Solution of fractional differential equations via $\alpha-\psi$-Geraghty type mappings

Abstract

Using fixed point results of $\alpha-\psi$-Geraghty contractive type mappings, we examine the existence of solutions for some fractional differential equations in b-metric spaces. By some concrete examples we illustrate the obtained results.

1 Introduction

In 2012, Samet et al. [11] presented the concepet of α-admissible mappings, which was expanded by several authors (see [5, 6, 9]). Baleanu, Rezapour, and Mohammadi [3] studied the existence of a solution for problem $D^{\nu}w(\xi)=h(\xi, w(\xi))$ $(\xi\in[0,1],1<\nu\leq2)$. Afshari, Aydi, and Karapinar [1, 2] considered generalized $\alpha-\psi$-Geraghty contractive mappings in b-metric spaces.

We investigate the existence of solutions for some fractional differential equations in b-metric spaces. We denote $I=[0,1]$.

Definition 1.1

([7, 10])

The Caputo derivative of order ν of a continuous function $h:[0,\infty)\rightarrow\mathbb{R}$ is defined by

$$^{c}D^{\nu}h(\xi)=\frac{1}{\Gamma(n-\nu)} \int_{0}^{\xi}(\xi-\zeta )^{n-\nu-1}h^{(n)}( \zeta)\,d\zeta, $$

where $n-1<\nu<n$, $n=[\nu]+1$, $[\nu]$ is the integer part of ν, and

$$\begin{aligned} \Gamma(z)= \int_{0}^{\infty}x^{z-1}e^{-x}\,dx. \end{aligned}$$

(1)

Definition 1.2

([7, 10])

The Riemann–Liouville derivative of a continuous function h is defined by

$$D^{\nu}h(\xi)=\frac{1}{\Gamma(n-\nu)}\biggl(\frac{d}{d\xi} \biggr)^{n} \int _{0}^{\xi}\frac{h(\zeta)}{ (\xi-\zeta)^{\nu-n-1}}\,d\zeta\quad \bigl(n=[ \nu]+1\bigr), $$

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

Let Ψ be the set of all increasing continuous functions $\psi: [0,\infty) \to\mathbb{[}0,\infty)$ such that $\psi(\lambda x)\leq \lambda\psi(x)\leq\lambda x$ for $\lambda>1$, and let $\mathcal{B}$ be the family of nondecreasing functions $\gamma: [0,\infty) \to[0,\frac{1}{s^{2}})$ for some $s\geq1$.

Definition 1.3

([1])

Let $(X,d)$ be a b-metric space (with constant s). A function $g:X\rightarrow X$ is a generalized $\alpha-\psi$-Geraghty contraction if there exists $\alpha:X\times X\to [0,\infty)$ such that

$$\begin{aligned} \alpha(z,t)\psi\bigl(s^{3} d(gz,gt)\bigr)\leq\gamma\bigl( \psi\bigl(d(z,t)\bigr)\bigr)\psi\bigl(d(z,t)\bigr) \end{aligned}$$

(2)

for all $z,t \in X$, where $\gamma\in\mathcal{B}$ and $\psi\in\Psi$.

Definition 1.4

([11])

Let $g: X\rightarrow X$ and $\alpha: X\times X\rightarrow[0,\infty)$ be given. Then g is called α-admissible if for $z,t\in X$,

$$ \alpha(z,t)\geq1 \quad \Longrightarrow\quad\alpha(gz,gt)\geq1. $$

(3)

Theorem 1.5

([1])

Let $(X,d)$ be a complete b-metric space, and let $f:X\rightarrow X$ be a generalized $\alpha-\psi$-Geraghty contraction such that

• (i) f is α-admissible;

• (ii) there exists $u_{0}\in X$ such that $\alpha(u_{0},fu_{0})\geq1$;

• (iii) if $\{u_{n}\}\subseteq X$, $u_{n}\rightarrow u$ in X, and $\alpha(u_{n},u_{n+1})\geq1$, then $\alpha(u_{n},u)\geq 1$.

Then f has a fixed point.

2 Main result

By $X=C(I)$ we denote the set of continuous functions. Let $d:X\times X\to[0,\infty)$ be given by

$$\begin{aligned} d(y,z)=\bigl\| (y-z)^{2} \bigr\| _{\infty}=\sup _{\xi\in I}\bigl(y(\xi)-z(\xi)\bigr)^{2}. \end{aligned}$$

(4)

Evidently, $(X,d)$ is a complete b-metric space with $s=2$ but is not a metric space.

