Boundary value problems for Caputo fractional differential equations with nonlocal and fractional integral boundary conditions

Abstract

In this paper, we study the existence and uniqueness of solutions for fractional differential equations with nonlocal and fractional integral boundary conditions. New existence and uniqueness results are established using the Banach contraction principle. Other existence results are obtained using O’Regan fixed point theorem and Burton and Kirk fixed point. In addition, an example is given to demonstrate the application of our main results.

1 Introduction

Fractional differential equations have gained much importance and attention due to the fact that they have been proved to be valuable tools in the modeling of many phenomena in engineering and sciences such as physics, mechanics, economics, biology, etc. (see, for example, [16, 19, 23, 24] and references therein). Actually, the concepts of fractional derivatives are not the only generalization of the ordinary derivatives, but also it has been found that they can efficiently and properly describe the behavior of many physical systems (real-life phenomena) more accurately than integer-order derivatives.

Different kind of fixed point theorems are widely used as fundamental tools to prove the existence and uniqueness of solutions for various classes of fractional differential equations; for example, we refer the reader to [1,2,3,4, 9, 12, 14, 15, 17, 18, 21, 25, 27] and the references cited therein. Moreover, some mathematicians considered Caputo fractional differential equations with a nonlinear term depending on the Caputo derivative (see Benchohra and Souid [5], Benchohra and Lazreg [6], Benchohra et al. [7], El-Sayed and Bin-Taher [10, 11], Guezane-Lakoud and Khaldi [12], Guezane-Lakoud and Bensebaa [13], Houas and Benbachir [14], Nieto et al. [20], and Yan et al. [27]).

Motivated by the above papers, in this paper, we establish various existence and uniqueness results of solutions for a boundary value problem of nonlinear fractional differential equations of order $\alpha \in (2,3]$ with nonlocal and fractional integral boundary conditions given by:

$$\begin{aligned} {\left\{ \begin{array}{ll} {^{c}\!D}_{0^+}^\alpha u(t)=f(t, u(t), {^{c}\!D}_{0^+}^{\beta }u(t)), t\in J:=[0,1] \\ u(0)=g(u), \\ u'(0)=a I_{0^+}^{\sigma _1}u(\eta _1),&{}\;\quad 0<\eta _1<1 \\ {^{c}\!D}_{0^+}^{\beta _1} u (1)=b I_{0^+}^{\sigma _2}u(\eta _2), &{}\;\quad 0<\eta _2<1, \end{array}\right. } \end{aligned}$$

(1)

where ${^{c}\!D}_{0^+}^\kappa$ is the Caputo fractional derivative of order $\kappa \in \{\alpha ,\beta , \beta _1\}$, such that $2<\alpha \le 3$, and $0<\beta , \beta _1 \le 1$, and $f: [0,1]\times \mathbb {R}^{2} \longrightarrow \mathbb {R}$ and $g : C( [0,1], \mathbb {R}) \longrightarrow \mathbb {R}$ are continuous functions, $I_{0^+}^{\sigma _i}$ is the Riemann–Liouville fractional integral of order $\sigma _i > 0, i = 1, 2, \eta _1, \eta _2, a, b$ are suitably chosen real constants, such that:

$$\begin{aligned} \varDelta&=\left[ \left( \frac{a\eta _1^{\sigma _1+1}}{\varGamma (\sigma _1+2)} -1 \right) \left( \frac{b\eta _2^{\sigma _2+2}}{\varGamma (\sigma _2+3)}-\frac{1}{\varGamma (3-\beta _1)}\right) \right. \nonumber \\&\quad \quad \left. -\left( \frac{b\eta _2^{\sigma _2+1}}{\varGamma (\sigma _2+2)}- \frac{1}{\varGamma (2-\beta _1)}\right) \left( \frac{a\eta _1^{\sigma _1+2}}{\varGamma (\sigma _1+3)}\right) \right] \ne 0. \end{aligned}$$

(2)

The paper is organized as follows. In Sect. 2, we collect some background material and basic results about fractional calculus that will be used later. In Sect. 3, we establish the existence of solutions for the boundary value problem (1) using Banach contraction principle, O’Regan fixed point theorem, and Burton and Kirk fixed point. In the fourth section, we give an example to demonstrate the application of our main results. Finally, we add a conclusion to the paper in the last section.

2 Preliminaries

In this section, we will recall some notions and definition which we will need in the sequel (details can be found, e.g., in [16, 23, 24]).

Definition 2.1

( [16]) The Riemann–Liouville fractional integral of order $\alpha >0$ of a function $u \in L^1([0, 1])$ is defined by:

$$\begin{aligned} I^{\alpha }_{0^+}u(t)=\frac{1}{\varGamma (\alpha )}\int _{0}^{t}(t-s)^{\alpha -1}u(s) \mathrm {d}s, \quad (t>0,\;\alpha >0). \end{aligned}$$

where $\varGamma (\cdot )$ is the (Euler’s) Gamma function $\varGamma (\alpha )=\int _{0}^{+\infty }e^{-t}t^{\alpha -1}\mathrm {d}t,\quad \alpha > 0.$ Moreover, for $\alpha = 0$, we set $I^{0}_{0^+}u := u$.

Definition 2.2

( [16, 23]) The Caputo fractional derivative of order $\alpha$ of a function $u \in AC^{n}([0, 1])$ is represented by:

$$\begin{aligned} {^{c}\!D}_{0^+}^{\alpha }u(t)= {\left\{ \begin{array}{ll} \frac{1}{\varGamma (n-\alpha )}\int _{0}^{t}(t-s)^{n-\alpha -1}u^{(n)}(s)\mathrm {d}s,&{}\quad \text {if}\; \alpha \notin \mathbb {N}, \nonumber \\ \\ u^{(n)}(t),&{}\quad \quad \text {if}\;\alpha \in \mathbb {N}, \end{array}\right. } \end{aligned}$$

where $u^{(n)}(t)=\frac{{\mathrm d}^n u(t)}{{\mathrm d}t^n}, \alpha >0, n=[\alpha ]+1$ and $[\alpha ]$ denotes the integer part of the real number $\alpha$.

