Theory and Modern Applications

# Positive solutions for a singular fractional nonlocal boundary value problem

## Abstract

We investigate a singular fractional differential equation with an infinite-point fractional boundary condition, where the nonlinearity $$f(t,x)$$ may be singular at $$x = 0$$, and $$g(t)$$ may also have singularities at $$t= 0$$ or $$t=1$$. We establish the existence of positive solutions using the fixed point index theory in cones.

## Introduction

We consider the existence of positive solutions for the following fractional nonlocal boundary value problem:

$$\textstyle\begin{cases} D^{\alpha }_{0^{+}}x(t)+\lambda g(t)f(t,x(t))=0,& t\in (0,1), \\ x(0)=x'(0)=\cdots =x^{(n-2)}(0)=0,& D^{\beta }_{0^{+}}x(1)=\sum_{i=1}^{\infty }\alpha_{i}D^{\gamma }_{0^{+}}x(\xi_{i}), \end{cases}$$
(1.1)

where $$\lambda >0$$ is a parameter, $$D^{\alpha }_{0^{+}}$$, $$D^{\beta } _{0^{+}}$$, and $$D^{\gamma }_{0^{+}}$$ denote the Riemann–Liouville fractional derivatives, $$2\leq n-1<\alpha \leq n$$, $$1\leq \beta \leq n-2, 0\leq \gamma \leq \beta$$, $$\alpha_{i}\geq 0$$, $$0<\xi_{1}<\xi_{2}< \cdots <\xi_{i-1}<\xi_{i}<\cdots <1$$ ($$i=1,2,\ldots$$), and $$\Gamma (\alpha -\gamma )>\Gamma (\alpha -\beta )\sum_{i=1}^{\infty } \alpha_{i}\xi_{i}^{\alpha -\gamma -1}$$. The function $$f(t,x)$$ may have singularity at $$x = 0$$, and $$g(t)$$ may be singular at $$t= 0$$ and/or $$t=1$$.

Fractional differential equations describe many phenomena in various fields of science and engineering . For the development of the fractional differential equations, see  and the references therein. Recently, the existence of positive solutions for fractional differential equation multipoint boundary value problems (BVPs) have been studied by many authors; see . Using the compression expansion fixed point theorem due to Krasnosel’skii, Henderson and Luca  studied the fractional BVP

$$\textstyle\begin{cases} D^{\alpha }_{0^{+}}x(t)+\lambda f(t,x(t))=0,& 0< t< 1, \\ x(0)=x'(0)=\cdots =x^{(n-2)}(0)=0,& D^{\beta }_{0^{+}}x(1)=\sum_{i=1}^{m}\alpha_{i}D^{\gamma }_{0^{+}}x(\xi_{i}), \end{cases}$$
(1.2)

where $$\lambda >0$$, $$2\leq n-1<\alpha \leq n$$, $$\alpha_{i}\geq 0$$, $$0<\xi _{1}<\xi_{2}<\cdots <\xi_{m}<1$$ ($$i=1,2,\ldots ,m$$), $$1\leq \beta \leq n-2$$, $$0\leq \gamma \leq \beta$$, and $$f(t,x)$$ may be singular at $$t= 0,1$$ and may change sign. In , for $$\lambda =1$$, the authors investigated the existence and multiplicity of positive solutions for BVP (1.2). In [29, 30], the authors discussed the following infinite-point BVP:

$$\textstyle\begin{cases} D^{\alpha }_{0^{+}}x(t)+f(t,x(t))=0,& 0< t< 1, \\ x(0)=x'(0)=\cdots =x^{(n-2)}(0)=0,& x^{(i)}(1)=\sum_{j=1}^{\infty }\alpha _{j}x(\xi_{j}), \end{cases}$$
(1.3)

where $$i\in \{1,2,\ldots ,n-2\}$$, and $$\sum_{j=1}^{\infty }\alpha_{j} \xi_{j}^{\alpha -1}<(\alpha -1)\cdots (\alpha -i)$$. The existence, uniqueness, and multiplicity of positive solutions for BVP (1.3) are established. Qiao and Zhou  discussed the singular BVP

