Theory and Modern Applications

# The existence of solutions for a nonlinear mixed problem of singular fractional differential equations

## Abstract

By using fixed point results on cones, we study the existence of solutions for the singular nonlinear fractional boundary value problem

$\begin{array}{c}{}^{c}D^{\alpha }u\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right){,}^{c}{D}^{\beta }u\left(t\right)\right),\hfill \\ u\left(0\right)=au\left(1\right),\phantom{\rule{2em}{0ex}}{u}^{\prime }\left(0\right)=b{\phantom{\rule{0.2em}{0ex}}}^{c}{D}^{\beta }u\left(1\right),\phantom{\rule{2em}{0ex}}{u}^{″}\left(0\right)={u}^{‴}\left(0\right)={u}^{\left(n-1\right)}\left(0\right)=0,\hfill \end{array}$

where $n\ge 3$ is an integer, $\alpha \in \left(n-1,n\right)$, $0<\beta <1$, $0, $0, f is an ${L}^{q}$-Caratheodory function, $q>\frac{1}{\alpha -1}$ and $f\left(t,x,y,z\right)$ may be singular at value 0 in one dimension of its space variables x, y, z. Here, cD stands for the Caputo fractional derivative.

## 1 Introduction

Fractional differential equations (see, for example, [16] and references therein) started to play an important role in several branches of science and engineering. There are some works about existence of solutions for the nonlinear mixed problems of singular fractional boundary value problem (see, for example, [711] and [12]). Also, there are different methods for solving distinct fractional differential equations (see, for example, [1318] and [19]). By using fixed point results on cones, we focus on the existence of positive solutions for a nonlinear mixed problem of singular fractional boundary value problem. For the convenience of the reader, we present some necessary definitions from fractional calculus theory (see, for example, [20]). The Caputo derivative of fractional order α for a function $f:\left[0,\mathrm{\infty }\right)\to \mathbb{R}$ is defined by

${}^{c}D^{\alpha }f\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(n-\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{n-\alpha -1}{f}^{\left(n\right)}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{1em}{0ex}}\left(n-1<\alpha

Let $q\ge 1$. As you know, ${L}^{q}\left[0,1\right]$ denotes the space of functions, whose q th powers of modulus are integrable on $\left[0,1\right]$, equipped with the norm ${\parallel x\parallel }_{q}={\left({\int }_{0}^{1}{|x\left(t\right)|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}$. We consider the sup norm

$\parallel x\parallel =sup\left\{|x\left(t\right)|:t\in \left[0,1\right]\right\}$

on the space $C\left[0,1\right]$. Also, $\mathit{AC}\left[0,1\right]$ is the set of absolutely continuous functions on $\left[0,1\right]$. Let B be a subset of ${\mathbb{R}}^{3}$. A function $f:\left[0,1\right]×B\to \mathbb{R}$ is called an ${L}^{q}$-Caratheodory function whenever the real-valued function $f\left(\cdot ,x,y,z\right)$ on $\left[0,1\right]$ is measurable for all $\left(x,y,z\right)\in B$, the function $f\left(t,\cdot ,\cdot ,\cdot \right):B\to \mathbb{R}$ is continuous for almost all $t\in \left(0,1\right]$, and for each compact set $U\subset B$, there exists a function ${\phi }_{u}\in {L}^{q}\left[0,1\right]$ such that $|f\left(t,x,y,z\right)|\le {\phi }_{u}\left(t\right)$ for almost all $t\in \left[0,1\right]$ and $\left(x,y,z\right)\in U$. Consider the nonlinear fractional boundary value problem

$\begin{array}{l}{D}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right),{\phantom{\rule{1em}{0ex}}}^{c}{D}^{\beta }u\left(t\right)\right),\\ u\left(0\right)=au\left(1\right),\phantom{\rule{1em}{0ex}}{u}^{\prime }\left(0\right)={b}^{c}{D}^{\beta }u\left(1\right),\phantom{\rule{1em}{0ex}}{u}^{″}\left(0\right)={u}^{‴}\left(0\right)={u}^{\left(n-1\right)}\left(0\right)=0,\end{array}$
(*)

where $n\ge 3$ is an integer, $\alpha \in \left(n-1,n\right)$, $0<\beta <1$, $0, $0 and $q>\frac{1}{\alpha -1}$. We say that the function $u:\left[0,1\right]\to \mathbb{R}$ is a positive solution for the problem whenever $u>0$ on $\left[0,1\right]$, ${}^{c}D^{\alpha }u$ is a function in ${L}^{q}\left[0,1\right]$, and u satisfies the boundary conditions almost everywhere on $\left[0,1\right]$. In this paper, we suppose that f is an ${L}^{q}$-Caratheodory function on $\left[0,1\right]×B$, where $B=\left(0,\mathrm{\infty }\right)×\left(0,\mathrm{\infty }\right)×\left(0,\mathrm{\infty }\right)$, there exists a positive constant m such that $m\le f\left(t,x,y,z\right)$ for almost all $t\in \left[0,1\right]$ and $\left(x,y,z\right)\in B$, f satisfies the estimate

$f\left(t,x,y,z\right)\le h\left(x\right)+r\left(|y|\right)+k\left(|z|\right)+\gamma \left(t\right)w\left(x,|y|,|z|\right),$

where $h,r,k\in C\left(0,\mathrm{\infty }\right)$ are positive and non-increasing, $\gamma \in {L}^{q}\left[0,1\right]$ and $w\in C\left(\left[0,\mathrm{\infty }\right)×\left[0,\mathrm{\infty }\right)×\left[0,\mathrm{\infty }\right)\right)$ are positive, w is non-decreasing in all its variables, ${\int }_{0}^{1}{h}^{q}\left({s}^{\alpha }\right)\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty }$, ${\int }_{0}^{1}{r}^{q}\left({s}^{\alpha -1}\right)\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty }$, ${\int }_{0}^{1}{k}^{q}\left({s}^{\alpha -\beta }\right)\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty }$, and ${lim}_{x↦\mathrm{\infty }}\frac{w\left(x,x,x\right)}{x}=0$. Since we suppose that problem () is singular, that is, $f\left(t,x,y,z\right)$ may be singular at the value 0 of its space variables x, y, z, we use regularization and sequential techniques for the existence of positive solutions of the problem. In this way, for each natural number n define the function ${f}_{n}$ by

${f}_{n}\left(t,x,y,z\right)=f\left(t,{\chi }_{n}^{+}\left(x\right),{\chi }_{n}^{+}\left(y\right),{\chi }_{n}^{+}\left(z\right)\right)$

for all $t\in \left[0,1\right]$ and $\left(x,y,z\right)\in {\mathbb{R}}^{3}$, where

