{5})\) does not hold so that there exists \(\ell _{1} > \ell _{0}\) such that
$$\begin{aligned} \int _{\ell _{1}}^{\infty }\left[ \frac{1}{a(s)} \int _{s}^{\infty }\Lambda (\theta )\Delta \theta \right] ^{1/\alpha }\Delta s < \infty . \end{aligned}$$
(12)
Letting
$$\begin{aligned} \chi =\left\{ u: u\in C_{rd}\left( [\ell _{0},\infty )_{{\mathbb{T}}}, {\mathbb{R}}\right) \left \sup _{\ell \in [\ell _{0},\infty )_{{\mathbb{T}}}}u(\ell )<\infty \right. \right\} , \end{aligned}$$
we see that \(\chi\) is a Banach space with the norm \(\Vert u\Vert =\sup _{\ell \in [\ell _{0},\infty )_{{\mathbb{T}}}}u(\ell )\). Choose \(\varsigma _{1}>0\) and \(\varsigma _{2} >0\) so that \(\varsigma _{1}q_{1} \varsigma _{2} < \varsigma _{2}\) and consider \(\Omega _{\varsigma _{1}, \varsigma _{2}} \subset \chi\) to be
$$\begin{aligned} \Omega _{\varsigma _{1}, \varsigma _{2}} =\{u \in \chi : \varsigma _{1}\le u(\ell )\le \varsigma _{2}, \ \ \ell \in [\ell _{0},\infty )_{{\mathbb{T}}}\}. \end{aligned}$$
By (12), we can find \(\ell ^{*} > \ell _{1}\) and \(\varsigma _{3} >0\) such that \(\varsigma _{1}<\varsigma _{3}<(1+q_{1})\varsigma _{2}\) and
$$\begin{aligned} \int _{\ell ^{*}}^{\infty }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )\Delta \theta \right] ^{1/\alpha }\Delta s\le \frac{(1+q_{1})\varsigma _{2}\varsigma _{3}}{\varsigma _{2}^{\beta /\alpha }}. \end{aligned}$$
(13)
Define two maps \(\Gamma _{1}\) and \(\Gamma _{2}\) on \(\Omega\) by
$$\begin{aligned} (\Gamma _{1}u)(\ell )= \left\{ \begin{array}{ll} (\Gamma _{1}u)(\ell ^{*}),&{}\quad \ell \in [\ell _{0}, \ell ^{*})_{{\mathbb{T}}},\\ \varsigma _{3}q(\ell )u(m(\ell )),&{}\quad \ell \in [\ell ^{*}, \infty )_{{\mathbb{T}}} \end{array}\right. \end{aligned}$$
and
$$\begin{aligned} (\Gamma _{2}u)(\ell )= \left\{ \begin{array}{ll} (\Gamma _{2}u)(\ell ^{*}),&{}\quad \ell \in [\ell _{0}, \ell ^{*})_{{\mathbb{T}}},\\ \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s, &{}\quad \ell \in [\ell ^{*}, \infty )_{{\mathbb{T}}}. \end{array}\right. \end{aligned}$$
To show that \(\Gamma _{1} + \Gamma _{2} : \Omega \rightarrow \Omega\), let \(u_{1}\), \(u_{2}\in \Omega\). Then from (13),
$$\begin{aligned} (\Gamma _{1}u_{1})(\ell )+(\Gamma _{2}u_{2})(\ell )&=\varsigma _{3} q(\ell )u_{1}(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u_{2}^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s\\&\le \varsigma _{3}q_{1}\varsigma _{2}+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u_{2}^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s\\&\le \varsigma _{3}q_{1}\varsigma _{2}+ \varsigma _{2}^{\beta /\alpha }\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )\Delta \theta \right] ^{1/\alpha }\Delta s\\&\le \varsigma _{2} \end{aligned}$$
and
$$\begin{aligned} (\Gamma _{1}u_{1})(\ell )+(\Gamma _{2}u_{2})(\ell ) \ge \varsigma _{3} q(\ell )u_{1}(m(\ell )) \ge \varsigma _{3}\ge \varsigma _{1} \end{aligned}$$
for \(\ell \ge \ell ^{*}\). Hence, \(\Gamma _{1}u_{1}+ \Gamma _{2}u_{2} \in \Omega _{\varsigma _{1}, \varsigma _{2}}\).
To see that \(\Gamma _{1}\) is a contraction, let \(u_{1}\), \(u_{2}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\) and \(\ell \ge \ell ^{*}\). Then,
$$\begin{aligned} (\Gamma _{1}u_{1})(\ell )(\Gamma _{1}u_{2})(\ell )&\le q(\ell ) u_{1}(m(\ell ))u_{2}(m(\ell ))\le q_{1}u_{1}(m(\ell ))u_{2}(m(\ell )), \end{aligned}$$
so
$$\begin{aligned} \Vert \Gamma _{1}u_{1}\Gamma _{1}u_{2}\Vert \le q_{1}\Vert u_{1}u_{2}\Vert , \end{aligned}$$
i.e., \(\Gamma _{1}\) is a contraction mapping.
