1 Introduction

The boundary value problems of fractional differential equations have attracted the attention of many authors. Fractional differential equations are used in mathematical modelling, engineering, biology, chemistry, and many other fields of science; see the references. However, the impulsive fractional differential equations has become a new topic, therefore more researchers interest focused on the field of impulsive problems for fractional differential equations; see [128] and the references therein.

Tian and Bai, [7] used the Banach fixed point theorem and Schauder’s fixed point theorem to obtain the existence of the solutions of the problem which is given as follows:

$$\begin{aligned}& ^{c}D_{0+}^{\alpha}u(t) =f\bigl(t,u(t)\bigr), \\& \Delta u ( t) _{t=t_{k}} =I_{k} \bigl( u(t) \bigr),\quad k=1,2,\ldots,m, \\& \Delta u^{\prime} ( t) _{t=t_{k}} =\bar{I}_{k} \bigl( u(t) \bigr),\quad k=1,2,\ldots,m, \\& u(0)+u^{\prime} ( 0 ) =0, \\& u(1)+u^{\prime} ( \xi) =0. \end{aligned}$$

The existence and uniqueness of the solutions for an anti-periodic BVP of nonlinear impulsive differential equations of order \(\alpha\in ( 2,3 ] \) were obtained, in 2010 [17], given in the following:

$$\begin{aligned}& \bigl( ^{c}D_{0+}^{\alpha}u(t) \bigr) =f\bigl(t,u(t) \bigr),\quad 2< \alpha\leq3 \\& \Delta u ( t_{k} ) =Q_{k} \bigl( u ( t_{k} ) \bigr),\quad k=1,2,\ldots,p, \\& \Delta u^{\prime} ( t_{k} ) =I_{k} \bigl( u ( t_{k} ) \bigr),\quad k=1,2,\ldots,p, \\& \Delta u^{\prime\prime} ( t_{k} ) =I_{k}^{\ast} \bigl( u ( t_{k} ) \bigr),\quad k=1,2,\ldots,p, \\& u(0) =-u(1), \\& u^{\prime}(0) =-u^{\prime}(1), \\& u^{\prime\prime}(0) =-u^{\prime\prime}(1), \end{aligned}$$

with the Caputo fractional derivative \({}^{c}D_{0+}^{\alpha}\), \(f\in C ( J\times\mathbb{R} , \mathbb{R} ) \) and \(Q_{k},I_{k},I_{k}^{\ast}\in C ( \mathbb{R} \times \mathbb{R} )\), \(0=t_{0}< t_{1}<\cdots<t_{k}<\cdots<t_{p}<t_{p+1}=1\).

In 2011, Cao and Chen, [6], studied the following problem to give some existence results and a continuous version of Filippov’s theorem of a fractional differential inclusion:

$$\begin{aligned}& \bigl( ^{C}D_{0+}^{\alpha}u(t) \bigr) \in f \bigl(t,u(t)\bigr), \quad \mbox{a.e. }t\in J \\& \Delta u ( t) _{t=t_{k}} =I_{k} \bigl( u(t) \bigr),\quad k=1,2,\ldots,m, \\& \bigl( \Delta D_{0+}^{\beta}u ( t) _{t=t_{k}} \bigr) = \bar{I}_{k} \bigl( u(t) \bigr),\quad k=1,2,\ldots,m, \\& u(0)+D_{0+}^{\beta}u ( 0 ) =A, \\& u(1)+D_{0+}^{\beta}u ( \zeta) =B. \end{aligned}$$

Here, \({}^{C}D_{0+}^{\alpha}\) is the Caputo fractional derivative and multi-valued map with compact values \(F:J\times \mathbb{R} \rightarrow P(\mathbb{R} )\) where \(P(\mathbb{R} )\) is the family of all nonempty subsets of \(\mathbb{R} \), \(1<\alpha\leq2\) and \(0<\beta<\alpha-1\) with real numbers A, B.

In 2012, the contraction mapping principle, Krasnoselskii’s theorem, Schaefer’s theorem, and the Leray-Schauder alternative were used, in [18], to find the existence of the solutions of the following problem:

$$\begin{aligned}& ^{C}D_{0+}^{q}u(t) =f\bigl(t,u(t)\bigr), \\& \Delta u ( t_{k} ) =y_{k}, \\& \Delta u^{\prime} ( t_{k} ) =\bar{y}_{k},\quad k=1,\ldots,m \\& u(0) =u_{0},\qquad u^{\prime}(0)=\bar{u}_{0};\quad y_{k},\bar{y}_{k},u_{0}, \bar{u}_{0}\in \mathbb{R} . \end{aligned}$$

By using fixed point theorems, the existence and uniqueness solutions for an impulsive mixed boundary value problem of nonlinear differential equations of fractional order were studied in 2016, [1], which is given as

$$\begin{aligned}& ^{C}D_{0+}^{q}u(t) =f\bigl(t,u(t)\bigr),\quad t\in J^{\prime} \\& \Delta u ( t_{k} ) =I_{k} \bigl( u ( t_{k} ) \bigr),\qquad \Delta u^{\prime} ( t_{k} ) =J_{k} \bigl( u ( t_{k} ) \bigr), \\& u ( 0 ) +u^{\prime} ( 0 ) =0,\qquad u ( 1) +u^{\prime} ( 1) =0, \end{aligned}$$

where \(q\in ( 1,2) \) and \({}^{C}D_{0+}^{q}\) is the Caputo derivative of order q.

