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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.

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

  1. (i)

    f is α-admissible;

  2. (ii)

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

  3. (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.

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. (1)

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

  2. (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

  1. (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 \(\theta(c,d)\geq0\);

  2. (ii)

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

  3. (iii)

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

    $$\theta \biggl( \int_{0}^{1}G(\xi,\zeta)h\bigl(\zeta,y(\zeta)\bigr)d \zeta, \int _{0}^{1}G(\xi,\zeta)h\bigl(\zeta,z(\xi)\bigr)d \zeta \biggr)\geq 0; $$
  4. (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.

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Afshari, H., Kalantari, S. & Baleanu, D. Solution of fractional differential equations via \(\alpha-\psi\)-Geraghty type mappings. Adv Differ Equ 2018, 347 (2018). https://doi.org/10.1186/s13662-018-1807-4

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Keywords

  • Fractional differential equation
  • Normal cone
  • \(\alpha-\psi\)-Geraghty contractive type mapping