# Positive solution for a fractional singular boundary value problem with p-Laplacian operator

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## Abstract

In this paper, we consider a fractional singular three-point boundary value problem with p-Laplacian operator. The nonlinearity $$f(t,u)$$ may be singular at $$t = 0,1$$ and $$u = 0$$. Some properties of the associated Green function are obtained. By using the upper and lower solutions method and a fixed point theorem, the existence result of positive solution is established.

## Introduction

In this paper, we investigate the following fractional three-point boundary value problem (BVP) with p-Laplacian operator:

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=f(t,u(t)),\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0,\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad D^{\beta}_{0^{+}}u(1)=bD^{\beta}_{0^{+}}u(\eta), \end{cases}$$
(1.1)

where $$\alpha\in(1,2]$$, $$\beta\in(3,4]$$, $$D^{\alpha}_{0^{+}}$$ and $$D^{\beta}_{0^{+}}$$ are the standard Riemann–Liouville derivatives, $$\varphi_{p}(s)= \vert s \vert ^{p-2}s$$, $$p>1$$, $$\varphi_{p}^{-1}=\varphi_{q}$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\eta\in (0,1)$$, $$b\in (0,\eta^{\frac{1-\alpha}{p-1}} )$$, $$f(t,u): (0,1)\times(0,+\infty)\to [0,+\infty)$$ is continuous and may be singular at $$t=0, 1$$ and $$u=0$$.

The differential equations with p-Laplacian operator have deep background in physics. In recent years, boundary value problems of fractional differential equations with or without p-Laplacian operator have been widely studied. By means of nonlinear analysis theory and methods, many existence and multiplicity results of solutions or positive solutions have been obtained, see  and the references therein.

In , Xu and Dong considered three-point BVP (1.1), but their nonlinearity $$f:[0,1]\times [0,+\infty)\to [0,+\infty)$$ is continuous, the existence and uniqueness of positive solutions were obtained by using the upper and lower solutions method and Schauder’s fixed point theorem.

By means of the lower and upper solutions method and monotone iterative technique, Liu et al.  investigated the existence of positive solutions for mixed fractional BVP with p-Laplacian operator

$$\textstyle\begin{cases} D^{\alpha}_{0^{+}}(\varphi_{p}(^{c}D^{\beta}_{0^{+}}u(t)))=f(t,u(t),^{c}D_{0^{+}}^{\beta}u(t)),\quad 0< t< 1, \\ ^{c}D_{0^{+}}^{\beta}u(0)=u'(0)=0,\qquad u(1)=r_{1}u(\eta),\qquad ^{c}D_{0^{+}}^{\beta}u(1)=r_{2}^{c}D_{0^{+}}^{\beta}u(\xi), \end{cases}$$

where $$\alpha, \beta\in(1,2]$$, $$D^{\alpha}_{0^{+}}$$ and $$^{c}D^{\beta}_{0^{+}}$$ are the Riemann–Liouville fractional derivative and Caputo fractional derivative, respectively.

By using upper and lower solutions method, Wang and Xiang  established existence results of positive solution for a fractional BVP with p-Laplacian operator

$$\textstyle\begin{cases} D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=f(t,u(t)),\quad 0< t< 1, \\ u(0) = 0,\qquad u(1)=au(\xi),\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad D^{\beta}_{0^{+}}u(1)=bD^{\beta}_{0^{+}}u(\eta), \end{cases}$$

where $$\alpha,\beta \in(1,2]$$, $$a,b \in(0,1]$$, $$\xi,\eta \in(0,1)$$, $$D^{\alpha}_{0^{+}}$$ and $$D^{\beta}_{0^{+}}$$ are the Riemann–Liouville fractional derivatives.

In , Zhang et al. studied the integral BVP of fractional differential equations with parameter and p-Laplacian operator

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=\lambda f(t,u(t),\quad 0< t< 1, \\ u(0)=0,\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad u(1)=\int^{1}_{0}u(s)dA(s), \end{cases}$$

where $$\alpha\in(0,1]$$, $$\beta\in(1,2]$$, $$D^{\alpha}_{0^{+}}$$ and $$D^{\beta}_{0^{+}}$$ are the Riemann–Liouville fractional derivatives, $$\int^{1}_{0}u(s)dA(s)$$ is the Riemann–Stieltjes integral, $$f(t,u):(0,1)\times(0,+\infty)\rightarrow [0,+\infty)$$ is continuous.

