In this section, we present some necessary definitions from the fractional calculus theory which can be found in the literature [1–5].
Definition 2.1
[1]
The Riemann-Liouville fractional integral of order \(\alpha>0\) of a function \(h:(a,+\infty)\mapsto R\) is given by
$$ I_{a^{+}}^{\alpha}h(t)=\frac{1}{\Gamma(\alpha)} \int_{a}^{t}(t-s)^{\alpha-1}h(s)\,ds, $$
(7)
provided that the right-hand side exists.
Definition 2.2
[1]
The Riemann-Liouville fractional derivative of order \(\alpha>0\) of a function \(h:(a,+\infty)\mapsto R\) is given by
$$ D_{a^{+}}^{\alpha}h(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}} \int_{a}^{t}\frac{h(s)}{(t-s)^{\alpha-n+1}}\,ds, $$
(8)
where \(n-1\le\alpha< n\), provided that the right-hand side exists.
Lemma 2.1
(Schauder fixed point theorem)
Let Ω be a closed convex subset of the Banach space
X. Suppose
\(T:\Omega\mapsto \Omega\)
and
T
is compact (i.e., bounded sets in Ω are mapped into relatively compact sets). Then
T
has a fixed point in Ω.
Remark 2.1
Let \(a< b\). From (2.106) in [5], we have \(D_{a^{+}}^{\gamma}I_{a^{+}}^{\gamma}h(t)=h(t)\), \(t\in[a,b]\), \(\gamma>0\) if f satisfies some suitable assumptions. For example we know \((\int _{a}^{t}h(s)\,ds )'=h(t)\) for all continuous function \(h:[a,b]\to \mathbb {R}\). From (2.108), in [5], for a function \(h:[a,b]\mapsto \mathbb {R}\), if \(D_{a^{+}}^{\alpha}h(t)\) (\(D_{a^{+}}^{\delta}h(t)\), \(D_{a^{+}}^{\alpha-1} h(t)\)) is integrable, then we have \(A,B,C,D\in \mathbb {R}\) such that
$$\begin{aligned}& I_{a^{+}}^{\alpha}D_{a^{+}}^{\alpha}h(t)=h(t)+A(t-a)^{\alpha-1}+B(t-a)^{\alpha -2},\quad t\in(a,b], \\& I_{a^{+}}^{\delta}D_{a^{+}}^{\delta}h(t)=h(t)+C(t-a)^{\delta-1},\quad t\in(a,b], \\& I_{a^{+}}^{\alpha-1} D_{a^{+}}^{\alpha-1} h(t)=h(t)+D(t-a)^{\alpha-2},\quad t\in(a,b]. \end{aligned}$$
These results are generalizations of \(\int h'(s)\,ds=h(t)+C\) and \([\int h(t)\,dt ]'=h(t)\) if h is an absolutely continuous function. If h is a piecewise continuous function, what are the results? We give the following lemma.
Lemma 2.2
Let
\(\alpha\in(1,2)\). Suppose that
\(h:(0,1]\mapsto \mathbb {R}\)
satisfies
\(h|_{(t_{i},t_{i+1}]}\), \(D_{0^{+}}^{\alpha-1}h\in C^{0}(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)), \(\lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha}h(t)\), and
\(\lim _{t\to t_{i}^{+}}D_{0^{+}}^{\alpha-1}h(t)\)
exist for all
\(i\in \mathbb {N}_{0}^{m}\). We can prove that there exist constants
\(c_{i,j}\in \mathbb {R}\) (\(i\in N_{0}^{m}\), \(j=1,2\)) such that
$$ I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(t)=h(t)+ \sum _{j=0}^{i} \bigl(c_{1,j}(t-t_{j})^{\alpha-1}+c_{2,j}(t-t_{j})^{\alpha-2} \bigr),\quad t\in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, $$
(9)
and
$$ D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}h(t)=h(t),\quad t \in(0,1]. $$
(10)
Proof
Step 1. We first of all prove (10). Since \(\lim _{t\to0^{+}}t^{2-\alpha}h(t)=A_{0}\) exists, for any \(\epsilon>0\), we have \(A_{0}-\epsilon< t^{2-\alpha}h(t)< A_{0}+\epsilon\) for sufficiently small \(t\in(0,t_{1}]\). Then there exists \(A>0\) such that \(|h(t)|\le At^{\alpha-2}\) for all \(t\in(t_{0},t_{1}]\). We have for \(t\in(t_{0},t_{1}]\)
$$\begin{aligned} \biggl\vert \int_{0}^{t}(t-v)^{\alpha-1}h(v)\,dv\biggr\vert \le& \int_{0}^{t}(t-v)^{\alpha -1}\bigl|h(v)\bigr|\,dv \\ \le& A \int_{0}^{t}(t-v)^{\alpha-1}v^{\alpha-2}\,dv \quad\hbox{by }\frac{v}{t}=w \\ =&At^{2\alpha-2} \int_{0}^{1}(1-w)^{\alpha-1}w^{\alpha-2}\,dw. \end{aligned}$$
Then \(\int_{0}^{t}(t-v)^{\alpha-1}h(v)\,dv\) is convergent for all \(t\in (t_{0},t_{1}]\). So for \(t\in(0,t_{1}]\), we have
$$\begin{aligned} D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}h(t) =& \frac{1}{\Gamma(2-\alpha)\Gamma (\alpha)} \biggl[ \int_{0}^{t}(t-s)^{1-\alpha} \int_{0}^{s}(s-v)^{\alpha -1}h(v)\,dv\,ds \biggr]'' \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)} \biggl[ \int_{0}^{t} \int _{s}^{t}(t-s)^{1-\alpha}(s-v)^{\alpha-1}\,dsh(v)\,dv \biggr]'' \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)} \biggl[ \int_{0}^{t}(t-v) \int _{0}^{1}(1-w)^{1-\alpha}w^{\alpha-1}\,dwh(v)\,dv \biggr]'' \\ =& \biggl[ \int_{0}^{t}(t-v)h(v)\,dv \biggr]''=h(t). \end{aligned}$$
Then \(D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}h(t)=h(t)\), \(t\in(t_{0},t_{1}]\). For \(t\in(t_{i+1},t_{i+2}]\) (\(i\ge0\)), similar to above discussion, use that \(\lim _{t\to t_{j}^{+}}(t-t_{j})^{2-\alpha}h(t)\) exists, we know that \(\int_{t_{j}}^{t_{j+1}}(t-v)^{\alpha-1}h(v)\,dv\) and \(\int _{t_{i+1}}^{t}(t-v)^{\alpha-1}h(v)\,dv\) are convergent. So \(\int _{0}^{t}(t-v)^{\alpha-1}h(v)\,dv\) is convergent. Then similarly to the above discussion \(D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}h(t)=h(t)\). The proof of (10) is complete.