To show that \(\Gamma _{2}\) is completely continuous, we begin by letting \(u_{k}\in \Omega _{\varsigma _{1}, \varsigma _{2}}\) be such that \(u_{k}(\ell )\rightarrow u(\ell )\) as \(k\rightarrow \infty\). Since \(\Omega _{\varsigma _{1}, \varsigma _{2}}\) is closed, \(u(\ell )\in \Omega _{\varsigma _{1}, \varsigma _{2}}\). Now
$$\begin{aligned} (\Gamma _{2}u_{k})(\ell )(\Gamma _{2}u)(\ell )&\le \int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)} \int _{s}^{\infty } \Lambda (\theta ) u_{k}^{\beta }(\tau (\theta ))u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s. \end{aligned}$$
Since \(u_{k}^{\beta }(\tau (\theta ))u^{\beta }(\tau (\theta ))\rightarrow 0\) as \(k\rightarrow \infty\), an application of Lebesgue’s dominated convergence theorem implies \(\lim _{k\rightarrow \infty }(\Gamma _{2}u_{k})(\ell )(\Gamma _{2}u)(\ell )\rightarrow 0\). Hence, \(\Gamma _{2}u\) is continuous. To show that \(\Gamma _{2}u\) is relatively compact, it suffices to show that the family of functions \(\{\Gamma _{2}u:u\in \Omega _{\varsigma _{1}, \varsigma _{2}}\}\) is uniformly bounded and equicontinuous on \([\ell ^{*},\infty )_{{\mathbb{T}}}\). The uniform boundedness is clear.
To show \(\Gamma _{2}u\) is equicontinuous, let \(\epsilon >0\) be given and choose \(\delta > 0\) such that \(\ell _{3}>\ell _{2}\ge \ell ^{*}\) and \(\ell _{2}  \ell _{1} < \delta\) implies
$$\begin{aligned} \int _{\ell _{2}}^{\ell _{3}}\left[ \frac{1}{a(s)} \int _{s}^{\infty }\Lambda (\theta )\Delta \theta \right] ^{1/\alpha }\Delta s < \frac{\epsilon }{\varsigma _{2}^{\beta / \alpha }}. \end{aligned}$$
Then,
$$\begin{aligned}&(\Gamma _{2}u)(\ell _{3})(\Gamma _{2}u)(\ell _{2})\\&\quad = \left \int _{\ell ^{*}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s  \int _{\ell ^{*}}^{\ell _{2}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s \right \\&\quad = \left \int _{\ell _{2}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s \right \\&\quad \le \varsigma _{2}^{\beta /\alpha } \int _{\ell _{2}}^{\ell _{3}}\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )\Delta \theta \right] ^{1/\alpha }\Delta s < \epsilon . \end{aligned}$$
Therefore, \(\Gamma _{2}u\) is relatively compact, and by Krasnosel’skii’s fixed point theorem [, Lemma 5], 29\(\Gamma _{1} + \Gamma _{2}\) has a unique fixed point \(u\in \Omega _{\varsigma _{1}, \varsigma _{2}}\). It follows that
$$\begin{aligned} u(\ell )= \varsigma _{3}q(\ell )u(m(\ell ))+\int _{\ell ^{*}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty }\Lambda (\theta )u^{\beta }(\tau (\theta ))\Delta \theta \right] ^{1/\alpha }\Delta s, \ \ell \in [\ell ^{*},\infty )_{{\mathbb{T}}} \end{aligned}$$
is a nonoscillatory solution of (3).
Sufficiency: Let u be a nonoscillatory solution of (3) with Lemma 3 holding for \(\ell \in [\ell _{1},\infty )_{{\mathbb{T}}}\). We again consider two cases.
