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.

1 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 [1–4]. For the development of the fractional differential equations, see [5–23] 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 [24–33]. Using the compression expansion fixed point theorem due to Krasnosel’skii, Henderson and Luca [27] 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 [28], 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 [31] 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 $t=1$.

2 Preliminaries and lemmas

Definition 2.1

([1–4])

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

([1–4])

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:

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

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

• (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 [36]. Using the Krein–Rutman theorem (see Theorem 19.2 on p. 226 of [37]), 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

([38])

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

• (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 $i(T,K_{r},K)=0$.

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

3 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). □

4 An example

Let $\alpha =\frac{7}{2}$, $\beta =\frac{3}{2}$, $\gamma =\frac{1}{2}, \alpha_{i}=\frac{2}{i^{2}}$, $\xi_{i}=1-\frac{1}{i+1} (i=1,2,\ldots )$, $g(t)=\frac{1}{\sqrt[4]{t(1-t)}}$,$f(t,x)=\sqrt{2-t+\vert \ln x \vert }$. Consider the following fractional BVP:

$$ \textstyle\begin{cases} D^{\frac{7}{2}}_{0^{+}}x(t)+\lambda \frac{1}{\sqrt[4]{t(1-t)}}\sqrt{2-t+\vert \ln x(t) \vert } =0, & t\in (0,1), \\ x(0)=x'(0)=x''(0)=0,& D^{\frac{3}{2}}_{0^{+}}x(1)=\sum_{i=1}^{\infty }\frac{2}{i^{2}}D^{ \frac{1}{2}}_{0^{+}}x ( 1-\frac{1}{i+1} ) . \end{cases} $$

(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 $\lambda \in (0,+\infty )$.

5 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 $t=1$.