Existence results for hybrid fractional integro-differential equations

Abstract

In this paper we study existence results for initial value problems for hybrid fractional integro-differential equations. A couple of hybrid fixed point theorems for the sum of three operators are used for proving the main results. Examples illustrating the results are also presented.

1 Introduction

Fractional differential equations arise in the mathematical modeling of systems and processes occurring in many engineering and scientific disciplines such as physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, etc. [1–6]. For some recent developments on the topic, see [7–17] and the references therein.

Hybrid fractional differential equations have also been studied by several researchers. This class of equations involves the fractional derivative of an unknown function hybrid with the nonlinearity depending on it. Some recent results on hybrid differential equations can be found in a series of papers [18–26].

In this paper we study existence results for initial value problems for hybrid fractional integro-differential equations. In Section 3 we consider the following initial value problem for hybrid fractional integro-differential equations:

$$ \left \{ \textstyle\begin{array}{l} D^{\alpha} [\frac{x(t)-\sum_{i=1}^{m}I^{\beta_{i}}h_{i}(t,x(t))}{f(t, x(t))} ]=g(t,x(t)),\quad t\in J:=[0,T], \\ x(0)=0, \end{array}\displaystyle \right . $$

(1.1)

where $D^{\alpha}$ denotes the Riemann-Liouville fractional derivative of order α, $0<\alpha\leq1$, $I^{\phi}$ is the Riemann-Liouville fractional integral of order $\phi>0$, $\phi\in\{ \beta_{1},\beta_{2},\ldots,\beta_{m}\}$, $f\in C( J\times{\mathbb{R}},{\mathbb{R}}\setminus\{0\})$, $g\in C(J\times\mathbb{R}, \mathbb {R})$ and $h_{i}\in C(J\times\mathbb{R}, \mathbb{R})$ with $h_{i}(0,0)=0$, $i=1,2,\ldots,m$. An existence result is obtained for the initial value problem (1.1) by using a hybrid fixed point theorem for three operators in a Banach algebra due to Dhage [27].

As a second problem we discuss in Section 4 an initial value problem for hybrid fractional sequential integro-differential equations,

$$ \left \{ \textstyle\begin{array}{l} D^{\alpha} [\frac{D^{\omega}x(t)-\sum_{i=1}^{m}I^{\beta_{i}}h_{i}(t,x(t))}{f(t, x(t))} ]=g(t,x(t),I^{\gamma }x(t)), \quad t\in J, \\ x(0)=0,\qquad D^{\omega}x(0)=0, \end{array}\displaystyle \right . $$

(1.2)

where $0<\alpha,\omega\leq1$, $1<\alpha+\omega\leq2$, functions f, h, and constants $\beta_{1},\beta_{2},\ldots,\beta_{m}$ are defined as in problem (1.1), $g\in C(J\times\mathbb{R}^{2}, \mathbb {R})$, and $I^{\gamma}$ is the Riemann-Liouville fractional integral of order γ. By using a useful generalization of Krasnoselskii’s fixed point theorem due to Dhage [28], we prove an existence result for the initial value problem (1.2). Note that if $m=1$, $\beta_{1}=\gamma=0$, and $h_{1}(t,x(t))=-\lambda x(t)$, λ is a constant, then the first equation of (1.2) is reduced to the hybrid fractional Langevin equation (first formulated by Langevin in 1908) of the form

$$ D^{\alpha} \biggl[\frac{(D^{\omega}+\lambda)x(t)}{f(t, x(t))} \biggr]=g\bigl(t,x(t)\bigr),\quad t\in J, $$

(1.3)

which is a generalization of the well-known classical results in [29].

The rest of the paper is organized as follows: In Section 2 we recall some useful preliminaries. In Section 3 we study the existence of the initial value problem (1.1), while in Section 4 we deal with the initial value problem (1.2). Examples illustrating the obtained results are presented in Section 5.

2 Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [1, 5] and present preliminary results needed in our proofs later.

Definition 2.1

The Riemann-Liouville fractional derivative of order $q > 0$ of a continuous function $f:(0,\infty)\rightarrow\mathbb{R}$ is defined by

$$ D^{q} f(t)=\frac{1}{\Gamma(n-q)} \biggl(\frac{d}{dt} \biggr)^{n} \int_{0}^{t}(t-s)^{n-q-1}f(s) \,ds, \quad n-1< q< n, $$

where $n=[q]+1$, $[q]$ denotes the integer part of a real number q, provided the right-hand side is point-wise defined on $(0,\infty)$, where Γ is the gamma function defined by $\Gamma(q)=\int_{0}^{\infty} e^{-s}s^{q-1}\,ds$.

Definition 2.2

The Riemann-Liouville fractional integral of order $p > 0$ of a continuous function $f:(0,\infty)\rightarrow \mathbb{R}$ is defined by

$$ I^{p} f(t)=\frac{1}{\Gamma(p)}\int_{0}^{t}(t-s)^{p-1}f(s) \,ds, $$

provided the right-hand side is point-wise defined on $(0,\infty)$.

Lemma 2.1

[1]

Let $q > 0$ and $x\in C(0,T)\cap L(0,T)$. Then the fractional differential equation

$$ D^{q} x(t)=0 $$

has a unique solution

$$ x(t) = k_{1}t^{q-1}+k_{2}t^{q-2}+\cdots+ k_{n}t^{q-n}, $$

where $k_{i} \in\mathbb{R}$, $i=1,2,\ldots,n$, and $n-1< q< n$.

Lemma 2.2

[1]

Let $q > 0$. Then for $x\in C(0,T)\cap L(0,T)$ we have

$$ I^{q}D^{q} x(t)=x(t)-\sum_{j=1}^{n} \frac{(I^{n-q}x)^{(n-j)}(0)}{\Gamma (q-j+1)}t^{q-j}, $$

where $n-1< q< n$.

