1 Introduction

Fractional calculus has emerged as an effective and powerful tool in the modeling of many physical and engineering phenomena. One of the reasons for the popularity of this subject is that fractional derivatives naturally describe memory and hereditary properties of several materials and processes. Specific applications of fractional calculus can be found in a variety of disciplines such as physics, chemistry, statistics, economics, biology, biophysics, blood flow phenomena, control theory, signal and image processing, etc. [9, 22, 28, 30]. For some recent development in the theory, methods, and applications of fractional calculus, we refer the reader to [2, 4, 5, 7, 8, 10, 18, 19, 23, 32] and the references therein.

The concept of nonlocal Cauchy problems introduced by Byszewski [12] is found to be more practical than the classical Cauchy problems with the initial conditions [13, 14]. In the last few decades, several kinds of nonlocal problems have been studied. More recently, the topic of nonlocal-integral boundary conditions has attracted a considerable attention. In most of the work dealing with nonlocal boundary value problems, the contribution expressed in terms of the integral is related to the value of the unknown function at a fixed point (left/right end-point of the interval under consideration), for instance, see [1, 3, 25, 34] and references therein.

In the present study, we introduce a more general variant of nonlocal-integral boundary conditions, which relates the integral contribution due to a strip of arbitrary length with the value of the unknown function at an arbitrary (nonlocal) point of the interval instead of its value at a fixed point. Precisely, we formulate this variant of nonlocal-integral boundary conditions as follows: \(x(\xi ) \propto \int _{\eta }^{1}x(s)\mathrm{d}s, ~0<\xi < \eta <1\) or \(x(\xi ) =a \int _{\eta }^{1}x(s)\mathrm{d}s\)   (a is constant of proportionality). Thus, we consider the following nonlocal boundary value problem of fractional differential equations with new integral boundary conditions:

$$\begin{aligned} \left\{ \begin{array}{ll} ^cD^{q}x(t)=f(t,x(t)), \quad t\in [0,1], \quad 1<q\le 2,\\ \displaystyle x(0) =x_0+g(x),\quad x(\xi ) =a \int _{\eta }^{1}x(s)\mathrm{d}s, \quad 0<\xi <\eta <1, \quad x_0 \in \mathbb {R},\\ \end{array} \right. \end{aligned}$$
(1.1)

where \(^cD^{q}\) denotes the Caputo fractional derivative of order q, \(f : [0,1] \times \mathbb {R} \rightarrow \mathbb {R}\) and \(g : C([0, 1], \mathbb {R}) \rightarrow \mathbb {R}\) are given continuous functions, and a is real constant.

As a second problem, we study the multivalued analog of problem (1.1) given by

$$\begin{aligned} \left\{ \begin{array}{ll} ^cD^{q}x(t)\in F(t,x(t)), \quad t\in [0,1], \quad 1<q\le 2,\\ \displaystyle x(0) =x_0+g(x),\quad x(\xi ) =a \int _{\eta }^{1}x(s)\mathrm{d}s, \quad 0<\xi <\eta <1, \quad x_0 \in \mathbb {R}, \\ \end{array} \right. \end{aligned}$$
(1.2)

where \(F{:}\, [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) is a multivalued map, \({\mathcal {P}}(\mathbb {R})\) is the family of all nonempty subsets of \(\mathbb {R},\) and g(x) and a are the same as defined in the problem (1.1). Moreover, g (x) considered in the problems (1.1) and (1.2) may be understood as \(g(x)=\sum _{j=1}^p \alpha _j x(t_j)\) where \(\alpha _j, j=1,\ldots , p,\) are given constants and \(0<t_1 <\cdots <t_p \le 1.\) For more details we refer to the work by Byszewski [12, 13]. For some real-world problems and engineering applications involving the strip conditions similar to the ones considered in the present study , we refer the reader to the works [6, 15, 29, 31, 33].

The paper is organized as follows. In Sect. 2, we recall some basic definitions from fractional calculus and establish a lemma which plays a pivotal role in the sequel. Section 3 deals with the existence results for the problem (1.1) which are shown by applying Banach’s contraction principle and a fixed point theorem due to D. O’Regan. In Sect. 4, we discuss the existence of solutions for the problem (1.2) by means of the nonlinear alternative for contractive maps and the combination of the nonlinear alternative of Leray–Schauder type for single-valued maps and a selection theorem due to Bressan and Colombo for lower semi-continuous multivalued maps with nonempty closed and decomposable values. Some examples are constructed for the illustration of main results.

2 Preliminaries

In this section, some basic definitions on fractional calculus and an auxiliary lemma are presented [22, 28].

Definition 2.1

The Riemann–Liouville fractional integral of order q for a continuous function g is defined as

$$\begin{aligned} I^q g(t)=\frac{1}{\Gamma (q)}\int _0^t \frac{g(s)}{(t-s)^{1-q}}\mathrm{d}s, \quad q>0, \end{aligned}$$

provided the integral exists.

Definition 2.2

For at least n-times continuously differentiable function \(g: [0,\infty ) \rightarrow \mathbb {R},\) the Caputo derivative of fractional order q is defined as

$$\begin{aligned} ^cD^q g(t)=\frac{1}{\Gamma (n-q)}\int _{0}^t(t-s)^{n-q-1}g^{(n)}(s)\mathrm{d}s, \quad n-1 < q < n, n=[q]+1, \end{aligned}$$

where [q] denotes the integer part of the real number q.

Lemma 2.1

For any \(y \in L[0, 1]\) the unique solution of the linear fractional boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} ^cD^{q}x(t)=y(t), \quad 1<q\le 2,\\ \displaystyle x(0) =x_0+g(x), \quad x(\xi ) =a \int _{\eta }^{1}x(s)\mathrm{d}s, \quad t\in [0,1]\\ \end{array} \right. \end{aligned}$$
(2.1)

is

$$\begin{aligned} x(t)= & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}y(s)\mathrm{d}s +\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}y(s)\mathrm{d}s\nonumber \\&-\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}y(s)\mathrm{d}s -\int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}y(s)\mathrm{d}s\Bigg \}\\&+\,\Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)),\nonumber \end{aligned}$$
(2.2)

where

$$\begin{aligned} A=\xi -\frac{a}{2}(1-\eta ^2)\ne 0. \end{aligned}$$
(2.3)

Proof

It is well known that the general solution of the fractional differential equation in (2.1) can be written as

$$\begin{aligned} x(t)= c_0+c_1t+\int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}y(s)\mathrm{d}s, \end{aligned}$$
(2.4)

where \(c_0, c_1 \in \mathbb {R}\) are arbitrary constants.