Lemma 2.3

Let $\alpha , \beta >0,\, n=[\alpha ]+1$, and then, the following relation holds:

$$\begin{aligned} {^{c}\!D}_{0^+}^{\alpha }t^\beta = {\left\{ \begin{array}{ll} \frac{\varGamma (\beta +1)}{\varGamma (\beta -\alpha +1)}t^{\beta -\alpha }, &{}\quad (\beta \in \mathbb {N} \;\text {and}\;\beta \ge n \;\text {or}\; \beta \notin \mathbb {N} \;\text {and}\; \beta > n-1),\\ \\ 0, &{}\quad \quad \beta \in \{0, \dots , n-1\}. \end{array}\right. } \end{aligned}$$

Lemma 2.4

Let $\alpha>\beta >0$, and $u \in L^1([0, 1])$. Then, we have:

• (1) $I_{0^+}^{\alpha } I_{0^+}^{\beta }u(t)= I_{0^+}^{\alpha +\beta }u(t)$,

• (2) ${^{c}\!}D_{0^+}^{\alpha }I_{0^+}^{\alpha }u(t)=u(t)$,

• (3) ${^{c}\!}D_{0^+}^{\beta }I_{0^+}^{\alpha }u(t)=I_{0^+}^{\alpha - \beta }u(t)$.

Lemma 2.5

Let $\alpha >0$. Then, the differential equation

$$\begin{aligned} ({^{c}\!}D_{0^+}^{\alpha }u)(t)=0, \end{aligned}$$

has solutions

$$\begin{aligned} u(t)=\sum _{j=0}^{n-1}c_jt^j, \quad c_j \in \mathbb {R}, j=0\dots n-1, \end{aligned}$$

where $n-1<\alpha < n$

Lemma 2.6

Let $\alpha >0$. Then:

$$\begin{aligned} I_{0^+}^{\alpha }({^{c}}\!D_{0^+}^{\alpha }u(t))=u(t)+\sum _{j=0}^{n-1}c_jt^j, \;\; \end{aligned}$$

for some $c_j\in \mathbb {R}, j= 0, 1, 2,\dots , m -1$, where $n = [\alpha ] + 1$.

Lemma 2.7

Let $\alpha >0, u \in L^1([0, 1],\mathbb {R})$. Then, for all $t \in [0, 1]$, we have:

$$\begin{aligned} I_{0^+}^{{\alpha +1}}u(t) \le \Vert I_{0^+}^{{\alpha }}u\Vert _{L^1}. \end{aligned}$$

Proof

Let $u \in L^1([0, 1],\mathbb {R})$, and from Lemma 2.4, we have:

$$\begin{aligned} I_{0^+}^{\alpha +1}u(t)=I_{0^+}^{1} I_{0^+}^{\alpha }u(t)=\int _{0}^{t}I_{0^+}^{{\alpha }}u(s)ds\le \int _{0}^{1}|I_{0^+}^{{\alpha }}u(s)|\mathrm{d}s=\Vert I_{0^+}^{\alpha }u\Vert _{L^{1}}. \end{aligned}$$

$\square$

Lemma 2.8

The fractional integral $I_{0^+}^{{\alpha }}$, $\alpha >0$ is bounded in $L^1([0, 1],\mathbb {R})$ with

$$\begin{aligned} \Vert I_{0^+}^{{\alpha }}u\Vert _{L^1}\le \frac{\Vert u\Vert _{L^{1}}}{\varGamma (\alpha +1)}. \end{aligned}$$

Proof

Let $u \in L^1([0, 1],\mathbb {R})$, and then:

$$\begin{aligned} \Vert I_{0^+}^{\alpha }u\Vert _{L^{1}}&=\int _{0}^{1}|I_{0^+}^{{\alpha }}u(t)|\mathrm{d}t \le \frac{1}{\varGamma (\alpha )} \int _{0}^{1}\int _{0}^{t}(t-s)^{\alpha -1}|u(s)|\mathrm{d}s\mathrm{d}t\\&\le \frac{1}{\varGamma (\alpha )} \int _{0}^{1}|u(s)|\mathrm{d}s\int _{s}^{1}(t-s)^{\alpha -1}\mathrm{d}t\\&\le \frac{1}{\varGamma (\alpha +1)} \int _{0}^{1}(t-s)^{\alpha }|u(s)|\mathrm{d}s\\&\le \frac{\Vert u\Vert _{L^{1}}}{\varGamma (\alpha +1)}. \end{aligned}$$

$\square$

Let us now introduce the space $E=\big \{u :u \in C([0, 1], \mathbb {R}) : {^{c}\!D}_{0^+}^{\beta }u \in C([0, 1], \mathbb {R}) \big \}$ equipped with the norm:

$$\begin{aligned} \Vert u\Vert _{E}=\Vert u\Vert _\infty +\Vert ^{c}\!D_{0^+}^{\beta }u\Vert _\infty =\sup _{t \in J}|u(t)|+\sup _{t \in J}|{^{c}\!D}_{0^+}^{\beta }u(t)|. \end{aligned}$$

Clearly, $(E, \Vert \cdot \Vert _{E})$ is a Banach space [26].

3 Main results

For the existence of solutions for the problem (1), we need the following lemma:

Lemma 3.1

For a given $h\in C(J,\mathbb {R})$, the unique solution of the linear fractional boundary value problem:

$$\begin{aligned} {^{c}\!}D_{0^+}^\alpha u(t)=h(t), \; 2<\alpha \le 3, \end{aligned}$$

(3)

supplemented with boundary conditions

$$\begin{aligned} u(0)=g(u),\;\; u'(0)=aI_{0^+}^{\sigma _1}u(\eta _1),\;\;{^{c}\!}D_{0^+}^{\beta _1} u(1)= bI_{0^+}^{\sigma _2}u(\eta _2), \end{aligned}$$

(4)

is given by:

$$\begin{aligned} u(t)&=I_{0^+}^{\alpha }h(t)+(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}h(\eta _1)+ (v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}h(\eta _2)\nonumber \\&\quad \quad +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}h(1)+(1-v_3t-v_7t^2)g(u), \end{aligned}$$

