Let \(\mathcal{C}=C([a, T],\mathbb{R})\) denote the Banach space of all continuous functions from \([a, T]\) to \(\mathbb{R}\) endowed with the norm defined by \(\|x\|=\sup_{t\in [a, T]}|x(t)|\). Throughout this paper, for convenience, the expression \({{}_{a}}\mathfrak{I}^{b,\rho }f(s,x(s))(c)\) means
$$ {{}_{a}}\mathfrak{I}^{b,\rho } f\bigl(s,x(s)\bigr) (c)= \frac{1}{\varGamma (b)} \int _{a}^{c} \biggl(\frac{(c-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr)^{b-1}\frac{f(s,x(s))\,ds}{(s-a)^{1- \rho }},\quad t\in [a,T], $$
where \(b\in \{\alpha ,\alpha +\beta \}\) and \(c\in \{t,T,\xi ,\sigma \}\).
In view of Lemma 2.4, we define an operator \(\mathcal{F}: \mathcal{C}\rightarrow \mathcal{C}\) by
$$\begin{aligned} (\mathcal{F}x) (t) =& {{}_{a}}\mathfrak{I}^{\alpha ,\rho } f \bigl(s,x(s)\bigr) (t)+\lambda _{1}(t)\, {{}_{a}} \mathfrak{I}^{\alpha ,\rho }f\bigl(s,x(s)\bigr) (\xi ) \\ &{} +\lambda _{2}(t)\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }f \bigl(s,x(s)\bigr) (T)-\lambda _{3}(t)\, {{}_{a}} \mathfrak{I}^{\alpha +\beta ,\rho }f\bigl(s,x(s)\bigr) (\sigma )+ \lambda _{4}(t). \end{aligned}$$
(3.1)
It should be noticed that problem (1.1) has solutions if and only if the operator \(\mathcal{F}\) has fixed points. In the following subsections we prove existence, as well as existence and uniqueness results, for the boundary value problem (1.1) by using a variety of fixed point theorems. In addition, we set
$$\begin{aligned}& \bigl\vert \lambda _{1}(t) \bigr\vert \le M_{1}:=\frac{|\mu _{1}|}{|\mathcal{J}|} \bigl\{ | \eta _{2}|+(T-a)^{\rho }| \eta _{1}| \bigr\} , \end{aligned}$$
(3.2)
$$\begin{aligned}& \bigl\vert \lambda _{2}(t) \bigr\vert \le M_{2}:= \frac{1}{|\mathcal{J}|} \bigl\{ (T-a)^{ \rho }\bigl(1+ \vert \mu _{1} \vert \bigr)+|\mu _{1}|(\xi -a)^{\rho } \bigr\} , \end{aligned}$$
(3.3)
$$\begin{aligned}& \bigl\vert \lambda _{3}(t) \bigr\vert \le M_{3}:= \frac{|\lambda |}{|\mathcal{J}|} \bigl\{ | \mu _{1}|(\xi -a)^{\rho }+(T-a)^{\rho } \bigl(1+ \vert \mu _{1} \vert \bigr) \bigr\} , \end{aligned}$$
(3.4)
$$\begin{aligned}& \bigl\vert \lambda _{4}(t) \bigr\vert \le M_{4}:= \frac{1}{|\mathcal{J}|} \bigl\{ |\mu _{2}|\bigl(| \eta _{2}|+(T-a)^{\rho }| \eta _{1}|\bigr) \bigr\} , \end{aligned}$$
(3.5)
and
$$ \varPhi = \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M_{1}\frac{( \xi -a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M_{2} \frac{(T-a)^{ \rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha +\beta }\varGamma (\alpha + \beta +1)}. $$
(3.6)
Existence and uniqueness result via Banach’s fixed point theorem
The first existence and uniqueness result is based on the Banach contraction mapping principle (Banach’s fixed point theorem).