Motivated by the papers mentioned above, in this paper, we study the p-Laplacian fractional differential equation three-point BVP (1.1). The existence of positive solution is obtained by using the upper and lower solutions method and a fixed point theorem. It is worth pointing out that $$f(t,u)$$ may be singular at $$t = 0,1$$ and $$u = 0$$.

## Preliminaries and lemmas

Let $$\varphi_{p}(D^{\beta}_{0^{+}}u(t))=v(t)$$, then $$v(0)=0$$, $$v(1)=b^{p-1}v(\eta)$$. We now consider the following BVP:

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}v(t)=y(t),\quad 0< t< 1, \\ v(0)=0,\qquad v(1)=b^{p-1}v(\eta). \end{cases}$$
(2.1)

### Lemma 2.1

()

If $$y\in C[0,1]$$, then BVP (2.1) has a unique solution

$$v(t)= \int_{0}^{1} H(t,s)y(s)\,ds,$$

where

\begin{aligned} &H(t,s)=h(t,s)+\frac{b^{p-1}t^{\alpha-1}}{1-b^{p-1}\eta^{\alpha-1}}h(\eta,s), \\ &h(t,s)=\frac{1}{\Gamma(\alpha)} \textstyle\begin{cases} t^{\alpha-1}(1-s)^{\alpha-1}, & 0\leq t\leq s\leq 1, \\ t^{\alpha-1}(1-s)^{\alpha-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq 1. \end{cases}\displaystyle \end{aligned}

From the above analysis, the BVP

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=y(t),\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0,\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad D^{\beta}_{0^{+}}u(1)=bD^{\beta}_{0^{+}}u(\eta) \end{cases}$$

is equal to

$$\textstyle\begin{cases} D^{\beta}_{0^{+}}u(t)=\varphi_{q} (\int_{0}^{1} H(t,s)y(s)\,ds ),\quad 0< t< 1,\\ u(0)=u(1)=u'(0)=u'(1)=0. \end{cases}$$
(2.2)

### Lemma 2.2

()

If $$y\in C[0,1]$$, BVP (2.2) has a unique solution

$$u(t)= \int_{0}^{1} G(t,s)\varphi_{q} \biggl( \int_{0}^{1} H(s,\tau)y(\tau)d\tau \biggr)\,ds,$$

where

$$G(t,s)=\frac{1}{\Gamma(\beta)} \textstyle\begin{cases} t^{\beta-2}(1-s)^{\beta-2}[(s-t)+(\beta-2)(1-t)s],& 0\leq t\leq s\leq 1,\\ t^{\beta-2}(1-s)^{\beta-2}[(s-t)+(\beta-2)(1-t)s]+(t-s)^{\beta-1},& 0\leq s\leq t\leq 1. \end{cases}$$

### Lemma 2.3

The functions $$H, G\in C([0,1]\times[0,1],[0,+\infty))$$ have the following properties:

1. (1)
$$H(t,s)\leq d_{1} (1-s)^{\alpha-1},\quad t,s\in(0,1),$$

where $$d_{1}=\frac{1}{(1-b^{p-1}\eta^{\alpha-1})\Gamma(\alpha)}$$.

2. (2)
$$(\beta-2)k(t)q(s)\leq\Gamma(\beta)G(t,s)\leq M_{0}q(s),\quad t,s \in (0,1),$$

where

$$k(t)=t^{\beta-2}(1-t)^{2},\qquad q(s)=s^{2}(1-s)^{\beta-2}, \qquad M_{0}=\max\bigl\{ \beta-1,(\beta-2)^{2}\bigr\} .$$
3. (3)
$$G(t,s)\geq \frac{\beta-2}{M_{0}}t^{\beta-2}(1-t)^{2}G(t_{0},s), \quad t, s, t_{0}\in(0,1).$$

### Proof

1. (1)

For any $$t,s\in (0,1)$$,

$$h(t,s)\leq\frac{1}{\Gamma(\alpha)}\bigl[t(1-s)\bigr]^{\alpha-1}\leq \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)},$$

then

$$h(\eta,s)\leq\frac{1}{\Gamma(\alpha)}\bigl[\eta(1-s)\bigr]^{\alpha-1}.$$

Therefore

\begin{aligned} H(t,s) \leq&\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+\frac{b^{p-1}}{1-b^{p-1}\eta^{\alpha-1}} \frac{\eta^{\alpha-1}(1-s)^{\alpha-1}}{\Gamma(\alpha)} \\ =&\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)(1-b^{p-1}\eta^{\alpha-1})}=d_{1}(1-s)^{\alpha-1}. \end{aligned}
2. (2)

See Lemma 2.4(2) of .