Now we study the problem

$$ \frac{D^{\nu}}{D\xi}w(\xi)=h\bigl(\xi,w(\xi)\bigr),\quad \xi\in I, 3< \nu\leq4, $$

(5)

under the conditions

$$ w(0)=w^{\prime}(0)=w(1)=w^{\prime}(1)=0, $$

(6)

where $D^{\nu}$ is the Riemann–Liouville derivative, and $h:I\times X\rightarrow\mathbb{R}$ is continuous.

Lemma 2.1

([13])

Given $h\in C(I\times X,\mathbb{R})$ and $3 <\nu\leq4$, the unique solution of

$$ \frac{D^{\nu}}{D\xi}w(\xi)=h\bigl(\xi,w(\xi)\bigr),\quad \xi \in I,3< \nu\leq4, $$

(7)

where

$$ w(0)=w^{\prime}(0)=w(1)=w^{\prime }(1)=0, $$

(8)

is given by $w(\xi)=\int_{0}^{1} G(\xi,\zeta)h(s,w(s))\,ds$, where

$$\begin{aligned} G(\xi,\zeta)= \textstyle\begin{cases} \frac{(\xi-1)^{\nu-1}+(1-\zeta)^{\nu-2}\xi^{\nu-2}[(\zeta-\xi )+(\nu-2)(1-\xi)\zeta]}{\Gamma(\nu)},&0\leq\zeta\leq\xi\leq1,\\ \frac{(1-\zeta)^{\nu-2}\xi^{\nu-2}[(\zeta-\xi)+(\nu-2)(1-\xi )\zeta]}{\Gamma(\nu)},&0\leq\xi\leq\zeta\leq1. \end{cases}\displaystyle \end{aligned}$$

(9)

If $h(\xi,w(\xi))=1$, then the unique solution of (7)–(8) is given by

$$f(\xi)= \int_{0}^{1}G(\xi,\zeta)\,ds=\frac{1}{\Gamma(\nu+1)} \xi^{\nu -2}(1-\xi)^{2}. $$

Lemma 2.2

([13])

In Lemma 2.1, $G(\xi,\zeta)$ given in (9) satisfies the following conditions:

• (1) $G(\xi, \zeta) > 0$, and $G(\xi,\zeta)$ is continuous for $\xi,\zeta\in I$;

• (2) $\frac{(\nu-2)\sigma(\xi)\rho(\zeta)}{\Gamma(\nu)}\leq G(\xi,\zeta)\leq\frac{r_{0}\rho(\zeta)}{\Gamma(\nu)}$,

where

$$r_{0}=\max\bigl\{ \nu-1,(\nu-2)^{2}\bigr\} ,\qquad\sigma(\xi)= \xi^{\nu-2}(1-\xi )^{2},\quad\textit{and}\quad\rho(\zeta)= \zeta^{2}(1-\zeta)^{\nu-2}. $$

Theorem 2.3

Suppose

• (i) there exist $\theta:\mathbb{R}^{2}\rightarrow\mathbb{R}$ and $\psi\in\Psi$ such that

  $$\bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert \leq \frac{1}{2\sqrt{2}} \frac{\Gamma(\nu+1)}{4\nu} \frac{\psi( \vert c-d \vert ^{2})}{\sqrt{4\| (c-d)^{2}\|_{\infty}+1}} $$

  for $\xi\in I$ and $c,d\in\mathbb{R}$ with $\theta(c,d)\geq0$;

• (ii) there exists $y_{0}\in C(I)$ such that $\theta(y_{0}(\xi),\int _{0}^{1}G(\xi,\zeta)h(\zeta,y_{0}(\xi))\,d\zeta)\geq0$, $\xi\in I$;

• (iii) for $\xi\in I$ and $y,z\in C(I)$, $\theta(y(\xi),z(\xi))\geq0$ implies

  $$\theta \biggl( \int_{0}^{1}G(\xi,\zeta)h\bigl(\zeta,y(\zeta)\bigr)d \zeta, \int _{0}^{1}G(\xi,\zeta)h\bigl(\zeta,z(\xi)\bigr)d \zeta \biggr)\geq 0; $$

• (iv) if $\{y_{n}\}\subseteq C(I)$, $y_{n}\rightarrow y$ in $C(I)$, and $\theta(y_{n},y_{n+1})\geq0$, then $\theta(y_{n},y)\geq0$.

