# Solutions for a class of fractional Langevin equations with integral and anti-periodic boundary conditions

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## Abstract

In this paper, we consider a class of fractional Langevin equations with integral and anti-periodic boundary conditions. By using some fixed point theorems and the Leray–Schauder degree theory, several new existence results of solutions are obtained.

## Introduction

In this paper, we consider the existence of solutions for the following fractional Langevin equation with integral and anti-periodic conditions:

\begin{aligned} \textstyle\begin{cases} ^{c}D^{\beta }_{0^{+}}(^{c}D^{\alpha }_{0^{+}}+\lambda)x(t)=f(t,x(t)),\quad 0< t< 1, \\ x(0)=0,\qquad x(1)=\mu \int_{0}^{1}x(s)\,ds,\qquad {}^{c}D^{\alpha }_{0^{+}}x(0)+^{c}D ^{\alpha }_{0^{+}}x(1)=0, \end{cases}\displaystyle \end{aligned}
(1)

where $$0<\alpha <1$$, $$1<\beta \leq 2$$, $$\lambda >0$$, and $$\mu >0$$ are real numbers, $${}^{c}D^{\alpha }_{0^{+}}x(t)$$ and $${}^{c}D^{\beta }_{0^{+}}x(t)$$ are the Caputo fractional derivatives, and $$f\in C([0,1]\times R,R)$$.

Theory of fractional differential equations has important application in many areas. It has become a new research field in differential equations . There are a lot of good research results on boundary value problems of fractional differential equations . Recently fractional Langevin equations have been studied by some scholars (see, for example, ).

In , via fixed point theorems, Ahmad et al. discussed the existence of solutions for fractional Langevin equations with three-point nonlocal boundary value conditions.

In , Li et al. investigated the infinite-point boundary value problem of fractional Langevin equations. By means of the nonlinear alternative and Leray–Schauder degree theory, they got some existence results for the boundary value problem.

To our knowledge, for fractional Langevin equation, the boundary value problem with integral and anti-periodic boundary conditions is rarely studied, so the research of this paper is new.

The rest of this paper is organized as follows. In Sect. 2, we present some preliminaries and lemmas. In Sect. 3, we adopt some fixed point theorems and the Leray–Schauder degree theory to obtain the existence of solutions for boundary value problem (1).

## Relevant lemmas

Some necessary definitions from fractional calculus theory can be found in [2, 3]. We omit them here.

### Lemma 2.1

Let $$h\in C(0,1)\cap L(0,1)$$, $$0<\alpha <1$$, $$1<\beta \leq 2$$, then the following problem

\begin{aligned} \textstyle\begin{cases} ^{c}D^{\beta }_{0^{+}}(^{c}D^{\alpha }_{0^{+}}+\lambda)x(t)=h(t),\quad t \in (0,1) \\ x(0)=0,\qquad x(1)=\mu \int_{0}^{1}x(s)\,ds,\qquad {}^{c}D^{\alpha }_{0^{+}}x(0)+^{c}D ^{\alpha }_{0^{+}}x(1)=0 \end{cases}\displaystyle \end{aligned}
(2)

has a solution $$x(t)$$ satisfying

\begin{aligned} x(t) =&\frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha + \beta -1}h(s)\,ds - \frac{t^{\alpha }(2t-\alpha -1)}{(1-\alpha)\Gamma ( \alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1}h(s)\,ds \\ &{}+ \frac{\lambda t^{\alpha }(2t-\alpha -1)}{(1-\alpha)\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\ &{} -\frac{t^{\alpha }(1-t)}{(1- \alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}h(s)\,ds \\ &{} +\frac{\lambda \mu t^{\alpha }(1-t)}{(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1}x(s)\,ds \\ &{} +\frac{\mu t^{\alpha }(2t-\alpha -1)}{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{ \alpha -1}x(s)\,ds. \end{aligned}

### Proof

From the relevant lemma in , it follows that

\begin{aligned}& \bigl(^{c}D^{\alpha }_{0^{+}}+\lambda\bigr)x(t)=I_{0^{+}}^{\beta }h(t)+c_{0}+c _{1}t, \\& {}^{c}D^{\alpha }_{0^{+}}x(t)=I_{0^{+}}^{\beta }h(t)+c_{0}+c_{1}t- \lambda x(t)=\frac{1}{\Gamma (\beta)} \int_{0}^{t}(t-s)^{\beta -1}h(s)\,ds +c_{0}+c_{1}t-\lambda x(t), \\& x(t)=I_{0^{+}}^{\alpha +\beta }h(t)+I_{0^{+}}^{\alpha }c_{0}+I_{0^{+}} ^{\alpha }c_{1}t-I_{0^{+}}^{\alpha }\lambda x(t)+c_{2} \\& \hphantom{x(t)}=\frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha +\beta -1}h(s)\,ds + \frac{c_{0}}{\Gamma (\alpha +1)}t^{\alpha } +\frac{c_{1}}{\Gamma ( \alpha +2)}t^{\alpha +1} \\& \hphantom{x(t)=} -\frac{\lambda }{\Gamma (\alpha)} \int_{0} ^{t}(t-s)^{\alpha -1}x(s) \,ds+c_{2}. \end{aligned}

