\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 20, 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 (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/20\hfil Positive solutions] {Positive solutions for a class of nonresonant boundary-value problems} \author[X. Zhang\hfil EJDE-2007/20\hfilneg] {Xuemei Zhang} \address{ Xuemei Zhang \newline Department of Mathematics and Physics, North China Electric Power University, Beijing 102206, China} \email{zxm74@sina.com} \thanks{Submitted November 28, 2006. Published January 27, 2007.} \thanks{Supported by grant 10626018 from the National Nature Science Foundation of China and \hfill\break\indent the Science Foundation for Young Teachers of North China Electric Power University.} \subjclass[2000]{34B15} \keywords{Positive solution; fixed point theorem; existence; complete continuity} \begin{abstract} This paper concerns the existence and multiplicity of positive solutions to the nonresonant second-order boundary-value problem $$Lx=\lambda w(t)f(t,x).$$ We are interested in the operator $Lx:=-x''+\rho qx$ when $w$ is in $L^{p}$ for $1\leq p \leq +\infty$. Our arguments are based on fixed point theorems in a cone and H\"older's inequality. The nonexistence of positive solutions is also studied. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} Consider the second-order boundary-value problem (BVP) $$\begin{gathered} Lx= \lambda w(t) f(t,x),\quad 0< t < 1 , \\ x(0)=x(1)=0, \end{gathered} \label{e1.1}$$ where $\lambda$ is a positive parameter and $L$ denotes the linear operator $$Lx:=-x''+\rho qx,$$ where $q\in C([0,1],[0,\infty))$ and $\rho>0$ such that \begin{gather*} Lx= 0,\quad 0< t < 1 , \\ x(0)=x(1)=0, \end{gather*} has only the trivial solution. For the classical case $Lx=-x''$ and $f(t,x)=f(x)$, several results are available in the literature. Bandle [1] and Lin [12] established the existence of positive solutions under the assumption that $f$ is superlinear, i.e., $f_{0}=\lim_{x\to 0}\frac{f(x)}{x}=0$, $f_{\infty}=\lim_{x\to \infty}\frac{f(x)}{x}=\infty$. Wang [14] established the existence of positive solutions under the assumption that f is sublinear, i.e., $f_{0}=\infty$ and $f_{\infty}=0$. Eloe and Henderson [3] and Henderson and Wang [7] obtained the existence of positive solutions under the assumption that $f_{0}$ and $f_{\infty}$ exist. In the case $Lx=|x'|^{p-2}x'$, $p>1$, i.e., the one-dimensional p-Laplacian, Jiang [9] obtained existence and multiplicity results under the assumption that f may be semilinear or superlinear at $x=\infty$ and change sign. Wang and Gao [16] established the existence of positive solutions under the assumption that $f(t,x)=f(x)$ is positive, right continuous, nonincreasing in $(0,+\infty)$ and $f_{0}=\infty$. In recent papers, when $Lx=|x'|^{p-2}x'$, $p>1$ and $Lx=-x'', f(t,x)=f(x)$, where $x=(x_{1},\ldots,x_{n})$. If $01$, $g\in L^{q}[a,b]$ with $q>1$, and $\frac{1}{p}+\frac{1}{q}=1$. Then $fg\in L^{1}[a,b]$ and $\|fg\|_{1}\leq\|f\|_{p}\|g\|_{q}$. Let $f\in L^{1}[a,b]$, $g\in L^{\infty}[a,b]$. Then $fg\in L^{1}[a,b]$ and $\|fg\|_{1}\leq\|f\|_{1}\|g\|_{\infty}$. \end{lemma} This paper is organized as follows: In Section 2, we provide some necessary background. In particular, we state some properties of the Green's function associated with BVP (1.1). In Section 3, the main result will be stated and proved. Finally some examples illustrate our main results. \section{Preliminaries} Let $J=[0,1]$. The basic space used in this paper is $E=C[0,1]$. It is well known that $E$ is a real Banach space with the norm $\| \cdot \|$ defined by $\|x\|=\max_{t\in J}|x(t)|$. Let $K$ be a cone of $E$, $K_{r}=\{x\in K :\|x\|\leq r\}$, $\partial K_{r}=\{x\in K:\|x\|=r\}$, $\bar K_{r,R}=\{x\in K:r\leq \|x\|\leq R\}$, where $00$ such that $w(t)\geq m$ a.e. on $[0,1]$; \item[(H2)] $f\in C([0,1]\times [0,+\infty),[0,+\infty))$. \end{itemize} In this paper, the Green's function of the corresponding homogeneous BVP is $$G(t,s) =\frac{1}{\Delta} \begin{cases} \phi(s)\psi(t) &\text{if } 0\leq s \leq t \leq 1,\\ \phi(t)\psi(s) &\text{if } 0\leq t \leq s \leq 1. \end{cases} \label{e2.1}$$ Where $\phi$ and $\psi$ satisfy \begin{gather} L\phi=0,\quad \phi(0)=0,\quad \phi'(0)=1, \label{e2.2}\\ L\psi=0,\quad \psi(1)=0,\quad \psi'(1)=-1. \label{e2.3} \end{gather} From [17,18], it is not difficult to show that $\Delta= -(\phi(t)\psi'(t)-\phi'(t)\psi(t))>0$ and $\phi'(t)>0$ on $(0,1]$ and $\psi'(t)<0$ on $[0,1)$. It is easy to prove that $G(t,s)$ has the following properties; \begin{itemize} \item %{\bf Property 2.1.} For $t,s \in (0,1)$, we have $$G(t,s)>0. \label{e2.4}$$ \item % {\bf Property 2.2.} For $t,s\in J$, we have $$0\leq G(t,s)\leq G(s,s).\label{e2.5}$$ \item % {\bf Property 2.3.} Let $\theta \in (0,\frac{1}{2})$ and define $J_{\theta}=[\theta,1-\theta]$. Then for all $t \in J_{\theta},s\in J$ we have $$G(t,s)\geq \sigma G(s,s),\label{e2.6}$$ where $$\sigma(=\sigma(\theta))= \min\big\{\frac{\psi(1-\theta)}{\psi(0)}, \frac{\phi(\theta)}{\phi(1)}\big\}. \label{e2.7}$$ In fact, for $t\in [\theta,1-\theta]$, we have $$\frac{G(t,s)}{G(s,s)}\geq \min\big\{\frac{\psi(1-\theta)}{\psi(s)}, \frac{\phi(\theta)}{\phi(s)}\big\} \geq\min\big\{\frac{\psi(1-\theta)}{\psi(0)}, \frac{\phi(\theta)}{\phi(1)}\big\} =:\sigma.$$ It is easy to see that $0<\sigma<1$. \end{itemize} For the sake of applying Lemma 1.1 and Lemma 1.2, we construct a cone in $E=C[0,1]$ by $$K=\{x\in C[0,1]:x\geq 0, \; \min_{t\in J_{\theta}}x(t) \geq \sigma\|x\|\} .\label{e2.8}$$ It is easy to see $K$ is a closed convex cone of $E$ and $\bar K_{r,R}\subset K$. Define an operator $T_{\lambda}:\bar K_{r,R}\to K$ by $$T_{\lambda} x(t) =\lambda\int_{0}^{1}G(t,s)w(s)f(s,x(s))ds, \label{e2.9}$$ From the above equality, it is well known that (1.1) has a positive solution $x$ if and only if $x\in \bar K_{r,R}$ is a fixed point of $T_{\lambda}$. \begin{lemma} \label{lem2.1} Let (H1) and (H2) hold. Then $T_{\lambda}\bar K_{r,R}\subset K$ and $T_{\lambda}:\bar K_{r,R}\to K$ is completely continuous. \end{lemma} \begin{proof} For $x\in K$, by (2.9), we have $T_{\lambda}x(t)\geq 0$ and $$\|T_{\lambda}x\|\leq\lambda \int_{0}^{1}G(s,s)w(s)f(s,x(s))ds,\label{e2.10}$$ On the other hand, by (2.9), (2.10) and (2.6), we obtain \begin{align*} \min_{t\in J_{\theta}}T_{\lambda}x(t) &=\min_{t\in J_{\theta}}\lambda \int_{0}^{1}G(t,s)w(s)f(s,x(s))ds\\ & \geq \lambda \sigma\int_{0}^{1}G(s,s)w(s)f(s,x(s))ds \\ & \geq \sigma\|T_{\lambda}x\|. \end{align*} Therefore $T_{\lambda}x\in K$, i.e., $T_{\lambda}K\subset K$. Also we have $T_{\lambda}\bar K_{r,R}\subset K$ by $\bar K_{r,R}\subset K$. Hence we have $T_{\lambda}:\bar K_{r,R}\to K$. Next by standard methods and Ascoli-Arzela theorem one can prove $T_{\lambda}:\bar K_{r,R}\to K$ is completely continuous. So it is omitted. \end{proof} \section{Main results} Write $$f^{\beta}=\limsup_ {x \to \beta}\max_{t\in J} \frac{f(t,x)}{x}, \quad f_{\beta}=\liminf_ {x \to \beta}\min_{t\in J} \frac{f(t,x)}{x},$$ where $\beta$ denotes $0$ or $\infty$. In this section, we apply the Lemmas 1.1-1.3 to establish the existence of positive solutions for BVP (1.1). We consider the following three cases for $w\in L^{p}[0,1]$; $p> 1$, $p=1$ and $p=\infty$. Case $p>1$ is treated in the following theorem. \begin{theorem} \label{thm3.1} Assume that (H1) and (H2) hold. In addition, letting $f_{0}=\infty$ and $f^{\infty}=0$ be satisfied, then, for all $\lambda >0$, BVP (1.1) has at least one positive solution $x^{*}(t)$. \end{theorem} \begin{proof} Let $T_{\lambda}$ be cone preserving, completely continuous operator that was defined by (2.9). Considering $f_{0}=\infty$, there exists $r_{1}>0$ such that $f(t,x)\geq \varepsilon_{1}x$, for $00$ satisfies $\lambda\sigma^2\frac{1}{\Delta} m \phi(\theta)\psi(1-\theta)\varepsilon_{1}\geq 1$. So, for $x\in \partial K_{r_{1}}$, $t \in J$, from (2.6), we have \begin{aligned} (T_{\lambda}x)(t) &= \lambda \int_{0}^{1}G(t,s)w(s)f(s,x(s))ds\\ & \geq\lambda \varepsilon_{1}\int_{0}^{1}G(t,s)w(s)x(s)ds\\ & \geq \lambda m\varepsilon_{1}\min_{t\in J_{\theta}}\int_{0}^{1}G(t,s)x(s)ds\\ &\geq \lambda m\sigma\varepsilon_{1}\int_{0}^{1}G(s,s)x(s)ds\\ & \geq \lambda m\sigma^2\varepsilon_{1}\|x\|\int_{\theta}^{1-\theta}G(s,s)ds\\ & \geq \lambda\sigma^2\frac{1}{\Delta} m \phi(\theta) \psi(1-\theta)\varepsilon_{1}\|x\| \\ & \geq\|x\|. \end{aligned} \label{e3.1} Consequently, for $x \in\partial K_{r_{1}}$, we have $$\|T_{\lambda}x\|\geq\|x\|. \label{e3.2}$$ Next, turning to $f^{\infty}=0$, there exists $\bar r_{2}>0$ such that $f(t,x)\leq \varepsilon_{2}x$, for $x\geq \bar r_{2},\ \ t \in J$, where $\varepsilon_{2}>0$ satisfies $\varepsilon_{2}\lambda \|G\|_{q}\|w\|_{p}\leq \frac{1}{2}$. Let $$M=\lambda\sup_{x\in \partial K_{\bar r_{2},t\in J}}f(t,x) \int_{0}^{1}G(t,t)w(t)dt.