3. (3)

For any $$t, s, t_{0}\in(0,1)$$, we have

\begin{aligned} G(t,s) \geq&\frac{(\beta-2)k(t)q(s)}{\Gamma(\beta)}\\ =&\frac{(\beta-2)k(t)}{M_{0}\Gamma(\beta)}M_{0}q(s) \\ \geq&\frac{(\beta-2)k(t)}{M_{0}\Gamma(\beta)}\Gamma(\beta)G(t_{0},s) \\ =&\frac{\beta-2}{M_{0}}t^{\beta-2}(1-t)^{2}G(t_{0},s). \end{aligned}

This completes the proof.

□

### Remark 2.1

By Lemmas 2.2 and 2.3, if $$D^{\beta}_{0^{+}}u\geq0$$ and $$u(0)=u(1)=u'(0)=u'(1)=0$$, we conclude that $$u(t)\geq0$$, $$t\in[0,1]$$.

u is said to be a lower solution for BVP (1.1) if u satisfies the following inequality system:

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))\leq f(t,u(t)),\quad 0< t< 1, \\ u(0)\leq0,\qquad u(1)\leq0,\qquad u'(0)\leq0,\qquad u'(1)\leq0,\\ -D^{\beta}_{0^{+}}u(0)\leq0,\qquad -D^{\beta}_{0^{+}}u(1)\leq -bD^{\beta}_{0^{+}}u(\eta). \end{cases}$$

Similarly, we define the upper solution for BVP (1.1) by replacing “least or equal” by “greater or equal”.

## Main result

### Theorem 3.1

Assume that the following conditions $$(\mathrm{H}_{1})$$$$(\mathrm{H}_{3})$$ are satisfied:

$$(\mathrm{H}_{1})$$ :

$$f(t,u)\in C((0,1)\times(0,+\infty),[0,+\infty))$$ and $$f(t,u)$$ is nonincreasing relative to u.

$$(\mathrm{H}_{2})$$ :

For any constant $$\lambda>0$$,

$$0< \int^{1}_{0} (1-s)^{\alpha-1}f\bigl(s,\lambda s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty.$$
$$(\mathrm{H}_{3})$$ :

There exist a function $$a\in C[0,1]$$ and a constant $$k>0$$ such that $$a(t)\geq kt^{\beta-2}(1-t)^{2}$$, $$t\in[0,1]$$, and

\begin{aligned} & \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,a(s)\bigr)\,ds \biggr)\,dr=b(t)\geq a(t), \\ & \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,b(s)\bigr)\,ds \biggr)\,dr \geq a(t). \end{aligned}

Then BVP (1.1) has at least one positive solution w which satisfies $$w(t)\geq mt^{\beta-2}\times(1-t)^{2}$$ for some $$m>0$$.

### Proof

Let

$$P= \bigl\{ u\in C[0,1]: \mbox{there exists }k_{u}>0\mbox{ such that }u(t) \geq k_{u}t^{\beta-2}(1-t)^{2}, t\in[0,1] \bigr\} .$$

Define an operator T by

$$Tu(t)= \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr, \quad u\in P.$$

For $$u\in P$$, there exists $$k_{u}>0$$ such that $$u(t)\geq k_{u}t^{\beta-2}(1-t)^{2}$$, $$t\in[0,1]$$. By $$(H_{1})$$ and $$(H_{2})$$, we have

$$\int^{1}_{0}H(t,s)f\bigl(s,u(s)\bigr)\,ds\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty.$$

Hence

\begin{aligned} Tu(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& \int^{1}_{0}\frac{M_{0}}{\Gamma(\beta)}q(r) \varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)\,dr \\ =&\frac{M_{0}}{\Gamma(\beta)}d_{1}^{q-1} \int^{1}_{0}q(r)\,dr\varphi_{q} \biggl( \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< + \infty. \end{aligned}
(3.1)

On the other hand, choose $$t_{0}\in(0,1)$$ such that $$Tu(t_{0})=k_{T_{u}}>0$$. It follows from Lemma 2.3 that

\begin{aligned} Tu(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \geq&\frac{\beta-2}{\Gamma(\beta)}k(t) \int^{1}_{0}q(r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \geq&\frac{\beta-2}{M_{0}}k(t)Tu(t_{0}) =\frac{\beta-2}{M_{0}}k_{T_{u}}t^{\beta-2}(1-t)^{2}, \quad t\in[0,1]. \end{aligned}
(3.2)

It follows from (3.1) and (3.2) that T is well defined and $$T(P)\subset P$$.