By the boundary value conditions $$x(0)=0$$, $$x(1)=\mu \int_{0}^{1}x(s)\,ds$$, and $${}^{c}D^{\alpha }_{0^{+}}x(0)+$$ $${}^{c}D^{\alpha }_{0^{+}}x(1)=0$$, we can get

\begin{aligned}& c_{2}=0, \\& \frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1}h(s)\,ds + \frac{c_{0}}{\Gamma (\alpha +1)} +\frac{c_{1}}{\Gamma (\alpha +2)} -\frac{ \lambda }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\& \quad =\mu \int_{0}^{1}x(s)\,ds, \end{aligned}

and

$$c_{0}+\frac{1}{\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}h(s)\,ds +c _{0}+c_{1}-\lambda \mu \int_{0}^{1}x(s)\,ds=0.$$

By direct computation, we have

\begin{aligned} c_{0} =&\frac{\Gamma (\alpha +2)}{(1-\alpha)\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1}h(s)\,ds - \frac{\lambda \alpha (1+ \alpha)}{1-\alpha } \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\ &{} +\frac{\lambda \mu }{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{\mu \Gamma (\alpha +2)}{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{1}{(1- \alpha)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}h(s)\,ds, \end{aligned}

and

\begin{aligned} c_{1} =&\frac{2\lambda \alpha (1+\alpha)}{1-\alpha } \int_{0}^{1}(1-s)^{ \alpha -1}x(s)\,ds + \frac{1+\alpha }{(1-\alpha)\Gamma (\beta)} \int _{0}^{1}(1-s)^{\beta -1}h(s)\,ds \\ &{} -\frac{\lambda \mu (1+\alpha)}{1- \alpha } \int_{0}^{1}x(s)\,ds \\ &{} +\frac{2\mu \Gamma (\alpha +2)}{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{2 \Gamma (\alpha +2)}{(1-\alpha)\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{ \alpha +\beta -1}h(s)\,ds. \end{aligned}

Thus, the solution of (2) satisfies

\begin{aligned} x(t) =&\frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha + \beta -1}h(s)\,ds - \frac{t^{\alpha }(2t-\alpha -1)}{(1-\alpha)\Gamma ( \alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1}h(s)\,ds \\ &{} + \frac{\lambda t^{\alpha }(2t-\alpha -1)}{(1-\alpha)\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\ &{} -\frac{t^{\alpha }(1-t)}{(1- \alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}h(s)\,ds \\ &{} +\frac{\lambda \mu t^{\alpha }(1-t)}{(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1}x(s)\,ds +\frac{\mu t^{\alpha }(2t-\alpha -1)}{1-\alpha } \int_{0}^{1}x(s)\,ds \\ &{} -\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{ \alpha -1}x(s)\,ds. \end{aligned}

This completes the proof. □

## Main results

Let $$E=C[0,1]$$. Obviously, the space E is a Banach space if it is endowed with the norm as follows:

$$\Vert x \Vert =\max_{t\in [0,1]} \bigl\vert x(t) \bigr\vert .$$

Define an operator $$T:E\rightarrow E$$ as

\begin{aligned} (Tx) (t) =&\frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha + \beta -1}f\bigl(s,x(s)\bigr) \,ds -\frac{t^{\alpha }(2t-\alpha -1)}{(1-\alpha) \Gamma (\alpha +\beta)} \\ &{}\times \int_{0}^{1}(1-s)^{\alpha +\beta -1}f\bigl(s,x(s)\bigr) \,ds +\frac{\lambda t^{ \alpha }(2t-\alpha -1)}{(1-\alpha)\Gamma (\alpha)} \int_{0}^{1}(1-s)^{ \alpha -1}x(s)\,ds \\ &{} - \frac{t^{\alpha }(1-t)}{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}f\bigl(s,x(s)\bigr) \,ds+\frac{\lambda \mu t^{\alpha }(1-t)}{(1- \alpha)\Gamma (\alpha +1)} \int_{0}^{1}x(s)\,ds \\ &{} +\frac{\mu t^{\alpha }(2t-\alpha -1)}{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{ \lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}x(s)\,ds,\quad t \in [0,1]. \end{aligned}
(3)

It is easy to see that the solution for (1) is equivalent to the fixed point of T.

