Positive solutions for the singular nonlocal boundary value problems involving nonlinear integral conditions

Abstract

In this paper, using the theory of fixed point index on a cone and the Leray-Schauder fixed point theorem, we present the multiplicity of positive solutions for the singular nonlocal boundary-value problems involving nonlinear integral conditions and the existence of at least one positive solution for the singular nonlocal boundary-value problems with sign-changed nonlinearities.

MSC:34B10, 34B15, 34B18.

1 Introduction

Nonlocal boundary-value problems with linear and nonlinear integral conditions have seen a great deal of study lately (see [1–16], and references therein) because of their interesting theory and their applications to various problems, such as heat flow in a bar of finite length [4, 11]. In this paper, we consider the existence of positive solutions of the nonlinear boundary-value problem (BVP) of the form

$$-y^{″}=q(t)f(t,y(t)),t\in (0,1)$$

(1.1)

with integral boundary conditions

$$y(0)=H(\varphi (y)),y(1)=0,$$

(1.2)

where $\varphi (y)$ is a linear functional on $C[0,1]$ given by

$$\varphi (y)={\int }_0^{1}y(s)d\alpha (s)$$

involving a Stieltjes integral with a signed measure.

In [2], Goodrich considered the following problem:

$$-y^{″}=\lambda g(t,y(t)),t\in (0,1)$$

(1.3)

with integral boundary conditions

$$y(0)=H(\varphi (y)),y(1)=0$$

(1.4)

and deduced the existence of at least one positive solution to the BVP (1.3)-(1.4) in which $H(\varphi (y))$ has either asymptotically sublinear or asymptotically superlinear growth, and in [3] Goodrich demonstrated that if the nonlinear functional $H(\varphi (y))$ satisfies a certain asymptotic behavior, then the BVP (1.3)-(1.4) possesses at least one positive solution. For the case that H is linear and $\varphi (y)={\int }_0^{1}y(s)d\alpha (s)$ involves a signed measure, Webb and Infante discussed the multiplicity of positive solutions for nonlocal boundary-value problems [12–14]. For the case that H is linear and the Borel measure associated with the Lebesgue-Stieltjes integral is positive, we can find some results on the existence of positive solutions [7, 8, 16, 17]. The results in the above literature are obtained under the condition that $f(t,x)$ is continuous on $(0,1)\times [0,+\mathrm{\infty })$, i.e., f has no singularity at $x=0$. And it is well known that study of singular two-point boundary-value problems for the second-order differential equation (1.1) (singular in the dependent variable) is very important and there are many results on the existence of positive solutions [15, 18–24]. But there are fewer results on the existence of positive solutions for the singular BVP (1.1)-(1.2) [5, 6]. One goal in this paper is to consider the existence of positive solutions under the condition that $f(t,x)$ is singular at $x=0$. Our paper has the following features.

Firstly, in order to overcome the difficulties of the singularity of f we establish a new cone and get the new condition (3.13) which is different from that in [5, 6]. Moreover, we get a multiplicity of positive solutions for BVP (1.1)-(1.2) different from that in [2, 3, 12–14] under the condition that $H(y)$ or $f(t,y)$ is superlinear at $y=+\mathrm{\infty }$.

Secondly, when f is singular and sign-changed, we get the existence of at least one positive solution to the BVP (1.1)-(1.2) which is different from that in [2, 3, 5, 6, 12–14] where f is nonnegative and continuous at $x=0$. Moreover, the results are different from that in [7, 8, 16, 17] where integral boundary conditions are linear and the Borel measure is positive.

Our paper is organized as follows. In Section 2, we present some lemmas and preliminaries. Section 3 discusses the existence of multiple positive solutions for the BVP (1.1)-(1.2) when f is positive. In Section 4, we discuss the existence of at least one positive solution of BVP (1.1)-(1.2) when f is singular and sign-changed.

2 Preliminaries

In this paper, the following lemmas are needed.

Lemma 2.1 (see [25])

Let Ω be a bounded open set in real Banach space E, P a cone of $E,\theta \in \mathrm{\Omega }$ and $A:\overline{\mathrm{\Omega }}\cap P\to P$ continuous and compact. Suppose $\lambda Ax\ne x$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap P$, $\lambda \in (0,1]$. Then

$$i(A,\mathrm{\Omega }\cap P,P)=1.$$

Lemma 2.2 (see [25])

Let Ω be a bounded open set in real Banach space E, P a cone of $E,\theta \in \mathrm{\Omega }$ and $A:\overline{\mathrm{\Omega }}\cap P\to P$ continuous and compact. Suppose $Ax\nleqq x$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap P$. Then

$$i(A,\mathrm{\Omega }\cap P,P)=0.$$

Lemma 2.3 (see [25, 26])

Let E be a Banach space, $R>0$, $B_R=\{x\in E:\parallel x\parallel \le R\}$, and $F:B_R\to E$ a continuous compact operator. If $x\ne \lambda F(x)$ for any $x\in E$ with $\parallel x\parallel =R$ and $0<\lambda <1$, then F has a fixed point in $B_R$.

Let us begin by stating the hypotheses which we shall impose on the BVP (1.1)-(1.2).

• (C₁) Assume that there are three linear functionals $\varphi ,{\varphi }_1,{\varphi }_2:C([0,1])\to R$ such that

  $$\varphi (y)={\varphi }_1(y)+{\varphi }_2(y).$$

  Moreover, assume that there exists a constant ${\epsilon }_0>0$ such that

  $${\varphi }_2(y)\ge {\epsilon }_0\parallel y\parallel$$

  holds for each $y\in P$, where P is the cone introduced in (2.1) below [2].

• (C₂) The functionals ${\varphi }_1(y)$ and ${\varphi }_2(y)$ are linear and, in particular, have the form

  $${\varphi }_1(y):={\int }_0^{1}y(t)d{\alpha }_1(t),{\varphi }_2(y):={\int }_0^{1}y(t)d{\alpha }_2(t),$$

  where ${\alpha }_1,{\alpha }_2:[0,1]\to R$ satisfy ${\alpha }_1,{\alpha }_2\in BV([0,1])$ with

  $${\int }_0^1(1-t)d{\alpha }_1(t)\ge 0,{\int }_0^1(1-t)d{\alpha }_2(t)\ge 0$$

  and

  $${\int }_0^{1}k(t,s)d{\alpha }_1(t)\ge 0,{\int }_0^{1}k(t,s)d{\alpha }_2(t)\ge 0$$

  hold, where the latter holds for each $s\in [0,1]$ and $k(t,s)$ is defined in (3.2) below [2].

• (C₃) Let $H:R\to R$ be a real-valued, continuous function. Moreover, $H:(0,+\mathrm{\infty })\to (0,+\mathrm{\infty })$.

• (C₄)

  $$\{\begin{array}{c}f:[0,1]\times (0,\mathrm{\infty })\to (0,\mathrm{\infty })\text{ is continuous}\hfill \\ \text{and there exists a function }{\psi }_1\hfill \\ \text{continuous on }[0,1]\text{ and positive on }(0,1)\text{ such that}\hfill \\ f(t,y)\ge {\psi }_1(t)\text{ on }(0,1)\times (0,1].\hfill \end{array}$$

• (C₅)

  $$q\in C(0,1),q>0\text{on }(0,1)\text{and}{\int }_0^1(1-t)q(t)dt<\mathrm{\infty }.$$

Let $C[0,1]=\{y:[0,1]\to R:y(t)\text{ is continuous on }[0,1]\}$ with norm $\parallel y\parallel ={max}_{t\in [0,1]}|y(t)|$. It is easy to see that $C[0,1]$ is a Banach space.

Assume that (C₂) hold. Define

$$\begin{array}{rl}P=& \{y\in C[0,1]:y\text{ is concave on }[0,1]\text{ with }y(t)\ge 0\text{ for all }t\in [0,1],\\ {\varphi }_1(y)\ge 0,{\varphi }_2(y)\ge 0\}.\end{array}$$

(2.1)

It is easy to prove P is a cone of $C[0,1]$ and we have the following lemma.

Lemma 2.4 (see [20])

Let $y\in P$ (defined in (2.1)). Then

$$y(t)\ge t(1-t)\parallel y\parallel \mathit{\text{for }}t\in [0,1].$$

3 Multiplicity of positive solutions for the singular boundary-value problems with positive nonlinearities

In this section, we consider the existence of multiple positive solutions for the BVP (1.1)-(1.2). To show that the BVP (1.1)-(1.2) has a solution, for $y\in P$, we define

$$\begin{array}{r}(T_{ϵ}y)(t)=(1-t)H(\varphi (y))+{\int }_0^{1}k(t,s)q(s)f(s,max\{ϵ,y(s)\})ds,\\ t\in [0,1],1\ge ϵ>0,\end{array}$$

(3.1)

where

$$k(t,s)=\{\begin{array}{ll}(1-t)s,& 0\le s\le t\le 1,\\ (1-s)t,& 0\le t\le s\le 1.\end{array}$$

(3.2)

Lemma 3.1 Suppose (C₁)-(C₅) hold. Then $T_{ϵ}:P\to P$ is continuous and compact for all $1\ge ϵ>0$.