${\chi }_{n}^{+}\left(u\right)=\left\{\begin{array}{cc}u,\hfill & u\ge \frac{1}{n},\hfill \\ \frac{1}{n},\hfill & u<\frac{1}{n}.\hfill \end{array}$

It is easy to see that each ${f}_{n}$ is an ${L}^{q}$-Caratheodory function on $\left[0,1\right]×{\mathbb{R}}^{3}$, $m\le {f}_{n}\left(t,x,y,z\right)$,

${f}_{n}\left(t,x,y,z\right)\le h\left(\frac{1}{n}\right)+r\left(\frac{1}{n}\right)+k\left(\frac{1}{n}\right)+\gamma \left(t\right)w\left(1+x,1+|y|,1+|z|\right)$

and

${f}_{n}\left(t,x,y,z\right)\le h\left(x\right)+r\left(|y|\right)+k\left(|z|\right)+\gamma \left(t\right)w\left(1+x,1+|y|,1+|z|\right)$

for almost all $t\in \left[0,1\right]$ and all $\left(x,y,z\right)\in B$. In 2012, Agarwal et al. proved the following result.

Lemma 1.1 [7]

Let $\rho \in {L}^{q}\left[0,1\right]$ and $0\le {t}_{1}<{t}_{2}\le 1$. Then we have $|{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\rho \left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\le {\left(\frac{{t}^{d}}{d}\right)}^{1/p}{\parallel \rho \parallel }_{q}$ for all $t\in \left[0,1\right]$ and

$\begin{array}{r}|{\int }_{0}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}\rho \left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -2}\rho \left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le {\left(\frac{{t}_{1}^{d}+{\left({t}_{2}-{t}_{1}\right)}^{d}-{t}_{2}^{d}}{d}\right)}^{1/p}{\parallel \rho \parallel }_{q}+{\left(\frac{{\left({t}_{2}-{t}_{1}\right)}^{d}}{d}\right)}^{1/p}{\parallel \rho \parallel }_{q},\end{array}$

where $d=\left(\alpha -2\right)p+1$.

## 2 Main results

Now, we are ready to investigate the problem in regular and singular cases. First, we give the following result.

Lemma 2.1 Let $y\in C\left[0,1\right]$. Then the boundary value problem

$\begin{array}{c}{}^{c}D^{\alpha }u\left(t\right)=y\left(t\right)\phantom{\rule{1em}{0ex}}\left(t\in \left(0,1\right)\right),\hfill \\ u\left(0\right)=au\left(1\right),\phantom{\rule{2em}{0ex}}{u}^{\prime }\left(0\right)=b{\phantom{\rule{0.2em}{0ex}}}^{c}{D}^{\beta }u\left(1\right),\phantom{\rule{2em}{0ex}}{u}^{″}\left(0\right)={u}^{‴}\left(0\right)={u}^{\left(n-1\right)}\left(0\right)=0\hfill \end{array}$

is equivalent to the fractional integral equation $u\left(t\right)={\int }_{0}^{1}G\left(t,s\right)y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$, where

$\begin{array}{rl}G\left(t,s\right)=& \frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\\ +\frac{a\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right){\left(1-s\right)}^{\alpha -1}+b\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2-\beta \right)\left(a+t-at\right){\left(1-s\right)}^{\alpha -\beta -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}\end{array}$

whenever $0\le s\le t\le 1$ and

$G\left(t,s\right)=\frac{a\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right){\left(1-s\right)}^{\alpha -1}+b\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2-\beta \right)\left(a+t-at\right){\left(1-s\right)}^{\alpha -\beta -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}$

whenever $0\le t\le s\le 1$.

Proof From ${}^{c}D^{\alpha }u\left(t\right)=y\left(t\right)$ and the boundary conditions, we obtain

$\begin{array}{rl}u\left(t\right)& ={I}^{\alpha }y\left(t\right)+u\left(0\right)+{u}^{\prime }\left(0\right)t+\frac{{u}^{″}\left(0\right)}{2!}{t}^{2}+\cdots +\frac{{u}^{\left(n-1\right)}\left(0\right)}{\left(n-1\right)!}{t}^{n-1}\\ =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+u\left(0\right)+{u}^{\prime }\left(0\right)t.\end{array}$

By properties of the Caputo derivative, we get

$\begin{array}{rl}{}^{c}D^{\beta }u\left(t\right)& ={I}^{\alpha -\beta }y\left(t\right){+}^{c}{D}^{\beta }\left(u\left(0\right)+{u}^{\prime }\left(0\right)t\right)\\ =\frac{1}{\mathrm{\Gamma }\left(\alpha -\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -\beta -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{{u}^{\prime }\left(0\right){t}^{1-\beta }}{\mathrm{\Gamma }\left(2-\beta \right)}.\end{array}$

Thus, $u\left(1\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+u\left(0\right)+{u}^{\prime }\left(0\right)$ and

${}^{c}D^{\beta }u\left(1\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -\beta \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -\beta -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{{u}^{\prime }\left(0\right)}{\mathrm{\Gamma }\left(2-\beta \right)}.$

By using the boundary conditions $u\left(0\right)=au\left(1\right)$ and ${u}^{\prime }\left(0\right)=b{\phantom{\rule{0.2em}{0ex}}}^{c}{D}^{\beta }u\left(1\right)$, we get $u\left(0\right)=a\left(\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+u\left(0\right)+{u}^{\prime }\left(0\right)\right)$ and

${u}^{\prime }\left(0\right)=b\left(\frac{1}{\mathrm{\Gamma }\left(\alpha -\beta \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -\beta -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\frac{{u}^{\prime }\left(0\right)}{\mathrm{\Gamma }\left(2-\beta \right)}\right).$

Hence, ${u}^{\prime }\left(0\right)=\frac{b\mathrm{\Gamma }\left(2-\beta \right)}{\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -\beta -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ and

$\begin{array}{rl}u\left(0\right)=& \frac{a}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\frac{ab\mathrm{\Gamma }\left(2-\beta \right)}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}{\int }_{0}^{1}{\left(1-s\right)}^{\alpha -\beta -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

Thus,

$\begin{array}{rl}u\left(t\right)=& \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -1}y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+u\left(0\right)+{u}^{\prime }\left(0\right)t\\ =& {\int }_{0}^{t}\left(\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}+\frac{a\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right){\left(1-s\right)}^{\alpha -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}\\ +\frac{b\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2-\beta \right)\left(a+t-at\right){\left(1-s\right)}^{\alpha -\beta -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}\right)y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ +{\int }_{t}^{1}\left(\frac{a\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right){\left(1-s\right)}^{\alpha -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}\\ +\frac{b\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2-\beta \right)\left(a+t-at\right){\left(1-s\right)}^{\alpha -\beta -1}}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}\right)y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ =& {\int }_{0}^{1}G\left(t,s\right)y\left(s\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