### Lemma 3.1

1. (i)

$$\max_{t\in [0,1]}|t^{\alpha }(2t-\alpha -1)|=\max \{(\frac{ \alpha }{2})^{\alpha },1-\alpha \}$$;

2. (ii)

$$\max_{t\in [0,1]}t^{\alpha }(1-t)=\frac{\alpha^{\alpha }}{(1+ \alpha)^{1+\alpha }}$$.

### Proof

(i) Let $$g(t)=t^{\alpha }(2t-\alpha -1)$$, $$0\leq t\leq 1$$, then $$g'(t)=(\alpha +1)t^{\alpha -1}(2t-\alpha)$$, $$g(0)=0$$, $$g(1)=1-\alpha >0$$.

When $$0\leq t<\frac{\alpha }{2}$$, $$g'(t)\leq 0$$; when $$\frac{\alpha }{2}< t\leq 1$$, $$g'(t)\geq 0$$. In conclusion, $$|g(t)|_{ \max }=\max \{|g(\frac{\alpha }{2})|,g(1))\}=\max \{( \frac{\alpha }{2})^{\alpha },1-\alpha \}$$.

(ii) Let $$g(t)=t^{\alpha }(1-t)$$, $$0\leq t\leq 1$$, then $$g'(t)=t^{ \alpha -1}[\alpha -(\alpha +1)t]$$.

When $$0\leq t<\frac{\alpha }{1+\alpha }$$, $$g'(t)\geq 0$$; when $$\frac{\alpha }{1+\alpha }< t\leq 1$$, $$g'(t)\leq 0$$. So $$g(t)_{\max }=g(\frac{ \alpha }{1+\alpha })=\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }}$$.

The proof is completed. □

Let $$\eta =\frac{1}{1-\alpha }\max \{(\frac{\alpha }{2})^{\alpha },1- \alpha \}$$.

### Lemma 3.2

$$T:E\rightarrow E$$ is completely continuous.

### Proof

Since the continuity of f, $$T:E\rightarrow E$$ is continuous. For any bounded set $$D\subset E$$, there exists $$K>0$$ such that $$\forall x\in D$$, $$\|x\|\leq K$$. There exists a constant $$L_{1}>0$$ such that $$|f(t,x)|\leq L_{1}$$ for any $$t\in [0,1]$$ and $$x\in [-K,K]$$. Then $$\forall x\in D$$, it follows that

\begin{aligned} \bigl\vert (Tx) (t) \bigr\vert \leq& \frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{ \alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\eta }{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \eta }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma ( \alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds + \mu \eta \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ \leq& \frac{L_{1}}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{ \alpha +\beta -1}\,ds + \frac{L_{1}\eta }{\Gamma (\alpha +\beta)} \int _{0}^{1}(1-s)^{\alpha +\beta -1}\,ds \\ &{} +\frac{\eta \lambda \Vert x \Vert }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1}\,ds \\ &{} +\frac{\alpha^{\alpha }L_{1}}{(1+\alpha)^{1+\alpha }(1-\alpha) \Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}\,ds \\ &{} +\frac{ \lambda \mu \alpha^{\alpha } \Vert x \Vert }{(1+\alpha)^{1+\alpha }(1-\alpha) \Gamma (\alpha +1)} \\ &{} +\eta \mu \Vert x \Vert +\frac{\lambda \Vert x \Vert }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{ \alpha -1}\,ds \\ =&\frac{L_{1}t^{\alpha +\beta }}{\Gamma (\alpha +\beta +1)} +\frac{ \eta L_{1}}{\eta \Gamma (\alpha +\beta +1)} +\frac{\lambda \Vert x \Vert }{ \Gamma (\alpha +1)} \\ &{} +\frac{\alpha^{\alpha }L_{1}}{(1+\alpha)^{1+ \alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha } \Vert x \Vert }{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu \Vert x \Vert +\frac{\lambda \Vert x \Vert t^{ \alpha }}{\Gamma (\alpha +1)} \\ \leq& \frac{(1+\eta)L_{1}}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda \Vert x \Vert }{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }L_{1}}{(1+ \alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha } \Vert x \Vert }{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu \Vert x \Vert \\ \leq& \frac{(1+\eta)L_{1}}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda K}{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }L_{1}}{(1+ \alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha }K}{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu K, \end{aligned}

which implies that TD is uniformly bounded.