(5)

where

$$\begin{aligned} v_1&=\frac{a}{\varDelta }\left( \frac{ b\eta _2^{\sigma _2+2}}{\varGamma (\sigma _2+3)}-\frac{1}{\varGamma (3-\beta _1)}\right) ,\;v_2=\frac{ab}{\varDelta }\left( \frac{\eta _1^{\sigma _1+2}}{\varGamma (\sigma _1+3)}\right) ,\nonumber \\ v_3&=\frac{\eta _1^{\sigma _1} v_1}{\varGamma (\sigma _1+1)}-\frac{\eta _2^{\sigma _2} v_2}{\varGamma (\sigma _2+1)},\;v_4= \frac{ v_2}{b},\;v_5=\frac{b}{2\varDelta }\left( \frac{a\eta _1^{\sigma _1+1}}{\varGamma (\sigma _1+2)} -1 \right) ,\\ v_6&=\frac{a}{2\varDelta }\left( \frac{b\eta _2^{\sigma _2+1}}{\varGamma (\sigma _2+2)}-\frac{1}{\varGamma (2-\beta _1)}\right) ,\;v_7=\frac{\eta _2^{\sigma _2}v_5}{\varGamma (\sigma _2+1)}-\frac{\eta _1^{\sigma _1} v_6}{\varGamma (\sigma _1+1)},\nonumber \\ v_8&=\frac{v_5}{b}\nonumber , \end{aligned}$$

(6)

and $\varDelta$ is defined by (2).

Proof

Applying the Riemann–Liouville fractional integral operator of order $\alpha$ to both sides of (3) and using Lemma 2.6, we have:

$$\begin{aligned} u(t) =I_{0^+}^{\alpha }h(t)-c_0-c_1t-c_2t^2,\quad c_0, c_1 ,c_2\in \mathbb {R}. \end{aligned}$$

(7)

Applying the boundary conditions (4) in (7), we find that:

$$\begin{aligned} c_0=-g(u). \end{aligned}$$

Using the boundary conditions of (4) in (7), we find a system of equations:

$$\begin{aligned} \left( \frac{a\eta _1^{\sigma _1+1}}{\varGamma (\sigma _1+2)} -1 \right) c_1 +2\left( \frac{a\eta _1^{\sigma _1+2}}{\varGamma (\sigma _1+3)}\right) c_2 = a I_{0^+}^{\alpha +\sigma _1}h(\eta _1)+\left( \frac{a\eta _1^{\sigma _1}}{\varGamma (\sigma _1+1)}\right) g(u). \end{aligned}$$

(8)

$$\begin{aligned}&\left( \frac{b\eta _2^{\sigma _2+1}}{\varGamma (\sigma _2+2)}-\frac{1}{\varGamma (2-\beta _1)}\right) c_1 +2\left( \frac{b\eta _2^{\sigma _2+2}}{\varGamma (\sigma _2+3)}-\frac{1}{\varGamma (3-\beta _1)}\right) c_2=b I_{0^+}^{\alpha +\sigma _2}h(\eta _2)\nonumber \\&\quad +\left( \frac{b\eta _2^{\sigma _2}}{\varGamma (\sigma _2+1)}\right) g(u)-I_{0^+} ^{\alpha -\beta _1}h(1). \end{aligned}$$

(9)

Solving the system of equations (8) and (9) together with notations (6), we find that:

$$\begin{aligned} c_1= & {} v_1I_{0^+}^{\alpha +\sigma _1}h(\eta _1)-v_2I_{0^+}^{\alpha +\sigma _2}h(\eta _2)+v_3g(u)+v_4I_{0^+} ^{\alpha -\beta _1}h(1),\\ c_2= & {} v_5I_{0^+}^{\alpha +\sigma _2}h(\eta _2)-v_6I_{0^+}^{\alpha +\sigma _1}h(\eta _1)+v_7g(u)-v_8I_{0^+} ^{\alpha -\beta _1}h(1). \end{aligned}$$

Substituting the values of $c_0, c_1, c_2$ in (7), we obtain the solution (5). This completes the proof. $\square$

For computational convenience, we introduce the notations:

$$\begin{aligned} I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)=\frac{1}{\varGamma (\alpha )}\int _{0}^{t}(t- s)^{\alpha -1}f(s, u(s), {^{c}}\!D_{0^+}^{\beta }u(s))\mathop {}\mathopen {}\mathrm {d}s,\; \forall t\in J. \end{aligned}$$

Define the integral operator $T: E \longrightarrow E$ by:

$$\begin{aligned} T(u)(t)&=I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t) \nonumber \\&\quad +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\nonumber \\&\quad +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\nonumber \\&\quad +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\nonumber \\&\quad +(1-v_3t-v_7t^2)g(u). \end{aligned}$$

(10)

Next, we introduce two operators $T_{1, 2}: E \longrightarrow E$ as follows:

$$\begin{aligned} T_1(u)(t)&=I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)\nonumber \\&\quad +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\nonumber \\&\quad +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\nonumber \\&\quad +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1), \end{aligned}$$

(11)

and

$$\begin{aligned} T_2u(t)=(1-v_3t-v_7t^2)g(u). \end{aligned}$$

(12)

Clearly:

$$\begin{aligned} Tu(t)= T_1u(t)+T_2u(t), t \in J. \end{aligned}$$

(13)

Remark 3.2

In our work, we assume that the restriction $\varDelta \ne 0$ is satisfied, because in the case when $\varDelta =0$, the problem (1) will be under the resonant conditions, and hence, the previously mentioned theorems are not applicable, and then, the problem can be managed using the method of Mawhin continuation theorem.

Lemma 3.3

Let $f \in C ([0, 1] \times \mathbb {R} \times \mathbb {R}, \mathbb R)$. Then, $u \in E$ is a solution of the fractional boundary value problem (1) if and only if $T u(t) = u(t), \forall t \in [0, 1]$.