Proof It is easy to prove that $T_{ϵ}$ is well defined and $(T_{ϵ}y)(t)\ge 0$ for all $t\in P$. For $y\in P$, we have

$$\{\begin{array}{l}{(T_{ϵ}y)}^{″}(t)\le 0\text{on }(0,1),\\ (T_{ϵ}y)(0)=H(\varphi (y)),(T_{ϵ}y)(1)=0,\end{array}$$

and so

$$(T_{ϵ}y)(t)\text{ is concave on }[0,1].$$

(3.3)

Moreover, from (C₂), we may estimate

$$\begin{array}{rl}{\varphi }_1(T_{ϵ}y)& ={\int }_0^1(1-t)d{\alpha }_1(t)H(\varphi (y))+{\int }_0^1{\int }_0^{1}k(t,s)q(s)f(s,max\{ϵ,y(s)\})dsd{\alpha }_1(t)\\ ={\int }_0^1(1-t)d{\alpha }_1(t)H(\varphi (y))+{\int }_0^{1}q(s)f(s,max\{ϵ,y(s)\}){\int }_0^{1}k(t,s)d{\alpha }_1(t)ds\\ \ge 0\end{array}$$

(3.4)

and

$$\begin{array}{rl}{\varphi }_2(T_{ϵ}y)& ={\int }_0^1(1-t)d{\alpha }_2(t)H(\varphi (y))+{\int }_0^1{\int }_0^{1}k(t,s)q(s)f(s,max\{ϵ,y(s)\})dsd{\alpha }_2(t)\\ ={\int }_0^1(1-t)d{\alpha }_2(t)H(\varphi (y))+{\int }_0^{1}q(s)f(s,max\{ϵ,y(s)\}){\int }_0^{1}k(t,s)d{\alpha }_2(t)ds\\ \ge 0.\end{array}$$

(3.5)

Combining (3.3), (3.4), and (3.5), $T_{ϵ}:P\to P$. A standard argument shows that $T_{ϵ}:P\to P$ is continuous and compact [9, 18, 26]. □

Define

$$\begin{array}{rl}{\mathrm{\Phi }}_r:=& \{x\in P\cap C^2((0,1),R):\parallel x\parallel \le r\text{ and }x\text{ satisfies}\\ x^{″}(t)+q(t)f(t,max\{ϵ,x(t)\})=0,0<t<1,x(0)=H(\varphi (x)),x(1)=0,\mathrm{\forall }1\ge ϵ>0\}.\end{array}$$

Lemma 3.2 If ${\mathrm{\Phi }}_r\ne \mathrm{\varnothing }$ and (C₂) hold, there exists a ${\delta }_r>0$ such that

$$x(0)\ge {\delta }_{r}t(1-t),\mathrm{\forall }t\in [0,1],x\in {\mathrm{\Phi }}_r.$$

Proof Suppose $x\in {\mathrm{\Phi }}_r$. There are two cases to consider.

(1)$\parallel x\parallel >1$. Lemma 2.4 implies that

$$x(t)\ge t(1-t)\parallel x\parallel \ge t(1-t),t\in [0,1].$$

(3.6)

(2) $0<\parallel x\parallel \le 1$. Condition (C₄) guarantees that

$$\begin{array}{rl}x(t)& =(1-t)H(\varphi (x))+{\int }_0^{1}k(t,s)q(s)f(s,max\{ϵ,x(s)\})ds\\ \ge {\int }_0^{1}k(t,s)q(s){\psi }_1(s)ds:={\gamma }_0(t),t\in [0,1].\end{array}$$

Since ${\gamma }_0^{″}(t)\ge 0$, ${\gamma }_0(0)=0$, and ${\gamma }_0(1)=0$, we know that ${\gamma }_0$ is concave on $[0,1]$ and ${\gamma }_0(t)\ge 0$ for all $t\in [0,1]$. And from (C₂), a similar argument as (3.4) and (3.5) shows that ${\varphi }_1({\gamma }_0)\ge 0$ and ${\varphi }_2({\gamma }_0)\ge 0$. Then ${\gamma }_0\in P$ and Lemma 2.4 implies that

$${\gamma }_0(t)\ge t(1-t)\parallel {\gamma }_0\parallel ,\mathrm{\forall }t\in [0,1].$$

(3.7)

Let ${\delta }_1=min\{1,\parallel {\gamma }_0\parallel \}$. From (3.6) and (3.7), one has

$$x(t)\ge {\delta }_{1}t(1-t),\mathrm{\forall }t\in [0,1],$$

which means that

$$r\ge \parallel x\parallel \ge {\delta }_1.$$

Thus

$$\varphi (x)={\int }_0^{1}x(s)d{\alpha }_1(s)+{\int }_0^{1}x(s)d{\alpha }_2(s)\le c_0\parallel x\parallel \le c_{0}r,$$

where

$$c_0\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}.}{=}{\int }_0^1|d{\alpha }_1(s)|+{\int }_0^1|d{\alpha }_2(s)|$$

and (C₁) guarantees that

$$\varphi (x)\ge {\varphi }_2(x)\ge {\epsilon }_0\parallel x\parallel .$$

And so

$$x(0)=H(\varphi (x))\ge \underset{s\in [{\epsilon }_0{\delta }_1,c_{0}r]}{min}H(s):={\delta }_r>0.$$

The concavity $x(t)$ yields

$$x(t)\ge {\delta }_r(1-t)>0,\mathrm{\forall }t\in [0,1],x\in {\mathrm{\Phi }}_r.$$

The proof is complete. □

For $R>0$, let

$${\mathrm{\Omega }}_R=\{x\in C[0,1]:\parallel x\parallel <R\}.$$

We have the following lemmas.

Lemma 3.3 Suppose that (C₁)-(C₅) hold and there exists an $a\in (0,\frac{1}{2})$ such that

$$\underset{y\to +\mathrm{\infty }}{lim}\frac{f(t,y)}{y}=+\mathrm{\infty }$$

(3.8)

uniformly on $[a,1-a]$. Then, there exists an $R^{\prime }>1$ such that for all $R\ge R^{\prime }$

$$i(T_{ϵ},{\mathrm{\Omega }}_R\cap P,P)=0,\mathrm{\forall }0<ϵ\le 1.$$

Proof From (3.8), there exists an $R_1>1$ such that

$$f(t,y)\ge N^{\ast }y,\mathrm{\forall }y\ge R_1,$$

(3.9)

where

$$N^{\ast }>\frac{2}{a^2{\int }_a^{1-a}k(a,s)q(s)ds}.$$

Let $R^{\prime }=\frac{R_1}{a^2}$ and

$${\mathrm{\Omega }}_R:=\{x\in C[0,1]:\parallel x\parallel <R\},\mathrm{\forall }R\ge R^{\prime }.$$

Now we show

$$T_{ϵ}y\nleqq y\text{for }y\in P\cap \partial {\mathrm{\Omega }}_R,\mathrm{\forall }0<ϵ\le 1.$$

(3.10)

Suppose that there exists a $y_0\in P\cap \partial {\mathrm{\Omega }}_R$ with $T_{ϵ}{y}_0\le y_0$. Then, $\parallel y_0\parallel =R$. Since $y_0(t)$ is concave on $[0,1]$ (since $y_0\in P$) we find from Lemma 2.4 that $y_0(t)\ge t(1-t)\parallel y_0\parallel \ge t(1-t)R$ for $t\in [0,1]$. For $t\in [a,1-a]$, one has

$$y_0(t)\ge a^{2}R\ge a^2{R}^{\prime }=R_1,\mathrm{\forall }t\in [a,1-a],$$

which together with (3.9) yields

$$f(t,max\{ϵ,y_0(t)\})=f(t,y_0(t))\ge N^{\ast }{y}_0(t)\ge N^{\ast }{a}^{2}R,\mathrm{\forall }t\in [a,1-a].$$

(3.11)

Then we have, using (3.11),

$$\begin{array}{rcl}{y}_0(a)& \ge & T_{ϵ}{y}_0(a)=(1-a)H(\varphi (y_0))+{\int }_0^{1}k(a,s)q(s)f(s,max\{ϵ,y_0(s)\})ds\\ \ge & {\int }_a^{1-a}k(a,s)q(s)f(s,max\{ϵ,y_0(s)\})ds\\ =& {\int }_a^{1-a}k(a,s)q(s)f(s,y_0(s))ds\\ \ge & N^{\ast }R{a}^2{\int }_a^{1-a}k(a,s)q(s)ds\\ >& R=\parallel y_0\parallel ,\end{array}$$

which is a contradiction. Hence equation (3.10) is true. Lemma 2.2 guarantees that

$$i(T_{ϵ},{\mathrm{\Omega }}_R\cap P,P)=0,\mathrm{\forall }0<ϵ\le 1.$$

The proof is complete. □

Lemma 3.4 Suppose that (C₁)-(C₅) hold and

$$\underset{s\to +\mathrm{\infty }}{lim}\frac{H(s)}{s}=+\mathrm{\infty }.$$

(3.12)

Then, there exists an $R^{\prime }>1$ such that for all $R\ge R^{\prime }$

$$i(T_{ϵ},{\mathrm{\Omega }}_R\cap P,P)=0,\mathrm{\forall }0<ϵ\le 1.$$

Proof From equation (3.12), there exists an $R_1>1$ such that

$$H(s)\ge N^{\ast }s,\mathrm{\forall }s\ge R_1,$$

(3.13)

where

$$N^{\ast }>\frac{2}{{\epsilon }_0}\left({\epsilon }_0\text{ defined in (C}{}_1\text{)}\right).$$

Let $R^{\prime }=\frac{R_1}{{\epsilon }_0}$ and

$${\mathrm{\Omega }}_R=\{x\in C[0,1]:\parallel x\parallel <R\},\mathrm{\forall }R\ge R^{\prime }.$$

Now we show

$$T_{ϵ}y\nleqq y\text{for }y\in P\cap \partial {\mathrm{\Omega }}_R,\mathrm{\forall }0<ϵ\le 1.$$

(3.14)

Suppose that there exists a $y_0\in P\cap \partial {\mathrm{\Omega }}_R$ with $T_{ϵ}{y}_0\le y_0$. Then, $\parallel y_0\parallel =R$. Now (C₁) guarantees that

$$\varphi (y_0)={\varphi }_1(y_0)+{\varphi }_2(y_0)\ge {\epsilon }_0\parallel y_0\parallel ={\epsilon }_{0}R\ge R_1,$$

which together with equation (3.13) implies that

$$y_0(0)\ge T_{ϵ}{y}_0(0)=H(\varphi (y_0))\ge N^{\ast }\varphi (y_0)>\frac{2}{{\epsilon }_0}{\epsilon }_0\parallel y_0\parallel >\parallel y_0\parallel .$$

