\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 04, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/04\hfil Monotone iterative method] {Monotone iterative method and regular singular nonlinear BVP in the presence of reverse ordered upper and lower solutions} \author[A. K. Verma\hfil EJDE-2012/04\hfilneg] {Amit K. Verma} \address{Amit K. Verma \newline Department of Mathematics, BITS Pilani Pilani - 333031, Rajasthan, India\newline Phone +919413789285; fax: +911596244183} \email{amitkverma02@yahoo.co.in, akverma@bits-pilani.ac.in} \thanks{Submitted October 19, 2011. Published January 9, 2012.} \subjclass[2000]{34B16} \keywords{Monotone iterative technique; lower and upper solutions; \hfill\break\indent Neumann boundary conditions} \begin{abstract} Monotone iterative technique is employed for studying the existence of solutions to the second-order nonlinear singular boundary value problem $$ -\big(p(x)y'(x)\big)'+p(x)f\big(x,y(x),p(x)y'(x)\big)=0 $$ for $00$ in $(0,1)$. \item[(ii)] $p\in C[0,1]\cap C^1(0,1)$ and \item[(iii)] for some $r>1$, $x\frac{p'(x)}{p(x)}$ is analytic in $\{z:|z|0$ for $i=0,1,2,\dots$. Where $-\lambda_0$ is the first negative zero of $z_0'(1,\lambda)$ or in other words first negative eigenvalue of \eqref{2.3}--\eqref{2.4}. Since $z_0(x,\lambda)$ does not change its sign for $-\lambda_0<\lambda<0$ and $z_0(0,\lambda)=1$ therefore $z_0(x,\lambda)>0$ for all $x\in[0,1]$ and for all $-\lambda_0<\lambda<0$. \end{remark} \begin{remark}\label{RM1}\rm Using \eqref{2.12} and the fact that $z_1(x)=z_0(1-x)$ it is easy to prove that if $\lambda>0$ then for all $x\in(0,1]$, $z_0(x)>1$ and $z'_0(x)>0$ and for all $x\in[0,1)$ we have $z_1(x)>1$ and $z_1'(x)<0$. \end{remark} \begin{remark}\label{RM2}\rm Using Remark \ref{RM0}, $z_1(x)=z_0(1-x)$ and the differential equation \eqref{2.13} it is easy to prove that if $-\lambda_0<\lambda<0$ for all $x\in[0,1)$, $z_0(x)>0$ and $z'_1(x)>0$ and for all $x\in(0,1]$ we have $z_0'(x)<0$ and $z_1(x)>0$. \end{remark} \begin{remark}\label{R1} \rm Let $\lambda>0$ and $h\in C[0,1]$. If $h\geq0$ $(or~h\leq0)$ then \[ \int_{0}^{x}\frac{p(t)h(t)z_0(t)}{W_{p}(z_1,z_0)}dt \quad\text{and}\quad \int_x^1 \frac{p(t)h(t)z_1(t)}{W_{p}(z_1,z_0)}dt \] are non-negative (or non-positive). \end{remark} \begin{remark}\label{R2} \rm Let $-\lambda_0<\lambda<0$ and $h\in C[0,1]$. If $h\geq0$ (or $h\leq0$) then \[ \int_{0}^{x}\frac{p(t)h(t)z_0(t)}{W_{p}(z_1,z_0)}dt\quad\text{and}\quad \int_x^1\frac{p(t)h(t)z_1(t)}{W_{p}(z_1,z_0)}dt \] are non-positive (or non-negative). \end{remark} \begin{proposition}[Maximum~Principle]\label{MP} Let $\lambda>0$. If $A\leq 0$, $B\geq0$ (or $A\geq 0$, $B\leq0$) and $h\in C[0,1]$ such that $h\geq0$ (or $h\leq0$), then $w(x)\geq0$ (or $w(x)\leq0$), where $w(x)$ is the solution of \eqref{2.1}-\eqref{2.2}. \end{proposition} \begin{proposition}[Anti-maximum~Principle]\label{AMP} Let $-\lambda_0<\lambda<0$. If $A\leq 0$, $B\geq0$ (or $A\geq 0$, $B\leq0$) and $h\in C[0,1]$ such that $h\geq0$ (or $h\leq0$), then $w(x)\leq0$ (or $w(x)\geq0$), where $w(x)$ is the solution of \eqref{2.1}-\eqref{2.2}. \end{proposition} Now we derive conditions on $\lambda$ which will help us to prove the monotonicity of the solutions generated by iterative scheme \eqref{1.5}-\eqref{1.6}. \begin{lemma}\label{DI1} Let $M$ and $N$ $\in\mathbb{R}^{+}$. If $\lambda>0$ such that \[ \lambda\geq M \Big(1-N\int_0^1p(t)dt\Big)^{-1}, \] then for all $x\in [0,1]$, \begin{equation} \label{2.14} (M-\lambda)z_0(x)+Np(x)z'_0(x)\leq0. \end{equation} \end{lemma} \begin{proof} Integrating \eqref{2.12} from $0$ to $x$ and using that $z_0'(x)>0$ in $(0,1]$ we obtain \[ p(x)z_0'(x)\leq \lambda z_0(x)\int_0^1p(t)dt. \] Therefore we obtain $(M-\lambda)z_0(x)+Np(x)z'_0(x)\leq (M-\lambda)z_0+N\lambda z_0(x)\int_0^1p(t)dt$. Hence \eqref{2.14} will hold if $(M-\lambda)+N\lambda\int_0^1p(t)dt\leq 0$. Hence the result. \end{proof} \begin{lemma}\label{DI2} Let $M$ and $N$ $\in\mathbb{R}^{+}$. If $-\lambda_0<\lambda<0$ is such that $$ -\Big(\int_0^1\frac{1}{p(x)}\int_0^xp(t)\,dt\,dx\Big)^{-1}<\lambda<-M $$ and \[ (M+\lambda)\Big(1+\lambda\int_0^1\frac{1}{p(x)}\int_0^xp(t)\,dt\,dx\big) -N\lambda \int_0^1p(x)dx\leq0 \] then for all $x\in [0,1]$, \begin{equation} \label{2.15} (M+\lambda)z_0(x)-Np(x)z'_0(x)\leq0. \end{equation} \end{lemma} \begin{proof} Using \eqref{2.12} and Remark \ref{RM2} it can be deduced that $z_0(x)$ and $p(x)z_0'(x)$ are decreasing functions of $x$ for $-\lambda_0<\lambda<0$, thus \[ (M+\lambda)z_0(x)-Np(x)z'_0(x)\leq (M+\lambda)z_0(1)-Np(1)z'_0(1). \] Now using \eqref{2.12} we obtain \[ -p(1)z_0'(1)\leq (-\lambda)\int_0^1p(x)dx\quad\text{and}\quad z_0(1)>1+\lambda \int_0^1\frac{1}{p(x)}\int_0^xp(t)\,dt\,dx. \] Which completes the proof. \end{proof} \section{Well-ordered upper and lower solutions}\label{wouls} Let us define upper and lower solutions: A function $u_0\in C^2[0,1]$ is an upper solution of \eqref{1.3}-\eqref{1.4} if \begin{equation} \label{3.1} \begin{gathered} -(pu_0')'+p(x)f(x,u_0,pu_0')\geq0,\quad 00$. If $u_n$ is an upper solution of \eqref{1.3}-\eqref{1.4} and $u_{n+1}$ is defined by \eqref{1.5}-\eqref{1.6} then $u_{n+1}\leq u_{n}$. \end{lemma} \begin{proof} Let $w_{n}=u_{n+1}-u_{n}$, then \begin{gather*} -(pw_{n}')'+\lambda pw_{n} =(pu'_n)'-pf(x,u_n,pu'_n)\leq0,\\ w_{n}'(0)\geq0,\quad w_{n}'(1)\leq0, \end{gather*} and using Proposition \ref{MP} we have $u_{n+1}\leq u_n$. \end{proof} For the next proposition we use the following assumptions: \begin{itemize} \item[(H1)] there exists upper solution $(u_0)$ and lower solution $(v_0)$ in $C^2[0,1]$ such that $v_0\leq u_0$ for all $x\in[0,1]$; \item[(H2)] the function $f:D\to \mathbb{R}$ is continuous on \[ D:=\{(x,y,py')\in[0,1]\times R\times R: v_0\leq y\leq u_0\}; \] \item[(H3)] there exists $M\geq0$ such that for all $(x,\tau,pv'), (x,\sigma,pv')\in D$, \[ f(x,\tau,pv')-f(x,\sigma,pv')\geq M(\tau-\sigma),\quad (\tau\leq \sigma); \] \item[(H4)] there exist $N\geq0$ such that for all $(x,u_,pv'_1),(x,u,pv'_2)\in D$, \[ |f(x,u,pv'_1)-f(x,u,pv'_2)|\leq N|pv'_2-pv'_1|. \] \end{itemize} \begin{proposition}\label{U-L-2} Assume {\em (H1)--(H4)}, and let $\lambda>0$ such that $$ \lambda\geq M \big(1-N\int_0^1p(t)dt\big)^{-1}. $$ Then the functions $u_{n+1}$ defined recursively by \eqref{1.5}-\eqref{1.6} are such that for all $n\in \mathbb{N}$, \begin{itemize} \item[(i)] $u_n$ is an upper solution of \eqref{1.3}-\eqref{1.4}. \item[(ii)] $u_{n+1}\leq u_{n}$. \end{itemize} \end{proposition} \begin{proof} We prove the claims by the principle of mathematical induction. Since $u_0$ is an upper solution and by Lemma \ref{U-L-1} $u_0\geq u_1$, therefore both the claims are true for $n=0$. Further, let the claims be true for $n-1$; i.e., $u_{n-1}$ is an upper solution and $u_{n-1}\geq u_{n}$. Now we are required to prove that $u_{n}$ is an upper solution and $u_{n+1}\leq u_{n}$. To prove this let $w=u_n-u_{n-1}$, then we have \[ -(pu_n')'+pf(x,u_n,pu_n')\geq p[(M-\lambda)w-N (\operatorname{sign}w') pw']. \] Thus to prove that $u_n$ is an upper solution we require to prove that \begin{equation} \label{ineq} (M-\lambda)w-N (\operatorname{sign}w') pw'\geq 0. \end{equation} Now, since $w$ satisfies \[ -(pw')'+\lambda pw=(pu'_{n-1})'-pf(x,u_{n-1},pu_{n-1}')\leq 0, \quad w'(0)\geq0,\; w'(1)\leq0, \] from Proposition \ref{MP} we have $w\leq0$ for $\lambda>0$. Now, putting the value of $w$ from \eqref{2.11} in \eqref{ineq} and in view of $h=(pu'_{n-1})'-pf(x,u_{n-1},pu_{n-1}')\leq 0$ we deduce that to prove \eqref{ineq} it is sufficient to prove that \[ (M-\lambda)z_0-N(sign~w')pz'_0\leq0 \] and \[ (M-\lambda)z_1-N(sign~w')pz'_1\leq0 \] for all $x\in[0,1]$. Since $z_1=z_0(1-x)$, using Remark \ref{R1}, above inequalities will be true if for all $x\in[0,1]$ we have \[ (M-\lambda)z_0(x)+N p(x)z_0'(x)\leq0, \] and which is true (by Lemma \ref{DI1}). Therefore \eqref{ineq} holds and hence $u_n$ is an upper solution. Now applying Lemma \ref{U-L-1} we deduce that $u_{n+1}\leq u_n$. This completes the proof. \end{proof} Similarly we can prove the following two results (Lemma \ref{L-L-1}, Proposition \ref{L-L-2}) for lower solutions. \begin{lemma}\label{L-L-1} Let $\lambda>0$. If $v_n$ is a lower solution of \eqref{1.3}-\eqref{1.4} and $v_{n+1}$ is defined by \eqref{1.5}-\eqref{1.6} then $v_{n}\leq v_{n+1}$. \end{lemma} \begin{proposition}\label{L-L-2} Assume that {\rm (H1)--(H4)} hold and let $\lambda>0$ be such that $\lambda\geq M \big(1-N\int_0^1p(t)dt\big)^{-1}$. Then the functions $v_{n+1}$ defined recursively by \eqref{1.5}-\eqref{1.6} are such that for all $n\in \mathbb{N}$, \begin{itemize} \item[(i)] $v_n$ is a lower solution of \eqref{1.3}-\eqref{1.4}. \item[(ii)] $v_{n}\leq v_{n+1}$. \end{itemize} \end{proposition} In the next result we prove that upper solution $u_n$ is larger than lower solution $v_n$ for all $n$. \begin{proposition}\label{U-L-L-WO} Assume that {\rm (H1)--(H4)} hold and let $\lambda>0$ such