\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 49, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/49\hfil Exact asymptotic behavior] {Exact asymptotic behavior of the positive solutions for some singular Dirichlet problems on the half line} \author[H. M\^aagli, R. Alsaedi, N. Zeddini \hfil EJDE-2016/49\hfilneg] {Habib M\^aagli, Ramzi Alsaedi, Noureddine Zeddini} \address{Habib M\^aagli \newline Department of Mathematics, College of Sciences and Arts \\ King Abdulaziz University, Rabigh Campus \\ P.O. Box 344, Rabigh 21911, Saudi Arabia} \email{habib.maagli@fst.rnu.tn} \address{Ramzi Alsaedi \newline Department of Mathematics, Faculty of Sciences, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia} \email{ramzialsaedi@yahoo.co.uk} \address{Noureddine Zeddini \newline Department of Mathematics, College of Sciences and Arts \\ King Abdulaziz University, Rabigh Campus \\ P.O. Box 344, Rabigh 21911, Saudi Arabia} \email{noureddine.zeddini@ipein.rnu.tn} \thanks{Submitted December 8, 2015. Published February 17, 2016.} \subjclass[2010]{34B16, 34B18, 34D05} \keywords{Singular nonlinear boundary value problems; positive solution; \hfill\break\indent exact asymptotic behavior; Karamata regular variation theory} \begin{abstract} In this article, we give an exact behavior at infinity of the unique solution to the following singular boundary value problem \begin{gather*} -\frac{1}{A}(Au')'=q(t)g(u), \quad t \in (0,\infty), \\ u>0, \quad \lim_{t\to 0}A(t)u'(t)=0, \quad \lim_{t\to \infty}u(t)=0. \end{gather*} Here $A$ is a nonnegative continuous function on $[0,\infty)$, positive and differentiable on $(0,\infty)$ such that \[ \lim_{t\to \infty}\frac{tA'(t)}{A(t)}=\alpha>1, \quad g \in C^1((0,\infty),(0,\infty)) \] is non-increasing on $(0,\infty)$ with $\lim_{t\to 0}g'(t)\int_0^t\frac{ds}{g(s)}=-C_g\leq 0$ and the function $q$ is a nonnegative continuous, satisfying \[ 00$ and $y$ is continuous on $[ 1,\infty)$ such that $\lim_{t\to \infty}y(t)=0$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article, we give the exact asymptotic behavior at infinity of the unique positive solution to the singular problem \begin{equation}\label{P} \begin{gathered} \frac{1}{A}(Au')'=-q(t)g(u)\,,\quad t \in (0,\infty) , \\ u>0\,,\quad \text{in } (0,\infty)\\ \lim_{t\to 0^+}A(t)u'(t)=0,\quad \lim_{t \to \infty}u(t)=0, \end{gathered} \end{equation} where the functions $A$, $q$ and $g$ satisfy the following assumptions. \begin{itemize} \item[(H1)] $A$ is a continuous function on $[0,\infty)$, positive and differentiable on $(0,\infty)$ such that \begin{equation*} \lim_{t\to \infty}\frac{t\,A'(t)}{A(t)}=\alpha>1. \end{equation*} \item[(H2)] $q$ is a nonnegative continuous function on $(0,\infty)$ satisfying \begin{equation*} 00$. \end{itemize} Using that $g$ is non-increasing, for $t>0$, we obtain \begin{equation*} 00$, \begin{equation*} \int_0^tg'(s)\int_0^s\frac{1}{g(r)}\,dr\,ds=g(t)\int_0^t\frac{1}{g(s)}ds-t, \end{equation*} we obtain \begin{equation}\label{eq 1.2} \lim_{t\to 0}\frac{g(t)}{t}\int_0^t\frac{1}{g(s)}\,ds=1-C_g. \end{equation} This implies that $0\leq C_g\leq 1$. The functions $t^{-1}\log(1+t)$, $\log(\log(e+\frac{1}{t}))$, $t^{-\nu}\log(1+\frac{1}{t})$, $\exp\{(\log(1+\frac{1}{t}))^{\nu}\}$, $\nu \in (0,1)$ satisfy the assumption (H3), as well as the function \begin{gather*} t^2 e^{1/t}\,,\quad \text{if } 00$ and $y\in C( [ 1,\infty )) $ such that $\lim_{t\to \infty}y(t)=0$. \begin{remark} \label{remark1} \rm It is clear that a function $L$ is in $\mathcal{K}$ if and only if $L$ is a positive function in $C^{1}( [1,\infty ))$ such that \begin{equation}\label{tLprimesurL-infty} \lim_{t\to \infty}\frac{t\,L^{\prime }(t)}{L(t)}=0. \end{equation} \end{remark} Throughout this paper, we denote by $\psi_g$ the unique solution of the equation \begin{equation}\label{eq psi g} \int_0^{\psi_g(t)}\frac{ds}{g(s)}=t\,,\quad \text{for }t\in [0,\infty), \end{equation} and we mention that \begin{equation}\label{eq 1.3} \lim_{t\to 0} t g'(\psi_g(t))=-C_g. \end{equation} \begin{theorem} \label{result1} Assume {\rm (H1)--(H4)}. Then problem \eqref{P} has a unique solution $u\in C^{2}((0,\infty))\cap C([0,\infty))$ satisfying \begin{itemize} \item[(i)] If $\lambda>2$, \begin{equation*} \Big(\frac{\xi_1}{\lambda-2}\Big)^{1-C_g} \leq \liminf_{t \to \infty} \frac{u(x)}{\psi_g(t^{2-\lambda}\,L(t))} \leq \limsup_{t \to \infty} \frac{u(x)}{\psi_g(t^{2-\lambda}\,L(t))} \leq \Big(\frac{\xi_2}{\lambda-2}\Big)^{1-C_g}\,, \end{equation*} where $\xi_i=\frac{a_i}{\alpha+1-\lambda+(\lambda-2)C_g}$ for $i\in \{1\,2\}$. \item[(ii)]If $\lambda=2$, \begin{equation*} {\xi_1}^{1-C_g} \leq \liminf_{t \to \infty} \frac{u(t)}{\psi_g(\int_t^{\infty}\frac{L(s)}{s}\,ds)} \leq \limsup_{t \to \infty} \frac{u(t)}{\psi_g(\int_t^{\infty}\frac{L(s)}{s}\,ds)} \leq {\xi_2}^{1-C_g} \end{equation*} \end{itemize} \end{theorem} An immediate consequence of Theorem \ref{result1} is the following result. \begin{corollary} \label{Corol1} Let $u$ be the unique solution of \eqref{P}. Then, we have the following exact asymptotic behavior: \\ (a) When $C_g=1$, we have \begin{itemize} \item[(i)] $\lim_{t \to \infty} \frac{u(t)}{\psi_g(t^{2-\lambda}\,L(t))}=1$ if $\lambda >2$; \item[(ii)] $\lim_{t \to \infty} \frac{u(t)}{\psi_g(\int_t^{\infty}\frac{L(s)}{s}\,ds)}=1$ if $\lambda =2$. \end{itemize} (b) When $C_g<1$ and $a_1=a_2=a_0$, we have \begin{itemize} \item[(i)] $\lim_{t \to \infty}\frac{u(t)}{\psi_g(t^{2-\lambda}\,L(t))}= (\frac{a_0}{(\lambda-2)(\alpha+1-\lambda+(\lambda-2)C_g)})^{1-C_g}$ if $\lambda >2$; \item[(ii)] $\lim_{t \to \infty}\frac{u(t)}{\psi_g(\int_t^{\infty} \frac{L(s)}{s}\,ds)}=(\frac{a_0}{\alpha-1})^{1-C_g}$ if $\lambda =2$. \end{itemize} \end{corollary} \begin{remark} \rm In the hypothesis (H3), we do not need the monotonicity of the function $g$ on $(0,\infty)$, but only the fact that $g$ is non-increasing in a neighborhood of zero. \end{remark} \begin{example} \rm Let $g$ be the function \begin{equation*} g(t)=\begin{cases} t^2 e^{1/t}\,, & \text{if } 02$; \item[(ii)] $\lim_{t \to \infty}u(t) \log\big(\frac{1}{\int_t^{\infty} \frac{L(s)}{s}\,ds}\big)=1$ if $\lambda =2$. \end{itemize} \end{example} To establish our second result, we consider the special case where $g(t)=t^{-\gamma}$ with $\gamma\geq 0$, and $\lambda=\alpha+1+\gamma(\alpha-1)$. Note that in this case $C_g=\frac{\gamma}{\gamma+1}$ and $(\alpha+1-\lambda)+(\lambda-2)C_g=0$. We assume that $A$ and $q$ satisfy the following hypotheses: \begin{itemize} \item[(H5)] $A$ is a continuous function on $(0,\infty)$ such that $A(t)=t^{\alpha}B(t)$ with $\alpha>1$ and $\frac{t^{\nu}B'(t)}{B(t)}$ is bounded for $t$ large and $\nu \in (0,1)$. \item[(H6)] $q$ is a nonnegative continuous function in $(0,\infty)$ and satisfies \begin{equation*} 00$, then \begin{equation*} \lim_{t\to \infty}t^{-\varepsilon }L(t)=0. \end{equation*} (ii) Let $L_{1},L_{2}\in \mathcal{K}$ and $p\in \mathbb{R}$. Then $L_{1}+L_{2}\in \mathcal{K}$, $L_{1}L_{2}\in \mathcal{K}$ and $L_{1}^{p}\in \mathcal{K}$ . \end{lemma} Applying Karamata's theorem (see \cite{M,Sen}), we get the following result. \begin{lemma} \label{lemma 2.3} Let $\gamma \in {\mathbb{R}}$, $L$ be a function in $\mathcal{K}$ defined on $[1, \infty)$. We have \begin{itemize} \item[(i)] If $\gamma <-1$, then $\int_{1}^{\infty }s^{\gamma}L(s)ds$ converges. Moreover \[ \int_{t}^{\infty }s^{\gamma }L(s)ds \sim_{t\to \infty}-\frac{ t^{1+\gamma }L(t)}{\gamma+1}. \] \item[(ii)] If $\gamma>-1$, then $\int_{1}^{\infty }s^{\gamma }L(s)ds$ diverges. Moreover \[ \int_{1}^{t }s^{\gamma }L(s)ds \sim_{t\to \infty} \frac{t^{1+\gamma }L(t)}{\gamma+1}. \] \end{itemize} \end{lemma} \begin{lemma}[\cite{CDM,MMZ}] \label{lemma 2.4} Let $L\in \mathcal{K}$ be defined on $[1,\infty)$. Then \begin{equation}\label{lemma 2.4 eq1-infty} \lim_{t\to \infty}\frac{L( t) }{\int_{1}^{t } \frac{L(s)}{s}ds}=0. \end{equation} If further $\int_{1}^{\infty }\frac{L(s)}{s}ds$ converges, then \begin{equation}\label{lemma 2.4 eq2-infty} \lim_{t\to \infty}\frac{L( t) }{\int_{t}^{\infty} \frac{L(s)}{s}ds}=0. \end{equation} \end{lemma} \begin{remark} \rm Let $L \in \mathcal{K}$, then using Remark \ref{remark1} and \eqref{lemma 2.4 eq1-infty}, we deduce that \begin{equation*} t\to \int_1^{t}\frac{L(s)}{s}\,ds \in {\mathcal{K}}. \end{equation*} If further $\int_1^{\infty}\frac{L(s)}{s}\,ds$ converges, then $t \to \int_t^{\infty}\frac{L(s)}{s}\,ds\in {\mathcal{K}}$. \end{remark} \begin{definition}\label{NRVI-class} \rm A positive measurable function $k$ is called normalized regularly varying at infinity with index $\rho \in {\mathbb{R}}$ and we write $k \in NRVI_{\rho}$ if $k(s)=s^{\rho}L(s)$ for $s \in [1,\infty)$ with $ L \in {\mathcal{K}}$. \end{definition} Using the definition of the class ${\mathcal{K}}$ and the above Lemmas we obtain the following lemma. \begin{lemma}[\cite{AMZ}] \label{lem 2.5} \begin{itemize} \item[(i)] If $k \in NRVI_{\rho}$, then $\lim_{t \to \infty} \frac{k(\xi t)}{k(t)}=\xi^{\rho}$, uniformly for $\xi \in [c_1, c_2] \subset (0,\infty)$. \item[(ii)] A positive measurable function $k$ belongs to the class $NRVI_{\rho}$ if and only if $\lim_{t \to \infty} \frac{tk'(t)}{k(t)}=\rho$. \item[(iii)] Let $L\in {\mathcal{K}}$ and assume that $\int_1^{\infty}s^{1-\lambda}\,L(s)\,ds<\infty$. Then the function $\theta(t)=\int_t^{\infty}s^{1-\lambda}\,L(s)\,ds$ belongs to $NRVI_{(2-\lambda)}$. \item[(iv)] The function $\psi_g\circ \theta \in NRVI_{(2-\lambda)(1-C_g)}$. \item[(v)] Let $m_1$, $m_2$ be positive functions on $(0,\infty)$ such that $\lim_{t\to \infty}m_1(t)= \lim_{t\to \infty}m_2(t)=0$ and $\lim_{t\to \infty}\frac{m_1(t)}{m_2(t)}=1$. Then $\lim_{t \to \infty}\frac{\psi_g(m_1(t))}{\psi_g(m_2(t))}=1$. \end{itemize} \end{lemma} \section{Proofs of Theorems \ref{result1} and \ref{result2}} In the sequel, we denote by $$ v_0(t)=\int_t^{\infty}\frac{1}{A(s)}\,ds\quad \text{for } t\in (0,\infty). $$ Since the function $A$ satisfies (H1), then using Definition \ref{NRVI-class} and assertion (ii) of Lemma \ref{lem 2.5}, we deduce that there exists $L_0\in {\mathcal{K}}$ such that $A(t)=t^{\alpha}\,L_0(t)$, for $t>1$. Hence, using Lemma \ref{lemma 2.1}, we deduce that $1/A$ is integrable near infinity. So the function $v_0$ is well defined, and by Lemma \ref{lemma 2.3} we have \begin{equation}\label{v0 comparable t 1-alpha sur L0} v_0(t)=\int_t^{\infty}\frac{1}{A(s)}\,ds \sim \frac{t^{1-\alpha }}{(\alpha-1)\,L_0(t)}\quad \text{as } t \to \infty. \end{equation} In the sequel, we denote also by $L_Au:=\frac{1}{A}(Au')'=u''+\frac{A'}{A}u'$ and we remark that $L_Av_0=0$. \begin{proof}[Proof of Theorem \ref{result1}] Let $\varepsilon \in (0,a_1/2)$. Put \[ \xi_i=\frac{a_i}{(\alpha+1-\lambda)+(\lambda-2)C_g}\quad\text{for } i\in\{1,2\}, \] $\tau_1=\xi_1-\varepsilon \frac{\xi_1}{a_1}$ and $\tau_2=\xi_2+\varepsilon \frac{\xi_2}{a_2}$. Clearly, we have $\frac{\xi_1}{2}<\tau_1<\tau_2<\frac{3}{2}\xi_2$. Let $\theta(t)=\int_t^{\infty}s^{1-\lambda}\,L(s)\,ds$ and put $$ \omega_i(t)=\psi_g\Big(\tau_i\int_t^{\infty}s^{1-\lambda}\,L(s)\,ds\Big) =\psi_g(\tau_i\,\theta(t)), \quad \text{for } t>0. $$ By a simple calculus, for $i\in\{1,2\}$ we obtain \begin{align*} &L_A \omega_i(t)+q(t)g(\omega_i(t)) \\ &= g(\omega_i(t))t^{-\lambda}\,L(t)) \Big[\tau_i(\tau_i t^{{2-\lambda}}L(t)g'(\omega_i(t))+(\lambda-2)C_g)\\ &\quad - \tau_i\Big(\frac{t\,A'(t)}{A(t)}-\alpha+\frac{t\,L'(t)}{L(t)}\Big) -\tau_i((\alpha+1-\lambda+(\lambda-2)C_g)+a_i \\ &\quad +\Big(\frac{q(t)}{t^{-\lambda}\,L(t)}-a_i\Big)\Big]. \end{align*} So, for the fixed $\varepsilon>0$, there exists $M_{\varepsilon}>1$ such that for $t> M_{\varepsilon}$ and $i\in\{1,2\}$, we have \begin{gather*} \tau_i\big|\frac{t\,A'(t)}{A(t)}-\alpha+\frac{t\,L'(t)}{L(t)}\big| \leq \frac{3}{2}\xi_2\Big(\big|\frac{t\,A'(t)}{A(t)}-\alpha\big| +\big|\frac{t\,L'(t)}{L(t)}\big|\Big) \leq \frac{\varepsilon}{4}\,, \\ a_1-\frac{\varepsilon}{2} \leq \frac{a(t)}{t^{-\mu}\,L(t)}\leq a_2 +\frac{\varepsilon}{2} \\ |\tau_i(\tau_i\,t^{2-\lambda}\,L(t)g'(\omega_i(t))+(\lambda-2)C_g)| \leq \frac{3}{2}\xi_2|\tau_i\,t^{2-\lambda}\,L(t)g'(\omega_i(t)) +(\lambda-2)C_g | \leq \frac{\varepsilon}{4}. \end{gather*} Indeed, the last inequality follows from \eqref{eq 1.3} and the fact that from Lemmas \ref{lemma 2.3} and \ref{lemma 2.4}, we have \[ \lim_{t \to \infty} \frac{t^{2-\lambda}\,L(t)}{\int_t^{\infty}s^{1-\lambda} \,L(s)\,ds}=2-\lambda, \] for all $\lambda\geq 2$. This implies that for each $t> M_{\varepsilon}$, we have \begin{equation*} L_A \omega_1(t)+q(t)g(\omega_1(t)) \geq g(\omega_1(t))t^{-\lambda} L(t)[-\varepsilon +a_1-\tau_1((\alpha+1-\lambda)+(\lambda-2)C_g)]=0 \end{equation*} and \begin{equation*} L_A \omega_2(t)+q(t)g(\omega_2(t)) \leq g(\omega_2(t))t^{-\lambda}\,L(t)[\varepsilon+a_2- \tau_2((\alpha+1-\lambda)+(\lambda-2)C_g)]=0. \end{equation*} Let $u \in C^{2}((0,\infty))\cap C([0,\infty))$ be the unique solution of \eqref{P} (see \cite{MM}). Then, there exists $B>0$ such that \begin{equation}\label{eq-t=M-epsilon} \omega_1(M_{\varepsilon})-B\,v_0(M_{\varepsilon}) \leq u(M_{\varepsilon})\leq \omega_2(M_{\varepsilon})+B\,v_0(M_{\varepsilon})\,. \end{equation} We claim that \begin{equation} \label{eq-t sup M-epsilon} \omega_1(t)-B\,v_0(t)\leq u(t)\leq \omega_2(t)+B\,v_0(t)\quad \text{for all } t>M_{\varepsilon}. \end{equation} Assume for instance that the right inequality of \eqref{eq-t sup M-epsilon} is not true. Then the function $h(t)= u(t)- \omega_2(t)-B\,v_0(t)$ for $t>M_{\varepsilon}$ is not negative. Consequently, there exists $t_1>M_{\varepsilon}$ such that $h(t_1)=\max_{M_{\varepsilon}\leq t<\infty}h(t)>0$. Since $h$ is continuous on $[M_{\varepsilon},\infty)$, $h(M_{\varepsilon})\leq 0$ and $\lim_{t \to \infty}h(t)=0$, then $h'(t_1)=0$ and $h(t)>0$ for $t \in (t_1-\delta,t_1+\delta)$ for some $\delta>0$, sufficiently small. Namely $u(t)>\omega_2(t)+B\,v_0(t)$ for $t \in (t_1-\delta,t_1+\delta)$. Since $g$ is non-increasing on $(0,\infty)$, then \[ \frac{1}{A(t)}(A(t)h'(t))' =-q(t)g(u(t))-\frac{1}{A(t)}(A(t)\omega_2'(t))' \geq q(t)(g(\omega_2(t))-g(u(t)))\geq0, \] for $t\in (t_1-\delta,t_1+\delta)$. Which implies $h'(t)\leq h'(t_1)=0$ for $t\in (t_1-\delta,t_1)$ and $h'(t)\geq h'(t_1)=0$ for $t\in (t_1,t_1+\delta)$. This implies that $h$ has a local minimum at $t_1$. Which contradicts the fact that $h$ a global maximum at $t_1$ on $[M_{\varepsilon}, \infty)$. This proves that \begin{equation*} u(t)\leq \omega_2(t)+B\,v_0(t)\quad \text{for all } t>M_{\varepsilon}. \end{equation*} Similarly, we show that \begin{equation*} \omega_1(t)-B\,v_0(t)\leq u(t)\quad \text{for all } t>M_{\varepsilon}. \end{equation*} This proves \eqref{eq-t sup M-epsilon}. Now, since $\psi_g \circ \theta \in NRVI_{(2-\lambda)(1-C_g)}$, there exists $\hat{L}\in {\mathcal{K}}$ such that $\psi_g \circ \theta=t^{(2-\lambda)(1-C_g)}\, \hat{L}(t)$ for $t\in [1,\infty)$. Moreover since $(\alpha-1)-(\lambda-2)(1-C_g)>0$, it follows by Lemma \ref{lemma 2.1} that \[ \lim_{t\to \infty}\frac{t^{1-\alpha}}{t^{(2-\lambda)(1-C_g)}\hat{L}(t)} =0. \] This implies that \[ \lim_{t\to \infty} \frac{t^{1-\alpha}}{\psi_g(\tau_i\int_t^{\infty}s^{1-\lambda} L(s)\,ds)} =\lim_{t\to \infty} \frac{t^{1-\alpha}}{\psi_g(\tau_i\theta(t))} =\lim_{t\to \infty} \frac{\psi_g(\theta(t))}{\psi_g(\tau_i\theta(t))} \frac{t^{1-\alpha}}{\psi_g(\theta(t))}=0 \] uniformly in $\tau_i \in [\frac{\xi_1}{2},\frac{3}{2}\xi_2] \subset (0,\infty)$. This together with \eqref{v0 comparable t 1-alpha sur L0} implies \begin{equation*} \lim_{t \to \infty}\frac{v_0(t)}{\psi_g(\tau_1\theta(t))} =\lim_{t \to \infty}\;\frac{v_0(t)}{\psi_g(\tau_2\theta(t))}=0. \end{equation*} So, we obtain \begin{equation*} \limsup_{t \to \infty}\frac{u(t)}{\omega_2(t)} \leq 1\leq \liminf_{t \to \infty} \frac{u(t)}{\omega_1(t)}. \end{equation*} Using this fact and assertions (i) and (iv) of Lemma \ref{lem 2.5}, we deduce that \[ \liminf_{t \to \infty}\frac{u(t)}{\psi_g(\theta(t))} =\liminf_{t \to \infty} \frac{u(t)}{\omega_1(t)}\,\frac{\omega_1(t)}{\psi_g(\theta(t))} \geq \lim_{t \to \infty}\,\frac{\psi_g(\tau_1 \theta(t))}{\psi_g(\theta(t))}= {\tau_1}^{1-C_g}. \] By letting $\varepsilon$ approach zero, we obtain \begin{equation*} {\xi_1}^{1-C_g} \leq \liminf_{t \to \infty}\;\frac{u(t)}{\psi_g(\theta(t))}. \end{equation*} Similarly, we obtain \begin{equation*} \limsup_{t \to \infty} \frac{u(t)}{\psi_g(\theta(t))} \leq {\xi_2}^{1-C_g}. \end{equation*} This proves in particular the exact behavior at infinity in the case $\lambda=2$. Now, for $\lambda>2$, we have by Lemma \ref{lemma 2.3} that $\theta(t)\sim_{t\to \infty}\frac{t^{2-\lambda}}{\lambda-2}L(t)$. Hence it follows by assertions (i), (iv) and (v) of Lemma \ref{lem 2.5} that for $\lambda>2$, we have $$ \lim_{t \to \infty}\frac{\psi_g(\theta(t))}{\psi_g((t)^{2-\lambda}L(t))} =\lim_{t \to \infty}\frac{\psi_g(\theta(t))}{\psi_g((\lambda-2)\theta(t))} \frac{\psi_g((\lambda-2)\theta(t))}{\psi_g((t)^{2-\lambda}L(t))} =\frac{1}{(\lambda-2)^{1-C_g}}. $$ This achieves the proof of the Theorem. \end{proof} \begin{proof}[Proof of Theorem \ref{result2}] We recall that $g(t)=t^{-\gamma}$, $\lambda=\alpha+1+(\alpha-1)\gamma$ and $C_g=\frac{\gamma}{1+\gamma}$. Let $\varepsilon \in (0,\frac{a_1}{2})$ and put $\tau_1=(\gamma+1)(a_1-\varepsilon)$ and $\tau_2=(\gamma+1)(a_2+\varepsilon)$. Put $k(t)=\int_1^{t}\frac{L(s)}{s}\,ds$ and \[ \omega_i(t)=\Big((1+\gamma)\tau_i \int_{t}^{\infty}s^{1-\lambda}\,k(s)\,ds \Big)^{\frac{1}{1+\gamma}}\quad\text{for }i\in\{1,2\}, \] where $L$ is the function given in hypothesis $({{H}}_6)$. Then, by a simple computation, we have \begin{align*} &L_A \omega_i(t)+q(t)g(\omega_i(t))\\ &= g(\omega_i(t))t^{-\lambda}\,L(t) \Big[\tau_i\Big(\frac{k(t)}{L(t)} (\tau_i t^{{2-\lambda}}k(t)g'(\omega_i) +(\lambda-1-\alpha) )-\frac{\gamma}{\gamma+1}\Big) \\ &\quad -\tau_i \frac{k(t)}{L(t)} \Big(\frac{t\,A'(t)}{A(t)}-\alpha\Big)+\frac{\gamma}{\gamma+1}\tau_i -\tau_i+a_i +\Big(\frac{q(t)}{t^{-\lambda}\,L(t)}-a_i \Big)\Big]\\ &= g(\omega_i(t))t^{-\lambda}\,L(t) \Big[\tau_i\Big(\frac{k(t)}{L(t)} (\tau_i t^{{2-\lambda}}k(t) g'(\omega_i)+(\alpha-1)\gamma )-\frac{\gamma}{\gamma+1}\Big)\\ & \quad -\tau_i\,\frac{k(t)}{L(t)}\,\frac{t\,B'(t)}{B(t)} -\frac{\tau_i}{\gamma+1}+a_i +\Big(\frac{q(t)}{t^{-\lambda}\, L(t)}-a_i \Big)\Big]. \end{align*} Since $g(t)=t^{-\gamma}$ and $\lambda=\alpha+1+(\alpha-1)\gamma$, integrating by parts, we obtain \begin{align*} &\tau_i t^{{2-\lambda}}k(t)g'(\omega_i(t))+(\alpha-1)\gamma \\ &= -\gamma \tau_i\, t^{2-\lambda}k(t) (\omega_i(t))^{-(1+\gamma)}+(\alpha-1)\gamma\\ &= \gamma \Big((\alpha-1)-\frac{t^{2-\alpha}k(t)}{(\gamma+1) \int_{t}^{\infty}s^{1-\lambda}k(s)\,ds}\Big)\\ &=\gamma \Big( \frac{(\alpha-1)(1+\gamma) \int_{t}^{\infty}s^{1-\lambda} k(s) ds -t^{2-\lambda}k(t)} {(1+\gamma) \int_{t}^{\infty}s^{1-\lambda} k(s) ds}\Big)\\ &= \frac{\gamma}{\gamma+1} \frac{\int_{t}^{\infty}s^{1-\lambda}L(s)\,ds} {\int_{t}^{\infty}s^{1-\lambda} k(s)ds}. \end{align*} This gives \begin{align*} &\frac{k(t)}{L(t)}(\tau_i t^{{2-\lambda}}k(t)g'(\omega_i(t))+(\alpha-1)\gamma) -\frac{\gamma}{\gamma+1}\\ &=\frac{\gamma}{\gamma+1} \Big[\frac{\int_{t}^{\infty}s^{1-\lambda} L(s)\,ds}{t^{2-\lambda}\,L(t)}\frac{t^{2-\lambda}\,k(t)} {\int_{t}^{\infty}s^{1-\lambda} k(s)ds}-1\Big]. \end{align*} This together with Lemma \ref{lemma 2.3} and the fact that $k$ and $L$ are in ${\mathcal{K}}$, implies \begin{equation*} \lim_{t \to \infty} \frac{k(t)}{L(t)}(\tau_i t^{{2-\lambda}}k(t) g'(\omega_i(t))+(\alpha-1)\gamma)-\frac{\gamma}{\gamma+1}=0. \end{equation*} Now since $\frac{t^{\nu}\,B'(t)}{B(t)}$ is bounded for $t$ large and by Lemma \ref{lemma 2.1}, we have $\frac{k}{L}\in {\mathcal{K}}$ and $\lim_{t\to \infty}\frac{t^{1-\nu}\,k(t)}{L(t)}=0$, we deduce that $$ \lim_{t\to \infty}\frac{k(t)}{L(t)}\Big(\frac{t\,B'(t)}{B(t)}\Big) = \lim_{t\to \infty}\frac{t^{1-\nu}\,k(t)}{L(t)}\Big(\frac{t^{\nu}\,B'(t)}{B(t)}\Big)=0. $$ So, for the fixed $\varepsilon>0$, there exists $M_{\varepsilon} >1$ such that for $t\geq M_{\varepsilon}$, we have \begin{gather*} L_A \omega_2(t)+q(t)g(\omega_2(t)) \leq g(\omega_2(t))t^{-\lambda}L(t) \Big[\frac{\varepsilon}{3}+\frac{\varepsilon}{3}-\frac{\tau_2}{\gamma+1}+ a_2+\frac{\varepsilon}{3}\Big]=0, \\ L_A \omega_1(t)+q(t)g(\omega_1(t)) \geq g(\omega_1(t))t^{-\mu}\,L(t)[-\frac{\varepsilon}{3} -\frac{\varepsilon}{3}-\frac{\tau_1}{\gamma+1}+a_1-\frac{\varepsilon}{3}\Big]=0. \end{gather*} Let $u \in C([0,\infty))\cap C^2((0,\infty))$ be the unique solution of \eqref{P2}. As in the proof of Theorem \ref{result1}, we choose $C>0$ such that $$ \omega_1(t)-C v_0(t)\leq u(t) \leq \omega_2(t)+C v_0(t) \quad \text{for } t\geq M_{\varepsilon}. $$ Moreover, thanks to (H6), we have $\lim_{t\to \infty}k(t)=\infty$. So, using Lemma \ref{lemma 2.3}, we obtain \begin{align*} \lim_{t\to \infty }\frac{1}{t^{\alpha-1} \Big((1+\gamma)\tau_1 \int_{t}^{\infty}s^{1-\lambda}\,k(s)\,ds \Big)^{\frac{1}{1+\gamma}}} &= \lim_{t\to \infty} \frac{1}{t^{\alpha-1}\big(t^{(2-\lambda)} \frac{\tau_1\,k(t)}{\alpha-1}\big)^{\frac{1}{1+\gamma}}}\\ &= \lim_{t\to \infty} \Big(\frac{\alpha-1}{\tau_1\,k(t)}\Big)^{\frac{1}{1+\gamma}}=0. \end{align*} This and \eqref{v0 comparable t 1-alpha sur L0} gives $\lim_{t\to \infty}\frac{v_0(t)}{\omega_1(t)}=0$. Similarly, we obtain $\lim_{t \to \infty} \frac{v_0(t)}{\omega_2(t)}=0$. So we have \[ \limsup_{t \to \infty} \frac{u(t)}{\omega_2(t)}\leq 1 \leq \liminf_{t \to \infty} \frac{u(t)}{\omega_1(t)}. \] This implies that $$ \liminf_{t \to \infty} \frac{u(t)}{\Big((1+\gamma) \int_{t}^{\infty}s^{1-\lambda}\,k(s) \,ds\Big)^{\frac{1}{1+\gamma}}}\geq \tau_1^{\frac{1}{1+\gamma}}. $$ Now, as $\varepsilon$ tends to zero, we obtain $$ \liminf_{t \to \infty} \frac{u(t)}{\Big((1+\gamma) \int_{t}^{\infty}s^{1-\lambda}\,k(s)\,ds \Big)^{\frac{1}{1+\gamma}}}\geq ((\gamma+1)a_1)^{\frac{1}{1+\gamma}}. $$ Similarly, we obtain $$ \limsup_{t \to \infty} \frac{u(t)}{\Big((1+\gamma) \int_{t}^{\infty}s^{1-\lambda}\,k(s)\,ds\Big)^{\frac{1}{1+\gamma}}} \leq ((\gamma+1)a_2)^{\frac{1}{1+\gamma}}. $$ Now, since $(\gamma+1) \int_{t}^{\infty}s^{1-\lambda}\,k(s)\,ds \sim_{t \to \infty }\frac{t^{2-\lambda}\,k(t)}{\alpha-1} =\frac{t^{(1-\alpha)(\gamma+1)}}{\alpha}\int_1^{t}\frac{L(s)}{s}\,ds$, we deduce that \begin{align*} \Big(\frac{(\gamma +1)a_1}{\alpha-1}\Big)^{\frac{1}{1+\gamma}} & \leq \liminf_{t \to \infty} \frac{u(t)}{t^{1-\alpha}\big(\int_1^{t} \frac{L(s)}{s}\,ds\big)^{\frac{1}{1+\gamma}}}\\ &\leq \limsup_{t \to \infty} \frac{u(t)}{t^{1-\alpha} \big(\int_1^{t}\frac{L(s)}{s}\,ds\big)^{\frac{1}{1+\gamma}}}\\ &\leq (\frac{(\gamma +1)a_2}{\alpha})^{\frac{1}{1+\gamma}}. \end{align*} In particular, if $a_1=a_2$, we obtain \begin{equation*} \lim_{t \to \infty} \frac{u(t)}{t^{1-\alpha}\big(\int_1^{t}\frac{L(s)}{s}\,ds\big)^{\frac{1}{1+\gamma}}} = \Big(\frac{(\gamma+1)a_1}{\alpha-1}\Big)^{\frac{1}{1+\gamma}}. \end{equation*} \end{proof} \section{Applications} \subsection*{Application 