Motivated by the above mentioned work, we focus on the existence of solutions of fractional differential equation:

$$ ^{C}D_{0+}^{q}u(t)=f\bigl(t,u(t)\bigr),\quad t\in J^{\prime}, $$
(1)

with boundary conditions;

$$\begin{aligned}& \Delta u ( t_{k} ) =I_{k} \bigl( u ( t_{k} ) \bigr) =u \bigl( t_{k}^{+} \bigr) -u \bigl( t_{k}^{-} \bigr), \\& \Delta u^{\prime} ( t_{k} ) =J_{k} \bigl( u ( t_{k} ) \bigr) =u^{\prime} \bigl( t_{k}^{+} \bigr) -u^{\prime} \bigl( t_{k}^{-} \bigr);\quad k=1,\ldots,p \\& u ( 0 ) +\mu_{1}u^{\prime} ( 1) =\sigma _{1}, \\& u^{\prime} ( 0 ) +\mu_{2}u ( 1) =\sigma _{2}, \end{aligned}$$
(2)

where \({}^{C}D_{0+}^{q}\) is the Caputo derivative of order \(q\in ( 1,2)\), \(J= [ 0,1 ] \), \(J^{\prime}=J\backslash \{ t_{1},t_{2},\ldots,t_{p} \} \), \(0=t_{0}< t_{1}<\cdots<t_{p}<t_{p+1}=1\), \(\Delta u ( t_{k} ) =u ( t_{k}^{+} ) -u ( t_{k}^{-} ) \) and \(\Delta u^{\prime} ( t_{k} ) =u^{\prime} ( t_{k}^{+} ) -u^{\prime} ( t_{k}^{-} ) \). Here, respectively, the right and the left limits of \(u ( t) \) at \(t=t_{k}^{+}\) are represented by \(u ( t_{k}^{+} ) \) and \(u ( t_{k}^{-} ) \).

2 Preliminaries

In this section, we introduce preliminary facts which are used throughout this paper. We have

$$\begin{aligned} \mathit{PC}( J ) =& \bigl\{ u: [ 0,1 ] \rightarrow R,u\in C \bigl( J^{\prime} \bigr),\text{and }u \bigl( t_{k}^{+} \bigr),u \bigl( t_{k}^{-} \bigr) \text{ exists}, \\ &\mbox{and }u \bigl( t_{k}^{-} \bigr) =u ( t_{k} ),1\leq k\leq p \bigr\} . \end{aligned}$$

Obviously, \(\mathit{PC}( J ) \) is a Banach space with the norm

$$\Vert u\Vert _{\mathit{PC}}=\sup_{0\leq t\leq1}\bigl\vert u ( t) \bigr\vert . $$

Definition 1

The Riemann-Liouville fractional integral of order \(\alpha>0\) for a function \(f:[ 0,+\infty) \rightarrow R\) is defined as

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

provided that the right hand side of the integral is pointwise defined on \((0,+\infty)\) and Γ is the gamma function.

Definition 2

The Caputo derivative of order \(\alpha>0\) for a function \(f: [ 0,+\infty) \rightarrow R\) is written as

$$D_{0+}^{\alpha}f(t)=\frac{1}{\Gamma(n-\alpha)} \int _{0}^{t}(t-s)^{n-\alpha-1}f^{ ( n) }(s)\,ds, $$

where \(n= [ \alpha ] +1\), \([ \alpha ] \) is the integral part of α.

Lemma 3

Let \(\alpha>0\). Then the differential equation \(D_{0+}^{\alpha}f(t)=0\) has solutions

$$f(t)=k_{0}+k_{1}t+k_{2}t^{2}+\cdots+k_{n-1}t^{n-1} $$

and

$$I_{0+}^{\alpha}D_{0+}^{\alpha}f(t)=f(t)+k_{0}+k_{1}t+k_{2}t^{2}+\cdots+k_{n-1}t^{n-1}, $$

where \(k_{i}\in R\) and \(i=1,2,\ldots\) , \(n= [ \alpha ] +1\).

Lemma 4

[9]

The set \(F\subset \mathit{PC}( [ 0,1 ] ,R^{n} ) \) is relatively compact if and only if F is bounded, that is, \(\Vert x\Vert \leq C\) for each \(x\in F\) and some \(C>0\), and/or F is quasi-equicontinuous in \([ 0,1 ] \). That is to say, for any \(\varepsilon>0\) there exists \(\delta>0\) such that if \(x\in F\), \(k\in N\); \(\tau_{1},\tau_{2}\in( t_{k-1},t_{k} ] \) and \(\vert \tau _{1}-\tau_{2}\vert <\delta\), we have \(\vert x ( \tau _{1} ) -x ( \tau_{2} ) \vert <\varepsilon\).

Lemma 5

[19]

Let M be a closed, convex, and nonempty subset of Banach space X, and the operators A and B be such that

  1. (i)

    \(Ax+By\in M\) whenever \(x,y\in M\);

  2. (ii)

    A is compact and continuous;

  3. (iii)

    B is contraction mapping.

Then there exists \(z\in M\) such that \(z=Az+Bz\).

Lemma 6

For \(q\in ( 1,2) \), and the continuous function \(f:J\rightarrow R\), we have the following impulsive fractional boundary value problem:

$$\begin{aligned}& ^{C}D_{0+}^{q}u(t) =f\bigl(t,u(t)\bigr), \\& \Delta u ( t_{k} ) =I_{k} \bigl( u ( t_{k} ) \bigr) =u \bigl( t_{k}^{+} \bigr) -u \bigl( t_{k}^{-} \bigr), \\& \Delta u^{\prime} ( t_{k} ) =J_{k} \bigl( u ( t_{k} ) \bigr) =u^{\prime} \bigl( t_{k}^{+} \bigr) -u^{\prime} \bigl( t_{k}^{-} \bigr); \quad k=1,\ldots,p, \\& u ( 0 ) +\mu_{1}u^{\prime} ( 1) =\sigma_{1}, \\& u^{\prime} ( 0 ) +\mu_{2}u ( 1) =\sigma_{2}, \end{aligned}$$