Then problem (7) has at least one solution.

Proof

By Lemma 2.1 $y\in C(I)$ is a solution of (7) if and only if it is a solution of $y(\xi)=\int_{0}^{1} G(\xi,\zeta)h(\zeta,y(\zeta))\,d\zeta$, and we define $A : C(I)\rightarrow C(I)$ by $Ay(\xi)=\int_{0}^{1} G(\xi,\zeta)h(\zeta,y(\zeta))\,d\zeta$ for $\xi\in I$. For this purpose, we find a fixed point of A. Let $y,z\in C(I)$ be such that $\theta(y(\xi),z(\xi))\geq0$ for $\xi\in I$. Using (i), we get

$$\begin{aligned} \bigl\vert Ay(\xi)-Az(\xi) \bigr\vert ^{2}&= \biggl\vert \int_{0}^{1} G(\xi,\zeta) \bigl(h\bigl(\zeta,y(\zeta )\bigr)-h\bigl(\zeta,z(\zeta)\bigr)\bigr)\,d\zeta \biggr\vert ^{2} \\ &\leq \biggl[ \int_{0}^{1} G(\xi,\zeta) \bigl\vert h\bigl( \zeta,y(\zeta)\bigr)-h\bigl(\zeta ,z(\zeta)\bigr) \bigr\vert \,d\zeta \biggr]^{2} \\ & \leq \biggl[ \int_{0}^{1} G(\xi,\zeta)\frac{1}{2\sqrt{2}} \frac {\Gamma(\nu+1)}{4\nu} \frac{\psi( \vert y(\zeta)-z(\zeta) \vert ^{2})}{\sqrt{4\| (y-z)^{2}\|_{\infty}+1}}\,d\zeta \biggr]^{2} \\ &\leq\frac{1}{8}\frac{(\psi(\|(y-z)^{2}\|_{\infty }))^{2}}{4\| (y-z)^{2}\|_{\infty}+1}. \end{aligned}$$

Hence, for $y,z\in C(I)$ and $\xi\in I$ with $\theta(y(\xi),z(\xi ))\geq 0$, we have

$$\begin{aligned} \|(Ay-Az)^{2}\|_{\infty}\leq\frac{1}{8} \frac {(\psi(\|(y-z)^{2}\|_{\infty}))^{2}}{4\| (y-z)^{2}\|_{\infty}+1}. \end{aligned}$$

Let $\alpha: C(I)\times C(I)\rightarrow[0,\infty)$ be defined by

$$\begin{aligned} \alpha(y,z)= \textstyle\begin{cases} 1,&\theta(y(\xi),z(\xi))\geq0, \xi\in I,\\ 0&\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$

Define $\gamma: [0,\infty)\rightarrow[0,\frac{1}{4})$ by $\gamma (q)=\frac{q}{4q+1}$ and $s=2$.

So

$$\begin{aligned} \alpha(y,z)\psi\bigl(8d(Ay,Az)\bigr)&\leq 8\alpha(y,z)\psi\bigl(d(Ay,Az)\bigr) \leq\frac{(\psi (d(y,z)))^{2}}{4d(y,z)+1} \\ &\leq\frac{(\psi(d(y,z)))^{2}}{4\psi(d(y,z))+1} \\ &= \frac{1}{\gamma(\psi(d(y,z)))}\gamma\bigl(\psi\bigl(d(y,z)\bigr)\bigr)\frac{(\psi (d(y,z)))^{2}}{4\psi(d(y,z))+1} \\ & \leq\gamma\bigl(\psi\bigl(d(y,z)\bigr)\bigr)\psi\bigl(d(y,z)\bigr),\quad\gamma\in\mathcal{B}. \end{aligned}$$

Then A is an $\alpha-\psi-$contractive mapping. From (iii) and the definition of α we have

$$\begin{aligned} \alpha(y,z)\geq1&\quad\Rightarrow\quad\theta\bigl(y(\xi),z(\xi)\bigr)\geq 0\\ &\quad \Rightarrow\quad \theta\bigl(A(y),A(z)\bigr)\geq0\\ &\quad \Rightarrow\quad\alpha\bigl(A(y),A(z)\bigr)\geq 1, \end{aligned}$$

for $y,z\in C(I)$. Thus, A is α-admissible. By (ii) there exists $y_{0}$ $\in C(I)$ such that $\alpha(y_{0},Ay_{0})\geq1$. By (iv) and Theorem 1.5 there is $y^{*}\in C(I)$ such that $y^{*}=Ay^{*}$. Hence $y^{*}$ is a solution of the problem. □

