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# Existence and iterative solutions of a new kind of Sturm-Liouville-type boundary value problem with one-dimensional p-Laplacian

## Abstract

We study a kind of Sturm-Liouville-type four-point boundary value problems. The main tool is monotone iteration theory.

## Introduction

In this paper, we are concerned with the following Sturm-Liouville-type four-point boundary value problem with one-dimensional p-Laplacian:

$$\textstyle\begin{cases} (\phi_{p}(x'(t)))'+h(t)f(t,x(t),x'(t))=0,\quad 0< t< 1, \\ x'(0)-\alpha x(\xi)=0,\qquad x'(1)+\beta x(\eta)=0, \end{cases}$$
(1.1)

where $$\phi_{p}(s)=\vert s\vert ^{p-2}s$$, $$p>1$$, $$0<\alpha\leq\frac{ 1}{ \xi}$$, $$0<\beta\leq\frac{ 1}{ 1-\eta}$$, $$0<\xi<\eta<1$$. By applying the monotone iterative technique, we not only prove the existence of positive solutions for the problem, but also establish iterative schemes for approximating the solutions.

We will assume throughout:

(C1):

$$h(t)\in L(0,1)$$ is nonnegative on $$(0,1)$$ and is not identically zero on any subset of $$(0,1)$$.

(C2):

$$f\in C([0,1]\times[0,+\infty)\times R, [0,+\infty))$$, $$f(t,0,0)\not\equiv0$$ for $$0\leq t\leq1$$.

Boundary value problems (BVPs) have been studied for a long period. At the beginning, most researchers focused on two-point BVPs with four classical boundary conditions (BCs) of Dirichlet type $$u(0)=u(1)=0$$, Neumann type $$u'(0)=u'(1)=0$$, Robin type $$u(0)=u'(1)=0$$ or $$u'(0)=u(1)=0$$, and Sturm-Liouville type $$\alpha u(0)-\beta u'(0)=0$$, $$\gamma u(1)+\delta u'(1)=0$$. Later, in order to meet the requirements of various applications, some researchers began to pay their attentions on multipoint BVPs, such as three-point BC $$u(0)=\alpha u(\eta)$$, $$u(1)=0$$ or $$u'(0)=0$$, $$u(1)=\alpha u(\eta)$$, and so on. Although the points involved are larger than that involved in two-point BC, the difficulties remain similar. However, when we study this kind of four-point BVPs, difficulties have a qualitative leap.

Recently, some research articles on the theory of positive solutions to multipoint BVPs have appeared . More recently, in , BVPs subject to the boundary conditions

$$\alpha x(0)-\beta x'(\xi)=0,\qquad \gamma x(1)+\delta x'( \eta)=0$$
(1.2)

(Sturm-Liouville-type BC) were studied. Notice that BC in equation (1.1) can also be seen as a Sturm-Liouville-type BC. However, to the best knowledge of the authors, such a kind of BVPs has been rarely considered up to now. The reason is that it is not easy to convert BVP (1.1) to its equivalent integral equation. In this paper, we overcome this difficulty and also get its iterative solutions. The main tool is the monotone iterative technique. For more references, we refer the readers to .

## Background material

In the following, there are some lemmas.

### Definition 2.1

A map α is said to be a nonnegative concave continuous function if α: $$P\rightarrow [0,\infty)$$ is continuous and

$$\alpha \bigl(\lambda x+(1-\lambda)y \bigr)\geqslant\lambda\alpha(x)+(1-\lambda) \alpha(y)$$

for all $$x, y\in P$$ and $$0\leqslant\lambda\leqslant1$$.

By $$\phi_{q}$$ we denote the inverse to $$\phi_{p}$$, where $$\frac { 1}{ p}+\frac{1}{ q}=1$$. Consider the following BVP:

$$\textstyle\begin{cases} (\phi_{p}(x'(t)))'+v(t)=0,\quad 0< t< 1, \\ x'(0)-\alpha x(\xi)=0,\qquad x'(1)+\beta x(\eta)=0. \end{cases}$$
(2.1)

Let

\begin{aligned}& B_{1}(t)=\frac{1}{\alpha} \phi_{q} \biggl( \int_{0}^{t}v(s)\,\mathrm{d} s \biggr)+ \int _{\xi}^{t}\phi_{q} \biggl( \int_{s}^{t}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s, \\& B_{2}(t)=\frac{1}{\beta}\phi_{q} \biggl( \int_{t}^{1}v(s)\,\mathrm{d} s \biggr)+ \int _{t}^{\eta}\phi_{q} \biggl( \int_{t}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s. \end{aligned}

### Lemma 2.1

Suppose that $$v\in L[0,1]$$, $$v(t)\geq0$$, and $$v(t)\not\equiv0$$ on any subinterval of $$[0,1]$$. Then BVP (2.1) has the unique solution

$$x(t)= \textstyle\begin{cases} \frac{1}{\alpha}\phi_{q}(\int_{0}^{\sigma_{x}}v(s)\,\mathrm{d} s)+ \int_{\xi}^{t}\phi_{q}(\int_{s}^{\sigma_{x}}v(\tau)\,\mathrm{d} \tau)\,\mathrm{d} s, & t\in[0,\sigma_{x}],\qquad\hphantom{1} (2.2_{1}) \\ \frac{1}{\beta}\phi_{q}(\int_{\sigma_{x}}^{1}v(s)\,\mathrm{d} s)+\int _{t}^{\eta}\phi_{q}(\int_{\sigma_{x}}^{s} v(\tau)\,\mathrm{d} \tau)\,\mathrm{d} s, & t\in[\sigma_{x},1],\qquad\hphantom{0} (2.2_{2}) \end{cases}$$
(2.2)

where $$\sigma_{x}$$ is a solution of the equation

$$B_{1}(t)-B_{2}(t)=0,\quad t\in[0,1].$$
(2.3)

### Proof

We first prove that the solution of (2.1) can be expressed as (2.2). Let x be a solution of BVP (2.1). Then $$(\phi_{p}(x'(t)))'=-v(t)\leq0$$ means that $$x'(t)$$ is nonincreasing. We show that $$x'(0)>0>x'(1)$$, which implies that there exists a point $$\sigma\in(0,1)$$ such that $$x'(\sigma)=0$$.