This is a contradiction. Hence (3.14) is true. Lemma 2.2 guarantees that

$$i(T_{ϵ},{\mathrm{\Omega }}_R\cap P,P)=0,\mathrm{\forall }0<ϵ\le 1.$$

The proof is complete. □

Theorem 3.1 Suppose (C₁)-(C₅) hold and the following conditions are satisfied:

$$\{\begin{array}{c}0\le f(t,y)\le g(y)+h(y)\mathit{\text{ on }}[0,1]\times (0,\mathrm{\infty })\mathit{\text{ with}}\hfill \\ g>0\mathit{\text{ continuous and nonincreasing on }}(0,\mathrm{\infty }),\hfill \\ h\ge 0\mathit{\text{ continuous on }}[0,\mathrm{\infty }),\mathit{\text{ and}}\hfill \\ \frac{h}{g}\mathit{\text{ nondecreasing on }}(0,\mathrm{\infty })\hfill \end{array}$$

(3.15)

and

$$\underset{r\in (0,+\mathrm{\infty })}{sup}min\{\frac{1}{1+\frac{h(r)}{g(r)}}{\int }_0^r\frac{dy}{g(y)},\frac{r}{{max}_{y\in [0,c_{0}r]}H(y)}\}>max\{1,b_0\}$$

(3.16)

hold; here

$$b_0={\int }_0^1(1-s)q(s)ds,c_0={\int }_0^1|d{\alpha }_1(s)|+{\int }_0^1|d{\alpha }_2(s)|.$$

Then the BVP (1.1)-(1.2) has at least one positive solution.

Proof From equation (3.16), choose $ϵ>0$ and $r>0$ with $ϵ<min\{1,r\}$ such that

$$min\{\frac{1}{1+\frac{h(r)}{g(r)}}{\int }_0^r\frac{dy}{g(y)},\frac{r}{{max}_{y\in [0,c_{0}r]}H(y)}\}>max\{1,b_0\}.$$

(3.17)

Let

$${\mathrm{\Omega }}_1=\{y\in C[0,1]:\parallel y\parallel <r\},$$

and $n_0>\frac{1}{ϵ}$. For $n\in \{n_0,n_0+1,\dots \}$, we define $T_{\frac{1}{n}}$ as in equation (3.1). Lemma 3.1 guarantees that $T_{\frac{1}{n}}:P\to P$ is continuous and compact.

Now we show that

$$y\ne \lambda T_{\frac{1}{n}}y,\mathrm{\forall }y\in \partial {\mathrm{\Omega }}_1\cap P,\lambda \in (0,1].$$

(3.18)

Suppose that there is a $y_0\in \partial {\mathrm{\Omega }}_1\cap P$ and ${\lambda }_0\in [0,1]$ with $y_0={\lambda }_0{T}_{\frac{1}{n}}{y}_0$, i.e., $y_0$ satisfies

$$\{\begin{array}{l}{y}_0^{″}(t)+{\lambda }_{0}q(t)f(t,max\{\frac{1}{n},y_0(t)\})=0,0<t<1,\\ y_0(0)={\lambda }_{0}H(\varphi (y)),y_0(1)=0.\end{array}$$

(3.19)

Then $y_0^{″}(t)\le 0$ on $(0,1)$. From equation (3.17), we have $y_0(0)={\lambda }_{0}H(\varphi (y_0))\le {max}_{y\in [0,c_{0}r]}H(y)<r$, which together with $y_0(1)=0<r$ implies that there exists a $t_0\in (0,1)$ with $y_0(t_0)=\parallel y_0\parallel =r$, $y_0^{\prime }(t_0)=0$ and $y_0^{\prime }(t)\le 0$ for all $t\in (t_0,1)$. For $t\in (0,1)$, from equations (3.15) and (3.19), we have

$$\begin{array}{rl}-y_0^{″}(t)& \le g(max\{\frac{1}{n},y_0(t)\})\{1+\frac{h(max\{\frac{1}{n},y_0(t)\})}{g(max\{\frac{1}{n},y_0(t)\})}\}q(t)\\ \le g(max\{\frac{1}{n},y_0(t)\})\{1+\frac{h(r)}{g(r)}\}q(t).\end{array}$$

(3.20)

We integrate equation (3.20) from $t_0$ ($t_0<t$) to t to obtain

$$\begin{array}{rl}-y_0^{\prime }(t)& \le g(max\{\frac{1}{n},y_0(t)\})\{1+\frac{h(r)}{g(r)}\}{\int }_{t_0}^{t}q(s)ds\\ \le g(y_0(t))\{1+\frac{h(r)}{g(r)}\}{\int }_{t_0}^{t}q(s)ds\end{array}$$

(3.21)

and then integrate equation (3.21) from $t_0$ to 1 to obtain

$$\begin{array}{rl}{\int }_{y_0(1)}^{y_0(t_0)}\frac{dy}{g(y)}& \le \{1+\frac{h(r)}{g(r)}\}{\int }_{t_0}^1{\int }_{t_0}^{s}q(\tau )d\tau ds\\ =\{1+\frac{h(r)}{g(r)}\}{\int }_{t_0}^1(1-s)q(s)ds\\ \le \{1+\frac{h(r)}{g(r)}\}{\int }_0^1(1-s)q(s)ds,\end{array}$$

i.e.,

$${\int }_0^r\frac{dy}{g(y)}\le \{1+\frac{h(r)}{g(r)}\}{\int }_0^1(1-s)q(s)ds,$$

which contradicts equation (3.17). Therefore, equation (3.18) is true. Lemma 2.1 implies that

$$i(T_{\frac{1}{n}},{\mathrm{\Omega }}_1\cap P,P)=1,$$

which yields the result that there exists a $y_n\in {\mathrm{\Omega }}_1\cap P$ such that

$$T_{\frac{1}{n}}{y}_n=y_n,$$

i.e., ${\mathrm{\Phi }}_r\ne \mathrm{\varnothing }$ in Lemma 3.2. Now Lemma 3.2 guarantees that there exists a ${\delta }_r>0$ such that

$$y_n(0)\ge {\delta }_r,y_n(t)\ge {\delta }_r(1-t),\mathrm{\forall }t\in [0,1],x\in \{n_0,n_0+1,\dots \}.$$

(3.22)

Now we consider the set ${\{y_n\}}_{n=n_0}^{\mathrm{\infty }}$. Obviously, $\parallel y_n\parallel \le r$ means that

$$\text{the functions belonging to }\{y_n(t)\}\text{ are uniformly bounded on }[0,1].$$

(3.23)

Now we show that

$$\text{the functions belonging to }\{y_n(t)\}\text{ are equicontinuous on }[0,1].$$

(3.24)

There are two cases to consider.

(1) There exists a subsequence $\{y_{n_i}\}$ of $\{y_n\}$ with $y_{n_i}(0)=H(\varphi (y_{n_i}))<\parallel y_{n_i}\parallel$. Without loss of generality, we assume that $y_n(0)=H(\varphi (y_n))<\parallel y_n\parallel$, $n\in \{n_0,n_0+1,\dots \}$, which together with $y_n(1)=0$ implies that there exists a $t_n$ satisfying that $y_n^{\prime }(t_n)=0$ with $y_n^{\prime }(t)\ge 0$ for $t\in (0,t_n)$ and $y_n^{\prime }(t)\le 0$ for $t\in (t_n,1)$. Let $t^{\prime }=sup\{t_n,n\ge n_0\}$. Now we show that $t^{\prime }<1$. To the contrary, suppose that $t^{\prime }=1$. Then there exists a subsequence $\{n_i\}$ of $\{n\}$ such that $t_{n_i}\to 1$ as $n_i\to +\mathrm{\infty }$. From equation (3.21), using $y_n$ in place of $y_0$, we have

$${\int }_0^{y_{n_i}(t_{n_i})}\frac{1}{g(y)}dy\le (1+\frac{h(r)}{g(r)}){\int }_{t_{n_i}}^1(1-s)q(s)ds,$$

which implies that

$$y_{n_i}(t_{n_i})\to 0,\text{as }{n}_i\to +\mathrm{\infty }.$$

This contradicts $y_{n_i}(t)\ge {\delta }_r(1-t)$ for all $t\in [0,1]$.

Let $t_0\in (t^{\prime },1)$. From equation (3.22), we have

$$y_n(t)\ge k_0:=\underset{t\in [0,t_0]}{min}{\delta }_r(1-t),t\in [0,t_0].$$

Similarly as the proof in equation (3.21), one has

$$y_n^{\prime }(t)\le g(k_0)(1+\frac{h(r)}{g(r)}){\int }_0^{1}q(s)ds,$$

which means that

$$\text{the functions belonging to }\{y_n(t)\}\text{ are equicontinuous on }[0,t_0].$$

(3.25)

For $t_1,t_2\in [t_0,1)$, from equation (3.21), using $y_n$ in place of $y_0$, we have

$$\left|{\int }_{y_n(t_1)}^{y_n(t_2)}\frac{1}{g(y)}dy\right|\le (1+\frac{h(r)}{g(r)}){\int }_0^{1}q(s)ds|t_1-t_2|,$$

which yields

$$\text{the functions belonging to }\{y_n(t)\}\text{ are equicontinuous on }[t_0,1].$$

(3.26)

Combining equations (3.25) and (3.26), we find that equation (3.24) holds.

