Theory and Modern Applications

# Boundary value problems of the nonlinear multiple base points impulsive fractional differential equations with constant coefficients

## Abstract

In this paper, the nonlinear multiple base points boundary value problems of the impulsive fractional differential equations are studied. By using the fixed point theorem and the Mittag-Leffler functions, the sufficient conditions for the existence of the solutions to the given problems are formulated. An example is presented to illustrate the result.

## Introduction

The application of fractional calculus is very broad, including characterization of me- chanics and electricity, earthquake analysis, the memory of many kinds of material, elec- tronic circuits, electrolysis chemical, etc. ([15]). In recent years, there has been a signif- icant development in solving differential equations involving fractional derivatives ([614] and the references therein).

In the left and right fractional derivatives $${}^{c} D_{a^{+}}^{\alpha}x$$ and $${}^{c} D_{b^{-}}^{\alpha}x$$, a is called a left base point and b a right base point. Both a and b are called base points of the fractional derivatives. A fractional differential equation (FDE) containing more than one base point is called a multiple base points FDE ([10]). In this paper, we study the following boundary value problem (BVP) of nonlinear multiple base points fractional differential equations with impulses:

where $$\alpha,\gamma,\delta\in (0,1)$$, $$\alpha>\gamma$$, $$\alpha>\delta$$, $$\lambda>0$$. $${}^{c}D^{\cdot}_{*}$$ is the standard Caputo fractional derivative at the base points $$t=t_{k}$$ ($$k=0,1,2,\ldots,m$$), that is, $${}^{c}D^{\cdot}_{*}|_{(t_{k},\,t_{k+1}]}x(t)={}^{c}D^{\cdot}_{t^{+}_{k}}x(t)$$ for all $$t\in (t_{k},\,t_{k+1}]$$, $$I_{0^{+}}^{\gamma}$$ denotes the fractional integral of order γ, $$f: J \times\mathbb{R}\rightarrow\mathbb{R}$$ is an appropriate function to be specified later. The impulsive moments $$\{t_{k}\}$$ are given such that $$0 < t_{1} < \cdots< t_{m-1} < 1$$, $$\Delta x(t_{k})$$ represents the jump of function x at $$t_{k}$$, which is defined by $$\Delta x(t_{k}) = x(t_{k}^{+})- x(t_{k}^{-})$$, where $$x(t_{k}^{+})$$, $$x(t_{k}^{-})$$ represent the right and left limits of $$x(t)$$ at $$t=t_{k}$$ respectively, the constant $$I_{k}$$ denotes the size of the jump.

In 1954, Barrett ([6]) applied the method of successive approximations to derive the existence of solutions to the fractional differential equations of order $$\alpha\in(0, 1)$$ with constant coefficients. Recently, as mentioned in [13, 14] and the references therein, the existence results of the impulsive fractional differential equations with anti-periodic boundary conditions involving the Caputo differential operator of order $$\alpha\in(0,1)$$ are obtained by the Mittag-Leffler functions. Inspired by the work of the above papers, the aim of the present paper is to establish some simple criteria for the existence of solutions of BVP (1.1)-(1.3).

The paper is organized as follows. In Section 2, we present some basic concepts, the notations about the fractional calculus and the properties of the Mittag-Leffler functions. In Section 3, we present the definition of solution for (1.1)-(1.3). In Section 4, by applying Krasnoselskii’s fixed point theorem, we verify the existence of solutions for problem (1.1)-(1.3). We give an example to illustrate the result in Section 5.

## Preliminaries

In this paper, we denote by $$L^{p}(J,\mathbb{R})$$ the Banach space of all Lebesgue measurable functions $$l: J\rightarrow\mathbb{R}$$ with the norm $$\Vert l \Vert _{L^{p}}= (\int_{J} \vert l(t) \vert ^{p}\,dt )^{\frac {1}{p}}<\infty$$ and by $$\operatorname{AC}([a,b], \mathbb{R})$$ the space of all the absolutely continuous functions defined on $$[a,b]$$.

### Definition 2.1

[2, 3]

The fractional integral of order q with the lower limit a for a function $$g(t)\in L^{1}([a, +\infty), \mathbb{R})$$ is defined as

$$\bigl(I_{a^{+}}^{q}g\bigr) (t)=\frac{1}{\Gamma(q)} \int_{a}^{t} (t-s)^{q-1}g(s)\,ds, \quad t>a, q>0,$$

where $$\Gamma(\cdot)$$ is the gamma function.

### Definition 2.2

[2, 3]

If $$g(t)\in \operatorname{AC}^{n}([a,b], \mathbb{R})$$, then the Riemann-Liouville fractional derivative $$({}^{L} D_{a^{+}}^{q}g)(t)$$ of order q exists almost everywhere on $$[a, b]$$ and can be written as

$$\bigl({}^{L} D_{a^{+}}^{q}g\bigr) (t)= \frac{1}{\Gamma(n-q)}\frac{d^{n}}{dt^{n}} \int _{a}^{t}(t-s)^{n-q-1}g(s)\,ds,\quad t> a, n-1< q< n.$$

### Definition 2.3

[2, 3]

If $$g(t)\in \operatorname{AC}^{n}([a,b], \mathbb{R})$$, then the Caputo derivative $$({}^{c}D_{a^{+}}^{q}g)(t)$$ of order q exists almost everywhere on $$[a, b]$$ and can be written as

$$\bigl({}^{c}D_{a^{+}}^{q}g\bigr) (t)= \Biggl( {}^{L} D_{a^{+}}^{q} \Biggl[g(s)-\sum _{k=0}^{n-1}\frac{g^{(k)}(a)}{k!}(s-a)^{k} \Biggr] \Biggr) (t), \quad t> a, n-1< q< n,$$

moreover, if $$g(a)=g'(a)=\cdots=g^{(n-1)}(a)=0$$, then $$({}^{c}D_{a^{+}}^{q}g)(t)=({}^{L} D_{a^{+}}^{q}g)(t)$$.

### Remark 2.4

[2, 3]

The Caputo fractional derivative of order q for a function $$g\in C^{n}([a,b], \mathbb{R})$$ is defined by

$$\bigl({}^{c}D_{a^{+}}^{q}g\bigr) (t)= \frac{1}{\Gamma(n-q)} \int_{a}^{t}\frac {g^{(n)}(s)}{(t-s)^{q-n+1}}\,ds, \quad t> a, n-1< q< n.$$

### Definition 2.5

[2, 3]

For $$\alpha, \beta> 0$$, $$z\in\mathbb{C}$$, the classical Mittag-Leffler function $$E_{\alpha}(z)$$ and the generalized Mittag-Leffler functions $$E_{\alpha, \beta}(z)$$ are defined by

$$\begin{gathered} E_{\alpha}(z)=\sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma(\alpha k+1)},\qquad E_{\alpha, \beta}(z)=\sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma(\alpha k+\beta)},\\ E^{\rho}_{\alpha,\beta}(z)=\sum _{k=0}^{\infty}\frac{z^{k}}{\Gamma(\alpha k+\beta)}\frac{(\rho)_{k}}{k!}, \end{gathered}$$

where $$(\rho)_{0} =1$$ and $$(\rho)_{k}=\rho(\rho+ 1) \cdots(\rho+k -1)$$ for $$k\in \mathbb{N}$$.

Clearly, $$E_{\alpha,1}(z)=E_{\alpha}(z)$$.