has a unique solution, and Green’s function is given by

$$u(t)= \textstyle\begin{cases} \int _{0}^{t}\frac{ ( t-s) ^{q-1}}{\Gamma ( q) }f(s)\,ds-\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac {(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ \quad {}-\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+\sigma_{1}\omega_{1} ( t) +\sigma_{2}\omega_{2} ( t),\quad t\in [ 0,t_{1} ],\\\int _{0}^{t}\frac{ ( t-s) ^{q-1}}{\Gamma ( q) }f(s)\,ds-\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac {(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ \quad {}-\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+\sigma_{1}\omega_{1} ( t) +\sigma_{2}\omega_{2} ( t) \\ \quad {}-\omega_{1} ( t) \sum_{j=1}^{p}J_{j} ( u(t_{j})) t_{j}-\omega_{2} ( t) \sum_{j=1}^{p}J_{j} ( u(t_{j})) \\ \quad {}+\omega_{1} ( t) \sum_{j=1}^{p}I_{j} ( u(t_{j})) \\ \quad {}+\sum_{j=k+1}^{p}J_{j} ( u(t_{j})) ( t_{j}-t) -\sum_{j=k+1}^{p}I_{j} ( u(t_{j}));\quad t\in [ t_{k},t_{k+1} ] \\ \int _{0}^{t}\frac{ ( t-s) ^{q-1}}{\Gamma ( q) }f(s)\,ds-\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac {(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ \quad {}-\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+\sigma_{1}\omega_{1} ( t) +\sigma_{2}\omega_{2} ( t) \\ \quad {}-\omega_{1} ( t) \sum_{j=1}^{p}J_{j} ( u(t_{j})) t_{j}-\omega_{2} ( t) \sum_{j=1}^{p}J_{j} ( u(t_{j})) \\ \quad {}+\omega_{1} ( t) \sum_{j=1}^{p}I_{j} ( u(t_{j})),\quad t\in [ t_{p},t_{p+1} ] ,\end{cases} $$

where

$$\omega_{1} ( t) =\frac{1+\mu_{2}-\mu_{2}t}{1+\mu_{2}-\mu_{1}\mu_{2}}\quad \textit{and}\quad \omega_{2} ( t) =\frac{t-\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}}. $$

Proof

A general solution \({}^{C}D_{0+}^{q}u(t)=f(t,u(t))\), on \(( t_{k},t_{k+1} ] \), \(k=1,\ldots,p\),

$$u(t)= \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+a_{k}+b_{k}t,\quad \text{for }t\in ( t_{k},t_{k+1} ] , $$

where \(t_{0}=0\), \(t_{p+1}=1\) and taking the derivative,

$$u^{\prime}(t)= \int _{0}^{t}\frac{(t-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+b_{k},\quad \text{for }t\in ( t_{k},t_{k+1} ] . $$

We use the boundary conditions \(u ( 0 ) +\mu_{1}u^{\prime} ( 1) =\sigma_{1}\) and \(u^{\prime} ( 0 ) +\mu_{2}u ( 1) =\sigma_{2}\) to get

$$a_{0}+\mu_{1} \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+ \mu_{1}b_{p}=\sigma_{1} $$

and

$$b_{0}+\mu_{2} \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+ \mu_{2}a_{p}+\mu_{2}b_{p}= \sigma_{2}, $$

where

$$\begin{aligned}& u(0) =a_{0},\qquad u^{\prime}(0)=b_{0}, \\& u(1) = \int _{0}^{t}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+a_{p}+b_{p}, \\& u^{\prime}(1) = \int _{0}^{t}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+b_{p}. \end{aligned}$$

That is,

$$\begin{aligned}& \begin{aligned} \Delta u^{\prime} ( t_{k} ) & =J_{k} \bigl( u ( t_{k} ) \bigr) \\ & =u^{\prime} \bigl( t_{k}^{+} \bigr) -u^{\prime} \bigl( t_{k}^{-} \bigr) \\ & =b_{k}-b_{k-1}, \end{aligned} \\& b_{k} =b_{k-1}+J_{k} \bigl( u ( t_{k} ) \bigr), \\& b_{k+1} =b_{k}+J_{k+1} \bigl( u ( t_{k}+1) \bigr), \\& b_{p} =b_{k-1}+\sum_{j=k}^{p}J_{j} \bigl( u ( t_{j} ) \bigr), \\& b_{k} =b_{p}-\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$

and

$$\begin{aligned}& \Delta u ( t_{k} ) =I_{k} \bigl( u ( t_{k} ) \bigr) =u \bigl( t_{k}^{+} \bigr) -u \bigl( t_{k}^{-} \bigr), \\& a_{k}+b_{k}t_{k} =a_{k-1}+b_{k-1}t_{k}+I_{k} \bigl( u ( t_{k} ) \bigr). \end{aligned}$$

Since \(b_{k}=b_{k-1}+J_{k} ( u ( t_{k} )) \), we have

$$\begin{aligned}& a_{k}+ \bigl( b_{k-1}+J_{k} \bigl( u ( t_{k} ) \bigr) \bigr) t_{k} =a_{k-1}+b_{k-1}t_{k}+I_{k} \bigl( u ( t_{k} ) \bigr), \\& a_{k}+b_{k-1}t_{k}+J_{k} \bigl( u ( t_{k} ) \bigr) t_{k} =a_{k-1}+b_{k-1}t_{k}+I_{k} \bigl( u ( t_{k} ) \bigr), \\& a_{k}+J_{k} \bigl( u ( t_{k} ) \bigr) t_{k} =a_{k-1}+I_{k} \bigl( u ( t_{k} ) \bigr), \\& a_{k} =a_{k-1}-J_{k} \bigl( u ( t_{k} ) \bigr) t_{k}+I_{k} \bigl( u ( t_{k} ) \bigr), \\& a_{k} =a_{p}+\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\sum _{j=k+1}^{p}I_{j} \bigl( u ( tj) \bigr). \end{aligned}$$