Next, we determine upper and lower solutions of BVP (1.1). In fact, by simple computations, we have

\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\bigl(Tu(t)\bigr)\bigr)\bigr)=f\bigl(t,u(t) \bigr),\quad t\in(0,1), \end{aligned}
(3.3)
\begin{aligned} & \textstyle\begin{cases} (Tu)(0)=(Tu)(1)=(Tu)'(0)=(Tu)'(1)=0,\\ D^{\beta}_{0^{+}}(Tu)(0)=0,\qquad D^{\beta}_{0^{+}}(Tu)(1)=bD^{\beta}_{0^{+}}(Tu)(\eta). \end{cases}\displaystyle \end{aligned}
(3.4)

Let $$b(t)=Ta(t)$$, then by $$(H_{1})$$ and $$(H_{3})$$, we have

$$a(t)\leq Ta(t)=b(t),\qquad b(t)=Ta(t)\geq Tb(t),\quad t\in[0,1].$$
(3.5)

Since $$a(t)\in P$$, from (3.2), we obtain $$Ta(t),Tb(t)\in P$$. Thus, by (3.3) and (3.5),

\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Tb) (t)\bigr)\bigr)-f\bigl(t,(Tb) (t)\bigr) \leq-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Tb) (t)\bigr)\bigr)-f\bigl(t,b(t)\bigr)=0, \end{aligned}
(3.6)
\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr)\bigr)-f\bigl(t,(Ta) (t)\bigr) \geq-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr)\bigr)-f\bigl(t,a(t)\bigr)=0. \end{aligned}
(3.7)

Meanwhile, (3.4) implies that $$Ta(t), Tb(t)$$ satisfy the boundary conditions of BVP (1.1). Then, from (3.5)–(3.7), $$\varphi(t)=Tb(t)$$ and $$\psi(t)=Ta(t)$$ are lower and upper solutions of BVP (1.1), respectively.

Next, we shall show that the BVP

$$\textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=g(t,u(t)),\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0,\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad D^{\beta}_{0^{+}}u(1)=bD^{\beta}_{0^{+}}u(\eta) \end{cases}$$
(3.8)

has a positive solution, where

$$g\bigl(t,u(t)\bigr)= \textstyle\begin{cases} f(t,\varphi(t)),& u(t)< \varphi(t), \\ f(t,u(t)),& \varphi(t)\leq u(t)\leq \psi(t),\\ f(t,\psi(t)),& u(t)>\psi(t). \end{cases}$$
(3.9)

To see this, we consider the operator $$A:C[0,1]\rightarrow C[0,1]$$ defined as follows:

$$Au(t)= \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr.$$

It is well known that a fixed point of the operator A is a solution of BVP (3.8).

It is clear that A is continuous. Since $$\varphi(t)\in P$$, there exists $$k_{\varphi}>0$$ such that $$\varphi(t)\geq k_{\varphi}t^{\beta-2}(1-t)^{2}$$, $$t\in[0,1]$$. It follows from $$(H_{2})$$ that

\begin{aligned} \int^{1}_{0}H(t,s)g\bigl(s,u(s)\bigr)\,ds&\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f\bigl(s,\varphi(s) \bigr)\,ds \\ &\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty. \end{aligned}
(3.10)

Consequently, for $$u\in C[0,1]$$ and $$t\in [0,1]$$, by (3.9) and (3.10), we have

\begin{aligned} Au(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& M_{0} \int^{1}_{0}q(r)\varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}g \bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& M_{0}d_{1}^{q-1} \int^{1}_{0} q(r)\,dr\varphi_{q} \biggl( \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< + \infty, \end{aligned}

which implies that A is uniformly bounded.