### Lemma 2.6

[2]

Let $$\nu, \beta, \alpha>0$$. The usual derivatives of $$E_{\alpha}(z)$$, $$E_{\alpha,\beta}(z)$$ and the Riemann-Liouville integration of $$E_{\alpha}(-\lambda t^{\alpha})$$ are expressed by

1. (1)

$${ (\frac{d}{dz} )^{n}[E_{\alpha,\beta}(z)]=n! E^{n+1}_{\alpha, \beta+\alpha n}(z)}$$, $$n\in\mathbb{N}$$;

2. (2)

$${ (\frac{d}{dz} )^{n}[E_{\alpha}(z)]=n! E^{n+1}_{\alpha , 1+\alpha n}(z)}$$, $$n\in\mathbb{N}$$;

3. (3)

$${ (\frac{d}{dt} )^{n}[t^{\beta-1}E_{\alpha, \beta }(-\lambda t^{\alpha})]=t^{\beta-n-1} E_{\alpha, \beta-n}(-\lambda t^{\alpha})}$$, $$n\geq1$$;

4. (4)

$${[I_{0^{+}}^{\beta}(s^{\nu-1}E_{\alpha, \nu}(-\lambda s^{\alpha }))](t):=\frac{1}{\Gamma(\beta)}\int_{0}^{t} (t-s)^{\beta-1}s^{\nu -1}E_{\alpha, \nu}(-\lambda s^{\alpha})\,ds=t^{\beta+\nu-1}E_{\alpha, \beta+\nu}(-\lambda t^{\alpha})}$$.

As mentioned in ([14]), $$E_{\alpha}(-\lambda t^{\alpha})$$ and $$E_{\alpha, \alpha}(-\lambda t^{\alpha})$$ can be represented by

\begin{aligned}& E_{\alpha}\bigl(-\lambda t^{\alpha}\bigr) = \int_{0}^{\infty}e^{-\lambda t^{\alpha}\theta}\phi(\theta)\,d \theta, \end{aligned}
(2.1)
\begin{aligned}& E_{\alpha,\alpha}\bigl(-\lambda t^{\alpha}\bigr) = \alpha \int_{0}^{\infty}\theta e^{-\lambda t^{\alpha}\theta}\phi(\theta) \,d\theta, \end{aligned}
(2.2)

where

$$\phi(\theta)=\frac{1}{\pi}\sum_{n=1}^{\infty}(-1)^{n-1} \theta^{\alpha n-1}\frac{\Gamma(n\alpha+1)}{n!}\sin(n\pi\alpha),\quad0< \alpha< 1, \theta>0.$$

Moreover,

\begin{aligned} \int_{0}^{\infty}\theta^{\xi}\phi(\theta)\,d \theta=\frac{\Gamma (\xi+1)}{\Gamma(\alpha\xi+1)}\quad (\xi\geq0). \end{aligned}
(2.3)

### Lemma 2.7

For $$\lambda>0$$, $$\alpha, \beta, \theta_{1}, \theta_{2}\in(0, 1)$$, $$\alpha \geq\theta_{2}$$, the generalized Mittag-Leffler functions have the following properties:

1. (1)

$${\frac{d}{dt}[E_{\alpha}(-\lambda t^{\alpha})]=-\lambda t^{\alpha-1}E_{\alpha, \alpha}(-\lambda t^{\alpha})}$$;

2. (2)

$${E_{\alpha,\alpha+\beta}(-\lambda t^{\alpha})=\frac{1}{\Gamma (\beta)}\int_{0}^{1} E_{\alpha,\alpha} (-\lambda t^{\alpha}u^{\alpha })u^{\alpha-1}(1-u)^{\beta-1}\,du}$$;

3. (3)

$${E_{\alpha,\beta}(-\lambda t^{\alpha})=\frac{1}{\Gamma(\beta )}-\lambda t^{\alpha}E_{\alpha, \alpha+\beta}(-\lambda t^{\alpha})}$$;

4. (4)

$${E_{\alpha,\theta_{1}+1}(-\lambda t^{\alpha})=\frac{1}{\Gamma (\theta_{1})}\int_{0}^{1} E_{\alpha}(-\lambda t^{\alpha}u^{\alpha })(1-u)^{\theta_{1}-1}\,du}$$;

5. (5)

$${[{}^{c} D_{a^{+}}^{\theta_{2}}E_{\alpha}(-\lambda(s-a)^{\alpha})](t)=-\lambda(t-a)^{\alpha-\theta_{2}}E_{\alpha, \alpha-\theta _{2}+1}(-\lambda(t-a)^{\alpha})}$$.

In particular, when $$\alpha=\theta_{2}$$, $$[{}^{c} D_{a^{+}}^{\alpha}E_{\alpha}(-\lambda(s-a)^{\alpha})](t)=-\lambda E_{\alpha}(-\lambda(t-a)^{\alpha})$$.

### Proof

We denote the beta function by $$\mathbb{B}(\cdot, \cdot)$$. From Lemma 2.6(2),

\begin{aligned} \begin{aligned}[b]\frac{d}{dt}\bigl[E_{\alpha}\bigl(-\lambda t^{\alpha}\bigr)\bigr]&=-\lambda \alpha t^{\alpha-1}E^{2}_{\alpha, 1+\alpha} \bigl(-\lambda t^{\alpha}\bigr) \\ &=-\lambda\alpha t^{\alpha-1}\sum_{k=0}^{\infty} \frac{(-\lambda t^{\alpha})^{k}(1+k)}{\Gamma(\alpha k+1+\alpha)} \\ &=-\lambda t^{\alpha-1}\sum_{k=0}^{\infty} \frac{(-\lambda t^{\alpha })^{k}}{\Gamma(\alpha k+\alpha)} \\ &=-\lambda t^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha} \bigr). \end{aligned} \end{aligned}

From [14], the second result holds. Moreover,

\begin{aligned}& \begin{aligned} E_{\alpha,\beta}\bigl(-\lambda t^{\alpha}\bigr)&=\sum _{k=0}^{\infty}\frac{(-\lambda t^{\alpha})^{k}}{\Gamma(\alpha k+\beta)} =\frac{1}{\Gamma(\beta)}- \lambda t^{\alpha}\sum_{k=1}^{\infty}\frac {(-\lambda t^{\alpha})^{k-1}}{\Gamma(\alpha k+\beta)} \\ &=\frac{1}{\Gamma(\beta)}-\lambda t^{\alpha}E_{\alpha, \alpha+\beta }\bigl(-\lambda t^{\alpha}\bigr), \end{aligned} \\& \begin{aligned} E_{\alpha,\theta_{1}+1}\bigl(-\lambda t^{\alpha}\bigr)&=\sum _{k=0}^{\infty}\frac {(-\lambda t^{\alpha})^{k}}{\Gamma(\alpha k+\theta_{1}+1)} =\frac{1}{\Gamma(\theta_{1})}\sum _{k=0}^{\infty}\frac{(-\lambda t^{\alpha })^{k} \mathbb{B}(\alpha k+1, \theta_{1})}{\Gamma(\alpha k+1)} \\ &=\frac{1}{\Gamma(\theta_{1})} \int_{0}^{1}\sum_{k=0}^{\infty}\frac{(-\lambda t^{\alpha}u^{\alpha})^{k}}{\Gamma(\alpha k+1)}(1-u)^{\theta_{1}-1}\,du \\ &=\frac{1}{\Gamma(\theta_{1})} \int_{0}^{1} E_{\alpha}\bigl(-\lambda t^{\alpha}u^{\alpha}\bigr) (1-u)^{\theta_{1}-1}\,du. \end{aligned} \end{aligned}

Applying Remark 2.4 and the fact $${\int _{a}^{t}(t-s)^{m_{1}-1}(s-a)^{m_{2}-1}\,ds=(t-a)^{m_{1}+m_{2}-1}\mathbb{B}(m_{1}, m_{2})}$$, we have

\begin{aligned} \bigl[{}^{c} D_{a^{+}}^{\theta_{2}}E_{\alpha}\bigl(- \lambda(s-a)^{\alpha}\bigr)\bigr](t) =&\frac {1}{\Gamma(1-\theta_{2})} \int_{a}^{t} (t-s)^{-\theta_{2}}\frac{d}{ds} \Biggl(\sum_{k=0}^{\infty} \frac{(-\lambda(s-a)^{\alpha})^{k})}{\Gamma(\alpha k+1)} \Biggr)\,ds \\ =&\frac{1}{\Gamma(1-\theta_{2})}\sum_{k=1}^{\infty} \frac{(-\lambda )^{k}}{\Gamma(\alpha k)} \int_{a}^{t} (t-s)^{-\theta_{2}}(s-a)^{\alpha k-1} \,ds \\ =&-\lambda(t-a)^{\alpha-\theta_{2}}\sum_{k=1}^{\infty} \frac{(-\lambda )^{k-1}(t-a)^{\alpha(k-1)}}{\Gamma(\alpha k+1-\theta_{2})} \\ =&-\lambda(t-a)^{\alpha-\theta_{2}}E_{\alpha, \alpha-\theta _{2}+1}\bigl(-\lambda(t-a)^{\alpha}\bigr). \end{aligned}

□

### Lemma 2.8

[3]

If $$0<\alpha<2$$, β is an arbitrary real number, $$\frac{\pi\alpha}{2} <\mu<\min\{\pi, \pi\alpha\}$$, then

$$\bigl\vert E_{\alpha, \beta}(z) \bigr\vert \leq\frac{\mathcal {C}}{1+ \vert z \vert },\qquad\mu \leq \bigl\vert \operatorname{arg}(z) \bigr\vert \leq\pi,\qquad \vert z \vert \geq0,$$

where $$\mathcal{C}$$ is a positive constant.