This completes the proof. □

Put ${k}_{1}=\frac{\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)+b\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(2-\beta \right)}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}$ and ${k}_{2}=\frac{ab\mathrm{\Gamma }\left(2-\beta \right)}{\left(1-a\right)\mathrm{\Gamma }\left(\alpha -\beta \right)\left(\mathrm{\Gamma }\left(2-\beta \right)-b\right)}$. It is easy to check that the Green function G in the last result belongs to $C\left(\left[0,1\right]×\left[0,1\right]\right)$, $G\left(t,s\right)>0$ for all $\left(t,s\right)\in \left[0,1\right)×\left[0,1\right)$,

$G\left(t,s\right)\le {k}_{1}{\left(1-s\right)}^{\alpha -\beta -1}\le 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}G\left(t,s\right)\ge {k}_{2}{\left(1-s\right)}^{\alpha -\beta -1}$

for all $\left(t,s\right)\in \left[0,1\right]×\left[0,1\right]$. Consider the Banach space $X={C}^{1}\left[0,1\right]$ with the norm ${\parallel x\parallel }_{\ast }=max\left\{\parallel x\parallel ,\parallel {x}^{\prime }\parallel \right\}$ and the cone

For each natural number n, define the operator ${Q}_{n}$ on P by

$\left({Q}_{n}x\right)\left(t\right)={\int }_{0}^{1}G\left(t,s\right){f}_{n}\left(s,u\left(s\right),{u}^{\prime }\left(s\right){,}^{c}{D}^{\beta }u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

Now, we prove that ${Q}_{n}$ is a completely continuous operator (see [2]).

Lemma 2.2 The operator ${Q}_{n}$ is a completely continuous operator.

Proof Let $x\in P$. Then, ${}^{c}D^{\beta }x\in C\left[0,1\right]$ and ${}^{c}D^{\beta }x\ge 0$. Now, define $\rho \left(t\right)={f}_{n}\left(t,u\left(t\right),{u}^{\prime }\left(t\right){,}^{c}{D}^{\beta }u\left(t\right)\right)$ for almost all $t\in \left[0,1\right]$. Then $\rho \in {L}^{q}\left[0,1\right]$ and $\rho \left(t\right)\ge m$ for almost all $t\in \left[0,1\right]$. By using the properties of fractional integral ${I}^{\alpha }$, it is easy to see that ${Q}_{n}x\in C\left[0,1\right]$, $\left({Q}_{n}x\right)\left(t\right)\ge 0$ and

${\left({Q}_{n}x\right)}^{\prime }\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\rho \left(s\right)\phantom{\rule{0.2em}{0ex}}ds$

for all $t\in \left[0,1\right]$. This implies that ${\left({Q}_{n}x\right)}^{\prime }\in C\left[0,1\right]$ and ${\left({Q}_{n}x\right)}^{\prime }\ge 0$ on $\left[0,1\right]$. Consequently, ${Q}_{n}$ maps P into P. In order to prove that ${Q}_{n}$ is a continuous operator, let ${x}_{m}$ be a convergent sequence in P and ${lim}_{m↦\mathrm{\infty }}{x}_{m}=x$. Thus, ${lim}_{m↦\mathrm{\infty }}{x}_{m}^{\left(j\right)}\left(t\right)={x}^{\left(j\right)}\left(t\right)$ uniformly on $\left[0,1\right]$ for $j=0,1$. Since

$\begin{array}{rl}{}^{c}D^{\beta }x\left(t\right)& =\frac{1}{\mathrm{\Gamma }\left(1-\beta \right)}\frac{d}{dt}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }\left(x\left(s\right)-x\left(0\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ =\frac{1}{\mathrm{\Gamma }\left(1-\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{x}^{\prime }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds,\end{array}$

we get ${|}^{c}{D}^{\beta }{x}_{m}\left(t\right){-}^{c}{D}^{\beta }x\left(t\right)|\le \frac{\parallel {x}_{m}^{\prime }-{x}^{\prime }\parallel }{\mathrm{\Gamma }\left(1-\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }\phantom{\rule{0.2em}{0ex}}ds\le \frac{{\parallel {x}_{m}-x\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}$ and $lim_{m↦\mathrm{\infty }}{}^{c}{D}^{\beta }{x}_{m}\left(t\right){=}^{c}{D}^{\beta }x\left(t\right)$ uniformly on $\left[0,1\right]$. Also, we have ${|}^{c}{D}^{\beta }{x}_{m}\left(t\right)|\le \frac{{x}_{m}^{\prime }}{\mathrm{\Gamma }\left(\beta \right)}$ on $\left[0,1\right]$, and so ${\parallel }^{c}{D}^{\beta }{x}_{m}\parallel \le \frac{\parallel {x}_{m}^{\prime }\parallel }{\mathrm{\Gamma }\left(\beta \right)}$. Now, put

${\rho }_{m}\left(t\right)={f}_{n}\left(t,{x}_{m}\left(t\right),{x}_{m}^{\prime }\left(t\right){,}^{c}{D}^{\beta }{x}_{m}\left(t\right)\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\rho \left(t\right)={f}_{n}\left(t,x\left(t\right),{x}^{\prime }\left(t\right){,}^{c}{D}^{\beta }x\left(t\right)\right).$

Then, it is easy to see that ${lim}_{m↦\mathrm{\infty }}{\rho }_{m}\left(t\right)=\rho \left(t\right)$ for almost all $t\in \left[0,1\right]$, and there exists $\beta \in {L}^{q}\left[0,1\right]$ such that $0\le {\rho }_{m}\left(t\right)\le \beta \left(t\right)$ for almost all $t\in \left[0,1\right]$ and all $m\ge 1$. Since ${f}_{n}$ is an ${L}^{q}$-Caratheodory function, $\left\{{x}_{m}\right\}$ is bounded in ${C}^{1}\left[0,1\right]$, and ${\left\{}^{c}{D}^{\beta }{x}_{m}\right\}$ is bounded in $C\left[0,1\right]$. Therefore, ${lim}_{m↦\mathrm{\infty }}\left({Q}_{n}{x}_{m}\right)\left(t\right)=\left({Q}_{n}x\right)\left(t\right)$ uniformly on $\left[0,1\right]$. Since $\left\{{\rho }_{m}\right\}$ is ${L}^{q}$-convergent on $\left[0,1\right]$,