Proof

Let u be a solution of (1). Then, using the same method as used in Lemma 3.1, we can prove that:

$$\begin{aligned} u(t)&= T(u)(t)=I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t) \nonumber \\&\quad +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\nonumber \\&\quad +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\nonumber \\&\quad +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\nonumber \\&\quad +(1-v_3t-v_7t^2)g(u). \end{aligned}$$

Conversely, u satisfies:

$$\begin{aligned} u(t)&=I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t) \nonumber \\&\quad +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\nonumber \\&\quad +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\nonumber \\&\quad +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\nonumber \\&\quad +(1-v_3t-v_7t^2)g(u), \end{aligned}$$

and denotes the right-hand side of the equation by v(t). Then, by Lemma 2.4, we obtain:

$$\begin{aligned} {^{c}\!}D_{0^+}^{\alpha }v(t)&={^{c}\!}D_{0^+}^{\alpha }I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)\\&\quad +{^{c}\!}D_{0^+}^{\alpha }(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\nonumber \\&\quad +{^{c}\!}D_{0^+}^{\alpha }(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\\&\quad +{^{c}\!}D_{0^+}^{\alpha }(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\nonumber \\&\quad +{^{c}\!}D_{0^+}^{\alpha }(1-v_3t-v_7t^2)g(u)\\&=f(t, u(t),{^{c}\!}D_{0^+}^{\beta }u(t)). \end{aligned}$$

Hence, v(t) is a solution of the fractional differential Eq. (1). Also, it is easy to verify by direct computation that v satisfies conditions (4). This completes the proof. $\square$

To simplify the computations in the main results, we set some notations:

$$\begin{aligned} M_1&=\frac{1}{\varGamma (\alpha )}+\frac{|v_6|+|v_1|}{\varGamma (\alpha +\sigma _1)} + \frac{|v_2|+|v_5|}{\varGamma (\alpha +\sigma _2)}+\frac{|v_8|+|v_4|}{\varGamma (\alpha - \beta _1)},\nonumber \\ M_2&= \frac{1}{\varGamma (\alpha - \beta )}+\frac{2|v_6|+(2-\beta )|v_1|}{\varGamma (3-\beta )\varGamma (\alpha +\sigma _1)}+\frac{2|v_5|+(2-\beta )|v_2|}{\varGamma (3-\beta )\varGamma (\alpha +\sigma _2)}\nonumber \\&\quad +\frac{2|v_8|+(2-\beta )|v_4|}{\varGamma (3-\beta )\varGamma (\alpha -\beta _1)},\nonumber \\ N_1&=(1+|v_3|+|v_7|),\quad N_2 =\frac{2|v_7|+(2-\beta )|v_3|}{\varGamma (3-\beta )}. \end{aligned}$$

(14)

In the sequel, we need the following assumptions.

• (H1) The function $f: J\times \mathbb {R}\times \mathbb {R} \longrightarrow \mathbb {R}$ is continuous.

• (H2)
  $$\begin{aligned} |f(t,u_1,v_1)-f(t,u_2,v_2)|\le g(t)\left( |u_1-u_2|+|v_1-v_2|\right) . \end{aligned}$$

  $\text {for all}\; u_1,\,u_2,\,v_1,\,v_2 \in \mathbb {R}; g \in L^1 (J,\mathbb {R}_+)\;\text {and}\; t \in J.$

• (H3) There exist a positive constant $\omega < 1$ and a continuous function $\phi : [0, \infty ) \longrightarrow [0, \infty )$, such that $\phi (v) \le \omega v,\; \omega <\frac{1}{N_1+N_2}$ and $|g(u) - g(v)| \le \phi (\Vert u - v\Vert )$, for all $u,v \in C (J,\mathbb {R} )$.

• (H4) $g(0) = 0$.

• (H5) There exist continuous nondecreasing function $\psi :[0, \infty ) \longrightarrow (0, \infty )$ and function $p \in L^1([0, 1],\mathbb {R}^+),$ such that:

  $$\begin{aligned} |f(t, u, v)| \le p(t)\psi (|u|+|v|), \end{aligned}$$

  (15)

  for each $(t, u, v) \in [0, 1] \times \mathbb {R} \times \mathbb {R}$.

• (H16
  $$\begin{aligned} \sup _{r \in (0, \infty )} \frac{r}{(M_1+M_2)\Vert p\Vert _{L^1}\psi (r)} > \frac{1}{1-(N_1+N_2)\omega }. \end{aligned}$$

  (16)

• (H7) There exists non-negative function $p\in L^1 (J,\mathbb {R}_+)$, such that:

  $$\begin{aligned} |f(t,u,v)|\le p(t)(1+|u|+|v|), \end{aligned}$$

  (17)

  for all $u,v \in \mathbb {R}$ and $t \in J$.

Our first results is based on the Banach contraction principle.

Theorem 3.4

Assume $(H_1) ,(H_2)$, and $(H_3)$ hold. If

$$\begin{aligned} \gamma _1=(M_1+M_2)\Vert g\Vert _{L^1}+(N_1+N_2)\omega )< 1. \end{aligned}$$

(18)

holds, then the problem (1) has a unique solution on J.

Proof

Transform the problem (1) into a fixed point problem. Clearly, the fixed points of the operator T defined by (10) are solutions of the problem (1).

Let $u, v \in E$ and $t\in J$, and from the definition of T, we can write:

$$\begin{aligned} |&(Tu)(t) - (Tv)(t)|=I_{0^+}^{\alpha }|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))- f(s, v(s),{^{c}\!}D_{0^+}^{\beta }v(s))|(t)\nonumber \\&\quad +|v_6t^2-v_1t|I_{0^+}^{\alpha +\sigma _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))- f(s, v(s),{^{c}\!}D_{0^+}^{\beta }v(s))|(\eta _1)\nonumber \\&\quad +|v_2t-v_5t^2|I_{0^+}^{\alpha +\sigma _2}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))- f(s, v(s),{^{c}\!}D_{0^+}^{\beta }v(s))|(\eta _2)\nonumber \\&\quad +|v_8t^2-v_4t|I_{0^+}^{\alpha -\beta _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))- f(s, v(s),{^{c}\!}D_{0^+}^{\beta }v(s))|(1)\\&\quad +|1-v_3t-v_7t^2||g(u)-g(v)|. \end{aligned}$$