### Lemma 2.9

Let $$\alpha, \beta\in(0, 1)$$, $$\lambda>0$$. Then the functions $$E_{\alpha}$$, $$E_{\alpha,\alpha}$$ and $$E_{\alpha,\alpha+\beta}$$ are nonnegative and have the following properties:

1. (i)

For any $$t\in J$$, $$E_{\alpha}(-\lambda t^{\alpha})\leq1$$, $$E_{\alpha,\alpha}(-\lambda t^{\alpha})\leq\frac{1}{\Gamma(\alpha)}$$, $$E_{\alpha,\alpha+\beta}(-\lambda t^{\alpha})\leq\frac{1}{\Gamma(\alpha +\beta)}$$, $$E_{\alpha,\beta}(-\lambda t^{\alpha})\leq\frac{1}{\Gamma(\beta)}$$, moreover, $$E_{\alpha}(0)=1$$. In particular,

\begin{aligned} E_{\alpha,\alpha-\delta}\bigl(-\lambda t^{\alpha}\bigr)\leq \frac {1}{\Gamma(\alpha-\delta)},\qquad \bigl\vert E_{\alpha,\alpha-\delta }\bigl(-\lambda t^{\alpha}\bigr) \bigr\vert \leq\mathcal{C}. \end{aligned}
(2.4)
2. (ii)

For any $$t_{1}, t_{2}\in J$$,

\begin{aligned}& \bigl\vert E_{\alpha}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha}\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert = O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\textit{as } t_{2}\rightarrow t_{1}, \\& \bigl\vert E_{\alpha,\alpha}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha,\alpha }\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert = O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\textit{as } t_{2}\rightarrow t_{1}, \\& \bigl\vert E_{\alpha,\alpha-\delta}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha ,\alpha-\delta}\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert = O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\textit{as } t_{2}\rightarrow t_{1}. \end{aligned}

### Proof

(i) From (2.1), we get $$E_{\alpha}(-\lambda t^{\alpha})=\int_{0}^{\infty}e^{-\lambda t^{\alpha}\theta}\phi(\theta)\,d\theta\leq \int_{0}^{\infty}\phi(\theta)\,d\theta=1$$.

By (2.2), we find $$E_{\alpha,\alpha}(-\lambda t^{\alpha})=\alpha \int_{0}^{\infty}\theta e^{-\lambda t^{\alpha}\theta}\phi(\theta)\,d\theta\leq \frac{1}{\Gamma(\alpha)}$$.

Using Lemma 2.7(2), one sees

\begin{aligned} E_{\alpha,\alpha+\beta}\bigl(-\lambda t^{\alpha}\bigr)=\frac{1}{\Gamma(\beta)} \int _{0}^{1} E_{\alpha,\alpha} \bigl(-\lambda t^{\alpha}u^{\alpha}\bigr)u^{\alpha -1}(1-u)^{\beta-1}\,du \leq\frac{1}{\Gamma(\alpha+\beta)}. \end{aligned}

Noting $$E_{\alpha,\alpha+\beta}(-\lambda t^{\alpha})>0$$ and Lemma 2.7(3), we have $$E_{\alpha,\beta}(-\lambda t^{\alpha})\leq\frac {1}{\Gamma(\beta)}$$. Taking $$\beta=\alpha-\delta$$ in $$E_{\alpha,\beta}(-\lambda t^{\alpha})\leq \frac{1}{\Gamma(\beta)}$$, we obtain $$E_{\alpha,\alpha-\delta}(-\lambda t^{\alpha})\leq\frac{1}{\Gamma(\alpha-\delta)}$$. By Lemma 2.8, we get $$\vert E_{\alpha,\alpha-\delta}(-\lambda t^{\alpha}) \vert \leq\mathcal{C}$$.

(ii) For $$0\leq t_{1}< t_{2}\leq1$$, using the Lagrange mean value theorem and the fact $$\vert t_{2}^{\alpha}-t_{1}^{\alpha} \vert \leq (t_{2}-t_{1})^{\alpha}$$, (2.1), (2.2) and (2.3), we find

\begin{aligned}& \begin{aligned} \bigl\vert E_{\alpha}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha}\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert &= \int_{0}^{\infty}\bigl\vert e^{-\lambda t_{2}^{\alpha}\theta }-e^{-\lambda t_{1}^{\alpha}\theta} \bigr\vert \phi(\theta)\,d\theta\leq \lambda(t_{2}-t_{1})^{\alpha} \int_{0}^{\infty}\theta\phi(\theta)\,d\theta \\ &=\frac{\lambda(t_{2}-t_{1})^{\alpha}}{\Gamma(\alpha+1)}:=O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\text{as } t_{2}\rightarrow t_{1}, \end{aligned} \\& \begin{aligned} \bigl\vert E_{\alpha,\alpha}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha,\alpha }\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert &=\alpha \int_{0}^{\infty}\bigl\vert e^{-\lambda t_{2}^{\alpha}\theta}-e^{-\lambda t_{1}^{\alpha}\theta} \bigr\vert \theta\phi(\theta)\,d\theta \\ &\leq\frac{2\lambda\alpha(t_{2}-t_{1})^{\alpha}}{\Gamma(2\alpha +1)}:=O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\text{as } t_{2}\rightarrow t_{1}, \end{aligned} \end{aligned}

by Lemma 2.7(3), Lemma 2.9(i) and Lemma 2.7(2), one has

\begin{aligned}& \bigl\vert E_{\alpha, \alpha-\delta}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha, \alpha-\delta}\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert \\& \quad = \lambda \bigl\vert t_{2}^{\alpha}E_{\alpha, 2\alpha-\delta}\bigl(- \lambda t_{2}^{\alpha}\bigr)-t_{1}^{\alpha}E_{\alpha, 2\alpha-\delta} \bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert \\& \quad \leq \lambda \bigl[ \vert t_{2}-t_{1} \vert ^{\alpha}E_{\alpha, 2\alpha-\delta}\bigl(-\lambda t_{2}^{\alpha}\bigr)+t_{1}^{\alpha} \bigl\vert E_{\alpha, 2\alpha-\delta}\bigl(-\lambda t_{2}^{\alpha}\bigr)-E_{\alpha, 2\alpha-\delta }\bigl(-\lambda t_{1}^{\alpha}\bigr) \bigr\vert \bigr] \\& \quad \leq \frac{\lambda}{\Gamma(2\alpha-\delta)} \vert t_{2}-t_{1} \vert ^{\alpha} \\& \qquad {}+\frac{\lambda}{\Gamma(\alpha-\delta)} \int_{0}^{1} \bigl\vert E_{\alpha, \alpha}\bigl(- \lambda t_{2}^{\alpha}u^{\alpha}\bigr)-E_{\alpha, \alpha } \bigl(-\lambda t_{1}^{\alpha}u^{\alpha}\bigr) \bigr\vert u^{\alpha-1}(1-u)^{\alpha -\delta-1}\,du \\& \quad := O\bigl( \vert t_{2}-t_{1} \vert ^{\alpha}\bigr),\quad\text{as } t_{2}\rightarrow t_{1}. \\ & \end{aligned}
(2.5)

□

### Lemma 2.10

[2]