$$\textstyle\begin{cases} D^{\alpha }_{0^{+}}x(t)+g(t) f(t,x(t))=0, & 0< t< 1, \\ x(0)=x'(0)= \cdots =x^{(n-2)}(0)=0,& D^{\beta }_{0^{+}}x(1)=\sum_{i=1}^{\infty }\alpha_{i} x(\xi_{i}), \end{cases}$$
(1.4)

where $$\beta \in [1,\alpha -1]$$, and $$\Gamma (\alpha )>\Gamma (\alpha -\beta )\sum_{i=1}^{\infty }\alpha_{i}\xi_{i}^{\alpha -1}$$. For more results on the fractional infinite-point BVPs, see [24, 25, 32, 33] and the references therein.

In the present paper, we investigate the existence of positive solutions for the singular fractional infinite-point BVP (1.1) using the fixed point index theory in cones. Note that $$f(t,x)$$ may be singular at $$x = 0$$ and $$g(t)$$ may be singular at $$t= 0$$ or $$t=1$$.

## Preliminaries and lemmas

### Definition 2.1

()

The Riemann–Liouville fractional integral of order $$\alpha > 0$$ of a function $$h: (0,+ \infty )\rightarrow \mathbb{R}$$ is given by

$$I_{0^{+}}^{\alpha }h(t)=\frac{1}{\Gamma (\alpha )} \int_{0}^{t} (t-s)^{ \alpha -1}h(s)\,ds,$$

provided that the right-hand side is defined pointwise on $$(0,+\infty )$$.

### Definition 2.2

()

The Riemann–Liouville fractional derivative of order $$\alpha > 0$$ of a function $$h: (0,+ \infty )\rightarrow \mathbb{R}$$ is given by

$$D^{\alpha }_{0^{+}}h(t)=\frac{1}{\Gamma (n-\alpha )} \biggl( \frac{d}{dt} \biggr) ^{n} \int_{0}^{t} {\frac{h(s)}{(t-s)^{\alpha -n+1}}}\,ds,$$

where n is the smallest integer not less than α, provided that the right-hand side is defined pointwise on $$(0,+\infty )$$.

By arguments similar to those in [30, 31], we have the following two lemmas.

### Lemma 2.1

Given $$y\in C(0,1)\cap L^{1}(0,1)$$, the solution of the BVP

$$\textstyle\begin{cases} D^{\alpha }_{0^{+}}x(t)+y(t)=0,& t\in (0,1), \\ x(0)=x'(0)=\cdots =x^{(n-2)}(0)=0,& D^{\beta }_{0^{+}}x(1)=\sum_{i=1}^{\infty } \alpha_{i}D^{\gamma }_{0^{+}}x(\xi_{i}), \end{cases}$$

is

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

where $$G(t,s)$$ is the Green’s function given by

$$G(t,s)=\frac{1}{\Gamma (\alpha )q(0)} \textstyle\begin{cases} q(s)(1-s)^{\alpha -\beta -1}t^{\alpha -1}-q(0)(t-s)^{\alpha -1},& 0\leq s\leq t\leq 1, \\ q(s)(1-s)^{\alpha -\beta -1}t^{\alpha -1},& 0\leq t\leq s\leq 1, \end{cases}$$

and

$$q(s)=\frac{1}{\Gamma (\alpha -\beta )}-\frac{1}{\Gamma (\alpha - \gamma )}\sum_{s\leq \xi_{i}} \alpha_{i} \biggl( \frac{\xi_{i}-s}{1-s} \biggr) ^{\alpha -\gamma -1}(1-s)^{\beta -\gamma }.$$

### Lemma 2.2

The functions q and G given in Lemma 2.1 have the following properties:

1. (i)

$$q\in C([0,1],(0,+\infty ))$$ is nondecreasing;

2. (ii)

$$G(t,s)\in C([0,1]\times [0,1],[0,+\infty ))$$;

3. (iii)

$$p(t)G(1,s)\leq G(t,s)\leq G(1,s), t,s\in [0,1]$$, where $$p(t)=t^{\alpha -1}$$.