If not, then, for example, $$x'(0)\le0$$. Then $$x'(t)\le0$$ on $$[0,1]$$ and $$x'(1)<0$$ at the same time. Considering $$\xi<\eta$$, we have $$x(\eta)\leq0$$. Then from the boundary condition in (2.1) we have $$x'(1)\geq0$$, a contradiction.

Integrating both sides of

$$- \bigl(\phi_{p} \bigl(x'(t) \bigr) \bigr)'=v(t)$$
(2.4)

from σ to t, we get

$$\phi_{p} \bigl(x'(t) \bigr)=- \int_{\sigma}^{t}v(s)\,\mathrm{d} s.$$

Then

$$x'(t)=-\phi_{q} \biggl( \int_{\sigma}^{t}v(s)\,\mathrm{d} s \biggr),$$
(2.5)

where q is given by $$\frac{1}{p}+\frac{1}{q}=1$$.

Integrating both sides of (2.5) from t to 1, we have

$$x(t)=x(1)+ \int_{t}^{1}\phi_{q} \biggl( \int_{\sigma}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s.$$
(2.6)

By (2.5) and (2.6) we have

\begin{aligned}& x'(1)=-\phi_{q} \biggl( \int_{\sigma}^{1}v(s)\,\mathrm{d} s \biggr), \\& x(\eta)=x(1)+ \int_{\eta}^{1}\phi_{q} \biggl( \int_{\sigma}^{s}v(\tau )\,\mathrm{d} \tau \biggr)\, \mathrm{d} s. \end{aligned}

Considering the BC in (2.1), we have

$$x(1)=\frac{1}{\beta} \biggl(\phi_{q} \biggl( \int_{\sigma }^{1}v(s)\,\mathrm{d} s \biggr) \biggr)- \int_{\eta}^{1}\phi_{q} \biggl( \int_{\sigma}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s.$$
(2.7)

Substituting (2.7) into (2.6), we obtain

\begin{aligned} x(t) =&\frac{1}{\beta} \biggl( \phi_{q} \biggl( \int_{\sigma }^{1}v(s)\,\mathrm{d} s \biggr) \biggr)- \int_{\eta}^{1}\phi_{q} \biggl( \int_{\sigma}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s+ \int_{t}^{1}\phi_{q} \biggl( \int _{\sigma}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \\ =&\frac{1}{\beta} \biggl(\phi_{q} \biggl( \int_{\sigma}^{1}v(s)\,\mathrm{d} s \biggr) \biggr)+ \int_{t}^{\eta}\phi_{q} \biggl( \int_{\sigma}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s ,\quad t \in[0,1]. \end{aligned}
(2.8)

By a similar argument we have

$$x(t)=\frac{1}{\alpha} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma }v(s)\,\mathrm{d} s \biggr) \biggr)+ \int_{\xi}^{t}\phi_{q} \biggl( \int_{s}^{\sigma} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s,\quad t\in[0,1].$$
(2.9)

Let $$t=\sigma$$ in (2.8) and (2.9). Then $$B_{1}(\sigma)=B_{2}(\sigma)$$, that is, σ can be determined by $$B_{1}(t)-B_{2}(t)=0$$. Next, we show that such a σ is unique.

Clearly, $$B_{1}(t)-B_{2}(t)$$ is increasing on $$t\in[0,1]$$. It can be easily seen that $$B_{1}(0)-B_{2}(0)<0$$ and $$B_{1}(1)-B_{2}(1)>0$$. Indeed,

\begin{aligned} B_{1}(0)&= \int_{\xi}^{0}\phi_{q} \biggl( \int_{s}^{0}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s = \int_{0}^{\xi}\phi_{q} \biggl( \int_{0}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \\ &< \int_{0}^{\eta}\phi_{q} \biggl( \int_{0}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \leq B_{2}(0). \end{aligned}

Thus, $$B_{1}(0)-B_{2}(0)<0$$. Similarly, $$B_{1}(1)-B_{2}(1)>0$$. Therefore, $$B_{1}(t)$$ and $$B_{2}(t)$$ must intersect at one point in $$(0,1)$$, which solves (2.3), that is, σ exists and is unique. This also implies that $$x(t)$$ defined by (2.1) is continuous at σ.

Since σ has something to do with x, we denote σ by $$\sigma_{x}$$.

Hence, for $$t\in[0,1]$$, the solution of (2.1) can be expressed as (2.2), which completes the proof. □

### Remark 2.1

In fact, for any $$t\in[0,1]$$, the solution of (2.1) can be expressed both by (2.21) and (2.22), but just for convenience, we write it in two parts.

### Lemma 2.2

Let $$v(t)$$ satisfy all the conditions in Lemma  2.1. Then the solution $$x(t)$$ of BVP (2.1) is concave on $$t\in[0,1]$$. Moreover, $$x(t)\geq0$$.