In addition, for $$x\in D$$, $$0\leq t_{1}< t_{2}\leq 1$$, we have

\begin{aligned}& \bigl\vert (Tx) (t_{2})-(Tx) (t_{1}) \bigr\vert \\& \quad =|\frac{1}{\Gamma (\alpha +\beta)} \int_{0} ^{t_{2}}(t_{2}-s)^{\alpha +\beta -1}f \bigl(s,x(s)\bigr)\,ds -\frac{t_{2}^{\alpha }(2t_{2}-\alpha -1)}{(1-\alpha)\Gamma (\alpha +\beta)} \\& \qquad {}\times \int_{0}^{1}(1-s)^{\alpha +\beta -1}f\bigl(s,x(s)\bigr) \,ds +\frac{\lambda t_{2} ^{\alpha }(2t_{2}-\alpha -1)}{(1-\alpha)\Gamma (\alpha)} \int_{0} ^{1}(1-s)^{\alpha -1}x(s)\,ds \\& \qquad {} -\frac{t_{2}^{\alpha }(1-t_{2})}{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1}f\bigl(s,x(s)\bigr) \,ds +\frac{\lambda \mu t_{2}^{\alpha }(1-t_{2})}{(1-\alpha)\Gamma (\alpha +1)} \int_{0} ^{1}x(s)\,ds \\& \qquad {} +\frac{\mu t_{2}^{\alpha }(2t_{2}-\alpha -1)}{1-\alpha } \int_{0}^{1}x(s)\,ds -\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t_{2}}(t_{2}-s)^{\alpha -1}x(s) \,ds \\& \qquad {} -\frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t_{1}}(t_{1}-s)^{\alpha + \beta -1}f \bigl(s,x(s)\bigr)\,ds \\& \qquad {} +\frac{t_{1}^{\alpha }(2t_{1}-\alpha -1)}{(1- \alpha)\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1}f\bigl(s,x(s)\bigr) \,ds \\& \qquad {} -\frac{\lambda t_{1}^{\alpha }(2t_{1}-\alpha -1)}{(1-\alpha)\Gamma ( \alpha)} \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\& \qquad {} +\frac{t_{1}^{\alpha }(1-t _{1})}{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{ \beta -1}f\bigl(s,x(s)\bigr) \,ds \\& \qquad {} -\frac{\lambda \mu t_{1}^{\alpha }(1-t_{1})}{(1-\alpha)\Gamma ( \alpha +1)} \int_{0}^{1}x(s)\,ds \\& \qquad {} -\frac{\mu t_{1}^{\alpha }(2t_{1}- \alpha -1)}{1-\alpha } \int_{0}^{1}x(s)\,ds +\frac{\lambda }{\Gamma ( \alpha)} \int_{0}^{t_{1}}(t_{1}-s)^{\alpha -1}x(s) \,ds \\& \quad = \biggl\vert \int_{0}^{t_{1}}\frac{f(s,x(s))}{\Gamma (\alpha +\beta)} \bigl[(t_{2}-s)^{ \alpha +\beta -1}-(t_{1}-s)^{\alpha +\beta -1}\bigr] \,ds + \int_{t_{1}}^{t _{2}}\frac{f(s,x(s))}{\Gamma (\alpha +\beta)}(t_{2}-s)^{\alpha + \beta -1} \,ds \\& \qquad {} +\frac{(\alpha +1)(t_{2}^{\alpha }-t_{1}^{\alpha })+2(t_{1}^{\alpha +1}-t _{2}^{\alpha +1})}{(1-\alpha)\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{ \alpha +\beta -1}f\bigl(s,x(s)\bigr) \,ds \\& \qquad {} +\frac{\lambda (\alpha +1)(t_{1}^{\alpha }-t_{2}^{\alpha })+2\lambda (t_{2}^{\alpha +1} -t_{1}^{\alpha +1})}{(1-\alpha)\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1}x(s)\,ds \\& \qquad {} +\frac{(t_{2}^{\alpha +1}-t_{1}^{\alpha +1})+(t_{1}^{\alpha }-t_{2} ^{\alpha })}{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta)} \int_{0} ^{1}(1-s)^{\beta -1}f\bigl(s,x(s)\bigr) \,ds \\& \qquad {} +\frac{\lambda \mu (t_{1}^{\alpha +1}-t_{2}^{\alpha +1})+\lambda \mu (t_{2}^{\alpha }-t_{1}^{\alpha })}{(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1}x(s)\,ds \\& \qquad {} +\frac{2\mu (t_{2}^{\alpha +1}-t_{1}^{\alpha +1})+\mu (\alpha +1)(t _{1}^{\alpha }-t_{2}^{\alpha })}{1-\alpha } \int_{0}^{1}x(s)\,ds \\& \qquad {} + \int_{0}^{t_{1}}\frac{\lambda x(s)}{\Gamma (\alpha)}\bigl[(t_{1}-s)^{ \alpha -1}-(t_{2}-s)^{\alpha -1} \bigr]\,ds - \int_{t_{1}}^{t_{2}}\frac{\lambda x(s)}{\Gamma (\alpha)}(t_{2}-s)^{\alpha -1} \,ds \biggr\vert \\& \quad \leq \frac{L_{1} \vert t_{2}^{\alpha +\beta }-t_{1}^{\alpha +\beta } \vert }{ \Gamma (\alpha +\beta +1)} +\frac{L_{1} \vert (\alpha +1)(t_{2}^{\alpha }-t _{1}^{\alpha })+2(t_{1}^{\alpha +1}-t_{2}^{\alpha +1}) \vert }{(1-\alpha) \Gamma (\alpha +\beta +1)} \\& \qquad {} +\frac{ \vert \lambda (\alpha +1)(t_{1}^{\alpha }-t_{2}^{\alpha }) +2\lambda (t_{2}^{\alpha +1}-t_{1}^{\alpha +1}) \vert \Vert x \Vert }{(1-\alpha)\Gamma ( \alpha +1)} \\& \qquad {} +\frac{L_{1} \vert (t_{2}^{\alpha +1}-t_{1}^{\alpha +1})+(t_{1}^{\alpha }-t _{2}^{\alpha }) \vert }{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} +\frac{ \vert \lambda \mu (t_{1}^{\alpha +1}-t_{2}^{\alpha +1})+\lambda \mu (t_{2} ^{\alpha }-t_{1}^{\alpha }) \vert \Vert x \Vert }{(1-\alpha)\Gamma (\alpha +1)} \\& \qquad {} +\frac{ \vert 2\mu (t_{2}^{\alpha +1}-t_{1}^{\alpha +1})+\mu (\alpha +1)(t _{1}^{\alpha }-t_{2}^{\alpha }) \vert \Vert x \Vert }{1-\alpha } +\frac{\lambda \Vert x \Vert \vert t_{1}^{\alpha }-t_{2}^{\alpha } \vert }{\Gamma (\alpha +1)} \\& \quad \leq \frac{L_{1}}{\Gamma (\alpha +\beta +1)}\bigl(t_{2}^{\alpha +\beta }-t _{1}^{\alpha +\beta }\bigr) +\biggl[\frac{2L_{1}}{(1-\alpha)\Gamma (\alpha + \beta +1)} + \frac{2\lambda K}{(1-\alpha)\Gamma (\alpha +1)} \\& \qquad {} +\frac{L_{1}}{(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} +\frac{ \lambda \mu K}{(1-\alpha)\Gamma (\alpha +1)} +\frac{2\mu K}{1-\alpha }\biggr] \bigl(t_{2}^{\alpha +1}-t_{1}^{\alpha +1}\bigr) \\& \qquad {} +\biggl[\frac{L_{1}(\alpha +1)}{(1-\alpha)\Gamma (\alpha +\beta +1)} +\frac{ \lambda K(\alpha +1)}{(1-\alpha)\Gamma (\alpha +1)} +\frac{L_{1}}{(1- \alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\& \qquad {} +\frac{\lambda \mu K}{(1-\alpha)\Gamma (\alpha +1)}+\frac{\mu K( \alpha +1)}{1-\alpha } +\frac{\lambda K}{\Gamma (\alpha +1)}\biggr] \bigl(t_{2} ^{\alpha }-t_{1}^{\alpha }\bigr), \end{aligned}