(2) There exists a $k_1>0$ such that $y_n(0)=\parallel y_n\parallel$ and $y_n(t)$ is nonincreasing on $[0,1]$ for all $n>k_1$. From $y_n(0)=H(\varphi (y_n))=\parallel y_n\parallel$ and $y_n(1)=0$, there exists $t_n\in (0,1)$ such that $y_n^{\prime }(t_n)=-H(\varphi (y_n))$. Now $y_n^{″}(t)\le 0$ implies that $y_n^{\prime }(0)\ge y_n^{\prime }(t_n)=-H(\varphi (y_n))$. Hence, from equation (3.20), using $y_n$ in place of $y_0$, we have

$$-y_n^{\prime }(t)+y_n^{\prime }(0)\le g(y_n(t))(1+\frac{h(r)}{g(r)}){\int }_0^{t}q(s)ds,t\in (0,1)$$

and so

$$\begin{array}{rl}-\frac{y_n^{\prime }(t)}{g(y_n(t))}& \le (1+\frac{h(r)}{g(r)}){\int }_0^{t}q(s)ds-\frac{y_0^{\prime }(0)}{g(y_n(t))}\\ \le (1+\frac{h(r)}{g(r)}){\int }_0^{t}q(s)ds+\frac{H(\varphi (y_n))}{g(y_n(t))}\\ \le (1+\frac{h(r)}{g(r)}){\int }_0^{t}q(s)ds+\frac{1}{g(r)}\underset{s\in [0,c_{0}r]}{max}H(r),t\in (0,1).\end{array}$$

Then

$$\begin{array}{rl}\left|{\int }_{y_n(t_1)}^{y_n(t_2)}\frac{1}{g(y)}dy\right|& =\left|{\int }_{t_1}^{t_2}\frac{y_n^{\prime }(s)}{g(y_n(s))}ds\right|\\ \le (1+\frac{h(r)}{g(r)})|{\int }_{t_1}^{t_2}{\int }_0^{s}q(\tau )d\tau ds|+\frac{1}{g(r)}\underset{s\in [0,c_{0}r]}{max}H(r)|t_1-t_2|,\\ \phantom{\le }\mathrm{\forall }{t}_1,t_2\in [0,1],\end{array}$$

which implies that (3.24) hold.

Now Arzela-Ascoli theorem guarantees that $\{y_n(t)\}$ has a convergent subsequence. Without loss of generality, we assume that there is a $y_{\ast }\in C[0,1]$ such that

$$\underset{n\to +\mathrm{\infty }}{lim}{y}_n=y_{\ast },$$

which together with equation (3.22) and $y_n(1)=0$ implies that

$$y_{\ast }(1)=0,y_{\ast }(t)\ge {\delta }_r(1-t),\mathrm{\forall }t\in [0,1].$$

(3.27)

Since $y_n$ ($n\in \mathbb{N}$) satisfies $y_n=T_{\frac{1}{n}}{y}_n$, we have

$$y_n^{″}(t)=-q(t)f(t,max\{\frac{1}{n},y_n(t)\})=0,0<t<1.$$

We integrate the above equation from $\frac{1}{2}$ to t to yield

$$y_n^{\prime }(t)=y_n^{\prime }\left(\frac{1}{2}\right)-{\int }_{\frac{1}{2}}^{t}q(s)f(s,max\{\frac{1}{n},y_n(s)\})ds,$$

and so

$$\begin{array}{rl}{y}_n(t)& =y_n\left(\frac{1}{2}\right)+y_n^{\prime }\left(\frac{1}{2}\right)(t-\frac{1}{2})-{\int }_{\frac{1}{2}}^t{\int }_{\frac{1}{2}}^{s}q(\tau )f(\tau ,max\{\frac{1}{n},y_n(\tau )\})d\tau ds\\ =y_n\left(\frac{1}{2}\right)+y_n^{\prime }\left(\frac{1}{2}\right)(t-\frac{1}{2})+{\int }_{\frac{1}{2}}^t(s-t)q(s)f(s,max\{\frac{1}{n},y_n(s)\})ds\end{array}$$

for $t\in (0,1)$ and

$$y_n(0)=H(\varphi (y_n))=H({\int }_0^1{y}_n(s)d{\alpha }_1(s)+{\int }_0^1{y}_n(s)d{\alpha }_2(s)),$$

and the Lebesgue Dominated Convergent theorem together with equation (3.27) implies that

$$\begin{array}{rl}{y}_{\ast }(t)& =\underset{n\to +\mathrm{\infty }}{lim}{y}_n(t)\\ =\underset{n\to +\mathrm{\infty }}{lim}[y_n\left(\frac{1}{2}\right)+y_n^{\prime }\left(\frac{1}{2}\right)(t-\frac{1}{2})+{\int }_{\frac{1}{2}}^t(s-t)q(s)f(s,max\{\frac{1}{n},y_n(s)\})ds]\\ =y_{\ast }\left(\frac{1}{2}\right)+y_{\ast }^{\prime }\left(\frac{1}{2}\right)(t-\frac{1}{2})+{\int }_{\frac{1}{2}}^t(s-t)q(s)f(s,y_{\ast }(s))ds\end{array}$$

(3.28)

for $t\in (0,1)$ and

$$\begin{array}{rl}{y}_{\ast }(0)& =\underset{n\to +\mathrm{\infty }}{lim}{y}_n(0)\\ =\underset{n\to +\mathrm{\infty }}{lim}H(\varphi (y_n))\\ =\underset{n\to +\mathrm{\infty }}{lim}H({\int }_0^1{y}_n(s)d{\alpha }_1(s)+{\int }_0^1{y}_n(s)d{\alpha }_2(s))\\ =H({\varphi }_1(y_{\ast })+{\varphi }_2(y_{\ast }))\\ =H(\varphi (y_{\ast })).\end{array}$$

(3.29)

We differentiate equation (3.28) to get

$$y_{\ast }^{″}(t)+q(t)f(t,y_{\ast }(t))=0,t\in (0,1),$$

which together with equations (3.27) and (3.29) means that the BVP (1.1)-(1.2) has at least one positive solution. The proof is complete. □

Theorem 3.2 Suppose the conditions of Theorem  3.1 hold and there exists an $a\in (0,\frac{1}{2})$ such that

$$\underset{y\to +\mathrm{\infty }}{lim}\frac{f(t,y)}{y}=+\mathrm{\infty }$$

uniformly on $[a,1-a]$. Then the BVP (1.1)-(1.2) has at least two positive solutions.

Proof Choose $r>0$ as in (3.17), $n_0>0$ with $\frac{1}{n_0}<min\{1,r\}$, and $R>max\{r,R^{\prime }\}$ in Lemma 3.3. Set ${\mathbb{N}}_{n_0}=\{n_0,n_0+1,\dots \}$, and

$$\begin{array}{c}{\mathrm{\Omega }}_1=\{y\in C[0,1]:\parallel y\parallel <r\},\hfill \\ {\mathrm{\Omega }}_2=\{y\in C[0,1]:\parallel y\parallel <R\}.\hfill \end{array}$$

By the proof of Theorem 3.1 and Lemma 3.3, we have

$$i(T_{\frac{1}{n}},{\mathrm{\Omega }}_1\cap P,P)=1$$

and

$$i(T_{\frac{1}{n}},{\mathrm{\Omega }}_2\cap P,P)=0,$$

which implies that

$$i(T_{\frac{1}{n}},({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)\cap P,P)=-1.$$

Then, there exist $x_{1,n}\in {\mathrm{\Omega }}_1\cap P$ and $x_{2,n}\in ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)\cap P$ such that

$$T_{\frac{1}{n}}{x}_{1,n}=x_{1,n},T_{\frac{1}{n}}{x}_{2,n}=x_{2,n}.$$

By the proof of Theorem 3.1, there exist a subsequence $\{x_{1,n_i}\}$ of $\{x_{1,n}\}$ and $x_1\in P$ such that

$$\underset{n_i\to +\mathrm{\infty }}{lim}{x}_{1,n_i}(t)=x_1(t),t\in [0,1].$$

And moreover, $x_1(t)$ is a positive solution to the BVP (1.1)-(1.2) with $r>x_1(t)\ge {\delta }_r(1-t)$, $\mathrm{\forall }t\in [0,1]$.

A similar argument shows that there exist a subsequence $\{x_{2,n_j}\}$ of $\{x_{2,n}\}$ and $x_2\in P\cap ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)$ such that

$$\underset{n_i\to +\mathrm{\infty }}{lim}{x}_{2,n_j}(t)=x_2(t),t\in [0,1].$$

And moreover, $x_2(t)$ is a positive solution to the BVP (1.1)-(1.2) and equation (3.18) guarantees that $\parallel x_2\parallel >r$. Hence, $x_1(t)$ and $x_2(t)$ are two positive solutions for the BVP (1.1)-(1.2). The proof is complete. □

Theorem 3.3 Suppose the conditions of Theorem  3.1 hold and

$$\underset{s\to +\mathrm{\infty }}{lim}\frac{H(s)}{s}=+\mathrm{\infty }.$$

Then the BVP (1.1)-(1.2) has at least two positive solutions.