### Proof

Since $$(\phi_{p}(x'(t)))'=-v(t)\leq0$$, we have $$x''(t)\leq0$$, so $$x(t)$$ is concave on $$t\in[0,1]$$.

Next, we prove that $$x(t)\geq0$$. By Lemma 2.1 we know that $$x(t)$$ can be expressed as (2.2). When $$t\in[\sigma_{x},1]$$, since $$0<\beta\leq\frac{ 1}{ 1-\eta}$$, that is, $$\frac{ 1}{ \beta}\geq1-\eta$$, we have

\begin{aligned} (2.2_{2})&=\frac{1}{\beta} \phi_{q} \biggl( \int_{\sigma_{x}}^{1}v(s)\,\mathrm{d} s \biggr)+ \int_{t}^{\eta}\phi_{q} \biggl( \int_{\sigma_{x}}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \\ &=\frac{1}{\beta}\phi_{q} \biggl( \int_{\sigma_{x}}^{1}v(s)\,\mathrm{d} s \biggr)- \int _{\eta}^{1}\phi_{q} \biggl( \int_{\sigma_{x}}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s+ \int_{t}^{1}\phi_{q} \biggl( \int_{\sigma_{x}}^{s}v(\tau )\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \\ &\geq \int_{\eta}^{1}\phi_{q} \biggl( \int_{\sigma_{x}}^{1}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s- \int_{\eta}^{1}\phi_{q} \biggl( \int_{\sigma _{x}}^{1}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s+ \int_{t}^{1}\phi_{q} \biggl( \int_{\sigma_{x}}^{s}v(\tau )\,\mathrm{d} \tau \biggr)\, \mathrm{d} s \\ &\geq \int_{t}^{1}\phi_{q} \biggl( \int_{\sigma_{x}}^{s}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s\geq0. \end{aligned}

Similarly, when $$t\in[0,\sigma_{x}]$$ and $$0<\alpha\leq\frac { 1}{ \xi}$$, we get $$(2.2_{1})\geq0$$. Thus, $$x(t)\geq0$$ for all $$t\in[0,1]$$. The proof is complete. □

Let $$X=C^{1}[0,1]$$ be endowed with the maximum norm, $$\Vert x\Vert =\max\{\Vert x\Vert _{1},\Vert x'\Vert _{1}\}$$, where $$\Vert x\Vert _{1}=\max_{0\leq t\leq1} \vert x(t)\vert$$. Define the cone $$P\subset X$$ as

\begin{aligned} P =& \bigl\{ x\in X: x \mbox{ is concave on } t\in [0,1], \\ &{} \mbox{and there exists one point } \sigma_{x}\in (0,1)\mbox{ such that } x'(\sigma_{x})=0 \bigr\} . \end{aligned}

For $$x, y\in P$$, by $$x\le y$$ we mean that $$x(t)\le y(t)$$ and $$\vert x'(t)\vert \le \vert y'(t)\vert$$ for $$t\in[0,1]$$.

Define $$T:P\rightarrow X$$ as follows:

$$(Tx) (t)= \textstyle\begin{cases} \frac{1}{\alpha}\phi_{q}(\int_{0}^{\sigma _{x}}h(s)f(s,x(s),x'(s))\,\mathrm{d} s) \\ \quad{} +\int_{\xi}^{t}\phi_{q}(\int_{s}^{\sigma_{x}}h(\tau)f(\tau,x(\tau),x'(\tau))\,\mathrm{d} \tau )\,\mathrm{d} s, & t\in[0,\sigma_{x}],\\ \frac{1}{\beta}\phi_{q}(\int_{\sigma _{x}}^{1}h(s)f(s,x(s),x'(s))\,\mathrm{d} s) \\ \quad{}+\int_{t}^{\eta}\phi_{q}(\int_{\sigma_{x}}^{s}h(\tau)f(\tau,x(\tau),x'(\tau))\,\mathrm{d} \tau )\,\mathrm{d} s, & t\in[\sigma_{x},1]. \end{cases}$$
(2.10)

### Lemma 2.3

For $$x\in P$$, $$x(t)\geq\min\{ t,1-t\}\max_{0\leq t\leq1}\vert x(t)\vert$$.

### Lemma 2.4

Suppose that (C1) and (C2) hold. Then $$T:P\rightarrow P$$ is completely continuous.

### Proof

We divide the proof into three steps.

Step 1. We first show that $$T:P\rightarrow P$$ is well defined. Let $$x\in P$$. Then Tx is concave on $$t\in[0,1]$$. Indeed, by (2.4),

$$(Tx)'(t)= \textstyle\begin{cases} \phi_{q}(\int_{t}^{\sigma_{x}}h(s)f(s,x(s),x'(s))\,\mathrm{d} s), & t\in[0,\sigma_{x}],\\ -\phi_{q}(\int_{\sigma_{x}}^{t}h(s)f(s,x(s),x'(s))\,\mathrm{d} s), & t\in[\sigma_{x},1]. \end{cases}$$
(2.11)

Obviously, $$(Tx)''(t)\leq0$$, that is, Tx is concave on $$t\in[0,1]$$. Further, $$(Tx)'(t)\geq0$$ on $$t\in[0,\sigma_{x}]$$, $$(Tx)'(t)\leq0$$ on $$t\in[\sigma_{x},1]$$, and $$(Tx)'(\sigma_{x})=0$$. Thus, $$T:P\rightarrow P$$ is well defined.