Then

$$ a_{0}+\mu_{1}b_{p}+\mu_{1} \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds= \sigma_{1} $$
(3)

and

$$ b_{0}+\mu_{2}a_{p}+\mu_{2}b_{p}+ \mu_{2} \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds= \sigma_{2}. $$
(4)

Also we get

$$\begin{aligned}& \begin{aligned} &b_{k} =b_{k-1}+J_{k} \bigl( u ( t_{k} ) \bigr), \\ &b_{k} =b_{p}-\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned} \end{aligned}$$
(5)

and

$$\begin{aligned}& \begin{aligned} &a_{k} =a_{k-1}-J_{k} \bigl( u ( t_{k} ) \bigr) t_{k}+I_{k} \bigl( u ( t_{k} ) \bigr), \\ &a_{k} =a_{p}+\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned} \end{aligned}$$
(6)

By combining (3), (4), (5), and (6)

$$\begin{aligned}& a_{0}+\mu_{1}b_{p}+\mu_{1} \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds = \sigma_{1}, \\& a_{p}+\mu_{1}b_{p}+\mu_{1} \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds+\sum _{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) =\sigma_{1}, \end{aligned}$$

and

$$\begin{aligned}& b_{0}+\mu_{2}a_{p}+\mu_{2}b_{p}+ \mu_{2} \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds = \sigma_{2}, \\& \Biggl[ b_{p}-\sum_{j=1}^{p}J_{k} \bigl( u ( t_{j} ) \bigr) \Biggr] +\mu_{2}a_{p}+ \mu_{2}b_{p}+\mu_{2} \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds =\sigma_{2}, \\& \mu_{2}a_{p}+ ( 1+\mu_{2} ) b_{p}+ \mu_{2} \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds- \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) =\sigma_{2}. \end{aligned}$$

Then

$$\begin{aligned}& a_{p} =\sigma_{1}-\mu_{1}b_{p}- \mu_{1} \int _{0}^{1}\frac {(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds-\sum _{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}+\sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr), \\& a_{p} =\frac{\sigma_{2}}{\mu_{2}}-\frac{ ( 1+\mu_{2} ) }{\mu_{2}}b_{p}- \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+ \frac{1}{\mu_{2}}\sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned}$$

Also we have

$$\begin{aligned}& \sigma_{1}-\mu_{1}b_{p}-\mu_{1} \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds-\sum _{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}+\sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \\& \quad =\frac{\sigma_{2}}{\mu_{2}}-\frac{ ( 1+\mu_{2} ) }{\mu_{2}}b_{p}- \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+ \frac{1}{\mu_{2}}\sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned}$$

Therefore \(a_{p}\) and \(b_{p}\) are found as follows:

$$\begin{aligned} b_{p} =&- \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma _{1}+ \biggl( \frac{1}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma_{2} \\ &{} - \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ &{}+ \biggl( \frac{\mu_{1}\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds \\ &{} + \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}+ \biggl( \frac{1}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} - \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \end{aligned}$$
(7)

and

$$\begin{aligned} a_{p} =& \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma_{1}- \biggl( \frac{\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma _{2} \\ &{} + \biggl( \frac{\mu_{1}\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ &{} -\mu_{1} \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds \\ &{} - \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}- \biggl( \frac{\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} + \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned}$$
(8)

By (5), (6), (7), and (8) since

$$\begin{aligned}& b_{k} =b_{p}-\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr), \\& a_{k} =a_{p}+\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$

are known. By (5),

$$\begin{aligned} b_{k} =&- \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma_{1}+ \biggl( \frac{1}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma _{2} \\ &{} - \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+ \biggl( \frac{\mu_{1}\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds \\ &{} + \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}+ \biggl( \frac{1}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} - \biggl( \frac{\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) -\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$
(9)

with the help of (6),

$$\begin{aligned} a_{k} = &\biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma_{1}- \biggl( \frac{\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sigma _{2} \\ &{} + \biggl( \frac{\mu_{1}\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ &{} -\mu_{1} \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds \\ &{}- \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}- \biggl( \frac{\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} + \biggl( \frac{1+\mu_{2}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr) \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) +\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$
(10)

for \(k=0,1,\ldots,p-1\). By using (9) and (10), we get

$$\begin{aligned} a_{k}+b_{k}t =& \biggl[ \frac{1+\mu_{2}-\mu_{2}t}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \sigma_{1}+ \biggl[ \frac{t-\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \sigma_{2} \\ &{} + \biggl[ \frac{-\mu_{2} ( t-\mu_{1} ) }{1+\mu_{2}-\mu_{1}\mu _{2}} \biggr] \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ &{} + \biggl[ \frac{-\mu_{1} ( 1+\mu_{2}-\mu_{2}t) }{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds \\ &{} - \biggl[ \frac{1+\mu_{2} ( 1-t) }{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j} \\ &{} + \biggl[ \frac{-\mu_{1}+t}{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} + \biggl[ \frac{1+\mu_{2} ( 1-t) }{1+\mu_{2}-\mu_{1}\mu_{2}} \biggr] \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} +\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) ( t_{j}-t) -\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned}$$

Thus

$$\begin{aligned} u(t) =& \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }f(s)\,ds+ \omega_{1} ( t) \sigma_{1}+\omega_{2} ( t) \sigma_{2} \\ &{} -\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f(s)\,ds \\ &{} -\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f(s)\,ds- \omega_{1} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j} \\ &{} -\omega_{2} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) +\omega_{1} ( t) \sum _{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} +\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) ( t_{j}-t) -\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$

where

$$\omega_{1} ( t) =\frac{1+\mu_{2}-\mu_{2}t}{1+\mu_{2}-\mu_{1}\mu_{2}}\quad \text{and}\quad \omega_{2} ( t) =\frac{t-\mu_{1}}{1+\mu _{2}-\mu_{1}\mu_{2}}. $$