Proof Choose $r>0$ as in (3.17), $n_0>0$ with $\frac{1}{n_0}<min\{1,r\}$, and $R>max\{r,R^{\prime }\}$ in Lemma 3.4. Set ${\mathbb{N}}_{n_0}=\{n_0,n_0+1,\dots \}$, and

$$\begin{array}{c}{\mathrm{\Omega }}_1=\{y\in C[0,1]:\parallel y\parallel <r\},\hfill \\ {\mathrm{\Omega }}_2=\{y\in C[0,1]:\parallel y\parallel <R\}.\hfill \end{array}$$

By the proof of Theorem 3.1 and Lemma 3.4, we have

$$i(T_{\frac{1}{n}},{\mathrm{\Omega }}_1\cap P,P)=1$$

and

$$i(T_{\frac{1}{n}},{\mathrm{\Omega }}_2\cap P,P)=0,$$

which implies that

$$i(T_{\frac{1}{n}},({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)\cap P,P)=-1.$$

Then, there exist $x_{1,n}\in {\mathrm{\Omega }}_1\cap P$ and $x_{2,n}\in ({\mathrm{\Omega }}_2-{\overline{\mathrm{\Omega }}}_1)\cap P$ such that

$$T_{\frac{1}{n}}{x}_{1,n}=x_{1,n},T_{\frac{1}{n}}{x}_{2,n}=x_{2,n}.$$

A similar argument to that in Theorem 3.2 shows that the BVP (1.1)-(1.2) has at least two positive solutions. The proof is complete. □

Example 3.1 Consider

$$y^{″}(t)+\mu \frac{1}{\sqrt{1-t}}(\frac{1}{200}+\frac{1}{300}sin{t}^2+\frac{1}{100}{y}^{-{\delta }_1}(t)+\frac{1}{100}{y}^{{\delta }_2}(t))=0,0<t<1,$$

(3.30)

with

$$y(0)=H(\varphi (y)),y(1)=0,$$

(3.31)

where

$$H(t)=\frac{1}{2}t+\frac{1}{3}{t}^{\frac{1}{3}},\varphi (y)={\varphi }_1(y)+{\varphi }_2(y)={\int }_0^{1}y(s)d{\alpha }_1(s)+{\int }_0^{1}y(s)d{\alpha }_2(s),$$

with

$$\begin{array}{r}d{\alpha }_1(s)=\frac{1}{8}cos2\pi sds,d{\alpha }_2(s)=\frac{1}{8}d{e}^s,\\ {\delta }_1>0,{\delta }_2>1,\frac{100}{({\delta }_1+1)3}>1.\end{array}$$

(3.32)

Then equations (3.30)-(3.31) have at least two positive solutions.

To prove that the BVP (3.30)-(3.31) has at least two positive solutions, we use Theorem 3.2. Let $q(t)=\mu \frac{1}{\sqrt{1-t}}$, $f(t,y)=\frac{1}{200}+\frac{1}{300}sin{t}^2+\frac{1}{100}{y}^{-{\delta }_1}+\frac{1}{100}{y}^{{\delta }_2}$, $g(y)=\frac{1}{100}{y}^{-{\delta }_1}$, $h(y)=\frac{1}{100}+\frac{1}{100}{y}^{{\delta }_2}$, $c_0={\int }_0^1|d{\alpha }_1(s)|+{\int }_0^1|d{\alpha }_2(s)|=\frac{1}{4\pi }+\frac{e-1}{8}$, $b_0=\frac{2}{3}\mu$. For $y\in P$ (defined in (2.1)), we have

$${\varphi }_2(y)={\int }_0^{1}y(t)\frac{1}{8}{e}^{s}ds\ge \parallel y\parallel {\int }_0^{1}s(1-s)\frac{1}{8}{e}^{s}ds,$$

which means that (C₁) holds. Since

$$\begin{array}{c}{\int }_0^1(1-t)d{\alpha }_1(t)=0,{\int }_0^1(1-t)d{\alpha }_2(t)>0,\hfill \\ {\int }_0^{1}k(t,s)d{\alpha }_1(t)=(1-s){\int }_0^{s}td{\alpha }_1(t)+s{\int }_s^1(1-t)d{\alpha }_1(t)=\frac{1-cos2\pi s}{32{\pi }^2}\ge 0,\hfill \end{array}$$

and

$${\int }_0^{1}k(t,s)d{\alpha }_2(t)=(1-s){\int }_0^{s}td{\alpha }_2(t)+s{\int }_s^1(1-t)d{\alpha }_2(t)\ge 0,$$

(C₂) is true. Since $c_0<1$, we have ${max}_{y\in [0,c_{0}r]}H(y)=\frac{1}{2}{c}_{0}r+\frac{1}{3}{(c_{0}r)}^{\frac{1}{3}}\le \frac{1}{2}r+\frac{1}{3}{r}^{\frac{1}{3}}$. Then

$$\frac{1}{{max}_{y\in [0,c_{0}1]}H(y)}=\frac{1}{\frac{1}{2}{c}_{0}1+\frac{1}{3}{(c_{0}1)}^{\frac{1}{3}}}>1.$$

Equation (3.32) guarantees that

$$\frac{1}{1+\frac{h(1)}{g(1)}}{\int }_0^1\frac{1}{g(y)}dy=\frac{100}{3(1+{\delta }_1)}>1.$$

Letting ${\mu }_0<3$, we have

$$\underset{r\in (0,+\mathrm{\infty })}{sup}min\{\frac{1}{1+\frac{h(r)}{g(r)}}{\int }_0^r\frac{dy}{g(y)},\frac{r}{{max}_{y\in [0,c_{0}r]}H(y)}\}>max\{1,b_0\},$$

for all $\mu \le {\mu }_0$, which means that equations (3.15)-(3.16) hold. Since

$$f(t,x)\ge \frac{1}{200}+\frac{1}{300}sin{t}^2,\mathrm{\forall }(t,x)\in [0,1]\times (0,1],$$

we get (C₄). Moreover, since

$$\underset{y\to +\mathrm{\infty }}{lim}\frac{f(t,y)}{y}=+\mathrm{\infty }$$

uniformly on $[0,1]$, all conditions of Theorem 3.2 hold, which implies that equations (3.30)-(3.31) have at least two positive solutions.

Example 3.2 Consider

$$y^{″}(t)+\mu y^{-{\delta }_1}(t)=0,0<t<1,$$

(3.33)

with

$$y(0)=H(\varphi (y)),y(1)=0,$$

(3.34)

where

$$H(t)=\frac{1}{2}{t}^3+\frac{1}{3}{t}^{\frac{1}{3}},\varphi (y)={\varphi }_1(y)+{\varphi }_2(y)={\int }_0^{1}y(s)d{\alpha }_1(s)+{\int }_0^{1}y(s)d{\alpha }_2(s),$$

with

$$d{\alpha }_1(s)=\frac{1}{8}cos2\pi sds,d{\alpha }_2(s)=\frac{1}{8}d{e}^s,{\delta }_1>0.$$

Then equations (3.33)-(3.34) have at least two positive solutions.

To prove that the BVP (3.33)-(3.34) has at least two positive solutions, we use Theorem 3.3. Let $q(t)=\mu$, $f(t,y)=y^{-{\delta }_1}$, $g(y)=y^{-{\delta }_1}$, $h(y)=0$, $c_0=\frac{1}{4\pi }+\frac{e-1}{8}$, $b_0=\frac{1}{2}\mu$. Since $c_0<1$, we have ${max}_{y\in [0,c_{0}r]}H(y)=\frac{1}{2}{(c_{0}r)}^3+\frac{1}{3}{(c_{0}r)}^{\frac{1}{3}}\le \frac{1}{2}{r}^3+\frac{1}{3}{r}^{\frac{1}{3}}$. Then

$$\frac{1}{{max}_{y\in [0,c_{0}1]}H(y)}=\frac{1}{\frac{1}{2}{(c_{0}1)}^3+\frac{1}{3}{(c_{0}1)}^{\frac{1}{3}}}>1.$$

Also we have

$$\underset{r\to +\mathrm{\infty }}{lim}{\int }_0^r\frac{dy}{g(y)}{(1+\frac{h(r)}{g(r)})}^{-1}=+\mathrm{\infty }.$$

Then, letting ${\mu }_0\le 2$, we get

$$\underset{r\in (0,+\mathrm{\infty })}{sup}min\{\frac{1}{1+\frac{h(r)}{g(r)}}{\int }_0^r\frac{dy}{g(y)},\frac{r}{{max}_{y\in [0,c_{0}r]}H(y)}\}>max\{1,b_0\},$$

for all $\mu \le {\mu }_0$, which means that equations (3.15)-(3.16) hold. Since

$$f(t,x)\ge 1,\mathrm{\forall }(t,x)\in [0,1]\times (0,1],$$

we get (C₄). Obviously, (C₁)-(C₃), and (C₅) hold. Moreover, since

$$\underset{y\to +\mathrm{\infty }}{lim}\frac{H(s)}{s}=+\mathrm{\infty }$$

uniformly on $[0,1]$, all conditions of Theorem 3.3 hold, which implies that equations (3.30)-(3.31) have at least two positive solutions.

4 Positive solutions for singular boundary-value problems with sign-changing nonlinearities

• (H₁) Assume that there are three linear functionals $\varphi ,{\varphi }_1,{\varphi }_2:C([0,1])\to R$

  $$\varphi (y)={\varphi }_1(y)+{\varphi }_2(y),{\varphi }_1(y):={\int }_0^{1}y(t)d{\alpha }_1(t),{\varphi }_2(y):={\int }_0^{1}y(t)d{\alpha }_2(t),$$

  where ${\alpha }_1,{\alpha }_2:[0,1]\to R$ satisfy ${\alpha }_1,{\alpha }_2\in BV([0,1])$;

• (H₂) $a(t)\in C([0,1],(0,+\mathrm{\infty }))$, $(1-t)q(t)\in L^1((0,1])$;

• (H₃) Let $H:R\to [0,+\mathrm{\infty })$ be a real-valued, continuous function. Moreover, $H:(0,+\mathrm{\infty })\to (0,+\mathrm{\infty })$;

• (H₄) $f(t,y)\in C([0,1]\times (0,+\mathrm{\infty }),(-\mathrm{\infty },+\mathrm{\infty }))$, there exists a decreasing function $F(y)\in C((0,+\mathrm{\infty }),(0,+\mathrm{\infty }))$, and a nonnegative function $G(y)\in C([0,+\mathrm{\infty }),[0,+\mathrm{\infty }))$ such that $f(t,y)\le F(y)+G(y)$ and there exists a $b\in C((0,1),(0,+\mathrm{\infty }))$ such that

  $$f(t,y)\ge a(t),\mathrm{\forall }0<y\le b(t),t\in (0,1);$$

• (H₅) there exist $R>1$ such that

  $${\int }_0^R\frac{dy}{F(y)}\cdot {(1+\frac{\overline{G}(R)}{F(R)})}^{-1}>{\int }_0^1(1-s)q(s)ds$$

  and

  $$\underset{y\in [0,r{c}_0]}{max}H(y)<r,\mathrm{\forall }R\ge r>0,\text{where }{c}_0={\int }_0^1|d{\alpha }_1(s)|+{\int }_0^1|d{\alpha }_2(s)|,$$

  where $\overline{G}(R)={max}_{s\in [0,R]}G(s)$.