Step 2. T is continuous. Let $$x_{n}\to x_{0}$$ in P. Similarly to Lemma 2.1, there exists a unique $$\sigma_{x_{n}}$$ such that $$W_{1,n}(\sigma_{x_{n}})=W_{2,n}(\sigma_{x_{n}})$$, where

\begin{aligned}& W_{1,n}(t)=\frac{1}{\alpha} \phi_{q} \biggl( \int_{0}^{\sigma _{x_{n}}}v(s)\,\mathrm{d} s \biggr)+ \int_{\xi}^{t}\phi_{q} \biggl( \int_{s}^{\sigma _{x_{n}}}v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s, \\& W_{2,n}(t)=\frac{1}{\beta}\phi_{q} \biggl( \int_{\sigma _{x_{n}}}^{1}v(s)\,\mathrm{d} s \biggr)+ \int_{\sigma_{x_{n}}}^{\eta}\phi_{q} \biggl( \int_{t}^{s} v(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s. \end{aligned}

Meanwhile, we can obtain that $$\sigma_{x_{n}}\to \sigma_{x_{0}}$$ ($$n\to+\infty$$), $$W_{i,n}\to W_{i,0}$$ ($$n\to+\infty$$), $$i=1,2$$. Let $$\underline{\sigma}_{n}=\min\{\sigma_{x_{n}},\sigma_{x_{0}}\}$$ and $$\overline{\sigma}_{n}=\max\{\sigma_{x_{n}},\sigma_{x_{0}}\}$$, $$n=1,2,\ldots$$ . Obviously, when $$t\in\Delta_{n}=[\underline{\sigma}_{n},\overline{\sigma}_{n}]$$, $$t-\sigma_{x_{0}}\to0$$ as $$n\to+\infty$$. Noticing that

\begin{aligned} \max_{t\in\Delta_{n}} \bigl\vert W_{i,n}(t)-W_{j,0}(t) \bigr\vert \le&\max _{t\in\Delta _{n}} \bigl\vert W_{i,n}(t)-W_{i,n}( \sigma_{x_{n}}) \bigr\vert + \bigl\vert W_{j,n}( \sigma_{x_{n}})-W_{i,n}( \sigma_{x_{n}}) \bigr\vert \\ &{}+\max_{t\in\Delta_{n}} \bigl\vert W_{j,0}( \sigma_{x_{0}})-W_{j,0}(t) \bigr\vert \quad\text{as } n\to+ \infty,i,j=1,2,i\neq j, \end{aligned}

we have

\begin{aligned} & \max_{t\in[0,1]}\vert Tx_{n}-Tx_{0}\vert \\ &\quad =\max \bigl\{ \vert W_{1,n}-W_{1,0}\vert _{[0,\underline {\sigma}_{n}]}, \vert W_{2,n}-W_{1,0}\vert _{\Delta_{n}},\vert W_{1,n}-W_{2,0}\vert _{\Delta _{n}},\vert W_{2,n}-W_{2,0}\vert _{[\overline{\sigma}_{n},1]} \bigr\} \\ &\quad\to0\quad\text{as } n\to+\infty. \end{aligned}

Similarly, by (2.11) and the continuity of $$\phi_{q}$$ we can prove that

$$\max_{t\in[0,1]} \bigl\vert (Tx_{n})'-(Tx_{0})' \bigr\vert \to0\quad\text{as } n\to +\infty.$$

Thus, T is continuous.

It is easy to prove that $$T(D)$$ is bounded and equicontinuous, where $$D\subset P$$ is a bounded set. By the Arzelà-Ascoli theorem, $$T(D)$$ is relatively compact. So $$T: P\to P$$ is completely continuous. □

### Lemma 2.5

Suppose that (C1) and (C2) hold. Then T is increasing with respect to $$x\in P$$.

### Proof

Suppose $$x_{1},x_{2}\in P$$, $$x_{1}\le x_{2}$$. Then $$x_{1}(t)\le x_{2}(t)$$ and $$\vert x_{1}'(t)\vert \le \vert x_{2}'(t)\vert$$. Let us prove that $$Tx_{1}\le Tx_{2}$$. According to the definition of P, we know that there exists $$\sigma_{x_{2}}\in(0,1)$$ such that $$x_{2}'(\sigma_{x_{2}})=0$$, and considering $$\vert x_{1}'(t)\vert \le \vert x_{2}'(t)\vert$$, we have $$x_{1}'(\sigma_{x_{2}})=0$$, which means that $$\sigma_{x_{1}}=\sigma_{x_{2}}$$. In what follows, we try to prove that $$Tx_{1}\le Tx_{2}$$.