 □

2.1 Existence and uniqueness results

In this section, we state and prove existence and uniqueness results of the fractional BVP (1)-(2) by using the Banach fixed point theorem. We use the following notations throughout this paper:

$$\omega_{1} ( t) =\frac{1+\mu_{2}-\mu_{2}t}{1+\mu_{2}-\mu_{1}\mu_{2}},\qquad \omega_{2} ( t) = \frac{t-\mu_{1}}{1+\mu_{2}-\mu_{1}\mu_{2}}$$

and

$$\omega_{1} ( t) \leq\omega_{1}:=\frac{1+2\vert \mu _{2}\vert }{\vert 1+\mu_{2}-\mu_{1}\mu_{2}\vert },\qquad \omega_{2} ( t) \leq\omega_{2}:=\frac{1+\vert \mu _{1}\vert }{\vert 1+\mu_{2}-\mu_{1}\mu_{2}\vert }. $$

By using the following conditions, we state and prove our first result.

  1. (A1)

    The function \(f: [ 0,1 ] \times R\rightarrow R\) is jointly continuous.

  2. (A2)

    There exist positive constants \(L_{1}\), \(L_{2}\), \(L_{3}\), \(M_{1}\), \(M_{2}\) such that

    $$\begin{aligned}& \bigl\vert f ( t,x) -f ( t,y) \bigr\vert \leq L_{1}\vert x-y\vert ,\quad t\in [ 0,1 ] , x,y\in R; \\& \bigl\vert I_{k} ( x) -I_{k} ( y) \bigr\vert \leq L_{2}\vert x-y\vert ,\qquad \bigl\vert J_{k} ( x) -J_{k} ( y) \bigr\vert \leq L_{3}\vert x-y\vert , \\& \bigl\vert I_{k} ( x) \bigr\vert \leq M_{1}, \qquad \bigl\vert J_{k} ( x) \bigr\vert \leq M_{2}. \end{aligned}$$

Also it is clear that

$$\begin{aligned} \bigl\vert f ( t,x) \bigr\vert & \leq\bigl\vert f ( t,x) -f ( t,0 ) \bigr\vert +\bigl\vert f ( t,0 ) \bigr\vert \\ & \leq L_{1}\vert x\vert +M, \end{aligned}$$

where \(\sup_{t\in [ 0,1 ] }\vert f ( t,0 ) \vert =M\).

Theorem 7

Assume (A1)-(A2) holds. If

$$\begin{aligned} & L_{1} \biggl( \frac{1+\vert \mu_{2}\vert \omega_{2}}{\Gamma ( q+1) }+\frac{\vert \mu_{1}\vert \omega_{1}}{\Gamma ( q) } \biggr) + ( \omega_{1}+1) p ( L_{2}+L_{3} ) + \omega_{2}pL_{3} < 1, \end{aligned}$$
(11)

then our boundary value problem (1)-(2) has a unique solution on \([ 0,1 ] \).

Proof

By using (11) r can be chosen as follows:

$$\begin{aligned} r >& \biggl( 1-\frac{L_{1}}{\Gamma ( q+1) } \bigl( 1+\vert \mu_{2}\vert \omega_{2} \bigr) -\vert \mu_{1}\vert \omega_{1}\frac{L_{1}}{\Gamma ( q) } \biggr) ^{-1} \biggl\{ \frac{M}{\Gamma ( q+1) }+\omega_{1}\vert \sigma _{1}\vert + \omega_{2}\vert \sigma_{2}\vert \\ &{} +\vert \mu_{2}\vert \omega_{2} \frac{M}{\Gamma ( q+1) }+\vert \mu_{1}\vert \omega_{1} \frac{M}{\Gamma ( q) } \\ &{}+ ( \omega_{1}+2) p ( M_{1}+M_{2} ) + \omega_{2}pM_{2} \biggr\} . \end{aligned}$$

Define an operator \(\mathcal{T}:\mathit{PC}( [ 0,1 ] ,R ) \rightarrow \mathit{PC}( [ 0,1 ] ,R ) \) to transform (1)-(2) into the fixed point problem

$$\begin{aligned} ( \mathcal{T} u) (t) =& \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }f\bigl(s,u ( s) \bigr)\,ds+\omega_{1} ( t) \sigma_{1}+\omega _{2} ( t) \sigma_{2} \\ &{} -\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f\bigl(s,u ( s) \bigr)\,ds \\ &{} -\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f\bigl(s,u ( s) \bigr)\,ds \\ &{} -\omega_{1} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}-\omega_{2} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} +\omega_{1} ( t) \sum_{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} +\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) ( t_{j}-t) -\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr), \end{aligned}$$

where \(t_{k}< t< t_{k+1}\), \(k=0,\ldots,p\). Then

$$\begin{aligned} \bigl\vert Tu(t)\bigr\vert \leq& \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl(s,u ( s) \bigr)\bigr\vert \,ds+\bigl\vert \omega _{1} ( t) \bigr\vert \vert \sigma_{1}\vert +\bigl\vert \omega_{2} ( t) \bigr\vert \vert \sigma_{2}\vert \\ &{} +\vert \mu_{2}\vert \bigl\vert \omega_{2} ( t) \bigr\vert \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl(s,u ( s) \bigr)\bigr\vert \,ds \\ &{} +\vert \mu_{1}\vert \bigl\vert \omega_{1} ( t) \bigr\vert \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }\bigl\vert f \bigl(s,u ( s) \bigr)\bigr\vert \,ds \\ &{} +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert +\bigl\vert \omega_{2} ( t) \bigr\vert \sum_{j=1}^{p} \bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert \\ &{} +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert I_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert \\ &{} +2\sum_{j=k+1}^{p}\bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert +\sum _{j=k+1}^{p}\bigl\vert I_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert \end{aligned}$$