For $n>3$, let $b_n=min\{\frac{1}{n},{min}_{t\in [\frac{1}{n},1-\frac{1}{n}]}b(t)\}$. Obviously, $b_n>0$. For $y\in C_n=C[\frac{1}{n},1-\frac{1}{n}]$, we define $T_n$ as

$$\begin{array}{r}(T_{n}y)(t)=(1-\frac{1}{n}-t)H({\varphi }_n(y))+b_n+{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{k}_n(t,s)q(s)f(s,max\{b_n,y(s)\})ds,\\ t\in [\frac{1}{n},1-\frac{1}{n}],\end{array}$$

where

$$k_n(t,s)=\{\begin{array}{ll}(s-\frac{1}{n})(1-\frac{1}{n}-t),& \frac{1}{n}\le s\le t\le 1-\frac{1}{n},\\ (t-\frac{1}{n})(1-\frac{1}{n}-s),& \frac{1}{n}\le t\le s\le 1-\frac{1}{n}\end{array}$$

and

$${\varphi }_n(y)={\int }_{\frac{1}{n}}^{1-\frac{1}{n}}y(s)d{\alpha }_1(s)+{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}y(s)d{\alpha }_2(s).$$

From a standard argument (see [18, 25, 26]), we have the following result.

Lemma 4.1 Suppose (H₁)-(H₄) hold. Then the operator $T_n$ is continuous and compact from $C_n$ to $C_n$.

From (H₃) and (H₅), there exists ${ϵ}_0>0$ such that

$$\begin{array}{r}{\int }_{{ϵ}_0}^R\frac{dy}{F(y)}\cdot {(1+\frac{\overline{G}(R)}{F(R)})}^{-1}>{\int }_0^1(1-s)q(s)ds,\\ \underset{y\in [0,c_{0}R]}{max}H(y)+{ϵ}_0<R.\end{array}$$

(4.1)

Choose $n_0>3$ with $\frac{1}{n_0}<{ϵ}_0$ and let ${\mathbb{N}}_{n_0}=\{n_0,n_0+1,\dots \}$. Now we have the following lemmas.

Lemma 4.2 Suppose (H₁)-(H₅) hold. Then, for $n\in {\mathbb{N}}_0$, there exists a $x_n\in C_n$ with $b_n\le x_n(t)\le R$ such that

$$x_n(t)=(1-\frac{1}{n}-t)H({\varphi }_n(x_n))+b_n+{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{k}_n(t,s)q(s)f(s,x_n(s))ds,t\in [\frac{1}{n},1-\frac{1}{n}].$$

Proof Let $\mathrm{\Omega }=\{y\in C_n:\parallel y\parallel <R\}$. For $y\in \partial \mathrm{\Omega }$, we now prove that

$$\begin{array}{rl}y(t)\ne & \lambda (T_{n}y)(t)=\lambda ((1-\frac{1}{n}-t)H({\varphi }_n(y))+b_n)\\ +\lambda {\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{k}_n(t,s)q(s)f(s,max\{b_n,y(s)\})ds,t\in [\frac{1}{n},1-\frac{1}{n}]\end{array}$$

(4.2)

for any $\lambda \in (0,1]$.

Suppose equation (4.2) is not true. Then there exists $y\in C[\frac{1}{n},1-\frac{1}{n}]$ with $\parallel y\parallel =R$ and $0<\lambda <1$ such that

$$\begin{array}{rl}y(t)=& \lambda (Ty)(t)=\lambda ((1-\frac{1}{n}-t)H({\varphi }_n(y))+b_n)\\ +\lambda {\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{k}_n(t,s)q(s)f(s,max\{b_n,y(s)\})ds,t\in [\frac{1}{n},1-\frac{1}{n}].\end{array}$$

(4.3)

We first claim that $y(t)\ge \lambda b_n$ for any $t\in [\frac{1}{n},1-\frac{1}{n}]$.

Suppose there exists a $\eta \in (0,1)$ with $y(\eta )<\lambda b_n$. Let ${\gamma }_0=inf\{t_1:y(s)<\lambda b_n,\mathrm{\forall }s\in [t_1,\eta ]\}$ and ${\gamma }_1=sup\{t_1:y(s)<\lambda b_n,\mathrm{\forall }s\in [\eta ,t_1]\}$. Since $y(\frac{1}{n})\ge \lambda b_n$ and $y(1-\frac{1}{n})=\lambda b_n$, we have ${\gamma }_0\ge \frac{1}{n}$, ${\gamma }_1\le 1-\frac{1}{n}$, $y({\gamma }_0)=y({\gamma }_1)=\lambda b_n$, and $y(t)<\lambda b_n$ for all $t\in ({\gamma }_0,{\gamma }_1)$, which implies that

$$y^{″}(t)=-\lambda q(t)f(t,b_n)<0,t\in ({\gamma }_0,{\gamma }_1)$$

and so $y(t)$ is concave down on $[{\gamma }_0,{\gamma }_1]$. This is a contradiction.

Now (H₅) guarantees that

$$y\left(\frac{1}{n}\right)=\lambda ((1-\frac{2}{n})H({\varphi }_n(y))+b_n)\le \underset{r\in [0,c_{0}R]}{max}h(r)+{ϵ}_0<R,$$

which together with $y(1-\frac{1}{n})=\lambda b_n<R$ means that there is a $t\in (\frac{1}{n},1-\frac{1}{n})$ with $y^{\prime }(t)=0$ and $y(t)=R$. Let $t^{\ast }=sup\{t:y(t)=R,y^{\prime }(t)=0\}$ and $t_{\ast }=inf\{t:y(t)=R,y^{\prime }(t)=0\}$. Obviously, $\frac{1}{n}<t_{\ast }\le t^{\ast }<1-\frac{1}{n}$, $y(t_{\ast })=R$, $y^{\prime }(t_{\ast })=0$, $y(t^{\ast })=R$, $y^{\prime }(t^{\ast })=0$, $y(t)<R$ for all $t\in (t^{\ast },1-\frac{1}{n}]$ and $y(t)<R$ for all $t\in (\frac{1}{n},t_{\ast }]$. Let $t_1=inf\{t^{\ast }<t\le 1-\frac{1}{n}:y(t)=\lambda y(1-\frac{1}{n})\}$ and $t_1^{\prime }=sup\{t<t_{\ast }\le 1-\frac{1}{n}:y(t)=\lambda y(\frac{1}{n})\}$. It is easy to see that $t^{\ast }<t_1\le 1-\frac{1}{n}$, $y(t)>y(t_1)$ for all $t\in (t^{\ast },t_1)$, $t_1^{\prime }<t_{\ast }$ and $y(t)>y(t_1^{\prime })$ for all $t\in (t_1^{\prime },t_{\ast })$.

Now we consider the properties of y on $(t^{\ast },t_1)$. We get a countable set $\{t_i\}$ of $(t^{\ast },t_1]$ such that

• 1. $t^{\ast }>\cdots \ge t_{2m}>t_{2m-1}>\cdots >t_5\ge t_4>t_3\ge t_2>t_1=t_1$, $t_{2m}\to t^{\ast }$,

• 2. $y(t_{2i})=y(t_{2i+1})$, $y^{\prime }(t_{2i})=0$, $i=1,2,3,\dots$ ,

• 3. $y(t)$ is strictly decreasing in $[t_{2i},t_{2i-1}]$, $i=1,2,3,\dots$ (if $y(t)$ is strictly decreasing in $[t^{\ast },t_1]$, put $m=1$; i.e, $[t_2,t_1]=[t^{\ast },t_1]$).

Differentiating equation (4.3) and using the assumptions (H₂) and (H₄), we obtain

$$\begin{array}{rl}-y^{″}(t)& =\lambda q(t)f(t,max\{b_n,y(t)\})\\ \le \lambda q(t)(F(max\{b_n,y(t)\})+G(max\{b_n,y(t)\}))\\ =\lambda q(t)F(max\{b_n,y(t)\})(1+\frac{G(max\{b_n,y(t)\})}{F(max\{b_n,y(t)\})})\\ <q(t)F(max\{b_n,y(t)\})(1+\frac{\overline{G}(R)}{F(R)})\\ \le q(t)F(y(t))(1+\frac{\overline{G}(R)}{F(R)}),t\in [t_{2i},t_{2i-1}),i=1,2,3,\dots .\end{array}$$

(4.4)

Integrating (4.4) from $t_{2i}$ to t, we have, by the decreasing property of $F(y)$,

$$-{\int }_{t_{2i}}^t{y}^{″}(s)ds\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_{2i}}^{t}q(s)F(y(s))ds\le F(y(t))(1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_{2i}}^{t}q(s)ds,$$

for $t\in [t_{2i},t_{2i-1})$, $i=1,2,3,\dots$ ; that is to say,

$$-y^{\prime }(t)\le F(y(t))(1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_{2i}}^{t}q(s)ds,t\in [t_{2i},t_{2i-1}),i=1,2,3,\dots .$$

(4.5)

It follows from equation (4.5) that

$$-\frac{y^{\prime }(t)}{F(y(t))}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_{2i}}^{t}q(s)ds\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_0^{t}q(s)ds,$$

(4.6)

for $t\in [t_{2i},t_{2i-1})$, $i=1,2,3,\dots$ .

