Existence of solutions for fractional differential equations with nonlocal and average type integral boundary conditions

Abstract

In this paper, we establish sufficient conditions for the existence and uniqueness of solutions for a boundary value problem of fractional differential equations with nonlocal and average type integral boundary conditions. The Leray–Schauder nonlinear alternative, Krasnoselskii’s fixed point theorem and Banach’s fixed point theorem together with Hölder inequality are applied to construct proofs for the main results. Examples illustrating the obtained results are also presented.

1 Introduction

In this paper, we study a boundary value problem of nonlinear fractional differential equations supplemented with nonlocal and average type integral boundary conditions of the form:

$$\begin{aligned} \left\{ \begin{array} {ll} \displaystyle ^C_0D^{\alpha }_tx(t)= f(t,x(t), ^C_0D^{\beta }_tx(t)), \quad 0<t<1, \\ \displaystyle x(0)+x'(0)= h(x),~ \int _0^\eta x(t)dt=\xi , ~~ 0<\eta <1,\\ x''(0)=x'''(0)=\ldots =x^{(n-1)}(0)=0, \end{array} \right. \end{aligned}$$

(1.1)

where $^C_0D^{\alpha }_t, ^C_0D^{\beta }_t$ denote the Caputo fractional derivatives of order $\alpha \in (n-1, n), n\ge 2$ and $\beta \in (0, 1),$ $f: [0,1]\times {\mathbb R}\times {\mathbb R}\rightarrow {\mathbb R},$ $h: C([0,1],{\mathbb R})\rightarrow {\mathbb R}$ are given continuous functions and $\xi \in {\mathbb R}$ is a real constant.

Fractional differential equations have recently been investigated by several scientists and modelers. It has been mainly due to the extensive application of the subject in a variety of disciplines such as biological sciences, ecology, aerodynamics, control theory, viscoelasticity, electro-dynamics of complex medium, electron-analytical chemistry, environmental issues, etc. Fractional calculus modelling techniques have revolutionized the modelling methodology based on integer-order calculus due to the nonlocal characteristic of fractional-order differential and integral operators which are capable of tracing the past history of many materials and processes, for instance, see [1–7] and the references cited therein.

Fractional-order boundary value problems involving classical, nonlocal, multi-point, periodic/anti-periodic, fractional-order, and integral boundary conditions have extensively been studied by many researchers and a variety of results can be found in recent literature on the topic [8–24].

Nonlocal conditions, dated back to the works [25–27], are considered to be more practical and plausible than the classical conditions as they can describe some peculiar phenomena associated with physical, chemical or other processes occurring at the interior positions of the domain. Integral boundary conditions are found to be of great help in computational fluid dynamics (CFD) studies of blood flow problems, for instance, see [28] and the references cited therein. Also, integral boundary conditions are applied in regularizing ill-posed parabolic backward problems in time partial differential equations such as mathematical models for bacterial self-regularization [29].

In a recent paper [20], the authors studied a boundary value problem analog to (1.1) with an integral condition of the form $\int _0^1 x(t)dt=\xi .$ In the present work, we have considered a more flexible problem by introducing a nonlocal integral condition $\int _0^{\eta } x(t)dt=\xi , \eta \in (0,1),$ which reduces to the one assumed in [20] in the limit $\eta \rightarrow 1^{-}.$ Existence and uniqueness results in [20] were obtained by using contraction mapping principle and Schauder’s fixed-point theorem. We can apply the tools of fixed point theory employed by the authors in [20] (a brief description of application of these tools to (1.1) has been given in Sect. 3). However, in order to broaden the sphere of existence theory for problem (1.1) as well as for its special case addressed in [20], we make use of Leray–Schauder nonlinear alternative, Krasnoselskii’s fixed point theorem and Banach’s fixed point theorem coupled with Hölder inequality.

The rest of the paper is organized as follows. In Sect. 2, we give some preliminary concepts of fractional calculus and establish an auxiliary lemma. Section 3 contains the main results for the problem (1.1), while the illustrative examples for these results are presented in Sect. 4.

2 Preliminaries

This section is devoted to some basic definitions and relations for fractional-order derivatives and integrals [2]. An auxiliary lemma dealing with the linear variant of problem (1.1) is also presented.

Definition 2.1

The Riemann-Liouville fractional integral of order q for a function $g\in L^1_\mathrm{loc}[0,1]$ is defined as

$$\begin{aligned} I^q_{0+} g(t)=\frac{1}{\Gamma (q)}\int _0^t \frac{g(s)}{(t-s)^{1-q}}ds, ~~q>0, \end{aligned}$$

provided the integral exists.

Definition 2.2

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

$$\begin{aligned} ^C_0D^q_t g(t)=\frac{1}{\Gamma (n-q)}\int _{0}^t(t-s)^{n-q-1}g^{(n)}(s)ds, ~~n-1 < q < n,~ n=[q]+1, \end{aligned}$$

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

Lemma 2.3

(see [2]) (i)   If $\alpha > 0, \beta > 0, \beta >\alpha , f\in L[0, 1],$ then

$$\begin{aligned}&I^{\alpha }_{0+} I^{\beta }_{0+} f(t) = I^{\alpha +\beta }_{0+} f(t),\;\;^C_0D^{\alpha }_t I^{\alpha }_{0+} f(t) = f(t),\;^C_0D^{\alpha }_t I^{\beta }_{0+} f(t)=I^{\beta -\alpha }_{0+} f(t).\\&(ii)\qquad ^C_0D^{\alpha }_tt^{\lambda -1}=\frac{\Gamma (\lambda )}{\Gamma (\lambda -\alpha )}t^{\lambda -\alpha -1}, ~~\lambda >[\alpha ] ~~\text{ and }~~ ^C_0D^{\alpha }_tt^{\lambda -1}=0, ~~\lambda <[\alpha ]. \end{aligned}$$

In order to define the solution for problem (1.1), we present the following lemma.