Applying the given boundary conditions, we find that \(c_0=x_0+g(x)\), and

$$\begin{aligned} c_1= & {} \frac{1}{A}\left( a(1-\eta )-1\right) (x_0+g(x))+\frac{1}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}y(s)\mathrm{d}s\nonumber \\&-\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}y(s)\mathrm{d}s-\int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}y(s)\mathrm{d}s\Bigg \}. \end{aligned}$$
(2.5)

Substituting the values of \(c_0, c_1\) in (2.4), we get (2.2). This completes the proof. \(\square \)

3 Existence results for single-valued problem (1.1)

We denote by \(\mathcal{{C}}=C([0,1], \mathbb {R})\) the Banach space of all continuous functions from \([0,1] \rightarrow \mathbb {R}\) endowed with a topology of uniform convergence with the norm defined by \(\Vert x\Vert =\sup \{|x(t)|: t\in [0,1]\}.\) Also by \(L^1([0,1],\mathbb {R})\) we denote the Banach space of measurable functions \(x : [0,1] \rightarrow \mathbb {R}\) which are Lebesgue integrable and normed by \(\displaystyle \Vert x\Vert _{L^1} = \int _0^1 |x(t)|dt.\)

In view of Lemma 2.1, we define an operator \(\mathcal {P}:\mathcal{{C}} \rightarrow \mathcal{{C}}\) by

$$\begin{aligned} (\mathcal {P}x)(t)= & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s + \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s\nonumber \\&-\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s \quad - \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s\Bigg \}\nonumber \\&+\,\Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)), \quad t\in [0,1]. \end{aligned}$$
(3.1)

Let us define \(\mathcal {P}_{1, 2:}\, \mathcal{{C}} \rightarrow \mathcal{{C}}\) by

$$\begin{aligned} (\mathcal {P}_1 x)(t)= & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s\nonumber \\&-\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s\Bigg \},\qquad \end{aligned}$$
(3.2)

and

$$\begin{aligned} (\mathcal {P}_2 x)(t)=\displaystyle \left[ 1+\frac{t}{A}\left( a(1-\eta )-1\right) \right] (x_0+g(x)). \end{aligned}$$
(3.3)

Clearly

$$\begin{aligned} (\mathcal {P} x)(t)=(\mathcal {P}_1 x)(t)+(\mathcal {P}_2 x)(t), \quad t \in [0,1]. \end{aligned}$$
(3.4)

For convenience we introduce the notations:

$$\begin{aligned} p_0:= \frac{1}{\Gamma (q+1)}\left[ 1+\frac{1}{|A|}\left\{ \frac{|a|(1+\eta ^{q+1})}{q+1}+ \xi ^q \right\} \right] , \end{aligned}$$
(3.5)

and

$$\begin{aligned} k_0:=\left| 1+\frac{1}{A}\left( a(1-\eta )-1\right) \right| . \end{aligned}$$
(3.6)

Theorem 3.1

Let \(f : [0,1]\times {\mathbb R} \rightarrow {\mathbb R}\) be a continuous function. Assume that

\({(A_{1})}\) :

\(|f(t,x)-f(t,y)| \le L \Vert x-y\Vert , \forall t \in [0,1], ~L>0, ~x, y\in {\mathbb R}\);

\({(A_{2})}\) :

\(g : C([0, 1], \mathbb {R}) \rightarrow \mathbb {R}\) is a continuous function satisfying the condition:

$$\begin{aligned} |g(u)-g(v)| \le \ell \Vert u-v\Vert , ~ \ell <k_0^{-1}, \quad \forall ~u,v\in C([0, 1], \mathbb {R}), ~\ell >0; \end{aligned}$$
\({(A_{3})}\) :

\(\displaystyle \gamma :=Lp_0+k_0\ell <1.\)

Then the boundary value problem (1.1) has a unique solution.

Proof

For \(x, y \in \mathcal{{C}}\) and for each \(t \in [0,1],\) from the definition of \(\mathcal {P}\) and assumptions \({(A_{1})}\) and \({(A_{2})},\) we obtain

$$\begin{aligned} |(\mathcal {P}x)(t)-(\mathcal {P}y)(t)|\le & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}\big |f(s,x(s))-f(s,y(s))\big |\mathrm{d}s\\&+\,\frac{1}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}\big |f(s,x(s))-f(s,y(s))\big |\mathrm{d}s\\&\displaystyle +\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}\big |f(s,x(s))-f(s,y(s))\big |\mathrm{d}s\\&+\, \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}|f(s,x(s))-f(s,y(s))|\mathrm{d}s\Bigg \}\\&+\,\Big |1+\frac{1}{A}\big (a(1-\eta )-1\Big )\Big ||g(x)-g(y)|\\\le & {} L\Vert x-y\Vert \Bigg [\int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}\mathrm{d}s+\frac{1}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}\mathrm{d}s\\&+\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}\mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}\mathrm{d}s\Bigg \}\Bigg ]\\&+\,\Big |1+\frac{1}{A}\big (a(1-\eta )-1\Big )\Big |\ell \Vert x-y\Vert \\\le & {} L\Vert x-y\Vert \Bigg [\frac{1}{\Gamma (q+1)}+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1})}{(q+1)\Gamma (q+1)}+\frac{\xi ^q}{\Gamma (q+1)}\Bigg \}\Bigg ]\\&+\,\Big |1+\frac{1}{A}\big (a(1-\eta )-1\Big )\Big |\ell \Vert x-y\Vert \\= & {} \frac{L}{\Gamma (q+1)}\Bigg (\Bigg [1+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1})}{q+1} + \xi ^q \Bigg \}\Bigg ]\\&+\,\Big |1+\frac{1}{A}\Big (a(1-\eta )-1\Big )\Big |\ell \Bigg ) \Vert x-y\Vert \\= & {} ( Lp_0+k_0\ell )\Vert x-y\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \Vert (\mathcal {P}x)-(\mathcal {P}y)\Vert \le \gamma \Vert x-y\Vert . \end{aligned}$$

As \(\gamma < 1\) by \((A_3),\) the operator \(\mathcal {P}\) is a contraction map from the Banach space \({\mathcal C}\) into itself. Hence the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem). \(\square \)

Example 3.1

Consider the following fractional boundary value problem

$$\begin{aligned} \left\{ \begin{array} {ll} \displaystyle ^cD^{3/2}x(t) = \frac{1}{t+16}(\sin x +x)+1, \quad t\in [0,1], \\ \displaystyle x(0)=\frac{1}{2}+\frac{1}{12} \tan ^{-1}(x(1/8)), \quad x\left( \frac{1}{4}\right) =\frac{1}{4}\int _{1/3}^1 x(s)\mathrm{d}s. \end{array} \right. \end{aligned}$$
(3.7)