Let $E=C(J,\mathbb{R})$ be the space of continuous real-valued functions defined on $J=[0,T]$. Define a norm $\|\cdot\|$ and a multiplication in E by

$$ \|x\|=\sup_{t\in J} \bigl\vert x(t)\bigr\vert \quad \text{and}\quad (xy) (t)=x(t)y(t),\quad \forall t\in J. $$

Clearly E is a Banach algebra with respect to above supremum norm and the multiplication in it.

3 Hybrid fractional integro-differential equations

In this section we consider the initial value problem (1.1). The following hybrid fixed point theorem for three operators in a Banach algebra E, due to Dhage [27], will be used to prove the existence result for the initial value problem (1.1).

Lemma 3.1

Let S be a nonempty, closed convex and bounded subset of a Banach algebra E and let $A,C:E\to E$ and $B:S\to E$ be three operators satisfying:

(a₁):

A and C are Lipschitzian with Lipschitz constants δ and ρ, respectively,

(b₁):

B is compact and continuous,

(c₁):

$x=AxBy+Cx\Rightarrow x \in S$ for all $y\in S$,

(d₁):

$\delta M+\rho<1$, where $M=\|B(S)\|$.

Then the operator equation $x=AxBx+Cx$ has a solution.

Lemma 3.2

Suppose that $0<\alpha\leq1$ and functions f, g, $h_{i}$, $i=1,2,\ldots,m$ satisfy problem (1.1). Then the unique solution of the hybrid fractional integro-differential problem (1.1) is given by

$$ x(t)=\frac{f(t,x(t))}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha -1}g \bigl(s,x(s)\bigr)\,ds+\sum_{i=1}^{m}I^{\beta_{i}}h_{i} \bigl(t,x(t)\bigr), \quad t\in J. $$

(3.1)

Proof

Applying the Riemann-Liouville fractional integral of order α to both sides of (1.1) and using Lemma 2.2, we have

$$\begin{aligned} \begin{aligned} &\biggl[\frac{x(t)-\sum_{i=1}^{m}I^{\beta_{i}}h_{i}(t,x(t))}{f(t, x(t))} \biggr]-\frac{t^{\alpha-1}}{\Gamma(\alpha)}I^{1-\alpha } \biggl[ \frac{x(t)-\sum_{i=1}^{m}I^{\beta_{i}}h_{i}(t,x(t))}{f(t, x(t))} \biggr]_{t=0} \\ &\quad =I^{\alpha}g\bigl(t,x(t)\bigr). \end{aligned} \end{aligned}$$

Since $x(0)=0$, $h(0,0)=0$, and $f(0,0)\neq0$, it follows that

$$ x(t)=f\bigl(t, x(t)\bigr)I^{\alpha}g\bigl(t,x(t)\bigr)+\sum _{i=1}^{m}I^{\beta_{i}}h_{i}\bigl(t,x(t) \bigr). $$

Thus (3.1) holds. The proof is completed. □

Theorem 3.1

Assume that:

(H₁):

The functions $f:J\times\mathbb{R}\to\mathbb {R}\setminus\{0\}$ and $h_{i}:J\times\mathbb{R}\to\mathbb{R}$, $h_{i}(0,0)=0$, $i=1,2,\ldots,m$, are continuous and there exist two positive functions ϕ, $\psi_{i}$, $i=1,2,\ldots,m$ with bound $\| \phi\|$ and $\|\psi_{i}\|$, $i=1,2,\ldots,m$, respectively, such that

$$ \bigl\vert f\bigl(t,x(t)\bigr)-f\bigl(t,y(t)\bigr)\bigr\vert \leq\phi(t)\bigl\vert x(t)-y(t)\bigr\vert $$

(3.2)

and

$$ \bigl\vert h_{i}\bigl(t,x(t)\bigr)-h_{i} \bigl(t,y(t)\bigr)\bigr\vert \leq\psi_{i}(t)\bigl\vert x(t)-y(t) \bigr\vert ,\quad i=1,2,\ldots,m, $$

(3.3)

for $t\in J$ and $x,y\in\mathbb{R}$.

(H₂):

There exist a function $p\in C(J,\mathbb{R}^{+})$ and a continuous nondecreasing function $\Psi:[0,\infty)\to(0,\infty)$ such that

$$ \bigl\vert g\bigl(t,x(t)\bigr)\bigr\vert \leq p(t)\Psi\bigl( \vert x\vert \bigr),\quad (t,x)\in J\times\mathbb{R}. $$

(3.4)

(H₃):

There exists a number $r>0$ such that

$$ r\ge\frac{ F_{0}\|p\|\Psi(r)\frac{T^{\alpha}}{\Gamma (\alpha+1)}+K_{0}\sum_{i=1}^{m}\frac{T^{\beta_{i}}}{\Gamma(\beta _{i}+1)}}{ 1-\|\phi\|\|p\|\Psi(r)\frac{T^{\alpha }}{\Gamma(\alpha+1)}-\sum_{i=1}^{m}\frac{\|\psi_{i}\|T^{\beta _{i}}}{\Gamma(\beta_{i}+1)}}, $$

(3.5)

where $F_{0}=\sup_{t\in J}|f(t,0)|$ and $K_{0}=\sup_{t\in J}|h_{i}(t,0)|$, $i=1,2,\ldots,m$, and

$$ \|\phi\| \|p\|\Psi(r)\frac{T^{\alpha}}{\Gamma(\alpha+1)} +\sum _{i=1}^{m}\frac{\|\psi_{i}\|T^{\beta_{i}}}{\Gamma(\beta_{i}+1)}< 1. $$

(3.6)

Then problem (1.1) has at least one solution on J.