Step 2. We prove (9). Let \(D_{0^{+}}^{\alpha}h(t)=H(t)\). Since \(\lim _{t\to0^{+}}t^{2-\alpha}h(t)=A_{0}\) exists, for any \(\epsilon >0\), we have \(A_{0}-\epsilon< t^{2-\alpha}h(t)< A_{0}+\epsilon\) for sufficiently small \(t\in(0,t_{1}]\). Then for sufficiently small \(t\in (0,t_{1}]\), we have
$$\begin{aligned} (A_{0}-\epsilon)\mathbf{B}(2-\alpha,\alpha-1) \le&(A_{0}- \epsilon) \int _{0}^{t}(t-s)^{1-\alpha}s^{\alpha-2}\,ds< \int_{0}^{t}(t-s)^{1-\alpha}h(s)\,ds \\ < &(A_{0}+\epsilon) \int_{0}^{t}(t-s)^{1-\alpha}s^{\alpha-2}\,ds=(A_{0}+ \epsilon )\mathbf{B}(2-\alpha,\alpha-1). \end{aligned}$$
Then
$$\begin{aligned} (A_{0}-\epsilon)\mathbf{ B}(2-\alpha,\alpha-1) \le& \mathop{\underline{\lim}}_{t\to0^{+}} \int_{0}^{t}(t-s)^{1-\alpha}h(s)\,ds \\ \le& \mathop{\overline{\lim}} _{t\to0^{+}} \int_{0}^{t}(t-s)^{1-\alpha}h(s)\,ds\le (A_{0}+\epsilon)\mathbf{B}(2-\alpha,\alpha-1). \end{aligned}$$
Let \(\epsilon\to0\), we get \(\lim _{t\to0^{+}}\int _{0}^{t}(t-s)^{1-\alpha}h(s)\,ds=A_{0}\mathbf{ B}(2-\alpha,\alpha-1)\). Furthermore, since \(\lim _{t\to0^{+}}D_{0^{+}}^{\alpha-1}h(t)=B\) exists, \(\lim _{t\to0^{+}} (\int_{0}^{t}(t-s)^{1-\alpha}h(s)\,ds )'=\Gamma (2-\alpha)B\) exists by Definition 2.2. So
$$\begin{aligned} I_{0^{+}}^{\alpha}H(t) =&I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(t)\\ =&\frac{1}{\Gamma (2-\alpha)\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1} \biggl( \int _{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)''\,ds \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)} \biggl[ \int_{0}^{t}(t-s)^{\alpha -1}\,d \biggl( \int_{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)' \biggr] \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[ (t-s)^{\alpha-1} \biggl( \int _{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)'\bigg|_{0}^{t} \\ &{}+(\alpha-1) \int_{0}^{t}(t-s)^{\alpha-2} \biggl( \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du \biggr)'\,ds\biggr] \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[-t^{\alpha-1}\lim _{s\to0^{+}} \biggl( \int_{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)' \\ &{}+ \biggl( \int_{0}^{t}(t-s)^{\alpha-1} \biggl( \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du \biggr)'\,ds \biggr)'\biggr] \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[-t^{\alpha-1}\lim _{s\to0^{+}} \biggl( \int_{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)' \\ &{}+ \biggl[ \int_{0}^{t}(t-s)^{\alpha-1}\,d \biggl( \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du \biggr) \biggr]'\biggr] \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[-t^{\alpha-1}\lim _{s\to0^{+}} \biggl( \int_{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)' \\ &{}- \biggl(t^{\alpha-1}\lim _{s\to0^{+}} \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du \biggr)' \\ &{}+(\alpha-1) \biggl( \int_{0}^{t}(t-s)^{\alpha-2} \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du\,ds \biggr)'\biggr] \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[-t^{\alpha-1}\lim _{s\to0^{+}} \biggl( \int_{0}^{s}(s-u)^{1-\alpha}h(u)\,du \biggr)' \\ &{}-(\alpha-1)t^{\alpha-2}\lim _{s\to0^{+}} \int_{0}^{s}(s-u)^{1-\alpha }h(u)\,du \\ &{}+(\alpha-1) \biggl( \int_{0}^{t} \int_{s}^{t}(t-s)^{\alpha-2}(s-u)^{1-\alpha }\,dsh(u)\,du \biggr)'\biggr]\quad\hbox{by }\frac{s-u}{t-u} = w \\ =&\frac{1}{\Gamma(2-\alpha)\Gamma(\alpha)}\biggl[-t^{\alpha-1}\Gamma(2-\alpha )B -( \alpha-1)t^{\alpha-2}A_{0}\mathbf{B}(2-\alpha,\alpha-1) \\ &{}+(\alpha-1) \biggl( \int_{0}^{t} \int_{0}^{1}(1-w)^{\alpha-2}w^{1-\alpha }\,dwh(u)\,du \biggr)'\biggr] \\ =:&c_{10}t^{\alpha-1}+c_{20}t^{\alpha-2}+h(t). \end{aligned}$$
So
$$ I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(t)=I_{0^{+}}^{\alpha}H(t)=h(t)+c_{10}t^{\alpha-1}+c_{20}t^{\alpha-2},\quad t\in(0,t_{1}]. $$
(11)
It follows from (11) that (9) holds for \(i=0\). Now we suppose that (9) holds for \(i=0,1,2,\ldots,n< m\). We will prove that (9) holds for \(i=n+1\). Then by the method of mathematical induction, we know (9) holds for all \(i\in \mathbb {N}_{0}^{m}\).
In fact, suppose that
$$ I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(t)=h(t)+ \sum _{j=0}^{n} \bigl(c_{1,j}(t-t_{j})^{\alpha-1}+c_{2,j}(t-t_{j})^{\alpha-2} \bigr)+\Phi (t),\quad t\in(t_{n+1},t_{n+2}]. $$
(12)
Use the assumption, we get for \(t\in(t_{n+1},t_{n+2}]\)
$$\begin{aligned} H(t) =&D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}H(t)=D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(t)\\ =&\frac{1}{\Gamma(2-\alpha)} \biggl[ \int _{0}^{t}(t-s)^{1-\alpha}I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(s)\,ds \biggr]'' \\ =&\frac{1}{\Gamma(2-\alpha)} \Biggl[\sum _{j=0}^{n} \int _{t_{j}}^{t_{j+1}}(t-s)^{1-\alpha}I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(s)\,ds+ \int_{t_{n+1}}^{t}(t-s)^{1-\alpha}I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}h(s)\,ds \Biggr]'' \\ =&\frac{1}{\Gamma(2-\alpha)}\Biggl[\sum _{j=0}^{n} \int _{t_{j}}^{t_{j+1}}(t-s)^{1-\alpha} \Biggl( h(s)+ \sum _{\nu=0}^{j} \bigl(c_{1,\nu}(s-t_{\nu})^{\alpha-1}+c_{2,\nu }(s-t_{\nu})^{\alpha-2} \bigr) \Biggr)\,ds \\ &{}+ \int_{t_{n+1}}^{t}(t-s)^{1-\alpha} \Biggl(h(s)+\sum _{\nu=0}^{n} \bigl(c_{1,\nu}(s-t_{\nu})^{\alpha-1}+c_{2,\nu}(s-t_{j})^{\alpha-2} \bigr)+\Phi(s) \Biggr)\,ds\Biggr]'' \\ =&D_{t_{n+1}^{+}}^{\alpha}\Phi(t)+D_{0^{+}}^{\alpha}h(t) +\frac{1}{\Gamma(2-\alpha)}\Biggl[\sum _{j=0}^{n}\sum _{\nu =0}^{j}c_{1,\nu} \int_{t_{j}}^{t_{j+1}}(t-s)^{1-\alpha} (s-t_{\nu})^{\alpha-1}\,ds \\ &{}+\sum _{j=0}^{n}\sum _{\nu=0}^{j}c_{2,\nu} \int _{t_{j}}^{t_{j+1}}(t-s)^{1-\alpha} (s-t_{\nu})^{\alpha-2}\,ds \\ &{}+\sum _{\nu=0}^{n}c_{1,\nu} \int_{t_{n+1}}^{t}(t-s)^{1-\alpha}(s-t_{\nu})^{\alpha-1}\,ds+\sum _{\nu=0}^{n} c_{2,\nu} \int_{t_{n+1}}^{t}(t-s)^{1-\alpha}(s-t_{j})^{\alpha-2}\,ds \Biggr]'' \\ =&D_{t_{n+1}^{+}}^{\alpha}\Phi(t)+D_{0^{+}}^{\alpha}h(t) +\frac{1}{\Gamma(2-\alpha)}\Biggl[\sum _{j=0}^{n}\sum _{\nu =0}^{j}c_{1,\nu}(t-t_{\nu}) \int_{\frac{t_{j}-t_{\nu}}{t-t_{\nu}}}^{\frac {t_{j+1}-t_{\nu}}{t-t_{\nu}} }(1-w)^{1-\alpha} w^{\alpha-1}\,dw \\ &{}+\sum _{j=0}^{n}\sum _{\nu=0}^{j}c_{2,\nu} \int_{\frac {t_{j}-t_{\nu}}{t-t_{\nu}}}^{\frac{t_{j+1}-t_{\nu}}{t-t_{\nu}} }(1-w)^{1-\alpha} w^{\alpha-2}\,dw \\ &{}+\sum _{\nu=0}^{n}c_{1,\nu}(t-t_{\nu}) \int_{\frac{t_{n+1}-t_{\nu}}{t-t_{\nu}}}^{1}(1-w)^{1-\alpha}w^{\alpha-1}\,dw \\ &{}+ \sum _{\nu=0}^{n} c_{2,\nu} \int_{\frac{t_{n+1}-t_{\nu}}{t-t_{\nu}}}^{1}(1-w)^{1-\alpha}w^{\alpha-2}\,dw \Biggr]'' \\ =&D_{t_{n+1}^{+}}^{\alpha}\Phi(t)+D_{0^{+}}^{\alpha}h(t) +\frac{1}{\Gamma(2-\alpha)}\Biggl[\sum _{\nu=0}^{n}c_{1,\nu}(t-t_{\nu})\sum _{j=\nu}^{n} \int_{\frac{t_{j}-t_{\nu}}{t-t_{\nu}}}^{\frac {t_{j+1}-t_{\nu}}{t-t_{\nu}} }(1-w)^{1-\alpha} w^{\alpha-1}\,dw \\ &{}+\sum _{\nu=0}^{n}c_{2,\nu}\sum _{j=\nu}^{n} \int_{\frac {t_{j}-t_{\nu}}{t-t_{\nu}}}^{\frac{t_{j+1}-t_{\nu}}{t-t_{\nu}} }(1-w)^{1-\alpha} w^{\alpha-2}\,dw \\ &{}+\sum _{\nu=0}^{n}c_{1,\nu}(t-t_{\nu}) \int_{\frac{t_{n+1}-t_{\nu}}{t-t_{\nu}}}^{1}(1-w)^{1-\alpha}w^{\alpha-1}\,dw \\ &{}+ \sum _{\nu=0}^{n} c_{2,\nu} \int_{\frac{t_{n+1}-t_{\nu}}{t-t_{\nu}}}^{1}(1-w)^{1-\alpha}w^{\alpha -2}\,dw \Biggr]'' \\ =&D_{t_{n+1}^{+}}^{\alpha}\Phi(t)+H(t) +\frac{1}{\Gamma(2-\alpha)}\Biggl[\sum _{\nu=0}^{n}c_{1,\nu}(t-t_{\nu}) \int_{1}^{1}(1-w)^{1-\alpha} w^{\alpha-1}\,dw \\ &{}+\sum _{\nu=0}^{n}c_{2,\nu} \int_{0}^{1}(1-w)^{1-\alpha} w^{\alpha-2}\,dw\Biggr]'' \\ =&D_{t_{n+1}^{+}}^{\alpha}\Phi(t)+H(t). \end{aligned}$$
It follows that \(D_{t_{n+1}^{+}}^{\alpha}\Phi(t)=0\) on \((t_{n+1},t_{n+2}]\). Similarly to the proof of (11), there exist constants \(c_{1,n+1},c_{2,n+1}\in R\) such that \(\Phi (t)=c_{1,n+1}(t-t_{n+1})^{\alpha-1}+c_{2,n+1}(t-t_{n+1})^{\alpha-2}\). Substituting \(\Phi(t)\) into (12), we know that (9) holds for \(i=n+1\). This completes the proof of (9). □
Remark 2.2
Let \(\delta\in(0,1)\). Suppose that \(h:(0,1]\to \mathbb {R}\) satisfies \(h|_{(t_{i},t_{i+1}]}\in C^{0}(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)) and \(\lim _{t\to t_{i}^{+}}(t-t_{i})^{1-\delta}h(t)\) exists for all \(i\in \mathbb {N}_{0}^{m}\). We can prove that there exist constants \(c_{i}\in \mathbb {R}\) (\(i\in \mathbb {N}_{0}^{m}\)) such that
$$\begin{aligned}& I_{0^{+}}^{\delta}D_{0^{+}}^{\delta}h(t)=h(t)+ \sum _{j=0}^{i}c_{j}(t-t_{j})^{\delta-1},\quad t \in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, \\& D_{0^{+}}^{\delta}I_{0^{+}}^{\delta}h(t)=h(t),\quad t \in(0,1]. \end{aligned}$$
Proof
The proof is similar to that of Lemma 2.2 and is omitted. □
Definition 2.3
Let \(b>0\), \(c,d\in R\) with \(c< d\) be fixed. A function \(h:(c,d)\to \mathbb {R}\) is called a b-well integrable function on \((c,d)\) if both \(\int_{0}^{t}(t-s)^{b-1}|h(s)|\,ds\) and \(\int _{0}^{t}(t-s)^{b-1}h(s)\,ds\) are convergent for all \(t\in[c,d]\).
Definition 2.4
Let \(b\in(1,2)\) and \(a\in(0,b-1)\) and \(\mu :(0,1)\to R\). \(h:(0,1)\times \mathbb {R}^{3}\to \mathbb {R}\) is called a \((a,b;\mu )\)-Carathéodory function if
-
(i)
\(t\mapsto\mu(t)h (t,\frac{x_{1}}{(t-t_{i})^{2-b}},\frac {x_{2}}{(t-t_{i})^{2+a-b}},x_{3} )\) is a b-well integrable function on \((0,1]\) for every \((x_{1},x_{2},x_{3})\in \mathbb {R}^{3}\),
-
(ii)
\((x_{1},x_{2},x_{3})\mapsto h (t,\frac {x_{1}}{(t-t_{i})^{2-b}},\frac{x_{2}}{(t-t_{i})^{2+a-b}},x_{3} )\) is continuous on \(\mathbb {R}^{3}\) for each \(t\in(t_{i},t_{i+1})\) (\(i\in \mathbb {N}_{0}^{m}\)),
-
(iii)
for each \(r>0\), there exists \(M_{r}>0\) such that \(|x_{i}|\le r\) (\(i=1,2,3\)) imply that
$$\biggl\vert h \biggl(t,\frac{x_{1}}{(t-t_{i})^{2-b}},\frac {x_{2}}{(t-t_{i})^{2+a-b}},x_{3} \biggr)\biggr\vert \le M_{r},\quad t\in(t_{i},t_{i+1}),i \in \mathbb {N}_{0}^{m}. $$
Definition 2.5
Let \(b\in(1,2)\) and \(a\in(0,b-1)\). \(I:\{ t_{i}:i\in \mathbb {N}_{1}^{m}\}\times \mathbb {R}^{3}\to \mathbb {R}\) is a \((a,b)\)-Carathéodory function if
-
(i)
\((x_{1},x_{2},x_{3})\mapsto I (t_{i},\frac {x_{1}}{(t_{i}-t_{i-1})^{2-b}},\frac{x_{2}}{(t_{i}-t_{i-1})^{2+a-b}},x_{3} )\) is continuous on \(\mathbb {R}^{3}\) for each \(i\in N_{1}^{m}\),
-
(ii)
for each \(r>0\), there exists \(M_{I,r}>0\) such that \(|x_{i}|\le r\) (\(i=1,2,3\)) imply that
$$\biggl\vert I \biggl(t_{i},\frac{x_{1}}{(t_{i}-t_{i-1})^{2-b}},\frac {x_{2}}{(t_{i}-t_{i-1})^{2+a-b}},x_{3} \biggr)\biggr\vert \le M_{I,r},\quad i\in \mathbb {N}_{1}^{m}. $$
Choose
$$\begin{aligned} X_{\delta,\alpha} =& \Bigl\{ u:(0,1]\to R , u, D^{\delta}u, D^{\alpha-1}u \in C^{0}(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m},\\ &\hbox{the following limits exist: } \lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha}u(t), i\in \mathbb {N}_{0}^{m},\\ & \lim _{t\to t_{i}^{+}}(t-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}u(t),i\in \mathbb {N}_{0}^{m}, \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\alpha-1}u(t),i \in \mathbb {N}_{0}^{m} \Bigr\} . \end{aligned}$$
For \(u\in X_{\delta,\alpha}\), define
$$\|u\|=:\|u\|_{\delta,\alpha}=\max \left \{ \textstyle\begin{array}{l}\displaystyle\sup _{t\in(t_{i},t_{i+1}]}(t-t_{i})^{2-\alpha }\bigl|u(t)\bigr|\\ \displaystyle\sup _{t\in(t_{i},t_{i+1}]}(t-t_{i})^{2+\delta-\alpha}\bigl|D_{0^{+}}^{\delta}u(t)\bigr|\\ \displaystyle\sup _{t\in(t_{i},t_{i+1}]}\bigl|D_{0^{+}}^{\alpha-1}u(t)\bigr| \end{array}\displaystyle :i\in \mathbb {N}_{0}^{m} \right \}. $$
Lemma 2.3
\(X_{\delta,\alpha}\)
is a Banach space with the norm defined.