Here, \(q=3/2, x_0=1/2, ~a=1/4, ~\xi =1/4, ~ \eta =1/3,\) \(f(t,x)=\displaystyle \frac{1}{t+16}(\sin x +x)+1,\) and \(g(x)=\frac{1}{12} \tan ^{-1}(x(1/8)).\) Since \(\displaystyle |f(t,x)-f(t,y)| \le \frac{1}{8} \Vert x-y\Vert , |g(x)-g(y)| \le \frac{1}{12} \Vert x-y\Vert ,\) therefore, \((A_{1})\) and \((A_{2})\) are respectively satisfied with \(L=1/8\) and \(\ell =1/12.\) Using the given values, it is found that \(A=5/36, ~p_0 \approx 2.00565, k_0=5.\) Clearly \( \gamma =Lp_0+k_0\ell \approx 0.667373<1.\) Thus, the conclusion of Theorem 3.1 applies, and the boundary value problem (3.7) has a solution on [0, 1].

Our next existence result relies on a fixed point theorem due to O’Regan in [26].

Lemma 3.1

Denote by U an open set in a closed, convex set C of a Banach space E. Assume \(0\in U.\) Also assume that \(F({\bar{U}})\) is bounded and that \(F: {\bar{U}}\rightarrow C\) is given by \(F = F_1 + F_2,\) in which \(F_1: {\bar{U}}\rightarrow E\) is continuous and completely continuous and \(F_2: {\bar{U}}\rightarrow E\) is a nonlinear contraction (i.e., there exists a nonnegative nondecreasing function \(\phi : [0,\infty )\rightarrow [0,\infty )\) satisfying \(\phi (z)<z\) for \(z > 0,\) such that \(\Vert F_2(x)-F_2(y) \Vert \le \phi (\Vert x-y\Vert )\) for all \(x, y \in {\bar{U}}).\) Then, either

(C1):

F has a fixed point \(u\in {\bar{U}}\) or

(C2):

there exist a point \(u\in \partial U\) and \(\lambda \in (0,1)\) with \(u = \lambda F(u),\) where \({\bar{U}}\) and \( \partial U,\) respectively, represent the closure and boundary of U.

Let

$$\begin{aligned} \Omega _r=\left\{ x\in C([0,1],{\mathbb R}): \Vert x\Vert <r\right\} \end{aligned}$$

and denote the maximum number by

$$\begin{aligned} M_r=\max \left\{ |f(t,x)|: (t,x)\in [0,1]\times [-r,r]\right\} . \end{aligned}$$

Theorem 3.2

Let \(f{:}\,\, [0,1]\times {\mathbb R} \rightarrow {\mathbb R}\) be a continuous function. Suppose that \((A_1), (A_2)\) hold. In addition we assume that

\((A_4)\) :

\(g(0)=0;\)

\((A_5)\) :

there exist a nonnegative function \(p\in C([0,1], {\mathbb R})\) and a nondecreasing function \(\psi : [0,\infty )\rightarrow (0,\infty )\) such that

$$\begin{aligned} |f(t,u)|\le p(t)\psi (|u|) \quad \text{ for } \text{ any } \quad (t,u)\in [0,1]\times {\mathbb R}; \end{aligned}$$
\((A_6)\) :

\(\displaystyle \sup _{r\in (0,\infty )}\frac{r}{k_0 |x_0|+p_0\psi (r)\Vert p\Vert }>\frac{1}{1-k_0\ell },\) where \(p_0\) and \(k_0\) are defined in (3.5) and (3.6), respectively.

Then the boundary value problem (1.1) has at least one solution on [0, 1].

Proof

By the assumption \((A_6),\) there exists a number \(r_0>0\) such that

$$\begin{aligned} \displaystyle \frac{r_0}{k_0|x_0|+p_0\psi (r_0)\Vert p\Vert }>\frac{1}{1-k_0\ell }. \end{aligned}$$
(3.8)

We shall show that the operators \(\mathcal {P}_1\) and \( \mathcal {P}_2\) defined by (3.2) and (3.3), respectively, satisfy all the conditions of Lemma 3.1.

Step 1. The operator \(\mathcal {P}_1\) is continuous and completely continuous. We first show that \( \mathcal {P}_1({\bar{\Omega }_{r_{0}}})\) is bounded. For any \(x\in {\bar{\Omega }_{r_{0}}},\) we have

$$\begin{aligned} \Vert \mathcal {P}_1x\Vert\le & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}\big |f(s,x(s))\big |\mathrm{d}s+\frac{1}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}\big |f(s,x(s))\big |\mathrm{d}s\\&\displaystyle +\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}\big |f(s,x(s))\big |\mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}\big |f(s,x(s))\big |\mathrm{d}s\Bigg \}\\\le & {} \frac{M_r}{\Gamma (q+1)} \Bigg [1+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1} )}{q+1} + \xi ^q \Bigg \}\Bigg ]\Vert p\Vert =M_r p_0\Vert p\Vert . \end{aligned}$$

Thus, the operator \(\mathcal {P}_1({\bar{\Omega }_{r_{0}}})\) is uniformly bounded. For any \(t_1, t_2\in [0,1], t_1<t_2,\) we have

$$\begin{aligned}&|(\mathcal {P}_1 x)(t_2)-(\mathcal {P}_1 x)(t_1)|\\&\quad \le \frac{1}{\Gamma (q)}\int _0^{t_1}\left[ (t_2-s)^{q-1}-(t_1-s)^{q-1}\right] |f(s,x(s))|\mathrm{d}s \\&\qquad +\,\frac{1}{\Gamma (q)}\int _{t_1}^{t_2}(t_2-s)^{q-1}|f(s,x(s))|\mathrm{d}s \\&\qquad +\,\frac{|t_2-t_1|}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}|f(s,x(s))|\mathrm{d}s\\&\qquad +\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}\big |f(s,x(s))\big |\mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}\big |f(s,x(s))\big |\mathrm{d}s\Bigg \}\\&\quad \le \frac{M_r}{\Gamma (q)}\int _0^{t_1}\left[ (t_2-s)^{q-1}-(t_1-s)^{q-1}\right] \mathrm{d}s + \frac{M_r}{\Gamma (q)}\int _{t_1}^{t_2}(t_2-s)^{q-1} \mathrm{d}s \\&\qquad +\,\frac{M_r|t_2-t_1|}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)} \mathrm{d}s +|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)} \mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)} \mathrm{d}s\Bigg \}\\&\quad \le \frac{M_r}{\Gamma (q)}\int _0^{t_1}\left[ (t_2-s)^{q-1}-(t_1-s)^{q-1}\right] \mathrm{d}s + \frac{M_r}{\Gamma (q)}\int _{t_1}^{t_2}(t_2-s)^{q-1} \mathrm{d}s \\&\qquad +\frac{M_r|t_2-t_1|}{|A|\Gamma (q+1)}\Bigg \{\frac{|a|(1+\eta ^{q+1})}{q+1} + \xi ^q \Bigg \}, \end{aligned}$$

