## 1 Preliminaries

It is well known that we can make better exact models for most natural phenomena by using fractional differential equations. Most researchers are working on fractional integro-differential equations (see, for example, [1, 2, 5,6,7,8, 10,11,12,13,14,15,16,17,18,19,20,21]).

In 2010, Agarwal et al. reviewed the existence of solutions $$D^{\alpha } u(t)+ f(t , u(t) )=0$$ with boundary conditions $$u'(0) = \cdots = u^{(n-1)} = 0$$ and $$u(1)=\int _{0} ^{1} u(s) \,d\mu (s)$$, where $$n \geq 2$$, $$\alpha \in (n-1,n)$$, $$\mu (s)$$ is a functional of bounded variation, f may have singularity at $$t=0$$ and $$\int _{0} ^{1} d\mu (s) < 1$$ [3]. In 2012, Agarwal et al. studied positive solutions for the integral value problem $$D^{\alpha } u_{i}(t)+ f_{i}(t , u_{1}(t) , u_{2}(t) )=0$$ with boundary conditions $$u_{i}(0)=u'_{i}(0)=0$$ and $$u_{i}(1) = \int _{0} ^{1} u_{i}(t) \,d\eta (t)$$ for $$i=1,2$$, where $$t \in (0,1)$$, $$\alpha \in (2,3]$$, $$D^{\alpha }$$ is the Riemann–Liouville fractional derivative of order α, $$f_{i}$$ is a real valued continuous map on $$[0,1] \times \mathbb{R} ^{+} \times \mathbb{R}^{+}$$ and $$\int _{0} ^{1} u_{i}(t) \,d\eta (t)$$ denotes the Riemann–Stieltjes integral [4]. In 2013, the singular fractional problem $$D^{\alpha } u+ f(t , u , D^{ \gamma } u, D^{\mu } u )+ g(t , u , D^{\gamma } u, D^{\mu } u)=0$$ with boundary conditions $$u(0)=u'(0)=u''(0)=u'''(0)=0$$ was reviewed, where $$3< \alpha < 4$$, $$0< \gamma <1$$, $$1<\mu <2$$, $$D^{\alpha }$$ is the Caputo fractional derivative and f is a Caratheodory function on $$[0,1] \times (0 , \infty )^{3}$$ [9].

Recently, the authors introduced a new model for investigating the fractional differential equations called three step crisis integro-differential equations [11]. By using the idea, we investigate the existence of solutions for the three step crisis integro-differential equation

\begin{aligned} D^{\alpha } x(t)+ f\biggl(t , x(t), x'(t), D^{\beta }x(t), \int _{0}^{t} h( \xi ) x(\xi ) \,d\xi \biggr)=0 \end{aligned}
(1)

with boundary conditions $$x(0)=x'(T_{0})$$, $$x(1)=x'(T_{1})$$ and $$x''(0)=x^{(n)}(0)=0$$, where $$\alpha >1$$ with $$n=[\alpha ]-1$$, $$T_{0},T_{1}, \beta , \lambda , \mu \in (0,1)$$, $$h \in L^{1}[0,1]$$, $$D^{\alpha }$$ is the Caputo fractional derivative of order α, $$f(t,x_{1}(t),\ldots, x_{5}(t))=f_{1}(t,x_{1}(t),\ldots, x_{4}(t))$$ on $$[0,\lambda )$$, $$f(t,x_{1}(t),\ldots, x_{5}(t))=f_{2}(t,x_{1}(t),\ldots, x _{4}(t))$$ on $$[\lambda ,\mu ]$$ and $$f(t,x_{1}(t),\ldots, x_{5}(t))=f(t,x _{1}(t),\ldots, x_{4}(t))$$ on $$(\mu ,1]$$ in which $$f_{1}(t,\cdot,\cdot,\cdot,\cdot)$$ and $$f_{3}(t,\cdot,\cdot,\cdot,\cdot)$$ are continuous on $$[0,\lambda )$$ and $$(\mu ,1]$$, respectively, and $$f_{2}(t,\cdot,\cdot,\cdot,\cdot)$$ is singular at some points $$t \in [\lambda ,\mu ]$$. In this case, we use the symbol $$f=[f_{1},f _{2},f_{3},\lambda ,\mu ]$$ [11].

As is well known, the Caputo fractional derivative of order $$\alpha >0$$ of a function $$f:(0,\infty )\to \mathbb{R}$$ is defined by $${}^{c}D^{\alpha }f(t)=\frac{1}{\varGamma (n-\alpha )} \int _{0}^{t}\!\frac{f^{n}(s)}{(t-s)^{\alpha +1-n}}\,ds$$, where $$n=[\alpha ]+1$$ (see, for example, [13]). Let Ψ be the family of nondecreasing functions $$\psi :[0,\infty ) \to [0,\infty )$$ such that $$\sum_{n=1}^{\infty } \psi ^{n}(t)<\infty$$ for all $$t> 0$$ [22]. One can check that $$\psi (t)< t$$ for all $$t>0$$ [22]. Let $$T:X \to X$$ and $$\alpha :X \times X \to [0,\infty )$$ be two maps. Then T is called an α-admissible map whenever $$\alpha (x,y) \geq 1$$ implies $$\alpha (Tx,Ty) \geq 1$$ [22]. Let $$(X,d)$$ be a complete metric space, $$\psi \in \varPsi$$ and $$\alpha :X \times X \to [0,\infty )$$ a map. A self-map $$T:X \to X$$ is called an α-ψ-contraction whenever $$\alpha (x,y) d(Tx,Ty) \leq \psi (d(x,y))$$ for all $$x,y \in X$$ [22]. We need the following results.

### Lemma 1

([23])

Let $$0< n-1\leq \alpha < n$$. Then $$I^{\alpha } D^{\alpha }x(t)=x(t)+ \sum_{i=0}^{n-1} c_{i}t^{i}$$ for some constants $$c_{0},\dots ,c_{n-1}$$.

### Lemma 2

([24])

If E is a closed, bounded and convex subset of a Banach space X and $$T : E \to E$$ is completely continuous, then T has a fixed point in E.

### Lemma 3

([22])

Let $$(X,d)$$ be a complete metric space, $$\psi \in \varPsi$$, $$\alpha :X \times X \to [0,\infty )$$ a map and $$T:X \to X$$ an α-admissible α-ψ-contraction. If T is continuous and there exists $$x_{0} \in X$$ such that $$\alpha (x _{0}, Tx_{0}) \geq 1$$, then T has a fixed point.

## 2 Main results

Now, we are ready to state and prove our main results.

### Lemma 4

Let $$\alpha > 1$$, $$n=[\alpha ] +1$$, $$T_{0}, T_{1} \in (0,1)$$ and $$f \in L^{1}[0,1]$$. Then $$x(t)= \int ^{1}_{0} G(t,s) f(s) \,ds$$ is the solution of the pointwise defined equation $$D^{\alpha }x(t) +f(t) = 0$$ with boundary conditions $$x(0)=x'(T_{0})$$, $$x(1)=x'(T_{1})$$ and $$x''(0) = \cdots =x^{(n-1)}(0)=0$$, where $$G(t,s)=\frac{-(t-s)^{\alpha -1}}{\varGamma (\alpha )}+ \frac{(1+t)(1-s)^{\alpha -1}}{\varGamma (\alpha )} - \frac{(1+t)(T_{1} -s)^{\alpha -2}}{\varGamma (\alpha -1 )} - \frac{t(T _{0} -s)^{\alpha -2}}{\varGamma (\alpha -1 )}$$ whenever $$0\leq s \leq t$$, $$s \leq T_{0}$$, $$G(t,s)=\frac{-(t-s)^{\alpha -1}}{\varGamma (\alpha )}+ \frac{(1+t)(1-s)^{ \alpha -1}}{\varGamma (\alpha )} - \frac{(1+t)(T_{1} -s)^{\alpha -2}}{ \varGamma (\alpha -1 )}$$ whenever $$0\leq T_{0} \leq s \leq t$$, $$s \leq T _{1}$$, $$G(t,s)=\frac{(1+t)(1-s)^{\alpha -1}}{\varGamma (\alpha )} - \frac{(1+t)(T _{1} -s)^{\alpha -2}}{\varGamma (\alpha -1 )}$$ whenever $$0\leq t \leq s \leq T_{1}$$, $$s \geq T_{0}$$, $$G(t,s)=\frac{(1+t)(1-s)^{\alpha -1}}{ \varGamma (\alpha )} - \frac{(1+t)(T_{1} -s)^{\alpha -2}}{\varGamma (\alpha -1 )} - \frac{t(T_{0} -s)^{\alpha -2}}{\varGamma (\alpha -1 )}$$ whenever $$0\leq t \leq s \leq T_{0} \leq T_{1}$$, $$G(t,s)=\frac{-(t-s)^{\alpha -1}}{\varGamma (\alpha )}+ \frac{(1+t)(1-s)^{\alpha -1}}{\varGamma (\alpha )}$$ whenever $$0\leq T_{0} \leq T_{1} \leq s \leq t$$ and $$G(t,s)=\frac{(1+t)(1-s)^{ \alpha -1}}{\varGamma (\alpha )}$$ whenever $$0 \leq t \leq s$$, $$s \geq T _{1}$$.

