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
-
(i)
\(Ax+By\in M\)
whenever
\(x,y\in M\);
-
(ii)
A
is compact and continuous;
-
(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.
-
(A1)
The function
\(f: [ 0,1 ] \times R\rightarrow R\)
is jointly continuous.
-
(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 \([ 0,1 ] \).