which is independent of x and tends to zero as \(t_2-t_1\rightarrow 0.\) Thus, \(\mathcal {P}_1\) is equicontinuous. Hence, by the Arzelá-Ascoli Theorem, \(\mathcal {P}_1({\bar{\Omega }_{r_{0}}})\) is a relatively compact set. Now, let \(x_n\subset {\bar{\Omega }_{r_{0}}}\) with \(\Vert x_n-x\Vert \rightarrow 0.\) Then the limit \(\Vert x_n(t)-x(t)\Vert \rightarrow 0\) is uniformly valid on [0, 1]. From the uniform continuity of f(t, x) on the compact set \([0,1]\times [-r_0,r_0],\) it follows that \(\Vert f(t,x_n(t))-f(t,x(t))\Vert \rightarrow 0\) is uniformly valid on [0, 1]. Hence \(\Vert \mathcal {P}_1x_n-\mathcal {P}_1x\Vert \rightarrow 0\) as \(n\rightarrow \infty \) which proves the continuity of \(\mathcal {P}_1.\) This completes the proof of Step 1.

Step 2. The operator \({\mathcal {P}_2}: {\bar{\Omega }_{r_{0}}}\rightarrow C([0,1],{\mathbb R})\) is contractive. This is a consequence of \((A_2).\)

Step 3. The set \(\mathcal {P}({\bar{\Omega }_{r_{0}}})\) is bounded. The assumptions \((A_2)\) and \((A_4)\) imply that

$$\begin{aligned} \Vert {\mathcal {P}_2}(x)\Vert \le k_0(|x_0|+\ell r_0), \end{aligned}$$

for any \(x\in {\bar{\Omega }_{r_{0}}}.\) This, with the boundedness of the set \(\mathcal {P}_1({\bar{\Omega }_{r_{0}}})\) implies that the set \(\mathcal {P}({\bar{\Omega }_{r_{0}}})\) is bounded.

Step 4. Finally, it will be shown that the case (C2) in Lemma 3.1 does not hold. On the contrary, we suppose that (C2) holds. Then, we have that there exist \(\lambda \in (0,1)\) and \(x\in \partial \Omega _{r_{0}}\) such that \(x=\lambda \mathcal {P}x.\) So, we have \(\Vert x\Vert =r_0\) and

$$\begin{aligned} x(t)= & {} \displaystyle \lambda \Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x))+ \lambda \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s\\&+ \displaystyle \,\lambda \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s -a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s,x(s))\mathrm{d}s\\&- \displaystyle \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s,x(s))\mathrm{d}s\Bigg \}, \quad t \in [0,1]. \end{aligned}$$

Using the assumptions \((A_4)-(A_6),\) we get

$$\begin{aligned} r_0\le & {} \displaystyle \psi (r_0) \Bigg [\frac{1}{\Gamma (q)}\int _0^1 (t-s)^{q-1}p(s)\mathrm{d}s +\,\frac{1}{|A|}\Bigg \{\frac{|a|}{\Gamma (q+1)} \int _0^1(1-s)^{q}p(s) \mathrm{d}s \\&\displaystyle +\frac{|a|}{\Gamma (q+1)}\int _0^{\eta } (\eta -s)^{q}p(s) \mathrm{d}s+\frac{1}{\Gamma (q)}\int _0^{\xi } (\xi -s)^{q}p(s) \mathrm{d}s\Bigg \}\Bigg ]\\&+\,\Big |1+\frac{1}{A}\Big (a(1-\eta )-1\Big )\Big |(|x_0|+\ell r_0), \end{aligned}$$

which yields

$$\begin{aligned} r_0\le k_0|x_0|+p_0\psi (r_0)\Vert p\Vert + k_0\ell r_0. \end{aligned}$$

Thus, we get a contradiction:

$$\begin{aligned} \frac{r_0}{k_0|x_0|+p_0\psi (r_0)\Vert p\Vert }\le \frac{1}{1-k_0\ell }. \end{aligned}$$

Thus, the operators \(\mathcal {P}_1\) and \( \mathcal {P}_2\) satisfy all the conditions of Lemma 3.1. Hence, the operator \(\mathcal {P}\) has at least one fixed point \(x\in \bar{\Omega }_{r_{0}},\) which is the solution of the problem (1.1). This completes the proof. \(\square \)

Example 3.2

Consider the following boundary value problem

$$\begin{aligned} \displaystyle \left\{ \begin{array} {cc} ^cD^{3/2}x(t) =\displaystyle \frac{1}{3} \Big (\sqrt{t}+\frac{1}{18}\Big ) \sin x, \quad 0<t<1, \\ \displaystyle x(0)=\frac{1}{18} x\left( \frac{1}{9}\right) , \quad x\left( \frac{1}{4}\right) =\frac{1}{4} \int _{1/3}^1 x(s)\mathrm{d}s. \end{array} \right. \end{aligned}$$
(3.9)

Observe that \(\displaystyle q=3/2, f(t,x)=\frac{1}{3} \Big (\sqrt{t}+\frac{1}{18}\Big ) \sin x, ~x_0=0, g(x)=\frac{x(1/9)}{12}, ~\ell =1/18, a=1/4, \xi =1/4, \eta =1/3.\) Clearly \((A_1), (A_2)\), and \((A_4)\) are satisfied. Further \(\displaystyle |f(t,x)|\le \frac{1}{3} \Big (\sqrt{t}+\frac{1}{18}\Big ) x, ~\text{ for } \text{ any }~ (t,x)\in [0,1]\times {\mathbb R},\) and hence \(\displaystyle p(t)= \sqrt{t}+\frac{1}{18}\) and \(\displaystyle \psi (x)=\frac{1}{3}x.\) Thus, the condition \((A_5)\) is satisfied. With the given data, \(p_0\approx 2.00565, k_0=5,\)

$$\begin{aligned} \displaystyle \sup _{r\in (0,\infty )}\frac{r}{k_0|x_0|+p_0\psi (r)\Vert p\Vert }\approx 1.417049, \quad \frac{1}{1-k_{0} \ell } \approx 1.38461, \end{aligned}$$

it is found that \((A_6)\) holds. Therefore, all the conditions of Theorem 3.2 are satisfied and hence by its conclusion, the problem (3.9) has at least one solution on [0, 1].