### Proof

Suppose that the equation $$D^{\alpha }x(t) +f(t) = 0$$ holds for all $$t \in E \subset [0,1]$$, where $$m(E^{c})=0$$ and m is the Lebesgue measure on $$\mathbb{R}$$. Let $$f_{0}$$ be a function such that $$f_{0}=f$$ on E. It is easy to check that $$I^{\alpha }(f(t))=I^{ \alpha }(f_{0}(t))$$ for all $$t\in [0,1]$$. This implies that $$I^{\alpha }(D^{\alpha }x(t))= I^{\alpha }(-f_{0}(t))$$ and by using Lemma 1 we get $$x(t)= - \frac{1}{\varGamma (\alpha )} \int ^{t} _{0} (t-s)^{\alpha - 1} f(s) \,ds +c_{0}+ c_{1} t$$ for some constants $$c_{0}$$ and $$c_{1}$$. By using the boundary conditions, we obtain $$x(0)=c_{0}$$ and

$$x'(T_{0})= - \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{0}}_{0} (T_{0}-s)^{ \alpha - 1} f(s) \,ds +c_{1}.$$

Thus, $$c_{1}-c_{0}= - \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{0}}_{0} (T_{0}-s)^{\alpha - 1} f(s) \,ds$$. Since $$x(1)= x'(T_{1})$$, we get

$$c_{0}=\frac{1}{\varGamma (\alpha - 1)} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s) \,ds - \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 1} f(s) \,ds$$

and so

\begin{aligned} c_{1} =&\frac{1}{\varGamma (\alpha )} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s) \,ds \\ &{}- \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 2} f(s) \,ds- \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{0}}_{0} (T_{0}-s)^{ \alpha - 2} f(s) \,ds. \end{aligned}

Hence,

\begin{aligned} x(t) =& - \frac{1}{\varGamma (\alpha )} \int ^{t}_{0} (t-s)^{\alpha - 1} f(s) \,ds - \frac{1}{\varGamma (\alpha )} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s) \,ds \\ &{}- \frac{1}{\varGamma (\alpha - 1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 2} f(s) \,ds +\frac{t}{\varGamma (\alpha )} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s) \,ds \\ &{}- \frac{t}{\varGamma (\alpha - 1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 2} f(s) \,ds -\frac{t}{\varGamma (\alpha -1)} \int ^{T_{0}}_{0} (T_{0}-s)^{ \alpha - 2} f(s) \,ds \\ =& - \frac{1}{\varGamma (\alpha )} \int ^{t}_{0} (t-s)^{\alpha - 1} f(s) \,ds - \frac{1+t}{\varGamma (\alpha )} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s) \,ds \\ &{}-\frac{1+t}{\varGamma (\alpha -1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 2} f(s) \,ds -\frac{t}{\varGamma (\alpha -1)} \int ^{T_{0}}_{0} (T_{0}-s)^{ \alpha - 2} f(s) \,ds. \end{aligned}

Now it is easy to check that $$x(t)= \int _{0}^{1} G(t,s) f(s)\,ds$$, where G is the given Green function. □

By using some usual calculations, we find that $$|G(t,s)| \leq \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} (1-s)^{\alpha -2}$$ for all $$t,s \in [0,1]$$ and $$|\frac{\partial G}{\partial t}(t,s)| \leq \frac{3 \alpha }{\varGamma (\alpha )} (1-s)^{\alpha -2}$$ for all $$t,s \in [0,1]$$. Also, it is easy to see that $$D^{\mu } x \in C[0,1]$$ and $$|D^{\mu } x| \leq \frac{ \Vert x' \Vert }{\varGamma (2-\mu )}$$ whenever $$x \in C^{1}[0,1]$$. Here, $$0 \leq \mu \leq 1$$. Now, consider the Banach space $$X= C^{1}[0,1]$$ with the norm $$\|x\|_{*} = \max \{ \|x\|, \|x' \| \}$$, $$\| \cdot \|$$ is the sup norm on $$C[0,1]$$. Assume that $$f=[f_{1},f_{2},f_{3},\lambda ,\mu ]$$. Define $$T:X \to X$$ by

\begin{aligned} T_{x}(t) =& \int _{0}^{1} G(t,s) f\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi , \phi \bigl(x(s) \bigr)\biggr) \,ds \\ =& - \frac{1}{\varGamma (\alpha )} \int ^{t}_{0} (t-s)^{\alpha - 1} f(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi , \phi \bigl(x(s) \bigr) \,ds \\ &{}- \frac{1+t}{\varGamma (\alpha )} \int ^{1}_{0} (1-s)^{\alpha - 1} f(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi , \phi \bigl(x(s) \bigr) \,ds \\ &{}-\frac{1+t}{\varGamma (\alpha -1)} \int ^{T_{1}}_{0} (T_{1}-s)^{\alpha - 2} f(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi , \phi \bigl(x(s) \bigr) \,ds \\ &{}-\frac{t}{\varGamma (\alpha -1)} \int ^{T_{0}}_{0} (T_{0}-s)^{\alpha - 2} f(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi , \phi \bigl(x(s) \bigr) \,ds \end{aligned}

for all $$x\in X$$ and $$t \in [0,1]$$. Note that the singular pointwise defined problem (1) has a solution if and only if T has a fixed point in X. We are going to investigate the singular pointwise defined problem (1) under two different conditions. Here. we present first one. In our second result we denote the map T by F.

### Theorem 5

Let $$f=[f_{1},f_{2},f_{3},\lambda ,\mu ]$$, $$f_{1}(t,0,0,0,0,0)=0$$, $$f_{2}(s,0,0,0,0,0)=0$$ and $$f_{3}(u,0,0,0, 0,0)=0$$ for all $$t \in [0, \lambda ]$$, $$s \in [0,\lambda ]$$ and $$u \in [\mu , 1]$$. Assume that there are nondecreasing maps $$\varLambda , \varLambda ': X \to [0, \infty )$$ and mappings $$a_{1}, a_{2}, a_{3}, a_{4} :(\lambda , \mu )\to [0, \infty )$$ such that $$\lim_{z\to 0^{+}} \frac{\varLambda (z)}{z} =q< \infty$$, $$\lim_{z\to 0^{+}} \frac{\varLambda '(z)}{z} =q< \infty$$ and $$\hat{a_{1}}, \hat{a_{2}}, \hat{a_{3}}, \hat{a_{4}} \in L^{1}[\lambda , \mu ]$$, where $$\hat{a_{i}}= (1-s)^{\alpha -2}$$ for $$i=1,2,3,4$$. Suppose that $$|f_{1}(t, x_{1}, \ldots, x_{5}) - f_{1}(t, y_{1}, \ldots, y _{5})| \leq \sum _{i=1}^{4} \varLambda (|x_{i} - y_{i}|)$$,

$$\bigl\vert f_{2}(t, x_{1}, \ldots, x_{5}) - f_{2}(t, y_{1}, \ldots, y_{5}) \bigr\vert \leq \sum _{i=1}^{4} a_{i}(t) \vert x_{i} - y_{i} \vert$$

and $$|f_{3}(t, x_{1}, \ldots, x_{5}) - f_{3}(t, y_{1}, \ldots, y_{5})| \leq \sum _{i=1}^{4} \varLambda ' (|x_{i} - y_{i}|)$$ for almost all $$t \in [0,1]$$ and every $$x_{1}, x_{2}, \ldots, x_{5}, y_{1}, y_{2}, \ldots, y_{5} \in X$$. If

$$\frac{4q}{\alpha -1} \bigl(1-(1-\lambda )^{\alpha -1}\bigr)+ \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + \frac{4q'}{\alpha -1} (1-\mu )^{ \alpha -1}< \frac{\varGamma (\alpha )}{l \theta _{0}},$$

then the pointwise defined equation (1) with boundary conditions has a solution, where $$\|h\|_{1}= m_{0}$$, $$l = \max \{ 1, \frac{1}{ \varGamma (2 - \beta )}, m_{0} \}$$ and $$\theta _{0}= \max \{ 3 \alpha , 2+ \alpha + T_{0} \}$$.

### Proof

Let $$x_{1}, x_{2} \in X$$ and $$t \in [0,1]$$. Then we have

\begin{aligned} \bigl\vert T_{x_{1}}(t)- T_{x_{2}}(t) \bigr\vert \leq{}& \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f\biggl(s, x_{1}(s), x'_{1}(s), D^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}(\xi )\,d \xi \biggr) \\ &{}- f\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h(\xi ) x _{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ \leq{}& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{1}\biggl(s, x_{1}(s), x'_{1}(s), D ^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}(\xi )\,d\xi \biggr) \\ &{}- f_{1}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{2}\biggl(s, x_{1}(s), x'_{1}(s), D ^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}(\xi )\,d\xi \biggr) \\ &{}- f_{2}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{3}\biggl(s, x_{1}(s), x'_{1}(s), D^{\beta }x _{1}(s), \int _{0}^{s} h(\xi ) x_{1}(\xi )\,d\xi \biggr) \\ &{}\times f_{3}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ \leq{}& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl( \bigl\vert x_{1}(s) - x_{2}(s) \bigr\vert \bigr)+ \varLambda \bigl( \bigl\vert x'_{1}(s) - x'_{2}(s) \bigr\vert \bigr) \\ &{}+ \varLambda \bigl( \bigl\vert D^{\beta } (x_{1} - x_{2}) (s) \bigr\vert \bigr) +\varLambda \biggl( \biggl\vert \int _{0}^{s} h( \xi ) \bigl(x_{1}(\xi ) - x_{2}(\xi )\bigr) \,d\xi \biggr\vert \biggr) \biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[a_{1}(s) \bigl\vert x_{1}(s) - x_{2}(s) \bigr\vert + a _{2}(s) \bigl\vert x'_{1}(s) - x'_{2}(s) \bigr\vert \\ &{}+a_{3}(s) \bigl\vert D^{\beta } (x_{1} - x_{2}) (s) \bigr\vert +a_{4}(s) \biggl\vert \int _{0}^{s} h( \xi ) \bigl(x_{1}(\xi ) - x_{2}(\xi )\bigr) \,d\xi \biggr\vert \biggr] \,ds \\ &{}+ \int _{\mu }^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda ' \bigl( \bigl\vert x_{1}(s) - x_{2}(s) \bigr\vert \bigr)+ \varLambda ' \bigl( \bigl\vert x'_{1}(s) - x'_{2}(s) \bigr\vert \bigr) \\ &{}+ \varLambda ' \bigl( \bigl\vert D^{\beta } (x_{1} - x_{2}) (s) \bigr\vert \bigr) +\varLambda ' \biggl( \biggl\vert \int _{0} ^{s} h(\xi ) \bigl(x_{1}(\xi ) - x_{2}(\xi )\bigr) \,d\xi \biggr\vert \biggr) \biggr] \,ds \\ \leq{}& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl( \Vert x_{1} - x_{2} \Vert \bigr)+ \varLambda \bigl( \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \bigr) \\ &{}+ \varLambda \bigl( \bigl\Vert D^{\beta } (x_{1} - x_{2}) \bigr\Vert \bigr) +\varLambda \biggl( \int _{0}^{s} |h( \xi ) \bigl\vert \Vert x_{1}- x_{2} \Vert \,d\xi \bigr\vert \biggr) \biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[a_{1}(s) \Vert x_{1} - x_{2} \Vert + a _{2}(s) \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \\ &{}+ a_{3}(s) \bigl\Vert D^{\beta } (x_{1} - x_{2}) \bigr\Vert +a_{4}(s) \int _{0}^{s} |h( \xi ) \bigl\vert \Vert x_{1} - x_{2} \Vert \,d\xi \bigr\vert \biggr] \,ds \\ &{}+ \int _{\mu }^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda ' \bigl( \Vert x_{1} - x_{2} \Vert \bigr)+ \varLambda ' \bigl( \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \bigr) \\ &{}+ \varLambda ' \bigl(|D^{\beta } \Vert x_{1} - x_{2} \Vert \bigr) +\varLambda ' \biggl( \int _{0}^{s} |h(\xi ) \bigl\vert \Vert x_{1} - x_{2} \Vert \,d\xi \bigr\vert \biggr) \biggr] \,ds \\ \leq{}& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl( \Vert x_{1} - x_{2} \Vert \bigr)+ \varLambda \bigl( \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \bigr) \\ &{}+ \varLambda \biggl( \frac{ \Vert x'_{1} - x'_{2} \Vert }{\varGamma (2- \beta )}\biggr) +\varLambda \bigl(m_{0} \Vert x_{1}- x_{2} \Vert \bigr) \biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[a_{1}(s) \Vert x_{1} - x_{2} \Vert + a _{2}(s) \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \\ &{}+ a_{3}(s) \frac{ \Vert x'_{1} - x'_{2} \Vert }{\varGamma (2- \beta )} +a_{4}(s) m _{0} \Vert x_{1} - x_{2} \Vert \biggr] \,ds \\ &{}+ \int _{\mu }^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda ' \bigl( \Vert x_{1} - x_{2} \Vert \bigr)+ \varLambda ' \bigl( \bigl\Vert x'_{1} - x'_{2} \bigr\Vert \bigr) \\ &{}+ \varLambda ' \biggl(\frac{ \Vert x'_{1} - x'_{2} \Vert }{\varGamma (2- \beta )}\biggr) +\varLambda ' \bigl( m_{0} \Vert x_{1} - x_{2} \Vert \bigr) \biggr] \,ds \\ \leq{}& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl(l \Vert x_{1} - x_{2} \Vert _{*}\bigr)+ \varLambda \bigl(l \Vert x_{1} - x_{2} \Vert _{*}\bigr) \\ &{}+ \varLambda \biggl( \frac{l \Vert x_{1} - x_{2} \Vert _{*}}{\varGamma (2- \beta )}\biggr) + \varLambda \bigl(m_{0} l \Vert x_{1} - x_{2} \Vert _{*} \bigr) \biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[a_{1}(s) l \Vert x_{1} - x_{2} \Vert _{*} + a_{2}(s) l \Vert x_{1} - x_{2} \Vert _{*} \\ &{}+ a_{3}(s) \frac{l \Vert x_{1} - x_{2} \Vert _{*}}{\varGamma (2- \beta )} +a_{4}(s) m_{0} l \Vert x_{1} - x_{2} \Vert _{*}\biggr] \,ds \\ &{}+ \int _{\mu }^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda ' \bigl(l \Vert x_{1} - x_{2} \Vert _{*}\bigr)+ \varLambda ' \bigl(l \Vert x_{1} - x_{2} \Vert _{*}\bigr) \\ &{}+ \varLambda ' \biggl(\frac{l \Vert x_{1} - x_{2} \Vert _{*}}{\varGamma (2- \beta )}\biggr) + \varLambda ' \bigl( m_{0} l \Vert x_{1} - x_{2} \Vert _{*}\bigr) \biggr] \,ds, \end{aligned}
(*)