$\underset{m↦\mathrm{\infty }}{lim}{\left({Q}_{n}{x}_{m}\right)}^{\prime }\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\underset{m↦\mathrm{\infty }}{lim}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}{\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds={\left({Q}_{n}x\right)}^{\prime }\left(t\right)$

uniformly on $\left[0,1\right]$. Hence, ${Q}_{n}$ is a continuous operator. Now, we have to show that for each bounded sequence $\left\{{x}_{m}\right\}$ in P, the sequence $\left\{{Q}_{n}{x}_{m}\right\}$ is relatively compact in $C\left[0,1\right]$. Choose a positive constant k such that $\parallel {x}_{m}\parallel \le k$ and $\parallel {x}_{m}^{\prime }\parallel \le k$ for all m. Note that ${\parallel }^{c}{D}^{\beta }{x}_{m}\parallel \le \frac{k}{\mathrm{\Gamma }\left(\beta \right)}$ and $|{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}{\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\le {\left({\int }_{0}^{t}{\left(t-s\right)}^{\left(\alpha -2\right)p}\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{p}}{\left({\int }_{0}^{t}{|{\rho }_{m}\left(s\right)|}^{q}ds\right)}^{\frac{1}{q}}\le {\left(\frac{{t}^{d}}{d}\right)}^{\frac{1}{p}}{\parallel {\rho }_{m}\parallel }_{q}$ for all m, where $d=\left(\alpha -2\right)p+1$. But we have

$0\le \left({Q}_{n}{x}_{m}\right)\left(t\right)={\int }_{0}^{1}G\left(t,s\right){\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{0}^{1}G\left(t,s\right)\beta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\le \frac{{\parallel \beta \parallel }_{1}}{\mathrm{\Gamma }\left(\alpha \right)}$

and

$\begin{array}{rl}0& \le {\left({Q}_{n}{x}_{m}\right)}^{\prime }\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}{\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\beta \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\left(\frac{1}{\left(\alpha -2\right)p+1}\right)}^{\frac{1}{p}}{\parallel \beta \parallel }_{q}\end{array}$

for all $t\in \left[0,1\right]$ and m. This implies that $\left\{{Q}_{n}{x}_{m}\right\}$ is bounded in ${C}^{1}\left[0,1\right]$. Also, we have

$\begin{array}{r}|{\left({Q}_{n}{x}_{m}\right)}^{\prime }\left({t}_{2}\right)-{\left({Q}_{n}{x}_{m}\right)}^{\prime }\left({t}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}|{\int }_{0}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}{\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -2}{\rho }_{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\parallel {\rho }_{m}\parallel }_{q}}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\left(\frac{{t}_{1}^{d}+{\left({t}_{2}-{t}_{1}\right)}^{d}-{t}_{2}^{d}}{d}\right)}^{\frac{1}{p}}+{\left(\frac{{\left({t}_{2}-{t}_{1}\right)}^{d}}{d}\right)}^{\frac{1}{p}}\right)\\ \phantom{\rule{1em}{0ex}}\le \frac{{\parallel \beta \parallel }_{q}}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\left(\frac{{t}_{1}^{d}+{\left({t}_{2}-{t}_{1}\right)}^{d}-{t}_{2}^{d}}{d}\right)}^{\frac{1}{p}}+{\left(\frac{{\left({t}_{2}-{t}_{1}\right)}^{d}}{d}\right)}^{\frac{1}{p}}\right)\end{array}$

for all $0\le {t}_{1}\le {t}_{2}\le 1$, where $d=\left(\alpha -2\right)p+1$. Hence, $\left\{{\left({Q}_{n}{x}_{m}\right)}^{\prime }\right\}$ is equicontinuous on $\left[0,1\right]$. Thus, $\left\{{Q}_{n}{x}_{m}\right\}$ is relatively compact in ${C}^{1}\left[0,1\right]$ by the Arzela-Ascoli theorem. Hence, ${Q}_{n}$ is a completely continuous operator. □

We need the following result (see [2] and [21]).

Lemma 2.3 [21]

Let Y be a Banach space, P a cone in Y and ${\mathrm{\Omega }}_{1}$ and ${\mathrm{\Omega }}_{2}$ bounded open balls in Y centered at the origin with ${\overline{\mathrm{\Omega }}}_{1}\subset {\mathrm{\Omega }}_{2}$. Suppose that $T:P\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)\to P$ is a completely continuous operator such that $\parallel Tx\parallel \ge \parallel x\parallel$ for all $x\in P\cap \partial {\mathrm{\Omega }}_{1}$ and $\parallel Tx\parallel \le \parallel x\parallel$ for all $x\in P\cap \partial {\mathrm{\Omega }}_{2}$. Then T has a fixed point in $P\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)$.

Theorem 2.4 For each natural number n, problem () has a solution ${u}_{n}\in P$ such that ${u}_{n}\ge \frac{m{k}_{2}}{\alpha -\beta }$, ${u}_{n}^{\prime }\left(t\right)\ge \frac{m{t}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}$ and ${}^{c}D^{\beta }{u}_{n}\left(t\right)\ge \frac{m{t}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}$ for all $t\in \left[0,1\right]$.

Proof Let $n\ge 1$. It is sufficient to show that ${Q}_{n}$ has a fixed point ${u}_{n}$ in P with the desired conditions. In this way, note that

$\begin{array}{rl}\left({Q}_{n}x\right)\left(t\right)& ={\int }_{0}^{1}G\left(t,s\right){f}_{n}\left(s,x\left(s\right),{x}^{\prime }\left(s\right){,}^{c}{D}^{\beta }x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge m{\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds\ge m{\int }_{0}^{1}{k}_{2}{\left(1-s\right)}^{\alpha -\beta -1}\phantom{\rule{0.2em}{0ex}}ds=\frac{m{k}_{2}}{\alpha -\beta },\end{array}$

and so ${\parallel {Q}_{n}x\parallel }_{\ast }\ge \parallel {Q}_{n}\left(x\right)\parallel \ge \frac{m{k}_{2}}{\alpha -\beta }$. Put ${\mathrm{\Omega }}_{1}=\left\{x\in X:{\parallel x\parallel }_{\ast }<\frac{m{k}_{2}}{\alpha -\beta }\right\}$. Then ${\parallel {Q}_{n}x\parallel }_{\ast }\ge {\parallel x\parallel }_{\ast }$ for all $x\in P\cap \partial {\mathrm{\Omega }}_{1}$. If ${v}_{n}=h\left(\frac{1}{n}\right)+r\left(\frac{1}{n}\right)+k\left(\frac{1}{n}\right)$, then

