\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 08, 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 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/08\hfil Superlinear second-order differential equations] {Existence and uniqueness for superlinear second-order differential equations on the half-line} \author[I. Bachar, H. M\^aagli \hfil EJDE-2015/08\hfilneg] {Imed Bachar, Habib M\^aagli} % in alphabetical order \address{Imed Bachar \newline King Saud University, College of Science, Mathematics Department, P.O.Box 2455 Riyadh 11451, Saudi Arabia} \email{abachar@ksu.edu.sa} \address{Habib M\^aagli \newline King Abdulaziz University, College of Sciences and Arts, Rabigh Campus, Department of Mathematics P.O. Box 344, Rabigh 21911, Saudi Arabia} \email{habib.maagli@fst.rnu.tn, abobaker@kau.edu.sa} \thanks{Submitted October 11, 2014. Published January 5, 2015.} \subjclass[2000]{34B15, 34B18, 34B27} \keywords{Second order differential equation; boundary value problem; \hfill\break\indent half-line; Green's function; positive solution} \begin{abstract} We prove the existence and uniqueness, and study the global behavior of a positive continuous solution to the superlinear second-order differential equation \begin{gather*} \frac{1}{A(t)}(A(t)u'(t))'=u(t)g(t,u(t)),\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \end{gather*} where $a,b$ are nonnegative constants such that $a+b>0$, $A$ is a continuous function on $[0,\infty)$, positive and continuously differentiable on $(0,\infty )$ such that $1/A$ is integrable on $[0,1]$ and $\int_0^{\infty }1/A(t)\,dt=\infty $. Here $\rho (t)=\int_0^t 1/A(s)\,ds$, for $t\geq 0$ and $g(t,s)$ is a nonnegative continuous function satisfying suitable integrability condition. Our Approach is based on estimates of the Green's function and a perturbation argument. Finally two illustrative examples are given. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} We are concerned with the existence, uniqueness and global behavior of a positive continuous solution to the second-order differential equation \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'=u(t)g(t,u(t)),\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \end{gathered} \label{e1.1} \end{equation} where $a,b$ are nonnegative constants such that $a+b>0$, $A$ is a continuous function on $[0,\infty )$, positive and continuously differentiable on $(0,\infty )$ such that $\frac{1}{A}$ is integrable on $[0,1]$ and $\int_0^{\infty } 1/A(t)\,dt=\infty $. Here $\rho (t)=\int_0^t 1/A(s)\,ds$, for $t\geq 0$. The nonnegative nonlinearity $g$ is required to satisfy an appropriate condition related to the class $\mathcal{K}$, defined next. \begin{definition}\label{def1.1} \rm A Borel measurable function $q$ in $(0,\infty )$ belongs to the class $\mathcal{K}$ if \begin{equation} \| q\| :=\int_0^{\infty }A(r)\rho (r)|q(r)|dr<\infty . \label{e1.2} \end{equation} \end{definition} The motivation for the present work stems from both practical and theoretical aspects. In fact, boundary value problems on the half-line arise quite naturally in the study of radially symmetric solutions of nonlinear elliptic equations, see for instance \cite{Ba,KYY}, and various physical phenomena \cite{GGLO,I}, such as unsteady flow of gas through a semi-infinite, porous media and the theory of drain flows. Note that boundary value problems for second-order differential equations have been considering widely and there are many results on the existence of solutions, see for example \cite{AO,B,CMMZ,gherad,gherad2,R}. Zhao \cite{Zh} considered the second-order differential equation \begin{equation} \frac{1}{A(t)}(A(t)u'(t))'+h(t,u(t))=0,\quad t\in (0,\infty ), \label{e1.2.1} \end{equation} subject to the boundary conditions \begin{equation} u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \label{e1.2.2} \end{equation} where $A(t)\equiv 1$, $a=0$ and $h$ is a measurable function on $(0,\infty )\times (0,\infty )$, dominated by a convex positive function. Then he showed that there exists $\mu >0$ such that for each $b\in (0,\mu ]$, there exists a positive continuous solution $u$ of \eqref{e1.2.1}--\eqref{e1.2.2}. This result has been generalized by M\^{a}agli and Masmoudi \cite{MM}. On the other hand, Yan \cite{Y} studied equation \eqref{e1.2.1} subject to the boundary condition $u(0)=a\geq 0$, $\lim_{t\to \infty} A(t)u'(t)=b\geq 0$, where $A$ is a continuous function satisfying some appropriate conditions and \[ q_0(t)k(s)\leq h(t,s)\leq q_0(t)\widetilde{k}(s), \] where $q_0,k$ and $\widetilde{k}$ are nonnegative continuous functions on $(0,\infty )$ such that \[ \int_0^{\infty }A(r)|q_0(r)|dr<\infty ,\quad \lim_{s\to 0^{+}} \frac{k(s)}{s}=\infty \] and $\widetilde{k}$ satisfying some growth condition. By using fixed-point index theory, the existence of at least one nonnegative nonzero solution is established. Recently, in \cite{BM}, we have studied problem \eqref{e1.2.1}-\eqref{e1.2.2} where $A$ is a continuous function satisfying some appropriate conditions, $h(t,s)=q(t)s^{\sigma }$, with $\sigma <1$, $a=b=0$ and $q(t)$ is a nonnegative continuous function that is required to satisfy some assumptions related to Karamata regular variation theory. Using monotonicity methods, we established the existence, uniqueness and the global asymptotic behavior of a positive continuous solution; see also \cite{ciradulescu}. Here, we shall use estimates of the Green's function and a perturbation argument to address existence, uniqueness and global behavior of a positive continuous solution to problem \eqref{e1.1}. Throughout this paper, and without loss of generality, we assume that $\rho (1)=1$. We let $a,b\geq 0$ such that $a+b>0$ and we denote by $\omega (t):=a+b\rho (t)$, $t\geq 0$, the unique solution of the problem \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'=0,\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b. \end{gathered} \label{e1.3} \end{equation} We denote by $G(t,s)$ the Green's function of the operator $u\mapsto -\frac{1}{A}(Au')'$ on $(0,\infty )$ with the Dirichlet conditions $u(0)=0$ and $\lim_{t\to\infty}\frac{u(t)}{\rho (t)}=0$, which is given by \begin{equation} G(t,s)=A(s)\min (\rho (t),\rho (s)). \label{e1.3.a} \end{equation} The outline of the article is as follows. In Section 2, we give some sharp estimates on the Green's function $G(t,s)$, including the following 3G-inequality: for each $t,s,r\in (0,\infty )$, \[ \frac{G(t,r)G(r,s)}{G(t,s)}\leq A(r)\rho (r). \] In particular, we derive from this 3G-inequality that for each $q\in \mathcal{K}$, we have \begin{equation} \alpha _{q}:=\sup_{t,s\in (0,\infty )}\int_0^{\infty }\frac{G(t,r)G(r,s)}{G(t,s)}| q(r)| dr=\| q\| <\infty . \label{e1.4} \end{equation} In Section 3, for a given nonnegative function $q$ in $\mathcal{K}\cap C((0,\infty ))$, we prove that the Green's function $G_{q}(t,s) $ of the operator $u\mapsto -\frac{1}{A}(Au')'+qu$ on $(0,\infty )$ with the Dirichlet conditions $u(0)=0$ and $\lim_{t\to \infty} \frac{u(t)}{\rho (t)}=0$ is given by \[ G_{q}( t,s) =A(s)\rho (t)\rho (s)\varphi (t)\varphi (s)\int_{\max (t,s)}^{\infty }\frac{dr}{A(r)\rho ^2(r)\varphi ^2(r)}, \] where $\varphi $ is the unique positive solution in $C([0,\infty ))\cap C^2((0,\infty ))$ of the equation \[ \frac{1}{A(t)\rho ^2(t)} (A(t)\rho ^2(t)u'(t))'-q(t)u(t)=0 \] $\lim_{t\to0} (A\rho ^2\varphi ')(t)=0$ and $\varphi (0)=1$. In particular, we deduce the comparison result, \[ e^{-2\| q\| }G( t,s) \leq G_{q}( t,s) \leq G( t,s) ,\quad \text{for }t,s\geq 0. \] Moreover, we establish the resolvent equality \[ Vf=V_{q}f+V_{q}( qVf) =V_{q}f+V( qV_{q}f) ,\quad \text{for } f\in \mathcal{B}^{+}( (0,\infty )) , \] where $\mathcal{B}^{+}( (0,\infty )) $ is the set of nonnegative Borel measurable functions in $(0,\infty )$ and the kernels $V$ and $V_{q}$ are defined on $\mathcal{B} ^{+}( (0,\infty )) $ by \[ Vf( t) :=\int_0^{\infty }G( t,s) f(s)ds, \quad