where $$m_{0}= \int _{0}^{1}|h(\xi )|\,d\xi$$ and $$l= \max \{1, \frac{1 }{ \varGamma (2-\beta )}, m_{0}, \theta _{0}+\theta _{1} \}$$. On the other hand, $$\lim_{z \to 0^{+}} \frac{\varLambda (z)}{z}=q$$ and so for each $$\epsilon >0$$ there exists $$0<\delta _{\varLambda }= \delta (\epsilon , \varLambda )$$ such that $$| \frac{\varLambda (z)}{z}- q| <\epsilon$$ for all $$0< z \leq \delta _{\varLambda }$$. Thus, $$0< z \leq \delta _{\varLambda }$$ implies $$| \frac{\varLambda (z)}{z}| -q \leq | \frac{\varLambda (z)}{z} -q|< \epsilon$$. Hence, $$| \varLambda (z)|< (\epsilon +q)|z|$$. By choosing $$0< z \leq \delta _{1} := \min \{ \delta _{\varLambda }, \epsilon \}$$, we have

\begin{aligned} \bigl\vert \varLambda (z) \bigr\vert < (\epsilon +q) \vert z \vert < (\epsilon +q) \epsilon . \end{aligned}
(2)

For $$\varLambda '$$ we have similar conclusion, that is,

\begin{aligned} \bigl\vert \varLambda '(z) \bigr\vert < \bigl( \epsilon +q'\bigr) \epsilon \end{aligned}
(3)

for all $$0< z \leq \delta _{1} := \min \{ \delta _{\varLambda '}, \epsilon \}$$. Let $$\epsilon >0$$ be given, $$l \|x_{1} - x_{2}\|_{*} < \min \{ \delta _{1}, \delta _{2} \}$$ and $$x_{1} \to x_{2}$$. By using (2) and (3), we get $$\varLambda ( l \|x_{1} - x_{2}\|_{*} )< ( \epsilon +q) \epsilon$$ and $$\varLambda '( l \|x_{1} - x_{2}\|_{*} )< ( \epsilon +q') \epsilon$$. Now by using (*), we obtain

\begin{aligned} &\bigl\vert T_{x_{1}}(t)- T_{x_{2}}(t) \bigr\vert \\ &\quad \leq 4(q+ \epsilon )\epsilon \int _{0}^{ \lambda } \bigl\vert G(t,s) \bigr\vert \,ds + \epsilon \int _{\lambda }^{\mu } \bigl[a_{1}(s)+ \cdots+ a_{4}(s)\bigr] \bigl\vert G(t,s) \bigr\vert \,ds \\ &\qquad {}+ 4\bigl(q'+\epsilon \bigr)\epsilon \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \,ds \leq 4(q+\epsilon )\epsilon \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{0}^{\lambda } (1-s)^{\alpha -2}\,ds \\ &\qquad {}+ \frac{\epsilon (2+ \alpha + T_{0})}{\varGamma (\alpha )} \sum _{i=1} ^{4} \int _{\lambda }^{\mu } a_{i}(s) (1-s)^{\alpha -2}\,ds+ 4\bigl(q'+\epsilon \bigr)\epsilon \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{\mu }^{1} (1-s)^{ \alpha -2}\,ds \\ &\quad = \epsilon \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \Biggl[4(q+\epsilon ) \cdot \frac{1}{\alpha -1} \bigl(1-(1-\lambda )^{\alpha -1} \bigr)\\ &\qquad {}+ \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + 4\bigl(q'+ \epsilon \bigr) \cdot \frac{1}{ \alpha -1} (1- \mu )^{\alpha -1} \Biggr]. \end{aligned}

Hence,

\begin{aligned} \Vert T_{x_{1}}- T_{x_{2}} \Vert \leq& \epsilon \frac{2+ \alpha + T_{0}}{ \varGamma (\alpha )} \Biggl[4(q+\epsilon ) \cdot \frac{1}{\alpha -1} \bigl(1-(1-\lambda )^{\alpha -1} \bigr) \\ &{}+ \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + 4\bigl(q'+\epsilon \bigr) \cdot \frac{1}{\alpha -1} (1- \mu )^{\alpha -1} \Biggr]. \end{aligned}

In a similar way, we get

\begin{aligned} &\bigl\vert T'_{x_{1}}(t)- T'_{x_{2}}(t) \bigr\vert \\ &\quad \leq \int _{0}^{1} \biggl\vert \frac{\partial G}{ \partial t} (t,s) \biggr\vert \biggl\vert f\biggl(s, x_{1}(s), x'_{1}(s), D^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}(\xi )\,d\xi \biggr) \\ &\qquad {}- f\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h(\xi ) x _{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &\quad \leq \int _{0}^{\lambda } \biggl\vert \frac{\partial G}{\partial t} (t,s) \biggr\vert \biggl\vert f_{1}\biggl(s, x_{1}(s), x'_{1}(s), D^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}( \xi )\,d\xi \biggr) \\ &\qquad {}- f_{1}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &\qquad {}+ \int _{\lambda }^{\mu } \biggl\vert \frac{\partial G}{\partial t} (t,s) \biggr\vert \biggl\vert f_{2}\biggl(s, x_{1}(s), x'_{1}(s), D^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}( \xi )\,d\xi \biggr) \\ &\qquad {}- f_{2}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &\qquad {}+ \int _{\mu }^{1} \biggl\vert \frac{\partial G}{\partial t} (t,s) \biggr\vert \biggl\vert f_{3}\biggl(s, x _{1}(s), x'_{1}(s), D^{\beta }x_{1}(s), \int _{0}^{s} h(\xi ) x_{1}( \xi )\,d\xi \biggr) \\ &\qquad {}- f_{3}\biggl(s, x_{2}(s), x'_{2}(s), D^{\beta }x_{2}(s), \int _{0}^{s} h( \xi ) x_{2}(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &\quad \leq 4(q+\epsilon )\epsilon \int _{0}^{\lambda } \biggl\vert \frac{\partial G}{ \partial t} (t,s) \biggr\vert \,ds + \epsilon \int _{\lambda }^{\mu } \bigl[a_{1}(s)+ \cdots+ a_{4}(s)\bigr] \bigl\vert G(t,s) \bigr\vert \,ds \\ &\qquad {}+ 4\bigl(q'+\epsilon \bigr)\epsilon \int _{\mu }^{1} \biggl\vert \frac{\partial G}{\partial t} (t,s) \biggr\vert \,ds \leq 4(q+\epsilon ) \frac{3 \epsilon \alpha }{\varGamma ( \alpha )} \int _{0}^{\lambda } (1-s)^{\alpha -2}\,ds \\ &\qquad {}+ \frac{3 \epsilon \alpha }{\varGamma (\alpha )} \sum _{i=1}^{4} \int _{\lambda }^{\mu } a_{i}(s) (1-s)^{\alpha -2}\,ds + 4\bigl(q'+\epsilon \bigr) \frac{3 \epsilon \alpha }{\varGamma (\alpha )} \int _{\mu }^{1} (1-s)^{ \alpha -2}\,ds \\ &\quad = \frac{3 \epsilon \alpha }{\varGamma (\alpha )} \Biggl[4(q+\epsilon ) \cdot \frac{1}{ \alpha -1} \bigl(1-(1- \lambda )^{\alpha -1} \bigr) \\ &\qquad {}+ \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + 4\bigl(q'+\epsilon \bigr) \cdot \frac{1}{\alpha -1} (1- \mu )^{\alpha -1} \Biggr] \end{aligned}

and so

\begin{aligned} \bigl\Vert T'_{x_{1}}- T'_{x_{2}} \bigr\Vert \leq & \frac{3 \epsilon \alpha }{\varGamma (\alpha )} \Biggl[4(q+\epsilon ) \cdot \frac{1}{\alpha -1} \bigl(1-(1-\lambda )^{ \alpha -1} \bigr) \\ &{} + \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + 4\bigl(q'+ \epsilon \bigr) \cdot \frac{1}{\alpha -1} (1- \mu )^{\alpha -1} \Biggr]. \end{aligned}