By (H2), we can find that:

$$\begin{aligned}&| (Tu)(t) - (Tv)(t)| \le \left( \Vert u-v\Vert _\infty +\Vert {^{c}\!}D_{0^+}^{\beta }u-{^{c}\!}D_{0^+}^{\beta }v\Vert _\infty \right) \left[ I_{0^+}^{\alpha }g(s)(t)\right. \\&\quad \left. +(|v_6|+|v_1|)I_{0^+}^{\alpha +\sigma _1}g(s)(\eta _1)+(|v_2|+|v_5|)I_{0^+}^{\alpha +\sigma _2}g(s)(\eta _2)\right. \\&\quad \left. +\,(|v_8|+|v_4|)I_{0^+}^{\alpha -\beta _1}g(s)(1)\right] \\&\quad +(1+|v_3|+|v_7|)|g(u)-g(v)|. \end{aligned}$$

According to the Lemmas 2.7, 2.8, and (H3), we have:

$$\begin{aligned} \Vert Tu- Tv\Vert _\infty&\le \Vert u-v\Vert _E\left( \frac{1}{\varGamma (\alpha )}+\frac{|v_6|+|v_1|}{\varGamma (\alpha +\sigma _1)} + \frac{|v_2|+|v_5|}{\varGamma (\alpha +\sigma _2)}+\frac{|v_8|+|v_4|}{\varGamma (\alpha - \beta _1)}\right) \Vert g\Vert _{L^1}\nonumber \\&\quad +(1+|v_3|+|v_7|)\omega \Vert u-v\Vert _\infty . \end{aligned}$$

Thus:

$$\begin{aligned} \Vert Tu - Tv\Vert _{\infty }\le (M_1\Vert g\Vert _{L^1}+N_1\omega )\Vert u-v\Vert _{E}. \end{aligned}$$

(19)

Observe that:

$$\begin{aligned} {^{c}\!}D_{0^+}^{\beta }Tu(t)&=I_{0^+}^{\alpha -\beta }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)\\&\quad +\frac{2v_6t^{2-\beta }+(2-\beta )v_1t^{1-\beta }}{\varGamma (3-\beta )} I_{0^+}^{\alpha +\sigma _1} f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\\&\quad +\frac{2v_5t^{2-\beta }+(2-\beta )v_2t^{1-\beta }}{\varGamma (3-\beta )}I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\\&\quad +\frac{2v_8t^{2-\beta }+(2-\beta )v_4t^{1-\beta }}{\varGamma (3-\beta )}I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\\&\quad +\frac{2v_7t^{2-\beta }+(2-\beta )v_3t^{1-\beta }}{\varGamma (3-\beta )}g(u), \end{aligned}$$

using similar techniques as to get (19), it yields:

$$\begin{aligned} \Vert {^{c}\!}D_{0^+}^{\beta }Tu- {^{c}\!}D_{0^+}^{\beta }Tv\Vert _{\infty } \le (M_2 \Vert g\Vert _{L^1}+N_2\omega )\Vert u-v\Vert _{E}. \end{aligned}$$

(20)

Combining (19) and (20), we get:

$$\begin{aligned} \Vert Tu - Tv\Vert _{E}\le ((M_1+M_2)\Vert g\Vert _{L^1}+(N_1+N_2)\omega ) \Vert u-v\Vert _{E}. \end{aligned}$$

Therefore, we have:

$$\begin{aligned} \Vert Tu - Tv\Vert _{E}\le \gamma _1 \Vert u-v\Vert _{E}. \end{aligned}$$

In view of the given condition $\gamma _1=(M_1+M_2)\Vert g\Vert _{L^1}+(N_1+N_2)\omega )< 1$, it follows that the mapping T is a contraction. Hence, by the Banach fixed point theorem, T has a unique fixed point which is a unique solution of Problem (1). This completes the proof.$\square$

Our next result relies on a fixed point theorem of O’Regan [22]

Lemma 3.5

Denote by U an open set in a closed, convex set C of a Banach space E. Assume $0 \in U$. Also assume that $T(\overline{U} )$ is bounded and that $F : \overline{U} \longrightarrow C$ is given by $T = T_1 + T_2$, in which $T_1 :\overline{U} \longrightarrow E$ is continuous and completely continuous and $T_2 : \overline{U} \longrightarrow E$ is a nonlinear contraction (i.e., there exists a non-negative nondecreasing function $\phi : [0, \infty ) \longrightarrow [0, \infty )$ satisfying $\phi (z) < z$ for $z > 0$, such that $\Vert T_2(u) - T_2(v)\Vert \le \phi (\Vert u - v\Vert )$ for all $u,v \in \overline{U}$). Then, either

• (c1) T has a fixed point in $\overline{U}$, or

• (c2) there exists $u \in \partial U$ and $\lambda \in (0, 1)$, such that $u=\lambda Tu$, where $\overline{U}$ and $\partial U$, respectively, represent the closure and boundary of U.

Let

$$\begin{aligned} B_{r}=\left\{ u \in E : \Vert u\Vert _{E}< r \right\} , \end{aligned}$$

and denote the maximum number by:

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

Theorem 3.6

Suppose that (H1),(H3)–(H6) are satisfied. Then, the problem (1) has at least one solution on J.

Proof

From (H6), there exists a number $r_0 > 0$, such that:

$$\begin{aligned} \frac{r_0}{(M_1+M_2)\Vert p\Vert _{L^1}\psi (r_0)} > \frac{1}{1-(N_1+N_2)\omega }. \end{aligned}$$

(21)

We shall prove that $T_1$ and $T_2$ defined by (11) and (12), respectively, satisfy the conditions of Lemma 3.5. We split the proof into several steps.

Step 1 : The operator $T_1$ is completely continuous. It is clear that $T_1$ is continuous, since f is continuous.

