# On high order fractional integro-differential equations including the Caputo–Fabrizio derivative

## Abstract

By using the fractional Caputo–Fabrizio derivative, we introduce two types new high order derivations called CFD and DCF. Also, we study the existence of solutions for two such type high order fractional integro-differential equations. We illustrate our results by providing two examples.

## 1 Introduction

Fractional integro-differential equations have been studied by many researchers from different points of view during the last decades (see for example, [5, 10] and [1519]). In 2015, a new fractional derivation without singular kernel was introduced by Caputo and Fabrizio ([8]). Some researchers tried to use it for solving different equations (see, for example, [2, 9] and [14]). Recently, approximate solutions of some fractional differential equations have been reviewed (see, for example, [3, 4, 6, 12, 13] and [7]). Also, one is finding some new applications for fractional derivations (see, for example, [3]).

In this manuscript we consider $$b>0$$, $$x\in H^{1}(0,b)$$ and $$\alpha\in (0,1)$$. The expression of the Caputo–Fabrizio fractional derivative of order α for the function x has the form $${}^{\mathrm{CF}}D^{\alpha}x(t)=\frac{B(\alpha)}{1-\alpha}\int_{0}^{t}\exp(\frac {-\alpha}{1-\alpha}(t-s))x^{\prime}(s)\,ds$$, where $$t\geq0$$ ([1, 8] and [9]). $$B(\alpha)$$ is a normalization constant $$(B(1)=B(0)=1)$$. The fractional integral of order α for the function x is written as ([14]) $${}^{\mathrm{CF}}I^{\alpha} x(t)=\frac{ 1-\alpha}{B(\alpha)}x(t) +\frac{ \alpha}{B(\alpha)}\int_{0}^{t} x(s) \,ds$$, whenever $$0<\alpha<1$$. If $$n\geq1$$ and $$\alpha\in[0,1]$$, then the fractional derivative $${}^{\mathrm{CF}}D^{\alpha+n}$$ of order $$n+\alpha$$ is defined by $${}^{\mathrm{CF}}D^{\alpha+n}x:= {}^{\mathrm{CF}}D^{\alpha}( D^{n}x(t))$$ ([6] and [8]). If the function x is such that $$x^{(k)}=0$$ for $$k=1,2,3,\ldots,n$$, then $${}^{\mathrm{CF}}D^{\alpha}( D^{n}x(t))=D^{n} ({}^{\mathrm{CF}}D^{\alpha}) x(t)$$([8]). Here, D is the ordinary derivation.

### Lemma 1.1

([1] and [14])

Let $$0<\alpha<1$$. Then the unique solution for the problem $${}^{\mathrm{CF}}D^{\alpha}x(t)=y(t)$$ is given by $$x(t)=x(0)+\frac{ 1-\alpha}{B(\alpha)}y (t) +\frac{ \alpha}{B(\alpha )}\int_{0}^{t} x(s)\,ds$$.

### Theorem 1.2

([11])

Let $$(X,d)$$ be a complete metric space and $$F:X\to X$$ be a mapping such that $$\varphi (d(Fx,Fy))\leq\varphi(d(x,y))-\phi(d(x,y))$$, for all $$x,y \in X$$, where $$\varphi,\phi:[0,1]\to[0,1]$$ are continuous non-decreasing maps and $$\varphi(t)=\phi(t)=0$$ if and only if $$t=0$$. Then F has a unique fixed point.

## 2 Main result

Let n be a natural number, $$\alpha\in(0,1)$$ and $$x^{(n)}\in H^{1}(0,1)$$. Then the fractional CFD of order α and n is defined by

$${}^{\mathrm{CF}}D^{\alpha+n}x(t) = {}^{\mathrm{CF}}D^{\alpha } \bigl(D^{n}x(t)\bigr)=\frac{B(\alpha)}{1-\alpha} \int_{0}^{t}\exp\biggl(\frac{-\alpha }{1-\alpha}(t-s) \biggr)x^{(n+1)}(s)\,ds.$$

Also, the fractional DCF of order α and n is defined by

$$\bigl( ^{CF }D^{\alpha}\bigr)^{(n)}x(t) =D^{n}\bigl( {}^{\mathrm{CF}}D^{\alpha }x(t)\bigr)= \frac{B(\alpha)}{1-\alpha}\frac{d^{n}}{dt^{n}} \int_{0}^{t}\exp\biggl(\frac {-\alpha}{1-\alpha}(t-s) \biggr)x^{\prime} ( s)\,ds.$$

Here, D is the ordinary derivative.

### Lemma 2.1

Let n be a natural number and $$\alpha\in(0,1)$$. Then

$$\bigl( {}^{\mathrm{CF}}D^{\alpha}\bigr)^{(n)}x(t) = {}^{\mathrm{CF}}D^{\alpha +n}x(t) + \exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr) \sigma(\alpha,n,0),$$

where $$\sigma(\alpha,n,t)=\frac{B(\alpha)}{1-\alpha}\sum_{i=1}^{n} (\frac{-\alpha}{1-\alpha})^{n-i}x^{(i)}(t)$$.

### Proof

For each $$k\geq1$$, we have

\begin{aligned} \int_{0}^{t}\exp\biggl(\frac{-\alpha}{1-\alpha}(t-s) \biggr)x^{(k)} ( s)\,ds&=x^{(k-1)}(t) -\exp\biggl( \frac{-\alpha}{1-\alpha} t \biggr)x^{(k-1)} (0) \\ &\quad {}+\biggl(\frac{-\alpha}{1-\alpha}\biggr) \int_{0}^{t}\exp\biggl(\frac{-\alpha}{1-\alpha }(t-s) \biggr)x^{(k-1)} ( s)\,ds. \end{aligned}

Now by using repetition of the last relation, we get

\begin{aligned} {}^{\mathrm{CF}}D^{\alpha+n}x(t) &= {}^{\mathrm{CF}}D^{\alpha} \bigl( D^{n}x(t)\bigr)=\frac{B(\alpha)}{1-\alpha} \int_{0}^{t}\exp\biggl(\frac{-\alpha }{1-\alpha}(t-s) \biggr)x^{(n+1)} ( s)\,ds \\ &=\frac{B(\alpha)}{1-\alpha}\biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n} \int_{0}^{t}\exp \biggl(\frac{-\alpha}{1-\alpha}(t-s) \biggr)x^{\prime} ( s)\,ds\\ &\quad {}+\frac{B(\alpha )}{1-\alpha}\sum_{i=1}^{n} \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(t) \\ &\quad {}-\frac{B(\alpha)}{1-\alpha}\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\sum _{i=1}^{n} \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(0)\\ &=\frac{B(\alpha )}{1-\alpha}\biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n} \int_{0}^{t}\exp\biggl(\frac{-\alpha }{1-\alpha}(t-s) \biggr)x^{\prime}(s)\,ds \\ &\quad {}+\sigma(\alpha,n,t)-\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\sigma( \alpha,n,0)\\ & = \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n} {}^{\mathrm{CF}}D^{\alpha}x(t)+ \sigma (\alpha,n,t)-\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\sigma(\alpha,n,0). \end{aligned}

Also, we have

\begin{aligned} \bigl({}^{\mathrm{CF}}D^{\alpha}\bigr)^{(n)}x(t)&= D^{n}\bigl(D^{\alpha}x(t)\bigr)=\biggl(\frac {-\alpha}{1-\alpha} \biggr)^{n} {}^{\mathrm{CF}}D^{\alpha}x(t)+\frac{B(\alpha )}{1-\alpha} \sum_{i=1}^{n} \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(t) \\ &=\frac{B(\alpha)}{1-\alpha}\biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n} \int_{0}^{t}\exp \biggl(\frac{-\alpha}{1-\alpha}(t-s) \biggr)x^{\prime} ( s)\,ds\\ &\quad {}+\frac{B(\alpha )}{1-\alpha}\sum_{i=1}^{n} \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(t) \\ &=\biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n} {}^{\mathrm{CF}}D^{\alpha}x(t)+ \sigma (\alpha,n,t). \end{aligned}

Hence $$( {}^{\mathrm{CF}}D^{\alpha})^{(n)}x(t) = {}^{\mathrm{CF}}D^{\alpha+n}x(t) + \exp(\frac{-\alpha}{1-\alpha}t)\sigma(\alpha ,n,0)$$. □

By using Lemma 2.1, we conclude that $${}^{\mathrm{CF}}D^{\alpha +n}x(t)=({}^{\mathrm{CF}}D^{\alpha})^{(n)}x(t)$$ whenever $$x^{(k)}(0)=0$$ for $$0\leq k \leq n$$.