which implies that TD is equicontinuous. Thus, by the Arzel àAscoli theorem, $$T:E\rightarrow E$$ is completely continuous.

The proof is completed. □

### Theorem 3.1

Suppose that f satisfies the Lipschitz condition

$$\bigl\vert f(t,x)-f(t,y) \bigr\vert \leq L \vert x-y \vert ,\quad \forall t\in [0,1],x,y\in R,$$

and

\begin{aligned} A =&\frac{(1+\eta)L}{\Gamma (\alpha +\beta +1)} +\frac{(1+\eta)\lambda }{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }L}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{}+\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} +\eta \mu < 1. \end{aligned}

Then (1) has a unique solution.

### Proof

Define $$Q=\max_{t\in [0,1]}|f(t,0)|$$ and select $$r\geq \frac{\frac{2Q}{\Gamma (\alpha +\beta +1)} +\frac{\alpha^{ \alpha }Q}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)}}{1-A}$$, define a closed ball as $$B_{r}=\{x\in E:\|x\| \leq r\}$$, then for $$x\in B_{r}$$, we have

\begin{aligned} \bigl\vert (Tx) (t) \bigr\vert \leq& \frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{ \alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\eta }{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \eta }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma ( \alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds + \mu \eta \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ \leq& \frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha + \beta -1}\bigl( \bigl\vert f \bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) \,ds \\ &{} +\frac{\eta }{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha + \beta -1}\bigl( \bigl\vert f \bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) \,ds \\ &{} +\frac{\lambda \eta }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds +\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma ( \alpha +1)\Gamma (\beta)} \\ &{}\times \int_{0}^{1}(1-s)^{\beta -1}\bigl( \bigl\vert f\bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) \,ds \\ &{} +\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds + \mu \eta \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ \leq& \frac{(Lr+Q)t^{\alpha +\beta }}{\Gamma (\alpha +\beta +1)} +\frac{ \eta (Lr+Q)}{\Gamma (\alpha +\beta +1)} +\frac{\lambda r\eta }{\Gamma (\alpha +1)} \\ &{} +\frac{\alpha^{\alpha }(Lr+Q)}{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha }r}{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu r +\frac{\lambda rt^{\alpha }}{ \Gamma (\alpha +1)} \\ \leq& \frac{(1+\eta)(Lr+Q)}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda r}{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }(Lr+Q)}{(1+ \alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha }r}{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu r\leq r, \end{aligned}

which implies that $$\|Tx\|\leq r$$, that is, $$T(B_{r})\subset B_{r}$$. In what follows, for $$x,y\in E$$, for each $$t\in [0,1]$$, we can get that

\begin{aligned}& \bigl\vert (Tx) (t)-(Ty) (t) \bigr\vert \\& \quad \leq \frac{1}{\Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{ \alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert \,ds \\& \qquad {} +\frac{\eta }{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha + \beta -1} \bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert \,ds \\& \qquad {} +\frac{\lambda \eta }{\Gamma (\alpha)} \int_{0}^{1}(1-s)^{\alpha -1} \bigl\vert x(s)-y(s) \bigr\vert \,ds \\& \qquad {} +\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma ( \alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1} \bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert \,ds \\& \qquad {} +\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1} \bigl\vert x(s)-y(s) \bigr\vert \,ds +\mu \eta \int_{0}^{1} \bigl\vert x(s)-y(s) \bigr\vert \,ds \\& \qquad {} +\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1} \bigl\vert x(s)-y(s) \bigr\vert \,ds \\& \quad \leq \frac{(1+\eta)L \Vert x-y \Vert }{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda \Vert x-y \Vert }{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }L \Vert x-y \Vert }{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma ( \beta +1)} \\& \qquad {} +\frac{\lambda \mu \alpha^{\alpha } \Vert x-y \Vert }{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu \Vert x-y \Vert =A \Vert x-y \Vert , \end{aligned}

which implies T is a contraction. Thus, by Banach fixed point theorem , T has a unique fixed point, that is, (1) has a unique solution.

The proof is completed. □

### Theorem 3.2

Suppose there exist $$M>0$$ and $$c\geq 0$$ with $$\frac{(1+\eta)c}{ \Gamma (\alpha +\beta +1)} + \frac{(1+\eta)\lambda }{\Gamma (\alpha +1)} + \frac{\alpha^{\alpha }c}{(1+ \alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} +\frac{ \lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha) \Gamma (\alpha +1)} +\eta \mu <1$$ such that $$|f(t,x)|\leq c|x|+M$$ for $$t\in [0,1]$$ and $$x\in R$$, then (1) has at least one solution.

### Proof

First we analyze the a priori bound of solutions of the equation $$x=\sigma Tx$$ for some $$\sigma \in [0,1]$$.

If $$x=\sigma Tx$$ for some $$\sigma [0,1]$$, $$x\in E$$, then we get

\begin{aligned} \forall t\in [0,1],\quad \bigl\vert x(t) \bigr\vert =& \bigl\vert \sigma Tx(t) \bigr\vert \leq \bigl\vert Tx(t) \bigr\vert \leq \frac{1}{ \Gamma (\alpha +\beta)} \int_{0}^{t}(t-s)^{\alpha +\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\eta }{\Gamma (\alpha +\beta)} \int_{0}^{1}(1-s)^{\alpha + \beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \eta }{\Gamma (\alpha)} \int _{0}^{1}(1-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma ( \alpha +1)\Gamma (\beta)} \int_{0}^{1}(1-s)^{\beta -1} \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert \,ds \\ &{} +\frac{\lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)} \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds + \mu \eta \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \\ &{} +\frac{\lambda }{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{\alpha -1} \bigl\vert x(s) \bigr\vert \,ds \\ \leq& \frac{(1+\eta)2(c \Vert x \Vert +M)}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda \Vert x \Vert }{\Gamma (\alpha +1)} \\ &{}+\frac{\alpha^{\alpha }(c \Vert x \Vert +M)}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma ( \beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha } \Vert x \Vert }{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu \Vert x \Vert . \end{aligned}