4 Existence results for multivalued problem (1.2)

Let us recall some basic definitions on multivalued maps [16, 20].

For a normed space \((X, \Vert \cdot \Vert )\), let \(P_{cl}(X)=\{Y \in {\mathcal {P}}(X) : Y \,\text{ is } \text{ closed }\}\), \(P_{b}(X)=\{Y \in {\mathcal {P}}(X) : Y \,\text{ is } \text{ bounded }\}\), \(P_{cp}(X)=\{Y \in {\mathcal {P}}(X) : Y \,\text{ is } \text{ compact }\}\), and \(P_{cp, c}(X)=\{Y \in {\mathcal {P}}(X) : Y \,\text{ is } \text{ compact } \text{ and } \text{ convex }\}\). A multivalued map \(G : X \rightarrow {\mathcal {P}}(X)\) is convex (closed) valued if G(x) is convex (closed) for all \(x \in X.\) The map G is bounded on bounded sets if \(G(\mathbb {B}) = \cup _{x \in \mathbb {B}}G(x)\) is bounded in X for all \(\mathbb {B} \in P_{b}(X)\) (i.e., \(\sup _{x \in \mathbb {B}}\{\sup \{|y| : y \in G(x)\}\} < \infty ).\) G is called upper semi-continuous (u.s.c.) on X if for each \(x_0 \in X,\) the set \(G(x_0)\) is a nonempty closed subset of X, and if for each open set N of X containing \(G(x_0),\) there exists an open neighborhood \(\mathcal {N}_0\) of \(x_0\) such that \(G(\mathcal {N}_0) \subseteq N.\) G is said to be completely continuous if \(G(\mathbb {B})\) is relatively compact for every \(\mathbb {B} \in P_{b}(X)\). If the multivalued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph, i.e., \(x_n \rightarrow x_*, ~y_n \rightarrow y_*,\) \(y_n \in G(x_n)\) imply \(y_* \in G(x_*).\) G has a fixed point if there is \(x \in X\) such that \(x \in G(x)\). The fixed point set of the multivalued operator G will be denoted by \({ {Fix}} G\). A multivalued map \(G : [0;1] \rightarrow P_{cl}(\mathbb {R})\) is said to be measurable if for every \(y \in \mathbb {R}\), the function

$$\begin{aligned} t \longmapsto d(y,G(t)) = \inf \{|y-z|: z \in G(t)\} \end{aligned}$$

is measurable.

Definition 4.1

A function \(x \in AC^1([0,1], \mathbb R)\) is a solution of the problem (1.2) if \(\displaystyle x(0)=x_0+g(x), ~ x(\xi )=a \int _{\eta }^1x(s)\, \mathrm{d}s,\) and there exists a function \(f \in L^1([0,1],\mathbb {R})\) such that \(f(t) \in F(t, x(t))\) a.e. on [0, 1] and

$$\begin{aligned} \begin{array}{ll} \displaystyle x(t)&{}= \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s +\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s\\ &{}\displaystyle \quad -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)ds - \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)ds\Bigg \}\\ &{}\displaystyle \quad \,+\,\Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)). \end{array} \end{aligned}$$
(4.1)

Here \(AC^1([0,1], {\mathbb R})\) will denote the space of functions \(x{:}\, [0,1]\rightarrow {\mathbb R}\) that are absolutely continuous and whose first derivative is absolutely continuous.

4.1 The Carathéodory Case

Definition 4.2

A multivalued map \(F : [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) is said to be Carathéodory if

  1. (i)

    \(t \longmapsto F(t,x)\) is measurable for each \(x \in \mathbb {R}\);

  2. (ii)

    \(x \longmapsto F(t,x)\) is upper semi-continuous for almost all \(t\in [0,1]\);

Further a Carathéodory function F is called \(L^1-\)Carathéodory if

  1. (iii)

    for each \(\alpha > 0\), there exists \(\varphi _{\alpha } \in L^1([0,1],\mathbb {R}^+)\) such that

    $$\begin{aligned} \Vert F (t, x)\Vert = \sup \{|v| : v \in F (t, x)\} \le \varphi _{\alpha } (t) \end{aligned}$$

    for all \(\Vert x\Vert \le \alpha \) and for a. e. \(t \in [0,1].\)

For each \(y \in C([0,1], \mathbb {R})\), define the set of selections of F by

$$\begin{aligned} S_{F,y} := \{ v \in L^1([0,1],\mathbb {R}) : v (t) \in F (t, y(t)) ~\text{ for } \text{ a.e. } ~t \in [0,1]\}. \end{aligned}$$

The following lemma will be used in the sequel.

Lemma 4.1

([24]) Let X be a Banach space. Let \(F : [0, 1] \times \mathbb {R} \rightarrow {\mathcal P}_{cp,c}(X)\) be an \(L^1-\) Carathéodory multivalued map, and let \(\Theta \) be a linear continuous mapping from \(L^1([0,1],X)\) to C([0, 1], X). Then the operator

$$\begin{aligned} \Theta \circ S_F : C([0,1],X) \rightarrow {\mathcal P}_{cp,c} (C([0,1],X)), ~~ x \mapsto (\Theta \circ S_F) (x) = \Theta ( S_{F,x}) \end{aligned}$$

is a closed graph operator in \(C([0,1],X) \times C([0,1],X).\)

To prove our main result in this section, we use the following form of the Nonlinear Alternative for contractive maps [27, Corollary3.8].

Theorem 4.1

Let X be a Banach space and D a bounded neighborhood of \(0\in X.\) Let \(Z_1:X\rightarrow {\mathcal P}_{cp,c}(X)\) and \(Z_2:\bar{D}\rightarrow {\mathcal P}_{cp,c}(X)\) two multivalued operators satisfying

  1. (a)

    \(Z_1\) is contraction, and

  2. (b)

    \(Z_2\) is u.s.c and compact.

Then, if \(G=Z_1+Z_2,\) either

  1. (i)

    G has a fixed point in \(\bar{D}\) or

  2. (ii)

    there is a point \(u\in \partial D\) and \(\lambda \in (0,1)\) with \(u\in \lambda G(u)\).