Proof
In fact, it is easy to see that X is a normed linear space with the norm \(\|\cdot\|\). Let \(\{x_{u}\}\) be a Cauchy sequence in \(X_{\delta,\alpha}\). Then \(\|x_{u}-x_{v}\|\to0\), \(u,v \to+\infty\). It follows that
$$\begin{aligned}& \sup _{t\in(t_{i},t_{i+1}]}(t-t_{i})^{2-\alpha}\bigl|x_{u}(t)-x_{v}(t)\bigr| \to 0,\quad v,u\to+\infty,i\in \mathbb {N}_{0}^{m}, \\& \sup _{t\in(t_{i},t_{i+1}]}(t-t_{i})^{2+\delta-\alpha }\bigl|D_{0^{+}}^{\delta}x_{u}(t)-D_{0^{+}}^{\delta}x_{v}(t)\bigr| \to0,\quad v,u\to+\infty,i\in \mathbb {N}_{0}^{m}, \\& \sup _{t\in(t_{i},t_{i+1}]}\bigl|D_{0^{+}}^{\alpha -1}x_{u}(t)-D_{0^{+}}^{\alpha-1}x_{v}(t)\bigr| \to0,\quad v,u\to+\infty,i\in \mathbb {N}_{0}^{m}. \end{aligned}$$
Denote \(x_{u,i}=x_{u}|_{(t_{i},t_{i+1}]}\). Since
$$\lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha}x_{u}(t), \lim _{t\to t_{i}^{+}}(t-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}x(t), \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\alpha-1}x_{u}(t) $$
exist, we know that \(t\mapsto(t-t_{i})^{2-\alpha}x_{u,i}(t)\) are continuous on \([t_{i},t_{i+1}]\). Thus \(t\mapsto(t-t_{i})^{2-\alpha}x_{u,i}(t)\) are Cauchy sequences in \(C[t_{i},t_{i+1}]\). So \((t-t_{i})^{2-\alpha }x_{u,i}(t)\) uniformly converges to some \(x_{0,i}\) in \(C[t_{i},t_{i+1}]\) as \(u\to+\infty\). It follows that
$$\sup _{t\in[t_{i},t_{i+1}]}\bigl|(t-t_{i})^{2-\alpha }{x}_{u,i}(t)-x_{0,i}(t)\bigr| \to0,\quad u\to+\infty,i\in \mathbb {N}_{0}^{m}. $$
That is,
$$\sup _{t\in[t_{i},t_{i+1}]}(t-t_{i})^{2-\alpha }\bigl|{x}_{u,i}(t)-(t-t_{i})^{\alpha-1}x_{0,i}(t)\bigr| \to0,\quad u\to+\infty,i\in \mathbb {N}_{0}^{m}. $$
Let \(x_{0}(t)=(t-t_{i})^{\alpha-2}x_{0,i}(t)\) for \(t\in(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)). It is easy to see that \(x_{0}\in C^{0}(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)) and the limit \(\lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha}x_{0}(t)\) exists for all \(i\in \mathbb {N}_{0}^{m}\).
Similarly there exist \(y_{0,i},z_{0,i}:(0,1]\to \mathbb {R}\) such that
$$\begin{aligned}& \sup _{t\in[t_{i},t_{i+1}]}\bigl|(t-t_{i})^{2+\delta-\alpha }D_{0^{+}}^{\delta}{x}_{u,i}(t)-y_{0,i}(t)\bigr| \to0,\quad u\to+\infty,i\in \mathbb {N}_{0}^{m}, \\& \sup _{t\in[t_{i},t_{i+1}]}\bigl|D_{0^{+}}^{\alpha -1}x_{u,i}(t)-z_{0,i}(t)\bigr| \to0,\quad u\to+\infty,i\in \mathbb {N}_{0}^{m}. \end{aligned}$$
Let \(y_{0}(t)= (t-t_{i})^{\alpha-\delta-2}y_{0,i}(t)\) and \(z_{0}(t)=z_{0,i}(t)\) for \(t\in(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)). Then \(y_{0},z_{0}\in C^{0}(t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)) and the limits \(\lim _{t\to t_{i}^{+}}(t-t_{i})^{2+\delta-\alpha}y_{0}(t)\) and \(\lim _{t\to t_{i}^{+}}z_{0}(t)\) exist for all \(i\in \mathbb {N}_{0}^{m}\).
Furthermore, using Lemma 2.2, for \(t\in(t_{i},t_{i+1}]\) there exists \(c_{uj}\in \mathbb {R}\) such that
$$\begin{aligned}& \Biggl\vert x_{u}(t)+\sum _{j=0}^{i}c_{uj}(t-t_{j})^{\delta -1}-I_{0^{+}}^{\delta}y_{0}(t)\Biggr\vert \\& \quad= \bigl|I_{0^{+}}^{\delta}D_{0^{+}}^{\delta}x_{u}(t)-I_{0^{+}}^{\delta}y_{0}(t)\bigr| \\& \quad\le \int_{0}^{t}\frac{(t-s)^{\delta-1}}{\Gamma(\delta)}\bigl\vert D_{0^{+}}^{\delta}x_{u}(s)- y_{0}(s)\bigr\vert \,ds \\& \quad= \sum _{j=0}^{i-1} \int_{t_{j}}^{t_{j+1}}\frac{(t-s)^{\delta -1}}{\Gamma(\delta)}(s-t_{j})^{\alpha-\delta-2} \bigl\vert (s-t_{j})^{2+\delta -\alpha} D_{0^{+}}^{\delta}x_{u}(s)- (s-t_{j})^{2+\delta-\alpha}y_{0}(s)\bigr\vert \, ds \\& \quad\quad{}+ \int_{t_{i}}^{t}\frac{(t-s)^{\delta-1}}{\Gamma(\delta)}(s-t_{i})^{\alpha -\delta-2} \bigl\vert (s-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}x_{u}(s)- (s-t_{i})^{2+\delta-\alpha}y_{0}(s)\bigr\vert \, ds \\& \quad= \sum _{j=0}^{i-1} \int_{t_{j}}^{t_{j+1}}\frac{(t-s)^{\delta -1}}{\Gamma(\delta)}(s-t_{j})^{\alpha-\delta-2} \bigl\vert (s-t_{j})^{2+\delta -\alpha} D_{0^{+}}^{\delta}x_{u}(s)- y_{0,j}(s)\bigr\vert \,ds \\& \quad\quad{}+ \int_{t_{i}}^{t}\frac{(t-s)^{\delta-1}}{\Gamma(\delta)}(s-t_{i})^{\alpha -\delta-2} \bigl\vert (s-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}x_{u}(s)- y_{0,i}(s)\bigr\vert \, ds \\& \quad\le \Biggl( \sum _{j=0}^{i-1} \int_{t_{j}}^{t_{j+1}}\frac {(t-s)^{\delta-1}}{\Gamma(\delta)}(s-t_{j})^{\alpha-\delta-2}\,ds + \int_{t_{i}}^{t}\frac{(t-s)^{\delta-1}}{\Gamma(\delta)}(s-t_{i})^{\alpha -\delta-2}\,ds \Biggr) \\& \quad\quad{}\times \max \Bigl\{ \sup _{t\in(t_{i},t_{i+1}]}\bigl\vert (t-t_{i})^{2+\delta -\alpha}D_{0^{+}}^{\delta}x_{u}(t)- y_{0,i}(t)\bigr\vert :i\in N_{0} \Bigr\} \\& \quad= \Biggl( \sum _{j=0}^{i-1}(t-t_{j})^{\alpha-2} \int_{0}^{\frac {t_{j+1}-t_{j}}{t-t_{j}}}\frac{(1-w)^{\delta-1}}{\Gamma(\delta)}w^{\alpha -\delta-2}\,dw +(t-t_{i})^{\alpha-2} \int_{0}^{1}\frac{(1-w)^{\delta-1}}{\Gamma(\delta )}w^{\alpha-\delta-2}\,dw \Biggr) \\& \quad\quad{}\times \max \Bigl\{ \sup _{t\in(t_{i},t_{i+1}]}\bigl\vert (t-t_{i})^{2+\delta -\alpha}D_{0^{+}}^{\delta}x_{u}(t)- y_{0,i}(t)\bigr\vert :i\in N_{0} \Bigr\} \\& \quad\le\frac{\mathbf{B}(\delta,\alpha-\delta-1)}{\Gamma(\delta)}\sum _{j=0}^{i}(t-t_{j})^{\alpha-2} \max \Bigl\{ \sup _{t\in(t_{i},t_{i+1}]}\bigl\vert (t-t_{i})^{2+\delta -\alpha}D_{0^{+}}^{\delta}x_{u}(t)- y_{0,i}(t)\bigr\vert :i\in N_{0} \Bigr\} \\& \quad\to0\quad\hbox{as }u\to+\infty. \end{aligned}$$
Then \(\lim _{u\to+\infty} [x_{u}(t)+\sum _{j=0}^{i}c_{uj}(t-t_{j})^{\delta-1} ]=I_{0^{+}}^{\delta}y_{0}(t)\). So \((t-t_{i})^{\alpha-2}x_{0,i}(t)+\sum _{j=0}^{i}c_{0j}(t-t_{j})^{\delta -1}=I_{0^{+}}^{\delta}y_{0}(t)\) (\(i\in \mathbb {N}_{0}^{m}\)). It follows that \(x_{0}(t)+\sum _{j=0}^{i}c_{0j}(t-t_{j})^{\delta-1}=I_{0^{+}}^{\delta}y_{0}(t)\). Thus \(y_{0}(t)=D_{0^{+}}^{\delta}x_{0}(t)\) for \(t\in(t_{i},t_{i+1}]\).