Lemma 2.4

For $y\in L[0,1],$ let $x\in AC^n[0,1]$ be a solution of the linear boundary value problem

$$\begin{aligned} \left\{ \begin{array} {ll} \displaystyle ^C_0D^{\alpha }_{t}x(t)=y(t), \quad 0<t<1, \\ \displaystyle x(0)+x'(0)= h(x),~ \int _0^\eta x(t)dt=\xi ,\quad 0<\eta <1, \\ x''(0)=x'''(0)=\ldots =x^{(n-1)}(0)=0. \end{array} \right. \end{aligned}$$

(2.1)

Then

$$\begin{aligned} \begin{array}{ll} x(t) &{}= \displaystyle \int _{0}^{t} \frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )} y(s)ds+\frac{2(1-t)}{\eta (2-\eta )}\xi +\frac{(2t-\eta )}{2-\eta }h(x)\\ &{}\quad \displaystyle +\frac{2(t-1)}{\eta (2-\eta )}\int _0^\eta \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}y(s)ds. \end{array} \end{aligned}$$

(2.2)

Proof

We know that the general solution of the fractional differential equation in (2.1) can be written as

$$\begin{aligned} x(t)=\int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}y(s) ds+c_0+c_1t+c_2t^2 +\cdots +c_{n-2}t^{n-2}+c_{n-1}t^{n-1}, \end{aligned}$$

(2.3)

where $c_0,c_1,\ldots ,c_{n-1}\in \mathbb {R}$ are arbitrary constants. Using the initial conditions: $x''(0)=x'''(0)=\ldots =x^{(n-1)}(0)=0,$ we find that $c_2=\cdots =c_{n-1}=0.$ Thus (2.3) becomes

$$\begin{aligned} x(t)=\int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}y(s) ds+c_0+c_1 t. \end{aligned}$$

(2.4)

Next, by the conditions: $x(0)+x'(0)= h(x), \, \int _0^\eta x(t)dt=\xi ,$ we obtain

$$\begin{aligned} c_{0}= & {} \frac{2}{\eta (2-\eta )}\Big (\xi -\frac{\eta ^2}{2} h(x)-\int _0^\eta \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}y(s)ds\Big ),\\ c_{1}= & {} \frac{2}{\eta (2-\eta )}\Big (\int _0^\eta \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}y(s)ds-\xi \Big ) +\frac{2h(x)}{2-\eta }. \end{aligned}$$

Substituting the values of $c_0$ and $c_1$ in (2.4), we obtain the solution (2.2). This completes the proof. $\square$

3 Existence and uniqueness results

Let $X = \left\{ x|x\in C([0,1],{\mathbb R})\;\;\text{ and }\; \; ^C_0D^{\beta }_tx \in C([0,1],{\mathbb R})\right\}$ denote a Banach space equipped with the norm $\Vert x\Vert _{X}=\Vert x\Vert +\Vert ^C_0D^{\beta }_tx\Vert =\sup _{t\in [0,1]}|x(t)|+\sup _{t\in [0,1]} |^C_0D^{\beta }_tx(t)|.$

In relation to problem (1.1), we introduce an operator $F: X \rightarrow X$ as

$$\begin{aligned} \begin{array}{ll} F(x)(t) &{}= \displaystyle \int _{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}f(s, x(s), ^C_0D^{\beta }_tx(s))ds+\frac{2(1-t)}{\eta (2-\eta )} \xi +\frac{(2t-\eta )}{2-\eta }h(x)\\ &{}\quad \displaystyle +\frac{2(t-1)}{\eta (2-\eta )}\int _{0}^{\eta } \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}f(s, x(s), ^C_0D^{\beta }_tx(s))ds. \end{array} \end{aligned}$$

(3.1)

Observe that problem (1.1) has solutions if the operator F has fixed points.

Here we remark that the existence and uniqueness result for problem (1.1) analog to Theorem 3.1 of [20] can be established with the aid of the operator (3.1) together with the conditions $(H_1), (H_2)$ of [20] and modified form of $\theta$ in $(H_3)$ of [20] given by $\tilde{\theta }=\max \{\theta _1, \theta _2\},$ where

$$\begin{aligned}&\displaystyle \quad \theta _1=\ell \left( \frac{2}{\Gamma (\alpha +1)}+\frac{4 \eta ^{\alpha +1}}{\eta (2-\eta )\Gamma (\alpha +2)}\right) +\ell _1, \\&\displaystyle \quad \quad \quad \quad \theta _2=\frac{1}{\Gamma (1-\beta )}\left[ \ell \left( \frac{2}{\Gamma (\alpha )}+\frac{4 \eta ^{\alpha +1}}{\eta (2-\eta )\Gamma (\alpha +2)}\right) +\frac{2 \ell _1}{2-\eta }\right] . \end{aligned}$$

Notice that $\tilde{\theta }$ takes the form of $\theta$ in $(H_3)$ of [20] in the limit $\eta \rightarrow 1^{-}.$ The existence results for problem (1.1) analog to Theorems 3.2 and 3.3 of [20] can be proved as done in [20] with the aid of the operator (3.1).

In the sequel, we establish some more existence and uniqueness results for problem (1.1) which are based on Leray–Schauder nonlinear alternative, Krasnoselskii’s fixed point theorem and Banach’s fixed point theorem coupled with Hölder inequality.

3.1 Existence result via Leray–Schauder’s nonlinear alternative

Lemma 3.1

(Nonlinear alternative for single valued maps)[30]. Let E be a Banach space, C a closed, convex subset of E,  U an open subset of C and $0\in U.$ Suppose that $F:\overline{U}\rightarrow C$ is a continuous, compact (that is, $F(\overline{U})$ is a relatively compact subset of C) map. Then either

• (i) F has a fixed point in $\overline{U},$ or

• (ii) there is a $u\in \partial U$ (the boundary of U in C) and $\lambda \in (0,1)$ with $u=\lambda F(u).$

Theorem 3.2

Assume that:

${(A_{1})}$ :

$f: [0,1]\times {\mathbb R}\times {\mathbb R} \rightarrow {\mathbb R}$ is a continuous function and there exist a function $m\in L^{\frac{1}{\gamma }}([0,1], {\mathbb R}^+),$ $\gamma \in (0,1/(\alpha -1))$ and nondecreasing functions $\psi _i: {\mathbb {R}}^+\rightarrow { \mathbb {R}}^+, i=1,2$ such that $|f(t,x,y)| \le m(t)(\psi _1(|x|)+\psi _2(|y|)),$ $\forall (t,x,y) \in [0,1] \times {\mathbb R}\times {\mathbb R};$

${(A_{2})}$ :