Set $$E=C[0,1]$$ and $$\Vert x \Vert =\sup_{t\in [0,1]}\vert x(t) \vert$$. We define the cones

$$P= \bigl\{ x\in E: x(t)\geq 0, t\in [0,1] \bigr\} \quad \mbox{and}\quad K= \bigl\{ x \in P: x(t)\geq p(t)\Vert x \Vert , t\in [0,1] \bigr\} .$$

For $$0< r<+\infty$$, denote $$K_{r}=\{x\in K:\Vert x \Vert < r\}$$, $$\partial K_{r}= \{x\in K:\Vert x \Vert =r\}$$ and $$\overline{K}_{r}=\{x\in K:\Vert x \Vert \leq r\}$$. Define the operators $$A:\overline{K}_{R}\backslash K_{r}\rightarrow P$$ and $$L: E\rightarrow E$$ by

\begin{aligned}& Ax(t)=\lambda \int_{0}^{1}G(t,s)g(s)f \bigl(s,x(s) \bigr)\,ds,\quad t\in [0,1], \\& Lx(t)=\int_{0}^{1}G(t,s)g(s)x(s)\,ds,\quad t\in [0,1]. \end{aligned}

Clearly, $$L: K\rightarrow K$$ is a completely continuous linear operator. Moreover, if x is a fixed point of A, then x is a solution of BVP (1.1).

We further assume that:

$$(H_{1})$$ :

$$g\in C((0,1), [0,\infty ))$$ and $$0<\int_{0}^{1} G(1,s)g(s)\,ds <+\infty$$.

$$(H_{2})$$ :

$$f\in C([0,1]\times (0,\infty ), [0,\infty ))$$, and for any $$0< r< R<+\infty$$,

$$\lim_{m\rightarrow \infty }\sup_{u\in \overline{K}_{R}\backslash K _{r}} \int_{D(m)}g(s)f \bigl(s,x(s) \bigr)\,ds=0,$$

where $$D(m)=[0,\frac{1}{m}] \cup [\frac{m-1}{m},1]$$.

We obtain the following lemma using proofs similar to those in [34, 35].

### Lemma 2.3

Suppose that $$(H_{1})$$ and $$(H_{2})$$ hold. Then $$A: \overline{K}_{R}\backslash K_{r}\rightarrow K$$ is completely continuous.

By Lemma 2.2 we can show that the spectral radius $$r(L)>0$$; see, for example, Lemma 2.5 of . Using the Krein–Rutman theorem (see Theorem 19.2 on p. 226 of ), we have the following result.

### Lemma 2.4

Suppose that $$(H_{1})$$ and $$(H_{2})$$ are satisfied. Then the first eigenvalue of L is $$\lambda_{1}=(r(L))^{-1}>0$$, and there exists a positive eigenfunction $$\varphi_{1}$$ such that $$\varphi_{1}=\lambda_{1} L \varphi_{1}$$.

The main tool in the paper is the following fixed point index theorem.

### Lemma 2.5

()

Let K be a cone in a Banach space E, and let $$T:\overline{K}_{r} \rightarrow K$$ be a completely continuous operator.

1. (i)

If there exists $$u_{0}\in K\backslash \{\theta \}$$ such that $$u-Tu\neq\mu u_{0}$$ for any $$u\in \partial K_{r}$$ and $$\mu \geq 0$$, then $$i(T,K_{r},K)=0$$.

2. (ii)

If $$Tu\neq\mu u$$ for any $$u\in \partial K_{r}$$ and $$\mu \geq 1$$, then $$i(T,K_{r},K)=1$$.