So

\begin{aligned} \Vert x \Vert \leq& \frac{(1+\eta)(c \Vert x \Vert +M)}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda \Vert x \Vert }{\Gamma (\alpha +1)} + \frac{\alpha^{\alpha }(c \Vert x \Vert +M)}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma ( \beta +1)} \\ &{} +\frac{\lambda \mu \alpha^{\alpha } \Vert x \Vert }{(1+\alpha)^{1+\alpha }(1- \alpha)\Gamma (\alpha +1)} +\eta \mu \Vert x \Vert . \end{aligned}

This implies

\begin{aligned} \Vert x \Vert \leq& \frac{\frac{2M}{\Gamma (\alpha +\beta +1)} +\frac{ \alpha^{\alpha }M}{(1+\alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1) \Gamma (\beta +1)}}{1-(\frac{(1+\eta)c}{\Gamma (\alpha +\beta +1)} +\frac{(1+ \eta)\lambda }{\Gamma (\alpha +1)} +\frac{\alpha^{\alpha }c}{(1+ \alpha)^{1+\alpha }(1-\alpha)\Gamma (\alpha +1)\Gamma (\beta +1)} +\frac{ \lambda \mu \alpha^{\alpha }}{(1+\alpha)^{1+\alpha }(1-\alpha) \Gamma (\alpha +1)} +\eta \mu)} \\ :=& B. \end{aligned}

Let $$\gamma =B+1$$. Set $$P_{\gamma }=\{x\in E: \parallel x\parallel <\gamma \}$$, then $$\forall x\in \partial P_{\gamma }$$, $$\parallel x\parallel =\gamma >B$$. Now, we consider $$T: \overline{P_{\gamma }}\rightarrow E$$. By the above analysis, it follows that $$x\neq \sigma Tx$$ for $$\forall x\in \partial P_{\gamma }, \forall \sigma \in [0,1]$$.

Define operators $$H_{\sigma }: E\rightarrow E$$ ($$\forall \sigma \in [0,1]$$), as $$H_{\sigma }(x)=x-\sigma Tx$$. It is easy to see that

$$\forall \sigma \in [0,1], \forall x\in \partial P_{\gamma },\quad H_{\sigma }(x)=x-\sigma Tx\neq 0.$$

According to Lemma 3.2, T is completely continuous. This yields that $$\forall \sigma \in [0,1]$$, $$H_{\sigma }$$ is a completely continuous field.

Hence by the homotopy invariance of Leray–Schauder degree, we know that

$$\operatorname{deg}(H_{\sigma }, P_{\gamma }, 0)=\operatorname{deg}(H_{1}, P_{\gamma }, 0)=\operatorname{deg}(H_{0}, P_{\gamma }, 0)=\operatorname{deg}(I, P_{\gamma }, 0)=1\neq 0.$$

By the nonzero property of Leray–Schauder degree, the equation $$H_{1}(x)=x-Tx=0$$ has at least one solution in $$P_{\gamma }$$, that is, problem (1) has at least one solution. The proof is completed. □

## References

1. 1.

Podlubny, I.: Fractional Differential Equations. Mathematics in Science and Engineering, vol. 198. Academic Press, New Tork (1999)

2. 2.

Yukunthorn, W., Ntouyas, S.K., Tariboon, J.: Nonlinear fractional Caputo–Langevin equation with nonlocal Riemann–Liouville fractional integral conditions. Adv. Differ. Equ. 2014, 315 (2014)

3. 3.

Zhao, K.H., Gong, P.: Existence of positive solutions for a class of higher-order Caputo fractional differential equation. Qual. Theory Dyn. Syst. 14(1), 157–171 (2015)

4. 4.

Zhou, Y.: Basic Theory of Fractional Differential Equations. World Scientific, Singapore (2014)

5. 5.

Zhang, X.: Positive solutions for singular higher-order fractional differential equations with nonlocal conditions. J. Appl. Math. Comput. 49, 69–89 (2015)

6. 6.

Agarwal, R.P., Zhou, Y., He, Y.Y.: Existence of fractional neutral functional differential equations. Comput. Math. Appl. 59(3), 1095–1100 (2010)

7. 7.

Zhang, L., Wang, G., Ahmad, B., Agarwal, R.P.: Nonlinear fractional integro-differential equations on unbounded domains in a Banach space. J. Comput. Appl. Math. 249, 51–56 (2013)

8. 8.

Langevin, P.: On the theory of Brownian motion. C. R. Acad. Bulgare Sci. 146, 530 (1908)

9. 9.

Zhao, K.H.: Triple positive solutions for two classes of delayed nonlinear fractional FDEs with nonlinear integral boundary value conditions. Bound. Value Probl. 2015, 181 (2015)

10. 10.