Corollary 2.4

Suppose that there exist $\theta:\mathbb{R}^{2}\rightarrow\mathbb{R}$ and $\psi\in \Psi$ such that

$$\begin{aligned} \bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert \leq\frac{10^{3}}{4\sqrt{8}} \frac{\psi( \vert c-d \vert ^{2})}{\sqrt{4\| (c-d)^{2}\|_{\infty}+1}} \end{aligned}$$

(10)

for $\xi\in I$ and $c,d\in\mathbb{R}$ with $\theta(c,d)\geq0$. Also, suppose that conditions (ii)–(iv) from Theorem 2.3 hold for h, where $G(\xi,\zeta)$ is given in (9). Then the problem

$$ \frac{D^{\frac{7}{2}}}{D\xi}w(\xi)=h\bigl(\xi,w(\xi)\bigr),\quad \xi\in I, $$

(11)

where

$$ w(0)=w^{\prime}(0)=w(1)=w^{\prime }(1)=0, $$

has at least one solution.

Proof

By Lemma 2.2

$$\begin{aligned} \min \int_{0}^{1}G(\xi,\zeta)\,d\zeta=10^{-5} \quad\mbox{and}\quad \max \int_{0}^{1}G(\xi,\zeta)\,d\zeta=4 \times10^{-3}. \end{aligned}$$

(12)

Using (10) and (12), by Theorem 2.3 we obtain

$$\begin{aligned} & \bigl\vert Ay(\xi)-Az(\xi) \bigr\vert ^{2}\leq\frac{1}{8} \frac{(\psi( \vert y-z \vert ^{2}))^{2}}{{4\|(y-z)^{2}\|_{\infty}+1}}. \end{aligned}$$

The rest of the proof is according to Theorem 2.3. □

Lemma 2.5

([8])

If $h\in C(I\times X,\mathbb{R})$ and $h(\xi,w(\xi))\leq0$, then the problem

$$ \begin{aligned} &{-}D_{0+}^{\nu}w(\xi)=h\bigl(\xi,w(\xi)\bigr), \quad (0< \xi< 1, 3< \nu\leq4), \\ &w(0)=w^{\prime}(0)=w^{\prime\prime}(0)=w^{\prime\prime}(1)=0 \end{aligned} $$

(13)

has a unique positive solution

$$w(\xi)= \int_{0}^{1} G(\xi,\zeta)h\bigl(\zeta,w(\zeta) \bigr)\,d\zeta, $$

where $G(\xi,\zeta)$ is given by

$$\begin{aligned} G(\xi,\zeta)=\frac{1}{\Gamma(\nu)} \textstyle\begin{cases} \xi^{\nu-1}(1-\zeta)^{\nu-3}-(\xi-\zeta)^{\nu-1},&0\leq\zeta \leq\xi\leq1,\\ \xi^{\nu-1}(1-\zeta)^{\nu-3},&0\leq\xi\leq\zeta\leq1. \end{cases}\displaystyle \end{aligned}$$

(14)

Lemma 2.6

([12])

The function $G(\xi,\zeta)$ in Lemma 2.5 has the following property:

$$\frac{1}{\Gamma(\nu)}\zeta(2-\zeta) (1-\zeta)^{\nu-3}\xi^{\nu -1}\leq G(\xi,\zeta)\leq\frac{1}{\Gamma(\nu)}(1-\zeta)^{\nu-3}\xi^{\nu-1}, $$

where $\xi,\zeta\in I$ and $3<\nu\leq4$.

Based on Theorem 2.3, we get the following result.

Corollary 2.7

Assume that there exist $\theta:\mathbb {R}^{2}\rightarrow\mathbb{R}$ and $\psi\in\Psi$ such that

$$\bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert \leq \frac{1}{2\sqrt{2}M} \frac{\psi( \vert c-d \vert ^{2})}{\sqrt{4\| (c-d)^{2}\|_{\infty}+1}}, $$

where $M=\sup_{\xi\in I} \int_{0}^{1}G(\xi,\zeta)\,d\zeta$. Also, suppose that conditions (ii)–(iv) from Theorem 2.3 are satisfied, where $G(\xi,\zeta)$ is given in (14). Then problem (13) has at least one solution.