Theorem 3.1
Assume that
\(f:[a,T]\times \mathbb{R}\to \mathbb{R}\)
is a continuous function such that
- (H1):
-
there exists a constant
\(L>0\)
such that
\(|f(t,x)-f(t,y)| \leq L|x-y|\)
for each
\(t\in [a, T]\)
and
\(x, y\in \mathbb{R}\).
If
$$ L\varPhi < 1, $$
(3.7)
where
Φ
is defined by (3.6), then the boundary value problem (1.1) has a unique solution on
\([a, T]\).
Proof
We transform problem (1.1) into a fixed point problem, \(x=\mathcal{F}x\), where the operator \(\mathcal{F}\) is defined as in (3.1). Observe that the fixed points of the operator \(\mathcal{F}\) are solutions of problem (1.1). Applying the Banach contraction mapping principle, we shall show that \(\mathcal{F}\) has a unique fixed point. Now, we let \(\sup_{t \in [a,T]}|f(t,0)|=M< \infty \) and choose a positive constant r satisfying
$$ r\geq \frac{\varPhi M+M_{4}}{1-L\varPhi }. $$
Next, we show that \(\mathcal{F} B_{r} \subset B_{r}\), where \(B_{r}=\{x \in {\mathcal{{C}}}: \|x\|\le r \}\). For any \(x \in B_{r}\), we have
$$\begin{aligned}& \bigl\vert (\mathcal{F}x) (t) \bigr\vert \\& \quad = \bigl\vert {{}_{a}}\mathfrak{I}^{\alpha ,\rho } f\bigl(s,x(s) \bigr) (t)+\lambda _{1}(t)\, {{}_{a}}\mathfrak{I}^{\alpha ,\rho }f \bigl(s,x(s)\bigr) (\xi )+\lambda _{2}(t)\, {{}_{a}} \mathfrak{I}^{\alpha ,\rho }f\bigl(s,x(s)\bigr) (T) \\& \qquad {} -\lambda _{3}(t)\,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho }f \bigl(s,x(s)\bigr) ( \sigma )+\lambda _{4}(t) \bigr\vert \\& \quad \le {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (t)+ \bigl\vert \lambda _{1}(t) \bigr\vert \, {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (\xi )+ \bigl\vert \lambda _{2}(t) \bigr\vert \, {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (T) \\& \qquad {} + \bigl\vert \lambda _{3}(t) \bigr\vert \,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho } \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert ( \sigma )+ \bigl\vert \lambda _{4}(t) \bigr\vert \\& \quad \le {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl( \bigl\vert f \bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) (T) \\& \qquad {} +M_{1}\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl( \bigl\vert f\bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) (\xi ) \\& \qquad {} +M_{2}\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl( \bigl\vert f\bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) (T) \\& \qquad {} +M_{3}\,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho } \bigl( \bigl\vert f\bigl(s,x(s)\bigr)-f(s,0) \bigr\vert + \bigl\vert f(s,0) \bigr\vert \bigr) (\sigma )+M_{4} \\& \quad \le (Lr+M)\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }(1) (T)+(Lr+M)M_{1}\,{{}_{a}} \mathfrak{I}^{\alpha ,\rho }(1) ( \xi )+(Lr+M)M_{2}\,{{}_{a}}\mathfrak{I} ^{\alpha ,\rho }(1) (T) \\& \qquad {} +(Lr+M)M_{3}\,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho }(1) (\sigma )+M _{4} \\& \quad = (Lr+M) \biggl[\frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M_{1}\frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma ( \alpha +1)}+M_{2} \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma ( \alpha +1)} \\& \qquad {} +M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha + \beta }\varGamma (\alpha +\beta +1)} \biggr]+M_{4} \\& \quad = (Lr+M)\varPhi +M_{4}\leq r, \end{aligned}$$
which implies that \(\|{\mathcal{F}}x\|\le r\) and therefore \(\mathcal{F}B_{r}\subset B_{r}\).