### Lemma 2.2

Let n be a natural number, $$\alpha\in(0,1)$$ and $$y\in H^{1}(0,1)$$. Then the solution of the problem $${}^{\mathrm{CF}}D^{{\alpha+n}}x(t)=y(t)$$ is given by

$$x(t)=\frac{ 1-\alpha}{B(\alpha)}J^{n}y(t)+\frac{\alpha}{B(\alpha)} J^{n+1}y(t) + x(0)+tx^{\prime}(0)+t^{2} \frac{x^{\prime\prime }(0)}{2!}+\cdots+t^{n}\frac{x^{(n)}(0)}{n!}.$$

### Proof

By using Lemma 1.1 for the equation $${}^{\mathrm{CF}} D^{\alpha +n} x(t)= {}^{\mathrm{CF}} D^{\alpha} x^{(n)}(t)=y(t)$$, we get $$x^{(n)}(t)= x^{(n)}(0)+\frac{ 1-\alpha}{B(\alpha)} y(t) +\frac{ \alpha}{B(\alpha)} \int_{0}^{t} y(s)\,ds$$. By using an integration, we obtain

$$x^{(n-1)}(t)= x^{(n-1)}(0)+tx^{(n)}(0)+\frac{ 1-\alpha}{B(\alpha)} \int _{0}^{t}y(s)\,ds +\frac{ \alpha}{B(\alpha)} \int_{0}^{t} \int_{0}^{s} y(r)\,dr\,ds.$$

By repeating this method, we deduce that

\begin{aligned} x^{(n-2)}(t)&= x^{(n-2)}(0)+tx^{(n-1)}(0)+ \frac{t^{2}}{2}x^{(n)}(0) \\ &\quad {}+\frac{ 1-\alpha}{B(\alpha)} \int_{0}^{t} \int_{0}^{s} y(r)\,dr\,ds +\frac{ \alpha}{B(\alpha)} \int_{0}^{t} \int_{0}^{s} \int_{0}^{r} y(k)\,dk\,dr\,ds. \end{aligned}

By continuing the process, we conclude that

$$x(t)=\frac{ 1-\alpha}{B(\alpha)}J^{n}y(t)+\frac{ \alpha}{B(\alpha)} J^{n+1}y(t) + x(0)+tx^{\prime}(0)+t^{2} \frac{x^{\prime\prime }(0)}{2!}+\cdots+t^{n}\frac{x^{(n)}(0)}{n!}.$$

On the other hand, by using some calculation, one can find that the given map $$x(t)$$ is a solution for the problem $${}^{\mathrm{CF}}D^{{\alpha+n}}x(t)=y(t)$$. □

### Lemma 2.3

Let n be a natural number, $$\alpha\in(0,1)$$ and $$y\in H^{1}(0,1)$$. Then the solution of the problem $$({}^{\mathrm{CF}}D^{\alpha})^{(n)}x(t)=y(t)$$ is given by

\begin{aligned} x(t) &=\frac{ 1-\alpha}{B(\alpha)}J^{n}y(t)+\frac{ \alpha}{B(\alpha)} J^{n+1}y(t) + x(0)+tx^{\prime}(0)+\cdots\\ &\quad {}+t^{n} \frac{x^{(n)}(0)}{n!}- \frac{t^{n}}{n!}\sum_{i=1}^{n} \biggl( \frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(0). \end{aligned}

### Proof

By using Lemma 2.2 for $$( {}^{\mathrm{CF}}D^{\alpha })^{(n)}x(t)= {}^{\mathrm{CF}}D^{\alpha+n}x(t)+ \exp(\frac{-\alpha }{1-\alpha}t)\sigma(\alpha,n,0)$$, we get

\begin{aligned} x(t)& =\frac{ 1-\alpha}{B(\alpha)}J^{n}\biggl(y(t)- \exp\biggl( \frac{-\alpha }{1-\alpha}t\biggr)\sigma(\alpha,n,0)\biggr)\\ &\quad {}+\frac{ \alpha}{B(\alpha)} J^{n+1}\biggl(y(t)- \exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\sigma( \alpha,n,0)\biggr) \\ &\quad {}+ x(0)+tx^{\prime}(0)+t^{2}\frac{x^{\prime\prime}(0)}{2!}+ \cdots+t^{n}\frac {x^{(n)}(0)}{n!} \end{aligned}

or equivalently

\begin{aligned} x(t)& =\frac{ 1-\alpha}{B(\alpha)} J^{n}y(t)- \frac{ 1-\alpha}{B(\alpha )}\sigma(\alpha,n,0)J^{n}\exp\biggl(\frac{-\alpha}{1-\alpha}t \biggr)+\frac{ \alpha}{B(\alpha)} J^{n+1}y(t) \\ &\quad {}- \frac{ \alpha}{B(\alpha)}\sigma(\alpha,n,0)J^{n+1}\exp\biggl( \frac{-\alpha }{1-\alpha}t\biggr) + x(0)+tx^{\prime}(0)+t^{2} \frac{x^{\prime\prime }(0)}{2!}+\cdots+t^{n}\frac{x^{(n)}(0)}{n!}\\ & = \frac{ 1-\alpha}{B(\alpha )}J^{n}y(t) +\frac{ \alpha}{B(\alpha)} J^{n+1}y(t)- \sigma(\alpha,n,0)J^{n} \biggl[\frac{ 1-\alpha}{B(\alpha)}\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\\ &\quad {}+ \frac{ \alpha }{B(\alpha)}J^{1}\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr) \biggr] + x(0)+tx^{\prime }(0)+t^{2}\frac{x^{\prime\prime}(0)}{2!}+\cdots+t^{n}\frac{x^{(n)}(0)}{n!} \\ & =\frac{ 1-\alpha}{B(\alpha )}J^{n}y(t)+ \frac{ \alpha}{B(\alpha)} J^{n+1}y(t)- \sigma(\alpha ,n,0)J^{n}\biggl[ \frac{ 1-\alpha}{B(\alpha)}\exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr) \\ &\quad {}+\frac{ 1-\alpha}{B(\alpha)}\biggl(1- \exp\biggl(\frac{-\alpha}{1-\alpha}t\biggr)\biggr) \biggr] + x(0)+tx^{\prime}(0)+t^{2}\frac{x^{\prime\prime}(0)}{2!}+ \cdots+t^{n}\frac {x^{(n)}(0)}{n!} \\ &=\frac{ 1-\alpha}{B(\alpha)}J^{n}y(t)+\frac{ \alpha}{B(\alpha)} J^{n+1}y(t)+ x(0)+tx^{\prime}(0)+t^{2} \frac {x^{\prime\prime}(0)}{2!}+\cdots+t^{n}\frac{x^{(n)}(0)}{n!} \\ &\quad {}- \frac {t^{n}}{n!} \sum_{i=1}^{n} \biggl(\frac{-\alpha}{1-\alpha}\biggr)^{n-i}x^{(i)}(0). \end{aligned}

□

### Lemma 2.4

Let $$\alpha\in(0,1)$$, $$2< q=2+\alpha<3$$ and $$y\in H^{1}(0,1)$$. The fractional differential equation $$^{ {CF}} D^{q} x(t)=y(t)$$ with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime \prime}(0) =0$$ has the unique solution of the form $$x(t)=\int_{0}^{1} G(t,s)y(s)\,ds$$, where $$G(t,s)=\frac{- (1-\alpha)t}{2B(\alpha) }-\frac {\alpha t}{2B(\alpha)}$$ whenever $$0< t\leq s<1$$ and $$G(t,s)=\frac {1-\alpha}{B(\alpha)}(t-s) + \frac{\alpha}{2B(\alpha)}(t-s)^{2} -\frac {(1-\alpha)t}{2B(\alpha)}- \frac{\alpha t}{2B(\alpha)}(t-s)$$ whenever $$0< s\leq t<1$$.

### Proof

By using Lemma 2.2, we get $$x(t)=\frac{ 1-\alpha}{B(\alpha )}J^{2 }y(t)+\frac{ \alpha}{B(\alpha)} J^{3}y(t) +tx^{\prime}(0)$$. Hence, we obtain $$x^{\prime}(t)= \frac{ 1-\alpha}{B(\alpha)}J^{1}y(t)+\frac{ \alpha }{B(\alpha)} J^{2 }y(t) + x^{\prime}(0)$$. By using the boundary conditions $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime}(1)= \frac{ 1-\alpha}{B(\alpha)}J^{1}y(1)+\frac{ \alpha }{B(\alpha)} J^{2 }y(1) + x^{\prime}(0)$$, we have $$x(t)=\frac{ 1-\alpha}{B(\alpha)}J^{2 }y(t)+\frac{ \alpha}{B(\alpha)} J^{3}y(t)-\frac{ (1-\alpha) t}{2B(\alpha) }J^{1}y(1)-\frac{\alpha t}{2B(\alpha) }J^{2 }y(1)$$. Thus, $$x(t)= \frac{ 1-\alpha}{B(\alpha)} \int_{0}^{t}y(s)(t-s) \,ds+ \frac{\alpha}{2B(\alpha) }\int_{0}^{t}y(s)(t-s)^{2}\,ds -\frac{ (1-\alpha) t}{2B(\alpha)} \int_{0}^{1}y(s) \,ds-\frac{\alpha t}{2B(\alpha) } \int_{0}^{1}y(s)(t-s) \,ds =\int_{0}^{1} G(t,s)y(s)\,ds$$. Note that $${}^{\mathrm{CF}}D^{q}x(t)=0$$ if and only if $$x(t)=0$$. This implies that the given map $$x(t)$$ is a unique solution. □