Hence,

\begin{aligned} \Vert T_{x_{1}}- T_{x_{2}} \Vert _{*} \leq & \epsilon \Biggl[4(q+\epsilon ) \cdot \frac{1}{ \alpha -1} \bigl(1-(1-\lambda )^{\alpha -1} \bigr) + \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} \\ & {}+ 4\bigl(q'+\epsilon \bigr) \cdot \frac{1}{\alpha -1} (1- \mu )^{\alpha -1} \Biggr] \max \biggl\{ \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )}, \frac{3 \alpha }{ \varGamma (\alpha )} \biggr\} . \end{aligned}

This implies that $$\|T_{x_{1}}- T_{x_{2}}\|_{*} \to 0$$ as $$x_{1} \to x_{2}$$. Hence, T is continuous. Since $$\lim_{z\to 0^{+}} \frac{ \varLambda (z)}{z} =q$$ and Λ is nondecreasing, for each $$\epsilon >0$$ there exists $$\delta _{1}= \delta _{1}(\epsilon )>0$$ such that $$\frac{\varLambda (l z)}{ l z} < q+ \epsilon$$ for all $$z\in (0, \delta _{1}]$$. Thus, $$\varLambda (l z) < (q+ \epsilon ) l z$$. By using similar reason, there exists $$\delta _{2}(\epsilon )>0$$ such that $$\varLambda '(l z) < (q'+ \epsilon ) l z$$ for all $$z\in (0,\delta _{2}]$$. Put $$\delta = \delta ( \epsilon ) := \min \{ \delta _{1}( \epsilon ), \delta _{2}( \epsilon ) \}$$. Then $$\varLambda (l z) < (q+ \epsilon ) l z$$ and $$\varLambda '(l z) < (q'+ \epsilon ) l z$$ for all $$z\in (0,\delta ]$$. In particular, $$\varLambda (l \delta ) < (q+ \epsilon ) l \delta$$ and $$\varLambda '(l \delta ) < (q'+ \epsilon ) l \delta$$. On other hand, we have $$\frac{4q}{\alpha -1} (1-(1-\lambda )^{\alpha -1})+ \sum _{i=1} ^{4} \| \hat{a_{i}}\|_{[\lambda , \mu ]} + \frac{4q'}{\alpha -1} (1- \mu )^{\alpha -1}< \frac{\varGamma (\alpha )}{l \theta _{0}}$$. Choose $$\epsilon _{0}>0$$ such that $$\frac{4(q+\epsilon _{0})}{\alpha -1} (1-(1- \lambda )^{\alpha -1})+ \sum _{i=1}^{4} \| \hat{a_{i}}\|_{[\lambda , \mu ]} + \frac{4(q' +\epsilon _{0})}{\alpha -1} (1-\mu )^{\alpha -1}< \frac{ \varGamma (\alpha )}{l \theta _{0}}$$ and put $$\delta _{0}= \delta (\epsilon _{0})$$. Then $$\varLambda (l \delta _{0}) < (q+ \epsilon ) l \delta _{0}$$ and $$\varLambda '(l \delta _{0}) < (q'+ \epsilon ) l \delta _{0}$$. Now, assume that $$E= \{ x \in X : \|x\|_{*}< \delta _{0} \}$$, $$x \in E$$ and $$t \in [0,1]$$. Then we have

\begin{aligned} \bigl\vert T_{x}(t) \bigr\vert \leq& \int _{0}^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ =& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{3}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ =& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) -f_{1}(s,0,0,0,0,0) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) -f_{2}(s,0,0,0,0,0) \biggr\vert \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{3}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) -f_{3}(s,0,0,0,0,0) \biggr\vert \,ds \\ \leq& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl( \bigl\vert x(s) \bigr\vert \bigr) +\varLambda \bigl( \bigl\vert x'(s) \bigr\vert \bigr) +\varLambda \bigl( \bigl\vert D^{\beta }x(s) \bigr\vert \bigr) \\ &{}+ \varLambda \biggl( \int _{0}^{s} \bigl\vert h( \xi ) \bigr\vert \bigl\vert x(\xi ) \bigr\vert \,d\xi \biggr)\biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[ a_{1}(s) \bigl\vert x(s) \bigr\vert + a_{2}(s) \bigl\vert x'(s) \bigr\vert + a_{3}(s) \bigl\vert D^{\beta }x(s) \bigr\vert \\ &{} + a_{4}(s) \int _{0}^{s} \bigl\vert h(\xi ) \bigr\vert \bigl\vert x(\xi ) \bigr\vert \,d\xi \biggr] \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda '\bigl( \bigl\vert x(s) \bigr\vert \bigr) + \varLambda '\bigl( \bigl\vert x'(s) \bigr\vert \bigr) + \varLambda '\bigl( \bigl\vert D^{\beta }x(s) \bigr\vert \bigr) \\ &{} + \varLambda '\biggl( \int _{0}^{s} \bigl\vert h(\xi ) \bigr\vert \bigl\vert x( \xi ) \bigr\vert \,d\xi \biggr)\biggr] \,ds \\ \leq& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda \bigl( \Vert x \Vert \bigr) +\varLambda \bigl( \bigl\Vert x' \bigr\Vert \bigr) +\varLambda \biggl(\frac{ \Vert x' \Vert }{\varGamma (2- \beta )}\biggr) + \varLambda \bigl( m_{0} \Vert x \Vert \bigr)\biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl[ a_{1}(s) \Vert x \Vert + a_{2}(s) \bigl\Vert x' \bigr\Vert + a_{3}(s) \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}+ a_{4}(s) m_{0} \Vert x \Vert \biggr] \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl[ \varLambda '\bigl( \Vert x \Vert \bigr) +\varLambda '\bigl( \bigl\Vert x' \bigr\Vert \bigr) + \varLambda '\biggl(\frac{ \Vert x' \Vert }{\varGamma (2- \beta )}\biggr) + \varLambda '\bigl( m_{0} \Vert x \Vert \bigr)\biggr] \,ds \\ \leq& 4 \varLambda \bigl(l \Vert x \Vert _{*}\bigr) \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \,ds \\ &{}+ l \Vert x \Vert _{*} \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \sum _{i=1}^{4} a_{i}(s) \,ds + 4 \varLambda \bigl(l \Vert x \Vert _{*}\bigr) \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \,ds \\ \leq& 4 \varLambda (l \delta _{0}) \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{0}^{\lambda } (1-s)^{\alpha -2} \,ds \\ &{}+ l \delta _{0} \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \sum _{i=1} ^{4} \int _{\lambda }^{\mu } a_{i}(s) (1-s)^{\alpha -2} \,ds \\ &{}+ 4 \varLambda '(l \delta _{0}) \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{ \mu }^{1} (1-s)^{\alpha -2} \,ds \\ \leq& 4(q+\epsilon _{0}) l \delta _{0} \frac{(2+ \alpha + T_{0})}{\varGamma (\alpha )}\cdot \frac{1}{\alpha -1}\bigl[1-(1-\lambda )^{\alpha -1}\bigr] \\ &{}+ l \delta _{0} \frac{(2+ \alpha + T_{0})}{\varGamma (\alpha )} \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} \\ &{}+ 4(q+\epsilon _{0}) l \delta _{0} \frac{(2+ \alpha + T_{0})}{\varGamma ( \alpha )}\cdot \frac{1}{\alpha -1}\bigl[1-(1-\lambda )^{\alpha -1}\bigr] \\ =&\delta _{0} \frac{(2+ \alpha + T_{0}) l}{\varGamma (\alpha )}[ \frac{4(q+ \epsilon _{0})}{\alpha -1} \bigl(1-(1- \lambda )^{\alpha -1}\bigr)\\ &{}+ \sum _{i=1} ^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + \frac{4(q' +\epsilon _{0})}{ \alpha -1} (1-\mu )^{\alpha -1}< \delta _{0}. \end{aligned}

Hence, $$\|Tx\| \leq \delta _{0}$$. By using a similar method, we get

\begin{aligned} \bigl\vert T'_{x}(t) \bigr\vert \leq& \int _{0}^{1} \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl\vert f\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d \xi \biggr) \biggr\vert \,ds \\ =& \int _{0}^{\lambda } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggr\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggl\vert \,ds \\ &{}+ \int _{\mu }^{1} \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggr\vert f_{3}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr)\biggr| \,ds \\ \leq& \int _{0}^{\lambda } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ \varLambda \bigl( \bigl\vert x(s) \bigr\vert \bigr) + \varLambda \bigl( \bigl\vert x'(s) \bigr\vert \bigr) +\varLambda \bigl( \bigl\vert D^{\beta }x(s) \bigr\vert \bigr) \\ &{}+ \varLambda \biggl( \int _{0}^{s} \bigl\vert h(\xi ) \bigr\vert \bigl\vert x(\xi ) \bigr\vert \,d\xi \biggr)\biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ a_{1}(s) \bigl\vert x(s) \bigr\vert + a_{2}(s) \bigl\vert x'(s) \bigr\vert + a_{3}(s) \bigl\vert D^{\beta }x(s) \bigr\vert \\ &{} + a_{4}(s) \int _{0}^{s} \bigl\vert h(\xi ) \bigr\vert \bigl\vert x(\xi ) \bigr\vert \,d\xi \biggr] \,ds \\ &{}+ \int _{\mu }^{1} \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ \varLambda '\bigl( \bigl\vert x(s) \bigr\vert \bigr) +\varLambda '\bigl( \bigl\vert x'(s) \bigr\vert \bigr) +\varLambda '\bigl( \bigl\vert D^{\beta }x(s) \bigr\vert \bigr) \\ &{}+ \varLambda '\biggl( \int _{0}^{s} \bigl\vert h(\xi ) \bigr\vert \bigl\vert x(\xi ) \bigr\vert \,d\xi \biggr)\biggr] \,ds \\ \leq& \int _{0}^{\lambda } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ \varLambda \bigl( \Vert x \Vert \bigr) +\varLambda \bigl( \bigl\Vert x' \bigr\Vert \bigr) +\varLambda \biggl( \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}\biggr) + \varLambda \bigl( m_{0} \Vert x \Vert \bigr)\biggr] \,ds \\ &{}+ \int _{\lambda }^{\mu } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ a _{1}(s) \Vert x \Vert + a_{2}(s) \bigl\Vert x' \bigr\Vert + a_{3}(s) \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}+ a_{4}(s) m_{0} \Vert x \Vert \biggr] \,ds \\ &{}+ \int _{\mu }^{1} \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \biggl[ \varLambda '\bigl( \Vert x \Vert \bigr) + \varLambda '\bigl( \bigl\Vert x' \bigr\Vert \bigr) + \varLambda '\biggl( \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}\biggr) + \varLambda '\bigl( m_{0} \Vert x \Vert \bigr)\biggr] \,ds \\ \leq& 4 \varLambda \bigl(l \Vert x \Vert _{*}\bigr) \int _{0}^{\lambda } \biggl\vert \frac{ \partial G}{ \partial t}(t,s) \biggr\vert \,ds + l \Vert x \Vert _{*} \int _{\lambda }^{\mu } \biggl\vert \frac{ \partial G}{\partial t}(t,s) \biggr\vert \sum _{i=1}^{4} a_{i}(s) \,ds + 4 \varLambda \bigl(l \Vert x \Vert _{*}\bigr) \\ &{}\times \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \,ds\leq 4 \varLambda (l \delta _{0}) \frac{3 \alpha }{\varGamma (\alpha )} \int _{0}^{\lambda } (1-s)^{\alpha -2} \,ds \\ &{}+ l \delta _{0} \frac{3 \alpha }{\varGamma (\alpha )} \sum _{i=1}^{4} \int _{\lambda }^{\mu } a_{i}(s) (1-s)^{\alpha -2} \,ds+ 4 \varLambda '(l \delta _{0}) \frac{3 \alpha }{\varGamma (\alpha )} \int _{\mu }^{1} (1-s)^{ \alpha -2} \,ds \\ \leq& 4(q+\epsilon _{0}) l \delta _{0} \frac{3 \alpha }{\varGamma (\alpha )}\cdot \frac{1}{\alpha -1}\bigl[1-(1-\lambda )^{\alpha -1}\bigr] + l \delta _{0} \frac{3 \alpha }{\varGamma (\alpha )} \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} \\ &{}+ 4(q+\epsilon _{0}) l \delta _{0} \frac{3 \alpha }{\varGamma (\alpha )}\cdot \frac{1}{ \alpha -1}\bigl[1-(1-\lambda )^{\alpha -1}\bigr] \\ =&\delta _{0} \frac{3 \alpha l}{\varGamma (\alpha )}\Biggl[ \frac{4(q+\epsilon _{0})}{\alpha -1} \bigl(1-(1- \lambda )^{\alpha -1}\bigr)+ \sum _{i=1}^{4} \Vert \hat{a_{i}} \Vert _{[\lambda , \mu ]} + \frac{4(q' +\epsilon _{0})}{\alpha -1} (1-\mu )^{\alpha -1}\Biggr]\\ < & \delta _{0} \end{aligned}