The solution to the Cauchy problem

\begin{aligned} \textstyle\begin{cases} {}^{c} D_{a^{+}}^{\alpha}x(t)+\lambda x(t)=f(t),\\ x(a)=b_{1},\quad b_{1}\in\mathbb{R}, \end{cases}\displaystyle \end{aligned}

with $$0<\alpha<1$$ has the form

$$x(t)=b_{1}E_{\alpha}\bigl(-\lambda(t-a)^{\alpha}\bigr)+ \int_{a}^{t}(t-s)^{\alpha -1}E_{\alpha,\alpha} \bigl(-\lambda(t-s)^{\alpha}\bigr)f(s)\,ds.$$

### Theorem 2.11

Krasnoselskii’s fixed point theorem

Let $$\mathcal{M}$$ be a closed convex and nonempty subset of a Banach space X. Let $$\mathcal{A}$$, $$\mathcal{B}$$ be two operators such that (i) $$\mathcal{A}x+\mathcal{B}y\in\mathcal{M}$$ whenever $$x, y\in\mathcal{M}$$, (ii) $$\mathcal{A}$$ is compact and continuous, (iii) $$\mathcal{B}$$ is a contraction mapping. Then there exists a $$z\in\mathcal{M}$$ such that $$z=\mathcal{A}z+\mathcal{B}z$$.

## Solutions for BVP

Setting $$J_{0}=[0, t_{1}]$$, $$J_{k}=(t_{k}, t_{k+1}]$$, $$k=1,\ldots, m-1$$, $$J_{m}=[t_{m}, 1]$$, and we define $$X=\{x:[0,1]\rightarrow\mathbb{R}: x\vert_{J_{k}}\in C(J_{k}, \mathbb{R}) \text{ and there exist } x(t_{k}^{+}) \text{ and } x(t_{k}^{-}), \text{with } x(t_{k}^{-})=x(t_{k}), k=1,\ldots,m-1\}$$ with the norm

$$\Vert x \Vert _{1}:=\sup_{k=0, 1, \ldots,m}\sup _{t \in J_{k}} \bigl\vert x(t) \bigr\vert .$$

Obviously, X is a real Banach space.

In this paper, we consider the following assumption.

(H1):

$$f:J\times\mathbb{R}\rightarrow\mathbb{R}$$ satisfies $$f(\cdot ,x):J\rightarrow\mathbb{R}$$ is measurable for all $$x\in\mathbb{R}$$ and $$f(t,\cdot):\mathbb{R}\rightarrow\mathbb{R}$$ is continuous for a.e. $$t\in J$$, and there exists a function $$\mu\in L^{\frac {1}{q_{1}}}(J,\mathbb{R}^{+})$$ ($$0< q_{1}<\min\{\frac{\alpha}{2}, \alpha-\delta\}$$) such that $$\vert f(t,x) \vert \leq\mu(t)$$.

### Definition 3.1

A function $$x:J\rightarrow\mathbb{R}$$ is said to be a solution of (1.1)-(1.3) if

1. (1)

$$x\in \operatorname{AC}(J_{k}, \mathbb{R})$$;

2. (2)

x satisfies the equation $${}^{c} D_{t_{k}^{+}}^{\alpha} x(t)+\lambda x(t)= f(t, x(t))$$ on $$J_{k}$$;

3. (3)

for $$k=1,2,\ldots,m-1$$, $$\Delta x(t_{k})=I_{k}$$, and $${x(0)+I^{\gamma}_{0^{+}}x(\eta)=0}$$, $$x(1)+{}^{c} D_{t_{m}^{+}}^{\delta}x(1)=0$$.

Next, we present the following lemmas.

### Lemma 3.2

For any $$\tau_{2}, \tau_{1}\in J_{k}$$ ($$k=0,1,2,\ldots,m$$) and $$\tau_{2}<\tau_{1}$$,

$$\int_{t_{k}}^{\tau_{2}}\bigl[(\tau_{2}-s)^{\alpha-1}-( \tau_{1}-s)^{\alpha-1}\bigr]\mu (s)\,ds\rightarrow0, \quad\textit{as } \tau_{2}\rightarrow\tau_{1}.$$

### Proof

It follows from the Hölder inequality that

\begin{aligned}& \biggl\vert \int_{t_{k}}^{\tau_{2}}\bigl[(\tau_{2}-s)^{\alpha-1}-( \tau _{1}-s)^{\alpha-1}\bigr]\mu(s)\,ds \biggr\vert \\& \quad \leq \Vert \mu \Vert _{L^{\frac{1}{q_{1}}}} \biggl[ \int _{t_{k}}^{\tau_{2}} \bigl\vert (\tau_{2}-s)^{\alpha-1}-( \tau_{1}-s)^{\alpha -1} \bigr\vert ^{\frac{1}{1-q_{1}}}\,ds \biggr]^{1-q_{1}} \\& \quad = (1-\alpha) \Vert \mu \Vert _{L^{\frac{1}{q_{1}}}} \biggl( \int _{t_{k}}^{\tau_{2}} \biggl\vert \int_{\tau_{2}}^{\tau_{1}}(\zeta-s)^{\alpha -2}\,d\zeta \biggr\vert ^{\frac{1}{1-q_{1}}}\,ds \biggr) ^{1-q_{1}} \\& \quad \leq \overline{M} \biggl[ \int_{t_{k}}^{\tau_{2}}\bigl((\tau_{2}-s)^{\theta}-( \tau _{1}-s)^{\theta}\bigr)\,ds \biggr]^{1-q_{1}} \\& \quad = \frac{\overline{M}}{(1+\theta)^{1-q_{1}}} \bigl[(\tau_{1}-\tau_{2})^{1+\theta}-( \tau_{1}-t_{k})^{1+\theta} +(\tau_{2}-t_{k})^{1+\theta} \bigr]^{1-q_{1}} \\& \quad \rightarrow 0, \quad\text{as } \tau_{2}\rightarrow\tau_{1}, \end{aligned}

where $$\overline{M}>0$$ is a constant and $${\theta=\frac{\alpha-1-q_{1}}{1-q_{1}}\in(-1, 0)}$$. □

For $$y>q_{1}$$ and $$t_{i-1}\in J$$ ($$i=1,\ldots,m+1$$), from the Hölder inequality, we have

\begin{aligned} \int_{t_{i-1}}^{t_{i}}(t_{i}-s)^{y-1} \mu(s)\,ds \leq& \biggl( \int _{t_{i-1}}^{t_{i}}(t_{i}-s)^{\frac{y-1}{1-q_{1}}} \,ds \biggr)^{1-q_{1}} \Vert \mu \Vert _{L^{\frac{1}{q_{1}}}}= \zeta_{y}(t_{i}-t_{i-1})^{y-q_{1}}, \end{aligned}
(3.1)

where $${\zeta_{y}= (\frac{1-q_{1}}{y-q_{1}} )^{1-q_{1}} \Vert \mu \Vert _{L^{\frac{1}{q_{1}}}}}$$.

For brevity, we define

$$\bigl(Q_{k}^{\varsigma} x\bigr) (t):= \int_{t_{k}}^{t} (t-s)^{\varsigma-1} E_{\alpha ,\varsigma} \bigl(-\lambda(t-s)^{\alpha}\bigr) f\bigl(s,x(s)\bigr)\,ds,$$

then, for $$t\in(t_{k}, t_{k+1}]$$, from (3.1) and Lemma 2.9(i), we obtain

\begin{aligned}& \bigl\vert \bigl(Q_{k}^{\alpha}x\bigr) (t) \bigr\vert \leq \int_{t_{k}}^{t}\frac{(t-s)^{\alpha-1}\mu(s)}{\Gamma(\alpha)}\,ds \leq \zeta_{\alpha}\frac{(t-t_{k})^{\alpha-q_{1}}}{\Gamma(\alpha)}, \end{aligned}
(3.2)
\begin{aligned}& \bigl\vert \bigl(Q_{k}^{\alpha-\delta}x\bigr) (t) \bigr\vert \leq \mathcal{C} \int_{t_{k}}^{t}(t-s)^{\alpha-\delta-1}\mu(s)\,ds \leq \mathcal{C}\zeta_{\alpha-\delta}(t-t_{k})^{\alpha-\delta-q_{1}}, \end{aligned}
(3.3)

which means that $$(t-s)^{\alpha-1}E_{\alpha,\alpha}(-\lambda (t-s)^{\alpha})f(s, x(s))$$ and $$(t-s)^{\alpha-\delta-1}E_{\alpha,\alpha -\delta}(-\lambda(t-s)^{\alpha})f(s, x(s))$$ are Lebesgue integrable with respect to $$s\in[t_{k}, t_{k+1}]$$ for all $$t\in[t_{k}, t_{k+1}]$$ and $$x\in X$$.