V_{q}f( t) :=\int_0^{\infty }G_{q}( t,s) f(s)ds, \quad t\geq 0. \] To state our existence results, we use the following assumptions: \begin{itemize} \item[(H1)] $g$ is a nonnegative continuous function in $(0,\infty )\times [ 0,\infty )$. \item[(H2)] There exists a nonnegative function $q\in \mathcal{K}\cap C((0,\infty ))$ such that for each $t\in (0,\infty )$, the map $s\to s( q(t) -g( t,s\omega ( t) ) ) $ is nondecreasing on $[0,1]$. \item[(H3)] For each $t\in (0,\infty )$, the function $s\to sg( t,s) $ is nondecreasing on $[0,\infty )$. \end{itemize} Using properties of the Green's function $G_{q}(t,s) $ and using a perturbation argument, we prove the following result. \begin{theorem}\label{thm1.2} Assume {\rm (H1)--(H3)}. Then \eqref{e1.1} has a unique positive solution $u\in C([0,\infty ))\cap C^2((0,\infty ))$ satisfying \begin{equation} c\omega (t)\leq u( t) \leq \omega (t), \label{e1.7} \end{equation} where $c$ is a constant in $(0,1]$. \end{theorem} \begin{corollary} \label{coro1.3} Let $f$ be a nonnegative function in $C^{1}([0,\infty ))$ such that the map $s\to \theta (s)=sf(s)$ is nondecreasing on $[0,\infty )$. Let $p$ be a nonnegative continuous function on $(0,\infty )$ such that the function $t\to q(t):=p(t)\max_{0\leq \xi \leq \omega ( t) }\theta '(\xi )$ belongs to the class $\mathcal{K}$. Then the problem \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'=p(t)u(t)f(u(t)),\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \end{gathered} \label{e1.8} \end{equation} has a unique positive solution $u\in C([0,\infty ))\cap C^2((0,\infty ))$ satisfying \begin{equation} e^{-2\| q\| }\omega (t)\leq u( t) \leq \omega (t),\quad t\geq 0. \label{e1.9} \end{equation} \end{corollary} Observe that in Theorem \ref{thm1.2} we obtain a positive continuous solution $u$, to \eqref{e1.1}, which behavior is not affected by the perturbed term. That is, it behaves like the solution $\omega $ of the homogeneous problem \eqref{e1.3}. As a typical example of nonlinearity satisfying (H1)--(H3), we have $g( t,s) =p(t)s^{\sigma }$, for $\sigma \geq 0$, where $p$ is a positive continuous function on $(0,\infty )$ such that \[ \int_0^{\infty }A(r)\rho (r)(a+b\rho (r))^{\sigma }p(r)dr<\infty . \] \section{Estimates on the Green's function} In this section, we prove some estimates on the Green's function $G(t,s)$. \begin{proposition} \label{prop2.1} (i) For each $t,s\in [ 0,\infty )$, we have \begin{equation} A(s)\min (1,\rho (s))\min (1,\rho (t))\leq G(t,s)\leq A(s)\min (1,\rho (s))\max (1,\rho (t)). \label{e2.1} \end{equation} (ii) For $f\in \mathcal{B}^{+}( (0,\infty )) $, the function $t\to Vf(t)$ is continuous on $[0,\infty )$ if and only if the integral $\int_0^{\infty }A(s)\min (1,\rho (s))f(s)ds$ converges. \end{proposition} \begin{proof} (i) The inequalities in \eqref{e2.1} follow from \eqref{e1.3.a} and the fact that for $\alpha ,\beta \geq 0$, \[ \min (1,\alpha )\min (1,\beta )\leq \min (\alpha ,\beta )\leq \min (1,\alpha )\max (1,\beta ). \] (ii) Using \eqref{e1.3.a}, \eqref{e2.1} and the dominated convergence theorem, we obtain the required assertion. \end{proof} \begin{corollary} \label{coro2.3} Let $f\in \mathcal{B}^{+}( (0,\infty )) $ such that $s\mapsto A(s)\min (1,\rho (s))f(s)$ is continuous and integrable on $(0,\infty )$. Then $Vf$ is the unique continuous solution of the boundary-value problem \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'=-f,\quad \text{in }(0,\infty ), \\ u(0)=0,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)} =0. \end{gathered} \label{e2.2} \end{equation} \end{corollary} We have the following 3G-inequality. \begin{proposition} \label{prop2.4} For each $t,s,r\in (0,\infty )$, we have \begin{equation} \frac{G(t,r)G(r,s)}{G(t,s)}\leq A(r)\rho (r). \label{e2.3} \end{equation} \end{proposition} \begin{proof} Using \eqref{e1.3.a}, for each $t,s,r\in (0,\infty )$, we deduce that \[ \frac{G(t,r)G(r,s)}{G(t,s)}=\frac{A(r)\min (\rho (t),\rho (r))\min (\rho (r),\rho (s))}{\min (\rho (t),\rho (s))}. \] We claim that \[ \frac{\min (\rho (t),\rho (r))\min (\rho (r),\rho (s))}{\min (\rho (t),\rho (s))}\leq \rho (r). \] Indeed, by symmetry, we may assume that $t\leq s$. Therefore, we obtain \begin{align*} \frac{\min (\rho (t),\rho (r))\min (\rho (r),\rho (s))}{\min (\rho (t),\rho (s))} &= \frac{\min (\rho (t),\rho (r))\min (\rho (r),\rho (s))}{\rho (t)} \\ &\leq \frac{\rho (t)\rho (r)}{\rho (t)}=\rho (r). \end{align*} This completes the proof. \end{proof} In the sequel, we denote \[ \alpha _{q}=\sup_{t,s\in (0,\infty )}\int_0^{\infty }\frac{G(t,r)G(r,s)}{G(t,s)}|q(r)|\ dr,\quad \| q\| =\int_0^{\infty }A(r)\rho (r)| q(r)| dr. \] \begin{proposition} \label{prop2.5} Let $q$ be a nonnegative function in $\mathcal{K}$, then: $(i)$ For $t\in [ 0,\infty )$, we have \begin{equation} V(q)(t)\leq \alpha _{q}. \label{e2.4} \end{equation} In particular, \begin{equation} \alpha _{q}=\| q\| <\infty . \label{e2.5} \end{equation} $(ii)$ For $t\in [ 0,\infty )$, we have \begin{equation} V(q\rho )(t)\leq \alpha _{q}\rho (t). \label{e2.6} \end{equation} In particular for $t\in [ 0,\infty )$, we obtain \begin{equation} V(q\omega )(t)\leq \alpha _{q}\omega (t). \label{e2.7} \end{equation} $(iii)$ Let $f\in \mathcal{B}^{+}(0,\infty )$, then \begin{equation} V(qV(f))(t)\leq \alpha _{q}V(f)(t). \label{e2.8} \end{equation} \end{proposition} \begin{proof} Let $q$ be a nonnegative function in $\mathcal{K}$. (i) Since for each $t,s\in (0,\infty )$, we have $\lim_{r\to 0} \frac{G(s,r)}{G(t,r)}=1$, then by Fatou's lemma and \eqref{e1.4}, we deduce that \[ V(q)(t)=\int_0^{\infty }G(t,s)q(s)ds\leq \underset{r\to 0}{\lim \inf }\int_0^{\infty }G(t,s)\frac{G(s,r)}{G(t,r)}q(s)ds\leq \alpha _{q}. \] This proves \eqref{e2.4}. To prove \eqref{e2.5}, observe that $\| q\| =\| V(q)\| _{\infty }:=\sup_{t>0}|V(q)(t)| $. So it follows from \eqref{e2.4} that $\| q\| =\| V(q)\| _{\infty }\leq \alpha _{q}$. On the other hand, by using \eqref{e2.3}, for $t,s\in (0,\infty )$, we have \[ \int_0^{\infty }\frac{G(t,r)G(r,s)}{G(t,s)}q(r)dr\leq \int_0^{\infty }A(r)\rho (r)q(r)dr=\| q\| . \] Hence $\alpha _{q}\leq \| q\| <\infty$. Therefore $\alpha _{q}=\| q\| <\infty $. (ii) Since for each $t,s\in (0,\infty )$, we have $\lim_{r\to \infty } \frac{G(s,r)}{G(t,r)}=\frac{\rho (s)}{\rho (t)}$, then we deduce by Fatou's lemma and \eqref{e1.4}, that \[ \int_0^{\infty }\frac{G(t,s)}{\rho (t)}\rho (s)q(s)dr \leq \liminf_{r\to \infty } \int_0^{\infty }G(t,s)\frac{ G(s,r)}{G(t,r)}q(s)ds\leq \alpha _{q}. \] This proves \eqref{e2.6}. Inequality \eqref{e2.7} follows from inequalities \eqref{e2.4}, \eqref{e2.6} and the fact that $\omega (t)=a+b\rho (t)$. (iii) Using \eqref{e1.4} and Fubini-Tonelli's theorem, we obtain \begin{align*} V(qV(f))(t) &= \int_0^{\infty }[\int_0^{\infty }G(t,r)G(r,s)q(r)dr]f(s)ds\\ &\leq \int_0^{\infty }\alpha _{q}G(t,s)f(s)ds=\alpha _{q}V(f)(t). \end{align*} This completes the proof. \end{proof} \section{Proofs of main results} In this section, we prove Theorem \ref{thm1.2} and Corollary \ref{coro1.3}. First, for a given nonnegative function $q$ in $\mathcal{K}\cap C((0,\infty ))$, we aim at determining the Green's function $G_{q}(t,s)$ of the linear problem \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)u'(t))'-q(t)u(t)=-f(t),\quad t\in (0,\infty ), \\ u(0)=0,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=0. \end{gathered} \label{e3.0.3} \end{equation} Put $u(t):=\rho (t)v(t)$. It is easy to check that $u$ is a solution of \eqref{e3.0.3} if and only if $v$ is a solution of the problem \begin{equation} \begin{gathered} \frac{1}{A(t)\rho ^2(t)}(A(t)\rho ^2(t)v'(t))'-q(t)v(t) =\frac{-f(t)}{\rho (t)},\quad t\in (0,\infty ), \\ \lim_{t\to0} (A\rho ^2v')(t)=0,\quad \lim_{t\to\infty} v(t)=0. \end{gathered} \label{e3.0.4} \end{equation} Therefore, to obtain $G_{q}(t,s)$ it is sufficient to determine the Green's function $H_{q}(t,s)$ of the operator $u\mapsto \frac{-1}{A\rho ^2}(A\rho ^2v')'+qv$ on $(0,\infty )$ with the Dirichlet conditions $\lim_{t\to0} (A\rho ^2v')(t)=0$, $\lim_{t\to\infty} v(t)=0$. To this end, we need the following results. \begin{proposition}\label{prop3.0} Let $q$ be a nonnegative function in $\mathcal{K}\cap C((0,\infty ))$, then the problem \begin{equation} \begin{gathered} \frac{1}{A(t)\rho ^2(t)}(A(t)\rho ^2(t)u'(t))'-q(t)u(t)=0, \quad t\in (0,\infty ), \\ \lim_{t\to0} (A\rho ^2u')(t)=0,\quad u(0)=1, \end{gathered} \label{e3.0} \end{equation} has a unique positive solution $\varphi \in C([0,\infty ))\cap C^2((0,\infty ))$. Moreover, $\varphi $ is nondecreasing and for $t\geq 0$ satisfies \begin{equation} 1\leq \varphi (t)\leq \exp \Big(\int_0^t \frac{1}{A(s)\rho ^2(s)} (\int_0^{s}A(r)\rho ^2(r)q(r)dr)ds\Big)\leq \exp (\| q\| ). \label{e3.0.1} \end{equation} In particular, $\varphi (\infty ):=\lim_{t\to \infty}\varphi (t)$ exists and $1\leq \varphi (\infty )\leq \exp (\|q\| )$. \end{proposition} \begin{proof} (see \cite{Yaz}). Let $K$ be the operator defined on $C([0,\infty ))$ by \[ Kf(t):=\int_0^t \frac{1}{A(s)\rho ^2(s)}\Big( \int_0^{s}A(r)\rho ^2(r)q(r)f(r)dr\Big) ds,\quad t\in [ 0,\infty ). \] We put $K^{j}=K^{j-1}\circ K$ for any integer $j\geq 2$. Then we claim that for each $t\geq 0$ and $m\in \mathbb{N}$, we have \begin{equation} 0\leq K^{m}\mathbf{1}(t)\leq \frac{(K\mathbf{1)}^{m}(t)}{m!}. \label{e3.0.2} \end{equation} Indeed, if $m=0$ or $1$, \eqref{e3.0.2} is valid. Now for a given $m\in \mathbb{N}$, suppose \eqref{e3.0.2}, then we have \begin{align*} K^{m+1}\mathbf{1}(t) &= K(K^{m}\mathbf{1})(t) \\ &\leq \frac{1}{m!}K((K\mathbf{1})^{m})(t) \\ &= \frac{1}{m!}\int_0^t \frac{1}{A(s)\rho ^2(s)}\Big( \int_0^{s}A(r)\rho ^2(r)q(r)(K\mathbf{1})^{m}(r)\,dr\Big) ds. \end{align*} Since the function $K\mathbf{1}$ is nondecreasing, it follows that \begin{align*} K^{m+1}\mathbf{1}(t) &\leq \frac{1}{m!}\int_0^t (K\mathbf{1} )^{m}(s)\Big( \frac{1}{A(s)\rho ^2(s)}\int_0^{s}A(r)\rho ^2(r)q(r)\,dr\Big) \,ds \\ &= \frac{1}{m!}\int_0^t (K\mathbf{1})^{m}(s)(K\mathbf{1})'(s)\,ds\\ &=\frac{1}{(m+1)!}(K\mathbf{1})^{m+1}(t). \end{align*} Therefore, the series $\sum_{m=0}^\infty (K^{m}\mathbf{1})(t)$ converges locally uniformly to a function $\varphi \in C([0,\infty ))$ satisfying for each $t\geq 0$, \[ \varphi (t)=1+\int_0^t {\frac{1}{A(s)\rho ^2(s)}\Big( \int_0^{s}A(r)\rho ^2(r)q(r)\varphi (r)dr\Big) ds}. \] Hence $\varphi \in C([0,\infty ))\cap C^2((0,\infty ))$ and $\varphi $ is a positive solution of \eqref{e3.0}. Now, we show the uniqueness. Let $u,v\in C([0,\infty ))\cap C^2((0,\infty ))$ be two positive solutions of \eqref{e3.0}. Then for each $R\in (0,\infty )$ and $t\in [ 0,R]$ we have \[ | u(t)-v(t)| \leq K(| u-v| )(t). \] Since $K$ is a nondecreasing operator, we deduce by induction that for each $m\geq 0$, \begin{align*} | u(t)-v(t)| &\leq K^{m}(| u-v| )(t) \\ &\leq \sup_{r\in [ 0,R]} | u(r)-v(r)| K^{m}\mathbf{1}(R) \\ &\leq \sup_{r\in [ 0,R]} | u(r)-v(r)| \frac{(K\mathbf{1)}^{m}(R)}{m!}. \end{align*} Letting $m$ tend to infinity, we obtain $|u(t)-v(t)| =0$ for all $t\in [ 0,R]$. So $u=v$ on $[0,\infty )$. Finally \eqref{e3.0.1} follows from the fact that \begin{gather*} 1\leq \varphi (t)=\underset{m=0}{\overset{\infty }{\sum }}(K^{m}\mathbf{1} )(t)\leq \underset{m=0}{\overset{\infty }{\sum }}\frac{(K\mathbf{1)}^{m}(t)}{ m!}=\exp (K\mathbf{1}(t))\quad \forall t\geq 0, \\ K\mathbf{1}(t)\leq \int_0^{\infty } \frac{1}{A(s)\rho ^2(s)}\Big( \int_0^{s}A(r)\rho ^2(r)q(rdr)\Big) ds=\| q\| . \end{gather*} \end{proof} \begin{remark}\label{rem3.1} \rm Let $q$ be a nonnegative function in $\mathcal{K}\cap C((0,\infty ))$ and $\varphi $ be the solution of \eqref{e3.0}. It follows that the function $\psi $ defined on $(0,\infty )$ by \[ \psi (t):=\varphi (t)\int_{t}^{\infty }\frac{ds}{A(s)\rho ^2(s)\varphi ^2(s)}, \] is a second solution of the equation \[ \frac{1}{A(t)\rho ^2(t)}(A(t)\rho ^2(t)u'(t))'-q(t)u(t)=0, \quad \text{ on }(0,\infty ), \] such that $\varphi $ and $\psi $ are linearly independent. Furthermore, since for $t>0$, \begin{equation} \frac{1}{\varphi ^2(\infty )\rho (t)}\leq \psi (t)\leq \frac{1}{\varphi (t)}\int_{t}^{\infty }\frac{ds}{A(s)\rho ^2(s)}=\frac{1}{\varphi (t)\rho (t)}, \label{e3.0.6} \end{equation} it follows that $\lim_{t\to\infty} \psi (t)=0$ and also we have \[ \psi (t) \sim \int_{t}^{\infty }\frac{ds}{A(s)\rho ^2(s)}=\frac{1}{\rho (t)}\quad\text{as } t\to 0. \] Hence $\lim_{t\to0} \rho (t)\psi (t)=1$. \end{remark} Now, following \cite[Section 2, p.294]{MM1}, we deduce that $H_{q}(t,s)$ is given by \[ H_{q}(t,s)=\begin{cases} A(s)\rho ^2(s)\varphi (s)\psi (t), & \text{if }00$, we have \begin{align*} V_{q}f( t) &= \int_0^{\infty }G_{q}( t,s) f(s)ds \\ &= (\rho \psi )(t)\int_0^t A(s)\rho (s)\varphi (s)f(s)ds+(\rho \varphi )(t)\int_{t}^{\infty }A(s)\rho (s)\psi (s)f(s)ds. \end{align*} So $V_{q}f$ is differentiable on $(0,\infty )$ and we have for $t>0$, \[ (V_{q}f)'( t) =(\rho \psi )'(t)\int_0^t A(s)\rho (s)\varphi (s)f(s)ds +(\rho \varphi )'(t)\int_{t}^{\infty }A(s)\rho (s)\psi (s)f(s)ds. \] Therefore by using the fact that $\rho \varphi $ and $\rho \psi $ are solutions of the equation $\frac{1}{A}(Au')'-qu=0$ on $(0,\infty )$ and \eqref{e3.0.5}, we obtain \begin{align*} (A(V_{q}f)')'( t) &= (A(\rho \psi )')'(t)\int_0^t A(s)\rho (s)\varphi (s)f(s)ds \\ &\quad +(A(\rho \varphi )')'(t)\int_{t}^{\infty }A(s)\rho (s)\psi (s)f(s)ds \\ &\quad +A(t)f(t)[A(\rho \varphi )(\rho \psi )'-A(\rho \psi )(\rho \varphi )'](t) \\ &= A(t)q(t)V_{q}f( t) -A(t)f(t). \end{align*} So $V_{q}f$ is a solution of the equation $\frac{1}{A(t)} (A(t)u'(t))'-q(t)u(t)=-f(t)$. Now since $0\leq V_{q}f\leq Vf$, it follows by Corollary \ref{coro2.3}, that $V_{q}f(0)=0$ and $\lim_{t\to\infty}\frac{V_{q}f( t) }{\rho (t)}=0$. It remains to prove the uniqueness. Assume that there exist two positive solutions $u,v\in C([0,\infty ))\cap C^2((0,\infty ))$ to problem \eqref{e3.0.3}. Let $\theta :=u-v$, then $\theta \in C([0,\infty ))\cap C^2((0,\infty ))$ and satisfies \begin{gather*} \frac{1}{A(t)}(A(t)\theta '(t))'-q(t)\theta (t)=0\quad \text{on }(0,\infty ), \\ \theta (0)=0,\quad \lim_{t\to\infty} \frac{\theta(t)}{\rho (t)}=0. \end{gather*} Hence, there exists $\lambda ,\mu \in \mathbb{R}$, such that \[ \theta (t)=\lambda \rho (t)\varphi (t)+\mu \rho (t)\psi (t),\text{ for } t\geq 0. \] So using this fact, Proposition \ref{prop3.0}, Remark \ref{rem3.1} and that \[ \theta (0)=\lim_{t\to\infty} \frac{\theta (t) }{\rho (t)}=0, \] we deduce that $\lambda =\mu =0$. That is, $u=v$. This completes the proof. \end{proof} \begin{corollary}\label{coro3.2} Let $q$ be a nonnegative function in $\mathcal{K}\cap C((0,\infty ))$ and let $f\in \mathcal{B}^{+}( (0,\infty )) $ such that $s\to A(s)\min (1,\rho (s))f(s)$ is continuous and integrable on $(0,\infty )$. Then $V_{q}f$ satisfies the resolvent equation \begin{equation} Vf=V_{q}f+V_{q}( qVf) =V_{q}f+V( qV_{q}f) . \label{e3.5} \end{equation} In particular, if $V(qf)<\infty $, we have \begin{equation} (I-V_{q}(q\cdot) )(I+V(q\cdot) )f=(I+V(q\cdot) )(I-V_{q}(q\cdot) )f=f. \label{e3.5.1} \end{equation} \end{corollary} \begin{proof} Let $q$ be a nonnegative function in $\mathcal{K}\cap C((0,\infty ))$ and let $f\in \mathcal{B}^{+}( (0,\infty )) $ such that $s\to A(s)\min (1,\rho (s))f(s)$ is continuous and integrable on $(0,\infty )$. By Proposition \ref{prop2.1} it is clear that the function $t\mapsto q(t)Vf( t) $ is continuous on $(0,\infty )$ and there exists a nonnegative constant $c$ such that \begin{equation} Vf(t)\leq (1+\rho (t))\int_0^{\infty }A(s)\min (1,\rho (s))f(s)ds\leq c(1+\rho (t)). \label{e3.5.3} \end{equation} So we deduce by Proposition \ref{prop2.5} that \begin{align*} \int_0^{\infty }A(s)\min (1,\rho (s))q(s)Vf(s)ds &\leq c\int_0^{\infty}G(1,s)(1+\rho (s))q(s)ds \\ &\leq 2c\alpha _{q}<\infty . \end{align*} Let $\theta :=Vf-V_{q}f-V_{q}( qVf) $. By using Corollary \ref{coro2.3} and Proposition \ref{prop3.3}, the function $\theta $ is a solution of the problem \begin{equation} \begin{gathered} \frac{1}{A(t)}(A(t)\theta '(t))'-q(t)\theta (t)=0,\quad t\in (0,\infty ), \\ \theta (0)=0,\quad \lim_{t\to\infty} \frac{\theta (t)}{\rho (t)}=0. \end{gathered} \label{e3.5.4} \end{equation} From the uniqueness in Proposition \ref{prop3.3}, we deduce that $ \theta =0$. Now, by using Corollary \ref{coro3.0} and \eqref{e3.3}, we deduce that the function $t\mapsto q(t)V_{q}f( t) $ is continuous on $(0,\infty )$ and that \[ \int_0^{\infty }A(s)\min (1,\rho (s))q(s)V_{q}f(s)ds\leq \int_0^{\infty }A(s)\min (1,\rho (s))q(s)Vf(s)ds<\infty . \] So by similar arguments as above, we obtain $Vf-V_{q}f-V(qV_{q}f) =0$. This completes the proof. \end{proof} We recall that for $a,b\geq 0$ such that $a+b>0$, we have \[ \omega (t)=a+b\rho (t),\text{ }t\in [ 0,\infty ). \] The next Lemma will be useful for the proof of Theorem \ref{thm1.2}. \begin{lemma}\label{lem3.4} Let $q$ be a nonnegative function in $\mathcal{K}\cap C((0,\infty ))$, then we have \[ e^{-2\| q\| }\omega \leq \omega -V_{q}( q\omega ) \leq \omega . \] \end{lemma} \begin{proof} Let $\theta :=\omega -V_{q}( q\omega ) $. It is clear that $\theta \leq \omega $. Now since $\{\rho \varphi ,\rho \psi \}$ is a fundamental system of solutions of the equation \begin{equation} \frac{1}{A(t)}(A(t)u'(t))'-q(t)u(t)=0, \label{e3.5.5} \end{equation} and the function $\theta $ is also a solution of this equation with $\theta (0)=a$ and $\lim_{t\to\infty} \frac{\theta (t)}{\rho (t)}=b$, we deduce by using Proposition \ref{prop3.0} and Remark \ref{rem3.1} that \[ \theta (t)=\frac{b}{\varphi (\infty )}\rho (t)\varphi (t)+a\rho (t)\psi (t), \quad t>0. \] Using Proposition \ref{prop3.0} and \eqref{e3.0.6}, this implies that \[ \theta =\omega -V_{q}( q\omega ) \geq \frac{b}{\varphi (\infty )} \rho +\frac{a}{\varphi ^2(\infty )}\geq \frac{1}{\varphi ^2(\infty )} \omega \geq e^{-2\| q\| }\omega . \] The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] Since $g$ satisfies (H2), there exists a nonnegative continuous function $q$ in $\mathcal{K}$ such for each $t\in (0,\infty )$, the map $s\to s( q( t) -g(t,s\omega ( t) ) ) $ is nondecreasing on $[0,1]$. Let \[ \Lambda :=\big\{ u\in \mathcal{B}^{+}( (0,\infty )) :e^{-2\| q\| }\omega \leq u\leq \omega \big\} , \] and define the operator $T$ on $\Lambda $ by \[ Tu=\omega -V_{q}( q\omega ) +V_{q}(( q-g(\cdot,u)) u). \] By (H2), we have \begin{equation} 0\leq g(.,u)\leq q,\quad \text{for all }u\in \Lambda . \label{e3.7} \end{equation} We claim that $\Lambda $ is invariant under $T$. Indeed, since $g$ is nonnegative, we have for $u\in \Lambda $ \[ Tu\leq \omega -V_{q}( q\omega ) +V_{q}(qu)\leq \omega \] and by \eqref{e3.7} and Lemma \ref{lem3.4}, \[ Tu\geq \omega -V_{q}( q\omega ) \geq e^{-2\| q\| }\omega . \] Next, we will prove that the operator $T$ is nondecreasing on $ \Lambda $. Indeed, let $u,v\in \Lambda $ be such that $u\leq v$. Since for $t\in (0,\infty )$, the map $s\to s( q( t) -g( t,s\omega (t) ) ) $ is nondecreasing on $[0,1]$, then we obtain \[ Tv-Tu=V_{q}([ v( q-g(\cdot,v) ) -u(q-g(\cdot,u) )] ) \geq 0. \] Now, we consider the sequence $\{ u_{n}\} $ defined by $u_0=e^{-2\| q\| }\omega $ and $u_{n+1}=Tu_{n}$, for $n\in \mathbb{N}$. Since $\Lambda $ is invariant under $T$, we have $u_1=Tu_0\geq $ $u_0$ and by the monotonicity of $T$, we deduce that \[ e^{-2\| q\| }\omega =u_0\leq u_1\leq \dots \leq u_{n}\leq u_{n+1}\leq \omega . \] So the sequence $\{ u_{n}\} $ converges to a function $u\in \Lambda $. Using hypotheses (H1)--(H2) and the monotone convergence theorem, we deduce that \[ u=( I-V_{q}(q\cdot) ) \omega +V_{q}(( q-g(.,u) ) u). \] That is, \[ ( I-V_{q}(q\cdot) ) u=( I-V_{q}( q\cdot)) \omega -V_{q}( ug( \cdot,u) ) . \] On the other hand, since $u\leq \omega $, then by \eqref{e2.7}, we obtain $V( qu) \leq V( q\omega ) \leq \alpha_{q}\omega <\infty $. So by applying the