$h: C([0,1], {\mathbb R})\rightarrow {\mathbb R}$ is a continuous function and there exists a nondecreasing function $\Theta : {\mathbb R}^+\rightarrow {\mathbb R}^+$ such that $|h(x)|\le \Theta (\Vert x\Vert ), \forall x\in C([0,1], {\mathbb R});$

${(A_{3})}$ :

there exists a constant $M>0$ such that

$$\begin{aligned} \frac{M}{\displaystyle [\psi _1(M)+\psi _2(M)] \Vert m\Vert \, H+H_0} >1, \end{aligned}$$

where $\displaystyle \Vert m\Vert =\left( \int _0^1|m(s)|^{\frac{1}{\gamma }}ds\right) ^{\gamma },$

$$\begin{aligned} H= & {} \frac{1}{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\\&+ \frac{1}{\Gamma (\alpha -\beta )}\left( \frac{1-\gamma }{\alpha -\beta -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (2-\beta )\Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma },\\ H_0= & {} 2\frac{|\xi |}{\eta }\Bigg (1+\frac{2}{\Gamma (2-\beta )}\Bigg )+ \Theta (M)\Bigg (2+\eta +\frac{2}{\Gamma (2-\beta )}\Bigg ). \end{aligned}$$

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

Proof

We complete the proof in several steps. Firstly, we show that the operator F defined by (3.1) maps bounded sets into bounded sets in the space X. For $r>0,$ we consider a ball $B_r=\{x: x\in X\,\, \text{ and }\,\, \Vert x\Vert _X\le r\}.$ In view of the inequalities $\displaystyle \frac{2(1-t)}{\eta (2-\eta )}<\frac{2}{\eta },$ $\displaystyle \frac{|2t-\eta |}{2-\eta }<2+\eta ,$ $\displaystyle \frac{2|t-1|}{\eta (2-\eta )}<\frac{2}{\eta }$ and for any $x\in B_r,$ we have

$$\begin{aligned}&|(Fx)(t)|\\\le & {} \int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}m(s)\left( \psi _1(|x(s)|)+\psi _2\left( \left| ^C_0D^{\beta }_tx(s)\right| \right) \right) ds+\frac{2}{\eta }|\xi |+(2+\eta ) \Theta (\Vert x\Vert )\\&+\frac{2}{\eta } \int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\left( \psi _1(|x(s)|)+\psi _2\left( \left| ^C_0D^{\beta }_tx(s)\right| \right) \right) ds\\\le & {} \int _0^1 \frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}m(s)\left( \psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X)\right) ds+\frac{2}{\eta }|\xi |+(2+\eta ) \Theta \big (\Vert x\Vert _X\big ) \\&+ \frac{2}{\eta }\int _{0}^{1}\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\left( \psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X)\right) ds\\\le & {} \frac{1}{\Gamma (\alpha )}\left( \int _0^1\left( (1-s)^{\alpha -1}\right) ^{\frac{1}{1-\gamma }}ds\right) ^{1-\gamma }\left( \int _0^1(m(s))^{1/{\gamma }}ds\right) ^{\gamma }\left( \psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X)\right) \\&+\frac{2}{\eta }|\xi |+(2+\eta ) \Theta \big (\Vert x\Vert _X\big )\\&+\frac{2}{\eta \Gamma (\alpha +1)}\left( \int _0^1\left( (1-s)^{\alpha }\right) ^{\frac{1}{1-\gamma }}ds\right) ^{1-\gamma }\left( \int _0^1(m(s))^{1/{\gamma }}ds\right) ^{\gamma }\left( \psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X)\right) \\\le & {} \Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma } +\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \} [\psi _1(r)+\psi _2(r)]\\&+\frac{2}{\eta }|\xi |+(2+\eta ) \Theta (r). \end{aligned}$$

Thus

$$\begin{aligned} \Vert Fx\Vert\le & {} \Bigg \{\frac{1}{\Gamma (\alpha )} \left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\Vert m\Vert [\psi _1(r)+\psi _2(r)]\\&+\frac{2}{\eta }|\xi |+(2+\eta ) \Theta (r). \end{aligned}$$

Using Definition 2.2 with

$$\begin{aligned} (Fx)'(t)= & {} \int _{0}^{t}\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha -1)}f(s, x(s), D^{\beta }x(s))ds-\frac{2}{\eta (2-\eta )}\xi + \frac{2}{2-\eta } h(x)\\&+\frac{2}{\eta (2-\eta )} \int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}f(s, x(s), D^{\beta }x(s))ds, \end{aligned}$$

we get

$$\begin{aligned} ^C_0D^{\beta }_t(Fx)(t)= & {} \int _0^t\frac{(t-s)^{-\beta }}{\Gamma (1-\beta )}(Fx)'(s)ds\\= & {} \int _0^t\frac{(t-s)^{-\beta }}{\Gamma (1-\beta )}\Bigg (\int _0^s\frac{(s-\tau )^{\alpha -2}}{\Gamma (\alpha -1)}f(\tau , x(\tau ), ^C_0D^{\beta }_tx(\tau ))d\tau \\&-\frac{2}{\eta (2-\eta )}\xi + \frac{2}{2-\eta } h(x)+\frac{2}{\eta (2-\eta )}\int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}f(s, x(s), ^C_0D^{\beta }_tx(s))ds\Bigg )ds\\= & {} \int _0^t\frac{(t-s)^{\alpha -1-\beta }}{\Gamma (\alpha -\beta )}f(s,x(s),^C_0D^{\beta }_tx(s))ds+\frac{ t^{1-\beta }}{\Gamma (2-\beta )}\frac{2}{2-\eta }\left( h(x)-\frac{\xi }{\eta }\right) \\&+\frac{2}{\eta (2-\eta )}\frac{t^{1-\beta }}{\Gamma (2-\beta )}\int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}f(s,x(s),^C_0D^{\beta }_tx(s))ds, \end{aligned}$$