Note that $$|G(t,s)|\leq|\frac{ 1-\alpha}{B(\alpha)}|+ |\frac{\alpha }{2B(\alpha) }| +|\frac{-\alpha t}{2B(\alpha) } |+|\frac{-(1-\alpha) t}{2B(\alpha)}|<\frac{3}{2B(\alpha)}$$, for $$t \in[0,1]$$. Let $$\mu,\mu_{1},\mu_{2}, k_{1},k_{2}\in C^{1}[0,1]$$, $$m_{1}$$, $$m_{2}$$, h and g be bounded continuous functions on $$I:=[0,1]$$ with $$M_{1}=\sup_{t\in I}|\mu(t)|<\infty$$, $$M_{2}=\sup_{t\in I}|\mu_{1}(t)|<\infty$$, $$M_{3}=\sup_{t\in I}|\mu_{2}(t)|<\infty$$, $$M_{4}=\sup_{t\in I}|k_{1}(t)|<\infty$$, $$M_{5}=\sup_{t\in I}|k_{2}(t)|<\infty$$, $$M_{6}=\sup_{t\in I}|m_{1}(t)| <\infty$$, $$M_{7}=\sup_{t\in I}|m_{2}(t)| <\infty$$, $$M_{8}=\sup_{t\in I}|h(t)| <\infty$$, $$M_{9}=\sup_{t\in I}|g (t)| <\infty$$, $$N_{1}=\sup_{t\in I}|\mu^{\prime}(t)|<\infty$$, $$N_{2}=\sup_{t\in I}|\mu_{1}^{\prime }(t)|<\infty$$, $$N_{3}=\sup_{t\in I}|\mu_{2}^{\prime}(t)|<\infty$$, $$N_{4}=\sup_{t\in I}|K_{1}^{\prime}(t)|<\infty$$ and $$N_{5}=\sup_{t\in I}|K_{2}^{\prime}(t)|<\infty$$. Let $$\alpha\in(0,1)$$ and $$2< q= 2+\alpha<3$$. Now, we investigate the CFD fractional integro-differential problem

\begin{aligned}[b] {}^{\mathrm{CF}}D^{q}x(t)&=\mu(t) x(t)+\mu_{1}(t) x^{\prime}(t)+\mu_{2} (t)x^{\prime\prime}(t)+ k_{1}(t){}^{\mathrm{CF}}D^{\beta_{1}}x(t)+ k_{2}(t){}^{\mathrm{CF}}D^{\beta_{2}}x(t) \\ &\quad{}+ \int_{0}^{t} f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime \prime}(s), h(s){}^{\mathrm{CF}}D^{ \gamma}x(s) , g(s){}^{\mathrm{CF}}D^{ \nu}x(s) \bigr)\,ds, \end{aligned}
(1)

with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime\prime}(0) =0$$, where $$1<\beta_{1}< 2<\beta_{2}<3$$ and $$1<\gamma< 2<\nu<3$$.

### Theorem 2.5

Let $$\xi_{1}$$, $$\xi_{2}$$, $$\xi_{3}$$, $$\xi_{4}$$ and $$\xi_{5}$$ be nonnegative real numbers, $$f:[0,1]\times\Bbb {R}^{5}\to\Bbb{R}$$ an integrable function such that

\begin{aligned} &\bigl\vert f (t,x,y,z,v,w )-f \bigl(t,x^{\prime},y^{\prime} ,z^{\prime},v^{\prime},w^{\prime} \bigr) \bigr\vert \\ &\quad \leq \xi_{ 1} \bigl\vert x -x^{\prime} \bigr\vert + \xi_{ 2} \bigl\vert y -y^{\prime} \bigr\vert + \xi_{ 3} \bigl\vert z -z^{\prime} \bigr\vert + \xi_{ 4} \bigl\vert v -v^{\prime} \bigr\vert + \xi_{ 5} \bigl\vert v -v^{\prime} \bigr\vert , \end{aligned}

for all real numbers x, y, z, v, w, $$x^{\prime}$$, $$y^{\prime}$$, $$z^{\prime}$$, $$v^{\prime} ,w ^{\prime}\in\Bbb{R}$$ and $$t \in I$$. If $$\Delta<\frac {1}{2}$$, then the problem (1) has a unique solution, where $$\Delta :=\max\lbrace\Delta_{1},\Delta_{2},\Delta_{3},\Delta_{4}\rbrace$$, $$\Delta_{1}= \frac{3}{2B(\alpha)} [M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta _{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} ]$$, $$\Delta_{2}=\frac{3}{2}[ \frac{3+4\alpha}{2B(\alpha)} ] [M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu}]$$, $$\Delta_{3}= \frac{1+ \alpha}{B(\alpha)}[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} ]$$ and $$\Delta_{4}= \frac{\alpha}{B(\alpha)} [M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} ] +\frac{1-\alpha }{B(\alpha)}[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3} +B(\beta_{1}-1)[ \frac{ |1-\beta _{1}|M_{4} }{(2-\beta_{1})^{2}} +\frac{ N_{4} +M_{4}}{ 2-\beta_{1} } ] +B(\beta _{2}-2)[\frac{ |2-\beta_{2}|M_{5} }{(3-\beta_{2})^{2}} +\frac{ M_{5}+ N_{5} }{ 3-\beta _{2} } ] + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{M_{ 8}B(\gamma-1)}{ 2-\l\gamma} +\xi_{ 5} \frac{ M_{9}B(\nu-1)}{ 3-\nu}]$$.

### Proof

Consider the Banach space $$C^{3}_{\mathbb{R}}[0,1]$$ equipped with the norm $$\Vert x \Vert= \max_{t\in I}|x(t)|+ \max_{t\in I}| x^{\prime }(t) |+ \max_{t\in I}| x^{\prime\prime}(t) |+ \max_{t\in I}| x^{\prime \prime\prime}(t) |$$. Define the map $$F: C^{3}_{\mathbb{R}}[0,1]\to C^{3}_{\mathbb{R}}[0,1]$$ by

\begin{aligned} Fx(t) &= \int_{0}^{1} G(t,s)R(s)\,ds \\ &= \frac{ 1-\alpha}{B(\alpha)} \int _{0}^{t}R(s) (t-s) \,ds+ \frac{\alpha}{2B(\alpha) } \int_{0}^{t}R(s) (t-s)^{2}\,ds \\ &\quad {}-\frac{ (1-\alpha) t}{2B(\alpha)} \int_{0}^{1}R(s) \,ds-\frac{\alpha t}{2B(\alpha) } \int_{0}^{1}R(s) (t-s) \,ds, \end{aligned}

where

\begin{aligned} (Rx) (t)&=\mu(t) x(t)+\mu_{1} (t) x^{\prime}(t)+ \mu_{2}(t) x^{\prime\prime }(t)+ k_{1}(t) {}^{\mathrm{CF}}D^{\beta_{1}}x(t)+ k_{2}(t){}^{\mathrm{CF}}D^{\beta_{2}}x(t) \\ &\quad {}+ \int_{0}^{t} f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime\prime }(s), h(s) {}^{\mathrm{CF}}D^{ \gamma}x(s) , g(s) {}^{\mathrm{CF}}D^{ \nu}x(s) \bigr)\,ds \end{aligned}

and

\begin{aligned} \bigl(R^{\prime}x\bigr) (t)&=\mu(t) x ^{\prime}(t)+ \mu^{\prime}(t) x (t)+\mu _{1}^{\prime}(t) x^{\prime}(t)+\mu_{1} x^{\prime\prime}(t)\\ &\quad {}+\mu_{2}^{\prime }(t) x^{\prime\prime}(t)+\mu_{2} (t) x^{\prime\prime\prime}(t)+ k_{1}^{\prime}(t) {}^{\mathrm{CF}}D^{\beta_{1}}x(t) \\ &\quad {}+ k_{1}(t)\biggl[\frac{1-\beta_{1}}{2-\beta_{1}} {}^{\mathrm{CF}}D^{\beta _{1}}x(t)+ \frac{B(\beta_{1}-1)}{2-\beta_{1}}x^{\prime\prime}(t)\biggr]\\ &\quad {} + k_{2}(t)\biggl[ \frac{2-\beta_{2}}{3-\beta_{2}} {}^{\mathrm{CF}}D^{\beta_{2}}x(t)+\frac {B(\beta_{2}-2)}{3-\beta_{2}}x^{\prime\prime\prime}(t) \biggr] + k_{2}^{\prime}(t) {}^{\mathrm{CF}}D^{\beta_{2}}x(t) \\ &\quad {}+f \bigl(t,x(t), m _{1}(t) x^{\prime}(t) ,m _{2}(t) x^{\prime\prime}(t), h(t) {}^{\mathrm{CF}}D^{ \gamma}x(t) , g(t) {}^{\mathrm{CF}}D^{ \nu}x(t)\bigr). \end{aligned}