Next, we shall prove that the operator $T_1$ maps bounded sets into bounded sets in E. For $u\in \overline{B}_{r_0}$ and for each $t \in J$, we have:

$$\begin{aligned} |(T_1u)(t)|&\le I_{0^+}^{\alpha }|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(t)\\&\quad +(|v_6|+|v_1|)I_{0^+}^{\alpha +\sigma _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _1)\\&\quad +(|v_2|+|v_5|)I_{0^+}^{\alpha +\sigma _2}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _2)\\&\quad +(|v_8|+|v_4|)I_{0^+}^{\alpha -\beta _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(1). \end{aligned}$$

In view of (15), we obtain:

$$\begin{aligned} |(T_1u)(t)|&\le \psi (\Vert u\Vert )\left\{ I_{0^+}^{\alpha } p(s)(t)+(|v_6|+|v_1|)I_{0^+}^{\alpha +\sigma _1} p(s)(\eta _1)\right. \\&\quad +\left. (|v_2|+|v_5|)I_{0^+}^{\alpha +\sigma _2} p(s)(\eta _2) +(|v_8|+|v_4|)I_{0^+}^{\alpha -\beta _1} p(s)(1)\right\} . \end{aligned}$$

By Lemmas (2.7), ( 2.8), we obtain:

$$\begin{aligned} |(T_1u)(t)| \le M_1\Vert p\Vert _{L^1}\psi _1(r_0). \end{aligned}$$

Thus:

$$\begin{aligned} \Vert T_1u\Vert _\infty \le M_1\Vert p\Vert _{L^1}\psi _1(r_0). \end{aligned}$$

(22)

With the same arguments as in (22), we get:

$$\begin{aligned} \Vert {^{c}\!D}_{0^+}^{\beta }T_1u\Vert _\infty \le M_2\Vert p\Vert _{L^1}\psi _1(r_0). \end{aligned}$$

(23)

Combining (22) with (23), we get:

$$\begin{aligned} \Vert T_1u\Vert _E \le (M_1+M_2)\Vert p\Vert _{L^1}\psi _1(r_0), \end{aligned}$$

(24)

which shows that $T_1$ is uniform bounded.

Now, we will prove that $T_1(\overline{B}_{r_0})$ is equi-continuous. Let $t_1, t_2 \in [0, 1] , t_1 < t_2, u \in \overline{B}_{r_0}$:

$$\begin{aligned}&|(T_1u)(t_2) - (T_1u)(t_1)|\\&\le \frac{1}{\varGamma (\alpha )}\int _{0}^{t_1} \Big [(t_2- s)^{\alpha -1}- (t_1- s)^{\alpha -1} \Big ]|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|\mathrm{d}s\\&\quad +\frac{1}{\varGamma (\alpha )}\int _{t_1}^{t_2} (t_2- s)^{\alpha -1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|\mathrm{d}s\\&\quad +(|v_6|(t_2^2-t_1^2)+|v_1|(t_2-t_1))I_{0^+}^{\alpha +\sigma _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _1)\\&\quad +(|v_2|(t_2-t_1)+|v_5|(t_2^2-t_1^2))I_{0^+}^{\alpha +\sigma _2}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _2)\\&\quad +(|v_8|(t_2^2-t_1^2)+|v_4|(t_2-t_1))I_{0^+}^{\alpha -\beta _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(1). \end{aligned}$$

Some computations give:

$$\begin{aligned} |(T_1u)(t_2) - (T_1u)(t_1)|&\le M_r\left[ \frac{1}{\varGamma (\alpha )} +\frac{2|v_6|+|v_1|}{\varGamma (\alpha +\sigma _1+1)}\eta _1^{\alpha +\sigma _1}\frac{2|v_2|+|v_5|}{\varGamma (\alpha +\sigma _2+1)}\eta _2^{\alpha +\sigma _2}\right. \nonumber \\&\quad \left. +\frac{2|v_8|+|v_4|}{\varGamma (\alpha -\beta _1+1)}\right] (t_2-t_1). \end{aligned}$$

In a like manner:

$$\begin{aligned}&|{^{c}\!D}_{0^+}^\beta (T_1u)(t_2) - {^{c}\!D}_{0^+}^\beta (T_1u)(t_1)| \le M_r\;\left\{ \frac{t_2^{\alpha -\beta }-t_1^{\alpha -\beta } }{\varGamma (\alpha - \beta +1)}\right. \nonumber \\&\quad +\left. \frac{2|v_6||t_2^{2-\beta }-t_1^{2-\beta }|+(2-\beta )|v_1||t_2^{1-\beta }-t_1^{1-\beta }|}{\varGamma (3-\beta )\varGamma (\alpha +\sigma _1+1)}\eta _1^{\alpha +\sigma _1}\right. \nonumber \\&\quad +\left. \frac{2|v_5||t_2^{2-\beta }-t_1^{2-\beta }|+(2-\beta )|v_2||t_2^{1-\beta }-t_1^{1-\beta }|}{\varGamma (3-\beta )\varGamma (\alpha +\sigma _2+1)}\eta _2^{\alpha +\sigma _2}\right. \nonumber \\&\quad +\left. \frac{2|v_8||t_2^{2-\beta }-t_1^{2-\beta }|+(2-\beta )|v_4||t_2^{1-\beta }-t_1^{1-\beta }|}{\varGamma (3-\beta )\varGamma (\alpha -\beta _1+1)}\right\} . \end{aligned}$$

In consequence, we obtain:

$$\begin{aligned} \sup _{u\in \overline{B}_{r_0}}|(T_1u)(t_2) - (T_1u)(t_1)|+\sup _{u\in \overline{B}_{r_0}}|{^{c}\!D}_{0^+}^\beta (T_1u)(t_2) - {^{c}\!D}_{0^+}^\beta (T_1u)(t_1)|\rightarrow 0, \end{aligned}$$

as $t_1\rightarrow t_2$ independent of $u \in \overline{B}_{r_0}$. Therefore, the operator $T_1$ is equi-continuous and uniformly bounded. Hence, by the Arzelá–Ascoli theorem, it follows that $T_1(\overline{B}_{r_0})$ is relatively compact in E.