for all $$x \in E$$ and $$t \in [0,1]$$. Hence, $$\|Tx\| \leq \delta _{0}$$, and so $$\|Tx\|_{*} \leq \delta _{0}$$. Thus, T maps E into E. It is easy to check that T maps bounded sets into bounded sets. Assume that $$t_{1}, t_{2} \in [0,1]$$ and $$x \in E$$. Since $$G(t,s)$$ and $$\frac{ \partial G(t,s)}{\partial t}$$ are continuous with respect to t, we get

\begin{aligned} \lim_{t_{2} \to t_{1}} T'x(t_{2}) =&\lim _{t_{2} \to t_{1}} \int _{0} ^{1} \biggl\vert \frac{ \partial G}{\partial t}(t_{2}, s) \biggr\vert \biggl\vert f\biggl(s, x(s), x'(s), D ^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ =& \int _{0}^{1} \lim_{t_{2} \to t_{1}} \frac{ \partial G}{\partial t}(t _{2}, s) f\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x( \xi )\,d\xi \biggr) \,ds \\ =& \int _{0}^{1} \frac{ \partial G}{\partial t}(t_{1}, s) f\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \,ds\\ =& T'x(t _{1}). \end{aligned}

Hence, T is equi-continuous on E and so $$T: E \to E$$ is completely continuous. Now by using Lemma 2, T has a fixed point on E and so the problem (1) has a solution. □

### Example 1

Consider the pointwise defined equation

$$D^{\frac{7}{2}} x(t) +f\biggl(t, x(t), x'(t), D^{\frac{1}{2}} x(t), \int _{0}^{t} s x(s) \,ds\biggr)=0$$

with the boundary conditions in the last result, where

$$f(t, x_{1}, x_{2}, x_{3}, x_{4}, x_{5})= \textstyle\begin{cases} t \sum _{i=1}^{4} x_{i}, & 0 \leq t < 0.2, \\ d(t) \sum _{i=1}^{4} x_{i}, & 0.2 \leq t \leq 0.8, \\ t^{2} \sum _{i=1}^{4} x_{i}, & 0.8 < t \leq 1, \end{cases}$$

and $$d(t)=0$$ whenever $$t \in [0.1, 0.8] \cap Q$$ and $$d(t)=0.1$$ whenever $$t \in [0.1, 0.8] \cap Q^{c}$$. Now, put $$f_{1}(t, x_{1}, x_{2}, x_{3}, x_{4}, x_{5})= \frac{1}{2} t \sum _{i=1}^{4} x_{i}$$, $$f_{2}(t, x_{1}, x_{2}, x_{3}, x_{4}, x_{5})= d(t) \sum _{i=1}^{4} x_{i}$$ and $$f_{3}(t, x_{1}, x_{2}, x_{3}, x_{4}, x_{5})= t^{2} \sum _{i=1}^{4} x _{i}$$. Then we have $$f_{1}(t, 0, 0, 0, 0)= f_{2}(t, 0, 0, 0, 0) = f _{3}(t, 0, 0, 0, 0) =0$$,

\begin{aligned} \bigl\vert f_{1}(t, x_{1}, x_{2}, x_{3}, x_{4})- f_{1}(t, y_{1}, y_{2}, y_{3}, y _{4}) \bigr\vert \leq& t \sum _{i=1}^{4} \bigl\vert \vert x_{i} \vert - \vert y_{i} \vert \bigr\vert \\ \leq& t \sum _{i=1}^{4} \vert x_{i} - y_{i} \vert \leq 0.1 \sum _{i=1}^{4} \varLambda \bigl( \vert x_{i} - y_{i} \vert \bigr), \\ \bigl\vert f_{2}(t, x_{1}, x_{2}, x_{3}, x_{4})- f_{2}(t, y_{1}, y_{2}, y_{3}, y _{4}) \bigr\vert \leq& d(t) \sum _{i=1}^{4} \bigl\vert \vert x_{i} \vert - \vert y_{i} \vert \bigr\vert \\ \leq& d(t) \sum _{i=1}^{4} \vert x_{i} - y_{i} \vert , \end{aligned}

and

\begin{aligned} \bigl\vert f_{3}(t, x_{1}, x_{2}, x_{3}, x_{4})- f_{3}(t, y_{1}, y_{2}, y_{3}, y _{4}) \bigr\vert \leq& \frac{1}{2} t \sum _{i=1}^{4} \bigl\vert \vert x_{i} \vert - \vert y_{i} \vert \bigr\vert \\ \leq& \frac{1}{2} t \sum _{i=1}^{4} \vert x_{i} - y_{i} \vert \leq \frac{1}{2} \sum _{i=1}^{4} \varLambda \bigl( \vert x_{i} - y_{i} \vert \bigr), \end{aligned}

where $$\varLambda (x) = |x|$$ and $$\varLambda ' (x) =\frac{1}{2} |x|$$. Hence, $$\lim_{z \to 0^{+}} \frac{\varLambda ' (z)}{z} = 0.1:= q$$, $$\lim_{z \to 0^{+}} \frac{\varLambda ' (z)}{z} = \frac{1}{2}:= q'$$, $$\hat{a_{i}}= \hat{d} \in L^{1}[0.1, 0.8]$$, $$\sum _{i=1}^{4} \| \hat{a_{i}}\|_{[\lambda , \mu ]}< 0.092$$ and

\begin{aligned} & \Biggl[ \frac{4q (1-(1-\lambda )^{\alpha -1})}{\alpha -1} + \sum _{i=1} ^{4} \Vert \hat{a_{i}} \Vert + \frac{4q' }{\alpha -1} (1-\mu )^{\alpha -1}\Biggr] \\ &\quad < \biggl[ \frac{4 \times 0.1 (1-(1-0.1)^{\frac{5}{2}}}{\frac{5}{2}}+ 0.092 +\frac{4 \times 0.5 (1-0.9)^{\frac{5}{2}}}{\frac{5}{2}} \biggr] < \frac{ \varGamma (\alpha )}{3l \alpha }. \end{aligned}

Now by using Theorem 5, the problem has a solution.

Now, we present our second result by using different conditions.

### Theorem 6

Suppose that $$f=[f_{1},f_{2},f_{3},\lambda ,\mu ]$$, f is nonnegative on $$[0,1]$$ and there exist nonnegative functions $$a_{1},a_{2}, a_{3}, a_{4}:[0,\lambda ] \to \mathbb{R}^{+}$$, maps $$b_{1},\dots , b_{k_{0}}:[ \lambda , \mu ] \to \mathbb{R}^{+}$$ for some $$k_{0} \geq 1$$, and functions $$c_{1},c_{2}, c_{3}, c_{4}:[\mu , 1] \to \mathbb{R}^{+}$$ such that $$\hat{a_{i}} \in L^{1}[0,\lambda ]$$, $$\hat{b_{j}} \in L^{1}[ \lambda , \mu ]$$, $$\hat{c_{i}} \in L^{1}[\mu , 1]$$ and $$\hat{a_{1}}(s) =(1-s)^{\alpha -2} a_{i}(s)$$. Assuming that there are nonnegative and nondecreasing functions $$\phi _{i}, \varPhi _{i} : \mathbb{R}^{+} \to \mathbb{R}^{+}$$ and $$H_{j}: \mathbb{R_{+}}^{4} \to \mathbb{R}_{+}$$ such that $$\lim_{z \to 0^{+}} \frac{\phi _{i}(z)}{z^{\mu _{i}} }:= l_{\mu _{i}}< \infty$$, $$\lim_{z \to 0^{+}} \frac{\varPhi _{i}(z)}{z^{\gamma _{i}} }:= l_{\gamma _{i}}< \infty$$ and $$\lim_{z \to 0^{+}} \frac{H_{j}(z, z, z, z)}{z^{m} }:=q_{j}< \infty$$ for some $$\mu _{i}, \gamma _{i}, m \in [1, \infty )$$ and $$H_{j}$$ are nonnegative and nondecreasing with respect to all their components ($$1 \leq i \leq 4$$, $$1 \leq j \leq k _{0}$$),