$\begin{array}{rl}|\left({Q}_{n}x\right)\left(t\right)|& \le |{\int }_{0}^{1}G\left(t,s\right){f}_{n}\left(s,u\left(s\right),{u}^{\prime }\left(s\right){,}^{c}{D}^{\beta }u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \le {\int }_{0}^{1}|G\left(t,s\right)|\left({v}_{n}+\gamma \left(s\right)w\left(1+|x\left(s\right)|,1+|{x}^{\prime }\left(s\right)|,1+{|}^{c}{D}^{\beta }x\left(s\right)|\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le {k}_{1}\left({v}_{n}+w\left(1+\parallel x\parallel ,1+\parallel {x}^{\prime }\parallel ,1+{\parallel }^{c}{D}^{\beta }x\parallel \right){\parallel \gamma \parallel }_{1}\right)\end{array}$

and

$\begin{array}{r}|{\left({Q}_{n}x\right)}^{\prime }\left(t\right)|\\ \phantom{\rule{1em}{0ex}}=|\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}{f}_{n}\left(s,x\left(s\right),{x}^{\prime }\left(s\right){,}^{c}{D}^{\beta }x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\left({v}_{n}+\gamma \left(s\right)w\left(1+|x\left(s\right)|,1+|{x}^{\prime }\left(s\right)|,1+{|}^{c}{D}^{\beta }x\left(s\right)|\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left(\frac{{v}_{n}{t}^{\alpha -1}}{\alpha -1}+w\left(1+\parallel x\parallel ,1+\parallel {x}^{\prime }\parallel ,1+{\parallel }^{c}{D}^{\beta }x\parallel \right){\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\gamma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\end{array}$

for all $x\in P$ and $t\in \left[0,1\right]$, because w is non-decreasing in all its variables. Since $\parallel x\parallel \le {\parallel x\parallel }_{\ast }$, $\parallel {x}^{\prime }\parallel \le {\parallel x\parallel }_{\ast }$, ${\parallel }^{c}{D}^{\beta }x\parallel \le \frac{\parallel {x}^{\prime }\parallel }{\mathrm{\Gamma }\left(\beta \right)}\le \frac{{\parallel x\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}$ and

${\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\gamma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\le {\left(\frac{1}{d}\right)}^{1/p}{\parallel \gamma \parallel }_{q},$

where $d=\left(\alpha -2\right)p+1$, we have

$\parallel {Q}_{n}\left(x\right)\parallel \le {k}_{1}\left({v}_{n}+w\left(1+{\parallel x\parallel }_{\ast },1+{\parallel x\parallel }_{\ast },1+\frac{{\parallel x\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\parallel \gamma \parallel }_{1}\right)$

and

$\begin{array}{r}\parallel {\left({Q}_{n}x\right)}^{\prime }\parallel \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left(\frac{{v}_{n}}{\alpha -1}+w\left(1+{\parallel x\parallel }_{\ast },1+{\parallel x\parallel }_{\ast },1+\frac{{\parallel x\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\left(1/d\right)}^{1/p}{\parallel \gamma \parallel }_{q}\right).\end{array}$

Hence, ${\parallel {Q}_{n}x\parallel }_{\ast }\le M\left(\frac{{v}_{n}}{\alpha -1}+Nw\left(1+{\parallel x\parallel }_{\ast },1+{\parallel x\parallel }_{\ast },1+\frac{{\parallel x\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right)\right)$, where $N=max\left\{{\parallel \gamma \parallel }_{1},{\left(1/d\right)}^{1/p}{\parallel \gamma \parallel }_{q}\right\}$ and $M=max\left\{\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)},{k}_{1}\right\}$. Since

$\underset{v↦\mathrm{\infty }}{lim}\frac{1}{v}w\left(1+v,1+v,1+v\right)=0,$

there exists a positive constant L such that

$M\left(\frac{{v}_{n}}{\alpha -1}+Nw\left(1+v,1+v,\frac{v}{\mathrm{\Gamma }\left(\beta \right)}\right)\right)

for all $v\ge L$. Thus, ${\parallel {Q}_{n}x\parallel }_{\ast }<{\parallel x\parallel }_{\ast }$ for all $x\in P$ with ${\parallel x\parallel }_{\ast }\ge L$. Put ${\mathrm{\Omega }}_{2}=\left\{x\in X:{\parallel x\parallel }_{\ast }. Then ${\parallel {Q}_{n}x\parallel }_{\ast }<{\parallel x\parallel }_{\ast }$ for all $x\in P\cap \partial {\mathrm{\Omega }}_{2}$. By using last result, ${Q}_{n}$ has a fixed point ${u}_{n}$ in $P\cap \left({\overline{\mathrm{\Omega }}}_{2}\mathrm{\setminus }{\mathrm{\Omega }}_{1}\right)$. But ${u}_{n}=\left({Q}_{n}{u}_{n}\right)\left(t\right)\ge \frac{m{k}_{2}}{\alpha -\beta }$ and

$\begin{array}{rl}{\left({Q}_{n}x\right)}^{\prime }\left(t\right)& =\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}{f}_{n}\left(s,x\left(s\right),{x}^{\prime }\left(s\right){,}^{c}{D}^{\beta }x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge \frac{m}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds=\frac{m{t}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\end{array}$

for all $t\in \left[0,1\right]$ and $x\in P$. Since ${\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{s}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}ds=\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\beta \right)}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}{t}^{\alpha -\beta }$,

$\begin{array}{rl}{}^{c}D^{\beta }{u}_{n}\left(t\right)& =\frac{1}{\mathrm{\Gamma }\left(1-\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{u}_{n}^{\prime }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge \frac{m}{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(1-\beta \right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{s}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}ds=\frac{m{t}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\end{array}$

for all $t\in \left[0,1\right]$. This completes the proof. □

Now, we give our last result.

Theorem 2.5 Problem () has a solution u such that $u\left(t\right)\ge \frac{m{k}_{2}}{\alpha -\beta }$, ${u}^{\prime }\left(t\right)\ge \frac{m{t}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}$ and ${}^{c}D^{\beta }u\left(t\right)\ge \frac{m{t}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}$ for all $t\in \left[0,1\right]$.