By using Lemma 2.4, $$x_{0}$$ is a solution for the problem (1) if and only if $$x_{0}$$ is a fixed point of the operator F. Note that

\begin{aligned} & \bigl\vert (Rx) (t)-(Ry) (t) \bigr\vert \\ &\quad \leq \biggl\vert \mu(t) x(t)+\mu_{1} (t) x^{\prime}(t)+ \mu_{2}(t) x^{\prime\prime}(t)+ k_{1}(t) {}^{\mathrm{CF}}D^{\beta_{1}}x(t)+ k_{2}(t){}^{\mathrm{CF}}D^{\beta _{2}}x(t) \\ &\qquad {}+ \int_{0}^{t} f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime\prime }(s), h(s) {}^{\mathrm{CF}}D^{ \gamma}x(s) , g(s) {}^{\mathrm{CF}}D^{ \nu}x(s) \bigr)\,ds \\ &\qquad {}- \biggl(\mu(t) y(t)+\mu_{1} (t) y^{\prime}(t)+ \mu_{2}(t) y^{\prime\prime}(t)+ k_{1}(t){}^{\mathrm{CF}}D^{\beta_{1}}y(t)+ k_{2}(t){}^{\mathrm{CF}}D^{\beta _{2}}y (t) \\ &\qquad {}+ \int_{0}^{t} f \bigl(s,y(s), m _{1}(s) y^{\prime}(s) ,m _{2}(s) y^{\prime\prime }(s), h(s) {}^{\mathrm{CF}}D^{ \gamma}y(s) , g(s) {}^{\mathrm{CF}}D^{ \nu}y(s) \bigr)\,ds\biggr) \biggr\vert \\ &\quad \leq \bigl\vert \mu(t) \bigr\vert \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert \mu_{1}(t) \bigr\vert \bigl\vert x^{\prime}(t)-y^{\prime }(t) \bigr\vert + \bigl\vert \mu_{2}(t) \bigr\vert \bigl\vert x^{\prime\prime }(t)-y^{\prime\prime}(t) \bigr\vert \\ &\qquad {}+ \bigl\vert k_{1}(t) \bigr\vert {}^{\mathrm{CF}}D^{\beta_{1}} \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert k_{2}(t) \bigr\vert {}^{\mathrm{CF}}D^{\beta_{2}} \bigl\vert x(t)-y(t) \bigr\vert \\ &\qquad {}+ \int_{0}^{t} \bigl\vert f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime \prime}(s), h(s){}^{\mathrm{CF}}D^{ \gamma}x(s) , g(s){}^{\mathrm{CF}}D^{ \nu}x(s) \bigr)\,ds \\ &\qquad {}- f \bigl(s,y(s), m _{1}(s) y^{\prime}(s) ,m _{2}(s) y^{\prime\prime}(s), h(s) {}^{\mathrm{CF}}D^{ \gamma}y(s) , g(s){}^{\mathrm{CF}}D^{ \nu}y(s) \bigr) \bigr\vert \,ds \\ &\quad \leq M_{1} \Vert x-y \Vert + M_{2} \Vert x-y \Vert + M_{3} \Vert x-y \Vert \\ &\qquad {} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } \Vert x-y \Vert + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } \Vert x-y \Vert \\ &\qquad {}+ \biggl[\xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} + \xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} \biggr] \Vert x-y \Vert \\ &\quad \leq \biggl[M_{1} + M_{2} + M_{3}+ \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } \\ &\qquad {} + \xi_{ 1} + \xi_{ 2} M_{6}+\xi_{ 3} M_{7}+ \xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} \biggr] \Vert x-y \Vert \end{aligned}

and

\begin{aligned} &\bigl\vert R^{\prime}x(t)-R^{\prime}y(t) \bigr\vert \\ &\quad \leq \bigl\vert \mu^{\prime}(t) \bigr\vert \bigl\vert x(t)-y (t) \bigr\vert + \bigl\vert \mu(t)+\mu_{1}^{\prime}(t) \bigr\vert \bigl\vert x^{\prime }(t)-y^{\prime}(t) \bigr\vert \\ &\qquad {} + \bigl\vert \mu_{1}(t)+ \mu^{\prime}_{2}(t) \bigr\vert \bigl\vert x^{\prime\prime}(t)-y^{\prime\prime}(t) \bigr\vert + \bigl\vert \mu_{2}(t) \bigr\vert \bigl\vert x^{\prime\prime\prime }(t)-y^{\prime\prime\prime}(t) \bigr\vert \\ &\qquad {} + \biggl\vert \frac{1-\beta _{1}}{2-\beta_{1}} k_{1} (t)+k_{1}^{\prime}(t) \biggr\vert {}^{\mathrm{CF}}D^{\beta_{1}} \bigl\vert x (t)-y (t) \bigr\vert + \frac{ \vert k_{1} (t)B(\beta_{1}-1) \vert }{2-\beta_{1}} \bigl\vert x^{\prime\prime }(t)-y^{\prime\prime}(t) \bigr\vert \\ &\qquad {}+ \biggl\vert \frac{2-\beta_{2}}{3-\beta_{2}} k_{2}(t)+k_{2}^{\prime}(t) \biggr\vert {}^{\mathrm{CF}}D^{\beta_{2}} \bigl\vert x (t)-y (t) \bigr\vert +\frac{ \vert k_{2}^{\prime}(t)B(\beta_{2}-2) \vert }{3-\beta_{2}} \bigl\vert x^{\prime\prime\prime}(t)-y^{\prime\prime\prime}(t) \bigr\vert \\ &\qquad {}+ \bigl\vert f \bigl(t,x(t), m _{1}(t) x^{\prime}(t) ,m _{2}(t) x^{\prime\prime}(t), h(t){}^{\mathrm{CF}}D^{ \gamma}x(t) , g(t){}^{\mathrm{CF}}D^{ \nu}x(t) \bigr) \\ &\qquad {}- f \bigl(t,y(t), m _{1}(t) y^{\prime}(t) ,m _{2}(t) y^{\prime\prime}(t), h(t){}^{\mathrm{CF}}D^{ \gamma}y(t) , g(t){}^{\mathrm{CF}}D^{ \nu}y(t) \bigr) \bigl\vert . \end{aligned}

Hence, we get

\begin{aligned} &\bigl\vert R^{\prime}x(t)-R^{\prime}y(t) \bigr\vert \\ &\quad \leq \biggl(N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+B( \beta_{1}-1)\biggl[ \frac{ \vert 1-\beta_{1} \vert M_{4} }{(2-\beta_{1})^{2}} +\frac{ N_{4} +M_{4}}{ 2-\beta_{1} } \biggr] \\ &\qquad {}+B(\beta_{2}-2)\biggl[\frac{ \vert 2-\beta_{2} \vert M_{5} }{(3-\beta _{2})^{2}} +\frac{ M_{5}+ N_{5} }{ 3-\beta_{2} } \biggr] + \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}\\ &\qquad {}+\xi_{ 4} \frac{M_{ 8}B(\gamma-1)}{ 2-\l\gamma} + \xi_{ 5} \frac{ M_{9}B(\nu-1)}{ 3-\nu} \biggr) \Vert x-y \Vert . \end{aligned}

On the other hand, we have

\begin{aligned} \bigl\vert Fx(t)-Fy(t) \bigr\vert &\leq\frac{3}{2B(\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } \\ &\quad {}+ \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} + \xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} \biggr] \Vert x-y \Vert \\ & = \Delta_{1} \Vert x-y \Vert \end{aligned}

and

\begin{aligned} F^{\prime}x(t)-F^{\prime}y(t)&=\biggl[ \frac{1-\alpha}{B(\alpha)} +\frac{ \alpha }{B(\alpha)} t\biggr] \int_{0}^{t}\bigl(Rx(s)-Ry(s)\bigr)\,ds- \frac{ \alpha}{B(\alpha)} \int _{0}^{t} s\bigl(Rx(s)-Ry(s)\bigr)\,ds \\ &\quad {}-\frac{ (1-\alpha)}{2B(\alpha)} \int_{0}^{1} \bigl(Rx(s)-Ry(s)\bigr)\,ds- \frac{ \alpha t}{B(\alpha)} \int_{0}^{1} \bigl(Rx(s)-Ry(s)\bigr)\,ds\\ &\quad {}+ \frac{ \alpha}{2B(\alpha)} \int _{0}^{1} s\bigl(Rx(s)-Ry(s)\bigr)\,ds \\ & {} \leq\frac{3}{2}\biggl[ \frac{3+4\alpha}{2B(\alpha)} \biggr] \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta _{2} } \\ &\quad {}+ \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} + \xi_{ 5}\frac{ M_{9} B(\nu-2)}{ 3-\nu} \biggr] \Vert x-y \Vert \\ & = \Delta_{2} \Vert x-y \Vert \end{aligned}