and then

$$\begin{aligned}& \bigl\vert \mathcal{T} u(t)\bigr\vert \\& \quad \leq \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl(s,u ( s) \bigr)-f(s,0)\bigr\vert \,ds+ \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }\bigl\vert f(s,0)\bigr\vert \,ds \\& \qquad {} +\bigl\vert \omega_{1} ( t) \bigr\vert \vert \sigma _{1}\vert +\bigl\vert \omega_{2} ( t) \bigr\vert \vert \sigma_{2}\vert \\& \qquad {} +\vert \mu_{2}\vert \bigl\vert \omega_{2} ( t) \bigr\vert \biggl[ \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl(s,u ( s) \bigr)-f(s,0)\bigr\vert \,ds+ \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }\bigl\vert f(s,0) \bigr\vert \,ds \biggr] \\& \qquad {} +\vert \mu_{1}\vert \bigl\vert \omega_{1} ( t) \bigr\vert \biggl[ \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }\bigl\vert f \bigl(s,u ( s) \bigr)-f(s,0)\bigr\vert \,ds+ \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }\bigl\vert f(s,0) \bigr\vert \,ds \biggr] \\& \qquad {} +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert +\bigl\vert \omega_{2} ( t) \bigr\vert \sum_{j=1}^{p} \bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert I_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert \\& \qquad {} +\sum_{j=k+1}^{p}\bigl\vert J_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert +\sum _{j=k+1}^{p}\bigl\vert I_{j} \bigl( u ( t_{j} ) \bigr) \bigr\vert . \end{aligned}$$

Thus

$$\begin{aligned} \bigl\vert \mathcal{T} u(t)\bigr\vert \leq&\frac{L_{1}r}{\Gamma ( q+1) }+ \frac{M}{\Gamma ( q+1) }+\omega_{1}\vert \sigma_{1}\vert + \omega_{2}\vert \sigma_{2}\vert \\ &{} +\vert \mu_{2}\vert \omega_{2} \biggl[ \frac{L_{1}r}{\Gamma ( q+1) }+\frac{M}{\Gamma ( q+1) } \biggr] \\ &{} +\vert \mu_{1}\vert \omega_{1} \biggl[ \frac{L_{1}r}{\Gamma ( q) }+\frac{M}{\Gamma ( q) } \biggr] \\ &{} + ( \omega_{1}+2) p ( M_{1}+M_{2} ) +\omega _{2}pM_{2}< r. \end{aligned}$$

For \(t\in [ 0,1 ] \), the expression is well defined. The fixed point of the operator T is the solution of our boundary value problem (1)-(2). To show the existence and uniqueness of the solution, the Banach fixed point theorem is used and then it is shown that T is a contraction and we get

$$\begin{aligned}& \bigl\vert (\mathcal{T}x) (t)-(\mathcal{T}y) (t)\bigr\vert \\& \quad \leq \int _{0}^{t}\frac{(t-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl( s,x ( s) \bigr) -f \bigl( s,y ( s) \bigr) \bigr\vert \,ds \\& \qquad {} +\vert \mu_{2}\vert \bigl\vert \omega_{2} ( t) \bigr\vert \biggl[ \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }\bigl\vert f \bigl( s,x ( s) \bigr) -f \bigl( s,y ( s) \bigr) \bigr\vert \,ds \biggr] \\& \qquad {} +\vert \mu_{1}\vert \bigl\vert \omega_{1} ( t) \bigr\vert \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }\bigl\vert f \bigl( s,x ( s) \bigr) -f \bigl( s,y ( s) \bigr) \bigr\vert \,ds \\& \qquad {} +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert J_{j} \bigl( x ( t_{j} ) \bigr) -J_{j} \bigl( y ( t_{j} ) \bigr) \bigr\vert \\& \qquad {} +\bigl\vert \omega_{2} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert J_{j} \bigl( x ( t_{j} ) \bigr) -J_{j} \bigl( y ( t_{j} ) \bigr) \bigr\vert \\& \qquad {} +\bigl\vert \omega_{1} ( t) \bigr\vert \sum _{j=1}^{p}\bigl\vert I_{j} \bigl( x ( t_{j} ) \bigr) -I_{j} \bigl( y ( t_{j} ) \bigr) \bigr\vert \\& \qquad {} +2\sum_{j=k+1}^{p}\bigl\vert J_{j} \bigl( x ( t_{j} ) \bigr) -J_{j} \bigl( y ( t_{j} ) \bigr) \bigr\vert +\sum_{j=k+1}^{p} \bigl\vert I_{j} \bigl( x ( t_{j} ) \bigr) -I_{j} \bigl( y ( t_{j} ) \bigr) \bigr\vert . \end{aligned}$$

Thus

$$\begin{aligned}& \bigl\vert (\mathcal{T}x) (t)-(\mathcal{T}y) (t)\bigr\vert \\& \quad \leq \biggl[ L_{1} \biggl( \frac{1}{\Gamma ( q+1) }+ \frac{\vert \mu_{2}\vert \vert \omega_{2} ( t) \vert }{\Gamma ( q+1) }+\frac{\vert \mu_{1}\vert \vert \omega_{1} ( t) \vert }{\Gamma ( q) } \biggr) \\& \qquad {} + \bigl( \bigl\vert \omega_{1} ( t) \bigr\vert +1 \bigr) p ( L_{2}+L_{3} ) +\bigl\vert \omega_{2} ( t) \bigr\vert pL_{3} \biggr] \Vert x-y\Vert . \end{aligned}$$
(12)