Zhang, S.: Existence of positive solution for some class of nonlinear fractional differential equations. J. Math. Anal. Appl. 278(1), 136–148 (2003)

11. 11.

Cui, Y., Zou, Y.: Existence results and the monotone iterative technique for nonlinear fractional differential systems with coupled four-point boundary value problems. Abstr. Appl. Anal. 2014, Article ID 242591 (2014)

12. 12.

Ahmad, B., Nieto, J.J., Alsaedi, A., El-Shahed, M.: A study of nonlinear Langevin equation involving two fractional orders in different intervals. Nonlinear Anal. 13, 599–606 (2012)

13. 13.

Li, X., Sun, S., Sun, Y.: Existence of solutions for fractional Langevin equation with infinite-point boundary conditions. J. Appl. Math. Comput. 53(1), 1–10 (2016)

14. 14.

Kilbas, A., Srivastava, H., Trujillo, J.: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam (2006)

15. 15.

Xu, Y.F.: Fractional boundary value problems with integral and anti-periodic boundary conditions. Bull. Malays. Math. Sci. Soc. 39, 571–587 (2016)

16. 16.

Smart, D.R.: Fixed Point Theorems. Cambridge University Press, Cambridge (1980)

17. 17.

Lutz, E.: Fractional Langevin equation. Phys. Rev. E 64(5), 051106 (2011)

18. 18.

Henderson, J., Luca, R.: Existence and multiplicity of positive solutions for a system of fractional boundary value problems. Bound. Value Probl. 2014, 60 (2014)

19. 19.

Zhang, K.: On sign-changing solution for some fractional differential equations. Bound. Value Probl. 2017, 59 (2017)

20. 20.

Guo, L., Liu, L., Wu, Y.: Iterative unique positive solutions for singular p-Laplacian fractional differential equation system with several parameters. Nonlinear Anal., Model. Control 23(2), 182–203 (2018)

21. 21.

Zhang, X., Liu, L., Wu, Y., Wiwatanapataphee, B.: The spectral analysis for a singular fractional differential equation with a signed measure. Appl. Math. Comput. 257, 252–263 (2015)

22. 22.

Jiang, J., Liu, W., Wang, H.: Positive solutions to singular Dirichlet-type boundary value problems of nonlinear fractional differential equations. Adv. Differ. Equ. 2018, 169 (2018)

23. 23.

Jiang, J., Liu, L.: Existence of solutions for a sequential fractional differential system with coupled boundary conditions. Bound. Value Probl. 2016, 159 (2016)

24. 24.

Jiang, J., Liu, L., Wu, Y.: Positive solutions to singular fractional differential system with coupled boundary conditions. Commun. Nonlinear Sci. Numer. Simul. 18, 3061–3074 (2013)

25. 25.

Giacomoni, J., Mukherjee, T., Sreenadh, K.: Positive solutions of fractional elliptic equation with critical and singular nonlinearity. Adv. Nonlinear Anal. 6(3), 327–354 (2017)

26. 26.

Denton, Z., Ramírez, J.D.: Existence of minimal and maximal solutions to RL fractional integro-differential initial value problems. Opusc. Math. 37(5), 705–724 (2017)

27. 27.

Guan, Y., Zhao, Z., Lin, X.: On the existence of solutions for impulsive fractional differential equations. Adv. Math. Phys. 2017, Article ID 1207456 (2017)

28. 28.

Wang, Y., Jiang, J.: Existence and nonexistence of positive solutions for the fractional coupled system involving generalized p-Laplacian. Adv. Differ. Equ. 2017, 337 (2017)

29. 29.

Xiang, M., Zhang, B., Radulescu, V.D.: Multiplicity of solutions for a class of quasilinear Kirchhoff system involving the fractional p-Laplacian. Nonlinearity 29(10), 3186–3205 (2016)

30. 30.

Lyons, J.W., Neugebauer, J.T.: Positive solutions of a singular fractional boundary value problem with a fractional boundary condition. Opusc. Math. 37(3), 421–434 (2017)

### Acknowledgements

The authors would like to thank the referees for their careful reading and very constructive comments that have led to the present improved version of the original manuscript.

Not applicable.

### Funding

The research was supported by the National Natural Science Foundation of China (Grant No. 11371027), Anhui Provincial Natural Science Foundation (1608085MA12), University Science Research Key Project of Anhui Province (KJ2018A0565), and Research Fund Project of Hefei University (17ZR05ZDA).

## Author information

The authors have equal contributions to each part of this paper. All authors read and approved the final manuscript.

Correspondence to Zongfu Zhou.

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### Competing interests

The authors declare that they have no competing interests. 