On the other hand, for any $z\in (\frac{1}{n},1-\frac{1}{n})$ with $y(z)>\lambda b_n$, we can choose $i_0$ and $z^{\prime }\in (t^{\ast },t_1)$ such that $z^{\prime }\in [t_{2i_0},t_{2i_0-1})$, $y(z^{\prime })=y(z)$ and $z\le z^{\prime }$. Integrating equation (4.6) from $t_{2i}$ to $t_{2i-1}$, $i=1,2,3,\dots ,i_0-1$ and from $t_{2i_0}$ to $z^{\prime }$, we have

$${\int }_{y(t_{2i-1})}^{y(t_{2i})}\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_{2i}}^{t_{2i-1}}{\int }_0^{t}q(s)dsdt,i=1,2,3,\dots ,i_0-1,$$

(4.7)

and

$${\int }_{y(t_{2i_0})}^{y(z^{\prime })}\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{z^{\prime }}^{t_{2i_0}}{\int }_0^{t}q(s)dsdt.$$

(4.8)

Summing equation (4.7) from 1 to $i_0-1$, we have by equation (4.8) and $y(t_{2i})=y(t_{2i+1})$

$${\int }_{y(t_1)}^{y(z^{\prime })}\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{z^{\prime }}^{t_1}{\int }_0^{t}q(s)dsdt\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_z^{t_1}{\int }_0^{t}q(s)dsdt.$$

Since $y(z)=y(z^{\prime })$,

$${\int }_{y(t_1)}^{y(z)}\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_z^{t_1}{\int }_0^{t}q(s)dsdt.$$

(4.9)

For the properties of y on $(t_1^{\prime },t_{\ast })$, a similar argument shows that for any $z>t_1^{\prime }$

$${\int }_{y(t_1^{\prime })}^{y(z)}\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_1^{\prime }}^z{\int }_0^{t}q(s)dsdt.$$

(4.10)

Letting $z\to t^{\ast }$ in (4.9), we have

$$\begin{array}{rl}{\int }_{{ϵ}_0}^R\frac{dy}{F(y)}& \le {\int }_{y(t_1)}^R\frac{dy}{F(y)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t^{\ast }}^{t_1}{\int }_0^{t}q(s)dsdt\\ \le (1+\frac{\overline{G}(R)}{F(R)}){\int }_0^1{\int }_0^{t}q(s)dsdt\\ =(1+\frac{\overline{G}(R)}{F(R)}){\int }_0^1(1-s)q(s)ds,\end{array}$$

which contradicts equation (4.1). Hence equation (4.2) holds.

It follows from Lemma 3.2 that $T_n$ has a fixed point $x_n$ in $C_n$. Using $x_n$ and 1 in place of y and λ in (4.3), we obtain easily $b_n\le x_n(t)\le R$, $t\in [\frac{1}{n},1-\frac{1}{n}]$. And $x_n$ satisfies

$$\begin{array}{r}{x}_n(t)=(1-\frac{1}{n}-t)H({\varphi }_n(x_n))+b_n+{\int }_0^1{k}_n(t,s)q(s)f(s,x_n(s))ds,\\ t\in [\frac{1}{n},1-\frac{1}{n}].\end{array}$$

(4.11)

The proof is complete. □

Lemma 4.3 Suppose that all conditions of Lemma  4.2 hold and $x_n$ satisfies (4.11). For a fixed $h\in (0,min\{\frac{1}{2},\eta \})$, let $m_{n,h}=min\{x_n(t),t\in [h,1-h]\}$. Then $m_h=inf\{m_{n,h}\}>0$.

Proof Since $x_n(t)\ge b_n>0$, we get $m_h\ge 0$. For any fixed natural number n ($n>n_0$ defined in Lemma 4.2), let $t_n\in [h,1-h]$ such that $x_n(t_n)=min\{x_n(t),t\in [h,1-h]\}$. If $m_h=0$, there exists a countable set $\{n_i\}$ such that

$$\underset{n_i\to +\mathrm{\infty }}{lim}{x}_{n_i}(t_{n_i})=0.$$

So there exists $N_0$ such that $x_{n_i}(t_{n_i})\le min\{b(t),t\in [\frac{h}{2},1-h]\}$, $n_i>N_0$. Let ${\overline{\mathbb{N}}}_0=\{n_0>N_0:n\in {\mathbb{N}}_0\text{ with }{lim}_{n_i\to +\mathrm{\infty }}{x}_{n_i}(t_{n_i})=0\}$. Then we have two cases.

Case 1. There exist $n_k\in {\overline{\mathbb{N}}}_0$ and $t_{n_k}^{\ast }\in [\frac{h}{2},h]$ such that $x_{n_k}(t_{n_k}^{\ast })\ge x_{n_k}(t_{n_k})$. By the same argument in Lemma 4.2, we can get $t_{n_k}^{\prime },t_{n_k}^{″}\in [\frac{h}{2},1]$, $t_{n_k}^{\prime }<t_{n_k}^{″}$ such that

$$\begin{array}{r}{x}_{n_k}(t)\le min\{b(t),t\in [\frac{h}{2},1]\},t\in [t_{n_k}^{\prime },t_{n_k}^{″}],\\ x_{n_k}(t)\le x_{n_k}\left(t_{n_k}^{\prime }\right),x_{n_k}(t)\le x_{n_k}\left(t_{n_k}^{″}\right),t\in (t_{n_k}^{\prime },t_{n_k}^{″}),\end{array}$$

(4.12)

and

$$x_{n_k}^{″}(t)=-q(t)f(t,x_{n_k}(t))<0,t\in (t_{n_k}^{\prime },t_{n_k}^{″}).$$

(4.13)

The inequality (4.13) shows that $x_{n_k}(t)$ is concave down in $[t_{n_k}^{\prime },t_{n_k}^{″}]$, which contradicts equation (4.12).

Case 2. $x_{n_i}(t)<x_{n_i}(t_{n_i})$, $t\in [\frac{h}{2},h]$ for any $n_i\in {\overline{\mathbb{N}}}_0$. And so we have

$$\underset{n_i\to +\mathrm{\infty }}{lim}{x}_{n_i}(t)=0,t\in [\frac{h}{2},h].$$

(4.14)

On the other hand, for any $t\in [\frac{h}{2},h]$,

$$\begin{array}{rl}{x}_{n_i}(t)=& \frac{2}{h}{\int }_{\frac{h}{2}}^t(s-\frac{h}{2})(h-t)q(s)f(s,x_{n_i}(s))ds\\ +\frac{2}{h}{\int }_t^h(t-\frac{h}{2})(h-s)q(s)f(s,x_{n_i}(s))ds+x_{n_i}\left(\frac{h}{2}\right)+x_{n_i}(h)\\ \ge & \frac{2}{h}[{\int }_{\frac{h}{2}}^t(s-\frac{h}{2})(h-t)a(s)ds+{\int }_t^h(t-\frac{h}{2})(h-s)a(s)ds]>0,\end{array}$$

which contradicts equation (4.14). Hence, $m_h>0$. The proof is complete. □

Theorem 4.1 If (H₁)-(H₅) hold, then BVP (1.1)-(1.2) has at least one positive solution.

Proof For any natural number $n\in \mathbb{N}$ (defined in Lemma 4.2), it follows from Lemma 4.2 that there exist $x_n\in C_n$, $b_n\le x_n(t)\le R$ for all $t\in [\frac{1}{n},1-\frac{1}{n}]$ satisfying (4.11). Now we divide the proof into three steps.

Step 1. There exists a convergent subsequence of $\{x_n\}$ in $(0,1)$. For a natural number $k\ge n_0$ in Lemma 4.2, it follows from Lemma 4.3 that $0<m_{\frac{1}{k}}\le x_n(t)\le R$, $t\in [\frac{1}{k},1-\frac{1}{k}]$ for any natural numbers $n\in N$; i.e., $\{x_n\}$ is uniformly bounded in $[\frac{1}{k},1-\frac{1}{k}]$. Since $x_n$ also satisfies

$$\begin{array}{rl}{x}_n(t)=& \frac{1}{1-\frac{2}{k}}{\int }_{\frac{1}{k}}^t(s-\frac{1}{k})(1-\frac{1}{k}-t)q(s)f(s,x_n(s))ds\\ +\frac{1}{1-\frac{2}{k}}{\int }_t^{1-\frac{1}{k}}(t-\frac{1}{k})(1-\frac{1}{k}-s)q(s)f(s,x_n(s))ds+x_n\left(\frac{1}{k}\right)+x_n(1-\frac{1}{k}),\end{array}$$

we have

$$\begin{array}{rl}{x}_n^{\prime }(t)=& -\frac{1}{1-\frac{2}{k}}{\int }_{\frac{1}{k}}^t(s-\frac{1}{k})q(s)f(s,x_n(s))ds\\ +\frac{1}{1-\frac{2}{k}}{\int }_t^{1-\frac{1}{k}}(1-\frac{1}{k}-s)q(s)f(s,x_n(s))ds.\end{array}$$

Obviously

$$|x_n^{\prime }(t)|\le 2(1-\frac{2}{k})max\{q(t)\left|f(t,x_n(t))\right|:(t,x)\in [\frac{1}{k},1-\frac{1}{k}]\times [m_{\frac{1}{k}},R]\},$$