Proof By using Theorem 2.4, one gets that for each natural number n, problem () has a solution ${u}_{n}\in P$ with the desired conditions. Thus, $h\left({u}_{n}\left(t\right)\right)\le h\left(\frac{m{k}_{2}}{\alpha -\beta }\right)$, $r\left(|{u}_{n}^{\prime }\left(t\right)|\right)\le r\left(\frac{m{t}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\right)$ and $k\left({|}^{c}{D}^{\beta }{u}_{n}\left(t\right)|\right)\le k\left(\frac{m{t}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\right)$ for all $t\in \left[0,1\right]$ and n. Also, we have ${\parallel }^{c}{D}^{\beta }{u}_{n}\parallel \le \frac{\parallel {u}_{n}^{\prime }\parallel }{\mathrm{\Gamma }\left(\beta \right)}$. Suppose that

$S\left(t\right)=h\left(\frac{m{k}_{2}}{\alpha -\beta }\right)+r\left(\frac{m{t}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\right)+k\left(\frac{m{t}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\right).$

Then

$\begin{array}{rl}m& \le {f}_{n}\left(t,{u}_{n}\left(t\right),{u}_{n}^{\prime }\left(t\right){,}^{c}{D}^{\beta }{u}_{n}\left(t\right)\right)\\ \le S\left(t\right)+\gamma \left(t\right)w\left(1+\parallel {u}_{n}\parallel ,1+\parallel {u}_{n}^{\prime }\parallel ,1+{\parallel }^{c}{D}^{\beta }{u}_{n}\parallel \right)\\ \le S\left(t\right)+\gamma \left(t\right)w\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}^{\prime }\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right)\end{array}$

for almost all $t\in \left[0,1\right]$ and n. Since $0\le G\left(t,s\right)\le {k}_{1}$, we get

$\begin{array}{rl}0& \le {u}_{n}\left(t\right)\\ ={\int }_{0}^{1}G\left(t,s\right){f}_{n}\left(s,{u}_{n}\left(s\right),{u}_{n}^{\prime }\left(s\right){,}^{c}{D}^{\beta }{u}_{n}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le {k}_{1}\left({\int }_{0}^{1}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+w\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\parallel \gamma \parallel }_{1}\right)\end{array}$

and

$\begin{array}{rl}0\le & {u}_{n}^{\prime }\left(t\right)\\ \le & \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ +w\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}\gamma \left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right).\end{array}$

We show that ${\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ is bounded on $\left[0,1\right]$. Let $d=\left(\alpha -2\right)p+1$. Note that

$\begin{array}{c}\begin{array}{r}{\int }_{0}^{1}{\left(t-s\right)}^{\alpha -2}h\left(\frac{m{k}_{2}}{\alpha -\beta }\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}=h\left(\frac{m{k}_{2}}{\alpha -\beta }\right){\int }_{0}^{1}{\left(t-s\right)}^{\alpha -2}\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{\alpha -1}h\left(\frac{m{k}_{2}}{\alpha -\beta }\right)=:{\eta }_{1}<\mathrm{\infty },\end{array}\hfill \\ \begin{array}{r}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}r\left(\frac{m{s}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}={\left(\frac{1}{d}\right)}^{1/p}{\left(\frac{\mathrm{\Gamma }\left(\alpha \right)}{m}\right)}^{\frac{1}{\left(\alpha -1\right)q}}{\left({\int }_{0}^{{\left(\frac{m}{\mathrm{\Gamma }\left(\alpha \right)}\right)}^{\frac{1}{\alpha -1}}}{r}^{q}\left({s}^{\alpha -1}\right)\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/q}=:{\eta }_{2}<\mathrm{\infty }\end{array}\hfill \end{array}$

and

$\begin{array}{r}{\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}k\left(\frac{m{s}^{\alpha -\beta }}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}={\left(\frac{1}{d}\right)}^{1/p}{\left(\frac{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}{m}\right)}^{\frac{1}{\left(\alpha -\beta \right)q}}{\left({\int }_{0}^{{\left(\frac{m}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\right)}^{\frac{1}{\alpha -\beta }}}{k}^{q}\left({s}^{\alpha -\beta }\right)\phantom{\rule{0.2em}{0ex}}ds\right)}^{1/q}=:{\eta }_{3}<\mathrm{\infty }.\end{array}$

Thus, ${\int }_{0}^{t}{\left(t-s\right)}^{\alpha -2}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\le \eta$ for all $t\in \left[0,1\right]$, where $\eta ={\eta }_{1}+{\eta }_{2}+{\eta }_{3}$. Also, we have

$\begin{array}{r}{\int }_{0}^{1}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\alpha -1}h\left(\frac{m{k}_{2}}{\alpha -\beta }\right)+{\left(\frac{\mathrm{\Gamma }\left(\alpha \right)}{m}\right)}^{\frac{1}{\alpha -1}}{\int }_{0}^{{\left(\frac{m}{\mathrm{\Gamma }\left(\alpha \right)}\right)}^{\frac{1}{\alpha -1}}}r\left({s}^{\alpha -1}\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+{\left(\frac{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}{m}\right)}^{\frac{1}{\alpha -\beta }}{\int }_{0}^{{\left(\frac{m}{\mathrm{\Gamma }\left(\alpha -\beta +1\right)}\right)}^{\frac{1}{\alpha -\beta }}}k\left({s}^{\alpha -\beta }\right)\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty }.\end{array}$

Since

$\parallel {u}_{n}\parallel ={k}_{1}\left({\int }_{0}^{1}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+w\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\parallel \gamma \parallel }_{1}\right)$

and

$\parallel {u}_{n}^{\prime }\parallel \le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left(\eta +w\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right){\left(\frac{1}{d}\right)}^{1/p}{\parallel \gamma \parallel }_{q}\right),$

we get ${\parallel {u}_{n}\parallel }_{\ast }\le M\left(\mathrm{\Phi }+Kw\left(1+{\parallel {u}_{n}\parallel }_{\ast },1+{\parallel {u}_{n}\parallel }_{\ast },1+\frac{{\parallel {u}_{n}\parallel }_{\ast }}{\mathrm{\Gamma }\left(\beta \right)}\right)\right)$ for all n, where $\mathrm{\Phi }=max\left\{\eta ,{\int }_{0}^{1}S\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right\}$, $K=max\left\{{\parallel \gamma \parallel }_{1},{\left(\frac{1}{d}\right)}^{1/p}{\parallel \gamma \parallel }_{q}\right\}$ and also $M=max\left\{{k}_{1},\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\right\}$. On the other hand, there exists a positive constant L such that $M\left(\mathrm{\Phi }+Kw\left(1+v,1+v,1+\frac{v}{\mathrm{\Gamma }\left(\beta \right)}\right)\right) for all $v\ge L$, and so ${\parallel {u}_{n}\parallel }_{\ast } for all n. Thus, for almost all $t\in \left[0,1\right]$ and all n, we have ${f}_{n}\left(t,{u}_{n}\left(t\right),{u}_{n}^{\prime }\left(t\right){,}^{c}{D}^{\beta }{u}_{n}\left(t\right)\right)\le R\left(t\right)$, where