## Main results

### Theorem 3.1

Suppose that $$(H_{1})$$ and $$(H_{2})$$ are satisfied. If

$$0\leq f^{\infty }:=\limsup_{x\to +\infty }\max_{t\in [0,1]} \frac{f(t,x)}{x} < \lambda_{1}< f_{0}:=\liminf _{x\to 0}\min_{t\in [0, 1]} \frac{f(t,x)}{x}\leq + \infty ,$$

then BVP (1.1) has at least one positive solution for any

$$\lambda \in \biggl( \frac{\lambda_{1}}{ f_{0}}, \frac{\lambda_{1}}{ f ^{\infty }} \biggr) .$$
(3.1)

### Proof

By (3.1) we have $$f_{0}>\frac{\lambda_{1}}{\lambda }$$, and there exists $$r_{1}>0$$ such that $$f(t,x) \geq \frac{\lambda_{1}}{\lambda }x$$ for $$0< x\leq r_{1}$$ and $$0 \leq t \leq 1$$. For any $$x \in \partial K_{r _{1}}$$, we obtain

$$(Ax) (t)=\lambda \int_{0}^{1} G(t,s)g(s)f \bigl(s,x(s) \bigr)\,ds \geq \lambda_{1}(Lx) (t),\quad t\in [0, 1].$$

Suppose that $$\varphi_{1}$$ is the positive eigenfunction corresponding to $$\lambda_{1}$$ and that A has no fixed points on $$\partial K_{r_{1}}$$. We claim that

$$x-Ax\neq\mu \varphi_{1},\quad x\in \partial K_{r_{1}}, \mu\geq 0.$$
(3.2)

Otherwise, there would exist $$x_{1}\in \partial K_{r_{1}}$$ and $$\mu_{1}\geq 0$$ such that $$x_{1}-Ax_{1}=\mu_{1} \varphi_{1}$$. Then $$\mu_{1}> 0$$ and $$x_{1}=Ax_{1}+\mu_{1} \varphi_{1}\geq \mu_{1} \varphi _{1}$$. Denote $$\overline{\mu }=\sup \{\mu \mid x_{1}\geq \mu \varphi_{1}\}$$. Then $$\overline{\mu }\geq \mu_{1}$$, $$x_{1}\geq \overline{\mu } \varphi_{1}$$, and $$A x_{1}\geq \lambda_{1} \overline{\mu } L\varphi _{1}=\overline{\mu } \varphi_{1}$$. Thus

$$x_{1}=Ax_{1}+\mu_{1} \varphi_{1} \geq \overline{\mu } \varphi_{1}+\mu _{1} \varphi_{1}=(\overline{\mu }+\mu_{1}) \varphi_{1},$$

which contradicts to the definition of μ̅. It follows from (3.2) and Lemma 2.5(i) that

$$i(A, K_{r_{1}},K)=0.$$
(3.3)

On the other hand, by (3.1) we have $$f^{\infty }<\frac{\lambda_{1}}{ \lambda }$$, and there exist $$r_{2}>r_{1}$$ and $$0< \sigma <1$$ such that $$f(t,x)\leq \sigma \frac{\lambda_{1}}{\lambda }x$$ for $$x\geq r_{2}$$ and $$0 \leq t \leq 1$$. We define $$L_{1}u= \sigma \lambda_{1}Lu$$. Obviously, the linear operator $$L_{1}:E\rightarrow E$$ is bounded, and $$L_{1}(K) \subset K$$. From the definition of $$\lambda_{1}$$ and $$0< \sigma <1$$ it follows that

$$\bigl(r(L_{1}) \bigr)^{-1}=(\sigma \lambda_{1})^{-1} \bigl(r(L) \bigr)^{-1}=\sigma^{-1}>1.$$
(3.4)