and so $$|F^{\prime}x(t)-F^{\prime}y(t)|\leq \Delta_{2} \Vert x-y \Vert$$. Also, we have

\begin{aligned} \bigl\vert F^{\prime\prime}x(t)-F^{\prime\prime}y(t) \bigr\vert &\leq \frac{ \alpha}{B(\alpha)} \int_{0}^{t} \bigl\vert Rx(s)-Ry(s) \bigr\vert \,ds+\frac{1-\alpha}{B(\alpha)} \bigl\vert Rx(t)-Ry(t) \bigr\vert \\ &\quad {}+ \frac{ \alpha}{B(\alpha)} \int_{0}^{1} \bigl\vert Rx(s)-Ry(s) \bigr\vert \,ds\\ &\leq \frac{1+\alpha}{B(\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta _{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } \\ &\quad {}+ \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} + \xi_{ 5}\frac{ M_{9}B(\nu-2)}{ 3-\nu} \biggr]\\ & =\Delta_{3} \Vert x-y \Vert \end{aligned}

and

\begin{aligned} &\bigl\vert F^{\prime\prime\prime}x(t)-F^{\prime\prime\prime}y(t) \bigr\vert \\ &\quad =\frac{ \alpha}{B(\alpha)} \bigl\vert R x(t)-Ry(t) \bigr\vert + \frac{ 1-\alpha}{B(\alpha)} \bigl\vert R^{\prime}x(t)- R^{\prime}y(t) \bigr\vert \\ &\quad \leq \biggl(\frac{ \alpha}{B(\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta _{1}-1)}{ 2-\beta_{1} } + \frac{ M_{5}B(\beta_{2}-2)}{ 3-\beta_{2} } + \xi_{ 1} + \xi_{ 2} M_{6}+\xi_{ 3} M_{7} \\ &\qquad {}+\xi_{ 4} \frac{ M_{8}B(\gamma-1)}{ 2-\gamma} +\xi_{ 5} \frac{ M_{9}B(\nu -2)}{ 3-\nu} \biggr] +\frac{ 1-\alpha}{B(\alpha)}\biggl[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3} \\ &\qquad {}+B(\beta_{1}-1)\biggl[ \frac{ \vert 1-\beta_{1} \vert M_{4} }{(2-\beta _{1})^{2}} +\frac{ N_{4} +M_{4}}{ 2-\beta_{1} } \biggr] +B(\beta_{2}-2)\biggl[\frac{ \vert 2-\beta_{2} \vert M_{5} }{(3-\beta_{2})^{2}} +\frac{ M_{5}+ N_{5} }{ 3-\beta _{2} } \biggr] \\ &\qquad {}+ \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{M_{ 8}B(\gamma -1)}{ 2-\l\gamma} + \xi_{ 5} \frac{ M_{9}B(\nu-1)}{ 3-\nu} \biggr]\biggr) \Vert x-y \Vert \\ & \quad = \Delta_{4} \Vert x-y \Vert . \end{aligned}

Thus, $$\Vert Fx-Fy \Vert \leq\Delta \Vert x-y \Vert$$ for all $$x,y\in C^{3}_{\mathbb{R}}[0,1]$$. Put $$\varphi(t)=2t$$ and $$\phi(t)=t$$ for all t. Now by using Theorem 1.2, F has a unique fixed point which is the unique solution for the problem (1). □

### Lemma 2.6

Let $$\alpha\in(0,1)$$ and $$y\in H^{1}(0,1)$$. Then the fractional differential equation $${}^{\mathrm{CF}}D^{\alpha^{(2)}} x(t)=y(t)$$ with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime\prime}(0) =0$$ has the unique solution $$x(t)=\int_{0}^{1} G(t,s)y(s)\,ds$$, where $$G(t,s)=\frac{- (1-\alpha)t}{(2-\alpha)B(\alpha) } +\frac{- \alpha t}{(2-\alpha)B(\alpha) } (t-s)$$ whenever $$0< t\leq s<1$$ and $$G(t,s)=\frac{1-\alpha}{B(\alpha)}(t-s) + \frac{\alpha}{2B(\alpha) }(t-s)^{2} -\frac{ (1-\alpha) t}{B(\alpha)(2-\alpha)} -\frac{\alpha t}{B(\alpha)(2-\alpha) } (t-s)$$ whenever $$0< s\leq t<1$$.

### Proof

By using Lemma 2.3, we get $$x(t)=\frac{1-\alpha}{B(\alpha)}J^{2 }y(t)+\frac{ \alpha}{B(\alpha)} J^{3}y(t) + x^{\prime}(0)t+\frac{\alpha }{ 1-\alpha} x^{\prime}(0)t$$. Hence, $$x^{\prime}(t)= \frac{1-\alpha}{B(\alpha)}J^{1}y(t)+\frac{ \alpha }{B(\alpha)} J^{2 }y(t) + \frac{1}{ 1-\alpha} x^{\prime}(0)$$. By using the boundary conditions $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime}(1)= \frac{1-\alpha}{B(\alpha)}J^{1}y(1)+\frac{ \alpha }{B(\alpha)} J^{2 }y(1) + \frac{1}{ 1-\alpha} x^{\prime}(0)$$, we obtain $$x(t)=\frac{1-\alpha}{B(\alpha)}J^{2 }y(t)+\frac{ \alpha }{B(\alpha)} J^{3}y(t)-\frac{ (1-\alpha) t}{(2-\alpha)B(\alpha) }J^{1}y(1)-\frac{\alpha t}{(2-\alpha)B(\alpha) }J^{2 }y(1)$$. Thus, $$x(t)=\frac{1-\alpha}{B(\alpha)} \int_{0}^{t}y(s)(t-s) \,ds+\frac{ \alpha}{2B(\alpha)}\int_{0}^{t}y(s)(t-s)^{2}\,ds -\frac{(1-\alpha)t}{(2-\alpha )B(\alpha)} \int_{0}^{1}y(s) \,ds-\frac{\alpha t}{(2-\alpha)B(\alpha) }\int _{0}^{1}y(s)(t-s) \,ds =\int_{0}^{1} G(t,s)y(s)\,ds$$. Note that $$( {}^{\mathrm{CF}} D^{\alpha})^{(2)}x(t)=0$$ if and only if $$x(t)=0$$. This implies that the given map $$x(t)$$ is a unique solution. □

Note that $$|G(t,s)|\leq|\frac{1-\alpha}{B(\alpha)}|+ |\frac{\alpha }{2B(\alpha) }| +|\frac{-\alpha t}{ B(\alpha)} |+|\frac{-(1-\alpha) t}{ B(\alpha)}|<\frac{2}{B(\alpha)}$$, for $$t \in[0,1]$$. Let $$\alpha,\beta_{1},\beta_{2},\gamma,\nu\in(0,1)$$. Now, we investigate the DCF fractional integro-differential problem

\begin{aligned}[b] \bigl( {}^{\mathrm{CF}}D^{\alpha} \bigr)^{(2)}x(t)&=\mu(t) x(t)+\mu_{1}(t) x^{\prime }(t)+ \mu_{2} (t)x^{\prime\prime}(t)\\ &\quad {}+\lambda_{1} k_{1}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{1}}\bigr)^{(1)}x(t)+ \lambda_{2} k_{2}(t) \bigl( {}^{\mathrm{CF}}D^{\beta _{2} } \bigr)^{(2)}x(t) \\ &\quad{}+ \int_{0}^{t} f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime \prime}(s),\\ &\quad {} h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma}\bigr)^{(1)}x(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu}\bigr)^{(2)}x(s) \bigr)\,ds \end{aligned}
(2)

with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime\prime}(0)=0$$.