\begin{aligned}& \bigl\vert f_{1}(t, x_{1}, \ldots, x_{4}) - f_{1}(t, y_{1}, \ldots, y_{4}) \bigr\vert \leq \sum _{i=1}^{4} a_{i}(t) \phi _{i} \bigl( \vert x_{i} - y_{i} \vert \bigr), \\& \bigl\vert f_{2}(t, x_{1}, \ldots, x_{4}) - f_{2}(t, y_{1}, \ldots, y_{4}) \bigr\vert \leq \sum _{i=1}^{5} b_{j}(t) H_{j} \bigl( \vert x_{1} - y_{1} \vert , \ldots, \vert x_{4} - y _{4} \vert \bigr) \end{aligned}

and $$|f_{3}(t, x_{1}, \ldots, x_{4}) - f_{3}(t, y_{1}, \ldots, y_{4})| \leq \sum _{i=1}^{4} c_{i}(t) \varPhi _{i}(| x_{i} - y_{i} |)$$. Suppose that $$|f_{2}(t, x_{1}, \ldots, x_{4})| \leq \varTheta (t) \varLambda (x_{1}, \ldots, x_{4})$$, where Λ are nonnegative and nondecreasing with respect to all their components, $$\lim_{x \to 0^{+}} \frac{\varLambda (x, x, x, x)}{x }:=P_{2},< \infty$$, $$\hat{\varTheta } \in L^{1}[\lambda , \mu ]$$, $$\lim_{\max |x_{i}| \to 0} \frac{|f_{1}(t, x_{1}, \ldots, x_{4})|}{ \max |x_{i}|} =P_{1}(t)$$ and $$\lim_{\max |x_{i}| \to 0} \frac{|f_{3}(t, x_{1}, \ldots, x_{4})|}{\max |x_{i}|} =P_{3}(t)$$, where $$\hat{P_{1}} \in L^{1}[0, \lambda ]$$, $$\hat{P_{3}} \in L^{1}[\mu , 1]$$. If

\begin{aligned}& \begin{gathered} \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T_{0}}{ \varGamma (\alpha ) } \biggr\} \max \Biggl\{ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} ) + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} q_{j} \\ \quad {}+\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} ] ( l_{\gamma _{i}} ) , \max \biggl\{ 1, \frac{1}{\varGamma (2- \beta ) }, m_{0} \biggr\} \bigl[ \Vert \hat{P_{1}} \Vert _{[0, \lambda ]} + P_{2} \Vert \varTheta \Vert _{[\lambda , \mu ]} + \Vert \hat{P_{3}} \Vert _{[\mu , 1]}\bigr] \Biggr\} < 1, \end{gathered} \end{aligned}

then the pointwise defined equation (1) with boundary conditions has a solution.

### Proof

Let $$x,y \in X$$ and $$t \in [0,1]$$. Then we have

\begin{aligned}& \bigl\vert F_{x}(t)-F_{y}(t) \bigr\vert \\& \quad \leq \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \\& \qquad {}-f_{1}\biggl(s, y(s), y'(s), D^{\beta }y(s), \int _{0}^{s} h(\xi ) y(\xi )\,d \xi \biggr) \biggr\vert \,ds \\& \qquad {}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \\& \qquad {}-f_{2}\biggl(s, y(s), y'(s), D^{\beta }y(s), \int _{0}^{s} h(\xi ) y(\xi )\,d \xi \biggr) \biggr\vert \,ds \\& \qquad {}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{3}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \\& \qquad {}- f_{3}\biggl(s, y(s), y'(s), D^{\beta }y(s), \int _{0}^{s} h(\xi ) y(\xi )\,d \xi \biggr) \biggr\vert \,ds \\& \quad \leq \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert [a_{1}(s) \phi \bigl( \bigl\vert x(s) - y(s) \bigr\vert \bigr)+ a _{2}(s) \phi \bigl( \bigl\vert x'(s) - y'(s) \bigr\vert \bigr) \\& \qquad {}+ a_{3}(s) \phi \bigl( \bigl\vert D^{\beta }x(s) - D^{\beta }y(s) \bigr\vert \bigr) + a_{4}(s) \phi \biggl( \biggl\vert \int _{0}^{s} h(\xi ) \bigl( x(\xi ) - y(\xi ) \bigr) \,d\xi \biggr\vert \biggr) \,ds \\& \qquad {}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \sum _{i=1}^{k_{0}} b_{i}(s) H_{i}( \bigl\vert x(s) - y(s) \bigr\vert , \bigl\vert x'(s) - y'(s) \bigr\vert , \\& \qquad {}\bigl\vert D^{\beta }x(s) - D^{\beta }y(s) \bigr\vert , \biggl\vert \int _{0}^{s} h(\xi ) \bigl( x(\xi ) - y( \xi )\, d \xi \bigr)\biggr| \,ds \\& \qquad {}+ \int _{\mu }^{1} \bigl\vert G(t,s)| \biggl[c_{1}(s) \varPhi \bigl( \bigl\vert x(s) - y(s) \bigr\vert \bigr)+ c_{2}(s) \varPhi \bigl( \bigl\vert x'(s) - y'(s) \bigr\vert \bigr) \\& \qquad {}+ c_{3}(s) \varPhi \bigl( \bigl\vert D^{\beta }x(s) - D^{\beta }y(s) \bigr\vert \bigr) + c_{4}(s) \varPhi ( \biggl\vert \int _{0}^{s} h(\xi ) \bigl( x(\xi ) - y(\xi ) \bigr) \,d\xi \biggr\vert \biggr] \,ds \\& \quad \leq \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \int _{0}^{\lambda } (1-s)^{ \alpha -2} \bigl[a_{1}(s) \phi \bigl( \Vert x - y \Vert \bigr)+ a_{2}(s) \phi \bigl( \bigl\Vert x' - y' \bigr\Vert \bigr)\bigr] \\& \qquad {}+ a_{3}(s) \phi \biggl({\frac{ \Vert x' - y' \Vert }{\varGamma (2-\beta )}} \biggr) + a_{4}(s) \phi \bigl( m_{0} \Vert x - y \Vert \bigr) \,ds \\& \qquad {}+\frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{\lambda }^{\mu } (1-s)^{ \alpha -2}\\& \qquad {}\times \sum _{i=1}^{k_{0}} b_{i}(s) H_{i}\biggl( \Vert x - y \Vert , \bigl\Vert x' - y' \bigr\Vert , \frac{ \Vert x' - y' \Vert }{\varGamma (2-\beta )}, m_{0} \Vert x - y \Vert \biggr) \,ds \\& \qquad {}+ \frac{2+ \alpha + T_{0}}{\varGamma (\alpha )} \int _{\mu }^{1} (1-s)^{ \alpha -2} \biggl[c_{1}(s) \varPhi \bigl( \Vert x - y \Vert \bigr)+ c_{2}(s) \varPhi \bigl( \bigl\Vert x' - y' \bigr\Vert \bigr) \\& \qquad {}+ c_{3}(s) \varPhi \biggl({\frac{ \Vert x' - y' \Vert }{\varGamma (2-\beta )}}\biggr) + c_{4}(s) \varPhi \bigl(m_{0} \Vert x - y \Vert \bigr) \biggr] \,ds \\& \quad \leq \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \Biggl[ \sum _{i=1}^{4} \phi \bigl(l \Vert x - y \Vert _{*}\bigr) \int _{0}^{\lambda } (1-s)^{\alpha -2} a_{i}(s) \,ds \\& \qquad {}+ \sum _{i=1}^{k_{0}} H_{i}\bigl(l \Vert x - y \Vert _{*}, \ldots, l \Vert x - y \Vert _{*}\bigr) \int _{\lambda }^{\mu } (1-s)^{\alpha -2} b_{i}(s) \,ds \\& \qquad {}+\sum _{i=1}^{4} \varPhi \bigl(l \Vert x - y \Vert _{*}\bigr) \int _{\mu }^{1} (1-s)^{ \alpha -2} c_{i}(s) \,ds \Biggr] \\& \quad = \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \Biggl[ \sum _{i=1}^{5} \phi \bigl(l \Vert x - y \Vert _{*} \bigr) \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} \\& \qquad {}+ \sum _{i=1}^{k_{0}} H_{i}\bigl(l \Vert x - y \Vert _{*}, \ldots, l \Vert x - y \Vert _{*}\bigr) \Vert \hat{b_{i}} \Vert _{[\lambda , \mu ]} +\sum _{i=1}^{4} \varPhi \bigl(l \Vert x - y \Vert _{*}\bigr) \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr]. \end{aligned}

On the other hand, $$\lim_{z \to 0^{+}} \frac{\phi _{i}(z)}{z^{\mu _{i}}}= l_{\mu _{i}}$$ for $$1 \leq i \leq 4$$. This implies that for each $$\epsilon >0$$ there exists $$0< \delta _{i} = \delta _{i}(\epsilon ) < \epsilon$$ such that $$\frac{\phi _{i}(z)}{z ^{\mu _{i}}}< l_{\mu _{i}}+ \epsilon$$ for all $$z \in (0, \delta _{i}]$$. Hence, $$\phi _{i}(\delta _{i})< ( l_{\mu _{i}} +\epsilon ) \delta ^{\mu _{i}}_{i} < ( l_{\mu _{i}} +\epsilon ) \epsilon ^{\mu _{i}}$$. By using a similar method, we conclude that there exists $$0< \delta '_{i} = \delta '_{i}(\epsilon ) < \epsilon$$ such that $$\varPhi _{i}(\delta '_{i})< ( l_{\gamma _{i}} +\epsilon ) (\delta '_{i})^{\gamma _{i}} < ( l_{\gamma _{i}} +\epsilon ) \epsilon ^{\gamma _{i}}$$. Also, we have $$\lim_{z \to 0^{+}} \frac{H_{j}(z, z, z, z)}{z^{m}}= q_{j}$$ for $$1 \leq j \leq k_{0}$$ and so there exists $$0< \delta _{q_{j}} < \epsilon$$ such that $$\frac{H_{j}(z, z, z, z)}{z^{m}}< q_{j} + \epsilon$$ for all $$z \in (0, \delta _{q_{j}}]$$ and $$1 \leq j \leq k_{0}$$. Hence, $$H_{j}(z, z, z, z) < (q_{j} + \epsilon ) z^{m}$$ for all $$z \in (0, \delta _{q_{j}}]$$ and so $$H_{j}(\delta _{q_{j}}, \delta _{q_{j}}, \delta _{q_{j}}, \delta _{q_{j}}) < (q_{j} + \epsilon ) \delta _{q_{j}} ^{m} < (q_{j} + \epsilon ) \epsilon ^{m}$$. Let $$x \to y$$ in X. If $$l \|x -y\|_{*}< \delta := \min \{ \delta _{1}, \ldots, \delta _{4}, \delta '_{1}, \ldots, \delta '_{4}, \delta _{q_{1}}, \ldots, \delta _{q_{k _{0}}} \}$$, then $$\phi _{i}(\delta )< \phi _{i}(\delta _{i})< ( l_{\mu _{i}} +\epsilon ) \epsilon ^{\mu _{i}}$$, $$\varPhi _{i}(\delta )< \varPhi _{i}( \delta '_{i})< ( l_{\gamma _{i}} +\epsilon ) \epsilon ^{\gamma _{i}}$$ and $$H_{j}(\delta , \ldots, \delta ) < H_{j}(\delta _{q_{j}}, \ldots, \delta _{q _{j}}) < (q_{j} + \epsilon ) \epsilon ^{m}$$ for $$1 \leq i \leq 4$$ and $$1 \leq j \leq k_{0}$$. If $$l \|x -y\|_{*}< \delta$$, then $$|F_{x}(t)-F _{y}(t)| \leq \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } [ \sum _{i=1}^{5} \| \hat{a_{i}} \|_{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon ) \epsilon ^{\mu _{i}} + \sum _{j=1}^{k_{0}} \| \hat{b_{j}} \|_{[ \lambda , \mu ]} (q_{j} + \epsilon ) \epsilon ^{m} +\sum _{i=1}^{4} \| \hat{c_{i}} \|_{[\mu , 1]} ] ( l_{\gamma _{i}} +\epsilon ) \epsilon ^{\gamma _{i}}]$$ and so $$| F_{x}-F_{y} \| \leq \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) }\times [ \sum _{i=1}^{5} \| \hat{a_{i}} \|_{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon ) \epsilon ^{\mu _{i}} + \sum _{j=1}^{k_{0}} \| \hat{b_{j}} \|_{[\lambda , \mu ]} (q_{j} + \epsilon ) \epsilon ^{m} +\sum _{i=1}^{4} \| \hat{c_{i}} \|_{[\mu , 1]} ] ( l_{\gamma _{i}} +\epsilon ) \epsilon ^{\gamma _{i}}]$$. By a similar way, we get