and therefore

$$\begin{aligned}&\left| ^C_0D^{\beta }_t(Fx)(t)\right| \\\le & {} \int _0^t \frac{(t-s)^{\alpha -1-\beta }}{\Gamma (\alpha -\beta )} m(s)\left( \psi _1(|x(s)|)+\psi _2\left( \left| ^C_0D^{\beta }_tx(s)\right| \right) \right) ds\\&+\frac{2t^{1-\beta }}{(2-\eta )\Gamma (2-\beta )}\left( |h(x)|+ \frac{|\xi |}{\eta }\right) \\&+\frac{2t^{1-\beta }}{\eta (2-\eta )\Gamma (2-\beta )} \int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s) \left( \psi _1(|x(s)|)+\psi _2\left( \left| ^C_0D^{\beta }_tx(s)\right| \right) \right) ds\\\le & {} \int _0^t\frac{(t-s)^{\alpha -1-\beta }}{\Gamma (\alpha -\beta )}m(s) (\psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X))ds+\frac{2}{\Gamma (2-\beta )}\Bigg (\frac{|\xi |}{\eta }+ \Theta \big (\Vert x\Vert _X\big )\Bigg )\\&+\frac{2}{\eta \Gamma (2-\beta )}\int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\left( \psi _1(\Vert x\Vert _X)+\psi _2(\Vert x\Vert _X)\right) ds\\\le & {} \Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha -\beta )}\left( \frac{1-\gamma }{\alpha -\beta -\gamma }\right) ^{1-\gamma } +\frac{2\Vert m\Vert }{\eta \Gamma (2-\beta )\Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\\&\times [\psi _1(r)+\psi _2(r)] +\frac{2}{\Gamma (2-\beta )}\Bigg (\frac{|\xi |}{\eta }+ \Theta (r)\Bigg ). \end{aligned}$$

Consequently, we get

$$\begin{aligned}&\Big \Vert ^C_0D^{\beta }_tFx\Big \Vert \\ {}\le & {} \Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha -\beta )}\left( \frac{1-\gamma }{\alpha -\beta -\gamma }\right) ^{1-\gamma } +\frac{2\Vert m\Vert }{\eta \Gamma (2-\beta )\Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\times \\&\times [\psi _1(r)+\psi _2(r)] +\frac{2}{\Gamma (2-\beta )}\Bigg (\frac{|\xi |}{\eta }+ \Theta (r)\Bigg ). \end{aligned}$$

Thus it follows that

$$\begin{aligned} \Vert Fx\Vert _X\le & {} \Bigg \{\frac{1}{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \\&+\,\, \frac{1}{\Gamma (\alpha -\beta )}\left( \frac{1-\gamma }{\alpha -\beta -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (2-\beta )\Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\! \Bigg \}\Vert m\Vert \\&\times \,\, [\psi _1(r)+\psi _2(r)]+2\frac{|\xi |}{\eta } \Bigg (1+\frac{2}{\Gamma (2-\beta )}\Bigg )+ \Theta (r) \Bigg (2+\eta +\frac{2}{\Gamma (2-\beta )}\Bigg )\\= & {} [\psi _1(r)+\psi _2(r)]\Vert m\Vert \, H+H_0. \end{aligned}$$

Next we show that the operator F is equicontinuous on bounded sets of X. Observe that F is continuous in view of continuity of f and h. Further, we have that $|f(t,x(t), ^C_0D^{\beta }_tx(t))|\le N$ for any $x\in B_r$ and $t\in [0,1]$ and $|h(x)|\le N_1, \forall x\in B_r.$

Now let $0\le t_1<t_2\le 1.$ Then

$$\begin{aligned}&|(Fx)(t_2)-(Fx)(t_1)|\\\le & {} \Bigg |\frac{1}{\Gamma (\alpha )}\int _0^{t_1}\left[ (t_2-s)^{\alpha -1} -(t_1-s)^{\alpha -1}\right] f(s,x(s), ^C_0D^{\beta }_tx(s))ds\\&+ \frac{1}{\Gamma (\alpha )}\int _{t_1}^{t_2} (t_2-s)^{\alpha -1}f(s,x(s),^C_0D^{\beta }_tx(t))ds\Bigg |\\&+\frac{2|\xi |(t_2-t_1)}{\eta (2-\eta )}+\frac{2(t_2-t_1)}{2-\eta }|h(x)|\\&+\frac{2(t_2-t_1)}{\eta (2-\eta )}\int _{0}^{\eta } \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}\Big |f(s, x(s), ^C_0D^{\beta }_tx(s))\Big |ds\\\le & {} \frac{N}{\Gamma (\alpha +1)}(t_2^\alpha -t_1^\alpha )+ \frac{2|\xi |}{\eta }(t_2-t_1)+2N_1(t_2-t_1)+ \frac{2N(t_2-t_1)}{\eta \Gamma (\alpha +2)}, \end{aligned}$$

and

$$\begin{aligned}&\left| ^C_0D^{\beta }_t(Fx)(t_2)- ^C_0D^{\beta }_t(Fx)(t_1)\right| \\&\quad \le \frac{N}{\Gamma (\alpha -\beta +1)} \left( t_2^{\alpha -\beta -1}-t_1^{\alpha -\beta -1}\right) +\frac{2(|\xi |/\eta +N_1)\left( t_2^{1-\beta }-t_1^{1-\beta }\right) }{\Gamma (2-\beta )}\\&\qquad +\frac{2N}{\eta \Gamma (\alpha +2)\Gamma (2-\beta )} \left( t_2^{1-\beta }-t_1^{1-\beta }\right) . \end{aligned}$$

Clearly

$$\begin{aligned} \Vert (Fx)(t_2)-(Fx)(t_1)\Vert _X\rightarrow 0\,\,\, \text{ as }\,\,\, t_2\rightarrow t_1, \end{aligned}$$

and the limit is independent of $x\in B_r.$ Therefore the operator $F: B_r\rightarrow B_r$ is equicontinuous and uniformly bounded. Hence, the Arzelá-Ascoli theorem applies and thus $F(B_r)$ is relatively compact in X.