$R\left(t\right)=S\left(t\right)+\gamma \left(t\right)w\left(1+L,1+L,1+\frac{L}{\mathrm{\Gamma }\left(\beta \right)}\right).$

Note that $R\in {L}^{q}\left[0,1\right]$. We show that $\left\{{u}_{n}^{\prime }\right\}$ is equicontinuous on $\left[0,1\right]$. Let ${\rho }_{n}\left(t\right)={f}_{n}\left(t,{u}_{n}\left(t\right),{u}_{n}^{\prime }\left(t\right){,}^{c}{D}^{\beta }{u}_{n}\left(t\right)\right)$ and $0\le {t}_{1}<{t}_{2}\le T$. Then

$\begin{array}{r}|{u}_{n}^{\prime }\left({t}_{2}\right)-{u}_{n}^{\prime }\left({t}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}|{\int }_{0}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}{\rho }_{n}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds-{\int }_{0}^{{t}_{1}}{\left({t}_{1}-s\right)}^{\alpha -2}{\rho }_{n}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\int }_{0}^{{t}_{1}}\left({\left({t}_{1}-s\right)}^{\alpha -2}-{\left({t}_{2}-s\right)}^{\alpha -2}\right){\rho }_{n}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{t}_{1}}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}{\rho }_{n}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\int }_{0}^{{t}_{1}}\left({\left({t}_{1}-s\right)}^{\alpha -2}-{\left({t}_{2}-s\right)}^{\alpha -2}\right)R\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{t}_{1}}^{{t}_{2}}{\left({t}_{2}-s\right)}^{\alpha -2}R\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),\end{array}$

and so

$\begin{array}{r}|{u}_{n}^{\prime }\left({t}_{2}\right)-{u}_{n}^{\prime }\left({t}_{1}\right)|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\parallel R\parallel }_{q}}{\mathrm{\Gamma }\left(\alpha -1\right)}\left({\left(\frac{{t}_{1}^{d}+{\left({t}_{2}-{t}_{1}\right)}^{d}-{t}_{2}^{d}}{d}\right)}^{1/p}+{\left(\frac{{\left({t}_{2}-{t}_{1}\right)}^{d}}{d}\right)}^{1/p}\right).\end{array}$

Hence, $\left\{{u}_{n}^{\prime }\right\}$ is equicontinuous on $\left[0,1\right]$. Since $\left\{{u}_{n}\right\}$ is a bounded sequence in $C\left[0,1\right]$, by using the Arzela-Ascoli theorem, without loss of generality, we can assume that $\left\{{u}_{n}\right\}$ is convergent in $C\left[0,1\right]$. Let ${lim}_{n↦\mathrm{\infty }}{u}_{n}=u$. Then, it is easy to see that ${}^{c}D^{\beta }{u}_{n}\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{u}_{n}^{\prime }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$, and ${}^{c}D^{\beta }{u}_{n}\left(t\right)$ uniformly converges to $\frac{1}{\mathrm{\Gamma }\left(\alpha -1\right)}{\int }_{0}^{t}{\left(t-s\right)}^{-\beta }{u}^{\prime }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds$ on $\left[0,1\right]$. Thus, ${}^{c}D^{\beta }{u}_{n}$ converges to ${}^{c}D^{\beta }u$ in $C\left[0,1\right]$. Hence,

$\underset{n↦\mathrm{\infty }}{lim}{f}_{n}\left(t,{u}_{n}\left(t\right),{u}_{n}^{\prime }\left(t\right){,}^{c}{D}^{\beta }{u}_{n}\left(t\right)\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right){,}^{c}{D}^{\beta }u\left(t\right)\right)$

for almost all $t\in \left[0,1\right]$. Since $R\in {L}^{q}\left[0,1\right]$, by using the dominated convergence theorem on the relation

${u}_{n}\left(t\right)={\int }_{0}^{1}G\left(t,s\right){f}_{n}\left(s,{u}_{n}\left(s\right),{u}_{n}^{\prime }\left(s\right){,}^{c}{D}^{\beta }{u}_{n}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,$

we get $u\left(t\right)={\int }_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right){,}^{c}{D}^{\beta }u\left(s\right)\right)$ for all $t\in \left[0,1\right]$. This completes the proof. □

### 2.1 Examples for the problem

Example 2.1 Let ${\rho }_{1},{\rho }_{2}\in {L}^{q}\left[0,1\right]$, ${\rho }_{1}\left(t\right)\ge m>0$ for almost all t in $\left[0,1\right]$. Suppose that

$f\left(t,x,y,z\right)={\rho }_{1}\left(t\right)+\frac{1}{{x}^{\frac{1}{3}}-\lambda }+\frac{1}{{y}^{\frac{1}{4}}}+\frac{1}{{z}^{\frac{1}{4}}}+|{\rho }_{2}\left(t\right)|\left({x}^{\frac{1}{3}}+{y}^{\frac{1}{4}}+{z}^{\frac{1}{4}}\right)$

on $\left[0,1\right]×B$, $\lambda ={\left(au\left(1\right)\right)}^{\frac{1}{3}}$, $h\left(x\right)=\frac{1}{{x}^{\frac{1}{3}}-\lambda }$ whenever ${x}^{\frac{1}{3}}-\lambda \ge 0$ and $h\left(x\right)=0$ whenever ${x}^{\frac{1}{3}}-\lambda <0$, $r\left(x\right)=\frac{1}{{x}^{\frac{1}{4}}}$, $k\left(x\right)=\frac{1}{{x}^{\frac{1}{4}}}$, $w\left(x,y,z\right)=\left({x}^{\frac{1}{3}}+{y}^{\frac{1}{4}}+{z}^{\frac{1}{4}}+1\right)$ and $\gamma \left(t\right)={\rho }_{1}\left(t\right)+|{\rho }_{2}\left(t\right)|$. Then Theorem 2.5 guarantees that problem () has a positive solution.