### Theorem 2.7

Let $$\xi_{1}$$, $$\xi_{2}$$, $$\xi_{3}$$, $$\xi_{4}$$, and $$\xi_{5}$$ be nonnegative real numbers, $$f :[0,1]\times\Bbb {R}^{5}\to\Bbb{R}$$ an integrable function such that

\begin{aligned} &\bigl\vert f (t,x,y,z,v,w )-f \bigl(t,x^{\prime},y^{\prime} ,z^{\prime},v^{\prime},w^{\prime} \bigr) \bigr\vert \\ &\quad \leq \xi_{ 1} \bigl\vert x -x^{\prime} \bigr\vert + \xi_{ 2} \bigl\vert y -y^{\prime} \bigr\vert + \xi_{ 3} \bigl\vert z -z^{\prime} \bigr\vert + \xi_{ 4} \bigl\vert v -v^{\prime} \bigr\vert + \xi_{ 5} \bigl\vert v -v^{\prime} \bigr\vert \end{aligned}

for all real numbers x, y, z, v, w, $$x^{\prime}$$, $$y^{\prime}$$, $$z^{\prime}$$, $$v^{\prime}$$, $$w^{\prime}$$ and $$t \in I$$. If $$\Delta<\frac{1}{2}$$, then the problem (2) has a unique solution, where $$\Delta:=\max\lbrace \Delta_{1},\Delta_{2},\Delta_{3},\Delta_{4}\rbrace$$, $$\Delta_{1}=\frac{2}{(2-\alpha)B(\alpha)} [M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5} B(\beta _{2})}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac { M_{8} B(\gamma)}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu )}{(1-\nu)^{3}} ]$$, $$\Delta_{2}= \frac{3+\alpha }{ B(\alpha)(2-\alpha)} [M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta _{2}+1)M_{5} B(\beta_{2})}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8} B(\gamma)}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu ^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} ]$$, $$\Delta_{3}= \frac{2+\alpha }{B(\alpha)(2-\alpha)} [M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta _{2}+1)M_{5} B(\beta_{2})}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8} B(\gamma)}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu ^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} ]$$ and $$\Delta_{4}=\frac{\alpha}{B(\alpha)}[M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5} B(\beta _{2})}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac { M_{8} B(\gamma)}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu )}{(1-\nu)^{3}} ]+\frac{1-\alpha}{B(\alpha )}[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+M_{4}B(\beta_{1}) \frac{ ( \beta_{1}^{2}- \beta _{1}+1)}{(1-\beta_{1})^{3}} +\frac{ B(\beta_{1}) N_{4}}{(1-\beta_{1})^{2}} +M_{5}B(\beta_{2})\frac{ ( 2\beta_{2}^{2}-2 \beta_{2}+1)}{(1-\beta_{2})^{4}} +N_{5}B(\beta_{2}) \frac{ ( \beta_{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\frac{\xi_{ 4} M_{ 8}B(\gamma)}{(1-\l \gamma)^{2}} +\frac{(\nu^{2}- \nu+1)M_{9}B(\nu)}{(1-\nu)^{3}}]$$.

### Proof

Consider the Banach space $$C^{3}_{\mathbb{R}}[0,1]$$ equipped with the norm $$\Vert x \Vert= \max_{t\in I}|x(t)| + \max_{t\in I}| x^{\prime}(t) | + \max_{t\in I}| x^{\prime\prime}(t) |+ \max_{t\in I}| x^{\prime\prime\prime}(t) |$$. Define the map $$F: C^{3}_{\mathbb {R}}[0,1]\to C^{3}_{\mathbb{R}}[0,1]$$ by

\begin{aligned} Fx(t) &= \int_{0}^{1} G(t,s)R(s)\,ds \\ &= \frac{1-\alpha}{B(\alpha)} \int _{0}^{t}R(s) (t-s) \,ds+ \frac{\alpha}{2B(\alpha) } \int_{0}^{t}R(s) (t-s)^{2}\,ds \\ &\quad {}-\frac{ (1-\alpha) t}{B(\alpha)(2-\alpha)} \int_{0}^{1}R(s) \,ds-\frac {\alpha t}{B(\alpha)(2-\alpha) } \int_{0}^{1}R(s) (t-s) \,ds, \end{aligned}

where

\begin{aligned} (Rx) (t)&=\mu(t) x(t)+\mu_{1}(t) x^{\prime}(t)+ \mu_{2} (t)x^{\prime\prime }(t)\\ &\quad {}+\lambda_{1} k_{1}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{1}}\bigr)^{(1)}x(t)+\lambda _{2} k_{2}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{2} } \bigr)^{(2)}x(t) \\ &\quad {}+ \int_{0}^{t} f \bigl(s,x(s), m _{1}(s) x^{\prime}(s) ,m _{2}(s) x^{\prime\prime }(s), h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma}\bigr)^{(1)}x(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu}\bigr)^{(2)}x(s)\bigr)\,ds \end{aligned}

and

\begin{aligned} \bigl(R^{\prime}x\bigr) (t)&=\mu(t) x ^{\prime}(t)+ \mu^{\prime}(t) x (t)+\mu _{1}^{\prime}(t) x^{\prime}(t)+\mu_{1} x^{\prime\prime}(t) \\ &\quad {}+\mu_{2}^{\prime }(t) x^{\prime\prime}(t)+\mu_{2} (t) x^{\prime\prime\prime}(t) + k_{1}^{\prime}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{1}} \bigr)^{(1)}x(t)\\ &\quad {} + k_{1}(t)B(\beta_{1})\biggl[\biggl( \frac{ -\beta_{1}}{1-\beta_{1}}\biggr)^{2} {}^{\mathrm{CF}}D^{\beta_{1}}x(t) - \frac{\beta_{1}}{(1-\beta_{1})^{2}}x^{\prime}(t) +\frac {1}{ 1-\beta_{1} }x^{\prime\prime}(t) \biggr] \\ &\quad {}+ k_{2}(t)B(\beta_{2})\biggl[\biggl( \frac{ -\beta_{2}}{1-\beta_{2}}\biggr)^{3} {}^{\mathrm{CF}}D^{\beta_{2}}x(t)+ \frac{\beta^{2} _{2}}{(1-\beta_{2})^{3}}x^{\prime}(t)\\ &\quad {} -\frac{\beta_{2}}{( 1-\beta_{2})^{2} }x^{\prime\prime}(t)+ \frac{x^{\prime \prime\prime}(t)}{ 1-\beta_{2} }\biggr] + k_{2}^{\prime}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{2}} \bigr)^{(2)}x(t)\\ &\quad {} +f \bigl(t,x(t), m _{1}(t) x^{\prime}(t) ,m _{2}(t) x^{\prime\prime}(t), h(t) \bigl( {}^{\mathrm{CF}}D^{ \gamma} \bigr)^{(1)}x(t) , g(t) \bigl( {}^{\mathrm{CF}}D^{ \nu} \bigr)^{(2)}x(t)\bigr). \end{aligned}

By using Lemma 2.6, $$x_{0}$$ is a solution for the problem (2) if and only if $$x_{0}$$ is a fixed point of the operator F. Note that

\begin{aligned} &(Rx) (t)-(Ry) (t) \\ &\quad \leq\biggl[\mu(t) x(t)+\mu_{1} (t) x^{\prime}(t)+ \mu_{2}(t) x^{\prime\prime }(t)+\lambda_{1} k_{1}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{1}}\bigr)^{(1)}x(t)+ k_{2}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{2}}\bigr)^{(2)}x(t) \\ &\qquad {} + \int_{0}^{t} f \bigl(s,x(s), m _{1}(t) x^{\prime}(s) ,m _{2}(s) x^{\prime\prime }(s), h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma}\bigr)^{(1)}x(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu}\bigr)^{(2)}x(s) \bigr)\,ds\biggr] \\ &\qquad {} - \biggl[\mu(t) y(t)+\mu_{1} (t) y^{\prime}(t)+ \mu_{2}(t) y^{\prime\prime}(t)+ k_{1}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{1}}\bigr)^{(1)}y(t)+ k_{2}(t) \bigl( {}^{\mathrm{CF}}D^{\beta_{2}}\bigr)^{(2)}y(t) \\ &\qquad {} + \int_{0}^{t} f \bigl(s,y(s), m _{1}(t) y^{\prime}(s) ,m _{2}(s) y^{\prime\prime }(s), h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma}\bigr)^{(1)}y(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu}\bigr)^{(2)}y(s) \bigr)\,ds\biggr] \\ &\quad \leq \bigl\vert \mu(t) \bigr\vert \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert \mu_{1}(t) \bigr\vert \bigl\vert x^{\prime}(t)-y^{\prime }(t) \bigr\vert + \bigl\vert \mu_{2}(t) \bigr\vert \bigl\vert x^{\prime\prime}(t)-y^{\prime\prime}(t) \bigr\vert \\ &\qquad {} + \bigl\vert k_{1}(t) \bigr\vert \bigl( {}^{\mathrm{CF}}D^{\beta_{1}}\bigr)^{(1)} \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert k_{2}(t) \bigr\vert \bigl( {}^{\mathrm{CF}}D^{\beta_{2}}\bigr)^{(2)} \bigl\vert x(t)-y(t) \bigr\vert \\ &\qquad {} + \int_{0}^{t} \vert f \bigl(s,x(s), m _{1}(t) x^{\prime}(s) ,m _{2}(s) x^{\prime \prime}(s), h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma}\bigr)^{(1)}x(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu}\bigr)^{(2)}x(s) \bigr)\,ds \\ &\qquad {} - f \bigl(s,y(s), m _{1}(t) y^{\prime}(s) ,m _{2}(s) y^{\prime\prime}(s), h(s) \bigl( {}^{\mathrm{CF}}D^{ \gamma} \bigr)^{(1)}y(s) , g(s) \bigl( {}^{\mathrm{CF}}D^{ \nu } \bigr)^{(2)}y(s) \bigr) \vert \,ds \\ & \quad \leq M_{1} \Vert x-y \Vert + M_{2} \Vert x-y \Vert + M_{3} \Vert x-y \Vert \\ &\qquad {} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} \Vert x-y \Vert + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5} B(\beta _{2})}{(1-\beta_{2})^{3}} \Vert x-y \Vert \\ &\qquad {} + \biggl[\xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7} +\xi_{ 4} \frac{ M_{8} B(\gamma )}{(1-\gamma)^{2}} + \xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} \biggr] \Vert x-y \Vert \\ &\quad \leq \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta _{1})}{(1-\beta_{1})^{2}}+ \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5} B(\beta_{2})}{(1-\beta_{2})^{3}} \\ &\qquad {} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8} B(\gamma)}{(1-\gamma )^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} \biggr] \Vert x-y \Vert \end{aligned}