Proof

By Lemma 2.5 $y\in C(I)$ is a solution of (13) if and only if a solution of $y(\xi)=\int_{0}^{1} G(\xi,\zeta)h(\zeta,y(\zeta))\,d\zeta$. Define $A : C(I)\rightarrow C(I)$ by $Ay(\xi)=\int_{0}^{1} G(\xi,\zeta)h(\zeta,y(\zeta))\,d\zeta$ for $\xi\in I$. We find a fixed point of A. Let $y,z\in C(I)$ be such that $\theta(y(\xi),z(\xi))\geq0$ for $\xi\in I$. By (i) and Lemma 2.6 we get

$$\begin{aligned} &\bigl\vert Ay(\xi)-Az(\xi) \bigr\vert ^{2}\\ &\quad = \biggl\vert \int_{0}^{1} G(\xi,\zeta) \bigl(h\bigl(\zeta,y(\zeta )\bigr)-h\bigl(\zeta,z(\zeta)\bigr)\bigr)\,d\zeta \biggr\vert ^{2} \\ &\quad \leq \biggl[ \int_{0}^{1} G(\xi,\zeta) \bigl\vert h\bigl( \zeta,y(\zeta)\bigr)-h\bigl(\zeta ,z(\zeta)\bigr) \bigr\vert \,d\zeta \biggr]^{2} \\ &\quad \leq \biggl[ \int_{0}^{1} G(\xi,\zeta)\frac{1}{2\sqrt{2}M} \frac{\psi( \vert y(\zeta)-z(\zeta) \vert ^{2})}{\sqrt{4\| (y-z)^{2}\|_{\infty}+1}}\,d\zeta \biggr]^{2} \\ &\quad \leq \biggl[ \int_{0}^{1} G(\xi,\zeta)\frac{1}{2\sqrt{2}(\sup_{\xi \in I} \int_{0}^{1}G(\xi,\zeta)\,d\zeta)} \frac{\psi( \vert y(\zeta)-z(\zeta) \vert ^{2})}{\sqrt{4\| (y-z)^{2}\|_{\infty}+1}}\,d\zeta \biggr]^{2} \\ &\quad \leq \biggl[ \int_{0}^{1} G(\xi,\zeta)\frac{1}{2\sqrt{2}(\int _{0}^{1}G(\xi,\zeta)\,d\zeta)} \frac{\psi( \vert y(\zeta)-z(\zeta) \vert ^{2})}{\sqrt{4\| (y-z)^{2}\|_{\infty}+1}}\,d\zeta \biggr]^{2} \\ &\quad \leq \biggl[ \int_{0}^{1} \frac{1}{\Gamma(\nu)}(1- \zeta)^{\nu-3}\xi ^{\nu-1}\frac{\Gamma(\nu)}{2\sqrt{2}(\int_{0}^{1}\zeta(2-\zeta )(1-\zeta)^{\nu-3}\xi^{\nu-1}\,d\zeta)} \\ &\qquad {}\times\frac{\psi( \vert y(\zeta)-z(\zeta) \vert ^{2})}{\sqrt{4\| (y-z)^{2}\|_{\infty}+1}}\,d \zeta \biggr]^{2} \\ &\quad \leq\frac{1}{8}\frac{(\psi(\|(y-z)^{2}\|_{\infty }))^{2}}{4\| (y-z)^{2}\|_{\infty}+1}. \end{aligned}$$

Suppose that conditions (ii)–(iv) from Theorem 2.3 are satisfied, where $G(\xi,\zeta)$ is given in (14). By Theorem 2.3 problem (13) has at least one solution.

Let $(X,d)$ be given in (4). For the equation

$$\begin{aligned} {}^{c}D^{\nu}y(\xi)=h\bigl(\xi,y(\xi)\bigr), \quad (\xi\in I,1< \nu\leq2), \end{aligned}$$

(15)

via

$$y(0)=0,\qquad y(1)= \int_{0}^{\eta}y(\zeta)\,d\zeta\quad(0< \eta< 1), $$

where $h:I\times X\rightarrow\mathbb{R}$ is continuous, we have the following result. □

Theorem 2.8

Assume that there exist $\theta:\mathbb{R}^{2}\rightarrow\mathbb{R}$, $\gamma\in \mathcal{B}$, and $\psi\in\Psi$ such that

$$\bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert \leq \frac{\Gamma(\nu+1)}{5}\sqrt{ \frac{1}{8}\gamma\bigl(\psi \bigl( \vert c-d \vert ^{2}\bigr) \bigr)\psi\bigl( \vert c-d \vert ^{2}\bigr)}. $$