Step 2 : The operator $T_2: \overline{B}_{r_0} \longrightarrow E$ is contractive. Let $u, v \in$ and $t\in J$ by (H3), we can show that:

$$\begin{aligned} \Vert T_2(u) - T_2(v)\Vert _{E}&\le (N_1+N_2)\omega \big \Vert u-v\Vert _{E}, \end{aligned}$$

which, in view of (H3), implies that $T_2$ is a contraction.

Step 3: The set $T_2( \overline{B}_{r_0})$ is bounded. For $u \in \overline{B}_{r_0}$ and for each $t \in J$, and by (H3) and (H4), we obtain:

$$\begin{aligned}\Vert T_2u\Vert _{E}\le (N_1+N_2)\omega r_0.\end{aligned}$$

Hence, $T_2( \overline{B}_{r_0})$ is bounded.

Step 4 : Finally, we show that the case (c2) in Lemma 3.5 does not occur. To this end, we suppose that (c2) holds. Let $u \in \partial B_{r_0}$, and then, $u =\lambda Tu$, for some $0< \lambda < 1$. Therefore, we have $\Vert u\Vert _{E} = r_0$, and:

$$\begin{aligned} u(t)&=\lambda \left\{ I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)\right. \\&\quad \left. +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\right. \\&\quad \left. +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\right. \\&\quad \left. +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\right. \\&\quad \left. +(1-v_3t-v_7t^2)g(u)\right\} . \end{aligned}$$

With hypotheses (H3)–(H6), we can write:

$$\begin{aligned} r_0 \le (M_1+M_2)\Vert p\Vert _{L^1}\psi (r_0)+(N_1+N_2)\omega r_0, \end{aligned}$$

(25)

which implies that:

$$\begin{aligned} \frac{r_0}{ (M_1+M_2)\Vert p\Vert _{L^1}\psi (r_0)} \le \frac{1}{1-(N_1+N_2)\omega } , \end{aligned}$$

(26)

which contradicts (16). Consequently, we have proved that the operators $T_1$ and $T_2$ satisfy all the conditions in Lemma 3.5. Hence, the operator T has at least one fixed point $u \in \overline{B}_{r_0}$, which is the solution of the boundary value problem (1).$\square$

Our next result relies on the following fixed point theorem due to Burton and Kirk [8]

Theorem 3.7

Let E be a Banach space, and $A, B : E \longrightarrow E$ be two operators, such that A is a contraction and B is completely continuous. Then, either

• the operator equation $u = A(u) + B(u)$ has a solution, or

• the set $\varOmega = \big \{u \in E : \lambda A(\frac{u}{\lambda }) +\lambda B(u) = u \big \}$ is unbounded for $\lambda \in (0, 1)$.

Then, there exists $z\in U$, such that $z=Az+Bz.$

Theorem 3.8

Assume that (H1), (H3), (H4), and (H7) hold. In addition, we suppose that:

$$\begin{aligned} (M_1+M_2)\Vert p\Vert _{L^1}+(N_1+N_2)\omega <1. \end{aligned}$$

Then, the fractional boundary value problem (1) has at least one solution on J.

Proof

We define the map $T: E \longrightarrow E$ by:

$$\begin{aligned} Tu(t)= T_1u(t)+T_2u(t), t \in J, \end{aligned}$$

where $T_1$ and $T_2$ are defined by (11) and (12), respectively. Notice that Problem (1) is equivalent to the fixed point problem $T(u) = u$. We shall prove that $T_1$ and $T_2$ satisfy the conditions of Theorem 3.7. For clarity, we will divide the remain of the proof into several steps.

Step 1: $T_1$ is continuous. The continuity of f implies that the operator $T_1$ is continuous.

Step 2: The operator $T_1$ maps bounded sets into bounded sets in E. In a similar manner as previous proof of Theorem 3.6, we can find that:

$$\begin{aligned} \Vert T_1u\Vert _{E}\le (M_1+M_2)\Vert p\Vert _{L^1}(1+\Vert u\Vert _{E}). \end{aligned}$$

Step 3: The operator $T_1$ maps bounded sets into equi-continuous sets. By a discussion similar to that of Theorem 3.6, we can get:

$$\begin{aligned} |(T_1u)(t_2) - (T_1u)(t_1)| \rightarrow 0. \end{aligned}$$

And

$$\begin{aligned} | {^{c}\!D}_{0^+}^{\beta }T_1y(t_2)- {^{c}\!D}_{0^+}^{\beta }T_1y(t_1)| \rightarrow 0, \end{aligned}$$

as $t_2\rightarrow t_1$. This implies that:

$$\begin{aligned} \Vert (Tu)(t_2) - (Tu)(t_1)\Vert _E \rightarrow 0, \;\;\text {as}\;\; t_2\rightarrow t_1. \end{aligned}$$

Step 4: The operator $T_2$ is a contraction. This was established in Step 2 of Theorem 3.6.

Step 5: A priori bounds on solutions. Now, it remains to show that the set $\varOmega = \big \{u \in E : \lambda T_2\left( \frac{u}{\lambda }\right) +\lambda T_1(u) = u \big \}$ is unbounded for $\lambda \in (0, 1)$. Let $\lambda \in (0, 1)$, and $u \in E$ be a solution of the integral equation:

$$\begin{aligned} u(t)&= \lambda \left\{ I_{0^+}^{\alpha }f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(t)\right. \\&\quad \left. +(v_6t^2-v_1t)I_{0^+}^{\alpha +\sigma _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _1)\right. \\&\quad \left. +(v_2t-v_5t^2)I_{0^+}^{\alpha +\sigma _2}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(\eta _2)\right. \\&\quad \left. +(v_8t^2-v_4t)I_{0^+}^{\alpha -\beta _1}f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))(1)\right. \\&\quad \left. +(1-v_3t-v_7t^2)g\left( \frac{y}{\lambda }\right) \right\} . \end{aligned}$$