operator $( I+V(q\cdot) ) $ on both sides of the above equality and using \eqref{e3.5} and \eqref{e3.5.1}, we conclude that $u$ satisfies \begin{equation} u=\omega -V( ug( \cdot,u) ). \label{e3.8} \end{equation} Next we aim at proving that $u$ is a solution of problem \eqref{e1.1}. To this end, we remark by \eqref{e3.7} that \begin{equation} ug(\cdot,u) \leq q\omega . \label{e3.9} \end{equation} By Proposition \ref{prop2.1} (ii) and \eqref{e2.7}, this implies that the function $t\to V( ug(\cdot,u) ) (t)$ is continuous on $[0,\infty )$ and so by \eqref{e3.8}, $u$ is continuous on $[0,\infty )$. Now, since by $(H_1)$ and \eqref{e3.9}, the function $s\to A(s)\min ( 1,\rho (s)) u(s)g(s,u(s)) $ is continuous and integrable on $(0,\infty )$, we conclude by Corollary \ref{coro2.3} that $u$ is the required solution. It remains to prove that $u$ is the unique solution to \eqref{e1.1}. Assume that $v\in C([0,\infty ))\cap C^2((0,\infty ))$ is another nonnegative solution to problem \eqref{e1.1}. Then we have \begin{equation} v=\omega -V( vg(\cdot,v) ) . \label{e3.10} \end{equation} Now let $h$ be the function defined on $(0,\infty )$ by \[ h(t)=\begin{cases} \frac{v(t)g( t,v(t)) -u(t)g( t,u(t)) }{v(t)-u(t)} & \text{if }v(t)\neq u(t), \\ 0 & \text{if }v(t)=u(t). \end{cases} \] From (H3), we have $h\in \mathcal{B}^{+}( (0,\infty )) $ and by using \eqref{e3.8} and \eqref{e3.10}, we obtain \[ (I+V(h.))(v-u)=0. \] On the other hand, since by (H2), we have $h\leq q$, then by using \eqref{e2.7} we deduce that \[ V(h| v-u| )\leq 2V(q\omega )\leq 2\alpha _{q}\omega <\infty . \] Hence by \eqref{e3.5.1}, we conclude that $u=v$. This completes the proof. \end{proof} \begin{proof}[Proof of Corollary \ref{coro1.3}] Let $g( t,s) =p(t)f(s)$ and $\theta (s)=sf(s)$, and let $q(t)=p(t)\max_{0\leq \xi \leq \omega ( t) } \theta'(\xi )\in \mathcal{K}$. It is clear that hypotheses (H1) and (H3) are satisfied. Moreover, by a simple computation, we obtain \[ \frac{d}{ds}[s( q( t) -g( t,s\omega ( t) ) ) ]=q(t)-p(t)\theta '(s\omega ( t) )\geq 0 \quad \text{for }s\in [ 0,1]\text{ and }t>0. \] This implies that the function $g$ satisfies hypothesis (H2). So the result follows by Theorem \ref{thm1.2}. \end{proof} \begin{example} \rm Let $a\geq 0$ and $b\geq 0$ with $a+b>0$. Let $\sigma \geq 0$, and $p$ be a positive continuous function on $(0,\infty )$ such that \[ \int_0^{\infty }A(r)\rho (r)(\omega ( t) )^{\sigma}p(r)dr<\infty . \] Since the function $q(t):=(\sigma +1)p(t)(\omega ( t))^{\sigma }$ belongs to the class $\mathcal{K}$, the problem \begin{gather*} \frac{1}{A(t)}(A(t)u'(t))'=p(t)u^{\sigma +1}(t),\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \end{gather*} has a unique positive solution $u\in C([0,\infty ))\cap C^2((0,\infty ))$ satisfying \[ e^{-2\| q\| }\omega (t)\leq u( t) \leq \omega (t),\quad t\geq 0. \] \end{example} \begin{example} \rm Let $a\geq 0$ and $b\geq 0$ with $a+b>0$. Let $\sigma \geq 0$, $\gamma >0$ and $p$ be a positive continuous function on $(0,\infty )$ such that \[ \int_0^{\infty }A(r)\rho (r)(\omega ( t) )^{\sigma +\gamma }p(r)dr<\infty . \] Let $\theta (s)=s^{\sigma +1}\log (1+s^{\gamma })$. Since the function $q(t):=p(t)\max_{0\leq \xi \leq \omega ( t) }\theta '(\xi )$ belongs to the class $\mathcal{K}$, then the problem \begin{gather*} \frac{1}{A(t)}(A(t)u'(t))'=p(t)u^{\sigma +1}(t)\log (1+u^{\gamma }(t)),\quad t\in (0,\infty ), \\ u(0)=a,\quad \lim_{t\to\infty} \frac{u(t)}{\rho (t)}=b, \end{gather*} has a unique positive solution $u\in C([0,\infty ))\cap C^2((0,\infty ))$ satisfying \[ e^{-2\| q\| }\omega (t)\leq u( t) \leq \omega (t),\quad t\geq 0. \] \end{example} \subsection*{Acknowledgments} The authors would like to thank the anonymous referees for their careful reading of the paper. The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Research group No RG-1435-043. \begin{thebibliography}{99} \bibitem{AO} R. P. Agarwal, D. 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