Case a Let \(v>0\); then \(u(\ell )\ge v(\ell )\) for \(\ell \in [\ell _{1},\infty )_{{\mathbb{T}}}\). From the fact that \(v^{\Delta }(\ell )>0\) for \(\ell \in [\ell _{1},\infty )_{{\mathbb{T}}}\), it follows that \(v(\tau (\ell ))\ge v(\tau (\ell _{1}))={\mathcal{C}}\) for \(\ell \in [\ell _{2},\infty )_{{\mathbb{T}}}\) for some \(\ell _{2} \ge \ell _{1}\). Since \(\gamma <\beta\),
$$\begin{aligned} v^{\beta }(\tau (\ell ))=v^{\beta \gamma }(\tau (\ell ))v^{\gamma }(\tau (\ell ))\ge {\mathcal{C}}^{\beta \gamma } v^{\gamma }(\tau (\ell )). \end{aligned}$$
(14)
Using (14) in (3), we obtain
$$\begin{aligned} {[}a(\ell )(v^{\Delta }(\ell ))^{\alpha }]^{\Delta }+ {\mathcal{C}}^{\beta \gamma } \Lambda (\ell ) v^{\gamma }(\tau (\ell ))\le 0, \end{aligned}$$
and an integration from \(\ell\) to \(\infty\) gives
$$\begin{aligned} \lim _{t\rightarrow \infty }a(t)(v^{\Delta }(t))^{\alpha }a(\ell )(v^{\Delta }(\ell ))^{\alpha }+{\mathcal{C}}^{\beta \gamma } \int _{\ell }^{\infty } \Lambda (s) v^{\gamma }(\tau (s))\Delta s\le 0, \end{aligned}$$
that is,
$$\begin{aligned} {\mathcal{C}}^{\beta \gamma }\int _{\ell }^{\infty }\Lambda (s) v^{\gamma }(\tau (s))\Delta s \le a(\ell )(v^{\Delta }(\ell ))^{\alpha }\le a(\tau (\ell ))(v^{\Delta }(\tau (\ell )))^{\alpha }. \end{aligned}$$
Using the fact that \(a^{\Delta }(\ell )\ge 0\), we see that
$$\begin{aligned} (v^{\Delta }(\tau (\ell )))^{\alpha } \ge \frac{{\mathcal{C}}^{\beta \gamma }}{a(\ell )} \int _{\ell }^{\infty }\Lambda (s) v^{\gamma }(\tau (s))\Delta s, \end{aligned}$$
which implies
$$\begin{aligned} v^{\Delta }(\tau (\ell ))&\ge \frac{{\mathcal{C}}^{(\beta \gamma )/\alpha }}{a^{1/\alpha }(\ell )} \left[ \int _{\ell }^{\infty }\Lambda (s) v^{\gamma }(\tau (s))\Delta s \right] ^{1/\alpha }\\&\ge \frac{{\mathcal{C}}^{(\beta \gamma )/\alpha }}{a^{1/\alpha }(\ell )} \left[ \int _{\sigma (\ell )}^{\infty }\Lambda (s) v^{\gamma }(\tau (s))\Delta s \right] ^{1/\alpha }\\&\ge \frac{{\mathcal{C}}^{(\beta \gamma )/\alpha }}{a^{1/\alpha }(\ell )} \left[ \int _{\sigma (\ell )}^{\infty }\Lambda (s)\Delta s \right] ^{1/\alpha }(v^{\sigma }(\tau (\ell )))^{\gamma /\alpha }, \end{aligned}$$
that is
$$\begin{aligned} v^{\Delta }(\tau (\ell ))(v^{\sigma }(\tau (\ell )))^{\gamma / \alpha } \ge \frac{{\mathcal{C}}^{(\beta \gamma )/\alpha }}{a^{1/\alpha }(\ell )}\left[ \int _{\sigma (\ell )}^{\infty }\Lambda (s)\Delta s \right] ^{1/\alpha }. \end{aligned}$$
Since \(\alpha < \gamma\), by Lemma 2
$$\begin{aligned} \frac{{\mathcal{C}}^{(\beta \gamma )/\alpha }}{a^{1/\alpha }(\ell )}\left[ \int _{\sigma (\ell )}^{\infty }\Lambda (s)\Delta s \right] ^{1/\alpha } \le v^{\Delta }(\tau (\ell ))(v^{\sigma }(\tau (\ell )))^{\gamma / \alpha } \le \frac{[v^{1 \gamma / \alpha }(\tau (\ell ))]^{\Delta }}{1 \gamma /\alpha }. \end{aligned}$$
Integrating the preceding inequality from \(\ell _{2}\) to \(\ell\) gives
$$\begin{aligned} {\mathcal{C}}^{(\beta \gamma )/\alpha }\int _{\ell _{2}}^{\ell }\left[ \frac{1}{a(s)}\int _{s}^{\infty } \Lambda (\theta )\Delta \theta \right] ^{1/\alpha } \Delta s&\le \frac{1}{1 \gamma /\alpha }\int _{\ell _{2}}^{\ell } \big [v^{1 \gamma / \alpha }(\tau (s)))\big ]^{\Delta }\Delta s\\&= \frac{1}{\gamma /\alpha 1} \big [v^{1 \gamma / \alpha }(\tau (\ell _{2})))  v^{1 \gamma / \alpha }(\tau (\ell ))\big ]\\&\le \frac{1}{\gamma /\alpha 1}v^{1 \gamma / \alpha }(\tau (\ell _{2})), \end{aligned}$$
contradicting \(({\mathcal{H}}_{5})\).
Case b If \(v<0\), then \(u(\ell ) \rightarrow 0\) by Lemma 4. This proves the theorem. \(\square\)
The following corollary is analogous to Corollary 6.
Corollary 8
Under the assumption of Theorem 5, every unbounded solution of (3) oscillates if and only if \(({\mathcal{H}}_{5})\) holds.