Next, we let \(x, y\in \mathcal{C}\). Then, for \(t\in [a,T]\), we have
$$\begin{aligned} \bigl\vert (\mathcal{F}x) (t)-(\mathcal{F}y) (t) \bigr\vert \le & {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,x(s)\bigr)-f \bigl(s,y(s)\bigr) \bigr\vert (t) \\ &{}+ \bigl\vert \lambda _{1}(t) \bigr\vert \,{{}_{a}} \mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert ( \xi ) \\ &{}+ \bigl\vert \lambda _{2}(t) \bigr\vert \,{{}_{a}} \mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert (T) \\ &{}+ \bigl\vert \lambda _{3}(t) \bigr\vert \,{{}_{a}} \mathfrak{I}^{\alpha +\beta ,\rho } \bigl\vert f\bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr) \bigr\vert ( \sigma ) \\ \le & L\|x-y\|\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }(1) (T)+L\|x-y \|M_{1}\,{{}_{a}} \mathfrak{I}^{\alpha ,\rho }(1) (\xi ) \\ &{}+L\|x-y\|M_{2}\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }(1) (T)+L\|x-y\|M _{3}\,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho }(1) (\sigma ) \\ = & L \biggl(\frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M _{1}\frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M _{2}\frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)} \\ &{} +M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha + \beta }\varGamma (\alpha +\beta +1)} \biggr)\|x-y\| \\ = & L\varPhi \|x-y\|, \end{aligned}$$
which implies that \(\|\mathcal{F}x-\mathcal{F}y\|\leq L\varPhi \|x-y\|\). As \(L\varPhi <1\), \(\mathcal{F}\) is a contraction operator. Therefore, we deduce, by the Banach contraction mapping principle, that \(\mathcal{F}\) has a fixed point which is the unique solution of problem (1.1) on \([a,T]\). The proof is completed. □
Existence result via Krasnoselskii’s fixed point theorem
The next existence theorem is based on Krasnoselskii’s fixed point theorem.
Lemma 3.1
(Krasnoselskii’s fixed point theorem [46])
Let
M
be a closed, bounded, convex, and nonempty subset of a Banach space
X. Let
A, B
be the operators such that (a) \(Ax+By \in M\)
whenever
\(x, y \in M\); (b) A
is compact and continuous; (c) B
is a contraction mapping. Then there exists
\(z \in M\)
such that
\(z=Az+Bz\).
Theorem 3.2
Let
\(f : [a,T]\times {\mathbb{R}} \to \mathbb{R}\)
be a continuous function satisfying (H1). In addition we assume that
- (H2):
-
\(|f(t,x)|\le \delta (t)\), \(\forall (t,x) \in [a,T] \times {\mathbb{R}}\), and
\(\delta \in C([a,T], {\mathbb{R}}^{+})\).
Then the boundary value problem (1.1) has at least one solution on
\([a,T]\)
provided
$$ M_{1} \frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M _{2} \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)}+M _{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha +\beta } \varGamma (\alpha +\beta +1)}< 1. $$
(3.8)
Proof
Setting \(\sup_{t\in [a, T]}|\delta (t)|=\|\delta \|\) and choosing
$$ \overline{r}\geq \|\delta \|\varPhi +M_{4}, $$
(3.9)
where Φ is defined by (3.6), we consider \(B_{\overline{r}}=\{x\in \mathcal{C}:\|x\|\leq \overline{r}\}\). Let us define the operators \(\mathcal{F}_{1}\) and \(\mathcal{F}_{2}\) on \(B_{\overline{r}}\) by
$$\begin{aligned}& \mathcal{F}_{1}x(t) = {{}_{a}}\mathfrak{I}^{\alpha ,\rho } f \bigl(s,x(s)\bigr) (t),\quad t\in [a,T], \\& \begin{aligned} \mathcal{F}_{2}x(t) &= \lambda _{1}(t) \,{{}_{a}}\mathfrak{I}^{\alpha , \rho }f\bigl(s,x(s)\bigr) (\xi )+\lambda _{2}(t)\,{{}_{a}}\mathfrak{I}^{\alpha , \rho }f\bigl(s,x(s) \bigr) (T) \\ &\quad {}-\lambda _{3}(t)\,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho }f \bigl(s,x(s)\bigr) ( \sigma )+\lambda _{4}(t),\quad t\in [a,T]. \end{aligned} \end{aligned}$$