### Lemma 3.3

For any $$k=0,1,2,\ldots,m$$, $$(Q_{k}^{\alpha}x)(t)\in C(J_{k}, \mathbb{R})$$, $$(Q_{k}^{\alpha-\delta} x)(t)\in C(J_{k}, \mathbb{R})$$.

### Proof

For any $$h>0$$, $$t_{k}< t< t+h< t_{k+1}$$, by (H1), Lemma 2.9(i), (ii), Lemma 3.2 and (3.1), we get

\begin{aligned}& \bigl\vert \bigl(Q_{k}^{\alpha}x\bigr) (t+h)- \bigl(Q_{k}^{\alpha}x\bigr) (t) \bigr\vert \\& \quad \leq \int_{t_{k}}^{t} \bigl\vert (t+h-s)^{\alpha-1}-(t-s)^{\alpha-1} \bigr\vert E_{\alpha,\alpha}\bigl(-\lambda(t+h-s)^{\alpha}\bigr) \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,ds \\& \qquad {} + \int_{t_{k}}^{t}(t-s)^{\alpha-1} \bigl\vert E_{\alpha,\alpha}\bigl(-\lambda (t+h-s)^{\alpha}\bigr)-E_{\alpha,\alpha} \bigl(-\lambda(t-s)^{\alpha}\bigr) \bigr\vert \bigl\vert f\bigl(s,x(s) \bigr) \bigr\vert \,ds \\& \qquad {} + \int_{t}^{t+h}(t+h-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-\lambda (t+h-s)^{\alpha}\bigr) \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert \,ds \\& \quad \leq \int_{t_{k}}^{t}\frac{ \vert (t+h-s)^{\alpha-1}-(t-s)^{\alpha -1} \vert }{\Gamma(\alpha)}\mu(s)\,ds+O \bigl(h^{\alpha}\bigr) \int _{t_{k}}^{t}(t-s)^{\alpha-1}\mu(s)\,ds \\& \qquad {} + \int_{t}^{t+h}\frac{(t+h-s)^{\alpha-1}}{\Gamma(\alpha)}\mu(s)\,ds \\& \quad \rightarrow 0,\quad\text{as } h\rightarrow0. \end{aligned}

Similarly, noting (2.4) and (2.5), we find $$(Q_{k}^{\alpha -\delta} x)(t)\in C(J_{k}, \mathbb{R})$$. □

### Lemma 3.4

Assume that (H1) holds. Then $$(Q_{k}^{\alpha}x)(t)\in \operatorname{AC}([t_{k}, t_{k+1}], \mathbb{R})$$, for $$x\in X$$, $$k=0,1,\ldots,m$$.

### Proof

For every finite collection $$\{(a_{i}, b_{i})\}_{1\leq i\leq n}$$ on $$[t_{k}, t_{k+1}]$$ with $$\sum_{i=1}^{n}(b_{i}-a_{i})\rightarrow0$$, noting (3.1), Lemma 3.2 and Lemma 2.9(ii), we have

\begin{aligned}& \sum_{i=1}^{n} \bigl\vert \bigl(Q_{k}^{\alpha} x\bigr) (b_{i})- \bigl(Q_{k}^{\alpha} x\bigr) (a_{i}) \bigr\vert \\& \quad \leq \sum_{i=1}^{n} \biggl\vert \int_{a_{i}}^{b_{i}}(b_{i}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-\lambda(b_{i}-s)^{\alpha}\bigr)f\bigl(s, x(s)\bigr)\,ds \biggr\vert \\& \qquad {} + \sum_{i=1}^{n} \int_{t_{k}}^{a_{i}} \bigl\vert \bigl[(b_{i}-s)^{\alpha -1}-(a_{i}-s)^{\alpha-1} \bigr]E_{\alpha,\alpha}\bigl(-\lambda(b_{i}-s)^{\alpha}\bigr)f \bigl(s, x(s)\bigr) \bigr\vert \,ds \\& \qquad {} +\sum_{i=1}^{n} \int_{t_{k}}^{a_{i}}(a_{i}-s)^{\alpha-1} \bigl\vert E_{\alpha ,\alpha}\bigl(-\lambda(b_{i}-s)^{\alpha}\bigr)-E_{\alpha,\alpha}\bigl(-\lambda (a_{i}-s)^{\alpha}\bigr) \bigr\vert \bigl\vert f\bigl(s, x(s)\bigr) \bigr\vert \,ds \\& \quad \leq \sum_{i=1}^{n} \int_{a_{i}}^{b_{i}}\frac{(b_{i}-s)^{\alpha-1} \mu (s)}{\Gamma(\alpha)}\,ds+ \frac{1}{\Gamma(\alpha)}\sum_{i=1}^{n} \int _{t_{k}}^{a_{i}} \bigl[(a_{i}-s)^{\alpha-1}-(b_{i}-s)^{\alpha-1} \bigr] \mu(s)\,ds \\& \qquad {} +\sum_{i=1}^{n} \int_{t_{k}}^{a_{i}}(a_{i}-s)^{\alpha-1} \mu(s)\,ds\cdot O\bigl( \vert b_{i}-a_{i} \vert ^{\alpha}\bigr) \\& \quad \leq \frac{\zeta_{\alpha}}{\Gamma(\alpha)}\sum_{i=1}^{n}(b_{i}-a_{i})^{\alpha-q_{1}} +\frac{1}{\Gamma(\alpha)}\sum_{i=1}^{n} \int_{t_{k}}^{a_{i}} \bigl[(a_{i}-s)^{\alpha-1}-(b_{i}-s)^{\alpha-1} \bigr] \mu(s)\,ds \\& \qquad {} +\zeta_{\alpha}\sum_{i=1}^{n}O \bigl( \vert b_{i}-a_{i} \vert ^{\alpha}\bigr) \\& \quad \rightarrow 0. \end{aligned}

Hence, $$(Q_{k}^{\alpha}x)(t)$$ is absolutely continuous on $$[t_{k}, t_{k+1}]$$. Furthermore, for almost all $$t\in[t_{k}, t_{k+1}]$$, $$[ {}^{c} D_{t_{k}^{+}}^{\alpha }(Q_{k}^{\alpha}x)(s) ](t)$$ and $$[ {}^{c} D_{t_{k}^{+}}^{\delta }(Q_{k}^{\alpha}x)(s) ](t)$$ exist. □

### Lemma 3.5

Assume that (H1) holds. Then, for $$x\in X$$, $$k=0,1,\ldots,m$$,

\begin{aligned}& \bigl[ {}^{c} D_{t_{k}^{+}}^{\alpha}\bigl(Q_{k}^{\alpha}x \bigr) (s) \bigr](t) = f\bigl(t, x(t)\bigr)-\lambda\bigl(Q_{k}^{\alpha}x \bigr) (t), \quad\textit{a.e. } t\in J_{k}, \\& \bigl[ {}^{c} D_{t_{k}^{+}}^{\delta}\bigl(Q_{k}^{\alpha}x \bigr) (s) \bigr](t) = \bigl(Q_{k}^{\alpha -\delta}x\bigr) (t), \quad \textit{a.e. } t\in J_{k}. \end{aligned}