For convenience, we give the notation

$$F_{i}(t)=h(t)f \bigl(t,x_{i}(t),x_{i}'(t) \bigr),\quad i=1,2.$$

If $$t\in[0,\sigma_{x_{1}}(\sigma_{x_{2}})]$$, then, in view of (C2), we have

\begin{aligned}& (Tx_{2}) (t)-(Tx_{1}) (t) =\frac{1}{\alpha} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma _{x_{2}}}F_{2}(s)\,\mathrm{d} s \biggr) - \phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}}F_{1}(s)\,\mathrm{d} s \biggr) \biggr) \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{} + \int _{\xi}^{t} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{1}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)} =\frac{1}{\alpha} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma _{x_{2}}}F_{2}(s)\,\mathrm{d} s \biggr) - \phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}}F_{1}(s)\,\mathrm{d} s \biggr) \biggr) \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{} + \int _{\xi}^{t} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)} =\frac{1}{\alpha} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma _{x_{2}}}F_{2}(s)\,\mathrm{d} s \biggr) - \phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}}F_{1}(s)\,\mathrm{d} s \biggr) \biggr) \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{} - \int_{0}^{\xi} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{}+ \int_{0}^{t} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)}\ge \int_{0}^{\xi} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}}F_{2}(\tau )\,\mathrm{d} \tau \biggr) -\phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}}F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{}- \int_{0}^{\xi} \biggl(\phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{0}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)=}{}+ \int_{0}^{t} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s \\& \hphantom{(Tx_{2}) (t)-(Tx_{1}) (t)}= \int_{0}^{t} \biggl(\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s-\phi_{q} \biggl( \int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau \biggr) \biggr)\,\mathrm{d} s\ge0, \\& (Tx_{2})'(t)-(Tx_{1})'(t) = \phi_{q} \biggl( \int_{t}^{\sigma _{x_{2}}}F_{2}(s)\,\mathrm{d} s \biggr)- \phi_{q} \biggl( \int_{t}^{\sigma_{x_{1}}}F_{1}(s)\,\mathrm{d} s \biggr) \\& \hphantom{(Tx_{2})'(t)-(Tx_{1})'(t)} =\phi_{q} \biggl( \int_{t}^{\sigma_{x_{2}}}F_{2}(s)\,\mathrm{d} s \biggr)- \phi_{q} \biggl( \int _{t}^{\sigma_{x_{2}}}F_{1}(s)\,\mathrm{d} s \biggr) \ge0. \end{aligned}

If $$t\in[\sigma_{x_{1}}(\sigma_{x_{2}}),1]$$, then we can similarly prove that $$(Tx_{2})(t)-(Tx_{1})(t)\ge0$$ and $$(Tx_{2})'(t)-(Tx_{1})'(t)\le0$$.

To sum up, we have $$Tx_{1}\le Tx_{2}$$, which is the desired result. The proof is complete. □

### Remark 2.2

We can easily verify that $$\phi_{q}(\int_{s}^{\sigma_{x_{2}}} F_{2}(\tau)\,\mathrm{d} \tau)\,\mathrm{d} s-\phi_{q}(\int_{s}^{\sigma_{x_{2}}} F_{1}(\tau)\,\mathrm{d} \tau)$$ is nonincreasing with respect to $$s\in[0,\sigma_{x_{2}}]$$ by calculating its derivative.

## The existence of positive solutions

Let

\begin{aligned} \lambda =& \max \biggl\{ \frac{1}{\alpha} \phi_{q} \biggl( \int_{0}^{\frac {1}{2}}h(s)\,\mathrm{d} s \biggr)+ \int_{\xi}^{\frac{1}{2}} \phi_{q} \biggl( \int_{s}^{\frac{1}{2}}h(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s, \frac{1}{\beta}\phi_{q} \biggl( \int_{\frac{1}{2}}^{1}h(s)\,\mathrm{d} s \biggr) \\ &{}+ \int_{\frac{1}{2}}^{\eta} \phi_{q} \biggl( \int_{\frac{1}{2}}^{s}h(\tau)\,\mathrm{d} \tau \biggr)\, \mathrm{d} s, \phi_{q} \biggl( \int_{0}^{1}h(s)\,\mathrm{d} s \biggr) \biggr\} \cdot \biggl(\max \biggl\{ \frac{1}{\alpha}, \frac{1}{\beta} \biggr\} + \eta+ \frac{5}{4} \biggr). \end{aligned}

### Theorem 3.1

Assume that (C1) and (C2) hold. Further, suppose that there exists $$r>0$$ such that:

(C3):

$$f(t,u_{1},v_{1})\leq f(t,u_{2},v_{2})$$ for any $$0\leq t\leq1$$, $$0\leq u_{1} \leq u_{2}\leq r$$, $$0\leq \vert v_{1}\vert \leq \vert v_{2}\vert \leq r$$;

(C4):

$$\max_{t\in[0,1]}f(t,r,r)\le \phi_{p}(\frac{r}{\lambda})$$.

Then the boundary value problem (1.1) has at least two positive solutions $$w^{*}$$ and $$v^{*}$$ in P such that

$$0< w^{*}\leq r,\qquad 0< \bigl\vert \bigl(w^{*} \bigr)' \bigr\vert \leq r,$$

and

$$\lim_{n\rightarrow\infty}w_{n}=\lim _{n\rightarrow \infty}T^{n}w_{0}=w^{*},\qquad \lim_{n\rightarrow\infty}(w_{n})'=\lim _{n\rightarrow\infty } \bigl(T^{n}w_{0} \bigr)'= \bigl(w^{*} \bigr)',$$

where

$$w_{0}(t)= \frac{r}{\lambda} \biggl(\max \biggl\{ \frac{1}{\alpha}, \frac{1}{\beta} \biggr\} +\eta +t+1-t^{2} \biggr) \cdot \phi_{q} \biggl( \int_{0}^{1}h(s)\,\mathrm{d} s \biggr)$$

and

$$0< v^{*}\leq r,\qquad 0< \bigl\vert \bigl(v^{*} \bigr)' \bigr\vert \leq r,$$

and

$$\lim_{n\rightarrow\infty}v_{n}=\lim _{n\rightarrow \infty}T^{n}v_{0}=v^{*},\qquad \lim_{n\rightarrow\infty}(v_{n})'=\lim _{n\rightarrow\infty } \bigl(T^{n}v_{0} \bigr)'= \bigl(v^{*} \bigr)',$$

where $$v_{0}(t)=0$$, $$0\leq t\leq1$$.