Finally, for $\lambda \in (0,1),$ let $x=\lambda Fx.$ Then, as in the first step, we can obtain

$$\begin{aligned} \frac{\Vert x\Vert _X}{\displaystyle [\psi _1(\Vert x\Vert _X)+ \psi _2(\Vert x\Vert _X)]\Vert m\Vert \, H+H_0}\le 1. \end{aligned}$$

By the condition $(A_3)$, there exists M such that $\Vert x\Vert \ne M.$ Let us set

$$\begin{aligned} U = \{x \in C([0,1], {\mathbb R}) : \Vert x\Vert _X< M \}. \end{aligned}$$

Note that the operator $F:\overline{U} \rightarrow C([0,1], \mathbb {R})$ is continuous and completely continuous. From the choice of U,  there is no $x \in \partial U$ such that $x =\lambda Fx$ for some $\lambda \in (0,1).$ Consequently, by the nonlinear alternative of Leray–Schauder type (Lemma 3.1), we deduce that F has a fixed point $x \in \overline{U}$ which is a solution of the problem (1.1). This completes the proof.$\square$

Remark 3.3

The conclusion of Theorem 3.2 remains valid if we replace $(A_2)$ by one of the following conditions:

${(A_{2})'}$ :

$h: C([0,1], {\mathbb R})\rightarrow {\mathbb R}$ is a continuous function and there exists a constant $M_1>0$ such that $|h(x)|\le M_1,~~ \forall x\in C([0,1], {\mathbb R}).$

${(A_{2})''}$ :

$h: C([0,1], {\mathbb R})\rightarrow {\mathbb R}$ is a continuous function and there exist constants $h_0, h_1>0$ such that $|h(x)|\le h_0+h_1\Vert x\Vert , ~~\forall x\in C([0,1], {\mathbb R}).$

3.2 Existence result via Krasnoselskii’s fixed point theorem

Lemma 3.4

(Krasnoselskii’s fixed point theorem [31]). 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 3.5

Let $h: C([0,1], {\mathbb R})\rightarrow {\mathbb R}$ be a continuous function satisfying $(A_2)'.$ Moreover we assume that:

$(B_1)$ :

$f : [0,1]\times {\mathbb R} \rightarrow \mathbb R$ is a continuous function such that

$$\begin{aligned} |f(t,x_1,y_1)-f(t,x_2,y_2)|\le L(|x_1-x_2|+|y_1-y_2|), \end{aligned}$$

for $t\in [0,1], ~x_i,y_i\in {\mathbb R},~ i=1,2$ and $L>0$ is a constant;

$(B_{2})$ :

$|f(t,u,v)|\le \mu (t), ~~\forall (t,u,v) \in [0,1] \times {\mathbb R}\times {\mathbb R},$ and $\mu \in C\big ([0,1], {\mathbb R}^+\big );$

$(B_{3})$ :

there exists a constant $L_1>0$ such that

$$\begin{aligned} |h(x_1)-h(x_2)|\le L_1\Vert x-y\Vert , \end{aligned}$$

for each $x_1, x_2\in C([0,1], {\mathbb R}).$

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

$$\begin{aligned} \displaystyle (2+\eta )L_1+\frac{2L}{\eta \Gamma (\alpha +2)}+ \frac{2}{\eta \Gamma (2-\beta )}\Bigg (L_1+\frac{L}{\eta \Gamma (\alpha +2)} \Bigg )<1. \end{aligned}$$

(3.2)

Proof

Letting $\sup _{t \in [0,T]}|\mu (t)|=\Vert \mu \Vert ,$ we fix

$$\begin{aligned} \overline{r}\ge & {} \Vert \mu \Vert \Bigg \{\frac{1}{\Gamma (\alpha +1)} +\frac{2}{\eta \Gamma (\alpha +2)}+\frac{1}{\Gamma (\alpha -\beta +1)} +\frac{2}{\eta \Gamma (2-\beta )\Gamma (\alpha +2)}\Bigg \}\\&+\frac{2|\xi |}{\eta }\Bigg (1+\frac{1}{\Gamma (2-\beta )}\Bigg ) +\Bigg (2+\eta + \frac{2}{\Gamma (2-\beta )}\Big )M_1, \end{aligned}$$

and consider $B_{\overline{r}}=\{x \in X: \Vert x\Vert _X\le \overline{r} \}.$ We define the operators $\mathcal{{P}}$ and $\mathcal{{Q}}$ on $B_{\overline{r}}$ as

$$\begin{aligned} (\mathcal {P}x)(t)= & {} \displaystyle \int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}f(s,x(s), ^C_0D^{\beta }_tx(s))ds, \\ (\mathcal {Q}x)(t)= & {} \frac{2(1-t)}{\eta (2-\eta )}\xi +\frac{(2t-\eta )}{2-\eta }h(x) +\frac{2(t-1)}{\eta (2-\eta )}\int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}f(s, x(s), ^C_0D^{\beta }_tx(s))ds. \end{aligned}$$

For $x, y \in B_{\overline{r}},$ we find that

$$\begin{aligned} \Vert {\mathcal {P}} x+{\mathcal {Q}} y\Vert\le & {} \int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\mu (s)ds+ \frac{2|\xi |}{\eta }+(2+\eta )M_1 +\frac{2}{\eta } \int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}\mu (s)ds\\\le & {} \Vert \mu \Vert \Bigg \{\frac{1}{\Gamma (\alpha +1)}+ \frac{2}{\eta \Gamma (\alpha +2)}\Bigg \}+\frac{2|\xi |}{\eta }+(2+\eta )M_1, \end{aligned}$$

and

$$\begin{aligned}&\left\| ^C_0D^{\beta }_t{\mathcal {P}} x+^C_0D^{\beta }_t{\mathcal {Q}} y\right\| \\&\quad \le \int _0^t\frac{(t-s)^{\alpha -1-\beta }}{\Gamma (\alpha -\beta )} \mu (s)ds+\frac{2}{\Gamma (2-\beta )}\Bigg (\frac{|\xi |}{\eta }+ M_1\Bigg ) + \frac{2}{\eta } \int _0^1\frac{(1-s)^{\alpha }}{\Gamma (2-\beta )\Gamma (\alpha +1)} \mu (s)ds\\&\quad \le \Vert \mu \Vert \Bigg \{\frac{1}{\Gamma (\alpha -\beta +1)}+ \frac{2}{\eta \Gamma (2-\beta )\Gamma (\alpha +2)}\Bigg \}+ \frac{2}{\Gamma (2-\beta )}\Bigg (\frac{|\xi |}{\eta }+ M_1\Bigg ) . \end{aligned}$$