We have similarly for \(t\in(t_{i},t_{i+1}]\)
$$\begin{aligned}& \Biggl\vert x_{u}(t)+\sum _{j=0}^{i}c_{uj}(t-t_{j})^{\alpha -2}-I_{0^{+}}^{\alpha-1}z_{0}(t) \Biggr\vert \\& \quad = \bigl|I_{0^{+}}^{\alpha-1} D_{0^{+}}^{\alpha-1} x_{u}(t)-I_{0^{+}}^{\alpha -1}z_{0}(t)\bigr| \\& \quad\le \int_{0}^{t}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}\bigl\vert D_{0^{+}}^{\alpha-1} x_{u}(s)- z_{0}(s)\bigr\vert \,ds \\& \quad= \sum _{j=0}^{i-1} \int_{t_{j}}^{t_{j+1}}\frac{(t-s)^{\alpha -2}}{\Gamma(\alpha-1)}\bigl\vert D_{0^{+}}^{\alpha-1} x_{u,j}(s)- z_{0,j}(s)\bigr\vert \, ds \\& \quad\quad{}+ \int_{t_{i}}^{t}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}\bigl\vert D_{0^{+}}^{\alpha-1} x_{u,i}(s)- z_{0,i}(s)\bigr\vert \,ds \\& \quad\le \int_{0}^{t}\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}\,ds \max \Bigl\{ \sup _{t\in(t_{i},t_{i+1}]}\bigl\vert D_{0^{+}}^{\alpha-1}x_{u,i}(t)- z_{0,i}(t)\bigr\vert :i\in N_{0} \Bigr\} \\& \quad= \frac{t^{\alpha-1}}{\Gamma(\alpha)} \max \Bigl\{ \sup _{t\in (t_{i},t_{i+1}]}\bigl\vert D_{0^{+}}^{\alpha-1}x_{u,i}(t)- z_{0,i}(t)\bigr\vert :i\in N_{0} \Bigr\} \to0\quad\hbox{ as }u\to+\infty. \end{aligned}$$
Then \(\lim _{u\to+\infty} [x_{u}(t)+\sum _{j=0}^{i}c_{uj}(t-t_{j})^{\alpha-2} ]=I_{0^{+}}^{\alpha-1}z_{0}(t)\). So \((t-t_{i})^{\alpha-2}x_{0,i}(t)+\sum _{j=0}^{i}c_{0j}(t-t_{j})^{\alpha -2}=I_{0^{+}}^{\alpha-1}z_{0}(t)\) (\(i\in \mathbb {N}_{0}^{m}\)). It follows that \(x_{0}(t)+\sum _{j=0}^{i}c_{0j}(t-t_{j})^{\alpha-2}=I_{0^{+}}^{\alpha-1}z_{0}(t)\). Thus \(z_{0}(t)=D_{0^{+}}^{\alpha-1} x_{0}(t)\) for \(t\in(t_{i},t_{i+1}]\).
From the above discussion, we know that \(x_{u}\to x_{0}\) as \(u\to+\infty\) in \(X_{\delta,\alpha}\). It follows that \(X_{\delta,\alpha}\) is a Banach space. The proof is complete. □
Denote \(E=X_{\delta,\alpha}\times X_{\theta,\beta}\). Define \(\|(x,y)\| \max\{\|x\|_{\delta,\alpha},\|y\|_{\theta,\beta}\}\). Then E is a Banach space. For \(y\in X_{\theta,\beta}\) and \(x\in X_{\delta,\alpha}\), denote
$$\begin{aligned}& F_{y}(t)=p(t)f \bigl(t,y(t),D_{0^{+}}^{\theta}y(t),D_{0^{+}}^{\beta-1}y(t) \bigr), \\& G_{x}(t)=q(t)g\bigl(t,x(t),D_{0^{+}}^{\delta}x(t),D_{0^{+}}^{\alpha-1}x(t)\bigr), \\& I_{1y}(t_{i})=I_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta -1}y(t_{i})\bigr), \\& J_{1y}(t_{i})=J_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta -1}y(t_{i})\bigr), \\& I_{2x}(t_{i})=I_{2}\bigl(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha -1}x(t_{i})\bigr), \\& J_{2x}(t_{i})=J_{2}\bigl(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha-1}x(t_{i})\bigr). \end{aligned}$$
Lemma 2.4
Suppose that (a)-(d) hold and
\((x,y)\in E\). Then
\((u,v)\in E\)
is a solution of
$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha}u(t)=p(t)f (t,y(t),D_{0^{+}}^{\theta}y(t),D_{0^{+}}^{\beta-1}y(t) ),\quad t\in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, \\ D_{0^{+}}^{\beta}v(t)=q(t)g (t,x(t),D_{0^{+}}^{\delta}x(t),D_{0^{+}}^{\alpha-1}x(t) ),\quad t\in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, \\ D_{0^{+}}^{\alpha-1}u(0)=0, \quad\quad u(1)=0, \quad\quad D_{0^{+}}^{\beta-1}v(0)=0, \quad\quad v(1)=0,\\ \lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha }u(t)=I_{1}(t_{i},y(t_{i}),D_{0^{+}}^{\delta}y(t_{i}),D_{0^{+}}^{\beta -1}y(t_{i})),\quad i\in \mathbb {N}_{1}^{m},\\ \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\alpha-1}u(t)-D_{0^{+}}^{\alpha -1}u(t_{i})=J_{1}(t_{i},y(t_{i}),D_{0^{+}}^{\delta}y(t_{i}),D_{0^{+}}^{\beta -1}y(t_{i})),\quad i\in \mathbb {N}_{1}^{m},\\ \lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\beta }v(t)=I_{1}(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha -1}x(t_{i})),\quad i\in \mathbb {N}_{1}^{m},\\ \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\beta-1}v(t)-D_{0^{+}}^{\beta -1}v(t_{i})=J_{2}(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha -1}x(t_{i})),\quad i\in \mathbb {N}_{1}^{m}, \end{cases} $$
(13)
if and only if
$$\begin{aligned} u(t) =& \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds \\ &{}- \Biggl[ \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds+ \sum _{j=1}^{m}\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha )}J_{1y}(t_{j})+ \sum _{j=1}^{m} (1-t_{j})^{\alpha-2}I_{1y}(t_{j}) \Biggr]t^{\alpha-2} \\ &{}+\sum _{j=1}^{i}I_{1y}(t_{j}) (t-t_{j})^{\alpha-2}+\sum _{j=1}^{i} \frac{J_{1y}(t_{j}) }{\Gamma(\alpha)}(t-t_{j})^{\alpha-1},\quad t\in(t_{i},t_{i+1}],i \in \mathbb {N}_{0}^{m}, \end{aligned}$$
(14)
and
$$\begin{aligned} v(t) =& \int_{0}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}G_{x}(s)\,ds \\ &{}- \Biggl[ \int_{0}^{1}\frac{(1-s)^{\beta-1}}{\Gamma(\beta)}G_{x}(s)\,ds+ \sum _{j=1}^{m}I_{2x}(t_{j}) (1-t_{j})^{\beta-2} +\sum _{j=1}^{m} \frac{J_{2x}(t_{j})}{\Gamma(\beta)}(1-t_{j})^{\beta -1} \Biggr]t^{\beta-2} \\ &{}+\sum _{j=1}^{i}I_{2x}(t_{j}) (t-t_{j})^{\beta-2}+\sum _{j=1}^{i} \frac{J_{2x}(t_{j}) }{\Gamma(\beta)}(t-t_{j})^{\beta-1},\quad t\in(t_{i},t_{i+1}],i \in \mathbb {N}_{0}^{m}. \end{aligned}$$
(15)
Proof
From \(x\in X\) and \(y\in Y\), we see that there exists a constant \(r>0\) such that \(\|x\|=r<+\infty\). Then there exist constants \(M_{r,f},M_{r,g},M_{1,r,I},M_{2,r,I},M_{1,r,J},M_{2,r,J}\ge0\) such that
$$\begin{aligned} \bigl|F_{y}(t)\bigr| =&\bigl|p(t)\bigr|\bigl\vert f \bigl(t,y(t),D_{0^{+}}^{\theta}y(t),D_{0^{+}}^{\beta -1}y(t) \bigr)\bigr\vert \\ =&\bigl|p(t)\bigr|\biggl\vert f \biggl(t,\frac{(t-t_{i})^{2-\beta}y(t)}{(t-t_{i})^{2-\beta }},\frac{(t-t_{i})^{2+\theta-\beta}D_{0^{+}}^{\theta}y(t)}{(t-t_{i})^{2+\theta -\beta}},D_{0^{+}}^{\beta-1}x(t) \biggr)\biggr\vert \\ \le& M_{r,f}t^{k_{1}}(1-t)^{l_{1}},\quad t\in(t_{i},t_{i+1}),i \in \mathbb {N}_{0}^{m}, \end{aligned}$$
(16)
and similarly