### Proof

Let $$\overline{P}_{r}=\{u\in P\mid \Vert u\Vert \leq r\}$$. First, we prove that $$T: \overline{P}_{r}\rightarrow\overline{P}_{r}$$. For any $$u\in \overline{P}_{r}$$, $$\Vert u\Vert \leq r$$, we have

$$0\leq u(t)\leq\max_{0\leq t\leq1} \bigl\vert u(t) \bigr\vert \leq \Vert u\Vert \leq r,\qquad \bigl\vert u'(t) \bigr\vert \leq \max_{0\leq t\leq1} \bigl\vert u'(t) \bigr\vert \leq \Vert u\Vert \leq r.$$

Then considering (C1)-(C4), we get

$$0\leq f \bigl(t,u(t),u'(t) \bigr)\leq f(t,r,r)\leq\max _{0\leq t\leq 1}f(t,r,r)\leq\phi_{p} \biggl(\frac{r}{\lambda } \biggr).$$

By (2.4) and (2.5) we obtain

\begin{aligned} (Tu) (\sigma_{x}) =&\frac{1}{\alpha} \phi_{q} \biggl( \int_{0}^{\sigma _{x}}h(s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d} s \biggr) \\ &{}+ \int_{\xi}^{\sigma_{x}} \phi_{q} \biggl( \int_{s}^{\sigma_{x}}h(\tau)f \bigl(\tau,x( \tau),x'(\tau) \bigr)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s \\ \leq&\max \biggl\{ \frac{1}{\alpha}\phi_{q} \biggl( \int_{0}^{\frac {1}{2}}h(s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d} s \biggr) \\ &{}+ \int_{\xi}^{\frac{1}{2}} \phi_{q} \biggl( \int_{s}^{\frac{1}{2}}h(\tau)f \bigl(\tau,x( \tau),x'(\tau ) \bigr)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s, \frac{1}{\beta}\phi_{q} \biggl( \int_{\frac {1}{2}}^{1}h(s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d} s \biggr) \\ &{}+ \int_{\frac{1}{2}}^{\eta} \phi_{q} \biggl( \int_{\frac{1}{2}}^{s}h(\tau)f \bigl(\tau,x( \tau),x'(\tau ) \bigr)\,\mathrm{d} \tau \biggr)\,\mathrm{d} s \biggr\} \\ \leq&\frac{r}{\lambda}\cdot\lambda=r, \end{aligned}

and

\begin{aligned}& (Tu)'(0) =\phi_{q} \biggl( \int_{0}^{\sigma_{x}}h(s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d} s \biggr) \\& \hphantom{(Tu)'(0)}\leq\phi_{q} \biggl( \int_{0}^{1}h(s)f \bigl(s,x(s),x'( s) \bigr)\,\mathrm{d} s \biggr) \leq\frac{r}{\lambda}\cdot\lambda= r, \\& -(Tu)'(1)=\phi_{q} \biggl( \int_{\sigma_{x}}^{1}h(s)f \bigl(s,x(s),x'(s) \bigr)\,\mathrm{d} s \biggr) \\& \hphantom{-(Tu)'(1)} \leq\phi_{q} \biggl( \int_{0}^{1}h(s)f \bigl(s,x(s),x '(s) \bigr)\,\mathrm{d} s \biggr)\leq\frac{r}{\lambda}\cdot\lambda= r. \end{aligned}

Thus, we obtain that $$\Vert Tu\Vert \leq r$$. So, we have shown that $$T:\overline{P}_{r}\rightarrow\overline{P}_{r}$$.

Second, we will establish iterative schemes for approximating the solutions. Let

$$w_{0}(t)= \frac{r}{\lambda} \bigl(-t^{2}+t+c \bigr) \cdot \phi_{q} \biggl( \int_{0}^{1}h(s)\,\mathrm{d} s \biggr),$$

where $$c=\frac{1}{\alpha}+\frac{1}{\alpha}+1+\frac{1}{\beta}$$. Obviously, $$w_{0}(t)\in P$$ and $$w_{0}'(\frac{1}{2})=0$$. Let $$w_{1}(t)=Tw_{0}(t)$$. Then we have $$w_{1}\in\overline{P}_{r}$$. We denote $$w_{n+1}=Tw_{n}=T^{n}w_{0}$$, $$n=1,2,\ldots$$ . Then we have $$w_{n}\in\overline{P}_{r}$$. Since T is completely continuous, $$\{w_{n}\}_{n=1}^{\infty}$$ is a sequentially compact set. We have