For any \(x,y\in B_{\overline{r}}\), we have
$$\begin{aligned} \|\mathcal{F}_{1}x+\mathcal{F}_{2}y\| \le & \sup _{t\in [a,T]} \bigl\{ {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert (t)+ \bigl\vert \lambda _{1}(t) \bigr\vert \,{{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,y(s)\bigr) \bigr\vert (\xi ) \\ &{}+ \bigl\vert \lambda _{2}(t) \bigr\vert \,{{}_{a}} \mathfrak{I}^{\alpha ,\rho } \bigl\vert f\bigl(s,y(s)\bigr) \bigr\vert (T)+ \bigl\vert \lambda _{3}(t) \bigr\vert \,{{}_{a}} \mathfrak{I}^{\alpha +\beta ,\rho } \bigl\vert f\bigl(s,y(s)\bigr) \bigr\vert ( \sigma )+ \bigl\vert \lambda _{4}(t) \bigr\vert \bigr\} \\ \le & \|\delta \| \biggl[M_{1}\frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha } \varGamma (\alpha +1)}+M_{2} \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha } \varGamma (\alpha +1)}+M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{ \rho ^{\alpha +\beta }\varGamma (\alpha +\beta +1)} \biggr]+M_{4} \\ =&\|\delta \|\varPhi +M_{4} \le {\overline{r}}. \end{aligned}$$
This shows that \(\mathcal{F}_{1}x+\mathcal{F}_{2}y\in B_{\overline{r}}\) which satisfies condition (a) of Lemma 3.1. It is easy to see, using (3.8), that \(\mathcal{F}_{2}\) is a contraction mapping and also condition (c) of Lemma 3.1 holds.
To show that condition (b) of Lemma 3.1 is fulfilled, we apply the continuity of a function f, which leads to operator \(\mathcal{F}_{1}\) being continuous. Also, the set \(\mathcal{F}_{1} B _{\overline{r}}\) is uniformly bounded as
$$ \|\mathcal{F}_{1} x\| \le \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha } \varGamma (\alpha +1)}\|\delta \|. $$
Next, we prove the compactness of the operator \(\mathcal{F}_{1}\) by setting \(\sup_{(t,x) \in [a,T] \times B_{\overline{r}}}|f(t,x)|= \overline{f}< \infty \). Then, for \(a\le t_{1}\le t_{2}\le T\), we have
$$\begin{aligned} & \bigl\vert \mathcal{F}_{1}x(t_{2})- \mathcal{F}_{1}x(t_{1}) \bigr\vert \\ &\quad = \bigl\vert {{}_{a}}\mathfrak{I}^{\alpha ,\rho } f \bigl(s,x(s)\bigr) (t_{2})-{{}_{a}} \mathfrak{I}^{\alpha ,\rho } f\bigl(s,x(s)\bigr) (t_{1}) \bigr\vert \\ &\quad \le \biggl\vert \frac{1}{\varGamma (\alpha )} \biggl[ \int _{a}^{t_{1}} \biggl(\frac{(t _{2}-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr)^{\alpha -1}- \int _{a} ^{t_{1}} \biggl(\frac{(t_{1}-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr) ^{\alpha -1} \biggr] \\ &\qquad {}\times\frac{f(s,x(s))\,ds}{(s-a)^{1-\rho }}+\frac{1}{\varGamma (\alpha )} \int _{t_{1}}^{t_{2}} \biggl(\frac{(t _{2}-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr)^{\alpha -1}\frac{f(s,x(s))\,ds}{(s-a)^{1- \rho }} \biggr\vert \\ &\quad \le \frac{\overline{f}}{\rho ^{\alpha }\varGamma (\alpha +1)} \bigl[2 \bigl\vert (t_{2}-a)^{\rho \alpha }-(t_{1}-a)^{\rho \alpha } \bigr\vert +\bigl\vert (t _{2}-a)^{\rho \alpha }-(t_{1}-a)^{\rho \alpha }\bigr\vert \bigr], \end{aligned}$$
which is independent of x and tends to zero as \(t_{2}\to t_{1}\). Thus, the set \(\mathcal{F}_{1}B_{\overline{r}}\) is equicontinuous. So the set \(\mathcal{F}_{1}B_{\overline{r}}\) is relatively compact. Hence, by the Arzelá–Ascoli theorem, the operator \(\mathcal{F}_{1}\) is compact on \(B_{\overline{r}}\). Thus all the assumptions of Lemma 3.1 are satisfied. So, the conclusion of Lemma 3.1 implies that the boundary value problem (1.1) has at least one solution on \([a,T]\). The proof is completed. □
Existence result via Leray–Schauder’s nonlinear alternative
By using Leray–Schauder’s nonlinear alternative, we give in this subsection our last existence theorem.