Proof

Set $E=C(J,\mathbb{R})$ and define a subset S of E as

$$ S=\bigl\{ x\in E : \Vert x\Vert \leq r\bigr\} , $$

where r satisfies inequality (3.5).

Clearly S is closed, convex, and bounded subset of the Banach space E. By Lemma 3.2, problem (1.1) is equivalent to the integral equation (3.1). Now we define three operators; $\mathcal{A}:E\to E$ by

$$ \mathcal{A}x(t)=f\bigl(t,x(t)\bigr), \quad t\in J, $$

(3.7)

$\mathcal{B}:S\to E$ by

$$ \mathcal{B}x(t)= \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}g\bigl(s,x(s)\bigr)\,ds, \quad t\in J, $$

(3.8)

and $\mathcal{C}: E\to E$ by

$$ \mathcal{C}x(t) = \sum_{i=1}^{m}I^{\beta_{i}}h_{i} \bigl(t,x(t)\bigr) = \sum_{i=1}^{m}\int _{0}^{t}\frac{(t-s)^{\beta_{i}-1}}{\Gamma(\beta _{i})}h_{i} \bigl(s,x(s)\bigr)\,ds, \quad t\in J. $$

(3.9)

We shall show that the operators $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ satisfy all the conditions of Lemma 3.1. This will be achieved in the following series of steps.

Step 1. We first show that $\mathcal{A}$ and $\mathcal{C}$ are Lipschitzian on E.

Let $x,y\in E$. Then by (H₁), for $t\in J$ we have

$$\begin{aligned} \bigl\vert \mathcal{A}x(t)-\mathcal{A}y(t)\bigr\vert = & \bigl\vert f \bigl(t,x(t)\bigr)-f\bigl(t,y(t)\bigr)\bigr\vert \\ \leq& \phi(t)\bigl\vert x(t)-y(t)\bigr\vert \leq \|\phi\| \|x-y\|, \end{aligned}$$

which implies $\|\mathcal{A}x-\mathcal{A}y\|\leq\|\phi\| \|x-y\|$ for all $x,y\in E$. Therefore, $\mathcal{A}$ is a Lipschitzian on E with Lipschitz constant $\|\phi\|$.

Analogously, for any $x,y\in E$, we have

$$\begin{aligned} \bigl\vert \mathcal{C}x(t)-\mathcal{C}y(t)\bigr\vert = & \Biggl\vert \sum_{i=1}^{m}I^{\beta_{i}}h_{i} \bigl(t,x(t)\bigr)-\sum_{i=1}^{m}I^{\beta_{i}}h_{i} \bigl(t,y(t)\bigr)\Biggr\vert \\ \leq& \sum_{i=1}^{m} \int _{0}^{t}\frac{(t-s)^{\beta _{i}-1}}{\Gamma(\beta_{i})}\psi_{i}(s) \bigl\vert x(s)-y(s)\bigr\vert \,ds \\ \leq& \|x-y\|\sum_{i=1}^{m} \frac{\|\psi_{i}\| T^{\beta _{i}}}{\Gamma(\beta_{i}+1)}. \end{aligned}$$

This means that

$$ \|\mathcal{C}x-\mathcal{C}y\|\leq \sum_{i=1}^{m} \frac{\| \psi_{i}\|T^{\beta_{i}}}{\Gamma(\beta_{i}+1)}\|x-y\|. $$

Thus, $\mathcal{C}$ is a Lipschitzian on E with Lipschitz constant $\sum_{i=1}^{m}\frac{\|\psi_{i}\|T^{\beta_{i}}}{\Gamma (\beta_{i}+1)}$.

Step 2. The operator $\mathcal{B}$ is completely continuous on S.

We first show that the operator $\mathcal{B}$ is continuous on E. Let $\{x_{n}\}$ be a sequence in S converging to a point $x\in S$. Then by the Lebesgue dominated convergence theorem, for all $t\in J$, we obtain

$$\begin{aligned} \begin{aligned} \lim_{n\to\infty}\mathcal{B} x_{n}(t) & = \lim _{n\to\infty}\int_{0}^{t} \frac{(t-s)^{\alpha -1}}{\Gamma(\alpha)}g\bigl(s,x_{n}(s)\bigr)\,ds \\ & = \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}\lim _{n\to\infty}g\bigl(s,x_{n}(s)\bigr)\,ds \\ & = \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}g\bigl(s,x(s) \bigr)\,ds. \end{aligned} \end{aligned}$$

This implies that $\mathcal{B}$ is continuous on S.

Next we will prove that the set $\mathcal{B}(S)$ is a uniformly bounded in S. For any $x\in S$, we have

$$\begin{aligned} \bigl\vert \mathcal{B} x(t)\bigr\vert = & \biggl\vert \int _{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)}g\bigl(s,x(s)\bigr)\,ds \biggr\vert \\ \leq& \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )} p(s) \Psi(r) \,ds \\ \leq& \|p\|\Psi(r)\frac{T^{\alpha}}{\Gamma(\alpha +1)} =K_{1}, \end{aligned}$$

for all $t\in J$. Therefore, $\|\mathcal{B}\|\leq K_{1}$, which shows that $\mathcal{B}$ is uniformly bounded on S.