Theorem 4.2

Assume that \((A_2)\) holds. In addition we suppose that

\({(H_1)}\) :

\(F : [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}_{cp,c}(\mathbb {R})\) is \(L^1-\)Carathéodory multivalued map;

\({(H_2)}\) :

there exists a continuous nondecreasing function \(\psi : [0,\infty ) \rightarrow (0,\infty )\) and a function \(p \in C([0,1],\mathbb {R}^+)\) such that

$$\begin{aligned} \Vert F(t,x)\Vert _\mathcal {P}:=\sup \{|y|: y \in F(t,x)\}\le p(t)\psi (\Vert x\Vert ) ~~ \text{ for } \text{ each } \,\,\, (t,x) \in [0,1] \times \mathbb {R}; \end{aligned}$$
\({(H_3)}\) :

there exists a number \(M>0\) such that

$$\begin{aligned} \displaystyle \frac{(1-k_0\ell )M}{ \displaystyle \psi (M)p_0\Vert p\Vert +k_0|x_0|} > 1, \end{aligned}$$
(4.2)

where \(p_0\) and \(k_0\) are defined in (3.5) and (3.6), respectively.

Then the boundary value problem (1.2) has at least one solution on [0, 1].

Proof

To transform the problem (1.2) to a fixed point, we introduce an operator \({\mathcal N}:C([0,1],{\mathbb R})\longrightarrow {\mathcal P}(C([0,1],{\mathbb R}))\) defined by

$$\begin{aligned} {\mathcal N}(x)=\left\{ \begin{array}{ll} &{}h \in C([0,1], \mathbb {R}): \\ &{}h(t) =\left\{ \begin{array}{l} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s + \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s \\ \displaystyle \quad -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s -\, \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s\Bigg \}\\ \displaystyle \quad +\Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)), \end{array} \right. \end{array}\right\} \end{aligned}$$

for \(f \in S_{F,x}.\)

Now, we define two operators \({\mathcal A}: C([0,1],\mathbb {R})\longrightarrow C([0,1],\mathbb {R})\) by

$$\begin{aligned} {\mathcal A}x(t) = \displaystyle \Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)), \end{aligned}$$
(4.3)

and a multivalued operator \({\mathcal B}: C([0,1],\mathbb {R})\longrightarrow {\mathcal P}(C([0,1],\mathbb {R}))\) by

$$\begin{aligned} {\mathcal B}(x)=\left\{ \begin{array}{ll} &{}h \in C([0,1], \mathbb {R}): \\ &{}h(t) =\left\{ \begin{array}{l} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s + \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s \\ \displaystyle -a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s - \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s\Bigg \}. \end{array} \right. \end{array}\right\} \nonumber \\ \end{aligned}$$
(4.4)

Observe that \({\mathcal N}={\mathcal A}+{\mathcal B}\). We shall show that the operators \({\mathcal A}\) and \({\mathcal B}\) satisfy all the conditions of Theorem 4.1 on [0, 1]. The proof consists of several steps and claims.

Step 1: We show that \({\mathcal A}\) is a contraction on \(C([0,1],{\mathbb R})\). For \(x,y\in C([0,1],{\mathbb R}),\) we have

$$\begin{aligned} |{\mathcal A}x(t)-{\mathcal A}y(t)|= & {} \displaystyle \Big |1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ||g(x)-g(y)|\\\le & {} \Big |1+\frac{1}{A}\big (a(1-\eta )-1\Big )\Big ||g(x)-g(y)|,\\\le & {} k_0\ell \Vert x-y\Vert , \end{aligned}$$

which, on taking supremum over \(t \in [0, 1],\) yields

$$\begin{aligned} \Vert {\mathcal A}x - {\mathcal A}y\Vert \le L_0 \Vert x-y\Vert , \quad L_0=k_0\ell <1. \end{aligned}$$

This shows that \({\mathcal A}\) is a contraction as \(L_0<1\).

Step 2: \({\mathcal B}\) is compact and convex valued, and it is completely continuous. This will be established in several claims.

Claim I: \({\mathcal B}\) maps bounded sets into bounded sets in \(C([0,1],{\mathbb R})\). Let \(B_r = \{x \in C([0,1], \mathbb {R}): \Vert x\Vert \le r \}\) be a bounded set in \(C([0,1], \mathbb {R})\). Then, for each \(h \in {\mathcal B} (x), x \in B_{\rho }\), there exists \(f \in S_{F,x}\) such that

$$\begin{aligned} h(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s + \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s \\&\displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s - \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s\Bigg \}. \end{aligned}$$

Then, for \(t\in [0,1],\) we have

$$\begin{aligned} |h(t)|\le & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}|f(s)|\mathrm{d}s+\frac{1}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}|f(s)|\mathrm{d}s\\&\displaystyle +\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}|f(s)|\mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}|f(s)|\mathrm{d}s\Bigg \}\\\le & {} \frac{ \psi (\Vert x\Vert )}{\Gamma (q+1)} \Bigg [1+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1} )}{q+1} + \xi ^q \Bigg \}\Bigg ]\Vert p\Vert . \end{aligned}$$

Thus,

$$\begin{aligned} \Vert h\Vert \le \frac{ \psi (\Vert x\Vert )}{\Gamma (q+1)} \Bigg [1+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1} )}{q+1} + \xi ^q \Bigg \}\Bigg ]\Vert p\Vert . \end{aligned}$$

Claim II: \({\mathcal B}\) maps bounded sets into equicontinuous sets. Let \(t_1, t_2 \in [0,1]\) with \(t_1< t_2\) and \(x \in B_{\rho }.\) Then, for each \(h \in {\mathcal B}(x),\) we obtain

$$\begin{aligned}&|h(t_2)-h(t_1)|\\&\quad \le \frac{\psi (\Vert x\Vert )\Vert p\Vert }{\Gamma (q)}\int _0^{t_1}[(t_2{-}s)^{q-1}{-}(t_1-s)^{q-1}]\mathrm{d}s + \frac{\psi (\Vert x\Vert )\Vert p\Vert }{\Gamma (q)}\int _{t_1}^{t_2}(t_2-s)^{q-1}\mathrm{d}s\quad \\&\qquad +\,\frac{\psi (\Vert x\Vert )\Vert p\Vert |t_2-t_1|}{|A|}\Bigg \{|a|\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}\mathrm{d}s\\&\qquad +\,|a|\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}\mathrm{d}s+ \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}\mathrm{d}s\Bigg \}. \end{aligned}$$

Obviously the right-hand side of the above inequality tends to zero independently of \(x \in B_{\rho }\) as \(t_2- t_1 \rightarrow 0.\) Therefore it follows by the Ascoli-Arzelá theorem that \({\mathcal B}: C([0,1], \mathbb {R}) \rightarrow {\mathcal {P}}(C([0,1], \mathbb {R}))\) is completely continuous.