$$\begin{aligned}& \bigl|G_{x}(t)\bigr|\le M_{r,g}t^{k_{2}}(1-t)^{l_{2}},\quad t \in(t_{i},t_{i+1}),i\in \mathbb {N}_{0}^{m}, \\& \bigl|I_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta-1}y(t_{i})\bigr)\bigr|\le M_{1,r,I},\quad i\in \mathbb {N}_{1}^{m}, \\& \bigl|I_{2}\bigl(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha-1}x(t_{i})\bigr)\bigr|\le M_{2,r,I},\quad i\in \mathbb {N}_{1}^{m}, \\& \bigl|J_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta-1}y(t_{i})\bigr)\bigr|\le M_{1,r,J},\quad i\in \mathbb {N}_{1}^{m}, \\& \bigl|J_{2}\bigl(t_{i},x(t_{i}),D_{0^{+}}^{\delta}x(t_{i}),D_{0^{+}}^{\alpha-1}x(t_{i})\bigr)\bigr|\le M_{2,r,J},\quad i\in \mathbb {N}_{1}^{m}. \end{aligned}$$
(17)
Hence
$$\begin{aligned} \biggl\vert \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}F_{y}(s)\,ds \biggr\vert \le& \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl|F_{y}(s)\bigr|\,ds \\ \le& \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}s^{k_{1}}(1-s)^{l_{1}} M_{r,f}\,ds \\ \le &M_{r,f} \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}s^{k_{1}}(t-s)^{l_{1}}\,ds \\ =& M_{r,f}t^{\alpha+l_{1}+k_{1}} \int _{0}^{1}\frac{(1-w)^{\alpha +l_{1}-1}}{\Gamma(\alpha)}w^{k_{1}}\,dw \\ =& M_{r,f}t^{\alpha+l_{1}+k_{1}}\frac{\mathbf{B}(\alpha+l_{1},k_{1}+1)}{\Gamma (\alpha)}< \infty,\quad t\in(0,1]. \end{aligned}$$
This means \(F_{y}\) is an α-integral function. Similarly we get
$$\biggl\vert \int _{0}^{t}F_{y}(s)\,ds\biggr\vert \le M_{r,f}\mathbf{B}(l_{1}+1,k_{1}+1)< \infty,\quad t\in(0,1]. $$
From \(u\in X\) and (9) in Lemma 2.3, we know that there exist constants \(A_{i}\), \(B_{i}\) (\(i\in \mathbb {N}_{0}^{m}\)) such that
$$\begin{aligned}& u(t)= \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds +\sum _{j=0}^{i}\bigl[A_{j}(t-t_{j})^{\alpha-1}+B_{j}(t-t_{j})^{\alpha-2} \bigr], \\ & \quad t\in (t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}. \end{aligned}$$
(18)
So
$$\begin{aligned} D_{0^{+}}^{\delta}u(t) =& \int_{0}^{t}\frac{(t-s)^{\alpha-\delta-1}}{\Gamma(\alpha -\delta)}F_{y}(s)\,ds +\frac{\Gamma(\alpha)}{\Gamma(\alpha-\delta)}\sum _{j=0}^{i}A_{j}(t-t_{j})^{\alpha-\delta-1} \\ &{}+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)}\sum _{j=0}^{i} B_{j}(t-t_{j})^{\alpha-\delta-2},\quad t\in(t_{i},t_{i+1}],i \in \mathbb {N}_{0}^{m}, \end{aligned}$$
(19)
and
$$ D_{0^{+}}^{\alpha-1}u(t)= \int_{0}^{t}F_{y}(s)\,ds +\Gamma(\alpha) \sum _{j=0}^{i}A_{j},\quad t \in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}. $$
(20)
From \(D_{0^{+}}^{\alpha-1}u(0)=0 \) and \(u(1)=0\) imply that \(A_{0}=0\) and
$$ \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds +\sum _{j=0}^{m}\bigl[A_{j}(1-t_{j})^{\alpha-1}+B_{j}(1-t_{j})^{\alpha-2} \bigr]=0. $$
(21)
Now we get by using the impulse conditions
$$\begin{aligned}& B_{i}=I_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta-1}y(t_{i})\bigr),\quad i \in \mathbb {N}_{1}^{m}, \\& \Gamma(\alpha)A_{i}=J_{1}\bigl(t_{i},y(t_{i}),D_{0^{+}}^{\theta}y(t_{i}),D_{0^{+}}^{\beta -1}y(t_{i})\bigr),\quad i \in \mathbb {N}_{1}^{m}. \end{aligned}$$
Together with (21), we obtain
$$B_{0}=- \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds- \sum _{j=1}^{m} \biggl[\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)}J_{1y}(t_{j}) +(1-t_{j})^{\alpha-2}I_{1y}(t_{j}) \biggr]. $$
Substitute \(A_{i}\), \(B_{i}\) into (19), we get (14). Similarly we get (15).
Now we suppose that u satisfies (14) and v satisfies (15). We will prove that \(u\in X\) and \(v\in Y\), u, v is a solution of BVP (13).
It is easy to see that \(u\in X\), \(v\in Y\). Furthermore, by direct computation, we get
$$\begin{aligned}& D_{0^{+}}^{\alpha}u(t)=F_{y}(t),\quad t \in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, \\& D_{0^{+}}^{\alpha-1}u(0)=0,\quad\quad u(1)=0, \\& D_{0^{+}}^{\beta}v(t)=G_{x}(t),\quad t \in(t_{i},t_{i+1}],i\in \mathbb {N}_{0}^{m}, \\& D_{0^{+}}^{\beta-1}v(0)=0,\quad\quad v(1)=0, \\& \lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\alpha}u(t)=I_{1y}(t_{i}),\quad i \in \mathbb {N}_{1}^{m}, \\& \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\alpha-1}u(t)-D_{0^{+}}^{\alpha -1}u(t_{i})=J_{1y}(t_{i}),\quad i \in \mathbb {N}_{1}^{m}, \\& \lim _{t\to t_{i}^{+}}(t-t_{i})^{2-\beta}v(t)=I_{2y}(t_{i}),\quad i \in \mathbb {N}_{1}^{m}, \\& \lim _{t\to t_{i}^{+}}D_{0^{+}}^{\beta-1}v(t)-D_{0^{+}}^{\beta -1}v(t_{i})=J_{2y}(t_{i}),\quad i \in \mathbb {N}_{1}^{m}. \end{aligned}$$
Then \((u,v)\) is a solution of BVP (13). The proof is completed. □
For \((x,y)\in E\), define \(T(x,y)\) by \(T(x,y)(t)=((T_{1}y)(t),(T_{2}x)(t))\) with
$$\begin{aligned} (T_{1}y) (t) =& \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds \\ &{}- \Biggl[ \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}F_{y}(s)\,ds+ \sum _{j=1}^{m}\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha )}J_{1y}(t_{j})+ \sum _{j=1}^{m} (1-t_{j})^{\alpha-2}I_{1y}(t_{j}) \Biggr]t^{\alpha-2} \\ &{}+\sum _{j=1}^{i}I_{1y}(t_{j}) (t-t_{j})^{\alpha-2}+\sum _{j=1}^{i} \frac{J_{1y}(t_{j}) }{\Gamma(\alpha)}(t-t_{j})^{\alpha-1},\quad t\in(t_{i},t_{i+1}],i \in \mathbb {N}_{0}^{m}, \end{aligned}$$
(22)
and
$$\begin{aligned} (T_{2}x) (t) =& \int_{0}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}G_{x}(s)\,ds \\ &{}- \Biggl[ \int_{0}^{1}\frac{(1-s)^{\beta-1}}{\Gamma(\beta)}G_{x}(s)\,ds+ \sum _{j=1}^{m}I_{2x}(t_{j}) (1-t_{j})^{\beta-2} +\sum _{j=1}^{m} \frac{J_{2x}(t_{j})}{\Gamma(\beta)}(1-t_{j})^{\beta -1} \Biggr]t^{\beta-2} \\ &{}+\sum _{j=1}^{i}I_{2x}(t_{j}) (t-t_{j})^{\beta-2}+\sum _{j=1}^{i} \frac{J_{2x}(t_{j}) }{\Gamma(\beta)}(t-t_{j})^{\beta-1},\quad t\in(t_{i},t_{i+1}],i \in \mathbb {N}_{0}^{m}. \end{aligned}$$
(23)
Lemma 2.5
Suppose that (a)-(d) hold. Then
\(T:E\to E\)
is well defined and is completely continuous, \((x,y)\)
is a solution of BVP (6) if and only if
\((x,y)=T(x,y)\).