This implies by (H3), (H4), and (H7) that, for each $t \in J$ , we have:

$$\begin{aligned} |u(t)|&\le I_{0^+}^{\alpha }|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(t)\\&\quad +(|v_6|+|v_1|)I_{0^+}^{\alpha +\sigma _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _1)\\&\quad +(|v_2|+|v_5|)I_{0^+}^{\alpha +\sigma _2}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(\eta _2)\\&\quad +(|v_8|+|v_4|)I_{0^+}^{\alpha -\beta _1}|f(s, u(s),{^{c}\!}D_{0^+}^{\beta }u(s))|(1)\\&\quad +(1+|v_3|+|v_7|)|g\left( \frac{u(t)}{\lambda }\right) \\&\le (1+\Vert u\Vert _E)\left( I_{0^+}^{\alpha }p(s)(t)+(|v_6|+|v_1|)I_{0^+}^{\alpha +\sigma _1}p(s)(\eta _1)\right. \\&\quad \left. +(|v_2|+|v_5|)I_{0^+}^{\alpha +\sigma _2}p(s)(\eta _2)+(|v_8|+|v_4|)I_{0^+}^{\alpha -\beta _1}p(s)(1)\right) \\&\quad +(1+|v_3|+|v_7|)\omega \Vert u\Vert . \end{aligned}$$

According to the Lemmas 2.7 and 2.8, we have:

$$\begin{aligned} \Vert u\Vert _{\infty } \le M_1\Vert p\Vert _{L^1}(1+\Vert u\Vert _E)+N_1\omega \Vert u\Vert , \end{aligned}$$

(27)

for all $t \in J$. In a similar manner, we obtain:

$$\begin{aligned} \Vert {^{c}\!D}_{0^+}^{\beta }u\Vert _{\infty } \le M_2\Vert p\Vert _{L^1}(1+\Vert u\Vert _E)+N_2\omega \Vert u\Vert . \end{aligned}$$

(28)

Combining (27) with (28), we get:

$$\begin{aligned} \Vert u\Vert _{E} \le \frac{(M_1+M_2)\Vert p\Vert _{L^1}}{1-\left( (M_1+M_2)\Vert p\Vert _{L^1}+(N_1+N_2)\omega \right) }. \end{aligned}$$

This shows that the set $\varOmega$ is bounded, since $(M_1+M_2)\Vert p\Vert _{L^1}+(N_1+N_2)\omega <1$. Hence, T has a fixed point by Theorem 3.7, and consequently, the problem (1) has a solution. This completes the proof.$\square$

4 An example

Example 4.1

Let us consider Problem (1) with specific data:

$$\begin{aligned} \alpha&=\frac{5}{2}, \beta =\frac{1}{2},\beta _1=\frac{1}{4},\nonumber \\ a&=1,b=2, \sigma _1=\frac{5}{2},\\ \sigma _2&=\frac{7}{2}, \eta _1=\frac{1}{4},\eta _2=\frac{1}{2}.\nonumber \end{aligned}$$

(29)

Using the given values of the parameters in (6) and (14), by the Matlab program, we find that:

$$\begin{aligned} v_1&=-1.0062,\;v_2=0.0027,\;v_3=-0.0535,\nonumber \\ v_4&= 0.0014 ,\;v_5=-1.6060,\;v_6= -0.8803,\nonumber \\ v_7&= 0.0457,\; v_8 = -0.8030,\; M_1 =1.5542, \\ M_2&=2.1903, M_1+M_2 =3.7445,\; N_1=1.0993,\nonumber \\ N_2&=0.1292,\; N_1+N_2=1.2285,\; \varDelta = 0.6179,\nonumber . \end{aligned}$$

(30)

To illustrate Theorem 3.4, we take:

$$\begin{aligned} f(t, u(t), {^{c}\!}D_{0^+}^{\frac{1}{4}}u(t))&= \frac{(1-t)^2}{10}\biggl (\frac{1}{2}\left( x(t)+\sqrt{x^2(t)+1}\right) \nonumber \\&+ \frac{\left| {^{c}\!}D_{0^+}^{\frac{1}{4}}u(t)\right| }{1+\left| {^{c}\!}D_{0^+}^{\frac{1}{4}}u(t)\right| }\biggr ) +e^{-t},\\ g(u)&=\frac{1}{10}(1 -\cos u),\nonumber , \end{aligned}$$

(31)

in (1) and note that:

$$\begin{aligned} |f (t, u, v)-f (t, u_1, v_1)|&\le \frac{(1-t)^2}{10}\left( \left| \frac{1}{2}\left( u-u_1+\sqrt{1+u^2}-\sqrt{1+u_1^2}\right) \right| \right. \\&\quad \left. +\left| \frac{|v|}{1+|v|}-\frac{|v_1|}{1+|v_1|}\right| \right) \\&\le \frac{(1-t)^2}{10}\left( \left| \frac{1}{2}(u-u_1)\left( 1+\frac{u+u_1}{\sqrt{1+u^2}+\sqrt{1+u_1^2}}\right) \right| \right. \\&\quad \left. +\frac{|v-v_1|}{(1+|v|)(1+|v_1|)}\right) \\&\le \frac{(1-t)^2}{10}\left( |u-u_1|+|v-v_1|\right) , \end{aligned}$$

for each $(t, u, v) \in [0, 1] \times \mathbb {R} \times \mathbb {R}$ and

$$\begin{aligned} |g(u)-g(v)|&\le \frac{1}{10}|u-v|, \end{aligned}$$

for each $(u, v) \in \mathbb {R} \times \mathbb {R}$.

Here, $\omega =\frac{1}{10}, g(t)=\frac{(1-t)^2}{10}$ with $\Vert g\Vert _{L^1}=\frac{1}{30}$. Using the values of $M_1+M_2$ and $N_1+N_2$ given by (30), we find that $\gamma _1=(M_1+M_2)\Vert g\Vert _{L^1}+(N_1+N_2)\omega =0.2477<1.$ As all the conditions of Theorem 3.4 are satisfied, therefore, there exists a unique solution of Problem (1) with the data (29) and (31) on J.

5 Conclusion

We have established the existence and uniqueness of solutions for fractional boundary value problem under more general assumptions. Using the techniques of fixed point theorems such as Banach contraction principle, O’Regan fixed point theorem, and Burton and Kirk fixed point theorem, we establish the main results of the paper. Finally, we conclude the paper with an example to make our results clear.