On the other hand, since $$G(t,s)$$ is continuous on $$[0,1]\times[0,1]$$, it is uniformly continuous on $$[0,1]\times[0,1]$$. Thus, for $$s\in[0,1]$$ and for each $$\varepsilon>0$$, there exists $$\delta>0$$ such that $$\vert t_{1}-t_{2} \vert <\delta$$ implies

$$\bigl\vert G(t_{1},s)-G(t_{2},s) \bigr\vert < \frac{\varepsilon}{\varphi_{q} (d_{1}\int^{1}_{0}(1-s)^{\alpha-1}f(s,k_{\varphi}s^{\beta-2}(1-s)^{2})\,ds )}.$$

Furthermore, for $$u\in C[0,1]$$,

\begin{aligned} & \bigl\vert Au(t_{1})-Au(t_{2}) \bigr\vert \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r)) \bigr\vert \varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r) \bigr\vert \varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}f \bigl(s,\varphi(s)\bigr)\,ds \biggr)\,dr \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r) \bigr\vert \,dr\varphi_{q} \biggl(d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< \varepsilon, \end{aligned}

which implies that A is equicontinuous. Thus, the Ascoli–Arzela theorem guarantees A is a compact operator. It follows from Schauder’s fixed point theorem that A has a fixed point w, i.e., $$w=Aw$$. Consequently, (3.8) has a solution.

Finally, we will show that BVP (1.1) has at least one positive solution. In fact, we only need to prove that $$\varphi(t)\leq w(t)\leq\psi(t)$$, $$t\in[0,1]$$. By $$(H_{1})$$, we have

$$f\bigl(t,\psi(t)\bigr)\leq g\bigl(t,w(t)\bigr)\leq f\bigl(t, \varphi(t)\bigr),\quad t\in[0,1].$$
(3.11)

It follows from (3.5) and $$(H_{3})$$ that

$$f\bigl(t,b(t)\bigr)\leq g\bigl(t,w(t)\bigr)\leq f\bigl(t,a(t) \bigr),\quad t\in[0,1].$$
(3.12)

Since $$a(t)\in P$$, by (3.3), we have

$$-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)=-D^{\alpha}_{0^{+}} \bigl(\varphi_{p}\bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr) \bigr)=f\bigl(t,a(t)\bigr),\quad t\in(0,1).$$

By (3.4), (3.5), (3.11), and (3.12), we have

\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)-\bigl[-D^{\alpha}_{0^{+}} \bigl(\varphi_{p}\bigl(D^{\beta}_{0^{+}}w(t)\bigr)\bigr) \bigr]=f\bigl(t,a(t)\bigr)-g\bigl(t,w(t)\bigr)\geq0,\quad t\in[0,1], \\ &(\psi-w) (0)=(\psi-w) (1)=(\psi-w)'(0)=(\psi-w)'(1)=0, \\ &D^{\beta}_{0^{+}}(\psi-w) (0)=0,\qquad D^{\beta}_{0^{+}}( \psi-w) (1)=bD^{\beta}_{0^{+}}(\psi-w) (\eta). \end{aligned}

Setting $$z=\varphi_{p}(D^{\beta}_{0^{+}}\psi(t))-\varphi_{p}(D^{\beta}_{0^{+}}w(t))$$, then

\begin{aligned} &{-}D^{\alpha}_{0^{+}}z(t)=-D^{\alpha}_{0^{+}}\bigl( \varphi_{p}\bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)- \bigl[-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}w(t)\bigr)\bigr)\bigr]\geq0, \\ &z(0)=0,\quad z(1)=\varphi_{p}(b)z(\eta). \end{aligned}

Hence, by Lemma 2.1, we get $$z(t)\geq0$$, $$t\in[0,1]$$. Since $$\varphi_{p}$$ is monotone increasing, we have $$D^{\beta}_{0^{+}}\psi(t)\geq D^{\beta}_{0^{+}}w(t)$$, that is, $$D^{\beta}_{0^{+}}(\psi(t)-w(t))\geq0$$, $$t\in[0,1]$$. By Remark 2.1, we have $$w(t)\leq\psi(t)$$ for $$t\in[0,1]$$. Similarly, $$w(t)\geq\varphi(t)$$ on $$[0,1]$$. Therefore, $$w(t)$$ is a positive solution of BVP (1.1). And $$\varphi(t)\in P$$ implies that there exists $$m>0$$ such that $$w(t)\geq\varphi(t)\geq mt^{\beta-2}(1-t)^{2}$$, $$t\in[0,1]$$. This completes the proof. □