Proof
By Lemma 2.4, we know that \(T_{1}y\in X\) and \(T_{2}x\in Y\). Then \(T:E\to E\) is well defined. It is easy to show from Lemma 2.4 that \((x,y)\) is a solution of BVP (6) if and only if \((x,y)=T(x,y)\).
Now, we prove that T is completely continuous. It suffices to prove that \(T_{1}:Y\to X\) and \(T_{2}:X\to Y\) are completely continuous. We divide the proof of completely continuous property of \(T_{1}\) into four steps. Similarly we can prove the completely continuous property of \(T_{2}\).
Step 1. Prove that \(T_{1}\) is continuous.
Let \(y_{n}\in Y\) (\(n=0,1,2,\ldots\)) with \(y_{n}\to y_{0}\) as \(n\to+\infty\). We will prove that \(T_{1}y_{n}\to T_{1}y_{0}\) as \(n\to+\infty\). It is easy to show that there exists \(r>0\) such that \(\|y_{n}\|\le r\), \(n=0,1,2,\ldots\) , and \(\|y_{n}-y_{0}\|\to0\) as \(n\to+\infty\). Then there exist constants \(M_{r,f}, M_{1,r,I}, M_{1,r,J}\ge0\) such that
$$\begin{aligned}& \bigl|F_{y_{n}}(t)\bigr|\le M_{r,f}t^{k_{1}}(1-t)^{l_{1}},\quad t \in(t_{i},t_{i+1}), i\in \mathbb {N}_{0}^{m}0,n=0,1,2, \ldots, \\& \bigl|I_{1}\bigl(t_{i},y_{n}(t_{i}),D_{0^{+}}^{\theta}y_{n}(t_{i}),D_{0^{+}}^{\beta -1}y_{n}(t_{i}) \bigr)\bigr|\le M_{1,r,I},\quad i\in \mathbb {N}_{1}^{m}, n=0,1,2,\ldots, \\& \bigl|J_{1}\bigl(t_{i},y_{n}(t_{i}),D_{0^{+}}^{\theta}y_{n}(t_{i}),D_{0^{+}}^{\beta -1}y_{n}(t_{i}) \bigr)\bigr|\le M_{1,r,J},\quad i\in \mathbb {N}_{1}^{m}, n=0,1,2,\ldots. \end{aligned}$$
(24)
Using the definition in (22), one sees for \(t\in(t_{i},t_{i+1}]\) that
$$\begin{aligned}& (t-t_{i})^{2-\alpha}\bigl|(T_{1}y_{n}) (t)-(T_{1}{y_{0}}) (t)\bigr| \\& \quad\le(t-t_{i})^{2-\alpha} \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\bigl|F_{y_{n}}(s)-F_{y_{0}}(s)\bigr|\,ds \\& \quad\quad{}+(t-t_{i})^{2-\alpha}\Biggl[ \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha )}\bigl|F_{y_{n}}(s)-F_{y_{0}}(s)\bigr|\,ds +\sum _{j=1}^{m}(1-t_{j})^{\alpha-2}\bigl|I_{1y_{n}}(t_{j})-I_{1y_{0}}(t_{j})\bigr| \\& \quad\quad{}+\sum _{j=1}^{m}\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)} \bigl|J_{1y_{n}}(t_{j})-J_{1y_{0}}(t_{j})\bigr| \Biggr]t^{\alpha-2} +(t-t_{i})^{2-\alpha}\sum _{j=1}^{i}(t-t_{j})^{\alpha -2}\bigl|I_{1y_{n}}(t_{j})-I_{1y_{0}}(t_{j})\bigr| \\& \quad\quad{}+(t-t_{i})^{2-\alpha}\sum _{j=1}^{i} \frac {|J_{1y_{n}}(t_{j})-J_{1y_{0}}(t_{j})|}{\Gamma(\alpha)}(t-t_{j})^{\alpha-1} \\& \quad\le2M_{r,f}(t-t_{i})^{2-\alpha} \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}s^{k_{1}}(1-s)^{l_{1}}\,ds \\& \quad\quad{}+2M_{r,f} \int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}s^{k_{1}}(1-s)^{l_{1}}\,ds +2M_{1,r,I}\sum _{j=1}^{m}(1-t_{j})^{\alpha-2}+2M_{1,r,J} \sum _{j=1}^{m}\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)} \\& \quad\quad{}+2mM_{1,r,I}+2M_{1,r,J}\sum _{j=1}^{m} \frac{t_{j+1}-t_{j}}{\Gamma (\alpha)} \\& \quad\le2M_{r,f}(t-t_{i})^{2-\alpha}t^{\alpha+k_{1}+l_{1}} \int_{0}^{1}\frac {(1-w)^{\alpha+l_{1}-1}}{\Gamma(\alpha)}w^{k_{1}}\,dw \\& \quad\quad{}+2M_{r,f}\frac{\mathbf{B}(\alpha+l_{1},k_{1}+1)}{\Gamma(\alpha)} +2M_{1,r,I}\sum _{j=1}^{m}(1-t_{j})^{\alpha-2}+2M_{1,r,J} \sum _{j=1}^{m}\frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)} \\& \quad\quad{}+2mM_{1,r,I}+2M_{1,r,J}\sum _{j=1}^{m} \frac{t_{j+1}-t_{j}}{\Gamma (\alpha)} \\& \quad\le4M_{r,f}\frac{\mathbf{B}(\alpha+l_{1},k_{1}+1)}{\Gamma(\alpha )}+2M_{1,r,I} \Biggl( \sum _{j=1}^{m}(1-t_{j})^{\alpha-2}+m \Biggr) \\& \quad\quad{}+2M_{1,r,J}\sum _{j=1}^{m} \biggl( \frac{(1-t_{j})^{\alpha-1}}{\Gamma (\alpha)}+ \frac{t_{j+1}-t_{j}}{\Gamma(\alpha)} \biggr). \end{aligned}$$
Similarly we can prove for \(t\in(t_{i},t_{i+1}]\) that
$$\begin{aligned}& (t-t_{i})^{2+\delta-\alpha}\bigl|D_{0^{+}}^{\delta}(T_{1}y_{n}) (t)-D_{0^{+}}^{\delta}(T_{1}y_{0}) (t)\bigr| \\& \quad\le2M_{r,f} \biggl(\frac{\mathbf{B}(\alpha+l_{1}-\delta,k_{1}+1)}{\Gamma(\alpha -\delta)}+ \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \frac{\mathbf{B}(\alpha +l_{1},k_{1}+1)}{\Gamma(\alpha)} \biggr) \\& \quad\quad{}+ 2M_{1,r,J} \Biggl(\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \sum _{j=1}^{m} \frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)}+\frac {\Gamma(\alpha)}{\Gamma(\alpha-\delta)}\sum _{j=1}^{m} \frac{t_{j+1}-t_{j} }{\Gamma(\alpha)} \Biggr) \\& \quad\quad{}+2M_{1,r,I} \Biggl(\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \sum _{j=1}^{m} (1-t_{j})^{\alpha-2}+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta -1)}m \Biggr) \end{aligned}$$
and
$$\begin{aligned}& \bigl|D_{0^{+}}^{\alpha-1}(T_{1}y_{n}) (t)-D_{0^{+}}^{\alpha-1}(T_{1}{y_{0}}) (t)\bigr| \\& \quad\le \int_{0}^{t}\bigl|F_{y_{n}}(s)-F_{y_{0}}(s)\bigr|\,ds +\sum _{j=1}^{i}\bigl|J_{1y_{n}}(t_{j})-J_{1y_{0}}(t_{j})\bigr| \le2M_{r,f}\mathbf{B}(l_{1}+1,k_{1}+1)+2mM_{1,r,J}. \end{aligned}$$
By Lebesgue’s dominated convergence theorem, we can show that
$$\begin{aligned}& \lim _{n\to\infty}\sup _{t\in (t_{i},t_{i+1}]}(t-t_{i})^{2-\alpha}\bigl|(Tx_{n}) (t)-(T{x_{0}}) (t)\bigr|=0, \\ & \lim _{n\to\infty}\sup _{t\in (t_{i},t_{i+1}]}(t-t_{i})^{2+\delta-\alpha}\bigl|D_{0^{+}}^{\delta}(Tx_{n}) (t)-D_{0^{+}}^{\delta}(T{x_{0}}) (t)\bigr|=0, \\ & \lim _{n\to\infty}\sup _{t\in (t_{i},t_{i+1}]}\bigl|D_{0^{+}}^{\alpha-1}(Tx_{n}) (t)-D_{0^{+}}^{\alpha-1}(T{x_{0}}) (t)\bigr|=0. \end{aligned}$$
Hence \(\|Tx_{n}-Tx_{0}\|\to0\) as \(n\to\infty\). Then T is continuous.