Claim III: \({\mathcal B}\) has a closed graph. Let \(x_n \rightarrow x_*, h_n \in {\mathcal B} (x_n)\) and \(h_n \rightarrow h_*.\) Then we need to show that \(h_* \in {\mathcal B}(x_*).\) Associated with \(h_n \in {\mathcal B} (x_n),\) there exists \(f_n \in S_{F,x_n}\) such that for each \(t \in [0,1],\)

$$\begin{aligned} h_n(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f_n(s)\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f_n(s)\mathrm{d}s\\&\displaystyle -\,a\,\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f_n(s)\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f_n(s)\mathrm{d}s\Bigg \}. \end{aligned}$$

Thus, it suffices to show that there exists \(f_* \in S_{F,x_*}\) such that for each \(t \in [0,1],\)

$$\begin{aligned} h_*(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f_*(s)\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f_*(s)\mathrm{d}s\\&\displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f_*(s)\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f_*(s)\mathrm{d}s\Bigg \}. \end{aligned}$$

Let us consider the linear operator \(\Theta : L^1([0,1], \mathbb {R}) \rightarrow C([0,1], \mathbb {R})\) given by

$$\begin{aligned} f \mapsto \Theta (f)(t)= & {} \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s\\&\displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s\Bigg \}. \end{aligned}$$

Observe that

$$\begin{aligned}&\Vert h_n(t)-h_*(t)\Vert \\&\quad =\Bigg \Vert \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}(f_n(s)-f_*(s))\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}(f_n(s)-f_*(s))\mathrm{d}s\\&\qquad -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}(f_n(s)-f_*(s))\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}(f_n(s)-f_*(s))\mathrm{d}s\Bigg \}\Bigg \Vert \rightarrow 0, \end{aligned}$$

as \(n\rightarrow \infty .\)

Thus, it follows by Lemma 4.1 that \(\Theta \circ S_F\) is a closed graph operator. Further, we have \(h_n(t) \in \Theta (S_{F,x_n}).\) Since \(x_n \rightarrow x_*,\) therefore, we have

$$\begin{aligned} h_*(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f_*(s)\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f_*(s)\mathrm{d}s\\&\displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f_*(s)\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f_*(s)\mathrm{d}s\Bigg \}, \end{aligned}$$

for some \(f_* \in S_{F,x_*}\). Hence \({\mathcal B}\) has a closed graph (and therefore has closed values). In consequence, the operator \({\mathcal B}\) is compact valued.

Thus, the operators \({\mathcal A}\) and \({\mathcal B}\) satisfy all the conditions of Theorem 4.1, and hence, its conclusion implies either condition (i) or condition (ii) holds. We show that the conclusion (ii) is not possible. If \(x \in \lambda {\mathcal A} (x)+\lambda {\mathcal B}(x)\) for \(\lambda \in (0,1),\) then there exists \(f\in S_{F,x}\) such that

$$\begin{aligned} x(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s+\frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s\\&\displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(s)\mathrm{d}s- \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(s)\mathrm{d}s\Bigg \}\\&+\,\Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)), \quad t\in [0,1]. \end{aligned}$$

Following the method for proof of Claim I, we can obtain

$$\begin{aligned} |x(t)|\le & {} \frac{ \psi (\Vert x\Vert )}{\Gamma (q+1)} \Bigg [1+\frac{1}{|A|}\Bigg \{\frac{|a|(1+\eta ^{q+1} )}{q+1} + \xi ^q \Bigg \}\Bigg ]\Vert p\Vert \\&+\,\Big |1+\frac{1}{A}\big (a(1-\eta )-1\Big )\Big |[|x_0|+\ell \Vert x\Vert ]. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert x\Vert \le \psi (\Vert x\Vert )p_0\Vert p\Vert +k_0(|x_0|+\ell \Vert x\Vert ). \end{aligned}$$
(4.5)

If condition (ii) of Theorem 4.1 holds, then there exist \(\lambda \in (0, 1)\) and \(x\in \partial B_r\) with \(x = \lambda {\mathcal N}(x).\) Then, x is a solution of (1.2) with \(\Vert x\Vert = M.\) Now, by the inequality (4.5), we get

$$\begin{aligned} \displaystyle \frac{(1-k_0\ell )M}{ \displaystyle \psi (M)p_0\Vert p\Vert +k_0|x_0|} \le 1 \end{aligned}$$

which contradicts (4.2). Hence, \({\mathcal N}\) has a fixed point in [0, 1] by Theorem 4.1, and consequently the problem (1.2) has a solution. This completes the proof. \(\square \)

Example 4.1

Consider the following boundary value problem

$$\begin{aligned} \left\{ \begin{array} {ll} ^cD^{3/2}x(t) \in F(t, x), &{}\quad 0<t<1, \\ \displaystyle x(0)=\frac{1}{15}+\frac{1}{12} x(1/9), &{}\quad x\left( \frac{1}{4}\right) =\frac{1}{4} \int _{1/3}^1 x(s)\mathrm{d}s. \end{array} \right. \end{aligned}$$
(4.6)

Here, \(q=3/2, ~x_0=1/15, ~g(x)=\frac{1}{12} x(1/9), ~ a=1/4, \xi =1/4, \eta =1/3, \) and \(F: [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) is a multivalued map given by

$$\begin{aligned} x \rightarrow F(t,x)=\left[ \frac{1}{9}\frac{x^3}{x^3+3}+\frac{1}{8}(t+1),\frac{1}{2}\cos x\right] . \end{aligned}$$

For \(f \in F,\) we have

$$\begin{aligned} |f| \le \max \left[ \frac{1}{9}\frac{x^3}{x^3+3}+\frac{1}{8}(t+1),\frac{1}{2}\cos x\right] \le \frac{1}{2}. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert F(t,x)\Vert _\mathcal {P}:=\sup \{|y|: y \in F(t,x)\}\le p(t)\psi (\Vert x\Vert ), \quad x \in \mathbb {R}, \end{aligned}$$

with \(p(t)=1, ~\psi (\Vert x\Vert )=1/2.\) Furthermore, \(p_0\approx 2.00565, k_0=5\), and by the condition \((H_3),\) we have \(\ell =1/12.\) Obviously \(\ell <k_0^{-1}.\) With the given data, it is found that \(M>M_1, ~M_1\approx 2.29056.\) Clearly, all the conditions of Theorem 4.2 are satisfied, and hence the problem (4.6) has at least one solution on [0, 1].