\begin{aligned} w_{1}(t) =&Tw_{0}(t) \\ =& \textstyle\begin{cases} \frac{1}{\alpha}\phi_{q}(\int_{0}^{\sigma_{x}}h(s)f(s,w_{0}(s),w_{0}'(s))\,\mathrm{d} s) \\ \quad{} +\int_{\xi}^{t} \phi_{q}(\int_{s}^{\sigma_{x}}h(\tau)f(\tau,w_{0}(\tau),w_{0}'(\tau ))\,\mathrm{d} \tau)\,\mathrm{d} s, & t\in[0,\sigma_{x}], \\ \frac{1}{\beta}\phi_{q}(\int_{\sigma _{x}}^{1}h(s)f(s,w_{0}(s),w_{0}'(s))\,\mathrm{d} s) \\ \quad{} +\int_{t}^{\eta}\phi_{q}(\int_{\sigma_{x}}^{s}h(\tau)f(\tau,w_{0}(\tau ),w_{0}'(\tau))\,\mathrm{d} \tau)\,\mathrm{d} s, & t\in[\sigma_{x},1], \end{cases}\displaystyle \\ \leq& \textstyle\begin{cases} \frac{r}{\lambda}(\frac{1}{\alpha}+\xi-t)\cdot\phi_{q}(\int _{0}^{1}h(s)\,\mathrm{d} s), & 0 \leq t\leq\min\{\xi, \sigma_{x}\}\leq1,\\ \frac{r}{\lambda}(\frac{1}{\alpha}+t-\xi)\cdot\phi_{q}(\int _{0}^{1}h(s)\,\mathrm{d} s),& 0 \leq\xi\leq t\leq\sigma_{x}\leq1,\\ \frac{r}{\lambda}(\frac{1}{\beta}+\eta-t)\cdot\phi_{q}(\int _{0}^{1}h(s)\,\mathrm{d} s),& 0 \leq\sigma_{x}\leq t\leq\eta\leq1,\\ \frac{r}{\lambda}(\frac{1}{\beta}+t-\eta)\cdot\phi_{q}(\int _{0}^{1}h(s)\,\mathrm{d} s),& 0 \leq\max\{\eta, \sigma_{x}\}\leq t\leq1 \end{cases}\displaystyle \\ \leq&\frac{r}{\lambda} \bigl(-t^{2}+t+c \bigr) \cdot\phi_{q} \biggl( \int_{0}^{1}h(s)\,\mathrm{d} s \biggr) \\ =& w_{0}(t), \quad 0\leq t\leq1, \end{aligned}

and

\begin{aligned} \bigl\vert w_{1}'(t) \bigr\vert =& \bigl\vert (Tw_{0})'(t) \bigr\vert \\ =& \textstyle\begin{cases} \vert \phi_{q}(\int_{t}^{\sigma_{w_{0}}}h(s)f(s,w_{0}(s),w_{0}'(s))\,\mathrm{d} s)\vert ,& t\in[0,\sigma_{w_{0}}],\\ \vert -\phi_{q}(\int_{\sigma_{w_{0}}}^{t}h(s)f(s,w_{0}(s),w_{0}'(s))\,\mathrm{d} s)\vert ,& t\in[\sigma_{w_{0}},1], \end{cases}\displaystyle \\ \le& \textstyle\begin{cases} \frac{r}{\lambda} \vert \phi_{q}(\int_{t}^{\frac{1}{2}}h(s)\,\mathrm{d} s)\vert ,& t\in[0,\frac{1}{2}],\\ \frac{r}{\lambda} \vert -\phi_{q}(\int_{\frac{1}{2}}^{t}h(s)\,\mathrm{d} s)\vert ,& t\in[\frac{1}{2},1], \end{cases}\displaystyle \\ \le&\frac{r}{\lambda} \vert b-at\vert \phi_{q} \biggl( \int_{0}^{1}h(s)\,\mathrm{d} s \biggr),\quad t \in[0,1], \\ =& \bigl\vert w_{0}'(t) \bigr\vert ,\quad 0\leq t \leq1. \end{aligned}

Then we obtain that

$$w_{1}(t)\leq w_{0}(t),\qquad \bigl\vert w_{1}'(t) \bigr\vert \leq \bigl\vert w_{0}'(t) \bigr\vert ,\quad 0\leq t\leq1.$$

Hence, by Lemma 2.5 we have

\begin{aligned}& w_{2}(t)=(Tw_{1}) (t)\leq(Tw_{0}) (t)= w_{1}(t),\quad 0\leq t\leq1, \\& \bigl\vert w_{2}'(t) \bigr\vert = \bigl\vert (Tw_{1})'(t) \bigr\vert \leq \bigl\vert (Tw_{0})'(t) \bigr\vert = \bigl\vert w_{1}'(t) \bigr\vert ,\quad 0\leq t\leq1. \end{aligned}

Thus, by induction we get

$$w_{n+1}(t)\leq w_{n}(t),\qquad \bigl\vert w_{n+1}'(t) \bigr\vert \leq \bigl\vert w_{n}'(t) \bigr\vert ,\quad 0\leq t\leq1, n=1,2, \ldots.$$

So, there exists $$w^{*}\in\overline{P}_{r}$$ such that $$w_{n}\rightarrow w^{*}$$. Considering that T is completely continuous and $$w_{n+1}=Tw_{n}$$, we have $$Tw^{*}=w^{*}$$.

Let $$v_{0}(t)=0$$, $$0\leq t\leq1$$. Then $$v_{0}(t)\in\overline{P}_{r}$$. Let $$v_{1}=Tv_{0}$$; then $$v_{1}\in\overline{P}_{r}$$. We denote $$v_{n+1}=Tv_{n}=T^{n}v_{0}$$, $$n=1,2,\ldots$$ . Since $$T: \overline{P}_{r}\rightarrow\overline{P}_{r}$$, we get $$v_{n}\in T\overline{P}_{r}\subseteq \overline{P}_{r}$$, $$n=1,2,\ldots$$ . Since T is completely continuous, $$\{v_{n}\}_{n=1}^{\infty}$$ is a sequentially compact set. We have

\begin{aligned}& v_{1}(t)=(Tv_{0}) (t)=(T0) (t)\geq0,\quad 0\leq t \leq1, \\& \bigl\vert v_{1}'(t) \bigr\vert = \bigl\vert (Tv_{0})'(t) \bigr\vert = \bigl\vert (T0)'(t) \bigr\vert \geq0,\quad 0\leq t\leq1. \end{aligned}