Now, we will show that $\mathcal{B}(S)$ is an equicontinuous set in E. Let $\tau_{1},\tau_{2}\in J$ with $\tau_{1}<\tau_{2}$ and $x\in S$. Then we have

$$\begin{aligned} \bigl\vert \mathcal{B} x(\tau_{2})-\mathcal{B} x( \tau_{1})\bigr\vert = & \biggl\vert \int_{0}^{\tau_{2}} \frac{(\tau_{2}-s)^{\alpha -1}}{\Gamma(\alpha)}g\bigl(s,x(s)\bigr)\,ds -\int_{0}^{\tau_{1}} \frac{(\tau _{1}-s)^{\alpha-1}}{\Gamma(\alpha)}g\bigl(s,x(s)\bigr)\,ds\biggr\vert \\ \leq& \int_{0}^{\tau_{1}}\frac{\vert (\tau _{2}-s)^{\alpha-1}-(\tau_{1}-s)^{\alpha-1}\vert }{\Gamma(\alpha )}\bigl\vert g\bigl(s,x(s)\bigr)\bigr\vert \,ds \\ &{} +\int_{\tau_{1}}^{\tau_{2}}\frac{(\tau_{2}-s)^{\alpha -1}}{\Gamma(\alpha)}\bigl\vert g\bigl(s,x(s)\bigr)\bigr\vert \,ds \\ \le& \int_{0}^{\tau_{1}}\frac{\vert (\tau _{2}-s)^{\alpha-1}-(\tau_{1}-s)^{\alpha-1}\vert }{\Gamma(\alpha)}\| p\| \Psi(r) \,ds \\ &{}+\int_{\tau_{1}}^{\tau_{2}}\frac{(\tau_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} \|p\|\Psi(r) \,ds, \end{aligned}$$

which is independent of $x\in S$. As $\tau_{1}\to\tau_{2}$, the right-hand side of the above inequality tends to zero. Therefore, it follows from the Arzelá-Ascoli theorem that $\mathcal{B}$ is a completely continuous operator on S.

Step 3. The hypothesis (c₁) of Lemma 3.1 is satisfied.

Let $x\in E$ and $y\in S$ be arbitrary elements such that $x=\mathcal {A}x\mathcal{B}y+\mathcal{C}x$. Then we have

$$\begin{aligned} \bigl\vert x(t)\bigr\vert \leq& \bigl\vert \mathcal{A}x(t)\bigr\vert \bigl\vert \mathcal{B}y(t)\bigr\vert +\bigl\vert \mathcal {C}x(t)\bigr\vert \\ \leq& \bigl\vert f\bigl(t,x(t)\bigr)\bigr\vert \int_{0}^{t} \frac{(t-s)^{\alpha -1}}{\Gamma(\alpha)}\bigl\vert g\bigl(s,x(s)\bigr)\bigr\vert \,ds +\sum _{i=1}^{m} \int_{0}^{t} \frac{(t-s)^{\beta_{i}-1}}{\Gamma(\beta _{i})}\bigl\vert h_{i}\bigl(s,x(s)\bigr)\bigr\vert \,ds \\ \leq& \bigl(\bigl\vert f\bigl(t,x(t)\bigr)-f(t,0)\bigr\vert +\bigl\vert f(t,0)\bigr\vert \bigr) \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \|p\|\Psi(r) s \\ &{} +\sum_{i=1}^{m} \int _{0}^{t}\frac{ (t-s)^{\beta _{i}-1}}{\Gamma(\beta_{i})} \bigl(\bigl\vert h_{i}\bigl(s,x(s)\bigr)-h_{i}(s,0)\bigr\vert +\bigl\vert h_{i}(s,0)\bigr\vert \bigr)\,ds \\ \leq& \bigl(r\Vert \phi \Vert +F_{0}\bigr) \biggl(\|p\|\Psi(r) \frac {T^{\alpha}}{\Gamma(\alpha+1)} \biggr) +\sum_{i=1}^{m} \frac{(r\|\psi_{i}\|+K_{0})T^{\beta_{i}}}{\Gamma(\beta_{i}+1)}, \end{aligned}$$

which leads to

$$ \|x\| \leq \bigl(r\Vert \phi \Vert +F_{0}\bigr) \biggl(\|p\| \Psi(r)\frac {T^{\alpha}}{\Gamma(\alpha+1)} \biggr) +\sum_{i=1}^{m} \frac{(r\|\psi_{i}\|+K_{0})T^{\beta_{i}}}{\Gamma(\beta _{i}+1)} \leq r. $$

Therefore, $x\in S$.

Step 4. Finally we show that $\delta M+\rho<1$, that is, (d₁) of Lemma 3.1 holds.

Since

$$\begin{aligned} M = & \bigl\Vert \mathcal{B}(S)\bigr\Vert \\ =& \sup_{x\in S} \Bigl\{ \sup_{t\in J}\bigl\vert \mathcal {B}x(t)\bigr\vert \Bigr\} \\ \leq& \|p\|\Psi(r)\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \end{aligned}$$

(3.10)

and by (H₃) we have

$$ \|\phi\|M+\sum_{i=1}^{m} \frac{\|\psi_{i}\|T^{\beta_{i}}}{\Gamma(\beta _{i}+1)}< 1, $$

with $\delta=\|\phi\|$ and $\rho=\sum_{i=1}^{m}\frac {\|\psi_{i}\|T^{\beta_{i}}}{\Gamma(\beta_{i}+1)}$.

Thus all the conditions of Lemma 3.1 are satisfied and hence the operator equation $x=\mathcal{A}x\mathcal {B}x+\mathcal{C}x$ has a solution in S. In consequence, problem (1.1) has a solution on J. This completes the proof. □

4 Hybrid fractional sequential integro-differential equations

In this section we consider the initial value problem (1.2). An existence result will be proved by using the following fixed point theorem due to Dhage.

Lemma 4.1

[28]

Let M be a nonempty, closed, convex and bounded subset of the Banach space X and let $A: X\to X$ and $B:M\to X$ be two operators such that

• (i) A is a contraction,

• (ii) B is completely continuous, and

• (iii) $x=Ax+By$ for all $y\in M \Rightarrow x\in M$.

Then the operator equation $Ax+Bx=x$ has a solution.

