Proof of Theorem 3.1
Since q satisfies the condition (1.1), we obtain that the equation (E) has a strongly increasing solution.
Let x be arbitrary strongly increasing solution of (E) defined on \([T,\infty ), T\geq a\). First, we show that there exist positive constants m, M such that
$$ m X_{2}(t)\leq x(t)\leq M X_{1}(t),\quad \text{for large } t, $$
(5.1)
where \(X_{1}\) and \(X_{2}\) are given by (4.1) and (4.2), respectively. Integrating \(x'\) on \([T,t]\), we get
$$ x(t)=x(T)+ \int _{T}^{t} x'(s)\,ds\leq x(T)+ x'(t) (t-T), \quad t\geq T,$$
because \(x'\) is increasing. Hence, we find \(K_{1} > 0\) such that
$$ x(t)\leq K_{1} t x'(t),\quad t\geq T. $$
(5.2)
Since x is increasing, integration of (E) from T to t gives
$$ x'(t)=x'(T)+ \int _{T}^{t} q(s) x(s)^{\gamma }\,ds \leq x'(T)+x(t)^{\gamma } \int _{T}^{t} q(s) \,ds, \quad t\geq T,$$
implying, due to the fact \(\int _{T}^{t} q(s)\,ds \rightarrow \infty \) as \(t\to \infty \), that we find \(K_{2}>0\) such that
$$ x'(t) \leq K_{2} x(t)^{\gamma } \int _{T}^{t} q(s)\,ds,\quad t\geq T. $$
(5.3)
By combining (5.2) and (5.3), we have
$$ x(t) \leq K_{1} K_{2} t x(t)^{\gamma } \int _{T}^{t} q(s)\,ds, \quad t \geq T. $$
Thus, there exists \(M>0\) such that
$$ x(t) \leq M \biggl(t \int _{T}^{t} q(s)\,ds \biggr)^{\frac{1}{1-\gamma }},\quad t\geq T. $$
The right-hand side of the inequality (5.1) is proved.
Next, we prove the left-hand side of the inequality (5.1). Set \(w(t)=x(t)x'(t)\) and
$$ \nu =\frac{\gamma +1}{\gamma +3},\qquad \mu =\frac{2}{\gamma +3},\qquad \kappa =\frac{1-\gamma }{\gamma +3}. $$
(5.4)
An application of Young’s inequality gives
$$\begin{aligned} w'(t) =&w(t) \biggl(\frac{q(t)x(t)^{\gamma }}{x'(t)}+\frac{x'(t)}{x(t)} \biggr)\geq \frac{ w(t)}{\mu ^{\mu }\nu ^{\nu }} \biggl( \frac{q(t)x(t)^{\gamma }}{x'(t)} \biggr)^{\mu } \biggl(\frac{x'(t)}{x(t)} \biggr)^{\nu } \\ =& \frac{w(t)}{\mu ^{\mu }\nu ^{\nu }} x(t)^{\gamma \mu -\nu } x'(t)^{ \nu -\mu }q(t)^{\mu }. \end{aligned}$$
Since, \(\gamma \mu -\nu =\nu -\mu =-\kappa \), we get
$$ w'(t)\geq \frac{1}{\mu ^{\mu }\nu ^{\nu }} w(t)^{1-\kappa }q(t)^{\mu }. $$
(5.5)
After dividing (5.5) by \(w(t)^{1-\kappa }\) and integrating the obtained inequality on \([T,t]\), we get that there is \(k_{1}>0\) such that
$$ w(t)^{\kappa }\geq k_{1} \int _{T}^{t} q(s)^{\mu }\,ds, \quad t\geq T $$
or
$$ x(t) x'(t)\geq k_{1}^{\frac{1}{\kappa }} \biggl( \int _{T}^{t} q(s)^{\mu }\,ds \biggr)^{\frac{1}{\kappa }},\quad t\geq T. $$
(5.6)
Integrating (5.6) from T to t, we find \(k_{2}>0\) and \(T^{*}\geq T\) sufficiently large such that
$$ \frac{x(t)^{2}}{2}\geq k_{2} \int _{a}^{t} \biggl( \int _{a}^{s} q(r)^{\frac{2}{\gamma +3}} \,dr \biggr)^{\frac{\gamma +3}{1-\gamma }} \,ds,\quad t \geq T^{*}. $$
(5.7)
From (5.7), we obtain that there exists \(m>0\) such that the left-hand side of the inequality (5.1) is satisfied.
Next, we prove that x is a rapidly varying function of index ∞. Fix arbitrary \(\lambda >1\). Indeed, from (5.1) for sufficiently large t, we have
$$ m X_{2}(\lambda t) \leq x(\lambda t)\leq M X_{1}(\lambda t), $$
(5.8)
and
$$ \frac{1}{M X_{1}(t)}\leq \frac{1}{ x( t)}\leq \frac{1}{m X_{2}( t)}. $$
(5.9)
From (5.8) and (5.9), we obtain
$$ \frac{m}{M} \frac{ X_{2}(\lambda t)}{ X_{1}(t)}\leq \frac{x(\lambda t)}{ x( t)}\leq \frac{M}{m} \frac{X_{1}(\lambda t)}{X_{2}( t)} $$
(5.10)
for sufficiently large t. By Lemma 4.1 and Lemma 4.2, we have \(X_{1}(t)\stackrel {r}{\sim }X_{2}(t)\), \(t\to \infty \), which means
$$ \lim_{t\to \infty } \frac{ X_{2}(\lambda t)}{ X_{1}(t)}=\lim _{t \to \infty } \frac{X_{1}(\lambda t)}{X_{2}( t)}=\infty . $$
(5.11)
Since λ was arbitrary, combining (5.10) and (5.11) gives us \(\lim_{t\to \infty }\frac{x(\lambda t)}{x(t)}=\infty \) for all \(\lambda >1 \), that is, \(x\in \operatorname{RPV}(\infty )\).