Thus, it follows from the above inequalities that

$$\begin{aligned} \Vert {\mathcal {P}} x+{\mathcal {Q}} y\Vert _X= \Vert {\mathcal {P}} x+ {\mathcal {Q}} y\Vert +\left\| ^C_0D^{\beta }_t{\mathcal {P}} x +^C_0D^{\beta }_t{\mathcal {Q}} y\right\| \le \overline{r}, \end{aligned}$$

which means that ${\mathcal {P}} x+{\mathcal {Q}} y \in B_{\overline{r}}.$

Next we will show that ${\mathcal {Q}}$ is a contraction mapping. Let $x_1, x_2\in X.$ Then

$$\begin{aligned}&|{\mathcal {Q}}(x_1)(t)-{\mathcal {Q}}(x_2)(t)|\\\le & {} (2+\eta )|h(x_1)-h(x_2)|\\&+\frac{2}{\eta } \int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}|f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))|ds\\\le & {} (2+\eta ) L_1\Vert x_1-x_2\Vert +\frac{2}{\eta }\int _0^1 \frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}L\Big (|x_1(s)-x_2(s)| +\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds\\\le & {} (2+\eta ) L_1\Vert x_1-x_2\Vert +\frac{2}{\eta } \int _0^1 \frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}L\Big (\Vert x_1-x_2\Vert +\left\| ^C_0D^{\beta }_tx_1-^C_0D^{\beta }_tx_2\right\| \Big )ds\\\le & {} (2+\eta ) L_1\Vert x_1-x_2\Vert _X+\frac{2L}{\eta \Gamma (\alpha +2)}\Vert x_1-x_2\Vert _X, \end{aligned}$$

and thus

$$\begin{aligned} \Vert {\mathcal {Q}}(x_1)-{\mathcal {Q}}(x_2)\Vert \le \Bigg \{(2+\eta )L_1+\frac{2L}{\eta \Gamma (\alpha +2)}\Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

Also we have

$$\begin{aligned}&|{\mathcal {Q}}(x_1)'(t)-{\mathcal {Q}}(x_2)'(t)|\\&\quad \le 2| h(x_1)-h(x_2)| +\frac{2}{\eta } \int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}|f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))|ds\\&\quad \le 2L_1\Vert x_1-x_2\Vert + \frac{2}{\eta } \int _{0}^{1}\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}L\Big (|x_1(s)-x_2(s)|+\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds\\&\quad \le \Bigg \{2L_1+\frac{2L}{\eta \Gamma (\alpha +2)}\Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

Hence we obtain

$$\begin{aligned} \left\| ^C_0D^{\beta }_t({\mathcal {Q}}x_1)-^C_0D^{\beta }_t ({\mathcal {Q}}x_2)\right\|\le & {} \int _0^t\frac{(t-s)^{-\beta }}{\Gamma (1-\beta )}|({\mathcal {Q}}x_1)'(s)-({\mathcal {Q}}x_2)'(s)|ds\\\le & {} \frac{1}{\Gamma (2-\beta )}\Bigg \{2L_1+ \frac{2L}{\eta \Gamma (\alpha +2)}\Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

Consequently we have

$$\begin{aligned}&\Vert {\mathcal {Q}}(x_1)-{\mathcal {Q}}(x_2)\Vert _X\\&\quad \le \Bigg \{\displaystyle (2+\eta )L_1+\frac{2L}{\eta \Gamma (\alpha +2)}+\frac{2}{\Gamma (2-\beta )}\Bigg (L_1+\frac{L}{\eta \Gamma (\alpha +2)}\Bigg )\Bigg \}\Vert x_1-x_2\Vert _X, \end{aligned}$$

which, in view of (3.2)), implies that ${\mathcal {Q}}$ is a contraction.

Continuity of f implies that the operator ${\mathcal {P}}$ is continuous. Also, ${\mathcal {P}}$ is uniformly bounded on $B_{\overline{r}}$ as

$$\begin{aligned} \Vert {\mathcal {P}} x\Vert _X \le \Bigg \{ \frac{1}{\Gamma (\alpha +1)}+\frac{1}{\Gamma (\alpha -\beta +1)}\Bigg \}\Vert \mu \Vert . \end{aligned}$$

Now we establish the compactness of the operator ${\mathcal {P}}.$ Letting $\max _{t \in [0,1]} |f(t,x(t),$ $^C_0D^{\beta }_tx(t))|=\overline{f}< \infty$ and $0 < t_1 < t_2 <1,$ we obtain

$$\begin{aligned} |({\mathcal {P}} x)(t_2)-({\mathcal {P}} x)(t_1)|\le & {} \left| \int _0^{t_1}\frac{\Big [(t_2-s)^{\alpha -1}-(t_1-s)^{\alpha -1}\Big ]}{\Gamma (\alpha )}f(s,x(s), ^C_0D^{\beta }_tx(t))ds\right| \\&+\left| \int _{t_1}^{t_2}\frac{(t_2-s)^{\alpha -1}}{\Gamma (\alpha )}f\left( s,x(s), ^C_0D^{\beta }_tx(t)\right) ds\right| \\\le & {} \frac{\overline{f}\Big [2(t_2-t_1)^\alpha +(t_2^\alpha -t_1^\alpha )\Big ]}{\Gamma (\alpha +1)}, \end{aligned}$$

and

$$\begin{aligned} \left| ^C_0D^{\beta }_t({\mathcal {P}} x)(t_2)-^C_0D^{\beta }_t({\mathcal {P}} x)(t_1)\right| \le \frac{\overline{f}\Big [2(t_2-t_1)^{\alpha -\beta } +\left( t_2^{\alpha -\beta }-t_1^{\alpha -\beta }\right) \Big ]}{\Gamma ({\alpha -\beta }+1)}, \end{aligned}$$

which are independent of x. Hence

$$\begin{aligned} \Vert ({\mathcal {P}} x)(t_2)-({\mathcal {P}} x)(t_1)\Vert _X\rightarrow 0~~\text{ as }~~ t_2\rightarrow t_1. \end{aligned}$$

Thus, ${\mathcal {P}}$ is equicontinuous. So ${\mathcal {P}}$ is relatively compact on $B_{\overline{r}}.$ Hence, by the Arzelá-Ascoli theorem, ${\mathcal {P}}$ is compact on $B_{\overline{r}}.$ Thus all the assumptions of Lemma 3.4 are satisfied. So the conclusion of Lemma 3.4 implies that the boundary value problem (1.1) has at least one solution on [0, 1]. $\square$

Remark 3.6

In the above theorem we can interchange the roles of the operators ${\mathcal {P}}$ and ${\mathcal {Q}}$ to obtain a second result. In this case, the condition (3.2) is replaced with the following one:

$$\begin{aligned} L\Bigg \{\frac{1}{\Gamma (\alpha +1)}+\frac{1}{\Gamma (\alpha -\beta +1)}\Bigg \}<1. \end{aligned}$$