Lemma 4.2

Suppose that $0<\alpha,\omega\leq 1$, $1<\alpha+\omega\leq2$, $\gamma>0$, and the functions f, g, $h_{i}$, $i=1,2,\ldots,m$ satisfy problem (1.2). Then the unique solution of the hybrid fractional sequential integro-differential problem (1.2) is given by

$$\begin{aligned} x(t) =& \int_{0}^{t} \frac{(t-s)^{\omega-1}}{\Gamma(\omega )}f\bigl(s,x(s)\bigr)\int_{0}^{s} \frac{(s-u)^{\alpha-1}}{\Gamma(\alpha )}g\bigl(u,x(u),I^{\gamma}x(u)\bigr)\, du\,ds \\ &{} +\sum_{i=1}^{m}I^{\beta_{i}+\omega}h_{i} \bigl(t,x(t)\bigr),\quad t\in J. \end{aligned}$$

(4.1)

Proof

By using the method of proving of Lemma 3.2 and applying the semigroup property, i.e., $I^{\omega}I^{\beta_{i}}h_{i}=I^{\beta_{i}+\omega}h_{i}$, $i=1,2,\ldots ,m$, we obtain the desired integral equation in (4.1). □

Theorem 4.1

Assume that:

(A₁):

The functions $f:J\times\mathbb{R}\to\mathbb {R}\setminus\{0\}$ and $g:J\times\mathbb{R}^{2}\to\mathbb{R}$, are continuous and there exist two positive functions ϕ, χ with bound $\|\phi\|$ and $\|\chi\|$, respectively, such that

$$ \bigl\vert f\bigl(t,x(t)\bigr)-f\bigl(t,y(t)\bigr)\bigr\vert \leq\phi(t)\bigl\vert x(t)-y(t)\bigr\vert $$

(4.2)

and

$$ \bigl\vert g\bigl(t,x(t),y(t)\bigr)-g\bigl(t,\bar{x}(t), \bar{y}(t)\bigr)\bigr\vert \leq\chi (t) \bigl(\bigl\vert x(t)-y(t)\bigr\vert +\bigl\vert \bar{x}(t)-\bar{y}(t)\bigr\vert \bigr), $$

(4.3)

for $t\in J$ and $x,y, \bar{x}, \bar{y}\in\mathbb{R}$.

(A₂):

$|f(t,x)|\le\mu(t)$, $\forall(t,x)\in J\times {\mathbb{R}}$, $\mu\in C(J,{\mathbb{R}}^{+})$, $|g(t, x, y)|\le\nu(t)$, $\forall(t,x, y)\in J\times{\mathbb{R}}\times{\mathbb{R}}$, $\nu\in C(J,{\mathbb{R}}^{+})$, and $|h_{i}(t, x)|\le\theta_{i}(t)$, $\forall(t,x)\in J\times{\mathbb{R}}$, $\theta_{i}\in C(J,{\mathbb{R}}^{+})$, $i=1,2, \ldots, m$.

If

$$ \frac{T^{\alpha}}{\Gamma(\alpha+1)}\frac{T^{\omega}}{\Gamma (\omega+1)}\|\nu\|\|\phi\|+\|\mu\|\|\chi \| \biggl[\frac{T^{\omega }}{\Gamma(\omega+1)}+\frac{T^{\omega+\gamma}}{\Gamma(\omega +\gamma+1)} \biggr]< 1, $$

(4.4)

then problem (1.2) has at least one solution on J.

Proof

Setting $\sup_{t\in J}|\mu(t)|=\|\mu\|$, $\sup_{t\in J}|\nu(t)|=\|\nu\|$, $\sup_{t\in J}|\theta_{i}(t)|=\|\theta_{i}\|$, $i=1,2,\ldots, m$, and choosing

$$ R\ge \sum_{i=1}^{m}\frac{T^{\beta_{i}+\omega}}{\Gamma(\beta_{i}+\omega +1)}\| \theta_{i}\|+\frac{T^{\omega+\alpha}}{\Gamma(\omega+\alpha +1)}\|\mu\|\|\nu\|, $$

(4.5)

we consider $B_{R}=\{x\in C(J, {\mathbb{R}}): \|x\|\le R\}$. We define the operators $\mathcal{A}:E\to E$ as in (3.7), $\mathcal{D}: B_{R}\to E$ by

$$ \mathcal{D}x(t)= \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )}g\bigl(s,x(s),I^{\gamma}x(s)\bigr)\,ds, \quad t\in J $$

(4.6)

and

$$ \mathcal{Q}x(t)= \sum_{i=1}^{m} \frac{1}{\Gamma(\beta_{i}+\omega )}\int_{0}^{t}(t-s)^{\beta_{i}+\omega-1}h_{i} \bigl(s,x(s)\bigr)\,ds, \quad t\in J, $$

(4.7)

and

$$ \mathcal{T}x(t)= \frac{1}{\Gamma(\omega)}\int_{0}^{t}(t-s)^{\omega -1} \mathcal{A}x(s)\mathcal{D}x(s)\,ds, \quad t\in J. $$

(4.8)