Choose $$\varepsilon_{0}=\frac{1}{2}(1-r(L_{1}))$$. Then by Gelfand’s formula there exists a natural number $$N\geq 1$$ such that $$\Vert L^{k}_{1} \Vert \leq [r(L_{1})+\varepsilon_{0}]^{k}$$ for $$k\geq N$$. We now define

$$\Vert x \Vert ^{*}=\sum_{i=1}^{N} \bigl[r(L_{1})+\varepsilon_{0} \bigr]^{N-i} \bigl\Vert L^{i-1}_{1}x \bigr\Vert ,\quad x\in E,$$

where $$L^{0}_{1}=I$$ is the identity operator. Since $$L_{1}$$ is linear, it is easy to verify that $$\Vert x \Vert ^{*}$$ is a norm in E. Let $$M_{0}=\sup_{x\in \partial K_{r_{2}}}\lambda \int_{0}^{1}G(1,s)g(s)f(s,x(s))\,ds$$. Then $$M_{0} < + \infty$$. We define $$M_{0}^{*}=\Vert M_{0} \Vert ^{*}$$ and take $$r_{3}>\max \{r_{2}, 2M_{0}^{*}\varepsilon^{-1}_{0}\}$$. Noting that $$\Vert x \Vert ^{*}>[r(L_{1})+\varepsilon_{0}]^{N-1}\Vert x \Vert$$, we can find $$r_{4}>r_{3}$$ large enough such that $$\Vert x \Vert \geq r_{4}$$and thus $$\Vert x \Vert ^{*}>r_{3}$$.

We next prove that

$$Ax\neq\mu x,\quad x\in \partial K_{r_{4}}, \mu \geq 1.$$
(3.5)

Arguing indirectly, we get that there exist $$x_{2}\in \partial K_{r _{4}}$$ and $$\mu_{2} \geq 1$$ such that $$Ax_{2}=\mu_{2}x_{2}$$. We define $$\widetilde{x}(t)=\min \{x_{2}(t), r_{2}\}$$ for $$t\in [0,1]$$ and $$H(x_{2})=\{t\in [0,1]: x_{2}(t)>r_{2}\}$$. It is easy to see that $$\Vert \widetilde{x} \Vert =r_{2}$$. We have $$\widetilde{x}\in \partial K_{r _{2}}$$ since $$\widetilde{x}(t)=\min \{x_{2}(t), r_{2}\}\geq \min \{p(t)r _{4}, r_{2}\}\geq p(t)r_{2}$$, $$t\in [0,1]$$. It follows that

\begin{aligned} \mu_{2}x_{2}(t)&=(Ax_{2}) (t) \\ &=\lambda \int_{0}^{1}G(t,s)g(s)f \bigl(s,x_{2}(s) \bigr)\,ds \\ & \leq \lambda \int_{H(x_{2})}G(t,s)g(s)f \bigl(s,x_{2}(s) \bigr)\,ds + \lambda \int_{[0,1]\backslash H(x_{2})}G(1,s)g(s)f \bigl(s,x _{2}(s) \bigr)\,ds \\ & \leq \sigma \lambda_{1} \int_{0}^{1}G(t,s)g(s)x_{2}(s)\,ds + \lambda \int_{0}^{1}G(1,s)g(s)f \bigl(s,\widetilde{x}(s) \bigr)\,ds \\ & \leq (L_{1}x_{2}) (t)+M_{0},\quad t\in [0,1]. \end{aligned}

Since $$L_{1}(K)\subset K$$, we have $$0\leq (L^{j}_{1}(Ax_{2})(t)) \leq (L^{j}_{1}(L_{1}x_{2}+M_{0})(t))$$, $$j=0,1,2,\ldots , N-1$$. Then $$\Vert L^{j}_{1}(Ax_{2}) \Vert \leq \Vert L^{j}_{1}(L_{1}x_{2}+M_{0}) \Vert$$, $$j=0,1,2, \ldots , N-1$$, and hence

$$\Vert Ax_{2} \Vert ^{*}\leq \sum _{i=1}^{N} \bigl[r(L_{1})+ \varepsilon_{0} \bigr]^{N-i}\bigl\Vert L^{i-1}_{1}(L_{1}x_{2}+M_{0}) \bigr\Vert = \Vert L_{1}x_{2}+M_{0} \Vert ^{*}.$$