Suppose conditions (ii)–(iv) from Theorem 2.3 hold, where $A:C(I)\rightarrow C(I)$ is defined by

$$\begin{aligned} Ay(\xi):={}&\frac{1}{\Gamma(\nu)} \int_{0}^{1}(\xi-\zeta)^{\nu -1}h\bigl( \zeta,y(\zeta)\bigr) \,d\zeta-\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1}(1-\zeta)^{\nu -1}h\bigl(\zeta,y( \zeta)\bigr)\,d\zeta \\ &{}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta}\biggl( \int_{0}^{\zeta}(\zeta-n)^{\nu-1}h\bigl(n,y(n) \bigr)dn\biggr)\,d\zeta\quad (\xi\in I); \end{aligned}$$

Then (15) has at least one solution.

Proof

A function $y\in C(I)$ is a solution of (15) if and only if it is a solution of

$$\begin{aligned} y(\xi)={}&\frac{1}{\Gamma({\nu})} \int_{0}^{1}(\xi-\zeta)^{\nu -1}h\bigl( \zeta,y(\zeta)\bigr) \,d\zeta-\frac{2\xi}{(2-{\eta}^{2})\Gamma(\nu)} \int_{0}^{1}(1-\zeta )^{\nu-1}h\bigl(\zeta,y( \zeta)\bigr)\,d\zeta \\ &{}+\frac{2\xi}{(2-{\eta}^{2})\Gamma(\nu)} \int_{0}^{{\eta}}\biggl( \int _{0}^{\zeta}(\zeta-n)^{\nu-1}h\bigl(n,y(n) \bigr)dn\biggr)\,d\zeta\quad (\xi\in I). \end{aligned}$$

Then (15) is equivalent to finding $y^{*}\in C(I)$ that is a fixed point of A. Let $y,z\in C(I)$ with $\theta(y(\xi),z(\xi))\geq0$, $\xi\in I$. By (i) we have

$$\begin{aligned} &\bigl\vert Ay(\xi)-Az(\xi) \bigr\vert ^{2}\\ &\quad = \biggl\vert \frac{1}{\Gamma(\nu)} \int_{0}^{1}(\xi-\zeta )^{\nu-1}h\bigl( \zeta,y(\zeta)\bigr)\,d\zeta \\ &\qquad {}-\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1}(1-\zeta)^{\nu -1}h\bigl(\zeta,y( \zeta)\bigr)\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta}\biggl( \int_{0}^{\zeta}(\zeta-n)^{\nu-1}h\bigl(n,y(n) \bigr)dn\biggr)\,d\zeta \\ &\qquad {}-\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(\xi-\zeta)^{\nu-1}h\bigl(\zeta ,z(\zeta)\bigr)\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1}(1-\zeta)^{\nu -1}h\bigl(\zeta,z( \zeta)\bigr)\,d\zeta \\ &\qquad {}-\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta}\biggl( \int_{0}^{\zeta}(\zeta-n)^{\nu-1}h\bigl(n,z(n) \bigr)dn\biggr)\,d\zeta \biggr\vert ^{2} \\ &\quad \leq \biggl\vert \frac{1}{\Gamma(\nu)} \int_{0}^{1}|\xi-\zeta \biggr\vert ^{\nu-1}\bigl|h\bigl(\zeta ,y(\zeta)\bigr)-h\bigl(\zeta,z(\zeta)\bigr)\bigr|\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1} \vert 1-\zeta \vert ^{\nu -1} \bigl\vert h\bigl(\zeta,y(\zeta)\bigr)-h\bigl(\zeta,z(\zeta)\bigr) \bigr\vert \,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta}\biggl| \int_{0}^{\zeta} \vert \zeta-n \vert ^{\nu-1} \bigl\vert h\bigl(n,y(n)\bigr)-h\bigl(n,z(n)\bigr) \bigr\vert dn\vert \,d\zeta \biggr\vert ^{2} \\ &\quad \leq \biggl\vert \frac{1}{\Gamma(\nu)} \int_{0}^{1} \vert \xi-\zeta \vert ^{\nu-1}\frac{\Gamma (\nu+1)}{5}\sqrt{\frac{1}{8} \gamma\bigl(\psi\bigl( \bigl\vert y(\zeta)-z(\zeta) \bigr\vert ^{2}\bigr)\bigr)\psi\bigl( \bigl\vert y(\zeta)-z(\zeta ) \bigr\vert ^{2}\bigr)}\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1} \vert 1-\zeta \vert ^{\nu -1} \frac{\Gamma(\nu+1)}{5} \\ &\qquad {}\times\sqrt{\frac{1}{8}\gamma\bigl(\psi\bigl( \bigl\vert y( \zeta)-z(\zeta) \bigr\vert ^{2}\bigr)\bigr)\psi \bigl( \bigl\vert y( \zeta)-z(\zeta) \bigr\vert ^{2}\bigr)}\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta} \int_{0}^{\zeta} \vert \zeta-n \vert ^{\nu-1} \\ &\qquad {}\times\sqrt{\frac{1}{8}\gamma\bigl(\psi\bigl( \bigl\vert y(n)-z(n) \bigr\vert ^{2}\bigr)\bigr)\psi \bigl( \bigl\vert y(n)-z(n) \bigr\vert ^{2}\bigr)}\,d\zeta \biggr\vert ^{2} \\ &\quad \leq\biggl(\frac{\Gamma(\nu+1)}{5}\biggr)^{2}\frac{1}{8}\gamma \bigl(\psi\bigl(\| y-z\|_{\infty}^{2}\bigr)\bigr) \psi\bigl(\| y-z\|_{\infty}^{2}\bigr)\biggl[\sup \biggl( \int_{0}^{1} \vert \xi-\zeta \vert ^{\nu-1}\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{1} \vert 1-\zeta \vert ^{\nu -1}\,d\zeta \\ &\qquad {}+\frac{2\xi}{(2-\eta^{2})\Gamma(\nu)} \int_{0}^{\eta}\biggl( \int_{0}^{\zeta} \vert \zeta-n \vert ^{\nu-1}dn\biggr)\,d\zeta\biggr)\biggr]^{2}\\ &\quad \leq \frac{1}{8} \gamma\bigl(\psi\bigl(\| y-z\|_{\infty}^{2}\bigr) \bigr)\psi\bigl(\| y-z\|_{\infty}^{2}\bigr) \end{aligned}$$