For any $y\in B_{R}$, we have

$$\begin{aligned} \bigl\vert x(t)\bigr\vert =&\bigl\vert \mathcal{Q}x(t)+\mathcal{T}y(t) \bigr\vert \\ \leq& \sum_{i=1}^{m}\frac{1}{\Gamma(\beta_{i}+\omega )} \int_{0}^{t}(t-s)^{\beta_{i}+\omega-1}\bigl\vert h_{i}\bigl(s,x(s)\bigr)\bigr\vert \,ds \\ &{} +\frac{1}{\Gamma(\omega)}\int_{0}^{t}(t-s)^{\omega -1} \bigl\vert \mathcal{A}y(s)\bigr\vert \bigl\vert \mathcal{D}y(s)\bigr\vert \,ds \\ \leq& \sum_{i=1}^{m}\frac{1}{\Gamma(\beta_{i}+\omega )} \int_{0}^{t}(t-s)^{\beta_{i}+\omega-1}\bigl\vert \theta_{i}(s)\bigr\vert \,ds \\ &{}+ \frac{1}{\Gamma(\omega)}\int_{0}^{t}(t-s)^{\omega -1} \bigl\vert \mu(s)\bigr\vert \int_{0}^{s} \frac{(s-\tau)^{\alpha-1}}{\Gamma(\alpha )}\bigl\vert \nu(\tau)\bigr\vert \, d\tau \,ds \\ \leq& \sum_{i=1}^{m}\frac{T^{\beta_{i}+\omega}}{\Gamma(\beta _{i}+\omega+1)}\| \theta_{i}\|+\frac{T^{\omega+\alpha}}{\Gamma(\omega +\alpha+1)}\|\mu\|\|\nu\| \\ \le& R, \end{aligned}$$

and therefore $\|x\|\le R$, which means that $x\in B_{R}$. Hence, the condition (iii) of Lemma 4.1 holds.

Next we will show that $\mathcal{Q}$ satisfy the condition (ii) of Lemma 4.1. The operator $\mathcal{Q}$ is obviously continuous. Also, $\mathcal{Q}$ is uniformly bounded on $B_{R}$ as

$$ \|\mathcal{Q}x\|\leq \sum_{i=1}^{m} \frac{T^{\beta_{i}+\omega}}{\Gamma (\beta_{i}+\omega+1)}\|\theta_{i}\|. $$

Let $\tau_{1},\tau_{2}\in J$ with $\tau_{1}<\tau_{2}$ and $x\in B_{R}$. We define $\sup_{(t,x)\in J\times B_{R}}|h_{i}(t,x)|=\bar{h}_{i}<\infty$, $i=1,2,\ldots, m$. Then we have

$$\begin{aligned} \bigl\vert \mathcal{Q}x(\tau_{2})-\mathcal{Q}x(\tau_{1}) \bigr\vert =& \sum_{i=1}^{m} \frac{1}{\Gamma(\beta_{i}+\omega)}\biggl\vert \int_{0}^{\tau _{2}}( \tau_{2}-s)^{\beta_{i}+\omega-1}h_{i}\bigl(s,x_{i}(s) \bigr)\,ds \\ &{}-\int_{0}^{\tau_{1}}(\tau_{1}-s)^{\beta_{i}+\omega -1}h_{i} \bigl(s,x_{i}(s)\bigr)\,ds\biggr\vert \\ \le& \sum_{i=1}^{m}\frac{\bar{h}_{i}}{\Gamma(\beta _{i}+\omega)} \biggl\vert \int_{0}^{\tau_{1}}\bigl[( \tau_{2}-s)^{\beta_{i}+\omega -1}-(\tau_{1}-s)^{\beta_{i}+\omega-1}\bigr] \,ds \\ &{}+\int_{\tau_{1}}^{\tau_{2}}(\tau_{2}-s)^{\beta_{i}+\omega-1} \,ds\biggr\vert \\ \leq& \sum_{i=1}^{m}\frac{\bar{h}_{i}}{\Gamma(\beta _{i}+\omega+1)} \bigl\vert \tau_{2}^{\beta_{i}+\omega}-\tau_{1}^{\beta _{i}+\omega} \bigr\vert , \end{aligned}$$

which is independent of x and tends to zero as $\tau_{2}-\tau_{1}\to 0$. Thus, $\mathcal{Q}$ is equicontinuous. So $\mathcal{Q}$ is relatively compact on $B_{R}$. Hence, by the Arzelá-Ascoli theorem, $\mathcal{Q}$ is compact on $B_{R}$.

Now we show that ${\mathcal{T}}$ is a contraction mapping. Let $x, y\in B_{R}$. Then for $t\in J$ we have

$$\begin{aligned}& \bigl\vert {\mathcal{T}}x(t)-{\mathcal{T}}y(t)\bigr\vert \\& \quad = \biggl\vert \int_{0}^{t} \frac{(t-s)^{\omega-1}}{\Gamma(\omega )}\bigl[{\mathcal{A}}x(s){\mathcal{D}}x(s)-{\mathcal{A}}y(s){ \mathcal{D}}y(s)\bigr]\,ds \biggr\vert \\& \quad = \biggl\vert \int_{0}^{t} \frac{(t-s)^{\omega-1}}{\Gamma(\omega )}\bigl[{\mathcal{A}}x(s){\mathcal{D}}x(s)-{\mathcal{A}}y(s){ \mathcal{D}}x(s)+ {\mathcal{A}}y(s){\mathcal{D}}x(s)-{\mathcal{A}}y(s){ \mathcal{D}}y(s)\bigr]\,ds \biggr\vert \\& \quad \le \int_{0}^{t}\frac{(t-s)^{\omega-1}}{\Gamma(\omega)} \bigl\{ \bigl\vert {\mathcal{D}}x(s)\bigr\vert \bigl\vert {\mathcal{A}}x(s)-{ \mathcal{A}}y(s)\bigr|+\bigl|{\mathcal{A}}y(s)\bigr\vert \bigl\vert {\mathcal{D}}x(s)-{ \mathcal{D}}y(s)\bigr\vert \bigr\} \,ds \\& \quad \le \int_{0}^{t}\frac{(t-s)^{\omega-1}}{\Gamma(\omega)} \biggl\{ \frac {T^{\alpha}}{\Gamma(\alpha+1)}\|\nu\|\|\phi\|\|x-y\| \\& \qquad {}+\|\mu\|\|\chi\| \biggl[\|x-y\|+\|x-y\|\int_{0}^{s} \frac{(s-u)^{\gamma -1}}{\Gamma(\gamma)}\, du \biggr] \biggr\} \,ds \\& \quad \le \biggl\{ \frac{T^{\alpha}}{\Gamma(\alpha+1)}\frac{T^{\omega }}{\Gamma(\omega+1)}\|\nu\|\|\phi\|+\|\mu\|\| \chi\| \biggl[\frac {T^{\omega}}{\Gamma(\omega+1)} +\frac{T^{\omega+\gamma}}{\Gamma(\omega+\gamma+1)} \biggr] \biggr\} \|x-y\|. \end{aligned}$$

Hence, by (4.4), $\mathcal{T}$ is a contraction mapping, and thus the condition (i) of Lemma 4.1 is satisfied.