Thus,

\begin{aligned}& v_{2}(t)=(Tv_{1}) (t)\geq(Tv_{0}) (t)=v_{1}(t),\quad 0\leq t\leq1, \\& \bigl\vert v_{2}'(t) \bigr\vert = \bigl\vert (Tv_{1})'(t) \bigr\vert \geq \bigl\vert (T0)'(t) \bigr\vert = \bigl\vert v_{1}'(t) \bigr\vert ,\quad 0\leq t\leq1. \end{aligned}

Similarly, by induction we obtain

$$v_{n+1}(t)\geq v_{n}(t),\qquad \bigl\vert v_{n+1}'(t) \bigr\vert \geq \bigl\vert v_{n}'(t) \bigr\vert ,\quad 0\leq t\leq1, n=1,2, \ldots.$$

So, there exists $$v^{*}\in\overline{P}_{r}$$ such that $$v_{n}\rightarrow v^{*}$$. Considering that T is completely continuous and $$v_{n+1}=Tv_{n}$$, we have $$Tv^{*}=v^{*}$$.

Since $$f(t,0,0)\not\equiv0$$ for $$0\leq t\leq1$$, the zero function is not the solution of (1.1). Hence, since $$\max \vert v^{*}(t)\vert >0$$, we have $$v^{*}(t)\geq\min\{t,1-t\} \max_{0\leq t\leq1}\vert v^{*}(t)\vert$$, $$0\leq t\leq1$$.

As we all know, the fixed point of T is a solution of BVP (1.1). Hence, we have shown that $$w^{*}$$, $$v^{*}$$ are two positive solutions of problem (1.1).

The proof is complete. □

### Remark 3.1

We can see that $$w^{*}$$ and $$v^{*}$$ may be the same solution of BVP (1.1), but for convenience, we say that there exist at least two solutions.

### Corollary 3.2

Assume that (C1) and (C2) hold. Further, suppose that there exists $$r>0$$ such that:

(C5):

$$\lim_{l\rightarrow+\infty}\max_{0\leq t\leq1}\frac{f(t,l,r)}{l ^{p-1}}\leq\phi_{p}(\frac{1}{\lambda})$$ (particularly, $$\lim_{l\rightarrow+\infty}\max_{0\leq t\leq1}\frac{f(t,l,r)}{l^{p-1}}=0$$).

Then problem (1.1) has two positive solutions in P.

At the end of this paper, we give an example to illustrate our main result.

Consider the following four-point boundary value problem.

### Example 1

$$\textstyle\begin{cases} (\phi_{p}(x'))'+tf(t,x(t),x'(t))=0,\quad 0< t< 1,\\ x'(0)-3x(1/4)=0, \qquad x'(1)+x(2/3)=0, \end{cases}$$
(3.1)

where

$$f(t,u,v)=t^{2}+\frac{u}{6}+\frac{v^{2}}{100}.$$

We can see that $$h(t)=t$$, $$\xi=\frac{ 1}{ 4}$$, $$\eta=\frac { 2}{ 3}$$, $$\alpha=3$$, $$\beta=1$$. Let $$p=\frac{ 3}{ 2}$$, $$r=14$$. By direct calculation we obtain $$q=3$$, $$\lambda=\frac{ 35}{ 36}$$. Then the conditions of Theorem 3.1 are all satisfied. So BVP (3.1) has at least two positive solutions $$w^{*}$$, $$v^{*}$$, and there exists $$\sigma_{x}\in(0,1)$$ such that $$(w^{*})'(\sigma _{x})=0$$, $$(v^{*})'(\sigma_{x})=0$$. Further,

$$0\leq w^{*}\leq14,\qquad 0\leq \bigl\vert \bigl(w^{*} \bigr)' \bigr\vert \leq14,$$

and

$$\lim_{n\rightarrow\infty}w_{n}=\lim _{n\rightarrow \infty}T^{n}w_{0}=w^{*},\qquad \lim_{n\rightarrow\infty}(w_{n})'=\lim _{n\rightarrow\infty } \bigl(T^{n}w_{0} \bigr)'= \bigl(w^{*} \bigr)',$$

where

$$w_{0}(t)= \textstyle\begin{cases} 8(\frac{5}{3}-t),\quad 0\leq t\leq\frac{11}{24},\\ 8(t+\frac{3}{4}),\quad \frac{11}{24}\leq t\leq1. \end{cases}$$

At the same time, we have

$$0< v^{*}\leq14,\qquad 0< \bigl\vert \bigl(v^{*} \bigr)' \bigr\vert \leq14,$$

and

$$\lim_{n\rightarrow\infty}v_{n}=\lim _{n\rightarrow \infty}T^{n}v_{0}=v^{*},\qquad \lim_{n\rightarrow\infty}(v_{n})'=\lim _{n\rightarrow\infty } \bigl(T^{n}v_{0} \bigr)'= \bigl(v^{*} \bigr)',$$

where $$v_{0}(t)=0$$, $$0\leq t\leq1$$, and T is as defined in (2.4).

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## Acknowledgements

The authors were very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. The study was supported by the Fundamental Research Funds for the Central Universities (No. 2652015194) and Beijing Higher Education Young Elite Teacher Project.

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Correspondence to Junfang Zhao.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

JZ and BS conceived of the study and participated in its coordination. JZ drafted the manuscript, and YW proofread the manuscript. All authors read and approved the final manuscript.

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