Therefore we obtain

\begin{aligned} \mu_{2}\Vert x_{2} \Vert ^{*} &=\Vert Ax_{2} \Vert ^{*} \\ &\leq \Vert L_{1}x_{2} \Vert ^{*}+M_{0}^{*} \\ &= \sum _{i=1}^{N} \bigl[r(L_{1})+ \varepsilon_{0} \bigr]^{N-i}\bigl\Vert L^{i}_{1}x_{2} \bigr\Vert +M _{0}^{*} \\ & \leq \bigl[r(L_{1})+\varepsilon_{0} \bigr]\sum _{i=1}^{N-1} \bigl[r(L_{1})+ \varepsilon_{0} \bigr]^{N-i-1}\bigl\Vert L^{i}_{1}x_{2} \bigr\Vert + \bigl[r(L_{1})+\varepsilon_{0} \bigr]^{N} \Vert x_{2} \Vert +M_{0}^{*} \\ & = \bigl[r(L_{1})+\varepsilon_{0} \bigr]\sum _{i=1}^{N} \bigl[r(L _{1})+ \varepsilon_{0} \bigr]^{N-i}\bigl\Vert L^{i-1}_{1}x_{2} \bigr\Vert +M_{0}^{*} \\ & = \bigl[r(L _{1})+\varepsilon_{0} \bigr] \Vert x_{2} \Vert ^{*}+M_{0}^{*} \\ &\leq \bigl[r(L_{1})+ \varepsilon_{0} \bigr]\Vert x_{2} \Vert ^{*}+\frac{\varepsilon_{0}}{2}r_{3} \\ & < \bigl[r(L _{1})+\varepsilon_{0} \bigr] \Vert x_{2} \Vert ^{*}+\frac{\varepsilon_{0}}{2}\Vert x_{2} \Vert ^{*} \\ &= \biggl[ \frac{1}{4}r(L_{1})+ \frac{3}{4} \biggr] \Vert x_{2} \Vert ^{*}. \end{aligned}

Thus $$\frac{1}{4}r(L_{1})+\frac{3}{4}\geq 1$$, that is, $$r(L_{1}) \geq 1$$, which contradicts (3.4). It follows from (3.5) and Lemma 2.5(ii) that

$$i(A,K_{r_{4}},K)=1.$$
(3.6)

By (3.3), (3.6), and the additivity of the fixed point index we have

$$i(A,K_{r_{4}}\backslash \overline{K}_{r_{1}},K)=i(A,K_{r_{4}},K)-i(A, K_{r_{1}},K)=1.$$

Therefore A has at least one fixed point $$x^{*}\in K_{r_{4}}\backslash \overline{K}_{r_{1}}$$, which is a positive solution of BVP (1.1). □

## An example

Let $$\alpha =\frac{7}{2}$$, $$\beta =\frac{3}{2}$$, $$\gamma =\frac{1}{2}, \alpha_{i}=\frac{2}{i^{2}}$$, $$\xi_{i}=1-\frac{1}{i+1} (i=1,2,\ldots )$$, $$g(t)=\frac{1}{\sqrt{t(1-t)}}$$,$$f(t,x)=\sqrt{2-t+\vert \ln x \vert }$$. Consider the following fractional BVP:

$$\textstyle\begin{cases} D^{\frac{7}{2}}_{0^{+}}x(t)+\lambda \frac{1}{\sqrt{t(1-t)}}\sqrt{2-t+\vert \ln x(t) \vert } =0, & t\in (0,1), \\ x(0)=x'(0)=x''(0)=0,& D^{\frac{3}{2}}_{0^{+}}x(1)=\sum_{i=1}^{\infty }\frac{2}{i^{2}}D^{ \frac{1}{2}}_{0^{+}}x ( 1-\frac{1}{i+1} ) . \end{cases}$$
(4.1)

Direct computation shows that $$\Gamma (\alpha -\beta )=1, \Gamma (\alpha -\gamma )=2$$, $$\sum_{i=1}^{\infty }\alpha_{i}\xi_{i} ^{\alpha -\gamma -1}=2 ( \frac{\pi^{2}}{6}-1 )$$, and $$\frac{1}{\Gamma (\alpha -\beta )}-\frac{1}{\Gamma (\alpha -\gamma )} \sum_{i=1}^{\infty }\alpha_{i}\xi_{i}^{\alpha -\gamma -1}\approx 0.355>0$$.