Thus all the assumptions of Lemma 4.1 are satisfied. Therefore, the conclusion of Lemma 4.1 implies that problem (1.2) has at least one solution on J. □

5 Examples

In this section, we present two examples to illustrate our results.

Example 5.1

Consider the following hybrid fractional integro-differential equation:

$$ \left \{ \textstyle\begin{array}{l} D^{\frac{3}{4}} [\frac{x(t)- \sum_{i=1}^{4}I^{\beta_{i}}h_{i}(t,x(t))}{f(t,x(t))} ]=\frac {(t-1)^{2}+3}{35(13-t^{2})}(7|x(t)|+15), \\ x(0)=0,\quad t\in[0,1], \end{array}\displaystyle \right . $$

(5.1)

where

$$\begin{aligned} \sum_{i=1}^{4}I^{\beta_{i}}h_{i} \bigl(t,x(t)\bigr) =&I^{1/3}\frac {2te^{-3t}}{15(3+t)} \biggl(\frac{x^{2}(t)+9|x(t)|}{|x(t)|+5}+ \frac {12e^{3t}}{5} \biggr) \\ &{}+ I^{7/4}\frac{t\sin t}{7(4+e^{t})} \biggl(\frac {x^{2}(t)+4|x(t)|}{|x(t)|+3}+\cos t \biggr) \\ &{}+ I^{10/3}\frac{2\sin\pi t}{9+(5+t)^{2}} \biggl(\frac {x^{2}(t)+8|x(t)|}{|x(t)|+5}+ \frac{4}{3} \biggr) \\ &{}+ I^{29/6}\frac{3t\cos t+4t\sin t}{10(4-t)^{2}} \biggl(\frac{x^{2}(t)+5|x(t)|}{|x(t)|+4}+ \frac{t}{t+2} \biggr) \end{aligned}$$

and

$$ f\bigl(t,x(t)\bigr)=\frac{3(\cos\pi t+2t)}{5(2+10t)^{2}} \biggl(\frac{x^{2}(t)+5|x(t)|}{|x(t)|+3} \biggr)+ \frac{8-2^{2-t}}{27}. $$

Here $\alpha=3/4$, $T=1$, $m=4$, $\beta_{1}=1/3$, $\beta_{2}=7/4$, $\beta_{3}=10/3$, and $\beta_{4}=29/6$. We can show that

$$ \bigl\vert f(t,x)-f(t,y)\bigr\vert \leq \biggl(\frac{1+2t}{(2+10t)^{2}} \biggr)|x-y| $$

and

$$\begin{aligned}& \bigl\vert h_{1}(t,x)-h_{1}(t,y)\bigr\vert \leq \biggl(\frac{18t}{75(3+t)} \biggr)|x-y|, \\& \bigl\vert h_{2}(t,x)-h_{2}(t,y)\bigr\vert \leq \biggl( \frac{4t}{21(4+e^{t})} \biggr)|x-y|, \\& \bigl\vert h_{3}(t,x)-h_{3}(t,y)\bigr\vert \leq \biggl( \frac{16}{45+5(5+t)^{2}} \biggr)|x-y|, \\& \bigl\vert h_{4}(t,x)-h_{4}(t,y)\bigr\vert \leq \biggl( \frac{5t}{8(4-t)^{2}} \biggr)|x-y|. \end{aligned}$$

It follows that $\phi(t)=(1+2t)(2+10t)^{-2}$, $\psi _{1}(t)=18t(3+t)^{-1}/75$, $\psi_{2}(t)=4t(4+e^{t})^{-1}/21$, $\psi _{3}(t)=16(45+5(5+t)^{2})^{-1}$, and $\psi_{4}(t)=5t(4-t)^{-2}/8$, which give norms $\|\phi\|=1/4$, $\|\psi_{1}\|=3/50$, $\|\psi_{2}\| =4(28+7e)^{-1}/3$, $\|\psi_{3}\|=8/85$, and $\|\psi_{4}\|=5/72$.

Since

$$ \bigl\vert g(t,x,y)\bigr\vert = \biggl\vert \frac {(t-1)^{2}+3}{35(13-t^{2})}\bigl(7 \vert x\vert +15\bigr)\biggr\vert \leq \biggl(\frac {(t-1)^{2}+3}{13-t^{2}} \biggr) \biggl(\frac{|x|}{5}+\frac{3}{7} \biggr), $$

we set $p(t)=((t-1)^{2}+3)/(13-t^{2})$ and also the function Ψ as

$$ \Psi(x)=\frac{x}{5}+\frac{3}{7}. $$

It is easy to verify that $\|p\|=4/13$, $F_{0}=\sup_{t\in[0,2]}|f(t,0)|=4/27$, and $K_{0}=\sup_{t\in[0,2]}|h_{i}(t, 0)|=2/25$, $i=1,2,3,4$. We see that condition (H₃) is followed with a number $r\in[0.187454,131.292851]$. Consequently all conditions in Theorem 3.1 are satisfied. Therefore, problem (5.1) has at least one solution on $[0,1]$.