$$ It is not difficult to see that $M<+\infty$. Choosing $r_{2}>max\{r_{1},\bar r_{2},2M\}$, we get $M<\frac{1}{2}r_{2}$. Now, we choose $x\in \partial K_{r_{2}}$ arbitrary. Letting $\bar x(t)=\min\{x(t),\bar r_{2}\}$, we have $\bar x\in \partial K_{\bar r_{2}}$. In addition, writing $e(x)=\{t \in J:x(t)>\bar r_{2}\}$, for $t\in e(x)$, we get $\bar r_{2}0$, $t\in J_{\theta}$. Thus it follows that (1.1) has a positive solution $x^{*}$ for all $\lambda>0$. The proof is complete. \end{proof} The following theorem studies the case $p=\infty$. \begin{theorem} \label{thm3.2} Suppose the conditions of Theorem 3.1 hold. Then, for all $\lambda >0$, BVP (1.1) has at least one positive solution $x^{*}(t)$. \end{theorem} To prove the above theorem, let $\|G\|_{1}\|w\|_{\infty}$ replace $\|G\|_{p}\|w\|_{q}$ and repeat the argument above. Finally we consider the case of $p=1$. \begin{theorem} \label{thm3.3} Suppose the conditions of Theorem 3.1 hold. Then, for all $\lambda >0$, (1.1) has at least one positive solution $x^{*}(t)$. \end{theorem} \begin{proof} As in the proof of Theorem 3.1, choose $r_{2}=max\{r_{1},\bar r_{2},2M\}$. For $x\in \partial K_{r_{2}}$, from (2.5) we have \begin{aligned} (T_{\lambda}x)(t) &\leq\lambda \int_{0}^{1}G(s,s)w(s)f(s,x(s))ds\\ & = \lambda\int_{e(x)}G(s,s)w(s)f(s,x(s))ds + \lambda\int_{[0,1]\setminus e(x)}G(s,s)w(s)f(s,x(s))ds\\ & \leq \lambda \varepsilon_{2} r_{2}\int_{0}^{1}G(s,s)w(s)ds+ \lambda\int_{0}^{1}G(s,s)w(s)f(s,\bar x(s))ds\\ & \leq\lambda \varepsilon_{2} r_{2}\frac{1}{\Delta}\phi(1)\psi(0)\|w\|_{1} +M\\ & <\frac{1}{2}r_{2}+\frac{1}{2}r_{2}\\ &= r_{2}=\|x\|, \end{aligned} \label{e3.3'} where $\varepsilon_{2}>0$ satisfies $\varepsilon_{2}\lambda \frac{1}{\Delta}\phi(1)\psi(0)\|w\|_{1}\leq 1$. Consequently, from (3.5), for $x \in\partial K_{r_{2}}$, we have $\|T_{\lambda}x\|< \|x\|$. This and (3.2) complete the proof. \end{proof} \begin{corollary} \label{coro3.1} Let $f^{0}=0$ replace $f^{\infty}=0$ and $f_{\infty}=\infty$ replace $f_{0}=\infty$ in Theorems 3.1-3.3. Then the results still hold. \end{corollary} In the following theorems we only consider the case of $p > 1$. The existence theorems corresponding to the cases of $p=1$ and $p=\infty$ are similar and are omitted. \begin{theorem} \label{thm3.4} Assume (H1), (H2) and the following two conditions: \begin{itemize} \item[(i)] $f_{0}=\infty$ or $f_{\infty}=\infty$; \item[(ii)] There exist $\rho>0$ and $\delta>0$, for $00$ such that for all $0<\lambda <\lambda_{0}$, BVP (1.1) has at least one positive solution $x^{*}(t)$. \end{theorem} \begin{proof} Considering $f_{0}=\infty$, there exists $00$ satisfies $\varepsilon_{3}\lambda\sigma^2\frac{1}{\Delta} m \phi(\theta)\psi(1-\theta)\geq 1$. So, for $x\in \partial K_{r_{3}}$, from (2.6), we have \begin{aligned} (T_{\lambda}x)(t)&= \lambda \int_{0}^{1}G(t,s)w(s)f(s,x(s))ds\\ & \geq\lambda \varepsilon_{3}\int_{0}^{1}G(t,s)w(s)x(s)ds\\ & \geq \lambda m\varepsilon_{3}\min_{t\in J_{\theta}}\int_{0}^{1}G(t,s)x(s)ds\\ & \geq \lambda m\sigma\varepsilon_{3}\int_{0}^{1}G(s,s)x(s)ds\\ & \geq \lambda m\sigma\varepsilon_{3}\int_{\theta}^{1-\theta}G(s,s)x(s)ds\\ & \geq \lambda\sigma^2\frac{1}{\Delta} m \phi(\theta)\psi(1-\theta) \varepsilon_{1}\|x\| \\ & \geq\|x\|. \end{aligned} \label{e3.5} Consequently, for $x \in\partial K_{r_{3}}$, we have $$\|T_{\lambda}x\|\geq\|x\|. \label{e3.6}$$ If $f_{\infty}=\infty$, similar to the proof of (3.7), there exists $r_{4}>\rho$ such that $f(t,x)\geq \varepsilon_{4}x$, for $x\geq r_{4}$, $t \in J$, where $\varepsilon_{4}>0$ satisfies $\varepsilon_{4}\lambda\sigma^2\frac{1}{\Delta} m \phi(\theta)\psi(1-\theta)\geq 1$, and, for $x \in\partial K_{r_{4}}$, we have $$\|T_{\lambda}x\|\geq\|x\|. \label{e3.7}$$ On the other hand, from (ii), when a $\rho>0$ is fixed, then there exists a $\lambda_{0}>0$ such that $f(t,x)\leq \delta<\frac{1}{\lambda} [\|G\|_{q}\|w\|_{p}]^{-1}\rho$ for $0<\lambda<\lambda_{0}$, $x \in\partial K_{\rho}$. Therefore for $x \in\partial K_{\rho}$ and $t \in J$ we have \begin{align*} (T_{\lambda}x)(t) &= \lambda \int_{0}^{1}G(t,s) w(s)f(s,x(s))ds\\ & \leq \delta\lambda \int_{0}^{1}G(t,s) w(s)ds \\ & \leq \delta\lambda \|G\|_{q}\|w\|_{p}\\ & <\rho=\|x\|. \end{align*} Consequently, for $x \in\partial K_{\rho}$, we have $$\|T_{\lambda}x\|< \|x\|. \label{e3.8}$$ By Lemma 1.1, for all $0<\lambda <\lambda_{0}$, (3.7) and (3.9), (3.8) and (3.9), respectively, yield that $T_{\lambda}$ has a fixed point $x^{*}\in \bar K_{r_{3},\rho}$, $r_{3} \leq \|x^{*}\| < \rho$ and $x^{*}(t)\geq \sigma \|x^{*}\|>0,\ \ t\in J_{\theta}$ or $x^{*}\in \bar K_{\rho,r_{4}}$, $\rho_{1} < \|x^{*}\| \leq r_{4}$ and $x^{*}(t)\geq \sigma \|x^{*}\|>0,\ \ t\in J_{\theta}$. Thus it follows that BVP (1.1) has at least one positive solution $x^{*}$ for all $0<\lambda<\lambda_{0}$. \end{proof} \begin{theorem} \label{thm3.5} Assume (H1), (H2) and the following two conditions: \begin{itemize} \item[(i)] $f_{0}=\infty$ and $f_{\infty}=\infty$; \item[(ii)] There exist $\rho>0$, $\delta>0$, for $00$ such that for all $0<\lambda <\lambda_{0}$, BVP (1.1) has at least two positive solutions $x^{*}(t),x^{**}(t)$. \end{theorem} \begin{proof} The proof is similar to that of Theorem 3.5. Lemma 1.2, (3.7)-(3.9) yield that $T_{\lambda}$ has at least two fixed points $x^{*}$, $x^{**}$, where $x^{*}\in \bar K_{r_{3},\rho}$, $r_{3} \leq \|x^{*}\| < \rho$ and $x^{*}(t)\geq \sigma \|x^{*}\|>0$, $t\in J_{\theta}$, $x^{**}\in \bar K_{\rho,r_{4}}$, $\rho < \|x^{*}\| \leq r_{4}$ and $x^{**}(t)\geq \sigma \|x^{**}\|>0$, $t\in J_{\theta}$. Thus it follows that BVP 1.1 has at least two positive solutions $x^{*}$, $x^{**}$ for all $0<\lambda<\lambda_{0}$. \end{proof} \begin{corollary} \label{coro3.2} Assume (H1), (H2) and the following two conditions: \begin{itemize} \item[(i)] $f^{0}=0$ or $f^{\infty}=0$; \item[(ii)] There exist $\rho >0$, $\delta>0$, such that $f(t,x)\geq \delta$ for $x \geq \rho$ and $t \in J$. \end{itemize} Then there exists $\lambda_{0}>0$ such that for all $\lambda >\lambda_{0}$, BVP (1.1) has at least one positive solution $x^{*}(t)$. \end{corollary} \begin{corollary} \label{coro3.3} Assume (H1), (H2) and the following two conditions: \begin{itemize} \item[(i)] $f^{0}=0$ and $f^{\infty}=0$; \item[(ii)] There exist $\rho >0$ and $\delta>0$, such that $f(t,x)\geq \delta$ for $x \geq \rho$ and $t \in J$. \end{itemize} Then there exists $\lambda_{0}>0$ such that for all $\lambda >\lambda_{0}$, BVP (1.1) has at least two positive solutions $x^{*}(t),x^{**}(t)$. \end{corollary} Our last result corresponds to the case when (1.1) has