### Proof

According to Lemma 2.6(4), we can see that

\begin{aligned}& \begin{aligned} \int_{s}^{t}(t-\tau)^{-\alpha}( \tau-s)^{\alpha-1}E_{\alpha, \alpha }\bigl(-\lambda(\tau-s)^{\alpha}\bigr) \,d\tau&= \int_{0}^{t-s}(t-s-\tau)^{-\alpha } \tau^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda\tau^{\alpha}\bigr)\,d\tau \\ &=\Gamma(1-\alpha)E_{\alpha}\bigl(-\lambda(t-s)^{\alpha}\bigr), \end{aligned} \\& \begin{aligned} \int_{s}^{t}(t-\tau)^{-\delta}( \tau-s)^{\alpha-1}E_{\alpha, \alpha }\bigl(-\lambda(\tau-s)^{\alpha}\bigr) \,d\tau&= \int_{0}^{t-s}(t-s-\tau)^{-\delta } \tau^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda\tau^{\alpha}\bigr)\,d\tau \\ &=\Gamma(1-\delta) (t-s)^{\alpha-\delta}E_{\alpha, \alpha-\delta +1}\bigl(- \lambda(t-s)^{\alpha}\bigr). \end{aligned} \end{aligned}

Moreover, noting Lemma 2.6(1) and Lemma 2.7(1), we obtain

\begin{aligned}& \bigl[{}^{L} D_{t_{k}^{+}}^{\alpha} \bigl(Q_{k}^{\alpha}x\bigr) (s) \bigr](t) \\& \quad = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dt} \int_{t_{k}}^{t}(t-s)^{-\alpha} \biggl[ \int_{t_{k}}^{s}(s-\tau)^{\alpha-1}E_{\alpha, \alpha} \bigl(-\lambda(s-\tau )^{\alpha}\bigr)f\bigl(\tau, x(\tau)\bigr)\,d\tau \biggr]\,ds \\& \quad = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dt} \int_{t_{k}}^{t} f\bigl(\tau, x(\tau )\bigr)\,d\tau \int_{\tau}^{t}(t-s)^{-\alpha}(s- \tau)^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda(s-\tau)^{\alpha}\bigr) \,d\tau \\& \quad = \frac{d}{dt} \int_{t_{k}}^{t} E_{\alpha}\bigl(-\lambda(t- \tau)^{\alpha}\bigr)f\bigl(\tau , x(\tau)\bigr)\,d\tau \\& \quad = f\bigl(t, x(t)\bigr)-\lambda\bigl(Q_{k}^{\alpha}x\bigr) (t), \quad\text{a.e. } t\in [t_{k}, t_{k+1}], \end{aligned}
(3.4)

and by Lemma 2.6(3), one gets

\begin{aligned}& \bigl[{}^{L} D_{t_{k}^{+}}^{\delta} \bigl(Q_{k}^{\alpha}x\bigr) (s) \bigr](t) \\& \quad = \frac{1}{\Gamma(1-\delta)}\frac{d}{dt} \int_{t_{k}}^{t} f\bigl(\tau, x(\tau )\bigr)\,d\tau \int_{\tau}^{t}(t-s)^{-\delta}(s- \tau)^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda(s-\tau)^{\alpha}\bigr) \,ds \\& \quad = \frac{d}{dt} \int_{t_{k}}^{t} (t-\tau)^{\alpha-\delta}E_{\alpha, \alpha -\delta+1} \bigl(-\lambda(t-\tau)^{\alpha}\bigr)f\bigl(\tau, x(\tau)\bigr)\,d\tau \\& \quad = \int_{t_{k}}^{t} (t-\tau)^{\alpha-\delta-1}E_{\alpha, \alpha-\delta } \bigl(-\lambda(t-\tau)^{\alpha}\bigr)f\bigl(\tau, x(\tau)\bigr)\,d\tau \\& \quad = \bigl(Q_{k}^{\alpha -\delta} x\bigr) (t), \quad\text{a.e. } t \in[t_{k}, t_{k+1}]. \end{aligned}
(3.5)

Noting (3.2) and (3.3), we have $${ (Q_{k}^{\alpha}x)(t_{k}^{+})=0}$$ and $${ (Q_{k}^{\alpha-\delta}x)(t_{k}^{+})=0}$$. Then, from Definition 2.3, with $$g(t)$$ replaced by $$(Q_{k}^{\alpha}x)(t)$$ and $$(Q_{k}^{\alpha-\delta}x)(t)$$, and applying (3.4) and (3.5), we derive

$$\bigl[ {}^{c} D_{t_{k}^{+}}^{\alpha}\bigl(Q_{k}^{\alpha}x \bigr) (s) \bigr](t)= \bigl[ {}^{L} D_{t_{k}^{+}}^{\alpha} \bigl(Q_{k}^{\alpha}x\bigr) (s) \bigr](t)=f\bigl(t, x(t)\bigr)- \lambda \bigl(Q_{k}^{\alpha}x\bigr) (t)$$

and $$[ {}^{c} D_{t_{k}^{+}}^{\delta}(Q_{k}^{\alpha}x)(s) ](t)=(Q_{k}^{\alpha-\delta}x)(t)$$. This completes the proof. □

### Lemma 3.6

Assume that (H1) holds. Then $${ [I^{\gamma}_{0^{+}}(Q^{\alpha}_{0} x)(s) ](t) =(Q^{\alpha+\gamma}_{0} x)(t)}$$.

### Proof

It follows from (3.2) that $$(Q^{\alpha}_{0} x)(t)$$ is Lebesgue integrable, noting Lemma 2.6(4), we have

\begin{aligned}& \bigl[I^{\gamma}_{0^{+}}\bigl(Q^{\alpha}_{0} x \bigr) (s) \bigr](t) \\& \quad = \frac{1}{\Gamma(\gamma)} \int_{0}^{t}(t-s)^{\gamma-1} \biggl( \int_{0}^{s}(s-\tau )^{\alpha-1}E_{\alpha, \alpha} \bigl(-\lambda(s-\tau)^{\alpha}\bigr)f\bigl(\tau,x(\tau )\bigr)\,d\tau \biggr)\,ds \\& \quad = \frac{1}{\Gamma(\gamma)} \int_{0}^{t}f\bigl(\tau,x(\tau)\bigr)\,d\tau \int_{0}^{t-\tau }(t-\tau-s)^{\gamma-1}{s}^{\alpha-1}E_{\alpha, \alpha} \bigl(-\lambda {s}^{\alpha}\bigr)\,ds \\& \quad = \int_{0}^{t}(t-\tau)^{\alpha+\gamma-1}E_{\alpha, \alpha+\gamma} \bigl(-\lambda (t-\tau)^{\alpha}\bigr)f\bigl(\tau,x(\tau)\bigr)\,d\tau= \bigl(Q^{\alpha+\gamma}_{0} x\bigr) (t). \end{aligned}

□

As a consequence of Lemmas 3.4-3.6, by directly computation, we get the following result. For brevity, we define

\begin{aligned}& \widetilde{c} := -\frac{(Q_{0}^{\alpha+\gamma}x)(\eta)}{1+\eta^{\gamma}E_{\alpha,\gamma+1}(-\lambda\eta^{\alpha})}, \\& (P_{0}x) (t) := \widetilde{c}E_{\alpha}\bigl(-\lambda t^{\alpha}\bigr), \\& (P_{i}x) (t) := \bigl[(P_{i-1}x) (t_{i})+ \bigl(Q_{i-1}^{\alpha}x\bigr) (t_{i})+I_{i} \bigr]E_{\alpha}\bigl(-\lambda(t-t_{i})^{\alpha}\bigr),\quad i=1,\ldots,m-1, \\& (P_{m}x) (t) := -\frac{ [(Q_{m}^{\alpha}x)(1)+(Q_{m}^{\alpha-\delta} x)(1) ]E_{\alpha}(-\lambda(t-t_{m})^{\alpha})}{E_{\alpha}(-\lambda (1-t_{m})^{\alpha})-\lambda(1-t_{m})^{\alpha-\delta}E_{\alpha, \alpha-\delta +1}(-\lambda(1-t_{m})^{\alpha})}. \end{aligned}