\begin{aligned} \bigl\Vert F'_{x}-F'_{y} \bigr\Vert \leq{}& \frac{3 \alpha }{\varGamma (\alpha ) } \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} + \epsilon ) \epsilon ^{\mu _{i}} + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q_{j} + \epsilon ) \epsilon ^{m} \\ &{}+\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} + \epsilon ) \epsilon ^{\gamma _{i}}] \end{aligned}

and so $$\| F_{x}-F_{y} \|_{*} \leq \max \{ \frac{3 \alpha }{\varGamma ( \alpha ) }, \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \} [ \sum _{i=1}^{5} \| \hat{a_{i}} \|_{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon ) \epsilon ^{\mu _{i}} + \sum _{j=1}^{k_{0}} \| \hat{b_{j}} \|_{[ \lambda , \mu ]} (q_{j} + \epsilon ) \epsilon ^{m} +\sum _{i=1}^{4} \| \hat{c_{i}} \|_{[\mu , 1]} ] ( l_{\gamma _{i}} +\epsilon ) \epsilon ^{\gamma _{i}}]$$. Since $$\epsilon >0$$ was arbitrary, we conclude that $$\| F_{x}-F_{y} \|_{*} \to 0$$ as $$x \to y$$. This implies that F is continuous on X. Since $$\lim_{x \to 0^{+}} \frac{\varLambda (x, x, x, x)}{x} = P_{2}$$, $$\lim_{x \to 0^{+}} \frac{\varLambda (lx, lx, lx, lx)}{lx} = P_{2}$$, where $$l = \max \{ 1, \frac{1}{\varGamma (2- \beta )}, m_{0} \}$$. Thus for each $$\epsilon >0$$ there exists $$\delta _{1} = \delta _{1}(\epsilon )$$ such that $$\frac{\varLambda (lx, lx, lx, lx)}{lx} < P_{2} + \epsilon$$ for all $$x \in (0, \delta _{1}]$$. Hence,

\begin{aligned} \varLambda (lx, lx, lx, lx) < ( P_{2} + \epsilon ) l x \end{aligned}
(4)

for $$x \in (0, \delta _{1}]$$. Also, $$\lim_{|x_{i}| \to 0} \frac{|f_{1}(t, x_{1}, \ldots, x_{4})|}{\min |x_{i}|} = P_{1}(t)$$. Thus, there exists $$\delta _{2} = \delta _{2}(\epsilon )$$ such that

\begin{aligned} \bigl\vert f_{1}(t, x_{1}, \ldots, x_{4}) \bigr\vert < \bigl(P_{1}(t)+ \epsilon \bigr) \min \vert x_{i} \vert \end{aligned}
(5)

for all $$t \in [0,1]$$ and $$|x_{i}| \in (0, \delta _{2}]$$ for $$1 \leq i \leq 4$$. Similarly, there exists $$\delta _{3} = \delta _{3}( \epsilon )$$ such that

\begin{aligned} \bigl\vert f_{3}(t, x_{1}, \ldots, x_{4}) \bigr\vert < \bigl(P_{3}(t)+ \epsilon \bigr) \min \vert x_{i} \vert \end{aligned}
(6)

for all $$t \in [0,1]$$ and $$|x_{i}| \in (0, \delta _{3}]$$ for $$1 \leq i \leq 4$$. Since $$\| \hat{P_{1}} \|_{[0, \lambda ]} + P_{2} \| \varTheta \|_{[\lambda , \mu ]} + \| \hat{P_{3}} \|_{[\mu , 1]} < \frac{ \varGamma (\alpha )}{ l \theta _{0}}$$, we can choose $$\epsilon _{0} >0$$ such that $$\| \hat{P_{1}} \|_{[0, \lambda ]} + \frac{\epsilon _{0}}{\alpha -1} (1-(1-\lambda )^{\alpha -1} ) + (P_{2} + \epsilon _{0}) \| \varTheta \|_{[\lambda , \mu ]} + \| \hat{P_{3}} \|_{[\mu , 1]} + \frac{\epsilon _{0}}{\alpha -1} (1- \mu )^{\alpha -1} < \frac{\varGamma (\alpha )}{ l \theta _{0}}$$. Since

$$\max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T_{0}}{ \varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} ) + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} q_{j} +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[ \mu , 1]} \Biggr] ( l_{\gamma _{i}} ) ] < 1,$$

pick $$\epsilon _{1} \in (0,1)$$ such that

\begin{aligned} & \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T _{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \\ & \quad {} + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) ] < 1. \end{aligned}
(7)

Let $$r_{0} = \min \{ \delta _{1}(\epsilon _{0}), \delta _{2}(\epsilon _{0}), \delta _{3}(\epsilon _{0}), \frac{\epsilon _{1}}{2} \}$$, and $$C= \{ x\in X : \|x\|_{*}< r_{0} \}$$. Define the map α on $$X\times X$$ by

$$\alpha (x, y)= \textstyle\begin{cases} 1, & x,y \in C, \\ 0, & \mbox{otherwise}. \end{cases}$$

Let $$x, y \in X$$ and $$\alpha (x, y) \geq 1$$. Then $$x, y \in C$$ and so

\begin{aligned} \bigl\vert F_{x}(t) \bigr\vert \leq& \int _{0}^{\lambda } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\mu }^{1} \bigl\vert G(t,s) \bigr\vert \biggl\vert f_{3}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int_{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[ \int _{0}^{\lambda } (1-s)^{\alpha -2} \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } (1-s)^{\alpha -2} \biggl\vert f_{2}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} \biggl\vert 3_{2}\biggl(s, x(s), x'(s),D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \biggr] \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[ \int _{0}^{\lambda} (1-s)^{\alpha -2} \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } (1-s)^{\alpha -2} \varTheta (s) \varLambda \biggl( x(s),x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \,ds \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} \biggl\vert 3_{2}\biggl(s, x(s), x'(s),D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \biggr] \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[ \int _{0}^{\lambda} (1-s)^{\alpha -2} \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } (1-s)^{\alpha -2} \varTheta (s) \varLambda \biggl( \Vert x \Vert , \bigl\Vert x' \bigr\Vert , \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}, m_{0} \Vert x \Vert \biggr) \,ds \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} \biggl\vert 3_{2}\biggl(s, x(s), x'(s),D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \biggr] \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[ \int _{0}^{\lambda} (1-s)^{\alpha -2} \biggl\vert f_{1}\biggl(s, x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \\ &{}+ \int _{\lambda }^{\mu } (1-s)^{\alpha -2} \varTheta (s) \varLambda \bigl( l \Vert x \Vert _{*}, l \Vert x \Vert _{*}, l \Vert x \Vert _{*}, l \Vert x \Vert _{*}\bigr) \,ds \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} \biggl\vert 3_{2}\biggl(s, x(s), x'(s),D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr) \biggr\vert \,ds \biggr] \end{aligned}

for all $$t \in [0,1]$$. Since $$\|x\|_{*}< r_{0}$$, $$x \in [0, \min \{ \delta _{1}, \delta _{2}, \delta _{3} \} )$$ and so by using (4), (5) and (6) we conclude that