for all $y,z\in C(I)$ with $\theta(y(\xi),z(\xi))\geq 0$, $\xi\in I$, so that

$$\begin{aligned} \bigl\| (Ay-Az)^{2}\bigr\| _{\infty}\leq\frac{1}{8} \gamma\bigl(\psi \bigl(\| y-z\|_{\infty}^{2}\bigr) \bigr)\psi \bigl(\| y-z\|_{\infty}^{2}\bigr). \end{aligned}$$

Let $\alpha: C(I)\times C(I)\rightarrow[0,\infty)$ be defined by

$$\begin{aligned} \alpha(y,z)= \textstyle\begin{cases} 1&\theta(y(\xi),z(\xi))\geq0, \xi\in I,\\ 0&\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$

Then

$$\begin{aligned} \alpha(y,z)\psi\bigl(8d(Ay,Az)\bigr)&\leq8\alpha(y,z)\psi\bigl(d(Ay,Az)\bigr) \\ &\leq\alpha(y,z)\psi\bigl(\gamma\bigl(\psi\bigl(d(y,z)\bigr)\bigr)\psi \bigl(d(y,z)\bigr)\bigr)\\ &\leq \gamma\bigl(\psi\bigl(d(y,z)\bigr)\bigr)\psi \bigl(d(y,z)\bigr) \end{aligned}$$

for all $y,z\in C(I)$, and thus A is an $\alpha-\psi-$contractive mapping. From Theorem 1.5, based on the proof of Theorem 2.3, we can deduce the proof of Theorem 2.8. □

Here we find a positive solution for

$$ \frac{{}^{c}D^{\nu}}{D\xi}w(\xi)=h\bigl(\xi,w(\xi)\bigr),\quad 0< \nu\leq1, \xi\in I, $$

(16)

where

$$w(0)+ \int_{0}^{1}w(\zeta)\,d\zeta=w(1). $$

Note that $^{c}D^{\nu}$ is the Caputo derivative of order ν. We consider the Banach space of continuous functions on I endowed with the sup norm. We have the following lemma.