Lemma 3.2
(Nonlinear alternative for single-valued maps [47])
Let
E
be a Banach space, C
be a closed, convex subset of
E, X
be an open subset of
C, and
\(0\in X\). Suppose that
\(F:\overline{X}\to C\)
is a continuous, compact (that is, \(F( \overline{X})\)
is a relatively compact subset of
C) map. Then either
-
(i)
F
has a fixed point in
X̅, or
-
(ii)
there is
\(x\in \partial X\) (the boundary of
X
in
C) and
\(\lambda \in (0,1)\)
with
\(x=\lambda F(x)\).
Theorem 3.3
Assume that:
- (H3):
-
there exist a continuous nondecreasing function
\(\psi :[0,\infty )\to (0,\infty )\)
and a function
\(p\in C([a,T], \mathbb{R}^{+})\)
such that
$$ \bigl\vert f(t,x) \bigr\vert \le p(t)\psi \bigl( \Vert x \Vert \bigr) \quad \textit{for each } (t,x) \in [a,T]\times \mathbb{R}; $$
- (H4):
-
there exists a constant
\(N>0\)
such that
$$ \frac{N}{\|p\|\psi (N)\varPhi +M_{4}}> 1, $$
where
Φ
is defined by (3.6).
Then the boundary value problem (1.1) has at least one solution on
\([a,T]\).
Proof
Let the operator \(\mathcal{F}\) be defined by (3.1). Firstly, we shall show that \(\mathcal{F}\)
maps bounded sets (balls) into bounded sets in
\(\mathcal{C}\). For a number \(R>0\), let \(B_{R} = \{x \in \mathcal{C}: \|x\| \le R\}\) be a bounded ball in \(\mathcal{C}\). Then, for \(t\in [a,T]\), we have
$$\begin{aligned}& \bigl\vert (\mathcal{F}x) (t) \bigr\vert \\& \quad \le {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (t)+ \bigl\vert \lambda _{1}(t) \bigr\vert \, {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (\xi )+ \bigl\vert \lambda _{2}(t) \bigr\vert \, {{}_{a}}\mathfrak{I}^{\alpha ,\rho } \bigl\vert f \bigl(s,x(s)\bigr) \bigr\vert (T) \\& \qquad {} + \bigl\vert \lambda _{3}(t) \bigr\vert \,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho } \bigl\vert f\bigl(s,x(s)\bigr) \bigr\vert ( \sigma )+ \bigl\vert \lambda _{4}(t) \bigr\vert \\& \quad \le \Vert p \Vert \psi \bigl( \Vert x \Vert \bigr) \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma ( \alpha +1)}+ \Vert p \Vert \psi \bigl( \Vert x \Vert \bigr)M_{1}\frac{(\xi -a)^{\rho \alpha }}{ \rho ^{\alpha }\varGamma (\alpha +1)} \\& \qquad {}+ \Vert p \Vert \psi \bigl( \Vert x \Vert \bigr)M_{2}\frac{(T-a)^{ \rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)} + \Vert p \Vert \psi \bigl( \Vert x \Vert \bigr)M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{ \rho ^{\alpha +\beta }\varGamma (\alpha +\beta +1)}+M_{4}, \end{aligned}$$
which leads to