Example 5.2

Consider the following hybrid fractional sequential integro-differential equation:

$$ \left \{ \textstyle\begin{array}{l} D^{4/5} [\frac{D^{2/3}x(t)- \sum_{i=1}^{5}I^{\beta_{i}}h_{i}(t,x(t))}{f(t,x(t))} ] \\ \quad =\frac{1}{2}\arctan (\frac{2|x(t)|}{3+|x(t)|}\cos\frac{\pi t}{6} ) -\frac{1}{2}\arctan (\frac{2 \vert I^{8/3}x(t)\vert }{3+\vert I^{8/3}x(t)\vert }\cos\frac{\pi t}{6} ), \\ x(0)=0, \qquad D^{2/3}x(0)=0,\quad t\in[0,1], \end{array}\displaystyle \right . $$

(5.2)

where

$$\begin{aligned} \sum_{i=1}^{5}I^{\beta_{i}}h_{i} \bigl(t,x(t)\bigr) =&I^{1/6}\log \biggl( 1+\bigl|x(t)\bigr|\frac{e^{t}\sin t}{1+|x(t)|} \biggr) \\ &{}+ I^{3/5} \biggl(\frac{\cos t\sin x(t)}{\sqrt{2+t}}+\frac{\cos x(t) \sin t}{\sqrt{2-t}} \biggr) \\ &{} + I^{5/4}\bigl(t e^{t}\cos\bigl(x(t)+\pi t \bigr)+e^{-t}\sin\bigl(x(t)+\pi t\bigr)\bigr) \\ &{} + I^{7/3} \biggl(\frac{2t^{2}+e^{-|x(t)|}\sin t}{1+|x(t)|+t^{3}} \biggr) \\ &{} + I^{9/2} \biggl(\frac{te^{t}}{1+e^{t}}+\frac {t^{2}|x(t)|}{1+t|x(t)|} \biggr) \end{aligned}$$

and

$$ f\bigl(t,x(t)\bigr)= \biggl(\frac{|x(t)|+1}{|x(t)|+2} \biggr)\frac {(7-e^{t})}{2\sqrt{25-t^{2}}}+ \frac{2-t}{10}. $$

Here $\alpha=4/5$, $\omega=2/3$, $m=5$, $T=1$, $\beta_{1}=1/6$, $\beta _{2}=3/5$, $\beta_{3}=5/4$, $\beta_{4}=7/3$, $\beta_{5}=9/2$, $\gamma=8/3$. We can show that

$$ \bigl\vert f(t,x)-f(t,y)\bigr\vert \leq \biggl(\frac{7-e^{t}}{8\sqrt{25-t^{2}}} \biggr)|x-y| $$

and

$$ \bigl\vert g(t,x,y)-g(t,\bar{x},\bar{y})\bigr\vert \leq\frac{1}{3} \cos\frac{\pi t}{6}\bigl(\vert x-y\vert +|\bar{x}-\bar{y}|\bigr). $$

Therefore, we choose

$$ \phi(t)=\frac{7-e^{t}}{8\sqrt{25-t^{2}}} \quad \text{and}\quad \chi (t)=\frac{1}{3}\cos \frac{\pi t}{6}. $$

It is easy to see that

$$\begin{aligned}& \bigl\vert f(t,x)\bigr\vert \leq \frac{7-e^{t}}{2\sqrt{25-t^{2}}}+\frac{2-t}{10}=\mu (t), \\& \bigl\vert g(t,x,y)\bigr\vert \leq 2\cos\frac{\pi t}{6}=\nu(t). \end{aligned}$$

The functions $h_{i}(t,x(t))$, $i=1,2,\ldots,5$, are bounded by the corresponding positive functions $\theta_{i}(t)$, $i=1,2,\ldots,5$, as follows:

$$\begin{aligned}& \bigl\vert h_{1}(t,x)\bigr\vert \leq 1+e^{t}\sin t= \theta_{1}(t), \\& \bigl\vert h_{2}(t,x)\bigr\vert \leq \sqrt{\frac{2-t\cos2t}{4-t^{2}}}+ \frac{1}{2}=\theta _{2}(t), \\& \bigl\vert h_{3}(t,x)\bigr\vert \leq \sqrt{\frac{e^{4t}+1}{e^{2t}}}= \theta_{3}(t), \\& \bigl\vert h_{4}(t,x)\bigr\vert \leq \frac{2t^{2}+\sin t}{1+t^{3}}= \theta_{4}(t), \\& \bigl\vert h_{5}(t,x)\bigr\vert \leq \frac{t(1+2e^{t})}{1+e^{t}}= \theta_{5}(t). \end{aligned}$$

Hence the conditions (A₁) and (A₂) are satisfied. Putting $\|\phi\| =3/20$, $\|\chi\|=1/3$, $\|\mu\|=4/5$, and $\|\nu\|=2$, we obtain

$$ \frac{T^{\alpha}}{\Gamma(\alpha+1)}\frac{T^{\omega}}{\Gamma (\omega+1)}\|\nu\|\|\phi\|+\|\mu\|\|\chi\| \biggl[ \frac{T^{\omega }}{\Gamma(\omega+1)}+\frac{T^{\omega+\gamma}}{\Gamma(\omega +\gamma+1)} \biggr]\approx0.680993 < 1. $$

By Theorem 4.1, problem (5.2) has a solution on $[0,1]$.

6 Conclusions

In this paper, we have studied the existence of solutions for initial value problems of first and second order for hybrid fractional integro-differential equations. The first result has been obtained by using a hybrid fixed point theorem for three operators in a Banach algebra due to Dhage [27], while the second has been obtained by using a useful generalization of Krasnoselskii’s fixed point theorem due to Dhage [28]. The main results are well illustrated with the help of examples.