Example 2.2 Consider the nonlinear mixed problem of singular fractional boundary value problem

${D}^{\frac{7}{3}}u\left(t\right)=t=1=\frac{1}{u{\left(t\right)}^{\frac{1}{3}}-\rho }=\frac{1}{{u}^{\prime }{\left(t\right)}^{\frac{1}{4}}}=\frac{1}{{{\left(}^{c}{D}^{\frac{1}{4}}u\left(t\right)\right)}^{\frac{1}{4}}}=2\left(u{\left(t\right)}^{\frac{1}{3}}={u}^{\prime }{\left(t\right)}^{\frac{1}{4}}={{\left(}^{c}{D}^{\frac{1}{4}}u\left(t\right)\right)}^{\frac{1}{4}}=1\right)\right)$

via boundary value conditions $u\left(0\right)=\frac{1}{4}u\left(1\right)$, ${u}^{\prime }\left(0\right)=\frac{1}{3}{\left(}^{c}{D}^{\frac{1}{4}}\right)u\left(1\right)$ and ${u}^{″}\left(0\right)={u}^{‴}\left(0\right)=\cdots ={u}^{\left(n-1\right)}\left(0\right)=0$, where $\rho ={\left(\left(\frac{1}{4}\right)u\left(1\right)\right)}^{\frac{1}{3}}$. Let f

Then the map f is singular at $t=0$, and f satisfies the desired conditions, where $h\left(x\right)=\frac{1}{{x}^{\frac{1}{3}}-\rho }$ whenever ${x}^{\frac{1}{3}}-\rho \ge 0$ and $h\left(x\right)=0$ whenever ${x}^{\frac{1}{3}}-\rho <0$, $r\left(x\right)=\frac{1}{{x}^{\frac{1}{4}}}$, $k\left(x\right)=\frac{1}{{x}^{\frac{1}{4}}}$, $w\left(x,y,z\right)={x}^{\frac{1}{3}}+{y}^{\frac{1}{4}}+{z}^{\frac{1}{4}}+1$, ${\rho }_{1}\left(t\right)=t+1>1=m$, ${\rho }_{2}\left(t\right)=2$ and $\gamma \left(t\right)={\rho }_{1}\left(t\right)+|{\rho }_{2}\left(t\right)|$. Then Theorem 2.5 guarantees that this problem has a positive solution.

## 3 Conclusions

One of the most interesting branches is obtaining solutions of singular fractional differential via boundary value problems. Having these things in mind, we study the existence of solutions for a singular nonlinear fractional boundary value problem. Two illustrative examples illustrate the applicability of the proposed method. It seems that the obtained results could be extended to more general functional spaces. Finally, note that all calculations in proofs of the results depend on the definition of the fractional derivative.

## References

1. Bai Z, Lu H: Positive solutions for boundary value problem of nonlinear fractional differential equation. J. Math. Anal. Appl. 2005, 311: 495-505. 10.1016/j.jmaa.2005.02.052

2. Baleanu D, Mohammadi H, Rezapour S: Positive solutions of a boundary value problem for nonlinear fractional differential equations. Abstr. Appl. Anal. 2012., 2012: Article ID 837437

3. Baleanu D, Mohammadi H, Rezapour S: Some existence results on nonlinear fractional differential equations. Philos. Trans. R. Soc. Lond. A 2013., 371: Article ID 20120144

4. Baleanu D, Mohammadi H, Rezapour S: On a nonlinear fractional differential equation on partially ordered metric spaces. Adv. Differ. Equ. 2013., 2013: Article ID 83

5. Kilbas AA, Trujillo JJ: Differential equations of fractional order: methods, results and problems. Appl. Anal. 2002, 81(2):435-493. 10.1080/0003681021000022032

6. Zhang SQ: Existence of positive solution for some class of nonlinear fractional differential equations. J. Math. Anal. Appl. 2003, 278: 136-148. 10.1016/S0022-247X(02)00583-8

7. Agarwal RP, O’Regan D, Stanek S: Positive solutions for mixed problems of singular fractional differential equations. Math. Nachr. 2012, 285(1):27-41. 10.1002/mana.201000043

8. Cabrera IJ, Harjani J, Sadarangani KB: Existence and uniqueness of positive solutions for a singular fractional three-point boundary value problem. Abstr. Appl. Anal. 2012., 2012: Article ID 803417

9. Cao J, Chen H: Positive solution of singular fractional differential equation in Banach space. J. Appl. Math. 2011., 2011: Article ID 352341

10. Liu Y, Nieto JJ, Otero-Zarraquinos O: Existence results for a coupled system of nonlinear singular fractional differential equations with impulse effects. Math. Probl. Eng. 2013., 2013: Article ID 498781

11. Wei X, Lu X: The periodic solutions of the compound singular fractional differential system with delay. Int. J. Differ. Equ. 2010., 2010: Article ID 509286

12. Zhou WX, Chu YD, Baleanu D: Uniqueness and existence of positive solutions for a multi-point boundary value problem of singular fractional differential equations. Adv. Differ. Equ. 2013., 2013: Article ID 114

13. Baleanu D, Agarwal RP, Mohammadi H, Rezapour S: Some existence results for a nonlinear fractional differential equation on partially ordered Banach spaces. Bound. Value Probl. 2013., 2013: Article ID 112

14. Baleanu D, Bhrawy AH, Taha TM: Two efficient generalized Laguerre spectral algorithms for fractional initial value problems. Abstr. Appl. Anal. 2013., 2013: Article ID 546502

15. Baleanu D, Diethelm K, Scalas E, Trujillo JJ Series on Complexity, Nonlinearity and Chaos. In Fractional Calculus Models and Numerical Methods. World Scientific, Singapore; 2012.

16. Bhrawy AH, Alghamdi MA: A shifted Jacobi-Gauss-Lobatto collocation method for solving nonlinear fractional Langevin equation involving two fractional orders in different intervals. Bound. Value Probl. 2012., 2012: Article ID 62

17. Bhrawy AH, Alofi AS: The operational matrix of fractional integration for shifted Chebyshev polynomials. Appl. Math. Lett. 2013, 26: 25-31. 10.1016/j.aml.2012.01.027

18. Bhrawy AH, Al-Shomrani MM: A shifted Legendre spectral method for fractional-order multi-point boundary value problems. Adv. Differ. Equ. 2012., 2012: Article ID 8

19. Bhrawy AH, Tharwata MM, Yildirim A: A new formula for fractional integrals of Chebyshev polynomials: application for solving multi-term fractional differential equations. Appl. Math. Model. 2013, 37: 4245-4252. 10.1016/j.apm.2012.08.022

20. Podlubny I Mathematics in Science and Engineering 198. In Fractional Differential Equations. Academic Press, San Diego; 1999.

21. Krasnoselski MA: Positive Solutions of Operator Equations. Noordhoff, Groningen; 1964.

## Acknowledgements

Research of the second and third authors was supported by Azarbaijan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions, which improved the final version of this paper.

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Correspondence to Dumitru Baleanu.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors have equal contributions. All authors read and approved the final manuscript.

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Baleanu, D., Mohammadi, H. & Rezapour, S. The existence of solutions for a nonlinear mixed problem of singular fractional differential equations. Adv Differ Equ 2013, 359 (2013). https://doi.org/10.1186/1687-1847-2013-359

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• DOI: https://doi.org/10.1186/1687-1847-2013-359

### Keywords

• boundary value problem
• fixed point
• fractional differential equation
• Green function
• regularization
• singular