\begin{aligned} \bigl\vert F_{x}(t) \bigr\vert \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \\ &{}\times\biggl[ \int _{0}^{\lambda } (1-s)^{\alpha -2} \bigl(P_{1}(s)+ \epsilon \bigr) \min \biggl\{ x_{(}s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d\xi \biggr\} \,ds \\ &{}+ ( P_{2} + \epsilon _{0}) l \Vert x \Vert _{*} \int _{\lambda }^{\mu } (1-s)^{ \alpha -2} \varTheta (s) \,ds \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} | \bigl(P_{3}(s)+ \epsilon _{0}\bigr) \min \biggl\{ x(s), x'(s), D^{\beta }x(s), \int _{0}^{s} h(\xi ) x(\xi )\,d \xi \biggr\} \,ds \biggr] \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[ \int _{0}^{\lambda } (1-s)^{\alpha -2} \bigl(P_{1}(s)+ \epsilon \bigr) \min \biggl\{ \Vert x \Vert , \bigl\Vert x' \bigr\Vert , \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}, m_{0} \Vert x \Vert \biggr\} \,ds \\ &{}+ ( P_{2} + \epsilon _{0}) l \Vert x \Vert _{*} \Vert \hat{\varTheta } \Vert _{[\lambda , \mu ]} \\ &{}+ \int _{\mu }^{1} (1-s)^{\alpha -2} | \bigl(P_{3}(s)+ \epsilon _{0}\bigr) \min \biggl\{ \Vert x \Vert , \bigl\Vert x' \bigr\Vert , \frac{ \Vert x' \Vert }{\varGamma (2- \beta )}, m_{0} \Vert x \Vert \biggr\} \,ds \biggr] \\ \leq& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggl[l \Vert x \Vert _{*} \int _{0}^{\lambda } (1-s)^{\alpha -2} \bigl(P_{1}(s)+ \epsilon _{0}\bigr) \,ds + ( P _{2} + \epsilon ) l \Vert x \Vert _{*} \Vert \hat{\varTheta } \Vert _{[\lambda , \mu ]} \\ &{}+ l \Vert x \Vert _{*} \int _{\mu }^{1} (1-s)^{\alpha -2} \bigl(P_{3}(s)+ \epsilon _{0}\bigr) \,ds \biggr] \\ = &\frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } l \Vert x \Vert _{*} \biggl[ \int _{0} ^{\lambda } (1-s)^{\alpha -2} P_{1}(s) \,ds+ \epsilon _{0} \int _{0}^{ \lambda } (1-s)^{\alpha -2} \,ds \\ &{}+ ( P_{2} + \epsilon _{0}) \Vert \hat{\varTheta } \Vert _{[\lambda , \mu ]} + \int _{\mu }^{1} (1-s)^{\alpha -2} P_{3}(s) \,ds + \epsilon _{0} \int _{ \mu }^{1} (1-s)^{\alpha -2} \,ds\biggr] \\ =& \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } l \Vert x \Vert _{*} \biggl[ \Vert \hat{ P_{1}} \Vert _{[0, \lambda ]} +\frac{\epsilon _{0}}{\alpha -1} \bigl(1 - (1- \lambda )^{\alpha -1}\bigr) \\ &{}+ ( P_{2} + \epsilon _{0}) \Vert \hat{\varTheta } \Vert _{[\lambda , \mu ]} + \Vert \hat{ P_{3}} \Vert _{[\mu , 1]} + \frac{\epsilon _{0}}{\alpha -1} (1 - \mu )^{\alpha -1}\biggr] \\ \leq& \theta _{0} l \biggl[ \Vert \hat{ P_{1}} \Vert _{[0, \lambda ]} +\frac{\epsilon _{0}}{\alpha -1} \bigl(1 - (1-\lambda )^{\alpha -1}\bigr) \\ &{}+ ( P_{2} + \epsilon _{0}) \Vert \hat{\varTheta } \Vert _{[\lambda , \mu ]} + \Vert \hat{ P_{3}} \Vert _{[\mu , 1]} + \frac{\epsilon _{0}}{\alpha -1} (1 - \mu )^{\alpha -1}\biggr] \Vert x \Vert _{*} \\ \leq& \Vert x \Vert _{*} \end{aligned}

and so $$\|Fx \| \leq \|x\|_{*} < r_{0}$$. Also, we can conclude that $$\|F'x \| \leq \|x\|_{*} < r_{0}$$. Hence, $$\|Fx \| < r_{0}$$ and so $$F_{x} \in C$$. For the same reason, $$F_{y} \in C$$. Similar to (7), we conclude that

\begin{aligned} \Vert F_{x}-F_{y} \Vert _{*} \leq & \max \biggl\{ \frac{3 \alpha }{\varGamma ( \alpha ) }, \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \Vert x-y \Vert _{*}^{\mu _{i}} \\ &{} + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) \Vert x-y \Vert _{*}^{m} +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) \Vert x-y \Vert _{*}^{\gamma _{i}}] \\ \leq & \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \Vert x-y \Vert _{*} \\ &{} + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) \Vert x-y \Vert _{*} +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) \Vert x-y \Vert _{*}] \\ =& \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T _{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \\ & {}+ \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) ] \Vert x-y \Vert _{*} \end{aligned}

whenever $$\|x-y\|_{*}< \epsilon _{1}$$. Thus, $$\|x - y\|_{*} \leq \|x\| _{*} + \|y\|_{*} \leq \epsilon _{0}$$ whenever $$x, y \in C$$. Hence,

\begin{aligned} \Vert F_{x}-F_{y} \Vert _{*} \leq & \max \biggl\{ \frac{3 \alpha }{\varGamma ( \alpha ) }, \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \\ & {}+ \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) +\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) ] \Vert x-y \Vert _{*} \\ = & \psi \bigl( \Vert x-y \Vert _{*}\bigr), \end{aligned}

where

\begin{aligned} \psi (t) = & \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T_{0}}{\varGamma (\alpha ) } \biggr\} \Biggl[ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} +\epsilon _{1}) \\ & {}+ \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} (q _{j} + \epsilon _{1} ) \\ & {}+\sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} \Biggr] ( l_{\gamma _{i}} +\epsilon _{1}) ] t. \end{aligned}

Note that $$\psi \in \varPsi$$. Now by using Theorem 6, F has a fixed point and so the pointwise defined problem (1) has a solution. □

### Example 2

Consider the problem

$$D^{\frac{7}{2}} x(t) + f\biggl(t, x(t), x'(t), D^{\frac{1}{2}} x(t), \int _{0}^{t} x(\xi ) \,d\xi \biggr)=0$$

with boundary conditions $$x(0)=x'(\frac{1}{3})$$, $$x(1)=x'(\frac{1}{2})$$ and $$x''(0) =0$$, where

$$f(t, x_{1}, \ldots, x_{4})= \textstyle\begin{cases} f_{1}(t, x_{1}, \ldots, x_{4}):= t^{2} (\sum _{i=1}^{4} x_{i}(s) ), & t \in [0, 0.7), \\ f_{2}(t, x_{1}, \ldots, x_{4}):= \frac{0.1}{p(t)} \sum _{i=1}^{4} \frac{ \vert x_{i}(t) \vert }{1+ \vert x_{i}(t) \vert }, & \frac{t}{3} \in [0.7, 0.7], \\ f_{3}(t, x_{1}, \ldots, x_{4}):= t(\sum _{i=1}^{4} x_{i}(s) ), & t \in [0.9, 1], \end{cases}$$

and

$$p(t)= \textstyle\begin{cases} 0, & t \in [0.2, 0.9] \cap Q, \\ t, & t \in [0.2, 0.9] \cap Q^{c}. \end{cases}$$

Put $$a_{i}(t) = a(t)= t^{2}$$, $$b_{j}(t)= b(t)= \frac{0.1}{p(t)}$$ and $$c_{i}(t)= c(t) = t$$ (for $$1 \leq i \leq 4$$, $$k_{0}=1$$). Then we have $$|f_{1}(t, x_{1}, \ldots, x_{4}) - f_{1}(t, y_{1}, \ldots, y_{4})| \leq t ^{2} \sum _{i=1}^{4} |x_{i} - y_{i}| =a(t) \sum _{i=1}^{4} \phi (|x _{i} - y_{i}|)$$,

$$\bigl\vert f_{3}(t, x_{1}, \ldots, x_{4}) - f_{3}(t, y_{1}, \ldots, y_{4}) \bigr\vert \leq t \sum _{i=1}^{4} \vert x_{i} - y_{i} \vert =c(t) \sum _{i=1}^{4} \varPhi \bigl( \vert x_{i} - y_{i} \vert \bigr),$$

and

$$\bigl\vert f_{2}(t, x_{1}, \ldots, x_{4}) - f_{2}(t, y_{1}, \ldots, y_{4}) \bigr\vert \leq t ^{2} \sum _{i=1}^{4} \vert x_{i} - y_{i} \vert =b(t) \sum _{i=1}^{4} H\bigl( \vert x- y \vert , \ldots, \vert x- y \vert \bigr),$$

where $$\phi (z)=z$$, $$\varPhi (z)= z$$ and $$H(z_{1}, \ldots, z_{4})= z_{1}+ \cdots+ z_{4}$$. Put $$\mu _{i}=\gamma _{i}=m= 1$$. Then we have $$\lim_{z \to 0^{+}} \frac{\phi (z)}{z }=1$$, $$\lim_{z \to 0^{+}} \frac{\varPhi (z)}{z }=1$$ and $$\lim_{z \to 0^{+}} \frac{H_{j}(z, z, z, z)}{z }=1$$. Also, $$\lim_{\max |x_{i}| \to 0} \frac{|f_{1}(t, x_{1}, \ldots, x_{4})|}{\max |x _{i}|} =4 t^{2}=P_{1}(t)$$, $$\lim_{\max |x_{i}| \to 0} \frac{|f_{3}(t, x_{1}, \ldots, x_{4})|}{\max |x_{i}|} = 4t =P_{3}(t)$$ and $$|f_{2}(t, x _{1}, \ldots, x_{4})| \leq \varTheta (t) \varLambda (x_{1}, \ldots, x_{4})$$, where $$\varTheta (t) = p(t)$$ and $$\varLambda (x_{1}, \ldots, x_{4})= \sum _{i=1} ^{5} \frac{|x_{i}|}{1+|x_{i}|}$$. It is easy to see that ϕ, Φ, H and Λ satisfy the conditions of Theorem 6 and $$\lim_{x \to 0^{+}} \frac{\varLambda (x, x, x, x)}{x } =4 :=P_{2}$$. Also, we have

\begin{aligned}& \max \biggl\{ \frac{3 \alpha }{\varGamma (\alpha ) }, \frac{2+ \alpha + T_{0}}{ \varGamma (\alpha ) } \biggr\} \cdot \max \biggl\{ \sum _{i=1}^{5} \Vert \hat{a_{i}} \Vert _{[0, \lambda ]} ( l_{\mu _{i}} ) + \sum _{j=1}^{k_{0}} \Vert \hat{b_{j}} \Vert _{[\lambda , \mu ]} q_{j} \\& \qquad {}+ \sum _{i=1}^{4} \Vert \hat{c_{i}} \Vert _{[\mu , 1]} ] ( l_{\gamma _{i}} ) , \max \biggl\{ 1, \frac{1}{\varGamma (2- \beta ) }, m_{0} \biggr\} \bigl[ \Vert \hat{P_{1}} \Vert _{[0, \lambda ]} + P_{2} \Vert \varTheta \Vert _{[\lambda , \mu ]} + \Vert \hat{P_{3}} \Vert _{[\mu , 1]}\bigr] \biggr\} \\& \quad \leq \frac{\frac{21}{2} }{\frac{15 \sqrt{\pi }}{8} } \max \bigl\{ 0.19 + 0.005+ 0.9, 1.13[0.19 + 0.02+ 0.9] \bigr\} < 1. \end{aligned}

By using Theorem 6, the pointwise defined problem has a solution.

## 3 Conclusion

It is very important that we increase our abilities of natural phenomenon modeling. In this way, it is better we investigate different types of high order fractional integro-differential equations or new type model ones. One of the new models is described by the three step crisis fractional integro-differential equations which have been introduced recently. In this work, we reviewed the existence of solutions for a three step crisis fractional integro-differential equation under some boundary conditions.