Let \(\Omega_{1}\subseteq X\) and \(\Omega_{2}\subseteq Y\) be bounded sets of X and Y, respectively. Then there exists \(r>0\) such that \(\|x\|,\|y\|\le r\), \(x\in\Omega_{1}\), \(y\in\Omega_{2}\). So there exist constants \(M_{r,f} , M_{r,g} , M_{1,r,I} , M_{2,r,I} , M_{1,r,J} , M_{2,r,J}\ge0\) such that (16) and (17) hold for all \(x\in\Omega _{1}\), \(y\in\Omega_{2}\).
Step 2. Prove that \(\{T_{1}y:y\in\Omega\}\) is bounded.
We have similarly to Step 1 for \(t\in(t_{i},t_{i+1}]\)
$$\begin{aligned}& (t-t_{i})^{2-\alpha}\bigl|(Ty) (t)\bigr| \le2M_{r,f} \frac{\mathbf{B}(\alpha+l_{1},k_{1}+1)}{\Gamma(\alpha )}+M_{1,r,I} \Biggl( \sum _{j=1}^{m}(1-t_{j})^{\alpha-2}+m \Biggr) \\& \hphantom{(t-t_{i})^{2-\alpha}\bigl|(Ty) (t)\bigr|\le}{}+M_{1,r,J}\sum _{j=1}^{m} \biggl( \frac{(1-t_{j})^{\alpha-1}}{\Gamma (\alpha)}+ \frac{t_{j+1}-t_{j}}{\Gamma(\alpha)} \biggr), \\& (t-t_{i})^{2+\delta-\alpha}\bigl|D_{0^{+}}^{\delta}(T_{1}y) (t)\bigr| \\& \quad \le M_{r,f} \biggl(\frac{\mathbf{B}(\alpha+l_{1}-\delta,k_{1}+1)}{\Gamma(\alpha -\delta)}+ \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \frac{\mathbf{B}(\alpha +l_{1},k_{1}+1)}{\Gamma(\alpha)} \biggr) \\& \quad\quad{}+ M_{1,r,J} \Biggl(\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \sum _{j=1}^{m} \frac{(1-t_{j})^{\alpha-1}}{\Gamma(\alpha)}+\frac {\Gamma(\alpha)}{\Gamma(\alpha-\delta)}\sum _{j=1}^{m} \frac{t_{j+1}-t_{j} }{\Gamma(\alpha)} \Biggr) \\& \quad\quad{}+M_{1,r,I} \Biggl(\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta-1)} \sum _{j=1}^{m} (1-t_{j})^{\alpha-2}+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\delta -1)}m \Biggr), \\& \bigl|D_{0^{+}}^{\alpha-1}(T_{1}y) (t)\bigr|\le M_{r,f}\mathbf{B} (l_{1}+1,k_{1}+1)+mM_{1,r,J}. \end{aligned}$$
It follows that there exists a constant \(M>0\) such that \(\|T_{1}y\|\le M_{1}\) for all \(y\in\Omega_{2}\). Similarly we see that there exists a constant \(M_{1}>0\) such that \(\|T_{2}x\|\le M_{2}\) for all \(x\in\Omega_{1}\). From the above discussion, we see that \(\{T(x,y):x\in\Omega_{1},y\in\Omega_{2}\}\) is bounded.
Step 3. Prove that
$$\begin{aligned}& \bigl\{ t\to(t-t_{i})^{2-\alpha}(T_{1}y) (t):y\in\Omega \bigr\} , \\& \bigl\{ t\to(t-t)^{2+\delta-\alpha}D_{0^{+}}^{\delta}(T_{1}y) (t):x\in\Omega\bigr\} , \\& \bigl\{ t\to D_{0^{+}}^{\alpha-1}(T_{1}y) (t):y\in\Omega \bigr\} \end{aligned}$$
are equi-continuous on each \((t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)).
Since \(T:X\rightarrow X\), we can take
$$\begin{aligned}& (t-t_{i})^{2-\alpha}(Tx) (t)|_{t=t_{i}}=\lim {t\rightarrow t_{i}^{+}} (t-t_{i})^{2-\alpha}(Tx) (t), \\& (t-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}(Tx) (t)|_{t=t_{i}}=\lim _{t\rightarrow t_{i}^{+}}(t-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}(Tx) (t), \\& D_{0^{+}}^{\alpha-1}(Tx) (t)|_{t=t_{i}}=\lim _{t\rightarrow t_{i}^{+}} D_{0^{+}}^{\alpha-1}(Tx) (t). \end{aligned}$$
Then
$$(t-t_{i})^{2-\alpha}(Tx) (t), (t-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}(Tx) (t), D_{0^{+}}^{\alpha-1}(Tx) (t) $$
are continuous on \([t_{i},t_{i+1}]\) for each \(x\in\Omega\). So
$$t^{2-\alpha}(Tx) (t), t^{2+\delta-\alpha}D_{0^{+}}^{\delta}(Tx) (t), D_{0^{+}}^{\alpha-1}(Tx) (t) $$
are uniformly continuous, for any \(\varepsilon>0\), there exists \(\delta _{0}>0\), when \(s_{1}, s_{2}\in[t_{i},t_{i+1}]\), \(|s_{1}-s_{2}|<\delta_{0}\), and \(x\in \Omega\), we can get
$$\begin{aligned}& \bigl|(s_{1}-t_{i})^{2-\alpha}(Tx) (s_{1})-(s_{2}-t_{i}) ^{2-\alpha}(Tx) (s_{2})\bigr|< \varepsilon, \\& \bigl|(s_{1}-t_{i})^{2+\delta-\alpha}D_{0^{+}}^{\delta}(Tx) (s_{1})-( s_{2}-t_{i})^{2-\alpha}(Tx) (s_{2})\bigr|< \varepsilon, \\& \bigl|D_{0^{+}}^{\alpha-1}(Tx) (s_{1})-D_{0^{+}}^{\alpha-1}(Tx) (s_{2})\bigr|< \varepsilon. \end{aligned}$$
This shows us that
$$\begin{aligned}& \bigl\{ t\to t^{2-\alpha}(Tx) (t):x\in\Omega\bigr\} , \\& \bigl\{ t\to t^{2+\delta-\alpha }D_{0^{+}}^{\delta}(Tx) (t):x\in\Omega\bigr\} ,\\& \bigl\{ t\to D_{0^{+}}^{\alpha-1}(Tx) (t):x\in\Omega\bigr\} \end{aligned}$$
are equi-continuous on any closed subinterval of \([t_{i},t_{i+1}]\) (\(i\in \mathbb {N}_{0}^{m}\)).
It follows from Steps 1-3 that T is completely continuous. The proof is ended. □