T is contraction mapping. By condition (11), we have

$$\begin{aligned}& \Vert Tx-Ty\Vert \\& \quad \leq \biggl[ L_{1} \biggl( \frac{1+\vert \mu_{2}\vert \omega_{2}}{\Gamma ( q+1) }+\frac{\vert \mu_{1}\vert \omega_{1}}{\Gamma ( q) } \biggr) \\& \qquad {} + ( \omega_{1}+2) p ( L_{2}+L_{3} ) +\omega_{2}pL_{3} \biggr] \Vert x-y\Vert . \end{aligned}$$

Thus \(\mathcal{T}\) is a contraction mapping. \(\mathcal{T}\) has a fixed point, and that is the solution of the BVP by the Banach fixed point theorem. □

Theorem 8

Assume \(\vert f ( t,u) \vert \leq\rho ( t) \) for \(( t,u) \in J\times R\) where \(\rho\in L^{\frac{1}{\sigma}} ( J\times R ) \) and \(\sigma\in ( 0,q-1) \), moreover, there exist positive constants \(L_{1}\), \(L_{2}\), \(L_{3}\), \(M_{1}\), \(M_{2}\) and M such that

$$\begin{aligned}& \bigl\vert f ( t,x) -f ( t,y) \bigr\vert \leq L_{1}\vert x-y\vert ,\quad t\in [ 0,1 ] , x,y\in R; \\& \bigl\vert I_{k} ( x) -I_{k} ( y) \bigr\vert \leq L_{2}\vert x-y\vert ,\qquad \bigl\vert J_{k} ( x) -J_{k} ( y) \bigr\vert \leq L_{3}\vert x-y\vert , \\& \bigl\vert I_{k} ( x) \bigr\vert \leq M_{1},\qquad \bigl\vert J_{k} ( x) \bigr\vert \leq M_{2}, \end{aligned}$$

with

$$ ( \omega_{1}+2) p ( L_{2}+L_{3} ) + \omega_{2}pL_{3}< 1. $$
(13)

Then our boundary value problem has at least one solution on J.

Proof

Let us choose

$$\begin{aligned} r \geq& \biggl[ \Vert \rho \Vert _{L^{\frac{1}{\sigma}}} \biggl( \frac{ ( 1+\vert \mu_{2}\vert \omega_{2} ) }{\Gamma ( q) ( \frac{q-\sigma}{1-\sigma} ) ^{1-\sigma}}+ \frac{\vert \mu_{1}\vert \omega_{1}}{\Gamma ( q-1) ( \frac{q-\sigma-1}{1-\sigma} ) ^{1-\sigma}} \biggr) \\ &{} + ( \omega_{1}+2) p ( M_{1}+M_{2} ) +\omega_{2}pM_{2} \biggr] , \end{aligned}$$

and \(B_{r}= \{ u\in \mathit{PC}( J,R ) \mid \Vert u\Vert _{\mathit{PC}}\leq r \} \). The operators S and N on \(B_{r}\) are defined as

$$\begin{aligned} ( \mathit{Su}) ( t) =& \int _{0}^{t}\frac {(t-s)^{q-1}}{\Gamma ( q) }f \bigl( s,u ( s) \bigr)\,ds-\mu_{2}\omega_{2} ( t) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }f \bigl( s,u ( s) \bigr)\,ds \\ &{} -\mu_{1}\omega_{1} ( t) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }f \bigl( s,u ( s) \bigr)\,ds \end{aligned}$$

and

$$\begin{aligned} ( Nu) ( t) =&-\omega_{1} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) t_{j}- \omega_{2} ( t) \sum_{j=1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) +\omega_{1} ( t) \sum _{j=1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr) \\ &{} +\sum_{j=k+1}^{p}J_{j} \bigl( u ( t_{j} ) \bigr) ( t_{j}-t) -\sum _{j=k+1}^{p}I_{j} \bigl( u ( t_{j} ) \bigr). \end{aligned}$$

For any \(u,v\in B_{r}\) and \(t\in J\), by using \(\vert f ( t,u) \vert \leq\rho ( t) \) and the Hölder inequality,

$$\begin{aligned}& \begin{aligned} \frac{1}{\Gamma ( q) } \int _{0}^{t}\bigl\vert (t-s)^{q-1}f \bigl( s,u ( s) \bigr) \bigr\vert \,ds & \leq\frac{1}{\Gamma ( q) } \biggl( \int _{0}^{t}(t-s)^{\frac{q-1}{1-\sigma}}\,ds \biggr) ^{1-\sigma} \biggl( \int _{0}^{t} \bigl( \rho ( s) \bigr) ^{\frac{1}{\sigma}}\,ds \biggr) ^{\sigma} \\ & \leq\frac{\Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q) ( \frac{q-\sigma}{1-\sigma} ) ^{1-\sigma}}, \end{aligned} \\& \begin{aligned} \frac{1}{\Gamma ( q) } \int _{0}^{1}\bigl\vert (1-s)^{q-1}f \bigl( s,u ( s) \bigr) \bigr\vert \,ds & \leq\frac{1}{\Gamma ( q) } \biggl( \int _{0}^{1}(1-s)^{\frac{q-1}{1-\sigma}}\,ds \biggr) ^{1-\sigma} \biggl( \int _{0}^{1} \bigl( \rho ( s) \bigr) ^{\frac{1}{\sigma}}\,ds \biggr) ^{\sigma} \\ & \leq\frac{\Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q) ( \frac{q-\sigma}{1-\sigma} ) ^{1-\sigma}}, \end{aligned} \end{aligned}$$

and at last

$$\begin{aligned} \frac{1}{\Gamma ( q-1) } \int _{0}^{1}\bigl\vert (1-s)^{q-2}f \bigl( s,u ( s) \bigr) \bigr\vert \,ds & \leq\frac{1}{\Gamma ( q-1) } \biggl( \int _{0}^{1}(1-s)^{\frac{q-2}{1-\sigma}}\,ds \biggr) ^{1-\sigma} \biggl( \int _{0}^{1} \bigl( \rho ( s) \bigr) ^{\frac{1}{\sigma}}\,ds \biggr) ^{\sigma} \\ & \leq\frac{\Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q-1) ( \frac{q-\sigma-1}{1-\sigma} ) ^{1-\sigma}}. \end{aligned}$$