no positive solution. \begin{theorem} \label{thm3.6} Assume (H1), (H2), $f_{0}>0$ and $f_{\infty}>0$. Then there exists $\lambda_{0}>0$ such that for all $\lambda >\lambda_{0}$, BVP (1.1) has no positive solution. \end{theorem} \begin{proof} Since $f_{0}>0$ and $f_{\infty}>0$, then there exist $\eta_{1}>0$, $\eta_{2}>0$, $h_{1}>0$ and $h_{2}>0$ such that h_{1}0. $$Thus, for t \in J, x \geq \sigma h_{1}, we have $$f(t,x)\geq \eta x, \label{e3.11}$$ and for t \in J, x \leq h_{1}, we have $$f(t,x)\geq \eta x. \label{e3.12}$$ Assume y is a positive solution of (1.1). We will show that this leads to a contradiction for \lambda >\lambda_{0}=[\eta\sigma^{2} \int_{\theta}^{1-\theta}G(s,s)w(s)ds]^{-1}. In fact, if \|y\|\leq h_{1}, (3.13) implies that$$ f(t,y)\geq \eta y,\ for \ t\in J. On the other hand, if \|y\|>h_{1}, then \min_{t\in J_{\theta}}y(t) \geq \sigma\|y\|>\sigma h_{1}, which, together with (3.12), implies that, for t\in J_{\theta}, we get f(t,y)\geq \eta y. Since (Ty)(t)=y(t), it follows that, for \lambda > \lambda_{0}, t \in J, \begin{align*} \|y\| &=\|(T_{\lambda}y)\|\\ &= \max_{ t\in J}\lambda \int_{0}^{1}G(t,s) w(s)f(s,y(s))ds\\ &\geq \min_{ t \in J_{\theta}}\lambda \int_{\theta}^{1-\theta}G(t,s) w(s)\eta y(s)ds\\ &\geq \lambda \eta\sigma\|y\|\sigma \int_{\theta}^{1-\theta}G(s,s)w(s) ds \\ &\geq \lambda \eta\|y\|\sigma^{2} \int_{\theta}^{1-\theta}G(s,s)w(s) ds \\ &> \|y\|, \end{align*} which is a contradiction. The proof is complete. \end{proof} \begin{corollary} \label{coro3.4} Let f^{0}<\infty and f^{\infty}<\infty replace f_{0}>0 and f_{\infty}>0 in Theorem 3.9. Then the results are still valid. \end{corollary} \begin{remark} \label{rmk3.1} \rm We did not use H\"older's inequality in the proof of Theorem 3.3 and Theorem 3.9. \end{remark} It is clear that the results obtained here improve the results of [2,3,11,12,13,14,15]. To illustrate how our main results can be used in practice we present two examples. \begin{example} \label{exa3.1} \rm We set \omega(t)=|2t-\frac{1}{8}|^{-1/15}. Then w\in L^{p} for 11. It is not difficult to see that f_{0}=\liminf_{x\to 0} \min_{t\in [0,1]}\frac{f(t,x)}{x} =\infty,\quad f^{\infty}=\limsup_{x\to \infty} \max_{0\leq t\leq 1}\frac{f(t,x)}{x} =0. $$Hence the conditions of the Theorem 3.1 are satisfied. \end{example} \begin{example} \label{exa3.2} \rm We set \omega(t)=|t-\frac{1}{8}|^{-1/8}. Then w\in L^{p} for 11. It is not difficult to see that$$ f_{0}=\liminf_{x\to 0} \min_{t\in J}\frac{f(t,x)}{x} =\infty,\quad f_{\infty}=\liminf_{x\to \infty} \min_{t\in J}\frac{f(t,x)}{x} =\infty. $$Therefore, conditions (H1), (H2) and (i) of the Theorem 3.6 are satisfied. Finally we verify (ii) of the Theorem 3.6. Choosing q(t)=0, then we get \phi(t)=t,\psi(t)=1-t, \phi(0)=0, \psi(1)=0, \phi'(0)=1=a, \psi'(1)=-1=-c, \Delta=-(-t-(1-t))=1>0 and$$ G(t,s) = \begin{cases} s(1-t) &\text{if } 0\leq s \leq t \leq 1,\\ t(1-s) &\text{if } 0\leq t \leq s \leq 1. \end{cases} $$Clearly, for \rho=1, \delta=9/4, we obtain$$ f(t,x)\leq 2x^2+\frac{1}{4}x^{1/2}\leq 2+\frac{1}{4} =\frac{9}{4}=\delta$for$0