and

\begin{aligned} &\bigl\vert R^{\prime}x(t)-R^{\prime}y(t) \bigr\vert \\ &\quad \leq \biggl(N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+M_{4}B( \beta_{1}) \frac{ ( \beta_{1}^{2}- \beta _{1}+1)}{(1-\beta_{1})^{3}} +\frac{ B(\beta_{1}) N_{4}}{(1-\beta_{1})^{2}} \\ &\qquad{} +M_{5}B(\beta_{2})\frac{ ( 2\beta_{2}^{2}-2 \beta_{2}+1)}{(1-\beta_{2})^{4}} +N_{5}B(\beta_{2}) \frac{ ( \beta_{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7} \\ &\qquad{} + \frac{\xi_{ 4} M_{ 8}B(\gamma)}{(1-\l \gamma)^{2}}+\frac{(\nu^{2}- \nu+1)M_{9}B(\nu)}{(1-\nu)^{3}} \biggr) \Vert x-y \Vert . \end{aligned}

Thus,

\begin{aligned} \bigl\vert Fx(t)-Fy(t) \bigr\vert &\leq\frac{2}{(2-\alpha)B(\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta _{2}^{2}- \beta_{2}+1)M_{5} B(\beta_{2})}{(1-\beta_{2})^{3}} \\ &\quad{} + \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8} B(\gamma )}{(1-\gamma)^{2}} + \xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} \biggr] \Vert x-y \Vert \\ &= \Delta_{1} \Vert x-y \Vert \end{aligned}

and

\begin{aligned} \bigl\vert F^{\prime}x(t)-F^{\prime}y(t) \bigr\vert &\leq\biggl[ \frac {3+\alpha }{ B(\alpha)(2-\alpha)} \biggr] \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}}+ \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5} B(\beta _{2})}{(1-\beta_{2})^{3}} \\ &\quad{} + \xi_{ 1} +\xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8} B(\gamma )}{(1-\gamma)^{2}} + \xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} \biggr] \Vert x-y \Vert \\ &= \Delta_{2} \Vert x-y \Vert . \end{aligned}

Also, we have

\begin{aligned} \bigl\vert F^{\prime\prime}x(t)-F^{\prime\prime}y(t) \bigr\vert & \leq \frac{2+\alpha }{B(\alpha)(2-\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4}B(\beta_{1})}{(1-\beta_{1})^{2}} \\ &\quad{} + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5}B(\beta_{2})}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}\\ &\quad {}+\xi_{ 4} \frac{ M_{8}B(\gamma)}{(1-\gamma )^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9}B(\nu)}{(1-\nu)^{3}} \biggr]\\ &= \Delta_{3} \Vert x-y \Vert \end{aligned}

and

\begin{aligned} &\bigl\vert F^{\prime\prime\prime}x(t)-F^{\prime\prime\prime}y(t) \bigr\vert \\ &\quad \leq \frac{\alpha}{B(\alpha)} \bigl\vert R x(t)-Ry(t) \bigr\vert + \frac{1-\alpha}{B(\alpha)} \bigl\vert R^{\prime}x(t)- R^{\prime }y(t) \bigr\vert \\ &\quad \leq\biggl( \frac{\alpha}{B(\alpha)} \biggl[M_{1} + M_{2} + M_{3} + \frac{ M_{4} B(\beta_{1})}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta _{2}+1)M_{5} B(\beta_{2})}{(1-\beta_{2})^{3}} \\ &\qquad {}+ \xi_{ 1} + \xi_{ 2} M_{6}+\xi_{ 3} M_{7}+ \xi_{ 4} \frac{ M_{8} B(\gamma)}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu ^{2}- \nu+1)M_{9} B(\nu)}{(1-\nu)^{3}} \biggr] \\ &\qquad{} + \frac{1-\alpha}{B(\alpha)}\biggl[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+M_{4}B( \beta_{1}) \frac { ( \beta_{1}^{2}- \beta_{1}+1)}{(1-\beta_{1})^{3}} \\ &\qquad{} +\frac{ B(\beta_{1}) N_{4}}{(1-\beta_{1})^{2}}+M_{5}B(\beta_{2})\frac{ ( 2\beta_{2}^{2}-2 \beta_{2}+1)}{(1-\beta_{2})^{4}} +N_{5}B(\beta_{2}) \frac{ ( \beta_{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} \\ &\qquad{} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+ \frac{\xi_{ 4} M_{ 8}B(\gamma)}{(1-\l \gamma)^{2}}+\frac{(\nu^{2}- \nu+1)M_{9}B(\nu)}{(1-\nu)^{3}} \biggr]\biggr) \Vert x-y \Vert \\ &\quad = \Delta_{4} \Vert x-y \Vert . \end{aligned}

Hence, $$\Vert Fx-Fy \Vert \leq\Delta \Vert x-y \Vert$$ for all $$x,y\in C^{3}_{\mathbb{R}}[0,1]$$. Put $$\varphi(t)=2t$$ and $$\phi(t)=t$$ for all t. By using Theorem 1.2, F has a unique fixed point, which is the desired solution for the problem. □

Here, we provide three examples to illustrate our main results. Consider the bounded continuous functions $$\mu(t)=\frac{1}{100}\sin (t)$$, $$\mu_{1}(t)=\frac{3t-1}{20t+162}$$, $$\mu_{2}(t)= \frac {1}{100}e^{-6t}$$, $$k_{1}(t)=\frac{1}{300}t^{3}+\frac{1}{100}t+\frac {1}{50}$$, $$k_{2}(t)= \frac{1}{800}\cos(t)$$, $$m_{1}(t)= e^{2t}$$, $$m_{2}(t)=\frac{Ln(t+2)}{20}$$, $$h(t)=0$$ and $$g(t)=\frac{1}{t-900}$$ for all $$t\in I=[0,1]$$. Note that $$M_{1}=\sup_{t\in I}|\mu(t)|=\frac {1}{100}$$, $$M_{2}=\sup_{t\in I}|\mu_{1}(t)|=\frac{1}{91}$$, $$M_{3}=\sup_{t\in I}|\mu_{2}(t)|=\frac{1}{100e^{6}}$$, $$M_{4}=\sup_{t\in I}|k_{1}(t)|=\frac{1}{30}$$, $$M_{5}=\sup_{t\in I}|k_{2}(t)|=\frac{1}{800}$$, $$M_{6}=\sup_{t\in I}|m_{1}(t)| =e^{2}$$, $$M_{7}=\sup_{t\in I}|m_{2}(t)|=\frac {Ln(3)}{1200}$$, $$M_{8}=\sup_{t\in I}|h(t)| =0$$ and $$M_{9}=\sup_{t\in I}|g (t)| = \frac{1}{900 }$$. Also, $$N_{1}= \sup_{t\in I}|\mu^{\prime }(t)|=\frac{1}{100}$$, $$N_{2}=\sup_{t\in I}|\mu_{1}^{\prime}(t)|=\frac {506}{(162)^{2}}$$, $$N_{3}=\sup_{t\in I}|\mu_{2}^{\prime}(t)|=\frac{ 6}{100e^{ 6 }}$$, $$N_{4}=\sup_{t\in I}|k_{1}^{\prime}(t)|= \frac{1}{50}$$ and $$N_{5}=\sup_{t\in I}|k_{2}^{\prime}(t)|= \frac{1}{800}$$. Also, consider the function $$B(\alpha)=1$$ for $$\alpha\in(0,1)$$.

### Example 2.1

Let $$\alpha\in(0,1)$$. By using Lemma 2.4, the fractional differential equation $${}^{\mathrm{CF}}D^{2+\alpha}x_{1}(t)=t$$ with boundary conditions $$x_{1}(0)=0$$, $$x_{1}^{\prime}(1)+ x_{1}^{\prime}(0)=0$$ and $$x_{1}^{\prime\prime}(0) =0$$ has the unique solution $$x_{1}(t)$$. Also by using Lemma 2.6, the fractional differential equation $$({}^{\mathrm{CF}}D^{ \alpha})^{(2)}x_{2}(t)=t$$ with boundary conditions $$x_{2}(0)=0$$, $$x_{2}^{\prime}(1)+ x_{2}^{\prime}(0)=0$$ and $$x_{2}^{\prime\prime }(0) =0$$ has the unique solution $$x_{2}(t)$$. For $$\alpha=\frac {1}{100}$$, $$\alpha=\frac{1}{10}$$, $$\alpha=\frac{1}{5}$$, $$\alpha=\frac {1}{2}$$, $$\alpha=\frac{4}{5}$$ and $$\alpha=\frac{99}{100}$$ we compare the solutions $$x_{1}(t)$$, $$x_{2}(t)$$ and $$X(t)= x_{1}(t)-x_{2}(t)$$ in Fig. 1.