$$\begin{aligned} \|\mathcal{F}x\| \le & \|p\|\psi (R) \biggl\{ \frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha } \varGamma (\alpha +1)}+ M_{1}\frac{(\xi -a)^{\rho \alpha }}{\rho ^{ \alpha }\varGamma (\alpha +1)}+ M_{2}\frac{(T-a)^{\rho \alpha }}{ \rho ^{\alpha }\varGamma (\alpha +1)} \\ &{}+ M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha + \beta }\varGamma (\alpha +\beta +1)} \biggr\} +M_{4} \\ :=& K. \end{aligned}$$
Secondly, we show that
\(\mathcal{F}\)
maps bounded sets into equicontinuous sets of
\(\mathcal{C}\). Let \(v_{1}, v_{2} \in [a,T]\) with \(v_{1}< v_{2}\) and \(x \in B_{R}\). Then we have
$$\begin{aligned}& \bigl\vert (\mathcal{F}x) (v_{2})-(\mathcal{F}x) (v_{1}) \bigr\vert \\& \quad = \bigl\vert {{}_{a}}\mathfrak{I}^{\alpha ,\rho } f\bigl(s,x(s) \bigr) (v_{2})+\lambda _{1}(v _{2}) \,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }f\bigl(s,x(s)\bigr) (\xi )+\lambda _{2}(v _{2})\,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }f \bigl(s,x(s)\bigr) (T) \\& \qquad {}-\lambda _{3}(v_{2})\,{{}_{a}} \mathfrak{I}^{\alpha +\beta ,\rho }f\bigl(s,x(s)\bigr) ( \sigma )+\lambda _{4}(v_{2}) -{{}_{a}}\mathfrak{I}^{\alpha ,\rho } f\bigl(s,x(s)\bigr) (v _{1}) \\& \qquad {}-\lambda _{1}(v_{1}) \,{{}_{a}}\mathfrak{I}^{\alpha ,\rho }f\bigl(s,x(s)\bigr) ( \xi )-\lambda _{2}(v_{1})\,{{}_{a}} \mathfrak{I}^{\alpha ,\rho }f\bigl(s,x(s)\bigr) (T) \\& \qquad {}+ \lambda _{3}(v_{1}) \,{{}_{a}}\mathfrak{I}^{\alpha +\beta ,\rho }f\bigl(s,x(s)\bigr) ( \sigma )- \lambda _{4}(v_{1}) \bigr\vert \\& \quad \le \frac{\|p\|\psi (R)}{\varGamma (\alpha )} \biggl\vert \int _{a}^{v_{1}} \biggl[ \biggl(\frac{(v_{2}-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr) ^{\alpha -1}- \biggl(\frac{(v_{1}-a)^{\rho }-(s-a)^{\rho }}{\rho } \biggr) ^{\alpha -1} \biggr] \frac{ds}{(s-a)^{1-\rho }} \\& \qquad {}+ \int _{v_{1}}^{v_{2}} \biggl(\frac{(v_{2}-a)^{\rho }-(s-a)^{ \rho }}{\rho } \biggr)^{\alpha -1}\frac{ds}{(s-a)^{1-\rho }} \biggr\vert \\& \qquad {}+ \|p\|\psi (R) \biggl\{ \frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha } \varGamma (\alpha +1)} \bigl\vert \lambda _{1}(v_{2})-\lambda _{1}(v_{1}) \bigr\vert \\& \qquad {}+\frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)} \bigl\vert \lambda _{2}(v_{2})- \lambda _{2}(v_{1}) \bigr\vert +\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha +\beta }\varGamma (\alpha + \beta +1)} \bigl\vert \lambda _{3}(v_{2})-\lambda _{3}(v_{1}) \bigr\vert \biggr\} \\& \qquad {}+ \bigl\vert \lambda _{4}(v_{2})-\lambda _{4}(v_{1}) \bigr\vert \\& \quad \le \frac{\|p\|\psi (R)}{\rho ^{\alpha }\varGamma (\alpha +1)} \bigl[2 \bigl\vert (t_{2}-a)^{\rho \alpha }-(t_{1}-a)^{\rho \alpha } \bigr\vert + \bigl\vert (t _{2}-a)^{\rho \alpha }-(t_{1}-a)^{\rho \alpha } \bigr\vert \bigr] \\& \qquad {}+\|p\|\psi (R) \biggl\{ \frac{(\xi -a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)} \bigl\vert \lambda _{1}(v_{2})-\lambda _{1}(v_{1}) \bigr\vert \\& \qquad {}+\frac{(T-a)^{\rho \alpha }}{\rho ^{\alpha }\varGamma (\alpha +1)} \bigl\vert \lambda _{2}(v_{2})- \lambda _{2}(v_{1}) \bigr\vert +\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha +\beta }\varGamma (\alpha + \beta +1)} \bigl\vert \lambda _{3}(v_{2})-\lambda _{3}(v_{1}) \bigr\vert \biggr\} \\& \qquad {}+ \bigl\vert \lambda _{4}(v_{2})-\lambda _{4}(v_{1}) \bigr\vert . \end{aligned}$$
Obviously, the above inequality tends to zero independently of \(x\in B_{R}\) as \(v_{2} \to v_{1}\). Therefore it follows from the Arzelá–Ascoli theorem that \(\mathcal{F}: \mathcal{C} \to \mathcal{C}\) is completely continuous.
Finally, we show that there exists an open set
\(X\subseteq \mathcal{C}\)
with
\(x\ne \theta \mathcal{F}(x)\)
for
\(\theta \in (0,1)\)
and
\(x\in \partial X\).
Let \(x\in \mathcal{C}\) be a solution of \(x=\theta \mathcal{F}x\) for \(\theta \in [ 0,1]\). Then, for \(t\in [a,T]\), we have
$$\begin{aligned} \bigl\vert x(t) \bigr\vert =& \bigl\vert \nu (\mathcal{F}x) (t) \bigr\vert \\ \le &\|p\|\psi \bigl( \Vert x \Vert \bigr) \biggl\{ \frac{(T-a)^{\rho \alpha }}{\rho ^{ \alpha }\varGamma (\alpha +1)}+ M_{1}\frac{(\xi -a)^{\rho \alpha }}{ \rho ^{\alpha }\varGamma (\alpha +1)}+ M_{2}\frac{(T-a)^{\rho \alpha }}{ \rho ^{\alpha }\varGamma (\alpha +1)} \\ &{}+ M_{3}\frac{(\sigma -a)^{\rho (\alpha +\beta )}}{\rho ^{\alpha + \beta }\varGamma (\alpha +\beta +1)} \biggr\} +M_{4} \\ = & \|p\|\psi \bigl( \Vert x \Vert \bigr)\varPhi +M_{4}, \end{aligned}$$
which, on taking the norm for \(t \in [a,T]\), implies that
$$ \|x\|\le \|p\|\psi \bigl( \Vert x \Vert \bigr)\varPhi +M_{4}. $$
Consequently, we have
$$ \frac{\|x\|}{\|p\|\psi ( \Vert x \Vert )\varPhi +M_{4}}\leq 1. $$
In view of (H4), there exists N such that \(\|x\|\ne N\). Let us set
$$ X=\bigl\{ x\in \mathcal{C}: \Vert x \Vert < N\bigr\} \quad \text{and} \quad Y=X\cap B_{R}. $$
Note that the operator \(\mathcal{F}:\overline{Y}\rightarrow \mathcal{C}\) is continuous and completely continuous. From the choice of Y, there is no \(x\in \partial Y\) such that \(x=\theta \mathcal{F}x\) for some \(\theta \in (0,1)\). Consequently, by the nonlinear alternative of Leray–Schauder type (Lemma 3.2), we deduce that \(\mathcal{F}\) has a fixed point \(x\in \overline{Y}\) which is a solution of the boundary value problem (1.1). This completes the proof. □