### Lemma 3.7

A function x is a solution of (1.1)-(1.3) if and only if x is a solution of the following equation:

$$x(t)= \textstyle\begin{cases} (P_{0}x)(t)+(Q_{0}^{\alpha}x)(t),& \textit{for } t\in J_{0},\\ (P_{1}x)(t)+(Q_{1}^{\alpha}x)(t),& \textit{for } t\in J_{1},\\ \cdots\\ (P_{m-1}x)(t)+(Q_{m-1}^{\alpha}x)(t),& \textit{for } t\in J_{m-1}, \\ (P_{m}x)(t)+(Q_{m}^{\alpha}x)(t),& \textit{for } t\in J_{m}. \end{cases}$$
(3.6)

### Proof

(Necessity) For $$t\in J_{0}$$, it follows from Lemma 2.10 that $$x(t)=a_{0} E_{\alpha}(-\lambda t^{\alpha})+(Q_{0}^{\alpha}x)(t)$$. Obviously, $$x(0)=a_{0}$$. Moreover, from Lemma 2.6(4) (taking $$\beta:=\gamma$$, $$\nu:=1$$) and Lemma 3.6, we have

$$I^{\gamma}_{0^{+}}x(\eta)=a_{0}\eta^{\gamma}E_{\alpha,\gamma+1}\bigl(-\lambda\eta ^{\alpha}\bigr)+\bigl(Q_{0}^{\alpha+\gamma}x \bigr) (\eta).$$

Using the condition $$x(0)+I^{\gamma}_{0^{+}}x(\eta)=0$$, we obtain $$a_{0}=\widetilde{c}$$, then, for $$t\in J_{0}$$,

$$x(t)=(P_{0}x) (t)+\bigl(Q_{0}^{\alpha}x\bigr) (t).$$

For $$t\in J_{1}$$, $$x(t)=a_{1} E_{\alpha}(-\lambda(t-t_{1})^{\alpha})+(Q_{1}^{\alpha}x)(t)$$, since $${x(t_{1}^{+})=a_{1}= (P_{0}x)(t_{1})+(Q_{0}^{\alpha}x)(t_{1})+I_{1}}$$, then, for $$t\in J_{1}$$,

$$x(t)=(P_{1}x) (t)+\bigl(Q_{1}^{\alpha}x\bigr) (t).$$

Repeating the above process, we find

$$x(t)=(P_{k}x) (t)+\bigl(Q_{k}^{\alpha}x\bigr) (t),\quad t\in J_{k}, k=0,1,\ldots,m-1.$$

For $$t\in J_{m}=[t_{m},1]$$, $${x(t)=a_{m} E_{\alpha}(-\lambda(t-t_{m})^{\alpha})+(Q_{m}^{\alpha}x)(t)}$$.

Noting Lemma 2.7(5) and Lemma 3.5, we get

\begin{aligned} {}^{c} D_{t_{m}^{+}}^{\delta}x(t) =&-\lambda a_{m} (t-t_{m})^{\alpha-\delta} E_{\alpha, \alpha-\delta+1}\bigl(- \lambda(t-t_{m})^{\alpha}\bigr)+\bigl(Q_{m}^{\alpha-\delta}x \bigr) (t). \end{aligned}

From $$x(1)+{}^{c} D_{t_{m}^{+}}^{\delta}x(1)=0$$, one can obtain

$$a_{m}=-\frac{(Q_{m}^{\alpha}x)(1)+(Q_{m}^{\alpha-\delta} x)(1)}{E_{\alpha }(-\lambda(1-t_{m})^{\alpha})-\lambda(1-t_{m})^{\alpha-\delta}E_{\alpha, \alpha-\delta+1}(-\lambda(1-t_{m})^{\alpha})}.$$

Now, $$x(t)=(P_{m}x)(t)+(Q_{m}^{\alpha}x)(t)$$.

(Sufficiency) Let $$x(t)$$ satisfy (3.6). Noting Lemma 2.7(5) and Lemma 3.5, $$({}^{c}D_{t_{k}^{+}}^{\alpha}x)(t)$$ exists and $${}^{c}D_{t_{k}^{+}}^{\alpha}x(t)+\lambda x(t)=f(t, x(t))$$ for $$t\in J_{k}$$ ($$k=0,1,\ldots,m$$). Moreover, for $$k=1,2,\ldots,m-1$$,

\begin{aligned} x\bigl(t_{k}^{+}\bigr)-x\bigl(t_{k}^{-}\bigr) =&(P_{k} x) (t_{k})+\bigl(Q_{k}^{\alpha}x\bigr) (t_{k})-(P_{k-1} x) (t_{k})-\bigl(Q_{k-1}^{\alpha}x \bigr) (t_{k}) \\ =&(P_{k-1} x) (t_{k})+\bigl(Q_{k-1}^{\alpha}x \bigr) (t_{k})+I_{k}-(P_{k-1} x) (t_{k})- \bigl(Q_{k-1}^{\alpha}x\bigr) (t_{k}) \\ =&I_{k}. \end{aligned}

The boundary conditions of (1.3) are clearly satisfied, that is, $$x(t)$$ satisfies (1.1)-(1.3). □

## Existence result

In this section, we deal with the existence of solution for the problem (1.1)-(1.3). To this end, we consider the following assumption.

(H2):

There exists a function $$\psi\in L^{\frac{1}{q_{2}}}(J,\mathbb {R}^{+})$$ ($$q_{2}\in(0, \alpha)$$) such that

$$\bigl\vert f(t,x)-f(t,y) \bigr\vert \leq\psi(t) \vert x-y \vert .$$

For convenience, we introduce the following notation:

\begin{aligned}& c_{\alpha} = \frac{1}{\Gamma(\alpha)} \biggl(\frac{1-q_{1}}{\alpha -q_{1}} \biggr)^{1-q_{1}} \Vert \mu \Vert _{L^{\frac {1}{q_{1}}}},\qquad M_{\alpha}= \frac{1}{\Gamma(\alpha)} \biggl(\frac {1-q_{2}}{\alpha-q_{2}} \biggr)^{1-q_{2}} \Vert \psi \Vert _{L^{\frac {1}{q_{2}}}}, \\& T_{0} = \frac{c_{\alpha+\gamma}}{1+\eta^{\gamma}E_{\alpha,\gamma +1}(-\lambda\eta^{\alpha})}, \\& T_{i} = T_{i-1}+c_{\alpha}+ \vert I_{i} \vert ,\quad i=1,2,\ldots, m-1, \\& T_{m} = \frac{c_{\alpha}+\mathcal{C}\zeta_{\alpha-\delta}}{ \vert E_{\alpha}(-\lambda(1-t_{m})^{\alpha})-\lambda(1-t_{m})^{\alpha-\delta }E_{\alpha, \alpha-\delta+1}(-\lambda(1-t_{m})^{\alpha}) \vert }. \end{aligned}

Clearly, $$T_{0}< T_{1}<\cdots<T_{m-1}$$.

### Theorem 4.1

Assume that (H1) and (H2) are satisfied, then the problem (1.1)-(1.3) has at least a solution $$x\in X$$ if $$M_{\alpha}<1$$.

### Proof

Define an operator $$\mathcal{F}: X\rightarrow X$$ by

$$(\mathcal{F} x) (t)= \textstyle\begin{cases} (P_{0}x)(t)+(Q_{0}^{\alpha}x)(t),& t\in J_{0},\\ (P_{1}x)(t)+(Q_{1}^{\alpha}x)(t),& t\in J_{1},\\ \cdots\\ (P_{m-1}x)(t)+(Q_{m-1}^{\alpha}x)(t),& t\in J_{m-1},\\ (P_{m}x)(t)+(Q_{m}^{\alpha}x)(t),& t\in J_{m}. \end{cases}$$
(4.1)

From Lemma 2.9(ii) and Lemma 3.3, we see that $$\mathcal{F}:X\rightarrow X$$ is clearly well defined.