It remains to prove that the solution x satisfies the asymptotic relation (3.1). Fix arbitrary \(\lambda >1\). Let m and M be positive numbers, satisfying (5.1) for \(t\geq T_{1}\geq T\). By Lemma 4.1 and Lemma 4.2, we have (4.3) and (4.5), so there exists \(T_{2}=T_{2}(\lambda )\geq T_{1} \) such that
$$ M X_{1}(t)\leq X(\lambda t) \wedge X \biggl( \frac{t}{\lambda } \biggr)\leq m X_{2}(t), \quad t\geq T_{2}. $$
Therefore, from (5.1), we conclude that
$$ X \biggl( \frac{t}{\lambda } \biggr)\leq x(t)\leq X(\lambda t),\quad t\geq T_{2}, $$
(5.12)
implying \(x(t)\stackrel {\star }{\sim }X(t)\), \(t\to \infty \). This completes the proof of Theorem 3.1. □
Proof of Theorem 3.2
Assumption (1.2) ensures the existence of strongly decreasing solution of (E). Assume that x is the arbitrary strongly decreasing solution of (E) defined on \([T,\infty ), T\geq a\). First, we show that there exist positive constants m and M such that
$$ m Y_{2}(t)\leq x(t)\leq M Y_{1}(t),\quad \text{for large } t, $$
(5.13)
where \(Y_{1}\) and \(Y_{2}\) are given by (4.8) and (4.9), respectively. Since \(x'(t)\to 0\), \(t\to \infty \), and x is decreasing, integrating (E) from t to ∞, we get
$$ -x'(t)= \int _{t}^{\infty }q(s) x(s)^{\gamma }\,ds\leq x(t)^{\gamma } \int _{t}^{\infty }q(s) \,ds,\quad t\geq T. $$
(5.14)
Dividing (5.14) by \(x(t)^{\gamma }\) and then integrating from t to ∞, since \(x(t)\to 0\), \(t\to \infty \), we have
$$ \frac{1}{1-\gamma } x(t)^{1-\gamma }\leq \int _{t}^{\infty } \int _{s}^{\infty }q(r) \,dr\,ds,\quad t\geq T $$
implying that there exists \(M>0\) such that the right-hand side of the inequality (5.13) is satisfied.
Next, we prove the left-hand side of the inequality (5.13). Setting \(w(t)=x(t)|x'(t)|\) and ν, μ, κ as in (5.4), application of Young’s inequality gives
$$\begin{aligned} -w'(t) =&w(t) \biggl(\frac{q(t)x(t)^{\gamma }}{ \vert x'(t) \vert } + \frac{ \vert x'(t) \vert }{x(t)} \biggr)\geq \frac{ w(t)}{\mu ^{\mu }\nu ^{\nu }} \biggl(\frac{q(t)x(t)^{\gamma }}{ \vert x'(t) \vert } \biggr)^{\mu } \biggl( \frac{ \vert x'(t) \vert }{x(t)} \biggr)^{\nu} \\ =& \frac{w(t)}{\mu ^{\mu }\nu ^{\nu }} x(t)^{\gamma \mu -\nu } \bigl\vert x'(t) \bigr\vert ^{ \nu -\mu }q(t)^{\mu }= \frac{1}{\mu ^{\mu }\nu ^{\nu }}w(t)^{1-\kappa }q(t)^{\mu }, \end{aligned}$$
afterwards multiplying by \(w(t)^{\kappa -1}\) and integrating from t to ∞, we find \(k_{1}>0\) such that
$$ w(t)^{\kappa }\geq k_{1} \int _{t}^{\infty }q(s)^{\mu }\,ds,\quad t\geq T, $$
or
$$ -x(t) x'(t)\geq k_{1}^{\frac{1}{\kappa }} \biggl( \int _{t}^{\infty }q(s)^{\mu }\,ds \biggr)^{\frac{1}{\kappa }},\quad t\geq T. $$
(5.15)
Since \(x(t)\to 0\), \(t\to \infty \), integrating (5.15) from t to ∞ yields that there is \(k_{2}>0\) such that
$$ \frac{x(t)^{2}}{2}\ge k_{2} \int _{t}^{\infty } \biggl( \int _{s}^{\infty }q(r)^{\frac{2}{\gamma +3}} \,dr \biggr)^{\frac{\gamma +3}{1-\gamma }} \,ds. $$
(5.16)
From (5.16), we obtain that there exists \(m>0\) such that the left-hand side of the inequality (5.13) is satisfied.
That \(x\in \operatorname{RPV}(-\infty )\) and satisfies the asymptotic relation (3.3) can be proved in the same way as in the proof of Theorem 3.1, using Lemma 4.3 and Lemma 4.4. □