We get

$$\begin{aligned} \Vert \mathit{Su}+\mathit{Nv}\Vert \leq&\frac{ ( 1+\vert \mu _{2}\vert \omega_{2} ) \Vert \rho \Vert _{L^{\frac {1}{\sigma}}}}{\Gamma ( q) ( \frac{q-\sigma}{1-\sigma } ) ^{1-\sigma}}+\frac{\vert \mu_{1}\vert \omega _{1}\Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q-1) ( \frac{q-\sigma-1}{1-\sigma} ) ^{1-\sigma}} \\ &{} + ( \omega_{1}+2) p ( M_{1}+M_{2} ) +\omega _{2}pM_{2}. \end{aligned}$$

Thus \(\mathit{Su}+\mathit{Nv}\in B_{r}\). By (13), it is obvious that N is a contraction mapping. Moreover, the continuity of f implies S is continuous. And the operator S is uniformly bounded on \(B_{r}\) where

$$\Vert \mathit{Su}\Vert \leq\frac{ ( 1+\vert \mu_{2}\vert \omega_{2} ) \Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q) ( \frac{q-\sigma}{1-\sigma} ) ^{1-\sigma}}+\frac{\vert \mu_{1}\vert \omega_{1}\Vert \rho \Vert _{L^{\frac{1}{\sigma}}}}{\Gamma ( q-1) ( \frac{q-\sigma-1}{1-\sigma} ) ^{1-\sigma}}\leq r. $$

Here the quasi-equicontinuity of the operator S is proved. Let \(\Lambda=J\times B_{r}\), \(f_{\sup}= \sup_{ ( t,u) \in \Lambda} \vert f ( t,u) \vert \). For any \(t_{k}< t_{2}< t_{1}< t_{k+1}\), we have

$$\begin{aligned}& \bigl\vert ( \mathit{Su}) ( t_{2} ) - ( \mathit{Su}) ( t_{1} ) \bigr\vert \\& \quad \leq\frac{f_{\sup}}{\Gamma ( q) }\biggl\vert \int _{0}^{t_{2}} \bigl[ (t_{2}-s)^{q-1}-(t_{1}-s)^{q-1} \bigr]\,ds+ \int _{t_{1}}^{t_{2}}\frac{(t_{1}-s)^{q-1}}{\Gamma ( q) }\,ds\biggr\vert \\& \qquad {} +\biggl\vert \vert \mu_{2}\vert ( t_{2}-t_{1} ) \int _{0}^{1}\frac{(1-s)^{q-1}}{\Gamma ( q) }\,ds\biggr\vert \\& \qquad {} +\biggl\vert \vert \mu_{1}\vert \vert \mu_{2} \vert ( t_{2}-t_{1} ) \int _{0}^{1}\frac{(1-s)^{q-2}}{\Gamma ( q-1) }\,ds\biggr\vert \\& \quad \leq f_{\sup} \biggl[ \frac{ ( t_{2}-t_{1} ) ^{q}+t_{2}^{q}-t_{1}^{q}}{\Gamma ( q+1) }+\vert \mu_{2} \vert \frac{ [ ( t_{2}^{q}-t_{1}^{q} ) ] }{\Gamma ( q+1) } \\& \qquad {} + \vert \mu_{1}\vert \frac{\vert \mu_{2}\vert ( t_{2}^{q}-t_{1}^{q} ) }{\Gamma ( q) } \biggr] . \end{aligned}$$

It tends to zero as \(t_{2}\rightarrow t_{1}\). On the interval \(( t_{k},t_{k+1} ] \), S is quasi-equicontinuous. Also by lemma ( 4), S is compact and is relatively compact on \(B_{r}\). Therefore our BVP has at least one solution on \(J= [ 0,1 ] \). □

2.2 Examples

Example 9

Consider the following boundary value problem of fractional differential equation:

$$\textstyle\begin{cases} D_{0+}^{\frac{3}{2}}u ( t) =\frac{\cos u ( t) }{ ( t+10 ) ^{2} ( 1+u^{6} ( t)) },\\ \Delta u(\frac{1}{3})=\frac{\vert u(\frac{1}{3})\vert }{100+\vert u(\frac{1}{3})\vert },\\ \Delta u^{\prime}(\frac{1}{3})=\frac{\vert u(\frac{1}{3})\vert }{100+\vert u(\frac{1}{3})\vert },\\ u(0)+u\prime(1)=0,\\ u(1)+u\prime(0)=0. \end{cases} $$

Here \(t\in [ 0,1 ] \), let

$$\begin{aligned}& q =\frac{3}{2},\qquad t=\frac{1}{3},\qquad \mu_{1}= \mu_{2}=1, \\& \sigma_{1} =\sigma_{2}=0, \\& L_{1} =L_{2}=L_{3}=0.01, \end{aligned}$$

and since \(0.88<\Gamma ( 1.5) <0.89\) and \(1.33<\Gamma (2.5) <1.34\), we found

$$\omega_{1}\leq3 \qquad \omega_{2}\leq2. $$

Therefore,

$$\begin{aligned}& 0.01 \biggl( \frac{ ( 1+1.99) }{\Gamma ( 2.5) }+\frac{2.99}{\Gamma ( 1.5) } \biggr) + ( 0.01+0.01) ( 2.99+2) +0.01 ( 1.99) < 1, \\& 0.01 ( 2.24+3.38) + ( 0.02) ( 4.99) + ( 0.01) ( 1.99) < 1, \\& 0.0562+0.0998+0.0199 < 1, \\& 0.1759 < 1. \end{aligned}$$

Thus, by Theorem 7, the BVP has a unique solution on \([ 0,1 ] \).