1} We consider the Dirichlet problem \begin{equation}\label{P-beta} \begin{gathered} -\frac{1}{A}(Au')'+\frac{\beta}{u} (u')^2=q(t)g(u)\,,\quad t \in (0,\infty) , \\ u>0\,,\quad \text{in } (0,\infty)\,,\\ \lim_{t\to 0^+}A(t)u'(t)=0 \,\quad \lim_{t \to \infty}u(t)=0, \end{gathered} \end{equation} where $\beta <1$ and $\lim_{t \to \infty}\frac{q(t)}{t^{-\lambda}\,L(t)}=a_0>0$ with $\lambda \geq 2$ and $L\in {\mathcal{K}}$ with \\ $\int_1^{\infty}s^{1-\lambda}\,L(s)\,ds<\infty$. We assume that $A$ satisfies (H1) and $g$ satisfies the following hypotheses: \begin{itemize} \item[(A1)] The function $t \to t^{-\beta}\,g(t)$ is non-increasing from $(0,\infty)$ into $(0,\infty)$. \item[(A2)] $\lim_{t\to 0}g'(t)\int_0^t\frac{1}{g(s)}\,ds=-C_g$ with $\max(0,\frac{\beta}{\beta-1})\leq C_g\leq 1$. \item[(A3)] $(\alpha-1)-(\lambda-2)(1-\beta)(1-C_g)>0$ \end{itemize} Note that for $\gamma>0$ and $-\gamma<\beta<1$, the function $g(t)=t^{-\gamma}$ satisfies ${({A_1})}$ and (A2). Put $u=v^{\frac{1}{1-\beta}}$. Then $v$ satisfies \begin{equation}\label{Q-beta} \begin{gathered} -\frac{1}{A}(Av')'=(1-\beta)q(t)g(v^{\frac{1}{1-\beta}})v^{\frac{-\beta}{1-\beta}}\,, \quad t \in (0,\infty) , \\ v>0\,,\quad \text{in } (0,\infty)\;,\\ \lim_{t\to 0^+}A(t)v'(t)=0, \quad \lim_{t \to \infty}v(t)=0, \end{gathered} \end{equation} The function $f(r)=(1-\beta)g(r^{\frac{1}{1-\beta}})r^{-\frac{\beta}{1-\beta}}$ is non-increasing on $(0,\infty)$ and a simple computation shows that $\psi_g=(\psi_f)^{\frac{1}{1-\beta}}$ and \begin{equation*} \lim_{r\to 0}f'(r)\int_0^r\frac{1}{f(s)}\,ds=(1-\beta)(1-C_g)-1 =:-C_f \,,\quad \text{with } 0\leq C_f\leq 1. \end{equation*} Applying Corollary \ref{Corol1} to problem \eqref{Q-beta}, we deduce that there exists a unique solution $v$ to \eqref{Q-beta} such that\\ (a) When $C_f=1$, we have \begin{itemize} \item[(i)] $\lim_{t \to \infty}\frac{v(t)}{\psi_f(t^{2-\lambda}L(t))} =1$ if $\lambda >2$; \item[(ii)] $\lim_{t \to \infty}\frac{v(t)} {\psi_f\big(\int_{t}^{\infty}\frac{L(s)}{s}\,ds\big)}=1$ if $\lambda=2$; \end{itemize} (b) When $C_f<1$, we have: \begin{itemize} \item[(i)] $\lim_{t \to \infty}\frac{v(t)}{\psi_f(t^{2-\lambda}\,L(t))} =[\frac{a_0}{\alpha+1-\lambda+(\lambda-2)C_f}]^{1-C_f}$ if $ 2< \lambda <2+\frac{\alpha-1}{(1-\beta)(1-c_g)}$; \item[(ii)] $\lim_{t \to \infty}\frac{v(t)}{\psi_f(\int_{t}^{\infty} \frac{L(s)}{s}\,ds)}=[\frac{a_0}{\alpha-1}]^{1-C_f}$ if $\lambda=2$. \end{itemize} This implies that problem \eqref{P-beta} has a solution $u\in C([0,\infty))\cap C^2((0,\infty))$ satisfying the following exact behavior\\ (a) When $C_g=1$, we have: \begin{itemize} \item[(i)] if $\lambda >2$, then \[ \lim_{t \to \infty}\frac{u(t)}{\psi_g(t^{2-\lambda}\,L(t))}=1; \] \item[(ii)] if $\lambda=2$, then \[ \lim_{t \to \infty}\frac{u(t)}{\psi_g(\int_{t}^{\infty} \frac{L(s)}{s}\,ds)}=1; \] \end{itemize} (b) If $\max(0,\frac{\beta}{\beta-1})\leq C_g<1$, then: \begin{itemize} \item[(i)] if $ 2< \lambda <2+\frac{\alpha-1}{(1-\beta)(1-C_g)}$, then \[ \lim_{t\to \infty}\frac{u(t)}{\psi_g(t^{2-\lambda}\,L(t))} =[\frac{a_0}{\alpha-1-(\lambda-2)(1-\beta)(1-C_g)}]^{1-C_g} \] \item[(ii)] if $\lambda=2$, then \[ \lim_{|x| \to \infty}\frac{u(t)}{\psi_g(\int_{t}^{\infty} \frac{L(s)}{s}\,ds)}=[\frac{a_0}{\alpha-1}]^{1-C_g}. \] \end{itemize} \subsection*{Application 2} In this subsection, we assume that the function $A$ satisfy the following hypothesis \begin{itemize} \item[(A4)] $A$ is a continuous function on $[0,\infty)$, positive and differentiable on $(0,\infty)$ such that $\frac{1}{A}$ is integrable near 0 and $\lim_{t\to \infty}\frac{t\,A'(t)}{A(t)}=\sigma \in {\mathbb{R}}-\{1\}$. \end{itemize} We are interested in the exact behavior at infinity of the unique positive solution of the problem \begin{equation}\label{Applic2-prob} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'=-p(t)u^{-\gamma}\,,\quad t \in (0,\infty) , \\ u>0\,,\quad \text{in } (0,\infty)\;,\\ u(0)=0\,,\quad \lim_{t\to \infty}\frac{u(t)}{\rho(t)}=0 \,, \end{gathered} \end{equation} where $\gamma>0$ and $\rho(t)=\int_0^t\frac{ds}{A(s)}$. Let $u(t)=\rho(t)\,v(t)$ and $B(t)=A(t)\rho^2(t)$ for $t\in [0,\infty)$. Then $u$ is a positive solution of \eqref{Applic2-prob} if and only if $v$ is a positive solution of the problem \begin{equation}\label{Applic2-transf-prob} \begin{gathered} \frac{1}{B(t)}(B(t)v'(t))'=-\frac{p(t)}{(\rho(t))^{\gamma+1}}\,v^{-\gamma}\,,\quad t \in (0,\infty) , \\ v>0\,,\quad \text{in } (0,\infty)\;,\\ \lim_{t\to 0^+}B(t)v'(t)=0\,,\quad \lim_{t\to \infty}v(t)=0 \,. \end{gathered} \end{equation} First, we claim that if $A$ satisfies (A4), then \begin{equation}\label{lim-tBprime sur B} \lim_{t\to \infty}\frac{tB'(t)}{B(t)}=1+|\sigma-1|>1. \end{equation} Since $\frac{t\,B'(t)}{B(t)}=\frac{t\,A'(t)}{A(t)}+\frac{2t}{A(t)\,\rho(t)}$ and by Definition \ref{NRVI-class} and assertion (ii) of Lemma \ref{lem 2.5}, we have $A(t)=t^{\sigma}L_0(t)$ for $t\geq a>1$ with $L_0\in {\mathcal{K}}$, then we deduce from Lemma \ref{lemma 2.3} that \noindent$\bullet$ For $\sigma<1$, we have $\rho(\infty)=\infty$ and so \begin{equation*} \rho(t)=\int_0^a\frac{ds}{A(s)}+\int_a^t\frac{1}{s^{\sigma}\,L_0(s)}\,ds \sim \frac{1}{1-\sigma}\frac{t^{1-\sigma}}{L_0(t)}\quad \text{as } t\to \infty. \end{equation*} So \begin{equation*} \frac{2t}{A(t)\,\rho(t)}\sim (1-\sigma)\frac{L_0(t)}{t^{1-\sigma}} \frac{2t}{t^{\sigma} L_0(t)}=2(1-\sigma) \quad \text{as } t \to \infty. \end{equation*} Consequently in this case we have \begin{equation*} \lim_{t\to \infty}\frac{t\,B'(t)}{B(t)} =\sigma+2(1-\sigma)=2-\sigma=1+|\sigma-1|. \end{equation*} \noindent$\bullet$ For $\sigma>1$, we have $\rho(\infty)=\int_0^{\infty}\frac{ds}{A(s)}\,ds<\infty$. So \begin{equation*} \frac{2t}{A(t)\,\rho(t)}\sim \frac{2t}{t^{\sigma}L_0(t)\rho(\infty)}\to 0 \quad \text{as } t \to \infty. \end{equation*} In this case we have \begin{equation*} \lim_{t\to \infty}\frac{t\,B'(t)}{B(t)}=\sigma=1+|\sigma-1|. \end{equation*} This proves \eqref{lim-tBprime sur B}. Taking into account this fact, we assume that the function $p$ satisfies the following hypotheses \begin{itemize} \item[(A5)] $p$ is a nonnegative continuous function $(0,\infty)$ satisfying \begin{equation*} 00$. \end{itemize} Assume that $A$ and $p$ satisfy (A4)--(A6) and let $v$ be the unique positive solution of problem \eqref{Applic2-transf-prob}. Then $v$ has the following exact behavior at infinity \begin{itemize} \item[(i)] if $\lambda>2$, then \[ \lim_{t \to \infty}\frac{v(t)}{[(\gamma+1)t^{2-\lambda} L(t) ]^{\frac{1}{1+\gamma}}} =\Big[\frac{a_0}{(\lambda-2)(2+|\sigma-1| -\lambda+(\lambda-2)\frac{\gamma}{\gamma+1})}\Big]^{\frac{1}{1+\gamma}}. \] \item[(ii)] if $\lambda=2$, then \[ \lim_{t \to \infty}\frac{v(t)}{[(\gamma+1)\int_t^{\infty}\frac{L(s)}{s}ds ]^{\frac{1}{1+\gamma}}} =[\frac{a_0}{|\sigma-1|}\Big]^{\frac{1}{1+\gamma}} \] \end{itemize} Consequently, the unique positive solution $u$ of problem \eqref{Applic2-prob} has the following exact behavior at infinity \begin{itemize} \item[(i)] if $\lambda>2$, then \[ \lim_{t \to \infty}\frac{u(t)}{\rho(t)[(\gamma+1)t^{2-\lambda}\, L(t) ]^{\frac{1}{1+\gamma}}} =\Big[\frac{a_0}{(\lambda-2)(2+|\sigma-1|-\lambda+(\lambda-2) \frac{\gamma}{\gamma+1})} \Big]^{\frac{1}{1+\gamma}}\,. \] \item[ii)] if $\lambda=2$, then \[ \lim_{t \to \infty}\frac{u(t)}{\rho(t) [(\gamma+1)\int_t^{\infty}\frac{L(s)}{s}ds]^{\frac{1}{1+\gamma}}} =\Big[\frac{a_0}{|\sigma-1|}\Big]^{\frac{1}{1+\gamma}} \] \end{itemize} \subsection*{Acknowledgements} This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. (253-662-1436-G). 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