### Example 2.2

Consider the CFD fractional integro-differential problem

\begin{aligned}[b] {}^{\mathrm{CF}}D^{ \frac{12}{5 }}x(t)&= \frac{\sin(t)}{100}x(t)+\frac {3t-1}{20t+162} x^{\prime}(t)+ \frac{e^{-6t}}{100} x^{\prime\prime }(t)\\ &\quad {}+\biggl( \frac{t^{3}+3t+6}{300} \biggr) {}^{\mathrm{CF}}D^{\frac{3}{2}}x(t) + \frac{\cos(t)}{800} {}^{\mathrm{CF}}D^{\frac{5}{2}}x(t) \\ &\quad{}+ \int_{0}^{t} f \biggl(s,x(s),e^{2t} x^{\prime}(s),\frac{Ln(t+2)}{20} x^{\prime\prime}(s), 0 , \frac{1}{t-900} {}^{\mathrm{CF}}D^{ \frac {8}{3}}x(s) \biggr)\,ds \end{aligned}
(3)

with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime\prime}(0) =0$$, where $$1<\beta_{1}=\frac{3}{2}< 2<\beta _{2}=\frac{5}{2}<3$$ and $$1<\gamma=\frac{4}{3 }< 2<\nu=\frac{8}{3}<3$$. Put $$f(t,x,y,z,v,w)=\frac{2}{91}t+\frac{3}{604 }x+\frac{1}{200}y+\frac {1}{80 }z+\frac{1}{e^{18}}w+2v$$. Note that $$\Delta_{1}= \frac{3}{2} [M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{ 2-\beta_{1} } + \frac{ M_{5}}{ 3-\beta_{2} } + \xi_{ 1} + \xi_{ 2} M_{6}+ \xi_{ 3} M_{7}+ \xi _{ 4} \frac{ M_{8}}{ 2-\gamma } +\xi_{ 5}\frac{ M_{9}}{ 3-\nu} ]=0.02$$, $$\Delta_{2}=\frac{3}{2}[ \frac{3+4\alpha}{2} ] [M_{1} + M_{2} + M_{3} + \frac { M_{4}}{ 2-\beta_{1} } + \frac{ M_{5}}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}}{ 2-\gamma } +\xi_{ 5}\frac{ M_{9}}{ 3-\nu} ]=0.46$$, $$\Delta_{3}= (1+ \alpha) [M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{ 2-\beta_{1} } + \frac{ M_{5}}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}}{ 2-\gamma } +\xi_{ 5}\frac{ M_{9}}{ 3-\nu} ] =0.3 2$$ and $$\Delta_{4}= \alpha[ M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{ 2-\beta_{1} } + \frac{ M_{5}}{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7} +\xi _{ 4} \frac{ M_{8}}{ 2-\gamma } +\xi_{ 5}\frac{ M_{9}}{ 3-\nu} ] +(1-\alpha)[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+ \frac{ |1-\beta_{1}|M_{4} }{(2-\beta _{1})^{2}} +\frac{ N_{4} +M_{4}}{ 2-\beta_{1} } +\frac{ |2-\beta_{2}|M_{5} }{(3-\beta _{2})^{2}} +\frac{ M_{5}+ N_{5} }{ 3-\beta_{2} } + \xi_{ 1} +\xi_{ 2} M_{6}+\xi _{ 3} M_{7}+\xi_{ 4} \frac{M_{ 8}}{ 2-\l\gamma} +\xi_{ 5} \frac{ M_{6}}{ 3-\nu} ]=0.166$$. By using Theorem 2.5, the problem (3) has a unique solution.

### Example 2.3

Consider the DCF fractional integro-differential problem

\begin{aligned}[b] &\bigl( {}^{\mathrm{CF}}D^{\frac{2}{5} }\bigr) ^{(2)}x(t)\\ &\quad =\frac{\sin (t)}{100}x(t)+\frac{3t-1}{20t+162} x^{\prime}(t)+ \frac{e^{-6t}}{100} x^{\prime\prime}(t) \\ &\qquad{}+\biggl( \frac{t^{3}+3t+6}{300} \biggr) \bigl( {}^{\mathrm{CF}}D^{\frac {1}{2}} \bigr)^{(1)}x(t)+ \frac{\cos(t)}{800} \bigl( {}^{\mathrm{CF}}D^{\frac{2}{3} } \bigr)^{(2)}x(t) \\ &\qquad {}+ \int_{0}^{t} f \biggl(s,x(s),e^{-40t} x^{\prime}(s),\frac {Ln(t+2)}{20} x^{\prime\prime}(s), 0 , \frac{1}{t-900}\bigl( {}^{\mathrm{CF}}D^{ \frac{1}{5} }\bigr)^{(2)}x(s) \biggr)\,ds, \end{aligned}
(4)

with boundary conditions $$x(0)=0$$, $$x^{\prime}(1)+ x^{\prime}(0)=0$$ and $$x^{\prime\prime}(0) =0$$, where $$\alpha=\frac{2}{5}$$, $$\beta_{1}=\frac {1}{2}$$, $$\beta_{2}=\frac{2}{3}$$, $$\gamma=\frac{1}{3 }$$ and $$\nu=\frac {1}{5}$$. Put $$f(t,x,y,z,v,w)=\frac{2}{91}t+\frac{3}{604 }x+\frac {1}{200}y+\frac{1}{80 }z+\frac{1}{e^{18}}w+2v$$. Note that $$\Delta _{1}=\frac{2}{2-\alpha} [M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{(1-\beta _{1})^{2}} + M_{5}\frac{( \beta_{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}}{(1-\gamma)^{2}} +\xi _{ 5}\frac{( \nu^{2}- \nu+1)M_{9}}{(1-\nu)^{3}} ]<0.391$$, $$\Delta_{2}=[ \frac {3+\alpha }{ 2-\alpha} ] [M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{(1-\beta _{1})^{2}} + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5}}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9}}{(1-\nu)^{3}}]<0.225$$, $$\Delta_{3}= \frac {2+\alpha }{2-\alpha} [M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{(1-\beta_{1})^{2}} + \frac{( \beta_{2}^{2}- \beta_{2}+1)M_{5}}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi_{ 4} \frac{ M_{8}}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu+1)M_{9}}{(1-\nu)^{3}} ] <0.132$$ and $$\Delta_{4}=\alpha[M_{1} + M_{2} + M_{3} + \frac{ M_{4}}{(1-\beta_{1})^{2}} + M_{5}\frac{( \beta_{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+\xi _{ 4} \frac{ M_{8}}{(1-\gamma)^{2}} +\xi_{ 5}\frac{( \nu^{2}- \nu +1)M_{9}}{(1-\nu)^{3}} ]+(1-\alpha)[N_{1}+M_{1}+N_{2}+M_{2}+N_{3}+M_{3}+M_{4} \frac{ ( \beta_{1}^{2}- \beta_{1}+1)}{(1-\beta_{1})^{3}} +\frac{ N_{4}}{(1-\beta_{1})^{2}} +M_{5}\frac{ ( 2\beta_{2}^{2}-2 \beta_{2}+1)}{(1-\beta_{2})^{4}} + \frac{ N_{5}( \beta _{2}^{2}- \beta_{2}+1)}{(1-\beta_{2})^{3}} + \xi_{ 1} +\xi_{ 2} M_{6}+\xi_{ 3} M_{7}+ \frac{\xi_{ 4}M_{ 8}}{(1-\l\gamma)^{2}} +\xi_{ 5}\frac{(\nu^{2}- \nu +1)M_{9}}{(1-\nu)^{3}} ] < 0.493$$. By using Theorem 2.7, the problem (4) has a unique solution.

## 3 Conclusion

It is important that researchers have some methods available enabling them to review some high order fractional integro-differential equations. In this manuscript, we introduce two types of new fractional derivatives entitled CFD and DCF and by using those we investigate the existence of solutions for two high order fractional integro-differential equations of such a type including the new derivatives.

## Abbreviations

DFC:

Caputo–Fabrizio derivation followed by a differentiation

CFD:

Differentiation followed by Caputo–Fabrizio derivation

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### Acknowledgements

The third author was supported by Azarbaijan Shahid Madani University.

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Correspondence to Dumitru Baleanu.

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Aydogan, M.S., Baleanu, D., Mousalou, A. et al. On high order fractional integro-differential equations including the Caputo–Fabrizio derivative. Bound Value Probl 2018, 90 (2018). https://doi.org/10.1186/s13661-018-1008-9

• Accepted:

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• DOI: https://doi.org/10.1186/s13661-018-1008-9

• 34A08
• 34A99

### Keywords

• Caputo–Fabrizio derivative
• Fractional integro-differential equation
• High order derivation