Lemma 2.9

([4])

Let $0<\nu\leq1$ and $h\in C([0,T]\times X,\mathbb{R})$ be given. Then the equation

$$^{c}D^{\nu}w(\xi)=h\bigl(\xi,w(\xi)\bigr)\quad \bigl(\xi \in[0,T], T\geq1\bigr) $$

with

$$w(0)+ \int_{0}^{T}w(\zeta)\,d\zeta=w(T) $$

has a unique solution given by

$$w(\xi)= \int_{0}^{T}G(\xi,\zeta)h\bigl(\zeta,w(\zeta ) \bigr)\,d\zeta, $$

where $G(\xi,\zeta)$ is defined by

$$\begin{aligned} G(\xi,\zeta)= \textstyle\begin{cases} \frac{-(T-\zeta)^{\nu}+\nu T(\xi-\zeta)^{\nu-1}}{T\Gamma(\nu+1)}+\frac{(T-\zeta)^{\nu-1}}{T\Gamma(\nu)},&0\leq \zeta< \xi,\\ \frac{-(T-\zeta)^{\nu}}{T\Gamma(\nu+1)}+\frac{(T-\zeta)^{\nu-1}}{T\Gamma(\nu)},&\xi \leq \zeta< T. \end{cases}\displaystyle \end{aligned}$$

(17)

By Lemma 2.9 and Theorem 2.4 we get the following conclusion.

Corollary 2.10

Assume that there exist $\theta:\mathbb{R}^{2}\rightarrow\mathbb{R}$ and $\psi\in \Psi$ such that

$$\bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert \leq\frac{51}{80\sqrt{8}} \frac{\psi( \vert c-d \vert ^{2})}{\sqrt{4\| (c-d)^{2}\|_{\infty}+1}} $$

for $\xi\in I$ and $c,d\in\mathbb{R}$ with $\theta(c,d)\geq0$. Suppose conditions (ii)–(iv) from Theorem 2.3 are satisfied, where $G(\xi,\zeta)$ is given in (17). Then the following problem has at least one solution:

$${}^{c}D^{\frac{1}{2}}w(\xi)=h\bigl(\xi,w(\xi)\bigr),\quad \bigl(\xi \in[0,1]\bigr),\qquad w(0)+ \int _{0}^{1}w(\zeta)\,d\zeta=w(1). $$

Proof

It is easily that $\min_{t\in[0,1]}\int_{0}^{1} G(t,s)\,ds=\frac{1}{3}$ and $\max_{t\in[0,1]}\int_{0}^{1} G(t,s)\,ds=\frac{80}{51}$. By Theorem 2.3 we conclude the desired result. □

Example 2.11

Let $\psi(r)=r$, $\theta(x,z)=xz$, and $y_{n}(\xi)=\frac{\xi}{n^{2}+1}$. We consider $h:I\times[-2,2]\to[-2,2]$ and the periodic boundary value problem

$$ \frac{D^{\frac{7}{2}}}{D\xi}w(\xi)=h\bigl(\xi,w(\xi)\bigr)=w(\xi),\quad \xi \in I, $$

(18)

with

$$ w(0)=w^{\prime}(0)=w(1)=w^{\prime}(1)=0. $$

Then

$$\begin{aligned} \bigl\vert h(\xi,c)-h(\xi,d) \bigr\vert = \vert c-d \vert \leq \frac{10^{3}}{4\sqrt{8}}\frac{\psi( \vert c-d \vert ^{2})}{\sqrt{4\| (c-d)^{2}\|_{\infty}+1}} \end{aligned}$$

for $\xi\in I$ and $c,d\in[-2,2]$ with $\theta(c,d)\geq0$. Because $y_{0}(\xi)=\xi$, thus

$$\theta\biggl(y_{0}(\xi), \int_{0}^{1}G(\xi,\zeta)h\bigl(\zeta,y_{0}( \zeta)\bigr)\,d\zeta \biggr)\geq0 $$

for all $\xi\in I$. Also, $\theta(y(\xi),z(\xi))=y(\xi)z(\xi)\geq 0$ implies that

$$\theta\biggl( \int_{0}^{1}G(\xi,\zeta)h\bigl(\zeta,y_{(}\zeta) \bigr)\,d\zeta, \int _{0}^{1}G(\xi,\zeta)h\bigl(\zeta,z(\zeta) \bigr)\,d\zeta\biggr)\geq 0. $$

It is obvious that condition (iv) in Corollary (2.4) holds. Hence by Corollary 2.4 problem (18) has at least one solution.