3.3 Existence result via Banach’s fixed point theorem

Theorem 3.7

Let $h: C([0,1], {\mathbb R})\rightarrow {\mathbb R}$ be a continuous function satisfying $(B_3).$ In addition we assume that:

$(C_1)$ :

$f : [0,1]\times {\mathbb R} \times {\mathbb R}\rightarrow \mathbb R$ is a continuous function and there exists $m\in L^{\frac{1}{\gamma }}\left( [0,1],{\mathbb R}^+\right) ,$ $\gamma \in (0,1/(\alpha -1))$ such that

$$\begin{aligned} |f(t,x_1,y_1)-f(t,x_2,y_2)|\le m(t)(|x_1-x_2|+|y_1-y_2|), \end{aligned}$$

for $t\in [0,1], ~x_i,y_i\in {\mathbb R},~ i=1,2.$

If

$$\begin{aligned} Z:= & {} \Vert m\Vert \Bigg \{\frac{1}{\Gamma (\alpha )} \left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \\&+\frac{1}{\Gamma (2-\beta )}\Bigg \{\frac{1}{\Gamma (\alpha -1)}\left( \frac{1-\gamma }{\alpha -\gamma -1} \right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\\&+L_1\Bigg (2+\eta +\frac{2}{\Gamma (2-\beta )}\Bigg )<1, \end{aligned}$$

with $\displaystyle \Vert m\Vert =\left( \int _0^1|m(s)|^{\frac{1}{\gamma }}ds\right) ^{\gamma },$ then the boundary value problem (1.1) has a unique solution on [0, 1].

Proof

In order to show that the operator $F: X \rightarrow X$ defined by (3.1) is a contraction, let $x_1,x_2\in X.$ Then, for each $t\in [0,1],$ we have the following estimate by using Hölder inequality:

$$\begin{aligned}&|(Fx_1)(t)-(Fx_2)(t)|\\\le & {} \int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\left| f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))\right| ds\\&+(2+\eta )|h(x_1)-h(x_2)|\\&+\frac{2}{\eta } \int _0^\eta \frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}\left| f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))\right| ds\\\le & {} \int _0^t\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}m(s)\Big (|x_1(s)-x_2(s)|+\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds\\&+(2+\eta )L_1\Vert x_1-x_2\Vert \\&+\frac{2}{\eta }\int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\Big (|x_1(s)-x_2(s)|+\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds\\\le & {} \int _0^1\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}m(s)\Big (\Vert x_1 -x_2\Vert +\left\| ^C_0D^{\beta }_tx_1-^C_0D^{\beta }_tx_2\right\| \Big )ds+(2+\eta )L_1\Vert x_1-x_2\Vert \\&+\frac{2}{\eta }\int _0^1\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\Big (\Vert x_1-x_2\Vert +\left\| ^C_0D^{\beta }_tx_1-^C_0D^{\beta }_tx_2\right\| \Big )ds\\\le & {} \frac{\Vert x_1-x_2\Vert _X}{\Gamma (\alpha )}\left( \int _0^1\left( (1-s)^{\alpha -1}\right) ^{\frac{1}{1-\gamma }}ds\right) ^{1-\gamma }\left( \int _0^1(m(s))^{1/{\gamma }}ds\right) ^{\gamma }\\&+(2+\eta )L_1\Vert x_1-x_2\Vert _X\\&+ \frac{2\Vert x_1-x_2\Vert _X}{\eta \Gamma (\alpha +1)}\left( \int _0^1\left( (1-s)^{\alpha }\right) ^{\frac{1}{1-\gamma }}ds\right) ^{1-\gamma }\left( \int _0^1(m(s))^{1/{\gamma }}ds\right) ^{\gamma }\\\le & {} \frac{\Vert m\Vert ~\Vert x_1-x_2\Vert _X}{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma }+(2+\eta )L_1\Vert x_1-x_2\Vert _X\\&+ \frac{2\Vert m\Vert ~\Vert x_1-x_2\Vert _X}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\\= & {} \Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma }\right) ^{1-\gamma }+(2+\eta )L_1+\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

Also we have

$$\begin{aligned}&|(Fx_1)'(t)-(Fx_2)'(t)|\\\le & {} \int _{0}^{t}\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha -1)}|f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))|ds+ 2| h(x_1)-h(x_2)|\\&+\frac{2}{\eta } \int _{0}^{\eta }\frac{(\eta -s)^{\alpha }}{\Gamma (\alpha +1)}|f(s, x_1(s), ^C_0D^{\beta }_tx_1(s))-f(s,x_2(s), ^C_0D^{\beta }_tx_2(s))|ds\\\le & {} \int _{0}^{t}\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha -1)}m(s)\Big (|x_1(s)-x_2(s)|+\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds+2L_1\Vert x_1-x_2\Vert \\&+ \frac{2}{\eta }\int _{0}^{1}\frac{(1-s)^{\alpha }}{\Gamma (\alpha +1)}m(s)\Big (|x_1(s)-x_2(s)|+\left| ^C_0D^{\beta }_tx_1(s)-^C_0D^{\beta }_tx_2(s)\right| \Big )ds\\\le & {} \frac{\Vert m\Vert ~\Vert x_1-x_2\Vert _X}{\Gamma (\alpha -1)}\left( \frac{1-\gamma }{\alpha -\gamma -1}\right) ^{1-\gamma }+2L_1\Vert x_1-x_2\Vert _X\\&+ \frac{2\Vert m\Vert ~\Vert x_1-x_2\Vert _X}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\\= & {} \Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha -1)}\left( \frac{1-\gamma }{\alpha -\gamma -1}\right) ^{1-\gamma }+2L_1+\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

Hence we obtain

$$\begin{aligned} \left| ^C_0D^{\beta }_t(Fx_1)(t)-^C_0D^{\beta }_t(Fx_2)(t)\right|\le & {} \int _0^t\frac{(t-s)^{-\beta }}{\Gamma (1-\beta )}|(Fx_1)'(s)-(Fx_2)'(s)|ds\\\le & {} \frac{1}{\Gamma (2-\beta )}\Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha -1)}\left( \frac{1-\gamma }{\alpha -\gamma -1}\right) ^{1-\gamma }\\&+2L_1+\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\Vert x_1-x_2\Vert _X. \end{aligned}$$