## An example

### Example 4.1

Consider the following fractional singular BVP:

$$\textstyle\begin{cases} -D^{\frac{3}{2}}_{0^{+}}(\varphi_{p}(D^{\frac{7}{2}}_{0^{+}}u(t)))=(1-t)^{-\frac{1}{4}}u^{-\frac{1}{2}},\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0,\qquad D^{\frac{7}{2}}_{0^{+}}u(0)=0,\qquad D^{\frac{7}{2}}_{0^{+}}u(1)=\frac{1}{4}D^{\frac{7}{2}}_{0^{+}}u (\frac{1}{2} ), \end{cases}$$
(4.1)

where $$\varphi_{p}(t)= \vert t \vert ^{p-2}t$$, $$p>1$$. Then BVP (4.1) has a positive solution $$w(t)\geq m t^{\frac {3}{2}}(1-t)^{2}$$ for some $$m>0$$.

In fact, let $$\alpha=\frac{3}{2}$$, $$\beta=\frac{7}{2}$$, $$f(t,u)=(1-t)^{-\frac{1}{4}}u^{-\frac{1}{2}}$$, $$t\in(0,1)$$. Obviously, $$f(t,u)$$ is singular at $$t =1$$ and $$u = 0$$. It is easy to check that $$(H_{1})$$ in Theorem 3.1 is satisfied. For any constant $$\lambda>0$$,

\begin{aligned} 0 < & \int^{1}_{0} (1-s)^{\alpha-1}f\bigl(s,\lambda s^{\beta-2}(1-s)^{2}\bigr)\,ds \\ =& \int^{1}_{0} (1-s)^{\frac{1}{2}}(1-s)^{-\frac{1}{4}} \bigl[\lambda s^{\frac{3}{2}}(1-s)^{2}\bigr]^{-\frac{1}{2}}\,ds \\ =&\lambda^{-\frac{1}{2}} \int^{1}_{0} s^{-\frac{3}{4}}(1-s)^{-\frac{3}{4}}\,ds \\ =&\lambda^{-\frac{1}{2}}B \biggl(\frac{1}{4},\frac{1}{4} \biggr)= \lambda^{-\frac{1}{2}}\frac{\Gamma^{2}(\frac{1}{4})}{\sqrt{\pi}}< +\infty, \end{aligned}

so $$(H_{2})$$ in Theorem 3.1 is satisfied.

Set $$\mu=\frac{1}{2}$$, then $$f(t,u)\leq f(t,ru)\leq r^{-\mu}f(t,u)$$ for any $$r\in (0,1)$$. Since $$e(t)=t^{\frac{3}{2}}(1-t)^{2}\in P$$, by (3.2) we know $$Te\in P$$, $$T^{2}e\in P$$, then there exist positive numbers k and l such that $$Te\geq ke$$ and $$T^{2}e\geq le$$. Take $$0< r_{0}< \min \{1,k,l^{\frac{1}{1-\mu^{2}}} \}$$, then

$$T(r_{0}e)\geq Te\geq ke \geq r_{0}e,\qquad T^{2}(r_{0}e)\geq r_{0}^{\mu^{2}}T^{2}e \geq r_{0}^{\mu^{2}}le\geq r_{0}e.$$

If we take $$a(t)=r_{0}t^{\frac{3}{2}}(1-t)^{2}$$, then condition $$(H_{3})$$ of Theorem 3.1 is satisfied. Consequently, the above conclusion is guaranteed by Theorem 3.1.

## Conclusion

In this paper, we consider the fractional singular three-point boundary value problem with p-Laplacian operator. It is worth pointing out that $$f(t,u)$$ may be singular at $$t = 0,1$$ and $$u = 0$$. Some properties of the associated Green function are obtained. By using the upper and lower solutions method and a fixed point theorem, the existence result of positive solution is established.

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### Acknowledgements

The authors would like to thank the referees for their pertinent comments and valuable suggestions.

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### Funding

This work is supported financially by the National Natural Science Foundation of China (11501318), the China Postdoctoral Science Foundation (2017M612230) and the Natural Science Foundation of Shandong Province of China (ZR2017MA036).

## Author information

All authors contributed equally to the writing of this paper. The authors read and approved the final manuscript.

Correspondence to Xinan Hao.

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