(4.15)

for $t\in [\frac{1}{k},1-\frac{1}{k}]$. It follows from inequality (4.15) that $\{x_n\}$ is equicontinuous in $[\frac{1}{k},1-\frac{1}{k}]$. The Ascoli-Arzela theorem guarantees that there exists a subsequence of $\{x_n(t)\}$ which converges uniformly on $[\frac{1}{k},1-\frac{1}{k}]$. Then, for $k=n_0$, we choose a convergent subsequence of $\{x_n\}$ on $[\frac{1}{n_0},1-\frac{1}{n_0}]$,

$$x_{n_1(n_0)}(t),x_{n_2(n_0)}(t),x_{n_3(n_0)}(t),\dots ,x_{n_k(n_0)}(t),\dots ;$$

for $k=n_0+1$, we choose a convergent subsequence of $\{x_{n_k(n_0)}\}$ on $[\frac{1}{n_0+1},1-\frac{1}{n_0+1}]$,

$$x_{n_1(n_0+1)}(t),x_{n_2(n_0+1)}(t),x_{n_3(n_0+1)}(t),\dots ,x_{n_k(n_0+1)}(t),\dots ;$$

for $k=n_0+2$, we choose a convergent subsequence of $\{x_{n_k(n_0+1)}\}$ on $[\frac{1}{n_0+2},1-\frac{1}{n_0+2}]$,

$$\begin{array}{c}{x}_{n_1(n_0+2)}(t),x_{n_2(n_0+2)}(t),x_{n_3(n_0+2)}(t),\dots ,x_{n_k(n_0+2)}(t),\dots ;\hfill \\ \dots ,\dots ,\dots ,\dots ;\hfill \end{array}$$

for $k=n_0+j$, we choose a convergent subsequence of $\{x_{n_k(n_0+j-1)}\}$ on $[\frac{1}{n_0+j},1-\frac{1}{n_0+j}]$,

$$\begin{array}{c}{x}_{n_1(n_0+j)}(t),x_{n_2(n_0+j)}(t),x_{n_3(n_0+j)}(t),\dots ,x_{n_k(n_0+j)}(t),\dots ;\hfill \\ \dots ,\dots ,\dots ,\dots .\hfill \end{array}$$

We may choose the diagonal sequence $\{x_{n_{k+1}(n_0+k)}(t)\}$ which converges everywhere in $(0,1)$ and it is easy to verify that $\{x_{n_{k+1}(n_0+k)}(t)\}$ converges uniformly on any interval $[c,d]\subseteq (0,1)$. Without loss of generality, let $\{x_{n_{k+1}(n_0+k)}(t)\}$ be $\{x_n(t)\}$ in the rest. Putting $x(t)={lim}_{n\to +\mathrm{\infty }}{x}_n(t)$, $t\in (0,1)$, we have $x(t)$ continuous in $(0,1)$ and $x(t)\ge m_h>0$, $t\in [h,1-h]$ for any $h\in (0,\frac{1}{2})$ by Lemma 4.3.

Step 2. $x(t)$ satisfies equation (1.1). Fixed $t\in (0,1)$, we may choose $h\in (0,\frac{1}{2})$ such that $t\in (h,1-h)$ and

$$\begin{array}{rl}{x}_n(t)=& \frac{1}{1-2h}{\int }_h^t(s-h)(1-h-t)q(s)f(s,x_n(s))ds\\ +\frac{1}{1-2h}{\int }_t^{1-h}(t-h)(1-h-s)q(s)f(s,x_n(s))ds+x_n(h)+x_n(1-h).\end{array}$$

Letting $n\to +\mathrm{\infty }$ in above equation, we have

$$\begin{array}{rl}x(t)=& \frac{1}{1-2h}{\int }_h^t(s-h)(1-h-t)q(s)f(s,x(s))ds\\ +\frac{1}{1-2h}{\int }_t^{1-h}(t-h)(1-h-s)q(s)f(s,x(s))ds+x(h)+x(1-h).\end{array}$$

(4.16)

Differentiating equation (4.16), we get the desired result.

Step 3. $x(t)$ satisfies equation (1.2). Let

$$t_n=sup\{t:x_n(t)=\parallel x_n\parallel ,x_n^{\prime }(t)=0,t\in [\frac{1}{n},1-\frac{1}{n}]\}$$

and

$$t_n^{\prime }=inf\{t:x_n(t)=\parallel x_n\parallel ,x_n^{\prime }(t)=0,t\in [\frac{1}{n},1-\frac{1}{n}]\},$$

where $\parallel x_n\parallel ={max}_{\frac{1}{n}\le t\le 1-\frac{1}{n}}{x}_n(t)\le R$. Then

$$t_n,t_n^{\prime }\in [\frac{1}{n},1-\frac{1}{n}],x_n(t_n)=x_n\left(t_n^{\prime }\right)=\parallel x_n\parallel ,x_n^{\prime }(t_n)=x_n^{\prime }\left(t_n^{\prime }\right)=0.$$

Using $x_n(t)$, 1, $t_n$ in place of $y(t)$, λ and $t^{\ast }$ in Lemma 4.2, from equation (4.9); we have

$${\int }_{b_n}^{\parallel x_n\parallel }\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{t_n}^{1-\frac{1}{n}}{\int }_0^{t}q(s)dsdt$$

and using $x_n(t)$, 1, $t_n^{\prime }$ in place of $y(t)$, λ and $t_{\ast }$ in Lemma 4.2, from equation (4.10), we obtain easily

$${\int }_{x_n(\frac{1}{n})+b_n}^{\parallel x_n\parallel }\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{\frac{1}{n}}^{t_n^{\prime }}{\int }_0^{t}q(s)dsdt.$$

It follows from the above inequalities that $a=inf\{t_n^{\prime }\}>0$ and $b=sup\{t_n\}<1$.

• (1) Fixing $z\in (b,1)$, we get $b_n<x_n(z)<\parallel x_n\parallel \le R$. From equation (4.9) of the proof in Lemma 4.2, one easily has

  $${\int }_{b_n}^{x_n(z)}\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_z^{1-\frac{1}{n}}{\int }_0^{t}q(s)dsdt,z\in (b,1).$$

Letting $n\to +\mathrm{\infty }$ in the above inequality and noticing $b_n\to 0$, we have

$${\int }_0^{x(z)}\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_z^1{\int }_0^{t}q(s)dsdt,z\in (b,1).$$

(4.17)

It follows from equation (4.17) that $x(1)={lim}_{z\to 1^{-}}x(z)=0$.

• (2) Fixing $z\in (0,a)$, we get $x_n(\frac{1}{n})+b_n<x_n(z)<\parallel x_n\parallel \le R$. From equation (4.10) in the proof of Lemma 4.2, we easily get

  $${\int }_{x_n(\frac{1}{n})+b_n}^{x_n(z)}\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_{\frac{1}{n}}^z{\int }_0^{t}q(s)dsdt,z\in (0,a).$$

  (4.18)

Since ${lim}_{n\to +\mathrm{\infty }}{x}_n(t)=x(t)$ and $\parallel x_n\parallel \le R$, the Lebesgue Dominated Convergent theorem guarantees that

$$\underset{n\to +\mathrm{\infty }}{lim}{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{x}_n(t)d{\alpha }_1(t)={\int }_0^{1}x(t)d{\alpha }_1(t),\underset{n\to +\mathrm{\infty }}{lim}{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{x}_n(t)d{\alpha }_2(t)={\int }_0^{1}x(t)d{\alpha }_2(t).$$

Since H is continuous, we have

$$\underset{n\to +\mathrm{\infty }}{lim}{x}_n\left(\frac{1}{n}\right)=\underset{n\to +\mathrm{\infty }}{lim}(1-\frac{2}{n})H({\varphi }_n(x_n))=H(\varphi (x)).$$

(4.19)

Letting $n\to +\mathrm{\infty }$ in equation (4.18) and noticing $b_n\to 0$ and equation (4.19), we have

$${\int }_{H(\varphi (x))}^{x(z)}\frac{dx}{F(x)}\le (1+\frac{\overline{G}(R)}{F(R)}){\int }_0^z{\int }_0^{t}q(s)dsdt,z\in (0,a).$$

(4.20)

It follows from equation (4.20) that $x(0)={lim}_{z\to 0+}x(z)=H(\varphi (x))$. This complete the proof. □

Example 4.1 Consider

$$y^{″}(t)+\frac{1}{8}(\frac{1}{217}{y}^2(t)+\frac{1}{100}(\frac{1}{y^2(t)}-\frac{y^3(t)}{t^{10}}-\frac{3}{t^4}))=0,0<t<1,$$

(4.21)

with boundary conditions

$$y(0)=\frac{1}{100}|{\int }_0^{1}y(s)d{\alpha }_1(s)+{\int }_0^{1}y(s)d{\alpha }_2(s){|}^3,y(1)=0,$$

(4.22)

where

$$d{\alpha }_1(s)=-\frac{1}{10}cos4\pi sds,d{\alpha }_2(s)=\frac{1}{9}(e^s-2)ds.$$

Then the BVP (4.21)-(4.22) has at least one positive solution.

Let $q(t)=\frac{1}{8}$, $f(t,y)=\frac{1}{217}{y}^2+\frac{1}{100}(\frac{1}{y^2}-\frac{y^3}{t^{10}}-\frac{3}{t^4})$, $G(y)=\frac{1}{217}{y}^2$, $F(y)=\frac{1}{100y^2}$, $b(t)=\frac{1}{2}{t}^2$, $a(t)=\frac{7}{8t^4}$. Let $R=2$ and $H(y)=\frac{1}{100}{|y|}^3$. We have

$$\begin{array}{c}{\int }_0^2\frac{1}{F(y)}dy{(1+\frac{G(2)}{F(2)})}^{-1}>\frac{200}{9}>\frac{1}{16}={\int }_0^1(1-s)q(s)ds,\hfill \\ \underset{y\in [0,c_{0}r]}{max}H(r)=\frac{1}{100}{(c_{0}r)}^3<r,\mathrm{\forall }r\in (0,2],\hfill \end{array}$$

where $c_0={\int }_0^1|d{\alpha }_1(s)|+{\int }_0^1|d{\alpha }_2(s)|<1$ and

$$f(t,y)\ge a(t),\mathrm{\forall }0<y\le b(t),t\in (0,1).$$

Then (H₁)-(H₅) hold. Now Theorem 4.1 guarantees that the BVP (4.21)-(4.22) has at least one positive solution.