Let $$K=\{ x\in C[0,1]: x(t)\geq p(t)\Vert x \Vert , t\in [0,1] \}$$, where $$p(t)=t^{\frac{5}{2}}$$. For $$x\in \overline{K}_{R}\backslash K_{r}$$, we obtain $$\vert \ln x(t) \vert \leq \vert \ln rp(t) \vert +\vert \ln R \vert$$. Due to $$\int_{0}^{1}\vert \ln p(t) \vert dt =\frac{5}{2}$$, we have $$\lim_{m\rightarrow \infty }\int_{D(m)}\vert \ln p(t) \vert \,dt=0$$. Since $$0\leq G(t,s)\leq G(1,s) \leq \frac{1}{\Gamma (\frac{7}{2})(2-\frac{\pi^{2}}{6})}$$, it follows that $$\int_{0}^{1}G(1,s)g(s)\,ds\leq \frac{1}{\Gamma (\frac{7}{2})(2-\frac{\pi^{2}}{6})} \int_{0}^{1}g(s)\,ds=\frac{2[\Gamma (\frac{3}{4})]^{2}}{\Gamma (\frac{7}{2})(2-\frac{\pi^{2}}{6})\sqrt{\pi }}$$. For $$x\in \overline{K}_{R}\backslash K_{r}$$, we have

$$\int_{0}^{1}f^{2} \bigl(s,x(s) \bigr) \,ds \leq \int_{0}^{1} \bigl(2-s+\vert \ln r \vert + \vert \ln R \vert +\bigl\vert \ln p(s) \bigr\vert \bigr)\,ds=4+\vert \ln r \vert +\vert \ln R \vert .$$

Therefore

\begin{aligned} &\lim_{m\rightarrow \infty }\sup_{x\in \overline{K}_{R}\backslash K _{r}} \int_{D(m)}g(s)f \bigl(s,x(s) \bigr)\,ds \\ & \quad \leq \lim_{m\rightarrow \infty }\sup_{x\in \overline{K}_{R}\backslash K _{r}} \biggl( \int_{D(m)}g^{2}(s)\,ds \biggr) ^{\frac{1}{2}} \biggl( \int_{D(m)}f ^{2} \bigl(s,x(s) \bigr)\,ds \biggr) ^{\frac{1}{2}} \\ & \quad \leq \lim_{m\rightarrow \infty }\sqrt{\pi } \biggl( \int_{D(m)} \bigl(2-s+\vert \ln r \vert +\vert \ln R \vert +\bigl\vert \ln p(s) \bigr\vert \bigr)\,ds \biggr) ^{\frac{1}{2}}=0. \end{aligned}

Direct computation yields $$f^{\infty }=0$$ and $$f_{0}=+\infty$$. Using Theorem 3.1, we can conclude that the BVP (4.1) has at least one positive solution for any $$\lambda \in (0,+\infty )$$.

## Conclusions

We established the existence of positive solutions for the singular fractional differential equation infinite-point BVP (1.1) using the fixed point index theory in cones. Note that the nonlinearity may possess singularities, that is, $$f(t,x)$$ may have a singularity at $$x = 0$$, and $$g(t)$$ may be singular at $$t= 0$$ or $$t=1$$.

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## Funding

Supported financially by the National Natural Science Foundation of China (11501318, 11871302), the China Postdoctoral Science Foundation (2017M612230), and the Natural Science Foundation of Shandong Province of China (ZR2017MA036).

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### Contributions

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

### Corresponding author

Correspondence to Xinan Hao.

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### Competing interests

The authors declare that they have no competing interests.

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