Similar to (3.2) and (3.3), combining with Lemma 2.9(i) and (2.4), one can get

$$\begin{gathered} \bigl\vert \bigl(Q_{0}^{\alpha+\gamma}x\bigr) (t) \bigr\vert \leq c_{\alpha+\gamma},\qquad \bigl\vert \bigl(Q_{m}^{\alpha-\delta}x \bigr) (t) \bigr\vert \leq\mathcal{C}\zeta_{\alpha-\delta},\\ \bigl\vert \bigl(Q_{k}^{\alpha }x\bigr) (t) \bigr\vert \leq c_{\alpha}, \quad k=0,1,\ldots,m. \end{gathered}$$
(4.2)

Setting $$B_{r}=\{x\in X: \Vert x \Vert _{1}\leq r\}$$, where $$r\geq \max\{T_{m}, T_{m-1}\}+c_{\alpha}$$, we shall prove $$(P_{i}x)(t)+(Q_{i}^{\alpha}y)(t)\in B_{r}$$ for any $$x, y\in B_{r}$$ and $$t\in J_{i}$$ ($$i=0,1,\ldots, m$$).

By Lemma 2.9(i) and (4.2), we have

\begin{aligned} \bigl\vert (P_{0}x) (t)+\bigl(Q_{0}^{\alpha}y \bigr) (t) \bigr\vert \leq& \frac {c_{\alpha+\gamma}}{1+\eta^{\gamma}E_{\alpha,\gamma+1}(-\lambda\eta ^{\alpha})}+c_{\alpha}= T_{0}+c_{\alpha}\leq r. \end{aligned}

For $$t\in J_{1}$$, one has

\begin{aligned} \bigl\vert (P_{1}x) (t)+\bigl(Q_{1}^{\alpha}y\bigr) (t) \bigr\vert \leq& \bigl\vert (P_{0} x) (t_{1})+ \bigl(Q_{0}^{\alpha}x\bigr) (t_{1})+I_{1} \bigr\vert + \bigl\vert \bigl(Q_{1}^{\alpha }y\bigr) (t) \bigr\vert \\ \leq&T_{0}+c_{\alpha}+ \vert I_{1} \vert +c_{\alpha}= T_{1}+c_{\alpha}\leq r. \end{aligned}

Repeating the above process, for $$t\in J_{i}$$ ($$i=2,\ldots,m-1$$), we find

$$\bigl\vert (P_{i}x) (t)+\bigl(Q_{i}^{\alpha}y \bigr) (t) \bigr\vert \leq T_{i}+c_{\alpha}\leq r.$$

For $$t\in J_{m}$$, one sees

$$\bigl\vert (P_{m}x) (t)+\bigl(Q_{m}^{\alpha}y \bigr) (t) \bigr\vert \leq T_{m}+c_{\alpha }\leq r.$$

Now, we can see that $$(P_{i}x)(t)+(Q_{i}^{\alpha}y)(t)\in B_{r}$$ for any $$t\in J_{i}$$ ($$i=0,1,\ldots, m$$) and $$x, y\in B_{r}$$.

Similar to (3.1), for $$t\in J_{i}$$, $$i=0,1,\ldots,m$$, one gets

\begin{aligned} \bigl\vert \bigl(Q_{i}^{\alpha}x\bigr) (t)- \bigl(Q_{i}^{\alpha}y\bigr) (t) \bigr\vert \leq& \int _{t_{i}}^{t} (t-s)^{\alpha-1} E_{\alpha,\alpha}\bigl(-\lambda(t-s)^{\alpha}\bigr) \bigl\vert f \bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert \,ds \\ \leq& \frac{1}{\Gamma(\alpha)} \int_{t_{i}}^{t} (t-s)^{\alpha-1} \psi (s)\,ds \Vert x-y \Vert _{1}\leq M_{\alpha} \Vert x-y \Vert _{1}. \end{aligned}

This implies that $$Q_{i}^{\alpha}$$ ($$i=0,1,\ldots,m$$) is a contraction mapping.

Let $$\{x_{n}\}$$ be a sequence such that $$x_{n}\rightarrow x$$ in X, then there exists $$\varepsilon>0$$ such that $$\Vert x_{n}-x \Vert _{1}\leq\varepsilon$$ for n sufficiently large. By (H2), we obtain

$$\bigl\vert f\bigl(t,x_{n}(t)\bigr)-f\bigl(t,x(t)\bigr) \bigr\vert \leq \psi(t)\varepsilon.$$

Moreover, f satisfies (H1), for almost every $$t\in J$$, we get $$f(t,x_{n}(t))\rightarrow f(t,x(t))$$ as $$n\rightarrow\infty$$. It follows from the Lebesgue dominated convergence theorem that

$$\bigl\Vert (P_{i} x_{n})-(P_{i} x) \bigr\Vert _{1}\rightarrow0,\quad \text{as } n\rightarrow\infty.$$

Now we can see that $$P_{i}$$ ($$i=0,1,\ldots,m$$) is continuous.

Moreover, by Lemma 2.9(ii) and (4.2), $$\{P_{i}x: x\in B_{r}\}$$ is an equicontinuous and uniformly bounded set. Therefore, $$P_{i}$$ is a completely continuous operator on $$B_{r}\vert_{J_{i}}$$ ($$i=0,1,\ldots,m$$). Now, it follows from Theorem 2.11 that problem (1.1)-(1.3) has at least a solution $$x\in B_{r}$$. □

## Application

In this section, we give an example to illustrate the usefulness of our main result.

### Example 5.1

Consider the following impulsive boundary problem of fractional order:

$$\textstyle\begin{cases} {}^{c}D_{*}^{\frac{1}{2}}x(t)+5x(t)={\frac{1}{6\sqrt[14]{t}}\sin(3+ \vert x(t) \vert ),\quad\text{a.e. } t\in(0,1]\setminus\{\frac {1}{4}\}},\\ {\Delta x (\frac{1}{4} )=2},\\ {x(0)+I^{\frac{1}{3}}_{0^{+}}x(\frac{1}{10})=0},\qquad {x(1)+{}^{c} D_{{\frac{1}{3}}^{+}}^{\frac{1}{4}}x(1)=0}. \end{cases}$$
(5.1)

Corresponding to (1.1)-(1.3), we have $$\alpha=\frac {1}{2}$$, $$\gamma=\frac{1}{3}$$, $$\delta=\frac{1}{4}$$, $$\lambda=5$$, $$m=2$$, $$t_{1}=\frac{1}{4}$$, $$t_{2}=\frac{1}{3}$$, $$\eta=\frac{1}{10}$$, $$f(t, x(t))=\frac{1}{6\sqrt[14]{t}}\sin(3+ \vert x(t) \vert )$$, $$I_{1}=2$$.

It is easy to see that $$\vert f(t, x(t)) \vert \leq \nu(t)$$ and $$\vert f(t, x(t))-f(t, y(t)) \vert \leq\psi (t) \vert x(t)-y(t) \vert$$, where $${\nu(t)=\psi(t)=\frac{1}{6\sqrt[14]{t}}\in L^{\frac{1}{q}} ([0, 1])}(q=\frac{1}{7})$$ and $$\Vert \psi \Vert _{L^{7}}= \frac {2^{\frac{1}{7}}}{6}$$. By direct computation, we find that

\begin{aligned} M_{\alpha}=\frac{1}{\Gamma(\alpha)} \biggl(\frac{1-q}{\alpha-q} \biggr)^{1-q} \Vert \psi \Vert _{L^{\frac{1}{q}}}=\frac{ 1}{3\sqrt {\pi}} \biggl(\frac{6}{5}\biggr)^{\frac{6}{7}}\approx0.22< 1. \end{aligned}

Now, due to the fact that all the assumptions of Theorem 4.1 hold, problem (5.1) has at least a solution.

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## Acknowledgements

This work was supported partly by the Natural Science Foundation of China (11561077, 11471227) and the Reserve Talents of Young and Middle-Aged Academic and Technical Leaders of the Yunnan Province.

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Correspondence to Fang Li.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All the authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Miao, Y., Li, F. Boundary value problems of the nonlinear multiple base points impulsive fractional differential equations with constant coefficients. Adv Differ Equ 2017, 190 (2017). https://doi.org/10.1186/s13662-017-1249-4

• Accepted:

• Published:

• DOI: https://doi.org/10.1186/s13662-017-1249-4

• 34A08
• 34A37
• 34B10

### Keywords

• nonlinear multiple base points fractional differential equations
• boundary value problem
• impulsive condition
• Mittag-Leffler functions