that $\lambda\geq M \big(1-N\int_0^1p(t)dt\big)^{-1}$ and for all $x\in [0,1]$ \[ f(x,v_0,pv_0')-f(x,u_0,pu_0')+\lambda(u_0-v_0)\geq0. \] Then for all $n\in\mathbb{N}$, the functions $u_n$ and $v_n$ defined recursively by \eqref{1.5}-\eqref{1.6} satisfy $v_n\leq u_n$. \end{proposition} \begin{proof} We define a function \[ h_i(x)=f(x,v_ipv'_i)-f(x,u_i,pu'_i)+\lambda (u_i-v_i),\quad i\in\mathbb{N}. \] It is easy to see that for all $i\in\mathbb{N}$, $w_i=u_i-v_i$ satisfies the differential equation \begin{align*} -(pw_i')'+\lambda p w_i &=p\{f(x,v_{i-1},pv'_{i-1})-f(x,u_{i-1},pu'_{i-1}) +\lambda (u_{i-1}-v_{i-1})\}\\ &= p h_{i-1}\,. \end{align*} To prove this proposition we use again the principle of mathematical induction. For $i=1$ we have $h_0\geq0$ and $w_1$ is the solution of \eqref{2.1}-\eqref{2.2} with $A=0$ and $B=0$. Using Proposition \ref{MP} we deduce that $w_1\geq0$; i.e., $u_1\geq v_1$. Now, let $n\geq2$, $h_{n-2}\geq0$ and $u_{n-1}\geq v_{n-1}$, then we are required to prove that $h_{n-1}\geq0$ and $u_{n}\geq v_{n}$. First we show that for all $x\in [0,1]$ the function $h_{n-1}$ is non-negative. Indeed we have \begin{align*} h_{n-1}&= f(x,v_{n-1},pv'_{n-1})-f(x,u_{n-1},pu'_{n-1}) +\lambda (u_{n-1}-v_{n-1})\\ &\geq -[(M-\lambda)w_{n-1}+N(sign~w'_{n-1})pw'_{n-1}]. \end{align*} Here $w_{n-1}$ is a solution of \eqref{2.1} with $h(x)=h_{n-2}\geq0$, $A=0$ and $B=0$. Arguments similar to the Proposition \ref{U-L-2} can be used to prove that $h_{n-1}\geq0$. Now, we have $h_{n-1}\geq0$, $w'_n(0)=0$ and $w'_{n}(1)=0$ thus from Proposition \ref{MP} we deduce that $w_n\geq0$, i.e., $u_n\geq v_n$. \end{proof} For the next lemma we use the assumption \begin{itemize} \item[(H5)] For all $(x,u,pu')\in D$, $|f(x,u,pu')|\leq \varphi(|pu'|)$ where $\varphi:[0,\infty)\to(0,\infty)$ is continuous and satisfies \[ \int_{0}^{\infty}\frac{ds}{\varphi(s)}>\int_0^1p(x)dx. \] \end{itemize} \begin{lemma}\label{U-L-3} If $f(x,u,pu')$ satisfies {\rm(H1), (H2), (H5)}, then there exists $R_0>0$ such that any solution of \[ -(pu')'+pf(x,u,pu')\geq0,\quad 00$ such that any solution of \[ -(pv')'+pf(x,v,pv')\leq0,\quad 00$ be such that \[ \lambda\geq M \Big(1-N\int_0^1p(t)dt\Big)^{-1} \] and for all the $x\in [0,1]$, \[ f(x,v_0,pv_0')-f(x,u_0,pu_0')+\lambda(u_0-v_0)\geq0. \] Then the sequences $u_n$ and $v_n$ defined by \eqref{1.5}--\eqref{1.6} converge monotonically to solutions $\widetilde{u}(x)$ and $\widetilde{v}(x)$ of \eqref{1.3}-\eqref{1.4}. Any solution $z(x)$ of \eqref{1.3}-\eqref{1.4} in $D$ satisfies \[ \widetilde{v}(x)\leq z(x)\leq \widetilde{u}(x). \] \end{theorem} \begin{proof} Using Lemma \ref{U-L-1} to Lemma \ref{L-L-3} and Proposition \ref{U-L-2} to Proposition \ref{U-L-L-WO} we deduce that sequences $\{u_n\}$ and $\{v_n\}$ are monotonic $(u_0\geq u_1\geq u_2\dots \geq u_n\geq v_n\dots \geq v_2\geq v_1\geq v_0)$ and are bounded by $v_0$ and $u_0$ in $C[0,1]$ and by Dini's theorem they converge uniformly to $\widetilde{u}$ and $\widetilde{v}$ (say). We can also deduce that the sequences $\{pu'_n\}$ and $\{pv'_n\}$ are uniformly bounded and equi-continuous in $C[0,1]$ and by Arzela-Ascoli theorem there exists uniformly convergent subsequences $\{pu'_{n_{k}}\}$ and $\{pv'_{n_{k}}\}$ in $C[0,1]$. It is easy to observe that $u_n\to \widetilde{u}$ and $v_n \to \widetilde{v}$ implies $pu'_n\to p\widetilde{u}'$ and $p\widetilde{v}'_n\to p\widetilde{v}'$. Solution of \eqref{1.5}-\eqref{1.6} is given by \eqref{2.11} where $h(x)=-pf(x,y_{n-1},py_{n-1}')+\lambda py_{n-1}$. Since the sequences are uniformly convergent taking limit as $n\to \infty$ we obtain $\widetilde{u}$ and $\widetilde{v}$ as the solutions of the nonlinear boundary value problem \eqref{1.3}-\eqref{1.4}. Any solution $z(x)$ in $D$ plays the role of $u_0$. Hence $z(x)\geq \widetilde{v}(x)$. Similarly one concludes that $z(x)\leq \widetilde{u}(x)$. \end{proof} \begin{remark} \rm When the source function is derivative independent; i.e., $N=0$. In this case we can choose $\lambda=M$. \end{remark} \section{Upper and lower solutions in reverse order}\label{rouls} In this section we consider the case when the upper and lower solutions are in reverse order; i.e., $u_0(x)\leq v_0(x)$. For this we require opposite one-sided Lipschitz condition and we assume that \begin{itemize} \item[(F1)] there exists upper solution ($u_0$) and lower solution ($v_0$) in $C^2[0,1]$ such that $u_0\leq v_0$ for all $x\in[0,1]$; \item[(F2)] the function $f:D_0\to \mathbb{R}$ is continuous on \[ D_0:=\{(x,y,py')\in[0,1]\times R\times R: u_0\leq y\leq v_0\}; \] \item[(F3)] there exists $M\geq0$ such that for all $(x,\widetilde{\tau},pv'),~(x,\widetilde{\sigma},pv')\in D_0$, \[ f(x,\widetilde{\sigma},pv')-f(x,\widetilde{\tau},pv') \geq -M(\widetilde{\sigma}-\widetilde{\tau}), \quad (\widetilde{\tau}\leq \widetilde{\sigma}); \] \item[(F4)] there exist $N\geq0$ such that for all $(x,u_,pv'_1), (x,u,pv'_2)\in D_0$, \[ |f(x,u,pv'_1)-f(x,u,pv'_2)|\leq N|pv'_2-pv'_1|. \] \end{itemize} Here we define the approximation scheme by \eqref{1.5}-\eqref{1.6} and use Anti-maximum principle. We make a good choice of $\lambda$ so that the sequences thus generated converge to the solution of the nonlinear problem. Similar to the Section \ref{wouls} we require the following Lemmas and Propositions. \begin{lemma}\label{R-U-L-1} Let $-\lambda_0<\lambda<0$. If $u_n$ is an upper solution of \eqref{1.3}-\eqref{1.4} and $u_{n+1}$ is defined by \eqref{1.5}-\eqref{1.6} then $u_{n+1}\geq u_{n}$. \end{lemma} \begin{proof} Let $w_{n}=u_{n+1}-u_{n}$, then \begin{gather*} -(pw_{n}')'+\lambda pw_{n} =(pu'_n)'-pf(x,u_n,pu'_n)\leq0,\\ w_{n}'(0)\geq0,\quad w_{n}'(1)\leq0, \end{gather*} and using Proposition \ref{MP} we have $u_{n+1}\geq u_n$. \end{proof} \begin{proposition}\label{R-U-L-2} Assume that {\rm (F1)--(F4)} hold. Let $-\lambda_0<\lambda<0$ be such that $M+\lambda\leq0$ and $(M+\lambda)\big(1+\lambda\int_0^1\frac{1}{p(x)}\int_0^xp(t) \,dt\,dx\big)-N\lambda \int_0^1p(x)dx\leq0$. Then the functions $u_{n+1}$ defined recursively by \eqref{1.5}-\eqref{1.6} are such that for all $n\in \mathbb{N}$, \begin{itemize} \item[(i)] $u_n$ is an upper solution of \eqref{1.3}-\eqref{1.4}. \item[(ii)] $u_{n+1}\geq u_{n}$. \end{itemize} \end{proposition} \begin{proof} Using Remark \ref{RM2}, Remark \ref{R2}, Lemma \ref{DI2}, Lemma \ref{R-U-L-1} and on the lines of the proof of Proposition \ref{U-L-2} this proposition can be deduced easily. \end{proof} In the same way we can prove the following results for the lower solutions. \begin{lemma}\label{R-L-L-1} Let $-\lambda_0<\lambda<0$. If $v_n$ is a lower solution of \eqref{1.3}-\eqref{1.4} and $v_{n+1}$ is defined by \eqref{1.5}-\eqref{1.6} then $v_{n}\geq v_{n+1}$. \end{lemma} \begin{proposition}\label{R-L-L-2} Assume that {\rm (F1)--(F4)} hold. Let $-\lambda_0<\lambda<0$ be such that $M+\lambda\leq0$ and $(M+\lambda)\big(1+\lambda\int_0^1\frac{1}{p(x)} \int_0^xp(t)\,dt\,dx\big)-N\lambda \int_0^1p(x)dx\leq0$. Then the functions $v_{n+1}$ defined recursively by \eqref{1.5}-\eqref{1.6} are such that for all $n\in \mathbb{N}$, \begin{itemize} \item[(i)] $v_n$ is a lower solution of \eqref{1.3}-\eqref{1.4}. \item[(ii)] $v_{n}\geq v_{n+1}$. \end{itemize} \end{proposition} In the next result we prove that lower solution $v_n$ is larger than upper solution $u_n$ for all $n$. \begin{proposition}\label{R-U-L-L-WO} Assume that {\rm (F1)--(F4)} hold. Let $-\lambda_0<\lambda<0$ be such that $M+\lambda\leq0$ and \[ (M+\lambda)\Big(1+\lambda\int_0^1\frac{1}{p(x)}\int_0^xp(t) \,dt\,dx\Big)-N\lambda \int_0^1p(x)dx\leq0 \] and for all $x\in [0,1]$ \[ f(x,v_0,pv_0')-f(x,u_0,pu_0')+\lambda(u_0-v_0)\geq0. \] Then for all $n\in\mathbb{N}$, the functions $u_n$ and $v_n$ defined recursively by \eqref{1.5}-\eqref{1.6} satisfy $v_n\geq u_n$. \end{proposition} Now similar to the Lemma \ref{U-L-3} and Lemma \ref{L-L-3} we state the following two results. These results establish a bound on $p(x) u'(x)$ and $p(x) v'(x)$. We use the assumption \begin{itemize} \item[(F5)] For all $(x,u,pu')\in D_0$, $|f(x,u,pu')|\leq \varphi(|pu'|)$ where $\varphi:[0,\infty)\to(0,\infty)$ is continuous and satisfies \[ \int_{0}^{\infty}\frac{ds}{\varphi(s)}>\int_0^1p(x)dx. \] \end{itemize} \begin{lemma}\label{R-U-L-3} If $f(x,u,pu')$ satisfies {\rm (F1), (F2), (F5)}, then there exists $R_0>0$ such that any solution of \[ -(pu')'+pf(x,u,pu')\geq0,\quad 00$ such that any solution of \[ -(pv')'+pf(x,v,pv')\leq0,\quad 00$ in $(0,1)$. \begin{thebibliography}{0} \bibitem{DOR-MAEG-IMAJAM-2008} D. O'Regan, M. A. El-Gebeily; Existence, upper and lower solutions and quasilinearization for singular differential equations, {\it IMA J. Appl. Math.}, Vol. 73 (2008) 323--344. \bibitem{MC-CDC-PH-AMC-2001} M. Cherpion, C. D. Coster, P. Habets; A constructive monotone iterative method for second-order BVP in the presence of lower and upper solutions, {\it Appl. Math. 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