From the above inequalities, it follows that

$$\begin{aligned} \Vert Fx_1-Fx_2\Vert _X\le & {} \Bigg [ \frac{\Vert m\Vert }{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma } \right) ^{1-\gamma }+(2+\eta )L_1+\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\\&+\frac{1}{\Gamma (2-\beta )}\Bigg \{\frac{\Vert m\Vert }{\Gamma (\alpha -1)} \left( \frac{1-\gamma }{\alpha -\gamma -1}\right) ^{1-\gamma }\\&+2L_1+\frac{2\Vert m\Vert }{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \} \Bigg ]\Vert x_1-x_2\Vert _X\\= & {} Z\Vert x-y\Vert _X. \end{aligned}$$

Since $Z<1,$ the operator F is a contraction. Hence the conclusion of Banach’s fixed point theorem implies that the operator F has a unique fixed point which corresponds to a unique solution of problem (1.1). This completes the proof. $\square$

Remark 3.8

By taking the limit $\eta \rightarrow 1^{-},$ we obtain the new results for the problem considered in [20].

Remark 3.9

For $\xi =\eta ,$ our results correspond to the boundary conditions with a unit average value of the unknown function over an interval of arbitrary size, that is, $\displaystyle \frac{1}{\eta }\int _0^\eta x(t)dt=1, ~0< \eta <1.$

4 Examples

Example 4.1

Consider the problem

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle ^C_0D^{7/2}_tx(t)=\frac{e^{-t}}{\sqrt{4+t^2}}\left( \frac{2}{7}\sin x(t)+\frac{1}{2 \pi } \arctan \left( ^C_0D^{1/2}_tx(t)\right) +\frac{1}{4}\right) , ~ 0<t<1,\\ \displaystyle x(0)+x'(0)=\frac{2}{25}x\Big (\frac{1}{4}\Big ), ~~ \int _0^{1/2}x(s)ds=\frac{1}{6},\\ x''(0)=x'''(0)=0. \end{array}\right. \end{aligned}$$

(4.1)

Here $\alpha =7/2, ~n=4, ~\beta =1/2, ~\eta =1/2, ~\gamma =1/4 \in (0, 2/5), ~\xi =1/6.$

Observe that

$$\begin{aligned} |f(t,x,^C_0D^{1/2}_tx(t))|= & {} \left| \frac{e^{-t}}{\sqrt{4+t^2}} \left( \frac{2}{7}\sin x(t)+\frac{1}{2 \pi } \arctan \left( ^C_0D^{1/2}_tx(t)\right) +\frac{1}{4}\right) \right| \\\le & {} m(t)(\psi _1(|x|)+\psi _2\left( \left| ^C_0D^{1/2}_tx(t)\right| \right) ), \end{aligned}$$

with $\displaystyle m(t)= \frac{e^{-t}}{2}, ~\psi _1(|x|)=\frac{2}{7}|x|, ~\psi _2\left( \left| ^C_0D^{1/2}_tx(t)\right| \right) )=\frac{1}{2}.$ Further,

$$\begin{aligned} H= & {} \frac{1}{\Gamma (\alpha )}\left( \frac{1-\gamma }{\alpha -\gamma } \right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma }\\&+\, \frac{1}{\Gamma (\alpha -\beta )}\left( \frac{1-\gamma }{\alpha -\beta -\gamma }\right) ^{1-\gamma } \quad +\frac{2}{\eta \Gamma (2-\beta )\Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \\\simeq & {} 0.4881660854,\\ H_0= & {} 2\frac{|\xi |}{\eta }\Big (1+\frac{2}{\Gamma (2-\beta )}\Big )+ \Theta (M)\left( 2+\eta +\frac{2}{\Gamma (2-\beta )}\right) \\\simeq & {} 2.170880207+0.3805056249M. \end{aligned}$$

Using the above values in the condition $(A_3),$ it is found that $M>M_1\simeq 3.956418218.$ Thus all the conditions of Theorem 3.2 are satisfied. In consequence, the conclusion of Theorem 3.2 implies that the problem (4.1) has a solution.

Example 4.2

Consider the fractional boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle ^C_0D^{13/2}_tx(t)=\frac{a e^{-t^2}}{\sqrt{4+t}}\left( \arctan (x(t))+\frac{\left| ^C_0D^{1/2}_tx(t)\right| }{1+\left| ^C_0D^{1/2}_tx(t)\right| }\right) +\frac{1}{3}, ~ 0<t<1,\\ \displaystyle x(0)+x'(0)=\frac{1}{15}\frac{|x|}{(1+|x|)}, ~~ \int _0^{1/2}x(s)ds=\frac{1}{6},\\ x^{(i)}(0)=0, ~i=2, 3, 4, 5, 6, \end{array}\right. \end{aligned}$$

(4.2)

where $\alpha =13/2, ~n=7, ~\beta =3/4, ~\eta =1/2, ~\xi =1/6,$ and a is a real constant to be determined later. By the conditions $(B_1)$ and $(B_3),$ of Theorem 3.5, we find that $L=a$ and $L_1=1/15$ respectively. Using the given data, we find that the condition (3.2) holds for $0< a < [25 \Gamma (5/4)-8]\Gamma (17/2)/120.$ Thus, by the conclusion of Theorem 3.5, there exists a solution for the problem (4.2) on [0, 1].

Example 4.3

Let us consider the problem (4.1).

Using the given values and $L_1=2/25,$ we find that

$$\begin{aligned} Z:= & {} \Vert m\Vert \Bigg \{\frac{1}{\Gamma (\alpha )} \left( \frac{1-\gamma }{\alpha -\gamma } \right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)} \left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \\&+\frac{1}{\Gamma (2-\beta )}\Bigg \{\frac{1}{\Gamma (\alpha -1)} \left( \frac{1-\gamma }{\alpha -\gamma -1}\right) ^{1-\gamma } +\frac{2}{\eta \Gamma (\alpha +1)}\left( \frac{1-\gamma }{\alpha +1-\gamma }\right) ^{1-\gamma } \Bigg \}\\&+L_1\Bigg (2+\eta +\frac{2}{\Gamma (2-\beta )}\Bigg ) \simeq 0.6169772240<1. \end{aligned}$$

Clearly the hypothesis of Theorem 3.7 is satisfied. Thus the problem (4.1) has a unique solution on [0, 1].