4.2 The Lower Semi-continuous Case

In this section, we study the case when F is not necessarily convex valued by applying the nonlinear alternative of Leray-Schauder type and a selection theorem due to Bressan and Colombo [11] for lower semi-continuous maps with decomposable values.

Let us mention some auxiliary facts. Let X be a nonempty closed subset of a Banach space E and \(G : X \rightarrow {\mathcal {P}}(E)\) be a multivalued operator with nonempty closed values. G is lower semi-continuous (l.s.c.) if the set \(\{y \in X : G(y)\cap B \ne \emptyset \}\) is open for any open set B in E. Let A be a subset of \([0,1]\times \mathbb {R}\). A is \(\mathcal {L}\otimes \mathcal {B}\) measurable if A belongs to the \(\sigma \)-algebra generated by all sets of the form \(\mathcal {J} \times \mathcal {D}\), where \(\mathcal {J}\) is Lebesgue measurable in [0, 1] and \(\mathcal {D}\) is Borel measurable in \(\mathbb {R}\). A subset \(\mathcal {A}\) of \(L^1([0,1], \mathbb {R})\) is decomposable if for all \(u, v \in \mathcal {A}\) and measurable \(\mathcal {J} \subset [0,1]=J\), the function \(u \chi _{\mathcal {J}}+v \chi _{J-\mathcal {J}} \in \mathcal {A}\), where \(\chi _{\mathcal {J}}\) stands for the characteristic function of \(\mathcal {J}\).

Definition 4.3

Let Y be a separable metric space and let \(N : Y \rightarrow {\mathcal {P}}(L^1([0,1],\mathbb {R}))\) be a multivalued operator. We say N has a property (BC) if N is lower semi-continuous (l.s.c.) and has nonempty closed and decomposable values.

Let \(F{:}\, [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) be a multivalued map with nonempty compact values. Define a multivalued operator \(\mathcal {F} : C([0,1] \times \mathbb {R}) \rightarrow {\mathcal {P}}(L^1([0,1],\mathbb {R}))\) associated with F as

$$\begin{aligned} \mathcal {F}(x)=\{w \in L^1([0,1],\mathbb {R}) : w(t) \in F(t,x(t)) ~\text{ for } \text{ a.e. } \quad t \in [0,1]\}, \end{aligned}$$

which is called the Nemytskii operator associated with F.

Definition 4.4

Let \(F{:} [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) be a multivalued function with nonempty compact values. We say F is of lower semi-continuous type (l.s.c. type) if its associated Nemytskii operator \(\mathcal {F }\) is lower semi-continuous and has nonempty closed and decomposable values.

Lemma 4.2

([17]) Let Y be a separable metric space, and let \(N : Y \rightarrow {\mathcal {P}}(L^1([0,1],\mathbb {R}))\) be a multivalued operator satisfying the property (BC). Then N has a continuous selection, that is, there exists a continuous function (single-valued) \(g : Y \rightarrow L^1([0,1],\mathbb {R})\) such that \(g(x) \in N(x)\) for every \(x \in Y\).

Theorem 4.3

Assume that \((A_2), (H_2), (H_3)\) and the following condition hold:

\((H_4)\) :

\(F : [0,1] \times \mathbb {R} \rightarrow {\mathcal {P}}(\mathbb {R})\) is a nonempty compact-valued multivalued map such that

(a):

\((t,x) \longmapsto F(t,x)\) is \(\mathcal {L}\otimes \mathcal {B}\) measurable;

(b):

\( x \longmapsto F(t,x)\) is lower semi-continuous for each \(t \in [0,1].\)

Then the boundary value problem (1.2) has at least one solution on [0, 1].

Proof

It follows from \((H_2)\) and \((H_4)\) that F is of l.s.c. type. Then, by Lemma 4.2, there exists a continuous function \(f : C([0,1],\mathbb {R}) \rightarrow L^1([0,1],\mathbb {R})\) such that \(f (x) \in \mathcal {F}(x)\) for all \(x \in C([0,1],\mathbb {R})\).

Consider the problem

$$\begin{aligned} \left\{ \begin{array} {ll} ^cD^q x(t) = f(x(t)), \quad 0<t<1, \quad 1 < q \le 2, \\ \displaystyle x(0)=x_0+g(x), \quad x(\xi ) =a \int _{\eta }^{1}x(s)\mathrm{d}s, \quad 0<\xi <\eta . \end{array} \right. \end{aligned}$$
(4.7)

Observe that if \(x \in AC^1([0,1], {\mathbb R})\) is a solution of (4.7), then x is a solution to the problem (1.2). Now, we define two operators, namely \({\mathcal A'}: C([0,1],{\mathbb R})\longrightarrow C([0,1],{\mathbb R})\) by

$$\begin{aligned} {\mathcal A'}x(t) = \displaystyle \displaystyle \Big [1+\frac{t}{A}\big (a(1-\eta )-1\Big )\Big ](x_0+g(x)), \end{aligned}$$
(4.8)

and \({\mathcal B'}: C([0,1],{\mathbb R})\longrightarrow C([0,1],{\mathbb R})\) by

$$\begin{aligned} {\mathcal B'}x(t)=\left\{ \begin{array}{ll} \displaystyle \int _0^t\frac{(t-s)^{q-1}}{\Gamma (q)}f(x(s))\mathrm{d}s + \frac{t}{A}\Bigg \{a\int _0^1\frac{(1-s)^{q}}{\Gamma (q+1)}f(x(s))\mathrm{d}s \\ \displaystyle -\,a\int _0^\eta \frac{(\eta -s)^{q}}{\Gamma (q+1)}f(x(s))\mathrm{d}s - \int _0^\xi \frac{(\xi -s)^{q-1}}{\Gamma (q)}f(x(s))\mathrm{d}s\Bigg \}. \end{array} \right. \end{aligned}$$
(4.9)

Clearly \({\mathcal A'},{\mathcal B'}{:}\, C([0,1],{\mathbb R}) \rightarrow C([0,1],{\mathbb R})\) are continuous. The arguments used in the proof of Theorem 4.2 apply and hence guarantee that \({\mathcal A'}\) and \({\mathcal B'}\) satisfy all the conditions of the Nonlinear Alternative for contractive maps in the single-valued setting [21], and hence the problem (4.7) has a solution. \(\square \)